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Topics In Analysis Kuttler March 19, 2009 2 Contents I Review Of Advanced Calculus 1 Set 1.1 1.2 1.3 1.4 Theory B...

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Topics In Analysis Kuttler March 19, 2009

2

Contents I

Review Of Advanced Calculus

1 Set 1.1 1.2 1.3 1.4

Theory Basic Definitions . . . . . . . . . The Schroder Bernstein Theorem Equivalence Relations . . . . . . Partially Ordered Sets . . . . . .

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19 19 22 25 26

2 Continuous Functions Of One Variable 2.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Theorems About Continuous Functions . . . . . . . . . . . . . . . .

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3 The 3.1 3.2 3.3 3.4 3.5 3.6

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Riemann Stieltjes Integral Upper And Lower Riemann Stieltjes Sums . Exercises . . . . . . . . . . . . . . . . . . . Functions Of Riemann Integrable Functions Properties Of The Integral . . . . . . . . . . Fundamental Theorem Of Calculus . . . . . Exercises . . . . . . . . . . . . . . . . . . .

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35 35 39 40 43 47 51

4 Some Important Linear Algebra 4.1 Algebra in Fn . . . . . . . . . . . . . . . . . 4.2 Exercises . . . . . . . . . . . . . . . . . . . 4.3 Subspaces Spans And Bases . . . . . . . . . 4.4 An Application To Matrices . . . . . . . . . 4.5 The Mathematical Theory Of Determinants 4.6 Exercises . . . . . . . . . . . . . . . . . . . 4.7 The Cayley Hamilton Theorem . . . . . . . 4.8 An Identity Of Cauchy . . . . . . . . . . . . 4.9 Block Multiplication Of Matrices . . . . . . 4.10 Exercises . . . . . . . . . . . . . . . . . . . 4.11 Shur’s Theorem . . . . . . . . . . . . . . . . 4.12 The Right Polar Decomposition . . . . . . .

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53 55 56 57 61 63 76 76 78 79 83 85 91

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4

CONTENTS

5 Multi-variable Calculus 5.1 Continuous Functions . . . . . . . . . . . . . . . 5.1.1 Distance In Fn . . . . . . . . . . . . . . . 5.2 Open And Closed Sets . . . . . . . . . . . . . . . 5.3 Continuous Functions . . . . . . . . . . . . . . . 5.3.1 Sufficient Conditions For Continuity . . . 5.4 Exercises . . . . . . . . . . . . . . . . . . . . . . 5.5 Limits Of A Function . . . . . . . . . . . . . . . 5.6 Exercises . . . . . . . . . . . . . . . . . . . . . . 5.7 The Limit Of A Sequence . . . . . . . . . . . . . 5.7.1 Sequences And Completeness . . . . . . . 5.7.2 Continuity And The Limit Of A Sequence 5.8 Properties Of Continuous Functions . . . . . . . 5.9 Exercises . . . . . . . . . . . . . . . . . . . . . . 5.10 Proofs Of Theorems . . . . . . . . . . . . . . . . 5.11 The Space L (Fn , Fm ) . . . . . . . . . . . . . . . 5.11.1 The Operator Norm . . . . . . . . . . . . 5.12 The Frechet Derivative . . . . . . . . . . . . . . . 5.13 C 1 Functions . . . . . . . . . . . . . . . . . . . . 5.14 C k Functions . . . . . . . . . . . . . . . . . . . . 5.15 Mixed Partial Derivatives . . . . . . . . . . . . . 5.16 Implicit Function Theorem . . . . . . . . . . . . 5.16.1 More Continuous Partial Derivatives . . . 5.17 The Method Of Lagrange Multipliers . . . . . . . 6 Metric Spaces And General Topological 6.1 Metric Space . . . . . . . . . . . . . . . 6.2 Compactness In Metric Space . . . . . . 6.3 Some Applications Of Compactness . . . 6.4 Ascoli Arzela Theorem . . . . . . . . . . 6.5 The Tietze Extension Theorem . . . . . 6.6 General Topological Spaces . . . . . . . 6.7 Connected Sets . . . . . . . . . . . . . .

Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7 Weierstrass Approximation Theorem 7.1 The Bernstein Polynomials . . . . . . . . . . . 7.2 Stone Weierstrass Theorem . . . . . . . . . . . 7.2.1 The Case Of Compact Sets . . . . . . . 7.2.2 The Case Of Locally Compact Sets . . . 7.2.3 The Case Of Complex Valued Functions 7.3 Exercises . . . . . . . . . . . . . . . . . . . . .

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97 97 97 100 102 102 103 104 108 108 110 111 112 113 113 118 118 120 124 129 130 132 136 137

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139 139 141 145 147 150 153 158

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163 163 167 167 170 171 172

CONTENTS

II

5

Real And Abstract Analysis

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8 Abstract Measure And Integration 8.1 σ Algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3 The Abstract Lebesgue Integral . . . . . . . . . . . . . . . . . . . . . 8.3.1 Preliminary Observations . . . . . . . . . . . . . . . . . . . . 8.3.2 Definition Of The Lebesgue Integral For Nonnegative Measurable Functions . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.3 The Lebesgue Integral For Nonnegative Simple Functions . . 8.3.4 Simple Functions And Measurable Functions . . . . . . . . . 8.3.5 The Monotone Convergence Theorem . . . . . . . . . . . . . 8.3.6 Other Definitions . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.7 Fatou’s Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . 8.3.8 The Righteous Algebraic Desires Of The Lebesgue Integral . 8.4 The Space L1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Vitali Convergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 8.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

177 177 188 190 190

9 The 9.1 9.2 9.3 9.4 9.5 9.6 9.7 9.8 9.9

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9.12 9.13

Construction Of Measures Outer Measures . . . . . . . . . . . . . . . . . . Regular Measures . . . . . . . . . . . . . . . . . Urysohn’s lemma . . . . . . . . . . . . . . . . . Positive Linear Functionals . . . . . . . . . . . One Dimensional Lebesgue Measure . . . . . . One Dimensional Lebesgue Stieltjes Measure . The Distribution Function . . . . . . . . . . . . Product Measures . . . . . . . . . . . . . . . . Alternative Treatment Of Product Measure . . 9.9.1 Monotone Classes And Algebras . . . . 9.9.2 Product Measure . . . . . . . . . . . . . Completion Of Measures . . . . . . . . . . . . . Another Version Of Product Measures . . . . . 9.11.1 General Theory . . . . . . . . . . . . . . 9.11.2 Completion Of Product Measure Spaces Disturbing Examples . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . .

10 Lebesgue Measure 10.1 Basic Properties . . . . . . . . . . . . . . 10.2 The Vitali Covering Theorem . . . . . . . 10.3 The Vitali Covering Theorem (Elementary 10.4 Vitali Coverings . . . . . . . . . . . . . . . 10.5 Change Of Variables For Linear Maps . . 10.6 Change Of Variables For C 1 Functions . .

191 193 196 200 202 202 205 205 212 214

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219 219 224 227 234 244 245 247 251 263 263 266 271 275 275 279 281 283

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285 285 289 291 294 298 301

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CONTENTS 10.7 Mappings Which Are Not One To One . . . 10.8 Lebesgue Measure And Iterated Integrals . 10.9 Spherical Coordinates In Many Dimensions 10.10The Brouwer Fixed Point Theorem . . . . . 10.11The Brouwer Fixed Point Theorem Another

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308 309 311 313 317

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321 321 323 326 333

Lp Spaces Basic Inequalities And Properties . . . . . . . Density Considerations . . . . . . . . . . . . . Separability . . . . . . . . . . . . . . . . . . . Continuity Of Translation . . . . . . . . . . . Mollifiers And Density Of Smooth Functions Exercises . . . . . . . . . . . . . . . . . . . .

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337 337 345 347 349 350 355

13 Banach Spaces 13.1 Theorems Based On Baire Category . . . . . . . . . . . . . . 13.1.1 Baire Category Theorem . . . . . . . . . . . . . . . . . 13.1.2 Uniform Boundedness Theorem . . . . . . . . . . . . . 13.1.3 Open Mapping Theorem . . . . . . . . . . . . . . . . . 13.1.4 Closed Graph Theorem . . . . . . . . . . . . . . . . . 13.2 Hahn Banach Theorem . . . . . . . . . . . . . . . . . . . . . . 13.2.1 Partially Ordered Sets . . . . . . . . . . . . . . . . . . 13.2.2 Gauge Functions And Hahn Banach Theorem . . . . . 13.2.3 The Complex Version Of The Hahn Banach Theorem 13.2.4 The Dual Space And Adjoint Operators . . . . . . . . 13.3 Approximation With Sums . . . . . . . . . . . . . . . . . . . 13.4 Weak And Weak ∗ Topologies . . . . . . . . . . . . . . . . . . 13.4.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . 13.4.2 Banach Alaoglu Theorem . . . . . . . . . . . . . . . . 13.4.3 Eberlein Smulian Theorem . . . . . . . . . . . . . . . 13.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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361 361 361 365 366 368 370 370 371 373 375 378 382 382 383 385 388

14 Locally Convex Topological Vector Spaces 14.1 Fundamental Considerations . . . . . . . . . 14.2 Separation Theorems . . . . . . . . . . . . . 14.2.1 Convex Functionals . . . . . . . . . 14.2.2 More Separation Theorems . . . . . 14.3 The Weak And Weak∗ Topologies . . . . . 14.4 The Tychonoff Fixed Point Theorem . . . . 14.5 Set-Valued Maps . . . . . . . . . . . . . . .

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393 393 397 400 403 406 407 412

11 Some Extension Theorems 11.1 Caratheodory Extension Theorem 11.2 The Tychonoff Theorem . . . . . . 11.3 Kolmogorov Extension Theorem . 11.4 Exercises . . . . . . . . . . . . . . 12 The 12.1 12.2 12.3 12.4 12.5 12.6

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CONTENTS

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15 Hilbert Spaces 15.1 Basic Theory . . . . . . . . . . . . . . . . . 15.2 The Hilbert Space L (U ) . . . . . . . . . . . 15.3 Approximations In Hilbert Space . . . . . . 15.4 The M¨ untz Theorem . . . . . . . . . . . . . 15.5 Orthonormal Sets . . . . . . . . . . . . . . . 15.6 Fourier Series, An Example . . . . . . . . . 15.7 Compact Operators . . . . . . . . . . . . . 15.7.1 Compact Operators In Hilbert Space 15.7.2 Nuclear Operators . . . . . . . . . . 15.7.3 Hilbert Schmidt Operators . . . . . 15.8 Compact Operators In Banach Space . . . . 15.9 The Fredholm Alternative . . . . . . . . . . 15.10Square Roots . . . . . . . . . . . . . . . . . 15.11Quotient Spaces . . . . . . . . . . . . . . . 15.12General Theory Of Continuous Semigroups 15.12.1 An Evolution Equation . . . . . . . 15.12.2 Adjoints, Hilbert Space . . . . . . . 15.12.3 Adjoints, Reflexive Banach Space . .

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421 421 427 432 436 440 441 443 443 448 451 456 457 459 464 468 477 480 484

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489 489 495 503 510 513 514 516 521

17 Integrals And Derivatives 17.1 The Fundamental Theorem Of Calculus . . . . . . . . . . . . . . 17.2 Absolutely Continuous Functions . . . . . . . . . . . . . . . . . . 17.3 Differentiation Of Measures With Respect To Lebesgue Measure 17.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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525 525 530 535 541

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16 Representation Theorems 16.1 Radon Nikodym Theorem . . . . . . . . . . . . . 16.2 Vector Measures . . . . . . . . . . . . . . . . . . 16.3 Representation Theorems For The Dual Space Of 16.4 The Dual Space Of C (X) . . . . . . . . . . . . . 16.5 The Dual Space Of C0 (X) . . . . . . . . . . . . . 16.6 More Attractive Formulations . . . . . . . . . . . 16.7 Sequential Compactness In L1 . . . . . . . . . . . 16.8 Exercises . . . . . . . . . . . . . . . . . . . . . .

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18 Orlitz Spaces 545 18.1 Basic Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 545 18.2 Dual Spaces In Orlitz Space . . . . . . . . . . . . . . . . . . . . . . . 560 19 Hausdorff Measure 19.1 Definition Of Hausdorff Measures . . . . 19.1.1 Properties Of Hausdorff Measure 19.2 Hn And mn . . . . . . . . . . . . . . . . 19.3 Technical Considerations . . . . . . . . . 19.3.1 Steiner Symmetrization . . . . .

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565 565 566 569 571 573

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CONTENTS 19.3.2 The Isodiametric Inequality . . . . . . . . . . . . 19.4 The Proper Value Of β (n) . . . . . . . . . . . . . . . . . 19.4.1 A Formula For α (n) . . . . . . . . . . . . . . . . 19.4.2 Hausdorff Measure And Linear Transformations . 19.5 The Area Formula . . . . . . . . . . . . . . . . . . . . . 19.5.1 Preliminary Results . . . . . . . . . . . . . . . . 19.5.2 The Area Formula . . . . . . . . . . . . . . . . . 19.6 The Area Formula Alternate Version . . . . . . . . . . . 19.6.1 Preliminary Results . . . . . . . . . . . . . . . . 19.6.2 The Area Formula . . . . . . . . . . . . . . . . . 19.7 The Divergence Theorem . . . . . . . . . . . . . . . . .

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575 576 577 579 581 581 589 592 592 599 601

20 Differentiation With Respect To General Radon Measures 20.1 Besicovitch Covering Theorem . . . . . . . . . . . . . . . . . 20.2 Fundamental Theorem Of Calculus For Radon Measures . . . 20.3 Slicing Measures . . . . . . . . . . . . . . . . . . . . . . . . . 20.4 Vitali Coverings . . . . . . . . . . . . . . . . . . . . . . . . . . 20.5 Differentiation Of Radon Measures . . . . . . . . . . . . . . . 20.6 The Radon Nikodym Theorem For Radon Measures . . . . .

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615 615 620 624 630 632 635

21 Fourier Transforms 21.1 An Algebra Of Special Functions . . . . . . . . . . 21.2 Fourier Transforms Of Functions In G . . . . . . . 21.3 Fourier Transforms Of Just About Anything . . . . 21.3.1 Fourier Transforms Of G ∗ . . . . . . . . . . 21.3.2 Fourier Transforms Of Functions In L1 (Rn ) 21.3.3 Fourier Transforms Of Functions In L2 (Rn ) 21.3.4 The Schwartz Class . . . . . . . . . . . . . 21.3.5 Convolution . . . . . . . . . . . . . . . . . . 21.4 Exercises . . . . . . . . . . . . . . . . . . . . . . .

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637 637 638 641 641 645 648 652 654 656

22 Fourier Analysis In Rn An Introduction 22.1 The Marcinkiewicz Interpolation Theorem 22.2 The Calderon Zygmund Decomposition . 22.3 Mihlin’s Theorem . . . . . . . . . . . . . . 22.4 Singular Integrals . . . . . . . . . . . . . . 22.5 Helmholtz Decompositions . . . . . . . . .

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661 661 664 666 679 689

23 The Bochner Integral 23.1 Strong And Weak Measurability . . . . . . . . . . . . 23.2 The Bochner Integral . . . . . . . . . . . . . . . . . . . 23.2.1 Definition And Basic Properties . . . . . . . . 23.2.2 Taking A Closed Operator Out Of The Integral 23.3 Operator Valued Functions . . . . . . . . . . . . . . . 23.3.1 Review Of Hilbert Schmidt Theorem . . . . . . 23.3.2 Measurable Compact Operators . . . . . . . . .

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697 697 705 705 710 715 717 721

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CONTENTS 23.4 23.5 23.6 23.7 23.8 23.9

Fubini’s Theorem For Bochner Integrals The Spaces Lp (Ω; X) . . . . . . . . . . Measurable Representatives . . . . . . . Vector Measures . . . . . . . . . . . . . The Riesz Representation Theorem . . . Maximal Monotone Operators . . . . . . 23.9.1 Basic Theory . . . . . . . . . . . 23.9.2 Evolution Inclusions . . . . . . . 23.10Exercises . . . . . . . . . . . . . . . . .

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721 724 732 734 739 745 745 752 757

24 The Yankov von Neumann Aumann theorem 759 24.1 Multifunctions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 770

III

Complex Analysis

773

25 The Complex Numbers 775 25.1 The Extended Complex Plane . . . . . . . . . . . . . . . . . . . . . . 777 25.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 778 26 Riemann Stieltjes Integrals 779 26.1 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 789 27 Fundamentals Of Complex Analysis 27.1 Analytic Functions . . . . . . . . . . . . . . . 27.1.1 Cauchy Riemann Equations . . . . . . 27.1.2 An Important Example . . . . . . . . 27.2 Exercises . . . . . . . . . . . . . . . . . . . . 27.3 Cauchy’s Formula For A Disk . . . . . . . . . 27.4 Exercises . . . . . . . . . . . . . . . . . . . . 27.5 Zeros Of An Analytic Function . . . . . . . . 27.6 Liouville’s Theorem . . . . . . . . . . . . . . 27.7 The General Cauchy Integral Formula . . . . 27.7.1 The Cauchy Goursat Theorem . . . . 27.7.2 A Redundant Assumption . . . . . . . 27.7.3 Classification Of Isolated Singularities 27.7.4 The Cauchy Integral Formula . . . . . 27.7.5 An Example Of A Cycle . . . . . . . . 27.8 Exercises . . . . . . . . . . . . . . . . . . . . 28 The 28.1 28.2 28.3 28.4

Open Mapping Theorem A Local Representation . . . . . . . . . . . Branches Of The Logarithm . . . . . . . . . Maximum Modulus Theorem . . . . . . . . Extensions Of Maximum Modulus Theorem 28.4.1 Phragmên Lindelöf Theorem . . . .

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791 791 793 795 796 797 804 807 809 810 810 813 814 817 824 828

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831 831 833 835 837 837

10

CONTENTS 28.4.2 Hadamard Three Circles Theorem . . . . . . . 28.4.3 Schwarz’s Lemma . . . . . . . . . . . . . . . . . 28.4.4 One To One Analytic Maps On The Unit Ball Exercises . . . . . . . . . . . . . . . . . . . . . . . . . Counting Zeros . . . . . . . . . . . . . . . . . . . . . . An Application To Linear Algebra . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . .

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839 840 841 842 844 848 852

29 Residues 29.1 Rouche’s Theorem And The Argument Principle . . . 29.1.1 Argument Principle . . . . . . . . . . . . . . . 29.1.2 Rouche’s Theorem . . . . . . . . . . . . . . . . 29.1.3 A Different Formulation . . . . . . . . . . . . . 29.2 Singularities And The Laurent Series . . . . . . . . . . 29.2.1 What Is An Annulus? . . . . . . . . . . . . . . 29.2.2 The Laurent Series . . . . . . . . . . . . . . . . 29.2.3 Contour Integrals And Evaluation Of Integrals 29.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . .

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855 858 858 861 862 863 863 866 870 879

30 Some Important Functional Analysis Applications 30.1 The Spectral Radius Of A Bounded Linear Transformation 30.2 Analytic Semigroups . . . . . . . . . . . . . . . . . . . . . . 30.2.1 Sectorial Operators And Analytic Semigroups . . . . 30.2.2 The Numerical Range . . . . . . . . . . . . . . . . . 30.2.3 An Interesting Example . . . . . . . . . . . . . . . . 30.2.4 Fractional Powers Of Sectorial Operators . . . . . . 30.2.5 A Scale Of Banach Spaces . . . . . . . . . . . . . . .

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881 881 884 884 895 897 900 915

31 Complex Mappings 31.1 Conformal Maps . . . . . . . . . . . . . . . 31.2 Fractional Linear Transformations . . . . . 31.2.1 Circles And Lines . . . . . . . . . . 31.2.2 Three Points To Three Points . . . . 31.3 Riemann Mapping Theorem . . . . . . . . . 31.3.1 Montel’s Theorem . . . . . . . . . . 31.3.2 Regions With Square Root Property 31.4 Analytic Continuation . . . . . . . . . . . . 31.4.1 Regular And Singular Points . . . . 31.4.2 Continuation Along A Curve . . . . 31.5 The Picard Theorems . . . . . . . . . . . . 31.5.1 Two Competing Lemmas . . . . . . 31.5.2 The Little Picard Theorem . . . . . 31.5.3 Schottky’s Theorem . . . . . . . . . 31.5.4 A Brief Review . . . . . . . . . . . . 31.5.5 Montel’s Theorem . . . . . . . . . .

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919 919 920 920 922 923 924 926 930 930 932 933 935 938 939 943 945

28.5 28.6 28.7 28.8

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CONTENTS

11

31.5.6 The Great Big Picard Theorem . . . . . . . . . . . . . . . . . 946 31.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 948 32 Approximation By Rational Functions 32.1 Runge’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1.1 Approximation With Rational Functions . . . . . . . . . . . . 32.1.2 Moving The Poles And Keeping The Approximation . . . . . 32.1.3 Merten’s Theorem. . . . . . . . . . . . . . . . . . . . . . . . . 32.1.4 Runge’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 The Mittag-Leffler Theorem . . . . . . . . . . . . . . . . . . . . . . . 32.2.1 A Proof From Runge’s Theorem . . . . . . . . . . . . . . . . 32.2.2 A Direct Proof Without Runge’s Theorem . . . . . . . . . . . b . . . . . . . . . . . . . . . . . . 32.2.3 Functions Meromorphic On C 32.2.4 A Great And Glorious Theorem About Simply Connected Regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

951 951 951 953 953 958 960 960 962 964

33 Infinite Products 33.1 Analytic Function With Prescribed Zeros . . . . . . . . . . 33.2 Factoring A Given Analytic Function . . . . . . . . . . . . . 33.2.1 Factoring Some Special Analytic Functions . . . . . 33.3 The Existence Of An Analytic Function With Given Values 33.4 Jensen’s Formula . . . . . . . . . . . . . . . . . . . . . . . . 33.5 Blaschke Products . . . . . . . . . . . . . . . . . . . . . . . 33.5.1 The M¨ untz-Szasz Theorem Again . . . . . . . . . . . 33.6 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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969 973 978 980 982 986 989 992 994

34 Elliptic Functions 34.1 Periodic Functions . . . . . . . . . . . . . . . 34.1.1 The Unimodular Transformations . . . 34.1.2 The Search For An Elliptic Function . 34.1.3 The Differential Equation Satisfied By 34.1.4 A Modular Function . . . . . . . . . . 34.1.5 A Formula For λ . . . . . . . . . . . . 34.1.6 Mapping Properties Of λ . . . . . . . 34.1.7 A Short Review And Summary . . . . 34.2 The Picard Theorem Again . . . . . . . . . . 34.3 Exercises . . . . . . . . . . . . . . . . . . . .

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1003 . 1004 . 1008 . 1011 . 1014 . 1016 . 1022 . 1024 . 1032 . 1036 . 1037

IV

Sobolev Spaces

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964 968

1039

35 Weak Derivatives 1041 35.1 Weak ∗ Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . 1041 35.2 Test Functions And Weak Derivatives . . . . . . . . . . . . . . . . . 1042 35.3 Weak Derivatives In Lploc . . . . . . . . . . . . . . . . . . . . . . . . . 1046

12

CONTENTS 35.4 Morrey’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1049 35.5 Rademacher’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 1052 35.6 Change Of Variables Formula Lipschitz Maps . . . . . . . . . . . . . 1055

36 The 36.1 36.2 36.3 36.4

Area And Coarea Formulas The Area Formula Again . . . . . . Mappings That Are Not One To One The Coarea Formula . . . . . . . . . A Nonlinear Fubini’s Theorem . . .

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1065 . 1065 . 1068 . 1072 . 1083

37 Integration On Manifolds 1085 37.1 Partitions Of Unity . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1085 37.2 Integration On Manifolds . . . . . . . . . . . . . . . . . . . . . . . . 1089 37.3 Comparison With Hn . . . . . . . . . . . . . . . . . . . . . . . . . . 1095 38 Basic Theory Of Sobolev Spaces 38.1 Embedding Theorems For W m,p (Rn ) . 38.2 An Extension Theorem . . . . . . . . . 38.3 General Embedding Theorems . . . . . 38.4 More Extension Theorems . . . . . . .

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1097 . 1106 . 1120 . 1127 . 1130

39 Sobolev Spaces Based On L2 39.1 Fourier Transform Techniques . . . . . . . . . . 39.2 Fractional Order Spaces . . . . . . . . . . . . . 39.3 An Intrinsic Norm . . . . . . . . . . . . . . . . 39.4 Embedding Theorems . . . . . . . . . . . . . . 39.5 The Trace On The Boundary Of A Half Space 39.6 Sobolev Spaces On Manifolds . . . . . . . . . . 39.6.1 General Theory . . . . . . . . . . . . . . 39.6.2 The Trace On The Boundary . . . . . .

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1137 . 1137 . 1142 . 1144 . 1151 . 1152 . 1160 . 1160 . 1165

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40 Weak Solutions 1171 40.1 The Lax Milgram Theorem . . . . . . . . . . . . . . . . . . . . . . . 1171 41 Korn’s Inequality 1177 41.1 A Fundamental Inequality . . . . . . . . . . . . . . . . . . . . . . . . 1177 41.2 Korn’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1183 42 Elliptic Regularity And Nirenberg Differences 1185 42.1 The Case Of A Half Space . . . . . . . . . . . . . . . . . . . . . . . . 1185 42.2 The Case Of Bounded Open Sets . . . . . . . . . . . . . . . . . . . . 1195 43 Interpolation In Banach Space 43.1 An Assortment Of Important Theorems . . . . . . . . . . . . . . . 43.1.1 Weak Vector Valued Derivatives . . . . . . . . . . . . . . . 43.1.2 Some Imbedding Theorems . . . . . . . . . . . . . . . . . .

1205 . 1205 . 1205 . 1215

CONTENTS

13

43.2 The K Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1220 43.3 The J Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1225 43.4 Duality And Interpolation . . . . . . . . . . . . . . . . . . . . . . . . 1231 44 Trace Spaces 1243 44.1 Definition And Basic Theory Of Trace Spaces . . . . . . . . . . . . . 1243 44.2 Trace And Interpolation Spaces . . . . . . . . . . . . . . . . . . . . . 1250 45 Traces Of Sobolev Spaces And Fractional Order Spaces 45.1 Traces Of Sobolev Spaces On The Boundary Of A Half Space 45.2 A Right Inverse For The Trace For A Half Space . . . . . . . 45.3 Intrinsic Norms . . . . . . . . . . . . . . . . . . . . . . . . . . 45.4 Fractional Order Sobolev Spaces . . . . . . . . . . . . . . . .

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1257 . 1257 . 1260 . 1262 . 1283

46 Sobolev Spaces On Manifolds 1289 46.1 Basic Definitions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1289 46.2 The Trace On The Boundary Of An Open Set . . . . . . . . . . . . . 1291

V

Topics In Probability

1295

47 Basic Probability 47.1 Random Variables And Independence . . . . . . . . . . . . . . . . 47.1.1 Independence . . . . . . . . . . . . . . . . . . . . . . . . . . 47.1.2 Independence For Banach Space Valued Random Variables 47.1.3 0, 1 Laws . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47.1.4 Kolmogorov’s Inequality, Strong Law Of Large Numbers . . 47.2 The Characteristic Function . . . . . . . . . . . . . . . . . . . . . . 47.3 Conditional Probability . . . . . . . . . . . . . . . . . . . . . . . . 47.4 Characteristic Functions And Independence . . . . . . . . . . . . . 47.5 Characteristic Functions For Measures . . . . . . . . . . . . . . . . 47.6 Characteristic Functions And Independence In Banach Space . . . 47.7 Convolution And Sums . . . . . . . . . . . . . . . . . . . . . . . . . 47.8 The Convergence Of Sums Of Symmetric Random Variables . . . . 47.9 The Multivariate Normal Distribution . . . . . . . . . . . . . . . . 47.10Use Of Characteristic Functions To Find Moments . . . . . . . . . 47.11The Central Limit Theorem . . . . . . . . . . . . . . . . . . . . . . 47.12Characteristic Functions Of Probability Measures . . . . . . . . . . 47.13Positive Definite Functions, Bochner’s Theorem . . . . . . . . . . .

1297 . 1297 . 1301 . 1304 . 1309 . 1312 . 1318 . 1319 . 1323 . 1326 . 1329 . 1331 . 1337 . 1343 . 1350 . 1351 . 1359 . 1363

48 Conditional Expectation And Martingales 48.1 Conditional Expectation . . . . . . . . . . . . . . 48.2 Discrete Martingales . . . . . . . . . . . . . . . . 48.2.1 Upcrossings . . . . . . . . . . . . . . . . . 48.2.2 The Submartingale Convergence Theorem 48.3 Optional Sampling And Stopping Times . . . . .

1371 . 1371 . 1374 . 1376 . 1378 . 1379

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14

CONTENTS 48.3.1 Stopping Times And Their Properties . . 48.4 Optional Stopping Times And Martingales . . . . 48.4.1 Stopping Times And Their Properties . . 48.4.2 Maximal Inequalities . . . . . . . . . . . . 48.4.3 The Upcrossing Estimate . . . . . . . . . 48.5 The Submartingale Convergence Theorem . . . . 48.6 A Reverse Submartingale Convergence Theorem 48.7 Strong Law Of Large Numbers . . . . . . . . . .

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49 Probability In Infinite Dimensions 49.1 Conditional Expectation In Banach Spaces . . . . 49.2 Probability Measures And Tightness . . . . . . . . 49.3 A Major Existence And Convergence Theorem . . 49.4 Bochner’s Theorem In Infinite Dimensions . . . . . 49.5 The Multivariate Normal Distribution . . . . . . . 49.6 Gaussian Measures . . . . . . . . . . . . . . . . . . 49.6.1 Definitions And Basic Properties . . . . . . 49.6.2 Fernique’s Theorem . . . . . . . . . . . . . 49.7 Gaussian Measures For A Separable Hilbert Space 49.8 Abstract Wiener Spaces . . . . . . . . . . . . . . . 49.9 White Noise . . . . . . . . . . . . . . . . . . . . . . 49.10Existence Of Abstract Wiener Spaces . . . . . . .

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1403 . 1403 . 1407 . 1414 . 1420 . 1425 . 1430 . 1430 . 1434 . 1439 . 1449 . 1463 . 1463

50 Stochastic Processes 50.1 Fundamental Definitions And Properties . . . . . . . . . . 50.2 Filtrations . . . . . . . . . . . . . . . . . . . . . . . . . . . 50.3 Martingales . . . . . . . . . . . . . . . . . . . . . . . . . . 50.3.1 Some Maximal Estimates . . . . . . . . . . . . . . 50.4 Optional Sampling Theorems . . . . . . . . . . . . . . . . 50.4.1 Stopping Times And Their Properties . . . . . . . 50.4.2 Doob Optional Sampling Theorem . . . . . . . . . 50.5 Doob Optional Sampling Continuous Case . . . . . . . . . 50.5.1 Stopping Times . . . . . . . . . . . . . . . . . . . . 50.5.2 The Optional Sampling Theorem Continuous Case 50.6 Right Continuity Of Submartingales . . . . . . . . . . . . 50.7 Some Maximal Inequalities . . . . . . . . . . . . . . . . . 50.8 Continuous Submartingale Convergence Theorem . . . . . 50.9 Hitting This Before That . . . . . . . . . . . . . . . . . . 50.10The Space MpT (E) . . . . . . . . . . . . . . . . . . . . . .

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1469 . 1469 . 1475 . 1479 . 1480 . 1484 . 1484 . 1487 . 1492 . 1492 . 1495 . 1503 . 1510 . 1513 . 1517 . 1520

51 The 51.1 51.2 51.3 51.4

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1525 . 1525 . 1536 . 1539 . 1542

Quadratic Variation Of A Martingale How To Recognize A Martingale . . . . . . The Burkholder Davis Gundy Inequality . . The Case Of M2T (H) . . . . . . . . . . . . The Case Of Local Martingales . . . . . . .

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1379 1383 1383 1389 1391 1393 1397 1399

CONTENTS

15

51.5 The Quadratic Variation And Stochastic Integration . . . . . . . . . 1550 51.6 Doob Meyer Decomposition . . . . . . . . . . . . . . . . . . . . . . . 1558 51.7 Levy’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1578 52 Wiener Processes 52.1 Real Wiener Processes . . . . . . . . . . . . . . . . . . 52.2 Nowhere Differentiability Of Wiener Processes . . . . 52.3 Wiener Processes In Separable Banach Space . . . . . 52.4 An Example Of Martingales, Independent Increments 52.5 Hilbert Space Valued Wiener Processes . . . . . . . . 52.6 Levy’s Theorem In Hilbert Space . . . . . . . . . . . .

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1587 . 1587 . 1591 . 1592 . 1598 . 1602 . 1616

53 Stochastic Integration 53.1 Integrals Of Elementary Processes . . 53.2 Integrals Of More General Processes . 53.3 Progressively Measurable . . . . . . . 53.4 Localization . . . . . . . . . . . . . . . 53.5 The Integral For A More General Case 53.6 Taking Out A Linear Transformation . 53.7 The Case Where Q Is Not Trace Class

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1621 . 1621 . 1634 . 1641 . 1645 . 1651 . 1655 . 1657

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54 The Ito Formula 1663 54.1 The Case Of A Q Wiener Process . . . . . . . . . . . . . . . . . . . . 1663 54.2 The Case Where Q Is Not Trace Class . . . . . . . . . . . . . . . . . 1675 55 Stochastic Differential Equations 55.1 Theorems Based On Contraction Mapping . . . . . . . . . . . 55.1.1 Predictable And Stochastic Continuity . . . . . . . . . 55.1.2 Existence And Uniqueness For A Stochastic Equation 55.2 Gelfand Triples . . . . . . . . . . . . . . . . . . . . . . . . . . 55.3 An Ito Formula . . . . . . . . . . . . . . . . . . . . . . . . . . 55.4 Finite To Infinite Dimensions . . . . . . . . . . . . . . . . . . 55.4.1 Some Nonlinear Operators. . . . . . . . . . . . . . . . 55.4.2 The Situation . . . . . . . . . . . . . . . . . . . . . . . 55.4.3 The Finite Dimensional Problem, Estimates . . . . . . 55.4.4 Passing To The Limit . . . . . . . . . . . . . . . . . . 55.5 Finite Dimensional Problems . . . . . . . . . . . . . . . . . . 55.5.1 The Ito Formula . . . . . . . . . . . . . . . . . . . . . 55.5.2 An Unbelievably Technical Lemma . . . . . . . . . . . 55.5.3 An Existence Theorem . . . . . . . . . . . . . . . . . . 55.6 Infinite Dimensional Problems . . . . . . . . . . . . . . . . . .

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1679 . 1679 . 1679 . 1683 . 1687 . 1689 . 1707 . 1707 . 1710 . 1712 . 1714 . 1722 . 1722 . 1723 . 1737 . 1746

16 56 Another Approach, Semigroups 56.1 Stochastic Fubini Theorem . . . . . . . . . . . . . . 56.2 Correlation Operators . . . . . . . . . . . . . . . . . 56.3 Weak Solutions To Stochastic Differential Equations 56.3.1 Convolution With A Semigroup . . . . . . . . 56.3.2 Continuity Of WA (t) . . . . . . . . . . . . . . 56.3.3 Temporal Regularity Analytic Semigroups . . 56.3.4 Spacial Regularity . . . . . . . . . . . . . . . 56.3.5 Weak Solutions . . . . . . . . . . . . . . . . .

CONTENTS

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1751 . 1751 . 1756 . 1761 . 1762 . 1772 . 1787 . 1790 . 1797

A The Hausdorff Maximal Theorem 1807 A.1 The Hamel Basis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1811 A.2 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1812 c 2004, Copyright °

Part I

Review Of Advanced Calculus

17

Set Theory 1.1

Basic Definitions

A set is a collection of things called elements of the set. For example, the set of integers, the collection of signed whole numbers such as 1,2,-4, etc. This set whose existence will be assumed is denoted by Z. Other sets could be the set of people in a family or the set of donuts in a display case at the store. Sometimes parentheses, { } specify a set by listing the things which are in the set between the parentheses. For example the set of integers between -1 and 2, including these numbers could be denoted as {−1, 0, 1, 2}. The notation signifying x is an element of a set S, is written as x ∈ S. Thus, 1 ∈ {−1, 0, 1, 2, 3}. Here are some axioms about sets. Axioms are statements which are accepted, not proved. 1. Two sets are equal if and only if they have the same elements. 2. To every set, A, and to every condition S (x) there corresponds a set, B, whose elements are exactly those elements x of A for which S (x) holds. 3. For every collection of sets there exists a set that contains all the elements that belong to at least one set of the given collection. 4. The Cartesian product of a nonempty family of nonempty sets is nonempty. 5. If A is a set there exists a set, P (A) such that P (A) is the set of all subsets of A. This is called the power set. These axioms are referred to as the axiom of extension, axiom of specification, axiom of unions, axiom of choice, and axiom of powers respectively. It seems fairly clear you should want to believe in the axiom of extension. It is merely saying, for example, that {1, 2, 3} = {2, 3, 1} since these two sets have the same elements in them. Similarly, it would seem you should be able to specify a new set from a given set using some “condition” which can be used as a test to determine whether the element in question is in the set. For example, the set of all integers which are multiples of 2. This set could be specified as follows. {x ∈ Z : x = 2y for some y ∈ Z} . 19

20

SET THEORY

In this notation, the colon is read as “such that” and in this case the condition is being a multiple of 2. Another example of political interest, could be the set of all judges who are not judicial activists. I think you can see this last is not a very precise condition since there is no way to determine to everyone’s satisfaction whether a given judge is an activist. Also, just because something is grammatically correct does not mean it makes any sense. For example consider the following nonsense. S = {x ∈ set of dogs : it is colder in the mountains than in the winter} . So what is a condition? We will leave these sorts of considerations and assume our conditions make sense. The axiom of unions states that for any collection of sets, there is a set consisting of all the elements in each of the sets in the collection. Of course this is also open to further consideration. What is a collection? Maybe it would be better to say “set of sets” or, given a set whose elements are sets there exists a set whose elements consist of exactly those things which are elements of at least one of these sets. If S is such a set whose elements are sets, ∪ {A : A ∈ S} or ∪ S signify this union. Something is in the Cartesian product of a set or “family” of sets if it consists of a single thing taken from each set in the family. Thus (1, 2, 3) ∈ {1, 4, .2} × {1, 2, 7} × {4, 3, 7, 9} because it consists of exactly one element from each of the sets which are separated by ×. Also, this is the notation for the Cartesian product of finitely many sets. If S is a set whose elements are sets, Y A A∈S

signifies the Cartesian product. The Cartesian product is the set of choice functions, a choice function being a function which selects exactly one element of each set of S. You may think the axiom of choice, stating that the Cartesian product of a nonempty family of nonempty sets is nonempty, is innocuous but there was a time when many mathematicians were ready to throw it out because it implies things which are very hard to believe, things which never happen without the axiom of choice. A is a subset of B, written A ⊆ B, if every element of A is also an element of B. This can also be written as B ⊇ A. A is a proper subset of B, written A ⊂ B or B ⊃ A if A is a subset of B but A is not equal to B, A 6= B. A ∩ B denotes the intersection of the two sets, A and B and it means the set of elements of A which are also elements of B. The axiom of specification shows this is a set. The empty set is the set which has no elements in it, denoted as ∅. A ∪ B denotes the union of the two sets, A and B and it means the set of all elements which are in either of the sets. It is a set because of the axiom of unions.

1.1. BASIC DEFINITIONS

21

The complement of a set, (the set of things which are not in the given set ) must be taken with respect to a given set called the universal set which is a set which contains the one whose complement is being taken. Thus, the complement of A, denoted as AC ( or more precisely as X \ A) is a set obtained from using the axiom of specification to write AC ≡ {x ∈ X : x ∈ / A} The symbol ∈ / means: “is not an element of”. Note the axiom of specification takes place relative to a given set. Without this universal set it makes no sense to use the axiom of specification to obtain the complement. Words such as “all” or “there exists” are called quantifiers and they must be understood relative to some given set. For example, the set of all integers larger than 3. Or there exists an integer larger than 7. Such statements have to do with a given set, in this case the integers. Failure to have a reference set when quantifiers are used turns out to be illogical even though such usage may be grammatically correct. Quantifiers are used often enough that there are symbols for them. The symbol ∀ is read as “for all” or “for every” and the symbol ∃ is read as “there exists”. Thus ∀∀∃∃ could mean for every upside down A there exists a backwards E. DeMorgan’s laws are very useful in mathematics. Let S be a set of sets each of which is contained in some universal set, U . Then © ª C ∪ AC : A ∈ S = (∩ {A : A ∈ S}) and

© ª C ∩ AC : A ∈ S = (∪ {A : A ∈ S}) .

These laws follow directly from the definitions. Also following directly from the definitions are: Let S be a set of sets then B ∪ ∪ {A : A ∈ S} = ∪ {B ∪ A : A ∈ S} . and: Let S be a set of sets show B ∩ ∪ {A : A ∈ S} = ∪ {B ∩ A : A ∈ S} . Unfortunately, there is no single universal set which can be used for all sets. Here is why: Suppose there were. Call it S. Then you could consider A the set of all elements of S which are not elements of themselves, this from the axiom of specification. If A is an element of itself, then it fails to qualify for inclusion in A. Therefore, it must not be an element of itself. However, if this is so, it qualifies for inclusion in A so it is an element of itself and so this can’t be true either. Thus the most basic of conditions you could imagine, that of being an element of, is meaningless and so allowing such a set causes the whole theory to be meaningless. The solution is to not allow a universal set. As mentioned by Halmos in Naive set theory, “Nothing contains everything”. Always beware of statements involving quantifiers wherever they occur, even this one.

22

SET THEORY

1.2

The Schroder Bernstein Theorem

It is very important to be able to compare the size of sets in a rational way. The most useful theorem in this context is the Schroder Bernstein theorem which is the main result to be presented in this section. The Cartesian product is discussed above. The next definition reviews this and defines the concept of a function. Definition 1.2.1 Let X and Y be sets. X × Y ≡ {(x, y) : x ∈ X and y ∈ Y } A relation is defined to be a subset of X × Y . A function, f, also called a mapping, is a relation which has the property that if (x, y) and (x, y1 ) are both elements of the f , then y = y1 . The domain of f is defined as D (f ) ≡ {x : (x, y) ∈ f } , written as f : D (f ) → Y . It is probably safe to say that most people do not think of functions as a type of relation which is a subset of the Cartesian product of two sets. A function is like a machine which takes inputs, x and makes them into a unique output, f (x). Of course, that is what the above definition says with more precision. An ordered pair, (x, y) which is an element of the function or mapping has an input, x and a unique output, y,denoted as f (x) while the name of the function is f . “mapping” is often a noun meaning function. However, it also is a verb as in “f is mapping A to B ”. That which a function is thought of as doing is also referred to using the word “maps” as in: f maps X to Y . However, a set of functions may be called a set of maps so this word might also be used as the plural of a noun. There is no help for it. You just have to suffer with this nonsense. The following theorem which is interesting for its own sake will be used to prove the Schroder Bernstein theorem. Theorem 1.2.2 Let f : X → Y and g : Y → X be two functions. Then there exist sets A, B, C, D, such that A ∪ B = X, C ∪ D = Y, A ∩ B = ∅, C ∩ D = ∅, f (A) = C, g (D) = B. The following picture illustrates the conclusion of this theorem. X

Y A

B = g(D) ¾

f

g

- C = f (A)

D

1.2. THE SCHRODER BERNSTEIN THEOREM

23

Proof: Consider the empty set, ∅ ⊆ X. If y ∈ Y \ f (∅), then g (y) ∈ / ∅ because ∅ has no elements. Also, if A, B, C, and D are as described above, A also would have this same property that the empty set has. However, A is probably larger. Therefore, say A0 ⊆ X satisfies P if whenever y ∈ Y \ f (A0 ) , g (y) ∈ / A0 . A ≡ {A0 ⊆ X : A0 satisfies P}. Let A = ∪A. If y ∈ Y \ f (A), then for each A0 ∈ A, y ∈ Y \ f (A0 ) and so g (y) ∈ / A0 . Since g (y) ∈ / A0 for all A0 ∈ A, it follows g (y) ∈ / A. Hence A satisfies P and is the largest subset of X which does so. Now define C ≡ f (A) , D ≡ Y \ C, B ≡ X \ A. It only remains to verify that g (D) = B. Suppose x ∈ B = X \ A. Then A ∪ {x} does not satisfy P and so there exists y ∈ Y \ f (A ∪ {x}) ⊆ D such that g (y) ∈ A ∪ {x} . But y ∈ / f (A) and so since A satisfies P, it follows g (y) ∈ / A. Hence g (y) = x and so x ∈ g (D) and this proves the theorem. Theorem 1.2.3 (Schroder Bernstein) If f : X → Y and g : Y → X are one to one, then there exists h : X → Y which is one to one and onto. Proof: Let A, B, C, D be the sets of Theorem1.2.2 and define ½ f (x) if x ∈ A h (x) ≡ g −1 (x) if x ∈ B Then h is the desired one to one and onto mapping. Recall that the Cartesian product may be considered as the collection of choice functions. Definition 1.2.4 Let I be a set and let Xi be a set for each i ∈ I. f is a choice function written as Y f∈ Xi i∈I

if f (i) ∈ Xi for each i ∈ I. The axiom of choice says that if Xi 6= ∅ for each i ∈ I, for I a set, then Y Xi 6= ∅. i∈I

Sometimes the two functions, f and g are onto but not one to one. It turns out that with the axiom of choice, a similar conclusion to the above may be obtained. Corollary 1.2.5 If f : X → Y is onto and g : Y → X is onto, then there exists h : X → Y which is one to one and onto.

24

SET THEORY

Proof: For each y ∈ Y , f −1 (y)Q≡ {x ∈ X : f (x) = y} 6= ∅. Therefore, by the axiom of choice, there exists f0−1 ∈ y∈Y f −1 (y) which is the same as saying that for each y ∈ Y , f0−1 (y) ∈ f −1 (y). Similarly, there exists g0−1 (x) ∈ g −1 (x) for all x ∈ X. Then f0−1 is one to one because if f0−1 (y1 ) = f0−1 (y2 ), then ¡ ¢ ¡ ¢ y1 = f f0−1 (y1 ) = f f0−1 (y2 ) = y2 . Similarly g0−1 is one to one. Therefore, by the Schroder Bernstein theorem, there exists h : X → Y which is one to one and onto. Definition 1.2.6 A set S, is finite if there exists a natural number n and a map θ which maps {1, · · · , n} one to one and onto S. S is infinite if it is not finite. A set S, is called countable if there exists a map θ mapping N one to one and onto S.(When θ maps a set A to a set B, this will be written as θ : A → B in the future.) Here N ≡ {1, 2, · · · the natural numbers. S is at most countable if there exists a map θ : N →S which is onto. The property of being at most countable is often referred to as being countable because the question of interest is normally whether one can list all elements of the set, designating a first, second, third etc. in such a way as to give each element of the set a natural number. The possibility that a single element of the set may be counted more than once is often not important. Theorem 1.2.7 If X and Y are both at most countable, then X ×Y is also at most countable. If either X or Y is countable, then X × Y is also countable. Proof: It is given that there exists a mapping η : N → X which is onto. Define η (i) ≡ xi and consider X as the set {x1 , x2 , x3 , · · · }. Similarly, consider Y as the set {y1 , y2 , y3 , · · · }. It follows the elements of X × Y are included in the following rectangular array. (x1 , y1 ) (x1 , y2 ) (x1 , y3 ) · · · (x2 , y1 ) (x2 , y2 ) (x2 , y3 ) · · · (x3 , y1 ) (x3 , y2 ) (x3 , y3 ) · · · .. .. .. . . .

← Those which have x1 in first slot. ← Those which have x2 in first slot. ← Those which have x3 in first slot. . .. .

Follow a path through this array as follows. (x1 , y1 ) → (x1 , y2 ) . % (x2 , y1 ) (x2 , y2 ) ↓ % (x3 , y1 )

(x1 , y3 ) →

Thus the first element of X × Y is (x1 , y1 ), the second element of X × Y is (x1 , y2 ), the third element of X × Y is (x2 , y1 ) etc. This assigns a number from N to each element of X × Y. Thus X × Y is at most countable.

1.3. EQUIVALENCE RELATIONS

25

It remains to show the last claim. Suppose without loss of generality that X is countable. Then there exists α : N → X which is one to one and onto. Let β : X × Y → N be defined by β ((x, y)) ≡ α−1 (x). Thus β is onto N. By the first part there exists a function from N onto X × Y . Therefore, by Corollary 1.2.5, there exists a one to one and onto mapping from X × Y to N. This proves the theorem. Theorem 1.2.8 If X and Y are at most countable, then X ∪Y is at most countable. If either X or Y are countable, then X ∪ Y is countable. Proof: As in the preceding theorem, X = {x1 , x2 , x3 , · · · } and Y = {y1 , y2 , y3 , · · · } . Consider the following array consisting of X ∪ Y and path through it. x1 y1

→ . →

x2

x3

→

% y2

Thus the first element of X ∪ Y is x1 , the second is x2 the third is y1 the fourth is y2 etc. Consider the second claim. By the first part, there is a map from N onto X × Y . Suppose without loss of generality that X is countable and α : N → X is one to one and onto. Then define β (y) ≡ 1, for all y ∈ Y ,and β (x) ≡ α−1 (x). Thus, β maps X × Y onto N and this shows there exist two onto maps, one mapping X ∪ Y onto N and the other mapping N onto X ∪ Y . Then Corollary 1.2.5 yields the conclusion. This proves the theorem.

1.3

Equivalence Relations

There are many ways to compare elements of a set other than to say two elements are equal or the same. For example, in the set of people let two people be equivalent if they have the same weight. This would not be saying they were the same person, just that they weighed the same. Often such relations involve considering one characteristic of the elements of a set and then saying the two elements are equivalent if they are the same as far as the given characteristic is concerned. Definition 1.3.1 Let S be a set. ∼ is an equivalence relation on S if it satisfies the following axioms. 1. x ∼ x

for all x ∈ S. (Reflexive)

2. If x ∼ y then y ∼ x. (Symmetric)

26

SET THEORY

3. If x ∼ y and y ∼ z, then x ∼ z. (Transitive) Definition 1.3.2 [x] denotes the set of all elements of S which are equivalent to x and [x] is called the equivalence class determined by x or just the equivalence class of x. With the above definition one can prove the following simple theorem. Theorem 1.3.3 Let ∼ be an equivalence class defined on a set, S and let H denote the set of equivalence classes. Then if [x] and [y] are two of these equivalence classes, either x ∼ y and [x] = [y] or it is not true that x ∼ y and [x] ∩ [y] = ∅.

1.4

Partially Ordered Sets

Definition 1.4.1 Let F be a nonempty set. F is called a partially ordered set if there is a relation, denoted here by ≤, such that x ≤ x for all x ∈ F. If x ≤ y and y ≤ z then x ≤ z. C ⊆ F is said to be a chain if every two elements of C are related. This means that if x, y ∈ C, then either x ≤ y or y ≤ x. Sometimes a chain is called a totally ordered set. C is said to be a maximal chain if whenever D is a chain containing C, D = C. The most common example of a partially ordered set is the power set of a given set with ⊆ being the relation. It is also helpful to visualize partially ordered sets as trees. Two points on the tree are related if they are on the same branch of the tree and one is higher than the other. Thus two points on different branches would not be related although they might both be larger than some point on the trunk. You might think of many other things which are best considered as partially ordered sets. Think of food for example. You might find it difficult to determine which of two favorite pies you like better although you may be able to say very easily that you would prefer either pie to a dish of lard topped with whipped cream and mustard. The following theorem is equivalent to the axiom of choice. For a discussion of this, see the appendix on the subject. Theorem 1.4.2 (Hausdorff Maximal Principle) Let F ordered set. Then there exists a maximal chain.

be a nonempty partially

Continuous Functions Of One Variable There is a theorem about the integral of a continuous function which requires the notion of uniform continuity. This is discussed in this section. Consider the function f (x) = x1 for x ∈ (0, 1) . This is a continuous function because, it is continuous at every point of (0, 1) . However, for a given ε > 0, the δ needed in the ε, δ definition of continuity becomes very small as x gets close to 0. The notion of uniform continuity involves being able to choose a single δ which works on the whole domain of f. Here is the definition. Definition 2.0.3 Let f : D ⊆ R → R be a function. Then f is uniformly continuous if for every ε > 0, there exists a δ depending only on ε such that if |x − y| < δ then |f (x) − f (y)| < ε. It is an amazing fact that under certain conditions continuity implies uniform continuity. Definition 2.0.4 A set, K ⊆ R is sequentially compact if whenever {an } ⊆ K is a sequence, there exists a subsequence, {ank } such that this subsequence converges to a point of K. The following theorem is part of the Heine Borel theorem. Theorem 2.0.5 Every closed interval, [a, b] is sequentially compact. ¤ £ ¤ £ and a+b Proof: Let {xn } ⊆ [a, b] ≡ I0 . Consider the two intervals a, a+b 2 2 ,b each of which has length (b − a) /2. At least one of these intervals contains xn for infinitely many values of n. Call this interval I1 . Now do for I1 what was done for I0 . Split it in half and let I2 be the interval which contains xn for infinitely many values of n. Continue this way obtaining a sequence of nested intervals I0 ⊇ I1 ⊇ I2 ⊇ I3 · · · where the length of In is (b − a) /2n . Now pick n1 such that xn1 ∈ I1 , n2 such that n2 > n1 and xn2 ∈ I2 , n3 such that n3 > n2 and xn3 ∈ I3 , etc. (This can be done because in each case the intervals contained xn for infinitely many values of n.) By 27

28

CONTINUOUS FUNCTIONS OF ONE VARIABLE

the nested interval lemma there exists a point, c contained in all these intervals. Furthermore, |xnk − c| < (b − a) 2−k and so limk→∞ xnk = c ∈ [a, b] . This proves the theorem. Theorem 2.0.6 Let f : K → R be continuous where K is a sequentially compact set in R. Then f is uniformly continuous on K. Proof: If this is not true, there exists ε > 0 such that for every δ > 0 there exists a pair of points, xδ and yδ such that even though |xδ − yδ | < δ, |f (xδ ) − f (yδ )| ≥ ε. Taking a succession of values for δ equal to 1, 1/2, 1/3, · · · , and letting the exceptional pair of points for δ = 1/n be denoted by xn and yn , |xn − yn |
0 such that whenever |x − y| < δ 1 , it follows |f (x) − f (y)| < 2(|a|+|b|+1) and there exists δ 2 > 0 such that whenever |x − y| < δ 2 , it follows that |g (x) − g (y)| < ε 2(|a|+|b|+1) . Then let 0 < δ ≤ min (δ 1 , δ 2 ) . If |x − y| < δ, then everything happens at once. Therefore, using the triangle inequality |af (x) + bf (x) − (ag (y) + bg (y))| ≤ |a| |f (x) − f (y)| + |b| |g (x) − g (y)| µ ¶ µ ¶ ε ε < |a| + |b| < ε. 2 (|a| + |b| + 1) 2 (|a| + |b| + 1) Now consider 2.) There exists δ 1 > 0 such that if |y − x| < δ 1 , then |f (x) − f (y)| < 1. Therefore, for such y, |f (y)| < 1 + |f (x)| . It follows that for such y, |f g (x) − f g (y)| ≤ |f (x) g (x) − g (x) f (y)| + |g (x) f (y) − f (y) g (y)|

30


≤ |g (x)| |f (x) − f (y)| + |f (y)| |g (x) − g (y)| ≤ (1 + |g (x)| + |f (y)|) [|g (x) − g (y)| + |f (x) − f (y)|] . Now let ε > 0 be given. There exists δ 2 such that if |x − y| < δ 2 , then ε |g (x) − g (y)| < , 2 (1 + |g (x)| + |f (y)|) and there exists δ 3 such that if |x−y| < δ 3 , then |f (x) − f (y)|
0 be such that for |x−y| < δ 0 , |g (x) − g (y)| < |g (x)| /2 and so by the triangle inequality, − |g (x)| /2 ≤ |g (y)| − |g (x)| ≤ |g (x)| /2 which implies |g (y)| ≥ |g (x)| /2, and |g (y)| < 3 |g (x)| /2. Then if |x−y| < min (δ 0 , δ 1 ) , ¯ ¯ ¯ ¯ ¯ f (x) f (y) ¯ ¯ f (x) g (y) − f (y) g (x) ¯ ¯ ¯ ¯ ¯ ¯ g (x) − g (y) ¯ = ¯ ¯ g (x) g (y) |f (x) g (y) − f (y) g (x)| ³ ´ ≤ 2 |g(x)| 2

= ≤ ≤

2 2

|g (x)| 2

2 |f (x) g (y) − f (y) g (x)| |g (x)|

2

[|f (x) g (y) − f (y) g (y) + f (y) g (y) − f (y) g (x)|]

2 [|g (y)| |f (x) − f (y)| + |f (y)| |g (y) − g (x)|] |g (x)| · ¸ 3 2 |g (x)| |f (x) − f (y)| + (1 + |f (x)|) |g (y) − g (x)| ≤ 2 |g (x)| 2 2 ≤ 2 (1 + 2 |f (x)| + 2 |g (x)|) [|f (x) − f (y)| + |g (y) − g (x)|] |g (x)| ≡ M [|f (x) − f (y)| + |g (y) − g (x)|]

2.2. THEOREMS ABOUT CONTINUOUS FUNCTIONS

31

where M is defined by M≡

2 |g (x)|

2

(1 + 2 |f (x)| + 2 |g (x)|)

Now let δ 2 be such that if |x−y| < δ 2 , then |f (x) − f (y)|
0 such that if |y−f (x)| < η and y ∈ D (g) , it follows that |g (y) − g (f (x))| < ε. From continuity of f at x, there exists δ > 0 such that if |x−z| < δ and z ∈ D (f ) , then |f (z) − f (x)| < η. Then if |x−z| < δ and z ∈ D (g ◦ f ) ⊆ D (f ) , all the above hold and so |g (f (z)) − g (f (x))| < ε. This proves part 3.) To verify part 4.), let ε > 0 be given and let δ = ε. Then if |x−y| < δ, the triangle inequality implies |f (x) − f (y)| = ||x| − |y|| ≤ |x−y| < δ = ε. This proves part 4.) and completes the proof of the theorem. Next here is a proof of the intermediate value theorem. Theorem 2.2.2 Suppose f : [a, b] → R is continuous and suppose f (a) < c < f (b) . Then there exists x ∈ (a, b) such that f (x) = c. and consider the intervals [a, d] and [d, b] . If f (d) ≥ c, Proof: Let d = a+b 2 then on [a, d] , the function is ≤ c at one end point and ≥ c at the other. On the other hand, if f (d) ≤ c, then on [d, b] f ≥ 0 at one end point and ≤ 0 at the

32


other. Pick the interval on which f has values which are at least as large as c and values no larger than c. Now consider that interval, divide it in half as was done for the original interval and argue that on one of these smaller intervals, the function has values at least as large as c and values no larger than c. Continue in this way. Next apply the nested interval lemma to get x in all these intervals. In the nth interval, let xn , yn be elements of this interval such that f (xn ) ≤ c, f (yn ) ≥ c. Now |xn − x| ≤ (b − a) 2−n and |yn − x| ≤ (b − a) 2−n and so xn → x and yn → x. Therefore, f (x) − c = lim (f (xn ) − c) ≤ 0 n→∞

while f (x) − c = lim (f (yn ) − c) ≥ 0. n→∞

Consequently f (x) = c and this proves the theorem. Lemma 2.2.3 Let φ : [a, b] → R be a continuous function and suppose φ is 1 − 1 on (a, b). Then φ is either strictly increasing or strictly decreasing on [a, b] . Proof: First it is shown that φ is either strictly increasing or strictly decreasing on (a, b) . If φ is not strictly decreasing on (a, b), then there exists x1 < y1 , x1 , y1 ∈ (a, b) such that (φ (y1 ) − φ (x1 )) (y1 − x1 ) > 0. If for some other pair of points, x2 < y2 with x2 , y2 ∈ (a, b) , the above inequality does not hold, then since φ is 1 − 1, (φ (y2 ) − φ (x2 )) (y2 − x2 ) < 0. Let xt ≡ tx1 + (1 − t) x2 and yt ≡ ty1 + (1 − t) y2 . Then xt < yt for all t ∈ [0, 1] because tx1 ≤ ty1 and (1 − t) x2 ≤ (1 − t) y2 with strict inequality holding for at least one of these inequalities since not both t and (1 − t) can equal zero. Now define h (t) ≡ (φ (yt ) − φ (xt )) (yt − xt ) . Since h is continuous and h (0) < 0, while h (1) > 0, there exists t ∈ (0, 1) such that h (t) = 0. Therefore, both xt and yt are points of (a, b) and φ (yt ) − φ (xt ) = 0 contradicting the assumption that φ is one to one. It follows φ is either strictly increasing or strictly decreasing on (a, b) . This property of being either strictly increasing or strictly decreasing on (a, b) carries over to [a, b] by the continuity of φ. Suppose φ is strictly increasing on (a, b) , a similar argument holding for φ strictly decreasing on (a, b) . If x > a, then pick y ∈ (a, x) and from the above, φ (y) < φ (x) . Now by continuity of φ at a, φ (a) = lim φ (z) ≤ φ (y) < φ (x) . x→a+

Therefore, φ (a) < φ (x) whenever x ∈ (a, b) . Similarly φ (b) > φ (x) for all x ∈ (a, b). This proves the lemma.

2.2. THEOREMS ABOUT CONTINUOUS FUNCTIONS

33

Corollary 2.2.4 Let f : (a, b) → R be one to one and continuous. Then f (a, b) is an open interval, (c, d) and f −1 : (c, d) → (a, b) is continuous. Proof: Since f is either strictly increasing or strictly decreasing, it follows that f (a, b) is an open interval, (c, d) . Assume f is decreasing. Now let x ∈ (a, b). Why is f −1 is continuous at f (x)? Since f is decreasing, if f (x) < f (y) , then y ≡ f −1 (f (y)) < x ≡ f −1 (f (x)) and so f −1 is also decreasing. Let ε > 0 be given. Let ε > η > 0 and (x − η, x + η) ⊆ (a, b) . Then f (x) ∈ (f (x + η) , f (x − η)) . Let δ = min (f (x) − f (x + η) , f (x − η) − f (x)) . Then if |f (z) − f (x)| < δ, it follows so

z ≡ f −1 (f (z)) ∈ (x − η, x + η) ⊆ (x − ε, x + ε) ¯ −1 ¯ ¯ ¯ ¯f (f (z)) − x¯ = ¯f −1 (f (z)) − f −1 (f (x))¯ < ε.

This proves the theorem in the case where f is strictly decreasing. The case where f is increasing is similar.

34


The Riemann Stieltjes Integral The integral originated in attempts to find areas of various shapes and the ideas involved in finding integrals are much older than the ideas related to finding derivatives. In fact, Archimedes1 was finding areas of various curved shapes about 250 B.C. using the main ideas of the integral. What is presented here is a generalization of these ideas. The main interest is in the Riemann integral but if it is easy to generalize to the so called Stieltjes integral in which the length of an interval, [x, y] is replaced with an expression of the form F (y) − F (x) where F is an increasing function, then the generalization is given. However, there is much more that can be written about Stieltjes integrals than what is presented here. A good source for this is the book by Apostol, [3].

3.1

Upper And Lower Riemann Stieltjes Sums

The Riemann integral pertains to bounded functions which are defined on a bounded interval. Let [a, b] be a closed interval. A set of points in [a, b], {x0 , · · · , xn } is a partition if a = x0 < x1 < · · · < xn = b. Such partitions are denoted by P or Q. For f a bounded function defined on [a, b] , let Mi (f ) ≡ sup{f (x) : x ∈ [xi−1 , xi ]}, mi (f ) ≡ inf{f (x) : x ∈ [xi−1 , xi ]}. 1 Archimedes 287-212 B.C. found areas of curved regions by stuffing them with simple shapes which he knew the area of and taking a limit. He also made fundamental contributions to physics. The story is told about how he determined that a gold smith had cheated the king by giving him a crown which was not solid gold as had been claimed. He did this by finding the amount of water displaced by the crown and comparing with the amount of water it should have displaced if it had been solid gold.

35

36

THE RIEMANN STIELTJES INTEGRAL

Definition 3.1.1 Let F be an increasing function defined on [a, b] and let ∆Fi ≡ F (xi ) − F (xi−1 ) . Then define upper and lower sums as U (f, P ) ≡

n X

Mi (f ) ∆Fi and L (f, P ) ≡

i=1

n X

mi (f ) ∆Fi

i=1

respectively. The numbers, Mi (f ) and mi (f ) , are well defined real numbers because f is assumed to be bounded and R is complete. Thus the set S = {f (x) : x ∈ [xi−1 , xi ]} is bounded above and below. In the following picture, the sum of the areas of the rectangles in the picture on the left is a lower sum for the function in the picture and the sum of the areas of the rectangles in the picture on the right is an upper sum for the same function which uses the same partition. In these pictures the function, F is given by F (x) = x and these are the ordinary upper and lower sums from calculus.

y = f (x)

x0

x1

x2

x3

x0

x1

x2

x3

What happens when you add in more points in a partition? The following pictures illustrate in the context of the above example. In this example a single additional point, labeled z has been added in.

y = f (x)

x0

x1

x2

z

x3

x0

x1

x2

z

x3

Note how the lower sum got larger by the amount of the area in the shaded rectangle and the upper sum got smaller by the amount in the rectangle shaded by dots. In general this is the way it works and this is shown in the following lemma. Lemma 3.1.2 If P ⊆ Q then U (f, Q) ≤ U (f, P ) , and L (f, P ) ≤ L (f, Q) .

3.1. UPPER AND LOWER RIEMANN STIELTJES SUMS

37

Proof: This is verified by adding in one point at a time. Thus let P = {x0 , · · · , xn } and let Q = {x0 , · · · , xk , y, xk+1 , · · · , xn }. Thus exactly one point, y, is added between xk and xk+1 . Now the term in the upper sum which corresponds to the interval [xk , xk+1 ] in U (f, P ) is sup {f (x) : x ∈ [xk , xk+1 ]} (F (xk+1 ) − F (xk ))

(3.1.1)

and the term which corresponds to the interval [xk , xk+1 ] in U (f, Q) is sup {f (x) : x ∈ [xk , y]} (F (y) − F (xk )) + sup {f (x) : x ∈ [y, xk+1 ]} (F (xk+1 ) − F (y))

(3.1.2)

≡ M1 (F (y) − F (xk )) + M2 (F (xk+1 ) − F (y)) All the other terms in the two sums coincide. Now sup {f (x) : x ∈ [xk , xk+1 ]} ≥ max (M1 , M2 ) and so the expression in 3.1.2 is no larger than sup {f (x) : x ∈ [xk , xk+1 ]} (F (xk+1 ) − F (y)) + sup {f (x) : x ∈ [xk , xk+1 ]} (F (y) − F (xk )) = sup {f (x) : x ∈ [xk , xk+1 ]} (F (xk+1 ) − F (xk )) , the term corresponding to the interval, [xk , xk+1 ] and U (f, P ) . This proves the first part of the lemma pertaining to upper sums because if Q ⊇ P, one can obtain Q from P by adding in one point at a time and each time a point is added, the corresponding upper sum either gets smaller or stays the same. The second part about lower sums is similar and is left as an exercise. Lemma 3.1.3 If P and Q are two partitions, then L (f, P ) ≤ U (f, Q) . Proof: By Lemma 3.1.2, L (f, P ) ≤ L (f, P ∪ Q) ≤ U (f, P ∪ Q) ≤ U (f, Q) . Definition 3.1.4 I ≡ inf{U (f, Q) where Q is a partition} I ≡ sup{L (f, P ) where P is a partition}. Note that I and I are well defined real numbers.

38


Theorem 3.1.5 I ≤ I. Proof: From Lemma 3.1.3, I = sup{L (f, P ) where P is a partition} ≤ U (f, Q) because U (f, Q) is an upper bound to the set of all lower sums and so it is no smaller than the least upper bound. Therefore, since Q is arbitrary, I = sup{L (f, P ) where P is a partition} ≤ inf{U (f, Q) where Q is a partition} ≡ I where the inequality holds because it was just shown that I is a lower bound to the set of all upper sums and so it is no larger than the greatest lower bound of this set. This proves the theorem. Definition 3.1.6 A bounded function f is Riemann Stieltjes integrable, written as f ∈ R ([a, b]) if I=I and in this case,

Z

b

f (x) dF ≡ I = I. a

When F (x) = x, the integral is called the Riemann integral and is written as Z b f (x) dx. a

Thus, in words, the Riemann integral is the unique number which lies between all upper sums and all lower sums if there is such a unique number. Recall the following Proposition which comes from the definitions. Proposition 3.1.7 Let S be a nonempty set and suppose sup (S) exists. Then for every δ > 0, S ∩ (sup (S) − δ, sup (S)] 6= ∅. If inf (S) exists, then for every δ > 0, S ∩ [inf (S) , inf (S) + δ) 6= ∅. This proposition implies the following theorem which is used to determine the question of Riemann Stieltjes integrability. Theorem 3.1.8 A bounded function f is Riemann integrable if and only if for all ε > 0, there exists a partition P such that U (f, P ) − L (f, P ) < ε.

(3.1.3)

3.2. EXERCISES

39

Proof: First assume f is Riemann integrable. Then let P and Q be two partitions such that U (f, Q) I − ε/2. Then since I = I, U (f, Q ∪ P ) − L (f, P ∪ Q) ≤ U (f, Q) − L (f, P ) 0 there exists a partition such that 3.1.3 holds. Then for given ε and partition P corresponding to ε I − I ≤ U (f, P ) − L (f, P ) ≤ ε. Since ε is arbitrary, this shows I = I and this proves the theorem. The condition described in the theorem is called the Riemann criterion . Not all bounded functions are Riemann integrable. For example, let F (x) = x and ½ 1 if x ∈ Q f (x) ≡ (3.1.4) 0 if x ∈ R \ Q Then if [a, b] = [0, 1] all upper sums for f equal 1 while all lower sums for f equal 0. Therefore the Riemann criterion is violated for ε = 1/2.

3.2

Exercises

1. Prove the second half of Lemma 3.1.2 about lower sums. 2. Verify that for f given in 3.1.4, the lower sums on the interval [0, 1] are all equal to zero while the upper sums are all equal to one. © ª 3. Let f (x) = 1 + x2 for x ∈ [−1, 3] and let P = −1, − 31 , 0, 21 , 1, 2 . Find U (f, P ) and L (f, P ) for F (x) = x and for F (x) = x3 . 4. Show that if f ∈ R ([a, b]) for F (x) = x, there exists a partition, {x0 , · · · , xn } such that for any zk ∈ [xk , xk+1 ] , ¯Z ¯ n ¯ b ¯ X ¯ ¯ f (x) dx − f (zk ) (xk − xk−1 )¯ < ε ¯ ¯ a ¯ k=1

Pn

This sum, k=1 f (zk ) (xk − xk−1 ) , is called a Riemann sum and this exercise shows that the Riemann integral can always be approximated by a Riemann sum. For the general Riemann Stieltjes case, does anything change? ª © 5. Let P = 1, 1 14 , 1 12 , 1 34 , 2 and F (x) = x. Find upper and lower sums for the function, f (x) = x1 using this partition. What does this tell you about ln (2)? 6. If f ∈ R ([a, b]) with F (x) = x and f is changed at finitely many points, show the new function is also in R ([a, b]) . Is this still true for the general case where F is only assumed to be an increasing function? Explain.

40


7. In the case where F (x) = x, define a “left sum” as n X

f (xk−1 ) (xk − xk−1 )

k=1

and a “right sum”,

n X

f (xk ) (xk − xk−1 ) .

k=1

Also suppose that all partitions have the property that xk − xk−1 equals a constant, (b − a) /n so the points in the partition are equally spaced, and define the integral to be the number these right and left sums R x get close to as n gets larger and larger. Show that for f given in 3.1.4, 0 f (t) dt = 1 if x Rx is rational and 0 f (t) dt = 0 if x is irrational. It turns out that the correct answer should always equal zero for that function, regardless of whether x is rational. This is shown when the Lebesgue integral is studied. This illustrates why this method of defining the integral in terms of left and right sums is total nonsense. Show that even though this is the case, it makes no difference if f is continuous.

3.3

Functions Of Riemann Integrable Functions

It is often necessary to consider functions of Riemann integrable functions and a natural question is whether these are Riemann integrable. The following theorem gives a partial answer to this question. This is not the most general theorem which will relate to this question but it will be enough for the needs of this book. Theorem 3.3.1 Let f, g be bounded functions and let f ([a, b]) ⊆ [c1 , d1 ] , g ([a, b]) ⊆ [c2 , d2 ] . Let H : [c1 , d1 ] × [c2 , d2 ] → R satisfy, |H (a1 , b1 ) − H (a2 , b2 )| ≤ K [|a1 − a2 | + |b1 − b2 |] for some constant K. Then if f, g ∈ R ([a, b]) it follows that H ◦ (f, g) ∈ R ([a, b]) . Proof: In the following claim, Mi (h) and mi (h) have the meanings assigned above with respect to some partition of [a, b] for the function, h. Claim: The following inequality holds. |Mi (H ◦ (f, g)) − mi (H ◦ (f, g))| ≤ K [|Mi (f ) − mi (f )| + |Mi (g) − mi (g)|] . Proof of the claim: By the above proposition, there exist x1 , x2 ∈ [xi−1 , xi ] be such that H (f (x1 ) , g (x1 )) + η > Mi (H ◦ (f, g)) ,

3.3. FUNCTIONS OF RIEMANN INTEGRABLE FUNCTIONS

41

and H (f (x2 ) , g (x2 )) − η < mi (H ◦ (f, g)) . Then |Mi (H ◦ (f, g)) − mi (H ◦ (f, g))| < 2η + |H (f (x1 ) , g (x1 )) − H (f (x2 ) , g (x2 ))| < 2η + K [|f (x1 ) − f (x2 )| + |g (x1 ) − g (x2 )|] ≤ 2η + K [|Mi (f ) − mi (f )| + |Mi (g) − mi (g)|] . Since η > 0 is arbitrary, this proves the claim. Now continuing with the proof of the theorem, let P be such that n X

(Mi (f ) − mi (f )) ∆Fi
0 be given and let µ ¶ b−a xi = a + i , i = 0, · · · , n. n Since F is continuous, it follows from Corollary 2.0.7 on Page 28 that it is uniformly continuous. Therefore, if n is large enough, then for all i, F (xi ) − F (xi−1 )
0 is given, there exists a δ > 0 such that if |xi − xi−1 | < δ, then Mi − mi < F (b)−Fε (a)+1 . Let P ≡ {x0 , · · · , xn } be a partition with |xi − xi−1 | < δ. Then U (f, P ) − L (f, P )
−f (x) dF + Mi (f ) ∆Fi ≥ a

n X

Z

b

b

−f (x) dF +

a

i=1

Mi (f ) ∆Fi

i=1

f (x) dF. a

Thus, since ε is arbitrary, Z

Z

b

−f (x) dF ≤ −

f (x) dF

a

a

whenever f ∈ R ([a, b]) . It follows Z b Z b Z −f (x) dF ≤ − f (x) dF = − a

a

and this proves the lemma.

b

Z

b

− (−f (x)) dF ≤ a

b

−f (x) dF a

44


Theorem 3.4.3 The integral is linear, Z

Z

b

(αf + βg) (x) dF = α

Z

b

b

f (x) dF + β

a

a

g (x) dF. a

whenever f, g ∈ R ([a, b]) and α, β ∈ R. Proof: First note that by Theorem 3.3.1, αf + βg ∈ R ([a, b]) . To begin with, consider the claim that if f, g ∈ R ([a, b]) then Z

Z

b

(f + g) (x) dF = a

Z

b

f (x) dF + a

b

g (x) dF.

(3.4.7)

a

Let P1 ,Q1 be such that U (f, Q1 ) − L (f, Q1 ) < ε/2, U (g, P1 ) − L (g, P1 ) < ε/2. Then letting P ≡ P1 ∪ Q1 , Lemma 3.1.2 implies U (f, P ) − L (f, P ) < ε/2, and U (g, P ) − U (g, P ) < ε/2. Next note that mi (f + g) ≥ mi (f ) + mi (g) , Mi (f + g) ≤ Mi (f ) + Mi (g) . Therefore, L (g + f, P ) ≥ L (f, P ) + L (g, P ) , U (g + f, P ) ≤ U (f, P ) + U (g, P ) . For this partition, Z

b

(f + g) (x) dF ∈ [L (f + g, P ) , U (f + g, P )] a

⊆ [L (f, P ) + L (g, P ) , U (f, P ) + U (g, P )] and Z

Z

b

f (x) dF + a

b

g (x) dF ∈ [L (f, P ) + L (g, P ) , U (f, P ) + U (g, P )] . a

Therefore, ¯Z ÃZ !¯ Z b ¯ b ¯ b ¯ ¯ (f + g) (x) dF − f (x) dF + g (x) dF ¯ ≤ ¯ ¯ a ¯ a a U (f, P ) + U (g, P ) − (L (f, P ) + L (g, P )) < ε/2 + ε/2 = ε. This proves 3.4.7 since ε is arbitrary.

3.4. PROPERTIES OF THE INTEGRAL It remains to show that

Z

45

Z

b

α

f (x) dF = a

b

αf (x) dF. a

Suppose first that α ≥ 0. Then Z b αf (x) dF ≡ sup{L (αf, P ) : P is a partition} = a

Z

b

α sup{L (f, P ) : P is a partition} ≡ α

f (x) dF. a

If α < 0, then this and Lemma 3.4.2 imply Z b Z b αf (x) dF = (−α) (−f (x)) dF a

a

Z

Z

b

= (−α)

b

(−f (x)) dF = α a

f (x) dF. a

This proves the theorem. In the next theorem, suppose F is defined on [a, b] ∪ [b, c] . Theorem 3.4.4 If f ∈ R ([a, b]) and f ∈ R ([b, c]) , then f ∈ R ([a, c]) and Z c Z b Z c f (x) dF = f (x) dF + f (x) dF. (3.4.8) a

a

b

Proof: Let P1 be a partition of [a, b] and P2 be a partition of [b, c] such that U (f, Pi ) − L (f, Pi ) < ε/2, i = 1, 2. Let P ≡ P1 ∪ P2 . Then P is a partition of [a, c] and U (f, P ) − L (f, P ) = U (f, P1 ) − L (f, P1 ) + U (f, P2 ) − L (f, P2 ) < ε/2 + ε/2 = ε.

(3.4.9)

Thus, f ∈ R ([a, c]) by the Riemann criterion and also for this partition, Z b Z c f (x) dF + f (x) dF ∈ [L (f, P1 ) + L (f, P2 ) , U (f, P1 ) + U (f, P2 )] a

b

= [L (f, P ) , U (f, P )] and

Z

c

f (x) dF ∈ [L (f, P ) , U (f, P )] . a

Hence by 3.4.9, ¯Z ÃZ !¯ Z c ¯ c ¯ b ¯ ¯ f (x) dF − f (x) dF + f (x) dF ¯ < U (f, P ) − L (f, P ) < ε ¯ ¯ a ¯ a b which shows that since ε is arbitrary, 3.4.8 holds. This proves the theorem.

46


Corollary 3.4.5 Let F be continuous and let [a, b] be a closed and bounded interval and suppose that a = y1 < y2 · · · < yl = b and that f is a bounded function defined on [a, b] which has the property that f is either increasing on [yj , yj+1 ] or decreasing on [yj , yj+1 ] for j = 1, · · · , l − 1. Then f ∈ R ([a, b]) . Proof: This Rfollows from Theorem 3.4.4 and Theorem 3.3.2. b The symbol, a f (x) dF when a > b has not yet been defined. Definition 3.4.6 Let [a, b] be an interval and let f ∈ R ([a, b]) . Then Z

Z

a

b

f (x) dF ≡ − b

f (x) dF. a

Note that with this definition, Z a Z f (x) dF = − a

and so

a

f (x) dF

a

Z

a

f (x) dF = 0. a

Theorem 3.4.7 Assuming all the integrals make sense, Z

Z

b a

Z

c

f (x) dF +

c

f (x) dF = b

f (x) dF. a

Proof: This follows from Theorem 3.4.4 and Definition 3.4.6. For example, assume c ∈ (a, b) . Then from Theorem 3.4.4, Z

Z

c

f (x) dF + a

Z

b

b

f (x) dF =

f (x) dF

c

a

and so by Definition 3.4.6, Z

Z

c

f (x) dF = a

b

f (x) dF − a

Z =

f (x) dF c

Z

b

f (x) dF + a

The other cases are similar.

Z

b

c

f (x) dF. b

3.5. FUNDAMENTAL THEOREM OF CALCULUS

47

The following properties of the integral have either been established or they follow quickly from what has been shown so far. If f ∈ R ([a, b]) then if c ∈ [a, b] , f ∈ R ([a, c]) , Z

b

α dF = α (F (b) − F (a)) , Z

a

Z

b

Z

f (x) dF + a

Z a

a

Z

c

f (x) dF = b

(3.4.11)

b

f (x) dF + β

Z

b

Z

b

(αf + βg) (x) dF = α a

(3.4.10)

g (x) dF,

(3.4.12)

a c

f (x) dF,

(3.4.13)

a

b

f (x) dF ≥ 0 if f (x) ≥ 0 and a < b,

(3.4.14)

¯Z ¯ ¯Z ¯ ¯ b ¯ ¯ b ¯ ¯ ¯ ¯ ¯ f (x) dF ¯ ≤ ¯ |f (x)| dF ¯ . ¯ ¯ a ¯ ¯ a ¯

(3.4.15)

The only one of these claims which may not be completely obvious is the last one. To show this one, note that |f (x)| − f (x) ≥ 0, |f (x)| + f (x) ≥ 0. Therefore, by 3.4.14 and 3.4.12, if a < b, Z

Z

b

b

|f (x)| dF ≥ a

and

Z

f (x) dF a

Z

b

b

|f (x)| dF ≥ − a

Therefore,

Z a

f (x) dF. a

b

¯Z ¯ ¯ b ¯ ¯ ¯ |f (x)| dF ≥ ¯ f (x) dF ¯ . ¯ a ¯

If b < a then the above inequality holds with a and b switched. This implies 3.4.15.

3.5

Fundamental Theorem Of Calculus

In this section F (x) = x so things are specialized to the ordinary Riemann integral. With these properties, it is easy to prove the fundamental theorem of calculus2 . 2 This theorem is why Newton and Liebnitz are credited with inventing calculus. The integral had been around for thousands of years and the derivative was by their time well known. However the connection between these two ideas had not been fully made although Newton’s predecessor, Isaac Barrow had made some progress in this direction.

48


Let f ∈ R ([a, b]) . Then by 3.4.10 f ∈ R ([a, x]) for each x ∈ [a, b] . The first version of the fundamental theorem of calculus is a statement about the derivative of the function Z x x→

f (t) dt. a

Theorem 3.5.1 Let f ∈ R ([a, b]) and let Z x F (x) ≡ f (t) dt. a

Then if f is continuous at x ∈ (a, b) , F 0 (x) = f (x) . Proof: Let x ∈ (a, b) be a point of continuity of f and let h be small enough that x + h ∈ [a, b] . Then by using 3.4.13, Z h−1 (F (x + h) − F (x)) = h−1

x+h

f (t) dt. x

Also, using 3.4.11,

Z f (x) = h−1

x+h

f (x) dt. x

Therefore, by 3.4.15, ¯ ¯ Z ¯ ¯ −1 ¯ ¯¯ −1 x+h ¯ ¯h (F (x + h) − F (x)) − f (x)¯ = ¯h (f (t) − f (x)) dt¯ ¯ ¯ x ¯ ¯ Z x+h ¯ ¯ ¯ −1 ¯ ≤ ¯h |f (t) − f (x)| dt¯ . ¯ ¯ x Let ε > 0 and let δ > 0 be small enough that if |t − x| < δ, then |f (t) − f (x)| < ε. Therefore, if |h| < δ, the above inequality and 3.4.11 shows that ¯ −1 ¯ ¯h (F (x + h) − F (x)) − f (x)¯ ≤ |h|−1 ε |h| = ε. Since ε > 0 is arbitrary, this shows lim h−1 (F (x + h) − F (x)) = f (x)

h→0

and this proves the theorem. Note this gives existence for the initial value problem, F 0 (x) = f (x) , F (a) = 0

3.5. FUNDAMENTAL THEOREM OF CALCULUS

49

whenever f is Riemann integrable and continuous.3 The next theorem is also called the fundamental theorem of calculus. Theorem 3.5.2 Let f ∈ R ([a, b]) and suppose there exists an antiderivative for f, G, such that G0 (x) = f (x) for every point of (a, b) and G is continuous on [a, b] . Then Z

b

f (x) dx = G (b) − G (a) .

(3.5.16)

a

Proof: Let P = {x0 , · · · , xn } be a partition satisfying U (f, P ) − L (f, P ) < ε. Then G (b) − G (a) = G (xn ) − G (x0 ) n X = G (xi ) − G (xi−1 ) . i=1

By the mean value theorem, G (b) − G (a) =

n X

G0 (zi ) (xi − xi−1 )

i=1

=

n X

f (zi ) ∆xi

i=1

where zi is some point in [xi−1 , xi ] . It follows, since the above sum lies between the upper and lower sums, that G (b) − G (a) ∈ [L (f, P ) , U (f, P )] , and also

Z

b

f (x) dx ∈ [L (f, P ) , U (f, P )] . a

Therefore, ¯ ¯ Z b ¯ ¯ ¯ ¯ f (x) dx¯ 0 is arbitrary, 3.5.16 holds. This proves the theorem. 3 Of course it was proved that if f is continuous on a closed interval, [a, b] , then f ∈ R ([a, b]) but this is a hard theorem using the difficult result about uniform continuity.

50


The following notation is often used in this context. Suppose F is an antiderivative of f as just described with F continuous on [a, b] and F 0 = f on (a, b) . Then Z

b a

f (x) dx = F (b) − F (a) ≡ F (x) |ba .

Definition 3.5.3 Let f be a bounded function defined on a closed interval [a, b] and let P ≡ {x0 , · · · , xn } be a partition of the interval. Suppose zi ∈ [xi−1 , xi ] is chosen. Then the sum n X f (zi ) (xi − xi−1 ) i=1

is known as a Riemann sum. Also, ||P || ≡ max {|xi − xi−1 | : i = 1, · · · , n} . Proposition 3.5.4 Suppose f ∈ R ([a, b]) . Then there exists a partition, P ≡ {x0 , · · · , xn } with the property that for any choice of zk ∈ [xk−1 , xk ] , ¯Z ¯ n ¯ b ¯ X ¯ ¯ f (x) dx − f (zk ) (xk − xk−1 )¯ < ε. ¯ ¯ a ¯ k=1

Rb Proof: Choose P such that U (f, P ) − L (f, P ) < ε and then both a f (x) dx Pn and k=1 f (zk ) (xk − xk−1 ) are contained in [L (f, P ) , U (f, P )] and so the claimed inequality must hold. This proves the proposition. It is significant because it gives a way of approximating the integral. The definition of Riemann integrability given in this chapter is also called Darboux integrability and the integral defined as the unique number which lies between all upper sums and all lower sums which is given in this chapter is called the Darboux integral . The definition of the Riemann integral in terms of Riemann sums is given next. Definition 3.5.5 A bounded function, f defined on [a, b] is said to be Riemann integrable if there exists a number, I with the property that for every ε > 0, there exists δ > 0 such that if P ≡ {x0 , x1 , · · · , xn } is any partition having ||P || < δ, and zi ∈ [xi−1 , xi ] , ¯ ¯ n ¯ ¯ X ¯ ¯ f (zi ) (xi − xi−1 )¯ < ε. ¯I − ¯ ¯ i=1

The number

Rb a

f (x) dx is defined as I.

Thus, there are two definitions of the Riemann integral. It turns out they are equivalent which is the following theorem of of Darboux.

3.6. EXERCISES

51

Theorem 3.5.6 A bounded function defined on [a, b] is Riemann integrable in the sense of Definition 3.5.5 if and only if it is integrable in the sense of Darboux. Furthermore the two integrals coincide. The proof of this theorem is left for the exercises in Problems 10 - 12. It isn’t essential that you understand this theorem so if it does not interest you, leave it out. Note that it implies that given a Riemann integrable function f in either sense, it can be approximated by Riemann sums whenever ||P || is sufficiently small. Both versions of the integral are obsolete but entirely adequate for most applications and as a point of departure for a more up to date and satisfactory integral. The reason for using the Darboux approach to the integral is that all the existence theorems are easier to prove in this context.

3.6

Exercises

1. Let F (x) = 2. Let F (x) = it does.

R x3

t5 +7 x2 t7 +87t6 +1

Rx

1 2 1+t4

dt. Find F 0 (x) .

dt. Sketch a graph of F and explain why it looks the way

3. Let a and b be positive numbers and consider the function, Z

ax

F (x) = 0

1 dt + a2 + t2

Z b

a/x

1 dt. a2 + t2

Show that F is a constant. 4. Solve the following initial value problem from ordinary differential equations which is to find a function y such that y 0 (x) =

x6

x7 + 1 , y (10) = 5. + 97x5 + 7

R 5. If F, G ∈ f (x) dx for all x ∈ R, show F (x) = G (x) + C for some constant, C. Use this to give a different proof of the fundamental theorem of calculus Rb which has for its conclusion a f (t) dt = G (b) − G (a) where G0 (x) = f (x) . 6. Suppose f is Riemann integrable on [a, b] and continuous. (In fact continuous implies Riemann integrable.) Show there exists c ∈ (a, b) such that 1 f (c) = b−a

Z

b

f (x) dx. a

Rx Hint: You might consider the function F (x) ≡ a f (t) dt and use the mean value theorem for derivatives and the fundamental theorem of calculus.

52


7. Suppose f and g are continuous functions on [a, b] and that g (x) 6= 0 on (a, b) . Show there exists c ∈ (a, b) such that Z f (c)

Z

b

g (x) dx = a

b

f (x) g (x) dx. a

Rx Rx Hint: Define F (x) ≡ a f (t) g (t) dt and let G (x) ≡ a g (t) dt. Then use the Cauchy mean value theorem on these two functions. 8. Consider the function ½ f (x) ≡

¡ ¢ sin x1 if x 6= 0 . 0 if x = 0

Is f Riemann integrable? Explain why or why not. 9. Prove the second part of Theorem 3.3.2 about decreasing functions. 10. Suppose f is a bounded function defined on [a, b] and |f (x)| < M for all x ∈ [a, b] . Now let Q be a partition having n points, {x∗0 , · · · , x∗n } and let P be any other partition. Show that |U (f, P ) − L (f, P )| ≤ 2M n ||P || + |U (f, Q) − L (f, Q)| . Hint: Write the sum for U (f, P ) − L (f, P ) and split this sum into two sums, the sum of terms for which [xi−1 , xi ] contains at least one point of Q, and terms for which [xi−1 , xi ] does not contain any£points of¤Q. In the latter case, [xi−1 , xi ] must be contained in some interval, x∗k−1 , x∗k . Therefore, the sum of these terms should be no larger than |U (f, Q) − L (f, Q)| . 11. ↑ If ε > 0 is given and f is a Darboux integrable function defined on [a, b], show there exists δ > 0 such that whenever ||P || < δ, then |U (f, P ) − L (f, P )| < ε. 12. ↑ Prove Theorem 3.5.6.

Some Important Linear Algebra This chapter contains some important linear algebra as distinguished from that which is normally presented in undergraduate courses consisting mainly of uninteresting things you can do with row operations. The notation, Cn refers to the collection of ordered lists of n complex numbers. Since every real number is also a complex number, this simply generalizes the usual notion of Rn , the collection of all ordered lists of n real numbers. In order to avoid worrying about whether it is real or complex numbers which are being referred to, the symbol F will be used. If it is not clear, always pick C. Definition 4.0.1 Define Fn ≡ {(x1 , · · · , xn ) : xj ∈ F for j = 1, · · · , n} . (x1 , · · · , xn ) = (y1 , · · · , yn ) if and only if for all j = 1, · · · , n, xj = yj . When (x1 , · · · , xn ) ∈ Fn , it is conventional to denote (x1 , · · · , xn ) by the single bold face letter, x. The numbers, xj are called the coordinates. The set {(0, · · · , 0, t, 0, · · · , 0) : t ∈ F} for t in the ith slot is called the ith coordinate axis. The point 0 ≡ (0, · · · , 0) is called the origin. Thus (1, 2, 4i) ∈ F3 and (2, 1, 4i) ∈ F3 but (1, 2, 4i) 6= (2, 1, 4i) because, even though the same numbers are involved, they don’t match up. In particular, the first entries are not equal. The geometric significance of Rn for n ≤ 3 has been encountered already in calculus or in precalculus. Here is a short review. First consider the case when n = 1. Then from the definition, R1 = R. Recall that R is identified with the points of a line. Look at the number line again. Observe that this amounts to 53

54

SOME IMPORTANT LINEAR ALGEBRA

identifying a point on this line with a real number. In other words a real number determines where you are on this line. Now suppose n = 2 and consider two lines which intersect each other at right angles as shown in the following picture. · (2, 6)

6 (−8, 3) ·

3 2

−8 Notice how you can identify a point shown in the plane with the ordered pair, (2, 6) . You go to the right a distance of 2 and then up a distance of 6. Similarly, you can identify another point in the plane with the ordered pair (−8, 3) . Go to the left a distance of 8 and then up a distance of 3. The reason you go to the left is that there is a − sign on the eight. From this reasoning, every ordered pair determines a unique point in the plane. Conversely, taking a point in the plane, you could draw two lines through the point, one vertical and the other horizontal and determine unique points, x1 on the horizontal line in the above picture and x2 on the vertical line in the above picture, such that the point of interest is identified with the ordered pair, (x1 , x2 ) . In short, points in the plane can be identified with ordered pairs similar to the way that points on the real line are identified with real numbers. Now suppose n = 3. As just explained, the first two coordinates determine a point in a plane. Letting the third component determine how far up or down you go, depending on whether this number is positive or negative, this determines a point in space. Thus, (1, 4, −5) would mean to determine the point in the plane that goes with (1, 4) and then to go below this plane a distance of 5 to obtain a unique point in space. You see that the ordered triples correspond to points in space just as the ordered pairs correspond to points in a plane and single real numbers correspond to points on a line. You can’t stop here and say that you are only interested in n ≤ 3. What if you were interested in the motion of two objects? You would need three coordinates to describe where the first object is and you would need another three coordinates to describe where the other object is located. Therefore, you would need to be considering R6 . If the two objects moved around, you would need a time coordinate as well. As another example, consider a hot object which is cooling and suppose you want the temperature of this object. How many coordinates would be needed? You would need one for the temperature, three for the position of the point in the object and one more for the time. Thus you would need to be considering R5 . Many other examples can be given. Sometimes n is very large. This is often the case in applications to business when they are trying to maximize profit subject to constraints. It also occurs in numerical analysis when people try to solve hard problems on a computer.

4.1. ALGEBRA IN FN

55

There are other ways to identify points in space with three numbers but the one presented is the most basic. In this case, the coordinates are known as Cartesian coordinates after Descartes1 who invented this idea in the first half of the seventeenth century. I will often not bother to draw a distinction between the point in n dimensional space and its Cartesian coordinates. The geometric significance of Cn for n > 1 is not available because each copy of C corresponds to the plane or R2 .

4.1

Algebra in Fn

There are two algebraic operations done with elements of Fn . One is addition and the other is multiplication by numbers, called scalars. In the case of Cn the scalars are complex numbers while in the case of Rn the only allowed scalars are real numbers. Thus, the scalars always come from F in either case. Definition 4.1.1 If x ∈ Fn and a ∈ F, also called a scalar, then ax ∈ Fn is defined by ax = a (x1 , · · · , xn ) ≡ (ax1 , · · · , axn ) . (4.1.1) This is known as scalar multiplication. If x, y ∈ Fn then x + y ∈ Fn and is defined by x + y = (x1 , · · · , xn ) + (y1 , · · · , yn ) ≡ (x1 + y1 , · · · , xn + yn )

(4.1.2)

With this definition, the algebraic properties satisfy the conclusions of the following theorem. Theorem 4.1.2 For v, w ∈ Fn and α, β scalars, (real numbers), the following hold. v + w = w + v,

(4.1.3)

(v + w) + z = v+ (w + z) ,

(4.1.4)

the commutative law of addition,

the associative law for addition, v + 0 = v,

(4.1.5)

the existence of an additive identity, v+ (−v) = 0,

(4.1.6)

1 Ren´ e Descartes 1596-1650 is often credited with inventing analytic geometry although it seems the ideas were actually known much earlier. He was interested in many different subjects, physiology, chemistry, and physics being some of them. He also wrote a large book in which he tried to explain the book of Genesis scientifically. Descartes ended up dying in Sweden.

56


the existence of an additive inverse, Also α (v + w) = αv+αw,

(4.1.7)

(α + β) v =αv+βv,

(4.1.8)

α (βv) = αβ (v) ,

(4.1.9)

1v = v.

(4.1.10)

In the above 0 = (0, · · · , 0). You should verify these properties all hold. For example, consider 4.1.7 α (v + w) = α (v1 + w1 , · · · , vn + wn ) = (α (v1 + w1 ) , · · · , α (vn + wn )) = (αv1 + αw1 , · · · , αvn + αwn ) = (αv1 , · · · , αvn ) + (αw1 , · · · , αwn ) = αv + αw. As usual subtraction is defined as x − y ≡ x+ (−y) .

4.2

Exercises

1. Verify all the properties 4.1.3-4.1.10. 2. Compute 5 (1, 2 + 3i, 3, −2) + 6 (2 − i, 1, −2, 7) . 3. Draw a picture of the points in R2 which are determined by the following ordered pairs. (a) (1, 2) (b) (−2, −2) (c) (−2, 3) (d) (2, −5) 4. Does it make sense to write (1, 2) + (2, 3, 1)? Explain. 5. Draw a picture of the points in R3 which are determined by the following ordered triples. (a) (1, 2, 0) (b) (−2, −2, 1) (c) (−2, 3, −2)

4.3. SUBSPACES SPANS AND BASES

4.3

57

Subspaces Spans And Bases

Definition 4.3.1 Let {x1 , · · · , xp } be vectors in Fn . A linear combination is any expression of the form p X c i xi i=1

where the ci are scalars. The set of all linear combinations of these vectors is called span (x1 , · · · , xn ) . If V ⊆ Fn , then V is called a subspace if whenever α, β are scalars and u and v are vectors of V, it follows αu + βv ∈ V . That is, it is “closed under the algebraic operations of vector addition and scalar multiplication”. A linear combination of vectors is said to be trivial if all the scalars in the linear combination equal zero. A set of vectors is said to be linearly independent if the only linear combination of these vectors which equals the zero vector is the trivial linear combination. Thus {x1 , · · · , xn } is called linearly independent if whenever p X

ck xk = 0

k=1

it follows that all the scalars, ck equal zero. A set of vectors, {x1 , · · · , xp } , is called linearly dependent if it is not linearly independent. Thus the set of vectors is Pplinearly dependent if there exist scalars, ci , i = 1, · · · , n, not all zero such that k=1 ck xk = 0. Lemma 4.3.2 A set of vectors {x1 , · · · , xp } is linearly independent if and only if none of the vectors can be obtained as a linear combination of the others. Proof: Suppose first that {x1 , · · · , xp } is linearly independent. If X xk = cj xj , j6=k

then 0 = 1xk +

X

(−cj ) xj ,

j6=k

a nontrivial linear combination, contrary to assumption. This shows that if the set is linearly independent, then none of the vectors is a linear combination of the others. Now suppose no vector is a linear combination of the others. Is {x1 , · · · , xp } linearly independent? If it is not there exist scalars, ci , not all zero such that p X

ci xi = 0.

i=1

Say ck 6= 0. Then you can solve for xk as X xk = (−cj ) /ck xj j6=k

58


contrary to assumption. This proves the lemma. The following is called the exchange theorem. Theorem 4.3.3 (Exchange Theorem) Let {x1 , · · · , xr } be a linearly independent set of vectors such that each xi is in span(y1 , · · · , ys ) . Then r ≤ s. Proof: Define span{y1 , · · · , ys } ≡ V, it follows there exist scalars, c1 , · · · , cs such that s X x1 = ci yi . (4.3.11) i=1

Not all of these scalars can equal zero because if this were the case, it would follow that x1 = 0 and Prso {x1 , · · · , xr } would not be linearly independent. Indeed, if x1 = 0, 1x1 + i=2 0xi = x1 = 0 and so there would exist a nontrivial linear combination of the vectors {x1 , · · · , xr } which equals zero. Say ck 6= 0. Then solve (4.3.11) for yk and obtain   s-1 vectors here }| { z yk ∈ span x1 , y1 , · · · , yk−1 , yk+1 , · · · , ys  . Define {z1 , · · · , zs−1 } by {z1 , · · · , zs−1 } ≡ {y1 , · · · , yk−1 , yk+1 , · · · , ys } Therefore, span {x1 , z1 , · · · , zs−1 } = V because if v ∈ V, there exist constants c1 , · · · , cs such that s−1 X v= ci zi + cs yk . i=1

Now replace the yk in the above with a linear combination of the vectors, {x1 , z1 , · · · , zs−1 } to obtain v ∈ span {x1 , z1 , · · · , zs−1 } . The vector yk , in the list {y1 , · · · , ys } , has now been replaced with the vector x1 and the resulting modified list of vectors has the same span as the original list of vectors, {y1 , · · · , ys } . Now suppose that r > s and that span (x1 , · · · , xl , z1 , · · · , zp ) = V where the vectors, z1 , · · · , zp are each taken from the set, {y1 , · · · , ys } and l+p = s. This has now been done for l = 1 above. Then since r > s, it follows that l ≤ s < r

4.3. SUBSPACES SPANS AND BASES

59

and so l + 1 ≤ r. Therefore, xl+1 is a vector not in the list, {x1 , · · · , xl } and since span {x1 , · · · , xl , z1 , · · · , zp } = V, there exist scalars, ci and dj such that xl+1 =

l X

ci xi +

i=1

p X

dj zj .

(4.3.12)

j=1

Now not all the dj can equal zero because if this were so, it would follow that {x1 , · · · , xr } would be a linearly dependent set because one of the vectors would equal a linear combination of the others. Therefore, (4.3.12) can be solved for one of the zi , say zk , in terms of xl+1 and the other zi and just as in the above argument, replace that zi with xl+1 to obtain 

 p-1 vectors here z }| { span x1 , · · · xl , xl+1 , z1 , · · · zk−1 , zk+1 , · · · , zp  = V. Continue this way, eventually obtaining span (x1 , · · · , xs ) = V. But then xr ∈ span (x1 , · · · , xs ) contrary to the assumption that {x1 , · · · , xr } is linearly independent. Therefore, r ≤ s as claimed. Definition 4.3.4 A finite set of vectors, {x1 , · · · , xr } is a basis for Fn if span (x1 , · · · , xr ) = Fn and {x1 , · · · , xr } is linearly independent. Corollary 4.3.5 Let {x1 , · · · , xr } and {y1 , · · · , ys } be two bases2 of Fn . Then r = s = n. Proof: From the exchange theorem, r ≤ s and s ≤ r. Now note the vectors, 1 is in the ith slot

z }| { ei = (0, · · · , 0, 1, 0 · · · , 0) for i = 1, 2, · · · , n are a basis for Fn . This proves the corollary. Lemma 4.3.6 Let {v1 , · · · , vr } be a set of vectors. Then V ≡ span (v1 , · · · , vr ) is a subspace. 2 This is the plural form of basis. We could say basiss but it would involve an inordinate amount of hissing as in “The sixth shiek’s sixth sheep is sick”. This is the reason that bases is used instead of basiss.

60


Proof: Suppose α, β are two scalars and let elements of V. What about α

r X

ck vk + β

r X

Pr

k=1 ck vk

and

Pr k=1

dk vk are two

dk vk ?

k=1

k=1

Is it also in V ? α

r X k=1

ck vk + β

r X k=1

dk vk =

r X

(αck + βdk ) vk ∈ V

k=1

so the answer is yes. This proves the lemma. Definition 4.3.7 A finite set of vectors, {x1 , · · · , xr } is a basis for a subspace, V of Fn if span (x1 , · · · , xr ) = V and {x1 , · · · , xr } is linearly independent. Corollary 4.3.8 Let {x1 , · · · , xr } and {y1 , · · · , ys } be two bases for V . Then r = s. Proof: From the exchange theorem, r ≤ s and s ≤ r. Therefore, this proves the corollary. Definition 4.3.9 Let V be a subspace of Fn . Then dim (V ) read as the dimension of V is the number of vectors in a basis. Of course you should wonder right now whether an arbitrary subspace even has a basis. In fact it does and this is in the next theorem. First, here is an interesting lemma. Lemma 4.3.10 Suppose v ∈ / span (u1 , · · · , uk ) and {u1 , · · · , uk } is linearly independent. Then {u1 , · · · , uk , v} is also linearly independent. Pk Proof: Suppose i=1 ci ui + dv = 0. It is required to verify that each ci = 0 and that d = 0. But if d 6= 0, then you can solve for v as a linear combination of the vectors, {u1 , · · · , uk }, k ³ ´ X ci v=− ui d i=1 Pk contrary to assumption. Therefore, d = 0. But then i=1 ci ui = 0 and the linear independence of {u1 , · · · , uk } implies each ci = 0 also. This proves the lemma. Theorem 4.3.11 Let V be a nonzero subspace of Fn . Then V has a basis. Proof: Let v1 ∈ V where v1 6= 0. If span {v1 } = V, stop. {v1 } is a basis for V . Otherwise, there exists v2 ∈ V which is not in span {v1 } . By Lemma 4.3.10 {v1 , v2 } is a linearly independent set of vectors. If span {v1 , v2 } = V stop, {v1 , v2 } is a basis for V. If span {v1 , v2 } = 6 V, then there exists v3 ∈ / span {v1 , v2 } and

4.4. AN APPLICATION TO MATRICES

61

{v1 , v2 , v3 } is a larger linearly independent set of vectors. Continuing this way, the process must stop before n + 1 steps because if not, it would be possible to obtain n + 1 linearly independent vectors contrary to the exchange theorem. This proves the theorem. In words the following corollary states that any linearly independent set of vectors can be enlarged to form a basis. Corollary 4.3.12 Let V be a subspace of Fn and let {v1 , · · · , vr } be a linearly independent set of vectors in V . Then either it is a basis for V or there exist vectors, vr+1 , · · · , vs such that {v1 , · · · , vr , vr+1 , · · · , vs } is a basis for V. Proof: This follows immediately from the proof of Theorem 47.9.4. You do exactly the same argument except you start with {v1 , · · · , vr } rather than {v1 }. It is also true that any spanning set of vectors can be restricted to obtain a basis. Theorem 4.3.13 Let V be a subspace of Fn and suppose span (u1 · · · , up ) = V where the ui are nonzero vectors. Then there exist vectors, {v1 · · · , vr } such that {v1 · · · , vr } ⊆ {u1 · · · , up } and {v1 · · · , vr } is a basis for V . Proof: Let r be the smallest positive integer with the property that for some set, {v1 · · · , vr } ⊆ {u1 · · · , up } , span (v1 · · · , vr ) = V. Then r ≤ p and it must be the case that {v1 · · · , vr } is linearly independent because if it were not so, one of the vectors, say vk would be a linear combination of the others. But then you could delete this vector from {v1 · · · , vr } and the resulting list of r − 1 vectors would still span V contrary to the definition of r. This proves the theorem.

4.4

An Application To Matrices

The following is a theorem of major significance. Theorem 4.4.1 Suppose A is an n × n matrix. Then A is one to one if and only if A is onto. Also, if B is an n × n matrix and AB = I, then it follows BA = I. Proof: First suppose A is one to one. Consider the vectors, {Ae1 , · · · , Aen } where ek is the column vector which is all zeros except for a 1 in the k th position. This set of vectors is linearly independent because if n X

ck Aek = 0,

k=1

then since A is linear,

Ã A

n X

k=1

! ck ek

=0

62


and since A is one to one, it follows n X

ck ek = 03

k=1

which implies each ck = 0. Therefore, {Ae1 , · · · , Aen } must be a basis for Fn because if not there would exist a vector, y ∈ / span (Ae1 , · · · , Aen ) and then by Lemma 4.3.10, {Ae1 , · · · , Aen , y} would be an independent set of vectors having n + 1 vectors in it, contrary to the exchange theorem. It follows that for y ∈ Fn there exist constants, ci such that Ã n ! n X X y= ck Aek = A ck ek k=1

k=1

showing that, since y was arbitrary, A is onto. Next suppose A is onto. This means the span of the columns of A equals Fn . If these columns are not linearly independent, then by Lemma 4.3.2 on Page 57, one of the columns is a linear combination of the others and so the span of the columns of A equals the span of the n−1 other columns. This violates the exchange theorem because {e1 , · · · , en } would be a linearly independent set of vectors contained in the span of only n − 1 vectors. Therefore, the columns of A must be independent and this equivalent to saying that Ax = 0 if and only if x = 0. This implies A is one to one because if Ax = Ay, then A (x − y) = 0 and so x − y = 0. Now suppose AB = I. Why is BA = I? Since AB = I it follows B is one to one since otherwise, there would exist, x 6= 0 such that Bx = 0 and then ABx = A0 = 0 6= Ix. Therefore, from what was just shown, B is also onto. In addition to this, A must be one to one because if Ay = 0, then y = Bx for some x and then x = ABx = Ay = 0 showing y = 0. Now from what is given to be so, it follows (AB) A = A and so using the associative law for matrix multiplication, A (BA) − A = A (BA − I) = 0. But this means (BA − I) x = 0 for all x since otherwise, A would not be one to one. Hence BA = I as claimed. This proves the theorem. This theorem shows that if an n×n matrix, B acts like an inverse when multiplied on one side of A it follows that B = A−1 and it will act like an inverse on both sides of A. The conclusion of this theorem pertains to square matrices only. For example, let   µ ¶ 1 0 1 0 0 A =  0 1 , B = (4.4.13) 1 1 −1 1 0 Then

µ BA =

1 0 0 1

¶

4.5. THE MATHEMATICAL THEORY OF DETERMINANTS but

4.5



1 AB =  1 1

0 1 0

63

 0 −1  . 0

The Mathematical Theory Of Determinants

It is assumed the reader is familiar with matrices. However, the topic of determinants is often neglected in linear algebra books these days. Therefore, I will give a fairly quick and grubby treatment of this topic which includes all the main results. Two books which give a good introduction to determinants are Apostol [3] and Rudin [57]. A recent book which also has a good introduction is Baker [8] Let (i1 , · · · , in ) be an ordered list of numbers from {1, · · · , n} . This means the order is important so (1, 2, 3) and (2, 1, 3) are different. The following Lemma will be essential in the definition of the determinant. Lemma 4.5.1 There exists a unique function, sgnn which maps each list of n numbers from {1, · · · , n} to one of the three numbers, 0, 1, or −1 which also has the following properties. sgnn (1, · · · , n) = 1 (4.5.14) sgnn (i1 , · · · , p, · · · , q, · · · , in ) = − sgnn (i1 , · · · , q, · · · , p, · · · , in )

(4.5.15)

In words, the second property states that if two of the numbers are switched, the value of the function is multiplied by −1. Also, in the case where n > 1 and {i1 , · · · , in } = {1, · · · , n} so that every number from {1, · · · , n} appears in the ordered list, (i1 , · · · , in ) , sgnn (i1 , · · · , iθ−1 , n, iθ+1 , · · · , in ) ≡ n−θ

(−1)

sgnn−1 (i1 , · · · , iθ−1 , iθ+1 , · · · , in )

(4.5.16)

where n = iθ in the ordered list, (i1 , · · · , in ) . Proof: To begin with, it is necessary to show the existence of such a function. This is clearly true if n = 1. Define sgn1 (1) ≡ 1 and observe that it works. No switching is possible. In the case where n = 2, it is also clearly true. Let sgn2 (1, 2) = 1 and sgn2 (2, 1) = −1 while sgn2 (2, 2) = sgn2 (1, 1) = 0 and verify it works. Assuming such a function exists for n, sgnn+1 will be defined in terms of sgnn . If there are any repeated numbers in (i1 , · · · , in+1 ) , sgnn+1 (i1 , · · · , in+1 ) ≡ 0. If there are no repeats, then n + 1 appears somewhere in the ordered list. Let θ be the position of the number n + 1 in the list. Thus, the list is of the form (i1 , · · · , iθ−1 , n + 1, iθ+1 , · · · , in+1 ) . From 4.5.16 it must be that sgnn+1 (i1 , · · · , iθ−1 , n + 1, iθ+1 , · · · , in+1 ) ≡ n+1−θ

(−1)

sgnn (i1 , · · · , iθ−1 , iθ+1 , · · · , in+1 ) .

64


It is necessary to verify this satisfies 4.5.14 and 4.5.15 with n replaced with n + 1. The first of these is obviously true because sgnn+1 (1, · · · , n, n + 1) ≡ (−1)

n+1−(n+1)

sgnn (1, · · · , n) = 1.

If there are repeated numbers in (i1 , · · · , in+1 ) , then it is obvious 4.5.15 holds because both sides would equal zero from the above definition. It remains to verify 4.5.15 in the case where there are no numbers repeated in (i1 , · · · , in+1 ) . Consider ³ ´ r s sgnn+1 i1 , · · · , p, · · · , q, · · · , in+1 , where the r above the p indicates the number, p is in the rth position and the s above the q indicates that the number, q is in the sth position. Suppose first that r < θ < s. Then µ ¶ θ r s sgnn+1 i1 , · · · , p, · · · , n + 1, · · · , q, · · · , in+1 ≡ n+1−θ

(−1) while

³ ´ r s−1 sgnn i1 , · · · , p, · · · , q , · · · , in+1

µ ¶ θ r s sgnn+1 i1 , · · · , q, · · · , n + 1, · · · , p, · · · , in+1 = (−1)

n+1−θ

³ ´ r s−1 sgnn i1 , · · · , q, · · · , p , · · · , in+1

and so, by induction, a switch of p and q introduces a minus sign in the result. Similarly, if θ > s or if θ < r it also follows that 4.5.15 holds. The interesting case is when θ = r or θ = s. Consider the case where θ = r and note the other case is entirely similar. ³ ´ r s sgnn+1 i1 , · · · , n + 1, · · · , q, · · · , in+1 = n+1−r

(−1) while

³ ´ s−1 sgnn i1 , · · · , q , · · · , in+1

(4.5.17)

³ ´ s r sgnn+1 i1 , · · · , q, · · · , n + 1, · · · , in+1 = n+1−s

(−1)

³ ´ r sgnn i1 , · · · , q, · · · , in+1 .

(4.5.18)

By making s − 1 − r switches, move the q which is in the s − 1th position in 4.5.17 to the rth position in 4.5.18. By induction, each of these switches introduces a factor of −1 and so ³ ´ ³ ´ s−1 r s−1−r sgnn i1 , · · · , q , · · · , in+1 = (−1) sgnn i1 , · · · , q, · · · , in+1 .

4.5. THE MATHEMATICAL THEORY OF DETERMINANTS

65

Therefore, ³ ´ r s sgnn+1 i1 , · · · , n + 1, · · · , q, · · · , in+1 ³ ´ s−1 n+1−r (−1) sgnn i1 , · · · , q , · · · , in+1

=

n+1−r

= (−1)

s−1−r

(−1)

³ ´ r sgnn i1 , · · · , q, · · · , in+1

³ ´ r sgnn i1 , · · · , q, · · · , in+1 ³ ´ r 2s−1 n+1−s = (−1) (−1) sgnn i1 , · · · , q, · · · , in+1 n+s

= (−1)

³ ´ s r = − sgnn+1 i1 , · · · , q, · · · , n + 1, · · · , in+1 . This proves the existence of the desired function. To see this function is unique, note that you can obtain any ordered list of distinct numbers from a sequence of switches. If there exist two functions, f and g both satisfying 4.5.14 and 4.5.15, you could start with f (1, · · · , n) = g (1, · · · , n) and applying the same sequence of switches, eventually arrive at f (i1 , · · · , in ) = g (i1 , · · · , in ) . If any numbers are repeated, then 4.5.15 gives both functions are equal to zero for that ordered list. This proves the lemma. In what follows sgn will often be used rather than sgnn because the context supplies the appropriate n. Definition 4.5.2 Let f be a real valued function which has the set of ordered lists of numbers from {1, · · · , n} as its domain. Define X

f (k1 · · · kn )

(k1 ,··· ,kn )

to be the sum of all the f (k1 · · · kn ) for all possible choices of ordered lists (k1 , · · · , kn ) of numbers of {1, · · · , n} . For example, X (k1 ,k2 )

f (k1 , k2 ) = f (1, 2) + f (2, 1) + f (1, 1) + f (2, 2) .

66


Definition 4.5.3 Let (aij ) = A denote an n × n matrix. The determinant of A, denoted by det (A) is defined by X det (A) ≡ sgn (k1 , · · · , kn ) a1k1 · · · ankn (k1 ,··· ,kn )

where the sum is taken over all ordered lists of numbers from {1, · · · , n}. Note it suffices to take the sum over only those ordered lists in which there are no repeats because if there are, sgn (k1 , · · · , kn ) = 0 and so that term contributes 0 to the sum. Let A be an n × n matrix, A = (aij ) and let (r1 , · · · , rn ) denote an ordered list of n numbers from {1, · · · , n}. Let A (r1 , · · · , rn ) denote the matrix whose k th row is the rk row of the matrix, A. Thus X det (A (r1 , · · · , rn )) = sgn (k1 , · · · , kn ) ar1 k1 · · · arn kn (4.5.19) (k1 ,··· ,kn )

and A (1, · · · , n) = A. Proposition 4.5.4 Let (r1 , · · · , rn ) be an ordered list of numbers from {1, · · · , n}. Then sgn (r1 , · · · , rn ) det (A) =

X

sgn (k1 , · · · , kn ) ar1 k1 · · · arn kn

(4.5.20)

(k1 ,··· ,kn )

= det (A (r1 , · · · , rn )) .

(4.5.21)

Proof: Let (1, · · · , n) = (1, · · · , r, · · · s, · · · , n) so r < s. det (A (1, · · · , r, · · · , s, · · · , n)) = X

(4.5.22)

sgn (k1 , · · · , kr , · · · , ks , · · · , kn ) a1k1 · · · arkr · · · asks · · · ankn ,

(k1 ,··· ,kn )

and renaming the variables, calling ks , kr and kr , ks , this equals X = sgn (k1 , · · · , ks , · · · , kr , · · · , kn ) a1k1 · · · arks · · · askr · · · ankn (k1 ,··· ,kn )

 =

X (k1 ,··· ,kn )

− sgn k1 , · · · ,

These got switched

z }| { kr , · · · , ks

 , · · · , kn  a1k1 · · · askr · · · arks · · · ankn

4.5. THE MATHEMATICAL THEORY OF DETERMINANTS = − det (A (1, · · · , s, · · · , r, · · · , n)) .

67 (4.5.23)

Consequently, det (A (1, · · · , s, · · · , r, · · · , n)) = − det (A (1, · · · , r, · · · , s, · · · , n)) = − det (A) Now letting A (1, · · · , s, · · · , r, · · · , n) play the role of A, and continuing in this way, switching pairs of numbers, p

det (A (r1 , · · · , rn )) = (−1) det (A) where it took p switches to obtain(r1 , · · · , rn ) from (1, · · · , n). By Lemma 4.5.1, this implies p

det (A (r1 , · · · , rn )) = (−1) det (A) = sgn (r1 , · · · , rn ) det (A) and proves the proposition in the case when there are no repeated numbers in the ordered list, (r1 , · · · , rn ). However, if there is a repeat, say the rth row equals the sth row, then the reasoning of 4.5.22 -4.5.23 shows that A (r1 , · · · , rn ) = 0 and also sgn (r1 , · · · , rn ) = 0 so the formula holds in this case also. Observation 4.5.5 There are n! ordered lists of distinct numbers from {1, · · · , n} . To see this, consider n slots placed in order. There are n choices for the first slot. For each of these choices, there are n − 1 choices for the second. Thus there are n (n − 1) ways to fill the first two slots. Then for each of these ways there are n − 2 choices left for the third slot. Continuing this way, there are n! ordered lists of distinct numbers from {1, · · · , n} as stated in the observation. With the above, it is possible to give a more symmetric ¡ ¢ description of the determinant from which it will follow that det (A) = det AT . Corollary 4.5.6 The following formula for det (A) is valid. det (A) = X

X

1 · n!

sgn (r1 , · · · , rn ) sgn (k1 , · · · , kn ) ar1 k1 · · · arn kn .

(4.5.24)

(r1 ,··· ,rn ) (k1 ,··· ,kn )

¡ ¢ And also det AT = det (A) where AT is the transpose of A. (Recall that for ¢ ¡ AT = aTij , aTij = aji .) Proof: From Proposition 4.5.4, if the ri are distinct, X det (A) = sgn (r1 , · · · , rn ) sgn (k1 , · · · , kn ) ar1 k1 · · · arn kn . (k1 ,··· ,kn )

Summing over all ordered lists, (r1 , · · · , rn ) where the ri are distinct, (If the ri are not distinct, sgn (r1 , · · · , rn ) = 0 and so there is no contribution to the sum.) n! det (A) =

68


X

X

sgn (r1 , · · · , rn ) sgn (k1 , · · · , kn ) ar1 k1 · · · arn kn .

(r1 ,··· ,rn ) (k1 ,··· ,kn )

This proves the corollary since the formula gives the same number for A as it does for AT . Corollary 4.5.7 If two rows or two columns in an n × n matrix, A, are switched, the determinant of the resulting matrix equals (−1) times the determinant of the original matrix. If A is an n × n matrix in which two rows are equal or two columns are equal then det (A) = 0. Suppose the ith row of A equals (xa1 + yb1 , · · · , xan + ybn ). Then det (A) = x det (A1 ) + y det (A2 ) where the ith row of A1 is (a1 , · · · , an ) and the ith row of A2 is (b1 , · · · , bn ) , all other rows of A1 and A2 coinciding with those of A. In other words, det is a linear function of each row A. The same is true with the word “row” replaced with the word “column”. Proof: By Proposition 4.5.4 when two rows are switched, the determinant of the resulting matrix is (−1) times the determinant of the original matrix. By Corollary 4.5.6 the same holds for columns because the columns of the matrix equal the rows of the transposed matrix. Thus if A1 is the matrix obtained from A by switching two columns, ¡ ¢ ¡ ¢ det (A) = det AT = − det AT1 = − det (A1 ) . If A has two equal columns or two equal rows, then switching them results in the same matrix. Therefore, det (A) = − det (A) and so det (A) = 0. It remains to verify the last assertion. X sgn (k1 , · · · , kn ) a1k1 · · · (xaki + ybki ) · · · ankn det (A) ≡ (k1 ,··· ,kn )

=x

X

sgn (k1 , · · · , kn ) a1k1 · · · aki · · · ankn

(k1 ,··· ,kn )

+y

X

sgn (k1 , · · · , kn ) a1k1 · · · bki · · · ankn

(k1 ,··· ,kn )

≡ x det (A1 ) + y det (A2 ) . ¡ ¢ The same is true of columns because det AT = det (A) and the rows of AT are the columns of A. Definition 4.5.8 A vector, w, is a linear combination of the vectors {v1 , · · · , vr } Pr if there exists scalars, c1 , · · · cr such that w = k=1 ck vk . This is the same as saying w ∈ span {v1 , · · · , vr } . The following corollary is also of great use.


69

Corollary 4.5.9 Suppose A is an n × n matrix and some column (row) is a linear combination of r other columns (rows). Then det (A) = 0. ¡ ¢ Proof: Let A = a1 · · · an be the columns of A and suppose the condition that one column is a linear combination of r of the others is satisfied. Then by using Corollary 4.5.7 you may rearrange the columns have the nth column a linear Pto r combination of the first r columns. Thus an = k=1 ck ak and so ¡ ¢ Pr det (A) = det a1 · · · ar · · · an−1 . k=1 ck ak By Corollary 4.5.7 r X

det (A) =

ck det

¡

a1

···

ar

···

an−1

ak

¢

= 0.

k=1

¡ ¢ The case for rows follows from the fact that det (A) = det AT . This proves the corollary. Recall the following definition of matrix multiplication. Definition 4.5.10 If A and B are n × n matrices, A = (aij ) and B = (bij ), AB = (cij ) where n X cij ≡ aik bkj . k=1

One of the most important rules about determinants is that the determinant of a product equals the product of the determinants. Theorem 4.5.11 Let A and B be n × n matrices. Then det (AB) = det (A) det (B) . Proof: Let cij be the ij th entry of AB. Then by Proposition 4.5.4, det (AB) = X

sgn (k1 , · · · , kn ) c1k1 · · · cnkn

(k1 ,··· ,kn )

=

X

sgn (k1 , · · · , kn )

(k1 ,··· ,kn )

=

X

X

Ã X r1

! a1r1 br1 k1

Ã ···

X

! anrn brn kn

rn

sgn (k1 , · · · , kn ) br1 k1 · · · brn kn (a1r1 · · · anrn )

(r1 ··· ,rn ) (k1 ,··· ,kn )

=

X

sgn (r1 · · · rn ) a1r1 · · · anrn det (B) = det (A) det (B) .

(r1 ··· ,rn )

This proves the theorem.

70


Lemma 4.5.12 Suppose a matrix is of the form µ ¶ A ∗ M= 0 a or

µ M=

A ∗

0 a

(4.5.25)

¶ (4.5.26)

where a is a number and A is an (n − 1) × (n − 1) matrix and ∗ denotes either a column or a row having length n − 1 and the 0 denotes either a column or a row of length n − 1 consisting entirely of zeros. Then det (M ) = a det (A) . Proof: Denote M by (mij ) . Thus in the first case, mnn = a and mni = 0 if i 6= n while in the second case, mnn = a and min = 0 if i 6= n. From the definition of the determinant, X det (M ) ≡ sgnn (k1 , · · · , kn ) m1k1 · · · mnkn (k1 ,··· ,kn )

Letting θ denote the position of n in the ordered list, (k1 , · · · , kn ) then using the earlier conventions used to prove Lemma 4.5.1, det (M ) equals µ ¶ X θ n−1 n−θ (−1) sgnn−1 k1 , · · · , kθ−1 , kθ+1 , · · · , kn m1k1 · · · mnkn (k1 ,··· ,kn )

Now suppose 4.5.26. Then if kn 6= n, the term involving mnkn in the above expression equals zero. Therefore, the only terms which survive are those for which θ = n or in other words, those for which kn = n. Therefore, the above expression reduces to X sgnn−1 (k1 , · · · kn−1 ) m1k1 · · · m(n−1)kn−1 = a det (A) . a (k1 ,··· ,kn−1 )

To get the assertion in the situation of 4.5.25 use Corollary 4.5.6 and 4.5.26 to write µµ T ¶¶ ¡ ¢ ¡ ¢ A 0 det (M ) = det M T = det = a det AT = a det (A) . ∗ a This proves the lemma. In terms of the theory of determinants, arguably the most important idea is that of Laplace expansion along a row or a column. This will follow from the above definition of a determinant. Definition 4.5.13 Let A = (aij ) be an n × n matrix. Then a new matrix called the cofactor matrix, cof (A) is defined by cof (A) = (cij ) where to obtain cij delete the ith row and the j th column of A, take the determinant of the (n − 1) × (n − 1) matrix which results, (This is called the ij th minor of A. ) and then multiply this i+j number by (−1) . To make the formulas easier to remember, cof (A)ij will denote the ij th entry of the cofactor matrix.


71

The following is the main result. Earlier this was given as a definition and the outrageous totally unjustified assertion was made that the same number would be obtained by expanding the determinant along any row or column. The following theorem proves this assertion. Theorem 4.5.14 Let A be an n × n matrix where n ≥ 2. Then det (A) =

n X

aij cof (A)ij =

j=1

n X

aij cof (A)ij .

(4.5.27)

i=1

The first formula consists of expanding the determinant along the ith row and the second expands the determinant along the j th column. Proof: Let (ai1 , · · · , ain ) be the ith row of A. Let Bj be the matrix obtained from A by leaving every row the same except the ith row which in Bj equals (0, · · · , 0, aij , 0, · · · , 0) . Then by Corollary 4.5.7, det (A) =

n X

det (Bj )

j=1

Denote by Aij the (n − 1) × (n − 1) matrix obtained by deleting the ith row and the ¡ ¢ i+j j th column of A. Thus cof (A)ij ≡ (−1) det Aij . At this point, recall that from Proposition 4.5.4, when two rows or two columns in a matrix, M, are switched, this results in multiplying the determinant of the old matrix by −1 to get the determinant of the new matrix. Therefore, by Lemma 4.5.12, µµ ij ¶¶ A ∗ n−j n−i det (Bj ) = (−1) (−1) det 0 aij µµ ij ¶¶ A ∗ i+j = (−1) det = aij cof (A)ij . 0 aij Therefore, det (A) =

n X

aij cof (A)ij

j=1

which is the formula for expanding det (A) along the ith row. Also, det (A)

=

n ¡ ¢ X ¡ ¢ det AT = aTij cof AT ij j=1

=

n X

aji cof (A)ji

j=1

which is the formula for expanding det (A) along the ith column. This proves the theorem. Note that this gives an easy way to write a formula for the inverse of an n × n matrix.

72


4.5.15 A−1 exists if and only if det(A) 6= 0. If det(A) 6= 0, then A−1 = ¡Theorem ¢ −1 aij where −1 a−1 cof (A)ji ij = det(A) for cof (A)ij the ij th cofactor of A. Proof: By Theorem 4.5.14 and letting (air ) = A, if det (A) 6= 0, n X

air cof (A)ir det(A)−1 = det(A) det(A)−1 = 1.

i=1

Now consider

n X

air cof (A)ik det(A)−1

i=1 th

when k 6= r. Replace the k column with the rth column to obtain a matrix, Bk whose determinant equals zero by Corollary 4.5.7. However, expanding this matrix along the k th column yields −1

0 = det (Bk ) det (A)

=

n X

−1

air cof (A)ik det (A)

i=1

Summarizing,

n X

−1

air cof (A)ik det (A)

= δ rk .

i=1

Using the other formula in Theorem 4.5.14, and similar reasoning, n X

arj cof (A)kj det (A)

−1

= δ rk

j=1

¡ ¢ This proves that if det (A) 6= 0, then A−1 exists with A−1 = a−1 ij , where a−1 ij = cof (A)ji det (A)

−1

.

Now suppose A−1 exists. Then by Theorem 4.5.11, ¡ ¢ ¡ ¢ 1 = det (I) = det AA−1 = det (A) det A−1 so det (A) 6= 0. This proves the theorem. The next corollary points out that if an n × n matrix, A has a right or a left inverse, then it has an inverse. Corollary 4.5.16 Let A be an n × n matrix and suppose there exists an n × n matrix, B such that BA = I. Then A−1 exists and A−1 = B. Also, if there exists C an n × n matrix such that AC = I, then A−1 exists and A−1 = C.


73

Proof: Since BA = I, Theorem 4.5.11 implies det B det A = 1 and so det A 6= 0. Therefore from Theorem 4.5.15, A−1 exists. Therefore, ¢ ¡ A−1 = (BA) A−1 = B AA−1 = BI = B. The case where CA = I is handled similarly. The conclusion of this corollary is that left inverses, right inverses and inverses are all the same in the context of n × n matrices. Theorem 4.5.15 says that to find the inverse, take the transpose of the cofactor matrix and divide by the determinant. The transpose of the cofactor matrix is called the adjugate or sometimes the classical adjoint of the matrix A. It is an abomination to call it the adjoint although you do sometimes see it referred to in this way. In words, A−1 is equal to one over the determinant of A times the adjugate matrix of A. In case you are solving a system of equations, Ax = y for x, it follows that if A−1 exists, ¡ ¢ x = A−1 A x = A−1 (Ax) = A−1 y thus solving the system. Now in the case that A−1 exists, there is a formula for A−1 given above. Using this formula, xi =

n X

a−1 ij yj =

j=1

n X j=1

1 cof (A)ji yj . det (A)

By the formula for the expansion of a determinant along a column,   ∗ · · · y1 · · · ∗ 1  .. ..  , xi = det  ... . .  det (A) ∗ · · · yn · · · ∗ T

where here the ith column of A is replaced with the column vector, (y1 · · · ·, yn ) , and the determinant of this modified matrix is taken and divided by det (A). This formula is known as Cramer’s rule. Definition 4.5.17 A matrix M , is Thus such a matrix equals zero below as shown.  ∗   0   .  ..

upper triangular if Mij = 0 whenever i > j. the main diagonal, the entries of the form Mii  ∗ ··· ∗ .  .. . ..  ∗   .. .. . . ∗  0 ··· 0 ∗

A lower triangular matrix is defined similarly as a matrix for which all entries above the main diagonal are equal to zero.

74


With this definition, here is a simple corollary of Theorem 4.5.14. Corollary 4.5.18 Let M be an upper (lower) triangular matrix. Then det (M ) is obtained by taking the product of the entries on the main diagonal. Definition 4.5.19 A submatrix of a matrix A is the rectangular array of numbers obtained by deleting some rows and columns of A. Let A be an m × n matrix. The determinant rank of the matrix equals r where r is the largest number such that some r × r submatrix of A has a non zero determinant. The row rank is defined to be the dimension of the span of the rows. The column rank is defined to be the dimension of the span of the columns. Theorem 4.5.20 If A has determinant rank, r, then there exist r rows of the matrix such that every other row is a linear combination of these r rows. Proof: Suppose the determinant rank of A = (aij ) equals r. If rows and columns are interchanged, the determinant rank of the modified matrix is unchanged. Thus rows and columns can be interchanged to produce an r × r matrix in the upper left corner of the matrix which has non zero determinant. Now consider the r + 1 × r + 1 matrix, M,   a11 · · · a1r a1p  ..  .. ..  .  . .    ar1 · · · arr arp  al1 · · · alr alp where C will denote the r × r matrix in the upper left corner which has non zero determinant. I claim det (M ) = 0. There are two cases to consider in verifying this claim. First, suppose p > r. Then the claim follows from the assumption that A has determinant rank r. On the other hand, if p < r, then the determinant is zero because there are two identical columns. Expand the determinant along the last column and divide by det (C) to obtain r X cof (M )ip alp = − aip . det (C) i=1 Now note that cof (M )ip does not depend on p. Therefore the above sum is of the form r X alp = mi aip i=1

which shows the lth row is a linear combination of the first r rows of A. Since l is arbitrary, this proves the theorem. Corollary 4.5.21 The determinant rank equals the row rank.


75

Proof: From Theorem 4.5.20, the row rank is no larger than the determinant rank. Could the row rank be smaller than the determinant rank? If so, there exist p rows for p < r such that the span of these p rows equals the row space. But this implies that the r × r submatrix whose determinant is nonzero also has row rank no larger than p which is impossible if its determinant is to be nonzero because at least one row is a linear combination of the others. Corollary 4.5.22 If A has determinant rank, r, then there exist r columns of the matrix such that every other column is a linear combination of these r columns. Also the column rank equals the determinant rank. Proof: This follows from the above by considering AT . The rows of AT are the columns of A and the determinant rank of AT and A are the same. Therefore, from Corollary 4.5.21, column rank of A = row rank of AT = determinant rank of AT = determinant rank of A. The following theorem is of fundamental importance and ties together many of the ideas presented above. Theorem 4.5.23 Let A be an n × n matrix. Then the following are equivalent. 1. det (A) = 0. 2. A, AT are not one to one. 3. A is not onto. Proof: Suppose det (A) = 0. Then the determinant rank of A = r < n. Therefore, there exist r columns such that every other column is a linear combination of these columns by Theorem 4.5.20. In particular, it follows that for some¡ m, the mth column is a ¢linear combination of all the others. Thus letting A = a1 · · · am · · · an where the columns are denoted by ai , there exists scalars, αi such that X α k ak . am = k6=m

¡

α1 · · · −1 · · · X Ax = −am + αk ak = 0.

Now consider the column vector, x ≡

αn

¢T

. Then

k6=m

Since also A0 = 0, it follows A is not one to one. Similarly, AT is not one to one by the same argument applied to AT . This verifies that 1.) implies 2.). Now suppose 2.). Then since AT is not one to one, it follows there exists x 6= 0 such that AT x = 0. Taking the transpose of both sides yields xT A = 0

76


where the 0 is a 1 × n matrix or row vector. Now if Ay = x, then ¡ ¢ 2 |x| = xT (Ay) = xT A y = 0y = 0 contrary to x 6= 0. Consequently there can be no y such that Ay = x and so A is not onto. This shows that 2.) implies 3.). Finally, suppose 3.). If 1.) does not hold, then det (A) 6= 0 but then from Theorem 4.5.15 A−1 exists and so for every y ∈ Fn there exists a unique x ∈ Fn such that Ax = y. In fact x = A−1 y. Thus A would be onto contrary to 3.). This shows 3.) implies 1.) and proves the theorem. Corollary 4.5.24 Let A be an n × n matrix. Then the following are equivalent. 1. det(A) 6= 0. 2. A and AT are one to one. 3. A is onto. Proof: This follows immediately from the above theorem.

4.6

Exercises

1. Let m < n and let A be an m × n matrix. Show that A is not one to one. Hint: Consider the n × n matrix, A1 which is of the form µ ¶ A A1 ≡ 0 where the 0 denotes an (n − m) × n matrix of zeros. Thus det A1 = 0 and so A1 is not one to one. Now observe that A1 x is the vector, µ ¶ Ax A1 x = 0 which equals zero if and only if Ax = 0.

4.7

The Cayley Hamilton Theorem

Definition 4.7.1 Let A be an n×n matrix. The characteristic polynomial is defined as pA (t) ≡ det (tI − A) and the solutions to pA (t) = 0 are called eigenvalues. For A a matrix and p (t) = tn + an−1 tn−1 + · · · + a1 t + a0 , denote by p (A) the matrix defined by p (A) ≡ An + an−1 An−1 + · · · + a1 A + a0 I. The explanation for the last term is that A0 is interpreted as I, the identity matrix.

4.7. THE CAYLEY HAMILTON THEOREM

77

The Cayley Hamilton theorem states that every matrix satisfies its characteristic equation, that equation defined by PA (t) = 0. It is one of the most important theorems in linear algebra. The following lemma will help with its proof. Lemma 4.7.2 Suppose for all |λ| large enough, A0 + A1 λ + · · · + Am λm = 0, where the Ai are n × n matrices. Then each Ai = 0. Proof: Multiply by λ−m to obtain A0 λ−m + A1 λ−m+1 + · · · + Am−1 λ−1 + Am = 0. Now let |λ| → ∞ to obtain Am = 0. With this, multiply by λ to obtain A0 λ−m+1 + A1 λ−m+2 + · · · + Am−1 = 0. Now let |λ| → ∞ to obtain Am−1 = 0. Continue multiplying by λ and letting λ → ∞ to obtain that all the Ai = 0. This proves the lemma. With the lemma, here is a simple corollary. Corollary 4.7.3 Let Ai and Bi be n × n matrices and suppose A0 + A1 λ + · · · + Am λm = B0 + B1 λ + · · · + Bm λm for all |λ| large enough. Then Ai = Bi for all i. Consequently if λ is replaced by any n × n matrix, the two sides will be equal. That is, for C any n × n matrix, A0 + A1 C + · · · + Am C m = B0 + B1 C + · · · + Bm C m . Proof: Subtract and use the result of the lemma. With this preparation, here is a relatively easy proof of the Cayley Hamilton theorem. Theorem 4.7.4 Let A be an n × n matrix and let p (λ) ≡ det (λI − A) be the characteristic polynomial. Then p (A) = 0. Proof: Let C (λ) equal the transpose of the cofactor matrix of (λI − A) for |λ| large. (If |λ| is large enough, then λ cannot be in the finite list of eigenvalues of A −1 and so for such λ, (λI − A) exists.) Therefore, by Theorem 4.5.15 −1

C (λ) = p (λ) (λI − A)

.

Note that each entry in C (λ) is a polynomial in λ having degree no more than n−1. Therefore, collecting the terms, C (λ) = C0 + C1 λ + · · · + Cn−1 λn−1

78


for Cj some n × n matrix. It follows that for all |λ| large enough, ¡ ¢ (A − λI) C0 + C1 λ + · · · + Cn−1 λn−1 = p (λ) I and so Corollary 4.7.3 may be used. It follows the matrix coefficients corresponding to equal powers of λ are equal on both sides of this equation. Therefore, if λ is replaced with A, the two sides will be equal. Thus ¡ ¢ 0 = (A − A) C0 + C1 A + · · · + Cn−1 An−1 = p (A) I = p (A) . This proves the Cayley Hamilton theorem.

4.8

An Identity Of Cauchy

There is a very interesting identity for determinants due to Cauchy. Theorem 4.8.1 The following identity holds. ¯ ¯ 1 1 ¯ a +b ¯ · · · a1 +b 1 1 n ¯ ¯ Y Y ¯ ¯ . . .. .. (ai + bj ) ¯ (ai − aj ) (bi − bj ) . ¯= ¯ ¯ 1 1 i,j j 2. Therefore, there are exactly n − 1 factors which contain a2 . Therefore, a2 has an exponent of n − 1. Similarly, each ak is raised to the n − 1 power and the same holds for the bk as well. Therefore, the right side of 4.8.28 is of the form can−1 an−1 · · · an−1 bn−1 · · · bn−1 n n 1 2 1 where c is some constant. Now consider the left side of 4.8.28. This is of the form 1 Y (ai + bj ) n! i,j i

X

sgn (i1 · · · in ) sgn (j1 · · · jn ) ·

1 ···in ,j1 ,···jn

1 1 1 ··· . ai1 + bj1 ai2 + bj2 ain + bjn For a given i1 · · · in , j1 , · · · jn , let S (i1 · · · in , j1 , · · · jn ) ≡ {(i1 , j1 ) , (i2 , j2 ) · · · , (in , jn )} . This equals 1 n! i

X 1 ···in ,j1 ,···jn

sgn (i1 · · · in ) sgn (j1 · · · jn )

Y (i,j)∈{(i / 1 ,j1 ),(i2 ,j2 )··· ,(in ,jn )}

(ai + bj )

4.9. BLOCK MULTIPLICATION OF MATRICES

79

where you can assume the ik are all distinct and the jk are also all distinct because otherwise sgn will produce a 0. Therefore, in Y (ai + bj ) , (i,j)∈{(i / 1 ,j1 ),(i2 ,j2 )··· ,(in ,jn )}

there are exactly n − 1 factors which contain ak for each k and similarly, there are exactly n − 1 factors which contain bk for each k. Therefore, the left side of 4.8.28 is of the form da1n−1 a2n−1 · · · ann−1 bn−1 · · · bn−1 n 1 and it remains to verify that c = d. Using the properties of determinants, the left side of 4.8.28 is of the form ¯ a1 +b1 +b1 ¯ ¯ ¯ · · · aa11+b 1 n ¯ ¯ a2 +b2 a1 +b2 a2 +b2 ¯ ¯ 1 · · · a2 +bn ¯ Y ¯ a2 +b1 (ai + bj ) ¯ ¯ . . .. .. .. .. ¯ ¯ . . i6=j ¯ ¯ ¯ ¯ an +bn an +bn · · · 1 an +b1

an +b2

Q

Let ak → −bk . Then this converges to i6=j (−bi + bj ) . The right side of 4.8.28 converges to Y Y (−bi + bj ) (bi − bj ) = (−bi + bj ) . j

i6=j

Therefore, d = c and this proves the identity.

4.9

Block Multiplication Of Matrices

Consider the following problem µ

A C

B D

¶µ

You know how to do this. You get µ AE + BG CE + DG

E G

F H

¶

AF + BH CF + DH

¶ .

Now what if instead of numbers, the entries, A, B, C, D, E, F, G are matrices of a size such that the multiplications and additions needed in the above formula all make sense. Would the formula be true in this case? I will show below that this is true. Suppose A is a matrix of the form   A11 · · · A1m  ..  .. A =  ... (4.9.29) . .  Ar1

···

Arm

80


where Aij is a si × pj matrix where si is constant for j = 1, · · · , m for each i = 1, · · · , r. Such a matrix is called a block matrix, also a partitioned matrix. How do you get the block Aij ? Here is how for A an m × n matrix:

¡z

si ×m

}|

0

Isi ×si

z

n×pj

}| { 0 {¢ 0 A Ipj ×pj . 0

(4.9.30)

In the block column matrix on the right, you need to have cj − 1 rows of zeros above the small pj × pj identity matrix where the columns of A involved in Aij are cj , · · · , cj + pj and in the block row matrix on the left, you need to have ri − 1 columns of zeros to the left of the si ×si identity matrix where the rows of A involved in Aij are ri , · · · , ri + si . An important observation to make is that the matrix on the right specifies columns to use in the block and the one on the left specifies the rows used. There is no overlap between the blocks of A. Thus the identity n × n identity matrix corresponding to multiplication on the right of A is of the form   Ip1 ×p1 0   ..   . 0

Ipm ×pm

these little identity matrices don’t overlap. A similar conclusion follows from consideration of the matrices Isi ×si . Next consider the question of multiplication of two block matrices. Let B be a block matrix of the form   B11 · · · B1p  .. ..  .. (4.9.31)  . . .  Br1 and A is a block matrix of the form  A11  ..  . Ap1

···

Brp

··· .. . ···

 A1m ..  .  Apm

(4.9.32)

and that for all i, j, it makes sense to multiply Bis Asj for all s ∈ {1, · · · , p}. (That is the two matrices, Bis and Asj are conformable.) and that for fixed P ij, it follows Bis Asj is the same size for each s so that it makes sense to write s Bis Asj . The following theorem says essentially that when you take the product of two matrices, you can do it two ways. One way is to simply multiply them forming BA. The other way is to partition both matrices, formally multiply the blocks to get another block matrix and this one will be BA partitioned. Before presenting this theorem, here is a simple lemma which is really a special case of the theorem.

4.9. BLOCK MULTIPLICATION OF MATRICES

81

Lemma 4.9.1 Consider the following product.   0 ¡ ¢  I  0 I 0 0 where the first is n × r and the second is r × n. The small identity matrix I is an r × r matrix and there are l zero rows above I and l zero columns to the left of I in the right matrix. Then the product of these matrices is a block matrix of the form   0 0 0  0 I 0  0 0 0 Proof: From the definition of the way you multiply matrices, the product is               0 0 0 0 0 0   I  0 · · ·  I  0  I  e1 · · ·  I  er  I  0 · · ·  I  0  0 0 0 0 0 0 which yields the claimed result. In the formula ej referrs to the column vector of length r which has a 1 in the j th position. This proves the lemma. Theorem 4.9.2 Let B be a q × p block matrix as in 4.9.31 and let A be a p × n block matrix as in 4.9.32 such that Bis is conformable with Asj and each product, Bis Asj for s = 1, · · · , p is of the same size so they can be added. Then BA can be obtained as a block matrix such that the ij th block is of the form X Bis Asj . (4.9.33) s

Proof: From 4.9.30 Bis Asj =

¡

0

Iri ×ri

 0

¢

0



B  Ips ×ps  0

 ¡

0

Ips ×ps

0

¢

0



A  Iqj ×qj  0

where here it is assumed Bis is ri × ps and Asj is ps × qj . The product involves the sth block in the ith row of blocks for B and the sth block in the j th column of A. Thus there are the same number of rows above the Ips ×ps as there are columns to the left of Ips ×ps in those two inside matrices. Then from Lemma 4.9.1     0 0 0 0 ¡ ¢  Ips ×ps  0 Ips ×ps 0 =  0 Ips ×ps 0  0 0 0 0 Since the blocks of small identity matrices do not overlap,     I 0 p1 ×p1 0 0 0 X  ..  0 Ips ×ps 0  =   =I . s 0 0 0 0 Ipp ×pp

82


and so

X

Bis Asj =

s

 X¡

0

=

¡

0

Iri ×ri

0



 ¡

¢

0



B  Ips ×ps  0 Ips ×ps 0 A  Iqj ×qj  0 0     0 0 ¡ ¢ ¢ 0 BIA  Iqj ×qj  = 0 Iri ×ri 0 BA  Iqj ×qj  0 0

Iri ×ri

s

¢

0

Hence the ij th block of BA equals the formal multiplication according to matrix multiplication, X Bis Asj . s

This proves the theorem. Example 4.9.3 Let an n × n matrix have the form µ ¶ a b A= c P where P is n − 1 × n − 1. Multiply it by µ ¶ p q B= r Q where B is also an n × n matrix and Q is n − 1 × n − 1. You use block multiplication µ ¶µ ¶ µ ¶ a b p q ap + br aq + bQ = c P r Q pc + P r cq + P Q Note that this all makes sense. For example, b = 1 × n − 1 and r = n − 1 × 1 so br is a 1 × 1. Similar considerations apply to the other blocks. Here is an interesting and significant application of block multiplication. In this theorem, pM (t) denotes the characteristic polynomial, det (tI − M ) . Thus the zeros of this polynomial are the eigenvalues of the matrix, M . Theorem 4.9.4 Let A be an m×n matrix and let B be an n×m matrix for m ≤ n. Then pBA (t) = tn−m pAB (t) , so the eigenvalues of BA and AB are the same including multiplicities except that BA has n − m extra zero eigenvalues.

4.10. EXERCISES

83

Proof: Use block multiplication to write µ ¶µ ¶ µ AB 0 I A = B 0 0 I µ ¶µ ¶ µ I A 0 0 = 0 I B BA

AB B

ABA BA

AB B

ABA BA

¶ ¶ .

Therefore, µ

¶ 0 BA µ ¶ µ ¶ 0 0 AB 0 Since the two matrices above are similar it follows that and B BA B 0 have the same characteristic polynomials. Therefore, noting that BA is an n × n matrix and AB is an m × m matrix, I 0

A I

¶−1 µ

AB B

0 0

¶µ

I 0

A I

¶

µ

=

0 B

tm det (tI − BA) = tn det (tI − AB) and so det (tI − BA) = pBA (t) = tn−m det (tI − AB) = tn−m pAB (t) . This proves the theorem.

4.10

Exercises

1. Show that matrix multiplication is associative. That is, (AB) C = A (BC) . 2. Show the inverse of a matrix, if it exists, is unique. Thus if AB = BA = I, then B = A−1 . 3. In the proof of Theorem 4.5.15 it was claimed that det (I) = 1. Here I = (δ ij ) . Prove this assertion. Also prove Corollary 4.5.18. 4. Let v1 , · · · , vn be vectors in Fn and let M (v1 , · · · , vn ) denote the matrix whose ith column equals vi . Define d (v1 , · · · , vn ) ≡ det (M (v1 , · · · , vn )) . Prove that d is linear in each variable, (multilinear), that d (v1 , · · · , vi , · · · , vj , · · · , vn ) = −d (v1 , · · · , vj , · · · , vi , · · · , vn ) , (4.10.34) and d (e1 , · · · , en ) = 1

(4.10.35)

n

where here ej is the vector in F which has a zero in every position except the j th position in which it has a one. 5. Suppose f : Fn × · · · × Fn → F satisfies 4.10.34 and 4.10.35 and is linear in each variable. Show that f = d.

84


6. Show that if you replace a row (column) of an n×n matrix A with itself added to some multiple of another row (column) then the new matrix has the same determinant as the original one. P 7. If A = (aij ) , show det (A) = (k1 ,··· ,kn ) sgn (k1 , · · · , kn ) ak1 1 · · · akn n . 8. Use the result of Problem 6 to evaluate by hand the determinant   1 2 3 2  −6 3 2 3   det   5 2 2 3 . 3 4 6 4 9. Find the inverse if it exists of the matrix,   t e cos t sin t  et − sin t cos t  . et − cos t − sin t 10. Let Ly = y (n) +an−1 (x) y (n−1) +· · ·+a1 (x) y 0 +a0 (x) y where the ai are given continuous functions defined on a closed interval, (a, b) and y is some function which has n derivatives so it makes sense to write Ly. Suppose Lyk = 0 for k = 1, 2, · · · , n. The Wronskian of these functions, yi is defined as   y1 (x) ··· yn (x)  y10 (x) ··· yn0 (x)    W (y1 , · · · , yn ) (x) ≡ det  .. ..    . . (n−1) (n−1) (x) y1 (x) · · · yn Show that for W (x) = W (y1 , · · · , yn ) (x) to save space,  y1 (x) · · · yn (x)  y10 (x) · · · yn0 (x)  W 0 (x) = det  .. ..  . . (n)

y1 (x) · · ·

   . 

(n)

yn (x)

Now use the differential equation, Ly = 0 which is satisfied by each of these functions, yi and properties of determinants presented above to verify that W 0 + an−1 (x) W = 0. Give an explicit solution of this linear differential equation, Abel’s formula, and use your answer to verify that the Wronskian of these solutions to the equation, Ly = 0 either vanishes identically on (a, b) or never. 11. Two n × n matrices, A and B, are similar if B = S −1 AS for some invertible n × n matrix, S. Show that if two matrices are similar, they have the same characteristic polynomials.

4.11. SHUR’S THEOREM

85

12. Suppose the characteristic polynomial of an n × n matrix, A is of the form tn + an−1 tn−1 + · · · + a1 t + a0 and that a0 6= 0. Find a formula A−1 in terms of powers of the matrix, A. Show that A−1 exists if and only if a0 6= 0. 13. In constitutive modeling P of the stress and strain tensors, one sometimes con∞ siders sums of the form k=0 ak Ak where A is a 3×3 matrix. Show using the Cayley Hamilton theorem that if such a thing makes any sense, you can always obtain it as a finite sum having no more than n terms.

4.11

Shur’s Theorem

Every matrix is related to an upper triangular matrix in a particularly significant way. This is Shur’s theorem and it is the most important theorem in the spectral theory of matrices. Lemma 4.11.1 Let {x1 , · · · , xn } be a basis for Fn . Then there exists an orthonormal basis for Fn , {u1 , · · · , un } which has the property that for each k ≤ n, span (x1 , · · · , xk ) = span (u1 , · · · , uk ) . Proof: Let {x1 , · · · , xn } be a basis for Fn . Let u1 ≡ x1 / |x1 | . Thus for k = 1, span (u1 ) = span (x1 ) and {u1 } is an orthonormal set. Now suppose for some k < n, u1 , · · · , uk have been chosen such that (uj · ul ) = δ jl and span (x1 , · · · , xk ) = span (u1 , · · · , uk ). Then define Pk xk+1 − j=1 (xk+1 · uj ) uj ¯, uk+1 ≡ ¯¯ (4.11.36) Pk ¯ ¯xk+1 − j=1 (xk+1 · uj ) uj ¯ where the denominator is not equal to zero because the xj form a basis and so xk+1 ∈ / span (x1 , · · · , xk ) = span (u1 , · · · , uk ) Thus by induction, uk+1 ∈ span (u1 , · · · , uk , xk+1 ) = span (x1 , · · · , xk , xk+1 ) . Also, xk+1 ∈ span (u1 , · · · , uk , uk+1 ) which is seen easily by solving 4.11.36 for xk+1 and it follows span (x1 , · · · , xk , xk+1 ) = span (u1 , · · · , uk , uk+1 ) .

86


If l ≤ k,  (uk+1 · ul )

= C (xk+1 · ul ) −

k X

 (xk+1 · uj ) (uj · ul )

j=1

 = C (xk+1 · ul ) −

k X

 (xk+1 · uj ) δ lj 

j=1

= C ((xk+1 · ul ) − (xk+1 · ul )) = 0. n

The vectors, {uj }j=1 , generated in this way are therefore an orthonormal basis because each vector has unit length. The process by which these vectors were generated is called the Gram Schmidt process. Recall the following definition. Definition 4.11.2 An n × n matrix, U, is unitary if U U ∗ = I = U ∗ U where U ∗ is defined to be the transpose of the conjugate of U. Theorem 4.11.3 Let A be an n × n matrix. Then there exists a unitary matrix, U such that U ∗ AU = T, (4.11.37) where T is an upper triangular matrix having the eigenvalues of A on the main diagonal listed according to multiplicity as roots of the characteristic equation. Proof: Let v1 be a unit eigenvector for A . Then there exists λ1 such that Av1 = λ1 v1 , |v1 | = 1. Extend {v1 } to a basis and then use Lemma 4.11.1 to obtain {v1 , · · · , vn }, an orthonormal basis in Fn . Let U0 be a matrix whose ith column is vi . Then from the above, it follows U0 is unitary. Then U0∗ AU0 is of the form   λ1 ∗ · · · ∗  0     ..   .  A 1

0 where A1 is an n − 1 × n − 1 matrix. Repeat the process for the matrix, A1 above. e1 such that U e ∗ A1 U e1 is of the form There exists a unitary matrix U 1   λ2 ∗ · · · ∗  0     .. .  .  A 2

0


87

Now let U1 be the n × n matrix of the form µ ¶ 1 0 e1 . 0 U This is also a unitary matrix because by block multiplication, µ ¶∗ µ ¶ µ ¶µ ¶ 1 0 1 0 1 0 1 0 = e1 e1 e∗ e1 0 U 0 U 0 U 0 U 1 ¶ µ ¶ µ 1 0 1 0 = = e e ∗U 0 I 0 U 1 1 Then using block multiplication, U1∗ U0∗ AU0 U1 is of the form   λ1 ∗ ∗ ··· ∗  0 λ2 ∗ · · · ∗      0 0    ..  ..   . . A2 0 0 where A2 is an n − 2 × n − 2 matrix. Continuing in this way, there exists a unitary matrix, U given as the product of the Ui in the above construction such that U ∗ AU = T where T is some upper triangular matrix. Since the matrix is upper triangular, the Qn characteristic equation is i=1 (λ − λi ) where the λi are the diagonal entries of T. Therefore, the λi are the eigenvalues. What if A is a real matrix and you only want to consider real unitary matrices? Theorem 4.11.4 Let A be a real n × n matrix. Then there exists a real unitary matrix, Q and a matrix T of the form   P1 · · · ∗  .  .. T = (4.11.38) . ..  0 Pr where Pi equals either a real 1 × 1 matrix or Pi equals a real 2 × 2 matrix having two complex eigenvalues of A such that QT AQ = T. The matrix, T is called the real Schur form of the matrix A. Proof: Suppose Av1 = λ1 v1 , |v1 | = 1 where λ1 is real. Then let {v1 , · · · , vn } Let Q0 be a matrix whose ith column is  λ1 ∗  0   ..  . 0

be an orthonormal basis of vectors in Rn . vi . Then Q∗0 AQ0 is of the form  ··· ∗     A1

88


where A1 is a real n − 1 × n − 1 matrix. This is just like the proof of Theorem 4.11.3 up to this point. Now in case λ1 = α + iβ, it follows since A is real that v1 = z1 + iw1 and that v1 = z1 − iw1 is an eigenvector for the eigenvalue, α − iβ. Here z1 and w1 are real vectors. It is clear that {z1 , w1 } is an independent set of vectors in Rn . Indeed,{v1 , v1 } is an independent set and it follows span (v1 , v1 ) = span (z1 , w1 ) . Now using the Gram Schmidt theorem in Rn , there exists {u1 , u2 } , an orthonormal set of real vectors such that span (u1 , u2 ) = span (v1 , v1 ) . Now let {u1 , u2 , · · · , un } be an orthonormal basis in Rn and let Q0 be a unitary matrix whose ith column is ui . Then Auj are both in span (u1 , u2 ) for j = 1, 2 and so uTk Auj = 0 whenever k ≥ 3. It follows that Q∗0 AQ0 is of the form   ∗ ∗ ··· ∗  ∗ ∗     0     ..   .  A 1

0 e 1 an n − 2 × n − 2 where A1 is now an n − 2 × n − 2 matrix. In this case, find Q matrix to put A1 in an appropriate form as above and come up with A2 either an n − 4 × n − 4 matrix or an n − 3 × n − 3 matrix. Then the only other difference is to let   1 0 0 ··· 0  0 1 0 ··· 0      Q1 =  0 0   .. ..  e  . .  Q1 0 0 thus putting a 2 × 2 identity matrix in the upper left corner rather than a one. Repeating this process with the above modification for the case of a complex eigenvalue leads eventually to 4.11.38 where Q is the product of real unitary matrices Qi above. Finally,   λI1 − P1 · · · ∗   .. .. λI − T =   . . 0

λIr − Pr

where Ik is the 2 × 2 identity matrix in the case that Pk is 2 × 2 and is the number Qr 1 in the case where Pk is a 1 × 1 matrix. Now, it follows that det (λI − T ) = k=1 det (λIk − Pk ) . Therefore, λ is an eigenvalue of T if and only if it is an eigenvalue of some Pk . This proves the theorem since the eigenvalues of T are the same as those of A because they have the same characteristic polynomial due to the similarity of A and T. Definition 4.11.5 When a linear transformation, A, mapping a linear space, V to V has a basis of eigenvectors, the linear transformation is called non defective.


89

Otherwise it is called defective. An n×n matrix, A, is called normal if AA∗ = A∗ A. An important class of normal matrices is that of the Hermitian or self adjoint matrices. An n × n matrix, A is self adjoint or Hermitian if A = A∗ . The next lemma is the basis for concluding that every normal matrix is unitarily similar to a diagonal matrix. Lemma 4.11.6 If T is upper triangular and normal, then T is a diagonal matrix. Proof: Since T is normal, T ∗ T = T T ∗ . Writing this in terms of components and using the description of the adjoint as the transpose of the conjugate, yields the following for the ik th entry of T ∗ T = T T ∗ . X X X X tij t∗jk = tij tkj = t∗ij tjk = tji tjk . j

j

j

j

Now use the fact that T is upper triangular and let i = k = 1 to obtain the following from the above. X X 2 2 2 |t1j | = |tj1 | = |t11 | j

j

You see, tj1 = 0 unless j = 1 due to the assumption that T is upper triangular. This shows T is of the form   ∗ 0 ··· 0  0 ∗ ··· ∗     .. . . . . ..  . . . ..  0 ··· 0 ∗ Now do the same thing only this time take i = k = 2 and use the result just established. Thus, from the above, X X 2 2 2 |t2j | = |tj2 | = |t22 | , j

j

showing that t2j = 0 if j > 2 which means  ∗ 0 0  0 ∗ 0   0 0 ∗   .. .. . .  . . . 0

0

0

T has the form  ··· 0 ··· 0   ··· ∗  . ..  .. . .  0 ∗

Next let i = k = 3 and obtain that T looks like a diagonal matrix in so far as the first 3 rows and columns are concerned. Continuing in this way it follows T is a diagonal matrix. Theorem 4.11.7 Let A be a normal matrix. Then there exists a unitary matrix, U such that U ∗ AU is a diagonal matrix.

90


Proof: From Theorem 4.11.3 there exists a unitary matrix, U such that U ∗ AU equals an upper triangular matrix. The theorem is now proved if it is shown that the property of being normal is preserved under unitary similarity transformations. That is, verify that if A is normal and if B = U ∗ AU, then B is also normal. But this is easy. B∗B

U ∗ A∗ U U ∗ AU = U ∗ A∗ AU U ∗ AA∗ U = U ∗ AU U ∗ A∗ U = BB ∗ .

= =

Therefore, U ∗ AU is a normal and upper triangular matrix and by Lemma 4.11.6 it must be a diagonal matrix. This proves the theorem. Corollary 4.11.8 If A is Hermitian, then all the eigenvalues of A are real and there exists an orthonormal basis of eigenvectors. Proof: Since A is normal, there exists unitary, U such that U ∗ AU = D, a diagonal matrix whose diagonal entries are the eigenvalues of A. Therefore, D∗ = U ∗ A∗ U = U ∗ AU = D showing D is real. Finally, let ¡ ¢ U = u1 u2 · · · un where the ui denote the columns of U and  λ1  .. D=

0 .

0

  

λn

The equation, U ∗ AU = D implies AU

= =

¡

Au1

UD =

¡

Au2 λ1 u1

···

Aun

λ2 u2

¢

···

λn u n

¢

where the entries denote the columns of AU and U D respectively. Therefore, Aui = λi ui and since the matrix is unitary, the ij th entry of U ∗ U equals δ ij and so δ ij = uTi uj = uTi uj = ui · uj . This proves the corollary because it shows the vectors {ui } form an orthonormal basis. Corollary 4.11.9 If A is a real symmetric matrix, then A is Hermitian and there exists a real unitary matrix, U such that U T AU = D where D is a diagonal matrix. Proof: This follows from Theorem 4.11.4 and Corollary 4.11.8.

4.12. THE RIGHT POLAR DECOMPOSITION

4.12

91

The Right Polar Decomposition

The right polar decomposition involves writing a matrix as a product of two other matrices, one which preserves distances and the other which stretches and distorts. First here are some lemmas. Lemma 4.12.1 Let A be a Hermitian matrix such that all its eigenvalues are nonnegative. Then there exists a Hermitian matrix, A1/2 such that A1/2 has all non¡ ¢2 negative eigenvalues and A1/2 = A. Proof: Since A is Hermitian, there exists a diagonal matrix D having all real nonnegative entries and a unitary matrix U such that A = U ∗ DU. Then denote by D1/2 the matrix which is obtained by replacing each diagonal entry of D with its square root. Thus D1/2 D1/2 = D. Then define A1/2 ≡ U ∗ D1/2 U. Then

³

Since D1/2 is real, ³

A1/2

´2

= U ∗ D1/2 U U ∗ D1/2 U = U ∗ DU = A.

U ∗ D1/2 U

´∗

³ ´∗ ∗ = U ∗ D1/2 (U ∗ ) = U ∗ D1/2 U

so A1/2 is Hermitian. This proves the lemma. There is also a useful observation about orthonormal sets of vectors which is stated in the next lemma. Lemma 4.12.2 Suppose {x1 , x2 , · · · , xr } is an orthonormal set of vectors. Then if c1 , · · · , cr are scalars, ¯ ¯2 r r ¯X ¯ X ¯ ¯ 2 ck xk ¯ = |ck | . ¯ ¯ ¯ k=1

k=1

Proof: This follows from the definition. From the properties of the dot product and using the fact that the given set of vectors is orthonormal,  ¯ r ¯2  r r ¯X ¯ X X ¯ ¯ ck xk ¯ =  ck xk , cj xj  ¯ ¯ ¯ k=1

=

k=1

X k,j

ck cj (xk , xj ) =

j=1

r X

2

|ck | .

k=1

This proves the lemma. Next it is helpful to recall the Gram Schmidt algorithm and observe a certain property stated in the next lemma.

92


Lemma 4.12.3 Suppose {w1 , · · · , wr , vr+1 , · · · , vp } is a linearly independent set of vectors such that {w1 , · · · , wr } is an orthonormal set of vectors. Then when the Gram Schmidt process is applied to the vectors in the given order, it will not change any of the w1 , · · · , wr . Proof: Let {u1 , · · · , up } be the orthonormal set delivered by the Gram Schmidt process. Then u1 = w1 because by definition, u1 ≡ w1 / |w1 | = w1 . Now suppose uj = wj for all j ≤ k ≤ r. Then if k < r, consider the definition of uk+1 . wk+1 −

Pk+1

j=1 (wk+1 , uj ) uj ¯ uk+1 ≡ ¯¯ Pk+1 ¯ ¯wk+1 − j=1 (wk+1 , uj ) uj ¯

By induction, uj = wj and so this reduces to wk+1 / |wk+1 | = wk+1 . This proves the lemma. This lemma immediately implies the following lemma. Lemma 4.12.4 Let V be a subspace of dimension p and let {w1 , · · · , wr } be an orthonormal set of vectors in V . Then this orthonormal set of vectors may be extended to an orthonormal basis for V, {w1 , · · · , wr , yr+1 , · · · , yp } Proof: First extend the given linearly independent set {w1 , · · · , wr } to a basis for V and then apply the Gram Schmidt theorem to the resulting basis. Since {w1 , · · · , wr } is orthonormal it follows from Lemma 4.12.3 the result is of the desired form, an orthonormal basis extending {w1 , · · · , wr }. This proves the lemma. Here is another lemma about preserving distance. Lemma 4.12.5 Suppose R is an m × n matrix with m > n and R preserves distances. Then R∗ R = I. Proof: Since R preserves distances, |Rx| = |x| for every x. Therefore from the axioms of the dot product, 2

2

|x| + |y| + (x, y) + (y, x) =

2

|x + y|

= (R (x + y) , R (x + y)) = (Rx,Rx) + (Ry,Ry) + (Rx, Ry) + (Ry, Rx) =

2

2

|x| + |y| + (R∗ Rx, y) + (y, R∗ Rx)

and so for all x, y, (R∗ Rx − x, y) + (y,R∗ Rx − x) = 0 Hence for all x, y, Re (R∗ Rx − x, y) = 0


93

Now for a x, y given, choose α ∈ C such that α (R∗ Rx − x, y) = |(R∗ Rx − x, y)| Then 0 = Re (R∗ Rx − x,αy) = Re α (R∗ Rx − x, y) = |(R∗ Rx − x, y)| Thus |(R∗ Rx − x, y)| = 0 for all x, y because the given x, y were arbitrary. Let y = R∗ Rx − x to conclude that for all x, R∗ Rx − x = 0 which says R∗ R = I since x is arbitrary. This proves the lemma. With this preparation, here is the big theorem about the right polar decomposition. Theorem 4.12.6 Let F be an m × n matrix where m ≥ n. Then there exists a Hermitian n × n matrix, U which has all nonnegative eigenvalues and an m × n matrix, R which preserves distances and satisfies R∗ R = I such that F = RU. Proof: Consider F ∗ F. This is a Hermitian matrix because ∗

∗

(F ∗ F ) = F ∗ (F ∗ ) = F ∗ F Also the eigenvalues of the n × n matrix F ∗ F are all nonnegative. This is because if x is an eigenvalue, λ (x, x) = (F ∗ F x, x) = (F x,F x) ≥ 0. Therefore, by Lemma 4.12.1, there exists an n × n Hermitian matrix, U having all nonnegative eigenvalues such that U 2 = F ∗ F. Consider the subspace U (Fn ). Let {U x1 , · · · , U xr } be an orthonormal basis for U (Fn ) ⊆ Fn . Note that U (Fn ) might not be all of Fn . Using Lemma 4.12.4, extend to an orthonormal basis for all of Fn , {U x1 , · · · , U xr , yr+1 , · · · , yn } . Next observe that {F x1 , · · · , F xr } is also an orthonormal set of vectors in Fm . This is because ¡ ¢ (F xk , F xj ) = (F ∗ F xk , xj ) = U 2 xk , xj = (U xk , U ∗ xj ) = (U xk , U xj ) = δ jk

94


Therefore, from Lemma 4.12.4 again, this orthonormal set of vectors can be extended to an orthonormal basis for Fm , {F x1 , · · · , F xr , zr+1 , · · · , zm } Thus there are at least as many zk as there are yj . Now for x ∈ Fn , since {U x1 , · · · , U xr , yr+1 , · · · , yn } is an orthonormal basis for Fn , there exist unique scalars, c1 · · · , cr , dr+1 , · · · , dn such that x=

r X

ck U xk +

Rx ≡

r X

dk yk

j=r+1

k=1

Define

n X

n X

ck F xk +

dk zk

(4.12.39)

j=r+1

k=1

Then also there exist scalars bk such that Ux =

r X

bk U xk

k=1

and so from 4.12.39,applied to U x in place of x Ã r ! r X X bk xk RU x = bk F xk = F k=1

k=1

Pr Is F ( k=1 bk xk ) = F (x)? Ã Ã r ! Ã r ! ! X X F bk xk − F (x) , F bk xk − F (x) k=1

k=1

Ã

Ã ∗

=

(F F ) Ã

=

Ã U

Ã =

U Ã

=

2

r X

! Ã bk xk − x ,

k=1 r X

k=1 Ã r X

! Ã

bk xk − x , !

bk xk − x , U

k=1 r X k=1

bk U xk − U x,

r X

k=1

bk xk − x

k=1 r X

!!

bk xk − x

k=1 Ã r X k=1

r X

!!

!! bk xk − x !

bk U xk − U x

=0


95

Pr Therefore, F ( k=1 bk xk ) = F (x) and this shows RU x = F x. From 4.12.39 and Lemma 4.12.2 R preserves distances. Therefore, by Lemma 4.12.5 R∗ R = I. This proves the theorem.

96


Multi-variable Calculus 5.1

Continuous Functions

In what follows, F will denote either R or C. It turns out it is more efficient to not make a distinction. However, the main interest is in R so if you like, you can think R whenever you see F.

5.1.1

Distance In Fn

It is necessary to give a generalization of the dot product for vectors in Cn . This definition reduces to the usual one in the case the components of the vector are real. Definition 5.1.1 Let x, y ∈ Cn . Thus x = (x1 , · · · , xn ) where each xk ∈ C and a similar formula holding for y. Then the dot product of these two vectors is defined to be X x·y ≡ xj yj ≡ x1 y1 + · · · + xn yn . j

Notice how you put the conjugate on the entries of the vector, y. It makes no difference if the vectors happen to be real vectors but with complex vectors you must do it this way. The reason for this is that when you take the dot product of a vector with itself, you want to get the square of the length of the vector, a positive number. Placing the conjugate on the components of y in the above definition assures this will take place. Thus X X 2 x·x= xj xj = |xj | ≥ 0. j

j

If you didn’t place a conjugate as in the above definition, things wouldn’t work out correctly. For example, 2 (1 + i) + 22 = 4 + 2i and this is not a positive number. The following properties of the dot product follow immediately from the definition and you should verify each of them. Properties of the dot product: 97

98

MULTI-VARIABLE CALCULUS

1. u · v = v · u. 2. If a, b are numbers and u, v, z are vectors then (au + bv) · z = a (u · z) + b (v · z) . 3. u · u ≥ 0 and it equals 0 if and only if u = 0. The norm is defined in the usual way. Definition 5.1.2 For x ∈ Cn , Ã |x| ≡

n X

!1/2 2

1/2

|xk |

= (x · x)

k=1

Here is a fundamental inequality called the Cauchy Schwarz inequality which is stated here in Cn . First here is a simple lemma. Lemma 5.1.3 If z ∈ C there exists θ ∈ C such that θz = |z| and |θ| = 1. Proof: Let θ = 1 if z = 0 and otherwise, let θ =

z . Recall that for z = |z|

2

x + iy, z = x − iy and zz = |z| . Theorem 5.1.4 (Cauchy Schwarz)The following inequality holds for xi and yi ∈ C. ¯ ¯ Ã n !1/2 Ã n !1/2 n ¯X ¯ X X ¯ ¯ 2 2 |(x · y)| = ¯ |xi | |yi | xi y i ¯ ≤ = |x| |y| ¯ ¯ i=1

i=1

i=1

Proof: Let θ ∈ C such that |θ| = 1 and ¯ ¯ n n ¯X ¯ X ¯ ¯ θ xi y i = ¯ xi y i ¯ ¯ ¯ i=1

Thus θ

n X

xi y i =

i=1

n X

i=1

Consider p (t) ≡ 0

¡

xi θyi

¢

i=1

¯ n ¯ ¯X ¯ ¯ ¯ =¯ xi y i ¯ . ¯ ¯ i=1

´ ¡ ¢³ xi + tθyi where t ∈ R. i=1 xi + tθyi

Pn

≤

p (t) =

n X

Ã |xi | + 2t Re θ

i=1

=

2

n X

i=1 ¯ ¯ n ¯X ¯ ¯ ¯ 2 2 |x| + 2t ¯ xi y i ¯ + t2 |y| ¯ ¯ i=1

! xi y i

+t

2

n X i=1

|yi |

2

(5.1.1)

5.1. CONTINUOUS FUNCTIONS

99

If |y| = 0 then 5.1.1 is obviously true because both sides equal zero. Therefore, assume |y| 6= 0 and then p (t) is a polynomial of degree two whose graph opens up. Therefore, it either has no zeroes, two zeros or one repeated zero. If it has two zeros, the above inequality must be violated because in this case the graph must dip below the x axis. Therefore, it either has no zeros or exactly one. From the quadratic formula this happens exactly when ¯ n ¯2 ¯X ¯ ¯ ¯ 2 2 xi y i ¯ − 4 |x| |y| ≤ 0 4¯ ¯ ¯ i=1

and so

¯ n ¯ ¯X ¯ ¯ ¯ xi y i ¯ ≤ |x| |y| ¯ ¯ ¯ i=1

as claimed. This proves the inequality. By analogy to the case of Rn , length or magnitude of vectors in Cn can be defined. Definition 5.1.5 Let z ∈ Cn . Then |z| ≡ (z · z) be referred to as scalars.

1/2

. Also numbers in F will often

Theorem 5.1.6 For length defined in Definition 5.1.5, the following hold. |z| ≥ 0 and |z| = 0 if and only if z = 0

(5.1.2)

If α is a scalar, |αz| = |α| |z|

(5.1.3)

|z + w| ≤ |z| + |w| .

(5.1.4)

Proof: The first two claims are left as exercises. To establish the third, |z + w|

2

= (z + w, z + w) = z·z+w·w+w·z+z·w =

2

2

2

2

2

2

|z| + |w| + 2 Re w · z

≤ |z| + |w| + 2 |w · z| 2

≤ |z| + |w| + 2 |w| |z| = (|z| + |w|) . The main difference between Cn and Rn is that the scalars are complex numbers. Definition 5.1.7 Suppose you have a vector space, V and for z, w ∈ V and α a scalar a norm is a way of measuring distance or magnitude which satisfies the properties 5.1.2 - 5.1.4. Thus a norm is something which does the following. ||z|| ≥ 0 and ||z|| = 0 if and only if z = 0

(5.1.5)

If α is a scalar, ||αz|| = |α| ||z||

(5.1.6)

||z + w|| ≤ ||z|| + ||w|| .

(5.1.7)

Here is is understood that for all z ∈ V, ||z|| ∈ [0, ∞). Note that |·| provides a norm on Fn from the above.

100

5.2


Open And Closed Sets

Eventually, one must consider functions which are defined on subsets of Fn and their properties. The next definition will end up being quite important. It describe a type of subset of Fn with the property that if x is in this set, then so is y whenever y is close enough to x. Definition 5.2.1 Let U ⊆ Fn . U is an open set if whenever x ∈ U, there exists r > 0 such that B (x, r) ⊆ U. More generally, if U is any subset of Fn , x ∈ U is an interior point of U if there exists r > 0 such that x ∈ B (x, r) ⊆ U. In other words U is an open set exactly when every point of U is an interior point of U . If there is something called an open set, surely there should be something called a closed set and here is the definition of one. Definition 5.2.2 A subset, C, of Fn is called a closed set if Fn \ C is an open set. They symbol, Fn \ C denotes everything in Fn which is not in C. It is also called the complement of C. The symbol, S C is a short way of writing Fn \ S. To illustrate this definition, consider the following picture.

qx B(x, r)

U

You see in this picture how the edges are dotted. This is because an open set, can not include the edges or the set would fail to be open. For example, consider what would happen if you picked a point out on the edge of U in the above picture. Every open ball centered at that point would have in it some points which are outside U . Therefore, such a point would violate the above definition. You also see the edges of B (x, r) dotted suggesting that B (x, r) ought to be an open set. This is intuitively clear but does require a proof. This will be done in the next theorem and will give examples of open sets. Also, you can see that if x is close to the edge of U, you might have to take r to be very small. It is roughly the case that open sets don’t have their skins while closed sets do. Here is a picture of a closed set, C.

5.2. OPEN AND CLOSED SETS

101

C

qx B(x, r)

Note that x ∈ / C and since Fn \ C is open, there exists a ball, B (x, r) contained n entirely in F \ C. If you look at Fn \ C, what would be its skin? It can’t be in Fn \ C and so it must be in C. This is a rough heuristic explanation of what is going on with these definitions. Also note that Fn and ∅ are both open and closed. Here is why. If x ∈ ∅, then there must be a ball centered at x which is also contained in ∅. This must be considered to be true because there is nothing in ∅ so there can be no example to show it false1 . Therefore, from the definition, it follows ∅ is open. It is also closed because if x ∈ / ∅, then B (x, 1) is also contained in Fn \ ∅ = Fn . Therefore, ∅ is both open and closed. From this, it follows Fn is also both open and closed. Theorem 5.2.3 Let x ∈ Fn and let r ≥ 0. Then B (x, r) is an open set. Also, D (x, r) ≡ {y ∈ Fn : |y − x| ≤ r} is a closed set. Proof: Suppose y ∈ B (x,r) . It is necessary to show there exists r1 > 0 such that B (y, r1 ) ⊆ B (x, r) . Define r1 ≡ r − |x − y| . Then if |z − y| < r1 , it follows from the above triangle inequality that |z − x| = ≤
r and defining δ ≡ |x − y| − r, it follows that if z ∈ B (y, δ) , then by the triangle inequality, |x − z| ≥ |x − y| − |y − z| > |x − y| − δ = |x − y| − (|x − y| − r) = r and this shows that B (y, δ) ⊆ Fn \ D (x, r) . Since y was an arbitrary point in Fn \ D (x, r) , it follows Fn \ D (x, r) is an open set which shows from the definition that D (x, r) is a closed set as claimed. A picture which is descriptive of the conclusion of the above theorem which also implies the manner of proof is the following. 6 r rq q1 x y B(x, r)

5.3

6 r q x

rq1 y

D(x, r)

Continuous Functions

With the above definition of the norm in Fp , it becomes possible to define continuity. Definition 5.3.1 A function f : D (f ) ⊆ Fp → Fq is continuous at x ∈ D (f ) if for each ε > 0 there exists δ > 0 such that whenever y ∈ D (f ) and |y − x| < δ it follows that |f (x) − f (y)| < ε. f is continuous if it is continuous at every point of D (f ) . Note the total similarity to the scalar valued case.

5.3.1

Sufficient Conditions For Continuity

The next theorem is a fundamental result which will allow us to worry less about the ε δ definition of continuity. Theorem 5.3.2 The following assertions are valid. 1. The function, af + bg is continuous at x whenever f , g are continuous at x ∈ D (f ) ∩ D (g) and a, b ∈ F.

5.4. EXERCISES

103

2. If f is continuous at x, f (x) ∈ D (g) ⊆ Fp , and g is continuous at f (x) ,then g ◦ f is continuous at x. 3. If f = (f1 , · · · , fq ) : D (f ) → Fq , then f is continuous if and only if each fk is a continuous F valued function. 4. The function f : Fp → F, given by f (x) = |x| is continuous. The proof of this theorem is in the last section of this chapter. Its conclusions are not surprising. For example the first claim says that (af + bg) (y) is close to (af + bg) (x) when y is close to x provided the same can be said about f and g. For the second claim, if y is close to x, f (x) is close to f (y) and so by continuity of g at f (x), g (f (y)) is close to g (f (x)) . To see the third claim is likely, note that closeness in Fp is the same as closeness in each coordinate. The fourth claim is immediate from the triangle inequality. For functions defined on Fn , there is a notion of polynomial just as there is for functions defined on R. Definition 5.3.3 Let α be an n dimensional multi-index. This means α = (α1 , · · · , αn ) where each αi is a natural number or zero. Also, let |α| ≡

n X

|αi |

i=1

The symbol, xα ,means

αn 1 α2 xα ≡ xα 1 x2 · · · x3 .

An n dimensional polynomial of degree m is a function of the form X p (x) = dα xα. |α|≤m

where the dα are complex or real numbers. The above theorem implies that polynomials are all continuous.

5.4

Exercises

1. Let f (t) = (t, sin t) . Show f is continuous at every point t. 2. Suppose |f (x) − f (y)| ≤ K |x − y| where K is a constant. Show that f is everywhere continuous. Functions satisfying such an inequality are called Lipschitz functions. α

3. Suppose |f (x) − f (y)| ≤ K |x − y| where K is a constant and α ∈ (0, 1). Show that f is everywhere continuous.

104


4. Suppose f : F3 → F is given by f (x) = 3x1 x2 + 2x23 . Use Theorem 5.3.2 to verify that f is continuous. Hint: You should first verify that the function, π k : F3 → F given by π k (x) = xk is a continuous function. 5. Generalize the previous problem to the case where f : Fq → F is a polynomial. 6. State and prove a theorem which involves quotients of functions encountered in the previous problem.

5.5

Limits Of A Function

As in the case of scalar valued functions of one variable, a concept closely related to continuity is that of the limit of a function. The notion of limit of a function makes sense at points, x, which are limit points of D (f ) and this concept is defined next. Definition 5.5.1 Let A ⊆ Fm be a set. A point, x, is a limit point of A if B (x, r) contains infinitely many points of A for every r > 0. Definition 5.5.2 Let f : D (f ) ⊆ Fp → Fq be a function and let x be a limit point of D (f ) . Then lim f (y) = L y→x

if and only if the following condition holds. For all ε > 0 there exists δ > 0 such that if 0 < |y − x| < δ, and y ∈ D (f ) then, |L − f (y)| < ε. Theorem 5.5.3 If limy→x f (y) = L and limy→x f (y) = L1 , then L = L1 . Proof: Let ε > 0 be given. There exists δ > 0 such that if 0 < |y − x| < δ and y ∈ D (f ) , then |f (y) − L| < ε, |f (y) − L1 | < ε. Pick such a y. There exists one because x is a limit point of D (f ) . Then |L − L1 | ≤ |L − f (y)| + |f (y) − L1 | < ε + ε = 2ε. Since ε > 0 was arbitrary, this shows L = L1 . As in the case of functions of one variable, one can define what it means for limy→x f (x) = ±∞. Definition 5.5.4 If f (x) ∈ F, limy→x f (x) = ∞ if for every number l, there exists δ > 0 such that whenever |y − x| < δ and y ∈ D (f ) , then f (x) > l. The following theorem is just like the one variable version presented earlier.

5.5. LIMITS OF A FUNCTION

105

Theorem 5.5.5 Suppose limy→x f (y) = L and limy→x g (y) = K where K, L ∈ Fq . Then if a, b ∈ F, lim (af (y) + bg (y)) = aL + bK,

(5.5.8)

lim f · g (y) = LK

(5.5.9)

y→x

y→x

and if g is scalar valued with limy→x g (y) = K 6= 0, lim f (y) g (y) = LK.

y→x

(5.5.10)

Also, if h is a continuous function defined near L, then lim h ◦ f (y) = h (L) .

y→x

(5.5.11)

Suppose limy→x f (y) = L. If |f (y) − b| ≤ r for all y sufficiently close to x, then |L − b| ≤ r also. Proof: The proof of 5.5.8 is left for you. It is like a corresponding theorem for continuous functions. Now 5.5.9is to be verified. Let ε > 0 be given. Then by the triangle inequality, |f · g (y) − L · K| ≤ |fg (y) − f (y) · K| + |f (y) · K − L · K| ≤ |f (y)| |g (y) − K| + |K| |f (y) − L| . There exists δ 1 such that if 0 < |y − x| < δ 1 and y ∈ D (f ) , then |f (y) − L| < 1, and so for such y, the triangle inequality implies, |f (y)| < 1 + |L| . Therefore, for 0 < |y − x| < δ 1 , |f · g (y) − L · K| ≤ (1 + |K| + |L|) [|g (y) − K| + |f (y) − L|] .

(5.5.12)

Now let 0 < δ 2 be such that if y ∈ D (f ) and 0 < |x − y| < δ 2 , |f (y) − L|
0 given, there exists η > 0 such that if |y − L| < η, then |h (y) −h (L)| < ε

106


Now since limy→x f (y) = L, there exists δ > 0 such that if 0 < |y − x| < δ, then |f (y) −L| < η. Therefore, if 0 < |y − x| < δ, |h (f (y)) −h (L)| < ε. It only remains to verify the last assertion. Assume |f (y) − b| ≤ r for all y close enough to x. It is required to show that |L − b| ≤ r. If this is not true, then |L − b| > r. Consider B (L, |L − b| − r) . Since L is the limit of f , it follows f (y) ∈ B (L, |L − b| − r) whenever y ∈ D (f ) is close enough to x. Thus, by the triangle inequality, |f (y) − L| < |L − b| − r and so r

< |L − b| − |f (y) − L| ≤ ||b − L| − |f (y) − L|| ≤ |b − f (y)| ,

a contradiction to the assumption that |b − f (y)| ≤ r. Theorem 5.5.6 For f : D (f ) → Fq and x ∈ D (f ) a limit point of D (f ) , f is continuous at x if and only if lim f (y) = f (x) .

y→x

Proof: First suppose f is continuous at x a limit point of D (f ) . Then for every ε > 0 there exists δ > 0 such that if |y − x| < δ and y ∈ D (f ) , then |f (x) − f (y)| < ε. In particular, this holds if 0 < |x − y| < δ and this is just the definition of the limit. Hence f (x) = limy→x f (y) . Next suppose x is a limit point of D (f ) and limy→x f (y) = f (x) . This means that if ε > 0 there exists δ > 0 such that for 0 < |x − y| < δ and y ∈ D (f ) , it follows |f (y) − f (x)| < ε. However, if y = x, then |f (y) − f (x)| = |f (x) − f (x)| = 0 and so whenever y ∈ D (f ) and |x − y| < δ, it follows |f (x) − f (y)| < ε, showing f is continuous at x. The following theorem is important. Theorem 5.5.7 Suppose f : D (f ) → Fq . Then for x a limit point of D (f ) , lim f (y) = L

(5.5.13)

lim fk (y) = Lk

(5.5.14)

y→x

if and only if y→x

where f (y) ≡ (f1 (y) , · · · , fp (y)) and L ≡ (L1 , · · · , Lp ) .

5.5. LIMITS OF A FUNCTION

107

Proof: Suppose 5.5.13. Then letting ε > 0 be given there exists δ > 0 such that if 0 < |y − x| < δ, it follows |fk (y) − Lk | ≤ |f (y) − L| < ε which verifies 5.5.14. Now suppose 5.5.14 holds. Then letting ε > 0 be given, there exists δ k such that if 0 < |y − x| < δ k , then ε |fk (y) − Lk | < √ . p Let 0 < δ < min (δ 1 , · · · , δ p ) . Then if 0 < |y − x| < δ, it follows Ã |f (y) − L|

=

p X

!1/2 |fk (y) − Lk |

k=1

Ã
0 such that B (x, δ) possibly x itself. Argue that 33.3 is a In other words the concept is totally

The Limit Of A Sequence

As in the case of real numbers, one can consider the limit of a sequence of points in Fp . ∞

Definition 5.7.1 A sequence {an }n=1 converges to a, and write lim an = a or an → a

n→∞

if and only if for every ε > 0 there exists nε such that whenever n ≥ nε , |an −a| < ε. In words the definition says that given any measure of closeness, ε, the terms of the sequence are eventually all this close to a. There is absolutely no difference between this and the definition for sequences of numbers other than here bold face is used to indicate an and a are points in Fp . Theorem 5.7.2 If limn→∞ an = a and limn→∞ an = a1 then a1 = a.

5.7. THE LIMIT OF A SEQUENCE

109

Proof: Suppose a1 6= a. Then let 0 < ε < |a1 −a| /2 in the definition of the limit. It follows there exists nε such that if n ≥ nε , then |an −a| < ε and |an −a1 | < ε. Therefore, for such n, |a1 −a|

≤ |a1 −an | + |an −a| < ε + ε < |a1 −a| /2 + |a1 −a| /2 = |a1 −a| ,

a contradiction. As in the case of a vector valued function, it suffices to consider the components. This is the content of the next theorem. ¡ ¢ Theorem 5.7.3 Let an = an1 , · · · , anp ∈ Fp . Then limn→∞ an = a ≡ (a1 , · · · , ap ) if and only if for each k = 1, · · · , p, lim ank = ak .

(5.7.15)

n→∞

Proof: First suppose limn→∞ an = a. Then given ε > 0 there exists nε such that if n > nε , then |ank − ak | ≤ |an − a| < ε which establishes 5.7.15. Now suppose 5.7.15 holds for each k. Then letting ε > 0 be given there exist nk such that if n > nk , √ |ank − ak | < ε/ p. Therefore, letting nε > max (n1 , · · · , np ) , it follows that for n > nε , Ã |an − a| =

n X

!1/2 |ank

− ak |

2

Ã
0 be given and choose n1 such that if n ≥ n1 then |an −a| < 1. Then for such n, the triangle inequality and Cauchy Schwarz inequality imply |an · bn −a · b| ≤ |an · bn −an · b| + |an · b − a · b| ≤ |an | |bn −b| + |b| |an −a| ≤ (|a| + 1) |bn −b| + |b| |an −a| . Now let n2 be large enough that for n ≥ n2 , |bn −b|
max (n1 , n2 ) . For n ≥ nε , |an · bn −a · b|

≤ (|a| + 1) |bn −b| + |b| |an −a| ε ε < (|a| + 1) + |b| ≤ ε. 2 (|a| + 1) 2 (|b| + 1)

This proves 5.7.16. The proof of 5.7.17 is entirely similar and is left for you.

5.7.1

Sequences And Completeness

Recall the definition of a Cauchy sequence. Definition 5.7.6 {an } is a Cauchy sequence if for all ε > 0, there exists nε such that whenever n, m ≥ nε , |an −am | < ε. A sequence is Cauchy means the terms are “bunching up to each other” as m, n get large. ∞

Theorem 5.7.7 Let {an }n=1 be a Cauchy sequence in Fp . Then there exists a unique a ∈ Fp such that an → a. ¡ ¢ Proof: Let an = an1 , · · · , anp . Then |ank − am k | ≤ |an − am | ∞

which shows for each k = 1, · · · , p, it follows {ank }n=1 is a Cauchy sequence in F. This requires that both the real and imaginary parts of ank are Cauchy sequences ∞ in R which means the real and imaginary parts converge in R. This shows {ank }n=1

5.7. THE LIMIT OF A SEQUENCE

111

must converge to some ak . That is limn→∞ ank = ak . Letting a = (a1 , · · · , ap ) , it follows from Theorem 5.7.3 that lim an = a.

n→∞

This proves the theorem. Theorem 5.7.8 The set of terms in a Cauchy sequence in Fp is bounded in the sense that for all n, |an | < M for some M < ∞. Proof: Let ε = 1 in the definition of a Cauchy sequence and let n > n1 . Then from the definition, |an −an1 | < 1. It follows that for all n > n1 , |an | < 1 + |an1 | . Therefore, for all n, |an | ≤ 1 + |an1 | +

n1 X

|ak | .

k=1

This proves the theorem. Theorem 5.7.9 If a sequence {an } in Fp converges, then the sequence is a Cauchy sequence. Proof: Let ε > 0 be given and suppose an → a. Then from the definition of convergence, there exists nε such that if n > nε , it follows that |an −a|
0 be given. By continuity, there exists δ > 0 such that if |y − x| < δ, then |f (x) − f (y)| < ε. However, there exists nδ such that if n ≥ nδ , then |xn −x| < δ and so for all n this large, |f (x) −f (xn )| < ε which shows f (xn ) → f (x) . Now suppose the condition about taking convergent sequences to convergent sequences holds at x. Suppose f fails to be continuous at x. Then there exists ε > 0 and xn ∈ D (f ) such that |x − xn | < n1 , yet |f (x) −f (xn )| ≥ ε. But this is clearly a contradiction because, although xn → x, f (xn ) fails to converge to f (x) . It follows f must be continuous after all. This proves the theorem.

5.8

Properties Of Continuous Functions

Functions of p variables have many of the same properties as functions of one variable. First there is a version of the extreme value theorem generalizing the one dimensional case. Theorem 5.8.1 Let C be closed and bounded and let f : C → R be continuous. Then f achieves its maximum and its minimum on C. This means there exist, x1 , x2 ∈ C such that for all x ∈ C, f (x1 ) ≤ f (x) ≤ f (x2 ) . There is also the long technical theorem about sums and products of continuous functions. These theorems are proved in the next section. Theorem 5.8.2 The following assertions are valid 1. The function, af + bg is continuous at x when f , g are continuous at x ∈ D (f ) ∩ D (g) and a, b ∈ F. 2. If and f and g are each F valued functions continuous at x, then f g is continuous at x. If, in addition to this, g (x) 6= 0, then f /g is continuous at x. 3. If f is continuous at x, f (x) ∈ D (g) ⊆ Fp , and g is continuous at f (x) ,then g ◦ f is continuous at x. 4. If f = (f1 , · · · , fq ) : D (f ) → Fq , then f is continuous if and only if each fk is a continuous F valued function. 5. The function f : Fp → F, given by f (x) = |x| is continuous.

5.9. EXERCISES

5.9

113

Exercises

1. f : D ⊆ Fp → Fq is Lipschitz continuous or just Lipschitz for short if there exists a constant, K such that |f (x) − f (y)| ≤ K |x − y| for all x, y ∈ D. Show every Lipschitz function is uniformly continuous which means that given ε > 0 there exists δ > 0 independent of x such that if |x − y| < δ, then |f (x) − f (y)| < ε. 2. If f is uniformly continuous, does it follow that |f | is also uniformly continuous? If |f | is uniformly continuous does it follow that f is uniformly continuous? Answer the same questions with “uniformly continuous” replaced with “continuous”. Explain why.

5.10

Proofs Of Theorems

This section contains the proofs of the theorems which were just stated without proof. Theorem 5.10.1 The following assertions are valid 1. The function, af + bg is continuous at x when f , g are continuous at x ∈ D (f ) ∩ D (g) and a, b ∈ F. 2. If and f and g are each F valued functions continuous at x, then f g is continuous at x. If, in addition to this, g (x) 6= 0, then f /g is continuous at x. 3. If f is continuous at x, f (x) ∈ D (g) ⊆ Fp , and g is continuous at f (x) ,then g ◦ f is continuous at x. 4. If f = (f1 , · · · , fq ) : D (f ) → Fq , then f is continuous if and only if each fk is a continuous F valued function. 5. The function f : Fp → F, given by f (x) = |x| is continuous. Proof: Begin with 1.) Let ε > 0 be given. By assumption, there exist δ 1 > 0 ε such that whenever |x − y| < δ 1 , it follows |f (x) − f (y)| < 2(|a|+|b|+1) and there exists δ 2 > 0 such that whenever |x − y| < δ 2 , it follows that |g (x) − g (y)| < ε 2(|a|+|b|+1) . Then let 0 < δ ≤ min (δ 1 , δ 2 ) . If |x − y| < δ, then everything happens at once. Therefore, using the triangle inequality |af (x) + bf (x) − (ag (y) + bg (y))|

114


≤ |a| |f (x) − f (y)| + |b| |g (x) − g (y)| µ ¶ µ ¶ ε ε < |a| + |b| < ε. 2 (|a| + |b| + 1) 2 (|a| + |b| + 1) Now begin on 2.) There exists δ 1 > 0 such that if |y − x| < δ 1 , then |f (x) − f (y)| < 1. Therefore, for such y, |f (y)| < 1 + |f (x)| . It follows that for such y, |f g (x) − f g (y)| ≤ |f (x) g (x) − g (x) f (y)| + |g (x) f (y) − f (y) g (y)| ≤ |g (x)| |f (x) − f (y)| + |f (y)| |g (x) − g (y)| ≤ (1 + |g (x)| + |f (y)|) [|g (x) − g (y)| + |f (x) − f (y)|] . Now let ε > 0 be given. There exists δ 2 such that if |x − y| < δ 2 , then |g (x) − g (y)|
0 such that if |x − z| < δ and z ∈ D (f ) , then |f (z) − f (x)| < η. Then if |x − z| < δ and z ∈ D (g ◦ f ) ⊆ D (f ) , all the above hold and so |g (f (z)) − g (f (x))| < ε. This proves part 3.) Part 4.) says: If f = (f1 , · · · , fq ) : D (f ) → Fq , then f is continuous if and only if each fk is a continuous F valued function. Then |fk (x) − fk (y)| ≤ |f (x) − f (y)| Ã ≡

q X

!1/2 |fi (x) − fi (y)|

2

i=1

≤

q X

|fi (x) − fi (y)| .

(5.10.18)

i=1

Suppose first that f is continuous at x. Then there exists δ > 0 such that if |x − y| < δ, then |f (x) − f (y)| < ε. The first part of the above inequality then shows that for each k = 1, · · · , q, |fk (x) − fk (y)| < ε. This shows the only if part. Now suppose each function, fk is continuous. Then if ε > 0 is given, there exists δ k > 0 such that whenever |x − y| < δ k |fk (x) − fk (y)| < ε/q. Now let 0 < δ ≤ min (δ 1 , · · · , δ q ) . For |x − y| < δ, the above inequality holds for all k and so the last part of 5.10.18 implies |f (x) − f (y)| ≤
0 be given and let δ = ε. Then if |x − y| < δ, the triangle inequality implies |f (x) − f (y)| = ||x| − |y|| ≤ |x − y| < δ = ε. This proves part 5.) and completes the proof of the theorem. Here is a multidimensional version of the nested interval lemma. The following definition is similar to that given earlier. It defines what is meant by a sequentially compact set in Fp .

5.10. PROOFS OF THEOREMS

117

Definition 5.10.2 A set, K ⊆ Fp is sequentially compact if and only if whenever ∞ {xn }n=1 is a sequence of points in K, there exists a point, x ∈ K and a subsequence, ∞ {xnk }k=1 such that xnk → x. It turns out the sequentially compact sets in Fp are exactly those which are closed and bounded. Only half of this result will be needed in this book and this is proved next. First note that C can be considered as R2 . Therefore, Cp may be considered as R2p . Theorem 5.10.3 Let C ⊆ Fp be closed and bounded. Then C is sequentially compact. ¢ ¡ Proof: Let {an } ⊆ C. Then let an = an1 , · · · , anp . It follows the real and © ª∞ imaginary parts of the terms of the sequence, anj n=1 are each contained in some sufficiently large closed bounded interval. By©Theorem 2.0.5 on Page 27, there is a ª∞ subsequence of the sequence of real parts of anj n=1 which converges. Also there © ª∞ is a further subsequence of the imaginary parts of anj n=1 which converges. Thus there is a subsequence, nk with the property that anj k converges to a point, aj ∈ F. Taking further subsequences, one obtains the existence of a subsequence, still called nk such that for each r = 1, · · · , p, anr k converges to a point, ar ∈ F as k → ∞. Therefore, letting a ≡ (a1 , · · · , ap ) , limk→∞ ank = a. Since C is closed, it follows a ∈ C. This proves the theorem. Here is a proof of the extreme value theorem. Theorem 5.10.4 Let C be closed and bounded and let f : C → R be continuous. Then f achieves its maximum and its minimum on C. This means there exist, x1 , x2 ∈ C such that for all x ∈ C, f (x1 ) ≤ f (x) ≤ f (x2 ) . Proof: Let M = sup {f (x) : x ∈ C} . Recall this means +∞ if f is not bounded above and it equals the least upper bound of these values of f if f is bounded above. Then there exists a sequence, {xn } such that f (xn ) → M. Since C is sequentially compact, there exists a subsequence, xnk , and a point, x ∈ C such that xnk → x. But then since f is continuous at x, it follows from Theorem 5.7.10 on Page 111 that f (x) = limk→∞ f (xnk ) = M. This proves f achieves its maximum and also shows its maximum is less than ∞. Let x2 = x. The case of a minimum is handled similarly. Recall that a function is uniformly continuous if the following definition holds. Definition 5.10.5 Let f : D (f ) → Fq . Then f is uniformly continuous if for every ε > 0 there exists δ > 0 such that whenever |x − y| < δ, it follows |f (x) − f (y)| < ε. Theorem 5.10.6 Let f :C → Fq be continuous where C is a closed and bounded set in Fp . Then f is uniformly continuous on C.

118


Proof: If this is not so, there exists ε > 0 and pairs of points, xn and yn satisfying |xn − yn | < 1/n but |f (xn ) − f (yn )| ≥ ε. Since C is sequentially compact, there exists x ∈ C and a subsequence, {xnk } satisfying xnk → x. But |xnk − ynk | < 1/k and so ynk → x also. Therefore, from Theorem 5.7.10 on Page 111, ε ≤ lim |f (xnk ) − f (ynk )| = |f (x) − f (x)| = 0, k→∞

a contradiction. This proves the theorem.

5.11

The Space L (Fn , Fm )

Definition 5.11.1 The symbol, L (Fn , Fm ) will denote the set of linear transformations mapping Fn to Fm . Thus L ∈ L (Fn , Fm ) means that for α, β scalars and x, y vectors in Fn , L (αx + βy) = αL (x) + βL (y) . It is convenient to give a norm for the elements of L (Fn , Fm ) . This will allow the consideration of questions such as whether a function having values in this space of linear transformations is continuous.

5.11.1

The Operator Norm

How do you measure the distance between linear transformations defined on Fn ? It turns out there are many ways to do this but I will give the most common one here. Definition 5.11.2 L (Fn , Fm ) denotes the space of linear transformations mapping Fn to Fm . For A ∈ L (Fn , Fm ) , the operator norm is defined by ||A|| ≡ max {|Ax|Fm : |x|Fn ≤ 1} < ∞. Theorem 5.11.3 Denote by |·| the norm on either Fn or Fm . Then L (Fn , Fm ) with this operator norm is a complete normed linear space of dimension nm with ||Ax|| ≤ ||A|| |x| . Here Completeness means that every Cauchy sequence converges. Proof: It is necessary to show the norm defined on L (Fn , Fm ) really is a norm. This means it is necessary to verify ||A|| ≥ 0 and equals zero if and only if A = 0. For α a scalar, ||αA|| = |α| ||A|| , and for A, B ∈ L (Fn , Fm ) , ||A + B|| ≤ ||A|| + ||B||

¡

5.11. THE SPACE L FN , FM

¢

119

The first two properties are obvious but you should verify them. It remains to verify the norm is well defined and also to verify the triangle inequality above. First if |x| ≤ 1, and (Aij ) is the matrix of the linear transformation with respect to the usual basis vectors, then Ã  !1/2  X  2 ||A|| = max |(Ax)i | : |x| ≤ 1   i   ¯ ¯2 1/2    X ¯X ¯    ¯ ¯   ¯ ¯ = max  A x : |x| ≤ 1 ij j ¯  ¯    ¯   i ¯ j  which is a finite number by the extreme value theorem. It is clear that a basis for L (Fn , Fm ) consists of linear transformations whose matrices are of the form Eij where Eij consists of the m × n matrix having all zeros except for a 1 in the ij th position. In effect, this considers L (Fn , Fm ) as Fnm . Think of the m × n matrix as a long vector folded up. If x 6= 0, ¯ ¯ ¯ x¯ 1 |Ax| = ¯¯A ¯¯ ≤ ||A|| (5.11.19) |x| |x| It only remains to verify completeness. Suppose then that {Ak } is a Cauchy sequence in L (Fn , Fm ) . Then from 5.11.19 {Ak x} is a Cauchy sequence for each x ∈ Fn . This follows because |Ak x − Al x| ≤ ||Ak − Al || |x| which converges to 0 as k, l → ∞. Therefore, by completeness of Fm , there exists Ax, the name of the thing to which the sequence, {Ak x} converges such that lim Ak x = Ax.

k→∞

Then A is linear because A (ax + by)

≡ =

lim Ak (ax + by)

k→∞

lim (aAk x + bAk y)

k→∞

= a lim Ak x + b lim Ak y k→∞

k→∞

= aAx + bAy. By the first part of this argument, ||A|| < ∞ and so A ∈ L (Fn , Fm ) . This proves the theorem. Proposition 5.11.4 Let A (x) ∈ L (Fn , Fm ) for each x ∈ U ⊆ Fp . Then letting (Aij (x)) denote the matrix of A (x) with respect to the standard basis, it follows Aij is continuous at x for each i, j if and only if for all ε > 0, there exists a δ > 0 such that if |x − y| < δ, then ||A (x) − A (y)|| < ε. That is, A is a continuous function having values in L (Fn , Fm ) at x.

120


Proof: Suppose first the second condition holds. Then from the material on linear transformations, |Aij (x) − Aij (y)|

= |ei · (A (x) − A (y)) ej | ≤ |ei | |(A (x) − A (y)) ej | ≤ ||A (x) − A (y)|| .

Therefore, the second condition implies the first. Now suppose the first condition holds. That is each Aij is continuous at x. Let |v| ≤ 1. ¯ ¯2 1/2 ¯X ¯ X ¯ ¯   ¯ ¯ (A (x) − A (y)) v  ij ij j¯  ¯ ¯ i ¯ j  |(A (x) − A (y)) (v)| =

(5.11.20)

 2 1/2 X X    |Aij (x) − Aij (y)| |vj |  .  

≤

j

i

By continuity of each Aij , there exists a δ > 0 such that for each i, j ε |Aij (x) − Aij (y)| < √ n m whenever |x − y| < δ. Then from 5.11.20, if |x − y| < δ, 

|(A (x) − A (y)) (v)|

 2 1/2 X X ε    √ |v|  <  n m i j 

 2 1/2 X X ε    √   ≤  =ε n m i j This proves the proposition.

5.12

The Frechet Derivative

Let U be an open set in Fn , and let f : U → Fm be a function. Definition 5.12.1 A function g is o (v) if g (v) =0 |v|→0 |v| lim

(5.12.21)

5.12. THE FRECHET DERIVATIVE

121

A function f : U → Fm is differentiable at x ∈ U if there exists a linear transformation L ∈ L (Fn , Fm ) such that f (x + v) = f (x) + Lv + o (v) This linear transformation L is the definition of Df (x). This derivative is often called the Frechet derivative. . Usually no harm is occasioned by thinking of this linear transformation as its matrix taken with respect to the usual basis vectors. The definition 5.12.21 means that the error, f (x + v) − f (x) − Lv converges to 0 faster than |v|. Thus the above definition is equivalent to saying lim

|v|→0

|f (x + v) − f (x) − Lv| =0 |v|

(5.12.22)

or equivalently, |f (y) − f (x) − Df (x) (y − x)| = 0. y→x |y − x| lim

(5.12.23)

Now it is clear this is just a generalization of the notion of the derivative of a function of one variable because in this more specialized situation, |f (x + v) − f (x) − f 0 (x) v| = 0, |v| |v|→0 lim

due to the definition which says f 0 (x) = lim

v→0

f (x + v) − f (x) . v

For functions of n variables, you can’t define the derivative as the limit of a difference quotient like you can for a function of one variable because you can’t divide by a vector. That is why there is a need for a more general definition. The term o (v) is notation that is descriptive of the behavior in 5.12.21 and it is only this behavior that is of interest. Thus, if t and k are constants, o (v) = o (v) + o (v) , o (tv) = o (v) , ko (v) = o (v) and other similar observations hold. The sloppiness built in to this notation is useful because it ignores details which are not important. It may help to think of o (v) as an adjective describing what is left over after approximating f (x + v) by f (x) + Df (x) v. Theorem 5.12.2 The derivative is well defined.

122


Proof: First note that for a fixed vector, v, o (tv) = o (t). Now suppose both L1 and L2 work in the above definition. Then let v be any vector and let t be a real scalar which is chosen small enough that tv + x ∈ U . Then f (x + tv) = f (x) + L1 tv + o (tv) , f (x + tv) = f (x) + L2 tv + o (tv) . Therefore, subtracting these two yields (L2 − L1 ) (tv) = o (tv) = o (t). Therefore, dividing by t yields (L2 − L1 ) (v) = o(t) t . Now let t → 0 to conclude that (L2 − L1 ) (v) = 0. Since this is true for all v, it follows L2 = L1 . This proves the theorem. Lemma 5.12.3 Let f be differentiable at x. Then f is continuous at x and in fact, there exists K > 0 such that whenever |v| is small enough, |f (x + v) − f (x)| ≤ K |v| Proof: From the definition of the derivative, f (x + v)−f (x) = Df (x) v+o (v). Let |v| be small enough that o(|v|) |v| < 1 so that |o (v)| ≤ |v|. Then for such v, |f (x + v) − f (x)| ≤ |Df (x) v| + |v| ≤ (|Df (x)| + 1) |v| This proves the lemma with K = |Df (x)| + 1. Theorem 5.12.4 (The chain rule) Let U and V be open sets, U ⊆ Fn and V ⊆ Fm . Suppose f : U → V is differentiable at x ∈ U and suppose g : V → Fq is differentiable at f (x) ∈ V . Then g ◦ f is differentiable at x and D (g ◦ f ) (x) = D (g (f (x))) D (f (x)) . Proof: This follows from a computation. Let B (x,r) ⊆ U and let r also be small enough that for |v| ≤ r, it follows that f (x + v) ∈ V . Such an r exists because f is continuous at x. For |v| < r, the definition of differentiability of g and f implies g (f (x + v)) − g (f (x)) =

= =

Dg (f (x)) (f (x + v) − f (x)) + o (f (x + v) − f (x)) Dg (f (x)) [Df (x) v + o (v)] + o (f (x + v) − f (x)) D (g (f (x))) D (f (x)) v + o (v) + o (f (x + v) − f (x)) .

(5.12.24)

It remains to show o (f (x + v) − f (x)) = o (v). By Lemma 5.12.3, with K given there, letting ε > 0, it follows that for |v| small enough, |o (f (x + v) − f (x))| ≤ (ε/K) |f (x + v) − f (x)| ≤ (ε/K) K |v| = ε |v| .

5.12. THE FRECHET DERIVATIVE

123

Since ε > 0 is arbitrary, this shows o (f (x + v) − f (x)) = o (v) because whenever |v| is small enough, |o (f (x + v) − f (x))| ≤ ε. |v| By 5.12.24, this shows g (f (x + v)) − g (f (x)) = D (g (f (x))) D (f (x)) v + o (v) which proves the theorem. The derivative is a linear transformation. What is the matrix of this linear transformation taken with respect to the usual basis vectors? Let ei denote the vector of Fn which has a one in the ith entry and zeroes elsewhere. Then the matrix of the linear transformation is the matrix whose ith column is Df (x) ei . What is this? Let t ∈ R such that |t| is sufficiently small. f (x + tei ) − f (x)

= Df (x) tei + o (tei ) = Df (x) tei + o (t) .

Then dividing by t and taking a limit, Df (x) ei = lim

t→0

∂f f (x + tei ) − f (x) ≡ (x) . t ∂xi

Thus the matrix of Df (x) with respect to the usual basis vectors is the matrix of the form   f1,x1 (x) f1,x2 (x) · · · f1,xn (x)   .. .. ..  . . . . fm,x1 (x) fm,x2 (x) · · · fm,xn (x) As mentioned before, there is no harm in referring to this matrix as Df (x) but it may also be referred to as Jf (x) . This is summarized in the following theorem. Theorem 5.12.5 Let f : Fn → Fm and suppose f is differentiable at x. Then all the i (x) partial derivatives ∂f∂x exist and if Jf (x) is the matrix of the linear transformation j with respect to the standard basis vectors, then the ij th entry is given by fi,j or ∂fi ∂xj (x). What if all the partial derivatives of f exist? Does it follow that f is differentiable? Consider the following function. ½ xy x2 +y 2 if (x, y) 6= (0, 0) . f (x, y) = 0 if (x, y) = (0, 0) Then from the definition of partial derivatives, lim

h→0

f (h, 0) − f (0, 0) 0−0 = lim =0 h→0 h h

124


and

f (0, h) − f (0, 0) 0−0 = lim =0 h→0 h h However f is not even continuous at (0, 0) which may be seen by considering the behavior of the function along the line y = x and along the line x = 0. By Lemma 5.12.3 this implies f is not differentiable. Therefore, it is necessary to consider the correct definition of the derivative given above if you want to get a notion which generalizes the concept of the derivative of a function of one variable in such a way as to preserve continuity whenever the function is differentiable. lim

h→0

5.13

C 1 Functions

However, there are theorems which can be used to get differentiability of a function based on existence of the partial derivatives. Definition 5.13.1 When all the partial derivatives exist and are continuous the function is called a C 1 function. Because of Proposition 5.11.4 on Page 119 and Theorem 5.12.5 which identifies the entries of Jf with the partial derivatives, the following definition is equivalent to the above. Definition 5.13.2 Let U ⊆ Fn be an open set. Then f : U → Fm is C 1 (U ) if f is differentiable and the mapping x →Df (x) , is continuous as a function from U to L (Fn , Fm ). The following is an important abstract generalization of the familiar concept of partial derivative. Definition 5.13.3 Let g : U ⊆ Fn × Fm → Fq , where U is an open set in Fn × Fm . Denote an element of Fn × Fm by (x, y) where x ∈ Fn and y ∈ Fm . Then the map x → g (x, y) is a function from the open set in Fn , {x : (x, y) ∈ U } to Fq . When this map is differentiable, its derivative is denoted by D1 g (x, y) , or sometimes by Dx g (x, y) . Thus, g (x + v, y) − g (x, y) = D1 g (x, y) v + o (v) . A similar definition holds for the symbol Dy g or D2 g. The special case seen in beginning calculus courses is where g : U → Fq and gxi (x) ≡

∂g (x) g (x + hei ) − g (x) ≡ lim . h→0 ∂xi h

5.13. C 1 FUNCTIONS

125

The following theorem will be very useful in much of what follows. It is a version of the mean value theorem. You might call it the mean value inequality. Theorem 5.13.4 Suppose U is an open subset of Fn and f : U → Fm has the property that Df (x) exists for all x in U and that, x+t (y − x) ∈ U for all t ∈ [0, 1]. (The line segment joining the two points lies in U .) Suppose also that for all points on this line segment, ||Df (x+t (y − x))|| ≤ M. Then |f (y) − f (x)| ≤ M |y − x| . Proof: Let S ≡ {t ∈ [0, 1] : for all s ∈ [0, t] , |f (x + s (y − x)) − f (x)| ≤ (M + ε) s |y − x|} . Then 0 ∈ S and by continuity of f , it follows that if t ≡ sup S, then t ∈ S and if t < 1, |f (x + t (y − x)) − f (x)| = (M + ε) t |y − x| . (5.13.25) ∞

If t < 1, then there exists a sequence of positive numbers, {hk }k=1 converging to 0 such that |f (x + (t + hk ) (y − x)) − f (x)| > (M + ε) (t + hk ) |y − x| which implies that |f (x + (t + hk ) (y − x)) − f (x + t (y − x))| + |f (x + t (y − x)) − f (x)| > (M + ε) (t + hk ) |y − x| . By 5.13.25, this inequality implies |f (x + (t + hk ) (y − x)) − f (x + t (y − x))| > (M + ε) hk |y − x| which yields upon dividing by hk and taking the limit as hk → 0, |Df (x + t (y − x)) (y − x)| ≥ (M + ε) |y − x| . Now by the definition of the norm of a linear operator, M |y − x| ≥

||Df (x + t (y − x))|| |y − x|

≥ |Df (x + t (y − x)) (y − x)| ≥ (M + ε) |y − x| , a contradiction. Therefore, t = 1 and so |f (x + (y − x)) − f (x)| ≤ (M + ε) |y − x| . Since ε > 0 is arbitrary, this proves the theorem. The next theorem proves that if the partial derivatives exist and are continuous, then the function is differentiable.

126


Theorem 5.13.5 Let g : U ⊆ Fn × Fm → Fq . Then g is C 1 (U ) if and only if D1 g and D2 g both exist and are continuous on U . In this case, Dg (x, y) (u, v) = D1 g (x, y) u+D2 g (x, y) v. Proof: Suppose first that g ∈ C 1 (U ). Then if (x, y) ∈ U , g (x + u, y) − g (x, y) = Dg (x, y) (u, 0) + o (u) . Therefore, D1 g (x, y) u =Dg (x, y) (u, 0). Then |(D1 g (x, y) − D1 g (x0 , y0 )) (u)| = |(Dg (x, y) − Dg (x0 , y0 )) (u, 0)| ≤ ||Dg (x, y) − Dg (x0 , y0 )|| |(u, 0)| . Therefore, |D1 g (x, y) − D1 g (x0 , y0 )| ≤ ||Dg (x, y) − Dg (x0 , y0 )|| . A similar argument applies for D2 g and this proves the continuity of the function, (x, y) → Di g (x, y) for i = 1, 2. The formula follows from Dg (x, y) (u, v)

= Dg (x, y) (u, 0) + Dg (x, y) (0, v) ≡ D1 g (x, y) u+D2 g (x, y) v.

Now suppose D1 g (x, y) and D2 g (x, y) exist and are continuous. g (x + u, y + v) − g (x, y) = g (x + u, y + v) − g (x, y + v) +g (x, y + v) − g (x, y) = g (x + u, y) − g (x, y) + g (x, y + v) − g (x, y) + [g (x + u, y + v) − g (x + u, y) − (g (x, y + v) − g (x, y))] = D1 g (x, y) u + D2 g (x, y) v + o (v) + o (u) + [g (x + u, y + v) − g (x + u, y) − (g (x, y + v) − g (x, y))] .

(5.13.26)

Let h (x, u) ≡ g (x + u, y + v) − g (x + u, y). Then the expression in [ ] is of the form, h (x, u) − h (x, 0) . Also D2 h (x, u) = D1 g (x + u, y + v) − D1 g (x + u, y) and so, by continuity of (x, y) → D1 g (x, y), ||D2 h (x, u)|| < ε

5.13. C 1 FUNCTIONS

127

whenever ||(u, v)|| is small enough. By Theorem 5.13.4 on Page 125, there exists δ > 0 such that if ||(u, v)|| < δ, the norm of the last term in 5.13.26 satisfies the inequality, ||g (x + u, y + v) − g (x + u, y) − (g (x, y + v) − g (x, y))|| < ε ||u|| .

(5.13.27)

Therefore, this term is o ((u, v)). It follows from 5.13.27 and 5.13.26 that g (x + u, y + v) = g (x, y) + D1 g (x, y) u + D2 g (x, y) v+o (u) + o (v) + o ((u, v)) = g (x, y) + D1 g (x, y) u + D2 g (x, y) v + o ((u, v)) Showing that Dg (x, y) exists and is given by Dg (x, y) (u, v) = D1 g (x, y) u + D2 g (x, y) v. The continuity of (x, y) → Dg (x, y) follows from the continuity of (x, y) → Di g (x, y). This proves the theorem. Not surprisingly, it can be generalized to many more factors. Qn Definition 5.13.6 Let g : U ⊆ i=1 Fri → Fq , where U is an open set. Then the map xi → g (x) is a function from the open set in Fri , ©

¡ ¢ ª x : x = x1 , · · · , xi−1 , x, xi+1 , · · · , xn ∈ U

to Fq . When this map is differentiable, its Qnderivative is denoted by Di g (x). To aid in the notation, for v ∈ Fri , let θi v ∈ Q i=1 Fri be the vector (0, · · · , v, · · · , 0) n where the v is in the ith slot and for v ∈ i=1 Fri , let vi denote the entry in the ith slot of v. Thus, by saying xi → g (x) is differentiable is meant that for v ∈ Fri sufficiently small, g (x + θi v) − g (x) = Di g (x) v + o (v) . Note Di g (x) ∈ L (Fri ,

Qn i=1

Fri ) .

Here is a generalization of Theorem 5.13.5. Qn Theorem 5.13.7 Let g, U, i=1 Fri , be given as in Definition 5.13.6. Then g is C 1 (U ) if and only if Di g exists and is continuous on U for each i. In this case, Dg (x) (v) =

X k

where v = (v1 , · · · , vn ) .

Dk g (x) vk

(5.13.28)

128


and is continuous for each P i. Note that PkProof: Suppose then that Di g exists P n 0 j=1 θ j vj = (v1 , · · · , vk , 0, · · · , 0). Thus j=1 θ j vj = v and define j=1 θ j vj ≡ 0. Therefore,      n k k−1 X X X g x+ g (x + v) − g (x) = θj vj  − g x + θj vj  (5.13.29) j=1

k=1

j=1

Consider the terms in this sum.     k−1 k X X θj vj  = g (x+θk vk ) − g (x) + θj vj  − g x + g x+   g x+

k X

(5.13.30)

j=1

j=1





 

θj vj  − g (x+θk vk ) − g x +

j=1

k−1 X





θj vj  − g (x)

(5.13.31)

j=1

and the expression in 5.13.31 is of the form h (vk ) − h (0) where for small w ∈ Frk ,   k−1 X h (w) ≡ g x+ θj vj + θk w − g (x + θk w) . j=1

Therefore,  Dh (w) = Dk g x+

k−1 X

 θj vj + θk w − Dk g (x + θk w)

j=1

and by continuity, ||Dh (w)|| < ε provided |v| is small enough. Therefore, by Theorem 5.13.4, whenever |v| is small enough, |h (vk ) − h (0)| ≤ ε |vk | ≤ ε |v| which shows that since ε is arbitrary, the expression in 5.13.31 is o (v). Now in 5.13.30 g (x+θk vk ) − g (x) = Dk g (x) vk + o (vk ) = Dk g (x) vk + o (v) . Therefore, referring to 5.13.29, g (x + v) − g (x) =

n X

Dk g (x) vk + o (v)

k=1

which shows Dg exists and equals the formula given in 5.13.28. Next suppose g is C 1 . I need to verify that Dk g (x) exists and is continuous. Let v ∈ Frk sufficiently small. Then g (x + θk v) − g (x) = =

Dg (x) θk v + o (θk v) Dg (x) θk v + o (v)

5.14. C K FUNCTIONS

129

since |θk v| = |v|. Then Dk g (x) exists and equals Dg (x) ◦ θk Qn Since x → Dg (x) is continuous and θk : Frk → i=1 Fri is also continuous, this proves the theorem The way this is usually used is in the following corollary, a case of Theorem 5.13.7 obtained by letting Frj = F in the above theorem. Corollary 5.13.8 Let U be an open subset of Fn and let f :U → Fm be C 1 in the sense that all the partial derivatives of f exist and are continuous. Then f is differentiable and f (x + v) = f (x) +

n X ∂f (x) vk + o (v) . ∂xk

k=1

5.14

C k Functions

Recall the notation for partial derivatives in the following definition. Definition 5.14.1 Let g : U → Fn . Then gxk (x) ≡

∂g g (x + hek ) − g (x) (x) ≡ lim h→0 ∂xk h

Higher order partial derivatives are defined in the usual way. gxk xl (x) ≡

∂2g (x) ∂xl ∂xk

and so forth. To deal with higher order partial derivatives in a systematic way, here is a useful definition. Definition 5.14.2 α = (α1 , · · · , αn ) for α1 · · · αn positive integers is called a multiindex. For α a multi-index, |α| ≡ α1 + · · · + αn and if x ∈ Fn , x = (x1 , · · · , xn ), and f a function, define αn α 1 α2 xα ≡ xα 1 x2 · · · xn , D f (x) ≡

∂ |α| f (x) . n · · · ∂xα n

α2 1 ∂xα 1 ∂x2

The following is the definition of what is meant by a C k function. Definition 5.14.3 Let U be an open subset of Fn and let f : U → Fm . Then for k a nonnegative integer, f is C k if for every |α| ≤ k, Dα f exists and is continuous.

130

5.15


Mixed Partial Derivatives

Under certain conditions the mixed partial derivatives will always be equal. This astonishing fact is due to Euler in 1734. Theorem 5.15.1 Suppose f : U ⊆ F2 → R where U is an open set on which fx , fy , fxy and fyx exist. Then if fxy and fyx are continuous at the point (x, y) ∈ U , it follows fxy (x, y) = fyx (x, y) . Proof: Since U is open, there exists r > 0 such that B ((x, y) , r) ⊆ U. Now let |t| , |s| < r/2, t, s real numbers and consider h(t)

h(0)

}| { z }| { 1 z ∆ (s, t) ≡ {f (x + t, y + s) − f (x + t, y) − (f (x, y + s) − f (x, y))}. st

(5.15.32)

Note that (x + t, y + s) ∈ U because |(x + t, y + s) − (x, y)| = ≤

¡ ¢1/2 |(t, s)| = t2 + s2 µ 2 ¶1/2 r r2 r + = √ < r. 4 4 2

As implied above, h (t) ≡ f (x + t, y + s) − f (x + t, y). Therefore, by the mean value theorem from calculus and the (one variable) chain rule, ∆ (s, t) = =

1 1 (h (t) − h (0)) = h0 (αt) t st st 1 (fx (x + αt, y + s) − fx (x + αt, y)) s

for some α ∈ (0, 1) . Applying the mean value theorem again, ∆ (s, t) = fxy (x + αt, y + βs) where α, β ∈ (0, 1). If the terms f (x + t, y) and f (x, y + s) are interchanged in 5.15.32, ∆ (s, t) is unchanged and the above argument shows there exist γ, δ ∈ (0, 1) such that ∆ (s, t) = fyx (x + γt, y + δs) . Letting (s, t) → (0, 0) and using the continuity of fxy and fyx at (x, y) , lim

(s,t)→(0,0)

∆ (s, t) = fxy (x, y) = fyx (x, y) .

This proves the theorem. The following is obtained from the above by simply fixing all the variables except for the two of interest.

5.15. MIXED PARTIAL DERIVATIVES

131

Corollary 5.15.2 Suppose U is an open subset of Fn and f : U → R has the property that for two indices, k, l, fxk , fxl , fxl xk , and fxk xl exist on U and fxk xl and fxl xk are both continuous at x ∈ U. Then fxk xl (x) = fxl xk (x) . By considering the real and imaginary parts of f in the case where f has values in F you obtain the following corollary. Corollary 5.15.3 Suppose U is an open subset of Fn and f : U → F has the property that for two indices, k, l, fxk , fxl , fxl xk , and fxk xl exist on U and fxk xl and fxl xk are both continuous at x ∈ U. Then fxk xl (x) = fxl xk (x) . Finally, by considering the components of f you get the following generalization. Corollary 5.15.4 Suppose U is an open subset of Fn and f : U → F m has the property that for two indices, k, l, fxk , fxl , fxl xk , and fxk xl exist on U and fxk xl and fxl xk are both continuous at x ∈ U. Then fxk xl (x) = fxl xk (x) . It is necessary to assume the mixed partial derivatives are continuous in order to assert they are equal. The following is a well known example [5]. Example 5.15.5 Let ( f (x, y) =

xy (x2 −y 2 ) x2 +y 2

if (x, y) 6= (0, 0) 0 if (x, y) = (0, 0)

From the definition of partial derivatives it follows immediately that fx (0, 0) = fy (0, 0) = 0. Using the standard rules of differentiation, for (x, y) 6= (0, 0) , fx = y

x4 − y 4 + 4x2 y 2 (x2 +

2 y2 )

, fy = x

x4 − y 4 − 4x2 y 2 2

(x2 + y 2 )

Now fxy (0, 0) ≡ =

fx (0, y) − fx (0, 0) y −y 4 lim = −1 y→0 (y 2 )2 lim

y→0

while fyx (0, 0) ≡ =

fy (x, 0) − fy (0, 0) x x4 lim =1 x→0 (x2 )2 lim

x→0

showing that although the mixed partial derivatives do exist at (0, 0) , they are not equal there.

132

5.16


Implicit Function Theorem

The implicit function theorem is one of the greatest theorems in mathematics. There are many versions of this theorem. However, I will give a very simple proof valid in finite dimensional spaces. Theorem 5.16.1 (implicit function theorem) Suppose U is an open set in Rn ×Rm . Let f : U → Rn be in C 1 (U ) and suppose −1

f (x0 , y0 ) = 0, D1 f (x0 , y0 )

∈ L (Rn , Rn ) .

(5.16.33)

Then there exist positive constants, δ, η, such that for every y ∈ B (y0 , η) there exists a unique x (y) ∈ B (x0 , δ) such that f (x (y) , y) = 0.

(5.16.34)

Furthermore, the mapping, y → x (y) is in C 1 (B (y0 , η)). Proof: Let

   f (x, y) =  

f1 (x, y) f2 (x, y) .. .

   . 

fn (x, y) ¡ 1 ¢ n Define for x , · · · , xn ∈ B (x0 , δ) and y ∈ B (y0 , η) the following matrix. ¡ ¢ ¡ ¢   f1,x1 x1 , y · · · f1,xn x1 , y ¡ ¢   .. .. J x1 , · · · , xn , y ≡  . . . fn,x1 (xn , y) · · · fn,xn (xn , y) Then by the assumption of continuity of all the partial derivatives, there ¡ exists δ 0¢ > 0 and η 0 > 0 such that if δ < δ 0 and η < η 0 , it follows that for all x1 , · · · , xn ∈ n B (x0 , δ) and y ∈ B (y0 , η) , ¡ ¡ ¢¢ det J x1 , · · · , xn , y > r > 0. (5.16.35) and B (x0 , δ 0 )× B (y0 , η 0 ) ⊆ U . Pick y ∈ B (y0 , η) and suppose there exist x, z ∈ B (x0 , δ) such that f (x, y) = f (z, y) = 0. Consider fi and let h (t) ≡ fi (x + t (z − x) , y) . Then h (1) = h (0) and so by the mean value theorem, h0 (ti ) = 0 for some ti ∈ (0, 1) . Therefore, from the chain rule and for this value of ti , h0 (ti ) = Dfi (x + ti (z − x) , y) (z − x) = 0. Then denote by xi the vector, x + ti (z − x) . It follows from 5.16.36 that ¡ ¢ J x1 , · · · , xn , y (z − x) = 0

(5.16.36)

5.16. IMPLICIT FUNCTION THEOREM

133

and so from 5.16.35 z − x = 0. Now it will be shown that if η is chosen sufficiently small, then for all y ∈ B (y0 , η) , there exists a unique x (y) ∈ B (x0 , δ) such that f (x (y) , y) = 0. 2 Claim: If η is small enough, then the function, hy (x) ≡ |f (x, y)| achieves its minimum value on B (x0 , δ) at a point of B (x0 , δ) . Proof of claim: Suppose this is not the case. Then there exists a sequence η k → 0 and for some yk having |yk −y0 | < η k , the minimum of hyk occurs on a point of the boundary of B (x0 , δ), xk such that |x0 −xk | = δ. Now taking a subsequence, still denoted by k, it can be assumed that xk → x with |x − x0 | = δ and yk → y0 . Let ε > 0. Then for k large enough, hyk (x0 ) < ε because f (x0 , y0 ) = 0. Therefore, from the definition of xk , hyk (xk ) < ε. Passing to the limit yields hy0 (x) ≤ ε. Since ε > 0 is arbitrary, it follows that hy0 (x) = 0 which contradicts the first part of the argument in which it was shown that for y ∈ B (y0 , η) there is at most one point, x of B (x0 , δ) where f (x, y) = 0. Here two have been obtained, x0 and x. This proves the claim. Choose η < η 0 and also small enough that the above claim holds and let x (y) denote a point of B (x0 , δ) at which the minimum of hy on B (x0 , δ) is achieved. Since x (y) is an interior point, you can consider hy (x (y) + tv) for |t| small and conclude this function of t has a zero derivative at t = 0. Thus T

Dhy (x (y)) v = 0 = 2f (x (y) , y) D1 f (x (y) , y) v for every vector v. But from 5.16.35 and the fact that v is arbitrary, it follows f (x (y) , y) = 0. This proves the existence of the function y → x (y) such that f (x (y) , y) = 0 for all y ∈ B (y0 , η) . It remains to verify this function is a C 1 function. To do this, let y1 and y2 be points of B (y0 , η) . Then as before, consider the ith component of f and consider the same argument using the mean value theorem to write 0 = fi (x (y1 ) , y1 ) − fi (x (y2 ) , y2 ) = fi (x (y¡1 ) , y1 )¢− fi (x (y2 ) , y1 ) + fi (x (y¡2 ) , y1 ) − f¢i (x (y2 ) , y2 ) = D1 fi xi , y1 (x (y1 ) − x (y2 )) + D2 fi x (y2 ) , yi (y1 − y2 ) . Therefore, ¡ ¢ J x1 , · · · , xn , y1 (x (y1 ) − x (y2 )) = −M (y1 − y2 )

(5.16.37)

¡ ¢ where M is the matrix whose ith row is D2 fi x (y2 ) , yi . Then from 5.16.35 there exists a constant, C independent of the choice of y ∈ B (y0 , η) such that ¯¯ ¡ ¢−1 ¯¯¯¯ ¯¯ ¯¯J x1 , · · · , xn , y ¯¯ < C ¢ ¡ n whenever x1 , · · · , xn ∈ B (x0 , δ) . By continuity of the partial derivatives of f it also follows there exists a constant, C1 such that ||D2 fi (x, y)|| < C1 whenever, (x, y) ∈ B (x0 , δ) × B (y0 , η) . Hence ||M || must also be bounded independent of the

134


choice of y1 and y2 in B (y0 , η) . From 5.16.37, it follows there exists a constant, C such that for all y1 , y2 in B (y0 , η) , |x (y1 ) − x (y2 )| ≤ C |y1 − y2 | .

(5.16.38)

It follows as in the proof of the chain rule that o (x (y + v) − x (y)) = o (v) .

(5.16.39)

Now let y ∈ B (y0 , η) and let |v| be sufficiently small that y + v ∈ B (y0 , η) . Then 0 = =

f (x (y + v) , y + v) − f (x (y) , y) f (x (y + v) , y + v) − f (x (y + v) , y) + f (x (y + v) , y) − f (x (y) , y)

= D2 f (x (y + v) , y) v + D1 f (x (y) , y) (x (y + v) − x (y)) + o (|x (y + v) − x (y)|) = =

D2 f (x (y) , y) v + D1 f (x (y) , y) (x (y + v) − x (y)) + o (|x (y + v) − x (y)|) + (D2 f (x (y + v) , y) v−D2 f (x (y) , y) v) D2 f (x (y) , y) v + D1 f (x (y) , y) (x (y + v) − x (y)) + o (v) .

Therefore, x (y + v) − x (y) = −D1 f (x (y) , y)

−1

D2 f (x (y) , y) v + o (v)

−1

which shows that Dx (y) = −D1 f (x (y) , y) D2 f (x (y) , y) and y →Dx (y) is continuous. This proves the theorem. −1 In practice, how do you verify the condition, D1 f (x0 , y0 ) ∈ L (Fn , Fn )?   f1 (x1 , · · · , xn , y1 , · · · , yn )   .. f (x, y) =  . . fn (x1 , · · · , xn , y1 , · · · , yn ) The matrix of the linear transformation, D1 f (x0 , y0 ) is then  ∂f (x ,··· ,x ,y ,··· ,y ) ,xn ,y1 ,··· ,yn ) 1 1 n 1 n · · · ∂f1 (x1 ,··· ∂x ∂x1 n  .. ..  . .  ,xn ,y1 ,··· ,yn ) ∂fn (x1 ,··· ,xn ,y1 ,··· ,yn ) · · · ∂fn (x1 ,···∂x ∂x1 n −1

and from linear algebra, D1 f (x0 , y0 ) has an inverse. In other words when  ∂f (x ,··· ,x ,y ,··· ,y )  det  

∂x1 ∂fn (x1 ,··· ,xn ,y1 ,··· ,yn ) ∂x1

···

1

n

1

n

  

∈ L (Fn , Fn ) exactly when the above matrix ···

1



.. .

∂f1 (x1 ,··· ,xn ,y1 ,··· ,yn ) ∂xn

.. .

∂fn (x1 ,··· ,xn ,y1 ,··· ,yn ) ∂xn

   6= 0 

5.16. IMPLICIT FUNCTION THEOREM

135

at (x0 , y0 ). The above determinant is important enough that it is given special notation. Letting z = f (x, y) , the above determinant is often written as ∂ (z1 , · · · , zn ) . ∂ (x1 , · · · , xn ) Of course you can replace R with F in the above by applying the above to the situation in which each F is replaced with R2 . Corollary 5.16.2 (implicit function theorem) Suppose U is an open set in Fn ×Fm . Let f : U → Fn be in C 1 (U ) and suppose −1

f (x0 , y0 ) = 0, D1 f (x0 , y0 )

∈ L (Fn , Fn ) .

(5.16.40)


(5.16.41)

Furthermore, the mapping, y → x (y) is in C 1 (B (y0 , η)). The next theorem is a very important special case of the implicit function theorem known as the inverse function theorem. Actually one can also obtain the implicit function theorem from the inverse function theorem. It is done this way in [46] and in [3]. Theorem 5.16.3 (inverse function theorem) Let x0 ∈ U ⊆ Fn and let f : U → Fn . Suppose f is C 1 (U ) , and Df (x0 )−1 ∈ L(Fn , Fn ). (5.16.42) Then there exist open sets, W , and V such that x0 ∈ W ⊆ U,

(5.16.43)

f : W → V is one to one and onto,

(5.16.44)

f −1 is C 1 .

(5.16.45)

Proof: Apply the implicit function theorem to the function F (x, y) ≡ f (x) − y where y0 ≡ f (x0 ). Thus the function y → x (y) defined in that theorem is f −1 . Now let W ≡ B (x0 , δ) ∩ f −1 (B (y0 , η)) and V ≡ B (y0 , η) . This proves the theorem.

136

5.16.1


More Continuous Partial Derivatives

Corollary 5.16.2 will now be improved slightly. If f is C k , it follows that the function which is implicitly defined is also in C k , not just C 1 . Since the inverse function theorem comes as a case of the implicit function theorem, this shows that the inverse function also inherits the property of being C k . Theorem 5.16.4 (implicit function theorem) Suppose U is an open set in Fn ×Fm . Let f : U → Fn be in C k (U ) and suppose −1

f (x0 , y0 ) = 0, D1 f (x0 , y0 )

∈ L (Fn , Fn ) .

(5.16.46)


(5.16.47)

Furthermore, the mapping, y → x (y) is in C k (B (y0 , η)). Proof: From Corollary 5.16.2 y → x (y) is C 1 . It remains to show it is C k for k > 1 assuming that f is C k . From 5.16.47 ∂x −1 ∂f = −D1 (x, y) . l ∂y ∂y l Thus the following formula holds for q = 1 and |α| = q. X Dα x (y) = Mβ (x, y) Dβ f (x, y)

(5.16.48)

|β|≤q

where Mβ is a matrix whose entries are differentiable functions of Dγ (x) for |γ| < q −1 and Dτ f (x, y) for |τ | ≤ q. This follows easily from the description of D1 (x, y) in terms of the cofactor matrix and the determinant of D1 (x, y). Suppose 5.16.48 holds for |α| = q < k. Then by induction, this yields x is C q . Then X ∂Mβ (x, y) ∂Dα x (y) ∂Dβ f (x, y) β = D f (x, y) + M (x, y) . β ∂y p ∂y p ∂y p |β|≤|α|

∂M (x,y)

β By the chain rule is a matrix whose entries are differentiable functions of ∂y p τ D f (x, y) for |τ | ≤ q + 1 and Dγ (x) for |γ| < q + 1. It follows since y p was arbitrary that for any |α| = q + 1, a formula like 5.16.48 holds with q being replaced by q + 1. By induction, x is C k . This proves the theorem. As a simple corollary this yields an improved version of the inverse function theorem.

Theorem 5.16.5 (inverse function theorem) Let x0 ∈ U ⊆ Fn and let f : U → Fn . Suppose for k a positive integer, f is C k (U ) , and Df (x0 )−1 ∈ L(Fn , Fn ).

(5.16.49)

5.17. THE METHOD OF LAGRANGE MULTIPLIERS

137

Then there exist open sets, W , and V such that x0 ∈ W ⊆ U,

(5.16.50)

f : W → V is one to one and onto,

(5.16.51)

f

5.17

−1

k

is C .

(5.16.52)

The Method Of Lagrange Multipliers

As an application of the implicit function theorem, consider the method of Lagrange multipliers from calculus. Recall the problem is to maximize or minimize a function subject to equality constraints. Let f : U → R be a C 1 function where U ⊆ Rn and let gi (x) = 0, i = 1, · · · , m (5.17.53) be a collection of equality constraints with m < n. Now consider the system of nonlinear equations f (x) = gi (x) =

a 0, i = 1, · · · , m.

x0 is a local maximum if f (x0 ) ≥ f (x) for all x near x0 which also satisfies the constraints 5.17.53. A local minimum is defined similarly. Let F : U × R → Rm+1 be defined by   f (x) − a  g1 (x)    F (x,a) ≡  (5.17.54) . ..   . gm (x) Now consider the m + 1 × n Jacobian matrix,  fx1 (x0 ) · · · fxn (x0 )  g1x1 (x0 ) · · · g1xn (x0 )   .. ..  . . gmx1 (x0 ) · · ·

   . 

gmxn (x0 )

If this matrix has rank m + 1 then some m + 1 × m + 1 submatrix has nonzero determinant. It follows from the implicit function theorem that there exist m + 1 variables, xi1 , · · · , xim+1 such that the system F (x,a) = 0

(5.17.55)

specifies these m + 1 variables as a function of the remaining n − (m + 1) variables and a in an open set of Rn−m . Thus there is a solution (x,a) to 5.17.55 for some x close to x0 whenever a is in some open interval. Therefore, x0 cannot be either a local minimum or a local maximum. It follows that if x0 is either a local maximum

138


or a local minimum, then the above matrix must have rank less than m + 1 which requires the rows to be linearly dependent. Thus, there exist m scalars, λ1 , · · · , λ m , and a scalar µ, not all zero such that       fx1 (x0 ) g1x1 (x0 ) gmx1 (x0 )       .. .. .. µ  = λ1   + · · · + λm  . . . . fxn (x0 ) g1xn (x0 ) gmxn (x0 )

(5.17.56)

If the column vectors    gmx1 (x0 ) g1x1 (x0 )     .. ..   ,··· . . gmxn (x0 ) g1xn (x0 ) 

(5.17.57)

are linearly independent, then, µ 6= 0 and dividing by µ yields an expression of the form       fx1 (x0 ) g1x1 (x0 ) gmx1 (x0 )       .. .. .. (5.17.58)   = λ1   + · · · + λm   . . . fxn (x0 ) g1xn (x0 ) gmxn (x0 ) at every point x0 which is either a local maximum or a local minimum. This proves the following theorem. Theorem 5.17.1 Let U be an open subset of Rn and let f : U → R be a C 1 function. Then if x0 ∈ U is either a local maximum or local minimum of f subject to the constraints 5.17.53, then 5.17.56 must hold for some scalars µ, λ1 , · · · , λm not all equal to zero. If the vectors in 5.17.57 are linearly independent, it follows that an equation of the form 5.17.58 holds.

Metric Spaces And General Topological Spaces 6.1

Metric Space

Definition 6.1.1 A metric space is a set, X and a function d : X × X → [0, ∞) which satisfies the following properties. d (x, y) = d (y, x) d (x, y) ≥ 0 and d (x, y) = 0 if and only if x = y d (x, y) ≤ d (x, z) + d (z, y) . You can check that Rn and Cn are metric spaces with d (x, y) = |x − y| . However, there are many others. The definitions of open and closed sets are the same for a metric space as they are for Rn . Definition 6.1.2 A set, U in a metric space is open if whenever x ∈ U, there exists r > 0 such that B (x, r) ⊆ U. As before, B (x, r) ≡ {y : d (x, y) < r} . Closed sets are those whose complements are open. A point p is a limit point of a set, S if for every r > 0, B (p, r) contains infinitely many points of S. A sequence, {xn } converges to a point x if for every ε > 0 there exists N such that if n ≥ N, then d (x, xn ) < ε. {xn } is a Cauchy sequence if for every ε > 0 there exists N such that if m, n ≥ N, then d (xn , xm ) < ε. Lemma 6.1.3 In a metric space, X every ball, B (x, r) is open. A set is closed if and only if it contains all its limit points. If p is a limit point of S, then there exists a sequence of distinct points of S, {xn } such that limn→∞ xn = p. Proof: Let z ∈ B (x, r). Let δ = r − d (x, z) . Then if w ∈ B (z, δ) , d (w, x) ≤ d (x, z) + d (z, w) < d (x, z) + r − d (x, z) = r. Therefore, B (z, δ) ⊆ B (x, r) and this shows B (x, r) is open. The properties of balls are presented in the following theorem. 139

140

METRIC SPACES AND GENERAL TOPOLOGICAL SPACES

Theorem 6.1.4 Suppose (X, d) is a metric space. Then the sets {B(x, r) : r > 0, x ∈ X} satisfy ∪ {B(x, r) : r > 0, x ∈ X} = X (6.1.1) If p ∈ B (x, r1 ) ∩ B (z, r2 ), there exists r > 0 such that B (p, r) ⊆ B (x, r1 ) ∩ B (z, r2 ) .

(6.1.2)

Proof: Observe that the union of these balls includes the whole space, X so 6.1.1 is obvious. Consider 6.1.2. Let p ∈ B (x, r1 ) ∩ B (z, r2 ). Consider r ≡ min (r1 − d (x, p) , r2 − d (z, p)) and suppose y ∈ B (p, r). Then d (y, x) ≤ d (y, p) + d (p, x) < r1 − d (x, p) + d (x, p) = r1 and so B (p, r) ⊆ B (x, r1 ). By similar reasoning, B (p, r) ⊆ B (z, r2 ). This proves the theorem. Let K be a closed set. This means K C ≡ X \ K is an open set. Let p be a limit point of K. If p ∈ K C , then since K C is open, there exists B (p, r) ⊆ K C . But this contradicts p being a limit point because there are no points of K in this ball. Hence all limit points of K must be in K. Suppose next that K contains its limit points. Is K C open? Let p ∈ K C . Then p is not a limit point of K. Therefore, there exists B (p, r) which contains at most finitely many points of K. Since p ∈ / K, it follows that by making r smaller if necessary, B (p, r) contains no points of K. That is B (p, r) ⊆ K C showing K C is open. Therefore, K is closed. Suppose now that p is a limit point of S. Let x1 ∈ (S \ {p}) ∩ B (p, 1) . If x1 , · · · , xk have been chosen, let ½ ¾ 1 rk+1 ≡ min d (p, xi ) , i = 1, · · · , k, . k+1 Let xk+1 ∈ (S \ {p}) ∩ B (p, rk+1 ) . This proves the lemma. Lemma 6.1.5 If {xn } is a Cauchy sequence in a metric space, X and if some subsequence, {xnk } converges to x, then {xn } converges to x. Also if a sequence converges, then it is a Cauchy sequence. Proof: Note first that nk ≥ k because in a subsequence, the indices, n1 , n2 , · · · are strictly increasing. Let ε > 0 be given and let N be such that for k > N, d (x, xnk ) < ε/2 and for m, n ≥ N, d (xm , xn ) < ε/2. Pick k > n. Then if n > N, d (xn , x) ≤ d (xn , xnk ) + d (xnk , x)
N, then d (xn , x) < ε/2. it follows that for m, n > N, d (xn , xm ) ≤ d (xn , x) + d (x, xm )
0 is arbitrary, this proves the lemma.

6.2

Compactness In Metric Space

Many existence theorems in analysis depend on some set being compact. Therefore, it is important to be able to identify compact sets. The purpose of this section is to describe compact sets in a metric space. Definition 6.2.1 Let A be a subset of X. A is compact if whenever A is contained in the union of a set of open sets, there exists finitely many of these open sets whose union contains A. (Every open cover admits a finite subcover.) A is “sequentially compact” means every sequence has a convergent subsequence converging to an element of A. In a metric space compact is not the same as closed and bounded! Example 6.2.2 Let X be any infinite set and define d (x, y) = 1 if x 6= y while d (x, y) = 0 if x = y. You should verify the details that this is a metric space because it satisfies the axioms of a metric. The set X is closed and bounded because its complement is ∅ which is clearly open because every point of ∅ is an interior point. (There are none.) Also © ¡X is ¢bounded ªbecause X = B (x, 2). However, X is clearly not compact because B x, 21 : x ∈ X is a collection of open sets whose union contains X but

142


since they are all disjoint and there is no finite subset of these whose ¡ nonempty, ¢ union contains X. In fact B x, 21 = {x}. From this example it is clear something more than closed and bounded is needed. If you are not familiar with the issues just discussed, ignore them and continue. Definition 6.2.3 In any metric space, a set E is totally bounded if for every ε > 0 there exists a finite set of points {x1 , · · · , xn } such that E ⊆ ∪ni=1 B (xi , ε). This finite set of points is called an ε net. The following proposition tells which sets in a metric space are compact. First here is an interesting lemma. Lemma 6.2.4 Let X be a metric space and suppose D is a countable dense subset of X. In other words, it is being assumed X is a separable metric space. Consider the open sets of the form B (d, r) where r is a positive rational number and d ∈ D. Denote this countable collection of open sets by B. Then every open set is the union of sets of B. Furthermore, if C is any collection of open sets, there exists a countable subset, {Un } ⊆ C such that ∪n Un = ∪C. Proof: Let U be an open set and let x ∈ U. Let B (x, δ) ⊆ U. Then by density of D, there exists d ∈ D∩B (x, δ/4) . Now pick r ∈ Q∩(δ/4, 3δ/4) and consider B (d, r) . Clearly, B (d, r) contains the point x because r > δ/4. Is B (d, r) ⊆ B (x, δ)? if so, this proves the lemma because x was an arbitrary point of U . Suppose z ∈ B (d, r) . Then δ 3δ δ d (z, x) ≤ d (z, d) + d (d, x) < r + < + =δ 4 4 4 Now let C be any collection of open sets. Each set in this collection is the union of countably many sets of B. Let B 0 denote the sets of B which are contained in some set of C. Thus ∪B0 = ∪C. Then for each B ∈ B 0 , pick UB ∈ C such that B ⊆ UB . Then {UB : B ∈ B 0 } is a countable collection of sets of C whose union equals ∪C. Therefore, this proves the lemma. Proposition 6.2.5 Let (X, d) be a metric space. Then the following are equivalent. (X, d) is compact,

(6.2.3)

(X, d) is sequentially compact,

(6.2.4)

(X, d) is complete and totally bounded.

(6.2.5)

Proof: Suppose 6.2.3 and let {xk } be a sequence. Suppose {xk } has no convergent subsequence. If this is so, then no value of the sequence is repeated more than finitely many times. Also {xk } has no limit point because if it did, there would exist a subsequence which converges. To see this, suppose p is a limit point of {xk } . Then in B (p, 1) there are infinitely many points of {xk } . Pick one called xk1 . Now if

6.2. COMPACTNESS IN METRIC SPACE

143

xk1 , xk2 , · · · , xkn have been picked with xki ∈ B (p, 1/i) , consider B (p, 1/ (n + 1)) . There are infinitely many points of {xk } in this ball also. Pick xkn+1 such that ∞ kn+1 > kn . Then {xkn }n=1 is a subsequence which converges to p and it is assumed this does not happen. Thus {xk } has no limit points. It follows the set Cn = ∪{xk : k ≥ n} is a closed set because it has no limit points and if Un = CnC , then

X = ∪∞ n=1 Un

but there is no finite subcovering, because no value of the sequence is repeated more than finitely many times. This contradicts compactness of (X, d). Note xk is not in Un whenever k > n. Thus 6.2.3 implies 6.2.4. Now suppose 6.2.4 and let {xn } be a Cauchy sequence. Is {xn } convergent? By sequential compactness xnk → x for some subsequence. By Lemma 6.1.5 it follows that {xn } also converges to x showing that (X, d) is complete. If (X, d) is not totally bounded, then there exists ε > 0 for which there is no ε net. Hence there exists a sequence {xk } with d (xk , xl ) ≥ ε for all l 6= k. By Lemma 6.1.5 again, this contradicts 6.2.4 because no subsequence can be a Cauchy sequence and so no subsequence can converge. This shows 6.2.4 implies 6.2.5. n Now suppose 6.2.5. What about 6.2.4? Let {pn } be a sequence and let {xni }m i=1 be a 2−n net for n = 1, 2, · · · . Let ¡ ¢ Bn ≡ B xnin , 2−n be such that Bn contains pk for infinitely many values of k and Bn ∩ Bn+1 6= ∅. To do this, suppose Bn contains infinitely many values of k. Then one of ¡ pk for ¢ the sets which intersect Bn , B xn+1 , 2−(n+1) must contain pk for infinitely many i values of k because all these indices of points¡from {pn } contained in Bn must be ¢ accounted for in one of finitely many sets, B xn+1 , 2−(n+1) . Thus there exists a i strictly increasing sequence of integers, nk such that pnk ∈ Bk . Then if k ≥ l, d (pnk , pnl ) ≤

k−1 X

¡ ¢ d pni+1 , pni

i=l

max (|xi | : i = 1, · · · , p) . If z ∈ A, then z ∈ B (xj , 1) for some j and so by the triangle inequality, |z − 0| ≤ |z − xj | + |xj | < 1 + r. Thus A ⊆ B (0,r + 1) and so A is bounded. Now suppose A is bounded and suppose A is not totally bounded. Then there exists ε > 0 such that there is no ε net for A. Therefore, there exists a sequence of points {ai } with |ai − aj | ≥ ε if i 6= j. Since A is bounded, there exists r > 0 such that A ⊆ [−r, r)n. (x ∈[−r, r)n means xi ∈ [−r, r) for each i.) Now define S to be all cubes of the form n Y

[ak , bk )

k=1

where ak = −r + i2−p r, bk = −r + (i + 1) 2−p r,

6.3. SOME APPLICATIONS OF COMPACTNESS

145

¢n ¡ for i ∈ {0, 1, · · · , 2p+1 − 1}. Thus S is a collection of 2p+1 non overlapping √ cubes whose union equals [−r, r)n and whose diameters are all equal to 2−p r n. Now choose p large enough that the diameter of these cubes is less than ε. This yields a contradiction because one of the cubes must contain infinitely many points of {ai }. This proves the lemma. The next theorem is called the Heine Borel theorem and it characterizes the compact sets in Rn . Theorem 6.2.7 A subset of Rn is compact if and only if it is closed and bounded. Proof: Since a set in Rn is totally bounded if and only if it is bounded, this theorem follows from Proposition 6.2.5 and the observation that a subset of Rn is closed if and only if it is complete. This proves the theorem.

6.3

Some Applications Of Compactness

The following corollary is an important existence theorem which depends on compactness. Corollary 6.3.1 Let X be a compact metric space and let f : X → R be continuous. Then max {f (x) : x ∈ X} and min {f (x) : x ∈ X} both exist. Proof: First it is shown f (X) is compact. Suppose C is a set of open sets whose −1 union contains © −1f (X). Thenª since f is continuous f (U ) is open for all U ∈ C. Therefore, f (U ) : U ∈ C is a collection of open sets ©whose union contains X. ª Since X is compact, it follows finitely many of these, f −1 (U1 ) , · · · , f −1 (Up ) contains X in their union. Therefore, f (X) ⊆ ∪pk=1 Uk showing f (X) is compact as claimed. Now since f (X) is compact, Theorem 6.2.7 implies f (X) is closed and bounded. Therefore, it contains its inf and its sup. Thus f achieves both a maximum and a minimum. Definition 6.3.2 Let X, Y be metric spaces and f : X → Y a function. f is uniformly continuous if for all ε > 0 there exists δ > 0 such that whenever x1 and x2 are two points of X satisfying d (x1 , x2 ) < δ, it follows that d (f (x1 ) , f (x2 )) < ε. A very important theorem is the following. Theorem 6.3.3 Suppose f : X → Y is continuous and X is compact. Then f is uniformly continuous. Proof: Suppose this is not true and that f is continuous but not uniformly continuous. Then there exists ε > 0 such that for all δ > 0 there exist points, pδ and qδ such that d (pδ , qδ ) < δ and yet d (f (pδ ) , f (qδ )) ≥ ε. Let pn and qn be the points which go with δ = 1/n. By Proposition 6.2.5 {pn } has a convergent

146


subsequence, {pnk } converging to a point, x ∈ X. Since d (pn , qn ) < that qnk → x also. Therefore,

1 n,

it follows

ε ≤ d (f (pnk ) , f (qnk )) ≤ d (f (pnk ) , f (x)) + d (f (x) , f (qnk )) but by continuity of f , both d (f (pnk ) , f (x)) and d (f (x) , f (qnk )) converge to 0 as k → ∞ contradicting the above inequality. This proves the theorem. Another important property of compact sets in a metric space concerns the finite intersection property. Definition 6.3.4 If every finite subset of a collection of sets has nonempty intersection, the collection has the finite intersection property. Theorem 6.3.5 Suppose F is a collection of compact sets in a metric space, X which has the finite intersection property. Then there exists a point in their intersection. (∩F 6= ∅). Proof: First I show each compact set is closed. Let K be a nonempty compact set and suppose p ∈ / K. Then for each x ∈ K, let Vx = B (x, d (p, x) /3) and Ux = B (p, d (p, x) /3) so that Ux and Vx have empty intersection. Then since V is compact, there are finitely many Vx which cover K say Vx1 , · · · , Vxn . Then let U = ∩ni=1 Uxi . It follows p ∈ U and U has empty intersection with K. In fact U has empty intersection with ∪ni=1 Vxi . Since U is an open set and p ∈ K C is arbitrary, it follows K C is an open set. © ª Consider now the claim about the intersection. ©If this were not so, ∪ F C : F ∈ F = ª X and so, in particular, picking some F0 ∈ F , F C : F ∈ F would be an open C exists. But cover of F0 . Since F0 is compact, some finite subcover, F1C , · · · , Fm m C m then F0 ⊆ ∪k=1 Fk which means ∩k=0 Fk = ∅, contrary to the finite intersection property. To see this, note that if x ∈ F0 , then it must fail to be in some Fk and so m it is not in ∩m k=0 Fk . Since this is true for every x it follows ∩k=0 Fk = ∅. Qm Theorem 6.3.6 Let Xi be a compact metric space with metric di . Then i=1 Xi is also a compact metric space with respect to the metric, d (x, y) ≡ maxi (di (xi , yi )). © ª∞ Proof: This is most easily seen from sequential compactness. Let xk k=1 Qm th k k be a sequence © k ª of points in i=1 Xi . Consider the i component of x , xi . It follows xi is a sequence of points in Xi and so it has a convergent subsequence. © ª Compactness of X1 implies there exists a subsequence of xk , denoted by xk1 such that lim xk11 → x1 ∈ X1 . k1 →∞

© ª Now there exists a further subsequence, denoted by xk2 such that in addition to this,ª xk22 → x2 ∈ X2 . After taking m such subsequences, there exists a subsequence, © l xl such that Qm liml→∞ xi = xi ∈ Xi for each i. Therefore, letting x ≡ (x1 , · · · , xm ), l x → x in i=1 Xi . This proves the theorem.

6.4. ASCOLI ARZELA THEOREM

6.4

147

Ascoli Arzela Theorem

Definition 6.4.1 Let (X, d) be a complete metric space. Then it is said to be locally compact if B (x, r) is compact for each r > 0. Thus if you have a locally compact metric space, then if {an } is a bounded sequence, it must have a convergent subsequence. Let K be a compact subset of Rn and consider the continuous functions which have values in a locally compact metric space, (X, d) where d denotes the metric on X. Denote this space as C (K, X) . Definition 6.4.2 For f, g ∈ C (K, X) , where K is a compact subset of Rn and X is a locally compact complete metric space define ρK (f, g) ≡ sup {d (f (x) , g (x)) : x ∈ K} . Then ρK provides a distance which makes C (K, X) into a metric space. The Ascoli Arzela theorem is a major result which tells which subsets of C (K, X) are sequentially compact. Definition 6.4.3 Let A ⊆ C (K, X) for K a compact subset of Rn . Then A is said to be uniformly equicontinuous if for every ε > 0 there exists a δ > 0 such that whenever x, y ∈ K with |x − y| < δ and f ∈ A, d (f (x) , f (y)) < ε. The set, A is said to be uniformly bounded if for some M < ∞, and a ∈ X, f (x) ∈ B (a, M ) for all f ∈ A and x ∈ K. Uniform equicontinuity is like saying that the whole set of functions, A, is uniformly continuous on K uniformly for f ∈ A. The version of the Ascoli Arzela theorem I will present here is the following. Theorem 6.4.4 Suppose K is a nonempty compact subset of Rn and A ⊆ C (K, X) is uniformly bounded and uniformly equicontinuous. Then if {fk } ⊆ A, there exists a function, f ∈ C (K, X) and a subsequence, fkl such that lim ρK (fkl , f ) = 0.

l→∞

To give a proof of this theorem, I will first prove some lemmas. ∞

Lemma 6.4.5 If K is a compact subset of Rn , then there exists D ≡ {xk }k=1 ⊆ K such that D is dense in K. Also, for every ε > 0 there exists a finite set of points, {x1 , · · · , xm } ⊆ K, called an ε net such that ∪m i=1 B (xi , ε) ⊇ K.

148


m Proof: For m ∈ N, pick xm 1 ∈ K. If every point of K is within 1/m of x1 , stop. Otherwise, pick m xm 2 ∈ K \ B (x1 , 1/m) . m If every point of K contained in B (xm 1 , 1/m) ∪ B (x2 , 1/m) , stop. Otherwise, pick m m xm 3 ∈ K \ (B (x1 , 1/m) ∪ B (x2 , 1/m)) . m m If every point of K is contained in B (xm 1 , 1/m) ∪ B (x2 , 1/m) ∪ B (x3 , 1/m) , stop. Otherwise, pick m m m xm 4 ∈ K \ (B (x1 , 1/m) ∪ B (x2 , 1/m) ∪ B (x3 , 1/m))

Continue this way until the process stops, say at N (m). It must stop because if it didn’t, there would be a convergent subsequence due to the compactness of K. Ultimately all terms of this convergent subsequence would be closer than 1/m, N (m) m violating the manner in which they are chosen. Then D = ∪∞ m=1 ∪k=1 {xk } . This is countable because it is a countable union of countable sets. If y ∈ K and ε > 0, then for some m, 2/m < ε and so B (y, ε) must contain some point of {xm k } since otherwise, the process stopped too soon. You could have picked y. This proves the lemma. Lemma 6.4.6 Suppose D is defined above and {gm } is a sequence of functions of A having the property that for every xk ∈ D, lim gm (xk ) exists.

m→∞

Then there exists g ∈ C (K, X) such that lim ρ (gm , g) = 0.

m→∞

Proof: Define g first on D. g (xk ) ≡ lim gm (xk ) . m→∞

Next I show that {gm } converges at every point of K. Let x ∈ K and let ε > 0 be given. Choose xk such that for all f ∈ A, d (f (xk ) , f (x))
0 there exists δ > 0 such that ρ (f (x) , f (z)) < ε whenever d (x, z) < δ. As is usual in such definitions, f is said to be continuous if it is continuous at every point of X. The following lemma gives an important example of a continuous real valued function defined on a metric space, (X, d) . Lemma 6.5.2 Let (X, d) be a metric space and let S ⊆ X be a nonempty subset. Define dist (x, S) ≡ inf {d (x, y) : y ∈ S} . Then x → dist (x, S) is a continuous function satisfying the inequality, |dist (x, S) − dist (y, S)| ≤ d (x, y) .

(6.5.7)

Proof: The continuity of x → dist (x, S) is obvious if the inequality 6.5.7 is established. So let x, y ∈ X. Without loss of generality, assume dist (x, S) ≥ dist (y, S) and pick z ∈ S such that d (y, z) − ε < dist (y, S) . Then |dist (x, S) − dist (y, S)|

= dist (x, S) − dist (y, S) ≤ d (x, z) − (d (y, z) − ε) ≤ d (z, y) + d (x, y) − d (y, z) + ε = d (x, y) + ε.

Since ε is arbitrary, this proves 6.5.7. Lemma 6.5.3 Let H, K be two nonempty disjoint closed subsets of a metric space, (X, d) . Then there exists a continuous function, g : X → [−1, 1] such that g (H) = −1/3, g (K) = 1/3, g (X) ⊆ [−1/3, 1/3] . Proof: Let f (x) ≡

dist (x, H) . dist (x, H) + dist (x, K)

The denominator is never equal to zero because if dist (x, H) = 0, then x ∈ H becasue H is closed. (To see this, pick hk ∈ B (x, 1/k) ∩ H. Then hk → x and since H is closed, x ∈ H.) Similarly, if dist (x, K) = 0, then x ∈ K and so the denominator is never zero as claimed. Hence, by Lemma 6.5.2, f is continuous and ¡ ¢ from its definition, f = 0 on H and f = 1 on K. Now let g (x) ≡ 23 f (x) − 12 . Then g has the desired properties. Definition 6.5.4 For f a real or complex valued bounded continuous function defined on a metric space, M ||f ||M ≡ sup {|f (x)| : x ∈ M } . Lemma 6.5.5 Suppose M is a closed set in X where (X, d) is a metric space and suppose f : M → [−1, 1] is continuous at every point of M. Then there exists a function, g which is defined and continuous on all of X such that ||f − g||M < 23 .

152


Proof: Let H = f −1 ([−1, −1/3]) , K = f −1 ([1/3, 1]) . Thus H and K are disjoint closed subsets of M. Suppose first H, K are both nonempty. Then by Lemma 6.5.3 there exists g such that g is a continuous function defined on all of X and g (H) = −1/3, g (K) = 1/3, and g (X) ⊆ [−1/3, 1/3] . It follows ||f − g||M < 2/3. If H = ∅, then f has all its values in [−1/3, 1] and so letting g ≡ 1/3, the desired condition is obtained. If K = ∅, let g ≡ −1/3. This proves the lemma. Lemma 6.5.6 Suppose M is a closed set in X where (X, d) is a metric space and suppose f : M → [−1, 1] is continuous at every point of M. Then there exists a function, g which is defined and continuous on all of X such that g = f on M and g has its values in [−1, 1] . Proof: Let g1 be such that g1 (X) ⊆ [−1/3, 1/3] and ||f − g1 ||M ≤ 32 . Suppose g1 , · · · , gm have been chosen such that gj (X) ⊆ [−1/3, 1/3] and ¯¯ ¯¯ µ ¶m m µ ¶i−1 ¯¯ ¯¯ X 2 2 ¯¯ ¯¯ f − g < . (6.5.8) ¯¯ i ¯¯ ¯¯ ¯¯ 3 3 i=1 M

Then

¯¯µ ¶ Ã !¯¯ m µ ¶i−1 ¯¯ 3 m ¯¯ X 2 ¯¯ ¯¯ f− gi ¯¯ ≤ 1 ¯¯ ¯¯ 2 ¯¯ 3 i=1 M ¡ 3 ¢m ³ Pm ¡ 2 ¢i−1 ´ and so 2 f − i=1 3 gi can play the role of f in the first step of the proof. Therefore, there exists gm+1 defined and continuous on all of X such that its values are in [−1/3, 1/3] and ¯¯µ ¶ Ã ¯¯ ! m µ ¶i−1 ¯¯ 3 m ¯¯ X 2 2 ¯¯ ¯¯ f− gi − gm+1 ¯¯ ≤ . ¯¯ ¯¯ 2 ¯¯ 3 3 i=1 M

Hence

¯¯Ã ¯¯ ! µ ¶ m µ ¶i−1 m ¯¯ ¯¯ X 2 2 ¯¯ ¯¯ gi − gm+1 ¯¯ ¯¯ f − ¯¯ ¯¯ 3 3 i=1

M

µ ¶m+1 2 . ≤ 3

It follows there exists a sequence, {gi } such that each has its values in [−1/3, 1/3] and for every m 6.5.8 holds. Then let ∞ µ ¶i−1 X 2 g (x) ≡ gi (x) . 3 i=1 It follows

¯ ¯ ∞ µ ¶i−1 m µ ¶i−1 ¯X ¯ X 1 2 2 ¯ ¯ |g (x)| ≤ ¯ ≤1 gi (x)¯ ≤ ¯ ¯ 3 3 3 i=1 i=1

and since convergence is uniform, g must be continuous. The estimate 6.5.8 implies f = g on M . The following is the Tietze extension theorem.

6.6. GENERAL TOPOLOGICAL SPACES

153

Theorem 6.5.7 Let M be a closed nonempty subset of a metric space (X, d) and let f : M → [a, b] is continuous at every point of M. Then there exists a function, g continuous on all of X which coincides with f on M such that g (X) ⊆ [a, b] . 2 Proof: Let f1 (x) = 1 + b−a (f (x) − b) . Then f1 satisfies the conditions of Lemma 6.5.6 and so there exists g1 : X → [−1, ¡ 1]¢ such that g is continuous on X and equals f1 on M . Let g (x) = (g1 (x) − 1) b−a + b. This works. 2

6.6

General Topological Spaces

It turns out that metric spaces are not sufficiently general for some applications. This section is a brief introduction to general topology. In making this generalization, the properties of balls which are the conclusion of Theorem 6.1.4 on Page 140 are stated as axioms for a subset of the power set of a given set which will be known as a basis for the topology. More can be found in [45] and the references listed there. Definition 6.6.1 Let X be a nonempty set and suppose B ⊆ P (X). Then B is a basis for a topology if it satisfies the following axioms. 1.) Whenever p ∈ A ∩ B for A, B ∈ B, it follows there exists C ∈ B such that p ∈ C ⊆ A ∩ B. 2.) ∪B = X. Then a subset, U, of X is an open set if for every point, x ∈ U, there exists B ∈ B such that x ∈ B ⊆ U . Thus the open sets are exactly those which can be obtained as a union of sets of B. Denote these subsets of X by the symbol τ and refer to τ as the topology or the set of open sets. Note that this is simply the analog of saying a set is open exactly when every point is an interior point. Proposition 6.6.2 Let X be a set and let B be a basis for a topology as defined above and let τ be the set of open sets determined by B. Then ∅ ∈ τ, X ∈ τ,

(6.6.9)

If C ⊆ τ , then ∪ C ∈ τ

(6.6.10)

If A, B ∈ τ , then A ∩ B ∈ τ .

(6.6.11)

Proof: If p ∈ ∅ then there exists B ∈ B such that p ∈ B ⊆ ∅ because there are no points in ∅. Therefore, ∅ ∈ τ . Now if p ∈ X, then by part 2.) of Definition 6.6.1 p ∈ B ⊆ X for some B ∈ B and so X ∈ τ . If C ⊆ τ , and if p ∈ ∪C, then there exists a set, B ∈ C such that p ∈ B. However, B is itself a union of sets from B and so there exists C ∈ B such that p ∈ C ⊆ B ⊆ ∪C. This verifies 6.6.10. Finally, if A, B ∈ τ and p ∈ A ∩ B, then since A and B are themselves unions of sets of B, it follows there exists A1 , B1 ∈ B such that A1 ⊆ A, B1 ⊆ B, and

154


p ∈ A1 ∩ B1 . Therefore, by 1.) of Definition 6.6.1 there exists C ∈ B such that p ∈ C ⊆ A1 ∩ B1 ⊆ A ∩ B, showing that A ∩ B ∈ τ as claimed. Of course if A ∩ B = ∅, then A ∩ B ∈ τ . This proves the proposition. Definition 6.6.3 A set X together with such a collection of its subsets satisfying 6.6.9-6.6.11 is called a topological space. τ is called the topology or set of open sets of X. Definition 6.6.4 A topological space is said to be Hausdorff if whenever p and q are distinct points of X, there exist disjoint open sets U, V such that p ∈ U, q ∈ V . In other words points can be separated with open sets.

U

V ·

p

·

q

Hausdorff Definition 6.6.5 A subset of a topological space is said to be closed if its complement is open. Let p be a point of X and let E ⊆ X. Then p is said to be a limit point of E if every open set containing p contains a point of E distinct from p. Note that if the topological space is Hausdorff, then this definition is equivalent to requiring that every open set containing p contains infinitely many points from E. Why? Theorem 6.6.6 A subset, E, of X is closed if and only if it contains all its limit points. Proof: Suppose first that E is closed and let x be a limit point of E. Is x ∈ E? If x ∈ / E, then E C is an open set containing x which contains no points of E, a contradiction. Thus x ∈ E. Now suppose E contains all its limit points. Is the complement of E open? If x ∈ E C , then x is not a limit point of E because E has all its limit points and so there exists an open set, U containing x such that U contains no point of E other than x. Since x ∈ / E, it follows that x ∈ U ⊆ E C which implies E C is an open set because this shows E C is the union of open sets. Theorem 6.6.7 If (X, τ ) is a Hausdorff space and if p ∈ X, then {p} is a closed set. Proof: If x 6= p, there exist open sets U and V such that x ∈ U, p ∈ V and C U ∩ V = ∅. Therefore, {p} is an open set so {p} is closed. Note that the Hausdorff axiom was stronger than needed in order to draw the conclusion of the last theorem. In fact it would have been enough to assume that if x 6= y, then there exists an open set containing x which does not intersect y.


155

Definition 6.6.8 A topological space (X, τ ) is said to be regular if whenever C is a closed set and p is a point not in C, there exist disjoint open sets U and V such that p ∈ U, C ⊆ V . Thus a closed set can be separated from a point not in the closed set by two disjoint open sets. U

V ·

p

C Regular

Definition 6.6.9 The topological space, (X, τ ) is said to be normal if whenever C and K are disjoint closed sets, there exist disjoint open sets U and V such that C ⊆ U, K ⊆ V . Thus any two disjoint closed sets can be separated with open sets.

U

V C

K Normal

Definition 6.6.10 Let E be a subset of X. E is defined to be the smallest closed set containing E. Lemma 6.6.11 The above definition is well defined. Proof: Let C denote all the closed sets which contain E. Then C is nonempty because X ∈ C. © ª C (∩ {A : A ∈ C}) = ∪ AC : A ∈ C , an open set which shows that ∩C is a closed set and is the smallest closed set which contains E. Theorem 6.6.12 E = E ∪ {limit points of E}. Proof: Let x ∈ E and suppose that x ∈ / E. If x is not a limit point either, then there exists an open set, U ,containing x which does not intersect E. But then U C is a closed set which contains E which does not contain x, contrary to the definition that E is the intersection of all closed sets containing E. Therefore, x must be a limit point of E after all. Now E ⊆ E so suppose x is a limit point of E. Is x ∈ E? If H is a closed set containing E, which does not contain x, then H C is an open set containing x which contains no points of E other than x negating the assumption that x is a limit point of E. The following is the definition of continuity in terms of general topological spaces. It is really just a generalization of the ε - δ definition of continuity given in calculus.

156


Definition 6.6.13 Let (X, τ ) and (Y, η) be two topological spaces and let f : X → Y . f is continuous at x ∈ X if whenever V is an open set of Y containing f (x), there exists an open set U ∈ τ such that x ∈ U and f (U ) ⊆ V . f is continuous if f −1 (V ) ∈ τ whenever V ∈ η. You should prove the following. Proposition 6.6.14 In the situation of Definition 6.6.13 f is continuous if and only if f is continuous at every point of X. Qn Definition 6.6.15 Let (Xi , τ i ) be topological spaces. i=1 Xi is the Cartesian Q n product. Define a product topology as follows. Let B = i=1 Ai where Ai ∈ τ i . Then B is a basis for the product topology. Theorem 6.6.16 The set B of Definition 6.6.15 is a basis for a topology. Qn Qn Proof: Suppose x ∈ i=1 Ai ∩ i=1 Bi where Ai and Bi are open sets. Say x = (x1 , · · · , xn ) . Qn Qn Then Qn xi ∈ Ai ∩ Bi for each i. Therefore, x ∈ i=1 Ai ∩ Bi ∈ B and i=1 Ai ∩ Bi ⊆ i=1 Ai . The definition of compactness is also considered for a general topological space. This is given next. Definition 6.6.17 A subset, E, of a topological space (X, τ ) is said to be compact if whenever C ⊆ τ and E ⊆ ∪C, there exists a finite subset of C, {U1 · · · Un }, such that E ⊆ ∪ni=1 Ui . (Every open covering admits a finite subcovering.) E is precompact if E is compact. A topological space is called locally compact if it has a basis B, with the property that B is compact for each B ∈ B. In general topological spaces there may be no concept of “bounded”. Even if there is, closed and bounded is not necessarily the same as compactness. However, in any Hausdorff space every compact set must be a closed set. Theorem 6.6.18 If (X, τ ) is a Hausdorff space, then every compact subset must also be a closed set. Proof: Suppose p ∈ / K. For each x ∈ X, there exist open sets, Ux and Vx such that x ∈ Ux , p ∈ Vx , and Ux ∩ Vx = ∅. If K is assumed to be compact, there are finitely many of these sets, Ux1 , · · · , Uxm which cover K. Then let V ≡ ∩m i=1 Vxi . It follows that V is an open set containing p which has empty intersection with each of the Uxi . Consequently, V contains no points of K and is therefore not a limit point of K. This proves the theorem. A useful construction when dealing with locally compact Hausdorff spaces is the notion of the one point compactification of the space.


157

Definition 6.6.19 Suppose (X, τ ) is a locally compact Hausdorff space. Then let e ≡ X ∪ {∞} where ∞ is just the name of some point which is not in X which is X e is called the point at infinity. A basis for the topology e τ for X © ª τ ∪ K C where K is a compact subset of X . e and so the open sets, K C are basic open The complement is taken with respect to X sets which contain ∞. The reason this is called a compactification is contained in the next lemma. ³ ´ e e Lemma 6.6.20 If (X, τ ) is a locally compact Hausdorff space, then X, τ is a compact Hausdorff space. Also if U is an open set of e τ , then U \ {∞} is an open set of τ . ³ ´ e e Proof: Since (X, τ ) is a locally compact Hausdorff space, it follows X, τ is a Hausdorff topological space. The only case which needs checking is the one of p ∈ X and ∞. Since (X, τ ) is locally compact, there exists an open set of τ , U C having compact closure which contains p. Then p ∈ U and ∞ ∈ U and these are disjoint open sets containing the points, p and ∞ respectively. Now let C be an e with sets from e open cover of X τ . Then ∞ must be in some set, U∞ from C, which must contain a set of the form K C where K is a compact subset of X. Then there e is exist sets from C, U1 , · · · , Ur which cover K. Therefore, a finite subcover of X U1 , · · · , Ur , U∞ . To see the last claim, suppose U contains ∞ since otherwise there is nothing to show. Notice that if C is a compact set, then X \ C is an open set. Therefore, if e \ C is a basic open set contained in U containing ∞, then x ∈ U \ {∞} , and if X e it is also in the open set X \ C ⊆ U \ {∞} . If x if x is in this basic open set of X, e \ C then x is contained in an open set of is not in any basic open set of the form X τ which is contained in U \ {∞}. Thus U \ {∞} is indeed open in τ . Definition 6.6.21 If every finite subset of a collection of sets has nonempty intersection, the collection has the finite intersection property. Theorem 6.6.22 Let K be a set whose elements are compact subsets of a Hausdorff topological space, (X, τ ). Suppose K has the finite intersection property. Then ∅ 6= ∩K. Proof: Suppose to the contrary that ∅ = ∩K. Then consider © ª C ≡ KC : K ∈ K . It follows C is an open cover of K0 where K0 is any particular element of K. But then there are finitely many K ∈ K, K1 , · · · , Kr such that K0 ⊆ ∪ri=1 KiC implying that ∩ri=0 Ki = ∅, contradicting the finite intersection property.

158


Lemma 6.6.23 Let (X, τ ) be a topological space and let B be a basis for τ . Then K is compact if and only if every open cover of basic open sets admits a finite subcover. Proof: Suppose first that X is compact. Then if C is an open cover consisting of basic open sets, it follows it admits a finite subcover because these are open sets in C. Next suppose that every basic open cover admits a finite subcover and let C be an open cover of X. Then define Ce to be the collection of basic open sets which are contained in some set of C. It follows Ce is a basic open cover of X and so it admits a finite subcover, {U1 , · · · , Up }. Now each Ui is contained in an open set of C. Let Oi be a set of C which contains Ui . Then {O1 , · · · , Op } is an open cover of X. This proves the lemma. In fact, much more can be said than Lemma 6.6.23. However, this is all which I will present here.

6.7

Connected Sets

Stated informally, connected sets are those which are in one piece. More precisely, Definition 6.7.1 A set, S in a general topological space is separated if there exist sets, A, B such that S = A ∪ B, A, B 6= ∅, and A ∩ B = B ∩ A = ∅. In this case, the sets A and B are said to separate S. A set is connected if it is not separated. One of the most important theorems about connected sets is the following. Theorem 6.7.2 Suppose U and V are connected sets having nonempty intersection. Then U ∪ V is also connected. Proof: Suppose U ∪ V = A ∪ B where A ∩ B = B ∩ A = ∅. Consider the sets, A ∩ U and B ∩ U. Since ¡ ¢ (A ∩ U ) ∩ (B ∩ U ) = (A ∩ U ) ∩ B ∩ U = ∅, It follows one of these sets must be empty since otherwise, U would be separated. It follows that U is contained in either A or B. Similarly, V must be contained in either A or B. Since U and V have nonempty intersection, it follows that both V and U are contained in one of the sets, A, B. Therefore, the other must be empty and this shows U ∪ V cannot be separated and is therefore, connected.

6.7. CONNECTED SETS

159

The intersection of connected sets is not necessarily connected as is shown by the following picture. U V

Theorem 6.7.3 Let f : X → Y be continuous where X and Y are topological spaces and X is connected. Then f (X) is also connected. Proof: To do this you show f (X) is not separated. Suppose to the contrary that f (X) = A ∪ B where A and B separate f (X) . Then consider the sets, f −1 (A) and f −1 (B) . If z ∈ f −1 (B) , then f (z) ∈ B and so f (z) is not a limit point of A. Therefore, there exists an open set, U containing f (z) such that U ∩ A = ∅. But then, the continuity of f implies that f −1 (U ) is an open set containing z such that f −1 (U ) ∩ f −1 (A) = ∅. Therefore, f −1 (B) contains no limit points of f −1 (A) . Similar reasoning implies f −1 (A) contains no limit points of f −1 (B). It follows that X is separated by f −1 (A) and f −1 (B) , contradicting the assumption that X was connected. An arbitrary set can be written as a union of maximal connected sets called connected components. This is the concept of the next definition. Definition 6.7.4 Let S be a set and let p ∈ S. Denote by Cp the union of all connected subsets of S which contain p. This is called the connected component determined by p. Theorem 6.7.5 Let Cp be a connected component of a set S in a general topological space. Then Cp is a connected set and if Cp ∩ Cq 6= ∅, then Cp = Cq . Proof: Let C denote the connected subsets of S which contain p. If Cp = A ∪ B where A ∩ B = B ∩ A = ∅, then p is in one of A or B. Suppose without loss of generality p ∈ A. Then every set of C must also be contained in A also since otherwise, as in Theorem 6.7.2, the set would be separated. But this implies B is empty. Therefore, Cp is connected. From this, and Theorem 6.7.2, the second assertion of the theorem is proved. This shows the connected components of a set are equivalence classes and partition the set.

160


A set, I is an interval in R if and only if whenever x, y ∈ I then (x, y) ⊆ I. The following theorem is about the connected sets in R. Theorem 6.7.6 A set, C in R is connected if and only if C is an interval. Proof: Let C be connected. If C consists of a single point, p, there is nothing to prove. The interval is just [p, p] . Suppose p < q and p, q ∈ C. You need to show (p, q) ⊆ C. If x ∈ (p, q) \ C let C ∩ (−∞, x) ≡ A, and C ∩ (x, ∞) ≡ B. Then C = A ∪ B and the sets, A and B separate C contrary to the assumption that C is connected. Conversely, let I be an interval. Suppose I is separated by A and B. Pick x ∈ A and y ∈ B. Suppose without loss of generality that x < y. Now define the set, S ≡ {t ∈ [x, y] : [x, t] ⊆ A} and let l be the least upper bound of S. Then l ∈ A so l ∈ / B which implies l ∈ A. But if l ∈ / B, then for some δ > 0, (l, l + δ) ∩ B = ∅ contradicting the definition of l as an upper bound for S. Therefore, l ∈ B which implies l ∈ / A after all, a contradiction. It follows I must be connected. The following theorem is a very useful description of the open sets in R. Theorem 6.7.7 Let U be an open set in R. Then there exist countably many ∞ disjoint open sets, {(ai , bi )}i=1 such that U = ∪∞ i=1 (ai , bi ) . Proof: Let p ∈ U and let z ∈ Cp , the connected component determined by p. Since U is open, there exists, δ > 0 such that (z − δ, z + δ) ⊆ U. It follows from Theorem 6.7.2 that (z − δ, z + δ) ⊆ Cp . This shows Cp is open. By Theorem 6.7.6, this shows Cp is an open interval, (a, b) where a, b ∈ [−∞, ∞] . There are therefore at most countably many of these connected components because each must contain a rational number and the rational ∞ numbers are countable. Denote by {(ai , bi )}i=1 the set of these connected components. This proves the theorem. Definition 6.7.8 A topological space, E is arcwise connected if for any two points, p, q ∈ E, there exists a closed interval, [a, b] and a continuous function, γ : [a, b] → E such that γ (a) = p and γ (b) = q. E is locally connected if it has a basis of connected open sets. E is locally arcwise connected if it has a basis of arcwise connected open sets.

6.7. CONNECTED SETS

161

An example of an arcwise connected topological space would be the any subset of Rn which is the continuous image of an interval. Locally connected is not the same as connected. A well known example is the following. ½µ ¶ ¾ 1 : x ∈ (0, 1] ∪ {(0, y) : y ∈ [−1, 1]} (6.7.12) x, sin x You can verify that this set of points considered as a metric space with the metric from R2 is not locally connected or arcwise connected but is connected. Proposition 6.7.9 If a topological space is arcwise connected, then it is connected. Proof: Let X be an arcwise connected space and suppose it is separated. Then X = A ∪ B where A, B are two separated sets. Pick p ∈ A and q ∈ B. Since X is given to be arcwise connected, there must exist a continuous function γ : [a, b] → X such that γ (a) = p and γ (b) = q. But then we would have γ ([a, b]) = (γ ([a, b]) ∩ A) ∪ (γ ([a, b]) ∩ B) and the two sets, γ ([a, b]) ∩ A and γ ([a, b]) ∩ B are separated thus showing that γ ([a, b]) is separated and contradicting Theorem 6.7.6 and Theorem 6.7.3. It follows that X must be connected as claimed. Theorem 6.7.10 Let U be an open subset of a locally arcwise connected topological space, X. Then U is arcwise connected if and only if U if connected. Also the connected components of an open set in such a space are open sets, hence arcwise connected. Proof: By Proposition 6.7.9 it is only necessary to verify that if U is connected and open in the context of this theorem, then U is arcwise connected. Pick p ∈ U . Say x ∈ U satisfies P if there exists a continuous function, γ : [a, b] → U such that γ (a) = p and γ (b) = x. A ≡ {x ∈ U such that x satisfies P.} If x ∈ A, there exists, according to the assumption that X is locally arcwise connected, an open set, V, containing x and contained in U which is arcwise connected. Thus letting y ∈ V, there exist intervals, [a, b] and [c, d] and continuous functions having values in U , γ, η such that γ (a) = p, γ (b) = x, η (c) = x, and η (d) = y. Then let γ 1 : [a, b + d − c] → U be defined as ½ γ (t) if t ∈ [a, b] γ 1 (t) ≡ η (t + c − b) if t ∈ [b, b + d − c] Then it is clear that γ 1 is a continuous function mapping p to y and showing that V ⊆ A. Therefore, A is open. A 6= ∅ because there is an open set, V containing p which is contained in U and is arcwise connected. Now consider B ≡ U \ A. This is also open. If B is not open, there exists a point z ∈ B such that every open set containing z is not contained in B. Therefore, letting V be one of the basic open sets chosen such that z ∈ V ⊆ U, there exist

162


points of A contained in V. But then, a repeat of the above argument shows z ∈ A also. Hence B is open and so if B 6= ∅, then U = B ∪ A and so U is separated by the two sets, B and A contradicting the assumption that U is connected. It remains to verify the connected components are open. Let z ∈ Cp where Cp is the connected component determined by p. Then picking V an arcwise connected open set which contains z and is contained in U, Cp ∪ V is connected and contained in U and so it must also be contained in Cp . This proves the theorem. As an application, consider the following corollary. Corollary 6.7.11 Let f : Ω → Z be continuous where Ω is a connected open set. Then f must be a constant. Proof: Suppose not. Then it achieves two different values, k and l 6= k. Then Ω = f −1 (l) ∪ f −1 ({m ∈ Z : m 6= l}) and these are disjoint nonempty open sets which separate Ω. To see they are open, note µ ¶¶ µ 1 1 f −1 ({m ∈ Z : m 6= l}) = f −1 ∪m6=l m − , m + 6 6 which is the inverse image of an open set.

Weierstrass Approximation Theorem 7.1

The Bernstein Polynomials

This short chapter is on the important Weierstrass approximation theorem. It is about approximating an arbitrary continuous function uniformly by a polynomial. It will be assumed only that f has values in C and that all scalars are in C. First here is some notation. Definition 7.1.1 α = (α1 , · · · , αn ) for α1 · · · αn positive integers is called a multiindex. For α a multi-index, |α| ≡ α1 + · · · + αn and if x ∈ Rn , x = (x1 , · · · , xn ) , and f a function, define αn 1 α2 xα ≡ xα 1 x2 · · · xn .

A polynomial in n variables of degree m is a function of the form p (x) =

X

aα xα .

|α|≤m

Here α is a multi-index as just described. The following estimate will be the basis for the Weierstrass approximation theorem. It is actually a statement about the variance of a binomial random variable. Lemma 7.1.2 The following estimate holds for x ∈ [0, 1]. m µ ¶ X m k=0

k

2

m−k

(k − mx) xk (1 − x)

163

≤

1 m 4

164

WEIERSTRASS APPROXIMATION THEOREM

Proof: By the Binomial theorem, m µ ¶ X ¡ ¢m m ¡ t ¢k m−k e x (1 − x) = 1 − x + et x . k

(7.1.1)

k=0

Differentiating both sides with respect to t and then evaluating at t = 0 yields m µ ¶ X m m−k kxk (1 − x) = mx. k k=0

Now doing two derivatives of 7.1.1 with respect to t yields Pm ¡m¢ 2 t k m−2 2t 2 m−k = m (m − 1) (1 − x + et x) e x k=0 k k (e x) (1 − x) m−1 t t +m (1 − x + e x) xe . Evaluating this at t = 0, m µ ¶ X m 2 k m−k k (x) (1 − x) = m (m − 1) x2 + mx. k k=0

Therefore, m µ ¶ X m 2 m−k (k − mx) xk (1 − x) k

=

m (m − 1) x2 + mx − 2m2 x2 + m2 x2

=

¡ ¢ 1 m x − x2 ≤ m. 4

k=0

This proves the lemma. n Now for x = (x1 , · · · , xn ) ∈ [0, 1] consider the polynomial, µ ¶ m m µ ¶µ ¶ X X m m m k1 m−k2 m−k1 k2 pm (x) ≡ ··· x2 (1 − x2 ) ··· x (1 − x1 ) k1 k2 kn 1 k1 =1

kn =1

µ · · · xknn

(1 − xn )

m−kn

f

kn k1 ,··· , m m

¶ .

(7.1.2)

Also define if I is a set in Rn ||h||I ≡ sup {|h (x)| : x ∈ I} . Thus pm converges uniformly to f on a set, I if lim ||pm − f ||I = 0.

m→∞

Also to simplify the notation, let k = (k1 , · · · , kn ) where each ki ∈ [0, m], ¡ k1 ¢ kn m , · · · , m , and let µ ¶ µ ¶µ ¶ µ ¶ m m m m ≡ ··· . k k1 k2 kn

k m

≡

7.1. THE BERNSTEIN POLYNOMIALS

165

Also define ||k||∞ ≡ max {ki , i = 1, 2, · · · , n} xk (1 − x)

m−k

≡ xk11 (1 − x1 )

m−k1

m−k2

xk22 (1 − x2 )

m−kn

· · · xknn (1 − xn )

.

Thus in terms of this notation, X

pm (x) =

||k||∞ ≤m

µ ¶ µ ¶ m k k m−k x (1 − x) f m k

n

n

Lemma 7.1.3 For x ∈ [0, 1] , f a continuous function defined on [0, 1] , and pm n given in 7.1.2, pm converges uniformly to f on [0, 1] as m → ∞. Proof: The function, f is uniformly continuous because it is continuous on a compact set. Therefore, there exists δ > 0 such that if |x − y| < δ, then |f (x) − f (y)| < ε. √ 2 Denote by G the set of k such that (ki − mxi ) < η 2 m2 for each η = δ/ n. ¯ k i where ¯ Note this condition is equivalent to saying that for each i, ¯ mi − xi ¯ < η. By the binomial theorem, X µm¶ m−k xk (1 − x) =1 k ||k||∞ ≤m

n

and so for x ∈ [0, 1] , |pm (x) − f (x)| ≤

X ||k||∞ ≤m

¯ µ ¶ ¯ µ ¶ ¯ m k k m−k ¯¯ x (1 − x) f − f (x)¯¯ ¯ k m

¯ ¯ µ ¶ X µm¶ ¯ k m−k ¯¯ k ¯ ≤ x (1 − x) ¯f m − f (x)¯ k k∈G

¯ µ ¶ ¯ X µm¶ ¯ k m−k ¯¯ k + x (1 − x) f − f (x)¯¯ ¯ k m C

(7.1.3)

k∈G

Now for k ∈ G it follows that for each i ¯ ¯ ¯ ki ¯ ¯ − xi ¯ < √δ (7.1.4) ¯m ¯ n ¯ ¡k¢ ¯ ¯k ¯ and so ¯f m − f (x)¯ < ε because the above implies ¯ m − x¯ < δ. Therefore, the first sum on the right in 7.1.3 is no larger than X µm¶ X µm¶ m−k m−k xk (1 − x) ε≤ xk (1 − x) ε = ε. k k k∈G

||k||∞ ≤m

166

WEIERSTRASS APPROXIMATION THEOREM n

Letting M ≥ max {|f (x)| : x ∈ [0, 1] } it follows |pm (x) − f (x)| X µm¶ m−k ≤ ε + 2M xk (1 − x) k C k∈G ¶n X µ ¶ Y µ n m 1 2 m−k (kj − mxj ) xk (1 − x) ≤ ε + 2M k η 2 m2 j=1 k∈GC ¶n X µ ¶ Y µ n 1 m 2 m−k ≤ ε + 2M (kj − mxj ) xk (1 − x) η 2 m2 k j=1 ||k||∞ ≤m

because on GC ,

2

(kj − mxj ) < 1, j = 1, · · · , n. η 2 m2 Now by Lemma 7.1.2, µ |pm (x) − f (x)| ≤ ε + 2M

1 2 η m2

¶n ³

m ń . 4

Therefore, since the right side does not depend on x, it follows lim sup ||pm − f ||[0,1]n ≤ ε m→∞

and since ε is arbitrary, this shows limm→∞ ||pm − f ||[0,1]n = 0. This proves the lemma. The following is not surprising. n

Lemma 7.1.4 Let f be a continuous function defined on [−M, M ] . Then there n exists a sequence of polynomials, {pm } converging uniformly to f on [−M, M ] . Proof: Let h (t) = −M + 2M t so h : [0, 1] → [−M, M ] and let h (t) ≡ n (h (t1 ) , · · · , h (tn )) . Therefore, f ◦h is a continuous function defined on [0, 1] . From 1 Lemma 7.1.3 there exists a polynomial, p (t) such that ||pm − f ◦ h||[0,1]n < m . Now ¡ ¢ n −1 −1 −1 for x ∈ [−M, M ] , h (x) = h (x1 ) , · · · , h (xn ) and so ¯¯ ¯¯ ¯¯pm ◦ h−1 − f ¯¯

[−M,M ]n

= ||pm − f ◦ h||[0,1]n
0, there exists an open set U (y) containing y such that fxy (z) > h (z) − ε if z ∈ U (y). Since A is compact, let U (y1 ) , · · · , U (yl ) cover A. Let fx ≡ max (fxy1 , fxy2 , · · · , fxyl ). Then fx ∈ A and fx (z) > h (z) − ε for all z ∈ A and fx (x) = h (x). This implies that for each x ∈ A there exists an open set V (x) containing x such that for z ∈ V (x), fx (z) < h (z) + ε. Let V (x1 ) , · · · , V (xm ) cover A and let f ≡ min (fx1 , · · · , fxm ). Therefore, f (z) < h (z) + ε for all z ∈ A and since fx (z) > h (z) − ε for all z ∈ A, it follows f (z) > h (z) − ε also and so |f (z) − h (z)| < ε for all z. Since ε is arbitrary, this shows h ∈ A and proves A = C (A; R). This proves the theorem.

170

7.2.2


The Case Of Locally Compact Sets

Definition 7.2.7 Let (X, τ ) be a locally compact Hausdorff space. C0 (X) denotes the space of real or complex valued continuous functions defined on X with the property that if f ∈ C0 (X) , then for each ε > 0 there exists a compact set K such that |f (x)| < ε for all x ∈ / K. Define ||f ||∞ = sup {|f (x)| : x ∈ X}. Lemma 7.2.8 For (X, τ ) a locally compact Hausdorff space with the above norm, C0 (X) is a complete space. ³ ´ e e Proof: Let X, τ be the one point compactification described in Lemma 6.6.20. n ³ ´ o e : f (∞) = 0 . D≡ f ∈C X ³ ´ e . For f ∈ C0 (X) , Then D is a closed subspace of C X ½ fe(x) ≡

f (x) if x ∈ X 0 if x = ∞

and let θ : C0 (X) → D be given by θf = fe. Then θ is one to one and onto and also satisfies ||f ||∞ = ||θf ||∞ . Now D is complete because it is a closed subspace of a complete space and so C0 (X) with ||·||∞ is also complete. This proves the lemma. The above refers to functions which have values in C but the same proof works for functions which have values in any complete normed linear space. In the case where the functions in C0 (X) all have real values, I will denote the resulting space by C0 (X; R) with similar meanings in other cases. With this lemma, the generalization of the Stone Weierstrass theorem to locally compact sets is as follows. Theorem 7.2.9 Let A be an algebra of functions in C0 (X; R) where (X, τ ) is a locally compact Hausdorff space which separates the points and annihilates no point. Then A is dense in C0 (X; R). ³ ´ e e Proof: Let X, τ be the one point compactification as described in Lemma 6.6.20. Let Ae denote all finite linear combinations of the form ( n ) X ci fei + c0 : f ∈ A, ci ∈ R i=1

where for f ∈ C0 (X; R) , ½ fe(x) ≡

f (x) if x ∈ X . 0 if x = ∞

7.2. STONE WEIERSTRASS THEOREM

171

³ ´ e R . It separates points because Then Ae is obviously an algebra of functions in C X; this is true of A. Similarly, it annihilates no point because of the inclusion of c0 an e arbitrary element ³ óf R in the definition above. Therefore from³Theorem ´ 7.2.5, A is e e e dense in C X; R . Letting f ∈ C0 (X; R) , it follows f ∈ C X; R and so there exists a sequence {hn } ⊆ Ae such that hn converges uniformly to fe. Now hn is of Pn n n n e the form i=1 cni ff i + c0 and since f (∞) = 0, you can take each c0 = 0 and so this has shown the existence of a sequence of functions in A such that it converges uniformly to f . This proves the theorem.

7.2.3

The Case Of Complex Valued Functions

What about the general case where C0 (X) consists of complex valued functions and the field of scalars is C rather than R? The following is the version of the Stone Weierstrass theorem which applies to this case. You have to assume that for f ∈ A it follows f ∈ A. Such an algebra is called self adjoint. Theorem 7.2.10 Suppose A is an algebra of functions in C0 (X) , where X is a locally compact Hausdorff space, which separates the points, annihilates no point, and has the property that if f ∈ A, then f ∈ A. Then A is dense in C0 (X). Proof: Let Re A ≡ {Re f : f ∈ A}, Im A ≡ {Im f : f ∈ A}. First I will show that A = Re A + i Im A = Im A + i Re A. Let f ∈ A. Then f=

¢ 1¡ ¢ 1¡ f +f + f − f = Re f + i Im f ∈ Re A + i Im A 2 2

and so A ⊆ Re A + i Im A. Also ´ ¢ i³ 1 ¡ f= if + if − if + (if ) = Im (if ) + i Re (if ) ∈ Im A + i Re A 2i 2 This proves one half of the desired equality. Now suppose h ∈ Re A + i Im A. Then h = Re g1 + i Im g2 where gi ∈ A. Then since Re g1 = 21 (g1 + g1 ) , it follows Re g1 ∈ A. Similarly Im g2 ∈ A. Therefore, h ∈ A. The case where h ∈ Im A + i Re A is similar. This establishes the desired equality. Now Re A and Im A are both real algebras. I will show this now. First consider Im A. It is obvious this is a real vector space. It only remains to verify that the product of two functions in Im A is in Im A. Note that from the first part, Re A, Im A are both subsets of A because, for example, if u ∈ Im A then u + 0 ∈ Im A + i Re A = A. Therefore, if v, w ∈ Im A, both iv and w are in A and so Im (ivw) = vw and ivw ∈ A. Similarly, Re A is an algebra. Both Re A and Im A must separate the points. Here is why: If x1 6= x2 , then there exists f ∈ A such that f (x1 ) 6= f (x2 ) . If Im f (x1 ) 6= Im f (x2 ) , this shows there is a function in Im A, Im f which separates these two points. If Im f fails to separate the two points, then Re f must separate the points and so you could consider Im (if ) to get a function in Im A which separates these points. This shows Im A separates the points. Similarly Re A separates the points.

172


Neither Re A nor Im A annihilate any point. This is easy to see because if x is a point there exists f ∈ A such that f (x) 6= 0. Thus either Re f (x) 6= 0 or Im f (x) 6= 0. If Im f (x) 6= 0, this shows this point is not annihilated by Im A. If Im f (x) = 0, consider Im (if ) (x) = Re f (x) 6= 0. Similarly, Re A does not annihilate any point. It follows from Theorem 7.2.9 that Re A and Im A are dense in the real valued functions of C0 (X). Let f ∈ C0 (X) . Then there exists {hn } ⊆ Re A and {gn } ⊆ Im A such that hn → Re f uniformly and gn → Im f uniformly. Therefore, hn + ign ∈ A and it converges to f uniformly. This proves the theorem.

7.3

Exercises

1. Let (X, τ ) , (Y, η) be topological spaces and let A ⊆ X be compact. Then if f : X → Y is continuous, show that f (A) is also compact. 2. ↑ In the context of Problem 1, suppose R = Y where the usual topology is placed on R. Show f achieves its maximum and minimum on A. 3. Let V be an open set in Rn . Show there is an increasing sequence of compact sets, Km , such that V = ∪∞ m=1 Km . Hint: Let ½ ¾ ¡ ¢ 1 n C Cm ≡ x ∈ R : dist x,V ≥ m where dist (x,S) ≡ inf {|y − x| such that y ∈ S}. Consider Km ≡ Cm ∩ B (0,m). 4. Let B (X; Rn ) be the space of functions f , mapping X to Rn such that sup{|f (x)| : x ∈ X} < ∞. Show B (X; Rn ) is a complete normed linear space if ||f || ≡ sup{|f (x)| : x ∈ X}. 5. Let α ∈ [0, 1]. Define, for X a compact subset of Rp , C α (X; Rn ) ≡ {f ∈ C (X; Rn ) : ρα (f ) + ||f || ≡ ||f ||α < ∞} where ||f || ≡ sup{|f (x)| : x ∈ X} and ρα (f ) ≡ sup{

|f (x) − f (y)| : x, y ∈ X, x 6= y}. α |x − y|

Show that (C α (X; Rn ) , ||·||α ) is a complete normed linear space.

7.3. EXERCISES

173

α n p 6. Let {fn }∞ n=1 ⊆ C (X; R ) where X is a compact subset of R and suppose

||fn ||α ≤ M for all n. Show there exists a subsequence, nk , such that fnk converges in C (X; Rn ). The given sequence is called precompact when this happens. (This also shows the embedding of C α (X; Rn ) into C (X; Rn ) is a compact embedding.) 7. Use the general Stone Weierstrass approximation theorem to prove Theorem 7.1.6. 8. Let (X, d) be a metric space where d is a bounded metric. Let C denote the collection of closed subsets of X. For A, B ∈ C, define ρ (A, B) ≡ inf {δ > 0 : Aδ ⊇ B and Bδ ⊇ A} where for a set S, Sδ ≡ {x : dist (x, S) ≡ inf {d (x, s) : s ∈ S} ≤ δ} . Show x → dist (x, S) is continuous and that therefore, Sδ is a closed set containing S. Also show that ρ is a metric on C. This is called the Hausdorff metric. 9. ↑Suppose (X, d) is a compact metric space. Show (C, ρ) is a complete metric space. Hint: Show first that if Wn ↓ W where Wn is closed, then ρ (Wn , W ) → 0. Now let {An } be a Cauchy sequence in C. Then if ε > 0 there exists N such that when m, n ≥ N, then ρ (An , Am ) < ε. Therefore, for each n ≥ N, (An )ε ⊇∪∞ k=n Ak . ∞ Let A ≡ ∩∞ n=1 ∪k=n Ak . By the first part, there exists N1 > N such that for n ≥ N1 , ¡ ¢ ∞ ρ ∪∞ k=n Ak , A < ε, and (An )ε ⊇ ∪k=n Ak .

Therefore, for such n, Aε ⊇ Wn ⊇ An and (Wn )ε ⊇ (An )ε ⊇ A because (An )ε ⊇ ∪∞ k=n Ak ⊇ A. 10. ↑ Let X be a compact metric space. Show (C, ρ) is compact. Hint: Let Dn be a 2−n net for X. Let Kn denote finite unions of sets of the form B (p, 2−n ) where p ∈ Dn . Show Kn is a 2−(n−1) net for (C, ρ) .

174


Part II

Real And Abstract Analysis

175

Abstract Measure And Integration 8.1

σ Algebras

This chapter is on the basics of measure theory and integration. A measure is a real valued mapping from some subset of the power set of a given set which has values in [0, ∞]. Many apparently different things can be considered as measures and also there is an integral defined. By discussing this in terms of axioms and in a very abstract setting, many different topics can be considered in terms of one general theory. For example, it will turn out that sums are included as an integral of this sort. So is the usual integral as well as things which are often thought of as being in between sums and integrals. Let Ω be a set and let F be a collection of subsets of Ω satisfying ∅ ∈ F, Ω ∈ F ,

(8.1.1)

E ∈ F implies E C ≡ Ω \ E ∈ F , ∞ If {En }∞ n=1 ⊆ F, then ∪n=1 En ∈ F .

(8.1.2)

Definition 8.1.1 A collection of subsets of a set, Ω, satisfying Formulas 8.1.18.1.2 is called a σ algebra. As an example, let Ω be any set and let F = P(Ω), the set of all subsets of Ω (power set). This obviously satisfies Formulas 8.1.1-8.1.2. Lemma 8.1.2 Let C be a set whose elements are σ algebras of subsets of Ω. Then ∩C is a σ algebra also. Be sure to verify this lemma. It follows immediately from the above definitions but it is important for you to check the details. Example 8.1.3 Let τ denote the collection of all open sets in Rn and let σ (τ ) ≡ intersection of all σ algebras that contain τ . σ (τ ) is called the σ algebra of Borel sets . In general, for a collection of sets, Σ, σ (Σ) is the smallest σ algebra which contains Σ. 177

178

ABSTRACT MEASURE AND INTEGRATION

This is a very important σ algebra and it will be referred to frequently as the Borel sets. Attempts to describe a typical Borel set are more trouble than they are worth and it is not easy to do so. Rather, one uses the definition just given in the example. Note, however, that all countable intersections of open sets and countable unions of closed sets are Borel sets. Such sets are called Gδ and Fσ respectively. Definition 8.1.4 Let F be a σ algebra of sets of Ω and let µ : F → [0, ∞]. µ is called a measure if ∞ ∞ [ X µ( Ei ) = µ(Ei ) (8.1.3) i=1

i=1

whenever the Ei are disjoint sets of F. The triple, (Ω, F, µ) is called a measure space and the elements of F are called the measurable sets. (Ω, F, µ) is a finite measure space when µ (Ω) < ∞. Note that the above definition immediately implies that if Ei ∈ F and the sets Ei are not necessarily disjoint, µ(

∞ [

Ei ) ≤

i=1

∞ X

µ (Ei ) .

i=1

To see this, let F1 ≡ E1 , F2 ≡ E2 \ E1 , · · · , Fn ≡ En \ ∪n−1 i=1 Ei , then the sets Fi are disjoint sets in F and µ(

∞ [ i=1

Ei ) = µ(

∞ [

i=1

Fi ) =

∞ X

µ (Fi ) ≤

i=1

∞ X

µ(Ei )

i=1

because of the fact that each Ei ⊇ Fi and so µ (Ei ) = µ (Fi ) + µ (Ei \ Fi ) which implies µ (Ei ) ≥ µ (Fi ) . The following theorem is the basis for most of what is done in the theory of measure and integration. It is a very simple result which follows directly from the above definition. Theorem 8.1.5 Let {Em }∞ m=1 be a sequence of measurable sets in a measure space (Ω, F, µ). Then if · · · En ⊆ En+1 ⊆ En+2 ⊆ · · · , µ(∪∞ i=1 Ei ) = lim µ(En ) n→∞

(8.1.4)

and if · · · En ⊇ En+1 ⊇ En+2 ⊇ · · · and µ(E1 ) < ∞, then µ(∩∞ i=1 Ei ) = lim µ(En ). n→∞

(8.1.5)

Stated more succinctly, Ek ↑ E implies µ (Ek ) ↑ µ (E) and Ek ↓ E with µ (E1 ) < ∞ implies µ (Ek ) ↓ µ (E).

8.1. σ ALGEBRAS

179

C C ∞ ∈ F so ∩∞ Proof: First note that ∩∞ i=1 Ei is measurable. i=1 Ei = (∪i=1 Ei ) ¡ C ¢C Also note that for A and B sets of F, A \ B ≡ A ∪ B ∈ F . To show 8.1.4, note that 8.1.4 is obviously true if µ(Ek ) = ∞ for any k. Therefore, assume µ(Ek ) < ∞ for all k. Thus µ(Ek+1 \ Ek ) + µ(Ek ) = µ(Ek+1 )

and so µ(Ek+1 \ Ek ) = µ(Ek+1 ) − µ(Ek ). Also,

∞ [

Ek = E1 ∪

k=1

∞ [

(Ek+1 \ Ek )

k=1

and the sets in the above union are disjoint. Hence by 8.1.3, µ(∪∞ i=1 Ei ) = µ(E1 ) +

∞ X

µ(Ek+1 \ Ek ) = µ(E1 )

k=1

+

∞ X

µ(Ek+1 ) − µ(Ek )

k=1

= µ(E1 ) + lim

n X

n→∞

µ(Ek+1 ) − µ(Ek ) = lim µ(En+1 ). n→∞

k=1

This shows part 8.1.4. To verify 8.1.5, ∞ µ(E1 ) = µ(∩∞ i=1 Ei ) + µ(E1 \ ∩i=1 Ei ) n ∞ since µ(E1 ) < ∞, it follows µ(∩∞ i=1 Ei ) < ∞. Also, E1 \ ∩i=1 Ei ↑ E1 \ ∩i=1 Ei and so by 8.1.4, ∞ n µ(E1 ) − µ(∩∞ i=1 Ei ) = µ(E1 \ ∩i=1 Ei ) = lim µ(E1 \ ∩i=1 Ei ) n→∞

= µ(E1 ) − lim µ(∩ni=1 Ei ) = µ(E1 ) − lim µ(En ), n→∞

n→∞

Hence, subtracting µ (E1 ) from both sides, lim µ(En ) = µ(∩∞ i=1 Ei ).

n→∞

This proves the theorem. It is convenient to allow functions to take the value +∞. You should think of +∞, usually referred to as ∞ as something out at the right end of the real line and its only importance is the notion of sequences converging to it. xn → ∞ exactly when for all l ∈ R, there exists N such that if n ≥ N, then xn > l.

180


This is what it means for a sequence to converge to ∞. Don’t think of ∞ as a number. It is just a convenient symbol which allows the consideration of some limit operations more simply. Similar considerations apply to −∞ but this value is not of very great interest. In fact the set of most interest is the complex numbers or some vector space. Therefore, this topic is not considered. Lemma 8.1.6 Let f : Ω → (−∞, ∞] where F is a σ algebra of subsets of Ω. Then the following are equivalent. f −1 ((d, ∞]) ∈ F for all finite d, f −1 ((−∞, d)) ∈ F for all finite d, f −1 ([d, ∞]) ∈ F for all finite d, f −1 ((−∞, d]) ∈ F for all finite d, f −1 ((a, b)) ∈ F for all a < b, −∞ < a < b < ∞. Proof: First note that the first and the third are equivalent. To see this, observe −1 ((d − 1/n, ∞]), f −1 ([d, ∞]) = ∩∞ n=1 f

and so if the first condition holds, then so does the third. −1 f −1 ((d, ∞]) = ∪∞ ([d + 1/n, ∞]), n=1 f

and so if the third condition holds, so does the first. Similarly, the second and fourth conditions are equivalent. Now f −1 ((−∞, d]) = (f −1 ((d, ∞]))C so the first and fourth conditions are equivalent. Thus the first four conditions are equivalent and if any of them hold, then for −∞ < a < b < ∞, f −1 ((a, b)) = f −1 ((−∞, b)) ∩ f −1 ((a, ∞]) ∈ F . Finally, if the last condition holds, ¡ ¢C −1 f −1 ([d, ∞]) = ∪∞ ((−k + d, d)) ∈ F k=1 f and so the third condition holds. Therefore, all five conditions are equivalent. This proves the lemma. This lemma allows for the following definition of a measurable function having values in (−∞, ∞]. Definition 8.1.7 Let (Ω, F, µ) be a measure space and let f : Ω → (−∞, ∞]. Then f is said to be measurable if any of the equivalent conditions of Lemma 8.1.6 hold. When the σ algebra, F equals the Borel σ algebra, B, the function is called Borel measurable. More generally, if f : Ω → X where X is a topological space, f is said to be measurable if f −1 (U ) ∈ F whenever U is open.

8.1. σ ALGEBRAS

181

Theorem 8.1.8 Let fn and f be functions mapping Ω to (−∞, ∞] where F is a σ algebra of measurable sets of Ω. Then if fn is measurable, and f (ω) = limn→∞ fn (ω), it follows that f is also measurable. (Pointwise limits of measurable functions are measurable.) ¡ ¢ 1 1 Proof: First is is shown f −1 ((a, b)) ∈ F. Let Vm ≡ a + m ,b − m and £ ¤ 1 1 Vm = a+ m ,b − m . Then for all m, Vm ⊆ (a, b) and ∞ (a, b) = ∪∞ m=1 Vm = ∪m=1 V m .

Note that Vm 6= ∅ for all m large enough. Since f is the pointwise limit of fn , f −1 (Vm ) ⊆ {ω : fk (ω) ∈ Vm for all k large enough} ⊆ f −1 (V m ). You should note that the expression in the middle is of the form −1 ∞ ∪∞ n=1 ∩k=n fk (Vm ).

Therefore, −1 −1 ∞ ∞ f −1 ((a, b)) = ∪∞ (Vm ) ⊆ ∪∞ m=1 f m=1 ∪n=1 ∩k=n fk (Vm ) −1 ⊆ ∪∞ (V m ) = f −1 ((a, b)). m=1 f

It follows f −1 ((a, b)) ∈ F because it equals the expression in the middle which is measurable. This shows f is measurable. The following theorem considers the case of functions which have values in a metric space. Its proof is similar to the proof of the above. Theorem 8.1.9 Let {fn } be a sequence of measurable functions mapping Ω to (X, d) where (X, d) is a metric space and (Ω, F) is a measure space. Suppose also that f (ω) = limn→∞ fn (ω) for all ω. Then f is also a measurable function. Proof: It is required to show f −1 (U ) is measurable for all U open. Let ½ ¾ ¡ ¢ 1 C Vm ≡ x ∈ U : dist x, U > . m Thus

½ Vm ⊆

¡ ¢ 1 x ∈ U : dist x, U C ≥ m

¾

and Vm ⊆ Vm ⊆ Vm+1 and ∪m Vm = U. Then since Vm is open, −1 ∞ f −1 (Vm ) = ∪∞ n=1 ∩k=n fk (Vm )

and so f −1 (U )

−1 = ∪∞ (Vm ) m=1 f ∞ ∞ = ∪m=1 ∪n=1 ∩∞ f −1 (Vm ) ¡ k=n ¢ k −1 ∞ −1 Vm = f (U ) ⊆ ∪m=1 f

which shows f −1 (U ) is measurable. This proves the theorem. Now here is a simple observation.

182


Observation 8.1.10 Let f : Ω → X where X is some topological space. Suppose f (ω) =

m X

xk XAk (ω)

k=1

where each xk ∈ X and the Ak are disjoint measurable sets. (Such functions are often referred to as simple functions.) The sum means the function has value xk on set Ak . Then f is measurable. Proof: Letting U be open, f −1 (U ) = ∪ {Ak : xk ∈ U } , a finite union of measurable sets. There is also a very interesting theorem due to Kuratowski [44] which is presented next. Theorem 8.1.11 Let E be a compact metric space and let (Ω, F) be a measure space. Suppose ψ : E × Ω → R has the property that x → ψ (x, ω) is continuous and ω → ψ (x, ω) is measurable. Then there exists a measurable function, f having values in E such that ψ (f (ω) , ω) = sup ψ (x, ω) . x∈E

Furthermore, ω → ψ (f (ω) , ω) is measurable. Proof: Let C1 be a 2−1 net of E. Suppose C1 , · · · , Cm have © ¡been chosen¢ such ª that Ck is a 2−k net and Ci+1 ⊇ Ci for all i. Then consider E\∪ B x, 2−(m+1) : x ∈ Cm . r If this set is empty, let Cm+1 = Cm . If it is nonempty, let {yi }i=1 be a 2−(m+1) net r ∞ for this compact set. Then let Cm+1 = Cm ∪ {yi }i=1 . It follows {Cm }m=1 satisfies Cm is a 2−m net and Cm ⊆ Cm+1 . © ªm(1) Let x1k k=1 equal C1 . Let ¾ ½ ¡ 1 ¢ ¡ 1 ¢ 1 A1 ≡ ω : ψ x1 , ω = max ψ xk , ω k

For ω ∈ A11 , define s1 (ω) ≡ x11 . Next let ½ ¾ ¡ 1 ¢ ¡ 1 ¢ 1 1 A2 ≡ ω ∈ / A1 : ψ x2 , ω = max ψ xk , ω k

and let s1 (ω) ≡ x12 on A12 . Continue in this way to obtain a simple function, s1 such that ψ (s1 (ω) , ω) = max {ψ (x, ω) : x ∈ C1 } and s1 has values in C1 . Suppose s1 (ω) , s2 (ω) , · · · , sm (ω) are simple functions with the property that if m > 1, |sk (ω) − sk+1 (ω)| < 2−k , ψ (sk (ω) , ω) = max {ψ (x, ω) : x ∈ Ck } sk has values in Ck

8.1. σ ALGEBRAS

183

for each k + 1 ≤ m, only the second and third assertions holding if m = 1. Letting N Cm = {xk }k=1 , it follows sm (ω) is of the form sm (ω) =

N X

xk XAk (ω) , Ai ∩ Aj = ∅.

(8.1.6)

k=1 n

1 Denote by {y1i }i=1 those points of Cm+1 which are contained in B (x1 , 2−m ) . Letting Ak play the role of Ω in the first step in which s1 was constructed, for each ω ∈ A1 let sm+1 (ω) be a simple function which has one of the values y1i and satisfies

ψ (sm+1 (ω) , ω) = max ψ (y1i , ω) i≤n1

n2 for each ω ∈ A1 . Next let {y2i }i=1 be which are contained in B (x2 , 2−m ). n2 taken from {y2i }i=1 and

n

1 those points of Cm+1 different than {y1i }i=1 Then define sm+1 (ω) on A2 to have values

ψ (sm+1 (ω) , ω) = max ψ (y2i , ω) i≤n2

for each ω ∈ A2 . Continuing this way defines sm+1 on all of Ω and it satisfies |sm (ω) − sm+1 (ω)| < 2−m for all ω ∈ Ω

(8.1.7)

ψ (sm+1 (ω) , ω) = max {ψ (x, ω) : x ∈ Cm+1 } .

(8.1.8)

It remains to verify

To see this is so, pick ω ∈ Ω. Let max {ψ (x, ω) : x ∈ Cm+1 } = ψ (yj , ω)

(8.1.9)

where yj ∈ Cm+1 and out of all the balls B (xl , 2−m ) , the first one which contains yj is B (xk , 2−m ). Then by the construction, sm+1 (ω) = yj . This and 8.1.9 verifies 8.1.8. From 8.1.7 it follows sm (ω) converges uniformly on Ω to a measurable function, f (ω) . Then from the construction, ψ (f (ω) , ω) ≥ ψ (sm (ω) , ω) for all m and ω. Now pick ω ∈ Ω and let z be such that ψ (z, ω) = maxx∈E ψ (x, ω). Letting yk → z where yk ∈ Ck , it follows from continuity of ψ in the first argument that max ψ (x, ω) x∈E

= ψ (z, ω) = lim ψ (yk , ω) k→∞

≤

lim ψ (sm (ω) , ω) = ψ (f (ω) , ω) ≤ max ψ (x, ω) .

m→∞

x∈E

To show ω → ψ (f (ω) , ω) is measurable, note that since E is compact, there exists a countable dense subset, D. Then using continuity of ψ in the first argument, ψ (f (ω) , ω)

=

sup ψ (x, ω) x∈E

=

sup ψ (x, ω) x∈D

which equals a measurable function of ω because D is countable. This proves the theorem.

184


Theorem 8.1.12 Let B consist of open cubes of the form Qx ≡

n Y

(xi − δ, xi + δ)

i=1

where δ is a positive rational number and x ∈ Qn . Then every open set in Rn can be written as a countable union of open cubes from B. Furthermore, B is a countable set. Proof: Let U be an open set and√let y ∈ U. Since U is open, B (y, r) ⊆ U for some r > 0 and it can be assumed r/ n ∈ Q. Let µ x ∈ B y,

r √ 10 n

¶ ∩ Qn

and consider the cube, Qx ∈ B defined by Qx ≡

n Y

(xi − δ, xi + δ)

i=1

√ where δ = r/4 n. The following picture is roughly illustrative of what is taking place.

B(y, r) Qx

qxq y

Then the diameter of Qx equals Ã µ ¶2 !1/2 r r √ n = 2 2 n and so, if z ∈ Qx , then |z − y|

|z − x| + |x − y| r r < + = r. 2 2

≤

8.1. σ ALGEBRAS

185

Consequently, Qx ⊆ U. Now also, Ã n X

!1/2 2

(xi − yi )

α} ≡ f −1 ((α, ∞]) with obvious modifications for the symbols [α ≤ f ] , [α ≥ f ] , [α ≥ f ≥ β], etc. Definition 8.1.16 For a set E, ½ XE (ω) =

1 if ω ∈ E, 0 if ω ∈ / E.

This is called the characteristic function of E. Sometimes this is called the indicator function which I think is better terminology since the term characteristic function has another meaning. Note that this “indicates” whether a point, ω is contained in E. It is exactly when the function has the value 1. Theorem 8.1.17 (Egoroff ) Let (Ω, F, µ) be a finite measure space, (µ(Ω) < ∞) and let fn , f be complex valued functions such that Re fn , Im fn are all measurable and lim fn (ω) = f (ω) n→∞

for all ω ∈ / E where µ(E) = 0. Then for every ε > 0, there exists a set, F ⊇ E, µ(F ) < ε, such that fn converges uniformly to f on F C .

8.1. σ ALGEBRAS

187

Proof: First suppose E = ∅ so that convergence is pointwise everywhere. It follows then that Re f and Im f are pointwise limits of measurable functions and are therefore measurable. Let Ekm = {ω ∈ Ω : |fn (ω) − f (ω)| ≥ 1/m for some n > k}. Note that q 2 2 |fn (ω) − f (ω)| = (Re fn (ω) − Re f (ω)) + (Im fn (ω) − Im f (ω)) and so, By Theorem 8.1.13,

·

¸ 1 m is measurable. Hence Ekm is measurable because · ¸ 1 Ekm = ∪∞ |f − f | ≥ . n n=k+1 m |fn − f | ≥

For fixed m, ∩∞ k=1 Ekm = ∅ because fn converges to f . Therefore, if ω ∈ Ω there 1 exists k such that if n > k, |fn (ω) − f (ω)| < m which means ω ∈ / Ekm . Note also that Ekm ⊇ E(k+1)m . Since µ(E1m ) < ∞, Theorem 8.1.5 on Page 178 implies 0 = µ(∩∞ k=1 Ekm ) = lim µ(Ekm ). k→∞

Let k(m) be chosen such that µ(Ek(m)m ) < ε2−m and let F =

∞ [

Ek(m)m .

m=1

Then µ(F ) < ε because µ (F ) ≤

∞ X

∞ ¡ ¢ X µ Ek(m)m < ε2−m = ε

m=1

m=1

C Now let η > 0 be given and pick m0 such that m−1 0 < η. If ω ∈ F , then

ω∈

∞ \

C Ek(m)m .

m=1

Hence ω ∈

C Ek(m 0 )m0

so |fn (ω) − f (ω)| < 1/m0 < η

for all n > k(m0 ). This holds for all ω ∈ F C and so fn converges uniformly to f on F C. ∞ Now if E 6= ∅, consider {XE C fn }n=1 . Each XE C fn has real and imaginary parts measurable and the sequence converges pointwise to XE f everywhere. Therefore, from the first part, there exists a set of measure less than ε, F such that on C F C , {XE C fn } converges uniformly to XE C f. Therefore, on (E ∪ F ) , {fn } converges uniformly to f . This proves the theorem. Finally here is a comment about notation.

188


Definition 8.1.18 Something happens for µ a.e. ω said as µ almost everywhere, if there exists a set E with µ(E) = 0 and the thing takes place for all ω ∈ / E. Thus f (ω) = g(ω) a.e. if f (ω) = g(ω) for all ω ∈ / E where µ(E) = 0. A measure space, (Ω, F, µ) is σ finite if there exist measurable sets, Ωn such that µ (Ωn ) < ∞ and Ω = ∪∞ n=1 Ωn .

8.2

Exercises

1. Let Ω = N ={1, 2, · · · }. Let F = P(N) and let µ(S) = number of elements in S. Thus µ({1}) = 1 = µ({2}), µ({1, 2}) = 2, etc. Show (Ω, F, µ) is a measure space. It is called counting measure. What functions are measurable in this case? 2. Let Ω be any uncountable set and let F = {A ⊆ Ω : either A or AC is countable}. Let µ(A) = 1 if A is uncountable and µ(A) = 0 if A is countable. Show (Ω, F, µ) is a measure space. This is a well known bad example. 3. Let F be a σ algebra of subsets of Ω and suppose F has infinitely many elements. Show that F is uncountable. Hint: You might try to show there exists a countable sequence of disjoint sets of F, {Ai }. It might be easiest to verify this by contradiction if it doesn’t exist rather than a direct construction. Once this has been done, you can define a map, θ, from P (N) into F which is one to one by θ (S) = ∪i∈S Ai . Then argue P (N) is uncountable and so F is also uncountable. 4. Prove Lemma 8.1.2. 5. g is Borel measurable if whenever U is open, g −1 (U ) is Borel. Let f : Ω → Rn and let g : Rn → R and F is a σ algebra of sets of Ω. Suppose f is measurable and g is Borel measurable. Show g ◦ f is measurable. To say g is Borel measurable means g −1 (open set) = (Borel set) where a Borel set is one of those sets in the smallest σ algebra containing the open sets of Rn . See Lemma 8.1.2. Hint: You should show, using Theorem 8.1.12 that f −1 (open set) ∈ F. Now let © ª S ≡ E ⊆ Rn : f −1 (E) ∈ F By what you just showed, S contains the open sets. Now verify S is a σ algebra. Argue that from the definition of the Borel sets, it follows S contains the Borel sets. 6. Let (Ω, F) be a measure space and suppose f : Ω → C. Then f is said to be mesurable if f −1 (open set) ∈ F . Show f is measurable if and only if Re f and Im f are measurable real-valued functions. Thus it suffices to define a complex valued function to be measurable if the real and imaginary parts are measurable. Hint: Argue that

8.2. EXERCISES

189 −1

−1

f −1 (((a, b) + i (c, d))) = (Re f ) ((a, b)) ∩ (Im f ) ((c, d)) . Then use Theorem 8.1.12 to verify that if Re f and Im f are measurable, it follows f is. −1 Conversely, argue that (Re f ) ((a, b)) = f −1 ((a, b) + iR) with a similar formula holding for Im f. 7. Let (Ω, F, µ) be a measure space. Define µ : P(Ω) → [0, ∞] by µ(A) = inf{µ(B) : B ⊇ A, B ∈ F}. Show µ satisfies µ(∅) = µ(∪∞ i=1 Ai ) ≤

0, if A ⊆ B, µ(A) ≤ µ(B), ∞ X µ(Ai ), µ (A) = µ (A) if A ∈ F. i=1

If µ satisfies these conditions, it is called an outer measure. This shows every measure determines an outer measure on the power set. 8. Let {Ei } be a sequence of measurable sets with the property that ∞ X

µ(Ei ) < ∞.

i=1

Let S = {ω ∈ Ω such that ω ∈ Ei for infinitely many values of i}. Show µ(S) = 0 and S is measurable. This is part of the Borel Cantelli lemma. Hint: Write S in terms of intersections and unions. Something is in S means that for every n there exists k > n such that it is in Ek . Remember the tail of a convergent series is small. 9. ↑ Let fn , f be measurable functions. fn converges in measure if lim µ(x ∈ Ω : |f (x) − fn (x)| ≥ ε) = 0

n→∞

for each fixed ε > 0. Prove the theorem of F. Riesz. If fn converges to f in measure, then there exists a subsequence {fnk } which converges to f a.e. Hint: Choose n1 such that µ(x : |f (x) − fn1 (x)| ≥ 1) < 1/2. Choose n2 > n1 such that µ(x : |f (x) − fn2 (x)| ≥ 1/2) < 1/22, n3 > n2 such that µ(x : |f (x) − fn3 (x)| ≥ 1/3) < 1/23, etc. Now consider what it means for fnk (x) to fail to converge to f (x). Then use Problem 8.

190

8.3 8.3.1


The Abstract Lebesgue Integral Preliminary Observations

This section is on the Lebesgue integral and the major convergence theorems which are the reason for studying it. In all that follows µ will be a measure defined on a σ algebra F of subsets of Ω. 0 · ∞ = 0 is always defined to equal zero. This is a meaningless expression and so it can be defined arbitrarily but a little thought will soon demonstrate that this is the right definition in the context of measure theory. To see this, consider the zero function defined on R. What should the integral of this function equal? Obviously, by an analogy with the Riemann integral, it should equal zero. Formally, it is zero times the length of the set or infinity. This is why this convention will be used. Lemma 8.3.1 Let f (a, b) ∈ [−∞, ∞] for a ∈ A and b ∈ B where A, B are sets. Then sup sup f (a, b) = sup sup f (a, b) . a∈A b∈B

b∈B a∈A

Proof: Note that for all a, b, f (a, b) ≤ supb∈B supa∈A f (a, b) and therefore, for all a, sup f (a, b) ≤ sup sup f (a, b) . b∈B a∈A

b∈B

Therefore, sup sup f (a, b) ≤ sup sup f (a, b) . a∈A b∈B

b∈B a∈A

Repeating the same argument interchanging a and b, gives the conclusion of the lemma. Lemma 8.3.2 If {An } is an increasing sequence in [−∞, ∞], then sup {An } = limn→∞ An . The following lemma is useful also and this is a good place to put it. First ∞ {bj }j=1 is an enumeration of the aij if ∪∞ j=1 {bj } = ∪i,j {aij } . ∞

In other words, the countable set, {aij }i,j=1 is listed as b1 , b2 , · · · . P∞ P∞ P∞ P∞ ∞ Lemma 8.3.3 Let aij ≥ 0. Then i=1 j=1 aij = j=1 i=1 aij . Also if {bj }j=1 P∞ P∞ P∞ is any enumeration of the aij , then j=1 bj = i=1 j=1 aij . Proof: First note there is no trouble in defining these sums because the aij are all nonnegative. If a sum diverges, it only diverges to ∞ and so ∞ is written as the answer. ∞ X ∞ ∞ X n m X n X X X aij ≥ sup aij = sup lim aij j=1 i=1

n

j=1 i=1

n m→∞

j=1 i=1

8.3. THE ABSTRACT LEBESGUE INTEGRAL = sup lim

n m→∞

n X m X

aij = sup n

i=1 j=1

n X ∞ X

191 aij =

∞ ∞ X X

aij .

(8.3.10)

i=1 j=1

i=1 j=1

Interchanging the i and j in the above argument the first part of the lemma is proved. Finally, note that for all p, p X

bj ≤

j=1

and so

P∞

j=1 bj

≤

P∞ P∞ i=1

j=1

∞ X ∞ X

aij

i=1 j=1

aij . Now let m, n > 1 be given. Then m X n X

aij ≤

i=1 j=1

p X

bj

j=1

where p is chosen large enough that {b1 , · · · , bp } ⊇ {aij : i ≤ m and j ≤ n} . Therefore, since such a p exists for any choice of m, n,it follows that for any m, n, m X n X

aij ≤

i=1 j=1

∞ X

bj .

j=1

Therefore, taking the limit as n → ∞, m X ∞ X

aij ≤

i=1 j=1

∞ X

bj

j=1

and finally, taking the limit as m → ∞, ∞ X ∞ X

aij ≤

i=1 j=1

∞ X

bj

j=1

proving the lemma.

8.3.2

Definition Of The Lebesgue Integral For Nonnegative Measurable Functions

The following picture illustrates the idea used to define the Lebesgue integral to be like the area under a curve.

3h 2h

hµ([3h < f ])

h

hµ([2h < f ]) hµ([h < f ])

192


You can see that by following the procedure illustrated in the picture and letting h get smaller, you would expect to obtain better approximations to the area under the curve1 although all these approximations would likely be too small. Therefore, define Z f dµ ≡ sup

∞ X

hµ ([ih < f ])

h>0 i=1

Lemma 8.3.4 The following inequality holds. ∞ X

hµ ([ih < f ]) ≤

i=1

µ· ¸¶ ∞ X h h µ i 0 be given and let δ1

p X i=1

µ (Ei ) < ε.

194


Pick δ < δ 1 such that for h < δ it is also true that h
kh]) =

k=1

∞ ∞ X X h µ ([ih < s ≤ (i + 1) h]) k=1

= =

i=k

∞ X i X

hµ ([ih < s ≤ (i + 1) h])

i=1 k=1 ∞ X

ihµ ([ih < s ≤ (i + 1) h]) .

(8.3.13)

i=1

Because of the choice of h there exist positive integers, ik such that i1 < i2 < · · · , < ip and i1 h < a1 ≤ (i1 + 1) h < · · · < i2 h < a2 < < (i2 + 1) h < · · · < ip h < ap ≤ (ip + 1) h Then in the sum of 8.3.13 the only terms which are nonzero are those for which i ∈ {i1 , i2 · · · , ip }. To see this, you might consider the following picture.

a3 a2 a1

i3 h i2 h i1 h

When ih and (i + 1) h are both in between two of the ai the set [ih < s ≤ (i + 1) h] must be empty because the only values of the function are one of the ai . At an ik , ik h is smaller than ak while (ik + 1) h is at least as large. Therefore, the set [ih < s ≤ (i + 1) h] equals Ek and so µ ([ik h < s ≤ (ik + 1) h]) = µ (Ek ) . Therefore, ∞ X k=1

hµ ([s > kh]) =

p X k=1

ik hµ (Ek ) .

8.3. THE ABSTRACT LEBESGUE INTEGRAL

195

It follows that for all h this small, 0

< =

p X k=1 p X

ak µ (Ek ) − ak µ (Ek ) −

k=1

∞ X k=1 p X

hµ ([s > kh]) ik hµ (Ek ) ≤ h

k=1

p X

µ (Ek ) < ε.

k=1

Taking the inf for h this small and using Lemma 8.3.4, 0

≤ =

p X k=1 p X

ak µ (Ek ) − sup

δ>h>0

Z ak µ (Ek ) −

∞ X

hµ ([s > kh])

k=1

sdµ ≤ ε.

k=1

Since ε > 0 is arbitrary, this proves the first part. To verify 8.3.12 Note the formula is obvious if λ = 0 because then [ih < λf ] = ∅ for all i > 0. Assume λ > 0. Then Z λf dµ

≡ = =

sup

∞ X

h>0 i=1 ∞ X

sup

hµ ([ih/λ < f ])

h>0 i=1 ∞ X

sup λ h>0

λ

(h/λ) µ ([i (h/λ) < f ])

i=1

Z

=

hµ ([ih < λf ])

f dµ.

This proves the lemma. Lemma 8.3.7 Let the nonnegative simple function, s be defined as s (ω) =

n X

ci XEi (ω)

i=1

where the ci are not necessarily distinct but the Ei are disjoint. It follows that Z s=

n X

ci µ (Ei ) .

i=1

Proof: Let the values of s be {a1 , · · · , am }. Therefore, since the Ei are disjoint, each ai equal to one of the cj . Let Ai ≡ ∪ {Ej : cj = ai }. Then from Lemma 8.3.6

196


it follows that Z s

=

m X

ai µ (Ai ) =

i=1

=

m X

m X i=1

X

X

ai

µ (Ej )

{j:cj =ai }

cj µ (Ej ) =

i=1 {j:cj =ai }

n X

ci µ (Ei ) .

i=1

This proves the R lemma. R Note that s could equal +∞ if µ (Ak ) = ∞ and ak > 0 for some k, but s is well defined because s ≥ 0. Recall that 0 · ∞ = 0. Lemma 8.3.8 If a, b ≥ 0 and if s and t are nonnegative simple functions, then Z Z Z as + bt = a s + b t. Proof: Let s(ω) =

n X

αi XAi (ω), t(ω) =

i=1

m X

β j XBj (ω)

i=1

where αi are the distinct values of s and the β j are the distinct values of t. Clearly as + bt is a nonnegative simple function because it is measurable and has finitely many values. Also, (as + bt)(ω) =

m X n X

(aαi + bβ j )XAi ∩Bj (ω)

j=1 i=1

where the sets Ai ∩ Bj are disjoint. By Lemma 8.3.7, Z as + bt = =

m X n X

(aαi + bβ j )µ(Ai ∩ Bj )

j=1 i=1 n X

m X

i=1

j=1

a

αi µ(Ai ) + b

Z =

a

s+b

Z

β j µ(Bj )

t.

This proves the lemma.

8.3.4

Simple Functions And Measurable Functions

There is a fundamental theorem about the relationship of simple functions to measurable functions given in the next theorem.


197

Theorem 8.3.9 Let f ≥ 0 be measurable. Then there exists a sequence of nonnegative simple functions {sn } satisfying 0 ≤ sn (ω)

(8.3.14)

· · · sn (ω) ≤ sn+1 (ω) · · · f (ω) = lim sn (ω) for all ω ∈ Ω. n→∞

(8.3.15)

If f is bounded the convergence is actually uniform. Proof : Letting I ≡ {ω : f (ω) = ∞} , define n

2 X k tn (ω) = X[k/n≤f 0 be given and pick ω ∈ Ω. Then there exists xn ∈ D such that d (xn , f (ω)) < ε. It follows from the construction that d (fn (ω) , f (ω)) ≤ d (xn , f (ω)) < ε. This proves the first half. Now suppose the existence of the sequence of simple functions as described above. Each fn is a measurable function because fn−1 (U ) = ∪ {Ak : xk ∈ U }. Therefore, the conclusion that f is measurable follows from Theorem 8.1.9 on Page 181. In the context of this more general notion of measurable function having values in a metric space, here is a version of Egoroff’s theorem. Theorem 8.3.11 (Egoroff ) Let (Ω, F, µ) be a finite measure space, (µ(Ω) < ∞) and let fn , f be X valued measurable functions where X is a separable metric space and for all ω ∈ / E where µ(E) = 0 fn (ω) → f (ω) Then for every ε > 0, there exists a set, F ⊇ E, µ(F ) < ε, such that fn converges uniformly to f on F C .


199

Proof: First suppose E = ∅ so that convergence is pointwise everywhere. Let Ekm = {ω ∈ Ω : d (fn (ω) , f (ω)) ≥ 1/m for some n > k}. ¤ £ 1 Claim: ω : d (fn (ω) , f (ω)) ≥ m is measurable. ∞ Proof of claim: Let {xk }k=1 be a countable dense subset of X and let r denote a positive rational number, Q+ . Then ¶¶ µ µ 1 −r ∪k∈N,r∈Q+ fn−1 (B (xk , r)) ∩ f −1 B xk , m · ¸ 1 = d (f, fn ) < (8.3.19) m Here is why. If ω is in the set on the left, then d (fn (ω) , xk ) < r and d (f (ω) , xk )
k, |fn (ω) − f (ω)| < m which means ω ∈ / Ekm . Note also that Ekm ⊇ E(k+1)m . Since µ(E1m ) < ∞, Theorem 8.1.5 on Page 178 implies 0 = µ(∩∞ k=1 Ekm ) = lim µ(Ekm ). k→∞

200


Let k(m) be chosen such that µ(Ek(m)m ) < ε2−m and let F =

∞ [

Ek(m)m .

m=1

Then µ(F ) < ε because µ (F ) ≤

∞ X

∞ ¡ ¢ X µ Ek(m)m < ε2−m = ε

m=1

m=1

C Now let η > 0 be given and pick m0 such that m−1 0 < η. If ω ∈ F , then

ω∈

∞ \

C . Ek(m)m

m=1 C Hence ω ∈ Ek(m so 0 )m0

d (f (ω) , fn (ω)) < 1/m0 < η for all n > k(m0 ). This holds for all ω ∈ F C and so fn converges uniformly to f on F C. ∞ Now if E 6= ∅, consider {XE C fn }n=1 . Then XE C fn is measurable and the sequence converges pointwise to XE f everywhere. Therefore, from the first part, there exists a set of measure less than ε, F such that on F C , {XE C fn } converges uniformly C to XE C f. Therefore, on (E ∪ F ) , {fn } converges uniformly to f . This proves the theorem.

8.3.5

The Monotone Convergence Theorem

The following is called the monotone convergence theorem. This theorem and related convergence theorems are the reason for using the Lebesgue integral. Theorem 8.3.12 (Monotone Convergence theorem) Let f have values in [0, ∞] and suppose {fn } is a sequence of nonnegative measurable functions having values in [0, ∞] and satisfying lim fn (ω) = f (ω) for each ω.

n→∞

· · · fn (ω) ≤ fn+1 (ω) · · · Then f is measurable and Z

Z f dµ = lim

n→∞

fn dµ.


201

Proof: From Lemmas 8.3.1 and 8.3.2, Z f dµ

≡ sup

∞ X

hµ ([ih < f ])

h>0 i=1

= sup sup h>0

k

k X

= sup sup sup h>0

k

= sup sup

hµ ([ih < f ])

i=1

m

∞ X

m h>0 i=1

k X

hµ ([ih < fm ])

i=1

hµ ([ih < fm ])

Z

≡ sup

fm dµ Z lim fm dµ.

m

=

m→∞

The third equality follows from the observation that lim µ ([ih < fm ]) = µ ([ih < f ])

m→∞

which follows from Theorem 8.1.5 since the sets, [ih < fm ] are increasing in m and their union equals [ih < f ]. This proves the theorem. To illustrate what goes wrong without the Lebesgue integral, consider the following example. Example 8.3.13 Let {rn } denote the rational numbers in [0, 1] and let ½ fn (t) ≡

1 if t ∈ / {r1 , · · · , rn } 0 otherwise

Then fn (t) ↑ f (t) where f is the function which is one on the rationals and zero on the irrationals. Each fn is Riemann but f is not Riemann R integrable (why?) R integrable. Therefore, you can’t write f dx = limn→∞ fn dx. A meta-mathematical observation related to this type of example is this. If you can choose your functions, you don’t need the Lebesgue integral. The Riemann integral is just fine. It is when you can’t choose your functions and they come to you as pointwise limits that you really need the superior Lebesgue integral or at least something more general than the Riemann integral. The Riemann integral is entirely adequate for evaluating the seemingly endless lists of boring problems found in calculus books.

202

8.3.6


Other Definitions

To review and summarize the above, if f ≥ 0 is measurable, Z f dµ ≡ sup

∞ X

hµ ([f > ih])

(8.3.20)

h>0 i=1

R another way to get the same thing for f dµ is to take an increasing sequence of nonnegative simple functions, {sn } with sn (ω) → f (ω) and then by monotone convergence theorem, Z Z f dµ = lim

n→∞

where if sn (ω) =

Pm

j=1 ci XEi

sn

(ω) , Z sn dµ =

m X

ci m (Ei ) .

i=1

Similarly this also shows that for such nonnegative measurable function, ½Z ¾ Z f dµ = sup s : 0 ≤ s ≤ f, s simple which is the usual way of defining the Lebesgue integral for nonnegative simple functions in most books. I have done it differently because this approach led to an easier proof of the Monotone convergence theorem. Here is an equivalent definition of the integral. The fact it is well defined has been discussed above. Definition 8.3.14 For s a nonnegative simple function, s (ω) =

n X k=1

Z ck XEk (ω) ,

s=

n X

ck µ (Ek ) .

k=1

For f a nonnegative measurable function, ½Z ¾ Z f dµ = sup s : 0 ≤ s ≤ f, s simple .

8.3.7

Fatou’s Lemma

Sometimes the limit of a sequence does not exist. There are two more general notions known as lim sup and lim inf which do always exist in some sense. These notions are dependent on the following lemma. Lemma 8.3.15 Let {an } be an increasing (decreasing) sequence in [−∞, ∞] . Then limn→∞ an exists.


203

Proof: Suppose first {an } is increasing. Recall this means an ≤ an+1 for all n. If the sequence is bounded above, then it has a least upper bound and so an → a where a is its least upper bound. If the sequence is not bounded above, then for every l ∈ R, it follows l is not an upper bound and so eventually, an > l. But this is what is meant by an → ∞. The situation for decreasing sequences is completely similar. Now take any sequence, {an } ⊆ [−∞, ∞] and consider the sequence {An } where An ≡ inf {ak : k ≥ n} . Then as n increases, the set of numbers whose inf is being taken is getting smaller. Therefore, An is an increasing sequence and so it must converge. Similarly, if Bn ≡ sup {ak : k ≥ n} , it follows Bn is decreasing and so {Bn } also must converge. With this preparation, the following definition can be given. Definition 8.3.16 Let {an } be a sequence of points in [−∞, ∞] . Then define lim inf an ≡ lim inf {ak : k ≥ n} n→∞

n→∞

and lim sup an ≡ lim sup {ak : k ≥ n} n→∞

n→∞

In the case of functions having values in [−∞, ∞] , ³ ´ lim inf fn (ω) ≡ lim inf (fn (ω)) . n→∞

n→∞

A similar definition applies to lim supn→∞ fn . Lemma 8.3.17 Let {an } be a sequence in [−∞, ∞] . Then limn→∞ an exists if and only if lim inf an = lim sup an n→∞

n→∞

and in this case, the limit equals the common value of these two numbers. Proof: Suppose first limn→∞ an = a ∈ R. Then, letting ε > 0 be given, an ∈ (a − ε, a + ε) for all n large enough, say n ≥ N. Therefore, both inf {ak : k ≥ n} and sup {ak : k ≥ n} are contained in [a − ε, a + ε] whenever n ≥ N. It follows lim supn→∞ an and lim inf n→∞ an are both in [a − ε, a + ε] , showing ¯ ¯ ¯ ¯ ¯lim inf an − lim sup an ¯ < 2ε. ¯ ¯ n→∞

n→∞

Since ε is arbitrary, the two must be equal and they both must equal a. Next suppose limn→∞ an = ∞. Then if l ∈ R, there exists N such that for n ≥ N, l ≤ an

204


and therefore, for such n, l ≤ inf {ak : k ≥ n} ≤ sup {ak : k ≥ n} and this shows, since l is arbitrary that lim inf an = lim sup an = ∞. n→∞

n→∞

The case for −∞ is similar. Conversely, suppose lim inf n→∞ an = lim supn→∞ an = a. Suppose first that a ∈ R. Then, letting ε > 0 be given, there exists N such that if n ≥ N, sup {ak : k ≥ n} − inf {ak : k ≥ n} < ε therefore, if k, m > N, and ak > am , |ak − am | = ak − am ≤ sup {ak : k ≥ n} − inf {ak : k ≥ n} < ε showing that {an } is a Cauchy sequence. Therefore, it converges to a ∈ R, and as in the first part, the lim inf and lim sup both equal a. If lim inf n→∞ an = lim supn→∞ an = ∞, then given l ∈ R, there exists N such that for n ≥ N, inf an > l.

n>N

Therefore, limn→∞ an = ∞. The case for −∞ is similar. This proves the lemma. The next theorem, known as Fatou’s lemma is another important theorem which justifies the use of the Lebesgue integral. Theorem 8.3.18 (Fatou’s lemma) Let fn be a nonnegative measurable function with values in [0, ∞]. Let g(ω) = lim inf n→∞ fn (ω). Then g is measurable and Z Z gdµ ≤ lim inf fn dµ. n→∞

In other words,

Z ³

Z ´ lim inf fn dµ ≤ lim inf fn dµ n→∞

n→∞

Proof: Let gn (ω) = inf{fk (ω) : k ≥ n}. Then −1 gn−1 ([a, ∞]) = ∩∞ k=n fk ([a, ∞]) ∈ F.

Thus gn is measurable by Lemma 8.1.6 on Page 180. Also g(ω) = limn→∞ gn (ω) so g is measurable because it is the pointwise limit of measurable functions. Now the functions gn form an increasing sequence of nonnegative measurable functions so the monotone convergence theorem applies. This yields Z Z Z gdµ = lim gn dµ ≤ lim inf fn dµ. n→∞

n→∞

The last inequality holding because Z Z gn dµ ≤ fn dµ. R (Note that it is not known whether limn→∞ fn dµ exists.) This proves the Theorem.

8.4. THE SPACE L1

8.3.8

205

The Righteous Algebraic Desires Of The Lebesgue Integral

The monotone convergence theorem shows the integral wants to be linear. This is the essential content of the next theorem. Theorem 8.3.19 Let f, g be nonnegative measurable functions and let a, b be nonnegative numbers. Then Z Z Z (af + bg) dµ = a f dµ + b gdµ. (8.3.21) Proof: By Theorem 8.3.9 on Page 197 there exist sequences of nonnegative simple functions, sn → f and tn → g. Then by the monotone convergence theorem and Lemma 8.3.8, Z Z (af + bg) dµ = lim asn + btn dµ n→∞ µ Z ¶ Z = lim a sn dµ + b tn dµ n→∞ Z Z = a f dµ + b gdµ. As long as you are allowing functions to take the value +∞, you cannot consider something like f + (−g) and so you can’t very well expect a satisfactory statement about the integral being linear until you restrict yourself to functions which have values in a vector space. This is discussed next.

8.4

The Space L1

The functions considered here have values in C, a vector space. Definition 8.4.1 Let (Ω, S, µ) be a measure space and suppose f : Ω → C. Then f is said to be measurable if both Re f and Im f are measurable real valued functions. Definition 8.4.2 A complex simple function will be a function which is of the form s (ω) =

n X

ck XEk (ω)

k=1

where ck ∈ C and µ (Ek ) < ∞. For s a complex simple function as above, define I (s) ≡

n X k=1

ck µ (Ek ) .

206


Lemma 8.4.3 The definition, 8.4.2 is well defined. Furthermore, I is linear on the vector space of complex simple functions. Also the triangle inequality holds, |I (s)| ≤ I (|s|) . Pn

Proof: Suppose k=1 ck XEk (ω) = 0. Does it follow that The supposition implies n X

Re ck XEk (ω) = 0,

k=1

n X

P

k ck µ (Ek )

Im ck XEk (ω) = 0.

= 0?

(8.4.22)

k=1

Choose λ large and positive so that λ + Re ck ≥ 0. Then adding sides of the first equation above, n X

(λ + Re ck ) XEk (ω) =

k=1

n X

k

λXEk to both

λXEk

k=1

and by Lemma 8.3.8 on Page 196, it follows upon taking n X

P

(λ + Re ck ) µ (Ek ) =

k=1

n X

R

of both sides that

λµ (Ek )

k=1

Pn Pn which k=1 Im ck µ (Ek ) = 0 and so k=1 Re ck µ (Ek ) = 0. Similarly, Pn implies k=1 ck µ (Ek ) = 0. Thus if X X cj XEj = dk XFk j

P

k

P

+ k (−dk ) XFk = 0 and so the result just established verifies then j cj XEjP P c µ (E ) − j k dk µ (Fk ) = 0 which proves I is well defined. j j That I is linear is now obvious. It only remains to verify the triangle inequality. Let s be a simple function, X s= cj XEj j

Then pick θ ∈ C such that θI (s) = |I (s)| and |θ| = 1. Then from the triangle inequality for sums of complex numbers, X |I (s)| = θI (s) = I (θs) = θcj µ (Ej ) j

=

¯ ¯ ¯ ¯ ¯X ¯ X ¯ θcj µ (Ej )¯¯ ≤ |θcj | µ (Ej ) = I (|s|) . ¯ ¯ j ¯ j

This proves the lemma. With this lemma, the following is the definition of L1 (Ω) .

8.4. THE SPACE L1

207

Definition 8.4.4 f ∈ L1 (Ω) means there exists a sequence of complex simple functions, {sn } such that sn (ω) → f (ω) for all ωR ∈ Ω limm,n→∞ I (|sn − sm |) = limn,m→∞ |sn − sm | dµ = 0

(8.4.23)

Then I (f ) ≡ lim I (sn ) .

(8.4.24)

n→∞

Lemma 8.4.5 Definition 8.4.4 is well defined. Proof: There are several things which need to be verified. First suppose 8.4.23. Then by Lemma 8.4.3 |I (sn ) − I (sm )| = |I (sn − sm )| ≤ I (|sn − sm |) and for m, n large enough this last is given to be small so {I (sn )} is a Cauchy sequence in C and so it converges. This verifies the limit in 8.4.24 at least exists. It remains to consider another sequence {tn } having the same properties as {sn } and verifying I (f ) determined by this other sequence is the same. By Lemma 8.4.3 and Fatou’s lemma, Theorem 8.3.18 on Page 204, Z |I (sn ) − I (tn )| ≤ I (|sn − tn |) = |sn − tn | dµ Z ≤ |sn − f | + |f − tn | dµ Z Z ≤ lim inf |sn − sk | dµ + lim inf |tn − tk | dµ < ε k→∞

k→∞

whenever n is large enough. Since ε is arbitrary, this shows the limit from using the tn is the same as the limit from using Rsn . This proves the lemma. What if f has values in [0, ∞)? Earlier f dµ was defined for such functions and now I (f ) has R been defined. Are they the same? If so, I can be regarded as an extension of dµ to a larger class of functions. Lemma 8.4.6 Suppose f has values in [0, ∞) and f ∈ L1 (Ω) . Then f is measurable and Z I (f ) = f dµ. Proof: Since f is the pointwise limit of a sequence of complex simple functions, {sn } having the properties described in Definition 8.4.4, it follows f (ω) = limn→∞ Re sn (ω) and so f is measurable. Also Z ¯ Z Z ¯ ¯ + +¯ ¯(Re sn ) − (Re sm ) ¯ dµ ≤ |Re sn − Re sm | dµ ≤ |sn − sm | dµ

208


where x+ ≡ 12 (|x| + x) , the positive part of the real number, x. 2 Thus there is no loss of generality in assuming {sn } is a sequence of complex simple functions having R values in [0, ∞). Then since for such complex simple functions, I (s) = sdµ, ¯ ¯ ¯Z ¯ Z Z ¯ ¯ ¯ ¯ ¯I (f ) − f dµ¯ ≤ |I (f ) − I (sn )| + ¯ sn dµ − f dµ¯ ¯ ¯ ¯ ¯

0 such that for all f ∈ S Z | f dµ| < ε whenever µ(E) < δ. E

Lemma 8.5.2 If S is uniformly integrable, then |S| ≡ {|f | : f ∈ S} is uniformly integrable. Also S is uniformly integrable if S is finite. Proof: Let ε > 0 be given and suppose S is uniformly integrable. First suppose the functions are real valued. Let δ be such that if µ (E) < δ, then ¯Z ¯ ¯ ¯ ¯ f dµ¯ < ε ¯ ¯ 2 E

8.5. VITALI CONVERGENCE THEOREM

213

for all f ∈ S. Let µ (E) < δ. Then if f ∈ S, Z Z Z |f | dµ ≤ (−f ) dµ + f dµ E E∩[f ≤0] E∩[f >0] ¯ ¯Z ¯ ¯Z ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ f dµ¯ f dµ¯ + ¯ = ¯ ¯ ¯ ¯ E∩[f >0] ¯ E∩[f ≤0] ε ε < + = ε. 2 2 In general, if S is a uniformly integrable set of complex valued functions, the inequalities, ¯ ¯ ¯Z ¯ ¯Z ¯ ¯Z ¯Z ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ Re f dµ¯ ≤ ¯ f dµ¯ , ¯ Im f dµ¯ ≤ ¯ f dµ¯ , ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ E

E

E

E

imply Re S ≡ {Re f : f ∈ S} and Im S ≡ {Im f : f ∈ S} are also uniformly integrable. Therefore, applying the above result for real valued functions to these sets of functions, it follows |S| is uniformly integrable also. For the last part, is suffices to verify a single function in L1 (Ω) is uniformly integrable. To do so, note that from the dominated convergence theorem, Z lim |f | dµ = 0. R→∞

[|f |>R]

Let ε > 0 be given and choose R large enough that ε µ (E) < 2R . Then Z

Z |f | dµ

E

R [|f |>R]

|f | dµ
R]

ε ε ε < + = ε. 2 2 2

This proves the lemma. The following theorem is Vitali’s convergence theorem. Theorem 8.5.3 Let {fn } be a uniformly integrable set of complex valued functions, µ(Ω) < ∞, and fn (x) → f (x) a.e. where f is a measurable complex valued function. Then f ∈ L1 (Ω) and Z lim |fn − f |dµ = 0. (8.5.25) n→∞

Ω

Proof: First it will be shown that f ∈ L1 (Ω). By uniform integrability, there exists δ > 0 such that if µ (E) < δ, then Z |fn | dµ < 1 E

214


for all n. By Egoroff’s theorem, there exists a set, E of measure less than δ such that on E C , {fn } converges uniformly. Therefore, for p large enough, and n > p, Z |fp − fn | dµ < 1 EC

which implies

Z

Z EC

|fn | dµ < 1 +

|fp | dµ. Ω

Then since there are only finitely many functions, fn with n ≤ p, there exists a constant, M1 such that for all n, Z |fn | dµ < M1 . EC

But also,

Z

Z |fm | dµ

Z

= EC

Ω

|fm | dµ +

|fm | E

≤ M1 + 1 ≡ M. Therefore, by Fatou’s lemma, Z Z |f | dµ ≤ lim inf |fn | dµ ≤ M, n→∞

Ω

showing that f ∈ L1 as hoped. Now R S∪{f } is uniformly integrable so there exists δ 1 > 0 such that if µ (E) < δ 1 , then E |g| dµ < ε/3 for all g ∈ S ∪ {f }. By Egoroff’s theorem, there exists a set, F with µ (F ) < δ 1 such that fn converges uniformly to f on F C . Therefore, there exists N such that if n > N , then Z ε |f − fn | dµ < . 3 C F It follows that for n > N , Z Z |f − fn | dµ ≤ Ω

r] . Note that l (x, y) = x − y is not continuous on (−∞, ∞] so the obvious idea doesn’t work.

216


9. Let {fn } be a sequence of real or complex valued measurable functions. Let S = {ω : {fn (ω)} converges}. Show S is measurable. Hint: You might try to exhibit the set where fn converges in terms of countable unions and intersections using the definition of a Cauchy sequence. 10. Let (Ω, S, µ) be a measure space and let f be a nonnegative measurable function defined on Ω. Also let φ : [0, ∞) → [0, ∞) be strictly increasing and have a continuous derivative and φ (0) = 0. Suppose f is bounded and that 0 ≤ φ (f (ω)) ≤ M for some number, M . Show that Z ∞ Z φ (f ) dµ = φ0 (s) µ ([s < f ]) ds, Ω

0

where the integral on the right is the ordinary improper Riemann integral. Hint: First note that s → φ0 (s) µ ([s < f ]) is Riemann integrable because φ0 is continuous and s → µ ([s < f ]) is a nonincreasing function, hence Riemann integrable. From the second description of the Lebesgue integral and the assumption that φ (f (ω)) ≤ M , argue that for [M/h] the greatest integer less than M/h, Z

[M/h]

φ (f ) dµ =

sup

X

hµ ([ih < φ (f )])

h>0 i=1

Ω

[M/h]

=

sup

X

hµ

¡£

φ−1 (ih) < f

¤¢

h>0 i=1 [M/h]

X h∆i ¡£ ¤¢ µ φ−1 (ih) < f ∆ i h>0 i=1 ¡ −1 ¢ where ∆i = φ (ih) − φ−1 ((i − 1) h) . Now use the mean value theorem to write ¡ ¢0 ∆i = φ−1 (ti ) h 1 ¡ ¢h = φ0 φ−1 (ti ) =

sup

for some ti between (i − 1) h and ih. Therefore, the right side is of the form [M/h]

sup h

¡

X

¡ ¢ ¡£ ¤¢ φ0 φ−1 (ti ) ∆i µ φ−1 (ih) < f

i=1

¢ where φ (ti ) ∈ φ ((i − 1) h) , φ−1 (ih) . Argue that if ti were replaced with ih, this would be a Riemann sum for the Riemann integral Z φ−1 (M ) Z ∞ 0 φ (t) µ ([t < f ]) dt = φ0 (t) µ ([t < f ]) dt. −1

0

−1

0

8.6. EXERCISES

217

11. Let (Ω, F, µ) be a measure space and suppose fn converges uniformly to f and that fn is in L1 (Ω). When is Z Z lim fn dµ = f dµ? n→∞

12. Suppose un (t) is a differentiable function for t ∈ (a, b) and suppose that for t ∈ (a, b), |un (t)|, |u0n (t)| < Kn P∞ where n=1 Kn < ∞. Show (

∞ X

un (t))0 =

n=1

∞ X

u0n (t).

n=1

Hint: This is an exercise in the use of the dominated convergence theorem and the mean value theorem. P∞ 13. Show that { i=1 2−n µ ([i2−n < f ])} for f a nonnegative measurable function is an increasing sequence. Could you define Z f dµ ≡ lim

n→∞

∞ X

2−n µ

¡£

i2−n < f

¤¢

i=1

and would it be equivalent to the above definitions of the Lebesgue integral? 14. Suppose {fn } is a sequence of nonnegative measurable functions defined on a measure space, (Ω, S, µ). Show that Z X ∞ k=1

fk dµ =

∞ Z X

fk dµ.

k=1

Hint: Use the monotone convergence theorem along with the fact the integral is linear.

218


The Construction Of Measures 9.1

Outer Measures

What are some examples of measure spaces? In this chapter, a general procedure is discussed called the method of outer measures. It is due to Caratheodory (1918). This approach shows how to obtain measure spaces starting with an outer measure. This will then be used to construct measures determined by positive linear functionals. Definition 9.1.1 Let Ω be a nonempty set and let µ : P(Ω) → [0, ∞] satisfy µ(∅) = 0, If A ⊆ B, then µ(A) ≤ µ(B), µ(∪∞ i=1 Ei ) ≤

∞ X

µ(Ei ).

i=1

Such a function is called an outer measure. For E ⊆ Ω, E is µ measurable if for all S ⊆ Ω, µ(S) = µ(S \ E) + µ(S ∩ E). (9.1.1) To help in remembering 9.1.1, think of a measurable set, E, as a process which divides a given set into two pieces, the part in E and the part not in E as in 9.1.1. In the Bible, there are four incidents recorded in which a process of division resulted in more stuff than was originally present.1 Measurable sets are exactly 1 1 Kings 17, 2 Kings 4, Mathew 14, and Mathew 15 all contain such descriptions. The stuff involved was either oil, bread, flour or fish. In mathematics such things have also been done with sets. In the book by Bruckner Bruckner and Thompson there is an interesting discussion of the Banach Tarski paradox which says it is possible to divide a ball in R3 into five disjoint pieces and assemble the pieces to form two disjoint balls of the same size as the first. The details can be found in: The Banach Tarski Paradox by Wagon, Cambridge University press. 1985. It is known that all such examples must involve the axiom of choice.

219

220

THE CONSTRUCTION OF MEASURES

those for which no such miracle occurs. You might think of the measurable sets as the nonmiraculous sets. The idea is to show that they form a σ algebra on which the outer measure, µ is a measure. First here is a definition and a lemma. Definition 9.1.2 (µbS)(A) ≡ µ(S ∩ A) for all A ⊆ Ω. Thus µbS is the name of a new outer measure, called µ restricted to S. The next lemma indicates that the property of measurability is not lost by considering this restricted measure. Lemma 9.1.3 If A is µ measurable, then A is µbS measurable. Proof: Suppose A is µ measurable. It is desired to to show that for all T ⊆ Ω, (µbS)(T ) = (µbS)(T ∩ A) + (µbS)(T \ A). Thus it is desired to show µ(S ∩ T ) = µ(T ∩ A ∩ S) + µ(T ∩ S ∩ AC ).

(9.1.2)

But 9.1.2 holds because A is µ measurable. Apply Definition 9.1.1 to S ∩ T instead of S. If A is µbS measurable, it does not follow that A is µ measurable. Indeed, if you believe in the existence of non measurable sets, you could let A = S for such a µ non measurable set and verify that S is µbS measurable. The next theorem is the main result on outer measures. It is a very general result which applies whenever one has an outer measure on the power set of any set. This theorem will be referred to as Caratheodory’s procedure in the rest of the book. Theorem 9.1.4 The collection of µ measurable sets, S, forms a σ algebra and If Fi ∈ S, Fi ∩ Fj = ∅, then µ(∪∞ i=1 Fi ) =

∞ X

µ(Fi ).

(9.1.3)

i=1

If · · · Fn ⊆ Fn+1 ⊆ · · · , then if F = ∪∞ n=1 Fn and Fn ∈ S, it follows that µ(F ) = lim µ(Fn ). n→∞

(9.1.4)

If · · · Fn ⊇ Fn+1 ⊇ · · · , and if F = ∩∞ n=1 Fn for Fn ∈ S then if µ(F1 ) < ∞, µ(F ) = lim µ(Fn ). n→∞

(9.1.5)

Also, (S, µ) is complete. By this it is meant that if F ∈ S and if E ⊆ Ω with µ(E \ F ) + µ(F \ E) = 0, then E ∈ S.

9.1. OUTER MEASURES

221

Proof: First note that ∅ and Ω are obviously in S. Now suppose A, B ∈ S. I will show A \ B ≡ A ∩ B C is in S. To do so, consider the following picture. S T T S AC B C

S S

T

A

T

BC S

T

A

T

T

AC

T

B

B B

A Since µ is subadditive, ¡ ¢ ¡ ¢ ¡ ¢ µ (S) ≤ µ S ∩ A ∩ B C + µ (A ∩ B ∩ S) + µ S ∩ B ∩ AC + µ S ∩ AC ∩ B C . Now using A, B ∈ S, ¡ ¢ ¡ ¢ ¡ ¢ µ (S) ≤ µ S ∩ A ∩ B C + µ (S ∩ A ∩ B) + µ S ∩ B ∩ AC + µ S ∩ AC ∩ B C ¡ ¢ = µ (S ∩ A) + µ S ∩ AC = µ (S) It follows equality holds in the above. Now observe using the picture if you like that ¡ ¢ ¡ ¢ (A ∩ B ∩ S) ∪ S ∩ B ∩ AC ∪ S ∩ AC ∩ B C = S \ (A \ B) and therefore, ¡ ¢ ¡ ¢ ¡ ¢ µ (S) = µ S ∩ A ∩ B C + µ (A ∩ B ∩ S) + µ S ∩ B ∩ AC + µ S ∩ AC ∩ B C ≥

µ (S ∩ (A \ B)) + µ (S \ (A \ B)) .

Therefore, since S is arbitrary, this shows A \ B ∈ S. Since Ω ∈ S, this shows that A ∈ S if and only if AC ∈ S. Now if A, B ∈ S, A ∪ B = (AC ∩ B C )C = (AC \ B)C ∈ S. By induction, if A1 , · · · , An ∈ S, then so is ∪ni=1 Ai . If A, B ∈ S, with A ∩ B = ∅, µ(A ∪ B) = µ((A ∪ B) ∩ A) + µ((A ∪ B) \ A) = µ(A) + µ(B).

222


By induction, if Ai ∩ Aj = ∅ and Ai ∈ S, µ(∪ni=1 Ai ) = Now let A = ∪∞ i=1 Ai where Ai ∩ Aj = ∅ for i 6= j. ∞ X

µ(Ai ) ≥ µ(A) ≥ µ(∪ni=1 Ai ) =

i=1

Pn

n X

i=1

µ(Ai ).

µ(Ai ).

i=1

Since this holds for all n, you can take the limit as n → ∞ and conclude, ∞ X

µ(Ai ) = µ(A)

i=1

which establishes 9.1.3. Part 9.1.4 follows from part 9.1.3 just as in the proof of Theorem 8.1.5 on Page 178. That is, letting F0 ≡ ∅, use part 9.1.3 to write µ (F ) =

µ (∪∞ k=1 (Fk \ Fk−1 )) =

∞ X

µ (Fk \ Fk−1 )

k=1

=

lim

n→∞

n X

(µ (Fk ) − µ (Fk−1 )) = lim µ (Fn ) . n→∞

k=1

In order to establish 9.1.5, let the Fn be as given there. Then, since (F1 \ Fn ) increases to (F1 \ F ), 9.1.4 implies lim (µ (F1 ) − µ (Fn )) = µ (F1 \ F ) .

n→∞

Now µ (F1 \ F ) + µ (F ) ≥ µ (F1 ) and so µ (F1 \ F ) ≥ µ (F1 ) − µ (F ). Hence lim (µ (F1 ) − µ (Fn )) = µ (F1 \ F ) ≥ µ (F1 ) − µ (F )

n→∞

which implies lim µ (Fn ) ≤ µ (F ) .

n→∞

But since F ⊆ Fn , µ (F ) ≤ lim µ (Fn ) n→∞

and this establishes 9.1.5. Note that it was assumed µ (F1 ) < ∞ because µ (F1 ) was subtracted from both sides. It remains to show S is closed under countable unions. Recall that if A ∈ S, then n AC ∈ S and S is closed under finite unions. Let Ai ∈ S, A = ∪∞ i=1 Ai , Bn = ∪i=1 Ai . Then µ(S) =

µ(S ∩ Bn ) + µ(S \ Bn )

= (µbS)(Bn ) +

(9.1.6)

(µbS)(BnC ).

By Lemma 9.1.3 Bn is (µbS) measurable and so is BnC . I want to show µ(S) ≥ µ(S \ A) + µ(S ∩ A). If µ(S) = ∞, there is nothing to prove. Assume µ(S) < ∞.

9.1. OUTER MEASURES

223

Then apply Parts 9.1.5 and 9.1.4 to the outer measure, µbS in 9.1.6 and let n → ∞. Thus Bn ↑ A, BnC ↓ AC and this yields µ(S) = (µbS)(A) + (µbS)(AC ) = µ(S ∩ A) + µ(S \ A). Therefore A ∈ S and this proves Parts 9.1.3, 9.1.4, and 9.1.5. It remains to prove the last assertion about the measure being complete. Let F ∈ S and let µ(E \ F ) + µ(F \ E) = 0. Consider the following picture.

S

E

F

Then referring to this picture and using F ∈ S, µ(S) ≤ ≤ ≤ =

µ(S ∩ E) + µ(S \ E) µ (S ∩ E ∩ F ) + µ ((S ∩ E) \ F ) + µ (S \ F ) + µ (F \ E) µ (S ∩ F ) + µ (E \ F ) + µ (S \ F ) + µ (F \ E) µ (S ∩ F ) + µ (S \ F ) = µ (S)

Hence µ(S) = µ(S ∩ E) + µ(S \ E) and so E ∈ S. This shows that (S, µ) is complete and proves the theorem. Completeness usually occurs in the following form. E ⊆ F ∈ S and µ (F ) = 0. Then E ∈ S. Where do outer measures come from? One way to obtain an outer measure is to start with a measure µ, defined on a σ algebra of sets, S, and use the following definition of the outer measure induced by the measure. Definition 9.1.5 Let µ be a measure defined on a σ algebra of sets, S ⊆ P (Ω). Then the outer measure induced by µ, denoted by µ is defined on P (Ω) as µ(E) = inf{µ(F ) : F ∈ S and F ⊇ E}. A measure space, (S, Ω, µ) is σ finite if there exist measurable sets, Ωi with µ (Ωi ) < ∞ and Ω = ∪∞ i=1 Ωi . You should prove the following lemma. Lemma 9.1.6 If (S, Ω, µ) is σ finite then there exist disjoint measurable sets, {Bn } such that µ (Bn ) < ∞ and ∪∞ n=1 Bn = Ω. The following lemma deals with the outer measure generated by a measure which is σ finite. It says that if the given measure is σ finite and complete then no new measurable sets are gained by going to the induced outer measure and then considering the measurable sets in the sense of Caratheodory.

224


Lemma 9.1.7 Let (Ω, S, µ) be any measure space and let µ : P(Ω) → [0, ∞] be the outer measure induced by µ. Then µ is an outer measure as claimed and if S is the set of µ measurable sets in the sense of Caratheodory, then S ⊇ S and µ = µ on S. Furthermore, if µ is σ finite and (Ω, S, µ) is complete, then S = S. Proof: It is easy to see that µ is an outer measure. Let E ∈ S. The plan is to show E ∈ S and µ(E) = µ(E). To show this, let S ⊆ Ω and then show µ(S) ≥ µ(S ∩ E) + µ(S \ E).

(9.1.7)

This will verify that E ∈ S. If µ(S) = ∞, there is nothing to prove, so assume µ(S) < ∞. Thus there exists T ∈ S, T ⊇ S, and µ(S) > µ(T ) − ε = µ(T ∩ E) + µ(T \ E) − ε ≥ µ(T ∩ E) + µ(T \ E) − ε ≥ µ(S ∩ E) + µ(S \ E) − ε. Since ε is arbitrary, this proves 9.1.7 and verifies S ⊆ S. Now if E ∈ S and V ⊇ E with V ∈ S, µ(E) ≤ µ(V ). Hence, taking inf, µ(E) ≤ µ(E). But also µ(E) ≥ µ(E) since E ∈ S and E ⊇ E. Hence µ(E) ≤ µ(E) ≤ µ(E). Next consider the claim about not getting any new sets from the outer measure in the case the measure space is σ finite and complete. Suppose first F ∈ S and µ (F ) < ∞. Then there exists E ∈ S such that E ⊇ F and µ (E) = µ (F ) . Since µ (F ) < ∞, µ (E \ F ) = µ (E) − µ (F ) = 0. Then there exists D ⊇ E \ F such that D ∈ S and µ (D) = µ (E \ F ) = 0. Then by completeness of S, it follows E \ F ∈ S and so E = (E \ F ) ∪ F Hence F = E \ (E \ F ) ∈ S. In the general case where µ (F ) is not known to be finite, let µ (Bn ) < ∞, with Bn ∩ Bm = ∅ for all n 6= m and ∪n Bn = Ω. Apply what was just shown to F ∩ Bn , obtaining each of these is in S. Then F = ∪n F ∩ Bn ∈ S. This proves the Lemma.

9.2

Regular Measures

Usually Ω is not just a set. It is also a topological space. It is very important to consider how the measure is related to this topology.

9.2. REGULAR MEASURES

225

Definition 9.2.1 Let µ be a measure on a σ algebra S, of subsets of Ω, where (Ω, τ ) is a topological space. µ is a Borel measure if S contains all Borel sets. µ is called outer regular if µ is Borel and for all E ∈ S, µ(E) = inf{µ(V ) : V is open and V ⊇ E}. µ is called inner regular if µ is Borel and µ(E) = sup{µ(K) : K ⊆ E, and K is compact}. If the measure is both outer and inner regular, it is called regular. There is an interesting situation in which regularity is obtained automatically. To save on words, let B (E) denote the σ algebra of Borel sets in E, a closed subset of Rn . It is a very interesting fact that every finite measure on B (E) must be regular. Lemma 9.2.2 Let µ be a finite measure defined on B (E) where E is a closed subset of Rn . Then for every F ∈ B (E) , µ (F ) = sup {µ (K) : K ⊆ F, K is closed } µ (F ) = inf {µ (V ) : V ⊇ F, V is open} Proof: For convenience, I will call a measure which satisfies the above two conditions “almost regular”. It would be regular if closed were replaced with compact. First note every open set is the countable union of compact sets and every closed set is the countable intersection of open sets. Here is why. Let V be an open set and let © ¡ ¢ ª Kk ≡ x ∈ V : dist x, V C ≥ 1/k . Then clearly the union of the Kk equals V and each is closed because x → dist (x, S) is always a continuous function whenever S is any nonempty set. Next, for K closed let Vk ≡ {x ∈ E : dist (x, K) < 1/k} . Clearly the intersection of the Vk equals K. Therefore, letting V denote an open set and K a closed set, µ (V ) = sup {µ (K) : K ⊆ V and K is closed} µ (K) = inf {µ (V ) : V ⊇ K and V is open} . Also since V is open and K is closed, µ (V ) = inf {µ (U ) : U ⊇ V and V is open} µ (K) = sup {µ (L) : L ⊆ K and L is closed} In words, µ is almost regular on open and closed sets. Let F ≡ {F ∈ B (E) such that µ is almost regular on F } .

226


Then F contains the open sets. I want to show F is a σ algebra and then it will follow F = B (E). First I will show F is closed with respect to complements. Let F ∈ F . Then since µ is finite and F is inner regular, there¢ exists K ⊆ F such that µ (F \ K) < ε. ¡ But K C \ F C = F \ K and so µ K C \ F C < ε showing that F C is outer regular. I have just approximated the measure of F C with the measure of K C , an open set containing F C . A similar argument works to show F C is inner regular. You start with F C \ V C = V \ F, and then conclude ¡ CV ⊇ CF¢ such that µ (V \ F ) < ε, note C < ε, thus approximating F with the closed subset, V C . µ F \V Next I will show F is closed with respect to taking countable unions. Let {Fk } be a sequence of sets in F. Then since Fk ∈ F , there exist {Kk } such that Kk ⊆ Fk and µ (Fk \ Kk ) < ε/2k+1 . First choose m large enough that m µ ((∪∞ k=1 Fk ) \ (∪k=1 Fk ))
a} = ∩{V t : t > a}

which is a closed set. If a = 1, f −1 ([0, 1]) = f −1 ([0, a]) = X. Therefore, f −1 ((a, 1]) = X \ f −1 ([0, a]) = open set. It follows f is continuous. Clearly f (x) = 0 on H. If x ∈ U C , then x ∈ / Vt for any t ∈ D so f (x) = 1 on U C. Let g (x) = 1 − f (x). This proves the theorem. In any metric space there is a much easier proof of the conclusion of Urysohn’s lemma which applies. Lemma 9.3.2 Let S be a nonempty subset of a metric space, (X, d) . Define f (x) ≡ dist (x, S) ≡ inf {d (x, y) : y ∈ S} . Then f is continuous.

9.3. URYSOHN’S LEMMA

229

Proof: Consider |f (x) − f (x1 )|and suppose without loss of generality that f (x1 ) ≥ f (x) . Then choose y ∈ S such that f (x) + ε > d (x, y) . Then |f (x1 ) − f (x)|

= ≤

f (x1 ) − f (x) ≤ f (x1 ) − d (x, y) + ε d (x1 , y) − d (x, y) + ε

≤ =

d (x, x1 ) + d (x, y) − d (x, y) + ε d (x1 , x) + ε.

Since ε is arbitrary, it follows that |f (x1 ) − f (x)| ≤ d (x1 , x) and this proves the lemma. Theorem 9.3.3 (Urysohn’s lemma for metric space) Let H be a closed subset of an open set, U in a metric space, (X, d) . Then there exists a continuous function, g : X → [0, 1] such that g (x) = 1 for all x ∈ H and g (x) = 0 for all x ∈ / U. Proof: If x ∈ / C, a closed set, then dist (x, C) > 0 because if not, there would exist a sequence of points of¡ C converging to x and it would follow that x ∈ C. ¢ Therefore, dist (x, H) + dist x, U C > 0 for all x ∈ X. Now define a continuous function, g as ¡ ¢ dist x, U C g (x) ≡ . dist (x, H) + dist (x, U C ) It is easy to see this verifies the conclusions of the theorem and this proves the theorem. Theorem 9.3.4 Every compact Hausdorff space is normal. Proof: First it is shown that X, is regular. Let H be a closed set and let p ∈ / H. Then for each h ∈ H, there exists an open set Uh containing p and an open set Vh containing h such that Uh ∩ Vh = ∅. Since H must be compact, it follows there are finitely many of the sets Vh , Vh1 · · · Vhn such that H ⊆ ∪ni=1 Vhi . Then letting U = ∩ni=1 Uhi and V = ∪ni=1 Vhi , it follows that p ∈ U , H ∈ V and U ∩ V = ∅. Thus X is regular as claimed. Next let K and H be disjoint nonempty closed sets.Using regularity of X, for every k ∈ K, there exists an open set Uk containing k and an open set Vk containing H such that these two open sets have empty intersection. Thus H ∩U k = ∅. Finitely many of the Uk , Uk1 , · · · , Ukp cover K and so ∪pi=1 U ki is a closed set which has ¡ ¢C empty intersection with H. Therefore, K ⊆ ∪pi=1 Uki and H ⊆ ∪pi=1 U ki . This proves the theorem. A useful construction when dealing with locally compact Hausdorff spaces is the notion of the one point compactification of the space discussed earler. However, it is reviewed here for the sake of convenience or in case you have not read the earlier treatment. Definition 9.3.5 Suppose (X, τ ) is a locally compact Hausdorff space. Then let e ≡ X ∪ {∞} where ∞ is just the name of some point which is not in X which is X

230


e is called the point at infinity. A basis for the topology e τ for X © C ª τ ∪ K where K is a compact subset of X . e and so the open sets, K C are basic open The complement is taken with respect to X sets which contain ∞. The reason this is called a compactification is contained in the next lemma. ³ ´ e e Lemma 9.3.6 If (X, τ ) is a locally compact Hausdorff space, then X, τ is a compact Hausdorff space. Also if U is an open set of e τ , then U \ {∞} is an open set of τ . ³ ´ e e Proof: Since (X, τ ) is a locally compact Hausdorff space, it follows X, τ is a Hausdorff topological space. The only case which needs checking is the one of p ∈ X and ∞. Since (X, τ ) is locally compact, there exists an open set of τ , U C having compact closure which contains p. Then p ∈ U and ∞ ∈ U and these are disjoint open sets containing the points, p and ∞ respectively. Now let C be an e with sets from e open cover of X τ . Then ∞ must be in some set, U∞ from C, which must contain a set of the form K C where K is a compact subset of X. Then there e is exist sets from C, U1 , · · · , Ur which cover K. Therefore, a finite subcover of X U1 , · · · , Ur , U∞ . To see the last claim, suppose U contains ∞ since otherwise there is nothing to show. Notice that if C is a compact set, then X \ C is an open set. Therefore, if e \ C is a basic open set contained in U containing ∞, then x ∈ U \ {∞} , and if X e it is also in the open set X \ C ⊆ U \ {∞} . If x if x is in this basic open set of X, e \ C then x is contained in an open set of is not in any basic open set of the form X τ which is contained in U \ {∞}. Thus U \ {∞} is indeed open in τ . Theorem 9.3.7 Let X be a locally compact Hausdorff space, and let K be a compact subset of the open set V . Then there exists a continuous function, f : X → [0, 1], such that f equals 1 on K and {x : f (x) 6= 0} ≡ spt (f ) is a compact subset of V . e be the space just described. Then K and V are respectively Proof: Let X closed and open in e τ . By Theorem 9.3.4 there exist open sets in e τ , U , and W such that K ⊆ U, ∞ ∈ V C ⊆ W , and U ∩ W = U ∩ (W \ {∞}) = ∅. VC K

U

W


231

Thus W \ {∞} is an open set in the original topological space which contains V C , U is an open set in the original topological space which contains K, and W \{∞} and U are disjoint. Now for each x ∈ K, let Ux be a basic open set whose closure is compact and such that x ∈ Ux ⊆ U. Thus Ux must have empty intersection with V C because the open set, W \ {∞} contains no points of Ux . Since K is compact, there are finitely many of these sets, Ux1 , Ux2 , · · · , Uxn which cover K. Now let H ≡ ∪ni=1 Uxi . Claim: H = ∪ni=1 Uxi Proof of claim: Suppose p ∈ H. If p ∈ / ∪ni=1 Uxi then if follows p ∈ / Uxi for each i. Therefore, there exists an open set, Ri containing p such that Ri contains no other points of Uxi . Therefore, R ≡ ∩ni=1 Ri is an open set containing p which contains no other points of ∪ni=1 Uxi = W, a contradiction. Therefore, H ⊆ ∪ni=1 Uxi . On the other hand, if p ∈ Uxi then p is obviously in H so this proves the claim. From the claim, K ⊆ H ⊆ H ⊆ V and H is compact because it is the finite union of compact sets. By Urysohn’s lemma, there exists f1 continuous on H which has values in [0, 1] such that f1 equals 1 on K and equals 0 off H. Let f denote the function which extends f1 to be 0 off H. Then for α > 0, the continuity of f1 implies there exists U open in the topological space such that f −1 ((−∞, α)) = f1−1 ((−∞, α)) ∪ H

C

¡ ¢ C C = U ∩H ∪H =U ∪H

an open set. If α ≤ 0, f −1 ((−∞, α)) = ∅ an open set. If α > 0, there exists an open set U such that f −1 ((α, ∞)) = f1−1 ((α, ∞)) = U ∩ H = U ∩ H because U must be a subset of H since by definition f = 0 off H. If α ≤ 0, then f −1 ((α, ∞)) = X, an open set. Thus f is continuous and spt (f ) ⊆ H, a compact subset of V. This proves the theorem. In fact, the conclusion of the above theorem could be used to prove that the topological space is locally compact. However, this is not needed here. In case you would like a more elementary proof which does not use the one point compactification idea, here is such a proof. Theorem 9.3.8 Let X be a locally compact Hausdorff space, and let K be a compact subset of the open set V . Then there exists a continuous function, f : X → [0, 1], such that f equals 1 on K and {x : f (x) 6= 0} ≡ spt (f ) is a compact subset of V .

232


Proof: To begin with, here is a claim. This claim is obvious in the case of a metric space but requires some proof in this more general case. Claim: If k ∈ K then there exists an open set Uk containing k such that Uk is contained in V. Proof of claim: Since X is locally compact, there exists a basis of open sets whose closures are compact, U. Denote by C the set of all U ∈ U which contain k and let C 0 denote the set of all closures of these sets of C intersected with the closed set V C . Thus C 0 is a collection of compact sets. I will argue that there are finitely many of the sets of C 0 which have empty intersection. If not, then C 0 has the finite intersection property and so there exists a point p in all of them. Since X is a Hausdorff space, there exist disjoint basic open sets from U, A, B such that k ∈ A and p ∈ B. Therefore, p ∈ / A contrary to the above requirement that p be in all such sets. It follows there are sets A1 , · · · , Am in C such that V C ∩ A1 ∩ · · · ∩ Am = ∅ Let Uk = A1 ∩ · · · ∩ Am . Then Uk ⊆ A1 ∩ · · · ∩ Am and so it has empty intersection with V. Thus it is contained in V . Also Uk is a closed subset of the compact set A1 so it is compact. This proves the claim. Now to complete the proof of the theorem, since K is compact, there are finitely many Uk of the sort just described which cover K, Uk1 , · · · , Ukr . Let H = ∪ri=1 Uki so it follows H = ∪ri=1 Uki and so K ⊆ H ⊆ H ⊆ V and H is a compact set. By Urysohn’s lemma, there exists f1 continuous on H which has values in [0, 1] such that f1 equals 1 on K and equals 0 off H. Let f denote the function which extends f1 to be 0 off H. Then for α > 0, the continuity of f1 implies there exists U open in the topological space such that f −1 ((−∞, α)) = f1−1 ((−∞, α)) ∪ H

C

¡ ¢ C C = U ∩H ∪H =U ∪H

an open set. If α ≤ 0, f −1 ((−∞, α)) = ∅ an open set. If α > 0, there exists an open set U such that f −1 ((α, ∞)) = f1−1 ((α, ∞)) = U ∩ H = U ∩ H because U must be a subset of H since by definition f = 0 off H. If α ≤ 0, then f −1 ((α, ∞)) = X, an open set. Thus f is continuous and spt (f ) ⊆ H, a compact subset of V. This proves the theorem.


233

Definition 9.3.9 Define spt(f ) (support of f ) to be the closure of the set {x : f (x) 6= 0}. If V is an open set, Cc (V ) will be the set of continuous functions f , defined on Ω having spt(f ) ⊆ V . Thus in Theorem 9.3.7 or 9.3.8, f ∈ Cc (V ). Definition 9.3.10 If K is a compact subset of an open set, V , then K ≺ φ ≺ V if φ ∈ Cc (V ), φ(K) = {1}, φ(Ω) ⊆ [0, 1], where Ω denotes the whole topological space considered. Also for φ ∈ Cc (Ω), K ≺ φ if φ(Ω) ⊆ [0, 1] and φ(K) = 1. and φ ≺ V if φ(Ω) ⊆ [0, 1] and spt(φ) ⊆ V. Theorem 9.3.11 (Partition of unity) Let K be a compact subset of a locally compact Hausdorff topological space satisfying Theorem 9.3.7 or 9.3.8 and suppose K ⊆ V = ∪ni=1 Vi , Vi open. Then there exist ψ i ≺ Vi with

n X

ψ i (x) = 1

i=1

for all x ∈ K. Proof: Let K1 = K \ ∪ni=2 Vi . Thus K1 is compact and K1 ⊆ V1 . Let K1 ⊆ W1 ⊆ W 1 ⊆ V1 with W 1 compact. To obtain W1 , use Theorem 9.3.7 or 9.3.8 to get f such that K1 ≺ f ≺ V1 and let W1 ≡ {x : f (x) 6= 0} . Thus W1 , V2 , · · · Vn covers K and W 1 ⊆ V1 . Let K2 = K \ (∪ni=3 Vi ∪ W1 ). Then K2 is compact and K2 ⊆ V2 . Let K2 ⊆ W2 ⊆ W 2 ⊆ V2 W 2 compact. Continue this way finally obtaining W1 , · · · , Wn , K ⊆ W1 ∪ · · · ∪ Wn , and W i ⊆ Vi W i compact. Now let W i ⊆ Ui ⊆ U i ⊆ Vi , U i compact.

Wi

Ui

Vi

By Theorem 9.3.7 or 9.3.8, let U i ≺ φi ≺ Vi , ∪ni=1 W i ≺ γ ≺ ∪ni=1 Ui . Define Pn Pn ½ γ(x)φi (x)/ j=1 φj (x) if j=1 φj (x) 6= 0, P ψ i (x) = n 0 if j=1 φj (x) = 0. Pn If x is such that j=1 φj (x) = 0, then x ∈ / ∪ni=1 U i . Consequently γ(y) = 0 for all y near x and so ψ i (y) = 0 for all y near x. Hence ψ i is continuous at such x.

234


Pn If j=1 φj (x) 6= 0, this situation persists near x and so ψ i is continuous at such Pn points. Therefore ψ i is continuous. If x ∈ K, then γ(x) = 1 and so j=1 ψ j (x) = 1. Clearly 0 ≤ ψ i (x) ≤ 1 and spt(ψ j ) ⊆ Vj . This proves the theorem. The following corollary won’t be needed immediately but is of considerable interest later. Corollary 9.3.12 If H is a compact subset of Vi , there exists a partition of unity such that ψ i (x) = 1 for all x ∈ H in addition to the conclusion of Theorem 9.3.11. fj ≡ Vj \ H. Now in the proof Proof: Keep Vi the same but replace Vj with V above, applied to this modified collection of open sets, if j 6= i, φj (x) = 0 whenever x ∈ H. Therefore, ψ i (x) = 1 on H.

9.4

Positive Linear Functionals

Definition 9.4.1 Let (Ω, τ ) be a topological space. L : Cc (Ω) → C is called a positive linear functional if L is linear, L(af1 + bf2 ) = aLf1 + bLf2 , and if Lf ≥ 0 whenever f ≥ 0. Theorem 9.4.2 (Riesz representation theorem) Let (Ω, τ ) be a locally compact Hausdorff space and let L be a positive linear functional on Cc (Ω). Then there exists a σ algebra S containing the Borel sets and a unique measure µ, defined on S, such that

µ(K)
µ(Vi ). µ(E) ≤ µ(∪∞ i=1 Vi ) ≤

∞ X

µ(Vi ) ≤ ε +

i=1

Since ε was arbitrary, µ(E) ≤

P∞ i=1

∞ X

µ(Ei ).

i=1

µ(Ei ) which proves the lemma.

Lemma 9.4.5 Let K be compact, g ≥ 0, g ∈ Cc (Ω), and g = 1 on K. Then µ(K) ≤ Lg. Also µ(K) < ∞ whenever K is compact. Proof: Let α ∈ (0, 1) and Vα = {x : g(x) > α} so Vα ⊇ K and let h ≺ Vα .

K

Vα

g>α Then h ≤ 1 on Vα while gα−1 ≥ 1 on Vα and so gα−1 ≥ h which implies L(gα−1 ) ≥ Lh and that therefore, since L is linear, Lg ≥ αLh. Since h ≺ Vα is arbitrary, and K ⊆ Vα , Lg ≥ αµ (Vα ) ≥ αµ (K) .

236


Letting α ↑ 1 yields Lg ≥ µ(K). This proves the first part of the lemma. The second assertion follows from this and Theorem 9.3.7. If K is given, let K≺g≺Ω and so from what was just shown, µ (K) ≤ Lg < ∞. This proves the lemma. Lemma 9.4.6 If A and B are disjoint compact subsets of Ω, then µ(A ∪ B) = µ(A) + µ(B). Proof: By Theorem 9.3.7 or 9.3.8, there exists h ∈ Cc (Ω) such that A ≺ h ≺ B C . Let U1 = h−1 (( 12 , 1]), V1 = h−1 ([0, 12 )). Then A ⊆ U1 , B ⊆ V1 and U1 ∩V1 = ∅.

A

U1

B

V1

From Lemma 9.4.5 µ(A ∪ B) < ∞ and so there exists an open set, W such that W ⊇ A ∪ B, µ (A ∪ B) + ε > µ (W ) . Now let U = U1 ∩ W and V = V1 ∩ W . Then U ⊇ A, V ⊇ B, U ∩ V = ∅, and µ(A ∪ B) + ε ≥ µ (W ) ≥ µ(U ∪ V ). Let A ≺ f ≺ U, B ≺ g ≺ V . Then by Lemma 9.4.5, µ(A ∪ B) + ε ≥ µ(U ∪ V ) ≥ L(f + g) = Lf + Lg ≥ µ(A) + µ(B). Since ε > 0 is arbitrary, this proves the lemma. From Lemma 9.4.5 the following lemma is obtained. Lemma 9.4.7 Let f ∈ Cc (Ω), f (Ω) ⊆ [0, 1]. Then µ(spt(f )) ≥ Lf . Also, every open set, V satisfies µ (V ) = sup {µ (K) : K ⊆ V } . Proof: Let V ⊇ spt(f ) and let spt(f ) ≺ g ≺ V . Then Lf ≤ Lg ≤ µ(V ) because f ≤ g. Since this holds for all V ⊇ spt(f ), Lf ≤ µ(spt(f )) by definition of µ.

spt(f )

V

Finally, let V be open and let l < µ (V ) . Then from the definition of µ, there exists f ≺ V such that L (f ) > l. Therefore, l < µ (spt (f )) ≤ µ (V ) and so this shows the claim about inner regularity of the measure on an open set.

9.4. POSITIVE LINEAR FUNCTIONALS

237

Lemma 9.4.8 If K is compact there exists V open, V ⊇ K, such that µ(V \ K) ≤ ε. If V is open with µ(V ) < ∞, then there exists a compact set, K ⊆ V with µ(V \ K) ≤ ε. Proof: Let K be compact. Then from the definition of µ, there exists an open set U , with µ(U ) < ∞ and U ⊇ K. If µ (U \ K) ≤ ε, stop. The open set has been found. If µ (U \ K) > ε, then from Lemma 9.4.7 there exists K1 ⊆ U \ K with µ (K1 ) > ε. Now let U \ K1 play the role of U in the above argument. If µ ((U \ K1 ) \ K) ≤ ε, stop. Otherwise µ ((U \ K1 ) \ K) > ε and there exists K2 ⊆ (U \ K1 ) \ K such that µ (K2 ) > ε. continue this way. You obtain a sequence of disjoint compact sets contained in U such that µ (Kj ) > ε. U

K1 K2

K

K3

Therefore, the process eventually stops and you obtain µ ε since otherwise, by Lemma 9.4.6, µ(U ) ≥ µ(K ∪

∪ri=1 Ki )

= µ(K) +

r X

¡¡

¢ ¢ U \ ∪m j=1 Kj \ K ≤

µ(Ki ) ≥ rε

i=1

for all r, contradicting µ(U ) < ∞. This demonstrates the first part of the lemma. To show the second part, employ a similar construction. Suppose µ(V \ K) > ε for all K ⊆ V . Then µ(V ) > ε so there exists K1 ⊆ V with µ(K1 ) > ε. If K1 · · · Kn , disjoint, compact subsets of V have been chosen, µ(V \ ∪ni=1 Ki ) > ε and so there exists Kn+1 ⊆ V \ ∪ni=1 Ki such that µ (Kn+1 ) > ε. In this way there exists a sequence of disjoint compact subsets of V , {Ki } with µ(Ki ) > ε. Thus for any m, K1 · · · Km are all contained in V and are disjoint and compact. By Lemma 9.4.6 m X µ(V ) ≥ µ(∪m µ(Ki ) > mε i=1 Ki ) = i=1

for all m, a contradiction to µ(V ) < ∞. This proves the second part. Lemma 9.4.9 Let S be the σ algebra of µ measurable sets in the sense of Caratheodory. Then S ⊇ Borel sets and µ is inner regular on every open set and for every E ∈ S with µ(E) < ∞.

238


Proof: Define S1 = {E ⊆ Ω : E ∩ K ∈ S} for all compact K. First it will be shown the compact sets are in S. From this it will follow the closed sets are in S1 . Then you show S1 is a σ algebra and so it contains the Borel sets. Finally, it is shown that S1 = S and then the inner regularity assertion is established. Let V be an open set with µ (V ) < ∞. I will show that µ (V ) ≥ µ(V \ C) + µ(V ∩ C). By Lemma 9.4.8, there exists an open set U containing C and a compact subset of V , K, such that µ(V \ K) < ε and µ (U \ C) < ε. U

V

C

K

Then by Lemma 9.4.6, µ(V )

≥ µ(K) ≥ µ((K \ U ) ∪ (K ∩ C)) = µ(K \ U ) + µ(K ∩ C) ≥ µ(V \ C) + µ(V ∩ C) − 3ε

(9.4.11)

To see the last step, µ (V \ C)

≤ µ (K \ U ) + µ (V \ K) + µ (U \ C) ≤ µ (K \ U ) + 2ε

while µ(V ∩ C)

≤ µ(K ∩ C) + µ (V \ K) ≤ µ(K ∩ C) + ε

Since ε is arbitrary, it follows from 9.4.11 µ(V ) = µ(V \ C) + µ(V ∩ C)

(9.4.12)

whenever C is compact and V is open. (If µ (V ) = ∞, it is obvious that µ (V ) ≥ µ(V \ C) + µ(V ∩ C) and it is always the case that µ (V ) ≤ µ (V \ C) + µ (V ∩ C) .)


239

Of course 9.4.12 is exactly what needs to be shown for arbitrary S in place of V . It suffices to consider only S having µ (S) < ∞. If S ⊆ Ω, with µ(S) < ∞, let V ⊇ S, µ(S) + ε > µ(V ). Then from what was just shown, if C is compact, ε + µ(S) > µ(V ) = µ(V \ C) + µ(V ∩ C) ≥ µ(S \ C) + µ(S ∩ C). Since ε is arbitrary, this shows the compact sets are in S. As discussed above, this verifies the closed sets are in S1 . Therefore, if E ∈ S and K is a compact set, it follows K ∩ E ∈ S and so S1 ⊇ S. To see that S1 is closed with respect to taking complements, let E ∈ S1 and K a compact set. K = (E C ∩ K) ∪ (E ∩ K). Then from the fact, just established, that the compact sets are in S, E C ∩ K = K \ (E ∩ K) ∈ S. S1 is closed under countable unions because if K is a compact set and En ∈ S1 , ∞ K ∩ ∪∞ n=1 En = ∪n=1 K ∩ En ∈ S

because it is a countable union of sets of S. Thus S1 is a σ algebra which contains the Borel sets since it contains the closed sets. The next task is to show S1 = S. Let E ∈ S1 and let V be an open set with µ(V ) < ∞ and choose K ⊆ V such that µ(V \ K) < ε. Then since E ∈ S1 , it follows E ∩ K ∈ S and µ(V ) = ≥ because

µ(V \ (K ∩ E)) + µ(V ∩ (K ∩ E)) µ(V \ E) + µ(V ∩ E) − ε µ (V ) . and so µ (V \ (U \ F )) = µ (V ) − µ (U \ F ) < ε. Also, V \ (U \ F ) = = = ⊇

¡ ¢C V ∩ U ∩ FC £ ¤ V ∩ UC ∪ F ¡ ¢ (V ∩ F ) ∪ V ∩ U C V ∩F

and so µ(V ∩ F ) ≤ µ (V \ (U \ F )) < ε. C

Since V ⊇ U ∩ F , V ⊆ U C ∪ F so U ∩ V C ⊆ U ∩ F = F . Hence U ∩ V C is a subset of F . Now let K ⊆ U, µ(U \ K) < ε. Thus K ∩ V C is a compact subset of F and µ(F ) =
0 is arbitrary,

Z Lf ≤

f dµ

(9.4.16)

R for all f ∈ C R c (Ω), f real. HenceR equality holds in 9.4.16 because L(−f ) ≤ − f dµ so L(f ) ≥ f dµ. Thus Lf = f dµ for all f ∈ Cc (Ω). Just apply the result for real functions to the real and imaginary parts of f . This proves the Lemma. This gives the existence part of the Riesz representation theorem. It only remains to prove uniqueness. Suppose both µ1 and µ2 are measures on S satisfying the conclusions of the theorem. Then if K is compact and V ⊇ K, let K ≺ f ≺ V . Then Z Z µ1 (K) ≤ f dµ1 = Lf = f dµ2 ≤ µ2 (V ). Thus µ1 (K) ≤ µ2 (K) for all K. Similarly, the inequality can be reversed and so it follows the two measures are equal on compact sets. By the assumption of inner regularity on open sets, the two measures are also equal on all open sets. By outer regularity, they are equal on all sets of S. This proves the theorem. An important example of a locally compact Hausdorff space is any metric space in which the closures of balls are compact. For example, Rn with the usual metric is an example of this. Not surprisingly, more can be said in this important special case. Theorem 9.4.11 Let (Ω, τ ) be a metric space in which the closures of the balls are compact and let L be a positive linear functional defined on Cc (Ω) . Then there exists a measure representing the positive linear functional which satisfies all the conclusions of Theorem 9.3.7 or 9.3.8 and in addition the property that µ is regular. The same conclusion follows if (Ω, τ ) is a compact Hausdorff space. Theorem 9.4.12 Let (Ω, τ ) be a metric space in which the closures of the balls are compact and let L be a positive linear functional defined on Cc (Ω) . Then there exists a measure representing the positive linear functional which satisfies all the conclusions of Theorem 9.3.7 or 9.3.8 and in addition the property that µ is regular. The same conclusion follows if (Ω, τ ) is a compact Hausdorff space.


243

Proof: Let µ and S be as described in Theorem 9.4.2. The outer regularity comes automatically as a conclusion of Theorem 9.4.2. It remains to verify inner regularity. Let F ∈ S and let l < k < µ (F ) . Now let z ∈ Ω and Ωn = B (z, n) for n ∈ N. Thus F ∩ Ωn ↑ F. It follows that for n large enough, k < µ (F ∩ Ωn ) ≤ µ (F ) . Since µ (F ∩ Ωn ) < ∞ it follows there exists a compact set, K such that K ⊆ F ∩ Ωn ⊆ F and l < µ (K) ≤ µ (F ) . This proves inner regularity. In case (Ω, τ ) is a compact Hausdorff space, the conclusion of inner regularity follows from Theorem 9.4.2. This proves the theorem. The proof of the above yields the following corollary. Corollary 9.4.13 Let (Ω, τ ) be a locally compact Hausdorff space and suppose µ defined on a σ algebra, S represents the positive linear functional L where L is defined on Cc (Ω) in the sense of Theorem 9.3.7 or 9.3.8. Suppose also that there exist Ωn ∈ S such that Ω = ∪∞ n=1 Ωn and µ (Ωn ) < ∞. Then µ is regular. The following is on the uniqueness of the σ algebra in some cases. Definition 9.4.14 Let (Ω, τ ) be a locally compact Hausdorff space and let L be a positive linear functional defined on Cc (Ω) such that the complete measure defined by the Riesz representation theorem for positive linear functionals is inner regular. Then this is called a Radon measure. Thus a Radon measure is complete, and regular. Corollary 9.4.15 Let (Ω, τ ) be a locally compact Hausdorff space which is also σ compact meaning Ω = ∪∞ n=1 Ωn , Ωn is compact, and let L be a positive linear functional defined on Cc (Ω) . Then if (µ1 , S1 ) , and (µ2 , S2 ) are two Radon measures, together with their σ algebras which represent L then the two σ algebras are equal and the two measures are equal. Proof: Suppose (µ1 , S1 ) and (µ2 , S2 ) both work. It will be shown the two measures are equal on every compact set. Let K be compact and let V be an open set containing K. Then let K ≺ f ≺ V. Then Z Z Z µ1 (K) = dµ1 ≤ f dµ1 = L (f ) = f dµ2 ≤ µ2 (V ) . K

Therefore, taking the infimum over all V containing K implies µ1 (K) ≤ µ2 (K) . Reversing the argument shows µ1 (K) = µ2 (K) . This also implies the two measures are equal on all open sets because they are both inner regular on open sets. It is being assumed the two measures are regular. Now let F ∈ S1 with µ1 (F ) < ∞. Then there exist sets, H, G such that H ⊆ F ⊆ G such that H is the countable

244


union of compact sets and G is a countable intersection of open sets such that µ1 (G) = µ1 (H) which implies µ1 (G \ H) = 0. Now G \ H can be written as the countable intersection of sets of the form Vk \Kk where Vk is open, µ1 (Vk ) < ∞ and Kk is compact. From what was just shown, µ2 (Vk \ Kk ) = µ1 (Vk \ Kk ) so it follows µ2 (G \ H) = 0 also. Since µ2 is complete, and G and H are in S2 , it follows F ∈ S2 and µ2 (F ) = µ1 (F ) . Now for arbitrary F possibly having µ1 (F ) = ∞, consider F ∩ Ωn . From what was just shown, this set is in S2 and µ2 (F ∩ Ωn ) = µ1 (F ∩ Ωn ). Taking the union of these F ∩Ωn gives F ∈ S2 and also µ1 (F ) = µ2 (F ) . This shows S1 ⊆ S2 . Similarly, S2 ⊆ S1 . The following lemma is often useful. Lemma 9.4.16 Let (Ω, F, µ) be a measure space where Ω is a topological space. Suppose µ is a Radon measure and f is measurable with respect to F. Then there exists a Borel measurable function, g, such that g = f a.e. Proof: Assume without loss of generality that f ≥ 0. Then let sn ↑ f pointwise. Say Pn X sn (ω) = cnk XEkn (ω) k=1 n where Ek ∈ F. By the outer regularity of µ, there that µ (Fkn ) = µ (Ekn ). In fact Fkn can be assumed

tn (ω) ≡

Pn X

exists a Borel set, Fkn ⊇ Ekn such to be a Gδ set. Let

cnk XFkn (ω) .

k=1

Then tn is Borel measurable and tn (ω) = sn (ω) for all ω ∈ / Nn where Nn ∈ F is N . Then N is a set of measure zero a set of measure zero. Now let N ≡ ∪∞ n n=1 and if ω ∈ / N , then tn (ω) → f (ω). Let N 0 ⊇ N where N 0 is a Borel set and µ (N 0 ) = 0. Then tn X(N 0 )C converges pointwise to a Borel measurable function, g, and g (ω) = f (ω) for all ω ∈ / N 0 . Therefore, g = f a.e. and this proves the lemma.

9.5

One Dimensional Lebesgue Measure

To obtain one dimensional Lebesgue measure, you use the positive linear functional L given by Z Lf =

f (x) dx

whenever f ∈ Cc (R) . Lebesgue measure, denoted by m is the measure obtained from the Riesz representation theorem such that Z Z f dm = Lf = f (x) dx. From this it is easy to verify that m ([a, b]) = m ((a, b)) = b − a.

(9.5.17)

9.6. ONE DIMENSIONAL LEBESGUE STIELTJES MEASURE

245

This will be done in general a little later but for now, consider the following picture of functions, f k and g k . Note that f k ≤ X(a,b) ≤ X[a,b] ≤ g k .

1

a + 1/k £ £ @£ @ R £ a

B

fk b − 1/k B ¡ B ª ¡ B b

1 £ a − 1/k £ @ £ @ R£ @

B

a

b

gk B

b + 1/k ¡ B ¡ ª B

Then considering lower sums and upper sums in the inequalities on the ends, µ ¶ 2 b−a− ≤ k =

Z

Z k

f k dm ≤ m ((a, b)) ≤ m ([a, b]) µ ¶ Z Z Z 2 k k X[a,b] dm ≤ g dm = g dx ≤ b − a + . k f dx =

From this the claim in 9.5.17 follows.

9.6

One Dimensional Lebesgue Stieltjes Measure

This is just a generalization of Lebesgue measure. Instead of the functional, Z Lf ≡

f (x) dx, f ∈ Cc (R) ,

you use the functional Z Lf ≡

f (x) dF (x) f ∈ Cc (R) ,

where F is an increasing function defined on R. By Theorem 3.3.4 this functional is easily seen to be well defined. Therefore, by the Riesz representation theorem there exists a unique Radon measure µ representing the functional. Thus Z

Z f dµ =

f dF

R

for all f ∈ Cc (R) . Now consider what this measure does to intervals. To begin with, consider what it does to the closed interval, [a, b] . The following picture may help.

246


1 £

B

£ £

fn B

B B b bn

£ an a

In this picture {an } increases to a and bn decreases to b. Also suppose a, b are points of continuity of F . Therefore, Z fn dµ ≤ F (bn ) − F (an ) F (b) − F (a) ≤ Lfn = R

Passing to the limit and using the dominated convergence theorem, this shows µ ([a, b]) = F (b) − F (a) = F (b+) − F (a−) . Next suppose a, b are arbitrary, maybe not points of continuity of F. Then letting an and bn be as in the above picture which are points of continuity of F, µ ([a, b])

= =

lim µ ([an , bn ]) = lim F (bn ) − F (an )

n→∞

n→∞

F (b+) − F (a−) .

In particular µ (a) = F (a+) − F (a−) and so µ ((a, b)) = =

F (b+) − F (a−) − (F (a+) − F (a−)) − (F (b+) − F (b−)) F (b−) − F (a+)

This shows what µ does to intervals. This is stated as the following proposition. Proposition 9.6.1 Let µ be the measure representing the functional Z Lf ≡ f dF, f ∈ Cc (R) for F an increasing function defined on R. Then µ ([a, b]) = F (b+) − F (a−) µ ((a, b)) = F (b−) − F (a+) µ (a) = F (a+) − F (a−) . Observation 9.6.2 Note that all the above would work as well if Z Lf ≡ f dF, f ∈ Cc ([0, ∞)) where F is continuous at 0 and ν is the measure representing this functional. This is because you could just extend F (x) to equal F (0) for x ≤ 0 and apply the above to the extended F . In this case, ν ([0, b]) = F (b+) − F (0).

9.7. THE DISTRIBUTION FUNCTION

9.7

247

The Distribution Function

There is an interesting connection between the Lebesgue integral of a nonnegative function with something called the distribution function. Definition 9.7.1 Let f ≥ 0 and suppose f is measurable. The distribution function is the function defined by t → µ ([t < f ]) . Lemma 9.7.2 If {fn } is an increasing sequence of functions converging pointwise to f then µ ([f > t]) = lim µ ([fn > t]) n→∞

Proof: The sets, [fn > t] are increasing and their union is [f > t] because if f (ω) > t, then for all n large enough, fn (ω) > t also. Therefore, from Theorem 8.1.5 on Page 178 the desired conclusion follows. Lemma 9.7.3 Suppose s ≥ 0 is a measurable simple function, s (ω) ≡

n X

ak XEk (ω)

k=1

where the ak are the distinct nonzero values of s, a1 < a2 < · · · < an . Suppose φ is a C 1 function defined on [0, ∞) which has the property that φ (0) = 0, φ0 (t) > 0 for all t. Then Z ∞ Z φ0 (t) µ ([s > t]) dm =

φ (s) dµ.

0

Proof: First note that if µ (Ek ) = ∞ for any k then both sides equal ∞ and so without loss of generality, assume µ (Ek ) < ∞ for all k. Letting a0 ≡ 0, the left side equals n Z ak n Z ak n X X X φ0 (t) µ ([s > t]) dm(t) = φ0 (t) µ (Ei ) dm k=1

ak−1

= =

k=1 ak−1 n X n X

i=k ak

Z

φ0 (t) dm

µ (Ei )

k=1 i=k n X n X

ak−1

µ (Ei ) (φ (ak ) − φ (ak−1 ))

k=1 i=k

= =

n X i=1 n X

µ (Ei )

i X

(φ (ak ) − φ (ak−1 ))

k=1

µ (Ei ) φ (ai ) =

Z φ (s) dµ.

i=1

This proves the lemma. With this lemma the next theorem which is the main result follows easily.

248


Theorem 9.7.4 Let f ≥ 0 be measurable and let φ be a C 1 function defined on [0, ∞) which satisfies φ0 (t) > 0 for all t > 0 and φ (0) = 0. Then Z

Z

∞

φ (f ) dµ =

φ0 (t) µ ([f > t]) dm.

0

Proof: By Theorem 8.3.9 on Page 197 there exists an increasing sequence of nonnegative simple functions, {sn } which converges pointwise to f. By the monotone convergence theorem and Lemma 9.7.2, Z Z Z ∞ φ (f ) dµ = lim φ (sn ) dµ = lim φ0 (t) µ ([sn > t]) dm n→∞ n→∞ 0 Z ∞ = φ0 (t) µ ([f > t]) dm 0

This proves the theorem. This theorem can be generalized to a situation in which φ is only increasing and continuous. In the generalization I will replace the symbol φ with F to coincide with earlier notation. Lemma 9.7.5 Suppose s ≥ 0 is a measurable simple function, s (ω) ≡

n X

ak XEk (ω)

k=1

where the ak are the distinct nonzero values of s, a1 < a2 < · · · < an . Suppose F is an increasing function defined on [0, ∞), F (0) = 0, F being continuous at 0 from the right and continuous at every ak . Then letting µ be a measure and (Ω, F, µ) a measure space, Z Z µ ([s > t]) dν = F (s) dµ. (0,∞]

Ω

where the integral on the left is the Lebesgue integral for the measure ν given as the Radon measure representing the functional Z ∞ gdF 0

for g ∈ Cc ([0, ∞)) . Proof: This follows from the following computation and Proposition 9.6.1. Since F is continuous at 0 and the values ak , Z

∞

µ ([s > t]) dν (t) = 0

n Z X k=1

(ak−1 ,ak ]

µ ([s > t]) dν (t)

9.7. THE DISTRIBUTION FUNCTION

=

n Z X k=1

=

n X

n X

249

µ (Ej ) dF (t) =

(ak−1 ,ak ] j=k

µ (Ej )

j=1

j X

n X

µ (Ej )

j=1

(F (ak ) − F (ak−1 )) =

j X

ν ((ak−1 , ak ])

k=1 n X

Z µ (Ej ) F (aj ) ≡

j=1

k=1

F (s) dµ Ω

This proves the lemma. Now here is the generalization to nonnegative measurable f . Theorem 9.7.6 Let f ≥ 0 be measurable with respect to F where (Ω, F, µ) a measure space, and let F be an increasing continuous function defined on [0, ∞) and F (0) = 0. Then Z Z F (f ) dµ =

µ ([f > t]) dν (t) (0,∞]

Ω

where ν is the Radon measure representing Z ∞ Lg = gdF 0

for g ∈ Cc ([0, ∞)) . Proof: By Theorem 8.3.9 on Page 197 there exists an increasing sequence of nonnegative simple functions, {sn } which converges pointwise to f. By the monotone convergence theorem and Lemma 9.7.5, Z Z Z F (f ) dµ = lim F (sn ) dµ = lim µ ([sn > t]) dν n→∞ Ω n→∞ (0,∞] Ω Z = µ ([f > t]) dν (0,∞]

This proves the theorem. Note that the function t → µ ([f > t]) is a decreasing function. R ∞ Therefore, one can make sense of an improper Riemann Stieltjes integral 0 µ ([f > t]) dF (t) . With more work, one can have this equal to the corresponding Lebesgue integral above. There is a very interesting and important inequality called the good lambda inequality (I am not sure if there is a bad lambda inequality.) which follows from the above theory of distribution functions. It involves the inequality µ ([f > βλ] ∩ [g ≤ δλ]) ≤ φ (δ) µ ([f > λ]) for β > 1, nonnegative functions f, g and is supposed to hold for all small positive δ and φ (δ) → 0 as δ → 0. Note the left side is small when g is large and f is small. The inequality involves dominating an integral involving f with one involving g as described below.

250


Theorem 9.7.7 Let (Ω, F, µ) be a finite measure space and let F be a continuous increasing function defined on [0, ∞) such that F (0) = 0. Suppose also that for all α > 1, there exists a constant Cα such that for all x ∈ [0, ∞), F (αx) ≤ Cα F (x) . Also suppose f, g are nonnegative measurable functions and there exists β > 1 such that for all λ > 0 and 1 > δ > 0, µ ([f > βλ] ∩ [g ≤ δλ]) ≤ φ (δ) µ ([f > λ])

(9.7.18)

where limδ→0+ φ (δ) = 0 and φ is increasing. Under these conditions, there exists a constant C depending only on β, φ such that Z Z F (f (ω)) dµ (ω) ≤ C F (g (ω)) dµ (ω) . Ω

Ω

Proof: This follows from Theorem 9.7.6. By assumption there exists a constant Cβ such that for all y ≥ 0, µ ¶ µ ¶ y y ≤ Cβ−1 F F (y) = F β β β and so for all y ≥ 0,

µ ¶ y F ≥ Cβ F (y) . β

(9.7.19)

Let ν be the measure representing the functional Lg = Integrating both sides gives Z Z µ ([f > βλ] ∩ [g ≤ δλ]) dν (λ) ≤ φ (δ) [0,∞)

R∞ 0

gdF for g ∈ Cc ([0, ∞)) .

µ ([f > λ]) dν (λ)

[0,∞)

and so by Theorem 9.7.6 applied to the right side, Z Z µ ([f /β > λ] ∩ [g/δ ≤ λ]) dν (λ) ≤ φ (δ) F (f (ω)) dµ (ω) [0,∞)

Ω

Consider the left side. Z Z

[0,∞)

= [0,∞)

µ ([f /β > λ] ∩ [g/δ ≤ λ]) dν (λ) Z X([f /β>λ]∩[g/δ≤λ]) (ω) dµdν Ω

Consider the function k (λ, ω) ≡ X([f /β>λ]∩[g/δ≤λ]) (ω)

(9.7.20)

9.8. PRODUCT MEASURES

251

this is product measurable because you can approximate f, g with increasing sequences of simple functions and replacing f, g respectively with the corresponding function from the sequence, the resulting function is obviously product measurable. Therefore, this function, being the limit of those functions, must also be product measurable. Thus it makes sense to interchange the order of integration in 9.7.20 which implies 9.7.20 equals Z Z X([f /β>λ]∩[g/δ≤λ]) (ω) dνdµ Ω

[0,∞)

Z ÃZ

!

f (ω)/β

=

dν Ω

g(ω)/δ

dµ. +

By Proposition 9.6.1 this reduces to ¶ µ ¶¶ Z µ µ f (ω) g (ω) F −F dµ β δ Ω + where the + indicates the positive part of the expression. Then by 9.7.19 µ ¶ µ ¶ Z Z f (ω) g (ω) ≥ F dµ − F dµ β δ Ω Ω µ ¶ Z Z g (ω) ≥ Cβ F (f (ω)) dµ − F dµ. δ Ω Ω It follows

Z

φ (δ)

Z F (f (ω)) dµ (ω) ≥ Cβ

Ω

µ

Z F (f (ω)) dµ −

Ω

F Ω

g (ω) δ

¶ dµ

and so if δ is picked so small that Cβ /2 > φ (δ) , it follows µ ¶ Z Z Z g (ω) Cβ F (f (ω)) dµ ≤ F dµ ≤ C1/δ F (g (ω)) dµ 2 Ω δ Ω Ω and this proves the theorem with C = 2C1/δ /Cβ .

9.8

Product Measures

Let (X, S, µ) and (Y, T , ν) be two complete measure spaces. In this section consider the problem of defining a product measure, µ × ν which is defined on a σ algebra of sets of X × Y such that (µ × ν) (E × F ) = µ (E) ν (F ) whenever E ∈ S and F ∈ T . I found the following approach to product measures in [24] and they say they got it from [26]. Definition 9.8.1 Let R denote the set of countable unions of sets of the form A × B, where A ∈ S and B ∈ T (Sets of the form A × B are referred to as measurable rectangles) and also let ρ (A × B) = µ (A) ν (B)

(9.8.21)

252


More generally, define

Z Z ρ (E) ≡

XE (x, y) dµdν

(9.8.22)

whenever E is such that x → XE (x, y) is µ measurable for all y and

(9.8.23)

Z y→

XE (x, y) dµ is ν measurable.

(9.8.24)

Note that if E = A × B as above, then Z Z Z Z XE (x, y) dµdν = XA×B (x, y) dµdν Z Z = XA (x) XB (y) dµdν = µ (A) ν (B) = ρ (E) and so there is no contradiction between 9.8.22 and 9.8.21. The first goal is to show that for Q ∈ R, 9.8.23 Rand 9.8.24 both hold. That is, x → XQ (x, y) is µ measurable for all y and y → XQ (x, y) dµ is ν measurable. This is done so that it is possible to speak of ρ (Q) . The following lemma will be the fundamental result which will make this possible. First here is a picture. A B D C n

Lemma 9.8.2 Given C × D and {Ai × Bi }i=1 , there exist finitely many disjoint p rectangles, {Ci0 × Di0 }i=1 such that none of these sets intersect any of the Ai × Bi , each set is contained in C × D and (∪ni=1 Ai × Bi ) ∪ (∪pk=1 Ck0 × Dk0 ) = (C × D) ∪ (∪ni=1 Ai × Bi ) . Proof: From the above picture, you see that (C × D) \ (A1 × B1 ) = C × (D \ B1 ) ∪ (C \ A1 ) × (D ∩ B1 ) and these last two sets are disjoint, have empty intersection with A1 × B1 , and (C × (D \ B1 ) ∪ (C \ A1 ) × (D ∩ B1 )) ∪ (∪ni=1 Ai × Bi ) = (C × D) ∪ (∪ni=1 Ai × Bi )

9.8. PRODUCT MEASURES Now suppose disjoint sets, of C × D such that

253

n om ei × D ei have been obtained, each being a subset C i=1

³ ´ n e e (∪ni=1 Ai × Bi ) ∪ ∪m k=1 Ck × Dk = (∪i=1 Ai × Bi ) ∪ (C × D) ek × D e k has empty intersection with each set of {Ai × Bi }p . Then and for all k, C i=1 ek × D e k with finitely many disjoint using the same procedure, replace each of C rectangles such that none of these intersect Ap+1 × Bp+1 while preserving the union of all the sets involved. The process stops when you have gotten to n. This proves the lemma. Lemma 9.8.3 If Q = ∪∞ i=1 Ai × Bi ∈ R, then there exist disjoint sets, of the form 0 0 0 0 A0i × Bi0 such that Q = ∪∞ i=1 Ai × Bi , each Ai × Bi is a subset of some Ai × Bi , and 0 0 Ai ∈ S while Bi ∈ T . Also, the intersection of finitely many sets of R is a set of R. For ρ defined in 9.8.22, it follows that 9.8.23 and 9.8.24 hold for any element of R. Furthermore, X X ρ (Q) = µ (A0i ) ν (Bi0 ) = ρ (A0i × Bi0 ) . i

i

Proof: Let Q be given as above. Let A01 × B10 = A1 × B1 . By Lemma 9.8.2, it m2 is possible to replace A2 × B2 with finitely many disjoint rectangles, {A0i × Bi0 }i=2 0 0 such that none of these rectangles intersect A1 × B1 , each is a subset of A2 × B2 , and m2 0 0 ∞ ∪∞ i=1 Ai × Bi = (∪i=1 Ai × Bi ) ∪ (∪k=3 Ak × Bk ) m

p Now suppose disjoint rectangles, {A0i × Bi0 }i=1 have been obtained such that each rectangle is a subset of Ak × Bk for some k ≤ p and ¡ mp 0 ¢ ¡ ∞ ¢ 0 ∪∞ i=1 Ai × Bi = ∪i=1 Ai × Bi ∪ ∪k=p+1 Ak × Bk .

m

p+1 By Lemma 9.8.2 again, there exist disjoint rectangles {A0i × Bi0 }i=m such that p +1 mp 0 each is contained in Ap+1 × Bp+1 , none have intersection with any of {Ai × Bi0 }i=1 and ¢ ¡ mp+1 0 ¢ ¡ ∞ 0 ∪∞ i=1 Ai × Bi = ∪i=1 Ai × Bi ∪ ∪k=p+2 Ak × Bk .

m

p Note that no change is made in {A0i × Bi0 }i=1 . Continuing this way proves the existence of the desired sequence of disjoint rectangles, each of which is a subset of at least one of the original rectangles and such that

0 0 Q = ∪∞ i=1 Ai × Bi .

It remains to verify x → XQ (x, y) is µ measurable for all y and Z y → XQ (x, y) dµ

254


is ν measurable whenever Q ∈ R. Let Q ≡ ∪∞ i=1 Ai × Bi ∈ R. Then by the first ∞ part of this lemma, there exists {A0i × Bi0 }i=1 such that the sets are disjoint and 0 0 ∪∞ i=1 Ai × Bi = Q. Therefore, since the sets are disjoint, XQ (x, y) =

∞ X

XA0i ×Bi0 (x, y) =

∞ X

XA0i (x) XBi0 (y) .

i=1

i=1

It follows x → XQ (x, y) is measurable. Now by the monotone convergence theorem, Z XQ (x, y) dµ =

Z X ∞

XA0i (x) XBi0 (y) dµ

i=1

= =

∞ X i=1 ∞ X

Z

XBi0 (y)

XA0i (x) dµ

XBi0 (y) µ (A0i ) .

i=1

It follows y → theorem again,

R

XQ (x, y) dµ is measurable and so by the monotone convergence Z Z XQ (x, y) dµdν

=

Z X ∞

XBi0 (y) µ (A0i ) dν

i=1

=

∞ Z X

XBi0 (y) µ (A0i ) dν

i=1

=

∞ X

ν (Bi0 ) µ (A0i ) .

(9.8.25)

i=1

This shows the measurability conditions, 9.8.23 and 9.8.24 hold for Q ∈ R and also establishes the formula for ρ (Q) , 9.8.25. If ∪i Ai × Bi and ∪j Cj × Dj are two sets of R, then their intersection is ∪i ∪j (Ai ∩ Cj ) × (Bi ∩ Dj ) a countable union of measurable rectangles. Thus finite intersections of sets of R are in R. This proves the lemma. Now note that from the definition of R if you have a sequence of elements of R then their union is also in R. The next lemma will enable the definition of an outer measure. ∞

Lemma 9.8.4 Suppose {Ri }i=1 is a sequence of sets of R then ρ (∪∞ i=1 Ri ) ≤

∞ X i=1

ρ (Ri ) .


255 ∞

i i 0 0 Proof: Let Ri = ∪∞ j=1 Aj × Bj . Using Lemma 9.8.3, let {Am × Bm }m=1 be a i sequence of disjoint rectangles each of which is contained in some Aj × Bji for some i, j such that ∞ 0 0 ∪∞ i=1 Ri = ∪m=1 Am × Bm .

Now define

© ª 0 Si ≡ m : A0m × Bm ⊆ Aij × Bji for some j .

It is not important to consider whether some m might be in more than one Si . The important thing to notice is that 0 i i ⊆ ∪∞ ∪m∈Si A0m × Bm j=1 Aj × Bj = Ri .

Then by Lemma 9.8.3, ρ (∪∞ i=1 Ri ) = ≤

X m ∞ X

0 ρ (A0m × Bm )

X

0 ρ (A0m × Bm )

i=1 m∈Si

≤

∞ X

0 ρ (∪m∈Si A0m × Bm )≤

i=1

∞ X

ρ (Ri ) .

i=1

This proves the lemma. So far, there is no measure and no σ algebra. However, the next step is to define an outer measure which will lead to a measure on a σ algebra of measurable sets from the Caratheodory procedure. When this is done, it will be shown that this measure can be computed using ρ which implies the important Fubini theorem. Now it is possible to define an outer measure. Definition 9.8.5 For S ⊆ X × Y, define (µ × ν) (S) ≡ inf {ρ (R) : S ⊆ R, R ∈ R} .

(9.8.26)

The following proposition is interesting but is not needed in the development which follows. It gives a different description of (µ × ν) . P∞ Proposition 9.8.6 (µ × ν) (S) = inf { i=1 µ (Ai ) ν (Bi ) : S ⊆ ∪∞ i=1 Ai × Bi } P∞ Proof: Let λ (S) ≡ inf { i=1 µ (Ai ) ν (Bi ) : S ⊆ ∪∞ i=1 Ai × Bi } . Suppose S ⊆ ∞ ∪i=1 Ai × Bi ≡ Q ∈ R. Then by Lemma 9.8.3, QP = ∪i A0i × Bi0 where these rectan∞ gles are disjoint. Thus by this lemma, ρ (Q) = i=1 µ (A0i ) ν (Bi0 ) ≥ λ (S) and so λ (S) ≤ (µ × ν) (S) . If λ (S) = ∞, this shows λ (S) = (µ × ν) (S) . Suppose then P∞ that λ (S) < ∞ and λ (S) + ε > i=1 µ (Ai ) ν (Bi ) where Q = ∪∞ i=1 Ai × Bi ⊇ S. ∞ 0 0 Then by Lemma 9.8.3 again, ∪∞ i=1 Ai ×Bi = ∪i=1 Ai ×Bi where the primed rectangles are disjoint, each is a subset of some Ai × Bi and so λ (S) + ε ≥

∞ X i=1

µ (Ai ) ν (Bi ) ≥

∞ X

µ (A0i ) ν (Bi0 ) = ρ (Q) ≥ (µ × ν) (S) .

i=1

Since ε is arbitrary, this shows λ (S) ≥ (µ × ν) (S) and this proves the proposition.

256


Lemma 9.8.7 µ × ν is an outer measure on X × Y and for R ∈ R (µ × ν) (R) = ρ (R) .

(9.8.27)

Proof: First consider 9.8.27. Since R ⊇ R, it follows ρ (R) ≥ (µ × ν) (R) . On the other hand, if Q ∈ R and Q ⊇ R, then ρ (Q) ≥ ρ (R) and so, taking the infimum on the left yields (µ × ν) (R) ≥ ρ (R) . This shows 9.8.27. It is necessary to show that if S ⊆ T, then (µ × ν) (S) ≤ (µ × ν) (T ) , (µ × ν) (∪∞ i=1 Si ) ≤

∞ X

(9.8.28)

(µ × ν) (Si ) .

(9.8.29)

i=1

To do this, note that 9.8.28 is obvious. To verify 9.8.29, note that it is obvious if (µ × ν) (Si ) = ∞ for any i. Therefore, assume (µ × ν) (Si ) < ∞. Then letting ε > 0 be given, there exist Ri ∈ R such that (µ × ν) (Si ) +

ε > ρ (Ri ) , Ri ⊇ Si . 2i

Then by Lemma 9.8.4, 9.8.27, and the observation that ∪∞ i=1 Ri ∈ R, (µ × ν) (∪∞ i=1 Si ) ≤ =

(µ × ν) (∪∞ i=1 Ri ) ∞ X ρ (∪∞ R ) ≤ ρ (Ri ) i=1 i i=1

≤

∞ ³ X

(µ × ν) (Si ) +

i=1

=

Ã∞ X

! (µ × ν) (Si )

ε´ 2i + ε.

i=1

Since ε is arbitrary, this proves the lemma. By Caratheodory’s procedure, it follows there is a σ algebra of subsets of X × Y, denoted here by S × T such that (µ × ν) is a complete measure on this σ algebra. The first thing to note is that every rectangle is in this σ algebra. Lemma 9.8.8 Every rectangle is (µ × ν) measurable. Proof: Let S ⊆ X × Y. The following inequality must be established. (µ × ν) (S) ≥ (µ × ν) (S ∩ (A × B)) + (µ × ν) (S \ (A × B)) . The following claim will be used to establish this inequality. Claim: Let P, A × B ∈ R. Then ρ (P ∩ (A × B)) + ρ (P \ (A × B)) = ρ (P ) .

(9.8.30)


257

0 0 0 0 Proof of the claim: From Lemma 9.8.3, P = ∪∞ i=1 Ai × Bi where the Ai × Bi are disjoint. Therefore,

P ∩ (A × B) =

∞ [

(A ∩ A0i ) × (B ∩ Bi0 )

i=1

while P \ (A × B) =

∞ [

(A0i \ A) × Bi0 ∪

i=1

∞ [

(A ∩ A0i ) × (Bi0 \ B) .

i=1

Since all of the sets in the above unions are disjoint, ρ (P ∩ (A × B)) + ρ (P \ (A × B)) = Z Z X ∞ i=1

X(A∩A0 ) (x) XB∩Bi0 (y) dµdν +

Z Z X ∞

i

+

Z Z X ∞

i=1

X(A0 \A) (x) XBi0 (y) dµdν i

XA∩A0i (x) XBi0 \B (y) dµdν

i=1

=

∞ X

µ (A ∩ A0i ) ν (B ∩ Bi0 ) + µ (A0i \ A) ν (Bi0 ) + µ (A ∩ A0i ) ν (Bi0 \ B)

i=1

=

∞ X

µ (A ∩ A0i ) ν (Bi0 ) + µ (A0i \ A) ν (Bi0 ) =

i=1

∞ X

µ (A0i ) ν (Bi0 ) = ρ (P ) .

i=1

This proves the claim. Now continuing to verify 9.8.30, without loss of generality, (µ × ν) (S) can be assumed finite. Let P ⊇ S for P ∈ R and (µ × ν) (S) + ε > ρ (P ) . Then from the claim, (µ × ν) (S) + ε

> ρ (P ) = ρ (P ∩ (A × B)) + ρ (P \ (A × B)) ≥ (µ × ν) (S ∩ (A × B)) + (µ × ν) (S \ (A × B)) .

Since ε > 0 this shows A × B is µ × ν measurable as claimed. Lemma 9.8.9 Let R1 be defined as the set of all countable intersections of sets of R. Then if S ⊆ X × Y, there exists R ∈ R1 for which it makes sense to write ρ (R) because 9.8.23 and 9.8.24 hold such that (µ × ν) (S) = ρ (R) . Also, every element of R1 is µ × ν measurable.

(9.8.31)

258


Proof: Consider 9.8.31. Let S ⊆ X × Y. If (µ × ν) (S) = ∞, let R = X × Y and it follows ρ (X × Y ) = ∞ = (µ × ν) (S) . Assume then that (µ × ν) (S) < ∞. Therefore, there exists Pn ∈ R such that Pn ⊇ S and (µ × ν) (S) ≤ ρ (Pn ) < (µ × ν) (S) + 1/n.

(9.8.32)

Let Qn = ∩ni=1 Pi ∈ R. Define P ≡ ∩∞ i=1 Qi ⊇ S. Then 9.8.32 holds with Qn in place of Pn . It is clear that x → XP (x, y) is µ measurable because this function is the pointwise R limit of functions for which this is so. It remains to consider whether y → XP (x, y) dµ is ν measurable. First observe Qn ⊇ Qn+1 , XQi ≤ XPi , and Z Z ρ (Q1 ) = ρ (P1 ) = XP1 (x, y) dµdν < ∞. (9.8.33) Therefore, there exists a set of ν measure 0, N, such that if y ∈ / N, then Z XP1 (x, y) dµ < ∞. It follows from the dominated convergence theorem that Z Z lim XN C (y) XQn (x, y) dµ = XN C (y) XP (x, y) dµ n→∞

and so

Z y → XN C (y)

XP (x, y) dµ

is also measurable. By completeness of ν, Z y → XP (x, y) dµ must also be ν measurable and so it makes sense to write Z Z XP (x, y) dµdν for every P ∈ R1 . Also, by the dominated convergence theorem, Z Z Z Z XP (x, y) dµdν = XN C (y) XP (x, y) dµdν Z Z = lim XN C (y) XQn (x, y) dµdν n→∞ Z Z = lim XQn (x, y) dµdν n→∞

=

lim ρ (Qn ) ∈ [(µ × ν) (S) , (µ × ν) (S) + 1/n]

n→∞


259

for all n. Therefore, Z Z ρ (P ) ≡

XP (x, y) dµdν = (µ × ν) (S) .

The sets of R1 are µ × ν measurable because these sets are countable intersections of countable unions of rectangles and Lemma 9.8.8 verifies the rectangles are µ × ν measurable. This proves the Lemma. The following theorem is the main result. Theorem 9.8.10 Let E ⊆ X × Y be µ × ν measurable and suppose (µ × ν) (E) < ∞. Then x → XE (x, y) is µ measurable a.e. y. Modifying XE on a set of measure zero, it is possible to write Z XE (x, y) dµ. The function,

Z y→

is ν measurable and

XE (x, y) dµ Z Z

(µ × ν) (E) = Similarly,

XE (x, y) dµdν. Z Z

(µ × ν) (E) =

XE (x, y) dνdµ.

Proof: By Lemma 9.8.9, there exists R ∈ R1 such that ρ (R) = (µ × ν) (E) , R ⊇ E. Therefore, since R is µ × ν measurable and ρ (R) = (µ × ν) (R), it follows (µ × ν) (R \ E) = 0. By Lemma 9.8.9 again, there exists P ⊇ R \ E with P ∈ R1 and ρ (P ) = (µ × ν) (R \ E) = 0. Thus

Z Z XP (x, y) dµdν = 0.

(9.8.34)

Since P ∈ R1 Lemma 9.8.9 implies x → XP (x, y) is µ measurable and it follows from the above there exists a set of ν measure zero, N such that if y ∈ / N, then R XP (x, y) dµ = 0. Therefore, by completeness of ν, x → XN C (y) XR\E (x, y)

260


is µ measurable and

Z XN C (y) XR\E (x, y) dµ = 0.

(9.8.35)

Now also XN C (y) XR (x, y) = XN C (y) XR\E (x, y) + XN C (y) XE (x, y)

(9.8.36)

and this shows that x → XN C (y) XE (x, y) is µ measurable because it is the difference of two functions with this property. Then by 9.8.35 it follows Z Z XN C (y) XE (x, y) dµ = XN C (y) XR (x, y) dµ. The right side of this equation equals a ν measurable function and so the left side which equals R it is also a ν measurable function. It follows from completeness of ν that y → XE (x,Ry) dµ is ν measurable because for y outside of a set of ν measure zero, N it equals XR (x, y) dµ. Therefore, Z Z Z Z XE (x, y) dµdν = XN C (y) XE (x, y) dµdν Z Z = XN C (y) XR (x, y) dµdν Z Z = XR (x, y) dµdν =

ρ (R) = (µ × ν) (E) .

In all the above there would be no change in writing dνdµ instead of dµdν. The same result would be obtained. This proves the theorem. Now let f : X × Y → [0, ∞] be µ × ν measurable and Z f d (µ × ν) < ∞. (9.8.37) Pm Let s (x, y) ≡ i=1 ci XEi (x, y) be a nonnegative simple function with ci being the nonzero values of s and suppose 0 ≤ s ≤ f. Then from the above theorem, Z Z Z sd (µ × ν) = sdµdν In which

Z

Z sdµ =

XN C (y) sdµ


261

R for N a set of ν measure zero such that y → XN C (y) sdµ is ν measurable. This follows because 9.8.37 implies (µ × ν) (Ei ) < ∞. Now let sn ↑ f where sn is a nonnegative simple function and Z Z Z sn d (µ × ν) = XNnC (y) sn (x, y) dµdν where

Z y→

XNnC (y) sn (x, y) dµ

is ν measurable. Then let N ≡ ∪∞ n=1 Nn . It follows N is a set of ν measure zero. Thus Z Z Z sn d (µ × ν) = XN C (y) sn (x, y) dµdν and letting n → ∞, the monotone convergence theorem implies Z Z Z f d (µ × ν) = XN C (y) f (x, y) dµdν Z Z = f (x, y) dµdν because of completeness of the measures, µ and ν. This proves Fubini’s theorem. Theorem 9.8.11 (Fubini) Let (X, S, µ) and (Y, T , ν) be complete measure spaces and let ½Z Z ¾ (µ × ν) (E) ≡ inf XR (x, y) dµdν : E ⊆ R ∈ R 2 where Ai ∈ S and Bi ∈ T . Then µ × ν is an outer measure on the subsets of X × Y and the σ algebra of µ × ν measurable sets, S × T , contains all measurable rectangles. If f ≥ 0 is a µ × ν measurable function satisfying Z f d (µ × ν) < ∞, (9.8.38) X×Y

then

Z

Z Z f d (µ × ν) = X×Y

f dµdν, Y

X

where the iterated integral Ron the right makes sense because for ν a.e. y, x → f (x, y) is µ measurable and y → f (x, y) dµ is ν measurable. Similarly, Z Z Z f d (µ × ν) = f dνdµ. X×Y 2 Recall

this is the same as inf

(∞ X

X

Y

) µ (Ai ) ν (Bi ) : E ⊆ ∪∞ i=1 Ai × Bi

i=1

in which the Ai and Bi are measurable.

262


In the case where (X, S, µ) and (Y, T , ν) are both σ finite, it is not necessary to assume 9.8.38. Corollary 9.8.12 (Fubini) Let (X, S, µ) and (Y, T , ν) be complete measure spaces such that (X, S, µ) and (Y, T , ν) are both σ finite and let ½Z Z ¾ (µ × ν) (E) ≡ inf XR (x, y) dµdν : E ⊆ R ∈ R where Ai ∈ S and Bi ∈ T . Then µ × ν is an outer measure. If f ≥ 0 is a µ × ν measurable function then Z Z Z f dµdν, f d (µ × ν) = X×Y

Y

X

where the iterated integral Ron the right makes sense because for ν a.e. y, x → f (x, y) is µ measurable and y → f (x, y) dµ is ν measurable. Similarly, Z Z Z f d (µ × ν) = f dνdµ. X×Y

∪∞ n=1 Xn

X

Y

∪∞ n=1 Yn

= Y where Xn ∈ S, Yn ∈ T , Xn ⊆ = X and Proof: Let Xn+1 , Yn ⊆ Yn+1 for all n and µ (Xn ) < ∞, ν (Yn ) < ∞. From Theorem 9.8.11 applied to Xn , Yn and fm ≡ min (f, m) , Z Z Z fm dµdν fm d (µ × ν) = Xn ×Yn

Yn

Xn

Now take m → ∞ and use the monotone convergence theorem to obtain Z Z Z f dµdν. f d (µ × ν) = Xn ×Yn

Yn

Xn

Then use the monotone convergence theorem again letting n → ∞ to obtain the desired conclusion. The argument for the other order of integration is similar. Corollary 9.8.13 If f ∈ L1 (X × Y ) , then Z Z Z Z Z f d (µ × ν) = f (x, y) dµdν = f (x, y) dνdµ. If µ and R Rν are σ finite, thenRifR f is µ × ν measurableR having complex values and either |f | dµdν < ∞ or |f | dνdµ < ∞, then |f | d (µ × ν) < ∞ so f ∈ L1 (X × Y ) . Proof: Without loss of generality, it can be assumed that f has real values. Then |f | + f − (|f | − f ) f= 2

9.9. ALTERNATIVE TREATMENT OF PRODUCT MEASURE

263

and both f + ≡ |f |+f and f − ≡ |f |−f are nonnegative and are less than |f |. There2 2 R fore, gd (µ × ν) < ∞ for g = f + and g = f − so the above theorem applies and Z Z Z f d (µ × ν) ≡ f + d (µ × ν) − f − d (µ × ν) Z Z Z Z = f + dµdν − f − dµdν Z Z = f dµdν. It remains to verify the last claim. Suppose s is a simple function, s (x, y) ≡

m X

ci XEi ≤ |f | (x, y)

i=1

where the ci are the nonzero values of s. Then sXRn ≤ |f | XRn where Rn ≡ Xn × Yn where Xn ↑ X and Yn ↑ Y with µ (Xn ) < ∞ and ν (Yn ) < ∞. It follows, since the nonzero values of sXRn are achieved on sets of finite measure, Z Z Z sXRn d (µ × ν) = sXRn dµdν. Letting n → ∞ and applying the monotone convergence theorem, this yields Z Z Z sdµdν. (9.8.39) sd (µ × ν) = Now let sn ↑ |f | where sn is a nonnegative simple function. From 9.8.39, Z Z Z sn d (µ × ν) = sn dµdν. Letting n → ∞ and using the monotone convergence theorem, yields Z Z Z |f | d (µ × ν) = |f | dµdν < ∞

9.9 9.9.1

Alternative Treatment Of Product Measure Monotone Classes And Algebras

Measures are defined on σ algebras which are closed under countable unions. It is for this reason that the theory of measure and integration is so useful in dealing with limits of sequences. However, there is a more basic notion which involves only finite unions and differences.

264


Definition 9.9.1 A is said to be an algebra of subsets of a set, Z if Z ∈ A, ∅ ∈ A, and when E, F ∈ A, E ∪ F and E \ F are both in A. It is important to note that if A is an algebra, then it is also closed under finite intersections. This is because E ∩ F = (E C ∪ F C )C ∈ A since E C = Z \ E ∈ A and F C = Z \ F ∈ A. Note that every σ algebra is an algebra but not the other way around. Something satisfying the above definition is called an algebra because union is like addition, the set difference is like subtraction and intersection is like multiplication. Furthermore, only finitely many operations are done at a time and so there is nothing like a limit involved. How can you recognize an algebra when you see one? The answer to this question is the purpose of the following lemma. Lemma 9.9.2 Suppose R and E are subsets of P(Z)3 such that E is defined as the set of all finite disjoint unions of sets of R. Suppose also ∅, Z ∈ R A ∩ B ∈ R whenever A, B ∈ R, A \ B ∈ E whenever A, B ∈ R. Then E is an algebra of sets of Z. Proof: Note first that if A ∈ R, then AC ∈ E because AC = Z \ A. Now suppose that E1 and E2 are in E, n E1 = ∪m i=1 Ri , E2 = ∪j=1 Rj

where the Ri are disjoint sets in R and the Rj are disjoint sets in R. Then n E1 ∩ E2 = ∪m i=1 ∪j=1 Ri ∩ Rj

which is clearly an element of E because no two of the sets in the union can intersect and by assumption they are all in R. Thus by induction, finite intersections of sets of E are in E. Consider the difference of two elements of E next. If E = ∪ni=1 Ri ∈ E, E C = ∩ni=1 RiC = finite intersection of sets of E which was just shown to be in E. Now, if E1 , E2 ∈ E, E1 \ E2 = E1 ∩ E2C ∈ E from what was just shown about finite intersections. 3 Set

of all subsets of Z


265

Finally consider finite unions of sets of E. Let E1 and E2 be sets of E. Then E1 ∪ E2 = (E1 \ E2 ) ∪ E2 ∈ E because E1 \ E2 consists of a finite disjoint union of sets of R and these sets must be disjoint from the sets of R whose union yields E2 because (E1 \ E2 ) ∩ E2 = ∅. This proves the lemma. The following corollary is particularly helpful in verifying the conditions of the above lemma. Corollary 9.9.3 Let (Z1 , R1 , E1 ) and (Z2 , R2 , E2 ) be as described in Lemma 9.9.2. Then (Z1 × Z2 , R, E) also satisfies the conditions of Lemma 9.9.2 if R is defined as R ≡ {R1 × R2 : Ri ∈ Ri } and E ≡ { finite disjoint unions of sets of R}. Consequently, E is an algebra of sets. Proof: It is clear ∅, Z1 × Z2 ∈ R. Let A × B and C × D be two elements of R. A×B∩C ×D =A∩C ×B∩D ∈R by assumption. A × B \ (C × D) = ∈E2

∈E1

∈R2

z }| { z }| { z }| { A × (B \ D) ∪ (A \ C) × (D ∩ B) = (A × Q) ∪ (P × R) where Q ∈ E2 , P ∈ E1 , and R ∈ R2 . C D B A Since A × Q and P × R do not intersect, it follows the above expression is in E because each of these terms are. This proves the corollary. Definition 9.9.4 M ⊆ P(Z) is called a monotone class if a.) · · · En ⊇ En+1 · · · , E = ∩∞ n=1 En, and En ∈ M, then E ∈ M. b.) · · · En ⊆ En+1 · · · , E = ∪∞ n=1 En , and En ∈ M, then E ∈ M. (In simpler notation, En ↓ E and En ∈ M implies E ∈ M. En ↑ E and En ∈ M implies E ∈ M.)

266


Theorem 9.9.5 (Monotone Class theorem) Let A be an algebra of subsets of Z and let M be a monotone class containing A. Then M ⊇ σ(A), the smallest σ-algebra containing A. Proof: Consider all monotone classes which contain A, and take their intersection. The result is still a monotone class which contains A and is therefore the smallest monotone class containing A. Therefore, assume without loss of generality that M is the smallest monotone class containing A because if it is shown the smallest monotone class containing A contains σ (A), then the given monotone class does also. To avoid more notation, let M denote this smallest monotone class. The plan is to show M is a σ-algebra. It will then follow M ⊇ σ(A) because σ (A) is defined as the intersection of all σ algebras which contain A. For A ∈ A, define MA ≡ {B ∈ M such that A ∪ B ∈ M}. Clearly MA is a monotone class containing A. Hence MA ⊇ M because M is the smallest such monotone class. But by construction, MA ⊆ M. Therefore, M = MA . This shows that A ∪ B ∈ M whenever A ∈ A and B ∈ M. Now pick B ∈ M and define MB ≡ {D ∈ M such that D ∪ B ∈ M}. It was just shown that A ⊆ MB . It is clear that MB is a monotone class. Thus by a similar argument, MB = M and it follows that D ∪ B ∈ M whenever D ∈ M and B ∈ M. This shows M is closed under finite unions. Next consider the diference of two sets. Let A ∈ A MA ≡ {B ∈ M such that B \ A and A \ B ∈ M}. Then MA , is a monotone class containing A. As before, M = MA . Thus B \ A and A \ B are both in M whenever A ∈ A and B ∈ M. Now pick A ∈ M and consider MA ≡ {B ∈ M such that B \ A and A \ B ∈ M}. It was just shown MA contains A. Now MA is a monotone class and so MA = M as before. Thus M is both a monotone class and an algebra. Hence, if E ∈ M then Z \ E ∈ M. Next consider the question of whether M is a σ-algebra. If Ei ∈ M and Fn = ∪ni=1 Ei , then Fn ∈ M and Fn ↑ ∪∞ i=1 Ei . Since M is a monotone class, ∪∞ i=1 Ei ∈ M and so M is a σ-algebra. This proves the theorem.

9.9.2

Product Measure

Definition 9.9.6 Let (X, S, µ) and (Y, F, λ) be two measure spaces. A measurable rectangle is a set A×B ⊆ X ×Y where A ∈ S and B ∈ F. An elementary set will be any subset of X × Y which is a finite union of disjoint measurable rectangles. S × F will denote the smallest σ algebra of sets in P(X × Y ) containing all elementary sets.


267

Example 9.9.7 It follows from Lemma 9.9.2 or more easily from Corollary 9.9.3 that the elementary sets form an algebra. Definition 9.9.8 Let E ⊆ X × Y, Ex = {y ∈ Y : (x, y) ∈ E}, E y = {x ∈ X : (x, y) ∈ E}. These are called the x and y sections. Y

Ex x

X

Theorem 9.9.9 If E ∈ S × F, then Ex ∈ F and E y ∈ S for all x ∈ X and y ∈ Y . Proof: Let M = {E ⊆ S × F such that for all x ∈ X, Ex ∈ F, and for all y ∈ Y, E y ∈ S.} Then M contains all measurable rectangles. If Ei ∈ M, ∞ (∪∞ i=1 Ei )x = ∪i=1 (Ei )x ∈ F.

Similarly,

y

y ∞ (∪∞ i=1 Ei ) = ∪i=1 Ei ∈ S.

It follows M is closed under countable unions. If E ∈ M, ¡ C¢ C E x = (Ex ) ∈ F. ¡ ¢y Similarly, E C ∈ S. Thus M is closed under complementation. Therefore M is a σ-algebra containing the elementary sets. Hence, M ⊇ S × Fbecause S × F is the smallest σ algebra containing these elementary sets. But M ⊆ S × Fby definition and so M = S × F. This proves the theorem. It follows from Lemma 9.9.2 that the elementary sets form an algebra because clearly the intersection of two measurable rectangles is a measurable rectangle and (A × B) \ (A0 × B0 ) = (A \ A0 ) × B ∪ (A ∩ A0 ) × (B \ B0 ), an elementary set.

268


Theorem 9.9.10 If (X, S, µ) and (Y, F, λ) are both finite measure spaces (µ(X), λ(Y ) < ∞), then for every E ∈ S × F, a.) Rx → λ(Ex ) is µR measurable, y → µ(E y ) is λ measurable b.) X λ(Ex )dµ = Y µ(E y )dλ. Proof: Let M = {E ∈ S × F such that both a.) and b.) hold} . Since µ and λ are both finite, the monotone convergence and dominated convergence theorems imply M is a monotone class. Next I will argue M contains the elementary sets. Let E = ∪ni=1 Ai × Bi where the measurable rectangles, Ai × Bi are disjoint. Then Z λ (Ex )

=

XE (x, y) dλ =

Z X n

Y

=

n Z X i=1

Y

Y i=1

XAi ×Bi (x, y) dλ

XAi ×Bi (x, y) dλ =

n X

XAi (x) λ (Bi )

i=1

which is clearly µ measurable. Furthermore, Z

Z X n

λ (Ex ) dµ = X

Similarly,

X i=1

XAi (x) λ (Bi ) dµ =

Z µ (E y ) dλ = Y

n X

µ (Ai ) λ (Bi ) .

i=1

n X

µ (Ai ) λ (Bi )

i=1

and y → µ (E y ) is λ measurable and this shows M contains the algebra of elementary sets. By the monotone class theorem, M = S × F. This proves the theorem. One can easily extend this theorem to the case where the measure spaces are σ finite. Theorem 9.9.11 If (X, S, µ) and (Y, F, λ) are both σ finite measure spaces, then for every E ∈ S × F, a.) Rx → λ(Ex ) is µR measurable, y → µ(E y ) is λ measurable. b.) X λ(Ex )dµ = Y µ(E y )dλ. ∞ Proof: Let X = ∪∞ n=1 Xn , Y = ∪n=1 Yn where,

Xn ⊆ Xn+1 , Yn ⊆ Yn+1 , µ (Xn ) < ∞, λ(Yn ) < ∞. Let Sn = {A ∩ Xn : A ∈ S}, Fn = {B ∩ Yn : B ∈ F}.


269

Thus (Xn , Sn , µ) and (Yn , Fn , λ) are both finite measure spaces. Claim: If E ∈ S × F, then E ∩ (Xn × Yn ) ∈ Sn × Fn . Proof: Let Mn = {E ∈ S × F : E ∩ (Xn × Yn ) ∈ Sn × Fn } . Clearly Mn contains the algebra of elementary sets. It is also clear that Mn is a monotone class. Thus Mn = S × F. Now let E ∈ S × F. By Theorem 9.9.10, Z Z λ((E ∩ (Xn × Yn ))x )dµ = µ((E ∩ (Xn × Yn ))y )dλ (9.9.40) Xn

Yn

where the integrands are measurable. Also (E ∩ (Xn × Yn ))x = ∅ if x ∈ / Xn and a similar observation holds for the second integrand in 9.9.40 if y∈ / Yn . Therefore, Z Z λ((E ∩ (Xn × Yn ))x )dµ = λ((E ∩ (Xn × Yn ))x )dµ X Xn Z = µ((E ∩ (Xn × Yn ))y )dλ Yn Z = µ((E ∩ (Xn × Yn ))y )dλ. Y

Then letting n → ∞, the monotone convergence theorem implies b.) and the measurability assertions of a.) are valid because λ (Ex ) = y

µ (E ) =

lim λ((E ∩ (Xn × Yn ))x )

n→∞

lim µ((E ∩ (Xn × Yn ))y ).

n→∞

This proves the theorem. This theorem makes it possible to define product measure. Definition 9.9.12 For E ∈ S × F and (X, S, µ), (Y, F, λ) σ finite, (µ × λ)(E) ≡ R R y λ(E )dµ = µ(E )dλ. x X Y This definition is well defined because of Theorem 9.9.11. Theorem 9.9.13 If A ∈ S, B ∈ F, then (µ × λ)(A × B) = µ(A)λ(B), and µ × λ is a measure on S × F called product measure. Proof: The first assertion about the measure of a measurable rectangle was ∞ established above. Now suppose {Ei }i=1 is a disjoint collection of sets of S × F.

270


Then using the monotone convergence theorem along with the observation that (Ei )x ∩ (Ej )x = ∅, Z ∞ (µ × λ) (∪i=1 Ei ) = λ((∪∞ i=1 Ei )x )dµ Z

X

= X

= =

λ (∪∞ i=1 (Ei )x ) dµ =

∞ Z X i=1 ∞ X

X

Z X ∞ X i=1

λ ((Ei )x ) dµ

λ ((Ei )x ) dµ

(µ × λ) (Ei )

i=1

This proves the theorem. The next theorem is one of several theorems due to Fubini and Tonelli. These theorems all have to do with interchanging the order of integration in a multiple integral. Theorem 9.9.14 Let f : X × Y → [0, ∞] be measurable with respect to S × F and suppose µ and λ are σ finite. Then Z Z Z Z Z f d(µ × λ) = f (x, y)dλdµ = f (x, y)dµdλ (9.9.41) X×Y

X

Y

Y

X

and all integrals make sense. Proof: For E ∈ S × F, Z Z XE (x, y)dλ = λ(Ex ), XE (x, y)dµ = µ(E y ). Y

X

Thus from Definition 9.9.12, 9.9.41 holds if f = XE . It follows that 9.9.41 holds for every nonnegative simple function. By Theorem 8.3.9 on Page 197, there exists an increasing sequence, {fn }, of simple functions converging pointwise to f . Then Z Z f (x, y)dλ = lim fn (x, y)dλ, Z

n→∞

Y

Y

Z f (x, y)dµ = lim

n→∞

X

fn (x, y)dµ. X

This follows from the monotone convergence theorem. Since Z x→ fn (x, y)dλ Y

R is measurable with respect to S, it follows that x → Y f (x, y)dλR is also measurable with respect to S. A similar conclusion can be drawn about y → X f (x, y)dµ. Thus the two iterated integrals make sense. Since 9.9.41 holds for fn, another application of the Monotone Convergence theorem shows 9.9.41 holds for f . This proves the theorem.

9.10. COMPLETION OF MEASURES

271

Corollary 9.9.15R Let R R R f : X × Y → C be S × F measurable. Suppose either |f | dλdµ or Y X |f | dµdλ < ∞. Then f ∈ L1 (X × Y, µ × λ) and X Y Z

Z Z

Z Z

f d(µ × λ) = X×Y

f dλdµ = X

Y

f dµdλ Y

(9.9.42)

X

with all integrals making sense. Proof : Suppose first that f is real valued. Apply Theorem 9.9.14 to f + and f . 9.9.42 follows from observing that f = f + − f − ; and that all integrals are finite. If f is complex valued, consider real and imaginary parts. This proves the corollary. Suppose f is product measurable. From the above discussion, and breaking f down into a sum of positive and negative parts of real and imaginary parts and then using Theorem 8.3.9 on Page 197 on approximation by simple functions, it follows that whenever f is S × F measurable, x → f (x, y) is µ measurable, y → f (x, y) is λ measurable. −

9.10

Completion Of Measures

Suppose (Ω, F, µ) is a measure space. Then it is always possible to enlarge the¢ σ ¡ algebra and define a new measure µ on this larger σ algebra such that Ω, F, µ is a complete measure space. Recall this means that if N ⊆ N 0 ∈ F and µ (N 0 ) = 0, then N ∈ F. The following theorem is the main result. The new measure space is called the completion of the measure space. Theorem 9.10.1 Let ¡(Ω, F, µ)¢ be a σ finite measure space. Then there exists a unique measure space, Ω, F, µ satisfying ¡ ¢ 1. Ω, F, µ is a complete measure space. 2. µ = µ on F 3. F ⊇ F 4. For every E ∈ F there exists G ∈ F such that G ⊇ E and µ (G) = µ (E) . 5. For every E ∈ F there exists F ∈ F such that F ⊆ E and µ (F ) = µ (E) . Also for every E ∈ F there exist sets G, F ∈ F such that G ⊇ E ⊇ F and µ (G \ F ) = µ (G \ F ) = 0

(9.10.43)

Proof: First consider the claim about uniqueness. Suppose (Ω, F1 , ν 1 ) and (Ω, F2 , ν 2 ) both work and let E ∈ F1 . Also let µ (Ωn ) < ∞, · · · Ωn ⊆ Ωn+1 · · · , and ∪∞ n=1 Ωn = Ω. Define En ≡ E ∩ Ωn . Then pick Gn ⊇ En ⊇ Fn such that µ (Gn ) =

272


µ (Fn ) = ν 1 (En ). It follows µ (Gn \ Fn ) = 0. Then letting G = ∪n Gn , F ≡ ∪n Fn , it follows G ⊇ E ⊇ F and µ (G \ F )

≤ µ (∪n (Gn \ Fn )) X ≤ µ (Gn \ Fn ) = 0. n

It follows that ν 2 (G \ F ) = 0 also. Now E \ F ⊆ G \ F and since (Ω, F2 , ν 2 ) is complete, it follows E \ F ∈ F2 . Since F ∈ F2 , it follows E = (E \ F ) ∪ F ∈ F2 . Thus F1 ⊆ F2 . Similarly F2 ⊆ F1 . Now it only remains to verify ν 1 = ν 2 . Thus let E ∈ F1 = F2 and let G and F be as just described. Since ν i = µ on F, µ (F ) ≤ = ≤ =

ν 1 (E) ν 1 (E \ F ) + ν 1 (F ) ν 1 (G \ F ) + ν 1 (F ) ν 1 (F ) = µ (F )

Similarly ν 2 (E) = µ (F ) . This proves uniqueness. The construction has also verified 9.10.43. Next define an outer measure, µ on P (Ω) as follows. For S ⊆ Ω, µ (S) ≡ inf {µ (E) : E ∈ F} . Then it is clear µ is increasing. It only remains to verify µ is subadditive. Then let S = ∪∞ i=1 Si . If any µ (Si ) = ∞, there is nothing to prove so suppose µ (Si ) < ∞ for each i. Then there exist Ei ∈ F such that Ei ⊇ Si and µ (Si ) + ε/2i > µ (Ei ) . Then µ (S)

= µ (∪i Si ) ≤ µ (∪i Ei ) ≤

X

µ (Ei )

i

≤

X¡ ¢ X µ (Si ) + ε/2i = µ (Si ) + ε. i

i

Since ε is arbitrary, this verifies µ is subadditive and is an outer measure as claimed. Denote by F the σ algebra of measurable sets in the sense of Caratheodory. Then it follows from the Caratheodory procedure, Theorem 9.1.4, on Page 220 that ¡ ¢ Ω, F, µ is a complete measure space. This verifies 1. Now let E ∈ F . Then from the definition of µ, it follows µ (E) ≡ inf {µ (F ) : F ∈ F and F ⊇ E} ≤ µ (E) . If F ⊇ E and F ∈ F, then µ (F ) ≥ µ (E) and so µ (E) is a lower bound for all such µ (F ) which shows that µ (E) ≡ inf {µ (F ) : F ∈ F and F ⊇ E} ≥ µ (E) .

9.10. COMPLETION OF MEASURES

273

This verifies 2. Next consider 3. Let E ∈ F and let S be a set. I must show µ (S) ≥ µ (S \ E) + µ (S ∩ E) . If µ (S) = ∞ there is nothing to show. Therefore, suppose µ (S) < ∞. Then from the definition of µ there exists G ⊇ S such that G ∈ F and µ (G) = µ (S) . Then from the definition of µ, µ (S) ≤ µ (S \ E) + µ (S ∩ E) ≤ µ (G \ E) + µ (G ∩ E) = µ (G) = µ (S) This verifies 3. Claim 4 comes by the definition of µ as used above. The only other case is when µ (S) = ∞. However, in this case, you can let G = Ω. It only remains to verify 5. Let the Ωn be as described above and let E ∈ F such that E ⊆ Ωn . By 4 there exists H ∈ F such that H ⊆ Ωn , H ⊇ Ωn \ E, and µ (H) = µ (Ωn \ E) .

(9.10.44)

Then let F ≡ Ωn ∩ H C . It follows F ⊆ E and E\F

¡ ¢ = E ∩ F C = E ∩ H ∪ ΩC n = E ∩ H = H \ (Ωn \ E)

Hence from 9.10.44 µ (E \ F ) = µ (H \ (Ωn \ E)) = 0. It follows µ (E) = µ (F ) = µ (F ) . In the case where E ∈ F is arbitrary, not necessarily contained in some Ωn , it follows from what was just shown that there exists Fn ∈ F such that Fn ⊆ E ∩ Ωn and µ (Fn ) = µ (E ∩ Ωn ) . Letting F ≡ ∪n Fn µ (E \ F ) ≤ µ (∪n (E ∩ Ωn \ Fn )) ≤

X

µ (E ∩ Ωn \ Fn ) = 0.

n

Therefore, µ (E) = µ (F ) and this proves 5. This proves the theorem. Now here is an interesting theorem about complete measure spaces. Theorem 9.10.2 Let (Ω, F, µ) be a complete measure space and let f ≤ g ≤ h be functions having values in [0, ∞] . Suppose also that f (ω) ¡ = h (ω) ¢ a.e. ω and that f and h are measurable. Then g is also measurable. If Ω, F, µ is the completion

274


of a σ finite measure space (Ω, F, µ) as described above in Theorem 9.10.1 then if f is measurable with respect to F having values in [0, ∞] , it follows there exists g measurable with respect to F , g ≤ f, and a set N ∈ F with µ (N ) = 0 and g = f on N C . There also exists h measurable with respect to F such that h ≥ f, and a set of measure zero, M ∈ F such that f = h on M C . Proof: Let α ∈ R. [f > α] ⊆ [g > α] ⊆ [h > α] Thus [g > α] = [f > α] ∪ ([g > α] \ [f > α]) and [g > α] \ [f > α] is a measurable set because it is a subset of the set of measure zero, [h > α] \ [f > α] . Now consider the last assertion. By Theorem 8.3.9 on Page 197 there exists an increasing sequence of nonnegative simple functions, {sn } measurable with respect to F which converges pointwise to f . Letting sn (ω) =

mn X

cnk XEkn (ω)

(9.10.45)

k=1

be one of these simple functions, it follows from Theorem 9.10.1 there exist sets, Fkn ∈ F such that Fkn ⊆ Ekn and µ (Fkn ) = µ (Ekn ) . Then let tn (ω) ≡

mn X

cnk XFkn (ω) .

k=1

Thus tn = sn off a set of measure zero, Nn ∈ F, tn ≤ sn . Let N 0 ≡ ∪n Nn . Then by Theorem 9.10.1 again, there exists N ∈ F such that N ⊇ N 0 and µ (N ) = 0. Consider the simple functions, s0n (ω) ≡ tn (ω) XN C (ω) . It is an increasing sequence so let g (ω) = limn→∞ sn0 (ω) . It follows g is mesurable with respect to F and equals f off N . Finally, to obtain the function, h ≥ f, in 9.10.45 use Theorem 9.10.1 to obtain the existence of Fkn ∈ F such that Fkn ⊇ Ekn and µ (Fkn ) = µ (Ekn ). Then let tn (ω) ≡

mn X

cnk XFkn (ω) .

k=1

Thus tn = sn off a set of measure zero, Mn ∈ F, tn ≥ sn , and tn is measurable with respect to F. Then define s0n = max tn . k≤n

9.11. ANOTHER VERSION OF PRODUCT MEASURES

275

It follows s0n is an increasing sequence of F measurable nonnegative simple functions. Since each s0n ≥ sn , it follows that if h (ω) = limn→∞ s0n (ω) ,then h (ω) ≥ f (ω) . Also if h (ω) > f (ω) , then ω ∈ ∪n Mn ≡ M 0 , a set of F having measure zero. By Theorem 9.10.1, there exists M ⊇ M 0 such that M ∈ F and µ (M ) = 0. It follows h = f off M. This proves the theorem.

9.11

Another Version Of Product Measures

9.11.1

General Theory

Given two finite measure spaces, (X, F, µ) and (Y, S, ν) , there is a way to define a σ algebra of subsets of X × Y , denoted by F × S and a measure, denoted by µ × ν defined on this σ algebra such that µ × ν (A × B) = µ (A) ν (B) whenever A ∈ F and B ∈ S. This is naturally related to the concept of iterated integrals similar to what is used in calculus to evaluate a multiple integral. The approach is based on something called a π system, [18]. Definition 9.11.1 Let (X, F, µ) and (Y, S, ν) be two measure spaces. A measurable rectangle is a set of the form A × B where A ∈ F and B ∈ S. Definition 9.11.2 Let Ω be a set and let K be a collection of subsets of Ω. Then K is called a π system if ∅, Ω ∈ K and whenever A, B ∈ K, it follows A ∩ B ∈ K. Obviously an example of a π system is the set of measurable rectangles because A × B ∩ A0 × B 0 = (A ∩ A0 ) × (B ∩ B 0 ) . The following is the fundamental lemma which shows these π systems are useful. This lemma is due to Dynkin. Lemma 9.11.3 Let K be a π system of subsets of Ω, a set. Also let G be a collection of subsets of Ω which satisfies the following three properties. 1. K ⊆ G 2. If A ∈ G, then AC ∈ G ∞

3. If {Ai }i=1 is a sequence of disjoint sets from G then ∪∞ i=1 Ai ∈ G. Then G ⊇ σ (K) , where σ (K) is the smallest σ algebra which contains K. Proof: First note that if H ≡ {G : 1 - 3 all hold}

276


then ∩H yields a collection of sets which also satisfies 1 - 3. Therefore, I will assume in the argument that G is the smallest collection satisfying 1 - 3. Let A ∈ K and define GA ≡ {B ∈ G : A ∩ B ∈ G} . I want to show GA satisfies 1 - 3 because then it must equal G since G is the smallest collection of subsets of Ω which satisfies 1 - 3. This will give the conclusion that for A ∈ K and B ∈ G, A ∩ B ∈ G. This information will then be used to show that if A, B ∈ G then A ∩ B ∈ G. From this it will follow very easily that G is a σ algebra which will imply it contains σ (K). Now here are the details of the argument. Since K is given to be a π system, K ⊆ G A . Property 3 is obvious because if {Bi } is a sequence of disjoint sets in GA , then ∞ A ∩ ∪∞ i=1 Bi = ∪i=1 A ∩ Bi ∈ G

because A ∩ Bi ∈ G and the property 3 of G. It remains to verify Property 2 so let B ∈ GA . I need to verify that B C ∈ GA . In other words, I need to show that A ∩ B C ∈ G. However, ¡ ¢C A ∩ B C = AC ∪ (A ∩ B) ∈ G Here is why. Since B ∈ GA , A ∩ B ∈ G and since A ∈ K ⊆ G it follows AC ∈ G by assumption 2. It follows from assumption 3 the union of the disjoint sets, AC and (A ∩ B) is in G and then from 2 the complement of their union is in G. Thus GA satisfies 1 - 3 and this implies since G is the smallest such, that GA ⊇ G. However, GA is constructed as a subset of G. This proves that for every B ∈ G and A ∈ K, A ∩ B ∈ G. Now pick B ∈ G and consider GB ≡ {A ∈ G : A ∩ B ∈ G} . I just proved K ⊆ GB . The other arguments are identical to show GB satisfies 1 - 3 and is therefore equal to G. This shows that whenever A, B ∈ G it follows A∩B ∈ G. This implies G is a σ algebra. To show this, all that is left is to verify G is closed under countable unions because then it follows G is a σ algebra. Let {Ai } ⊆ G. Then let A01 = A1 and A0n+1

≡ = =

An+1 \ (∪ni=1 Ai ) ¡ ¢ An+1 ∩ ∩ni=1 AC i ¡ ¢ ∩ni=1 An+1 ∩ AC ∈G i

because finite intersections of sets of G are in G. Since the A0i are disjoint, it follows ∞ 0 ∪∞ i=1 Ai = ∪i=1 Ai ∈ G

Therefore, G ⊇ σ (K) and this proves the Lemma. With this lemma, it is easy to define product measure.


277

Let (X, F, µ) and (Y, S, ν) be two finite measure spaces. Define K to be the set of measurable rectangles, A × B, A ∈ F and B ∈ S. Let ½ ¾ Z Z Z Z G ≡ E ⊆X ×Y : XE dµdν = XE dνdµ (9.11.46) Y

X

X

Y

where in the above, part of the requirement is for all integrals to make sense. Then K ⊆ G. This is obvious. Next I want to show that if E ∈ G then E C ∈ G. Observe XE C = 1 − XE and so Z Z Z Z (1 − XE ) dµdν XE C dµdν = Y X ZY ZX = (1 − XE ) dνdµ ZX ZY = XE C dνdµ X

Y

C

which shows that if E ∈ G, then E ∈ G. Next I want to show G is closed under countable unions of disjoint sets of G. Let {Ai } be a sequence of disjoint sets from G. Then Z Z Z Z X ∞ X∪∞ dµdν = XAi dµdν i=1 Ai Y

X

Y

= = =

X i=1

Z X ∞ Z Y i=1 ∞ Z X

X

XAi dµdν

Z

X

Y

Z X ∞ Z Y

X i=1

=

XAi dµdν

Z

i=1 Y ∞ Z X i=1

=

X

Z Z X ∞ X

Y i=1

X

Y

XAi dνdµ XAi dνdµ XAi dνdµ

Z Z =

X∪∞ dνdµ, i=1 Ai

(9.11.47)

the interchanges between the summation and the integral depending on the monotone convergence theorem. Thus G is closed with respect to countable disjoint unions. From Lemma 9.11.3, G ⊇ σ (K) . Also the computation in 9.11.47 implies that on σ (K) one can define a measure, denoted by µ × ν and that for every E ∈ σ (K) , Z Z Z Z (µ × ν) (E) = XE dµdν = XE dνdµ. (9.11.48) Y

X

X

Y

278


Now here is Fubini’s theorem. Theorem 9.11.4 Let f : X × Y → [0, ∞] be measurable with respect to the σ algebra, σ (K) just defined and let µ × ν be the product measure of 9.11.48 where µ and ν are finite measures on (X, F) and (Y, S) respectively. Then Z Z Z Z Z f d (µ × ν) = f dµdν = f dνdµ. X×Y

Y

X

X

Y

Proof: Let {sn } be an increasing sequence of σ (K) measurable simple functions which converges pointwise to f. The above equation holds for sn in place of f from what was shown above. The final result follows from passing to the limit and using the monotone convergence theorem. This proves the theorem. The symbol, F × S denotes σ (K). Of course one can generalize right away to measures which are only σ finite. Theorem 9.11.5 Let f : X × Y → [0, ∞] be measurable with respect to the σ algebra, σ (K) just defined and let µ × ν be the product measure of 9.11.48 where µ and ν are σ finite measures on (X, F) and (Y, S) respectively. Then Z Z Z Z Z f d (µ × ν) = f dµdν = f dνdµ. X×Y

Y

X

X

Y

Proof: Since the measures are σ finite, there exist increasing sequences of sets, {Xn } and {Yn } such that µ (Xn ) < ∞ and ν (Yn ) < ∞. Then µ and ν restricted to Xn and Yn respectively are finite. Then from Theorem 9.11.4, Z Z Z Z f dµdν = f dνdµ Yn

Xn

Xn

Yn

Passing to the limit yields Z Z

Z Z f dµdν =

Y

X

f dνdµ X

Y

whenever f is as above. In particular, you could take f = XE where E ∈ F × S and define Z Z Z Z (µ × ν) (E) ≡ XE dµdν = XE dνdµ. Y

X

X

Y

Then just as in the proof of Theorem 9.11.4, the conclusion of this theorem is obtained. This proves the theorem. Qn It is also useful to note that all the above holds for i=1 Xi in place of X × Y. You would simply modify the definition of G in 9.11.46 including allQpermutations n for the iterated integrals and for K you would use sets of the form i=1 Ai where Ai is measurable. Everything goes through exactly as above. Thus the following is obtained.


279

Qn n Theorem 9.11.6 Let {(Xi , Fi , µi )}i=1 be σ finite measure spaces and let i=1 Fi denote the smallest σ algebra which contains the measurable boxesQof the form Qn n there exists a measure, λ defined on i=1 Fi such i=1 Ai where Qn Ai ∈ Fi . Then Q n that if f : i=1 Xi → [0, ∞] is i=1 Fi measurable, and (i1 , · · · , in ) is any permutation of (1, · · · , n) , then Z Z Z f dλ = ··· f dµi1 · · · dµin Xin

9.11.2

Xi1

Completion Of Product Measure Spaces

Using Theorem 9.10.2 it is easy to give a generalization to yield a theorem for the completion of product spaces. Qn n Theorem 9.11.7 Let {(Xi , Fi , µi )}i=1 be σ finite measure spaces and let i=1 Fi denote the smallest σ algebra which contains the measurable boxesQof the form Qn n there exists a measure, λ defined on i=1 Fi such i=1 Ai where Qn Ai ∈ Fi . Then Q n that if f : i=1 Xi → [0, ∞] is i=1 Fi measurable, and (i1 , · · · , in ) is any permutation of (1, · · · , n) , then Z Z Z f dλ = ··· f dµi1 · · · dµin Xin

Let let

³Q

n i=1

Xi ,

Qn i=1

Xi1

´ Fi , λ denote the completion of this product measure space and f:

n Y

Xi → [0, ∞]

i=1

Qn Qn Fi such that λ (N ) = 0 and a be i=1 Fi measurable. Then there exists N ∈ i=1 Q n nonnegative function, f1 measurable with respect to i=1 Fi such that f1 = f off N and if (i1 , · · · , in ) is any permutation of (1, · · · , n) , then Z Z Z f dλ = ··· f1 dµi1 · · · dµin . Xin

Xi1

Furthermore, f1 may be chosen to satisfy either f1 ≤ f or f1 ≥ f. Proof: This follows immediately from Theorem 9.11.6 and Theorem 9.10.2. By the second theorem, there Qn exists a function f1 ≥ f such that f1 = f for all (x1 , · · · , xn ) ∈ / N, a set of i=1 Fi having measure zero. Then by Theorem 9.10.1 and Theorem 9.11.6 Z Z Z Z f dλ = f1 dλ = ··· f1 dµi1 · · · dµin . Xin

Xi1

To get f1 ≤ f, just use that part of Theorem 9.10.2.

280


Since f1 = f off a set of measure zero, I will dispense with the subscript. Also it is customary to write λ = µ1 × · · · × µn and λ = µ1 × · · · × µn . Thus in more standard notation, one writes Z Z Z f d (µ1 × · · · × µn ) = ··· Xin

Xi1

f dµi1 · · · dµin

This theorem is often referred to as Fubini’s theorem. The next theorem is also called this. ³Q ´ Qn n Corollary 9.11.8 Suppose f ∈ L1 where each i=1 Xi , i=1 Fi , µ1 × · · · × µn Xi is a σ finite measure space. Then if (i1 , · · · , in ) is any permutation of (1, · · · , n) , it follows Z Z Z f d (µ1 × · · · × µn ) = ··· f dµi1 · · · dµin . Xin

Xi1

Proof: Just apply Theorem 9.11.7 to the positive and negative parts of the real and imaginary parts of f. This proves the theorem. Here is another easy corollary. Corollary 9.11.9 Suppose in the situation of Corollary 9.11.8, f = f1 off N, a set Qn of i=1 Fi having µ1 ×· · ·×µQ n measure zero and that f1 is a complex valued function n measurable with respect to i=1 Fi . Suppose also that for some permutation of (1, 2, · · · , n) , (j1 , · · · , jn ) Z Z ··· |f1 | dµj1 · · · dµjn < ∞. Xjn

Then

Xj1

Ã f ∈L

1

n Y

Xi ,

i=1

n Y

! Fi , µ1 × · · · × µn

i=1

and the conclusion of Corollary 9.11.8 holds. Qn Proof: Since |f1 | is i=1 Fi measurable, it follows from Theorem 9.11.6 that Z Z ∞ > ··· |f1 | dµj1 · · · dµjn Xjn

Z =

Xj1

|f1 | d (µ1 × · · · × µn ) Z |f1 | d (µ1 × · · · × µn )

= Z =

|f | d (µ1 × · · · × µn ) .

9.12. DISTURBING EXAMPLES

281

³Q ´ Qn n Thus f ∈ L1 X , F , µ × · · · × µ i i 1 n as claimed and the rest follows from i=1 i=1 Corollary 9.11.8. This proves the corollary. The following lemma is also useful. Lemma 9.11.10 Let (X, F, µ) and (Y, S, ν) be σ finite complete measure spaces and suppose f ≥ 0 is F × S measurable. Then for a.e. x, y → f (x, y) is S measurable. Similarly for a.e. y, x → f (x, y) is F measurable. Proof: By Theorem 9.10.2, there exist F × S measurable functions, g and h and a set, N ∈ F × S of µ × λ measure zero such that g ≤ f ≤ h and for (x, y) ∈ / N, it follows that g (x, y) = h (x, y) . Then Z Z Z Z gdνdµ = hdνdµ X

Y

and so for a.e. x,

X

Z

Y

Z gdν = Y

hdν. Y

Then it follows that for these values of x, g (x, y) = h (x, y) and so by Theorem 9.10.2 again and the assumption that (Y, S, ν) is complete, y → f (x, y) is S measurable. The other claim is similar. This proves the lemma.

9.12

Disturbing Examples

There are examples which help to define what can be expected of product measures and Fubini type theorems. Three such examples are given in Rudin [58]. Some of the theorems given above are more general than those in this reference but the same examples are still useful for showing that the hypotheses of the above theorems are all necessary. Example 9.12.1 Let {an } be an increasing sequence R of numbers in (0, 1) which converges to 1. Let gn ∈ Cc (an , an+1 ) such that gn dx = 1. Now for (x, y) ∈ [0, 1) × [0, 1) define f (x, y) ≡

∞ X

gn (y) (gn (x) − gn+1 (x)) .

k=1

Note this is actually a finite sum for each such (x, y) . Therefore, this is a continuous function on [0, 1) × [0, 1). Now for a fixed y, Z 1 Z 1 ∞ X gn (y) (gn (x) − gn+1 (x)) dx = 0 f (x, y) dx = 0

k=1

0

282


showing that Z

R1R1 0

0

f (x, y) dxdy =

1

f (x, y) dy = 0

∞ X

R1 0

0dy = 0. Next fix x. Z

1

gn (y) dy = g1 (x) .

(gn (x) − gn+1 (x)) 0

k=1

R1 R1R1 Hence 0 0 f (x, y) dydx = 0 g1 (x) dx = 1. The iterated integrals are not equal. Note theR function, g is not nonnegative R 1 R 1 even though it is measurable. In addition, 1R1 neither 0 0 |f (x, y)| dxdy nor 0 0 |f (x, y)| dydx is finite and so you can’t apply Corollary 9.8.13. The problem here is the function is not nonnegative and is not absolutely integrable. Example 9.12.2 This time let µ = m, Lebesgue measure on [0, 1] and let ν be counting measure on [0, 1] , in this case, the σ algebra is P ([0, 1]) . Let l denote the line segment in [0, 1] × [0, 1] which goes from (0, 0) to (1, 1). Thus l = (x, x) where x ∈ [0, 1] . Consider the outer measure of l in m × ν. Let l ⊆ ∪k Ak × Bk where Ak is Lebesgue measurable and Bk is a subset of [0, 1] . Let B ≡ {k ∈ N : ν (Bk ) = ∞} . If m (∪k∈B Ak ) has measure zero, then there are uncountably many points of [0, 1] outside of ∪k∈B Ak . For p one of these points, (p, p) ∈ Ai × Bi and i ∈ / B. Thus each of these points is in ∪i∈B B , a countable set because these B are each finite. But i i / this is a contradiction because there need to be uncountably many of these points as just indicated. Thus m (Ak ) > 0 for some k ∈ B and so mR× ν (Ak × Bk ) = ∞. It follows m × ν (l) = ∞ and so l is m × ν measurable. Thus Xl (x, y) d m × ν = ∞ and so you cannot apply Fubini’s theorem, Theorem 9.8.11. Since ν is not σ finite, you cannot apply the corollary to this theorem either. Thus there is no contradiction to the above theorems in the following observation. Z Z Z Z Z Z Xl (x, y) dνdm = 1dm = 1, Xl (x, y) dmdν = 0dν = 0. The problem here is that you have neither spaces.

R

f d m × ν < ∞ not σ finite measure

The next example is far more exotic. It concerns the case where both iterated integrals make perfect sense but are unequal. In 1877 Cantor conjectured that the cardinality of the real numbers is the next size of infinity after countable infinity. This hypothesis is called the continuum hypothesis and it has never been proved or disproved4 . Assuming this continuum hypothesis will provide the basis for the following example. It is due to Sierpinski. Example 9.12.3 Let X be an uncountable set. It follows from the well ordering theorem which says every set can be well ordered which is presented in the appendix that X can be well ordered. Let ω ∈ X be the first element of X which is 4 In 1940 it was shown by Godel that the continuum hypothesis cannot be disproved. In 1963 it was shown by Cohen that the continuum hypothesis cannot be proved. These assertions are based on the axiom of choice and the Zermelo Frankel axioms of set theory. This topic is far outside the scope of this book and this is only a hopefully interesting historical observation.

9.13. EXERCISES

283

preceded by uncountably many points of X. Let Ω denote {x ∈ X : x < ω} . Then Ω is uncountable but there is no smaller uncountable set. Thus by the continuum hypothesis, there exists a one to one and onto mapping, j which maps [0, 1] onto nΩ. Thus, for x ∈ [0, 1] , j (x) o is preceeded by countably many points. Let 2 Q ≡ (x, y) ∈ [0, 1] : j (x) < j (y) and let f (x, y) = XQ (x, y) . Then Z

Z

1

1

f (x, y) dy = 1, 0

f (x, y) dx = 0 0

In each case, the integrals make sense. In the first, for fixed x, f (x, y) = 1 for all but countably many y so the function of y is Borel measurable. In the second where y is fixed, f (x, y) = 0 for all but countably many x. Thus Z 1Z 1 Z 1Z 1 f (x, y) dydx = 1, f (x, y) dxdy = 0. 0

0

0

0

The problem here must be that f is not m × m measurable.

9.13

Exercises

1. Let Ω = N, the natural numbers and let d (p, q) = |p − q|, the usual distance in R. Show that (Ω, d) the closures of the balls are compact. Now let P∞ Λf ≡ k=1 f (k) whenever f ∈ Cc (Ω). Show this is a well defined positive linear functional on the space Cc (Ω). Describe the measure of the Riesz representation theorem which results from this positive linear functional. What if Λ (f ) = f (1)? What measure would result from this functional? Which functions are measurable? 2. Verify that µ defined in Lemma 9.1.7 is an outer measure. R 3. Let F : R → R be increasing and right continuous. Let Λf ≡ f dF where the integral is the Riemann Stieltjes integral of f . Show the measure µ from the Riesz representation theorem satisfies µ ([a, b]) µ ([a, a])

= =

F (b) − F (a−) , µ ((a, b]) = F (b) − F (a) , F (a) − F (a−) .

4. Let Ω be a metric space with the closed balls compact and suppose µ is a measure defined on the Borel sets of Ω which is finite on compact sets. Show there exists a unique Radon measure, µ which equals µ on the Borel sets. 5. ↑ Random vectors are measurable functions, X, mapping a probability space, (Ω, P, F) to Rn . Thus X (ω) ∈ Rn for each ω ∈ Ω and P is a probability measure defined on the sets of F, a σ algebra of subsets of Ω. For E a Borel set in Rn , define ¡ ¢ µ (E) ≡ P X−1 (E) ≡ probability that X ∈ E.

284


Show this is a well defined measure on the Borel sets of Rn and use Problem 4 to obtain a Radon measure, λX defined on a σ algebra of sets of Rn including the Borel sets such that for E a Borel set, λX (E) =Probability that (X ∈E). 6. Suppose X and Y are metric spaces having compact closed balls. Show (X × Y, dX×Y ) is also a metric space which has the closures of balls compact. Here dX×Y ((x1 , y1 ) , (x2 , y2 )) ≡ max (d (x1 , x2 ) , d (y1 , y2 )) . Let A ≡ {E × F : E is a Borel set in X, F is a Borel set in Y } . Show σ (A), the smallest σ algebra containing A contains the Borel sets. Hint: Show every open set in a metric space which has closed balls compact can be obtained as a countable union of compact sets. Next show this implies every open set can be obtained as a countable union of open sets of the form U × V where U is open in X and V is open in Y . 7. Suppose (Ω, S, µ) is a measure space ¡ which¢may not be complete. Could you obtain a complete measure space, Ω, S, µ1 by simply letting S consist of all sets of the form E where there exists F ∈ S such that (F \ E) ∪ (E \ F ) ⊆ N for some N ∈ S which has measure zero and then let µ (E) = µ1 (F )? 8. If µ and ν are Radon measures defined on Rn and Rm respectively, show µ × ν is also a radon measure on Rn+m . Hint: Show the µ × ν measurable sets include the open sets using the observation that every open set in Rn+m is the countable union of sets of the form U × V where U and V are open in Rn and Rm respectively. Next verify outer regularity by considering A × B for A, B measurable. Argue sets of R defined above have the property that they can be approximated in measure from above by open sets. Then verify the same is true of sets of R1 . Finally conclude using an appropriate lemma that µ × ν is inner regular as well. 9. Let (Ω, S, µ) be a σ finite measure space and let f : Ω → [0, ∞) be measurable. Define A ≡ {(x, y) : y < f (x)} Verify that A is µ × m measurable. Show that Z Z Z Z f dµ = XA (x, y) dµdm = XA dµ × m.

Lebesgue Measure 10.1

Basic Properties

Definition 10.1.1 Define the following positive linear functional for f ∈ Cc (Rn ) . Z ∞ Z ∞ Λf ≡ ··· f (x) dx1 · · · dxn . −∞

−∞

Then the measure representing this functional is Lebesgue measure. The following lemma will help in understanding Lebesgue measure. Lemma 10.1.2 Every open set in Rn is the countable disjoint union of half open boxes of the form n Y (ai , ai + 2−k ] i=1 −k

where ai = l2 for some integers, l, k. The sides of these boxes are of equal length. One could also have half open boxes of the form n Y

[ai , ai + 2−k )

i=1

and the conclusion would be unchanged. Proof: Let Ck = {All half open boxes

n Y

(ai , ai + 2−k ] where

i=1

ai = l2−k for some integer l.} Thus Ck consists of a countable disjoint collection of boxes whose union is Rn . This is sometimes called a tiling of Rn . Think of tiles on the floor of a bathroom and 285

286

LEBESGUE MEASURE

√ you will get the idea. Note that each box has diameter no larger than 2−k n. This is because if n Y (ai , ai + 2−k ], x, y ∈ i=1 −k

then |xi − yi | ≤ 2

. Therefore, Ã n !1/2 X¡ ¢ √ 2 |x − y| ≤ 2−k = 2−k n. i=1

Let U be open and let B1 ≡ all sets of C1 which are contained in U . If B1 , · · · , Bk have been chosen, Bk+1 ≡ all sets of Ck+1 contained in ¡ ¢ U \ ∪ ∪ki=1 Bi . Let B∞ = ∪∞ i=1 Bi . In fact ∪B∞ = U . Clearly ∪B∞ ⊆ U because every box of every Bi is contained in U . If p ∈ U , let k be the smallest integer such that p is contained in a box from Ck which is also a subset of U . Thus p ∈ ∪Bk ⊆ ∪B∞ . Hence B∞ is the desired countable disjoint collection of half open boxes whose union is U . The last assertion about the other type of half open rectangle is obvious. This proves the lemma. Qn Now what does Lebesgue measure do to a rectangle, i=1 (ai , bi ]? Qn Qn Lemma 10.1.3 Let R = i=1 [ai , bi ], R0 = i=1 (ai , bi ). Then mn (R0 ) = mn (R) =

n Y

(bi − ai ).

i=1

Proof: Let k be large enough that ai + 1/k < bi − 1/k for i = 1, · · · , n and consider functions gik and fik having the following graphs.

1

ai + ££ @£ R @ £ ai 1 k

fik B bi − B ¡ B ª ¡ B bi

Let g k (x) =

1 k

1

ai − k1 @

@ R£ @

n Y i=1

£

£ £

gik (xi ), f k (x) =

B ai n Y i=1

gik B

bi

fik (xi ).

bi + ¡ B ¡ B ª

1 k

10.1. BASIC PROPERTIES

287

Then by elementary calculus along with the definition of Λ, n Y

Z (bi − ai + 2/k) ≥ Λg k =

g k dmn ≥ mn (R) ≥ mn (R0 )

i=1

Z ≥

f k dmn = Λf k ≥

n Y

(bi − ai − 2/k).

i=1

Letting k → ∞, it follows that mn (R) = mn (R0 ) =

n Y

(bi − ai ).

i=1

This proves the lemma. Lemma 10.1.4 Let U be an open or closed set. Then mn (U ) = mn (x + U ) . Proof: By Lemma 10.1.2 there is a sequence of disjoint half open rectangles, {Ri } such that ∪i Ri = U. Therefore, x + U = ∪i (x + Ri ) and the x + Ri are also disjoint rectangles which arePidentical to the Ri but translated. From Lemma P 10.1.3, mn (U ) = i mn (Ri ) = i mn (x + Ri ) = mn (x + U ) . It remains to verify the lemma for a closed set. Let H be a closed bounded set first. Then H ⊆ B (0,R) for some R large enough. First note that x + H is a closed set. Thus mn (B (x, R)) = = = = =

mn (x + H) + mn ((B (0, R) + x) \ (x + H)) mn (x + H) + mn ((B (0, R) \ H) + x) mn (x + H) + mn ((B (0, R) \ H)) mn (B (0, R)) − mn (H) + mn (x + H) mn (B (x, R)) − mn (H) + mn (x + H)

the last equality because of the first part of the lemma which implies mn (B (x, R)) = mn (B (0, R)) . Therefore, mn (x + H) = mn (H) as claimed. If H is not bounded, consider Hm ≡ B (0, m) ∩ H. Then mn (x + Hm ) = mn (Hm ) . Passing to the limit as m → ∞ yields the result in general. Theorem 10.1.5 Lebesgue measure is translation invariant. That is mn (E) = mn (x + E) for all E Lebesgue measurable. Proof: Suppose mn (E) < ∞. By regularity of the measure, there exist sets G, H such that G is a countable intersection of open sets, H is a countable union of compact sets, mn (G \ H) = 0, and G ⊇ E ⊇ H. Now mn (G) = mn (G + x) and

288

LEBESGUE MEASURE

mn (H) = mn (H + x) which follows from Lemma 10.1.4 applied to the sets which are either intersected to form G or unioned to form H. Now x+H ⊆x+E ⊆x+G and both x + H and x + G are measurable because they are either countable unions or countable intersections of measurable sets. Furthermore, mn (x + G \ x + H) = mn (x + G) − mn (x + H) = mn (G) − mn (H) = 0 and so by completeness of the measure, x + E is measurable. It follows mn (E) = ≤

mn (H) = mn (x + H) ≤ mn (x + E) mn (x + G) = mn (G) = mn (E) .

If mn (E) is not necessarily less than ∞, consider Em ≡ B (0, m) ∩ E. Then mn (Em ) = mn (Em + x) by the above. Letting m → ∞ it follows mn (E) = mn (E + x). This proves the theorem. Corollary 10.1.6 Let D be an n × n diagonal matrix and let U be an open set. Then mn (DU ) = |det (D)| mn (U ) . Proof: If any of the diagonal entries of D equals 0 there is nothing to prove because then both sides equal zero. Therefore, it can be assumed none are equal to zero. Suppose these diagonal entries are k1 , · · · , kn . From Lemma 10.1.2 there exist half open boxes, i . Suppose Qi R Qn{Ri } having all sides equal such that U = ∪ n one of these is Ri = j=1 (aj , bj ], where bj − aj = li . Then DRi = j=1 Ij where Ij = (kj aj , kj bj ] if kj > 0 and Ij = [kj bj , kj aj ) if kj < 0. Then the rectangles, DRi are disjoint because D is one to one and their union is DU. Also, mn (DRi ) =

n Y

|kj | li = |det D| mn (Ri ) .

j=1

Therefore, mn (DU ) =

∞ X i=1

mn (DRi ) = |det (D)|

∞ X

mn (Ri ) = |det (D)| mn (U ) .

i=1

and this proves the corollary. From this the following corollary is obtained. Corollary 10.1.7 Let M > 0. Then mn (B (a, M r)) = M n mn (B (0, r)) . Proof: By Lemma 10.1.4 there is no loss of generality in taking a = 0. Let D be the diagonal matrix which has M in every entry of the main diagonal so |det (D)| = M n . Note that DB (0, r) = B (0, M r) . By Corollary 10.1.6 mn (B (0, M r)) = mn (DB (0, r)) = M n mn (B (0, r)) .

10.2. THE VITALI COVERING THEOREM

289

There are many norms on Rn . Other common examples are ||x||∞ ≡ max {|xk | : x = (x1 , · · · , xn )} or ||x||p ≡

Ã n X

!1/p |xi |

p

.

i=1

With ||·|| any norm for Rn you can define a corresponding ball in terms of this norm. B (a, r) ≡ {x ∈ Rn such that ||x − a|| < r} It follows from general considerations involving metric spaces presented earlier that these balls are open sets. Therefore, Corollary 10.1.7 has an obvious generalization. Corollary 10.1.8 Let ||·|| be a norm on Rn . Then for M > 0, mn (B (a, M r)) = M n mn (B (0, r)) where these balls are defined in terms of the norm ||·||.

10.2

The Vitali Covering Theorem

The Vitali covering theorem is concerned with the situation in which a set is contained in the union of balls. You can imagine that it might be very hard to get disjoint balls from this collection of balls which would cover the given set. However, it is possible to get disjoint balls from this collection of balls which have the property that if each ball is enlarged appropriately, the resulting enlarged balls do cover the set. When this result is established, it is used to prove another form of this theorem in which the disjoint balls do not cover the set but they only miss a set of measure zero. Recall the Hausdorff maximal principle, Theorem 1.4.2 on Page 26 which is proved to be equivalent to the axiom of choice in the appendix. For convenience, here it is: Theorem 10.2.1 (Hausdorff Maximal Principle) Let F be a nonempty partially ordered set. Then there exists a maximal chain. I will use this Hausdorff maximal principle to give a very short and elegant proof of the Vitali covering theorem. This follows the treatment in Evans and Gariepy [24] which they got from another book. I am not sure who first did it this way but it is very nice because it is so short. In the following lemma and theorem, the balls will be either open or closed and determined by some norm on Rn . When pictures are drawn, I shall draw them as though the norm is the usual norm but the results are unchanged for any norm. Also, I will write (in this section only) B (a, r) to indicate a set which satisfies {x ∈ Rn : ||x − a|| < r} ⊆ B (a, r) ⊆ {x ∈ Rn : ||x − a|| ≤ r} b (a, r) to indicate the usual ball but with radius 5 times as large, and B {x ∈ Rn : ||x − a|| < 5r} .

290

LEBESGUE MEASURE

Lemma 10.2.2 Let ||·|| be a norm on Rn and let F be a collection of balls determined by this norm. Suppose ∞ > M ≡ sup{r : B(p, r) ∈ F} > 0 and k ∈ (0, ∞) . Then there exists G ⊆ F such that if B(p, r) ∈ G then r > k,

(10.2.1)

if B1 , B2 ∈ G then B1 ∩ B2 = ∅,

(10.2.2)

G is maximal with respect to 10.2.1 and 10.2.2. Note that if there is no ball of F which has radius larger than k then G = ∅. Proof: Let H = {B ⊆ F such that 10.2.1 and 10.2.2 hold}. If there are no balls with radius larger than k then H = ∅ and you let G =∅. In the other case, H 6= ∅ because there exists B(p, r) ∈ F with r > k. In this case, partially order H by set inclusion and use the Hausdorff maximal principle (see the appendix on set theory) to let C be a maximal chain in H. Clearly ∪C satisfies 10.2.1 and 10.2.2 because if B1 and B2 are two balls from ∪C then since C is a chain, it follows there is some element of C, B such that both B1 and B2 are elements of B and B satisfies 10.2.1 and 10.2.2. If ∪C is not maximal with respect to these two properties, then C was not a maximal chain because then there would exist B ! ∪C, that is, B contains C as a proper subset and {C, B} would be a strictly larger chain in H. Let G = ∪C. Theorem 10.2.3 (Vitali) Let F be a collection of balls and let A ≡ ∪{B : B ∈ F}. Suppose ∞ > M ≡ sup{r : B(p, r) ∈ F } > 0. Then there exists G ⊆ F such that G consists of disjoint balls and b : B ∈ G}. A ⊆ ∪{B Proof: Using Lemma 10.2.2, there exists G1 ⊆ F ≡ F0 which satisfies B(p, r) ∈ G1 implies r >

M , 2

B1 , B2 ∈ G1 implies B1 ∩ B2 = ∅, G1 is maximal with respect to 10.2.3, and 10.2.4. Suppose G1 , · · · , Gm have been chosen, m ≥ 1. Let Fm ≡ {B ∈ F : B ⊆ Rn \ ∪{G1 ∪ · · · ∪ Gm }}.

(10.2.3) (10.2.4)

10.3. THE VITALI COVERING THEOREM (ELEMENTARY VERSION)

291

Using Lemma 10.2.2, there exists Gm+1 ⊆ Fm such that B(p, r) ∈ Gm+1 implies r >

M , 2m+1

B1 , B2 ∈ Gm+1 implies B1 ∩ B2 = ∅,

(10.2.5) (10.2.6)

Gm+1 is a maximal subset of Fm with respect to 10.2.5 and 10.2.6. Note it might be the case that Gm+1 = ∅ which happens if Fm = ∅. Define G ≡ ∪∞ k=1 Gk . b : B ∈ G} covers A. Thus G is a collection of disjoint balls in F. I must show {B Let x ∈ B(p, r) ∈ F and let M M < r ≤ m−1 . 2m 2 Then B (p, r) must intersect some set, B (p0 , r0 ) ∈ G1 ∪ · · · ∪ Gm since otherwise, Gm would fail to be maximal. Then r0 > 2Mm because all balls in G1 ∪ · · · ∪ Gm satisfy this inequality.

¾

r0

r p0

r

xr rp ?

Then for x ∈ B (p, r) , the following chain of inequalities holds because r ≤ and r0 > 2Mm

M 2m−1

|x − p0 | ≤

|x − p| + |p − p0 | ≤ r + r0 + r 2M 4M ≤ + r0 = m + r0 < 5r0 . 2m−1 2

b (p0 , r0 ) and this proves the theorem. Thus B (p, r) ⊆ B

10.3

The Vitali Covering Theorem (Elementary Version)

The proof given here is from Basic Analysis [45]. It first considers the case of open balls and then generalizes to balls which may be neither open nor closed or closed.

292

LEBESGUE MEASURE

Lemma 10.3.1 Let F be a countable collection of balls satisfying ∞ > M ≡ sup{r : B(p, r) ∈ F} > 0 and let k ∈ (0, ∞) . Then there exists G ⊆ F such that If B(p, r) ∈ G then r > k,

(10.3.7)

If B1 , B2 ∈ G then B1 ∩ B2 = ∅,

(10.3.8)

G is maximal with respect to 10.3.7 and 10.3.8.

(10.3.9)

Proof: If no ball of F has radius larger than k, let G = ∅. Assume therefore, that ∞ some balls have radius larger than k. Let F ≡ {Bi }i=1 . Now let Bn1 be the first ball in the list which has radius greater than k. If every ball having radius larger than k intersects this one, then stop. The maximal set is just Bn1 . Otherwise, let Bn2 be the next ball having radius larger than k which is disjoint from Bn1 . Continue ∞ this way obtaining {Bni }i=1 , a finite or infinite sequence of disjoint balls having radius larger than k. Then let G ≡ {Bni }. To see that G is maximal with respect to 10.3.7 and 10.3.8, suppose B ∈ F, B has radius larger than k, and G ∪ {B} satisfies 10.3.7 and 10.3.8. Then at some point in the process, B would have been chosen because it would be the ball of radius larger than k which has the smallest index. Therefore, B ∈ G and this shows G is maximal with respect to 10.3.7 and 10.3.8. e the open ball, For the next lemma, for an open ball, B = B (x, r) , denote by B B (x, 4r) . Lemma 10.3.2 Let F be a collection of open balls, and let A ≡ ∪ {B : B ∈ F} . Suppose ∞ > M ≡ sup {r : B(p, r) ∈ F } > 0. Then there exists G ⊆ F such that G consists of disjoint balls and e : B ∈ G}. A ⊆ ∪{B Proof: Without loss of generality assume F is countable. This is because there is a countable subset of F, F 0 such that ∪F 0 = A. To see this, consider the set of balls having rational radii and centers having all components rational. This is a countable set of balls and you should verify that every open set is the union of balls of this form. Therefore, you can consider the subset of this set of balls consisting of those which are contained in some open set of F, G so ∪G = A and use the axiom of choice to define a subset of F consisting of a single set from F containing each set of G. Then this is F 0 . The union of these sets equals A . Then consider F 0 instead of F. Therefore, assume at the outset F is countable. By Lemma 10.3.1, there exists G1 ⊆ F which satisfies 10.3.7, 10.3.8, and 10.3.9 with k = 2M 3 .

10.3. THE VITALI COVERING THEOREM (ELEMENTARY VERSION)

293

Suppose G1 , · · · , Gm−1 have been chosen for m ≥ 2. Let union of the balls in these Gj

Fm = {B ∈ F : B ⊆ Rn \

z }| { ∪{G1 ∪ · · · ∪ Gm−1 } }

and using Lemma 10.3.1, let Gm be a maximal collection¡ of¢ disjoint balls from Fm m M. Let G ≡ ∪∞ with the property that each ball has radius larger than 23 k=1 Gk . Let x ∈ B (p, r) ∈ F. Choose m such that µ ¶m µ ¶m−1 2 2 M M. 3 Consider the picture, in which w ∈ B (p0 , r0 ) ∩ B (p, r) .

¾

r0 p0

w ·

r

·x p ?

Then 0. Then there exists G ⊆ F such that G consists of disjoint balls and b : B ∈ G}. A ⊆ ∪{B Proof: For ¢ B one of these balls, say B (x, r) ⊇ B ⊇ B (x, r), denote by B1 , the ¡ ball B x, 5r 4 . Let F1 ≡ {B1 : B ∈ F } and let A1 denote the union of the balls in F1 . Apply Lemma 10.3.2 to F1 to obtain f1 : B1 ∈ G1 } A1 ⊆ ∪{B where G1 consists of disjoint balls from F1 . Now let G ≡ {B ∈ F : B1 ∈ G1 }. Thus G consists of disjoint balls from F because they are contained in the disjoint open balls, G1 . Then f1 : B1 ∈ G1 } = ∪{B b : B ∈ G} A ⊆ A1 ⊆ ∪{B ¡ ¢ f b because for B1 = B x, 5r 4 , it follows B1 = B (x, 5r) = B. This proves the theorem.

10.4

Vitali Coverings

There is another version of the Vitali covering theorem which is also of great importance. In this one, balls from the original set of balls almost cover the set,leaving out only a set of measure zero. It is like packing a truck with stuff. You keep trying to fill in the holes with smaller and smaller things so as to not waste space. It is remarkable that you can avoid wasting any space at all when you are dealing with balls of any sort provided you can use arbitrarily small balls. Definition 10.4.1 Let F be a collection of balls that cover a set, E, which have the property that if x ∈ E and ε > 0, then there exists B ∈ F, diameter of B < ε and x ∈ B. Such a collection covers E in the sense of Vitali. In the following covering theorem, mn denotes the outer measure determined by n dimensional Lebesgue measure. Theorem 10.4.2 Let E ⊆ Rn and suppose 0 < mn (E) < ∞ where mn is the outer measure determined by mn , n dimensional Lebesgue measure, and let F be a collection of closed balls of bounded radii such that F covers E in the sense of Vitali. Then there exists a countable collection of disjoint balls from F, {Bj }∞ j=1 , B ) = 0. such that mn (E \ ∪∞ j j=1

10.4. VITALI COVERINGS

295

Proof: From the definition of outer measure there exists a Lebesgue measurable set, E1 ⊇ E such that mn (E1 ) = mn (E). Now by outer regularity of Lebesgue measure, there exists U , an open set which satisfies mn (E1 ) > (1 − 10−n )mn (U ), U ⊇ E1 .

U

E1

Each point of E is contained in balls of F of arbitrarily small radii and so there exists a covering of E with balls of F which are themselves contained in U . ∞ Therefore, by the Vitali covering theorem, there exist disjoint balls, {Bi }i=1 ⊆ F such that b E ⊆ ∪∞ j=1 Bj , Bj ⊆ U. Therefore, mn (E1 ) = =

´ X ³ ´ ³ bj ≤ b B m mn (E) ≤ mn ∪∞ n Bj j=1 5

n

X

j

¡ ¢ mn (Bj ) = 5 mn ∪∞ j=1 Bj n

j

Then E1 and ∪∞ j=1 Bj are contained in U and so mn (E1 ) > (1 − 10−n )mn (U ) ∞ ≥ (1 − 10−n )[mn (E1 \ ∪∞ j=1 Bj ) + mn (∪j=1 Bj )] =mn (E1 )

z }| { −n ≥ (1 − 10−n )[mn (E1 \ ∪∞ mn (E) ]. j=1 Bj ) + 5 and so ¡

¡ ¢ ¢ 1 − 1 − 10−n 5−n mn (E1 ) ≥ (1 − 10−n )mn (E1 \ ∪∞ j=1 Bj )

which implies mn (E1 \ ∪∞ j=1 Bj ) ≤

(1 − (1 − 10−n ) 5−n ) mn (E1 ) (1 − 10−n )

296

LEBESGUE MEASURE

Now a short computation shows 0
0 be given. Now by outer regularity, there exists an open set, V , containing Tk which is contained in U such that mn (V ) < ε. Let x ∈ Tk . Then by differentiability, h (x + v) = h (x) + Dh (x) v + o (v) and so there exist arbitrarily small rx < 1 such that B (x,5rx ) ⊆ V and whenever |v| ≤ rx , |o (v)| < k |v| . Thus h (B (x, rx )) ⊆ B (h (x) , 2krx ) . From the Vitali covering theorem there exists a countable disjoint sequence of n o∞ ∞ ∞ c these sets, {B (xi , ri )}i=1 such that {B (xi , 5ri )}i=1 = Bi covers Tk Then i=1 letting mn denote the outer measure determined by mn , ´´ ³ ³ b mn (h (Tk )) ≤ mn h ∪∞ i=1 Bi ≤

∞ X

∞ ³ ³ ´´ X bi mn h B ≤ mn (B (h (xi ) , 2krxi ))

i=1

=

∞ X

i=1 n

mn (B (xi , 2krxi )) = (2k)

i=1

≤

∞ X

mn (B (xi , rxi ))

i=1 n

n

(2k) mn (V ) ≤ (2k) ε.

Since ε > 0 is arbitrary, this shows mn (h (Tk )) = 0. Now mn (h (T )) = lim mn (h (Tk )) = 0. k→∞


10.5. CHANGE OF VARIABLES FOR LINEAR MAPS

299

Lemma 10.5.2 Let h satisfy 10.5.11. If S is a Lebesgue measurable subset of U , then h (S) is Lebesgue measurable. Proof: Let Sk = S ∩ B (0, k) , k ∈ N. By inner regularity of Lebesgue measure, there exists a set, F , which is the countable union of compact sets and a set T with mn (T ) = 0 such that F ∪ T = Sk . Then h (F ) ⊆ h (Sk ) ⊆ h (F ) ∪ h (T ). By continuity of h, h (F ) is a countable union of compact sets and so it is Borel. By Lemma 10.5.1, mn (h (T )) = 0 and so h (Sk ) is Lebesgue measurable because of completeness of Lebesgue measure. Now h (S) = ∪∞ k=1 h (Sk ) and so it is also true that h (S) is Lebesgue measurable. This proves the lemma. In particular, this proves the following corollary. Corollary 10.5.3 Suppose A is an n × n matrix. Then if S is a Lebesgue measurable set, it follows AS is also a Lebesgue measurable set. Lemma 10.5.4 Let R be unitary (R∗ R = RR∗ = I) and let V be a an open or closed set. Then mn (RV ) = mn (V ) . Proof: First assume V is a bounded open set. By Corollary 10.4.6 there is a disjoint sequence of closed balls, {Bi } such that V = ∪∞ i=1 Bi ∪N where mn (N ) = 0. Denote by xP i the center of Bi and let ri be the radius of Bi . Then by Lemma 10.5.1 ∞ of translation of Lebesgue measure, mn (RV ) = P i=1 mn (RBi ) . Now by invariance P∞ ∞ this equals i=1 mn (RBi − Rxi ) = i=1 mn (RB (0, ri )) . Since R is unitary, it preserves all distances and so RB (0, ri ) = B (0, ri ) and therefore, mn (RV ) =

∞ X i=1

mn (B (0, ri )) =

∞ X

mn (Bi ) = mn (V ) .

i=1

This proves the lemma in the case that V is bounded. Suppose now that V is just an open set. Let Vk = V ∩ B (0, k) . Then mn (RVk ) = mn (Vk ) . Letting k → ∞, this yields the desired conclusion. This proves the lemma in the case that V is open. Suppose now that H is a closed and bounded set. Let B (0,R) ⊇ H. Then letting B = B (0, R) for short, mn (RH)

= mn (RB) − mn (R (B \ H)) = mn (B) − mn (B \ H) = mn (H) .

In general, let Hm = H ∩ B (0,m). Then from what was just shown, mn (RHm ) = mn (Hm ) . Now let m → ∞ to get the conclusion of the lemma in general. This proves the lemma. Lemma 10.5.5 Let E be Lebesgue measurable set in Rn and let R be unitary. Then mn (RE) = mn (E) .

300

LEBESGUE MEASURE

Proof: First suppose E is bounded. Then there exist sets, G and H such that H ⊆ E ⊆ G and H is the countable union of closed sets while G is the countable intersection of open sets such that mn (G \ H) = 0. By Lemma 10.5.4 applied to these sets whose union or intersection equals H or G respectively, it follows mn (RG) = mn (G) = mn (H) = mn (RH) . Therefore, mn (H) = mn (RH) ≤ mn (RE) ≤ mn (RG) = mn (G) = mn (E) = mn (H) . In the general case, let Em = E ∩ B (0, m) and apply what was just shown and let m → ∞. Lemma 10.5.6 Let V be an open or closed set in Rn and let A be an n × n matrix. Then mn (AV ) = |det (A)| mn (V ). Proof: Let RU be the right polar decomposition (Theorem 4.12.6 on Page 93) of A and let V be an open set. Then from Lemma 10.5.5, mn (AV ) = mn (RU V ) = mn (U V ) . Now U = Q∗ DQ where D is a diagonal matrix such that |det (D)| = |det (A)| and Q is unitary. Therefore, mn (AV ) = mn (Q∗ DQV ) = mn (DQV ) . Now QV is an open set and so by Corollary 10.1.6 on Page 288 and Lemma 10.5.4, mn (AV ) = |det (D)| mn (QV ) = |det (D)| mn (V ) = |det (A)| mn (V ) . This proves the lemma in case V is open. Now let H be a closed set which is also bounded. First suppose det (A) = 0. Then letting V be an open set containing H, mn (AH) ≤ mn (AV ) = |det (A)| mn (V ) = 0 which shows the desired equation is obvious in the case where det (A) = 0. Therefore, assume A is one to one. Since H is bounded, H ⊆ B (0, R) for some R > 0. Then letting B = B (0, R) for short, mn (AH)

= mn (AB) − mn (A (B \ H)) = |det (A)| mn (B) − |det (A)| mn (B \ H) = |det (A)| mn (H) .

If H is not bounded, apply the result just obtained to Hm ≡ H ∩ B (0, m) and then let m → ∞. With this preparation, the main result is the following theorem.

10.6. CHANGE OF VARIABLES FOR C 1 FUNCTIONS

301

Theorem 10.5.7 Let E be Lebesgue measurable set in Rn and let A be an n × n matrix. Then mn (AE) = |det (A)| mn (E) . Proof: First suppose E is bounded. Then there exist sets, G and H such that H ⊆ E ⊆ G and H is the countable union of closed sets while G is the countable intersection of open sets such that mn (G \ H) = 0. By Lemma 10.5.6 applied to these sets whose union or intersection equals H or G respectively, it follows mn (AG) = |det (A)| mn (G) = |det (A)| mn (H) = mn (AH) . Therefore, |det (A)| mn (E) = ≤

|det (A)| mn (H) = mn (AH) ≤ mn (AE) mn (AG) = |det (A)| mn (G) = |det (A)| mn (E) .

In the general case, let Em = E ∩ B (0, m) and apply what was just shown and let m → ∞.

10.6

Change Of Variables For C 1 Functions

In this section theorems are proved which generalize the above to C 1 functions. More general versions can be seen in Kuttler [45], Kuttler [46], and Rudin [58]. There is also a very different approach to this theorem given in [45]. The more general version in [45] follows [58] and both are based on the Brouwer fixed point theorem and a very clever lemma presented in Rudin [58]. This same approach will be used later in this book to prove a different sort of change of variables theorem in which the functions are only Lipschitz. The proof will be based on a sequence of easy lemmas. Lemma 10.6.1 Let U and V be bounded open sets in Rn and let h, h−1 be C 1 functions such that h (U ) = V . Also let f ∈ Cc (V ) . Then Z Z f (y) dmn = f (h (x)) |det (Dh (x))| dmn V

U

Proof: First note h−1 (spt (f )) is a closed subset of the bounded set, U and so it is compact. Thus x → f (h (x)) |det (Dh (x))| is bounded and continuous. Let x ∈ U. By the assumption that h and h−1 are C 1 , h (x + v) − h (x) = = =

Dh (x) v + o (v) ¡ ¢ Dh (x) v + Dh−1 (h (x)) o (v) Dh (x) (v + o (v))

and so if r > 0 is small enough then B (x, r) is contained in U and h (B (x, r)) − h (x) =

302

LEBESGUE MEASURE

h (x+B (0,r)) − h (x) ⊆ Dh (x) (B (0, (1 + ε) r)) .

(10.6.12)

Making r still smaller if necessary, one can also obtain |f (y) − f (h (x))| < ε

(10.6.13)

for any y ∈ h (B (x, r)) and also |f (h (x1 )) |det (Dh (x1 ))| − f (h (x)) |det (Dh (x))|| < ε

(10.6.14)

whenever x1 ∈ B (x, r) . The collection of such balls is a Vitali cover of U. By Corollary 10.4.6 there is a sequence of disjoint closed balls {Bi } such that U = ∪∞ i=1 Bi ∪ N where mn (N ) = 0. Denote by xi the center of Bi and ri the radius. Then by Lemma 10.5.1, the monotone convergence theorem, and 10.6.12 - 10.6.14, R P∞ R f (y) dmn = i=1 h(Bi ) f (y) dmn V P∞ R ≤ εmn (V ) + i=1 h(Bi ) f (h (xi )) dmn P∞ ≤ εmn (V ) + i=1 f (h (xi )) mn (h (Bi )) P∞ ≤ εmn (V ) + i=1 f (h (xi )) mn (Dh (xi ) (B (0, (1 + ε) ri ))) n P∞ R = εmn (V ) + (1 + ε) i=1 Bi f (h (xi )) |det (Dh (xi ))| dmn ³ ´ R n P∞ ≤ εmn (V ) + (1 + ε) f (h (x)) |det (Dh (x))| dm + εm (B ) n n i i=1 Bi n P∞ R n ≤ εmn (V ) + (1 + ε) i=1 Bi f (h (x)) |det (Dh (x))| dmn + (1 + ε) εmn (U ) nR n = εmn (V ) + (1 + ε) U f (h (x)) |det (Dh (x))| dmn + (1 + ε) εmn (U ) Since ε > 0 is arbitrary, this shows Z Z f (y) dmn ≤ f (h (x)) |det (Dh (x))| dmn V

(10.6.15)

U

whenever f ∈ Cc (V ) . Now x →f (h (x)) |det (Dh (x))| is in Cc (U ) and so using the same argument with U and V switching roles and replacing h with h−1 , Z f (h (x)) |det (Dh (x))| dmn U Z ¡ ¡ ¢¢ ¯ ¡ ¡ ¢¢¯ ¯ ¡ ¢¯ ≤ f h h−1 (y) ¯det Dh h−1 (y) ¯ ¯det Dh−1 (y) ¯ dmn ZV = f (y) dmn V

by the chain rule. This with 10.6.15 proves the lemma. The next task is to relax the assumption that f is continuous.


303

Corollary 10.6.2 Let U and V be bounded open sets in Rn and let h, h−1 be C 1 functions such that h (U ) = V and |det (Dh (x))| is bounded. Also let E ⊆ V be measurable. Then Z Z XE (y) dmn = XE (h (x)) |det (Dh (x))| dmn . V

U

Proof: By regularity, there exist compact sets, Kk and open sets Gk such that Kk ⊆ E ⊆ Gk and mn (Gk \ Kk ) < 2−k . By Theorem 9.3.7, there exist fk such that Kk ≺ fk ≺ Gk . Then fk (y) → XE (y) a.e. because if y is such that convergence fails, it must P be the case that y is in Gk \ Kk for infinitely many k and k mn (Gk \ Kk ) < ∞. This set equals ∞ N = ∩∞ m=1 ∪k=m Gk \ Kk and so for each m ∈ N mn (N )

≤ mn (∪∞ k=m Gk \ Kk ) ∞ ∞ X X ≤ mn (Gk \ Kk ) < 2−k = 2−(m−1) k=m

k=m

showing mn (N ) = 0. Then fk (h (x)) must converge to XE (h (x)) for all x ∈ / h−1 (N ) , a set of measure zero by Lemma 10.5.1. Thus Z Z fk (y) dmn = fk (h (x)) |det (Dh (x))| dmn . V

U

By the dominated convergence theorem using a dominating function, XV in the integral on the left and XU |det (Dh)| on the right, it follows Z Z XE (y) dmn = XE (h (x)) |det (Dh (x))| dmn . V

U

This proves the corollary. You don’t need to assume the open sets are bounded. Corollary 10.6.3 Let U and V be open sets in Rn and let h, h−1 be C 1 functions such that h (U ) = V . Also let E ⊆ V be measurable. Then Z Z XE (y) dmn = XE (h (x)) |det (Dh (x))| dmn . V

U

Proof: For each x ∈ U, there exists rx such that B (x, rx ) ⊆ U and rx < 1. Then by the mean value inequality Theorem 5.13.4, it follows h (B (x, rx )) is also

304

LEBESGUE MEASURE

bounded. This is a Vitali cover of U and so by Corollary 10.4.6 there is a sequence of these balls, {Bi } such that they are disjoint, h (Bi ) is also bounded and mn (U \ ∪i Bi ) = 0. It follows from Lemma 10.5.1 that h (U \ ∪i Bi ) also has measure zero. Then from Corollary 10.6.2 Z XZ XE (y) dmn = XE∩h(Bi ) (y) dmn V

=

i

h(Bi )

i

Bi

XZ Z

=

XE (h (x)) |det (Dh (x))| dmn

XE (h (x)) |det (Dh (x))| dmn . U

This proves the corollary. With this corollary, the main theorem follows. Theorem 10.6.4 Let U and V be open sets in Rn and let h, h−1 be C 1 functions such that h (U ) = V. Then if g is a nonnegative Lebesgue measurable function, Z

Z g (y) dy =

V

g (h (x)) |det (Dh (x))| dx.

(10.6.16)

U

Proof: From Corollary 10.6.3, 10.6.16 holds for any nonnegative simple function in place of g. In general, let {sk } be an increasing sequence of simple functions which converges to g pointwise. Then from the monotone convergence theorem Z Z Z g (y) dy = lim sk dy = lim sk (h (x)) |det (Dh (x))| dx k→∞ V k→∞ U V Z = g (h (x)) |det (Dh (x))| dx. U

This proves the theorem. This is a pretty good theorem but it isn’t too hard to generalize it. In particular, it is not necessary to assume h−1 is C 1 . Lemma 10.6.5 Suppose V is an n − 1 dimensional subspace of Rn and K is a compact subset of V . Then letting Kε ≡ ∪x∈K B (x,ε) = K + B (0, ε) , it follows that n−1

mn (Kε ) ≤ 2n ε (diam (K) + ε)

.


305

Proof: Let an orthonormal basis for V be {v1 , · · · , vn−1 } and let {v1 , · · · , vn−1 , vn } be an orthonormal basis for Rn . Now define a linear transformation, Q by Qvi = ei . Thus QQ∗ = Q∗ Q = I and Q preserves all distances because ¯ ¯2 ¯2 ¯ ¯2 ¯ ¯X ¯ ¯X ¯ ¯ X ¯ X ¯ ¯ ¯ ¯ ¯ ¯ 2 ai ei ¯ . ai vi ¯ = |ai | = ¯ a i ei ¯ = ¯ ¯Q ¯ ¯ ¯ ¯ ¯ ¯ i

i

i

i

Letting k0 ∈ K, it follows K ⊆ B (k0 , diam (K)) and so, QK ⊆ B n−1 (Qk0 , diam (QK)) = B n−1 (Qk0 , diam (K)) where B n−1 refers to the ball taken with respect to the usual norm in Rn−1 . Every point of Kε is within ε of some point of K and so it follows that every point of QKε is within ε of some point of QK. Therefore, QKε ⊆ B n−1 (Qk0 , diam (QK) + ε) × (−ε, ε) , To see this, let x ∈ QKε . Then there exists k ∈ QK such that |k − x| < ε. Therefore, |(x1 , · · · , xn−1 ) − (k1 , · · · , kn−1 )| < ε and |xn − kn | < ε and so x is contained in the set on the right in the above inclusion because kn = 0. However, the measure of the set on the right is smaller than n−1

[2 (diam (QK) + ε)]

n−1

(2ε) = 2n [(diam (K) + ε)]

ε.

This proves the lemma. Note this is a very sloppy estimate. You can certainly do much better but this estimate is sufficient to prove Sard’s lemma which follows. Definition 10.6.6 In any metric space, if x is a point of the metric space and S is a nonempty subset, dist (x,S) ≡ inf {d (x, s) : s ∈ S} . More generally, if T, S are two nonempty sets, dist (S, T ) ≡ inf {d (t, s) : s ∈ S, t ∈ T } . Lemma 10.6.7 The function x → dist (x,S) is continuous. Proof: Let x, y be given. Suppose dist (x, S) ≥ dist (y, S) and pick s ∈ S such that dist (y, S) + ε ≥ d (y, s) . Then 0

≤ dist (x, S) − dist (y, S) ≤ dist (x, S) − (d (y, s) − ε) ≤ d (x, s) − d (y, s) + ε ≤ d (x, y) + d (y, s) − d (y, s) + ε = d (x, y) + ε.

Since ε > 0 is arbitrary, this shows |dist (x, S) − dist (y, S)| ≤ d (x, y) . This proves the lemma.

306

LEBESGUE MEASURE

Lemma 10.6.8 Let h be a C 1 function defined on an open set, U and let K be a compact subset of U. Then if ε > 0 is given, there exits r1 > 0 such that if |v| ≤ r1 , then for all x ∈ K, |h (x + v) − h (x) − Dh (x) v| < ε |v| . ¡ ¢ Proof: Let 0 < δ < dist K, U C . Such a positive number exists because if there exists a sequence of points in K, {kk } and points in U C , {sk } such that |kk − sk | → 0, then you could take a subsequence, still denoted by k such that kk → k ∈ K and then sk → k also. But U C is closed so k ∈ K ∩ U C , a contradiction. Then ¯R ¯ ¯ ¯ 1 Dh (x + tv) vdt − Dh (x) v ¯ ¯ 0 |h (x + v) − h (x) − Dh (x) v| ≤ |v| |v| R1 |Dh (x + tv) v − Dh (x) v| dt 0 ≤ . |v| Now from uniform continuity of Dh on the compact set, {x : dist (x,K) ≤ δ} it follows there exists r1 < δ such that if |v| ≤ r1 , then ||Dh (x + tv) − Dh (x)|| < ε for every x ∈ K. From the above formula, it follows that if |v| ≤ r1 , R1 |Dh (x + tv) v − Dh (x) v| dt |h (x + v) − h (x) − Dh (x) v| 0 ≤ |v| |v| R1 ε |v| dt 0 < = ε. |v| This proves the lemma. A different proof of the following is in [45]. See also [46]. Lemma 10.6.9 (Sard) Let U be an open set in Rn and let h : U → Rn be C 1 . Let Z ≡ {x ∈ U : det Dh (x) = 0} . Then mn (h (Z)) = 0. ∞

Proof: Let {Uk }k=1 be an increasing sequence of open sets whose closures are compact and © whose union ¡ equals ¢ U and ª let Zk ≡ Z ∩Uk . To obtain such a sequence, let Uk = x ∈ U : dist x,U C < k1 ∩ B (0, k) . First it is shown that h (Zk ) has measure zero. Let W be an open set contained in Uk+1 which contains Zk and satisfies mn (Zk ) + ε > mn (W ) where here and elsewhere, ε < 1. Let ¡ ¢ C r = dist Uk , Uk+1 and let r1 > 0 be a constant as in Lemma 10.6.8 such that whenever x ∈ Uk and 0 < |v| ≤ r1 , |h (x + v) − h (x) − Dh (x) v| < ε |v| . (10.6.17)


307

Now the closures of balls which are contained in W and which have the property that their diameters are less than r1 yield a Vitali covering of W. by o n Therefore, ei such that Corollary 10.4.6 there is a disjoint sequence of these closed balls, B e W = ∪∞ i=1 Bi ∪ N where N is a set of measure zero. Denote by {Bi } those closed balls in this sequence which have nonempty intersection with Zk , let di be the diameter of Bi , and let zi be a point in Bi ∩ Zk . Since zi ∈ Zk , it follows Dh (zi ) B (0,di ) = Di where Di is contained in a subspace, V which has dimension n − 1 and the diameter of Di is no larger than 2Ck di where Ck ≥ max {||Dh (x)|| : x ∈ Zk } Then by 10.6.17, if z ∈ Bi , h (z) − h (zi ) ∈ Di + B (0, εdi ) ⊆ Di + B (0,εdi ) . Thus h (Bi ) ⊆ h (zi ) + Di + B (0,εdi ) By Lemma 10.6.5 n−1

≤ 2n (2Ck di + εdi ) εdi ³ ´ n−1 n n ≤ di 2 [2Ck + ε] ε

mn (h (Bi ))

≤ Cn,k mn (Bi ) ε. Therefore, by Lemma 10.5.1 mn (h (Zk )) ≤

X

mn (h (Bi )) ≤ Cn,k ε

i

≤

X

mn (Bi )

i

εCn,k mn (W ) ≤ εCn,k (mn (Zk ) + ε)

Since ε is arbitrary, this shows mn (h (Zk )) = 0 and so 0 = limk→∞ mn (h (Zk )) = mn (h (Z)). With this important lemma, here is a generalization of Theorem 10.6.4. Theorem 10.6.10 Let U be an open set and let h be a 1 − 1, C 1 function with values in Rn . Then if g is a nonnegative Lebesgue measurable function, Z Z g (y) dy = g (h (x)) |det (Dh (x))| dx. (10.6.18) h(U )

U

Proof: Let Z = {x : det (Dh (x)) = 0} . Then by the inverse function theorem, h−1 is C 1 on h (U \ Z) and h (U \ Z) is an open set. Therefore, from Lemma 10.6.9 and Theorem 10.6.4, Z Z Z g (y) dy = g (y) dy = g (h (x)) |det (Dh (x))| dx h(U ) h(U \Z) U \Z Z = g (h (x)) |det (Dh (x))| dx. U

308

LEBESGUE MEASURE

This proves the theorem. Of course the next generalization considers the case when h is not even one to one.

10.7

Mappings Which Are Not One To One

Now suppose h is only C 1 , not necessarily one to one. For U+ ≡ {x ∈ U : |det Dh (x)| > 0} and Z the set where |det Dh (x)| = 0, Lemma 10.6.9 implies mn (h(Z)) = 0. For x ∈ U+ , the inverse function theorem implies there exists an open set Bx such that x ∈ Bx ⊆ U+ , h is one to one on Bx . Let {Bi } be a countable subset of {Bx }x∈U+ such that U+ = ∪∞ i=1 Bi . Let E1 = B1 . If E1 , · · · , Ek have been chosen, Ek+1 = Bk+1 \ ∪ki=1 Ei . Thus ∪∞ i=1 Ei = U+ , h is one to one on Ei , Ei ∩ Ej = ∅, and each Ei is a Borel set contained in the open set Bi . Now define ∞ X n(y) ≡ Xh(Ei ) (y) + Xh(Z) (y). i=1

The set, h (Ei ) , h (Z) are measurable by Lemma 10.5.2. Thus n (·) is measurable. Lemma 10.7.1 Let F ⊆ h(U ) be measurable. Then Z Z n(y)XF (y)dy = XF (h(x))| det Dh(x)|dx. h(U )

U

Proof: Using Lemma 10.6.9 and the Monotone Convergence Theorem or Fubini’s Theorem,   mn (h(Z))=0 Z Z ∞ z }| {  X n(y)XF (y)dy = Xh(Ei ) (y) + Xh(Z) (y)  XF (y)dy  h(U )

h(U )

= = = =

∞ Z X i=1 h(U ) ∞ Z X

Xh(Ei ) (y)XF (y)dy

i=1 h(Bi ) ∞ Z X i=1 Bi ∞ Z X i=1

=

i=1

U

Z X ∞ U i=1

Xh(Ei ) (y)XF (y)dy

XEi (x)XF (h(x))| det Dh(x)|dx

XEi (x)XF (h(x))| det Dh(x)|dx XEi (x)XF (h(x))| det Dh(x)|dx

10.8. LEBESGUE MEASURE AND ITERATED INTEGRALS Z

309

Z

=

XF (h(x))| det Dh(x)|dx = U+

XF (h(x))| det Dh(x)|dx. U

This proves the lemma. Definition 10.7.2 For y ∈ h(U ), define a function, #, according to the formula #(y) ≡ number of elements in h−1 (y). Observe that #(y) = n(y) a.e.

(10.7.19)

because n(y) = #(y) if y ∈ / h(Z), a set of measure 0. Therefore, # is a measurable function. Theorem 10.7.3 Let g ≥ 0, g measurable, and let h be C 1 (U ). Then Z Z #(y)g(y)dy = g(h(x))| det Dh(x)|dx. h(U )

(10.7.20)

U

Proof: From 10.7.19 and Lemma 10.7.1, 10.7.20 holds for all g, a nonnegative simple function. Approximating an arbitrary measurable nonnegative function, g, with an increasing pointwise convergent sequence of simple functions and using the monotone convergence theorem, yields 10.7.20 for an arbitrary nonnegative measurable function, g. This proves the theorem.

10.8

Lebesgue Measure And Iterated Integrals

The following is the main result. Theorem 10.8.1R Let f ≥ 0 and suppose f is a Lebesgue measurable function defined on Rn and Rn f dmn < ∞. Then Z Z Z f dmn = f dmn−k dmk . Rn

Rk

Rn−k

This will be accomplished by Fubini’s theorem, Theorem 9.8.11 and the following lemma. Lemma 10.8.2 mk × mn−k = mn on the mn measurable sets. Qn Proof: First of all, let R = i=1 (ai , bi ] be a measurable rectangle and let Qk Qn Rk = i=1 (ai , bi ], Rn−k = i=k+1 (ai , bi ]. Then by Fubini’s theorem, Z Z Z XR d (mk × mn−k ) = XRk XRn−k dmk dmn−k k n−k ZR R Z = XRk dmk XRn−k dmn−k Rk Rn−k Z = XR dmn

310

LEBESGUE MEASURE

and so mk × mn−k and mn agree on every half open rectangle. By Lemma 10.1.2 these two measures agree on every open set. ¡Now¢ if K is a compact set, then 1 K = ∩∞ k=1 © Uk where Uk is the ª open set, K + B 0, k . Another way of saying this 1 is Uk ≡ x : dist (x,K) < k which is obviously open because x → dist (x,K) is a continuous function. Since K is the countable intersection of these decreasing open sets, each of which has finite measure with respect to either of the two measures, it follows that mk × mn−k and mn agree on all the compact sets. Now let E be a bounded Lebesgue measurable set. Then there are sets, H and G such that H is a countable union of compact sets, G a countable intersection of open sets, H ⊆ E ⊆ G, and mn (G \ H) = 0. Then from what was just shown about compact and open sets, the two measures agree on G and on H. Therefore, mn (H)

= mk × mn−k (H) ≤ mk × mn−k (E) ≤ mk × mn−k (G) = mn (E) = mn (H)

By completeness of the measure space for mk × mn−k , it follows E is mk × mn−k measurable and mk × mn−k (E) = mn (E) . This proves the lemma. You could also show that the two σ algebras are the same. However, this is not needed for the lemma or the theorem. Proof of Theorem 10.8.1: By the lemma and Fubini’s theorem, Theorem 9.8.11, Z Z Z Z f dmn = f d (mk × mn−k ) = f dmn−k dmk . Rn

Rn

Rk

Rn−k

Corollary 10.8.3 Let f be a nonnegative real valued measurable function. Then Z Z Z f dmn = f dmn−k dmk . Rn

Rk

Rn−k

Rk

Rn−k

R Proof: Let Sp ≡ {x ∈ Rn : 0 ≤ f (x) ≤ p} ∩ B (0, p) . Then Rn f XSp dmn < ∞. Therefore, from Theorem 10.8.1, Z Z Z f XSp dmn = XSp f dmn−k dmk . Rn

Now let p → ∞ and use the Monotone convergence theorem and the Fubini Theorem 9.8.11 on Page 261. Not surprisingly, the following corollary follows from this. Corollary 10.8.4 Let f ∈ L1 (Rn ) where the measure is mn . Then Z Z Z f dmn = f dmn−k dmk . Rn

Rk

Rn−k

Proof: Apply Corollary 10.8.3 to the postive and negative parts of the real and imaginary parts of f .

10.9. SPHERICAL COORDINATES IN MANY DIMENSIONS

10.9

311

Spherical Coordinates In Many Dimensions

Sometimes there is a need to deal with spherical coordinates in more than three dimensions. In this section, this concept is defined and formulas are derived for these coordinate systems. Recall polar coordinates are of the form y1 = ρ cos θ y2 = ρ sin θ where ρ > 0 and θ ∈ [0, 2π). Here I am writing ρ in place of r to emphasize a pattern which is about to emerge. I will consider polar coordinates as spherical coordinates in two dimensions. I will also simply refer to such coordinate systems as polar coordinates regardless of the dimension. This is also the reason I am writing y1 and y2 instead of the more usual x and y. Now consider what happens when you go to three dimensions. The situation is depicted in the following picture. r(x1 , x2 , x3 )

R

¶¶

φ1 ¶ ¶ρ ¶

¶

R2

From this picture, you see that y3 = ρ cos φ1 . Also the distance between (y1 , y2 ) and (0, 0) is ρ sin (φ1 ) . Therefore, using polar coordinates to write (y1 , y2 ) in terms of θ and this distance, y1 = ρ sin φ1 cos θ, y2 = ρ sin φ1 sin θ, y3 = ρ cos φ1 . where φ1 ∈ [0, π] . What was done is to replace ρ with ρ sin φ1 and then to add in y3 = ρ cos φ1 . Having done this, there is no reason to stop with three dimensions. Consider the following picture: r(x1 , x2 , x3 , x4 )

R

¶¶

φ2 ¶ ¶ρ ¶

¶

R3

From this picture, you see that y4 = ρ cos φ2 . Also the distance between (y1 , y2 , y3 ) and (0, 0, 0) is ρ sin (φ2 ) . Therefore, using polar coordinates to write (y1 , y2 , y3 ) in

312

LEBESGUE MEASURE

terms of θ, φ1 , and this distance, y1 y2 y3 y4

= ρ sin φ2 sin φ1 cos θ, = ρ sin φ2 sin φ1 sin θ, = ρ sin φ2 cos φ1 , = ρ cos φ2

where φ2 ∈ [0, π] . Continuing this way, given spherical coordinates in Rn , to get the spherical coordinates in Rn+1 , you let yn+1 = ρ cos φn−1 and then replace every occurrence of ρ with ρ sin φn−1 to obtain y1 · · · yn in terms of φ1 , φ2 , · · · , φn−1 ,θ, and ρ. It is always the case that ρ measures the distance from the point in Rn to the origin in Rn , 0. Each φi ∈ [0, π] , and θ ∈ [0, 2π). It can be shown using math Qn−2 induction that these coordinates map i=1 [0, π] × [0, 2π) × (0, ∞) one to one onto Rn \ {0} . Theorem 10.9.1 Let y = h (φ, θ, ρ) be the spherical coordinate transformations in Qn−2 Rn . Then letting A = i=1 [0, π] × [0, 2π), it follows h maps A × (0, ∞) one to one onto Rn \ {0} . Also |det Dh (φ, θ, ρ)| will always be of the form |det Dh (φ, θ, ρ)| = ρn−1 Φ (φ, θ) .

(10.9.21)

where Φ is a continuous function of φ and θ.1 Furthermore whenever f is Lebesgue measurable and nonnegative, Z Z ∞ Z n−1 f (y) dy = ρ f (h (φ, θ, ρ)) Φ (φ, θ) dφ dθdρ (10.9.22) Rn

0

A

where here dφ dθ denotes dmn−1 on A. The same formula holds if f ∈ L1 (Rn ) . Proof: Formula 10.9.21 is obvious from the definition of the spherical coordinates. The first claim is also clear from Qn−2the definition and math induction. It remains to verify 10.9.22. Let A0 ≡ i=1 (0, π) × (0, 2π) . Then it is clear that (A \ A0 ) × (0, ∞) ≡ N is a set of measure zero in Rn . Therefore, from Lemma 10.5.1 it follows h (N ) is also a set of measure zero. Therefore, using the change of variables theorem, Corollary 10.8.3, and Sard’s lemma, Z Z Z f (y) dy = f (y) dy = f (y) dy Rn Rn \{0} Rn \({0}∪h(N )) Z = f (h (φ, θ, ρ)) ρn−1 Φ (φ, θ) dmn A0 ×(0,∞) Z = XA×(0,∞) (φ, θ, ρ) f (h (φ, θ, ρ)) ρn−1 Φ (φ, θ) dmn µZ ¶ Z ∞ n−1 = ρ f (h (φ, θ, ρ)) Φ (φ, θ) dφ dθ dρ. 0 1 Actually

A

it is only a function of the first but this is not important in what follows.

10.10. THE BROUWER FIXED POINT THEOREM

313

Now the claim about f ∈ L1 follows routinely from considering the positive and negative parts of the real and imaginary parts of f in the usual way. This proves the theorem. Notation 10.9.2 Often this is written differently. Note that from the spherical coordinate formulas, f (h (φ, θ, ρ)) = f (ρω) where |ω| = 1. Letting S n−1 denote the unit sphere, {ω ∈ Rn : |ω| = 1} , the inside integral in the above formula is sometimes written as Z f (ρω) dσ S n−1

where σ is a measure on S n−1 . See [45] for another description of this measure. It isn’t an important issue here. Later in the book when integration on manifolds is discussed, more general considerations will be dealt with. Either 10.9.22 or the formula ¶ µZ Z ∞

ρn−1

f (ρω) dσ dρ S n−1

0

will be referred ¡ ¢ toRas polar coordinates and is very useful in establishing estimates. Here σ S n−1 ≡ A Φ (φ, θ) dφ dθ. ³ ´s R 2 Example 10.9.3 For what values of s is the integral B(0,R) 1 + |x| dy bounded independent of R? Here B (0, R) is the ball, {x ∈ Rn : |x| ≤ R} . I think you can see immediately that s must be negative but exactly how negative? It turns out it depends on n and using polar coordinates, you can find just exactly what is needed. From the polar coordinats formula above, Z

³ 1 + |x|

2

´s

Z dy

B(0,R)

R

Z

¡ ¢s 1 + ρ2 ρn−1 dσdρ

= 0

Z

S n−1 R

= Cn

¡ ¢s 1 + ρ2 ρn−1 dρ

0

Now the very hard problem has been reduced to considering an easy one variable problem of finding when Z R ¡ ¢s ρn−1 1 + ρ2 dρ 0

is bounded independent of R. You need 2s + (n − 1) < −1 so you need s < −n/2.

10.10

The Brouwer Fixed Point Theorem

This seems to be a good place to present a short proof of one of the most important of all fixed point theorems. There are many approaches to this but one of the easiest and shortest I have ever seen is the one in Dunford and Schwartz [22]. This is what is presented here. In Evans [25] there is a different proof which depends on

314

LEBESGUE MEASURE

integration theory. A good reference for an introduction to various kinds of fixed point theorems is the book by Smart [61]. This book also gives an entirely different approach to the Brouwer fixed point theorem. The proof given here is based on the following lemma. Recall that the mixed partial derivatives of a C 2 function are equal. In the following lemma, and elsewhere, a comma followed by an index indicates the partial derivative with respect to the ∂f indicated variable. Thus, f,j will mean ∂x . Also, write Dg for the Jacobian matrix j which is the matrix of Dg taken with respect to the usual basis vectors in Rn . Recall that for A an n × n matrix, cof (A)ij is the determinant of the matrix which results i+j

from deleting the ith row and the j th column and multiplying by (−1)

.

Lemma 10.10.1 Let g : U → Rn be C 2 where U is an open subset of Rn . Then n X

cof (Dg)ij,j = 0,

j=1

where here (Dg)ij ≡ gi,j ≡

∂gi ∂xj .

Also, cof (Dg)ij =

∂ det(Dg) . ∂gi,j

Proof: From the cofactor expansion theorem, det (Dg) =

n X

gi,j cof (Dg)ij

i=1

and so

∂ det (Dg) = cof (Dg)ij ∂gi,j

(10.10.23)

which shows the last claim of the lemma. Also X δ kj det (Dg) = gi,k (cof (Dg))ij

(10.10.24)

i

because if k 6= j this is just the cofactor expansion of the determinant of a matrix in which the k th and j th columns are equal. Differentiate 10.10.24 with respect to xj and sum on j. This yields X r,s,j

δ kj

X X ∂ (det Dg) gr,sj = gi,kj (cof (Dg))ij + gi,k cof (Dg)ij,j . ∂gr,s ij ij

Hence, using δ kj = 0 if j 6= k and 10.10.23, X X X gr,ks (cof (Dg))rs + gi,k cof (Dg)ij,j . (cof (Dg))rs gr,sk = rs

rs

ij

Subtracting the first sum on the right from both sides and using the equality of mixed partials,   X X gi,k  (cof (Dg))ij,j  = 0. i

j

10.10. THE BROUWER FIXED POINT THEOREM If det (gi,k ) 6= 0 so that (gi,k ) is invertible, this shows det (Dg) = 0, let gk = g + εk I

315 P j

(cof (Dg))ij,j = 0. If

where εk → 0 and det (Dg + εk I) ≡ det (Dgk ) 6= 0. Then X X (cof (Dg))ij,j = lim (cof (Dgk ))ij,j = 0 k→∞

j

j

and this proves the lemma. To prove the Brouwer fixed point theorem, first consider a version of it valid for C 2 mappings. This is the following lemma. Lemma 10.10.2 Let Br = B (0,r) and suppose g is a C 2 function defined on Rn which maps Br to Br . Then g (x) = x for some x ∈ Br . Proof: Suppose not. Then |g (x) − x| must be bounded away from zero on Br . Let a (x) be the larger of the two roots of the equation, 2

|x+a (x) (x − g (x))| = r2 .

(10.10.25)

Thus r − (x, (x − g (x))) + a (x) =

³ ´ 2 2 2 (x, (x − g (x))) + r2 − |x| |x − g (x)| 2

|x − g (x)|

(10.10.26) The expression under the square root sign is always nonnegative and it follows from the formula that a (x) ≥ 0. Therefore, (x, (x − g (x))) ≥ 0 for all x ∈ Br . The reason for this is that a (x) is the larger zero of a polynomial of the form 2 2 p (z) = |x| + z 2 |x − g (x)| − 2z (x, x − g (x)) and from the formula above, it is nonnegative. −2 (x, x − g (x)) is the slope of the tangent line to p (z) at z = 0. If 2 x 6= 0, then |x| > 0 and so this slope needs to be negative for the larger of the two zeros to be positive. If x = 0, then (x, x − g (x)) = 0. Now define for t ∈ [0, 1], f (t, x) ≡ x+ta (x) (x − g (x)) . The important properties of f (t, x) and a (x) are that a (x) = 0 if |x| = r.

(10.10.27)

|f (t, x)| = r for all |x| = r

(10.10.28)

and These properties follow immediately from 10.10.26 and the above observation that for x ∈ Br , it follows (x, (x − g (x))) ≥ 0. Also from 10.10.26, a is a C 2 function near Br . This is obvious from 10.10.26 as long as |x| < r. However, even if |x| = r it is still true. To show this, it suffices to

316

LEBESGUE MEASURE

verify the expression under the square root sign is positive. If this expression were not positive for some |x| = r, then (x, (x − g (x))) = 0. Then also, since g (x) = 6 x, ¯ ¯ ¯ g (x) + x ¯ ¯ ¯

µ ¶ 2 g (x) + x 1 r2 |x| r2 x, = (x, g (x)) + = + = r2 , 2 2 2 2 2

a contradiction. Therefore, the expression under the square root in 10.10.26 is always positive near Br and so a is a C 2 function near Br as claimed because the square root function is C 2 away from zero. Now define Z I (t) ≡ det (D2 f (t, x)) dx. Br

Then

Z I (0) =

dx = mn (Br ) > 0.

(10.10.29)

Br

Using the dominated convergence theorem one can differentiate I (t) as follows. Z X ∂ det (D2 f (t, x)) ∂fi,j I 0 (t) = dx ∂fi,j ∂t Br ij Z X ∂ (a (x) (xi − gi (x))) = cof (D2 f )ij dx. ∂xj Br ij Now from 10.10.27 a (x) = 0 when |x| = r and so integration by parts and Lemma 10.10.1 yields Z X ∂ (a (x) (xi − gi (x))) 0 I (t) = cof (D2 f )ij dx ∂xj Br ij Z X = − cof (D2 f )ij,j a (x) (xi − gi (x)) dx = 0. Br ij

Therefore, I (1) = I (0). However, from 10.10.25 it follows that for t = 1, X fi fi = r2 i

P

and so, i fi,j fi = 0 which implies since |f (1, x)| = r by 10.10.25, that det (fi,j ) = det (D2 f (1, x)) = 0 and so I (1) = 0, a contradiction to 10.10.29 since I (1) = I (0). This proves the lemma. The following theorem is the Brouwer fixed point theorem for a ball. Theorem 10.10.3 Let Br be the above closed ball and let f : Br → Br be continuous. Then there exists x ∈ Br such that f (x) = x.

10.11. THE BROUWER FIXED POINT THEOREM ANOTHER PROOF Proof: Let fk (x) ≡

f (x) 1+k−1 .

Thus ||fk − f ||
0 but

Z I (1) =

Z det (Dh (x)) dmn =

B(0,1)

# (y) dmn = 0 ∂B(0,1)

because from polar coordinates or other elementary reasoning, mn (∂B (0, 1)) = 0. This proves the lemma. The following is the Brouwer fixed point theorem for C 2 maps. ³ ´ Lemma 10.11.4 If h ∈ C 2 B (0, R) and h : B (0, R) → B (0, R), then h has a fixed point, x such that h (x) = x. Proof: Suppose the lemma is not true. Then for all x, |x − h (x)| = 6 0. Then define x − h (x) t (x) g (x) = h (x) + |x − h (x)|

10.11. THE BROUWER FIXED POINT THEOREM ANOTHER PROOF

319

where t (x) is nonnegative and is chosen such that g (x) ∈ ∂B (0, R) . This mapping is illustrated in the following picture.

tf (x)

g(x)

xt ¡ ¡ ¡ ¡ t

If x →t (x) is C 2 near B (0, R), it will follow g is a C 2 retraction onto ∂B (0, R) contrary to Lemma 10.11.3. Thus t (x) is the nonnegative solution to µ ¶ x − h (x) 2 H (x, t) = |h (x)| + 2 h (x) , t + t2 = R 2 (10.11.30) |x − h (x)| Then

µ ¶ x − h (x) Ht (x, t) = 2 h (x) , + 2t. |x − h (x)|

If this is nonzero for all x near B (0, R), it follows from the implicit function theorem that t is a C 2 function of x. Then from 10.11.30 µ ¶ x − h (x) 2t = −2 h (x) , |x − h (x)| s µ ¶2 ³ ´ x − h (x) 2 ± 4 h (x) , − 4 |h (x)| − R2 |x − h (x)| and so

µ ¶ x − h (x) Ht (x, t) = 2t + 2 h (x) , |x − h (x)| s ¶2 µ ³ ´ x − h (x) 2 2 = ± 4 R − |h (x)| + 4 h (x) , |x − h (x)|

If |h (x)| < R, this is nonzero. If |h (x)| = R, then it is still nonzero unless (h (x) , x − h (x)) = 0. But this cannot happen because the angle between h (x) and x − h (x) cannot be π/2. Alternatively, if the above equals zero, you would need 2

(h (x) , x) = |h (x)| = R2 which cannot happen unless x = h (x) which is assumed not to happen. Therefore, x → t (x) is C 2 near B (0, R) and so g (x) given above contradicts Lemma 10.11.3. This proves the lemma. Now it is easy to prove the Brouwer fixed point theorem.

320

LEBESGUE MEASURE

Theorem 10.11.5 Let f : B (0, R) → B (0, R) be continuous. Then f has a fixed point. Proof: If this is not so, there exists ε > 0 such that for all x ∈ B (0, R), |x − f (x)| > ε. By the Weierstrass approximation theorem, there exists h, a polynomial such that n o ε max |h (x) − f (x)| : x ∈ B (0, R) < . 2 Then for all x ∈ B (0, R), |x − h (x)| ≥ |x − f (x)| − |h (x) − f (x)| > ε − contradicting Lemma 10.11.4. This proves the theorem.

ε ε = 2 2

Some Extension Theorems 11.1

Caratheodory Extension Theorem

The Caratheodory extension theorem is a fundamental result which makes possible the consideration of measures on infinite products among other things. The idea is that if a finite measure defined only on an algebra is trying to be a measure, then in fact it can be extended to a measure. Definition 11.1.1 Let E be an algebra of sets of Ω and let µ0 be a finite measure on E. This means µ0 is finitely additive and if Ei , E are sets of E with the Ei disjoint and E = ∪∞ i=1 Ei , then µ0 (E) =

∞ X

µ0 (Ei )

i=1

while µ0 (Ω) < ∞. In this definition, µ0 is trying to be a measure and acts like one whenever possible. Under these conditions, µ0 can be extended uniquely to a complete measure, µ, defined on a σ algebra of sets containing E such that µ agrees with µ0 on E. The following is the main result. Theorem 11.1.2 Let µ0 be a measure on an algebra of sets, E, which satisfies µ0 (Ω) < ∞. Then there exists a complete measure space (Ω, S, µ) such that µ (E) = µ0 (E) for all E ∈ E. Also if ν is any such measure which agrees with µ0 on E, then ν = µ on σ (E), the σ algebra generated by E. Proof: Define an outer measure as follows. (∞ ) X ∞ µ (S) ≡ inf µ0 (Ei ) : S ⊆ ∪i=1 Ei , Ei ∈ E i=1

321

322

SOME EXTENSION THEOREMS

Claim 1: µ is an outer measure. ∞ Proof of Claim 1: Let S ⊆ ∪∞ i=1 Si and let Si ⊆ ∪j=1 Eij , where ∞

X ε µ (Si ) + i ≥ µ (Eij ) . 2 j=1 Then µ (S) ≤

XX i

µ (Eij ) =

j

X³

µ (Si ) +

i

ε´ X = µ (Si ) + ε. 2i i

Since ε is arbitrary, this shows µ is an outer measure as claimed. By the Caratheodory procedure, there exists a unique σ algebra, S, consisting of the µ measurable sets such that (Ω, S, µ) is a complete measure space. It remains to show µ extends µ0 . Claim 2: If S is the σ algebra of µ measurable sets, S ⊇ E and µ = µ0 on E. Proof of Claim 2: First observe that if A ∈ E, then µ (A) ≤ µ0 (A) by definition. Letting µ (A) + ε >

∞ X

µ0 (Ei ) , ∪∞ i=1 Ei ⊇A, Ei ∈ E,

i=1

it follows µ (A) + ε >

∞ X

µ0 (Ei ∩ A) ≥ µ0 (A)

i=1

since A = ∪∞ i=1 Ei ∩ A. Therefore, µ = µ0 on E. Consider the assertion that E ⊆ S. Let A ∈ E and let S ⊆ Ω be any set. There exist sets {Ei } ⊆ E such that ∪∞ i=1 Ei ⊇ S but µ (S) + ε >

∞ X

µ (Ei ) .

i=1

Then µ (S) ≤ µ (S ∩ A) + µ (S \ A) ∞ ≤ µ (∪∞ i=1 Ei \ A) + µ (∪i=1 (Ei ∩ A)) ∞ ∞ ∞ X X X ≤ µ (Ei \A) + µ (Ei ∩ A) = µ (Ei ) < µ (S) + ε. i=1

i=1

i=1

Since ε is arbitrary, this shows A ∈ S. This has proved the existence part of the theorem. To verify uniqueness, Let M ≡ {E ∈ σ (E) : µ (E) = ν (E)} . Then M is given to contain E and is obviously a monotone class. Therefore by Theorem 9.9.5 on monotone classes, M = σ (E) and this proves the lemma. The following lemma is also very significant.

11.2. THE TYCHONOFF THEOREM

323

Lemma 11.1.3 Let M be a metric space with the closed balls compact and suppose µ is a measure defined on the Borel sets of M which is finite on compact sets. Then there exists a unique Radon measure, µ which equals µ on the Borel sets. In particular µ must be both inner and outer regular on all Borel sets. R Proof: Define a positive linear functional, Λ (f ) = f dµ. Let µ be the Radon measure which comes from the Riesz representation theorem for positive linear functionals. Thus for all f continuous, Z Z f dµ = f dµ. If V is an open set, let {fn } be a sequence of continuous functions which is increasing and converges to XV pointwise. Then applying the monotone convergence theorem, Z Z XV dµ = µ (V ) = XV dµ = µ (V ) and so the two measures coincide on all open sets. Every compact set is a countable intersection of open sets and so the two measures coincide on all compact sets. Now let B (a, n) be a ball of radius n and let E be a Borel set contained in this ball. Then by regularity of µ there exist sets F, G such that G is a countable intersection of open sets and F is a countable union of compact sets such that F ⊆ E ⊆ G and µ (G \ F ) = 0. Now µ (G) = µ (G) and µ (F ) = µ (F ) . Thus µ (G \ F ) + µ (F ) = =

µ (G) µ (G) = µ (G \ F ) + µ (F )

and so µ (G \ F ) = µ (G \ F ) . It follows µ (E) = µ (F ) = µ (F ) = µ (G) = µ (E) . If E is an arbitrary Borel set, then µ (E ∩ B (a, n)) = µ (E ∩ B (a, n)) and letting n → ∞, this yields µ (E) = µ (E) .

11.2

The Tychonoff Theorem

Sometimes it is necessary to consider infinite Cartesian products of topological spaces. When you have finitely many topological spaces in the product and each is compact, it can be shown that the Cartesian product is compact with the product topology. It turns out that the same thing holds for infinite products but you have to be careful how you define the topology. The first thing likely to come to mind by analogy with finite products is not the right way to do it. First recall the Hausdorff maximal principle.

324


Theorem 11.2.1 (Hausdorff maximal principle) Let F be a nonempty partially ordered set. Then there exists a maximal chain. The main tool in the study of products of compact topological spaces is the Alexander subbasis theorem which is presented next. Recall a set is compact if every basic open cover admits a finite subcover. This was pretty easy to prove. However, there is a much smaller set of open sets called a subbasis which has this property. The proof of this result is much harder. Definition 11.2.2 S ⊆ τ is called a subbasis for the topology τ if the set B of finite intersections of sets of S is a basis for the topology, τ . Theorem 11.2.3 Let (X, τ ) be a topological space and let S ⊆ τ be a subbasis for τ . Then if H ⊆ X, H is compact if and only if every open cover of H consisting entirely of sets of S admits a finite subcover. Proof: The only if part is obvious because the subasic sets are themselves open. By Lemma 6.6.23 on Page 6.6.23, if every basic open cover admits a finite subcover then the set in question is compact. Suppose then that H is a subset of X having the property that subbasic open covers admit finite subcovers. Is H compact? Assume this is not so. Then what was just observed about basic covers implies there exists a basic open cover of H, O, which admits no finite subcover. Let F be defined as {O : O is a basic open cover of H which admits no finite subcover}. The assumption is that F is nonempty. Partially order F by set inclusion and use the Hausdorff maximal principle to obtain a maximal chain, C, of such open covers and let D = ∪C. If D admits a finite subcover, then since C is a chain and the finite subcover has only finitely many sets, some element of C would also admit a finite subcover, contrary to the definition of F. Therefore, D admits no finite subcover. If D0 % D and D0 is a basic open cover of H, then D0 has a finite subcover of H since otherwise, C would fail to be a maximal chain, being properly contained in C∪ {D0 }. Every set of D is of the form U = ∩m i=1 Bi , Bi ∈ S because they are all basic open sets. If it is the case that for all U ∈ D one of the Bi is found in D, then replace each such U with the subbasic set from D containing it. But then this would be a subbasic open cover of H which by assumption would admit a finite subcover contrary to the properties of D. Therefore, one of the sets of D, denoted by U , has the property that U = ∩m i=1 Bi , Bi ∈ S

11.2. THE TYCHONOFF THEOREM

325

and no Bi is in D. Thus D ∪ {Bi } admits a finite subcover, for each of the above Bi because it is strictly larger than D. Let this finite subcover corresponding to Bi be denoted by V1i , · · · , Vmi i , Bi Consider

{U, Vji , j = 1, · · · , mi , i = 1, · · · , m}.

If p ∈ H \ ∪{Vji }, then p ∈ Bi for each i and so p ∈ U . This is therefore a finite subcover of D contradicting the properties of D. Therefore, F must be empty and by Lemma 6.6.23, this proves the theorem. Let I be a set and suppose for each i ∈ I, (Xi , τ i )Qis a nonempty topological space. The Cartesian product of the Xi , denoted by i∈I Xi , consists of the set of allQchoice functions defined on I which select a single element of each XQ i . Thus f ∈ i∈I Xi means for every i ∈ I, f (i) ∈ Xi . The axiom of choice says i∈I Xi is nonempty. Let Y Pj (A) = Bi i∈I

where Bi = Xi if i 6= j and Bj = A. A subbasis for a topology on the product space consists of all sets Pj (A) where A ∈ τ j . (These sets have an open set from the topology of Xj in the j th slot and the whole space in the other slots.) Thus a basis consists of finite intersections of these sets. Note that theQintersection of two of these basic sets is another basic set and their union yields i∈I Xi . Therefore, they satisfy the condition needed for a collection of sets to serve as a basis Q for a topology. This topology is called the product topology and is denoted by Qτ i . It is tempting to define a basis for a topology to be sets of the form i∈I Ai where Ai is open in Xi . This is not the same thing at all. Note that the basis just described has at most finitely many slots filled with an open set which is not the whole space. The thing just mentioned in which every slot may be filled by a proper open set is called the box topology and there exist people who are interested in it. The Alexander subbasis theorem is used to prove the Tychonoff theorem which says Q that if each Xi is a compact topological space, then in the product topology, i∈I Xi is also compact. Q Q Theorem 11.2.4 If (Xi τ i ) is compact, then so is ( i∈I Xi , τ i ). Proof: By the Alexander subbasis theorem, the theorem will be proved if every subbasic Q open cover admits a finite subcover. Therefore, let O be a subbasic open cover of i∈I Xi . Let Oj = {Q ∈ O : Q = Pj (A) for some A ∈ τ j }. Thus Oj consists of those sets of O which have a possibly proper subset of Xi only in the slot i = j. Let π j Oj = {A : Pj (A) ∈ Oj }. Thus π j Oj picks out those proper open subsets of Xj which occur in Oj .

326


If no π j Oj covers Xj , then by the axiom of choice, there exists f∈

Y

Xi \ ∪π i Oi

i∈I

Q Therefore, f (j) ∈ / ∪π j Oj for each j ∈ I. Now f is a point of i∈I Xi and so f ∈ Pk (A) ∈ O for some k. However, this is a contradiction it was shown that f (k) is not an element of A. (A is one of the sets whose union makes up ∪π k Ok .) This contradiction shows that for some j, π j Oj covers Xj . Thus Xj = ∪π j Oj and so by compactness of Xj , there exist A1 , · · · , Am , Q sets in τ j such that Xj ⊆ m ∪m A and P (A ) ∈ O. Therefore, {P (A )} covers i j i j i i=1 i=1 i∈I Xi . By the Alexander Q subbasis theorem this proves i∈I Xi is compact.

11.3

Kolmogorov Extension Theorem n

Here Mt ≡ [−∞, ∞] t where nt is a positive integer. Let At denote finite disjoint unions the Qnntt dimensional boxes. Thus At denotes finite disjoint unions of sets of Ek where Ek is an interval. By Corollary 9.9.3 on Page 265 this is an the form k=1 algebra of sets. This algebra has the property that every set in At is the countable union of an increasing sequence of compact sets each of which is in At . I will denote a totally ordered Q index set, and the interest will be in building a measure on the product space, t∈I Mt . By the well ordering principle, you can always put an order on the index set so this order is no restriction. Also for X a topological space, B (X) willQdenote the Borel sets. Let J ⊆ I. Then if E ≡ t∈I Et , define γJ E ≡

Y

Ft

t∈I

where

½ Ft =

Et if t ∈ J Mt if t ∈ /J

Thus γ J E leaves alone Et for t ∈ J and changes the other Et into Mt . If γ J E = E, thenQthis means Et = Mt for all t ∈ / J. Also define for J a subset of I and x ∈ t∈I Mt , Y πJ x ≡ xt so π J is a continuous mapping from

t∈J

Q t∈I

πJ E ≡

Mt to Y t∈J

Et .

Q t∈J

Mt .

11.3. KOLMOGOROV EXTENSION THEOREM

327

Definition 11.3.1 Now define for J a finite subset of I, ) ( Y RJ ≡ E= Et : γ J E = E, Et ∈ At t∈I

R ≡

∪ {RJ : J ⊆ I, and J finite}

Q Thus R consists of those sets of t∈I Mt for which every slot is filled with Mt except for a finite set, J ⊆ I where the slots are filled with Et , a set of At . Define E as finite disjoint unions of sets of R and EJ is defined as finite disjoint unions of sets of RJ . First I show E is an algebra of sets. Lemma 11.3.2 The sets, E ,EJ defined above form an algebra of sets of

Q t∈I

Mt .

Proof: First consider RJ . If A, B ∈ RJ , then A ∩ B ∈ RJ also. Is A \ B a finite disjoint union of sets of RJ ? Using induction on the size of J and Corollary 9.9.3 on Page 265 this follows. Thus Q finite disjoint unions of sets of RJ form an algebra since it is clear that ∅ and t∈I Mt are both in RJ . This shows EJ is an algebra. Now suppose A, B ∈ R. Then for some finite set, J, γ J A = A, γ J B = B. Then from what was just shown, A \ B ∈ E, A ∩ B ∈ R. By Lemma 9.9.2 on Page 264 this shows E is an algebra. With this preparation here is the Kolmogorov extension theorem. In the statement and proof of the theorem, Fi , Gi , and Ei will denote Borel sets. Any list of indices from I will always be assumed to be taken in order. Thus, if J ⊆ I and J = (t1 , · · · , tn ) , it will always be assumed t1 < t2 < · · · < tn . Theorem 11.3.3 For each finite set J = (t1 , · · · , tn ) ⊆ I, supposeQthere exists a Borel probability measure, ν J = ν t1 ···tn defined on the Borel sets of t∈J Mt such that the following consistency condition holds. If (t1 , · · · , tn ) ⊆ (s1 , · · · , sp ) , then

¡ ¢ ν t1 ···tn (Ft1 × · · · × Ftn ) = ν s1 ···sp Gs1 × · · · × Gsp

(11.3.1)

328


where if si = tj , then Gsi = Ftj and if si is not equal to any of the indices, tk , then Gsi = Msi . Then for E defined in Definition 11.3.1, there exists a probability measure, P and a σ algebra F = σ (E) such that Ã ! Y Mt , P, F t∈I

is a probability space. Also there exist measurable functions, Xs : defined as Xs x ≡ xs

Q t∈I

Mt → Ms

ν t1 ···tn (Ft1 × · · · × Ftn ) = P ([Xt1 ∈ Ft1 ] ∩ · · · ∩ [Xtn ∈ Ftn ])   Ã ! n Y Y   Ft = P (Xt1 , · · · , Xtn ) ∈ Ftj = P

(11.3.2)

for each s ∈ I such that for each (t1 · · · tn ) ⊆ I,

j=1

t∈I

where Ft = Mt for every t ∈ / {t1 · · · tn } and Fti is a Borel set. Also if f is a nonnegative function of finitely many variables, xt1 , · · · , xtn , measurable with respect to ´ ³Q n B j=1 Mtj , then f is also measurable with respect to F and Z f (xt1 , · · · , xtn ) dν t1 ···tn Z =

Mt1 ×···×Mtn

Q t∈I

Mt

f (xt1 , · · · , xtn ) dP

(11.3.3)

Proof: Let E be the algebra of sets defined in Definition 11.3.1. I want to define 2 a measure on E. Suppose E = E1 ∪ ∪ · · · ¢∪ Em where γ Jk Ek = Ek and the sets, ¡E k k E are disjoint. Thus E ∈ E. Let t1 · · · tkmk = Jk . Then letting J = (s1 , · · · , sp ) ⊇ ∪m k=1 Jk define P0 (E) ≡

m X

³ ´ ν s1 ···sp Gks1 × · · · × Gksp

k=1

where

Gksi

=

Etkk j

in case si =

tkj

and Msi otherwise. By the consistency condition,

11.3.1 this is well defined and equals m X

³ ν tk1 ···tkm

k=1

k

Etkk × · · · × Etkkm 1

´ .

k

P0 is clearly finitely additive because the ν J are measures and one can pick J as large as desired. Also, from the definition, Ã ! Y P0 Mt = ν t1 (Mt1 ) = 1. t∈I


329

Next I will show P0 is a finite measure on E. After this it is only a matter of using the Caratheodory extension theorem to get the existence of the desired probability measure, P. Claim: If En ↓ ∅, then P0 (En ) ↓ 0. Proof of the claim: If not, there exists a sequence such that although En ↓ ∅, P0 (En ) ↓ ε > 0. Suppose first En =

Y

Etn

t∈I

where Etn = Mt except for t ∈ Jn , a finite subset of I. By the assumption that every set of At is the countable union of compact sets, there exists a compact set, Q n t∈Jn Kt such that Y Y Ktn ⊆ Etn t∈Jn

and

Ã ν Jn

Y

t∈Jn

! Ktn

Ã + η > ν Jn

t∈Jn

Y

! Etn

t∈Jn

where η is a small positive number. Now let Y Y Kn = Kt ⊆ Mt t∈I

t∈I

be such that for t ∈ / Jn , Kt = Mt . Doing this for each of the finitely many sets of the above form whose union is En and choosing η small enough, it follows there exists Kn ∈ E such that Kn ⊆ E n and n

P0 (K ) +

ε 2n+2

Ã =

ν Jn

Y

! Ktn

t∈Jn

Ã > ν Jn

Y

+

ε 2n+2

! Etn

≡ P0 (En )

t∈Jn

By Tychonoff, theorem, Kn is compact. The interesting thing about these Kn is they have the finite intersection property. Here is why. ¡ ¢ ¡ ¢ k k ε ≤ P0 ∩m + P0 Em \ ∩m k=1 K k=1 K ¢ ¢ ¡ ¡ k k k + P0 ∪m ≤ P0 ∩m k=1 E \ K k=1 K ∞ ¡ ¢ ¡ ¢ X ε k k < P0 ∩m + ε/2 < P0 ∩m K + k=1 K k=1 k+2 2 k=1

330


¡ ¢ k and so P0 ∩m > ε/2. Now this yields a contradiction because this finite ink=1 K tersection property implies the intersection of all the Kn is nonempty, contradicting En ↓ ∅ since each Kn is contained in En . ∞ With the claim, it follows P0 is a measure on E. Here is why: If E = ∪k=1 Ek n k where E, E ∈ E, then (E \ ∪k=1 Ek ) ↓ ∅ and so P0 (∪nk=1 Ek ) → P0 (E) . Pn Hence if the Ek are disjoint, P0 (∪nk=1 Ek ) = k=1 P0 (Ek ) → P0 (E) . Now to conclude the proof, apply the Caratheodory extension theorem to obtain P a probability measure which extends P0 to σ (E) the sigma algebra generated by E. ¡Q ¢ Let F ≡ σ (E) . It follows t∈I Mt , F, P is a probability measure space with the property that when γ J (E) = E for J = (t1 · · · tn ) a finite subset of I, P (E) = P0 (E) = ν t1 ···tn (Et1 × · · · × Etn ) . ¡Q ¢ Q Next, let be the probability space and for x ∈ t∈I Mt let t∈I Mt , F, P Xt (x) = xt , the tth entry of x. (xt = π t x). It follows Xt is measurable because if U is open in Mt , then Xt−1 (U ) has a U in the tth slot and Ms everywhere else for s 6= t. Thus inverse images of open sets are measurable. Also, letting (t1 · · · tn ) be a finite subset of I and Ft1 , · · · , Ftn be Borel sets in Mt1 · · · Mtn respectively, P ([Xt1 ∈ Ft1 ] ∩ [Xt2 ∈ Ft2 ] ∩ · · · ∩ [Xtn ∈ Ftn ]) = Ã ! Y P ((Xt1 , Xt2 , · · · , Xtn ) ∈ Ft1 × · · · × Ftn ) = P Ft t∈I

= ν t1 ···tn (Ft1 × · · · × Ftn ) where Ft ≡ Mt if t ∈ / {t1 · · · tn }. Finally consider the claim Q about the integrals. Suppose f (xt1 , · · · , xtn ) = XF where F is a Borel set of t∈J Mt where J = {t1 , · · · , tn }. To begin with suppose F = Ft1 × · · · × Ftn

(11.3.4)

where each Ftj is in Atj . Then Z XF (xt1 , · · · , xtn ) dν t1 ···tn = ν t1 ···tn (Ft1 × · · · × Ftn ) Mt1 ×···×Mtn

=

P

Ã Y

Z =

! Ft

t∈I

Q t∈I

Mt

Z =

Q t∈I

Mt

XQt∈I Ft (x) dP

XF (xt1 , · · · , xtn ) dP

(11.3.5)


331

where Ft = Mt if t ∈ / J. Let K denote sets, F of¡Q the sort ¢in 11.3.4. It is clearly a π system. Now let G denote those sets, F in B t∈J Mt such that 11.3.5 holds. Thus G ⊇ K. It is clear that G is closed with respect disjoint unions ¡Qto countable ¢ and complements. Hence G ⊇ σ (K) but σ (K) = B M because every open t t∈J Q set in t∈J Mt is the ¡countable¢ union of rectangles like 11.3.4. Therefore, 11.3.5 Q holds for every F ∈ B t∈J Mt . Passing to simple functions and then using the monotone convergence theorem yields the final claim of the theorem. This proves the theorem. n The next task is to consider the case where Mt = (−∞, ∞) t . To consider this case, here is a lemma which will allow this case to be deduced from the above theorem. In this lemma, n Mt0 ≡ [−∞, ∞] t . Q Lemma 11.3.4 Let J be a finite subset ofQI. Then U is a Borel set in Qt∈J Mt if and only if there exists a Borel set, U0 in t∈J Mt0 such that U = U0 ∩ t∈J Mt . Borel is replaced with open. If U is an open Q set in0 Q Proof: First suppose Q 0 M from the definition of the open sets of t t∈J Mt t∈J t∈J Mt , then it is open in Q which consist of open sets of t∈J Mt along with complements of compact sets 0 which have the points Q let U0 = U in this case. Next Q ±∞ added in. 0 Thus you can 0 suppose U = U ∩ t∈J Mt where U is open in t∈J Mt . I need show U is open Q in Qt∈J Mt . Letting x ∈ U, it follows xt 6= ±∞ for each t. It follows from U0 open in t∈J Mt0 that xt ∈ Vt0 where Vt0 is open in Mt0 and x∈

Y

Vt0 ⊆ U0

t∈J

Letting Vt = Vt0 \ {∞, −∞} , it follows Vt is open in Mt and x∈

Y t∈J

Vt ⊆ U0 ∩

Y

Mt .

t∈J

Q This shows that U is an open set in t∈J Mt . Now let ( ) Y Y Y G ≡ F Borel in Mt0 such that F ∩ Mt is Borel in Mt t∈J

t∈J

t∈J

then from what was just shown G contains Q the open sets. It is also clearly a σ algebra. Hence G equals the Borel sets of t∈J Mt0Q . It only remains to verify that any Borel set in t∈J Mt is the intersection of a Q Q Borel set of t∈J Mt0 with t∈J Mt . Let ( H≡

F Borel in

Y t∈J

0

Mt such that F = F ∩

Y t∈J

0

Mt , F Borel in

Y t∈J

) Mt0

332


From the first part of the argument, n } be a Q H contains the 0 open sets. Now Q let {F 0 0 sequence in H. Thus F = F ∩ M where F is Borel in M . Then n t n n t t∈J t∈J Q Q ∪n Fn = ∪n F0n ∩ t∈J Mt and ∪n F0n is Borel inQ t∈J Mt0 . Thus H is Q closed under countable unions. Next let¡ F ∈ H and F¢ = F0 ∩ t∈J Mt . Then F0C ≡ t∈J Mt0 \F0 Q Q Q 0C is Borel in t∈J Mt0 and ∩ t∈J Mt . Thus t∈J Mt \ F = F Q H is a σ algebra containing the open sets and so H equals the Borel sets in t∈J Mt . This proves the lemma. Now here is the Kolmogorov extension theorem. Theorem 11.3.5 (Kolmogorov extension theorem) For each finite set J = (t1 , · · · , tn ) ⊆ I, supposeQthere exists a Borel probability measure, ν J = ν t1 ···tn defined on the Borel sets of t∈J Mt for Mt = Rnt for nt an integer, such that the following consistency condition holds. If (t1 , · · · , tn ) ⊆ (s1 , · · · , sp ) , then

¡ ¢ ν t1 ···tn (Ft1 × · · · × Ftn ) = ν s1 ···sp Gs1 × · · · × Gsp

(11.3.6)

where if si = tj , then Gsi = Ftj and if si is not equal to any of the indices, tk , then Gsi = Msi . Then for E defined as in Definition 11.3.1, adjusted so that ±∞ never appears as any endpoint of any interval, there exists a probability measure, P and a σ algebra F = σ (E) such that Ã ! Y Mt , P, F t∈I

Q t∈I

Mt → Ms

ν t1 ···tn (Ft1 × · · · × Ftn ) = P ([Xt1 ∈ Ft1 ] ∩ · · · ∩ [Xtn ∈ Ftn ])   Ã ! n Y Y   = P (Xt1 , · · · , Xtn ) ∈ Ftj = P Ft

(11.3.7)

is a probability space. Also there exist measurable functions, Xs : defined as Xs x ≡ xs for each s ∈ I such that for each (t1 · · · tn ) ⊆ I,

j=1

t∈I

where Ft = Mt for every t ∈ / {t1 · · · tn } and Fti is a Borel set. Also if f is a nonnegative function of finitely many variables, xt1 , · · · , xtn , measurable with respect to ³Q ´ n B j=1 Mtj , then f is also measurable with respect to F and Z f (xt1 , · · · , xtn ) dν t1 ···tn Z =

Mt1 ×···×Mtn

Q t∈I

Mt

f (xt1 , · · · , xtn ) dP

(11.3.8)

11.4. EXERCISES

333

0 Proof: Using Lemma 11.3.4, extend each measure, ν¡J to M ¢ by adding Qt , defined E ∩ for all E ∈ in¡the points ±∞ at the ends, by letting ν (E) ≡ ν M J J t t∈I ¢ Q 0 M . Then apply Theorem B 11.3.3 to these extended measures and use the t t∈I definition of the extensions of each ν J to replace each Mt0 with Mt everywhere it occurs. This proves the theorem. As a special case, you can obtain a version of product measure for possibly infinitely many factors. Suppose in the context of the above theorem that ν t is a probability measure defined on the Borel sets of Mt ≡ Rnt forQnt a positive integer, n and let the measures, ν t1 ···tn be defined on the Borel sets of i=1 Mti by

ν t1 ···tn (E) ≡ (ν t1 × · · · × ν tn ) (E) . Then these measures satisfy the necessary consistency condition and so the Kolmogorov extension theorem to obtain a measure, P ¡Q ¢ given above can be applied Q defined on a t∈I Mt , F and measurable functions Xs : t∈I Mt → Ms such that for Fti a Borel set in Mti , Ã P

(Xt1 , · · · , Xtn ) ∈

n Y

! Fti

= ν t1 ···tn (Ft1 × · · · × Ftn )

i=1

= ν t1 (Ft1 ) · · · ν tn (Ftn ) .

(11.3.9)

In particular, P (Xt ∈ Ft ) = ν t (Ft ) . Then P in the resulting probability space, Ã

Y

! Mt , F, P

t∈I

Q will be denoted as t∈I ν t . This proves the following theorem which describes an infinite product measure. Theorem 11.3.6 Let Mt for t ∈ I be given as in Theorem 11.3.5 and let ν t be a Borel probability measure defined on the Borel sets of Mt . Then there exists a measure, Q P and a σ algebra F = σ (E) where E is given in Definition 11.3.1 such that ( t Mt , F, P ) is a probability space satisfying 11.3.9 wheneverQeach Fti is a Borel set of Mti . This probability measure is sometimes denoted as t ν t .

11.4

Exercises

1. Let (X, S, µ) and (Y, F, λ) be two finite measure spaces. A subset of X × Y is called a measurable rectangle if it is of the form A × B where A ∈ S and B ∈ F. A subset of X × Y is called an elementary set if it is a finite disjoint union of measurable rectangles. Denote this set of functions by E. Show that E is an algebra of sets.

334


2. ↑For A ∈ σ (E) , the smallest σ algebra containing E, show that x → XA (x, y) is µ measurable and that Z y → XA (x, y) dµ is λ measurable. Show similar assertions hold for y → XA (x, y) and Z x → XA (x, y) dλ and that

Z Z

Z Z XA (x, y) dµdλ =

XA (x, y) dλdµ.

(11.4.10)

Hint: Let M ≡ {A ∈ σ (E) : 11.4.10 holds} along with all relevant measurability assertions. Show M contains E and is a monotone class. Then apply the Theorem 9.9.5. RR 3. ↑For A ∈ σ (E) define (µ × λ) (A) ≡ XA (x, y) dµdλ. Show that (µ × λ) is a measur on σ (E) and that whenever f ≥ 0 is measurable with respect to σ (E) , Z Z Z Z Z f d (µ × λ) = f (x, y) dµdλ = f (x, y) dλdµ. X×Y

This is a common approach to Fubini’s theorem. 4. ↑Generalize the above version of Fubini’s theorem to the case where the measure spaces are only σ finite. 5. ↑Suppose now that µ and λ are both complete σ finite measures. Let (µ × λ) denote the completion´ of this measure. Let the larger measure space be ³ X × Y, σ (E), (µ × λ) . Thus if E ∈ σ (E), it follows there exists a set A ∈ σ (E) such that E ∪ N = A where (µ × λ) (N ) = 0. Now argue that for λ a.e. y, x → XN (x, y) is measurable because it is equal to zero µ a.e. and µ is complete. Therefore, Z Z XN (x, y) dµdλ makes sense and equals zero.R Use to argue that for λ a.e. y, x → XE (x, y) is R µ measurable and equals XA (x, y) dµ. Then by completeness of λ, y → XE (x, y) dµ is λ measurable and Z Z Z Z XA (x, y) dµdλ = XE (x, y) dµdλ = (µ × λ) (E) . Similarly

Z Z XE (x, y) dλdµ = (µ × λ) (E) .

11.4. EXERCISES

335

Use this to give a generalization of the above Fubini theorem. Prove that if f is measurable with respect to the σ algebra, σ (E) and nonnegative, then Z Z Z Z Z f d(µ × λ) = f (x, y) dµdλ = f (x, y) dλdµ X×Y

where the iterated integrals make sense.

336


The Lp Spaces 12.1

Basic Inequalities And Properties

One of the main applications of the Lebesgue integral is to the study of various sorts of functions space. These are vector spaces whose elements are functions of various types. One of the most important examples of a function space is the space of measurable functions whose absolute values are pth power integrable where p ≥ 1. These spaces, referred to as Lp spaces, are very useful in applications. In the chapter (Ω, S, µ) will be a measure space. Definition 12.1.1 Let 1 ≤ p < ∞. Define Z Lp (Ω) ≡ {f : f is measurable and

|f (ω)|p dµ < ∞} Ω

In terms of the distribution function, Z Lp (Ω) = {f : f is measurable and

∞

ptp−1 µ ([|f | > t]) dt < ∞}

0

For each p > 1 define q by

1 1 + = 1. p q

Often one uses p0 instead of q in this context. Lp (Ω) is a vector space and has a norm. This is similar to the situation for Rn but the proof requires the following fundamental inequality. . Theorem 12.1.2 (Holder’s inequality) If f and g are measurable functions, then if p > 1, µZ ¶ p1 µZ ¶ q1 Z p q |f | |g| dµ ≤ |f | dµ |g| dµ . (12.1.1) Proof: First here is a proof of Young’s inequality . Lemma 12.1.3 If

p > 1, and 0 ≤ a, b then ab ≤ 337

ap p

+

bq q .

THE LP SPACES

338 Proof: Consider the following picture:

b

x x = tp−1 t = xq−1 a

t

From this picture, the sum of the area between the x axis and the curve added to the area between the t axis and the curve is at least as large as ab. Using beginning calculus, this is equivalent to the following inequality. Z a Z b ap bq ab ≤ tp−1 dt + xq−1 dx = + . p q 0 0 The above picture represents the situation which occurs when p > 2 because the graph of the function is concave up. If 2 ≥ p > 1 the graph would be concave down or a straight line. You should verify that the same argument holds in these cases just as well. In fact, the only thing which matters in the above inequality is that the function x = tp−1 be strictly increasing. Note equality occurs when ap = bq . Here is an alternate proof. Lemma 12.1.4 For a, b ≥ 0, ab ≤

ap bq + p q

and equality occurs when if and only if ap = bq . Proof: If b = 0, the inequality is obvious. Fix b > 0 and consider f (a) ≡

ap bq + − ab. p q

Then f 0 (a) = ap−1 − b. This is negative when a < b1/(p−1) and is positive when a > b1/(p−1) . Therefore, f has a minimum when a = b1/(p−1) . In other words, when ap = bp/(p−1) = bq since 1/p + 1/q = 1. Thus the minimum value of f is bq bq + − b1/(p−1) b = bq − bq = 0. p q It follows f ≥ 0 and this yields the desired inequality. R R Proof of Holder’s inequality: If either |f |p dµ or |g|p dµ equals R ∞, the inequality 12.1.1 is obviously valid because ∞ ≥ anything. If either |f |p dµ or

12.1. BASIC INEQUALITIES AND PROPERTIES

339

R

|g|p dµ equals 0, then f = 0 a.e. or that g = 0 a.e. and so in this case the left side of theRinequality equals 0 and so the inequality is therefore true. Therefore assume R both |f |p dµ and |g|p dµ are less than ∞ and not equal to 0. Let µZ

¶1/p p

|f | dµ and let

¡R

¢1/q |g|p dµ = I (g). Then using the lemma, Z

Hence,

= I (f )

1 |f | |g| dµ ≤ I (f ) I (g) p

Z

|f |p 1 p dµ + q I (f ) µZ

Z

Z

|g|q q dµ = 1. I (g)

¶1/p µZ p

|f | |g| dµ ≤ I (f ) I (g) =

¶1/q q

|f | dµ

|g| dµ

.

This proves Holder’s inequality. The following lemma will be needed. Lemma 12.1.5 Suppose x, y ∈ C. Then p

p

p

|x + y| ≤ 2p−1 (|x| + |y| ) . Proof: The function f (t) = tp is concave up for t ≥ 0 because p > 1. Therefore, the secant line joining two points on the graph of this function must lie above the graph of the function. This is illustrated in the following picture.

¡

¡ ¡

¡

¡

(|x| + |y|)/2 = m

¡

|x|

m

|y|

Now as shown above, µ

|x| + |y| 2

¶p

p

≤

|x| + |y| 2

p

which implies p

p

p

p

|x + y| ≤ (|x| + |y|) ≤ 2p−1 (|x| + |y| ) and this proves the lemma. Note that if y = φ (x) is any function for which the graph of φ is concave up, you could get a similar inequality by the same argument.

THE LP SPACES

340 Corollary 12.1.6 (Minkowski inequality) Let 1 ≤ p < ∞. Then µZ p

¶1/p

|f + g| dµ

µZ ≤

¶1/p µZ ¶1/p p p |f | dµ + |g| dµ .

(12.1.2)

Proof: If p = 1, this is obvious because it is just the triangle inequality. Let p > 1. Without loss of generality, assume µZ

¶1/p ¶1/p µZ p p 0 there exists N such that if n, m ≥ N , then ||fn − fm ||p < ε. Now select a subsequence as follows. Let n1 be such that ||fn − fm ||p < 2−1 whenever n, m ≥ n1 . 1 These spaces are named after Stefan Banach, 1892-1945. Banach spaces are the basic item of study in the subject of functional analysis and will be considered later in this book. There is a recent biography of Banach, R. Katu˙za, The Life of Stefan Banach, (A. Kostant and W. Woyczy´ nski, translators and editors) Birkhauser, Boston (1996). More information on Banach can also be found in a recent short article written by Douglas Henderson who is in the department of chemistry and biochemistry at BYU. Banach was born in Austria, worked in Poland and died in the Ukraine but never moved. This is because borders kept changing. There is a rumor that he died in a German concentration camp which is apparently not true. It seems he died after the war of lung cancer. He was an interesting character. He hated taking examinations so much that he did not receive his undergraduate university degree. Nevertheless, he did become a professor of mathematics due to his important research. He and some friends would meet in a cafe called the Scottish cafe where they wrote on the marble table tops until Banach’s wife supplied them with a notebook which became the ”Scotish notebook” and was eventually published.

THE LP SPACES

342

Let n2 be such that n2 > n1 and ||fn − fm ||p < 2−2 whenever n, m ≥ n2 . If n1 , · · · , nk have been chosen, let nk+1 > nk and whenever n, m ≥ nk+1 , ||fn − fm ||p < 2−(k+1) . The subsequence just mentioned is {fnk }. Thus, ||fnk −fnk+1 ||p < 2−k . Let gk+1 = fnk+1 − fnk . Then by the corollary to Minkowski’s inequality, ∞>

∞ X

||gk+1 ||p ≥

k=1

for all m. It follows that Z ÃX m

m X k=1

¯¯ ¯¯ m ¯¯ ¯¯ X ¯¯ ¯¯ |gk+1 |¯¯ ||gk+1 ||p ≥ ¯¯ ¯¯ ¯¯

!p

k=1

Ã

|gk+1 |

dµ ≤

k=1

∞ X

p

!p ||gk+1 ||p

n. Thus fn is uniformly bounded and product measurable. By the above lemma, Z

µZ

¶ p1 µZ Z |fn (x, y)| dλ dµ ≥ ( p

Xm

Yk

Yk

¶ p1 |fn (x, y)| dµ) dλ . p

(12.1.9)

Xm

Now observe that |fn (x, y)| increases in n and the pointwise limit is |f (x, y)|. Therefore, using the monotone convergence theorem in 12.1.9 yields the same inequality with f replacing fn . Next let k → ∞ and use the monotone convergence theorem again to replace Yk with Y . Finally let m → ∞ in what is left to obtain 12.1.8. This proves the theorem.

12.2. DENSITY CONSIDERATIONS

345

Note that the proof of this theorem depends on two manipulations, the interchange of the order of integration and Holder’s inequality. Note that there is nothing to check in the case of double sums. Thus if aij ≥ 0, it is always the case that  

Ã X X j

!p 1/p aij

i



≤

X i

 1/p X p  aij  j

because the integrals in this case are just sums and (i, j) → aij is measurable. The Lp spaces have many important properties.

12.2

Density Considerations

Theorem 12.2.1 Let p ≥ 1 and let (Ω, S, µ) be a measure space. Then the simple functions are dense in Lp (Ω). Proof: Recall that a function, f , having values in R can be written in the form f = f + − f − where f + = max (0, f ) , f − = max (0, −f ) . Therefore, an arbitrary complex valued function, f is of the form ¡ ¢ f = Re f + − Re f − + i Im f + − Im f − . If each of these nonnegative functions is approximated by a simple function, it follows f is also approximated by a simple function. Therefore, there is no loss of generality in assuming at the outset that f ≥ 0. Since f is measurable, Theorem 8.3.9 implies there is an increasing sequence of simple functions, {sn }, converging pointwise to f (x). Now |f (x) − sn (x)| ≤ |f (x)|. By the Dominated Convergence theorem, Z 0 = lim |f (x) − sn (x)|p dµ. n→∞

Thus simple functions are dense in Lp . Recall that for Ω a topological space, Cc (Ω) is the space of continuous functions with compact support in Ω. Also recall the following definition. Definition 12.2.2 Let (Ω, S, µ) be a measure space and suppose (Ω, τ ) is also a topological space. Then (Ω, S, µ) is called a regular measure space if the σ algebra of Borel sets is contained in S and for all E ∈ S, µ(E) = inf{µ(V ) : V ⊇ E and V open}

THE LP SPACES

346 and if µ (E) < ∞, µ(E) = sup{µ(K) : K ⊆ E and K is compact } and µ (K) < ∞ for any compact set, K.

For example Lebesgue measure is an example of such a measure. Lemma 12.2.3 Let Ω be a metric space in which the closed balls are compact and let K be a compact subset of V , an open set. Then there exists a continuous function f : Ω → [0, 1] such that f (x) = 1 for all x ∈ K and spt(f ) is a compact subset of V . That is, K ≺ f ≺ V. Proof: Let K ⊆ W ⊆ W ⊆ V and W is compact. To obtain this list of inclusions consider a point in K, x, and take B (x, rx ) a ball containing x such that B (x, rx ) is a compact subset of V . Next use the fact that K is compact to obtain m the existence of a list, {B (xi , rxi /2)}i=1 which covers K. Then let ³ r ´ xi W ≡ ∪m . i=1 B xi , 2 It follows since this is a finite union that ³ r ´ x W = ∪m B xi , i i=1 2 and so W , being a finite union of compact sets is itself a compact set. Also, from the construction W ⊆ ∪m i=1 B (xi , rxi ) . Define f by f (x) =

dist(x, W C ) . dist(x, K) + dist(x, W C )

It is clear that f is continuous if the denominator is always nonzero. But this is clear because if x ∈ W C there must be a ball B (x, r) such that this ball does not intersect K. Otherwise, x would be a limit point of K and since K is closed, x ∈ K. However, x ∈ / K because K ⊆ W . It is not necessary to be in a metric space to do this. You can accomplish the same thing using Urysohn’s lemma. Theorem 12.2.4 Let (Ω, S, µ) be a regular measure space as in Definition 12.2.2 where the conclusion of Lemma 12.2.3 holds. Then Cc (Ω) is dense in Lp (Ω). Proof: First consider a measurable set, E where µ (E) < ∞. Let K ⊆ E ⊆ V where µ (V \ K) < ε. Now let K ≺ h ≺ V. Then Z Z p |h − XE | dµ ≤ XVp \K dµ = µ (V \ K) < ε.

12.3. SEPARABILITY

347

It follows that for each s a simple function in Lp (Ω) , there exists h ∈ Cc (Ω) such that ||s − h||p < ε. This is because if s(x) =

m X

ci XEi (x)

i=1

is a simple function in Lp where the ci are the distinct nonzero values of s each µ (Ei ) < ∞ since otherwise s ∈ / Lp due to the inequality Z p p |s| dµ ≥ |ci | µ (Ei ) . By Theorem 12.2.1, simple functions are dense in Lp (Ω), and so this proves the Theorem.

12.3

Separability

Theorem 12.3.1 For p ≥ 1 and µ a Radon measure, Lp (Rn , µ) is separable. Recall this means there exists a countable set, D, such that if f ∈ Lp (Rn , µ) and ε > 0, there exists g ∈ D such that ||f − g||p < ε. Proof: Let Q be all functions of the form cX[a,b) where [a, b) ≡ [a1 , b1 ) × [a2 , b2 ) × · · · × [an , bn ), and both ai , bi are rational, while c has rational real and imaginary parts. Let D be the set of all finite sums of functions in Q. Thus, D is countable. In fact D is dense in Lp (Rn , µ). To prove this it is necessary to show that for every f ∈ Lp (Rn , µ), there exists an element of D, s such that ||s − f ||p < ε. If it can be shown that for every g ∈ Cc (Rn ) there exists h ∈ D such that ||g − h||p < ε, then this will suffice because if f ∈ Lp (Rn ) is arbitrary, Theorem 12.2.4 implies there exists g ∈ Cc (Rn ) such that ||f − g||p ≤ 2ε and then there would exist h ∈ Cc (Rn ) such that ||h − g||p < 2ε . By the triangle inequality, ||f − h||p ≤ ||h − g||p + ||g − f ||p < ε. Therefore, assume at the outset that f ∈ Cc (Rn ).Q n Let Pm consist of all sets of the form [a, b) ≡ i=1 [ai , bi ) where ai = j2−m and −m bi = (j + 1)2 for j an integer. Thus Pm consists of a tiling of Rn into half open 1 rectangles having diameters 2−m n 2 . There are countably many of these rectangles; n ∞ m so, let Pm = {[ai , bi )}∞ i=1 and R = ∪i=1 [ai , bi ). Let ci be complex numbers with rational real and imaginary parts satisfying −m |f (ai ) − cm , i | 0 there exists δ > 0 such that if |x − y| < δ, |f (x) − f (y)| < ε/2. Now let m be large enough that every box in Pm has diameter less than δ and also that 2−m < ε/2. Then if [ai , bi ) is one of these boxes of Pm , and x ∈ [ai , bi ), |f (x) − f (ai )| < ε/2 and

−m < ε/2. |f (ai ) − cm i | 0, there exists g ∈ D such that ||f − g||p < ε. Proof: Let P denote the set of all polynomials which have rational coefficients. n n Then P is countable. Let τ k ∈ Cc ((− (k + 1) , (k + 1)) ) such that [−k, k] ≺ τ k ≺ n (− (k + 1) , (k + 1)) . Let Dk denote the functions which are of the form, pτ k where p ∈ P. Thus Dk is also countable. Let D ≡ ∪∞ k=1 Dk . It follows each function in D is in Cc (Rn ) and so it in Lp (Rn , µ). Let f ∈ Lp (Rn , µ). By regularity of µ there exists n g ∈ Cc (Rn ) such that ||f − g||Lp (Rn ,µ) < 3ε . Let k be such that spt (g) ⊆ (−k, k) . Now by the Weierstrass approximation theorem there exists a polynomial q such that ||g − q||[−(k+1),k+1]n

≡
0 be given. Choose g ∈ Cc (Rn ) such that ||f − g||p < 2ε . This can be done by using Theorem 12.2.4. Now let Z Z gm (x) = g ∗ ψ m (x) ≡ g (x − y) ψ m (y) dmn (y) = g (y) ψ m (x − y) dmn (y) where {ψ m } is a mollifier. It follows from Lemma 12.5.7 gm ∈ Cc∞ (Rn ). It vanishes 1 if x ∈ / spt(g) + B(0, m ). ¶ p1 µZ Z p ||g − gm ||p = |g(x) − g(x − y)ψ m (y)dmn (y)| dmn (x) ≤

µZ Z ¶ p1 p ( |g(x) − g(x − y)|ψ m (y)dmn (y)) dmn (x) Z µZ p

≤

¶ p1

|g(x) − g(x − y)| dmn (x) Z

= 1 B(0, m )

||g − gy ||p ψ m (y)dmn (y)
0 and

1 p

+ 1q = 1. Show

directly that f (x, y) ≥ 1 for all such x, y and show this implies xy ≤

xp p

q

+ yq .

5. Give an example of a sequence of functions in Lp (R) which converges to zero in Lp but does not converge pointwise to 0. Does this contradict the proof of the theorem that Lp is complete? 6. Let K be a bounded subset of Lp (Rn ) and suppose that there exists G such that G is compact with Z p |u (x)| dx < εp Rn \G

and for all ε > 0, there exist a δ > 0 and such that if |h| < δ, then Z p |u (x + h) − u (x)| dx < εp

THE LP SPACES

356

for all u ∈ K. Show that K is precompact in Lp (Rn ). Hint: Let φk be a mollifier and consider Kk ≡ {u ∗ φk : u ∈ K} . Verify the conditions of the Ascoli Arzela theorem for these functions defined on G and show there is an ε net for each ε > 0. Can you modify this to let an arbitrary open set take the place of Rn ? 7. Let (Ω, d) be a metric space and suppose also that (Ω, S, µ) is a regular measure space such that µ (Ω) < ∞ and let f ∈ L1 (Ω) where f has complex values. Show that for every ε > 0, there exists an open set of measure less than ε, denoted here by V and a continuous function, g defined on Ω such that f = g on V C . Thus, aside from a set of small measure, f is continuous. If |f (ω)| ≤ M , show that it can be assumed that |g (ω)| ≤ M . This is called Lusin’s theorem. Hint: Use Theorems 12.2.4 and 12.1.10 to obtain a sequence of functions in Cc (Ω) , {gn } which converges pointwise a.e. to f and then use Egoroff’s theorem to obtain a small set, W of measure less than ε/2 such that convergence on W C . Now let F be a closed subset of W C such ¡ C is uniform ¢ that µ W \ F < ε/2. Let V = F C . Thus µ (V ) < ε and on F = V C , the convergence of {gn } is uniform showing that the restriction of f to V C is continuous. Now use the Tietze extension theorem. R 8. Let φm ∈ Cc∞ (Rn ), φm (x) ≥ 0,and Rn φm (y)dy = 1 with lim sup {|x| : x ∈ spt (φm )} = 0.

m→∞

Show if f ∈ Lp (Rn ), limm→∞ f ∗ φm = f in Lp (Rn ). 9. Let φ : R → R be convex. This means φ(λx + (1 − λ)y) ≤ λφ(x) + (1 − λ)φ(y) whenever λ ∈ [0, 1]. Verify that if x < y < z, then

φ(y)−φ(x) y−x

≤

φ(z)−φ(y) z−y

and that φ(z)−φ(x) ≤ φ(z)−φ(y) . Show if s ∈ R there exists λ such that z−x z−y φ(s) ≤ φ(t) + λ(s − t) for all t. Show that if φ is convex, then φ is continuous. 10. ↑ Prove Jensen’s inequality. If φ R: R → R is convex, µ(Ω) = 1,R and f : Ω → R R is in L1 (Ω), then φ( Ω f du) ≤ Ω φ(f )dµ. Hint: Let s = Ω f dµ and use Problem 9. 0

11. Let p1 + p10 = 1, p > 1, let f ∈ Lp (R), g ∈ Lp (R). Show f ∗ g is uniformly continuous on R and |(f ∗ g)(x)| ≤ ||f ||Lp ||g||Lp0 . Hint: You need to consider why f ∗ g exists and then this follows from the definition of convolution and continuity of translation in Lp . R1 R∞ 12. B(p, q) = 0 xp−1 (1 − x)q−1 dx, Γ(p) = 0 e−t tp−1 dt for p, q > 0. The first of these is called the beta function, while the second is the gamma function. Show a.) Γ(p + 1) = pΓ(p); b.) Γ(p)Γ(q) = B(p, q)Γ(p + q).

12.6. EXERCISES

357

13. Let f ∈ Cc (0, ∞) and define F (x) = ||F ||Lp (0,∞) ≤

1 x

Rx 0

f (t)dt. Show

p ||f ||Lp (0,∞) whenever p > 1. p−1

Hint: Argue there isR no loss of generality in assuming f ≥ 0 and then assume ∞ this is so. Integrate 0 |F (x)|p dx by parts as follows: Z

∞ 0

show = 0

Z z }| { p ∞ F dx = xF |0 − p

∞

p

xF p−1 F 0 dx.

0

Now show xF 0 = f − F and use this in the last integral. Complete the argument by using Holder’s inequality and p − 1 = p/q. 14. ↑ Now supposeRf ∈ Lp (0, ∞), p > 1, and f not necessarily in Cc (0, ∞). Show x that F (x) = x1 0 f (t)dt still makes sense for each x > 0. Show the inequality of Problem 13 is still valid. This inequality is called Hardy’s inequality. Hint: To show this, use the above inequality along with the density of Cc (0, ∞) in Lp (0, ∞). 15. Suppose f, g ≥ 0. When does equality hold in Holder’s inequality? 16. Prove Vitali’s Convergence theorem: Let {fn } be uniformly integrable and complex valued, µ(Ω) R < ∞, fn (x) → f (x) a.e. where f is measurable. Then f ∈ L1 and limn→∞ Ω |fn − f |dµ = 0. Hint: Use Egoroff’s theorem to show {fn } is a Cauchy sequence in L1 (Ω). This yields a different and easier proof than what was done earlier. See Theorem 8.5.3 on Page 213. 17. ↑ Show the Vitali Convergence theorem implies the Dominated Convergence theorem for finite measure spaces but there exist examples where the Vitali convergence theorem works and the dominated convergence theorem does not. 18. ↑ Suppose µ(Ω) < ∞, {fn } ⊆ L1 (Ω), and Z h (|fn |) dµ < C Ω

for all n where h is a continuous, nonnegative function satisfying lim

t→∞

h (t) = ∞. t

Show {fn } is uniformly integrable. In applications, this often occurs in the form of a bound on ||fn ||p . 19. ↑ Sometimes, especially in books on probability, a different definition of uniform integrability is used than that presented here. A set of functions, S,

THE LP SPACES

358

defined on a finite measure space, (Ω, S, µ) is said to be uniformly integrable if for all ε > 0 there exists α > 0 such that for all f ∈ S, Z |f | dµ ≤ ε. [|f |≥α]

Show that this definition is equivalent to the definition of uniform integrability given earlier in Definition 8.5.1 on Page 212 with the addition of the condition that there is a constant, C < ∞ such that Z |f | dµ ≤ C for all f ∈ S. 20. f ∈ L∞ (Ω, µ) if there exists a set of measure zero, E, and a constant C < ∞ such that |f (x)| ≤ C for all x ∈ / E. ||f ||∞ ≡ inf{C : |f (x)| ≤ C a.e.}. Show || ||∞ is a norm on L∞ (Ω, µ) provided f and g are identified if f (x) = g(x) a.e. Show L∞ (Ω, µ) is complete. Hint: You might want to show that [|f | > ||f ||∞ ] has measure zero so ||f ||∞ is the smallest number at least as large as |f (x)| for a.e. x. Thus ||f ||∞ is one of the constants, C in the above. 21. Suppose f ∈ L∞ ∩ L1 . Show limp→∞ ||f ||Lp = ||f ||∞ . Hint: Z p p (||f ||∞ − ε) µ ([|f | > ||f ||∞ − ε]) ≤ |f | dµ ≤ [|f |>||f ||∞ −ε] Z Z Z p p−1 p−1 |f | dµ = |f | |f | dµ ≤ ||f ||∞ |f | dµ. Now raise both ends to the 1/p power and take lim inf and lim sup as p → ∞. You should get ||f ||∞ − ε ≤ lim inf ||f ||p ≤ lim sup ||f ||p ≤ ||f ||∞ 22. Suppose µ(Ω) < ∞. Show that if 1 ≤ p < q, then Lq (Ω) ⊆ Lp (Ω). Hint Use Holder’s inequality. 23. Show L1 (R)√* L2 (R) and L2 (R) * L1 (R) if Lebesgue measure is used. Hint: Consider 1/ x and 1/x. 24. Suppose that θ ∈ [0, 1] and r, s, q > 0 with 1 θ 1−θ = + . q r s show that

Z Z Z ( |f |q dµ)1/q ≤ (( |f |r dµ)1/r )θ (( |f |s dµ)1/s )1−θ.

12.6. EXERCISES

359

If q, r, s ≥ 1 this says that ||f ||q ≤ ||f ||θr ||f ||1−θ . s Using this, show that ³ ´ ln ||f ||q ≤ θ ln (||f ||r ) + (1 − θ) ln (||f ||s ) . Hint:

Z

Z |f |q dµ =

Now note that 1 =

θq r

+

q(1−θ) s

|f |qθ |f |q(1−θ) dµ.

and use Holder’s inequality.

25. Suppose f is a function in L1 (R) and f is infinitely differentiable. Is f 0 ∈ L1 (R)? Hint: What if φ ∈ Cc∞ (0, 1) and f (x) = φ (2n (x − n)) for x ∈ (n, n + 1) , f (x) = 0 if x < 0?

360

THE LP SPACES

Banach Spaces 13.1

Theorems Based On Baire Category

13.1.1

Baire Category Theorem

Some examples of Banach spaces that have been discussed up to now are Rn , Cn , and Lp (Ω). Theorems about general Banach spaces are proved in this chapter. The main theorems to be presented here are the uniform boundedness theorem, the open mapping theorem, the closed graph theorem, and the Hahn Banach Theorem. The first three of these theorems come from the Baire category theorem which is about to be presented. They are topological in nature. The Hahn Banach theorem has nothing to do with topology. Banach spaces are all normed linear spaces and as such, they are all metric spaces because a normed linear space may be considered as a metric space with d (x, y) ≡ ||x − y||. You can check that this satisfies all the axioms of a metric. As usual, if every Cauchy sequence converges, the metric space is called complete. Definition 13.1.1 A complete normed linear space is called a Banach space. The following remarkable result is called the Baire category theorem. To get an idea of its meaning, imagine you draw a line in the plane. The complement of this line is an open set and is dense because every point, even those on the line, are limit points of this open set. Now draw another line. The complement of the two lines is still open and dense. Keep drawing lines and looking at the complements of the union of these lines. You always have an open set which is dense. Now what if there were countably many lines? The Baire category theorem implies the complement of the union of these lines is dense. In particular it is nonempty. Thus you cannot write the plane as a countable union of lines. This is a rather rough description of this very important theorem. The precise statement and proof follow. Theorem 13.1.2 Let (X, d) be a complete metric space and let {Un }∞ n=1 be a sequence of open subsets of X satisfying Un = X (Un is dense). Then D ≡ ∩∞ n=1 Un is a dense subset of X. 361

362

BANACH SPACES

Proof: Let p ∈ X and let r0 > 0. I need to show D ∩ B(p, r0 ) 6= ∅. Since U1 is dense, there exists p1 ∈ U1 ∩ B(p, r0 ), an open set. Let p1 ∈ B(p1 , r1 ) ⊆ B(p1 , r1 ) ⊆ U1 ∩ B(p, r0 ) and r1 < 2−1 . This is possible because U1 ∩ B (p, r0 ) is an open set and so there exists r1 such that B (p1 , 2r1 ) ⊆ U1 ∩ B (p, r0 ). But B (p1 , r1 ) ⊆ B (p1 , r1 ) ⊆ B (p1 , 2r1 ) because B (p1 , r1 ) = {x ∈ X : d (x, p) ≤ r1 }. (Why?)

¾ r0 p p· 1 There exists p2 ∈ U2 ∩ B(p1 , r1 ) because U2 is dense. Let p2 ∈ B(p2 , r2 ) ⊆ B(p2 , r2 ) ⊆ U2 ∩ B(p1 , r1 ) ⊆ U1 ∩ U2 ∩ B(p, r0 ). and let r2 < 2−2 . Continue in this way. Thus rn < 2−n, B(pn , rn ) ⊆ U1 ∩ U2 ∩ ... ∩ Un ∩ B(p, r0 ), B(pn , rn ) ⊆ B(pn−1 , rn−1 ). The sequence, {pn } is a Cauchy sequence because all terms of {pk } for k ≥ n are contained in B (pn , rn ), a set whose diameter is no larger than 2−n . Since X is complete, there exists p∞ such that lim pn = p∞ .

n→∞

Since all but finitely many terms of {pn } are in B(pm , rm ), it follows that p∞ ∈ B(pm , rm ) for each m. Therefore, ∞ p∞ ∈ ∩∞ m=1 B(pm , rm ) ⊆ ∩i=1 Ui ∩ B(p, r0 ).

This proves the theorem. The following corollary is also called the Baire category theorem. Corollary 13.1.3 Let X be a complete metric space and suppose X = ∪∞ i=1 Fi where each Fi is a closed set. Then for some i, interior Fi 6= ∅. Proof: If all Fi has empty interior, then FiC would be a dense open set. Therefore, from Theorem 13.1.2, it would follow that C

∞ C ∅ = (∪∞ i=1 Fi ) = ∩i=1 Fi 6= ∅.

13.1. THEOREMS BASED ON BAIRE CATEGORY

363

The set D of Theorem 13.1.2 is called a Gδ set because it is the countable intersection of open sets. Thus D is a dense Gδ set. Recall that a norm satisfies: a.) ||x|| ≥ 0, ||x|| = 0 if and only if x = 0. b.) ||x + y|| ≤ ||x|| + ||y||. c.) ||cx|| = |c| ||x|| if c is a scalar and x ∈ X. From the definition of continuity, it follows easily that a function is continuous if lim xn = x n→∞

implies lim f (xn ) = f (x).

n→∞

Theorem 13.1.4 Let X and Y be two normed linear spaces and let L : X → Y be linear (L(ax + by) = aL(x) + bL(y) for a, b scalars and x, y ∈ X). The following are equivalent a.) L is continuous at 0 b.) L is continuous c.) There exists K > 0 such that ||Lx||Y ≤ K ||x||X for all x ∈ X (L is bounded). Proof: a.)⇒b.) Let xn → x. It is necessary to show that Lxn → Lx. But (xn − x) → 0 and so from continuity at 0, it follows L (xn − x) = Lxn − Lx → 0 so Lxn → Lx. This shows a.) implies b.). b.)⇒c.) Since L is continuous, L is continuous at 0. Hence ||Lx||Y < 1 whenever ||x||X ≤ δ for some δ. Therefore, suppressing the subscript on the || ||, µ ¶ δx ||L || ≤ 1. ||x|| Hence ||Lx|| ≤

1 ||x||. δ

c.)⇒a.) follows from the inequality given in c.). Definition 13.1.5 Let L : X → Y be linear and continuous where X and Y are normed linear spaces. Denote the set of all such continuous linear maps by L(X, Y ) and define ||L|| = sup{||Lx|| : ||x|| ≤ 1}. (13.1.1) This is called the operator norm.

364

BANACH SPACES

Note that from Theorem 13.1.4 ||L|| is well defined because of part c.) of that Theorem. The next lemma follows immediately from the definition of the norm and the assumption that L is linear. Lemma 13.1.6 With ||L|| defined in 13.1.1, L(X, Y ) is a normed linear space. Also ||Lx|| ≤ ||L|| ||x||. Proof: Let x 6= 0 then x/ ||x|| has norm equal to 1 and so ¯¯ µ ¶¯¯ ¯¯ x ¯¯¯¯ ¯¯L ≤ ||L|| . ¯¯ ||x|| ¯¯ Therefore, multiplying both sides by ||x||, ||Lx|| ≤ ||L|| ||x||. This is obviously a linear space. It remains to verify the operator norm really is a norm. First ³of all, ´ x if ||L|| = 0, then Lx = 0 for all ||x|| ≤ 1. It follows that for any x 6= 0, 0 = L ||x|| and so Lx = 0. Therefore, L = 0. Also, if c is a scalar, ||cL|| = sup ||cL (x)|| = |c| sup ||Lx|| = |c| ||L|| . ||x||≤1

||x||≤1

It remains to verify the triangle inequality. Let L, M ∈ L (X, Y ) . ||L + M ||

≡

sup ||(L + M ) (x)|| ≤ sup (||Lx|| + ||M x||) ||x||≤1

≤

||x||≤1

sup ||Lx|| + sup ||M x|| = ||L|| + ||M || . ||x||≤1

||x||≤1

This shows the operator norm is really a norm as hoped. This proves the lemma. For example, consider the space of linear transformations defined on Rn having values in Rm . The fact the transformation is linear automatically imparts continuity to it. You should give a proof of this fact. Recall that every such linear transformation can be realized in terms of matrix multiplication. Thus, in finite dimensions the algebraic condition that an operator is linear is sufficient to imply the topological condition that the operator is continuous. The situation is not so simple in infinite dimensional spaces such as C (X; Rn ). This explains the imposition of the topological condition of continuity as a criterion for membership in L (X, Y ) in addition to the algebraic condition of linearity. Theorem 13.1.7 If Y is a Banach space, then L(X, Y ) is also a Banach space. Proof: Let {Ln } be a Cauchy sequence in L(X, Y ) and let x ∈ X. ||Ln x − Lm x|| ≤ ||x|| ||Ln − Lm ||. Thus {Ln x} is a Cauchy sequence. Let Lx = lim Ln x. n→∞


365

Then, clearly, L is linear because if x1 , x2 are in X, and a, b are scalars, then L (ax1 + bx2 ) = = =

lim Ln (ax1 + bx2 )

n→∞

lim (aLn x1 + bLn x2 )

n→∞

aLx1 + bLx2 .

Also L is continuous. To see this, note that {||Ln ||} is a Cauchy sequence of real numbers because |||Ln || − ||Lm ||| ≤ ||Ln −Lm ||. Hence there exists K > sup{||Ln || : n ∈ N}. Thus, if x ∈ X, ||Lx|| = lim ||Ln x|| ≤ K||x||. n→∞


13.1.2

Uniform Boundedness Theorem

The next big result is sometimes called the Uniform Boundedness theorem, or the Banach-Steinhaus theorem. This is a very surprising theorem which implies that for a collection of bounded linear operators, if they are bounded pointwise, then they are also bounded uniformly. As an example of a situation in which pointwise bounded does not imply uniformly bounded, consider the functions fα (x) ≡ X(α,1) (x) x−1 for α ∈ (0, 1). Clearly each function is bounded and the collection of functions is bounded at each point of (0, 1), but there is no bound for all these functions taken together. One problem is that (0, 1) is not a Banach space. Therefore, the functions cannot be linear. Theorem 13.1.8 Let X be a Banach space and let Y be a normed linear space. Let {Lα }α∈Λ be a collection of elements of L(X, Y ). Then one of the following happens. a.) sup{||Lα || : α ∈ Λ} < ∞ b.) There exists a dense Gδ set, D, such that for all x ∈ D, sup{||Lα x|| α ∈ Λ} = ∞. Proof: For each n ∈ N, define Un = {x ∈ X : sup{||Lα x|| : α ∈ Λ} > n}. Then Un is an open set because if x ∈ Un , then there exists α ∈ Λ such that ||Lα x|| > n But then, since Lα is continuous, this situation persists for all y sufficiently close to x, say for all y ∈ B (x, δ). Then B (x, δ) ⊆ Un which shows Un is open. Case b.) is obtained from Theorem 13.1.2 if each Un is dense. The other case is that for some n, Un is not dense. If this occurs, there exists x0 and r > 0 such that for all x ∈ B(x0 , r), ||Lα x|| ≤ n for all α. Now if y ∈

366

BANACH SPACES

B(0, r), x0 + y ∈ B(x0 , r). Consequently, for all such y, ||Lα (x0 + y)|| ≤ n. This implies that for all α ∈ Λ and ||y|| < r, ||Lα y|| ≤ n + ||Lα (x0 )|| ≤ 2n. ¯¯ r ¯¯ Therefore, if ||y|| ≤ 1, ¯¯ 2 y ¯¯ < r and so for all α, ³r ´ ||Lα y || ≤ 2n. 2 Now multiplying by r/2 it follows that whenever ||y|| ≤ 1, ||Lα (y)|| ≤ 4n/r. Hence case a.) holds.

13.1.3

Open Mapping Theorem

Another remarkable theorem which depends on the Baire category theorem is the open mapping theorem. Unlike Theorem 13.1.8 it requires both X and Y to be Banach spaces. Theorem 13.1.9 Let X and Y be Banach spaces, let L ∈ L(X, Y ), and suppose L is onto. Then L maps open sets onto open sets. To aid in the proof, here is a lemma. Lemma 13.1.10 Let a and b be positive constants and suppose B(0, a) ⊆ L(B(0, b)). Then L(B(0, b)) ⊆ L(B(0, 2b)). Proof of Lemma 13.1.10: Let y ∈ L(B(0, b)). There exists x1 ∈ B(0, b) such that ||y − Lx1 || < a2 . Now this implies 2y − 2Lx1 ∈ B(0, a) ⊆ L(B(0, b)). Thus 2y − 2Lx1 ∈ L(B(0, b)) just like y was. Therefore, there exists x2 ∈ B(0, b) such that ||2y − 2Lx1 − Lx2 || < a/2. Hence ||4y − 4Lx1 − 2Lx2 || < a, and there exists x3 ∈ B (0, b) such that ||4y − 4Lx1 − 2Lx2 − Lx3 || < a/2. Continuing in this way, there exist x1 , x2 , x3 , x4 , ... in B(0, b) such that ||2n y −

n X

2n−(i−1) L(xi )|| < a

i=1

which implies ||y −

n X i=1

2

−(i−1)

L(xi )|| = ||y − L

Ã n X i=1

! −(i−1)

2

(xi ) || < 2−n a

(13.1.2)

13.1. THEOREMS BASED ON BAIRE CATEGORY P∞

Now consider the partial sums of the series, ||

n X

2−(i−1) xi || ≤ b

∞ X

i=1

367

2−(i−1) xi .

2−(i−1) = b 2−m+2 .

i=m

i=m

Therefore, these P partial sums form a Cauchy sequence and so since X is complete, ∞ there exists x = i=1 2−(i−1) xi . Letting n → ∞ in 13.1.2 yields ||y −Lx|| = 0. Now ||x|| = lim || n→∞

≤ lim

n→∞

n X i=1

n X

2−(i−1) xi ||

i=1

2−(i−1) ||xi || < lim

n→∞

n X

2−(i−1) b = 2b.

i=1

This proves the lemma. Proof of Theorem 13.1.9: Y = ∪∞ n=1 L(B(0, n)). By Corollary 13.1.3, the set, L(B(0, n0 )) has nonempty interior for some n0 . Thus B(y, r) ⊆ L(B(0, n0 )) for some y and some r > 0. Since L is linear B(−y, r) ⊆ L(B(0, n0 )) also. Here is why. If z ∈ B(−y, r), then −z ∈ B(y, r) and so there exists xn ∈ B (0, n0 ) such that Lxn → −z. Therefore, L (−xn ) → z and −xn ∈ B (0, n0 ) also. Therefore z ∈ L(B(0, n0 )). Then it follows that B(0, r) ⊆ ≡ ⊆

B(y, r) + B(−y, r) {y1 + y2 : y1 ∈ B (y, r) and y2 ∈ B (−y, r)} L(B(0, 2n0 ))

The reason for the last inclusion is that from the above, if y1 ∈ B (y, r) and y2 ∈ B (−y, r), there exists xn , zn ∈ B (0, n0 ) such that Lxn → y1 , Lzn → y2 . Therefore, ||xn + zn || ≤ 2n0 and so (y1 + y2 ) ∈ L(B(0, 2n0 )). By Lemma 13.1.10, L(B(0, 2n0 )) ⊆ L(B(0, 4n0 )) which shows B(0, r) ⊆ L(B(0, 4n0 )). Letting a = r(4n0 )−1 , it follows, since L is linear, that B(0, a) ⊆ L(B(0, 1)). It follows since L is linear, L(B(0, r)) ⊇ B(0, ar). (13.1.3) Now let U be open in X and let x + B(0, r) = B(x, r) ⊆ U . Using 13.1.3, L(U ) ⊇ L(x + B(0, r)) = Lx + L(B(0, r)) ⊇ Lx + B(0, ar) = B(Lx, ar).

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BANACH SPACES

Hence Lx ∈ B(Lx, ar) ⊆ L(U ). which shows that every point, Lx ∈ LU , is an interior point of LU and so LU is open. This proves the theorem. This theorem is surprising because it implies that if |·| and ||·|| are two norms with respect to which a vector space X is a Banach space such that |·| ≤ K ||·||, then there exists a constant k, such that ||·|| ≤ k |·| . This can be useful because sometimes it is not clear how to compute k when all that is needed is its existence. To see the open mapping theorem implies this, consider the identity map id x = x. Then id : (X, ||·||) → (X, |·|) is continuous and onto. Hence id is an open map which implies id−1 is continuous. Theorem 13.1.4 gives the existence of the constant k.

13.1.4

Closed Graph Theorem

Definition 13.1.11 Let f : D → E. The set of all ordered pairs of the form {(x, f (x)) : x ∈ D} is called the graph of f . Definition 13.1.12 If X and Y are normed linear spaces, make X × Y into a normed linear space by using the norm ||(x, y)|| = max (||x||, ||y||) along with component-wise addition and scalar multiplication. Thus a(x, y) + b(z, w) ≡ (ax + bz, ay + bw). There are other ways to give a norm for X × Y . For example, you could define ||(x, y)|| = ||x|| + ||y|| Lemma 13.1.13 The norm defined in Definition 13.1.12 on X × Y along with the definition of addition and scalar multiplication given there make X × Y into a normed linear space. Proof: The only axiom for a norm which is not obvious is the triangle inequality. Therefore, consider ||(x1 , y1 ) + (x2 , y2 )|| = ||(x1 + x2 , y1 + y2 )|| = max (||x1 + x2 || , ||y1 + y2 ||) ≤ max (||x1 || + ||x2 || , ||y1 || + ||y2 ||) ≤ =

max (||x1 || , ||y1 ||) + max (||x2 || , ||y2 ||) ||(x1 , y1 )|| + ||(x2 , y2 )|| .

It is obvious X × Y is a vector space from the above definition. This proves the lemma. Lemma 13.1.14 If X and Y are Banach spaces, then X × Y with the norm and vector space operations defined in Definition 13.1.12 is also a Banach space.


369

Proof: The only thing left to check is that the space is complete. But this follows from the simple observation that {(xn , yn )} is a Cauchy sequence in X × Y if and only if {xn } and {yn } are Cauchy sequences in X and Y respectively. Thus if {(xn , yn )} is a Cauchy sequence in X × Y , it follows there exist x and y such that xn → x and yn → y. But then from the definition of the norm, (xn , yn ) → (x, y). Lemma 13.1.15 Every closed subspace of a Banach space is a Banach space. Proof: If F ⊆ X where X is a Banach space and {xn } is a Cauchy sequence in F , then since X is complete, there exists a unique x ∈ X such that xn → x. However this means x ∈ F = F since F is closed. Definition 13.1.16 Let X and Y be Banach spaces and let D ⊆ X be a subspace. A linear map L : D → Y is said to be closed if its graph is a closed subspace of X × Y . Equivalently, L is closed if xn → x and Lxn → y implies x ∈ D and y = Lx. Note the distinction between closed and continuous. If the operator is closed the assertion that y = Lx only follows if it is known that the sequence {Lxn } converges. In the case of a continuous operator, the convergence of {Lxn } follows from the assumption that xn → x. It is not always the case that a mapping which is closed is necessarily continuous. Consider the function f (x) = tan (x) if x is not an odd multiple of π2 and f (x) ≡ 0 at every odd multiple of π2 . Then the graph is closed and the function is defined on R but it clearly fails to be continuous. Of course this function is not linear. You could also consider the map, ª d © : y ∈ C 1 ([0, 1]) : y (0) = 0 ≡ D → C ([0, 1]) . dx where the norm is the uniform norm on C ([0, 1]) , ||y||∞ . If y ∈ D, then Z

x

y (x) =

y 0 (t) dt.

0

Therefore, if

dyn dx

→ f ∈ C ([0, 1]) and if yn → y in C ([0, 1]) it follows that yn (x) = ↓ y (x) =

Rx 0

dyn (t) dx dt

Rx ↓ f (t) dt 0

and so by the fundamental theorem of calculus f (x) = y 0 (x) and so the mapping is closed. It is obviously not continuous because it takes y (x) and y (x) + n1 sin (nx) to two functions which are far from each other even though these two functions are very close in C ([0, 1]). Furthermore, it is not defined on the whole space, C ([0, 1]). The next theorem, the closed graph theorem, gives conditions under which closed implies continuous.

370

BANACH SPACES

Theorem 13.1.17 Let X and Y be Banach spaces and suppose L : X → Y is closed and linear. Then L is continuous. Proof: Let G be the graph of L. G = {(x, Lx) : x ∈ X}. By Lemma 13.1.15 it follows that G is a Banach space. Define P : G → X by P (x, Lx) = x. P maps the Banach space G onto the Banach space X and is continuous and linear. By the open mapping theorem, P maps open sets onto open sets. Since P is also one to one, this says that P −1 is continuous. Thus ||P −1 x|| ≤ K||x||. Hence ||Lx|| ≤ max (||x||, ||Lx||) ≤ K||x|| By Theorem 13.1.4 on Page 363, this shows L is continuous and proves the theorem. The following corollary is quite useful. It shows how to obtain a new norm on the domain of a closed operator such that the domain with this new norm becomes a Banach space. Corollary 13.1.18 Let L : D ⊆ X → Y where X, Y are a Banach spaces, and L is a closed operator. Then define a new norm on D by ||x||D ≡ ||x||X + ||Lx||Y . Then D with this new norm is a Banach space. Proof: If {xn } is a Cauchy sequence in D with this new norm, it follows both {xn } and {Lxn } are Cauchy sequences and therefore, they converge. Since L is closed, xn → x and Lxn → Lx for some x ∈ D. Thus ||xn − x||D → 0.

13.2

Hahn Banach Theorem

The closed graph, open mapping, and uniform boundedness theorems are the three major topological theorems in functional analysis. The other major theorem is the Hahn-Banach theorem which has nothing to do with topology. Before presenting this theorem, here are some preliminaries about partially ordered sets.

13.2.1

Partially Ordered Sets

Definition 13.2.1 Let F be a nonempty set. F is called a partially ordered set if there is a relation, denoted here by ≤, such that x ≤ x for all x ∈ F. If x ≤ y and y ≤ z then x ≤ z. C ⊆ F is said to be a chain if every two elements of C are related. This means that if x, y ∈ C, then either x ≤ y or y ≤ x. Sometimes a chain is called a totally ordered set. C is said to be a maximal chain if whenever D is a chain containing C, D = C.

13.2. HAHN BANACH THEOREM

371

The most common example of a partially ordered set is the power set of a given set with ⊆ being the relation. It is also helpful to visualize partially ordered sets as trees. Two points on the tree are related if they are on the same branch of the tree and one is higher than the other. Thus two points on different branches would not be related although they might both be larger than some point on the trunk. You might think of many other things which are best considered as partially ordered sets. Think of food for example. You might find it difficult to determine which of two favorite pies you like better although you may be able to say very easily that you would prefer either pie to a dish of lard topped with whipped cream and mustard. The following theorem is equivalent to the axiom of choice. For a discussion of this, see the appendix on the subject. Theorem 13.2.2 (Hausdorff Maximal Principle) Let F be a nonempty partially ordered set. Then there exists a maximal chain.

13.2.2

Gauge Functions And Hahn Banach Theorem

Definition 13.2.3 Let X be a real vector space ρ : X → R is called a gauge function if ρ(x + y) ≤ ρ(x) + ρ(y), ρ(ax) = aρ(x) if a ≥ 0.

(13.2.4)

Suppose M is a subspace of X and z ∈ / M . Suppose also that f is a linear real-valued function having the property that f (x) ≤ ρ(x) for all x ∈ M . Consider the problem of extending f to M ⊕ Rz such that if F is the extended function, F (y) ≤ ρ(y) for all y ∈ M ⊕ Rz and F is linear. Since F is to be linear, it suffices to determine how to define F (z). Letting a > 0, it is required to define F (z) such that the following hold for all x, y ∈ M . f (x)

z }| { F (x) + aF (z) = F (x + az) ≤ ρ(x + az), f (y)

z }| { F (y) − aF (z) = F (y − az) ≤ ρ(y − az).

(13.2.5)

Now if these inequalities hold for all y/a, they hold for all y because M is given to be a subspace. Therefore, multiplying by a−1 13.2.4 implies that what is needed is to choose F (z) such that for all x, y ∈ M , f (x) + F (z) ≤ ρ(x + z), f (y) − ρ(y − z) ≤ F (z) and that if F (z) can be chosen in this way, this will satisfy 13.2.5 for all x, y and the problem of extending f will be solved. Hence it is necessary to choose F (z) such that for all x, y ∈ M f (y) − ρ(y − z) ≤ F (z) ≤ ρ(x + z) − f (x).

(13.2.6)

372

BANACH SPACES

Is there any such number between f (y) − ρ(y − z) and ρ(x + z) − f (x) for every pair x, y ∈ M ? This is where f (x) ≤ ρ(x) on M and that f is linear is used. For x, y ∈ M , ρ(x + z) − f (x) − [f (y) − ρ(y − z)] = ρ(x + z) + ρ(y − z) − (f (x) + f (y)) ≥ ρ(x + y) − f (x + y) ≥ 0. Therefore there exists a number between sup {f (y) − ρ(y − z) : y ∈ M } and inf {ρ(x + z) − f (x) : x ∈ M } Choose F (z) to satisfy 13.2.6. This has proved the following lemma. Lemma 13.2.4 Let M be a subspace of X, a real linear space, and let ρ be a gauge function on X. Suppose f : M → R is linear, z ∈ / M , and f (x) ≤ ρ (x) for all x ∈ M . Then f can be extended to M ⊕ Rz such that, if F is the extended function, F is linear and F (x) ≤ ρ(x) for all x ∈ M ⊕ Rz. With this lemma, the Hahn Banach theorem can be proved. Theorem 13.2.5 (Hahn Banach theorem) Let X be a real vector space, let M be a subspace of X, let f : M → R be linear, let ρ be a gauge function on X, and suppose f (x) ≤ ρ(x) for all x ∈ M . Then there exists a linear function, F : X → R, such that a.) F (x) = f (x) for all x ∈ M b.) F (x) ≤ ρ(x) for all x ∈ X. Proof: Let F = {(V, g) : V ⊇ M, V is a subspace of X, g : V → R is linear, g(x) = f (x) for all x ∈ M , and g(x) ≤ ρ(x) for x ∈ V }. Then (M, f ) ∈ F so F 6= ∅. Define a partial order by the following rule. (V, g) ≤ (W, h) means V ⊆ W and h(x) = g(x) if x ∈ V. By Theorem 13.2.2, there exists a maximal chain, C ⊆ F . Let Y = ∪{V : (V, g) ∈ C} and let h : Y → R be defined by h(x) = g(x) where x ∈ V and (V, g) ∈ C. This is well defined because if x ∈ V1 and V2 where (V1 , g1 ) and (V2 , g2 ) are both in the chain, then since C is a chain, the two element related. Therefore, g1 (x) = g2 (x). Also h is linear because if ax + by ∈ Y , then x ∈ V1 and y ∈ V2 where (V1 , g1 ) and (V2 , g2 ) are elements of C. Therefore, letting V denote the larger of the two Vi ,


373

and g be the function that goes with V , it follows ax + by ∈ V where (V, g) ∈ C. Therefore, h (ax + by)

= g (ax + by) = ag (x) + bg (y) = ah (x) + bh (y) .

Also, h(x) = g (x) ≤ ρ(x) for any x ∈ Y because for such x, x ∈ V where (V, g) ∈ C. Is Y = X? If not, there exists z ∈ X \ Y and there exists an extension of h to Y ⊕ Rz using Lemma 13.2.4. Letting¡ h denote this ¢ extended function, contradicts the maximality of C. Indeed, C ∪ { Y ⊕ Rz, h } would be a longer chain. This proves the Hahn Banach theorem. This is the original version of the theorem. There is also a version of this theorem for complex vector spaces which is based on a trick.

13.2.3

The Complex Version Of The Hahn Banach Theorem

Corollary 13.2.6 (Hahn Banach) Let M be a subspace of a complex normed linear space, X, and suppose f : M → C is linear and satisfies |f (x)| ≤ K||x|| for all x ∈ M . Then there exists a linear function, F , defined on all of X such that F (x) = f (x) for all x ∈ M and |F (x)| ≤ K||x|| for all x. Proof: First note f (x) = Re f (x) + i Im f (x) and so Re f (ix) + i Im f (ix) = f (ix) = if (x) = i Re f (x) − Im f (x). Therefore, Im f (x) = − Re f (ix), and f (x) = Re f (x) − i Re f (ix). This is important because it shows it is only necessary to consider Re f in understanding f . Now it happens that Re f is linear with respect to real scalars so the above version of the Hahn Banach theorem applies. This is shown next. If c is a real scalar Re f (cx) − i Re f (icx) = cf (x) = c Re f (x) − ic Re f (ix). Thus Re f (cx) = c Re f (x). Also, Re f (x + y) − i Re f (i (x + y)) = =

f (x + y) f (x) + f (y)

= Re f (x) − i Re f (ix) + Re f (y) − i Re f (iy). Equating real parts, Re f (x + y) = Re f (x) + Re f (y). Thus Re f is linear with respect to real scalars as hoped.

374

BANACH SPACES

Consider X as a real vector space and let ρ(x) ≡ K||x||. Then for all x ∈ M , | Re f (x)| ≤ |f (x)| ≤ K||x|| = ρ(x). From Theorem 13.2.5, Re f may be extended to a function, h which satisfies h(ax + by) = ah(x) + bh(y) if a, b ∈ R h(x) ≤ K||x|| for all x ∈ X. Actually, |h (x)| ≤ K ||x|| . The reason for this is that h (−x) = −h (x) ≤ K ||−x|| = K ||x|| and therefore, h (x) ≥ −K ||x||. Let F (x) ≡ h(x) − ih(ix). By arguments similar to the above, F is linear. F (ix) = h (ix) − ih (−x) = ih (x) + h (ix) = i (h (x) − ih (ix)) = iF (x) . If c is a real scalar, F (cx)

= h(cx) − ih(icx) = ch (x) − cih (ix) = cF (x)

Now F (x + y)

= h(x + y) − ih(i (x + y)) = h (x) + h (y) − ih (ix) − ih (iy) = F (x) + F (y) .

Thus F ((a + ib) x) = = =

F (ax) + F (ibx) aF (x) + ibF (x) (a + ib) F (x) .

This shows F is linear as claimed. Now wF (x) = |F (x)| for some |w| = 1. Therefore must equal zero

|F (x)|

= wF (x) = h(wx) −

z }| { ih(iwx)

= |h(wx)| ≤ K||wx|| = K ||x|| . This proves the corollary.

= h(wx)


13.2.4

375

The Dual Space And Adjoint Operators

Definition 13.2.7 Let X be a Banach space. Denote by X 0 the space of continuous linear functions which map X to the field of scalars. Thus X 0 = L(X, F). By Theorem 13.1.7 on Page 364, X 0 is a Banach space. Remember with the norm defined on L (X, F), ||f || = sup{|f (x)| : ||x|| ≤ 1} X 0 is called the dual space. Definition 13.2.8 Let X and Y be Banach spaces and suppose L ∈ L(X, Y ). Then define the adjoint map in L(Y 0 , X 0 ), denoted by L∗ , by L∗ y ∗ (x) ≡ y ∗ (Lx) for all y ∗ ∈ Y 0 . The following diagram is a good one to help remember this definition. X0 X

L∗ ← → L

Y0 Y

This is a generalization of the adjoint of a linear transformation on an inner product space. Recall (Ax, y) = (x, A∗ y) What is being done here is to generalize this algebraic concept to arbitrary Banach spaces. There are some issues which need to be discussed relative to the above definition. First of all, it must be shown that L∗ y ∗ ∈ X 0 . Also, it will be useful to have the following lemma which is a useful application of the Hahn Banach theorem. Lemma 13.2.9 Let X be a normed linear space and let x ∈ X \ V where V is a closed subspace of X. Then there exists x∗ ∈ X 0 such that x∗ (x) = ||x||, x∗ (V ) = {0}, and 1 ||x∗ || ≤ dist (x, V ) In the case that V = {0} , ||x∗ || = 1. Proof: Let f : Fx+V → F be defined by f (αx+v) = α||x||. First it is necessary to show f is well defined and continuous. If α1 x + v1 = α2 x + v2 then if α1 6= α2 , then x ∈ V which is assumed not to happen so f is well defined. It remains to show f is continuous. Suppose then that αn x + vn → 0. It is necessary to show αn → 0. If this does not happen, then there exists a subsequence, still denoted by αn such that |αn | ≥ δ > 0. Then x + (1/αn ) vn → 0 contradicting the assumption

376

BANACH SPACES

that x ∈ / V and V is a closed subspace. Hence f is continuous on Fx + V. Being a little more careful, ||f || =

sup

|f (αx + v)| =

||αx+v||≤1

sup

|α| ||x|| =

|α|||x+(v/α)||≤1

1 ||x|| dist (x, V )

By the Hahn Banach theorem, there exists x∗ ∈ X 0 such that x∗ = f on Fx + V. Thus x∗ (x) = ||x|| and also ||x∗ || ≤ ||f || =

1 dist (x, V )

In case V = {0} , the result follows from the above or alternatively, ||f || ≡ sup |f (αx)| = ||αx||≤1

sup

|α| ||x|| = 1

|α|≤1/||x||

and so, in this case, ||x∗ || ≤ ||f || = 1. Since x∗ (x) = ||x|| it follows ¯ µ ¶¯ ¯ ∗ x ¯¯ ||x|| ∗ ¯ ||x || ≥ ¯x = 1. = ||x|| ¯ ||x|| Thus ||x∗ || = 1 and this proves the lemma. Theorem 13.2.10 Let L ∈ L(X, Y ) where X and Y are Banach spaces. Then a.) L∗ ∈ L(Y 0 , X 0 ) as claimed and ||L∗ || = ||L||. b.) If L maps one to one onto a closed subspace of Y , then L∗ is onto. c.) If L maps onto a dense subset of Y , then L∗ is one to one. Proof: It is routine to verify L∗ y ∗ and L∗ are both linear. This follows immediately from the definition. As usual, the interesting thing concerns continuity. ||L∗ y ∗ || = sup |L∗ y ∗ (x)| = sup |y ∗ (Lx)| ≤ ||y ∗ || ||L|| . ||x||≤1

||x||≤1

Thus L∗ is continuous as claimed and ||L∗ || ≤ ||L|| . By Lemma 13.2.9, there exists yx∗ ∈ Y 0 such that ||yx∗ || = 1 and yx∗ (Lx) = ||Lx|| .Therefore, ||L∗ || =

sup ||L∗ y ∗ || = sup ||y ∗ ||≤1

=

sup |y ∗ (Lx)| = sup

sup ||y ∗ ||≤1

≥

sup ||x||≤1

sup |L∗ y ∗ (x)|

||y ∗ ||≤1 ||x||≤1

||x||≤1 |yx∗ (Lx)|

sup |y ∗ (Lx)|

||x||≤1 ||y ∗ ||≤1

= sup ||Lx|| = ||L|| ||x||≤1

showing that ||L∗ || ≥ ||L|| and this shows part a.). If L is one to one and onto a closed subset of Y , then L (X) being a closed subspace of a Banach space, is itself a Banach space and so the open mapping theorem implies L−1 : L(X) → X is continuous. Hence ¯¯ ¯¯ ||x|| = ||L−1 Lx|| ≤ ¯¯L−1 ¯¯ ||Lx||


377

Now let x∗ ∈ X 0 be given. Define f ∈ L(L(X), C) by f (Lx) = x∗ (x). The function, f is well defined because if Lx1 = Lx2 , then since L is one to one, it follows x1 = x2 and so f (L (x1 )) = x∗ (x1 ) = x∗ (x2 ) = f (L (x1 )). Also, f is linear because f (aL (x1 ) + bL (x2 ))

=

f (L (ax1 + bx2 ))

≡ = =

x∗ (ax1 + bx2 ) ax∗ (x1 ) + bx∗ (x2 ) af (L (x1 )) + bf (L (x2 )) .

In addition to this, ¯¯ ¯¯ |f (Lx)| = |x∗ (x)| ≤ ||x∗ || ||x|| ≤ ||x∗ || ¯¯L−1 ¯¯ ||Lx|| ¯¯ ¯¯ and so the norm of f on L (X) is no larger than ||x∗ || ¯¯L−1 ¯¯. By the Hahn Banach ∗ 0 ∗ theorem, ¯¯ there ¯¯ exists an extension of f to an element y ∈ Y such that ||y || ≤ ∗ ¯¯ −1 ¯¯ ||x || L . Then L∗ y ∗ (x) = y ∗ (Lx) = f (Lx) = x∗ (x) so L∗ y ∗ = x∗ because this holds for all x. Since x∗ was arbitrary, this shows L∗ is onto and proves b.). Consider the last assertion. Suppose L∗ y ∗ = 0. Is y ∗ = 0? In other words is y ∗ (y) = 0 for all y ∈ Y ? Pick y ∈ Y . Since L (X) is dense in Y, there exists a sequence, {Lxn } such that Lxn → y. But then by continuity of y ∗ , y ∗ (y) = limn→∞ y ∗ (Lxn ) = limn→∞ L∗ y ∗ (xn ) = 0. Since y ∗ (y) = 0 for all y, this implies y ∗ = 0 and so L∗ is one to one. Corollary 13.2.11 Suppose X and Y are Banach spaces, L ∈ L(X, Y ), and L is one to one and onto. Then L∗ is also one to one and onto. There exists a natural mapping, called the James map from a normed linear space, X, to the dual of the dual space which is described in the following definition. Definition 13.2.12 Define J : X → X 00 by J(x)(x∗ ) = x∗ (x). Theorem 13.2.13 The map, J, has the following properties. a.) J is one to one and linear. b.) ||Jx|| = ||x|| and ||J|| = 1. c.) J(X) is a closed subspace of X 00 if X is complete. Also if x∗ ∈ X 0 , ||x∗ || = sup {|x∗∗ (x∗ )| : ||x∗∗ || ≤ 1, x∗∗ ∈ X 00 } . Proof: J (ax + by) (x∗ ) ≡ = =

x∗ (ax + by) ax∗ (x) + bx∗ (y) (aJ (x) + bJ (y)) (x∗ ) .

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BANACH SPACES

Since this holds for all x∗ ∈ X 0 , it follows that J (ax + by) = aJ (x) + bJ (y) and so J is linear. If Jx = 0, then by Lemma 13.2.9 there exists x∗ such that x∗ (x) = ||x|| and ||x∗ || = 1. Then 0 = J(x)(x∗ ) = x∗ (x) = ||x||. This shows a.). To show b.), let x ∈ X and use Lemma 13.2.9 to obtain x∗ ∈ X 0 such that ∗ x (x) = ||x|| with ||x∗ || = 1. Then ||x|| ≥

sup{|y ∗ (x)| : ||y ∗ || ≤ 1}

= sup{|J(x)(y ∗ )| : ||y ∗ || ≤ 1} = ||Jx|| ≥ |J(x)(x∗ )| = |x∗ (x)| = ||x|| Therefore, ||Jx|| = ||x|| as claimed. Therefore, ||J|| = sup{||Jx|| : ||x|| ≤ 1} = sup{||x|| : ||x|| ≤ 1} = 1. This shows b.). To verify c.), use b.). If Jxn → y ∗∗ ∈ X 00 then by b.), xn is a Cauchy sequence converging to some x ∈ X because ||xn − xm || = ||Jxn − Jxm || and {Jxn } is a Cauchy sequence. Then Jx = limn→∞ Jxn = y ∗∗ . Finally, to show the assertion about the norm of x∗ , use what was just shown applied to the James map from X 0 to X 000 still referred to as J. ||x∗ || = sup {|x∗ (x)| : ||x|| ≤ 1} = sup {|J (x) (x∗ )| : ||Jx|| ≤ 1} ≤ sup {|x∗∗ (x∗ )| : ||x∗∗ || ≤ 1} = sup {|J (x∗ ) (x∗∗ )| : ||x∗∗ || ≤ 1} ≡ ||Jx∗ || = ||x∗ ||. This proves the theorem. Definition 13.2.14 When J maps X onto X 00 , X is called reflexive. It happens the Lp spaces are reflexive whenever p > 1. This is shown later.

13.3

Approximation With Sums

Theorem 13.3.1 Let X be a Banach space and let V = span (x1 , · · · , xn ) . Then V is a closed subspace of X.

13.3. APPROXIMATION WITH SUMS

379

Proof: Without loss of generality, it can be assumed {x1 , · · · , xn } is linearly independent. Otherwise, delete those vectors which are in the span of the others till a linearly independent set is obtained. Let x = lim

p→∞

n X

cpk xk ∈ V .

(13.3.7)

k=1

First suppose cp ≡ (cp1 , · · · , cpn ) is not bounded in Fn . Then dp ≡ cp / |cp |Fn is a unit vector in Fn and so there exists a subsequence, still denoted by dp which converges to d where |d| = 1. Then n

n

k=1

k=1

X p X x = lim d x = dk xk k k p→∞ ||cp || p→∞

0 = lim P

2

where k |dk | = 1 in contradiction to the linear independence of the {x1 , · · · , xn }. Hence it must be the case that cp is bounded in Fn . Then taking a subsequence, still denoted as p, it can be assumed cp → c and then in 13.3.7 it follows x=

n X

ck xk ∈ span (x1 , · · · , xn ) .

k=1

This proves the Theorem. Lemma 13.3.2 Let E be a separable Banach space. Then there exists an increasing sequence of subspaces, {Fn } such that dim (Fn+1 ) − dim (Fn ) ≤ 1 and equals 1 for all n if the dimension of E is infinite. Also ∪∞ n=1 Fn is dense in E. In the case where E is infinite dimensional, Fn = span (e1 , · · · , en ) where for each n dist (en+1 , Fn ) ≥

1 2

(13.3.8)

and defining, Gk ≡ span ({ej : j 6= k}) dist (ek , Gk ) ≥

1 . 4

(13.3.9)

Proof: Since E is separable, so is ∂B (0, 1) , the boundary of the unit ball. Let ∞ {wk }k=1 be a countable dense subset of ∂B (0, 1). Let e1 = w1 . Let F1 = Fe1 . Suppose Fn has been obtained and equals span (e1 , · · · , en ) where {e1 , · · · , en } is independent, ||ek || = 1, and dist (en , span (e1 , · · · , en−1 )) ≥

1 . 2

For each n, Fn is closed by Theorem 13.3.1. ∞ If Fn contains {wk }k=1 , let Fm = Fn for all m > n. Otherwise, pick w ∈ {wk } ∞ to be the point of {wk }k=1 having the smallest subscript which is not contained in

380

BANACH SPACES

Fn . Then w is at a positive distance, λ from Fn because Fn is closed. Therefore, w−y there exists y ∈ Fn such that λ ≤ ||y − w|| ≤ 2λ. Let en+1 = ||w−y|| . It follows w = ||w − y|| en+1 + y ∈ span (e1 , · · · , en+1 ) ≡ Fn+1 Then if x ∈ span (e1 , · · · , en ) ,

¯¯ ¯¯ ¯¯ w − y ¯¯ ¯ ¯ ||en+1 − x|| = ¯¯ − x¯¯¯¯ ||w − y|| ¯¯ ¯¯ ¯¯ w − y ||w − y|| x ¯¯¯¯ ¯ ¯ = ¯¯ − ||w − y|| ||w − y|| ¯¯ 1 ≥ ||w − y − ||w − y|| x|| 2λ λ 1 ≥ = . 2λ 2 This has shown the existence of an increasing sequence of subspaces, {Fn } as described above. It remains to show the union of these subspaces is dense. First ∞ note that the union of these subspaces must contain the {wk }k=1 because if wm is th missing, then it would contradict the construction at the m step. That one should ∞ have been chosen. However, {wk }k=1 is dense in ∂B (0, 1). If x ∈ E and x 6= 0, x then ||x|| ∈ ∂B (0, 1) then there exists ∞

¯¯ ¯¯ such that ¯¯wm −

wm ∈ {wk }k=1 ⊆ ∪∞ n=1 Fn

¯¯

x ¯¯ ||x|| ¯¯

k. Suppose 13.3.9 does not hold for some such y so that ¯¯  ¯¯ ¯¯ ¯¯ k−1 n X X ¯¯ ¯¯ 1 ¯¯e k −   (13.3.10) cj ej + cj ej ¯¯¯¯ < . ¯¯ ¯¯ ¯¯ 4 j=1 j=k+1 Then from the construction, ¯¯   ¯¯ ¯¯ ¯¯ k−1 n−1 X X ¯ ¯ ¯¯ 1 > |cn | ¯¯¯¯ek −  (cj /cn ) ej + (cj /cn ) ej + en ¯¯¯¯ 4 ¯¯ ¯¯ j=1 j=k+1

≥

|cn |

1 2

13.3. APPROXIMATION WITH SUMS

381

and so |cn | < 1/2. Consider the left side of 13.3.10. By the construction ¯¯   ¯¯ ¯¯ ¯¯ k−1 n−1 X X ¯¯ ¯¯ ¯¯cn (ek − en ) + (1 − cn ) ek −  cj ej + cj ej ¯¯¯¯ ¯¯ ¯¯ ¯¯ j=1 j=k+1 ¯¯  ¯¯ ¯¯ ¯¯ k−1 n−1 X X ¯¯ ¯¯ ¯ ¯ (cj /cn ) ej + (cj /cn ) ej ¯¯¯¯ ≥ |1 − cn | − |cn | ¯¯(ek − en ) −  ¯¯ ¯¯ j=1 j=k+1 1 3 31 1 ≥ 1 − |cn | > 1 − = , 2 2 22 4 a contradiction. This proves the desired estimate. This proves the lemma. The following is the main result. ≥ |1 − cn | − |cn |

Theorem 13.3.3 Let X be a separable Banach space. Then there exists a sequence, ∞ {e∗k }k=1 ⊆ X 0 and {ek } as in Lemma 13.3.2 such that for every x ∈ X , x=

∞ X

e∗k (x) ek .

k=1

Proof: The case where X is finite dimensional is left to the reader. Assume then that dim (X) = ∞. Then letting {Fn } be the sequence of closed finite dimensional subspaces described in Lemma 13.3.2, Fn = span (e1 , · · · , en ) . Also let Gk be as defined there, Gk ≡ span ({ej : j 6= k}) . Then, as shown in Lemma 13.3.2, ¡ ¢ dist ek , Gk ≥ 1/4. By Lemma 13.2.9 there exists e∗k ∈ X 0 such that e∗k (e1 ) = ||ek ||X , e∗k (Gk ) = {0} . Now let x ∈ X be given. By Lemma 13.3.2 there exist scalars {ck } such that x=

∞ X

ck ek .

k=1

Then doing e∗m to both sides, yields e∗m (x) =

∞ X k=1


ck e∗m (ek ) = cm

382

BANACH SPACES

13.4

Weak And Weak ∗ Topologies

13.4.1

Basic Definitions

Let X be a Banach space and let X 0 be its dual space.1 For A0 a finite subset of X 0 , denote by ρA0 the function defined on X ρA0 (x) ≡ max |x∗ (x)| ∗ 0 x ∈A

(13.4.11)

and also let BA0 (x, r) be defined by BA0 (x, r) ≡ {y ∈ X : ρA0 (y − x) < r}

(13.4.12)

Then certain things are obvious. First of all, if a ∈ F and x, y ∈ X, ρA0 (x + y) ≤ ρA0 (ax) =

ρA0 (x) + ρA0 (y) , |a| ρA0 (x) .

Similarly, letting A be a finite subset of X, denote by ρA the function defined on X 0 ρA (x∗ ) ≡ max |x∗ (x)| (13.4.13) x∈A

∗

and let BA (x , r) be defined by BA (x∗ , r) ≡ {y ∗ ∈ X 0 : ρA (y ∗ − x∗ ) < r} .

(13.4.14)

It is also clear that ρA (x∗ + y ∗ ) ≤ ρA (ax∗ ) =

ρ (x∗ ) + ρA (y ∗ ) , |a| ρA (x∗ ) .

Lemma 13.4.1 The sets, BA0 (x, r) where A0 is a finite subset of X 0 and x ∈ X form a basis for a topology on X known as the weak topology. The sets BA (x∗ , r) where A is a finite subset of X and x∗ ∈ X 0 form a basis for a topology on X 0 known as the weak ∗ topology. Proof: The two assertions are very similar. I will verify the one for the weak topology. The union of these sets, BA0 (x, r) for x ∈ X and r > 0 is all of X. Now suppose z is contained in the intersection of two of these sets. Say z ∈ BA0 (x, r) ∩ BA01 (x1 , r1 ) Then let C 0 = A0 ∪ A01 and let ´ ³ 0 < δ ≤ min r − ρA0 (z − x) , r1 − ρA01 (z − x1 ) . 1 Actually,

all this works in much more general settings than this.

13.4. WEAK AND WEAK ∗ TOPOLOGIES

383

Consider y ∈ BC 0 (z, δ) . Then r − ρA0 (z − x) ≥ δ > ρC 0 (y − z) ≥ ρA0 (y − z) and so r > ρA0 (y − z) + ρA0 (z − x) ≥ ρA0 (y − x) which shows y ∈ BA0 (x, r) . Similar reasoning shows y ∈ BA01 (x1 , r1 ) and so BC 0 (z, δ) ⊆ BA0 (x, r) ∩ BA01 (x1 , r1 ) . Therefore, the weak topology consists of the union of all sets of the form BA (x, r).

13.4.2

Banach Alaoglu Theorem

Why does anyone care about these topologies? The short answer is that in the weak ∗ topology, closed unit ball in X 0 is compact. This is not true in the normal topology. This wonderful result is the Banach Alaoglu theorem. First recall the notion of the product topology, and the Tychonoff theorem, Theorem 11.2.4 on Page 325 which are stated here for convenience. Definition 13.4.2 Let I be a set and suppose for each i ∈ I, (Xi , Q τ i ) is a nonempty topological space. The Cartesian product of the Xi , denoted by i∈I Xi , consists of the set of allQchoice functions defined on I which select a single element of each X Qi . Thus f ∈ i∈I Xi means for every i ∈ I, f (i) ∈ Xi . The axiom of choice says i∈I Xi is nonempty. Let Y Pj (A) = Bi i∈I

where Bi = Xi if i 6= j and Bj = A. A subbasis for a topology on the product space consists of all sets Pj (A) where A ∈ τ j . (These sets have an open set from the topology of Xj in the j th slot and the whole space in the other slots.) Thus a basis consists of finite intersections of these sets. Note that theQintersection of two of these basic sets is another basic set and their union yields i∈I Xi . Therefore, they satisfy the condition needed for a collection of sets to serve as a Q basis for a topology. This topology is called the product topology and is denoted by τ i . Q Q Theorem 13.4.3 If (Xi , τ i ) is compact, then so is ( i∈I Xi , τ i ). The Banach Alaoglu theorem is as follows. Theorem 13.4.4 Let B 0 be the closed unit ball in X 0 . Then B 0 is compact in the weak ∗ topology. Proof: By the Tychonoff theorem, Theorem 13.4.3 Y P ≡ B (0, ||x||) x∈X

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BANACH SPACES

is compact in the product topology where the topology on B (0, ||x||) is the usual topology of F. Recall P is the set of functions which map a point, x ∈ X to a point in B (0, ||x||). Therefore, B 0 ⊆ P. Also the basic open sets in the weak ∗ topology on B 0 are obtained as the intersection of basic open sets in the product topology of P to B 0 and so it suffices to show B 0 is a closed subset of P. Suppose then that f ∈ P \ B 0 . It follows f cannot be linear. There are two ways this can happen. One way is that for some x, y f (x + y) 6= f (x) + f (y) for some x, y ∈ X. However, if g is close enough to f at the three points, x + y, x, and y, the above inequality will hold for g in place of f. In other words there is a basic open set containing f such that for all g in this basic open set, g ∈ / B0. A similar consideration applies in case f (λx) 6= λf (x) for some scalar, λ and x. Since P \ B 0 is open, it follows B 0 is a closed subset of P and is therefore, compact. This proves the theorem. Sometimes one can consider the weak ∗ topology in terms of a metric space. Theorem 13.4.5 If K ⊆ X 0 is compact in the weak ∗ topology and X is separable then there exists a metric, d, on K such that if τ d is the topology on K induced by d and if τ is the topology on K induced by the weak ∗ topology of X 0 , then τ = τ d . Thus one can consider K with the weak ∗ topology as a metric space. Proof: Let D = {xn } be the dense countable subset. The metric is d (f, g) ≡

∞ X n=1

2−n

ρxn (f − g) 1 + ρxn (f − g)

where ρxn (f ) = |f (xn )|. Clearly d (f, g) = d (g, f ) ≥ 0. If d (f, g) = 0, then this requires f (xn ) = g (xn ) for all xn ∈ D. Since f and g are continuous and D is dense, this requires that f (x) = g (x) for all x. It is routine to verify the triangle inequality from the easy to establish inequality, x y x+y + ≥ , 1+x 1+y 1+x+y valid whenever x, y ≥ 0. Therefore this is a metric. Now for each n g→

ρxn (f − g) 1 + ρxn (f − g)

is a continuous function from (K, τ ) to [0, ∞) and also the above sum defining d converges uniformly. It follows g → d (f, g) is continuous. Therefore, the ball with respect to d, Bd (f, r) ≡ {g ∈ K : d (g, f ) < r}


385

is open in τ which implies τ d ⊆ τ . Now suppose U ∈ τ . Then K \ U is closed in K. Hence, K \ U is compact in τ because it is a closed subset of the compact set K. It follows that K \ U is compact with respect to τ d because τ d ⊆ τ . But (K, τ d ) is a Hausdorff space and so K \ U must be closed with respect to τ d . This implies U ∈ τ d . Thus τ ⊆ τ d and this proves τ = τ d . This proves the theorem. The fact that this set with the weak ∗ topology can be considered a metric space is very significant because if a point is a limit point in a metric space, one can extract a convergent sequence. Corollary 13.4.6 If X is separable and K ⊆ X 0 is compact in the weak ∗ topology, then K is sequentially compact. That is, if {fn }∞ n=1 ⊆ K, then there exists a subsequence fnk and f ∈ K such that for all x ∈ X, lim fnk (x) = f (x).

k→∞

Proof: By Theorem 13.4.5, K is a metric space for the metric described there and it is compact. Therefore by the characterization of compact metric spaces, Proposition 6.2.5 on Page 142, K is sequentially compact. This proves the corollary.

13.4.3

Eberlein Smulian Theorem

Next consider the weak topology. The most interesting results have to do with a reflexive Banach space. The following lemma ties together the weak and weak ∗ topologies in the case of a reflexive Banach space. Lemma 13.4.7 Let J : X → X 00 be the James map Jx (f ) ≡ f (x) and let X be reflexive so that J is onto. Then J is a homeomorphism of (X, weak topology) and (X 00 , weak ∗ topology).This means J is one to one, onto, and both J and J −1 are continuous. Proof: Let f ∈ X 0 and let Bf (x, r) ≡ {y : |f (x) − f (y)| < r}. Thus Bf (x, r) is a subbasic set for the weak topology on X. Now by the definition of J, y ∈ Bf (x, r) if and only if |Jy (f ) − Jx (f )| < r if and only if Jy ∈ Bf (Jx, r) ≡ {y

∗∗

∈ X 00 : |y ∗∗ (f ) − J (x) (f )| < r},

a subbasic set for the weak ∗ topology on X 00 . Since J −1 and J are one to one and onto and map subbasic sets to subbasic sets, it follows that J is a homeomorphism. This proves the Lemma. The following is an easy corollary.

386

BANACH SPACES

Corollary 13.4.8 If X is a reflexive Banach space, then the closed unit ball is weakly compact. Proof: Let B be the closed unit ball. Then B = J −1 (B ∗∗ ) where B ∗∗ is the unit ball in X 00 which is compact in the weak ∗ topology. Therefore B is weakly compact because J −1 is continuous. Corollary 13.4.9 Let X be a reflexive Banach space. If K ⊆ X is compact in the weak topology and X 0 is separable then there exists a metric d, on K such that if τ d is the topology on K induced by d and if τ is the topology on K induced by the weak topology of X, then τ = τ d . Thus one can consider K with the weak topology as a metric space. Proof: This follows from Theorem 13.4.5 and Lemma 13.4.7. Lemma 13.4.7 implies J (K) is compact in X 00 . Then since X 0 is separable, there is a metric, d00 on J (K) which delivers the weak ∗ topology. Let d (x, y) ≡ d00 (Jx, Jy) . Then J

J −1

id

(K, τ d ) → (J (K) , τ d00 ) → (J (K) , τ weak ∗ ) → (K, τ weak ) and all the maps are homeomorphisms. Next is the Eberlein Smulian theorem which states that a Banach space is reflexive if and only if the closed unit ball is weakly sequentially compact. Actually, only half the theorem is proved here, the more useful only if part. The book by Yoshida [67] has the complete theorem discussed. First here is an interesting lemma for its own sake. Lemma 13.4.10 A closed subspace of a reflexive Banach space is reflexive. Proof: Let Y be the closed subspace of the reflexive space, X. Consider the following diagram i∗∗ 1-1 Y 00 → X 00 Y0 Y

i∗ onto

← i →

X0 X

This diagram follows from Theorem 13.2.10 on Page 376, the theorem on adjoints. Now let y ∗∗ ∈ Y 00 . Then i∗∗ y ∗∗ = JX (y) because X is reflexive. I want to show that y ∈ Y . If it is not in Y then since Y is closed, there exists x∗ ∈ X 0 such that x∗ (y) 6= 0 but x∗ (Y ) = 0. Then i∗ x∗ = 0. Hence 0 = y ∗∗ (i∗ x∗ ) = i∗∗ y ∗∗ (x∗ ) = J (y) (x∗ ) = x∗ (y) 6= 0, a contradiction. Hence y ∈ Y . Letting JY denote the James map from Y to Y 00 and x∗ ∈ X 0 , y ∗∗ (i∗ x∗ ) = =

i∗∗ y ∗∗ (x∗ ) = JX (y) (x∗ ) x∗ (y) = x∗ (iy) = i∗ x∗ (y) = JY (y) (i∗ x∗ )

Since i∗ is onto, this shows y ∗∗ = JY (y) and this proves the lemma.


387

Theorem 13.4.11 (Eberlein Smulian) The closed unit ball in a reflexive Banach space X, is weakly sequentially compact. By this is meant that if {xn } is contained in the closed unit ball, there exists a subsequence, {xnk } and x ∈ X such that for all x∗ ∈ X 0 , x∗ (xnk ) → x∗ (x) . Proof: Let {xn } ⊆ B ≡ B (0, 1). Let Y be the closure of the linear span of {xn }. Thus Y is a separable. It is reflexive because it is a closed subspace of a reflexive space so the above lemma applies. By the Banach Alaoglu theorem, the closed unit ball B ∗ in Y 0 is weak ∗ compact. Also by Theorem 13.4.5, B ∗ is a metric space with a suitable metric. Thus B ∗ is complete and totally bounded with respect to this metric and it follows that B ∗ with the weak ∗ topology is separable. This implies Y 0 is also separable in the weak ∗ topology. To see this, let {yn∗ } ≡ D be a weak ∗ dense set in B ∗ and let y ∗ ∈ Y 0 . Let p be a large enough positive rational number that y ∗ /p ∈ B ∗ . Then if A is any finite set from Y, there exists yn∗ ∈ D such that ρA (y ∗ /p − yn∗ ) < pε . It follows pyn∗ ∈ BA (y ∗ , ε) showing that rational multiples of D are weak ∗ dense in Y 0 . Since Y is reflexive, the weak and weak ∗ topologies on Y 0 coincide and so Y 0 is weakly separable. Since Y 0 is separable, Corollary 13.4.6 implies B ∗∗ , the closed unit ball in Y 00 is weak ∗ sequentially compact. Then by Lemma 13.4.7 B, the unit ball in Y , is weakly sequentially compact. It follows there exists a subsequence xnk , of the sequence {xn } and a point x ∈ Y , such that for all f ∈ Y 0 , f (xnk ) → f (x). Now if x∗ ∈ X 0 , and i is the inclusion map of Y into X, x∗ (xnk ) = i∗ x∗ (xnk ) → i∗ x∗ (x) = x∗ (x). which shows xnk converges weakly and this shows the unit ball in X is weakly sequentially compact. Corollary 13.4.12 Let {xn } be any bounded sequence in a reflexive Banach space, X. Then there exists x ∈ X and a subsequence, {xnk } such that for all x∗ ∈ X 0 , lim x∗ (xnk ) = x∗ (x)

k→∞

Proof: If a subsequence, xnk has ||xnk || → 0, then the conclusion follows. Simply let x = 0. Suppose then that ||xn || is bounded away from 0. That is, ||xn || ∈ [δ, C]. Take a subsequence such that ||xnk || → a. Then consider xnk / ||xnk ||. By the Smulian theorem, this subsequence has a further subsequence, ¯¯ Eberlein ¯¯ ¯¯ ¯¯ xnkj / ¯¯xnkj ¯¯ which converges weakly to x ∈ B where B is the closed unit ball. It follows from routine considerations that xnkj → ax weakly. This proves the corollary.

388

BANACH SPACES

13.5

Exercises

1. Is N a Gδ set? What about Q? What about a countable dense subset of a complete metric space? 2. ↑ Let f : R → C be a function. Define the oscillation of a function in B (x, r) by ω r f (x) = sup{|f (z) − f (y)| : y, z ∈ B(x, r)}. Define the oscillation of the function at the point, x by ωf (x) = limr→0 ω r f (x). Show f is continuous at x if and only if ωf (x) = 0. Then show the set of points where f is continuous is a Gδ set (try Un = {x : ωf (x) < n1 }). Does there exist a function continuous at only the rational numbers? Does there exist a function continuous at every irrational and discontinuous elsewhere? Hint: Suppose ∞ D is any countable set, D = {di }i=1 , and define the function, fn (x) to equal zero for P every x ∈ / {d1 , · · · , dn } and 2−n for x in this finite set. Then consider ∞ g (x) ≡ n=1 fn (x). Show that this series converges uniformly. 3. Let f ∈ C([0, 1]) and suppose f 0 (x) exists. Show there exists a constant, K, such that |f (x) − f (y)| ≤ K|x − y| for all y ∈ [0, 1]. Let Un = {f ∈ C([0, 1]) such that for each x ∈ [0, 1] there exists y ∈ [0, 1] such that |f (x) − f (y)| > n|x − y|}. Show that Un is open and dense in C([0, 1]) where for f ∈ C ([0, 1]), ||f || ≡ sup {|f (x)| : x ∈ [0, 1]} . Show that ∩n Un is a dense Gδ set of nowhere differentiable continuous functions. Thus every continuous function is uniformly close to one which is nowhere differentiable. P∞ 4. ↑Suppose f (x) = k=1 uk (x) where the convergence is uniform P∞ and each uk is a polynomial. Is it reasonable to conclude that f 0 (x) = k=1 u0k (x)? The answer is no. Use Problem 3 and the Weierstrass approximation theorem do show this. 5. Let X be a normed linear space. We say A ⊆ X is “weakly bounded” if for each x∗ ∈ X 0 , sup{|x∗ (x)| : x ∈ A} < ∞, while A is bounded if sup{||x|| : x ∈ A} < ∞. Show A is weakly bounded if and only if it is bounded. 6. Let X and Y be two Banach spaces. Define the norm |||(x, y)||| ≡ ||x||X + ||y||Y . Show this is a norm on X × Y which is equivalent to the norm given in the chapter for X × Y . Can you do the same for the norm defined for p > 1 by p

p 1/p

|(x, y)| ≡ (||x||X + ||y||Y )

?

7. Let f be a 2π periodic locally integrable function on R. The Fourier series for f is given by ∞ X k=−∞

ak eikx ≡ lim

n→∞

n X k=−n

ak eikx ≡ lim Sn f (x) n→∞

13.5. EXERCISES where

389 Z

1 ak = 2π

Show

π

e−ikx f (x) dx.

−π

Z

π

Dn (x − y) f (y) dy

Sn f (x) = −π

where Dn (t) = Verify that

Rπ −π

sin((n + 12 )t) . 2π sin( 2t )

Dn (t) dt = 1. Also show that if g ∈ L1 (R) , then Z g (x) sin (ax) dx = 0.

lim

a→∞

R

This last is called the Riemann Lebesgue lemma. Hint: For the last part, assume first that g ∈ Cc∞ (R) and integrate by parts. Then exploit density of the set of functions in L1 (R). 8. ↑It turns out that the Fourier series sometimes converges to the function pointwise. Suppose f is 2π periodic and Holder continuous. That is |f (x) − f (y)| ≤ θ K |x − y| where θ ∈ (0, 1]. Show that if f is like this, then the Fourier series converges to f at every point. Next modify your argument to show that if θ at every point, x, |f (x+) − f (y)| ≤ K |x − y| for y close enough to x and θ larger than x and |f (x−) − f (y)| ≤ K |x − y| for every y close enough to x (x−) and smaller than x, then Sn f (x) → f (x+)+f , the midpoint of the jump 2 of the function. Hint: Use Problem 7. 9. ↑ Let Y = {f such that f is continuous, defined on R, and 2π periodic}. Define ||f ||Y = sup{|f (x)| : x ∈ [−π, π]}. Show that (Y, || ||Y ) is a Banach space. Let x ∈ R and define Ln (f ) = Sn f (x). Show Ln ∈ Y 0 but limn→∞ ||Ln || = ∞. Show that for each x ∈ R, there exists a dense Gδ subset of Y such that for f in this set, |Sn f (x)| is unbounded. Finally, show there is a dense Gδ subset of Y having the property that |Sn f (x)| is unbounded on the rational numbers. Hint: To do the first part, let f (y) approximate sgn(Dn (x−y)). Here sgn r = 1 if r > 0, −1 if r < 0 and 0 if r = 0. This rules out one possibility of the uniform boundedness principle. After this, show the countable intersection of dense Gδ sets must also be a dense Gδ set. 10. Let α ∈ (0, 1]. Define, for X a compact subset of Rp , C α (X; Rn ) ≡ {f ∈ C (X; Rn ) : ρα (f ) + ||f || ≡ ||f ||α < ∞} where ||f || ≡ sup{|f (x)| : x ∈ X}

390

BANACH SPACES

and ρα (f ) ≡ sup{

|f (x) − f (y)| : x, y ∈ X, x 6= y}. α |x − y|

Show that (C α (X; Rn ) , ||·||α ) is a complete normed linear space. This is called a Holder space. What would this space consist of if α > 1? 11. ↑Now recall Problem 10 about the Holder spaces. Let X be the Holder functions which are periodic of period 2π. Define Ln f (x) = Sn f (x) where Ln : X → Y for Y given in Problem 9. Show ||Ln || is bounded independent of n. Conclude that Ln f → f in Y for all f ∈ X. In other words, for the Holder continuous and 2π periodic functions, the Fourier series converges to the function uniformly. Hint: Ln f (x) is given by Z π Ln f (x) = Dn (y) f (x − y) dy −π α

where f (x − y) = f (x) + g (x, y) where |g (x, y)| ≤ C |y| . Use the fact the Dirichlet kernel integrates to one to write ¯Z ¯ ¯ ¯

π

−π

¯ ¯zZ ¯ ¯ Dn (y) f (x − y) dy ¯¯ ≤ ¯¯

¯Z ¯ +C ¯¯

=|f (x)|

π

−π

}|

{¯ ¯ Dn (y) f (x) dy ¯¯

¯ ¶ ¶ µµ π ¯ 1 y (g (x, y) / sin (y/2)) dy ¯¯ sin n+ 2 −π

Show the functions, y → g (x, y) / sin (y/2) are bounded in L1 independent of x and get a uniform bound on ||Ln ||. Now use a similar argument to show {Ln f } is equicontinuous in addition to being uniformly bounded. If Ln f fails to converge to f uniformly, then there exists ε > 0 and a subsequence, nk such that ||Lnk f − f ||∞ ≥ ε where this is the norm in Y or equivalently the sup norm on [−π, π]. By the Arzela Ascoli theorem, there is a further subsequence, Lnkl f which converges uniformly on [−π, π]. But by Problem 8 Ln f (x) → f (x). 12. Let X be a normed linear space and let M be a convex open set containing 0. Define x ρ(x) = inf{t > 0 : ∈ M }. t Show ρ is a gauge function defined on X. This particular example is called a Minkowski functional. It is of fundamental importance in the study of locally convex topological vector spaces. A set, M , is convex if λx + (1 − λ)y ∈ M whenever λ ∈ [0, 1] and x, y ∈ M . 13. ↑ The Hahn Banach theorem can be used to establish separation theorems. Let M be an open convex set containing 0. Let x ∈ / M . Show there exists x∗ ∈ X 0 ∗ ∗ such that Re x (x) ≥ 1 > Re x (y) for all y ∈ M . Hint: If y ∈ M, ρ(y) < 1.

13.5. EXERCISES

391

Show this. If x ∈ / M, ρ(x) ≥ 1. Try f (αx) = αρ(x) for α ∈ R. Then extend f to the whole space using the Hahn Banach theorem and call the result F , show F is continuous, then fix it so F is the real part of x∗ ∈ X 0 . 14. A Banach space is said to be strictly convex if whenever ||x|| = ||y|| and x 6= y, then ¯¯ ¯¯ ¯¯ x + y ¯¯ ¯¯ ¯¯ ¯¯ 2 ¯¯ < ||x||. F : X → X 0 is said to be a duality map if it satisfies the following: a.) ||F (x)|| = ||x||. b.) F (x)(x) = ||x||2 . Show that if X 0 is strictly convex, then such a duality map exists. The duality map is an attempt to duplicate some of the features of the Riesz map in Hilbert space which is discussed in the chapter on Hilbert space. Hint: For an arbitrary Banach space, let n o 2 F (x) ≡ x∗ : ||x∗ || ≤ ||x|| and x∗ (x) = ||x|| Show F (x) 6= ∅ by using the Hahn Banach theorem on f (αx) = α||x||2 . Next show F (x) is closed and convex. Finally show that you can replace the inequality in the definition of F (x) with an equal sign. Now use strict convexity to show there is only one element in F (x). 15. Prove the following theorem which is an improved version of the open mapping theorem, [20]. Let X and Y be Banach spaces and let A ∈ L (X, Y ). Then the following are equivalent. AX = Y, A is an open map. There exists a constant M such that for every y ∈ Y , there exists x ∈ X with y = Ax and ||x|| ≤ M ||y||. Note this gives the equivalence between A being onto and A being an open map. The open mapping theorem says that if A is onto then it is open. 16. Suppose D ⊆ X and D is dense in X. Suppose L : D → Y is linear and e defined ||Lx|| ≤ K||x|| for all x ∈ D. Show there is a unique extension of L, L, e e is linear. You do not get uniqueness on all of X with ||Lx|| ≤ K||x|| and L when you use the Hahn Banach theorem. Therefore, in the situation of this problem, it is better to use this result. 17. ↑ A Banach space is uniformly convex if whenever ||xn ||, ||yn || ≤ 1 and ||xn + yn || → 2, it follows that ||xn − yn || → 0. Show uniform convexity implies strict convexity (See Problem 14). Hint: Suppose it ¯is ¯ not strictly ¯¯ n ¯¯ convex. Then there exist ||x|| and ||y|| both equal to 1 and ¯¯ xn +y = 1 2 consider xn ≡ x and yn ≡ y, and use the conditions for uniform convexity to get a contradiction. It can be shown that Lp is uniformly convex whenever ∞ > p > 1. See Hewitt and Stromberg [31] or Ray [56].

392

BANACH SPACES

18. Show that a closed subspace of a reflexive Banach space is reflexive. Hint: The proof of this is an exercise in the use of the Hahn Banach theorem. Let Y be the closed subspace of the reflexive space X and let y ∗∗ ∈ Y 00 . Then i∗∗ y ∗∗ ∈ X 00 and so i∗∗ y ∗∗ = Jx for some x ∈ X because X is reflexive. Now argue that x ∈ Y as follows. If x ∈ / Y , then there exists x∗ such that x∗ (Y ) = 0 but x∗ (x) 6= 0. Thus, i∗ x∗ = 0. Use this to get a contradiction. When you know that x = y ∈ Y , the Hahn Banach theorem implies i∗ is onto Y 0 and for all x∗ ∈ X 0 , y ∗∗ (i∗ x∗ ) = i∗∗ y ∗∗ (x∗ ) = Jx (x∗ ) = x∗ (x) = x∗ (iy) = i∗ x∗ (y). 19. We say that xn converges weakly to x if for every x∗ ∈ X 0 , x∗ (xn ) → x∗ (x). xn * x denotes weak convergence. Show that if ||xn − x|| → 0, then xn * x. 20. ↑ Show that if X is uniformly convex, then if xn * x and ||xn || → ||x||, it follows ||xn −x|| → 0. Hint: Use Lemma 13.2.9 to obtain f ∈ X 0 with ||f || = 1 and f (x) = ||x||. See Problem 17 for the definition of uniform convexity. Now by the weak convergence, you can argue that if x 6= 0, f (xn / ||xn ||) → f (x/ ||x||). You also might try to show this in the special case where ||xn || = ||x|| = 1. 21. Suppose L ∈ L (X, Y ) and M ∈ L (Y, Z). Show M L ∈ L (X, Z) and that ∗ (M L) = L∗ M ∗ . 22. Let X and Y be Banach spaces and suppose f ∈ L (X, Y ) is compact. Recall this means that if B is a bounded set in X, then f (B) has compact closure in Y. Show that f ∗ is also a compact map. Hint: Take a bounded subset of Y 0 , S. You need to show f ∗ (S) is totally bounded. You might consider using the Ascoli Arzela theorem on the functions of S applied to f (B) where B is the closed unit ball in X.

Locally Convex Topological Vector Spaces 14.1

Fundamental Considerations

The right context to consider certain topics like separation theorems is in locally convex topological vector spaces, a generalization of normed linear spaces. Let X be a vector space and let Ψ be a collection of functions defined on X such that if ρ ∈ Ψ, ρ(x + y) ≤ ρ(x) + ρ(y), ρ(ax) = |a| ρ(x) if a ∈ F, ρ(x) ≥ 0, where F denotes the field of scalars, either R or C, assumed to be C unless otherwise specified. These functions are called seminorms because it is not necessarily true that x = 0 when ρ (x) = 0. A basis for a topology, B, is defined as follows. Definition 14.1.1 For A a finite subset of Ψ and r > 0, BA (x, r) ≡ {y ∈ X : ρ(x − y) < r for all ρ ∈ A}. Then B ≡ {BA (x, r) : x ∈ X, r > 0, and A ⊆ Ψ, A finite}. That this really is a basis is the content of the next theorem. Theorem 14.1.2 B is the basis for a topology. Proof: I need to show that if BA (x, r1 ) and BB (y, r2 ) are two elements of B and if z ∈ BA (x, r1 ) ∩ BB (y, r2 ), then there exists U ∈ B such that z ∈ U ⊆ BA (x, r1 ) ∩ BB (y, r2 ). Let r = min (min{(r1 − ρ(z − x)) : ρ ∈ A}, 393

394

LOCALLY CONVEX TOPOLOGICAL VECTOR SPACES

min{(r2 − ρ(z − y)) : ρ ∈ B}) and consider BA∪B (z, r). If w belongs to this set, then for ρ ∈ A, ρ(w − z) < r1 − ρ(z − x). Hence ρ (w − x) ≤ ρ(w − z) + ρ(z − x) < r1 for each ρ ∈ A and so BA∪B (z, r) ⊆ BA (x, r1 ). Similarly, BA∪B (z, r) ⊆ BB (y, r2 ). This proves the theorem. Let τ be the topology consisting of unions of all subsets of B. Then (X, τ ) is a locally convex topological vector space. Theorem 14.1.3 The vector space operations of addition and scalar multiplication are continuous. More precisely, + : X × X → X, · : F×X→X are continuous. Proof: It suffices to show +−1 (B) is open in X × X and ·−1 (B) is open in F×X if B is of the form B = {y ∈ X : ρ(y − x) < r} because finite intersections of such sets form the basis B. (This collection of sets is a subbasis.) Suppose u + v ∈ B where B is described above. Then ρ(u + v − x) < λr for some λ < 1. Consider Bρ (u, δ) × Bρ (v, δ). If (u1 , v1 ) is in this set, then ρ(u1 + v1 − x) ≤
0 be small enough that δ < 1 and also λr + δ (ρ (z) + 1) + δ |α| < r.

14.1. FUNDAMENTAL CONSIDERATIONS

395

Then consider (β, w) ∈ B (α, δ) × Bρ (z, δ). ρ (βw − x) − ρ (αz − x) ≤ ρ (βw − αz) ≤ |β − α| ρ (w) + ρ (w − z) |α| ≤ |β − α| (ρ (z) + 1) + ρ (w − z) |α|
0. Suppose ρA (x − y) < r (CA + 1)

−1

. Then

|f (x) − f (y)| = |f (x − y)| ≤ CA ρA (y − x) < r. Hence

³ ³ ´´ −1 f BA x, r (CA + 1) ⊆ B(f (x), r) ⊆ V.

Thus f is continuous at x. This proves the theorem. What are some examples of locally convex topological vector spaces? It is obvious that any normed linear space is such an example. More generally, here is a theorem which shows how to make any vector space into a locally convex topological vector space. Theorem 14.1.7 Let X be a vector space and let Y be a vector space of linear functionals defined on X. For each y ∈ Y , define ρy (x) ≡ |y (x)|. Then the collection of seminorms {ρy }y∈Y defined on X makes X into a locally convex topological vector space and Y = X 0 .

14.2. SEPARATION THEOREMS

397

Proof: Clearly {ρy }y∈Y is a collection of seminorms defined on X; so, X supplied with the topology induced by this collection of seminorms is a locally convex topological vector space. Is Y = X 0 ? Let y ∈ Y , let U ⊆ F be open and let x ∈ y −1 (U ). Then B (y (x) , r) ⊆ U for some r > 0. Letting A = {y}, it is easy to see from the definition that BA (x, r) ⊆ y −1 (U ) and so y −1 (U ) is an open set as desired. Thus, Y ⊆ X 0 . Now suppose z ∈ X 0 . Then by 14.1.2, there exists a finite subset of Y, A ≡ {y1 , · · · , yn }, such that |z (x)| ≤ CρA (x). Let π (x) ≡ (y1 (x) , · · · , yn (x)) and let f be a linear map from π (X) to F defined by f (πx) ≡ z (x). (This is well defined because if π (x) = π (x1 ), then yi (x) = yi (x1 ) for i = 1, · · · , n and so ρA (x − x1 ) = 0. Thus, |z (x1 ) − z (x)| = |z (x1 − x)| ≤ CρA (x − x1 ) = 0.) Extend f to all of Fn and denote the resulting linear map by F . Then there exists a vector α = (α1 , · · · , αn ) ∈ Fn with αi = F (ei ) such that F (β) = α · β. Hence for each x ∈ X, z (x) = f (πx) = F (πx) =

n X

αi yi (x)

i=1

and so z=

n X

αi yi ∈ Y.

i=1


14.2

Separation Theorems

It will always be assumed that X is a locally convex topological vector space. A set, K, is said to be convex if whenever x, y ∈ K, λx + (1 − λ) y ∈ K for all λ ∈ [0, 1].

398


Definition 14.2.1 Let U be an open convex set containing 0 and define m (x) ≡ inf{t > 0 : x/t ∈ U }. This is called a Minkowski functional. Proposition 14.2.2 Let X be a locally convex topological vector space. Then m is defined on X and satisfies m (x + y) ≤ m (x) + m (y)

(14.2.4)

m (λx) = λm (x) if λ > 0.

(14.2.5)

Thus, m is a gauge function on X. Proof: Let x ∈ X be arbitrary. There exists A ⊆ Ψ such that 0 ∈ BA (0, r) ⊆ U. Then

rx ∈ BA (0, r) ⊆ U 2ρA (x)

which implies

2ρA (x) ≥ m (x) . r

(14.2.6)

Thus m (x) is defined on X. Let x/t ∈ U, y/s ∈ U . Then since U is convex, µ ¶³ ´ µ ¶³ ´ x+y t x s y = + ∈ U. t+s t+s t t+s s It follows that m (x + y) ≤ t + s. Choosing s, t such that t − ε < m (x) and s − ε < m (y), m (x + y) ≤ m (x) + m (y) + 2ε. Since ε is arbitrary, this shows 14.2.4. It remains to show 14.2.5. Let x/t ∈ U . Then if λ > 0, λx ∈U λt and so m (λx) ≤ λt. Thus m (λx) ≤ λm (x) for all λ > 0. Hence ¡ ¢ m (x) = m λ−1 λx ≤ λ−1 m (λx) ≤ λ−1 λm (x) = m (x) and so λm (x) = m (λx) . This proves the proposition.


399

Lemma 14.2.3 Let U be an open convex set containing 0 and let q ∈ / U . Then there exists f ∈ X 0 such that Re f (q) > Re f (x) for all x ∈ U . Proof: Let m be the Minkowski functional just defined and let F (cq) = cm (q) for c ∈ R. If c > 0 then F (cq) = m (cq) while if c ≤ 0, F (cq) = cm (q) ≤ 0 ≤ m (cq) . By the Hahn Banach theorem, F has an extension, g, defined on all of X satisfying g (x + y) = g (x) + g(y), g (cx) = cg (x) for all c ∈ R, and g (x) ≤ m (x). Thus, g (−x) ≤ m (−x) and so −m (−x) ≤ g (x) ≤ m (x). It follows as in 14.2.6 that for some A ⊆ Ψ, A finite, and r > 0, |g (x)| ≤ m (x) + m (−x) ≤

2 4 2 ρ (x) + ρA (−x) = ρA (x) r A r r

because ρA (−x) = |−1| ρA (x) = ρA (x). Hence g is continuous by Theorem 14.1.6. Now define f (x) ≡ g (x) − ig (ix). Thus f is linear and continuous so f ∈ X 0 and Re f (x) = g (x). But for x ∈ U, Theorem 14.1.3 implies that x/t ∈ U for some t < 1 and so m (x) < 1. Since U is convex and 0 ∈ U , it follows q/t ∈ / U if t < 1 because if it were, ³q ´ q=t + (1 − t) 0 ∈ U. t Therefore, m (q) ≥ 1 and for x ∈ U , Re f (x) = g (x) ≤ m (x) < 1 ≤ m (q) = g (q) = Re f (q) and this proves the lemma.

400


Theorem 14.2.4 Let K be closed and convex in a locally convex topological vector space and let p ∈ / K. Then there exists a real number, c, and f ∈ X 0 such that Re f (p) > c > Re f (k) for all k ∈ K. Proof: Since K is closed, and p ∈ / K, there exists a finite subset of Ψ, A, and a positive r > 0 such that K ∩ BA (p, 2r) = ∅. Pick k0 ∈ K and let U = K + BA (0, r) − k0 , q = p − k0 . It follows that U is an open convex set containing 0 and q ∈ / U . Therefore, by Lemma 14.2.3, there exists f ∈ X 0 such that Re f (p − k0 ) = Re f (q) > Re f (k + e − k0 )

(14.2.7)

for all k ∈ K and e ∈ BA (0, r). If Re f (e) = 0 for all e ∈ BA (0, r), then Re f = 0 and 14.2.7 could not hold. Therefore, Re f (e) > 0 for some e ∈ BA (0, r) and so, Re f (p) > Re f (k) + Re f (e) for all k ∈ K. Let c1 ≡ sup {Re f (k) : k ∈ K}. Then for all k ∈ K, Re f (p) ≥ c1 + Re f (e) > c1 + Let c = c1 +

Re f (e) . 2

Re f (e) > Re f (k). 2


K {x : Ref (x) = c}

³ ³³

³ ³³ ³³ · p

Note that if the field of scalars comes from R rather than C there is no essential change to the above conclusions. Just eliminate all references to the real part.

14.2.1

Convex Functionals

As an important application, this theorem gives the basis for proving something about lower semicontinuity of functionals. Definition 14.2.5 Let X be a Banach space and let φ : X → (0, ∞] be convex and lower semicontinuous. This means whenever x ∈ X and limn→∞ xn = x, φ (x) ≤ lim inf φ (xn ) . n→∞

Also assume φ is not identically equal to ∞.


401

Lemma 14.2.6 Let X, Y be two Banach spaces. Then letting ||(x, y)|| ≡ max (||x||X , ||y||Y ) , 0

it follows X × Y is a Banach space and φ ∈ (X × Y ) if and only if there exist x∗ ∈ X 0 and y ∗ ∈ Y 0 such that φ ((x, y)) = x∗ (x) + y ∗ (y) . The topology coming from this norm is called the strong topology. Proof: Most of these conclusions are obvious. In particular it is clear X × Y is 0 a Banach space with the given norm. Let φ ∈ (X × Y ) . Also let π X (x, y) ≡ (x, 0) and π Y (x, y) ≡ (0, y) . Then each of π X and π Y is continuous and φ ((x, y)) = =

φ (π X + π Y ) ((x, y)) φ ((x, 0)) + φ ((0, y)) .

Thus φ ◦ π X and φ ◦ π Y are both continuous and their sum equals φ. Let x∗ (x) ≡ φ◦π X (x, 0) and let y ∗ ≡ φ◦π Y (x, 0) . Then it is clear both x∗ and y ∗ are continuous and linear defined on X and Y respectively. Also, if (x∗ , y ∗ ) ∈ X 0 × Y 0 , then if 0 φ ((x, y)) ≡ x∗ (x) + y ∗ (y) , it follows φ ∈ (X × Y ) . This proves the lemma. Lemma 14.2.7 Let φ be a functional as described in Definition 14.2.5. Then φ is lower semicontinuous if and only if the epigraph of φ is closed in X × R with the strong topology. Here the epigraph is defined as epi (φ) ≡ {(x, y) : y ≥ φ (x)} . In this case the functional is called strongly lower semicontinuous. Proof: First suppose epi (φ) is closed and suppose xn → x. Let l < φ (x) . Then (x, l) ∈ / epi (φ) and so there exists δ > 0 such that if |x − y| < δ and |α − l| < δ, then α < φ (y) . This implies that if |x − y| < δ and α < l + δ, then the above holds. Therefore, (xn , φ (xn )), being in epi (φ) cannot satisfy both conditions, |xn − x| < δ, φ (xn ) < l + δ. However, for all n large enough, the first condition is satisfied. Consequently, for all n large enough, φ (xn ) ≥ l + δ ≥ l. Thus lim inf φ (xn ) ≥ l n→∞

and since l < φ (x) is arbitrary, it follows lim inf φ (xn ) ≥ φ (x) . n→∞

402


Next suppose the condition about the lim inf. If epi (φ) is not closed, then there exists (x, l) ∈ / epi (φ) which is a limit point of points of epi (φ) , Thus there exists (xn , ln ) ∈ epi (φ) such that (xn , ln ) → (x, l) and so l = lim inf ln ≥ lim inf φ (xn ) ≥ φ (x) , n→∞

n→∞

contradicting (x, l) ∈ / epi (φ). This proves the lemma. Definition 14.2.8 Let φ be convex and defined on X, a Banach space. Then φ is said to be weakly lower semicontinuous if epi (φ) is closed in X × R where a basis for the topology of X × R consists of sets of the form U × (a, b) for U a weakly open set in X. Theorem 14.2.9 Let φ be a lower semicontinuous convex functional as described in Definition 14.2.5 and let X be a real Banach space. Then φ is also weakly lower semicontinuous. Proof: By Lemma 14.2.7 epi (φ) is closed in X × R with the strong topology as well as being convex. Letting (z, l) ∈ / epi (φ) , it follows from Theorem 14.2.4 and Lemma 14.2.6 there exists (x∗ , α) ∈ X 0 × R such that for some c x∗ (z) + αl > c > x∗ (x) + αβ whenever β ≥ φ (x) . Consider B{(x∗ ,α)} ((z, l) , r) where r is chosen so small that if (y, γ) ∈ B{(x∗ ,α)} ((z, l) , r) , then x∗ (y) + αγ > c. This shows that the complement of epi (φ) is weakly open and this proves the theorem. Corollary 14.2.10 Let φ be a lower semicontinuous convex functional as described in Definition 14.2.5 and let X be a real Banach space. Then if xn converges weakly to x, it follows that φ (x) ≤ lim inf φ (xn ) . n→∞

Proof: Let l < φ (x) so that (x, l) ∈ / epi (φ). Then by Theorem 14.2.9 there exists B × (−∞, l + δ) such that B is a weakly open set in X containing x and C

B × (−∞, l + δ) ⊆ epi (φ) . Thus (xn , φ (xn )) ∈ / B×(−∞, l + δ) for all n. However, xn ∈ B for all n large enough. Therefore, for those values of n, it must be the case that φ (xn ) ∈ / (−∞, l + δ) and so lim inf φ (xn ) ≥ l + δ ≥ l n→∞

which shows, since l < φ (x) is arbitrary that lim inf φ (xn ) ≥ φ (x) . n→∞

This proves the corollary.


14.2.2

403

More Separation Theorems

There are other separation theorems which can be proved in a similar way. The next theorem considers the separation of an open convex set from a convex set. Theorem 14.2.11 Let A and B be disjoint, convex and nonempty sets with B open. Then there exists f ∈ X 0 such that Re f (a) < Re f (b) for all a ∈ A and b ∈ B. Proof: Let b0 ∈ B, a0 ∈ A. Then the set B − A + a0 − b0 is open, convex, contains 0, and does not contain a0 − b0 . By Lemma 14.2.3 there exists f ∈ X 0 such that Re f (a0 − b0 ) > Re f (b − a + a0 − b0 ) for all a ∈ A and b ∈ B. Therefore, for all a ∈ A, b ∈ B, Re f (b) > Re f (a) . Before giving another separation theorem, here is a lemma. Lemma 14.2.12 If B is convex, then int (B) ≡ union of all open sets contained in B is convex. Also, if int (B) 6= ∅, then B ⊆ int (B). Proof: Suppose x, y ∈ int (B). Then there exists r > 0 and a finite set A ⊆ Ψ such that BA (x, r) , BA (y, r) ⊆ B. Let V ≡ ∪λ∈[0,1] λBA (x, r) + (1 − λ) BA (y, r). Then V is open, V ⊆ B, and if λ ∈ [0, 1], then λx + (1 − λ) y ∈ V ⊆ B. Therefore, int (B) is convex as claimed. Now let y ∈ B and x ∈ int (B). Let x ∈ BA (x, r) ⊆ int (B) and let xλ ≡ (1 − λ) x + λy. Define the open cone, C ≡ ∪λ∈[0,1] BA (xλ , (1 − λ) r).

404


Thus C is represented in the following picture.

BA (x, r) • x

B

C

• xλ

•y

I claim C ⊆ B as suggested in the picture. To see this, let z ∈ BA (xλ , (1 − λ) r) , λ ∈ (0, 1). Then ρA (z − xλ ) < (1 − λ) r and so

µ ρA

Therefore,

z λy −x− 1−λ 1−λ

¶ < r.

z λy − ∈ BA (x, r) ⊆ B. 1−λ 1−λ

It follows

µ (1 − λ)

z λy − 1−λ 1−λ

¶ + λy = z ∈ B

and so C ⊆ B as claimed. Now this shows xλ ∈ int (B) and limλ→1 xλ = y. Thus, y ∈ int (B) and this proves the lemma. Corollary 14.2.13 Let A, B be convex, nonempty sets. Suppose int (B) 6= ∅ and A ∩ int (B) = ∅. Then there exists f ∈ X 0 , f 6= 0, such that for all a ∈ A and b ∈ B, Re f (b) ≥ Re f (a). Proof: By Theorem 14.2.11, there exists f ∈ X 0 such that for all b ∈ int (B), and a ∈ A, Re f (b) > Re f (a) . Thus, in particular, f 6= 0. By Lemma 14.2.12, if b ∈ B and a ∈ A, Re f (b) ≥ Re f (a) . This proves the theorem.


405

Lemma 14.2.14 If X is a topological Hausdorff space then compact implies closed. Proof: Let K be compact and suppose K C is not open. Then there exists p ∈ K C such that Vp ∩ K 6= ∅ for all open sets Vp containing p. Let ¡ ¢C C ={ V p : Vp is an open set containing p}. Then C is an open cover of K because if q ∈ K, there exist disjoint open sets Vp ¡ ¢C and Vq containing p and q respectively. Thus q ∈ V p . This is an example of an open cover of K which has no finite subcover, contradicting the assumption that K is compact. This proves the lemma. Lemma 14.2.15 If X is a locally convex topological vector space, and if every point is a closed set, then the seminorms and X 0 separate the points. This means if x 6= y, then for some ρ ∈ Ψ, ρ (x − y) 6= 0 and for some f ∈ X 0 , f (x) 6= f (y) . In this case, X is a Hausdorff space. Proof: Let x 6= y. Then by Theorem 14.2.4, there exists f ∈ X 0 such that f (x) 6= f (y). Thus X 0 separates the points. Since f ∈ X 0 , Theorem 14.1.6 implies |f (z)| ≤ CρA (z) for some A a finite subset of Ψ. Thus 0 < |f (x − y)| ≤ CρA (x − y) and so ρ (x − y) 6= 0 for some ρ ∈ A ⊆ Ψ. Now to show X is Hausdorff, let 0 < r < ρ (x − y) 2−1. Then the two disjoint open sets containing x and y respectively are Bρ (x, r) and Bρ (y, r). This proves the lemma.

406

14.3


The Weak And Weak∗ Topologies

The weak and weak ∗ topologies are examples which make the underlying vector space into a topological vector space. This section gives a description of these topologies. Unless otherwise specified, X is a locally convex topological vector space. For G a finite subset of X 0 define δ G : X → [0, ∞) by δ G (x) = max{|f (x)| : f ∈ G}. Lemma 14.3.1 The functions δ G for G a finite subset of X 0 are seminorms and the sets BG (x, r) ≡ {y ∈ X : δ G (x − y) < r} form a basis for a topology on X. Furthermore, X with this topology is a locally convex topological vector space. If each point in X is a closed set, then the same is true of X with respect to this new topology. Proof: It is obvious that the functions δ G are seminorms and therefore the proof that the sets BG (x, r) form a basis for a topology is the same as in Theorem 14.1.2. To see every point is a closed set in this new topology, assuming this is true for X with the original topology, use Lemma 14.2.15 to assert X 0 separates the points. Let x ∈ X and let y 6= x. There exists f ∈ X 0 such that f (x) 6= f (y). Let G = {f } and consider BG (y, |f (x − y)| /2). Then this open set does not contain x. Thus {x}C is open and so {x} is closed. This proves the Lemma. This topology for X is called the weak topology for X. For F a finite subset of X, define γ F : X 0 → [0, ∞) by γ F (f ) = max{|f (x) | : x ∈ F }. Lemma 14.3.2 The functions γ F for F a finite subset of X are seminorms and the sets BF (f, r) ≡ {g ∈ X 0 : γ F (f − g) < r} form a basis for a topology on X 0 . Furthermore, X 0 with this topology is a locally convex topological vector space having the property that every point is a closed set. Proof: The proof is similar to that of Lemma 14.3.1 but there is a difference in the part where every point is shown to be a closed set. Let f ∈ X 0 and let g 6= f . Thus there exists x ∈ X such that f (x) 6= g (x). Let F = {x}. Then BF (g, | (f − g) (x) |/2) contains g but not f . Thus {f }C is open and so {f } is closed. This proves the lemma. Note that it was not necessary to assume points in X are closed sets to get this.

14.4. THE TYCHONOFF FIXED POINT THEOREM

407

The topology for X 0 just described is called the weak ∗ topology. In terms of Theorem 14.1.7 the weak topology is obtained by letting Y = X 0 in that theorem while the weak ∗ topology is obtained by letting Y = X with the understanding that X is a vector space of linear functionals on X 0 defined by x (x∗ ) ≡ x∗ (x). By Theorem 14.2.4, there is a useful result which follows immediately. Theorem 14.3.3 Let K be closed and convex in a Banach space X. Then it is also weakly closed. Furthermore, if p ∈ / K, there exists f ∈ X 0 such that Re f (p) > c > Re f (k)

(14.3.8)

for all k ∈ K. If K ∗ is closed and convex in the dual of a Banach space, X 0 , then it is also weak ∗ closed. Proof: By Theorem 14.2.4 there exists f ∈ X 0 such that 14.3.8 holds. Therefore, letting A = {f } , it follows that for r small enough, BA (p, r) ∩ K = ∅. Thus K is weakly closed. This establishes the first part. For the second part, the seminorms for the weak ∗ toplogy are determined from X and the continuous linear functionals are of the form x∗ → x∗ (x) where x ∈ X. Thus if p∗ ∈ / K ∗ , it follows from Theorem 14.2.4 there exists x ∈ X such that Re p∗ (x) > c > Re k ∗ (x) for all k ∗ ∈ K ∗ . Therefore, letting A = {x} , BA (p∗ , r) ∩ K ∗ = ∅ whenever r is small enough and this shows K ∗ is weak ∗ closed. This proves the theorem.

14.4

The Tychonoff Fixed Point Theorem

The objective in this section is to present a proof of the Tychonoff fixed point theorem, one of the most useful of the generalizations of the Brouwer fixed point theorem. In this section, X will be a locally convex topological vector space in which every point is a closed set. Let B be the basis described earlier and let B0 consist of all sets of B which are of the form BA (0, r) where A is a finite subset of Ψ as described earlier. Note that for U ∈ B0 , U = −U and U is convex. Also, if U ∈ B0 , there exists V ∈ B0 such that V +V ⊆U where V + V ≡ {v1 + v2 : vi ∈ V }. To see this, note BA (0, r/2) + BA (0, r/2) ⊆ B(0, r). We let K be a closed convex subset of X and let f be continuous, f : K → K, and f (K) is compact.

408


Lemma 14.4.1 For each U ∈ B0 , there exists a finite set of points {y1 · · · yn } ⊆ f (K) and continuous functions ψ i defined on f (K) such that for x ∈ f (K), n X

ψ i (x) = 1,

(14.4.9)

i=1

ψ i (x) = 0 if x ∈ / yi + U, ψ i (x) > 0 if x ∈ yi + U. If fU (x) ≡

n X

yi ψ i (f (x)),

(14.4.10)

i=1

then whenever x ∈ K, f (x) − fU (x) ∈ U. Proof: Let U = BA (0, r) . Using the compactness of f (K), there exists {y1 · · · yn } ⊆ f (K) such that

n

{yi + U }i=1 covers f (K). Let

+

φi (y) ≡ (r − ρA (y − yi )) . Thus φi (y) > 0 if y ∈ yi + U and φi (y) = 0 if y ∈ / yi + U . For x ∈ f (K), let  −1 n X ψ i (x) ≡ φi (x)  φj (x) . j=1

Then 14.4.9 is satisfied. Now let fU be given by 14.4.10 for x ∈ K. For such x, X f (x) − fU (x) = (f (x) − yi ) ψ i (f (x)) {i:f (x)−yi ∈U }

+

X

(f (x) − yi ) ψ i (f (x))

{i:f (x)−yi ∈U / }

= X {i:f (x)−yi ∈U }

X

(f (x) − yi ) ψ i (f (x)) =

{i:f (x)−yi ∈U }

(f (x) − yi ) ψ i (f (x)) +

X

0ψ i (f (x)) ∈ U

{i:f (x)−yi ∈U / }

because 0 ∈ U , U is convex, and 14.4.9. This proves the lemma. We think of fU as an approximation to f .


409

Lemma 14.4.2 For each U ∈ B0 , there exists xU ∈ convex hull of f (K) such that fU (xU ) = xU . Proof: If fU (xU ) = xU and xU =

n X

ai yi

i=1

for

Pn i=1

ai = 1, we need n X

Ã Ã n !! n X X yj ψ j f ai yi = aj yj .

j=1

i=1

j=1

This will be satisfied if for each j = 1, · · · , n, Ã Ã n !! X ai yi ; aj = ψ j f

(14.4.11)

i=1

so, let

( Σn−1 ≡

n

a ∈R :

n X

) ai = 1, ai ≥ 0

i=1

and let h : Σn−1 → Σn−1 be given by Ã Ã h (a)j ≡ ψ j

f

n X

!! ai yi

.

i=1

Since h is continuous, the Brouwer fixed point theorem applies and we see there exists a fixed point for h which is a solution to 14.4.11. This proves the lemma. Theorem 14.4.3 Let K be a closed and convex subset of X, a locally convex topological vector space in which every point is closed. Let f : K → K be continuous and suppose f (K) is compact. Then f has a fixed point. Proof: First consider the following claim which will yield a candidate for the fixed point. Claim: There exists x ∈ f (K) with the property that if V ∈ B0 , there exists U ⊆ V , U ∈ B0 , such that f (xU ) ∈ x + V. Proof of the claim: If no such x exists, then for each x ∈ f (K), there exists Vx ∈ B0 such that whenever U ⊆ Vx , with U ∈ B0 , f (xU ) ∈ / x + Vx .

410


Since f (K) is compact, there exist x1 , · · · , xn ∈ f (K) such that n

{xi + Vxi }i=1 cover f (K). Let U ∈ B0 , U ⊆ ∩ni=1 Vxi and consider xU . f (xU ) ∈ xi + Vxi for some i but U ⊆ Vxi , a contradiction. This shows the claim. Now I show x is the desired fixed point. Let W ∈ B0 and let V ∈ B0 with V + V + V ⊆ W. Since f is continuous at x, there exists V0 ∈ B0 such that V0 + V0 ⊆ V and if y − x ∈ V0 + V0 , then f (x) − f (y) ∈ V. Using the claim, let U ∈ B0 , U ⊆ V0 , such that f (xU ) ∈ x + V0 . Then x − xU = x − f (xU ) + f (xU ) − fU (xU ) ∈ V0 + U ⊆ V0 + V0 ⊆ V and so f (x) − x = = ⊆

f (x) − f (xU ) + f (xU ) − fU (xU ) + fU (xU ) − x f (x) − f (xU ) + f (xU ) − fU (xU ) + xU − x V + U + V ⊆ W.

Since W ∈ B0 is arbitrary, it follows from Lemma 14.2.15 that f (x) − x = 0. This proves the theorem. In the case where X is normed linear space, this fixed point theorem is called the Schauder fixed point theorem. As an example of the usefulness of this fixed point theorem, consider the following application to the theory of ordinary differential equations. In the context of this theorem, X = C ([0, T ] ; Rn ), a Banach space with norm given by ||x|| ≡ max {|x (t)| : t ∈ [0, T ]} .


411

Theorem 14.4.4 Let f : [0, T ] × Rn → Rn be continuous and suppose there exists L > 0 such that for all λ ∈ (0, 1), if x0 = λf (t, x) , x (0) = x0

(14.4.12)

for all t ∈ [0, T ], then ||x|| < L. Then there exists a solution to x0 = f (t, x) , x (0) = x0

(14.4.13)

for t ∈ [0, T ]. Proof: Let

Z N x (t) ≡

t

f (s, x (s)) ds. 0

Thus a solution to the initial value problem exists if there exists a solution to x0 + N (x) = x. Let

n o m ≡ max |f (t, x)| : (t, x) ∈ [0, T ] × B (0, L) , M ≡ |x0 | + mT

and let K ≡ {x ∈ C (0, T ; Rn ) such that x (0) = x0 and ||x|| ≤ M } . Now define

( Ax ≡

x0 + N x if ||N x|| ≤ M − |x0 |, 0 |)N x x0 + (M −|x if ||N x|| > M − |x0 |. ||N x||

Then A is continuous and maps X to K. Also A (K) is equicontinuous because Z

t

x0 + N x (t) − (x0 + N x (t1 )) =

f (s, x (s)) ds t1

and the integrand is bounded. Thus A (K) is a compact set in X by the Ascoli Arzela theorem. By the Schauder fixed point theorem, A has a fixed point, x ∈K. If ||N (x)|| > M − |x0 |, then x0 + λN (x) = x where λ=

(M − |x0 |) 0 such that if y ∈ B (x,δ) , then Ay ⊆O. (14.5.15) This last condition is sometimes refered to as upper semicontinuity. In words, A is upper semicontinuous and has values which are compact and convex. Lemma 14.5.1 Let A satisfy 14.5.14 and 14.5.15. Then AK is a subset of a compact set whenever K is compact. Also the graph of A is closed. Proof: Let x ∈ K. Then Ax is compact and contained in some open set whose closure is compact, Ux . By assumption 14.5.15 there exists an open set Vx containing x such that if y ∈Vx , then Ay ⊆Ux . Let Vx1 , · · · , Vxm cover K. Then AK ⊆ ∪m k=1 U xk , a compact set. To see the graph of A is closed, let xk → x, yk → y where yk ∈ Axk . Then letting O = Ax+B (0, r) it follows from 14.5.15 that yk ∈ Axk ⊆ O for all k large enough. Therefore, y ∈ Ax+B (0, 2r) and since r > 0 is arbitrary and Ax is closed it follows y ∈ Ax. The next lemma is an application of the Brouwer fixed point theorem. First define an n simplex, denoted by [x0 , · · · , xn ], to be the convex hull of the n + 1 n points, {x0 , · · · , xn } where {xi − x0 }i=1 are independent. Thus ( n ) n X X [x0 , · · · , xn ] ≡ ti x i : ti = 1, ti ≥ 0 . i=1

i=1

14.5. SET-VALUED MAPS

413

If n ≤ 2, the simplex is a triangle, line segment, or point. If n ≤ 3, it is a tetrahedron, triangle, line segment or point. A collection of simplicies is a tiling of Rn if Rn is contained in their union and if S1 , S2 are two simplicies in the tiling, with h i Sj = xj0 , · · · , xjn , then S1 ∩ S2 = [xk0 , · · · , xkr ] where

© ª © ª {xk0 , · · · , xkr } ⊆ x10 , · · · , x1n ∩ x20 , · · · , x2n

or else the two simplicies do not intersect. The collection of simplicies is said to be locally finite if, for every point, there exists a ball containing that point which also intersects only finitely many of the simplicies in the collection. It is left to the reader to verify that for each ε > 0, there exists a locally finite tiling of Rn which is composed of simplicies which have diameters less than ε. Lemma 14.5.2 Suppose A : Cn → P (Cn ) satisfies 14.5.14 and 14.5.15 and K is a nonempty compact convex set in Cn . Then if y ∈ Cn there exists (x, w) ∈ G (A) such that x ∈ K and Re (y − w, z − x) ≤ 0 for all z ∈ K. Proof: Tile Cn with 2n simplicies such that the collection is locally finite and each simplex has diameter less than ε < 1. This collection of simplicies is determined by a countable collection of vertices. For each vertex, x, pick Aε x ∈ Ax and define Aε on all of Cn by the following rule. If x ∈ [x0 , · · · , x2n ], so x =

P2n

i=0 ti xi ,

then Aε x ≡

2n X

tk Aε xk .

k=0

Thus Aε is a continuous map defined on Cn thanks to the local finiteness of the collection of simplicies. Let PK denote the projection on the convex set K. By the Brouwer fixed point theorem, there exists a fixed point, xε ∈ K such that PK (y−Aε xε + xε ) = xε . By Corollary 15.1.9 this requires Re (y−Aε xε , z − xε ) ≤ 0 for all z ∈ K. P2n Suppose xε ∈ [xε0 , · · · , xε2n ] so xε = k=0 tεk xεk . Then since xε is contained in K, a compact set, and the diameter of each simplex is less than 1, it follows that

414


Aε xεk is contained in A(K + B (0,1)), which is contained in a compact set thanks to Lemma 14.5.1. From the Heine Borel theorem, there exists a sequence ε → 0 such that tεk → tk , xε → x,Aε xεk → yk for k = 0, · · · , 2n. Since the diameter of the simplex containing xε converges to 0, it follows xεk → x, Aε xεk → yk . Since the graph of A is closed and Aε xεk ∈ Axεk , this implies yk ∈ Ax. Since Ax is convex, 2n X tk yk ∈ Ax. k=1

Hence for all z ∈ K, ! Ã ! Ã 2n 2n X X ε ε tk Aε xk , z − xε Re y− tk yk , z − x = lim Re y− ε→0

k=1

k=1

= lim Re (y−Aε xε , z − xε ) ≤ 0. Let w =

P2n

k=1 tk yk .

ε→0


Lemma 14.5.3 Suppose in addition to 14.5.14 and 14.5.15, A is coercive, ½ ¾ Re (y, x) lim inf : y ∈ Ax = ∞. |x| |x|→∞ Then A is onto. Proof: Let y ∈ Cn and let Kr ≡ B (0,r). By Lemma 14.5.2 there exists xr ∈ Kr and wr ∈ Axr such that Re (y − wr , z − xr ) ≤ 0 (14.5.16) for all z ∈ Kr . Letting z = 0, Re (wr , xr ) ≤ Re (y, xr ). Therefore,

½ inf

Re (w, xr ) : w ∈ Axr |xr |

¾ ≤ |y| .

It follows from the assumption of coercivity that |xr | is bounded independent of r. Therefore, picking r strictly larger than this bound, 14.5.16 implies Re (y − wr , v) ≤ 0 for all v in some open ball containing 0. Therefore, for all v in this ball Re (y − wr , v) = 0 and hence this holds for all v ∈ Cn and so y = wr ∈ Axr . This proves the lemma.


415

Lemma 14.5.4 Let F be a finite dimensional Banach space of dimension n, and let T be a mapping from F to P (F 0 ) such that 14.5.14 and 14.5.15 both hold for F 0 in place of Cn . Then if T is also coercive, ½ ¾ Re y∗ (u) ∗ lim inf : y ∈ T u = ∞, (14.5.17) ||u|| ||u||→∞ it follows T is onto. Proof: Let |·| be an equivalent norm for F such that there is an isometry of Cn and F, θ. Now define A : Cn → P (Cn ) by Ax ≡ θ∗ T θx. θ∗

P (F 0 ) → Cn T ↑ ◦ ↑A F

θ

←

Cn

Thus y ∈ Ax means that there exists z∗ ∈ T θx such that (w, y)Cn = z∗ (θw) for all w ∈ Cn . Then A satisfies the conditions of Lemma 14.5.3 and so A is onto. Consequently T is also onto. With these lemmas, it is possible to prove a very useful result about a class of mappings which map a reflexive Banach space to the power set of its dual space. For more theorems about these mappings and their applications, see [52]. In the discussion below, we will use the symbol, *, to denote weak convergence. Definition 14.5.5 Let V be a Reflexive Banach space. We say T : V → P (V 0 ) is pseudomonotone if the following conditions hold. T u is closed, nonempty, convex, and bounded.

(14.5.18)

If F is a finite dimensional subspace of V , then if u ∈ F and W ⊇ T u for W a weakly open set in V 0 , then there exists δ > 0 such that v ∈ B (u, δ) ∩ F implies T v ⊆ W .

(14.5.19)

If uk * u and if u∗k ∈ T uk is such that lim sup Re u∗k (uk − u) ≤ 0, k→∞

then for all v ∈ V , there exists u∗ (v) ∈ T u such that lim inf Re u∗k (uk − v) ≥ Re u∗ (v) (u − v). k→∞

(14.5.20)

We say T is coercive if ½ lim inf

||v||→∞

Re z ∗ (v) ∗ : z ∈ Tv ||v||

¾ = ∞.

(14.5.21)

416


Theorem 14.5.6 Let V be a reflexive Banach space and let T : V → P (V 0 ) be pseudomonotone and coercive. Then T is onto. Proof: Let F be the set of finite subspaces of V and let F ∈ F. Then define TF as TF ≡ i∗F T iF where here iF is the identity map from F to V. Then TF satisfies the conditions of Lemma 14.5.4 and so TF is onto P (F 0 ). Let w∗ ∈ V 0 . Then since TF is onto, there exists uF ∈ F such that i∗F w∗ ∈ i∗F T iF uF . Thus for each finite dimensional subspace, F , there exists uF ∈ F such that for all v ∈ F, w∗ (v) = u∗F (v) , u∗F ∈ T uF . (14.5.22) Replacing v with uF , in 14.5.22, u∗F (uF ) w∗ (uF ) = ≤ ||w∗ ||. ||uF || ||uF || Therefore, the assumption that T is coercive implies {uF : F ∈ F} is bounded in V . Now define WF ≡ ∪ {uF 0 : F 0 ⊇ F } . Then WF is bounded and if WF ≡ weak closure of WF , then © ª WF : F ∈ F is a collection of nonempty weakly compact (since V is reflexive) sets having the finite intersection property because WF 6= ∅ for each F . Thus there exists © ª u ∈ ∩ WF : F ∈ F . I will show w∗ ∈ T u. If w∗ ∈ / T u, a closed convex set, there exists v ∈ V such that Re w∗ (u − v) < Re u∗ (u − v) (14.5.23) for all u∗ ∈ T u. This follows from the separation theorems. (These theorems imply there exists z ∈ V such that Re w∗ (z) < Re u∗ (z) for all u∗ ∈ T u. Define u − v ≡ z.) Now let F ⊇ {u, v}. Since u ∈ WF , a weakly sequentially compact set, there exists a sequence, {uk }, such that uk * u, uk ∈ WF . Then since F ⊇ {u, v}, there exists u∗k ∈ T uk such that u∗k (uk − u) = w∗ (uk − u) .


417

Therefore, lim sup Re u∗k (uk − u) = lim sup Re w∗ (uk − u) = 0. k→∞

k→∞

It follows by the assumption T is pseudomonotone; the following holds for the v defined above in 14.5.23. lim inf Re u∗k (uk − v) ≥ Re u∗ (v) (u − v) , u∗ (v) ∈ T u. k→∞

But since v ∈ F, Re u∗k (uk − v) = Re w∗ (uk − v) and so lim inf Re u∗k (uk − v) = lim inf Re w∗ (uk − v) = Re w∗ (u − v), k→∞

so from 14.5.23,

k→∞

Re w∗ (u − v) = lim inf Re u∗k (uk − v) k→∞

∗

≥ Re u (v) (u − v) > Re w∗ (u − v), a contradiction. Thus, w∗ ∈ T u and this proves the theorem. This is an interesting theorem, but one might wonder if there are easy to verify examples of such possibly set valued mappings. In what follows consider only real spaces because the essential ideas are included in this case which is also the case of most use in applications. Definition 14.5.7 Let V be a real reflexive Banach space and let f : V → R be a locally Lipschitz function, meaning that f is Lipschitz near every point of V although f need not be Lipschitz on all of V. Under these conditions, f 0 (x, y) ≡ lim

f (x + h + µy) − f (x + h) µ µ→0+ h→0 sup

and ∂f (x) ⊆ X 0 is defined by © ª ∂f (x) ≡ x∗ ∈ X 0 : x∗ (y) ≤ f 0 (x, y) for all y ∈ X .

(14.5.24)

(14.5.25)

The set just described is called the generalized gradient. In 14.5.24 we mean the following by the right hand side. ½ ¾ f (x + h + µy) − f (x + h) lim sup : µ ∈ (0, r) , h ∈ B (0, δ) µ (r,δ)→(0,0) I will show, following [52], that these generalized gradients of locally Lipschitz functions are sometimes pseudomonotone. First here is a lemma. Lemma 14.5.8 Let f be as described in the above definition. Then ∂f (x) is a closed, bounded, convex, and non empty subset of V 0 . Furthermore, for x∗ ∈ ∂f (x) , ||x∗ || ≤ Lipx (f ) .

(14.5.26)

418


Proof: It is left as an exercise to verify the assertions that ∂f (x) is closed, and convex. It follows directly from the definition. To verify this set is bounded, let Lipx (f ) denote a Lipschitz constant valid near x ∈ V and let x∗ ∈ ∂f (x) . Then choosing y with ||y|| = 1 and x∗ (y) ≥ 12 ||x∗ || , 1 ∗ ||x || = x∗ (y) ≤ f 0 (x, y) . 2

(14.5.27)

Also, for small µ and h, ¯ ¯ ¯ f (x + h + µy) − f (x + h) ¯ ¯ ¯ ≤ Lipx (f ) ||y|| = Lipx (f ) . ¯ ¯ µ Therefore, f 0 (x, y) ≤ Lipx (f ) and so 14.5.27 shows ||x∗ || ≤ 2Lipx (f ) . The interesting part of this Lemma is that ∂f (x) 6= ∅. To verify this first note that the definition of f 0 implies that y → f 0 (x, y) is a gauge function. Now fix y ∈ V and define on Ry a linear map x∗0 by x∗0 (αy) ≡ αf 0 (x, y) . Then if α ≥ 0, x∗0 (αy) = αf 0 (x, y) = f 0 (x, αy) . If α < 0, x∗0 (αy) ≡ αf 0 (x, y) = lim

inf

µ→0+ h→0

(−α) lim

(−α) f (x + h) − (−α) f (x + h + µy) = µ inf

µ→0+ h→0

f (x + h − µy) − f (x + h) ≤ µ

(−α) f 0 (x, −y) = f 0 (x, αy) . Therefore, x∗0 (αy) ≤ f 0 (x, αy) for all α. By the Hahn Banach theorem there is an extention of x∗0 to all of V, x∗ which satisfies, x∗ (y) ≤ f 0 (x, y) for all y. It remains to verify x∗ is continuous. This follows easily from |x∗ (y)| = max (x∗ (−y) , x∗ (y)) ≤ ¡ ¢ max f 0 (x, y) , f 0 (x, −y) ≤ Lipx (f ) ||y|| , which verifies 14.5.26 and proves the lemma. This lemma has verified the first condition needed in the definition of pseudomonotone. The next lemma verifies that these generalized subgradients satisfy the second of the conditions needed in the definition. In fact somewhat more than is needed in the definition is shown.


419

Lemma 14.5.9 Let U be weakly open in V 0 and suppose ∂f (x) ⊆ U. Then ∂f (z) ⊆ U whenever z is close enough to x. Proof: Suppose to the contrary there exists zn → x but zn∗ ∈ ∂f (zn ) \ U. From the first lemma, we may assert that ||zn∗ || ≤ 2Lip (f ) for all n large enough. Therefore, there is a subsequence, still denoted by n such that zn∗ converges weakly to z ∗ ∈ / U. Claim: f 0 (x, y) ≥ lim supn→∞ f 0 (xn , y) . Proof of the claim: There exists δ > 0 such that if µ, ||h|| < δ, then ε + f 0 (x, y) ≥

f (x + h + µy) − f (x + h) . µ

Thus, for ||h|| < δ, ε + f 0 (x, y) ≥

f (xn + (x − xn ) + h + µy) − f (xn + (x − xn ) + h) . µ

Now let ||h0 || < 2δ and let n be so large that ||x − xn || < 2δ . Suppose||h0 || < 2δ . Then choosing h ≡ h0 − (x − xn ) , it follows the above inequality holds because ||h|| < δ. Therefore, if ||h0 || < 2δ , and n is sufficiently large, ε + f 0 (x, y) ≥

f (xn + h0 + µy) − f (xn + h0 ) . µ

Consequently, for all n large enough, ε + f 0 (x, y) ≥ f 0 (xn , y) which proves the claim. Now with the claim, z ∗ (y) = lim sup zn∗ (y) ≤ lim sup f 0 (xn , y) ≤ f 0 (x, y) n→∞

n→∞

so z ∗ ∈ ∂f (x) contradicting the assumption that z ∗ ∈ / U. This proves the lemma. It is necessary to assume more on f 0 in order to obtain the third axiom defining pseudomonotone. The following theorem describes the situation. Theorem 14.5.10 Let f : V → V 0 be locally Lipschitz and suppose it satisfies the condition that whenever xn converges weakly to x and

lim sup f 0 (xn , x − xn ) ≥ 0 n→∞

it follows that

lim sup f 0 (xn , z − xn ) ≤ f 0 (x, z − x) n→∞

for all z ∈ V. Then ∂f is pseudomonotone.

420


Proof: 14.5.18 and 14.5.19 both are satisfied thanks to Lemmas 14.5.7 and 14.5.8. It remains to verify 14.5.20. To do so, I will adopt the convention that x∗ ∈ ∂f (x) . Suppose lim sup x∗n (xn − x) ≤ 0. (14.5.28) n→∞

This implies

lim inf n→∞ x∗n

(x − xn ) ≥ 0. Thus,

0 ≤ lim inf x∗n (x − xn ) ≤ lim inf f 0 (xn , x − xn ) n→∞

≤ lim sup f 0 (xn , x − xn ) , n→∞

which implies, by the above assumption that for all z, lim sup x∗n (z − xn ) ≤ lim sup f 0 (xn , z − xn ) ≤ f 0 (x, z − x) .

(14.5.29)

In particular, this holds for z = x and this implies lim sup x∗n (x − xn ) ≤ 0 which along with 14.5.28 yields lim x∗n (xn − x) = 0 (14.5.30) n→∞

. Now let z be arbitrary. There exists a subsequence, nk , depending on z such that lim x∗nk (xnk − z) = lim inf x∗nk (xnk − z) . k→∞

Now from Lemma 14.5.8 and its proof, the ||x∗n || are all bounded by Lipx (f ) whenever n is large enough. Therefore, there is a further subsequence, still denoted by nk such that x∗nk converges weakly to x∗ (z) . We need to verify that x∗ (z) ∈ ∂f (x) . To do so, let y be arbitrary. Then from the definition, x∗n (y − xn ) ≤ f 0 (xn , y − xn ) . (14.5.31) From 14.5.30, we can take the lim sup of both sides and obtain, using 14.5.29 x∗ (z) (y − x) ≤ lim sup f 0 (xn , y − xn ) ≤ f 0 (x, y − x) . Since y is arbitrary, this shows x∗ (z) ∈ ∂f (x) and proves the theorem.

Hilbert Spaces 15.1

Basic Theory

Definition 15.1.1 Let X be a vector space. An inner product is a mapping from X × X to C if X is complex and from X × X to R if X is real, denoted by (x, y) which satisfies the following. (x, x) ≥ 0, (x, x) = 0 if and only if x = 0,

(15.1.1)

(x, y) = (y, x).

(15.1.2)

(ax + by, z) = a(x, z) + b(y, z).

(15.1.3)

For a, b ∈ C and x, y, z ∈ X,

Note that 15.1.2 and 15.1.3 imply (x, ay + bz) = a(x, y) + b(x, z). Such a vector space is called an inner product space. The Cauchy Schwarz inequality is fundamental for the study of inner product spaces. Theorem 15.1.2 (Cauchy Schwarz) In any inner product space |(x, y)| ≤ ||x|| ||y||. Proof: Let ω ∈ C, |ω| = 1, and ω(x, y) = |(x, y)| = Re(x, yω). Let F (t) = (x + tyω, x + tωy). If y = 0 there is nothing to prove because (x, 0) = (x, 0 + 0) = (x, 0) + (x, 0) and so (x, 0) = 0. Thus, it can be assumed y 6= 0. Then from the axioms of the inner product, F (t) = ||x||2 + 2t Re(x, ωy) + t2 ||y||2 ≥ 0. 421

422

HILBERT SPACES

This yields

||x||2 + 2t|(x, y)| + t2 ||y||2 ≥ 0.

Since this inequality holds for all t ∈ R, it follows from the quadratic formula that 4|(x, y)|2 − 4||x||2 ||y||2 ≤ 0. This yields the conclusion and proves the theorem. 1/2

Proposition 15.1.3 For an inner product space, ||x|| ≡ (x, x) norm.

does specify a

Proof: All the axioms are obvious except the triangle inequality. To verify this, ||x + y||

2

2

2

≡

(x + y, x + y) ≡ ||x|| + ||y|| + 2 Re (x, y)

≤

||x|| + ||y|| + 2 |(x, y)|

≤

||x|| + ||y|| + 2 ||x|| ||y|| = (||x|| + ||y||) .

2

2

2

2

2

The following lemma is called the parallelogram identity. Lemma 15.1.4 In an inner product space, ||x + y||2 + ||x − y||2 = 2||x||2 + 2||y||2. The proof, a straightforward application of the inner product axioms, is left to the reader. Lemma 15.1.5 For x ∈ H, an inner product space, ||x|| = sup |(x, y)|

(15.1.4)

||y||≤1

Proof: By the Cauchy Schwarz inequality, if x 6= 0, µ ¶ x ||x|| ≥ sup |(x, y)| ≥ x, = ||x|| . ||x|| ||y||≤1 It is obvious that 15.1.4 holds in the case that x = 0. Definition 15.1.6 A Hilbert space is an inner product space which is complete. Thus a Hilbert space is a Banach space in which the norm comes from an inner product as described above. In Hilbert space, one can define a projection map onto closed convex nonempty sets. Definition 15.1.7 A set, K, is convex if whenever λ ∈ [0, 1] and x, y ∈ K, λx + (1 − λ)y ∈ K.

15.1. BASIC THEORY

423

Theorem 15.1.8 Let K be a closed convex nonempty subset of a Hilbert space, H, and let x ∈ H. Then there exists a unique point P x ∈ K such that ||P x − x|| ≤ ||y − x|| for all y ∈ K. Proof: Consider uniqueness. Suppose that z1 and z2 are two elements of K such that for i = 1, 2, ||zi − x|| ≤ ||y − x|| (15.1.5) for all y ∈ K. Also, note that since K is convex, z1 + z2 ∈ K. 2 Therefore, by the parallelogram identity, z1 − x z2 − x 2 z1 + z2 − x||2 = || + || 2 2 2 z1 − x 2 z2 − x 2 z1 − z2 2 = 2(|| || + || || ) − || || 2 2 2 1 1 z1 − z2 2 2 2 = ||z1 − x|| + ||z2 − x|| − || || 2 2 2 z1 − z2 2 ≤ ||z1 − x||2 − || || , 2

||z1 − x||2

≤ ||

where the last inequality holds because of 15.1.5 letting zi = z2 and y = z1 . Hence z1 = z2 and this shows uniqueness. Now let λ = inf{||x − y|| : y ∈ K} and let yn be a minimizing sequence. This means {yn } ⊆ K satisfies limn→∞ ||x − yn || = λ. Now the following follows from properties of the norm. 2

||yn − x + ym − x|| = 4(||

yn + ym − x||2 ) 2

Then by the parallelogram identity, and convexity of K,

yn +ym 2

z

2

|| (yn − x) − (ym − x) ||

∈ K, and so =||yn −x+ym −x||2

}| { yn + ym 2 = 2(||yn − x|| + ||ym − x|| ) − 4(|| − x|| ) 2 ≤ 2(||yn − x||2 + ||ym − x||2 ) − 4λ2. 2

2

Since ||x − yn || → λ, this shows {yn − x} is a Cauchy sequence. Thus also {yn } is a Cauchy sequence. Since H is complete, yn → y for some y ∈ H which must be in K because K is closed. Therefore ||x − y|| = lim ||x − yn || = λ. n→∞

Let P x = y.

424

HILBERT SPACES

Corollary 15.1.9 Let K be a closed, convex, nonempty subset of a Hilbert space, H, and let x ∈ H. Then for z ∈ K, z = P x if and only if Re(x − z, y − z) ≤ 0

(15.1.6)

for all y ∈ K. Before proving this, consider what it says in the case where the Hilbert space is Rn. yX yXXrθ z

K

- x

Condition 15.1.6 says the angle, θ, shown in the diagram is always obtuse. Remember from calculus, the sign of x · y is the same as the sign of the cosine of the included angle between x and y. Thus, in finite dimensions, the conclusion of this corollary says that z = P x exactly when the angle of the indicated angle is obtuse. Surely the picture suggests this is reasonable. The inequality 15.1.6 is an example of a variational inequality and this corollary characterizes the projection of x onto K as the solution of this variational inequality. Proof of Corollary: Let z ∈ K and let y ∈ K also. Since K is convex, it follows that if t ∈ [0, 1], z + t(y − z) = (1 − t) z + ty ∈ K. Furthermore, every point of K can be written in this way. (Let t = 1 and y ∈ K.) Therefore, z = P x if and only if for all y ∈ K and t ∈ [0, 1], ||x − (z + t(y − z))||2 = ||(x − z) − t(y − z)||2 ≥ ||x − z||2 for all t ∈ [0, 1] and y ∈ K if and only if for all t ∈ [0, 1] and y ∈ K 2

2

2

||x − z|| + t2 ||y − z|| − 2t Re (x − z, y − z) ≥ ||x − z|| If and only if for all t ∈ [0, 1], 2

t2 ||y − z|| − 2t Re (x − z, y − z) ≥ 0.

(15.1.7)

Now this is equivalent to 15.1.7 holding for all t ∈ (0, 1). Therefore, dividing by t ∈ (0, 1) , 15.1.7 is equivalent to 2

t ||y − z|| − 2 Re (x − z, y − z) ≥ 0 for all t ∈ (0, 1) which is equivalent to 15.1.6. This proves the corollary. Corollary 15.1.10 Let K be a nonempty convex closed subset of a Hilbert space, H. Then the projection map, P is continuous. In fact, |P x − P y| ≤ |x − y| .

15.1. BASIC THEORY

425

Proof: Let x, x0 ∈ H. Then by Corollary 15.1.9, Re (x0 − P x0 , P x − P x0 ) ≤ 0, Re (x − P x, P x0 − P x) ≤ 0 Hence 0

≤ Re (x − P x, P x − P x0 ) − Re (x0 − P x0 , P x − P x0 ) = Re (x − x0 , P x − P x0 ) − |P x − P x0 |

and so

2

2

|P x − P x0 | ≤ |x − x0 | |P x − P x0 | . This proves the corollary. The next corollary is a more general form for the Brouwer fixed point theorem. Corollary 15.1.11 Let f : K → K where K is a convex compact subset of Rn . Then f has a fixed point. Proof: Let K ⊆ B (0, R) and let P be the projection map onto K. Then consider the map f ◦ P which maps B (0, R) to B (0, R) and is continuous. By the Brouwer fixed point theorem for balls, this map has a fixed point. Thus there exists x such that f ◦ P (x) = x Now the equation also requires x ∈ K and so P (x) = x. Hence f (x) = x. Definition 15.1.12 Let H be a vector space and let U and V be subspaces. U ⊕V = H if every element of H can be written as a sum of an element of U and an element of V in a unique way. The case where the closed convex set is a closed subspace is of special importance and in this case the above corollary implies the following. Corollary 15.1.13 Let K be a closed subspace of a Hilbert space, H, and let x ∈ H. Then for z ∈ K, z = P x if and only if (x − z, y) = 0

(15.1.8)

for all y ∈ K. Furthermore, H = K ⊕ K ⊥ where K ⊥ ≡ {x ∈ H : (x, k) = 0 for all k ∈ K} and

2

2

2

||x|| = ||x − P x|| + ||P x|| .

(15.1.9)

Proof: Since K is a subspace, the condition 15.1.6 implies Re(x − z, y) ≤ 0 for all y ∈ K. Replacing y with −y, it follows Re(x − z, −y) ≤ 0 which implies Re(x − z, y) ≥ 0 for all y. Therefore, Re(x − z, y) = 0 for all y ∈ K. Now let

426

HILBERT SPACES

|α| = 1 and α (x − z, y) = |(x − z, y)|. Since K is a subspace, it follows αy ∈ K for all y ∈ K. Therefore, 0 = Re(x − z, αy) = (x − z, αy) = α (x − z, y) = |(x − z, y)|. This shows that z = P x, if and only if 15.1.8. For x ∈ H, x = x − P x + P x and from what was just shown, x − P x ∈ K ⊥ and P x ∈ K. This shows that K ⊥ + K = H. Is there only one way to write a given element of H as a sum of a vector in K with a vector in K ⊥ ? Suppose y + z = y1 + z1 where z, z1 ∈ K ⊥ and y, y1 ∈ K. Then (y − y1 ) = (z1 − z) and so from what was just shown, (y − y1 , y − y1 ) = (y − y1 , z1 − z) = 0 which shows y1 = y and consequently z1 = z. Finally, letting z = P x, ||x||

2

2

2

= (x − z + z, x − z + z) = ||x − z|| + (x − z, z) + (z, x − z) + ||z|| 2

2

= ||x − z|| + ||z||

This proves the corollary. The following theorem is called the Riesz representation theorem for the dual of a Hilbert space. If z ∈ H then define an element f ∈ H 0 by the rule (x, z) ≡ f (x). It follows from the Cauchy Schwarz inequality and the properties of the inner product that f ∈ H 0 . The Riesz representation theorem says that all elements of H 0 are of this form. Theorem 15.1.14 Let H be a Hilbert space and let f ∈ H 0 . Then there exists a unique z ∈ H such that f (x) = (x, z) (15.1.10) for all x ∈ H. Proof: Letting y, w ∈ H the assumption that f is linear implies f (yf (w) − f (y)w) = f (w) f (y) − f (y) f (w) = 0 which shows that yf (w) − f (y)w ∈ f −1 (0), which is a closed subspace of H since f is continuous. If f −1 (0) = H, then f is the zero map and z = 0 is the unique element of H which satisfies 15.1.10. If f −1 (0) 6= H, pick u ∈ / f −1 (0) and let w ≡ u − P u 6= 0. Thus Corollary 15.1.13 implies (y, w) = 0 for all y ∈ f −1 (0). In particular, let y = xf (w) − f (x)w where x ∈ H is arbitrary. Therefore, 0 = (f (w)x − f (x)w, w) = f (w)(x, w) − f (x)||w||2. Thus, solving for f (x) and using the properties of the inner product, f (x) = (x,

f (w)w ) ||w||2

Let z = f (w)w/||w||2 . This proves the existence of z. If f (x) = (x, zi ) i = 1, 2, for all x ∈ H, then for all x ∈ H, then (x, z1 − z2 ) = 0 which implies, upon taking x = z1 − z2 that z1 = z2 . This proves the theorem. If R : H → H 0 is defined by Rx (y) ≡ (y, x) , the Riesz representation theorem above states this map is onto. This map is called the Riesz map. It is routine to show R is linear and |Rx| = |x|.

15.2. THE HILBERT SPACE L (U )

15.2

427

The Hilbert Space L (U )

Let L ∈ L (U, H) . Then one can consider the image of L, L (U ) as a Hilbert space. This is another interesting application of Theorem 15.1.8. First here is a definition which involves abominable and atrociously misleading notation which nevertheless seems to be well accepted. Definition 15.2.1 Let L ∈ L (U, H), the bounded linear maps from U to H for U, H Hilbert spaces. For y ∈ L (U ) , let L−1 y denote the unique vector in {x : Lx = y} ≡ My which is closest in U to 0.

L−1 (y) r¡ ¡@ ¡ ¡

¡ ¡ I ¡ @ ¡ {x : Lx = y}

Note this is a good definition because {x : Lx = y} is closed thanks to the continuity of L and it is obviously convex. Thus Theorem 15.1.8 applies. With this definition define an inner product on L (U ) as follows. For y, z ∈ L (U ) , ¡ ¢ (y, z)L(U ) ≡ L−1 y, L−1 z U The notation is abominable because L−1 (y) is the normal notation for My . With the above definition, here is the main result. Theorem 15.2.2 Let U, H be Hilbert spaces and let L ∈ L (U, H) . Then Definition 15.2.1 makes L (U ) into a Hilbert space. Furthermore there is a constant C independent of x ∈ U such that C ||Lx||L(U ) ≥ ||Lx||H ¡ ¢ If U is separable, so is L (U ). Also L−1 (y) , x = 0 for all x ∈ ker (L) . Proof: First it is necessary to verify this is an inner product. It is clear (y, z)L(U ) = (z, y)L(U ) . What about the property of being linear in the first argument? Is (ay1 + by2 , z)L(U ) = a (y1 , z)L(U ) + b (y2 , z)L(U ) ?

(15.2.11)

By Corollary 15.1.9 ¡ ¢ Re 0 − L−1 (y) , z − L−1 (y) U ≤ 0

(15.2.12)

428

HILBERT SPACES

for all z ∈ L−1 (y). Also it is clear that My = L−1 (y) + ker (L) (This is because if z ∈ My so that Lz = y, then ¡ ¢ L L−1 (y) − z = 0 and so L−1 (y) − z ∈ ker (L).) Thus from 15.2.12, ¡ ¢ Re 0 − L−1 (y) , x U ≤ 0 for all x ∈ ker (L) and so in fact ¡ −1 ¢ L (y) , x U = 0 for all x ∈ ker (L).Therefore, returning to 15.2.11 ¡ ¢ (ay1 + by2 , z)L(U ) ≡ L−1 (ay1 + by2 ) , L−1 z U Also

¡ ¢ L−1 (ay1 + by2 ) − aL−1 (y1 ) + bL−1 (y2 ) ∈ ker (L)

(15.2.13)

(15.2.14) (15.2.15)

and so the right side of 15.2.14 equals =0

z¡ }| ¡ ¢ ¢{ L−1 (ay1 + by2 ) − aL−1 (y1 ) + bL−1 (y2 ) , L−1 z U +

¡¡

¢ ¢ aL−1 (y1 ) + bL−1 (y2 ) , L−1 z U

which by 15.2.15 and 15.2.13 equals ¡¡ −1 ¢ ¢ aL (y1 ) + bL−1 (y2 ) , L−1 z U ¡ ¢ ¡ ¢ ≡ a L−1 (y1 ) , L−1 (z) + b L−1 (y2 ) , L−1 (z) which shows the proposed inner product is linear in the first argument. If (y, y)L(U ) = 0, does it follow y = 0? This is easy because the inner product is defined as ¡ ¢ (y, y)L(U ) = L−1 y, L−1 y U = 0 and so L−1 y = 0 which requires y = 0. Why is L (U ) a Hilbert© space? Let ª {yn } be a Cauchy sequence in L (U ) . Why does it converge? I claim L−1 (yn ) is a Cauchy sequence in U. ¡ −1 ¢ L (yn ) − L−1 (ym ) , L−1 (yn ) − L−1 (ym ) U ¡ ¡ ¢ ¢ = L−1 (yn ) − L−1 (ym ) , L−1 (yn ) − L−1 (ym ) − L−1 (yn − ym ) U ¡ ¢ + L−1 (yn ) − L−1 (ym ) , L−1 (yn − ym ) U

(15.2.16)


429

¡ ¢ Since L−1 (yn ) − L−1 (ym ) − L−1 (yn − ym ) ∈ ker (L) , it follows from 15.2.13 this term in 15.2.16 equals 0. Then also ¡ −1 ¢ ¡ ¢ L (yn ) − L−1 (ym ) , L−1 (yn − ym ) U = L−1 (yn − ym ) , L−1 (yn − ym ) U ¡ ¢ + L−1 (yn ) − L−1 (ym ) − L−1 (yn − ym ) , L−1 (yn − ym ) U

(15.2.17)

and by 15.2.13 again this last term equals 0. Therefore, ¡ −1 ¢ L (yn ) − L−1 (ym ) , L−1 (yn ) − L−1 (ym ) U ¡ ¢ = L−1 (yn − ym ) , L−1 (yn − ym ) U 2

≡ ||yn − ym ||L(U )

(15.2.18)

© ª and it follows L−1 (yn ) is a Cauchy sequence as claimed. Therefore, there exists z ∈ U such that L−1 (yn ) → z, (15.2.19) the convergence in U and therefore, by continuity of L, yn → Lz ∈ L (U ) , the convergence taking place in H. Does yn also converge to Lz in the inner product of L (U )? By the same reasoning which led to 15.2.18 ¡ ¢ 2 ||yn − Lz||L(U ) ≡ L−1 (yn − L (z)) , L−1 (yn − L (z)) U = =

¡ −1 ¢ L (yn ) − L−1 (L (z)) , L−1 (yn ) − L−1 (L (z)) U ¡ −1 ¢ L (yn ) − z, L−1 (yn ) − z U

and this last expression converges to 0 by 15.2.19. Also assuming L 6= 0, ¡ ¢ 2 2 ||Lx||L(U ) ≡ L−1 (Lx) , L−1 (Lx) U = ||x||U ≥

1

2

2

||L||

||Lx||H

and this proves the claim about the norm. If L = 0 there is nothing to prove in this case. Next suppose U is separable. Why is L (U ) separable? Let {xn } be a countable dense subset of U . I claim {Lxn } is a countable dense subset in L (U ) . This is because ¡ ¢ 2 ||Lx − Lxn ||L(U ) ≡ L−1 (Lx − Lxn ) , L−1 (Lx − Lxn ) U and by 15.2.13, this equals ¡ −1 ¢ L (Lx − Lxn ) − (x − xn ) , L−1 (Lx − Lxn ) U ¡ ¢ + x − xn , L−1 (Lx − Lxn )

430

HILBERT SPACES

¡ ¢ = x − xn , L−1 (Lx − Lxn ) ¡ ¢ = x − xn , L−1 (Lx − Lxn ) − (x − xn ) + (x − xn , x − xn )U Thus the density of {xn } implies the density of {Lxn } in L (U ) . This proves the theorem. Note what it really says is that the norm in L (U ) corresponds to the norm in U. Now here are some other very interesting results. I am following [55]. ³ ´ Lemma 15.2.3 Let L ∈ L (U, H) . Then L B (0, r) is closed and convex. Proof: It is clear this is convex since L is linear. Why is it closed? B (0, r) is compact in the weak topology by the Banach Alaoglu theorem, Theorem 13.4.4 on Page 383. Furthermore, ³ ´ L is continuous with respect to the weak topologies on U and H so L B (0, r) is weakly compact because it is the continuous image of a compact set. Therefore, it must also be weakly closed because the weak topology Hausdorff because in this topology which comes from seminorms each point is a closed set thanks to Lemma 14.3.2 on Page 406 and so you can apply the separation theorem, Theorem 14.2.4 on Page 400 to obtain a separating functional. Thus if x 6= y, there exists f ∈ H 0 such that Re f (y) > c > Re f (x) and so taking 2r < min (c − Re f (x) , Re f (y) − c) , Bf (x, r) ∩ Bf (y, r) = ∅ where Bf (x, r) ≡ {y ∈ H : |f (x − y)| < r} is an example of a basic ³open setín the weak topology. Now suppose p ∈ / L B (0, r) . Since the set is weakly closed and convex, it follows by Theorem 14.2.4 and the Riesz representation theorem for Hilbert space there exists z ∈ H such that Re (p, z) > c > Re (Lx, z) for all x ∈ B (0, r). Therefore, p cannot be a strong limit point because if it were, there would exist xn ∈ B (0, r) such that Lxn → p which would require Re (Lxn , z) → Re (p, z) which is prevented by the above inequality. This proves the lemma. Now here is a very interesting result about showing that T1 (U1 ) = T2 (U2 ) where Ui is a Hilbert space and Ti ∈ L (Ui , H).


431

Theorem 15.2.4 Let Ui , i = 1, 2 and H be Hilbert spaces and let Ti ∈ L (Ui , H). If there exists c ≥ 0 such that for all x ∈ H

then

||T1∗ x||1 ≤ c ||T2∗ x||2

(15.2.20)

³ ´ ³ ´ T1 B (0, 1) ⊆ T2 B (0, c)

(15.2.21)

and so T1 (U1 ) ⊆ T2 (U2 ). If ||T1∗ x||1 = ||T ∗ x||2 for all x ∈ H, then T1 (U1 ) = T2 (U2 ) and in addition to this, ¯¯ −1 ¯¯ ¯¯ ¯¯ ¯¯T x¯¯ = ¯¯T −1 x¯¯ (15.2.22) 1 2 1 2 for all x ∈ T1 (U1 ) = T2 (U2 ). In this theorem, Ti−1 refers to Definition 15.2.1. Proof: Consider the first claim. If it is not so, then there exists u0 , ||u0 ||1 ≤ 1 but ³ ´ T1 (u0 ) ∈ / T2 B (0, c) the latter set being a closed convex nonempty set thanks to Lemma 15.2.3. Then by the separation theorem, Theorem 14.2.4 there exists z ∈ H such that Re (T1 (u0 ) , z)H > 1 > Re (T2 (v) , z)H for all ||v||2 ≤ c. Therefore, replacing v with vθ where θ is a suitable complex number having modulus 1, it follows ¯ ¯ ||T1∗ z|| > 1 > ¯(v, T2∗ z)U2 ¯ (15.2.23) for all ||v||2 ≤ c. If c = 0 this gives a contradiction immediately because of 15.2.20. Assume then that c > 0. Then from 15.2.23, ¯ ¯ 1 1 ||T2∗ z||U2 = sup ¯(v, T2∗ z)U2 ¯ ≤ < ||T1∗ z|| c c ||v||≤1 which contradicts 15.2.20. Therefore, it is clear T1 (U1 ) ⊆ T2 (U2 ). Now consider the second claim. The first part shows T1 (U1 ) = T2 (U2 ). Denote by ui ∈ Ui , the point Ti−1 x. Without loss of generality, it can be assumed x 6= 0 because if x = 0, then the definition of Ti−1 gives Ti−1 (x) = 0. Thus for x 6= 0 neither ui can equal 0. I need to verify that ||u1 ||1 = ||u2 ||2 . Suppose then that this is not so. Say ||u1 ||1 > ||u2 ||2 > 0. µ ¶ ³ ´ u2 x = T2 ∈ T2 B (0, 1) ||u2 ||2 ||u2 ||2 ³ ´ But from the first part of the theorem this equals T1 B (0, 1) and so there exists u01 ∈ B (0, 1) such that

x = T1 u01 ||u2 ||2

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HILBERT SPACES

Hence

µ T1

u01

u1 − ||u2 ||2

¶ =

x x − = 0. ||u2 ||2 ||u2 ||2

From Theorem 15.2.2 this implies µ ¶ ||u1 ||1 u1 0 0 = u1 , u1 − ≤ ||u1 ||1 ||u01 ||1 − ||u1 ||1 ||u2 ||2 ||u2 ||2 ¶ µ ¶ µ ||u ||u || 1 ||1 1 1 0 ≤ ||u1 ||1 1 − = ||u1 ||1 ||u1 ||1 − ||u2 ||2 ||u2 ||2 which is a contradiction because it was assumed

15.3

||u1 ||1 ||u2 ||2

> 1. This proves the theorem.

Approximations In Hilbert Space

The Gram Schmidt process applies in any Hilbert space. Theorem 15.3.1 Let {x1 , · · · , xn } be a basis for M a subspace of H a Hilbert space. Then there exists an orthonormal basis for M, {u1 , · · · , un } which has the property that for each k ≤ n, span(x1 , · · · , xk ) = span (u1 , · · · , uk ) . Also if {x1 , · · · , xn } ⊆ H, then span (x1 , · · · , xn ) is a closed subspace. Proof: Let {x1 , · · · , xn } be a basis for M. Let u1 ≡ x1 / |x1 | . Thus for k = 1, span (u1 ) = span (x1 ) and {u1 } is an orthonormal set. Now suppose for some k < n, u1 , · · · , uk have been chosen such that (uj · ul ) = δ jl and span (x1 , · · · , xk ) = span (u1 , · · · , uk ). Then define xk+1 −

Pk

j=1 (xk+1 · uj ) uj ¯, uk+1 ≡ ¯¯ Pk ¯ ¯xk+1 − j=1 (xk+1 · uj ) uj ¯

(15.3.24)

where the denominator is not equal to zero because the xj form a basis and so xk+1 ∈ / span (x1 , · · · , xk ) = span (u1 , · · · , uk ) Thus by induction, uk+1 ∈ span (u1 , · · · , uk , xk+1 ) = span (x1 , · · · , xk , xk+1 ) . Also, xk+1 ∈ span (u1 , · · · , uk , uk+1 ) which is seen easily by solving 15.3.24 for xk+1 and it follows span (x1 , · · · , xk , xk+1 ) = span (u1 , · · · , uk , uk+1 ) .

15.3. APPROXIMATIONS IN HILBERT SPACE

433

If l ≤ k,  (uk+1 · ul ) =

C (xk+1 · ul ) −

 (xk+1 · uj ) (uj · ul )

j=1

 =

k X

C (xk+1 · ul ) −

k X

 (xk+1 · uj ) δ lj 

j=1

=

C ((xk+1 · ul ) − (xk+1 · ul )) = 0.

n

The vectors, {uj }j=1 , generated in this way are therefore an orthonormal basis because each vector has unit length. Consider the second claim about finite dimensional subspaces. Without loss of generality, assume {x1 , · · · , xn } is linearly independent. If it is not, delete vectors until a linearly independent set is obtained. Then by the first part, span (x1 , · · · , xn ) = span (u1 , · · · , un ) ≡ M where the ui are an orthonormal set of vectors. Suppose {yk } ⊆ M and yk → y ∈ H. Is y ∈ M ? Let yk ≡

n X

ckj uj

j=1

¢T ¡ Then let ck ≡ ck1 , · · · , ckn . Then ¯ k ¯ ¯c − cl ¯2

  n n n X X X ¯ k ¯ ¡ ¢ ¡ ¢ 2 ¯cj − clj ¯ =  ≡ ckj − clj uj , ckj − clj uj  j=1

j=1

j=1

2

= ||yk − yl || ª which shows ck is a Cauchy sequence in Fn and so it converges to c ∈ Fn . Thus ©

y = lim yk = lim k→∞

n X

k→∞

ckj uj =

j=1

n X

cj uj ∈ M.

j=1

This completes the proof. Theorem 15.3.2 Let M be the span of {u1 , · · · , un } in a Hilbert space, H and let y ∈ H. Then P y is given by Py =

n X

(y, uk ) uk

(15.3.25)

k=1

and the distance is given by v u n u 2 X 2 t|y| − |(y, uk )| . k=1

(15.3.26)

434

HILBERT SPACES

Proof: Ã y−

n X

! (y, uk ) uk , up

=

(y, up ) −

n X

(y, uk ) (uk , up )

k=1

k=1

= It follows that

Ã y−

n X

(y, up ) − (y, up ) = 0 !

(y, uk ) uk , u

=0

k=1

for all u ∈ M and so by Corollary 15.1.13 this verifies 15.3.25. The square of the distance, d is given by Ã ! n n X X 2 d = y− (y, uk ) uk , y − (y, uk ) uk =

2

k=1 n X

|y| − 2

k=1

k=1 2

|(y, uk )| +

n X

|(y, uk )|

2

k=1

and this shows 15.3.26. What if the subspace is the span of vectors which are not orthonormal? There is a very interesting formula for the distance between a point of a Hilbert space and a finite dimensional subspace spanned by an arbitrary basis. Definition 15.3.3 Let {x1 , · · · , xn } ⊆ H, a Hilbert space. Define   (x1 , x1 ) · · · (x1 , xn )   .. .. G (x1 , · · · , xn ) ≡   . . (xn , x1 ) · · · (xn , xn )

(15.3.27)

Thus the ij th entry of this matrix is (xi , xj ). This is sometimes called the Gram matrix. Also define G (x1 , · · · , xn ) as the determinant of this matrix, also called the Gram determinant. ¯ ¯ ¯ (x1 , x1 ) · · · (x1 , xn ) ¯ ¯ ¯ ¯ ¯ .. .. G (x1 , · · · , xn ) ≡ ¯ (15.3.28) ¯ . . ¯ ¯ ¯ (xn , x1 ) · · · (xn , xn ) ¯ The theorem is the following. Theorem 15.3.4 Let M = span (x1 , · · · , xn ) ⊆ H, a Real Hilbert space where {x1 , · · · , xn } is a basis and let y ∈ H. Then letting d be the distance from y to M, d2 =

G (x1 , · · · , xn , y) . G (x1 , · · · , xn )

(15.3.29)

15.3. APPROXIMATIONS IN HILBERT SPACE

435

Proof: By Theorem 15.3.1 M is a closed subspace of H. Let element of M which is closest to y. Then by Corollary 15.1.13, Ã ! n X y− αk xk , xp = 0

Pn k=1

αk xk be the

k=1

for each p = 1, 2, · · · , n. This yields the system of equations, (y, xp ) =

n X

(xp , xk ) αk , p = 1, 2, · · · , n

(15.3.30)

k=1

Also by Corollary 15.1.13, d2

z }| { ¯¯2 ¯¯ ¯¯2 ¯¯ n n ¯¯ X ¯¯ ¯¯ ¯¯ X ¯¯ ¯¯ ¯¯ ¯¯ 2 αk xk ¯¯ ||y|| = ¯¯y − αk xk ¯¯ + ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ k=1

k=1

and so, using 15.3.30, 2

||y||

=

2

d +

Ã X X j

=

d2 +

X

! αk (xk , xj ) αj

k

(y, xj ) αj

(15.3.31)

j

≡ in which

d2 + yxT α

yxT ≡ ((y, x1 ) , · · · , (y, xn )) , αT ≡ (α1 , · · · , αn ) .

Then 15.3.30 and 15.3.31 imply the following system ¶µ ¶ µ ¶ µ yx G (x1 , · · · , xn ) 0 α = 2 1 d2 yxT ||y|| By Cramer’s rule, µ

d2

=

= =

¶ G (x1 , · · · , xn ) yx det 2 yxT ||y|| ¶ µ G (x1 , · · · , xn ) 0 det yxT 1 µ ¶ G (x1 , · · · , xn ) yx det 2 yxT ||y|| det (G (x1 , · · · , xn )) det (G (x1 , · · · , xn , y)) G (x1 , · · · , xn , y) = det (G (x1 , · · · , xn )) G (x1 , · · · , xn )

and this proves the theorem.

(15.3.32)

436

15.4

HILBERT SPACES

The M¨ untz Theorem

Recall the polynomials are dense in C ([0, 1]) . This is a consequence of the Weierstrass approximation theorem. Now consider finite linear combinations of the functions, tpk where {p0 , p1 , p2 , · · · } is a sequence of nonnegative real numbers, p0 ≡ 0. The M¨ untz theorem P∞ says this set, S of finite linear combinations is dense in C ([0, 1]) exactly when k=1 p1k = ∞. There are two versions of this theorem, one for density of S in L2 (0, 1) and one for C ([0, 1]) . The presentation follows Cheney [17]. Recall the Cauchy identity presented earlier, Theorem 4.8.1 on Page 78 which is stated here for convenience. Theorem 15.4.1 The following identity holds. ¯ ¯ 1 1 ¯ ¯ a +b · · · a1 +b 1 1 n ¯ Y ¯ Y ¯ ¯ . . .. .. (ai − aj ) (bi − bj ) . (ai + bj ) ¯ ¯= ¯ ¯ 1 1 j ε > 0. 5. Give an example of a sequence of functions in L∞ (−π, π) which converges weak ∗ to zero but which does not converge pointwise a.e. to zero.

524

REPRESENTATION THEOREMS

Integrals And Derivatives 17.1

The Fundamental Theorem Of Calculus

The version of the fundamental theorem of calculus found in Calculus has already been referred to frequently. It says that if f is a Riemann integrable function, the function Z x x→

f (t) dt, a

has a derivative at every point where f is continuous. It is natural to ask what occurs for f in L1 . It is an amazing fact that the same result is obtained asside from a set of measure zero even though f , being only in L1 may fail to be continuous anywhere. Proofs of this result are based on some form of the Vitali covering theorem presented above. In what follows, the measure space is (Rn , S, m) where m is n-dimensional Lebesgue measure although the same theorems can be proved for arbitrary Radon measures [46]. To save notation, m is written in place of mn . By Lemma 9.1.7 on Page 224 and the completeness of m, the Lebesgue measurable sets are exactly those measurable in the sense of Caratheodory. Also, to save on notation m is also the name of the outer measure defined on all of P(Rn ) which is determined by mn . Recall B(p, r) = {x : |x − p| < r}.

(17.1.1)

b = B(p, 5r). If B = B(p, r), then B

(17.1.2)

Also define the following.

The first version of the Vitali covering theorem presented above will now be used to establish the fundamental theorem of calculus. The space of locally integrable functions is the most general one for which the maximal function defined below makes sense. Definition 17.1.1 f ∈ L1loc (Rn ) means f XB(0,R) ∈ L1 (Rn ) for all R > 0. For f ∈ L1loc (Rn ), the Hardy Littlewood Maximal Function, M f , is defined by Z 1 |f (y)|dy. M f (x) ≡ sup r>0 m(B(x, r)) B(x,r) 525

526

INTEGRALS AND DERIVATIVES

Theorem 17.1.2 If f ∈ L1 (Rn ), then for α > 0, m([M f > α]) ≤

5n ||f ||1 . α

(Here and elsewhere, [M f > α] ≡ {x ∈ Rn : M f (x) > α} with other occurrences of [ ] being defined similarly.) Proof: Let S ≡ [M f > α]. For x ∈ S, choose rx > 0 with Z 1 |f | dm > α. m(B(x, rx )) B(x,rx ) The rx are all bounded because 1 m(B(x, rx )) < α

Z |f | dm < B(x,rx )

1 ||f ||1 . α

By the Vitali covering theorem, there are disjoint balls B(xi , ri ) such that S ⊆ ∪∞ i=1 B(xi , 5ri ) and

1 m(B(xi , ri ))

Z |f | dm > α. B(xi ,ri )

Therefore m(S) ≤

∞ X i=1

≤ ≤

m(B(xi , 5ri )) = 5n

∞ Z 5n X |f | dm α i=1 B(xi ,ri ) Z 5n |f | dm, α Rn

∞ X

m(B(xi , ri ))

i=1

the last inequality being valid because the balls B(xi , ri ) are disjoint. This proves the theorem. Note that at this point it is unknown whether S is measurable. This is why m(S) and not m (S) is written. The following is the fundamental theorem of calculus from elementary calculus. Lemma 17.1.3 Suppose g is a continuous function. Then for all x, Z 1 g(y)dy = g(x). lim r→0 m(B(x, r)) B(x,r)

17.1. THE FUNDAMENTAL THEOREM OF CALCULUS Proof: Note that g (x) = and so

1 m(B(x, r))

527

Z g (x) dy B(x,r)

¯ ¯ Z ¯ ¯ 1 ¯ ¯ g(y)dy ¯ ¯g (x) − ¯ ¯ m(B(x, r)) B(x,r) = ≤

¯ ¯ Z ¯ ¯ 1 ¯ ¯ (g(y) − g (x)) dy ¯ ¯ ¯ m(B(x, r)) B(x,r) ¯ Z 1 |g(y) − g (x)| dy. m(B(x, r)) B(x,r)

Now by continuity of g at x, there exists r > 0 such that if |x − y| < r, |g (y) − g (x)| < ε. For such r, the last expression is less than Z 1 εdy < ε. m(B(x, r)) B(x,r) This proves the lemma. ¡ ¢ Definition 17.1.4 Let f ∈ L1 Rk , m . A point, x ∈ Rk is said to be a Lebesgue point if Z 1 lim sup |f (y) − f (x)| dm = 0. r→0 m (B (x, r)) B(x,r) Note that if x is a Lebesgue point, then Z 1 lim f (y) dm = f (x) . r→0 m (B (x, r)) B(x,r) and so the symmetric derivative exists at all Lebesgue points. Theorem 17.1.5 (Fundamental Theorem of Calculus) Let f ∈ L1 (Rk ). Then there exists a set of measure 0, N , such that if x ∈ / N , then Z 1 lim |f (y) − f (x)|dy = 0. r→0 m(B(x, r)) B(x,r) ¡ k¢ ¡ k ¢ 1 Proof: ¡ k ¢Let λ > 0 and let ε > 0. By density of Cc R in L R , m there exists g ∈ Cc R such that ||g − f ||L1 (Rk ) < ε. Now since g is continuous, Z 1 lim sup |f (y) − f (x)| dm r→0 m (B (x, r)) B(x,r) Z 1 |f (y) − f (x)| dm = lim sup r→0 m (B (x, r)) B(x,r) Z 1 − lim |g (y) − g (x)| dm r→0 m (B (x, r)) B(x,r)

528


Ã = lim sup r→0

≤

lim sup r→0

≤

lim sup r→0

≤

Ã

lim sup r→0

≤

Ã

Ã

1 m (B (x, r)) 1 m (B (x, r)) 1 m (B (x, r)) 1 m (B (x, r))

!

Z |f (y) − f (x)| − |g (y) − g (x)| dm B(x,r)

!

Z ||f (y) − f (x)| − |g (y) − g (x)|| dm B(x,r)

!

Z |f (y) − g (y) − (f (x) − g (x))| dm B(x,r)

!

Z

|f (y) − g (y)| dm

+ |f (x) − g (x)|

B(x,r)

M ([f − g]) (x) + |f (x) − g (x)| .

Therefore, "

⊆

# Z 1 x : lim sup |f (y) − f (x)| dm > λ r→0 m (B (x, r)) B(x,r) ¸ · ¸ · λ λ ∪ |f − g| > M ([f − g]) > 2 2

Now

Z ε > ≥

Z |f − g| dm ≥

[|f −g|> λ2 ] µ· ¸¶ λ λ m |f − g| > 2 2

|f − g| dm

This along with the weak estimate of Theorem 17.1.2 implies Ã" #! Z 1 m x : lim sup |f (y) − f (x)| dm > λ r→0 m (B (x, r)) B(x,r) µ ¶ 2 k 2 < 5 + ||f − g||L1 (Rk ) λ λ µ ¶ 2 k 2 < 5 + ε. λ λ Since ε > 0 is arbitrary, it follows Ã" #! Z 1 mn x : lim sup |f (y) − f (x)| dm > λ = 0. r→0 m (B (x, r)) B(x,r) Now let "

1 N = x : lim sup r→0 m (B (x, r))

#

Z |f (y) − f (x)| dm > 0 B(x,r)

17.1. THE FUNDAMENTAL THEOREM OF CALCULUS and

"

1 Nn = x : lim sup r→0 m (B (x, r))

Z

1 |f (y) − f (x)| dm > n B(x,r)

529 #

It was just shown that m (Nn ) = 0. Also, N = ∪∞ n=1 Nn . Therefore, m (N ) = 0 also. It follows that for x ∈ / N, Z 1 |f (y) − f (x)| dm = 0 lim sup r→0 m (B (x, r)) B(x,r) and this proves a.e. point is a Lebesgue point. ¡ ¢ Of course it is sufficient to assume f is only in L1loc Rk . Corollary 17.1.6 (Fundamental Theorem of Calculus) Let f ∈ L1loc (Rk ). Then there exists a set of measure 0, N , such that if x ∈ / N , then Z 1 lim |f (y) − f (x)|dy = 0. r→0 m(B(x, r)) B(x,r) Consider B (0, n) where n is a positive integer. Then fn ≡ f XB(0,n) ∈ ¡Proof: ¢ L1 Rk and so there exists a set of measure 0, Nn such that if x ∈ B (0, n) \ Nn , then Z 1 lim |fn (y) − fn (x)|dy r→0 m(B(x, r)) B(x,r) Z 1 = lim |f (y) − f (x)|dy = 0. r→0 m(B(x, r)) B(x,r) / N, the above equation holds. Let N = ∪∞ n=1 Nn . Then if x ∈ Corollary 17.1.7 If f ∈ L1loc (Rn ), then Z 1 lim f (y)dy = f (x) a.e. x. r→0 m(B(x, r)) B(x,r) Proof:

(17.1.3)

¯ ¯ Z ¯ ¯ 1 ¯ ¯ f (y)dy − f (x)¯ ¯ ¯ m(B(x, r)) B(x,r) ¯ Z 1 |f (y) − f (x)| dy ≤ m(B(x, r)) B(x,r)

and the last integral converges to 0 a.e. x. Definition 17.1.8 For N the set of Theorem 17.1.5 or Corollary 17.1.6, N C is called the Lebesgue set or the set of Lebesgue points. The next corollary is a one dimensional version of what was just presented.

530


Corollary 17.1.9 Let f ∈ L1 (R) and let Z x F (x) = f (t)dt. −∞

Then for a.e. x, F 0 (x) = f (x). Proof: For h > 0 Z Z x+h 1 1 x+h |f (y) − f (x)|dy ≤ 2( ) |f (y) − f (x)|dy h x 2h x−h By Theorem 17.1.5, this converges to 0 a.e. Similarly Z 1 x |f (y) − f (x)|dy h x−h converges to 0 a.e. x. ¯ ¯ Z ¯ F (x + h) − F (x) ¯ 1 x+h ¯ ¯ − f (x)¯ ≤ |f (y) − f (x)|dy ¯ h h x and

¯ ¯ Z ¯ F (x) − F (x − h) ¯ 1 x ¯ ¯ − f (x)¯ ≤ |f (y) − f (x)|dy. ¯ h h x−h

(17.1.4)

(17.1.5)

Now the expression on the right in 17.1.4 and 17.1.5 converges to zero for a.e. x. Therefore, by 17.1.4, for a.e. x the derivative from the right exists and equals f (x) while from 17.1.5 the derivative from the left exists and equals f (x) a.e. It follows F (x + h) − F (x) = f (x) a.e. x h→0 h lim


17.2

Absolutely Continuous Functions

Definition 17.2.1 Let [a, b] be a closed and bounded interval and let f : [a, b] → R. Then fP is said to be absolutely continuous if for every ε > 0 there exists δ > 0 such Pm m that if i=1 |yi − xi | < δ, then i=1 |f (yi ) − f (xi )| < ε. Definition 17.2.2 A finite subset, P of [a, b] is called a partition of [x, y] ⊆ [a, b] if P = {x0 , x1 , · · · , xn } where x = x0 < x1 < · · · , < xn = y. For f : [a, b] → R and P = {x0 , x1 , · · · , xn } define VP [x, y] ≡

n X i=1

|f (xi ) − f (xi−1 )| .

17.2. ABSOLUTELY CONTINUOUS FUNCTIONS

531

Denoting by P [x, y] the set of all partitions of [x, y] define V [x, y] ≡

sup

VP [x, y] .

P ∈P[x,y]

For simplicity, V [a, x] will be denoted by V (x) . It is called the total variation of the function, f. There are some simple facts about the total variation of an absolutely continuous function, f which are contained in the next lemma. Lemma 17.2.3 Let f be an absolutely continuous function defined on [a, b] and let V be its total variation function as described above. Then V is an increasing bounded function. Also if P and Q are two partitions of [x, y] with P ⊆ Q, then VP [x, y] ≤ VQ [x, y] and if [x, y] ⊆ [z, w] , V [x, y] ≤ V [z, w]

(17.2.6)

If P = {x0 , x1 , · · · , xn } is a partition of [x, y] , then V [x, y] =

n X

V [xi , xi−1 ] .

(17.2.7)

i=1

Also if y > x, V (y) − V (x) ≥ |f (y) − f (x)|

(17.2.8)

and the function, x → V (x) − f (x) is increasing. The total variation function, V is absolutely continuous. Proof: The claim that V is increasing is obvious as is the next claim about P ⊆ Q leading to VP [x, y] ≤ VQ [x, y] . To verify this, simply add in one point at a time and verify that from the triangle inequality, the sum involved gets no smaller. The claim that V is increasing consistent with set inclusion of intervals is also clearly true and follows directly from the definition. Now let t < V [x, y] where P0 = {x0 , x1 , · · · , xn } is a partition of [x, y] . There exists a partition, P of [x, y] such that t < VP [x, y] . Without loss of generality it can be assumed that {x0 , x1 , · · · , xn } ⊆ P since if not, you can simply add in the points of P0 and the resulting sum for the total variation will get no smaller. Let Pi be those points of P which are contained in [xi−1 , xi ] . Then t < Vp [x, y] =

n X

VPi [xi−1 , xi ] ≤

i=1

n X

V [xi−1 , xi ] .

i=1

Since t < V [x, y] is arbitrary, V [x, y] ≤

n X i=1

V [xi , xi−1 ]

(17.2.9)

532


Note that 17.2.9 does not depend on f being absolutely continuous. Suppose now that f is absolutely continuous. Let δ correspond to ε = 1. Then if [x, y] is an interval of length no larger than δ, the definition of absolute continuity implies V [x, y] < 1. Then from 17.2.9 V [a, nδ] ≤

n X

V [a + (i − 1) δ, a + iδ]
V [xi−1 , xi ] −

ε n

Then letting P = ∪Pi , −ε +

n X i=1

V [xi−1 , xi ]
0 be given and letP δ correspond to ε/2 in the definition Pn of absolute contin |y − x | < δ and consider nuity applied to f . Suppose i i i=1 |V (yi ) − V (xi )|. Pn i=1 By 17.2.9 this last equals i=1 V [xi , yi ] . Now let Pi be a partition of [xi , yi ] such ε that VPi [xi , yi ] + 2n > V [xi , yi ] . Then by the definition of absolute continuity, n X

|V (yi ) − V (xi )| =

i=1

≤

n X i=1 n X

V [xi , yi ] VPi [xi , yi ] + η < ε/2 + ε/2 = ε.

i=1

and shows V is absolutely continuous as claimed. Lemma 17.2.4 Suppose f : [a, b] → R is absolutely continuous and increasing. Then f 0 exists a.e., is in L1 ([a, b]) , and Z x f (x) = f (a) + f 0 (t) dt. a

17.2. ABSOLUTELY CONTINUOUS FUNCTIONS

533

Proof: Define L, a positive linear functional on C ([a, b]) by Z Lg ≡

b

gdf a

where this integral is the Riemann Stieltjes integral with respect to the integrating function, f. By the Riesz representation theorem for positive linear functionals, R there exists a unique Radon measure, µ such that Lg = gdµ. Now consider the following picture for gn ∈ C ([a, b]) in which gn equals 1 for x between x + 1/n and y.

¥ ¥ ¥ ¥ ¥

¥

D

¥ ¥ x x + 1/n

D D D D D D

D y y + 1/n

Then gn (t) → X(x,y] (t) pointwise. Therefore, by the dominated convergence theorem, Z µ ((x, y]) = lim gn dµ. n→∞

However,

≤

µ µ ¶¶ 1 f (y) − f x + n ¶ ¶ µ µ Z Z b 1 − f (y) gn dµ = gn df ≤ f y + n a µ µ ¶¶ µ µ ¶ ¶ 1 1 + f (y) − f x + + f x+ − f (x) n n

and so as n → ∞ the continuity of f implies µ ((x, y]) = f (y) − f (x) . Similarly, µ (x, y) = f (y) − f (y) and µ ([x, y]) = f (y) − f (x) , the argument used to establish this being very similar to the above. It follows in particular that Z f (x) − f (a) = dµ. [a,x]

Note that up till now, no referrence has been made to the absolute continuity of f. Any increasing continuous function would be fine.

534


Now if E is a Borel set such that m (E) = 0, Then the outer regularity of m implies there exists an open set, V containing E such that m (V ) < δ where δ corresponds to ε in the definition of absolute continuity of P f. Then letting {Ik } be the connected components of V it follows E ⊆ ∪∞ k=1 Ik with k m (Ik ) = m (V ) < δ. Therefore, from absolute continuity of f, it follows that for Ik = (ak , bk ) and each n n n X X µ (∪nk=1 Ik ) = µ (Ik ) = |f (bk ) − f (ak )| < ε k=1

k=1

and so letting n → ∞, µ (E) ≤ µ (V ) =

∞ X

|f (bk ) − f (ak )| ≤ ε.

k=1

Since ε is arbitrary, it follows µ (E) = 0. Therefore, µ ¿ m and so by the Radon Nikodym theorem there exists a unique h ∈ L1 ([a, b]) such that Z µ (E) = hdm. E

In particular,

Z µ ([a, x]) = f (x) − f (a) =

hdm. [a,x]

From the fundamental theorem of calculus f 0 (x) = h (x) at every Lebesgue point of h. Therefore, writing in usual notation, Z x f (x) = f (a) + f 0 (t) dt a

as claimed. This proves the lemma. With the above lemmas, the following is the main theorem about absolutely continuous functions. Theorem 17.2.5 Let f : [a, b] → R be absolutely continuous if and only if f 0 (x) exists a.e., f 0 ∈ L1 ([a, b]) and Z x f (x) = f (a) + f 0 (t) dt. a

Proof: Suppose first that f is absolutely continuous. By Lemma 17.2.3 the total variation function, V is absolutely continuous and f (x) = V (x) − (V (x) − f (x)) where both V and V − f are increasing and absolutely continuous. By Lemma 17.2.4 f (x) − f (a)

= V (x) − V (a) − [(V (x) − f (x)) − (V (a) − f (a))] Z x Z x 0 = V 0 (t) dt − (V − f ) (t) dt. a

a

17.3. DIFFERENTIATION OF MEASURES WITH RESPECT TO LEBESGUE MEASURE535 Now f 0 exists and is in L1 becasue f = V −(V − f ) and V and V −f have derivatives 0 in L1 . Therefore, (V − f ) = V 0 − f 0 and so the above reduces to Z x f 0 (t) dt. f (x) − f (a) = a

This proves one half of the theorem. Rx Now suppose f 0 ∈ L1 and f (x) = f (a) + a f 0 (t) dt. It is necessary to verify that f is absolutely continuous. But this follows easily from Lemma 8.5.2 on Page 0 212 which implies P that a single function, f is uniformly integrable. This lemma implies that if i |yi − xi | is sufficiently small then ¯ X X ¯¯Z yi ¯ 0 ¯ ¯= f (t) dt |f (yi ) − f (xi )| < ε. ¯ ¯ i

xi

i

The following simple corollary is a case of Rademacher’s theorem. Corollary 17.2.6 Suppose f : [a, b] → R is Lipschitz continuous, |f (x) − f (y)| ≤ K |x − y| . Then f 0 (x) exists a.e. and Z f (x) = f (a) +

x

f 0 (t) dt.

a

Proof: It is easy to see that f is absolutely continuous. Therefore, Theorem 17.2.5 applies.

17.3

Differentiation Of Measures With Respect To Lebesgue Measure

Recall the Vitali covering theorem in Corollary 10.4.5 on Page 297. Corollary 17.3.1 Let E ⊆ Rn and let F, be a collection of open balls of bounded radii such that F covers E in the sense of Vitali. Then there exists a countable ∞ collection of disjoint balls from F, {Bj }∞ j=1 , such that m(E \ ∪j=1 Bj ) = 0. Definition 17.3.2 Let µ be a Radon mesure defined on Rn . Then µ (B (x, r)) dµ (x) ≡ lim r→0 m (B (x, r)) dm whenever this limit exists. It turns out this limit exists for m a.e. x. To verify this here is another definition.

536


Definition 17.3.3 Let f (r) be a function having values in [−∞, ∞] . Then lim sup f (r) ≡ r→0+

lim inf f (r) ≡ r→0+

lim (sup {f (t) : t ∈ [0, r]})

r→0

lim (inf {f (t) : t ∈ [0, r]})

r→0

This is well defined because the function r → inf {f (t) : t ∈ [0, r]} is increasing and r → sup {f (t) : t ∈ [0, r]} is decreasing. Also note that limr→0+ f (r) exists if and only if lim sup f (r) = lim inf f (r) r→0+

r→0+

and if this happens lim f (r) = lim inf f (r) = lim sup f (r) .

r→0+

r→0+

r→0+

The claims made in the above definition follow immediately from the definition of what is meant by a limit in [−∞, ∞] and are left for the reader. Theorem 17.3.4 Let µ be a Borel measure on Rn then m a.e.

dµ dm

(x) exists in [−∞, ∞]

Proof:Let p < q and let p, q be rational numbers. Define ½ µ (B (x, r)) >q Npq (M ) ≡ x ∈ Rn such that lim sup r→0+ m (B (x, r)) ¾ µ (B (x, r)) > p > lim inf ∩ B (0, M ) , r→0+ m (B (x, r)) ½ µ (B (x, r)) Npq ≡ x ∈ Rn such that lim sup >q r→0+ m (B (x, r)) ¾ µ (B (x, r)) > p > lim inf , r→0+ m (B (x, r)) ½ µ (B (x, r)) N ≡ x ∈ Rn such that lim sup > r→0+ m (B (x, r)) ¾ µ (B (x, r)) lim inf . r→0+ m (B (x, r)) I will show m (Npq (M )) = 0. Use outer regularity to obtain an open set, V containing Npq (M ) such that m (Npq (M )) + ε > m (V ) . From the definition of Npq (M ) , it follows that for each x ∈ Npq (M ) there exist arbitrarily small r > 0 such that µ (B (x, r)) < p. m (B (x, r))

17.3. DIFFERENTIATION OF MEASURES WITH RESPECT TO LEBESGUE MEASURE537 Only consider those r which are small enough to be contained in B (0, M ) so that the collection of such balls has bounded radii. This is a Vitali cover of Npq (M ) and ∞ so by Corollary 17.3.1 there exists a sequence of disjoint balls of this sort, {Bi }i=1 such that µ (Bi ) < pm (Bi ) , m (Npq (M ) \ ∪∞ (17.3.10) i=1 Bi ) = 0. Now for x ∈ Npq (M ) ∩ (∪∞ i=1 Bi ) (most of Npq (M )), there exist arbitrarily small ∞ balls, B (x, r) , such that B (x, r) is contained in some set of {Bi }i=1 and µ (B (x, r)) > q. m (B (x, r)) This is a Vitali cover Npq (M )∩(∪∞ i=1 Bi ) and so there exists a sequence of disjoint © 0ofª∞ balls of this sort, Bj j=1 such that ¢ ¡ 0¢ ¡ 0¢ ¡ ∞ 0 m (Npq (M ) ∩ (∪∞ i=1 Bi )) \ ∪j=1 Bj = 0, µ Bj > qm Bj .

(17.3.11)

It follows from 17.3.10 and 17.3.11 that ¡ ∞ 0¢ m (Npq (M )) ≤ m ((Npq (M ) ∩ (∪∞ i=1 Bi ))) ≤ m ∪j=1 Bj Therefore, X ¡ ¢ µ Bj0 >

q

X

j

¡ ¢ m Bj0 ≥ qm (Npq (M ) ∩ (∪i Bi )) = qm (Npq (M ))

j

≥ ≥

pm (Npq (M )) ≥ p (m (V ) − ε) ≥ p X

(17.3.12)

µ (Bi ) − pε ≥

i

X ¡ ¢ µ Bj0 − pε.

X

m (Bi ) − pε

i

j

It follows pε ≥ (q − p) m (Npq (M )) Since ε is arbitrary, m (Npq (M )) = 0. Now Npq ⊆ ∪∞ M =1 Npq (M ) and so m (Npq ) = 0. Now N = ∪p.q∈Q Npq and since this is a countable union of sets of measure zero, m (N ) = 0 also. This proves the theorem. From Theorem 16.2.4 on Page 497 it follows that if µ is a complex measure then |µ| is a finite measure. This makes possible the following definition. Definition 17.3.5 Let µ be a real measure. Define the following measures. For E a measurable set, µ+ (E)

≡

µ− (E)

≡

1 (|µ| + µ) (E) , 2 1 (|µ| − µ) (E) . 2

538


These are measures thanks to Theorem 16.2.3 on Page 496 and µ+ −µ− = µ. These measures have values in [0, ∞). They are called the positive and negative parts of µ respectively. For µ a complex measure, define Re µ and Im µ by ´ 1³ µ (E) + µ (E) Re µ (E) ≡ 2 ´ 1 ³ Im µ (E) ≡ µ (E) − µ (E) 2i Then Re µ and Im µ are both real measures. Thus for µ a complex measure, ¡ ¢ µ = Re µ+ − Re µ− + i Im µ+ − Im µ− = ν 1 − ν 1 + i (ν 3 − ν 4 ) where each ν i is a real measure having values in [0, ∞). Then there is an obvious corollary to Theorem 17.3.4. Corollary 17.3.6 Let µ be a complex Borel measure on Rn . Then a.e.

dµ dm

(x) exists

Proof: Letting ν i be defined in Definition 17.3.5. By Theorem 17.3.4, for m i a.e. x, dν dm (x) exists. This proves the corollary because µ is just a finite sum of these ν i . Theorem 16.1.2 on Page 489, the Radon Nikodym theorem, implies that if you have two finite measures, µ and λ, you can write λ as the sum of a measure absolutely continuous with respect to µ and one which is singular to µ in a unique way. The next topic is related to this. It has to do with the differentiation of a measure which is singular with respect to Lebesgue measure. Theorem 17.3.7 Let µ be a Radon measure on Rn and suppose there exists a µ measurable set, N such that for all Borel sets, E, µ (E) = µ (E ∩ N ) where m (N ) = 0. Then dµ (x) = 0 m a.e. dm Proof: For k ∈ N, let ½ ¾ µ (B (x, r)) 1 Bk (M ) ≡ x ∈ N C : lim sup > ∩ B (0,M ) , k r→0+ m (B (x, r)) ½ ¾ µ (B (x, r)) 1 Bk ≡ x ∈ N C : lim sup > , k r→0+ m (B (x, r)) ¾ ½ µ (B (x, r)) >0 . B ≡ x ∈ N C : lim sup r→0+ m (B (x, r)) Let ε > 0. Since µ is regular, there exists H, a compact set such that H ⊆ N ∩ B (0, M ) and µ (N ∩ B (0, M ) \ H) < ε.

17.3. DIFFERENTIATION OF MEASURES WITH RESPECT TO LEBESGUE MEASURE539

B(0, M )

N ∩ B(0, M ) H

Bi Bk (M )

For each x ∈ Bk (M ) , there exist arbitrarily small r > 0 such that B (x, r) ⊆ B (0, M ) \ H and µ (B (x, r)) 1 > . (17.3.13) m (B (x, r)) k Two such balls are illustrated in the above picture. This is a Vitali cover of Bk (M ) ∞ and so there exists a sequence of disjoint balls of this sort, {Bi }i=1 such that m (Bk (M ) \ ∪i Bi ) = 0. Therefore, X X m (Bk (M )) ≤ m (Bk (M ) ∩ (∪i Bi )) ≤ m (Bi ) ≤ k µ (Bi ) = k

X

µ (Bi ∩ N ) = k

i

X

i

i

µ (Bi ∩ N ∩ B (0, M ))

i

≤ kµ (N ∩ B (0, M ) \ H) < εk Since ε was arbitrary, this shows m (Bk (M )) = 0. Therefore, ∞ X m (Bk ) ≤ m (Bk (M )) = 0 M =1

P and m (B) ≤ k m (Bk ) = 0. Since m (N ) = 0, this proves the theorem. It is easy to obtain a different version of the above theorem. This is done with the aid of the following lemma. Lemma 17.3.8 Suppose µ is a Borel measure on Rn having values in [0, ∞). Then there exists a Radon measure, µ1 such that µ1 = µ on all Borel sets. Proof: By assumption, µ (Rn ) < ∞ and so it is possible to define a positive linear functional, L on Cc (Rn ) by Z Lf ≡ f dµ.

540


By the Riesz representation theorem for positive linear functionals of this sort, there exists a unique Radon measure, µ1 such that for all f ∈ Cc (Rn ) , Z

Z f dµ1 = Lf =

f dµ.

¢ ª © ¡ Now let V be an open set and let Kk ≡ x ∈ V : dist x, V C ≤ 1/k ∩ B (0,k). Then {Kk } is an incresing sequence of compact sets whose union is V. Let Kk ≺ fk ≺ V. Then fk (x) → XV (x) for every x. Therefore, Z µ1 (V ) = lim

Z fk dµ1 = lim

k→∞

fk dµ = µ (V )

k→∞

and so µ = µ1 on open sets. Now if K is a compact set, let Vk ≡ {x ∈ Rn : dist (x, K) < 1/k} . Then Vk is an open set and ∩k Vk = K. Letting K ≺ fk ≺ Vk , it follows that fk (x) → XK (x) for all x ∈ Rn . Therefore, by the dominated convergence theorem with a dominating function, XRn Z µ1 (K) = lim

k→∞

Z fk dµ1 = lim

k→∞

fk dµ = µ (K)

and so µ and µ1 are equal on all compact sets. It follows µ = µ1 on all countable unions of compact sets and countable intersections of open sets. Now let E be a Borel set. By regularity of µ1 , there exist sets, H and G such that H is the countable union of an increasing sequence of compact sets, G is the countable intersection of a decreasing sequence of open sets, H ⊆ E ⊆ G, and µ1 (H) = µ1 (G) = µ1 (E) . Therefore, µ1 (H) = µ (H) ≤ µ (E) ≤ µ (G) = µ1 (G) = µ1 (E) = µ1 (H) . therefore, µ (E) = µ1 (E) and this proves the lemma. Corollary 17.3.9 Suppose µ is a complex Borel measure defined on Rn for which there exists a µ measurable set, N such that for all Borel sets, E, µ (E) = µ (E ∩ N ) where m (N ) = 0. Then dµ (x) = 0 m a.e. dm Proof: Each of Re µ+ , Re µ− , Im µ+ , and Im µ− are real measures having values in [0, ∞) and so by Lemma 17.3.8 each is a Radon measure having the same property that µ has in terms of being supported on a set of m measure zero. Therefore, for dν ν equal to any of these, dm (x) = 0 m a.e. This proves the corollary.

17.4. EXERCISES

17.4

541

Exercises

1. Suppose A and B are sets of positive Lebesgue measure in Rn . Show that A − B must contain B (c, ε) for some c ∈ Rn and ε > 0. A − B ≡ {a − b : a ∈ A and b ∈ B} . Hint: First assume both sets are bounded. This creates no loss of generality. Next there exist a0 ∈ A, b0 ∈ B and δ > 0 such that Z Z 3 3 XB (t) dt > m (B (b0 , δ)) . XA (t) dt > m (B (a0 , δ)) , 4 4 B(b0 ,δ) B(a0 ,δ) Now explain why this implies m (A − a0 ∩ B (0,δ)) >

3 m (B (0, δ)) 4

m (B − b0 ∩ B (0,δ)) >

3 m (B (0, δ)) . 4

and Explain why m ((A − a0 ) ∩ (B − b0 )) > Let

1 m (B (0, δ)) > 0. 2

Z f (x) ≡

XA−a0 (x + t) XB−b0 (t) dt.

Explain why f (0) > 0. Next explain why f is continuous and why f (x) > 0 for all x ∈ B (0, ε) for some ε > 0. Thus if |x| < ε, there exists t such that x + t ∈ A − a0 and t ∈ B − b0 . Subtract these. 2. Show M f is Borel measurable by verifying that R [M f > λ] ≡ Eλ is actually an open set. Hint: If x ∈ Eλ then for some r, B(x,r) |f | dm > λm (B (x, r)) . R Then for δ a small enough positive number, B(x,r) |f | dm > λm (B (x, r + 2δ)) . Now pick y ∈ B (x, δ) and argue that B (y, δ + r) ⊇ B (x, r) . Therefore show that, Z Z |f | dm > |f | dm > λB (x, r + 2δ) ≥ λm (B (y, r + δ)) . B(y,δ+r)

B(x,r)

Thus B (x, δ) ⊆ Eλ . 3. Consider ¤ nested sequence of compact sets, {Pn }.Let P1 = [0, 1], £ the ¤ following £ P2 = 0, 31 ∪ 23 , 1 , etc. To go from Pn to Pn+1 , delete the open interval which is the middle third of each closed interval in Pn . Let P = ∩∞ n=1 Pn . By the finite intersection property of compact sets, P 6= ∅. Show m(P ) = 0. If you feel ambitious also show there is a one to one onto mapping of [0, 1]

542


to P . The set P is called the Cantor set. Thus, although P has measure zero, it has the same number of points in it as [0, 1] in the sense that there is a one to one and onto mapping from one to the other. Hint: There are various ways of doing this last part but the most enlightenment is obtained by exploiting the topological properties of the Cantor set rather than some silly representation in terms of sums of powers of two and three. All you need to do is use the Schroder Bernstein theorem and show there is an onto map from the Cantor set to [0, 1]. If you do this right and remember the theorems about characterizations of compact metric spaces, Proposition 6.2.5 on Page 142, you may get a pretty good idea why every compact metric space is the continuous image of the Cantor set. 4. Consider the sequence of functions defined in the following way. Let f1 (x) = x on [0, 1]. To get from fn to fn+1 , let fn+1 = fn on all intervals where fn is constant. If fn is nonconstant on [a, b], let fn+1 (a) = fn (a), fn+1 (b) = fn (b), fn+1 is piecewise linear and equal to 12 (fn (a) + fn (b)) on the middle third of [a, b]. Sketch a few of these and you will see the pattern. The process of modifying a nonconstant section of the graph of this function is illustrated in the following picture.

¡

¡

¡

Show {fn } converges uniformly on [0, 1]. If f (x) = limn→∞ fn (x), show that f (0) = 0, f (1) = 1, f is continuous, and f 0 (x) = 0 for all x ∈ / P where P is the Cantor set of Problem 3. This function is called the Cantor function.It is a very important example to remember. Note it has derivative equal to zero a.e. and yet it succeeds in climbing from 0 to 1. Explain why this interesting function is not absolutely continuous although it is continuous. Hint: This isn’t too hard if you focus on getting a careful estimate on the difference between two successive functions in the list considering only a typical small interval in which the change takes place. The above picture should be helpful. 5. A function, f : [a, b] → R is Lipschitz if |f (x) − f (y)| ≤ K |x − y| . Show that every Lipschitz function is absolutely continuous. Thus R y every Lipschitz function is differentiable a.e., f 0 ∈ L1 , and f (y) − f (x) = x f 0 (t) dt. 6. Suppose f, g are both absolutely continuous on [a, b] . Show the product of 0 these functions is also absolutely continuous. Explain why (f g) = f 0 g + g 0 f and show the usual integration by parts formula Z b Z b f (b) g (b) − f (a) g (a) − f g 0 dt = f 0 gdt. a

a

7. In Problem 4 f 0 failed to give the expected result for 1 In

this example, you only know that

f0

exists a.e.

Rb a

f 0 dx

1

but at least

17.4. EXERCISES

543

f 0 ∈ L1 . Suppose f 0 exists for f a continuous function defined on [a, b] . Does it follow that f 0 is measurable? Can you conclude f 0 ∈ L1 ([a, b])? 8. A sequence of sets, {Ei } containing the point x is said to shrink to x nicely if there exists a sequence of positive numbers, {ri } and a positive constant, α such that ri → 0 and m (Ei ) ≥ αm (B (x, ri )) , Ei ⊆ B (x, ri ) . Show the above theorems about differentiation of measures with respect to Lebesgue measure all have a version valid for Ei replacing B (x, r) . Rx 9. Suppose F (x) = a f (t) dt. Using the concept of nicely shrinking sets in Problem 8 show F 0 (x) = f (x) a.e. 10. A random variable, X is a measurable real valued function defined on a measure space, (Ω, S, P ) where P is just a measure with P (Ω) = 1 called a probability measure. The distribution function for X is the function, F (x) ≡ P ([X ≤ x]) in words, F (x) is the probability that X has values no larger than x. Show that F is a right continuous increasing function with the property that limx→−∞ F (x) = 0 and limx→∞ F (x) = 1. 11. Suppose F is an increasing right continuous function. Rb (a) Show that Lf ≡ a f dF is a well defined positive linear functional on Cc (R) where here [a, b] is a closed interval containing the support of f ∈ Cc (R) . (b) Using the Riesz representation theorem for positive linear functionals on Cc (R) , let µ denote the Radon measure determined by L. Show that µ ((a, b]) = F (b) − F (a) and µ ({b}) = F (b) − F (b−) where F (b−) ≡ limx→b− F (x) . (c) Review Corollary 16.1.4 on Page 494 at this point. Show that the conditions of this corollary hold for µ and m. Consider µ⊥ + µ|| , the Lebesgue decomposition of µ where µ|| ¿ m and there exists a set of m measure zero, R x N such that µ⊥ (E) = µ⊥ (E ∩ N ) . Show µ ((0, x]) = µ⊥ ((0, x]) + 0 h (t) dt for some h ∈ L1 (m) . Using Theorem 17.3.7 show Rx h (x) = F 0 (x) m a.e. Explain why F (x) = F (0) + S (x) + 0 F 0 (t) dt for some function, S (x) which is increasing but has S 0 (x) = 0 a.e. Note this shows in particular that a right continuous increasing function has a derivative a.e. 12. Suppose now that G is just an increasing function defined on R. Show that G0 (x) exists a.e. Hint: You can mimic the proof of Theorem 17.3.4. The Dini derivates are defined as G (x + h) − G (x) , D+ G (x) ≡ lim inf h→0+ h G (x + h) − G (x) D+ G (x) ≡ lim sup h h→0+

544


G (x) − G (x − h) , h G (x) − G (x − h) D− G (x) ≡ lim sup . h h→0+ D− G (x) ≡ lim inf

h→0+

When D+ G (x) = D+ G (x) the derivative from the right exists and when D− G (x) = D− G (x) , then the derivative from the left exists. Let (a, b) be an open interval and let © ª Npq ≡ x ∈ (a, b) : D+ G (x) > q > p > D+ G (x) . Let V ⊆ (a, b) be an open set containing Npq such that n (V ) < m (Npq ) + ε. Show using a Vitali covering theorem there is a disjoint sequence of intervals ∞ contained in V , {(xi , xi + hi )}i=1 such that G (xi + hi ) − G (xi ) < p. hi ¢ª∞ ©¡ Next show there is a disjoint sequence of intervals x0i , x0j + h0j j=1 such that each of these is contained in one of the former intervals and ¢ ¡ ¢ ¡ X G x0j + h0j − G x0j > q, h0j ≥ m (Npq ) . h0j j Then qm (Npq )

≤ q

X j

≤ p

X

h0j ≤

X

¡ ¢ ¡ ¢ X G x0j + h0j − G x0j ≤ G (xi + hi ) − G (xi )

j

i

hi ≤ pm (V ) ≤ p (m (Npq ) + ε) .

i

Since ε was arbitrary, this shows m (Npq ) = 0. Taking a union of all Npq for p, q rational, shows the derivative from the right exists a.e. Do a similar argument to show the derivative from the left exists a.e. and then show the derivative from the left equals the derivative from the right a.e. using a simlar argument. Thus G0 (x) exists on (a, b) a.e. and so it exists a.e. on R because (a, b) was arbitrary.

Orlitz Spaces 18.1

Basic Theory

All the theorems about the Lp spaces have generalizations to something called an Orlitz space. [1], [49] Instead of the convex function, A (t) = tp /p, one considers a more general convex increasing function called an N function. Definition 18.1.1 A : [0, ∞) → [0, ∞) is an N function if the following two conditions hold. A is convex and strictly increasing (18.1.1) A (t) A (t) = 0, lim = ∞. t→∞ t t

(18.1.2)

e (s) ≡ max {st − A (t) : t ≥ 0} . A

(18.1.3)

lim

t→0+

For A an N function,

As an example see the following picture of a typical N function.

A(t)

e must Note that from the assumption, 18.1.2 the maximum in the definition of A exist. This is because for t 6= 0 (s − A (t) /t) t is negative for all t large enough. On the other hand, it equals 0 when t = 0 and so it suffices to consider only t in a compact set. 545

546

ORLITZ SPACES

Lemma 18.1.2 Let φ : R → R be a convex function. Then φ is Lipschitz continuous on [a, b] . Proof: Since it is convex, the difference quotients, φ (t) − φ (a) t−a are increasing because by convexity, if a < t < x µ ¶ t−a t−a φ (x) + 1 − φ (a) ≥ φ (t) x−a x−a and this reduces to

φ (x) − φ (a) φ (t) − φ (a) ≤ . t−a x−a Also these difference quotients are bounded below by φ (a) − φ (a − 1) = φ (a) − φ (a − 1) . 1 Let

½ A ≡ inf

¾ φ (t) − φ (a) : t ∈ (a, b) . t−a

Then A is some finite real number. Similarly there exists a real number B such that for all t ∈ (a, b) , φ (b) − φ (t) B≥ . b−t Now let a ≤ s < t ≤ b. Then φ (t) − φ (s) φ (t) − θφ (a) − (1 − θ) φ (t) ≥ t−s t−s where θ is such that θa + (1 − θ) t = s. Thus θ=

t−s t − t1

and so the above implies φ (t) − φ (s) t − s φ (t) − φ (a) φ (t) − φ (a) ≥ = ≥ A. t−s t − t1 t−s t − t1 Similarly, φ (t) − φ (s) t−s

≤ =

θφ (b) + (1 − θ) φ (s) − φ (s) t−s t − s φ (b) − φ (s) ≤ B. b−s t−s

18.1. BASIC THEORY

547

It follows |φ (t) − φ (s)| ≤ (|A| + |B|) |t − s| and this proves the lemma. The following is like the inequality, st ≤ tp /p + sq /q, important in the study of p L spaces. e and Proposition 18.1.3 If A is an N function, then so is A n o e (s) : s ≥ 0 , A (t) = max ts − A ee so A = A. Also

e (s) for all s, t ≥ 0 st ≤ A (t) + A

and for all s > 0,

Ã A

e (s) A s

(18.1.4)

(18.1.5)

! e (s) . ≤A

(18.1.6)

e is convex. Let λ ∈ [0, 1] . Proof: First consider the claim A e (λs1 + (1 − λ) s2 ) ≡ max {[s1 λ + (1 − λ) s2 ] t − A (t) : t ≥ 0} A ≤ λ max {s1 t − A (t) : t ≥ 0} + (1 − λ) max {s2 t − A (t) : t ≥ 0} e (s1 ) + (1 − λ) A e (s2 ) . = λA e is stictly increasing because st is strictly increasing in s. Next It is obvious A consider 18.1.2. For s > 0 let ts denote the number where the maximum is achieved. That is, e (s) ≡ sts − A (ts ) . A Thus

e (s) A A (ts ) = ts − ≥ 0. s s

(18.1.7)

It follows from this that lim ts = 0

s→0+

since otherwise, a contradiction results to 18.1.7, the expression becoming negative for small enough s. Thus e (s) A ≥0 ts ≥ s and this shows e (s) A lim = 0. s→0+ s which shows 18.1.2.

548

ORLITZ SPACES

To verify the second part of 18.1.2, let ts be as just described. Then for any t>0 e (s) A A (ts ) A (t) = ts − ≥t− s s s It follows lim inf

s→∞

e (s) A ≥ t. s

Since t is arbitrary, this proves the second part of 18.1.2. e (s) . The inequality 18.1.5 follows from the definition of A Next consider 18.1.4. It must be shown that n o e (s) : s ≥ 0 . A (t0 ) = max t0 s − A To do so, first note e (s) = max {st − A (t) : t ≥ 0} ≥ st0 − A (t0 ) . A Hence n o e (s) : s ≥ 0 ≤ max {t0 s − [st0 − A (t0 )]} = A (t0 ) . max t0 s − A Now let

½ s0 ≡ inf

¾ A (t) − A (t0 ) : t > t0 . t − t0

By convexity, the above difference quotients are nondecreasing in t and so s0 (t − t0 ) ≤ A (t) − A (t0 ) for all t 6= t0 . Hence for all t, s0 t − A (t) ≤ s0 t0 − A (t0 ) and so e (s0 ) = s0 t0 − A (t0 ) A implying n o e (s0 ) ≤ max st0 − A e (s) : s ≥ 0 ≤ A (t0 ) . A (t0 ) = s0 t0 − A Therefore, 18.1.4 holds. Consider 18.1.6 next. To do so, let a = A0 so that Z

t

A (t) =

a (r) dr, a increasing. 0

18.1. BASIC THEORY

549

This is possible by Rademacher’s theorem, Corollary 17.2.6 and the fact that since A is convex, it is locally Lipshitz found in Lemma 18.1.2 above. That a is increasing follows from convexity of A. Here is why. For a.e. s, t ≥ 0, and letting λ ∈ [0, 1] , A (s + λ (t − s)) − A (s) λ

(1 − λ) A (s) + λA (t) − A (s) λ A (t) − A (s)

≤ =

Then passing to a limit as λ → 0+, a (s) (t − s) ≤ A (t) − A (s) . Similarly a (t) (s − t) ≤ A (s) − A (t) and so (a (t) − a (s)) (t − s) ≥ 0. (If you like, you can simply assume from the beginning that A (t) is given this way as an integral of a positive increasing function, a, and verify directly that such an A is convex and satisfies the properties of an N function. There is no loss of generality in doing so.) Thus geometrically, A (t) equals the area under the curve defined by e (s) let ts be a and above the x axis from x = 0 to x = t. In the definition of A the point where the maximum is achieved. Then e (s) = sts − A (ts ) A e (s) + A (ts ) = sts . This means that A e (s) is the area to the and so at this point, A left of the graph of a which is to the right of the y axis for y between 0 and a (ts ) and that in fact a (ts ) = s. The following picture illustrates the reasoning which follows. s e A(s)

@ I @ A(t0 )

@

@

@ t0

@

graph of a

Therefore, e (s) A s

= =

Z A (ts ) 1 ts ts − = ts − a (r) dr s s 0 µ ¶ Z ts Z ts 1 1 a (r) dr = ts s − a (r) dr ts − a (ts ) 0 a (ts ) 0

550

ORLITZ SPACES

and so Ã A

e (s) A s

!

Z

Z

e A(s)/s

=

1 a(ts )

a (r) dr = Z

0

(s−a(r))dr

a (τ ) dτ

0

0

≤

R ts

ts

e (s) . s − a (r) dr = sts − A (ts ) = A

0

The inequality results from replacing a (τ ) with a (ts ) in the last integral on the top line. p An example of an N function is A (t) = tp for t ≥ 0 and p > 1. For this example, e (s) = A

0

sp p0

where

1 p

+

1 p0

= 1.

Definition 18.1.4 Let A be an N function and let (Ω, S, µ) be a measure space. Define ½ ¾ Z KA (Ω) ≡ u measurable such that A (|u|) dµ < ∞ . (18.1.8) Ω

This is called the Orlitz class. Also define LA (Ω) ≡ {λu : u ∈ KA (Ω) and λ ∈ F}

(18.1.9)

where F is the field of scalars, assumed to be either R or C. The pair (A, Ω) is called ∆ regular if either of the following conditions hold. A (rx) ≤ Kr A (x) for all x ∈ [0, ∞)

(18.1.10)

or µ (Ω) < ∞ and for all r > 0, there exists Mr and Kr > 0 such that A (rx) ≤ Kr A (x) for all x ≥ Mr .

(18.1.11)

Note there are N functions which are not ∆ regular. For example, consider 2

A (x) ≡ ex − 1. It can’t be ∆ regular because 2

2

er x − 1 = ∞. 2 r→∞ ex − 1 lim

However, functions like xp /p for p > 1 are ∆ regular. Then the following proposition is important. Proposition 18.1.5 If (A, Ω) is ∆ regular, then KA (Ω) = LA (Ω) . In any case, LA (Ω) is a vector space and KA (Ω) ⊆ LA (Ω) .

18.1. BASIC THEORY

551

Proof: Suppose (A, Ω) is ∆ regular. Then I claim KA (Ω) is a vector space. This will verify KA (Ω) = LA (Ω) . Let f, g ∈ KA (Ω) and suppose 18.1.10. Then µ µ ¶¶ µ ¶ 1 |f + g| |f + g| A (|f + g|) = A 2 ≤ K2 A ≤ K2 [A (|f |) + A (|g|)] 2 2 2 so f + g ∈ KA (Ω) in this case. Now suppose 18.1.11 Z Z Z A (|f + g|) dµ + A (|f + g|) dµ = [|f +g|≤M2 ]

Ω

Z ≤ A (M2 ) µ (Ω) + Ω

A (|f + g|) dµ [|f +g|>M2 ]

K2 (A (|f |) + A (|g|)) dµ < ∞. 2

Thus f + g ∈ KA (Ω) in this case also. Next consider scalar multiplication. First consider the case of 18.1.10. If f ∈ KA (Ω) and α ∈ F, Z Z A (|α| |f |) dµ ≤ K|α| A (f ) dµ Ω

Ω

so in the case of 18.1.10 αf ∈ KA (Ω) whenever f ∈ KA (Ω) . In the case of 18.1.11, Z Z Z A (|α| |f |) dµ = A (|α| |f |) dµ + A (|α| |f |) dµ Ω [|α||f |≤M|α| ] [|α||f |>M|α| ] Z ¡ ¢ ≤ A M|α| µ (Ω) + K|α| A (|f |) dµ < ∞. Ω

This establishes the first part of the proposition. Next consider the claim that LA (Ω) is always a vector space. First note KA (Ω) is always convex due to convexity of A. Let λu, αv ∈ LA (Ω) where u, v ∈ KA (Ω) and let a, b be scalars in F. Then aλu + bαv = |aλ| ωu + |bα| θv where |ω| = |θ| = 1. Then µ = (|aλ| + |bα|)

|aλ| ωu + |bα| θv |aλ| + |bα|

¶

which exhibits aλu + bαv as a multiple of a convex combination of two elements of KA (Ω) , ωu and θv. Thus LA (Ω) is closed with respect to linear combinations. This shows it is a vector space. This proves the proposition. The following norm for LA (Ω) is due to Luxemburg [49]. You might compare this to the definition of a Minkowski functional. The definition of LA (Ω) above was cooked up so that the following norm does make sense. Definition 18.1.6 Define ½ ||u||A = ||u||A,Ω ≡ inf t > 0 :

µ

Z A Ω

|u (x)| t

¶

¾ dµ ≤ 1 .

552

ORLITZ SPACES

If two functions of LA (Ω) are equal a.e. they are considered to be the same in the usual way. Proposition 18.1.7 The number defined in Definition 18.1.6 is a norm on LA (Ω) . Also, if Ω1 ⊆ Ω, then ||u||A,Ω1 ≤ ||u||A,Ω . Proof: Clearly ||u||A ≥ 0. Is ||u||A finite for u ∈ LA (Ω)? Let u ∈ LA (Ω) so u = λv where v ∈ KA (Ω) . Then for s > 0 µ µ ¶ ¶ Z Z |u| |v| A A dµ = dµ < ∞ s |λ| s Ω Ω whenever s > 1. Therefore, from the dominated convergence theorem, if s is large enough, µ ¶ Z |u| A dµ ≤ 1 s |λ| Ω and this shows there are values of t > 0 such that µ ¶ Z |u (x)| A dµ ≤ 1. t Ω Thus ||u||A is finite as hoped. Now suppose ||u||A = 0 and let ½ En ≡

x : |u (x)| ≥

1 n

¾ .

Then for arbitrarily small values of t, µ ¶ µ ¶ Z Z (1/n) |u (x)| A dµ ≤ A dµ ≤ 1 t t En Ω and so for arbitrarily small values of t, µ ¶ (1/n) A µ (En ) ≤ 1. t Letting t → 0+ yields a contradiction unless µ (En ) = 0. Now µ ([|u (x)| > 0]) ≤

∞ X

µ (En ) = 0.

n=1

Thus u = 0 as claimed. Consider the other axioms of a norm. Let u, v ∈ LA (Ω) and let α, β be scalars. Then ¶ ¾ ½ µ Z |u (x) + v (x)| dµ ≤ 1 ||αu + βv||A ≡ inf t > 0 : A t Ω

18.1. BASIC THEORY

553

Without loss of generality ||u||A , ||v||A < ∞ since otherwise there is nothing to prove. ½ ¶ ¾ µ Z |αu (x) + βv (x)| dµ ≤ 1 . ||u + v||A ≡ inf t > 0 : A t Ω ¶ ½ µ ¾ Z |α| |u| + |β| |v| ≤ inf t > 0 : A dµ ≤ 1 t Ω ( Ã ! ) Z + |β| (|α|+|β|)|v| |α| (|α|+|β|)|u| t t = inf t > 0 : A dµ ≤ 1 (|α| + |β|) Ω

≤

¶ ¾ ½ µ Z |α| |u| dµ ≤ 1 inf t > 0 : A (|α| + |β|) Ω t/ (|α| + |β|) ½ µ ¶ ¾ Z |β| |v| + inf t > 0 : A dµ ≤ 1 (|α| + |β|) Ω t/ (|α| + |β|) ½

=

¶ ¾ |u| |α| inf t/ (|α| + |β|) > 0 : A dµ ≤ 1 t/ (|α| + |β|) Ω ½ µ ¶ ¾ Z |v| + |β| inf t/ (|α| + |β|) > 0 : A dµ ≤ 1 t/ (|α| + |β|) Ω Z

µ

= |α| ||u||A + |β| ||v||A . Now let Ω1 ⊆ Ω. ||u||A,Ω1

½ µ ¶ ¾ Z |u (x)| ≡ inf t > 0 : A dµ ≤ 1 t Ω ¶ ¾ ½ Z 1 µ |u (x)| dµ ≤ 1 ≡ ||u||A,Ω . ≤ inf t > 0 : A t Ω

This occurs because if t is in the second set, then it is in the first so the infimum of the second is no smaller than that of the first. This proves the proposition. Next it is shown that LA (Ω) is a Banach space. Theorem 18.1.8 LA (Ω) is a Banach space and every Cauchy sequence has a subsequence which also converges pointwise a.e. Proof: Let {fn } be a Cauchy sequence in LA (Ω) and select a subsequence {fnk } such that ¯¯ ¯¯ ¯¯f n − fnk ¯¯A ≤ 2−k . k+1 Thus fnm (x) = fn1 (x) +

m−1 X k=1

fnk+1 (x) − fnk (x) .

554

ORLITZ SPACES

Let gm (x) ≡ |fn1 (x)| +

m−1 X

¯ ¯f n

k+1

¯ (x) − fnk (x)¯ .

k=1

Then ||gm ||A ≤ ||fn1 ||A +

∞ X

2−k ≡ K < ∞.

k=1

Let g (x) ≡ lim gm (x) ≡ |fn1 (x)| + m→∞

∞ X ¯ ¯f n

k+1

¯ (x) − fnk (x)¯ .

k=1

Now K > ||gm ||A so

µ

Z 1≥

A Ω

|gm (x)| K

¶ dµ.

By the monotone convergence theorem, µ ¶ Z |g (x)| 1≥ A dµ K Ω showing g (x) < ∞ a.e., say for all x ∈ / E where E is a measurable set having measure zero. Let ! Ã ∞ X ¡ ¢ fnk+1 (x) − fnk (x) f (x) ≡ XE C (x) fn1 (x) + k=1

=

lim XE C (x) fnm (x) .

m→∞

Thus f is measurable and fnm (x) → f (x) a.e. as m → ∞. For l > k, 1 ||fnk − fnl ||A < k−2 2 and so Ã ! Z |fnl (x) − fnk (x)| ¡ 1 ¢ 1≥ A dµ. Ω

2k−2

By Fatou’s lemma, let l → ∞ and obtain ! Ã Z |f (x) − fnk (x)| ¡ 1 ¢ dµ 1≥ A Ω

2k−2

and so (f − fnk ) 2k−2 ∈ KA (Ω) and so f − fnk ∈ LA (Ω) , fnk ∈ LA (Ω) . Since LA (Ω) is a vector space, this shows f ∈ LA (Ω) . Also ||f − fnk ||A ≤

1 , 2k−2

18.1. BASIC THEORY

555

showing that fnk → f in LA (Ω) . Since a subsequence converges in LA (Ω) , it follows the original Cauchy sequence also converges to f in LA (Ω). This proves the theorem. Next consider the space, EA (Ω) which will be a subspace of the Orlitz class, KA (Ω) just as LA (Ω) is a vector space containing the Orlitz class. Definition 18.1.9 Let S denote the set of simple functions, s, such that µ ({x : s (x) 6= 0}) < ∞. Then define EA (Ω) ≡ the closure in LA (Ω) of S. Proposition 18.1.10 EA (Ω) ⊆ KA (Ω) ⊆ LA (Ω) and they are all equal if (A, Ω) is ∆ regular. Proof: First note that S ⊆ KA (Ω) ∩ EA (Ω) . Let f ∈ EA (Ω) . Then by the definition of EA (Ω) , there exists sn ∈ S such that ||sn − f ||A → 0. Therefore, for n large enough, ||sn − f ||A < and so

Ã

Z A Ω

Since S ⊆ KA (Ω) ,

!

|f − sn | ¡1¢

1 2

Z dµ =

A (|2f − 2sn |) dµ < ∞. Ω

2

Z A (2 |sn |) dµ < ∞. Ω

Therefore, 2f − 2sn ∈ KA (Ω) and 2sn ∈ KA (Ω) and so, since KA (Ω) is convex, 2f − 2sn 2sn + = f ∈ KA (Ω) . 2 2 This shows EA (Ω) ⊆ KA (Ω) . Next consider the claim these spaces are all equal in the case that (A, Ω) is ∆ regular. It was already shown in Proposition 18.1.5 that in this case, KA (Ω) = LA (Ω) so it remains to show EA (Ω) = KA (Ω). Is every f ∈ KA (Ω) the limit in LA (Ω) of functions from S? First suppose µ (Ω) = ∞. Then A (r |f |) ≤ Kr A (|f |) and so A (r |f |) ∈ L1 (Ω) for any r. Let ε > 0 be given and let Ωδ ≡ {x : |f (x)| ≥ δ}

556

ORLITZ SPACES

Then by the dominated convergence theorem, µ ¶ Z |f | lim A dµ = 0. δ→0+ Ω\Ω ε δ Choose δ such that

µ

Z A Ω\Ωδ

|f | ε

¶ dµ
0 is arbitrary. Now suppose µ (Ω) < ∞. In this case, only assume A (rt) ≤ Kr A (t) for t large enough, say for t ≥ Mr . However, this is enough to conclude A (r |f |) ∈ L1 (Ω) for any r > 0 because µ (Ω) < ∞ and f ∈ KA (Ω) . Let sn → f pointwise with |sn | ≤ |f | , and s simple. Then µ ¶ µ ¶ |f − sn | 2 A ≤A |f | ∈ L1 (Ω) ε ε and so the dominated convergence theorem implies µ ¶ Z |f − sn | lim A dµ = 0. n→∞ Ω ε Hence

µ

Z A Ω

|f − sn | ε

¶ dµ < 1

for all n large enough and so for such n, ||f − sn ||A ≤ ε which proves the proposition. It turns out EA (Ω) is the largest linear subspace of KA (Ω) .

18.1. BASIC THEORY

557

Proposition 18.1.11 EA (Ω) is the maximal linear subspace of KA (Ω) . Proof: Let M be a subspace of KA (Ω) . Is M ⊆ EA (Ω)? For f ∈ M, f /ε ∈ KA (Ω) for all ε > 0 because of the fact that M is a subspace and f ∈ M. Thus A (|f | /ε) is in L1 (Ω). Let ε > 0 be given, choose δ > 0 and let Fδ ≡ {x : |f (x)| ≤ δ} By the dominated convergence theorem there exists δ small enough that ¶ µ Z 1 2 |f | A dµ < . ε 2 Fδ Let |sn | ≤ |f | XFδC and sn → f XFδC pointwise for sn a simple function. Thus sn = 0 ¡ ¢ on Fδ and so sn ∈ S because µ FδC < ∞. Now µ ¶ µ ¶ ¶ µ Z Z Z |f − sn | |f | |f − sn | A dµ = A dµ + dµ A ε ε ε Ω Fδ FδC µ ¶ Z 1 |f − sn | < + A dµ. 2 ε FδC | The integrand in the last integral is no larger than 2|f ε and so by the dominated convergence theorem, this integral converges to 0 as n → ∞. In particular, it is eventually less than 12 . Therefore, for such n,

||f − sn ||A ≤ r. Since r is arbitrary, this shows that f ∈ EA (Ω) which proves the proposition. Next is a comparison of these function spaces for different choices of the N function. The notation X ,→ Y for two normed linear spaces means X is a subset of Y and the identity map is continuous. Proposition 18.1.12 LB (Ω) ,→ LA (Ω) if either B (t) ≥ A (t) for all t ≥ 0

(18.1.12)

B (t) ≥ A (t) for all t > M

(18.1.13)

or if and µ (Ω) < ∞. Proof: Let f ∈ LB (Ω) and let µ ¶ Z |f | B dµ ≤ 1. t Ω Then if 18.1.12 holds, it follows µ

Z A Ω

|f | t

¶ dµ ≤ 1.

558

ORLITZ SPACES

Thus if t ≥ ||f ||B then t ≥ ||f ||A which implies ||f ||B ≥ ||f ||A . Now suppose 18.1.13 holds and µ (Ω) < ∞. Then max (A, B) is an N function dominating both A and B for all t. By what was just shown Lmax(A,B) (Ω) ,→ LB (Ω) . Then let f ∈ LB (Ω) and let µ ¶ Z |f | B dµ < 1. t Ω Then

µ

Z max (A, B)

|f | t

¶

Z

µ

¶

B dµ [ |ft | >M ] µ ¶ Z |f | + max (A, B) dµ |f | t [ t ≤M ] µ ¶ Z |f | ≤ B dµ + µ (Ω) max (A, B) (M ) < ∞. t Ω Ω

dµ =

|f | t

It follows |ft | ∈ Kmax(A,B) (Ω) and so f ∈ Lmax(A,B) (Ω) . Hence LB (Ω) = Lmax(A,B) (Ω) and the identity map from Lmax(A,B) (Ω) to LB (Ω) is continuous. Therefore, by the open mapping theorem, the norms || ||B and || ||max(A,B) are equivalent. Hence for f ∈ LB (Ω) , ||f ||A ≤ ||f ||max(A,B) ≤ C ||f ||B . This proves the proposition. Corollary 18.1.13 Suppose there exists C > 0, a constant such that either CB (t) ≥ A (t) for all t ≥ 0 or CB (t) ≥ A (t) for all t > M and µ (Ω) < ∞. Then LB (Ω) ,→ LA (Ω) . Proof: If f ∈ LB (Ω) then f = λu where u ∈ KB (Ω) = KCB (Ω) . Hence LCB (Ω) = LB (Ω) and the two norms on LB (Ω) , || ||CB , and || ||B , are equivalent norms by the open mapping theorem. Hence by the Proposition 18.1.12, if f ∈ LB (Ω) , ||f ||A ≤ C1 ||f ||CB ≤ C2 ||f ||B which proves the corollary.

18.1. BASIC THEORY

559

Definition 18.1.14 A increases essentially more slowly than B if for all a > 0, lim

t→∞

A (at) =0 B (t)

The next theorem gives added information on how these spaces are related in case that one N function increases essentially more slowly than the other. Theorem 18.1.15 Suppose µ (Ω) < ∞ and A increases essentially more slowly than B. Then LB (Ω) ,→ EA (Ω) Proof: Let f ∈ LB (Ω) . Then there exists λ > 0 such that µ ¶ Z |f | B dµ ≤ 1. λ Ω Let r be such that for t ≥ r, A (|λ| t) ≤ B (t) . Then

Z

Z A (|f |) dµ

Ω

Z

=

A (|f |) dµ + [|f |≥r]

¶ |f | dµ + A (r) µ (Ω) |λ| Ω < 1 + A (r) µ (Ω) . Z

≤

µ

A (|f |) dµ [|f | 0, |u| |v| ∈ KAe (Ω) , ∈ KA (Ω) . ||v||Ae + ε ||u||A + ε Therefore, Z µ Ω

|u| ||u||A + ε

¶µ

|v| ||v||Ae + ε µ

Z e A

+ Ω

¶

µ

Z dµ ≤

A Ω

|v| ||v||Ae + ε

|u| ||u||A + ε

¶ dµ

¶ dµ ≤ 2

and so uv ∈ L1 (Ω) and ¯Z ¯ Z ¯ ¯ ¡ ¢ ¯ uvdµ¯ ≤ |u| |v| dµ ≤ 2 ||v||Ae + ε (||u||A + ε) . ¯ ¯ Ω

Ω

Since ε is arbitrary this shows ¯Z ¯ Z ¯ ¯ ¯ uvdµ¯ ≤ |u| |v| dµ ≤ 2 ||v||Ae ||u||A . ¯ ¯ Ω

(18.2.14)

Ω

e by Defining Lv for v ∈ A Z Lv (u) ≡

uvdµ, Ω

0

it follows Lv ∈ LA (Ω) . From now on assume the measure space is σ finite. That is, there exist measurable sets, Ωk satisfying the following: Ω = ∪∞ k=1 Ωk , µ (Ωk ) < ∞, Ωk ⊆ Ωk+1 . Then Proposition 18.2.1 For v ∈ LAe (Ω) , the following inequality holds. ||v||Ae ≤ ||Lv || ≤ 2 ||v||Ae . 0

0

Here Lv is considered as either an element of EA (Ω) or LA (Ω) and ||Lv || refers to the operator norm in either dual space.

18.2. DUAL SPACES IN ORLITZ SPACE

561

Proof: The inequality 18.2.14 implies ||Lv || ≤ 2 ||v||Ae . It remains to show the other half of the inequality. If Lv = 0 there is nothing to show because this would imply that v = 0 so assume ||Lv || > 0. Define a measurable function, u, as follows. Letting r ∈ (0, 1) , ´ ( ³ e r|v(x)| / v(x) if v (x) 6= 0 A ||Lv || ||Lv || (18.2.15) u (x) ≡ 0 if v (x) = 0. Now let Fn ≡ {x : |u (x)| ≤ n} ∩ Ωn ∩ {x : v (x) 6= 0}

(18.2.16)

un (x) ≡ u (x) XFn (x) .

(18.2.17)

and define Thus un is bounded and equals zero off a set which has finite measure. It follows that µ ¶ |un | A ∈ L1 (Ω) α for all α > 0. I claim that ||un ||A ≤ 1. If not, there exists ε > 0 such that ||un ||A −ε > 1. Then since A is convex, µ ¶ Z Z |un | 1 1< A dµ ≤ A (|un |) dµ. ||un ||A − ε ||un ||A − ε Ω Ω Taking ε → 0+, using 18.1.6, and convexity of A along with 18.2.15 and 18.2.17, µ µ ¶ ¶ Z Z r |v (x)| rv (x) e ||un ||A ≤ A (|un |) dµ = A rA / dµ ||Lv || ||Lv || Ω Fn µ µ ¶ ¶ µ ¶ Z Z r |v (x)| rv (x) r |v (x)| e e ≤ r A A / dµ ≤ r A dµ ||Lv || ||Lv || ||Lv || Fn Fn Z 1 1 = r un (x) v (x) dµ ≤ r ||un ||A ||Lv || = r ||un ||A , ||Lv || Ω ||Lv || a contradiction since r < 1. Therefore, from 18.2.15, Z Z ||Lv || ≥ |Lv (un )| ≡ v (x) u (x) dµ = ||Lv ||

µ ¶ r |v (x)| e A dµ ||Lv || Fn

Fn

and so

µ ¶ r |v (x)| e 1≥ dµ A ||Lv || Fn Z

By the monotone convergence theorem, letting n → ∞, µ ¶ Z r |v (x)| e 1≥ A dµ ||Lv || Ω showing that ||v||Ae ≤

||Lv || . r

562

ORLITZ SPACES

Since this holds for all r ∈ (0, 1), it follows ||Lv || ≥ ||v||Ae as claimed. This proves the proposition. Now what follows is the Riesz representation theorem for the dual space of EA (Ω) . 0

Theorem 18.2.2 Suppose µ (Ω) < ∞ and suppose L ∈ EA (Ω) . Then the map 0 v → Lv from LAe (Ω) to EA (Ω) is one to one continuous, linear, and onto. If (Ω, A) is ∆ regular then v → Lv is one to one, linear, onto and continuous as a 0 map from LAe (Ω) to LA (Ω) . Proof: It is obvious this map is linear. From Proposition 18.2.1 it is continuous 0 and one to one. It remains only to verify that it is onto. Let L ∈ EA (Ω) and define a complex valued function, λ, mapping the measurable sets to C as follows. λ (F ) ≡ L (XF ) . In case µ (F ) 6= 0, µ µ Z A XF (x) A−1 Ω

1 µ (F )

¶¶

Z dµ

=

µ µ A A−1

F

Z

F

||XF ||A ≤

1 ³ A−1

1 µ(F )

¶¶ dµ

1 dµ = 1 µ (F )

= and so

1 µ (F )

´.

(18.2.18)

In fact, λ is actually a complex measure. To see this, suppose Fi ↑ F. Then from the formula just derived, ¯¯ ¯¯ ||XFi − XF ||A = ¯¯XF \Fi ¯¯A ≤

³ A−1

1 1 µ(F \Fi )

´

which converges to zero as i → ∞. Therefore, if the Fi are disjoint and F = ∪∞ i=1 Fi , let Sm ≡ ∪m F so that S ↑ F. Then since X → X in E (Ω) and L is i m S F A m i=1 continuous, λ (F )

≡ L (XF ) = lim L (XSm ) m→∞

=

lim

m→∞

m X i=1

L (XFi ) =

∞ X

λ (Fi ) .

i=1

Next observe that λ is absolutely continuous with respect to µ. To see this, suppose µ (F ) = 0. Then if t > 0, ¶ µ Z XF (x) dµ = 0 < 1 A t Ω

18.2. DUAL SPACES IN ORLITZ SPACE

563

for all t > 0 and so ||XF ||A = 0. Therefore, λ (F ) ≡ L (XF ) = 0. It follows by the Radon Nikodym theorem there exists v ∈ L1 (Ω) such that Z L (XF ) = λ (F ) = vdµ. F

Therefore, for all s ∈ S,

Z L (s) =

svdµ.

(18.2.19)

F

I need to show that v is actually in LAe (Ω) . If v = 0 a.e., there is nothing to prove so assume this is not so. Let u be defined by. ´ ( ³ e r|v(x)| / v(x) if v (x) 6= 0 A ||L|| ||L|| u (x) ≡ (18.2.20) 0 if v (x) = 0. for r ∈ (0, 1) . Now let Fn ≡ {x : |u (x)| ≤ n} ∩ {x : v (x) 6= 0}

(18.2.21)

un (x) ≡ u (x) XFn (x) .

(18.2.22)

and define I claim ||un ||A ≤ 1. It is clear that since µ (Ω) < ∞, un ∈ EA (Ω) . If ||un ||A > 1, Then for ε small enough, ||un ||A − ε > 1 and so, by convexity of A and the fact that A (0) = 0, µ ¶ Z Z |un (x)| 1 1< A dµ ≤ A (|un (x)|) dµ ||un ||A − ε ||un ||A − ε Ω Ω and so, letting ε → 0+ and using 18.1.6 and convexity of A as in the proof of the preceeding proposition, ¶ ¶ µ µ Z Z e r |v (x)| / r |v (x)| dµ ||un ||A ≤ A (|un (x)|) dµ ≤ A rA ||L|| ||L|| Ω F µ µ ¶n ¶ µ ¶ Z Z e r |v (x)| / r |v (x)| dµ ≤ r e r |v (x)| dµ ≤ r A A A ||L|| ||L|| ||L|| Fn Fn Z Z 1 r ≤ r u (x) v (x) dµ = un vdµ. (18.2.23) ||L|| Fn ||L|| Ω Now by Theorem 8.3.9 applied to the positive and negative parts of real and imaginary parts, there exists a uniformly bounded sequence of simple functions, {sk } converging uniformly to un , implying convergence in EA (Ω) , and so Z Z Lun = lim Lsk = lim sk vdµ = un vdµ. (18.2.24) k→∞

k→∞

Ω

Ω

564

ORLITZ SPACES

therefore, from 18.2.23, ||un ||A ≤

r ||L||

Z un vdµ = Ω

r r L (un ) ≤ ||L|| ||un ||A , ||L|| ||L||

which is a contradiction since r < 1. Therefore, ||un ||A ≤ 1 and from 18.2.23, Z un vdµ ||L|| ≥ ||L|| ||un ||A ≥ |Lun | = µ ¶ Ω Z e r |v (x)| dµ. ≥ ||L|| A ||L|| Fn Letting n → ∞ the monotone convergence theorem and the above imply ¶ µ Z r |v (x)| e dµ ≤ 1 A ||L|| Ω which shows that v ∈ LAe (Ω) and ||v||Ae ≤ ||L|| e ≤ r for all r ∈ (0, 1) . Therefore, ||v||A ||L|| . Since v ∈ LAe (Ω) it follows Lv = L on S and so Lv = L because S is dense in EA (Ω) . The last assertion follows from Proposition 18.1.10. This completes the proof.

Hausdorff Measure 19.1

Definition Of Hausdorff Measures

This chapter is on Hausdorff measures. First I will discuss some outer measures. In all that is done here, α (n) will be the volume of the ball in Rn which has radius 1. Definition 19.1.1 For a set, E, denote by r (E) the number which is half the diameter of E. Thus r (E) ≡

1 1 sup {|x − y| : x, y ∈ E} ≡ diam (E) 2 2

Let E ⊆ Rn . ∞ X β(s)(r (Cj ))s : E ⊆ ∪∞ Hδs (E) ≡ inf{ j=1 Cj , diam(Cj ) ≤ δ} j=1

Hs (E) ≡ lim Hδs (E). δ→0

In the above definition, β (s) is an appropriate positive constant depending on s. It will turn out that for n an integer, β (n) = α (n) where α (n) is the Lebesgue measure of the unit ball, B (0, 1) where the usual norm is used to determine this ball. Lemma 19.1.2 Hs and Hδs are outer measures. Proof: It is clear that Hs (∅) = 0 and if A ⊆ B, then Hs (A) ≤ Hs (B) with s similar assertions valid for Hδs . Suppose E = ∪∞ i=1 Ei and Hδ (Ei ) < ∞ for each i. i ∞ Let {Cj }j=1 be a covering of Ei with ∞ X

β(s)(r(Cji ))s − ε/2i < Hδs (Ei )

j=1

565

566

HAUSDORFF MEASURE

and diam(Cji ) ≤ δ. Then Hδs (E)

≤

∞ ∞ X X

β(s)(r(Cji ))s

i=1 j=1

≤

∞ X

Hδs (Ei ) + ε/2i

i=1

≤ ε+

∞ X

Hδs (Ei ).

i=1

It follows that since ε > 0 is arbitrary, Hδs (E) ≤

∞ X

Hδs (Ei )

i=1

Hδs

which shows is an outer measure. Now notice that Hδs (E) is increasing as δ → 0. Picking a sequence δ k decreasing to 0, the monotone convergence theorem implies Hs (E) ≤

∞ X

Hs (Ei ).

i=1

This proves the lemma. The outer measure Hs is called s dimensional Hausdorff measure when restricted to the σ algebra of Hs measurable sets. Next I will show the σ algebra of Hs measurable sets includes the Borel sets. This is done by the following very interesting condition known as Caratheodory’s criterion.

19.1.1

Properties Of Hausdorff Measure

Definition 19.1.3 For two sets, A, B in a metric space, we define dist (A, B) ≡ inf {d (x, y) : x ∈ A, y ∈ B} . Theorem 19.1.4 Let µ be an outer measure on the subsets of (X, d), a metric space. If µ(A ∪ B) = µ(A) + µ(B) whenever dist(A, B) > 0, then the σ algebra of measurable sets contains the Borel sets. Proof: It suffices to show that closed sets are in S, the σ-algebra of measurable sets, because then the open sets are also in S and consequently S contains the Borel sets. Let K be closed and let S be a subset of Ω. Is µ(S) ≥ µ(S ∩ K) + µ(S \ K)? It suffices to assume µ(S) < ∞. Let Kn ≡ {x : dist(x, K) ≤

1 } n

19.1. DEFINITION OF HAUSDORFF MEASURES

567

By Lemma 6.1.7 on Page 141, x → dist (x, K) is continuous and so Kn is closed. By the assumption of the theorem, µ(S) ≥ µ((S ∩ K) ∪ (S \ Kn )) = µ(S ∩ K) + µ(S \ Kn )

(19.1.1)

since S ∩ K and S \ Kn are a positive distance apart. Now µ(S \ Kn ) ≤ µ(S \ K) ≤ µ(S \ Kn ) + µ((Kn \ K) ∩ S).

(19.1.2)

If limn→∞ µ((Kn \ K) ∩ S) = 0 then the theorem will be proved because this limit along with 19.1.2 implies limn→∞ µ (S \ Kn ) = µ (S \ K) and then taking a limit in 19.1.1, µ(S) ≥ µ(S ∩ K) + µ(S \ K) as desired. Therefore, it suffices to establish this limit. Since K is closed, a point, x ∈ / K must be at a positive distance from K and so Kn \ K = ∪∞ k=n Kk \ Kk+1 . Therefore µ(S ∩ (Kn \ K)) ≤

∞ X

µ(S ∩ (Kk \ Kk+1 )).

(19.1.3)

k=n

If

∞ X

µ(S ∩ (Kk \ Kk+1 )) < ∞,

(19.1.4)

k=1

then µ(S ∩ (Kn \ K)) → 0 because it is dominated by the tail of a convergent series so it suffices to show 19.1.4. M X

µ(S ∩ (Kk \ Kk+1 )) =

k=1

X

X

µ(S ∩ (Kk \ Kk+1 )) +

k even, k≤M

µ(S ∩ (Kk \ Kk+1 )).

(19.1.5)

k odd, k≤M

By the construction, the distance between any pair of sets, S ∩ (Kk \ Kk+1 ) for different even values of k is positive and the distance between any pair of sets, S ∩ (Kk \ Kk+1 ) for different odd values of k is positive. Therefore, X X µ(S ∩ (Kk \ Kk+1 )) + µ(S ∩ (Kk \ Kk+1 )) ≤ k even, k≤M

µ(

[ k even

S ∩ (Kk \ Kk+1 )) + µ(

k odd, k≤M

[

S ∩ (Kk \ Kk+1 )) ≤ 2µ (S) < ∞

k odd

PM and so for all M, k=1 µ(S ∩ (Kk \ Kk+1 )) ≤ 2µ (S) showing 19.1.4 and proving the theorem. With the above theorem, the following theorem is easy to obtain.

568

HAUSDORFF MEASURE

Theorem 19.1.5 The σ algebra of Hs measurable sets contains the Borel sets and Hs has the property that for all E ⊆ Rn , there exists a Borel set F ⊇ E such that Hs (F ) = Hs (E). Proof: Let dist(A, B) = 2δ 0 > 0. Is it the case that Hs (A) + Hs (B) = Hs (A ∪ B)? This is what is needed to use Caratheodory’s criterion. Let {Cj }∞ j=1 be a covering of A ∪ B such that diam(Cj ) ≤ δ < δ 0 for each j and Hδs (A ∪ B) + ε >

∞ X

β(s)(r (Cj ))s.

j=1

Thus

Hδs (A ∪ B )˙ + ε >

X

β(s)(r (Cj ))s +

j∈J1

X

β(s)(r (Cj ))s

j∈J2

where J1 = {j : Cj ∩ A 6= ∅}, J2 = {j : Cj ∩ B 6= ∅}. Recall dist(A, B) = 2δ 0 , J1 ∩ J2 = ∅. It follows Hδs (A ∪ B) + ε > Hδs (A) + Hδs (B). Letting δ → 0, and noting ε > 0 was arbitrary, yields Hs (A ∪ B) ≥ Hs (A) + Hs (B). Equality holds because Hs is an outer measure. By Caratheodory’s criterion, Hs is a Borel measure. To verify the second assertion, note first there is no loss of generality in letting Hs (E) < ∞. Let E ⊆ ∪∞ j=1 Cj , r(Cj ) < δ, and Hδs (E) + δ >

∞ X

β(s)(r (Cj ))s.

j=1

Let

Fδ = ∪∞ j=1 Cj .

Thus Fδ ⊇ E and Hδs (E) ≤

Hδs (Fδ ) ≤

∞ X

¡ ¢ β(s)(r Cj )s

j=1

=

∞ X j=1

β(s)(r (Cj ))s < δ + Hδs (E).

19.2. HN AND MN

569

Let δ k → 0 and let F = ∩∞ k=1 Fδ k . Then F ⊇ E and Hδsk (E) ≤ Hδsk (F ) ≤ Hδsk (Fδ ) ≤ δ k + Hδsk (E). Letting k → ∞, Hs (E) ≤ Hs (F ) ≤ Hs (E) This proves the theorem. A measure satisfying the conclusion of Theorem 19.1.5 is called a Borel regular measure.

19.2

Hn And mn

Next I will compare Hn and mn . To do this, recall the following covering theorem which is a summary of Corollaries 10.4.5 and 10.4.4 found on Page 297. Theorem 19.2.1 Let E ⊆ Rn and let F, be a collection of balls of bounded radii such that F covers E in the sense of Vitali. Then there exists a countable collection ∞ of disjoint balls from F, {Bj }∞ j=1 , such that mn (E \ ∪j=1 Bj ) = 0. In the next lemma, the balls are the usual balls taken with respect to the usual distance in Rn . Lemma 19.2.2 If mn (S) = 0 then Hn (S) = Hδn (S) = 0. Also, there exists a constant, k such that Hn (E) ≤ kmn (E) for all E Borel. Also, if Q0 ≡ [0, 1)n , the unit cube, then Hn ([0, 1)n ) > 0. Proof: Suppose first mn (S) = 0. Without loss of generality, S is bounded. Then by outer regularity, there exists a bounded open V containing S ³ and m ´ n (V ) < c c ε. For each x ∈ S, there exists a ball Bx such that Bx ⊆ V and δ > r Bx . By the n o ck Vitali covering theorem there is a sequence of disjoint balls {Bk } such that B n covers S. Then letting α (n) be the Lebesgue measure of the unit ball in R Hδn (S) ≤

X k

≤

³ ń X n ck = β (n) 5n α (n) r (Bk ) β (n) r B α (n) k

β (n) n β (n) n 5 mn (V ) < 5 ε α (n) α (n)

Since ε is arbitrary, this shows Hδn (S) = 0 and now it follows Hn (S) = 0. Letting U be an open set and δ > 0, consider all balls, B contained in U which have diameters less than δ. This is a Vitali covering of U and therefore by Theorem 19.2.1, there exists {Bi } , a sequence of disjoint balls of radii less than δ contained

570

HAUSDORFF MEASURE

in U such that ∪∞ i=1 Bi differs from U by a set of Lebesgue measure zero. Let α (n) be the Lebesgue measure of the unit ball in Rn . Then from what was just shown, Hδn (U )

= Hδn (∪i Bi ) ≤

∞ X

∞

n

β (n) r (Bi ) =

i=1

=

∞ β (n) X

α (n)

mn (Bi ) =

i=1

β (n) X n α (n) r (Bi ) α (n) i=1

β (n) mn (U ) ≡ kmn (U ) . α (n)

Now letting E be Borel, it follows from the outer regularity of mn there exists a decreasing sequence of open sets, {Vi } containing E such such that mn (Vi ) → mn (E) . Then from the above, Hδn (E) ≤ lim Hδn (Vi ) ≤ lim kmn (Vi ) = kmn (E) . i→∞

i→∞

Since δ > 0 is arbitrary, it follows that also Hn (E) ≤ kmn (E) . This proves the first part of the lemma. To verify the second part, note that it is obvious Hδn and Hn are translation invariant because diameters of sets do not change when translated. Therefore, if Hn ([0, 1)n ) = 0, it follows Hn (Rn ) = 0 because Rn is the countable union of translates of Q0 ≡ [0, 1)n . Since each Hδn is no larger than Hn , the same must hold for Hδn . Therefore, there exists a sequence of sets, {Ci } each having diameter less than δ such that the union of these sets equals Rn but 1>

∞ X

n

β (n) r (Ci ) .

i=1

Now let Bi be a ball having radius equal to diam (Ci ) = 2r (Ci ) which contains Ci . It follows α (n) 2n n n mn (Bi ) = α (n) 2n r (Ci ) = β (n) r (Ci ) β (n) which implies 1>

∞ X

n

β (n) r (Ci ) =

i=1

∞ X β (n) mn (Bi ) = ∞, α (n) 2n i=1

a contradiction. This proves the lemma. Theorem 19.2.3 By choosing β (n) properly, one can obtain Hn = mn on all Lebesgue measurable sets. Proof: I will show Hn is a positive multiple of mn for any choice of β (n) . Define k=

mn (Q0 ) Hn (Q0 )

19.3. TECHNICAL CONSIDERATIONS

571

where Q0 = [0, 1)n is the half open unit cube in Rn . I will show kHn (E) = mn (E) for any Lebesgue measurable set. When this is done, it will follow that by adjusting β (n) the multiple Qn can be taken to be 1. Let Q¡= ¢ i=1 [ai , ai + 2−k ) be a half open box where ai = l2−k . Thus Q0 is the n union of 2k of these identical half open boxes. By translation invariance, of Hn and mn ¡ k ¢n n 1 1 ¡ k ¢n 2 H (Q) = Hn (Q0 ) = mn (Q0 ) = 2 mn (Q) . k k Therefore, kHn (Q) = mn (Q) for any such half open box and by translation invariance, for the translation of any such half open box. It follows from Lemma 10.1.2 that kHn (U ) = mn (U ) for all open sets. It follows immediately, since every compact set is the countable intersection of open sets that kHn = mn on compact sets. Therefore, they are also equal on all closed sets because every closed set is the countable union of compact sets. Now let F be an arbitrary Lebesgue measurable set. I will show that F is Hn measurable and that kHn (F ) = mn (F ). Let Fl = B (0, l) ∩ F. Then there exists H a countable union of compact sets and G a countable intersection of open sets such that H ⊆ Fl ⊆ G

(19.2.6)

and mn (G \ H) = 0 which implies by Lemma 19.2.2 mn (G \ H) = kHn (G \ H) = 0.

(19.2.7)

To do this, let {Gi } be a decreasing sequence of bounded open sets containing Fl and let {Hi } be an increasing sequence of compact sets contained in Fl such that kHn (Gi \ Hi ) = mn (Gi \ Hi ) < 2−i Then letting G = ∩i Gi and H = ∪i Hi this establishes 19.2.6 and 19.2.7. Then by completeness of Hn it follows Fl is Hn measurable and kHn (Fl ) = kHn (H) = mn (H) = mn (Fl ) . Now taking l → ∞, it follows F is Hn measurable and kHn (F ) = mn (F ). Therefore, adjusting β (n) it can be assumed the constant, k is 1. This proves the theorem. The exact determination of β (n) is more technical.

19.3

Technical Considerations

Let α(n) be the volume of the unit ball in Rn . Thus the volume of B(0, r) in Rn is α(n)rn from the change of variables formula. There is a very important and interesting inequality known as the isodiametric inequality which says that if A is any set in Rn , then n

m(A) ≤ α(n)(2−1 diam(A))n = α (n) r (A) .

572

HAUSDORFF MEASURE

This inequality may seem obvious at first but it is not really. The reason it is not is that there are sets which are not subsets of any sphere having the same diameter as the set. For example, consider an equilateral triangle. Lemma 19.3.1 Let f : Rn−1 → [0, ∞) be Borel measurable and let S = {(x,y) :|y| < f (x)}. Then S is a Borel set in Rn . Proof: Set sk be an increasing sequence of Borel measurable functions converging pointwise to f . Nk X sk (x) = ckm XEm k (x). m=1

Let k k k k Sk = ∪N m=1 Em × (−cm , cm ).

Then (x,y) ∈ Sk if and only if f (x) > 0 and |y| < sk (x) ≤ f (x). It follows that Sk ⊆ Sk+1 and S = ∪∞ k=1 Sk . But each Sk is a Borel set and so S is also a Borel set. This proves the lemma. Let Pi be the projection onto span (e1 , · · ·, ei−1 , ei+1 , · · · , en ) where the ek are the standard basis vectors inP Rn , ek being the vector having a 1 th in the k slot and a 0 elsewhere. Thus Pi x ≡ j6=i xj ej . Also let APi x ≡ {xi : (x1 , · · · , xi , · · · , xn ) ∈ A}

APi x

¡@ ¡ @ ¡x @ Q ´ Q ´ Q´ Pi x ∈ span{e1 , · · ·, ei−1 ei+1 , · · ·, en }.

Lemma 19.3.2 Let A ⊆ Rn be a Borel set. Then Pi x → m(APi x ) is a Borel measurable function defined on Pi (Rn ). Qn Proof: Let K be the π system consisting of sets of the form j=1 Aj where Ai is Borel. Also let G denote those Borel sets of Rn such that if A ∈ G then Pi x → m((A ∩ Rk )Pi x ) is Borel measurable.


573

where Rk = (−k, k)n . Thus K ∈ G. If A ∈ G ³¡ ¢ ´ Pi x → m AC ∩ Rk Pi x is Borel measurable because it is of the form ¡ ¢ ¡ ¢ m (Rk )Pi x − m (A ∩ Rk )Pi x and these are Borel measurable functions of Pi x. Also, if {Ai } is a disjoint sequence of sets in G then ¢ ¢ X ¡ ¡ m (∪i Ai ∩ Rk )Pi x = m (Ai ∩ Rk )Pi x i

and each function of Pi x is Borel measurable. Thus by the lemma on π systems, Lemma 9.11.3, G = B (Rn ) and this proves the lemma. Now let A ⊆ Rn be Borel. Let Pi be the projection onto ¡ ¢ span e1 , · · · , ei−1 , ei+1 , · · · , en and as just described, APi x = {y ∈ R : Pi x + yei ∈ A} Thus for x = (x1 , · · · , xn ), APi x = {y ∈ R : (x1 , · · · , xi−1 , y, xi+1 , · · · , xn ) ∈ A}. Since A is Borel, it follows from Lemma 19.3.1 that Pi x → m(APi x ) is a Borel measurable function on Pi Rn = Rn−1.

19.3.1

Steiner Symmetrization

Define S(A, ei ) ≡ {x =Pi x + yei : |y| < 2−1 m(APi x )} Lemma 19.3.3 Let A be a Borel subset of Rn . Then S(A, ei ) satisfies Pi x + yei ∈ S(A, ei ) if and only if Pi x − yei ∈ S(A, ei ), S(A, ei ) is a Borel set in Rn, mn (S(A, ei )) = mn (A),

(19.3.8)

diam(S(A, ei )) ≤ diam(A).

(19.3.9)

574

HAUSDORFF MEASURE

Proof : The first assertion is obvious from the definition. The Borel measurability of S(A, ei ) follows from the definition and Lemmas 19.3.2 and 19.3.1. To show Formula 19.3.8, Z mn (S(A, ei )) = Z

Z Pi R n

= Pi R n

2−1 m(APi x )

−2−1 m(APi x )

dxi dx1 · · · dxi−1 dxi+1 · · · dxn

m(APi x )dx1 · · · dxi−1 dxi+1 · · · dxn

= m(A). Now suppose x1 and x2 ∈ S(A, ei ) x1 = Pi x1 + y1 ei , x2 = Pi x2 + y2 ei . For x ∈ A define l(x) = sup{y : Pi x+yei ∈ A}. g(x) = inf{y : Pi x+yei ∈ A}. Then it is clear that l(x1 ) − g(x1 ) ≥ m(APi x1 ) ≥ 2|y1 |,

(19.3.10)

l(x2 ) − g(x2 ) ≥ m(APi x2 ) ≥ 2|y2 |.

(19.3.11)

Claim: |y1 − y2 | ≤ |l(x1 ) − g(x2 )| or |y1 − y2 | ≤ |l(x2 ) − g(x1 )|. Proof of Claim: If not, 2|y1 − y2 | > |l(x1 ) − g(x2 )| + |l(x2 ) − g(x1 )| ≥ |l(x1 ) − g(x1 ) + l(x2 ) − g(x2 )| = l(x1 ) − g(x1 ) + l(x2 ) − g(x2 ). ≥ 2 |y1 | + 2 |y2 | by 19.3.10 and 19.3.11 contradicting the triangle inequality. Now suppose |y1 − y2 | ≤ |l(x1 ) − g(x2 )|. From the claim, |x1 − x2 |

= (|Pi x1 − Pi x2 |2 + |y1 − y2 |2 )1/2 ≤ (|Pi x1 − Pi x2 |2 + |l(x1 ) − g(x2 )|2 )1/2 ≤ (|Pi x1 − Pi x2 |2 + (|z1 − z2 | + 2ε)2 )1/2 √ ≤ diam(A) + O( ε)

where z1 and z2 are such that Pi x1 + z1 ei ∈ A, Pi x2 + z2 ei ∈ A, and |z1 − l(x1 )| < ε and |z2 − g(x2 )| < ε. If |y1 − y2 | ≤ |l(x2 ) − g(x1 )|, then we use the same argument but let |z1 − g(x1 )| < ε and |z2 − l(x2 )| < ε,


575

Since x1 , x2 are arbitrary elements of S(A, ei ) and ε is arbitrary, this proves 19.3.9. The next lemma says that if A is already symmetric with respect to the j th direction, then this symmetry is not destroyed by taking S (A, ei ). Lemma 19.3.4 Suppose A is a Borel set in Rn such that Pj x + ej xj ∈ A if and only if Pj x+(−xj )ej ∈ A. Then if i 6= j, Pj x + ej xj ∈ S(A, ei ) if and only if Pj x+(−xj )ej ∈ S(A, ei ). Proof : By definition, Pj x + ej xj ∈ S(A, ei ) if and only if

|xi | < 2−1 m(APi (Pj x+ej xj ) ).

Now xi ∈ APi (Pj x+ej xj ) if and only if xi ∈ APi (Pj x+(−xj )ej ) by the assumption on A which says that A is symmetric in the ej direction. Hence Pj x + ej xj ∈ S(A, ei ) if and only if

|xi | < 2−1 m(APi (Pj x+(−xj )ej ) )

if and only if Pj x+(−xj )ej ∈ S(A, ei ). This proves the lemma.

19.3.2

The Isodiametric Inequality

The next theorem is called the isodiametric inequality. It is the key result used to compare Lebesgue and Hausdorff measures. Theorem 19.3.5 Let A be any Lebesgue measurable set in Rn . Then mn (A) ≤ α(n)(r (A))n. Proof: Suppose first that A is Borel. Let A1 = S(A, e1 ) and let Ak = S(Ak−1 , ek ). Then by the preceding lemmas, An is a Borel set, diam(An ) ≤ diam(A), mn (An ) = mn (A), and An is symmetric. Thus x ∈ An if and only if −x ∈ An . It follows that An ⊆ B(0, r (An )). (If x ∈ An \ B(0, r (An )), then −x ∈ An \ B(0, r (An )) and so diam (An ) ≥ 2|x| > diam(An ).) Therefore, mn (An ) ≤ α(n)(r (An ))n ≤ α(n)(r (A))n .

576

HAUSDORFF MEASURE

It remains to establish this inequality for arbitrary measurable sets. Letting A be such a set, let {Kn } be an increasing sequence of compact subsets of A such that m(A) = lim m(Kk ). k→∞

Then lim m(Kk ) ≤ lim sup α(n)(r (Kk ))n

m(A) =

k→∞

k→∞

α(n)(r (A))n.

≤ This proves the theorem.

19.4

The Proper Value Of β (n)

I will show that the proper determination of β (n) is α (n), the volume of the unit ball. Since β (n) has been adjusted such that k = 1, mn (B (0, 1)) = Hn (B (0, 1)). ∞ There exists a covering of B (0,1) of sets of radii less than δ, {Ci }i=1 such that X n Hδn (B (0, 1)) + ε > β (n) r (Ci ) i

Then by Theorem 19.3.5, the isodiametric inequality, Hδn (B (0, 1)) + ε >

X

n

β (n) r (Ci ) =

i

≥

¡ ¢n β (n) X α (n) r C i α (n) i

¡ ¢ β (n) β (n) X β (n) n mn C i ≥ mn (B (0, 1)) = H (B (0, 1)) α (n) i α (n) α (n)

Now taking the limit as δ → 0, Hn (B (0, 1)) + ε ≥

β (n) n H (B (0, 1)) α (n)

and since ε > 0 is arbitrary, this shows α (n) ≥ β (n). By the Vitali covering theorem, there exists a sequence of disjoint balls, {Bi } n such that B (0, 1) = (∪∞ i=1 Bi )∪N where mn (N ) = 0. Then Hδ (N ) = 0 can be conn n cluded because Hδ ≤ H and Lemma 19.2.2. Using mn (B (0, 1)) = Hn (B (0, 1)) again, Hδn (B (0, 1)) =

Hδn (∪i Bi ) ≤

∞ X

n

β (n) r (Bi )

i=1

= =

∞ β (n) X

α (n)

i=1

∞

n

α (n) r (Bi ) =

β (n) X mn (Bi ) α (n) i=1

β (n) β (n) β (n) n mn (∪i Bi ) = mn (B (0, 1)) = H (B (0, 1)) α (n) α (n) α (n)

19.4. THE PROPER VALUE OF β (N )

577

which implies α (n) ≤ β (n) and so the two are equal. This proves that if α (n) = β (n) , then the Hn = mn on the measurable sets of Rn . This gives another way to think of Lebesgue measure which is a particularly nice way because it is coordinate free, depending only on the notion of distance. For s < n, note that Hs is not a Radon measure because it will not generally be finite on compact sets. For example, let n = 2 and consider H1 (L) where L is a line segment joining (0, 0) to (1, 0). Then H1 (L) is no smaller than H1 (L) when L is considered a subset of R1 , n = 1. Thus by what was just shown, H1 (L) ≥ 1. Hence H1 ([0, 1] × [0, 1]) = ∞. The situation is this: L is a one-dimensional object inside R2 and H1 is giving a one-dimensional measure of this object. In fact, Hausdorff measures can make such heuristic remarks as these precise. Define the Hausdorff dimension of a set, A, as dim(A) = inf{s : Hs (A) = 0}

19.4.1

A Formula For α (n)

What is α(n)? Recall the gamma function which makes sense for all p > 0. Z Γ (p) ≡

∞

e−t tp−1 dt.

0

Lemma 19.4.1 The following identities hold. pΓ(p) = Γ(p + 1), µZ Γ(p)Γ(q) =

1

¶ xp−1 (1 − x)q−1 dx Γ(p + q),

0

Γ

µ ¶ √ 1 = π 2

Proof: Using integration by parts, Z Γ (p + 1)

= 0

=

∞

Z e−t tp dt = −e−t tp |∞ 0 +p

pΓ (p)

0

∞

e−t tp−1 dt

578

HAUSDORFF MEASURE

Next Z

Z ∞ e−t tp−1 dt e−s sq−1 ds 0 0 Z ∞Z ∞ e−(t+s) tp−1 sq−1 dtds = 0 0 Z ∞Z ∞ p−1 q−1 = e−u (u − s) s duds Z0 ∞ Zs u p−1 q−1 = e−u (u − s) s dsdu

Γ (p) Γ (q)

∞

=

Z

0

0

∞

Z

1

= Z

0

p−1

e−u (u − ux)

q−1

(ux)

udxdu

0

∞Z

1

p−1

e−u up+q−1 (1 − x) xq−1 dxdu 0 0 µZ 1 ¶ p−1 q−1 = Γ (p + q) x (1 − x) dx .

=

0

It remains to find Γ

¡1¢ 2

.

µ ¶ Z ∞ Z ∞ Z ∞ 2 1 −t −1/2 −u2 1 Γ = 2udu = 2 e t dt = e−u du e 2 u 0 0 0 Now µZ

∞

e

−x2

¶2 dx

Z

Z

∞

=

e

0

Z

0 ∞

−x2

and so Γ

dx

e

−y 2

Z

π/2

2

e−r rdθdr =

0

∞

Z

dy =

0

= 0

Z

∞

0

∞

e−(x

2

+y 2 )

dxdy

0

1 π 4

µ ¶ Z ∞ √ 2 1 =2 e−u du = π 2 0

This proves the lemma. Next let n be a positive integer. Theorem 19.4.2 α(n) = π n/2 (Γ(n/2 + 1))−1 where Γ(s) is the gamma function Z ∞ Γ(s) = e−t ts−1 dt. 0

Proof: First let n = 1. 3 1 Γ( ) = Γ 2 2

µ ¶ √ 1 π = . 2 2

19.4. THE PROPER VALUE OF β (N ) Thus

579

2 √ π 1/2 (Γ(1/2 + 1))−1 = √ π = 2 = α (1) . π

and this shows the theorem is true if n = 1. Assume the theorem is true for n and let Bn+1 be the unit ball in Rn+1 . Then by the result in Rn , Z 1 mn+1 (Bn+1 ) = α(n)(1 − x2n+1 )n/2 dxn+1 −1

Z

1

= 2α(n)

(1 − t2 )n/2 dt.

0

Doing an integration by parts and using Lemma 19.4.1 Z 1 = 2α(n)n t2 (1 − t2 )(n−2)/2 dt 0

Z 1 1 1/2 = 2α(n)n u (1 − u)n/2−1 du 2 0 Z 1 = nα(n) u3/2−1 (1 − u)n/2−1 du 0

= nα(n)Γ(3/2)Γ(n/2)(Γ((n + 3)/2))−1 = nπ n/2 (Γ(n/2 + 1))−1 (Γ((n + 3)/2))−1 Γ(3/2)Γ(n/2) = nπ n/2 (Γ(n/2)(n/2))−1 (Γ((n + 1)/2 + 1))−1 Γ(3/2)Γ(n/2) = 2π n/2 Γ(3/2)(Γ((n + 1)/2 + 1))−1 = π (n+1)/2 (Γ((n + 1)/2 + 1))−1 . This proves the theorem. From now on, in the definition of Hausdorff measure, it will always be the case that β (s) = α (s) . As shown above, this is the right thing to have β (s) to equal if s is a positive integer because this yields the important result that Hausdorff measure is the same as Lebesgue measure. Note the formula, π s/2 (Γ(s/2+1))−1 makes sense for any s ≥ 0.

19.4.2

Hausdorff Measure And Linear Transformations

Hausdorff measure makes possible a unified development of n dimensional area. As in the case of Lebesgue measure, the first step in this is to understand ¡ ¢basic considerations related to linear transformations. Recall that for L ∈ L Rk , Rl , L∗ is defined by (Lu, v) = (u, L∗ v) . Also recall Theorem 4.12.6 on Page 93 which is stated here for convenience. This theorem says you can write a linear transformation as the composition of two linear transformations, one which preserves length and the other which distorts, the right

580

HAUSDORFF MEASURE

polar decomposition. The one which distorts is the one which will have a nontrivial interaction with Hausdorff measure while the one which preserves lengths does not change Hausdorff measure. These ideas are behind the following theorems and lemmas. Theorem 19.4.3 Let F be an n × m matrix where m ≥ n. Then there exists an m × n matrix R and a n × n matrix U such that F = RU, U = U ∗ , all eigenvalues of U are non negative, U 2 = F ∗ F, R∗ R = I, and |Rx| = |x|. Lemma 19.4.4 Let R ∈ L(Rn , Rm ), n ≤ m, and R∗ R = I. Then if A ⊆ Rn , Hn (RA) = Hn (A). In fact, if P : Rn → Rm satisfies |P x − P y| = |x − y| , then Hn (P A) = Hn (A) . Proof: Note that 2

2

|R(x − y)| = (R (x − y) , R (x − y)) = (R∗ R (x − y) , x − y) = |x − y|

Thus R preserves lengths. Now let P be an arbitrary mapping which preserves lengths and let A be bounded, P (A) ⊆ ∪∞ j=1 Cj , r(Cj ) < δ, and Hδn (P A) + ε >

∞ X

α(n)(r(Cj ))n .

j=1

Since P preserves lengths, it follows P is one to one on P (Rn ) and P −1 also preserves lengths on P (Rn ) . Replacing each Cj with Cj ∩ (P A), Hδn (P A) + ε

> =

∞ X j=1 ∞ X

α(n)r(Cj ∩ (P A))n ¡ ¢n α(n)r P −1 (Cj ∩ (P A))

j=1

≥ Hδn (A). Thus Hδn (P A) ≥ Hδn (A).

19.5. THE AREA FORMULA

581

Now let A ⊆ ∪∞ j=1 Cj , diam(Cj ) ≤ δ, and Hδn (A) + ε ≥

∞ X

n

α(n) (r (Cj ))

j=1

Then Hδn (A)

+ε ≥

∞ X

α(n) (r (Cj ))

n

j=1

= ≥

∞ X

α(n) (r (P Cj ))

n

j=1 Hδn (P A).

Hence Hδn (P A) = Hδn (A). Letting δ → 0 yields the desired conclusion in the case where A is bounded. For the general case, let Ar = A ∩ B (0, r). Then Hn (P Ar ) = Hn (Ar ). Now let r → ∞. This proves the lemma. Lemma 19.4.5 Let F ∈ L(Rn , Rm ), n ≤ m, and let F = RU where R and U are described in Theorem 4.12.6 on Page 93. Then if A ⊆ Rn is Lebesgue measurable, Hn (F A) = det(U )mn (A). Proof: Using Theorem 10.5.7 on Page 301 and Theorem 19.2.3, Hn (F A) = Hn (RU A) = Hn (U A) = mn (U A) = det(U )mn (A). Definition 19.4.6 Define J to equal det(U ). Thus J = det((F ∗ F )1/2 ) = (det(F ∗ F ))1/2.

19.5

The Area Formula

19.5.1

Preliminary Results

It was shown in Lemma 19.4.5 that Hn (F A) = det(U )mn (A) where F = RU with R preserving distances and U a symmetric matrix having all positive eigenvalues. The area formula gives a generalization of this simple relationship to the case where F is replaced by a nonlinear mapping, h. It contains as a special case the earlier change of variables formula. There are two parts to this development. The first part is to generalize Lemma 19.4.5 to the case of nonlinear maps. When this is done, the area formula can be presented.

582

HAUSDORFF MEASURE

In this section, U will be an open set in Rn on which h is defined and A ⊆ U will be a Lebesgue measurable set. Assume m ≥ n and h : U → Rm is continuous,

(19.5.12)

Dh (x) exists for all x ∈ A,

(19.5.13)

Also assume that for every x ∈ A, there exists Rx and Lx such that for all y, z ∈ B (x, Rx ) , |h (z) − h (y)| ≤ Lx |x − y| (19.5.14) This last condition is weaker than saying h is Lipschitz. Instead, it is an assumption that h is locally Lipshitz, the Lipschitz constant depending on the point considered. An interesting case in which this would hold would be when h is differentiable on U and ||Dh (x)|| is uniformly bounded near each point x. Actually, it is the case that 19.5.14 will suffice to obtain 19.5.13 on all but a subset of measure zero of A but this has not been shown yet. Also, the condition 19.5.12 is redundant because you can simply replace U with the union of the sets B (x, Rx ) for x ∈ A. I think it is easiest to retain this condition because differentiability is defined for functions whose domains are open sets. To make this more formal, here is a definition. Definition 19.5.1 Let h be defined in some open set containing a set, A. Then h is locally Lipschitz on A if for every x ∈ A there exits Rx > 0 and a constant, Lx such that whenever y, z ∈ B (x, Rx ) , |h (z) − h (y)| ≤ Lx |z − y| . Lemma 19.5.2 If T ⊆ A and mn (T ) = 0, then Hn (h (T )) = 0. Proof: Let Tk ≡ {x ∈ T : ||Dh (x)|| < k} . Thus T = ∪k Tk . I will show h (Tk ) has Hn measure zero and then it will follow that h (T ) = ∪∞ k=1 h (Tk ) must also have measure zero. Let ε > 0 be given. By outer regularity, there exists an open set, V , containing Tk which is contained in U such that mn (V ) < knε6n . For x ∈ Tk it follows from differentiability, h (x + v) = h (x) + Dh (x) v + o (v) and so whenever rx is small enough, B (x,5rx ) ⊆ V and whenever |v| < 5rx , |o (v)| < krx . Therefore, if |v| < 5rx , Dh (x) v + o (v) ∈ B (0, 5krx ) + B (0,krx ) ⊆ B (0, 6krx ) and so h (B (x, 5rx )) ⊆ B (h (x) , 6krx ).


583

Letting δ > 0 be given, the Vitali covering theorem implies there exists a sequence of disjoint balls {Bi }, Bi =n B o (xi , rxi ), which are contained in V such that the bi , having the same center but 5 times the radius, sequence of enlarged balls, B covers Tk and 6krxi < δ. Then ³ ³ ´´ b Hδn (h (Tk )) ≤ Hδn h ∪∞ i=1 Bi ≤

∞ X

∞ ³ ³ ´´ X bi Hδn h B ≤ Hδn (B (h (xi ) , 6krxi ))

i=1

≤

i=1 ∞ X

n

n

α (n) (6krxi ) = (6k)

i=1

∞ X

α (n) rxni

i=1

∞ X n = (6k) mn (B (xi , rxi )) i=1

≤

n

n

(6k) mn (V ) ≤ (6k)

ε = ε. k n 6n

Since ε > 0 is arbitrary, this shows Hδn (h (Tk )) = 0. Since δ is arbitrary, this implies Hn (h (Tk )) = 0. Now Hn (h (T )) = lim Hn (h (Tk )) = 0. k→∞

This proves the lemma. Lemma 19.5.3 If S is a Lebesgue measurable subset of A, then h (S) is Hn measurable. Proof: Let Sk = S ∩ B (0, k) , k ∈ N. By inner regularity of Lebesgue measure, there exists a set, F , which is the countable union of compact sets and a set T with mn (T ) = 0 such that F ∪ T = Sk . Then h (F ) ⊆ h (Sk ) ⊆ h (F ) ∪ h (T ). By continuity of h, h (F ) is a countable union of compact sets and so it is Borel. By Lemma 19.5.2, Hn (h (T )) = 0 and so h (Sk ) is Hn measurable because of completeness of Hausdorff measure, which comes from Hn being obtained from an outer measure. Now h (S) = ∪∞ k=1 h (Sk ) and so it is also true that h (S) is Hn measurable. This proves the lemma. The following lemma, depending on the Brouwer fixed point theorem and found in Rudin [58], will be important for the following arguments. The idea is that if a continuous function mapping a ball in Rk to Rk doesn’t move any point very much, then the image of the ball must contain a slightly smaller ball. Lemma 19.5.4 Let B = B (0, r), a ball in Rk and let F : B → Rk be continuous and suppose for some ε < 1, |F (v) −v| < εr (19.5.15)

584

HAUSDORFF MEASURE

for all v ∈ B. Then F (B) ⊇ B (0, r (1 − ε)) . Proof: Suppose a ∈ B (0, r (1 − ε)) \ F (B) . I claim that a 6= F (v) for all v ∈ B. Here is why. By assumption, if F (v) = a, then |v| = r and so |F (v) − v| = |a − v| ≥ |v| − |a| > r − r (1 − ε) = rε, a contradiction to 19.5.15. Now letting G :B → B, be defined by G (v) ≡

r (a − F (v)) , |a − F (v)|

it follows G is continuous. Then by the Brouwer fixed point theorem, G (v) = v for some v ∈ B. Using the formula for G, it follows |v| = r. Taking the inner product with v, r 2 (G (v) , v) = |v| = r2 = (a − F (v) , v) |a − F (v)| r = (a − v + v − F (v) , v) |a − F (v)| r = [(a − v, v) + (v − F (v) , v)] |a − F (v)| h i r 2 = (a, v) − |v| + (v − F (v) , v) |a − F (v)| £ 2 ¤ r ≤ r (1 − ε) − r2 +r2 ε = 0, |a − F (v)| a contradiction to |v| = r. Therefore, B (0, r (1 − ε)) \ F (B) = ∅ and this proves the lemma. By Theorem 4.12.6 on Page 93, when Dh (x) exists, Dh (x) = R (x) U (x) where (U (x) u, v) = (U (x) v, u) , (U (x) u, u) ≥ 0 and R∗ R = I. Lemma 19.5.5 In this situation, |R∗ u| ≤ |u|. Proof: First note that (u−RR∗ u,RR∗ u)

=

(u,RR∗ u) − |RR∗ u| 2

2

2

= |R∗ u| − |R∗ u| = 0, and so 2

|u|


2

=

|u−RR∗ u+RR∗ u|

=

|u−RR∗ u| + |RR∗ u|

=

|u−RR∗ u| + |R∗ u| .

2 2

2

2


585

Lemma 19.5.6 If |P x − P y| ≤ L |x − y| , then for E a set, Hn (P E) ≤ Ln Hn (E) . Proof: Without loss of generality, assume Hn (E) < ∞. Let δ > 0 and let ∞ {Ci }i=1 be a covering of E such that r (Ci ) ≤ δ for each i and ∞ X

n

α (n) r (Ci ) ≤ Hδn (E) + ε.

i=1

Then

∞ {P Ci }i=1

is a covering of P E such that r (P Ci ) ≤ Lδ. Therefore,

n HLδ (P E)

≤

∞ X

n

α (n) r (P Ci )

i=1

≤ Ln

∞ X

n

α (n) r (Ci ) ≤ Ln Hδn (E) + Ln ε

i=1

≤ Ln Hn (E) + Ln ε. Letting δ → 0,

Hn (P E) ≤ Ln Hn (E) + Ln ε

and since ε > 0 is arbitrary, this proves the Lemma. Then the following corollary follows from Lemma 19.5.5. Corollary 19.5.7 Let T ⊆ Rm . Then Hn (T ) ≥ Hn (RR∗ T ) = Hn (R∗ T ). Definition 19.5.8 Let E be a Lebesgue measurable set. x ∈ E is a point of density if mn (E ∩ B(x, r)) lim = 1. r→0 mn (B(x, r)) Recall that from the fundamental theorem of calculus applied to XE almost every point of E is a point of density. −1

Lemma 19.5.9 Let x ∈ A be a point where U (x) following hold for all r small enough.

exists. Then if ε ∈ (0, 1) the

h (B (x, r)) ⊆ h (x) + R (x) U (x) B (0, r (1 + ε)), Hn (h (B (x,r))) ≤ mn (U (x) B (0, r (1 + ε))). ∗

∗

R (x) h (B (x, r)) ⊇ R (x) h (x) + U (x) B (0, r (1 − ε)), n

H (h (B (x,r))) ≥ mn (U (x) B (0, r (1 − ε))),

(19.5.16) (19.5.17) (19.5.18) (19.5.19)

If x is also a point of density of A, then Hn (h (B (x, r) ∩ A)) = 1. r→0 Hn (h (B (x, r))) lim

(19.5.20)

586

HAUSDORFF MEASURE

Proof: Since Dh (x) exists, h (x + v) = h (x) + Dh (x) v+o (|v|).

(19.5.21)

Consequently, when r is small enough, 19.5.16 holds. Using the fact R (x) preserves all distances, and Theorem 19.2.3 which says Hn = mn on the Borel sets of Rn implies, Hn (h (B (x,r))) ≤ Hn (R (x) U (x) B (0, r (1 + ε))) = Hn (U (x) B (0, r (1 + ε))) = mn (U (x) B (0, r (1 + ε))) which shows 19.5.17. From 19.5.21, R∗ (x) h (x + v) = R∗ (x) h (x) + U (x) (v+o (|v|)). Thus, from the assumption that U (x) −1

F (v) ≡ U (x)

−1

exists and letting F (v) be given by

R∗ (x) h (x + v) − U (x)

−1

R∗ (x) h (x)

(19.5.22)

It follows F (v) − v = o (|v|) and so Lemma 19.5.4 implies that for all r small enough, F (B (0, r)) ≡ ⊇

−1

U (x) R∗ (x) h (x+B (0,r)) − U (x) B (0, (1 − ε) r).

−1

R∗ (x) h (x)

Therefore, R∗ (x) h (B (x,r)) ⊇ R∗ (x) h (x) + U (x) B (0, (1 − ε) r) which proves 19.5.18. Therefore, R (x) R∗ (x) h (B (x, r)) ⊇ R (x) R∗ (x) h (x) + R (x) U (x) B (0, r (1 − ε)). From Lemma 19.5.7, this implies Hn (h (B (x,r))) ≥ =

Hn (R∗ (x) h (B (x, r))) Hn (R (x) R∗ (x) h (B (x, r)))

≥ Hn (R (x) U (x) B (0, r (1 − ε))) = Hn (U (x) B (0, r (1 − ε))) = mn (U (x) B (0, r (1 − ε))) which shows 19.5.19.


587

Now suppose that x is also a point of density of A. Then whenever r is small enough, mn (A ∩ B (x, r)) > 1 − ε. mn (B (x, r)) Consequently, for such r, 1

= >

and so

mn (A ∩ B (x, r)) mn (B (x,r) \ A) + mn (B (x, r)) mn (B (x, r)) mn (B (x,r) \ A) 1−ε+ α (n) rn mn (B (x,r) \ A) < εα (n) rn.

(19.5.23)

Also, h (B (x, r) ∩ A) ∪ h (B (x, r) \ A) = h (B (x, r)) and so ≥

Hn (h (B (x, r) ∩ A)) + Hn (h (B (x, r) \ A)) Hn (h (B (x, r)))

Then also letting r also be smaller than Rx mentioned in 19.5.14, it follows from Lemmas 19.5.6, 19.5.3, and 19.5.23, 19.5.18 that 1

Hn (h (B (x, r) ∩ A)) Hn (h (B (x, r))) Hn (h (B (x, r))) − Hn (h (B (x,r) \ A)) ≥ Hn (h (B (x, r))) Ln mn (B (x,r) \ A) ≥ 1− nx ∗ H (R (x) h (B (x, r))) Lnx mn (B (x,r) \ A) ≥ 1− mn (U (x) B (0, r (1 − ε))) Lnx εα (n) rn ≥ 1− n det (U (x)) α (n) rn (1 − ε) = 1 − g (ε) ≥

where limε→0 g (ε) = 0. Since ε is arbitrary, this proves 19.5.20. The next theorem is the generalization of Lemma 19.4.5 to nonlinear maps. Theorem 19.5.10 Let h : U → Rm where U is an open set in Rn for n ≤ m and suppose h is locally Lipschitz at every point of a Lebesgue measurable subset, A of U. Also suppose that for every x ∈ A, Dh (x) exists. Then for x ∈ A, Hn (h (B (x, r))) , r→0 mn (B (x,r)) ¡ ¢1/2 ∗ . where J (x) ≡ det (U (x)) = det Dh (x) Dh (x) J (x) = lim

(19.5.24)

588

HAUSDORFF MEASURE −1

Proof: Suppose first that U (x) of variables formula for linear maps,

exists. Using 19.5.19, 19.5.17 and the change

n

J (x) (1 − ε) Hn (h(B (x, r))) mn (U (x) B (0,r (1 − ε))) ≤ = mn (B (x, r)) mn (B (x, r)) mn (U (x) B (0,r (1 + ε))) n ≤ = J (x) (1 + ε) mn (B (x, r)) whenever r is small enough. It follows that since ε > 0 is arbitrary, 19.5.24 holds. −1 Now suppose U (x) does not exist. The first part shows that the conclusion of the theorem holds when J (x) 6= 0. I will apply this to a modified function. Let k : Rn → Rm × Rn be defined as

µ k (x) ≡

Then

h (x) εx

∗

¶ .

∗

Dk (x) Dk (x) = Dh (x) Dh (x) + ε2 In

and so Jk (x)

2

≡ =

¡ ¢ ∗ det Dh (x) Dh (x) + ε2 In ¡ ¢ det Q∗ DQ + ε2 In ∗

where D is a diagonal matrix having the nonnegative eigenvalues of Dh (x) Dh (x) down the main diagonal. Thus, since one of these eigenvalues equals 0, letting λ2i denote the ith eigenvalue, there exists a constant, C independent of ε such that 2

0 < Jk (x) =

n Y ¡

¢ λ2i + ε2 ≤ C 2 ε2.

i=1

Therefore, what was just shown applies to k. Let n o T T ≡ (h (w) , 0) : w ∈ B (x,r) , Tε

≡

n o T (h (w) , εw) : w ∈ B (x,r)

≡

k (B (x,r)),

then T = P Tε where P is the projection map defined by µ ¶ µ ¶ x x P ≡ . y 0

(19.5.25)


589

Since P decreases distances, it follows from Lemma 19.5.6 Hn (h (B (x,r))) = Hn (T ) = Hn (P Tε ) ≤ Hn (Tε ) = Hn (k (B (x,r))) . It follows from 19.5.25 and the first part of the proof applied to k that Cε

Hn (k (B (x, r))) r→0 mn (B (x,r)) Hn (h (B (x, r))) . lim sup mn (B (x,r)) r→0

≥

Jk (x) = lim

≥

−1

Since ε is arbitrary, this establishes 19.5.24 in the case where U (x) and completes the proof of the theorem. Define the following set for future reference. −1

S ≡ {x ∈ A : U (x)

19.5.2

does not exist}

does not exist

(19.5.26)

The Area Formula

Assume h : A → Rm is one to one in addition to 19.5.12 - 19.5.14. Since h is one to one, Lemma 19.5.3 implies one can define a measure, ν, on the σ− algebra of Lebesgue measurable sets as follows. ν (E) ≡ Hn (h (E ∩ A)). By Lemma 19.5.3, this is a measure and ν ¿ m. Therefore by the corollary to the Radon Nikodym theorem, Corollary 16.1.3 on Page 492, there exists f ∈ L1loc (Rn ) , f ≥ 0, f (x) = 0 if x ∈ / A, and Z Z ν (E) = f dm = f dm. E

A∩E

What is f ? I will show that f (x) = J (x) = det (U (x)) a.e. Define E ≡ {x ∈ A : x is not a point of density of A} ∪ {x ∈ A : x is not a Lebesgue point of f }. Then E is a set of measure zero and if x ∈ (A \ E), Lemma 19.5.9 and Theorem 19.5.10 imply Z 1 f (y) dm f (x) = lim r→0 mn (B (x,r)) B(x,r) = = =

Hn (h (B (x,r) ∩ A)) r→0 mn (B (x,r)) Hn (h (B (x,r) ∩ A)) Hn (h (B (x,r))) lim r→0 Hn (h (B (x,r))) mn (B (x,r)) J (x). lim

590

HAUSDORFF MEASURE

Therefore, f (x) = J (x) a.e., whenever x ∈ A \ E. Now let F be a Borel set in Rm . Recall this implies F is Hn measurable. Then Z Z n XF (y) dH = XF ∩h(A) (y) dHn h(A)

¡ ¡ ¢¢ Hn h h−1 (F ) ∩ A Z ¡ ¢ ν h−1 (F ) = XA∩h−1 (F ) (x) J (x) dm

= = Z =

XF (h (x)) J (x) dm.

(19.5.27)

A

Note there are no measurability questions in the above formula because h−1 (F ) is a Borel set due to the continuity of h. The Borel measurability of J (x) also follows from the observation that h is continuous and therefore, the partial derivatives are Borel measurable, being the limit of continuous functions. Then J (x) is just a continuous function of these partial derivatives. However, things are not so clear if E is only assumed Hn measurable. Is there a similar formula for F only Hn measurable? First consider the case where E is only Hn measurable but Hn (E ∩ h (A)) = 0. By Theorem 19.1.5 on Page 568, there exists a Borel set F ⊇ E ∩ h (A) such that Hn (F ) = Hn (E ∩ h (A)) = 0. Then from 19.5.27, XA∩h−1 (F ) (x) J (x) = 0 a.e. But 0 ≤ XA∩h−1 (E) (x) J (x) ≤ XA∩h−1 (F ) (x) J (x)

(19.5.28)

which shows the two functions in 19.5.28 are equal a.e. Therefore XA∩h−1 (E) (x) J (x) is Lebesgue measurable and so from 19.5.27, Z Z 0 = XE∩h(A) (y) dHn = XF ∩h(A) (y) dHn Z =

Z XA∩h−1 (F ) (x) J (x) dmn =

XA∩h−1 (E) (x) J (x) dmn ,

(19.5.29)

which shows 19.5.27 holds in this case where Hn (E ∩ h (A)) = 0. Now let AR ≡ A ∩ B (0,R) where R is large enough that AR 6= ∅ and let E be Hn measurable. By Theorem 19.1.5, there exists F ⊇ E ∩ h (AR ) such that F is Borel and Hn (F \ (E ∩ h (AR ))) = 0. (19.5.30)


591

Then (E ∩ h (AR )) ∪ (F \ (E ∩ h (AR )) ∩ h (AR )) = F ∩ h (AR ) and so XAR ∩h−1 (F ) J = XAR ∩h−1 (E) J + XAR ∩h−1 (F \(E∩h(AR ))) J where from 19.5.30 and 19.5.29, the second function on the right of the equal sign is Lebesgue measurable and equals zero a.e. Therefore, the first function on the right of the equal sign is also Lebesgue measurable and equals the function on the left a.e. Thus, Z Z XE∩h(AR ) (y) dHn = XF ∩h(AR ) (y) dHn Z Z = XAR ∩h−1 (F ) (x) J (x) dmn = XAR ∩h−1 (E) (x) J (x) dmn . (19.5.31) Letting R → ∞ yields 19.5.31 with A replacing AR and the function x → XAR ∩h−1 (E) (x) J (x) is Lebesgue measurable. Writing this in a more familiar form yields Z Z XE (y) dHn = XE (h (x)) J (x) dmn . h(A)

(19.5.32)

A

From this, it follows that if s is a nonnegative Hn measurable simple function, 19.5.32 continues to be valid with s in place of XE . Then approximating an arbitrary nonnegative Hn measurable function, g, by an increasing sequence of simple functions, it follows that 19.5.32 holds with g in place of XE and there are no measurability problems because x → g (h (x)) J (x) is Lebesgue measurable. This proves the area formula. Theorem 19.5.11 Let g : h (A) → [0, ∞] be Hn measurable where h is a continuous function and A is a Lebesgue measurable set which satisfies 19.5.12 - 19.5.14. That is, U is an open set in Rn on which h is defined and A ⊆ U is a Lebesgue measurable set, m ≥ n, and h : A → Rm is continuous,

(19.5.33)


(19.5.34)

Also assume that for every x ∈ A, there exists Rx and Lx such that for all y, z ∈ B (x, Rx ) , |h (z) − h (y)| ≤ Lx |x − y| (19.5.35) Then x → (g ◦ h) (x) J (x) is Lebesgue measurable and Z

Z n

g (y) dH = h(A)

g (h (x)) J (x) dm A

¡ ¢1/2 ∗ where J (x) = det (U (x)) = det Dh (x) Dh (x) .

592

HAUSDORFF MEASURE

19.6

The Area Formula Alternate Version

19.6.1

Preliminary Results

It was shown in Lemma 19.4.5 that Hn (F A) = det(U )mn (A) where F = RU with R preserving distances and U a symmetric matrix having all positive eigenvalues. The area formula gives a generalization of this simple relationship to the case where F is replaced by a nonlinear mapping, h. It contains as a special case the earlier change of variables formula. There are two parts to this development. The first part is to generalize Lemma 19.4.5 to the case of nonlinear maps. When this is done, the area formula can be presented. In this section, U will be an open set in Rn on which h is defined and A ⊆ U will be a Lebesgue measurable set. Assume m ≥ n and h : U → Rm is continuous,

(19.6.36)


(19.6.37)

n

H (h (U \ A)) = 0

(19.6.38)

These conditions are different than the ones considered earlier. Here no Lipschitz assumption is needed on h. In this sense, these conditions are more general than those considered earlier. However, they are not really more general because of 19.6.38 which says that A is essentially an open set. This was not necessary earlier. The area formula which results from the above conditions is a generalization of the change of variables formula given in Rudin [58] and the proof is essentially the same as the proof given in this book with modifications to account for the Hausdorff measure. Lemma 19.6.1 If T ⊆ A and mn (T ) = 0, then Hn (h (T )) = 0. Proof: Let Tk ≡ {x ∈ T : ||Dh (x)|| < k} . Thus T = ∪k Tk . I will show h (Tk ) has Hn measure zero and then it will follow that h (T ) = ∪∞ k=1 h (Tk ) must also have measure zero. Let ε > 0 be given. By outer regularity, there exists an open set, V , containing Tk which is contained in U such that mn (V ) < knε6n . For x ∈ Tk it follows from differentiability, h (x + v) = h (x) + Dh (x) v + o (v) and so whenever rx is small enough, B (x,5rx ) ⊆ V and whenever |v| < 5rx , |o (v)| < krx . Therefore, if |v| < 5rx , Dh (x) v + o (v) ∈ B (0, 5krx ) + B (0,krx ) ⊆ B (0, 6krx )

19.6. THE AREA FORMULA ALTERNATE VERSION

593

and so h (B (x, 5rx )) ⊆ B (h (x) , 6krx ). Letting δ > 0 be given, the Vitali covering theorem implies there exists a sequence of disjoint balls {Bi }, Bi =n B o (xi , rxi ), which are contained in V such that the b sequence of enlarged balls, Bi , having the same center but 5 times the radius, covers Tk and 6krxi < δ. Then ³ ³ ´´ b Hδn (h (Tk )) ≤ Hδn h ∪∞ i=1 Bi ≤

∞ X

∞ ³ ³ ´´ X bi Hδn h B ≤ Hδn (B (h (xi ) , 6krxi ))

i=1

≤

i=1 ∞ X

n

n

α (n) (6krxi ) = (6k)

i=1

∞ X

α (n) rxni

i=1

∞ X n = (6k) mn (B (xi , rxi )) i=1

≤

n

n

(6k) mn (V ) ≤ (6k)

ε k n 6n

= ε.

Since ε > 0 is arbitrary, this shows Hδn (h (Tk )) = 0. Since δ is arbitrary, this implies Hn (h (Tk )) = 0. Now Hn (h (T )) = lim Hn (h (Tk )) = 0. k→∞

This proves the lemma. Lemma 19.6.2 If S is a Lebesgue measurable subset of A, then h (S) is Hn measurable. Proof: Let Sk = S ∩ B (0, k) , k ∈ N. By inner regularity of Lebesgue measure, there exists a set, F , which is the countable union of compact sets and a set T with mn (T ) = 0 such that F ∪ T = Sk . Then h (F ) ⊆ h (Sk ) ⊆ h (F ) ∪ h (T ). By continuity of h, h (F ) is a countable union of compact sets and so it is Borel. By Lemma 19.6.1, Hn (h (T )) = 0 and so h (Sk ) is Hn measurable because of completeness of Hausdorff measure, which comes from Hn being obtained from an outer measure. Now h (S) = ∪∞ k=1 h (Sk ) and so it is also true that h (S) is Hn measurable. This proves the lemma. The following lemma, depending on the Brouwer fixed point theorem and found in Rudin [58], will be important for the following arguments. The idea is that if a continuous function mapping a ball in Rk to Rk doesn’t move any point very much, then the image of the ball must contain a slightly smaller ball.

594

HAUSDORFF MEASURE

Lemma 19.6.3 Let B = B (0, r), a ball in Rk and let F : B → Rk be continuous and suppose for some ε < 1, |F (v) −v| < εr (19.6.39) for all v ∈ B. Then F (B) ⊇ B (0, r (1 − ε)) . Proof: Suppose a ∈ B (0, r (1 − ε)) \ F (B) . I claim that a 6= F (v) for all v ∈ B. Here is why. By assumption, if F (v) = a, then |v| = r and so |F (v) − v| = |a − v| ≥ |v| − |a| > r − r (1 − ε) = rε, a contradiction to 19.6.39. Now letting G :B → B, be defined by G (v) ≡

r (a − F (v)) , |a − F (v)|

it follows G is continuous. Then by the Brouwer fixed point theorem, G (v) = v for some v ∈ B. Using the formula for G, it follows |v| = r. Taking the inner product with v, (G (v) , v) = = = = ≤

2

|v| = r2 =

r (a − F (v) , v) |a − F (v)|

r (a − v + v − F (v) , v) |a − F (v)| r [(a − v, v) + (v − F (v) , v)] |a − F (v)| h i r 2 (a, v) − |v| + (v − F (v) , v) |a − F (v)| £ 2 ¤ r r (1 − ε) − r2 +r2 ε = 0, |a − F (v)|

a contradiction to |v| = r. Therefore, B (0, r (1 − ε)) \ F (B) = ∅ and this proves the lemma. By Theorem 4.12.6 on Page 93, when Dh (x) exists, Dh (x) = R (x) U (x) where (U (x) u, v) = (U (x) v, u) , (U (x) u, u) ≥ 0 and R∗ R = I. Lemma 19.6.4 In this situation, |R∗ u| ≤ |u|. Proof: First note that (u−RR∗ u,RR∗ u)

=

(u,RR∗ u) − |RR∗ u| 2

2

2

= |R∗ u| − |R∗ u| = 0,


595

and so 2

|u|

2

=

|u−RR∗ u+RR∗ u|

=

|u−RR∗ u| + |RR∗ u|

=

|u−RR∗ u| + |R∗ u| .

2 2

2

2

This proves the lemma. Lemma 19.6.5 If |P x − P y| ≤ L |x − y| , then for E a set, Hn (P E) ≤ Ln Hn (E) . Proof: Without loss of generality, assume Hn (E) < ∞. Let δ > 0 and let ∞ {Ci }i=1 be a covering of E such that diam (Ci ) ≤ δ for each i and ∞ X

n

α (n) r (Ci ) ≤ Hδn (E) + ε.

i=1 ∞

Then {P Ci }i=1 is a covering of P E such that diam (P Ci ) ≤ Lδ. Therefore, n HLδ (P E)

≤

∞ X

n

α (n) r (P Ci )

i=1

≤ Ln

∞ X

n

α (n) r (Ci ) ≤ Ln Hδn (E) + Ln ε

i=1 n

≤ H (E) + ε. Letting δ → 0, Hn (P E) ≤ Ln Hn (E) + Ln ε and since ε > 0 is arbitrary, this proves the Lemma. Then the following corollary follows from Lemma 19.6.4. Corollary 19.6.6 Let T ⊆ Rm . Then Hn (T ) ≥ Hn (RR∗ T ) = Hn (R∗ T ). Definition 19.6.7 Let E be a Lebesgue measurable set. x ∈ E is a point of density if mn (E ∩ B(x, r)) lim = 1. r→0 mn (B(x, r)) Recall that from the fundamental theorem of calculus applied to XE almost every point of E is a point of density.

596

HAUSDORFF MEASURE −1

Lemma 19.6.8 Let x ∈ A be a point where U (x) following hold for all r small enough.

exists. Then if ε ∈ (0, 1) the

h (B (x, r)) ⊆ h (x) + R (x) U (x) B (0, r (1 + ε)),

(19.6.40)

Hn (h (B (x,r))) ≤ mn (U (x) B (0, r (1 + ε))). ∗

∗

R (x) h (B (x, r)) ⊇ R (x) h (x) + U (x) B (0, r (1 − ε)), Hn (h (B (x,r))) ≥ mn (U (x) B (0, r (1 − ε))),

(19.6.41) (19.6.42) (19.6.43)

If x is a point of A, then Hn (h (B (x, r) ∩ A)) = 1. r→0 Hn (h (B (x, r)))

(19.6.44)

lim

Proof: Since Dh (x) exists, h (x + v) = h (x) + Dh (x) v+o (|v|).

(19.6.45)

Consequently, when r is small enough, 19.6.40 holds. Using the fact R (x) preserves all distances, and Theorem 19.2.3 which says Hn = mn on the Borel sets of Rn ,this implies, Hn (h (B (x,r))) ≤ Hn (R (x) U (x) B (0, r (1 + ε))) = Hn (U (x) B (0, r (1 + ε))) = mn (U (x) B (0, r (1 + ε))) which shows 19.6.41. From 19.6.45, R∗ (x) h (x + v) = R∗ (x) h (x) + U (x) (v+o (|v|)). Thus, from the assumption that U (x) −1

F (v) ≡ U (x)

−1

exists and letting F (v) be given by

R∗ (x) h (x + v) − U (x)

−1

R∗ (x) h (x)

(19.6.46)

Since h is continuous near A, it follows F (v) − v = o (|v|) and so Lemma 19.6.3 implies that for all r small enough, F (B (0, r)) ≡ ⊇

U (x)

−1

R∗ (x) h (x+B (0,r)) − U (x)

−1

R∗ (x) h (x)

B (0, (1 − ε) r).

Therefore, R∗ (x) h (B (x,r)) ⊇ R∗ (x) h (x) + U (x) B (0, (1 − ε) r)


597

which proves 19.6.42. Therefore, R (x) R∗ (x) h (B (x, r)) ⊇ R (x) R∗ (x) h (x) + R (x) U (x) B (0, r (1 − ε)). From Lemma 19.6.6, this implies Hn (h (B (x,r))) ≥ =

Hn (R∗ (x) h (B (x, r))) Hn (R (x) R∗ (x) h (B (x, r)))

≥ Hn (R (x) U (x) B (0, r (1 − ε))) = Hn (U (x) B (0, r (1 − ε))) = mn (U (x) B (0, r (1 − ε))) which shows 19.6.43. Let x ∈ A. Choosing r small enough that B (x, r) ⊆ U, h (B (x, r) ∩ A) ∪ h (B (x, r) \ A) = h (B (x, r)) and so

≥

Hn (h (B (x, r) ∩ A)) + Hn (h (B (x, r) \ A)) Hn (h (B (x, r)))

Now by assumption 19.6.38, Hn (h (B (x, r) \ A)) = 0 and so for all r small enough, Hn (h (B (x, r) ∩ A)) = Hn (h (B (x, r))) . This establishes 19.6.44. The next theorem is the generalization of Lemma 19.4.5 to nonlinear maps. Theorem 19.6.9 Let h : U → Rm where U is an open set in Rn , n ≤ m, h is continuous on U, h is differentiable on A ⊆ U, and Hn (U \ A) = 0. Then for x ∈ A, Hn (h (B (x, r))) J (x) = lim , (19.6.47) r→0 mn (B (x,r)) ¡ ¢1/2 ∗ where J (x) ≡ det (U (x)) = det Dh (x) Dh (x) . −1

Proof: Suppose first that U (x) of variables formula for linear maps, n

exists. Using 19.6.43, 19.6.41 and the change

J (x) (1 − ε) Hn (h(B (x, r))) mn (U (x) B (0,r (1 − ε))) ≤ = mn (B (x, r)) mn (B (x, r)) mn (U (x) B (0,r (1 + ε))) n ≤ = J (x) (1 + ε) mn (B (x, r))

598

HAUSDORFF MEASURE

whenever r is small enough. It follows that since ε > 0 is arbitrary, 19.6.47 holds. −1 Now suppose U (x) does not exist. The first part shows that the conclusion of the theorem holds when J (x) 6= 0. I will apply this to a modified function. Let k : Rn → Rm × Rn be defined as

µ k (x) ≡

h (x) εx

¶ .

Then ∗

∗

Dk (x) Dk (x) = Dh (x) Dh (x) + ε2 In and so Jk (x)

2

≡ =

¡ ¢ ∗ det Dh (x) Dh (x) + ε2 In ¡ ¢ det Q∗ DQ + ε2 In ∗

where D is a diagonal matrix having the nonnegative eigenvalues of Dh (x) Dh (x) down the main diagonal. Thus, since one of these eigenvalues equals 0, letting λ2i denote the ith eigenvalue, there exists a constant, C independent of ε such that 2

0 < Jk (x) =

n Y ¡

¢ λ2i + ε2 ≤ C 2 ε2.

i=1

Therefore, what was just shown applies to k. Let n o T T ≡ (h (w) , 0) : w ∈ B (x,r) ,

Tε

≡

n o T (h (w) , εw) : w ∈ B (x,r)

≡

k (B (x,r)),

then T = P Tε where P is the projection map defined by µ ¶ µ ¶ x x P ≡ . y 0 Since P decreases distances, it follows from Lemma 19.6.5 Hn (h (B (x,r))) =

Hn (T ) = Hn (P Tε )

≤ Hn (Tε ) = Hn (k (B (x,r))) .

(19.6.48)


599

It follows from 19.6.48 and the first part of the proof applied to k that Cε

Hn (k (B (x, r))) r→0 mn (B (x,r)) Hn (h (B (x, r))) lim sup . mn (B (x,r)) r→0

≥

Jk (x) = lim

≥

−1

Since ε is arbitrary, this establishes 19.6.47 in the case where U (x) and completes the proof of the theorem.

19.6.2

does not exist

The Area Formula

Assume h : A → Rm is one to one in addition to 19.6.36 - 19.6.38. Since h is one to one on A, Lemma 19.6.2 implies one can define a measure, ν, on the σ− algebra of Lebesgue measurable sets as follows. ν (E) ≡ Hn (h (E ∩ A)). By Lemma 19.6.2, this is a measure and ν ¿ m. Therefore by the corollary to the Radon Nikodym theorem, Corollary 16.1.3 on Page 492, there exists f ∈ L1loc (Rn ) , f ≥ 0, f (x) = 0 if x ∈ / A, and Z Z ν (E) = f dmn = f dmn . E

A∩E

What is f ? I will show that f (x) = J (x) = det (U (x)) a.e. Let x be a Lebesgue point of f. Then by Lemma 19.6.8 and Theorem 19.6.9 Z 1 f (x) = lim f (y) dm r→0 mn (B (x,r)) B(x,r) = = =

Hn (h (B (x,r) ∩ A)) r→0 mn (B (x,r)) n H (h (B (x,r) ∩ A)) Hn (h (B (x,r))) lim r→0 Hn (h (B (x,r))) mn (B (x,r)) J (x). lim

Therefore, f (x) = J (x) a.e., whenever x is a Lebesgue point of f . Now let F be a Borel set in Rm . Recall this implies F is Hn measurable. Then Z Z XF (y) dHn = XF ∩h(A) (y) dHn h(A)

¡ ¡ ¢¢ Hn h h−1 (F ) ∩ A Z ¡ ¢ ν h−1 (F ) = XA∩h−1 (F ) (x) J (x) dmn

= = Z =

XF (h (x)) J (x) dmn . A

(19.6.49)

600

HAUSDORFF MEASURE

Note there are no measurability questions in the above formula because h−1 (F ) is a Borel set due to the continuity of h. The Borel measurability of J (x) also follows from the observation that h is continuous and therefore, the partial derivatives are Borel measurable, being the limit of continuous functions. Then J (x) is just a continuous function of these partial derivatives. However, things are not so clear if E is only assumed Hn measurable. Is there a similar formula for F only Hn measurable? First consider the case where E is only Hn measurable but Hn (E ∩ h (A)) = 0. By Theorem 19.1.5 on Page 568, there exists a Borel set F ⊇ E ∩ h (A) such that Hn (F ) = Hn (E ∩ h (A)) = 0. Then from 19.6.49, XA∩h−1 (F ) (x) J (x) = 0 a.e. But 0 ≤ XA∩h−1 (E) (x) J (x) ≤ XA∩h−1 (F ) (x) J (x)

(19.6.50)

which shows the two functions in 19.6.50 are equal a.e. Therefore XA∩h−1 (E) (x) J (x) is Lebesgue measurable and so from 19.6.49, Z Z n 0 = XE∩h(A) (y) dH = XF ∩h(A) (y) dHn Z =

Z XA∩h−1 (F ) (x) J (x) dmn =

XA∩h−1 (E) (x) J (x) dmn ,

(19.6.51)

which shows 19.6.49 holds in this case where Hn (E ∩ h (A)) = 0. Now let AR ≡ A ∩ B (0,R) where R is large enough that AR 6= ∅ and let E be Hn measurable. By Theorem 19.1.5, there exists F ⊇ E ∩ h (AR ) such that F is Borel and Hn (F \ (E ∩ h (AR ))) = 0. (19.6.52) Then (E ∩ h (AR )) ∪ (F \ (E ∩ h (AR )) ∩ h (AR )) = F ∩ h (AR ) and so XAR ∩h−1 (F ) J = XAR ∩h−1 (E) J + XAR ∩h−1 (F \(E∩h(AR ))) J where from 19.6.52 and 19.6.51, the second function on the right of the equal sign is Lebesgue measurable and equals zero a.e. Therefore, the first function on the right of the equal sign is also Lebesgue measurable and equals the function on the left a.e. Thus, Z Z XE∩h(AR ) (y) dHn = XF ∩h(AR ) (y) dHn

19.7. THE DIVERGENCE THEOREM Z =

601 Z

XAR ∩h−1 (F ) (x) J (x) dmn =

XAR ∩h−1 (E) (x) J (x) dmn .

(19.6.53)

Letting R → ∞ yields 19.6.53 with A replacing AR and the function x → XAR ∩h−1 (E) (x) J (x) is Lebesgue measurable. Writing this in a more familiar form yields Z Z XE (y) dHn = XE (h (x)) J (x) dmn . h(A)

(19.6.54)

A

From this, it follows that if s is a nonnegative Hn measurable simple function, 19.6.54 continues to be valid with s in place of XE . Then approximating an arbitrary nonnegative Hn measurable function, g, by an increasing sequence of simple functions, it follows that 19.6.54 holds with g in place of XE and there are no measurability problems because x → g (h (x)) J (x) is Lebesgue measurable. This proves the area formula. Theorem 19.6.10 Let g : h (A) → [0, ∞] be Hn measurable where h is a continuous function and A is a Lebesgue measurable set which satisfies 19.6.36 - 19.6.38. That is, U is an open set in Rn on which h is defined and continuous and A ⊆ U is a Lebesgue measurable set, m ≥ n, and h : U → Rm is continuous, h one to one on A,

(19.6.55)


(19.6.56)

n

H (h(U \ A)) = 0,

(19.6.57)

Then x → (g ◦ h) (x) J (x) is Lebesgue measurable and Z

Z g (y) dHn =

h(A)


¡ ¢1/2 ∗ where J (x) = det (U (x)) = det Dh (x) Dh (x) .

19.7

The Divergence Theorem

As an important application of the area formula I will give a general version of the divergence theorem. It will always be assumed n ≥ 2. Actually it is not necessary to make this assumption but what results in the case where n = 1 is nothing more than the fundamental theorem of calculus and the considerations necessary to draw this conclusion seem unneccessarily tedious. You have to consider H0 , zero dimensional Hausdorff measure. It is left as an exercise but I will not present it. It will be convenient to have some lemmas and theorems in hand before beginning the proof. First recall the Tietze extension theorem on Page 153. It is stated next for convenience.

602

HAUSDORFF MEASURE

Theorem 19.7.1 Let M be a closed nonempty subset of a metric space (X, d) and let f : M → [a, b] be continuous at every point of M. Then there exists a function, g continuous on all of X which coincides with f on M such that g (X) ⊆ [a, b] . The next topic needed is the concept of an infinitely differentiable partition of unity. Definition 19.7.2 Let C be a set whose elements are subsets of Rn .1 Then C is said to be locally finite if for every x ∈ Rn , there exists an open set, Ux containing x such that Ux has nonempty intersection with only finitely many sets of C. Lemma 19.7.3 Let C be a set whose elements are open subsets of Rn and suppose ∞ ∪C ⊇ H, a closed set. Then there exists a countable list of open sets, {Ui }i=1 such ∞ that each Ui is bounded, each Ui is a subset of some set of C, and ∪i=1 Ui ⊇ H. Proof: Let Wk ≡ B (0, k) , W0 = W−1 = ∅. For each x ∈ H ∩ Wk there exists an open set, Ux such that Ux is a subset of some set of C and Ux ⊆ Wk+1 \ Wk−1 . © ªm(k) Then since H ∩ Wk is compact, there exist finitely many of these sets, Uik i=1 whose union contains H ∩ Wk . If H ∩ Wk = ∅, let m (k) = 0 and there are no such © k ªm(k) sets obtained.The desired countable list of open sets is ∪∞ k=1 Ui i=1 . Each open set in this list is bounded. Furthermore, if x ∈ Rn , then x ∈ Wk where k is the first positive integer with x ∈ Wk . Then Wk \ Wk−1 is an open set containing x and © ªm(k) this open set can have nonempty intersection only with with a set of Uik i=1 ∪ © k−1 ªm(k−1) © k ªm(k) Ui , a finite list of sets. Therefore, ∪∞ k=1 Ui i=1 is locally finite. i=1 ∞ The set, {Ui }i=1 is said to be a locally finite cover of H. The following lemma gives some important reasons why a locally finite list of sets is so significant. First ∞ of all consider the rational numbers, {ri }i=1 each rational number is a closed set. ∞

∞ Q = {ri }i=1 = ∪∞ i=1 {ri } 6= ∪i=1 {ri } = R

The set of rational numbers is definitely not locally finite. Lemma 19.7.4 Let C be locally finite. Then © ª ∪C = ∪ H : H ∈ C . Next suppose the elements of C are open sets and that for each U ∈ C, there exists a differentiable function, ψ U having spt (ψ U ) ⊆ U. Then you can define the following finite sum for each x ∈ Rn X f (x) ≡ {ψ U (x) : x ∈ U ∈ C} . Furthermore, f is also a differentiable function2 and X Df (x) = {Dψ U (x) : x ∈ U ∈ C} . 1 The

definition applies with no change to a general topological space in place of Rn . each ψ U were only continuous, one could conclude f is continuous. Here the main interest is differentiable. 2 If

19.7. THE DIVERGENCE THEOREM

603

Proof: Let p be a limit point of ∪C and let W be an open set which intersects only finitely many sets ofªC. Then p must©be a limit ª point of one of these sets. It © follows p ∈ ∪ H : H ∈ C and so ∪C ⊆ ∪ H : H ∈ C . The inclusion in the other direction is obvious. Now consider the second assertion. Letting x ∈ Rn , there exists an open set, W intersecting only finitely many open sets of C, U1 , U2 , · · · , Um . Then for all y ∈ W, f (y) =

m X

ψ Ui (y)

i=1

and so the desired result is obvious. It merely says that a finite sum of differentiable functions is differentiable. Recall the following definition. Definition 19.7.5 Let K be a closed subset of an open set, U. K ≺ f ≺ U if f is continuous, has values in [0, 1] , equals 1 on K, and has compact support contained in U . Lemma 19.7.6 Let U be a bounded open set and let K be a closed subset of U. Then there exist an open set, W, such that W ⊆ W ⊆ U and a function, f ∈ Cc∞ (U ) such that K ≺ f ≺ U . Proof: The set, K is compact so is at a positive distance from U C . Let © ¡ ¢ª W ≡ x : dist (x, K) < 3−1 dist K, U C . Also let

© ¡ ¢ª W1 ≡ x : dist (x, K) < 2−1 dist K, U C

Then it is clear K ⊆ W ⊆ W ⊆ W1 ⊆ W1 ⊆ U Now consider the function, ¢ ¡ dist x, W1C ¡ ¢ ¡ ¢ h (x) ≡ dist x, W1C + dist x, W Since W is compact it is at a positive distance from W1C and so h is a well defined continuous function which has compact support contained in W 1 , equals 1 on W, and has values in [0, 1] . Now let φk be a mollifier. Letting ¢ ¡ ¢¢ ¡ ¡ k −1 < min dist K, W C , 2−1 dist W 1 , U C , it follows that for such k,the function, h ∗ φk ∈ Cc∞ (U ) , has values in [0, 1] , and equals 1 on K. Let f = h ∗ φk . The above lemma is used repeatedly in the following.

604

HAUSDORFF MEASURE ∞

Lemma 19.7.7 Let K be a closed set and let {Vi }i=1 be a locally finite list of bounded open sets whose union contains K. Then there exist functions, ψ i ∈ Cc∞ (Vi ) such that for all x ∈ K, 1=

∞ X

ψ i (x)

i=1

and the function f (x) given by f (x) =

∞ X

ψ i (x)

i=1

is in C ∞ (Rn ) . Proof: Let K1 = K \ ∪∞ i=2 Vi . Thus K1 is compact because K1 ⊆ V1 . Let K 1 ⊆ W 1 ⊆ W 1 ⊆ V1 Thus W1 , V2 , · · · , Vn covers K and W 1 ⊆ V1 . Suppose W1 , · · · , Wr have been defined such that Wi ⊆ Vi for each i, and W1 , · · · , Wr , Vr+1 , · · · , Vn covers K. Then let ¡ ¢ ¡ r ¢ Kr+1 ≡ K \ ( ∪∞ i=r+2 Vi ∪ ∪j=1 Wj ). It follows Kr+1 is compact because Kr+1 ⊆ Vr+1 . Let Wr+1 satisfy Kr+1 ⊆ Wr+1 ⊆ W r+1 ⊆ Vr+1 ∞

Continuing this way defines a sequence of open sets, {Wi }i=1 with the property W i ⊆ Vi , K ⊆ ∪ ∞ i=1 Wi . ∞

∞

Note {Wi }i=1 is locally finite because the original list, {Vi }i=1 was locally finite. Now let Ui be open sets which satisfy W i ⊆ Ui ⊆ U i ⊆ Vi . ∞

Similarly, {Ui }i=1 is locally finite.

Wi

∞

Ui

Vi

∞ Since the set, {Wi }i=1 is locally finite, it follows ∪∞ i=1 Wi = ∪i=1 Wi and so it is possible to define φi and γ, infinitely differentiable functions having compact support such that ∞ U i ≺ φi ≺ Vi , ∪∞ i=1 W i ≺ γ ≺ ∪i=1 Ui .

19.7. THE DIVERGENCE THEOREM Now define

½ ψ i (x) =

605

P∞ P∞ γ(x)φi (x)/ j=1 φj (x) if j=1 φj (x) 6= 0, P∞ 0 if j=1 φj (x) = 0.

P∞ If x is such that j=1 φj (x) = 0, then x ∈ / ∪∞ i=1 Ui because φi equals one on Ui . Consequently γ (y) = 0 for all y near x thanks to the fact that ∪∞ i=1 Ui is closed and so ψ (y) = 0 for all y near x. Hence ψ is infinitely differentiable at such x. If i i P∞ j=1 φj (x) 6= 0, this situation persists near x because each φj is continuous and so 19.7.4. Therefore ψ i is infinitely differentiable at such points also thanks to Lemma P ∞ ψ i is infinitely differentiable. If x ∈ K, then γ (x) = 1 and so j=1 ψ j (x) = 1. Clearly 0 ≤ ψ i (x) ≤ 1 and spt(ψ j ) ⊆ Vj . This proves the theorem. The functions, {ψ i } are called a C ∞ partition of unity. The method of proof of this lemma easily implies the following useful corollary. Corollary 19.7.8 If H is a compact subset of Vi for some Vi there exists a partition of unity such that ψ i (x) = 1 for all x ∈ H in addition to the conclusion of Lemma 37.1.6. fj ≡ Vj \ H. Now in the proof Proof: Keep Vi the same but replace Vj with V above, applied to this modified collection of open sets, if j 6= i, φj (x) = 0 whenever x ∈ H. Therefore, ψ i (x) = 1 on H. Lemma 19.7.9 Let Ω be a metric space with the closed balls compact and suppose µ is a measure defined on the Borel sets of Ω which is finite on compact sets. Then there exists a unique Radon measure, µ which equals µ on the Borel sets. In particular µ must be both inner and outer regular on all Borel sets. R Proof: Define a positive linear functional, Λ (f ) = f dµ. Let µ be the Radon measure which comes from the Riesz representation theorem for positive linear functionals. Thus for all f continuous, Z Z f dµ = f dµ. If V is an open set, let {fn } be a sequence of continuous functions which is increasing and converges to XV pointwise. Then applying the monotone convergence theorem, Z Z XV dµ = µ (V ) = XV dµ = µ (V ) and so the two measures coincide on all open sets. Every compact set is a countable intersection of open sets and so the two measures coincide on all compact sets. Now let B (a, n) be a ball of radius n and let E be a Borel set contained in this ball. Then by regularity of µ there exist sets F, G such that G is a countable intersection of open sets and F is a countable union of compact sets such that F ⊆ E ⊆ G and µ (G \ F ) = 0. Now µ (G) = µ (G) and µ (F ) = µ (F ) . Thus µ (G \ F ) + µ (F ) = =

µ (G) µ (G) = µ (G \ F ) + µ (F )

606

HAUSDORFF MEASURE

and so µ (G \ F ) = µ (G \ F ) . Thus µ (E) = µ (F ) = µ (F ) = µ (G) = µ (E) . If E is an arbitrary Borel set, then µ (E ∩ B (a, n)) = µ (E ∩ B (a, n)) and letting n → ∞, this yields µ (E) = µ (E) . One more lemma will be useful. Lemma ¡ ¢ 19.7.10 Let V be a bounded open set and let X be the closed subspace of C V , the space of continuous functions defined on V , which is given by the following. ¡ ¢ X = {u ∈ C V : u (x) = 0 on ∂V }. Then Cc∞ (V ) is dense in X with respect to the norm given by © ª ||u|| = max |u (x)| : x ∈ V ¡ ¢ Proof: Let O ⊆ O ⊆ W ⊆ W ⊆ V be such that dist O, V C < η and let ψ δ (·) be a mollifier. Let u ∈ X and consider XW u ∗ ψ δ . Let ε > 0 be given and let η be small enough that |u (x) | < ε/2 whenever x ∈ V \ O. Then if δ is small enough |XW u ∗ ψ δ (x) − u (x) | < ε for all x ∈ O and XW u ∗ ψ δ is in Cc∞ (V ). For x ∈ V \ O, |XW u ∗ ψ δ (x) | ≤ ε/2 and so for such x, |XW u ∗ ψ δ (x) − u (x) | ≤ ε. This proves the lemma since ε was arbitrary. Definition 19.7.11 A bounded open set, U ⊆ Rn is said to have a Lipschitz boundary and to lie on one side of its boundary if the following conditions hold. There exist open boxes, Q1 , · · · , QN , Qi =

n Y ¡ i i¢ aj , bj j=1

such that ∂U ≡ U \ U is contained in their union. Also, for each Qi , there exists k and a Lipschitz function, gi such that U ∩ Qi is of the form  k−1  Y¡ ¢ x : (x1 , · · · , xk−1 , xk+1 , · · · , xn ) ∈ aij , bij ×  j=1  n  Y ¡ i i¢ aj , bj and aik < xk < gi (x1 , · · · , xk−1 , xk+1 , · · · , xn ) (19.7.58)  j=k+1


607

or else of the form  k−1  Y¡ ¢ x : (x1 , · · · , xk−1 , xk+1 , · · · , xn ) ∈ aij , bij ×  j=1

n Y ¡

  and gi (x1 , · · · , xk−1 , xk+1 , · · · , xn ) < xk < bij (19.7.59) . 

¢ i

aij , bj

j=k+1

The function, gi has a derivative on Ai ⊆ 

k−1 Y

mn−1 

¡ i i¢ Qk−1 ¡ i i ¢ Qn j=1 aj , bj × j=k+1 aj , bj where

 n Y ¡ i i¢ ¡ i i¢ aj , bj × aj , bj \ Ai  = 0.

j=1

j=k+1

Also, there exists an open set, Q0 such that Q0 ⊆ Q0 ⊆ U and U ⊆ Q0 ∪Q1 ∪· · ·∪QN . Note that since there are only finitely many Qi and each gi is Lipschitz, it follows from an application of Lemma 19.5.6 that Hn−1 (∂U ) < ∞. Also from Lemma 19.7.9 Hn−1 is inner and outer regular on ∂U . Lemma 19.7.12 Suppose U is¡a bounded¢ open set as described above. Then there n exists a unique function in L∞ ∂U, Hn−1 , n (y) for y ∈ ∂U such that |n (y)| = n−1 1, n is H measurable, (meaning each component of n is Hn−1 measurable) and for every w ∈Rn satisfying |w| = 1, and for every f ∈ Cc1 (Rn ) , Z Z f (x + tw) − f (x) lim dx = f (n · w) dHn−1 t→0 U t ∂U N

∞ Proof: Let U ⊆ V ⊆ V ⊆ ∪N partition of unity i=0 Qi and let {ψ i }i=0 be a C on V such that spt (ψ i ) ⊆ Qi . Then for all t small enough and x ∈ U , N

f (x + tw) − f (x) 1X = ψ f (x + tw) − ψ i f (x) . t t i=0 i Thus using the dominated convergence theorem, Z f (x + tw) − f (x) lim dx t→0 U t ! Z Ã X N 1 = lim ψ f (x + tw) − ψ i f (x) dx t→0 U t i=0 i =

Z X N X n U i=0 j=1

Dj (ψ i f ) (x) wj dx

608

HAUSDORFF MEASURE

=

Z X n

Dj (ψ 0 f ) (x) wj dx +

U j=1

N Z X n X U j=1

i=1

Dj (ψ i f ) (x) wj dx

(19.7.60)

Since spt (ψ 0 ) ⊆ Q0 , it follows the first term in the above equals zero. In the second term, fix i. Without loss of generality, suppose the k in the above definition equals n and 19.7.58 holds. This just makes things a little easier to write. Thus gi is a function of n−1 Y¡ ¢ (x1 , · · · , xn−1 ) ∈ aij , bij ≡ Bi j=1

Then Z X n U j=1

Z

Z

= Z

= Bi

n gi (x1 ,··· ,xn−1 ) X

ain

Bi

Z

Dj (ψ i f ) (x) wj dx Dj (ψ i f ) (x) wj dxn dx1 · · · dxn−1

j=1

n gi (x1 ,··· ,xn−1 ) X −∞

Dj (ψ i f ) (x) wj dxn dx1 · · · dxn−1

j=1

Letting xn = y + gi (x1 , · · · , xn−1 ) and changing the variable, this equals Z

Z

0

= Bi

=

n X

−∞ j=1

Dj (ψ i f ) (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 )) ·

wj dydx1 · · · dxn−1 Z Z 0 X n Dj (ψ i f ) (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 )) · Ai

−∞ j=1

wj dydx1 · · · dxn−1 Recall Dj refers to the partial derivative taken with respect to the entry in the j th slot. In the nth slot is found not just xn but y +gi (x1 , · · · , xn−1 ) so a differentiation with respect to xj will not be the same as Dj . In fact, it will introduce another term involving gi,j . Thus from the chain rule, Z

Z

= Ai

0

n−1 X

−∞ j=1

∂ (ψ f (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 ))) wj − ∂xj i

Dn (ψ i f ) (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 )) · gi,j (x1 , · · · , xn−1 ) wj dydx1 · · · dxn−1 Z Z 0 + Dn (ψ i f ) (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 )) · Ai

−∞

wn dydx1 · · · dxn−1

(19.7.61)

19.7. THE DIVERGENCE THEOREM Consider the term Z Z

0

n−1 X

−∞ j=1

Ai

609

∂ (ψ f (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 ))) · ∂xj i

wj dydx1 · · · dxn−1 . This equals Z

Z

Bi

0

n−1 X

−∞ j=1

∂ (ψ f (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 ))) · ∂xj i

wj dydx1 · · · dxn−1 , and now interchanging the order of integration and using the fact that spt (ψ i ) ⊆ Qi , it follows this term equals zero. (The reason this is valid is that xj → ψ i f (x1 , · · · , xn−1 , y + gi (x1 , · · · , xn−1 )) is the composition of Lipschitz functions and is therefore Lipschitz. Therefore, this function is absolutely continuous and can be recovered by integrating its derivative.) Then, changing the variable back to xn it follows 19.7.61 reduces to  Z Z gi (x1 ,··· ,xn−1 ) n−1 X  =− Dn (ψ i f ) (x1 , · · · , xn−1 , xn ) · Ai

−∞

j=1

gi,j (x1 , · · · , xn−1 ) wj ) dxn dx1 · · · dxn−1 Z

Z

gi (x1 ,··· ,xn−1 )

+ Ai

−∞

Dn (ψ i f (x1 , · · · , xn−1 , xn )) wn dxn dx1 · · · dxn−1

Doing the integrals using the observation that gi,j (x1 , · · · , xn−1 ) does not depend on xn , this reduces further to Z (ψ i f ) (x1 , · · · , xn−1 , xn ) Ni (x1 , · · · , xn−1 , gi (x1 , · · · , xn−1 )) · wdmn−1 Ai

(19.7.62)

where Ni (x1 , · · · , xn−1 , gi (x1 , · · · , xn−1 )) is given by (−gi,1 (x1 , · · · , xn−1 ) , −gi,2 (x1 , · · · , xn−1 ) , · · · , −gi,n−1 (x1 , · · · , xn−1 ) , 1) . (19.7.63) At this point I need a technical lemma which will allow the use of the area formula. The part of the boundary of U which is contained in Qi is the image of the map, hi (x1 , · · · , xn−1 ) given by (x1 , · · · , xn−1 , gi (x1 , · · · , xn−1 )) for (x1 , · · · , xn−1 ) ∈ Ai . I need a formula for ¡ ¢1/2 ∗ det Dhi (x1 , · · · , xn−1 ) Dhi (x1 , · · · , xn−1 ) . To avoid interupting the argument, I will state the lemma here and prove it later.

610

HAUSDORFF MEASURE

Lemma 19.7.13

=

¡ ¢1/2 ∗ det Dhi (x1 , · · · , xn−1 ) Dhi (x1 , · · · , xn−1 ) v u n−1 X u 2 t1 + gi,j (x1 , · · · , xn−1 ) ≡ Ji (x1 , · · · , xn−1 ) . j−1

For y = (x1 , · · · , xn−1 , gi (x1 , · · · , xn−1 )) ∈ ∂U ∩ Qi and n defined by ni (y) =

1 Ni (y) Ji (x1 , · · · , xn−1 )

it follows from the description of Ji (x1 , · · · , xn−1 ) given in the above lemma, that ni is a unit vector. All components of ni are continuous functions of limits of continuous functions. Therefore, ni is Borel measurable and so it is Hn−1 measurable. Now 19.7.62 reduces to Z (ψ i f ) (x1 , · · · , xn−1 , gi (x1 , · · · , xn−1 )) × Ai

ni (x1 , · · · , xn−1 , gi (x1 , · · · , xn−1 )) · wJi (x1 , · · · , xn−1 ) dmn−1 . By the area formula this equals Z h(Ai )

ψ i f (y) ni (y) · wdHn−1 .

Now by Lemma 19.5.6 and the equality of mn−1 and Hn−1 on Rn−1 , the above integral equals Z Z n−1 ψ i f (y) ni (y) · wdH = ψ i f (y) ni (y) · wdHn−1 . ∂U ∩Qi

∂U

Returning to 19.7.60 similar arguments apply to the other terms and therefore, Z f (x + tw) − f (x) lim dmn t→0 U t N Z X = ψ i f (y) ni (y) · wdHn−1 i=1

∂U

Z =

f (y)

N X

∂U

i=1

Z

f (y) n (y) · wdHn−1

= ∂U

Then let n (y) ≡

PN i=1

ψ i (y) ni (y) · wdHn−1

ψ i (y) ni (y) .

(19.7.64)


611

I need to show first there is no other n which satisfies 19.7.64 and then I need to show that |n (y)| = 1. Note that it is clear |n (y)| ≤ 1 because each ni is a unit vector ¡ and this ¢ is just a convex combination of these. Suppose then that n1 ∈ L∞ ∂U, Hn−1 also works in 19.7.64. Then for all f ∈ Cc1 (Rn ) , Z Z n−1 f (y) n (y) · wdH = f (y) n1 (y) · wdHn−1 . ∂U

∂U

Suppose h ∈ C (∂U ) . Then by the Tietze extension theorem, there exists f ∈ Cc (Rn ) such that the restriction of f to ∂U equals h. Now by Lemma 19.7.10 applied to a bounded open set containing the support of f, there exists a sequence {fm } of functions in Cc1 (Rn ) converging uniformly to f . Therefore, Z h (y) n (y) · wdHn−1 ∂U Z = lim fm (y) n (y) · wdHn−1 m→∞ ∂U Z = lim fm (y) ni (y) · wdHn−1 m→∞ ∂U Z = h (y) ni (y) · wdHn−1 . ∂U

Now Hn−1 is a ¡Radon measure on ∂U and so the continuous functions on ∂U ¢ ∞ are dense in L1 ∂U, Hn−1 . It follows n · w = ni · w a.e. Now let {wm }m=1 be a countable dense subset of the unit sphere. From what was just shown, n · wm = ni · wm except for a set of measure zero, Nm . Letting N = ∪m Nm , it follows that for y ∈ / N, n (y) ·wm = ni (y) · wm for all m. Since the set is dense, it follows n (y) ·w = ni (y) · w for all y ∈ / N and for all w a unit vector. Therefore, n (y) = ni (y) for all y ∈ / N and this shows n is unique. In particular, although it appears to depend on the partition of unity {ψ i } from its definition, this is not the case. It only remains to verify |n (y)| = 1 a.e. I will do this by showing how to compute n. In particular, I will show that n = ni a.e. on ∂U ∩ Qi . Let W ⊆ W ⊆ Qi ∩ ∂U where W is open in ∂U. Let O be an open set such that O ∩ ∂U = W and O ⊆ Qi . Using Corollary 19.7.8 there exists a C ∞ partition of unity {ψ m } such that ψ i = 1 on O. Therefore, if m 6= i, ψ m = 0 on O. Then if f ∈ Cc1 (O) , Z f w · ndHn−1 W Z Z n−1 = f w · ndH = ∇f · wdmn U Z∂U = ∇ (ψ i f ) · wdmn U

which by the first part of the argument given above equals Z Z ψ i f ni · wdHn−1 = f w · ni dHn−1 . W

W

612

HAUSDORFF MEASURE

Thus for all f ∈ Cc1 (O) , Z

Z f w · ndH

n−1

f w · ni dHn−1

=

W

(19.7.65)

W

Since Cc1 (O) is dense in Cc (O) , the above equation is also true for all f ∈ Cc (O). Now ¡ ¢letting h ∈ Cc (W ) , the Tietze extension theorem implies there exists f1 ∈ C O whose restriction to W equals h. Let f be defined by ¡ ¢ dist x, OC f1 (x) = f (x) . dist (x, spt (h)) + dist (x, OC ) Then f = h on W and so this has shown that for all h ∈ Cc (W ) , 19.7.65 holds n−1 for h in place of f. But as observed is outer and inner regular on ¡ earlier, ¢H 1 n−1 ∂U and so Cc (W ) is dense in L W, H which implies w · n (y) = w · ni (y) for a.e. y. Considering a countable dense subset of the unit sphere as above, this implies n (y) = ni (y) a.e. y. This proves |n (y)| = 1 a.e. and in fact n (y) can be computed by using the formula for ni (y). This proves the lemma. It remains to prove Lemma 19.7.13. T Proof of Lemma 19.7.13: Let h (x) = (x1 , · · · , xn−1 , g (x1 , · · · , xn−1 ))   1 0  ..  . ..   . .. Dh (x) =  .   0  1 g,x1 · · · g,xn−1 Therefore,

¡ ¡ ¢¢1/2 ∗ J (x) = det Dh (x) Dh (x) .

Therefore, J (x) is the square root of the determinant of the following n × n matrix.   1 + (g,x1 )2 g,x1 g,x2 · · · g,x1 g,xn−1  g,x2 g,x1  1 + (g,x2 )2 · · · g,x2 g,xn−1   (19.7.66)  .. . . ..  .  . .. g,xn−1 g,x1

g,xn−1 g,x2

···

1 + (g,xn−1 )2

I need to show the determinant of the above matrix equals 1+

n−1 X

2

(g,xi (x)) .

i=1

This is implied by the following claim. To simplify the notation I will replace n − 1 with n. Claim: Let a1 , · · · , an be real numbers and let A (a1 , · · · , an ) be the matrix which has 1 + a2i in the iith slot and ai aj in the ij th slot when i 6= j. Then det A = 1 +

n X i=1

a2i.


613

Proof of the claim: The matrix, A (a1 , · · · , an ) is of the form   1 + a21 a1 a2 · · · a1 an  a1 a2 1 + a22 a2 an    A (a1 , · · · , an ) =   .. .. . .   . . . 2 a1 an a2 an · · · 1 + an Now consider the product  1 0 ···  0 1   .. ..  . .   0 −a1 −a2 · · ·

of a matrix and its  0 a1 1  0 0 a2   ..   ..  .   . 1 an   0 −an 1 a1

transpose, B T B below.  0 · · · 0 −a1 1 0 −a2   ..  .. . .   1 −an  a2 · · · an 1

(19.7.67)

This product equals a matrix of the form ¶ µ A (a1 , · · · , an ) P0n 0 1 + i=1 a2i ¢ ¡ ¢2 ¡ Pn 2 Therefore, 1 + i=1 a2i det (A (a1 , · · · , an )) = det (B) = det B T . However, using row operations,   1 0 ··· 0 a1  0 1  0 a2     .. .. det B T = det  ...  . .    0  1 a n Pn 2 0 0 · · · 0 1 + i=1 ai n X = 1+ a2i i=1

and therefore, Ã 1+

n X

! a2i

Ã det (A (a1 , · · · , an )) =

i=1

1+

n X

!2 a2i

i=1

¡ ¢ Pn which shows det (A (a1 , · · · , an )) = 1 + i=1 a2i . This proves the claim. Now the above lemma implies the divergence theorem. Theorem 19.7.14 Let U be a bounded open set with a Lipschitz boundary which lies on one side of its boundary. Then if f ∈ Cc1 (Rn ) , Z Z f,k (x) dmn = f nk dHn−1 (19.7.68) U

∂U

614

HAUSDORFF MEASURE

where n = (n1 , · · · , nn ) is the Hn−1 measurable unit vector of Lemma 19.7.12. Also, if F is a vector field such that each component is in Cc1 (Rn ) , then Z Z ∇ · F (x) dmn = F · ndHn−1 . (19.7.69) U

∂U

Proof: To obtain 19.7.68 apply Lemma 19.7.12 to w = ek . Then to obtain 19.7.69 from this, Z ∇ · F (x) dmn U

=

n Z X j=1

Z =

Fj,j dmn = U

n X

∂U j=1

n Z X j=1

Fj nj dH

n−1

Fj nj dHn−1

∂U

Z F · ndHn−1 .

= ∂U

This proves the theorem. What is the geometric significance of the vector, n? Recall that in the part of the boundary contained in Qi , this vector points in the same direction as the vector Ni (x1 , · · · , xn−1 , gi (x1 , · · · , xn−1 )) given by (−gi,1 (x1 , · · · , xn−1 ) , −gi,2 (x1 , · · · , xn−1 ) , · · · , −gi,n−1 (x1 , · · · , xn−1 ) , 1) (19.7.70) in the case where k = n. This vector is the gradient of the function, xn − gi (x1 , · · · , xn−1 ) and so is perpendicular to the level surface given by xn − gi (x1 , · · · , xn−1 ) = 0 in the case where gi is C 1 . It also points away from U so the vector n is the unit outer normal. The other cases work similarly. The divergence theorem is valid in situations more general than for Lipschitz boundaries. What you need is essentially the ability to say that the functions, gi above can be differentiated a.e. and more importantly that these functions can be recovered by integrating their partial derivatives. In other words, you need absolute continuity in each variable. Later in the chapter on weak derivatives, examples of such functions which are more general than Lipschitz functions will be discussed. However, the Lipschitz functions are pretty general and will suffice for now.

Differentiation With Respect To General Radon Measures This is a brief chapter on certain important topics on the differentiation theory for general Radon measures. For different proofs and some results which are not discussed here, a good source is [24] which is where I first read some of these things.

20.1

Besicovitch Covering Theorem

The fundamental theorem of calculus presented above for Lebesgue measures can be generalized to arbitrary Radon measures. It turns out that the same approach works if a different covering theorem is employed instead of the Vitali theorem. This covering theorem is the Besicovitch covering theorem of this section. It is necessary because for a general Radon measure µ, it is no longer the case that ³ ´the measure b in terms of is translation invariant. This implies there is no way to estimate µ B µ (B) and thus the Vitali covering theorem is of no use. In the Besicovitch covering theorem the balls in the covering are not enlarged as they are in the Vitali theorem. In this theorem they can also be either open or closed or neither open nor closed. The balls can also be taken with respect to any norm on Rn . The notation, B (x,r) in the above argument will denote any set which satisfies

{y : ||y − x|| < r} ⊆ B (x, r) ⊆ {y : ||y − x|| < r} and the norm, ||·|| is just some norm on Rn . The following picture is a distorted picture of the situation described in the following lemma. 615

616

DIFFERENTIATION WITH RESPECT TO GENERAL RADON MEASURES

ry 0r rx

Lemma 20.1.1 Let 10 ≤ rx ≤ ry and suppose B (x, rx ) and B (y, ry ) both have nonempty intersection with B (0, 1) but neither of these balls contains 0. Suppose also that ||x − y|| ≥ ry so that neither ball contains both centers in its interior. Then ¯¯ ¯¯ ¯¯ x y ¯¯¯¯ 4 ¯¯ − ¯¯ ||x|| ||y|| ¯¯ ≥ 5 . Proof: By hypothesis, ||x|| ≥ rx ≥ ||x|| − 1, ||y|| ≥ ry ≥ ||y|| − 1. Then

¯¯ ¯¯ ¯¯ ¯¯ ¯¯ x y ¯¯¯¯ ¯¯¯¯ x ||y|| − ||x|| y ¯¯¯¯ ¯¯ ¯¯ ||x|| − ||y|| ¯¯ = ¯¯ ||x|| ||y|| ¯¯ ¯¯ ¯¯ ¯¯ x ||y|| − y ||y|| + y ||y|| − ||x|| y ¯¯ ¯¯ ¯ ¯ = ¯¯ ¯¯ ||x|| ||y|| ≥

||x − y|| |||y|| − ||x||| − ||x|| ||x||

Now there are two cases. First suppose ||y|| ≥ ||x|| . Then the above is larger than ry ||y|| ry (ry + 1) − +1≥ − +1 ||x|| ||x|| ||x|| ||x|| 1 1 1 9 = 1− ≥1− ≥1− = . ||x|| rx 10 10 ≥

(20.1.1)

20.1. BESICOVITCH COVERING THEOREM

617

Next suppose ||x|| ≥ ||y|| . Then 20.1.1 is at least as large as

= ≥ ≥ =

||x|| − ||y|| ry − ||x|| ||x|| ry ||y|| −1+ ||x|| ||x|| 2ry 2ry −1≥ −1 ||x|| rx + 1 2rx 20 −1≥ −1 rx + 1 10 + 1 . 818 18

This proves the lemma. m

Lemma 20.1.2 There exists Ln depending on dimension, n, such that for {xk }k=1 distinct points on ∂B (0, 1) , if m ≥ Ln , then the distance between some pair of points m of {xk }k=1 is less than 4/5. L −1

m

n Proof: Let {zj }j=1 be a 1/3 net on ∂B (0, 1) . Then for m ≥ Ln , if {xk }k=1 is a set of m distinct points on ∂B (0, 1) , there must exist xi and xj for i 6= j such that both xi and xj are contained in some B (zk , 1/3) . This follows from the pigeon hole principle. There are more xi than there are B (zk , 1/3) and so one of these must have more than one xk in it. But then

||xi − xj || ≤ ||xi − zk || + ||zk − xj || ≤

2 4 < 3 5

This proves the lemma. Corollary 20.1.3 Let B0 = B (0,1) and let Bj = B (xj , rj ) for j = 1, · · · , K such that rj ≥ 10, 0 ∈ / Bj for all j > 0, Bj ∩ B0 6= ∅, and for all i 6= j, ||xi − xj || ≥ max (ri , rj ) . That is, no Bj contains two centers in its interior. Then K ≤ Ln , the constant of the above lemma. Proof: By¯¯ Lemma 20.1.2, ¯¯ if K > Ln , there exist two of the centers, xi and ¯¯ x i x j ¯¯ xj such that ¯¯ ||xi || − ||xj || ¯¯ < 4/5. By Lemma 20.1.1, ||xi − xj || < max (ri , rj ) contrary to the assumptions of the corollary. Hence K ≤ Ln as claimed. Theorem 20.1.4 There exists a constant Nn , depending only on n with the following property. If F is any collection of nonempty balls in Rn with sup {diam (B) : B ∈ F} < D < ∞ and if A is the set of centers of the balls in F, then there exist subsets of F, G1 , · · · , GNn , such that each Gi is a countable collection of disjoint balls from F and n A ⊆ ∪N i=1 {B : B ∈ Gi }.

618


Lemma 20.1.5 In the situation of Theorem 20.1.4, suppose the set of centers A is J bounded. Then there exists a sequence of balls from F, {Bj }j=1 where J ≤ ∞ such that 3 r (B1 ) ≥ sup {r (B) : B ∈ F} (20.1.2) 4 and if Am ≡ A \ (∪m (20.1.3) i=1 Bi ) 6= ∅, then

3 sup {r : B (a, r) ∈ F , a ∈ Am } . 4 Letting Bj = B (aj , rj ) , this sequence satisfies r (Bm+1 ) ≥

A ⊆ ∪Ji=1 Bi , r (Bk ) ≤

(20.1.4)

4 J r (Bj ) for j < k, {B (aj , rj /3)}j=1 are disjoint. 3 (20.1.5)

Proof: Pick B1 satisfying 20.1.2. If B1 , · · · , Bm have been chosen, and Am is given in 20.1.3, then if it equals ∅, it follows A ⊆ ∪m i=1 Bi . Set J = m. If Am 6= ∅, pick Bm+1 to satisfy 20.1.4. This defines the desired sequence. It remains to verify the claims in 20.1.5. Consider the second claim. Letting A0 ≡ A, Ak ⊆ Aj−1 and so sup {r : B (a, r) ∈ F , a ∈ Aj−1 } ≥ rk . Hence rj ≥ (3/4) rk . This proves the second claim of 20.1.5. Consider the third claim of 20.1.5. Suppose to the contrary that x ∈ B (aj , rj /3)∩ B (ai , ri /3) where i < j. Then ||ai − aj || ≤ < =

||ai − x|| + ||x − aj || µ ¶ 1 1 4 (ri + rj ) ≤ ri + ri 3 3 3 7 ri < ri 9

contrary to the construction which requires aj ∈ / B (ai , ri ). Finally consider the first claim of 20.1.5. It is true if J < ∞. This follows from the construction. If J = ∞, then since A is bounded and the balls, B (aj , rj /3) are disjoint, it must be the case that limi→∞ ri = 0. Suppose J = ∞ so that Am 6= ∅ for all m. If a0 fails to be covered, then a0 ∈ Ak for all k. Let a0 ∈ B (a0 , r0 ) ∈ F 1 for some ball B (a0 , r0 ) . Then for i large enough, ri < 10 r0 and so since a0 ∈ Ai−1 , 3 1 3 r0 ≤ sup {r : B (a, r) ∈ F, a ∈ Ai−1 } ≤ ri < r0 , 4 4 10 a contradiction. This proves the lemma. Lemma 20.1.6 There exists a constant Mn depending only on n such that for each 1 ≤ k ≤ J, Mn exceeds the number of sets Bj for j < k which have nonempty intersection with Bk .

20.1. BESICOVITCH COVERING THEOREM

619

Proof: These sets Bj which intersect Bk are of two types. Either they have large radius, rj > 10rk , or they have small radius, rj ≤ 10rk . In this argument let card (S) denote the number of elements in the set S. Define for fixed k, I ≡ {j : 1 ≤ j < k, Bj ∩ Bk 6= ∅, rj ≤ 10rk }, K ≡ {j : 1 ≤ j < k, Bj ∩ Bk 6= ∅, rj > 10rk }. r ¢ Claim 1: B aj , 3j ⊆ B (ak , 15rk ) for j ∈ I. Proof: Let j ∈ I. Then Bj ∩ Bk 6= ∅ and rj ≤ 10rk . Now if ³ rj ´ x ∈ B aj , , 3 ¡

then since rj ≤ 10rk , ||x − ak ||

≤

||x − aj || + ||aj − ak || ≤

rj + rj + rk = 3

4 4 43 rj + rk ≤ (10rk ) + rk = rk < 15rk . 3 3 3 ¡ r ¢ Therefore, B aj , 3j ⊆ B (ak , 15rk ). Claim 2: card (I) ≤ 60n . Proof: Recall r (Bk ) ≤ 34 r (Bj ) . Then letting α (n) rn be the Lebesgue measure of the n dimensional ball of radius r, (Note this α (n) depends on the norm used.) ³ ³ X rj ´´ α (n) 15n rkn ≡ mn (B (ak , 15rk )) ≥ m n B aj , 3 j∈I µ ¶ ´ ³ ´ ³ X rk n 4 rj n X ≥ α (n) since rk ≤ rj = α (n) 3 4 3 j∈I j∈I ³ r ń k = card (I) α (n) 4 and so it follows card (I) ≤ 60n as claimed. Claim 3: card (K) ≤ Ln where Ln is the constant of Corollary 20.1.3. Proof: Consider {Bj : j ∈ K} and Bk . Let f (x) ≡ rk−1 (x − xk ) . Then f (Bk ) = B (0, 1) and µ ¶ xj − xk −1 f (Bj ) = rk B (xj − xk , rj ) = B , rj /rk . rk Then rj /rk ≥ 10 because j ∈ K. None of the balls, f (Bj ) contain 0 but all these balls intersect B (0, 1) and as just noted, each of these balls has radius ≥ 10 and none of them contains two centers on its interior. By Corollary 20.1.3, it follows there are no more than Ln of them. This proves the claim. A constant which will satisfy the desired conditions is Mn ≡ Ln + 60n + 1.

620


This completes the proof of Lemma 20.1.6. J Next subdivide the balls {Bi }i=1 into Mn subsets G1 , · · · , GMn each of which consists of disjoint balls. This is done in the following way. Let B1 ∈ G1 . If B1 , · · · , Bk have each been assigned to one of the sets G1 , · · · , GMn , let Bk+1 ∈ Gr where r is the smallest index having the property that Bk+1 does not intersect any of the balls already in Gr . There must exist such an index r ∈ {1, · · · , Mn } because otherwise Bk+1 ∩ Bj 6= ∅ for at least Mn values of j < k + 1 contradicting Lemma 20.1.6. By Lemma 20.1.5 J n A ⊆ ∪M i=1 {B : B ∈ Gi } = ∪j=1 Bj .

This proves Theorem 20.1.4 in the case where A is bounded. To complete the proof of this theorem, the restriction that A is bounded must be removed. Define Al ≡ A ∩ {x ∈ Rn : 10 (l − 1) D ≤ ||x|| < 10lD} , l = 1, 2, · · · and Fl = {B (a,r) : B (a,r) ∈ F and a ∈ Al }. Then since D is an upper bound for all the diameters of these balls, (∪Fl ) ∩ (∪Fm ) = ∅

(20.1.6)

whenever m ≥ l + 2. Therefore, applying what was just shown to the pair (Al , Fl ), l such that each Gil is a countable collection of there exist subsets of Fl , G1l · · · GM n disjoint balls of Fl ⊆ F and © ª l n Al ⊆ ∪M i=1 B : B ∈ Gi . 2l−1 2l Now let Gj ≡ ∪∞ for 1 ≤ j ≤ Mn and for 1 ≤ j ≤ Mn , let Gj+Mn ≡ ∪∞ l=1 Gj l=1 Gj . Thus, letting Nn ≡ 2Mn , Nn ∞ A = ∪∞ l=1 A2l ∪ ∪l=1 A2l−1 ⊆ ∪j=1 {B : B ∈ Gj }

and by 20.1.6, each Gj is a countable set of disjoint balls of F. This proves the Besicovitch covering theorem.

20.2

Fundamental Theorem Of Calculus For Radon Measures

In this section the Besicovitch covering theorem will be used to give a generalization of the Lebesgue differentiation theorem to general Radon measures. In what follows, µ will be a Radon measure, Z ≡ {x ∈ Rn : µ (B (x,r)) = 0 for some r > 0},

20.2. FUNDAMENTAL THEOREM OF CALCULUS FOR RADON MEASURES 621 ½

Z −

f dµ ≡

0 if x ∈ Z, R 1 µ(B(x,r))

B(x,r)

B(x,r)

f dµ if x ∈Z, /

and the maximal function M f : Rn → [0, ∞] is given by Z M f (x) ≡ sup − |f | dµ. r≤1

B(x,r)

Lemma 20.2.1 Z is measurable and µ (Z) = 0. Proof: For each x ∈ Z, there exists a ball B (x,r) with µ (B (x,r)) = 0. Let C be the collection of these balls. Since Rn has a countable basis, a countable subset, e of C also covers Z. Let C, ∞ Ce = {Bi }i=1 . Then letting µ denote the outer measure determined by µ, µ (Z) ≤

∞ X

µ (Bi ) =

i=1

∞ X

µ (Bi ) = 0

i=1

Therefore, Z is measurable and has measure zero as claimed. Theorem 20.2.2 Let µ be a Radon measure and let f ∈ L1 (Rn , µ). Then Z lim − |f (y) − f (x)| dµ (y) = 0 r→0

B(x,r)

for µ a.e. x ∈ Rn . Proof: First consider the following claim which is a weak type estimate of the same sort used when differentiating with respect to Lebesgue measure. Claim 1: µ ([M f > ε]) ≤ Nn ε−1 ||f ||1 Proof: First note A ∩ Z = ∅. For each x ∈ A there exists a ball Bx = B (x,rx ) with rx ≤ 1 and Z −1 µ (Bx ) |f | dµ > ε. B(x,rx )

Let F be this collection of balls so that A is the set of centers of balls of F. By the Besicovitch covering theorem, n A ⊆ ∪N i=1 {B : B ∈ Gi }

where Gi is a collection of disjoint balls of F. Now for some i, µ (A) /Nn ≤ µ (∪ {B : B ∈ Gi })

622


because if this is not so, then µ (A) ≤

Nn X

µ (∪ {B : B ∈ Gi })
ε

(20.2.8)

B(x,r)

Z x : lim sup −

ε µ |f (y) − g (y)| dµ (y) > 2 r→0 B(x,r) ³h ε i´ +µ x : |g (x) − f (x)| > . 2

#!

20.2. FUNDAMENTAL THEOREM OF CALCULUS FOR RADON MEASURES 623 ≤µ Now

³h ³h ε i´ ε i´ M (f − g) > + µ |f − g| > 2 2

Z [|f −g|> 2ε ]

|f − g| dµ ≥

(20.2.9)

ε ³h ε i´ µ |f − g| > 2 2

and so from Claim 1 20.2.9 and hence 20.2.8 is dominated by µ ¶ 2 Nn + ||f − g||L1 (Rn ,µ) . ε ε But by regularity of Radon measures, Cc (Rn ) is dense in L1 (Rn , µ) and so since g in the above is arbitrary, this shows 20.2.8 equals 0. Now Ã" #! Z µ

x : lim sup − r→0

|f (y) − f (x)| dµ (y) > 0

B(x,r)

Ã"

#!! 1 = µ ∪∞ |f (y) − f (x)| dµ (y) > m=1 m r→0 B(x,r) #! Ã" Z ∞ X 1 ≤ µ x : lim sup − |f (y) − f (x)| dµ (y) > = 0. m r→0 B(x,r) m=1 Ã

Z x : lim sup −

By Lemma 20.2.1 the set Z is a set of measure zero and so if " # Z x∈ / lim sup − |f (y) − f (·)| dµ (y) > 0 ∪ Z r→0

B(·,r)

the above has shown 0

Z ≤ lim inf − |f (y) − f (x)| dµ (y) r→0 B(x,r) Z ≤ lim sup − |f (y) − f (x)| dµ (y) = 0 r→0

B(x,r)

which proves the theorem. The following corollary is the main result referred to as the Lebesgue Besicovitch Differentiation theorem. Corollary 20.2.3 If f ∈ L1loc (Rn , µ), Z |f (y) − f (x)| dµ (y) = 0 µ a.e. x. lim − r→0

(20.2.10)

B(x,r)

Proof: If f is replaced by f XB(0,k) then the conclusion 20.2.10 holds for all x ∈F / k where Fk is a set of µ measure 0. Letting k = 1, 2, · · · , and F ≡ ∪∞ k=1 Fk , it follows that F is a set of measure zero and for any x ∈ / F , and k ∈ {1, 2, · · · }, 20.2.10

624


holds if f is replaced by f XB(0,k) . Picking any such x, and letting k > |x| + 1, this shows Z lim −

r→0

Z = lim − r→0

B(x,r)

|f (y) − f (x)| dµ (y)

B(x,r)

¯ ¯ ¯f XB(0,k) (y) − f XB(0,k) (x)¯ dµ (y) = 0.


20.3

Slicing Measures

Let µ be a finite Radon measure. I will show here that a formula of the following form holds. Z Z Z µ (F ) = XF (x, y) dν x (y) dα (x) dµ = F

Rn

Rm

where α (E) = µ (E × Rm ). When this is done, the measures, ν x , are called slicing measures and this shows that an integral with respect to µ can be written as an iterated integral in terms of the measure α and the slicing measures, ν x . This is like going backwards in the construction of product measure. One starts with a measure, µ, defined on the Cartesian product and produces α and an infinite family of slicing measures from it whereas in the construction of product measure, one starts with two measures and obtains a new measure on a σ algebra of subsets of the Cartesian product of two spaces. First here are two technical lemmas. Lemma 20.3.1 The space Cc (Rm ) with the norm ||f || ≡ sup {|f (y)| : y ∈ Rm } is separable. Proof: Let Dl consist of all functions which are of the form ³ ³ ´ńα X C aα yα dist y,B (0,l + 1) |α|≤N

where aα ∈ Q, α is a multi-index, and nα is a positive integer. Then Dl is countable, separates the points of B (0,l) and annihilates ³no point´ of B (0,l). By the Stone Weierstrass theorem Dl is dense in the space C B (0,l) and so ∪ {Dl : l ∈ N} is a countable dense subset of Cc (Rm ). From the regularity of Radon measures, the following lemma follows. Lemma 20.3.2 If µ and ν are two Radon measures defined on σ algebras, Sµ and Sν , of subsets of Rn and if µ (V ) = ν (V ) for all V open, then µ = ν and Sµ = Sν .

20.3. SLICING MEASURES

625

Proof: Every compact set is a countable intersection of open sets so the two measures agree on every compact set. Hence it is routine that the two measures agree on every Gδ and Fσ set. (Recall Gδ sets are countable intersections of open sets and Fσ sets are countable unions of closed sets.) Now suppose E ∈ Sν is a bounded set. Then by regularity of ν there exists G a Gδ set and F, an Fσ set such that F ⊆ E ⊆ G and ν (G \ F ) = 0. Then it is also true that µ (G \ F ) = 0. Hence E = F ∪ (E \ F ) and E \ F is a subset of G \ F, a set of µ measure zero. By completeness of µ, it follows E ∈ Sµ and µ (E) = µ (F ) = ν (F ) = ν (E) . If E ∈ Sν not necessarily bounded, let Em = E ∩ B (0, m) and then Em ∈ Sµ and µ (Em ) = ν (Em ) . Letting m → ∞, E ∈ Sµ and µ (E) = ν (E) . Similarly, Sµ ⊆ Sν and the two measures are equal on Sµ . The main result in the section is the following theorem. Theorem 20.3.3 Let µ be a finite Radon measure on Rn+m defined on a σ algebra, F. Then there exists a unique finite Radon measure, α, defined on a σ algebra, S, of sets of Rn which satisfies α (E) = µ (E × Rm )

(20.3.11)

for all E Borel. There also exists a Borel set of α measure zero, N , such that for each x ∈ / N , there exists a Radon probability measure ν x such that if f is a nonnegative µ measurable function or a µ measurable function in L1 (µ), y → f (x, y) is ν x measurable α a.e. Z x→ f (x, y) dν x (y) is α measurable

(20.3.12)

Rm

and

Z

µZ

Z f (x, y) dµ = Rn+m

Rn

Rm

¶ f (x, y) dν x (y) dα (x).

(20.3.13)

If νbx is any other collection of Radon measures satisfying 20.3.12 and 20.3.13, then νbx = ν x for α a.e. x. Proof: First consider the uniqueness of α. Suppose α1 is another Radon measure satisfying 20.3.11. Then in particular, α1 and α agree on open sets and so the two measures are the same by Lemma 20.3.2. To establish the existence of α, define α0 on Borel sets by α0 (E) = µ (E × Rm ). Thus α0 is a finite Borel measure and so it is finite on compact sets. Lemma 11.1.3 on Page 11.1.3 implies the existence of the Radon measure α extending α0 . Next consider the uniqueness of ν x . Suppose ν x and νbx satisfy all conclusions b respectively. Then, of the theorem with exceptional sets denoted by N and N

626


b , one may also assume, using Lemma 20.2.1, that for x ∈ b, enlarging N and N / N ∪N α (B (x,r)) > 0 whenever r > 0. Now let A=

m Y

(ai , bi ]

i=1

where ai and bi are rational. Thus there are countably many such sets. Then from b, the conclusion of the theorem, if x0 ∈ / N ∪N Z Z 1 XA (y) dν x (y) dα α (B (x0 , r)) B(x0 ,r) Rm Z Z 1 = XA (y) db ν x (y) dα, α (B (x0 , r)) B(x0 ,r) Rm and by the Lebesgue Besicovitch Differentiation theorem, there exists a set of α b , then the limit in the above exists measure zero, EA , such that if x0 ∈ / EA ∪ N ∪ N as r → 0 and yields ν x0 (A) = νbx0 (A). Letting E denote the union of all the sets EA for A as described above, it follows b then ν x (A) = νbx (A) for that E is a set of measure zero and if x0 ∈ / E∪N ∪N 0 0 all such sets A. But every open set can be written as a disjoint union of sets of this form and so for all such x0 , ν x0 (V ) = νbx0 (V ) for all V open. By Lemma 20.3.2 this shows the two measures are equal and proves the uniqueness assertion for ν x . It remains to show the existence of the measures ν x . This will be done with the aid of the following lemma. The idea is to define a positive linear functional which will yield the desired measure, ν x and this lemma will help in making this definition. Lemma 20.3.4 There exists a set N of α measure 0, independent of f ∈ Cc (Rn+m ) such that if x ∈ / N , α (B (x,r)) > 0 for all r > 0 and Z 1 lim f dµ = gf (x) r→0 α (B (x,r)) B(x,r)×Rm where gf is a α measurable function with the property that Z Z gf (x) dα = f dµ. Rn

Rn ×Rm

Proof: Let f ∈ Cc (Rn+m ) and let Z η f (E) ≡

f dµ. E×Rm

Then η f is a finite measure defined on the Borel sets with η f ¿ α. By the Radon Nikodym theorem, there exists a Borel measurable function gef such that for all Borel E, Z Z η f (E) ≡

f dµ = E×Rm

gef dα. E

(20.3.14)


627

By the theory of differentiation for Radon measures, there exists a set of α measure zero, Nf such that if x ∈N / f , then α (B (x,r)) > 0 for all r > 0 and Z Z 1 lim f dµ = lim − gef dα = gef (x). r→0 α (B (x,r)) B(x,r)×Rm r→0 B(x,r) Let D be a countable dense subset of Cc (Rn+m ) and let N ≡ ∪ {Nf : f ∈ D}. Then if f ∈ Cc (Rn+m ) is arbitrary, and x ∈N / , referring to 20.3.14, it follows there exists g ∈ D close enough to f such that for all r1 , r2 , ¯ ¯ Z Z ¯ ¯ 1 1 ¯ ¯ f dµ − f dµ¯ < ¯ ¯ ¯ α (B (x,r1 )) B(x,r1 )×Rm α (B (x,r2 )) B(x,r2 )×Rm ¯ ¯ Z Z ¯ ε ¯¯ 1 1 ¯ +¯ gdµ − gdµ¯ ¯ 2 ¯ α (B (x,r1 )) B(x,r1 )×Rm α (B (x,r2 )) B(x,r2 )×Rm Therefore, taking ri small enough, the right side is less than ε. Since ε is arbitrary, this shows the limit as r → 0 exists. Define this limit which exists for all f ∈ Cc (Rn+m ) and x ∈ / N as gf (x). By the first part of the argument, gf (x) = gef (x) a.e. Thus, gf (x) is α measurable because it equals a Borel measurable function α a.e. The final formula follows from Z Z Z gf (x) dα = gef (x) dα ≡ η f (Rn ) ≡ f dµ. Rn

Rn

Rn ×Rm

This proves the lemma. Continuing with the proof of the theorem, let x ∈ / N and let f (z, y) ≡ ψ (z) φ (y) where ψ and φ are continuous functions with compact support in Rn and Rm respectively. Suppose first that ψ (x) = 1. Then define a positive linear functional Z 1 Lx φ ≡ lim ψ (z) φ (y) dµ (z, y). r→0 α (B (x,r)) B(x,r)×Rm This functional may appear to depend on the choice of ψ satisfying ψ (x) = 1 but this is not the case because all such ψ 0 s used in the definition of Lx are continuous. Let ν x be the Radon measure representing Lx . Thus replacing an arbitrary ψ ψ ∈ Cc (Rn ) with ψ(x) , in the case when ψ (x) 6= 0, Z ψ (x) Lx (φ)

= ψ (x)

φ (y) dν x (y) Z ψ ψ (x) φdµ = lim r→0 α (B (x,r)) B(x,r)×Rm ψ (x) Z 1 = lim ψφdµ r→0 α (B (x,r)) B(x,r)×Rm Rm

628


By Lemma 20.3.4, Z Rn

Z

Z

Rm

ψφdν x dα =

Z

Rn

gψφ dα =

ψφdµ. Rn ×Rm

Letting ψ k , φk increase to 1 pointwise, the monotone convergence theorem implies Z Z dµ = µ (Rn × Rm ) < ∞ (20.3.15) ν x (Rm ) dα = Rn ×Rm

Rn

showing that x →ν x (Rm ) is a function in L1 (α). In particular, this function is finite α a.e. Summarizing, the above has shown that whenever ψ ∈ Cc (Rn ) and φ ∈ Cc (Rm ) Z Z Z Rn

Rm

ψφdν x dα =

ψφdµ.

(20.3.16)

Rn ×Rm

Also ν x is a finite measure α a.e. and ν x and α are Radon measures. Next it is shown that ψφ can be replaced by XE where E is an arbitrary µ measurable set. To do so, let R1 ≡

n Y

(ai , bi ], R2 ≡

i=1

m Y

(ci , di ]

(20.3.17)

i=1

and let ψ k be a sequence of functions in Cc (Rn ) which is bounded, piecewise linear in each variable, and converging pointwise to XR1 . Also let φk be a similar sequence converging pointwise to XR2 . Then by the dominated convergence theorem, Z Z XR1 (x) XR2 (y) dµ = lim ψ k (x) φk (y) dµ k→∞

Rn ×Rm

= lim

k→∞

Rn

Rn ×Rm

µZ

Z ψ k (x)

Rm

¶ φk (y) dν x dα.

(20.3.18)

Since ν x is finite α a.e., it follows that for α a.e. x, Z Z lim φk (y) dν x = XR2 (y) dν x . k→∞

Rm

Rm

Since the φk are uniformly bounded, 20.3.15 implies the existence of a dominating function for the integrand in 20.3.18. Therefore, one can take the limit inside the integrals and obtain Z Z Z XR1 (x) XR2 (y) dµ = XR1 (x) XR2 (y) dν x dα Rn ×Rm

Rn

Rm

Every open set, V in Rn+m is a countable disjoint union of such half open rectangles and so the monotone convergence theorem implies for all V open in Rn+m , Z Z Z XV dµ = XV dν x dα. (20.3.19) Rn ×Rm

Rn

Rm


629

Since every compact set is the countable intersection of open sets, the above formula holds for V replaced with K where K is compact. Then it follows from the dominated convergence and monotone convergence theorems that whenever H is either a Gδ (countable intersection of open sets) or a Fσ (countable union of closed sets) Z Z Z Rn ×Rm

XH dµ =

Rn

Rm

XH dν x dα.

Now let E be µ measurable. Using the regularity of µ there exists F, G such that F is Fσ , G is Gδ , µ (G \ F ) = 0, and F ⊆ E ⊆ G. Also a routine application of the dominated convergence theorem and 20.3.19 shows Z Z Z X(G\F ) dµ = X(G\F ) dν x dα Rn ×Rm

Rn

Rm

and (G \ F )x ≡ {y : (x, y) ∈ G \ F } is ν x measurable for α a.e. x, wherever ν x is a finite measure, and for α a.e. x, ν x (G \ F )x = 0. Therefore, for α a.e. x, Ex is ν x measurable because Ex = Fx + Sx where Sx ⊆ (G \ F )x , a set of ν x measure zero and Fx is an Fσ set which is measurable because ν x is a Radon measure coming as it does from a positive linear functional. Therefore, Z Z Z Z XE dµ = XF dµ = XF dν x dα n m Rn ×Rm Rn Rm ZR ×R Z = XE dν x dα. (20.3.20) Rn

Rm

It follows from 20.3.20 that one can replace XE with an arbitrary nonnegative µ measurable simple function, s. Letting f be a nonnegative µ measurable function, it follows there is an increasing sequence of nonnegative simple functions converging to f pointwise and so by the monotone convergence theorem, Z Z Z f dµ = f dν x dα Rn ×Rm

Rn

Rm

R where y → f (x, y) is ν x measurable for α a.e. x and x → Rm f dν x is α measurable so the iterated integral makes sense. To see ν x is a probability measure for a.e. x, Z Z 1 dν x dα α (B (x, r)) B(x,r) Rm 1 = µ (B (x, r) × Rm ) = 1 α (B (x, r)) and so, using the fundamental theorem of calculus it follows that upon passing to a limit as r → 0, it follows that for α a.e. x Z m ν x (R ) = dν x = 1 Rm

Due to the regularity of the measures, all sets of measure zero may be taken to be Borel. In the case of f ∈ L1 (µ) , one applies the above to the positive and negative parts of the real and imaginary parts. This proves the theorem.

630

20.4


Vitali Coverings

There is another covering theorem which may also be referred to as the Besicovitch covering theorem. As before, the balls can be taken with respect to any norm on Rn . At first, the balls will be closed but this assumption will be removed. Definition 20.4.1 A collection of balls, F covers a set, E in the sense of Vitali if whenever x ∈ E and ε > 0, there exists a ball B ∈ F whose center is x having diameter less than ε. I will give a proof of the following theorem. Theorem 20.4.2 Let µ be a Radon measure on Rn and let E be a set with µ (E) < ∞. Where µ is the outer measure determined by µ. Suppose F is a collection of closed balls which cover E in the sense of Vitali. Then there exists a sequence of disjoint balls, {Bi } ⊆ F such that ¡ ¢ µ E \ ∪∞ j=1 Bj = 0. Proof: Let Nn be the constant of the Besicovitch covering theorem. Choose r > 0 such that ¶ µ 1 −1 (1 − r) ≡ λ < 1. 1− 2Nn + 2 If µ (E) = 0, there is nothing to prove so assume µ (E) > 0. Let U1 be an open set containing E with (1 − r) µ (U1 ) < µ (E) and 2µ (E) > µ (U1 ) , and let F1 be those sets of F which are contained in U1 whose centers are in E. Thus F1 is also a Vitali cover of E. Now by the Besicovitch covering theorem proved earlier, there exist balls, B, of F1 such that n E ⊆ ∪N i=1 {B : B ∈ Gi }

where Gi consists of a collection of disjoint balls of F1 . Therefore, µ (E) ≤

Nn X X

µ (B)

i=1 B∈Gi

and so, for some i ≤ Nn , X

(Nn + 1)

µ (B) > µ (E) .

B∈Gi

It follows there exists a finite set of balls of Gi , {B1 , · · · , Bm1 } such that (Nn + 1)

m1 X i=1

µ (Bi ) > µ (E)

(20.4.21)

20.4. VITALI COVERINGS

631

and so (2Nn + 2)

m1 X

µ (Bi ) > 2µ (E) > µ (U1 ) .

i=1

Since 2µ (E) ≥ µ (U1 ) , 20.4.21 implies m

1 X 2µ (E) µ (E) µ (U1 ) ≤ = < µ (Bi ) . 2N2 + 2 2N2 + 2 N2 + 1 i=1

Also U1 was chosen such that (1 − r) µ (U1 ) < µ (E) , and so µ λµ (E) ≥ λ (1 − r) µ (U1 ) =

≥ µ (U1 ) −

m1 X

1−

1

¶

2Nn + 2

µ (U1 )

¡ 1 ¢ µ (Bi ) = µ (U1 ) − µ ∪m j=1 Bj

i=1

¡

¢ ¡ ¢ m1 1 = µ U1 \ ∪m j=1 Bj ≥ µ E \ ∪j=1 Bj . Since the balls are closed, you can consider the sets of F which have empty intersecm1 1 tion with ∪m j=1 Bj and this new collection of sets will be a Vitali cover of E \∪j=1 Bj . Letting this collection of balls play the role of F in the above argument and letting 1 E \ ∪m j=1 Bj play the role of E, repeat the above argument and obtain disjoint sets of F, {Bm1 +1 , · · · , Bm2 } , such that ¡ ¢ ¡¡ ¢ ¢ ¡ ¢ m2 m2 1 1 λµ E \ ∪m E \ ∪m j=1 Bj > µ j=1 Bj \ ∪j=m1 +1 Bj = µ E \ ∪j=1 Bj , and so

¡ ¢ 2 λ2 µ (E) > µ E \ ∪m j=1 Bj .

Continuing in this way, yields a sequence of disjoint balls {Bi } contained in F and ¡ ¢ ¡ ¢ mk k µ E \ ∪∞ j=1 Bj ≤ µ E \ ∪j=1 Bj < λ µ (E ) ¡ ¢ for all k. Therefore, µ E \ ∪∞ j=1 Bj = 0 and this proves the Theorem. It is not necessary to assume µ (E) < ∞. Corollary 20.4.3 Let µ be a Radon measure on Rn . Letting µ be the outer measure determined by µ, suppose F is a collection of closed balls which cover E in the sense of Vitali. Then there exists a sequence of disjoint balls, {Bi } ⊆ F such that ¡ ¢ µ E \ ∪∞ j=1 Bj = 0.

632


Proof: Since µ is a Radon measure it is finite on compact sets. Therefore, ∞ there are at most countably many numbers, {bi }i=1 such that µ (∂B (0, bi )) > 0. It ∞ follows there exists an increasing sequence of positive numbers, {ri }i=1 such that limi→∞ ri = ∞ and µ (∂B (0, ri )) = 0. Now let D1 · · · , Dm

≡ {x : ||x|| < r1 } , D2 ≡ {x : r1 < ||x|| < r2 } , ≡ {x : rm−1 < ||x|| < rm } , · · · .

Let Fm denote those closed balls of F which are contained in Dm . Then letting Em denote E ∩ Dm , Fm is a Vitali cover of Em , µ (Em ) < ∞,© andªso by Theorem ∞ 20.4.2, there exists a countable sequence of balls from Fm Bjm j=1 , such that ª∞ ¡ ¢ © m = 0. Then consider the countable collection of balls, Bjm j,m=1 . µ Em \ ∪∞ j=1 Bj ¡ ¢ ∞ m µ E \ ∪∞ ≤ m=1 ∪j=1 Bj ∞ X ¡ ¢ m + µ Em \ ∪∞ = j=1 Bj

¡ ¢ µ ∪∞ j=1 ∂B (0, ri ) + 0

m=1

This proves the corollary. You don’t need to assume the balls are closed. In fact, the balls can be open closed or anything in between and the same conclusion can be drawn. Corollary 20.4.4 Let µ be a Radon measure on Rn . Letting µ be the outer measure determined by µ, suppose F is a collection of balls which cover E in the sense of Vitali, open closed or neither. Then there exists a sequence of disjoint balls, {Bi } ⊆ F such that ¢ ¡ µ E \ ∪∞ j=1 Bj = 0. Proof: Let x ∈ E. Thus x is the center of arbitrarily small balls from F. Since µ is a Radon measure, at most countably many radii, r of these balls can have the property that µ (∂B (0, r)) = 0. Let F 0 denote the closures of the balls of F, B (x, r) with the property that µ (∂B (x, r)) = 0. Since for each x ∈ E there are only countably many exceptions, F 0 is still a Vitali cover of©E. ªTherefore, by Corollary ∞ 20.4.3 there is a disjoint sequence of these balls of F 0 , Bi i=1 for which ¡ ¢ µ E \ ∪∞ j=1 Bj = 0 However, since their boundaries have µ measure zero, it follows ¡ ¢ µ E \ ∪∞ j=1 Bj = 0. This proves the corollary.

20.5

Differentiation Of Radon Measures

This section is a generalization of earlier material in which a measure was differentiated with respect to Lebesgue measure. Here an arbitrary Radon measure will

20.5. DIFFERENTIATION OF RADON MEASURES

633

be differentiated with respect to another arbitrary Radon measure. In this section, B (x, r) will denote a ball with center x and radius r. Also, let λ and µ be Radon measures and as above, Z will denote a µ measure zero set off of which µ (B (x, r)) > 0 for all r > 0. Definition 20.5.1 For x ∈Z, / define the upper and lower symmetric derivatives as λ (B (x, r)) λ (B (x, r)) , Dµ λ (x) ≡ lim inf . r→0 µ (B (x, r)) r→0 µ (B (x, r))

Dµ λ (x) ≡ lim sup respectively. Also define

Dµ λ (x) ≡ Dµ λ (x) = Dµ λ (x) in the case when both the upper and lower derivatives are equal. Lemma 20.5.2 Let ª λ and µ be Radon measures. If A is a bounded subset of © x∈ / Z : Dµ λ (x) ≥ a , then λ (A) ≥ aµ (A) ª © and if A is a bounded subset of x ∈ / Z : Dµ λ (x) ≤ a , then λ (A) ≤ aµ (A) © ª Proof: Suppose first that A is a bounded subset of x ∈ / Z : Dµ λ (x) ≥ a , let ε > 0, and let V be a bounded open set with V ⊇ A and λ (V ) − ε < λ (A) , µ (V ) − ε < µ (A) . Then if x ∈ A, λ (B (x, r)) > a − ε, B (x, r) ⊆ V, µ (B (x, r)) for infinitely many values of r which are arbitrarily small. Thus the collection of such balls constitutes a Vitali cover for A. By Corollary 20.4.4 there is a disjoint sequence of these balls {Bi } such that µ (A \ ∪∞ i=1 Bi ) = 0. Therefore, (a − ε)

∞ X

µ (Bi )
0 was arbitrary. © ª Now suppose A is a bounded subset of x ∈ / Z : Dµ λ (x) ≤ a and let V be a bounded open set containing A with µ (V ) − ε < µ (A) . Then if x ∈ A, λ (B (x, r)) < a + ε, B (x, r) ⊆ V µ (B (x, r)) for values of r which are arbitrarily small. Therefore, by Corollary 20.4.4 again, there exists a disjoint sequence of these balls, {Bi } satisfying this time, λ (A \ ∪∞ i=1 Bi ) = 0. Then by arguments similar to the above, λ (A) ≤

∞ X

λ (Bi ) < (a + ε) µ (V ) < (a + ε) (µ (A) + ε) .

i=1

Since ε was arbitrary, this proves the lemma. Theorem 20.5.3 There exists a set of measure zero, N containing Z such that for x ∈ / N, Dµ λ (x) exists and also XN C (·) Dµ λ (·) is a µ measurable function. Furthermore, Dµ λ (x) < ∞ µ a.e. Proof: First I show Dµ λ (x) exists a.e. Let 0 ≤ a b > a > Dµ λ (x) . By Lemma 20.5.2, aµ (A) ≥ λ (A) ≥ bµ (A) and so µ (A) = 0 and A is µ measurable. It follows µ (N (a, b)) = 0 because µ (N (a, b)) ≤

∞ X

µ (N (a, b) ∩ B (0, m)) = 0.

m=1

Define

ª © N0 ≡ x ∈ / Z : Dµ λ (x) > Dµ λ (x) .

Thus µ (N0 ) = 0 because N0 ⊆ ∪ {N (a, b) : 0 ≤ a < b, and a, b ∈ Q}

20.6. THE RADON NIKODYM THEOREM FOR RADON MEASURES

635

Therefore, N0 is also µ measurable and has µ measure zero. Letting N ≡ N0 ∪ Z, it follows Dµ λ (x) exists on N C . It remains to verify XN C (·) Dµ λ (·) is finite a.e. and is µ measurable. Let I = {x : Dµ λ (x) = ∞} . Then by Lemma 20.5.2 λ (I ∩ B (0, m)) ≥ aµ (I ∩ B (0, m)) for all a and since λ is finite on bounded sets, the above implies µ (I ∩ B (0, m)) = 0 for each m which implies that I is µ measurable and has µ measure zero since I = ∪∞ m=1 Im . Letting η be an arbitrary Radon measure, let r > 0, and suppose η (∂B (x, r)) = 0. (Since η is finite on every ball, there are only countably many r such that η (∂B (x, r)) > 0.) and let V be an open set containing B (x, r). Then whenever y is close enough to x, it follows that B (y, r) is also a subset of V. Since V is an arbitrary open set containing B (x, r), it follows ³ ´ η (B (x, r)) = η B (x, r) ≥ lim sup η (B (y, r)) y→x

and so y → η (B (y, r)) an upper semicontinuous real valued function of x, one which satisfies f (x) ≥ lim sup f (xn ) n→∞

whenever xn → x. Now it is routine to verify that a function f is upper semicontinuous if and only if f −1 ([−∞, a)) is open for all a ∈ R. Therefore, f −1 ([−∞, a)) is a Borel set for all a ∈ R and so f is Borel measurable by Lemma 8.1.6. Now the measurability of XN C (·) Dµ λ (·) follows from λ (B (x, ri )) XN C (x) ri →0 µ (B (x, ri ))

XN C (x) Dµ λ (x) = lim

where ri is such that ∂B (x, ri ) has µ and λ measure zero.

20.6

The Radon Nikodym Theorem For Radon Measures

The above theory can be used to give an alternate treatment of the Radon Nikodym theorem which exhibits the Radon Nikodym derivative as a specific limit. Theorem 20.6.1 Let λ and µ be Radon measures and suppose λ ¿ µ. Then for all E a µ measurable set, Z λ (E) = (Dµ λ) dµ. E

636


Proof: Let t > 1 and let E be a µ measurable set which is bounded and a subset of N C where N is the exceptional set of µ measure zero in Theorem 20.5.3 off of which µ (B (x,r)) > 0 for all r > 0 and Dµ λ (x) exists. Consider © ª Em ≡ E ∩ x ∈ N C : tm ≤ Dµ λ (x) < tm+1 for m ∈ Z, the integers. First note that © ª E ∩ x ∈ N C : Dµ λ (x) = 0 has λ measure zero because by Lemma 20.5.2, ¡ © ª¢ λ E ∩ x ∈ N C : Dµ λ (x) = 0 ≤ aµ (E ) for all a > 0 and µ (E ) is finite due to the assumption that E is bounded and µ is a Radon measure. Therefore, by Lemma 20.5.2, X X X λ (E) = λ (Em ) ≤ tm+1 µ (Em ) = t tm µ (Em ) m∈Z

m∈Z

XZ

≤t

m∈Z

m∈Z

Z Dµ λ (x) dµ = t

Em

Dµ λ (x) dµ. E

Also by this same lemma, X X X tm+1 µ (Em ) λ (E) = λ (Em ) ≥ tm µ (Em ) = t−1 m∈Z

≥ t−1

m∈Z

Thus,

m∈Z

m∈Z

XZ

Z Dµ λ (x) dµ = t−1

Dµ λ (x) dµ.

Em

E

Z

Z

t

Dµ λ (x) dµ ≥ λ (E) ≥ t E

−1

Dµ λ (x) dµ E

and letting t → 1, it follows Z λ (E) =

Dµ λ (x) dµ.

(20.6.25)

E

Now if E is an arbitrary measurable set, contained in N C , this formula holds with E replaced with E ∩ B (0,k) . Letting k → ∞ and using the monotone convergence theorem, the above formula holds for all E ⊆ N C . Since N is a set of µ measure zero, it follows N is also a set of λ measure zero due to the assumption of absolute continuity. Therefore 20.6.25 continues to hold for arbitrary µ measurable sets, E. This proves the theorem.

Fourier Transforms 21.1

An Algebra Of Special Functions

First recall the following definition of a polynomial. Definition 21.1.1 α = (α1 , · · · , αn ) for α1 · · · αn nonnegative integers is called a multi-index. For α a multi-index, |α| ≡ α1 + · · · + αn and if x ∈ Rn , x = (x1 , · · · , xn ) , and f a function, define αn 1 α2 xα ≡ xα 1 x2 · · · xn .

A polynomial in n variables of degree m is a function of the form X p (x) = aα xα . |α|≤m

Here α is a multi-index as just described and aα ∈ C. (α1 , · · · , αn ) a multi-index Dα f (x) ≡

Also define for α =

∂ |α| f . n · · · ∂xα n

α2 1 ∂xα 1 ∂x2

2

Definition 21.1.2 Define G1 to be the functions of the form p (x) e−a|x| where a > 0 and p (x) is a polynomial. Let G be all finite sums of functions in G1 . Thus G is an algebra of functions which has the property that if f ∈ G then f ∈ G. It is always assumed, unless stated otherwise that the measure will be Lebesgue measure. Lemma 21.1.3 G is dense in C0 (Rn ) with respect to the norm, ||f ||∞ ≡ sup {|f (x)| : x ∈ Rn } 637

638

FOURIER TRANSFORMS

Proof: By the Weierstrass approximation theorem, it suffices to show G separates the points and annihilates no point. It was already observed in the above definition that f ∈ G whenever f ∈ G. If y1 6= y2 suppose first that |y1 | 6= |y2 | . 2 Then in this case, you can let f (x) ≡ e−|x| and f ∈ G and f (y1 ) 6= f (y2 ). If |y1 | = |y2 | , then suppose y1k 6= y2k . This must happen for some k because y1 6= y2 . 2 2 Then let f (x) ≡ xk e−|x| . Thus G separates points. Now e−|x| is never equal to zero and so G annihilates no point of Rn . This proves the lemma. These functions are clearly quite specialized. Therefore, the following theorem is somewhat surprising. Theorem 21.1.4 For each p ≥ 1, p < ∞, G is dense in Lp (Rn ). Proof: Let f ∈ Lp (Rn ) . Then there exists g ∈ Cc (Rn ) such that ||f − g||p < ε. Now let b > 0 be large enough that Z ³ ´ 2 p e−b|x| dx < εp . Rn 2

Then x → g (x) eb|x| is in Cc (Rn ) ⊆ C0 (Rn ) . Therefore, from Lemma 21.1.3 there exists ψ ∈ G such that ¯¯ ¯¯ ¯¯ b|·|2 ¯¯ − ψ ¯¯ < 1 ¯¯ge ∞

Therefore, letting φ (x) ≡ e

−b|x|2

ψ (x) it follows that φ ∈ G and for all x ∈ Rn ,

|g (x) − φ (x)| < e−b|x|

2

Therefore, µZ p

|g (x) − φ (x)| dx

¶1/p

µZ ≤

³

−b|x|2

e

Rn

´p

¶1/p dx

< ε.

Rn

It follows ||f − φ||p ≤ ||f − g||p + ||g − φ||p < 2ε. Since ε > 0 is arbitrary, this proves the theorem. The following lemma is also interesting even if it is obvious. Lemma 21.1.5 For ψ ∈ G , p a polynomial, and α, β multiindices, Dα ψ ∈ G and pψ ∈ G. Also sup{|xβ Dα ψ(x)| : x ∈ Rn } < ∞

21.2

Fourier Transforms Of Functions In G

Definition 21.2.1 For ψ ∈ G Define the Fourier transform, F and the inverse Fourier transform, F −1 by Z e−it·x ψ(x)dx, F ψ(t) ≡ (2π)−n/2 Rn

21.2. FOURIER TRANSFORMS OF FUNCTIONS IN G

639

Z F −1 ψ(t) ≡ (2π)−n/2

eit·x ψ(x)dx. Rn

Pn where t · x ≡ i=1 ti xi .Note there is no problem with this definition because ψ is in L1 (Rn ) and therefore, ¯ it·x ¯ ¯e ψ(x)¯ ≤ |ψ(x)| , an integrable function. One reason for using the functions, G is that it is very easy to compute the Fourier transform of these functions. The first thing to do is to verify F and F −1 map G to G and that F −1 ◦ F (ψ) = ψ. Lemma 21.2.2 The following formulas are true √ Z Z π −c(x+it)2 −c(x−it)2 e e dx = dx = √ , c R R

(21.2.1)

µ √ ¶n π √ , c Rn Rn √ Z Z 2 π −ct2 −ist −ct2 ist − s4c √ , e e dt = e e dt = e c R R µ √ ¶n Z Z |s|2 π −c|t|2 −is·t −c|t|2 is·t − 4c √ . e e dt = e e dt = e c Rn Rn

Z

Z

e−c(x+it)·(x+it) dx =

e−c(x−it)·(x−it) dx =

(21.2.2) (21.2.3) (21.2.4)

Proof: Consider the first one. Simple manipulations yield Z Z 2 2 2 H (t) ≡ e−c(x+it) dx = ect e−cx cos (2cxt) dx. R

R

Now using the dominated convergence theorem to justify passing derivatives inside the integral where necessary and using integration by parts, Z Z 2 0 ct2 −cx2 ct2 H (t) = 2cte e cos (2cxt) dx − e e−cx sin (2cxt) 2xcdx R R Z ct2 −cx2 = 2ctH (t) − e 2ct e cos (2cxt) dx = 2ct (H (t) − H (t)) = 0 R

and so H (t) = H (0) =

R R

Z 2

I =

e R2

√

√

2

e−cx dx ≡ I. Thus

−c(x2 +y 2 )

Z

∞

Z

2π

dxdy = 0

0

2

e−cr rdθdr =

π . c

Therefore, I = π/ c. Since the sign of t is unimportant, this proves 21.2.1. This also proves 21.2.2 after writing as iterated integrals.

640

FOURIER TRANSFORMS

Consider 21.2.3. Z 2 e−ct eist dt =

Z e

R

´ ³ 2 is 2 −c t2 − ist − s4c c +( 2c )

e

R

=

2

Z

e

2

is −c(t− 2c )

− s4c

e

dt = e

dt 2

− s4c

R

√ π √ . c

Changing the variable t → −t gives the other part of 21.2.3. Finally 21.2.4 follows from using iterated integrals. With these formulas, it is easy to verify F, F −1 map G to G and F ◦ F −1 = −1 F ◦ F = id. Theorem 21.2.3 Each of F and F −1 map G to G. Also F −1 ◦ F (ψ) = ψ and F ◦ F −1 (ψ) = ψ. Proof: The first claim will be shown if it is shown that F ψ ∈ G for ψ (x) = x e because an arbitrary function of G is a finite sum of scalar multiples of functions such as ψ. Using Lemma 21.2.2, µ ¶n/2 Z 2 1 F ψ (t) ≡ e−it·x xα e−b|x| dx 2π Rn µZ ¶ µ ¶n/2 1 −|α| α −it·x −b|x|2 (i) Dt e e dx = 2π Rn µ ¶n/2 µ µ √ ¶n ¶ |t|2 1 π −|α| √ = (i) Dtα e− 2b 2π b α −b|x|2

|t|2

and this is clearly in G because it equals a polynomial times e− 2b . It remains to verify the other assertion. As in the first case, it suffices to consider ψ (x) = 2 xα e−b|x| . F −1 ◦ F (ψ) (s) µ ¶n/2 Z µ ¶n/2 Z 2 1 1 eis·t e−it·x xα e−b|x| dxdt 2π 2π Rn Rn

≡ µ = µ =

1 2π 1 2π

= =

µZ e

is·t

(−i)

Rn ¶n/2 Z

µ eis·t

Rn

−|α|

1 2π

Dtα

¶n/2

¶ e

−it·x −b|x|2

Rn −|α|

(−i)

e

dxdt

µ µ √ ¶n ¶ |t|2 π √ Dtα e− 4b dt b

¶n µ √ ¶n µ ¶ Z |t|2 π −|α| is·t α − 4b √ (−i) e Dt e dt b Rn µ ¶n µ √ ¶n Z |t|2 1 π −|α| |α| |α| √ (−i) (−1) sα (i) eis·t e− 4b dt 2π b Rn µ ¶n µ √ ¶n Z |t|2 1 π √ eis·t e− 4b dt sα 2π b Rn µ

=

¶n Z

1 2π

21.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING Ã √ !n ¶n µ √ ¶n |s|2 π π α − 4(1/(4b)) √ p s e b 1/ (4b) µ ¶n µ √ ¶n ³ ń √ √ 2 2 1 π √ sα e−b|s| π2 b = sα e−b|s| = ψ (s) . 2π b µ

= =

641

1 2π

This little computation proves the theorem. The other case is entirely similar.

21.3

Fourier Transforms Of Just About Anything

21.3.1

Fourier Transforms Of G ∗

Definition 21.3.1 Let G ∗ denote the vector space of linear functions defined on G which have values in C. Thus T ∈ G ∗ means T : G →C and T is linear, T (aψ + bφ) = aT (ψ) + bT (φ) for all a, b ∈ C, ψ, φ ∈ G Let ψ ∈ G. Then define Tψ ∈ G ∗ by Z Tψ (φ) ≡

ψ (x) φ (x) dx Rn

Lemma 21.3.2 The following is obtained for all φ, ψ ∈ G. ¡ ¢ TF ψ (φ) = Tψ (F φ) , TF −1 ψ (φ) = Tψ F −1 φ Also if ψ ∈ G and Tψ = 0, then ψ = 0. Proof: Z TF ψ (φ)

≡

F ψ (t) φ (t) dt Rn

Z =

Rn

Z =

µ

1 2π

ψ(x) Rn

Z =

Rn

¶n/2 Z e−it·x ψ(x)dxφ (t) dt µ

1 2π

Rn ¶n/2

Z e−it·x φ (t) dtdx Rn

ψ(x)F φ (x) dx ≡ Tψ (F φ)

The other claim is similar. Suppose now Tψ = 0. Then Z ψφdx = 0 Rn

for all φ ∈ G. Therefore, this is true for φ = ψ and so ψ = 0. This proves the lemma.

642

FOURIER TRANSFORMS

From now on regard G ⊆ G ∗ and for ψ ∈ G write ψ (φ) instead of Tψ (φ) . It was just shown that with this interpretation1 , ¡ ¢ F ψ (φ) = ψ (F (φ)) , F −1 ψ (φ) = ψ F −1 φ . This lemma suggests a way to define the Fourier transform of something in G ∗ . Definition 21.3.3 For T ∈ G ∗ , define F T, F −1 T ∈ G ∗ by ¡ ¢ F T (φ) ≡ T (F φ) , F −1 T (φ) ≡ T F −1 φ Lemma 21.3.4 F and F −1 are both one to one, onto, and are inverses of each other. Proof: First note F and F −1 are both linear. This follows directly from the definition. Suppose now F T = 0. Then F T (φ) = T (F¡φ) = 0 for ¢ all φ ∈ G. But F and F −1 map G onto G because if ψ ∈ G, then ψ = F F −1 (ψ) . Therefore, T = 0 and so F is one to one. Similarly F −1 is one to one. Now ¡ ¢ ¡ ¡ ¢¢ F −1 (F T ) (φ) ≡ (F T ) F −1 φ ≡ T F F −1 (φ) = T φ. Therefore, F −1 ◦ F (T ) = T. Similarly, F ◦ F −1 (T ) = T. Thus both F and F −1 are one to one and onto and are inverses of each other as suggested by the notation. This proves the lemma. Probably the most interesting things in G ∗ are functions of various kinds. The following lemma has to do with this situation. R Lemma 21.3.5 If f ∈ L1loc (Rn ) and Rn f φdx = 0 for all φ ∈ Cc (Rn ), then f = 0 a.e. Proof: First suppose f ≥ 0. Let E ≡ {x :f (x) ≥ r}, ER ≡ E ∩ B (0,R). Let Km be an increasing sequence of compact sets and let Vm be a decreasing sequence of open sets satisfying Km ⊆ ER ⊆ Vm , mn (Vm ) ≤ mn (Km ) + 2−m , V1 ⊆ B (0,R) . Therefore,

mn (Vm \ Km ) ≤ 2−m .

Let φm ∈ Cc (Vm ) , Km ≺ φm ≺ Vm . 1 This is not all that different from what was done with the derivative. Remember when you consider the derivative of a function of one variable, in elementary courses you think of it as a number but thinking of it as a linear transformation acting on R is better because this leads to the concept of a derivative which generalizes to functions of many variables. So it is here. You can think of ψ ∈ G as simply an element of G but it is better to think of it as an element of G ∗ as just described.

21.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING

643

Then φm (x) → XER (x) a.e. because the set where φm (x) fails to converge to this set is contained in the set of all x which are in infinitely many of the sets Vm \ Km . This set has measure zero because ∞ X

mn (Vm \ Km ) < ∞

m=1

and so, by the dominated convergence theorem, Z Z Z f φm dx = lim f φm dx = 0 = lim m→∞

m→∞

Rn

V1

f dx ≥ rm (ER ).

ER

Thus, mn (ER ) = 0 and therefore mn (E) = limR→∞ mn (ER ) = 0. Since r > 0 is arbitrary, it follows ¡£ ¤¢ mn ([f > 0]) = ∪∞ f > k −1 = 0. k=1 mn Now suppose f has values in R. Let E+ = [f ≥ 0] and E− = [f < 0] . Thus these are two measurable sets. As in the first part, let Km and Vm be sequences of compact and open sets such that Km ⊆ E+ ∩ B (0, R) ⊆ Vm ⊆ B (0, R) and let Km ≺ φm ≺ Vm with mn (Vm \ Km ) < 2−m . Thus φm ∈ Cc (Rn ) and the sequence converges pointwise to XE+ ∩B(0,R) . Then by the dominated convergence theorem, if ψ is any function in Cc (Rn ) Z Z 0 = f φm ψdmn → f ψXE+ ∩B(0,R) dmn . Hence, letting R → ∞, Z

Z f ψXE+ dmn =

f+ ψdmn = 0

Since ψ is arbitrary, the first part of the argument applies to f+ and implies f+ = 0. Similarly f− = 0. Finally, if f is complcx valued, the assumptions mean Z Z Re (f ) φdmn = 0, Im (f ) φdmn = 0 for all φ ∈ Cc (Rn ) and so both Re (f ) , Im (f ) equal zero a.e. This proves the lemma. Corollary 21.3.6 Let f ∈ L1 (Rn ) and suppose Z f (x) φ (x) dx = 0 Rn

for all φ ∈ G. Then f = 0 a.e.

644

FOURIER TRANSFORMS

Proof: Let ψ ∈ Cc (Rn ) . Then by the Stone Weierstrass approximation theorem, there exists a sequence of functions, {φk } ⊆ G such that φk → ψ uniformly. Then by the dominated convergence theorem, Z Z f ψdx = lim f φk dx = 0. k→∞

By Lemma 21.3.5 f = 0. The next theorem is the main result of this sort. Theorem 21.3.7 Let f ∈ Lp (Rn ) , p ≥ 1, or suppose f is measurable and has polynomial growth, ³ ´m 2 |f (x)| ≤ K 1 + |x| for some m ∈ N. Then if

Z f ψdx = 0

for all ψ ∈ G then it follows f = 0. Proof: The case where f ∈ L1 (Rn ) was dealt with in Corollary 21.3.6. Suppose 0 f ∈ Lp (Rn ) forR p > 1. Then by Holder’s inequality and the density of G in Lp (Rn ) , p0 n it follows that f gdx = 0 for all g ∈ L (R ) . By the Riesz representation theorem, f = 0. It remains to consider the case where f has polynomial growth. Thus x → 2 f (x) e−|x| ∈ L1 (Rn ) . Therefore, for all ψ ∈ G, Z 2 0 = f (x) e−|x| ψ (x) dx 2

2

because e−|x| ψ (x) ∈ G. Therefore, by the first part, f (x) e−|x| = 0 a.e. The following theorem shows that you can consider most functions you are likely to encounter as elements of G ∗ . Theorem 21.3.8 Let f be a measurable function with polynomial growth, ³ Ń 2 |f (x)| ≤ C 1 + |x| for some N, or let f ∈ Lp (Rn ) for some p ∈ [1, ∞]. Then f ∈ G ∗ if Z f (φ) ≡ f φdx. Proof: Let f have polynomial growth first. Then the above integral is clearly well defined and so in this case, f ∈ G ∗ . Next suppose f ∈ Lp (Rn ) with ∞ > p ≥ 1. Then it is clear again that the above integral is well defined because of the fact that φ is a sum of polynomials

21.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING 2

645

0

times exponentials of the form e−c|x| and these are in Lp (Rn ). Also φ → f (φ) is clearly linear in both cases. This proves the theorem. This has shown that for nearly any reasonable function, you can define its Fourier 0 transform as described above. Also you should note that G ∗ includes C0 (Rn ) , the space of complex measures whose total variation are Radon measures. It is especially interesting when the Fourier transform yields another function of some sort.

21.3.2

Fourier Transforms Of Functions In L1 (Rn )

First suppose f ∈ L1 (Rn ) . Theorem 21.3.9 Let f ∈ L1 (Rn ) . Then F f (φ) = µ g (t) = and F −1 f (φ) =

R Rn

1 2π

R Rn

gφdt where

¶n/2 Z e−it·x f (x) dx Rn

gφdt where g (t) =

¡

¢ R 1 n/2 2π Rn

eit·x f (x) dx. In short,

Z F f (t) ≡ (2π)

−n/2

e−it·x f (x)dx, Rn

Z F −1 f (t) ≡ (2π)−n/2

eit·x f (x)dx. Rn

Proof: From the definition and Fubini’s theorem, Z F f (φ) ≡

µ

Z

¶n/2 Z

f (t) F φ (t) dt = f (t) e−it·x φ (x) dxdt Rn Rn Z Ãµ ¶n/2 Z 1 f (t) e−it·x dt φ (x) dx. 2π n R Rn Rn

=

1 2π !

Since φ ∈ G is arbitrary, it follows from Theorem 21.3.7 that F f (x) is given by the claimed formula. The case of F −1 is identical. Here are interesting properties of these Fourier transforms of functions in L1 . Theorem 21.3.10 If f ∈ L1 (Rn ) and ||fk − f ||1 → 0, then F fk and F −1 fk converge uniformly to F f and F −1 f respectively. If f ∈ L1 (Rn ), then F −1 f and F f are both continuous and bounded. Also, lim F −1 f (x) = lim F f (x) = 0.

|x|→∞

|x|→∞

(21.3.5)

Furthermore, for f ∈ L1 (Rn ) both F f and F −1 f are uniformly continuous.

646

FOURIER TRANSFORMS

Proof: The first claim follows from the following inequality. Z ¯ −it·x ¯ ¯e |F fk (t) − F f (t)| ≤ (2π)−n/2 fk (x) − e−it·x f (x)¯ dx n ZR |fk (x) − f (x)| dx = (2π)−n/2 Rn

−n/2

= (2π)

||f − fk ||1 .

which a similar argument holding for F −1 . Now consider the second claim of the theorem. Z ¯ ¯ 0 ¯ −it·x ¯ |F f (t) − F f (t0 )| ≤ (2π)−n/2 − e−it ·x ¯ |f (x)| dx ¯e Rn

The integrand is bounded by 2 |f (x)|, a function in L1 (Rn ) and converges to 0 as t0 → t and so the dominated convergence theorem implies F f is continuous. To see F f (t) is uniformly bounded, Z −n/2 |F f (t)| ≤ (2π) |f (x)| dx < ∞. Rn

A similar argument gives the same conclusions for F −1 . It remains to verify 21.3.5 and the claim that F f and F −1 f are uniformly continuous. ¯ ¯ Z ¯ ¯ |F f (t)| ≤ ¯¯(2π)−n/2 e−it·x f (x)dx¯¯ n R

Now let ε > 0 be given and let g ∈ Then |F f (t)|

≤

≤

Cc∞

−n/2

n

(R ) such that (2π)

||g − f ||1 < ε/2.

Z

−n/2

(2π) |f (x) − g (x)| dx Rn ¯ ¯ Z ¯ ¯ −n/2 −it·x ¯ + ¯(2π) e g(x)dx¯¯ n R ¯ ¯ Z ¯ ¯ −n/2 −it·x ¯ ε/2 + ¯(2π) e g(x)dx¯¯ . Rn

Now integrating by parts, it follows that for ||t||∞ ≡ max {|tj | : j = 1, · · · , n} > 0 ¯ ¯ ¯ ¯ ¯ Z X n ¯ ¯ ¯ ¯ ¯ 1 ∂g (x) ¯ ¯ dx¯ |F f (t)| ≤ ε/2 + (2π)−n/2 ¯¯ (21.3.6) ¯ ¯ ¯ ¯ ||t||∞ Rn j=1 ∂xj ¯ and this last expression converges to zero as ||t||∞ → ∞. The reason for this is that if tj 6= 0, integration by parts with respect to xj gives Z Z ∂g (x) −it·x −n/2 1 −n/2 e g(x)dx = (2π) (2π) e−it·x dx. −itj Rn ∂xj Rn


647

Therefore, choose the j for which ||t||∞ = |tj | and the result of 21.3.6 holds. Therefore, from 21.3.6, if ||t||∞ is large enough, |F f (t)| < ε. Similarly, lim||t||→∞ F −1 (t) = 0. Consider the claim about uniform continuity. Let ε > 0 be given. Then there exists R such that if ||t||∞ > R, then |F f (t)| < 2ε . Since F f is continuous, it is n uniformly continuous on the compact set, [−R − 1, R + 1] . Therefore, there exists n δ 1 such that if ||t − t0 ||∞ < δ 1 for t0 , t ∈ [−R − 1, R + 1] , then |F f (t) − F f (t0 )| < ε/2.

(21.3.7)

Now let 0 < δ < min (δ 1 , 1) and suppose ||t − t0 ||∞ < δ. If both t, t0 are contained n n n in [−R, R] , then 21.3.7 holds. If t ∈ [−R, R] and t0 ∈ / [−R, R] , then both are n contained in [−R − 1, R + 1] and so this verifies 21.3.7 in this case. The other case n is that neither point is in [−R, R] and in this case, |F f (t) − F f (t0 )| ≤

|F f (t)| + |F f (t0 )| ε ε + = ε. < 2 2

This proves the theorem. There is a very interesting relation between the Fourier transform and convolutions. n/2

Theorem 21.3.11 Let f, g ∈ L1 (Rn ). Then f ∗g ∈ L1 and F (f ∗g) = (2π) Proof: Consider

Z

F f F g.

Z |f (x − y) g (y)| dydx.

Rn

Rn

The function, (x, y) → |f (x − y) g (y)| is Lebesgue measurable and so by Fubini’s theorem, Z Z Z Z |f (x − y) g (y)| dydx = |f (x − y) g (y)| dxdy Rn

Rn

Rn

=

Rn

||f ||1 ||g||1 < ∞.

R It follows that for a.e. R x, Rn |f (x − y) g (y)| dy < ∞ and for each of these values of x, it follows that Rn f (x − y) g (y) dy exists and equals a function of x which is in L1 (Rn ) , f ∗ g (x). Now F (f ∗ g) (t) Z −n/2 ≡ (2π) e−it·x f ∗ g (x) dx n ZR Z −n/2 = (2π) e−it·x f (x − y) g (y) dydx n Rn ZR Z −n/2 −it·y = (2π) e g (y) e−it·(x−y) f (x − y) dxdy Rn

= (2π)

n/2

Rn

F f (t) F g (t) .

There are many other considerations involving Fourier transforms of functions in L1 (Rn ).

648

21.3.3

FOURIER TRANSFORMS

Fourier Transforms Of Functions In L2 (Rn )

Consider F f and F −1 f for f ∈ L2 (Rn ). First note that the formula given for F f and F −1 f when f ∈ L1 (Rn ) will not work for f ∈ L2 (Rn ) unless f is also in L1 (Rn ). Recall that a + ib = a − ib. Theorem 21.3.12 For φ ∈ G, ||F φ||2 = ||F −1 φ||2 = ||φ||2 . Proof: First note that for ψ ∈ G, F (ψ) = F −1 (ψ) , F −1 (ψ) = F (ψ).

(21.3.8)

This follows from the definition. For example, Z e−it·x ψ (x) dx F ψ (t) = (2π)−n/2 Rn

Z =

(2π)−n/2

eit·x ψ (x) dx Rn

Let φ, ψ ∈ G. It was shown above that Z Z (F φ)ψ(t)dt = Rn

Similarly,

φ(F ψ)dx. Rn

Z

Z φ(F −1 ψ)dx =

(F −1 φ)ψdt.

Rn

(21.3.9)

Rn

Now, 21.3.8 - 21.3.9 imply Z

Z |φ|2 dx

φF −1 (F φ)dx

=

Rn

Z

Rn

=

φF (F φ)dx n

ZR =

F φ(F φ)dx Z

Rn

|F φ|2 dx.

= Rn

Similarly

||φ||2 = ||F −1 φ||2 .

This proves the theorem. Lemma 21.3.13 Let f ∈ L2 (Rn ) and let φk → f in L2 (Rn ) where φk ∈ G. (Such a sequence exists because of density of G in L2 (Rn ).) Then F f and F −1 f are both in L2 (Rn ) and the following limits take place in L2 . lim F (φk ) = F (f ) , lim F −1 (φk ) = F −1 (f ) .

k→∞

k→∞

21.3. FOURIER TRANSFORMS OF JUST ABOUT ANYTHING Proof: Let ψ ∈ G be given. Then Z F f (ψ) ≡ f (F ψ) ≡ f (x) F ψ (x) dx Rn Z Z φk (x) F ψ (x) dx = lim = lim k→∞

k→∞

Rn

Rn

649

F φk (x) ψ (x) dx.

∞

Also by Theorem 21.3.12 {F φk }k=1 is Cauchy in L2 (Rn ) and so it converges to some h ∈ L2 (Rn ). Therefore, from the above, Z F f (ψ) = h (x) ψ (x) Rn

which shows that F (f ) ∈ L2 (Rn ) and h = F (f ) . The case of F −1 is entirely similar. This proves the lemma. Since F f and F −1 f are in L2 (Rn ) , this also proves the following theorem. Theorem 21.3.14 If f ∈ L2 (Rn ), F f and F −1 f are the unique elements of L2 (Rn ) such that for all φ ∈ G, Z Z F f (x)φ(x)dx = f (x)F φ(x)dx, (21.3.10) Rn

Rn

Z

Z F −1 f (x)φ(x)dx =

Rn

f (x)F −1 φ(x)dx.

(21.3.11)

Rn

Theorem 21.3.15 (Plancherel) ||f ||2 = ||F f ||2 = ||F −1 f ||2 .

(21.3.12)

Proof: Use the density of G in L2 (Rn ) to obtain a sequence, {φk } converging to f in L2 (Rn ). Then by Lemma 21.3.13 ||F f ||2 = lim ||F φk ||2 = lim ||φk ||2 = ||f ||2 . k→∞

k→∞

Similarly, ||f ||2 = ||F −1 f ||2. This proves the theorem. The following corollary is a simple generalization of this. To prove this corollary, use the following simple lemma which comes as a consequence of the Cauchy Schwarz inequality. Lemma 21.3.16 Suppose fk → f in L2 (Rn ) and gk → g in L2 (Rn ). Then Z Z lim fk gk dx = f gdx k→∞

Rn

Rn

650 Proof:

FOURIER TRANSFORMS

¯Z ¯ ¯ ¯

Rn

Z fk gk dx −

Rn

¯ ¯Z ¯ ¯ f gdx¯¯ ≤ ¯¯

Z

Rn

¯Z ¯ ¯ ¯

Z

Rn

fk gdx −

fk gk dx − ¯ ¯ f gdx¯¯ n

Rn

¯ ¯ fk gdx¯¯ +

R

≤ ||fk ||2 ||g − gk ||2 + ||g||2 ||fk − f ||2 . Now ||fk ||2 is a Cauchy sequence and so it is bounded independent of k. Therefore, the above expression is smaller than ε whenever k is large enough. This proves the lemma. Corollary 21.3.17 For f, g ∈ L2 (Rn ), Z Z Z f gdx = F f F gdx = Rn

Rn

F −1 f F −1 gdx.

Rn

Proof: First note the above formula is obvious if f, g ∈ G. To see this, note Z Z Z 1 F f F gdx = F f (x) e−ix·t g (t) dtdx n/2 Rn Rn Rn (2π) Z Z 1 = eix·t F f (x) dxg (t)dt n/2 n n R (2π) R Z ¡ −1 ¢ = F ◦ F f (t) g (t)dt n ZR = f (t) g (t)dt. Rn

The formula with F −1 is exactly similar. Now to verify the corollary, let φk → f in L2 (Rn ) and let ψ k → g in L2 (Rn ). Then by Lemma 21.3.13 Z Z F f F gdx = lim F φk F ψ k dx k→∞ Rn Rn Z = lim φk ψ k dx k→∞ Rn Z = f gdx Rn

A similar argument holds for F −1 .This proves the corollary. How does one compute F f and F −1 f ? Theorem 21.3.18 For f ∈ L2 (Rn ), let fr = f XEr where Er is a bounded measurable set with Er ↑ Rn . Then the following limits hold in L2 (Rn ) . F f = lim F fr , F −1 f = lim F −1 fr . r→∞

r→∞


651

Proof: ||f − fr ||2 → 0 and so ||F f − F fr ||2 → 0 and ||F −1 f − F −1 fr ||2 → 0 by Plancherel’s Theorem. This proves the theorem. What are F fr and F −1 fr ? Let φ ∈ G Z Z F fr φdx = fr F φdx Rn Rn Z Z −n 2 fr (x)e−ix·y φ(y)dydx = (2π) Rn Rn Z Z n [(2π)− 2 = fr (x)e−ix·y dx]φ(y)dy. Rn

Rn

Since this holds for all φ ∈ G, a dense subset of L2 (Rn ), it follows that Z −n 2 fr (x)e−ix·y dx. F fr (y) = (2π) Rn

Similarly

Z F

−1

fr (y) = (2π)

−n 2 Rn

fr (x)eix·y dx.

This shows that to take the Fourier transform of a function in L2 (Rn ), it suffices R 2 n −n 2 to take the limit as r → ∞ in L (R ) of (2π) f (x)e−ix·y dx. A similar Rn r procedure works for the inverse Fourier transform. Note this reduces to the earlier definition in case f ∈ L1 (Rn ). Now consider the convolution of a function in L2 with one in L1 . Theorem 21.3.19 Let h ∈ L2 (Rn ) and let f ∈ L1 (Rn ). Then h ∗ f ∈ L2 (Rn ), n/2

F −1 (h ∗ f ) = (2π)

F −1 hF −1 f,

n/2

F (h ∗ f ) = (2π)

F hF f,

and ||h ∗ f ||2 ≤ ||h||2 ||f ||1 .

(21.3.13)

Proof: An application of Minkowski’s inequality yields ÃZ

µZ |h (x − y)| |f (y)| dy

Rn

Hence

R

!1/2

¶2 dx

Rn

≤ ||f ||1 ||h||2 .

|h (x − y)| |f (y)| dy < ∞ a.e. x and Z x → h (x − y) f (y) dy

is in L2 (Rn ). Let Er ↑ Rn , m (Er ) < ∞. Thus, hr ≡ XEr h ∈ L2 (Rn ) ∩ L1 (Rn ),

(21.3.14)

652

FOURIER TRANSFORMS

and letting φ ∈ G,

Z F (hr ∗ f ) (φ) dx

Z ≡

(hr ∗ f ) (F φ) dx Z Z Z −n/2 = (2π) hr (x − y) f (y) e−ix·t φ (t) dtdydx ¶ Z Z µZ −n/2 = (2π) hr (x − y) e−i(x−y)·t dx f (y) e−iy·t dyφ (t) dt Z n/2 = (2π) F hr (t) F f (t) φ (t) dt. Since φ is arbitrary and G is dense in L2 (Rn ), F (hr ∗ f ) = (2π)

n/2

F hr F f.

Now by Minkowski’s Inequality, hr ∗ f → h ∗ f in L2 (Rn ) and also it is clear that hr → h in L2 (Rn ) ; so, by Plancherel’s theorem, you may take the limit in the above and conclude n/2 F (h ∗ f ) = (2π) F hF f. The assertion for F −1 is similar and 21.3.13 follows from 21.3.14.

21.3.4

The Schwartz Class

The problem with G is that it does not contain Cc∞ (Rn ). I have used it in presenting the Fourier transform because the functions in G have a very specific form which made some technical details work out easier than in any other approach I have seen. The Schwartz class is a larger class of functions which does contain Cc∞ (Rn ) and also has the same nice properties as G. The functions in the Schwartz class are infinitely differentiable and they vanish very rapidly as |x| → ∞ along with all their partial derivatives. This is the description of these functions, not a specific 2 form involving polynomials times e−α|x| . To describe this precisely requires some notation. Definition 21.3.20 f ∈ S, the Schwartz class, if f ∈ C ∞ (Rn ) and for all positive integers N , ρN (f ) < ∞ where

2

ρN (f ) = sup{(1 + |x| )N |Dα f (x)| : x ∈ Rn , |α| ≤ N }.

Thus f ∈ S if and only if f ∈ C ∞ (Rn ) and sup{|xβ Dα f (x)| : x ∈ Rn } < ∞ for all multi indices α and β.

(21.3.15)


653

Also note that if f ∈ S, then p(f ) ∈ S for any polynomial, p with p(0) = 0 and that S ⊆ Lp (Rn ) ∩ L∞ (Rn ) for any p ≥ 1. To see this assertion about the p (f ), it suffices to consider the case of the product of two elements of the Schwartz class. If f, g ∈ S, then Dα (f g) is a finite sum of derivatives of f times derivatives of g. Therefore, ρN (f g) < ∞ for all N . You may wonder about examples of things in S. Clearly any function in 2 Cc∞ (Rn ) is in S. However there are other functions in S. For example e−|x| is in S as you can verify for yourself and so is any function from G. Note also that the density of Cc (Rn ) in Lp (Rn ) shows that S is dense in Lp (Rn ) for every p. Recall the Fourier transform of a function in L1 (Rn ) is given by Z −n/2 e−it·x f (x)dx. F f (t) ≡ (2π) Rn

Therefore, this gives the Fourier transform for f ∈ S. The nice property which S has in common with G is that the Fourier transform and its inverse map S one to one onto S. This means I could have presented the whole of the above theory as well as what follows in terms of S and its algebraic dual, S∗ rather than in terms of G and G ∗ . However, it is more technical. Nevertheless, letting S play the role of G in the above is convenient in certain applications because it is easier to reduce to S than G. I will make use of this simple observation whenever it will simplify a presentation. The fundamental result which makes it possible is the following. Theorem 21.3.21 If f ∈ S, then F f and F −1 f are also in S. Proof: To begin with, let α = ej = (0, 0, · · · , 1, 0, · · · , 0), the 1 in the j th slot. F −1 f (t + hej ) − F −1 f (t) = (2π)−n/2 h Consider the integrand in 21.3.16. ¯ ¯ ihxj ¯ it·x − 1 ¯¯ ¯e f (x)( e )¯ = ¯ h = ≤

Z eit·x f (x)( Rn

eihxj − 1 )dx. h

(21.3.16)

¯ ¯ i(h/2)x j ¯ e − e−i(h/2)xj ¯¯ ¯ )¯ |f (x)| ¯( h ¯ ¯ ¯ i sin ((h/2) xj ) ¯ ¯ |f (x)| ¯¯ ¯ (h/2) |f (x)| |xj |

and this is a function in L1 (Rn ) because f ∈ S. Therefore by the Dominated Convergence Theorem, Z ∂F −1 f (t) eit·x ixj f (x)dx = (2π)−n/2 ∂tj Rn Z eit·x xej f (x)dx. = i(2π)−n/2 Rn

654

FOURIER TRANSFORMS

Now xej f (x) ∈ S and so one can continue in this way and take derivatives indefinitely. Thus F −1 f ∈ C ∞ (Rn ) and from the above argument, Z α Dα F −1 f (t) =(2π)−n/2 eit·x (ix) f (x)dx. Rn

To complete showing F −1 f ∈ S, Z a

tβ Dα F −1 f (t) =(2π)−n/2

eit·x tβ (ix) f (x)dx. Rn

Integrate this integral by parts to get Z a tβ Dα F −1 f (t) =(2π)−n/2 i|β| eit·x Dβ ((ix) f (x))dx.

(21.3.17)

Rn

Here is how this is done. Z β α eitj xj tj j (ix) f (x)dxj R

=

eitj xj β j t (ix)α f (x) |∞ −∞ + itj j Z β −1 i eitj xj tj j Dej ((ix)α f (x))dxj R

where the boundary term vanishes because f ∈ S. Returning to 21.3.17, use the fact that |eia | = 1 to conclude Z a |tβ Dα F −1 f (t)| ≤C |Dβ ((ix) f (x))|dx < ∞. Rn

It follows F −1 f ∈ S. Similarly F f ∈ S whenever f ∈ S. Theorem 21.3.22 Let ψ ∈ S. Then (F ◦ F −1 )(ψ) = ψ and (F −1 ◦ F )(ψ) = ψ whenever ψ ∈ S. Also F and F −1 map S one to one and onto S. Proof: The first claim follows from the fact that F and F −1 are inverses ¡ of each ¢ other which was established above. For the second, let ψ ∈ S. Then ψ = F F −1 ψ . Thus F maps S onto S. If F ψ = 0, then do F −1 to both sides to conclude ψ = 0. Thus F is one to one and onto. Similarly, F −1 is one to one and onto. Note the above equations involving F and F −1 hold pointwise everywhere because F ψ and F −1 ψ are continuous.

21.3.5

Convolution

To begin with it is necessary to discuss the meaning of φf where f ∈ G ∗ and φ ∈ G. What should it mean? First suppose f ∈ Lp (Rn ) or measurable with polynomial growth. RThen φf alsoR has these properties. Hence, it should be the case that φf (ψ) = Rn φf ψdx = Rn f (φψ) dx. This motivates the following definition.


655

Definition 21.3.23 Let T ∈ G ∗ and let φ ∈ G. Then φT ≡ T φ ∈ G ∗ will be defined by φT (ψ) ≡ T (φψ) . The next topic is that of convolution. It was just shown that n/2

F (f ∗ φ) = (2π)

n/2

F φF f, F −1 (f ∗ φ) = (2π)

F −1 φF −1 f

whenever f ∈ L2 (Rn ) and φ ∈ G so the same definition is retained in the general case because it makes perfect sense and agrees with the earlier definition. Definition 21.3.24 Let f ∈ G ∗ and let φ ∈ G. Then define the convolution of f with an element of G as follows. f ∗ φ ≡ (2π)

n/2

F −1 (F φF f ) ∈ G ∗

There is an obvious question. With this definition, is it true that F −1 (f ∗ φ) = n/2 −1 (2π) F φF −1 f as it was earlier? Theorem 21.3.25 Let f ∈ G ∗ and let φ ∈ G. F (f ∗ φ) = (2π)

n/2

n/2

F −1 (f ∗ φ) = (2π)

F φF f,

(21.3.18)

F −1 φF −1 f.

(21.3.19)

Proof: Note that 21.3.18 follows from Definition 21.3.24 and both assertions hold for f ∈ G. Consider 21.3.19. Here is a simple formula involving a pair of functions in G. ¡ ¢ ψ ∗ F −1 F −1 φ (x) µZ Z Z =

¶ ψ (x − y) e

iy·y1 iy1 ·z

e

n

φ (z) dzdy1 dy (2π) µZ Z Z ¶ n −iy·˜ y1 −i˜ y1 ·z = ψ (x − y) e e φ (z) dzd˜ y1 dy (2π) = (ψ ∗ F F φ) (x) .

Now for ψ ∈ G, n/2

(2π)

¡ ¢ ¢ n/2 ¡ −1 F F −1 φF −1 f (ψ) ≡ (2π) F φF −1 f (F ψ) ≡

n/2

(2π)

¢¢ ¡ ¢ n/2 ¡ −1 ¡ −1 F φF ψ = F −1 f F −1 φF ψ ≡ (2π) f F ³ ¢ ¢´ n/2 −1 ¡¡ f (2π) F F F −1 F −1 φ (F ψ) ≡ ¡ ¢ f ψ ∗ F −1 F −1 φ = f (ψ ∗ F F φ)

(21.3.20)

656 Also

FOURIER TRANSFORMS

¡ ¢ n/2 F −1 (F φF f ) (ψ) ≡ (2π) (F φF f ) F −1 ψ ≡ ¡ ¢ ¢¢ n/2 n/2 ¡ ¡ (2π) F f F φF −1 ψ ≡ (2π) f F F φF −1 ψ = ³ ³ ¢´´ n/2 ¡ = f F (2π) F φF −1 ψ ³ ³ ¢´´ ¡ ¡ ¢¢ n/2 ¡ −1 = f F (2π) F F F φF −1 ψ = f F F −1 (F F φ ∗ ψ) (2π)

n/2

f (F F φ ∗ ψ) = f (ψ ∗ F F φ) . The last line follows from the following. Z F F φ (x − y) ψ (y) dy =

(21.3.21)

Z F φ (x − y) F ψ (y) dy Z

=

F ψ (x − y) F φ (y) dy Z

=

ψ (x − y) F F φ (y) dy.

From 21.3.21 and 21.3.20 , since ψ was arbitrary, ¡ ¢ n/2 n/2 −1 (2π) F F −1 φF −1 f = (2π) F (F φF f ) ≡ f ∗ φ which shows 21.3.19.

21.4

Exercises

1. For f ∈ L1 (Rn ), show that if F −1 f ∈ L1 or F f ∈ L1 , then f equals a continuous bounded function a.e. 2. Suppose f, g ∈ L1 (R) and F f = F g. Show f = g a.e. 3. Show that if f ∈ L1 (Rn ) , then lim|x|→∞ F f (x) = 0. 4. ↑ Suppose f ∗ f = f or f ∗ f = 0 and f ∈ L1 (R). Show f = 0. R∞ Rr 5. For this problem define a f (t) dt ≡ limr→∞ a f (t) dt. Note this coincides with the Lebesgue integral when f ∈ L1 (a, ∞). Show R∞ π (a) 0 sin(u) u du = 2 R ∞ sin(ru) du = 0 whenever δ > 0. (b) limr→∞ δ u R 1 (c) If f ∈ L (R), then limr→∞ R sin (ru) f (u) du = 0. R∞ Hint: For the first two, use u1 = 0 e−ut dt and apply Fubini’s theorem to RR R sin u R e−ut dtdu. For the last part, first establish it for f ∈ Cc∞ (R) and 0 then use the density of this set in L1 (R) to obtain the result. This is sometimes called the Riemann Lebesgue lemma.

21.4. EXERCISES

657

6. ↑Suppose that g ∈ L1 (R) and that at some x > 0, g is locally Holder continuous from the right and from the left. This means lim g (x + r) ≡ g (x+)

r→0+

exists, lim g (x − r) ≡ g (x−)

r→0+

exists and there exist constants K, δ > 0 and r ∈ (0, 1] such that for |x − y| < δ, r |g (x+) − g (y)| < K |x − y| for y > x and

r

|g (x−) − g (y)| < K |x − y|

for y < x. Show that under these conditions, µ ¶ Z 2 ∞ sin (ur) g (x − u) + g (x + u) lim du r→∞ π 0 u 2 =

g (x+) + g (x−) . 2

7. ↑ Let g ∈ L1 (R) and suppose g is locally Holder continuous from the right and from the left at x. Show that then Z R Z ∞ 1 g (x+) + g (x−) lim eixt e−ity g (y) dydt = . R→∞ 2π −R 2 −∞ This is very interesting. If g ∈ L2 (R), this shows F −1 (F g) (x) = g(x+)+g(x−) , 2 the midpoint of the jump in g at the point, x. In particular, if g ∈ G, F −1 (F g) = g. Hint: Show the left side of the above equation reduces to µ ¶ Z 2 ∞ sin (ur) g (x − u) + g (x + u) du π 0 u 2 and then use Problem 6 to obtain the result. 8. ↑ A measurable function g defined on (0, ∞) has exponential growth if |g (t)| ≤ Ceηt for some η. For Re (s) > η, define the Laplace Transform by Z ∞ Lg (s) ≡ e−su g (u) du. 0

Assume that g has exponential growth as above and is Holder continuous from the right and from the left at t. Pick γ > η. Show that Z R g (t+) + g (t−) 1 eγt eiyt Lg (γ + iy) dy = . lim R→∞ 2π −R 2

658

FOURIER TRANSFORMS

This formula is sometimes written in the form Z γ+i∞ 1 est Lg (s) ds 2πi γ−i∞ and is called the complex inversion integral for Laplace transforms. It can be used to find inverse Laplace transforms. Hint: Z R 1 eγt eiyt Lg (γ + iy) dy = 2π −R 1 2π

Z

R

Z γt iyt

e e −R

∞

e−(γ+iy)u g (u) dudy.

0

Now use Fubini’s theorem and do the integral from −R to R to get this equal to Z eγt ∞ −γu sin (R (t − u)) e g (u) du π −∞ t−u where g is the zero extension of g off [0, ∞). Then this equals Z eγt ∞ −γ(t−u) sin (Ru) g (t − u) e du π −∞ u which equals 2eγt π

Z 0

∞

g (t − u) e−γ(t−u) + g (t + u) e−γ(t+u) sin (Ru) du 2 u

and then apply the result of Problem 6. 9. Suppose f ∈ S. Show F (fxj )(t) = itj F f (t). 10. Let f ∈ S and let k be a positive integer. X ||Dα f ||22 )1/2. ||f ||k,2 ≡ (||f ||22 + |α|≤k

One could also define

Z |||f |||k,2 ≡ (

|F f (x)|2 (1 + |x|2 )k dx)1/2. Rn

Show both || ||k,2 and ||| |||k,2 are norms on S and that they are equivalent. These are Sobolev space norms. For which values of k does the second norm make sense? How about the first norm? 11. ↑ Define H k (Rn ), k ≥ 0 by f ∈ L2 (Rn ) such that Z 1 ( |F f (x)|2 (1 + |x|2 )k dx) 2 < ∞,

21.4. EXERCISES

659 Z 1 |||f |||k,2 ≡ ( |F f (x)|2 (1 + |x|2 )k dx) 2.

Show H k (Rn ) is a Banach space, and that if k is a positive integer, H k (Rn ) ={ f ∈ L2 (Rn ) : there exists {uj } ⊆ G with ||uj − f ||2 → 0 and {uj } is a Cauchy sequence in || ||k,2 of Problem 10}. This is one way to define Sobolev Spaces. Hint: One way to do the second part of this is to define a new measure, µ by Z ³ ´k 2 µ (E) ≡ 1 + |x| dx. E

Then show µ is a Radon measure and show there exists {gm } such that gm ∈ G and gm → F f in L2 (µ). Thus gm = F fm , fm ∈ G because F maps G onto G. Then by Problem 10, {fm } is Cauchy in the norm || ||k,2 . 12. ↑ If 2k > n, show that if f ∈ H k (Rn ), then f equals a bounded continuous function a.e. Hint: Show that for k this large, F f ∈ L1 (Rn ), and then use Problem 1. To do this, write k

|F f (x)| = |F f (x)|(1 + |x|2 ) 2 (1 + |x|2 ) So

Z

−k 2

,

Z |F f (x)|dx =

k

|F f (x)|(1 + |x|2 ) 2 (1 + |x|2 )

−k 2

dx.

Use the Cauchy Schwarz inequality. This is an example of a Sobolev imbedding Theorem. 13. Let u ∈ G. Then F u ∈ G and so, in particular, it makes sense to form the integral, Z F u (x0 , xn ) dxn R 0

n

where (x , xn ) = x ∈ R . For u ∈ G, define γu (x0 ) ≡ u (x0 , 0). Find a constant such that F (γu) (x0 ) equals this constant times the above integral. Hint: By the dominated convergence theorem Z Z 2 0 F u (x , xn ) dxn = lim e−(εxn ) F u (x0 , xn ) dxn . R

ε→0

R

Now use the definition of the Fourier transform and Fubini’s theorem as required in order to obtain the desired relationship. 14. Recall the Fourier series of a function in L2 (−π, π) converges to the function in L2 (−π, π). Prove a similar theorem with L2 (−π, π) replaced by L2 (−mπ, mπ) and the functions o n −(1/2) inx (2π) e n∈Z

660

FOURIER TRANSFORMS

used in the Fourier series replaced with n o n −(1/2) i m (2mπ) e x

n∈Z

Now suppose f is a function in L2 (R) satisfying F f (t) = 0 if |t| > mπ. Show that if this is so, then ¶ µ 1X −n sin (π (mx + n)) f (x) = . f π m mx + n n∈Z

Here m is a positive integer. This is sometimes called the Shannon sampling theorem.Hint: First note that since F f ∈ L2 and is zero off a finite interval, it follows F f ∈ L1 . Also Z mπ 1 eitx F f (x) dx f (t) = √ 2π −mπ and you can conclude from this that f has all derivatives and they are all bounded. Thus f is a very nice function. You can replace F f with its Fourier series. ¡ ¢ Then consider carefully the Fourier coefficient of F f. Argue it equals f −n or at least an appropriate constant times this. When you get this the m rest will fall quickly into place if you use F f is zero off [−mπ, mπ].

Fourier Analysis In Rn An Introduction The purpose of this chapter is to present some of the most important theorems on Fourier analysis in Rn . These theorems are the Marcinkiewicz interpolation theorem, the Calderon Zygmund decomposition, and Mihlin’s theorem. They are all fundamental results whose proofs depend on the methods of real analysis.

22.1

The Marcinkiewicz Interpolation Theorem

Let (Ω, µ, S) be a measure space. Definition 22.1.1 Lp (Ω) + L1 (Ω) will denote the space of measurable functions, f , such that f is the sum of a function in Lp (Ω) and L1 (Ω). Also, if T : Lp (Ω) + L1 (Ω) → space of measurable functions, T is subadditive if |T (f + g) (x)| ≤ |T f (x)| + |T g (x)|. T is of type (p, p) if there exists a constant independent of f ∈ Lp (Ω) such that ||T f ||p ≤ A kf kp , f ∈ Lp (Ω). T is weak type (p, p) if there exists a constant A independent of f such that µ ¶p A µ ([x : |T f (x)| > α]) ≤ ||f ||p , f ∈ Lp (Ω). α The following lemma involves writing a function as a sum of a functions whose values are small and one whose values are large. Lemma 22.1.2 If p ∈ [1, r], then Lp (Ω) ⊆ L1 (Ω) + Lr (Ω). Proof: Let λ > 0 and let f ∈ Lp (Ω) ½ ½ f (x) if |f (x)| ≤ λ f (x) if |f (x)| > λ f1 (x) ≡ , f2 (x) ≡ . 0 if |f (x)| > λ 0 if |f (x)| ≤ λ 661

FOURIER ANALYSIS IN RN AN INTRODUCTION

662 Thus f (x) = f1 (x) + f2 (x). Z

Z

Z

r

r

p

|f (x)| dµ ≤ λr−p

|f1 (x)| dµ = [|f |≤λ]

|f (x)| dµ < ∞.

[|f |≤λ]

Therefore, f1 ∈ Lr (Ω). Z

Z |f2 (x)| dµ =

|f (x)| dµ ≤ µ [|f | > λ]

1/p0

µZ p

|f | dµ

¶1/p < ∞.

[|f |>λ]

This proves the lemma since f = f1 + f2 , f1 ∈ Lr and f2 ∈ L1 . For f a function having nonnegative real values, α → µ ([f > α]) is called the distribution function. Lemma 22.1.3 Let φ (0) = 0, φ is strictly increasing, and C 1 . Let f : Ω → [0, ∞) be measurable. Then Z Z ∞ (φ ◦ f ) dµ = φ0 (α) µ [f > α] dα. (22.1.1) Ω

0

Proof: First suppose f=

m X

ai XEi

i=1

where ai > 0 and the ai are all distinct nonzero values of f , the sets, Ei being disjoint. Thus, Z m X (φ ◦ f ) dµ = φ (ai ) µ (Ei ). Ω

i=1

Suppose without loss of generality a1 < a2 < · · · < am . Observe α → µ ([f > α]) is constant on the intervals [0, a1 ), [a1 , a2 ), · · · . For example, on [ai , ai+1 ), this function has the value m X µ (Ej ). j=i+1

The function equals zero on [am , ∞). Therefore, α → φ0 (α) µ ([|f | > α])

22.1. THE MARCINKIEWICZ INTERPOLATION THEOREM

663

is Lebesgue measurable and letting a0 = 0, the second integral in 22.1.1 equals Z

∞

φ0 (α) µ ([f > α]) dα

=

0

m Z X i=1

=

ai

φ0 (α) µ ([f > α]) dα

ai−1

m X m X

Z

=

j m X X

φ0 (α) dα

ai−1

i=1 j=i

=

ai

µ (Ej )

µ (Ej ) (φ (ai ) − φ (ai−1 ))

j=1 i=1 m X

Z (φ ◦ f ) dµ

µ (Ej ) φ (aj ) =

j=1

Ω

and so this establishes 22.1.1 in the case when f is a nonnegative simple function. Since every measurable nonnegative function may be written as the pointwise limit of such simple functions, the desired result will follow by the Monotone convergence theorem and the next claim. Claim: If fn ↑ f , then for each α > 0, µ ([f > α]) = lim µ ([fn > α]). n→∞

Proof of the claim: [fn > α] ↑ [f > α] because if f (x) > α then for large enough n, fn (x) > α and so µ ([fn > α]) ↑ µ ([f > α]). This proves the lemma. (Note the importance of the strict inequality in [f > α] in proving the claim.) The next theorem is the main result in this section. It is called the Marcinkiewicz interpolation theorem. Theorem 22.1.4 Let (Ω, µ, S) be a σ finite measure space, 1 < r < ∞, and let T : L1 (Ω) + Lr (Ω) → space of measurable functions be subadditive, weak (r, r), and weak (1, 1). Then T is of type (p, p) for every p ∈ (1, r) and ||T f ||p ≤ Ap ||f ||p where the constant Ap depends only on p and the constants in the definition of weak (1, 1) and weak (r, r). Proof: Let α > 0 and let f1 and f2 be defined as in Lemma 22.1.2, ½ ½ f (x) if |f (x)| ≤ α f (x) if |f (x)| > α f1 (x) ≡ , f2 (x) ≡ . 0 if |f (x)| > α 0 if |f (x)| ≤ α


664

Thus f = f1 + f2 where f1 ∈ Lr and f2 ∈ L1 . Since T is subadditive , [|T f | > α] ⊆ [|T f1 | > α/2] ∪ [|T f2 | > α/2] . Let p ∈ (1, r). By Lemma 22.1.3, Z

Z p

∞

|T f | dµ ≤ p

αp−1 µ ([|T f1 | > α/2]) dα+

0

Z

∞

+p

αp−1 µ ([|T f2 | > α/2]) dα.

0

Therefore, since T is weak (1, 1) and weak (r, r), Z

Z p

µ

∞

αp−1

|T f | dµ ≤ p 0

2Ar ||f1 ||r α

¶r

Z

∞

0

Therefore, the right side of 22.1.2 equals Z ∞ Z Z r r p (2Ar ) αp−1−r |f1 | dµdα + 2A1 p 0

Ω

Z Z p (2Ar )

α Ω

∞

2A1 ||f2 ||1 dα. (22.1.2) α

Z αp−2

0

p−1−r

r

|f1 | dαdµ + 2A1 p

0

Ω

|f2 | dµdα = Ω

Z Z

∞

r

αp−1

dα + p

∞

αp−2 |f2 | dαdµ.

0

Now f1 (x) = 0 unless |f1 (x)| ≤ α and f2 (x) = 0 unless |f2 (x)| > α so this equals Z r

p (2Ar )

|f (x)| Ω

which equals

Z r

Z

∞

α

p−1−r

|f (x)|

dαdµ + 2A1 p

Z |f (x)|

Ω

|f (x)|

αp−2 dαdµ

0

Z 2pA1 p |f (x)| dµ + |f (x)| dµ p − 1 Ω Ω ¶ µ r r 2 Ar p 2pA1 p , ||f ||Lp (Ω) ≤ max r−p p−1

2r Arr p r−p

Z

p


22.2

The Calderon Zygmund Decomposition

For a given nonnegative integrable function, Rn can be decomposed into a set where the function is small and a set which is the union of disjoint cubes on which the average of the function is under some control. The measure in this section will always be Lebesgue measure on Rn . This theorem depends on the Lebesgue theory of differentiation.

22.2. THE CALDERON ZYGMUND DECOMPOSITION

665

R Theorem 22.2.1 Let f ≥ 0, f dx < ∞, and let α be a positive constant. Then there exist sets F and Ω such that Rn = F ∪ Ω, F ∩ Ω = ∅

(22.2.3)

f (x) ≤ α a.e. on F

(22.2.4)

Ω = ∪∞ k=1 Qk where the interiors of the cubes are disjoint and for each cube, Qk , Z 1 α< f (x) dx ≤ 2n α. (22.2.5) m (Qk ) Qk Proof: Let S0 be a tiling of Rn into cubes having sides of length M where M is chosen large enough that if Q is one of these cubes, then Z 1 f dm ≤ α. (22.2.6) m (Q) Q Suppose S0 , · · · , Sm have been chosen. To get Sm+1 , replace each cube of Sm by the 2n cubes obtained by bisecting the sides. Then Sm+1 consists of exactly those cubes of Sm for which 22.2.6 holds and let Tm+1 consist of the bisected cubes from Sm for which 22.2.6 does not hold. Now define F ≡ {x : x is contained in some cube from Sm for all m} , Ω ≡ Rn \ F = ∪∞ m=1 ∪ {Q : Q ∈ Tm } Note that the cubes from Tm have pair wise disjoint interiors and also the interiors of cubes from Tm have empty intersections with the interiors of cubes of Tk if k 6= m. Let x be a point of Ω and let x be in a cube of Tm such that m is the first index for which this happens. Let Q be the cube in Sm−1 containing x and let Q∗ be the cube in the bisection of Q which contains x. Therefore 22.2.6 does not hold for Q∗ . Thus ≤α z }|Z { Z 1 m (Q) 1 α< f dx ≤ f dx ≤ 2n α m (Q∗ ) Q∗ m (Q∗ ) m (Q) Q which shows Ω is the union of cubes having disjoint interiors for which 22.2.5 holds. Now a.e. point of F is a Lebesgue point of f . Let x be such a point of F and suppose x ∈ Qk for Qk ∈ Sk . Let dk ≡ diameter of Qk . Thus dk → 0. Z Z 1 1 |f (y) − f (x)| dy ≤ |f (y) − f (x)| dy m (Qk ) Qk m (Qk ) B(x,dk ) Z m (B (x,dk )) 1 = |f (x) − f (y)| dy m (Qk ) m (B (x,dk )) B(x,dk ) Z 1 ≤ Kn |f (x) − f (y)| dy m (B (x,dk )) B(x,dk )


666

where Kn is a constant which depends on n and measures the ratio of the volume of a ball with diamiter 2d and a cube with diameter d. The last expression converges to 0 because x is a Lebesgue point. Hence Z 1 f (x) = lim f (y) dy ≤ α k→∞ m (Qk ) Q k and this shows f (x) ≤ α a.e. on F . This proves the theorem.

22.3

Mihlin’s Theorem

In this section, the Marcinkiewicz interpolation theorem and Calderon Zygmund decomposition will be used to establish a remarkable theorem of Mihlin, a generalization of Plancherel’s theorem to the Lp spaces. It is of fundamental importance in the study of elliptic partial differential equations and can also be used to give proofs for the theory of singular integrals. Mihlin’s theorem involves a conclusion which is of the form ¯¯ −1 ¯¯ ¯¯F ρ ∗ φ¯¯ ≤ Ap ||φ|| (22.3.7) p p for p > 1 and φ ∈ G. Thus F −1 ρ∗ extends to a continuous linear map defined on Lp because of the density of G. It is proved by showing various weak type estimates and then applying the Marcinkiewicz Interpolation Theorem to get an estimate like the above. Recall that by Corollary 21.3.19, if f ∈ L2 (Rn ) and if φ ∈ G, then f ∗φ ∈ L2 (Rn ) and n/2 F (f ∗ φ) (x) = (2π) F φ (x) F f (x). The next lemma is essentially a weak (1, 1) estimate. The inequality 22.3.7 is established under the condition, 22.3.8 and then it is shown there exist conditions which are easier to verify which imply condition 22.3.8. I think the approach used here is due to Hormander [34] and is found in Berg and Lofstrom [10]. For many more references and generalizations, you might look in Triebel [64]. A different proof based on singular integrals is in Stein [63]. Functions, ρ which yield an inequality of the sort in 22.3.7 are called Lp multipliers. Lemma 22.3.1 Suppose ρ ∈ L∞ (Rn ) ∩ L2 (Rn ) and suppose also there exists a constant C1 such that Z ¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x)¯ dx ≤ C1 . (22.3.8) |x|≥2|y|

Then there exists a constant A depending only on C1 , ||ρ||∞ , and n such that m for all φ ∈ G.

¯ ¡£ ¯ −1 ¤¢ A x : ¯F ρ∗φ (x)¯ > α ≤ ||φ||1 α

22.3. MIHLIN’S THEOREM

667

Proof: Let φ ∈ G and use the Calderon decomposition to write Rn = E ∪ Ω where Ω is a union of cubes, {Qi } with disjoint interiors such that Z αm (Qi ) ≤ |φ (x)| dx ≤ 2n αm (Qi ) , |φ (x)| ≤ α a.e. on E. (22.3.9) Qi

The proof is accomplished by writing φ as the sum of a good function and a bad function and establishing a similar weak inequality for these two functions separately. Then this information is used to obtain the desired conclusion. ½ φ (x) if R x∈E g (x) = , g (x) + b (x) = φ (x). (22.3.10) 1 m(Qi ) Qi φ (x) dx if x ∈ Qi ⊆ Ω Thus Z

Z b (x) dx

Z

=

Qi

φ (x) dx −

Qi

b (x) = Claim:

Z

(φ (x) − g (x)) dx = Qi

φ (x) dx =(22.3.11) 0, Qi

0 if x ∈Ω. /

(22.3.12)

2

||g||2 ≤ α (1 + 4n ) ||φ||1 , ||g||1 ≤ ||φ||1 .

Proof of claim:

2

2

(22.3.13)

2

||g||2 = ||g||L2 (E) + ||g||L2 (Ω) . Thus 2

||g||L2 (Ω)

=

XZ i

≤

XZ i

≤

Z 2 ||g||L2 (E)

4n α2

µ

Qi

XZ i

≤

2

|g (x)| dx

Qi

1 m (Qi )

|φ (y)| dy

2

Qi

X

m (Qi )

i

Z

1X α i

dx

Qi

(2n α) dx ≤ 4n α2

Qi

|φ (x)| dx ≤ 4n α ||φ||1 . Z

2

=

¶2

Z

|φ (x)| dx ≤ α E

E

|φ (x)| dx = α ||φ||1 .

Now consider the second of the inequalities in 22.3.13. Z Z ||g||1 = |g (x)| dx + |g (x)| dx E Ω Z XZ = |φ (x)| dx + |g| dx E

i

Z ≤

|φ (x)| dx + E

Z =

|φ (x)| dx + E

Qi

XZ i

Qi

i

Qi

XZ

1 m (Qi )

Z |φ (x)| dm (x) dm Qi

|φ (x)| dm (x) = ||φ||1


668

This proves the claim. From the claim, it follows that b ∈ L2 (Rn ) ∩ L1 (Rn ) . Because of 22.3.13, g ∈ L1 (Rn ) and so F −1 ρ ∗ g ∈ L2 (Rn ). (Since ρ ∈ L2 , it follows F −1 ρ ∈ L2 and so this convolution is indeed in L2 .) By Plancherel’s theorem, ¯¯ −1 ¯¯ ¯¯ ¡ ¢¯¯ ¯¯F ρ ∗ g ¯¯ = ¯¯F F −1 ρ ∗ g ¯¯ . 2

2

By Corollary 21.3.19 on Page 651, the expression on the right equals n/2

(2π) and so

||ρF g||2

¯¯ −1 ¯¯ ¯¯F ρ ∗ g ¯¯ = (2π)n/2 ||ρF g|| ≤ Cn ||ρ|| ||g|| . 2 ∞ 2 2

From this and 22.3.13 m

¯ ¡£¯ −1 ¤¢ ¯F ρ ∗ g ¯ ≥ α/2

2

Cn ||ρ||∞ α (1 + 4n ) ||φ||1 = Cn α−1 ||φ||1 . (22.3.14) α2 This ¡£¯ is what ¯is wanted ¤¢ so far as g is concerned. Next it is required to estimate m ¯F −1 ρ ∗ b¯ ≥ α/2 . ∗ If Q is one of the cubes whose √ union is Ω, let Q be the cube with the same center as Q but whose sides are 2 n times as long. ≤

Q∗i rQi yi

Let

∗ Ω ∗ ≡ ∪∞ i=1 Qi

and let

E ∗ ≡ Rn \ Ω∗.

Thus E ∗ ⊆ E. Let x ∈ E ∗ . Then because of 22.3.11, Z F −1 ρ (x − y) b (y) dy Qi

Z = Qi

£ −1 ¤ F ρ (x − y) − F −1 ρ (x − yi ) b (y) dy,

(22.3.15)

√ where yi is the center of Qi . Consequently if the sides of Qi have length 2t/ n, 22.3.15 implies ¯ Z ¯Z ¯ ¯ −1 ¯ F ρ (x − y) b (y) dy ¯¯ dx ≤ (22.3.16) ¯ ∗ E

Qi

22.3. MIHLIN’S THEOREM Z

Z

E∗

Z

Qi

Z

= Qi

Z

E∗

¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x − yi )¯ |b (y)| dydx ¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x − yi )¯ dx |b (y)| dy

Z

≤ Qi

669

|x−yi |≥2t

(22.3.17)

¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x − yi )¯ dx |b (y)| dy (22.3.18)

since if x ∈ E ∗ , then |x − yi | ≥ 2t. Now for y ∈ Qi , 

 ¶2 1/2 n µ X t  √ |y − yi | ≤  = t. n j=1 From 22.3.8 and the change of variables u = x − yi 22.3.16 - 22.3.18 imply ¯ Z ¯Z Z ¯ ¯ −1 ¯ ¯ dx ≤ C1 F ρ (x − y) b (y) dy |b (y)| dy. (22.3.19) ¯ ¯ ∗ E

Qi

Qi

Now from 22.3.19, and the fact that b = 0 off Ω, ¯ Z Z ¯Z ¯ −1 ¯ ¯ ¯ −1 ¯F ρ ∗ b (x)¯ dx = ¯ F ρ (x − y) b (y) dy ¯¯ dx ¯ E∗ E∗ Rn ¯ Z ¯¯X ∞ Z ¯ ¯ ¯ −1 = F ρ (x − y) b (y) dy ¯ dx ¯ ¯ ¯ ∗ E i=1 Qi ¯ ¯ Z X Z ∞ ¯ ¯ ¯ ≤ F −1 ρ (x − y) b (y) dy ¯¯ dx ¯ ∗

= ≤

E i=1 ∞ XZ i=1 ∞ X

E

¯Z ¯ ¯ ¯ ∗

Qi

F Qi

−1

¯ ¯ ρ (x − y) b (y) dy ¯¯ dx

Z C1 Qi

i=1

|b (y)| dy = C1 ||b||1 .

Thus, by 22.3.13, Z

¯ ¯ −1 ¯F ρ ∗ b (x)¯ dx

≤

E∗

Consequently, m

C1 ||b||1

≤

C1 [||φ||1 + ||g||1 ]

≤ ≤

C1 [||φ||1 + ||φ||1 ] 2C1 ||φ||1 .

³h¯ ´ i ¯ ¯F −1 ρ ∗ b¯ ≥ α ∩ E ∗ ≤ 4C1 ||φ|| . 1 2 α


670

From 22.3.10, 22.3.14, and 22.3.9, h¯ h¯ ¯ ¯ αi ¯ αi £¯ ¤ + m ¯F −1 ρ ∗ b¯ ≥ m ¯F −1 ρ ∗ φ¯ > α ≤ m ¯F −1 ρ ∗ g ¯ ≥ 2 2 ³h¯ ´ ¯ αi Cn ≤ ||φ||1 + m ¯F −1 ρ ∗ b¯ ≥ ∩ E ∗ + m (Ω∗ ) α 2 4C1 A Cn ||φ||1 + ||φ||1 + Cn m (Ω) ≤ ||φ||1 ≤ α α α because m (Ω) ≤ α−1 ||φ||1 by 22.3.9. This proves the lemma. The next lemma extends this lemma by giving a weak (2, 2) estimate and a (2, 2) estimate. Lemma 22.3.2 Suppose ρ ∈ L∞ (Rn ) ∩ L2 (Rn ) and suppose also that there exists a constant C1 such that Z ¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x)¯ dx ≤ C1 . (22.3.20) |x|>2|y|

Then F −1 ρ∗ maps L1 (Rn ) + L2 (Rn ) to measurable functions and there exists a constant A depending only on C1 , n, ||ρ||∞ such that ¯ ¡£¯ −1 ¤¢ ¯F ρ ∗ f ¯ > α ≤ A ||f ||1 if f ∈ L1 (Rn ), α ¶2 µ ¯ ¡£¯ ¤¢ ||f ||2 m ¯F −1 ρ ∗ f ¯ > α ≤ A if f ∈ L2 (Rn ). α m

(22.3.21) (22.3.22)

Thus, F −1 ρ∗ is weak type (1, 1) and weak type (2, 2). Also ¯¯ −1 ¯¯ ¯¯F ρ ∗ f ¯¯ ≤ A ||f || if f ∈ L2 (Rn ). 2 2

(22.3.23)

Proof: By Plancherel’s theorem F −1 ρ is in L2 (Rn ). If f ∈ L1 (Rn ), then by Minkowski’s inequality, F −1 ρ ∗ f ∈ L2 (Rn ) . Now let g ∈ L2 (Rn ). By Holder’s inequality, Z

¯ −1 ¯ ¯F ρ (x − y)¯ |g (y)| dy ≤

µZ

¯ −1 ¯ ¯F ρ (x − y)¯2 dy

¶1/2 µZ

and so the following is well defined a.e. Z F −1 ρ ∗ g (x) ≡ F −1 ρ (x − y) g (y) dy

2

|g (y)| dy

¶1/2 α]

≥ αm

¯ ¡£¯ −1 ¤¢ ¯F ρ ∗ f ¯ > α 1/2

and so 22.3.22 follows. It remains to prove 22.3.21 which holds for all f ∈ G by Lemma 22.3.1. Let f ∈ L1 (Rn ) and let φk → f in L1 (Rn ) , φk ∈ G. Without loss of generality, assume that both f and F −1 ρ are Borel measurable. Therefore, by Minkowski’s inequality, and Plancherel’s theorem, ¯¯ −1 ¯¯ ¯¯F ρ ∗ φk − F −1 ρ ∗ f ¯¯ 2


672

≤

ÃZ ¯Z ¯2 !1/2 ¯ ¯ ¯ F −1 ρ (x − y) (φk (y) − f (y)) dy ¯ dx ¯ ¯

≤

||φk − f ||1 ||ρ||2

which shows that F −1 ρ ∗ φk converges to F −1 ρ ∗ f in L2 (Rn ). Therefore, there exists a subsequence such that the convergence is pointwise a.e. Then, denoting the subsequence by k, X[|F −1 ρ∗f |>α] (x) ≤ lim inf X[|F −1 ρ∗φk |>α] (x) a.e. x. k→∞

Thus by Lemma 22.3.1 and Fatou’s lemma, there exists a constant, A, depending on C1 , n, and ||ρ||∞ such that ¯ ¯ ¡£¯ ¤¢ ¡£¯ ¤¢ m ¯F −1 ρ ∗ f ¯ > α ≤ lim inf m ¯F −1 ρ ∗ φk ¯ > α k→∞

≤

lim inf A k→∞

||φk ||1 ||f ||1 =A . α α

This shows 22.3.21 and proves the lemma. Theorem 22.3.3 Let ρ ∈ L2 (Rn ) ∩ L∞ (Rn ) and suppose Z ¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x)¯ dx ≤ C1 . |x|≥2|y|

Then for each p ∈ (1, ∞), there exists a constant, Ap , depending only on p, n, ||ρ||∞ , and C1 such that for all φ ∈ G, ¯¯ −1 ¯¯ ¯¯F ρ ∗ φ¯¯ ≤ Ap ||φ|| . p p Proof: From Lemma 22.3.2, F −1 ρ∗ is weak (1, 1), weak (2, 2), and maps L1 (Rn ) + L2 (Rn ) to measurable functions. Therefore, by the Marcinkiewicz interpolation theorem, there exists a constant Ap depending only on p, C1 , n, and ||ρ||∞ for p ∈ (1, 2], such that for f ∈ Lp (Rn ), and p ∈ (1, 2], ¯¯ −1 ¯¯ ¯¯F ρ ∗ f ¯¯ ≤ Ap ||f || . p p Thus the theorem is proved for these values of p. Now suppose p > 2. Then p0 < 2 where 1 1 + 0 = 1. p p


673

By Plancherel’s theorem and Theorem 21.3.25, Z Z n/2 F −1 ρ ∗ φ (x) ψ (x) dx = (2π) ρ (x) F φ (x) F ψ (x) dx Z ¡ ¢ = F F −1 ρ ∗ ψ F φdx Z ¡ −1 ¢ F ρ ∗ ψ (φ) dx. = Thus by the case for p ∈ (1, 2) and Holder’s ¯Z ¯ ¯ ¯ ¯ F −1 ρ ∗ φ (x) ψ (x) dx¯ = ¯ ¯

inequality, ¯Z ¯ ¯ ¡ −1 ¯ ¢ ¯ F ρ ∗ ψ (φ) dx¯¯ ¯ ¯¯ ¯¯ ≤ ¯¯F −1 ρ ∗ ψ ¯¯p0 ||φ||p ≤ Ap0 ||ψ||p0 ||φ||p .

R

0

0

F −1 ρ ∗ φ (x) ψ (x) dx, this shows L ∈ Lp (Rn ) and ||L||(Lp0 )0 ≤ Ap0 ||φ||p which implies by the Riesz representation theorem that F −1 ρ∗φ represents L and ¯¯ ¯¯ ||L||(Lp0 )0 = ¯¯F −1 ρ ∗ φ¯¯Lp ≤ Ap0 ||φ||p Letting Lψ ≡

Since p0 = p/ (p − 1), this proves the theorem. It is possible to give verifiable conditions on ρ which imply 22.3.20. The condition on ρ which is presented here is the existence of a constant, C0 such that C0 ≥ sup{|x|

|α|

|Dα ρ (x)| : |α| ≤ L, x ∈ Rn \ {0}}, L > n/2.

(22.3.25)

ρ ∈ C L (Rn \ {0}) where L is an integer. Pn Here α is a multi-index and |α| = i=1 αi . The condition says roughly that ρ is pretty smooth away from 0 and all the partial derivatives vanish pretty fast as |x| → ∞. Also recall the notation αn 1 xα ≡ xα 1 · · · xn

where α = (α1 · · · αn ). For more general conditions, see [34]. Lemma 22.3.4 Let 22.3.25 hold and suppose ψ ∈ Cc∞ (Rn \ {0}). Then for each α, |α| ≤ L, there exists a constant C ≡ C (α, n, ψ) independent of k such that sup |x| x ∈Rn

|α|

¯ α¡ ¡ ¢¢¯ ¯D ρ (x) ψ 2k x ¯ ≤ CC0 .

Proof: |x|

|α|

X ¯ ¯ α¡ ¯ ¯ ¡ ¢¢¯ ¡ ¢¯ ¯D ρ (x) ψ 2k x ¯ ≤ |x||α| ¯Dβ ρ (x)¯ 2k|γ| ¯Dγ ψ 2k x ¯ β+γ=α


674 X

=

|x|

|β|

¯ β ¯¯ ¯ ¯ ¡ ¢¯ ¯D ρ (x)¯ ¯2k x¯|γ| ¯Dγ ψ 2k x ¯

β+γ=α

X

≤ C0 C (α, n)

|γ|

|Dγ ψ (z)| : z ∈ Rn } = C0 C (α, n, ψ)

sup{|z|

|γ|≤|α|

and this proves the lemma. Lemma 22.3.5 There exists φ ∈ Cc∞

¡£

and

¤¢ x :4−1 < |x| < 4 , φ (x) ≥ 0, ∞ X

¡ ¢ φ 2k x = 1

k=−∞

for each x 6= 0. Proof: Let

£ ¤ ψ ≥ 0, ψ = 1 on 2−1 ≤ |x| ≤ 2 , £ ¤ spt (ψ) ⊆ 4−1 < |x| < 4 .

Consider g (x) =

∞ X

¡ ¢ ψ 2k x .

k=−∞

Then for each x, only finitely many terms are not equal to 0. Also, g (x) 0 for £ l> l+2 ¤ all x 6= 0. To verify this last claim, note that for some k an integer, |x| ∈ 2 , 2 . £ ¤ −1 Therefore, choose k an integer such £that 2k |x| ∈ 2 , 2 . For example, let k = ¤ £ l−l−1 l+2−l−1 ¤ £ −1 ¤ l k l+2 k −l − 1. This works because 2k |x| ∈ 2 2 , 2 2 = 2 ,2 = 2 ,2 . ¡ k ¢ Therefore, for this value of k, ψ 2 x = 1 so g (x) > 0. Now notice that g (2r x) =

∞ X

∞ X ¡ ¢ ¡ ¢ ψ 2k 2r x = ψ 2k x = g (x).

k=−∞ −1

Let φ (x) ≡ ψ (x) g (x) ∞ X k=−∞

k=−∞

. Then

¡ ¢ ∞ ∞ X X ¡ ¢ ¢ ψ 2k x −1 = g (x) ψ 2k x = 1 φ 2 x = k g (2 x) ¡

k

k=−∞

k=−∞

for each x 6= 0. This proves the lemma. Now define ρm (x) ≡

m X k=−m

¡ ¢ ¡ ¢ ρ (x) φ 2k x , γ k (x) ≡ ρ (x) φ 2k x .


675

Let t > 0 and let |y| ≤ t. Consider the problem of estimating Z ¯ −1 ¯ ¯F γ k (x − y) − F −1 γ k (x)¯ dx.

(22.3.26)

|x|≥2t

In the following estimates, C (a, b, · · · , d) will denote a generic constant depending only on the indicated objects, a, b, · · · , d. For the first estimate, note that since |y| ≤ t, 22.3.26 is no larger than Z Z ¯ −1 ¯ −1 ¯ ¯ ¯ ¯F γ k (x)¯ |x|−L |x|L dx ¯ F γ k (x) dx = 2 2 |x|≥t

|x|≥t

ÃZ

!1/2 ÃZ

≤2

|x|

−2L

dx

|x|

|x|≥t

2L

|x|≥t

¯ −1 ¯ ¯F γ k (x)¯2 dx

!1/2

Using spherical coordinates and Plancherel’s theorem, ³R ¯2 ´1/2 2L ¯ |x| ¯F −1 γ k (x)¯ dx ≤ C (n, L) tn/2−L ³R P ¯2 ´1/2 n 2L ¯¯ −1 ≤ C (n, L) tn/2−L F γ k (x)¯ dx j=1 |xj | ³P ´1/2 ¯ R ¯ −1 L n ¯F D γ k (x)¯2 dx ≤ C (n, L) tn/2−L j j=1 ³P ´1/2 ¯ R ¯ L n n/2−L ¯D γ k (x)¯2 dx = C (n, L) t j j=1 Sk where

£ ¤ Sk ≡ x :2−2−k < |x| < 22−k ,

(22.3.27)

(22.3.28)

a set containing the support of γ k . Now from the definition of γ k , ¯ L ¯ ¯ ¡ ¡ ¢¢¯ ¯Dj γ k (z)¯ = ¯DjL ρ (z) φ 2k z ¯. By Lemma 22.3.4, this is no larger than −L

C (L, n, φ) C0 |z|

.

(22.3.29)

It follows, using polar coordinates, that the last expression in 22.3.27 is no larger than µZ ¶1/2 −2L C (n, L, φ, C0 ) tn/2−L |z| dz ≤ C (n, L, φ, C0 ) tn/2−L· (22.3.30) Sk

ÃZ

!1/2

22−k

ρ

n−1−2L

2−2−k

dρ

≤ C (n, L, φ, C0 ) tn/2−L 2k(L−n/2).

Now estimate 22.3.26 in another way. The support of γ k is in Sk , a bounded set, and so F −1 γ k is differentiable. Therefore, Z ¯ −1 ¯ ¯F γ k (x − y) − F −1 γ k (x)¯ dx = |x|≥2t


676

¯ ¯ ¯Z 1 X ¯ n ¯ ¯ −1 ¯ ¯ dx D F γ (x−sy) y ds j j k ¯ ¯ |x|≥2t ¯ 0 j=1 ¯

Z

Z

Z

≤ t |x|≥2t

n 1X ¯

0 j=1

¯ ¯Dj F −1 γ k (x−sy)¯ dsdx

Z X n ¯ ¯ ¯Dj F −1 γ k (x)¯ dx ≤ t j=1

≤ t

n µZ ³ X j=1

µZ ³ ·

≤ C (n, L) t2kn/2

¯ ¯2 ´−L 1 + ¯2−k x¯ dx

¶1/2

¶1/2 ¯ −k ¯2 ´L ¯ ¯2 −1 ¯ ¯ ¯ ¯ 1+ 2 x Dj F γ k (x) dx

n µZ ³ X

¶1/2 ¯ ¯2 ´ L ¯ ¯ ¯Dj F −1 γ k (x)¯2 dx . 1 + ¯2−k x¯

(22.3.31)

j=1

Now consider the j th term in the last sum in 22.3.31. R³

¯ ¯2 ´ L ¯ ¯ ¯Dj F −1 γ k (x)¯2 dx ≤ 1 + ¯2−k x¯ ¯ ¯2 RP −2k|α| 2α ¯ C (n, L) x Dj F −1 γ k (x)¯ dx |α|≤L 2 ¯ ¯2 R P = C (n, L) |α|≤L 2−2k|α| x2α ¯F −1 (π j γ k ) (x)¯ dx

(22.3.32)

where π j (z) ≡ zj . This last assertion follows from Z Z Dj e−ix·y γ k (y) dy = (−i) e−ix·y yj γ k (y) dy. Therefore, a similar computation and Plancherel’s theorem implies 22.3.32 equals Z X ¯ ¯2 = C (n, L) 2−2k|α| ¯F −1 Dα (π j γ k ) (x)¯ dx |α|≤L

= C (n, L)

X |α|≤L

Z 2

−2k|α|

2

Sk

|Dα (zj γ k (z))| dz

where Sk is given in 22.3.28. Now |Dα (zj γ k (z))|

¯ ¡ ¡ ¢¢¯ = 2−k ¯Dα ρ (z) zj 2k φ 2k z ¯ ¯ ¡ ¡ ¢¢¯ = 2−k ¯Dα ρ (z) ψ j 2k z ¯

(22.3.33)


677

where ψ j (z) ≡ zj φ (z). By Lemma 22.3.4, this is dominated by 2−k C (α, n, φ, j, C0 ) |z|

−|α|

.

Therefore, 22.3.33 is dominated by C (L, n, φ, j, C0 )

X

Z

C (L, n, φ, j, C0 )

X

2−2k |z|

dz

Sk

|α|≤L

≤

−2|α|

2−2k|α|

¡ ¢(−2|α|) ¡ 2−k ¢n 2−2k|α| 2−2k 2−2−k 2

|α|≤L

≤

C (L, n, φ, j, C0 )

X

2−kn−2k

|α|≤L

≤ C (L, n, φ, j, C0 ) 2−kn 2−2k. It follows that 22.3.31 is no larger than C (L, n, φ, C0 ) t2kn/2 2−kn/2 2−k = C (L, n, φ, C0 ) t2−k .

(22.3.34)

It follows from 22.3.34 and 22.3.30 that if |y| ≤ t, Z ¯ −1 ¯ ¯F γ k (x − y) − F −1 γ k (x)¯ dx ≤ |x|≥2t

³ ¡ ¢n/2−L ´ C (L, n, φ, C0 ) min t2−k , 2−k t . With this inequality, the next lemma which is the desired result can be obtained. Lemma 22.3.6 There exists a constant depending only on the indicated objects, C1 = C (L, n, φ, C0 ) such that when |y| ≤ t, Z ¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x)¯ dx ≤ C1 |x|≥2t

Z |x|≥2t

¯ −1 ¯ ¯F ρm (x − y) − F −1 ρm (x)¯ dx ≤ C1 .

(22.3.35)

Proof: F −1 ρ = limm→∞ F −1 ρm in L2 (Rn ). Let mk → ∞ be such that convergence is pointwise a.e. Then if |y| ≤ t, Fatou’s lemma implies Z ¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x)¯ dx ≤ |x|≥2t

Z lim inf

l→∞

|x|≥2t

¯ ¯ −1 ¯F ρm (x − y) − F −1 ρm (x)¯ dx l l


678 ≤ lim inf

l→∞

Z ml X |x|≥2t

k=−ml

∞ X

≤ C (L, n, φ, C0 )

¯ −1 ¯ ¯F γ k (x − y) − F −1 γ k (x)¯ dx

³ ¡ ¢n/2−L ´ . min t2−k , 2−k t

(22.3.36)

k=−∞

Now consider the sum in 22.3.36, ∞ X

³ ¡ ¢n/2−L ´ min t2−k , 2−k t .

(22.3.37)

k=−∞

³ ¡ ¢n/2−L ´ t2j = min t2j , 2j t exactly when t2j ≤ 1. This occurs if and only if j ≤ − ln (t) / ln (2). Therefore 22.3.37 is no larger than X

X

2j t +

j≤− ln(t)/ ln(2)

Letting a = L − n/2, this equals X t

¡ j ¢n/2−L 2 t .

j≥− ln(t)/ ln(2)

X

2−k + t−α

¡ −a ¢j 2

j≥− ln(t)/ ln(2)

k≥ln(t)/ ln(2)

µ ¶ln(t)/ ln(2) µ ¶− ln(t)/ ln(2) 1 1 ≤ 2t + t−a 2 2a µ ¶log2 (t) µ ¶− log2 (t) 1 1 = 2t + t−a 2 2a = 2 + 1 = 3. Similarly, 22.3.35 holds. This proves the lemma. Now it is possible to prove Mihlin’s theorem. Theorem 22.3.7 (Mihlin’s theorem) Suppose ρ satisfies |α|

C0 ≥ sup{|x|

|Dα ρ (x)| : |α| ≤ L, x ∈ Rn \ {0}},

where L is an integer greater than n/2 and ρ ∈ C L (Rn \ {0}). Then for every p > 1, there exists a constant Ap depending only on p, C0 , φ, n, and L, such that for all ψ ∈ G, ¯¯ −1 ¯¯ ¯¯F ρ ∗ ψ ¯¯ ≤ Ap ||ψ|| . p

p

Proof: Since ρm satisfies 22.3.35, and is obviously in L2 (Rn ) ∩ L∞ (Rn ), Theorem 22.3.3 implies there exists a constant Ap depending only on p, n, ||ρm ||∞ , and C1 such that for all ψ ∈ G and p ∈ (1, ∞), ¯¯ −1 ¯¯ ¯¯F ρm ∗ ψ ¯¯ ≤ Ap ||ψ|| . p p

22.4. SINGULAR INTEGRALS

679

Now ||ρm ||∞ ≤ ||ρ||∞ because m X

|ρm (x)| ≤ |ρ (x)|

¡ ¢ φ 2k x ≤ |ρ (x)|.

(22.3.38)

k=−m

Therefore, since C1 = C1 (L, n, φ, C0 ) and C0 ≥ ||ρ||∞ , ¯¯ −1 ¯¯ ¯¯F ρm ∗ ψ ¯¯ ≤ Ap (L, n, φ, C0 , p) ||ψ|| . p p In particular, Ap does not depend on m. Now, by 22.3.38, the observation that ρ ∈ L∞ (Rn ), limm→∞ ρm (y) = ρ (y) and the dominated convergence theorem, it follows that for θ ∈ G. ¯ ¯ Z ¯¡ −1 ¯ ¢ ¯ ¯ ¯ F ρ ∗ ψ (θ)¯ ≡ ¯(2π)n/2 ρ (x) F ψ (x) F −1 θ (x) dx¯ ¯ ¯ ¯¡ ¯¯ ¯¯ ¢ ¯ = lim ¯ F −1 ρm ∗ ψ (θ)¯ ≤ lim sup ¯¯F −1 ρm ∗ ψ ¯¯p ||θ||p0 m→∞

m→∞

≤ Ap (L, n, φ, C0 , p) ||ψ||p ||θ||p0 . ¯¯ ¯¯ Hence F −1 ρ ∗ ψ ∈ Lp (Rn ) and ¯¯F −1 ρ ∗ ψ ¯¯p ≤ Ap ||ψ||p . This proves the theorem.

22.4

Singular Integrals

If K ∈ L1 (Rn ) then when p > 1, ||K ∗ f ||p ≤ ||f ||p . It turns out that some meaning can be assigned to K ∗ f for some functions K which are not in L1 . This involves assuming a certain form for K and exploiting cancellation. The resulting theory of singular integrals is very useful. To illustrate, an application will be given to the Helmholtz decomposition of vector fields in the next section. Like Mihlin’s theorem, the theory presented here rests on Theorem 22.3.3, restated here for convenience. Theorem 22.4.1 Let ρ ∈ L2 (Rn ) ∩ L∞ (Rn ) and suppose Z ¯ −1 ¯ ¯F ρ (x − y) − F −1 ρ (x)¯ dx ≤ C1 . |x|≥2|y|

Then for each p ∈ (1, ∞), there exists a constant, Ap , depending only on p, n, ||ρ||∞ , and C1 such that for all φ ∈ G, ¯¯ −1 ¯¯ ¯¯F ρ ∗ φ¯¯ ≤ Ap ||φ|| . p p


680 Lemma 22.4.2 Suppose

K ∈ L2 (Rn ) , ||F K||∞ ≤ B < ∞, and

(22.4.39)

Z |K (x − y) − K (x)| dx ≤ B. |x|>2|y|

Then for all p > 1, there exists a constant, A (p, n, B), depending only on the indicated quantities such that ||K ∗ f ||p ≤ A (p, n, B) ||f ||p for all f ∈ G. Proof: Let F K = ρ so F −1 ρ = K. Then from 22.4.39 ρ ∈ L2 (Rn ) ∩ L∞ (Rn ) and K = F −1 ρ. By Theorem 22.3.3 listed above, ¯¯ ¯¯ ||K ∗ f || = ¯¯F −1 ρ ∗ f ¯¯ ≤ A (p, n, B) ||f || p

p

p

for all f ∈ G. This proves the lemma. The next lemma provides a situation in which the above conditions hold. Lemma 22.4.3 Suppose

−n

Z

|K (x)| ≤ B |x|

,

(22.4.40)

K (x) dx = 0,

(22.4.41)

|K (x − y) − K (x)| dx ≤ B.

(22.4.42)

a2|y|

and ||F Kε ||∞ ≤ C (n) B.

(22.4.45)

Proof: In the argument, C (n) will denote a generic constant depending only on n. Consider 22.4.44 first. The integral is broken up according to whether |x| , |x − y| > ε. |x| |x − y|

>ε >ε

>ε 0 be given, we choose N such that ∞ X

2−k
M, 2−i

di (xni , xi ) ε < n 1 + di (xi , xi ) 2 (N + 1) 759

760

THE YANKOV VON NEUMANN AUMANN THEOREM

for all i = 1, 2, · · · , N. Then letting x = {xi } , ρ (x, xn ) ≤

∞ X εN ε ε + 2−k < + = ε. 2 (N + 1) 2 2 k=N

We need ª∞verify that X is separable. Let Di denote a countable dense set in © to Xi , Di ≡ rki k=1 . Then let ª ª © © Dk ≡ D1 × · · · × Dk × r1k+1 × r1k+2 × · · · Thus Dk is a countable subset of X. Let D ≡ ∪∞ k=1 Dk . Then D is countable and we can see DQis dense in X as follows. The projection of Dk onto the first k entries is k dense in i=1 Xi and for k large enough the remaining component’s contribution to the metric, ρ is very small. Therefore, obtaining d ∈ D close to x ∈ X may be accomplished by finding d ∈ D such that d isQclose to x in the first k components ∞ for k large enough. Note that we do not use k=1 Dk ! Definition 24.0.3 A complete separable metric space is called a polish space. Theorem 24.0.4 Let X be a polishQspace. Then there exists f : NN → X which ∞ is onto and continuous. Here NN ≡ i=1 N and a metric is given according to the N above theorem. Thus for n, m ∈ N , ρ (n, m) ≡

∞ X i=1

2−i

|ni − mi | . 1 + |ni − mi |

Proof: Since X is polish, there exists a countable covering of X by closed ∞ sets having diameters no larger than 2−1 , {B (i)}i=1 . Each of these closed sets is also a polish space and so there exists a countable covering of B (i) by a countable ∞ collection of closed sets, {B (i, j)}j=1 each having diameter no larger than 2−2 where B (i, j) ⊆ B (i) 6= ∅ for all j. Continue this way. Thus B (n1 , n2 , · · · , nm ) = ∪∞ i=1 B (n1 , n2 , · · · , nm , i) and each of B (n1 , n2 , · · · , nm , i) is a closed set contained in B (n1 , n2 , · · · , nm ) whose diameter is at most half of the diameter of B (n1 , n2 , · · · , nm ) . Now we define ∞ our mapping from NN to X. If n = {nk }k=1 ∈ NN , we let f (n) ≡ ∩∞ m=1 B (n1 , n2 , · · · , nm ) . Since the diameters of these sets converge to 0, there exists a unique point in this countable intersection and this is f (n) . We need to verify f is continuous. Let n ∈ NN be given and suppose m is very close to n. The only way this can occur is for nk to coincide with mk for many k. Therefore, both f (n) and f (m) must be contained in B (n1 , n2 , · · · , nm ) for some fairly large m. This implies, from the above construction that f (m) is as close to f (n) as 2−m , proving f is continuous. To see that f is onto, note that from the construction, if x ∈ X, then x ∈ B (n1 , n2 , · · · , nm ) for some choice of n1 , · · · , nm for each m. Note nothing is said about f being one to one. It probably is not one to one.

761 Definition 24.0.5 We call a topological space, X a Suslin space if X is a Hausdorff space and there exists a polish space, Z and a continuous function f which maps Z onto X. These are also called analytic sets in some contexts but we will use the term Suslin space in referring to them. Corollary 24.0.6 X is a Suslin space, if and only if there exists a continuous mapping from NN onto X. Proof: We know there exists a polish space Z and a continuous function, h : Z → X which is onto. By the above theorem there exists a continuous map, g : NN → Z which is onto. Then h ◦ g is a continuous map from NN onto X. The “if” part of this theorem is accomplished by noting that NN is a polish space. Lemma 24.0.7 Let X be a Suslin space and suppose Xi is a subspace of X which ∞ is also a Suslin space. Then ∪∞ i=1 Xi and ∩i=1 Xi are also Suslin spaces. Also every Borel set in X is a Suslin space. Proof: Let fi : Zi → Xi where Zi is a polish space and fi is continuous and onto. Without loss of generality we may assume the spaces Zi are disjoint because if not, we could replace Zi with Zi × {i} . Now we define a metric, ρ, for Z ≡ ∪∞ i=1 Zi as follows. ρ (x, y) ≡ 1 if x ∈ Zi , y ∈ Zk , i 6= k ρ (x, y) ≡

di (x, y) if x, y ∈ Zi . 1 + di (x, y)

Here di is the metric on Zi . It is easy to verify ρ is a metric and that (Z, ρ) is a polish space. Now we define f : Z → ∪∞ i=1 Xi as follows. For x ∈ Zi , f (x) ≡ fi (x) . This is well defined because the Zi are disjoint. If y is very close to x it must be that x and y are in the same Zi otherwise this could not happen. Therefore, continuity of f follows from continuity of fi . This shows countable unions of Suslin subspaces of a Suslin space are Suslin spaces. If H ⊆ X is a closed subset, then, letting f : Z → X be onto and continuous, it follows f : f −1 (H) → H is onto and continuous. Since f −1 (H) is closed, it follows f −1 (H) is a polish space. Therefore, H is a Suslin space. Now we show countable Q∞ intersections of Suslin spaces are Suslin. It is clear Q∞ Z → by θ (z) ≡ x = {xi } where xi = fi (zi ) that θ : i=1 i i=1 Xi given Q ∞ is continuous and onto. Therefore, i=1 Xi is a Suslin space. Now let P ≡ Q∞ {y ∈ i=1 fi (Zi ) : yi = yj for all i, j} . Then P is a closed subspace of a Suslin space and so it is Suslin. Then we define h : P → ∩∞ i=1 Xi by h (y) ≡ fi (yi ) . This shows ∩∞ X is Suslin because h is continuous and onto. (h ◦ θ : θ−1 (P ) → ∩∞ i i=1 i=1 Xi is −1 continuous and θ (P ) being a closed subset of a polish space is polish.) Next let U be an open subset of X. Then f −1 (U ) , being an open subset of a polish space, can be obtained as an increasing limit of closed sets, Kn . Therefore, U = ∪∞ n=1 f (Kn ) . Each f (Kn ) is a Suslin space because it is the continuous image of a polish space, Kn . Therefore, by the first part of the lemma, U is a Suslin space. Now let © ª F ≡ E ⊆ X : both E C and E are Suslin .

762


We see that F is closed with respect to taking complements. The first part of this lemma shows F is closed with respect to countable unions. Therefore, F is a σ algebra and so, since it contains the open sets, must contain the Borel sets. It turns out that Suslin spaces tend to be measurable sets. In order to develop this idea, we need a technical lemma. Lemma 24.0.8 Let (Ω, F, µ) be a measure space and denote by µ∗ the outer measure generated by µ. Thus µ∗ (S) ≡ inf {µ (E) : E ⊇ S, E ∈ F} . Then µ∗ is regular, meaning that for every S, there exists E ∈ F such that E ⊇ S and µ (E) = µ∗ (S) . If Sn ↑ S, it follows that µ∗ (Sn ) ↑ µ∗ (S) . Also if µ (Ω) < ∞, then a set, E is measurable if and only if µ∗ (Ω) ≥ µ∗ (E) + µ∗ (Ω \ E) . Proof: First we verify that µ∗ is regular. If µ∗ (S) = ∞, let E = Ω. Then µ (S) = µ (E) and E ⊇ S. On the other hand, if µ∗ (S) < ∞, then we can obtain En ∈ F such that µ∗ (S) + n1 ≥ µ (En ) and En ⊇ S. Now let Fn = ∩ni=1 Ei . Then Fn ⊇ S and so µ∗ (S) + n1 ≥ µ (Fn ) ≥ µ∗ (S) . Therefore, letting F = ∩∞ k=1 Fk ∈ F it follows µ (F ) = limn→∞ µ (Fn ) = µ∗ (S) . Let En ⊇ Sn be such that En ∈ F and µ (En ) = µ∗ (Sn ) . Also let E∞ ⊇ S such that µ (E∞ ) = µ∗ (S) and E∞ ∈ F. Now consider Bn ≡ ∪nk=1 Ek . We claim ∗

µ (Bn ) = µ (Sn ) .

(24.0.1)

Here is why: µ (E1 \ E2 ) = µ (E1 ) − µ (E1 ∩ E2 ) = µ∗ (S1 ) − µ∗ (S1 ) = 0. Therefore, µ (B2 ) = µ (E1 ∪ E2 ) = µ (E1 \ E2 ) + µ (E2 ) = µ (E2 ) = µ∗ (S2 ) . Continuing in this way we see that 24.0.1 holds. Now let Bn ∩ E∞ ≡ Cn . Then ∗ Cn ↑ C ≡ ∪∞ k=1 Cn ∈ F and µ (Cn ) = µ (Sn ) . Since Sn ↑ S and each Cn ⊇ Sn , it follows C ⊇ S and therefore, µ∗ (S) ≤ µ (C) = lim µ (Cn ) = lim µ∗ (Sn ) ≤ µ∗ (S) . n→∞

n→∞

Now we verify the second claim of the lemma. It is clear the formula holds whenever E is measurable. Suppose now that the formula holds. Let S be an arbitrary set. We need to verify that µ∗ (S) ≥ µ∗ (S ∩ E) + µ∗ (S \ E) . Let F ⊇ S, F ∈ F, and µ (F ) = µ∗ (S) . Then since µ∗ is subadditive, ¡ ¢ µ∗ (Ω \ F ) ≤ µ∗ (E \ F ) + µ∗ Ω ∩ E C ∩ F C .

(24.0.2)

763 Since F is measurable,

and

µ∗ (E) = µ∗ (E ∩ F ) + µ∗ (E \ F )

(24.0.3)

¡ ¢ µ∗ (Ω \ E) = µ∗ (F \ E) + µ∗ Ω ∩ E C ∩ F C

(24.0.4)

and by the hypothesis, µ∗ (Ω) ≥ µ∗ (E) + µ∗ (Ω \ E) .

(24.0.5)

Therefore, µ (Ω)

≥ = = ≥ ≥

µ∗ (E) + µ∗ (Ω \ E) µ∗ (E ∩ F ) + µ∗ (E \ F ) + µ∗ (Ω \ E) ¡ ¢ µ∗ (E ∩ F ) + µ∗ (E \ F ) + µ∗ (F \ E) + µ∗ Ω ∩ E C ∩ F C µ∗ (Ω \ F ) + µ∗ (F \ E) + µ∗ (E ∩ F ) µ∗ (Ω \ F ) + µ∗ (F ) = µ (Ω)

showing that all the inequalities must be equal signs. Hence, referring to the top and fourth lines above, µ (Ω) = µ∗ (Ω \ F ) + µ∗ (F \ E) + µ∗ (E ∩ F ) . Subtracting µ∗ (Ω \ F ) = µ (Ω \ F ) from both sides gives µ∗ (S) = µ (F ) = µ∗ (F \ E) + µ∗ (E ∩ F ) ≥ µ∗ (S \ E) + µ∗ (E ∩ S) , This proves the lemma. The next theorem is a major result. It states that the Suslin subsets are measurable under appropriate conditions. Theorem 24.0.9 Let Ω be a metric space and let (Ω, F, µ) be a complete Borel measure space with µ (Ω) < ∞. Denote by µ∗ the outer measure generated by µ. Then if A is a Suslin subset of Ω, it follows that A is µ∗ measurable. Proof: We need to verify that µ∗ (Ω) ≥ µ∗ (A) + µ∗ (Ω \ A) . We know from Corollary 24.0.6, there exists a continuous map, f : NN → A which is onto. Let © ª E (k) ≡ n ∈ NN : n1 ≤ k . Then E (k) ↑ NN and so from Lemma 24.0.8 we know µ∗ (f (E (k))) ↑ µ∗ (A) . Therefore, there exists m1 such that ε µ∗ (f (E (m1 ))) > µ∗ (A) − . 2

764


Now E (k) is clearly not compact but it is trying to be as far as the first component is concerned. Now we let © ª E (m1 , k) ≡ n ∈ NN : n1 ≤ m1 and n2 ≤ k . Thus E (m1 , k) ↑ E (m1 ) and so we can pick m2 such that µ∗ (f (E (m1 , m2 ))) > µ∗ (f (E (m1 ))) −

ε . 22

We continue in this way obtaining a decreasing list of sets, f (E (m1 , m2 , · · · , mk−1 , mk )) , such that ε µ∗ (f (E (m1 , m2 , · · · , mk−1 , mk ))) > µ∗ (f (E (m1 , m2 , · · · , mk−1 ))) − k . 2 Therefore, µ∗ (f (E (m1 , m2 , · · · , mk−1 , mk ))) − µ∗ (A) >

k X

−

l=1

³ε´ 2l

> −ε.

Now define a closed set, C ≡ ∩∞ k=1 f (E (m1 , m2 , · · · , mk−1 , mk )). The sets f (E (m1 , m2 , · · · , mk−1 , mk )) are decreasing as k → ∞ and so ³ ´ µ∗ (C) = lim µ∗ f (E (m1 , m2 , · · · , mk−1 , mk )) ≥ µ∗ (A) − ε. k→∞

We wish to verify that C ⊆ A. If we can do this we will be done because C, being a closed set, is measurable and so µ∗ (Ω) = µ∗ (C) + µ∗ (Ω \ C) ≥ µ∗ (A) − ε + µ∗ (Ω \ A) . Since ε is arbitrary, this will conclude the proof. Therefore, we only need to verify that C ⊆ A. What we know is that each f (E (m1 , m2 , · · · , mk−1 , mk )) is contained in A. We ∞ do not know their closures are contained in A. We let m ≡ {mi }i=1 where the mi are defined above. Then letting © ª K ≡ n ∈ NN : ni ≤ mi for all i , we see that K is a closed, hence complete subset of NN which is also totally bounded due to the definition of the distance. Therefore, K is compact and so f (K) is also compact, hence closed due to the assumption that Ω is a Hausdorff space and we know that f (K) ⊆ A. We verify that C = f (K) . We know f (K) ⊆ C. Suppose therefore, p ∈ C. From the definition of C, we know there exists rk ∈ ¡ ¡ ¢ ¢ E (m1 , m2 , · · · , mk−1 , mk ) such that d f rk , p < k1 . Denote by rek the element of NN which consists of modifying rk by taking all components after the k th equal to

765 n o rek is in a compact set and so taking a subsequence we ³ ´ 1 can have rek → r ∈ K. But from the metric on NN , it follows that ρ rek , rk < 2k−2 . ¡ ¢ k k Therefore, r → r also and so f r → f (r) = p. Therefore, p ∈ f (K) and this proves the theorem. Note we could have proved this under weaker assumptions. If we had assumed only that every point has a countable basis (first axiom of countability) and Ω is Hausdorff, the same argument would work. We will need the following definition. one. Thus rek ∈ K. Now

Definition 24.0.10 Let F be a σ algebra of sets from Ω and let µ denote a finite measure defined on F. We let Fµ denote the completion of F with respect to µ. Thus we let µ∗ be the outer measure determined by µ and Fµ will be the σ algebra of µ∗ measurable subsets of Ω. We also define Fb by Fb ≡ ∩ {Fµ : µ is a finite measure defined on F} . Also, if X is a topological space, we will denote by B (X) the Borel sets of X. With this notation, we can give the following simple corollary of Theorem 24.0.9. Corollary 24.0.11 Let Ω be a compact metric space and let A be a Suslin subset \ of Ω. Then A ∈ B (Ω). Proof: Let µ be a finite measure defined on B (Ω) . By Theorem 24.0.9 A ∈ \ B (Ω)µ . Since this is true for every finite measure, µ, it follows A ∈ B (Ω) as claimed. This proves the corollary. We give another technical lemma about the completion of measure spaces. Lemma 24.0.12 Let µ be a finite measure on a σ algebra, Σ. Then A ∈ Σµ if and only if there exists A1 ∈ Σ and N1 such that A = A1 ∪ N1 where there exists N ∈ Σ such that µ (N ) = 0 and N1 ⊆ N. Proof: Suppose first A = A1 ∪ N1 where these sets are as described. Let S ∈ P (Ω) and let µ∗ denote the outer measure determined by µ. Then since A1 ∈ Σ ⊆ Σµ µ∗ (S) ≤ ≤ =

µ∗ (S \ A) + µ∗ (S ∩ A) µ∗ (S \ A1 ) + µ∗ (S ∩ A1 ) + µ∗ (N1 ) µ∗ (S \ A1 ) + µ∗ (S ∩ A1 ) = µ∗ (S)

showing that A ∈ Σµ . ∗ Now suppose A ∈ Σµ . Then there exists B1 ⊇ A such that ¡µ∗ (B ¢ 1 ) = ∗µ¡ (A) ¢, C C C C and B1 ∈ Σ. Also there exists A1 ∈ Σ with A1 ⊇ A and µ A1 = µ AC . Then A1 ⊆ A ⊆ B1 A ⊆ A1 ∪ (B1 \ A1 ) .

766 Now


¡ ¢ ¡ ¢ µ (A1 ) + µ∗ AC = µ (A1 ) + µ AC 1 = µ (Ω)

and so µ (B1 \ A1 )

= µ∗ (B1 \ A1 ) = µ∗ (B1 \ A) + µ∗ (A \ A1 ) = µ∗ (B1 ) − µ∗ (A) + µ∗ (A) − µ∗ (A1 ) ¡ ¢¢ ¡ = µ∗ (A) − µ (Ω) − µ∗ AC = 0 N1

z }| { because A ∈ Σµ implying A = A1 ∪ (B1 \ A1 ) ∩ A and N1 ⊆ N ≡ (B1 \ A1 ) ∈ Σ with µ (N ) = 0. This proves the lemma. Next we need another definition. Definition 24.0.13 We say (Ω, Σ), where Σ is a σ algebra of subsets of Ω, is ∞ separable if there exists a sequence {An }n=1 ⊆ Σ such that σ ({An }) = Σ and if 0 w 6= w , then there exists A ∈ Σ such that XA (ω) 6= XA (ω 0 ) . This last condition is referred to by saying {An } separates the points of Ω. Given two measure spaces, (Ω, Σ) and (Ω0 , Σ0 ) , we say they are isomorphic if there exists a function, f : Ω → Ω0 which is one to one and f (E) ∈ Σ0 whenever E ∈ Σ and f −1 (F ) ∈ Σ whenever F ∈ Σ0 . The interesting thing about separable measure spaces is that they are isomorphic to a very simple sort of measure space in which topology plays a significant role. N

Lemma 24.0.14 Let (Ω, Σ) be separable. Then there exists E ∈ {0, 1} such that (Ω, Σ) and (E, B (E)) are isomorphic. Proof: First we show {An } separates the points. We already know Σ separates the points. If this is not so, there exists ω, ω 1 ∈ Ω such that for all n, XAn (ω) = XAn (ω 1 ) . Then let F ≡ {F ∈ Σ : XF (ω) = XF (ω 1 )} Thus An ∈ F for all n. It is also clear that F is a σ algebra and so F = Σ contradicting the assumption that Σ separates points. Now we define a function N from Ω to {0, 1} as follows. ∞

f (ω) ≡ {XAn (ω)}n=1 We also let E ≡ f (Ω) . Since the {An } separate the points, we see that f is one Q∞ N to one. A subbasis for the topology of {0, 1} consists of sets of the form i=1 Hi where Hi = {0, 1} for all i except one, when i = j and Hj equals either {0} or {1} . Therefore, f −1 (subbasic open set) ∈ Σ because if Hj is the exceptional set then this equals Aj if Hj = {1} and AC j if Hj = {0} . Intersections of these subbasic sets with E gives a countable subbasis for E and so the inverse image of all sets

767 in a countable subbasis for E are in Σ, showing that f −1 (open set) ∈ Σ. Now we consider f (An ) . ∞ f (An ) ≡ {{λk }k=1 : λn = 1} ∩ E, an open set in E. Hence f (An ) ∈ B (E) . Now letting F ≡ {G ⊆ Ω : f (G) ∈ B (E)} , ∞

we see that F is a σ algebra which contains {An }n=1 and so F ⊇ σ ({An }) ≡ Σ. Thus f (F ) ∈ B (E) for all A ∈ Σ. This proves the lemma. Lemma 24.0.15 Let φ : (Ω1 , Σ1 ) → (Ω2 , Σ2 ) where φ−1 (U ) ∈ Σ1 for all U ∈ Σ2 . b 2 , it follows φ−1 (F ) ∈ Σ b 1. Then if F ∈ Σ Proof: Let µ be a finite measure on Σ1 and define a measure φ (µ) on Σ2 by the rule ¡ ¢ φ (µ) (F ) ≡ µ φ−1 (F ) . Now let A ∈ Σ2φ(µ) . Then by Lemma 24.0.12, A = A1 ∪ N1 where there exists N ∈ Σ2 with ¡ φ (µ) (N ¢ ) = 0 and A1 ∈ Σ2 . Therefore, from the definition of φ (µ) , we have µ φ−1 (N ) = 0 and therefore, φ−1 (A) = φ−1 (A1 ) ∪ φ−1 (N1 ) where ¡ ¢ φ−1 (N1 ) ⊆ φ−1 (N ) ∈ Σ1 and µ φ−1 (N ) = 0. Therefore, φ−1 (A) ∈ Σ1µ and so if b 2 , then F ∈Σ © ª F ∈ ∩ {Σ2ν : ν is a finite measure on Σ2 } ⊆ ∩ Σ2φ(µ) : µ is a finite measure on Σ1 , b 1. and so φ−1 (F ) ∈ Σ1µ . Since µ is arbitrary, this shows φ−1 (F ) ∈ Σ The next lemma is a special case of the Yankov von Neumann Aumann projection theorem. It contains the main idea of the proof of the more general theorem. Lemma 24.0.16 Let (Ω, Σ) be separable and let X be a Suslin space. Let G ∈ Σ × B (X) . (Recall Σ × B (X) is the σ algebra of product measurable sets, the smallest σ algebra containing the measurable rectangles.) Then b projΩ (G) ∈ Σ. Proof: Let f : (Ω, Σ) → (E, B (E)) be the isomorphism of Lemma 24.0.14. We have the following claim. Claim: f × idX maps Σ × B (X) to B (E) × B (X) . Proof of the claim: First of all, assume A × B is a measurable rectangle where A ∈ Σ and B ∈ B (X) . Then by the assumption that f is an isomorphism, f (A) ∈ B (E) and so f × idX (A × B) ∈ B (E) × B (X) . Now let F ≡ {P ∈ Σ × B (X) : f × idX (P ) ∈ B (E) × B (X)} .

768


Then we see that F is a σ algebra and contains the elementary sets. (F is closed with respect to complements because f is one to one.) Therefore, F = Σ × B (X) and this proves the claim. Therefore, since G ∈ Σ × B (X) , we see f × idX (G) ∈ B (E) × B (X) ⊆ B (E × X) . The set inclusion follows from the observation that if A ∈ B (E) and B ∈ B (X) then A × B is in B (E × X) and the collection of sets in B (E) × B (X) which are in B (E × X) is a σ algebra. Therefore, there exists D, a Borel set in E × X such that f × idX (G) = D ∩ (E × X) . Now from this it follows from Lemma 24.0.7 that D is a Suslin space. N Letting Y be {0, 1} , it follows that projY (D) is a Suslin space in Y. By Corollary \ 24.0.11, we see that projY (D) ∈ B (Y ). Now projΩ (G) = {ω ∈ Ω : there exists x ∈ X with (ω, x) ∈ G} = {ω ∈ Ω : there exists x ∈ X with (f (ω) , x) ∈ f × idX (G)} = f −1 ({y ∈ Y : there exists x ∈ X with (y, x) ∈ D}) = f −1 (projY (D)) . \ b This Now projY (D) ∈ B (Y ) and so Lemma 24.0.15 shows f −1 (projY (D)) ∈ Σ. proves the lemma. Now we are ready to prove the Yankov von Neumann Aumann projection theorem. First we must present another technical lemma. Lemma 24.0.17 Let X be a Hausdorff space and let G ∈ Σ × B (X) where Σ is a σ algebra of sets of Ω. Then there exists Σ0 ⊆ Σ a countably generated σ algebra such that G ∈ Σ0 × B (X) . Proof: First suppose G is a measurable rectangle,©G = A × Bª where A ∈ Σ and B ∈ B (X) . Letting Σ0 be the finite σ algebra, ∅, A, AC , Ω , we see that G ∈ Σ0 × B (X) . Similarly, if G equals an elementary set, then the conclusion of the lemma holds for G. Let F ≡ {H ∈ Σ × B (X) : H ∈ Σ0 × B (X)} for some countably generated σ algebra, Σ0 . We just saw that F contains the elementary sets. If H ∈ F , then H C ∈ Σ0 × B (X) for the same Σ0 and so F is closed with respect to complements. Now suppose Hn ∈ F. Then for each n, there exists a countably generated σ algebra, Σ0n such that Hn ∈ Σ0n × B (X) . Then ∪∞ n=1 Hn ∈ σ ({Σ0n × B (X)}) . We will be done when we show ∞

∞

σ ({Σ0n × B (X)}n=1 ) ⊆ σ ({Σ0n }n=1 ) × B (X) ∞

∞

because it is clear that σ ({Σ0n }n=1 ) is countably generated. We see that σ ({Σ0n × B (X)}n=1 ) is generated by sets of the form A × B where A ∈ Σ0n and B ∈ B (X) . But each

769 ∞

such set is also contained in σ ({Σ0n }n=1 ) × B (X) and so the desired inclusion is obtained. Therefore, F is a σ algebra and so since F was shown to contain the measurable rectangles, this verifies F = Σ × B (X) and this proves the lemma. b × B (X) where X Theorem 24.0.18 Let (Ω, Σ) be a measure space and let G ∈ Σ is a Suslin space. Then b projΩ (G) ∈ Σ. Proof: By the previous lemma, G ∈ Σ0 × B (X) where Σ0 is countably generated. If (Ω, Σ0 ) were separable, we could then apply Lemma 24.0.16 and be done. Unfortunately, we don’t know Σ0 separates the points of Ω. Therefore, we define an equivalence class on the points of Ω as follows. We say ω v ω 1 if and only if XA (ω) = XA (ω 1 ) for all A ∈ Σ0 . Now the nice thing to notice about this equivalence relation is that if ω ∈ A ∈ Σ0 , and if ω v ω 1 , then 1 = XA (ω) = XA (ω 1 ) implying ω 1 ∈ A also. Therefore, every set of Σ0 is the union of equivalence classes. It follows that for A ∈ Σ0 , and π the map given by πω ≡ [ω] where [ω] is the equivalence class determined by ω, π (A) ∩ π (Ω \ A) = ∅. Suppose now that Hn ∈ Σ0 × B (X). If ([ω] , x) ∈ ∩∞ n=1 π × idX (Hn ) , then for each n, ([w] , x) = (πwn , x) for some (ω n , x) ∈ Hn . But this implies ω v ω n and so from the above observation that the sets of Σ0 are unions of equivalence classes, it follows that (ω, x) ∈ Hn . ∞ Therefore, (ω, x) ∈ ∩∞ n=1 Hn and so ([ω] , x) = π × idX (ω, x) where (ω, x) ∈ ∩n=1 Hn . This shows that ∞ π × idX (∩∞ n=1 Hn ) ⊇ ∩n=1 π × idX (Hn ) .

In fact these two sets are equal because the other inclusion is obvious. We will denote by Ω1 the set of equivalence classes and Σ1 will be the subsets, S1 , of Ω1 such that S1 = {[ω] : ω ∈ S ∈ Σ0 } . Then (Ω1 , Σ1 ) is clearly a measure space which is separable. Let © ¡ ¢ ª F ≡ H ∈ Σ0 × B (X) : π × idX (H) , π × idX H C ∈ Σ1 × B (X) . We see that the measurable rectangles, A × B where A ∈ Σ0 and B ∈ B (X) are in F, that from the above observation on countable intersections, F is closed with respect to countable unions and closed with respect to complements. Therefore, F is a σ algebra and so F = Σ0 × B (X) . By Lemma 24.0.14 (Ω1 , Σ1 ) is isomorphic N to (E, B (E)) where E is a subspace of {0, 1} . Denoting the isomorphism by h, it follows as in Lemma 24.0.16 that h × idX maps Σ1 × B (X) to B (E) × B (X) . Therefore, we see f ≡ h ◦ π is a mapping from Ω to E which has the property that f × idX maps Σ0 × B (X) to B (E) × B (X) . Now from the proof of Lemma 24.0.16 b 0 . However, if µ is a finite measure on Σ, b starting with the claim, we see that G ∈ Σ ³ ´ ³d´ b = Σµ and so Σ b0 ⊆ Σ b ⊆ Σ. b This proves the theorem. then Σ µ

770


24.1

Multifunctions

Let X be a separable complete metric space and let (T, C, µ) be a set, a σ algebra of subsets of T, and a measure µ such that this is a complete σ finite measure space. Also let Γ : T → PF (X) , the closed subsets of X. Definition 24.1.1 We define Γ− (S) ≡ {t ∈ T : Γ (t) ∩ S 6= ∅} We will consider a theory of measurability of set valued functions. The following theorem is the main result in the subject. In this theorem the third condition is what we will refer to as measurable. Theorem 24.1.2 The following are equivalent. 1. For all B a Borel set in X, Γ− (B) ∈ C. 2. For all F closed in X, Γ− (F ) ∈ C 3. For all U open in X, Γ− (U ) ∈ C 4. There exists a sequence, {σ n } of measurable functions satisfying σ n (t) ∈ Γ (t) such that for all t ∈ T, Γ (t) = {σ n (t) : n ∈ N} These functions are called measurable selections. 5. For all x ∈ X, t → dist (x, Γ (t)) is a measurable real valued function. 6. G (Γ) ≡ {(t, x) : x ∈ Γ (t)} ⊆ C × B (X) . Proof: It is obvious that 1.) ⇒ 2.). To see that 2.) ⇒ 3.) note that Γ− (∪∞ i=1 Fi ) = (Fi ) . Since any open set in X can be obtained as a countable union of closed sets, this implies 2.) ⇒ 3.). ∞ Now we verify that 3.) ⇒ 4.). Let {xn }n=1 be a countable dense subset of X. For t ∈ T, let ψ 1 (t) = xn where n is the smallest integer such that Γ (t) ∩ B (xn , 1) 6= ∅. Therefore, ψ 1 (t) has countably many values, xn1 , xn2 , · · · where n1 < n2 < · · · . Now {t : ψ 1 = xn } = − ∪∞ i=1 Γ

{t : Γ (t) ∩ B (xn , 1) 6= ∅} ∩ [T \ ∪k Nε , then ||fn (z) − g (z)|| < ε P∞ for all z ∈ S. The infinite sum k=1 fn converges uniformly on S if the partial sums converge uniformly on S. Here ||·|| refers to the norm in X, the Banach space in which f has its values. The following proposition is also a routine application of the above definition. Neither the definition nor this proposition say anything new. Proposition 25.0.7 A sequence of functions, {fn } defined on a set S, converges uniformly to some function, g if and only if for all ε > 0 there exists Nε such that whenever m, n > Nε , ||fn − fm ||∞ < ε. Here ||f ||∞ ≡ sup {||f (z)|| : z ∈ S} . Just as in the case of functions of a real variable, one of the important theorems is the Weierstrass M test. Again, there is nothing new here. It is just a review of earlier material. Theorem 25.0.8 Let {fn } be a sequence of complex valued functions defined on P S ⊆ C. Suppose there exists M such that ||f || < M and M converges. n n n n ∞ P Then fn converges uniformly on S.

25.1. THE EXTENDED COMPLEX PLANE

777

Proof: Let z ∈ S. Then letting m < n ¯¯ n ¯¯ m n ∞ ¯¯ X ¯¯ X X X ¯¯ ¯¯ f (z) − f (z) ≤ ||f (z)|| ≤ Mk < ε ¯¯ ¯¯ k k k ¯¯ ¯¯ k=1

k=1

k=m+1

k=m+1

whenever m is large enough. Therefore, the sequence of partial sums is uniformly P∞ Cauchy on S and therefore, converges uniformly to k=1 fk (z) on S.

25.1

The Extended Complex Plane

The set of complex numbers has already been considered along with the topology of C which is nothing but the topology of R2 . Thus, for zn = xn + iyn , zn → z ≡ x + iy if and only if xn → x and yn → y. The norm in C is given by 1/2

|x + iy| ≡ ((x + iy) (x − iy))

¡ ¢1/2 = x2 + y 2

which is just the usual norm in R2 identifying (x, y) with x + iy. Therefore, C is a complete metric space topologically like R2 and so the Heine Borel theorem that compact sets are those which are closed and bounded is valid. Thus, as far as topology is concerned, there is nothing new about C. b , consists of the complex plane, C The extended complex plane, denoted by C along with another point not in C known as ∞. For example, ∞ could be any point in R3 . A sequence of complex numbers, zn , converges to ∞ if, whenever K is a compact set in C, there exists a number, N such that for all n > N, zn ∈ / K. Since compact sets in C are closed and bounded, this is equivalent to saying that for all R > 0, there exists N such that if n > N, then zn ∈ / B (0, R) which is the same as saying limn→∞ |zn | = ∞ where this last symbol has the same meaning as it does in calculus. A geometric way of understanding this in terms of more familiar objects involves a concept known as the Riemann sphere. 2 Consider the unit sphere, S 2 given by (z − 1) + y 2 + x2 = 1. Define a map from the complex plane to the surface of this sphere as follows. Extend a line from the point, p in the complex plane to the point (0, 0, 2) on the top of this sphere and let θ (p) denote the point of this sphere which the line intersects. Define θ (∞) ≡ (0, 0, 2). (0, 0, s 2) @ @ @ sθ(p) s (0, 0, 1) @ @ p @ @s

C

778

THE COMPLEX NUMBERS

Then θ−1 is sometimes called sterographic projection. The mapping θ is clearly continuous because it takes converging sequences, to converging sequences. Furthermore, it is clear that θ−1 is also continuous. In terms of the extended complex b a sequence, zn converges to ∞ if and only if θzn converges to (0, 0, 2) and plane, C, a sequence, zn converges to z ∈ C if and only if θ (zn ) → θ (z) . b In fact this makes it easy to define a metric on C. b including possibly w = ∞. Then let d (x, w) ≡ Definition 25.1.1 Let z, w ∈ C |θ (z) − θ (w)| where this last distance is the usual distance measured in R3 . ³ ´ b d is a compact, hence complete metric space. Theorem 25.1.2 C, b This means {θ (zn )} is a sequence in Proof: Suppose {zn } is a sequence in C. S 2 which is compact. Therefore, there exists a subsequence, {θznk } and a point, z ∈ S 2 such that θznk → θz in S 2 which implies immediately that d (znk , z) → 0. A compact metric space must be complete.

25.2

Exercises

1. Prove the root test for series of complex numbers. If ak ∈ C and r ≡ 1/n lim supn→∞ |an | then  ∞  converges absolutely if r < 1 X diverges if r > 1 ak  test fails if r = 1. k=0 2. Does limn→∞ n

¡ 2+i ¢n

exist? Tell why and find the limit if it does exist. Pn 3. Let A0 = 0 and let An ≡ k=1 ak if n > 0. Prove the partial summation formula, q−1 q X X ak bk = Aq bq − Ap−1 bp + Ak (bk − bk+1 ) . k=p

3

k=p

Now using this formula, suppose {bn } is a sequence of real numbers which converges toP0 and is decreasing. Determine those values of ω such that ∞ |ω| = 1 and k=1 bk ω k converges. 4. Let f : U ⊆ C → C be given by f (x + iy) = u (x, y) + iv (x, y) . Show f is continuous on U if and only if u : U → R and v : U → R are both continuous.

Riemann Stieltjes Integrals In the theory of functions of a complex variable, the most important results are those involving contour integration. I will base this on the notion of Riemann Stieltjes integrals as in [16], [50], and [32]. The Riemann Stieltjes integral is a generalization of the usual Riemann integral and requires the concept of a function of bounded variation. Definition 26.0.1 Let γ : [a, b] → C be a function. Then γ is of bounded variation if ( n ) X |γ (ti ) − γ (ti−1 )| : a = t0 < · · · < tn = b ≡ V (γ, [a, b]) < ∞ sup i=1

where the sums are taken over all possible lists, {a = t0 < · · · < tn = b} . The set of points γ ([a, b]) will also be denoted by γ ∗ . The idea is that it makes sense to talk of the length of the curve γ ([a, b]) , defined as V (γ, [a, b]) . For this reason, in the case that γ is continuous, such an image of a bounded variation function is called a rectifiable curve. Definition 26.0.2 Let γ : [a, b] → C be of bounded variation and let f : γ ∗ → X. Letting P ≡ {t0 , · · · , tn } where a = t0 < t1 < · · · < tn = b, define ||P|| ≡ max {|tj − tj−1 | : j = 1, · · · , n} and the Riemann Steiltjes sum by S (P) ≡

n X

f (γ (τ j )) (γ (tj ) − γ (tj−1 ))

j=1

where τ j ∈ [tj−1 , tj ] . (Note this notation is a little sloppy because it does not identify the R specific point, τ j used. It is understood that this point is arbitrary.) Define f dγ as the unique number which satisfies the following condition. For all ε > 0 γ there exists a δ > 0 such that if ||P|| ≤ δ, then ¯Z ¯ ¯ ¯ ¯ f dγ − S (P)¯ < ε. ¯ ¯ γ

779

780

RIEMANN STIELTJES INTEGRALS

Sometimes this is written as Z f dγ ≡ lim S (P) . ||P||→0

γ

The set of points in the curve, γ ([a, b]) will be denoted sometimes by γ ∗ . Then γ ∗ is a set of points in C and as t moves from a to b, γ (t) moves from γ (a) to γ (b) . Thus γ ∗ has a first point and a last point. If φ : [c, d] → [a, b] is a continuous nondecreasing function, then γ ◦ φ : [c, d] → C is also of bounded variation and yields the same set of points in C with the same first and last points. Theorem 26.0.3 Let φ and γ be as just described. Then assuming that Z f dγ γ

exists, so does

Z f d (γ ◦ φ) γ◦φ

and

Z

Z f dγ =

γ

f d (γ ◦ φ) .

(26.0.1)

γ◦φ

Proof: There exists δ > 0 such that if P is a partition of [a, b] such that ||P|| < δ, then ¯Z ¯ ¯ ¯ ¯ f dγ − S (P)¯ < ε. ¯ ¯ γ

By continuity of φ, there exists σ > 0 such that if Q is a partition of [c, d] with ||Q|| < σ, Q = {s0 , · · · , sn } , then |φ (sj ) − φ (sj−1 )| < δ. Thus letting P denote the points in [a, b] given by φ (sj ) for sj ∈ Q, it follows that ||P|| < δ and so ¯ ¯ ¯Z ¯ n X ¯ ¯ ¯ f dγ − f (γ (φ (τ j ))) (γ (φ (sj )) − γ (φ (sj−1 )))¯¯ < ε ¯ ¯ γ ¯ j=1 where τ j ∈ [sj−1 , sj ] . Therefore, from the definition 26.0.1 holds and Z f d (γ ◦ φ) γ◦φ

exists. R This theorem shows that γ f dγ is independent of the particular γ used in its computation to the extent that if φ is any nondecreasing continuous function from another interval, [c, d] , mapping to [a, b] , then the same value is obtained by replacing γ with γ ◦ φ. The fundamental result in this subject is the following theorem.

781 Theorem 26.0.4 Let f : γ ∗ → XRbe continuous and let γ : [a, b] → C be continuous and of bounded variation. Then γ f dγ exists. Also letting δ m > 0 be such that 1 |t − s| < δ m implies ||f (γ (t)) − f (γ (s))|| < m , ¯ ¯Z ¯ ¯ ¯ f dγ − S (P)¯ ≤ 2V (γ, [a, b]) ¯ ¯ m γ whenever ||P|| < δ m . Proof: The function, f ◦ γ , is uniformly continuous because it is defined on a compact set. Therefore, there exists a decreasing sequence of positive numbers, {δ m } such that if |s − t| < δ m , then |f (γ (t)) − f (γ (s))|
0 be given. Then there exists η : [a, b] → C such that η (a) = γ (a) , γ (b) = η (b) , η ∈ C 1 ([a, b]) , and ||γ − η|| < ε, ¯Z ¯ Z ¯ ¯ ¯ f (·, z) dγ − f (·, z) dη ¯ < ε, ¯ ¯

(26.0.9)

V (η, [a, b]) ≤ V (γ, [a, b]) ,

(26.0.10)

γ

(26.0.8)

η

where ||γ − η|| ≡ max {|γ (t) − η (t)| : t ∈ [a, b]} .

784


Proof: Extend γ to be defined on all R according to γ (t) = γ (a) if t < a and γ (t) = γ (b) if t > b. Now define γ h (t) ≡

1 2h

Z

2h t+ (b−a) (t−a)

γ (s) ds.

2h −2h+t+ (b−a) (t−a)

where the integral is defined in the obvious way. That is, Z

Z

b a

Z

b

α (t) + iβ (t) dt ≡

α (t) dt + i a

Therefore, 1 γ h (b) = 2h γ h (a) =

1 2h

Z

b

β (t) dt. a

b+2h

γ (s) ds = γ (b) , b

Z

a

γ (s) ds = γ (a) . a−2h

Also, because of continuity of γ and the fundamental theorem of calculus, ½ µ ¶µ ¶ 1 2h 2h γ 0h (t) = γ t+ (t − a) 1+ − 2h b−a b−a µ ¶µ ¶¾ 2h 2h γ −2h + t + (t − a) 1+ b−a b−a and so γ h ∈ C 1 ([a, b]) . The following lemma is significant. Lemma 26.0.8 V (γ h , [a, b]) ≤ V (γ, [a, b]) . Proof: Let a = t0 < t1 < · · · < tn = b. Then using the definition of γ h and changing the variables to make all integrals over [0, 2h] , n X

|γ h (tj ) − γ h (tj−1 )| =

j=1

¯ Z ¶ n ¯ 2h · µ X 2h ¯ 1 γ s − 2h + tj + (tj − a) − ¯ ¯ 2h 0 b−a j=1 ¶¸¯ µ ¯ 2h (tj−1 − a) ¯¯ γ s − 2h + tj−1 + b−a ¯ ¶ µ Z 2h X n ¯ 1 2h ¯ ≤ (tj − a) − γ s − 2h + tj + 2h 0 j=1 ¯ b−a ¶¯ µ ¯ 2h (tj−1 − a) ¯¯ ds. γ s − 2h + tj−1 + b−a

785 2h For a given s ∈ [0, 2h] , the points, s − 2h + tj + b−a (tj − a) for j = 1, · · · , n form an increasing list of points in the interval [a − 2h, b + 2h] and so the integrand is bounded above by V (γ, [a − 2h, b + 2h]) = V (γ, [a, b]) . It follows n X

|γ h (tj ) − γ h (tj−1 )| ≤ V (γ, [a, b])

j=1

which proves the lemma. With this lemma the proof of the theorem can be completed without too much ∗ trouble. Let H be open ¡ an ¢ set containing γ such that H is a compact subset of Ω. ∗ C Let 0 < ε < dist γ , H . Then there exists δ 1 such that if h < δ 1 , then for all t, |γ (t) − γ h (t)|

≤
0 is given, there exists η : [a, b] → Ω such that η (a) = γ (a) , η (b) = γ (b) , η is C 1 ([a, b]) , and ¯Z ¯ Z ¯ ¯ ¯ f (z, w) dz − f (z, w) dz ¯ < r, ||η − γ|| < r. ¯ ¯ γ

η

It will be very important to consider which functions have primitives. It turns out, it is not enough for f to be continuous in order to possess a primitive. This is in stark contrast to the situation for functions of a real variable in which the fundamental theorem of calculus will deliver a primitive for any continuous function. The reason for the interest in such functions is the following theorem and its corollary. Theorem 26.0.13 Let γ : [a, b] → C be continuous and of bounded variation. Also suppose F 0 (z) = f (z) for all z ∈ Ω, an open set containing γ ∗ and f is continuous on Ω. Then Z f (z) dz = F (γ (b)) − F (γ (a)) . γ

Proof: By Theorem 26.0.12 there exists η ∈ C 1 ([a, b]) such that γ (a) = η (a) , and γ (b) = η (b) such that ¯¯Z ¯¯ Z ¯¯ ¯¯ ¯¯ f (z) dz − f (z) dz ¯¯ < ε. ¯¯ ¯¯ γ

η

Then since η is in C 1 ([a, b]) , Z

Z f (z) dz

=

η

Z f (η (t)) η 0 (t) dt =

a

= Therefore,

b

b

a

dF (η (t)) dt dt

F (η (b)) − F (η (a)) = F (γ (b)) − F (γ (a)) .

¯¯ ¯¯ Z ¯¯ ¯¯ ¯¯(F (γ (b)) − F (γ (a))) − f (z) dz ¯¯ < ε ¯¯ ¯¯ γ

and since ε > 0 is arbitrary, this proves the theorem. Corollary 26.0.14 If γ : [a, b] → C is continuous, has bounded variation, is a closed curve, γ (a) = γ (b) , and γ ∗ ⊆ Ω where Ω is an open set on which F 0 (z) = f (z) , then Z f (z) dz = 0. γ

26.1. EXERCISES

26.1

789

Exercises

1. Let γ : [a, b] → R be increasing. Show V (γ, [a, b]) = γ (b) − γ (a) . 2. Suppose γ : [a, b] → C satisfies a Lipschitz condition, |γ (t) − γ (s)| ≤ K |s − t| . Show γ is of bounded variation and that V (γ, [a, b]) ≤ K |b − a| . 3. γ : [c0 , cm ] → C is piecewise smooth if there exist numbers, ck , k = 1, · · · , m such that c0 < c1 < · · · < cm−1 < cm such that γ is continuous and γ : [ck , ck+1 ] → C is C 1 . Show that such piecewise smooth functions are of bounded variation and give an estimate for V (γ, [c0 , cm ]) . 4. Let γ R: [0, 2π] → C be given by γ (t) = r (cos mt + i sin mt) for m an integer. Find γ dz z . 5. Show that if γ : [a, b] → C then there exists an increasing function h : [0, 1] → [a, b] such that γ ◦ h ([0, 1]) = γ ∗ . 6. Let γ : [a, b] → C be an arbitrary continuous curve having bounded variation and let f, g have continuous derivatives on some open set containing γ ∗ . Prove the usual integration by parts formula. Z Z 0 f g dz = f (γ (b)) g (γ (b)) − f (γ (a)) g (γ (a)) − f 0 gdz. γ

γ −(1/2)

θ

7. Let f (z) ≡ |z| e−i 2R where z = |z| eiθ . This function is called the principle −(1/2) branch of z . Find γ f (z) dz where γ is the semicircle in the upper half plane which goes from (1, 0) to (−1, 0) in the counter clockwise direction. Next do the integral in which γ goes in the clockwise direction along the semicircle in the lower half plane. 8. Prove an open set, U is connected if and only if for every two points in U, there exists a C 1 curve having values in U which joins them. 9. Let P, Q be two partitions of [a, b] with P ⊆ Q. Each of these partitions can be used to form an approximation to V (γ, [a, b]) as described above. Recall the total variation was the supremum of sums of a certain form determined by a partition. How is the sum associated with P related to the sum associated with Q? Explain. 10. Consider the curve, ½ γ (t) =

t + it2 sin 0 if t = 0

¡1¢ t

if t ∈ (0, 1]

.

Is γ a continuous curve having bounded variation? What if the t2 is replaced with t? Is the resulting curve continuous? Is it a bounded variation curve? R 11. Suppose γ : [a, b] → R is given by γ (t) = t. What is γ f (t) dγ? Explain.

790


Fundamentals Of Complex Analysis 27.1

Analytic Functions

Definition 27.1.1 Let Ω be an open set in C and let f : Ω → X. Then f is analytic on Ω if for every z ∈ Ω, lim

h→0

f (z + h) − f (z) ≡ f 0 (z) h

exists and is a continuous function of z ∈ Ω. Here h ∈ C. Note that if f is analytic, it must be the case that f is continuous. It is more common to not include the requirement that f 0 is continuous but it is shown later that the continuity of f 0 follows. What are some examples of analytic functions? In the case where X = C, the simplest example is any polynomial. Thus p (z) ≡

n X

ak z k

k=0

is an analytic function and p0 (z) =

n X

ak kz k−1 .

k=1

More generally, power series are analytic. This will be shown soon but first here is an important definition and a convergence theorem called the root test. P∞ Pn Definition 27.1.2 Let {ak } be a sequence in X. Then k=1 ak ≡ limn→∞ k=1 ak whenever this limit exists. When the limit exists, the series is said to converge. 791

792

FUNDAMENTALS OF COMPLEX ANALYSIS

P∞ 1/k Theorem 27.1.3 Consider k=1 ak and let ρ ≡ lim supk→∞ ||ak || . Then if ρ < 1, the series converges absolutely and if ρ > 1 the series diverges spectacularly in the P∞ k sense that limk→∞ ak 6= 0. If ρ = 1 the test fails. Also k=1 ak (z − a) converges on some disk B (a, R) . It converges absolutely if |z − a| < R and uniformly on P∞ k B (a, r1 ) whenever r1 < R. The function f (z) = k=1 ak (z − a) is continuous on B (a, R) . Proof: Suppose ρ < 1. Then there exists r ∈ P (ρ, 1) . Therefore, ||ak || ≤ rk for all k large enough and so by a comparison test, k ||ak || converges because the partial sums are bounded above. Therefore, the partial sums of the original series form a Cauchy sequence in X and so they also converge due to completeness of X. 1/k Now suppose ρ > 1. Then letting ρ > r > 1, it follows ||ak || ≥ r infinitely often. Thus ||ak || ≥ rk infinitely often. Thus there exists a subsequence for which ||ank || converges to ∞. Therefore, the series cannot converge. P∞ k Now consider k=1 ak (z − a) . This series converges absolutely if 1/k

lim sup ||ak ||

|z − a| < 1

k→∞

1/k

which is the same as saying |z − a| < 1/ρ where ρ ≡ lim supk→∞ ||ak || R = 1/ρ. Now suppose r1 < R. Consider |z − a| ≤ r1 . Then for such z,

. Let

k

||ak || |z − a| ≤ ||ak || r1k and

¡ ¢1/k r1 1/k 0. Then f is analytic on B (a, R) . So are all its derivatives.

27.1. ANALYTIC FUNCTIONS

793

P∞ k−1 on B (a, R) where R = ρ−1 as Proof: Consider g (z) = k=2 ak k (z − a) above. Let r1 < r < R. Then letting |z − a| < r1 and h < r − r1 , ¯¯ ¯¯ ¯¯ f (z + h) − f (z) ¯¯ ¯¯ − g (z)¯¯¯¯ ¯¯ h ¯ ¯ ∞ ¯ ¯ (z + h − a)k − (z − a)k X ¯ k−1 ¯ − k (z − a) ≤ ||ak || ¯ ¯ ¯ ¯ h k=2 ¯ ¯ Ã ! ∞ k µ ¶ ¯1 X ¯ X k ¯ k−i i k k−1 ¯ ≤ ||ak || ¯ (z − a) h − (z − a) − k (z − a) ¯ ¯h ¯ i i=0 k=2 ¯ ¯ Ã ! ∞ k µ ¶ ¯1 X ¯ X k ¯ k−i i k−1 ¯ = ||ak || ¯ (z − a) h − k (z − a) ¯ ¯ ¯h i i=1 k=2 ¯ Ã !¯ µ ¶ ∞ k ¯ X ¯ X k ¯ k−i i−1 ¯ ≤ ||ak || ¯ (z − a) h ¯ ¯ ¯ i i=2 k=2 Ã ! ∞ k−2 X Xµ k ¶ k−2−i i ≤ |h| ||ak || |z − a| |h| i+2 i=0 k=2 ! Ãk−2 µ ∞ X X k − 2¶ k (k − 1) k−2−i i |z − a| |h| = |h| ||ak || (i + 2) (i + 1) i i=0 k=2 Ãk−2 µ ! ¶ ∞ X k (k − 1) X k − 2 k−2−i i ≤ |h| ||ak || |z − a| |h| 2 i i=0 =

|h|

k=2 ∞ X

k=2

∞

||ak ||

X k (k − 1) k−2 k (k − 1) k−2 (|z − a| + |h|) < |h| ||ak || r . 2 2 k=2

Then

µ lim sup k→∞

and so

||ak ||

k (k − 1) k−2 r 2

¶1/k = ρr < 1

¯¯ ¯¯ ¯¯ f (z + h) − f (z) ¯¯ ¯¯ − g (z)¯¯¯¯ ≤ C |h| . ¯¯ h

therefore, g (z) = f 0 (z) . Now by Theorem 27.1.3 it also follows that f 0 is continuous. Since r1 < R was arbitrary, this shows that f 0 (z) is given by the differentiated series above for |z − a| < R. Now a repeat of the argument shows all the derivatives of f exist and are continuous on B (a, R).

27.1.1

Cauchy Riemann Equations

Next consider the very important Cauchy Riemann equations which give conditions under which complex valued functions of a complex variable are analytic.

794


Theorem 27.1.5 Let Ω be an open subset of C and let f : Ω → C be a function, such that for z = x + iy ∈ Ω, f (z) = u (x, y) + iv (x, y) . Then f is analytic if and only if u, v are C 1 (Ω) and ∂v ∂u ∂v ∂u = , =− . ∂x ∂y ∂y ∂x Furthermore, f 0 (z) =

∂u ∂v (x, y) + i (x, y) . ∂x ∂x

Proof: Suppose f is analytic first. Then letting t ∈ R, f (z + t) − f (z) = t→0 t

f 0 (z) = lim µ lim

t→0

u (x + t, y) + iv (x + t, y) u (x, y) + iv (x, y) − t t =

∂u (x, y) ∂v (x, y) +i . ∂x ∂x

But also f 0 (z) = lim µ lim

t→0

t→0

f (z + it) − f (z) = it

u (x, y + t) + iv (x, y + t) u (x, y) + iv (x, y) − it it µ ¶ 1 ∂u (x, y) ∂v (x, y) +i i ∂y ∂y =

¶

¶

∂u (x, y) ∂v (x, y) −i . ∂y ∂y

This verifies the Cauchy Riemann equations. We are assuming that z → f 0 (z) is continuous. Therefore, the partial derivatives of u and v are also continuous. To see this, note that from the formulas for f 0 (z) given above, and letting z1 = x1 + iy1 ¯ ¯ ¯ ∂v (x, y) ∂v (x1 , y1 ) ¯ ¯ ¯ ≤ |f 0 (z) − f 0 (z1 )| , − ¯ ∂y ¯ ∂y showing that (x, y) → ∂v(x,y) is continuous since (x1 , y1 ) → (x, y) if and only if ∂y z1 → z. The other cases are similar. Now suppose the Cauchy Riemann equations hold and the functions, u and v are C 1 (Ω) . Then letting h = h1 + ih2 , f (z + h) − f (z) = u (x + h1 , y + h2 )

27.1. ANALYTIC FUNCTIONS

795

+iv (x + h1 , y + h2 ) − (u (x, y) + iv (x, y)) We know u and v are both differentiable and so f (z + h) − f (z) = µ i

∂u ∂u (x, y) h1 + (x, y) h2 + ∂x ∂y

∂v ∂v (x, y) h1 + (x, y) h2 ∂x ∂y

¶ + o (h) .

Dividing by h and using the Cauchy Riemann equations, f (z + h) − f (z) = h

∂u ∂x

∂v (x, y) h1 + i ∂y (x, y) h2

h

∂v i ∂x (x, y) h1 +

∂u ∂y

(x, y) h2

h =

+

+

o (h) h

∂u h1 + ih2 ∂v h1 + ih2 o (h) (x, y) +i (x, y) + ∂x h ∂x h h

Taking the limit as h → 0, f 0 (z) =

∂u ∂v (x, y) + i (x, y) . ∂x ∂x

It follows from this formula and the assumption that u, v are C 1 (Ω) that f 0 is continuous. It is routine to verify that all the usual rules of derivatives hold for analytic functions. In particular, the product rule, the chain rule, and quotient rule.

27.1.2

An Important Example

An important example of an analytic function is ez ≡ exp (z) ≡ ex (cos y + i sin y) where z = x + iy. You can verify that this function satisfies the Cauchy Riemann equations and that all the partial derivatives are continuous. Also from the above 0 discussion, (ez ) = ex cos (y) + iex sin y = ez . Later I will show that ez is given by the usual power series. An important property of this function is that it can be used to parameterize the circle centered at z0 having radius r. Lemma 27.1.6 Let γ denote the closed curve which is a circle of radius r centered at z0 . Then a parameterization this curve is γ (t) = z0 + reit where t ∈ [0, 2π] . ¯ ¯ 2 Proof: |γ (t) − z0 | = ¯reit re−it ¯ = r2 . Also, you can see from the definition of the sine and cosine that the point described in this way moves counter clockwise over this circle.

796

27.2


Exercises

1. Verify all the usual rules of differentiation including the product and chain rules. 2. Suppose f and f 0 : U → C are analytic and f (z) = u (x, y) + iv (x, y) . Verify uxx + uyy = 0 and vxx + vyy = 0. This partial differential equation satisfied by the real and imaginary parts of an analytic function is called Laplace’s equation. We say these functions satisfying Laplace’s equation are harmonic Rfunctions. If u isR a harmonic function defined on B (0, r) show that y x v (x, y) ≡ 0 ux (x, t) dt − 0 uy (t, 0) dt is such that u + iv is analytic. 3. Let f : U → C be analytic and f (z) = u (x, y) + iv (x, y) . Show u, v and uv are all harmonic although it can happen that u2 is not. Recall that a function, w is harmonic if wxx + wyy = 0. 4. Define a function f (z) ≡ z ≡ x − iy where z = x + iy. Is f analytic? 5. If f (z) = u (x, y) + iv (x, y) and f is analytic, verify that µ ¶ ux uy 2 det = |f 0 (z)| . vx vy 6. Show that if u (x, y) + iv (x, y) = f (z) is analytic, then ∇u · ∇v = 0. Recall ∇u (x, y) = hux (x, y) , uy (x, y)i. 7. Show that every polynomial is analytic. 8. If γ (t) = x (t)+iy (t) is a C 1 curve having values in U, an open set of C, and if f : U → C is analytic, we can consider f ◦γ, another C 1 curve having values in 0 C. Also, γ 0 (t) and (f ◦ γ) (t) are complex numbers so these can be considered 2 as vectors in R as follows. The complex number, x + iy corresponds to the vector, hx, yi. Suppose that γ and η are two such C 1 curves having values in U and that γ (t0 ) = η (s0 ) = z and suppose that f : U → C is analytic. Show 0 0 that the angle between (f ◦ γ) (t0 ) and (f ◦ η) (s0 ) is the same as the angle 0 0 0 between γ (t0 ) and η (s0 ) assuming that f (z) 6= 0. Thus analytic mappings preserve angles at points where the derivative is nonzero. Such mappings are called isogonal. . Hint: To make this easy to show, first observe that hx, yi · ha, bi = 12 (zw + zw) where z = x + iy and w = a + ib. 9. Analytic functions are even better than what is described in Problem 8. In addition to preserving angles, they also preserve orientation. To verify this show that if z = x + iy and w = a + ib are two complex numbers, then hx, y, 0i and ha, b, 0i are two vectors in R3 . Recall that the cross product, hx, y, 0i × ha, b, 0i, yields a vector normal to the two given vectors such that the triple, hx, y, 0i, ha, b, 0i, and hx, y, 0i × ha, b, 0i satisfies the right hand rule

27.3. CAUCHY’S FORMULA FOR A DISK

797

and has magnitude equal to the product of the sine of the included angle times the product of the two norms of the vectors. In this case, the cross product either points in the direction of the positive z axis or in the direction of the negative z axis. Thus, either the vectors hx, y, 0i, ha, b, 0i, k form a right handed system or the vectors ha, b, 0i, hx, y, 0i, k form a right handed system. These are the two possible orientations. Show that in the situation of Problem 8 the orientation of γ 0 (t0 ) , η 0 (s0 ) , k is the same as the orientation of the 0 0 vectors (f ◦ γ) (t0 ) , (f ◦ η) (s0 ) , k. Such mappings are called conformal. If f is analytic and f 0 (z) 6= 0, then we know from this problem and the above that 0 f is a conformal map. Hint: You can do this by verifying that (f ◦ γ) (t0 ) × 2 0 (f ◦ η) (s0 ) = |f 0 (γ (t0 ))| γ 0 (t0 ) × η 0 (s0 ). To make the verification easier, you might first establish the following simple formula for the cross product where here x + iy = z and a + ib = w. (x, y, 0) × (a, b, 0) = Re (ziw) k. 10. Write the Cauchy Riemann equations in terms of polar coordinates. Recall the polar coordinates are given by x = r cos θ, y = r sin θ. This means, letting u (x, y) = u (r, θ) , v (x, y) = v (r, θ) , write the Cauchy Riemann equations in terms of r and θ. You should eventually show the Cauchy Riemann equations are equivalent to ∂u 1 ∂v ∂v 1 ∂u = , =− ∂r r ∂θ ∂r r ∂θ 11. Show that a real valued analytic function must be constant.

27.3

Cauchy’s Formula For A Disk

The Cauchy integral formula is the most important theorem in complex analysis. It will be established for a disk in this chapter and later will be generalized to much more general situations but the version given here will suffice to prove many interesting theorems needed in the later development of the theory. The following are some advanced calculus results. Lemma 27.3.1 Let f : [a, b] → C. Then f 0 (t) exists if and only if Re f 0 (t) and Im f 0 (t) exist. Furthermore, f 0 (t) = Re f 0 (t) + i Im f 0 (t) . Proof: The if part of the equivalence is obvious. Now suppose f 0 (t) exists. Let both t and t + h be contained in [a, b] ¯ ¯ ¯ ¯ ¯ ¯ f (t + h) − f (t) ¯ ¯ Re f (t + h) − Re f (t) 0 0 ¯ ¯ ¯ − Re (f (t))¯ ≤ ¯ − f (t)¯¯ ¯ h h

798


and this converges to zero as h → 0. Therefore, Re f 0 (t) = Re (f 0 (t)) . Similarly, Im f 0 (t) = Im (f 0 (t)) . Lemma 27.3.2 If g : [a, b] → C and g is continuous on [a, b] and differentiable on (a, b) with g 0 (t) = 0, then g (t) is a constant. Proof: From the above lemma, you can apply the mean value theorem to the real and imaginary parts of g. Applying the above lemma to the components yields the following lemma. Lemma 27.3.3 If g : [a, b] → Cn = X and g is continuous on [a, b] and differentiable on (a, b) with g 0 (t) = 0, then g (t) is a constant. If you want to have X be a complex Banach space, the result is still true. Lemma 27.3.4 If g : [a, b] → X and g is continuous on [a, b] and differentiable on (a, b) with g 0 (t) = 0, then g (t) is a constant. Proof: Let Λ ∈ X 0 . Then Λg : [a, b] → C . Therefore, from Lemma 27.3.2, for each Λ ∈ X 0 , Λg (s) = Λg (t) and since X 0 separates the points, it follows g (s) = g (t) so g is constant. Lemma 27.3.5 Let φ : [a, b] × [c, d] → R be continuous and let Z

b

g (t) ≡

φ (s, t) ds.

(27.3.1)

a

Then g is continuous. If

∂φ ∂t

exists and is continuous on [a, b] × [c, d] , then Z

b

g 0 (t) = a

∂φ (s, t) ds. ∂t

(27.3.2)

Proof: The first claim follows from the uniform continuity of φ on [a, b] × [c, d] , which uniform continuity results from the set being compact. To establish 27.3.2, let t and t + h be contained in [c, d] and form, using the mean value theorem, g (t + h) − g (t) h

= =

1 h 1 h Z

Z

b

[φ (s, t + h) − φ (s, t)] ds Z

= a

a b a

b

∂φ (s, t + θh) hds ∂t

∂φ (s, t + θh) ds, ∂t

where θ may depend on s but is some number between 0 and 1. Then by the uniform continuity of ∂φ ∂t , it follows that 27.3.2 holds.


799

Corollary 27.3.6 Let φ : [a, b] × [c, d] → C be continuous and let Z

b

g (t) ≡

φ (s, t) ds.

(27.3.3)

a


∂φ ∂t


b

g 0 (t) = a

∂φ (s, t) ds. ∂t

(27.3.4)

Proof: Apply Lemma 27.3.5 to the real and imaginary parts of φ. Applying the above corollary to the components, you can also have the same result for φ having values in Cn . Corollary 27.3.7 Let φ : [a, b] × [c, d] → Cn be continuous and let Z

b

g (t) ≡

φ (s, t) ds.

(27.3.5)

a


∂φ ∂t


b

0

g (t) = a

∂φ (s, t) ds. ∂t

(27.3.6)

If you want to consider φ having values in X, a complex Banach space a similar result holds. Corollary 27.3.8 Let φ : [a, b] × [c, d] → X be continuous and let Z

b

g (t) ≡

φ (s, t) ds.

(27.3.7)

a


∂φ ∂t

exists and is continuous on [a, b] × [c, d] , then Z 0

g (t) = a

b

∂φ (s, t) ds. ∂t

(27.3.8)

Proof: Let Λ ∈ X 0 . Then Λφ : [a, b] × [c, d] → C is continuous and and is continuous on [a, b] × [c, d] . Therefore, from 27.3.8, Z 0

0

Λ (g (t)) = (Λg) (t) = a

b

∂Λφ (s, t) ds = Λ ∂t

Z

and since X 0 separates the points, it follows 27.3.8 holds. The following is Cauchy’s integral formula for a disk.

a

b

∂φ (s, t) ds ∂t

∂Λφ ∂t

exists

800


Theorem 27.3.9 Let f : Ω → X be analytic on the open set, Ω and let B (z0 , r) ⊆ Ω. Let γ (t) ≡ z0 + reit for t ∈ [0, 2π] . Then if z ∈ B (z0 , r) , Z 1 f (w) f (z) = dw. 2πi γ w − z

(27.3.9)

Proof: Consider for α ∈ [0, 1] , ¡ ¢¢ Z 2π ¡ f z + α z0 + reit − z g (α) ≡ rieit dt. reit + z0 − z 0 If α equals one, this reduces to the integral in 27.3.9. The idea is to show g is a constant and that g (0) = f (z) 2πi. First consider the claim about g (0) . µZ 2π ¶ reit g (0) = dt if (z) reit + z0 − z 0 µZ 2π ¶ 1 = if (z) dt 0 1 − z−z 0 reit Z 2π X ∞ n = if (z) r−n e−int (z − z0 ) dt 0

n=0

¯ ¯ 0¯ because ¯ z−z < 1. Since this sum converges uniformly you can interchange the reit sum and the integral to obtain Z 2π ∞ X n −n g (0) = if (z) r (z − z0 ) e−int dt n=0

0

= 2πif (z) R 2π

because 0 e−int dt = 0 if n > 0. Next consider the claim that g is constant. By Corollary 27.3.7, for α ∈ (0, 1) , ¡ ¢¢ ¡ it ¢ Z 2π 0 ¡ f z + α z0 + reit − z re + z0 − z 0 g (α) = rieit dt reit + z0 − z 0 Z 2π ¡ ¡ ¢¢ = f 0 z + α z0 + reit − z rieit dt 0 µ ¶ Z 2π ¡ ¡ ¢¢ 1 d = f z + α z0 + reit − z dt dt α 0 ¡ ¡ ¢¢ 1 ¡ ¡ ¢¢ 1 = f z + α z0 + rei2π − z − f z + α z0 + re0 − z = 0. α α Now g is continuous on [0, 1] and g 0 (t) = 0 on (0, 1) so by Lemma 27.3.3, g equals a constant. This constant can only be g (0) = 2πif (z) . Thus, Z f (w) g (1) = dw = g (0) = 2πif (z) . w −z γ


801

This proves the theorem. This is a very significant theorem. A few applications are given next. Theorem 27.3.10 Let f : Ω → X be analytic where Ω is an open set in C. Then f has infinitely many derivatives on Ω. Furthermore, for all z ∈ B (z0 , r) , Z f (w) n! f (n) (z) = dw (27.3.10) 2πi γ (w − z)n+1 where γ (t) ≡ z0 + reit , t ∈ [0, 2π] for r small enough that B (z0 , r) ⊆ Ω. Proof: Let z ∈ B (z0 , r) ⊆ Ω and let B (z0 , r) ⊆ Ω. Then, letting γ (t) ≡ z0 + reit , t ∈ [0, 2π] , and h small enough, Z Z 1 f (w) 1 f (w) f (z) = dw, f (z + h) = dw 2πi γ w − z 2πi γ w − z − h Now

1 1 h − = w−z−h w−z (−w + z + h) (−w + z)

and so f (z + h) − f (z) h

= =

Z 1 hf (w) dw 2πhi γ (−w + z + h) (−w + z) Z 1 f (w) dw. 2πi γ (−w + z + h) (−w + z)

Now for all h sufficiently small, there exists a constant C independent of such h such that ¯ ¯ ¯ ¯ 1 1 ¯ ¯ − ¯ (−w + z + h) (−w + z) (−w + z) (−w + z) ¯ ¯ ¯ ¯ ¯ h ¯ ¯ = ¯ ¯ ≤ C |h| ¯ (w − z − h) (w − z)2 ¯ and so, the integrand converges uniformly as h → 0 to =

f (w) 2

(w − z)

Therefore, the limit as h → 0 may be taken inside the integral to obtain Z 1 f (w) 0 f (z) = dw. 2πi γ (w − z)2 Continuing in this way, yields 27.3.10. This is a very remarkable result. It shows the existence of one continuous derivative implies the existence of all derivatives, in contrast to the theory of functions of a real variable. Actually, more than what is stated in the theorem was shown. The above proof establishes the following corollary.

802


Corollary 27.3.11 Suppose f is continuous on ∂B (z0 , r) and suppose that for all z ∈ B (z0 , r) , Z 1 f (w) dw, f (z) = 2πi γ w − z where γ (t) ≡ z0 + reit , t ∈ [0, 2π] . Then f is analytic on B (z0 , r) and in fact has infinitely many derivatives on B (z0 , r) . Another application is the following lemma. Lemma 27.3.12 Let γ (t) = z0 + reit , for t ∈ [0, 2π], suppose fn → f uniformly on B (z0 , r), and suppose Z 1 fn (w) dw (27.3.11) fn (z) = 2πi γ w − z for z ∈ B (z0 , r) . Then f (z) =

1 2πi

Z γ

f (w) dw, w−z

(27.3.12)

implying that f is analytic on B (z0 , r) . Proof: From 27.3.11 and the uniform convergence of fn to f on γ ([0, 2π]) , the integrals in 27.3.11 converge to Z 1 f (w) dw. 2πi γ w − z Therefore, the formula 27.3.12 follows. Uniform convergence on a closed disk of the analytic functions implies the target function is also analytic. This is amazing. Think of the Weierstrass approximation theorem for polynomials. You can obtain a continuous nowhere differentiable function as the uniform limit of polynomials. The conclusions of the following proposition have all been obtained earlier in Theorem 27.1.4 but they can be obtained more easily if you use the above theorem and lemmas. Proposition 27.3.13 Let {an } denote a sequence in X. Then there exists R ∈ [0, ∞] such that ∞ X k ak (z − z0 ) k=0

converges absolutely if |z − z0 | < R, diverges if |z − z0 | > R and converges uniformly on B (z0 , r) for all r < R. Furthermore, if R > 0, the function, f (z) ≡

∞ X k=0

is analytic on B (z0 , R) .

ak (z − z0 )

k


803

Proof: The assertions about absolute convergence are routine from the root test if

µ ¶−1 1/n R ≡ lim sup |an | n→∞

with R = ∞ if the quantity in parenthesis equals zero. The root test can be used to verify absolute convergence which then implies convergence by completeness of X. The assertion about P∞uniform convergence follows from the Weierstrass M test and Mn ≡ |an | rn . ( n=0 |an | rn < ∞ by the root test). It only remains to verify the assertion about f (z) being analytic in the case where R > 0. Pn k Let 0 < r < R and define fn (z) ≡ k=0 ak (z − z0 ) . Then fn is a polynomial and so it is analytic. Thus, by the Cauchy integral formula above, fn (z) =

1 2πi

Z γ

fn (w) dw w−z

where γ (t) = z0 + reit , for t ∈ [0, 2π] . By Lemma 27.3.12 and the first part of this proposition involving uniform convergence, 1 f (z) = 2πi

Z γ

f (w) dw. w−z

Therefore, f is analytic on B (z0 , r) by Corollary 27.3.11. Since r < R is arbitrary, this shows f is analytic on B (z0 , R) . This proposition shows that all functions having values in X which are given as power series are analytic on their circle of convergence, the set of complex numbers, z, such that |z − z0 | < R. In fact, every analytic function can be realized as a power series. Theorem 27.3.14 If f : Ω → X is analytic and if B (z0 , r) ⊆ Ω, then f (z) =

∞ X

n

an (z − z0 )

(27.3.13)

n=0

for all |z − z0 | < r. Furthermore, an =

f (n) (z0 ) . n!

(27.3.14)

Proof: Consider |z − z0 | < r and let γ (t) = z0 + reit , t ∈ [0, 2π] . Then for w ∈ γ ([0, 2π]) , ¯ ¯ ¯ z − z0 ¯ ¯ ¯ ¯ w − z0 ¯ < 1

804


and so, by the Cauchy integral formula, Z 1 f (w) f (z) = dw 2πi γ w − z Z 1 f (w) ³ ´ dw = 2πi γ (w − z ) 1 − z−z0 0 w−z0 µ ¶n Z ∞ f (w) X z − z0 1 = dw. 2πi γ (w − z0 ) n=0 w − z0 Since the series converges uniformly, you can interchange the integral and the sum to obtain Ã ! Z ∞ X 1 f (w) n f (z) = (z − z0 ) n+1 2πi (w − z ) γ 0 n=0 ≡

∞ X

n

an (z − z0 )

n=0

By Theorem 27.3.10, 27.3.14 holds. Note that this also implies that if a function is analytic on an open set, then all of its derivatives are also analytic. This follows from Theorem 27.1.4 which says that a function given by a power series has all derivatives on the disk of convergence.

27.4

Exercises

¯P ∞ ¡ ¢¯ 1. Show that if |ek | ≤ ε, then ¯ k=m ek rk − rk+1 ¯ < ε if 0 ≤ r < 1. Hint: Let |θ| = 1 and verify that ¯ ¯ ∞ ∞ ∞ ¯X ¯ X X ¡ k ¢ ¡ ¢ ¡ ¢ ¯ ¯ θ ek r − rk+1 = ¯ ek rk − rk+1 ¯ = Re (θek ) rk − rk+1 ¯ ¯ k=m

k=m

k=m

where −ε < Re (θek ) < ε. P∞ n 2. Abel’s theorem saysPthat if n=0 an (z − a) Phas radius of convergence equal ∞ ∞ n to = = A. Hint: Show n=0¡ an , then ¢limr→1− n=0 an r P∞1 and kif A P ∞ k k+1 a r = A r − r where A denotes the k th partial sum k k k k=0 k=0 P of aj . Thus ∞ X k=0

k

ak r =

∞ X k=m+1

m ¡ k ¢ X ¡ ¢ k+1 Ak r − r Ak rk − rk+1 , + k=0

where |Ak − A| < ε for all k ≥ m. In the first sum, write Ak = A + ek and use P∞ k 1 Problem 1. Use this theorem to verify that arctan (1) = k=0 (−1) 2k+1 .

27.4. EXERCISES

805

3. Find the integrals using the Cauchy integral formula. R z it (a) γ sin z−i dz where γ (t) = 2e : t ∈ [0, 2π] . R 1 (b) γ z−a dz where γ (t) = a + reit : t ∈ [0, 2π] R cos z (c) γ z2 dz where γ (t) = eit : t ∈ [0, 2π] R 1 it (d) γ log(z) : t ∈ [0, 2π] and n = 0, 1, 2. In this z n dz where γ (t) = 1 + 2 e problem, log (z) ≡ ln |z| + i arg (z) where arg (z) ∈ (−π, π) and z = 0 |z| ei arg(z) . Thus elog(z) = z and log (z) = z1 . R z2 +4 4. Let γ (t) = 4eit : t ∈ [0, 2π] and find γ z(z 2 +1) dz. P∞ 5. Suppose f (z) = n=0 an z n for all |z| < R. Show that then 1 2π

Z

2π

∞ X ¯ ¡ iθ ¢¯2 2 ¯f re ¯ dθ = |an | r2n

0

n=0

for all r ∈ [0, R). Hint: Let fn (z) ≡

n X

ak z k ,

k=0

show 1 2π

Z

2π

n X ¯ ¡ iθ ¢¯2 2 ¯fn re ¯ dθ = |ak | r2k

0

k=0

and then take limits as n → ∞ using uniform convergence. 6. The Cauchy integral formula, marvelous as it is, can actually be improved upon. The Cauchy integral formula involves representing f by the values of f on the boundary of the disk, B (a, r) . It is possible to represent f by using only the values of Re f on the boundary. This leads to the Schwarz formula . Supply the details in the following outline. Suppose f is analytic on |z| < R and f (z) =

∞ X

an z n

(27.4.15)

n=0

with the series converging uniformly on |z| = R. Then letting |w| = R, 2u (w) = f (w) + f (w) and so 2u (w) =

∞ X k=0

ak w k +

∞ X k=0

k

ak (w) .

(27.4.16)

806


Now letting γ (t) = Reit , t ∈ [0, 2π] Z 2u (w) dw = w γ =

Z (a0 + a0 ) γ

1 dw w

2πi (a0 + a0 ) .

Thus, multiplying 27.4.16 by w−1 , Z u (w) 1 dw = a0 + a0 . πi γ w Now multiply 27.4.16 by w−(n+1) and integrate again to obtain Z 1 u (w) an = dw. πi γ wn+1 Using these formulas for an in 27.4.15, we can interchange the sum and the integral (Why can we do this?) to write the following for |z| < R. f (z)

= =

1 πi 1 πi

Z γ

1 X ³ z ´k+1 u (w) dw − a0 z w

γ

u (w) dw − a0 , w−z

Z

∞

k=0

R u(w) 1 dw and a0 = Re a0 − which is the Schwarz formula. Now Re a0 = 2πi γ w i Im a0 . Therefore, we can also write the Schwarz formula as Z 1 u (w) (w + z) f (z) = dw + i Im a0 . (27.4.17) 2πi γ (w − z) w 7. Take the real parts of the second form of the Schwarz formula to derive the Poisson formula for a disk, ¡ ¢¡ ¢ Z 2π ¡ iα ¢ u Reiθ R2 − r2 1 u re = dθ. (27.4.18) 2π 0 R2 + r2 − 2Rr cos (θ − α) 8. Suppose that u (w) is a given real continuous function defined on ∂B (0, R) and define f (z) for |z| < R by 27.4.17. Show that f, so defined is analytic. Explain why u given in 27.4.18 is harmonic. Show that ¡ ¢ ¡ ¢ lim u reiα = u Reiα . r→R−

Thus u is a harmonic function which approaches a given function on the boundary and is therefore, a solution to the Dirichlet problem.

27.5. ZEROS OF AN ANALYTIC FUNCTION

807

P∞ k 9. Suppose f (z) = k=0 ak (z − z0 ) for all |z − z0 | < R. Show that f 0 (z) = P∞ k−1 for all |z − z0 | < R. Hint: Let fn (z) be a partial sum k=0 ak k (z − z0 ) of f. Show that fn0 converges uniformly to some function, g on |z − z0 | ≤ r for any r < R. Now use the Cauchy integral formula for a function and its derivative to identify g with f 0 . ¡ ¢k P∞ 10. Use Problem 9 to find the exact value of k=0 k 2 13 . 11. Prove the binomial formula, α

(1 + z) = where

∞ µ ¶ X α n z n n=0

µ ¶ α α · · · (α − n + 1) ≡ . n n!

Can this be used to give a proof of the binomial formula, n µ ¶ X n n−k k n (a + b) = a b ? k k=0

Explain. 12. Suppose f is analytic on B (z0¯ , r) and¯continuous on B (z0 , r) and |f (z)| ≤ M on B (z0 , r). Show that then ¯f (n) (a)¯ ≤ Mrnn! .

27.5

Zeros Of An Analytic Function

In this section we give a very surprising property of analytic functions which is in stark contrast to what takes place for functions of a real variable. Definition 27.5.1 A region is a connected open set. It turns out the zeros of an analytic function which is not constant on some region cannot have a limit point. This is also a good time to define the order of a zero. Definition 27.5.2 Suppose f is an analytic function defined near a point, α where f (α) = 0. Thus α is a zero of the function, f. The zero is of order m if f (z) = m (z − α) g (z) where g is an analytic function which is not equal to zero at α. Theorem 27.5.3 Let Ω be a connected open set (region) and let f : Ω → X be analytic. Then the following are equivalent. 1. f (z) = 0 for all z ∈ Ω 2. There exists z0 ∈ Ω such that f (n) (z0 ) = 0 for all n.

808


3. There exists z0 ∈ Ω which is a limit point of the set, Z ≡ {z ∈ Ω : f (z) = 0} . Proof: It is clear the first condition implies the second two. Suppose the third holds. Then for z near z0 f (z) =

∞ X f (n) (z0 ) n (z − z0 ) n!

n=k

where k ≥ 1 since z0 is a zero of f. Suppose k < ∞. Then, k

f (z) = (z − z0 ) g (z) where g (z0 ) 6= 0. Letting zn → z0 where zn ∈ Z, zn 6= z0 , it follows k

0 = (zn − z0 ) g (zn ) which implies g (zn ) = 0. Then by continuity of g, we see that g (z0 ) = 0 also, contrary to the choice of k. Therefore, k cannot be less than ∞ and so z0 is a point satisfying the second condition. Now suppose the second condition and let n o S ≡ z ∈ Ω : f (n) (z) = 0 for all n . It is clear that S is a closed set which by assumption is nonempty. However, this set is also open. To see this, let z ∈ S. Then for all w close enough to z, f (w) =

∞ X f (k) (z) k=0

k!

k

(w − z) = 0.

Thus f is identically equal to zero near z ∈ S. Therefore, all points near z are contained in S also, showing that S is an open set. Now Ω = S ∪ (Ω \ S) , the union of two disjoint open sets, S being nonempty. It follows the other open set, Ω \ S, must be empty because Ω is connected. Therefore, the first condition is verified. This proves the theorem. (See the following diagram.) 1.) .% 2.)

& ←−

3.)

Note how radically different this is from the theory of functions of a real variable. Consider, for example the function ¡ ¢ ½ 2 x sin x1 if x 6= 0 f (x) ≡ 0 if x = 0

27.6. LIOUVILLE’S THEOREM

809

which has a derivative for all x ∈ R and for which 0 is a limit point of the set, Z, even though f is not identically equal to zero. Here is a very important application called Euler’s formula. Recall that ez ≡ ex (cos (y) + i sin (y)) Is it also true that ez =

(27.5.19)

P∞

zk k=0 k! ?

Theorem 27.5.4 (Euler’s Formula) Let z = x + iy. Then ez =

∞ X zk k=0

k!

.

Proof: It was already observed that ez given by 27.5.19 is analytic. So is P∞ k exp (z) ≡ k=0 zk! . In fact the power series converges for all z ∈ C. Furthermore the two functions, ez and exp (z) agree on the real line which is a set which contains a limit point. Therefore, they agree for all values of z ∈ C. This formula shows the famous two identities, eiπ = −1 and e2πi = 1.

27.6

Liouville’s Theorem

The following theorem pertains to functions which are analytic on all of C, “entire” functions. Definition 27.6.1 A function, f : C → C or more generally, f : C → X is entire means it is analytic on C. Theorem 27.6.2 (Liouville’s theorem) If f is a bounded entire function having values in X, then f is a constant. Proof: Since f is entire, pick any z ∈ C and write Z 1 f (w) f 0 (z) = dw 2πi γ R (w − z)2 where γ R (t) = z + Reit for t ∈ [0, 2π] . Therefore, ||f 0 (z)|| ≤ C

1 R

where C is some constant depending on the assumed bound on f. Since R is arbitrary, let R → ∞ to obtain f 0 (z) = 0 for any z ∈ C. It follows from this that f is constant for if zj j = 1, 2 are two complex numbers, let h (t) = f (z1 + t (z2 − z1 )) for t ∈ [0, 1] . Then h0 (t) = f 0 (z1 + t (z2 − z1 )) (z2 − z1 ) = 0. By Lemmas 27.3.2 27.3.4 h is a constant on [0, 1] which implies f (z1 ) = f (z2 ) .

810


With Liouville’s theorem it becomes possible to give an easy proof of the fundamental theorem of algebra. It is ironic that all the best proofs of this theorem in algebra come from the subjects of analysis or topology. Out of all the proofs that have been given of this very important theorem, the following one based on Liouville’s theorem is the easiest. Theorem 27.6.3 (Fundamental theorem of Algebra) Let p (z) = z n + an−1 z n−1 + · · · + a1 z + a0 be a polynomial where n ≥ 1 and each coefficient is a complex number. Then there exists z0 ∈ C such that p (z0 ) = 0. −1

Proof: Suppose not. Then p (z) is an entire function. Also ³ ´ n n−1 |p (z)| ≥ |z| − |an−1 | |z| + · · · + |a1 | |z| + |a0 | ¯ ¯ ¯ −1 ¯ and so lim|z|→∞ |p (z)| = ∞ which implies lim|z|→∞ ¯p (z) ¯ = 0. It follows that, −1

−1

since p (z) is bounded for z in any bounded set, we must have that p (z) is a −1 bounded entire function. But then it must be constant. However since p (z) → 0 1 is never equal to zero. This as |z| → ∞, this constant can only be 0. However, p(z) proves the theorem.

27.7

The General Cauchy Integral Formula

27.7.1

The Cauchy Goursat Theorem

This section gives a fundamental theorem which is essential to the development which follows and is closely related to the question of when a function has a primitive. First of all, if you have two points in C, z1 and z2 , you can consider γ (t) ≡ z1 + t (z2 − z1 ) for t ∈ [0, 1] to obtain a continuous bounded variation curve from z1 to z2 . More generally, if z1 , · · · , zm are points in C you can obtain a continuous bounded variation curve from z1 to zm which consists of first going from z1 to z2 and then from z2 to z3 and so on, till in the end one goes from zm−1 to zm . We denote this piecewise linear curve as γ (z1 , · · · , zm ) . Now let T be a triangle with vertices z1 , z2 and z3 encountered in the counter clockwise direction as shown. z3 ¡@ ¡ @ ¡ @ ¡ @ @ z2 z1¡ Denote by

R ∂T

f (z) dz, the expression,

R γ(z1 ,z2 ,z3 ,z1 )

f (z) dz. Consider the fol-

27.7. THE GENERAL CAUCHY INTEGRAL FORMULA

811

lowing picture. z3 ¡@ @ ¡ ª ¡ T11 @ I T ¡¾@ ¡@ ¡@ ¡ ª T21 R @ @ ¡ @ ¡ ª @ I ¡T31 @ ¡T41 @ I µ ¡ @ z @¡ z1¡ 2 By Lemma 26.0.11 Z f (z) dz = ∂T

4 Z X k=1

f (z) dz.

(27.7.20)

∂Tk1

On the “inside lines” the integrals cancel as claimed in Lemma 26.0.11 because there are two integrals going in opposite directions for each of these inside lines. Theorem 27.7.1 (Cauchy Goursat) Let f : Ω → X have the property that f 0 (z) exists for all z ∈ Ω and let T be a triangle contained in Ω. Then Z f (w) dw = 0. ∂T

Proof: Suppose not. Then ¯¯Z ¯¯ ¯¯ ¯¯

∂T

From 27.7.20 it follows

¯¯ ¯¯ f (w) dw¯¯¯¯ = α 6= 0.

¯¯ ¯¯ 4 ¯¯Z ¯¯ X ¯¯ ¯¯ α≤ f (w) dw¯¯ ¯¯ ¯¯ ∂T 1 ¯¯ k=1

k

Tk1 ,

and so for at least one of these ¯¯Z ¯¯ ¯¯ ¯¯

denoted from now on as T1 , ¯¯ ¯¯ α f (w) dw¯¯¯¯ ≥ . 4 ∂T1

Now let T1 play the same role as T , subdivide as in the above picture, and obtain T2 such that ¯¯Z ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ≥ α . f (w) dw ¯¯ ¯ ¯ 42 ∂T2

Continue in this way, obtaining a sequence of triangles, Tk ⊇ Tk+1 , diam (Tk ) ≤ diam (T ) 2−k ,

812


and

¯¯Z ¯¯ ¯¯ ¯¯

¯¯ ¯¯ α f (w) dw¯¯¯¯ ≥ k . 4 ∂Tk

0 Then let z ∈ ∩∞ k=1 Tk and note that by assumption, f (z) exists. Therefore, for all k large enough, Z Z f (w) dw = f (z) + f 0 (z) (w − z) + g (w) dw ∂Tk

∂Tk

where ||g (w)|| < ε |w − z| . Now observe that w → f (z) + f 0 (z) (w − z) has a primitive, namely, 2 F (w) = f (z) w + f 0 (z) (w − z) /2. Therefore, by Corollary 26.0.14. Z Z f (w) dw = ∂Tk

g (w) dw. ∂Tk

From the definition, of the integral, ¯¯Z ¯¯ ¯¯ ¯¯ α ¯ ¯ ≤ ¯¯ g (w) dw¯¯¯¯ ≤ ε diam (Tk ) (length of ∂Tk ) k 4 ∂Tk ≤ ε2−k (length of T ) diam (T ) 2−k , and so α ≤ ε (length of T ) diam (T ) .

R Since ε is arbitrary, this shows α = 0, a contradiction. Thus ∂T f (w) dw = 0 as claimed. This fundamental result yields the following important theorem. Theorem 27.7.2 (Morera1 ) Let Ω be an open set and let f 0 (z) exist for all z ∈ Ω. Let D ≡ B (z0 , r) ⊆ Ω. Then there exists ε > 0 such that f has a primitive on B (z0 , r + ε). Proof: Choose ε > 0 small enough that B (z0 , r + ε) ⊆ Ω. Then for w ∈ B (z0 , r + ε) , define Z F (w) ≡ f (u) du. γ(z0 ,w)

Then by the Cauchy Goursat theorem, and w ∈ B (z0 , r + ε) , it follows that for |h| small enough, Z 1 F (w + h) − F (w) = f (u) du h h γ(w,w+h) Z Z 1 1 1 = f (w + th) hdt = f (w + th) dt h 0 0 1 Giancinto

Morera 1856-1909. This theorem or one like it dates from around 1886


813

which converges to f (w) due to the continuity of f at w. This proves the theorem. The following is a slight generalization of the above theorem which is also referred to as Morera’s theorem. Corollary 27.7.3 Let Ω be an open set and suppose that whenever γ (z1 , z2 , z3 , z1 ) is a closed curve bounding a triangle T, which is contained in Ω, and f is a continuous function defined on Ω, it follows that Z f (z) dz = 0, γ(z1 ,z2 ,z3 ,z1 )

then f is analytic on Ω. Proof: As in the proof of Morera’s theorem, let B (z0 , r) ⊆ Ω and use the given condition to construct a primitive, F for f on B (z0 , r) . Then F is analytic and so by Theorem 27.3.10, it follows that F and hence f have infinitely many derivatives, implying that f is analytic on B (z0 , r) . Since z0 is arbitrary, this shows f is analytic on Ω.

27.7.2

A Redundant Assumption

Earlier in the definition of analytic, it was assumed the derivative is continuous. This assumption is redundant. Theorem 27.7.4 Let Ω be an open set in C and suppose f : Ω → X has the property that f 0 (z) exists for each z ∈ Ω. Then f is analytic on Ω. Proof: Let z0 ∈ Ω and let B (z0 , r) ⊆ Ω. By Morera’s theorem f has a primitive, F on B (z0 , r) . It follows that F is analytic because it has a derivative, f, and this derivative is continuous. Therefore, by Theorem 27.3.10 F has infinitely many derivatives on B (z0 , r) implying that f also has infinitely many derivatives on B (z0 , r) . Thus f is analytic as claimed. It follows a function is analytic on an open set, Ω if and only if f 0 (z) exists for z ∈ Ω. This is because it was just shown the derivative, if it exists, is automatically continuous. The same proof used to prove Theorem 27.7.2 implies the following corollary. Corollary 27.7.5 Let Ω be a convex open set and suppose that f 0 (z) exists for all z ∈ Ω. Then f has a primitive on Ω. Note that this implies that if Ω is a convex open set on which f 0 (z) exists and if γ : [a, b] → Ω is a closed, continuous curve having bounded variation, then letting F be a primitive of f Theorem 26.0.13 implies Z f (z) dz = F (γ (b)) − F (γ (a)) = 0. γ

814


Notice how different this is from the situation of a function of a real variable! It is possible for a function of a real variable to have a derivative everywhere and yet the derivative can be discontinuous. A simple example is the following. ¡ ¢ ½ 2 x sin x1 if x 6= 0 . f (x) ≡ 0 if x = 0 Then f 0 (x) exists for all x ∈ R. Indeed, if x 6= 0, the derivative equals 2x sin x1 −cos x1 which has no limit as x → 0. However, from the definition of the derivative of a function of one variable, f 0 (0) = 0.

27.7.3

Classification Of Isolated Singularities

First some notation. Definition 27.7.6 Let B 0 (a, r) ≡ {z ∈ C such that 0 < |z − a| < r}. Thus this is the usual ball without the center. A function is said to have an isolated singularity at the point a ∈ C if f is analytic on B 0 (a, r) for some r > 0. It turns out isolated singularities can be neatly classified into three types, removable singularities, poles, and essential singularities. The next theorem deals with the case of a removable singularity. Definition 27.7.7 An isolated singularity of f is said to be removable if there exists an analytic function, g analytic at a and near a such that f = g at all points near a. Theorem 27.7.8 Let f : B 0 (a, r) → X be analytic. Thus f has an isolated singularity at a. Suppose also that lim f (z) (z − a) = 0.

z→a

Then there exists a unique analytic function, g : B (a, r) → X such that g = f on B 0 (a, r) . Thus the singularity at a is removable. 2

Proof: Let h (z) ≡ (z − a) f (z) , h (a) ≡ 0. Then h is analytic on B (a, r) because it is easy to see that h0 (a) = 0. It follows h is given by a power series, h (z) =

∞ X

ak (z − a)

k

k=2

where a0 = a1 = 0 because of the observation above that h0 (a) = h (a) = 0. It follows that for |z − a| > 0 f (z) =

∞ X

ak (z − a)

k−2

≡ g (z) .

k=2

This proves the theorem. What of the other case where the singularity is not removable? This situation is dealt with by the amazing Casorati Weierstrass theorem.


815

Theorem 27.7.9 (Casorati Weierstrass) Let a be an isolated singularity and suppose for some r > 0, f (B 0 (a, r)) is not dense in C. Then either a is a removable singularity or there exist finitely many b1 , · · · , bM for some finite number, M such that for z near a, M X bk f (z) = g (z) + (27.7.21) k k=1 (z − a) where g (z) is analytic near a. Proof: Suppose B (z0 , δ) has no points of f (B 0 (a, r)) . Such a ball must exist if f (B 0 (a, r)) is not dense. Then for z ∈ B 0 (a, r) , |f (z) − z0 | ≥ δ > 0. It follows 1 from Theorem 27.7.8 that f (z)−z has a removable singularity at a. Hence, there 0 exists h an analytic function such that for z near a, 1 . f (z) − z0

h (z) =

(27.7.22)

P∞ k 1 There are two cases. First suppose h (a) = 0. Then k=1 ak (z − a) = f (z)−z 0 for z near a. If all the ak = 0, this would be a contradiction because then the left side would equal zero for z near a but the right side could not equal zero. Therefore, there is a first m such that am 6= 0. Hence there exists an analytic function, k (z) which is not equal to zero in some ball, B (a, ε) such that m

k (z) (z − a)

=

1 . f (z) − z0

Hence, taking both sides to the −1 power, f (z) − z0 =

∞ X 1 k bk (z − a) m (z − a) k=0

and so 27.7.21 holds. The other case is that h (a) 6= 0. In this case, raise both sides of 27.7.22 to the −1 power and obtain −1 f (z) − z0 = h (z) , a function analytic near a. Therefore, the singularity is removable. This proves the theorem. This theorem is the basis for the following definition which classifies isolated singularities. Definition 27.7.10 Let a be an isolated singularity of a complex valued function, f . When 27.7.21 holds for z near a, then a is called a pole. The order of the pole in 27.7.21 is M. If for every r > 0, f (B 0 (a, r)) is dense in C then a is called an essential singularity. In terms of the above definition, isolated singularities are either removable, a pole, or essential. There are no other possibilities.

816


Theorem 27.7.11 Suppose f : Ω → C has an isolated singularity at a ∈ Ω. Then a is a pole if and only if lim d (f (z) , ∞) = 0 z→a

b in C. Proof: Suppose first f has a pole at a. Then by definition, f (z) = g (z) + PM bk k=1 (z−a)k for z near a where g is analytic. Then |f (z)|

≥ =

|bM | |z − a|

M

Ã

1 |z − a|

− |g (z)| −

M

Ã

|bM | −

M −1 X

|bk | k

k=1

|z − a|

|g (z)| |z − a|

M

+

M −1 X

!! |bk | |z − a|

M −k

.

k=1

³ ´ PM −1 M M −k Now limz→a |g (z)| |z − a| + k=1 |bk | |z − a| = 0 and so the above inequality proves limz→a |f (z)| = ∞. Referring to the diagram on Page 778, you see this is the same as saying lim |θf (z) − (0, 0, 2)| = lim |θf (z) − θ (∞)| = lim d (f (z) , ∞) = 0

z→a

z→a

z→a

Conversely, suppose limz→a d (f (z) , ∞) = 0. Then from the diagram on Page 778, it follows limz→a |f (z)| = ∞ and in particular, a cannot be either removable or an essential singularity by the Casorati Weierstrass theorem, Theorem 27.7.9. The only case remaining is that a is a pole. This proves the theorem. Definition 27.7.12 Let f : Ω → C where Ω is an open subset of C. Then f is called meromorphic if all singularities are isolated and are either poles or removable and this set of singularities has no limit point. It is convenient to regard meromorphic b where if a is a pole, f (a) ≡ ∞. From now on, this functions as having values in C will be assumed when a meromorphic function is being considered. The usefulness of the above convention about f (a) ≡ ∞ at a pole is made clear in the following theorem. b be meromorphic. Theorem 27.7.13 Let Ω be an open subset of C and let f : Ω → C b Then f is continuous with respect to the metric, d on C. Proof: Let zn → z where z ∈ Ω. Then if z is a pole, it follows from Theorem 27.7.11 that d (f (zn ) , ∞) ≡ d (f (zn ) , f (z)) → 0. If z is not a pole, then f (zn ) → f (z) in C which implies |θ (f (zn )) − θ (f (z))| = d (f (zn ) , f (z)) → 0. Recall that θ is continuous on C.


27.7.4

817

The Cauchy Integral Formula

This section presents the general version of the Cauchy integral formula valid for arbitrary closed rectifiable curves. The key idea in this development is the notion of the winding number. This is the number also called the index, defined in the following theorem. This winding number, along with the earlier results, especially Liouville’s theorem, yields an extremely general Cauchy integral formula. Definition 27.7.14 Let γ : [a, b] → C and suppose z ∈ / γ ∗ . The winding number, n (γ, z) is defined by Z 1 dw n (γ, z) ≡ . 2πi γ w − z The main interest is in the case where γ is closed curve. However, the same notation will be used for any such curve. Theorem 27.7.15 Let γ : [a, b] → C be continuous and have bounded variation with γ (a) = γ (b) . Also suppose that z ∈ / γ ∗ . Define Z 1 dw n (γ, z) ≡ . (27.7.23) 2πi γ w − z Then n (γ, ·) is continuous and integer valued. Furthermore, there exists a sequence, η k : [a, b] → C such that η k is C 1 ([a, b]) , ||η k − γ||
r, it must be the case that z ∈ H and so for such z, the bottom description of g (z) found in 27.7.28 is valid. Therefore, it follows lim ||g (z)|| = 0

|z|→∞

and so g is bounded and entire. By Liouville’s theorem, g is a constant. Hence, from the above equation, the constant can only equal zero. For z ∈ Ω \ ∪m k=1 γ k ([ak , bk ]) , m

0 = h (z) =

1 X 2πi

Z

k=1 m

1 X 2πi

k=1


m

φ (z, w) dw = γk

Z γk

1 X 2πi

k=1 m

Z γk

f (w) − f (z) dw = w−z

X f (w) n (γ k , z) . dw − f (z) w−z k=1


823

Corollary 27.7.20 Let Ω be an open set and let γ k : [ak , bk ] → Ω, k = 1, · · · , m, be closed, continuous and of bounded variation. Suppose also that m X

n (γ k , z) = 0

k=1

for all z ∈ / Ω. Then if f : Ω → C is analytic, m Z X

f (w) dw = 0.

γk

k=1

Proof: This follows from Theorem 27.7.19 as follows. Let g (w) = f (w) (w − z) where z ∈ Ω \ ∪m k=1 γ k ([ak , bk ]) . Then by this theorem, 0=0

m X

n (γ k , z) = g (z)

k=1

m X

n (γ k , z) =

k=1

Z m m Z X 1 g (w) 1 X dw = f (w) dw. 2πi γ k w − z 2πi γk

k=1

k=1

Another simple corollary to the above theorem is Cauchy’s theorem for a simply connected region. Definition 27.7.21 An open set, Ω ⊆ C is a region if it is open and connected. A b \Ω is connected where C b is the extended complex region, Ω is simply connected if C plane. In the future, the term simply connected open set will be an open set which b \Ω is connected . is connected and C Corollary 27.7.22 Let γ : [a, b] → Ω be a continuous closed curve of bounded variation where Ω is a simply connected region in C and let f : Ω → X be analytic. Then Z f (w) dw = 0. γ

b ∗ . Thus ∞ ∈ C\γ b ∗. Proof: Let D denote the unbounded component of C\γ b b Then the connected set, C \ Ω is contained in D since every point of C \ Ω must be b ∗ and ∞ is contained in both C\Ω b in some component of C\γ and D. Thus D must b \ Ω. It follows that n (γ, ·) must be constant on be the component that contains C b \ Ω, its value being its value on D. However, for z ∈ D, C Z 1 1 n (γ, z) = dw 2πi γ w − z

824


and so lim|z|→∞ n (γ, z) = 0 showing n (γ, z) = 0 on D. Therefore this verifies the hypothesis of Theorem 27.7.19. Let z ∈ Ω ∩ D and define g (w) ≡ f (w) (w − z) . Thus g is analytic on Ω and by Theorem 27.7.19, Z Z 1 g (w) 1 f (w) dw. 0 = n (z, γ) g (z) = dw = 2πi γ w − z 2πi γ This proves the corollary. The following is a very significant result which will be used later. Corollary 27.7.23 Suppose Ω is a simply connected open set and f : Ω → X is analytic. Then f has a primitive, F, on Ω. Recall this means there exists F such that F 0 (z) = f (z) for all z ∈ Ω. Proof: Pick a point, z0 ∈ Ω and let V denote those points, z of Ω for which there exists a curve, γ : [a, b] → Ω such that γ is continuous, of bounded variation, γ (a) = z0 , and γ (b) = z. Then it is easy to verify that V is both open and closed in Ω and therefore, V = Ω because Ω is connected. Denote by γ z0 ,z such a curve from z0 to z and define Z F (z) ≡ f (w) dw. γ z0 ,z

Then F is well defined because if γ j , j = 1, 2 are two such curves, it follows from Corollary 27.7.22 that Z Z f (w) dw + f (w) dw = 0, γ1

implying that

−γ 2

Z

Z f (w) dw =

γ1

f (w) dw. γ2

Now this function, F is a primitive because, thanks to Corollary 27.7.22 Z 1 (F (z + h) − F (z)) h−1 = f (w) dw h γ z,z+h Z 1 1 = f (z + th) hdt h 0 and so, taking the limit as h → 0, F 0 (z) = f (z) .

27.7.5

An Example Of A Cycle

The next theorem deals with the existence of a cycle with nice properties. Basically, you go around the compact subset of an open set with suitable contours while staying in the open set. The method involves the following simple concept.


825

Definition 27.7.24 A tiling of R2 = C is the union of infinitely many equally spaced vertical and horizontal lines. You can think of the small squares which result as tiles. To tile the plane or R2 = C means to consider such a union of horizontal and vertical lines. It is like graph paper. See the picture below for a representation of part of a tiling of C.

Theorem 27.7.25 Let K be a compact subset of an open set, Ω. Then there exist m continuous, closed, bounded variation oriented curves {Γj }j=1 for which Γ∗j ∩ K = ∅ for each j, Γ∗j ⊆ Ω, and for all p ∈ K, m X

n (Γk , p) = 1.

k=1

while for all z ∈ / Ω

m X

n (Γk , z) = 0.

k=1

¡ ¢ Proof: Let δ = dist K, ΩC . Since K is compact, δ > 0. Now tile the plane with squares, each of which has diameter less than δ/2. Ω

K

K

826


Let S denote the set of all the closed squares in this tiling which have nonempty intersection with K.Thus, all the squares of S are contained in Ω. First suppose p is a point of K which is in the interior of one of these squares in the tiling. Denote by ∂Sk the boundary of Sk one of the squares in S, oriented in the counter clockwise direction and Sm denote the square of S which contains the point, p in its interior. n o4 Let the edges of the square, Sj be γ jk . Thus a short computation shows k=1

n (∂Sm , p) = 1 but n (∂Sj , p) = 0 for all j 6= m. The reason for this is that for z in Sj , the values {z − p : z ∈ Sj } lie in an open square, Q which is located at a b \ Q is connected and 1/ (z − p) is analytic on Q. positive distance from 0. Then C It follows from Corollary 27.7.23 that this function has a primitive on Q and so

Z ∂Sj

1 dz = 0. z−p

Similarly, if z ∈ / Ω, n (∂Sj , z) = 0. On the other direct computation will ³ hand, ´ a P P j verify that n (p, ∂Sm ) = 1. Thus 1 = = j,k n p, γ k Sj ∈S n (p, ∂Sj ) and if ³ ´ P P j z∈ / Ω, 0 = j,k n z, γ k = Sj ∈S n (z, ∂Sj ) . l∗ If γ j∗ k coincides with γ l , then the contour integrals taken over this edge are taken in opposite directions and so ³ the edge ´ the two squares have in common can P be deleted without changing j,k n z, γ jk for any z not on any of the lines in the tiling. For example, see the picture,

¾

¾

¾

?

? -

6 -

6

?

6 -

From the construction, if any of the γ j∗ k contains a point of K then this point is on one of the four edges of Sj and at this point, there is at least one edge of some Sl which also contains this ³ point. ´ As just discussed, this shared edge can be deleted P without changing i,j n z, γ jk . Delete the edges of the Sk which intersect K but not the endpoints of these edges. That is, delete the open edges. When this is done, m delete all isolated points. Let the resulting oriented curves be denoted by {γ k }k=1 . ∗ ∗ Note that you might have γ k = γ l . The construction is illustrated in the following picture.


827

Ω

K ?

? ¾

K

6

6

? -

Then as explained above, about the closed curves.

Pm k=1

n (p, γ k ) = 1. It remains to prove the claim

Each orientation on an edge corresponds to a direction of motion over that edge. Call such a motion over the edge a route. Initially, every vertex, (corner of a square in S) has the property there are the same number of routes to and from that vertex. When an open edge whose closure contains a point of K is deleted, every vertex either remains unchanged as to the number of routes to and from that vertex or it loses both a route away and a route to. Thus the property of having the same number of routes to and from each vertex is preserved by deleting these open edges.. The isolated points which result lose all routes to and from. It follows that upon removing the isolated points you can begin at any of the remaining vertices and follow the routes leading out from this and successive vertices according to orientation and eventually return to that end. Otherwise, there would be a vertex which would have only one route leading to it which does not happen. Now if you have used all the routes out of this vertex, pick another vertex and do the same process. Otherwise, pick an unused route out of the vertex and follow it to return. Continue this way till all routes are used exactly once, resulting in closed oriented curves, Γk . Then X k

n (Γk , p) =

X ¡ ¢ n γ j , p = 1. j

In case p ∈ K is on some line of the tiling, it is not on any of the Γk because Γ∗k ∩ K = ∅ and so the continuity of z → n (Γk , z) yields the desired result in this case also. This proves the lemma.

828


27.8

Exercises

1. If U is simply connected, f is analytic on U and f has no zeros in U, show there exists an analytic function, F, defined on U such that eF = f. 2. Let f be defined and analytic near the point a ∈ C. Show that then f (z) = P∞ k k=0 bk (z − a) whenever |z − a| < R where R is the distance between a and the nearest point where f fails to have a derivative. The number R, is called the radius of convergence and the power series is said to be expanded about a. 1 3. Find the radius of convergence of the function 1+z 2 expanded about a = 2. 1 Note there is nothing wrong with the function, 1+x2 when considered as a function of a real variable, x for any value of x. However, if you insist on using power series, you find there is a limitation on the values of x for which the power series converges due to the presence in the complex plane of a point, i, where the function fails to have a derivative.

4. Suppose f is analytic on all of C and satisfies |f (z)| < A + B |z| is constant.

1/2

. Show f

5. What if you defined an open set, U to be simply connected if C\U is connected. Would it amount to the same thing? Hint: Consider the outside of B (0, 1) . R 6. Let γ (t) = eit : t ∈ [0, 2π] . Find γ z1n dz for n = 1, 2, · · · . ¢2n ¡ 1 ¢ R 2π R ¡ 2n it 7. Show i 0 (2 cos θ) dθ = γ z + z1 z dz where γ (t) = e : t ∈ [0, 2π] . Then evaluate this integral using the binomial theorem and the previous problem. 8. Suppose that for some constants a, b 6= 0, a, b ∈ R, f (z + ib) = f (z) for all z ∈ C and f (z + a) = f (z) for all z ∈ C. If f is analytic, show that f must be constant. Can you generalize this? Hint: This uses Liouville’s theorem. 9. Suppose f (z) = u (x, y) + iv (x, y) is analytic for z ∈ U, an open set. Let g (z) = u∗ (x, y) + iv ∗ (x, y) where µ ∗ ¶ µ ¶ u u = Q v∗ v where Q is a unitary matrix. That is QQ∗ = Q∗ Q = I. When will g be analytic? 10. Suppose f is analytic on an open set, U, except for γ ∗ ⊂ U where γ is a one to one continuous function having bounded variation, but it is known that f is continuous on γ ∗ . Show that in fact f is analytic on γ ∗ also. Hint: Pick a point on γ ∗ , say γ (t0 ) and suppose for now that t0 ∈ (a, b) . Pick r > 0 such that B = B (γ (t0 ) , r) ⊆ U. Then show there exists t1 < t0 and t2 > t0 such

27.8. EXERCISES

829

that γ ([t1 , t2 ]) ⊆ B and γ (ti ) ∈ / B. Thus γ ([t1 , t2 ]) is a path across B going through the center of B which divides B into two open sets, B1 , and B2 along with γ ∗ . Let the boundary of Bk consist of γ ([t1 , t2 ]) and a circular arc, Ck . (w) over γ ∗ in two different directions Now letting z ∈ Bk , the line integral of fw−z R f (w) 1 cancels. Therefore, if z ∈ Bk , you can argue that f (z) = 2πi dw. By C w−z continuity, this continues to hold for z ∈ γ ((t1 , t2 )) . Therefore, f must be analytic on γ ((t1 , t1 )) also. This shows that f must be analytic on γ ((a, b)) . To get the endpoints, simply extend γ to have the same properties but defined on [a − ε, b + ε] and repeat the above argument or else do this at the beginning and note that you get [a, b] ⊆ (a − ε, b + ε) . 11. Let U be an open set contained in the upper half plane and suppose that there are finitely many line segments on the x axis which are contained in the boundary of U. Now suppose that f is defined, real, and continuous on e denote the these line segments and is defined and analytic on U. Now let U reflection of U across the x axis. Show that it is possible to extend f to a function, g defined on all of e ∪ U ∪ {the line segments mentioned earlier} W ≡U e , the reflection of U across the such that g is analytic in W . Hint: For z ∈ U e ∪ U and continuous on x axis, let g (z) ≡ f (z). Show that g is analytic on U the line segments. Then use Problem 10 or Morera’s theorem to argue that g is analytic on the line segments also. The result of this problem is know as the Schwarz reflection principle. 12. Show that rotations and translations of analytic functions yield analytic functions and use this observation to generalize the Schwarz reflection principle to situations in which the line segments are part of a line which is not the x axis. Thus, give a version which involves reflection about an arbitrary line.

830


The Open Mapping Theorem 28.1

A Local Representation

The open mapping theorem, is an even more surprising result than the theorem about the zeros of an analytic function. The following proof of this important theorem uses an interesting local representation of the analytic function. Theorem 28.1.1 (Open mapping theorem) Let Ω be a region in C and suppose f : Ω → C is analytic. Then f (Ω) is either a point or a region. In the case where f (Ω) is a region, it follows that for each z0 ∈ Ω, there exists an open set, V containing z0 and m ∈ N such that for all z ∈ V, m

f (z) = f (z0 ) + φ (z)

(28.1.1)

where φ : V → B (0, δ) is one to one, analytic and onto, φ (z0 ) = 0, φ0 (z) = 6 0 on V and φ−1 analytic on B (0, δ) . If f is one to one then m = 1 for each z0 and f −1 : f (Ω) → Ω is analytic. Proof: Suppose f (Ω) is not a point. Then if z0 ∈ Ω it follows there exists r > 0 such that f (z) 6= f (z0 ) for all z ∈ B (z0 , r) \ {z0 } . Otherwise, z0 would be a limit point of the set, {z ∈ Ω : f (z) − f (z0 ) = 0} which would imply from Theorem 27.5.3 that f (z) = f (z0 ) for all z ∈ Ω. Therefore, making r smaller if necessary and using the power series of f, ³ ´m ? m 1/m f (z) = f (z0 ) + (z − z0 ) g (z) (= (z − z0 ) g (z) ) for all z ∈ B (z0 , r) , where g (z) 6= 0 on B (z0 , r) . As implied in the above formula, one wonders if you can take the mth root of g (z) . g0 g is an analytic function on B (z0 , r) and so by Corollary 27.7.5 it has a primitive ¡ ¢0 on B (z0 , r) , h. Therefore by the product rule and the chain rule, ge−h = 0 and so there exists a constant, C = ea+ib such that on B (z0 , r) , ge−h = ea+ib . 831

832

THE OPEN MAPPING THEOREM

Therefore, g (z) = eh(z)+a+ib and so, modifying h by adding in the constant, a + ib, g (z) = eh(z) where h0 (z) = g 0 (z) g(z) on B (z0 , r) . Letting φ (z) = (z − z0 ) e

h(z) m

implies formula 28.1.1 is valid on B (z0 , r) . Now φ0 (z0 ) = e

h(z0 ) m

6= 0.

Shrinking r if necessary you can assume φ0 (z) 6= 0 on B (z0 , r). Is there an open set, V contained in B (z0 , r) such that φ maps V onto B (0, δ) for some δ > 0? Let φ (z) = u (x, y) + iv (x, y) where z = x + iy. Consider the mapping µ ¶ µ ¶ x u (x, y) → y v (x, y) where u, v are C 1 because φ is given (x, y) ∈ B (z0 , r) is ¯ ¯ ux (x, y) uy (x, y) ¯ ¯ vx (x, y) vy (x, y)

to be analytic. The Jacobian of this map at ¯ ¯ ¯ ¯ ux (x, y) ¯=¯ ¯ ¯ vx (x, y)

¯ −vx (x, y) ¯¯ ux (x, y) ¯

¯ ¯2 2 2 = ux (x, y) + vx (x, y) = ¯φ0 (z)¯ 6= 0. This follows from a use of the Cauchy Riemann equations. Also µ ¶ µ ¶ u (x0 , y0 ) 0 = v (x0 , y0 ) 0 Therefore, by the inverse function theorem there exists an open set, V, containing T z0 and δ > 0 such that (u, v) maps V one to one onto B (0, δ) . Thus φ is one to one onto B (0, δ) as claimed. Applying the same argument to other points, z of V and using the fact that φ0 (z) 6= 0 at these points, it follows φ maps open sets to open sets. In other words, φ−1 is continuous. It also follows that φm maps V onto B (0, δ m ) . Therefore, the formula 28.1.1 implies that f maps the open set, V, containing z0 to an open set. This shows f (Ω) is an open set because z0 was arbitrary. It is connected because f is continuous and Ω is connected. Thus f (Ω) is a region. It remains to verify that φ−1 is analytic on B (0, δ) . Since φ−1 is continuous, z1 − z 1 φ−1 (φ (z1 )) − φ−1 (φ (z)) = lim = 0 . z1 →z φ (z1 ) − φ (z) φ (z1 ) − φ (z) φ(z1 )→φ(z) φ (z) lim

Therefore, φ−1 is analytic as claimed.

28.2. BRANCHES OF THE LOGARITHM

833

It only remains to verify the assertion about the case where f is one to one. If 2πi m > 1, then e m 6= 1 and so for z1 ∈ V, e

2πi m

φ (z1 ) 6= φ (z1 ) .

(28.1.2)

2πi

But e m φ (z1 ) ∈ B (0, δ) and so there exists z2 6= z1 (since φ is one to one) such that 2πi φ (z2 ) = e m φ (z1 ) . But then ³ 2πi ´m m m φ (z2 ) = e m φ (z1 ) = φ (z1 ) implying f (z2 ) = f (z1 ) contradicting the assumption that f is one to one. Thus m = 1 and f 0 (z) = φ0 (z) 6= 0 on V. Since f maps open sets to open sets, it follows that f −1 is continuous and so ¡

¢0 f −1 (f (z))

= =

f −1 (f (z1 )) − f −1 (f (z)) f (z1 ) − f (z) f (z1 )→f (z) z1 − z 1 lim = 0 . z1 →z f (z1 ) − f (z) f (z) lim

This proves the theorem. One does not have to look very far to find that this sort of thing does not hold for functions mapping R to R. Take for example, the function f (x) = x2 . Then f (R) is neither a point nor a region. In fact f (R) fails to be open. Corollary 28.1.2 Suppose in the situation of Theorem 28.1.1 m > 1 for the local representation of f given in this theorem. Then there exists δ > 0 such that if w ∈ B (f (z0 ) , δ) = f (V ) for V an open set containing z0 , then f −1 (w) consists of m distinct points in V. (f is m to one on V ) Proof: Let w ∈ B (f (z0 ) , δ) . Then w = f (b zn) where zb o∈ V. Thus f (b z) = m 2kπi m m . Then each of f (z0 ) + φ (b z ) . Consider the m distinct numbers, e φ (b z) k=1

these numbers is in B (0, δ) and so since φ maps V one to one onto B (0, δ) , there 2kπi m are m distinct numbers in V , {zk }k=1 such that φ (zk ) = e m φ (b z ). Then ³ 2kπi ´m m f (zk ) = f (z0 ) + φ (zk ) = f (z0 ) + e m φ (b z) =

m

f (z0 ) + e2kπi φ (b z)

= f (z0 ) + φ (b z)

m

= f (b z) = w


28.2

Branches Of The Logarithm

The argument used in to prove the next theorem was used in the proof of the open mapping theorem. It is a very important result and deserves to be stated as a theorem.

834


Theorem 28.2.1 Let Ω be a simply connected region and suppose f : Ω → C is analytic and nonzero on Ω. Then there exists an analytic function, g such that eg(z) = f (z) for all z ∈ Ω. Proof: The function, f 0 /f is analytic on Ω and so by Corollary 27.7.23 there is a primitive for f 0 /f, denoted as g1 . Then ¡ −g1 ¢0 f0 e f = − e−g1 f + e−g1 f 0 = 0 f and so since Ω is connected, it follows e−g1 f equals a constant, ea+ib . Therefore, f (z) = eg1 (z)+a+ib . Define g (z) ≡ g1 (z) + a + ib. The function, g in the above theorem is called a branch of the logarithm of f and is written as log (f (z)). Definition 28.2.2 Let ρ be a ray starting at 0. Thus ρ is a straight line of infinite length extending in one direction with its initial point at 0. A special case of the above theorem is the following. Theorem 28.2.3 Let ρ be a ray starting at 0. Then there exists an analytic function, L (z) defined on C \ ρ such that eL(z) = z. This function, L is called a branch of the logarithm. This branch of the logarithm satisfies the usual formula for logarithms, L (zw) = L (z) + L (w) provided zw ∈ / ρ. Proof: C \ ρ is a simply connected region because its complement with respect b to C is connected. Furthermore, the function, f (z) = z is not equal to zero on C \ ρ. Therefore, by Theorem 28.2.1 there exists an analytic function L (z) such that eL(z) = f (z) = z. Now consider the problem of finding a description of L (z). Each z ∈ C \ ρ can be written in a unique way in the form z = |z| ei argθ (z) where argθ (z) is the angle in (θ, θ + 2π) associated with z. (You could of course have considered this to be the angle in (θ − 2π, θ) associated with z or in infinitely many other open intervals of length 2π. The description of the log is not unique.) Then letting L (z) = a + ib z = |z| ei argθ (z) = eL(z) = ea eib and so you can let L (z) = ln |z| + i argθ (z) . Does L (z) satisfy the usual properties of the logarithm? That is, for z, w ∈ C\ρ, is L (zw) = L (z) + L (w)? This follows from the usual rules of exponents. You know ez+w = ez ew . (You can verify this directly or you can reduce to the case where z, w are real. If z is a fixed real number, then the equation holds for all real w. Therefore,

28.3. MAXIMUM MODULUS THEOREM

835

it must also hold for all complex w because the real line contains a limit point. Now for this fixed w, the equation holds for all z real. Therefore, by similar reasoning, it holds for all complex z.) Now suppose z, w ∈ C \ ρ and zw ∈ / ρ. Then eL(zw) = zw, eL(z)+L(w) = eL(z) eL(w) = zw and so L (zw) = L (z) + L (w) as claimed. This proves the theorem. In the case where the ray is the negative real axis, it is called the principal branch of the logarithm. Thus arg (z) is a number between −π and π. Definition 28.2.4 Let log denote the branch of the logarithm which corresponds to the ray for θ = π. That is, the ray is the negative real axis. Sometimes this is called the principal branch of the logarithm.

28.3

Maximum Modulus Theorem

Here is another very significant theorem known as the maximum modulus theorem which follows immediately from the open mapping theorem. Theorem 28.3.1 (maximum modulus theorem) Let Ω be a bounded region and let f : Ω → C be analytic and f : Ω → C continuous. Then if z ∈ Ω, |f (z)| ≤ max {|f (w)| : w ∈ ∂Ω} .

(28.3.3)

If equality is achieved for any z ∈ Ω, then f is a constant. Proof: Suppose f is not a constant. Then f (Ω) is a region and so if z ∈ Ω, there exists r > 0 such that B (f ©(z) , r) ⊆ f (Ω)ª. It follows there exists z1 ∈ Ω with |f (z1 )| > |f (z)| . Hence max |f (w)| : w ∈ Ω is not achieved at any interior point of Ω. Therefore, the point at which the maximum is achieved must lie on the boundary of Ω and so © ª max {|f (w)| : w ∈ ∂Ω} = max |f (w)| : w ∈ Ω > |f (z)| for all z ∈ Ω or else f is a constant. This proves the theorem. You can remove the assumption that Ω is bounded and give a slightly different version. Theorem 28.3.2 Let f : Ω → C be analytic on a region, Ω and suppose B (a, r) ⊆ Ω. Then ©¯ ¡ ¢¯ ª |f (a)| ≤ max ¯f a + reiθ ¯ : θ ∈ [0, 2π] . Equality occurs for some r > 0 and a ∈ Ω if and only if f is constant in Ω hence equality occurs for all such a, r.

836


Proof: The claimed inequality holds by Theorem 28.3.1. Suppose equality in the above is achieved for some B (a, r) ⊆ Ω. Then by Theorem 28.3.1 f is equal to a constant, w on B (a, r) . Therefore, the function, f (·) − w has a zero set which has a limit point in Ω and so by Theorem 27.5.3 f (z) = w for all z ∈ Ω. Conversely, if f is constant, then the equality in the above inequality is achieved for all B (a, r) ⊆ Ω. Next is yet another version of the maximum modulus principle which is in Conway [16]. Let Ω be an open set. Definition 28.3.3 Define ∂∞ Ω to equal ∂Ω in the case where Ω is bounded and ∂Ω ∪ {∞} in the case where Ω is not bounded. Definition 28.3.4 Let f be a complex valued function defined on a set S ⊆ C and let a be a limit point of S. lim sup |f (z)| ≡ lim {sup |f (w)| : w ∈ B 0 (a, r) ∩ S} . z→a

r→0

The limit exists because {sup |f (w)| : w ∈ B 0 (a, r) ∩ S} is decreasing in r. In case a = ∞, lim sup |f (z)| ≡ lim {sup |f (w)| : |w| > r, w ∈ S} z→∞

r→∞

Note that if lim supz→a |f (z)| ≤ M and δ > 0, then there exists r > 0 such that if z ∈ B 0 (a, r) ∩ S, then |f (z)| < M + δ. If a = ∞, there exists r > 0 such that if |z| > r and z ∈ S, then |f (z)| < M + δ. Theorem 28.3.5 Let Ω be an open set in C and let f : Ω → C be analytic. Suppose also that for every a ∈ ∂∞ Ω, lim sup |f (z)| ≤ M < ∞. z→a

Then in fact |f (z)| ≤ M for all z ∈ Ω. Proof: Let δ > 0 and let H ≡ {z ∈ Ω : |f (z)| > M + δ} . Suppose H 6= ∅. Then H is an open subset of Ω. I claim that H is actually bounded. If Ω is bounded, there is nothing to show so assume Ω is unbounded. Then the condition involving the lim sup implies there exists r > 0 such that if |z| > r and z ∈ Ω, then |f (z)| ≤ M + δ/2. It follows H is contained in B (0, r) and so it is bounded. Now consider the components of Ω. One of these components contains points from H. Let this component be denoted as V and let HV ≡ H ∩ V. Thus HV is a bounded open subset of V. Let U be a component of HV . First suppose U ⊆ V . In this case, it follows that on ∂U, |f (z)| = M + δ and so by Theorem 28.3.1 |f (z)| ≤ M + δ for all z ∈ U contradicting the definition of H. Next suppose ∂U contains a point of ∂V, a. Then in this case, a violates the condition on lim sup . Either way you get a contradiction. Hence H = ∅ as claimed. Since δ > 0 is arbitrary, this shows |f (z)| ≤ M.

28.4. EXTENSIONS OF MAXIMUM MODULUS THEOREM

837

28.4

Extensions Of Maximum Modulus Theorem

28.4.1

Phragmˆ en Lindel¨ of Theorem

This theorem is an extension of Theorem 28.3.5. It uses a growth condition near the extended boundary to conclude that f is bounded. I will present the version found in Conway [16]. It seems to be more of a method than an actual theorem. There are several versions of it. Theorem 28.4.1 Let Ω be a simply connected region in C and suppose f is analytic on Ω. Also suppose there exists a function, φ which is nonzero and uniformly bounded on Ω. Let M be a positive number. Now suppose ∂∞ Ω = A ∪ B such that for every a ∈ A, lim supz→a |f (z)| ≤ M and for every b ∈ B, and η > 0, η lim supz→b |f (z)| |φ (z)| ≤ M. Then |f (z)| ≤ M for all z ∈ Ω. Proof: By Theorem 28.2.1 there exists log (φ (z)) analytic on Ω. Now define η g (z) ≡ exp (η log (φ (z))) so that g (z) = φ (z) . Now also η

|g (z)| = |exp (η log (φ (z)))| = |exp (η ln |φ (z)|)| = |φ (z)| . Let m ≥ |φ (z)| for all z ∈ Ω. Define F (z) ≡ f (z) g (z) m−η . Thus F is analytic and for b ∈ B, η

lim sup |F (z)| = lim sup |f (z)| |φ (z)| m−η ≤ M m−η z→b

z→b

while for a ∈ A, lim sup |F (z)| ≤ M. z→a

Therefore, for α ³∈ ∂∞ Ω,´ lim supz→α |F (z)| ≤ max (M, M η −η ) and so by Theorem mη 28.3.5, |f (z)| ≤ |φ(z)| max (M, M η −η ) . Now let η → 0 to obtain |f (z)| ≤ M. η In applications, it is often the case that B = {∞}. Now here is an interesting case of this theorem. It involves a particular form for © ª π Ω, in this case Ω = z ∈ C : |arg (z)| < 2a where a ≥ 12 .

Ω

Then ∂Ω equals the two slanted lines. Also on Ω you can define a logarithm, log (z) = ln |z| + i arg (z) where arg (z) is the angle associated with z between −π

838


and π. Therefore, if c is a real number you can define z c for such z in the usual way: zc

≡ =

exp (c log (z)) = exp (c [ln |z| + i arg (z)]) c c |z| exp (ic arg (z)) = |z| (cos (c arg (z)) + i sin (c arg (z))) .

If |c| < a, then |c arg (z)| < |exp (− (z c ))|

π 2

and so cos (c arg (z)) > 0. Therefore, for such c, c

= =

|exp (− |z| (cos (c arg (z)) + i sin (c arg (z))))| c |exp (− |z| (cos (c arg (z))))|

which is bounded since cos (c arg (z)) > 0. © ª π Corollary 28.4.2 Let Ω = z ∈ C : |arg (z)| < 2a where a ≥ 12 and suppose f is analytic on Ω and satisfies lim supz→a |f (z)| ≤ M on ∂Ω and suppose there are positive constants, P, b where b < a and ³ ´ b |f (z)| ≤ P exp |z| for all |z| large enough. Then |f (z)| ≤ M for all z ∈ Ω. Proof: Let b < c < a and let φ (z) ≡ exp (− (z c )) . Then as discussed above, φ (z) 6= 0 on Ω and |φ (z)| is bounded on Ω. Now η

c

|φ (z)| = |exp (− |z| η (cos (c arg (z))))| ³ ´ b P exp |z| η lim sup |f (z)| |φ (z)| ≤ lim sup =0≤M c z→∞ z→∞ |exp (|z| η (cos (c arg (z))))| and so by Theorem 28.4.1 |f (z)| ≤ M. The following is another interesting case. This case is presented in Rudin [58] Corollary 28.4.3 Let Ω be the open set consisting of {z ∈ C : a < Re z < b} and suppose f is analytic on Ω , continuous on Ω, and bounded on Ω. Suppose also that f (z) ≤ 1 on the two lines Re z = a and Re z = b. Then |f (z)| ≤ 1 for all z ∈ Ω. 1 Proof: This time let φ (z) = 1+z−a . Thus |φ (z)| ≤ 1 because Re (z − a) > 0 and η φ (z) 6= 0 for all z ∈ Ω. Also, lim supz→∞ |φ (z)| = 0 for every η > 0. Therefore, if a η is a point of the sides of Ω, lim supz→a |f (z)| ≤ 1 while lim supz→∞ |f (z)| |φ (z)| = 0 ≤ 1 and so by Theorem 28.4.1, |f (z)| ≤ 1 on Ω. This corollary yields an interesting conclusion.

Corollary 28.4.4 Let Ω be the open set consisting of {z ∈ C : a < Re z < b} and suppose f is analytic on Ω , continuous on Ω, and bounded on Ω. Define M (x) ≡ sup {|f (z)| : Re z = x} Then for x ∈ (a, b). b−x

x−a

M (x) ≤ M (a) b−a M (b) b−a .

28.4. EXTENSIONS OF MAXIMUM MODULUS THEOREM

839

Proof: Let ε > 0 and define b−z

z−a

g (z) ≡ (M (a) + ε) b−a (M (b) + ε) b−a

where for M > 0 and z ∈ C, M z ≡ exp (z ln (M )) . Thus g 6= 0 and so f /g is analytic on Ω and continuous on Ω. Also on the left side, ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ f (a + iy) ¯ ¯ f (a + iy) ¯ ¯¯ f (a + iy) ¯ ¯ ¯ ¯ ¯ b−a−iy ¯ = ¯ b−a ¯ ≤ 1 ¯ g (a + iy) ¯ = ¯¯ ¯ ¯ (M (a) + ε) b−a ¯ (M (a) + ε) b−a ¯ ¯ ¯ ¯ while on the right side a similar computation shows ¯ fg ¯ ≤ 1 also. Therefore, by Corollary 28.4.3 |f /g| ≤ 1 on Ω. Therefore, letting x + iy = z, ¯ ¯ ¯ ¯ z−a b−x x−a b−z ¯ ¯ ¯ ¯ |f (z)| ≤ ¯(M (a) + ε) b−a (M (b) + ε) b−a ¯ = ¯(M (a) + ε) b−a (M (b) + ε) b−a ¯ and so

b−x

x−a

M (x) ≤ (M (a) + ε) b−a (M (b) + ε) b−a . Since ε > 0 is arbitrary, it yields the conclusion of the corollary. Another way of saying this is that x → ln (M (x)) is a convex function. This corollary has an interesting application known as the Hadamard three circles theorem.

28.4.2

Hadamard Three Circles Theorem

Let 0 < R1 < R2 and suppose f is analytic on {z ∈ C : R1 < |z| < R2 } . Then letting R1 < a < b < R2 , note that g (z) ≡ exp (z) maps the strip {z ∈ C : ln a < Re z < b} onto {z ∈ C : a < |z| < b} and that in fact, g maps the line ln r + iy onto the circle reiθ . Now let M (x) be defined as above and m be defined by ¯ ¡ ¢¯ m (r) ≡ max ¯f reiθ ¯ . θ

Then for a < r < b, Corollary 28.4.4 implies ¯ ¡ ln b−ln r ln r−ln a ¢¯ m (r) = sup ¯f eln r+iy ¯ = M (ln r) ≤ M (ln a) ln b−ln a M (ln b) ln b−ln a y ln(b/r)/ ln(b/a)

= m (a) and so

ln(b/a)

m (r)

ln(r/a)/ ln(b/a)

m (b)

≤ m (a)

ln(b/r)

ln(r/a)

m (b)

.

Taking logarithms, this yields µ ¶ µ ¶ ³r´ b b ln ln (m (r)) ≤ ln ln (m (a)) + ln ln (m (b)) a r a which says the same as r → ln (m (r)) is a convex function of ln r. The next example, also in Rudin [58] is very dramatic. An unbelievably weak assumption is made on the growth of the function and still you get a uniform bound in the conclusion.

840


ª © Corollary 28.4.5 Let Ω = z ∈ C : |Im (z)| < π2 . Suppose f is analytic on Ω, continuous on Ω, and there exist constants, α < 1 and A < ∞ such that |f (z)| ≤ exp (A exp (α |x|)) for z = x + iy and

¯ ³ π ´¯¯ ¯ ¯f x ± i ¯ ≤ 1 2

for all x ∈ R. Then |f (z)| ≤ 1 on Ω. Proof: This time let φ (z) = [exp (A exp (βz)) exp (A exp (−βz))] β < 1. Then φ (z) 6= 0 on Ω and for η > 0 η

|φ (z)| =

−1

where α
0 because β < 1 and |y|
0. By uniform continuity there exists δ = b−a for p a positive integer such that if p ε |s − t| < δ, then |γ (s) − γ (t)| < 2 . Then ³ ε´ γ ([t, t + δ]) ⊆ B γ (t) , ⊆ Ω. 2 n ¡ ¢o Let C ≥ max |f 0 (z)| : z ∈ ∪pj=1 B γ (tj ) , 2ε where tj ≡ what was just shown, V (f ◦ γ, [a, b]) ≤

p−1 X

j p

(b − a) + a. Then from

V (f ◦ γ, [tj , tj+1 ])

j=0

≤

C

p−1 X

V (γ, [tj , tj+1 ]) < ∞

j=0

showing that f ◦ γ is bounded variation as claimed. Now from Theorem 27.7.15 there exists η ∈ C 1 ([a, b]) such that η (a) = γ (a) = γ (b) = η (b) , η ([a, b]) ⊆ Ω, and n (η, ak ) = n (γ, ak ) , n (f ◦ γ, α) = n (f ◦ η, α) for k = 1, · · · , m. Then n (f ◦ γ, α) = n (f ◦ η, α)

(28.6.7)

846


= = = =

1 2πi

Z f ◦η

dw w−α

Z b f 0 (η (t)) 0 1 η (t) dt 2πi a f (η (t)) − α Z 1 f 0 (z) dz 2πi η f (z) − α m X n (η, ak ) k=1

Pm By Theorem 28.6.1. By 28.6.7, this equals k=1 n (γ, ak ) which proves the theorem. The next theorem is incredible and is very interesting for its own sake. The following picture is descriptive of the situation of this theorem. t a3 ta

t a1

f

q

sz

t a2 t a4

sα

B(α, δ) B(a, ²) Theorem 28.6.3 Let f : B (a, R) → C be analytic and let m

f (z) − α = (z − a) g (z) , ∞ > m ≥ 1 where g (z) 6= 0 in B (a, R) . (f (z) − α has a zero of order m at z = a.) Then there exist ε, δ > 0 with the property that for each z satisfying 0 < |z − α| < δ, there exist points, {a1 , · · · , am } ⊆ B (a, ε) , such that f −1 (z) ∩ B (a, ε) = {a1 , · · · , am } and each ak is a zero of order 1 for the function f (·) − z. Proof: By Theorem 27.5.3 f is not constant on B (a, R) because it has a zero of order m. Therefore, using this theorem again, there exists ε > 0 such that B (a, 2ε) ⊆ B (a, R) and there are no solutions to the equation f (z) − α = 0 for z ∈ B (a, 2ε) except a. Also assume ε is small enough that for 0 < |z − a| ≤ 2ε, f 0 (z) 6= 0. This can be done since otherwise, a would be a limit point of a sequence of points, zn , having f 0 (zn ) = 0 which would imply, by Theorem 27.5.3 that f 0 = 0 on B (a, R) , contradicting the assumption that f − α has a zero of order m and is therefore not constant. Thus the situation is described by the following picture.

28.6. COUNTING ZEROS

847

f 0 6= 0 s 2ε ¡ ¡ ¡f − α 6= 0 ¡ ª Now pick γ (t) = a + εeit , t ∈ [0, 2π] . Then α ∈ / f (γ ∗ ) so there exists δ > 0 with B (α, δ) ∩ f (γ ∗ ) = ∅.

(28.6.8)

Therefore, B (α, δ) is contained on one component of C \ f (γ ([0, 2π])) . Therefore, n (f ◦ γ, α) = n (f ◦ γ, z) for all z ∈ B (α, δ) . Now consider f restricted to B (a, 2ε) . For z ∈ B (α, δ) , f −1 (z) must consist of a finite set of points because f 0 (w) 6= 0 for all w in B (a, 2ε) \ {a} implying that the zeros of f (·) − z in B (a, 2ε) have no limit point. Since B (a, 2ε) is compact, this means there are only finitely many. By Theorem 28.6.2, p X n (γ, ak ) (28.6.9) n (f ◦ γ, z) = k=1 −1

where {a1 , · · · , ap } = f (z) . Each point, ak of f −1 (z) is either inside the circle traced out by γ, yielding n (γ, ak ) = 1, or it is outside this circle yielding n (γ, ak ) = 0 because of 28.6.8. It follows the sum in 28.6.9 reduces to the number of points of f −1 (z) which are contained in B (a, ε) . Thus, letting those points in f −1 (z) which are contained in B (a, ε) be denoted by {a1 , · · · , ar } n (f ◦ γ, α) = n (f ◦ γ, z) = r. Also, by Theorem 28.6.1, m = n (f ◦ γ, α) because a is a zero of f − α of order m. Therefore, for z ∈ B (α, δ) m = n (f ◦ γ, α) = n (f ◦ γ, z) = r It is required to show r = m, the order of the zero of f − α. Therefore, r = m. Each of these ak is a zero of order 1 of the function f (·) − z because f 0 (ak ) 6= 0. This proves the theorem. This is a very fascinating result partly because it implies that for values of f near a value, α, at which f (·) − α has a zero of order m for m > 1, the inverse image of these values includes at least m points, not just one. Thus the topological properties of the inverse image changes radically. This theorem also shows that f (B (a, ε)) ⊇ B (α, δ) . Theorem 28.6.4 (open mapping theorem) Let Ω be a region and f : Ω → C be analytic. Then f (Ω) is either a point or a region. If f is one to one, then f −1 : f (Ω) → Ω is analytic.

848


Proof: If f is not constant, then for every α ∈ f (Ω) , it follows from Theorem 27.5.3 that f (·) − α has a zero of order m < ∞ and so from Theorem 28.6.3 for each a ∈ Ω there exist ε, δ > 0 such that f (B (a, ε)) ⊇ B (α, δ) which clearly implies that f maps open sets to open sets. Therefore, f (Ω) is open, connected because f is continuous. If f is one to one, Theorem 28.6.3 implies that for every α ∈ f (Ω) the zero of f (·) − α is of order 1. Otherwise, that theorem implies that for z near α, there are m points which f maps to z contradicting the assumption that f is one to one. Therefore, f 0 (z) 6= 0 and since f −1 is continuous, due to f being an open map, it follows ¡

¢0 f −1 (f (z))

= =

f −1 (f (z1 )) − f −1 (f (z)) f (z1 ) − f (z) f (z1 )→f (z) 1 z1 − z lim = 0 . z1 →z f (z1 ) − f (z) f (z) lim


28.7

An Application To Linear Algebra

Gerschgorin’s theorem gives a convenient way to estimate eigenvalues of a matrix from easy to obtain information. For A an n × n matrix, denote by σ (A) the collection of all eigenvalues of A. Theorem 28.7.1 Let A be an n × n matrix. Consider the n Gerschgorin discs defined as     X |aij | . Di ≡ λ ∈ C : |λ − aii | ≤   j6=i

Then every eigenvalue is contained in some Gerschgorin disc. This theorem says to add up the absolute values of the entries of the ith row which are off the main diagonal and form the disc centered at aii having this radius. The union of these discs contains σ (A) . Proof: Suppose Ax = λx where x 6= 0. Then for A = (aij ) X

aij xj = (λ − aii ) xi .

j6=i

Therefore, if we pick k such that |xk | ≥ |xj | for all xj , it follows that |xk | 6= 0 since |x| 6= 0 and X X |akj | ≥ |akj | |xj | ≥ |λ − akk | |xk | . |xk | j6=k

j6=k

Now dividing by |xk | we see that λ is contained in the k th Gerschgorin disc.

28.7. AN APPLICATION TO LINEAR ALGEBRA

849

More can be said using the theory about counting zeros. To begin with the distance between two n × n matrices, A = (aij ) and B = (bij ) as follows. 2

||A − B|| ≡

X

2

|aij − bij | .

ij

Thus two matrices are close if and only if their corresponding entries are close. Let A be an n × n matrix. Recall the eigenvalues of A are given by the zeros of the polynomial, pA (z) = det (zI − A) where I is the n × n identity. Then small changes in A will produce small changes in pA (z) and p0A (z) . Let γ k denote a very small closed circle which winds around zk , one of the eigenvalues of A, in the counter clockwise direction so that n (γ k , zk ) = 1. This circle is to enclose only zk and is to have no other eigenvalue on it. Then apply Theorem 28.6.1. According to this theorem Z 0 pA (z) 1 dz 2πi γ pA (z) is always an integer equal to the multiplicity of zk as a root of pA (t) . Therefore, small changes in A result in no change to the above contour integral because it must be an integer and small changes in A result in small changes in the integral. Therefore whenever every entry of the matrix B is close enough to the corresponding entry of the matrix A, the two matrices have the same number of zeros inside γ k under the usual convention that zeros are to be counted according to multiplicity. By making the radius of the small circle equal to ε where ε is less than the minimum distance between any two distinct eigenvalues of A, this shows that if B is close enough to A, every eigenvalue of B is closer than ε to some eigenvalue of A. The next theorem is about continuous dependence of eigenvalues. Theorem 28.7.2 If λ is an eigenvalue of A, then if ||B − A|| is small enough, some eigenvalue of B will be within ε of λ. Consider the situation that A (t) is an n × n matrix and that t → A (t) is continuous for t ∈ [0, 1] . Lemma 28.7.3 Let λ (t) ∈ σ (A (t)) for t < 1 and let Σt = ∪s≥t σ (A (s)) . Also let Kt be the connected component of λ (t) in Σt . Then there exists η > 0 such that Kt ∩ σ (A (s)) 6= ∅ for all s ∈ [t, t + η] . Proof: Denote by D (λ (t) , δ) the disc centered at λ (t) having radius δ > 0, with other occurrences of this notation being defined similarly. Thus D (λ (t) , δ) ≡ {z ∈ C : |λ (t) − z| ≤ δ} . Suppose δ > 0 is small enough that λ (t) is the only element of σ (A (t)) contained in D (λ (t) , δ) and that pA(t) has no zeroes on the boundary of this disc. Then by continuity, and the above discussion and theorem, there exists η > 0, t + η < 1, such that for s ∈ [t, t + η] , pA(s) also has no zeroes on the boundary of this disc and that

850


A (s) has the same number of eigenvalues, counted according to multiplicity, in the disc as A (t) . Thus σ (A (s)) ∩ D (λ (t) , δ) 6= ∅ for all s ∈ [t, t + η] . Now let H=

[

σ (A (s)) ∩ D (λ (t) , δ) .

s∈[t,t+η]

I will show H is connected. Suppose not. Then H = P ∪Q where P, Q are separated and λ (t) ∈ P. Let s0 ≡ inf {s : λ (s) ∈ Q for some λ (s) ∈ σ (A (s))} . There exists λ (s0 ) ∈ σ (A (s0 )) ∩ D (λ (t) , δ) . If λ (s0 ) ∈ / Q, then from the above discussion there are λ (s) ∈ σ (A (s)) ∩ Q for s > s0 arbitrarily close to λ (s0 ) . Therefore, λ (s0 ) ∈ Q which shows that s0 > t because λ (t) is the only element of σ (A (t)) in D (λ (t) , δ) and λ (t) ∈ P. Now let sn ↑ s0 . Then λ (sn ) ∈ P for any λ (sn ) ∈ σ (A (sn )) ∩ D (λ (t) , δ) and from the above discussion, for some choice of sn → s0 , λ (sn ) → λ (s0 ) which contradicts P and Q separated and nonempty. Since P is nonempty, this shows Q = ∅. Therefore, H is connected as claimed. But Kt ⊇ H and so Kt ∩σ (A (s)) 6= ∅ for all s ∈ [t, t + η] . This proves the lemma. The following is the necessary theorem. Theorem 28.7.4 Suppose A (t) is an n×n matrix and that t → A (t) is continuous for t ∈ [0, 1] . Let λ (0) ∈ σ (A (0)) and define Σ ≡ ∪t∈[0,1] σ (A (t)) . Let Kλ(0) = K0 denote the connected component of λ (0) in Σ. Then K0 ∩ σ (A (t)) 6= ∅ for all t ∈ [0, 1] . Proof: Let S ≡ {t ∈ [0, 1] : K0 ∩ σ (A (s)) 6= ∅ for all s ∈ [0, t]} . Then 0 ∈ S. Let t0 = sup (S) . Say σ (A (t0 )) = λ1 (t0 ) , · · · , λr (t0 ) . I claim at least one of these is a limit point of K0 and consequently must be in K0 which will show that S has a last point. Why is this claim true? Let sn ↑ t0 so sn ∈ S. Now let the discs, D (λi (t0 ) , δ) , i = 1, · · · , r be disjoint with pA(t0 ) having no zeroes on γ i the boundary of D (λi (t0 ) , δ) . Then for n large enough it follows from Theorem 28.6.1 and the discussion following it that σ (A (sn )) is contained in ∪ri=1 D (λi (t0 ) , δ). Therefore, K0 ∩ (σ (A (t0 )) + D (0, δ)) 6= ∅ for all δ small enough. This requires at least one of the λi (t0 ) to be in K0 . Therefore, t0 ∈ S and S has a last point. Now by Lemma 28.7.3, if t0 < 1, then K0 ∪ Kt would be a strictly larger connected set containing λ (0) . (The reason this would be strictly larger is that K0 ∩σ (A (s)) = ∅ for some s ∈ (t, t + η) while Kt ∩σ (A (s)) 6= ∅ for all s ∈ [t, t + η].) Therefore, t0 = 1 and this proves the theorem. The following is an interesting corollary of the Gerschgorin theorem.

28.7. AN APPLICATION TO LINEAR ALGEBRA

851

Corollary 28.7.5 Suppose one of the Gerschgorin discs, Di is disjoint from the union of the others. Then Di contains an eigenvalue of A. Also, if there are n disjoint Gerschgorin discs, then each one contains an eigenvalue of A. ¡ ¢ Proof: Denote by A (t) the matrix atij where if i 6= j, atij = taij and atii = aii . Thus to get A (t) we multiply all non diagonal terms by t. Let t ∈ [0, 1] . Then A (0) = diag (a11 , · · · , ann ) and A (1) = A. Furthermore, the map, t → A (t) is continuous. Denote by Djt the Gerschgorin disc obtained from the j th row for the matrix, A (t). Then it is clear that Djt ⊆ Dj the j th Gerschgorin disc for A. Then aii is the eigenvalue for A (0) which is contained in the disc, consisting of the single point aii which is contained in Di . Letting K be the connected component in Σ for Σ defined in Theorem 28.7.4 which is determined by aii , it follows by Gerschgorin’s theorem that K ∩ σ (A (t)) ⊆ ∪nj=1 Djt ⊆ ∪nj=1 Dj = Di ∪ (∪j6=i Dj ) and also, since K is connected, there are no points of K in both Di and (∪j6=i Dj ) . Since at least one point of K is in Di ,(aii ) it follows all of K must be contained in Di . Now by Theorem 28.7.4 this shows there are points of K ∩ σ (A) in Di . The last assertion follows immediately. Actually, this can be improved slightly. It involves the following lemma. Lemma 28.7.6 In the situation of Theorem 28.7.4 suppose λ (0) = K0 ∩ σ (A (0)) and that λ (0) is a simple root of the characteristic equation of A (0). Then for all t ∈ [0, 1] , σ (A (t)) ∩ K0 = λ (t) where λ (t) is a simple root of the characteristic equation of A (t) . Proof: Let S ≡ {t ∈ [0, 1] : K0 ∩ σ (A (s)) = λ (s) , a simple eigenvalue for all s ∈ [0, t]} . Then 0 ∈ S so it is nonempty. Let t0 = sup (S) and suppose λ1 6= λ2 are two elements of σ (A (t0 )) ∩ K0 . Then choosing η > 0 small enough, and letting Di be disjoint discs containing λi respectively, similar arguments to those of Lemma 28.7.3 imply Hi ≡ ∪s∈[t0 −η,t0 ] σ (A (s)) ∩ Di is a connected and nonempty set for i = 1, 2 which would require that Hi ⊆ K0 . But then there would be two different eigenvalues of A (s) contained in K0 , contrary to the definition of t0 . Therefore, there is at most one eigenvalue, λ (t0 ) ∈ K0 ∩ σ (A (t0 )) . The possibility that it could be a repeated root of the characteristic equation must be ruled out. Suppose then that λ (t0 ) is a repeated root of the characteristic equation. As before, choose a small disc, D centered at λ (t0 ) and η small enough that H ≡ ∪s∈[t0 −η,t0 ] σ (A (s)) ∩ D is a nonempty connected set containing either multiple eigenvalues of A (s) or else a single repeated root to the characteristic equation of A (s) . But since H is connected and contains λ (t0 ) it must be contained in K0 which contradicts the condition for

852


s ∈ S for all these s ∈ [t0 − η, t0 ] . Therefore, t0 ∈ S as hoped. If t0 < 1, there exists a small disc centered at λ (t0 ) and η > 0 such that for all s ∈ [t0 , t0 + η] , A (s) has only simple eigenvalues in D and the only eigenvalues of A (s) which could be in K0 are in D. (This last assertion follows from noting that λ (t0 ) is the only eigenvalue of A (t0 ) in K0 and so the others are at a positive distance from K0 . For s close enough to t0 , the eigenvalues of A (s) are either close to these eigenvalues of A (t0 ) at a positive distance from K0 or they are close to the eigenvalue, λ (t0 ) in which case it can be assumed they are in D.) But this shows that t0 is not really an upper bound to S. Therefore, t0 = 1 and the lemma is proved. With this lemma, the conclusion of the above corollary can be improved. Corollary 28.7.7 Suppose one of the Gerschgorin discs, Di is disjoint from the union of the others. Then Di contains exactly one eigenvalue of A and this eigenvalue is a simple root to the characteristic polynomial of A. Proof: In the proof of Corollary 28.7.5, first note that aii is a simple root of A (0) since otherwise the ith Gerschgorin disc would not be disjoint from the others. Also, K, the connected component determined by aii must be contained in Di because it is connected and by Gerschgorin’s theorem above, K ∩ σ (A (t)) must be contained in the union of the Gerschgorin discs. Since all the other eigenvalues of A (0) , the ajj , are outside Di , it follows that K ∩ σ (A (0)) = aii . Therefore, by Lemma 28.7.6, K ∩ σ (A (1)) = K ∩ σ (A) consists of a single simple eigenvalue. This proves the corollary. Example 28.7.8 Consider the matrix,  5 1  1 1 0 1

 0 1  0

The Gerschgorin discs are D (5, 1) , D (1, 2) , and D (0, 1) . Then D (5, 1) is disjoint from the other discs. Therefore, there should be an eigenvalue in D (5, 1) . The actual eigenvalues are not easy to find. They are the roots of the characteristic equation, t3 − 6t2 + 3t + 5 = 0. The numerical values of these are −. 669 66, 1. 423 1, and 5. 246 55, verifying the predictions of Gerschgorin’s theorem.

28.8

Exercises

1. Use Theorem 28.6.1 to give an alternate proof of the fundamental theorem of algebra. Hint: Take a contour of the form γ r = reit where t ∈ [0, 2π] . R 0 (z) dz and consider the limit as r → ∞. Consider γ pp(z) r

2. Let M be an n × n matrix. Recall that the eigenvalues of M are given by the zeros of the polynomial, pM (z) = det (M − zI) where I is the n × n identity. Formulate a theorem which describes how the eigenvalues depend on small

28.8. EXERCISES

853

changes in M. Hint: You could define a norm on the space of n × n matrices 1/2 as ||M || ≡ tr (M M ∗ ) where M ∗ is the conjugate transpose of M. Thus  1/2 X 2 ||M || =  |Mjk |  . j,k

Argue that small changes will produce small changes in pM (z) . Then apply Theorem 28.6.1 using γ k a very small circle surrounding zk , the k th eigenvalue. 3. Suppose that two analytic functions defined on a region are equal on some set, S which contains a limit point. (Recall p is a limit point of S if every open set which contains p, also contains infinitely many points of S. ) Show the two functions coincide. We defined ez ≡ ex (cos y + i sin y) earlier and we showed that ez , defined this way was analytic on C. Is there any other way to define ez on all of C such that the function coincides with ex on the real axis? 4. You know various identities for real valued functions. For example cosh2 x − z −z z −z and sinh z ≡ e −e , does it follow sinh2 x = 1. If you define cosh z ≡ e +e 2 2 that cosh2 z − sinh2 z = 1 for all z ∈ C? What about sin (z + w) = sin z cos w + cos z sin w? Can you verify these sorts of identities just from your knowledge about what happens for real arguments? 5. Was it necessary that U be a region in Theorem 27.5.3? Would the same conclusion hold if U were only assumed to be an open set? Why? What about the open mapping theorem? Would it hold if U were not a region? 6. Let f : U → C be analytic and one to one. Show that f 0 (z) 6= 0 for all z ∈ U. Does this hold for a function of a real variable? 7. We say a real valued function, u is subharmonic if uxx +uyy ≥ 0. Show that if u is subharmonic on a bounded region, (open connected set) U, and continuous on U and u ≤ m on ∂U, then u ≤ m on U. Hint: If not, u achieves its maximum at (x0 , y0 ) ∈ U. Let u (x0 , y0 ) > m + δ where δ > 0. Now consider uε (x, y) = εx2 + u (x, y) where ε is small enough that 0 < εx2 < δ for all (x, y) ∈ U. Show that uε also achieves its maximum at some point of U and that therefore, uεxx + uεyy ≤ 0 at that point implying that uxx + uyy ≤ −ε, a contradiction. 8. If u is harmonic on some region, U, show that u coincides locally with the real part of an analytic function and that therefore, u has infinitely many

854


derivatives on U. Hint: Consider the case where 0 ∈ U. You can always reduce to this case by a suitable translation. Now let B (0, r) ⊆ U and use the Schwarz formula to obtain an analytic function whose real part coincides with u on ∂B (0, r) . Then use Problem 7. 9. Show the solution to the Dirichlet problem of Problem 8 on Page 806 is unique. You need to formulate this precisely and then prove uniqueness.

Residues Definition 29.0.1 The residue of f at an isolated singularity α which is a pole, −1 written res (f, α) is the coefficient of (z − α) where f (z) = g (z) +

m X

bk k

k=1

(z − α)

.

Thus res (f, α) = b1 in the above. At this point, recall Corollary 27.7.20 which is stated here for convenience. Corollary 29.0.2 Let Ω be an open set and let γ k : [ak , bk ] → Ω, k = 1, · · · , m, be closed, continuous and of bounded variation. Suppose also that m X

n (γ k , z) = 0

k=1

for all z ∈ / Ω. Then if f : Ω → C is analytic, m Z X f (w) dw = 0. k=1

γk

The following theorem is called the residue theorem. Note the resemblance to Corollary 27.7.20. Theorem 29.0.3 Let Ω be an open set and let γ k : [ak , bk ] → Ω, k = 1, · · · , m, be closed, continuous and of bounded variation. Suppose also that m X

n (γ k , z) = 0

k=1

b is meromorphic such that no γ ∗ contains any poles for all z ∈ / Ω. Then if f : Ω → C k of f , m Z m X X 1 X n (γ k , α) (29.0.1) f (w) dw = res (f, α) 2πi γk k=1

α∈A

855

k=1

856

RESIDUES

where here A denotes the set of poles of f in Ω. The sum on the right is a finite sum. Proof: First note that there are at most finitely many α which are not in the unbounded component of C \ ∪m k=1 γ k ([ak , bk ]) . Thus there exists Pna finite set, {α1 , · · · , αN } ⊆ A such that these are the only possibilities for which k=1 n (γ k , α) might not equal zero. Therefore, 29.0.1 reduces to m Z N n X X 1 X f (w) dw = res (f, αj ) n (γ k , αj ) 2πi γk j=1 k=1

k=1

and it is this last equation which is established. Near αj , f (z) = gj (z) +

mj X r=1

bjr r ≡ gj (z) + Qj (z) . (z − αj )

where gj is analytic at and near αj . Now define G (z) ≡ f (z) −

N X

Qj (z) .

j=1

It follows that G (z) has a removable singularity at each αj . Therefore, by Corollary 27.7.20, m Z m Z N X m Z X X X 0= G (z) dz = f (z) dz − Qj (z) dz. γk

k=1

k=1

Now m Z X k=1

Qj (z) dz

=

γk

=

m Z X k=1 γ k m Z X k=1

γk

Ã

γk

j=1 k=1

mj

γk

X bj1 bjr + (z − αj ) r=2 (z − αj )r

! dz

m

X bj1 dz ≡ n (γ k , αj ) res (f, αj ) (2πi) . (z − αj ) k=1

Therefore, m Z X k=1

f (z) dz

=

γk

N X m Z X j=1 k=1

=

N X m X

Qj (z) dz γk

n (γ k , αj ) res (f, αj ) (2πi)

j=1 k=1

=

2πi

N X

res (f, αj )

j=1

=

(2πi)

X α∈A

m X

n (γ k , αj )

k=1 m X

res (f, α)

k=1

n (γ k , α)

857 which proves the theorem. The following is an important example. This example can also be done by real variable methods and there are some who think that real variable methods are always to be preferred to complex variable methods. However, I will use the above theorem to work this example. Example 29.0.4 Find limR→∞

RR −R

sin(x) x dx

Things are easier if you write it as 1 lim R→∞ i

ÃZ

−R−1

−R

eix dx + x

Z

R

R−1

! eix dx . x

This gives the same answer because cos (x) /x is odd. Consider the following contour in which the orientation involves counterclockwise motion exactly once around.

−R

−R−1

R−1

R

Denote by γ R−1 the little circle and γ R the big one. Then on the inside of this contour there are no singularities of eiz /z and it is contained in an open set with the property that the winding number with respect to this contour about any point not in the open set equals zero. By Theorem 27.7.22 1 i Now

ÃZ

−R−1

−R

eix dx + x

Z γ R−1

eiz dz + z

Z

R R−1

eix dx + x

Z γR

eiz dz z

! =0

(29.0.2)

¯Z ¯ ¯Z ¯ Z π ¯ eiz ¯¯ ¯¯ π R(i cos θ−sin θ) ¯¯ ¯ e−R sin θ dθ e idθ dz = ¯ ¯ ¯ ¯≤ ¯ γR z ¯ 0 0

and this last integral converges to 0 by the dominated convergence theorem. Now consider the other circle. By the dominated convergence theorem again, Z γ R−1

eiz dz = z

Z

0 π

eR

−1

(i cos θ−sin θ)

idθ → −iπ

858

RESIDUES

as R → ∞. Then passing to the limit in 29.0.2, Z R sin (x) dx lim R→∞ −R x ÃZ Z R ix ! −R−1 ix 1 e e = lim dx + dx R→∞ i x −R R−1 x Ã Z ! Z 1 eiz eiz −1 = lim − dz − dz = (−iπ) = π. R→∞ i z z i γ R−1 γR RR Example 29.0.5 Find limR→∞ −R eixt sinx x dx. Note this is essentially finding the inverse Fourier transform of the function, sin (x) /x. This equals Z

R

lim

R→∞

= = =

(cos (xt) + i sin (xt)) Z

−R R

lim

R→∞

Z

sin (x) dx x

cos (xt)

sin (x) dx x

−R R

lim

R→∞

cos (xt)

−R

1 lim R→∞ 2

Z

R −R

sin (x) dx x

sin (x (t + 1)) + sin (x (1 − t)) dx x

Let t 6= 1, −1. Then changing variables yields Ã Z ! Z 1 R(1+t) sin (u) 1 R(1−t) sin (u) lim du + du . R→∞ 2 −R(1+t) u 2 −R(1−t) u In case |t| < 1 Example 29.0.4 implies this limit is π. However, if t > 1 the limit equals 0 and this is also the case if t < −1. Summarizing, ½ Z R sin x π if |t| < 1 lim eixt dx = . 0 if |t| > 1 R→∞ −R x

29.1

Rouche’s Theorem And The Argument Principle

29.1.1

Argument Principle

A simple closed curve is just one which is homeomorphic to the unit circle. The Jordan Curve theorem states that every simple closed curve in the plane divides the plane into exactly two connected components, one bounded and the other unbounded. This is a very hard theorem to prove. However, in most applications the

29.1. ROUCHE’S THEOREM AND THE ARGUMENT PRINCIPLE

859

conclusion is obvious. Nevertheless, to avoid using this big topological result and to attain some extra generality, I will state the following theorem in terms of the winding number to avoid using it. This theorem is called the argument principle. m First recall that f has a zero of order m at α if f (z) = g (z) (z − α) where g is an analytic is not equal to zero at α. This is equivalent to having P∞ function which k f (z) = k=m ak (z − α) for z near α where am 6= 0. Also recall that f has a pole of order m at α if for z near α, f (z) is of the form f (z) = h (z) +

m X

bk

(29.1.3)

k

k=1

(z − α)

where bm 6= 0 and h is a function analytic near α. Theorem 29.1.1 (argument principle) Let f be meromorphic in Ω. Also suppose γ ∗ is a closed bounded variation curve containing none of the poles or zeros of f with the property that for all z ∈ / Ω, n (γ, z) = 0 and for all z ∈ Ω, n (γ, z) either equals 0 or 1. Now let {p1 , · · · , pm } and {z1 , · · · , zn } be respectively the poles and zeros for which the winding number of γ about these points equals 1. Let zk be a zero of order rk and let pk be a pole of order lk . Then Z 0 n m X X 1 f (z) dz = rk − lk 2πi γ f (z) k=1

k=1

Proof: This theorem follows from computing the residues of f 0 /f. It has residues at poles and zeros. I will do this now. First suppose f has a pole of order p at α. Then f has the form given in 29.1.3. Therefore, Pp kbk h0 (z) − k=1 (z−α) k+1 f 0 (z) = Pp b k f (z) h (z) + k=1 (z−α)k Pp−1 pbp −k−1+p p − (z−α) h0 (z) (z − α) − k=1 kbk (z − α) = Pp−1 p−k p + bp h (z) (z − α) + k=1 bk (z − α) This is of the form pbp

r (z) − (z−α) bp bp = = s (z) + bp bp s (z) + bp

µ

r (z) p − bp (z − α)

¶

where s (α) = r (α) = 0. From this, it is clear res (f 0 /f, α) = −p, the order of the pole. Next suppose f has a zero of order p at α. Then P∞ P∞ k−1 k−1−p f 0 (z) k=p ak k (z − α) k=p ak k (z − α) = P∞ = P∞ k k−p f (z) k=p ak (z − α) k=p ak (z − α) and from this it is clear res (f 0 /f ) = p, the order of the zero. The conclusion of this theorem now follows from Theorem 29.0.3.

860

RESIDUES

One can also generalize the theorem to the case where there are many closed curves involved. This is proved in the same way as the above. Theorem 29.1.2 (argument principle) Let f be meromorphic in Ω and let γ k : [ak , bk ] → Ω, k = 1, · · · , m, be closed, continuous and of bounded variation. Suppose also that m X n (γ k , z) = 0 k=1

Pm and for all z ∈ / Ω and for z ∈ Ω, k=1 n (γ k , z) either equals 0 or 1. Now let {p1 , · · · , pm } and {z1 , · · · , zn } be respectively the poles and zeros for which the above sum of winding numbers equals 1. Let zk be a zero of order rk and let pk be a pole of order lk . Then Z 0 n m X X 1 f (z) dz = rk − lk 2πi γ f (z) k=1

k=1

There is also a simple extension of this important principle which I found in [32]. Theorem 29.1.3 (argument principle) Let f be meromorphic in Ω. Also suppose γ ∗ is a closed bounded variation curve with the property that for all z ∈ / Ω, n (γ, z) = 0 and for all z ∈ Ω, n (γ, z) either equals 0 or 1. Now let {p1 , · · · , pm } and {z1 , · · · , zn } be respectively the poles and zeros for which the winding number of γ about these points equals 1 listed according to multiplicity. Thus if there is a pole of order m there will be this value repeated m times in the list for the poles. Also let g (z) be an analytic function. Then 1 2πi

Z g (z) γ

n

m

k=1

k=1

X X f 0 (z) dz = g (zk ) − g (pk ) f (z)

Proof: This theorem follows from computing the residues of g (f 0 /f ) . It has residues at poles and zeros. I will do this now. First suppose f has a pole of order m at α. Then f has the form given in 29.1.3. Therefore, f 0 (z) f (z) ³ ´ Pm kbk g (z) h0 (z) − k=1 (z−α) k+1 = Pm bk h (z) + k=1 (z−α) k Pm−1 m −k−1+m mbm 0 h (z) (z − α) − k=1 kbk (z − α) − (z−α) = g (z) Pm−1 m m−k h (z) (z − α) + k=1 bk (z − α) + bm g (z)

From this, it is clear res (g (f 0 /f ) , α) = −mg (α) , where m is the order of the pole. Thus α would have been listed m times in the list of poles. Hence the residue at this point is equivalent to adding −g (α) m times.

29.1. ROUCHE’S THEOREM AND THE ARGUMENT PRINCIPLE

861

Next suppose f has a zero of order m at α. Then P∞ P∞ k−1 k−1−m f 0 (z) k=m ak k (z − α) k=m ak k (z − α) g (z) = g (z) P = g (z) P∞ ∞ k k−m f (z) k=m ak (z − α) k=m ak (z − α) and from this it is clear res (g (f 0 /f )) = g (α) m, where m is the order of the zero. The conclusion of this theorem now follows from the residue theorem, Theorem 29.0.3. The way people usually apply these theorems is to suppose γ ∗ is a simple closed bounded variation curve, often a circle. Thus it has an inside and an outside, the outside being the unbounded component of C\γ ∗ . The orientation of the curve is such that you go around it once in the counterclockwise direction. Then letting rk and lk be as described, the conclusion of the theorem follows. In applications, this is likely the way it will be.

29.1.2

Rouche’s Theorem

With the argument principle, it is possible to prove Pm Rouche’s theorem . In the argument principle, denote by Z the quantity f k=1 rk and by Pf the quantity Pn l . Thus Z is the number of zeros of f counted according to the order of the k f k=1 zero with a similar definition holding for Pf . Thus the conclusion of the argument principle is. Z 0 1 f (z) dz = Zf − Pf 2πi γ f (z) Rouche’s theorem allows the comparison of Zh − Ph for h = f, g. It is a wonderful and amazing result. Theorem 29.1.4 (Rouche’s theorem)Let f, g be meromorphic in an open set Ω. Also suppose γ ∗ is a closed bounded variation curve with the property that for all z ∈ / Ω, n (γ, z) = 0, no zeros or poles are on γ ∗ , and for all z ∈ Ω, n (γ, z) either equals 0 or 1. Let Zf and Pf denote respectively the numbers of zeros and poles of f, which have the property that the winding number equals 1, counted according to order, with Zg and Pg being defined similarly. Also suppose that for z ∈ γ ∗ |f (z) + g (z)| < |f (z)| + |g (z)| . Then Zf − Pf = Zg − Pg . Proof: From the hypotheses, ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯1 + f (z) ¯ < 1 + ¯ f (z) ¯ ¯ ¯ ¯ g (z) g (z) ¯ which shows that for all z ∈ γ ∗ , f (z) ∈ C \ [0, ∞). g (z)

(29.1.4)

862

RESIDUES

Letting ³ ´ l denote a branch of the logarithm defined on C \ [0, ∞), it follows that (z) l fg(z) is a primitive for the function, 0

(f /g) f0 g0 = − . (f /g) f g Therefore, by the argument principle, ¶ Z Z µ 0 0 (f /g) 1 f g0 1 dz = − dz 0 = 2πi γ (f /g) 2πi γ f g =

Zf − Pf − (Zg − Pg ) .

This proves the theorem. Often another condition other than 29.1.4 is used. Corollary 29.1.5 In the situation of Theorem 29.1.4 change 29.1.4 to the condition, |f (z) − g (z)| < |f (z)| for z ∈ γ ∗ . Then the conclusion is the same. ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ g(z) ∗ / Proof: The new condition implies ¯1 − fg (z)¯ < ¯ fg(z) (z) ¯ on γ . Therefore, f (z) ∈ (−∞, 0] and so you can do the same argument with a branch of the logarithm.

29.1.3

A Different Formulation

In [60] I found this modification for Rouche’s theorem concerned with the counting of zeros of analytic functions. This is a very useful form of Rouche’s theorem because it makes no mention of a contour. Theorem 29.1.6 Let Ω be a bounded open set and suppose f, g are continuous on Ω and analytic on Ω. Also suppose |f (z)| < |g (z)| on ∂Ω. Then g and f +g have the same number of zeros in Ω provided each zero is counted according to multiplicity. © ª Proof: Let K = z ∈ Ω : |f (z)| ≥ |g (z)| . Then letting λ ∈ [0, 1] , if z ∈ / K, then |f (z)| < |g (z)| and so 0 < |g (z)| − |f (z)| ≤ |g (z)| − λ |f (z)| ≤ |g (z) + λf (z)| which shows that all zeros of g+λf are contained in K which must be a compact subset of Ω due to the assumption that |f (z)| < |g (z)| on ∂Ω. By Theorem 27.7.25 on Pn n cycle, {γ k }k=1 such that ∪nk=1 γ ∗k ⊆ Ω\K, k=1 n (γ k , z) = 1 Page 825 there exists aP n for every z ∈ K and k=1 n (γ k , z) = 0 for all z ∈ / Ω. Then as above, it follows from the residue theorem or more directly, Theorem 29.1.2, Z p n X X 1 λf 0 (z) + g 0 (z) dz = mj 2πi γ k λf (z) + g (z) j=1

k=1

29.2. SINGULARITIES AND THE LAURENT SERIES

863

where mj is the order of the j th zero of λf + g in K, hence in Ω. However, λ→

Z n X 1 λf 0 (z) + g 0 (z) dz 2πi γ k λf (z) + g (z)

k=1

is integer valued and continuous so it gives the same value when λ = 0 as when λ = 1. When λ = 0 this gives the number of zeros of g in Ω and when λ = 1 it is the number of zeros of f + g. This proves the theorem. Here is another formulation of this theorem. Corollary 29.1.7 Let Ω be a bounded open set and suppose f, g are continuous on Ω and analytic on Ω. Also suppose |f (z) − g (z)| < |g (z)| on ∂Ω. Then f and g have the same number of zeros in Ω provided each zero is counted according to multiplicity. Proof: You let f − g play the role of f in Theorem 29.1.6. Thus f − g + g = f and g have the same number of zeros. Alternatively, you can give a proof of this directly as follows. Let K = {z ∈ Ω : |f (z) − g (z)| ≥ |g (z)|} . Then if g (z) + λ (f (z) − g (z)) = 0 it follows 0 = |g (z) + λ (f (z) − g (z))| ≥ |g (z)| − λ |f (z) − g (z)| ≥ |g (z)| − |f (z) − g (z)| and so z ∈ K. Thus all zeros of g (z) + λ (f (z) − g (z)) are contained in K. By n n ∗ Theorem Pn {γ k }k=1 such that ∪k=1 γ k ⊆ Pn27.7.25 on Page 825 there exists a cycle, / Ω. Ω \ K, k=1 n (γ k , z) = 1 for every z ∈ K and k=1 n (γ k , z) = 0 for all z ∈ Then by Theorem 29.1.2, Z p n X X 1 λ (f 0 (z) − g 0 (z)) + g 0 (z) dz = mj 2πi γ k λ (f (z) − g (z)) + g (z) j=1

k=1

where mj is the order of the j th zero of λ (f − g) + g in K, hence in Ω. The left side is continuous as a function of λ and so the number of zeros of g corresponding to λ = 0 equals the number of zeros of f corresponding to λ = 1. This proves the corollary.

29.2

Singularities And The Laurent Series

29.2.1

What Is An Annulus?

In general, when you consider singularities, isolated or not, the fundamental tool is the Laurent series. This series is important for many other reasons also. In particular, it is fundamental to the spectral theory of various operators in functional analysis and is one way to obtain relationships between algebraic and analytical

864

RESIDUES

conditions essential in various convergence theorems. A Laurent series lives on an annulus. In all this f has values in X where X is a complex Banach space. If you like, let X = C. Definition 29.2.1 Define ann (a, R1 , R2 ) ≡ {z : R1 < |z − a| < R2 } . Thus ann (a, 0, R) would denote the punctured ball, B (a, R) \ {0} and when R1 > 0, the annulus looks like the following.

r a

The annulus is the stuff between the two circles. Here is an important lemma which is concerned with the situation described in the following picture. qz

qz qa

qa

Lemma 29.2.2 Let γ r (t) ≡ a + reit for t ∈ [0, 2π] and let |z − a| < r. Then n (γ r , z) = 1. If |z − a| > r, then n (γ r , z) = 0. Proof: For the first claim, consider for t ∈ [0, 1] , f (t) ≡ n (γ r , a + t (z − a)) . Then from properties of the winding number derived earlier, f (t) ∈ Z, f is continuous, and f (0) = 1. Therefore, f (t) = 1 for all t ∈ [0, 1] . This proves the first claim because f (1) = n (γ r , z) . For the second claim, Z 1 1 n (γ r , z) = dw 2πi γ r w − z Z 1 1 dw = 2πi γ r w − a − (z − a) Z 1 −1 1 ´ dw ³ = 2πi z − a γ r 1 − w−a z−a ¶k Z X ∞ µ −1 w−a = dw. 2πi (z − a) γ r z−a k=0


865

The series converges uniformly for w ∈ γ r because ¯ ¯ ¯w − a¯ r ¯ ¯ ¯z−a¯= r+c for some c > 0 due to the assumption that |z − a| > r. Therefore, the sum and the integral can be interchanged to give µ ¶k ∞ Z X w−a −1 dw = 0 n (γ r , z) = 2πi (z − a) z−a γr k=0

³

´k

because w → w−a has an antiderivative. This proves the lemma. z−a Now consider the following picture which pertains to the next lemma.

γr r a

Lemma 29.2.3 Let g be analyticR on ann (a, R1 , R2 ) . Then if γ r (t) ≡ a + reit for t ∈ [0, 2π] and r ∈ (R1 , R2 ) , then γ g (z) dz is independent of r. r

Proof: Let R1 < r1 < r2 < R2 and denote by −γ r (t) the curve, −γ r (t) ≡ a ¡+ rei(2π−t) for¡ t ∈ [0, ¢ ¢ 2π] . Then if z ∈ B (a, R1 ), Lemma 29.2.2 implies both n γ r2 , z and n γ r1 , z = 1 and so ¡ ¢ ¡ ¢ n −γ r1 , z + n γ r2 , z = −1 + 1 = 0. ³ ´ Also if z ∈ / B (a, R2 ) , then Lemma 29.2.2 implies n γ rj , z = 0 for j = 1, 2. Therefore, whenever z ∈ / ann (a, R1 , R2 ) , the sum of the winding numbers equals zero. Therefore, by Theorem 27.7.19 applied to the function, f (w) = g (z) (w − z) and z ∈ ann (a, R1 , R2 ) \ ∪2j=1 γ rj ([0, 2π]) , ¡ ¡ ¢ ¡ ¢¢ ¡ ¡ ¢ ¡ ¢¢ f (z) n γ r2 , z + n −γ r1 , z = 0 n γ r2 , z + n −γ r1 , z = Z Z 1 g (w) (w − z) g (w) (w − z) 1 dw − dw 2πi γ r w−z 2πi γ r w−z 2 1 Z Z 1 1 = g (w) dw − g (w) dw 2πi γ r 2πi γ r 2

which proves the desired result.

1

866

29.2.2

RESIDUES

The Laurent Series

The Laurent series is like a power series except it allows for negative exponents. First here is a definition of what is meant by the convergence of such a series. P∞ n Definition 29.2.4 n=−∞ an (z − a) converges if both the series, ∞ X

n

an (z − a) and

n=0

∞ X

a−n (z − a)

−n

n=1

P∞

converge. When this is the case, the symbol, ∞ X

n

an (z − a) +

n=0

n

n=−∞

∞ X

an (z − a) is defined as

a−n (z − a)

−n

.

n=1

Lemma 29.2.5 Suppose f (z) =

∞ X

an (z − a)

n

n=−∞

P∞ P∞ −n n for all |z − a| ∈ (R1 , R2 ) . Then both n=1 a−n (z − a) n=0 an (z − a) and converge absolutely and uniformly on {z : r1 ≤ |z − a| ≤ r2 } for any r1 < r2 satisfying R1 < r1 < r2 < R2 . P∞ −n converges Proof: Let R1 < |w − a| = r1 − δ < r1 . Then n=1 a−n (w − a) and so −n −n lim |a−n | |w − a| = lim |a−n | (r1 − δ) = 0 n→∞

n→∞

which implies that for all n sufficiently large, −n

|a−n | (r1 − δ)

< 1.

Therefore, ∞ X

|a−n | |z − a|

n=1

−n

=

∞ X

−n

|a−n | (r1 − δ)

n

−n

(r1 − δ) |z − a|

.

n=1

Now for |z − a| ≥ r1 , |z − a|

−n

≤

1 r1n

and so for all sufficiently large n n

(r1 − δ) . r1n P∞ −n Therefore, by the Weierstrass M test, the series, converges n=1 a−n (z − a) absolutely and uniformly on the set −n

|a−n | |z − a|

≤

{z ∈ C : |z − a| ≥ r1 } .

29.2. SINGULARITIES AND THE LAURENT SERIES Similar reasoning shows the series,

P∞ n=0

867

n

an (z − a) converges uniformly on the set

{z ∈ C : |z − a| ≤ r2 } . This proves the Lemma. Theorem 29.2.6 Let f be analytic on ann (a, R1 , R2 ) . Then there exist numbers, an ∈ C such that for all z ∈ ann (a, R1 , R2 ) , f (z) =

∞ X

n

an (z − a) ,

(29.2.5)

n=−∞

where the series converges absolutely and uniformly on ann (a, r1 , r2 ) whenever R1 < r1 < r2 < R2 . Also Z 1 f (w) an = dw (29.2.6) 2πi γ (w − a)n+1 where γ (t) = a + reit , t ∈ [0, 2π] for any r ∈ (R1 , R2 ) . Furthermore the series is unique in the sense that if 29.2.5 holds for z ∈ ann (a, R1 , R2 ) , then an is given in 29.2.6. Proof: Let R1 < r1 < r2 < R2 and define γ 1 (t) ≡ a + (r1 − ε) eit and γ 2 (t) ≡ a + (r2 + ε) eit for t ∈ [0, 2π] and ε chosen small enough that R1 < r1 − ε < r2 + ε < R2 .

γ2 qa

γ1 qz

Then using Lemma 29.2.2, if z ∈ / ann (a, R1 , R2 ) then n (−γ 1 , z) + n (γ 2 , z) = 0 and if z ∈ ann (a, r1 , r2 ) , n (−γ 1 , z) + n (γ 2 , z) = 1.

868

RESIDUES

Therefore, by Theorem 27.7.19, for z ∈ ann (a, r1 , r2 ) "Z # Z 1 f (w) f (w) f (z) = dw + dw 2πi −γ 1 w − z γ2 w − z  Z Z f (w) f (w) 1  h h i dw + = 2πi γ 1 (z − a) 1 − w−a γ 2 (w − a) 1 − z−a ¶n Z ∞ µ 1 f (w) X z − a = dw+ 2πi γ 2 w − a n=0 w − a ¶n Z ∞ µ 1 f (w) X w − a dw. 2πi γ 1 (z − a) n=0 z − a

 z−a w−a

i dw

(29.2.7)

From the formula 29.2.7, it follows that for z ∈ ann r1 , r2 ), the terms in the first ³ (a, ´ n r2 while those in the second sum are bounded by an expression of the form C r2 +ε ³ ń are bounded by one of the form C r1r−ε and so by the Weierstrass M test, the 1 convergence is uniform and so the integrals and the sums in the above formula may be interchanged and after renaming the variable of summation, this yields Ã ! Z ∞ X 1 f (w) n f (z) = n+1 dw (z − a) + 2πi (w − a) γ 2 n=0 −1 X

Ã

n=−∞

1 2πi

Z

!

f (w) (w − a)

γ1

n

(z − a) .

n+1

(29.2.8)

Therefore, by Lemma 29.2.3, for any r ∈ (R1 , R2 ) , Ã ! Z ∞ X f (w) 1 n f (z) = n+1 dw (z − a) + 2πi γ r (w − a) n=0 −1 X

Ã

n=−∞

and so f (z) =

∞ X n=−∞

1 2πi Ã

Z

n

n+1

(w − a)

γr

1 2πi

!

f (w)

Z

(z − a) . !

f (w)

γr

(29.2.9)

n+1 dw

(w − a)

n

(z − a) .

where r ∈ (R1 , R2 ) is arbitrary. This proves the existence part of the theorem. It remains to characterize an . P∞ n If f (z) = n=−∞ an (z − a) on ann (a, R1 , R2 ) let fn (z) ≡

n X k=−n

k

ak (z − a) .

(29.2.10)


869

This function is analytic in ann (a, R1 , R2 ) and so from the above argument, Ã ! Z ∞ X 1 fn (w) k fn (z) = dw (z − a) . (29.2.11) 2πi γ r (w − a)k+1 k=−∞

Also if k > n or if k < −n, Ã

and so fn (z) =

1 2πi

n X

Z

fn (w)

γr

Ã

k=−n

(w − a)

1 2πi

Z

k+1

! dw

fn (w)

γr

k+1

(w − a)

= 0. ! k

dw (z − a)

which implies from 29.2.10 that for each k ∈ [−n, n] , Z 1 fn (w) dw = ak 2πi γ r (w − a)k+1 However, from the uniform convergence of the series, ∞ X

n

an (w − a)

n=0

and

∞ X

−n

a−n (w − a)

n=1

ensured by Lemma 29.2.5 which allows the interchange of sums and integrals, if k ∈ [−n, n] , Z 1 f (w) dw 2πi γ r (w − a)k+1 P∞ Z P∞ −m m 1 m=1 a−m (w − a) m=0 am (w − a) + = dw k+1 2πi γ r (w − a) Z ∞ X 1 m−(k+1) = am (w − a) dw 2πi γ r m=0 Z ∞ X −m−(k+1) + a−m (w − a) dw =

γr

m=1 n X

am

m=0 n X

+

1 2πi

m−(k+1)

(w − a) Z

Z

γr

dw

γr

a−m

m=1

=

1 2πi

Z

(w − a) γr

fn (w) k+1

(w − a)

dw

−m−(k+1)

dw

870

RESIDUES

because if l > n or l < −n, Z

al (w − a)

γr

l

k+1

(w − a)

dw = 0

for all k ∈ [−n, n] . Therefore, ak =

1 2πi

Z

f (w)

γr

k+1

(w − a)

dw

and so this establishes uniqueness. This proves the theorem.

29.2.3

Contour Integrals And Evaluation Of Integrals

Here are some examples of hard integrals which can be evaluated by using residues. This will be done by integrating over various closed curves having bounded variation. Example 29.2.7 The first example we consider is the following integral. Z ∞ 1 dx 4 −∞ 1 + x One could imagine evaluating this integral by the method of partial fractions and it should work out by that method. However, we will consider the evaluation of this integral by the method of residues instead. To do so, consider the following picture. y

x

Let γ r (t) = reit , t ∈ [0, π] and let σ r (t) = t : t ∈ [−r, r] . Thus γ r parameterizes the top curve and σ r parameterizes the straight line from −r to r along the x axis. Denoting by Γr the closed curve traced out by these two, we see from simple estimates that Z 1 dz = 0. lim r→∞ γ 1 + z 4 r


871

This follows from the following estimate. ¯ ¯Z ¯ ¯ 1 1 ¯ ¯ dz πr. ¯ ¯≤ 4 4 ¯ r −1 ¯ γr 1 + z Therefore,

Z R

∞ −∞

1 dx = lim r→∞ 1 + x4

Z Γr

1 dz. 1 + z4

1 1+z 4 dz

using the method of residues. The only residues of the We compute Γr integrand are located at points, z where 1 + z 4 = 0. These points are 1√ 1 √ 1√ 2 − i 2, z = 2− 2 2 2 √ √ √ 1 1 1 2 + i 2, z = − 2+ 2 2 2

z

= −

z

=

1 √ i 2, 2 1 √ i 2 2

and it is only the last two which are found in the inside of Γr . Therefore, we need to calculate the residues at these points. Clearly this function has a pole of order one at each of these points and so we may calculate the residue at α in this list by evaluating 1 lim (z − α) z→α 1 + z4 Thus µ ¶ 1√ 1 √ Res f, 2+ i 2 2 2 µ µ ¶¶ 1√ 1 √ 1 = lim z − 2 + i 2 √ √ 1 1 2 2 1 + z4 z→ 2 2+ 2 i 2 1√ 1 √ = − 2− i 2 8 8 Similarly we may find the other residue in the same way µ ¶ 1√ 1 √ Res f, − 2+ i 2 2 2 µ µ ¶¶ 1√ 1 √ 1 = lim z− − 2+ i 2 √ √ 1 1 2 2 1 + z4 z→− 2 2+ 2 i 2 1 √ 1√ = − i 2+ 2. 8 8 Therefore, Z Γr

1 dz 1 + z4

= =

µ µ ¶¶ 1 √ 1√ 1√ 1 √ 2πi − i 2 + 2+ − 2− i 2 8 8 8 8 √ 1 π 2. 2

872

RESIDUES

√ R∞ 1 Thus, taking the limit we obtain 12 π 2 = −∞ 1+x 4 dx. Obviously many different variations of this are possible. The main idea being that the integral over the semicircle converges to zero as r → ∞. Sometimes we don’t blow up the curves and take limits. Sometimes the problem of interest reduces directly to a complex integral over a closed curve. Here is an example of this. Example 29.2.8 The integral is Z 0

π

cos θ dθ 2 + cos θ

This integrand is even and so it equals Z 1 π cos θ dθ. 2 −π 2 + cos θ ¡ ¢ For z on the unit circle, z = eiθ , z = z1 and therefore, cos θ = 21 z + z1 . Thus dz = ieiθ dθ and so dθ = dz iz . Note this is proceeding formally to get a complex integral which reduces to the one of interest. It follows that a complex integral which reduces to the one desired is ¡ ¢ Z Z 1 1 1 1 z2 + 1 2 z¡+ z ¢ dz = dz 2i γ 2 + 12 z + z1 z 2i γ z (4z + z 2 + 1) where γ ¡is the unit circle. Now the integrand has poles of order 1 at those points ¢ where z 4z + z 2 + 1 = 0. These points are √ √ 0, −2 + 3, −2 − 3. Only the first two are inside the unit circle. It is also clear the function has simple poles at these points. Therefore, µ ¶ z2 + 1 Res (f, 0) = lim z = 1. z→0 z (4z + z 2 + 1) ³ √ ´ Res f, −2 + 3 = ³ lim √

³ √ ´´ z − −2 + 3

z→−2+ 3

z2 + 1 2√ =− 3. 2 z (4z + z + 1) 3

It follows Z 0

π

cos θ dθ 2 + cos θ

Z

z2 + 1 dz 2 γ z (4z + z + 1) µ ¶ 1 2√ = 2πi 1 − 3 2i 3 µ ¶ 2√ = π 1− 3 . 3 =

1 2i


873

Other rational functions of the trig functions will work out by this method also. Sometimes you have to be clever about which version of an analytic function that reduces to a real function you should use. The following is such an example. Example 29.2.9 The integral here is Z ∞ 0

ln x dx. 1 + x4

The same curve used in the integral involving sinx x earlier will create problems with the log since the usual version of the log is not defined on the negative real axis. This does not need to be of concern however. Simply use another branch of the logarithm. Leave out the ray from 0 along the negative y axis and use Theorem 28.2.3 to define L (z) on this set. Thus L (z) = ln |z| + i arg1 (z) where arg1 (z) will iθ be the angle, θ, between − π2 and 3π 2 such that z = |z| e . Now the only singularities contained in this curve are 1√ 1 √ 1√ 1 √ 2 + i 2, − 2+ i 2 2 2 2 2 and the integrand, f has simple poles at these points. Thus using the same procedure as in the other examples, µ ¶ 1√ 1 √ Res f, 2+ i 2 = 2 2 1√ 1 √ 2π − i 2π 32 32 and

µ

¶

−1 √ 1 √ Res f, 2+ i 2 2 2

=

3 √ 3√ 2π + i 2π. 32 32 Consider the integral along the small semicircle of radius r. This reduces to Z

0

π

ln |r| + it ¡ it ¢ dt 4 rie 1 + (reit )

which clearly converges to zero as r → 0 because r ln r → 0. Therefore, taking the limit as r → 0, Z large semicircle

Z

R

lim

r→0+

r

L (z) dz + lim r→0+ 1 + z4

ln t dt = 2πi 1 + t4

µ

Z

−r

−R

ln (−t) + iπ dt+ 1 + t4

¶ 3√ 3 √ 1√ 1 √ 2π + i 2π + 2π − i 2π . 32 32 32 32

874

RESIDUES

Observing that

R

L(z) dz large semicircle 1+z 4

Z e (R) + 2 lim

r→0+

r

R

→ 0 as R → ∞,

ln t dt + iπ 1 + t4

Z

0

−∞

1 dt = 1 + t4

µ ¶ √ 1 1 − + i π2 2 8 4

where e (R) → 0 as R → ∞. From an earlier example this becomes Ã√ ! µ ¶ Z R √ ln t 1 1 2 e (R) + 2 lim π = − + i π 2 2. dt + iπ r→0+ r 1 + t4 4 8 4 Now letting r → 0+ and R → ∞, Z

∞

2 0

ln t dt = 1 + t4 =

and so

Z

∞

0

Ã√ ! ¶ µ √ 2 1 1 2 − + i π 2 − iπ π 8 4 4 1√ 2 − 2π , 8

1√ 2 ln t 2π , dt = − 4 1+t 16

which is probably not the first thing you would thing of. You might try to imagine how this could be obtained using elementary techniques. The next example illustrates the use of what is referred to as a branch cut. It includes many examples. Example 29.2.10 Mellin transformations are of the form Z ∞ dx f (x) xα . x 0 Sometimes it is possible to evaluate such a transform in terms of the constant, α. Assume f is an analytic function except at isolated singularities, none of which are on (0, ∞) . Also assume that f has the growth conditions, |f (z)| ≤

C |z|

b

,b > α

for all large |z| and assume that |f (z)| ≤

C0 b1

|z|

, b1 < α

for all |z| sufficiently small. It turns out there exists an explicit formula for this Mellin transformation under these conditions. Consider the following contour.


875

¾

−R

¾

In this contour the small semicircle in the center has radius ε which will converge to 0. Denote by γ R the large circular path which starts at the upper edge of the slot and continues to the lower edge. Denote by γ ε the small semicircular contour and denote by γ εR+ the straight part of the contour from 0 to R which provides the top edge of the slot. Finally denote by γ εR− the straight part of the contour from R to 0 which provides the bottom edge of the slot. The interesting aspect of this problem is the definition of f (z) z α−1 . Let z α−1 ≡ e(ln|z|+i arg(z))(α−1) = e(α−1) log(z) where arg (z) is the angle of z in (0, 2π) . Thus you use a branch of the logarithm which is defined on C\(0, ∞) . Then it is routine to verify from the assumed estimates that Z lim f (z) z α−1 dz = 0 R→∞

and

γR

Z f (z) z α−1 dz = 0.

lim

ε→0+

Also, it is routine to verify Z lim ε→0+

and

γ εR+

γε

Z f (z) z α−1 dz =

R

f (x) xα−1 dx

0

Z ε→0+

Z f (z) z α−1 dz = −ei2π(α−1)

lim

γ εR−

0

R

f (x) xα−1 dx.

876

RESIDUES

Therefore, letting ΣR denote the sum of the residues of f (z) z α−1 which are contained in the disk of radius R except for the possible residue at 0, ³ ´Z R e (R) + 1 − ei2π(α−1) f (x) xα−1 dx = 2πiΣR 0

where e (R) → 0 as R → ∞. Now letting R → ∞, Z R 2πi πe−πiα lim f (x) xα−1 dx = Σ Σ= i2π(α−1) R→∞ 0 sin (πα) 1−e where Σ denotes the sum of all the residues of f (z) z α−1 except for the residue at 0. The next example is similar to the one on the Mellin transform. In fact it is a Mellin transform but is worked out independently of the above to emphasize a slightly more informal technique related to the contour. R ∞ p−1 Example 29.2.11 0 x1+x dx, p ∈ (0, 1) . Since the exponent of x in the numerator is larger than −1. The integral does converge. However, the techniques of real analysis don’t tell us what it converges to. The contour to be used is as follows: From (ε, 0) to (r, 0) along the x axis and then from (r, 0) to (r, 0) counter clockwise along the circle of radius r, then from (r, 0) to (ε, 0) along the x axis and from (ε, 0) to (ε, 0) , clockwise along the circle of radius ε. You should draw a picture of this contour. The interesting thing about this is that z p−1 cannot be defined all the way around 0. Therefore, use a branch of z p−1 corresponding to the branch of the logarithm obtained by deleting the positive x axis. Thus z p−1 = e(ln|z|+iA(z))(p−1) where z = |z| eiA(z) and A (z) ∈ (0, 2π) . Along the integral which goes in the positive direction on the x axis, let A (z) = 0 while on the one which goes in the negative direction, take A (z) = 2π. This is the appropriate choice obtained by replacing the line from (ε, 0) to (r, 0) with two lines having a small gap joined by a circle of radius ε and then taking a limit as the gap closes. You should verify that the two integrals taken along the circles of radius ε and r converge to 0 as ε → 0 and as r → ∞. Therefore, taking the limit, Z ∞ p−1 Z 0 p−1 ³ ´ x x dx + e2πi(p−1) dx = 2πi Res (f, −1) . 1+x 0 ∞ 1+x Calculating the residue of the integrand at −1, and simplifying the above expression, ´ Z ∞ xp−1 ³ 1 − e2πi(p−1) dx = 2πie(p−1)iπ . 1+x 0 Upon simplification

Z

∞ 0

xp−1 π dx = . 1+x sin pπ


877

Example 29.2.12 The Fresnel integrals are Z ∞ Z ∞ ¡ 2¢ ¡ ¢ cos x dx, sin x2 dx. 0

0 2

To evaluate these integrals consider f (z) = eiz on the curve which goes³ from ´ √ the origin to the point r on the x axis and from this point to the point r 1+i 2 along a circle of radius r, and from there back to the origin as illustrated in the following picture. y

@ ¡

x

Thus the curve to integrate over is shaped like a slice of pie. Denote by γ r the curved part. Since f is analytic, ¶ Z Z r Z r ³ ³ 1+i ´´2 µ 1+i i t √2 iz 2 ix2 √ 0 = e dz + dt e dx − e 2 γr 0 0 µ ¶ Z Z r Z r 2 2 2 1+i √ = eiz dz + eix dx − e−t dt 2 γr 0 0 ¶ √ µ Z Z r 2 2 π 1+i √ = eiz dz + eix dx − + e (r) 2 2 γr 0 √ R∞ 2 where e (r) → 0 as r → ∞. Here we used the fact that 0 e−t dt = 2π . Now consider the first of these integrals. ¯ ¯ ¯Z π ¯Z ¯ ¯ ¯ ¯ 4 it 2 2 ¯ ¯ ¯ ¯ ei(re ) rieit dt¯ eiz dz ¯ = ¯ ¯ ¯ γr ¯ ¯ ¯ 0 Z π4 2 ≤ r e−r sin 2t dt 0

r ≤ 2

Z 0

r −(3/2)

Z

1

2

e−r u √ du 1 − u2 0 µZ 1 ¶ 1/2 r 1 1 √ √ du + e−(r ) 2 2 2 1−u 1−u 0 =

r 2

878

RESIDUES

which converges to zero as r → ∞. Therefore, taking the limit as r → ∞, ¶ Z ∞ √ µ 2 π 1+i √ = eix dx 2 2 0 and so

√ Z ∞ π sin x dx = √ = cos x2 dx. 2 2 0 0 The following example is one of the most interesting. By an auspicious choice of the contour it is possible to obtain a very interesting formula for cot πz known as the Mittag- Leffler expansion of cot πz. Z

∞

2

Example 29.2.13 Let γ N be the contour which goes from −N − 12 −N i horizontally to N + 21 − N i and from there, vertically to N + 21 + N i and then horizontally to −N − 12 + N i and finally vertically to −N − 12 − N i. Thus the contour is a large rectangle and the direction of integration is in the counter clockwise direction. Consider the following integral. Z π cos πz IN ≡ dz 2 2 γ N sin πz (α − z ) where α ∈ R is not an integer. This will be used to verify the formula of Mittag Leffler, ∞ X 2 π cot πα 1 + = . (29.2.12) α2 n=1 α2 − n2 α You should verify that cot πz is bounded on this contour and that therefore, IN → 0 as N → ∞. Now you compute the residues of the integrand at ±α and at n where |n| < N + 12 for n an integer. These are the only singularities of the integrand in this contour and therefore, you can evaluate IN by using these. It is left as an exercise to calculate these residues and find that the residue at ±α is −π cos πα 2α sin πα while the residue at n is α2

1 . − n2

Therefore, " 0 = lim IN = lim 2πi N →∞

N →∞

N X n=−N

1 π cot πα − α 2 − n2 α

which establishes the following formula of Mittag Leffler. lim

N →∞

N X n=−N

1 π cot πα = . α 2 − n2 α

Writing this in a slightly nicer form, yields 29.2.12.

#

29.3. EXERCISES

29.3

879

Exercises

1. Example 29.2.7 found the integral of a rational function of a certain sort. The technique used in this example typically works for rational functions of the (x) form fg(x) where deg (g (x)) ≥ deg f (x) + 2 provided the rational function has no poles on the real axis. State and prove a theorem based on these observations. 2. Fill in the missing details of Example 29.2.13 about IN → 0. Note how important it was that the contour was chosen just right for this to happen. Also verify the claims about the residues. 3. Suppose f has a pole of order m at z = a. Define g (z) by m

g (z) = (z − a) f (z) . Show Res (f, a) =

1 g (m−1) (a) . (m − 1)!

Hint: Use the Laurent series. 4. Give a proof of Theorem 29.1.1. Hint: Let p be a pole. Show that near p, a pole of order m, P∞ k −m + k=1 bk (z − p) f 0 (z) = P∞ k f (z) (z − p) + k=2 ck (z − p) Show that Res (f, p) = −m. Carry out a similar procedure for the zeros. 5. Use Rouche’s theorem to prove the fundamental theorem of algebra which says that if p (z) = z n + an−1 z n−1 · · · + a1 z + a0 , then p has n zeros in C. Hint: Let q (z) = −z n and let γ be a large circle, γ (t) = reit for r sufficiently large. 6. Consider the two polynomials z 5 + 3z 2 − 1 and z 5 + 3z 2 . Show that on |z| = 1, the conditions for Rouche’s theorem hold. Now use Rouche’s theorem to verify that z 5 + 3z 2 − 1 must have two zeros in |z| < 1. 7. Consider the polynomial, z 11 + 7z 5 + 3z 2 − 17. Use Rouche’s theorem to find a bound on the zeros of this polynomial. In other words, find r such that if z is a zero of the polynomial, |z| < r. Try to make r fairly small if possible. √ R∞ 2 8. Verify that 0 e−t dt = 2π . Hint: Use polar coordinates. 9. Use the contour described in Example 29.2.7 to compute the exact values of the following improper integrals. R∞ x (a) −∞ (x2 +4x+13) 2 dx

880

RESIDUES

(b) (c)

R∞ 0

R∞

x2 dx (x2 +a2 )2

dx , a, b −∞ (x2 +a2 )(x2 +b2 )

>0

10. Evaluate the following improper integrals. R∞ ax (a) 0 (xcos 2 +b2 )2 dx R∞ x dx (b) 0 (xx2 sin +a2 )2 11. Find the Cauchy principle value of the integral Z ∞ sin x dx 2 + 1) (x − 1) (x −∞ defined as µZ

1−ε

lim

ε→0+

−∞

sin x dx + 2 (x + 1) (x − 1)

12. Find a formula for the integral 13. Find

R∞ −∞

R∞

Z

∞

1+ε

dx −∞ (1+x2 )n+1

¶ sin x dx . (x2 + 1) (x − 1)

where n is a nonnegative integer.

sin2 x x2 dx.

14. If m < n for m and n integers, show Z ∞ x2m π 1 ¡ 2m+1 ¢ . dx = 2n 1 + x 2n sin 0 2n π 15. Find 16. Find

R∞

1 dx. −∞ (1+x4 )2

R∞ 0

ln(x) 1+x2 dx

= 0.

17. Suppose f has an isolated singularity at α. Show the singularity is essential if and only if the principal part series of f has infinitely many P∞of the Laurent k P∞ bk terms. That is, show f (z) = k=0 ak (z − α) + k=1 (z−α) k where infinitely many of the bk are nonzero. 18. Suppose Ω is a bounded open set and fn is analytic on Ω and continuous on Ω. Suppose also that fn → f uniformly on Ω and that f 6= 0 on ∂Ω. Show that for all n large enough, fn and f have the same number of zeros on Ω provided the zeros are counted according to multiplicity.

Some Important Functional Analysis Applications 30.1

The Spectral Radius Of A Bounded Linear Transformation

As a very important application of the theory of Laurent series, I will give a short description of the spectral radius. This is a fundamental result which must be understood in order to prove convergence of various important numerical methods such as the Gauss Seidel or Jacobi methods. Definition 30.1.1 Let X be a complex Banach space and let A ∈ L (X, X) . Then n o −1 r (A) ≡ λ ∈ C : (λI − A) ∈ L (X, X) This is called the resolvent set. The spectrum of A, denoted by σ (A) is defined as all the complex numbers which are not in the resolvent set. Thus σ (A) ≡ C \ r (A) Lemma 30.1.2 λ ∈ r (A) if and only if λI − A is one to one and onto X. Also if |λ| > ||A|| , then λ ∈ σ (A). If the Neumann series, ∞

1X λ

k=0

converges, then

∞

1X λ

k=0

µ ¶k A λ

µ ¶k A −1 = (λI − A) . λ

Proof: Note that to be in r (A) , λI − A must be one to one and map X onto −1 X since otherwise, (λI − A) ∈ / L (X, X) . 881

882

SOME IMPORTANT FUNCTIONAL ANALYSIS APPLICATIONS

By the open mapping theorem, if these two algebraic conditions hold, then −1 (λI − A) is continuous and so this proves the first part of the lemma. Now suppose |λ| > ||A|| . Consider the Neumann series ∞ µ ¶k 1X A . λ λ k=0

By the root test, Theorem 27.1.3 on Page 792 this series converges to an element of L (X, X) denoted here by B. Now suppose the series converges. Letting Bn ≡ Pn ¡ A ¢k 1 , k=0 λ λ (λI − A) Bn

n µ ¶k X A

=

Bn (λI − A) =

=

µ ¶n+1 A I− →I λ

k=0

λ

−

n µ ¶k+1 X A k=0

λ

as n → ∞ because the convergence of the series requires the nth term to converge to 0. Therefore, (λI − A) B = B (λI − A) = I which shows λI − A is both one to one and onto and the Neumann series converges −1 to (λI − A) . This proves the lemma. This lemma also shows that σ (A) is bounded. In fact, σ (A) is closed. ¯¯ ¯¯−1 ¯¯ −1 ¯¯ Lemma 30.1.3 r (A) is open. In fact, if λ ∈ r (A) and |µ − λ| < ¯¯(λI − A) ¯¯ , then µ ∈ r (A). Proof: First note

³

(µI − A) = =

´ −1 I − (λ − µ) (λI − A) (λI − A) ³ ´ −1 (λI − A) I − (λ − µ) (λI − A)

Also from the assumption about |λ − µ| , ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ −1 ¯¯ −1 ¯¯ ¯¯(λ − µ) (λI − A) ¯¯ ≤ |λ − µ| ¯¯(λI − A) ¯¯ < 1 and so by the root test, ∞ ³ X

(λ − µ) (λI − A)

−1

´k

k=0

converges to an element of L (X, X) . As in Lemma 30.1.2, ∞ ³ X k=0

−1

(λ − µ) (λI − A)

´k

³ ´−1 −1 = I − (λ − µ) (λI − A) .

(30.1.1) (30.1.2)

30.1. THE SPECTRAL RADIUS OF A BOUNDED LINEAR TRANSFORMATION883 Therefore, from 30.1.1, −1

(µI − A)

−1

³

= (λI − A)

I − (λ − µ) (λI − A)

−1

´−1

.

This proves the lemma. Corollary 30.1.4 σ (A) is a compact set. Proof: Lemma 30.1.2 shows σ (A) is bounded and Lemma 30.1.3 shows it is closed. Definition 30.1.5 The spectral radius, denoted by ρ (A) is defined by ρ (A) ≡ max {|λ| : λ ∈ σ (A)} . Since σ (A) is compact, this maximum exists. Note from Lemma 30.1.2, ρ (A) ≤ ||A||. There is a simple formula for the spectral radius. Lemma 30.1.6 If |λ| > ρ (A) , then the Neumann series, ∞

1X λ

k=0

µ ¶k A λ

converges. Proof: This follows directly from Theorem 29.2.6 on Page 867 and the obserP∞ ¡ ¢k −1 vation above that λ1 k=0 A = (λI − A) for all |λ| > ||A||. Thus the analytic λ −1 function, λ → (λI − A) has a Laurent expansion on |λ| > ρ (A) by Theorem P∞ ¡ ¢k 29.2.6 and it must coincide with λ1 k=0 A on |λ| > ||A|| so the Laurent expanP∞λ ¡ A ¢k −1 1 on |λ| > ρ (A) . This proves the sion of λ → (λI − A) must equal λ k=0 λ lemma. The theorem on the spectral radius follows. It is due to Gelfand. 1/n

Theorem 30.1.7 ρ (A) = limn→∞ ||An ||

.

Proof: If |λ| < lim sup ||An ||

1/n

n→∞

then by the root test, the Neumann series does not converge and so by Lemma 30.1.6 |λ| ≤ ρ (A) . Thus ρ (A) ≥ lim sup ||An || n→∞

1/n

.

884


Now let p be a positive integer. Then λ ∈ σ (A) implies λp ∈ σ (Ap ) because ¡ ¢ λp I − Ap = (λI − A) λp−1 + λp−2 A + · · · + Ap−1 ¡ ¢ = λp−1 + λp−2 A + · · · + Ap−1 (λI − A) It follows from Lemma 30.1.2 applied to Ap that for λ ∈ σ (A) , |λp | ≤ ||Ap || and so 1/p 1/p |λ| ≤ ||Ap || . Therefore, ρ (A) ≤ ||Ap || and since p is arbitrary, lim inf ||Ap || p→∞

1/p

1/n

≥ ρ (A) ≥ lim sup ||An || n→∞

.


30.2

Analytic Semigroups

30.2.1

Sectorial Operators And Analytic Semigroups

With the theory of functions of a complex variable, it is time to consider the notion of analytic semigroups. These are better than continuous semigroups. I am mostly following the presentation in Henry [30]. In what follows H will be a Banach space unless specified to be a Hilbert space. Definition 30.2.1 Let φ < π/2 and for a ∈ R, let Saφ denote the sector in the complex plane {z ∈ C\ {a} : |arg (z − a)| ≤ π − φ} This sector is as shown below.

Saφ φ

ba

A closed, densely defined linear operator, A is called sectorial if for some sector as described above, it follows that for all λ ∈ Saφ , −1

(λI − A)

∈ L (H, H)

30.2. ANALYTIC SEMIGROUPS

885

and for some M it satisfies ¯¯ ¯¯ ¯¯ −1 ¯¯ ¯¯(λI − A) ¯¯ ≤

M |λ − a|

To begin with it is interesting to have a perturbation theorem for sectorial operators. First note that for λ ∈ Saφ , −1

A (λI − A)

−1

= −I + λ (λI − A)

Proposition 30.2.2 Suppose A is a sectorial operator as defined above so it is a densely defined closed operator on D (A) ⊆ H which satisfies ¯¯ ¯¯ ¯¯ −1 ¯¯ (30.2.3) ¯¯A (λI − A) ¯¯ ≤ C whenever |λ| , λ ∈ Saφ , is sufficiently large and suppose B is a densely defined closed operator such that D (B) ⊇ D (A) and for all x ∈ D (A) , ||Bx|| ≤ ε ||Ax|| + K ||x||

(30.2.4)

where εC < 1. Then A + B is also sectorial. −1

Proof: I need to consider (λI − (A + B)) ³³

−1

I − B (λI − A)

´

. This equals

´−1 (λI − A) .

(30.2.5)

The issue is whether this makes any sense for all λ ∈ Sbφ for some b ∈ R. Let b > a be very large so that if λ ∈ Sbφ , then 30.2.3 holds. Then from 30.2.4, it follows that for ||x|| ≤ 1, ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ −1 ¯¯ −1 ¯¯ −1 ¯¯ ¯¯B (λI − A) x¯¯ ≤ ε ¯¯A (λI − A) x¯¯ + K ¯¯(λI − A) x¯¯ ≤ εC + K/ |λ − a| and so if b is made still larger, it follows this is less than r < 1 for all ||x|| ≤ 1. Therefore, for such b, ³ ´−1 −1 I − B (λI − A) exists and so for such b, the expression in 30.2.5 makes sense and equals −1

(λI − A)

³ ´−1 −1 I − B (λI − A)

and furthermore, ¯¯ ³ ´−1 ¯¯¯¯ ¯¯ ¯¯(λI − A)−1 I − B (λI − A)−1 ¯¯ ≤ ¯¯ ¯¯

1 M0 M ≤ |λ − a| 1 − r |λ − b|

886


by adjusting the constants because M |λ − b| |λ − a| 1 − r is bounded for λ ∈ Sbφ . This proves the proposition. It is an interesting proposition because when you have compact embeddings, such inequalities tend to hold. Definition 30.2.3 Let ε > 0 and for a sectorial operator as defined above, let the contour γ ε,φ be as shown next where the orientation is also as shown by the arrow.

Saφ φ

ra

The little circle has radius ε in the above contour. 0 Definition 30.2.4 For t ∈ S0(φ+π/2) the open sector shown in the following picture,

φ + π/2 b 0

S0(φ+π/2)

30.2. ANALYTIC SEMIGROUPS define

1 S (t) ≡ 2πi

887 Z −1

eλt (λI − A)

dλ

(30.2.6)

γ ε,φ

where ε is chosen such that t is a positive distance from the set of points included 0 in γ ε,φ . The situation is described by the following picture which shows S0(φ+π/2) and S0φ . Note how the dotted line is at right angles to the solid line.

t φ

b

S0(φ+π/2)

Also define S (0) ≡ I. It isn’t necessary that ε be small, just that γ ε,φ not contain t. This is because the integrand in 30.2.6 is analytic. Then it is necessary to show the above definition is well defined. 0 Lemma 30.2.5 The above definition is well defined for t ∈ S0(φ+π/2) . Also there is a constant, Mr such that ||S (t)|| ≤ Mr ¡ ¢ 0 for every t ∈ S0(φ+π/2) such that |arg t| ≤ r < π2 − φ .

Proof: In the definition of S (t) one can take ε = 1/ |t| . Then on the little circle which is part of γ ε,φ the contour integral equals µ ¶−1 Z π−φ i(θ+arg(t)) 1 iθ 1 1 iθ ee e −A e dθ 2π φ−π |t| |t| and by assumption the norm of the integrand is no larger than M 1 1/ |t| |t| and so the norm of this integral is dominated by Z M π−φ M dθ = (2π − 2φ) ≤ M 2π φ−π 2π

888


which is independent of t. Now consider the part of the contour used to define S (t) which is the top line segment. This equals Z ∞ 1 −1 eywt (ywI − A) wdy 2πi 1/|t| where w is a fixed complex number of unit length which gives a direction for motion along this segment, arg (w) = π − φ. Then the norm of this is dominated by Z ∞ Z ∞ ¯ ywt ¯ M 1 M ¯e ¯ dy = 1 exp (y |t| cos (arg (w) + arg (t))) dy 2π 1/|t| y 2π 1/|t| y By assumption |arg (t)| ≤ r
0. 2

It follows the integral dominated by Z ∞ 1 M exp (−c (r) |t| y) dy 2π 1/|t| y Z ∞ 1 M |t| 1 = exp (−c (r) x) dx 2π 1 x |t| Z ∞ 1 M = exp (−c (r) x) dx 2π 1 x where c (r) < 0 independent of |arg (t)| ≤ r. A similar estimate holds for the integral on the bottom segment. Thus for |arg (t)| ≤ r, ||S (t)|| is bounded. This proves the Lemma. Also note that if the contour is shifted to the right slightly, the integral over the shifted contour, γ 0ε,φ coincides with the integral over γ ε,φ thanks to the Cauchy integral formula and an approximation argument involving truncating the infinite contours and joining them at either end. Also note that in particular, ||S (t)|| is bounded for all positive real t. The following is the main result. Theorem 30.2.6 Let A be a sectorial operator as defined in Definition 30.2.1 for the sector S0φ . 0 1. Then S (t) given above in 30.2.6 is analytic for t ∈ S0(φ+π/2) .

2. For any x ∈ H and t > 0, then for n a positive integer, S (n) (t) x = An S (t) x


889

0 0 . That is, for all t, s ∈ S0(φ+π/2) , 3. S is a semigroup on the open sector, S0(φ+π/2)

S (t + s) = S (t) S (s) 4. t → S (t) x is continuous at t = 0 for all x ∈ H. 5. For some constants M, N such that if t is positive and real, ||S (t)|| ≤ M ||AS (t)|| ≤

N t

Proof: Consider the first claim. This follows right away from the formula. Z 1 −1 S (t) ≡ eλt (λI − A) dλ 2πi γ ε,φ The estimates for uniform convergence do not change for small changes in t and so the formula can be differentiated with respect to the complex variable t using the dominated convergence theorem to obtain Z 1 −1 S 0 (t) ≡ λeλt (λI − A) dλ 2πi γ ε,φ

= =

1 2πi 1 2πi

Z Z

³ ´ −1 eλt I + A (λI − A) dλ γ ε,φ −1

eλt A (λI − A)

dλ

γ ε,φ

because of the Cauchy integral theorem and an approximation result. Now approximating the infinite contour with a finite one and then the integral with Riemann sums, one can use the fact A is closed to take A out of the integral and write Ã ! Z 1 −1 0 λt S (t) = A e (λI − A) dλ = AS (t) 2πi γ ε,φ To get the higher derivatives, note S (t) has infinitely many derivatives due to t being a complex variable. Therefore, S 0 (t + h) − S 0 (t) S (t + h) − S (t) = lim A h→0 h→0 h h

S 00 (t) = lim and

S (t + h) − S (t) → AS (t) h

890


and so since A is closed, AS (t) ∈ D (A) and A2 S (t) Continuing this way yields the claims 1.) and 2.). Note this also implies S (t) x ∈ 0 . D (A) for each t ∈ S0(φ+π/2) 0 Next consider the semigroup property. Let s, t ∈ S0(φ+π/2) and let ε be sufficiently small that γ ε,φ is at a positive distance from both s and t. As described above let γ 0ε,φ denote the contour shifted slightly to the right, still at a positive distance from t. Then

µ S (t) S (s) =

1 2πi

¶2 Z

Z −1 µs

eλt (λI − A)

−1

e

(µI − A)

−1

− (µI − A)

dµdλ

γ 0ε,φ

γ ε,φ

At this point note that −1

(λI − A)

−1

(µI − A)

= (µ − λ)

−1

³ (λI − A)

−1

´ .

Then substituting this in the integrals above, it equals µ

1 2πi

¶2 Z

Z

γ ε,φ

³ ³ ´´ −1 −1 −1 eµs eλt (µ − λ) (λI − A) − (µI − A) dµdλ γ 0ε,φ

µ =

− µ +

1 2πi 1 2πi

¶2 Z

Z eλt

¶2 Z

γ ε,φ

Z

−1

(µI − A)

−1

(λI − A)

eµs (µ − λ) eµs eλt (µ − λ)

γ ε,φ

−1

dµdλ

−1

dµdλ

γ 0ε,φ

γ 0ε,φ

The order of integration can be interchanged because of the absolute convergence and Fubini’s theorem. Then this reduces to µ =

− µ +

1 2πi 1 2πi

¶2 Z

Z −1 µs

(µI − A) ¶2 Z

γ ε,φ

dλdµ

Z

−1

dµdλ

γ ε,φ

−1 λt

(λI − A)

−1

eλt (µ − λ)

e

γ 0ε,φ

eµs (µ − λ)

e

γ 0ε,φ

Now the following diagram might help in drawing some interesting conclusions.


891

The first iterated integral equals 0. This can be seen from the above picture. The inner integral taken over γ ε,φ is essentially equal to the integral over the closed contour in the above picture provided the radius of the part of the circle in the above closed contour is large enough. This closed contour integral equals 0 by the Cauchy integral theorem. The second iterated integral equals Z 1 −1 (λI − A) eλt eλs dλ 2πi γ ε,φ This can be seen from considering a similar closed contour, using the Cauchy integral formula. This verifies the semigroup identity. Next consider 4.), the continuity at t = 0. This requires showing lim S (t) x = x

t→0+

where the limit is taken through positive real values of t. It suffices to let x ∈ D (A) because by Lemma 30.2.5, ||S (t)|| is bounded for these values of t. Also in doing the computation, let γ ε,φ equal γ t−1 ,φ and it will be assumed t < 1. Then one must estimate ||S (t) x − x|| ≡ ¯¯ ¯¯ ¯¯ 1 Z ¯¯ ¯¯ ¯¯ −1 λt e (λI − A) xdλ − Ix¯¯ (30.2.7) ¯¯ ¯¯ 2πi γ t−1 ,φ ¯¯ By the Cauchy integral formula and approximating the integral with a contour integral over a finite closed contour, Z 1 eλt dλ = 1 2πi γ t−1 ,φ λ

892


and so 30.2.7 equals

=

¯¯ ¯¯ ¯¯ ¯¯ 1 Z ³ ´ ¯¯ ¯¯ −1 −1 λt e (λI − A) − Iλ xdλ¯¯ ¯¯ ¯¯ 2πi γ t−1 ,φ ¯¯ ¯¯ ¯¯ Z ¯¯ 1 ¯¯ ¯¯ ¯¯ −1 eλt λ−1 (λI − A) Axdλ¯¯ ¯¯ ¯¯ 2πi γ −1 ¯¯ t ,φ

Changing the variable letting λt = µ, the above equals ¯¯ ¯¯ ¯¯ 1 Z 1 ¯¯¯¯ ¯¯ −1 −1 µ e t (µ) ((µ/t) I − A) Ax dµ¯¯ ¯¯ ¯¯ 2πi γ 1,φ t ¯¯ which is dominated by Z 1 1 t2 M e|µ| cos ψ 2 ||Ax|| d |µ| = tC ||Ax|| . 2π γ 1,φ t |µ| Therefore, limt→0+ ||S (t) x − x|| = 0 whenever x ∈ D (A) . Since S (t) is bounded for positive t, if y ∈ H is arbitrary, choose x ∈ D (A) close enough to y such that ||S (t) y − y|| ≤ ||S (t) (y − x)|| + ||S (t) x − x|| + ||x − y|| ≤ ε/2 + ||S (t) x − x|| and now if t is close enough to 0 it follows the right side is less than ε. This proves 4.). Finally consider 5.). The first part follows from Lemma 30.2.5. It remains to show the second part. Let x ∈ H. First suppose t 6= 1. ¯¯ ¯¯ ¯¯ 1 Z ¯¯ ¯¯ ¯¯ −1 λt ||AS (t) x|| = ¯¯ e A (λI − A) xdλ¯¯ ¯¯ 2πi γ t−1 ,φ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ 1 Z ³ ´ ¯¯ ¯¯ −1 λt e −I + λ (λI − A) xdλ¯¯ = ¯¯ ¯¯ ¯¯ 2πi γ −1 t ,φ ¯¯ ¯¯ Z ¯¯ ¯¯ 1 ¯¯ ¯¯ −1 eλt λ (λI − A) xdλ¯¯ = ¯¯ ¯¯ ¯¯ 2πi γ −1 t ,φ ¯¯ ¯¯ ¯¯ 1 Z ´−1 1 ¯¯ ³ ¯¯ ¯¯ µµ µ I −A x dµ¯¯ e = ¯¯ ¯¯ 2πi γ 1,φ t t t ¯¯ and this is dominated by 1 ≤ 2π This proves the theorem.

Z e|µ| cos ψ M d |µ| γ 1,φ

N 1 = . t t


893

What if A is sectorial in the more general sense with Saφ taking the place of S0φ and the resolvent estimate being ¯¯ ¯¯ ¯¯ −1 ¯¯ ¯¯(λI − A) ¯¯ ≤

M , λ ∈ Saφ ? |λ − a|

Then Saφ = a + S0φ and so for λ ∈ S0φ , (λ + a) ∈ Saφ and so −1

(λI − (A − aI)) and

−1

= ((λ + a) I − A)

∈ L (H, H)

¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ −1 ¯¯ −1 ¯¯ ¯¯(λI − (A − aI)) ¯¯ = ¯¯((λ + a) − A) ¯¯ ≤

M M = |(λ + a) − a| |λ|

Therefore, letting Aa ≡ A − aI, it follows the result of Theorem 30.2.6 holds for this operator. There exists a semigroup, Sa (t) satisfying 0 1. Sa (t) is analytic for t ∈ S0(φ+π/2) .

2. For any x ∈ H and t > 0, then for n a positive integer, Sa(n) (t) x = Ana Sa (t) x 0 0 3. Sa is a semigroup on the open sector, S0(φ+π/2) . That is, for all t, s ∈ S0(φ+π/2) ,

Sa (t + s) = Sa (t) Sa (s) 4. t → Sa (t) x is continuous at t = 0 for all x ∈ H. 5. For some constants M, N such that if t is positive and real, ||Sa (t)|| ≤ M ||Aa Sa (t)|| ≤ Define

N t

S (t) ≡ eat Sa (t) .

This satisfies the semigroup identity, is continuous at 0, and satisfies the inequality ||S (t)|| ≤ M eat for all t real and nonnegative. What is its derivative? S 0 (t) = =

aeat Sa (t) + (A − aI) eat Sa (t) Aeat Sa (t) = AS (t)

894


and by continuing this way using A is closed as before, it follows S (n) (t) = An S (t) . Also for t > 0,

¯¯ ¯¯ ||AS (t)|| = ¯¯eat ASa (t)¯¯ ¯¯ ¯¯ = ¯¯eat (A − aI) Sa (t) + eat aSa (t)¯¯ ¯¯ ¯¯ = ¯¯eat Aa Sa (t) + eat aSa (t)¯¯ µ ¶ N N ≤ eat + aeat M = eat + aM . t t

It is clear S (t) x is continuous at 0 since this is true of Sa (t). This proves the following corollary which is a useful generalization of the above major theorem. Corollary 30.2.7 Let A be sectorial satisfying −1

(λI − A) for φ < π/2 and

∈ L (H, H) for λ ∈ Saφ

¯¯ ¯¯ ¯¯ −1 ¯¯ ¯¯(λI − A) ¯¯ ≤

M |λ − a|

0 for all λ ∈ Saλ . Then there exists a semigroup, S (t) defined on S0(φ+π/2) which satisfies 0 1. S (t) is analytic for t ∈ S0(φ+π/2) .

2. For any x ∈ H and t > 0, then for n a positive integer, S (n) (t) x = An S (t) x 0 3. S (t) is a semigroup on the open sector, S0(φ+π/2) . That is, for all t, s ∈ 0 S0(φ+π/2) , S (t + s) = S (t) S (s)

4. t → S (t) x is continuous at t = 0 for all x ∈ H. 5. For some constants M, N such that if t is positive and real, ||S (t)|| ≤ M eat µ ||AS (t)|| ≤ e

at

N + aM t

¶


30.2.2

895

The Numerical Range

In Hilbert space, there is a useful easy to check criterion which implies an operator is sectorial. Definition 30.2.8 Let A be a closed densely defined operator A : D (A) → H for H a Hilbert space. The numerical range is the following set. {(Au, u) : u ∈ D (A)} −1

Also recall the resolvent set, r (A) consists of those λ ∈ C such that (λI − A) ∈ L (H, H) . Thus, to be in this set λI − A is one to one and onto with continuous inverse. Proposition 30.2.9 Suppose the numerical range of A,a closed densely defined operator A : D (A) → H for H a Hilbert space is contained in the set {z ∈ C : |arg (z)| ≥ π − φ} where 0 < φ < π/2 and suppose A−1 ∈ L (H, H) , (0 ∈ r (A)). Then A is sectorial with the sector S0,φ0 where π/2 > φ0 > φ. Proof: Here is a picture of the situation along with details used to motivate the proof.

JJ J @ J @J λ (T u, u) @J B @JB φ @ JB ¡ ¡ ¡ ¡ ¡ In the picture the angle which is a little larger than φ is φ0 . Let λ be as shown with |arg λ| ≤ π − φ0 . Then from the picture and trigonometry, if u ∈ D (A) , ¯ µ ¶¯ ¡ ¢ ¯ u u ¯¯ |λ| sin φ0 − φ < ¯¯λ − A , |u| |u| ¯ and so

¯µ ¶¯ ¡ 0 ¢ ¯ u ¯¯ ¯ |u| |λ| sin φ − φ < ¯ λu − Au, ≤ ||(λI − A) u|| |u| ¯

896


Hence for all λ such that |arg λ| ≤ π − φ0 and u ∈ D (A) , ! Ã 1 1 ¡ 0 ¢ |(λI − A) u| |u| < |λ| sin φ − φ ≡

M |(λI − A) u| |λ|

Thus (λI − A) is one to one on S0,φ0 and if λ ∈ r (A) , then ¯¯ ¯¯ M ¯¯ −1 ¯¯ . ¯¯(λI − A) ¯¯ < |λ| By assumption 0 ∈ r (A). Now if |µ| is small, −1

(µI − A) must exist because it equals ¡¡

¢ ¢−1 µA−1 − I A

¯¯ ¯¯ ¡ ¢−1 and for |µ| < ¯¯A−1 ¯¯ , µA−1 − I ∈ L (H, H) since the infinite series ∞ X

(−1)

k

¡ −1 ¢k µA

k=0

¡

¢−1 converges and must equal to µA−1 − I . Therefore, there exists µ ∈ S0,φ0 such that µ 6= 0 and µ ∈ r (A). Also if µ 6= 0 and µ ∈ S0,φ0 , then if |λ − µ| < −1 |µ| must exist because M , (λI − A) −1

(λI − A)

h³ =

(λ − µ) (µI − A)

−1

´ i−1 − I (µI − A)

³ ´−1 −1 where (λ − µ) (µI − A) − I exists because ¯¯ ¯¯ ¯¯ −1 ¯¯ ¯¯(λ − µ) (µI − A) ¯¯

=
φ, −1

((−aI + A) − µI)

¯¯ ¯¯ ¯¯ −1 ¯¯ ¯¯((−aI + A) − µI) ¯¯

∈ ≤

L (H, H) , M |µ|

Therefore, for µ ∈ S0,φ0 , µ + a ∈ r (A) . Therefore, if λ ∈ Sa,φ0 , λ − a ∈ S0,φ0 ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ −1 ¯¯ −1 ¯¯ ¯¯(A − λI) ¯¯ = ¯¯(A − aI − (λ − a) I) ¯¯ ≤

M |λ − a|


30.2.3

An Interesting Example

In this section related to this example, for V a Banach space, V 0 will denote the space of continuous conjugate linear functions defined on V . Usually the symbol has meant the space of continuous linear functions but here they will be conjugate linear. That is f ∈ V 0 means f (ax + by) = af (x) + bf (y) and f is continuous. Let Ω be a bounded open set in Rn and define © ¡ ¢ ª V0 ≡ u ∈ C ∞ Ω : u = 0 on Γ ¡ ¢ where Γ is some measurable subset of the boundary of Ω and C ∞ Ω denotes the restrictions of functions in Cc∞ (Rn ) to Ω. By Corollary 12.5.11 V0 is dense in L2 (Ω) . Now define the following for u, v ∈ V0 . Z Z A0 u (v) ≡ −a uvdx − a (x) ∇u · ∇vdx Ω

Ω

¡ ¢ where a > 0 and a (x) ≥ 0 is a C 1 Ω function. Also define the following inner product on V0 . Z ¡ ¢ (u, v)1 ≡ auv + a (x) ∇u · ∇v dx Ω

Let ||·||1 denote the corresponding norm. Of course V0 is not a Banach space because it fails to be complete. u ∈ V will mean that u ∈ L2 (Ω) and there exists a sequence {un } ⊆ V0 such that lim ||un − um ||1 = 0

m,n→∞

898


and lim |un − u|L2 (Ω) = 0.

n→∞

For u ∈ V, define ∇u to be that element of L2 (Ω; Cn , a (x) dmn ) , the space of vector valued L2 functions taken with respect to the measure a (x) dmn which satisfies |∇u − ∇un |L2 (Ω;Cn ,a(x)dmn ) → 0. Denote this space by W for simplicity of notation. Observation 30.2.11 V is a Hilbert space with inner product given by Z ¡ ¢ (u, v)1 ≡ auv + a (x) ∇u · ∇v dx Ω

Everything is obvious except completeness. Suppose then that {un } is a Cauchy sequence in V. Then there exists a unique u ∈ L2 (Ω) such that |un − u|L2 (Ω) → 0. Now let |wn − un |L2 (Ω) + |∇wn − ∇un |W < 1/2n It follows {∇wn } is also a Cauchy sequence in W while {wn } is a Cauchy sequence in L2 (Ω) converging to u. Thus the thing to which ∇wn converges in W is the definition of ∇u and u ∈ V. Thus ||un − u||1

≤ ||un − wn ||1 + ||wn − u||1 1 < + ||wn − u||1 2n

and the last term converges to 0. Hence V is complete as claimed. Then it is clear V is a Hilbert space. The next observation is a simple one involving the Riesz map. Definition 30.2.12 Let V be a Hilbert space and let V 0 be the space of continuous conjugate linear functions defined on V . Then define R : V → V 0 by Rx (y) ≡ (x, y) . This is called the Riesz map. Lemma 30.2.13 The Riesz map is one to one and onto and linear. Proof: It is obvious it is one to one and linear. The only challenge is to show it is onto. Let z ∗ ∈ V 0 . If z ∗ (V ) = {0} , then letting z = 0, it follows Rz = z ∗ . If z ∗ (V ) 6= 0, then ker (z ∗ ) ≡ {x ∈ V : z ∗ (x) = 0} is a closed subspace. It is closed because z ∗ is continuous and it is just z ∗−1 (0) . Since ker (z ∗ ) is not everything in V there exists ⊥

w ∈ ker (z ∗ ) ≡ {x : (x, y) = 0 for all y ∈ ker (z ∗ )}


899

and w 6= 0. Then ³ ´ z ∗ z ∗ (x)w − z ∗ (w)x = z ∗ (x) z ∗ (w) − z ∗ (w) z ∗ (x) = 0 and so z ∗ (x)w − z ∗ (w)x ∈ ker (z ∗ ) . Therefore, for any x ∈ V, ³ ´ 0 = w, z ∗ (x)w − z ∗ (w)x = z ∗ (x) (w, w) − z ∗ (w) (w, x) and so

Ã ∗

z (x) =

z ∗ (w) ||w||

2

! ,x

2

so let z = w/ ||w|| . Then Rz = z ∗ and so R is onto. This proves the lemma. Now for the V described above, Z ¢ ¡ Ru (v) = auv + a (x) ∇u · ∇v dx Ω

Also, as noted above V is dense in H ≡ L2 (Ω) and so if H is identified with H 0 , it follows V ⊆ H = H 0 ⊆ V 0. Let A : D (A) → H be given by D (A) ≡ {u ∈ V : Ru ∈ H} and A ≡ −R on D (A). Then the numerical range for A is contained in (−∞, −a] and so A is sectorial by Proposition 30.2.9 provided A is closed and densely defined. Why is D (A) dense? It is because it contains Cc∞ (Ω) which is dense in L2 (Ω) . This follows from integration by parts which shows that for u, v ∈ Cc∞ (Ω) , Z Z − auvdx − a (x) ∇u · ∇vdx Ω

Ω

Z

Z auvdx +

=− Ω

∇ · (a (x) ∇u) vdx Ω

and since Cc∞ (Ω) is dense in H, Au = −au + ∇ · (a (x) ∇u) ∈ L2 (Ω) = H. Why is A closed? If un ∈ D (A) and un → u in H while Aun → ξ in H, then it follows from the definition that Run → −ξ and {un } converges to u in V so for any v ∈ V, Ru (v) = lim Run (v) = lim (Run , v)H = (−ξ, v)H n→∞

n→∞

900


which shows Ru = −ξ ∈ H and so u ∈ D (A) and Au = ξ. Thus A is closed. This completes the example. Obviously you could follow identical reasoning to include many other examples of more complexity. What does it mean for u ∈ D (A)? It means that in a weak sense −au + ∇ · (a (x) ∇u) ∈ H. Since A is sectorial for S−a,φ for any 0 < φ < π/2, this has shown the existence of a weak solution to the partial differential equation along with appropriate boundary conditions, −au + ∇ · (a (x) ∇u) = f, u ∈ V. What are these appropriate boundary conditions? u = 0 on Γ is one. the other would be a variational boundary condition which comes from integration by parts. Letting v ∈ V, formally do the following using the divergence theorem. Z (f, v)H = (−au + ∇ · (a (x) ∇u)) vdx Ω

Z

Z

−auvdx + Ω Z = (f, v)H +

Z

=

(a (x) ∇uv) · nds − ∂Ω

a (x) ∇u (x) · ∇v (x) dx Ω

(a (x) ∇u) · nvds

∂Ω\Γ

and so the other boundary condition is ∂u = 0 on ∂Ω \ Γ. ∂n To what extent this weak solution is really a classical solution depends on more technical considerations. a (x)

30.2.4

Fractional Powers Of Sectorial Operators

It will always be assumed in this section that A is sectorial for the sector S−a,φ where a > 0. To begin with, here is a useful lemma which will be used in the presentation of these fractional powers. Lemma 30.2.14 The following holds for α ∈ (0, 1) and σ < t. Z t π α−1 −α (t − s) (s − σ) ds = sin (πα) σ In particular,

Z

1

α−1 −α

(1 − s)

s

ds =

0

π . sin (πα)

Also for α, β > 0 µZ Γ (α) Γ (β) =

1

α−1

x 0

β−1

(1 − x)

¶ dx Γ (α + β) .


901 −1

Proof: First change variables to get rid of the σ. Let y = (t − σ) Then the integral becomes Z 1 α−1 −α (t − [(t − σ) y + σ]) (t − σ) y −α (t − σ) dy Z

0 1

=

α−1

((t − σ) (1 − y)) Z

(s − σ) .

(t − σ)

−α

y −α (t − σ) dy

0 1

=

α−1

(1 − y)

y −α dy

0

Next let y = x2 . The integral is Z 1 ¡ ¢α−1 1−2α 2 1 − x2 x dx 0

Next let x = sin θ Z 12 π Z 2α−1 2 (cos (θ)) sin(1−2α) (θ) dθ = 2 0

1 2π

0

µ

cos (θ) sin (θ)

¶2α−1 dθ

Now change the variable again. Let u = cot (θ) . Then this yields Z ∞ 2α−1 u 2 du 1 + u2 0 This is fairly easy to evaluate using contour integrals. Consider the following contour called ΓR for large R. As R → ∞, the integral over the little circle converges to 0 and so does the integral over the big circle. There is one singularity at i.

−R

−R−1

R−1

R

Thus Z lim

R→∞

=

ΓR

e(ln|z|+i arg(z))(1−2α) dz = 1 + z2

Z ∞ 2α−1 u (1 + cos (1 − 2α) π) du 1 + u2 0 Z ∞ 2α−1 u +i sin ((1 − 2α) π) du 1 + u2 0

902


³ ³π ´ ³π ´´ = π cos (1 − 2α) + i sin (1 − 2α) 2 2 Then equating the imaginary parts yields Z ∞ 2α−1 ³π ´ u sin ((1 − 2α) π) du = π sin (1 − 2α) 1 + u2 2 0 and so using the trig identities for the sum of two angles, ¢¢ ¡ ¡ Z ∞ 2α−1 π sin π2 (1 − 2α) u ¢ ¡ ¢ ¡ du = 1 + u2 2 sin π2 (1 − 2α) cos π2 (1 − 2α) 0 π π ¡ ¢= = 2 sin (πα) 2 cos π2 (1 − 2α) It remains to verify the last identity. Z ∞Z ∞ Γ (α) Γ (β) ≡ tα−1 e−t sβ−1 e−s dsdt Z0 ∞ Z0 ∞ β−1 = tα−1 e−u (u − t) dudt 0 t Z ∞ Z u β−1 = e−u tα−1 (u − t) dtdu Z

0

0 1

=

Z

xα−1 (1 − x)

β−1

∞

dx

0

e−u uα+β−1 du

0

µZ

1

=

xα−1 (1 − x)

¶ dx Γ (α + β)

β−1

0

This proves the lemma. If it is not stated otherwise, in all that follows α > 0. Definition 30.2.15 Let A be a sectorial operator corresponding to the sector S−aφ where −a < 0. Then define for α > 0, Z ∞ 1 −α (−A) ≡ tα−1 S (t) dt Γ (α) 0 where S (t) is the analytic semigroup generated by A as in Corollary 30.2.7. Note that from the estimate, ||S (t)|| ≤ M e−at of this corollary, the integral is well defined and is in L (H, H). −α

Theorem 30.2.16 For (−A)

(−A) Also

−1

(−A)

as defined in Definition 30.2.15

−α

−β

(−A)

−(α+β)

= (−A)

(−A) = I, (−A) (−A)

−1

=I

(30.2.8) (30.2.9)


903

−α

and (−A) is one to one if α ≥ 0, defining A0 ≡ I. If α < β, then −β −α (−A) (H) ⊆ (−A) (H) .

(30.2.10)

If α ∈ (0, 1) , then −α

(−A)

sin (πα) π

=

Z

(−A)

−β

(−A)

−1

λ−α (λI − A)

dλ

(30.2.11)

0

Proof: Consider 30.2.8. −α

∞

1 ≡ Γ (α) Γ (β)

Z

∞

Z

0

∞

tα−1 sβ−1 S (t + s) dsdt

0

Changing variables and using Fubini’s theorem which is justified because of the abolute convergence of the iterated integrals, which follows from Corollary 30.2.7, this becomes Z ∞Z ∞ 1 β−1 tα−1 (u − t) S (u) dudt Γ (α) Γ (β) 0 t =

1 Γ (α) Γ (β)

=

1 Γ (α) Γ (β)

=

1 Γ (α) Γ (β)

=

1 Γ (α) Γ (β)

=

Z

∞

Z

0

Z

u

β−1

tα−1 (u − t)

0 ∞

S (u)

Z

1

S (u) dtdu

α−1

β−1

(ux)

(u − ux) udxdu 0 0 µZ 1 ¶Z ∞ β−1 xα−1 (1 − x) dx S (u) uα+β−1 du 0

µZ

1

¶ β−1

xα−1 (1 − x)

0 −(α+β)

dx Γ (α + β) (−A)

0

−(α+β)

(−A)

This proves the first part of the theorem. Consider 30.2.9. Since A is a closed operator, and approximating the integral with an appropriate sequence of Riemann sums, (−A) can be taken inside the integral and so Z ∞ Z ∞ 1 (−A) t1−1 S (t) dt = (−A) S (t) dt Γ (1) 0 0 Z ∞ d = − (S (t)) dt = S (0) = I. dt 0 Next let x ∈ D (−A) . Then Z ∞ Z ∞ 1 1−1 t S (t) dt (−A) x = − S (t) Axdt Γ (1) 0 0 Z ∞ Z ∞ d =− AS (t) xdt = − (S (t)) dt = Ix dt 0 0

904


This shows that the integral in which α = 1 deserves to be called A−1 so the definition is not bad notation. Also, by assumption, A−1 is one to one. Thus (−A)

−1

−1

(−A)

implies

−1

(−A)

x=0

x=0

−2

−m

hence x = 0 so that (−A) is also one to one. Similarly, (−A) is one to one for all positive integers m. −α From what was just shown, if (−A) x = 0 for α ∈ (0, 1) , then −1

(−A)

−(1−α)

x = (−A)

(−A)

−α

x=0

−α

and so x = 0. This shows (−A) is one to one for all α ∈ [0, 1] if is defined as 0 (−A) ≡ I. What about α > 1? For such α, it is of the form m + β where β ∈ [0, 1) and m is a positive integer. Therefore, if −(m+β)

(−A) then (−A)

−β

³ (−A)

and so from what was just shown, ³ (−A)

−m

x=0

−m

´ x=0

´ x=0

−α

and now this implies x = 0 so that (−A) is one to one for all α ≥ 0. Consider 30.2.10. It was shown above that (−A) Let x = (−A)

−(α+β)

−β

(−A)

−(α+β)

= (−A)

y. Then

−α

x = (−A)

−α

−β

(−A)

−α

y ⊆ (−A)

(−A)

−β

−β

−α

(H) ⊆ (−A)

(H) .

−α

This proves 30.2.10. If α < β, (−A) (H) ⊆ (−A) (H) . −α Now consider the problem of writing (−A) for α ∈ (0, 1) in terms of A, not mentioning S (t) . By Proposition 15.12.5, Z ∞ −1 (λI − A) x = e−λt S (t) xdt 0

Then

Z

∞

Z −1

λ−α (λI − A)

dλ

∞

=

0

Z

Z0 ∞

Z0 ∞ =

∞

λ−α S (t)

Z

0

=

0

Z

∞

S (t) 0

0

∞

e−λt S (t) dtdλ λ−α e−λt dλdt λβ−1 e−λt dλdt


905

where β ≡ 1 − α. Then using Lemma 30.2.14, this equals Z ∞ Z ∞ Z ∞ Z S (t) µβ−1 t1−β e−µ t−1 dµdt = t−β S (t) 0

0

0

Z

∞

µβ−1 e−µ dµdt

0

∞

−α

tα−1 S (t) dt = Γ (α) Γ (1 − α) (−A) Γ (1 − α) 0 µZ 1 ¶ π −α −α −α α−1 = (−A) x (1 − x) dx (−A) = sin (πα) 0

=

and so this gives the formula −α

(−A)

Z

sin (πα) = π

∞

−1

λ−α (λI − A)

dλ.

0

This proves 30.2.11. α

α

−α

Definition 30.2.17 For α ≥ 0, define (−A) on D ((−A) ) ≡ (−A)

(H) by

³ ´−1 α −α (−A) ≡ (−A) ³ ´ α+β Note that if α, β > 0, then if x ∈ D (−A) , α+β

(−A)

³ ´−1 −(α+β) x = (−A) x=

³ ´−1 −α −β β α (−A) (−A) x = (−A) (−A) x. (30.2.12) ³ ´ β Next let β > α > 0 and let x ∈ D (−A) . Then from what was just shown, α

β−α

(−A) (−A) and so (−A)

β−α

β

x = (−A) x

x = (−A)

−α

β

(−A) x ³ ´ ³ ´ β −α β −β If x ∈ D (−A) , does it follow that (−A) x ∈ D (−A) ? Note x = (−A) y and so ³ ´ −α −α −β −(α+β) α+β (−A) x = (−A) (−A) y = (−A) y ∈ D (−A) . Therefore, from 30.2.12, β−α

(−A)

β−α

x = (−A)

(−A)

α

³

−α

(−A)

´ β −α x = (−A) (−A) x.

906

SOME IMPORTANT FUNCTIONAL ANALYSIS APPLICATIONS α

α

Theorem 30.2.18 The definition of (−A) is well defined and (−A) is densely defined and closed. Also for any α > 0, Cα 1 −δt e δ tα

α

||(−A) S (t)|| ≤

(30.2.13)

where −δ > −a. Furthermore, Cα is bounded as α → 0+ and is bounded on compact α intervals of (0, ∞). Also for α ∈ (0, 1) and x ∈ D ((−A) ) , ||(S (t) − I) x|| ≤

C1−α α α t ||(−A) x|| αδ

(30.2.14)

There exists a constant C independent of α ∈ [0, 1) such that for x ∈ D (A) and ε > 0, α ||(−A) x|| ≤ ε ||(−A) x|| + Cε−α/(1−α) ||x|| (30.2.15) There exists a constant C 0 independent of α ∈ [0, 1] such that for x ∈ D (A) , α

α

1−α

||(−A) x|| ≤ C 0 ||(−A) x|| ||x||

(30.2.16)

The formula 30.2.16 is called an interpolation inequality. α

Proof: It is obvious (−A) is densely defined because its domain is at least as large as D (A) which was assumed to be dense. It is a closed operator because if α xn ∈ D ((−A) ) and α xn → x, (−A) xn → y, then

−α

(−A)

xn → (−A)

−α

x, xn = (−A)

and so

−α

(−A) α

−α

α

(−A) xn → (−A)

−α

y

y=x

−α

α

showing x ∈ D ((−A) ) and y = (−A) x. Thus (−A) is closed and densely defined. Let −δ > −a where the sector for A was S−a,φ , a > 0. Then recall from Corollary 30.2.7 there is a constant, N such that ||(−A) S (t)|| ≤

N −δt e t

α

What about ||(−A) S (t)||? First note that for α ∈ [0, 1) this at least makes sense because S (t) maps into D (A). For any α > 0, −α

S (t) (−A) −α

follows from the definiton of (−A) α

= (−A)

−α

S (t)

. Therefore, −α

(−A) S (t) (−A)

= S (t) .

(30.2.17)


907 α

Note this implies that on D ((−A) ) , α

α

(−A) S (t) = S (t) (−A) . Also −1

(−A)

S (t) = S (t) (−A)

−1

= S (t) (−A)

−α

−(1−α)

(−A)

and so −α

S (t) = (−A) S (t) (−A)

−(1−α)

(−A)

From 30.2.17 it follows α

(−A) S (t)

=

α

−α

(−A) (−A) S (t) (−A)

(−A)

−(1−α)

−(1−α)

= (−A) S (t) (−A)

(30.2.18)

Then with this formula, α

||(−A) S (t)||

¯¯ ¯¯ ¯¯ −(1−α) ¯¯ = ¯¯(−A) S (t) (−A) ¯¯ ¯¯ ¯¯ Z ∞ ¯¯ ¯¯ 1 1−α ¯ ¯ s (−A) S (t + s) ds¯¯¯¯ = ¯¯ Γ (1 − α) 0

N ≤ Γ (1 − α)

= ≤ ≤ ≡

N Γ (1 − α)

Z

Z

∞

0

∞

s1−α −δ(s+t) e ds (t + s) 1−α

(u − t) e−δu ds u t ¶1−α Z ∞µ N 1 −δu t e ds 1− Γ (1 − α) t u uα Z ∞ N 1 N 1 −δt e−δu ds = e α Γ (1 − α) t t Γ (1 − α) δ tα Cα 1 −δt e . δ tα

this establishes the formula when α ∈ [0, 1). Next suppose α = m, a positive integer.

m

||A S (t)|| = =

¯¯ µ ¶m ¯¯ ¯¯ m ¯¯ ¯¯A S t ¯¯ ¯¯ ¯¯ m ¯¯µ µ ¶¶m ¯¯ ¯¯ ¯¯ ¯¯ ≤ N mm . ¯¯ AS t ¯¯ t m ¯¯ m

This is why the above inequality holds.

908


If α, β > 0, ¯¯ α+β ¯¯ ¯¯A S (t)¯¯

¯¯ µ ¶ µ ¶¯¯ ¯¯ t ¯¯¯¯ t S = ¯¯¯¯Aα+β S 2 2 ¯¯ ¯¯ µ ¶ µ ¶¯¯ ¯¯ t t ¯¯¯¯ = ¯¯¯¯Aα S Aβ S 2 2 ¯¯ Cα Cβ −2δt C ≤ e = α+β e−δt α β t t t

Suppose now that α > 0. Then α=m+β where β ∈ [0, 1). Then from what was just shown, ¯¯ m+β ¯¯ ¯¯A S (t)¯¯ ≤

C −δt e . tm+β

Next consider 30.2.14. First note that whenever α > 0, (−A)

−α

−α

S (s) = S (s) (−A)

α

and so on D ((−A) ) , α

−α

S (s) = (−A) S (s) (−A)

α

α

, S (s) (−A) = (−A) S (s)

α

Now for x ∈ D ((−A) ) ,

=

¯¯ Z t ¯¯ ¯¯ ¯¯ ¯¯− ¯¯ (−A) S (s) xds ¯¯ ¯¯ 0 ¯¯ Z t ¯¯ ¯¯ ¯¯ 1−α α ¯¯− ¯¯ (−A) (−A) S (s) xds ¯¯ ¯¯ 0 ¯¯ Z t ¯¯ ¯¯ ¯¯ 1−α α ¯¯− (−A) S (s) (−A) xds¯¯¯¯ ¯¯

≤

Z t ¯¯ ¯¯ ¯¯ ¯¯ 1−α α S (s)¯¯ ds ||(−A) x|| ¯¯(−A)

||(S (t) − I) x|| = =

0

0

Z

t

≤ 0

C1−α 1 −δs α e ds ||(−A) x|| δ s1−α ≤

C1−α 1 α α t ||(−A) x|| δ α

and this shows 30.2.14. Next consider 30.2.15. Let x ∈ H and β ∈ (0, 1) . Then ¯¯Z ¯¯ ¯¯ ¯¯ ¯¯ 1 ¯¯¯¯ ∞ β−1 ¯¯ −β ¯¯ t S (t) xdt¯¯¯¯ ¯¯(−A) x¯¯ = ¯ ¯ Γ (β) 0


909

¯¯Z ¯¯ Z ∞ ¯¯ 1 ¯¯¯¯ η β−1 β−1 = t S (t) xdt + t S (t) xdt¯¯¯¯ Γ (β) ¯¯ 0 η ¯¯Z ¯¯ Z η ¯¯ 1 ¯¯¯¯ ∞ β−1 1 ¯¯ tβ−1 ||S (t) x|| dt + t S (t) xdt ≤ ¯ ¯ ¯¯ Γ (β) 0 Γ (β) η ¯ ¯ ¯ ¯ Z ¯¯ 1 ¯¯¯¯ ∞ β−1 C ηβ ||x|| + ≤ t S (t) xdt¯¯¯¯ ¯ ¯ Γ (β) β Γ (β) η C ηβ ||x|| + Γ (β) β ¯¯ ¯¯ Z ∞ ¯¯ 1 ¯¯¯¯ β−1 −1 β−2 −1 η S (η) A x + (1 − β) t S (t) A xdt¯¯¯¯ ¯ ¯ Γ (β) η

≤

≤ =

µ

¯¯ ¯¯ ¯¯ ¯¯ Cη β ||x|| + η β−1 ¯¯A−1 x¯¯ + (1 − β) ¯¯A−1 x¯¯ β µ β ¶ ¯¯ ¯¯ Cη 1 ||x|| + 2η β−1 ¯¯A−1 x¯¯ . Γ (β) β 1 Γ (β)

Z

∞

β−2

t

¶ dt

η

1−β

Now let δ = Cη β so η = C −1/β δ 1/β and η β−1 = C β δ (β−1)/β . Thus for all x ∈ H, µ ¶ ¯¯ ¯¯ ¯¯ ¯¯ 1−β δ 1 ¯¯ −β ¯¯ (β−1)/β ¯¯ −1 ¯¯ β ||x|| + 2C δ A x . ¯¯(−A) x¯¯ ≤ Γ (β) β Let ε =

δ βΓ(β)

=

δ Γ(1+β) .

¯¯ ¯¯ ¯¯ −β ¯¯ ¯¯(−A) x¯¯

Then the above is of the form 1−β

C β (β−1)/β ¯¯¯¯ −1 ¯¯¯¯ (εΓ (1 + β)) A x Γ (β) 1−β (β−1)/β ¯¯¯¯ −1 ¯¯¯¯ ≤ ε ||x|| + 2C β (εΓ (1 + β)) A x

≤ ε ||x|| + 2

because Γ is decreasing on (0, 1) . I need to verify that for β ∈ (0, 1) , (β−1)/β

Γ (1 + β)

is bounded. It is continuous on (0, 1] and so if I can show limβ→0+ Γ (1 + β) exists, then it will follow the function is bounded. It suffices to show

(β−1)/β

ln Γ (1 + β) β−1 ln Γ (1 + β) = − lim β→0+ β→0+ β β lim

exists. Consider this. By L’Hospital’s rule and dominated convergence theorem, this is R∞ Z ∞ ln (t) tβ e−t dt lim 0 = lim ln (t) tβ e−t dt β→0+ β→0+ 0 Γ (1 + β) Z ∞ = lim ln (t) e−t dt. β→0+

0

910


Thus the function is bounded independent of β ∈ (0, 1) . This shows there is a constant C which is independent of β ∈ (0, 1) such that for any x ∈ H, ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ −β ¯¯ (30.2.19) ¯¯(−A) x¯¯ ≤ ε ||x|| + Cε(β−1)/β ¯¯A−1 x¯¯ . Now let y ∈ D (A) = D ((−A)) and let x = (−A) y. Then the above becomes ¯¯ ¯¯ ¯¯ ¯¯ −β ¯¯(−A) (−A) y ¯¯ ≤ ε ||(−A) y|| + Cε(β−1)/β ||y|| I claim that (−A)

−β

1−β

(−A) y = (−A)

y.

The reason for this is as follows. β

1−β

(−A) (−A)

y = (−A) y

and so the desired result follows from multiplying on the left by (−A) ¯¯ ¯¯ ¯¯ 1−β ¯¯ y ¯¯ ≤ ε ||(−A) y|| + Cε(β−1)/β ||y|| ¯¯(−A)

−β

. Hence

Now let 1 − β = α and obtain α

||(−A) y|| ≤ ε ||(−A) y|| + Cε−α/(1−α) ||y|| This proves 30.2.15. Finally choose ε to minimize the right side of the above expression. Thus let µ ¶1−α α ||y|| C ε= ||(−A) y|| (1 − α) Then the above expression becomes µ

α

||(−A) y||

≤

¶1−α α ||y|| C ||(−A) y|| (1 − α) Ãµ ¶1−α !−α/(1−α) α ||y|| C +C ||y|| ||(−A) y|| (1 − α) ||(−A) y||

µ

=

= ≤

¶1−α αC (1 − α) ¶−α µ αC α 1−α + ||(−A) y|| ||y|| (1 − α) Ãµ ¶1−α µ ¶−α ! αC αC α 1−α + ||(−A) y|| ||y|| (1 − α) (1 − α) α

1−α

||(−A) y|| ||y||

α

C 0 ||(−A) y|| ||y||

1−α


911

where C 0 does not depend on α ∈ (0, 1) . To see such a constant exists, note µ ¶1−α αC lim =1 α→1 (1 − α) and

µ lim

α→1

while

µ lim

α→0

αC (1 − α)

αC (1 − α)

¶−α =0

¶1−α

µ = 0, lim

α→0

αC (1 − α)

¶−α =1

Of course C 0 depends on C but as shown above, this did not depend on α ∈ (0, 1) . This proves 30.2.16. The following corollary follows from the proof of the above theorem. Corollary 30.2.19 Let α ∈ (0, 1) . Then for all ε > 0, there exists a constant C (α, ε) such that ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ −α ¯¯ −1 ¯¯ ¯¯(−A) x¯¯ ≤ ε ||x|| + C (ε, α) ¯¯(−A) x¯¯ Also if A−1 is compact, then so is (−A)

−α

for all α ∈ (0, 1).

Proof: The first part is done in the above theorem. Let S be a bounded set and nlet η > 0. Then let ε > 0 be small enough that for all x ∈ S, ε ||x|| < η/4. o −1 −1 −α Let (−A) xn be a η/ (2 + 2C (ε, α)) net for (−A) (S) . Then if (−A) x ∈ (−A)

−α

S, there exists xn such that ¯¯ ¯¯ ¯¯ −1 −1 ¯¯ ¯¯(−A) xn − (−A) x¯¯
0. −α

Proof: First suppose (−A) −α

Γ (α) (−A)

is compact for all α > 0. Then Z t Z ∞ = sα−1 S (s) ds + sα−1 S (s) ds 0

α

=

t S (t) − α

Z

t t

s AS (s) ds + sα−1 S (s) A−1 |∞ t α

Z 0∞

− (α − 1)

α

sα−2 S (s) A−1 ds

t

Now

¯¯ α ¯¯ α−1 ¯¯ s ¯¯ ¯¯ AS (s)¯¯ ≤ C s ¯¯ α ¯¯ α and so the second integral satisfies ¯¯Z t α ¯¯ α ¯¯ ¯¯ s ¯¯ ¯¯ ≤ C t AS (s) ds ¯¯ ¯¯ α2 0 α µ α¶ t tα −α Γ (α) (−A) = O + S (t) α2 α Z α−1 −1 −t A S (t) − (α − 1)

∞

sα−2 S (s) dsA−1

t

It follows that for t > 0, and ε > 0 given, µ α ¶−1 ³ t −α α−1 S (t) = −t Γ (α) (−A) α µ α ¶¶ Z ∞ t + (α − 1) sα−2 S (s) dsA−1 + O 2 α t ¶−1 ³ µ α t −α Γ (α) (−A) − tα−1 = α ¶ µ ¶ Z ∞ 1 α−2 −1 + (α − 1) s S (s) dsA +O α µ t¶ 1 = Nα + O . α

914


where Nα is a compact operator. Now let B be a bounded set in H, ≤¯ M for all ¯¯ ||x|| ¡ ¢¯ η . x ∈ B and let η > 0 be given. Then choose α large enough that ¯¯O α1 ¯¯ < 4+4M N

N

Then there exists a η/2 net, {Nα xn }n=1 for Nα (B) . Then consider {S (t) xn }n=1 . For x ∈ B, there exists xn such that ||Nα xn − Nα x|| < η/2. Then ||S (t) x − S (t) xn || ≤

≤

||S (t) x − Nα x|| + ||Nα x − Nα xn || + ||Nα xn − S (t) xn ||

η η η M+ + M 0 and so S (t) is compact. Next suppose S (t) is compact for all t > 0. Then Z ∞ 1 −α (−A) = tα−1 S (t) dt Γ (α) 0 and the integral is a limit in norm of Riemann sums of the form m X

tα−1 S (tk ) ∆tk k

k=1 −α

and each of these operators is compact. Since (−A) is the limit in norm of compact operators, it must also be compact. This proves the proposition. Here are some observations which are listed in the book by Henry [30]. Like the above proposition, these are exercises in this book. Observation 30.2.22 For each x ∈ H, t → tAS (t) is continuous and limt→0+ tAS (t) x = 0. The reason for this is that if x ∈ D (A) , then tAS (t) x = |tS (t) Ax| → 0 as t → 0. Now suppose y ∈ H is arbitrary. Then letting x ∈ D (A) , |tAS (t) y|

≤ |tAS (t) (y − x)| + |tAS (t) x| ≤ ε + |tAS (t) x|

provided x is close enough to y. The last term converges to 0 and so lim sup |tAS (t) y| ≤ ε t→0+

where ε > 0 is arbitrary. Thus lim |tAS (t) y| = 0.

t→0+


915

Why is t → tAS (t) x continuous on [0, T ]? This is true if x ∈ D (A) because t → tS (t) Ax is continuous. If y ∈ H is arbitrary, let xn converge to y in H where xn ∈ D (A) . Then |tAS (t) y − tAS (t) xn | ≤ C |y − xn | and so the convergence is uniform. Thus t → tAS (t) y is continuous because it is the uniform limit of a sequence of continuous functions. Observation 30.2.23 If x ∈ H and A is sectorial for S−a,φ , −a < 0, then for any α ∈ [0, 1] , α lim tα ||(−A) S (t) x|| = 0. t→0+

This follows as above because you can verify this is true for x ∈ D (A) and then use the fact shown above that α

tα ||(−A) S (t)|| ≤ C to extend it to x arbitrary.

30.2.5

A Scale Of Banach Spaces

Next I will present an important and interesting theorem which can be used to prove equivalence of certain norms. Theorem 30.2.24 Let A, B be sectorial for S−a,φ where −a < 0 and suppose D (A) = D (B) . Also suppose −α

(A − B) (−A)

, (A − B) (−B)

−α

are both bounded on D (A) for some α ∈ (0, 1). Then for all β ∈ [0, 1] , β

−β

(−A) (−B)

β

−β

, (−B) (−A) ³ ´ ³ ´ β β are both bounded on D (A) = D (B). Also D (−A) = D (−B) . −α

−α

Proof: First of all it is a good idea to verify (A − B) (−A) , (A − B) (−B) −α make sense on D (A) . If x ∈ D (A) , then why is (−A) x ∈ D (A)? Here is why. Since x ∈ D (A) , −1 x = (−A) y for some y ∈ H. Then −α

(−A)

x = (−A) −α

The case of (A − B) (−B)

−α

−1

(−A)

is similar.

y = (−A)

−1

−α

(−A)

y ∈ D (A) .

916


Next for β ∈ (0, 1) and λ > 0, use 30.2.16 to write ¯¯ ¯¯ ¯¯ β −1 ¯¯ ¯¯(−A) (λI − A) x¯¯ ¯¯ ¯¯β ¯¯ ¯¯1−β ¯¯ −1 ¯¯ ¯¯ −1 ¯¯ ≤ C ¯¯(−A) (λI − A) x¯¯ ¯¯(λI − A) x¯¯ ¯¯1−β ¯¯ ¯¯β ¯¯ ¯¯ −1 ¯¯ −1 ¯¯ ¯¯ ||x|| C ¯¯(−A) (λI − A) ¯¯ ¯¯(λI − A) ¯¯ ¯¯β ¯¯ M ¯¯ −1 ¯¯ C ¯¯I − λ (λI − A) ¯¯ ||x|| 1−β (λ + δ) µ ¶β λ M C ||x|| ≡ ||x|| C 1+ 1−β 1−β (λ + δ) (λ + δ) (λ + δ)

≤ ≤ ≤

(30.2.20)

where −a < −δ < 0 where C denotes a generic constant. Similarly, for all β ∈ (0, 1) , ¯¯ ¯¯ ¯¯ β −1 ¯¯ ¯¯(−B) (λI − B) x¯¯ ≤

C 1−β

(λ + δ)

||x||

(30.2.21)

Now from Theorem 30.2.16 and letting β ∈ (0, 1) , Z ´ sin (πβ) ∞ −β ³ −β −β −1 −1 (−B) − (−A) = λ (λI − B) − (λI − A) dλ π 0 Z sin (πβ) ∞ −β −1 −1 = λ (λI − B) (A − B) (λI − A) dλ. (30.2.22) π 0 Therefore, letting x ∈ D (A) and letting C denote a generic constant which can be changed from line to line and using 30.2.20 and 30.2.21, ¯¯ ¯¯ ¯¯ β −β ¯¯ ¯¯x − (−B) (−A) x¯¯ Z ∞ ¯¯ 1 ¯¯¯¯ β −1 −1 ¯¯ (−B) (λI − B) (A − B) (λI − A) x ≤ C ¯ ¯ ¯¯ dλ λβ 0 β

The reason (−B) goes inside the integral is that it is a closed operator. Then the above Z ∞ ¯¯ ¯¯ 1 ¯¯ −α α −1 ¯¯ ≤ C (A − B) (−A) (−A) (λI − A) x ¯ ¯ ¯¯ dλ 1−β λβ (λ + δ) 0 Z ∞ ¯¯ ¯¯ 1 ¯¯ α −1 ¯¯ ≤ C (−A) (λI − A) x ¯ ¯ ¯¯ dλ 1−β λβ (λ + δ) 0 Z ∞ 1 1 ≤ C 1−α dλ ||x|| = C ||x|| . 1−β β (λ + δ) λ (λ + δ) 0 β

−β

It follows (−B) (−A)

is bounded on D (A).


917

Next reverse A and B in 30.2.22. This yields Z sin (πβ) ∞ −β −β −β −1 −1 λ (λI − A) (B − A) (λI − B) dλ. (−A) − (−B) = π 0 Letting x ∈ D (A) , ¯¯ ¯¯ ¯¯ β −β ¯¯ ¯¯x − (−A) (−B) x¯¯ Z ∞ ¯¯ ¯¯ ¯¯ β −1 −1 ¯¯ ≤ C λ−β ¯¯(−A) (λI − A) (B − A) (λI − B) x¯¯ dλ 0

Z ≤

∞

C

λβ (λ + δ)

0

Z ≤

1−β

∞

C

1 β

λ (λ + δ)

0 β

¯¯ ¯¯ ¯¯ −α α −1 ¯¯ dλ ¯¯(B − A) (−B) (−B) (λI − B) x¯¯ (30.2.23)

1

1−β

1−α

(λ + δ)

dλ ||x|| = C ||x||

(30.2.24)

−β

This shows (−A) (−B) is bounded on D (A) = D (B) . Note the assertion these are bounded refers to the norm ³ on H. ´ ³ ´ β β It remains to verify D (−A) = D (−B) . Since D (A) is dense in H β

−β

there exists a unique L (A, B) ∈ L (H, H) such that L (A, B) = (−A) (−B) on D (A). Let L (B, A) be defined similarly as a continuous linear map which equals β −β (−B) (−A) on D (A) . Then (−A)

−β

(−B)

−β

L (A, B) =

−β

(−B)

−β

L (B, A) = (−A) ³ ´ ³ ´ β β The first of these equations shows D (−B) ⊆ D (−A) and the second turns the inclusion around. Thus they are equal as claimed. Next consider the case where β = 1. In this case (A − B) B −α is bounded on D (A) and so (A − B) B −α B −1+α is also bounded on D (A) . But this equals (A − B) B −1 . Thus AB −1 is bounded on D (A) . Similarly you can show (B − A) A−1 is bounded which implies BA−1 is bounded on D (A). This proves the theorem.

918


Definition 30.2.25 Let A be sectorial for the sector Sa,φ . Let b > a so that A − bI is sectorial for S−δ,φ where δ = b − a. Then for each α ∈ [0, 1] , define a norm on α D ((bI − A) ) ≡ Hα by α ||x||α ≡ ||(bI − A) x|| The {Hα }α∈[0,1] is called a scale of Banach spaces. Proposition 30.2.26 The Hα above are Banach spaces and they decrease in α. Furthermore, if bi > a for i = 1, 2 then the two norms associated with the bi are equivalent. Proof: That the Hα are decreasing was shown above in Theorem 30.2.16. They α are Banach spaces because (bI − A) is a closed mapping which is also one to one. It only remains to verify the claim about the equivalence of the norms. Let b2 > b1 > a. Then if α ∈ (0, 1) ,

=

((b1 I − A) − (b2 I − A)) (b2 I − A) −α (b1 − b2 ) (b2 I − A) ∈ L (H, H)

−α

and so by Theorem 30.2.24, for each β ∈ [0, 1] , ³ ´ ³ ´ β β D (b1 I − A) = D (b2 I − A) so the spaces, Hβ are the same for either choice of b > a. Also from this theorem, β

−β

(b1 I − A) (b2 I − A)

β

, (b2 I − A) (b1 I − A)

−β

are both bounded on D (A) . Therefore, for x ∈ Hβ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ β ¯¯ β −β β ¯¯ ¯¯(b1 I − A) x¯¯ = ¯¯(b1 I − A) (b2 I − A) (b2 I − A) x¯¯ ¯¯ ¯¯ ¯¯ β ¯¯ ≤ C ¯¯(b2 I − A) x¯¯ β

−β

Similarly using the boundedness of (b2 I − A) (b1 I − A) , it follows ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ β ¯¯ β ¯¯ ¯¯(b2 I − A) x¯¯ ≤ C 0 ¯¯(b1 I − A) x¯¯ Thus showing the two norms are equivalent. This proves the proposition.

Complex Mappings 31.1

Conformal Maps

If γ (t) = x (t) + iy (t) is a C 1 curve having values in U, an open set of C, and if f : U → C is analytic, consider f ◦ γ, another C 1 curve having values in C. 0 Also, γ 0 (t) and (f ◦ γ) (t) are complex numbers so these can be considered as 2 vectors in R as follows. The complex number, x + iy corresponds to the vector, (x, y) . Suppose that γ and η are two such C 1 curves having values in U and that γ (t0 ) = η (s0 ) = z and suppose that f : U → C is analytic. What can be said about 0 0 the angle between (f ◦ γ) (t0 ) and (f ◦ η) (s0 )? It turns out this angle is the same 0 0 as the angle between γ (t0 ) and η (s0 ) assuming that f 0 (z) 6= 0. To see this, note (x, y) · (a, b) = 12 (zw + zw) where z = x + iy and w = a + ib. Therefore, letting θ 0 0 be the cosine between the two vectors, (f ◦ γ) (t0 ) and (f ◦ η) (s0 ) , it follows from calculus that

=

cos θ 0 0 (f ◦ γ) (t0 ) · (f ◦ η) (s0 ) ¯ ¯ ¯ ¯ ¯(f ◦ η)0 (s0 )¯ ¯(f ◦ γ)0 (t0 )¯

=

1 f 0 (γ (t0 )) γ 0 (t0 ) f 0 (η (s0 ))η 0 (s0 ) + f 0 (γ (t0 )) γ 0 (t0 )f 0 (η (s0 )) η 0 (s0 ) 2 |f 0 (γ (t0 ))| |f 0 (η (s0 ))|

=

1 f 0 (z) f 0 (z)γ 0 (t0 ) η 0 (s0 ) + f 0 (z)f 0 (z) γ 0 (t0 )η 0 (s0 ) 2 |f 0 (z)| |f 0 (z)|

=

1 γ 0 (t0 ) η 0 (s0 ) + η 0 (s0 ) γ 0 (t0 ) 2 1

which equals the angle between the vectors, γ 0 (t0 ) and η 0 (t0 ) . Thus analytic mappings preserve angles at points where the derivative is nonzero. Such mappings are called isogonal. . Actually, they also preserve orientations. If z = x + iy and w = a + ib are two complex numbers, then (x, y, 0) and (a, b, 0) are two vectors in R3 . Recall that the cross product, (x, y, 0) × (a, b, 0) , yields a vector normal to the two given vectors such that the triple, (x, y, 0) , (a, b, 0) , and (x, y, 0) × (a, b, 0) satisfies the right hand 919

920

COMPLEX MAPPINGS

rule and has magnitude equal to the product of the sine of the included angle times the product of the two norms of the vectors. In this case, the cross product will produce a vector which is a multiple of k, the unit vector in the direction of the z axis. In fact, you can verify by computing both sides that, letting z = x + iy and w = a + ib, (x, y, 0) × (a, b, 0) = Re (ziw) k. Therefore, in the above situation, 0

0

(f ◦ γ) (t0 ) × (f ◦ η) (s0 ) ´ ³ = Re f 0 (γ (t0 )) γ 0 (t0 ) if 0 (η (s0 ))η 0 (s0 ) k ³ ´ 2 = |f 0 (z)| Re γ 0 (t0 ) iη 0 (s0 ) k which shows that the orientation of γ 0 (t0 ), η 0 (s0 ) is the same as the orientation of 0 0 (f ◦ γ) (t0 ) , (f ◦ η) (s0 ). Mappings which preserve both orientation and angles are called conformal mappings and this has shown that analytic functions are conformal mappings if the derivative does not vanish.

31.2

Fractional Linear Transformations

31.2.1

Circles And Lines

These mappings map lines and circles to either lines or circles. Definition 31.2.1 A fractional linear transformation is a function of the form f (z) =

az + b cz + d

(31.2.1)

where ad − bc 6= 0. Note that if c = 0, this reduces to a linear transformation (a/d) z +(b/d) . Special cases of these are defined as follows. dilations: z → δz, δ 6= 0, inversions: z →

1 , z

translations: z → z + ρ. The next lemma is the key to understanding fractional linear transformations. Lemma 31.2.2 The fractional linear transformation, 31.2.1 can be written as a finite composition of dilations, inversions, and translations. Proof: Let d 1 (bc − ad) S1 (z) = z + , S2 (z) = , S3 (z) = z c z c2

31.2. FRACTIONAL LINEAR TRANSFORMATIONS and S4 (z) = z +

921

a c

in the case where c 6= 0. Then f (z) given in 31.2.1 is of the form f (z) = S4 ◦ S3 ◦ S2 ◦ S1 . Here is why.

µ ¶ d 1 S2 (S1 (z)) = S2 z + ≡ c z+

d c

=

c . zc + d

Now consider µ S3

c zc + d

¶

(bc − ad) ≡ c2

µ

c zc + d

¶ =

bc − ad . c (zc + d)

Finally, consider µ S4

bc − ad c (zc + d)

¶ ≡

bc − ad a b + az + = . c (zc + d) c zc + d

In case that c = 0, f (z) = ad z + db which is a translation composed with a dilation. Because of the assumption that ad − bc 6= 0, it follows that since c = 0, both a and d 6= 0. This proves the lemma. This lemma implies the following corollary. Corollary 31.2.3 Fractional linear transformations map circles and lines to circles or lines. Proof: It is obvious that dilations and translations map circles to circles and lines to lines. What of inversions? If inversions have this property, the above lemma implies a general fractional linear transformation has this property as well. Note that all circles and lines may be put in the form ¡ ¢ ¡ ¢ α x2 + y 2 − 2ax − 2by = r2 − a2 + b2 where α = 1 gives a circle centered at (a, b) with radius r and α = 0 gives a line. In terms of complex variables you may therefore consider all possible circles and lines in the form αzz + βz + βz + γ = 0, (31.2.2) To see this let β = β 1 + iβ 2 where β 1 ≡ −a and β 2 ≡ b. Note that even if α is not 0 or 1 the expression still corresponds to either a circle or a line because you can divide by α if α 6= 0. Now I verify that replacing z with z1 results in an expression of the form in 31.2.2. Thus, let w = z1 where z satisfies 31.2.2. Then ¡ ¢ ¢ 1 ¡ α + βw + βw + γww = αzz + βz + βz + γ = 0 zz

922

COMPLEX MAPPINGS

and so w also satisfies a relation like 31.2.2. One simply switches α with γ and β with β. Note the situation is slightly different than with dilations and translations. In the case of an inversion, a circle becomes either a line or a circle and similarly, a line becomes either a circle or a line. This proves the corollary. The next example is quite important. Example 31.2.4 Consider the fractional linear transformation, w =

z−i z+i .

First consider what this mapping does to the points of the form z = x + i0. Substituting into the expression for w, w=

x2 − 1 − 2xi x−i = , x+i x2 + 1

a point on the unit circle. Thus this transformation maps the real axis to the unit circle. The upper half plane is composed of points of the form x + iy where y > 0. Substituting in to the transformation, w=

x + i (y − 1) , x + i (y + 1)

which is seen to be a point on the interior of the unit disk because |y − 1| < |y + 1| which implies |x + i (y + 1)| > |x + i (y − 1)|. Therefore, this transformation maps the upper half plane to the interior of the unit disk. One might wonder whether the mapping is one to one and onto. The mapping w+1 is clearly one to one because it has an inverse, z = −i w−1 for all w in the interior of the unit disk. Also, a short computation verifies that z so defined is in the upper half plane. Therefore, this transformation maps {z ∈ C such that Im z > 0} one to one and onto the unit disk {z ∈ C such that |z| < 1} . ¯ ¯ ¯ ¯ A fancy way to do part of this is to use Theorem 28.3.5. lim supz→a ¯ z−i z+i ¯ ≤ 1 ¯ ¯ ¯ ¯ whenever a is the real axis or ∞. Therefore, ¯ z−i z+i ¯ ≤ 1. This is a little shorter.

31.2.2

Three Points To Three Points

There is a simple procedure for determining fractional linear transformations which map a given set of three points to another set of three points. The problem is as follows: There are three distinct points in the extended complex plane, z1 , z2 , and z3 and it is desired to find a fractional linear transformation such that zi → wi for i = 1, 2, 3 where here w1 , w2 , and w3 are three distinct points in the extended complex plane. Then the procedure says that to find the desired fractional linear transformation solve the following equation for w. z − z1 z2 − z3 w − w1 w2 − w3 · = · w − w3 w2 − w1 z − z3 z2 − z1

31.3. RIEMANN MAPPING THEOREM

923

The result will be a fractional linear transformation with the desired properties. If any of the points equals ∞, then the quotient containing this point should be adjusted. Why should this procedure work? Here is a heuristic argument to indicate why you would expect this to happen rather than a rigorous proof. The reader may want to tighten the argument to give a proof. First suppose z = z1 . Then the right side equals zero and so the left side also must equal zero. However, this requires w = w1 . Next suppose z = z2 . Then the right side equals 1. To get a 1 on the left, you need w = w2 . Finally suppose z = z3 . Then the right side involves division by 0. To get the same bad behavior, on the left, you need w = w3 . Example 31.2.5 Let Im ξ > 0 and consider the fractional linear transformation which takes ξ to 0, ξ to ∞ and 0 to ξ/ξ, . The equation for w is z−ξ ξ−0 w−0 ¡ ¢= · z−0 ξ−ξ w − ξ/ξ After some computations, w=

z−ξ . z−ξ

is always a point on the unit circle because Note that this has the property that x−ξ x−ξ it is a complex number divided by its conjugate. Therefore, this fractional linear transformation maps the real line to the unit circle. It also takes the point, ξ to 0 and so it must map the upper half plane to the unit disk. You can verify the mapping is onto as well. Example 31.2.6 Let z1 = 0, z2 = 1, and z3 = 2 and let w1 = 0, w2 = i, and w3 = 2i. Then the equation to solve is w −i z −1 · = · w − 2i i z−2 1 Solving this yields w = iz which clearly works.

31.3

Riemann Mapping Theorem

From the open mapping theorem analytic functions map regions to other regions or else to single points. The Riemann mapping theorem states that for every simply connected region, Ω which is not equal to all of C there exists an analytic function, f such that f (Ω) = B (0, 1) and in addition to this, f is one to one. The proof involves several ideas which have been developed up to now. The proof is based on the following important theorem, a case of Montel’s theorem. Before, beginning,

924

COMPLEX MAPPINGS

note that the Riemann mapping theorem is a classic example of a major existence theorem. In mathematics there are two sorts of questions, those related to whether something exists and those involving methods for finding it. The real questions are often related to questions of existence. There is a long and involved history for proofs of this theorem. The first proofs were based on the Dirichlet principle and turned out to be incorrect, thanks to Weierstrass who pointed out the errors. For more on the history of this theorem, see Hille [32]. The following theorem is really wonderful. It is about the existence of a subsequence having certain salubrious properties. It is this wonderful result which will give the existence of the mapping desired. The other parts of the argument are technical details to set things up and use this theorem.

31.3.1

Montel’s Theorem

Theorem 31.3.1 Let Ω be an open set in C and let F denote a set of analytic functions mapping Ω to B (0, M ) ⊆ C. Then there exists a sequence of functions (k) ∞ from F, {fn }n=1 and an analytic function, f such that fn converges uniformly to f (k) on every compact subset of Ω. Proof: First note there exists a sequence of compact sets, Kn such that Kn ⊆ int Kn+1 ⊆ Ω for all n where here int K denotes the interior of the set K, the ∞ union of all open in¢ K and © sets contained ¡ ª ∪n=1 Kn = Ω. In fact, you can verify 1 C that B (0, n) ∩ z ∈ Ω : dist z, Ω ≤ n works for Kn . Then there exist positive numbers, δ n such that if z ∈ Kn , then B (z, δ n ) ⊆ int Kn+1 . Now denote by Fn the set of restrictions of functions of F to Kn . Then let z ∈ Kn and let γ (t) ≡ z + δ n eit , t ∈ [0, 2π] . It follows that for z1 ∈ B (z, δ n ) , and f ∈ F , ¯ µ ¶ ¯ Z ¯ 1 ¯ 1 1 |f (z) − f (z1 )| = ¯¯ f (w) − dw¯¯ 2πi γ w−z w − z1 ¯Z ¯ ¯ 1 ¯¯ z − z1 dw¯ ≤ f (w) 2π ¯ (w − z) (w − z1 ) ¯ γ

Letting |z1 − z|
0 such that for all z ∈ Ω, ¯ z − ξ + a¯ > r > 0. Then consider ψ (z) ≡ √

r . z−ξ+a

(31.3.7)

¯√ ¯ This is one to one, analytic, and maps Ω into B (0, 1) (¯ z − ξ + a¯ > r). Thus F is not empty and this proves the claim. Claim 2: Let z0 ∈ Ω. There exists a finite positive real number, η, defined by ¯ ©¯ ª η ≡ sup ¯ψ 0 (z0 )¯ : ψ ∈ F (31.3.8) and an analytic function, h ∈ F such that |h0 (z0 )| = η. Furthermore, h (z0 ) = 0. Proof of Claim 2: First you show η < ∞. Let γ (t) = z0 + reit for t ∈ [0, 2π] and r is small enough that B (z0 , r) ⊆ Ω. Then for ψ ∈ F, the Cauchy integral formula for the derivative implies Z 1 ψ (w) ψ 0 (z0 ) = dw 2πi γ (w − z0 )2 ¯ ¯ ¡ ¢ and so ¯ψ 0 (z0 )¯ ≤ (1/2π) 2πr 1/r2 = 1/r. Therefore, η < ∞ as desired. For ψ defined above in 31.3.7 ¢−1 ¡√ −r (1/2) z0 − ξ −rφ0 (z0 ) 0 ψ (z0 ) = 6= 0. 2 = 2 (φ (z0 ) + a) (φ (z0 ) + a)

928

COMPLEX MAPPINGS

Therefore, η > 0. It remains to verify the existence of the function, h. By Theorem 31.3.1, there exists a sequence, {ψ n }, of functions in F and an analytic function, h, such that ¯ 0 ¯ ¯ψ n (z0 )¯ → η (31.3.9) and ψ n → h, ψ 0n → h0 ,

(31.3.10)

uniformly on all compact subsets of Ω. It follows ¯ ¯ |h0 (z0 )| = lim ¯ψ 0n (z0 )¯ = η

(31.3.11)

n→∞

and for all z ∈ Ω, |h (z)| = lim |ψ n (z)| ≤ 1.

(31.3.12)

n→∞

By 31.3.11, h is not a constant. Therefore, in fact, |h (z)| < 1 for all z ∈ Ω in 31.3.12 by the open mapping theorem. Next it must be shown that h is one to one in order to conclude h ∈ F. Pick z1 ∈ Ω and suppose z2 is another point of Ω. Since the zeros of h − h (z1 ) have no limit point, there exists a circular contour bounding a circle which contains z2 but not z1 such that γ ∗ contains no zeros of h − h (z1 ). ¾ γ t z1

?

t z2

6

Using the theorem on counting zeros, Theorem 28.6.1, and the fact that ψ n is one to one, 0

= =

Z 1 ψ 0n (w) dw n→∞ 2πi γ ψ n (w) − ψ n (z1 ) Z 1 h0 (w) dw, 2πi γ h (w) − h (z1 ) lim

which shows that h − h (z1 ) has no zeros in B (z2 , r) . In particular z2 is not a zero of h − h (z1 ) . This shows that h is one to one since z2 6= z1 was arbitrary. Therefore, h ∈ F. It only remains to verify that h (z0 ) = 0. If h (z0 ) 6= 0,consider φh(z0 ) ◦ h where φα is the fractional linear transformation defined in Lemma 31.3.3. By this lemma it follows φh(z0 ) ◦ h ∈ F. Now using the

31.3. RIEMANN MAPPING THEOREM

929

chain rule, ¯³ ¯ ´0 ¯ ¯ ¯ φh(z ) ◦ h (z0 )¯ 0 ¯ ¯

= = =

¯ ¯ ¯ 0 ¯ ¯φh(z0 ) (h (z0 ))¯ |h0 (z0 )| ¯ ¯ ¯ ¯ 1 ¯ ¯ 0 ¯ ¯ |h (z0 )| ¯ 1 − |h (z0 )|2 ¯ ¯ ¯ ¯ ¯ 1 ¯ ¯ ¯ ¯η > η 2 ¯ 1 − |h (z0 )| ¯

Contradicting the definition of η. This proves Claim 2. Claim 3: The function, h just obtained maps Ω onto B (0, 1). Proof of Claim 3: To show h is onto, use the fractional linear transformation of Lemma 31.3.3. Suppose h is not onto. Then there exists α ∈ B (0, 1) \ h (Ω) . Then 0 6= φα ◦ h (z) for all z ∈ Ω because φα ◦ h (z) =

h (z) − α 1 − αh (z)

and it is assumed α ∈ / h (Ω) . Therefore, p since Ω has the square root property, you can consider an analytic function z → φα ◦ h (z). This function is one to one because both φα and h are. Also, the values of this function are in B (0, 1) by Lemma 31.3.3 so it is in F. Now let p ◦ φ ◦ h. ψ ≡ φ√ (31.3.13) α

φα ◦h(z0 )

Thus

ψ (z0 ) = φ√φ

α ◦h(z0 )

◦

p

φα ◦ h (z0 ) = 0

and ψ is a one to one mapping of Ω into B (0, 1) so ψ is also in F. Therefore, ¯ ¯³ ´0 ¯ 0 ¯ ¯ ¯ p ¯ψ (z0 )¯ ≤ η, ¯ φα ◦ h (z0 )¯¯ ≤ η. (31.3.14) ¯ Define s (w) ≡ w2 . Then using Lemma 31.3.3, in particular, the description of φ−1 α = φ−α , you can solve 31.3.13 for h to obtain h (z)

= φ−α ◦ s ◦ φ−√φ ◦h(z ) ◦ ψ 0 α   ≡F z }| { = φ−α ◦ s ◦ φ−√φ ◦h(z ) ◦ ψ  (z) α

= (F ◦ ψ) (z)

0

(31.3.15)

Now F (0) = φ−α ◦ s ◦ φ−√φ

α ◦h(z0 )

(0) = φ−1 α (φα ◦ h (z0 )) = h (z0 ) = 0

930

COMPLEX MAPPINGS

and F maps B (0, 1) into B (0, 1). Also, F is not one to one because it maps B (0, 1) to B (0, 1) and has s in its definition. Thus there exists z1 ∈ B (0, 1) such that φ−√φ ◦h(z ) (z1 ) = − 21 and another point z2 ∈ B (0, 1) such that φ−√φ ◦h(z ) (z2 ) = 1 2.

α

0

α

0

However, thanks to s, F (z1 ) = F (z2 ). Since F (0) = h (z0 ) = 0, you can apply the Schwarz lemma to F . Since F is not one to one, it can’t be true that F (z) = λz for |λ| = 1 and so by the Schwarz lemma it must be the case that |F 0 (0)| < 1. But this implies from 31.3.15 and 31.3.14 that ¯ ¯ η = |h0 (z0 )| = |F 0 (ψ (z0 ))| ¯ψ 0 (z0 )¯ ¯ ¯ ¯ ¯ = |F 0 (0)| ¯ψ 0 (z0 )¯ < ¯ψ 0 (z0 )¯ ≤ η, a contradiction. This proves the theorem. The following lemma yields the usual form of the Riemann mapping theorem. Lemma 31.3.7 Let Ω be a simply connected region with Ω 6= C. Then Ω has the square root property. 0

both be analytic on Ω. Then ff is analytic on Ω so by 0 Corollary 27.7.23, there exists Fe, analytic on Ω such that Fe0 = ff on Ω. Then ³ ´0 e e e f e−F = 0 and so f (z) = CeF = ea+ib eF . Now let F = Fe + a + ib. Then F is Proof: Let f and

1 f

1

still a primitive of f 0 /f and f (z) = eF (z) . Now let φ (z) ≡ e 2 F (z) . Then φ is the desired square root and so Ω has the square root property. Corollary 31.3.8 (Riemann mapping theorem) Let Ω be a simply connected region with Ω 6= C and let z0 ∈ Ω. Then there exists a function, f : Ω → B (0, 1) such that f is one to one, analytic, and onto with f (z0 ) = 0. Furthermore, f −1 is also analytic. Proof: From Theorem 31.3.6 and Lemma 31.3.7 there exists a function, f : Ω → B (0, 1) which is one to one, onto, and analytic such that f (z0 ) = 0. The assertion that f −1 is analytic follows from the open mapping theorem.

31.4

Analytic Continuation

31.4.1

Regular And Singular Points

Given a function which is analytic on some set, can you extend it to an analytic function defined on a larger set? Sometimes you can do this. It was done in the proof of the Cauchy integral formula. There are also reflection theorems like those discussed in the exercises starting with Problem 10 on Page 828. Here I will give a systematic way of extending an analytic function to a larger set. I will emphasize simply connected regions. The subject of analytic continuation is much larger than the introduction given here. A good source for much more on this is found in Alfors

31.4. ANALYTIC CONTINUATION

931

[2]. The approach given here is suggested by Rudin [58] and avoids many of the standard technicalities. Definition 31.4.1 Let f be analytic on B (a, r) and let β ∈ ∂B (a, r) . Then β is called a regular point of f if there exists some δ > 0 and a function, g analytic on B (β, δ) such that g = f on B (β, δ) ∩ B (a, r) . Those points of ∂B (a, r) which are not regular are called singular.

rβ r a

Theorem 31.4.2 Suppose f is analytic on B (a, r) and the power series f (z) =

∞ X

k

ak (z − a)

k=0

has radius of convergence r. Then there exists a singular point on ∂B (a, r). Proof: If not, then for every z ∈ ∂B (a, r) there exists δ z > 0 and gz analytic on B (z, δ z ) such that gz = f on B (z, δ z ) ∩ B (a, r) . Since ∂B (a, r) is compact, there n exist z1 , · · · , zn , points in ∂B (a, r) such that {B (zk , δ zk )}k=1 covers ∂B (a, r) . Now define ½ f (z) if z ∈ B (a, r) g (z) ≡ gzk (z) if z ∈ B (zk , δ zk ) ¡ ¢ Is this well defined? If z ∈ B (zi , δ zi ) ∩ B zj , δ zj , is gzi (z) = gzj (z)? Consider the following picture representing this situation.

¡ ¢ ¡ ¢ You see that if z ∈ B (zi , δ zi ) ∩ B zj , δ zj then I ≡ B (zi , δ zi ) ∩ B zj , δ zj ∩ B (a, r) is a nonempty open set. ¡Both g¢zi and gzj equal f on I. Therefore, they must be equal on B (zi , δ zi ) ∩ B zj , δ zj because I has a limit point. Therefore, g is well defined and analytic on an open set containing B (a, r). Since g agrees

932

COMPLEX MAPPINGS

with f on B (a, r) , the power series for g is the same as the power series for f and converges on a ball which is larger than B (a, r) contrary to the assumption that the radius of convergence of the above power series equals r. This proves the theorem.

31.4.2

Continuation Along A Curve

Next I will describe what is meant by continuation along a curve. The following definition is standard and is found in Rudin [58]. Definition 31.4.3 A function element is an ordered pair, (f, D) where D is an open ball and f is analytic on D. (f0 , D0 ) and (f1 , D1 ) are direct continuations of each other if D1 ∩D0 6= ∅ and f0 = f1 on D1 ∩D0 . In this case I will write (f0 , D0 ) ∼ (f1 , D1 ) . A chain is a finite sequence, of disks, {D0 , · · · , Dn } such that Di−1 ∩Di 6= ∅. If (f0 , D0 ) is a given function element and there exist function elements, (fi , Di ) such that {D0 , · · · , Dn } is a chain and (fj−1 , Dj−1 ) ∼ (fj , Dj ) then (fn , Dn ) is called the analytic continuation of (f0 , D0 ) along the chain {D0 , · · · , Dn }. Now suppose γ is an oriented curve with parameter interval [a, b] and there exists a chain, {D0 , · · · , Dn } such that γ ∗ ⊆ ∪nk=1 Dk , γ (a) is the center of D0 , γ (b) is the center of Dn , and there is an increasing list of numbers in [a, b] , a = s0 < s1 · · · < sn = b such that γ ([si , si+1 ]) ⊆ Di and (fn , Dn ) is an analytic continuation of (f0 , D0 ) along the chain. Then (fn , Dn ) is called an analytic continuation of (f0 , D0 ) along the curve γ. (γ will always be a continuous curve. Nothing more is needed. ) In the above situation it does not follow that if Dn ∩ D0 6= ∅, that fn = f0 ! However, there are some cases where this will happen. This is the monodromy theorem which follows. This is as far as I will go on the subject of analytic continuation. For more on this subject including a development of the concept of Riemann surfaces, see Alfors [2]. Lemma 31.4.4 Suppose (f, B (0, r)) for r < 1 is a function element and (f, B (0, r)) can be analytically continued along every curve in B (0, 1) that starts at 0. Then there exists an analytic function, g defined on B (0, 1) such that g = f on B (0, r) . Proof: Let R

=

sup{r1 ≥ r such that there exists gr1 analytic on B (0, r1 ) which agrees with f on B (0, r) .}

Define gR (z) ≡ gr1 (z) where |z| < r1 . This is well defined because if you use r1 and r2 , both gr1 and gr2 agree with f on B (0, r), a set with a limit point and so the two functions agree at every point in both B (0, r1 ) and B (0, r2 ). Thus gR is analytic on B (0, R) . If R < 1, then by the assumption there are no singular points on B (0, R) and so Theorem 31.4.2 implies the radius of convergence of the power series for gR is larger than R contradicting the choice of R. Therefore, R = 1 and this proves the lemma. Let g = gR . The following theorem is the main result in this subject, the monodromy theorem.

31.5. THE PICARD THEOREMS

933

Theorem 31.4.5 Let Ω be a simply connected proper subset of C and suppose (f, B (a, r)) is a function element with B (a, r) ⊆ Ω. Suppose also that this function element can be analytically continued along every curve through a. Then there exists G analytic on Ω such that G agrees with f on B (a, r). Proof: By the Riemann mapping theorem, there exists h : Ω → B (0, 1) which is analytic, one to one and onto such that f (a) = 0. Since h is an open map, there exists δ > 0 such that B (0, δ) ⊆ h (B (a, r)) . It follows f ◦ h−1 can be analytically continued along every curve through 0. By Lemma 31.4.4 there exists g analytic on B (0, 1) which agrees¡ with f¢◦ h−1 on B (0, δ). Define G (z) ≡¡g (h (z))¢. For z = h−1 (w) , it follows G h−1 (w) = g (w) . If w ∈ B (0, δ) , then G h−1 (w) = f ◦ h−1 (w) and so G = f on h−1 (B (0, δ)) , an open set contained in B (a, r). Therefore, G = f on B (a, r) because h−1 (B (0, δ)) has a limit point. This proves the theorem. Actually, you sometimes want to consider the case where Ω = C. This requires a small modification to obtain from the above theorem. Corollary 31.4.6 Suppose (f, B (a, r)) is a function element with B (a, r) ⊆ C. Suppose also that this function element can be analytically continued along every curve through a. Then there exists G analytic on C such that G agrees with f on B (a, r). Proof: Let Ω1 ≡ {z ∈ C : a + it : t > a} and Ω2 ≡ {z ∈ C : a − it : t > a} . Here is a picture of Ω1 . Ω1 r a A picture of Ω2 is similar except the line extends down from the boundary of B (a, r). Thus B (a, r) ⊆ Ωi and Ωi is simply connected and proper. By Theorem 31.4.5 there exist analytic functions, Gi analytic on Ωi such that Gi = f on B (a, r). Thus G1 = G2 on B (a, r) , a set with a limit point. Therefore, G1 = G2 on Ω1 ∩ Ω2 . Now let G (z) = Gi (z) where z ∈ Ωi . This is well defined and analytic on C. This proves the corollary.

31.5

The Picard Theorems

The Picard theorem says that if f is an entire function and there are two complex numbers not contained in f (C) , then f is constant. This is certainly one of the most amazing things which could be imagined. However, this is only the little

934

COMPLEX MAPPINGS

Picard theorem. The big Picard theorem is even more incredible. This one asserts that to be non constant the entire function must take every value of C but two infinitely many times! I will begin with the little Picard theorem. The method of proof I will use is the one found in Saks and Zygmund [60], Conway [16] and Hille [32]. This is not the way Picard did it in 1879. That approach is very different and is presented at the end of the material on elliptic functions. This approach is much more recent dating it appears from around 1924. Lemma 31.5.1 Let f be analytic on a region containing B (0, r) and suppose |f 0 (0)| = b > 0, f (0) = 0,

³ 2 2´ b and |f (z)| ≤ M for all z ∈ B (0, r). Then f (B (0, r)) ⊇ B 0, r6M . Proof: By assumption, ∞ X

f (z) =

ak z k , |z| ≤ r.

(31.5.16)

k=0

Then by the Cauchy integral formula for the derivative, Z 1 f (w) ak = dw 2πi ∂B(0,r) wk+1 where the integral is in the counter clockwise direction. Therefore, Z 2π ¯¯ ¡ iθ ¢¯¯ f re 1 M |ak | ≤ dθ ≤ k . k 2π 0 r r In particular, br ≤ M . Therefore, from 31.5.16 ³ |f (z)|

≥ b |z| −

∞ X M k=2

rk

k

|z| = b |z| −

M

|z| r

1−

´2

|z| r

2

= b |z| − Suppose |z| =

r2 b 4M

M |z| − r |z|

r2

< r. Then this is no larger than

1 2 2 3M − br 1 3M − M r 2 b2 b r ≥ b2 r 2 = . 4 M (4M − br) 4 M (4M − M ) 6M Let |w|
0. Then there exists a ∈ B (z0 , R) such that µ ¶ |f 0 (z0 )| R f (B (z0 , R)) ⊇ B f (a) , . 24 Proof: You look at g (z) ≡ f (z0 + z) − f (z0 ) for z ∈ B (0, R) . Then g 0 (0) = f (z0 ) and so by Lemma 31.5.2 there exists a1 ∈ B (0, R) such that µ ¶ |f 0 (z0 )| R g (B (0, R)) ⊇ B g (a1 ) , . 24 0

Now g (B (0, R)) = f (B (z0 , R)) − f (z0 ) and g (a1 ) = f (a) − f (z0 ) for some a ∈ B (z0 , R) and so µ ¶ |f 0 (z0 )| R f (B (z0 , R)) − f (z0 ) ⊇ B g (a1 ) , 24 µ ¶ |f 0 (z0 )| R = B f (a) − f (z0 ) , 24

31.5. THE PICARD THEOREMS which implies

937

¶ µ |f 0 (z0 )| R f (B (z0 , R)) ⊇ B f (a) , 24

as claimed. This proves the lemma. No attempt was made to find the best number to multiply by R |f 0 (z0 )|. A discussion of this is given in Conway [16]. See also [32]. Much larger numbers than 1/24 are available and there is a conjecture due to Alfors about the best value. The conjecture is that 1/24 can be replaced with ¡ ¢ ¡ ¢ Γ 13 Γ 11 12 √ ¢1/2 ¡ 1 ¢ ≈ . 471 86 ¡ 1+ 3 Γ 4 You can see there is quite a gap between the constant for which this lemma is proved above and what is thought to be the best constant. Bloch’s lemma above gives the existence of a ball of a certain size inside the image of a ball. By contrast the next lemma leads to conditions under which the values of a function do not contain a ball of certain radius. It concerns analytic functions which do not achieve the values 0 and 1. Lemma 31.5.4 Let F denote the set of functions, f defined on Ω, a simply connected region which do not achieve the values 0 and 1. Then for each such function, it is possible to define a function analytic on Ω, H (z) by the formula "r # r log (f (z)) log (f (z)) H (z) ≡ log − −1 . 2πi 2πi There exists a constant C independent of f ∈ F such that H (Ω) does not contain any ball of radius C. Proof: Let f ∈ F . Then since f does not take the value 0, there exists g1 a primitive of f 0 /f . Thus d ¡ −g1 ¢ e f =0 dz so there exists a, b such that f (z) e−g1 (z) = ea+bi . Letting g (z) = g1 (z) + a + ib, it follows eg(z) = f (z). Let log (f (z)) = g (z). Then for n ∈ Z, the integers, log (f (z)) log (f (z)) , − 1 6= n 2πi 2πi (z)) because if equality held, then f (z) = 1 which does not happen. It follows log(f 2πi (z)) and log(f − 1 are never equal to zero. Therefore, using the same reasoning, you 2πi can define a logarithm of these two quantities and therefore, a square root. Hence there exists a function analytic on Ω, r r log (f (z)) log (f (z)) − − 1. (31.5.20) 2πi 2πi

938

COMPLEX MAPPINGS

√ √ For n a positive integer, this function cannot equal n ± n − 1 because if it did, then Ãr ! r √ √ log (f (z)) log (f (z)) − −1 = n± n−1 (31.5.21) 2πi 2πi and you could take reciprocals of both sides to obtain Ãr ! r √ √ log (f (z)) log (f (z)) + − 1 = n ∓ n − 1. 2πi 2πi Then adding 31.5.21 and 31.5.22 r 2

(31.5.22)

√ log (f (z)) =2 n 2πi

(z)) which contradicts the above observation that log(f is not equal to an integer. 2πi Also, the function of 31.5.20 is never equal to zero. Therefore, you can define the logarithm of this function also. It follows Ãr ! r √ ¢ ¡√ log (f (z)) log (f (z)) H (z) ≡ log − − 1 6= ln n ± n − 1 + 2mπi 2πi 2πi

where m is an arbitrary integer and n is a positive integer. Now √ ¡√ ¢ lim ln n + n − 1 = ∞ n→∞

¡√

¢ √ and limn→∞ ln n − n − 1 = −∞ and so C is covered by rectangles having ¢ ¡√ √ vertices at points ln n ± n − 1 + 2mπi as described above. Each of these rectangles has height equal to 2π and a short computation shows their widths are bounded. Therefore, there exists C independent of f ∈ F such that C is larger than the diameter of all these rectangles. Hence H (Ω) cannot contain any ball of radius larger than C.

31.5.2

The Little Picard Theorem

Now here is the little Picard theorem. It is easy to prove from the above. Theorem 31.5.5 If h is an entire function which omits two values then h is a constant. Proof: Suppose the two values omitted are a and b and that h is not constant. Let f (z) = (h (z) − a) / (b − a). Then f omits the two values 0 and 1. Let H be defined in Lemma 31.5.4. Then H (z) is clearly ¡√ not √ of the¢ form az +b because then it would have values equal to the vertices ln n ± n − 1 +2mπi or else be constant neither of which happen if h is not constant. Therefore, by Liouville’s theorem, H 0 must be unbounded. Pick ξ such that |H 0 (ξ)| > 24C where C is such that H (C)


939

contains no balls of radius larger than C. But by Lemma 31.5.3 H (B (ξ, 1)) must |H 0 (ξ)| contain a ball of radius 24 > 24C 24 = C, a contradiction. This proves Picard’s theorem. The following is another formulation of this theorem. Corollary 31.5.6 If f is a meromophic function defined on C which omits three distinct values, a, b, c, then f is a constant. b−c Proof: Let φ (z) ≡ z−a z−c b−a . Then φ (c) = ∞, φ (a) = 0, and φ (b) = 1. Now consider the function, h = φ ◦ f. Then h misses the three points ∞, 0, and 1. Since h is meromorphic and does not have ∞ in its values, it must actually be analytic. Thus h is an entire function which misses the two values 0 and 1. Therefore, h is constant by Theorem 31.5.5.

31.5.3

Schottky’s Theorem

Lemma 31.5.7 Let f be analytic on an open set containing B (0, R) and suppose that f does not take on either of the two values 0 or 1. Also suppose |f (0)| ≤ β. Then letting θ ∈ (0, 1) , it follows |f (z)| ≤ M (β, θ) for all z ∈ B (0, θR) , where M (β, θ) is a function of only the two variables β, θ. (In particular, there is no dependence on R.) Proof: Consider the function, H (z) used in Lemma 31.5.4 given by Ãr ! r log (f (z)) log (f (z)) H (z) ≡ log − −1 . (31.5.23) 2πi 2πi You notice there are two explicit uses of logarithms. Consider first the logarithm inside the radicals. Choose this logarithm such that log (f (0)) = ln |f (0)| + i arg (f (0)) , arg (f (0)) ∈ (−π, π].

(31.5.24)

You can do this because elog(f (0)) = f (0) = eln|f (0)| eiα = eln|f (0)|+iα and by replacing α with α + 2mπ for a suitable integer, m it follows the above equation still holds. Therefore, you can assume 31.5.24. Similar reasoning applies to the logarithm on the outside of the parenthesis. It can be assumed H (0) equals ¯r ¯ ! Ãr r ¯ log (f (0)) r log (f (0)) ¯ log (f (0)) log (f (0)) ¯ ¯ ln ¯ − − 1¯ + i arg − −1 ¯ ¯ 2πi 2πi 2πi 2πi (31.5.25)

940

COMPLEX MAPPINGS

where the imaginary part is no larger than π in absolute value. Now if ξ ∈ B (0, R) is a point where H 0 (ξ) 6= 0, then by Lemma 31.5.2 µ ¶ |H 0 (ξ)| (R − |ξ|) H (B (ξ, R − |ξ|)) ⊇ B H (a) , 24 where a is some point in B (ξ, R − |ξ|). But by Lemma 31.5.4 H (B (ξ, R − |ξ|)) contains no balls of radius C where ¡C√depended only ¢ on the maximum diameters of √ those rectangles having vertices ln n ± n − 1 + 2mπi for n a positive integer and m an integer. Therefore, |H 0 (ξ)| (R − |ξ|) 0 there exists a δ > 0 such that whenever x, y ∈ K with |x − y| < δ and f ∈ A, d (f (x) , f (y)) < ε. The set, A is said to be uniformly bounded if for some M < ∞, and a ∈ X, f (x) ∈ B (a, M ) for all f ∈ A and x ∈ K. The Ascoli Arzela theorem follows. Theorem 31.5.15 Suppose K is a nonempty compact subset of Rn and A ⊆ C (K, X) , is uniformly bounded and uniformly equicontinuous where X is a locally compact complete metric space. Then if {fk } ⊆ A, there exists a function, f ∈ C (K, X) and a subsequence, fkl such that lim ρK (fkl , f ) = 0.

l→∞

b with the metric defined above. In the cases of interest here, X = C

31.5.5

Montel’s Theorem

The following lemma is another version of Montel’s theorem. It is this which will make possible a proof of the big Picard theorem. Lemma 31.5.16 Let Ω be a region and let F be a set of functions analytic on Ω none of which achieve the two distinct values, a and b. If {fn } ⊆ F then one of the following hold: Either there exists a function, f analytic on Ω and a subsequence, {fnk } such that for any compact subset, K of Ω, lim ||fnk − f ||K,∞ = 0.

k→∞

(31.5.29)

or there exists a subsequence {fnk } such that for all compact subsets K, lim ρK (fnk , ∞) = 0.

k→∞

(31.5.30)

946

COMPLEX MAPPINGS

Proof: Let B (z0 , 2R) ⊆ Ω. There are two cases to consider. The first case is that there exists a subsequence, nk such that {fnk (z0 )} is bounded. The second case is that limn→∞ |fnk (z0 )| = ∞. Consider the first case. By Theorem 31.5.8 {fnk (z)} is uniformly bounded on B (z0 , R) because by and letting θ = 1/2 applied to B (z0 , 2R) , it fol¡ this theorem, ¢ lows |fnk (z)| ≤ M a, b, 21 , β where β is an upper bound to the numbers, |fnk©(z0 )|. ª The Cauchy integral formula implies the existence of a uniform bound on the fn0 k which implies the functions are equicontinuous and uniformly bounded. Therefore, by the Ascoli Arzela theorem there exists a further subsequence which converges uniformly on B (z0 , R) to a function, f analytic on B (z0 , R). Thus denoting this subsequence by {fnk } to save on notation, lim ||fnk − f ||B(z0 ,R),∞ = 0.

k→∞

(31.5.31)

Consider the second case. In this case, it follows {1/fn (z0 )} is bounded on B (z0 , R) and so by the same argument just given {1/fn (z)} is uniformly bounded on B (z0 , R). Therefore, a subsequence converges uniformly on B (z0 , R). But {1/fn (z)} converges to 0 and so this requires that {1/fn (z)} must converge uniformly to 0. Therefore, lim ρB(z0 ,R) (fnk , ∞) = 0. (31.5.32) k→∞

Now let {Dk } denote a countable set of closed balls, Dk = B (zk , Rk ) such that B (zk , 2Rk ) ⊆ Ω and ∪∞ k=1 int (Dk ) = Ω. Using a Cantor diagonal process, there exists a subsequence, {fnk } of {fn } such that for each Dj , one of the above two alternatives holds. That is, either lim ||fnk − gj ||Dj ,∞ = 0

(31.5.33)

lim ρDj (fnk , ∞) .

(31.5.34)

k→∞

or, k→∞

Let A = {∪ int (Dj ) : 31.5.33 holds} , B = {∪ int (Dj ) : 31.5.34 holds} . Note that the balls whose union is A cannot intersect any of the balls whose union is B. Therefore, one of A or B must be empty since otherwise, Ω would not be connected. If K is any compact subset of Ω, it follows K must be a subset of some finite collection of the Dj . Therefore, one of the alternatives in the lemma must hold. That the limit function, f must be analytic follows easily in the same way as the proof in Theorem 31.3.1 on Page 924. You could also use Morera’s theorem. This proves the lemma.

31.5.6

The Great Big Picard Theorem

The next theorem is the main result which the above lemmas lead to. It is the Big Picard theorem, also called the Great Picard theorem.Recall B 0 (a, r) is the deleted ball consisting of all the points of the ball except the center.


947

Theorem 31.5.17 Suppose f has an isolated essential singularity at 0. Then for every R > 0, and β ∈ C, f −1 (β) ∩ B 0 (0, R) is an infinite set except for one possible exceptional β. Proof: Suppose this is not true. Then there exists R1 > 0 and two points, α and β such that f −1 (β) ∩ B 0 (0, R1 ) and f −1 (α) ∩ B 0 (0, R1 ) are both finite sets. Then shrinking R1 and calling the result R, there exists B (0, R) such that f −1 (β) ∩ B 0 (0, R) = ∅, f −1 (α) ∩ B 0 (0, R) = ∅. © ª Now let A0 denote the annulus z ∈ C : 2R2 < |z| < 3R and let An denote the 22 © ª R annulus z ∈ C : 22+n < |z| < 23R . The reason for the 3 is to insure that An ∩ 2+n An+1 6= ∅. This follows from the observation that 3R/22+1+n > R/22+n . Now define a set of functions on A0 as follows: ³z ´ fn (z) ≡ f n . 2 By the choice of R, this set of functions missed the two points α and β. Therefore, by Lemma 31.5.16 there exists a subsequence such that one of the two options presented there holds. First suppose limk→∞ ||fnk − f ||K,∞ = 0 for all K a compact subset of A0 and f is analytic on A0 . In particular, this happens for γ 0 the circular contour having radius R/2. Thus fnk must be bounded on this contour. But this says the same thing as f (z/2nk ) is bounded for |z| = R/2, this holding for each k = 1, 2, · · · . Thus there exists a constant, M such that on each of a shrinking sequence of concentric circles whose radii converge to 0, |f (z)| ≤ M . By the maximum modulus theorem, |f (z)| ≤ M at every point between successive circles in this sequence. Therefore, |f (z)| ≤ M in B 0 (0, R) contradicting the Weierstrass Casorati theorem. The other option which might hold from Lemma 31.5.16 is that limk→∞ ρK (fnk , ∞) = 0 for all K compact subset of A0 . Since f has an essential singularity at 0 the zeros of f in B (0, R) are isolated. Therefore, for all k large enough, fnk has no zeros for |z| < 3R/22 . This is because the values of fnk are the values of f on Ank , a small anulus which avoids all the zeros of f whenever k is large enough. Only consider k this large. Then use the above argument on the analytic functions 1/fnk . By the assumption that limk→∞ ρK (fnk , ∞) = 0, it follows limk→∞ ||1/fnk − 0||K,∞ = 0 and so as above, there exists a shrinking sequence of concentric circles whose radii converge to 0 and a constant, M such that for z on any of these circles, |1/f (z)| ≤ M . This implies that on some deleted ball, B 0 (0, r) where r ≤ R, |f (z)| ≥ 1/M which again violates the Weierstrass Casorati theorem. This proves the theorem. As a simple corollary, here is what this remarkable theorem says about entire functions. Corollary 31.5.18 Suppose f is entire and nonconstant and not a polynomial. Then f assumes every complex value infinitely many times with the possible exception of one.

948

COMPLEX MAPPINGS

Proof: Since f is entire, f (z) = g (z) ≡ f

P∞ n=0

an z n . Define for z 6= 0,

µ ¶n µ ¶ X ∞ 1 1 = an . z z n=0

Thus 0 is an isolated essential singular point of g. By the big Picard theorem, Theorem 31.5.17 it follows g takes every complex number but possibly one an infinite number of times. This proves the corollary. Note the difference between this and the little Picard theorem which says that an entire function which is not constant must achieve every value but two.

31.6

Exercises

1. Prove that in Theorem 31.3.1 it suffices to assume F is uniformly bounded on each compact subset of Ω. 2. Find conditions on a, b, c, d such that the fractional linear transformation, az+b cz+d maps the upper half plane onto the upper half plane. 3. Let D be a simply connected region which is a proper subset of C. Does there exist an entire function, f which maps C onto D? Why or why not? 4. Verify the conclusion of Theorem 31.3.1 involving the higher order derivatives. 5. What if Ω = C? Does there exist an analytic function, f mapping Ω one to one and onto B (0, 1)? Explain why or why not. Was Ω 6= C used in the proof of the Riemann mapping theorem? 6. Verify that |φα (z)| = 1 if |z| = 1. Apply the maximum modulus theorem to conclude that |φα (z)| ≤ 1 for all |z| < 1. 7. Suppose that |f (z)| ≤ 1 for |z| = 1 and f (α) = 0 for |α| < 1. Show that |f (z)| ≤ |φα (z)| for all z ∈ B (0, 1) . Hint: Consider f (z)(1−αz) which has a z−α removable singularity at α. Show the modulus of this function is bounded by 1 on |z| = 1. Then apply the maximum modulus theorem. 8. Let U and V be open subsets of C and suppose u : U → R is harmonic while h is an analytic map which takes V one to one onto U . Show that u ◦ h is harmonic on V . 9. Show that for a harmonic function, u defined on B (0, R) , there exists an analytic function, h = u + iv where Z y Z x v (x, y) ≡ ux (x, t) dt − uy (t, 0) dt. 0

0

31.6. EXERCISES

949

10. Suppose Ω is a simply connected region and u is a real valued function defined on Ω such that u is harmonic. Show there exists an analytic function, f such that u = Re f . Show this is not true if Ω is not a simply connected region. Hint: You might use the Riemann mapping theorem and Problems 8 and 9. ¡ For the¢ second part it might be good to try something like u (x, y) = ln x2 + y 2 on the annulus 1 < |z| < 2. 1+z maps {z ∈ C : Im z > 0 and |z| < 1} to the first quadrant, 11. Show that w = 1−z {z = x + iy : x, y > 0} . a1 z+b1 12. Let f (z) = az+b cz+d and let g (z) = c1 z+d1 . Show that f ◦g (z) equals the quotient of two expressions, the numerator being the top entry in the vector µ ¶µ ¶µ ¶ a b a1 b1 z c d c1 d1 1

and the denominator being the bottom entry. Show that if you define µµ ¶¶ az + b a b φ ≡ , c d cz + d then φ (AB) = φ (A) ◦ φ (B) . Find an easy way to find the inverse of f (z) = az+b cz+d and give a condition on the a, b, c, d which insures this function has an inverse. 13. The modular group2 is the set of fractional linear transformations, az+b cz+d such that a, b, c, d are integers and ad − bc = 1. Using Problem 12 or brute force show this modular group is really a group with the group operation being dz−b composition. Also show the inverse of az+b cz+d is −cz+a . 14. Let Ω be a region and suppose f is analytic on Ω and that the functions fn are also analytic on Ω and converge to f uniformly on compact subsets of Ω. Suppose f is one to one. Can it be concluded that for an arbitrary compact set, K ⊆ Ω that fn is one to one for all n large enough? 15. The Vitali theorem says that if Ω is a region and {fn } is a uniformly bounded sequence of functions which converges pointwise on a set, S ⊆ Ω which has a limit point in Ω, then in fact, {fn } must converge uniformly on compact subsets of Ω to an analytic function. Prove this theorem. Hint: If the sequence fails to converge, show you can get two different subsequences converging uniformly on compact sets to different functions. Then argue these two functions coincide on S. 16. Does there exist a function analytic on B (0, 1) which maps B (0, 1) onto B 0 (0, 1) , the open unit ball in which 0 has been deleted?

2 This

is the terminology used in Rudin’s book Real and Complex Analysis.

950

COMPLEX MAPPINGS

Approximation By Rational Functions 32.1

Runge’s Theorem

Consider the function, z1 = f (z) for z defined on Ω ≡ B (0, 1) \ {0} = B 0 (0, 1) . Clearly f is analytic on Ω. Suppose you could approximate f uniformly by poly¡ ¢ nomials on ann 0, 12 , 34 , a compact of Ω. Then,¯ there would exist a suit¯ subset ¯ 1 R ¯ 1 able polynomial p (z) , such that ¯ 2πi γ f (z) − p (z) dz ¯ < 10 where here γ is a R 2 1 circle of radius 3 . However, this is impossible because 2πi γ f (z) dz = 1 while R 1 2πi γ p (z) dz = 0. This shows you can’t expect to be able to uniformly approximate analytic functions on compact sets using polynomials. This is just horrible! In real variables, you can approximate any continuous function on a compact set with a polynomial. However, that is just the way it is. It turns out that the ability to approximate an analytic function on Ω with polynomials is dependent on Ω being simply connected. All these theorems work for f having values in a complex Banach space. However, I will present them in the context of functions which have values in C. The changes necessary to obtain the extra generality are very minor. Definition 32.1.1 Approximation will be taken with respect to the following norm. ||f − g||K,∞ ≡ sup {||f (z) − g (z)|| : z ∈ K}

32.1.1

Approximation With Rational Functions

It turns out you can approximate analytic functions by rational functions, quotients of polynomials. The resulting theorem is one of the most profound theorems in complex analysis. The basic idea is simple. The Riemann sums for the Cauchy integral formula are rational functions. The idea used to implement this observation is that if you have a compact subset, of an open set, Ω there exists a cycle © ªK n composed of closed oriented curves γ j j=1 which are contained in Ω \ K such that 951

952

APPROXIMATION BY RATIONAL FUNCTIONS

Pn for every z ∈ K, k=1 n (γ k , z) = 1. One more ingredient is needed and this is a theorem which lets you keep the approximation but move the poles. To begin with, consider the part about the cycle of closed oriented curves. Recall Theorem 27.7.25 which is stated for convenience. Theorem 32.1.2 Let K be a compact subset of an open set, Ω. Then there exist © ªm continuous, closed, bounded variation oriented curves γ j j=1 for which γ ∗j ∩K = ∅ for each j, γ ∗j ⊆ Ω, and for all p ∈ K, m X

n (p, γ k ) = 1.

k=1

and

m X

n (z, γ k ) = 0

k=1

for all z ∈ / Ω. This theorem implies the following. Theorem 32.1.3 Let K ⊆ Ω where K is compact and Ω is open. Then there exist oriented closed curves, γ k such that γ ∗k ∩ K = ∅ but γ ∗k ⊆ Ω, such that for all z ∈ K, p Z 1 X f (w) f (z) = dw. (32.1.1) 2πi w −z γk k=1

Proof: This follows from Theorem 27.7.25 and the Cauchy integral formula. As shown in the proof, you can assume the γ k are linear mappings but this is not important. Next I will show how the Cauchy integral formula leads to approximation by rational functions, quotients of polynomials. Lemma 32.1.4 Let K be a compact subset of an open set, Ω and let f be analytic on Ω. Then there exists a rational function, Q whose poles are not in K such that ||Q − f ||K,∞ < ε. Proof: By Theorem 32.1.3 there are oriented curves, γ k described there such that for all z ∈ K, p Z 1 X f (w) f (z) = dw. (32.1.2) 2πi w −z γk k=1

(w) for (w, z) ∈ g (w, z) ≡ fw−z K and ∪k γ ∗k is positive that g

∪pk=1 γ ∗k × K, it follows since the distance Defining is uniformly continuous and so there exists between a δ > 0 such that if ||P|| < δ, then for all z ∈ K, ¯ ¯ ¯ ¯ p X n X ¯ f (γ k (τ j )) (γ k (ti ) − γ k (ti−1 )) ¯¯ ε ¯f (z) − 1 ¯ ¯ < 2. 2πi γ k (τ j ) − z ¯ ¯ k=1 j=1

32.1. RUNGE’S THEOREM

953

The complicated expression is obtained by replacing each integral in 32.1.2 with a Riemann sum. Simplifying the appearance of this, it follows there exists a rational function of the form R (z) =

M X k=1

Ak wk − z

where the wk are elements of components of C \ K and Ak are complex numbers or in the case where f has values in X, these would be elements of X such that ||R − f ||K,∞
n. Then cn =

∞ X

pnk ak bn−k+r .

k=r 1 Actually, it is only necessary to assume one of the series converges and the other converges absolutely. This is known as Merten’s theorem and may be read in the 1974 book by Apostol listed in the bibliography.

954


Also, ∞ X ∞ X

pnk |ak | |bn−k+r | =

k=r n=r

= = =

∞ X k=r ∞ X k=r ∞ X k=r ∞ X

|ak | |ak | |ak | |ak |

∞ X

pnk |bn−k+r |

n=r ∞ X n=k ∞ X n=k ∞ X

|bn−k+r | ¯ ¯ ¯bn−(k−r) ¯ |bm | < ∞.

m=r

k=r

Therefore, ∞ X

cn

=

n=r

= =

∞ X n X

ak bn−k+r =

n=r k=r ∞ ∞ X X

ak

k=r ∞ X

ak

k=r

∞ X ∞ X n=r k=r ∞ X

pnk bn−k+r =

n=r ∞ X

pnk ak bn−k+r ak

k=r

∞ X

bn−k+r

n=k

bm

m=r

and this proves the Ptheorem. ∞ It follows that n=r cn converges absolutely. Also, you can see by induction that you can multiply any number of absolutely convergent series together and obtain a series which is absolutely convergent. Next, here are some similar results related to Merten’s theorem. P∞ P∞ Lemma 32.1.6 Let n=0 an (z) and n=0 bn (z) be two convergent series for z ∈ K which satisfy the conditions of the Weierstrass M test. Thus there exist positive constants, An and P∞ P∞Bn such that |an (z)| ≤ An , |bn (z)| ≤ Bn for all z ∈ K and A < ∞, n=0 n n=0 Bn < ∞. Then defining the Cauchy product, cn (z) ≡

n X

an−k (z) bk (z) ,

k−0

P∞ it follows n=0 cn (z) also converges absolutely and uniformly on K because cn (z) satisfies the conditions of the Weierstrass M test. Therefore, Ã∞ !Ã ∞ ! ∞ X X X cn (z) = ak (z) bn (z) . (32.1.3) n=0

Proof: |cn (z)| ≤

k=0 n X k=0

n=0

|an−k (z)| |bk (z)| ≤

n X k=0

An−k Bk .

32.1. RUNGE’S THEOREM

955

Also, ∞ X n X

An−k Bk

=

n=0 k=0

=

∞ X ∞ X

An−k Bk

k=0 n=k ∞ ∞ X X

Bk

k=0

An < ∞.

n=0

The claim of 32.1.3 follows from Merten’s theorem. This proves the lemma. P∞ Corollary 32.1.7 Let P be a polynomial and let n=0 an (z) converge uniformly and absolutely on K such that the an satisfy P∞ the conditions P∞ of the Weierstrass M test. Then there exists a series for P ( n=0 an (z)) , n=0 cn (z) , which also converges absolutely and uniformly for z ∈ K because cn (z) also satisfies the conditions of the Weierstrass M test. The following picture is descriptive of the following lemma. This lemma says that if you have a rational function with one pole off a compact set, then you can approximate on the compact set with another rational function which has a different pole.

ub V

ua

K

Lemma 32.1.8 Let R be a rational function which has a pole only at a ∈ V, a component of C \ K where K is a compact set. Suppose b ∈ V. Then for ε > 0 given, there exists a rational function, Q, having a pole only at b such that ||R − Q||K,∞ < ε.

(32.1.4)

If it happens that V is unbounded, then there exists a polynomial, P such that ||R − P ||K,∞ < ε.

(32.1.5)

Proof: Say that b ∈ V satisfies P if for all ε > 0 there exists a rational function, Qb , having a pole only at b such that ||R − Qb ||K,∞ < ε Now define a set, S ≡ {b ∈ V : b satisfies P } .

956


Observe that S 6= ∅ because a ∈ S. I claim S is open. Suppose b1 ∈ S. Then there exists a δ > 0 such that ¯ ¯ ¯ b1 − b ¯ 1 ¯ ¯ (32.1.6) ¯ z−b ¯< 2 for all z ∈ K whenever b ∈ B (b1 , δ) . In fact, it suffices to take |b − b1 | < dist (b1 , K) /4 because then ¯ ¯ ¯ ¯ ¯ ¯ ¯ b1 − b ¯ ¯ < ¯ dist (b1 , K) /4 ¯ ≤ dist (b1 , K) /4 ¯ ¯ ¯ |z − b1 | − |b1 − b| ¯ z−b ¯ z−b dist (b1 , K) /4 1 1 ≤ ≤ < . dist (b1 , K) − dist (b1 , K) /4 3 2 Since b1 satisfies P, there exists a rational function Qb1 with the desired properties. It is shown next that you can approximate Qb1 with Qb thus yielding an approximation to R by the use of the triangle inequality, ||R − Qb1 ||K,∞ + ||Qb1 − Qb ||K,∞ ≥ ||R − Qb ||K,∞ . Since Qb1 has poles only at b1 , it follows it is a sum of functions of the form Therefore, it suffices to consider the terms of Qb1 or that Qb1 is of the special form 1 Qb1 (z) = n. (z − b1 ) αn (z−b1 )n .

However,

1 1 ³ n = n (z − b1 ) (z − b) 1 −

b1 −b z−b

ń

Now from the choice of b1 , the series ¶k ∞ µ X b1 − b 1 ´ =³ z−b 1 −b 1 − bz−b k=0 converges absolutely independent of the choice of z ∈ K because ¯µ ¯ ¯ b − b ¶k ¯ 1 ¯ 1 ¯ ¯ ¯ < k. ¯ z−b ¯ 2 1 . Thus a suitable 1 −b n (1− bz−b ) partial sum can be made uniformly on K as close as desired to (z−b1 1 )n . This shows that b satisfies P whenever b is close enough to b1 verifying that S is open. Next it is shown S is closed in V. Let bn ∈ S and suppose bn → b ∈ V. Then since bn ∈ S, there exists a rational function, Qbn such that

By Corollary 32.1.7 the same is true of the series for

||Qbn − R||K,∞
0, there exists a rational function, Q whose poles are all contained in the set, {bj } such that ||Q − f ||K,∞ < ε.

(32.1.9)

b \ K has only one component, then Q may be taken to be a polynomial. If C Proof: By Lemma 32.1.4 there exists a rational function of the form R (z) =

M X k=1

Ak wk − z

where the wk are elements of components of C \ K and Ak are complex numbers such that ε ||R − f ||K,∞ < . 2 k Consider the rational function, Rk (z) ≡ wA where wk ∈ Vj , one of the comk −z ponents of C \ K, the given point of Vj being bj . By Lemma 32.1.8, there exists a function, Qk which is either a rational function having its only pole at bj or a polynomial, depending on whether Vj is bounded such that

||Rk − Qk ||K,∞ < Letting Q (z) ≡

PM k=1

ε . 2M

Qk (z) , ||R − Q||K,∞
n} ∪

[ z ∈Ω /

µ ¶ 1 B z, . n

Thus {z : |z| > n} contains the point, ∞. Now let b \ Vn = C \ Vn ⊆ Ω. Kn ≡ C You should verify that 32.1.10 and 32.1.11 hold. It remains to show that every b \ Kn contains a component of C b \ Ω. Let D be a component of component of C b C \ K n ≡ Vn . If ∞ ∈ / D, then D contains no point of {z : |z| > n} because this set is connected and D is a component. (If it did contain ¢ of this set, it would have to S a ¡point contain the whole set.) Therefore, D ⊆ B z, n1 and so D contains some point z ∈Ω / ¡ ¢ of B z, n1 for some z ∈ / Ω. Therefore, since this ball is connected, it follows D must contain the whole ball and consequently D contains some point of ΩC . (The point z at the center of the ball will do.) Since D contains z ∈ / Ω, it must contain the component, Hz , determined by this point. The reason for this is that b \Ω⊆C b \ Kn Hz ⊆ C and Hz is connected. Therefore, Hz can only have points in one component of b \ Kn . Since it has a point in D, it must therefore, be totally contained in D. This C verifies the desired condition in the case where ∞ ∈ / D.

960


Now suppose that ∞ ∈ D. ∞ ∈ / Ω because Ω is given to be a set in C. Letting b \ Ω determined by ∞, it follows both D and H∞ H∞ denote the component of C contain ∞. Therefore, the connected set, H∞ cannot have any points in another b \ Kn and it is a set which is contained in C b \ Kn so it must be component of C contained in D. This proves the lemma. The following picture is a very simple example of the sort of thing considered by Runge’s theorem. The picture is of a region which has a couple of holes.

sa1

sa2

Ω

However, there could be many more holes than two. In fact, there could be infinitely many. Nor does it follow that the components of the complement of Ω need to have any interior points. Therefore, the picture is certainly not representative. Theorem 32.1.11 (Runge) Let Ω be an open set, and let A be a set which has one b \ Ω and let f be analytic on Ω. Then there exists a point in each component of C sequence of rational functions, {Rn } having poles only in A such that Rn converges uniformly to f on compact subsets of Ω. Proof: Let Kn be the compact sets of Lemma 32.1.10 where each component of b \ Kn contains a component of C b \ Ω. It follows each component of C b \ Kn contains C a point of A. Therefore, by Theorem 32.1.9 there exists Rn a rational function with poles only in A such that 1 ||Rn − f ||Kn ,∞ < . n It follows, since a given compact set, K is a subset of Kn for all n large enough, that Rn → f uniformly on K. This proves the theorem. Corollary 32.1.12 Let Ω be simply connected and f analytic on Ω. Then there exists a sequence of polynomials, {pn } such that pn → f uniformly on compact sets of Ω. b \ Ω is connected Proof: By definition of what is meant by simply connected, C b and so there are no bounded components of C\Ω. Therefore, in the proof of Theorem 32.1.11 when you use Theorem 32.1.9, you can always have Rn be a polynomial by Lemma 32.1.8.

32.2

The Mittag-Leffler Theorem

32.2.1

A Proof From Runge’s Theorem

This theorem is fairly easy to prove once you have Theorem 32.1.9. Given a set of complex numbers, does there exist a meromorphic function having its poles equal

32.2. THE MITTAG-LEFFLER THEOREM

961

to this set of numbers? The Mittag-Leffler theorem provides a very satisfactory answer to this question. Actually, it says somewhat more. You can specify, not just the location of the pole but also the kind of singularity the meromorphic function is to have at that pole. ∞

Theorem 32.2.1 Let P ≡ {zk }k=1 be a set of points in an open subset of C, Ω. Suppose also that P ⊆ Ω ⊆ C. For each zk , denote by Sk (z) a function of the form Sk (z) =

mk X

akj j

j=1

(z − zk )

.

Then there exists a meromorphic function, Q defined on Ω such that the poles of ∞ Q are the points, {zk }k=1 and the singular part of the Laurent expansion of Q at zk equals Sk (z) . In other words, for z near zk , Q (z) = gk (z) + Sk (z) for some function, gk analytic near zk . Proof: Let {Kn } denote the sequence of compact sets described in Lemma 32.1.10. Thus ∪∞ n=1 Kn = Ω, Kn ⊆ int (Kn+1 ) ⊆ Kn+1 · · · , and the components b \ Kn contain the components of C b \ Ω. Renumbering if necessary, you can of C assume each Kn 6= ∅. Also let K0 = ∅. Let Pm ≡ P ∩ (Km \ Km−1 ) and consider the rational function, Rm defined by X

Rm (z) ≡

Sk (z) .

zk ∈Km \Km−1

Since each Km is compact, it follows Pm is finite and so the above really is a rational function. Now for m > 1,this rational function is analytic on some open set containing Km−1 . There exists a set of points, A one point in each component b \ Ω. Consider C b \ Km−1 . Each of its components contains a component of C b \Ω of C b and so for each of these components of C \ Km−1 , there exists a point of A which is contained in it. Denote the resulting set of points by A0 . By Theorem 32.1.9 there exists a rational function, Qm whose poles are all contained in the set, A0 ⊆ ΩC such that 1 ||Rm − Qm ||Km−1,∞ < m . 2 The meromorphic function is Q (z) ≡ R1 (z) +

∞ X

(Rk (z) − Qk (z)) .

k=2

It remains to verify this function works. First consider K1 . Then on K1 , the above sum converges uniformly. Furthermore, the terms of the sum are analytic in some open set containing K1 . Therefore, the infinite sum is analytic on this open set and so for z ∈ K1 The function, f is the sum of a rational function, R1 , having poles at

962


P1 with the specified singular terms and an analytic function. Therefore, Q works on K1 . Now consider Km for m > 1. Then Q (z) = R1 (z) +

m+1 X

∞ X

(Rk (z) − Qk (z)) +

k=2

(Rk (z) − Qk (z)) .

k=m+2

As before, the infinite sum converges uniformly on Km+1 and hence on some open set, O containing Km . Therefore, this infinite sum equals a function which is analytic on O. Also, m+1 X R1 (z) + (Rk (z) − Qk (z)) k=2

is a rational function having poles at ∪m k=1 Pk with the specified singularities because the poles of each Qk are not in Ω. It follows this function is meromorphic because it is analytic except for the points in P. It also has the property of retaining the specified singular behavior.

32.2.2

A Direct Proof Without Runge’s Theorem

There is a direct proof of this important theorem which is not dependent on Runge’s theorem in the case where Ω = C. I think it is arguably easier to understand and the Mittag-Leffler theorem is very important so I will give this proof here. ∞

Theorem 32.2.2 Let P ≡ {zk }k=1 be a set of points in C which satisfies limn→∞ |zn | = 1 ∞. For each zk , denote by Sk (z) a polynomial in z−z which is of the form k Sk (z) =

mk X

akj j

j=1

(z − zk )

.

Then there exists a meromorphic function, Q defined on C such that the poles of Q ∞ are the points, {zk }k=1 and the singular part of the Laurent expansion of Q at zk equals Sk (z) . In other words, for z near zk , Q (z) = gk (z) + Sk (z) for some function, gk analytic in some open set containing zk . Proof: First consider the case where none of the zk = 0. Letting Kk ≡ {z : |z| ≤ |zk | /2} , there exists a power series for this set. Here is why: 1 = z − zk

1 z−zk

Ã

which converges uniformly and absolutely on

−1 1 − zzk

!

∞

1 −1 X = zk zk l=0

µ

z zk

¶l


963

and the Weierstrass M test can be applied because ¯ ¯ ¯z¯ 1 ¯ ¯< ¯ zk ¯ 2 1 on this set. Therefore, by Corollary 32.1.7, Sk (z) , being a polynomial in z−z , has k a power series which converges uniformly to Sk (z) on Kk . Therefore, there exists a polynomial, Pk (z) such that

||Pk − Sk ||B(0,|zk |/2),∞
N, then |zk | > 2 |z| Q (z) =

N X

(Sk (z) − Pk (z)) +

k=1

∞ X

(Sk (z) − Pk (z)) .

k=N +1

On Km , the second sum converges uniformly to a function analytic on int (Km ) (interior of Km ) while the first is a rational function having poles at z1 , · · · , zN . Since any compact set is contained in Km for large enough m, this shows Q (z) is meromorphic as claimed and has poles with the given singularities. ∞ Now consider the case where the poles are at {zk }k=0 with z0 = 0. Everything is similar in this case. Let Q (z) ≡ S0 (z) +

∞ X

(Sk (z) − Pk (z)) .

k=1

The series converges uniformly on every compact set because of the assumption that limn→∞ |zn | = ∞ which implies that any compact set is contained in Kk for k large enough. Choose N such that z ∈ int(KN ) and zn ∈ / KN for all n ≥ N + 1. Then Q (z) = S0 (z) +

N X k=1

(Sk (z) − Pk (z)) +

∞ X

(Sk (z) − Pk (z)) .

k=N +1

The last sum is analytic on int(KN ) because each function in the sum is analytic due PN to the fact that none of its poles are in KN . Also, S0 (z) + k=1 (Sk (z) − Pk (z)) is a finite sum of rational functions so it is a rational function and Pk is a polynomial so zm is a pole of this function with the correct singularity whenever zm ∈ int (KN ).

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32.2.3


b Functions Meromorphic On C

Sometimes it is useful to think of isolated singular points at ∞. Definition 32.2.3 Suppose f is analytic on {z ∈ C : |z| > r}¡ . ¢Then f is said to have a removable singularity at ∞ if the function, g (z) ≡ f z1 has a removable ¡ ¢ singularity at 0. f is said to have a pole at ∞ if the function, g (z) = f z1 has a b if all its singularities are isolated pole at 0. Then f is said to be meromorphic on C and either poles or removable. So what is f like for these cases? First suppose f has a removable singularity at ∞. Then zg (z) converges to 0 as z → 0. It follows g (z) must be analytic¡ near ¢ 0 and so can be given as a power series. Thus f (z) is of the form f (z) = g z1 = ¡ ¢ P∞ 1 n . Next suppose f has a pole at ∞. This means g (z) has a pole at 0 so n=0 an z Pm g (z) is of the form g (z) = k=1 zbkk +h (z) where h (z) is analytic near 0. Thus in the ¡ ¢ Pm ¡ ¢n P∞ case of a pole at ∞, f (z) is of the form f (z) = g z1 = k=1 bk z k + n=0 an z1 . b are all rational It turns out that the functions which are meromorphic on C b functions. To see this suppose f is meromorphic on C and note that there exists r > 0 such that f (z) is analytic for |z| > r. This is required if ∞ is to be isolated. Therefore, there are only finitely many poles of f for |z| ≤ r, {a1 , · · · , am } , because by assumption, these poles are isolated and this is P a compact set. Let the singular m part of f at ak be denoted by Sk (z) . Then f (z) − k=1 Sk (z) is analytic on all of C. Therefore, it is bounded on |z| ≤ r. In one P case, f has a removable singularity at ∞. In this case, f is bounded as z → ∞ andP k Sk also converges to 0 as z → ∞. m Therefore, by Liouville’s theorem, f (z) − k=1 Sk (z) equals a constant and so P case that f has f − k Sk is a constant. Thus f is a rational function. In¡the ¢ other P∞ Pm Pm Pm 1 n k a b z = S (z) − a pole at ∞, f (z) − − n k k n=0 k=1 k=1 k=1 Sk (z) . Now z P Pm m k b z is analytic on C and so is bounded on |z| ≤ r. But − f (z) − k=1 S¡k (z) k ¢n Pk=1 P∞ m now n=0 an z1 P− k=1 Sk (z) converges to 0 as z → ∞ and so by Liouville’s Pm m theorem, f (z) − k=1 Sk (z) − k=1 bk z k must equal a constant and again, f (z) equals a rational function.

32.2.4

A Great And Glorious Theorem About Simply Connected Regions

Here is given a laundry list of properties which are equivalent to an open set being simply connected. Recall Definition 27.7.21 on Page 823 which said that an open b \ Ω is connected. Recall also that this is not set, Ω is simply connected means C the same thing at all as saying C \ Ω is connected. Consider the outside of a disk for example. I will continue to use this definition for simply connected because it is the most convenient one for complex analysis. However, there are many other equivalent conditions. First here is an interesting lemma which is interesting for its own sake. Recall n (p, γ) means the winding number of γ about p. Now recall Theorem 27.7.25 implies the following lemma in which B C is playing the role of Ω in Theorem 27.7.25.


965

Lemma 32.2.4 Let K be a compact subset of B C , the complement of a closed set. m Then there exist continuous, closed, bounded variation oriented curves {Γj }j=1 for which Γ∗j ∩ K = ∅ for each j, Γ∗j ⊆ Ω, and for all p ∈ K, m X

n (Γk , p) = 1.

k=1

while for all z ∈ B

m X

n (Γk , z) = 0.

k=1

Definition 32.2.5 Let γ be a closed curve in an open set, Ω, γ : [a, b] → Ω. Then γ is said to be homotopic to a point, p in Ω if there exists a continuous function, H : [0, 1]×[a, b] → Ω such that H (0, t) = p, H (α, a) = H (α, b) , and H (1, t) = γ (t) . This function, H is called a homotopy. Lemma 32.2.6 Suppose γ is a closed continuous bounded variation curve in an open set, Ω which is homotopic to a point. Then if a ∈ / Ω, it follows n (a, γ) = 0. Proof: Let H be the homotopy described above. The problem with this is that it is not known that H (α, ·) is of bounded variation. There is no reason it should be. Therefore, it might not make sense to take the integral which defines the winding number. There are various ways around this. Extend H as follows. H (α, t) = H (α, a) for t < a, H (α, t) = H (α, b) for t > b. Let ε > 0. 1 Hε (α, t) ≡ 2ε

Z

2ε t+ (b−a) (t−a)

2ε −2ε+t+ (b−a) (t−a)

H (α, s) ds, Hε (0, t) = p.

Thus Hε (α, ·) is a closed curve which has bounded variation and when α = 1, this converges to γ uniformly on [a, b]. Therefore, for ε small enough, n (a, Hε (1, ·)) = n (a, γ) because they are both integers and as ε → 0, n (a, Hε (1, ·)) → n (a, γ) . Also, Hε (α, t) → H (α, t) uniformly on [0, 1] × [a, b] because of uniform continuity of H. Therefore, for small enough ε, you can also assume Hε (α, t) ∈ Ω for all α, t. Now α → n (a, Hε (α, ·)) is continuous. Hence it must be constant because the winding number is integer valued. But Z 1 1 lim dz = 0 α→0 2πi H (α,·) z − a ε because the length of Hε (α, ·) converges to 0 and the integrand is bounded because a∈ / Ω. Therefore, the constant can only equal 0. This proves the lemma. Now it is time for the great and glorious theorem on simply connected regions. The following equivalence of properties is taken from Rudin [58]. There is a slightly different list in Conway [16] and a shorter list in Ash [6]. Theorem 32.2.7 The following are equivalent for an open set, Ω.

966


1. Ω is homeomorphic to the unit disk, B (0, 1) . 2. Every closed curve contained in Ω is homotopic to a point in Ω. 3. If z ∈ / Ω, and if γ is a closed bounded variation continuous curve in Ω, then n (γ, z) = 0. b \ Ω is connected and Ω is connected. ) 4. Ω is simply connected, (C 5. Every function analytic on Ω can be uniformly approximated by polynomials on compact subsets. 6. For every f analytic on Ω and every closed continuous bounded variation curve, γ, Z f (z) dz = 0. γ

7. Every function analytic on Ω has a primitive on Ω. 8. If f, 1/f are both analytic on Ω, then there exists an analytic, g on Ω such that f = exp (g) . 9. If f, 1/f are both analytic on Ω, then there exists φ analytic on Ω such that f = φ2 . Proof: 1⇒2. Assume 1 and let γ be a closed ¡ ¡ curve in ¢¢Ω. Let h be the homeomorphism, h : B (0, 1) → Ω. Let H (α, t) = h α h−1 γ (t) . This works. 2⇒3 This is Lemma 32.2.6. b \ Ω is not connected, there exist 3⇒4. Suppose 3 but 4 fails to hold. Then if C disjoint nonempty sets, A and B such that A ∩ B = A ∩ B = ∅. It follows each of these sets must be closed because neither can have a limit point in Ω nor in the other. Also, one and only one of them contains ∞. Let this set be B. Thus A is a closed set which must also be bounded. Otherwise, there would exist a sequence of points in A, {an } such that limn→∞ an = ∞ which would contradict the requirement that no limit points of A can be in B. Therefore, A is a compact set contained in the open set, B C ≡ {z ∈ C : z ∈ / B} . Pick p ∈ A. By Lemma 32.2.4 m there exist continuous bounded variation closed curves {Γk }k=1 which are contained C in B , do not intersect A and such that 1=

m X

n (p, Γk )

k=1

However, if these curves do not intersect A and they also do not intersect B then they must be all contained in Ω. Since p ∈ / Ω, it follows by 3 that for each k, n (p, Γk ) = 0, a contradiction. 4⇒5 This is Corollary 32.1.12 on Page 960.


967

5⇒6 Every polynomial has a primitive and so the integral over any closed bounded variation curve of a polynomial equals 0. Let f be analytic on Ω. Then let {fn } be a sequence of polynomials converging uniformly to f on γ ∗ . Then Z Z fn (z) dz = f (z) dz. 0 = lim n→∞

γ

γ

6⇒7 Pick z0 ∈ Ω. Letting γ (z0 , z) be a bounded variation continuous curve joining z0 to z in Ω, you define a primitive for f as follows. Z F (z) = f (w) dw. γ(z0 ,z)

This is well defined by 6 and is easily seen to be a primitive. You just write the difference quotient and take a limit using 6. ÃZ ! Z F (z + w) − F (z) 1 lim = lim f (u) du − f (u) du w→0 w→0 w w γ(z0 ,z+w) γ(z0 ,z) Z 1 = lim f (u) du w→0 w γ(z,z+w) Z 1 1 = lim f (z + tw) wdt = f (z) . w→0 w 0 7⇒8 Suppose then that f, 1/f are both analytic. Then f 0 /f is analytic and so it has a primitive by 7. Let this primitive be g1 . Then ¡ −g1 ¢0 e f = e−g1 (−g10 ) f + e−g1 f 0 µ 0¶ f = −e−g1 f + e−g1 f 0 = 0. f Therefore, since Ω is connected, it follows e−g1 f must equal a constant. (Why?) Let the constant be ea+ibi . Then f (z) = eg1 (z) ea+ib . Therefore, you let g (z) = g1 (z) + a + ib. 8⇒9 Suppose then that f, 1/f are both analytic on Ω. Then by 8 f (z) = eg(z) . Let φ (z) ≡ eg(z)/2 . 9⇒1 There are two cases. First suppose Ω = C. This satisfies condition 9 because if f, 1/f are both analytic, then the same argument involved in 8⇒9 gives the existence of a square root. A homeomorphism is h (z) ≡ √ z 2 . It obviously 1+|z|

maps onto B (0, 1) and is continuous. To see it is 1 - 1 consider the case of z1 and z2 having different arguments. Then h (z1 ) 6= h (z2 ) . If z2 = tz1 for a positive t 6= 1, then it is also clear h (z1 ) 6= h (z2 ) . To show h−1 is continuous, note that if you have an open set in C and a point in this open set, you can get a small open set containing this point by allowing the modulus and the argument to lie in some open interval. Reasoning this way, you can verify h maps open sets to open sets. In the case where Ω 6= C, there exists a one to one analytic map which maps Ω onto B (0, 1) by the Riemann mapping theorem. This proves the theorem.

968


32.3

Exercises

1. Let a ∈ C. Show there exists a sequence of polynomials, {pn } such that pn (a) = 1 but pn (z) → 0 for all z 6= a. 2. Let l be a line in C. Show there exists a sequence of polynomials {pn } such that pn (z) → 1 on one side of this line and pn (z) → −1 on the other side of the line. Hint: The complement of this line is simply connected. 3. Suppose Ω is a simply connected region, f is analytic on Ω, f 6= 0 on Ω, and n n ∈ N. Show that there exists an analytic function, g such that g (z) = f (z) th for all z ∈ Ω. That is, you can take the n root of f (z) . If Ω is a region which contains 0, is it possible to find g (z) such that g is analytic on Ω and 2 g (z) = z? 4. Suppose Ω is a region (connected open set) and f is an analytic function defined on Ω such that f (z) 6= 0 for any z ∈ Ω. Suppose also that for every positive integer, n there exists an analytic function, gn defined on Ω such that gnn (z) = f (z) . Show that then it is possible to define an analytic function, L on f (Ω) such that eL(f (z)) = f (z) for all z ∈ Ω. 5. You know that φ (z) ≡ z−i z+i maps the upper half plane onto the unit ball. Its 1+z inverse, ψ (z) = i 1−z maps the unit ball onto the upper half plane. Also for z in the upper half plane, you can define a square root as follows. If z = |z| eiθ 1/2 where θ ∈ (0, π) , let z 1/2 ≡ |z| eiθ/2 so the square root maps the upper half plane to the first quadrant. Now consider Ã · µ ¶¸1/2 ! 1+z z → exp −i log i . (32.3.13) 1−z Show this is an analytic function which maps the unit ball onto an annulus. Is it possible to find a one to one analytic map which does this?

Infinite Products The Mittag-Leffler theorem gives existence of a meromorphic function which has specified singular part at various poles. It would be interesting to do something similar to zeros of an analytic function. That is, given the order of the zero at various points, does there exist an analytic function which has these points as zeros with the specified orders? You know that if you have the zeros of the polynomial, you can factor it. Can you do something similar with analytic functions which are just limits of polynomials? These questions involve the concept of an infinite product. Q∞ Qn Definition 33.0.1 n=1 (1 + un ) ≡ limn→∞ k=1 (1 + uk ) whenever this limit exists. If un = un (z) for zQ∈ H, we say the infinite product converges uniformly on n H if the partial products, k=1 (1 + uk (z)) converge uniformly on H. The main theorem is the following.

Theorem 33.0.2 Let H ⊆ C and suppose that on H where un (z) bounded on H. Then P (z) ≡

∞ Y

P∞ n=1

|un (z)| converges uniformly

(1 + un (z))

n=1

converges uniformly on H. If (n1 , n2 , · · · ) is any permutation of (1, 2, · · · ) , then for all z ∈ H, P (z) =

∞ Y

(1 + unk (z))

k=1

and P has a zero at z0 if and only if un (z0 ) = −1 for some n. 969

970

INFINITE PRODUCTS

Proof: First a simple estimate: n Y

(1 + |uk (z)|)

k=m

Ã Ã

= exp ln Ã ≤ exp

n Y

!! (1 + |uk (z)|)

k=m ∞ X

Ã = exp

! ln (1 + |uk (z)|)

k=m

!

|uk (z)|

n X

<e

k=m

P∞ for all z ∈ H provided m is large enough. Since k=1 |uk (z)| converges uniformly on H, |uk (z)| < 12 for all z ∈ H provided k is large enough. Thus you can take log (1 + uk (z)) . Pick N0 such that for n > m ≥ N0 ,

|um (z)|
0 be given and let N0 be such that if n > N0 , ¯ ¯ n ¯Y ¯ ¯ ¯ (1 + uk (z)) − P (z)¯ < ε ¯ ¯ ¯ k=1

© ª for all z ∈ H. Let {1, 2, · · · , n} ⊆ n1 , n2 , · · · , np(n) where p (n) is an increasing sequence. Then from 33.0.1 and 33.0.2,

≤

≤

≤

≤

≤

¯ ¯ ¯ ¯ p(n) Y ¯ ¯ ¯P (z) − (1 + unk (z))¯¯ ¯ ¯ ¯ k=1 ¯ ¯ ¯ ¯ ¯ p(n) n n ¯ ¯ ¯¯ Y Y Y ¯ ¯ ¯ ¯ (1 + uk (z))¯ + ¯ (1 + uk (z)) − (1 + unk (z))¯¯ ¯P (z) − ¯ ¯ ¯ ¯ k=1 k=1 k=1 ¯ ¯ ¯ n ¯ p(n) Y ¯Y ¯ ¯ ε+¯ (1 + uk (z)) − (1 + unk (z))¯¯ ¯k=1 ¯ k=1 ¯ n ¯¯ ¯ ¯Y ¯¯ ¯ Y ¯ ¯¯ ¯ ε+¯ (1 + |uk (z)|)¯ ¯1 − (1 + unk (z))¯ ¯ ¯¯ ¯ nk >n k=1 ¯N ¯¯ ¯ ¯ ¯ n 0 ¯Y ¯¯ Y ¯¯ ¯ Y ¯ ¯¯ ¯¯ ¯ ε+¯ (1 + |uk (z)|)¯ ¯ (1 + |uk (z)|)¯ ¯1 − (1 + unk (z))¯ ¯ ¯¯ ¯¯ ¯ nk >n k=1 k=N0 +1 ¯ ¯ ¯ ¯ ¯M (p(n)) ¯ ¯Y ¯ Y ¯ ¯ ¯ ¯ ε + Ce ¯ (1 + |unk (z)|) − 1¯ ≤ ε + Ce ¯¯ (1 + |unk (z)|) − 1¯¯ ¯ ¯ ¯ ¯ n >n k

k=n+1

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INFINITE PRODUCTS

ª © where M (p (n)) is the largest index in the permuted list, n1 , n2 , · · · , np(n) . then from 33.0.1, this last term is dominated by ¯  ¯  ¯ ¯ M (p(n)) Y ¯ ¯ (1 + |unk (z)|) − ln 1¯¯ ε + Ce2 ¯¯ln  ¯ ¯ k=n+1 ∞ X

ε + Ce2

≤

ln (1 + |unk |) ≤ ε + Ce2

∞ X

|unk | < 2ε

k=n+1

k=n+1

¯ ¯ Qp(n) ¯ ¯ for all n large enough uniformly in z ∈ H. Therefore, ¯P (z) − k=1 (1 + unk (z))¯ < 2ε whenever n is large enough. This proves the part about the permutation. It remains to verify the assertion about the points, z0 , where P (z0 ) = 0. Obviously, if un (z0 ) = −1, then P (z0 ) = 0. Suppose then that P (z0 ) = 0 and M > N0 . Then ¯ ¯ M ¯Y ¯ ¯ ¯ (1 + uk (z0 ))¯ = ¯ ¯ ¯ k=1

≤

≤

≤

≤

≤

≤

¯ ¯ M ∞ ¯Y ¯ Y ¯ ¯ (1 + uk (z0 )) − (1 + uk (z0 ))¯ ¯ ¯ ¯ k=1 k=1 ¯ ¯¯ ¯ M ∞ ¯Y ¯¯ ¯ Y ¯ ¯¯ ¯ (1 + uk (z0 ))¯ ¯1 − (1 + uk (z0 ))¯ ¯ ¯ ¯¯ ¯ k=1 k=M +1 ¯ ¯ ¯ ¯ M ∞ ¯Y ¯¯ Y ¯ ¯ ¯¯ ¯ (1 + uk (z0 ))¯ ¯ (1 + |uk (z0 )|) − 1¯ ¯ ¯ ¯¯ ¯ k=1 k=M +1 ¯ ¯ ¯ ¯ M ∞ ¯Y ¯¯ ¯ Y ¯ ¯¯ ¯ e¯ (1 + uk (z0 ))¯ ¯ln (1 + |uk (z0 )|) − ln 1¯ ¯ ¯¯ ¯ k=1 k=M +1 ¯ Ã ∞ !¯ M ¯Y ¯ X ¯ ¯ e ln (1 + |uk (z)|) ¯ (1 + uk (z0 ))¯ ¯ ¯ k=M +1 k=1 ¯ ¯ ∞ M ¯Y ¯ X ¯ ¯ e |uk (z)| ¯ (1 + uk (z0 ))¯ ¯ ¯ k=M +1 k=1 ¯M ¯ ¯ 1 ¯¯ Y ¯ (1 + uk (z0 ))¯ ¯ ¯ 2¯ k=1

whenever M is large enough. Therefore, for such M, M Y

(1 + uk (z0 )) = 0

k=1

and so uk (z0 ) = −1 for some k ≤ M. This proves the theorem.

33.1. ANALYTIC FUNCTION WITH PRESCRIBED ZEROS

33.1

973

Analytic Function With Prescribed Zeros

Suppose you are given complex numbers, {zn } and you want to find an analytic function, f such that these numbers are the zeros of f . How can you do it? The problem is easy if there are only finitely many of these zeros, {z1 , z2 , · · · , zm } . You just write (z − z1³) (z − z2´) · · · (z − zm ) . Now if none of the zk = 0 you could Qm also write it at k=1 1 − zzk and this might have a better chance of success in the case of infinitely ¯many prescribed zeros. However, you would need to verify P∞ ¯ z ¯¯ something like n=1 ¯ zn ¯ < ∞ which might not be so. The way around this is to ´ Q∞ ³ adjust the product, making it k=1 1 − zzk egk (z) where gk (z) is some analytic ³ ´ P n −1 ∞ function. Recall also that for |x| < 1, ln (1 − x) = n=1 xn . If you had x/xn ³ ³ ´´ Q∞ −1 small and real, then 1 = (1 − x/xn ) exp ln (1 − x/xn ) and k=1 1 of course converges but loses all the information about zeros. However, this is why it is not too unreasonable to consider factors of the form µ ¶ Pp ³ ´k 1 z k z 1− e k=1 zk k zk where pk is suitably chosen. First here are some estimates. Lemma 33.1.1 For z ∈ C, |ez − 1| ≤ |z| e|z| ,

(33.1.3)

and if |z| ≤ 1/2, ¯ ¯ ∞ m ¯X 2 1 1 z k ¯¯ 1 |z| ¯ m ≤ |z| ≤ . ¯ ¯≤ ¯ k ¯ m 1 − |z| m m 2m−1

(33.1.4)

k=m

Proof: Consider 33.1.3. ¯ ¯ ∞ ∞ k ¯X k¯ z ¯ ¯ X |z| |ez − 1| = ¯ = e|z| − 1 ≤ |z| e|z| ¯≤ ¯ k! ¯ k! k=1

k=1

the last inequality holding by the mean value theorem. Now consider 33.1.4. ¯ ¯ ∞ ∞ ∞ k ¯X X |z| 1 X k z k ¯¯ ¯ ≤ |z| ¯ ¯ ≤ ¯ k¯ k m k=m

k=m

k=m

m

=

2 1 1 1 |z| m ≤ |z| ≤ . m 1 − |z| m m 2m−1

This proves the lemma. The functions, Ep in the next definition are called the elementary factors.

974

INFINITE PRODUCTS

Definition 33.1.2 Let E0 (z) ≡ 1 − z and for p ≥ 1, µ ¶ z2 zp Ep (z) ≡ (1 − z) exp z + + ··· + 2 p In terms of this new symbol, here is another estimate. A sharper inequality is available in Rudin [58] but it is more difficult to obtain. Corollary 33.1.3 For Ep defined above and |z| ≤ 1/2, |Ep (z) − 1| ≤ 3 |z|

p+1

.

Proof: From elementary calculus, ln (1 − x) = − Therefore, for |z| < 1, log (1 − z) = −

P∞

xn n=1 n

for all |x| < 1.

∞ ∞ ³ ´ X X zn zn −1 , log (1 − z) = , n n n=1 n=1

P∞ n because the function log (1 − z) and the analytic function, − n=1 zn both are equal to ln (1 − x) on the real line segment (−1, 1) , a set which has a limit point. Therefore, using Lemma 33.1.1,

= =

=

≤

|Ep (z) − 1| ¯ ¯ ¶ µ p 2 ¯ ¯ ¯(1 − z) exp z + z + · · · + z ¯ − 1 ¯ ¯ 2 p ¯ ¯ ! Ã ∞ ¯ ¯ ³ ´ X zn ¯ ¯ −1 − 1¯ − ¯(1 − z) exp log (1 − z) ¯ ¯ n n=p+1 ¯ ¯ ! Ã ∞ ¯ ¯ X zn ¯ ¯ − 1¯ ¯exp − ¯ ¯ n n=p+1 ¯ ¯ ∞ ¯ X zn z n ¯¯ |− P∞ ¯ n=p+1 n | ¯e ¯− ¯ n¯ n=p+1

≤

1 p+1 p+1 · 2 · e1/(p+1) |z| . ≤ 3 |z| p+1

This proves the corollary. With this estimate, it is easy to prove the Weierstrass product formula. Theorem 33.1.4 Let {zn } be a sequence of nonzero complex numbers which have no limit point in C and let {pn } be a sequence of nonnegative integers such that ¶pn +1 ∞ µ X R N, |zn | > 2R. Therefore, for |z| < R and letting 0 < a = min {|zn | : n ≤ N } , ¯ µ ¶ ∞ ¯ X ¯ ¯ z ¯Ep ¯ − 1 ¯ n zn ¯

≤

¶pn +1 ∞ µ X R +3 2R

r for all zn and so |h (zn )| < r−1 . Thus the sequence, {h (zn )} is a bounded sequence in the open set h (Ω1 ) . It has no limit point in h (Ω1 ) because this is true of {zn } and Ω1 . By Lemma 33.1.6 there exist wn ∈ ∂ (h (Ω1 )) such that limn→∞ |wn − h (zn )| = 0. Consider for z ∈ Ω1 µ ¶ ∞ Y h (zn ) − wn f (z) ≡ En . (33.1.8) h (z) − wn n=1 Letting K be a compact subset of Ω1 , h (K) is a compact subset of h (Ω1 ) and so if z ∈ K, then |h (z) − wn | is bounded below by a positive constant. Therefore, there exists N large enough that for all z ∈ K and n ≥ N, ¯ ¯ ¯ h (zn ) − wn ¯ 1 ¯ ¯ ¯ h (z) − wn ¯ < 2

33.1. ANALYTIC FUNCTION WITH PRESCRIBED ZEROS and so by Corollary 33.1.3, for all z ∈ K and n ≥ N, ¯ µ ¯ ¶ µ ¶n ¯ ¯ ¯En h (zn ) − wn − 1¯ ≤ 3 1 . ¯ ¯ h (z) − wn 2 Therefore,

977

(33.1.9)

¯ µ ¶ ∞ ¯ X ¯ ¯ ¯En h (zn ) − wn − 1¯ ¯ ¯ h (z) − wn

n=1

³ ´ Q∞ n )−wn converges uniformly for z ∈ K. This implies n=1 En h(z also converges h(z)−wn uniformly for z ∈ K by Theorem 33.0.2. Since K is arbitrary, this shows f defined in 33.1.8 is analytic on Ω1 . Also if zn is listed m times so it is a zero of multiplicity m and wn is the point from ∂ (h (Ω1 )) closest to h (zn ) , then there are m factors in 33.1.8 which are of the form µ ¶ ¶ µ h (zn ) − wn h (zn ) − wn En = 1− egn (z) h (z) − wn h (z) − wn ¶ µ h (z) − h (zn ) gn (z) e = h (z) − wn µ ¶ zn − z 1 = egn (z) (z − w) (zn − w) h (z) − wn = (z − zn ) Gn (z) (33.1.10) where Gn is an analytic function which is not zero at and near zn . Therefore, f has a zero of order m at zn . This proves the theorem except for the point, w which has been left out of Ω1 . It is necessary to show f is analytic at this point also and right now, f is not even defined at w. The {wn } are bounded because {h (zn )} is bounded and limn→∞ |wn − h (zn )| = 0 which implies |wn − h (zn )| ≤ C for some constant, C. Therefore, there exists δ > 0 such that if z ∈ B 0 (w, δ) , then for all n, ¯ ¯ ¯ ¯ ¯ ¯ ¯ h (zn ) − w ¯ ¯ h (zn ) − wn ¯ 1 ¯³ ¯=¯ ¯ ´ ¯ 1 ¯ ¯ h (z) − wn ¯ < 2 . ¯ −w ¯ z−w

n

Thus 33.1.9 holds for all z ∈ B 0 (w, δ) and n so by Theorem 33.0.2, the infinite product in 33.1.8 converges uniformly on B 0 (w, δ) . This implies f is bounded in B 0 (w, δ) and so w is a removable singularity and f can be extended to w such that the result is analytic. It only remains to verify f (w) 6= 0. After all, this would not do because it would be another zero other than those in the given list. By 33.1.10, a partial product is of the form ¶ N µ Y h (z) − h (zn ) n=1

h (z) − wn

egn (z)

(33.1.11)

978

INFINITE PRODUCTS

where Ã gn (z) ≡

1 h (zn ) − wn + h (z) − wn 2

µ

h (zn ) − wn h (z) − wn

¶2

1 + ··· + n

µ

h (zn ) − wn h (z) − wn

¶n !

Each of the quotients in the definition of gn (z) converges to 0 as ³ z → w and ´ so h(z)−h(zn ) the partial product of 33.1.11 converges to 1 as z → w because →1 h(z)−wn as z → w. If f (w) = 0, then if z is close enough to w, it follows |f (z)| < 12 . Also, by the uniform convergence on B 0 (w, δ) , it follows that for some N, the partial product up to N must also be less than 1/2 in absolute value for all z close enough to w and as noted above, this does not occur because such partial products converge to 1 as z → w. Hence f (w) 6= 0. This proves the corollary. Recall the definition of a meromorphic function on Page 816. It was a function which is analytic everywhere except at a countable set of isolated points at which the function has a pole. It is clear that the quotient of two analytic functions yields a meromorphic function but is this the only way it can happen? Theorem 33.1.8 Suppose Q is a meromorphic function on an open set, Ω. Then there exist analytic functions on Ω, f (z) and g (z) such that Q (z) = f (z) /g (z) for all z not in the set of poles of Q. Proof: Let Q have a pole of order m (z) at z. Then by Corollary 33.1.7 there exists an analytic function, g which has a zero of order m (z) at every z ∈ Ω. It follows gQ has a removable singularity at the poles of Q. Therefore, there is an analytic function, f such that f (z) = g (z) Q (z) . This proves the theorem. Corollary 33.1.9 Suppose Ω is a region and Q is a meromorphic function defined on Ω such that the set, {z ∈ Ω : Q (z) = c} has a limit point in Ω. Then Q (z) = c for all z ∈ Ω. Proof: From Theorem 33.1.8 there are analytic functions, f, g such that Q = fg . Therefore, the zero set of the function, f (z) − cg (z) has a limit point in Ω and so f (z) − cg (z) = 0 for all z ∈ Ω. This proves the corollary.

33.2

Factoring A Given Analytic Function

The next theorem is the Weierstrass factorization theorem which can be used to factor a given analytic function f . If f has a zero of order m when z = 0, then you could factor out a z m and from there consider the factorization of what remains when you have factored out the z m . Therefore, the following is the main thing of interest.

33.2. FACTORING A GIVEN ANALYTIC FUNCTION

979

Theorem 33.2.1 Let f be analytic on C, f (0) 6= 0, and let the zeros of f, be {zk } ,listed according to order. (Thus if z is a zero of order m, it will be listed m times in the list, {zk } .) Choosing nonnegative integers, pn such that for all r > 0, ¶pn +1 ∞ µ X r < ∞, |zn | n=1 There exists an entire function, g such that f (z) = eg(z)

∞ Y

µ Epn

n=1

z zn

¶ .

(33.2.12)

Note that eg(z) 6= 0 for any z and this is the interesting thing about this function. Proof: {zn } cannot have a limit point because if there were a limit point of this sequence, it would follow from Theorem 27.5.3 that f (z) = 0 for all z, contradicting the hypothesis that f (0) 6= 0. Hence limn→∞ |zn | = ∞ and so ¶1+n−1 X ¶n ∞ µ ∞ µ X r r = 0. Hint: You have already shown that this is true for positive real numbers. Verify this formula for Re z > 0 yields an analytic function. 16. ↑Show Γ

¡1¢

17. Show that

2

=

R∞

√

e

π. Then find Γ

−s2 2

ds =

√

¡5¢ 2

.

2π. Hint: Denote this integral by I and observe

R −∞−(x2 +y2 )/2 e dxdy. R2

that I 2 = x = r cos (θ), y = r sin θ.

Then change variables to polar coordinates,

18. ↑ Now that you know what the gamma function is, consider in the formula for Γ (α + 1) the following change of variables. t = α + α1/2 s. Then in terms of the new variable, s, the formula for Γ (α + 1) is Z −α α+ 21

∞

e

α

=e

−α α+ 12

√ − α

Z α

e

√ − αs

∞

√ − α

Show the integrand converges to e−

s2 2

e

µ ¶α s 1+ √ ds α

h ³ ´ i α ln 1+ √sα − √sα

ds

. Show that then

Γ (α + 1) = α→∞ e−α αα+(1/2)

Z

∞

lim

e

−s2 2

ds =

√

2π.

−∞

Hint: You will need to obtain a dominating function for the integral so that you can use the dominated convergence theorem. You might try considering 2 √ √ s ∈ (− α, α) first and consider something like e1−(s /4) on this interval. √ Then look for another function for s > α. This formula is known as Stirling’s formula. 19. This and the next several problems develop the zeta function and give a relation between the zeta and the gamma function. Define for 0 < r < 2π Z Ir (z) ≡

2π

e(z−1)(ln r+iθ) iθ ire dθ + ereiθ − 1 0 Z r (z−1) ln t e + dt t ∞ e −1

Z

∞ r

e(z−1)(ln t+2πi) dt(33.6.24) et − 1

Show that Ir is an entire function. The reason 0 < r < 2π is that this prevents iθ ere − 1 from equaling zero. The above is just a precise description of the R z−1 contour integral, γ eww −1 dw where γ is the contour shown below.

998

INFINITE PRODUCTS

¾

¾

?

-

in which on the integrals along the real line, the argument is different in going from r to ∞ than it is in going from ∞ to r. Now I have not defined such contour integrals over contours which have infinite length and so have chosen to simply write out explicitly what is involved. You have to work with these integrals given above anyway but the contour integral just mentioned is the motivation for them. Hint: You may want to use convergence theorems from real analysis if it makes this more convenient but you might not have to. 20. ↑ In the context of Problem 19 define for small δ > 0 Z Irδ (z) ≡ γ r,δ

wz−1 dw ew − 1

where γ rδ is shown below.

¾ µ r¡ ¡

?

¾

s¡

- 2δ

x

Show that limδ→0 Irδ (z) = Ir (z) . Hint: Use the dominated convergence theorem if it makes this go easier. This is not a hard problem if you use these theorems but you can probably do it without them with more work. 21. ↑ In the context of Problem 20 show that for r1 < r, Irδ (z) − Ir1 δ (z) is a contour integral, Z wz−1 dw w γ r,r ,δ e − 1 1

where the oriented contour is shown below.

33.6. EXERCISES

999

¾ γ r,r1 ,δ ?6 ¾ In this contour integral, wz−1 denotes e(z−1) log(w) where log (w) = ln |w| + i arg (w) for arg (w) ∈ (0, 2π) . Explain why this integral equals zero. From Problem 20 it follows that Ir = Ir1 . Therefore, you can define an entire function, I (z) ≡ Ir (z) for all r positive but sufficiently small. Hint: Remember the Cauchy integral formula for analytic functions defined on simply connected regions. You could argue there is a simply connected region containing γ r,r1 ,δ . 22. ↑ In case Re z > 1, you can get an interesting formula for I (z) by taking the limit as r → 0. Recall that Z 2π (z−1)(ln r+iθ) Z ∞ (z−1)(ln t+2πi) e e iθ Ir (z) ≡ ire dθ + dt(33.6.25) iθ re et − 1 e −1 0 r Z r (z−1) ln t e dt + t ∞ e −1 and now it is desired to take a limit in the case where Re z > 1. Show the first integral above converges to 0 as r → 0. Next argue the sum of the two last integrals converges to ³ ´ Z ∞ e(z−1) ln(t) e(z−1)2πi − 1 dt. et − 1 0 Thus

¡ ¢ I (z) = ez2πi − 1

Z

∞ 0

e(z−1) ln(t) dt et − 1

(33.6.26)

when Re z > 1. 23. ↑ So what does all this have to do with the zeta function and the gamma function? The zeta function is defined for Re z > 1 by ∞ X 1 ≡ ζ (z) . z n n=1

By Problem 15, whenever Re z > 0, Z Γ (z) = 0

∞

e−t tz−1 dt.

1000

INFINITE PRODUCTS

Change the variable and conclude Γ (z)

Z

1 = nz

∞

e−ns sz−1 ds.

0

Therefore, for Re z > 1, ∞ Z X

ζ (z) Γ (z) =

∞

e−ns sz−1 ds.

n=1

0

Now show that you can interchange the order of the sum and the integral. This possibly most ¯ −ns z−1 ¯ easily done by using Fubini’s theorem. Show that P∞ isR ∞ ¯e ¯ ds < ∞ and then use Fubini’s theorem. I think you s n=1 0 could do it other ways though. It is possible to do it without any reference to Lebesgue integration. Thus Z ζ (z) Γ (z)

∞

= Z

∞ X

sz−1

0

e−ns ds

n=1 ∞

= 0

sz−1 e−s ds = 1 − e−s

Z 0

∞

sz−1 ds es − 1

By 33.6.26, I (z)

= = =

¡ ¡ ¡

¢

Z

ez2πi − 1 z2πi

e

2πiz

e

¢

0

∞

e(z−1) ln(t) dt et − 1

− 1 ζ (z) Γ (z) ¢ − 1 ζ (z) Γ (z)

whenever Re z > 1. 24. ↑ Now show there exists an entire function, h (z) such that ζ (z) =

1 + h (z) z−1

for Re z > 1. Conclude ζ (z) extends to a meromorphic function defined on all of C which has a simple pole at z = 1, namely, the right side of the above formula. Hint: Use Problem 10 to observe that Γ (z) is never equal to zero but has simple poles at every nonnegative integer. Then for Re z > 1, ζ (z) ≡

I (z) . (e2πiz − 1) Γ (z)

By 33.6.26 ζ has no poles for Re z > 1. The right side of the above equation is defined for all z. There are no poles except possibly when z is a nonnegative integer. However, these points are not poles either because of Problem 10 which states that Γ has simple poles at these points thus cancelling the simple

33.6. EXERCISES

1001

¡ ¢ zeros of e2πiz − 1 . The only remaining possibility for a pole for ζ is at z = 1. Show it has a simple pole at this point. You can use the formula for I (z) Z

2π

e(z−1)(ln r+iθ) iθ ire dθ + ereiθ − 1 0 Z r (z−1) ln t e + dt t ∞ e −1

I (z) ≡

Z r

∞

e(z−1)(ln t+2πi) dt(33.6.27) et − 1

Thus I (1) is given by Z I (1) ≡ 0

2π

Z

1 ereiθ

−1

∞

ireiθ dθ + r

1 dt + t e −1

Z

r

∞

et

1 dt −1

R

= γ ewdw −1 where γ r is the circle of radius r. This contour integral equals 2πi r by the residue theorem. Therefore, I (z) 1 = + h (z) (e2πiz − 1) Γ (z) z−1 where h (z) is an entire function. People worry a lot about where the zeros of ζ are located. In particular, the zeros for Re z ∈ (0, 1) are of special interest. The Riemann hypothesis says they are all on the line Re z = 1/2. This is a good problem for you to do next. 25. There is an important relation between prime numbers and the zeta function ∞ due to Euler. Let {pn }n=1 be the prime numbers. Then for Re z > 1, ∞ Y

1 = ζ (z) . 1 − p−z n n=1 To see this, consider a partial product. ¶j N X ∞ µ Y 1 1 n = . pzn 1 − p−z n n=1 n=1 j =1 N Y

n

Let SN denote all positive integers which P use only p1 , · · · , pN in their prime factorization. Then the above equals n∈SN n1z . Letting N → ∞ and using the fact that Re z > 1 so that the order in which you sum is not important (See Theorem 34.0.1 P∞ or recall advanced calculus. ) you obtain the desired equation. Show n=1 p1n = ∞.

1002

INFINITE PRODUCTS

Elliptic Functions This chapter is to give a short introduction to elliptic functions. There is much more available. There are books written on elliptic functions. What I am presenting here follows Alfors [2] although the material is found in many books on complex analysis. Hille, [32] has a much more extensive treatment than what I will attempt here. There are also many references and historical notes available in the book by Hille. Another good source for more having much the same emphasis as what is presented here is in the book by Saks and Zygmund [60]. This is a very interesting subject because it has considerable overlap with algebra. Before beginning, recall that an absolutely convergent series can be summed in any order and you always get the same answer. The easy way to see this is to think of the series as a Lebesgue integral with respect to counting measure and apply convergence theorems as needed. The following theorem provides the necessary results. P∞ and let θ, φ : N → N be one to one and Theorem 34.0.1 Suppose n=1 |an | < P∞ P∞ ∞ onto mappings. Then n=1 aφ(n) and n=1 aθ(n) both converge and the two sums are equal. Proof: By the monotone convergence theorem, ∞ X

|an | = lim

n→∞

n=1

n n X X ¯ ¯ ¯ ¯ ¯aφ(k) ¯ = lim ¯aθ(k) ¯ n→∞

k=1

k=1

¯ ¯ ¯ P∞ ¯ ¯aφ(k) ¯ and P∞ ¯aθ(k) ¯ respectively. Therefore, but these last two equal k=1 k=1 P∞ P∞ 1 k=1 aθ(k) and k=1 aφ(k) exist (n → aθ(n) is in L with respect to counting measure.) It remains to show the two are equal. There exists M such that if n > M then ∞ ∞ X X ¯ ¯ ¯ ¯ ¯aθ(k) ¯ < ε, ¯aφ(k) ¯ < ε k=n+1

k=n+1

¯ ¯ ∞ n ¯X ¯ X ¯ ¯ aφ(k) − aφ(k) ¯ < ε, ¯ ¯ ¯ k=1

k=1

¯ ¯ ∞ n ¯X ¯ X ¯ ¯ aθ(k) − aθ(k) ¯ < ε ¯ ¯ ¯ k=1

1003

k=1

1004

ELLIPTIC FUNCTIONS

Pick such an n denoted by n1 . Then pick n2 > n1 > M such that {θ (1) , · · · , θ (n1 )} ⊆ {φ (1) , · · · , φ (n2 )} . Then

n2 X k=1

Therefore,

aφ(k) =

n1 X k=1

aφ(k) .

φ(k)∈{θ(1),··· / ,θ(n1 )}

¯ ¯¯ ¯n n1 2 ¯ ¯ ¯X X ¯ ¯ aφ(k) − aθ(k) ¯ = ¯¯ ¯ ¯ ¯ ¯ k=1

X

aθ(k) +

k=1

¯ ¯ ¯ aφ(k) ¯¯ ¯ φ(k)∈{θ(1),··· / ,θ(n1 )},k≤n2 X

Now all of these φ (k) in the last sum are contained in {θ (n1 + 1) , · · · } and so the last sum above is dominated by ≤

∞ X ¯ ¯ ¯aθ(k) ¯ < ε. k=n1 +1

Therefore,

¯ ¯ ∞ ∞ ¯X ¯ X ¯ ¯ aφ(k) − aθ(k) ¯ ¯ ¯ ¯ k=1

≤

k=1

¯ ¯ n2 ∞ ¯X ¯ X ¯ ¯ aφ(k) − aφ(k) ¯ ¯ ¯ ¯ k=1 k=1 ¯n ¯ n1 2 ¯X ¯ X ¯ ¯ +¯ aφ(k) − aθ(k) ¯ ¯ ¯ k=1

k=1

¯n ¯ ∞ 1 ¯X ¯ X ¯ ¯ +¯ aθ(k) − aθ(k) ¯ < ε + ε + ε = 3ε ¯ ¯ k=1 k=1 P∞ P∞ and since ε is arbitrary, it follows k=1 aφ(k) = k=1 aθ(k) as claimed. This proves the theorem.

34.1

Periodic Functions

Definition 34.1.1 A function defined on C is said to be periodic if there exists w such that f (z + w) = f (z) for all z ∈ C. Denote by M the set of all periods. Thus if w1 , w2 ∈ M and a, b ∈ Z, then aw1 + bw2 ∈ M. For this reason M is called the module of periods.1 In all which follows it is assumed f is meromorphic. Theorem 34.1.2 Let f be a meromorphic function and let M be the module of periods. Then if M has a limit point, then f equals a constant. If this does not happen then either there exists w1 ∈ M such that Zw1 = M or there exist w1 , w2 ∈ M such that M = {aw1 + bw2 : a, b ∈ Z} and w1 /w2 is not real. Also if τ = w2 /w1 , |τ | ≥ 1, 1A

1 −1 ≤ Re τ ≤ . 2 2

module is like a vector space except instead of a field of scalars, you have a ring of scalars.

34.1. PERIODIC FUNCTIONS

1005

Proof: Suppose f is meromorphic and M has a limit point, w0 . By Theorem 33.1.8 on Page 978 there exist analytic functions, p, q such that f (z) = p(z) q(z) . Now pick z0 such that z0 is not a pole of f . Then letting wn → w0 where {wn } ⊆ M, f (z0 + wn ) = f (z0 ) . Therefore, p (z0 + wn ) = f (z0 ) q (z0 + wn ) and so the analytic function, p (z) − f (z0 ) q (z) has a zero set which has a limit point. Therefore, this function is identically equal to zero because of Theorem 27.5.3 on Page 807. Thus f equals a constant as claimed. This has shown that if f is not constant, then M is discreet. Therefore, there exists w1 ∈ M such that |w1 | = min {|w| : w ∈ M }. Suppose first that every element of M is a real multiple of w1 . Thus, if w ∈ M, it follows there exists a real number, x such that w = xw1 . Then there exist positive integers, k, k + 1 such that k ≤ x < k + 1. If x > k, then w − kw1 = (x − k) w1 is a period having smaller absolute value than |w1 | which would be a contradiction. Hence, x = k and so M = Zw1 . Now suppose there exists w2 ∈ M which is not a real multiple of w1 . You can let w2 be the element of M having this property which has smallest absolute value. Now let w ∈ M. Since w1 and w2 point in different directions, it follows w = xw1 + yw2 for some real numbers, x, y. Let |m − x| ≤ 21 and |n − y| ≤ 12 where m, n are integers. Therefore, w = mw1 + nw2 + (x − m) w1 + (y − n) w2 and so w − mw1 − nw2 = (x − m) w1 + (y − n) w2

(34.1.1)

Now since w2 /w1 ∈ / R, |(x − m) w1 + (y − n) w2 |

< |(x − m) w1 | + |(y − n) w2 | 1 1 = |w1 | + |w2 | . 2 2

Therefore, from 34.1.1, |w − mw1 − nw2 | =
k, then w − mw1 − nw2 − kw1 = (x − k) w1 which would contradict the choice of w1 as being the period having minimal absolute value because the expression on the left in the above is a period and it equals

1006

ELLIPTIC FUNCTIONS

something which has absolute value less than |w1 |. Therefore, x = k and w is an integer linear combination of w1 and w2 . It only remains to verify the claim about τ. From the construction, |w1 | ≤ |w2 | and |w2 | ≤ |w1 − w2 | , |w2 | ≤ |w1 + w2 | . Therefore, |τ | ≥ 1, |τ | ≤ |1 − τ | , |τ | ≤ |1 + τ | . The last two of these inequalities imply −1/2 ≤ Re τ ≤ 1/2. This proves the theorem. Definition 34.1.3 For f a meromorphic function which has the last of the above alternatives holding in which M = {aw1 + bw2 : a, b ∈ Z} , the function, f is called elliptic. This is also called doubly periodic. Theorem 34.1.4 Suppose f is an elliptic function which has no poles. Then f is constant. Proof: Since f has no poles it is analytic. Now consider the parallelograms determined by the vertices, mw1 + nw2 for m, n ∈ Z. By periodicity of f it must be bounded because its values are identical on each of these parallelograms. Therefore, it equals a constant by Liouville’s theorem. Definition 34.1.5 Define Pa to be the parallelogram determined by the points a + mw1 + nw2 , a + (m + 1) w1 + nw2 , a + mw1 + (n + 1) w2 , a + (m + 1) w1 + (n + 1) w2 Such Pa will be referred to as a period parallelogram. The sum of the orders of the poles in a period parallelogram which contains no poles or zeros of f on its boundary is called the order of the function. This is well defined because of the periodic property of f . Theorem 34.1.6 The sum of the residues of any elliptic function, f equals zero on every Pa if a is chosen so that there are no poles on ∂Pa . Proof: Choose a such that there are no poles of f on the boundary of Pa . By periodicity, Z f (z) dz = 0 ∂Pa

because the integrals over opposite sides of the parallelogram cancel out because the values of f are the same on these sides and the orientations are opposite. It follows from the residue theorem that the sum of the residues in Pa equals 0. Theorem 34.1.7 Let Pa be a period parallelogram for a nonconstant elliptic function, f which has order equal to m. Then f assumes every value in f (Pa ) exactly m times.


1007

Proof: Let c ∈ f (Pa ) and consider Pa0 such that f −1 (c) ∩ Pa0 = f −1 (c) ∩ Pa and Pa0 contains the same poles and zeros of f − c as Pa but Pa0 has no zeros of f (z) − c or poles of f on its boundary. Thus f 0 (z) / (f (z) − c) is also an elliptic function and so Theorem 34.1.6 applies. Consider Z 1 f 0 (z) dz. 2πi ∂Pa0 f (z) − c By the argument principle, this equals Nz −Np where Nz equals the number of zeros of f (z)−c and Np equals the number of the poles of f (z). From Theorem 34.1.6 this must equal zero because it is the sum of the residues of f 0 / (f − c) and so Nz = Np . Now Np equals the number of poles in Pa counted according to multiplicity. There is an even better theorem than this one. Theorem 34.1.8 Let f be a non constant elliptic function and suppose it has poles p1 , · · · , pm and zeros, z1 , · · · , zm in Pα , P listed according Pm to multiplicity where ∂Pα m contains no poles or zeros of f . Then z − k k=1 k=1 pk ∈ M, the module of periods. Proof: You can assume ∂Pa contains no poles or zeros of f because if it did, then you could consider a slightly shifted period parallelogram, Pa0 which contains no new zeros and poles but which has all the old ones but no poles or zeros on its boundary. By Theorem 29.1.3 on Page 860 Z m m X X 1 f 0 (z) z dz = zk − pk . (34.1.2) 2πi ∂Pa f (z) k=1

k=1

Denoting by γ (z, w) the straight oriented line segment from z to w, Z f 0 (z) z dz ∂Pa f (z) Z Z f 0 (z) f 0 (z) = z dz + z dz γ(a,a+w1 ) f (z) γ(a+w1 +w2 ,a+w2 ) f (z) Z Z f 0 (z) f 0 (z) + z dz + z dz γ(a+w1 ,a+w2 +w1 ) f (z) γ(a+w2 ,a) f (z) Z =

(z − (z + w2 )) γ(a,a+w1 )

Z

+

(z − (z + w1 )) γ(a,a+w2 )

Now near these line segments

f 0 (z) dz f (z)

f 0 (z) f (z)

f 0 (z) dz f (z)

is analytic and so there exists a primitive, gwi (z)

on γ (a, a + wi ) by Corollary 27.7.5 on Page 813 which satisfies egwi (z) = f (z). Therefore, = −w2 (gw1 (a + w1 ) − gw1 (a)) − w1 (gw2 (a + w2 ) − gw2 (a)) .

1008

ELLIPTIC FUNCTIONS

Now by periodicity of f it follows f (a + w1 ) = f (a) = f (a + w2 ) . Hence gwi (a + w1 ) − gwi (a) = 2mπi for some integer, m because egwi (a+wi ) − egwi (a) = f (a + wi ) − f (a) = 0. Therefore, from 34.1.2, there exist integers, k, l such that Z f 0 (z) 1 z dz 2πi ∂Pa f (z) 1 = [−w2 (gw1 (a + w1 ) − gw1 (a)) − w1 (gw2 (a + w2 ) − gw2 (a))] 2πi 1 = [−w2 (2kπi) − w1 (2lπi)] 2πi = −w2 k − w1 l ∈ M. From 34.1.2 it follows

m X k=1

zk −

m X

pk ∈ M.

k=1

This proves the theorem. Hille says this relation is due to Liouville. There is also a simple corollary which follows from the above theorem applied to the elliptic function, f (z) − c. Corollary 34.1.9 Let f be a non constant elliptic function and suppose the function, f (z) − c has poles p1 , · · · , pm and zeros, z1 , · · · , zm on Pα , listedPaccording m to multiplicity where ∂Pα contains no poles or zeros of f (z) − c. Then k=1 zk − Pm k=1 pk ∈ M, the module of periods.

34.1.1

The Unimodular Transformations

Definition 34.1.10 Suppose f is a nonconstant elliptic function and the module of periods is of the form {aw1 + bw2 } where a, b are integers and w1 /w2 is not real. Then by analogy with linear algebra, {w1 , w2 } is referred to as a basis. The unimodular transformations will refer to matrices of the form µ ¶ a b c d where all entries are integers and ad − bc = ±1. These linear transformations are also called the modular group. The following is an interesting lemma which ties matrices with the fractional linear transformations.


1009

Lemma 34.1.11 Define µµ φ

a b c d

¶¶ ≡

az + b . cz + d

Then φ (AB) = φ (A) ◦ φ (B) ,

(34.1.3)

φ (A) (z) = z if and only if A = cI −1 where I is the identity matrix and c 6= 0. Also if f (z) = az+b (z) exists cz+d , then f −1 if and only if ad − cb 6= 0. Furthermore it is easy to find f .

Proof: The equation in 34.1.3 µ is just ¶ a simple computation. Now suppose a b φ (A) (z) = z. Then letting A = , this requires c d az + b = z (cz + d) and so az + b = cz 2 + dz. Since this is to hold for all z it follows c = 0 = b and a = d. The other direction is obvious. Consider the claim about the existence of an inverse. Let ad − cb 6= 0 for f (z) = az+b cz+d . Then ¶¶ µµ a b f (z) = φ c d µ ¶−1 µ ¶ a b d −b 1 It follows exists and equals ad−bc . Therefore, c d −c a z

µµ ¶µ µ ¶¶¶ 1 a b d −b = φ (I) (z) = φ (z) c d −c a ad − bc ¶¶ µµ ¶¶¶ µµ µ 1 a b d −b ◦φ (z) = φ c d −c a ad − bc = f ◦ f −1 (z)

which shows f −1 exists and it is easy to find. Next suppose f −1 exists. I need to verify the condition ad − cb 6= 0. If f −1 exists, then from the process used to µ find it,¶you see that it must be a fractional a b linear transformation. Letting A = so φ (A) = f, it follows there exists c d a matrix B such that φ (BA) (z) = φ (B) ◦ φ (A) (z) = z. However, it was shown that this implies BA is a nonzero multiple of I which requires that A−1 must exist. Hence the condition must hold.

1010

ELLIPTIC FUNCTIONS

Theorem 34.1.12 If f is a nonconstant elliptic function with a basis {w1 , w2 } for 0 0 the module of periods, then {w µ 1 , w2 } ¶is another basis, if and only if there exists a a b unimodular transformation, = A such that c d µ 0 ¶ µ ¶µ ¶ w1 a b w1 = . (34.1.4) w20 c d w2 Proof: Since {w1 , w2 } is a basis, there exist integers, a, b, c, d such that 34.1.4 holds. It remains to show the transformation determined by the matrix is unimodular. Taking conjugates, µ 0 ¶ µ ¶µ ¶ w1 w1 a b = . c d w2 w20 Therefore,

µ

w10 w20

w10 w20

¶

µ =

a b c d

¶µ

w1 w2

w1 w2

¶

given to be a basis, there exits another matrix having Now since {w10 , w20 }µ is also ¶ e f all integer entries, such that g h µ

and

µ

Therefore,

µ

w10 w20

w10 w20

w1 w2 w1 w2

¶

¶

µ =

¶

µ =

µ =

a b c d

e g

f h

e g

f h

¶µ

¶µ

¶µ

e g

f h

¶

w10 w20 w10 w20

¶ .

¶µ

w10 w20

w10 w20

¶ .

However, since w10 /w20 is not real, it is routine to verify that µ 0 ¶ w1 w10 det 6= 0. w20 w20 Therefore,

µ µ

1 0 µ

0 1

¶

µ =

a b c d

¶µ

e g

f h

¶

¶ ¶ a b e f det = 1. But the two matrices have all integer c d g h entries and so both determinants must equal either 1 or −1. Next suppose µ 0 ¶ µ ¶µ ¶ w1 a b w1 = (34.1.5) w20 c d w2

and so det


1011

¶ a b where is unimodular. I need to verify that {w10 , w20 } is a basis. If w ∈ M, c d there exist integers, m, n such that µ ¶ ¡ ¢ w1 w = mw1 + nw2 = m n w2 µ

From 34.1.5

µ ±

and so w=±

d −b −c a ¡

m

n

¶µ

¢

µ

w10 w20

¶

µ =

d −b −c a

w1 w2

¶µ

¶

w10 w20

¶

which is an integer linear combination of {w10 , w20 } . It only remains to verify that w10 /w20 is not real. Claim: Let w1 and w2 be nonzero complex numbers. Then w2 /w1 is not real if and only if µ ¶ w1 w1 w1 w2 − w1 w2 = det 6= 0 w2 w2 Proof of the claim: Let λ = w2 /w1 . Then ¡ ¢ 2 w1 w2 − w1 w2 = λw1 w1 − w1 λw1 = λ − λ |w1 | ¡ ¢ Thus the ratio is not real if and only if λ − λ 6= 0 if and only if w1 w2 − w1 w2 6= 0. Now to verify w20 /w10 is not real, µ 0 ¶ w1 w10 det w20 w20 µµ ¶µ ¶¶ a b w1 w1 = det c d w2 w2 µ ¶ w1 w1 = ± det 6= 0 w2 w2 This proves the theorem.

34.1.2

The Search For An Elliptic Function

By Theorem 34.1.4 and 34.1.6 if you want to find a nonconstant elliptic function it must fail to be analytic and also have either no terms in its Laurent expansion −1 which are of the form b1 (z − a) or else these terms must cancel out. It is simplest to look for a function which simply does not have them. Weierstrass looked for a function of the form Ã ! X 1 1 1 ℘ (z) ≡ 2 + (34.1.6) 2 − w2 z (z − w) w6=0

1012

ELLIPTIC FUNCTIONS

where w consists of all numbers of the form aw1 + bw2 for a, b integers. Sometimes people write this as ℘ (z, w1 , w2 ) to emphasize its dependence on the periods, w1 and w2 but I won’t do so. It is understood there exist these periods, which are given. This is a reasonable thing to try. Suppose you formally differentiate the right side. Never mind whether this is justified for now. This yields X −2 −2 −2 X = ℘0 (z) = 3 − 3 3 z (z − w) w (z − w) w6=0 which is clearly periodic having both periods w1 and w2 . Therefore, ℘ (z + w1 ) − ℘ (z) and ℘ (z + w2 ) − ℘ (z) are both constants, c1 and c2 respectively. The reason for this is that since ℘0 is periodic with periods w1 and w2 , it follows ℘0 (z + wi ) − ℘0 (z) = 0 as long as z is not a period. From 34.1.6 you can see right away that ℘ (z) = ℘ (−z) Indeed ℘ (−z)

X 1 + z2

=

w6=0

X 1 + 2 z

=

w6=0

and so c1

Ã

Ã

1 2 − w2 (−z − w) 1

1 2 − w2 (−z + w) 1

!

! = ℘ (z) .

³ w ´ ³ w ´ 1 1 = ℘ − + w1 − ℘ − 2 2 ³w ´ ³ w ´ 1 1 = ℘ −℘ − =0 2 2

which shows the constant for ℘ (z + w1 ) − ℘ (z) must equal zero. Similarly the constant for ℘ (z + w2 ) − ℘ (z) also equals zero. Thus ℘ is periodic having the two periods w1 , w2 . Of course to justify this, you need to consider whether the series of 34.1.6 converges. Consider the terms of the series. ¯ ¯ ¯ ¯ ¯ ¯ 2w − z ¯ 1 1 ¯¯ ¯ ¯ ¯ − 2 ¯ = |z| ¯ ¯ ¯ ¯ (z − w)2 ¯ (z − w)2 w2 ¯ w ¯ If |w| > 2 |z| , this can be estimated more. For such w, ¯ ¯ ¯ 1 1 ¯¯ ¯ − 2¯ ¯ ¯ (z − w)2 w ¯ ¯ ¯ ¯ 2w − z ¯ (5/2) |w| ¯ ¯ = |z| ¯ ¯ ≤ |z| 2 2 2 2 ¯ (z − w) w ¯ |w| (|w| − |z|) ≤ |z|

(5/2) |w| 2

|w| ((1/2) |w|)

2

= |z|

10 |w|

3.


1013

It follows the series converges uniformly and absolutely on every compact P in 34.1.6 1 set, K provided w6=0 |w| 3 converges. This question is considered next. Claim: There exists a positive number, k such that for all pairs of integers, m, n, not both equal to zero, |mw1 + nw2 | ≥ k > 0. |m| + |n| Proof of claim: If not, there exists mk and nk such that nk mk w1 + w2 = 0 k→∞ |mk | + |nk | |mk | + |nk | ³ ´ nk k However, |mkm is a bounded sequence in R2 and so, taking a sub|+|nk | , |mk |+|nk | sequence, still denoted by k, you can have µ ¶ mk nk , → (x, y) ∈ R2 |mk | + |nk | |mk | + |nk | lim

and so there are real numbers, x, y such that xw1 + yw2 = 0 contrary to the assumption that w2 /w1 is not equal to a real number. This proves the claim. Now from the claim, X 1 3

|w| X

w6=0

=

1

(m,n)6=(0,0)

=

∞ 1 X k 3 j=1

|mw1 + nw2 |

X

3

(m,n)6=(0,0)

1 3

|m|+|n|=j

X

≤

(|m| + |n|)

=

1 k3

3

(|m| + |n|)

∞ 1 X 4j < ∞. k 3 j=1 j 3

Now consider the series in 34.1.6. Letting z ∈ B (0, R) , ! Ã X 1 1 1 ℘ (z) ≡ + 2 − w2 z2 (z − w) w6=0,|w|≤R Ã ! X 1 1 + 2 − w2 (z − w) w6=0,|w|>R and the last series converges uniformly on B (0, R) to an analytic function. Thus ℘ is a meromorphic function and also the argument given above involving differentiation of the series termwise is valid. Thus ℘ is an elliptic function as claimed. This is called the Weierstrass ℘ function. This has proved the following theorem. Theorem 34.1.13 The function ℘ defined above is an example of an elliptic function. On any compact set, ℘ equals a rational function added to a series which is uniformly and absolutely convergent on the compact set.

1014

34.1.3

ELLIPTIC FUNCTIONS

The Differential Equation Satisfied By ℘

For z not a pole, ℘0 (z) =

2 −2 X − 3 z3 (z − w) w6=0

Also since there are no poles of order 1 you can obtain a primitive for ℘, −ζ.2 To do so, recall Ã ! X 1 1 1 ℘ (z) ≡ 2 + 2 − w2 z (z − w) w6=0

where for |z| < R this is the sum of a rational function with a uniformly convergent series. Therefore, you can take the integral along any path, γ (0, z) from 0 to z which misses the poles of ℘. By the uniform convergence of the above integral, you can interchange the sum with the integral and obtain ζ (z) =

1 X 1 z 1 + + + z z − w w2 w

(34.1.7)

w6=0

This function is odd. Here is why. ζ (−z) =

X 1 1 z 1 + − + −z −z − w w2 w w6=0

while −ζ (z)

=

X −1 1 z 1 + − − −z z − w w2 w w6=0

=

X −1 1 z 1 + − + . −z z + w w2 w w6=0

Now consider 34.1.7. It will be used to find the Laurent expansion about the origin for ζ which will then be differentiated to obtain the Laurent expansion for ℘ at the origin. Since w 6= 0 and the interest is for z near 0 so |z| < |w| , 1 z 1 + 2+ z−w w w

= =

z 1 1 1 + − 2 w w w 1 − wz ∞ z 1 1 X ³ z ´k + − w2 w w w k=0

∞ 1 X ³ z ´k = − w w k=2

2 I don’t know why it is traditional to refer to this antiderivative as −ζ rather than ζ but I am following the convention. I think it is to minimize the number of minus signs in the next expression.


1015

From 34.1.7 1 X + z

ζ (z) =

1 − z

=

Ã

∞ X zk − wk+1

!

k=2 w6=0 ∞ k XX k=2 w6=0

∞

z 1 X X z 2k−1 = − k+1 w z w2k k=2 w6=0

because the sum over odd powers must be zero because for each w 6= 0, there exists 2k z 2k −w 6= 0 such that the two terms wz2k+1 and (−w) 2k+1 cancel each other. Hence ∞

ζ (z) =

1 X − Gk z 2k−1 z k=2

where Gk =

P

1 w6=0 w2k .

Now with this, ∞

−ζ 0 (z) =

℘ (z) =

X 1 + Gk (2k − 1) z 2k−2 2 z k=2

1 + 3G2 z 2 + 5G3 z 4 + · · · z2

= Therefore,

℘0 (z) = ℘0 (z)

2

=

4℘ (z)

3

=

−2 + 6G2 z + 20G3 z 3 + · · · z3 4 24G2 − 2 − 80G3 + · · · z6 z µ ¶3 1 2 4 4 + 3G2 z + 5G3 z · · · z2 4 36 + 2 G2 + 60G3 + · · · z6 z

= and finally

60G2 + 0 + ··· z2 where in the above, the positive powers of z are not listed explicitly. Therefore, 60G2 ℘ (z) =

0

2

3

℘ (z) − 4℘ (z) + 60G2 ℘ (z) + 140G3 =

∞ X

an z n

n=1

In deriving the equation it was assumed |z| < |w| for all w = aw1 +bw2 where a, b are integers not both zero. The left side of the above equation is periodic with respect to w1 and w2 where w2 /w1 is not a real number. The only possible poles of the left side are at 0, w1 , w2 , and w1 + w2 , the vertices of the parallelogram determined by w1 and w2 . This follows from the original formula for ℘ (z) . However, the above

1016

ELLIPTIC FUNCTIONS

equation shows the left side has no pole at 0. Since the left side is periodic with periods w1 and w2 , it follows it has no pole at the other vertices of this parallelogram either. Therefore, the left side is periodic and has no poles. Consequently, it equals a constant by Theorem 34.1.4. But the right side of the above equation shows this constant is 0 because this side equals zero when z = 0. Therefore, ℘ satisfies the differential equation, 2

3

℘0 (z) − 4℘ (z) + 60G2 ℘ (z) + 140G3 = 0. It is traditional to define 60G2 ≡ g2 and 140G3 ≡ g3 . Then in terms of these new quantities the differential equation is 2

3

℘0 (z) = 4℘ (z) − g2 ℘ (z) − g3 . Suppose e1 , e2 and e3 are zeros of the polynomial 4w3 − g2 w − g3 = 0. Then the above equation can be written in the form 2

℘0 (z) = 4 (℘ (z) − e1 ) (℘ (z) − e2 ) (℘ (z) − e3 ) .

34.1.4

(34.1.8)

A Modular Function

The next task is to find the ei in 34.1.8. First recall that ℘ is an even function. That is ℘ (−z) = ℘ (z). This follows from 34.1.6 which is listed here for convenience. ! Ã X 1 1 1 ℘ (z) ≡ 2 + (34.1.9) 2 − w2 z (z − w) w6=0

Thus ℘ (−z)

=

X 1 + 2 z

w6=0

=

X 1 + 2 z

w6=0

Ã

Ã

1

1 2 − w2 (−z − w) 1 2 − w2 (−z + w) 1

!

! = ℘ (z) .

Therefore, ℘ (w1 − z) = ℘ (z − w1 ) = ℘ (z) and so −℘0 (w1 − z) = ℘0 (z) . Letting z = w1 /2, it follows ℘0 (w1 /2) = 0. Similarly, ℘0 (w2 /2) = 0 and ℘0 ((w1 + w2 ) /2) = 0. Therefore, from 34.1.8 0 = 4 (℘ (w1 /2) − e1 ) (℘ (w1 /2) − e2 ) (℘ (w1 /2) − e3 ) . It follows one of the ei must equal ℘ (w1 /2) . Similarly, one of the ei must equal ℘ (w2 /2) and one must equal ℘ ((w1 + w2 ) /2). Lemma 34.1.14 The numbers, ℘ (w1 /2) , ℘ (w2 /2) , and ℘ ((w1 + w2 ) /2) are distinct.


1017

Proof: Choose Pa , a period parallelogram which contains the pole 0, and the points w1 /2, w2 /2, and (w1 + w2 ) /2 but no other pole of ℘ (z) . Also ∂Pa∗ does not contain any zeros of the elliptic function, z → ℘ (z) − ℘ (w1 /2). This can be done by shifting P0 slightly because the poles are only at the points aw1 + bw2 for a, b integers and the zeros of ℘ (z) − ℘ (w1 /2) are discreet.

a

w1 + w2 »s » » »£ » » » £ w2 »»» »»» » s £ »» » » £ £ »£» £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £s w1 » £ » £ £ »»» £ £0»»»»»»» £ £s £ » »»» £ »» s»

If ℘ (w2 /2) = ℘ (w1 /2) , then ℘ (z) − ℘ (w1 /2) has two zeros, w2 /2 and w1 /2 and since the pole at 0 is of order 2, this is the order of ℘ (z) − ℘ (w1 /2) on Pa hence by Theorem 34.1.7 on Page 1006 these are the only zeros of this function on Pa . It follows by Corollary 34.1.9 on Page 1008 which says the sum of the zeros minus the sum of the poles is in M , w21 + w22 ∈ M. Thus there exist integers, a, b such that w1 + w2 = aw1 + bw2 2 which implies (2a − 1) w1 + (2b − 1) w2 = 0 contradicting w2 /w1 not being real. Similar reasoning applies to the other pairs of points in {w1 /2, w2 /2, (w1 + w2 ) /2} . For example, consider (w1 + w2 ) /2 and choose Pa such that its boundary contains no zeros of the elliptic function, z → ℘ (z) − ℘ ((w1 + w2 ) /2) and Pa contains no poles of ℘ on its interior other than 0. Then if ℘ (w2 /2) = ℘ ((w1 + w2 ) /2) , it follows from Theorem 34.1.7 on Page 1006 w2 /2 and (w1 + w2 ) /2 are the only two zeros of ℘ (z) − ℘ ((w1 + w2 ) /2) on Pa and by Corollary 34.1.9 on Page 1008 w1 + w1 + w2 = aw1 + bw2 ∈ M 2 for some integers a, b which leads to the same contradiction as before about w1 /w2 not being real. The other cases are similar. This proves the lemma. Lemma 34.1.14 proves the ei are distinct. Number the ei such that e1 = ℘ (w1 /2) , e2 = ℘ (w2 /2) and e3 = ℘ ((w1 + w2 ) /2) . To summarize, it has been shown that for complex numbers, w1 and w2 with w2 /w1 not real, an elliptic function, ℘ has been defined. Denote this function as

1018

ELLIPTIC FUNCTIONS

℘ (z) = ℘ (z, w1 , w2 ) . This in turn determined numbers, ei as described above. Thus these numbers depend on w1 and w2 and as described above, ³w ´ ³w ´ 1 2 e1 (w1 , w2 ) = ℘ , w1 , w2 , e2 (w1 , w2 ) = ℘ , w1 , w2 2 µ2 ¶ w1 + w2 e3 (w1 , w2 ) = ℘ , w1 , w2 . 2 Therefore, using the formula for ℘, 34.1.9, Ã ! X 1 1 1 ℘ (z) ≡ 2 + 2 − w2 z (z − w) w6=0

you see that if the two periods w1 and w2 are replaced with tw1 and tw2 respectively, then ei (tw1 , tw2 ) = t−2 ei (w1 , w2 ) . Let τ denote the complex number which equals the ratio, w2 /w1 which was assumed in all this to not be real. Then ei (w1 , w2 ) = w1−2 ei (1, τ ) Now define the function, λ (τ ) λ (τ ) ≡

e3 (1, τ ) − e2 (1, τ ) e1 (1, τ ) − e2 (1, τ )

µ ¶ e3 (w1 , w2 ) − e2 (w1 , w2 ) = . e1 (w1 , w2 ) − e2 (w1 , w2 )

(34.1.10)

This function is meromorphic for Im τ > 0 or for Im τ < 0. However, since the denominator is never equal to zero the function must actually be analytic on both the upper half plane and the lower half plane. It never is equal to 0 because e3 6= e2 and it never equals 1 because e3 6= e1 . This is stated as an observation. Observation 34.1.15 The function, λ (τ ) is analytic for τ in the upper half plane and never assumes the values 0 and 1. This is a very interesting function. Consider what happens when µ 0 ¶ µ ¶µ ¶ w1 a b w1 = w20 c d w2 and the matrix is unimodular. By Theorem 34.1.12 on Page 1010 {w10 , w20 } is just another basis for the same module of periods. Therefore, ℘ (z, w1 , w2 ) = ℘ (z, w10 , w20 ) because both are defined as sums over the same values of w, just in different order which does not matter because of the absolute convergence of the sums on compact subsets of C. Since ℘ is unchanged, it follows ℘0 (z) is also unchanged and so the numbers, ei are also the same. However, they might be permuted in which case the function λ (τ ) defined above would change. What would it take for λ (τ ) to not change? In other words, for which unimodular transformations will λ be left


1019

unchanged? This happens takes place for the ei . This ³ 0 ´ if and ¡only¢ if no³permuting ´ ¡ ¢ w w0 occurs if ℘ w21 = ℘ 21 and ℘ w22 = ℘ 22 . If w10 w1 w0 w2 − ∈ M, 2 − ∈M 2 2 2 2

³ 0´ ¡ ¢ w then ℘ w21 = ℘ 21 and so e1 will be unchanged and similarly for e2 and e3 . This occurs exactly when 1 1 ((a − 1) w1 + bw2 ) ∈ M, (cw1 + (d − 1) w2 ) ∈ M. 2 2 This happens if a and d are odd and if b and c are even. Of course the stylish way to say this is a ≡ 1 mod 2, d ≡ 1 mod 2, b ≡ 0 mod 2, c ≡ 0 mod 2.

(34.1.11)

This has shown that for unimodular transformations satisfying 34.1.11 λ is unchanged. Letting τ be defined as above, w20 cw1 + dw2 c + dτ ≡ = . w10 aw1 + bw2 a + bτ µ ¶ a b Thus for unimodular transformations, satisfying 34.1.11, or more succ d cinctly, µ ¶ µ ¶ a b 1 0 ∼ mod 2 (34.1.12) c d 0 1 τ0 =

it follows that

µ λ

c + dτ a + bτ

¶ = λ (τ ) .

(34.1.13)

Furthermore, this is the only way this can happen. Lemma 34.1.16 λ (τ ) = λ (τ 0 ) if and only if τ0 =

aτ + b cτ + d

where 34.1.12 holds. Proof: It only remains to verify that if ℘ (w10 /2) = ℘ (w1 /2) then it is necessary that w1 w10 − ∈M 2 2 w0

with a similar requirement for w2 and w20 . If 21 − w21 ∈ / M, then there exist integers, m, n such that −w10 + mw1 + nw2 2

1020

ELLIPTIC FUNCTIONS

is in the interior of P0 , the period parallelogram whose vertices are 0, w1 , w1 + w2 , and w2 . Therefore, it is possible to choose small a such that Pa contains the pole, −w0 ∗ 0, w21 , and 2 1 + mw1 + nw¡2 but ¢ no other poles of ℘ and in addition, ∂Pa contains w1 no zeros of z → ℘ (z) − ℘ 2 . Then the order of this elliptic function is 2. By assumption, and the fact that ℘ is even, µ ¶ µ ¶ µ 0¶ ³w ´ −w10 −w10 w1 1 ℘ + mw1 + nw2 = ℘ =℘ =℘ . 2 2 2 2 ¡ ¢ −w0 It follows both 2 1 + mw1 + nw2 and w21 are zeros of ℘ (z) − ℘ w21 and so by Theorem 34.1.7 on Page 1006 these are the only two zeros of this function in Pa . Therefore, from Corollary 34.1.9 on Page 1008 w1 w0 − 1 + mw1 + nw2 ∈ M 2 2 w0

which shows w21 − 21 ∈ M. This completes the proof of the lemma. Note the condition in the lemma is equivalent to the condition 34.1.13 because you can relabel the coefficients. The message of either version is that the coefficient of τ in the numerator and denominator is odd while the constant in the numerator and denominator is¶ even. µ µ ¶ 1 0 1 0 Next, ∼ mod 2 and therefore, 2 1 0 1 µ ¶ 2+τ λ = λ (τ + 2) = λ (τ ) . (34.1.14) 1 Thus λ is periodic of period 2. Thus λ leaves invariant a certain subgroup of the unimodular group. According to the next definition, λ is an example of something called a modular function. Definition 34.1.17 When an analytic or meromorphic function is invariant under a group of linear transformations, it is called an automorphic function. A function which is automorphic with respect to a subgroup of the modular group is called a modular function or an elliptic modular function. Now consider what happens for some other unimodular matrices which are not congruent to the identity mod 2. This will yield other functional equations for λ in addition to the fact that λ is periodic of period 2. As before, these functional equations come about because ℘ is unchanged when you change the basis for M, the module of periods. In particular, consider the unimodular matrices µ ¶ µ ¶ 1 0 0 1 , . (34.1.15) 1 1 1 0 Consider the first of these. Thus µ 0 ¶ µ ¶ w1 w1 = w20 w1 + w2


1021

Hence τ 0 = w20 /w10 = (w1 + w2 ) /w1 = 1 + τ . Then from the definition of λ,

λ (τ 0 ) = =

= = = = =

=

=

=

λ (1 + τ ) ³ 0 0´ ³ 0´ w +w w ℘ 1 2 2 − ℘ 22 ³ 0´ ³ 0´ w w ℘ 21 − ℘ 22 ¡ ¢ ¡ ¢ 2 ℘ w1 +w22 +w1 − ℘ w1 +w ¡ w +w 2¢ ¡ w1 ¢ ℘ − ℘ 12 2 ¡ w2 2 ¢ ¡ ¢ 2 ℘ 2 + w1 − ℘ w1 +w 2 ¡ ¢ ¡ ¢ 2 ℘ w21 − ℘ w1 +w 2 ¡ ¢ ¡ ¢ 2 ℘ w22 − ℘ w1 +w 2 ¡ ¢ ¡ ¢ 2 ℘ w21 − ℘ w1 +w 2 ¡ ¢ ¡ ¢ 2 ℘ w1 +w − ℘ w22 2 ¡ ¢ ¡ ¢ − 2 ℘ w21 − ℘ w1 +w 2 ¡ ¢ ¡ ¢ 2 ℘ w1 +w − ℘ w22 2 ¡ ¢ ¡ ¢ ¡ ¢ − ¡ w1 ¢ 2 ℘ 2 − ℘ w22 + ℘ w22 − ℘ w1 +w 2 µ w +w ¶ w ℘( 1 2 2 )−℘( 22 ) w w ℘( 21 )−℘( 22 ) µ w ¶ − w +w ℘( 22 )−℘( 1 2 2 ) 1+ w1 w2 ℘( 2 )−℘( 2 ) µ w +w ¶ w ℘( 1 2 2 )−℘( 22 ) w1 w2 ℘( 2 )−℘( 2 ) µ w +w ¶ w ℘( 1 2 2 )−℘( 22 ) −1 w w ℘( 21 )−℘( 22 ) λ (τ ) . λ (τ ) − 1

(34.1.16)

Summarizing the important feature of the above,

λ (1 + τ ) =

λ (τ ) . λ (τ ) − 1

(34.1.17)

1022

ELLIPTIC FUNCTIONS

Next consider the other unimodular matrix in 34.1.15. In this case w10 = w2 and w20 = w1 . Therefore, τ 0 = w20 /w10 = w1 /w2 = 1/τ . Then λ (τ 0 ) = =

= = =

λ (1/τ ) ³ 0 0´ ³ 0´ w +w w ℘ 1 2 2 − ℘ 22 ³ 0´ ³ 0´ w w ℘ 21 − ℘ 22 ¡ ¢ ¡ ¢ 2 ℘ w1 +w − ℘ w21 2 ¡ ¢ ¡ ¢ ℘ w22 − ℘ w21 e3 − e1 e3 − e2 + e2 − e1 =− e2 − e1 e1 − e2 − (λ (τ ) − 1) = −λ (τ ) + 1.

(34.1.18)

You could try other unimodular matrices and attempt to find other functional equations if you like but this much will suffice here.

34.1.5

A Formula For λ

Recall the formula of Mittag-Leffler for cot (πα) given in 33.2.15. For convenience, here it is. ∞ 1 X 2α + = π cot πα. α n=1 α2 − n2 As explained in the derivation of this formula it can also be written as ∞ X

α = π cot πα. 2 − n2 α n=−∞ Differentiating both sides yields π 2 csc2 (πα)

= =

∞ X

2

n=−∞ ∞ X

=

(α2 − n2 ) 2

(α + n) − 2αn 2

n=−∞

=

α 2 + n2

∞ X

n=−∞

z

2

(α + n) 2

n=−∞ ∞ X

2

(α + n) (α − n)

2

(α + n) (α − n) 1

2.

(α − n)

−

=0 ∞ X

n=−∞

}|

{

2αn (α2 − n2 )

2

(34.1.19)

Now this formula can be used to obtain a formula for λ (τ ) . As pointed out above, λ depends only on the ratio w2 /w1 and so it suffices to take w1 = 1 and

34.1. PERIODIC FUNCTIONS w2 = τ . Thus

1023

¡ ¢ ¡ ¢ ℘ 1+τ − ℘ τ2 2 ¡ ¢ ¡ ¢ . λ (τ ) = ℘ 12 − ℘ τ2

From the original formula for ℘, µ ¶ ³τ ´ 1+τ ℘ −℘ 2 2 X 1 1 − ¡ ¢2 + = ¡ ¢ 2 1+τ τ 2

¡ (k,m)6=(0,0) k −

2

X

=

(k,m)∈Z2

(k,m)∈Z2

1 2

1 1 ¡ ¢ ¢2 − ¡ ¡ ¢ ¢2 1 + m− 2 τ k + m − 12 τ

¡ k−

1 2

1 1 ¡ ¢ ¢2 − ¡ ¡ ¢ ¢2 1 + m− 2 τ k + m − 12 τ

¡ k−

1 2

1 1 ¡ ¢ ¢2 − ¡ ¡ ¢ ¢2 1 + −m − 2 τ k + −m − 12 τ

X

=

(k,m)∈Z2

X

=

(k,m)∈Z2

1 1 ¡ ¢ ¢2 − ¡ ¡ ¢ ¢2 1 + m− 2 τ k + m − 12 τ

¡ k−

X

=

1 2

(34.1.20)

¡1 2

1 1 ¡ ¢ ¢2 − ¡¡ ¢ ¢2 . 1 1 + m+ 2 τ −k m+ 2 τ −k

(34.1.21)

Similarly,

= =

µ ¶ ³τ ´ 1 ℘ −℘ 2 2 1 1 ¡ ¢2 − ¡ ¢2 + 1 2

=

1 2

¡ k−

1 2

X (k,m)∈Z2

¡1 2

1 k−

1 2

+ mτ

¢2 − ¡

¡

1

k+ m−

1 2

¢ ¢2 τ

1

¡ k−

X (k,m)∈Z2

¡

(k,m)6=(0,0)

X (k,m)∈Z2

=

τ 2

X

1 ¢2 − ¡ ¡ ¢ ¢2 + mτ k + m − 21 τ

1

1 ¢2 − ¡ ¡ ¢ ¢2 − mτ k + −m − 12 τ

1 + mτ − k

¢2 − ¡¡

m+

1 ¢

1 2

τ −k

¢2 .

Now use 34.1.19 to sum these over k. This yields, ¶ µ ³τ ´ 1+τ −℘ ℘ 2 2 2 X π π2 ¢ ¢¢ − ¢ ¢ ¡ ¡1 ¡ ¡ ¡ = 2 2 1 sin π 2 + m + 2 τ sin π m + 12 τ m X π2 π2 ¡ ¡ ¢ ¢− ¡ ¡ ¢ ¢ = 2 1 2 cos π m + 2 τ sin π m + 12 τ m

(34.1.22)

1024 and

ELLIPTIC FUNCTIONS

µ ¶ ³τ ´ 1 ℘ −℘ 2 2

=

X m

=

X m

π2 π2 ¡ ¡ ¢¢ ¡ ¡ ¢ ¢ − sin2 π 12 + mτ sin2 π m + 12 τ π2 π2 ¡ ¡ ¢ ¢. − 2 cos2 (πmτ ) sin π m + 12 τ

The following interesting formula for λ results. P 1 1 m cos2 (π (m+ 1 )τ ) − sin2 (π (m+ 1 )τ ) 2 2 . λ (τ ) = P 1 1 − 2 1 2 m cos (πmτ ) sin (π (m+ 2 )τ ) From this it is obvious λ (−τ ) = λ (τ ) . Therefore, from 34.1.18, µ ¶ µ ¶ −1 1 − λ (τ ) + 1 = λ =λ τ τ

(34.1.23)

(34.1.24)

(It is good to recall that λ has been defined for τ ∈ / R.)

34.1.6

Mapping Properties Of λ

The two functional equations, 34.1.24 and 34.1.17 along with some other properties presented above are of fundamental importance. For convenience, they are summarized here in the following lemma. Lemma 34.1.18 The following functional equations hold for λ. µ ¶ λ (τ ) −1 λ (1 + τ ) = , 1 = λ (τ ) + λ λ (τ ) − 1 τ λ (τ + 2) = λ (τ ) ,

(34.1.25) (34.1.26)

λ (z) = λ (w) if and only if there exists a unimodular matrix, µ ¶ µ ¶ a b 1 0 ∼ mod 2 c d 0 1 such that w=

az + b cz + d

(34.1.27)

Consider the following picture. Ω l1

C

r 1 2

l2

r 1


1025

In this picture, l1 is¡ the¢y axis and l2 is the line, x = 1 while C is the top half of the circle centered at 12 , 0 which has radius 1/2. Note the above formula implies λ has real values on l1 which are between 0 and 1. This is because 34.1.23 implies P 1 1 m cos2 (π (m+ 1 )ib) − sin2 (π (m+ 1 )ib) 2 2 P λ (ib) = 1 1 − m cos2 (πmib) sin2 (π (m+ 21 )ib) P 1 1 m cosh2 (π (m+ 1 )b) + sinh2 (π (m+ 1 )b) 2 2 P = ∈ (0, 1) . (34.1.28) 1 1 + 2 m cosh (πmb) sinh2 (π (m+ 21 )b) This follows from the observation that cos (ix) = cosh (x) , sin (ix) = i sinh (x) . Thus it is clear from 34.1.28 that limb→0+ λ (ib) = 1. Next I need to consider the behavior of λ (τ ) as Im (τ ) → ∞. From 34.1.23 listed here for convenience, P 1 1 m cos2 (π (m+ 1 )τ ) − sin2 (π (m+ 1 )τ ) 2 2 λ (τ ) = P , (34.1.29) 1 1 − 2 1 2 m cos (πmτ ) sin (π (m+ 2 )τ ) it follows λ (τ ) = =

1 cos2 (π (− 12 )τ ) 2 cos2 (π ( 12 )τ )

−

−

1 sin2 (π (− 12 )τ )

2 sin2 (π ( 12 )τ )

+

1 cos2 (π 12 τ )

−

1 sin2 (π 21 τ )

+ A (τ )

1 + B (τ ) + A (τ )

1 + B (τ )

(34.1.30)

Where A (τ ) , B (τ ) → 0 as Im (τ ) → ∞. I took out the m = 0 term involving 1/ cos2 (πmτ ) in the denominator and the m = −1 and m = 0 terms in the numerator of 34.1.29. In fact, e−iπ(a+ib) A (a + ib) , e−iπ(a+ib) B (a + ib) converge to zero uniformly in a as b → ∞. Lemma 34.1.19 For A, B defined in 34.1.30, e−iπ(a+ib) C (a + ib) → 0 uniformly in a for C = A, B. Proof: From 34.1.23, e−iπτ A (τ ) =

X m6=0 m6=−1

e−iπτ e−iπτ ¡ ¡ ¢ ¢− ¡ ¡ ¢ ¢ 2 1 cos2 π m + 2 τ sin π m + 12 τ

¡ ¢ Now let τ = a + ib. Then letting αm = π m + 21 , cos (αm a + iαm b) = cos (αm a) cosh (αm b) − i sinh (αm b) sin (αm a) sin (αm a + iαm b) =

sin (αm a) cosh (αm b) + i cos (αm a) sinh (αm b)

1026

ELLIPTIC FUNCTIONS

Therefore, ¯ 2 ¯ ¯cos (αm a + iαm b)¯

Similarly, ¯ 2 ¯ ¯sin (αm a + iαm b)¯

=

cos2 (αm a) cosh2 (αm b) + sinh2 (αm b) sin2 (αm a)

≥

sinh2 (αm b) .

= sin2 (αm a) cosh2 (αm b) + cos2 (αm a) sinh2 (αm b) ≥

sinh2 (αm b) .

It follows that for τ = a + ib and b large ¯ −iπτ ¯ ¯e A (τ )¯ X 2eπb ¡ ¡ ¢ ¢ ≤ 2 sinh π m + 12 b m6=0 m6=−1

≤ =

∞ X

−2 X 2eπb 2eπb ¡ ¡ ¢ ¢ ¡ ¡ ¢ ¢ + sinh2 π m + 12 b sinh2 π m + 12 b m=−∞ m=1 ∞ X

∞ X 2eπb eπb ¢ ¢ ¢ ¢ ¡ ¡ ¡ ¡ 2 = 4 2 2 sinh π m + 12 b sinh π m + 12 b m=1 m=1

Now a short computation shows eπb sinh2 (π (m+1+ 12 )b) eπb sinh2 (π (m+ 12 )b)

¡ ¡ ¢ ¢ sinh2 π m + 12 b 1 ¡ ¡ ¢ ¢ ≤ 3πb . = 2 3 e sinh π m + 2 b

Therefore, for τ = a + ib, ¯ −iπτ ¯ ¯e A (τ )¯

≤

4

¶m ∞ µ X 1 eπb ¡ 3πb ¢ sinh 2 m=1 e3πb

≤

4

eπb 1/e3πb ¡ 3πb ¢ sinh 2 1 − (1/e3πb )

which converges to zero as b → ∞. Similar reasoning will establish the claim about B (τ ) . This proves the lemma. Lemma 34.1.20 limb→∞ λ (a + ib) e−iπ(a+ib) = 16 uniformly in a ∈ R. Proof: From 34.1.30 and Lemma 34.1.19, this lemma will be proved if it is shown Ã ! 2 2 ¡ ¡ ¢ ¢− ¡ ¡ ¢ ¢ e−iπ(a+ib) = 16 lim b→∞ cos2 π 12 (a + ib) sin2 π 12 (a + ib)


1027

uniformly in a ∈ R. Let τ = a + ib to simplify the notation. Then the above expression equals ! Ã 8 8 +¡ π e−iπτ ¡ iπτ π ¢2 π ¢2 e 2 + e−i 2 τ ei 2 τ − e−i 2 τ Ã ! 8eiπτ 8eiπτ = e−iπτ 2 + 2 (eiπτ + 1) (eiπτ − 1) 8 8 = 2 + 2 iπτ iπτ (e + 1) (e − 1) 1 + e2πiτ = 16 2. (1 − e2πiτ ) Now ¯ ¯ ¯ 1 + e2πiτ ¯ ¯ ¯ − 1 ¯ ¯ ¯ (1 − e2πiτ )2 ¯

¯ ¡ ¢2 ¯ ¯ 1 + e2πiτ 1 − e2πiτ ¯¯ ¯ − = ¯ 2¯ ¯ (1 − e2πiτ )2 (1 − e2πiτ ) ¯ ¯ 2πiτ ¯ ¯3e − e4πiτ ¯ 3e−2πb + e−4πb ≤ ≤ 2 2 (1 − e−2πb ) (1 − e−2πb )

and this estimate proves the lemma. Corollary 34.1.21 limb→∞ λ (a + ib) = 0 uniformly in a ∈ R. Also λ (ib) for b > 0 is real and is between 0 and 1, λ is real on the line, l2 and on the curve, C and limb→0+ λ (1 + ib) = −∞. Proof: From Lemma 34.1.20, ¯ ¯ ¯ ¯ ¯λ (a + ib) e−iπ(a+ib) − 16¯ < 1 for all a provided b is large enough. Therefore, for such b, |λ (a + ib)| ≤ 17e−πb . 34.1.28 proves the assertion about λ (−bi) real. By the first part, limb→∞ |λ (ib)| = 0. Now from 34.1.24 ¶¶ µ µ ¶¶ µ µ i −1 = lim 1 − λ = 1. lim λ (ib) = lim 1 − λ b→0+ b→0+ b→0+ ib b

(34.1.31)

by Corollary 34.1.21. Next consider the behavior of λ on line l2 in the above picture. From 34.1.17 and 34.1.28, λ (ib) λ (1 + ib) = 0 but which excludes w if Im (w) < 0. Theorem 34.1.22 Let Ω be the domain described above. Then λ maps Ω one to one and onto the upper half plane of C, {z ∈ C such that Im (z) > 0} . Also, the line λ (l1 ) = (0, 1) , λ (l2 ) = (−∞, 0) , and λ (C) = (1, ∞). Proof: Let Im (w) > 0 and denote by γ the oriented contour described above and illustrated in the above picture. Then the winding number of λ ◦ γ about w equals 1. Thus Z 1 1 dz = 1. 2πi λ◦γ z − w But, splitting the contour integrals into l2 ,the top line, l1 , C1 , C, and C2 and changing variables on each of these, yields Z 1 λ0 (z) 1= dz 2πi γ λ (z) − w and by the theorem on counting zeros, Theorem 28.6.1 on Page 844, the function, z → λ (z) − w has exactly one zero inside the truncated Ω. However, this shows this function has exactly one zero inside Ω because b0 was arbitrary as long as it is sufficiently large. Since w was an arbitrary element of the upper half plane, this verifies the first assertion of the theorem. The remaining claims follow from the above description of λ, in particular the estimate for λ on C2 . This proves the theorem. Note also that the argument in the above proof shows that if Im (w) < 0, then w is not in λ (Ω) . However, if you consider the reflection of Ω about the y axis, then it will follow that λ maps this set one to one onto the lower half plane. The argument will make significant use of Theorem 28.6.3 on Page 846 which is stated here for convenience.

1030

ELLIPTIC FUNCTIONS

Theorem 34.1.23 Let f : B (a, R) → C be analytic and let m

f (z) − α = (z − a) g (z) , ∞ > m ≥ 1 where g (z) 6= 0 in B (a, R) . (f (z) − α has a zero of order m at z = a.) Then there exist ε, δ > 0 with the property that for each z satisfying 0 < |z − α| < δ, there exist points, {a1 , · · · , am } ⊆ B (a, ε) , such that f −1 (z) ∩ B (a, ε) = {a1 , · · · , am } and each ak is a zero of order 1 for the function f (·) − z. Corollary 34.1.24 Let Ω be the region above. Consider the set of points, Q = Ω ∪ Ω0 \ {0, 1} described by the following picture. Ω0

Ω l1

C

r

r −1

1 2

l2

r 1

Then λ (Q) = C\ {0, 1} . Also λ0 (z) 6= 0 for every z in ∪∞ k=−∞ (Q + 2k) ≡ H. Proof: By Theorem 34.1.22, this will be proved if it can be shown that λ (Ω0 ) = {z ∈ C : Im (z) < 0} . Consider λ1 defined on Ω0 by λ1 (x + iy) ≡ λ (−x + iy). Claim: λ1 is analytic. Proof of the claim: You just verify the Cauchy Riemann equations. Letting λ (x + iy) = u (x, y) + iv (x, y) , λ1 (x + iy)

= u (−x, y) − iv (−x, y) ≡

u1 (x, y) + iv (x, y) .

Then u1x (x, y) = −ux (−x, y) and v1y (x, y) = −vy (−x, y) = −ux (−x, y) since λ is analytic. Thus u1x = v1y . Next, u1y (x, y) = uy (−x, y) and v1x (x, y) = vx (−x, y) = −uy (−x, y) and so u1y = −vx . Now recall that on l1 , λ takes real values. Therefore, λ1 = λ on l1 , a set with a limit point. It follows λ = λ1 on Ω0 ∪ Ω. By Theorem 34.1.22 λ maps Ω one to one onto the upper half plane. Therefore, from the definition of λ1 = λ, it follows λ maps Ω0 one to one onto the lower half plane as claimed. This has shown that λ


1031

is one to one on Ω ∪ Ω0 . This also verifies from Theorem 28.6.3 on Page 846 that λ0 6= 0 on Ω ∪ Ω0 . Now consider the lines l2 and C. If λ0 (z) = 0 for z ∈ l2 , a contradiction can be obtained. Pick such a point. If λ0 (z) = 0, then z is a zero of order m ≥ 2 of the function, λ − λ (z) . Then by Theorem 28.6.3 there exist δ, ε > 0 such that if w ∈ B (λ (z) , δ) , then λ−1 (w) ∩ B (z, ε) contains at least m points.

Ω

0

Ω l1

z1 r r z B(z, ε)

C

l2 λ(z1 ) r

r −1

r 1 2

λ(z) r

r 1

B(λ(z), δ)

In particular, for z1 ∈ Ω ∩ B (z, ε) sufficiently close to z, λ (z1 ) ∈ B (λ (z) , δ) and so the function λ − λ (z1 ) has at least two distinct zeros. These zeros must be in B (z, ε) ∩ Ω because λ (z1 ) has positive imaginary part and the points on l2 are mapped by λ to a real number while the points of B (z, ε) \ Ω are mapped by λ to the lower half plane thanks to the relation, λ (z + 2) = λ (z) . This contradicts λ one to one on Ω. Therefore, λ0 6= 0 on l2 . Consider C. Points on C are of the form 1 − τ1 where τ ∈ l2 . Therefore, using 34.1.33, µ ¶ 1 λ (τ ) − 1 λ 1− = . τ λ (τ ) Taking the derivative of both sides, µ ¶µ ¶ 1 1 λ0 (τ ) λ0 1 − = 2 6= 0. τ τ2 λ (τ ) Since λ is periodic of period 2 it follows λ0 (z) 6= 0 for all z ∈ ∪∞ k=−∞ (Q + 2k) . µ Lemma 34.1.25 If Im (τ ) > 0 then there exists a unimodular c + dτ a + bτ

a b c d

¶ such that

1032

ELLIPTIC FUNCTIONS

¯ ¯ ¯ c+dτ ¯ is contained in the interior of Q. In fact, ¯ a+bτ ¯ ≥ 1 and µ −1/2 ≤ Re

c + dτ a + bτ

¶ ≤ 1/2.

Proof: Letting a basis for the module of periods of ℘ be {1, τ } , it follows from Theorem 34.1.2 on Page 1004 that there exists a basis for the same module of periods, {w10 , w20 } with the property that for τ 0 = w20 /w10 |τ 0 | ≥ 1,

−1 1 ≤ Re τ 0 ≤ . 2 2

Since this µ is a basis ¶ for the same module of periods, there exists a unimodular a b matrix, such that c d µ

w10 w20

¶

µ =

a b c d

¶µ

¶

1 τ

.

Hence, τ0 =

w20 c + dτ = . 0 w1 a + bτ

Thus τ 0 is in the interior of H. In fact, it is on the interior of Ω0 ∪ Ω ≡ Q.

0 τ s

−1

34.1.7

s −1/2

0

s 1/2

1

A Short Review And Summary

With this lemma, it is easy to extend Corollary 34.1.24. First, a simple observation and review is a good idea. Recall that when you change the basis for the module of periods, the Weierstrass ℘ function does not change and so the set of ei used in defining λ also do not change. Letting the new basis be {w10 , w20 } , it was shown that µ 0 ¶ µ ¶µ ¶ w1 a b w1 = w20 c d w2


1033 µ

for some unimodular transformation, w20 /w10 τ0 = Now as discussed earlier

a b c d

=

. Letting τ = w2 /w1 and τ 0 =

c + dτ ≡ φ (τ ) a + bτ ³

λ (τ 0 ) =

¶

³ 0´ w − ℘ 22 ³ 0´ ³ 0´ λ (φ (τ )) ≡ w w ℘ 21 − ℘ 22 ³ ´ ³ 0´ 0 τ ℘ 1+τ − ℘ 2 2 ¡1¢ ¡ τ0 ¢ ℘ 2 −℘ 2 ℘

w10 +w20 2

´

¡ 1+τ ¢ ¡ τ ¢ These ¡ 1 ¢ numbers in the above fraction must be the same as ℘ 2 , ℘ 2 , and ℘ 2 but they might occur differently. This is because ℘ does not change and these numbers are the zeros of a polynomial having coefficients involving only numbers ³ ´ ¡ ¢ 1+τ 0 and ℘ (z) . It could happen for example that ℘ = ℘ τ2 in which case this 2 would change the value of λ. In effect, you can keep track of all possibilities by −e2 simply permuting the ei in the formula for λ (τ ) given by ee13 −e . Thus consider the 2 following permutation table. 1 2 3 2 3 1 3 1 2 . 2 1 3 1 3 2 3 2 1 Corresponding to this list of 6 permutations, all possible formulas for λ (φ (τ )) can be obtained as follows. Letting τ 0 = φ (τ ) where φ is a unimodular matrix corresponding to a change of basis, λ (τ 0 ) = λ (τ 0 ) =

e3 − e2 = λ (τ ) e1 − e2

e1 − e3 e3 − e2 + e2 − e1 1 λ (τ ) − 1 = =1− = e2 − e3 e3 − e2 λ (τ ) λ (τ ) λ (τ 0 )

= =

0

λ (τ ) = =

· ¸−1 e2 − e1 e3 − e2 − (e1 − e2 ) =− e3 − e1 e1 − e2 1 −1 − [λ (τ ) − 1] = 1 − λ (τ ) · ¸ e3 − e1 e3 − e2 − (e1 − e2 ) =− e2 − e1 e1 − e2 − [λ (τ ) − 1] = 1 − λ (τ )

(34.1.34) (34.1.35)

(34.1.36)

(34.1.37)

1034

ELLIPTIC FUNCTIONS

λ (τ 0 ) =

e2 − e3 e3 − e2 1 λ (τ ) = = = 1 e1 − e3 e3 − e2 − (e1 − e2 ) λ (τ )−1 1 − λ(τ ) λ (τ 0 ) =

e1 − e3 1 = e3 − e2 λ (τ )

(34.1.38) (34.1.39)

Corollary 34.1.26 λ0 (τ ) 6= 0 for all τ in the upper half plane, denoted by P+ . Proof: Let τ ∈ P+ . By Lemma 34.1.25 there exists φ a unimodular transformation and τ 0 in the interior of Q such that τ 0 = φ (τ ). Now from the definition of λ in terms of the ei , there is at worst a permutation of the ei and so it might be the case that λ (φ (τ )) 6= λ (τ ) but it is the case that λ (φ (τ )) = ξ (λ (τ )) where ξ 0 (z) 6= 0. Here ξ is one of the functions determined by 34.1.34 - 34.1.39. (Since λ (τ ) ∈ / {0, 1} , ξ 0 (λ (z)) 6= 0. This follows from the above possibilities for ξ listed above in 34.1.34 - 34.1.39.) All the possibilities are ξ (z) = z,

1 z 1 z−1 , , 1 − z, , z 1−z z−1 z

and these are the same as the possibilities for ξ −1 . Therefore, λ0 (φ (τ )) φ0 (τ ) = ξ 0 (λ (τ )) λ0 (τ ) and so λ0 (τ ) 6= 0 as claimed. Now I will present a lemma which is of major significance. It depends on the remarkable mapping properties of the modular function and the monodromy theorem from analytic continuation. A review of the monodromy theorem will be listed here for convenience. First recall the definition of the concept of function elements and analytic continuation. Definition 34.1.27 A function element is an ordered pair, (f, D) where D is an open ball and f is analytic on D. (f0 , D0 ) and (f1 , D1 ) are direct continuations of each other if D1 ∩D0 6= ∅ and f0 = f1 on D1 ∩D0 . In this case I will write (f0 , D0 ) ∼ (f1 , D1 ) . A chain is a finite sequence, of disks, {D0 , · · · , Dn } such that Di−1 ∩Di 6= ∅. If (f0 , D0 ) is a given function element and there exist function elements, (fi , Di ) such that {D0 , · · · , Dn } is a chain and (fj−1 , Dj−1 ) ∼ (fj , Dj ) then (fn , Dn ) is called the analytic continuation of (f0 , D0 ) along the chain {D0 , · · · , Dn }. Now suppose γ is an oriented curve with parameter interval [a, b] and there exists a chain, {D0 , · · · , Dn } such that γ ∗ ⊆ ∪nk=1 Dk , γ (a) is the center of D0 , γ (b) is the center of Dn , and there is an increasing list of numbers in [a, b] , a = s0 < s1 · · · < sn = b such that γ ([si , si+1 ]) ⊆ Di and (fn , Dn ) is an analytic continuation of (f0 , D0 ) along the chain. Then (fn , Dn ) is called an analytic continuation of (f0 , D0 ) along the curve γ. (γ will always be a continuous curve. Nothing more is needed. ) Then the main theorem is the monodromy theorem listed next, Theorem 31.4.5 and its corollary on Page 933. Theorem 34.1.28 Let Ω be a simply connected subset of C and suppose (f, B (a, r)) is a function element with B (a, r) ⊆ Ω. Suppose also that this function element can be analytically continued along every curve through a. Then there exists G analytic on Ω such that G agrees with f on B (a, r).


1035

Here is the lemma. Lemma 34.1.29 Let λ be the modular function defined on P+ the upper half plane. Let V be a simply connected region in C and let f : V → C\ {0, 1} be analytic and nonconstant. Then there exists an analytic function, g : V → P+ such that λ◦g = f. Proof: Let a ∈ V and choose r0 small enough that f (B (a, r0 )) contains neither 0 nor 1. You need only let B (a, r0 ) ⊆ V . Now there exists a unique point in Q, τ 0 such that λ (τ 0 ) = f (a). By Corollary 34.1.24, λ0 (τ 0 ) 6= 0 and so by the open mapping theorem, Theorem 28.6.3 on Page 846, There exists B (τ 0 , R0 ) ⊆ P+ such that λ is one to one on B (τ 0 , R0 ) and has a continuous inverse. Then picking r0 still smaller, it can be assumed f (B (a, r0 )) ⊆ λ (B (τ 0 , R0 )). Thus there exists a local inverse for λ, λ−1 defined on f (B (a, r0 )) having values in B (τ 0 , R0 ) ∩ 0 λ−1 (f (B (a, r0 ))). Then defining g0 ≡ λ−1 0 ◦ f, (g0 , B (a, r0 )) is a function element. I need to show this can be continued along every curve starting at a in such a way that each function in each function element has values in P+ . Let γ : [α, β] → V be a continuous curve starting at a, (γ (α) = a) and suppose that if t < T there exists a nonnegative integer m and a function element (gm , B (γ (t) , rm )) which is an analytic continuation of (g0 , B (a, r0 )) along γ where gm (γ (t)) ∈ P+ and each function in every function element for j ≤ m has values in P+ . Thus for some small T > 0 this has been achieved. Then consider f (γ (T )) ∈ C\ {0, 1} . As in the first part of the argument, there exists a unique τ T ∈ Q such that λ (τ T ) = f (γ (T )) and for r small enough there is −1 an analytic local inverse, λ−1 (f (B (γ (T ) , r))) ∩ T between f (B (γ (T ) , r)) and λ B (τ T , RT ) ⊆ P+ for some RT > 0. By the assumption that the analytic continuation can be carried out for t < T, there exists {t0 , · · · , tm = t} and function elements (gj , B (γ (tj ) , rj )) , j = 0, · · · , m as just described with gj (γ (tj )) ∈ P+ , λ ◦ gj = f on B (γ (tj ) , rj ) such that for t ∈ [tm , T ] , γ (t) ∈ B (γ (T ) , r). Let I = B (γ (tm ) , rm ) ∩ B (γ (T ) , r) . Then since λ−1 T is a local inverse, it follows for all z ∈ I ¡ ¢ λ (gm (z)) = f (z) = λ λ−1 T ◦ f (z) Pick z0 ∈ I . Then by Lemma 34.1.18 on Page 1024 there exists a unimodular mapping of the form az + b φ (z) = cz + d where µ ¶ µ ¶ a b 1 0 ∼ mod 2 c d 0 1 such that

¡ ¢ gm (z0 ) = φ λ−1 T ◦ f (z0 ) . ¡ ¢ Since both gm (z0 ) and φ λ−1 T ◦ f (z0 ) are in the upper half plane, it follows ad − cb = 1 and φ maps the upper half plane to the upper half plane. Note the pole of

1036

ELLIPTIC FUNCTIONS

φ is real and all the sets being considered are contained in the upper half plane so φ is analytic where it needs to be. Claim: For all z ∈ I, gm (z) = φ ◦ λ−1 T ◦ f (z) .

(34.1.40)

Proof: For z = z0 the equation holds. Let © ¡ ¢ª A = z ∈ I : gm (z) = φ λ−1 . T ◦ f (z) Thus z0 ∈ I. If z ∈ I and if w is close enough to z, then w ∈ I also and so both sides of 34.1.40 with w in place of z are in λ−1 m (f (I)) . But by construction, λ is one to one on this set and since λ is invariant with respect to φ, ¡ ¢ ¡ ¢ −1 λ (gm (w)) = λ λ−1 T ◦ f (w) = λ φ ◦ λT ◦ f (w) and consequently, w ∈ A. This shows A is open. But A is also closed in I because the functions are continuous. Therefore, A = I and so 34.1.40 is obtained. Letting f (z) ∈ f (B (γ (T )) , r) , ¡ ¡ ¢¢ ¡ ¢ λ φ λ−1 = λ λ−1 T (f (z)) T (f (z)) = f (z) and so φ ◦  λ−1 T is a local inverse forλ on f (B (γ (T )) , r) . Let the new function gm+1 z }| {   element be φ ◦ λ−1 T ◦ f , B (γ (T ) , r) . This has shown the initial function element can be continued along every curve through a. By the monodromy theorem, there exists g analytic on V such that g has values in P+ and g = g0 on B (a, r0 ) . By the construction, it also follows λ ◦ g = f . This last claim is easy to see because λ ◦ g = f on B (a, r0 ) , a set with a limit point so the equation holds for all z ∈ V . This proves the lemma.

34.2

The Picard Theorem Again

Having done all this work on the modular function which is important for its own sake, there is an easy proof of the Picard theorem. In fact, this is the way Picard did it in 1879. I will state it slightly differently since it is no trouble to do so, [32]. Theorem 34.2.1 Let f be meromorphic on C and suppose f misses three distinct points, a, b, c. Then f is a constant function. b−c Proof: Let φ (z) ≡ z−a z−c b−a . Then φ (c) = ∞, φ (a) = 0, and φ (b) = 1. Now consider the function, h = φ ◦ f. Then h misses the three points ∞, 0, and 1. Since h is meromorphic and does not have ∞ in its values, it must actually be analytic. Thus h is an entire function which misses the two values 0 and 1. If h is not constant, then by Lemma 34.1.29 there exists a function, g analytic on C which has values

34.3. EXERCISES

1037

in the upper half plane, P+ such that λ ◦ g = h. However, g must be a constant because there exists ψ an analytic map on the upper half plane which maps the upper half plane to B (0, 1) . You can use the Riemann mapping theorem or more simply, ψ (z) = z−i z+i . Thus ψ ◦ g equals a constant by Liouville’s theorem. Hence g is a constant and so h must also be a constant because λ (g (z)) = h (z) . This proves f is a constant also. This proves the theorem.

34.3

Exercises

1. Show the set of modular transformations is a group. Also show those modular transformations which are congruent mod 2 to the identity as described above is a subgroup. 2. Suppose f is an elliptic function with period module M. If {w1 , w2 } and {w10 , w20 } are two bases, show that the resulting period parallelograms resulting from the two bases have the same area. 3. Given a module of periods with basis {w1 , w2 } and letting a typical element of this module be denoted by w as described above, consider the product Y³ z ´ (z/w)+ 1 (z/w)2 2 σ (z) ≡ z e 1− . w w6=0

Show this product converges uniformly on compact sets, is an entire function, and satisfies σ 0 (z) /σ (z) = ζ (z) where ζ (z) was defined above as a primitive of ℘ (z) and is given by ζ (z) =

1 X 1 z 1 + + 2+ . z z−w w w w6=0

4. Show ζ (z + wi ) = ζ (z) + η i where η i is a constant. 5. Let Pa be the parallelogram shown in the following picture. » » »»» £ » » » » w2 »» » £ £ » s »»» » » £ £ »£» £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ £ » » £ £s w1 £ £ »»»£ £ » £ 0»»» » £s »»» £ » » » £ »» s» a

1038

ELLIPTIC FUNCTIONS

R 1 ζ (z) dz = 1 where the contour is taken once around the Show that 2πi ∂Pa parallelogram in the counter clockwise direction. Next evaluate this contour integral directly to obtain Legendre’s relation, η 1 w2 − η 2 w1 = 2πi. 6. For σ defined in Problem 3, 4 explain the following steps. For j = 1, 2 σ 0 (z + wj ) σ 0 (z) = ζ (z + wj ) = ζ (z) + η j = + ηj σ (z + wj ) σ (z) Therefore, there exists a constant, Cj such that σ (z + wj ) = Cj σ (z) eηj z . Next show σ is an odd function, (σ (−z) = −σ (z)) and then let z = −wj /2 η j wj to find Cj = −e 2 and so σ (z + wj ) = −σ (z) eηj (z+

wj 2

).

(34.3.41)

7. Show any even elliptic function, f with periods w1 and w2 for which 0 is neither a pole nor a zero can be expressed in the form f (0)

n Y ℘ (z) − ℘ (ak ) ℘ (z) − ℘ (bk )

k=1

where C is some constant. Here ℘ is the Weierstrass function which comes from the two periods, w1 and w2 . Hint: You might consider the above function in terms of the poles and zeros on a period parallelogram and recall that an entire function which is elliptic is a constant. 8. Suppose f is any elliptic function with {w1 , w2 } a basis for the module of periods. Using Theorem 34.1.8 and 34.3.41 show that there exists constants a1 , · · · , an and b1 , · · · , bn such that for some constant C, f (z) = C

n Y σ (z − ak ) . σ (z − bk )

k=1

Hint: You might try something like this: By Theorem 34.1.8, it follows that if {α P k } arePthe zeros and {bk } the poles in an appropriate period P parallelogram, P αk − bk equals a period. Replace αk with ak such that ak − bk = 0. Then use 34.3.41 to show that the given formula for f is bi periodic. Anyway, you try to arrange things such that the given formula has the same poles as f. Remember an entire elliptic function equals a constant. 9. Show that the map τ → 1 − τ1 maps l2 onto the curve, C in the above picture on the mapping properties of λ. 10. Modify the proof of Theorem 34.1.22 to show that λ (Ω)∩{z ∈ C : Im (z) < 0} = ∅.

Part IV

Sobolev Spaces

1039

Weak Derivatives 35.1

Weak ∗ Convergence

A very important sort of convergence in applications of functional analysis is the concept of weak or weak ∗ convergence. It is important because it allows you to assert the existence of a convergent subsequence of a given bounded sequence. The only problem is the convergence is very weak so it does not tell you as much as you would like. Nevertheless, it is a very useful concept. The big theorems in the subject are the Eberlein Smulian theorem and the Banach Alaoglu theorem about the weak or weak ∗ compactness of the closed unit balls in either a Banach space or its dual space. These theorems are proved in Yosida [67]. Here I will present a special case which turns out to be by far the most important in applications and it is not hard to get from the Riesz representation theorem for Lp . First I define weak and weak ∗ convergence. Definition 35.1.1 Let X 0 be the dual of a Banach space X and let {x∗n } be a sequence of elements of X 0 . Then x∗n converges weak ∗ to x∗ if and only if for all x ∈ X, lim x∗n (x) = x∗ (x) . n→∞

A sequence in X, {xn } converges weakly to x ∈ X if and only if for all x∗ ∈ X 0 lim x∗ (xn ) = x∗ (x) .

n→∞

The main result is contained in the following lemma. Lemma 35.1.2 Let X 0 be the dual of a Banach space, X and suppose X is separable. Then if {x∗n } is a bounded sequence in X 0 , there exists a weak ∗ convergent subsequence. Proof: Let D be a dense countable set in X. Then the sequence, {x∗n (x)} is bounded for all x and in particular for all x ∈ D. Use the Cantor diagonal process to obtain a subsequence, still denoted by n such that x∗n (d) converges for each d ∈ D. 1041

1042

WEAK DERIVATIVES

Now let x ∈ X be completely arbitrary. In fact {x∗n (x)} is a Cauchy sequence. Let ε > 0 be given and pick d ∈ D such that for all n |x∗n (x) − x∗n (d)|
Nε , |x∗n (d) − x∗m (d)|
ε} and let Em ≡ E ∩ B(0, m). Is m (Em ) = 0? If not, there exists an open set, V , and a compact set K satisfying K ⊆ Em ⊆ V ⊆ B (0, m) , m (V \ K) < 4−1 m (Em ) , Z |f | dx < ε4−1 m (Em ) . V \K

Let H and W be open sets satisfying K⊆H⊆H⊆W ⊆W ⊆V and let H≺g≺W where the symbol, ≺, in the above implies spt (g) ⊆ W, g has all values in [0, 1] , and g (x) = 1 on H. Then let φδ be a mollifier and let h ≡ g ∗ φδ for δ small enough that K ≺ h ≺ V. Thus

Z 0

= ≥

Z f hdx =

Z f dx +

K −1

f hdx V \K

≥

εm (K) − ε4 m (Em ) ¡ ¢ ε m (Em ) − 4−1 m (Em ) − ε4−1 m (Em )

≥

2−1 εm(Em ).

Therefore, m (Em ) = 0, a contradiction. Thus m (E) ≤

∞ X

m (Em ) = 0

m=1

and so, since ε > 0 is arbitrary, m ({x : f (x) > 0}) = 0. Similarly m ({x : f (x) < 0}) = 0. This proves the lemma. This lemma allows the following definition. Definition 35.3.4 Let U be an open subset of Rn and let u ∈ L1loc (U ) . Then Dα u ∈ L1loc (U ) if there exists a function g ∈ L1loc (U ), necessarily unique by Lemma 35.3.3, such that for all φ ∈ Cc∞ (U ), Z Z |α| gφdx = Dα u (φ) ≡ (−1) u (Dα φ) dx. U α

U

Then D u is defined to equal g when this occurs.

1048

WEAK DERIVATIVES

Lemma 35.3.5 Let u ∈ L1loc (Rn ) and suppose u,i ∈ L1loc (Rn ), where the subscript on the u following the comma denotes the ith weak partial derivative. Then if φε is a mollifier and uε ≡ u ∗ φε , it follows uε,i ≡ u,i ∗ φε . Proof: If ψ ∈ Cc∞ (Rn ), then Z u (x − y) ψ ,i (x) dx

Z =

u (z) ψ ,i (z + y) dz Z = − u,i (z) ψ (z + y) dz Z = − u,i (x − y) ψ (x) dx.

Therefore, Z uε,i (ψ) =

Z Z

−

uε ψ ,i = − u (x − y) φε (y) ψ ,i (x) d ydx Z Z − u (x − y) ψ ,i (x) φε (y) dxdy Z Z u,i (x − y) ψ (x) φε (y) dxdy Z u,i ∗ φε (x) ψ (x) dx.

= = =

The technical questions about product measurability in the use of Fubini’s theorem may be resolved by picking a Borel measurable representative for u. This proves the lemma. What about the product rule? Does it have some form in the context of weak derivatives? Lemma 35.3.6 Let U be an open set, ψ ∈ C ∞ (U ) and suppose u, u,i ∈ Lploc (U ). Then (uψ),i and uψ are in Lploc (U ) and (uψ),i = u,i ψ + uψ ,i . Proof: Let φ ∈ Cc∞ (U ) then (uψ),i (φ)

Z ≡ − ZU

uψφ,i dx

= − u[(ψφ),i − φψ ,i ]dx U Z ¡ ¢ = u,i ψφ + uψ ,i φ dx ZU ¡ ¢ = u,i ψ + uψ ,i φdx U


35.4. MORREY’S INEQUALITY

1049

Recall the notation for the gradient of a function. T

∇u (x) ≡ (u,1 (x) · · · u,n (x)) thus Du (x) v =∇u (x) · v.

35.4

Morrey’s Inequality

The following inequality will be called Morrey’s inequality. It relates an expression which is given pointwise to an integral of the pth power of the derivative. Lemma 35.4.1 Let u ∈ C 1 (Rn ) and p > n. Then there exists a constant, C, depending only on n such that for any x, y ∈ Rn , |u (x) − u (y)| ÃZ

!1/p p

≤C

|∇u (z) | dz

³ ´ | x − y|(1−n/p) .

(35.4.1)

B(x,2|x−y|)

Proof: In the argument C will be a generic constant which depends on n. Consider the following picture.

U xr W

ry V

This is a picture of two balls of radius r in Rn , U and V having centers at x and y respectively, which intersect in the set, W. The center of U is on the boundary of V and the center of V is on the boundary of U as shown in the picture. There exists a constant, C, independent of r depending only on n such that m (W ) m (W ) = = C. m (U ) m (V ) You could compute this constant if you desired but it is not important here. Define the average of a function over a set, E ⊆ Rn as follows. Z − f dx ≡ E

1 m (E)

Z f dx. E

1050

WEAK DERIVATIVES

Then Z |u (x) − u (y)| = − |u (x) − u (y)| dz ZW Z ≤ − |u (x) − u (z)| dz + − |u (z) − u (y)| dz W W ¸ ·Z Z C |u (x) − u (z)| dz + |u (z) − u (y)| dz = m (U ) W W ·Z ¸ Z ≤ C − |u (x) − u (z)| dz + − |u (y) − u (z)| dz U

V

Now consider these two terms. Using spherical coordinates and letting U0 denote the ball of the same radius as U but with center at 0, Z − |u (x) − u (z)| dz U Z 1 |u (x) − u (z + x)| dz m (U0 ) U0 Z r Z 1 n−1 ρ |u (x) − u (ρw + x)| dσ (w) dρ m (U0 ) 0 n−1 Z r ZS Z ρ 1 n−1 ρ |∇u (x + tw) · w| dtdσdρ m (U0 ) 0 n−1 Z r ZS Z0 ρ 1 ρn−1 |∇u (x + tw)| dtdσdρ m (U0 ) 0 n−1 0 Z rZ Z r S 1 C |∇u (x + tw)| dtdσdρ r 0 S n−1 0 Z Z Z r 1 r |∇u (x + tw)| n−1 C t dtdσdρ r 0 S n−1 0 tn−1

= = ≤ ≤ ≤ =

Z

= = ≤

Z

r

|∇u (x + tw)| n−1 t dtdσ tn−1 n−1 0 ZS |∇u (x + z)| C dz n−1 |z| U0 µZ ¶1/p µZ ¶1/p0 p p0 −np0 C |∇u (x + z)| dz |z| C

U0

µZ =

p

C

Z

r

|∇u (z)| dz µZ p

C

ρ S n−1

U

=

U

¶1/p µZ ¶1/p ÃZ

Z S n−1

ρ

0 r

|∇u (z)| dz U

p0 −np0 n−1

0

1 n−1

ρ p−1

¶(p−1)/p dρdσ

!(p−1)/p dρdσ

35.4. MORREY’S INEQUALITY µ =

C µ

=

C

p−1 p−n p−1 p−n

1051

¶(p−1)/p µZ

¶1/p

p

U

¶(p−1)/p µZ

p

n

r1− p

|∇u (z)| dz ¶1/p

|∇u (z)| dz

|x − y|

1− n p

U

Similarly, µ ¶(p−1)/p µZ ¶1/p Z p−1 1− n p |x − y| p − |u (y) − u (z)| dz ≤ C |∇u (z)| dz p−n V V Therefore, µ |u (x) − u (y)| ≤ C

¶(p−1)/p ÃZ

p−1 p−n

!1/p p

|∇u (z)| dz

1− n p

|x − y|

B(x,2|x−y|)

because B (x, 2 |x − y|) ⊇ V ∪ U. This proves the lemma. The following corollary is also interesting Corollary 35.4.2 Suppose u ∈ C 1 (Rn ) . Then |u (y) − u (x) − ∇u (x) · (y − x)| Ã ≤C

1 m (B (x, 2 |x − y|))

!1/p

Z p

|∇u (z) − ∇u (x) | dz

| x− y|. (35.4.2)

B(x,2|x−y|)

Proof: This follows easily from letting g (y) ≡ u (y) − u (x) − ∇u (x) · (y − x) . Then g ∈ C 1 (Rn ), g (x) = 0, and ∇g (z) = ∇u (z) − ∇u (x) . From Lemma 35.4.1, |u (y) − u (x) − ∇u (x) · (y − x)| = |g (y)| = |g (y) − g (x)| ÃZ !1/p 1− n ≤ C |∇u (z) − ∇u (x) |p dz |x − y| p B(x,2|x−y|)

Ã = C

1 m (B (x, 2 |x − y|))

!1/p

Z |∇u (z) − ∇u (x) |p dz

| x − y|.

B(x,2|x−y|)

This proves the corollary. It may be interesting at this point to recall the definition of differentiability on Page 121. If you knew the above inequality held for ∇u having components in L1loc (Rn ) , then at Lebesgue points of ∇u, the above would imply Du (x) exists. This is exactly the approach taken below.

1052

WEAK DERIVATIVES

35.5

Rademacher’s Theorem

The inequality of Corollary 35.4.2 can be extended to the case where u and u,i are in Lploc (Rn ) for p > n. This leads to an elegant proof of the differentiability a.e. of a Lipschitz continuous function as well as a more general theorem. Theorem 35.5.1 Suppose u and all its weak partial derivatives, u,i are in Lploc (Rn ). Then there exists a set of measure zero, E such that if x, y ∈ / E then inequalities 35.4.2 and 35.4.1 are both valid. Furthermore, u equals a continuous function a.e. Proof: Let u ∈ Lploc (Rn ) and ψ k ∈ Cc∞ (Rn ) , ψ k ≥ 0, and ψ k (z) = 1 for all z ∈ B (0, k). Then it is routine to verify that uψ k , (uψ k ),i ∈ Lp (Rn ). Here is why: Z (uψ k ),i (φ)

≡ − ZR

uψ k φ,i dx

n

Z

= −

uψ k φ,i dx −

Rn

Z Rn

uψ k,i φdx +

Z Z = − u (ψ k φ),i dx + uψ k,i φdx n Rn Z R ¡ ¢ = u,i ψ k + uψ k,i φdx

Rn

uψ k,i φdx

Rn

which shows (uψ k ),i = u,i ψ k + uψ k,i as expected. Let φε be a mollifier and consider (uψ k )ε ≡ uψ k ∗ φε . By Lemma 35.3.5 on Page 1048, (uψ k )ε,i = (uψ k ),i ∗ φε . Therefore (uψ k )ε,i → (uψ k ),i in Lp (Rn )

(35.5.3)

(uψ k )ε → uψ k in Lp (Rn )

(35.5.4)

and as ε → 0. By 35.5.4, there exists a subsequence ε → 0 such that for |z| < k and for each i = 1, 2, · · · , n (uψ k )ε,i (z) → (uψ k ),i (z) = u,i (z) a.e.

35.5. RADEMACHER’S THEOREM

1053

(uψ k )ε (z) → uψ k (z) = u (z) a.e.

(35.5.5)

Denoting the exceptional set by Ek , let ∞

x, y ∈ / ∪k=1 Ek ≡ E and let k be so large that B (0,k) ⊇ B (x,2|x − y|). Then by 35.4.1 and for x, y ∈ / E, |(uψ k )ε (x) − (uψ k )ε (y)| ÃZ

!1/p

≤C B(x,2|y−x|)

|∇(uψ k )ε |p dz

|x − y|

(1−n/p)

where C depends only on n. Similarly, by 35.4.2, |(uψ k )ε (x) − (uψ k )ε (y) − ∇ (uψ k )ε (x) · (y − x)| ≤ Ã C

1 m (B (x, 2 |x − y|))

!1/p

Z p

B(x,2|x−y|)

|∇ (uψ k )ε (z) − ∇ (uψ k )ε (x) | dz

| x − y|.

Now by 35.5.5 and 35.5.3 passing to the limit as ε → 0 yields ÃZ

!1/p p

|u (x) − u (y)| ≤ C

|∇u| dz

(1−n/p)

|x − y|

(35.5.6)

B(x,2|y−x|)

and |u (y) − u (x) − ∇u (x) · (y − x)| Ã ≤C

1 m (B (x, 2 |x − y|))

!1/p

Z p

|∇u (z) − ∇u (x) | dz

| x− y|. (35.5.7)

B(x,2|x−y|)

Redefining u on the set of mesure zero, E yields 35.5.6 for all x, y. This proves the theorem. Corollary 35.5.2 Let u, u,i ∈ Lploc (Rn ) for i = 1, · · · , n and p > n. Then the representative of u described in Theorem 35.5.1 is differentiable a.e. Proof: From Theorem 35.5.1 |u (y) − u (x) − ∇u (x) · (y − x)| Ã ≤C

1 m (B (x, 2 |x − y|))

!1/p

Z p

|∇u (z) − ∇u (x) | dz B(x,2|x−y|)

| x− y|. (35.5.8)

1054

WEAK DERIVATIVES

and at every Lebesgue point, x of ∇u Ã !1/p Z 1 p lim |∇u (z) − ∇u (x) | dz =0 y→x m (B (x, 2 |x − y|)) B(x,2|x−y|) and so at each of these points, |u (y) − u (x) − ∇u (x) · (y − x)| =0 y→x |x − y| lim

which says that u is differentiable at x and Du (x) (v) = ∇u (x) · (v) . See Page 121. This proves the corollary. Definition 35.5.3 Now suppose u is Lipschitz on Rn , |u (x) − u (y)| ≤ K |x − y| for some constant K. Define Lip (u) as the smallest value of K that works in this inequality. The following corollary is known as Rademacher’s theorem. It states that every Lipschitz function is differentiable a.e. Corollary 35.5.4 If u is Lipschitz continuous then u is differentiable a.e. and ||u,i ||∞ ≤ Lip (u). Proof: This is done by showing that Lipschitz continuous functions have weak derivatives in L∞ (Rn ) and then using the previous results. Let Dehi u (x) ≡ h−1 [u (x + hei ) − u (x)]. Then Dehi u is bounded in L∞ (Rn ) and ||Dehi u||∞ ≤ Lip (u). It follows that Dehi u is contained in a ball in L∞ (Rn ), the dual space of L1 (Rn ). By Theorem 35.1.3 on Page 1042, there is a subsequence h → 0 such that Dehi u * w, ||w||∞ ≤ Lip (u) where the convergence takes place in the weak ∗ topology of L∞ (Rn ). Let φ ∈ Cc∞ (Rn ). Then Z Z wφdx = lim Dehi uφdx h→0 Z (φ (x − hei ) − φ (x)) = lim u (x) dx h→0 h Z = − u (x) φ,i (x) dx. Thus w = u,i and u,i ∈ L∞ (Rn ) for each i. Hence u, u,i ∈ Lploc (Rn ) for all p > n and so u is differentiable a.e. by Corollary 35.5.2. This proves the corollary.

35.6. CHANGE OF VARIABLES FORMULA LIPSCHITZ MAPS

35.6

1055

Change Of Variables Formula Lipschitz Maps

With Rademacher’s theorem, one can give a general change of variables formula involving Lipschitz maps. Definition 35.6.1 Let E be a Lebesgue measurable set.x ∈ E is a point of density if m(E ∩ B(x, r)) lim = 1. r→0 m(B(x, r)) You see that if x were an interior point of E, then this limit will equal 1. However, it is sometimes the case that the limit equals 1 even when x is not an interior point. In fact, these points of density make sense even for sets that have empty interior. Lemma 35.6.2 Let E be a Lebesgue measurable set. Then there exists a set of measure zero, N , such that if x ∈ E \ N , then x is a point of density of E. Proof: Consider the function, f (x) = XE (x). This function is in L1loc (Rn ). Let N C denote the Lebesgue points of f . Then for x ∈ E \ N , Z 1 1 = XE (x) = lim XE (y) dmn r→0 mn (B (x, r)) B(x,r) =

lim

r→0

mn (B (x, r) ∩ E) . mn (B (x, r))

In this section, Ω will be a Lebesgue measurable set in Rn and h : Ω → Rn will be Lipschitz. Recall the following definition and theorems. See Page 10.4.2 for the proofs and more discussion. Definition 35.6.3 Let F be a collection of balls that cover a set, E, which have the property that if x ∈ E and ε > 0, then there exists B ∈ F, diameter of B < ε and x ∈ B. Such a collection covers E in the sense of Vitali. Theorem 35.6.4 Let E ⊆ Rn and suppose mn (E) < ∞ where mn is the outer measure determined by mn , n dimensional Lebesgue measure, and let F, be a collection of closed balls of bounded radii such that F covers E in the sense of Vitali. Then there exists a countable collection of disjoint balls from F, {Bj }∞ j=1 , such that mn (E \ ∪∞ B ) = 0. j j=1 Now this theorem implies a simple lemma which is what will be used. Lemma 35.6.5 Let V be an open set in Rr , mr (V ) < ∞. Then there exists a sequence of disjoint open balls {Bi } having radii less than δ and a set of measure 0, T , such that V = (∪∞ i=1 Bi ) ∪ T.

1056

WEAK DERIVATIVES

As in the proof of the change of variables theorem given earlier, the first step is to show that h maps Lebesgue measurable sets to Lebesgue measurable sets. In showing this the key result is the next lemma which states that h maps sets of measure zero to sets of measure zero. Lemma 35.6.6 If mn (T ) = 0 then mn (h (T )) = 0. Proof: Let V be an open set containing T whose measure is less than ε. Now using the Vitali covering theorem, there exists a sequence of disjoint balls {Bi }, B n i =o B (xi , ri ) which are contained in V such that the sequence of enlarged balls, bi , having the same center but 5 times the radius, covers T . Then B ³ ³ ´´ b mn (h (T )) ≤ mn h ∪∞ i=1 Bi ≤

∞ X

³ ³ ´´ bi mn h B

i=1 ∞ X

≤

n

α (n) (Lip (h)) 5n rin = 5n (Lip (h))

i=1

n

∞ X

mn (Bi )

i=1 n

n

≤ (Lip (h)) 5n mn (V ) ≤ ε (Lip (h)) 5n. Since ε is arbitrary, this proves the lemma. With the conclusion of this lemma, the next lemma is fairly easy to obtain. Lemma 35.6.7 If A is Lebesgue measurable, then h (A) is mn measurable. Furthermore, n mn (h (A)) ≤ (Lip (h)) mn (A). (35.6.9) Proof: Let Ak = A ∩ B (0, k) , k ∈ N. Let V ⊇ Ak and let mn (V ) < ∞. By Lemma 35.6.5, there is a sequence of disjoint balls {Bi } and a set of measure 0, T , such that V = ∪∞ i=1 Bi ∪ T, Bi = B(xi , ri ). By Lemma 35.6.6, mn (h (Ak )) ≤ mn (h (V )) ≤ ≤

mn (h (∪∞ i=1 Bi )) ∞ X

mn (h (Bi )) ≤

i=1

≤

∞ X i=1

Therefore,

+ mn (h (T )) = mn (h (∪∞ i=1 Bi )) ∞ X

mn (B (h (xi ) , Lip (h) ri ))

i=1 n

n

α (n) (Lip (h) ri ) = Lip (h)

∞ X

n

mn (Bi ) = Lip (h) mn (V ).

i=1 n

mn (h (Ak )) ≤ Lip (h) mn (V ).


1057

Since V is an arbitrary open set containing Ak , it follows from regularity of Lebesgue measure that n

mn (h (Ak )) ≤ Lip (h) mn (Ak ).

(35.6.10)

Now let k → ∞ to obtain 35.6.9. This proves the formula. It remains to show h (A) is measurable. By inner regularity of Lebesgue measure, there exists a set, F , which is the countable union of compact sets and a set T with mn (T ) = 0 such that F ∪ T = Ak . Then h (F ) ⊆ h (Ak ) ⊆ h (F ) ∪ h (T ). By continuity of h, h (F ) is a countable union of compact sets and so it is Borel. By 35.6.10 with T in place of Ak , mn (h (T )) = 0 and so h (T ) is mn measurable. Therefore, h (Ak ) is mn measurable because mn is a complete measure and this exhibits h (Ak ) between two mn measurable sets whose difference has measure 0. Now h (A) = ∪∞ k=1 h (Ak ) so h (A) is also mn measurable and this proves the lemma. The following lemma, depending on the Brouwer fixed point theorem and found in Rudin [58], will be important for the following arguments. A slightly more precise version was presented earlier on Page 583 but this version given below will suffice in this context. The idea is that if a continuous function mapping a ball in Rk to Rk doesn’t move any point very much, then the image of the ball must contain a slightly smaller ball. Lemma 35.6.8 Let B = B (0, r), a ball in Rk and let F : B → Rk be continuous and suppose for some ε < 1, |F (v) −v| < εr for all v ∈ B. Then

¡ ¢ F B ⊇ B (0, r (1 − ε)).

¡ ¢ Proof: Suppose a ∈ B (0, r (1 − ε)) \ F B and let G (v) ≡

r (a − F (v)) . |a − F (v)|

1058

WEAK DERIVATIVES

Then by the Brouwer fixed point theorem, G (v) = v for some v ∈ B. Using the formula for G, it follows |v| = r. Taking the inner product with v, 2

|v| = r2 =

(G (v) , v) =

r (a − F (v) , v) |a − F (v)|

r (a − v + v − F (v) , v) |a − F (v)| r = [(a − v, v) + (v − F (v) , v)] |a − F (v)| h i r 2 = (a, v) − |v| + (v − F (v) , v) |a − F (v)| £ 2 ¤ r r (1 − ε) − r2 +r2 ε = 0, ≤ |a − F (v)| ¡ ¢ a contradiction. Therefore, B (0, r (1 − ε)) \ F B = ∅ and this proves the lemma. Now let Ω be a Lebesgue measurable set and suppose h : Rn → Rn is Lipschitz continuous and one to one on Ω. Let =

N ≡ {x ∈ Ω : Dh (x) does not exist} n o −1 S ≡ x ∈ Ω \ N : Dh (x) does not exist

(35.6.11) (35.6.12)

Lemma 35.6.9 Let x ∈ Ω \ (S ∪ N ). Then if ε ∈ (0, 1) the following hold for all r small enough. ³ ³ ´´ mn h B (x,r) ≥ mn (Dh (x) B (0, r (1 − ε))), (35.6.13) h (B (x, r)) ⊆ h (x) + Dh (x) B (0, r (1 + ε)), ³ ³ ´´ mn h B (x,r) ≤ mn (Dh (x) B (0, r (1 + ε)))

(35.6.14) (35.6.15)

If x ∈ Ω \ (S ∪ N ) is also a point of density of Ω, then lim

r→0

mn (h (B (x, r) ∩ Ω)) = 1. mn (h (B (x, r)))

(35.6.16)

If x ∈ Ω \ N , then |det Dh (x)| = lim

r→0

Proof: Since Dh (x) h (x + v)

−1

mn (h (B (x, r))) a.e. mn (B (x,r))

(35.6.17)

exists,

= h (x) + Dh (x) v+o (|v|)   =o(|v|) }| {  z −1  Dh (x) o (|v|) = h (x) + Dh (x)  v+  

(35.6.18) (35.6.19)


1059

Consequently, when r is small enough, 35.6.14 holds. Therefore, 35.6.15 holds. −1 From 35.6.19, and the assumption that Dh (x) exists, Dh (x)

−1

−1

h (x + v) − Dh (x)

Letting F (v) = Dh (x)

−1

h (x) − v =o(|v|). −1

h (x + v) − Dh (x)

(35.6.20)

h (x),

apply Lemma 35.6.8 in 35.6.20 to conclude that for r small enough, whenever |v| < r, −1 −1 Dh (x) h (x + v) − Dh (x) h (x) ⊇ B (0, (1 − ε) r). Therefore,

³ ´ h B (x,r) ⊇ h (x) + Dh (x) B (0, (1 − ε) r)

which implies ³ ³ ´´ mn h B (x,r) ≥ mn (Dh (x) B (0, r (1 − ε))) which shows 35.6.13. Now suppose that x is a point of density of Ω as well as being a point where −1 Dh (x) and Dh (x) exist. Then whenever r is small enough, 1−ε< and so 1−ε

< ≤

which implies

mn (h (B (x, r) ∩ Ω)) ≤1 mn (h (B (x, r)))

¡ ¡ ¢¢ mn h B (x, r) ∩ ΩC mn (h (B (x, r) ∩ Ω)) + mn (h (B (x, r))) mn (h (B (x, r))) ¡ ¡ ¢¢ C mn h B (x, r) ∩ Ω + 1. mn (h (B (x, r))) mn (B (x,r) \ Ω) < εα (n) rn.

Then for such r, 1≥ ≥

(35.6.21)

mn (h (B (x, r) ∩ Ω)) mn (h (B (x, r)))

mn (h (B (x, r))) − mn (h (B (x,r) \ Ω)) . mn (h (B (x, r)))

From Lemma 35.6.7, 35.6.21, and 35.6.13, this is no larger than n

1−

Lip (h) εα (n) rn . mn (Dh (x) B (0, r (1 − ε)))

By the theorem on the change of variables for a linear map, this expression equals n

1−

Lip (h) εα (n) rn n ≡ 1 − g (ε) |det (Dh (x))| rn α (n) (1 − ε)

1060

WEAK DERIVATIVES

where limε→0 g (ε) = 0. Then for all r small enough, 1≥

mn (h (B (x, r) ∩ Ω)) ≥ 1 − g (ε) mn (h (B (x, r)))

which shows 35.6.16 since ε is arbitrary. It remains to verify 35.6.17. In case x ∈ S, for small |v| , h (x + v) = h (x) + Dh (x) v+o (|v|) where |o (|v|)| < ε |v| . Therefore, for small enough r, h (B (x, r)) − h (x) ⊆ K + B (0, rε) where K is a compact subset of an n−1 dimensional subspace contained in Dh (x) (Rn ) which has diameter no more than 2 ||Dh (x)|| r. By Lemma 10.6.5 on Page 304, mn (h (B (x, r)))

=

mn (h (B (x, r)) − h (x))

≤

2n εr (2 ||Dh (x)|| r + rε)

n−1

and so, in this case, letting r be small enough, mn (h (B (x, r))) 2n εr (2 ||Dh (x)|| r + rε) ≤ mn (B (x,r)) α (n) rn

n−1

≤ Cε.

Since ε is arbitrary, the limit as r → 0 of this quotient equals 0. If x ∈ / S, use 35.6.13 - 35.6.15 along with the change of variables formula for linear maps. This proves the Lemma. Since h is one to one, there exists a measure, µ, defined by µ (E) ≡ mn (h (E)) on the Lebesgue measurable subsets of Ω. By Lemma 35.6.7 µ ¿ mn and so by the Radon Nikodym theorem, there exists a nonnegative function, J (x) in L1loc (Rn ) such that whenever E is Lebesgue measurable, Z µ (E) = mn (h (E ∩ Ω)) = J (x) dmn . (35.6.22) E∩Ω

Extend J to equal zero off Ω. Lemma 35.6.10 The function, J (x) equals |det Dh (x)| a.e. Proof: Define Q ≡ {x ∈ Ω : x is not a point of density of Ω} ∪ N ∪ {x ∈ Ω : x is not a Lebesgue point of J}.


1061

Then Q is a set of measure zero and if x ∈ / Q, then by 35.6.17, and 35.6.16,

= = = =

|det Dh (x)| mn (h (B (x, r))) lim r→0 mn (B (x, r)) mn (h (B (x, r))) mn (h (B (x, r) ∩ Ω)) lim r→0 mn (h (B (x, r) ∩ Ω)) mn (B (x, r)) Z 1 lim J (y) dmn r→0 mn (B (x, r)) B(x,r)∩Ω Z 1 lim J (y) dmn = J (x) . r→0 mn (B (x, r)) B(x,r)

the last equality because J was extended to be zero off Ω. This proves the lemma. Here is the change of variables formula for Lipschitz mappings. It is a special case of the area formula. Theorem 35.6.11 Let Ω be a Lebesgue measurable set, let f ≥ 0 be Lebesgue measurable. Then for h a Lipschitz mapping defined on Rn which is one to one on Ω, Z Z f (y) dmn = f (h (x)) |det Dh (x)| dmn . (35.6.23) h(Ω)

Ω

Proof: Let F be a Borel set. It follows that h−1 (F ) is a Lebesgue measurable set. Therefore, by 35.6.22, ¡ ¡ ¢¢ mn h h−1 (F ) ∩ Ω (35.6.24) Z Z = XF (y) dmn = Xh−1 (F ) (x) J (x) dmn h(Ω) Ω Z = XF (h (x)) J (x) dmn . Ω

What if F is only Lebesgue measurable? Note there are no measurability problems with the above expression because x →XF (h (x)) is Borel measurable due to the assumption that h is continuous while J is given to be Lebesgue measurable. However, if F is Lebesgue measurable, not necessarily Borel measurable, then it is no longer clear that x → XF (h (x)) is measurable. In fact this is not always even true. However, x →XF (h (x)) J (x) is measurable and 35.6.24 holds. Let F be Lebesgue measurable. Then by inner regularity, F = H ∪ N where N has measure zero, H is the countable union of compact sets so it is a Borel set, and H ∩ N = ∅. Therefore, letting N 0 denote a Borel set of measure zero which contains N, b (x) ≡ XH (h (x)) J (x) ≤ XF (h (x)) J (x) =

XH (h (x)) J (x) + XN (h (x)) J (x)

≤ XH (h (x)) J (x) + XN 0 (h (x)) J (x) ≡ u (x)

1062

WEAK DERIVATIVES

Now since N 0 is Borel, Z Z (u (x) − b (x)) dmn = XN 0 (h (x)) J (x) dmn Ω

Ω

¡ ¡ ¢¢ = mn h h−1 (N 0 ) ∩ Ω = mn (N 0 ∩ h (Ω)) = 0 and this shows XH (h (x)) J (x) = XF (h (x)) J (x) except on a set of measure zero. By completeness of Lebesgue measure, it follows x →XF (h (x)) J (x) is Lebesgue measurable and also since h maps sets of measure zero to sets of measure zero, Z Z XF (h (x)) J (x) dmn = XH (h (x)) J (x) dmn Ω Ω Z XH (y) dmn = h(Ω) Z XF (y) dmn . = h(Ω)

It follows that if s is any nonnegative Lebesgue measurable simple function, Z Z s (h (x)) J (x) dmn = s (y) dmn (35.6.25) Ω

h(Ω)

and now, if f ≥ 0 is Lebesgue measurable, let sk be an increasing sequence of Lebesgue measurable simple functions converging pointwise to f . Then since 35.6.25 holds for sk , the monotone convergence theorem applies and yields 35.6.23. This proves the theorem. It turns out that a Lipschitz function defined on some subset of Rn always has a Lipschitz extension to all of Rn . The next theorem gives a proof of this. For more on this sort of theorem see Federer [26]. He gives a better but harder theorem than what follows. Theorem 35.6.12 If h : Ω → Rm is Lipschitz, then there exists h : Rn → Rm which extends h and is also Lipschitz. Proof: It suffices to assume m = 1 because if this is shown, it may be applied to the components of h to get the desired result. Suppose |h (x) − h (y)| ≤ K |x − y|.

(35.6.26)

h (x) ≡ inf{h (w) + K |x − w| : w ∈ Ω}.

(35.6.27)

Define If x ∈ Ω, then for all w ∈ Ω, h (w) + K |x − w| ≥ h (x) by 35.6.26. This shows h (x) ≤ h (x). But also you could take w = x in 35.6.27 which yields h (x) ≤ h (x). Therefore h (x) = h (x) if x ∈ Ω.


1063

¯ ¯ Now suppose x, y ∈ Rn and consider ¯h (x) − h (y)¯. Without loss of generality assume h (x) ≥ h (y) . (If not, repeat the following argument with x and y interchanged.) Pick w ∈ Ω such that h (w) + K |y − w| − ε < h (y). Then

¯ ¯ ¯h (x) − h (y)¯ = h (x) − h (y) ≤ h (w) + K |x − w| − [h (w) + K |y − w| − ε] ≤ K |x − y| + ε.

Since ε is arbitrary,

¯ ¯ ¯h (x) − h (y)¯ ≤ K |x − y|

and this proves the theorem. This yields a simple corollary to Theorem 35.6.11. Corollary 35.6.13 Let h : Ω → Rn be Lipschitz continuous and one to one where Ω is a Lebesgue measurable set. Then if f ≥ 0 is Lebesgue measurable, Z Z ¯ ¯ f (y) dmn = f (h (x)) ¯det Dh (x)¯ dmn . (35.6.28) h(Ω)

Ω

where h denotes a Lipschitz extension of h.

1064

WEAK DERIVATIVES

The Area And Coarea Formulas 36.1

The Area Formula Again

Recall the area formula presented earlier. For convenience, here it is. Theorem 36.1.1 Let g : h (A) → [0, ∞] be Hn measurable where h is a continuous function and A is a Lebesgue measurable set which satisfies 19.5.12 - 19.5.14. That is, U is an open set in Rn on which h is defined and A ⊆ U is a Lebesgue measurable set, m ≥ n, and h : U → Rm is continuous, (36.1.1) Dh (x) exists for all x ∈ A,

(36.1.2)

Also assume that for every x ∈ A, there exists rx and Lx such that for all y, z ∈ B (x, rx ) , |h (z) − h (y)| ≤ Lx |x − y| (36.1.3) Then x → (g ◦ h) (x) J (x) is Lebesgue measurable and Z

Z n

g (y) dH = h(A)


¡ ¢1/2 ∗ where J (x) = det (U (x)) = det Dh (x) Dh (x) . Obviously, one can obtain improved versions of this important theorem by using Rademacher’s theorem and condition 36.1.3. As mentioned earlier, a function which satisfies 36.1.3 is called locally Lipschitz at x. Here is a simple lemma which is in the spirit of similar lemmas presented in the chapter on Hausdorff measures. Lemma 36.1.2 Let U be an open set in Rn and let h : U → Rm where m ≥ n. Let A ⊆ U and let h be locally Lipschitz at every point of A. Then if N ⊆ A has Lebesgue measure zero, it follows that Hn (h (N )) = 0. 1065

1066

THE AREA AND COAREA FORMULAS

Proof: Let Nk be defined as Nk ≡ {x ∈ N : for some Rx > 0, |h (z) − h (y)| ≤ k |z − y| for all y, z ∈ B (x,Rx )} Thus Nk ↑ N. Let ε > 0 be given and let U ⊇ Vk ⊇ N be open and mn (Vk ) < 5nεkn . ¡δ ¢ Now fix δ > 0. For x ∈ Nk let B (x, 5rx ) ⊆ Vk such that rx < min 5k , Rx . By ∞ the Vitali covering n o∞ theorem, there exists a disjoint sequence of these balls, {Bi }i=1 ci , the corresponding sequence of balls having the same centers such that B i=1 ³ ´ n ³ ó∞ ci < 2δ/k. Hence h B ci but five times the radius covers Nk . Then diam B i=1 ³ ³ ´´ ci covers h (Nk ) and diam h B < 2δ. It follows n H2δ (h (Nk ))

≤

∞ X

³ ³ ´ń ci α (n) r h B

i=1

≤

∞ X

n

α (n) k n 5n r (Bi )

i=1

= 5n k n

∞ X

mn (Bi ) ≤ 5n k n mn (Vk ) < ε

i=1 n

Since δ was arbitrary, this shows H (h (Nk )) ≤ ε. Since k was arbitrary, this shows Hn (h (N )) = limk→∞ Hn (h (Nk )) ≤ ε. Since ε is arbitrary, this shows Hn (h (N )) = 0. This proves the lemma. Now with this lemma, here is one of many possible generalizations of the area formula. Theorem 36.1.3 Let U be an open set in Rn and h : U → Rm . Let h be locally Lipschitz and one to one on A, a Lebesgue measurable subset of U and let g : h (A) → R be a nonnegative Hn measurable function. Then x → (g ◦ h) (x) J (x) is Lebesgue measurable and Z Z g (y) dHn = g (h (x)) J (x) dmn h(A)

A

¡ ¢1/2 ∗ where J (x) = det (U (x)) = det Dh (x) Dh (x) . Proof: For x ∈ A, there exists a ball, Bx on which h is Lipschitz. By Rademacher’s theorem, h is differentiable a.e. on Bx . There is a countable cover of A consisting of such balls on which h is Lipschitz. Therefore, h is differentiable on A0 ⊆ A where mn (A \ A0 ) = 0. Then by the earlier area formula, Z Z n g (y) dH = g (h (x)) J (x) dmn h(A0 )

A0

36.1. THE AREA FORMULA AGAIN By Lemma 36.1.2 Z g (y) dHn

1067

Z g (y) dHn

=

h(A)

Z

h(A0 )

=

Z

g (h (x)) J (x) dmn = A0

g (h (x)) J (x) dmn A

This proves the theorem. Note how a special case of this occurs when h is one to one and C 1 . Of course this yields the earlier change of variables formula as a still more special case. In addition to this, recall the divergence theorem, Theorem 19.7.14 on Page 613. This theorem was stated for bounded open sets which have a Lipschitz boundary. This definition of Lipschitz boundary involved an assumption that certain Lipschitz mappings had a derivative a.e. Rademacher’s theorem makes this assumption redundant. Therefore, the statement of Theorem 19.7.14 remains valid with the following definition of a Lipschitz boundary. Definition 36.1.4 A bounded open set, U ⊆ Rn is said to have a Lipschitz boundary and to lie on one side of its boundary if the following conditions hold. There exist open boxes, Q1 , · · · , QN , Qi =

n Y ¡ i i¢ aj , bj j=1

such that ∂U ≡ U \ U is contained in their union. Also, for each Qi , there exists k and a Lipschitz function, gi such that U ∩ Qi is of the form  k−1  Y¡ ¢ x : (x1 , · · · , xk−1 , xk+1 , · · · , xn ) ∈ aij , bij ×  j=1  n  Y ¡ i i¢ aj , bj and aik < xk < gi (x1 , · · · , xk−1 , xk+1 , · · · , xn ) (36.1.4)  j=k+1

or else of the form  k−1  Y¡ ¢ x : (x1 , · · · , xk−1 , xk+1 , · · · , xn ) ∈ aij , bij ×  j=1

n Y ¡ j=k+1

¢ i

aij , bj

and gi (x1 , · · · , xk−1 , xk+1 , · · · , xn ) < xk < bij

  

.(36.1.5)

Also, there exists an open set, Q0 such that Q0 ⊆ Q0 ⊆ U and U ⊆ Q0 ∪Q1 ∪· · ·∪QN .

1068


36.2

Mappings That Are Not One To One

Next I will consider the case where h is not necessarily one to one. Recall the major theorem presented earlier on which the proof of the area formula depended, Theorem 19.5.10 on Page 587. Here it is. Theorem 36.2.1 Let h : U → Rm where U is an open set in Rn for n ≤ m and suppose h is locally Lipschitz at every point of a Lebesgue measurable subset, A of U. Also suppose that for every x ∈ A, Dh (x) exists. Then for x ∈ A, Hn (h (B (x, r))) , r→0 mn (B (x,r)) ¡ ¢1/2 ∗ where J (x) ≡ det (U (x)) = det Dh (x) Dh (x) . J (x) = lim

(36.2.6)

The next lemma is a version of Sard’s lemma. Lemma 36.2.2 Let h : U → Rm where U is an open set in Rn for n ≤ m and suppose h is locally Lipschitz at every point of a Lebesgue measurable subset, A of U. Let N ≡ {x ∈ A : Dh (x) does not exist} (36.2.7) and let S ≡ {x ∈ A0 ≡ A \ N : J (x) = 0}

(36.2.8)

n

Then H (h (S ∪ N )) = 0. Proof: By Rademacher’s theorem, N has measure 0. Therefore, Hn (h (N )) = 0 by Lemma 36.1.2. It remains to show Hn (h (S)) = 0. Let Sk = B (0, k) ∩ S for k a positive integer large enough that Sk 6= ∅. By Theorem 19.5.10 on Page 587 stated above, if x ∈ Sk , there exists rx such that 5rx < min (Rx , 1) and if r ≤ 5rx , Hn (h (B (x, r))) ε < n n , B (x, r) ⊆ B (0,k) ∩ U mn (B (x,r)) 5 k

(36.2.9)

Then by the Vitali covering theorem, there exists a sequence of disjoint balls of this ∞ sort, 5 times the radius but the same center, i }i=1 such that the n ³ballsóhaving n o{B ∞ ∞ c c covers h (Sk ) . Then from 36.2.9 Bi cover Sk . Then h Bi i=1

i=1

Hn (h (Sk ))

≤

∞ X i=1

≤

∞ ³ ³ ´´ X ci Hn h B ≤ 5n Hn (h (Bi )) i=1

∞ X

ε ε 5n n n mn (Bi ) ≤ n mn (B (0, k)) = εα (n) 5 k k i=1

Since ε > 0 is arbitrary, it follows Hn (h (Sk )) = 0 and now letting k → 0, it follows Hn (h (S)) = 0. This proves the lemma. The following very technical lemma provides the necessary theory to generalize to functions which are not one to one.

36.2. MAPPINGS THAT ARE NOT ONE TO ONE

1069

Lemma 36.2.3 Let h : U → Rm where U is an open set in Rn for n ≤ m and suppose h is locally Lipschitz at every point of a Lebesgue measurable subset, A of U. Let N ≡ {x ∈ A : Dh (x) does not exist} and let S ≡ {x ∈ A0 ≡ A \ N : J (x) = 0} ∞

Let B = A \ (S ∪ N ) . Then there exist measurable disjoint sets, {Ei }i=1 such that −1 is Lipschitz on h (Ei ) . A = ∪∞ i=1 Ei and h is one to one on Ei . Furthermore, h Proof: Let C be a dense countable subset of B and let F be a countable dense subset of the invertible elements of L (Rn , Rn ). For i a positive integer and T ∈ F, c ∈ C ½ µ ¶ ¾ 1 E (c, T, i) ≡ b ∈ B c, ∩ B : (a) , (b) both hold i where (a) , (b) are given by 2 |T v| ≤ |U (b) v| for all v 3 |h (a) − h (b) − Dh (b) (a − b)| ≤ ¢ ¡ for all a ∈ B b, 2i .

1 |T (a − b)| . 2

(a) (b)

6 2 i

br

rc 1 i¡

¡ ª

a

First I will show these sets, E (c, T, i) cover B and that they are measurable sets. To begin with consider the measurability question. Inequality (a) is the same as saying 2 |T v| ≤ |Dh (b) v| for all v 3 which is the same as saying ¯ ¯ 2 |v| ≤ ¯Dh (b) T −1 v¯ for all v. 3

1070


Let {vi } denote a dense countable subset of Rn . Letting ½ ¾ ¯ ¯ 2 −1 ¯ ¯ Si ≡ b : |vi | ≤ Dh (b) T vi 3 it follows easily that Si is measurable because the component functions of the matrix of Dh (b) are limits of difference quotients of continuous functions so they are Borel measurable. (Note that if B were Borel, then Si would also be Borel.) Now by continuity, ½ ¾ ¯ ¯ 2 −1 ¯ ¯ ∪∞ S = b : |v| ≤ Dh (b) T v for all v i=1 i 3 and so this set is measurable also. Inequality (b) also determines a measurable set by similar reasoning. It is the same as saying that for all |v| < 2/i, |h (b + v) − h (b) − Dh (b) (v)| ≤

1 |T (v)| 2

Use {vi } a countable dense subset of B (0,2/i) in a similar fashion to (a)¡ . ¢ Next¡ I need to¢ show these sets cover B. Let x ∈ B. Then pick ci ∈ B x, 1i and Ti ∈ B U (x) ¯,¯1i . I need¯¯to show that x ∈ E (ci , Ti , i) for i large enough. For i ¯¯ −1 ¯¯ large enough, ¯¯Ti U (x) ¯¯ < 32 . Therefore, for such i ¯ ¯ 3 ¯ ¯ −1 ¯Ti U (x) (v)¯ < |v| 2 for all v and so |Ti w|
0 such that |T v| ≥ r |v| for all v. Therefore, 6 |a − b| ≤ |h (a) − h (b)| . r In other words, ¯ −1 ¯ ¯h (h (a)) − h−1 (h (b))¯ = |a − b| ≤ 6 |h (a) − h (b)| . r which completes the proof. With these lemmas, here is the main theorem which is a generalization of Theorem 36.1.3. First remember that from Lemma 19.5.3 on Page 583 a locally Lipschitz function maps Lebesgue measurable sets to Hausdorff measurable sets. Theorem 36.2.4 Let U be an open set in Rn and h : U → Rm . Let h be locally Lipschitz on A, a Lebesgue measurable subset of U and let g : h (A) → R be a nonnegative Hn measurable function. Also let # (y) ≡ Number of elements of h−1 (y)

1072


Then # is Hn measurable, x → (g ◦ h) (x) J (x) is Lebesgue measurable, and Z Z # (y) g (y) dHn = g (h (x)) J (x) dmn h(A)

A

¡ ¢1/2 ∗ where J (x) = det (U (x)) = det Dh (x) Dh (x) . Proof: Let B = A \ (S ∪ N ) where S is the set of points where J (x) = 0 and N is the set of points, x of A where Dh (x) does not exist. Also from Lemma 36.2.3 ∞ there exists {Ei }i=1 , a sequence of disjoint measurable sets whose union equals B such that h is one to one on each Ei . Then from Theorem 36.1.3 Z g (h (x)) J (x) dmn Z

A

=

g (h (x)) J (x) dmn = B

=

∞ Z X i=1

h(Ei )

Z

∞ Z X i=1

n

g (y) dH = h(B)

¡P∞

g (h (x)) J (x) dmn

Ei

Ã∞ X

! Xh(Ei ) (y) g (y) dHn . (36.2.11)

i=1

¢ n Now # (y) = i=1 Xh(Ei ) (y) on h (B) and # differs from this H measurable n function only on h (S ∪ N ) , which by Lemma 36.2.2 is a set of H measure zero. Therefore, # is Hn measurable and the last term of 36.2.11 equals Ã∞ ! Z Z X n Xh(Ei ) (y) g (y) dH = # (y) g (y) dHn . h(A)

i=1

h(A)


36.3

The Coarea Formula

The coarea formula involves a function, h which maps a subset of Rn to Rm where m ≤ n instead of m ≥ n as in the area formula. The symbol, Lip (h) will denote the Lipschitz constant for h. It is possible to obtain the coarea formula as a computation involving the area formula and some simple linear algebra and this is the approach taken here. To begin with, here is the necessary linear algebra. Theorem 36.3.1 Let A be an m × n matrix and let B be an n × m matrix for m ≤ n. Then pBA (t) = tn−m pAB (t) , so the eigenvalues of BA and AB are the same including multiplicities except that BA has n − m extra zero eigenvalues.

36.3. THE COAREA FORMULA

1073

Proof: Use block multiplication to write µ ¶µ ¶ µ ¶ AB 0 I A AB ABA = B 0 0 I B BA µ ¶µ ¶ µ ¶ I A 0 0 AB ABA = . 0 I B BA B BA Therefore, µ ¶−1 µ ¶µ ¶ µ ¶ I A AB 0 I A 0 0 = 0 I B 0 0 I B BA µ ¶ µ ¶ 0 0 AB 0 It follows that and have the same characteristic polynoB BA B 0 mials because the two matrices are simlar. Thus µ ¶ µ ¶ tI − AB 0 tI 0 det = det −B tI −B tI − BA and so noting that BA is an n × n matrix and AB is an m × m matrix, tm det (tI − BA) = tn det (tI − AB) and so det (tI − BA) = pBA (t) = tn−m det (tI − AB) = tn−m pAB (t) . This proves the theorem. The following corollary is what will be used to prove the coarea formula. Corollary 36.3.2 Let A be an m × n matrix. Then det (I + AA∗ ) = det (I + A∗ A) . Proof: Assume m ≤ n. From Theorem 36.3.1 AA∗ and A∗ A have the eigenvalues, λ1 , · · .λm , necessarily nonnegative, with the same multiplicities and some zero eigenvalues which have differing multiplicities. The eigenvalues, λ1 , · · .λm are the zeros of pAA∗ (t) . Thus there is an orthogonal transformation, P such that   λ1   ..   . 0     ∗ λ m ∗  P . A A=P  0     . .   . 0 0 Therefore,

     I + A∗ A = P     



λ1 + 1 ..

.

    ∗ P    

0 λm + 1 1

0

..

. 1

1074


and so   det (I + A∗ A) = det 

λ1 + 1

0 ..

0

.

  ∗  = det (I + AA ) .

λm + 1

This proves the corollary. The other main ingredient is the following version of the chain rule. Theorem 36.3.3 Let h and g be locally Lipschitz mappings from Rn to Rn with h (g (x)) = x on A, a Lebesgue measurable set. Then for a.e. x ∈ A, Dg (h (x)), Dh (x), and D (h ◦ g) (x) all exist and I = D (g ◦ h) (x) = Dg (h (x)) Dh (x). The proof of this theorem is based on the following lemma. Lemma 36.3.4 If h : Rn → Rn is locally Lipschitz, then if h (x) = 0 for all x ∈ A, then det (Dh (x)) = 0 a.e. Proof: By the R case of the Area R formula which involves mappings which are not one to one, 0 = {0} # (y) dy = A |det (Dh (x))| dx and so det (Dh (x)) = 0 a.e. Proof of the theorem: On A, g (h (x)) − x = 0 and so by the lemma, there exists a set of measure zero, N1 such that if x ∈ / N1 , D (g ◦ h) (x) − I = 0. Let M be the set of measure zero of points in h (Rn ) where g fails to be differentiable and let N2 ≡ g (M ) ∩ A, also a set of measure zero because locally Lipschitz maps take sets of measure zero to sets of measure zero. Finally let N3 be the set of points where h fails to be differentiable. Then if x ∈ / N1 ∪ N2 ∪ N3 , the chain rule implies I = D (g ◦ h) (x) = Dg (h (x)) Dh (x). This proves the theorem. Lemma 36.3.5 Let h : Rp → Rm be Lipschitz continuous and δ > 0. Then if A ⊆ Rp is either open or compact, ¡ ¢ y → Hδs A ∩ h−1 (y) is Borel measurable. Proof: Suppose first that A is compact and suppose for δ > 0, ¢ ¡ Hδs A ∩ h−1 (y) < t Then there exist sets Si , satisfying diam (Si ) < δ, A ∩ h−1 (y) ⊆ ∪∞ i=1 Si , and

∞ X i=1

s

α (s) (r (Si )) < t.


1075

I claim these sets can be taken to be open sets. Choose λ > 1 but close enough to 1 that ∞ X s α (s) (λr (Si )) < t i=1

Replace Si with Si + B (0, η i ) where η i is small enough that diam (Si ) + 2η i < λ diam (Si ) . Then diam (Si + B (0, η i )) ≤ λ diam (Si ) and so r (Si + B (0, η i )) ≤ λr (Si ) . Thus ∞ X

s

α (s) r (Si + B (0, η i )) < t.

i=1

Hence you could replace Si with Si + B (0, η i ) and so one can assume the sets Si are open. Claim: If z is close enough to y, then A ∩ h−1 (z) ⊆ ∪∞ i=1 Si . Proof: If not, then there exists a sequence {zk } such that zk → y, and

xk ∈ (A ∩ h−1 (zk )) \ ∪∞ i=1 Si .

By compactness of A, there exists a subsequence still denoted by k such that zk → y, xk → x ∈ A \ ∪∞ i=1 Si . Hence h (x) = lim h (xk ) = lim zk = y. k→∞

k→∞

∪∞ i=1 Si

But x ∈ / contrary to the assumption that A ∩ h−1 (y) ⊆ ∪∞ i=1 Si . It follows from this claim that whenever z is close enough to y, ¡ ¢ Hδs A ∩ h−1 (z) < t. This shows

©

¡ ¢ ª z ∈ Rp : Hδs A ∩ h−1 (z) < t ¢ ¡ is an open set and so y →Hδs A ∩ h−1 (y) is Borel measurable whenever A is compact. Now let V be an open set and let Ak ↑ V, Ak compact. Then

¡ ¢ ¡ ¢ Hδs V ∩ h−1 (y) = lim Hδs Ak ∩ h−1 (y) k→∞ ¡ ¢ so y →Hδs V ∩ h−1 (y) is Borel measurable for all V open. This proves the lemma.

1076


Lemma 36.3.6 Let h : Rp → Rm be¡ Lipschitz continuous. Suppose A is either ¢ open or compact in Rp . Then y → Hs A ∩ h−1 (y) is also Borel measurable and Z ¡ ¢ m α (s) α (m) s+m Hs A ∩ h−1 (y) dy ≤ 2m (Lip (h)) H (A) α (s + m) m R In particular, if s = n − m and p = n Z ¡ ¢ m α (n − m) α (m) Hn−m A ∩ h−1 (y) dy ≤ 2m (Lip (h)) mn (A) α (n) Rm ¡ ¢ Proof: From Lemma 36.3.5 y → Hδs A ∩ h−1 (y) is Borel measurable for each δ > 0. Without loss of generality, Hs+m (A) < ∞. Now let Bi be closed sets with diam (Bi ) < δ, A ⊆ ∪∞ i=1 Bi , and Hδs+m (A) + ε >

∞ X

s+m

α (s + m) r (Bi )

.

i=1

¡ ¢ Note each Bi is compact so y → Hδs Bi ∩ h−1 (y) is Borel measurable. Thus Z ¡ ¢ Hδs A ∩ h−1 (y) dy m ZR X ¡ ¢ Hδs Bi ∩ h−1 (y) dy ≤ Rm

= ≤ =

XZ

i

¡ ¢ Hδs Bi ∩ h−1 (y) dy

i

Rm

i

h(Bi )

XZ X

Hδs (Bi ) dy

mm (h (Bi )) Hδs (Bi )

i

≤

X

m

m

s

(Lip (h)) 2m α (m) r (Bi ) α (s) r (Bi )

i

=

(Lip (h))

m

≤

(Lip (h))

m

α (m) α (s) m X m+s 2 α (s + m) r (Bi ) α (m + s) i ¢ α (m) α (s) m ¡ s+m 2 Hδ (A) + ε α (m + s)

Since ε is arbitrary, Z ¡ ¢ m α (m) α (s) m s+m Hδs A ∩ h−1 (y) dy ≤ (Lip (h)) 2 Hδ (A) α (m + s) m R Taking a limit as δ → 0 this proves the lemma. Next I will show that whenever A is Lebesgue measurable, ¡ ¢ y → Hn−m A ∩ h−1 (y) is mm measurable and the above estimate holds.


1077

Lemma 36.3.7 Let A be a Lebesgue measurable subset of Rn and let h : Rn → Rm be Lipschitz. Then ¡ ¢ y → Hn−m A ∩ h−1 (y) is Lebesgue measurable. Furthermore, for all A Lebesgue measurable, Z ¡ ¢ m α (n − m) α (m) Hn−m A ∩ h−1 (y) dy ≤ 2m (Lip (h)) mn (A) α (n) m R Proof: Let A be a bounded Lebesgue measurable set in Rn . Then by inner and outer regularity of Lebesgue measure there exists an increasing sequence of compact sets, {Kk } contained in A and a decreasing sequence of open sets, {Vk } containing A such that mn (Vk \ Kk ) < 2−k . Thus mn (V1 ) ≤ mn (A) + 1. By Lemma 36.3.6 Z ¡ ¢ m α (n − m) α (m) Hδn−m V1 ∩ h−1 (y) dy < 2m (Lip (h)) (mn (A) + 1) . α (n) Rm Then ¡ ¢ ¡ ¢ ¡ ¢ Hδn−m Kk ∩ h−1 (y) ≤ Hδn−m A ∩ h−1 (y) ≤ Hδn−m Vk ∩ h−1 (y) (36.3.12) By Lemma 36.3.6 Z ¡ n−m ¡ ¢ ¡ ¢¢ = Hδ Vk ∩ h−1 (y) − Hδn−m Kk ∩ h−1 (y) dy m ZR ¡ ¢ = Hδn−m (Vk − Kk ) ∩ h−1 (y) dy Rm

m

≤ 2m (Lip (h))

m

< 2m (Lip (h))

α (n − m) α (m) mn (Vk \ Kk ) α (n) α (n − m) α (m) −k 2 α (n)

Let the Borel measurable functions, g and f be defined by ¡ ¢ ¡ ¢ g (y) ≡ lim Hδn−m Vk ∩ h−1 (y) , f (y) ≡ lim Hδn−m Kk ∩ h−1 (y) k→∞

k→∞

It follows from the dominated convergence theorem and 36.3.12 that ¡ ¢ f (y) ≤ Hδn−m A ∩ h−1 (y) ≤ g (y) and

Z (g (y) − f (y)) dy = 0. Rm

¡ ¢ By completness of mm , this establishes y →Hδn−m A ∩ h−1 (y) is Lebesgue measurable. Then by Lemma 36.3.6 again, Z ¡ ¢ m α (n − m) α (m) Hδn−m A ∩ h−1 (y) dy ≤ 2m (Lip (h)) mn (A) . α (n) Rm

1078


Letting δ → 0 and ¡using the monotone convergence theorem yields the desired ¢ inequality for Hn−m A ∩ h−1 (y) . The case where A is not bounded can be handled by considering Ar = A∩B (0, r) and letting r → ∞. This proves the lemma. By fussing with the isodiametric inequality one can remove the factor of 2m in the above inequalities obtaining much more attractive formulas. This is done in [24]. See also [46] which follows [24] and [26]. This last reference probably has the most complete treatment of these topics. With these lemmas, it is now possible to give a proof of the coarea formula. Define Λ (n, m) as all possible ordered lists of m numbers taken from {1, 2, · · · , n} . Lemma 36.3.8 Let A be a measurable set in Rn and let h : Rn → Rm be a Lipschitz map where m ≤ n which is differentiable at every point of A and for which ¡ ∗ ¢1/2 Jh (x) ≡ det Dh (x) Dh (x) 6= 0. Then the following formula holds along with all measurability assertions needed for it to make sense. Z Z ¡ ¢ Hn−m A ∩ h−1 (y) dy = Jh (x) dx (36.3.13) Rm

A

Proof: For x ∈ Rn , and i ∈ Λ (n, m), with i = (i1 , · · · , im ), define xi ≡ (xi1 , · · · , xim ), and π i x ≡ xi . Also for i ∈ Λ (n, m), let ic ∈ Λ (n, n − m) consist of the remaining indices taken in order. For h : Rn → Rm where m ≤ n, define ¡ ∗ ¢1/2 Jh (x) ≡ det Dh (x) Dh (x) . For each i ∈ Λ (n, m), define µ i

h (x) ≡

h (x) xic

¶ .

© ª∞ By Lemma 36.2.3, there exist disjoint measurable sets Fji j=1 such that hi is one ¡ ¢−1 ¡ ¢ to one on Fji , hi is Lipschitz on hi Fji , and © ¡ i ¢ ª i ∪∞ j=1 Fj = x ∈ A : det Dh (x) 6= 0 . For ¡x ∈ A,¢ det (Dxi h (x)) = 6 0 for some i ∈ Λ (n, m). det Dhi (x) and so x ∈ Fji for some i and j. Hence

But det (Dxi h (x)) =

∪i,j Fji = A. © ª Now let Eji be measurable sets such that Eji ⊆ Fki for some k, the sets are disjoint, and their union coincides with ∪i,j Fji . Then Z Jh (x) dx = A

X

∞ Z X

i ∈Λ(n,m) j=1

¡ ∗ ¢1/2 det Dh (x) Dh (x) dx. Eji ∩A

(36.3.14)


1079

¡ ¢−1 Let g : Rn → Rn be a Lipschitz extension of hi so g ◦ hi (x) = x for all x ∈ i Ej . First, using Theorem 36.3.3, and the fact that Lipschitz mappings take sets of ei ⊆ E i measure zero to sets of measure zero, replace Eji with a measurable set, E j j e i has measure zero and such that Eji \ E j Dhi (g (y)) Dg (y) = I ³ ´ e i . Changing the variables using the area formula, the expression in 36.3.14 on hi E j equals Z Jh (x) dx = A

X

∞ Z X

i ∈Λ(n,m) j=1

e i ∩A) hi (E j

¯−1 ¡ ∗ ¢1/2 ¯¯ det Dh (g (y)) Dh (g (y)) det Dhi (g (y))¯ dy.

(36.3.15) Note the integrands are all Borel measurable functions because they are continuous functions of the entries of matrices which entries come from taking limits of difference quotients of continuous functions. Thus, Z ¡ ∗ ¢1/2 det Dh (x) Dh (x) dx = e i ∩A E j

Z Rn

Z

¯−1 ¡ ∗ ¢1/2 ¯¯ Xhi (Eei ∩A) (y) det Dh (g (y)) Dh (g (y)) det Dhi (g (y))¯ dy j

Z

= Rm

e i ∩A) π ic (h−1 (y1 )∩E j

¡ ∗ ¢1/2 −1 det Dh (g (y)) Dh (g (y)) |det Dhxi (g (y))| dy2 dy1 (36.3.16)

where y1 = h (x) and y2 = xic . Thus ¡ ¢ y2 = π ic g (y) = π ic g hi (x) = xic .

(36.3.17)

Now consider the inner integral in 36.3.16 in which y1 is fixed. The integrand equals det

· ¡

Dxi h (g (y))

Dxic h (g (y))

¢

µ

∗

Dxi h (g (y)) ∗ Dxic h (g (y))

¶¸1/2 |det Dhxi (g (y))|

−1

.

(36.3.18) I want to massage the above expression slightly. Since y1 is fixed, and y1 = h (π i g (y) , π ic g (y)) = h (g (y)), it follows from 36.3.17 that 0 = =

Dxi h (g (y)) Dy2 π i g (y) + Dxic h (g (y)) Dy2 π ic g (y) Dxi h (g (y)) Dy2 π i g (y) + Dxic h (g (y)).

1080


Letting A ≡ Dxi h (g (y)) and B ≡ Dy2 π i g (y) and using the above formula, 36.3.18 is of the form · µ ¶¸1/2 ¡ ¢ A∗ −1 A −AB det |det A| −B ∗ A∗ = det [AA∗ + ABB ∗ A∗ ]

1/2

1/2

= det [A (I + BB ∗ ) A∗ ] = (det (A) det (A∗ )) 1/2

= det (I + BB ∗ )

1/2

−1

|det A|

|det A|

−1

det (I + BB ∗ )

1/2

|det A|

−1

, 1/2

which, by Corollary 36.3.2, equals det (I + B ∗ B) . (Note the size of the identity changes in these two expressions, the first being an m × m matrix and the second being a n − m × n − m matrix.) By 36.3.17 π ic g (y) = y2 and so, · µ ¶¸1/2 ¡ ∗ ¢ B 1/2 B I det (I + B ∗ B) = det I = det

· ¡

Therefore, 36.3.16 reduces to Z e i ∩A E j

Z

π ic (

h−1 (y

∗

µ

¶¸1/2

¡ ∗ ¢1/2 det Dh (x) Dh (x) dx =

Z

Rm

¢

Dy2 π i g (y) Dy2 π i g (y) Dy2 π ic g (y) Dy2 π ic g (y) ¡ ¢1/2 ∗ = det Dy2 g (y) Dy2 g (y) . ∗

ei 1 )∩Ej ∩A)

¡ ¢1/2 ∗ det Dy2 g (y) Dy2 g (y) dy2 dy1 .

(36.3.19)

By the area formula applied to the inside integral, this integral equals ³ ´ eji ∩ A Hn−m h−1 (y1 ) ∩ E and so

Z Z

e i ∩A E j

= Rm

¡ ∗ ¢1/2 det Dh (x) Dh (x) dx

³ ´ e i ∩ A dy1 . Hn−m h−1 (y1 ) ∩ E j

ei Using Lemma 36.3.7, along with the inner regularity of Lebesgue measure, E j i can be replaced with Ej . Therefore, summing the terms over all i and j, Z Z ¡ ¡ ¢ ∗ ¢1/2 det Dh (x) Dh (x) dx = Hn−m h−1 (y) ∩ A dy. A

Rm


1081

This proves the lemma. Now the following is the Coarea formula. Corollary 36.3.9 Let A be a measurable set in Rn and let h : Rn → Rm be a Lipschitz map where m ≤ n. Then the following formula holds along with all measurability assertions needed for it to make sense. Z Z ¡ ¢ n−m −1 H A ∩ h (y) dy = Jh (x) dx (36.3.20) Rm

A

¡ ∗ ¢1/2 where Jh (x) ≡ det Dh (x) Dh (x) . Proof: By Lemma 36.3.7 again, this formula is true for all measurable A ⊆ Rn \ S. It remains to verify the formula for all measurable sets, A, whether or not they intersect S. Consider the case where A ⊆ S ≡ {x : J (Dh (x)) = 0}. ¡ ¢ Let A be compact so that by Lemma 36.3.6, y →Hn−m A ∩ h−1 (y) is Borel. For ε > 0, define k, p : Rn × Rm → Rm by k (x, y) ≡ h (x) + εy, p (x, y) ≡ y. Then Dk (x, y) = (Dh (x) , εI) = (U R, εI) where the dependence of U and R on x has been suppressed. Thus µ ∗ ¶ ¡ ¢ R U 2 = det U 2 + ε2 I Jk = det (U R, εI) εI ¡ ¢ ¡ ¢ = det Q∗ DQQ∗ DQ + ε2 I = det D2 + ε2 I =

m Y ¡ 2 ¢ λi + ε2 ∈ [ε2m , C 2 ε2 ]

(36.3.21)

i=1

since one of the λi equals 0. All the eigenvalues must be bounded independent of x, since ||Dh (x)|| is bounded independent of x due to the assumption that h is Lipschitz. Since Jk 6= 0, the first part of the argument implies ³ ´ Z εCmn+m A × B (0,1) ≥ |Jk| dmn+m A×B(0,1)

Z =

³

´ Hn k−1 (y) ∩ A × B (0,1) dy

Rm

Which by Lemma 36.3.6, Z Z ³ ´ ≥ Cnm Hn−m k−1 (y) ∩ p−1 (w) ∩ A × B (0,1) dwdy Rm

Rm

(36.3.22)

1082


where Cnm = Claim:

α(n) α(n−m)α(m) .

³ ´ Hn−m k−1 (y) ∩ p−1 (w) ∩ A × B (0,1) ¡ ¢ ≥ XB(0,1) (w) Hn−m h−1 (y − εw) ∩ A .

Proof of the claim: If w ∈ / B (0,1), there is nothing to prove so assume w ∈ B (0,1). For such w, (x, w1 ) ∈ k−1 (y) ∩ p−1 (w) ∩ A × B (0,1) if and only if h (x) + εw1 = y, w1 = w, and x ∈ A, if and only if (x, w1 ) ∈ h−1 (y − εw) ∩ A × {w}. Therefore for w ∈ B (0,1), ³ ´ Hn−m k−1 (y) ∩ p−1 (w) ∩ A × B (0,1) ¡ ¢ ¡ ¢ ≥ Hn−m h−1 (y − εw) ∩ A × {w} = Hn−m h−1 (y − εw) ∩ A . (Actually equality holds in the claim.) From the claim, 36.3.22 is at least as large as Z Cnm

Z Rm

¡ ¢ Hn−m h−1 (y − εw) ∩ A dwdy

(36.3.23)

B(0,1)

Z = Cnm B(0,1)

Z

¡ ¢ Hn−m h−1 (y − εw) ∩ A dydw

Rm

α (n) = α (n − m)

Z

¡ ¢ Hn−m h−1 (y) ∩ A dy.

(36.3.24)

Rm

The use of Fubini’s theorem is justified because the integrand is Borel measurable. Now by 36.3.24, it follows since ε > 0 is arbitrary, Z Z ¡ ¢ Hn−m A ∩ h−1 (y) dy = 0 = Jh (x) dx. Rm

A

Since this holds for arbitrary compact sets in S, it follows from Lemma 36.3.7 and inner regularity of Lebesgue measure that the equation holds for all measurable subsets of S. This completes the proof of the coarea formula.There is a simple corollary to this theorem in the case of locally Lipschitz maps. Corollary 36.3.10 Let h : Rn → Rm where m ≤ n and h is locally Lipschitz. Then the Coarea formula, 36.3.13, holds for h.

36.4. A NONLINEAR FUBINI’S THEOREM

1083

Proof: The assumption that h is locally Lipschitz implies that for each r > 0 it follows h is Lipschitz on B (0, r) . To see this, cover the compact set, B (0, r) with finitely many balls on which h is Lipschitz. Let A ⊆ B (0,r) and let hr be Lipschitz with h (x) = hr (x) for x ∈ B (0,r + 1) . Then Z Z Z J (Dh (x)) dx = J(Dhr (x))dx = A

A

Z =

H

n−m

¡

A∩

hr (A)

h−1 r

Z

¢ (y) dy =

Rm

Z

¡ ¢ Hn−m A ∩ h−1 r (y) dy

¡ ¢ Hn−m A ∩ h−1 (y) dy

h(A)

¡ ¢ Hn−m A ∩ h−1 (y) dy

= Rm

Now for arbitrary measurable A the above shows for k = 1, 2, · · · Z Z ¡ ¢ J (Dh (x)) dx = Hn−m A ∩ B (0,k) ∩ h−1 (y) dy. Rm

A∩B(0,k)

Use the monotone convergence theorem to obtain 36.3.13. From the definition of Hausdorff measure it follows H0 (E) equals the number of elements in E. Thus, if n = m, the Coarea formula implies Z Z Z ¡ ¢ J (Dh (x)) dx = H0 A ∩ h−1 (y) dy = # (y) dy A

h(A)

h(A)

This gives a version of Sard’s theorem by letting S = A.

36.4

A Nonlinear Fubini’s Theorem

Coarea formula holds for h : Rn → Rm , n ≥ m if whenever A is a Lebesgue measurable subset of Rn ,the following formula is valid. Z Z ¡ ¢ Hn−m A ∩ h−1 (y) dy = Jh (x) dx (36.4.25) Rm

A

Note this is the same as Z Z J(Dh (x))dx = A

¡ ¢ Hn−m A ∩ h−1 (y) dy

h(A)

because if y ∈ / h (A) , then h−1 (y) = ∅. Now let s (x) =

p X i=1

ci XEi (x)

1084


where Ei is measurable and ci ≥ 0. Then Z

p X

s (x) J ((Dh (x))) dx = Rn

i=1 p X

=

Z ci

J (Dh (x)) dx Ei

Z

¡ ¢ Hn−m Ei ∩ h−1 (y) dy

ci h(Ei )

i=1

Z

p X

= h(Rn )

Z

¡ ¢ ci Hn−m Ei ∩ h−1 (y) dy

i=1

"Z

# s dH

= h(Rn )

#

=

s dH h(Rn )

dy

h−1 (y)

"Z

Z

n−m

n−m

dy.

(36.4.26)

h−1 (y)

Theorem 36.4.1 Let g ≥ 0 be Lebesgue measurable and let h : Rn → Rm , n ≥ m satisfy the Coarea formula. For example, it could be locally Lipschitz. Then "Z # Z Z g dHn−m dy.

g (x) J ((Dh (x))) dx = Rn

h(Rn )

h−1 (y)

Proof: Let si ↑ g where si is a simple function satisfying 36.4.26. Then let i → ∞ and use the monotone convergence theorem to replace si with g. This proves the change of variables formula. Note that this formula is a nonlinear version of Fubini’s theorem. The “n − m dimensional surface”, h−1 (y), plays the role of Rn−m and Hn−m is like n − m dimensional Lebesgue measure. The term, J ((Dh (x))), corrects for the error occurring because of the lack of flatness of h−1 (y) .

Integration On Manifolds You can do integration on various manifolds by using the Hausdorff measure of an appropriate dimension. However, it is possible to discuss this through the use of the Riesz representation theorem and some of the machinery for accomplishing this is interesting for its own sake so I will present this alternate point of view.

37.1

Partitions Of Unity

This material has already been mostly discussed starting on Page 602. However, that was a long time ago and it seems like it might be good to go over it again and so, for the sake of convenience, here it is again. Definition 37.1.1 Let C be a set whose elements are subsets of Rn .1 Then C is said to be locally finite if for every x ∈ Rn , there exists an open set, Ux containing x such that Ux has nonempty intersection with only finitely many sets of C. Lemma 37.1.2 Let C be a set whose elements are open subsets of Rn and suppose ∞ ∪C ⊇ H, a closed set. Then there exists a countable list of open sets, {Ui }i=1 such that each Ui is bounded, each Ui is a subset of some set of C, and ∪∞ i=1 Ui ⊇ H. Proof: Let Wk ≡ B (0, k) , W0 = W−1 = ∅. For each x ∈ H ∩ Wk there exists an open set, Ux such that Ux is a subset of some set of C and Ux ⊆ Wk+1 \ Wk−1 . © ªm(k) Then since H ∩ Wk is compact, there exist finitely many of these sets, Uik i=1 whose union contains H ∩ Wk . If H ∩ Wk = ∅, let m (k) = 0 and there are no such © k ªm(k) sets obtained.The desired countable list of open sets is ∪∞ k=1 Ui i=1 . Each open set in this list is bounded. Furthermore, if x ∈ Rn , then x ∈ Wk where k is the first positive integer with x ∈ Wk . Then Wk \ Wk−1 is an open set containing x and © ªm(k) this open set can have nonempty intersection only with with a set of Uik i=1 ∪ © k ªm(k) © k−1 ªm(k−1) , a finite list of sets. Therefore, ∪∞ Ui k=1 Ui i=1 is locally finite. i=1 ∞ The set, {Ui }i=1 is said to be a locally finite cover of H. The following lemma gives some important reasons why a locally finite list of sets is so significant. First 1 The

definition applies with no change to a general topological space in place of Rn .

1085

1086

INTEGRATION ON MANIFOLDS ∞

of all consider the rational numbers, {ri }i=1 each rational number is a closed set. ∞

∞ Q = {ri }i=1 = ∪∞ i=1 {ri } 6= ∪i=1 {ri } = R

The set of rational numbers is definitely not locally finite. Lemma 37.1.3 Let C be locally finite. Then © ª ∪C = ∪ H : H ∈ C . Next suppose the elements of C are open sets and that for each U ∈ C, there exists a differentiable function, ψ U having spt (ψ U ) ⊆ U. Then you can define the following finite sum for each x ∈ Rn X f (x) ≡ {ψ U (x) : x ∈ U ∈ C} . Furthermore, f is also a differentiable function2 and X Df (x) = {Dψ U (x) : x ∈ U ∈ C} . Proof: Let p be a limit point of ∪C and let W be an open set which intersects only finitely many sets ofªC. Then p must©be a limit ª point of one of these sets. It © follows p ∈ ∪ H : H ∈ C and so ∪C ⊆ ∪ H : H ∈ C . The inclusion in the other direction is obvious. Now consider the second assertion. Letting x ∈ Rn , there exists an open set, W intersecting only finitely many open sets of C, U1 , U2 , · · · , Um . Then for all y ∈ W, f (y) =

m X

ψ Ui (y)

i=1

and so the desired result is obvious. It merely says that a finite sum of differentiable functions is differentiable. Recall the following definition. Definition 37.1.4 Let K be a closed subset of an open set, U. K ≺ f ≺ U if f is continuous, has values in [0, 1] , equals 1 on K, and has compact support contained in U . Lemma 37.1.5 Let U be a bounded open set and let K be a closed subset of U. Then there exist an open set, W, such that W ⊆ W ⊆ U and a function, f ∈ Cc∞ (U ) such that K ≺ f ≺ U . Proof: The set, K is compact so is at a positive distance from U C . Let © ¡ ¢ª W ≡ x : dist (x, K) < 3−1 dist K, U C . 2 If each ψ U were only continuous, one could conclude f is continuous. Here the main interest is differentiable.

37.1. PARTITIONS OF UNITY Also let

1087

¢ª © ¡ W1 ≡ x : dist (x, K) < 2−1 dist K, U C

Then it is clear K ⊆ W ⊆ W ⊆ W1 ⊆ W1 ⊆ U Now consider the function, ¡ ¢ dist x, W1C ¡ ¢ ¡ ¢ h (x) ≡ dist x, W1C + dist x, W Since W is compact it is at a positive distance from W1C and so h is a well defined continuous function which has compact support contained in W 1 , equals 1 on W, and has values in [0, 1] . Now let φk be a mollifier. Letting ¡ ¡ ¢ ¡ ¢¢ k −1 < min dist K, W C , 2−1 dist W 1 , U C , it follows that for such k,the function, h ∗ φk ∈ Cc∞ (U ) , has values in [0, 1] , and equals 1 on K. Let f = h ∗ φk . The above lemma is used repeatedly in the following. ∞

Lemma 37.1.6 Let K be a closed set and let {Vi }i=1 be a locally finite list of bounded open sets whose union contains K. Then there exist functions, ψ i ∈ Cc∞ (Vi ) such that for all x ∈ K, 1=

∞ X

ψ i (x)

i=1

and the function f (x) given by f (x) =

∞ X

ψ i (x)

i=1

is in C ∞ (Rn ) . Proof: Let K1 = K \ ∪∞ i=2 Vi . Thus K1 is compact because K1 ⊆ V1 . Let K 1 ⊆ W 1 ⊆ W 1 ⊆ V1 Thus W1 , V2 , · · · , Vn covers K and W 1 ⊆ V1 . Suppose W1 , · · · , Wr have been defined such that Wi ⊆ Vi for each i, and W1 , · · · , Wr , Vr+1 , · · · , Vn covers K. Then let ¡ ¢ ¡ r ¢ Kr+1 ≡ K \ ( ∪∞ i=r+2 Vi ∪ ∪j=1 Wj ). It follows Kr+1 is compact because Kr+1 ⊆ Vr+1 . Let Wr+1 satisfy Kr+1 ⊆ Wr+1 ⊆ W r+1 ⊆ Vr+1

1088

INTEGRATION ON MANIFOLDS ∞

Continuing this way defines a sequence of open sets, {Wi }i=1 with the property W i ⊆ Vi , K ⊆ ∪ ∞ i=1 Wi . ∞

∞

Note {Wi }i=1 is locally finite because the original list, {Vi }i=1 was locally finite. Now let Ui be open sets which satisfy W i ⊆ Ui ⊆ U i ⊆ Vi . ∞

Similarly, {Ui }i=1 is locally finite.

Wi

Ui

Vi

∞

∞ Since the set, {Wi }i=1 is locally finite, it follows ∪∞ i=1 Wi = ∪i=1 Wi and so it is possible to define φi and γ, infinitely differentiable functions having compact support such that ∞ U i ≺ φi ≺ Vi , ∪∞ i=1 W i ≺ γ ≺ ∪i=1 Ui .

Now define

½ ψ i (x) =

P∞ P∞ γ(x)φi (x)/ j=1 φj (x) if j=1 φj (x) 6= 0, P∞ 0 if j=1 φj (x) = 0.

P∞ / ∪∞ If x is such that j=1 φj (x) = 0, then x ∈ i=1 Ui because φi equals one on Ui . Consequently γ (y) = 0 for all y near x thanks to the fact that ∪∞ i=1 Ui is closed and so ψ (y) = 0 for all y near x. Hence ψ is infinitely differentiable at such x. If i i P∞ φ (x) = 6 0, this situation persists near x because each φ is continuous and so j j=1 j ψ i is infinitely differentiable at such points also thanks to Lemma P 37.1.3. Therefore ∞ ψ i is infinitely differentiable. If x ∈ K, then γ (x) = 1 and so j=1 ψ j (x) = 1. Clearly 0 ≤ ψ i (x) ≤ 1 and spt(ψ j ) ⊆ Vj . This proves the theorem. The method of proof of this lemma easily implies the following useful corollary. Corollary 37.1.7 If H is a compact subset of Vi for some Vi there exists a partition of unity such that ψ i (x) = 1 for all x ∈ H in addition to the conclusion of Lemma 37.1.6. fj ≡ Vj \ H. Now in the proof Proof: Keep Vi the same but replace Vj with V above, applied to this modified collection of open sets, if j 6= i, φj (x) = 0 whenever x ∈ H. Therefore, ψ i (x) = 1 on H. Theorem 37.1.8 Let H be any closed set and let C be any open cover of H. Then ∞ there exist functions {ψ i }i=1 such that spt (ψ i ) is contained in some of C and ψ i Pset ∞ is infinitely differentiable having values in [0, 1] such that on H, i=1 ψ i (x) = 1. P∞ Furthermore, the function, f (x) ≡ i=1 ψ i (x) is infinitely differentiable on Rn . ∞ Also, spt (ψ i ) ⊆ Ui where Ui is a bounded open set with the property that {Ui }i=1 is locally finite and each Ui is contained in some set of C.

37.2. INTEGRATION ON MANIFOLDS

1089

Proof: By Lemma 37.1.2 there exists an open cover of H composed of bounded open sets, Ui such that each Ui is a subset of some set of C and the collection, ∞ {Ui }i=1 is locally finite. Then the result follows from Lemma 37.1.6 and Lemma 37.1.3. m

Corollary 37.1.9 Let H be any closed set and let {Vi }i=1 be a finite open cover m of H. Then there exist functions {φi }i=1 such that spt i ) ⊆ Vi and φi is infinitely P(φ m differentiable having values in [0, 1] such that on H, i=1 φi (x) = 1. ∞

Proof: By Theorem 37.1.8 there exists a set of functions, {ψ i }i=1 having the m properties listed in this theorem relative ¡to the ¢ open covering, {Vi }i=1 . Let φ1 (x) equal the sum of all ψ j (x) such that spt ψ j ⊆ V1 . Next let φ2 (x) equal the sum ¡ ¢ of all ψ j (x) which have not already been included and for which spt ψ j ⊆ V2 . ∞ Continue in this manner. Since the open sets, {Ui }i=1 mentioned in Theorem 37.1.8 are locally finite, it follows from Lemma 37.1.3 that each φi is infinitely differentiable having support in Vi . This proves the corollary.

37.2

Integration On Manifolds

Manifolds are things which locally appear to be Rn for some n. The extent to which they have such a local appearance varies according to various analytical characteristics which the manifold possesses. Definition 37.2.1 Let U ⊆ Rn be an open set and let h : U → Rm . Then for r ∈ [0, 1), h ∈ C k,r (U ) for k a nonnegative integer means that Dα h exists for all |α| ≤ k and each Dα h is Holder continuous with exponent r. That is r

|Dα h (x) − Dα h (y)| ≤ K |x − y| . ¡ ¢ Also h ∈ C k,r U if it is the restriction of a function of C k,r (Rn ) to U . Definition 37.2.2 Let Γ be a closed subset of Rp where p ≥ n. Suppose Γ = ∪∞ i=1 Γi ∞ where Γi = Γ∩Wi for Wi a bounded open set. Suppose also {Wi }i=1 is locally finite. This means every bounded open set intersects only finitely many. Also suppose there are open bounded sets, Ui having Lipschitz boundaries and functions hi : Ui → Γi which are one to one, onto, and in C m,1 (Ui ) . Suppose also there exist functions, gi : Wi → Ui such that gi is C m,1 (Wi ) , and gi ◦ h©¡ i = id on ¢ªUi while hi ◦ gi = id on Γi . The collection of sets, Γj and mappings, gj , Γj , gj is called an atlas and ¡ ¢ an individual entry in the atlas is called a chart. Thus Γj , gj is a chart. Then Γ as just described is called a C m,1 manifold. The number, m is just a nonnegative integer. When m = 0 this would be called a Lipschitz manifold, the least smooth of the manifolds discussed here. For example, take p = n + 1 and let T

hi (u) = (u1 , · · · , ui , φi (u) , ui+1 , · · · , un )

1090

INTEGRATION ON MANIFOLDS T

for u = (u1 , · · · , ui , ui+1 , · · · , un ) ∈ Ui for φi ∈ C m,1 (Ui ) and gi : Ui × R → Ui given by gi (u1 , · · · , ui , y, ui+1 , · · · , un ) ≡ u for i = 1, 2, · · · , p. Then for u ∈ Ui , the definition gives gi ◦ hi (u) = gi (u1 , · · · , ui , φi (u) , ui+1 , · · · , un ) = u T

and for Γi ≡ hi (Ui ) and (u1 , · · · , ui , φi (u) , ui+1 , · · · , un ) ∈ Γi , hi ◦ gi (u1 , · · · , ui , φi (u) , ui+1 , · · · , un ) T

= hi (u) = (u1 , · · · , ui , φi (u) , ui+1 , · · · , un ) . This example can be used to describe the boundary of a bounded open set and since φi ∈ C m,1 (Ui ) , such an open set is said to have a C m,1 boundary. Note also that in this example, Ui could be taken to be Rn or if Ui is given, both hi and and gi can be taken as restrictions of functions defined on all of Rn and Rp respectively. The symbol, I will refer to an increasing list of n indices taken from {1, · · · , p} . Denote by Λ (p, n) the set of all such increasing lists of n indices. Let  Ã ¡ ¢ !2 1/2 X ∂ xi1 · · · xin  Ji (u) ≡  ∂ (u1 · · · un ) I∈Λ(p,n)

where here the sum is taken over all possible ¡ ¢ increasing lists of n indices, I, from {1, · · · , p} and x = hi u. Thus there are np terms in the sum. In this formula, ∂ (xi1 ···xin ) ∂(u1 ···un ) is defined to be the determinant of the following matrix.    

∂xi1 ∂u1

···

∂xin ∂u1

···

.. .

∂xi1 ∂un

.. .

  . 

∂xin ∂un

Note that if p = n there is only one term in the sum, the absolute value of the determinant of Dx (u). Define a positive linear functional, Λ on Cc (Γ) as follows: ∞ First let {ψ i } be a C P partition of unity subordinate to the open sets, {Wi } . Thus ∞ ψ i ∈ Cc (Wi ) and i ψ i (x) = 1 for all x ∈ Γ. Then Λf ≡

∞ Z X i=1

gi Γi

f ψ i (hi (u)) Ji (u) du.

(37.2.1)

Is this well defined? Lemma 37.2.3 The functional defined in 37.2.1 does not depend on the choice of atlas or the partition of unity.


1091

Proof: In 37.2.1, let {ψ i } be a C ∞ partition of unity which is associated with the atlas (Γi , gi ) and let {η i } be a C ∞ partition of unity associated in the same manner with the atlas (Γ0i , gi0 ). In the following argument, the local finiteness of the Γ ¡ i implies ¢ that all sums are finite. Using the change of variables formula with u = gi ◦ h0j v ∞ Z X ψ i f (hi (u)) Ji (u) du = (37.2.2) i=1

∞ X ∞ Z X gi Γi

i=1 j=1

gi Γi

η j ψ i f (hi (u)) Ji (u) du =

∞ X ∞ Z X i=1 j=1

gj0 (Γi ∩Γ0j )

·

¯ ¯ ¡ ¯ ∂ u1 · · · un ¢ ¯ ¢ ¡ ¢ ¡ ¢ ¯ ¯ η j h0j (v) ψ i h0j (v) f h0j (v) Ji (u) ¯ ¯ dv ¯ ∂ (v 1 · · · v n ) ¯ ∞ X ∞ Z X ¡ ¢ ¡ ¢ ¡ ¢ = η j h0j (v) ψ i h0j (v) f h0j (v) Jj (v) dv. 0 0 i=1 j=1 gj (Γi ∩Γj ) ¡

(37.2.3)

Thus the definition of Λf using (Γi , gi ) ≡ ∞ Z X ψ i f (hi (u)) Ji (u) du = i=1

∞ X ∞ Z X i=1 j=1

gj0 (Γi ∩Γ0j )

=

gi Γi

¡ ¢ ¡ ¢ ¡ ¢ η j h0j (v) ψ i h0j (v) f h0j (v) Jj (v) dv

∞ Z X gj0 (Γ0j )

j=1

¡ ¢ ¡ ¢ η j h0j (v) f h0j (v) Jj (v) dv

the definition of Λf using (Vi , gi0 ) . This proves the lemma. This lemma and the Riesz representation theorem for positive linear functionals implies the part of the following theorem which says the functional is well defined. Theorem 37.2.4 Let Γ be a C m,1 manifold. Then there exists a unique Radon measure, µ, defined on Γ such that whenever f is a continuous function having compact support which is defined on Γ and (Γi , gi ) denotes an atlas and {ψ i } a partition of unity subordinate to this atlas, Z ∞ Z X Λf = f dµ = ψ i f (hi (u)) Ji (u) du. (37.2.4) Γ

i=1

gi Γi

Also, a subset, A, of Γ is µ measurable if and only if for all r, gr (Γr ∩ A) is ν r measurable where ν r is the measure defined by Z ν r (gr (Γr ∩ A)) ≡ Jr (u) du gr (Γr ∩A)

1092

INTEGRATION ON MANIFOLDS

Proof: To begin, here is a claim. Claim : A set, S ⊆ Γi , has µ measure zero if and only if gi S has measure zero in gi Γi with respect to the measure, ν i . Proof of the claim: Let ε > 0 be given. By outer regularity, there exists a set, V ⊆ Γi , open3 in Γ such that µ (V ) < ε and S ⊆ V ⊆ Γi . Then gi V is open in Rn and contains gi S. Letting h ≺ gi V and h1 (x) ≡ h (gi (x)) for x ∈ Γi it follows h1 ≺ V . By Corollary 37.1.7 on Page 1088 there exists a partition of unity such that spt (h1 ) ⊆ {x ∈ Rp : ψ i (x) = 1}. Thus ψ j h1 (hj (u)) = 0 unless j = i when this reduces to h1 (hi (u)). It follows Z ε

Z

≥ µ (V ) ≥ =

V

∞ Z X gj Γj

j=1

Z =

h1 dµ

h1 dµ = Γ

ψ j h1 (hj (u)) Jj (u) du Z

h1 (hi (u)) Ji (u) du = Z

gi Γi

=

h (u) Ji (u) du gi Γi

h (u) Ji (u) du gi V

Now this holds for all h ≺ gi V and so Z Ji (u) du ≤ ε. gi V

Since ε is arbitrary, this shows gi V has measure no more than ε with respect to the measure, ν i . Since ε is arbitrary, gi S has measure zero. Consider the converse. Suppose gi S has ν i measure zero. Then there exists an open set, O ⊆ gi Γi such that O ⊇ gi S and Z Ji (u) du < ε. O

Thus hi (O) is open in Γ and contains S. Let h ≺ hi (O) be such that Z hdµ + ε > µ (hi (O)) ≥ µ (S)

(37.2.5)

Γ

As in the first part, Corollary 37.1.7 on Page 1088 implies there exists a partition of unity such that h (x) = 0 off the set, {x ∈ Rp : ψ i (x) = 1} 3 This means V is the intersection of an open set with Γ. Equivalently, it means that V is an open set in the traditional way regarding Γ as a metric space with the metric it inherits from Rm .


1093

and so as in this part of the argument, Z hdµ

∞ Z X

≡

Γ

j=1

Z =

gj Uj

ψ j h (hj (u)) Jj (u) du

h (hi (u)) Ji (u) du Z

gi Γi

Z

O∩gi Γi

=

h (hi (u)) Ji (u) du Ji (u) du < ε

≤

(37.2.6)

O

and so from 37.2.5 and 37.2.6 µ (S) ≤ 2ε. Since ε is arbitrary, this proves the claim. For the last part of the theorem, it suffices to let A ⊆ Γr because otherwise, the above argument would apply to A ∩ Γr . Thus let A ⊆ Γr be µ measurable. By the regularity of the measure, there exists an Fσ set, F and a Gδ set, G such that Γr ⊇ G ⊇ A ⊇ F and µ (G \ F ) = 0.(Recall a Gδ set is a countable intersection of open sets and an Fσ set is a countable union of closed sets.) Then since Γr is compact, it follows each of the closed sets whose union equals F is a compact set. ∞ Thus if F = ∪∞ k=1 Fk , gr (Fk ) is also a compact set and so gr (F ) = ∪k=1 gr (Fk ) is a Borel set. Similarly, gr (G) is also a Borel set. Now by the claim, Z Jr (u) du = 0. gr (G\F )

Since gr is one to one, gr G \ gr F = gr (G \ F ) and so gr (F ) ⊆ gr (A) ⊆ gr (G) where gr (G) \ gr (F ) has measure zero. By completeness of the measure, ν r , gr (A) is measurable. It follows that if A ⊆ Γ is µ measurable, then gr (Γr ∩ A) is ν r measurable for all r. The converse is entirely similar. This proves the theorem. Corollary 37.2.5 Let f ∈ L1 (Γ; µ) and suppose f (x) = 0 for all x ∈ / Γr where (Γr , gr ) is a chart. Then Z Z Z f dµ = f dµ = f (hr (u)) Jr (u) du. (37.2.7) Γ

Γr

gr Γr

Furthermore, if {(Γi , gi )} is an atlas and {ψ i } is a partition of unity as described earlier, then for any f ∈ L1 (Γ, µ), Z f dµ = Γ

∞ Z X r=1

gr Γr

ψ r f (hr (u)) Jr (u) du.

(37.2.8)

1094


Proof: Let f ∈ L1 (Γ, µ) with f = 0 off Γr . Without loss of generality assume f ≥ 0 because if the formulas can be established for this case, the same formulas are obtained for an arbitrary complex valued function by splitting it up into positive and negative parts of the real and imaginary parts in the usual way. Also, let K ⊆ Γr a compact set. Since µ is a Radon measure there exists a sequence of continuous functions, {fk } , fk ∈ Cc (Γr ), which converges to f in L1 (Γ, µ) and for µ a.e. x. Take the partition of unity, {ψ i } to be such that K ⊆ {x : ψ r (x) = 1} . Therefore, the sequence {fk (hr (·))} is a Cauchy sequence in the sense that Z lim |fk (hr (u)) − fl (hr (u))| Jr (u) du = 0 k,l→∞

gr (K)

It follows there exists g such that Z |fk (hr (u)) − g (u)| Jr (u) du → 0, gr (K)

and

g ∈ L1 (gr K; ν r ) .

By the pointwise convergence and the claim used in the proof of Theorem 37.2.4, g (u) = f (hr (u)) for µ a.e. hr (u) ∈ K. Therefore, Z Z Z f dµ = lim fk dµ = lim fk (hr (u)) Jr (u) du k→∞ K k→∞ g (K) K r Z Z = g (u) Jr (u) du = f (hr (u)) Jr (u) du. (37.2.9) gr (K)

gr (K)

∪∞ j=1 Kj

= Γr where Kj is compact for all j. Replace Now let · · · Kj ⊆ Kj+1 · · · and K in 37.2.9 with Kj and take a limit as j → ∞. By the monotone convergence theorem, Z Z f dµ = f (hr (u)) Jr (u) du. Γr

gr (Γr )

This establishes 37.2.7. To establish 37.2.8, let f ∈ L1 (Γ, µ) and let {(Γi , gi )} be an atlas and {ψ i } be a partition of unity. Then f ψ i ∈ L1 (Γ, µ) and is zero off Γi . Therefore, from what was just shown, Z ∞ Z X f ψ i dµ f dµ = Γ

=

i=1 Γi ∞ Z X r=1

gr (Γr )

ψ r f (hr (u)) Jr (u) du

37.3. COMPARISON WITH HN

37.3

1095

Comparison With Hn

The above gives a measure on a manifold, Γ. I will now show that the measure obtained is nothing more than Hn , the n dimensional Hausdorff measure. Recall Λ (p, n) was the set of all increasing lists of n indices taken from {1, 2, · · · , p} Recall  Ã ¡ ¢ !2 1/2 X ∂ xi1 · · · xin  Ji (u) ≡  ∂ (u1 · · · un ) I∈Λ(p,n)

where here the sum is taken over all possible increasing lists of n indices, I, from {1, · · · , p} and x = hi u and the functional was given as Λf ≡

∞ Z X gi Γi

i=1

f ψ i (hi (u)) Ji (u) du

(37.3.10)

∞

∞

where the {ψ i }i=1 was a partition of unity subordinate to the open sets, {Wi }i=1 as described above. I will show ¡ ¢1/2 ∗ Ji (u) = det Dh (u) Dh (u) and then use the area formula. The key result is really a special case of the Binet Cauchy theorem and this special case is presented in the next lemma. Lemma 37.3.1 Let A = (aij ) be a real p×n matrix in which p ≥ n. For I ∈ Λ (p, n) denote by AI the n × n matrix obtained by deleting from A all rows except for those corresponding to an element of I. Then X 2 det (AI ) = det (A∗ A) I∈Λ(p,n)

Proof: For (j1 , · · · , jn ) ∈ Λ (p, n) , define θ (jk ) ≡ k. Then let for {k1 , · · · , kn } = {j1 , · · · , jn } define sgn (k1 , · · · , kn ) ≡ sgn (θ (k1 ) , · · · , θ (kn )) . Then from the definition of the determinant and matrix multiplication, ∗

det (A A)

=

X

sgn (i1 , · · · , in )

i1 ,··· ,in p X

···

p X k1 =1

ak1 i1 ak1 1

p X

ak2 i2 ak2 2

k2 =1

akn in akn n

kn =1

=

X

X

X

J∈Λ(p,n) {k1 ,··· ,kn }=J i1 ,··· ,in

sgn (i1 , · · · , in ) ak1 i1 ak1 1 ak2 i2 ak2 2 · · · akn in akn n

1096 =


X

X

X

sgn (i1 , · · · , in ) ak1 i1 ak2 i2 · · · akn in · ak1 1 ak2 2 · · · akn n

J∈Λ(p,n) {k1 ,··· ,kn }=J i1 ,··· ,in

=

X

X

sgn (k1 , · · · , kn ) det (AJ ) ak1 1 ak2 2 · · · akn n

J∈Λ(p,n) {k1 ,··· ,kn }=J

=

X

det (AJ ) det (AJ )

J∈Λ(p,n)

and this proves the lemma. It follows from this lemma that ¡ ¢1/2 ∗ Ji (u) = det Dh (u) Dh (u) . From 37.3.10 and the area formula, the functional equals ∞ Z X f ψ i (hi (u)) Ji (u) du Λf ≡ =

i=1 gi Γi ∞ Z X i=1

Γi

Z

f ψ i (y) dHn =

f (y) dHn . Γ

n

Now H is a Borel measure defined on Γ which is finite on all compact subsets of Γ. This finiteness follows from the above formula. If K is a compact subset of Γ, then there exists an open set, W whose closure is compact and a continuous function R with compact support, f such that K ≺ f ≺ W . Then Hn (K) ≤ Γ f (y) dHn < ∞ because of the above formula. Lemma 37.3.2 µ = Hn on every µ measurable set. Proof: The Riesz representation theorem shows that Z Z f dµ = f dHn Γ

Γ

for every continuous function having compact support. Therefore, since every open set is the countable union of compact sets, it follows µ = Hn on all open sets. Since compact sets can be obtained as the countable intersection of open sets, these two measures are also equal on all compact sets. It follows they are also equal on all countable unions of compact sets. Suppose now that E is a µ measurable set of finite measure. Then there exist sets, F, G such that G is the countable intersection of open sets each of which has finite measure and F is the countable union of compact sets such that µ (G \ F ) = 0 and F ⊆ E ⊆ G. Thus Hn (G \ F ) = 0, Hn (G) = µ (G) = µ (F ) = Hn (F ) By completeness of Hn it follows E is Hn measurable and Hn (E) = µ (E) . If E is not of finite measure, consider Er ≡ E ∩ B (0, r) . This is contained in the compact set Γ ∩ B (0, r) and so µ (Er ) if finite. Thus from what was just shown, Hn (Er ) = µ (Er ) and so, taking r → ∞ Hn (E) = µ (E) . This shows you can simply use Hn for the measure on Γ.

Basic Theory Of Sobolev Spaces Definition 38.0.3 Let U be an open set of Rn . Define X m,p (U ) as the set of all functions in Lp (U ) whose weak partial derivatives up to order m are also in Lp (U ) where 1 ≤ p. The norm1 in this space is given by  1/p Z X p ||u||m,p ≡  |Dα u| dx . U |α|≤m

¡ ¢ P where α = (α1 , · · · , αn ) ∈ Nn and |α| ≡ αi . Here D0 u ≡ u. C ∞ U is defined to be the ¡set¢of functions which are restrictions to U of a function in Cc∞ (Rn ). Thus C ∞ U ⊆ W m,p (U ) . The Sobolev space, W m,p (U ) is defined to be the clo¡ ¢ sure of C ∞ U in X m,p (U ) with respect to the above norm. Denote this norm by ||u||W m,p (U ) , ||u||X m,p (U ) , or ||u||m,p,U when it is important to identify the open set, U. Also the following notation will be used pretty consistently. Definition 38.0.4 Let u be a function defined on U. Define ½ u (x) if x ∈ U u e (x) ≡ . 0 if x ∈ /U Theorem 38.0.5 Both X m,p (U ) and W m,p (U ) are separable reflexive Banach spaces provided p > 1. w

Proof: Define Λ : X m,p (U ) → Lp (U ) where w equals the number of multi w indices, α, such that |α| ≤ m as follows. Letting {αi }i=1 be the set of all multi indices with α1 = 0, Λ (u) ≡ (Dα1 u, Dα2 u, · · · , Dαw u) = (u, Dα2 u, · · · , Dαw u) . P α could also let the norm be given by ||u||m,p ≡ |α|≤m ||D u||p or ||u||m,p ≡ o α p max ||D u||p : |α| ≤ m because all norms are equivalent on R where p is the number of multi indices no larger than m. This is used whenever convenient. 1 You

n

1097

1098

BASIC THEORY OF SOBOLEV SPACES

Then Λ is one to one because one of the multi indices is 0. Also Λ (X m,p (U )) w

is a closed subspace of Lp (U ) . To see this, suppose (uk , Dα2 uk , · · · , Dαw uk ) → (f1 , f2 , · · · , fw ) w

in Lp (U ) . Then uk → f1 in Lp (U ) and Dαj uk → fj in Lp (U ) . Therefore, letting φ ∈ Cc∞ (U ) and letting k → ∞, R U

|α| R (Dαj uk ) φdx = (−1) u Dαj φdx U k ↓ ↓ R |α| R f φdx (−1) f Dαj φdx ≡ Dαj (f1 ) (φ) U j U 1

It follows Dαj (f1 ) = fj and so Λ (X m,p (U )) is closed as claimed. This is clearly w also a subspace of Lp (U ) and so it follows that Λ (X m,p (U )) is a reflexive Banach w space. This is because Lp (U ) , being the product of reflexive Banach spaces, is reflexive and any closed subspace of a reflexive Banach space is reflexive. Now Λ is an isometry of X m,p (U ) and Λ (X m,p (U )) which shows that X m,p (U ) is a reflexive Banach space. Finally, W m,p (U ) is a closed subspace of the reflexive Banach space, X m,p (U ) and so it is also reflexive. To see X m,p (U ) is separable, w note that Lp (U ) is separable because it is the finite product of the separable hence w completely separable metric space, Lp (U ) and Λ (X m,p (U )) is a subset of Lp (U ) . Therefore, Λ (X m,p (U )) is separable and since Λ is an isometry, it follows X m,p (U ) is separable also. Now W m,p (U ) must also be separable because it is a subset of X m,p (U ) . The following theorem is obvious but is worth noting because it says that if a function has a weak derivative in Lp (U ) on a large open set, U then the restriction of this weak derivative is also the weak derivative for any smaller open set. Theorem 38.0.6 Suppose U is an open set and U0 ⊆ U is another open set. Suppose also Dα u ∈ Lp (U ) . Then for all ψ ∈ Cc∞ (U0 ) , Z

Z |α|

(Dα u) ψdx = (−1) U0

u (Dα ψ) . U0

The following theorem is a fundamental approximation result for functions in X m,p (U ) . Theorem 38.0.7 Let¡ U be an ¢ open set and let U0 be an open subset of U with e denotes the zero the property that dist U0 , U C > 0. Then if u ∈ X m,p (U ) and u extention of u off U, lim ||e u ∗ φl − u||X m,p (U0 ) = 0. l→∞

1099 ¢ ¡ Proof: Always assume l is large enough that 1/l < dist U0 , U C . Thus for x ∈ U0 , Z u e ∗ φl (x) = u (x − y) φl (y) dy. (38.0.1) B (0, 1l ) The theorem is proved if it can be shown that Dα (e u ∗ φl ) → Dα u in Lp (U0 ) . Let ∞ ψ ∈ Cc (U0 ) Z |α| α D (e u ∗ φl ) (ψ) ≡ (−1) (e u ∗ φl ) (Dα ψ) dx U0 Z Z |α| = (−1) u e (y) φl (x − y) (Dα ψ) (x) dydx U0 Z Z |α| = (−1) u (y) φl (x − y) (Dα ψ) (x) dxdy. U

U0

Also, ³

´ αu ∗ φ g D l (ψ) ≡ = = = =

¶ α u (y) φ (x − y) dy ψ (x) dx g D l U ¶ Z 0 µZ Dα u (y) φl (x − y) dy ψ (x) dx U U ¶ Z 0 µZ u (y) (Dα φl ) (x − y) dy ψ (x) dx U U Z 0 Z u (y) (Dα φl ) (x − y) ψ (x) dxdy U U0 Z Z |α| (−1) u (y) φl (x − y) (Dα ψ) (x) dxdy. Z

µZ

U

³

αu ∗ φ g It follows that Dα (e u ∗ φl ) = D l Therefore,

||Dα (e u ∗ φl ) − Dα u||Lp (U0 )

U0

´

as weak derivatives defined on Cc∞ (U0 ) .

¯¯ ¯¯ ¯¯ g α u ∗ φ − D α u¯¯¯¯ = ¯¯D l Lp (U0 ) ¯¯ ¯¯ ¯¯ g ¯ ¯ α u¯¯ g ≤ ¯¯Dα u ∗ φl − D → 0. p n L (R )

This proves the theorem. As part of the proof of the theorem, the following corollary was established. Corollary 38.0.8 Let U0 and U be as in the above theorem. Then for all l large enough and φl a mollifier, ³ ´ αu ∗ φ g Dα (e u ∗ φl ) = D (38.0.2) l as distributions on Cc∞ (U0 ) .

1100


Definition 38.0.9 Let U be an open set. C ∞ (U ) denotes the set of functions which are defined and infinitely differentiable on U. Note that f (x) = x1 is a function in C ∞ (0, 1) . However, it is not equal to the restriction to (0, 1) of some function Cc¡∞ (R) ¡ ¢ which is in ∞ ¢ . This illustrates the ∞ ∞ distinction between C (U ) and C U . The set, C U is a subset of C ∞ (U ) . The following theorem is known as the Meyer Serrin theorem. Theorem 38.0.10 (Meyer Serrin) Let U be an open subset of Rn . Then if δ > 0 and u ∈ X m,p (U ) , there exists J ∈ C ∞ (U ) such that ||J − u||m,p,U < δ. Proof: Let · · · Uk ⊆ Uk ⊆ Uk+1 · · · be a sequence of open subsets of U whose union equals U such that Uk is compact for all k. Also let U−3 = U−2 = U−1 = U0 = ∞ ∅. Now define Vk ≡ Uk+1 \ Uk−1 . Thus {Vk }k=1 is an open cover of U. Note the open cover is locally finite and therefore, there exists a partition of unity subordinate to ∞ this open cover, {η k }k=1 such that each spt (η k ) ∈ Cc (Vk ) . Let ψ m denote the sum of all the η k which are non zero at some point of Vm . Thus ∞ X

spt (ψ m ) ⊆ Um+2 \ Um−2 , ψ m ∈ Cc∞ (U ) ,

ψ m (x) = 1

(38.0.3)

m=1

for all x ∈ U, and ψ m u ∈ W m,p (Um+2 ) . Now let φl be a mollifier and consider J≡

∞ X

uψ m ∗ φlm

(38.0.4)

m=0

where lm is chosen large enough that the following two conditions hold: ¡ ¢ spt uψ m ∗ φlm ⊆ Um+3 \ Um−3 , ¯¯ ¯¯ ¯¯(uψ m ) ∗ φl − uψ m ¯¯ m m,p,U

m+3

¯¯ ¯¯ = ¯¯(uψ m ) ∗ φlm − uψ m ¯¯m,p,U
10, some large value. ¯¯ ¯¯ N ¯¯ X ¡ ¢¯¯¯¯ ¯¯ ||J − u||m,p,UN −3 = ¯¯ uψ k ∗ φlk − uψ k ¯¯ ¯¯ ¯¯ k=0

≤

N X

m,p,UN −3

¯¯ ¯¯ ¯¯uψ k ∗ φl − uψ k ¯¯ k

k=0

≤

N X k=0

δ 2m+5

< δ.

m,p,UN −3

1101 Now apply the monotone convergence theorem to conclude that ||J − u||m,p,U ≤ δ. This proves the theorem. Note that J = 0 on ∂U. Later on, you will see that this is pathological. In the study of partial differential equations it is the space W m,p (U ) which ¡ ¢is of the most use, not the space X m,p (U ) . This is because of the density of C ∞ U . Nevertheless, for reasonable open sets, U, the two spaces coincide. Definition 38.0.11 An open set, U ⊆ Rn is said to satisfy the segment condition if for all z ∈ U , there exists an open set Uz containing z and a vector a such that U ∩ U z + ta ⊆ U for all t ∈ (0, 1) .

Uz u z

U Uz

T

U + ta

You can imagine open sets which do not satisfy the segment condition. For example, a pair of circles which are tangent at their boundaries. The condition in the above definition breaks down at their point of tangency. Here is a simple lemma which will be used in the proof of the following theorem. Lemma 38.0.12 If u ∈ W m,p (U ) and ψ ∈ Cc∞ (Rn ) , then uψ ∈ W m,p (U ) . Proof: Let |α| ≤ m and let φ ∈ Cc∞ (U ) . Then Z (Dxi (uψ)) (φ) ≡ − uψφ,xi dx U Z ³ ´ = − u (ψφ),xi − φψ ,xi dx U Z = (Dxi u) (ψφ) + uψ ,xi φdx U Z ¡ ¢ = ψDxi u + uψ ,xi φdx U

Therefore, Dxi (uψ) = ψDxi u + uψ ,xi ∈ Lp (U ) . In other words, the product rule holds. Now considering the terms in the last expression, you can do the same

1102


argument with each of these as long as they all have derivatives in Lp (U ) . Therefore, continuing this process the lemma is proved. Theorem 38.0.13 Let U be an open set and suppose there exists a locally finite ∞ covering2 of U which is of the form {Ui }i=1 such that each Ui is a bounded open set which satisfies the conditions of Definition 38.0.11. Thus there exist vectors, ai such that for all t ∈ (0, 1) , Ui ∩ U + tai ⊆ U. ¡ ¢ ∞ m,p Then C U is dense in X (U ) and so W m,p (U ) = X m,p (U ) . ∞

Proof: Let {ψ i }i=1 be a partition of unity subordinate to the given open cover with ψ i ∈ Cc∞ (Ui ) and let u ∈ X m,p (U ) . Thus u=

∞ X

ψ k u.

k=1

Consider Uk for some k. Let ak be the special vector associated with Uk such that tak + U ∩ Uk ⊆ U

(38.0.7)

for all t ∈ (0, 1) and consider only t small enough that spt (ψ k ) − tak ⊆ Uk

(38.0.8)

Pick l (t) > 1/t which is also large enough that µ µ ¶ ¶ 1 1 tak + U ∩ Uk + B 0, ⊆ U, spt (ψ k ) + B 0, − tak ⊆ Uk . (38.0.9) l (t) l (tk ) This can be done because tak + U ∩ Uk is a compact subset of U and so has positive distance to U C and spt (ψ k )−tak is a compact subset of Uk having positive distance to UkC . Let tk be such a value for t and for φl a mollifier, define Z vtk (x) ≡ u e (x + tk ak − y) ψ k (x + tk ak − y) φl(tk ) (y) dy (38.0.10) Rn

where as usual, u e is the zero extention of u ³off U. For ´ vtk (x) 6= 0, it is necessary 1 that x + tk ak − y ∈ spt (ψ k ) for some y ∈ B 0, l(tk ) . Therefore, using 38.0.9, for vtk (x) 6= 0, it is necessary that µ ¶ 1 x ∈ y − tk ak + U ∩ spt (ψ k ) ⊆ B 0, + spt (ψ k ) − tk ak l (tk ) 2 This is never a problem in Rn . In fact, every open covering has a locally finite subcovering in Rn or more generally in any metric space due to Stone’s theorem. These are issues best left to you in case you are interested. I am usually interested in bounded sets, U, and for these, there is a finite covering.

1103 µ ⊆ B 0,

1 l (tk )

¶ + spt (ψ k ) − tk ak ⊆ Uk

showing that vtk has compact support in Uk . Now change variables in 38.0.10 to obtain Z u e (y) ψ k (y) φl(tk ) (x + tk ak − y) dy. (38.0.11) vtk (x) ≡ Rn

For x ∈ U ∩ Uk , the above equals zero unless µ y − tk ak − x ∈ B 0,

1 l (tk )

¶

which implies by 38.0.9 that µ y ∈ tk ak + U ∩ Uk + B 0,

1 l (tk )

¶ ⊆U

Therefore, for such x ∈ U ∩ Uk ,38.0.11 reduces to Z vtk (x) = u (y) ψ k (y) φl(tk ) (x + tk ak − y) dy n ZR = u (y) ψ k (y) φl(tk ) (x + tk ak − y) dy. U

It follows that for |α| ≤ m, and x ∈ U ∩ Uk Z α D vtk (x) = u (y) ψ k (y) Dα φl(tk ) (x + tk ak − y) dy U Z = Dα (uψ k ) (y) φl(tk ) (x + tk ak − y) dy ZU = Dα^ (uψ k ) (y) φl(tk ) (x + tk ak − y) dy Rn Z = Dα^ (uψ k ) (x + tk ak − y) φl(tk ) (y) dy.

(38.0.12)

Rn

Actually, this formula holds for all x ∈ U. If x ∈ U but x ∈ / Uk , then the left side of the above formula equals zero because, as noted above, spt (vtk ) ⊆ Uk . The integrand of the right side equals zero unless µ ¶ 1 x ∈ B 0, + spt (ψ k ) − tk ak ⊆ Uk l (tk ) by 38.0.9 and here x ∈ / Uk . Next an estimate is obtained for ||Dα vtk − Dα (uψ k )||Lp (U ) . By 38.0.12, ||Dα vtk − Dα (uψ k )||Lp (U ) ≤

1104


µZ µZ Rn

U

Z ≤ Rn

≤

ε 2k

¶p ¶1/p ¯ ¯ ¯ α^ ¯ α ^ D (uψ ) (x + t a − y) − D (uψ ) (x) φ (y) dy dx ¯ ¯ l(tk ) k k k k

µZ ¯ ¯p ¶1/p ¯ α^ ¯ α ^ dy φl(tk ) (y) ¯D (uψ k ) (x + tk ak − y) − D (uψ k ) (x)¯ dx U

whenever tk is taken small enough. Pick tk this small and let wk ≡ vtk . Thus ||Dα wk − Dα (uψ k )||Lp (U ) ≤ and wk ∈ Cc∞ (Rn ) . Now let J (x) ≡

∞ X

ε 2k

wk .

k=1

Since the Uk are locally finite and spt (wk ) ⊆ Uk for each k, it follows Dα J =

∞ X

Dα wk

k=0

and the sum is always finite. Similarly, Dα

∞ X

(ψ k u) =

k=1

∞ X

Dα (ψ k u)

k=1

and the sum is always finite. Therefore, ¯¯ ¯¯ ∞ ¯¯ X ¯¯ ¯¯ ¯¯ α α α α ||D J − D u||Lp (U ) = ¯¯ D wk − D (ψ k u)¯¯ ¯¯ ¯¯ k=1

≤

∞ X

Lp (U )

||Dα wk − Dα (ψ k u)||Lp (U ) ≤

k=1

∞ X ε = ε. 2k

k=1

By choosing tk small enough, such an inequality can be obtained for ¯¯ β ¯¯ ¯¯D J − Dβ u¯¯ p L (U ) for each multi index, β such that |β| ≤ m. Therefore, there exists J ∈ Cc∞ (Rn ) such that ||J − u||W m,p (U ) ≤ εK where K equals the number of multi indices no larger than m. Since ε is arbitrary, this proves the theorem.

1105 Corollary 38.0.14 Let U be an open set which has the segment property. Then W m,p (U ) = X m,p (U ) . Proof: Start with an open covering of U whose sets satisfy the segment condition and obtain a locally finite refinement consisting of bounded sets which are of the sort in the above theorem. Now consider a situation where h : U → V where U and V are two open sets in Rn and Dα h exists and is continuous and bounded if |α| < m − 1 and Dα h is Lipschitz if |α| = m − 1. Definition 38.0.15 Whenever h : U → V, define h∗ mapping the functions which are defined on V to the functions which are defined on U as follows. h∗ f (x) ≡ f (h (x)) . h : U → V is bilipschitz if h is one to one, onto and Lipschitz and h−1 is also one to one, onto and Lipschitz. Theorem 38.0.16 Let h : U → V be one ¡to one ¢ and onto where U and V are two open sets. Also suppose that Dα h and Dα h−1 exist and are Lipschitz continuous if |α| ≤ m − 1 for m a positive integer. Then h∗ : W m,p (V ) → W m,p (U ) is continuous,¡linear, ¢∗ one to one, and has an inverse with the same properties, the inverse being h−1 . Proof: It is clear that h∗ is linear. It is required to show it is one to one and continuous. First suppose h∗ f = 0. Then Z p 0= |f (h (x))| dx V

and so f (h (x)) = 0 for a.e. x ∈ U. Since h is Lipschitz, it takes sets of measure zero to sets of measure zero. Therefore, f (y) = 0 a.e. This shows h∗ is one to one. By the Meyer Serrin theorem, Theorem 38.0.10, it suffices to verify that h∗ is continuous on functions in C ∞ (V ) . Let f be such a function. Then using the chain rule and product rule, (h∗ f ),i (x) = f,k (h (x)) hk,i (x) , (h∗ f ),ij (x)

=

(f,k (h (x)) hk,i (x)),j

= f,kl (h (x)) hl,j (x) hk,i (x) + f,k (h (x)) hk,ij (x) etc. In general, for |α| ≤ m − 1, succsessive applications of the product rule and chain rule yield that Dα (h∗ f ) (x) has the form X ¡ ¢ Dα (h∗ f ) (x) = h∗ Dβ f (x) gβ (x) |β|≤|α|

1106


where gβ is a bounded Lipschitz function with Lipschitz constant dependent on h and its derivatives. It only remains to take one more derivative of the functions, Dα f for |α| = m − 1. This can be done again but this time you have to use Rademacher’s theorem which assures you that the derivative of a Lipschitz function exists a.e. in order to take the partial derivative of the gβ (x) . When this is done, the above formula remains valid for all |α| ≤ m. Therefore, using the change of variables formula for multiple integrals, Corollary 35.6.13 on Page 1063, Z X Z ¯ ¡ ¯ ¢ p ¯h∗ Dβ f (x)¯p dx |Dα (h∗ f ) (x)| dx ≤ Cm,p,h U

= = ≤

|β|≤m

U

|β|≤m

U

|β|≤m

V

X Z ¯¡ ¯ ¢ ¯ Dβ f (h (x))¯p dx

Cm,p,h

X Z ¯¡ ¯ ¯ ¯ ¢ ¯ Dβ f (y)¯p ¯det Dh−1 (y)¯ dy

Cm,p,h

Cm,p,h,h−1 ||f ||m,p,V

This shows h∗ is continuous on C ∞ (V ) ∩ W m,p (U ) and since ¡ this ¢∗ set is dense, this proves h∗ is continuous. The same argument applies to h−1 and now the ¡ ¢∗ definitions of h∗ and h−1 show these are inverses.

38.1

Embedding Theorems For W m,p (Rn )

Recall Theorem 35.5.1 which is listed here for convenience. Theorem 38.1.1 Suppose u, u,i ∈ Lploc (Rn ) for i = 1, · · · , n and p > n. Then u has a representative, still denoted by u, such that for all x, y ∈Rn , ÃZ

!1/p p

|u (x) − u (y)| ≤ C

|∇u| dz

|x − y|

(1−n/p)

.

(38.1.13)

B(x,2|y−x|)

This amazing result shows that every u ∈ W m,p (Rn ) has a representative which is continuous provided p > n. Using the above inequality, one can give an important embedding theorem. Definition 38.1.2 Let X, Y be two Banach spaces and let f : X → Y be a function. Then f is a compact map if whenever S is a bounded set in X, it follows that f (S) is precompact in Y . n Theorem 38.1.3 Let U be a bounded open set and for u a function ¡ ¢defined on R , 1,p n let rU u (x) ≡ u (x) for x ∈ U . Then if p > n, rU : W (R ) → C U is continuous and compact.

¡

38.1. EMBEDDING THEOREMS FOR W M,P RN

¢

1107

Proof: First suppose uk → 0 in W 1,p (Rn ) . Then if rU uk does not converge to 0, it follows there exists a sequence, still denoted by k and ε > 0 such that uk → 0 in W 1,p (Rn ) but ||rU uk ||∞ ≥ ε. Selecting a further subsequence which is still denoted by k, you can also assume uk (x) → 0 a.e. Pick such an x0 ∈ U where this convergence takes place. Then from 38.1.13, for all x ∈ U , |uk (x)| ≤ |uk (x0 )| + C ||uk ||1,p,Rn diam (U ) showing that uk converges uniformly to 0 on U contrary to ||rU uk ||∞ ≥ ε. Therefore, rU is continuous as claimed. Next let S be a bounded subset of W 1,p (Rn ) with ||u||1,p < M for all u ∈ S. Then for u ∈ S Z p p r mn ([|u| > r] ∩ U ) ≤ |u| dmn ≤ M p [|u|>r]∩U

and so mn ([|u| > r] ∩ U ) ≤

Mp . rp

Now choosing r large enough, M p /rp < mn (U ) and so, for such r, there exists xu ∈ U such that |u (xu )| ≤ r. Therefore from 38.1.13, whenever x ∈ U, |u (x)| ≤ ≤

1−n/p

|u (xu )| + CM diam (U ) r + CM diam (U )

1−n/p

showing that {rU u : u ∈ S} is uniformly bounded. But also, for x, y ∈ U ,38.1.13 implies 1− n |u (x) − u (y)| ≤ CM |x − y| p showing that {rU u : u ∈ S} is equicontinuous. By the Ascoli Arzela theorem, it follows rU (S) is precompact and so rU is compact. Definition 38.1.4 Let α ∈ (0, 1] and K a compact subset of Rn C α (K) ≡ {f ∈ C (K) : ρα (f ) + ||f || ≡ ||f ||α < ∞} where ||f || ≡ ||f ||∞ ≡ sup{|f (x)| : x ∈ K} and

½ ρα (f ) ≡ sup

¾ |f (x) − f (y)| : x, y ∈ K, x = 6 y . α |x − y|

Then (C α (K) , ||·||α ) is a complete normed linear space called a Holder space. The verification that this is a complete normed linear space is routine and is left for you. More generally, one considers the following class of Holder spaces.

1108


Definition 38.1.5 Let K be a compact subset of Rn and let λ ∈ (0, 1]. C m,λ (K) denotes the set of functions, u which are restrictions of functions defined on Rn to Ksuch that for |α| ≤ m, Dα u ∈ C (K) and if |α| = m,

Dα u ∈ C λ (K) .

Thus C 0,λ (K) = C λ (K) . The norm of a function in C m,λ (K) is given by X ||u||m,λ ≡ sup ρλ (Dα u) + ||Dα u||∞ . |α|=m

|α|≤m

Lemma 38.1.6 Let m be a positive integer, K a compact subset of Rn , and let 0 < β < λ ≤ 1. Then the identity map from C m,λ (K) into C m,β (K) is compact. if

Proof: First note that the containment is obvious because for any function, f, ( ) |f (x) − f (y)| ρλ (f ) ≡ sup : x, y ∈ K, x 6= y < ∞, λ |x − y|

Then

( ρβ (f ) ≡

sup (

= sup sup

|x − y|

β

|f (x) − f (y)| λ

( ≤

|f (x) − f (y)|

|x − y|

|f (x) − f (y)| λ

|x − y|

) : x, y ∈ K, x 6= y ) |x − y|

λ−β

: x, y ∈ K, x 6= y )

λ−β

diam (K)

: x, y ∈ K, x 6= y

< ∞.

Suppose the identity map, id, is not compact. Then there exists ε > 0 and a ∞ sequence, {fk }k=1 ⊆ C m,λ (K) such that ||fk ||m,λ < M for all k but ||fk − fl ||β ≥ ε whenever k 6= l. By the Ascoli PArzela theorem, there exists a subsequence of this, still denoted by fk such that |α|≤m ||Dα (fl − fk )||∞ < δ where δ satisfies ¶ µ ³ ´³ ε ´β/(λ−β) ε ε , . (38.1.14) 0 < δ < min 2 8 8M Therefore, sup|α|=m ρβ (Dα (fk − fl )) ≥ ε−δ for all k 6= l. It follows that there exist pairs of points and a multi index, α with |α| = m, {xkl , ykl , α} such that ε−δ |(Dα fk − Dα fl ) (xkl ) − ((Dα fk − Dα fl ) (ykl ))| λ−β < ≤ 2M |xkl − ykl | β 2 |xkl − ykl | (38.1.15) and so considering the ends of the above inequality, µ ¶1/(λ−β) ε−δ < |xkl − ykl | . 4M

¡

38.1. EMBEDDING THEOREMS FOR W M,P RN Now also, since 38.1.15 that

P |α|≤m

¢

1109

||Dα (fl − fk )||∞ < δ, it follows from the first inequality in ε−δ 2δ 1. r Let φ ∈ Cc1 (Rn ) and consider |φ| where r > 1. Then a short computation shows r 1 n |φ| ∈ Cc (R ) and ¯ ¯ ¯ ¯ r¯ r−1 ¯¯ φ,i ¯ . ¯|φ| ¯ = r |φ| Cc1

,i

Therefore, from Lemma 38.1.9, µZ rn n−1

|φ| n

≤

r X √ n n i=1

≤

r X √ n n i=1

n

¶(n−1)/n dmn

Z |φ| µZ

r−1

¯ ¯ ¯φ,i ¯ dmn

¯ ¯p ¯φ,i ¯

¶1/p µZ ³ |φ|

r−1

´p/(p−1)

¶(p−1)/p dmn

.

Now choose r such that (r − 1) p rn = . p−1 n−1 That is, let r = µZ |φ| Also,

n−1 n

np n−p

−

p(n−1) n−p

> 1 and so

¶(n−1)/n dmn

p−1 p

µZ

= np

n−p np

np n−p ,

n

=

r X ≤ √ n n i=1

np n−p .

µZ

Then this reduces to

¯ ¯p ¯φ,i ¯

¶1/p µZ |φ|

np n−p

¶(p−1)/p dmn

and so, dividing both sides by the last term yields ¶ n−p np

|φ| n−p dmn Letting q =

rn n−1

it follows

1 q

n

µZ

1 p

−

r X ≤ √ n n i=1

=

n−p np

=

1 n

¯ ¯p ¯φ,i ¯

¶1/p

and

r ||φ||1,p,Rn . ||φ||q ≤ √ n n

r ||φ||1,p,Rn . ≤ √ n n

.

¡


¢

1113

Now let f ∈ W m,p (Rn ) and let ||φk − f ||1,p,Rn → 0 as k → ∞. Taking another subsequence, if necessary, you can also assume φk (x) → f (x) a.e. Therefore, by Fatou’s lemma, µZ ||f ||q

q

¶1/q

|φk (x)| dmn

≤

lim inf

≤

r lim inf √ ||φk ||1,p,Rn = ||f ||1,p,Rn . k→∞ n n

k→∞

Rn

This proves the theorem. Corollary 38.1.11 Suppose mp < n. Then W m,p (Rn ) ⊆ Lq (Rn ) where q = and the identity map, id : W m,p (Rn ) → Lq (Rn ) is continuous.

np n−mp

Proof: This is true if m = 1 according to Theorem 38.1.10. Suppose it is true for m − 1 where m > 1. If u ∈ W m,p (Rn ) and |α| ≤ 1, then Dα u ∈ W m−1,p (Rn ) so by induction, for all such α, np

Dα u ∈ L n−(m−1)p (Rn ) . Thus u ∈ W 1,q1 (Rn ) where q1 =

np n − (m − 1) p

By Theorem 38.1.10, it follows that u ∈ Lq (Rn ) where 1 n − (m − 1) p 1 n − mp = − = . q np n np This proves the corollary. There is another similar corollary of the same sort which is interesting and useful. Corollary 38.1.12 Suppose m ≥ 1 and j is a nonnegative integer satisfying jp < n. Then W m+j,p (Rn ) ⊆ W m,q (Rn ) for q≡

np n − jp

(38.1.16)

and the identity map is continuous. Proof: If |α| ≤ m, then Dα u ∈ W j,p (Rn ) and so by Corollary 38.1.11, Dα u ∈ L (Rn ) where q is given above. This means u ∈ W m,q (Rn ). The above corollaries imply yet another interesting corollary which involves embeddings in the Holder spaces. q

1114


Corollary 38.1.13 Suppose jp < n < (j + 1) p and let m be a positive integer. n Let U be any bounded open set ¡ in ¢ R . Then letting rU denote the restriction nto U , m+j,p n m−1,λ rU : W (R ) → C U is continuous for every λ ≤ λ0 ≡ (j + 1) − p and if λ < (j + 1) − np , then rU is compact. Proof: From Corollary 38.1.12 W m+j,p (Rn ) ⊆ W m,q (Rn ) where q is given by 38.1.16. Therefore, np >n n − jp ¡ ¢ and so by Corollary 38.1.7, W m,q (Rn ) ⊆ C m−1,λ U for all λ satisfying 0 0 such that if |h| < δ, then Z p

|e u (x + h) − u e (x)| dx < εp

(38.1.17)

Rn

Suppose also that for each ε > 0 there exists an open set, G ⊆ U such that G is compact and for all u ∈ K, Z p

|u (x)| dx < εp

(38.1.18)

U \G

Then K is precompact in Lp (Rn ). Proof: To save fussing first consider the case where U = Rn so that u e = u. Suppose the two conditions hold and let φk be a mollifier of the form φk (x) = k n φ (kx) where spt (φ) ⊆ B (0, 1) . Consider Kk ≡ {u ∗ φk : u ∈ K} . and verify the conditions for the Ascoli Arzela theorem for these functions defined on G. Say ||u||p ≤ M for all u ∈ K.

¡


¢

1115

First of all, for u ∈ K and x ∈ Rn , µZ ¶p p |u ∗ φk (x)| ≤ |u (x − y) φk (y)| dy µZ ¶p = |u (y) φk (x − y)| dy Z p ≤ |u (y)| φk (x − y) dy µ ¶Z µ ¶ ≤ sup φk (z) |u (y)| dy ≤ M sup φk (z) z∈Rn

z∈Rn

showing the functions in Kk are uniformly bounded. Next suppose x, x1 ∈ Kk and consider

≤

|u ∗ φk (x) − u ∗ φk (x1 )| Z |u (x − y) − u (x1 −y)| φk (y) dy µZ p

≤

¶1/p µZ

|u (x − y) − u (x1 −y)| dy

q

¶q

φk (y) dy

which by assumption 38.1.17 is small independent of the choice of u whenever |x − x1 | is small enough. ¡ ¢ Note that k is fixed in the above. Therefore, the set, Kk is precompact in C G thanks to the Ascoli Arzela theorem. Next consider how well u ∈ K is approximated by u ∗ φk in Lp (Rn ) . By Minkowski’s inequality, µZ ¶1/p p |u (x) − u ∗ φk (x)| dx µZ µZ ≤

¶p |u (x) − u (x − y)| φk (y) dy µZ

Z ≤

1 B (0, k )

φk (y)

¶1/p dx

¶1/p p |u (x) − u (x − y)| dx dy.

Now let η > 0 be given. From 38.1.17 there exists k large enough that for all u ∈ K, µZ ¶1/p Z Z p φk (y) |u (x) − u (x − y)| dx dy ≤ φk (y) η dy = η. 1 1 B (0, k B (0, k ) ) Now let ε > 0 be given and let δ and G correspond to ε as given in the hypotheses and let 1/k < δ and also k is large enough that for all u ∈ K, ||u − u ∗ φk ||p < ε as in the above inequality. By the Ascoli Arzela theorem there exists an Ã !1/p ε ¡ ¢ m G + B (0, 1)

1116


¡ ¢ m net for Kk in C G . That is, there exist {ui }i=1 ⊆ K such that for any u ∈ K, Ã ||u ∗ φk − uj ∗ φk ||∞
1 n o S ≡ u ∈ W 1,1 (U ) ∩ Lp (U ) : ||u||1,1,U + ||u||Lp (U ) ≤ M and let φ ∈ Cc∞ (U ) and

S1 ≡ {uφ : u ∈ S} .

(38.1.19) (38.1.20)

q

Then S1 is precompact in L (U ) where 1 ≤ q < p. Proof: This depends on Theorem 38.1.14. The second condition is satisfied by taking G ≡ spt (φ). Thus, for w ∈ S1 , Z q |w (x)| dx = 0 < εp . U \G

It remains to satisfy the first condition. It is necessary to verify there exists δ > 0 such that if |v| < δ, then Z ¯ ¯q ¯f f (x)¯¯ dx < εp . (38.1.21) ¯φu (x + v) − φu Rn

Let spt (φ) ∪ (spt (φ) − v) ≡ Gv . Now if h is any measurable function, and if θ ∈ (0, 1) is chosen small enough that θq < 1, Z Z q θq (1−θ)q |h| dx = |h| |h| dx Gv

Gv

µZ

¶θq µZ

≤

|h| dx µZ

Gv

=

³

(1−θ)q

|h| Gv

¶θq µZ |h| dx

|h|

Gv

(1−θ)q 1−θq

¶1−θq 1 ´ 1−θq

¶1−θq .

(38.1.22)

Gv

Now let θ also be small enough that there exists r > 1 such that r

(1 − θ) q =p 1 − θq

and use Holder’s inequality in the last factor of the right side of 38.1.22. Then 38.1.22 is dominated by ¶θq µZ

µZ |h| dx Gv

|h|

p

¶ 1−θq µZ r 1dx

Gv

´ µZ = C ||h||Lp (Gv ) , mn (Gv )

¶1/r0 Gv

³

Therefore, for u ∈ S, Z ¯ Z ¯q ¯f ¯ f ¯φu (x + v) − φu (x)¯ dx = Rn

|h| dx

¶θq .

Gv

q

|φu (x + v) − φu (x)| dx ≤ Gv

¡


¢

³ ´ µZ C ||φu (·+v) − φu (·)||Lp (Gv ) , mn (Gv )

1119 ¶θq |φu (x + v) − φu (x)| dx

Gv

³ ´ µZ ≤ C 2 ||φu (·)||Lp (U ) , mn (U )

¶θq |φu (x + v) − φu (x)| dx Gv

µZ

¶θq

≤ C (φ, M, mn (U ))

|φu (x + v) − φu (x)| dx µZ

= C (φ, M, mn (U ))

Gv

¯ ¯ ¶θq ¯f ¯ f φu (x + v) − φu (x) . ¯ ¯ dx

(38.1.23)

Rn

Now by Lemma 38.1.15, Z Rn

¯ ¯ ³ ´ ¯f f (x)¯¯ dx ≤ C φ, ||u|| ¯φu (x + v) − φu 1,1,U |v|

(38.1.24)

and so from 38.1.23 and 38.1.24, and adjusting the constants Z Rn

¯ ¯q ¯f f (x)¯¯ dx ¯φu (x + v) − φu

³ ³ ´ ´θq ≤ C (φ, M, mn (U )) C φ, ||u||1,1,U |v| = C (φ, M, mn (U )) |v|

θq

which verifies 38.1.21 whenever |v| is sufficiently small. This proves the lemma because the conditions of Theorem 38.1.14 are satisfied. Theorem 38.1.17 Let U be a bounded open set and define for p > 1 n o S ≡ u ∈ W 1,1 (U ) ∩ Lp (U ) : ||u||1,1,U + ||u||Lp (U ) ≤ M

(38.1.25)

Then S is precompact in Lq (U ) where 1 ≤ q < p. ∞

Proof: If suffices to show that every sequence, {uk }k=1 ⊆ S has a subsequence ∞ which converges in Lq (U ) . Let {Km }m=1 denote a sequence of compact subsets of U with the property that Km ⊆ Km+1 for all m and ∪∞ m=1 Km = U. Now let φm ∈ Cc∞ (U ) such that φm (x) ∈ [0, 1] and φm (x) = 1 for all x ∈ Km . Let ∞ Sm ≡ {φm u : u ∈ S}. By Lemma 38.1.16 there exists a subsequence of {uk }k=1 , ∞ ∞ q denoted here by {u1,k }k=1 such that {φ1 u1,k }k=1 converges in L (U ) . Now S2 is ∞ also precompact in Lq (U ) and so there exists a subsequence of {u1,k }k=1 , denoted ∞ ∞ 2 by {u2,k }k=1 such that {φ2 u2,k }k=1 converges in L (U ) . Thus it is also the case that ∞ {φ1 u2,k }k=1 converges in Lq (U ) . Continue taking subsequences in this manner such ∞ ∞ ∞ that for all l ≤ m, {φl um,k }k=1 converges in Lq (U ). Let {wm }m=1 = {um,m }m=1 so ∞ ∞ ∞ that {wk }k=m is a subsequence of {um,k }k=1 . Then it follows for all k, {φk wm }m=1

1120


must converge in Lq (U ) . For u ∈ S, Z ||u −

q φk u||Lq (U )

q

= U

q

|u| (1 − φk ) dx

µZ p

≤

¶q/p µZ

|u| dx U

U

µZ ≤

M U

(1 − φk )

qr

(1 − φk )

qr

¶1/r dx

¶1/r dx

where q/p + 1/r = 1. Now φl (x) → XU (x) and so the integrand in the last integral converges to 0 by the dominated convergence theorem. Therefore, k may be chosen large enough that for all u ∈ S, q

||u − φk u||Lq (U ) ≤

³ ε ´q 3

.

Fix such a value of k. Then ||wq − wp ||Lq (U ) ≤ ||wq − φk wq ||Lq (U ) + ||φk wq − φk wp ||Lq (U ) + ||wp − φk wp ||Lq (U ) ≤

2ε + ||φk wq − φk wp ||Lq (U ) . 3

∞

But {φk wm }m=1 converges in Lq (U ) and so the last term in the above is less than ∞ ε/3 whenever p, q are large enough. Thus {wm }m=1 is a Cauchy sequence and must q therefore converge in L (U ). This proves the theorem.

38.2

An Extension Theorem

Definition 38.2.1 An open subset, U , of Rn has a Lipschitz boundary if it satisfies the following conditions. For each p ∈ ∂U ≡ U \ U , there exists an open set, Q, containing p, an open interval (a, b), a bounded open box B ⊆ Rn−1 , and an orthogonal transformation R such that RQ = B × (a, b), b ∈ B, a < yn < g (b R (Q ∩ U ) = {y ∈ Rn : y y)} © ª where g is Lipschitz continuous on B, a < min g (x) : x ∈ B , and b ≡ (y1 , · · · , yn−1 ). y Letting W = Q ∩ U the following picture describes the situation.

(38.2.26) (38.2.27)

38.2. AN EXTENSION THEOREM

¶ZZ

Q ¶ ¶ ¶

¶ Z

1121

b R(Q)

Z

Z Z ¶ ¶

W ¶ Z xZ ¶ Z Z¶

R

-

R(W ) a

y

The following lemma is important.

Lemma 38.2.2 If U is an open subset of Rn which has a Lipschitz boundary, then it satisfies the segment condition and so X m,p (U ) = W m,p (U ) . Proof: For x ∈ ∂U, simply look at a single open set, Qx described in the above which contains x. Then consider an open set whose intersection with U is of the form RT ({y :b y ∈ B, g (b y ) − ε < yn < y)}) and ªa vector of the form εRT (−en ) © g (b where ε is chosen smaller than min g (x) : x ∈ B − a. There is nothing to prove for points of U. One way to extend many of the above theorems to more general open sets than Rn is through the use of an appropriate extension theorem. In this section, a fairly general one will be presented.

Lemma 38.2.3 Let B × (a, b) be as described in Definition 38.2.1 and let V − ≡ {(b y, yn ) : yn < g (b y)} , V + ≡ {(b y, yn ) : yn > g (b y)}, for g a Lipschitz function of the sort described in this definition. Suppose u+ and u− are Lipschitz functions defined on V + and V − respectively and suppose that b ∈ B. Let u+ (b y, g (b y)) = u− (b y, g (b y)) for all y ½ u (b y , yn ) ≡

u+ (b y, yn ) if (b y , yn ) ∈ V + − u (b y, yn ) if (b y , yn ) ∈ V −

and suppose spt (u) ⊆ B × (a, b). Then extending u to be 0 off of B × (a, b), u is continuous and the weak partial derivatives, u,i , are all in L∞ (Rn ) ∩ Lp (Rn ) for all p > 1 and u,i = (u+ ),i on V + and u,i = (u− ),i on V − . Proof: Consider the following picture which is descriptive of the situation.

1122


¡¡

b

a

spt(u)

¡

¡

B

Note ¡ firsti ¢that u is Lipschitz continuous. To see this, consider |u (y1 ) − u (y2 )| bi , yn = yi . There are various cases to consider depending on whether where y y2 ) . Then letting K ≥ yni is above g (b yi ) . Suppose yn1 < g (b y1 ) and yn2 > g (b + − max (Lip (u ) , Lip (u ) , Lip (g)) , ¯ ¡ ¢ ¡ ¢¯ ¯u y b1 , yn1 − u y b2 , yn2 ¯ ≤ ¯ ¡ ¯ ¢ ¡ ¢¯ ¯ ¡ ¢ ¯u y b1 , yn1 − u y b2 , yn1 ¯ + ¯u y b2 , yn1 − u (b y2 , g (b y2 ))¯ ¯ ¡ ¢¯ b2 , yn2 ¯ + ¯u (b y2 , g (b y2 )) − u y £ ¤ b2 | + K |g (b ≤ K |b y1 − y y2 ) − g (b y1 )| + g (b y1 ) − yn1 + yn2 − g (b y2 ) ¯ ¯ ¡ ¢ b2 | + K ¯yn1 − yn2 ¯ ≤ 2K + K 2 |b y1 − y ¯ ¯¢ ¡ ¡ ¢¡ ¢√ b2 | + ¯yn1 − yn2 ¯ ≤ 2K + K 2 2 |y1 − y2 | = 2K + K 2 |b y1 − y The other cases are similar. Thus u is a Lipschitz continuous function which has compact support. By Corollary 35.5.4 on Page 1054 it follows that u,i ∈ L∞ (Rn ) ∩ Lp (Rn ) for all p > 1. It remains to verify u,i = (u+ ),i on V + and u,i = (u− ),i on V − . The last claim is obvious from the definition of weak derivatives. ³ ´ Lemma 38.2.4 In the situation of Lemma 38.2.3 let u ∈ C 1 V − ∩Cc1 (B × (a, b))3 and define  b ∈ B and yn ≤ g (b y, yn ) if y y) ,  u (b b ∈ B and yn > g (b u (b y, 2g (b y) − yn ) , if y y) w (b y , yn ) ≡  b∈ 0 if y / B. Then w ∈ W 1,p (Rn ) and there exists a constant, C depending only on Lip (g) and dimension such that ||w||W 1,p (Rn ) ≤ C ||u||W 1,p (V − ) . Denote w by E0 u. Thus E0 (u) (y) = u (y) for all y ∈ V − but E0 u = w is defined on all of Rn . Also, E0 is a linear mapping. 3 This

³ ´ means that spt (u) ⊆ B × (a, b) and u ∈ C 1 V − .


1123

Proof: As in the previous lemma, w is Lipschitz continuous and has compact b ∈ B and support so it is clear w ∈ W 1,p (Rn ) . The main task is to find w,i for y yn > g (b y) and then to extract an estimate of the right sort. Denote by U the set b∈ b∈B of points of Rn with the property that (b y, yn ) ∈ U if and only if y / B or y and yn > g (b y) . Then letting φ ∈ Cc∞ (U ) , suppose first that i < n. Then Z w (b y, yn ) φ,i (y) dy U

¡ ¢ ¢ b − hen−1 b − hen−1 u y , 2g y − yn − u (b y, 2g (b y) − yn ) i i ≡ lim φ (y) dy h→0 U h (38.2.28) ½ Z £ ¡ ¢ −1 φ (y) D1 u (b y, 2g (b y) − yn ) hen−1 = lim i h→0 h U ¡ ¡ ¢ ¢¤ b − hen−1 +2D2 u (b y, 2g (b y) − yn ) g y − g (b y) dy i ¾ Z £ ¡ ¡ ¢ ¢ ¤ −1 b − hen−1 + φ (y) o g y − g (b y ) + o (h) dy i h U Z

¡

where en−1 is the unit vector in Rn−1 having all zeros except for a 1 in the ith i b and so except for position. Now by Rademacher’s theorem, y) exists ¡ ¡Dg (b ¢ for a.e.¢ y b − hen−1 a set of measure zero, the expression, o g y − g (b y) is o (h) and also for i b not in the exceptional set, y ¡ ¢ b − hen−1 g y − g (b y) = −hDg (b y) en−1 + o (h) . i i Therefore, since the integrand in 38.2.28 has compact support and because of the Lipschitz continuity of all the functions, the dominated convergence theorem may be applied to obtain Z U

Z U

w (b y, yn ) φ,i (y) dy =

£ ¡ ¢ ¡ ¢¤ φ (y) −D1 u (b y, 2g (b y) − yn ) en−1 + 2D2 u (b y, 2g (b y) − yn ) Dg (b y) en−1 dy i i · ¸ ∂u ∂u ∂g (b y) = φ (y) − (b y, 2g (b y) − yn ) + 2 (b y, 2g (b y) − yn ) dy ∂yi ∂yn ∂yi U Z

and so w,i (y) =

∂u ∂u ∂g (b y) (b y, 2g (b y ) − yn ) − 2 (b y, 2g (b y) − yn ) ∂yi ∂yn ∂yi

(38.2.29)

whenever i < n which is what you would expect from a formal application of the chain rule. Next suppose i = n. Z w (b y, yn ) φ,n (y) dy U

1124


Z

u (b y, 2g (b y) − (yn + h)) − u (b y, 2g (b y ) − yn ) φ (y) dy h U Z D2 u (b y, 2g (b y) − yn ) h + o (h) = lim φ (y) dy h→0 U h Z ∂u = (b y, 2g (b y) − yn ) φ (y) dy ∂y n U

= lim − h→0

showing that w,n (y) =

−∂u (b y, 2g (b y) − yn ) ∂yn

(38.2.30)

which is also expected. From the definnition, for y ∈ Rn \ U ≡ {(b y, yn ) : yn ≤ g (b y)} it follows w,i = u,i p and on U, w,i is given by 38.2.29 and 38.2.30. Consider ||w,i ||Lp (U ) for i < n. From 38.2.29 ¯p Z ¯ ¯ ∂u ∂u ∂g (b y) ¯¯ p ¯ ||w,i ||Lp (U ) = (b y , 2g (b y ) − y ) − 2 (b y , 2g (b y ) − y ) dy n n ¯ ∂yn ∂yi ¯ U ∂yi ¯p Z ¯ ¯ ¯ ∂u ¯ y, 2g (b y) − yn )¯¯ ≤2 ¯ ∂yi (b U ¯p ¯ ¯ ¯ p p ¯ ∂u (b y, 2g (b y) − yn )¯¯ Lip (g) dy +2 ¯ ∂yn ¯p Z ¯ ¯ ¯ ∂u p ¯ ¯ (b y , 2g (b y ) − y ) ≤ 4p (1 + Lip (g) ) n ¯ ¯ ∂yi U ¯p ¯ ¯ ¯ ∂u ¯ (b y, 2g (b y) − yn )¯¯ dy +¯ ∂yn ¯p Z Z ∞ ¯ ¯ ∂u ¯ p p ¯ y, 2g (b y) − yn )¯¯ = 4 (1 + Lip (g) ) ¯ ∂yi (b B g(b y) p−1

¯p ¯ ¯ ¯ ∂u ¯ +¯ (b y, 2g (b y) − yn )¯¯ dyn db y ∂yn Z Z p

p

g(b y)

= 4 (1 + Lip (g) ) B

−∞

¯p ¯ ¯p ¯ ¯ ¯ ∂u ¯ ¯ ∂u ¯ ¯ ¯ y , zn )¯ + ¯ (b y, zn )¯¯ dzn db y ¯ ∂yi (b ∂yn Z Z p

= 4p (1 + Lip (g) ) B

a

g(b y)

¯ ¯p ¯ ∂u ¯ ¯ ¯ (b y , z ) n ¯ ¯ ∂yi

¯ ¯p ¯ ∂u ¯ p p ¯ +¯ (b y, zn )¯¯ dzn db y ≤ 4p (1 + Lip (g) ) ||u||1,p,V − ∂yn


1125

Now by similar reasoning, p

||w,n ||Lp (U )

¯p Z ¯ ¯ −∂u ¯ ¯ ¯ (b y , 2g (b y ) − y ) n ¯ dy ¯ ∂yn U ¯p Z Z ∞ ¯ ¯ ¯ −∂u ¯ y, 2g (b y) − yn )¯¯ dyn db y ¯ ∂yn (b B g(b y) ¯p Z Z g(by) ¯ ¯ −∂u ¯ p ¯ ¯ y = ||u,n ||1,p,V − . (b y , z ) n ¯ dzn db ¯ ∂yn B a

= = =

It follows p

||w||1,p,Rn

and so

p

p

=

||w||1,p,U + ||u||1,p,V −

≤

4p n (1 + Lip (g) ) ||u||1,p,V − + ||u||1,p,V −

p

p

p

p

p

p

||w||1,p,Rn ≤ 4p n (2 + Lip (g) ) ||u||1,p,V −

which implies p 1/p

||w||1,p,Rn ≤ 4n1/p (2 + Lip (g) )

||u||1,p,V −

It is obvious that E0 is a continuous linear mapping. This proves the lemma. Now recall Definition 38.2.1, listed here for convenience. Definition 38.2.5 An open subset, U , of Rn has a Lipschitz boundary if it satisfies the following conditions. For each p ∈ ∂U ≡ U \ U , there exists an open set, Q, containing p, an open interval (a, b), a bounded open box B ⊆ Rn−1 , and an orthogonal transformation R such that RQ = B × (a, b),

(38.2.31)

b ∈ B, a < yn < g (b R (Q ∩ U ) = {y ∈ Rn : y y)} © ª where g is Lipschitz continuous on B, a < min g (x) : x ∈ B , and b ≡ (y1 , · · · , yn−1 ). y Letting W = Q ∩ U the following picture describes the situation.

¶ ZZ

Q ¶ ¶ ¶

¶ Z

b R(Q)

Z

W ¶ Z xZ ¶ Z Z¶

Z Z ¶ ¶

R

-

R(W ) a

y

(38.2.32)

1126


¡ ¢ Lemma 38.2.6 In the situation of Definition 38.2.1 let u ∈ C 1 U ∩ Cc1 (Q) and define ¡ ¢∗ Eu ≡ R∗ E0 RT u. ¡ ¢∗ where RT maps W 1,p (U ∩ Q) to W 1,p (R (W )) . Then E is linear and satisfies ||Eu||W 1,p (Rn ) ≤ C ||u||W 1,p (Q∩U ) , Eu (x) = u (x) for x ∈ Q ∩ U. where C depends only on the dimension and Lip (g) . Proof: This follows from Theorem 38.0.16 and Lemma 38.2.4. The following theorem is a general extension theorem for Sobolev spaces. Theorem 38.2.7 Let U be a bounded ¡ open set which has¢ Lipschitz boundary. Then for each p ≥ 1, there exists E ∈ L W 1,p (U ) , W 1,p (Rn ) such that Eu (x) = u (x) a.e. x ∈ U. Proof: Let ∂U ⊆ ∪pi=1 Qi Where the Qi are as described in Definition 38.2.5. Also let Ri be the orthogonal trasformation and gi the Lipschitz functions associated p with Qi as in this definition. Now let Q0 ⊆ 0 ⊆ U be such that U ⊆ ∪i=0 Qi ,¡ and ¢ PQ p ∞ let ψ i ∈ Cc (Qi ) with ψ i (x) ∈ [0, 1] and i=0 ψ i (x) = 1 on U . For u ∈ C ∞ U , let E 0 (ψ 0 u) ≡ ψ 0 u on Q0 and 0 off Q0 . Thus ¯¯ 0 ¯¯ ¯¯E (ψ 0 u)¯¯ = ||ψ 0 u||1,p,U . 1,p,Rn For i ≥ 1, let

¡ ¢∗ E i (ψ i u) ≡ Ri∗ E0 RT (ψ i u) .

Thus, by Lemma 38.2.6 ¯¯ 1 ¯¯ ¯¯E (ψ i u)¯¯

1,p,Rn

≤ C ||ψ i u||1,p,Qi ∩U

¡ ¢ where the constant depends on Lip (gi ) but is independent of u ∈ C ∞ U . Now define E as follows. p X E i (ψ i u) . Eu ≡ i=0

Thus for u ∈ C

∞

¡ ¢ U , it follows Eu (x) = u (x) for all x ∈ U. Also,

||Eu||1,p,Rn

≤ =

p p X ¯¯ i ¯¯ X ¯¯E (ψ i u)¯¯ ≤ Ci ||ψ i u|| i=0 p X

Ci ||ψ i u||1,p,U ≤

i=0

≤

1,p,Qi ∩U

i=0

(p + 1)

p X

Ci ||u||1,p,U

i=0 p X i=0

Ci ||u||1,p,U ≡ C ||u||1,p,U .

(38.2.33)

38.3. GENERAL EMBEDDING THEOREMS

1127

¡ ¢ where C depends on the ψ i and the gi but is independent of u ∈ C ∞ U . Therefore, ¡ ¢ by density of C ∞ U in W 1,p (U ) , E has a unique continuous extension to W 1,p (U ) still denoted by E satisfying the inequality determined by the ends of 38.2.33. It remains to verify that Eu (x) = u (x) a.e. for ¡ x¢ ∈ U . Let uk → u in W 1,p (U ) where uk ∈ C ∞ U . Therefore, by 38.2.33, Euk → Eu in W 1,p (Rn ) . Since Euk (x) = uk (x) for each k, ||u − Eu||Lp (U )

= =

lim ||uk − Euk ||Lp (U )

k→∞

lim ||Euk − Euk ||Lp (U ) = 0

k→∞

which shows u (x) = Eu (x) for a.e. x ∈ U as claimed. This proves the theorem. Definition 38.2.8 Let U be an open set. Then W0m,p (U ) is the closure of the set, Cc∞ (U ) in W m,p (U ) . Corollary 38.2.9 Let U be a bounded open set which has Lipschitz boundary and let³ W be an open set ´ containing U . Then for each p ≥ 1, there exists EW ∈ 1,p 1,p L W (U ) , W0 (W ) such that EW u (x) = u (x) a.e. x ∈ U. Proof: Let ψ ∈ Cc∞ (W ) and ψ = 1 on U. Then let EW u ≡ ψEu where E is the extension operator of Theorem 38.2.7. Extension operators of the above sort exist for many open sets, U, not just for bounded ones. In particular, the above discussion would apply to an open set, U, not necessarily bounded, if you relax the condition that the Qi must be bounded p but require the existence of a finite partition of unity {ψ i }i=1 having the property that ψ i and ψ i,j are uniformly bounded for all i, j. The proof would be identical to the above. My main interest is in bounded open sets so the above theorem will suffice. Such an extension operator will be referred to as a (1, p) extension operator.

38.3

General Embedding Theorems

With the extension theorem it is possible to give a useful theory of embeddings. Theorem 38.3.1 Let 1 ≤ p < n and 1q = p1 − n1 and let U be any open set for which there exists a (1, p) extension operator. Then if u ∈ W 1,p (U ) , there exists a constant independent of u such that ||u||Lq (U ) ≤ C ||u||1,p,U . If U is bounded and r < q, then id : W 1,p (U ) → Lr (U ) is also compact. Proof: Let E be the (1, p) extension operator. Then by Theorem 38.1.10 on Page 1112 ||u||Lq (U )

≤ ≤

1 (n − 1) p ||Eu||1,p,Rn ||Eu||Lq (Rn ) ≤ √ n n (n − p) C ||u||1,p,U .

1128


It remains to prove the assertion about compactness. If S ⊆ W 1,p (U ) is bounded then n o sup ||u||1,1,U + ||u||Lq (U ) < ∞ u∈S

and so by Theorem 38.1.17 on Page 1119, it follows S is precompact in Lr (U ) .This proves the theorem. Corollary 38.3.2 Suppose mp < n and U is an open set satisfying the segment condition which has a (1, p) extension operator for all p. Then id ∈ L (W m,p (U ) , Lq (U )) np where q = n−mp . Proof: This is true if m = 1 according to Theorem 38.3.1. Suppose it is true for m − 1 where m > 1. If u ∈ W m,p (U ) and |α| ≤ 1, then Dα u ∈ W m−1,p (U ) so by induction, for all such α, np

Dα u ∈ L n−(m−1)p (U ) . Thus, since U has the segment condition, u ∈ W 1,q1 (U ) where q1 =

np n − (m − 1) p

By Theorem 38.3.1, it follows u ∈ Lq (Rn ) where 1 n − (m − 1) p 1 n − mp = − = . q np n np This proves the corollary. There is another similar corollary of the same sort which is interesting and useful. Corollary 38.3.3 Suppose m ≥ 1 and j is a nonnegative integer satisfying jp < n. Also suppose U has a (1, p) extension operator for all p ≥ 1 and satisfies the segment condition. Then ¡ ¢ id ∈ L W m+j,p (U ) , W m,q (U ) where q≡

np . n − jp

(38.3.34)

If, in addition to the above, U is bounded and 1 ≤ r < q, then ¡ ¢ id ∈ L W m+j,p (U ) , W m,r (U ) and is compact. Proof: If |α| ≤ m, then Dα u ∈ W j,p (U ) and so by Corollary 38.3.2, Dα u ∈ L (U ) where q is given above. Since U has the segment property, this means u ∈ W m,q (U ). It remains to verify the assertion about compactness of id. Let S be bounded in W m+j,p (U ) . Then S is bounded in W m,q (U ) by the ∞ first part. Now let {uk }k=1 be any sequence in S. The corollary will be proved q

38.3. GENERAL EMBEDDING THEOREMS

1129

if it is shown that any such sequence has a convergent subsequence in W m,r (U ). Let {α1 , α2 , · · · , αh } denote the indices satisfying |α| ≤ m. Then for each of these indices, α, n o sup ||Dα u||1,1,U + ||Dα u||Lq (U ) < ∞ u∈S

and so for each such α, satisfying |α| ≤ m, it follows from Lemma 38.1.16 on Page 1117 that {Dα u : u ∈ S} is precompact in Lr (U ) . Therefore, there exists a subsequence, still denoted by uk such that Dα1 uk converges in Lr (U ) . Applying the same lemma, there exists a subsequence of this subsequence such that both Dα1 uk and Dα2 uk converge in Lr (U ) . Continue taking subsequences until you obtain a ∞ ∞ subsequence, {uk }k=1 for which {Dα uk }k=1 converges in Lr (U ) for all |α| ≤ m. But this must be a convergent subsequence in W m,r (U ) and this proves the corollary. Theorem 38.3.4 Let U be a bounded ¡open ¢ set having a (1, p) extension operator and let p > n. Then id : W 1,p (U ) → C U is continuous and compact. ¡ ¢ Proof: Theorem 38.1.3 on Page 38.1.3 implies rU : W 1,p (Rn ) → C U is continuous and compact. Thus ||u||∞,U = ||Eu||∞,U ≤ C ||Eu||1,p,Rn ≤ C ||u||1,p,U . This proves continuity. If S is a bounded set in W 1,p (U ) , then define S1 ≡ {Eu : u ∈ S} . Then S1 is a bounded set in W 1,p (Rn ) and so by Theorem 38.1.3 the set of restrictions to U, is precompact. However, the restrictions to U are just the functions of S. Therefore, id is compact as well as continuous. Corollary 38.3.5 Let p > n, let U be a bounded open set having a (1, p) extension operator which also satisfies the segment ¡ ¢condition, and let m be a nonnegative integer. Then id : W m+1,p (U ) → C m,λ U is continuous for all λ ∈ [0, 1 − np ] and id is compact if λ < 1 − np . Proof: Let uk → 0 in W m+1,p (U ) . Then it follows that for each |α| ≤ m, D uk → 0 in W 1,p (U ) . Therefore, α

E (Dα uk ) → 0 in W 1,p (Rn ) . Then from Morrey’s inequality, 38.1.13 on Page 1106, if λ ≤ 1 −

n p

1− n p −λ

ρλ (E (Dα uk )) ≤ C ||E (Dα uk )||1,p,Rn diam (U )

and |α| = m .

Therefore, ρλ (E (Dα uk )) = ρλ (Dα uk ) → 0. From Theorem 38.3.4 it follows that for |α| ≤ m, ||Dα uk ||∞ → 0 and so ||uk ||m,λ → 0. This proves the claim about continuity. The claim about compactness for λ < 1 − np follows from Lemma 38.1.6 ¢ id n ¡ id on Page 1108 and this. (Bounded in W m,p (U ) → Bounded in C m,1− p U → ¡ ¢ Compact in C m,λ U .)

1130


Theorem 38.3.6 Suppose jp < n < (j + 1) p and let m be a positive integer. Let U be any bounded open set in Rn which has ¡a (1, p) extension operator ¡ ¢¢ for each p ≥ 1 and the segment property. Then id ∈ L W m+j,p (U ) , C m−1,λ U for every λ ≤ λ0 ≡ (j + 1) − np and if λ < (j + 1) − np , id is compact. Proof: From Corollary 38.3.3 W m+j,p (U ) ⊆ W m,q (U ) where q is given by 38.3.34. Therefore, np >n n − jp ¡ ¢ and so by Corollary 38.3.5, W m,q (U ) ⊆ C m−1,λ U for all λ satisfying (n − jp) n p (j + 1) − n n = = (j + 1) − . np p p

0 0 be given, there exists v ∈ H m (Rn ) such that v|U = u and ||v||H m (Rn ) < ε. Therefore, ||u||0,U ≤ ||v||0,Rn ≤ ||v||H m (U ) < ε. Since ε > 0 is arbitrary, it follows u = 0 a.e. Next suppose ui ∈ H m (U ) for i = 1, 2. There exists vi ∈ H m (Rn ) such that ||vi ||H m (Rn ) < ||ui ||H m (U ) + ε. Therefore, ||u1 + u2 ||H m (U )

≤

||v1 + v2 ||H m (Rn ) ≤ ||v1 ||H m (Rn ) + ||v2 ||H m (Rn )

≤

||u1 ||H m (U ) + ||u2 ||H m (U ) + 2ε

and since ε > 0 is arbitrary, this shows the triangle inequality. The interesting question is the one about completeness. Suppose then {uk } is a Cauchy sequence in H m (U ) . There exists Nk such that if k, l ≥ Nk , it follows ||uk − ul ||H m (U ) < 21k and the numbers, Nk can be taken to be strictly increasing in k. Thus for l ≥ Nk , ||ul − uNk ||H m (U ) < 1/2l . Therefore, there exists wl ∈ H m (Rn ) such that 1 wl |U = ul − uNk , ||wl ||H m (Rn ) < l . 2 Also let vNk |U = uNk with vNk ∈ H m (Rn ) and ||vNk ||H m (Rn ) < ||uNk ||H m (U ) +

1 . 2k

Now for l > Nk , define vl by vl − vNk = wNk so that ||vl − vNk ||H m (Rn ) < 1/2k . In particular, ¯¯ ¯¯ ¯¯v N − vNk ¯¯H m (Rn ) < 1/2k k+1 ∞

which shows that {vNk }k=1 is a Cauchy sequence. Consequently it must converge to v ∈ H m (Rn ) . Let u = v|U . Then ||u − uNk ||H m (U ) ≤ ||v − vNk ||H m (Rn )

39.1. FOURIER TRANSFORM TECHNIQUES

1141

which shows the subsequence, {uNk }k converges to u. Since {uk } is a Cauchy sequence, it follows it too must converge to u. This proves the lemma. The main result is next. Theorem 39.1.8 Suppose U satisfies Assumption 39.1.2. Then for m a nonnegative integer, H m (U ) = W m,2 (U ) and the two norms are equivalent. Proof: Let u ∈ H m (U ) . Then there exists v ∈ H m (Rn ) such that v|U = u. Hence v ∈ W k,2 (Rn ) and so all its weak derivatives up to order m are in L2 (Rn ) . Therefore, the restrictions of these weak derivitves are in L2 (U ) . Since U satisfies the segment condition, it follows u ∈ W m,2 (U ) which shows H m (U ) ⊆ W m,2 (U ) . Next take u ∈ W m,2 (U ) . Then Eu ∈ W m,2 (Rn ) = H m (Rn ) and this shows u ∈ H m (U ) . This has shown the two spaces are the same. It remains to verify their norms are equivalent. Let u ∈ H m (U ) and let v|U = u where v ∈ H m (Rn ) and ||u||H m (U ) + ε > ||v||H m (Rn ) . Then recalling that ||·||H m (Rn ) and ||·||m,2,Rn are equivalent norms for H m (Rn ) , there exists a constant, C such that ||u||H m (U ) + ε > ||v||H m (Rn ) ≥ C ||v||m,2,Rn ≥ C ||u||m,2,U Now consider the two Banach spaces, ³ ´ ³ ´ H m (U ) , ||·||H m (U ) , W m,2 (U ) , ||·||m,2,U . ³ ´ The above inequality shows since ε > 0 is arbitrary that id : H m (U ) , ||·||H m (U ) → ³ ´ W m,2 (U ) , ||·||m,2,U is continuous. By the open mapping theorem, it follows id is continuous in the other direction. Thus there exists a constant, K such that ||u||H m (U ) ≤ K ||u||k,2,U . Hence the two norms are equivalent as claimed. Specializing Corollary 38.3.3 and Theorem 38.3.6 starting on Page 1128 to the case of p = 2 while also assuming more on U yields the following embedding theorems. Theorem 39.1.9 Suppose m ≥ 0 and j is a nonnegative integer satisfying 2j < n.¡ Also suppose U is an ¢ open set which satisfies Assumption 39.1.2. Then id ∈ L H m+j (U ) , W m,q (U ) where q≡

2n . n − 2j

(39.1.6)

If, in addition to the above, U is bounded and 1 ≤ r < q, then ¡ ¢ id ∈ L H m+j (U ) , W m,r (U ) and is compact. Theorem 39.1.10 Suppose for let m be a positive integer. Let Assumption 39.1.2. Then id ∈ (j + 1) − n2 and if λ < (j + 1) −

j a nonnegative integer, 2j < n < 2 (j + 1) and n U ¡be any bounded open¡ set ¢¢ in R which satisfies m+j m−1,λ L H (U ) , C U for every λ ≤ λ0 ≡ n , id is compact. 2

SOBOLEV SPACES BASED ON L2

1142

39.2

Fractional Order Spaces

What has been gained by all this? The main thing is that H m+s (U ) makes sense for any s ∈ (0, 1) and m an integer. You simply replace m with m + s in the above for s ∈ (0, 1). This gives what is meant by H m+s (Rn ) Definition 39.2.1 For m an integer and s ∈ (0, 1) , let H m+s (Rn ) ≡ ( ) µZ ³ ¶1/2 ´m+s 2 2 2 n u ∈ L (R ) : ||u||H m+s (Rn ) ≡ 1 + |x| |F u (x)| dx 1, the integrand is bounded by 4 |t| , and changing to polar coordinates shows Z Z ¢¯2 −n−2s ¯¯¡ it·z −n−2s |t| e − 1 ¯ dt ≤ 4 |t| dt < ∞. [|t|>1]

[|t|>1]

Now for α > 0, Z G (αz)

−n−2s

=

|t|

¯¡ it·αz ¢¯2 ¯ e − 1 ¯ dt

Z

¢¯2 −n−2s ¯¯¡ iαt·z |t| e − 1 ¯ dt Z ¯ ¯−n−2s ¯¡ ir·z ¢¯2 1 ¯r¯ ¯ e − 1 ¯ n dr = ¯ ¯ α α Z ¯ ¡ ¢¯ −n−2s ¯ eir·z − 1 ¯2 dr = α2s G (z) . = α2s |r| =

Also G is continuous and strictly positive. Letting 0 < m (s) = min {G (w) : |w| = 1} and M (s) = max {G (w) : |w| = 1} , it follows from this, and letting α = |z| , w ≡ z/ |z| , that ³ ´ 2s 2s G (z) ∈ m (s) |z| , M (s) |z| . More can be said but this will suffice. Also observe that for s ∈ (0, 1) and b > 0, s

s

(1 + b) ≤ 1 + bs , 21−s (1 + b) ≥ 1 + bs . In what follows, C (s) will denote a constant which depends on the indicated quantities which may be different on different lines of the argument. Then from 39.3.11, Z Z 2 −n−2s |Dα u (x) − Dα u (y)| |x − y| dxdy Z 2

2s

|F Dα u (z)| |z| dz

≤ M (s) Z = M (s)

2

2

2s

|F u (z)| |zα | |z| dz.


1146

No reference was made to |α| = m and so this establishes the top half of 39.3.10. Therefore, 2

2

|||u|||m+s

≡ ||u||m,2,Rn +

X Z Z

2

|Dα u (x) − Dα u (y)| |x − y|

−n−2s

dxdy

|α|=m

Z ³ Z ´m X 2 2 2 2 2s ≤ C 1 + |z| |F u (z)| dz + M (s) |F u (z)| |zα | |z| dz |α|=m

Recall that  X

z12α1

· · · zn2αn ≤ 1 +

n X

m zj2 

j=1

|α|≤m

X

≤ C (n, m)

z12α1 · · · zn2αn .

(39.3.12)

|α|≤m

Therefore, where C (n, m) is the largest of the multinomial coefficients obtained in the expansion, m  n X 1 + zj2  . j=1

Therefore, 2

|||u|||m+s Z ³ Z ´m X 2 2 2 2 2s ≤ C 1 + |z| |F u (z)| dz + M (s) |F u (z)| |zα | |z| dz Z ³ ≤ C

1 + |z|

2

1 + |z|

2

Z ³ ≤ C

´m+s ´m+s

|α|=m

Z 2

2

|F u (z)| dz + M (s)

|F u (z)|

³

1 + |z|

2

|F u (z)| dz = C ||u||H m+s (Rn ) .

It remains to show the other inequality. From 39.3.11, Z Z 2

|Dα u (x) − Dα u (y)| |x − y|

−n−2s

dxdy

Z 2

2s

|F Dα u (z)| |z| dz

≥ m (s) Z = m (s)

2

2

2s

|F u (z)| |zα | |z| dz.

2

´m

2s

|z| dz

39.3. AN INTRINSIC NORM

1147

No reference was made to |α| = m and so this establishes the bottom half of 39.3.10. Therefore, from 39.3.12, 2

≥

|||u|||m+s Z ³ Z ´m X 2 2 2 2 2s C 1 + |z| |F u (z)| dz + m (s) |F u (z)| |zα | |z| dz Z ³

≥

C

=

C

2

1 + |z| Z ³

2

1 + |z| Z ³

≥

C

=

C

2

1 + |z| Z ³

2

1 + |z|

´m

Z 2

|F u (z)| dz + C

´m ³ ´m ³

1 + |z|

2s

1 + |z|

2

´m+s

´

|F u (z)|

2

|α|=m

³

2

1 + |z|

´m

2s

|z| dz

2

|F u (z)| dz

´s

2

|F u (z)| dz

2

|F u (z)| dz = ||u||H m+s (Rn ) .

This proves the theorem. With the above intrinsic norm, it becomes possible to prove the following version of Theorem 39.2.7. n n α Lemma¡ 39.3.2 ¢ Let h : R → R be one to one and onto. Also suppose that D h and Dα h−1 exist and are Lipschitz continuous if |α| ≤ m for m a positive integer. Then h∗ : H m+s (Rn ) → H m+s (Rn )

is continuous,¡linear, ¢∗ one to one, and has an inverse with the same properties, the inverse being h−1 . Proof: Let u ∈ S. From Theorem 39.2.7 and the equivalence of the norms in W m,2 (Rn ) and H m (Rn ) , 2

||h∗ u||H m (Rn ) + 2

RRP

2

|α|=m

|Dα h∗ u (x) − Dα h∗ u (y)| |x − y|

RRP

2

−n−2s

dxdy

−n−2s

|Dα h∗ u (x) − Dα h∗ u (y)| |x − y| dxdy ¯ ¡ β(α) ¢ RRP ¯P 2 ∗ = C ||u||H m (Rn ) + D u gβ(α) (x) |α|=m ¯ |β(α)|≤m h

≤ C ||u||H m (Rn ) +

|α|=m

¯2 ¡ ¢ −n−2s −h∗ Dβ(α) u gβ(α) (y)¯ |x − y| dxdy ¯ ¡ ¢ R R P P 2 ¯ ∗ Dβ(α) u gβ(α) (x) ≤ C ||u||H m (Rn ) + C |α|=m |β(α)|≤m h ¯2 ¡ ¢ −n−2s −h∗ Dβ(α) u gβ(α) (y)¯ |x − y| dxdy (39.3.13) A single term in the last sum corresponding to a given α is then of the form, Z Z ¯ ∗¡ β ¢ ¯ ¡ ¢ ¯h D u gβ (x) − h∗ Dβ u gβ (y)¯2 |x − y|−n−2s dxdy (39.3.14)


1148 ·Z Z

¯ ∗¡ β ¢ ¯ ¡ ¢ ¯h D u (x) gβ (x) − h∗ Dβ u (y) gβ (x)¯2 |x − y|−n−2s dxdy +

≤ Z Z

¯ ∗¡ β ¢ ¯ ¡ ¢ ¯h D u (y) gβ (x) − h∗ Dβ u (y) gβ (y)¯2 |x − y|−n−2s dxdy

·

Z Z

≤ C (h) Z Z

¸

¯ ∗¡ β ¢ ¯ ¡ ¢ ¯h D u (x) − h∗ Dβ u (y)¯2 |x − y|−n−2s dxdy +

¸ ¯ ∗¡ β ¢ ¯ ¯h D u (y)¯2 |gβ (x) − gβ (y)|2 |x − y|−n−2s dxdy .

Changing variables, and then using the names of the old variables to simplify the notation, · Z Z ¯¡ β ¢ ¯ ¡ ¢ ¡ ¢ ¯ D u (x) − Dβ u (y)¯2 |x − y|−n−2s dxdy + ≤ C h, h−1 Z Z

¸ ¯ ∗¡ β ¢ ¯ ¯h D u (y)¯2 |gβ (x) − gβ (y)|2 |x − y|−n−2s dxdy .

By 39.3.10, Z

≤

¯2 2s 2¯ C (h) |F (u) (z)| ¯zβ ¯ |z| dz Z Z ¯ ∗¡ β ¢ ¯ ¯h D u (y)¯2 |gβ (x) − gβ (y)|2 |x − y|−n−2s dxdy. +

In the second term, let t = x − y. Then this term is of the form Z Z ¯ ∗¡ β ¢ ¯ ¯h D u (y)¯2 |gβ (y + t) − gβ (y)|2 |t|−n−2s dtdy (39.3.15) Z ¯ ¡ ¯2 ¢ 2 ≤ C ¯h∗ Dβ u (y)¯ dy ≤ C ||u||H m (Rn ) . (39.3.16) because the inside integral equals a constant which depends on the Lipschitz constants and bounds of the function, gβ and these things depend only on h. The reason this integral is finite is that for |t| ≤ 1, 2

|gβ (y + t) − gβ (y)| |t|

−n−2s

2

≤ K |t| |t|

−n−2s

and using polar coordinates, you see Z 2 −n−2s |gβ (y + t) − gβ (y)| |t| dt < ∞. [|t|≤1] −n−2s

Now for |t| > 1, the integrand in 39.3.15 is dominated by 4 |t| and using polar coordinates, this yields Z Z 2 −n−2s −n−2s |gβ (y + t) − gβ (y)| |t| dt ≤ 4 |t| dt < ∞. [|t|>1]

[|t|>1]

39.3. AN INTRINSIC NORM

1149

It follows 39.3.14 is dominated by an expression of the form Z ¯2 2s 2¯ 2 C (h) |F (u) (z)| ¯zβ ¯ |z| dz + C ||u||H m (Rn ) and so the sum in 39.3.13 is dominated by Z X ¯ ¯2 2 2s ¯zβ ¯ dz + C ||u||2 m n C (m, h) |F (u) (z)| |z| H (R ) Z ≤ C (m, h)

|F (u) (z)|

2

³

|β|≤m 2

1 + |z|

´s ³

2

1 + |z|

´m

2

dz + C ||u||H m (Rn )

2

≤ C ||u||H m+s (Rn ) . This proves the theorem because the assertion about h−1 is obvious. Just replace h with h−1 in the above argument. Next consider the case where U is an open set. Lemma 39.3.3 Let h (U ) ⊆ V where U and V are open subsets of Rn and suppose that h, h−1 : Rn → Rn are both functions in C m,1 (Rn ) . Recall this means Dα h and Dα h−1 exist and are Lipschitz continuous for all |α| ≤ m. Then h∗ ∈ L (H m+s (V ) , H m+s (U )). Proof: Let u ∈ H m+s (V ) and let v ∈ H m+s (Rn ) such that v|V = u. Then from the above, h∗ v ∈ H m+s (Rn ) and so h∗ u ∈ H m+s (U ) because h∗ u = h∗ v|U . Then by Lemma 39.3.2, ||h∗ u||H m+s (U ) ≤ ||h∗ v||H m+s (Rn ) ≤ C ||v||H m+s (Rn ) Since this is true for all v ∈ H m+s (Rn ) , it follows that ||h∗ u||H m+s (U ) ≤ C ||u||H m+s (V ) . With harder work, you don’t need to have h, h−1 defined on all of Rn but I don’t feel like including the details so this lemma will suffice. Another interesting application of the intrinsic norm is the following. Lemma 39.3.4 Let φ ∈ C m,1 (Rn ) and suppose spt (φ) is compact. Then there exists a constant, Cφ such that whenever u ∈ H m+s (Rn ) , ||φu||H m+s (Rn ) ≤ Cφ ||u||H m+s (Rn ) . Proof: It is a routine exercise in the product rule to verify that ||φu||H m (Rn ) ≤ Cφ ||u||H m (Rn ) . It only remains to consider the term involving the integral. A typical term is Z Z 2

−n−2s

|Dα φu (x) − Dα φu (y)| |x − y|

dxdy.


1150

This is a finite sum of terms of the form Z Z ¯ γ ¯ ¯D φ (x) Dβ u (x) − Dγ φ (y) Dβ u (y)¯2 |x − y|−n−2s dxdy where |γ| and |β| ≤ m. Z Z

¯2 2¯ −n−2s |Dγ φ (x)| ¯Dβ u (x) − Dβ u (y)¯ |x − y| dxdy Z Z ¯ β ¯ ¯D u (y)¯2 |Dγ φ (x) − Dγ φ (y)|2 |x − y|−n−2s dxdy +2

≤ 2

By 39.3.10 and the Lipschitz continuity of all the derivatives of φ, this is dominated by Z

=

≤ ≤

¯2 2s 2¯ CM (s) |F u (z)| ¯zβ ¯ |z| dz Z Z ¯ β ¯ ¯D u (y)¯2 |x − y|2 |x − y|−n−2s dxdy +K Z ¯2 2s 2¯ CM (s) |F u (z)| ¯zβ ¯ |z| dz Z Z ¯ ¯2 −n+2(1−s) +K ¯Dβ u (y)¯ |t| dtdy µZ ¶ Z ¯2 2s ¯ ¯2 2¯ C (s) |F u (z)| ¯zβ ¯ |z| dz + K ¯Dβ u (y)¯ dy Z ³ ´m+s 2 2 C (s) 1 + |y| |F u (y)| dy.

Since there are only finitely many such terms, this proves the lemma. Corollary 39.3.5 Let t = m + s for s ∈ [0, 1) and let U, V be open sets. Let φ ∈ Ccm,1 (V ). This means spt (φ) ⊆ V and φ ∈ C m,1 (Rn ) . Then if u ∈ H t (U ) it follows that uφ ∈ H t (U ∩ V ) and ||uφ||H t (U ∩V ) ≤ Cφ ||u||H t (U ) . Proof: Let v|U = u a.e. where v ∈ H t (Rn ) . Then by Lemma 39.3.4, φv ∈ H (Rn ) and φv|U ∩V = φu a.e. Therefore, φu ∈ H t (U ∩ V ) and t

||φu||H t (U ∩V ) ≤ ||φv||H t (Rn ) ≤ Cφ ||v||H t (Rn ) . Taking the infimum for all such v whose restrictions equal u, this yields ||φu||H t (U ∩V ) ≤ Cφ ||u||H t (U ) . This proves the corollary.

39.4. EMBEDDING THEOREMS

39.4

1151

Embedding Theorems

The Fourier transform description of Sobolev spaces makes possible fairly easy proofs of various embedding theorems. Definition 39.4.1 Let Cbm (Rn ) denote the functions which are m times continuously differentiable and for which sup sup |Dα u (x)| ≡ ||u||C m (Rn ) < ∞. b

|α|≤m x∈Rn

¡ ¢ For U an open set, C m U denotes the functions which are restrictions of Cbm (Rn ) to U. It is clear this is a Banach space, the proof being a simple exercise in the use of the fundamental theorem of calculus along with standard results about uniform convergence. Lemma 39.4.2 Let u ∈ S and let n2 + m < t. Then there exists C independent of u such that ||u||C m (Rn ) ≤ C ||u||H t (Rn ) . b

Proof: Using the fact that the Fourier transform maps S to S and the definition of the Fourier transform, |Dα u (x)|

≤ = ≤ ≤

C ||F Dα u||L1 (Rn ) Z C |xα | |F u (x)| dx Z ³ ´|α|/2 2 C 1 + |x| |F u (x)| dx Z ³ ´m/2 ³ ´−t/2 ³ ´t/2 2 2 2 C 1 + |x| 1 + |x| 1 + |x| |F u (x)| dx µZ ³

≤

C

1 + |x|

≤

C ||u||H t (Rn )

2

´m−t

¶1/2 µZ ³ ¶1/2 ´t 2 t dx 1 + |x| |F u (x)|

because for the given values of t and m the first integral is finite. This follows from a use of polar coordinates. Taking sup over all x ∈ Rn and |α| ≤ m, this proves the lemma. Corollary 39.4.3 Let u ∈ H t (Rn ) where t > m + n2 . Then u is a.e. equal to a function of Cbm (Rn ) still denoted by u. Furthermore, there exists a constant, C independent of u such that ||u||C m (Rn ) ≤ C ||u||H t (Rn ) . b


1152

Proof: This follows from the above lemma. Let {uk } be a sequence of functions of S which converges to u in H t and a.e. Then by the inequality of the above lemma, this sequence is also Cauchy in Cbm (Rn ) and taking the limit, ||u||C m (Rn ) = lim ||uk ||C m (Rn ) ≤ C lim ||uk ||H t (Rn ) = C ||u||H t (Rn ) . k→∞

b

k→∞

b

What about open sets, U ? Corollary 39.4.4 Let t > m + n2 and let U be an open set with u ∈ H t (U ) . Then ¡ ¢ u is a.e. equal to a function of C m U still denoted by u. Furthermore, there exists a constant, C independent of u such that ||u||C m (U ) ≤ C ||u||H t (U ) . Proof: Let u ∈ H t (U ) and let v ∈ H t (Rn ) such that v|U = u. Then ||u||C m (U ) ≤ ||v||C m (Rn ) ≤ C ||v||H t (Rn ) . b

Now taking the inf for all such v yields ||u||C m (U ) ≤ C ||u||H t (U ) .

39.5

The Trace On The Boundary Of A Half Space

It is important to consider the restriction of functions in a Sobolev space onto a smaller dimensional set such as the boundary of an open set. Definition 39.5.1 For u ∈ S, define γu a function defined on Rn−1 by γu (x0 ) ≡ u (x0 , 0) where x0 ∈ Rn−1 is defined by x = (x0 , xn ). The following elementary lemma featuring trig. substitutions is the basis for the proof of some of the arguments which follow. Lemma 39.5.2 Consider the integral, Z ¡ 2 ¢−t a + x2 dx. R

for a > 0 and t > 1/2. Then this integral is of the form Ct a−2t+1 where Ct is some constant which depends on t. Proof: Letting x = a tan θ, Z Z ¢−t ¡ 2 dx = a−2t+1 a + x2 R

π/2

cos2t−2 (θ) dθ

−π/2

and since t > 1/2 the last integral is finite. This yields the desired conclusion and proves the lemma.

39.5. THE TRACE ON THE BOUNDARY OF A HALF SPACE

1153

Lemma 39.5.3 Let u ∈ S. Then there exists a constant, Cn , depending on n but independent of u ∈ S such that Z 0 F γu (x ) = Cn F u (x0 , xn ) dxn . R

Proof: Using the dominated convergence theorem, Z Z 2 F u (x0 , xn ) dxn ≡ lim e−(εxn ) F u (x0 , xn ) dxn ε→0

R

µ

Z ≡

2

e−(εxn )

lim

ε→0

R

µ =

lim

ε→0

1 2π

¶n/2 Z

1 2π

¶n/2 Z

e−i(x ·y +xn yn ) u (y0 , yn ) dy 0 dyn dxn 0

Rn 0

Rn

R

u (y , yn ) e

0

Z

−ix0 ·y0

2

e−(εxn ) e−ixn yn dxn dy 0 dyn . R

¡ ¢2 y2 2 Now − (εxn ) − ixn yn = −ε2 xn + iy2n − ε2 4n and so the above reduces to an expression of the form Z Z Z 0 0 0 0 1 −ε2 yn2 4 lim Kn u (y0 , yn ) e−ix ·y dy 0 dyn = Kn u (y0 , 0) e−ix ·y dy 0 e ε→0 ε n−1 n R R R = Kn F γu (x0 ) and this proves the lemma with Cn ≡ Kn−1 . Earlier H t (Rn ) was defined and then for U an open subset of Rn , H t (U ) was defined to be the space of restrictions of functions of H t (Rn ) to U and a norm was given which made H t (U ) into a Banach space. The next task is to consider Rn−1 ×{0} , a smaller dimensional subspace of Rn and examine the functions defined on this set, denoted by Rn−1 for short which are restrictions of functions in H t (Rn ) . You note this is somewhat different because heuristically, the dimension of the domain of the function is changing. An open set in Rn is considered an n dimensional thing but Rn−1 is only n−1 dimensional. I realize this is vague because the standard definition of dimension requires a vector space and an open set is not a vector space. However, think in terms of fatness. An open set is fat in n directions whereas Rn−1 is only fat in n − 1 directions. Therefore, something interesting is likely to happen. Let S denote the Schwartz class of functions on Rn and S0 the Schwartz class of functions on Rn−1 . Also, y0 ∈ Rn−1 while y ∈ Rn . Let u ∈ S. Then from Lemma 39.5.3 and s > 0, Z ³ ´ 2 s 2 1 + |y0 | |F γu (y0 )| dy 0 Rn−1

Z

= Cn

³

Rn−1

´ 2 s

1 + |y0 |

¯Z ¯2 ¯ ¯ ¯ F u (y0 , yn ) dyn ¯ dy 0 ¯ ¯ R

¯2 Z ³ ´s ¯¯Z ³ ´t/2 ³ ´−t/2 ¯ 2 2 0 2 0 ¯ = Cn 1 + |y | F u (y , yn ) 1 + |y| 1 + |y| dyn ¯¯ dy 0 ¯ n−1 R

R


1154

Then by the Cauchy Schwarz inequality, Z ³ Z ³ ´ Z ³ ´t ´−t 2 s 2 2 2 ≤ Cn 1 + |y0 | |F u (y0 , yn )| 1 + |y| dyn 1 + |y| dyn dy 0 . Rn−1

Consider

R

Z ³ R

R

³

R

by Lemma 39.5.2 and taking a = 1 + |y0 | µ³ Ct

(39.5.17)

Z ³ ´−t ´−t 2 2 1 + |y| dyn = 1 + |y0 | + yn2 dyn

´ 2 1/2

¶−2t+1

1 + |y0 |

2

´1/2

, this equals

³ ´ 2 (−2t+1)/2 = Ct 1 + |y0 | .

Now using this in 39.5.17, Z ³ ´ 2 s 2 1 + |y0 | |F γu (y0 )| dy 0 Rn−1 Z ³ ´s Z ³ ´t 2 2 0 2 ≤ Cn,t 1 + |y | |F u (y0 , yn )| 1 + |y| dyn · Rn−1

=

R

³ ´ 2 (−2t+1)/2 1 + |y0 | dy 0 Z ³ ´s+(−2t+1)/2 Z ³ ´t 2 2 0 2 Cn,t 1 + |y | |F u (y0 , yn )| 1 + |y| dyn dy 0 . Rn−1

R 2

What is the correct choice of t so that the above reduces to ||u||H t (Rn ) ? It is clearly the one for which s + (−2t + 1) /2 = 0 which occurs when t = s + 12 . Then for this choice of t, the following inequality is obtained for any u ∈ S. ||γu||H t−1/2 (Rn−1 ) ≤ Cn,t ||u||H t (Rn ) .

(39.5.18)

This has proved part of the following theorem. Theorem 39.5.4 For each t > 1/2 there exists a unique mapping ³ ¡ ¢´ γ ∈ L H t (Rn ) , H t−1/2 Rn−1 0 which has the property that for u ∈ S, γu (x0 ) = u (x 0) . In to this, ¡ ,t−1/2 ¡ addition ¢ ¢ γ is onto. In fact, there exists a continuous map, ζ ∈ L H Rn−1 , H t (Rn ) such that γ ◦ ζ = id.

Proof: It only remains to verify that γ is onto and that the continuous map, ζ exists. Now define ³ ´t−1/2 2 1 + |y0 | φ (y) ≡ φ (y0 , yn ) ≡ ³ ´t . 2 1 + |y|


1155

Then for u ∈ S0 , let ζu (x) ≡ CF −1 (φF u) (x) = ³ ´t−1/2 2 1 + |y0 | 0 eiy·x ³ C ´t F u (y ) dy 2 Rn 1 + |y| Z

(39.5.19)

Here the inside Fourier transform is taken with respect to Rn−1 because u is only defined on Rn−1 and C will be chosen in such a way that γ ◦ ζ = id. First the existence of C such that γ ◦ ζ = id will be shown. Since u ∈ S0 it follows ³

´t−1/2 2 1 + |y0 | 0 y→ ³ ´t F u (y ) 2 1 + |y| is in S. Hence the inverse Fourier transform of this function is also in S and so for u ∈ S0 , it follows ζu ∈ S. Therefore, to check γ ◦ ζ = id it suffices to plug in xn = 0. From Lemma 39.5.2 this yields γ (ζu) (x0 , 0) Z =

0

eiy ·x

C Rn

Z =

³

C Z

=

CCt ZR

=

CCt

³

´t−1/2 2 1 + |y0 | 0 ³ ´t F u (y ) dy 2 1 + |y|

´t−1/2 0 0 0 2 iy ·x

Z

1 0 ³ ´t dyn dy R 1 + |y|2 ³ ´ ³ ´ −2t+1 2 2 t−1/2 iy0 ·x0 2 1 + |y0 | dy 0 e F u (y0 ) 1 + |y0 |

1 + |y | Rn−1

0

e

0

F u (y )

n−1 0

Rn−1

0

n/2

eiy ·x F u (y0 ) dy 0 = CCt (2π)

F −1 (F u) (x0 )

³ ´−1 n/2 and so the correct value of C is Ct (2π) to obtain γ ◦ ζ = id. It only remains to verify that ζ is continuous. From 39.5.19, and Lemma 39.5.2, 2

= =

||ζu||H t (Rn ) Z ³ ´t 2 2 1 + |x| |F ζu (x)| dx n R Z ³ ´t ¯ ¡ ¢¯ 2 ¯F F −1 (φF u) (x) ¯2 dx 1 + |x| C2 Rn


1156 Z

³ ´t 2 2 1 + |x| |φ (x) F u (x0 )| dx

C2

=

Rn

Z C2

=

Z =

C

ZR =

³

2 n

C2

¯³ ¯2 ´t−1/2 ¯ ¯ 2 0 ¯ ¯ ´t 1 + |x | ³ ¯ 2 0 ¯ F u (x ) 1 + |x| ¯ dx ¯ ³ ´t ¯ ¯ 2 Rn 1 + |x| ¯ ¯

¯2 ´−t ¯¯³ ´t−1/2 ¯ 0 2 0 ¯ ¯ 1 + |x| F u (x )¯ dx ¯ 1 + |x | Z ³ ³ ´ ´−t 2 2t−1 2 2 1 + |x0 | |F u (x0 )| 1 + |x| dxn dx0 2

Rn−1

Z =

C Ct Z

=

³

2

Rn−1

2

C Ct

³

R

´ 2 2t−1

1 + |x0 |

|F u (x0 )|

´ 2 t−1/2

1 + |x0 |

Rn−1

2

³

2

1 + |y0 |

2

´ −2t+1 2

dx0 2

|F u (x0 )| dx0 = C 2 Ct ||u||H t−1/2 (Rn−1 ) .

This proves the theorem because S is dense in Rn . Actually, the assertion that γu (x0 ) = u (x0 , 0) holds for more functions, u than just u ∈ S. I will make no effort to obtain the most general description of such functions but the following is a useful lemma which will be needed when the trace on the boundary of an open set is considered. Lemma 39.5.5 Suppose u is continuous and u ∈ H 1 (Rn ) . Then there exists a set ¡ ¢ of m1 measure zero, N such that if xn ∈ / N, then for every φ ∈ L2 Rn−1 Z (γu, φ)H +

0

xn

(u,n (·, t) , φ)H dt = (u (·, xn ) , φ)H

where here

Z (f, g)H ≡

f gdx0 , Rn−1

¡ ¢ just the inner product in L2 Rn−1 . Furthermore, u (·, 0) = γu a.e. x0 . Proof: Let {uk } be a sequence of functions from S which to u in ¡ converges ¢ H 1 (Rn ) and let {φk } denote a countable dense subset of L2 Rn−1 . Then ¡ ¢ γuk , φj H +

Z

xn 0

¡

uk,n (·, t) , φj

¢ H

¡ ¢ dt = uk (·, xn ) , φj H .

(39.5.20)


1157

Now µZ

∞

0

µZ

∞

= 0

µZ

¶1/2 ¯¡ ¢ ¡ ¢ ¯¯2 ¯ ¯ uk (·, xn ) , φj H − u (·, xn ) , φj H ¯ dxn ¶1/2 ¯¡ ¢ ¯¯2 ¯ ¯ uk (·, xn ) − u (·, xn ) , φj H ¯ dxn

∞

≤

|uk (·, xn ) − 0

=

¯ ¯2 ¯φj ¯

=

¯ ¯2 ¯φj ¯

µZ

H

∞

H

¯ ¯2 ¯φj ¯ dxn H

¶1/2 ¶1/2

2

|uk (·, xn ) − u (·, xn )|H dxn

0

µZ

2 u (·, xn )|H

∞

¶1/2

Z 2

Rn−1

0

|uk (x0 , xn ) − u (x0 , xn )| dx0 dxn

which converges to zero. Therefore, there exists a set of measure zero, Nj and a subsequence, still denoted by k such that if xn ∈ / Nj , then ¡ ¢ ¡ ¢ uk (·, xn ) , φj H → u (·, xn ) , φj H . ¡ ¢ Now by Theorem 39.5.4, γuk → γu in H = L2 Rn−1 . It only remains to consider the term of 39.5.20 which involves an integral. ¯Z x n ¯ Z xn ¯ ¯ ¡ ¢ ¡ ¢ ¯ ¯ u (·, t) , φ dt − u (·, t) , φ dt k,n ,n j H j H ¯ ¯ 0 0 Z xn ¯ ¯ ¢ ¯ ¯¡ ≤ ¯ uk,n (·, t) − u,n (·, t) , φj H ¯ dt Z0 xn ¯ ¯ ≤ |uk,n (·, t) − u,n (·, t)|H ¯φj ¯H dt 0

µZ

xn

≤

|uk,n (·, t) − 0

=

2 u,n (·, t)|H

¯ ¯ ¯φ j ¯ x1/2 n

µZ H

0

xn

Z

¶1/2 µZ dt 0

xn

¶1/2 ¯ ¯2 ¯φj ¯ dt H 2

Rn−1

|uk,n (x0 , t) − u,n (x0 , t)| dx0

¶1/2 dt

and this converges to zero as k → ∞. Therefore, using a diagonal sequence argument, there exists a subsequence, still denoted by k and a set of measure zero, 0 N ≡ ∪∞ / N, you can pass to the limit in 39.5.20 and obtain j=1 Nj such that for x ∈ that for all φj , Z xn ¡ ¢ ¡ ¢ ¡ ¢ γu, φj H + u,n (·, t) , φj H dt = u (·, xn ) , φj H . 0

¢ ¡ By density of φj , this equality holds for all φ ∈ L2 Rn−1 . In particular, the ¡ n−1 ¢ equality holds for every φ ∈ Cc R . Since u is uniformly continuous on the compact set, spt (φ) × [0, 1] , there exists a sequence, (xn )k → 0 such that the above ©

ª


1158

equality holds for xn replaced with (xn )k and φ in place of φj . Now taking k → ∞, this uniform continuity implies (γu, φ)H = (u (·, 0) , φ)H ¡ ¢ ¡ n−1 ¢ This implies since Cc R is dense in L2 Rn−1 that γu = u (·, 0) a.e. and this proves the lemma. Lemma 39.5.6 Suppose U is an open subset of Rn of the form U ≡ {u ∈ Rn : u0 ∈ U 0 and 0 < un < φ (u0 )} where U 0 is an open subset of Rn−1 and φ (u0 ) is a positive function such that φ (u0 ) ≤ ∞ and inf {φ (u0 ) : u0 ∈ U 0 } = δ > 0 Suppose v ∈ H t (Rn ) such that v = 0 a.e. on U. Then γv = 0 mn−1 a.e. point of U 0 . Also, if v ∈ H t (Rn ) and φ ∈ Cc∞ (Rn ) , then γvγφ = γ (φv) . Proof: First consider the second claim. Let v ∈ H t (Rn ) and let vk → v in H (Rn ) where vk ∈ S. Then from Lemma 39.3.4 and Theorem 39.5.4 t

||γ (φv) − γφγv||H t−1/2 (Rn−1 ) = lim ||γ (φvk ) − γφγvk ||H t−1/2 (Rn−1 ) = 0 k→∞

because each term in the sequence equals zero due to the observation that for vk ∈ S and φ ∈ Cc∞ (U ) , γ (φvk ) = γvk γφ. Now suppose v = 0 a.e. on U . Define for 0 < r < δ, vr (x) ≡ v (x0 , xn + r) . Claim: If u ∈ H t (Rn ) , then lim ||vr − v||H t (Rn ) = 0.

r→0

Proof of claim: First of all, let v ∈ S. Then v ∈ H m (Rn ) for all m and so by Lemma 39.2.5, θ

1−θ

||vr − v||H t (Rn ) ≤ ||vr − v||H m (Rn ) ||vr − v||H m+1 (Rn ) where t ∈ [m, m + 1] . It follows from continuity of translation in Lp (Rn ) that θ

1−θ

lim ||vr − v||H m (Rn ) ||vr − v||H m+1 (Rn ) = 0

r→0

and so the claim is proved if v ∈ S. Now suppose u ∈ H t (Rn ) is arbitrary. By density of S in H t (Rn ) , there exists v ∈ S such that ||u − v||H t (Rn ) < ε/3. Therefore, ||ur − u||H t (Rn )

≤

||ur − vr ||H t (Rn ) + ||vr − v||H t (Rn ) + ||v − u||H t (Rn )

=

2ε/3 + ||vr − v||H t (Rn ) .


1159

Now using what was just shown, it follows that for r small enough, ||ur − u||H t (Rn ) < ε and this proves the claim. Now suppose v ∈ H t (Rn ) . By the claim, ||vr − v||H t (Rn ) → 0 and so by continuity of γ, ¡ ¢ γvr → γv in H t−1/2 Rn−1 .

(39.5.21)

Note vr = 0 a.e. on Ur ≡ {u ∈ Rn : u0 ∈ U 0 and − r < un < φ (u0 ) − r} Let φ ∈ Cc∞ (Ur ) and consider φvr . Then it follows φvr = 0 a.e. on Rn .¢ Let ¡ n−1 t−1/2 w ≡ 0. Then w ∈ S and so γw = 0 = γ (φvr ) = γφγvr in H R . It follows that for mn−1 a.e. x0 ∈ [φ 6= 0] ∩ Rn−1 , γvr (x0 ) = 0. Now let U 0 = ∪∞ k=1 Kk where the Kk are compact sets such that Kk ⊆ Kk+1 and let φk ∈ Cc∞ (U ) such that φk has values in [0, 1] and φk (x0 ) = 1 if x0 ∈ Kk . Then from what was just shown, γvr = 0 for a.e. point of Kk . Therefore, γvr = 0 for mn−1 a.e. point in U 0 . Therefore, since each γvr = 0, it follows from 39.5.21 that γv = 0 also. This proves the lemma. Theorem 39.5.7 Let t > 1/2 and let U be of the form {u ∈ Rn : u0 ∈ U 0 and 0 < un < φ (u0 )} where U 0 is an open subset of Rn−1 and φ (u0 ) is a positive function such that φ (u0 ) ≤ ∞ and inf {φ (u0 ) : u0 ∈ U 0 } = δ > 0. Then there exists a unique ³ ´ γ ∈ L H t (U ) , H t−1/2 (U 0 ) which has the property that if u = v|U where v is continuous and also a function of H 1 (Rn ) , then γu (x0 ) = u (x0 , 0) for a.e. x0 ∈ U 0 . Proof: Let u ∈ H t (U ) . Then u = v|U for some v ∈ H t (Rn ) . Define γu ≡ γv|U 0 Is this well defined? The answer is yes because if vi |U = u a.e., then γ (v1 − v2 ) = 0 a.e. on U 0 which implies γv1 = γv2 a.e. and so the two different versions of γu differ only on a set of measure zero. If u = v|U where v is continuous and also a function of H 1 (Rn ) , then for a.e. 0 x ∈ Rn−1 , it follows from Lemma 39.5.5 on Page 1156 that γv (x0 ) = v (x0 , 0) . Hence, it follows that for a.e. x0 ∈ U 0 , γu (x0 ) ≡ u (x0 , 0).


1160

In particular, γ is determined by γu (x0 ) = u (x0 , 0) on S|U and the density of S|U and continuity of γ shows γ is unique. It only remains to show γ is continuous. Let u ∈ H t (U ) . Thus there exists v ∈ H t (Rn ) such that u = v|U . Then ||γu||H t−1/2 (U 0 ) ≤ ||γv||H t−1/2 (Rn−1 ) ≤ C ||v||H t (Rn ) for C independent of v. Then taking the inf for all such v ∈ H t (Rn ) which are equal to u a.e. on U, it follows ||γu||H t−1/2 (U 0 ) ≤ C ||v||H t (Rn ) and this proves γ is continuous.

39.6

Sobolev Spaces On Manifolds

39.6.1

General Theory

The type of manifold, Γ for which Sobolev spaces will be defined on is: Definition 39.6.1

1. Γ is a closed subset of Rp where p ≥ n.

2. Γ = ∪∞ i=1 Γi where Γi = Γ ∩ Wi for Wi a bounded open set. ∞

3. {Wi }i=1 is locally finite. 4. There are open bounded sets, Ui and functions hi : Ui → Γi which are one to one, onto, and in C m,1 (Ui ) . There exists a constant, C, such that C ≥ Lip hr for all r. 5. There exist functions, gi : Wi → Ui such that gi is C m,1 (Wi ) , and gi ◦hi = id on Ui while hi ◦ gi = id on Γi . This will be referred to as a C m,1 manifold. Lemma 39.6.2 Let gi , hi , Ui , Wi , and Γi be as defined above. Then −1 gi ◦ hk : Uk ∩ h−1 k (Γi ) → Ui ∩ hi (Γk )

is C m,1 . Furthermore, the inverse of this map is gk ◦ hi . Proof: First it is well to show it does indeed map the given open sets. Let x ∈ Uk ∩h−1 k (Γi ) . Then hk (x) ∈ Γk ∩Γi and so gi (hk (x)) ∈ Ui because hk (x) ∈ Γi . Now since hk (x) ∈ Γk , gi (hk (x)) ∈ hi−1 (Γk ) also and this proves the mappings do what they should in terms of mapping the two open sets. That gi ◦hk is C m,1 follows immediately from the chain rule and the assumptions that the functions gi and hk are C m,1 . The claim about the inverse follows immediately from the definitions of the functions. ∞ Let {ψ i }i=1 be a partition of unity subordinate to the open cover {Wi } satisfying ∞ ψ i ∈ Cc (Wi ) . Then the following definition provides a norm for H m+s (Γ) .

39.6. SOBOLEV SPACES ON MANIFOLDS

1161

Definition 39.6.3 Let s ∈ (0, 1) and m is a nonnegative integer. Also let µ denote the surface measure for Γ defined in the last section. A µ measurable function, u ∞ is in H m+s (Γ) if whenever {Wi , ψ i , Γi , Ui , hi , gi }i=1 is described above, h∗i (uψ i ) ∈ H m+s (Ui ) and Ã ||u||H m+s (Γ) ≡

∞ X

!1/2 ||h∗i

2 (uψ i )||H m+s (Ui )

< ∞.

i=1

Are there functions which are in H m+s (Γ)? The answer is yes. Just take the restriction to Γ of any function, u ∈ Cc∞ (Rm ) . Then each h∗i (uψ i ) ∈ H m+s (Ui ) and the sum is finite because spt u has nonempty intersection with only finitely many Wi . ª∞ © It is not at all obvious this norm is well defined. What if Wi0 , ψ 0i , Γ0i , Ui , h0i , gi0 i=1 is as described above? Would the two norms be equivalent? If they aren’t, then this is not a good way to define H m+s (Γ) because it would depend on the choice of partition of unity and functions, hi and choice of the open sets, Ui . To begin with ∞ pick a particular choice for {Wi , ψ i , Γi , Ui , hi , gi }i=1 . Lemma 39.6.4 H m+s (Γ) as just described, is a Banach space. ∞

∞

Proof: Let {uj }j=1 be a Cauchy sequence in H m+s (Γ) . Then {h∗i (uj ψ i )}j=1 is a Cauchy sequence in H m+s (Ui ) for each i. Therefore, for each i, there exists wi ∈ H m+s (Ui ) such that lim h∗i (uj ψ i ) = wi in H m+s (Ui ) .

j→∞

(39.6.22)

It is required to show there exists u ∈ H m+s (Γ) such that wi = h∗i (uψ i ) for each i. Now from Corollary 37.2.5 it follows easily by approximating with simple functions that for ever nonnegative µ measurable function, f, Z f dµ = Γ

∞ Z X r=1

gr Γr

ψ r f (hr (u)) Jr (u) du.

Therefore, Z 2

|uj − uk | dµ = Γ

≤

∞ Z X

2

r=1 gr Γr ∞ Z X

ψ r |uj − uk | (hr (u)) Jr (u) du

C

r=1 ∞ X

gr Γr

2

ψ r |uj − uk | (hr (u)) du 2

||h∗r (ψ r |uj − uk |)||0,2,Ur

=

C

≤

C ||uj − uk ||H m+s (Γ)

r=1


1162 and it follows there exists u ∈ L2 (Γ) such that

||uj − u||0,2,Γ → 0. and a subsequence, still denoted by uj such that uj (x) → u (x) for µ a.e. x ∈ Γ. It is required to show that u ∈ H m+s (Γ) such that wi = h∗i (uψ i ) for each i. First of all, u is measurable because it is the limit of measurable functions. The pointwise convergence just established and the fact that sets of measure zero on Γi correspond to sets of measure zero on Ui which was discussed in the claim found in the proof of Theorem 37.2.4 on Page 1091 shows that h∗i (uj ψ i ) (x) → h∗i (uψ i ) (x) a.e. x. Therefore,

h∗i (uψ i ) = wi

and this shows that h∗i (uψ i ) ∈ H m+s (Ui ) . It remains to verify that u ∈ H m+s (Γ) . This follows from Fatou’s lemma. From 39.6.22, 2

2

||h∗i (uj ψ i )||H m+s (Ui ) → ||h∗i (uψ i )||H m+s (Ui ) and so ∞ X

2

||h∗i (uψ i )||H m+s (Ui )

i=1

≤ lim inf

j→∞

∞ X

2

||h∗i (uj ψ i )||H m+s (Ui )

i=1 2

= lim inf ||uj ||H m+s (Γ) < ∞. j→∞

This proves the lemma. In fact any two such norms are equivalent. This follows from the open mapping theorem. Suppose ||·||1 and ||·||2 are two such norms and consider the norm ||·||3 ≡ max (||·||1 , ||·||2 ) . Then (H m+s (Γ) , ||·||3 ) is also a Banach space and the identity map from this Banach space to (H m+s (Γ) , ||·||i ) for i = 1, 2 is continuous. Therefore, by the open mapping theorem, there exist constants, C, C 0 such that for all u ∈ H m+s (Γ) , ||u||1 ≤ ||u||3 ≤ C ||u||2 ≤ C ||u||3 ≤ CC 0 ||u||1 Therefore,

||u||1 ≤ C ||u||2 , ||u||2 ≤ C 0 ||u||1 .

This proves the following theorem. Theorem 39.6.5 Let Γ be described above. Defining H t (Γ) as in Definition 39.6.3, any two norms like those given in this definition are equivalent. Suppose (Γ, Wi , Ui , Γi , hi , gi ) are as defined above where hi , gi are C m,1 functions. Take W , an open set in Rp and define Γ0 ≡ W ∩ Γ. Then letting Wi0 ≡ W ∩ Wi , Γ0i ≡ Wi0 ∩ Γ,


1163

and 0 Ui0 ≡ gi (Γ0i ) = h−1 i (Wi ∩ Γ) ,

it follows that Ui0 is an open set because hi is continuous and (Γ0 , Wi0 , Ui0 , Γ0i , h0i , gi0 ) is also a C m,1 manifold if you define h0i to be the restriction of hi to Ui0 and gi0 to be the restriction of gi to Wi0 . As a case of this, consider a C m,1 manifold, Γ where (Γ, Wi , Ui , Γi , hi , gi ) are as described in Definition 39.6.1 and the submanifold consisting of Γi . The next lemma shows there is a simple way to define a norm on H t (Γi ) which does not depend on dragging in a partition of unity. Lemma 39.6.6 Suppose Γ is a C m,1 manifold and (Γ, Wi , Ui , Γi , hi , gi ) are as described in Definition 39.6.1. Then for t ∈ [m, m + s), it follows that if u ∈ H t (Γ) , then u ∈ H t (Γk ) and the restriction map is continuous. Also an equivalent norm for H t (Γk ) is given by |||u|||t ≡ ||h∗k u||H t (Uk ) . Proof: Let u ∈ H t (Γ) and let (Γk , Wi0 , Ui0 , Γ0i , h0i , gi0 ) be the sets and functions which define what is meant by Γk being a C m,1 manifold as described in Definition © ª 39.6.1. Also let (Γ, Wi , Ui , Γi , hi , gi ) be pertain to Γ in the same way and let φj be a C ∞ partition of unity for the {Wj }. Since the {Wi0 } are locally finite, only finitely many can intersect Γk , say {W10 , · · · , Ws0 } . Also © only ª finitely many of the Wi can intersect Γk , say {W1 , · · · , Wq } . Then letting ψ 0i be a C ∞ partition of unity subordinate to the {Wi0 } . ∞ X ¯¯ 0∗ ¡ 0 ¢¯¯ ¯¯hi uψ i ¯¯ i=1

=

≤ = =

H t (Ui0 )

¯¯  ¯¯ ¯¯ q s ¯¯ X ¯¯ 0∗ X ¯¯ 0 ¯¯ ¯¯h i  φj uψ i ¯¯ ¯¯ ¯¯ i=1 ¯¯ j=1 q s X X i=1 j=1 q X s X j=1 i=1 q X s X

¯¯ 0∗ ¯¯ ¯¯hi φj uψ 0i ¯¯ ¯¯ 0∗ ¯¯ ¯¯hi φj uψ 0i ¯¯

H t (Ui0 )

H t (Ui0 )

H t (Ui0 )

¯¯ ¯¯ ¯¯(gj ◦ h0i )∗ h∗j φj uψ 0i ¯¯ t H

j=1 i=1

(Ui0 )

.

By Lemma 39.3.3 on page 1149, there exists a single constant, C such that the above is dominated by q X s X ¯¯ ∗ ¯¯ ¯¯hj φj uψ 0i ¯¯ t C . H (U ) j=1 i=1

j


1164

Now by Corollary 39.3.5 on Page 1150, this is no larger than

C

q X s X

Cψ0i

¯¯ ∗ ¯¯ ¯¯hj φj u¯¯

j=1 i=1

H t (Uj )

q X s X ¯¯ ∗ ¯¯ ¯¯hj φj u¯¯ t ≤ C H (U

≤ C

j=1 i=1 q X ¯¯ ∗ ¯¯ ¯¯hj φj u¯¯ t H (Uj ) j=1

j)

< ∞.

This shows that u restricted to Γk is in H t (Γk ). It also shows that the restriction map of H t (Γ) to H t (Γk ) is continuous. Now consider the norm |||·|||t . For u ∈ H t (Γk ) , let (Γk , Wi0 , Ui0 , Γ0i , h0i , gi0 ) be sets and functions which define an atlas for Γk . Since the {Wi0 } are locally finite, only finitely many can have nonempty intersection with Γk , say {W1 , · · · , Ws } . Thus i ≤ s for some finite s. The problem is to compare |||·|||t with ||·||H t (Γk ) . As above, © ª © ª let ψ 0i denote a C ∞ partition of unity subordinate to the Wj0 . Then

|||u|||t

¯¯ ¯¯ ¯¯ ¯¯ s ¯¯ ∗ X ¯¯ = ¯¯¯¯hk ψ 0j u¯¯¯¯ ¯¯ j=1 ¯¯

≡

||h∗k u||H t (Uk )

≤

s X ¯¯ ∗ ¡ 0 ¢¯¯ ¯¯hk ψ j u ¯¯

=

j=1 s ¯¯ X

H t (Uk )

¢∗ ¡ 0 ¢¯¯¯¯ ¯¯¡ 0 ¯¯ gj ◦ hk h0∗ j ψ j u ¯¯

j=1

≤

C

s X ¯¯ 0∗ ¡ 0 ¢¯¯ ¯¯hj ψ j u ¯¯ j=1

≤

H t (Uj0 )

 s X ¯¯ 0∗ ¡ 0 ¢¯¯2 ¯¯hj ψ j u ¯¯ C j=1

H t (Uk )

H t (Uk )

.

H t (Uj0 )

1/2 

= ||u||H t (Γk ) .


1165

where Lemma 39.3.3 on page 1149 was used in the last step. Now also, from Lemma 39.3.3 on page 1149 1/2  s X ¯¯ 0∗ ¡ 0 ¢¯¯2 ¯¯hj ψ j u ¯¯ t 0  ||u||H t (Γk ) =  H (Uj ) j=1

=

H t (Uj0 )

j=1



≤

s X ¯¯ ∗ ¡ 0 ¢¯¯2 ¯¯h k ψ j u ¯¯ C

H t (Uk )

j=1

 ≤

1/2

 s ¯¯ X ¢∗ ¡ ¢¯¯¯¯2 ¯¯¡  ¯¯ gk ◦ h0j h∗k ψ 0j u ¯¯

C

s X



1/2 

1/2 2

||h∗k u||H t (Uk ) 

= Cs ||h∗k u||H t (Uk ) = |||u|||t .

j=1


39.6.2

The Trace On The Boundary

Definition 39.6.7 A bounded open subset, Ω, of Rn has a C m,1 boundary if it satisfies the following conditions. For each p ∈ Γ ≡ Ω \ Ω, there exists an open set, W , containing p, an open interval (0, b), a bounded open box U 0 ⊆ Rn−1 , and an affine orthogonal transformation, RW consisting of a distance preserving linear transformation followed by a translation such that RW W = U 0 × (0, b),

(39.6.23)

RW (W ∩ Ω) = {u ∈ Rn : u0 ∈ U 0 , 0 < un < φW (u0 )} ≡ UW (39.6.24) ¡ ¢ where φW ∈ C m,1 U 0 meaning¡ φW ¢is the restriction to U 0 of a function, still denoted by φW which is in C m,1 Rn−1 and inf {φW (u0 ) : u0 ∈ U 0 } > 0 The following picture depicts the situation.

b

¶ ZZ ¶ W Z

Z Z ¶ ¶

¶

¶

¶ Z

Ω Z

T

Z

W

¶ Z Z¶

¶

R

-

RW (Ω 0

T

W)

u0 ∈ U 0


1166

For the situation described in the above definition, let hW : U 0 → Γ ∩ W be defined by 0

−1 −1 hW (u0 ) ≡ RW (u0 , φW (u0 )) , gW (x) ≡ (RW x) , HW (u) ≡ RW (u0 , φW (u0 ) − un ) .

where x0 ≡ (x1 , · · · , xn−1 ) for x = (x1 , · · · , xn ). Thus gW ◦ hW = id on U 0 and hW ◦ gW = id on Γ ∩ W. Also note that HW is defined on all of Rn is C m,1 , and has an inverse with the same properties. To see this, let GW (u) = (u0 , φW (u0 ) − un ) . −1 −1 −1 0 0 Then HW = RW ◦ GW and G−1 W = (u , φW (u ) − un ) and so HW = GW ◦ RW . Note also that as indicated in the picture, RW (W ∩ Ω) = {u ∈ Rn : u0 ∈ U 0 and 0 < un < φW (u0 )} . Since Γ = ∂Ω is compact, there exist finitely many of these open sets, W, denoted q by {Wi }i=1 such that Γ ⊆ ∪qi=1 Wi . Let the corresponding sets, U 0 be denoted by Ui0 q and let the functions, φ be denoted by φi . Also let hi = hWi etc. Now let {ψ i }i=1 q be a C ∞ partition of unity subordinate to the {Wi }i=1 . If u ∈ H t (Ω) , then by Corollary 39.3.5 on Page 1150 it follows that uψ i ∈ H t (Wi ∩ Ω) . Now Hi : Ui ≡ {u ∈ Rn : u0 ∈ Ui0 , 0 < un < φi (u0 )} → Wi ∩ Ω and Hi and its inverse are defined on Rn and are in C m,1 (Rn ) . Therefore, by Lemma 39.3.3 on Page 1149, ¡ ¢ H∗i ∈ L H t (Wi ∩ Ω) , H t (Ui ) . Provide t = m + s where s > 0. Now it is possible to define the trace on Γ ≡ ∂Ω. For u ∈ H t (Ω) , γu ≡

q X

gi∗ (γH∗i (uψ i )) .

(39.6.25)

i=1

I must show it satisfies what it should. Recall the definition of what it means for a function to be in H t−1/2 (Γ) where t = m + s. Definition 39.6.8 Let s ∈ (0, 1) and m is a nonnegative integer. Also let µ denote the surface measure for Γ. A µ measurable function, u is in H m+s (Γ) if whenever ∞ {Wi , ψ i , Γi , Ui , hi , gi }i=1 is described above, h∗i (uψ i ) ∈ H m+s (Ui ) and Ã ||u||H m+s (Γ) ≡

∞ X

!1/2 ||h∗i

2 (uψ i )||H m+s (Ui )

< ∞.

i=1

Recall that all these norms which are obtained from various partitions of unity and functions, hi and gi , are equivalent. Here there are only finitely many Wi so the sum is a finite sum. The theorem is the following.


1167

Theorem 39.6.9 Let Ω be a bounded open set having C m,1 boundary as discussed above in Definition 39.6.7. Then for t ≤ m + 1, there exists a unique ³ ´ γ ∈ L H t (Ω) , H t−1/2 (Γ) which has the property that for µ the measure on the boundary, γu (x) = u (x) for µ a.e. x ∈ Γwhenever u ∈ S|Ω . (39.6.26) ¡ t ¢ Proof: First consider the claim that γ ∈ L H (Ω) , H t−1/2 (Γ) . This involves first showing that for u ∈ H t (Ω) , γu ∈ H t−1/2 (Γ) . To do this, use the above definition. q X

¡ ¢ h∗j ψ j (γu) =

¡ ¢ h∗j ψ j gi∗ (γH∗i (uψ i ))

i=1 q X ¡

=

h∗j ψ j

¢ ¢¡ ∗ ∗ hj (gi (γH∗i (uψ i )))

i=1 q X ¡

=

¢ ∗ h∗j ψ j (gi ◦ hj ) (γH∗i (uψ i ))

(39.6.27)

i=1

First note that γH∗i (uψ i ) ∈ H t−1/2 (Ui0 )

¡ ¢ Now gi ◦ hj and its inverse, gj ◦ hi are both functions in C m,1 Rn−1 and gi ◦ hj : Uj0 → Ui0 . Therefore, by Lemma 39.3.3 on Page 1149, ¡ ¢ ∗ (gi ◦ hj ) (γH∗i (uψ i )) ∈ H t−1/2 Uj0 and

¯¯ ¯¯ ¯¯(gi ◦ hj )∗ (γH∗i (uψ i ))¯¯ h∗j ψ j

Also ∈C on Page 1150 and

m,1

H t−1/2 (Uj0 )

≤ Cij ||γH∗i (uψ i )||H t−1/2 (U 0 ) . i

¡ 0¢ Uj and has compact support in Uj0 and so by Corollary 39.3.5 ¡ ∗ ¢ ¡ ¢ ∗ hj ψ j (gi ◦ hj ) (γH∗i (uψ i )) ∈ H t−1/2 Uj0 ¯¯ ¯¯¡ ∗ ¢ ¯¯ hj ψ j (gi ◦ hj )∗ (γH∗i (uψ i ))¯¯

H t−1/2 (Uj0 )

¯¯ ¯¯ ∗ ≤ Cij ¯¯(gi ◦ hj ) (γH∗i (uψ i ))¯¯H t−1/2 (U 0 )

(39.6.28)

≤ Cij ||γH∗i (uψ i )||H t−1/2 (U 0 ) .

(39.6.29)

j

i


1168

¡ ¡ ¢ ¢ This shows γu ∈ H t−1/2 (Γ) because each h∗j ψ j (γu) ∈ H t−1/2 Uj0 . Also from 39.6.29 and 39.6.27 2

||γu||H t−1/2 (Γ) ≤

q X ¯¯ ∗ ¡ ¢¯¯ ¯¯hj ψ j (γu) ¯¯2

H t−1/2 (Uj0 )

j=1

=

q X ¯¯ ∗ ¡ ¢¯¯ ¯¯hj ψ j (γu) ¯¯2

H t−1/2 (Uj0 )

j=1

=

¯¯2 ¯¯ q q ¯¯X ¯¯ X ¡ ∗ ¢ ¯¯ ¯¯ ∗ hj ψ j (gi ◦ hj ) (γH∗i (uψ i ))¯¯ ¯¯ ¯¯ ¯¯

≤

Cq

H t−1/2 (Uj0 )

i=1

j=1

q q X X ¯¯¡ ∗ ¢ ¯¯ ¯¯ hj ψ j (gi ◦ hj )∗ (γH∗i (uψ i ))¯¯2 t−1/2 H j=1 i=1

≤ Cq ≤ Cq

q q X X j=1 i=1 q X i=1

≤

Cq

(Uj0 )

2

Cij ||(γH∗i (uψ i ))||H t−1/2 (U 0 ) i

2

||(γH∗i (uψ i ))||H t−1/2 (U 0 )

q X

i

2

||H∗i (uψ i )||H t (Ri (Wi ∩Ω))

i=1

≤

Cq

q X

2

2

||uψ i ||H t (Wi ∩Ω) ≤ Cq ||u||H t (Ω) .

i=1

Does γ satisfy 39.6.26? Let x ∈ Γ and u ∈ S|Ω . Let Ix ≡ {i ∈ {1, 2, · · · , q} : x = hi (u0i ) for some u0i ∈ Ui0 } . Then γu (x)

=

X

(γH∗i (uψ i )) (gi (x))

i∈Ix

=

X

(γH∗i (uψ i )) (gi (hi (u0i )))

i∈Ix

=

X

i∈Ix

(γH∗i (uψ i )) (u0i ) .


1169

Now because Hi is Lipschitz continuous and uψ ∈ S, it follows that H∗i (uψ i ) ∈ H 1 (Rn ) and is continuous and so by Theorem 39.5.7 on Page 1159 for a.e. u0i , X H∗i (uψ i ) (u0i , 0) = i∈Ix

=

X

h∗i (uψ i ) (u0i )

i∈Ix

=

X

(uψ i ) (hi (u0i )) = u (x) for µ a.e.x.

i∈Ix

This verifies 39.6.26 and completes the proof of the theorem.

(39.6.30)

1170


Weak Solutions 40.1

The Lax Milgram Theorem

The Lax Milgram theorem is a fundamental result which is useful for obtaining weak solutions to many types of partial differential equations. It is really a general theorem in functional analysis. Definition 40.1.1 Let A ∈ L (V, V 0 ) where V is a Hilbert space. Then A is said to be coercive if 2 A (v) (v) ≥ δ ||v|| for some δ > 0. Theorem 40.1.2 (Lax Milgram) Let A ∈ L (V, V 0 ) be coercive. Then A maps one to one and onto. Proof: The proof that A is onto involves showing A (V ) is both dense and closed. Consider first the claim that A (V ) is closed. Let Axn → y ∗ ∈ V 0 . Then 2

δ ||xn − xm ||V ≤ ||Axn − Axm ||V 0 ||xn − xm ||V . Therefore, {xn } is a Cauchy sequence in V. It follows xn → x ∈ V and since A is continuous, Axn → Ax. This shows A (V ) is closed. Now let R : V → V 0 denote the Riesz map defined by Rx (y) = (y, x) . Recall that the Riesz map is one to one, onto, and preserves norms. Therefore, R−1 (A (V )) ¡ ¢⊥ is a closed subspace of V. If there R−1 (A (V )) 6= V, then R−1 (A (V )) 6= {0} . ¡ −1 ¢⊥ Let x ∈ R (A (V )) and x 6= 0. Then in particular, ¡ ¢ ¡ ¢ 2 0 = x, R−1 Ax = R R−1 (A (x)) (x) = A (x) (x) ≥ δ ||x||V , a contradiction to x 6= 0. Therefore, R−1 (A (V )) = V and so A (V ) = R (V ) = V 0 . Since A (V ) is both closed and dense, A (V ) = V 0 . This shows A is onto. 2 If Ax = Ay, then 0 = A (x − y) (x − y) ≥ δ ||x − y||V , and this shows A is one to one. This proves the theorem. 1171

1172

WEAK SOLUTIONS

Here is a simple example which illustrates the use of the above theorem. In the example the repeated index summation convention is being used. That is, you sum over the repeated indices. Example 40.1.3 Let U be an open subset of Rn and let V be a closed subspace of H 1 (U ) . Let αij ∈ L∞ (U ) for i, j = 1, 2, · · · , n. Now define A : V → V 0 by Z ¡ ij ¢ α (x) u,i (x) v,j (x) + u (x) v (x) dx. A (u) (v) ≡ U

Suppose also that

αij vi vj ≥ δ |v|

2

whenever v ∈ Rn . Then A maps V to V 0 one to one and onto. Here is why. It is obvious that A is in L (V, V 0 ) . It only remains to verify that it is coercive. Z ¡ ij ¢ A (u) (u) ≡ α (x) u,i (x) u,j (x) + u (x) u (x) dx ZU 2 2 ≥ δ |∇u (x)| + |u (x)| dx U 2

≥ δ ||u||H 1 (U ) This proves coercivity and verifies the claim. What has been obtained in the above example? This depends on how you choose V. In Example 40.1.3 suppose U is a bounded open set with C 0,1 boundary and V = H01 (U ) where © ª H01 (U ) ≡ u ∈ H 1 (U ) : γu = 0 . Also suppose f ∈ L2 (U ) . Then you can consider F ∈ V 0 by defining Z F (v) ≡ f (x) v (x) dx. U

According to the Lax Milgram theorem and the verification of its conditions in Example 40.1.3, there exists a unique solution to the problem of finding u ∈ H01 (U ) such that for all v ∈ H01 (U ) , Z Z ¡ ij ¢ α (x) u,i (x) v,j (x) + u (x) v (x) dx = f (x) v (x) dx (40.1.1) U

U

Cc∞

In particular, this holds for all v ∈ (U ) . Thus for all such v, Z ³ ´ ¡ ¢ − αij (x) u,i (x) ,j + u (x) − f (x) v (x) dx = 0. U

Therefore, in terms of weak derivatives, ¡ ¢ − αij u,i ,j + u = f

40.1. THE LAX MILGRAM THEOREM

1173

and since u ∈ H01 (U ) , it must be the case that γu = 0 on ∂U. This is why the solution to 40.1.1 is referred to as a weak solution to the boundary value problem ¡ ¢ − αij (x) u,i (x) ,j + u (x) = f (x) , u = 0 on ∂U. Of course you then begin to ask the important question whether ¡ u really has¢ two derivatives. It is not immediately clear that just because − αij (x) u,i (x) ,j ∈ L2 (U ) it follows that the second derivatives of u exist. Actually this will often be true and is discussed somewhat in the next section. Next suppose you choose V = H 1 (U ) and let g ∈ H 1/2 (∂U ). Define F ∈ V 0 by Z Z g (x) γv (x) dµ. F (v) ≡ f (x) v (x) dx + U

∂U

Everything works the same way and you get the existence of a unique u ∈ H 1 (U ) such that for all v ∈ H 1 (U ) , Z Z Z ¡ ij ¢ α (x) u,i (x) v,j (x) + u (x) v (x) dx = f (x) v (x) dx + g (x) γv (x) dµ U

U

∂U

(40.1.2) is satisfied. It you pretend u has all second order derivatives in L2 (U ) and apply the divergence theorem, you find that you have obtained a weak solution to ¡ ¢ − αij u,i ,j + u = f, αij u,i nj = g on ∂U where nj is the j th component of n, the unit outer normal. Therefore, u is a weak solution to the above boundary value problem. The conclusion is that the Lax Milgram theorem gives a way to obtain existence and uniqueness of weak solutions to various boundary value problems. The following theorem is often very useful in establishing coercivity. To prove this theorem, here is a definition. Definition 40.1.4 Let U be an open set and δ > 0. Then © ¡ ¢ ª Uδ ≡ x ∈ U : dist x,U C > δ . Theorem 40.1.5 Let U be a connected bounded open set having C 0,1 boundary such that for some sequence, η k ↓ 0, U = ∪∞ k=1 Uη k

(40.1.3)

and Uηk is a connected open set. Suppose Γ ⊆ ∂U has positive surface measure and that © ª V ≡ u ∈ H 1 (U ) : γu = 0 a.e. on Γ . Then the norm |||·||| given by µZ 2

|||u||| ≡

|∇u| dx U

is equivalent to the usual norm on V.

¶1/2

1174

WEAK SOLUTIONS

Proof: First it is necessary to verify this is actually a norm. It clearly satisfies all the usual axioms of a norm except for the condition that |||u||| = 0 if and only if u = 0. Suppose then that |||u||| = 0. Let δ 0 = η k for one of those η k mentioned above and define Z uδ (x) ≡ B(0,δ)

u (x − y) φδ (y) dy

where φδ is a mollifier having support in B (0, δ) . Then changing the variables, it follows that for x ∈ Uδ0 Z Z uδ (x) = u (t) φδ (x − t) dt = u (t) φδ (x − t) dt B(x,δ)

U

and so uδ ∈ C ∞ (Uδ0 ) and Z Z ∇uδ (x) = u (t) ∇φδ (x − t) dt = U

B(0,δ)

∇u (x − y) φδ (y) dy = 0.

Therefore, uδ equals a constant on Uδ0 because Uδ0 is a connected open set and uδ is a smooth function defined on this set which has its gradient equal to 0. By Minkowski’s inequality, ÃZ

!1/2 2

|u (x) − uδ (x)| dx Uδ0

ÃZ

Z ≤ B(0,δ)

φδ (y)

!1/2 2

|u (x) − u (x − y)| dx

dy

Uδ0

and this converges to 0 as δ → 0 by continuity of translation in L2 . It follows there exists a sequence of constants, cδ ≡ uδ (x) such that {cδ } converges to u in L2 (Uδ0 ) . Consequently, a subsequence, still denoted by uδ , converges to u a.e. By Eggoroff’s theorem there exists a set, Nk having measure no more than 3−k mn (Uδ0 ) such to u uniformly on NkC . Thus u is constant on NkC . Now P that uδ converges 1 ∞ k mn (Nk ) ≤ 2 mn (Uδ 0 ) and so there exists x0 ∈ Uδ 0 \ ∪k=1 Nk . Therefore, if x∈ / Nk it follows u (x) = u (x0 ) and so, if u (x) 6= u (x0 ) it must be the case that x ∈ ∩∞ k=1 Nk , a set of measure zero. This shows that u equals a constant a.e. on Uδ0 = Uηk . Since k is arbitrary, 40.1.3 shows u is a.e. equal to a constant on U. Therefore, u equals the restriction of a function of S to U and so γu equals this constant in L2 (∂Ω) . Since the surface measure of Γ is positive, the constant must equal zero. Therefore, |||·||| is a norm. It remains to verify that it is equivalent to the usual norm. It is clear that |||u||| ≤ ||u||1,2 . What about the other direction? Suppose it is not true that for some constant, K, ||u||1,2 ≤ K |||u||| . Then for every k ∈ N, there exists uk ∈ V such that ||uk ||1,2 > k |||uk ||| .

40.1. THE LAX MILGRAM THEOREM

1175

Replacing uk with uk / ||uk ||1,2 , it can be assumed that ||uk ||1,2 = 1 for all k. Therefore, using the compactness of the embedding of H 1 (U ) into L2 (U ) , there exists a subsequence, still denoted by uk such that uk uk |||uk ||| uk

→ u weakly in V, → u strongly in L2 (U ) , → 0, → u weakly in (V, |||·|||) .

(40.1.4) (40.1.5) (40.1.6) (40.1.7)

From 40.1.6 and 40.1.7, it follows u = 0. Therefore, |uk |L2 (U ) → 0. This with 40.1.6 contradicts the fact that ||uk ||1,2 = 1 and this proves the equivalence of the two norms. The proof of the above theorem yields the following interesting corollary. Corollary 40.1.6 Let U be a connected open set with the property that for some sequence, η k ↓ 0, U = ∪∞ k=1 Uη k for Uηk a connected open set and suppose u ∈ W 1,p (U ) and ∇u = 0 a.e. Then u equals a constant a.e. Example 40.1.7 Let U be a bounded open connected subset of Rn and let V be a closed subspace of H 1 (U ) defined by © ª V ≡ u ∈ H 1 (U ) : γu = 0 on Γ where the surface measure of Γ is positive. Let αij ∈ L∞ (U ) for i, j = 1, 2, · · · , n and define A : V → V 0 by Z A (u) (v) ≡ αij (x) u,i (x) v,j (x) dx. U

for

αij vi vj ≥ δ |v|

2

whenever v ∈ Rn . Then A maps V to V 0 one to one and onto. This follows from Theorem 40.1.5 using the equivalent norm defined there. Define F ∈ V 0 by Z Z f (x) v (x) dx + g (x) γv (x) dx U 2

for f ∈ L (U ) and g ∈ H

∂U \Γ 1/2

(∂U ) . Then the equation, Au = F in V 0

which is equivalent to u ∈ V and for all v ∈ V, Z Z Z ij α (x) u,i (x) v,j (x) dx = f (x) v (x) dx + U

U

g (x) γv (x) dµ ∂U \Γ

1176

WEAK SOLUTIONS

is a weak solution for the boundary value problem, ¡ ¢ − αij u,i ,j = f in U, αij u,i nj = g on ∂U \ Γ, u = 0 on Γ as you can verify by using the divergence theorem formally.

Korn’s Inequality A fundamental inequality used in elasticity to obtain coercivity and then apply the Lax Milgram theorem or some other theorem is Korn’s inequality. The proof given here of this fundamental result follows [53] and [23].

41.1

A Fundamental Inequality

The proof of Korn’s inequality depends on a fundamental inequality involving negative Sobolev space norms. The theorem to be proved is the following. Theorem 41.1.1 Let f ∈ L2 (Ω) where Ω is a bounded Lipschitz domain. Then there exist constants, C1 and C2 such that Ã ! ¯¯ n ¯¯ X ¯¯ ∂f ¯¯ ¯¯ ¯¯ C1 ||f ||0,2,Ω ≤ ||f ||−1,2,Ω + ≤ C2 ||f ||0,2,Ω , ¯¯ ∂xi ¯¯ −1,2,Ω i=1 where here ||·||0,2,Ω represents the L2 norm and ||·||−1,2,Ω represents the norm in the dual space of H01 (Ω) , denoted by H −1 (Ω) . Similar conventions will apply for any domain in place of Ω. The proof of this theorem will proceed through the use of several lemmas. Lemma 41.1.2 Let U − denote the set, {(x,xn ) ∈ Rn : xn < g (x)} where g : Rn−1 → R is Lipschitz and denote by U + the set {(x,xn ) ∈ Rn : xn > g (x)} . Let f ∈ L2 (U − ) and extend f to all of Rn in the following way. f (x,xn ) ≡ −3f (x,2g (x) − xn ) + 4f (x, 3g (x) − 2xn ) . 1177

1178

KORN’S INEQUALITY

Then there is a constant, Cg , depending on g such that Ã ! ¯¯ ¯¯ n ¯¯ n ¯¯ X X ¯¯ ∂f ¯¯ ¯¯ ∂f ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ||f ||−1,2,Rn + ≤ Cg ||f ||−1,2,U − + . ¯¯ ∂xi ¯¯ ¯¯ ∂xi ¯¯ −1,2,Rn −1,2,U − i=1 i=1 Proof: Let φ ∈ Cc∞ (Rn ) . Then, Z Z ∂φ ∂φ f dx = [−3f (x,2g (x) − xn ) + 4f (x, 3g (x) − 2xn )] dx ∂x ∂x n + n n R U Z f

+ U−

∂φ dx. ∂xn

(41.1.1)

Consider the first integral on the right in 41.1.1. Changing the variables, letting yn = 2g (x) − xn in the first term of the integrand and 3g (x) − 2xn in the next, it equals Z ∂φ −3 (x,2g (x) − yn ) f (x,yn ) dyn dx U − ∂xn µ ¶ Z 3 yn ∂φ x, g (x) − f (x,yn ) dyn dx. +2 2 2 U − ∂xn For (x,yn ) ∈ U − , and defining µ ¶ 3 yn ψ (x,yn ) ≡ φ (x,yn ) + 3φ (x,2g (x) − yn ) − 4φ x, g (x) − , 2 2 it follows ψ = 0 when yn = g (x) and so Z Z ∂φ ∂ψ f dx = f (x,yn ) dxdyn . ∂x ∂y n − n n R U Now from the definition of ψ given above, ||ψ||1,2,U − ≤ Cg ||φ||1,2,U − ≤ Cg ||φ||1,2,Rn and so

¯¯ ¯¯ ¯¯ ∂f ¯¯ ¯¯ ¯¯ ¯¯ ∂xn ¯¯

≡ −1,2,Rn

¾ ∂φ dx : φ ∈ Cc∞ (Rn ) , ||φ||1,2,Rn ≤ 1 ≤ Rn ∂xn ¯ ¾ ½¯Z ¯ ¯ ¡ ¢ ∂ψ sup ¯¯ dxdyn ¯¯ : ψ ∈ H01 U − , ||ψ||1,2,U − ≤ Cg f U − ∂xn ¯¯ ¯¯ ¯¯ ∂f ¯¯ ¯¯ = Cg ¯¯¯¯ ∂xn ¯¯−1,2,U − ½Z

sup

f

(41.1.2)

41.1. A FUNDAMENTAL INEQUALITY

1179

It remains to establish a similar inequality for the case where the derivatives are taken with respect to xi for i < n. Let φ ∈ Cc∞ (Rn ) . Then Z Z ∂φ ∂φ dx = f dx f ∂x ∂x − n i i U R Z ∂φ [−3f (x,g (x) − xn ) + 4f (x, 3g (x) − 2xn )] dx. U + ∂xi Changing the variables as before, this last integral equals Z −3 Di φ (x,2g (x) − yn ) f (x,yn ) dyn dx U−

µ ¶ 3 yn Di φ x, g (x) − f (x,yn ) dyn dx. 2 2 U−

Z +2 Now let

µ ¶ 3 yn ψ 1 (x,yn ) ≡ φ (x,2g (x) − yn ) , ψ 2 (x,yn ) ≡ φ x, g (x) − . 2 2 Then

∂ψ 1 = Di φ (x,2g (x) − yn ) + Dn φ (x,2g (x) − yn ) 2Di g (x) , ∂xi µ ¶ µ ¶ ∂ψ 2 3 yn 3 yn 3 = Di φ x, g (x) − + Dn φ x, g (x) − Di g (x) . ∂xi 2 2 2 2 2

Also

∂ψ 1 (x,yn ) = −Dn φ (x,2g (x) − yn ) , ∂yn µ ¶ µ ¶ −1 3 yn ∂ψ 2 (x,yn ) = Dn φ x, g (x) − . ∂yn 2 2 2

Therefore, ∂ψ 1 ∂ψ (x,yn ) = Di φ (x,2g (x) − yn ) − 2 1 (x,yn ) Di g (x) , ∂xi ∂yn µ ¶ 3 yn ∂ψ ∂ψ 2 (x,yn ) = Di φ x, g (x) − − 3 2 (x,yn ) Di g (x) . ∂xi 2 2 ∂yn Using this in 41.1.3, the integrals in this expression equal ¸ Z · ∂ψ 1 ∂ψ 1 −3 (x,yn ) + 2 (x,yn ) Di g (x) f (x,yn ) dyn dx+ ∂yn U − ∂xi ¸ Z · ∂ψ 2 ∂ψ 2 2 (x,yn ) + 3 (x,yn ) Di g (x) f (x,yn ) dyn dx ∂yn U − ∂xi

(41.1.3)

1180

KORN’S INEQUALITY

· ¸ ∂ψ 1 (x,y) ∂ψ 2 (x,yn ) = −3 +2 f (x,yn ) dyn dx. ∂xi ∂xi U− Z

Therefore,

Z Rn

∂φ f dx = ∂xi

·

Z U−

¸ ∂φ ∂ψ 1 ∂ψ 2 −3 +2 f dxdyn ∂xi ∂xi ∂xi

and also φ (x,g (x)) − 3ψ 1 (x,g (x)) + 2ψ 2 (x,g (x)) = φ (x,g (x)) − 3φ (x,g (x)) + 2φ (x,g (x)) = 0 and so φ − 3ψ 1 + 2ψ 2 ∈ H01 (U − ) . It also follows from the definition of the functions, ψ i and the assumption that g is Lipschitz, that ||ψ i ||1,2,U − ≤ Cg ||φ||1,2,U − ≤ Cg ||φ||1,2,Rn . Therefore,

(41.1.4)

¯¯ ¯¯ ¯ ½¯Z ¾ ¯¯ ∂f ¯¯ ¯ ∂φ ¯¯ ¯¯ ¯ ¯¯ ≡ sup ¯ dx¯ : ||φ||1,2,Rn ≤ 1 f ¯¯ ∂xi ¯¯ Rn ∂xi −1,2,Rn ½¯Z · ¸ ¯ ¾ ¯ ¯ ∂φ ∂ψ 1 ∂ψ 2 ¯ = sup ¯ f −3 +2 dx¯¯ : ||φ||1,2,Rn ≤ 1 ∂xi ∂xi ∂xi U− ¯¯ ¯¯ ¯¯ ∂f ¯¯ ¯¯ ≤ Cg ¯¯¯¯ ∂xi ¯¯−1,2,U −

where Cg is a constant which depends on g. This inequality along with 41.1.2 yields Ã n ¯¯ ! ¯¯ ¯¯ n ¯¯ X X ¯¯ ∂f ¯¯ ¯¯ ∂f ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ≤ Cg . ¯¯ ∂xi ¯¯ ¯¯ ∂xi ¯¯ −1,2,Rn −1,2,U − i=1 i=1 The inequality, ||f ||−1,2,Rn ≤ Cg ||f ||−1,2,U − follows from 41.1.4 and the equation, Z Z Z f φdx = f φdx − 3 Rn

U−

U−

f (x,yn ) ψ 1 (x,yn ) dxdyn

Z +2 U−

f (x,yn ) ψ 2 (x,yn ) dxdyn

which results in the same way as before by changing variables using the definition of f off U − . This proves the lemma. The next lemma is a simple application of Fourier transforms. Lemma 41.1.3 If f ∈ L2 (Rn ) , then the following formula holds. ¯¯ n ¯¯ X ¯¯ ∂f ¯¯ ¯ ¯ ¯¯ Cn ||f ||0,2,Rn = + ||f ||−1,2,Rn ¯¯ ∂xi ¯¯ −1,2,Rn i=1

41.1. A FUNDAMENTAL INEQUALITY

1181

Proof: For φ ∈ Cc∞ (Rn ) µZ ||φ||1,2,Rn ≡

³ 1 + |t|

2

´

¶1/2 |F φ| dt 2

Rn

is an equivalent norm to the usual Sobolev space norm for H01 (Rn ) and is used in the³ following argument which depends on Plancherel’s theorem and the fact that ´ ∂φ F ∂xi = ti F (φ) . ¯¯ ¯¯ ¯ ½¯Z ¾ ¯ ¯¯ ∂f ¯¯ ¯ ∂φ ¯ ¯¯ ¯¯ ¯ : ||φ|| ≤ 1 ≡ sup f dx 1,2 ¯ n ∂xi ¯¯ ∂xi ¯¯ ¯ R −1,2,Rn ¯ ½¯Z ¾ ¯ ¯ ¯ ¯ = Cn sup ¯ ti (F φ) (F f )dt¯ : ||φ||1,2 ≤ 1 Rn

¯ ¯  ³ ´1/2 ¯ ¯   ¯Z ti (F φ) 1 + |t|2 ¯  ¯ ¯ = Cn sup ¯ (F f )dt : ||φ|| ≤ 1 ¯ ³ ´ 1,2 1/2  ¯  2 ¯¯ Rn  ¯ 1 + |t| 

Z

= Cn  Also,

1/2 2 |F f | t2i ³ ´ dt 2 1 + |t|

(41.1.5)

¯ ½¯Z ¾ ¯ ¯ ||f ||−1,2 ≡ sup ¯¯ φf dx¯¯ : ||φ||1,2 ≤ 1 Rn ¯ ½¯Z ¾ ¯ ¡ ¢ ¯ ¯ ¯ = Cn sup ¯ (F φ) F f dx¯ : ||φ||1,2 ≤ 1 Rn

¯ ¯  ³ ´1/2 ¯ ¯   ¯Z F φ 1 + |t|2 ¯  ¯ ¯ (F f )dt = Cn sup ¯ : ||φ|| ≤ 1 ¯ ³ ´ 1,2 1/2  ¯  2 ¯¯ Rn  ¯ 1 + |t|  = Cn 

Z ³ Rn

|F f |

2 2

1 + |t|

1/2 ´ dt

This along with 41.1.5 yields the conclusion of the lemma because ¯¯ Z n ¯¯ X ¯¯ ∂f ¯¯2 2 2 2 ¯¯ ¯¯ + ||f ||−1,2 = Cn |F f | dx = Cn ||f ||0,2 . ¯¯ ∂xi ¯¯ n i=1

−1,2

R

Now consider Theorem 41.1.1. First note that by Lemma 41.1.2 and U − defined there, Lemma 41.1.3 implies that for f extended as in Lemma 41.1.2, Ã ! ¯¯ n ¯¯ X ¯¯ ∂f ¯¯ ¯ ¯ ¯ ¯ ||f ||0,2,U − ≤ ||f ||0,2,Rn = Cn ||f ||−1,2,Rn + ¯¯ ∂xi ¯¯ −1,2,Rn i=1

1182

KORN’S INEQUALITY

Ã ≤ Cgn

||f ||−1,2,U −

¯¯ n ¯¯ X ¯¯ ∂f ¯¯ ¯¯ ¯¯ + ¯¯ ∂xi ¯¯ i=1

! .

(41.1.6)

−1,2,U −

Let Ω be a bounded open set having Lipschitz boundary which lies locally on p one side of its boundary. Let {Qi }i=0 be cubes of the sort used in the proof of the divergence theorem such that Q0 ⊆ Ω and the other cubes cover the boundary of Ω. Let {ψ i } be a C ∞ partition of unity with spt (ψ i ) ⊆ Qi and let f ∈ L2 (Ω) . Then for φ ∈ Cc∞ (Ω) and ψ one of these functions in the partition of unity, ¯¯ ¯Z ¯¯ ¯ ¯Z ¯ ¯¯ ∂ (f ψ) ¯¯ ¯ ¯ ¯ ¯ ¯¯ ¯ f ∂ (ψφ) dx¯ + sup ¯ f φ ∂ψ dx¯ ¯¯ ≤ sup ¯¯ ∂xi ¯¯ ¯ ¯ ¯ ∂xi ¯ ||φ||1,2 ≤1 ||φ||1,2 ≤1 Ω ∂xi Ω −1,2,Ω Now if ||φ||1,2 ≤ 1, then for a suitable constant, Cψ , ¯¯ ¯¯ ¯¯ ∂ψ ¯¯ ¯ ¯¯ ≤ C ψ . ¯ ||ψφ||1,2 ≤ Cψ ||φ||1,2 ≤ Cψ , ¯¯φ ∂xi ¯¯1,2 Therefore, ¯¯ ¯¯ ¯¯ ∂ (f ψ) ¯¯ ¯¯ ¯¯ ¯¯ ∂xi ¯¯

−1,2,Ω

¯ ¯Z ¯ ¯Z ¯ ¯ ¯ ∂η ¯¯ ¯ ¯ dx¯ + sup ¯ f ηdx¯¯ ≤ sup ¯ f ∂x i ||η||1,2 ≤Cψ ||η||1,2 ≤Cψ Ω Ω

Ã ¯¯ ! ¯¯ ¯¯ ∂f ¯¯ ¯¯ ≤ Cψ ¯¯¯¯ + ||f ||−1,2,Ω . ∂xi ¯¯−1,2,Ω

(41.1.7)

Now using 41.1.7 and 41.1.6  ¯¯ ¢ ¯¯ n ¯¯ ¡ X ¯¯ ¯¯ ¯¯ ¯¯ ∂ f ψ j ¯¯¯¯ ¯ ¯ ¯¯f ψ j ¯¯ ≤ Cg ¯¯f ψ j ¯¯−1,2,Ω + ¯¯ ¯¯ 0,2,Ω ¯¯ ∂xi ¯¯ i=1

Ã ≤ Cψ j Cg Therefore, letting C = ||f ||0,2,Ω

j=1

¯¯ n ¯¯ X ¯¯ ∂f ¯¯ ¯¯ ¯¯ ¯¯ ∂xi ¯¯ i=1

Pp j=1

p X ¯¯ ¯¯ ¯¯f ψ j ¯¯ ≤

||f ||−1,2,Ω +

  −1,2,Ω

! . −1,2,Ω

Cψj Cg , Ã

0,2,Ω

≤C

¯¯ n ¯¯ X ¯¯ ∂f ¯¯ ¯ ¯ ¯¯ ||f ||−1,2,Ω + ¯¯ ∂xi ¯¯ i=1

! . −1,2,Ω

This proves the hard half of the inequality of Theorem 41.1.1. To complete the proof, let f denote the zero extension of f off Ω. Then ¯¯ ¯¯ n ¯¯ n ¯¯ X X ¯¯ ∂f ¯¯ ¯¯ ∂f ¯¯ ¯¯ ¯¯ ¯ ¯ ¯ ¯ ¯ ¯¯ ¯ ¯ ¯ ¯ ¯ ||f ||−1,2,Ω + ≤ f −1,2,Rn + ¯¯ ∂xi ¯¯ ¯¯ ∂xi ¯¯ −1,2,Ω −1,2,Rn i=1 i=1 ¯¯ ¯¯ ≤ Cn ¯¯f ¯¯0,2,Rn = Cn ||f ||0,2,Ω . This along with 41.1.8 proves Theorem 41.1.1.

(41.1.8)

41.2. KORN’S INEQUALITY

41.2

1183

Korn’s Inequality

The inequality in this section is known as Korn’s second inequality. It is also known as coercivity of strains. For u a vector valued function in Rn , define εij (u) ≡

1 (ui,j + uj,i ) 2

This is known as the strain or small strain. Korn’s inequality says that the norm given by,  1/2 n n X n X X 2 2 |||u||| ≡  ||ui ||0,2,Ω + ||εij (u)||0,2,Ω  (41.2.9) i=1

i=1 j=1

is equivalent to the norm,  ||u|| ≡ 

n X

2 ||ui ||0,2,Ω

i=1

¯¯ n X n ¯¯ X ¯¯ ∂ui ¯¯2 ¯ ¯ ¯¯ + ¯¯ ∂xj ¯¯

1/2 

(41.2.10)

0,2,Ω

i=1 j=1

It is very significant because it is the strain as just defined which occurs in many of the physical models proposed in continuum mechanics. The inequality is far from obvious because the strains only involve certain combinations of partial derivatives. Theorem 41.2.1 (Korn’s second inequality) Let Ω be any domain for which the conclusion of Theorem 41.1.1 holds. Then the two norms in 41.2.9 and 41.2.10 are equivalent. Proof: Let u be such that ui ∈ H 1 (Ω) for each i = 1, · · · , n. Note that ∂ 2 ui ∂ ∂ ∂ = (εik (u)) + (εij (u)) − (εjk (u)) . ∂xj , ∂xk ∂xj ∂xk ∂xi Therefore, by Theorem 41.1.1, "¯¯ # ¯¯ ¯¯ ¯¯ ¯¯ n ¯¯ X ¯¯ ∂ui ¯¯ ¯¯ ∂ui ¯¯ ¯¯ ∂ 2 u i ¯¯ ¯¯ ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ≤ C ¯¯ + ¯¯ ∂xj ¯¯ ¯¯ ∂xj , ∂xk ¯¯ ∂xj ¯¯−1,2,Ω 0,2,Ω −1,2,Ω k=1

"¯¯ # ¯¯ ¯¯ X ¯¯¯¯ ∂εrs (u) ¯¯ ¯¯ ∂ui ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ + ≤ C ¯¯ ∂xj ¯¯−1,2,Ω r,s,p ¯¯ ∂xp ¯¯−1,2,Ω # "¯¯ ¯¯ X ¯¯ ∂ui ¯¯ ¯¯ ||εrs (u)||0,2,Ω . ≤ C ¯¯¯¯ + ∂xj ¯¯−1,2,Ω r,s But also by this theorem, ¯ X ¯¯¯ ∂ui ¯¯¯¯ ¯¯ ¯¯ ||ui ||−1,2,Ω + ¯¯ ∂xp ¯¯ p

−1,2,Ω

≤ C ||ui ||0,2,Ω

1184 and so

KORN’S INEQUALITY

" # ¯¯ ¯¯ X ¯¯ ∂ui ¯¯ ¯¯ ¯¯ ≤ C ||ui ||0,2,Ω + ||εrs (u)||0,2,Ω ¯¯ ∂xj ¯¯ 0,2,Ω r,s

This proves the theorem. Note that Ω did not need to be bounded. It suffices to be able to conclude the result of Theorem 41.1.1 which would hold whenever the boundary of Ω can be covered with finitely many boxes of the sort to which Lemma 41.1.2 can be applied.

Elliptic Regularity And Nirenberg Differences 42.1

The Case Of A Half Space

Regularity theorems are concerned with obtaining more regularity given a weak solution. This extra regularity is essential in order to obtain error estimates for various problems. In this section a regularity is given for weak solutions to various elliptic boundary value problems. To save on notation, I will use the repeated index summation convention. Thus you sum over repeated indices. Consider the following picture.

R

V

¡

¡

¡ ¡ ª

U1

Γ

U

Rn−1

Here V is an open set, U ≡ {y ∈ V : yn < 0} , Γ ≡ {y ∈ V : yn = 0} and U1 is an open set as shown for which U1 ⊆ V ∩ U. Assume also that V is bounded. Suppose f ∈ L2 (U ) , ¡ ¢ αrs ∈ C 0,1 U , (42.1.1) 1185

1186

ELLIPTIC REGULARITY AND NIRENBERG DIFFERENCES 2

αrs (y) vr vs ≥ δ |v| , δ > 0.

(42.1.2)

The following technical lemma gives the essential ideas. Lemma 42.1.1 Suppose w α

rs

hs f and

Z αrs (y) U

∈ H 1 (U ) , ¡ ¢ ∈ C 0,1 U ,

(42.1.3) (42.1.4)

∈ H 1 (U ) , ∈ L2 (U ) .

(42.1.5) (42.1.6)

∂w ∂z dy + ∂y r ∂y s

Z hs (y) U

∂z dy = ∂y s

Z f zdy

(42.1.7)

U

for all z ∈ H 1 (U ) having the property that spt (z) ⊆ V. Then w ∈ H 2 (U1 ) and for some constant C, independent of f, w, and g, the following estimate holds. Ã ! X 2 2 2 2 ||w||H 2 (U1 ) ≤ C ||w||H 1 (U ) + ||f ||L2 (U ) + ||hs ||H 1 (U ) . (42.1.8) s

Proof: Define for small real h, Dkh l (y) ≡

1 (l (y + hek ) − l (y)) . h

Let U1 ⊆ U1 ⊆ W ⊆ W ⊆ V and let η ∈ Cc∞ (W ) with η (y) ∈ [0, 1] , and η = 1 on U1 as shown in the following picture.

R

V

W

¡ ¡

¡ ¡ ª

U1

U

Γ

Rn−1

¡ ¢ For h small (3h < dist W , V C ), let 1 z (y) ≡ h

½

·

w (y) − w (y − hek ) η (y−hek ) h 2

¸

42.1. THE CASE OF A HALF SPACE ·

w (y + hek ) − w (y) −η (y) h ¡ 2 h ¢ −h ≡ −Dk η Dk w ,

1187 ¸¾

2

(42.1.9) (42.1.10)

where here k < n. Thus z can be used in equation 42.1.7. Begin by estimating the left side of 42.1.7. Z ∂w ∂z αrs (y) r s dy ∂y ∂y U ¡ ¢ Z ∂ η 2 Dkh w 1 ∂w rs = α (y + hek ) r (y + hek ) dy h U ∂y ∂y s ¡ ¢ Z 1 ∂w ∂ η 2 Dkh w rs − α (y) r dy h U ∂y ∂y s ¡ ¢ ¡ ¢ Z ∂ Dkh w ∂ η 2 Dkh w rs = α (y + hek ) dy+ ∂y r ∂y s U ¡ ¢ Z 1 ∂w ∂ η 2 Dkh w rs rs (α (y + hek ) − α (y)) r dy (42.1.11) h U ∂y ∂y s Now

¡ ¢ ¡ h ¢ ∂ η 2 Dkh w ∂η h 2 ∂ Dk w = 2η s Dk w + η . ∂y s ∂y ∂y s

(42.1.12)

therefore, Z =

+ 1 + h

¢ ¡ ¢ ¡ ∂ Dkh w ∂ Dkh w dy ∂y r ∂y s ¡ ¢ ∂ Dkh w ∂η rs 2η s Dkh wdy α (y + hek ) r ∂y ∂y W ∩U

η 2 αrs (y + hek ) U (Z

¡ ¢ ) ∂w ∂ η 2 Dkh w (α (y + hek ) − α (y)) r dy ≡ A. + {B.} . (42.1.13) ∂y ∂y s W ∩U

Z

rs

rs

Now consider these two terms. From 42.1.2, Z ¯ ¯2 A. ≥ δ η 2 ¯∇Dkh w¯ dy.

(42.1.14)

U

Using the Lipschitz continuity of αrs and 42.1.12, n¯¯ ¯¯ ¯¯ ¯¯ B. ≤ C (η, Lip (α) , α) ¯¯Dkh w¯¯L2 (W ∩U ) ¯¯η∇Dkh w¯¯L2 (W ∩U ;Rn ) + ¯¯ ¯¯ ||η∇w||L2 (W ∩U ;Rn ) ¯¯η∇Dkh w¯¯L2 (W ∩U ;Rn ) o ¯¯ ¯¯ + ||η∇w||L2 (W ∩U ;Rn ) ¯¯Dkh w¯¯L2 (W ∩U ) .

(42.1.15)

1188

ELLIPTIC REGULARITY AND NIRENBERG DIFFERENCES

³ ¯¯ ´ ¯¯2 2 ≤ C (η, Lip (α) , α) Cε ¯¯Dkh w¯¯L2 (W ∩U ) + ||η∇w||L2 (W ∩U ;Rn ) + ³ ¯¯ ´ ¯¯2 ¯¯ ¯¯2 εC (η, Lip (α) , α) ¯¯η∇Dkh w¯¯L2 (W ∩U ;Rn ) + ¯¯Dkh w¯¯L2 (W ∩U ) . Now

¯¯ h ¯¯ ¯¯Dk w¯¯

2

L2 (W )

≤ ||∇w||L2 (U ;Rn ) .

(42.1.16)

(42.1.17)

To see this, observe that if w is smooth, then ÃZ W

≤

 Z  W

ÃZ

h

!1/2 ¯ ¯ ¯ w (y + hek ) − w (y) ¯2 ¯ ¯ dy ¯ ¯ h ¯ Z ¯2 1/2 ¯1 h ¯ ¯ ¯ ∇w (y + tek ) · ek dt¯ dy  ¯ ¯h 0 ¯

µZ 2

≤

¶1/2

|∇w (y + tek ) · ek | dy 0

W

dt h

! ≤ ||∇w||L2 (U ;Rn )

so by density of such functions in H 1 (U ) , 42.1.17 holds. Therefore, changing ε, yields ¯¯ ¯¯2 2 B. ≤ Cε (η, Lip (α) , α) ||∇w||L2 (U ;Rn ) + ε ¯¯η∇Dkh w¯¯L2 (W ∩U ;Rn ) .

(42.1.18)

With 42.1.14 and 42.1.18 established, consider the other terms of 42.1.7. ¯Z ¯ ¯ ¯ ¯ f zdy ¯ ¯ ¯ U

¯Z ¯ ¯ ¡ ¢ ¯ −h 2 h ¯ ≤ ¯ f −Dk η Dk w dy ¯¯ U µZ ¶1/2 µZ ¶1/2 ¯ −h ¡ 2 h ¢¯2 2 ¯ ¯ ≤ |f | dy Dk η Dk w dy U ¯¯ ¡ 2 hU ¢¯¯ ≤ ||f ||L2 (U ) ¯¯∇ η Dk w ¯¯L2 (U ;Rn ) ³ ¯¯ ´ ¯¯ ¯¯ ¯¯ ≤ ||f ||L2 (U ) ¯¯2η∇ηDkh w¯¯L2 (U ;Rn ) + ¯¯η 2 ∇Dkh w¯¯L2 (U ;Rn ) ¯¯ ¯¯ ≤ C ||f ||L2 (U ) ||∇w||L2 (U ;Rn ) + ||f ||L2 (U ) ¯¯η∇Dkh w¯¯L2 (U ;Rn ) ³ ´ ¯¯ ¯¯2 2 2 ≤ Cε ||f ||L2 (U ) + ||∇w||L2 (U ;Rn ) + ε ¯¯η∇Dkh w¯¯L2 (U ;Rn ) (42.1.19)

42.1. THE CASE OF A HALF SPACE

1189

¯Z ¯ ¯ ¯ ¯ hs (y) ∂z dy ¯ ¯ ¯ s ∂y ¯Z U ¡ ¡ ¢¢ ¯¯ ¯ ∂ −Dk−h η 2 Dkh w ¯ ¯ dy ¯ ¯ hs (y) ¯ U ¯ ∂y s ¯Z ¯ ¡¡ ¢¢ ¯ ∂ η 2 Dkh w ¯¯ ¯ ¯ Dkh hs (y) ¯ ¯ U ¯ ∂y s Ã ¡ ¢ !¯¯ ¯ Z ¯¯ Z ¯ h ¯ h ¯ ¡ h ¢ ∂ D w ∂η ¯ ¯ k h ¯Dk hs 2η D w¯ dy + ¯ ηDk hs η ¯ dy ¯ s k ¯ s ¯ ¯ ∂y ∂y U U ³ ´ X ¯ ¯ ¯¯ C ||hs || 1 ||w|| 1 + ¯¯η∇Dkh w¯¯ 2 n

≤ ≤ ≤ ≤

H (U )

s

≤ Cε

X

H (U )

L (U ;R )

¯¯ ¯¯2 2 2 ||hs ||H 1 (U ) + ||w||H 1 (U ) + ε ¯¯η∇Dkh w¯¯L2 (U ;Rn ) .

(42.1.20)

s

The following inequalities in 42.1.14,42.1.18, 42.1.19and 42.1.20 are summarized here. Z ¯ ¯2 A. ≥ δ η 2 ¯∇Dkh w¯ dy, U

¯¯ ¯¯2 2 B. ≤ Cε (η, Lip (α) , α) ||∇w||L2 (U ;Rn ) + ε ¯¯η∇Dkh w¯¯L2 (W ∩U ;Rn ) , ¯Z ¯ ³ ´ ¯ ¯ ¯¯ ¯¯ 2 h ¯¯2 ¯ f zdy ¯ ≤ Cε ||f ||2 2 ¯¯ + ||∇w|| 2 n L (U ) L (U ;R ) + ε η∇Dk w L2 (U ;Rn ) ¯ ¯ U

¯ ¯Z ¯ ¯ ¯ hs (y) ∂z dy ¯ ¯ ¯ s ∂y

≤

Cε

U

X

2

||hs ||H 1 (U )

s

¯¯ ¯¯2 2 + ||w||H 1 (U ) + ε ¯¯η∇Dkh w¯¯L2 (U ;Rn ) . Therefore, ¯¯ ¯¯2 δ ¯¯η∇Dkh w¯¯L2 (U ;Rn )

≤

¯¯ ¯¯2 2 Cε (η, Lip (α) , α) ||∇w||L2 (U ;Rn ) + ε ¯¯η∇Dkh w¯¯L2 (U ;Rn )

+Cε

X

³

¯¯ ¯¯2 2 2 ||hs ||H 1 (U ) + ||w||H 1 (U ) + ε ¯¯η∇Dkh w¯¯L2 (U ;Rn )

s

´ ¯¯ ¯¯2 2 2 +Cε ||f ||L2 (U ) + ||∇w||L2 (U ;Rn ) + ε ¯¯η∇Dkh w¯¯L2 (U ;Rn ) . Letting ε be small enough and adjusting constants yields ¯¯ ¯¯ ¯¯ ¯¯2 ¯¯∇Dkh w¯¯2 2 ≤ ¯¯η∇Dkh w¯¯L2 (U ;Rn ) ≤ L (U ;Rn ) 1

Ã C

2 ||w||H 1 (U )

+

2 ||f ||L2 (U )

+ Cε

X s

! 2 ||hs ||H 1 (U )

1190


where the constant, C, depends on η, Lip (α) , α, δ. Since this holds for all h small ∂w 1 enough, it follows ∂y k ∈ H (U1 ) and ¯¯ ¯¯ ¯¯ ∂w ¯¯2 ¯¯ ¯¯∇ ¯¯ ∂y k ¯¯ 2

≤

L (U1 ;Rn )

Ã C

2 ||w||H 1 (U )

2 ||f ||L2 (U )

+

+ Cε

X

! 2 ||hs ||H 1 (U )

(42.1.21)

s

¯¯ 2 ¯¯2 ¯¯ ¯¯ for each k < n. It remains to estimate ¯¯ ∂∂yw2 ¯¯ 2 n

L (U1 )

Cc∞

. To do this return to 42.1.7

which must hold for all z ∈ (U1 ) . Therefore, using 42.1.7 it follows that for all z ∈ Cc∞ (U1 ) , Z Z Z ∂w ∂z ∂hs αrs (y) r s dy = − zdy + f zdy. s ∂y ∂y U U ∂y U Now from the Lipschitz assumption on αrs , it follows µ ¶ X ∂ rs ∂w F ≡ α ∂y s ∂y r r,s≤n−1 ¶ X X ∂ µ ∂hs ns ∂w + α − +f s n ∂y ∂y ∂y s s s≤n−1

∈

2

L (U1 )

and Ã ||F ||L2 (U1 ) ≤ C

2 ||w||H 1 (U )

+

2 ||f ||L2 (U )

+ Cε

X

! 2 ||hs ||H 1 (U )

.

(42.1.22)

s

Therefore, from density of Cc∞ (U1 ) in L2 (U1 ) , µ ¶ ∂ ∂w nn − n α (y) n = F, no sum on n ∂y ∂y and so −

∂αnn ∂w ∂2w − αnn 2 =F n n ∂y ∂y ∂ (y n )

By 42.1.2 αnn (y) ≥ δ and so it follows from 42.1.22 that there exists a constant,C depending on δ such that ¯ ¯ ¯ ∂2w ¯ ³ ´ ¯ ¯ ≤ C |F |L2 (U1 ) + ||w||H 1 (U ) ¯ ¯ 2 ¯ ∂ (y n ) ¯ 2 L (U1 )

42.1. THE CASE OF A HALF SPACE

1191

which with 42.1.21 and 42.1.22 implies the existence of a constant, C depending on δ such that Ã ! X 2 2 2 2 ||w||H 2 (U1 ) ≤ C ||w||H 1 (U ) + ||f ||L2 (U ) + Cε ||hs ||H 1 (U ) , s

proving the lemma. What if more regularity is known for f , hs , αrs and w? Could more be said about the regularity of the solution? The answer is yes and is the content of the next corollary. First here is some notation. For α a multi-index with |α| = k − 1, α = (α1 , · · · , αn ) define n Y ¡ h ¢ αk Dαh l (y) ≡ Dk l (y) . k=1

Also, for α and τ multi indices, τ < α means τ i < αi for each i. Corollary 42.1.2 Suppose in the context of Lemma 42.1.1 the following for k ≥ 1. w αrs hs f and

Z

∂w ∂z α (y) r s dy + ∂y ∂y U rs

∈ H k (U ) , ¡ ¢ ∈ C k−1,1 U , ∈ H k (U ) , ∈ H k−1 (U ) , Z

∂z hs (y) s dy = ∂y U

Z f zdy

(42.1.23)

U

for all z ∈ H 1 (U ) or H01 (U ) such that spt (z) ⊆ V. Then there exists C independent of w such that Ã ! X ||w||H k+1 (U1 ) ≤ C ||f ||H k−1 (U ) + ||hs ||H k (U ) + ||w||H k (U ) . (42.1.24) s

Proof: The proof involves the following claim which is proved using the conclusion of Lemma 42.1.1 on Page 1186. Claim : If α = (α0 , 0) where |α0 | ≤ k − 1, then there exists a constant independent of w such that Ã ! X α ||D w||H 2 (U1 ) ≤ C ||f ||H k−1 (U ) + ||hs ||H k (U ) + ||w||H k (U ) . (42.1.25) s

Proof of claim: First note that if |α| = 0, then 42.1.25 follows from Lemma 42.1.1 on Page 1186. Now suppose the conclusion of the claim holds for all |α| ≤ j−1 where j < k. Let |α| = j and α = (α0 , 0) . Then for z ∈ H 1 (U ) having compact support in V, it follows that for h small enough, ¡ ¢ Dα−h z ∈ H 1 (U ) , spt Dαh z ⊆ V.

1192


Therefore, you can replace z in 42.1.23 with Dα−h z. Now note that you can apply the following manipulation. Z Z p (y) Dα−h z (y) dy = Dαh p (y) z (y) dy U

U

and obtain µ ¶ ¶ Z Z µ ¡¡ h ¢ ¢ ∂z ∂w ∂z h Dα f z dy. Dαh αrs r (h ) dy = + D s α s s ∂y ∂y ∂y U U

(42.1.26)

Letting h → 0, this gives µ ¶ ¶ Z µ Z ∂z ∂z α rs ∂w α α D ((Dα f ) z) dy. + D (h ) dy = s ∂y r ∂y s ∂y s U U Now

µ ¶ α τ X rs ∂w rs ∂ (D w) α−τ rs ∂ (D w) D α = α + C (τ ) D (α ) ∂y r ∂y r ∂y r τ 0 independent of y ∈ U such that for all y ∈ U , 2

αrs (y) vr vs ≥ r |v| . Proof of the claim: If this is not so, there exist vectors, vn , |vn | = 1, and yn ∈ U such that αrs (yn ) vrn vsn ≤ n1 . Taking a subsequence, there exists y ∈ U and |v| = 1 such that αrs (y) vr vs = 0 contradicting 42.2.32.

1202


Therefore, by Corollary 42.1.2, there exists a constant, C, independent of f, g, and w such that Ã ! ¯¯ ¯¯2 X ¯¯¯¯ ¯¯¯¯2 ¯¯ e¯¯ 2 2 ||w||H k+1 (Φ−1 (W1 ∩Ω)) ≤ C ¯¯f ¯¯ k−1 + ||w||H k (U ) + . ¯¯hel ¯¯ k H

(U )

H (U )

l

Therefore, Ã 2 ||u||H k+1 (W1 ∩Ω)

≤

C

2 ||f ||H k−1 (W ∩Ω)

+

2 ||w||H k (W ∩Ω)

C

! 2 ||hs ||H k (W ∩Ω)

s

Ã ≤

+

X

2 ||f ||H k−1 (Ω)

+

2 ||w||H k (Ω)

+

X

2 ||hs ||H k (Ω)

! .

s

which proves the lemma. Now here is a theorem which generalizes the one above in the case where more regularity is known. Theorem 42.2.6 Let Ω be a bounded open set with C k,1 boundary as in Definition 42.2.1, let f ∈ H k−1 (Ω) , hs ∈ H k (Ω), and suppose that for all x ∈ Ω, 2

aij (x) vi vj ≥ δ |v| . Suppose also that u ∈ H k (Ω) and Z Z Z ij a (x) u,i (x) v,j (x) dx + hk (x) v,k (x) dx = f (x) v (x) dx Ω

Ω

Ω

for all v ∈ H k (Ω) . Then u ∈ H k+1 (Ω) and for some C independent of f, g, and u, Ã ! X 2 2 2 2 ||u||H k+1 (Ω) ≤ C ||f ||H k−1 (Ω) + ||u||H k (Ω) + ||hs ||H k (Ω) . s

Proof: Let the Wi for i = 1, · · · , l be as described in Definition 42.2.1. Thus ∂Ω ⊆ ∪lj=1 Wj . Then let C1 ≡ ∂Ω \ ∪li=2 Wi , a closed subset of W1 . Let D1 be an open set satisfying C1 ⊆ D1 ⊆ D1 ⊆ W1 . ¡ ¡ ¢¢ Then D1 , W2 , · · · , Wl cover ∂Ω. Let C2 = ∂Ω \ D1 ∪ ∪li=3 Wi . Then C2 is a closed subset of W2 . Choose an open set, D2 such that C2 ⊆ D2 ⊆ D2 ⊆ W2 . Thus D1 , D2 , W3 · · · , Wl covers ∂Ω. Continue in this way to get Di ⊆ Wi , and ∂Ω ⊆ ∪li=1 Di , and Di is an open set. Now let D0 ≡ Ω \ ∪li=1 Di .

42.2. THE CASE OF BOUNDED OPEN SETS

1203

Also, let Di ⊆ Vi ⊆ Vi ⊆ Wi . Therefore, D0 , V1 , · · · , Vl covers Ω. Then the same estimation process used above yields ! Ã X 2 2 2 ||u||H k+1 (D0 ) ≤ C ||f ||H k−1 (Ω) + ||u||H k (Ω) + ||hk ||H k (Ω) . k

From Lemma 42.2.5 Ã ||u||H k+1 (Vi ∩Ω) ≤ C

2 ||f ||H k−1 (Ω)

+

2 ||u||H k (Ω)

+

X

! 2 ||hk ||H k (Ω)

k

also. This proves the theorem since ||u||H k+1 (Ω) ≤

l X i=1

||u||H k+1 (Vi ∩Ω) + ||u||H k+1 (D0 ) .

1204


Interpolation In Banach Space 43.1

An Assortment Of Important Theorems

43.1.1

Weak Vector Valued Derivatives

In this section, several significant theorems are presented. Unless indicated otherwise, the measure will be Lebesgue measure. First here is a lemma. Lemma 43.1.1 Suppose g ∈ L1 ([a, b] ; X) where X is a Banach space. Then if Rb g (t) φ (t) dt = 0 for all φ ∈ Cc∞ (a, b) , then g (t) = 0 a.e. a Proof: Let E be a measurable subset of (a, b) and let K ⊆ E ⊆ V ⊆ (a, b) where K is compact, V is open and m (V \ K) < ε. Let K ≺ h ≺ V as in the proof of the Riesz representation theorem for positive linear functionals. Enlarging K slightly and convolving with a mollifier, it can be assumed h ∈ Cc∞ (a, b) . Then ¯Z ¯ ¯Z ¯ ¯ b ¯ ¯ b ¯ ¯ ¯ ¯ ¯ XE (t) g (t) dt¯ = ¯ (XE (t) − h (t)) g (t) dt¯ ¯ ¯ a ¯ ¯ a ¯ Z b ≤ |XE (t) − h (t)| ||g (t)|| dt Za ≤ ||g (t)|| dt. V \K

Now let Kn ⊆ E ⊆ Vn with m (Vn \ Kn ) < 2−n . Then from the above, ¯Z ¯ Z ¯ b ¯ b ¯ ¯ XE (t) g (t) dt¯ ≤ XVn \Kn (t) ||g (t)|| dt ¯ ¯ a ¯ a and the integrand of the last integral converges to 0 a.e. as n → ∞ because P n m (Vn \ Kn ) < ∞. By the dominated convergence theorem, this last integral 1205

1206

INTERPOLATION IN BANACH SPACE

converges to 0. Therefore, whenever E ⊆ (a, b) , Z

b

XE (t) g (t) dt = 0. a

Since the endpoints have measure zero, it also follows that for any measurable E, the above equation holds. Now g ∈ L1 ([a, b] ; X) and so it is measurable. Therefore, g ([a, b]) is separable. Let D be P a countable dense subset and let E denote the set of linear combinations of the form i ai di where ai is a rational point of F and di ∈ D. Thus E is countable. Denote by Y the closure of E in X. Thus Y is a separable closed subspace of X which contains all the values of g. ∞ Now let Sn ≡ g −1 (B (yn , ||yn || /2)) where E = {yn }n=1 . Therefore, ∪ń Sn = ³

g −1 (X \ {0}) . This follows because if x ∈ Y and x 6= 0, then in B x, ||x|| 4 ||yn || 2

||yn || > 34 ||x|| if each Sn has

3||x|| 8

there

||x|| 4

is a point of E, yn . Therefore, and so > > so x ∈ B (yn , ||yn || /2) . It follows that measure zero, then g (t) = 0 for a.e. t. Suppose then that for some n, the set, Sn has positive mesure. Then from what was shown above, ¯¯ ¯¯ ¯¯ ¯¯ Z Z ¯¯ 1 ¯¯ ¯¯ 1 ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ ||yn || = ¯¯ g (t) dt − yn ¯¯ = ¯¯ g (t) − yn dt¯¯¯¯ m (Sn ) Sn m (Sn ) Sn Z Z 1 1 ||g (t) − yn || dt ≤ ||yn || /2dt = ||yn || /2 ≤ m (Sn ) Sn m (Sn ) Sn and so yn = 0 which implies Sn = ∅, a contradiction to m (Sn ) > 0. This contradiction shows each Sn has measure zero and so as just explained, g (t) = 0 a.e. Definition 43.1.2 For f ∈ L1 (a, b; X) , define an extension, f defined on [2a − b, 2b − a] = [a − (b − a) , b + (b − a)] as follows.

  f (t) if t ∈ [a, b] f (2a − t) if t ∈ [2a − b, a] f (t) ≡  f (2b − t) if t ∈ [b, 2b − a]

Definition 43.1.3 Also if f ∈ Lp (a, b; X) and h > 0, define for t ∈ [a, b] , fh (t) ≡ f (t − h) for all h < b − a. Thus the map f → fh is continuous and linear on Lp (a, b; X) . It is continuous because Z

b

Z p

a

Z

a+h

||fh (t)|| dt =

p

b−h

||f (2a − t + h)|| dt + a

Z =

a+h

Z

b−h

p

||f (t)|| dt + a

p

||f (t)|| dt a

a

p

p

||f (t)|| dt ≤ 2 ||f ||p .

The following lemma is on continuity of translation in Lp (a, b; X) .

43.1. AN ASSORTMENT OF IMPORTANT THEOREMS

1207

Lemma 43.1.4 Let f be as defined in Definition 43.1.2. Then for f ∈ Lp (a, b; X) for p ∈ [1, ∞), Z b ¯¯ ¯¯ ¯¯f (t − δ) − f (t)¯¯p dt = 0. lim X δ→0

a

Proof: Regarding the measure space as (a, b) with Lebesgue measure, by Lemma 23.5.9 there exists g ∈ Cc (a, b; X) such that ||f − g||p < ε. Here the norm is the norm in Lp (a, b; X) . Therefore, ||fh − f ||p

≤ ||fh − gh ||p + ||gh − g||p + ||g − f ||p ³ ´ ≤ 21/p + 1 ||f − g||p + ||gh − g||p ³ ´ < 21/p + 1 ε + ε

whenever h is sufficiently small. This is because of the uniform continuity of g. Therefore, since ε > 0 is arbitrary, this proves the lemma. Definition 43.1.5 Let f ∈ L1 (a, b; X) . Then the distributional derivative in the sense of X valued distributions is given by Z 0

f (φ) ≡ −

b

f (t) φ0 (t) dt

a

Then f 0 ∈ L1 (a, b; X) if there exists h ∈ L1 (a, b; X) such that for all φ ∈ Cc∞ (a, b) , Z

b

0

f (φ) =

h (t) φ (t) dt. a

Then f 0 is defined to equal h. Here f and f 0 are considered as vector valued distributions in the same way as was done for scalar valued functions. Lemma 43.1.6 The above definition is well defined. Proof: Suppose both h and g work in the definition. Then for all φ ∈ Cc∞ (a, b) , Z

b

(h (t) − g (t)) φ (t) dt = 0. a

Therefore, by Lemma 43.1.1, h (t) − g (t) = 0 a.e. The other thing to notice about this is the following lemma. It follows immediately from the definition. Lemma 43.1.7 Suppose f, f 0 ∈ L1 (a, b; X) . Then if [c, d] ⊆ [a, b], it follows that ¡ ¢0 f |[c,d] = f 0 |[c,d] . This notation means the restriction to [c, d] .

1208


Recall that in the case of scalar valued functions, if you had both f and its weak derivative, f 0 in L1 (a, b) , then you were able to conclude that f is almost everywhere equal to a continuous function, still denoted by f and Z t f 0 (s) ds. f (t) = f (a) + a

In particular, you can define f (a) to be the initial value of this continuous function. It turns out that an identical theorem holds in this case. To begin with here is the same sort of lemma which was used earlier for the case of scalar valued functions. It says that if f 0 = 0 where the derivative is taken in the sense of X valued distributions, then f equals a constant. Lemma 43.1.8 Suppose f ∈ L1 (a, b; X) and for all φ ∈ Cc∞ (a, b) , Z b f (t) φ0 (t) dt = 0. a

Then there exists a constant, a ∈ X such that f (t) = a a.e. Rb Proof: Let φ0 ∈ Cc∞ (a, b) , a φ0 (x) dx = 1 and define for φ ∈ Cc∞ (a, b) ÃZ ! Z x

ψ φ (x) ≡

b

[φ (t) − a

a

Then ψ φ ∈ Cc∞ (a, b) and ψ 0φ = φ − Z

Z

b

f (t)

ÃZ

f a

ÃZ

b

(t) ψ 0φ ÃZ

(t) +

a

b

f (t) − a

a

!Z

b

(t) dt +

φ (y) dy a

!

!

!!

b

f (t) φ0 (t) dt

φ (t) dt

and so by Lemma 43.1.1, ÃZ f (t) − a


b

a

f (t) φ0 (t) dt φ (y) dy .

a

It follows that for all φ ∈ Cc∞ (a, b) , ÃZ Z Ã

!

φ (y) dy φ0 (t) dt

a

b

=

!

b

ÃZ

b

=

´ φ (y) dy φ0 . Then

ψ 0φ

a

Z

b a

Ã

b

f (t) (φ (t)) dt = a

³R

φ (y) dy φ0 (t)]dt

! f (t) φ0 (t) dt

= 0 a.e.

b

f (t) φ0 (t) dt


1209

Theorem 43.1.9 Suppose f, f 0 both are in L1 (a, b; X) where the derivative is taken in the sense of X valued distributions. Then there exists a unique point of X, denoted by f (a) such that the following formula holds a.e. t. Z t f (t) = f (a) + f 0 (s) ds a

Proof: ¶ Z t Z bµ Z f (t) − f 0 (s) ds φ0 (t) dt = a

a

b

Z

b

f (t) φ0 (t) dt −

a

a

Z

t

f 0 (s) φ0 (t) dsdt.

a

RbRt Now consider a a f 0 (s) φ0 (t) dsdt. Let Λ ∈ X 0 . Then it is routine from approximating f 0 with simple functions to verify ÃZ Z ! Z Z b t b t 0 0 Λ f (s) φ (t) dsdt = Λ (f 0 (s)) φ0 (t) dsdt. a

a

a

a

Now the ordinary Fubini theorem can be applied to obtain Z bZ b = Λ (f 0 (s)) φ0 (t) dtds a ÃZs Z ! b

=

b

Λ a

f 0 (s) φ0 (t) dtds .

s

Since X 0 separates the points of X, it follows Z bZ t Z bZ 0 0 f (s) φ (t) dsdt = a

a

a

b

f 0 (s) φ0 (t) dtds.

s

Therefore, Z

b

µ ¶ Z t f (t) − f 0 (s) ds φ0 (t) dt

a

Z

a b

=

Z

0

b

Z

f (t) φ (t) dt − a

Z

a

b

=

Z f (t) φ0 (t) dt −

a

Z =

Z f (t) φ0 (t) dt +

f 0 (s) φ0 (t) dtds

s b

a

b

b

Z

b

f 0 (s)

φ0 (t) dtds

s b

f 0 (s) φ (s) ds = 0.

a

a

Therefore, by Lemma 43.1.8, there exists a constant, denoted as f (a) such that Z t f (t) − f 0 (s) ds = f (a) a

and this proves the theorem. The integration by parts formula is also important.

1210


Corollary 43.1.10 Suppose f, f 0 ∈ L1 (a, b; X) and suppose φ ∈ C 1 ([a, b]) . Then the following integration by parts formula holds. Z

b

Z f (t) φ0 (t) dt = f (b) φ (b) − f (a) φ (a) −

a

b

f 0 (t) φ (t) dt.

a

Proof: From Theorem 43.1.9 Z Z

b a b

=

f (t) φ0 (t) dt µ

Z f (a) +

a

t

¶ f (s) ds φ0 (t) dt 0

a

Z

b

Z

t

= f (a) (φ (b) − φ (a)) + a

Z

b

= f (a) (φ (b) − φ (a)) +

Z

b

f 0 (s)

a

Z

f 0 (s) dsφ0 (t) dt

a

φ0 (t) dtds

s b

= f (a) (φ (b) − φ (a)) +

0

f (s) (φ (b) − φ (s)) ds a

Z

b

= f (a) (φ (b) − φ (a)) − a

f 0 (s) φ (s) ds + (f (b) − f (a)) φ (b) Z

= f (b) φ (b) − f (a) φ (a) −

b

f 0 (s) φ (s) ds.

a

The interchange in order of integration is justified as in the proof of Theorem 43.1.9. With this integration by parts formula, the following interesting lemma is obtained. This lemma shows why it was appropriate to define f as in Definition 43.1.2.

Lemma 43.1.11 Let f be given in Definition 43.1.2 and suppose f, f 0 ∈ L1 (a, b; X) . 0 Then f , f ∈ L1 (2a − b, 2b − a; X) also and  0  f (t) if t ∈ [a, b] 0 −f (2a − t) if t ∈ [2a − b, a] f (t) ≡  −f (2b − t) if t ∈ [b, 2b − a]

(43.1.1)

Proof: It is clear from the definition of f that f ∈ L1 (2a − b, 2b − a; X) and that in fact ¯¯ ¯¯ ¯¯f ¯¯ 1 ≤ 3 ||f ||L1 (a,b;X) . (43.1.2) L (2a−b,2b−a;X)


1211

Let φ ∈ Cc∞ (2a − b, 2b − a) . Then from the integration by parts formula, Z

2b−a

f (t) φ0 (t) dt

2a−b Z b

Z

2b−a

f (t) φ0 (t) dt +

= Z

a

Z

b

b

a

Z

= f (b) φ (b) − f (a) φ (a) − Z

b

Z

+f (a) φ (a) +

b

b

Z

f 0 (t) φ (t) dt +

a

f 0 (t) φ (t) dt − f (b) φ (b) + f (a) φ (2b − a)

b

Z f 0 (u) φ (2b − u) du +

a

b

Z f 0 (t) φ (t) dt −

a

Z

b

f 0 (u) φ (2a − u) du

a

= − Z

f (u) φ0 (2a − u) du

a

f 0 (u) φ (2b − u) du − f (b) φ (2a − b)

a

= −

f (2a − t) φ0 (t) dt

a

+

Z

b

f (u) φ0 (2b − u) du +

a

a 2a−b

Z

b

f (t) φ0 (t) dt +

=

Z f (2b − t) φ0 (t) dt +

a 2b−a

−f 0 (2b − t) φ (t) dt −

b 2b−a

= −

b

f 0 (u) φ (2a − u) du Z

a

−f 0 (2a − t) φ (t) dt

2a−b

0

f (t) φ (t) dt 2a−b

0

where f (t) is given in 43.1.1. This proves the lemma. Definition 43.1.12 Let V be a Banach space and let H be a Hilbert space. (Typically H = L2 (Ω)) Suppose V ⊆ H is dense in H meaning that the closure in H of V gives H. Then it is often the case that H is identified with its dual space, and then because of the density of V in H, it is possible to write V ⊆ H = H0 ⊆ V 0 When this is done, H is called a pivot space. Another notation which is often used is hf, gi to denote f (g) for f ∈ V 0 and g ∈ V. This may also be written as hf, giV 0 ,V The next theorem is an example of a trace theorem. In this theorem, f ∈ Lp (0, T ; V ) while f 0 ∈ Lp (0, T ; V 0 ) . It makes no sense to consider the initial values of f in V because it is not even continuous with values in V . However, because of the derivative of f it will turn out that f is continuous with values in a larger space and so it makes sense to consider initial values of f in this other space. This other space is called a trace space. Theorem 43.1.13 Let V and H be a Banach space and Hilbert space as described 0 in Definition 43.1.12. Suppose f ∈ Lp (0, T ; V ) and f 0 ∈ Lp (0, T ; V 0 ) . Then f is

1212


a.e. equal to a continuous function mapping [0, T ] to H. Furthermore, there exists f (0) ∈ H such that Z t 1 1 2 2 |f (t)|H − |f (0)|H = hf 0 (s) , f (s)i ds, (43.1.3) 2 2 0 and for all t ∈ [0, T ] ,

Z

t

f 0 (s) ds ∈ H,

(43.1.4)

0

and for a.e. t ∈ [0, T ] , Z f (t) = f (0) +

t

f 0 (s) ds in H,

(43.1.5)

0

Here f 0 is being taken in the sense of V 0 valued distributions and p ≥ 2.

1 p

+

1 p0

= 1 and

Proof: Let Ψ ∈ Cc∞ (−T, 2T ) satisfy Ψ (t) = 1 if t ∈ [−T /2, 3T /2] and Ψ (t) ≥ 0. For t ∈ R, define ½ f (t) Ψ (t) if t ∈ [−T, 2T ] b f (t) ≡ 0 if t ∈ / [−T, 2T ] and

Z

1/n

fn (t) ≡ −1/n

fb(t − s) φn (s) ds

(43.1.6)

where φn is a mollifier having support in (−1/n, 1/n) . Then by Minkowski’s inequality ¯¯ ¯¯ ¯¯ ¯¯ ¯¯fn − fb¯¯

=

≤

Lp (R;V )

¯¯p !1/p Ã Z ¯¯ Z 1/n ¯¯ ¯¯ ¯¯ b ¯¯ = fb(t − s) φn (s) ds¯¯ dt ¯¯f (t) − ¯ ¯ ¯¯ R −1/n V

¯¯p !1/p Ã Z ¯¯Z ¯¯ 1/n ³ ¯¯ ´ ¯¯ ¯¯ fb(t) − fb(t − s) φn (s) ds¯¯ dt ¯¯ ¯ ¯ ¯¯ R −1/n V ÃZ ÃZ !p !1/p ¯¯ 1/n ¯¯ ¯¯ b ¯¯ b ¯¯f (t) − f (t − s)¯¯ φ (s) ds dt R

Z

1/n

≤ −1/n

Z

V

−1/n

φn (s)

n

µZ ¯¯ ¯¯p ¶1/p ¯¯ b ¯¯ ds ¯¯f (t) − fb(t − s)¯¯ dt R

V

1/n

≤ −1/n

φn (s) εds = ε

provided n is large enough. This follows from Lemma 43.1.4 about continuity of translation. Since ε > 0 is arbitrary, it follows fn → fb in Lp (R; V ) . Similarly,


1213

fn → f in L2 (R; H). This follows because p ≥ 2 and the norm in V and norm in H are related by |x|H ≤ C ||x||V for some constant, C. Now  Ψ (t) f (t) if t ∈ [0, T ] ,    Ψ (t) f (2T − t) if t ∈ [T, 2T ] , b f (t) = Ψ (t) f (−t) if t ∈ [0, T ] ,    0 if t ∈ / [−T, 2T ] . An easy modification of the argument of Lemma 43.1.11 yields  0 Ψ (t) f (t) + Ψ (y) f 0 (t) if t ∈ [0, T ] ,    0 Ψ (t) f (2T − t) − Ψ (t) f 0 (2T − t) if t ∈ [T, 2T ] , fb0 (t) = . Ψ0 (t) f (−t) − Ψ (t) f 0 (−t) if t ∈ [−T, 0] ,    0 if t ∈ / [−T, 2T ] . Recall Z fn (t)

1/n

= Z

−1/n

= R

Z fb(t − s) φn (s) ds =

R

fb(t − s) φn (s) ds

fb(s) φn (t − s) ds.

Therefore, Z fn0 (t) =

Z

R

Z =

fb(s) φ0n (t − s) ds =

1 2T + n

1 −T − n

R

1 −T − n

fb(s) φ0n (t − s) ds

Z

fb0 (s) φn (t − s) ds =

Z =

1 2T + n

fb0 (t − s) φn (s) ds =

Z

1/n −1/n

R

fb0 (s) φn (t − s) ds

fb0 (t − s) φn (s) ds

and it follows from the first line above that fn0 is continuous with values in V for all t ∈ R. Also note that both fn0 and fn equal zero if t ∈ / [−T, 2T ] whenever n is large 0 enough. Exactly similar reasoning to the above shows that fn0 → fb0 in Lp (R; V 0 ) . ∞ Now let φ ∈ Cc (0, T ) . Z Z 2 |fn (t)|H φ0 (t) dt = (fn (t) , fn (t))H φ0 (t) dt (43.1.7) R R Z Z 0 = − 2 (fn (t) , fn (t)) φ (t) dt = − 2 hfn0 (t) , fn (t)i φ (t) dt R

Now

≤

R

¯Z ¯ Z ¯ ¯ 0 ¯ hfn0 (t) , fn (t)i φ (t) dt − ¯ hf (t) , f (t)i φ (t) dt ¯ ¯ R R Z (|hfn0 (t) − f 0 (t) , fn (t)i| + |hf 0 (t) , fn (t) − f (t)i|) φ (t) dt. R

1214


From the first part of this proof which showed that fn → fb in Lp (R; V ) and fn0 → fb0 0 in Lp (R; V 0 ) , an application of Holder’s inequality shows the above converges to 0 as n → ∞. Therefore, passing to the limit as n → ∞ in the 43.1.8, Z ¯ Z D ¯ E ¯ b ¯2 0 ¯f (t)¯ φ (t) dt = − 2 fb0 (t) , fb(t) φ (t) dt R

H

R

¯ ¯2 ¯ ¯ which shows t → ¯fb(t)¯ equals a continuous function a.e. and it also has a weak D H E derivative equal to 2 fb0 , fb . It remains to verify that fb is continuous on [0, T ] . Of course fb = f on this interval. Let N be large enough that fn (−T ) = 0 for all n > N. Then for m, n > N and t ∈ [−T, 2T ] Z t 2 0 |fn (t) − fm (t)|H = 2 (fn0 (s) − fm (s) , fn (s) − fm (s)) ds −T t

Z =

2 −T

0 hfn0 (s) − fm (s) , fn (s) − fm (s)iV 0 ,V ds

Z ≤

2 R

≤

0 ||fn0 (s) − fm (s)||V 0 ||fn (s) − fm (s)||V ds

2 ||fn − fm ||Lp0 (R;V 0 ) ||fn − fm ||Lp (R;V )

which shows from the above that {fn } is uniformly Cauchy on [−T, 2T ] with values in H. Therefore, there exists g a continuous function defined on [−T, 2T ] having values in H such that lim max {|fn (t) − g (t)|H ; t ∈ [−T, 2T ]} = 0.

n→∞

However, g = fb a.e. because fn converges to f in Lp (0, T ; V ) . Therefore, taking a subsequence, the convergence is a.e. It follows from the fact that V ⊆ H = H 0 ⊆ V 0 and Theorem 43.1.9 there exists f (0) ∈ V 0 such that for a.e. t, Z t f (t) = f (0) + f 0 (s) ds in V 0 0

Now g = f a.e. and g is continuous with values in H hence continuous with values in V 0 and so Z t g (t) = f (0) + f 0 (s) ds in V 0 0

for all t. Since g is continuous with values in H it is continuous with values in V 0 . Taking the limit as t ↓ 0 in the above, g (a) = limt→0+ g (t) = f (0) , showing that f (0) ∈ H. Therefore, for a.e. t, Z t Z t f (t) = f (0) + f 0 (s) ds in H, f 0 (s) ds ∈ H. 0

0


1215

This proves the theorem. 0 Note that if f ∈ Lp (0, T ; V ) and f 0 ∈ Lp (0, T ; V 0 ) , then you can consider the initial value of f and it will be in H. What if you start with something in H? Is 0 it an initial condition for a function f ∈ Lp (0, T ; V ) such that f 0 ∈ Lp (0, T ; V 0 )? This is worth thinking about. If it is not so, what is the space of initial values? How can you give this space a norm? What are its properties? It turns out that if V is a closed subspace of the Sobolev space, W 1,p (Ω) which contains W01,p (Ω) for p ≥ 2 and H = L2 (Ω) the answer to the above question is yes. Not surprisingly, there are many generalizations of the above ideas.

43.1.2

Some Imbedding Theorems

The next theorem is very useful in getting estimates in partial differential equations. It is called Erling’s lemma. Definition 43.1.14 Let E, W be Banach spaces such that E ⊆ W and the injection map from E into W is continuous. The injection map is said to be compact if every bounded set in E has compact closure in W. In other words, if a sequence is bounded in E it has a convergent subsequence converging in W . This is also referred to by saying that bounded sets in E are precompact in W. Theorem 43.1.15 Let E ⊆ W ⊆ X where the injection map is continuous from W to X and compact from E to W . Then for every ε > 0 there exists a constant, Cε such that for all u ∈ E, ||u||W ≤ ε ||u||E + Cε ||u||X Proof: Suppose not. Then there exists ε > 0 and for each n ∈ N, un such that ||un ||W > ε ||un ||E + n ||un ||X Now let vn = un / ||un ||E . Therefore, ||vn ||E = 1 and ||vn ||W > ε + n ||vn ||X It follows there exists a subsequence, still denoted by vn such that vn converges to v in W. However, the above inequality shows that ||vn ||X → 0. Therefore, v = 0. But then the above inequality would imply that ||vn || > ε and passing to the limit yields 0 > ε, a contradiction. Definition 43.1.16 Define C ([a, b] ; X) the space of functions continuous at every point of [a, b] having values in X. You should verify that this is a Banach space with norm © ª ||u||∞,X = max ||unk (t) − u (t)||X : t ∈ [a, b] . The following theorem is an infinite dimensional version of the Ascoli Arzela theorem.

1216


Theorem 43.1.17 Let q > 1 and let E ⊆ W ⊆ X where the injection map is continuous from W to X and compact from E to W . Let S be defined by n o u such that ||u (t)||E + ||u0 ||Lq ([a,b];X) ≤ R for all t ∈ [a, b] . Then S ⊆ C ([a, b] ; W ) and if {un } ⊆ S, there exists a subsequence, {unk } which converges to a function u ∈ C ([a, b] ; W ) in the following way. lim ||unk − u||∞,W = 0.

k→∞

Proof: First consider the issue of S being a subset of C ([a, b] ; W ) . By Theorem 43.1.9 on Page 1209 the following holds in X for u ∈ S. Z t u (t) − u (s) = u0 (r) dr. s

Thus S ⊆ C ([a, b] ; X) . Let ε > 0 be given. Then by Theorem 43.1.15 there exists a constant, Cε such that for all u ∈ W ε ||u||E + Cε ||u||X . ||u||W ≤ 4R Therefore, for all u ∈ S, ||u (t) − u (s)||W

≤ ≤ ≤

ε ||u (t) − u (s)||E + Cε ||u (t) − u (s)||X 6R ¯¯Z t ¯¯ ¯¯ ¯¯ ε 0 ¯ ¯ + C ε ¯¯ u (r) dr¯¯¯¯ 3 s X Z t ε ε 1/q + Cε ||u0 (r)||X dr ≤ + Cε R |t − s| .(43.1.8) 3 3 s

Since ε is arbitrary, it follows u ∈ C ([a, b] ; W ). ∞ Let D = Q ∩ [a, b] so D is a countable dense subset of [a, b]. Let D = {tn }n=1 . By compactness of the embedding of E into W, there exists a subsequence u(n,1) such that as n → ∞, u(n,1) (t1 ) converges to a point in W. Now take a subsequence of this, called (n, 2) such that as n → ∞, u(n,2) (t2 ) converges to a point in W. It follows that u(n,2) (t1 ) also converges to a point of W. Continue this way. Now consider the diagonal sequence, uk ≡ u(k,k) This sequence is a subsequence of u(n,l) whenever k > l. Therefore, uk (tj ) converges for all tj ∈ D. Claim: Let {uk } be as just defined, converging at every point of [a, b] . Then {uk } converges at every point of [a, b]. Proof of claim: Let ε > 0 be given. Let t ∈ [a, b] . Pick tm ∈ D ∩ [a, b] such that in 43.1.8 Cε R |t − tm | < ε/3. Then there exists N such that if l, n > N, then ||ul (tm ) − un (tm )||X < ε/3. It follows that for l, n > N, ||ul (t) − un (t)||X

≤ ≤

||ul (t) − ul (tm )|| + ||ul (tm ) − un (tm )|| + ||un (tm ) − un (t)|| 2ε ε 2ε + + < 2ε 3 3 3


1217

∞

Since ε was arbitrary, this shows {uk (t)}k=1 is a Cauchy sequence. Since W is complete, this shows this sequence converges. Now for t ∈ [a, b] , it was just shown that if ε > 0 there exists Nt such that if n, m > Nt , then ε ||un (t) − um (t)|| < . 3 Now let s 6= t. Then ||un (s) − um (s)|| ≤ ||un (s) − un (t)|| + ||un (t) − um (t)|| + ||um (t) − um (s)|| From 43.1.8 ||un (s) − um (s)|| ≤ 2

³ε 3

1/q

´

+ Cε R |t − s|

+ ||un (t) − um (t)||

and so it follows that if δ is sufficiently small and s ∈ B (t, δ) , then when n, m > Nt ||un (s) − um (s)|| < ε. p

Since [a, b] is compact, there are finitely many of these balls, {B (ti , δ)}i=1 , such that ©for s ∈ B (tiª, δ) and n, m > Nti , the above inequality holds. Let N > max Nt1 , · · · , Ntp . Then if m, n > N and s ∈ [a, b] is arbitrary, it follows the above inequality must hold. Therefore, this has shown the following claim. Claim: Let ε > 0 be given. Then there exists N such that if m, n > N, then ||un − um ||∞,W < ε. Now let u (t) = limk→∞ uk (t) . ||u (t) − u (s)||W ≤ ||u (t) − un (t)||W + ||un (t) − un (s)||W + ||un (s) − u (s)||W (43.1.9) Let N be in the above claim and fix n > N. Then ||u (t) − un (t)||W = lim ||um (t) − un (t)||W ≤ ε m→∞

and similarly, ||un (s) − u (s)||W ≤ ε. Then if |t − s| is small enough, 43.1.8 shows the middle term in 43.1.9 is also smaller than ε. Therefore, if |t − s| is small enough, ||u (t) − u (s)||W < 3ε. Thus u is continuous. Finally, let N be as in the above claim. Then letting m, n > N, it follows that for all t ∈ [a, b] , ||um (t) − un (t)|| < ε. Therefore, letting m → ∞, it follows that for all t ∈ [a, b] , ||u (t) − un (t)|| ≤ ε. and so ||u − un ||∞,W ≤ ε. Since ε is arbitrary, this proves the theorem. The next theorem is another such imbedding theorem. It is often used in partial differential equations.

1218


Theorem 43.1.18 Let E ⊆ W ⊆ X where the injection map is continuous from W to X and compact from E to W . Let p ≥ 1, let q > 1, and define S ≡ {u ∈ Lp ([a, b] ; E) : u0 ∈ Lq ([a, b] ; X) and ||u||Lp ([a,b];E) + ||u0 ||Lq ([a,b];X) ≤ R}. ∞

Then S is precompact in Lp ([a, b] ; W ). This means that if {un }n=1 ⊆ S, it has a subsequence {unk } which converges in Lp ([a, b] ; W ) . Proof: By Proposition 6.2.5 on Page 142 it suffices to show S has an η net in Lp ([a, b] ; W ) for each η > 0. If not, there exists η > 0 and a sequence {un } ⊆ S, such that ||un − um || ≥ η

(43.1.10)

for all n 6= m and the norm refers to Lp ([a, b] ; W ). Let a = t0 < t1 < · · · < tn = b, tk − tk−1 = T /k. Now define un (t) ≡

k X

uni X[ti−1 ,ti ) (t) , uni

i=1

1 ≡ ti − ti−1

Z

ti

un (s) ds. ti−1

The idea is to show that un approximates un well and then to argue that a subsequence of the {un } is a Cauchy sequence yielding a contradiction to 43.1.10. Therefore, Z ti k X 1 un (t) − un (t) = (un (t) − un (s)) dsX[ti−1 ,ti ) (t) . t − ti−1 ti−1 i=1 i It follows from Jensen’s inequality that p

||un (t) − un (t)||W ¯¯ ¯¯p Z ti k ¯¯ ¯¯ X 1 ¯¯ ¯¯ (un (t) − un (s)) ds¯¯ X[ti−1 ,ti ) (t) ¯¯ ¯¯ ti − ti−1 ti−1 ¯¯ i=1 W Z k ti X 1 p ||un (t) − un (s)||W dsX[ti−1 ,ti ) (t) t − t i i−1 t i−1 i=1

= ≤ and so

Z

b

a

Z ti 1 p ||un (t) − un (s)||W dsX[ti−1 ,ti ) (t) dt t − t i i−1 ti−1 a i=1 Z Z ti k t i X 1 p ||un (t) − un (s)||W dsdt. (43.1.11) t − t i i−1 t t i−1 i−1 i=1 Z

≤ =

p

||(un (t) − un (s))||W ds b

k X


1219

From Theorems 43.1.15 and 43.1.9, if ε > 0, there exists Cε such that p

p

p

||un (t) − un (s)||W ≤ ε ||un (t) − un (s)||E + Cε ||un (t) − un (s)||X ≤ ≤

¯¯Z t ¯¯p ¯¯ ¯¯ 0 ¯ ¯ 2 ε (||un (t)|| + ||un (s)|| ) + Cε ¯¯ un (r) dr¯¯¯¯ s X ¶p µZ t p p p−1 0 2 ε (||un (t)|| + ||un (s)|| ) + Cε ||un (r)||X dr p−1

p

p

p

p

p

p

s

≤

2p−1 ε (||un (t)|| + ||un (s)|| ) ÃµZ !p ¶1/q t q 1/q 0 0 +Cε ||un (r)||X dr |t − s| s

= 2

p−1

p/q 0

ε (||un (t)|| + ||un (s)|| ) + Cε Rp/q |t − s|

.

This is substituted in to 43.1.11 to obtain Z b p ||(un (t) − un (s))||W ds ≤ a

k X i=1

1 ti − ti−1

Z

ti ti−1

+Cε Rp/q |t − s| =

k X i=1

Z p

ti

2 ε ti−1

Z

p

b

= 2 ε

Z

p/q 0

ti

p

ti−1

dsdt + Cε R

||un (t)|| dt + Cε R

p/q

a

Z

p/q

k X i=1

b

p

2p−1 ε (||un (t)|| + ||un (s)|| )

´

p ||un (t)||W

p

¡

1 ti − ti−1

Z

ti

Z

ti

|t − s| ti−1

p/q 0

dsdt

ti−1

1 p/q 0 (ti − ti−1 ) (ti − ti−1 )

Z

ti

Z

ti

dsdt ti−1

ti−1

k X

1 p/q 0 2 (ti − ti−1 ) (ti − ti−1 ) (t − t ) i i−1 a i=1 µ ¶1+p/q0 k X T 1+p/q 0 p p p/q p p p/q ≤ 2 εR + Cε R (ti − ti−1 ) = 2 εR + Cε R k . k i=1 = 2p ε

p

||un (t)|| dt + Cε Rp/q

Taking ε so small that 2p εRp < η p /8p and then choosing k sufficiently large, it follows η ||un − un ||Lp ([a,b];W ) < . 4 Now use compactness of the embedding of E into W to obtain a subsequence such that {un } is Cauchy in Lp (a, b; W ) and use this to contradict 43.1.10. Suppose Pk un (t) = i=1 uni X[ti−1 ,ti ) (t) . Thus ||un (t)||E =

k X i=1

||uni ||E X[ti−1 ,ti ) (t)

1220


and so

Z

b

R≥ a

p

||un (t)||E dt =

k T X n p ||u || k i=1 i E

{uni }

Therefore, the are all bounded. It follows that after taking subsequences k times there exists a subsequence {unk } such that unk is a Cauchy sequence in Lp (a, b; W ) . You simply get a subsequence such that uni k is a Cauchy sequence in W for each i. Then denoting this subsequence by n, ||un − um ||Lp (a,b;W )

≤ ||un − un ||Lp (a,b;W ) + ||un − um ||Lp (a,b;W ) + ||um − um ||Lp (a,b;W ) η η ≤ + ||un − um ||Lp (a,b;W ) + < η 4 4

provided m, n are large enough, contradicting 43.1.10. This proves the theorem.

43.2

The K Method

This considers the problem of interpolating Banach spaces. The idea is to build a Banach space between two others in a systematic way, thus constructing a new Banach space from old ones. The first method of defining intermediate Banach spaces is called the K method. For more on this topic as well as the other topics on interpolation see [10] which is what I am following. See also [64]. There is far more on these subjects in these books than what I am presenting here! My goal is to present only enough to give an introduction to the topic and to use it in presenting more theory of Sobolev spaces. In what follows a topological vector space is a vector space in which vector addition and scalar multiplication are continuous. That is · : F × X → X is continuous and + : X × X → X is also continuous. A common example of a topological vector space is the dual space, X 0 of a Banach space, X with the weak ∗ topology. For S ⊆ X a finite set, define BS (x∗ , r) ≡ {y ∗ ∈ X 0 : |y ∗ (x) − x∗ (x)| < r for all x ∈ S} Then the BS (x∗ , r) for S a finite subset of X and r > 0 form a basis for the topology on X 0 called the weak ∗ topology. You can check that the vector space operations are continuous. Definition 43.2.1 Let A0 and A1 be two Banach spaces with norms ||·||0 and ||·||1 respectively, also written as ||·||A0 and ||·||A1 and let X be a topological vector space such that Ai ⊆ X for i = 1, 2, and the identity map from Ai to X is continuous. For each t > 0, define a norm on A0 + A1 by K (t, a) ≡ ||a||t ≡ inf {||a0 ||0 + t ||a1 ||1 : a0 + a1 = a}. This is short for K (t, a, A0 , A1 ) . Thus K (t, a, A1 , A0 ) will mean © ª K (t, a, A1 , A0 ) ≡ inf ||a1 ||A1 + t ||a0 ||A0 : a0 + a1 = a

43.2. THE K METHOD

1221

but the default is K (t, a, A0 , A1 ) if K (t, a) is written. The following lemma is an interesting exercise. Lemma 43.2.2 (A0 + A1 , K (t, ·)) is a Banach space and all the norms, K (t, ·) are equivalent. Proof: First, why is K (t, ·) a norm? It is clear that K (t, a) ≥ 0 and that if a = 0 then K (t, a) = 0. Is this the only way this can happen? Suppose K (t, a) = 0. Then there exist a0n ∈ A0 and a1n ∈ A1 such that ||a0n ||0 → 0, ||a1n ||1 → 0, and a = a0n + a1n . Since the embedding of Ai into X is continuous and since X is a topological vector space1 , it follows a = a0n + a1n → 0 and so a = 0. Let α be a nonzero scalar. Then K (t, αa) =

inf {||a0 ||0 + t ||a1 ||1 : a0 + a1 = αa} ¯¯ a ¯¯ a n ¯¯ a ¯¯ o a1 ¯¯ 0 ¯¯ ¯¯ 1 ¯¯ 0 = inf |α| ¯¯ ¯¯ + t |α| ¯¯ ¯¯ : + =a ¯¯ a ¯¯α 1a α a α o n¯¯ aα ¯¯0 ¯¯ 0 ¯¯ ¯¯ 1 ¯¯ 1 0 = |α| inf ¯¯ ¯¯ + t ¯¯ ¯¯ : + =a α 0 α 1 α α = |α| inf {||a0 ||0 + t ||a1 ||1 : a0 + a1 = a} = |α| K (t, a) .

It remains to verify the triangle inequality. Let ε > 0 be given. Then there exist a0 , a1 , b0 , and b1 in A0 , A1 , A0 , and A1 respectively such that a0 +a1 = a, b0 +b1 = b and ε + K (t, a) + K (t, b) > ||a0 ||0 + t ||a1 ||1 + ||b0 ||0 + t ||b1 ||1 ≥ ||a0 + b0 ||0 + t ||b1 + a1 ||1 ≥ K (t, a + b) . This has shown that K (t, ·) is at least a norm. Are all these norms equivalent? If 0 < s < t then it is clear that K (t, a) ≥ K (s, a) . To show there exists a constant, C such that CK (s, a) ≥ K (t, a) for all a, t K (s, a) ≡ s = = ≥

t inf {||a0 ||0 + s ||a1 ||1 : a0 + a1 = a} s ½ ¾ t t ||a0 ||0 + s ||a1 ||1 : a0 + a1 = a inf s s ½ ¾ t inf ||a0 ||0 + t ||a1 ||1 : a0 + a1 = a s inf {||a0 ||0 + t ||a1 ||1 : a0 + a1 = a} = K (t, a) .

Therefore, the two norms are equivalent as hoped. 1 Vector

addition is continuous is the property which is used here.

1222


Finally, it is required to verify that (A0 + A1 , K (t, ·)) is a Banach space. Since all these norms are equivalent, it suffices to only consider the norm, K (1, ·). Let ∞ {a0n + a1n }n=1 be a Cauchy sequence in A0 + A1 . Then for m, n large enough, K (1, a0n + a1n − (a0m + a1m )) < ε. It follows there exist xn ∈ A0 and yn ∈ A1 such that xn + yn = 0 for every n and whenever m, n are large enough, ||a0n + xn − (a0m + xm )||0 + ||a1n + yn − (a1m + ym )||1 < ε Hence {a1n + yn } is a Cauchy sequence in A1 and {a0n + xn } is a Cauchy sequence in A0 . Let a0n + xn a1n + yn

→ a 0 ∈ A0 → a 1 ∈ A1 .

Then K (1, a0n + a1n − (a0 + a1 ))

= K (1, a0n + xn + a1n + yn − (a0 + a1 )) ≤ ||a0n + xn − a0 ||0 + ||a1n + yn − a1 ||1

which converges to 0. Thus A0 + A1 is a Banach space as claimed. With this, there exists a method for constructing a Banach space which lies between A0 ∩ A1 and A0 + A1 . Definition 43.2.3 Let 1 ≤ q < ∞, 0 < θ < 1. Define (A0 , A1 )θ,q to be those elements of A0 + A1 , a, such that ·Z ||a||θ,q ≡

∞

0

¡ −θ ¢q dt t K (t, a, A0 , A1 ) t

¸1/q < ∞.

Theorem 43.2.4 (A0 , A1 )θ,q is a normed linear space satisfying A0 ∩ A1 ⊆ (A0 , A1 )θ,q ⊆ A0 + A1 ,

(43.2.12)

with the inclusion maps continuous, and ³ ´ (A0 , A1 )θ,q , ||·||θ,q is a Banach space.

(43.2.13)

If a ∈ A0 ∩ A1 , then µ ||a||θ,q ≤

1 qθ (1 − θ)

¶1/q

θ

1−θ

||a||1 ||a||0

.

If A0 ⊆ A1 with ||·||0 ≥ ||·||1 , then A0 ∩ A1 = A0 ⊆ (A0 , A1 )θ,q ⊆ A1 = A0 + A1 .

(43.2.14)

43.2. THE K METHOD

1223

Also, if bounded sets in A0 have compact closures in A1 then the same is true if A1 is replaced with (A0 , A1 )θ,q . Finally, if T ∈ L (A0 , B0 ) , T ∈ L (A1 , B1 ),

(43.2.15)

and T is a linear map from A0 + A1 to B0 + B1 where the Ai and Bi are Banach spaces with the properties described above, then it follows ³ ´ T ∈ L (A0 , A1 )θ,q , (B0 , B1 )θ,q (43.2.16) and if M is its norm, and M0 and M1 are the norms of T as a map in L (A0 , B0 ) and L (A1 , B1 ) respectively, then M ≤ M01−θ M1θ .

(43.2.17)

Proof: Suppose first a ∈ A0 ∩ A1 . Then Z r Z ∞ ¡ −θ ¢q dt ¡ −θ ¢q dt q ||a||θ,q ≡ t K (t, a) + t K (t, a) t t Z0 r Z r∞ ¡ −θ ¢q dt ¡ −θ ¢q dt ≤ t ||a||1 t + t ||a||0 t t 0 r Z r Z ∞ q q = ||a||1 tq(1−θ)−1 dt + ||a||0 t−1−θq dt 0

=

(43.2.18)

r

q−qθ −θq q r q r ||a||1 + ||a||0 0 and in particular for the value of r which minimizes the expression on the right in 43.2.19, r = ||a||0 / ||a||1 . Therefore, doing some calculus, q

||a||θ,q ≤

1 q(1−θ) qθ ||a||0 ||a||1 θq (1 + θ)

which shows 43.2.14. This also verifies that the first inclusion map is continuous in 43.2.12 because if an → 0 in A0 ∩ A1 , then an → 0 in A0 and in A1 and so the above shows an → 0 in (A0 , A1 )θ,q . Now consider the second inclusion in 43.2.12. The inclusion is obvious because (A0 , A1 )θ,q is given to be a subset of A0 + A1 defined by µZ 0

∞

¡

¢q dt t−θ K (t, a) t

¶1/q 0. (Recall all these norms K (t, ·) are equivalent.) Therefore, by Fatou’s lemma, µZ ∞ µZ ∞ ¶ ¶ ¡ −θ ¢q dt 1/q ¡ −θ ¢q dt 1/q t K (t, a) ≤ lim inf t K (t, an ) n→∞ t t 0 0 n o ≤ max ||an ||θ,q : n ∈ N < ∞ and so a ∈ (A0 , A1 )θ,q . Now µZ

||a − an ||θ,q

∞

¡ −θ ¢q dt ≤ lim inf t K (t, an − am ) m→∞ t 0 = lim inf ||an − am ||θ,q < ε

¶1/q

m→∞

whenever n is large enough. Thus (A0 , A1 )θ,q is complete as claimed. Next suppose A0 ⊆ A1 and the inclusion map is compact. In this case, A0 ∩A1 = A0 and so it has been shown above that A0 ⊆ (A0 , A1 )θ,q . It remains to show that every bounded subset, S, contained in A0 has an η net in (A0 , A1 )θ,q . Recall the inequality, 43.2.14 µ ¶1/q 1 θ 1−θ ||a||θ,q ≤ ||a||1 ||a||0 qθ (1 − θ) C θ 1−θ = ||a||1 ε ||a||0 . ε Now this implies µ ¶1/θ C ||a||θ,q ≤ θ ||a||1 + ε1/(1−θ) (1 − θ) ||a||0 ε By compactness of the embedding of A0 into A1 , it follows there exists an ε(1+θ)/θ net for S in A1 , {a1 , · · · , ap } . Then for a ∈ S, there exists k such that ||a − ak ||1 < ε(1+θ)/θ . It follows µ ¶1/θ C ||a − ak ||θ,q ≤ θ ||a − ak ||1 + ε1/(1−θ) (1 − θ) ||a − ak ||0 ε µ ¶1/θ C θε(1+θ)/θ + ε1/(1−θ) (1 − θ) 2M ≤ ε = C 1/θ θε + ε1/(1−θ) (1 − θ) 2M

43.3. THE J METHOD

1225

where M is large enough that ||a||0 ≤ M for all a ∈ S. Since ε is arbitrary, this shows the existence of a η net and proves the compactness of the embedding into (A0 , A1 )θ,q . It remains to verify the assertions 43.2.15-43.2.17. Let T ∈ L (A0 , B0 ) , T ∈ L (A1 , B1 ) with T a linear map from A0 + A1 to B0 + B1 . Let a ∈ (A0 , A1 )θ,q ⊆ A0 + A1 and consider T a ∈ B0 + B1 . Denote by K (t, ·) the norm described above for both A0 + A1 and B0 + B1 since this will cause no confusion. Then ¶ µZ ∞ ¡ −θ ¢q dt 1/q . (43.2.20) ||T a||θ,q ≡ t K (t, T a) t 0 Now let a0 + a1 = a and so T a0 + T a1 = T a K (t, T a)

≤ ||T a0 ||0 + t ||T a1 ||1 ≤ M0 ||a0 ||0 + M1 t ||a1 ||1 µ µ ¶ ¶ M1 ≤ M0 ||a0 ||0 + t ||a1 ||1 M0

and so, taking inf for all a0 + a1 = a, yields µ µ ¶ ¶ M1 K (t, T a) ≤ M0 K t ,a M0 It follows from 43.2.20 that µZ ∞ ¶ ¡ −θ ¢q dt 1/q ||T a||θ,q ≡ t K (t, T a) t 0 µZ ∞ µ µ µ ¶ ¶¶q ¶1/q M1 dt ≤ t−θ M0 K t ,a M t 0 0 µZ ∞ µ ¶ ¶¶q ¶1/q µ µ dt M1 = M0 ,a t−θ K t M t 0 0 ÃZ Ãµ !q !1/q ¶ −θ ∞ M0 ds = M0 s K (s, a) M s 1 0 µZ ∞ ¶ ¡ −θ ¢q ds 1/q (1−θ) (1−θ) = M1θ M0 s K (s, a) = M1θ M0 ||a||θ,q . s 0 ³ ´ This shows T ∈ L (A0 , A1 )θ,q , (B0 , B1 )θ,q and if M is the norm of T, M ≤ M01−θ M1θ as claimed. This proves the theorem.

43.3

The J Method

There is another method known as the J method. Instead of © ª K (t, a) ≡ inf ||a0 ||A0 + t ||a1 ||A1 : a0 + a1 = a

1226


for a ∈ A0 + A1 , this method considers a ∈ A0 ∩ A1 and J (t, a) defined below gives a norm on A0 ∩ A1 . Definition 43.3.1 For A0 and A1 Banach spaces as described above, and a ∈ A0 ∩ A1 , ¡ ¢ J (t, a) ≡ max ||a||A0 , t ||a||A1 . (43.3.21) this is short for J (t, a, A0 , A1 ). Thus

¡ ¢ J (t, a, A1 , A0 ) ≡ max ||a||A1 , t ||a||A0

but unless indicated otherwise, A0 will come first. Now for θ ∈ (0, 1) and q ≥ 1, define a space, (A0 , A1 )θ,q,J as follows. The space, (A0 , A1 )θ,q,J will consist of those elements, a, of A0 + A1 which can be written in the form Z ∞ Z 1 Z r dt dt dt ≡ lim + lim (43.3.22) a= u (t) u (t) u (t) r→∞ ε→0+ t t t 0 ε 1 the limits taking place in A0 + A1 with the norm ¡ ¢ K (1, a) ≡ inf ||a0 ||A0 + ||a1 ||A1 , a=a0 +a1

where u (t) is strongly measurable with values in A0 ∩ A1 and bounded on every compact subset of (0, ∞) such that µZ ∞ ¶ ¡ −θ ¢q dt 1/q t J (t, u (t) , A0 , A1 ) < ∞. (43.3.23) t 0 For such a ∈ A0 + A1 , define (µZ ¶ ) ∞¡ ¢q dt 1/q −θ ||a||θ,q,J ≡ inf t J (t, u (t) , A0 , A1 ) u t 0

(43.3.24)

where the infimum is taken over all u satisfying 43.3.22 and 43.3.23. Note that a norm on A0 × A1 would be ¡ ¢ ||(a0 , a1 )|| ≡ max ||a0 ||A0 , t ||a1 ||A1 and so J (t, ·) is the restriction of this norm to the subspace of A0 × A1 defined by {(a, a) : a ∈ A0 ∩ A1 }. Also for each t > 0 J (t, ·) is a norm on A0 ∩ A1 and furthermore, any two of these norms are equivalent. In fact, for 0 < t < s, ¢ ¡ J (t, a) = max ||a||A0 , t ||a||A1 ¡ ¢ ≥ max ||a||A0 , s ||a||A1 = J (s, a) ³s ´ ≥ max ||a||A0 , s ||a||A1 t ¡ ¢ s max ||a||A0 , t ||a||A1 = t s ≥ J (t, a) . t

43.3. THE J METHOD

1227

The following lemma is significant and follows immediately from the above definition. R∞ Lemma 43.3.2 Suppose a ∈ (A0 , A1 )θ,q,J and a = 0 u (t) dt t where u is described above. Then letting r > 1, ¢ ¡ ½ u (t) if t ∈ 1r , r ur (t) ≡ . 0 otherwise it follows that

Z

∞

dt ∈ A0 ∩ A1 . t 0 Rr Rr 1 Proof: The integral equals 1/r u (t) dt t . 1/r t dt = 2 ln r < ∞. Now ur is measurable in A0 ∩ A1 and bounded. Therefore, there exists a sequence of measurable simple functions, {sn } having values in A0 ∩ A1 which converges pointwise and uniformly to ur . It can also be assumed J (r, sn (t)) ≤ J (r, ur (t)) for all t ∈ [1/r, r]. Therefore, Z r dt = 0. lim J (r, sm − sn ) n,m→∞ 1/r t ur (t)

It follows from the definition of the Bochner integral that Z r Z r dt dt lim sn = ur ∈ A0 ∩ A1 . n→∞ 1/r t t 1/r This proves the lemma. The remarkable thing is that the two spaces, (A0 , A1 )θ,q and (A0 , A1 )θ,q,J coincide and have equivalent norms. The following important lemma, called the fundamental lemma of interpolation theory in [10] is used to prove this. This lemma is really incredible. Lemma 43.3.3 Suppose for a ∈ A0 +A1 , limt→0+ K (t, a) = 0 and limt→∞ 0. Then for any ε > 0, there is a representation, a=

∞ X i=−∞

ui =

lim

n,m→∞

n X

ui , ui ∈ A0 ∩ A1 ,

K(t,a) t

=

(43.3.25)

i=−m

the convergences taking place in A0 + A1 , such that ¡ ¢ ¡ ¢ J 2i , ui ≤ 3 (1 + ε) K 2i , a .

(43.3.26)

Proof: For each i, there exist a0,i ∈ A0 and a1,i ∈ A1 such that a = a0,i + a1,i , and

¡ ¢ (1 + ε) K 2i , a ≥ ||a0,i ||A0 + 2i ||a1,i ||A1 .

(43.3.27)

1228


This follows directly from the definition of K (t, a) . From the assumed limit conditions on K (t, a) , (43.3.28) lim ||a1,i ||A1 = 0, lim ||a0,i ||A0 = 0. i→−∞

i→∞

Then let ui ≡ a0,i − a0,i−1 = a1,i−1 − a1,i . The reason these are equal is a = a0,i + a1,i = a0,i−1 + a1,i−1 . Then n X

ui = a0,n − a0,−(m+1) = a1,−(m+1) − a1,n .

i=−m

¢ ¡ Pn It follows a − i=−m ui = a − a0,n − a0,−(m+1) = a0,−(m+1) + a1,n , and both terms converge to zero as m and n converge to ∞ by 43.3.28. Therefore, Ã ! n X ¯¯ ¯¯ K 1, a − ui ≤ ¯¯a0,−(m+1) ¯¯ + ||a1,n || i=−m

P∞

and so this shows a = i=−∞ ui which is one of the claims of the lemma. Also ¡ ¢ ¡ ¢ J 2i , ui ≡ max ||ui ||A0 , 2i ||ui ||A1 ≤ ||ui ||A0 + 2i ||ui ||A1 ³ ´ ≤2 ||a0,i−1 ||A +2i−1 ||a1,i−1 ||A 0

1

}| { z ≤ ||a0,i ||A0 + 2 ||a1,i ||A1 + ||a0,i−1 ||A0 + 2i ||a1,i−1 ||A1 ¡ ¢ ¡ ¢ ¡ ¢ ≤ (1 + ε) K 2i , a + 2 (1 + ε) K 2i−1 , a ≤ 3 (1 + ε) K 2i , a i

because t → K (t, a) is nondecreasing. This proves the lemma. ¢ ¡ Lemma 43.3.4 If a ∈ A0 ∩ A1 , then K (t, a) ≤ min 1, st J (s, a) . ¡ ¢ Proof: If s ≥ t, then min 1, st = st and so µ ¶ µ ¶ ¡ ¢ t t t min 1, J (s, a) = max ||a||A0 , s ||a||A1 ≥ s ||a||A1 s s s = t ||a||A1 ≥ K (t, a) . ¡ ¢ Now in case s < t, then min 1, st = 1 and so µ ¶ ¡ ¢ t min 1, J (s, a) = max ||a||A0 , s ||a||A1 ≥ ||a||A0 s ≥ K (t, a) . This proves the lemma. Theorem 43.3.5 Let A0 , A1 , K and J be as described above. Then for all q ≥ 1 and θ ∈ (0, 1) , (A0 , A1 )θ,q = (A0 , A1 )θ,q,J and furthermore, the norms are equivalent.

43.3. THE J METHOD

1229

Proof: Begin with a ∈ (A0 , A1 )θ,q . Thus Z q

||a||θ,q =

∞

¡ −θ ¢q dt t K (t, a) ||a0 ||A0 + t ||a1 ||A1 let s > t. Then ||a0 ||A0 + t ||a1 ||A1 ||a0 ||A0 + s ||a1 ||A1 K (t, a) + tε K (s, a) ≥ ≥ ≥ . t t s s Since ε is arbitrary, this proves the claim. . Is r = 0? Suppose to the contrary that r > 0. Then the Let r ≡ limt→∞ K(t,a) t integrand of 43.3.29, is at least as large as q−1

t−θq K (t, a)

K (t, a) q−1 ≥ t−θq K (t, a) r t

≥ t−θq (tr)

q−1

r ≥ rq tq(1−θ)−1

whose integral is infinite. Therefore, r = 0. P∞ Lemma 43.3.3, implies there exist ui ∈ A0 ∩ A1 such that a = i=−∞ ui , the convergence taking place in A0 + A1 with the inequality of that Lemma holding, ¡ ¢ ¡ ¢ J 2i , ui ≤ 3 (1 + ε) K 2i , a . For i an integer and t ∈ [2i−1 , 2i ), let u (t) ≡ ui / ln 2. Then a=

∞ X i=−∞

Z

∞

ui =

u (t) 0

dt . t

(43.3.30)

1230


Now Z

q ||a||θ,q,J

≤ = ≤ ≤

∞

¡ −θ ¢q dt t J (t, u (t)) t 0 i Z ∞ 2 ³ ³ X ui ´´q dt t−θ J t, ln 2 t i−1 i=−∞ 2 i µ ¶q X ∞ Z 2 ¡ −θ ¡ i ¢¢q dt 1 t J 2 , ui ln 2 i=−∞ 2i−1 t i µ ¶q X Z ∞ 2 ¡ ¡ ¢¢q dt 1 t−θ 3 (1 + ε) K 2i , a ln 2 i=−∞ 2i−1 t

K (2i ,a) Using the above claim, ≤ 2i fore, the above is no larger than

µ ≤ 2 µ

1 ln 2

¶q X ∞ Z i=−∞

K (2i−1 ,a) 2i−1

2i

¡

2i−1

¶q X ∞ Z

2i

¡ ¢ ¡ ¢ and so K 2i , a ≤ 2K 2i−1 , a . There-

¡ ¢¢q dt t−θ 3 (1 + ε) K 2i−1 , a t

¢q dt t−θ 3 (1 + ε) K (t, a) t i−1 i=−∞ 2 µ ¶q Z ∞ µ ¶q ¡ −θ ¢q dt 3 (1 + ε) 3 (1 + ε) q = 2 t K (t, a) ≡2 ||a||θ,q(43.3.31) . ln 2 t ln 2 0

≤ 2

1 ln 2

¡

This has shown that if a ∈ (A0 , A1 )θ,q , then by 43.3.30 and 43.3.31, a ∈ (A0 , A1 )θ,q,J and µ ¶q 3 (1 + ε) q q ||a||θ,q,J ≤ 2 ||a||θ,q . (43.3.32) ln 2 It remains to prove the other inclusion and norm inequality, both of which are much easier to obtain. Thus, let a ∈ (A0 , A1 )θ,q,J with Z

∞

a=

u (t) 0

dt t

(43.3.33)

where u is a strongly measurable function having values in A0 ∩ A1 and for which Z ∞ ¡ −θ ¢q t J (t, u (t)) dt < ∞. (43.3.34) 0

µ Z K (t, a) = K t,

∞ 0

ds u (s) s

¶

Z

∞

≤

K (t, u (s)) 0

ds . s

Now by Lemma 43.3.4, this is dominated by an expression of the form µ ¶ µ ¶ Z ∞ Z ∞ t ds 1 ds ≤ min 1, J (s, u (s)) = min 1, J (ts, u (ts)) s s s s 0 0

(43.3.35)

(43.3.36)

43.4. DUALITY AND INTERPOLATION

1231

where the equation follows from a change of variable. From Minkowski’s inequality and 43.3.36, µZ ||a||θ,q

¶ ¡ −θ ¢q dt 1/q t K (t, a) t 0 ¶ ¶q ¶1/q µZ ∞ µ µ Z ∞ ds 1 dt t−θ J (ts, u (ts)) min 1, s s t 0 0

≡ ≤ Z

∞

µZ

∞

∞

≤ 0

0

µ

µ ¶ ¶q ¶1/q 1 dt ds . t−θ min 1, J (ts, u (ts)) s t s

Now change the variable in the inside integral to obtain, letting t = τ s, Z

∞

≤ 0

¶ µZ ∞ ¶ µ ¡ −θ ¢q dt 1/q ds 1 min 1, t J (ts, u (ts)) s t s 0

µ ¶ µZ ∞ ¶ ¡ −θ ¢q dτ 1/q 1 θ ds min 1, s τ J (τ , u (τ )) s s τ 0 0 ¶ µZ ∞ ¶1/q µ ¡ −θ ¢q dτ 1 . = τ J (τ , u (τ )) (1 − θ) θ τ 0 Z

∞

=

This has shown that µ ||a||θ,q ≤

1 (1 − θ) θ

¶ µZ 0

∞

¡ −θ ¢q dτ τ J (τ , u (τ )) τ

¶1/q 0. Z

t

f (·, t) = f (0) + 0

f,xn (·, s) ds

But also, for a.e. x0 ,the following equation holds for a.e. t > 0. Z t f (x0 , t) = γf (x0 ) + f,xn (x0 , s) ds,

(45.1.1)

0

showing that ¡ ¢ 1 γf = f (0) ∈ W 1− p ,p Rn−1 ≡ T

µ ¶ 1 W 1,p (Ω) , Lp (Ω) , p, . p

¡ ¢ To see that 45.1.1 holds, approximate f with a sequence from C ∞ Rn+ and finally obtain an equation of the form ¸ Z Z ∞· Z t 0 0 0 f (x , t) − γf (x ) − f,xn (x , s) ds ψ (x0 , t) dtdx0 = 0, Rn−1

0

0

¡ ¢ which holds for all ψ ∈ Cc∞ Rn+ . This proves the lemma. Thus taking the trace on the boundary loses exactly p1 derivatives.

1260

TRACES OF SOBOLEV SPACES AND FRACTIONAL ORDER SPACES

45.2

A Right Inverse For The Trace For A Half Space

It is also important to show there is a continuous linear function, ¡ ¢ ¡ ¢ 1 R : W 1− p ,p Rn−1 → W 1,p Rn+ which has the property that γ (Rg) = g. Define this function as follows. µ 0 ¶ Z x − y0 1 0 g (y0 ) φ Rg (x0 , xn ) ≡ n−1 dy x n−1 x n n R

(45.2.2)

where φ is a mollifier having support in B (0, 1) . ¡ ¢ Lemma 45.2.1 Let R be defined in 45.2.2. Then Rg ∈ W 1,p Rn+ and is a contin¢ ¡ ¢ ¡ 1 uous linear map from W 1− p ,p Rn−1 to W 1,p Rn+ with the property that γRg = g. ¡ ¢ Proof: Let f ∈ W 1,p Rn+ be such that γf = g. Let ψ (xn ) ≡ (1 − xn )+ and assume f is Borel measurable by taking a Borel measurable representative. Then for a.e. x0 we have the following formula holding for a.e. xn . Rg (x0 , xn ) " Z Z 0 ψ (xn ) f (y , ψ (xn )) − = Rn−1

ψ(xn )

0

# µ ¶ x0 − y 0 0 (ψf ),n (y , t) dt φ x1−n dy 0 . n xn

Using the repeated index summation convention to save space, we obtain that in terms of weak derivatives, Rg,n (x0 , xn ) " Z Z 0 = ψ (xn ) f (y , ψ (xn )) − ·

Rn−1

µ

φ,k

x0 − y0 xn

¶µ

"

Z Rn−1

0

¶

µ

+φ Z

0

=

yk − xk xnn

0

µ 0

φ,k (z ) and so |Rg,n (x0 , xn )| ≤

yk − xk xnn

¶

# (ψf ),n (y0 , t) dt ·

x0 − y 0 xn

ψ(xn )

ψ (xn ) f (x − xn z , ψ (xn )) − ·

ψ(xn )

¶

¸ (1 − n) dy 0 xnn 0

0

#

(ψf ),n (x − xn z , t) dt ·

0

¸ (1 − n) n 0 zk + φ (z ) xn dz xnn 0

¯Z ¯ ¯ C (φ) ¯ [ψ (xn ) f (x0 − xn z0 , ψ (xn )) ¯ B(0,1) #¯ Z ψ(xn ) ¯ ¯ 0 0 − (ψf ),n (x − xn z , t) dt ¯ ¯ 0

45.2. A RIGHT INVERSE FOR THE TRACE FOR A HALF SPACE C (φ) xn−1 n Z +

≤

(Z |ψ (xn ) f (x0 + y0 , ψ (xn ))| dy 0 B(0,xn )

Z

B(0,xn )

Therefore, µZ ∞ Z 0

ÃZ

∞

Ã

Z

Ã

Z

+C (φ)

xn−1 n

Rn−1

0

0

≤ !p 0

0

|ψ (xn ) f (x + y , ψ (xn ))| dy

0

!1/p 0

dx dxn

B(0,xn )

Z

1

)

Z

1 xnn−1

Rn−1

0 ∞

0

¯ ¯ ¯ ¯ ¯(ψf ),n (x0 + y0 , t)¯ dtdy 0

|Rg,n (x , xn )| dx dxn

C (φ) ÃZ

ψ(xn )

¶1/p

p

0

Rn−1

1261

Z

B(0,xn )

0

ψ(xn )

¯ ¯ ¯ ¯ ¯(ψf ),n (x0 + y0 , t)¯ dtdy 0

!p

!1/p dx0 dxn

(45.2.3) Consider the first term on the right. We change variables, letting y0 = z0 xn . Then this term becomes ÃZ Z ÃZ !p !1/p 1

|ψ (xn ) f (x0 + xn z0 , ψ (xn ))| dz 0

C (φ) Rn−1

0

B(0,1)

µZ

Z

dx0 dxn

1

¶1/p

Z p

≤ C (φ) B(0,1)

Rn−1

0

|ψ (xn ) f (x0 + xn z0 , ψ (xn ))| dx0 dxn

dz 0

Now we change variables, letting t = ψ (xn ) . This yields µZ

Z

1

Z

= C (φ) B(0,1)

0

Rn−1

¶1/p p |tf (x0 + xn z0 , t)| dx0 dt dz 0 ≤ C (φ) ||f ||0,p,Rn . +

(45.2.4) Now we consider the second term on the right in 45.2.3. Using the same arguments which were used on the first term involving Minkowski’s inequality and changing the variables, we obtain the second term ¶1/p Z Z 1 µZ 1 Z ¯ ¯p ¯ ¯ 0 0 0 ≤ C (φ) (ψf ) (x + x z , t) dx dx dtdy 0 ¯ ¯ n n ,n B(0,1)

≤

0

0

Rn−1

C (φ) ||f ||1,p,Rn .

(45.2.5)

+

It is somewhat easier to verify that ||Rg,j ||0,p,Rn ≤ C (φ) ||f ||1,p,Rn . +

+

Therefore, we have shown that whenever γf = f (0) = g, ||Rg||1,p,Rn ≤ C (φ) ||f ||1,p,Rn . +

+

1262


Taking the infimum over all such f and using the definition of the norm in ¡ ¢ 1 W 1− p ,p Rn−1 , it follows ||Rg||1,p,Rn ≤ C (φ) ||g||1− 1 ,p,Rn−1 , +

p

showing that this map, R, is continuous as claimed. It is obvious that lim Rg (xn ) = g,

xn →0

¡ ¢ the convergence taking place in Lp Rn−1 because of general results about convolution with mollifiers. This proves the lemma.

45.3

Intrinsic Norms

The above presentation is very abstract, involving the trace of a function in W (A0 , A1 , p, θ) and a norm which was the infimum of norms of functions in W which have trace equal to the given function. It is very useful to have a description of the norm in these fractional order spaces which is defined in terms of the function itself rather than functions which have the given function as trace. This leads to something called an intrinsic norm. I am following Adams [1]. The following interesting lemma is called Young’s inequality. It holds more generally than stated. Lemma 45.3.1 Let g = f ∗ h where f ∈ L1 (R) , h ∈ Lp (R) , and f, h are all Borel measurable, p ≥ 1. Then g ∈ Lp (R) and ||g||Lp (R) ≤ ||f ||L1 (R) ||h||Lp (R) Proof: First of all it is good to show g is well defined. Using Minkowski’s inequality µZ µZ ¶p ¶1/p |h (t − s) f (s)| ds dt Z µZ p

≤

µZ

Z =

p

|h (t − s)| |f (s)| dt |f (s)|

= ||f ||L1 ||h||Lp

¶1/p ds

¶1/p |h (t − s)| dt ds p

45.3. INTRINSIC NORMS

1263

Therefore, for a.e. t, Z Z |h (t − s) f (s)| ds = |h (s) f (t − s)| ds < ∞ and so for all such t the convolution f ∗ h (t) makes sense. The above also shows ¯p ¶1/p µZ ¯Z ¯ ¯ ¯ ¯ ≤ ||f ||L1 ||h||Lp ≡ ¯ f (t − s) h (s) ds¯ dt

||g||Lp

and this proves the lemma. The following is a very interesting inequality of Hardy Littlewood and Pólya. Lemma 45.3.2 Let f be a real valued function defined a.e. on [0, ∞) and let α ∈ (−∞, 1) and Z 1 t g (t) = f (ξ) dξ (45.3.6) t 0 For 1 ≤ p < ∞ Z

∞

tαp |g (t)|

0

p

dt 1 ≤ p t (1 − α)

Z

∞

tαp |f (t)|

0

p

dt t

(45.3.7)

Proof: First it can be assumed the right side of 45.3.7 is finite since otherwise there is nothing to show. Changing the variables letting t = eτ , the above inequality takes the form Z ∞ Z ∞ 1 p p eτ pα |g (eτ )| dτ ≤ eτ pα |f (eτ )| dτ p (1 − α) −∞ −∞ Now from the definition of g it follows Z τ

g (e )

= e

−τ

eτ

f (ξ) dξ −∞ τ

Z =

e−τ

f (eσ ) eσ dσ

−∞

and so the left side equals Z ∞

τ p(α−1)

e −∞

¯Z ¯ ¯ ¯

τ −∞

¯p ¯ f (e ) e dσ ¯¯ dτ σ

σ

¯Z τ ¯p ¯ ¯ τα σ −(τ −σ) ¯ = e f (e ) e dσ ¯¯ dτ ¯ −∞ −∞ ¯p Z ∞ ¯Z τ ¯ ¯ (τ −σ)α −(τ −σ) σα σ ¯ = e e e f (e ) dσ ¯¯ dτ ¯ Z

∞

−∞

−∞

1264


Z

∞

= −∞

¯Z ¯ ¯ ¯

∞

−∞

X(−∞,0) (τ − σ) e

(τ −σ)(α−1) σα

e

¯p ¯ f (e ) dσ ¯¯ dτ σ

and by Lemma 45.3.1, µZ ≤ =

¶p Z ∞ p e(α−1)u du epσα |f (eσ )| dσ −∞ −∞ µ ¶p Z ∞ 1 p epσα |f (eσ )| dσ 1−α −∞ 0

which was to be shown. This proves the lemma. Next consider the case where G (t) , t > 0 is a continuous semigroup on A1 and A0 ≡ D (Λ) where Λ is the generator of this semigroup. Recall that from Proposition 15.12.5 on Page 469 Λ is a closed densely defined operator and so A0 is a Banach space if the norm is given by ||u||A0 ≡ ||u||A1 + ||Λu||A1 Also assume ||G (t)|| is uniformly bounded for t ∈ [0, ∞). I have in mind the case where A1 = Lp (Rn ) and G (t) u (x) = u (x + tei ) but it is notationally easier to discuss this in the general case. First here is a simple lemma. Lemma 45.3.3 Let A0 = D (Λ) as just described. Then for u ∈ A1 ||u||A1 +A0 = ||u||A1 Proof: D (Λ) ⊆ A1 . Now let u ∈ A1 . © ª ||u||A0 +A1 ≡ inf ||u0 ||A1 + ||Λu0 ||A1 + ||u1 ||A1 : u = u0 + u1 To make this as small as possible you should clearly take u1 = u because ||u0 ||A1 + ||Λu0 ||A1 + ||u1 ||A1

≥ =

||u0 + u1 ||A1 + ||Λu0 || ||u||A1 + ||Λu0 ||

Therefore, the result of the lemma follows. Lemma 45.3.4 Let Λ be the generator of G (t) and let t → g (t) be in C 1 (0, ∞; A1 ). Then there exists a unique solution to the initial value problem y 0 − Λy = g, y (0) = y0 ∈ D (Λ) and it is given by

Z

t

y (t) = G (t) y0 +

G (t − s) g (s) ds.

(45.3.8)

0

This solution is continuous having continuous derivative and has values in D (Λ).


1265

Proof: First R t I show the following claim. Claim: 0 G (t − s) g (s) ds ∈ D (Λ) and µZ

t

Λ

¶ Z t G (t − s) g (s) ds = G (t) g (0) − g (t) + G (t − s) g 0 (s) ds

0

0

Proof of the claim: µ ¶ Z t Z t 1 G (h) G (t − s) g (s) ds − G (t − s) g (s) ds h 0 0 µZ t ¶ Z t 1 = G (t − s + h) g (s) ds − G (t − s) g (s) ds h 0 0 ÃZ ! Z t t−h 1 G (t − s) g (s + h) ds − G (t − s) g (s) ds = h −h 0

=

1 h −

Z

Z

0

t−h

G (t − s) g (s + h) ds +

G (t − s)

−h

1 h

Z

0

g (s + h) − g (s) h

t

G (t − s) g (s) ds t−h

Using the estimate in Theorem 15.12.3 on Page 468 and the dominated convergence theorem the limit as h → 0 of the above equals Z

t

G (t) g (0) − g (t) +

G (t − s) g 0 (s) ds

0

which proves the claim. Since y0 ∈ D (Λ) , G (t) Λy0

= = =

G (h) y0 − y0 h G (t + h) − G (t) lim y0 h→0 h G (h) G (t) y0 − G (t) y0 lim h→0 h G (t) lim

h→0

(45.3.9)

Since this limit exists, the last limit in the above exists and equals ΛG (t) y0 and G (t) y0 ∈ D (Λ). Now consider 45.3.8. y (t + h) − y (t) G (t + h) − G (t) = y0 + h h

(45.3.10)

1266


1 h =

ÃZ

Z

t+h

G (t − s + h) g (s) ds − 0

!

t

G (t − s) g (s) ds 0

Z 1 t+h G (t + h) − G (t) y0 + G (t − s + h) g (s) ds h h t µ ¶ Z t Z t 1 + G (h) G (t − s) g (s) ds − G (t − s) g (s) ds h 0 0

From the claim and 45.3.9, 45.3.10 the limit of the right side is µZ t ¶ ΛG (t) y0 + g (t) + Λ G (t − s) g (s) ds 0 µ ¶ Z t = Λ G (t) y0 + G (t − s) g (s) ds + g (t) 0

Hence

y 0 (t) = Λy (t) + g (t)

and from the formula, y 0 is continuous since by the claim and 45.3.10 it also equals Z t G (t) Λy0 + g (t) + G (t) g (0) − g (t) + G (t − s) g 0 (s) ds 0

which is continuous. The claim and 45.3.10 also shows y (t) ∈ D (Λ). This proves the existence part of the lemma. It remains to prove the uniqueness part. It suffices to show that if y 0 − Λy = 0, y (0) = 0 and y is C 1 having values in D (Λ) , then y = 0. Suppose then that y is this way. Letting 0 < s < t, d (G (t − s) y (s)) ds y (s + h) − y (s) h→0 h G (t − s) y (s) − G (t − s − h) y (s) − h 0 provided the limit exists. Since y exists and y (s) ∈ D (Λ) , this equals ≡

lim G (t − s − h)

G (t − s) y 0 (s) − G (t − s) Λy (s) = 0. Let y ∗ ∈ A01 . This has shown that on the open interval (0, t) the function s → y ∗ (G (t − s) y (s)) has a derivative equal to 0. Also from continuity of G and y, this function is continuous on [0, t]. Therefore, it is constant on [0, t] by the mean value theorem. At s = 0, this function equals 0. Therefore, it equals 0 on [0, t]. Thus for fixed s > 0 and letting t > s, y ∗ (G (t − s) y (s)) = 0. Now let t decrease toward s. Then y ∗ (y (s)) = 0 and since y ∗ was arbitrary, it follows y (s) = 0. This proves uniqueness.


1267

Definition 45.3.5 Let G (t) be a uniformly bounded continuous semigroup defined on A1 and let Λ be its generator. Let the norm on D (Λ) be given by ||u||D(Λ) ≡ ||u||A1 + ||Λu||A1 so that by Lemma 45.3.3 the norm on A1 + D (Λ) is just ||·||A1 . Let ( T0 ≡

Z

u ∈ A1 :

p ||u||A1

∞

θp

+

t 0

) ¯¯ ¯¯ ¯¯ G (t) u − u ¯¯p dt p ¯¯ ¯¯ ≡ ||u||T0 < ∞ ¯¯ ¯¯ t A1 t

Theorem 45.3.6 T0 = T (D (Λ) , A1 , p, θ) ≡ T and the two norms are equivalent. Proof: Take u ∈ T (D (Λ) , A1 , p, θ) . I will show ||u||T0 ≤ C (θ, p) ||u||T . By the definition of the norm in T, there exists f ∈ W (D (Λ) , A1 , p, θ) such that p

p

||u||T + δ > ||f ||W , f (0) = u. Now by Lemma 44.1.4 there exists gr ∈ W such that ||gr − f ||W < r and gr ∈ C ∞ (0, ∞; D (Λ)) and gr0 ∈ C ∞ (0, ∞; A1 ). Thus for each ε > 0, gr (ε) ∈ D (Λ) although possibly gr (0) ∈ / D (Λ) . Then letting hr (t) be defined by gr0 (t) − Λgr (t) = hr (t) it follows hr ∈ C 1 (0, ∞; A1 ) and applying Lemma 45.3.4 on [ε, ∞) it follows Z

t

gr (t) = G (t − ε) gr (ε) +

G (t − s) hr (s) ds.

(45.3.11)

ε

By Lemma 44.1.4 again, gr (ε) converges to gr (0) in A1 . Thus Z ε

t

||G (t − s) hr (s)||A1 ds ≤ C

for some constant independent of ε. Thus s → G (t − s) hr (s) is in L1 (0, t; A1 ) and it is possible to pass to the limit in 45.3.11 as ε → 0 to conclude Z

t

gr (t) = G (t) gr (0) +

G (t − s) hr (s) ds 0

Now

G (t) gr (0) − gr (0) 1 = t t

Z 0

t

gr0 (s) ds −

1 t

Z

t

G (t − s) hr (s) ds 0

and so using the assumption that G (t) is uniformly bounded, ¯¯ ¯¯ Z ¯¯ G (t) gr (0) − gr (0) ¯¯ 1 t 0 ¯¯ ¯¯ ≤ ||gr ||A1 + M ||hr ||A1 ¯¯ ¯¯ t t 0

1268


≤ ≤

Z 1 t 0 ||gr || (M + 1) + M ||Λgr || ds t 0 Z M +1 t 0 ||gr ||A1 + ||gr ||D(Λ) ds t 0

Therefore, from Lemma 45.3.2 Z ∞ p dt tpθ−p ||G (t) gr (0) − gr (0)||A1 t 0 ¯¯ ¯¯ ¯¯ G (t) gr (0) − gr (0) ¯¯p dt ¯¯ ¯ ¯ = t ¯¯ ¯¯ t 0 A1 t ¯ ¯p Z ∞ Z t ¯ ¯ pθ ¯ M + 1 0 ≤ t ¯ ||gr ||A1 + ||gr ||D(Λ) ds¯¯ dt/t t 0 0 µ ¶p Z ∞ ³ ´ 1 p p p ≤ (M + 1) 2p−1 tpθ ||gr0 ||A1 + ||gr ||D(Λ) 1−θ 0 Z

∞

pθ

Now since gr → f in W, it follows from Lemma 44.1.8 that gr (0) → u in T and hence by Theorem 44.1.9 this also in A1 . Therefore, using Fatou’s lemma in the above along with the convergence of gr to f , Z ∞ p dt tpθ−p ||G (t) u − u||A1 t 0 µ ¶p Z ∞ ³ ´ 1 p p p ≤ (M + 1) 2p−1 tpθ ||f 0 ||A1 + ||f ||D(Λ) 1−θ 0 ¶p µ 1 p p (||u||T + δ) ≤ (M + 1) 2p−1 1−θ Since u ∈ T, Theorem 44.1.9 implies u ∈ A1 and ||u||A1 ≤ C ||u||T . Therefore, since δ was arbitrary, this has shown that u ∈ T0 and ||u||T0 ≤ C (θ, p) ||u||T . This shows T ⊆ T0 with continuous inclusion. Now it is necessary to take u ∈ T0 and show it is in T. Since u ∈ T0 ¯¯ ¯¯ Z ∞ ¯¯ G (t) u − u ¯¯p dt p p ¯¯ ∞ > ||u||A1 + tθp ¯¯¯¯ ¯¯ t ≡ ||u||T0 t 0 Let φ be a nonnegative decreasing infinitely differentiable function such that φ (0) = 1 and φ (t) = 0 for all t > 1. Then define f (t) ≡ φ (t)

1 t

Z

t

G (τ ) udτ . 0


1269

It is easy to see that f (t) ∈ D (Λ) . In fact, changing variables as needed, 1 h

µ ¶ Z t Z t G (h) G (τ ) udτ − G (τ ) udτ 0

= =

1 h 1 h

0

Z

t+h

Z

h t+h

1 G (τ ) udτ − h G (τ ) udτ −

t

1 h

Z

t

G (τ ) udτ 0

Z

h

G (τ ) udτ 0

which converges to G (t) u − u and so Z

t

G (τ ) udτ = G (t) u − u.

Λ

(45.3.12)

0

Thus Z

∞

t

pθ

0

p ||Λf ||A1

dt t

Z

∞

≤

t

pθ

0

¯¯ ¯¯ ¯¯ G (t) u − u ¯¯p dt ¯¯ ¯¯ ¯¯ ¯¯ t A1 t

p

≤ ||u||T0 Next it is necessary to consider Z

∞

0

p

tpθ ||f 0 ||A1

dt . t

Z 1 t f (t) = φ (t) G (τ ) udτ + t 0 µ ¶ Z 1 t 1 φ (t) − 2 G (τ ) udτ + G (t) u t 0 t µ Z t ¶ Z t 1 1 = φ0 (t) G (τ ) udτ + φ (t) 2 (G (t) u − G (τ ) u) dτ t 0 t 0 0

0

and so there is a constant C depending on φ and the uniform bound on ||G (t)|| such that µ ¶ Z 1 t 0 ||f (t)||A1 ≤ CX[0,1] (t) ||u||A1 + 2 ||G (t − τ ) u − u|| dτ t 0 µ ¶ Z 1 t = CX[0,1] (t) ||u||A1 + 2 ||G (τ ) u − u|| dτ t 0 Now

Z 0

∞

p

p

X[0,1] (t) tpθ ||u||A1 dt/t ≤ C (p, θ) ||u||T0

1270


and using Lemma 45.3.2, ¯p ¯ Z t Z ∞ ¯1 ¯ dt X[0,1] (t) ¯¯ 2 ||G (τ ) u − u|| dτ ¯¯ tpθ t t 0 0 ¯ ¯ Z ∞ Z t ¯p p(θ−1) dt ¯1 ¯ t ¯ ||G (τ ) u − u|| dτ ≤ ¯ ¯t t 0 0 Z ∞ 1 dt p ≤ ||G (τ ) u − u|| tp(θ−1) p t (1 − (θ − 1)) 0 ¯¯p Z ∞ ¯¯ ¯¯ G (τ ) u − u ¯¯ pθ dt 1 p ¯¯ t ¯¯ = ≤ C (θ, p) ||u||T0 p ¯¯ t t (2 − θ) 0 ¯¯ This proves the theorem. Of course the case of most interest here is where A1 = Lp (Rn ) and G (t) u (x) ≡ u (x + tei ) Thus Λu = ∂u/∂xi , the weak derivative. The trace space T (D (Λ) , Lp (Rn ) , p, 1 − θ) then is a space of functions in Lp (Rn ) which have a fractional order partial derivative with respect to xi . Recall from Definition 45.1.1 that for θ ∈ (0, 1) , ¡ ¢ W θ,p (Rn ) ≡ T W 1,p (Rn ) , Lp (Rn ) , p, 1 − θ ¡ ¢ Let f ∈ W W 1,p (Rn ) , Lp (Rn ) , p, 1 − θ . Then µZ ∞ ¶ Z ∞ dt p dt p (1−θ)p (1−θ)p 0 ||f ||W ≡ max t ||f (t)||W 1,p , t ||f (t)||Lp t 0 t 0 Letting Gi (t) u (x) ≡ u (x + tei ) and Λi its generator, W 1,p (Ω) = ∩ni=1 DΛi ∩ Lp (Rn ) with the norm given by p

p

||u|| = ||u||Lp +

n X

p

||Λi u||Lp

i=1

which is equivalent to the norm p

||u|| =

n X

p

||u||D(Λi ) .

i=1

Then by considering each of the Gi and repeating the above argument in Theorem 45.3.6, it follows an equivalent intrinsic norm is ¯¯ ¯¯p n Z ∞ X ¯¯ ¯¯ dt p p (1−θ)p ¯¯ Gi (t) u − u ¯¯ ||u||W θ,p (Rn ) = ||u||Lp (Rn ) + t ¯¯ ¯¯ p t t L i=1 0


1271

p ||u||Lp (Rn )

=

+

n Z X i=1

∞

t

(1−θ)p

0

¯¯ ¯¯ ¯¯ u (· + tei ) − u (·) ¯¯p dt ¯¯ ¯¯ ¯¯ ¯¯ p t t L

(45.3.13)

and u ∈ W θ,p (Rn ) when this norm is finite. The only new detail is that in showing that for u ∈ T0 it follows it is in T, you use the function Z Z t 1 t f (t) ≡ φ (t) n ··· G1 (τ 1 ) G2 (τ 2 ) · · · Gn (τ n ) udτ 1 · · · dτ n t 0 0 and the fact that these semigroups commute. To get this started, note that Z t Z t g (t) ≡ ··· G1 (τ 1 ) G2 (τ 2 ) · · · Gn (τ n ) udτ 1 · · · dτ n ∈ D (Λi ) 0

0

for each i. This follows from writing it as Z t Gi (τ i ) (wi ) dτ i 0 p

for wi ∈ L coming from the other integrals and then repeating the earlier argument to get Λi g (t) = Gi (t) wi − wi and then

Z

∞

0

p

tp(1−θ) ||Λi f ||Lp

dt t

¯¯ ¯¯ ¯¯ Gi (t) wi − wi ¯¯p ¯ ¯¯ ¯ t ¯¯ ¯¯ p t 0 ¯¯L ¯¯ Z ∞ ¯¯p ¯ ¯ G (t) u − u i ¯¯ C tp(1−θ) ¯¯¯¯ ¯¯ p t Z

≤ ≤

∞

p(1−θ)

0

L

dt t dt p ≤ C ||u||T0 t

Thus all is well as far as f is concerned and the proof will work as it did earlier in Theorem 45.3.6. What about f 0 ? As before, the only term which is problematic is µ φ (t)

1 tn

Z

Z

t

0

¶0

t

···

G1 (τ 1 ) G2 (τ 2 ) · · · Gn (τ n ) udτ 1 · · · dτ n 0

After enough massaging, it becomes Z n Y X 1 i=1 j6=i

t

t

Gj (τ j ) dτ j 0

1 t2

Z

t

(Gi (t) u − Gi (τ i ) u) dτ i 0

Rt Pn Q where the operator i=1 j6=i 1t 0 Gj (τ j ) dτ j is bounded. Thus similar arguments to those of Theorem 45.3.6 will work, the only difference being a sum.

1272


Theorem 45.3.7 An equivalent norm for W θ,p (Rn ) for θ ∈ (0, 1) is p

||u||W θ,p (Rn ) = ¯¯ ¯¯ ¯¯ Gi (t) u − u ¯¯p dt ¯ ¯¯ ¯ + t ¯¯ p t ¯¯ t 0 L i=1 ¯¯p ¯ ¯ Z n ∞ X ¯¯ dt ¯¯ p (1−θ)p ¯¯ u (· + tei ) − u (·) ¯¯ ||u||Lp (Rn ) + t ¯¯ p t ¯¯ t L i=1 0 p ||u||Lp (Rn )

=

n Z X

∞

(1−θ)p

(45.3.14)

Note it is obvious from 45.3.13 that a Lipschitz map takes W θ,p (Rn ) to W θ,p (Rn ) and is continuous. The above description in Theorem 45.3.7 also makes possible the following corollary. Corollary 45.3.8 W θ,p (Rn ) is reflexive. Proof: Let u ∈ W θ,p (Rn ). For each i = 1, 2, · · · , n, define for t > 0, u (x + tei ) − u (x) t

∆i u (t) (x) ≡ Then by Theorem 45.3.7,

∆i u ∈ Lp ((0, ∞) ; Lp (Rn ) , µ) ≡ Y where

Z t(1−θ)p t−1 dt.

µ (E) ≡ E

Clearly the measure space is σ finite and so Y is reflexive by Corollary 23.8.9 on Page 745. Also ∆i is a closed operator whose domain is W θ,p (Rn ). To see this, suppose un ∈ W θ,p (Rn ) and un → u in Lp (Rn ) while ∆i un → g in Y. Then in particular ||∆i un ||Y is bounded. Now by Fatou’s lemma, ¯¯ ¯¯ Z ∞ ¯¯ u (· + tei ) − u (·) ¯¯p dt ¯¯ t(1−θ)p ¯¯¯¯ ≤ ¯ ¯ t 0 Lp (Rn ) t Z

∞

lim inf

n→∞

0

¯¯ ¯¯ ¯¯ un (· + tei ) − un (·) ¯¯p dt ¯¯ t(1−θ)p ¯¯¯¯ ¯¯ p n t < ∞. t L (R )

~ ≡ (∆1 , ∆2 , · · · , ∆n ) , it follows from similar reasoning that ∆ ~ is a Letting ∆ closed operator mapping W θ,p (Rn ) to Y n . Therefore ³ ´¡ ¢ ~ W θ,p (Rn ) ⊆ Lp (Rn ) × Y n id, ∆ and is a closed subspace of the reflexive space Lp (Rn ) × Y n . With the norm in Lp (Rn ) × Y n given as the sum of the norms of the components, it follows the


1273

³ ´ ~ is a norm preserving isomorphism between W θ,p (Rn ) and this mapping id, ∆ closed subspace of Lp (Rn ) × Y n . Since Lp (Rn ) and³Y is reflexive, their product is ´¡ ¢ ~ W θ,p (Rn ) and hence reflexive. By Lemma 23.2.7 on Page 710 it follows id, ∆ W θ,p (Rn ) is reflexive. This proves the theorem. One can generalize this to find an intrinsic norm for W θ,p (Ω). The version given above will not do because it requires the function to be defined on all of Rn in order to make sense of the shift operators Gi . However, you can give a different version of this intrinsic norm which will make sense for Ω 6= Rn . Lemma 45.3.9 Let t 6= 0 be a number. Then there is a constant C (n, θ, p) depending on the indicated quantities such that Z 1 C (n, θ, p) ds = 1 ³ ´ 1+pθ (n+pθ) |t| 2 2 Rn−1 t2 + |s| Proof: Change the integral to polar coordinates. Thus the integral equals Z Z ∞ ρn−2 dρdσ 1 (n+pθ) S n−1 0 (t2 + ρ2 ) 2 Now change the variables, ρ = |t| u. Then the above integral becomes Z

n−2

∞

|t|

Cn

|t|

0

Z

1

= Cn

|t|

1+pθ

n+pθ

∞

1

(1 + u2 ) 2

(n+pθ)

un−2 (1 +

0

un−2 |t|

1

(n+pθ) u2 ) 2

du ≡

du C (n, θ, p) 1+pθ

|t|

.

This proves the lemma. p Now let u ∈ W θ,p (Rn ) . This means the norm of ||u||W θ,p can be taken as p

¯ ¯ ¯ u (x + tei ) − u (x) ¯p dt ¯ ¯ dx ¯ ¯ t |t| Rn i=1 0 ¯ ¯ Z Z n p ¯ u (x + tei ) − u (x) ¯ 1 X ∞ (1−θ)p ¯ ¯ dx dt + |t| ¯ ¯ 2 i=1 −∞ t |t| n R

||u||Lp + =

p

||u||Lp

n Z X

Z

∞

|t|

(1−θ)p

That integral over Rn can be massaged and one obtains the above equal to p

||u||Lp + n

1X 2 i=1

Z

∞

−∞

Z

Z Rn−i

Ri

1 t1+pθ

|u (x1 , · · · , xi + t, yi+1 , · · · , yn ) p

− u (x1 , · · · , xi , yi+1 , · · · , yn )| dxi dyn−i dt

1274


where dxi refers to the first i entries and dyn−i refers to the remaining entries. From Lemma 45.3.9, the complicated expression above equals Z Z Z n Z 1 1 1X ∞ ´ 1 (n+pθ) C (n, θ, p) 2 i=1 −∞ Rn−i Ri Rn−1 ³ 2 2 2 t + |s| |u (x1 , · · · , xi−1 , xi + t, yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| dsdxi dyn−i dt

p

Now Fubini this to get n

1 1X C (n, θ, p) 2 i=1

Z Rn−i

Z

Z

Ri

Z

Rn−1

∞

−∞

³ t2 + |s|

1 ´ 12 (n+pθ)

2

|u (x1 , · · · , xi−1 , xi + t, yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| dtdsdxi dyn−i

p

Changing the variable in the inside integral to t = yi − xi ,this equals Z Z Z ∞ n Z 1 1 1X ´ 1 (n+pθ) C (n, θ, p) 2 i=1 Rn−i Ri Rn−1 −∞ ³ 2 2 2 (yi − xi ) + |s| |u (x1 , · · · , xi−1 , yi , yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| dyi dsdxi dyn−i

p

Next let ≡

(s1 , · · · , sn−1 ) (y1 − x1 , · · · , yi−1 − xi−1 , xi+1 − yi+1 , · · · , xn − yn )

where the new variables of integration in the integral corresponding to ds are y1 , · · · , yi−1 and xi+1 , · · · , xn . Then changing the variables, the above reduces to Z Z Z ∞ n Z 1 1X 1 C (n, θ, p) 2 i=1 Rn−i Ri Rn−1 −∞ |x − y|(n+pθ) |u (x1 , · · · , xi−1 , yi , yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| dyi dy1 · · · dyi−1 dxi+1 · · · dxn dx1 · · · dxi dyi+1 · · · dyn

p

Then if you Fubini again, it reduces to the expression n

1 1X C (n, θ, p) 2 i=1 Z Rn

Z Rn

|u (x1 , · · · , xi−1 , yi , yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| |x − y|

(n+pθ)

p

dxdy


1275

Now taking the sum inside and adjusting the constants yields Z Z p |u (y) − u (x)| ≥ C (n, θ, p) dxdy (n+pθ) Rn Rn |x − y| Thus there exists a constant C (n, θ, p) such that Ã

Z

p ||u||Lp

||u||W θ,p (Rn ) ≥ C (n, θ, p)

Z

|u (y) − u (x)|

+ Rn

|x − y|

Rn

(n+pθ)

!1/p

p

dxdy

.

Next start with the right side of the above. It suffices to consider only the complicated term. First note that for a a vector, Ã

!p/2

n X

p

a2i

≥ |ai |

i=1

and so n X

Ã p

|ai | ≤ n

i=1

from which it follows

Then it follows Pn i=1

!p/2

n X

a2i

= n |a|

p

i=1 n

1X p p |ai | ≤ |a| n i=1

Z Rn

Z Rn

p

|u (y) − u (x)| |x − y|

(n+pθ)

dxdy ≥

1 n

Z

Z Rn

Rn

|u (x1 , · · · , xi−1 , yi , yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| |x − y|

(n+pθ)

p

dxdy (45.3.15)

Consider the ith term. By Fubini’s theorem it equals Z Z 1 1 ··· ³ ´ 1 (n+pθ) P n 2 2 2 (yi − xi ) + j6=i (yj − xj ) |u (x1 , · · · , xi−1 , yi , yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| dyi dx1 · · · dxi−1 dyi+1 · · · dyn dy1 · · · dyi−1 dxi · · · dxn

p

Let t = yi − xi . Then it reduces to Z Z 1 1 ··· ³ ´ 1 (n+pθ) P n 2 2 t2 + j6=i (yj − xj ) |u (x1 , · · · , xi−1 , xi + t, yi+1 , · · · , yn ) − u (x1 , · · · , xi , yi+1 , · · · , yn )| dtdx1 · · · dxi−1 dyi+1 · · · dyn dy1 · · · dyi−1 dxi · · · dxn

p

1276


Now let (s1 , · · · , sn−1 ) = (x1 − y1 , · · · , xi−1 − yi−1 , yi+1 − xi+1 , · · · , yn − xn ) on the next n−1 iterated integrals. Then using Fubini’s theorem again and changing the variables, it equals Z Z 1 1 ··· ³ ´ 1 (n+pθ) · n 2 2 t2 + |s| |u (s1 + y1 , · · · , yi−1 + si−1 , xi + t, xi+1 + si , · · · , xn + sn−1 ) − u (s1 + y1 , · · · , yi−1 + si−1 , xi , xi+1 + si , · · · , xn + sn−1 )| dy1 · · · dyi−1 dxi · · · dxn ds1 · · · dsi−1 dsi · · · dsn−1 dt

p

By translation invariance of the measure, the inside integrals corresponding to dy1 · · · dyi−1 dxi · · · dxn simplify and the expression can be written as Z Z 1 1 ··· ³ ´ 1 (n+pθ) · n 2 2 t2 + |s| p

|u (x1 , · · · , xi−1 , xi + t, xi+1 , · · · , xn ) − u (x1 , · · · , xi−1 , xi , xi+1 , · · · , xn )| dx1 · · · dxi−1 dxi · · · dxn ds1 · · · dsi−1 dsi · · · dsn−1 dt where I just renamed the variables. Use Fubini’s theorem again to get Z Z 1 1 ··· ³ ´ 1 (n+pθ) · n 2 2 t2 + |s|

p

|u (x1 , · · · , xi−1 , xi + t, xi+1 , · · · , xn ) − u (x1 , · · · , xi−1 , xi , xi+1 , · · · , xn )| ds1 · · · dsi−1 dsi · · · dsn−1 dx1 · · · dxi−1 dxi · · · dxn dt

Now from Lemma 45.3.9, the inside n−1 integrals corresponding to ds1 · · · dsi−1 dsi · · · dsn−1 can be replaced with C (n, θ, p) 1+pθ

|t| and this yields Z R

=

1 C (n, θ, p) 2

Z

1

C (n, θ, p) Z

|t|

0

p

pθ

∞

Rn

|u (x + tei ) − u (x)| dx

dt t

p

tp(1−θ)

||u (· + tei ) − u (·)||Lp (Rn ) dt tp

t


1277

Applying this to each term of the sum in 45.3.15 and adjusting the constant, it follows Z Z p |u (y) − u (x)| dxdy ≥ (n+pθ) Rn Rn |x − y| p n Z ∞ X ||u (· + tei ) − u (·)||Lp (Rn ) dt tp(1−θ) C (n, θ, p) tp t i=1 0 Therefore, ||u||W θ,p (Rn ) Ã ≥

C (n, θ, p)

Z p ||u||Lp (Rn )

Z

|u (y) − u (x)|

+ Rn

(n+pθ)

|x − y|

Rn

!1/p

p

dxdy

This has proved most of the following theorem about the intrinsic norm. Theorem 45.3.10 An equivalent norm for W θ,p (Rn ) is ||u|| = Ã

Z p ||u||Lp (Rn )

Z

+ Rn

!1/p

p

|u (y) − u (x)|

Rn

|x − y|

dxdy

(n+pθ)

.

Also for any open subset of Rn ||u|| = Ã

Z Z p ||u||Lp (Ω)

+ Ω

Ω

!1/p

p

|u (y) − u (x)| |x − y|

(n+pθ)

dxdy

.

(45.3.16)

is a norm. Proof: It only remains to verify this is a norm. Recall the lp norm on R2 given by p 1/p

p

|(x, y)|lp ≡ (|x| + |y| ) For u, v ∈ W θ,p denote by ρ (u) the expression ÃZ Z Ω

|u (y) − u (x)| Ω

(n+pθ)

|x − y|

!1/p

p

dxdy

a similar definition holding for v. Then it follows from the usual Minkowski inequality that ρ (u + v) ≤ ρ (u) + ρ (v). Then from 45.3.16 p

p 1/p

||u + v|| = (||u + v||Lp + ρ (u + v) ) p

p 1/p

≤ ((||u||Lp + ||v||Lp ) + (ρ (u) + ρ (v)) )

1278


= |(||u||Lp , ρ (u)) + (||v||Lp , ρ (v))|lp ≤ |(||u||Lp , ρ (u))|lp + |(||v||Lp , ρ (v))|lp p 1/p

p

= (||u||Lp + ρ (u) )

p

p 1/p

+ (||v||Lp + ρ (v) )

= ||u|| + ||v|| The other properties of a norm are obvious. This proves the theorem. As pointed out in the above theorem, this is a norm in 45.3.16. One could define a set of functions for which this norm is finite. In the case where Ω = Rn the conclusion of Theorem 45.3.10 is that this space of functions is the same as W θ,p (Rn ) and the norms are equivalent. Does this happen for other open subsets of Rn ? θ,p (U ) the functions in Lp (U ) for which the Definition 45.3.11 Denote by W^ norm of Theorem 45.3.10 is finite. Here θ ∈ (0, 1) .

Proposition 45.3.12 Let U be a bounded open set which has Lipschitz boundary ³ ´ ^ θ,p and θ ∈ (0, 1). Then for each p ≥ 1, there exists E ∈ L W (U ), W θ,p (Rn ) such that Eu (x) = u (x) a.e. x ∈ U. Proof: In proving this, I will use the equivalent norm of Theorem 45.3.10 as the norm of W θ,p (Rn ) Consider the following picture. spt(u) b ¡ U+ ¡ ¡ ¡ ¡

a

U ∩ B × (a, b) b r0 B

The wavy line signifies a part of the boundary of U and spt (u) is contained in the circle as shown. It is drawn as a circle but this is not important. Denote by U + the region above the part of the boundary which is shown. Also let the boundary b ∈ B ≡ B (b be given by xn = g (b x) for x y0 , r) ⊆ Rn−1 . Of course u is only defined on U so actually the support of u is contained in the intersection of the circle with U . Let the Lipschitz constant for g be very small and denote it by K. In fact, assume 8K 2 < 1. I will first show how to extend when this condition holds and then I will remove it with a simple trick. Define  x, xn ) if xn ≤ g (b x)  u (b u (b x, 2g (b x) − xn ) if xn > g (b x) Eu (b x, xn ) ≡  b∈ 0 if x /B


1279

I will write U instead of U ∩B ×(a, b) to save space but this does not matter because u is assumed to be zero outside the indicated region. Then Z Z p |Eu (b x, xn ) − Eu (b y, yn )| ¯ ¯(1/2)(n+pθ) dxdy 2 2¯ Rn Rn ¯ b |b x − y | + (x − y ) ¯ n n ¯ Z Z

p

= U

Z

Z U+

p

U

Z Z

Z

|u (b x, xn ) − Eu (b y, yn )| ¯ ¯(1/2)(n+pθ) dxdy+ ¯ 2 2¯ b−y b | + (xn − yn ) ¯ ¯|x |Eu (b x, xn ) − u (b y, yn )| ¯ ¯(1/2)(n+pθ) dxdy+ ¯ ¯ b |2 + (xn − yn )2 ¯ x−y ¯|b

Z

U+

(45.3.17)

p

U+

U

U

|u (b x, xn ) − u (b y, yn )| ¯ ¯(1/2)(n+pθ) dxdy+ ¯ 2 2¯ b x − y| + (xn − yn ) ¯ ¯|b

(45.3.18)

p

U+

|Eu (b x, xn ) − Eu (b y, yn )| ¯ ¯(1/2)(n+pθ) dxdy ¯ ¯ b |2 + (xn − yn )2 ¯ x−y ¯|b

(45.3.19)

Consider the second of the integrals on the right of the equal sign. Using Fubini’s theorem, it equals Z Z p |u (b x, xn ) − u (b y, 2g (b y) − yn )| ¯ ¯(1/2)(n+pθ) dydx ¯ U U+ ¯ b |2 + (xn − yn )2 ¯ x−y ¯|b Z Z Z

p

∞

= U

B

g(b y)

Z Z Z

|u (b x, xn ) − u (b y, 2g (b y) − yn )| y dx ¯ ¯(1/2)(n+pθ) dyn db ¯ ¯ b |2 + (xn − yn )2 ¯ x−y ¯|b

g(b y)

= U

B

−∞

p

|u (b x, xn ) − u (b y, zn )| y dx ¯ ¯(1/2)(n+pθ) dzn db ¯ 2¯ b |2 + (xn − (2g (b x−y y) − zn )) ¯ ¯|b

I need to estimate |xn − zn | . |xn − zn | ≤ |xn − g (b x)| + |g (b x) − g (b y)| + |g (b y) − zn | ≤ ≤

b | + yn − g (b g (b x) − xn + K |b x−y y) b| |yn − xn | + 2K |b x−y

and so |xn − zn |

2

2

b | + 2 |yn − xn | ≤ 8K 2 |b x−y 2

b | + 2 |yn − xn | ≤ |b x−y

2

2

1280


Thus,

1 1 2 b |2 |xn − zn | − |b x−y 2 2 and so, the above change of variables results in an expression which is dominated by Z Z p |u (b x, xn ) − u (b y, zn )| ¯ ¯(1/2)(n+pθ) dydx ¯ U U ¯1 b |2 + 12 (xn − zn )2 ¯ x−y ¯ 2 |b 2

(yn − xn ) ≥

where y refers to (b y, zn ) in the above formula. Hence there is a constant C (n.θ) p such that 45.3.17 is dominated by C (n.θ) ||u|| ^ . A similar inequality holds θ,p W

(U )

for the third term. Finally consider 45.3.19.This equals Z Z p |u (b x, 2g (b x) − xn ) − u (b y, 2g (b y) − yn )| dxdy ¯ ¯(1/2)(n+pθ) ¯ ¯ U+ U+ b |2 + (xn − yn )2 ¯ x−y ¯|b Changing variables, x0n = 2g (b x) − xn , yn0 = 2g (b y) − yn , it equals Z Z p y, yn0 )| |u (b x, x0n ) − u (b 0 0 ¯ ¯(1/2)(n+pθ) dx dy , 2 2¯ U+ U+ ¯ b | + (xn − yn ) ¯ x−y ¯|b

(45.3.20)

each of xn , yn being a function of x0n , yn0 where an estimate needs to be obtained on |x0n − yn0 | in terms of |xn − yn | . 2

2

x) − g (b y)) + yn − xn ) (x0n − yn0 ) = (2 (g (b =

2

(yn − xn ) + 4 (g (b x) − g (b y)) (yn − xn ) +4 (g (b x) − g (b y))

≤

2

2

(yn − xn ) + 2 (g (b x) − g (b y)) 2

2 2

+2 (yn − xn ) + 4 (g (b x) − g (b y)) and so

2

2

b| (x0n − yn0 ) ≤ 3 (yn − xn ) + 6K 2 |b x−y

2

which implies

1 0 2 b |2 . (x − yn0 ) − 2K 2 |b x−y 3 n Then substituting this in to 45.3.20, a short computation shows 45.3.19 is domip nated by an expression of the form C (n, θ) ||u|| ^ and this proves the existence θ,p 2

(yn − xn ) ≥

W

(U )

of an extension operator provided the Lipschitz constant is small enough. It is clear E is linear where E is defined above. Now this assumption on the smallness of K needs to be removed. For (b x, xn ) ∈ U define n ³ ´ o b0 : x b−b b∈U U 0 ≡ xb0 = λ x


1281

here this is centering at 0 and stretching B since λ will be large. Let h be the name of this mapping. Thus ³ ´ b0 , b−b h (b x) ≡ λ x k (b x) ≡

1 b0 b+b x λ

h−1 (b x) =

These mappings are defined on all of Rn . Now let u0 be defined on U 0 as follows. u0 (b x, xn ) ≡ k∗ u (b x, xn ) . Also let g 0 (b x) ≡ k∗ g (b x) . Thus g 0 (b x) ≡ g

³

1 b λx

´

b 0 = g (k (b +b x)) . Then choosing λ large enough the Lipschitz

condition for g 0 is as small as desired. ³ ´ Always assume λ has been chosen this large and also λ ≥ 1. Furthermore, g 0 xb0 = x0n describes the boundary in the same way θ,p (U 0 ). Consider as xn = g (b x). Now I need to consider whether u0 ∈ W^ ¯ ³ ´ ³ ´¯p ¯ 0 b0 ¯ Z Z ¯u x , xn − u0 yb0 , yn ¯ 0 0 µ¯ ¶p+nθ dx dy ¯ 2 0 0 U U ¯ b0 b0 ¯ 2 ¯x − y ¯ + (xn − yn )

Z

¯ ³ ´ ³ ´¯p ¯ ∗ ¯ ¯k u xb0 , xn − k∗ u yb0 , yn ¯ 0 0 µ¯ ¶p+nθ dx dy ¯ ¯ b0 b0 ¯2 2 ¯x − y ¯ + (xn − yn )

Z

= U0

U0

³ ´ b 0 = h (b b−b Then change the variables xb0 = λ x x) with a similar change for yb0 , the above expression equals Z Z p ¡ n−1 ¢2 |u (b x, xn ) − u (b y, yn )| λ ³ ´p+nθ dxdy 2 2 2 U U b | + (xn − yn ) λ |b x−y Thus ||u0 ||

θ,p (U 0 ) W^

≤ λn−1 ||u||

θ,p (U ) W^

1. Proof: From the theory of interpolation spaces, W σ,p (Ω) is reflexive. This is because it is an iterpolation space for the two reflexive spaces, Lp (Ω) and W 1,p (Ω) . (Alternatively, you could use Corollary 45.3.14 in the case where Ω is a bounded

1284


open set with Lipschitz boundary or you could use Corollary 45.3.8 in case Ω = Rn . In addition, the same ideas would work if Ω were any space for which there was a continuous extension map from W σ,p (Ω) to W σ,p (Rn ) .) Now the formula for the norm of an element in W s,p (Ω) shows this space is isometric to a closed subspace k of W m,p (Ω) × W σ,p (Ω) for suitable k. Therefore, from Corollary 23.2.8 on Page s,p 711, W (Ω) is also reflexive. ¡ ¢ ¡ ¢ 1 Theorem 45.4.3 The trace map, γ : W m,p Rn+ → W m− p ,p Rn−1 is continuous. ³ ´ Proof: Let f ∈ S, the Schwartz class. Let σ = 1− p1 so that m− p1 = m−1+σ. Then from the definition and using f ∈ S,  1/p X p p ||γf ||m− 1 ,p,Rn−1 = ||γf ||m−1,p,Rn−1 + ||Dα γf ||1− 1 ,p,Rn−1  p

p

|α|=m−1

1/p

 p = ||γf ||m−1,p,Rn−1 +

X

p

||γDα f ||1− 1 ,p,Rn−1 

|α|=m−1

p

and from 45.1.4, ¡ Lemma ¢ ¡ and ¢the fact that the trace is continuous as a map from W m,p Rn+ to W m−1,p Rn−1 ,  1/p X p ||γf ||m− 1 ,p,Rn−1 ≤ C1 ||f ||m,p,Rn + C2 ||Dα f ||1,p,Rn  p

+

|α|=m−1

≤ C ||f ||m,p,Rn +p . Then using density of S this implies the desired result. With the definition of W s,p (Ω) for s not an integer, here is a generalization of an earlier theorem. Theorem 45.4.4 Let h : U → V where U and V are two open sets and suppose h is bilipschitz and that Dα h and Dα h−1 exist and are Lipschitz continuous if |α| ≤ m where m = 0, 1, · · · .and s = m + σ where σ ∈ (0, 1) . Then h∗ : W s,p (V ) → W s,p (U ) is continuous,¡linear, ¢∗ one to one, and has an inverse with the same properties, the inverse being h−1 . Proof: In case m = 0, the conclusion of the theorem is immediate from the general theory of trace spaces. Therefore, assume m ≥ 1. It follows from the definition that  1/p X p p ||h∗ u||m+σ,p,U ≡ ||h∗ u||m,p,U + ||Dα (h∗ u)||σ,p,U  |α|=m

45.4. FRACTIONAL ORDER SOBOLEV SPACES

1285

Consider the case when m = 1. Then it is routine to verify that Dj h∗ u (x) = u,k (h (x)) hk,j (x) . Let Lk : W 1,p (V ) → W 1,p (U ) be defined by Lk v = h∗ (v) hk,j . Then Lk is continuous as a map from W 1,p (V ) to W 1,p (U ) and as a map from Lp (V ) to Lp (U ) and therefore, it follows that Lk is continuous as a map from W σ,p (V ) to W σ,p (U ) . Therefore, ||Lk (v)||σ,p,U ≤ Ck ||v||σ,p,U and so ||Dj (h∗ u)||σ,p,U

X

≤

||Lk (u,k )||σ,p,U

k

X

≤

k

Ck ||Dk u||σ,p,V

Ã

≤ C

X

!1/p p ||Dk u||σ,p,V

.

k

Therefore, it follows that 1/p

 ∗

||h u||1+σ,p,U

≤ ||h

∗

p u||1,p,U

+

X

C

p

j

" ≤ C

p ||u||1,p,V

+

X

p ||Dk u||σ,p,V



k

X

#1/p p ||Dk u||σ,p,V

= C ||u||1+σ,p,V .

k

The general case is similar except for the use of a more complicated linear operator in place of Lk . This proves the theorem. It is interesting to prove this theorem using Theorem 45.3.13 and the intrinsic norm. Now we prove an important interpolation inequality for Sobolev spaces. Theorem 45.4.5 Let Ω be an open set in Rn which has the segment property and let f ∈ W m+1,p (Ω) and σ ∈ (0, 1) . Then for some constant, C, independent of f, 1−σ

σ

||f ||m+σ,p,Ω ≤ C ||f ||m+1,p,Ω ||f ||m,p,Ω . Also,¡ if¢ L ∈ L (W m,p (Ω) , W m,p (Ω)) for all m = 0, 1, · · · , and L ◦ Dα = Dα ◦ L on C ∞ Ω , then L ∈ L (W m+σ,p (Ω) , W m+σ,p (Ω)) for any m = 0, 1, · · · .

1286


¡ ¢ Proof: Recall from above, W 1−θ,p (Ω) ≡ T W 1,p (Ω) , Lp (Ω) , p, θ . Therefore, from Theorem 44.1.9, if f ∈ W 1,p (Ω) , θ

1−θ

||f ||1−θ,p,Ω ≤ K ||f ||1,p,Ω ||f ||0,p,Ω Therefore,  ||f ||m+σ,p,Ω

≤

X

||f ||p m,p,Ω +

³

1−σ

σ

K ||Dα f ||1,p,Ω ||Dα f ||0,p,Ω

´p

1/p 

|α|=m

≤

h ³ ´p i1/p p 1−σ σ C ||f ||m,p,Ω + ||f ||m+1,p,Ω ||f ||m,p,Ω h³ ´p ³ ´p i1/p 1−σ σ 1−σ σ C ||f ||m+1,p,Ω ||f ||m,p,Ω + ||f ||m+1,p,Ω ||f ||m,p,Ω

≤

C ||f ||m+1,p,Ω ||f ||m,p,Ω .

≤

1−σ

σ

¡ ¢ This proves the first part. Now consider the second. Let φ ∈ C ∞ Ω  X

p

||Lφ||m+σ,p,Ω = ||Lφ||m,p,Ω +

1/p p

||Dα Lφ||σ,p,Ω 

|α|=m

 p

= ||Lφ||m,p,Ω +

X

1/p p

||LDα φ||T (W 1,p ,Lp ,p,1−σ) 

|α|=m

 p

= ||Lφ||m,p,Ω +

X h

1/p ³¯¯ íp ¯ ¯ ¯ ¯ ¯ ¯ σ 1−σ  inf ¯¯t1−σ Lfα ¯¯ ¯¯t1−σ Lfα0 ¯¯ 1

2

(45.4.23)

|α|=m

where

³ ¯¯ ¯¯σ ¯¯ ¯¯1−σ ´ inf ¯¯t1−σ Lfα ¯¯1 ¯¯t1−σ Lfα0 ¯¯2 = ´ ³ ¯¯ ¯¯ ¯¯1−σ ¯¯σ inf ¯¯t1−σ Lfα ¯¯Lp (0,∞; dt ;W 1,p (Ω)) ¯¯t1−σ Lfα0 ¯¯Lp (0,∞; dt ;Lp (Ω)) , t

t

fα (0) ≡ limt→0 fα (t) = Dα φ in W 1,p (Ω) + Lp (Ω) , and the infimum is taken over all such functions. Therefore, from 45.4.23, and letting ||L||1 denote the operator

45.4. FRACTIONAL ORDER SOBOLEV SPACES

1287

norm of L in W 1,p (Ω) and ||L||2 denote the operator norm of L in Lp (Ω) , ||Lφ||m+σ,p,Ω  ≤

1/p íp h ³ X ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ σ 1−σ σ 1−σ ||Lφ||p  inf ||L||1 ||L||2 ¯¯t1−σ fα ¯¯1 ¯¯t1−σ fα0 ¯¯2 m,p,Ω + |α|=m

 ≤

||Lφ||p m,p,Ω

1/p ³ ´p X h ³¯¯ íp ¯ ¯ ¯ ¯ ¯ ¯ σ 1−σ σ 1−σ  + ||L||1 ||L||2 inf ¯¯t1−σ fα ¯¯1 ¯¯t1−σ fα0 ¯¯2 |α|=m



≤

1/p h ip X p C ||φ||m,p,Ω + ||Dα φ||σ,p,Ω  = C ||φ||m+σ,p,Ω . |α|=m

¡ ¢ Since C ∞ Ω is dense in all the Sobolev spaces, this inequality establishes the desired result. 0

Definition 45.4.6 Define for s ≥ 0, W −s,p (Rn ) to be the dual space of W s,p (Rn ) . Here

1 p

+

1 p0

= 1.

Note that in the case of m = 0 this is consistent with the Riesz representation theorem for the Lp spaces.

1288


Sobolev Spaces On Manifolds 46.1

Basic Definitions

Consider the following situation. There exists a set, Γ ⊆ Rm where m > n, mappings, hi : Ui → Γi = Γ ∩ Wi for Wi an open set in Rm with Γ ⊆ ∪li=1 Wi and Ui is an open subset of Rn which hi one to one and onto. Assume hi is of the form hi (x) = Hi (x, 0)

(46.1.1)

where for some open set, Oi , Hi : Ui × Oi → Wi is bilipschitz having bilipschitz α inverse such that for G = Hi or H−1 i , D G is Lipschitz for |α| ≤ k. For example, let m = n + 1 and let µ ¶ x Hi (x,y) = φ (x) + y where φ is a Lipschitz function having Dα φ Lipschitz for all |α| ≤ k. This is an example of the sort of thing just described if x ∈ Ui ⊆ Rn and Oi = R, because it is obvious the inverse of Hi is given by µ ¶ x −1 Hi (x, y) = . y − φ (x) l

Also let {ψ i }i=1 be a partition of unity subordinate to the open cover {Wi } satisfying ψ i ∈ Cc∞ (Wi ) . Then the definition of W s,p (Γ) follows. l

Definition 46.1.1 u ∈ W s,p (Γ) if whenever {Wi , ψ i , Γi , Ui , hi , Hi }i=1 is described above with hi ∈ C k,1 , h∗i (uψ i ) ∈ W s,p (Ui ) . The norm is given by ||u||s,p,Γ ≡

l X

||h∗i (uψ i )||s,p,Ui

i=1

It is not at all obvious this norm is well defined. What if r

{Wi0 , φi , Γi , Vi , gi , Gi }i=1 is as described above. Would the two norms be equivalent? To begin with consider l the following lemma which involves a particular choice for {Wi , ψ i , Γi , Ui , hi , Hi }i=1 . 1289

1290

SOBOLEV SPACES ON MANIFOLDS

Lemma 46.1.2 W s,p (Γ) as just described, is a Banach space. If p > 1 then it is reflexive. Ql Proof: Let L : W s,p (Γ) → i=1 W s,p (Ui ) be defined by (Lu)i ≡ h∗i (uψ i ) . ∞ ∞ Let {uj }j=1 be a Cauchy sequence in W s,p (Γ) . Then {h∗i (uj ψ i )}j=1 is a Cauchy s,p sequence in W (Ui ) for each i. Therefore, for each i, there exists wi ∈ W s,p (Ui ) such that lim h∗i (uj ψ i ) = wi in W s,p (Ui ) . j→∞

But also, there exists a subsequence, still denoted by j such that for each i ∞

{h∗i (uj ψ i ) (x)}j=1 is a Cauchy sequence for a.e. x. Since hi is given to be Lipschitz, it maps sets of measure 0 to sets of n dimensional Hausdorff measure zero. Therefore, ∞

{uj ψ i (y)}j=1 is a Cauchy sequence for µ a.e. y ∈ Wi ∩ Γ where µ denotes the n dimensional ∞ Hausdorff measure. It follows that for µ a.e. y, {uj (y)}j=1 is a Cauchy sequence and so it converges to a function denoted as u (y). uj (y) → u (y) µ a.e. Therefore, wi (x) = h∗i (uψ i ) (x) a.e. and this shows h∗i (uψ i ) ∈ W s,p (Ui ) . Thus u ∈ W s,p (Γ) showing completeness. It is clear ||·||s,p,Γ is a norm. Thus L is an Ql isometry of W s,p (Γ) and a closed subspace of i=1 W s,p (Ui ). By Corollary 45.4.2, W s,p (Ui ) is reflexive which implies the product is reflexive. Closed subspaces of reflexive spaces are reflexive by Lemma 23.2.7 on Page 710 and so W s,p (Γ) is also reflexive. This proves the lemma. I now show are equivalent. © that any two such norms ªr l Suppose Wj0 , φj , Γj , Vj , gj , Gj j=1 and {Wi , ψ i , Γi , Ui , hi , Hi }i=1 both satisfy 1

the conditions described above. Let ||·||s,p,Γ denote the norm defined by 1

||u||s,p,Γ ≡

r X ¯¯ ∗ ¡ ¢¯¯ ¯¯gj uφj ¯¯ j=1

¯¯ Ã !¯¯ r ¯¯ l ¯¯ X ¯¯ ∗ X ¯¯ ≤ uφj ψ i ¯¯ ¯¯g j ¯¯ ¯¯ j=1

i=1

=

≤ s,p,Vj

X ¯¯ ¡ ¢¯¯ ¯¯gj∗ uφj ψ i ¯¯ j,i 1,g

s,p,Vj

X ¯¯ ¡ ¢¯¯ ¯¯gj∗ uφj ψ i ¯¯ j,i

s,p,gj−1 (Wi ∩Wj0 )

s,p,Vj

(46.1.2)

Now define a new norm ||u||s,p,Γ by the formula 46.1.2. This norm is determined by © 0 ª Wj ∩ Wi , ψ i φj , Γj ∩ Γi , Vj , gi,j , Gi,j

46.2. THE TRACE ON THE BOUNDARY OF AN OPEN SET

1291

³ ´ 1,g where gi,j = gj . Thus the identity map is continuous from W s,p (Γ) , ||·||s,p,Γ to ´ ³ 1,g 1 1 W s,p (Γ) , ||·||s,p,Γ . It follows the two norms, ||·||s,p,Γ and ||·||s,p,Γ , are equivalent 2,h

2

by the open mapping theorem. In a similar way, the norms, ||·||s,p,Γ and ||·||s,p,Γ are equivalent where l X 2 ||u||s,p,Γ ≡ ||h∗i (uψ i )||s,p,Ui j=1

and 2,h

||u||s,p,Γ ≡

X ¯¯ ¡ ¢¯¯ ¯¯h∗i uφj ψ i ¯¯ j,i

s,p,Ui

=

X ¯¯ ¡ ¢¯¯ ¯¯h∗i uφj ψ i ¯¯ j,i

0 s,p,h−1 i (Wi ∩Wj )

But from the assumptions on h and g, in particular the assumption that these are restrictions of functions which are defined on open subsets of Rm which have Lipschitz derivatives up to order k along with their inverses, Theorem 45.4.4 implies, there exist constants Ci , independent of u such that ¯¯ ∗ ¡ ¯¯ ¡ ¢¯¯ ¢¯¯ ¯¯hi uφj ψ i ¯¯ ≤ C1 ¯¯gj∗ uφj ψ i ¯¯s,p,g−1 W ∩W 0 0 s,p,h−1 W ∩W ( ) i i i j j ( j) and

¯¯ ∗ ¡ ¢¯¯ ¯¯gj uφj ψ i ¯¯

s,p,gj−1 (Wi ∩Wj0 ) 1,g

¯¯ ¡ ¢¯¯ ≤ C2 ¯¯h∗i uφj ψ i ¯¯s,p,h−1 i

(Wi ∩Wj0 )

.

2,h

Therefore, the two norms, ||·||s,p,Γ and ||·||s,p,Γ are equivalent. It follows that the 2 1 norms, ||·||s,p,Γ and ||·||s,p,Γ are equivalent. This proves the following theorem. Theorem 46.1.3 Let Γ be described above. Then any two norms for W s,p (Γ) as in Definition 39.6.3 are equivalent.

46.2

The Trace On The Boundary Of An Open Set

Next is a generalization of earlier theorems about the loss of boundary.

1 p

derivatives on the

Definition 46.2.1 Define bk ≡ (x1 , · · · , xk−1 , 0, xk+1 , · · · , xn ) . Rn−1 ≡ {x ∈ Rn : xk = 0} , x k An open set, Ω is C m,1 if there exist open sets, Wi , i = 0, 1, · · · , l such that Ω = ∪li=0 Wi with W0 ⊆ Ω, open sets Ui ⊆ Rn−1 for some k, and open intervals, (ai , bi ) containing k 0 such that for i ≥ 1, bk ∈ Ui } , ∂Ω ∩ Wi = {b xk + φi (b x k ) ek : x

1292


Ω ∩ Wi = {b xk + (φi (b xk ) + xk ) ek : (b x k , x k ) ∈ U i × Ii } , where φi is Lipschitz with partial derivatives up to order m also Lipschitz. Here Ii = (ai , 0) or (0, bi ) . The case of (ai , 0) is shown in the picture.

Wi

¡

ck + φi (c x xk )ek

¡ Ω ∩ Wi

Ui

Ω

Assume Ω is C m−1,1 . Define bk + φi (b bk + (φi (b hi (b xk ) = x xk ) ek , Hi (x) ≡ x xk ) + xk ) ek , P l and let ψ i ∈ Cc∞ (Wi ) with i=0 ψ i (x) = 1 on Ω. Thus l

{Wi , ψ i , ∂Ω ∩ Wi , Ui , hi , Hi }i=1

¡ ¢ satisfies all the conditions for defining W s,p (∂Ω) for s ≤ m. Let u ∈ C ∞ Ω and let hi be as just described. The trace, denoted by¡ γ¢is that operator which evaluates ¡ ¢ functions in C ∞ Ω on ∂Ω. Thus for u ∈ C ∞ Ω , and y ∈ ∂Ω, u (y) =

l X

(uψ i ) (y)

i=1

and so using the notation to suppress the reference to y, γu =

l X

γ (uψ i )

i=1

It is necessary to show this is a continuous map. Letting u ∈ W m,p (Ω) , it follows from Theorem 45.4.3, and Theorem 38.0.16, ||γu||m− 1 ,p,∂Ω = p

=

l X i=1

l X

||h∗i (γ (ψ i u))||m− 1 ,p,Ui p

i=1

||h∗i γ (ψ i u)||m− 1 ,p,Rn−1 ≤ C p

k

l X i=1

||H∗i (ψ i u)||m,p,Rn

+

46.2. THE TRACE ON THE BOUNDARY OF AN OPEN SET

≤C

l X

||H∗i

(ψ i u)||m,p,Ui ×(ai ,0) ≤ C

l X

||(ψ i u)||m,p,Wi ∩Ω

i=1

i=1

≤C

1293

l X

||(ψ i u)||m,p,Ω ≤ C

i=1

l X

||u||m,p,Ω ≤ Cl ||u||m,p,Ω .

i=1

¡ ¢ Now use the density of C ∞ Ω in W m,p (Ω) to see that γ extends to a continuous linear map defined on W m,p (Ω) still called γ such that for all u ∈ W m,p (Ω) , ||γu||m− 1 ,p,∂Ω ≤ Cl ||u||m,p,Ω .

(46.2.3)

p

1

1

Also, it can be shown that γ maps W m,p (Ω) onto W m− p (∂Ω) . Let g ∈ W m− p (∂Ω). By definition, this means 1

h∗i (ψ i g) ∈ W m− p (Ui ) , each i and so, using a cutoff function, there exists wi ∈ W m,p (Ui × Ii ) such that γwi = h∗i (ψ i g) = h∗i (γψ i g) ¡ ¢∗ Thus H−1 wi ∈ W mp (Ω ∩ Wi ) . Let i w≡

l X

¡ ¢∗ ψ i H−1 wi ∈ W mp (Ω) i

i=1

then γw

=

X

X ¡ ¢∗ ¡ ¢∗ γψ j γ H−1 wi = γψ j H−1 γwi i i

i

=

X

γψ j

¡

¢∗ H−1 i

i

h∗i

(γψ i g) = g

i

In addition to this, in the case where m = 1, Lemma 45.2.1 implies there exists a 1 linear map, R, from W 1− p ,p (∂Ω) to W 1,p (Ω) which has the property that γRg = g 1 1 for every g ∈ W 1− p ,p (∂Ω) . I show this now. Letting g ∈ W 1− p ,p (∂Ω) , g=

l X

ψ i g.

i=1

Then also,

¡ ¢ 1 h∗i (ψ i g) ∈ W 1− p ,p Rn−1

if extended to 0 off Ui . From Lemma 45.2.1, there exists an extension of ¡ equal ¢ this to W 1,p Rn+ , Rh∗i (ψ i g) . Without loss of generality, assume that Rh∗i (ψ i g) ∈

1294


W 1,p (Ui × (ai , 0)) . If not so, multiply by a suitable cut off function in the definition of R . Then the extension is Rg =

l X ¡ −1 ¢∗ Hi Rh∗i (ψ i g) . i=1

¡ ¢ This works because from the definition of γ on C ∞ Ω and continuity of the map ¡ −1 ¢∗ established above, γ and Hi commute and so γRg

≡

l X ¡ ¢∗ γ H−1 Rh∗i (ψ i g) i i=1

=

l X ¡

H−1 i

¢∗

γRh∗i (ψ i g)

i=1

=

l X ¡

H−1 i

¢∗

h∗i (ψ i g) = g.

i=1

This proves the following theorem about the trace. Theorem 46.2.2 Let Ω ∈ C m−1,1 . Then there exists a constant, C independent of 1 p ,p (∂Ω) such that u ∈ W m,p (Ω) and a continuous linear map, γ : W m,p (Ω) → W m− ¡ ¢ 46.2.3 holds. This map satisfies γu (x) = u (x) for all u ∈ C ∞ Ω and γ is onto. 1 In the case where m = 1, there exists a continuous linear map, R : W 1− p ,p (∂Ω) → 1 W 1,p (Ω) which has the property that γRg = g for all g ∈ W 1− p ,p (∂Ω). Of course more can be proved but this is all to be presented here.

Part V

Topics In Probability

1295

Basic Probability Caution: This material on probability and stochastic processes may be half baked in places. I have not yet rewritten it several times. This is not to say that nothing else is half baked. However, the probability is higher here.

47.1

Random Variables And Independence

Recall Lemma 11.1.3 on Page 323 which is stated here for convenience. Lemma 47.1.1 Let M be a metric space with the closed balls compact and suppose λ is a measure defined on the Borel sets of M which is finite on compact sets. Then there exists a unique Radon measure, λ which equals λ on the Borel sets. In particular λ must be both inner and outer regular on all Borel sets. Also recall from earlier the following fundamental result which is called the Borel Cantelli lemma. Lemma 47.1.2 Let (Ω, F, λ) be a measure space and let {Ai } be a sequence of measurable sets satisfying ∞ X λ (Ai ) < ∞. i=1

Then letting S denote the set of ω ∈ Ω which are in infinitely many Ai , it follows S is a measurable set and λ (S) = 0. ∞ Proof: S = ∩∞ k=1 ∪m=k Am . Therefore, S is measurable and also

λ (S) ≤ λ (∪∞ m=k Am ) ≤

∞ X

λ (Ak )

m=k

and this converges to 0 as k → ∞ because of the convergence of the series. This proves the lemma. 1297

1298

BASIC PROBABILITY

Definition 47.1.3 A probability space is a measure space, (Ω, F, P ) where P is a measure satisfying P (Ω) = 1. A random vector (variable) is a measurable function, X : Ω → Z where Z is some topological space. It is often the case that Z will equal Rp . Assume Z is a separable Banach space. Define the following σ algebra. © ª σ (X) ≡ HX ≡ X−1 (E) : E is Borel in Z Thus σ (X) = HX ⊆ F. For E a Borel set in Z define ¡ ¢ λX (E) ≡ P X−1 (E) . This is called the distribution of the random variable, X. If Z |X (ω)| dP < ∞ Ω

then define

Z E (X) ≡

XdP Ω

where the integral is defined in the obvious way componentwise. For random variables having values in a separable Banach space or even more generally for a separable metric space, much can be said about regularity of λX . Definition 47.1.4 A measure, µ defined on B (E) will be called inner regular if for all F ∈ B (E) , µ (F ) = sup {µ (K) : K ⊆ F and K is closed} A measure, µ defined on B (E) will be called outer regular if for all F ∈ B (E) , µ (F ) = inf {µ (V ) : V ⊇ F and V is open} When a measure is both inner and outer regular, it is called regular. For probability measures, the above definition of regularity tends to come free. Note it is a little weaker than the usual definition of regularity because K is only assumed to be closed, not compact. Lemma 47.1.5 Let µ be a finite measure defined on B (E) where E is a metric space. Then µ is regular. Proof: First note every open set is the countable union of closed sets and every closed set is the countable intersection of open sets. Here is why. Let V be an open set and let © ¡ ¢ ª Kk ≡ x ∈ V : dist x, V C ≥ 1/k . Then clearly the union of the Kk equals V. Next, for K closed let Vk ≡ {x ∈ E : dist (x, K) < 1/k} .

47.1. RANDOM VARIABLES AND INDEPENDENCE

1299

Clearly the intersection of the Vk equals K. Therefore, letting V denote an open set and K a closed set, µ (V ) = sup {µ (K) : K ⊆ V and K is closed} µ (K) = inf {µ (V ) : V ⊇ K and V is open} . Also since V is open and K is closed, µ (V ) = inf {µ (U ) : U ⊇ V and V is open} µ (K) = sup {µ (L) : L ⊆ K and L is closed} In words, µ is regular on open and closed sets. Let F ≡ {F ∈ B (E) such that µ is regular on F } . Then F contains the open sets. I want to show F is a σ algebra and then it will follow F = B (E). First I will show F is closed with respect to complements. Let F ∈ F . Then since µ is finite and F is inner regular, there¢ exists K ⊆ F such that µ (F \ K) < ε. ¡ But K C \ F C = F \ K and so µ K C \ F C < ε showing that F C is outer regular. I have just approximated the measure of F C with the measure of K C , an open set containing F C . A similar argument works to show F C is inner regular. You start with F C \ V C = V \ F, and then conclude ¡ CV ⊇ CF¢ such that µ (V \ F ) < ε, note C µ F \V < ε, thus approximating F with the closed subset, V C . Next I will show F is closed with respect to taking countable unions. Let {Fk } be a sequence of sets in F. Then µ is inner regular on each of these so there exist {Kk } such that Kk ⊆ Fk and µ (Fk \ Kk ) < ε/2k+1 . First choose m large enough that ε m µ ((∪∞ k=1 Fk ) \ (∪k=1 Fk )) < . 2 Then m µ ((∪m k=1 Fk ) \ (∪k=1 Kk ))

≤ ≤

µ (∪m k=1 Fk \ Kk ) m X ε ε < 2k+1 2

k=1

and so m µ ((∪∞ k=1 Fk ) \ (∪k=1 Kk ))

≤
k i>k,j>k Written more succinctly, ¯ ¯2 ¯ ¯2 ¯X ¯ ¯X ¯ X ¯ n ¯ ¯ k ¯ ¯ ¯ ¯ Xj ¯ = ¯ Xj ¯¯ + (Xi , Xj ) ¯ ¯ j=1 ¯ ¯ j=1 ¯ j>k or i>k Now multiply both sides by XAk and integrate. Suppose i ≤ k for one of the terms in the second sum. Then by Lemma 47.1.12 and Ak ∈ σ (X1 , · · · , Xk ), the two random vectors XAk Xi , Xj are independent, µZ ¶ Z Z XAk (Xi , Xj ) dP = XAk Xi dP, Xj dP = 0 Ω

Ω

Ω

the last equality holding because by assumption E (Xj ) = 0. Therefore, it can be assumed both i, j are larger than k and ¯ ¯2 ¯ ¯2 ¯ n ¯ ¯ k ¯ Z Z ¯X ¯ ¯X ¯ ¯ ¯ ¯ X Ak ¯ Xj ¯ dP = XAk ¯ Xj ¯¯ dP Ω Ω ¯ j=1 ¯ ¯ j=1 ¯ X Z + XAk (Xi , Xj ) dP (47.1.1) j>k,i>k

Ω

The last term on the right is interesting. Suppose i > j. Then i > k because one of these i, j is larger than k. The integral inside the sum is of the form Z (Xi , XAk Xj ) dP (47.1.2) Ω

The second factor in the inner product is in σ (X1 , · · · , Xk , Xj ) and Xi is not included in the list of random vectors. Thus by Lemma 47.1.12, the two random vectors Xi , XAk Xj are independent and so 47.1.2 reduces to µZ ¶ µ Z ¶ Z Xi dP, XAk Xj dP = 0, XAk Xj dP = 0. Ω

Ω

Ω


1315

A similar result holds if j > i. Thus the mixed terms in the last term of 47.1.1 are all equal to 0. Hence 47.1.1 reduces to ¯ ¯2 ¯ ¯ n ¯X ¯ ¯ Xj ¯¯ dP X Ak ¯ Ω ¯ j=1 ¯

¯ ¯2 ¯X ¯ ¯ k ¯ ¯ XAk ¯ Xj ¯¯ dP Ω ¯ j=1 ¯ Z X 2 XAk |Xi | dP +

Z

Z

=

i>k

and so Z

Ω

¯ ¯2 ¯ ¯2 ¯X ¯ ¯X ¯ Z ¯ n ¯ ¯ k ¯ ¯ ¯ ¯ XAk ¯ Xj ¯ dP ≥ X Ak ¯ Xj ¯¯ dP ≥ ε2 P (Ak ) . Ω Ω ¯ j=1 ¯ ¯ j=1 ¯

Now, summing these yields ¯ ¯2 ¯ ¯2 ¯ n ¯ Z ¯X n X ¯¯ ¯ ¯ ¯ ¯ ¯ ε2 P (A) ≤ XA ¯¯ Xj ¯¯ dP ≤ X j ¯ dP ¯ Ω Ω ¯ j=1 ¯ j=1 ¯ ¯ Z

=

XZ i,j

(Xi , Xj ) dP Ω

By independence of the random vectors the mixed terms of the above sum equal zero and so it reduces to n Z X 2 |Xi | dP i=1

Ω

and this proves the theorem. This theorem implies the following amazing result. ∞

Theorem 47.1.26 Let {Xk }k=1 be independent random vectors having values in a separable real Hilbert space and suppose E (|Xk |) < ∞ for each k and E (Xk ) = 0. Suppose also that ∞ ³ ´ X 2 E |Xj | < ∞. j=1

Then ∞ X j=1

converges a.e.

Xj

1316

BASIC PROBABILITY

Proof: Let ε > 0 be given. By Kolmogorov’s inequality, Theorem 47.1.25, it follows that for p ≤ m < n ¯ ¯   ¯X ¯ n ´ ¯ k ¯ 1 X ³ 2 P  max ¯¯ Xj ¯¯ ≥ ε ≤ E |X | j m≤k≤n ¯ ε2 j=p ¯ j=m ≤

∞ ´ 1 X ³ 2 E |X | . j ε2 j=p

Therefore, letting n → ∞ it follows that for all m, n such that p ≤ m ≤ n ¯ ¯   ¯X ¯ ∞ ´ ¯ n ¯ 1 X ³ 2 P  max ¯¯ Xj ¯¯ ≥ ε ≤ 2 E |Xj | . p≤m≤n ¯ ε j=p ¯ j=m It follows from the assumption ∞ X

³ ´ 2 E |Xj | < ∞

j=1

there exists a sequence, {pn } such that if m ≥ pn ¯ ¯  ¯ k ¯ ¯X ¯ P  max ¯¯ Xj ¯¯ ≥ 2−n  ≤ 2−n . k≥m≥pn ¯ ¯ j=m 

By the Borel Cantelli lemma, Lemma 47.1.2, there is a set of measure 0, N such that for ω ∈ / N, ω is in only finitely many of the sets, ¯ ¯   ¯ k ¯ ¯X ¯  max ¯ Xj ¯¯ ≥ 2−n  k≥m≥pn ¯¯ ¯ j=m and so for ω ∈ / N, it follows that for large enough n, ¯ ¯   ¯X ¯ ¯ k ¯  max ¯ Xj (ω)¯¯ < 2−n  k≥m≥pn ¯¯ ¯ j=m nP o∞ k However, this says the partial sums X (ω) are a Cauchy sequence. j j=1 k=1 Therefore, they converge and this proves the theorem. With this amazing result, there is a simple proof of the strong law of large numbers. In the following lemma, sk and aj could have values in any normed linear space.


1317

Lemma 47.1.27 Suppose sk → s. Then n

1X sk = s. n→∞ n lim

k=1

Also if

∞ X aj j=1

converges, then

j

n

1X aj = 0. n→∞ n j=1 lim

Proof: Consider the first part. Since sk → s, it follows there is some constant, C such that |sk | < C for all k and |s| < C also. Choose K so large that if k ≥ K, then for n > K, |s − sk | < ε/2. ¯ ¯ n n ¯ 1X 1 X ¯¯ ¯ sk ¯ ≤ |sk − s| ¯s − ¯ n ¯ n k=1

=

1 n

K X

k=1

|sk − s| +

k=1

n 1 X |sk − s| n k=K

2CK εn−K 2CK ε + < + n 2 n n 2 Therefore, whenever n is large enough, ¯ ¯ n ¯ 1 X ¯¯ ¯ sk ¯ < ε. ¯s − ¯ ¯ n ≤

k=1

Now consider the second claim. Let sk =

k X aj j=1

j

and s = limk→∞ sk Then by the first part, s = =

=

n

n

k=1 n X

k=1

k

1X 1 X X aj sk = lim n→∞ n n→∞ n j j=1 lim

n n 1 aj X 1 X aj (n − j) lim 1 = lim n→∞ n n→∞ n j j j=1 j=1 k=j   n n n X X aj 1 1X lim  − aj  = s − lim aj n→∞ n→∞ n j n j=1 j=1 j=1

1318

BASIC PROBABILITY

and this proves the second part. Now here is the strong law of large numbers. Theorem 47.1.28 Suppose {Xk } are independent random variables and E (|Xk |) < ∞ for each k and E (Xk ) = mk . Suppose also ∞ ´ X 1 ³ 2 E |Xj − mj | < ∞. 2 j j=1

Then

(47.1.3)

n

1X (Xj − mj ) = 0 n→∞ n j=1 lim

Proof: Consider the sum

∞ X Xj − mj . j j=1

This sum converges a.e.obecause of 47.1.3 and Theorem 47.1.26 applied to the n Xj −mj random vectors . Therefore, from Lemma 47.1.27 it follows that for a.e. j ω, n 1X lim (Xj (ω) − mj ) = 0 n→∞ n j=1 This proves the theorem. The next corollary is often called the strong law of large numbers. It follows immediately from the above theorem. ∞

Corollary 47.1.29 Suppose {Xj }j=1 are independent and identically distributed random vectors, λXi = λXj for all i, j having mean m and variance equal to Z 2 2 σ ≡ |Xj − m| dP < ∞. Ω

Then for a.e. ω ∈ Ω

n

1X Xj (ω) = m n→∞ n j=1 lim

47.2

The Characteristic Function

One of the most important tools in probability is the characteristic function. To begin with, assume the random variables have values in Rm . Definition 47.2.1 Let X be a random variable as above. The characteristic function is Z Z ¡ it·X ¢ it·X(ω) φX (t) ≡ E e ≡ e dP = eit·x dλX Ω

the last equation holding by Lemma 47.1.8.

Rp

47.3. CONDITIONAL PROBABILITY

1319

Recall the following fundamental lemma and definition, Lemma 21.3.4 on Page 642. Definition 47.2.2 For T ∈ G ∗ , define F T, F −1 T ∈ G ∗ by ¡ ¢ F T (φ) ≡ T (F φ) , F −1 T (φ) ≡ T F −1 φ Lemma 47.2.3 F and F −1 are both one to one, onto, and are inverses of each other. The main result on characteristic functions is the following. p Theorem Let ¡ it·X ¢ 47.2.4 ¡ it·Y ¢ X and Y beprandom vectors with values in R and suppose E e =E e for all t ∈ R . Then λX = λY . R R Proof: For ψ ∈ G, let λX (ψ) ≡ Rp ψdλX and λY (ψ) ≡ Rp ψdλY . Thus both λX and λY are in G ∗ . Then letting ψ ∈ G and using Fubini’s theorem, Z Z Z Z eit·y ψ (t) dtdλY = eit·y dλY ψ (t) dt Rp Rp Rp Rp Z ¡ ¢ = E eit·Y ψ (t) dt p ZR ¡ ¢ = E eit·X ψ (t) dt p ZR Z = eit·x dλX ψ (t) dt Rp Rp Z Z = eit·x ψ (t) dtdλX . Rp

Rp

¡ ¢ ¡ ¢ Thus λY F −1 ψ = λX F −1 ψ . Since ψ ∈ G is arbitrary and F −1 is onto, this implies λX = λY in G ∗ . But G is dense in C0 (Rp ) and so λX = λY as measures. This proves the theorem.

47.3

Conditional Probability

Here I will consider the concept of conditional probability depending on the theory of differentiation of general Radon measures and leading to the Doob Dynkin lemma. If X, Y are two random vectors defined on a probability space having values in Rp1 and Rp2 respectively, and if E is a Borel set in the appropriate space, then (X, Y) is a random vector with values in Rp1 × Rp2 and λ(X,Y) (E × Rp2 ) = λX (E), λ(X,Y) (Rp1 × E) = λY (E). Thus, by Theorem 20.3.3 on Page 625, there exist probability measures, denoted here by λX|y and λY|x , such that whenever E is a Borel set in Rp1 × Rp2 , Z Z Z XE dλ(X,Y) = XE dλY|x dλX , Rp1 ×Rp2

Rp1

Rp2

1320

BASIC PROBABILITY

and

Z

Z

Rp1 ×Rp2

XE dλ(X,Y) =

Z Rp2

Rp1

XE dλX|y dλY .

Definition 47.3.1 Let X and Y be two random vectors defined on a probability space. The conditional probability measure of Y given X is the measure λY|x in the above. Similarly the conditional probability measure of X given Y is the measure λX|y . More generally, one can use the theory of slicing measures to consider any finite list of random vectors, {Xi }, defined on space with Xi ∈ Rpi , and Qna probability pi write the following for E a Borel set in i=1 R . Z

Z Rp1 ×···×Rpn

XE dλ(X1 ,···,Xn ) =

Z

Z

Rp1

Rp2 ×···×Rpn

XE dλ(X2 ,··· ,Xn )|x1 dλX1

Z

= Rp1

Z

Rp2

Rp3 ×···×Rpn

XE dλ(X3 ,··· ,Xn )|x1 x2 dλX2 |x1 dλX1 .. .

Z =

Z ···

Rp1

Rpn

XE dλXn |x1 x2 ···xn−1 dλXn−1 |x1 ···xn−2 · · · dλX2 |x1 dλX1 .

(47.3.4)

Obviously, this could have been done in any order in the iterated integrals by simply modifying the “given” variables, those occurring after the symbol |, to be those which have been integrated in an outer level of the iterated integral. Definition 47.3.2 Let {X1 , · · · , Xn } be random vectors defined on a probability space having values in Fp1 , · · · , Fpn respectively. The random vectors are independent if for every E a Borel set in Fp1 × · · · × Fpn , Z XE dλ(X1 ,··· ,Xn ) Rp1 ×···×Rpn

Z

Z

=

··· Rp1

Rpn

XE dλXn dλXn−1 · · · dλX2 dλX1

(47.3.5)

and the iterated integration may be taken in any order. If A is any set of random vectors defined on a probability space, A is independent if any finite set of random vectors from A is independent. Thus, the random vectors are independent exactly when the dependence on the givens in 47.3.4 can be dropped. Does this amount to the same thing as discussed earlier? These two random vectors, X, Y were independent if whenever A ∈ σ (X) and B ∈ σ (Y) , P (A ∩ B) =

47.3. CONDITIONAL PROBABILITY

1321

P (A) P (B) . Suppose the above definition and A and B as described. Let A = X−1 (E) and B = Y−1 (F ) . Then P (A ∩ B) = = = =

P ((X, Y) ∈ E × F ) Z XE (x) XF (y) dλ(X,Y) Rp1 ×Rp2 Z Z XE (x) XF (y) dλY|x dλX p p ZR 1 ZR 2 XE (x) XF (y) dλY dλX Rp1

=

Rp2

λX (E) λY (F ) = P (A) P (B)

Next suppose P (A ∩ B) = P (A) P (B) where A ∈ σ (X) and B ∈ σ (Y), A = X−1 (E) and B = Y−1 (F ) . Can it be asserted λX|y = λX ? In this case, for all E Borel in Rp1 and F Borel in Rp2 , Z Z XE (x) XF (y) dλY dλX Rp1

= = =

Rp2

P (A) P (B) = P (A ∩ B) Z XE (x) XF (y) dλ(X,Y) Rp1 ×Rp2 Z Z XE (x) XF (y) dλY|x dλX Rp1

Rp2

and so, by uniqueness of the slicing measures, dλY|x = dλY . A similar argument shows dλX|y = dλX . The argument clearly extends to any finite set of random variables. Thus this amounts to the same thing discussed earlier. Proposition 47.3.3 Equations 47.3.5 and 47.3.4 hold with XE replaced by any nonnegative Borel measurable function and for any bounded continuous function. Proof: The two equations hold for simple functions in place of XE and so an application of the monotone convergence theorem applied to an increasing sequence of simple functions converging pointwise to a given nonnegative Borel measurable function yields the conclusion of the proposition in the case of the nonnegative Borel function. For a bounded continuous function, one can apply the result just established to the positive and negative parts of the real and imaginary parts of the function. Lemma 47.3.4 Let X1 , · · · , Xn be random vectors with values in Rp1 , · · · , Rpn respectively and let g : Rp1 ×· · ·×Rpn → Rk be Borel measurable. Then g (X1 , · · · , Xn ) is a random vector with values in Rk and if h : Rk → [0, ∞), then Z h (y) dλg(X1 ,··· ,Xn ) (y) = Rk

1322

BASIC PROBABILITY

Z Rp1 ×···×Rpn

h (g (x1 , · · · , xn )) dλ(X1 ,··· ,Xn ) .

(47.3.6)

If Xi is a random vector with values in Rpi , i = 1, 2, · · · and if gi : Rpi → Rki , where gi is Borel measurable, then the random vectors gi (Xi ) are also independent whenever the Xi are independent. Proof: First let E be a Borel set in Rk . From the definition, Z Z Xg−1 (E) dλ(X1 ,··· ,Xn ) XE dλg(X1 ,··· ,Xn ) = p1 pn Rk ZR ×···×R XE (g (x1 , · · · , xn )) dλ(X1 ,··· ,Xn ) . = Rp1 ×···×Rpn

This proves 47.3.6 in the case when h is XE . To prove it in the general case, approximate the nonnegative Borel measurable function with simple functions for which the formula is true, and use the monotone convergence theorem. It remains to prove the last assertion that functions of independent random vectors are also independent random vectors. Let E be a Borel set in Rk1 ×· · ·×Rkn . Then for π i (x1 , · · · , xn ) ≡ xi , Z XE dλ(g1 (X1 ),··· ,gn (Xn )) Rk1 ×···×Rkn

Z ≡

Rp1 ×···×Rpn

Z

XE ◦ (g1 ◦ π 1 , · · · , gn ◦ π n ) dλ(X1 ,··· ,Xn )

Z

=

··· p1

ZR =

ZR

pn

··· Rk1

Rkn

XE ◦ (g1 ◦ π 1 , · · · , gn ◦ π n ) dλXn · · · dλX1 XE dλgn (Xn ) · · · dλg1 (X1 )

and this proves the last assertion. Proposition 47.3.5 Let ν 1 , · · · , ν n be Radon probability measures defined on Rp . Then there exists a probability space and independent random vectors {X1 , · · · , Xn } defined on this probability space such that λXi = ν i . n

Proof: Let (Ω, S, P ) ≡ ((Rp ) , S1 × · · · × Sn , ν 1 × · · · × ν n ) where this is just the product σ algebra and product measure which satisfies the following for measurable rectangles. Ã n ! n Y Y (ν 1 × · · · × ν n ) Ei = ν i (Ei ). i=1

i=1

Now let Xi (x1 , · · · , xi , · · · , xn ) = xi . Then from the definition, if E is a Borel set in Rp , λXi (E) ≡ P {Xi ∈ E}

47.4. CHARACTERISTIC FUNCTIONS AND INDEPENDENCE

1323

= (ν 1 × · · · × ν n ) (Rp × · · · × E × · · · × Rp ) = ν i (E). n

Let M consist of all Borel sets of (Rp ) such that Z Z Z ··· XE (x1 , · · · , xn ) dλX1 · · · dλXn = Rp

(Rp )n

Rp

XE dλ(X1 ,··· ,Xn ) .

From what was just shown and the definition of (ν 1 × · · · × ν n ) that M contains Qn all sets of the form i=1 Ei where each Ei ∈ Borel sets of Rp . Therefore, M contains the algebra of all finite disjoint unions of such sets. It is also clear that M is a monotone class and so by the theorem on monotone classes, M equals the Borel sets. Therefore, the given random vectors are independent and this proves the proposition. The following Lemma was proved earlier in a different way. n

Lemma 47.3.6 If {Xi }i=1 are independent random variables having values in R, Ã n ! n Y Y E Xi = E (Xi ). i=1

i=1

Qn Proof: By Lemma 47.3.4 and denoting by P the product, i=1 Xi , Ã n ! Z Z n Y Y xi dλ(X1 ,··· ,Xn ) Xi = zdλP (z) = E R

i=1

Z =

··· R

47.4

Z Y n R i=1

R×R i=1

xi dλX1 · · · dλXn =

n Y

E (Xi ).

i=1

Characteristic Functions And Independence

There is a way to tell if random vectors are independent by using their characteristic functions. Proposition 47.4.1 If X1 and X2 are random vectors having values in Rp1 and Rp2 respectively, then the random vectors are independent if and only if 2 ¡ ¢ Y ¡ ¢ E eiP = E eitj ·Xj j=1

P2 where P ≡ j=1 tj · Xj for tj ∈ Rpj . More generally, if Xi is a random vector Pn having values in Rpi for i = 1, 2, · · · , n, and if P = j=1 tj · Xj , then the random vectors are independent if and only if n ¢ ¡ ¡ iP ¢ Y = E eitj ·Xj . E e j=1

1324

BASIC PROBABILITY

The proof of this proposition will depend on the following lemma. Lemma 47.4.2 Let Y be a random vector with values in Rp and let f be bounded and measurable with respect to the Radon measure, λY , and satisfy Z f (y) eit·y dλY = 0 for all t ∈ Rp . Then f (y) = 0 for λY a.e. y. Proof: The proof is just like the R proof of Theorem 47.2.4 on Page 1319 applied to the measure, f (y) dλY . Thus E f (y) dλY = 0 for all E Borel. Hence f (y) = 0 a.e. Proof of the proposition: If the Xj are independent, the formula follows from Lemma 47.3.6 and Lemma 47.3.4. Now suppose the formula holds. Then Z Z ¡ ¢ eit1 ·x1 eit2 ·x2 dλX1 dλX2 = E eiP Rp2

Rp1

Z

Z

= Rp2

Rp1

eit1 ·x1 eit2 ·x2 dλX1 |x2 dλX2 .

Now apply Lemma 47.4.2 to conclude that Z Z eit1 ·x1 dλX1 = Rp1

Rp1

eit1 ·x1 dλX1 |x2

(47.4.7)

for λX2 a.e. x2 , the exceptional set depending on t1 . Therefore, taking the union of all exceptional sets corresponding to t1 ∈ Qp1 , it follows by continuity and the dominated convergence theorem that 47.4.7 holds for all t1 whenever x2 is not an element of this exceptional set of measure zero. Therefore, for such x2 , Theorem 47.2.4 applies and it follows λX1 |x2 = λX1 forRλX2R a.e. x2 . Hence, if R E is a Borel set in Rp1 × Rp2 , Rp1 +p2 XE dλ(X1 ,X2 ) = Rp2 Rp1 XE dλX1 |x2 dλX2 R R = Rp2 Rp1 XE dλX1 dλX2 . A repeat of the above argument will give the iterated integral in the reverse order or else one could apply Fubini’s theorem to obtain this. The proposition also holds if 2 is replaced with n and the argument is a longer version of what was just presented. This proves the proposition. With this preparation, it is time to present the Doob Dynkin lemma.I am not entirely sure what the Doob Dynkin lemma says actually. What follows is a generalization of what is identified as a special case of this lemma in [54]. I am not sure I have the right generalization. However, it is a very interesting lemma regardless its name. Lemma 47.4.3 Suppose X, Y1 , Y2 , · · · , Yk are random vectors, X having values in Rn and Yj having values in Rpj and X, Yj ∈ L1 (Ω) .

47.4. CHARACTERISTIC FUNCTIONS AND INDEPENDENCE

1325

Suppose X is σ (Y1 , · · · , Yk ) measurable. Thus   k  Y © −1 ª  −1 X (E) : E Borel ⊆ (Y1 , · · · , Yk ) (F ) : F is Borel in Rpj   j=1

Then there exists a Borel function, g :

Qk j=1

Rpj → Rn such that

X = g (Y1 , Y2 , · · · , Yk ) . Proof: For the sake of brevity, denote by Y the vector (Y1 , · · · , Yk ) and by y Qk the vector (y1 , · · · , yk ) and let j=1 Rpj ≡ RP . For E a Borel set of Rn , Z Z XRn ×E (x, y) xdλ(X,Y) XdP = Rn ×RP Y −1 (E) Z Z = xdλX|y dλY . (47.4.8) Rn

E

Consider the function

Z y→ Rn

xdλX|y .

Since dλY is a Radon measure having inner and outer regularity, it follows the above function is equal to a Borel function for λY a.e. y. This function will be denoted by g. Then from 47.4.8 Z Z Z XdP = g (y) dλY = XE (y) g (y) dλY Y −1 (E) E RP Z = XE (Y (ω)) g (Y (ω)) dP Ω Z = g (Y (ω)) dP Y −1 (E)

and since Y−1 (E) is an arbitrary element of σ (Y) , this shows that since X is σ (Y) measurable, X = g (Y) P a.e. This proves the lemma. Later the concept of conditional expectation will be presented. The above argument shows g (Y) = E (X|σ (Y1 , · · · , Yk )) . Then if the symbol, E (X|y1 , · · · , yk ) is defined as the function g in the above, then for a.e. ω, E (X|σ (Y1 , Y2 , · · · , Yk )) (ω) = E (X|Y1 (ω) , Y2 (ω) , · · · , Yk (ω)) which is a fairly attractive formula.

1326

47.5

BASIC PROBABILITY

Characteristic Functions For Measures

Recall the characteristic function for a random variable having values in Rn . I will give a review of this to begin with. Then the concept will be generalized to random variables (vectors) which have values in a real separable Banach space. Definition 47.5.1 Let X be a random variable. The characteristic function is Z Z ¢ ¡ eit·X(ω) dP = eit·x dλX φX (t) ≡ E eit·X ≡ Rp

Ω

the last equation holding by Lemma 47.1.8 on Page 1300. Recall the following fundamental lemma and definition, Lemma 21.3.4 on Page 642. Definition 47.5.2 For T ∈ G ∗ , define F T, F −1 T ∈ G ∗ by ¡ ¢ F T (φ) ≡ T (F φ) , F −1 T (φ) ≡ T F −1 φ Lemma 47.5.3 F and F −1 are both one to one, onto, and are inverses of each other. The main result on characteristic functions is the following is in Theorem 47.2.4 on Page 1319 which is stated here for convenience. p Theorem Let ¡ it·X ¢ 47.5.4 ¡ it·Y ¢ X and Y beprandom vectors with values in R and suppose E e =E e for all t ∈ R . Then λX = λY .

I want to do something similar for random variables which have values in a separable real Banach space, E instead of Rp . Corollary 47.5.5 Let K be a π system of subsets of Ω and suppose two probability measures, µ and ν defined on σ (K) are equal on K. Then µ = ν. Proof: This follows from the Lemma 9.11.3 on Page 275. Let G ≡ {E ∈ σ (K) : µ (E) = ν (E)} Then K ⊆ G, since µ and ν are both probability measures, it follows that if E ∈ G, then so is E C . Since these are measures, if {Ai } is a sequence of disjoint sets from G then X X µ (∪∞ µ (Ai ) = ν (Ai ) = ν (∪∞ i=1 Ai ) = i=1 A) i

i

and so from Lemma 9.11.3, G = σ (K) . This proves the corollary. Next recall the following fundamental lemma used to prove Pettis’ theorem. It is proved on Page 699 but is stated here for convenience.

47.5. CHARACTERISTIC FUNCTIONS FOR MEASURES

1327

Lemma 47.5.6 If E is a separable Banach space with B 0 the closed unit ball in 0 0 E 0 , then there exists a sequence {fn }∞ n=1 ≡ D ⊆ B with the property that for every x ∈ E, ||x|| = sup |f (x)| f ∈D 0

Definition 47.5.7 Let E be a separable real Banach space. A cylindrical set is one which is of the form {x ∈ E : x∗i (x) ∈ Γi , i = 1, 2, · · · , m} where here x∗i ∈ E 0 and Γi is a Borel set in R. It is obvious that ∅ is a cylindrical set and that the intersection of two cylindrical sets is another cylindrical set. Thus the cylindrical sets form a π system. What is ∞ the smallest σ algebra containing the cylindrical sets? Letting {fn }n=1 = D0 be the sequence of Lemma 47.5.6 it follows that {x ∈ E : ||x − a|| ≤ δ} ( =

x ∈ E : sup |f (x − a)| ≤ δ (

=

)

f ∈D 0

)

x ∈ E : sup |f (x) − f (a)| ≤ δ f ∈D 0

n o = ∩∞ n=1 x ∈ E : fn (x) ∈ B (fn (a) , δ) which yields a countable intersection of cylindrical sets. It follows the smallest σ algebra containing the cylindrical sets contains the closed balls and hence the open balls and consequently the open sets and so it contains the Borel sets. However, each cylindrical set is a Borel set and so in fact this σ algebra equals B (E). From Corollary 47.5.5 it follows that two probability measures which are equal on the cylindrical sets are equal on the Borel sets, B (E). Definition 47.5.8 Let µ be a probability measure on B (E) where E is a real separable Banach space. Then for x∗ ∈ E 0 , Z ∗ ∗ φµ (x ) ≡ eix (x) dµ (x) . E

φµ is called the characteristic function for the measure µ. Note this is a little different than earlier when the symbol φX (t) was used and X was a random variable. Here the focus is more on the measure than a random variable, X such that L (X) = µ. It might appear this is a more general concept but in fact this is not the case. This is a consequence of Skorokhod’s theorem which will be presented later.

1328

BASIC PROBABILITY

Theorem 47.5.9 Let µ and ν be two probability measures on B (E) where E is a separable real Banach space. Suppose φµ (x∗ ) = φν (x∗ ) for all x∗ ∈ E 0 . Then µ = ν. Proof: Let x∗1 , · · · , x∗n be in E 0 and define for A a Borel set of Rn , ≡ µ ({x ∈ E : (x∗1 (x) , · · · , x∗n (x)) ∈ A}) , ≡ ν ({x ∈ E : (x∗1 (x) , · · · , x∗n (x)) ∈ A}) .

µ e (A) νe (A)

(47.5.9)

Note these sets in the parentheses are cylindrical sets. Letting λ ∈ Rn , consider in the definition of the characteristic function, λ1 x∗1 + · · · + λn x∗n ∈ E 0 . Thus Z Z ∗ ∗ ∗ ∗ ei(λ1 x1 (x)+···+λn xn (x)) dν ei(λ1 x1 (x)+···+λn xn (x)) dµ = E

E n

Now if F is a Borel measurable subset of R , Z XF (y) de µ (y) = µ e (F ) Rn

≡ =

µ ({x ∈ E : (x∗1 (x) , · · · , x∗n (x)) ∈ F }) Z XF (x∗1 (x) , · · · , x∗n (x)) dµ E

and using the usual approximations involving simple functions, it follows that for any f bounded and Borel measurable, Z Z f (y) de µ (y) = f ((x∗1 (x) , · · · , x∗n (x))) dµ (x) . Rn

Similarly,

E

Z

Z f (y) de ν (y) =

Rn

E

f ((x∗1 (x) , · · · , x∗n (x))) dν (x) ,

Therefore, Z

Z eiλ·y de µ (y) =

Rn

∗

∗

∗

∗

ei(λ1 x1 (x)+···+λn xn (x)) dµ ZE ei(λ1 x1 (x)+···+λn xn (x)) dν

= ZE

eiλ·y de ν (y)

= Rn

which shows from Theorem 47.5.4 that νe = µ e on the Borel sets of Rn . However, from the definition of these measures in 47.5.9 this says nothing more than µ = ν on any cylindrical set. Hence by Corollary 47.5.5 this shows µ = ν on B (E) . This proves the theorem.

47.6. CHARACTERISTIC FUNCTIONS AND INDEPENDENCE IN BANACH SPACE1329

47.6

Characteristic Functions And Independence In Banach Space

Finally, I will consider the relation between the characteristic function and independence of random variables. Recall an earlier proposition which relates independence of random vectors with characteristic functions. It is proved starting on Page 1323 in the case of two random variables and concludes with the observation that the general case is entirely similar but more tedious to write down. n

Proposition 47.6.1 Let {Xk }k=1 be random vectors such that Xk has values in Rpk . Then the random vectors are independent if and only if n ¡ ¢ Y ¡ ¢ E eiP = E eitj ·Xj j=1

where P ≡

Pn

j=1 tj

· Xj for tj ∈ Rpj .

It turns out there is a generalization of the above proposition to the case where the random variables have values in a real separable Banach space. Before proving this recall an earlier theorem which had to do with reducing to the case where the random variables had values in Rn , Theorem 47.1.18. It is restated here for convenience. Theorem 47.6.2 The random variables, {Xi }i∈I are independent if whenever {i1 , · · · , in } ⊆ I, mi1 , · · · , min are positive integers, and gmi1 , · · · , gmin are in m i1

(E 0 ) respectively,

n on gmij ◦ Xij

j=1

m in

, · · · , (E 0 )

are independent random vectors having values in Rmi1 , · · · , Rmin

respectively. Now here is the theorem about independence and the characteristic functions. n

Theorem 47.6.3 Let {Xk }k=1 be random variables such that Xk has values in Ek , a real separable Banach space. Then the random variables are independent if and only if n ´ ³ ∗ ¡ ¢ Y E eiP = E eitj (Xj ) where P ≡

Pn

∗ j=1 tj

j=1

(Xj ) for

t∗j

∈

Ej0 .

1330

BASIC PROBABILITY

Proof: If the random variables are independent, then so are the random variables, t∗j (Xj ) and so the equation follows. The interesting case is when the equation holds. It suffices to consider only the caseQwhere each Ek = E. This is because you n can consider each Xj to have values in k=1 Ek by letting Xj take its values in the j th component of the product and 0 in the other components. Can you draw the conclusion the random variables are independent? By Theorem 47.1.18, it suffices n to show the random variables {gmk ◦ Xk }k=1 are independent. This happens if whenever tmk ∈ Rmk and P =

n X

tmk · (gmk ◦ Xk ) ,

k=1

it follows

n ³ ´ ¡ ¢ Y E eiP = E eitmk ·(gmk ◦Xk ) .

(47.6.10)

k=1

Now consider one of these terms in the exponent on the right. tmk · (gmk ◦ Xk ) (ω)

= =

where y ∗ ≡

P mk

∗ j=1 tj xj .

mk X

tj x∗j (Xk (ω))

j=1 yk∗ (Xk

(ω))

Therefore, 47.6.10 reduces to

n ³ Pn ∗ ´ Y ³ ∗ ´ E ei k=1 yk (Xk ) = E eiyk (Xk ) k=1

which is assumed to hold. Therefore, the random variables are independent. This proves the theorem. There is an obvious corollary which is useful. n

Corollary 47.6.4 Let {Xk }k=1 be random variables such that Xk has values in Ek , a real separable Banach space. Then the random variables are independent if and only if n ³ ∗ ´ ¡ ¢ Y E eiP = E eitj (Xj ) j=1

where P ≡

Pn

∗ j=1 tj

(Xj ) for t∗j ∈ Mj where Mj is a dense subset of Ej0 .

Proof: The easy direction follows from Theorem 47.6.3.© Suppose then the ª above equation holds for all t∗j ∈ Mj . Then let t∗j ∈ E 0 and let t∗nj be a sequence in Mj such that lim t∗nj = t∗j in E 0 n→∞

47.7. CONVOLUTION AND SUMS Then define P ≡

n X

1331

t∗j Xj , Pn ≡

j=1

n X

t∗nj Xj .

j=1

It follows ¡ ¢ E eiP = = =

¡ ¢ lim E eiPn

n→∞

lim

n→∞ n Y

n Y

³ ∗ ´ E eitnj (Xj )

j=1

³ ∗ ´ E eitj (Xj )

j=1

and this proves the corollary.

47.7

Convolution And Sums

Lemma 47.1.6 on Page 1300 makes possible a definition of convolution of two probability measures defined on B (E) where E is a separable Banach space as well as some other interesting theorems which held earlier in the context of locally compact spaces. I will first show a little theorem about density of continuous functions in Lp (E) and then define the convolution of two finite measures. First here is a simple technical lemma. Lemma 47.7.1 Suppose K is a compact subset of U an open set in E a metric space. Then there exists δ > 0 such that ¡ ¢ dist (x, K) + dist x, U C ≥ δ for all x ∈ E. Proof: For each x ∈ K, there exists a ball, B (x, δ x ) such that B (x, 3δ x ) ⊆ U . m Finitely many of these balls cover K because K is compact, say {B (xi , δ xi )}i=1 . Let 0 < δ < min (δ xi : i = 1, 2, · · · , m) . Now pick any x ∈ K. Then x ∈ B (x¡i , δ xi ) ¢for some xi and so B (x, δ) ⊆ B (xi , 2δ xi ) ⊆ C U. Therefore, ≥ δ. If x ∈ B (xi , 2δ xi ) for some xi , it ¢ x ∈ K, dist x, U ¡ forC any ≥ δ because then B (x, δ) ⊆ B (xi , 3δ xi ) ⊆ U. If x ∈ / B (xi , 2δ xi ) follows dist x, U / B (y, δ) for any y ∈ K because all these sets are contained for any of the xi , then x ∈ in some B (xi , 2δ xi ) . Consequently dist (x, K) ≥ δ. This proves the lemma. From this lemma, there is an easy corollary. Corollary 47.7.2 Suppose K is a compact subset of U, an open set in E a metric space. Then there exists a uniformly continuous function f defined on all of E, having values in [0, 1] such that f (x) = 0 if x ∈ / U and f (x) = 1 if x ∈ K.

1332

BASIC PROBABILITY

Proof: Consider ¡ ¢ dist x, U C f (x) ≡ . dist (x, U C ) + dist (x, K) Then some algebra yields |f (x) − f (x0 )| ≤ ¡ ¢ ¡ ¢¯ ¢ 1 ¡¯¯ dist x, U C − dist x0 , U C ¯ + |dist (x, K) − dist (x0 , K)| δ where δ is the constant of Lemma 47.7.1. Now it is a general fact that |dist (x, S) − dist (x0 , S)| ≤ d (x, x0 ) . Therefore, |f (x) − f (x0 )| ≤

2 d (x, x0 ) δ

and this proves the corollary. Now suppose µ is a finite measure defined on the Borel sets of a separable Banach space, E. It was shown above that µ is inner and outer regular. Lemma 47.1.6 on Page 1300 shows that µ is inner regular in the usual sense with respect to compact sets. This makes possible the following theorem. Theorem 47.7.3 Let µ be a finite measure on B (E) where E is a separable Banach space and let f ∈ Lp (E; µ) . Then for any ε > 0, there exists a uniformly continuous, bounded g defined on E such that ||f − g||Lp (E) < ε. Proof: As usual in such situations, it suffices to consider only f ≥ 0. Then by Theorem 8.3.9 on Page 197 and an application of the monotone convergence theorem, there exists a simple measurable function, s (x) ≡

m X

ck XAk (x)

k=1

such that ||f − s||Lp (E) < ε/2. Now by regularity of µ there exist compact sets, Pm 1/p Kk and open sets, Vk such that 2 k=1 |ck | µ (Vk \ K) < ε/2 and by Corollary 47.7.2 there exist uniformly continuous functions gk having values in [0, 1] such that gk = 1 on Kk and 0 on VkC . Then consider g (x) =

m X k=1

ck gk (x) .

47.7. CONVOLUTION AND SUMS

1333

This function is bounded and uniformly continuous. Furthermore, ||s − g||Lp (E)

≤

≤

¯p !1/p ÃZ ¯ m m ¯X ¯ X ¯ ¯ ck gk (x)¯ dµ ck XAk (x) − ¯ ¯ ¯ E k=1 k=1 ÃZ Ã m !p !1/p X |ck | |XAk (x) − gk (x)| E

≤

m X

k=1

µZ

|ck | E

k=1

≤ =

m X

¶1/p |XAk (x) − gk (x)| dµ p

ÃZ

!1/p p

|ck |

k=1 m X

2

2 dµ Vk \Kk 1/p

|ck | µ (Vk \ K)

< ε/2.

k=1

Therefore, ||f − g||Lp ≤ ||f − s||Lp + ||s − g||Lp < ε/2 + ε/2. This proves the theorem. Lemma 47.7.4 Let A ∈ B (E) where µ is a finite measure on B (E) for E a separable Banach space. Also let xi ∈ E for i = 1, 2, · · · , m. Then for x ∈ E m , Ã ! Ã ! m m X X x →µ A + xi , x → µ A − xi i=1

i=1

are Borel measurable functions. Furthermore, the above functions are B (E) × · · · × B (E) measurable where the above denotes the product measurable sets as described in Theorem 9.11.6 on Page 279. Proof: First consider the case where A = U, an open set. Let ( Ã ! ) m X y ∈ x ∈ Em : µ U + xi > α

(47.7.11)

i=1

Pm Then from Lemma 47.1.6 on Page 1300 there exists a compact set, K ⊆ U + Pi=1 yi m such that µ (K) > α. Then if y0 is close enough to y, it follows K ⊆ U + i=1 yi0 also. Therefore, for all y0 close enough to y, Ã ! m X 0 µ U+ yi ≥ µ (K) > α. i=1

1334

BASIC PROBABILITY

Pm In other words the set described in 47.7.11 is an open set and so y → µ (U + i=1 yi ) is Borel measurable whenever U is an open set in E. Define a π system, K to consist of all open sets in E. Then define G as ! ) ( Ã m X yi is Borel measurable A ∈ σ (K) = B (E) : y → µ A + i=1

I just showed G ⊇ K. Now suppose A ∈ G. Then Ã ! Ã ! m m X X µ AC + yi = µ (E) − µ A + yi i=1

i=1

C

and so A ∈ G whenever A ∈ G. Next suppose {Ai } is a sequence of disjoint sets of G. Then      m m X X Ai + µ (∪∞ yj  = µ ∪∞ y j  i=1 Ai ) + i=1 j=1

=

∞ X i=1

 µ  Ai +

j=1 m X



yj 

j=1

∪∞ i=1 Ai

and so ∈ G because it is the sum of Borel measurable functions. By the lemma on π systems, 9.11.3 on Page 275, it follows G = σ (K) = B (E) . ´ ³ Lemma Pm Similarly, x → µ A − j=1 xj is also Borel measurable whenever A ∈ B (E). Finally note that B (E) × · · · × B (E) contains the open sets of E m because the separability of E implies the existence of a countable basis for the topology of E m consisting of sets of the form m Y

Ui

i=1

where the Ui come from a countable basis for E. Since every open set is the countable union of sets like the above, each being a measurable box, the open sets are contained in B (E) × · · · × B (E) which implies B (E m ) ⊆ B (E) × · · · × B (E) also. This proves the lemma. With this lemma, it is possible to define the convolution of two finite measures. Definition 47.7.5 Let µ and ν be two finite measures on B (E) , for E a separable Banach space. Then define a new measure, µ ∗ ν on B (E) as follows Z µ ∗ ν (A) ≡ ν (A − x) dµ (x) . E

This is well defined because of Lemma 47.7.4 which says that x → ν (A − x) is Borel measurable.

47.7. CONVOLUTION AND SUMS

1335

Here is an interesting theorem about convolutions. However, first here is a little lemma. The following picture is descriptive of the set described in the following lemma. @ @

E

@

@

@

@

@

@ @ SA @ @ @ @ A @ @ @ @

E

Lemma 47.7.6 For A a Borel set in E, a separable Banach space, define SA ≡ {(x, y) ∈ E × E : x + y ∈ A} Then SA ∈ B (E) × B (E) , the σ algebra of product measurable sets, the smallest σ algebra which contains all the sets of the form A × B where A and B are Borel. Proof: Let K denote the open sets in E. Then K is a π system. Let G ≡ {A ∈ σ (K) = B (E) : SA ∈ B (E) × B (E)} . Then K ⊆ G because if U ∈ K then SU is an open set in E × E and all open sets are in B (E) × B (E) because a countable basis for the topology of E × E are sets of the form B × C where B and C come from a countable basis for E. Therefore, K ⊆ G. Now let A ∈ G. For (x, y) ∈ E × E, either x + y ∈ A or x + y ∈ / A. Hence E × E = SA ∪ SAC which shows that if A ∈ G then so is AC . Finally if {Ai } is a sequence of disjoint sets of G S∪∞ = ∪∞ i=1 SAi i=1 Ai and this shows that G is also closed with respect to countable unions of disjoint sets. Therefore, by the lemma on π systems, Lemma 9.11.3 on Page 275 it follows G = σ (K) = B (E) . This proves the lemma. Theorem 47.7.7 Let µ, ν, and λ be finite measures on B (E) for E a separable Banach space. Then µ∗ν =ν∗µ (47.7.12) (µ ∗ ν) ∗ λ = µ ∗ (ν ∗ λ)

(47.7.13)

If µ is the distribution for an E valued random variable, X and if ν is the distribution for an E valued random variable, Y, and X and Y are independent, then µ ∗ ν is the distribution for the random variable, X + Y . Also the characteristic function of a convolution equals the product of the characteristic functions.

1336

BASIC PROBABILITY

Proof: First consider 47.7.12. Letting A ∈ B (E) , the following computation holds from Fubini’s theorem and Lemma 47.7.6 Z Z Z µ ∗ ν (A) ≡ ν (A − x) dµ (x) = XSA (x, y) dν (y) dµ (x) E E ZE Z = XSA (x, y) dµ (x) dν (y) = ν ∗ µ (A) . E

E

Next consider 47.7.13. Using 47.7.12 whenever convenient, Z (µ ∗ ν) ∗ λ (A) ≡ (µ ∗ ν) (A − x) dλ (x) ZE Z = ν (A − x − y) dµ (y) dλ (x) E

while

E

Z µ ∗ (ν ∗ λ) (A) ≡

(ν ∗ λ) (A − y) dµ (y) ZE Z ν (A − y − x) dλ (x) dµ (y) E E Z Z ν (A − y − x) dµ (y) dλ (x) .

= =

E

E

The necessary product measurability comes from Lemma 47.7.4. Recall Z (µ ∗ ν) (A) ≡ ν (A − x) dµ (x) . E Pn Therefore, if s is a simple function, s (x) = k=1 ck XAk (x) , Z Z n X sd (µ ∗ ν) = ck ν (Ak − x) dµ (x) E

E

k=1

=

Z X n

ck ν (Ak − x) dµ (x)

E k=1

Z Z =

s (x + y) dν (x) dµ (y) E

E

Approximating with simple functions it follows that whenever f is bounded and measurable or nonnegative and measurable, Z Z Z f d (µ ∗ ν) = f (x + y) dν (y) dµ (x) (47.7.14) E

E

E

Therefore, letting Z = X + Y, and λ the distribution of Z, it follows from independence of X and Y that for t∗ ∈ E 0 , ´ ³ ∗ ´ ³ ∗ ´ ³ ∗ ³ ∗ ´ φλ (t∗ ) ≡ E eit (Z) = E eit (X+Y ) = E eit (X) E eit (Y )

47.8. THE CONVERGENCE OF SUMS OF SYMMETRIC RANDOM VARIABLES1337 But also, it follows from 47.7.14 Z ∗ φ(µ∗ν) (t∗ ) = eit (z) d (µ ∗ ν) (z) ZE Z ∗ = eit (x+y) dν (y) dµ (x) ZE ZE ∗ ∗ = eit (x) eit (y) dν (y) dµ (x) E E µZ ¶ µZ ¶ ∗ ∗ = eit (y) dν (y) eit (x) dµ (x) ³E ∗ ´ ³ ∗ E ´ = E eit (X) E eit (Y ) Since φλ (t∗ ) = φ(µ∗ν) (t∗ ) , it follows λ = µ ∗ ν. Note the last part of this argument shows the characteristic function of a convolution equals the product of the characteristic functions. This proves the theorem.

47.8

The Convergence Of Sums Of Symmetric Random Variables

It turns out that when random variables have symmetric distributions, some remarkable things can be said about infinite sums of these random variables. Conditions are given here that enable one to conclude the convergence of the sequence of partial sums from the convergence of some subsequence of partial sums. The following lemma is like an earlier result but does not depend on a function being bounded. Definition 47.8.1 Let X be a random variable. L (X) = µ means λX = µ. This is called the law of X. It is the same as saying the distribution measure of X is µ. Lemma 47.8.2 Let (Ω, F, P ) be a probability space and let X : Ω → E be a random variable, where E is a real separable Banach space. Also let L (X) = µ, a probability measure defined on B (E) , the Borel sets of E. Suppose h : E → R is continuous and also suppose h ◦ X is in L1 (Ω) . Then Z Z (h ◦ X) dP = h (x) dµ. Ω

E

¡ ¢ Proof: Let be a countable dense subset of R. Let Bin ≡ B ai , n1 ⊆ R and define Borel sets, Ani ⊆ E as follows: ¡ n ¢ ¡ k ¢ An1 = h−1 (B1n ) , Ank+1 ≡ h−1 Bk+1 \ ∪i=1 Ani . ∞ {ai }i=1

∞

Thus {Ani }i=1 ¡are disjoint Borel sets, with h (Ani ) ⊆ Bin . Also let bni denote the ¢ 1 endpoint of B ai , n which is closer to 0. ½ n bi if h−1 (Bin ) 6= ∅ n hi ≡ 0 if h−1 (Bin ) = ∅

1338

BASIC PROBABILITY

Then define hn (x) ≡

∞ X

hni XAni (x)

i=1

Thus |hn (x)| ≤ |h (x)| for all x ∈ E and |hn (x) − h (x)| ≤ 1/n for all x ∈ E. Then hn ◦ X is in L1 (Ω) and for all n, |hn ◦ X (ω)| ≤ |h ◦ X (ω)| . Let hnk (x) ≡

k X

hni XAni (x) .

i=1 ∞

Then from the construction in which the {Ani }i=1 are disjoint, |hnk (X (ω))| ≤ |h (X (ω))| , |hnk (x)| ≤ |h (x)| and |hnk (x)| is increasing in k. Z Ω

|hnk (X (ω))| dP

=

Z X k Ω i=1

=

Z X k Ω i=1

=

k X

|hni | XAni (X (ω)) dP |hni | XX−1 (An ) (ω) dP i

¢ ¡ |hni | P X−1 (Ani )

i=1

=

k X

|hni | µ (Ani )

Zi=1 = E

|hnk (x)| dµ.

By the monotone convergence theorem, and letting k → ∞, Z Z n |h (X (ω))| dP = |hn (x)| dµ. Ω

E

Now by the uniform convergence in the construction, you can let n → ∞ and obtain Z Z |h (X (ω))| dP = |h (x)| dµ. Ω

E

Thus h ∈ L1 (E, µ). It is obviously Borel measurable, being the limit of a sequence of Borel measurable functions. Now similar reasoning to the above and using the dominated convergence theorem when necessary yields Z Z hn (X (ω)) dP = hn (x) dµ Ω

E

47.8. THE CONVERGENCE OF SUMS OF SYMMETRIC RANDOM VARIABLES1339 Now another application of the dominated convergence theorem yields Z Z h (X (ω)) dP = h (x) dµ. Ω

E

This proves the lemma. First is a simple definition and lemma about random variables whose distribution is symmetric. Definition 47.8.3 Let X be a random variable defined on a probability space, (Ω, F, P ) having values in a Banach space, E. Then it has a symmetric distribution if whenever A is a Borel set, P ([X ∈ A]) = P ([X ∈ −A]) In terms of the distribution, λX = λ−X . It is good to observe that if X, Y are independent random variables defined on a probability space, (Ω, F, P ) such that each has symmetric distribution, then X + Y also has symmetric distribution. Here is why. Let A be a Borel set in E. Then by Theorem 47.7.7 on Page 1335, Z λX+Y (A) = λX (A − z) dλY (z) ZE = λ−X (A − z) dλ−Y (z) E

=

λ−(X+Y ) (A) = λX+Y (−A)

By induction, it follows that if you have n independent random variables each having symmetric distribution, then their sum has symmetric distribution. Here is a simple lemma about random variables having symmetric distributions. It will depend on Lemma 47.8.2 on Page 1337. Lemma 47.8.4 Let X ≡ (X1 , · · · , Xn ) and Y be random variables defined on a probability space, (Ω, F, P ) such that Xi , i = 1, 2, · · · , n and Y have values in E a separable Banach space. Thus X has values in E n . Suppose also that {X1 , · · · , Xn , Y } are independent and that Y has symmetric distribution. Then if A ∈ B (E n ) , it follows ¯¯ Ã "¯¯ n #! ¯¯X ¯¯ ¯¯ ¯¯ P [X ∈ A] ∩ ¯¯ Xi + Y ¯¯ < r ¯¯ ¯¯ i=1 ¯¯ Ã "¯¯ n #! ¯¯X ¯¯ ¯¯ ¯¯ = P [X ∈ A] ∩ ¯¯ Xi − Y ¯¯ < r ¯¯ ¯¯ i=1

You can also change the inequalities in the obvious way, < to ≤ , > or ≥.

1340

BASIC PROBABILITY

Proof: Denote by λX and λY the distribution measures for X and Y respectively. Since the random variables are independent, the distribution for the random variable, (X, Y ) mapping into E n+1 is λX ×λY where this denotes product measure. Since the Banach space is separable, the Borel sets are contained in the product measurable sets. Then by symmetry of the distribution of Y ¯¯ #! Ã "¯¯ n ¯¯ ¯¯X ¯¯ ¯¯ Xi + Y ¯¯ < r P [X ∈ A] ∩ ¯¯ ¯¯ ¯¯ i=1 Ã n ! Z X = XA (x) XB(0,r) xi + y d (λX × λY ) (x,y) E n ×E

= En

E

XA (x) XB(0,r)

Z Z = En

E

XA (x) XB(0,r)

Z = E n ×E

Ã

=

P

i=1 n X

Ã

Z Z

XA (x) XB(0,r)

i=1 Ã n X i=1 Ã n X

! xi + y dλX dλY ! xi + y dλX dλ−Y ! xi + y d (λX × λ−Y ) (x,y)

i=1

¯¯ "¯¯ n #! ¯¯X ¯¯ ¯¯ ¯¯ [X ∈ A] ∩ ¯¯ Xi + (−Y )¯¯ < r ¯¯ ¯¯ i=1

This proves the lemma. Other cases are similar. Now here is a really interesting lemma. Lemma 47.8.5 Let E be a real separable Banach space. Assume ξ 1 , · · · , ξ N are independent random variables having values P in E, a separable Banach space which k have symmetric distributions. Also let Sk = i=1 ξ i . Then for any r > 0, µ· ¸¶ P sup ||Sk || > r ≤ 2P ([||SN || > r]) . k≤N

Proof: First of all, µ· ¸¶ P sup ||Sk || > r µ· =

P

k≤N

¸¶

sup ||Sk || > r and ||SN || > r

k≤N

µ· +P

sup ||Sk || > r and ||SN || ≤ r

k≤N −1

≤

¸¶

P ([||SN || > r]) + P

µ·

¸¶ sup ||Sk || > r and ||SN || ≤ r

k≤N −1

I need to estimate the second of these terms. Let A1 ≡ [||S1 || > r] , · · · , Ak ≡ [||Sk || > r, ||Sj || ≤ r for j < k] .

. (47.8.15)

47.8. THE CONVERGENCE OF SUMS OF SYMMETRIC RANDOM VARIABLES1341 Thus Ak consists of those ω where ||Sk (ω)|| > r for the first time at k. Thus ·

¸ −1 sup ||Sk || > r and ||SN || ≤ r = ∪N j=1 Aj ∩ [||SN || ≤ r]

k≤N −1

and the sets in the above union are disjoint. Consider Aj ∩ [||SN || ≤ r] . For ω in this set, ||Sj (ω)|| > r, ||Si (ω)|| ≤ r if i < j. Since ||SN (ω)|| ≤ r in this set, it follows ¯¯ ¯¯ ¯¯ ¯¯ N X ¯¯ ¯¯ ¯ ¯ ξ i (ω)¯¯¯¯ ≤ r ||SN (ω)|| = ¯¯Sj (ω) + ¯¯ ¯¯ i=j+1 Thus P (Aj ∩ [||SN || ≤ r]) ¯¯ ¯¯  ¯¯ ¯¯ N X ¯ ¯ ¯ ¯ ¯¯ = P ∩j−1 ξ i ¯¯¯¯ ≤ r i=1 [||Si || ≤ r] ∩ [||Sj || > r] ∩ ¯¯Sj + ¯¯ i=j+1 ¯¯

(47.8.16)



(47.8.17)

Now ∩j−1 i=1 [||Si || ≤ r] ∩ [||Sj || > r] is of the form £¡

¢ ¤ ξ1, · · · , ξj ∈ A

for some Borel set, A. Then letting Y = 47.8.17 equals 

PN i=j+1

ξ i in Lemma 47.8.4 and Xi = ξ i ,

¯¯ ¯¯  ¯¯ ¯¯ N X ¯ ¯ ¯ ¯ j−1 P ∩i=1 [||Si || ≤ r] ∩ [||Sj || > r] ∩ ¯¯¯¯Sj − ξ i ¯¯¯¯ ≤ r ¯¯ i=j+1 ¯¯ ³ ´ = P ∩j−1 [||S || ≤ r] ∩ [||S || > r] ∩ [||S − (S − S )|| ≤ r] i j j N j i=1 ³ ´ j−1 = P ∩i=1 [||Si || ≤ r] ∩ [||Sj || > r] ∩ [||2Sj − SN || ≤ r]

Now since ||Sj (ω)|| > r, [||2Sj − SN || ≤ r]

⊆

[2 ||Sj || − ||SN || ≤ r]

⊆ [2r − ||SN || < r] = [||SN || > r] and so, referring to 47.8.16, this has shown P (Aj ∩ [||SN || ≤ r])

1342

BASIC PROBABILITY

≤

´ ³ P ∩j−1 [||S || ≤ r] ∩ [||S || > r] ∩ [||2S − S || ≤ r] i j j N i=1 ³ ´ P ∩j−1 i=1 [||Si || ≤ r] ∩ [||Sj || > r] ∩ [||SN || > r]

=

P (Aj ∩ [||SN || > r]) .

=

It follows that µ· P

¸¶ sup ||Sk || > r and ||SN || ≤ r

=

N −1 X

k≤N −1

P (Aj ∩ [||SN || ≤ r])

i=1

≤

N −1 X

P (Aj ∩ [||SN || > r])

i=1

≤

P ([||SN || > r])

and using 47.8.15, this proves the lemma. This interesting lemma will now be used to prove the following which concludes a sequence of partial sums converges given a subsequence of the sequence of partial sums converges. Lemma 47.8.6 Let {ζ k } be a sequence of independent random variables having values inP a separable real Banach space, E whose distributions are symmetric. Letting k Sk ≡ i=1 ζ i , suppose {Snk } converges a.e. Also suppose that for every m > nk , ¡£ ¤¢ P ||Sm − Snk ||E > 2−k < 2−k . (47.8.18) Then in fact, Sk (ω) → S (ω) a.e.ω and off a set of measure zero, the convergence of Sk to S is uniform. Proof: Let nk ≤ l ≤ m. Then by Lemma 47.8.5 µ· ¸¶ ¡£ ¤¢ −k P sup ||Sl − Snk || > 2 ≤ 2P ||Sm − Snk || > 2−k nk 2 ≤ 2P ||Sm − Snk || > 2−k < 2−(k−1) nk 2−k nk 1, then X1 and (X2 , · · · , Xp ) are both normally distributed and the two random vectors are independent. Here mj ≡ E (Xj ) . More generally, if the covariance matrix is a diagonal matrix, the random variables, {X1 , · · · , Xp } are linearly independent. Proof: From Theorem 47.9.2 ¡ ∗¢ Σ = E (X − m) (X − m) . Then by assumption,

µ Σ=

σ 21 0

¶

0 Σp−1

.

(47.9.24)

I need to verify that if E ∈ HX1 (σ (X1 )) and F ∈ H(X2 ,··· ,Xp ) (σ (X2 , · · · , Xp )), then P (E ∩ F ) = P (E) P (F ) . Let E = X1−1 (A) and

−1

F = (X2 , · · · , Xp )

(B)

p−1

where A and B are Borel sets in R and R respectively. Thus I need to verify that P ([(X1 , (X2 , · · · , Xp )) ∈ (A, B)]) = µ(X1 ,(X2 ,··· ,Xp )) (A × B) = µX1 (A) µ(X2 ,··· ,Xp ) (B) . Using 47.9.24, Fubini’s theorem, and definitions, µ(X1 ,(X2 ,··· ,Xp )) (A × B) =

(47.9.25)

47.9. THE MULTIVARIATE NORMAL DISTRIBUTION Z Rp

XA×B (x)

1 (2π)

Z

det (Σ)

1/2

e

−1 ∗ −1 (x−m) 2 (x−m) Σ

dx

Z

=

XA (x1 )

Rp−1

R

(2π) e

p/2

1349

−1 2

(p−1)/2

√

∗

(x0 −m0 )

XB (X2 , · · · , Xp ) · 1 1/2

2π (σ 21 )

1/2

det (Σp−1 )

³ 0 ´ 0 Σ−1 p−1 x −m

e

−(x1 −m1 )2 2σ 2 1

·

dx0 dx1

where x0 = (x2 , · · · , xp ) and m0 = (m2 , · · · , mp ) . Now this equals Z XA (x1 ) p

e

1

e

−(x1 −m1 )2 2σ 2 1

2πσ 21 R ³ 0 ´ −1 0 0 ∗ −1 0 2 (x −m ) Σp−1 x −m

Z

1 B

(p−1)/2

(2π)

det (Σp−1 )

1/2

dx0 dx.

· (47.9.26) (47.9.27)

In case B = Rp−1 , the inside integral equals 1 and ¡ ¢ λX1 (A) = λ(X1 ,(X2 ,··· ,Xp )) A × Rp−1 Z −(x1 −m1 )2 1 2σ 2 1 = XA (x1 ) p e dx1 2 2πσ R 1 which shows X1 is normally distributed as claimed. Similarly, letting A = R, = =

λ(X2 ,··· ,Xp ) (B) λ(X1 ,(X2 ,··· ,Xp )) (R × B) Z 1 B

e (p−1)/2 1/2 (2π) det (Σp−1 )

−1 2

∗

(x0 −m0 )

³ 0 ´ 0 Σ−1 p−1 x −m

dx0

and (X2 , · · · , Xp ) is also normally distributed with mean m0 and covariance Σp−1 . Now from 47.9.26, 47.9.25 follows. In case the covariance matrix is diagonal, the above reasoning extends in an obvious way to prove the random variables, {X1 , · · · , Xp } are independent. However, another way to prove this is to use Proposition 47.4.1 on Page 1323 and consider the characteristic function. Let E (Xj ) = mj and P =

p X

tj Xj .

j=1

Then since X is normally distributed and the covariance is a diagonal,  2  σ1 0   .. D≡  . 0 σ 2p

1350

BASIC PROBABILITY

, ¡ ¢ E eiP

³ ´ 1 ∗ = E eit·(X−m) = eit·m e− 2 t Σt   p X 1 = exp  itj mj − t2j σ 2j  2 j=1 =

(47.9.28)

¶ µ 1 exp itj mj − t2j σ 2j 2 j=1 p Y

Also,  ¡

E e

¢ itj Xj

=



E exp itj Xj +

 X

i0Xk 

k6=j

µ ¶ 1 = exp itj mj − t2j σ 2j 2 With 47.9.28, this shows

p ¡ ¢ ¡ ¢ Y E eitj Xj E eiP = j=1

which shows by Proposition 47.4.1 that the random variables, {X1 , · · · , Xp } are independent. This proves the theorem.

47.10

Use Of Characteristic Functions To Find Moments

Let X be a random variable with characteristic function φX (t) ≡ E (exp (itX)) Then this can be used to find moments of the random variable assuming they exist. The k th moment is defined as ¡ ¢ E Xk . This can be done by using the dominated convergence theorem to differentiate the characteristic function with respect to t and then plugging in t = 0. For example, φ0X (t) = E (iX exp (itX)) and now plugging in t = 0 you get iE (X) . Doing another differentiation you obtain ¡ ¢ φ00X (t) = E −X 2 exp (itX)

47.11. THE CENTRAL LIMIT THEOREM

1351

¡ ¢ and plugging in t = 0 you get −E X 2 and so forth. An important case is where X is normally distributed with mean 0 and variance σ 2 . In this case, as shown above, the characteristic function is 1 2

e− 2 t

σ2

Also all moments exist when X is normally distributed. So what are these moments? ³ 1 2 2´ 1 2 2 Dt e− 2 t σ = −tσ 2 e− 2 t σ and plugging in t = 0 you find the mean equals 0 as expected. ³ ´ 1 2 2 1 2 2 1 2 2 Dt −tσ 2 e− 2 t σ = −σ 2 e− 2 t σ + t2 σ 4 e− 2 t σ and plugging in t = 0 you find the second moment is σ 2 . Then do it again. ³ ´ 1 2 2 1 2 2 1 2 2 1 2 2 Dt −σ 2 e− 2 t σ + t2 σ 4 e− 2 t σ = 3σ 4 te− 2 t σ − t3 σ 6 e− 2 t σ ¡ ¢ Then E X 3 = 0. ³ ´ 1 2 2 1 2 2 Dt 3σ 4 te− 2 t σ − t3 σ 6 e− 2 t σ 1 2

= 3σ 4 e− 2 t

σ2

1 2

− 6σ 6 t2 e− 2 t

σ2

1 2

+ t4 σ 8 e− 2 t

σ2

¡ ¢ and so E X 4 = 3σ 4 . By now you can see the pattern. If you continue this way, you find the odd moments are all 0 and ¡ ¢ ¡ ¢m E X 2m = Cm σ 2 . (47.10.29) This is an important observation.

47.11

The Central Limit Theorem

The central limit theorem is one of the most marvelous theorems in mathematics. It can be proved through the use of characteristic functions. Recall for x ∈ Rp , ||x||∞ ≡ max {|xj | , j = 1, · · · , p} . Also recall the definition of the distribution function for a random vector, X. FX (x) ≡ P (Xj ≤ xj , j = 1, · · · , p) . ∞

Definition 47.11.1 Let {Xn } be random vectors with values in Rp . Then {λXn }n=1 is called “tight” if for all ε > 0 there exists a compact set, Kε such that λXn ([x ∈ / Kε ]) < ε

1352

BASIC PROBABILITY

for all λXn . Similarly, if {µn } is a sequence of probability measures defined on the Borel sets of Rp , then this sequence is “tight” if for each ε > 0 there exists a compact set, Kε such that µn ([x ∈ / Kε ]) < ε for all µn . Lemma 47.11.2 If {Xn }is a sequence of random vectors with values in Rp such that lim φXn (t) = ψ (t) n→∞

∞

for all t, where ψ (0) = 1 and ψ is continuous at 0, then {λXn }n=1 is tight. Proof: Let ej be the j th standard unit basis vector. ¯ Z u ¯ ¯1 ¡ ¢ ¯ ¯ 1 − φXn (tej ) dt¯¯ ¯u −u ¯ Z uµ ¶ ¯ Z ¯1 ¯ = ¯¯ 1− eitxj dλXn dt¯¯ u Rp ¯ Z−u µZ ¶ ¯ ¯1 u ¯ ¡ ¢ = ¯¯ 1 − eitxj dλXn dt¯¯ u p ¯ ¯Z −u Z uR ¯ ¯ ¡ ¢ 1 1 − eitxj dtdλXn (x)¯¯ = ¯¯ p u ¯ ¯ RZ µ −u ¶ ¯ ¯ sin (uxj ) dλXn (x)¯¯ = ¯¯2 1− uxj Rp µ ¶ Z 1 1− dλXn (x) ≥ 2 |uxj | [|xj |≥ u2 ] µ ¶ Z 1 ≥ 2 1− dλXn (x) |u| (2/u) [|xj |≥ u2 ] Z = 1dλXn (x) [|xj |≥ u2 ] ¸¶ µ· 2 = λX n x : |xj | ≥ . u If ε > 0 is given, there exists r > 0 such that if u ≤ r, Z 1 u (1 − ψ (tej )) dt < ε/p u −u for all j = 1, · · · , p and so, by the dominated convergence theorem, the same is true with φXn in place of ψ provided n is large enough, say n ≥ N (u). Thus, if u ≤ r, and n ≥ N (u), ¸¶ µ· 2 < ε/p λX n x : |xj | ≥ u

47.11. THE CENTRAL LIMIT THEOREM

1353

for all j ∈ {1, · · · , p}. It follows that for u ≤ r and n ≥ N (u) , µ· ¸¶ 2 λX n x : ||x||∞ ≥ < ε. u because

· ¸ · ¸ 2 2 p x : ||x||∞ ≥ ⊆ ∪j=1 x : |xj | ≥ u u

This proves the lemma because there are only finitely many measures, λXn for n < N (u) and the compact set can be enlarged finitely many times to obtain a single compact set, Kε such that for all n, λXn ([x ∈ / Kε ]) < ε. This proves the lemma. Lemma 47.11.3 If φXn (t) → φX (t) for all t, then whenever ψ ∈ S, Z Z ψ (y) dλXn (y) → ψ (y) dλX (y) ≡ λX (ψ) λXn (ψ) ≡ Rp

Rp

as n → ∞. by

Proof: Recall that if X is any random vector, its characteristic function is given Z φX (y) ≡ eiy·x dλX (x) . Rp

Also remember the inverse Fourier transform. Letting ψ ∈ S, the Schwartz class, Z ¡ −1 ¢ −1 F (λX ) (ψ) ≡ λX F ψ ≡ F −1 ψdλX Rp Z Z 1 = eiy·x ψ (x) dxdλX (y) p/2 p p R R (2π) Z Z 1 = ψ (x) eiy·x dλX (y) dx p/2 Rp Rp (2π) Z 1 ψ (x) φX (x) dx = p/2 Rp (2π) and so, considered as elements of S∗ , −(p/2)

F −1 (λX ) = φX (·) (2π)

∈ L∞ .

By the dominated convergence theorem Z p/2

(2π)

F

−1

(λXn ) (ψ)

≡ ZR

p

→ Rp

=

φXn (t) ψ (t) dt φX (t) ψ (t) dt p/2

(2π)

F −1 (λX ) (ψ)

1354

BASIC PROBABILITY

whenever ψ ∈ S. Thus λXn (ψ) = ≡

F F −1 λXn (ψ) ≡ F −1 λXn (F ψ) → F −1 λX (F ψ) F F −1 λX (ψ) = λX (ψ).

This proves the lemma. Lemma 47.11.4 If φXn (t) → φX (t) , then if ψ is any bounded uniformly continuous function, Z Z ψdλXn = ψdλX . lim n→∞

Rp

Rp

Proof: Let ε > 0 be given, let ψ be a bounded function in C ∞ (Rp ). Now let p η ∈ Cc∞ (Qr ) where Qr ≡ [−r, r] satisfy the additional requirement that η = 1 on ∞ Qr/2 and η (x) ∈ [0, 1] for all x. By Lemma 47.11.2 the set, {λXn }n=1 , is tight and so if ε > 0 is given, there exists r sufficiently large such that for all n, Z ε |1 − η| |ψ| dλXn < , 3 / r/2 ] [x∈Q and

Z / r/2 ] [x∈Q

Thus,

|1 − η| |ψ| dλX
1, it is also true that ÃÃ !p !1/p E

sup M (t) t∈[S,T ]

1t

≤

µ

p p−1

→ M (t) (ω) is continuous from the right for a.e. ω.

¶ p 1/p

E (M (T ) )

50.3. MARTINGALES

1481

−m m m m m . Proof: Let S ≤ tm 0 < t1 < · · · < tNm = T where tj+1 − tj = (T − S) 2 First consider m = 1. © ¡ ¢ ª © ¡ ¢ ª At10 ≡ ω ∈ Ω : M t10 (ω) ≥ λ , At11 ≡ ω ∈ Ω : M t11 (ω) ≥ λ \ At10 ´ © ¡ ¢ ª ³ At12 ≡ ω ∈ Ω : M t12 (ω) ≥ λ \ At10 ∪ At10 .

Do this type of construction for m = 2, 3, 4, · · · yielding disjoint sets,

n o2m At m j

j=0

whose union equals

∪t∈Dm [M (t) ≥ λ] m ª 2 ∞ where Dm = tm j j=0 . Thus Dm ⊆ Dm+1 . Then also, D ≡ ∪m=1 Dm is dense and countable. From Lemma 50.3.3, ©

m

P (∪t∈Dm [M (t) ≥ λ])

2 X

=

³ ´ P Atm j

j=0 2 Z 1 X p X M (T ) dP λp j=0 Atm [supt∈Dm M (t)≥λ] j Z 1 p X M (T ) dP. λp Ω [supt∈D M (t)≥λ] m

≤ ≤ Let m → ∞ in the above to obtain 1 λp

P (∪t∈D [M (t) ≥ λ]) ≤

Z Ω

X[sup

t∈D

M (t)≥λ] M

p

(T ) dP.

(50.3.11)

From now on, assume that for a.e. ω ∈ Ω, t → M (t) (ω) is right continuous. Then with this assumption, the following claim holds. Claim:For λ > ε > 0, " # ∪t∈D [M (t) ≥ λ − ε] ⊇

sup M (t) ≥ λ t∈[S,T ]

h i Proof of the claim: Suppose ω ∈ supt∈[S,T ] M (t) ≥ λ . Then there exists s such that |M (s) (ω)| > λ − ε. If s < T, then by right continuity, this situation persists for all t > s for t close enough to s. In particular, it is true for some t ∈ D. If s = T, then s ∈ D. This proves the claim. Letting P 0 denote the outer measure determined by P it follows from the claim and 50.3.11 that Ã" #! P0

sup M (t) ≥ λ

≤ P (∪t∈D [M (t) ≥ λ − ε])

t∈[S,T ]

≤

1 p (λ − ε)

Z p

Ω

X[sup M (T ) dP. t∈D M (t)≥λ−ε]

1482

STOCHASTIC PROCESSES

Since ε is arbitrary, this shows Ã" P

0

#! sup M (t) ≥ λ

≤

t∈[S,T ]

Z

1 λp

p

Ω

X[sup M (T ) dP. t∈D M (t)≥λ]

(50.3.12)

h i It would be interesting to consider whether supt∈[S,T ] M (t) ≥ λ is measurable. However, for every ε ∈ (0, λ) it follows from the right continuity of M that ·

# · ¸ " ¸ sup M (t) ≥ λ − ε ⊇ sup M (t) ≥ λ ⊇ sup M (t) ≥ λ t∈D

t∈D

t∈[S,T ]

Now let εn be a sequence decreasing to 0 and then · ¸ ∩∞ sup M (t) ≥ λ − ε n n=1

· =

t∈D

¸

sup M (t) ≥ λ t∈D "

#

sup M (t) ≥ λ

⊇ · ⊇

t∈[S,T ]

¸

sup M (t) ≥ λ t∈D

h i Thus supt∈[S,T ] M (t) ≥ λ is a measurable set due to countability of D. Also from 50.3.12 this shows Ã" P

#! sup M (t) ≥ λ

1 λp

≤

t∈[S,T ]

Z Ω

X[sup

t∈[S,T ]

M (t)≥λ] M

p

(T ) dP

Now consider the other inequality. Using the distribution function technique and the above estimate obtained in the first part, ÃÃ E

!p ! sup M (t)

Z

∞

=

t∈[S,T ]

Ã" pαp−1 P

≤

Ã"

∞

pα 0

sup M (t) > α t∈[S,T ]

0

Z

#!

p−1

P

sup M (t) ≥ α t∈[S,T ]

dα #! dα

50.3. MARTINGALES Z

1483

∞

≤

pα Z Z

p

α

Ω

=

≤

=

Z

1 α

X[sup M t∈[S,T ] M (t)≥α] Ω supt∈[S,T ] M (t) p−2

0

=

p−1

p p−1 p p−1

Z Ã

!p−1 sup M (t)

M (T ) dP

t∈[S,T ]

Ω

ÃZ Ã

!p !1/p0 µZ p

sup M (t)

¶1/p

M (T )

t∈[S,T ]

Ω

Ω

ÃÃ

!p !1/p0 p 1/p

sup M (t)

E (M (T ) )

.

t∈[S,T ]

³³ Therefore, dividing both sides by E ÃÃ E

dαM (T ) dP

0

p E p−1

(T ) dP dα

´p ´1/p0 supt∈[S,T ] M (t) yields

!p !1/p sup M (t)

µ ≤

t∈[S,T ]

p p−1

¶ p 1/p

E (M (T ) )

This proves the theorem. With Theorem 50.3.4, here is an important maximal estimate for martingales having values in E, a real separable Banach space. Theorem 50.3.5 Let X (t) for t ∈ I = [0, T ] be an E valued right continuous martingale with respect to a filtration, Ft . Then for p ≥ 1, µ· ¸¶ 1 p P sup ||X (t)|| ≥ λ ≤ p E (||X (T )|| ) . (50.3.13) λ t∈I If p > 1, ÃÃ E

!p !1/p sup ||X (t)||

µ ≤

t∈[S,T ]

p p−1

¶ p 1/p

E (||X (T )|| )

(50.3.14)

Proof: By Proposition 50.3.2 ||X (t)|| , t ∈ I is a submartingale and so from Theorem 50.3.4, it follows 50.3.13 and 50.3.14 hold. This proves the theorem. Definition 50.3.6 Let K be a set of functions of L1 (Ω, F, P ). Then K is called equi integrable if Z |f | dP = 0. lim sup λ→∞ f ∈K

[|f |≥λ]

Recall that from Corollary 16.7.6 on Page 520 such an equi integrable set of functions is weakly sequentially precompact in L1 (Ω, F, P ) in the sense that if {fn } ⊆ K, there exists a subsequence, {fnk } and a function, f ∈ L1 (Ω, F, P ) such 0 that for all g ∈ L1 (Ω, F, P ) , g (fnk ) → g (f ) .

1484


50.4

Optional Sampling Theorems

50.4.1

Stopping Times And Their Properties

The optional sampling theorem is very useful in the study of martingales and submartingales as will be shown. First it is necessary to define the notion of a stopping time. ∞

Definition 50.4.1 Let (Ω, F, P ) be a probability space and let {Fn }n=1 be an increasing sequence of σ algebras each contained in F, called a discreet filtration. A stopping time is a measurable function, τ which maps Ω to N, τ −1 (A) ∈ F for all A ∈ P (N) , such that for all n ∈ N, [τ ≤ n] ∈ Fn . Note this is equivalent to saying [τ = n] ∈ Fn because [τ = n] = [τ ≤ n] \ [τ ≤ n − 1] . For τ a stopping time define Fτ as follows. Fτ ≡ {A ∈ F : A ∩ [τ ≤ n] ∈ Fn for all n ∈ N} These sets in Fτ are referred to as “prior” to τ . Proposition 50.4.2 For τ a stopping time, Fτ is a σ algebra and if Y (k) is Fk measurable for all k, Y (k) having values in a separable Banach space E, then ω → Y (τ (ω)) is Fτ measurable. Proof: Let An ∈ Fτ . I need to show ∪n An ∈ Fτ . In other words, I need to show that ∪n An ∩ [τ ≤ k] ∈ Fk The left side equals ∪n (An ∩ [τ ≤ k]) which is a countable union of sets of Fk and so Fτ is closed with respect to countable unions. Next suppose A ∈ Fτ . ¡ C ¢ A ∩ [τ ≤ k] ∪ (A ∩ [τ ≤ k]) = Ω ∩ [τ ≤ k]

50.4. OPTIONAL SAMPLING THEOREMS

1485

and Ω ∩ [τ ≤ k] ∈ Fk . Therefore, so is AC ∩ [τ ≤ k] . It remains to verify the last claim. Let B be an open set in E [Y (τ ) ∈ B] = ∪k [τ = k] ∩ [Y (k) ∈ B] Thus [Y (τ ) ∈ B] ∩ [τ ≤ l] = ∪k [τ = k] ∩ [Y (k) ∈ B] ∩ [τ ≤ l] Consider a term in the union. If l ≥ k the term reduces to [τ = k]∩[Y (k) ∈ B] ∈ Fk while if l < k, this term reduces to ∅, also a set of Fk . Therefore, Y (τ ) must be Fτ measurable. This proves the proposition. The following lemma contains the fundamental properties of stopping times for discreet filtrations. Lemma 50.4.3 In the situation of Definition 50.4.1, let σ, τ be two stopping times. Then 1. τ is Fτ measurable 2. Fσ ∩ [σ ≤ τ ] ⊆ Fσ∧τ = Fσ ∩ Fτ 3. Fτ = Fk on [τ = k] for all k. That is if A ∈ Fk , then A ∩ [τ = k] ∈ Fτ and if A ∈ Fτ , then A ∩ [τ = k] ∈ Fk . Also if A ∈ Fτ , and Y ∈ L1 (Ω; E) , Z Z E (Y |Fτ ) dP = E (Y |Fk ) dP A∩[τ =k]

A∩[τ =k]

and E (Y |Fτ ) = E (Y |Fk ) a.e. on [τ = k]. Proof: Consider the first claim. I need to show that [τ ≤ a] ∩ [τ ≤ k] ∈ Fk for every k. However, this is easy if a ≥ k because the left side is then [τ ≤ k] which is given to be in Fk since τ is a stopping time. If a < k, it is also easy because then the left side is [τ ≤ a] ∈ F[a] where [a] is the greatest integer less than or equal to a. Next consider the second claim. Let A ∈ Fσ . I want to show first that A ∩ [σ ≤ τ ] ∈ Fτ In other words, I want to show A ∩ [σ ≤ τ ] ∩ [τ ≤ k] ∈ Fk for all k. This will be done if I can show A ∩ [σ ≤ j] ∩ [τ ≤ k] ∈ Fk

(50.4.15)

1486


for each j ≤ k because the two sets on the left are equal. However, since σ is Fσ measurable, it follows A ∩ [σ ≤ j] ∈ Fj ⊆ Fk for each j ≤ k and so this has shown what I wanted to show, A ∩ [σ ≤ τ ] ∈ Fτ . Now replace the stopping time, τ with the stopping time τ ∧ σ in what was just shown. First note τ ∧ σ is a stopping time because [τ ∧ σ ≤ k] = [τ ≤ k] ∪ [σ ≤ k] ∈ Fk . Thus from 50.4.15 A ∩ [σ ≤ τ ∧ σ] ∈ Fτ ∧σ However the left side equals A ∩ [σ ≤ τ ] . Thus A ∩ [σ ≤ τ ] ∈ Fτ ∧σ This has shown the first part of 2.), Fσ ∩ [σ ≤ τ ] ⊆ Fτ ∧σ . Now 50.4.15 implies if A ∈ Fσ∧τ , all of Ω

z }| { A = A ∩ [σ ∧ τ ≤ τ ] ∈ Fτ and so Fσ∧τ ⊆ Fτ . Similarly, Fσ∧τ ⊆ Fσ which shows Fσ∧τ ⊆ Fτ ∩ Fσ . Next let A ∈ Fτ ∩ Fσ . Then is it in Fσ∧τ ? Is A ∩ [σ ∧ τ ≤ k] ∈ Fk ? Of course this is so because A ∩ [σ ∧ τ ≤ k] = A ∩ ([σ ≤ k] ∪ [τ ≤ k]) = (A ∩ [σ ≤ k]) ∪ (A ∩ [τ ≤ k]) ∈ Fk since both σ, τ are stopping times. This proves part 2.). Now consider part 3.). Let A ∈ Fk . I need to show A ∩ [τ = k] ∈ Fτ . Is A ∩ [τ = k] ∩ [τ ≤ l] ∈ Fl for all l? There are two cases. First suppose l ≥ k. Then the expression on the left reduces to A ∩ [τ = k] = A ∩ ([τ ≤ k] \ [τ ≤ (k − 1)]) ∈ Fk . If l < k, the left side equals ∅ ∈ Fl and so this shows Fk ∩ [τ = k] ⊆ Fτ . Now consider the other inclusion. Let A ∈ Fτ . Then I need to show A∩[τ = k] ∈ Fk . Since A ∈ Fτ , A ∩ [τ ≤ j] ∈ Fj for all j. Therefore, C

A ∩ [τ = k] = A ∩ [τ ≤ k] ∩ (A ∩ [τ ≤ k − 1]) ∈ Fk


1487

It only remains to verify the claim about the conditional expectations. Let A ∈ Fτ be arbitrary. Z Z Z E (Y |Fτ ) dP ≡ Y dP ≡ E (Y |Fk ) dP A∩[τ =k]

A∩[τ =k]

A∩[τ =k]

This is because the set, A ∩ [τ = k] is in both Fk and Fτ . Since A is arbitrary, it follows E (Y |Fτ ) = E (Y |Fk ) a.e. on [τ = k]. This proves the third claim and the Lemma.

50.4.2

Doob Optional Sampling Theorem

With this lemma, here is a major theorem, the optional sampling theorem of Doob. This one is for martingales having values in a Banach space. To begin with, consider the case of a martingale defined on a countable set. Theorem 50.4.4 Let {M (k)} be a martingale having values in E a separable real Banach space with respect to the increasing sequence of σ algebras, {Fk } and let σ, τ be two stopping times such that τ is bounded. Then M (τ ) defined as ω → M (τ (ω)) is integrable and M (σ ∧ τ ) = E (M (τ ) |Fσ ) . Proof: By Proposition 50.4.2 M (τ ) is Fτ measurable. Next note that since τ is bounded, Z ||M (τ (ω))|| dP ≤ Ω

l Z X i=1

||M (i)|| dP < ∞. [τ =i]

This proves the first assertion and makes possible the consideration of conditional expectation. Let l ≥ τ as described above. Then for k ≤ l, by Lemma 50.4.3, Fk ∩ [τ = k] = Fτ ∩ [τ = k] and so if A ∈ Fk ∩ [τ = k] = Fτ ∩ [τ = k] , Z Z E (M (l) |Fτ ) dP ≡ M (l) dP A ZA = E (M (l) |Fk ) dP A Z Z = M (k) dP = M (τ ) dP. A

A

1488


Therefore, since A was arbitrary, E (M (l) |Fτ ) = M (τ ) a.e. on [τ = k] for every k ≤ l. It follows E (M (l) |Fτ ) = M (τ ) a.e.

(50.4.16)

since it is true on each [τ = k] for all k ≤ l. Claim: [σ ≤ τ ] ∈ Fτ ∩ Fl ∩ Fσ . Proof of claim: First I show [σ ≤ τ ] ∈ Fl . [σ ≤ τ ] = ∪li=1 [τ = i] ∩ [σ ≤ i] ∈ Fl This shows it is in Fl . Next I show [σ ≤ τ ] ∈ Fτ . To do this I have to show [σ ≤ τ ] ∩ [τ ≤ k] ∈ Fk for all k. But [σ ≤ τ ] ∩ [τ ≤ k] = ∪∞ i=1 [τ = i] ∩ [σ ≤ i] ∩ [τ ≤ k] and for i > k the terms of the union equal ∅ while for i ≤ k, the terms reduce to [τ = i] ∩ [σ ≤ i] ∈ Fi ⊆ Fk . Thus [σ ≤ τ ] ∈ Fτ . Now consider the claim [σ ≤ τ ] ∈ Fσ . This occurs because [σ ≤ τ ]

= =

∪∞ i=1 [σ = i] ∩ [i ≤ τ ] l ∪i=1 [σ = i] ∩ [i ≤ τ ]

An individual term is in Fσ because [σ = i] ∩ [i ≤ τ ] ∩ [σ ≤ k] = [σ = i] ∩ [i ≤ τ ] ∈ Fi ⊆ Fk when k ≥ i and for k < i, this set, [σ = i] ∩ [i ≤ τ ] ∩ [σ ≤ k] = ∅ ∈ Fk . Thus [σ ≤ τ ] is the finite union of sets in Fσ so it is also in Fσ . I also claim that if A ∈ Fσ , then A ∩ [σ ≤ τ ] ∈ Fσ ∩ Fτ .

(50.4.17)

Here is why. It is only necessary to show A ∩ [σ ≤ τ ] is in Fτ because the above claim verifies [σ ≤ τ ] ∈ Fσ . A ∩ [σ ≤ τ ] ∩ [τ ≤ k] = ∪ki=1 A ∩ [σ ≤ i] ∩ [τ = i]


1489

and by assumption A ∩ [σ ≤ i] ∈ Fi ⊆ Fk while [τ = i] ∈ Fi ⊆ Fk . Thus 50.4.17 holds. Now let A ∈ Fσ . Then using 50.4.16 and Lemma 50.4.3, Z Z E (M (τ ) |Fσ ) dP = E (E (M (l) |Fτ ) |Fσ ) dP A∩[σ≤τ ]

=

A∩[σ≤τ ]

l Z X i=1

=

E (E (M (l) |Fi ) |Fσ ) dP = A∩[σ≤i]∩[τ =i]

=

E (M (i) |Fj ) dP = A∩[σ=j]∩[τ =i]

l X i Z X i=1 j=1

l X i Z X i=1 j=1

E (M (i) |Fσ ) dP A∩[σ≤i]∩[τ =i]

i=1

l X i Z X i=1 j=1

l Z X

M (σ) dP =

A∩[σ=j]∩[τ =i]

Z

Z

=

l Z X i=1

M (σ) dP

A∩[σ≤i]∩[τ =i]

M (σ) dP = A∩[σ≤τ ]

M (j) dP A∩[σ=j]∩[τ =i]

M (σ ∧ τ ) dP

(50.4.18)

A∩[σ≤τ ]

It follows that on [σ ≤ τ ] , E (M (τ ) |Fσ ) = M (σ ∧ τ ) a.e.

(50.4.19)

because M (σ ∧ τ ) is Fσ∧τ ⊆ Fσ measurable thanks to Proposition 50.4.2. Next consider the complement of this set, [σ > τ ] . I want to show that 50.4.18 for [σ > τ ] in place of [σ ≤ τ ]. Letting A ∈ Fσ , Z Z E (M (τ ) |Fσ ) dP = E (M (τ ∧ σ) |Fσ ) dP A∩[σ>τ ] A∩[σ>τ ] Z = M (τ ∧ σ) dP A∩[σ>τ ]

because by Lemma 50.4.3, M (τ ∧ σ) is Fσ∧τ measurable and this is a subset of Fσ . Thus 50.4.19 holds on [σ > τ ] also. It follows 50.4.19 holds a.e. This proves the theorem. ∞ What about submartingales? Recall {X (k)}k=1 is a submartingale if E (X (k + 1) |Fk ) ≥ X (k) where the Fk are an increasing sequence of σ algebras in the usual way. The following is a very interesting result. ∞

Lemma 50.4.5 Let {X (k)}k=0 be a submartingale adapted to the increasing se∞ quence of σ algebras, {Fk } . Then there exists a unique increasing process {A (k)}k=0 such that A (0) = 0 and A (k + 1) is Fk measurable for all k and a martingale, ∞ {M (k)}k=0 such that X (k) = A (k) + M (k) .

1490


Proof: First consider the uniqueness assertion. Suppose A is a process which does what is supposed to do. n−1 X

E (X (k + 1) − X (k) |Fk )

k=0

=

n−1 X

E (A (k + 1) − A (k) |Fk )

k=0

+

n−1 X

E (M (k + 1) − M (k) |Fk )

k=0

Then since {M (k)} is a martingale, n−1 X

E (X (k + 1) − X (k) |Fk ) =

k=0

n−1 X

A (k + 1) − A (k) = A (n)

k=0

This shows uniqueness and gives a formula for A (n) assuming it exists. It is only a matter of verifying this does work. Define A (n) ≡

n−1 X

E (X (k + 1) − X (k) |Fk ) .

k=0

Then A is increasing because from the definition, A (n + 1) − A (n) = E (X (n + 1) − X (n) |Fn ) ≥ 0. Also from the definition above, A (n) is Fn−1 measurable. Consider {X (k) − A (k)} . Why is this a martingale?

=

E (X (k + 1) − A (k + 1) |Fk ) E (X (k + 1) |Fk ) − A (k + 1)

=

E (X (k + 1) |Fk ) −

k X

E (X (j + 1) − X (j) |Fj )

j=0

=

E (X (k + 1) |Fk ) − E (X (k + 1) − X (k) |Fk ) −

k−1 X

E (X (j + 1) − X (j) |Fj )

j=0

=

X (k) −

k−1 X j=0

E (X (j + 1) − X (j) |Fj ) = X (k) − A (k)


1491

Letting M (k) ≡ X (k) − A (k) , this proves the lemma. Note the nonnegative integers could be replaced with any finite set or ordered countable set of numbers with no change in the conclusions of this lemma or the above optional sampling theorem. Next consider the case of a submartingale. Theorem 50.4.6 Let {X (k)} be a submartingale with respect to the increasing sequence of σ algebras, {Fk } and let σ, τ be two stopping times such that τ is bounded. Then X (τ ) defined as ω → X (τ (ω)) is integrable and X (σ ∧ τ ) ≤ E (X (τ ) |Fσ ) . Proof: The claim about X (τ ) being integrable is the same as in Theorem 50.4.4. By Lemma 50.4.5 there is a martingale, {M (k)} and an increasing process {A (k)} such that A (k + 1) is Fk measurable such that X (k) = M (k) + A (k) . Then using Theorem 50.4.4 on the martingale and the fact A is increasing E (X (τ ) |Fσ )

= E (M (τ ) + A (τ ) |Fσ ) = M (τ ∧ σ) + E (A (τ ) |Fσ ) ≥ M (τ ∧ σ) + E (A (τ ∧ σ) |Fσ ) = M (τ ∧ σ) + A (τ ∧ σ) = X (τ ∧ σ) .

This is so provided that for τ a stopping time, A (τ ) is Fτ measurable because in the above, it will follow E (A (τ ∧ σ) |Fσ ) = A (τ ∧ σ) . I need to verify that for each l, [A (τ ) ≤ l] ∩ [τ ≤ k] ∈ Fk for each k. However, this is easy to show. [A (τ ) ≤ l] = ∪i≥0 [A (i) ≤ l] ∩ [τ = i] . Therefore, [A (τ ) ≤ l] ∩ [τ ≤ k] = ∪i≥0 [A (i) ≤ l] ∩ [τ = i] ∩ [τ ≤ k] Consider a term. If k ≥ i [A (i) ≤ l] ∩ [τ = i] ∩ [τ ≤ k] = [A (i) ≤ l] ∩ [τ = i] ∈ Fi ⊆ Fk while if k < i the term equals ∅ also in Fk . Therefore, A (τ ) is Fτ measurable as hoped. This proves the theorem.

1492


50.5

Doob Optional Sampling Continuous Case

50.5.1

Stopping Times

As in the case of discreet martingales, there is something called a stopping time. Definition 50.5.1 Let (Ω, F, P ) be a probability space and let Ft be a filtration. A measurable function, τ : Ω → [0, ∞] is called a stopping time if [τ ≤ t] ∈ Ft for all t ≥ 0. Associated with a stopping time is the σ algebra, Fτ defined by Fτ ≡ {A ∈ F : A ∩ [τ ≤ t] ∈ Ft for all t} . These sets are also called those “prior” to τ . Proposition 50.5.2 Let B be an open subset of a Banach space, E and let X (t) be a right continuous Ft adapted stochastic process such that Ft is normal. Then define τ (ω) ≡ inf {t > 0 : X (t) (ω) ∈ B} . This is called the first hitting time. Then τ is a stopping time. If X (t) is continuous and adapted to Ft , an arbitrary filtration, then if H is a nonempty closed set, τ (ω) ≡ inf {t > 0 : X (t) (ω) ∈ H} is also a stopping time. Proof: Since X (t) is right continuous and B is open, τ (ω) = inf {t > 0, t ∈ Q∩ (0, ∞) : X (t) (ω) ∈ B} I need to verify that [τ ≤ a] ∈ Fa . First consider [τ < a] . Suppose that τ (ω) < a. It follows then that for each ε > 0, there exists t ∈ Q∩ (0, ∞) ∩ [0, a) ≡ Q+ ∩ [0, a) such that X (t) (ω) ∈ B That is,

−1

[τ < a] ≡ τ −1 ([0, a)) ⊆ ∪t∈Q+ ∩[0,a) X (t) −1

Also if ω ∈ ∪t∈Q+ ∩[0,a) X (t) Thus τ (ω) ≤ t < a and so

(B) ∈ Fa .

(B) then for some t ∈ Q+ ∩ [0, a), X (t) (ω) ∈ B.

[τ < a] = ∪t∈Q+ ∩[0,a) X (t)

−1

(B)

which shows τ is measurable. Also this shows [τ < a] ∈ Fa .

50.5. DOOB OPTIONAL SAMPLING CONTINUOUS CASE

1493

¡ ¢ −1 It remains to verify τ −1 ([0,¡ a]) ≡ [τ ≤ a] ∈ Fa . But τ −1 ([0, a]) = ∩∞ [0, a + k1 ) k=1 τ ¢ and it was just shown τ −1 [0, a + k1 ) ∈ Fa+1/k and so τ −1 ([0, a]) ∈ Fa because τ −1 ([0, a]) ∈ Fa+1/k for all k ∈ N and the intersection of these equals ∩s>t Fs ≡ Ft+ = Ft . Now consider the case of a closed set where X is continuous and you are hitting a closed set, H. Let H = ∩∞ n=1 Un where {Un } is a decreasing sequence of open sets, ¡ ¢ 1 dist H, UnC = . n Then τ (ω) ≤ t < ∞ means for all n there exists rn ∈ [0, t] such that X (rn ) (ω) ∈ Un . This is because if this condition holds then taking a subsequence still denoted by rn it can be assumed to converge to s ≤ t and X (s) (ω) ∈ H. Conversely, if X (s) (ω) ∈ H for some s ≤ t, then X (s) (ω) ∈ Un for all n and so you could let rn ≡ s. By continuity of X you can replace [0, t] with [0, t] ∩ Q. Thus [τ ≤ t] = ∩∞ n=1 ∪r∈Q∩[0,t] [X (r) ∈ Un ] ∈ Ft Now consider the case where τ (ω) = ∞. This is the case where X (t) (ω) is never in H for any t. In this case, there are infinitely many of the Un such that for all r ∈ Q∩ [0, t] , X (r) (ω) fails to be in any of these Un . Since they are decreasing, this set is of the form £ ¤ ∞ C [τ = ∞] = ∪∞ m=1 ∩n=m ∩r∈Q∩[0,t] X (r) ∈ Un ∈ Ft Now note that if t ∈ [0, T ] where T ≤ ∞, [τ < T ] = ∪∞ n=1 [τ < n] and this completes showing this type of hitting time is a stopping time. This proves the proposition. Thus there do exist stopping times, the first hitting time being an example. The following is a fundamental observation. Proposition 50.5.3 Let (Ω, F, P ) be a probability space and let σ ≤ τ be two stopping times with respect to a filtration, Ft . Then Fσ ⊆ Fτ . If X (t) is a right continuous stochastic process adapted to Ft , a normal filtration, and τ is a stopping time, ω → X (τ (ω)) is Fτ measurable. Proof: Let A ∈ Fσ . Then A ∩ [σ ≤ t] ∈ Ft for all t ≥ 0. I claim ¡ ¢ A ∩ [τ ≤ t] = ∩s≤t,s∈Q∪{t} A ∩ [σ ≤ s] ∩ [τ ≤ t] Suppose ω ∈ A ∩ [τ ≤ t] . First suppose τ (ω) = t. Then ω ∈ [τ ≤ t] and ω ∈ A ∩ [σ ≤ t] and so ω is in the right side of the above expression. If τ (ω) < t, then

1494


for some s ∈ Q∩ {t} , τ (ω) < s and so ω ∈ [σ ≤ s]∩A. Thus the left side is contained in the right side. Next suppose ω is in the right side. Then it is obviously in the left. This proves the first part because A ∩ [τ ≤ t] ∈ Ft for all t ≥ 0 and so A ∈ Fτ . Next consider X (τ ). Let U be an open set. Since X is right continuous, −1

X (τ )

(U ) ≡ ∪[τ =s,s≥0] X (s)

−1

−1

(U ) = ∪[τ =s,s∈Q+ ] X (s)

(U ) ∈ F

(50.5.20)

It remains to verify X (τ ) is Fτ measurable. To do this, I need to show that −1 X (τ ) (U ) ∈ Fτ for all U open. That is, −1

X (τ )

(U ) ∩ [τ ≤ t] ∈ Ft

for all U open. However, −1

X (τ )

−1

(U ) ≡ ∪[τ =s,s≥0] X (s)

−1

(U ) = ∪[τ =s,s∈Q+ ∪{t}] X (s)

(U )

and so letting [τ = s, s ∈ Q+ ∪ {t} , s ≤ t] equal © ª ω : τ (ω) = s for some s ∈ Q+ ∩ [0, t] ∪ {t} , −1

X (τ )

−1

(U ) ∩ [τ ≤ t] = ∪[τ =s,s∈Q+ ∪{t},s≤t] X (s)

(U ) ∈ Ft .

This proves the proposition. Now consider an increasing family of stopping times, τ (t) (ω → τ (t) (ω)). It turns out this is a submartingale. Example 50.5.4 Let {τ (t)} be an increasing family of stopping times. Then τ (t) is adapted to the σ algebras Fτ (t) and {τ (t)} is a submartingale adapted to these σ algebras. First I need to show that a stopping time, τ is Fτ measurable. Consider [τ ≤ s] . Is this in Fτ ? Is [τ ≤ s]∩[τ ≤ r] ∈ Fr for each r? This is obviously so if s ≤ r because the intersection reduces to [τ ≤ s] ∈ Fs ⊆ Fr . On the other hand, if s > r then the intersection reduces to [τ ≤ r] ∈ Fr and so it is clear that τ is Fτ measurable. It remains to verify it is a submartingale. Let s < t and let A ∈ Fτ (s) Z Z Z ¡ ¢ E τ (t) |Fτ (s) dP ≡ τ (t) dP ≥ τ (s) dP A

A

A

¢ ¡ and this shows E τ (t) |Fτ (s) ≥ τ (s) . Now here is an important example. Example 50.5.5 Let τ be a stopping time and let X be continuous and adapted to the filtration Ft . Then for a > 0, define σ as σ (ω) ≡ inf {t > τ (ω) : ||X (t) (ω) − X (τ (ω))|| = a} Then σ is also a stopping time.


1495

To see this is so, let ½ Y (t) (ω) =

||X (t) (ω) − X (τ (ω))|| if t ≥ τ (ω) 0 if t < τ (ω)

Then Y is continuous and adapted and σ (ω)

= inf {t > τ (ω) : ||X (t) (ω) − X (τ (ω))|| = a} = inf {t > 0 : Y (t) (ω) = a}

By Proposition 50.5.2 σ is a stopping time.

50.5.2

The Optional Sampling Theorem Continuous Case

Next I want a version of the Doob optional sampling theorem which applies to martingales defined on [0, L], L ≤ ∞. First here is a little lemma. Lemma 50.5.6 Let f ∈ L1 (Ω; E, F) where E is a separable Banach space. Then if G is a σ algebra G ⊆ F, ||E (f |G)|| ≤ E (||f || |G) . Proof: First suppose f is a simple function, f (ω) =

n X

ak XFk (ω)

k=1

Letting A ∈ G, !¯¯ Z ¯¯¯¯ ÃX n ¯¯ ¯¯ ¯¯ ||E (f |G)|| dP = ak XFk (ω) |G ¯¯ ¯¯E ¯¯ A A ¯¯

Z

k=1

= =

¯¯ Z ¯¯¯¯X Z X ¯¯ ¯¯ ¯¯ ak E (XFk |G)¯¯ dP ≤ ||ak || E (XFk |G) dP ¯¯ ¯¯ A ¯¯ k A k Ã ! Z Z X ||ak || XFk |G dP = E (||f || |G) dP E A

k

A

and since A is arbitrary, the lemma is proved when f is simple. The general case follows from this and an approximating sequence of simple functions. This proves the lemma. Here is a technical lemma. Lemma 50.5.7 Let X (t) be a right continuous adapted process having values in a separable Banach space E such that the filtration {Ft } is normal. Recall this includes Ft = ∩s>t Fs .

1496

STOCHASTIC PROCESSES ∞

Let {σ n } be a sequence of stopping times whose values are {kT 2−n }k=0 instead of the positive integers and suppose σ is a measurable function for which T 2−n ≥ σ n − σ ≥ 0

(50.5.21)

and X (σ n ) is Fσn measurable. Then σ is a stopping time and so X (σ) given by ω → X (σ (ω)) is Fσ measurable. Proof: I need to verify [σ ≤ t] ∈ Ft . Note that for each p

£ ¤ n −n [σ ≤ t] = ∩∞ n=p σ n ≤ kt T 2

(50.5.22)

ktn T 2−n

where is the first value of σ n at least as large as t. Here is why. If ω ∈ [σ ≤ t] , then for each n ≥ p, the assumed inequality in 50.5.21 implies σ n (ω) cannot be any larger than ktn T 2−n since if it were, it would have to be at least as large as (ktn + 1) T 2−n and the inequality would be violated. Hence the left side of 50.5.22 is contained in the right. If ω is in the right side, then for all n, T 2−n ≥ ktn T 2−n − σ (ω) ≥ 0 Since {ktn T 2−n } is a decreasing sequence converging to t, passing to a limit implies t − σ (ω) ≥ 0 and this shows the© two sidesª are equal. The sequence Fktn T 2−n is decreasing in n because {ktn T 2−n } is a decreasing sequence. Therefore, for each p £ ¤ n −n [σ ≤ t] = ∩∞ ∈ Fktp T 2−p n=p σ n ≤ kt T 2 Consequently [σ ≤ t] ∈ ∩p Fktp T 2−p = Ft because the filtration is normal. It follows from Proposition 50.5.3 that X (σ) is Fσ measurable. This proves the lemma. Theorem 50.5.8 Let {M (t)} be a right continuous martingale having values in E a separable real Banach space with respect to the increasing sequence of σ algebras, {Ft } which is assumed to be a normal filtration satisfying, Ft = ∩s>t Fs , for t ∈ [0, L] , L ≤ ∞ and let σ, τ be two stopping times with τ bounded. Then M (τ ) defined as ω → M (τ (ω)) is integrable and M (σ ∧ τ ) = E (M (τ ) |Fσ ) .


1497

Proof: To begin with I need to show the random variable M (τ ) is integrable given that τ is a stopping time having values in [0, T ], T being the bound for τ . Otherwise nothing makes sense. To do this, let X 2−n (k + 1) T Xτ −1 ((k2−n T,(k+1)T 2−n ]) (ω) . τ n (ω) ≡ k≥0

Then τ n is a stopping time for the filtration Ft because ¡ ¢ [τ n ≤ t] = τ −1 (0, (k + 1) T 2−n ] ∈ F(k+1)T 2−n where (k + 1) T 2−n ≤ t. Also τ n takes values at equally spaced points from 0 up to T and for each ω, |τ n (ω) − τ (ω)| < 2−n T 2n

for all ω ∈ Ω while τ n (ω) ≥ τ (ω) . Consider {M (kT 2−n )}k=0 . It follows easily 2n that this is a martingale on {kT 2−n }k=0 , the σ algebras being FkT 2−n . Also τ n is 2n a stopping time on {kT 2−n }k=0 with respect to these σ algebras since £ ¤ £ ¤ τ n ≤ (k + 1) T 2−n = τ ≤ (k + 1) T 2−n ∈ F(k+1)T 2−n This is because from the definition of τ n , τ n = (k + 1) T 2−n corresponds to τ ∈ (k2−n T, (k + 1) T 2−n ] and so forth. The function M (τ n ) is measurable because for B an open set, ¡ ¢−1 £ ¤ n −1 M (τ n ) (B) = ∪2j=0 M jT 2−n (B) ∩ τ n = jT 2−n ¡ ¢−1 £ ¤ n = ∪2j=0 M jT 2−n (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∈ FjT 2−n In addition, £ ¤ (B) ∩ τ n ≤ kT 2−n ¡ ¢−1 £ ¤ £ ¤ n = ∪2j=0 M jT 2−n (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∩ τ n ≤ kT 2−n ¡ ¢−1 £ ¤ £ ¤ n = ∪2j=0 M jT 2−n (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∩ τ ≤ kT 2−n −1

M (τ n )

which is in FkT 2−n . This follows from looking at the individual terms. If k ≥ j the term reduces to ¡ ¢−1 £ ¤ M jT 2−n (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∈ FjT 2−n ⊆ FkT 2−n while if k ≤ j − 1, the term reduces to ∅ which is still in FkT 2−n . Thus M (τ n ) is Fτ n measurable. Now consider the random variables M (τ n ) . Also let A ∈ FT . From Lemma 50.5.6, Z 2n Z X ¯¯ ¡ ¢¯¯ ¯¯M kT 2−n ¯¯ dP ||M (τ n )|| dP = A

k=0

[τ n =kT 2−n ]∩A

1498

STOCHASTIC PROCESSES 2 Z X n

=

k=0

[τ n =kT 2−n ]∩A

2 Z X

||E (M (T ) |FkT 2−n )|| dP

n

≤

k=0

[τ n =kT 2−n ]∩A

E (||M (T )|| |FkT 2−n ) dP Z

2 Z X n

=

k=0

||M (T )|| dP

||M (T )|| dP = [τ n =kT 2−n ]∩A

(50.5.23)

A

Letting A = Ω, this shows the random variables M (τ n ) are at least in L1 (Ω; E) . Since M is right continuous, ||M (τ )|| = lim ||M (τ n )|| n→∞

and so by Fatou’s lemma, Z Z ||M (τ )|| dP ≤ lim inf ||M (τ n )|| dP ≤ E (||M (T )||) < ∞. n→∞

Ω

Ω

This shows the random variable, M (τ ) is at least in L1 (Ω; E). However, more ∞ importantly, 50.5.23 also shows the random variables {M (τ n )}n=1 are uniformly integrable. Consider the main claim now. Letting σ, τ be stopping times with τ bounded, it follows that for σ n and τ n as above, it follows from Theorem 50.4.4 M (σ n ∧ τ n ) = E (M (τ n ) |Fσn ) Thus, taking A ∈ Fσ and recalling σ ≤ σ n so that by Proposition 50.5.3, Fσ ⊆ Fσn , Z Z Z M (σ n ∧ τ n ) dP = E (M (τ n ) |Fσn ) dP = M (τ n ) dP. A

A

A

Now passing to a limit as n → ∞, the Vitali convergence theorem, Theorem 8.5.3 on Page 213 and the right continuity of M implies one can pass to the limit in the above and conclude Z Z M (σ ∧ τ ) dP = M (τ ) dP. A

A

By Proposition 50.5.3, M (σ ∧ τ ) is Fσ∧τ ⊆ Fσ measurable showing E (M (τ ) |Fσ ) = M (σ ∧ τ ) . This proves the theorem. A similar theorem is available for submartingales defined on [0, L] , L ≤ ∞. The hard part is showing the uniform integrablity of the approximate processes. Theorem 50.5.9 Let {X (t)} be a right continuous submartingale with respect to the increasing sequence of σ algebras, {Ft } which is assumed to be a normal filtration, Ft = ∩s>t Fs ,


1499

for t ∈ [0, L] , L ≤ ∞ and let σ, τ be two stopping times with τ bounded. Then X (τ ) defined as ω → X (τ (ω)) is integrable and X (σ ∧ τ ) ≤ E (X (τ ) |Fσ ) . Proof: To begin with I need to show the random variable X (τ ) is integrable given that τ is a stopping time having values in [0, T ], T being the bound for τ . Otherwise nothing makes sense. To do this, let X τ n (ω) ≡ 2−n (k + 1) T Xτ −1 ((k2−n T,(k+1)T 2−n ]) (ω) . k≥0

Then τ n is a stopping time for the filtration Ft because ¡ ¢ [τ n ≤ t] = τ −1 (0, (k + 1) T 2−n ] ∈ F(k+1)T 2−n where (k + 1) T 2−n ≤ t. Also τ n takes values at equally spaced points from 0 up to T and for each ω, |τ n (ω) − τ (ω)| < 2−n T 2n

for all ω ∈ Ω while τ n (ω) ≥ τ (ω) . Consider {X (kT 2−n )}k=0 . It follows easily that 2n this is a submartingale on {kT 2−n }k=0 , the σ algebras being FkT 2−n . Also τ n is a n 2 stopping time on {kT 2−n }k=0 with respect to these σ algebras since £ ¤ £ ¤ τ n ≤ (k + 1) T 2−n = τ ≤ (k + 1) T 2−n ∈ F(k+1)T 2−n This is because from the definition of τ n , τ n = (k + 1) T 2−n corresponds to τ ∈ (k2−n T, (k + 1) T 2−n ] and so forth. The function X (τ n ) is measurable because for B an open set, −1

X (τ n )

¡ ¢−1 £ ¤ n (B) ∩ τ n = jT 2−n (B) = ∪2j=0 X jT 2−n

¡ ¢−1 £ ¤ n = ∪2j=0 X jT 2−n (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∈ FjT 2−n In addition, £ ¤ (B) ∩ τ n ≤ kT 2−n ¡ ¢−1 £ ¤ £ ¤ n ∪2j=0 X jT 2−n (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∩ τ n ≤ kT 2−n £ ¤ £ ¤ ¡ ¢−1 n (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∩ τ ≤ kT 2−n ∪2j=0 X jT 2−n −1

X (τ n ) = =

which is in FkT 2−n . This follows from looking at the individual terms. If k ≥ j the term reduces to £ ¤ ¡ ¢−1 (B) ∩ τ ∈ ((j − 1) 2−n T, jT 2−n ] ∈ FjT 2−n ⊆ FkT 2−n X jT 2−n

1500


while if k ≤ j − 1, the term reduces to ∅ which is still in FkT 2−n . Thus X (τ n ) is Fτ n measurable. Now consider the random variables X (τ n ) . Each is in L1 (Ω) thanks to the assumption that τ is bounded. Z 2n Z X ¯ ¡ ¢¯ ¯X kT 2−n ¯ dP < ∞. |X (τ n )| dP = (50.5.24) Ω

k=0

[τ n =kT 2−n ]

The challenge is to show {X (τ n )} are uniformly integrable. The idea is that X (τ ) is the limit of these due to the right continuity of X. Then the uniform integrability can be used to transfer integrability of these random variables to X (τ ). By the discreet optional sampling theorem, Theorem 50.4.6, ¡ ¢ E X (τ n ) |Fτ n+1 ≥ X (τ n+1 ) and so

¡ ¡ ¢¢ E (X (τ n )) = E E X (τ n ) |Fτ n+1 ≥ E (X (τ n+1 ))

Also since τ n ≥ 0, it also follows from the same reasoning and the discreet optional sampling theorem, E (X (τ n )) ≥ E (X (0)) . Now let ε > 0 be given. Since {E (X (τ n ))} is decreasing and bounded below, there exists fixed k large enough, depending on ε such that ¯ ¯ ¯ ¯ (50.5.25) ¯E (X (τ k )) − lim E (X (τ m ))¯ < ε. m→∞

In what follows n > k. Let λ > 0. Then Z Z Z |X (τ n )| dP = X (τ n ) dP + [|X(τ n )|≥λ]

[X(τ n )≥λ]

Z =

Z =

Z

X (τ n ) dP + [X(τ n )≥λ]

Z

[X(τ n )≥λ]

(−X (τ n )) dP [−Xn ≥λ]

X (τ n ) dP +

(−X (τ n )) dP [X(τ n )≤−λ]

Z

(−X (τ n )) dP + Ω

X (τ n ) dP [−X(τ n ) 0 is given, there exists λ large enough that for all n > 0, Z |X (τ n )| dP < ε. [|X(τ n )|≥λ] ∞

Thus the functions {X (τ n )}n=1 are equiintegrable and so they are uniformly integrable. It follows from this that it is also the case that for any σ a stopping time, ∞ bounded or not, {X (σ n ∧ τ n )}n=1 are uniformly integrable. You just repeat the argument with σ ∧ τ in place of τ . Consider the main claim now. Letting σ, τ be stopping times with τ bounded, it follows that for σ n and τ n as above and the optional sampling theorem for discreet submartingales, X (σ n ∧ τ n ) ≤ E (X (τ n ) |Fσn ) Thus, taking A ∈ Fσ and recalling σ ≤ σ n so that Fσ ⊆ Fσn , both σ and σ n being considered as stopping times for Ft , Z Z Z X (σ n ∧ τ n ) dP ≤ E (X (τ n ) |Fσn ) dP ≡ X (τ n ) dP. A

A

A

1502


Now passing to a limit as n → ∞, the Vitali convergence theorem, Theorem 8.5.3 on Page 213 and the right continuity of X implies one can pass to the limit in the above and conclude Z Z Z X (σ ∧ τ ) dP ≤ X (τ ) dP = E (X (τ ) |Fσ ) dP A

A

A

By Proposition 50.5.3, X (σ ∧ τ ) is Fσ∧τ ⊆ Fσ measurable which shows that since A ∈ Fσ was arbitrary, E (X (τ ) |Fσ ) ≥ X (σ ∧ τ ) . This proves the theorem. Note that a function defined on a countable ordered set such as the integers or equally spaced points is right continuous. Here is a useful lemma which was essentially proved in the above. However, it is interesting and so I will list it again and also give a proof. Lemma 50.5.10 Suppose E (|Xn |) < ∞ for all n, Xn is Fn measurable, Fn+1 ⊆ Fn for all n ∈ N, and there exist X∞ F∞ measurable such that F∞ ⊆ Fn for all n and X0 F0 measurable such that F0 ⊇ Fn for all n such that for all n ∈ {0, 1, · · · } , E (Xn |Fn+1 ) ≥ Xn+1 , E (Xn |F∞ ) ≥ X∞ . Then {Xn : n ∈ N} is uniformly integrable. Proof: E (Xn+1 ) ≤ E (E (Xn |Fn+1 )) = E (Xn ) Therefore, the sequence E (Xn ) is a decreasing sequence bounded below by E (X∞ ) so it has a limit. Let k be large enough that ¯ ¯ ¯ ¯ (50.5.27) ¯E (Xk ) − lim E (Xm )¯ < ε m→∞

and suppose n > k. Then if λ > 0, Z Z |Xn | dP = [|Xn |≥λ]

Z Xn dP +

[Xn ≥λ]

Z

Z

=

Z

Xn dP + Z

[Xn ≥λ]

=

Ω

Z Xn dP −

[Xn ≥λ]

(−Xn ) dP [Xn ≤−λ]

Ω

(−Xn ) dP − Z Xn dP +

(−Xn ) dP [−Xn b > a > lim

inf

r→s+,r∈Q

X (r, ω) .

(50.6.29)

50.6. RIGHT CONTINUITY OF SUBMARTINGALES

1505

If this is so, then in (s, t) ∩ Q there must be infinitely many values of r ∈ Q such that X (r, ω) ≥ b as well as infinitely many values of r ∈ Q such that X (r, ω) ≤ a. Note this involves the consideration of a limit from one side. Thus, since it is a limit from one side only, there are an arbitrarily large number of upcrossings between s and t. Therefore, letting M be a large positive number, it follows that for all n sufficiently large, n [0, t] (ω) ≥ M U[a,b] which implies

h i n ω ∈ U[a,b] [0, t] ≥ M

which from 50.6.28 is a set of measure no more than µ ³ ´¶ 1 1 + E (X (t) − a) . M b−a This has shown that the set of ω such that for some s ∈ [0, t) 50.6.29 holds is contained in the set h i ∞ n N[a,b] ≡ ∩∞ M =1 ∪n=1 U[a,b] [0, t] ≥ M Now the sets,

h i n U[a,b] [0, t] ≥ M

are increasing in n and each has measure less than µ ³ ´¶ 1 1 + E (X (t) − a) M b−a and so

µ ³ h i´ ³ ´¶ 1 1 + n P ∪∞ U [0, t] ≥ M ≤ E (X (t) − a) . n=1 [a,b] M b−a

which shows that

µ ³ ´¶ ¡ ¢ 1 1 + P N[a,b] ≤ E (X (t) − a) M b−a ¡ ¢ for every M and therefore, P N[a,b] = 0. Therefore, corresponding to a < b, there exists a set of measure 0, N[a,b] such that for ω ∈ / N[a,b] 50.6.29 is not true for any s ∈ [0, t). Let N ≡ ∪a,b∈Q N[a,b] , a set of measure 0 with the property that if ω ∈ / N, then 50.6.29 fails to hold for any pair of rational numbers, a < b for any s ∈ [0, t). Thus for ω ∈ / N, lim

r→s+,r∈Q

X (r, ω)

exists for all s ∈ [0, t). Similar reasoning applies to show the existence of the limit lim

r→s−,r∈Q

X (r, ω) .

1506


for all s ∈ (0, t] whenever ω is outside of a set of measure zero. Of course, this exceptional set depends on t. However, if this exceptional set is denoted as Nt , one could consider N ≡ ∪∞ n=1 Nn . It is obvious there is no change if Q is replaced with any countable dense subset. This proves the theorem. Of course the above theorem does not say the left and right limits are equal, just that they exist in some way for ω not in some set of measure zero. Also it has not been shown that limr→s+,r∈Q X (r, ω) = X (r, ω) for a.e. ω. Corollary 50.6.2 In the situation of Theorem 50.6.1, let s > 0 and let D1 and D2 be two countable dense subsets of R. Then lim

X (r, ω) =

lim

X (r, ω) =

r→s−,r∈D1

r→s+,r∈D1

lim

X (r, ω) a.e. ω

lim

X (r, ω) a.e. ω

r→s−,r∈D2

r→s+,r∈D2

© ª Proof: Let rni be an increasing sequence from Di converging to s and let N be the exceptional set corresponding to the countable dense set D1 ∪ D2 . Then for ω∈ / N, and i = 1, 2, ¢ ¡ lim X (r, ω) lim X (r, ω) = lim X rni , ω = r→s−,r∈D1 ∪D2

n→∞

r→s−,r∈Di

The other claim is similar. This proves the corollary. Now here is an impressive lemma about submartingales and uniform integrability. Lemma 50.6.3 Let X (t) be a submartingale adapted to a filtration Ft . Let {rk } ⊆ ∞ [s, t) be a decreasing sequence converging to s. Then {X (rj )}j=1 is uniformly integrable. Proof: First I will show the sequence is equiintegrable. I need to show that for all ε > 0 there exists λ large enough that for all n Z |X (rn )| dP < ε. [|X(rn )|≥λ]

Let ε > 0 be given. Since {X (r)}r≥0 is a submartingale, E (X (rn )) is a decreasing sequence bounded below by E (X (s)). This is because for rn < rk , E (X (rn )) ≤ E (E (X (rk ) |Fn )) = E (X (rk )) Pick k such that E (X (rk )) − lim E (X (rn )) n→∞ ¯ ¯ ¯ ¯ = ¯E (X (rk )) − lim E (X (rn ))¯ < ε/2. n→∞

50.6. RIGHT CONTINUITY OF SUBMARTINGALES Then for n > k, Z

Z

Z

|X (rn )| dP = [|X(rn )|≥λ]

X (rn ) dP + [X(rn )≥λ]

Z =

1507

−X (rn ) dP [X(rn )≤−λ]

Z

Z

X (rn ) dP +

X (rn ) dP Z X (rn ) dP + E (X (rk ) |Fn ) dP − X (rn ) dP [X(rn )≥λ] [X(rn )>−λ] Ω Z Z Z X (rk ) dP + X (rk ) dP − X (rk ) dP + ε/2 [X(rn )≥λ] [X(rn )>−λ] Ω Z Z (−X (rk )) dP + ε/2 X (rk ) dP + [X(rn )≥λ] [X(rn )≤−λ] Z |X (rk )| dP + ε/2 [|X(rn )|≥λ] Z |X (rk )| dP + ε/2 (50.6.30) [sup{|X(r)|≥λ:r∈{rj }∞ j=1 }] [X(rn )≥λ]

≤ ≤ = = ≤

X (rn ) dP −

[X(rn )>−λ]

Z

From maximal inequalities of Theorem 48.4.5 Ã" #! P

Ω

Z

sup

|X (r)| ≥ λ

r∈{rn ,rn−1 ,··· ,r1 }

≤

2E (|X (t)| + |X (0)|) C ≡ λ λ

and so, letting n → ∞, Ã" P

#! sup

r∈{rn }∞ n=1

|X (r)| ≥ λ

≤

C . λ

It follows that for λ sufficiently large the first term in 50.6.30 is smaller than ε/2 because k is fixed. Now this shows there is a choice of λ such that for all n > k, Z |X (rn )| dP < ε [|X(rn )|≥λ]

There are only finitely many rn for n ≤ k and by choosing λ sufficiently large the above formula can be made to hold for these also, thus showing {X (rn )} is equi integrable. Now this implies the sequence of random variables is uniformly integrable as well. Let ε > 0 be given and choose λ large enough that for all n, Z |X (rn )| dP < ε/2 [|X(rn )|≥λ]

1508


Then let A be a measurable set. Z Z Z |X (rn )| dP = |X (rn )| dP + |X (rn )| dP A A∩[|X(rn )|≥λ] A∩[|X(rn )| λ]) ≤

In case p > 1 then for each t ≤ M, µZ ¶1/p p ≤ |X ∗ (t)| dP Ω

1511

Z p

X[X ∗ (M )>λ] X (M ) dP

p p−1

µZ

(50.7.32)

¶1/p

p

(50.7.33)

X (M ) dP Ω

Proof: First note that from the right continuity of X, X ∗ (t) is a random variable. This is because this right continuity makes it possible to replace (0, t) with (0, t) ∩ Q in the definition of X ∗ (t). Define the stopping time T ≡ min (inf {t > 0 : X (t) > λ} , M ) . (The infimum over an empty set will equal ∞.)This is a stopping time because it is a continuous function of the stopping time of Proposition 50.5.2. Since f (x) = xp p is convex and increasing {X (t) } is also a submartingale. Then by the optional p p p sampling theorem, X (0) , X (T ) , X (M ) is a submartingale and so p

p

E (X (T ) ) ≤ E (X (M ) ) which implies Z

Z

Z

p

p

X (T ) dP +

p

X (T ) dP ≤

[T <M ]

[T =M ]

X (M ) dP. Ω

On [T < M ] , then since X (t) is right continuous, X (T ) ≥ λ. On [T = M ] , X (T ) = X (M ) and so

Z p

λp P ([T < M ]) ≤

X[T <M ] X (M ) dP

Also [T < M ] = [X ∗ (M ) > λ]. Therefore, this yields 50.7.32. Next if p > 1, then from the first part, Z Z Z ∞ p p ∗ |X (t)| dP ≤ |X (M )| dP = pλp−1 P ([X ∗ (M ) > λ]) dλ Ω

Ω

Z

≤ =

∞

0

1 λ p λ 0 Z Z p X (M ) Ω

p

X ∗ (M )

λp−2 dλdP

p−1

X (M ) Ω

≤

X[X ∗ (M )>λ] X (M ) dP dλ

0

Z =

Z

p−1

p p−1

µZ

X ∗ (M ) p−1 ∗

p

dP ¶1/p0 µZ

p

X (M ) dP Ω

X (M ) dP Ω

¶1/p

1512


This establishes 50.7.33. Here is another sort of maximal inequality in which X (t) is not assumed nonnegative. Theorem 50.7.2 Let {X (t)} be a submartingale adapted to the normal filtration Ft for t ∈ [0, M ] and X ∗ (t) defined as in Theorem 50.7.1 P ([X ∗ (M ) > λ]) ≤

1 E (|X (M )|) λ

(50.7.34)

For t > 0, let X∗ (t) = inf {X (s) : s < t} . Then P ([X∗ (M ) < −λ]) ≤

1 E (|X (M )| + |X (0)|) λ

(50.7.35)

Also P ([sup {|X (s)| : s < M } > λ]) ≤

2 E (|X (M )| + |X (0)|) λ

(50.7.36)

Proof: First let the stopping time T be given by T = min (inf {t : X (t) > λ} , M ) Then by optional sampling theorem X (0) , X (T ) , X (M ) is a submartingale and so E (X (T )) ≤ E (X (M )) which implies Z

Z

Z

X (T ) dP + [T <M ]

X (T ) dP ≤ [T =M ]

X (M ) dP. Ω

On [T < M ] , X (T ) ≥ λ due to right continuity. Hence Z ³ ´ + λP ([T < M ]) ≤ X (M ) dP ≤ E X (M ) . [T <M ]

It follows P ([X ∗ (M ) > λ]) ≤

1 E (|X (M )|) . λ

which verifies 50.7.34. Next let T = min (inf {t : X (t) < −λ} , M ) then as before, Z

Z X (T ) dP +

[T <M ]

Z X (T ) dP ≥

[T =M ]

X (0) dP Ω

50.8. CONTINUOUS SUBMARTINGALE CONVERGENCE THEOREM and so

Z

Z

−λ

Z

dP ≥ − [T <M ]

1513

X (M ) dP + [T =M ]

X (0) dP Ω

and so since [T < M ] = [X∗ (M ) < −λ] P ([X∗ (M ) < −λ])

= ≤ ≤

P ([T < M ]) Z Z 1 1 X (M ) dP − X (0) dP λ [T =M ] λ Ω 1 E (|X (M )| + |X (0)|) λ

and this proves 50.7.35. Finally, combining the above two inequalities, P ([sup {|X (s)| : s < M } > λ]) = P ([X∗ (M ) < −λ]) + P ([X ∗ (M ) > λ]) 2 E (|X (M )| + |X (0)|) λ This verifies 50.7.36 and proves the theorem. ≤

50.8

Continuous Submartingale Convergence Theorem

In this section, {X (t)} will be a right continuous submartingale. Lemma 50.8.1 Let {X (t)} be a right continuous submartingale adapted to a normal filtration, Ft and let S be a stopping time. Let T ≡ inf {t > S : X (t) ∈ B} where B is an open set. Then T is also a stopping time. Proof: By the right continuity of X, T (ω) = inf {t > S, t ∈ Q : X (t) ∈ B} Therefore for p ∈ N, · T 0 : X (t) < a} , M ) , T1 ≡ min (inf {t > T0 : X (t) > b} , M ) , ≡ min (inf {t > T1 : X (t) < a} , M ) , T3 ≡ min (inf {t > T2 : X (t) > b} , M ) ,

etc. where the infimum of an empty set is ∞. Then by Lemma 50.8.1 this is an increasing sequence of stopping times. Let nM U[a,b]

≡ ≤ ≤

lim

ε→0

n X

X (T2k+1 ) − X (T2k ) ε + |X (T2k+1 ) − X (T2k )|

k=0 n X

1 b−a 1 b−a

k=0 n X

X (T2k+1 ) − X (T2k ) X (T2k+1 ) − X (T2k )

k=0

nM Note that an upcrossing occurs when T2k+1 > T2k . Then U[a,b] is clearly a random variable and it records the number of upcrossings occuring for t ≤ M using only n of the stopping times. Then by taking the expected value of both sides and using the optional sampling theorem which implies E (X (T2k )) − E (X (T2k−1 )) ≥ 0 due to the fact {X (Tk )} is a submartingale,

³ ´ nM E U[a,b] ≤

n

1 X E (X (T2k+1 )) − E (X (T2k )) b−a k=0

1 ≤ b−a

n X k=0

n

1 X E (X (T2k+1 )) − E (X (T2k )) + E (X (T2k )) − E (X (T2k−1 )) b−a k=1

≤ ≤

1 (E (X (T2n+1 )) − E (X (T0 ))) b−a 1 (E (X (M )) − E (X (0))) b−a

which does not depend on n. Let n → ∞ to obtain ³ ´ 1 M E U[a,b] ≤ (E (X (M )) − E (X (0))) b−a M where U[a,b] is the number of upcrossings of {X (t)} on [0, M ] . This proves the following interesting upcrossing estimate.

50.8. CONTINUOUS SUBMARTINGALE CONVERGENCE THEOREM

1515

Lemma 50.8.2 Let {X (t)} be a right continuous submartingale adapted to a norM mal filtration Ft for t ∈ [0, M ] . Then if U[a,b] is defined to be the number of upcrossings of {X (t)} for t ∈ [0, M ] , then this is a random variable and ³ ´ 1 M E U[a,b] ≤ (E (X (M )) − E (X (0))) . b−a With this it is easy to prove a continuous submartingale convergence theorem. Theorem 50.8.3 Let {X (t)} be a right continuous submartingale such that sup {E (|X (t)|)} = C < ∞. Then there exists X∞ ∈ L1 (Ω) such that lim X (t) (ω) = X∞ (ω) a.e. ω.

t→∞

Proof: Let U[a,b] denote the upcrossings for t ≥ 0. Then M U[a,b] = lim U[a,b] . M →∞

From Lemma 50.8.2 and the assumption of this theorem, there exists a constant, C independent of M such that ³ ´ 2C M E U[a,b] ≤ . b−a Letting M → ∞, it follows from monotone convergence theorem that ¡ ¢ 2C E U[a,b] ≤ b−a also. Therefore, there exists a set of measure 0, Nab such that if ω ∈ / Nab , then U[a,b] (ω) < ∞. That is, there are only finitely many upcrossings. Now let N = ∪ {Nab : a, b ∈ Q} . It follows that for ω ∈ / N, it cannot happen that lim sup X (t) (ω) − lim inf X (t) (ω) > 0 t→∞

t→∞

because if this expression is positive, there would be arbitrarily large values of t where X (t) (ω) > b and arbitrarily large values of t where X (t) (ω) < a where a, b are rational numbers chosen such that lim sup X (t) (ω) > b > a > lim inf X (t) (ω) t→∞

t→∞

Thus there would be infinitely many upcrossings which is not allowed for ω ∈ / N. Therefore, the limit limt→∞ X (t) (ω) exists for a.e. ω. Let X∞ (ω) equal this limit

1516


for ω ∈ / N and let X∞ (ω) = 0 for ω ∈ N . Then X∞ is measurable and by Fatou’s lemma, Z Z |X∞ (ω)| dP ≤ lim inf |X (n) (ω)| dP < C n→∞

Ω

Ω

This proves the theorem. Now here is an interesting result due to Doob. Theorem 50.8.4 Let {M (t)} be a right continuous real martingale adapted to the normal filtration Ft . Then the following are equivalent. 1. The random variables M (t) are equiintegrable. 2. There exists M (∞) ∈ L1 (Ω) such that limt→∞ ||M (∞) − M (t)||L1 (Ω) = 0. In this case, M (t) = E (M (∞) |Ft ) and convergence also takes place pointwise. Proof: Suppose the equiintegrable condition. Then there exists λ large enough that for all t, Z |M (t)| dt < 1. [|M (t)|≥λ]

It follows that for all t, Z |M (t)| dP

Z

Z

=

Ω

|M (t)| dt +

|M (t)| dt

[|M (t)|≥λ]

≤

[|M (t)| T. In this case, the equiintegrability of the M (t) follows because for t < T, Z Z |M (t)| dP = |E (M (T ) |Ft )| dP [|M (t)|>λ] [|M (t)|>λ] Z ≤ |M (T )| dP [|M (t)|>λ]

and from Theorem 50.7.2, P (|M (t)| > λ) ≤ P ([M ∗ (t) > λ]) ≤

1 λ

Z |M (T )| dP. Ω

Definition 50.9.1 Let M be a process adapted to the filtration Ft and let τ be a stopping time. Then M τ , called the stopped process is defined by M τ (t) ≡ M (τ ∧ t) . With this definition, here is a simple lemma. Lemma 50.9.2 Let M be a martingale adapted to the normal filtration Ft and let τ be a stopping time. Then M τ is also a martingale adapted to the filtration Ft . Proof:Let s < t. By the Doob optional sampling theorem, E (M τ (t) |Fs ) ≡ E (M (τ ∧ t) |Fs ) = M (τ ∧ t ∧ s) = M τ (s) . Theorem 50.9.3 Let {M (t)} be a continuous real valued martingale adapted to the normal filtration Ft and let M ∗ ≡ sup {|M (t)| : t ≥ 0} and M (0) = 0. Letting τ x ≡ inf {t > 0 : M (t) = x}

1518


Then if a < 0 0]) ≥ (b − a) P ([τ b < τ a ]) and (b − a) P ([τ a < τ b ]) ≤ bP ([M ∗ > 0]) ≤ (b − a) P ([τ a ≤ τ b ]) . Proof: For x ∈ R, define τ x ≡ inf {t ∈ R such that M (t) = x} with the usual convention that inf (∅) = ∞. Let a < 0 0. Then by the Doob optional sampling theorem, E (M (τ ∧ t)) =

E (E (M (t) |Fτ )) = E (M (t))

(50.9.38)

= E (E (M (t) |F0 )) = E (M (0)) = 0.

(50.9.39)

Observe the martingale M τ must be bounded. There are two cases according to whether τ = ∞. If τ = ∞, then M (t) never hits a or b so M (t) has values between a and b. In this case M τ (t) = M (t) ∈ [a, b] . On the other hand, you could have τ < ∞. Then in this case M τ (t) is eventually equal to either a or b depending on which it hits first. In either case, the martingale M τ is bounded and lim M τ (t) (ω) = M τ (∞) (ω) = M (τ ) (ω)

t→∞

and since the M τ (t) are bounded, the dominated convergence theorem implies E (M (τ )) = lim E (M (τ ∧ t)) = 0. t→∞

This proves the claim. Recall M ∗ (ω) ≡ sup {|M (t) (ω)| : t ∈ [0, ∞]} . Also note that [τ a = τ b ] = [τ = ∞]. Now from the claim, Z 0 = E (M (τ )) = M (τ ) dP [τ a 0]

Consider this last term. By the definition, [τ a = τ b ] corresponds to M (t) never hitting either a or b. Since M (0) = 0, this can only happen if M (t) has values in [a, b] . Therefore, this last term satisfies aP ([τ a = τ b ] ∩ [M ∗ > 0]) Z ≤

M (∞) dP [τ a =τ b ]∩[M ∗ >0]

≤

bP ([τ a = τ b ] ∩ [M ∗ > 0])

(50.9.42)

Thus from 50.9.41 0 ≤ aP ([τ a < τ b ]) + bP ([τ b < τ a ]) + bP ([τ a = τ b ] ∩ [M ∗ > 0]) Note that [τ b < τ a ] ⊆ [M ∗ > 0]. Thus the above reduces to 0 ≤ aP ([τ a < τ b ]) + bP ([τ b ≤ τ a ] ∩ [M ∗ > 0]) and so 0 ≤ aP ([τ a < τ b ]) + b (P ([M ∗ > 0]) − P ([τ a < τ b ])) which implies (b − a) P ([τ a < τ b ]) ≤ bP ([M ∗ > 0])

(50.9.43)

This gives an estimate for P ([τ a < τ b ]) . Next I will use the bottom half of 50.9.42 to get an estimate for P ([τ a ≤ τ b ]) . Thus from 50.9.41, 0

≥ aP ([τ a < τ b ]) + bP ([τ b < τ a ]) + aP ([τ a = τ b ] ∩ [M ∗ > 0]) = aP ([τ a ≤ τ b ] ∩ [M ∗ > 0]) + bP ([τ b < τ a ]) = aP ([τ a ≤ τ b ] ∩ [M ∗ > 0]) + b (P ([M ∗ > 0]) − P ([τ a ≤ τ b ] ∩ [M ∗ > 0]))

which shows (b − a) P ([τ a ≤ τ b ]) ≥ ≥

(b − a) P ([τ a ≤ τ b ] ∩ [M ∗ > 0]) bP ([M ∗ > 0])

Combining these inequalities gives (b − a) P ([τ a < τ b ]) ≤

(50.9.44)

1520


bP ([M ∗ > 0]) ≤ (b − a) P ([τ a ≤ τ b ]) .

(50.9.45)

Note that if you knew that it is not possible that τ a could ever equal τ b this would yield an equality bP ([M ∗ > 0]) = (b − a) P ([τ a < τ b ]) . Using the same reasoning, it is possible to estimate P ([τ b < τ a ]) , P ([τ b ≤ τ a ]) . Begin with 50.9.41 again and use the inequality 50.9.42 again. Thus 0 ≤ aP ([τ a < τ b ]) + bP ([τ b < τ a ]) + bP ([τ a = τ b ] ∩ [M ∗ > 0]) = a (P ([M ∗ > 0]) − P ([τ b ≤ τ a ] ∩ [M ∗ > 0])) + bP ([τ b ≤ τ a ] ∩ [M ∗ > 0]) and so (b − a) P ([τ b ≤ τ a ]) ≥ (b − a) P ([τ b ≤ τ a ] ∩ [M ∗ > 0]) ≥ −aP ([M ∗ > 0])

(50.9.46)

Using the other half of the inequality 50.9.42 this yields 0

≥ aP ([τ a < τ b ]) + bP ([τ b < τ a ]) + aP ([τ a = τ b ] ∩ [M ∗ > 0]) = aP ([τ a ≤ τ b ] ∩ [M ∗ > 0]) + bP ([τ b < τ a ]) = a (P ([M ∗ > 0]) − P ([τ b < τ a ])) + bP ([τ b < τ a ])

and so aP ([M ∗ > 0]) ≤ − (b − a) P ([τ b < τ a ]) which implies −aP ([M ∗ > 0]) ≥ (b − a) P ([τ b < τ a ]) combining these yields (b − a) P ([τ b ≤ τ a ]) ≥ −aP ([M ∗ > 0]) ≥ (b − a) P ([τ b < τ a ]) This proves Theorem 50.9.3. Note P ([τ a < τ b ]) means M (t) hits a before it hits b with other occurrences of similar expressions being defined similarly.

50.10 Here p > 1.

The Space MpT (E)

50.10. THE SPACE MPT (E)

1521

Definition 50.10.1 Let M be an E valued martingale. Then M ∈ MpT (E) if t → M (t) (ω) is continuous for a.e. ω and Ã ! p

E

sup ||M (t)||

0 : |M (t)| > C} ∧ l where inf (∅) ≡ ∞ and denote the stopped martingale M τ (t) ≡ M (t ∧ τ ) . Then I claim this is also a martingale with respect to the filtration Ft because by Doob’s optional sampling theorem for martingales, if s < t, E (M τ (t) |Fs ) ≡ E (M (τ ∧ t) |Fs ) = M (τ ∧ t ∧ s) = M (τ ∧ s) = M τ (s)

51.1. HOW TO RECOGNIZE A MARTINGALE

1529

Note the bounded stopping time is τ ∧ t and the other one is σ = s in this theorem. Then M τ is a continuous martingale which is also uniformly bounded. It equals Aτ − B τ . Since τ is also bounded being no larger than l, it must be the case that since A and B are both increasing, Aτ , B τ are bounded uniformly also. Let D be a large enough constant that it bounds all of |M τ (t)| , B τ (t) , Aτ (t) on [0, l] . Now n let Pn ≡ {tk }k=1 be a uniform partition of [0, l] and let M τ (Pn ) denote n

max {|M τ (ti+1 ) − M τ (ti )|}i=1 . Then

³

2

E M τ (l)

´

Ã =E

n−1 X

!2  M τ (tk+1 ) − M τ (tk ) 

k=0

Now consider a mixed term in the sum where j < k. E ((M τ (tk+1 ) − M τ (tk )) (M τ (tj+1 ) − M τ (tj ))) = E (E ((M τ (tk+1 ) − M τ (tk )) (M τ (tj+1 ) − M τ (tj )) |Ftk )) = E ((M τ (tj+1 ) − M τ (tj )) E ((M τ (tk+1 ) − M τ (tk )) |Ftk )) = E ((M τ (tj+1 ) − M τ (tj )) (M τ (tk ) − M τ (tk ))) = 0 It follows ³

2

τ

´

E M (l)

= E

Ãn−1 X

! τ

2

τ

(M (tk+1 ) − M (tk ))

k=0

≤ E

Ãn−1 X

! τ

τ

τ

M (Pn ) |M (tk+1 ) − M (tk )|

k=0

≤ E

Ãn−1 X Ã

≤ E

! τ

τ

τ

τ

τ

τ

τ

τ

τ

M (Pn ) (|A (tk+1 ) − A (tk )| + |B (tk+1 ) − B (tk )|)

k=0 τ

M (Pn )

n−1 X

! (|A (tk+1 ) − A (tk )| + |B (tk+1 ) − B (tk )|)

k=0

≤ E (M τ (Pn ) 2D) the last step holding because A and B are increasing. Now letting n → ∞, the right side converges to 0 by the dominated convergence theorem and the observation that for a.e. ω, lim M τ (Pn ) (ω) = 0 n→∞

τ

and M (Pn ) ≤ 2D. Thus for τ = τ C given above, M (l ∧ τ C ) = 0 a.e.

1530

THE QUADRATIC VARIATION OF A MARTINGALE

Since M is a continuous martingale, it follows that for a.e. ω, |M (t) (ω)| ≤ C for all t ∈ [0, l] provided C is large enough and so, M (l ∧ τ C ) reduces to M (l). Thus M (l) = 0 a.e. This proves the lemma. The following proposition is the main result. One can generalize it in many ways. Proposition 51.1.6 Let {M (t)} be a continuous martingale having values in H a separable Hilbert space adapted to the normal filtration {Ft } such that ||M (t) (ω)|| is bounded and M (0) = 0. Then there exists a unique continuous, increasing, nonnegative, submartingale {[M ] (t)} called the quadratic variation such that 2

||M (t)|| − [M ] (t) is a real martingale and [M ] (0) = 0. Here t ∈ [0, T ] . Also letting 2

N (t) ≡ ||M (t)|| − [M ] (t) , it follows ||N (t)||L2 (Ω) ≤ 2 ||M (t)||L2 (Ω) .

(51.1.4)

Proof: Define stopping times τ n0 ≡ η n0 ≡ 0. Then © ª η nk+1 ≡ inf T ≥ s ≥ η nk : ||M (s) − M (η nk )|| = 2−n , τ nk ≡ η nk ∧ T where inf ∅ ≡ ∞. These are stopping times by Example 50.5.5 on Page 1494. Then for t > 0 and fixed ω, eventually for large enough k, t ∈ (τ nk (ω) , τ nk+1 (ω)]. Here is ∞ why. The sequence {η nk (ω)}k=1 eventually equals ∞. This is because if it did not, it ∞ would converge, being bounded above and then by continuity of M, {M (η nk (ω))}k=1 would be a Cauchy sequence contrary to the requirement that ¯¯ ¡ n ¯¯ ¢ ¯¯M η k+1 (ω) − M (η nk (ω))¯¯ = 2−n . Thus M (t) ≡

X

¡ ¢ M t ∧ τ nk+1 − M (t ∧ τ nk )

(51.1.5)

k≥0

and the terms of the series are eventually 0, as soon as η nk = ∞. Therefore, ¯¯ ¯¯2 ¯¯ ¯¯ X ¯ ¯ ¯¯ ¡ ¢ 2 n n ¯¯ ¯ ¯ ||M (t)|| = ¯¯ M t ∧ τ k+1 − M (t ∧ τ k )¯¯ ¯¯k≥0 ¯¯ Then this equals X ¯¯ ¡ ¯¯ ¢ ¯¯M t ∧ τ nk+1 − M (t ∧ τ nk )¯¯2 = k≥0

+

X ¡¡ j6=k

¡ ¢ ¢ ¡ ¡ ¢ ¡ ¢¢¢ M t ∧ τ nk+1 − M (t ∧ τ nk ) , M t ∧ τ nj+1 − M t ∧ τ nj


1531

Consider the second sum. It equals 2

X k−1 X ¡¡

¡ ¢ ¢ ¡ ¡ ¢ ¡ ¢¢¢ M t ∧ τ nk+1 − M (t ∧ τ nk ) , M t ∧ τ nj+1 − M t ∧ τ nj

k≥0 j=0

= 2

X

  k−1 X¡ ¡ ¢ ¢ ¢ ¡ ¢¢ ¡ ¡  M t ∧ τ nk+1 − M (t ∧ τ nk ) , M t ∧ τ nj+1 − M t ∧ τ nj 

k≥0

= 2

X ¡¡

j=0

¡ ¢ ¢ ¢ M t ∧ τ nk+1 − M (t ∧ τ nk ) , M (t ∧ τ nk )

k≥0

Now for t ∈ (τ nm , τ nm+1 ], M (t ∧ τ nk ) = M (τ nk ) whenever k ≤ m. However, for k ≥ m + 1, ¢ ¡ M t ∧ τ nk+1 − M (t ∧ τ nk ) = M (t) − M (t) = 0 and so this last sum equals 2

X¡

¡ ¡ ¢ ¢¢ M (τ nk ) , M t ∧ τ nk+1 − M (t ∧ τ nk ) ≡ Pn (t)

k≥0

© ª By adding in the values of τ n+1 Pn (t) can be written in the form k 2

X¡

¡ ¢ ¡ ¡ ¢ ¡ ¢¢¢ n+1 M τ n+10 , M t ∧ τ n+1 k k+1 − M t ∧ τ k

k≥0

where τ n+10 has some repeats. From the definition, k ¯¯ ¡ n+10 ¢ ¡ ¢¯¯ ¯¯M τ ¯¯ ≤ 2−n − M τ n+1 k k and so from Proposition 51.1.4 ³ ´ ³ ´ 2 2 E ||Pn (t) − Pn+1 (t)|| ≤ 2−2n E ||M (t)|| . Note this shows that since P0 (t) = 0, ||Pn (t)||L2 (Ω)

≤

n−1 X

||Pk+1 (t) − Pk (t)||L2 (Ω)

k=0

≤

n−1 X

2−k ||M (t)||L2 (Ω) ≤ 2 ||M (t)||L2 (Ω)

(51.1.6)

k=0

Now t → Pn (t) is continuous because it is a finite sum of continuous functions. It is also the case that {Pn (t)} is a martingale. To see this use Lemma 51.1.1. Let

1532


σ be a stopping time having two values. Then using Corollary 51.1.3 and the Doob optional sampling theorem, Theorem 50.5.8   X¡ ¡ ¡ ¢ ¢¢ E M (τ nk ) , M σ ∧ τ nk+1 − M (σ ∧ τ nk )  k≥0

=

X

E

¡¡

¡ ¡ ¢ ¢¢¢ M (τ nk ) , M σ ∧ τ nk+1 − M (σ ∧ τ nk )

k≥0

=

X

E

k≥0

=

X

E

k≥0

=

X

E

¡ ¡ ¢ ¢¢ ¢¢ ¡¡ ¡ E M (τ nk ) , M σ ∧ τ nk+1 − M (σ ∧ τ nk ) |Fτ nk ¡¡

¡ ¡ ¢ ¢ ¢¢ M (τ nk ) , E M σ ∧ τ nk+1 − M (σ ∧ τ nk ) |Fτ nk

¡¡

¢¢¢ ¢ ¡ ¡ =0 M (τ nk ) , E M σ ∧ τ nk+1 ∧ τ nk − M (σ ∧ τ nk )

k≥0

Note the Doob theorem applies because σ ∧ τ nk+1 is a bounded stopping time due to the fact σ has only two values. You don’t need to use the fact that τ nk is bounded. Similarly   X¡ ¢¢ ¡ ¡ ¢ E M (τ nk ) , M t ∧ τ nk+1 − M (t ∧ τ nk )  k≥0

=

X

E

¡¡

¢¢¢ ¡ ¡ ¢ M (τ nk ) , M t ∧ τ nk+1 − M (t ∧ τ nk )

k≥0

=

X

E

k≥0

=

X

E

k≥0

=

X

E

¡¡ ¡ ¡ ¡ ¢ ¢¢ ¢¢ E M (τ nk ) , M t ∧ τ nk+1 − M (t ∧ τ nk ) |Fτ nk ¡¡

¡ ¡ ¢ ¢ ¢¢ M (τ nk ) , E M t ∧ τ nk+1 − M (t ∧ τ nk ) |Fτ nk

¡¡

¡ ¡ ¢ ¢¢¢ M (τ nk ) , E M t ∧ τ nk+1 ∧ τ nk − M (t ∧ τ nk ) =0

k≥0

Note the Doob theorem applies because t ∧ τ nk+1 is a bounded stopping time. Thus {Pn (t)} is a Cauchy sequence in M2T (R) . Therefore, by Proposition 50.10.2 there exists {N (t)} ∈ M2T (R) such that Pn → N in M2T (H) . That is Ã lim E

n→∞

!1/2 sup |Pn (t) − N (t)|

2

t∈[0,T ]

Also from 51.1.6 ||N (t)||L2 (Ω) ≤ 2 ||M (t)||L2 (Ω) .

= 0.


1533

Since {N (t)} ∈ M2T (R) it is a continuous martingale and N (t) ∈ L2 (Ω) , and N (0) = 0 because this is true of each Pn (0) . From the above, 2

||M (t)|| = Qn (t) + Pn (t) where Qn (t) =

(51.1.7)

X ¯¯ ¡ ¯¯ ¢ ¯¯M t ∧ τ nk+1 − M (t ∧ τ nk )¯¯2 k≥0

and Pn (t) is a martingale. Then from 51.1.7, Qn (t) is a submartingale and converges for each t to something, denoted as [M ] (t) in L1 (Ω) uniformly in t ∈ [0, T ]. 2 This is because Pn (t) converges in L2 (Ω) to N (t) and ||M (t)|| converges to 2 ||M (t)|| in L1 (Ω) . Then also [M ] is a submartingale which equals 0 at 0 because this is true of Qn and because if A ∈ Fs where s < t, Z Z ³ Z ´ 2 E ([M ] (t) |Fs ) dP ≡ [M ] (t) dP = lim ||M (t)|| − Pn (t) dP n→∞

A

A

Z = lim

n→∞

A

Z ³ ´ 2 2 E ||M (t)|| − Pn (t) |Fs dP ≥ lim inf ||M (s)|| − Pn (s) dP

A

Z

Z

= lim inf

n→∞

n→∞

Qn (s) dP = A

A

[M ] (s) dP. A

Note that Qn (t) is increasing because as t increases, the definition allows for the possibility of more nonzero terms in the sum. Therefore, [M ] (t) is also increasing 2 in t. The function t → [M ] (t) is continuous because ||M (t)|| = [M ] (t) + N (t) and 2 t → N (t) is continuous as is t → ||M (t)|| . That is, off a set of measure zero, these are both continuous functions of t and so the same is true of [M ] . It remains to consider the claim about the uniqueness. Suppose then there are two which work, [M ] , and [M ]1 . Then [M ] − [M ]1 equals a martingale, G which is 0 when t = 0. Thus the uniqueness assertion follows from Lemma 51.1.5. This proves the proposition. Here is a corollary which tells how to manipulate stopping times. Corollary 51.1.7 In the situation of Proposition 51.1.6 let τ be a stopping time. Then τ [M τ ] = [M ] . Proof: ³ ´τ τ 2 2 [M ] (t) + N1 (t) = ||M || (t) = ||M τ || (t) = [M τ ] (t) + N2 (t) where Ni is a martingale. Therefore, τ

[M ] (t) − [M τ ] (t) = N2 (t) − N1 (t) , τ

a martingale. Therefore, by Lemma 51.1.5, this shows [M ] (t) − [M τ ] (t) = 0. This proves the corollary.

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Definition 51.1.8 The covariation of two continuous H valued martingales for H a separable Hilbert space M, N is defined as follows. [M, N ] ≡

1 ([M + N ] − [M − N ]) 4

Thus [M ] = [M, M ]. In writing this formula, it is only assumed M is such that [M ] has been defined. Corollary 51.1.9 Let M, N be two continuous bounded martingales, M (0) = N (0) = 0, as in Proposition 51.1.6. Then [M, N ] is of bounded variation and (M, N )H − [M, N ] is a martingale. Also for τ a stopping time, τ

[M, N ] = [M τ , N τ ] = [M τ , N ] = [M, N τ ] . In addition to this, [M − M τ ] = [M ] − [M τ ] ≤ [M ] and also (M, N ) → [M, N ] is bilinear and symmetric. Proof: Since [M, N ] is the difference of increasing functions, it is of bounded variation. ´ 1³ 2 2 (M, N )H − [M, N ] = ||M + N || − ||M − N || 4 1 − ([M + N ] − [M − N ]) 4 which equals a martingale from the definition of [M + N ] and [M − N ]. It remains to verify the claim about the stopping time. Using Corollary 51.1.7 τ

[M, N ] =

= =

1 τ ([M + N ] − [M − N ]) 4

1 τ τ ([M + N ] − [M − N ] ) 4 1 ([M τ + N τ ] − [M τ − N τ ]) ≡ [M τ , N τ ] . 4

The really interesting part is the next equality. This will involve Lemma 51.1.5. τ

[M, N ] − [M τ , N ] = [M τ , N τ ] − [M τ , N ]

51.1. HOW TO RECOGNIZE A MARTINGALE = =

1535

1 1 ([M τ + N τ ] − [M τ − N τ ]) − ([M τ + N ] − [M τ − N ]) 4 4 1 1 τ τ τ ([M + N ] + [M − N ]) − ([M τ + N ] + [M τ − N τ ]) , (51.1.8) 4 4

the difference of two increasing adapted processes. Also, this equals martingale + (M τ , N ) − (M τ , N τ ) I claim (M τ , N ) − (M τ , N τ ) = (M τ , N − N τ ) is a martingale. This follows from Lemma 51.1.1. Let σ be a stopping time with two values. By the Doob optional sampling theorem, E ((M τ (σ) , N (σ) − N τ (σ))) = E (E ((M τ (σ) , N (σ) − N τ (σ)) |Fτ )) =

E (M τ (σ) , E (N (σ) − N τ (σ) |Fτ ))

=

E (M τ (σ) , N (σ ∧ τ ) − N (τ ∧ σ)) = 0

while E ((M τ (t) , N (t) − N τ (t))) = E (E ((M τ (t) , N (t) − N τ (t)) |Fτ )) = =

E ((M τ (t) , E (N (t) − N τ (t) |Fτ ))) E ((M τ (t) , E (N (t ∧ τ ) − N (t ∧ τ )))) = 0

This shows the claim is true. Now from 51.1.8 and Lemma 51.1.5, τ

[M, N ] − [M τ , N ] = 0. Similarly

τ

[M, N ] − [M, N τ ]

Now consider the last claim. From the definition, it follows [M − M τ ] − ([M ] + [M τ ] − 2 [M, M τ ]) = a martingale. By the first part of the corollary, the left side is the difference of two increasing adapted processes and so the left side equals 0. Thus from what was just shown, [M − M τ ] = [M ] + [M τ ] − 2 [M, M τ ] = [M ] + [M τ ] − 2 [M τ , M τ ] = [M ] + [M τ ] − 2 [M τ ] =

[M ] − [M τ ] ≤ [M ]

Finally consider the claim that [M, N ] is bilinear. From the definition, letting M1 , M2 , N be bounded H valued martingales, (aM1 + bM2 , N )H

=

[aM1 + bM2 , N ] + martingale

a (M1 , N ) + b (M2 , N )H

=

a [M1 , N ] + b [M2 , N ] + martingale

1536


Hence [aM1 + bM2 , N ] − (a [M1 , N ] + b [M2 , N ]) = martingale. The left side can be written as the difference of two increasing functions and so by Lemma 51.1.5 it equals 0. [M, N ] is obviously symmetric. This proves the corollary.

51.2

The Burkholder Davis Gundy Inequality

The next topic is the Burkholder Davis Gundy inequality. Before presenting this, here is the good lambda inequality, Theorem 9.7.7 on Page 250 listed here for convenience. Theorem 51.2.1 Let (Ω, F, µ) be a finite measure space and let F be a continuous increasing function defined on [0, ∞) such that F is continuous at 0 and F (0) = 0. Suppose also that for all α > 1, there exists a constant Cα such that for all x ∈ [0, ∞), F (αx) ≤ Cα F (x) . Also suppose f, g are nonnegative measurable functions and there exists β > 1 such that for all λ > 0 and 1 > δ > 0, µ ([f > βλ] ∩ [g ≤ δλ]) ≤ φ (δ) µ ([f > λ])

(51.2.9)

where limδ→0+ φ (δ) = 0 and φ is increasing. Under these conditions, there exists a constant C depending only on β, φ such that Z Z F (f (ω)) dµ (ω) ≤ C F (g (ω)) dµ (ω) . Ω

Ω

Then the Burkholder Davis Gundy inequality is as follows. Generalizations will be presented later. Theorem 51.2.2 Let {M (t)} be a continuous H valued martingale which is uniformly bounded, M (0) = 0, where H is a separable Hilbert space and t ∈ [0, T ] . Then if F is a function of the sort described in the good lambda inequality above, there are constants, C and c independent of such martingales M such that Z Z Z ³ ´ ³ ´ 1/2 1/2 c F [M ] (T ) dP ≤ F (M ∗ ) dP ≤ C F [M ] (T ) dP Ω

Ω

Ω

where M ∗ (ω) ≡ sup {||M (t) (ω)|| : t ∈ [0, T ]} . Proof: Using Corollary 51.1.9, let N (t)

2

≡ ||M (t) − M τ (t)|| − [M − M τ ] (t) 2

τ

= ||M (t) − M τ (t)|| − [M ] (t) + [M ] (t)

51.2. THE BURKHOLDER DAVIS GUNDY INEQUALITY

1537

where τ ≡ inf {t ∈ [0, T ] : ||M (t)|| > λ} Thus N is a martingale and N (0) = 0. In fact N (t) = 0 as long as t ≤ τ . As usual inf (∅) ≡ ∞. Note [τ < ∞] = [M ∗ > λ] ⊇ [N ∗ > 0] . This is because to say τ < ∞ is to say there exists t < T such that ||M (t)|| > λ which is the same as saying M ∗ > λ. Thus the first two sets are equal. If τ = ∞, then from the formula for N (t) above, N (t) = 0 for all t ∈ [0, T ] and so it can’t happen that N ∗ > 0. Thus the third set is contained in [τ < ∞] as claimed. Let β > 2 and let δ ∈ (0, 1) . Then β−1>1>δ >0 Consider

h i 1/2 Sr ≡ [M ∗ > βλ] ∩ [M ] (T ) ≤ rδλ

where 0 < r < 1. 2 Claim: For ω ∈ Sr , N (t) hits (β − 1) λ2 − δ 2 λ2 . Proof of claim: For ω ∈ Sr , there exists a t < T such that ||M (t)|| > βλ and so using Corollary 51.1.9, N (t) ≥ ≥

2

2

|||M (t)|| − ||M τ (t)||| − [M − M τ ] (t) ≥ |βλ − λ| − [M ] (t) 2

(β − 1) λ2 − δ 2 λ2 2

which shows that N (t) hits (β − 1) λ2 − δ 2 λ2 for ω ∈ Sr . This proves the claim. Claim: N (t) (ω) never hits −δ 2 λ2 for ω ∈ Sr . Proof of claim: Suppose t is the first time N (t) reaches −δ 2 λ2 . Then 2

0 ≤ ||M (t) − M τ (t)|| = [M − M τ ] (t) − δ 2 λ2 and by Corollary 51.1.9, ≤ [M ] (t) − δ 2 λ2 ≤ r2 δ 2 λ2 − δ 2 λ2 < 0, a contradiction. This proves the claim. 2 Therefore, for all ω ∈ Sr , N (t) (ω) reaches (β − 1) λ2 − δ 2 λ2 before it reaches 2 2 −δ λ . It follows ³ ´ 2 P (Sr ) ≤ P N (t) reaches (β − 1) λ2 − δ 2 λ2 first. and because of Theorem 50.9.3 this is no larger than P ([N ∗ > 0]) ³

δ 2 λ2

´

¡ 2 (β − 1) λ2 − δ 2 λ2 − −δ 2 λ

∗ ¢ = P ([N > 0]) 2

δ2

2.

(β − 1)

1538


Since this is true for each r < 1, it follows ³ h i´ 1/2 P [M ∗ > βλ] ∩ [M ] (T ) ≤ δλ ≤ P ([N ∗ > 0])

δ2 2

(β − 1)

and as noted above, this implies ³ h i´ 1/2 P [M ∗ > βλ] ∩ [M ] (T ) ≤ δλ ≤ P ([M ∗ > λ])

δ2 (β − 1)

2

By the good lambda inequality, Z Z ³ ´ 1/2 F [M ] (T ) dP F (M ∗ ) dP ≤ C Ω

Ω

which is one half the inequality. Now consider the other half. This time define the stopping time τ by n o 1/2 τ ≡ inf t ∈ [0, T ] : [M ] (t) > λ and let

h i 1/2 Sr ≡ [M ] (T ) > βλ ∩ [2M ∗ ≤ rδλ]

Then there exists t < T such that [M ] (t) > β 2 λ2 . This time, let 2

N (t) = [M ] (t) − [M τ ] (t) − ||M (t) − M τ (t)|| This is still a martingale since by Corollary 51.1.9 [M ] (t) − [M τ ] (t) = [M − M τ ] (t) 2

Claim: N (t) (ω) hits (β − 1) λ2 − δ 2 λ2 for some t < T for ω ∈ Sr . Proof of claim: Fix such a ω ∈ Sr . Let t < T be such that [M ] (t) > β 2 λ2 . Then t > τ and so for that ω, N (t) >

2

β 2 λ2 − λ2 − ||M (t) − M (τ )|| 2

2

≥

(β − 1) λ2 − (||M (t)|| + ||M (τ )||)

≥

(β − 1) λ2 − r2 δ 2 λ2 ≥ (β − 1) λ2 − δ 2 λ2

2

2

This proves the claim. Claim: N (t) (ω) never hits −δ 2 λ2 for ω ∈ Sr . Proof of claim: By Corollary 51.1.9, if it did at t, then t > τ and 0

2

≤ [M ] (t) − [M τ ] (t) = ||M (t) − M (τ )|| − δ 2 λ2 2

≤ (||M (t)|| + ||M (τ )||) − δ 2 λ2 ≤ r2 δ 2 λ2 − δ 2 λ2 < 0, a contradiction. This proves the claim.

51.3. THE CASE OF M2T (H)

1539

It follows that for each r ∈ (0, 1) , ³ ´ 2 P (Sr ) ≤ P N (t) hits (β − 1) λ2 − δ 2 λ2 before − δ 2 λ2 By Theorem 50.9.3 this is no larger than P ([N ∗ > 0])

δ 2 λ2 2

(β − 1) λ2 δ2

≤ P ([τ < ∞])

2

=P

= P ([N ∗ > 0])

δ2 2

(β − 1)

³h i´ 1/2 [M ] (T ) > λ

δ2 2

(β − 1) (β − 1) Since the inequality holds for all r < 1, it follows that the same inequality holds letting r → 1. Thus ³h i ´ ³h i´ δ2 1/2 1/2 P [M ] (T ) > βλ ∩ [2M ∗ ≤ δλ] ≤ P [M ] (T ) > λ 2 (β − 1) Now by the good lambda inequality, there is a constant k independent of M such that Z Z Z ³ ´ 1/2 F [M ] (T ) dP ≤ k F (2M ∗ ) dP ≤ kC2 F (M ∗ ) dP Ω

Ω

Ω

by the assumptions about F . Therefore, combining this result with the first part, Z Z ³ ´ −1 1/2 (kC2 ) F [M ] (T ) dP ≤ F (M ∗ ) dP Ω Ω Z ³ ´ 1/2 ≤ C F [M ] (T ) dP Ω

and this proves the inequality.

51.3

The Case Of M2T (H)

Now of course it would be more interesting if the assumption that {M (t)} is uniformly bounded were dropped. In particular, it would be better to have the above result for M ∈ M2T (H) . Theorem 51.3.1 Let {M (t)} be a continuous martingale in M2T (H) where H is a separable Hilbert space adapted to the filtration {Ft } such that M (0) = 0. Then there exists a unique continuous, increasing, nonnegative, submartingale {[M ] (t)} , [M ] (0) = 0 called the quadratic variation such that 2

||M (t)|| − [M ] (t) ≡ N (t) is in M2T (R). Here t ∈ [0, T ] . Also ||N (t)||L2 (Ω) ≤ 2 ||M (t)||L2 (Ω) .

1540


Proof: Since t → M (t) (ω) is continuous, there exists an increasing sequence of stopping times of the form σ m ≡ inf {t : ||M (t)|| > m} ∧ T such that σ m = T for all m large enough. Here inf ∅ ≡ ∞. Now from Proposition 51.1.6 there exists a martingale Nm and a submartingale [M σm ] satisfying the conditions of that proposition such that 2

||M σm (t)|| = [M σm ] (t) + Nm (t) where M σm (t) denotes the stopped martingale given by M σm (t) ≡ M (t ∧ σ m ) Recall this is a martingale with respect to the same filtration because if s < t, it follows from the Doob optional sampling theorem for martingales, Theorem 50.5.8 E (M σm (t) |Fs )

= E (M (t ∧ σ m ) |Fs ) = M (t ∧ σ m ∧ s) = M (σ m ∧ s) = M σm (s) .

Also from Proposition 51.1.6 and the Doob optional sampling theorem and Jensen’s inequality, ³ ´ 2 2 ||Nm (t)||L2 (Ω) ≤ 2E ||M σm (t)|| Z Z 2 2 =2 ||M (σ m ∧ t)|| dP = 2 ||E (M (t) |Fσm )|| dP Z ≤2 Ω

and so

Ω

³

2

´

Ω

E ||M (t)|| |Fσm ∧t dP = 2

Z 2

||M (t)|| dP

(51.3.10)

Ω

³ ´ ³ ´ 2 2 E |Nm (t)| ≤ 2E ||M (t)||

Also from the uniqueness part of Proposition 51.1.6, pointwise convergence takes place as m → ∞ uniformly in t for each fixed ω. This is because for each ω, eventually σ m (ω) is so large that ||M (t) (ω)|| < m for all t ∈ [0, T ] and so no further change can take place as a function of m. Thus Nm (t) converges pointwise to a function, N (t) . From Fatou’s lemma ³ ´ ³ ´ 2 2 E |N (t)| ≤ 2E ||M (t)|| (51.3.11) n o∞ 3/2 3/2 also. From 51.3.10 the set |Nm (t)| , |N (t)|

m=1

is uniformly integrable and

so the pointwise convergence uniformly in t ∈ [0, T ] implies Nm (t) → N (t)

(51.3.12)

uniformly on [0, T ] in L6/5 (Ω) . Here is why. First note that C ([0, T ]) is separable from the Weierstrass approximation theorem. By Egoroff’s theorem, Theorem

51.3. THE CASE OF M2T (H)

1541

8.3.11 on Page 198 applied to X = C ([0, T ]) there is a subsequence, still denoted by m and a set of measure ε called Bε such that for ω ∈ / Bε then for each η > 0 there exists Nη such that if m > Nη , then sup |Nm (t) − N (t)| < η t∈[0,T ]

Then for all t ∈ [0, T ] , Z Z 6/5 |Nm (t) − N (t)| dP ≤ 21/5 Ω

³

6/5

|Nm (t)|

6/5

´

+ |N (t)|

dP

Bε

Z + BεC

|Nm (t) − N (t)|

"µZ ≤

1/5

2

Ω

XBε dP

XBε

BεC

¶3/5

|Nm (t)| dP Ω 2

¶3/5 #

|N (t)| Ω

Z +

2

¶2/5 µZ Ω

dP

¶2/5 µZ

µZ +

6/5

Z |Nm (t) − N (t)|

6/5

dP +

|Nm (t) − N (t)|

BεC

³ ³ ´´ Z 2 ≤ 4 · 21/5 ε2/5 E ||M (t)|| + ≤ 4·2

ε

2 ||M ||M2 (H) T

dP

6/5

BεC

Z 1/5 2/5

6/5

|Nm (t) − N (t)|

dP

ε6/5 dP

+ BεC

and so since ε and t ∈ [0, T ] are arbitrary, this shows Ã ! Z 6/5 lim sup |Nm (t) − N (t)| dP = 0 m→∞

t∈[0,T ]

Ω 6/5

It follows from Proposition 50.10.2 that N ∈ MT (R) and so, in particular it is a continuous martingale. In fact it is in M2T (R) because of the above estimate 51.3.11. Letting [M ] (t) be the pointwise limit of [M σm ] (t) 2

||M (t)|| = [M ] (t) + N (t) 2

and N is a martingale. It follows [M ] must be a submartingale since ||M || is a submartingale. It is increasing because it is a limit of increasing functions of t for each ω. This proves the theorem.

1542

51.4


The Case Of Local Martingales

There is also the notion of a local martingale which can be used to generalize the above results. Definition 51.4.1 Let {M (t)} be adapted to the normal filtration Ft for t > 0. Then {M (t)} is a local martingale if there exist stopping times τ n increasing to infinity such that for each n, the process M τ n (t) − M (0) is a martingale with respect to the given filtration. The sequence of stopping times is called a localizing sequence. As in the proof of Lemma 51.1.5 if {M (t)} is a martingale, then M τ n (t) − M (0) is a martingale with respect to the σ algebras, Ft as shown by the following computation; E (M τ (t) − M (0) |Fs )

≡ = = =

E (M (τ ∧ t) − M (0) |Fs ) M (τ ∧ t ∧ s) − M (0) M (τ ∧ s) − M (0) M τ (s) − M (0) ;

so the idea of a local martingale is just an adapted process which preserves this property having to do with stopping times coming from a localizing sequence. The above reasoning is used in the following simple proposition which states you can assume the localizing sequence can be assumed such that {M τ n (t) − M (0)} is a bounded martingale. Proposition 51.4.2 If M is a local martingale, there exists a localizing sequence {σ m } such that M σm − M (0) is a bounded martingale. Proof: Let δ n ≡ inf {t : ||M (t) − M (0)|| > n} Thus M δn − M (0) is a bounded adapted process, bounded by n. Also δ n → ∞ because of continuity of M. (If ω is fixed and b > 0, eventually ||M (t) (ω) − M (0)|| < n for all t ∈ [0, b] and so δ n (ω) > b.) Then let σ n = δ n ∧τ n . Thus M δn ∧τ n −M (0) = δ (M τ n − M (0)) n , a martingale since M τ n − M (0) is given to be one. This proves the proposition. One can also give a generalization of Lemma 51.1.5 to conclude a local martingale must be constant or else they must fail to be of bounded variation. Corollary 51.4.3 Let Ft be a normal filtration and let A (t) , B (t) be adapted to Ft , continuous, and increasing with A (0) = B (0) = 0 and suppose A (t)−B (t) ≡ M (t) is a local martingale. Then A (t) − B (t) = 0 a.e. for all t.

51.4. THE CASE OF LOCAL MARTINGALES

1543

Proof: Let {τ n } be a localizing sequence for M . For given n consider the martingale, M τ n (t) = Aτ n (t) − B τ n (t) Also let

σ ≡ σ C ≡ inf {t ∈ [0, l] : |Aτ n (t)| + |B τ n (t)| > C}

As usual inf (∅) ≡ ∞. Letting N = M τ n , n−1 X

N σ (l) =

(N σ (tk+1 ) − N σ (tk ))

k=0 m

where {tk }k=0 is a uniform partition of [0, l]. Then since the mixed terms have 0 expectation due to the fact N σ is a martingale, Ã !2  m−1 ³ ´ X 2 E N σ (l) = E (N σ (tk+1 ) − N σ (tk ))  k=0

=

m−1 X

³ ´ 2 E (N σ (tk+1 ) − N σ (tk ))

k=0

≤

m−1 XZ Ω

k=0

∆σm (ω) |N σ (tk+1 ) − N σ (tk )| dP

Z =

Ω

where

∆σm (ω)

m−1 X

|N σ (tk+1 ) − N σ (tk )| dP

k=0

∆σm (ω) ≡ max {|N σ (tk+1 ) (ω) − N σ (tk ) (ω)|} k

Since |N σ (tk+1 ) (ω) − N σ (tk ) (ω)| σ

σ

σ

σ

σ

σ

= |(Aτ n ) (tk+1 ) − (B τ n ) (tk+1 ) σ σ − ((Aτ n ) (tk ) − (B τ n ) (tk ))|

≤ (Aτ n ) (tk+1 ) − (Aτ n ) (tk ) + (B τ n ) (tk+1 ) − (B τ n ) (tk ) , it follows

m−1 X

σ

σ

|N σ (tk+1 ) − N σ (tk )| ≤ (Aτ n ) (l) + (B τ n ) (l)

k=0

and so ³ ´ 2 E N σ (l)

Z σ

≤ ZΩ ≤ Ω

σ

∆σm (ω) ((Aτ n ) (l) + (B τ n ) (l)) dP ∆σm (ω) (Aσ (l) + B σ (l)) dP.

1544


Since N σ is uniformly bounded, the dominated convergence theorem yields this converges to 0 as m → ∞ due to the continuity of N σ . Thus N σC (l) = 0 for all C > 0 and letting C → ∞, it follows from continuity of M that for each ω, eventually σ C (ω) > l and so the above reduces to N (l) ≡ M τ n (l) = 0. Since n is arbitrary and τ n → ∞, it follows for large enough n, τ n > l and so M (l) = M τ n (l) = 0. This proves the corollary. The next task is to generalize everything to the case of local martingales. Theorem 51.4.4 Let {M (t)} be a continuous local martingale adapted to the normal filtration Ft for t ∈ [0, T ] , M (t) having values in a separable Hilbert space H, M (0) = 0. Then there exists a unique adapted process, [M ] (t) which is increasing, equal to 0 at t = 0 such that n o 2 ||M || (t) − [M ] (t) is a local martingale adapted to Ft . Furthermore, if σ is any stopping time σ

[M ] (t) = [M σ ] (t) . Also there exists an increasing localizing sequence {σ m } for the local martingale n o 2 ||M || (t) − [M ] (t) such that {M σm (t)} is a uniformly bounded martingale and [M ] (t) = lim [M σm ] (t) m→∞

∞

Proof: Let {τ n }n=1 be a localizing sequence for {M (t)} so that {M τ n (t)} is a martingale. Also let σ be a stopping time. Thus {M σ∧τ n } is a martingale and ∞ {τ n }n=1 is a localizing sequence for M σ . Lemma 51.4.5 Let σ m ≡ inf {t : ||M (t)|| > m} Then if σ is any stopping time, {M σ (t)} a local martingale and {σ m } is a localizing sequence for it. Also ½³ ¾ ´σ∧σm 2 σ∧σ m ||M || (t) − [M ] (t) is a martingale for each σ m .


1545

Proof: It follows from {τ n } is a localizing sequence that {M σ∧τ n ∧σm (t)} is a bounded martingale and so by Proposition 51.1.6 there exists a unique continuous increasing submartingale [M σ∧τ n ∧σm ] such that [M σ∧τ n ∧σm ] (0) = 0 and ½³ ¾ ´σ∧τ n ∧σm 2 σ∧τ n ∧σ m ||M || (t) − [M ] (t) is a martingale. By Corollary 51.1.7, for each t it equals ³

2

||M ||

´σ∧τ n ∧σm

(t) − [M σ∧σm ]

τn

(t)

Since τ n → ∞, it follows that for each ω, it converges to ³ ´σ∧σm 2 ||M || (t) − [M σ∧σm ] (t) because eventually τ n > t. Why is this a martingale? This is easy to see because convergence is pointwise and the functions are bounded so the dominated convergence theorem applies. Thus letting s < t and A ∈ Fs , µ³ ¶ Z ´σ∧σm 2 σ∧σ m E ||M || (t) − [M ] (t) |Fs dP ZA ³ ´σ∧σm 2 = ||M || (t) − [M σ∧σm ] (t) dP A Z ³ ´σ∧τ n ∧σm τ 2 = lim ||M || (t) − [M σ∧σm ] n (t) dP n→∞ A µ³ ¶ Z ´σ∧τ n ∧σm 2 σ∧σ m τ n = lim E ||M || (t) − [M ] (t) |Fs dP n→∞ A Z ³ ´σ∧τ n ∧σm τ 2 = lim ||M || (s) − [M σ∧σm ] n (s) dP n→∞ A Z ³ ´σ∧σm 2 = ||M || (s) − [M σ∧σm ] (s) dP (51.4.13) A

Also Z

Z E (M

σ∧σ m

(t) |Fs ) dP

=

A

= =

Z τ

σ∧σ m

M (t) dP = lim (M σ∧σm ) n (t) dP n→∞ A A Z ¡ σ∧σm τ n ¢ lim E (M ) (t) |Fs dP n→∞ A Z Z σ∧σ m τ n lim (M ) (s) dP = M σ∧σm (s) dP n→∞

A

A

Thus {σ m } is also a localizing sequence for {M σ (t)} . This proves the lemma. Note that if you take σ = ∞, the argument has shown M σm is a bounded martingale.

1546


Now it was just shown ³ ´σ∧σm 2 ||M || (t) = [M σ∧σm ] (t) + N1mσ (t) where N1mσ is a martingale. For each ω the left side is eventually constant in m because of continuity of M. Therefore, the same is true of the right side. Define [M σ ] (t) + N1σ (t) ≡ lim ([M σ∧σm ] (t) + N1mσ (t)) m→∞

(51.4.14)

as a pointwise limit. Note the limit of the [M σ∧σm ] (t) must exist because for each ω these are increasing in m. Therefore, the limit of the N1m (t) also exists and equals something referred to here as N1σ (t). If you let σ ≡ ∞, this has defined [M ] (t) ≡ lim [M σm ] (t) .

(51.4.15)

m→∞

However, from Corollary 51.1.7, the right side of the 51.4.14 equals σ

lim ([M σm ] (t) + N1mσ (t)) .

(51.4.16)

m→∞

Using 51.4.15, σ

lim [M σm ] (t) ≡

m→∞

= and so 51.4.16 equals

lim [M σm ] (σ ∧ t)

m→∞

σ

[M ] (σ ∧ t) ≡ [M ] (t)

σ

[M ] (t) + N1σ (t) and so comparing this with 51.4.14, σ

[M ] (t) = [M σ ] (t) as claimed. Letting σ = ∞ in the above lemma and using the definition of [M ] given above, 2 this has shown ||M || (t) − [M ] (t) is a local martingale with localizing sequence {σ m } . Uniqueness follows from Corollary 51.4.3 as before. If [M ]1 works then [M ] (t) − [M ]1 (t) = a local martingale and so is equal to 0. This proves the theorem. With this theorem, a similar one can be established for the covariation. Theorem 51.4.6 Let M, N be local martingales having values in a separable Hilbert space, H for t ∈ [0, T ] with respect to the normal filtration Ft , and let M (0) = N (0) = 0. Then there exists a unique bounded variation process adapted to the filtration [M, N ] such that [M, N ] (0) = 0 and (M, N )H − [M, N ]


1547

is a local martingale. Also if σ is a stopping time σ

[M, N ] = [M σ , N σ ] = [M σ , N ] Also

0 ≤ [M − M σ ] = [M ] − [M σ ] .

For {σ m } a sequence of stopping times converging to ∞ such that M σm and N σm are uniformly bounded martingales, [M, N ] (t) = [M σm , N σm ] (t) The map (M, N ) → [M, N ] is also bilinear and symmetric. Proof: Define

1 ([M + N ] − [M − N ]) (51.4.17) 4 where [M + N ] and [M − N ] come from Theorem 51.4.4. Then it is routine to verify this satisfies (M, N )H −[M, N ] is a local martingale and [M, N ] is of bounded variation. Uniqueness is also easy to establish. If [M, N ]1 is another one, [M, N ] (t) ≡

[M, N ]1 − [M, N ] is a local martingale and since this expression can be written as the difference of two increasing functions which are equal to 0 at t = 0, it follows from Corollary 51.4.3 this expression equals 0. It only remains to verify the claims about the stopping time. σ

σ

= (M, N )H = (M σ , N σ )H = [M σ , N σ ] + local martingale

[M, N ] + local martingale

and so by Corollary 51.4.3 again σ

[M, N ] = [M σ , N σ ] The other claim is more surprising. This will involve a use of Lemma 51.1.1 which tells how to recognize a martingale when you see one. First note there exists a single localizing sequence {τ n } which works for M and N simultaneously. You can get it by simply taking τ n = σ n ∧ δ n where the σ n and the δ n are localizing sequences for M and N respectively. Claim: (M τ n ∧σ , N τ n ∧σ − N τ n )H is a martingale and so (M σ , N σ − N )H is a local martingale. Proof of the claim: Let α be a stopping time with two values. Then using the Doob optional sampling theorem, E ((M τ n ∧σ (α) , N τ n ∧σ (α) − N τ n (α))H ) = E (E ((M τ n ∧σ (α) , N τ n ∧σ (α) − N τ n (α))H |Fσ ))

1548


= E ((M τ n ∧σ (α) , E (N τ n ∧σ (α) − N τ n (α) |Fσ ))) = E ((M τ n ∧σ (α) , N τ n ∧σ (α ∧ σ) − N τ n (α ∧ σ))) = 0 Also

E ((M τ n ∧σ (t) , N τ n ∧σ (t) − N τ n (t))H ) = E (E ((M τ n ∧σ (t) , N τ n ∧σ (t) − N τ n (t))H |Fσ )) = E ((M τ n ∧σ (t) , E (N τ n ∧σ (t) − N τ n (t) |Fσ ))) = E ((M τ n ∧σ (t) , N τ n ∧σ (t ∧ σ) − N τ n (t ∧ σ))) = 0

Therefore by Lemma 51.1.1 this proves the claim. Now with the claim, [M σ , N σ ] − [M σ , N ] = (M σ , N σ ) − (M σ , N ) + local martingale = (M σ , N σ − N ) + local martingale which by the claim is a local martingale. Thus the result follows from Corollary 51.4.3. Finally, consider the last claim. By Theorem 51.4.4 [M σ ] (t) = lim [M σ∧σm ] (t) m→∞

where {σ m } is an increasing localizing sequence of stopping times for which M σm is bounded. Similarly [M − M σ ] (t) = lim [M σm − M σ∧σm ] (t) m→∞

By Corollary 51.1.9 = lim [M σm ] (t) − [M σm ∧σ ] (t) = [M ] (t) − [M σ ] (t) . m→∞

The formula follows from Theorem 51.4.4 and the definition, 51.4.17. It remains to prove the claim about [M, N ] being bilinear. From the first part, letting M1 , M2 , N be H valued continuous local martingales equal to 0 at t = 0, (aM1 + bM2 , N )H a (M1 , N ) + b (M2 , N )H

= =

[aM1 + bM2 , N ] + local martingale a [M1 , N ] + b [M2 , N ] + local martingale

Hence [aM1 + bM2 , N ] − (a [M1 , N ] + b [M2 , N ]) = local martingale. It follows from Corollary 51.4.3, the left side, being the difference of two increasing functions equals 0. [M, N ] is obviously symmetric thanks to the definition. This proves the theorem. Now here is a generalization of the Burkholder Gundy inequality to local martingales.


1549

Theorem 51.4.7 Let {M (t)} be a continuous H valued local martingale, M (0) = 0, where H is a separable Hilbert space and t ∈ [0, T ] . Then if F is a function of the sort described in the good lambda inequality, that is, F (0) = 0, F continuous, F increasing, F (αx) ≤ cα F (x) , there are constants, C and c independent of such local martingales M such that Z Z Z ³ ´ ³ ´ 1/2 1/2 c F [M ] (T ) dP ≤ F (M ∗ ) dP ≤ C F [M ] (T ) dP Ω

Ω

where

Ω

M ∗ (ω) ≡ sup {||M (t) (ω)|| : t ∈ [0, T ]} .

Proof: Let {τ n } be a localizing sequence for M such that M τ n is uniformly bounded. Such a localizing sequence exists from Theorem 51.4.4. Then from Theorem 51.2.2 there exist constants c, C independent of τ n such that Z Z ³ ´ ¡ 1/2 ∗¢ c F [M τ n ] (T ) dP ≤ F (M τ n ) dP Ω Ω Z ³ ´ 1/2 ≤ C F [M τ n ] (T ) dP Ω

By Theorem 51.4.4, this implies Z ³ ´ τ 1/2 c F ([M ] n ) (T ) dP

Z ≤

Ω

≤

¡ ∗¢ F (M τ n ) dP Ω Z ³ ´ τ 1/2 C F ([M ] n ) (T ) dP Ω

τ

1/2

∗

1/2

and now note that ([M ] n ) (T ) and (M τ n ) increase in n to [M ] (T ) and M ∗ respectively. Then the result follows from the monotone convergence theorem. Here is a corollary [55]. Corollary 51.4.8 Let {M (t)} be a continuous H valued local martingale and let ε, δ ∈ (0, ∞) . Then there is a constant independent of ε, δ such that Ã" #! ´ ³ ´ C ³ 1/2 1/2 P sup ||M (t)|| ≥ ε ≤ E [M ] (T ) ∧ δ + P [M ] (T ) > δ ε t∈[0,T ] Proof: Let the stopping time τ be defined by n o 1/2 τ ≡ inf t ≥ 0 : [M ] (t) > δ and if the set is empty, τ ≡ T . P ([M ∗ ≥ ε]) = P ([M ∗ ≥ ε] ∩ [τ = T ]) + P ([M ∗ ≥ ε] ∩ [τ < T ])

1550


On the set where [τ = T ] , M τ = M and so 1 ≤ ε

Z

³ h i´ ∗ 1/2 (M τ ) dP + P [M ∗ ≥ ε] ∩ [M ] (T ) > δ Ω

By Theorem 51.4.7 and Theorem 51.4.4 Z ³ h i´ C 1/2 1/2 ≤ [M τ ] (T ) dP + P [M ∗ ≥ ε] ∩ [M ] (T ) > δ ε Ω Z ³ h i´ C τ 1/2 1/2 ([M ] ) (T ) dP + P [M ∗ ≥ ε] ∩ [M ] (T ) > δ ≤ ε Ω Z ³ h i´ C 1/2 1/2 ≤ [M ] (T ) ∧ δdP + P [M ∗ ≥ ε] ∩ [M ] (T ) > δ ε Ω C ≤ ε

Z [M ]

1/2

(T ) ∧ δdP + P

³h i´ 1/2 [M ] (T ) > δ

Ω


51.5

The Quadratic Variation And Stochastic Integration

Let Ft be a normal filtration and let {M (t)} be a local martingale adapted to Ft having values in U a separable real Hilbert space. Definition 51.5.1 Let Ft be a normal filtration and let f (t) ≡

n−1 X

fk X(tk ,tk+1 ] (t)

k=0 n

where {tk }k=0 is a partition of [0, T ] and each fk is Ftk measurable, fk M ∗ ∈ L2 (Ω) where M ∗ (ω) ≡ sup ||M (t) (ω)|| t∈[0,T ]

Such a function is called an elementary function. Also let {M (t)} be a local martingale adapted to Ft which has values in a separable real Hilbert space, U such that M (0) = 0. For such an elementary real valued function define Z

t

f dM ≡ 0

n−1 X

fk (M (t ∧ tk+1 ) − M (t ∧ tk )) .

k=0

Then with this definition, here is a wonderful lemma.

51.5. THE QUADRATIC VARIATION AND STOCHASTIC INTEGRATION

1551

o nR t Lemma 51.5.2 For f an elementary function as above, 0 f dM is a continuous local martingale and Ã¯¯Z ¯¯2 ! Z Z t ¯¯ t ¯¯ 2 ¯ ¯ E ¯¯ f dM ¯¯¯¯ = f (s) d [M ] (s) dP. (51.5.18) 0

Ω

U

0

If N is another continuous local martingale adapted to Ft and both f, g are elementary functions such that for each k, fk M ∗ , fk N ∗ , gk M ∗ , gk N ∗ ∈ L2 (Ω) , then

µµZ E

Z

t

f dM,

¶ ¶

t

gdN

0

0

Z Z

t

=

f gd [M, N ] Ω

U

(51.5.19)

0

and both sides make sense. Proof: Let {τ l } be a localizing sequence for M such that M τ l is a bounded martingale. Then from the definition, for each ω µZ t ¶τ l Z t Z t τl f dM = lim f dM = lim f dM 0

and it is clear that

l→∞

nR t

f dM τ l

o

l→∞

0

0

is a martingale because it is just the sum of Rt some martingales. Thus {τ l } is a localizing sequence for 0 f dM . It is also clear Rt f dM is continuous because it is a finite sum of continuous random variables. 0 0

Next consider the formula which is really a version of the Ito isometry. There is no loss of generality in assuming the mesh points are the same for the two elementary functions because if not, one can simply add in points to make this happen. It suffices to consider 51.5.19 because the other formula is a special case. To begin with, let {τ l } be a localizing sequence which makes both M τ l and N τ l into bounded martingales. Consider the stopped process. µµZ t ¶ ¶ Z t E f dM τ l , gdN τ l 0

=

E

ÃÃn−1 X

0

U

fk (M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) ,

k=0 n−1 X

gk (N

!! τl

(t ∧ tk+1 ) − N

τl

(t ∧ tk ))

k=0

 X = E fk gj ((M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) , k,j

´ (N τ l (t ∧ tj+1 ) − N τ l (t ∧ tj )))

1552


Now consider one of the mixed terms with j < k. E (fk (M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) , gj (N τ l (t ∧ tj+1 ) − N τ l (t ∧ tj ))) = E (E (fk (M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) , gj (N τ l (t ∧ tj+1 ) − N τ l (t ∧ tj ))) |Ftk ) = E ((gj (N τ l (t ∧ tj+1 ) − N τ l (t ∧ tj )) , fk E ((M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) |Ftk ))) but E ((M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) |Ftk ) = 0 by the Doob optional sampling theorem. Thus µµZ ¶ ¶ Z t

E 0

n−1 X

=

t

f dM τ l , 0

gdN τ l

=

(51.5.20)

U

E (fk gk ((M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) , (N τ l (t ∧ tk+1 ) − N τ l (t ∧ tk ))))

k=0 n−1 X

=

³ ³ ´ ´ t t t t E fk gk (M τ l ) k+1 − (M τ l ) k , (N τ l ) k+1 − (N τ l ) k (t)

k=0

=

n−1 X

³ ³h i ´´ t t t t E fk gk (M τ l ) k+1 − (M τ l ) k , (N τ l ) k+1 − (N τ l ) k (t) + Nk (t)

k=0

(51.5.21) where Nk (t) is a martingale such that Nk (t) = 0 for t ≤ tk since the martingale t t (N τ l ) k+1 − (N τ l ) k equals 0 for such t; and so E (Nk (t)) = 0. Now use Theorems 51.4.4 and 51.4.6 as needed to do the following computations on the right side of the above which equals =

n−1 X

³ ³h i ´´ t t t t E fk gk (M τ l ) k+1 − (M τ l ) k , (N τ l ) k+1 − (N τ l ) k (t) + Nk (t)

k=0

(51.5.22)

Now E (fk gk Nk (t)) = E (fk gk E (Nk (t) |Ftk )) = 0 because Nk is martingale which equals 0 when t = tk . Thus 51.5.22 equals =

n−1 X

³ ³h i ´´ t t t t E fk gk (M τ l ) k+1 − (M τ l ) k , (N τ l ) k+1 − (N τ l ) k (t)

k=0

which equals =

n−1 i ³h 1X ³ t t t t E fk gk (M τ l ) k+1 − (M τ l ) k + (N τ l ) k+1 − (N τ l ) k 4 k=0 í´´ ³ h t t t t − (M τ l ) k+1 − (M τ l ) k − (N τ l ) k+1 − (N τ l ) k

51.5. THE QUADRATIC VARIATION AND STOCHASTIC INTEGRATION n−1 ³ í ³h 1X ³ t t t t E fk gk (M τ l ) k+1 + (N τ l ) k+1 − (M τ l ) k + (N τ l ) k 4 k=0 h ³ í´´ t t t t − (M τ l ) k+1 − (N τ l ) k+1 − (M τ l ) k − (N τ l ) k

=

=

n−1 ³h i h i 1X ³ t t t t E fk gk (M τ l ) k+1 + (N τ l ) k+1 − (M τ l ) k + (N τ l ) k 4 k=0 h i h i´´ t t t t − (M τ l ) k+1 − (N τ l ) k+1 − (M τ l ) k − (N τ l ) k

=

n−1 ³ 1X ³ t t E fk gk [(M τ l ) + (N τ l )] k+1 − [(M τ l ) + (N τ l )] k 4 k=0 ´´ t t − [(M τ l ) − (N τ l )] k+1 − [(M τ l ) − (N τ l )] k

n−1 ¡£¡ t ¢ ¡ t ¢¤τ l £¡ t ¢ ¡ t ¢¤τ l 1X ¡ E fk gk M k+1 + N k+1 − M k + N k 4 k=0 £¡ ¢ ¡ ¢¤τ £¡ ¢ ¡ ¢¤τ ¢¢ − M tk+1 − N tk+1 l − M tk − N tk l

=

Letting l → ∞ this reduces to =

n−1 ¡£¡ t ¢ ¡ t ¢¤ £¡ t ¢ ¡ t ¢¤ 1X ¡ E fk gk M k+1 + N k+1 − M k + N k 4 k=0 £¡ ¢ ¡ ¢¤ £¡ ¢ ¡ ¢¤¢¢ − M tk+1 − N tk+1 − M tk − N tk

= =

µZ Z t ¶ 1 f g (d [M + N ] − d [M − N ]) 4 Ω 0 Z Z t f gd [M, N ] Ω

0

Now consider the left side of 51.5.21. ¶ ¶ µµZ t Z t gdN τ l E f dM τ l , 0

≡

Z X Ω k,j τl

(N

0

U

fk gj ((M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) ,

(t ∧ tj+1 ) − N τ l (t ∧ tj ))) dP

1553

1554


Then for each ω, the integrand converges as l → ∞ to X fk gj ((M (t ∧ tk+1 ) − M (t ∧ tk )) , (N (t ∧ tj+1 ) − N (t ∧ tj ))) k,j

But also you can do a sloppy estimate which will allow the use of the dominated convergence theorem. ¯¯ ¯¯X ¯¯ ¯¯ fk gj ((M τ l (t ∧ tk+1 ) − M τ l (t ∧ tk )) ¯¯ ¯¯ k,j ¯¯ ¯¯ , (N τ l (t ∧ tj+1 ) − N τ l (t ∧ tj )))¯¯ ≤

X

|fk | |gj | 4M ∗ N ∗ ∈ L1 (Ω)

k,j

by assumption. Thus the left side of 51.5.21 converges as l → ∞ to Z X fk gj ((M (t ∧ tk+1 ) − M (t ∧ tk )) , (N (t ∧ tj+1 ) − N (t ∧ tj ))) dP Ω k,j

Z µZ

Z

t

=

f dM, Ω

0

¶

t

gdN 0

dP U

This proves the lemma. Note for each ω, the inside integral in 51.5.18 is just a Stieltjes integral taken with respect to the increasing integrating function [M ]. Of course, with this estimate it is obvious how to extend the integral to a larger class of functions. Definition 51.5.3 Let ν (ω) denote the Radon measure representing the functional Z T Λ (ω) (g) ≡ gd [M ] (t) (ω) 0

(t → [M ] (t) (ω) is a continuous increasing function and ν (ω) is the measure representing the Stieltjes integral, one for each ω.) Then let GM denote ¡ functions f (s, ω) ¢ which are the limit of such elementary functions in the space L2 Ω; L2 ([0, T ] , ν (·)) , the norm of such functions being Z Z t 2 2 ||f ||G ≡ f (s) d [M ] (s) dP Ω

For f ∈ G just defined,

Z

0

Z

t

t

f dM ≡ lim 0

n→∞

fn dM 0

where {fn } is a sequence of elementary functions converging to f in ¡ ¢ L2 Ω; L2 ([0, T ] , ν (·)) .


1555

Now here is an interesting lemma. Lemma 51.5.4 Let M, N be continuous local martingales, M (0) = N (0) = 0 having values in a separable Hilbert space, U. Then ³ ´ 1/2 1/2 1/2 [M + N ] ≤ [M ] + [N ] (51.5.23) [M + N ] ≤ 2 ([M ] + [N ])

(51.5.24)

Also, letting ν M +N denote the measure obtained from the increasing function [M + N ] and ν N , ν M defined similarly, ν M +N ≤ 2 (ν M + ν N )

(51.5.25)

on all Borel sets. Proof: Since (M, N ) → [M, N ] is bilinear and satisfies [M, N ] = [N, M ] [aM + bM1 , N ] = a [M, N ] + b [M1 , N ] [M, M ] ≥ 0 which follows from Theorem 51.4.6, the usual Cauchy Schwarz inequality holds and so 1/2 1/2 |[M, N ]| ≤ [M ] [N ] Thus [M + N ]

≡ ≤

[M + N, M + N ] = [M, M ] + [N, N ] + 2 [M, N ] ³ ´2 1/2 1/2 1/2 1/2 [M ] + [N ] + 2 [M ] [N ] = [M ] + [N ]

This proves 51.5.23. Now the right side is no larger than 2 ([M ] + [N ]) and this shows 51.5.24. Now consider the claim about the measures. It was just shown that s

[(M + N ) − (M + N ) ] ≤ 2 ([M − M s ] + [N − N s ]) and from Theorem 51.4.6 this implies that for t > s

=

[M + N ] (t) − [M + N ] (s ∧ t) [M + N ] (t) − [M + N ] (s)

≤ =

2 ([M − M s ] + [N − N s ]) (t) 2 [M − M s ] (t) + 2 [N − N s ] (t)

≤

2 ([M ] (t) − [M ] (s)) + 2 ([N ] (t) − [N ] (s))

1556


Thus ν M +N ([s, t]) ≤ 2 (ν M ([s, t]) + ν N ([s, t])) By regularity of the measures, this continues to hold with any Borel set, F in place of [s, t]. This proves the lemma. Theorem 51.5.5 The integral is well defined and has a continuous version which is a local martingale. Furthermore it satisfies the Ito isometry, Ã¯¯Z ¯¯2 ! Z Z t ¯¯ ¯¯ t 2 = f (s) d [M ] (s) dP f dM ¯¯¯¯ E ¯¯¯¯ 0

Ω

U

0

Let the norm on GN ∩ GM be the maximum of the norms on GN and GM and denote by EN and EM the elementary functions corresponding to the martingales N and M respectively. Define GN M as the closure in GN ∩ GM of EN ∩ EM . Then for f, g ∈ GN M , µµZ t ¶¶ Z Z t Z t E f dM, gdN = f gd [M, N ] (51.5.26) 0

0

Ω

0

Proof: It is clear the definition is well defined because if¡ {fn } and {gn } are¢ two sequences of elementary functions converging to f in L2 Ω; L2 ([0, T ] , ν (·)) R 1t and if 0 f dM is the integral which comes from {gn } , ¯¯2 Z ¯¯Z 1t Z t ¯¯ ¯¯ ¯¯ f dM − f dM ¯¯¯¯ dP ¯¯ Ω 0 0 ¯¯2 Z ¯¯Z t Z t ¯¯ ¯¯ ¯¯ ¯¯ dP = lim g dM − f dM n n ¯¯ ¯¯ n→∞

Ω

0

Z Z

T

≤ lim

n→∞

0

2

||gn − fn || dνdP = 0 Ω

0

Consider the claim the integral has a continuous version. Recall Theorem 50.7.1, part of which is listed here for convenience. Theorem 51.5.6 Let {X (t)} be a right continuous nonnegative submartingale adapted to the normal filtration Ft for t ∈ [0, T ]. Let p ≥ 1. Define X ∗ (t) ≡ sup {X (s) : 0 < s < t} , X ∗ (0) ≡ 0. Then for λ > 0 P ([X ∗ (T ) > λ]) ≤

1 λp

Z p

X (T ) dP Ω

Let {fn } be a sequence of elementary functions converging to f in ¡ ¢ L2 Ω; L2 ([0, T ] , ν (·)) .

(51.5.27)


1557

Then letting ¯¯Z t ¯¯ ¯¯ ¯¯ = ¯¯¯¯ (fn − fm ) dM τ l ¯¯¯¯ , 0 ¯¯Z t ¯¯ U ¯¯ ¯¯ = ¯¯¯¯ (fn − fm ) dM ¯¯¯¯ 0 U ¯¯ ¯¯Z t Z t ¯¯ ¯¯ fn dM − fm dM ¯¯¯¯ = ¯¯¯¯

τl Xn,m (t)

Xn,m (t)

0

τl Xn,m

It follows just listed,

0

U

is a continuous nonnegative submartingale and from Theorem 50.7.1 Z ¡£ τ ∗ ¤¢ 1 2 τl l P Xn,m (T ) > λ ≤ 2 Xn,m (T ) dP λ Ω ≤ ≤

Z Z T 1 2 |fn − fm | d [M τ l ] dP λ2 Ω 0 Z Z T 1 2 |fn − fm | d [M ] dP 2 λ Ω 0

Letting l → ∞, P

¡£ ∗ ¤¢ 1 Xn,m (T ) > λ ≤ 2 λ

Z Z

T

2

|fn − fm | d [M ] dP Ω

0

Therefore, there exists a subsequence, still denoted by {fn } such that ¡£ ∗ ¤¢ P Xn,n+1 (T ) > 2−n < 2−n Then by the Borel Cantelli lemma, the ω in infinitely many of the sets £ ∗ ¤ Xn,n+1 (T ) > 2−n has measure 0. Denoting this exceptional set as N, it follows that for ω ∈ / N, there exists n (ω) such that for n > n (ω) , ¯¯Z t ¯¯ Z t ¯¯ ¯¯ ¯ ¯ sup ¯¯ fn dM − fn+1 dM ¯¯¯¯ ≤ 2−n t∈[0,T ]

0

0

and this implies uniform convergence of

nR t 0

o fn dM . Letting

Z

t

G (t) = lim

n→∞

fn dM, 0

for ω ∈ / N and G (t) = 0 for ω ∈ N, it follows for each t, the continuous adapted nR that o Rt t process G (t) equals 0 f dM a.e. Thus 0 f dM has a continuous version.

1558


It suffices to verify 51.5.26. Let {fn } and {gn } be sequences of elementary functions converging to f and g in GM ∩GN . By Lemma 51.5.2, µµZ t ¶ ¶ Z Z t Z t E fn dM, gn dN = fn gn d [M, N ] 0

0

Ω

U

0

Then by the Holder inequality and the above definition, µµZ t ¶ ¶ µµZ t ¶ ¶ Z t Z t lim E fn dM, gn dN =E f dM, gdN n→∞

0

0

0

U

0

U

Consider the right side which equals Z Z t Z Z t 1 1 fn gn d [M + N ] dP − fn gn d [M − N ] dP 4 Ω 0 4 Ω 0 Now from Lemma 51.5.4, ¯Z Z t ¯ Z Z t ¯ ¯ ¯ fn gn d [M + N ] dP − f gd [M + N ] dP ¯¯ ¯ Ω 0 Ω 0 ¯Z Z t ¯ Z Z t ¯ ¯ ¯ = ¯ fn gn dν M +N dP − f gdν M +N dP ¯¯ Ω

Z Z

0

Ω

Z Z

t

≤

0

|fn gn − f g| dν M dP + Ω

0

t

|fn gn − f g| dν N dP Ω

0

and by the choice of the fn and gn , these both converge to 0. Similar considerations apply to ¯Z Z t ¯ Z Z t ¯ ¯ ¯ ¯ f g d [M − N ] dP − f gd [M − N ] dP n n ¯ ¯ Ω

0

and show

Ω

Z Z lim

n→∞

0

Z Z

t

fn gn d [M, N ] = Ω

0

t

f gd [M, N ] Ω

0


51.6

Doob Meyer Decomposition

This section is on the Doob Meyer decomposition which is a way of starting with a submartingale and writing it as the sum of a martingale and an increasing adapted stochastic process of a certain form. This is more general than what was done above 2 with the submartingales ||M (t)|| for M (t) ∈ MT2 (H) where M is a continuous martingale. There are two forms for this theorem, one for discreet martingales and one for martingales defined on an interval of the real line which is much harder. According to [37], this material is found in [41] however, I am following [37] for the continuous version of this theorem.

51.6. DOOB MEYER DECOMPOSITION

1559

Theorem 51.6.1 Let {Xn } be a submartingale. Then there exists a unique stochastic process, {An } and martingale, {Mn } such that 1. An (ω) ≤ An+1 (ω) , A1 (ω) = 0, 2. An is Fn−1 adapted for all n ≥ 1 where F0 ≡ F1 . and also Xn = Mn + An . Proof: Let A1 ≡ 0 and define An ≡

n X

E (Xk − Xk−1 |Fk−1 ) .

k=2

It follows An is Fn−1 measurable. Since {Xk } is a submartingale, An is increasing because An+1 − An = E (Xn+1 − Xn |Fn ) ≥ 0 (51.6.28) It is a submartingale because E (An |Fn−1 ) = An ≥ An−1 . Now let Mn be defined by Xn = Mn + An . Then from 51.6.28, E (Mn+1 |Fn ) = E (Xn+1 |Fn ) − E (An+1 |Fn ) = = = = =

E (Xn+1 |Fn ) − E (An+1 − An |Fn ) − An E (Xn+1 |Fn ) − E (E (Xn+1 − Xn |Fn ) |Fn ) − An E (Xn+1 |Fn ) − E (Xn+1 − Xn |Fn ) − An E (Xn |Fn ) − An Xn − An ≡ Mn

This proves the existence part. It remains to verify uniqueness. Suppose then that Xn = Mn + An = Mn0 + A0n where {An } and {A0n } both satisfy the conditions of the theorem and {Mn } and {Mn0 } are both martingales. Then Mn − Mn0 = A0n − An and so, since A0n − An is Fn−1 measurable and {Mn − Mn0 } is a martingale, 0 Mn−1 − Mn−1

= E (Mn − Mn0 |Fn−1 ) = E (A0n − An |Fn−1 ) = A0n − An = Mn − Mn0 .

Continuing this way shows Mn − Mn0 is a constant. However, since A01 − A1 = 0 = M1 − M10 , it follows Mn = Mn0 and this proves uniqueness. This proves the theorem.

1560


Definition 51.6.2 A stochastic process, {An } which satisfies the conditions of Theorem 51.6.1, An (ω) ≤ An+1 (ω) and An is Fn−1 adapted for all n ≥ 1 where F0 ≡ F1 is said to be natural. The Doob Meyer theorem needs to be extended to continuous submartingales and this will require another description of what it means for a stochastic process to be natural. To get an idea of what this condition should be, here is a lemma. Lemma 51.6.3 Let a stochastic process, {An } be natural. Then for every martingale, {Mn } ,   n−1 X E (Mn An ) = E  Mj (Aj+1 − Aj ) j=1

Proof: Start with the right side.     n−1 n n−1 X X X E Mj (Aj+1 − Aj ) = E  Mj−1 Aj − Mj Aj  j=1

 E

=

j=2 n−1 X

j=1



Aj (Mj−1 − Mj ) + E (Mn−1 An )

j=2

Then the first term equals zero because since Aj is Fj−1 measurable, Z Z Z Z Aj Mj−1 dP − Aj Mj = Aj E (Mj |Fj−1 ) dP − Aj Mj dP Ω Ω ZΩ ZΩ = E (Aj Mj |Fj−1 ) dP − Aj Mj dP Ω ZΩ Z = Aj Mj dP − Aj Mj dP = 0. Ω

The last term equals Z Mn−1 An dP

Z =

Ω

E (Mn |Fn−1 ) An dP ZΩ

=

E (Mn An |Fn−1 ) dP = E (Mn An ) . Ω


Ω


1561

Definition 51.6.4 Let A be an increasing function defined on R. By Theorem 3.3.4 on Page 42 there exists a positive linear functional, L defined on Cc (R) given by Z b f dA where spt (f ) ⊆ [a, b] Lf ≡ a

where the integral is just the Riemann Stieltjes integral. Then by the Riesz representation theorem, Theorem 9.4.2 on Page 234, there exists a unique Radon measure, µ which extends this functional, as described in the Riesz representation theorem. Then for B a measurable set, I will write either Z Z f dA f dµ or B

B

to denote the Lebesgue integral,

Z XB f dµ.

Lemma 51.6.5 Let f be right continuous. Then f is Borel measurable. Also, if the limit from the left exists, then f− (x) ≡ f (x)− ≡ limy→x− f (y) is also Borel measurable. If A is an increasing right continuous function and f is right continuous and f− , the left limit function exists, then if f is bounded, on [a, b], and if n o∞ xp0 , · · · , xpnp

p=1

is a sequence of partitions of [a, b] such that ¯ ©¯ ª lim max ¯xpk − xpk−1 ¯ : k = 1, 2, · · · , np = 0 p→∞

then

Z (a,b]

np X ¡ ¢¡ ¡ ¢¢ f− dA = lim f xpk−1 A (xpk ) − A xpk−1

More generally, let

p→∞

(51.6.29)

(51.6.30)

k=1

n o∞ p p D ≡ ∪∞ x , · · · , x p=1 np 0

p=1

and f− (t) =

lim

s→t−,s∈D

f (s) .

Then 51.6.30 holds. Proof: For x ∈ f −1 ((a, ∞)) , denote by Ix the union of all intervals containing x such that f (y) is larger than a for all y in the interval. Since f is right continuous, each Ix has positive length. Now if Ix and Iy are two of these intervals, then either they must have empty intersection or they are the same interval. Thus f −1 ((a, ∞))

1562


is of the form ∪x∈f −1 ((a,∞)) Ix and there can only be countably many distinct intervals because each has positive length and R is separable. Hence f −1 ((a, ∞)) equals the countable union of intervals and is therefore, Borel measurable. Now f− (x) = lim f (x − rn ) ≡ lim frn (x) n→∞

n→∞

where rn is a decreasing sequence converging to 0. Now each frn is Borel measurable by the first part of the proof because it is right continuous and so it follows the same is true of f− . Finally consider the claim about the integral. Since A is right continuous, a simple argument involving the dominated convergence theorem and approximating (c, d] with a piecewise linear continuous function nonzero only on (c, d + h) which approximates X(c,d] will show that for µ the measure of Definition 51.6.4 µ ((c, d]) = A (d) − A (c) . Therefore, the sum in 51.6.30 is of the form Z np X ¡ p ¢ f xk−1 µ ((xk−1 , xk ]) =

np X ¡ ¢ f xpk−1 X(xk−1 ,xk ] dµ

(a,b] k=1

k=1

and by 51.6.29 lim

p→∞

np X ¡ ¢ f xpk−1 X(xk−1 ,xk ] (x) = f− (x) k=1

for each x ∈ (a, b]. Therefore, since f is bounded, 51.6.30 follows from the dominated convergence theorem. The last claim follows the same way. This proves the lemma. Definition 51.6.6 An increasing stochastic process, {A (t)} which is right continuous is said to be natural if A (0) = 0 and whenever {ξ (t)} is a bounded right continuous martingale, ÃZ ! E (A (t) ξ (t)) = E (0,t]

ξ − (s) dA (s) .

(51.6.31)

Here ξ − (s, ω) ≡

lim

r→s−,r∈D

ξ (r, ω)

a.e. where D is a countable dense subset of [0, t] . By Corollary 50.6.2 the right side of 51.6.31 is not dependent on the choice of D since if ξ − is computed using two different dense subsets, the two random variables are equal a.e. Some discussion is in order for this definition. Pick ω ∈ Ω. Then since A is right continuous, the function t → A (t, ω) is increasing and right continuous. Therefore, one can do the Lebesgue Stieltjes integral defined in Definition 51.6.4 for each ω whenever f is Borel measurable and bounded. Now it is assumed {ξ (t)} is bounded


1563

and right continuous. By Lemma 51.6.5 ξ − (t) ≡ limr→t−,r∈D ξ (r) is measurable and by this lemma, Z (0,t]

ξ − (s) dA (s) = lim

p→∞

np X ¡ ¢¡ ¡ ¢¢ ξ tpk−1 A (tpk ) − A tpk−1 k=1

n

p where {tpk }k=1 is a sequence of partitions of [0, t] such that ¯ ©¯ ª lim max ¯tpk − tpk−1 ¯ : k = 1, 2, · · · , np = 0.

(51.6.32)

p→∞

n

n

p p p and D ≡ ∪∞ p=1 ∪k=1 {tk }k=1 . Also, if t → A (t, ω) is right continuous, hence Borel measurable, then for ξ (t) the above bounded right continuous martingale, it follows it makes sense to write Z ξ (s) dA (s) . (0,t]

Consider the right sum, np X

¡ ¡ ¢¢ ξ (tpk ) A (tpk ) − A tpk−1

k=1

This equals

Z

np X

(0,t] k=1

ξ (tpk ) X(tpk−1 ,tpk ] (s) dA (s)

and by right continuity, it follows lim

p→∞

np X

ξ (tpk ) X(tpk−1 ,tpk ] (s) = ξ (s)

k=1

and so the dominated convergence theorem applies and it follows lim

p→∞

np X

ξ

¡ (tpk ) A (tpk )

¡ ¢¢ − A tpk−1 =

k=1

where this is a random variable. Thus ÃZ ! Z Ã Z E ξ (s) dA (s) = lim (0,t]

Ω

p→∞

np X

Z ξ (s) dA (s) (0,t]

! ξ

(0,t] k=1

(tpk ) X(tpk−1 ,tpk ]

(s) dA (s) dP (51.6.33)

Now as mentioned above, Z

np X

(0,t] k=1

ξ (tpk ) X(tpk−1 ,tpk ] (s) dA (s) =

np X k=1

¡ ¡ ¢¢ ξ (tpk ) A (tpk ) − A tpk−1

1564


and since A is increasing, this is bounded above by an expression of the form CA (t) , a function in L1 . Therefore, by the dominated convergence theorem, 51.6.33 reduces to Z Z np X ξ (tpk ) X(tpk−1 ,tpk ] (s) dA (s) dP lim p→∞

=

=

lim

p→∞

lim

p→∞

Ω

(0,t] k=1

Z X np

Ω k=1

¡ ¡ ¢¢ ξ (tpk ) A (tpk ) − A tpk−1 dP

Z ÃX np Ω

lim

p→∞

X

k=1

ξ

−

k=1

np −1 Z

=

np −1

(tpk ) A (tpk )

Ω

X

! ¡p ¢ p ξ tk+1 A (tk ) dP

k=0

¡ p ¡ ¢¢ ξ (tk ) − ξ tpk+1 A (tpk ) dP +

Since ξ is a martingale, Z ¡ ¢ ξ tpk+1 A (tpk ) dP

Z =

Ω

ZΩ = ZΩ = Ω

Z ξ (t) A (t) dP.(51.6.34) Ω

³ ¡ ´ ¢ E ξ tpk+1 A (tpk ) |Ftpk dP ´ ³ ¡ ¢ A (tpk ) E ξ tpk+1 |Ftpk dP A (tpk ) ξ (tpk ) dP

and so in 51.6.34 the term with the sum equals 0 and it reduces to E (ξ (t) A (t)) . This is sufficiently interesting to state as a lemma. Lemma 51.6.7 Let A be an increasing adapted stochastic process which is right continuous. Also let ξ (t) be a bounded right continuous martingale. Then ÃZ ! E (ξ (t) A (t)) = E ξ (s) dA (s) (0,t]

and A is natural, if and only if for all such bounded right continuous martingales, ÃZ ! ÃZ ! E (ξ (t) A (t)) = E

ξ (s) dA (s)

=E

(0,t]

(0,t]

ξ − (s) dA (s)

Lemma 51.6.8 Let (Ω, F, P ) be a probability space and let G be a σ algebra contained in F. Suppose also that {fn } is a sequence in L1 (Ω) which converges weakly to f in L1 (Ω) . That is, for every h ∈ L∞ (Ω) , Z Z fn hdP → f hdP. Ω

Ω 1

Then E (fn |G) converges weakly in L (Ω) to E (f |G).


1565

Proof:First note that if h ∈ L∞ (Ω, F) , then E (h|G) ∈ L∞ (Ω, G) because if A ∈ G, Z Z Z |E (h|G)| dP ≤ E (|h| |G) dP = |h| dP A

A

A

and so if A = [|E (h|G)| > ||h||∞ ] , then if P (A) > 0, Z ||h||∞ P (A)
0, the set of random variables, {X (T )} for T a stopping time bounded by a is equi integrable. Example 51.6.10 Let {M (t)} be a continuous martingale. Then {M (t)} is of class DL. To show this, let a > 0 be given and let T be a stopping time bounded by a. Then by the optional sampling theorem, M (0) , M (T ) , M (a) is a martingale and so E (M (a) |FT ) = M (T ) and so by Jensen’s inequality, |M (T )| ≤ E (|M (a)| |FT ) . Therefore, Z Z |M (T )| dP ≤ E (|M (a)| |FT ) dP [|M (T )|≥λ] [|M (T )|≥λ] Z = |M (a)| dP.

(51.6.35)

[|M (T )|≥λ]

Now by Theorem 50.3.5, P ([|M (T )| ≥ λ]) ≤

1 E (|M (a)|) λ

and so since a given L1 function is uniformly integrable, there exists δ such that if P (A) < δ then Z |M (a)| dP < ε. A


1567

Now choose λ large enough that 1 E (|M (a)|) < δ. λ Then for such λ, it follows from 51.6.35 that for any stopping time bounded by a, Z |M (T )| dP < ε. [|M (T )|≥λ]

This shows M is DL. Example 51.6.11 Let {X (t)} be a nonnegative submartingale with t → E (X (t)) right continuous so {X (t)} can be considered right continuous. Then {X (t)} is DL. To show this, let T be a stopping time bounded by a > 0. Then by the optional sampling theorem, Z Z X (T ) dP ≤ X (a) dP [X(T )≥λ]

[X(T )≥λ]

and now by Theorem 48.4.5 on Page 1389 P ([X (T ) ≥ λ]) ≤

1 ¡ +¢ E Xa . λ

Thus if ε > 0 is given, there exists λ large enough that for any stopping time, T ≤ a, Z X (T ) dP ≤ ε [X(T )≥λ]

Thus the submartingale is DL. Now with this preparation, here is the Doob Meyer decomposition. Theorem 51.6.12 Let {X (t)} be a submartingale of class DL. Then there exists a martingale, {M (t)} and an increasing submartingale, {A (t)} such that for each t, X (t) = M (t) + A (t) . If {A (t)} is chosen to be natural and A (0) = 0, then with this condition, {M (t)} and {A (t)} are unique. Proof: First I will show uniqueness. Suppose then that X (t) = M (t) + A (t) = M 0 (t) + A0 (t) where M, M 0 and A, A0 satisfy the given conditions. Let t > 0 and consider s ∈ [0, t] . Then A (s) − A0 (s) = M 0 (s) − M (s)

1568


Since A, A0 are natural, it follows that for ξ (t) a right continuous bounded martingale, ÃZ ! ÃZ ! 0 0 E (ξ (t) (A (t) − A (t))) = E ξ − (s) dA (s) − E ξ − (s) dA (s) (0,t]

Ã =E

(0,t]

mn mn X ¡ ¢¡ ¡ ¢¢ X ¢¡ ¡ ¢¢ ¡ lim ξ tnk−1 A (tnk ) − A tnk−1 − ξ tnk−1 A0 (tnk ) − A0 tnk−1

n→∞

k=1

!

k=1

mn {tnk }k=0

where is a sequence of partitions of [0, t] such that these are equally spaced © ªmn+1 mn points, limn→∞ tnk+1 − tnk = 0, and {tnk }k=0 ⊆ tn+1 . Then since A (t) and k k=0 A0 (t) are increasing, the absolute value of each sum is bounded above by an expression of the form CA (t) or CA0 (t) and so the dominated convergence theorem can be applied to get the above expression to equal Ãm ! mn n X ¡ n ¢¡ ¡ n ¢¢ X ¡ n ¢¡ 0 n ¡ ¢¢ n 0 n lim E ξ tk−1 A (tk ) − A tk−1 − ξ tk−1 A (tk ) − A tk−1 n→∞

k=1

k=1

Now using X = A + M and X = A0 + M 0 Ãm ! mn n X ¡ n ¢¡ ¡ n ¢¢ X ¡ n ¢¡ 0 n ¡ ¢¢ n 0 n = lim E ξ tk−1 M (tk ) − M tk−1 − ξ tk−1 M (tk ) − M tk−1 . n→∞

k=1

k=1

Both terms in the above equal 0. Here is why. ³ ³ ¡ ´´ ¢ ¢ ¡ ¡ ¢ E ξ tnk−1 M (tnk ) = E E ξ tnk−1 M (tnk ) |Ftnk−1 ³ ¡ ´´ ¢ ³ = E ξ tnk−1 E M (tnk ) |Ftnk−1 ¡ ¡ ¢ ¡ ¢¢ = E ξ tnk−1 M tnk−1 . Thus the expected value of the first sum equals 0. Similarly, the expected value of the second sum equals 0. Hence this has shown that for any bounded right continuous martingale, {ξ (s)} and t > 0, E (ξ (t) (A (t) − A0 (t))) = 0. Now let ξ be a bounded random variable and let ξ (t) be a right continuous version of the martingale E (ξ|Ft ) . Then 0 = E (E (ξ|Ft ) (A (t) − A0 (t))) = E (E (ξ (A (t) − A0 (t)) |Ft )) = E (ξ (A (t) − A0 (t))) and since ξ is arbitrary, it follows that A (t) = A0 (t) a.e. which proves uniqueness.


1569

Because of the uniqueness assertion, it suffices to prove the theorem on an arbitrary interval, [0, a]. Without loss of generality, it can be assumed X (0) = 0 since otherwise, you could simply consider X (t) − X (0) in its place and then at the end, add X (0) to mn M (t) . Let {tnk }k=0 be a sequence of partitions of [0, a] such that these are equally © ªmn+1 mn spaced points, limn→∞ tnk+1 − tnk = 0, and {tnk }k=0 ⊆ tn+1 . Then consider k k=0 mn the submartingale, {X (tnk )}k=0 . Theorem 51.6.1 implies there exists a unique marmn mn tingale, and increasing submartingale, {M (tnk )}k=0 and {A (tnk )}k=0 respectively such that M (0) = 0 = A (0) , X (tnk ) = M n (tnk ) + An (tnk ) . and An (tnk ) is Ftnk−1 measurable. Recall how these were defined. An (tnk ) =

k X

³ ¡ ¢ ´ ¡ ¢ E X tnj − X tnj−1 |Ftnj−1 , An (0) = 0

j=1

M n (tnk ) = X (tnk ) − An (tnk ) . I want to show that {An (a)} is equi integrable. From this there will be a weakly convergent subsequence and nice things will happen. Define T n (ω) to equal tnj−1 ¡ ¢ where tnj is the first time where An tnj , ω ≥ λ or T n (ω) = a if this never happens. ¤ £ I want to say that T n is a stopping time and so I need to verify that T n ≤ tnj ∈ Ftnj ¤ £ for each j. If ω ∈ T n ≤ tnj , then this means the first time, tnk , where An (tnk , ω) ≥ λ is such that tnk ≤ tnj+1 . Since Ank is increasing in k, £ n ¤ T ≤ tnj = =

n n ∪j+1 k=0 [A (tk ) ≥ λ] £ n¡ n ¢ ¤ A tj+1 ≥ λ ∈ Ftnj .

Note T n only has the values tnk . Thus for t ∈ [tnj−1 , tnj ), £ ¤ [T n ≤ t] = T n ≤ tnj−1 ∈ Ftnj−1 ⊆ Ft . Thus T n is one of those stopping times bounded by a. Since {X (t)} is DL, this shows {X (T n )} is equi integrable. Now from the definition of T n , it follows An (T n ) ≤ λ ¡ ¢ Recall T n (ω) = tnj−1 where tnj is the first time where An tnj , ω ≥ λ or T n (ω) = a if this never happens. Thus T n is such that it is before An gets larger than λ. Thus, Z Z 1 n A (a) dP ≤ (An (a) − λ) dP [An (a)≥2λ] 2 [An (a)≥2λ] Z ≤ (An (a) − An (T n )) dP [An (a)≥2λ]

1570


Z (An (a) − An (T n )) dP

≤ Ω

Z =

(X (a) − M n (a) − (X (T n )) − M n (T n )) dP Ω Z = (X (a) − X (T n )) dP Ω

Because by the discreet optional sampling theorem, Z (M n (a) − M n (T n )) dP = 0. Ω

Remember Z

mn {M n (tnk )}k=0

was a martingale. Z (X (a) − X (T n )) dP = (X (a) − X (T n )) dP Ω [An (a)≥λ] Z + (X (a) − X (T n )) dP. [An (a) t ÃZ

!

E ÃZ =

+ (A (a) − A (t)) E (ξ (t))

ξ (s) dA (s) (0,t]

! t

E

ξ (s) dA (s) (0,a]

From what was shown above, ÃZ =

E (0,a]

! ξ t−

(s) dA (s)

ÃZ =

E (0,t]

Z ξ − (s) dA (s) +

ÃZ =

E (0,t]

! ξ − (s) dA (s)

! ξ (t) sA (s)

(t,a]

+ (A (a) − A (t)) E (ξ (t))

1574


and so

ÃZ

!

E

ξ (s) dA (s) (0,t]

ÃZ

!

=E (0,t]

ξ − (s) dA (s)

which shows A is natural by Lemma 51.6.7. This proves the theorem. There is another interesting variation of the above theorem. It involves the following definition. Definition 51.6.13 A submartingale, {X (t)} is said to be D if {XT : T < ∞ is a stopping time} is equi integrable. In this case, you can consider partitions of the entire positive real line and the ∞ ∞ martingales, {M (tnk )}k=0 and {A (tnk )}k=0 as before. This time you don’t stop at mn . By the submartingale convergence theorem, you can argue there exists An∞ = limk→∞ A (tnk ) . Then repeat the above argument using An∞ in place of An (a) . This time you get {A (t)} equi integrable. Thus the following corollary is obtained. Corollary 51.6.14 Let {X (t)} be a right continuous submartingale of class D. Then there exists a right continuous martingale, {M (t)} and a right continuous increasing submartingale, {A (t)} such that for each t, X (t) = M (t) + A (t) . If {A (t)} is chosen to be natural and A (0) = 0, then with this condition, {M (t)} and {A (t)} are unique. Furthermore {M (t)} and {A (t)} are equi integrable on [0, ∞). In the above theorem, {X (t)} was a submartingale and so it has a right continuous version. What if {X (t)} is actually continuous? Can one conclude that A (t) and M (t) are also continuous? The answer is yes. Theorem 51.6.15 Let {X (t)} be a right continuous submartingale of class DL. Then there exists a right continuous martingale, {M (t)} and a right continuous increasing submartingale, {A (t)} such that for each t, X (t) = M (t) + A (t) . If {A (t)} is chosen to be natural and A (0) = 0, then with this condition, {M (t)} and {A (t)} are unique. Also, if {X (t)} is continuous, (t → X (t, ω) is continuous for a.e. ω) then the same is true of {A (t)} and {M (t)} . Proof: The first part is done above. Let {X (t)} be continuous. As before, mn let {tnk }k=0 be a sequence of partitions of [0, a] such that these are equally spaced © ªmn+1 mn where here a > 0 is an points, limn→∞ tnk+1 − tnk = 0, and {tnk }k=0 ⊆ tn+1 k k=0 arbitrary positive number and let λ > 0 be an arbitrary positive number. Define ¡ ¡ ¡ ¢¢ ¢ ξ n (t) ≡ E min λ, A tnj |Ft for tnj−1 < t ≤ tnj


1575

Thus on (tnj−1 , tnj ] ξ n (t) is a bounded martingale. Assuming we are dealing with a right continuous version of this martingale so there are no measurability questions, it follows since A is natural, ! ! ÃZ ÃZ ξ n (s) dA (s)

E

=E n (tn j−1 ,tj ]

n (tn j−1 ,tj ]

where here

ξ n− (s, ω) ≡

lim

r→s−,r∈D

ξ n− (s) dA (s)

ξ n (s, ω) a.e.

m

mn n n n n for D ≡ ∪∞ n=1 ∪k=1 {tk }k=0 . Thus, adding these up for all the intervals, (tj−1 , tj ] yields ÃZ ! ÃZ ! n n E ξ (s) dA (s) = E ξ − (s) dA (s) (0,a]

(0,a]

I want to show that for a.e. ω, ξ

nk

(t, ω) converges uniformly to

min (λ, A (t, ω)) ≡ λ ∧ A (t, ω) on (0, a]. From this it will follow ÃZ E

!

λ ∧ A (s, ω) dA (s) (0,a]

ÃZ

!

=E

λ ∧ A− (s, ω) dA (s) (0,a]

Now since s → A (s, ω) is increasing, there is no problem in writing A− (s, ω) and the above equation will suffice to show with simple considerations that for a.e. ω, s → A (s, ω) is left continuous. Since {A (s)} is a submartingale already, it has a right continuous version which we are using in the above. Thus for a.e. ω it must be the case that s → A (s, ω) is continuous. Let t ∈ (tnj−1 , tnj ]. Then since λ ∧ A (t) is Ft measurable, ¡ ¡ ¢ ¢ ξ n (t) − λ ∧ A (t) ≡ E λ ∧ A tnj − λ ∧ A (t) |Ft ≥ 0 because A (t) is increasing. Now define a stopping time, T n (ε) for ε > 0 by letting T n (ε) be the infimum of all t ∈ [0, a] with the property that ξ n (t) − λ ∧ A (t) > ε or if this does not happen, then T n (ε) = a. Thus T n (ε) (ω) = a ∧ inf {t ∈ [0, a] : ξ n (t, ω) − λ ∧ A (t, ω) > ε} I need to verify T n (ε) really is a stopping time. Letting s < a, it follows that if ω ∈ [T n (ε) ≤ s] , then for each N, there exists t ∈ [s, s + N1 ) such that ξ n (t, ω) − λ ∧ A (t, ω) > ε. Then by right continuity it follows there exists r ∈ D ∩ [s, s + N1 ) such that ξ n (r, ω) − λ ∧ A (r, ω) > ε

1576 Thus


n 1 [T n (ε) ≤ s] = ∩∞ ) [ξ (r, ω) − λ ∧ A (r, ω) > ε] N =1 ∪r∈D∩[s,s+ N

and each ∪r∈D∩[s,s+ N1 ) [ξ n (r, ω) − λ ∧ A (r, ω) > ε] ∈ Fs+1/N and so [T n (ε) ≤ s] ∈ ∩r∈D,r≥s Fr = Fs+ = Fs due to the assumption that the filtration is normal. What if s ≥ a? Then from the definition, [T n (ε) ≤ a] = Ω ∈ Fa . Thus this really is a stopping time. £ ¤ Now let Bj ≡ tnj−1 < T n (ε) ≤ tnj . Note that T n (ε)∧tnj is also a stopping time. Z Ω

ξ nT n (ε) dP =

mn Z X Bj

j=1

=

mn Z X j=1

Bj

ξ nT n (ε) dP =

mn Z X j=1

Bj

ξ nT n (ε)∧tnj dP

³ ´ E ξ nT n (ε)∧tnj |FT n (ε)∧tnj dP.

This is because Bj ∈ FT n (ε)∧tnj . Thus from the definition, the above equals = = =

mn Z X j=1 Bj mn Z X j=1 Bj mn Z X j=1

Bj

³ ³ ´ ´ ¡ ¢ E E λ ∧ A tnj |FT n (ε)∧tnj |FT n (ε)∧tnj dP ³ ´ ¡ ¢ E λ ∧ A tnj |FT n (ε)∧tnj dP ¡ ¢ λ ∧ A tnj dP =

Z λ ∧ A (dT n (ε)e) dP

(51.6.41)

Ω

where on (tnj−1 , tnj ], dT n (ε)e ≡ tnj . Now dT n (ε)e is also a bounded stopping time. Here is why. Suppose s ∈ (tnj−1 , tnj ]. Then £ ¤ [dT n (ε)e ≤ s] = T n (ε) ≤ tnj−1 ∈ Ftnj−1 ⊆ Fs . Now let

Qn ≡ sup |ξ n (t) − λ ∧ A (t)| . t∈[0,a]

Then first note that " [Qn > ε] =

# n

sup |ξ (t) − λ ∧ A (t)| > ε t∈[0,a)

because Qn (a) = 0 follows from the definition of ξ n (t) as ¡ ¡ ¢ ¢ E λ ∧ A tnj |Ft for tnj−1 < t ≤ tnj and so

ξ n (a) = E (λ ∧ A (a) |Fa ) = λ ∧ A (a) .


1577

Thus it suffices to take the supremum over the half open interval, [0, a). It follows [Qn > ε] = [T n (ε) < a] By right continuity,

ξ n (T n (ε)) − λ ∧ A (T n (ε)) ≥ ε

on [Qn > ε] . εP ([Qn > ε])

εP ([T n (ε) < a]) Z (ξ n (T n (ε)) − λ ∧ A (T n (ε))) dP [Q >ε] Z n (ξ n (T n (ε)) − λ ∧ A (T n (ε))) dP

= ≤ ≤

Ω

Therefore, from 51.6.41, Z 1 (λ ∧ A (dT n (ε)e) − λ ∧ A (T n (ε))) dP ε Ω Z 1 (A (dT n (ε)e) − A (T n (ε))) dP (51.6.42) ε Ω

P ([Qn > ε]) ≤ ≤

By optional sampling theorem, E (M (T n (ε))) = E (M (0)) = 0 and also

E (M (dT n (ε)e)) = E (M (0)) = 0.

Therefore, 51.6.42 reduces to 1 P ([Qn > ε]) ≤ ε

Z (X (dT n (ε)e) − X (T n (ε))) dP Ω

By the assumption that {X (t)} is DL, it follows the functions in the above integrand are equi integrable and so since limn→∞ X (dT n (ε)e) − X (T n (ε)) = 0, the above integral converges to 0 as n → ∞ by Vitali’s convergence theorem, Theorem 8.5.3 on Page 213. It follows that there is a subsequence, nk such that ¡£ ¤¢ P Qnk > 2−k ≤ 2−k and so from the definition of Qn , lim sup |ξ nk (t) − λ ∧ A (t)|

k→∞ t∈[0,a]

giving uniform convergence. Now recall that ! ÃZ ÃZ E

ξ (0,a]

nk

(s) dA (s)

=E (0,a]

! ξ n−k

(s) dA (s)

1578


and so passing to the limit as k → ∞ with the uniform convergence yields ÃZ ! ÃZ ! E λ ∧ A (s) dA (s) = E λ ∧ A− (s) dA (s) (0,a]

(0,a]

Now let λ → ∞. Then from the monotone convergence theorem, ! ! ÃZ ÃZ A− (s) dA (s) A (s) dA (s) = E E (0,a]

and so for a.e. ω,

(0,a]

Z (A (s) − A− (s)) dA (s) = 0. (0,a]

Thus letting the measure associated with this Lebesgue integral be denoted by µ, A (s) − A− (s) = 0 µ a.e. Suppose then that A (s) − A− (s) > 0. Then µ ({s}) = 0 = A (s) − A (s−) , a contradiction. Hence A (s) − A− (s) = 0 for all s. It is already the case that s → A (s) is right continuous. Therefore, this proves the theorem. Example 51.6.16 Suppose {M (t)} is a continuous martingale. Assume sup ||M (t)||L2 (Ω) < ∞

t∈[0,a]

n o 2 Then {||M (t)||} is a submartingale and so is ||M (t)|| . By Example 51.6.11, this is DL. Then there exists a unique Doob Meyer decomposition, 2

||M (t)|| = Y (t) + h||M (t)||i where Y (t) is a martingale and {h||M (t)||i} is a submartingale which is continuous, natural, increasing and equal to 0 when t = 0. This submartingale is called the quadratic variation.

51.7

Levy’s Theorem

This remarkable theorem has to do with when a martingale is a Wiener process. The proof I am giving here follows [21]. Definition 51.7.1 Let W (t) be a stochastic process which has the properties that whenever t1 < t2 < · · · < tm , the increments {W (ti ) − W (ti−1 )} are independent and whenever s < t, it follows W (t) − W (s) is normally distributed with variance t − s and mean 0. Also W (0) = 0. This is called a Wiener process. First here is a lemma.

51.7. LEVY’S THEOREM

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Lemma 51.7.2 Let {X (t)} be a real martingale³adapted ∈ ´ to the filtration n Ft for t o 2 2 [a, b] some interval such that for all t ∈ [a, b] , E X (t) < ∞. Then X (t) − t is also a martingale if and only if whenever s < t, ³ ´ 2 E (X (t) − X (s)) |Fs = t − s. n o 2 Proof: Suppose first X (t) − t is a real martingale. Then since {X (t)} is a martingale, ³ ´ ³ ´ 2 2 2 E (X (t) − X (s)) |Fs = E X (t) − 2X (t) X (s) + X (s) |Fs ³ ´ 2 2 = E X (t) |Fs − 2E (X (t) X (s) |Fs ) + X (s) ³ ´ 2 2 = E X (t) |Fs − 2X (s) E (X (t) |Fs ) + X (s) ³ ´ 2 2 2 = E X (t) |Fs − 2X (s) + X (s) ³ ´ 2 2 = E X (t) − t|Fs + t − X (s) 2

2

= X (s) − s + t − X (s) = t − s ³ ´ 2 Next suppose E (X (t) − X (s)) |Fs = t − s. Then since {X (t)} is a martingale, ³ ´ 2 2 t − s = E X (t) − X (s) |Fs ³ ´ 2 2 = E X (t) − t|Fs + t − X (s) and so

³ ´ ³ ´ 2 2 0 = E X (t) − t|Fs − X (s) − s

which proves the converse. Theorem 51.7.3 Suppose {X (t)} is a real stochastic process which satisfies all the conditions of a real Wiener process n o except the requirement that it be continuous. 2 Then both {X (t)} and X (t) − t are martingales. Proof: First define the filtration to be Ft ≡ σ (X (s) − X (r) : r ≤ s ≤ t) . Claim: If A ∈ Fs , then Z Z XA (X (t) − X (s)) dP = P (A) (X (t) − X (s)) dP. Ω

Ω

1580


Proof of claim: Let G denote those sets of Fs for which the above formula holds. Then it is clear that G is closed with respect to countable unions of disjoint sets and complements. Let K denote those sets which are finite intersections of sets −1 of the form (X (u) − X (r)) (B) where B is a Borel set and r ≤ u ≤ s. Say a set, A of K is of the form −1 ∩m (Bi ) i=1 (X (ui ) − X (ri )) Then since disjoint increments are independent, linear combinations of the random variables, X (ui ) − X (ri ) are normally distributed. Consequently, (X (u1 ) − X (r1 ) , · · · , X (um ) − X (rm ) , X (t) − X (s)) is multivariate normal. The covariance matrix is of the form µ ¶ A 0 0 t−s and so the random vector, (X (u1 ) − X (r1 ) , · · · , X (um ) − X (rm )) and the random variable X (t) − X (s) are independent. Consequently, XA is independent of X (t) − X (s) for any A ∈ K. Then by the lemma on π systems, Lemma 9.11.3 on Page 275, Fs ⊇ G ⊇ σ (K) = Fs . This proves the claim. Thus Z Z (X (t) − X (s)) dP = (X (t) − X (s)) XA dP A Ω Z = P (A) (X (t) − X (s)) dP = 0 Ω

which shows that since A ∈ Fs was arbitrary, E (X (t) |Fs ) = X (s) and {X (t)} is a martingale. n o 2 Now consider whether X (t) − t is a martingale. By assumption, L (X (t) − X (s)) = L (X (t − s)) = N (0, t − s) . Then for A ∈ Fs , the independence of XA and X (t) − X (s) shows Z Z ³ ´ 2 2 E (X (t) − X (s)) |Fs dP = (X (t) − X (s)) dP A A Z = P (A) (t − s) = (t − s) dP A

and since A ∈ Fs is arbitrary, ³ ´ 2 E (X (t) − X (s)) |Fs = t − s and so the result follows from Lemma 51.7.2. This proves the theorem. The next lemma is the main result from which Levy’s theorem will be established.


1581

Lemma 51.7.4 Let {X (t)} be a real continuous martingale³adapted ´ to the filtration 2 Ft for t ∈ [a, b] some interval such that for all t ∈ [a, b] , E X (t) < ∞. Suppose n o 2 also that X (t) − t is a martingale. Then for λ real, ³ ´ ³ ´ λ2 E eiλX(b) = E eiλX(a) e−(b−a) 2 Proof: Let λ ∈ [−p, p] where for most of the proof, p is fixed but arbitrary. 2n Let {tnk }k=0 be uniform partitions such that tnk − tnk−1 = δ n ≡ (b − a) /2n . Now for ε > 0 define a stopping time τ ε,n to be the first time, t such that there exist s1 , s2 ∈ [a, t] with |s1 − s2 | < δ n but |X (s1 ) − X (s2 )| = ε. If no such time exists, then τ ε,n ≡ b. Then τ ε,n really is a stopping time because from continuity of X (t) and denoting by r, r1 elements of Q, then [τ ε,n > t] =

∞ [

\

m=1 0≤r1 ,r2 ≤t,|r1 −r2 |≤δ n

·

¸ 1 ∈ Ft |X (r1 ) − X (r2 )| ≤ ε − m

because to be in [τ ε,n > t] it means that by t the absolute value of the differences must always be less than ε. Hence [τ ε,n ≤ t] = Ω \ [τ ε,n > t] ∈ Ft . Now consider [τ ε,n = b] for various n. By continuity, it follows that for each ω ∈ Ω, τ ε,n (ω) = b for all n large enough. Thus ∅ = ∩∞ n=1 [τ ε,n < b] , the sets in the intersection decreasing. Thus there exists n (ε) such that ¡£ ¤¢ P τ ε,n(ε) < b < ε. (51.7.43) Denote τ ε,n(ε) as τ ε for short and it will always be assumed that n (ε) is at least this large and that limε→0+ n (ε) = ∞. In addition to this, n (ε) will also be large enough that λ2 1 − δ n(ε) > 0 2 for all λ ∈ [−p, p] . To save on notation, tj will take the place of tnj . Then consider the stopping times τ ε ∧ tj for j = 0, 1, · · · , 2n(ε) . Let yj ≡ X (τ ε ∧ tj )−X (τ ε ∧ tj−1 ) , it follows from the definition of the stopping time that |yj | ≤ ε (51.7.44)

1582


because both τ ε ∧ tj and τ ε ∧ tj−1 are less than τ ε and closer together than δ n(ε) and so if |yj | > ε, then τ ε ≤ tj , tj−1 and so yj would need to equal 0. By the optional stopping theorem, {X (τ ε ∧ tj )}j is a martingale as is also {X (τ ε ∧ tj ) − τ ε ∧ tj }j . Thus for A ∈ Fτ ε ∧tj−1 , Z Z ³ ´ ¡ 2 ¢ 2 E yj |Fτ ε ∧tj−1 dP = E (X (τ ε ∧ tj ) − X (τ ε ∧ tj−1 )) |Fτ ε ∧tj−1 dP A

A

Z

³ ´ 2 2 E X (τ ε ∧ tj ) |Fτ ε ∧tj−1 + X (τ ε ∧ tj−1 ) A ¡ ¢ −2X (τ ε ∧ tj−1 ) E X (τ ε ∧ tj ) |Fτ ε ∧tj−1 dP Z Z ³ ´ ¡ ¢ 2 = E X (τ ε ∧ tj ) − τ ε ∧ tj |Fτ ε ∧tj−1 dP + E τ ε ∧ tj |Fτ ε ∧tj−1 dP =

A

Z 2

+

2

X (τ ε ∧ tj−1 ) dP − 2 A

X (τ ε ∧ tj−1 ) dP A

Z =

A

Z

Z

Z

¡ ¢ X (τ ε ∧ tj−1 ) dP − τ ε ∧ tj−1 dP + E τ ε ∧ tj |Fτ ε ∧tj−1 dP A A A Z Z 2 2 + X (τ ε ∧ tj−1 ) dP − 2 X (τ ε ∧ tj−1 ) dP 2

A

A

Z = A

¡ ¢ E τ ε ∧ tj |Fτ ε ∧tj−1 dP −

Z =

Z τ ε ∧ tj−1 dP A

Z (τ ε ∧ tj − τ ε ∧ tj−1 ) dP ≤

A

tj − tj−1 dP. A

Thus, since A is arbitrary, Z σ 2j ≡

A

¡ ¢ E yj2 |Fτ ε ∧tj−1 dP =

³

´ 2 E (X (τ ε ∧ tj ) − X (τ ε ∧ tj−1 )) |Fτ ε ∧tj−1 ≤ tj − tj−1 = δ n(ε)

(51.7.45)

Also, ¡ ¢ ¡ ¢ E yj |Fτ ε ∧tj−1 = E X (τ ε ∧ tj ) − X (τ ε ∧ tj−1 ) |Fτ ε ∧tj−1 = 0.

(51.7.46)

¡ ¢ Now it is time to find E eiλX(τ ε ∧tj ) . ´ ´ ³ ³ E eiλX(τ ε ∧tj ) = E eiλ(X(τ ε ∧tj−1 )+yj )


1583

³ ¢´ ¡ = E eiλX(τ ε ∧tj−1 ) E eiλyj |Fτ ε ∧tj−1 .

(51.7.47)

Now let o (1) denote any quantity which converges to 0 as ε → 0 for all λ ∈ [−p, p] and O (1) is a quantity which is bounded as ε → 0. Then from 51.7.46 and 51.7.47 you can consider the power series for eiλyj which converges uniformly due to 51.7.44 and write 51.7.47 as µ µ ¶¶ λ2 E eiλX(τ ε ∧tj−1 ) 1 − σ 2j (1 + o (1)) . 2 then noting that from 51.7.45 which shows σ 2j is o (1) ,it is routine to verify 1−

λ2 2 λ2 2 σ j (1 + o (1)) = e− 2 σj (1+o(1)) . 2

Now this shows ³ ´ ³ ´ λ2 2 E eiλX(τ ε ∧tj ) = E eiλX(τ ε ∧tj−1 ) e− 2 σj (1+o(1)) Recall that σ 2j ≤ δ n = tj − tj−1 . Consider ¯ ³ ´ ³ ´¯ λ2 ¯ ¯ ¯E eiλX(τ ε ∧tj ) − E eiλX(τ ε ∧tj−1 ) e− 2 δn ¯ ¯ ³ ´ ³ ´¯ λ2 λ2 2 ¯ ¯ = ¯E eiλX(τ ε ∧tj−1 ) e− 2 σj (1+o(1)) − E eiλX(τ ε ∧tj−1 ) e− 2 δn ¯ ¯ ³ ´´¯ ³ λ2 2 λ2 ¯ ¯ = ¯E eiλX(τ ε ∧tj−1 ) e− 2 σj (1+o(1)) − e− 2 δn ¯ ¯ ³ ³ λ2 2 ´´¯ λ2 λ2 ¯ ¯ = ¯E eiλX(τ ε ∧tj−1 ) e− 2 δn e− 2 σj (1+o(1))+ 2 δn − 1 ¯ ¯´ ³¯ λ2 2 2 ¯ ¯ ≤ E ¯e 2 (δn −σj )+σj o(1) − 1¯ Everything in the exponent is o (1) and so the above expression is bounded by ¯¶ µ¯ 2 ¯λ ¡ ¯ ¢ O (1) E ¯¯ δ n − σ 2j + σ 2j o (1)¯¯ 2 ¡¡ ¢ ¢ ≤ O (1) E δ n − σ 2j + δ n |o (1)| £ ¡ ¢ ¤ = O (1) δ n − E yj2 + δ n o (1) . Therefore,

≤

¯ ³ ´ ³ ´¯ λ2 ¯ ¯ ¯E eiλX(τ ε ∧tj ) − E eiλX(τ ε ∧tj−1 ) e− 2 δn ¯ £ ¡ ¢ ¤ O (1) δ n − E yj2 + δ n o (1)

(51.7.48)

1584


and so it also follows ¯ ³ ´ λ2 ³ ´¯ λ2 ¯ ¯ ¯E eiλX(τ ε ∧tj ) e 2 tj − E eiλX(τ ε ∧tj−1 ) e 2 tj−1 ¯ £ ¡ ¢ ¤ ≤ O (1) δ n − E yj2 + δ n o (1) Now also remember yj = X (τ ε ∧ tj ) − X (τ ε ∧ tj−1 ) and that {X (τ ε ∧ tj )}j is a martingale. Therefore it is routine to show, ³ ´ ³ ´ ¡ ¢ 2 2 E yj2 = E X (τ ε ∧ tj ) − E X (τ ε ∧ tj−1 ) . and so ¯ ³ ´ λ2 ³ ´¯ λ2 ¯ ¯ ¯E eiλX(τ ε ∧tj ) e 2 tj − E eiλX(τ ε ∧tj−1 ) e 2 tj−1 ¯ h ³ ³ ´ ³ ´´ i 2 2 ≤ O (1) δ n − E X (τ ε ∧ tj ) − E X (τ ε ∧ tj−1 ) + δ n o (1) and so, summing over all j = 1, · · · , 2n(ε) , ¯ ³ ´ λ2 ³ ´¯ λ2 ¯ ¯ ¯E eiλX(τ ε ∧b) e 2 b − E eiλX(a) e 2 a ¯ ³ ³ ³ ´ ³ ´´´ 2 2 ≤ O (1) (1 + o (1)) (b − a) − E X (τ ε ∧ b) − E X (a) (51.7.49) . Now recall 51.7.43 which said P ([τ ε < b]) < ε. Let εk ≡ 2−k and then by the Borel Cantelli lemma, X (τ ε ∧ b) → X (b) a.e. since if ω is such that convergence does not take place, ω must be in infinitely many of the sets, [τ εk < b] , a set of measure 0. Also since {X (τ ε ∧ tj )}j is a martinn o 2 2 2 gale, it follows from optional sampling theorem that X (a) , X (τ ε ∧ b) , X (b) is a submartingale and so Z Z 2 2 X (τ ε ∧ b) dP ≤ X (b) dP 2 2 X(τ ∧b) ≥α X(τ ∧b) ≥α [ ε ] [ ε ] and also from the maximal inequalities, Theorem 48.4.5 on Page 1389 it follows P

³h i´ ´ 1 ³ 2 2 X (τ ε ∧ b) ≥ α ≤ E X (b) α

51.7. LEVY’S THEOREM and so the functions,

1585

o n 2 X (τ εk ∧ b)

εk

are uniformly integrable which implies by

the Vitali convergence theorem, Theorem 8.5.3 on Page 213, that you can pass to the limit as εk → 0 in the inequality, 51.7.49 and conclude ¯ ³ ´ λ2 ³ ´¯ λ2 ¯ ¯ ¯E eiλX(b) e 2 b − E eiλX(a) e 2 a ¯ ³ ³ ³ ´ ³ ´´´ 2 2 ≤ O (1) (b − a) − E X (b) − E X (a) = 0. Therefore,

³ ´ ³ ´ λ2 E eiλX(b) = E eiλX(a) e− 2 (b−a)

This proves the lemma because p was arbitrary. Now from this lemma, it is not hard to establish Levy’s theorem. Theorem 51.7.5 Let {X (t)} be a real continuous martingale adapted to´the fil³ 2 tration Ft for t ∈ [0, a] some interval such that for all t ∈ [0, a] , E X (t) < ∞. n o 2 Suppose also that X (t) − t is a martingale. Then for s < t, X (t)−X (s) is normally distributed with mean 0 and variance t−s. Also if 0 ≤ t0 < t1 < · · · < tm ≤ b, then the increments {X (tj ) − X (tj−1 )} are independent. Proof: Let the tj be as described above and consider the interval [tm−1 , tm ] in place of [a, b] in Lemma 51.7.4. Also let λk for k = 1, 2, · · · , m be given. For t ∈ [tm−1 , tm ] , and λm 6= 0, m−1 1 X Zλm (t) = λj (X (tj ) − X (tj−1 )) + (X (t) − X (tm−1 )) λm j=1

Then it is clear (t)} is a martingale on [tm−1 , tm ] . What is possibly less n that {Zλm o 2 clear is that Zλm (t) − t is also a martingale. Note that Zλm (t) = X (t) + Y where Y is measurable in Ftm−1 . Therefore, for s < t, s ∈ [tm−1 , tm ] , ³ ´ ³ ´ 2 2 E Zλm (t) − t|Fs = E X (t) + 2X (t) Y + Y 2 − t|Fs 2

= X (s) − s + 2E (X (t) Y |Fs ) + Y 2 2

2

= X (s) − s + 2Y X (s) + Y 2 = Zλm (s) − s and so Lemma 51.7.4 can be applied to conclude ´ λ2 ´ ³ ³ E eiλZλm (tm ) = E eiλZλm (tm−1 ) e− 2 (tm −tm−1 ) . Now letting λ = λm , ³ Pm ´ ³ Pm−1 ´ λ2m E ei j=1 λj (X(tj )−X(tj−1 )) = E ei j=1 λj (X(tj )−X(tj−1 )) e− 2 (tm −tm−1 ) .

1586


By continuity, this equation continues to hold for λm = 0. Then iterate this, using a similar argument on the first factor of the right side to eventually obtain m ³ Pm ´ Y λ2 j E ei j=1 λj (X(tj )−X(tj−1 )) = e− 2 (tj −tj−1 ) . j=1

Then letting all but one λj equal zero, this shows the increment, X (tj ) − X (tj−1 ) is a random variable which is normally distributed having variance tj − tj−1 and mean 0. The above formula also shows from Proposition 47.4.1 on Page 1323 that the increments are independent. This proves the theorem.

Wiener Processes 52.1

Real Wiener Processes

Do Wiener processes exist? Yes, they do. First here is a simple lemma which has really been done before. It depends on the Kolmogorov extension theorem, Theorem 11.3.5 on Page 332. ∞

Lemma 52.1.1 There exists a sequence, {ξ k }k=1 of random variables such that L (ξ k ) = N (0, 1) ∞

and {ξ k }k=1 is independent. Proof: Let i1 < i2 · · · < in be positive integers and define Z 2 1 µi1 ···in (F1 × · · · × Fn ) ≡ ¡√ ¢n e−|x| /2 dx. 2π F1 ×···×Fn Then for the index set equal to N the measures satisfy the necessary consistency condition for the Kolmogorov theorem. Therefore, there exists a probability space, (Ω, P, F) and measurable functions, ξ k : Ω → R such that ¡£ ¤ £ ¤ £ ¤¢ P ξ i1 ∈ Fi1 ∩ ξ i2 ∈ Fi2 · · · ∩ ξ in ∈ Fin = µi1 ···in (F1 × · · · × Fn ) ¡£ ¤¢ ¡£ ¤¢ = P ξ i1 ∈ Fi1 · · · P ξ in ∈ Fin which shows the random variables are independent as well as normal with mean 0 and variance 1. This proves the Lemma. Recall that the sum of independent normal random variables is normal. The Wiener process is just an infinite weighted sum of the above independent normal random variables, the weights depending on t. Therefore, if the sum converges, it is not too surprising that the result will be normally distributed and the variance will depend on t. This is the idea behind the following theorem. 1587

1588

WIENER PROCESSES

Theorem 52.1.2 There exists a real Wiener process as defined in Definition 51.7.1. Furthermore, the distribution of W (t) − W (s) is the same as the distribution of W (t − s) and W is Holder continuous with exponent γ for any γ < 1/2.

∞

Proof: Let {gm }m=1 be a complete orthonormal set in L2 (0, ∞) . Thus, if f ∈ L2 (0, ∞) , f=

∞ X

(f, gi )L2 gi .

i=1

The Wiener process is defined as ∞ X ¡ ¢ X(0,t) , gi L2 ξ i (ω)

W (t, ω) ≡

i=1

where the random variables, {ξ i } are as described in Lemma 52.1.1. The series converges in L2 (Ω) where (Ω, F, P ) is the probability space on which the random variables, ξ i are defined. This will first be shown. Note first that from the independence of the ξ i , Z Ω

ξ i ξ j dP = 0

Therefore, ¯2 Z ¯¯ X n ¯ ¡ ¢ ¯ ¯ X(0,t) , gi L2 ξ i (ω)¯ dP ¯ ¯ ¯ Ω

=

i=m

=

n X ¡ i=m n X

¡

X(0,t) , gi X(0,t) , gi

¢2 L2

Z 2

Ω

|ξ i | dP

¢2 L2

i=m

which converges to 0 as m, n → ∞. Thus the partial sums are a Cauchy sequence in L2 (Ω, P ) . It just remains to verify this definition satisfies the desired conditions. First I will show that ω → W (t, ω) is normally distributed with mean 0 and variance t. Selecting a suitable subsequence, {nk } it can be assumed

W (t, ω) = lim

k→∞

nk X ¡ ¢ X(0,t) , gi L2 ξ i (ω) a.e. i=1

52.1. REAL WIENER PROCESSES

1589

and so from the dominated convergence theorem and the independence of the ξ i ,  nk X ¢ ¡ lim E exp iλ X(0,t) , gj L2 ξ j (ω) 

E (exp (iλW (t)))

=



k→∞

 lim E 

=

k→∞

=

lim

k→∞

=

lim

k→∞

j=1

nk Y

 ¢ ¡ ¢ exp iλ X(0,t) , gj L2 ξ j (ω)  ¡

j=1 nk Y j=1 nk Y

¡ ¡ ¡ ¢ ¢¢ E exp iλ X(0,t) , gj L2 ξ j (ω) 2

1 2 e− 2 λ (X(0,t) ,gj )L2

j=1

  nk X ¡ ¢ 1 2 lim exp  − λ2 X(0,t) , gj L2  k→∞ 2 j=1 µ ¶ ¶ µ ¯¯2 1 ¯¯ 1 exp − λ2 ¯¯X(0,t) ¯¯L2 = exp − λ2 t , 2 2

= =

the characteristic function of a normally distributed random variable having variance t and mean 0.. It is clear W (0) = 0. It remains to verify the increments are independent. First note that for a, b scalars and r < s < t, the above definition shows a (W (t) − W (s)) + b (W (s) − W (r))

=

X¡ ¢ aX(s,t) + bX(r,s) , gj L2 ξ j j

and by reasoning identical to the above this is normal. Since this holds for every linear combination, it follows the vector (W (s) − W (r) , W (t) − W (s)) is multivariate normal. The increments, W (s)−W (r) , W (t)−W (s) will be independent if the covariance matrix of this multivariate normal distribution is diagonal. However, the off diagonal terms are of the form E ((W (s) − W (r)) (W (t) − W (s))) ÃÃ =E

∞ X ¡

k=1

X(r,s) , gk

¢ L2

!Ã ξ k (ω)

∞ X ¡ ¢ X(s,t) , gi L2 ξ i (ω) i=1

!!

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WIENER PROCESSES

 = E =

 X¡

X¡

X(r,s) , gk

i,k

X(r,s) , gk

i,k

=

X¡

X(r,s) , gi

¢

¢ L2

¡

L2

¢ L2

¡

X(s,t) , gi

X(s,t) , gi

¢ L2

¢ L2

ξk ξi

E (ξ k ξ i )

¡ ¢ ¡ ¢ X(s,t) , gi L2 = X(r,s) , X(s,t) L2 = 0

i

and so this shows the increments are independent. Similarly for 0 ≤ t0 < t1 < m · · · < tm , the random variables {W (ti ) − W (ti−1 )}i=1 are independent. You observe in the same way as above that (W (t1 ) − W (t0 ) , · · · W (tm ) − W (tm−1 )) is multivariate normal and by the same reasoning as above the covariance matrix is diagonal. From the definition, if t > s W (t − s) =

∞ X ¡

X(0,t−s) , gk

¢ L2

ξk

k=1

while W (t) − W (s) =

∞ X ¡

X(s,t) , gk

¢ L2

ξk .

k=1

Then the same argument given above involving the characteristic function to show W (t) is normally distributed shows both of these random variables are normally distributed with mean 0 and variance t − s. Finally note the distribution of W (t − s) is the same as the distribution of W (1) (t − s)

1/2

=

∞ X ¡

X(0,1) , gk

¢

1/2

L2

ξ k (t − s)

k=1

because the characteristic function of this last random variable is the same as the 1 2 characteristic function of W (t − s) which is e− 2 λ (t−s) which follows from a simple computation. Hence for any positive α, α

E (|W (t) − W (s)| ) =

α

=

E (|W (t − s)| ) ¯α ´ ³¯ ¯ ¯ 1/2 E ¯(t − s) W (1)¯

=

|t − s|

α/2

α

E (|W (1)| )

(52.1.1)

It follows from Theorem 50.1.4 that W (t) is Holder continuous with exponent γ where γ is any positive number less than β/α where α/2 = 1 + β. Thus γ is any constant less than α 2 −1 . α Thus γ is any constant less than 12 . This proves the theorem. The proof of the theorem also implies the following corollary.

52.2. NOWHERE DIFFERENTIABILITY OF WIENER PROCESSES

1591

∞

Corollary 52.1.3 Let {ξ i }i=1 be independent random variables each normal with mean 0 and variance 1. Then W (t, ω) ≡

∞ X ¡

X[0,t] , gi

¢ L2

ξ i (ω)

i=1

is a real Wiener process.

52.2

Nowhere Differentiability Of Wiener Processes

If W (t) is a Wiener process, it turns out that t → W (t, ω) is nowhere differentiable for a.e. ω. This fact is based on the independence of the increments and the fact that these increments are normally distributed. 1/2 First note that W (t) − W (s) has the same distribution as (t − s) W (1) . This is because they have the same characteristic function. Next it follows that because of the independence of the increments and what was just noted that, ¡ ¢ P ∩5r=1 [|W (t + rδ) − W (t + (r − 1) δ)| ≤ Kδ] =

5 Y

P ([|W (t + rδ) − W (t + (r − 1) δ)| ≤ Kδ])

r=1

= ≤

5 Y

¯ ³h¯ i´ ¯ ¯ P ¯δ 1/2 W (1)¯ ≤ Kδ =

r=1 5/2

Cδ

Ã

1 √ 2π

Z

K

!5

√ δ

−K

√ δ

e

− 12 t2

dt

.

(52.2.2)

With this observation, here is the proof which follows [62] and according to this reference is due to Payley, Wiener and Zygmund and the proof is like one given by Dvoretsky, Erdös and Kakutani. Theorem 52.2.1 Let W (t) be a Wiener process. Then there exists a set of measure 0, N such that for all ω ∈ / N, t → W (t, ω) is nowhere differentiable. Proof: Let [0, a] be an interval. If for some ω, t → W (t, ω) is differentiable at some s, then for some n, p > 0, ¯ ¯ ¯ W (t, ω) − W (s, ω) ¯ ¯ ¯≤p ¯ ¯ t−s whenever |t − s| < 5a2−n ≡ 5δ n . Define Cnp by ¯ ¯ ½ ¾ ¯ W (t, ω) − W (s, ω) ¯ ¯ ¯ ω : for some s ∈ [0, a), ¯ ¯ ≤ p if |t − s| ≤ 5δ n . t−s

(52.2.3)

1592

WIENER PROCESSES

Thus ∪n,p∈N Cnp contains the set of ω such that t → W (t, ω) is differentiable for some s ∈ [0, a). 2n Now define uniform partitions of [0, a), {tnk }k=0 such that ¯ ¯n ¯tk − tnk−1 ¯ = a2−n ≡ δ n Let ¢ ¡ n Dnp ≡ ∪2i=0−1 ∩5r=1 [|W (tni + rδ n ) − W (tni + (r − 1) δ n )| ≤ 10pδ n ] If ω ∈ Cnp , then for some s ∈ [0, a), the condition of 52.2.3 holds. Suppose k is the number such that s ∈ [tnk−1 , tnk ). Then for r ∈ {1, 2, 3, 4, 5} , ¯ ¡n ¢ ¡ ¢¯ ¯W tk−1 + rδ n , ω − W tnk−1 + (r − 1) δ n , ω ¯ ¯ ¯ ¯ ¡ ¢ ¡ ¢¯ ≤ ¯W tnk−1 + rδ n , ω − W (s, ω)¯ + ¯W (s, ω) − W tnk−1 + (r − 1) δ n , ω ¯ ≤ 5pδ n + 5pδ n = 10pδ n Thus Cnp ⊆ Dnp . Now from 52.2.2, ¡ ¢5/2 ¡√ ¢5 − 3 n 5/2 n P (Dnp ) ≤ 2n Cδ 5/2 2 2−n =C a 2 2 n = Ca

(52.2.4)

Let ∞ ∞ ∞ Cp = ∪∞ n=1 ∩k=n Ckp ⊆ ∪n=1 ∩k=n Dkp .

It was just shown in 52.2.4 that P (∩∞ k=n Dkp ) = 0 and so Cp has measure 0. C , the set of points, ω where t → W (t, ω) could have a derivative has Thus ∪∞ p=1 p measure 0. Taking the union of the exceptional sets corresponding to intervals [0, n) for n ∈ N, this proves the theorem. This theorem on nowhere differentiability is very important because it shows it R is doubtful one can define an integral f (s) dW (s) by simply fixing ω and then doing some sort of Stieltjes integral in time. The reason for this is that the nowhere differentiability of W implies it is also not of bounded variation on any interval since if it were, it would equal the difference of two increasing functions and would therefore have a derivative at a.e. point. I have presented the theorem on nowhere differentiability for one dimensional Wiener processes but the same proof holds with minor modifications if you have defined the Wiener process in Rn or you could simply consider the components and apply the above result.

52.3

Wiener Processes In Separable Banach Space

Here is an important lemma on which the existence of Wiener processes will be based.

52.3. WIENER PROCESSES IN SEPARABLE BANACH SPACE

1593 ∞

Lemma 52.3.1 There exists a sequence of real Wiener processes, {ψ k (t)}k=1 which have the following properties. Let t0 < t1 < · · · < tn be an arbitrary sequence. Then the random variables {ψ k (tq ) − ψ k (tq−1 ) : (q, k) ∈ (1, 2, · · · , n) × (k1 , · · · , km )} are independent. Also each ψ k is Holder continuous with exponent γ for any γ < 1/2 and for each m ∈ N there exists a constant Cm independent of k such that Z 2m m |ψ k (t) − ψ k (s)| dP ≤ Cm |t − s| Ω

© ª © ª Proof: First, there exists a sequence ξ ij (i,j)∈N×N such that the ξ ij are independent and each normally distributed with mean 0 and variance 1. This follows from Lemma 52.1.1. (Let θ be a one to one and onto map from N to N × N. Then define ξ ij ≡ ξ θ−1 (i,j) .) Let ∞ X ¡ ¢ ψ k (t) = X[0,t] , gj L2 ξ kj j=1

where {gj } is a orthonormal basis for L2 (0, ∞). By Corollary 52.1.3 this defines a real Wiener process. It remains to show the sequence of these Wiener processes has the desired properties. From the definition of ψ k (t) , ψ k (t) − ψ k (s) =

∞ X ¡

X[s,t] , gj

¢ L2

ξ kj

(52.3.5)

j=1

and this series converges in L2 (Ω) to a normal random variable having variance t − s and mean 0. The convergence follows from ¯ ¯2 ¯ Z ¯X ¯ n ¡ ¯ ¢ ¯ ¯ X , g ξ [s,t] j L2 kj ¯ dP ¯ Ω ¯j=m ¯ Z X ¡ ¢ ¡ ¢ = X[s,t] , gj L2 ξ kj X[s,t] , gl L2 ξ kl dP =

Ω j,l n X

¡

X[s,t] , gj

¢2 L2

j=m

which converges to 0 as m, n → ∞ due to X[s,t] ∈ L2 (0, ∞). Thus Z 2

Ω

|ψ k (t) − ψ k (s)| dP

=

∞ X ¡

X[s,t] , gj

¢2 L2

j=1

=

∞ X ¡ j=1

X[s,t] , gj

¢2 L2

Z Ω

ξ 2kj dP

= (t − s)

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WIENER PROCESSES

The rest of the claim that ψ k (t) − ψ k (s) is normally distributed with mean 0 and variance (t − s) follows from routine considerations involving partial sums and passing to a limit in the characteristic function of a partial sum. It will also follow from the next part of the argument. Given this is so it follows from considering the characteristic function as in 47.10.29 that an inequality of the following sort holds. Z 2m m |ψ k (t) − ψ k (s)| dP ≤ Cm |t − s| Ω

ˇ By Theorem 50.1.4, the Kolmogorov Centsov theorem, it follows ψ k (t) is Holder continuous with exponent γ for γ = (m − 1) /2m. Why are the random variables {ψ k (tq ) − ψ k (tq−1 ) : (q, k) ∈ (1, 2, · · · , n) × (k1 , · · · , km )}

(52.3.6)

linearly independent? Let P =

n X m X

¡ ¢ sqr ψ kr (tq ) − ψ kr (tq−1 )

q=1 r=1

¡ ¢ and consider E ¡eiP ¢. I want to use Proposition 47.4.1 on Page 1323. To do this I need to show E eiP equals n Y m Y

¡ ¡ ¡ ¢¢¢ E exp isqr ψ kr (tq ) − ψ kr (tq−1 ) .

q=1 r=1

¡ ¢ Using 52.3.5 E eiP equals 



E exp i

n X m X

 ∞ X ¡ ¢ sqr X[tq−1 ,tq ] , gj L2 ξ kr j 

q=1 r=1





= lim E exp i N →∞

j=1

n X m X q=1 r=1

sqr

N X ¡

 X[tq−1 ,tq ] , gj

¢ L2

ξ kr j 

j=1

Now the ξ kr j are independent by construction. Therefore, the above equals = lim

N →∞

n Y m Y N Y

³ ³ ´´ ¡ ¢ E exp isqr X[tq−1 ,tq ] , gj L2 ξ kr j

q=1 r=1 j=1

¶ µ ¢2 1 2 ¡ = lim exp − sqr X[tq−1 ,tq ] , gj L2 N →∞ 2 q=1 r=1 j=1 n Y m Y N Y


1595



=

 N X ¡ ¢ 1 2 lim exp − s2qr X[tq−1 ,tq ] , gj L2  N →∞ 2 q=1 r=1 j=1

=

µ ¶ 1 exp − s2qr (tq − tq−1 ) 2 q=1 r=1

m n Y Y

n Y m Y

Letting m = n = 1 this shows the earlier claim about ψ k (t) − ψ k (s) being normally distributed with variance (t − s). Now the above equals =

m n Y Y

¡ ¡ ¡ ¢¢¢ E exp isqr ψ kr (tq ) − ψ kr (tq−1 )

q=1 r=1

and so this verifies n Y m ¡ ¢ Y ¡ ¡ ¡ ¢¢¢ E eiP = E exp isqr ψ kr (tq ) − ψ kr (tq−1 ) q=1 r=1

and by Proposition 47.4.1 on Page 1323 it follows the random variables in 52.3.6 are independent. Note that as a special case, this also shows the random variables, ∞ {ψ k (t)}k=1 are independent due to the fact ψ k (0) = 0. This proves the lemma. Recall Corollary 49.10.4 which is stated here for convenience. Corollary 52.3.2 Let E be any real separable Banach space. Then there exists a sequence, {ek } ⊆ E such that for any {ξ k } a sequence of independent random variables such that L (ξ k ) = N (0, 1), it follows X (ω) ≡

∞ X

ξ k (ω) ek

k=1

converges a.e. and P its law is a Gaussian measure defined on B (E). Furthermore, ||ek ||E ≤ λk where k λk < ∞. Now let {ψ k (t)} be the sequence of Wiener processes described in Lemma 52.3.1. Then define a process with values in E by W (t) ≡

∞ X

ψ k (t) ek

(52.3.7)

k=1

√ Then ψ k (t) / t is N (0, 1) and so by Corollary 49.10.4 the law of ∞ ³ X √ √´ W (t) / t = ψ k (t) / t ek k=1

is a Gaussian measure. Therefore, the same is true of W (t) . Similar reasoning applies to the increments, W (t) − W (s) to conclude the law of each of these is

1596

WIENER PROCESSES

Gaussian. Consider the question whether the increments are independent. Let 0 ≤ t0 < t1 < · · · < tm and let φj ∈ E 0 . Then by the dominated convergence theorem and the properties of the {ψ k } ,    m X E exp i φj (W (tj ) − W (tj−1 ))  =

E exp i 

=

j=1



E

Ã∞ m X X j=1

Ã

m Y

exp i

j=1

lim E 

n→∞

=

=

m Y

E

=

=

m Y

=

Ã

m Y

m Y j=1 m Y

Ã E

j=1

(ψ k (tj ) − ψ k (tj−1 )) φj (ek ) ! (ψ k (tj ) − ψ k (tj−1 )) φj (ek ) 

¡ ¡ ¢¢ E exp i (ψ k (tj ) − ψ k (tj−1 )) φj (ek ) Ã

exp i

n X

!! (ψ k (tj ) − ψ k (tj−1 )) φj (ek )

k=1

Ã exp i

Ã E

!!

k=1

Ã

E

n X

exp i

j=1

m Y

n X k=1

j=1 k=1

j=1

=

exp i

m Y n Y

lim

n→∞

Ã

j=1

lim

n→∞

¡ ¡ ¢¢ E exp i (ψ k (tj ) − ψ k (tj−1 )) φj (ek )

Ã

lim E 

=

! (ψ k (tj ) − ψ k (tj−1 )) φj (ek ) 

k=1

j=1

n→∞

n X

exp i

j=1 k=1

 =

! (ψ k (tj ) − ψ k (tj−1 )) φj (ek ) 

Ã

m Y

m Y n Y

lim

n→∞

∞ X

j=1

lim

n→∞

k=1

k=1

 =

! (ψ k (tj ) − ψ k (tj−1 )) φj (ek ) 

Ã

∞ X

!! (ψ k (tj ) − ψ k (tj−1 )) φj (ek )

k=1

exp iφj

Ã

∞ X

!!! (ψ k (tj ) − ψ k (tj−1 )) ek

k=1

¡ ¡ ¢¢ E exp iφj (W (tj ) − W (tj−1 ))


1597 m

which shows by Theorem 47.6.3 on Page 1329 that the random vectors, {W (tj ) − W (tj−1 )}j=1 are independent. It is also routine to verify using properties of the ψ k and characteristic functions that L (W (t) − W (s)) = L (W (t − s)). To see this, let φ ∈ E 0 E (exp (iφ (W (t) − W (s)))) ÃÃ Ã ∞ !!! X =E exp iφ (ψ k (t) − ψ k (s)) ek ÃÃ = lim E

exp iφ

n→∞

= = =

k=1

Ã

n X

!!! (ψ k (t) − ψ k (s)) ek

k=1

lim

n→∞

n Y

E (exp (iφ (ek ) (ψ k (t) − ψ k (s))))

k=1 n Y

¶¶ µ µ 1 2 E exp − φ (ek ) (t − s) n→∞ 2 k=1 Ã ¶! µ n X 1 2 lim E exp − φ (ek ) (t − s) n→∞ 2 lim

k=1

which is the same as the result for E (exp (iφ (W (t − s)))) and

¡ ¡ ¡√ ¢¢¢ E exp iφ t − sW (1) .

This has proved the following lemma. Lemma 52.3.3 Let E be a real separable Banach space. Then there exists an E valued stochastic process, W (t) such that L (W (t)) and L (W (t) − W (s)) are Gaussian measures and the increments, {W (t) − W (s)} are independent. Furthermore, the increment W (t) − W√(s) has the same distribution as W (t − s) and W (t) has the same distribution as tW (1). Now I want to consider the question of Holder continuity of the functions, t → W (t, ω). Z Z α α ||W (t) − W (s)|| dP = ||x|| dµW (t)−W (s) Ω E Z Z α α = ||x|| dµW (t−s) = ||x|| dµ√t−sW (1) E ZE ¯¯α ¯¯√ ¯¯ t − sW (1)¯¯ dP = Ω Z α/2 α α/2 = |t − s| ||W (1)|| dP = Cα |t − s| Ω

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ˇ by Fernique’s theorem, Theorem 49.6.5. From the Kolmogorov Centsov theo50.1.4, it follows {W (t)} is Holder continuous with exponent γ < Theorem ¡rem, ¢ α 2 − 1 /α. This completes the proof of the following theorem. Theorem 52.3.4 Let E be a separable real Banach space. Then there exists a stochastic process, {W (t)} such that the distribution of W (t) and every increment, W (t) − W (s) is Gaussian. Furthermore, the increments corresponding ¡√ to disjoint ¢ intervals are independent, L (W (t) − W (s)) = L (W (t − s)) = L t − sW (1) . Also for a.e. ω, t → W (t, ω) is Holder continuous with exponent γ < 1/2.

52.4

An Example Of Martingales, Independent Increments

Here is an interesting lemma. Lemma 52.4.1 Let W (t) be a stochastic process having values in E a real separable Banach space which has independent increments. Let A ∈ Hs ≡ σ (W (u) − W (r) : 0 ≤ r < u ≤ s) Suppose g (W (t) − W (s)) ∈ L1 (Ω; E) . Then g (W (t) − W (s)) and XA are independent and Z Z XA g (W (t) − W (s)) dP = P (A) g (W (t) − W (s)) dP (52.4.8) Ω

Ω

Proof: Let G denote the set, of all A ∈ Hs such that 52.4.8 holds. Then it is obvious G is closed with respect to complements and countable disjoint unions. Let K denote those sets which are finite intersections of the form A = ∩m i=1 Ai where each Ai is in a set of σ (W (ui ) − W (ri )) for some 0 ≤ ri < ui ≤ s. For such A, it follows A ∈ σ (W (ui ) − W (ri ) , i = 1, · · · , m) . Now consider the random vector having values in E m+1 , (W (u1 ) − W (r1 ) , · · · , W (um ) − W (rm ) , g (W (t) − W (s))) m

Let t∗ ∈ (E 0 )

and s∗ ∈ E 0 . t∗ · (W (u1 ) − W (r1 ) , · · · , W (um ) − W (rm ))

can be written in¢ the form g∗ · (W (τ 1 ) − W (η 1 ) , · · · , W (τ l ) − W (η l )) where the ¡ intervals, τ j , η j are disjoint and each η j ≤ s. Therefore, by independence of the increments, E (exp i (t∗ · (W (u1 ) − W (r1 ) , · · · , W (um ) − W (rm )) + s∗ (g (W (t) − W (s)))))

52.4. AN EXAMPLE OF MARTINGALES, INDEPENDENT INCREMENTS

1599

= E (exp i (g∗ · (W (τ 1 ) − W (η 1 ) , · · · , W (τ l ) − W (η l )) + s∗ (g (W (t) − W (s))))) =

l Y

¡ ¡ ¡ ¡ ¢¢¢¢ E exp igj W (τ j ) − W η j E (exp (is∗ (g (W (t) − W (s)))))

j=1

=

E (exp (i (t∗ · (W (u1 ) − W (r1 ) , · · · , W (um ) − W (rm ))))) · E (exp (is∗ (g (W (t) − W (s))))) .

By Theorem 47.6.3, it follows the vector (W (u1 ) − W (r1 ) , · · · , W (um ) − W (rm )) is independent of the random variable g (W (t) − W (s)) which shows that for A ∈ K, XA , measurable in σ (W (u1 ) − W (r1 ) , · · · , W (um ) − W (rm )) is independent of g (W (t) − W (s)) . Therefore, Z Z Z XA g (W (t) − W (s)) dP = XA dP g (W (t) − W (s)) dP Ω Ω Ω Z = P (A) g (W (t) − W (s)) dP Ω

Thus K ⊆ G and so by the lemma on π systems, Lemma 9.11.3 on Page 275, it follows G ⊇ σ (K) ⊇ Fs ⊇ G. This proves the lemma. Lemma 52.4.2 Let {W (t)} be a stochastic process having values in a separable Banach space which has the property that if t1 < t2 · · · < tn , then the increments, {W (tk ) − W (tk−1 )} are independent and integrable and E (W (t) − W (s)) = 0. Suppose also that W (t) is right continuous meaning that for ω off a set of measure zero, t → W (t) (ω) is right continuous. Also suppose that for some q > 1 ||W (t) − W (s)||Lq (Ω) is bounded independent of s ≤ t. Then it is also a martingale with respect to the normal filtration defined by Fs ≡ ∩t>s σ (W (u) − W (r) : 0 ≤ r s and A ∈ Fs it follows that if g is a continuous function such that g (W (t) − W (s)) ∈ L1 (Ω; E) ||g (W (t) − W (s))||Lq (Ω) is bounded independent of s ≤ t for some q > 1 then Z Z XA g (W (t) − W (s)) dP = P (A) g (W (t) − W (s)) dP. Ω

Ω

(52.4.9)

(52.4.10)

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Proof: Consider first the claim, 52.4.10. To begin with I show that if A ∈ Fs then for all ε small enough that t > s + ε, Z Z XA g (W (t) − W (s + ε)) dP = P (A) g (W (t) − W (s + ε)) dP (52.4.11) Ω

Ω

This will happen if XA and g (W (t) − W (s + ε)) are independent. First note that from the definition A ∈ σ (W (u) − W (r) : 0 ≤ r < u ≤ s + ε) and so from the process of completion of a measure space, there exists B ∈ σ (W (u) − W (r) : 0 ≤ r < u ≤ s + ε) such that B ⊇ A and P (B \ A) = 0. Therefore, letting φ ∈ E 0 , E (exp (itXA + iφ (g (W (t) − W (s + ε))))) = E (exp (itXB + iφ (g (W (t) − W (s + ε))))) = E (exp (itXB )) E (exp (iφ (g (W (t) − W (s + ε))))) because XB is independent of g (W (t) − W (s + ε)) by Lemma 52.4.1 above. Then the above equals = E (exp (itXA )) E (exp (iφ (g (W (t) − W (s + ε))))) Now by Theorem 47.6.3, 52.4.11 follows. Next pass to the limit in both sides of 52.4.11 as ε → 0. One can do this because of 52.4.9 which implies the functions in the integrands are uniformly integrable and Vitali’s convergence theorem, Theorem 23.5.7. This yields 52.4.10. Now consider the part about the stochastic process being a martingale. If A ∈ Fs , the assumptions imply Z Z E (W (t) |Fs ) − W (s) dP = XA E ((W (t) − W (s)) |Fs ) dP A Ω Z = E (XA (W (t) − W (s)) |Fs ) dP Z

Z XA (W (t) − W (s)) dP =

Ω

Ω

Z

XA dP Ω

W (t) − W (s) dP = 0 Ω

and so since A is arbitrary, E (W (t) |Fs ) = W (s). This proves the lemma. Note this implies immediately from Lemma 51.1.5 that Wiener process is not of bounded variation on any interval. This is because this lemma implies if it were of bounded variation, then it would be constant which is not the case due to ¢ ¡√ L (W (t) − W (s)) = L (W (t − s)) = L t − sW (1) . Here is an interesting theorem about approximation.

52.4. AN EXAMPLE OF MARTINGALES, INDEPENDENT INCREMENTS

1601

Theorem 52.4.3 Let {W (t)} be a Wiener process having values in a separable Banach space as described in Theorem 52.3.4. There exists a set of measure 0, N such that for ω ∈ / N, the sum in 52.3.7 converges uniformly to W (t, ω) on any interval, [0, T ] . That is, for each ω not in a set of measure zero, the partial sums of the sum in that formula converge uniformly to t → W (t, ω) on [0, T ]. Proof: By Lemma 52.4.2 the independence of the increments imply n X

ψ k (t) ek

k=m

is a martingale and so by Theorem 50.3.5, ¯¯ n ¯¯ ¯¯ #! Ã" Z ¯¯¯¯ X n ¯¯ X ¯¯ ¯¯ 1 ¯¯ ¯¯ ¯¯ ¯¯ ψ k (t) ek ¯¯ ≥ α ≤ ψ k (T ) ek ¯¯ dP P sup ¯¯ ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯¯ α t∈[0,T ] Ω k=m

k=m

From Corollary 52.3.2 ¯¯ Z ¯¯¯¯ X n ¯¯ ¯¯ ¯¯ ψ k (T ) ek ¯¯ dP ¯¯ ¯ ¯ ¯¯ Ω

≤

k=m

≤

n Z X k=m n X

Ω

|ψ k (T )| dP λk

λk

k=m

which shows that there exists a subsequence, ml such that whenever n > ml , ¯¯ ¯¯ Ã" #! n ¯¯ X ¯¯ ¯¯ ¯¯ sup ¯¯ P ψ k (t) ek ¯¯ ≥ 2−k ≤ 2−k . ¯¯ t∈[0,T ] ¯¯ k=ml

Recall Lemma 47.8.6 stated below for convenience. Lemma 52.4.4 Let {ζ k } be a sequence of random variables having values in a separable real Banach space, E whose distributions are symmetric. Letting Sk ≡ P k i=1 ζ i , suppose {Snk } converges a.e. Also suppose that for every m > nk , ¡£ ¤¢ P ||Sm − Snk ||E > 2−k < 2−k . (52.4.12) Then in fact, Sk (ω) → S (ω) a.e.ω

(52.4.13)

Apply this lemma to the situation in which the Banach space, E is C ([0, T ] ; E) and ζ k = ψ k ek . Then you can conclude uniform convergence of the partial sums, m X k=1


ψ k (t) ek .

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WIENER PROCESSES 2n

Why is C ([0, T ] ; E) separable? Consider uniform partitions {tnk }k=0 such that n tk − tnk−1 = T 2−n . Let S denote all the step functions which are constant on the subintervals of one of these partitions which also have values in a countable dense subset of E. If f ∈ C ([0, T ] ; E) , there exists something in S which is uniformly close to f . Now let {ψ n } be a mollifier and let Sn denote the convolution of elements of S with ψ n . Then ∪∞ n=1 Sn is a countable dense subset of C ([0, T ] ; E) . You could also develop a proof based on the Bernstein polynomials as in Lemma 7.1.3 on Page 165.

52.5

Hilbert Space Valued Wiener Processes

Next I will consider the case of Hilbert space valued Wiener processes. This will include the case of Rn valued Wiener processes. I will present this material independent of the more general case of E valued Wiener processes. Definition 52.5.1 Let W (t) be a stochastic process with values in H, a real separable Hilbert space which has the properties that t → W (t, ω) is continuous, whenever t1 < t2 < · · · < tm , the increments {W (ti ) − W (ti−1 )} are independent, W (0) = 0, and whenever s < t, L (W (t) − W (s)) = N (0, (t − s) Q) which means that whenever h ∈ H, L ((h, W (t) − W (s))) = N (0, (t − s) (Qh, h)) Also E ((h1 , W (t) − W (s)) (h2 , W (t) − W (s))) = (Qh1 , h2 ) (t − s) . Here Q is a nonnegative trace class operator. Recall this means Q=

∞ X

λi e i ⊗ e i

i=1

where {ei } is a complete orthonormal basis, λi ≥ 0, and ∞ X

λi < ∞

i=1

Such a stochastic process is called a Q Wiener process. Note the characteristic function of a Q Wiener process is ´ ³ 1 2 E ei(h,W (t)) = e− 2 t (Qh,h)

(52.5.14)

Note that by Theorem 49.7.5 it you simply say that the distribution measure of W (t) is Gaussian, then it follows there exists a trace class operator Qt and mt ∈ H

52.5. HILBERT SPACE VALUED WIENER PROCESSES

1603

such that this measure is N (mt , Qt ) . Thus for W (t) a Wiener process, Qt = tQ and mt = 0. In addition, the increments are independent so this is much more specific than the earlier definition of a Gaussian measure. What is a Q Wiener process if the Hilbert space is Rn ? In particular, what is Q? It is given that L ((h, W (t) − W (s))) = N (0, (t − s) (Qh, h)) In this case everything is a vector in Rn and so for h ∈ Rn , ³ ´ 1 2 E eiλ(h,W (t)−W (s)) = e− 2 λ (t−s)(Qh,h) In particular, letting λ = 1 this shows W (t) − W (s) is normally distributed with 1 ∗ covariance (t − s) Q because its characteristic function is e− 2 h (t−s)Qh . With this and definition, one can describe Hilbert space valued Wiener processes in a fairly general setting. Theorem 52.5.2 Let U be a real separable Hilbert space and let J : U0 → U be a Hilbert Schmidt operator where U0 is a real separable Hilbert space. Then let {gk } be a complete orthonormal basis for U0 and define for t ∈ [0, T ] W (t) ≡

∞ X

ψ k (t) Jgk

k=1

Then W (t) is a Q Wiener process for Q = JJ ∗ as in Definition 52.5.1. Furthermore, the distribution of W (t) − W (s) is the same as the distribution of W (t − s) and W is Holder continuous with exponent γ for any γ < 1/2. There also is a subsequence denoted by N such that the convergence of the series N X

ψ k (t) Jgk

k=1

is uniform for all ω not in some set of measure zero. Proof: First it is necessary to show the series converges in L2 (Ω; U ) for each t. For convenience I will consider the series for W (t)−W (s) . Then since ψ k (t)−ψ k (s) is normal with mean 0 and variance (t − s) and ψ k (t) − ψ k (s) and ψ l (t) − ψ l (s) are independent, ¯2 Z ¯¯ X n ¯ ¯ ¯ (ψ k (t) − ψ k (s)) Jgk ¯ dP ¯ ¯ ¯ Ω k=m U Z X n = ((ψ k (t) − ψ k (s)) Jgk , (ψ l (t) − ψ l (s)) Jgl ) Ω k,l=m

= (t − s)

n X k=m

(Jgk , Jgk ) = (t − s)

n X k=m

2

||Jgk ||U

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WIENER PROCESSES

which converges to 0 as m, n → ∞ thanks to the assumption that J is Hilbert Schmidt. It follows the above sum converges in L2 (Ω; U ) for each t. Now letting m < n, it follows by the maximal estimate, Theorem 50.3.5, and the above ¯ #! Ã"¯ m n ¯ ¯X X ¯ ¯ ψ k (t) Jgk − ψ k (t) Jgk ¯ ≥ λ P ¯ ¯ ¯ k=1 k=1 U ¯  ¯ n n ¯ ¯ 1 ¯ X 1 X ¯ 2 ≤ E ¯ ψ k (T ) Jgk ¯  ≤ T ||Jgk ||U ¯ ¯ λ λ k=m+1

k=m

U

and so there exists a subsequence nl such that for all p ≥ 0, ¯ Ã"¯ n #! nX l +p l ¯X ¯ ¯ ¯ −l ψ k (t) Jgk − P ψ k (t) Jgk ¯ ≥ 2 < 2−l ¯ ¯ ¯ k=1

k=1

U

Therefore, by Borel Cantelli lemma, there is a set of measure zero such that for ω not in this set, nl ∞ X X lim ψ k (t) Jgk = ψ k (t) Jgk l→∞

k=1

k=1

is uniform. From now on denote this subsequence by N to save on notation. I need to consider the characteristic function of (h, W (t) − W (s))U for h ∈ U. Then since convergence is in L2 (Ω; U ) ,

=

E (exp (ir (h, (W (t) − W (s)))U ))     N X ¡ ¢ lim E exp ir  ψ j (t) − ψ j (s) (h, Jgj ) N →∞

j=1

 = lim E  N →∞

N Y

 eir(ψj (t)−ψj (s))(h,Jgj ) 

j=1

Since the random variables ψ j (t) − ψ j (s) are independent, = lim

N Y

N →∞

³ ´ E eir(h,Jgj )(ψj (t)−ψj (s))

j=1

Since ψ j (t) − ψ j (s) is a Gaussian random variable having mean 0 and variance (t − s), the above equals = lim

N →∞

N Y j=1

1

e− 2 r

2

(h,Jgj )2 (t−s)

52.5. HILBERT SPACE VALUED WIENER PROCESSES 

1605

 1 2 − r2 (h, Jgj ) (t − s) lim exp  N →∞ 2 j=1   ∞ X 1 2 (h, Jgj )U  exp − r2 (t − s) 2 j=1   ∞ X 1 2 exp − r2 (t − s) (J ∗ h, gj )U0  2 j=1 ¶ µ ¶ µ 1 2 1 2 2 ∗ ∗ exp − r (t − s) ||J h||U = exp − r (t − s) (JJ h, h)U 2 2 µ ¶ 1 exp − r2 (t − s) (Qh, h)U (52.5.15) 2

=

=

= = =

N X

which shows (h, W (t) − W (s))U is normally distributed with mean 0 and variance (t − s) (Qh, h) where Q ≡ JJ ∗ . It is obvious from the definition that W (0) = 0. Note that Q is of trace class because if {ek } is an orthonormal basis for U, X

(Qek , ek )U

=

k

X

2

||J ∗ ek ||U0 =

XX

k

=

XX l

k 2 (ek , Jgl )U

k

=

2

(J ∗ ek , gl )U0

l

X

2

||Jgl ||U < ∞

l

To find the covariance, consider E ((h1 , W (t) − W (s)) (h2 , W (t) − W (s))) , and use 52.5.19 to obtain this is equal to  E

∞ X

 ∞ X ¡ ¢ (ψ k (t) − ψ k (s)) (h1 , Jgk ) ψ j (t) − ψ j (s) (h2 , Jgj ) . j=1

k=1

Since the series converge in L2 (Ω; U ) , the above equals

 = lim E  n→∞

n X

k=1

 n X ¡ ¢ (ψ k (t) − ψ k (s)) (h1 , Jgk ) ψ j (t) − ψ j (s) (h2 , Jgj ) j=1

1606

WIENER PROCESSES

=

lim (t − s)

n→∞

=

lim (t − s)

n→∞

n X k=1 n X

(h1 , Jgk ) (h2 , Jgk ) (J ∗ h1 , gk )U0 (J ∗ h2 , gk )U0

k=1 ∞ X

(J ∗ h1 , gk )U0 (J ∗ h2 , gk )U0

=

(t − s)

=

(t − s) (J ∗ h1 , J ∗ h2 ) = (t − s) (Qh1 , h2 ) .

k=1

Next consider the claim that the increments are independent. Let W N (t) be m given by the appropriate partial sum and let {hj }j=1 be a finite list of vectors of U . Then from the independence properties of ψ j explained above, 

 m X ¡ ¢ E exp i hj , W N (tj ) − W N (tj−1 ) U  j=1

 E exp

Ã i hj ,

j=1

 =

m X

E exp

N X

!  Jgk (ψ k (tj ) − ψ k (tj−1 )) 

k=1

m X N X

U



i (hj , Jgk )U (ψ k (tj ) − ψ k (tj−1 ))

j=1 k=1

  Y ¡ ¢ = E  exp i (hj , Jgk )U (ψ k (tj ) − ψ k (tj−1 ))  j,k

=

Y

¡ ¡ ¢¢ E exp i (hj , Jgk )U (ψ k (tj ) − ψ k (tj−1 ))

j,k

This can be done because of the independence of the random variables {ψ k (tj ) − ψ k (tj−1 )}j,k . Thus the above equals Y j,k

µ ¶ 1 2 exp − (hj , Jgk )U (tj − tj−1 ) 2 Ã

! N 1X 2 = exp − (hj , Jgk )U (tj − tj−1 ) 2 j=1 m Y

k=1


1607

because ψ k (tj ) − ψ k (tj−1 ) is normally distributed having variance tj − tj−1 . Now letting N → ∞, this implies  E exp

m X

 i (hj , W (tj ) − W (tj−1 ))U 

j=1

=

= = = =

Ã

! ∞ 1X 2 exp − (hj , Jgk )U (tj − tj−1 ) 2 j=1 k=1 ! Ã ∞ m X Y 1 2 ∗ exp − (tj − tj−1 ) (J hj , gk )U0 2 j=1 k=1 µ ¶ m Y 1 2 exp − (tj − tj−1 ) ||J ∗ hj ||U0 2 j=1 ¶ µ m Y 1 exp − (tj − tj−1 ) (Qh, h)U 2 j=1 m Y

m Y

¡ ¢ exp i (hj , W (tj ) − W (tj−1 ))U

(52.5.16)

j=1

Now from 52.5.15, letting r = 1, the above equals m Y

¡ ¢ E i (hj , W (tj ) − W (tj−1 ))U

j=1

By Theorem 47.6.3 on Page 1329, this shows the increments are independent. It remains to verify the Holder continuity. Recall

W (t) =

∞ X

Jgk ψ k (t)

k=1

where ψ k is a real Wiener process. Let m > 2 and let W N (t) denote the appropriate partial sum up to N in the definition of W (t). Also let N be a subsequence such that in addition to convergence in L2 (Ω; U ) , pointwise convergence takes place for both W N (t) and W N (s). Then using 52.1.1 for ψ k and Jensen’s inequality along with Lemma 52.3.1,  ¯ ¯2m 1/m N ¯X ¯ ³ ³¯ ´´ 1/m ¯ 2m ¯ ¯ E ¯W N (t) − W N (s)¯U = E ¯ Jgk (ψ k (t) − ψ k (s))¯  ¯ ¯ k=1

U

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WIENER PROCESSES

=

m 1/m   N  X E   (Jgk Jgl )U (ψ k (t) − ψ k (s)) (ψ l (t) − ψ l (s))   k,l=1

≤

N X

m

m

m

1/m

m

m

1/m

(E (|(Jgk Jgl )U | |ψ k (t) − ψ k (s)| |ψ l (t) − ψ l (s)| ))

k,l=1

=

N X

m

|(Jgk Jgl )U | (E (|ψ k (t) − ψ k (s)| |ψ l (t) − ψ l (s)| ))

k,l=1

=

N X

2m

||Jgk ||U

³ ³ ´´1/m 2m E |ψ k (t) − ψ k (s)|

k=1

+

X

m

³ ³ ´´1/2m 2m E |ψ k (t) − ψ k (s)| ·

m

||Jgk || ||Jgl ||

k6=l

³ ³ ´´1/2m 2m E |ψ l (t) − ψ l (s)|

≤ Cm |t − s|

N X

2m

||Jgk ||U

k=1

+Cm

XX k

m 1/2m

m

||Jgk || (|t − s| )

m 1/2m

m

||Jgl || (|t − s| )

l

≤ Cm |t − s|

∞ X

2m

||Jgk ||U +

k=1

Cm |t − s|

∞ X

m

||Jgk ||

k=1

∞ X

m

||Jgl ||

0 ≤ Cm |t − s|

l=1

It follows an expression of the following form holds. ³¯ ¯2m ´ m E ¯W N (t) − W N (s)¯U ≤ Cm |t − s| , m > 2 where Cm is some constant independent of N . Now an application of Fatou’s lemma as N → ∞ yields ³ ´ 2m

E |W (t) − W (s)|U

m

≤ Cm |t − s| .

ˇ By the Kolmogorov Centsov Theorem, Theorem 50.1.4, it follows that off a set of measure 0, t → W (t, ω) is Holder continuous with exponent γ such that γ
2. 2m


1609

Finally, from 52.5.15with r = 1, ¶ µ 1 E (exp i (h, W (t) − W (s))U ) = exp − (t − s) (Qh, h) 2 which is the same as E (exp i (h, W (t − s))U ) due to the fact W (0) = 0. This proves the theorem. Now suppose you start with a nonnegative trace class operator Q. Then in this case also one can define a Q Wiener process. It is possible to get this theorem from Theorem 52.5.2 but this will not be done here. Theorem 52.5.3 Let U be a real separable Hilbert space and let Q be a nonnegative trace class operator defined on U . Then there exists a Q Wiener process as defined in Definition 52.5.1. Furthermore, the distribution of W (t) − W (s) is the same as the distribution of W (t − s) and W is Holder continuous with exponent γ for any γ < 1/2. Proof: One can obtain this theorem as a corollary of Theorem 52.5.2 but this will not be done here. Let ∞ X Q= λi e i ⊗ e i i=1

where {ei } is a complete orthonormal set and λi ≥ 0 and definition of the Q Wiener process is W (t) ≡

P

λi < ∞. Now the

∞ p X

λk ek ψ k (t)

(52.5.17)

k=1

where {ψ k (t)} are the real Wiener processes defined in Lemma 52.3.1. Now consider 52.5.17. From this formula, if s < t W (t) − W (s) =

∞ p X

λk ek (ψ k (t) − ψ k (s))

(52.5.18)

k=1

First it is necessary to show this sum converges. Since ψ j (t) is a Wiener process,

=

¯2 ¯ ¯ Z ¯X ¯ n p ¡ ¢ ¯ ¯ λj ψ j (t) − ψ j (s) ej ¯¯ dP ¯ Ω ¯j=m ¯ U Z X n ¡ ¢2 λj ψ j (t) − ψ j (s) dP Ω j=m

=

(t − s)

n X j=m

λj

1610

WIENER PROCESSES

and this converges to 0 as m, n → ∞ because it was given that ∞ X

λj < ∞

j=1

so the series in 52.5.18 converges in L2 (Ω; U ) . Therefore, there exists a subsequence (N ) Xp λk ek (ψ k (t) − ψ k (s)) k=1

which converges pointwise a.e. to W (t) − W (s) as well as in L2 (Ω; U ) as N → ∞. Then letting h ∈ U, (h, W (t) − W (s))U =

∞ p X

λk (ψ k (t) − ψ k (s)) (h, ek )

(52.5.19)

k=1

Then by the dominated convergence theorem,

=

E (exp (ir (h, (W (t) − W (s)))U ))     N X p ¡ ¢ lim E exp ir  λj ψ j (t) − ψ j (s) (h, ej ) N →∞

j=1

 = lim E  N →∞

N Y

eir

√

 λj (ψ j (t)−ψ j (s))(h,ej ) 

j=1

Since the random variables ψ j (t) − ψ j (s) are independent, = lim

N Y

N →∞

´ ³ √ E eir λj (ψj (t)−ψj (s))(h,ej )

j=1

Since ψ j (t) is a real Wiener process, = lim

N →∞

N Y

1

e− 2 r

2

λj (t−s)(h,ej )2

j=1

  N X 1 2 = lim exp  − r2 λj (t − s) (h, ej )  N →∞ 2 j=1   ∞ X 1 2 = exp − r2 (t − s) λj (h, ej )  2 j=1 ¶ µ 1 2 = exp − r (t − s) (Qh, h) 2

(52.5.20)


1611

Thus (h, W (t) − W (s)) is normally distributed with mean 0 and variance (t − s) (Qh, h). It is obvious from the definition that W (0) = 0. Also to find the covariance, consider E ((h1 , W (t) − W (s)) (h2 , W (t) − W (s))) , and use 52.5.19 to obtain this is equal to   ∞ p ∞ X X p ¡ ¢ E λk (ψ k (t) − ψ k (s)) (h1 , ek ) λj ψ j (t) − ψ j (s) (h2 , ej ) j=1

k=1

 = lim E  n→∞

 ¡ ¢ λj ψ j (t) − ψ j (s) (h2 , ej )

n p X

n X p

k=1

j=1

λk (ψ k (t) − ψ k (s)) (h1 , ek )

= lim (t − s) n→∞

n X

λk (h1 , ek ) (h2 , ej ) = (t − s) (Qh1 , h2 )

k=1

P (Recall Q ≡ k λk ek ⊗ ek .) Next I show the increments are independent. Let N be the subsequence defined m above and let W N (t) be given by the appropriate partial sum and let {hj }j=1 be a finite list of vectors of U . Then from the independence properties of ψ j explained above,   E exp

m X ¡ ¢ i hj , W N (tj ) − W N (tj−1 ) U  j=1

 E exp

m X

Ã i hj ,

j=1

 = E exp

!  λk ek (ψ k (tj ) − ψ k (tj−1 )) 

N p X k=1

U



m X N p X i λk (hj , ek )U (ψ k (tj ) − ψ k (tj−1 )) j=1 k=1

  ³p ´ Y = E  exp i λk (hj , ek )U (ψ k (tj ) − ψ k (tj−1 ))  j,k

=

Y

³ ³p ´´ E exp i λk (hj , ek )U (ψ k (tj ) − ψ k (tj−1 ))

j,k

This can be done because of the independence of the random variables {ψ k (tj ) − ψ k (tj−1 )}j,k .

1612

WIENER PROCESSES

Thus the above equals m Y

Ã

N

1X 2 λk (hj , ek )U (tj − tj−1 ) exp − = 2 j=1

!

k=1

because ψ k (tj ) − ψ k (tj−1 ) is normally distributed having variance tj − tj−1 and mean 0. Now letting N → ∞, this implies   m X E exp i (hj , W (tj ) − W (tj−1 ))U  j=1

= = =

m Y

Ã

∞

X 1 2 exp − (tj − tj−1 ) λk (hj , ek )U 2 j=1 k=1 µ ¶ m Y 1 exp − (tj − tj−1 ) (Qh, h)U 2 j=1 m Y

!

¢ ¡ exp i (hj , W (tj ) − W (tj−1 ))U

(52.5.21)

j=1

because of the fact shown above that (h, W (t) − W (s)) is normally distributed with mean 0 and variance (t − s) (Qh, h). By Theorem 47.6.3 on Page 1329, this shows the increments are independent. Next consider the continuity assertion. Recall W (t) =

∞ p X

λk ek ψ k (t)

k=1

where ψ k is a real Wiener process. Therefore, letting 2m > 2, m ∈ N and using 52.1.1 for ψ k and Jensen’s inequality along with Lemma 52.3.1, ¯ ¯2m  ∞ p ¯X ¯ ³ ´ ¯ ¯ 2m E |W (t) − W (s)| = E ¯ λk ek (ψ k (t) − ψ k (s))¯  ¯ ¯ k=1 ÃÃ ∞ !m ! X 2 λk |ψ k (t) − ψ k (s)| = E k=1

Ã  !m−1 ∞ ∞ X X 2m ≤ E λk λk |ψ k (t) − ψ k (s)|  k=1

≤ Cm

∞ X

k=1

³ λk E |ψ k (t) − ψ k (s)|

2m

´ (52.5.22)

k=1

≤ Cm |t − s|

m

(52.5.23)


1613

ˇ By the Kolmogorov Centsov Theorem, Theorem 50.1.4, it follows that off a set of measure 0, t → W (t, ω) is Holder continuous with exponent γ such that γ
2 < 2−k ¯ t∈[0,T ] ¯ k=ml

and so by the Borel Cantelli lemma, off a set of measure 0 the partial sums (m ) l X (W (t) , ek )U ek k=1

1616

WIENER PROCESSES

converge uniformly on [0, T ] . This is very interesting but more can be said. In fact the original partial sums converge. Recall Lemma 47.8.6 stated below for convenience. Lemma 52.5.5 Let {ζ k } be a sequence of random variables having values in a separable real Banach space, E whose distributions are symmetric. Letting Sk ≡ P k i=1 ζ i , suppose {Snk } converges a.e. Also suppose that for every m > nk , ¤¢ ¡£ P ||Sm − Snk ||E > 2−k < 2−k . (52.5.29) Then in fact, Sk (ω) → S (ω) a.e.ω

(52.5.30)

Apply this lemma to the situation in which the Banach space, E is C ([0, T ] ; U ) . Then you can conclude uniform convergence of the partial sums, m X

(W (t) , ek )U ek .

k=1


52.6

Levy’s Theorem In Hilbert Space

Recall the concept of quadratic variation. Let W (t) be a Q Wiener process. Does it follow {W (t)} ∈ M2T (H)? The Wiener process is continuous. Furthermore, ³ ´ 2 E |W (t)|H < ∞ for each t ∈ [0, T ] . Since {W (t)} is a martingale, Theorem 50.3.5 can be applied to conclude Ã !2 1/2 ³ ´1/2 ³ ´1/2 2 2 E |W (t)|H ≤ E  sup |W (t)|  ≤ 2E |W (T )|H t∈[0,T ]

and so {W (t)} ∈ M2T (H) . Therefore, by the Doob Meyer decomposition, Theorem 51.6.15, there exists an increasing natural process, A (t) and a martingale, Y (t) such that 2 |W (t)|H = Y (t) + A (t) . What is A (t)? Consider the process 2

|W (t)| From Theorem 52.5.4 this equals ∞ X k=1

2

λk ψ k (t)

52.6. LEVY’S THEOREM IN HILBERT SPACE

1617

where ψ k (t) is a one dimensional Wiener process and Q=

∞ X

λk e k ⊗ e k ,

k=1

∞ X

λk < ∞.

k=1

By Lemma 52.4.2, {W (t)} is a martingale. Therefore, for s < t and A ∈ Fs , it follows since XA is independent of W (t) − W (s) as in the proof of Lemma 52.4.2 that the following holds. Z ³ ´ 2 2 E |W (t)| |Fs − |W (s)| dP A

Z

³ ´ 2 2 E |W (t)| + |W (s)| − 2W (t) · W (s) |Fs dP

= A

Z

Z ³ ´ 2 2 E |W (t) − W (s)| |Fs dP = |W (t) − W (s)| dP

= A

A

Z =

2

P (A)

|W (t) − W (s)| dP Ω

=

P (A)

∞ X

³ ´ 2 λk E (ψ k (t) − ψ k (s))

k=1

=

P (A) (t − s)

∞ X

λk = P (A) (t − s) tr (Q) .

k=1

Therefore, Z

³ ´ ³ ´ 2 2 E |W (t)| − t tr (Q) |Fs − |W (s)| − s tr (Q) dP = 0

A

and since A ∈ Fs is arbitrary, this shows

n o 2 |W (t)| − t tr (Q) is a martingale. 2

Hence the Doob Meyer decomposition for |W (t)| is 2

|W (t)| = Y (t) + t tr (Q) where Y (t) is a martingale. There is a generalization of Levy’s theorem to Hilbert space valued Wiener processes. Theorem 52.6.1 Let {W (t)} ∈ M2T (H) , E (W (t)) = 0, where H is a real separable Hilbert space. Then for Q a nonnegative symmetricntrace class operator, {Wo(t)} 2

is a Q Wiener process if and only if both {W (t)} and (W (t) , h) − t (Qh, h) are martingales for every h ∈ H.

1618

WIENER PROCESSES

Proof: First suppose {W (t)} is a Q Wiener process. Then defining the filtration to be Ft ≡ σ (W (s) − W (u) : u ≤ s ≤ t) , it follows from Lemma 52.4.2 that {W (t)} is a martingale. Consider n o 2 (W (t) , h) − t (Qh, h) . Let A ∈ Fs where s ≤ t. Then using the fact {W (t)} is a martingale, Z ³ ´ 2 E (W (t) − W (s) , h) |Fs dP Z = Z =

A

³

´ 2 2 E (W (t) , h) + (W (s) , h) − 2 (W (t) , h) (W (s) , h) |Fs dP

A

³ ´ 2 2 E (W (t) , h) |Fs + (W (s) , h) − E (2 (W (t) , h) (W (s) , h) |Fs ) dP

A

Z

Z ³ ´ 2 2 E (W (t) , h) |Fs dP + (W (s) , h) dP A A Z − (W (s) , h) E (2 (W (t) , h) |Fs ) dP

=

Z =

A

Z ´ 2 2 E (W (t) , h) |Fs dP − (W (s) , h) dP. ³

A

A 2

Also since XA is independent of (W (t) − W (s) , h) as in the proof of Lemma 52.4.2, and {W (t)} is a Q Wiener process, Z ³ ´ 2 E (W (t) − W (s) , h) |Fs dP ZA 2 = (W (t) − W (s) , h) dP A Z 2 = P (A) (W (t) − W (s) , h) dP Ω

=

P (A) (t − s) (Qh, h) .

Thus, this has shown that for all A ∈ Fs , Z Z ³ ´ 2 2 E (W (t) , h) |Fs dP − (W (s) , h) dP A A Z = P (A) (t − s) (Qh, h) = (t − s) (Qh, h) dP A

and since A ∈ Fs is arbitrary, this proves ³ ´ 2 2 E (W (t) , h) − t (Qh, h) |Fs = (W (s) , h) − s (Qh, h)

52.6. LEVY’S THEOREM IN HILBERT SPACE

1619

This proves one half of the theorem. n o 2 Next suppose both {W (t)} and (W (t) , h) − t (Qh, h) are martingales for n o 2 any h ∈ H. It follows that both {(W (t) , h)} and (W (t) , h) − t (Qh, h) are martingales also. Therefore, by Levy’s theorem, Theorem 51.7.5, {(W (t) , h)} is a Wiener process with the property that its variance at t equals (Qh, h) t instead of t. Thus the time increments are normal and independent. I need to verify that {W (t)} is a Q Wiener process. One of the things which needs to be shown is that E ((W (t) − W (s) , h1 ) (W (t) − W (s) , h2 )) = (Qh1 , h2 ) (t − s) .

(52.6.31)

I have just shown ³ ´ 2 E (W (t) − W (s) , h) = (t − s) (Qh, h)

(52.6.32)

which follows from Levy’s theorem which concludes {(W (t) , h)} is a Wiener process. Therefore, E ((W (t) − W (s) , h1 + h2 ) (W (t) − W (s) , h2 + h1 )) = (Q (h1 + h2 ) , (h1 + h2 )) (t − s) Now using 52.6.32, it follows from this that E ((W (t) − W (s) , h1 ) (W (t) − W (s) , h2 )) = (Qh1 , h2 ) (t − s) which shows 52.6.31. This completes the proof.

1620

WIENER PROCESSES

Stochastic Integration 53.1

Integrals Of Elementary Processes

Stochastic integration starts with a Q Wiener process having values in a separable Hilbert space, U . Thus it satisfies the following definition. Definition 53.1.1 Let W (t) be a stochastic process with values in U, a real separable Hilbert space which has the properties that t → W (t, ω) is continuous. Whenever t1 < t2 < · · · < tm , the increments {W (ti ) − W (ti−1 )} are independent, W (0) = 0, and whenever s < t, L (W (t) − W (s)) = N (0, (t − s) Q) which means that whenever h ∈ H, L ((h, W (t) − W (s))) = N (0, (t − s) (Qh, h)) Also E ((h1 , W (t) − W (s)) (h2 , W (t) − W (s))) = (Qh1 , h2 ) (t − s) . Here Q is a nonnegative trace class operator. Recall this means Q=

∞ X

λi e i ⊗ e i

i=1

where {ei } is a complete orthonormal basis, λi ≥ 0, and ∞ X

λi < ∞

i=1

Such a stochastic process is called a Q Wiener process. Recall the definition of L2 (U, H) ≡ L2 the space of Hilbert Schmidt operators. Ψ ∈ L2 (U, H) means Ψ has the property that for some (equivalently all) orthonormal basis of U {ek } , it follows ∞ X

2

||Ψ (ek )|| < ∞

k=1

1621

1622

STOCHASTIC INTEGRATION

and the inner product for two of these, Ψ, Φ is given by (Ψ, Φ)L2 ≡

X

(Ψ (ek ) , Φ (ek ))

k

Then for such a Hilbert Schmidt operator, the norm in L2 is given by Ã

∞ X

!1/2 2

||Ψ (ek )||

≡ ||Ψ||L2 .

k=1

Note this is the same as  

∞ X ∞ X

1/2 2 (Ψ (ek ) , fj ) 

(53.1.1)

k=1 j=1

where {fj } is an orthonormal basis for H. This is the analog of the Frobenius norm for matrices obtained as 1/2

trace (M M ∗ )

=

Ã X

!1/2 (M M ∗ )ii

1/2  X 2 = Mij

i

i,j

Also 53.1.1 shows right away that if Ψ ∈ L2 (U, H) , then 2

||Ψ||L2 (U,H)

= =

∞ X ∞ X k=1 j=1 ∞ X ∞ X

2

(Ψek , fj )H 2

2

(ek , Ψ∗ fj )U = ||Ψ∗ ||L2 (H,U )

k=1 j=1

The filtration will continue to be denoted by Ft . It will be defined as the following normal filtration in which σ (W (s) − W (r) : 0 ≤ r < s ≤ r) is the completion of σ (W (s) − W (r) : 0 ≤ r < s ≤ r). Ft ≡ ∩r>t σ (W (s) − W (r) : 0 ≤ r < s ≤ r).

(53.1.2)

and σ (W (s) − W (r) : 0 ≤ r < s ≤ r) denotes the σ algebra of all sets of the form (W (s) − W (r)) where 0 ≤ r < s ≤ t.

−1

(Borel)

53.1. INTEGRALS OF ELEMENTARY PROCESSES

1623

Definition 53.1.2 Let Φ (t) ∈ L (U, H) be constant on each interval, (tm , tm+1 ] determined by a partition of [0, T ] , 0 = t0 < t1 · · · < tn = T. Then Φ (t) is said to be elementary if also Φ (tm ) is Ftm measurable and Φ (tm ) takes on only finitely many values. What does this mean? It means that if O is an open (Borel) set in −1 the topological space L (U, H), Φ (tm ) (O) ∈ Ftm . Thus an elementary function is of the form n−1 X Φ (t) = Φ (tk ) X(tk ,tk+1 ] (t) . k=0

Then for Φ elementary, the stochastic integral is defined by Z

t

Φ (s) dW (s) ≡ 0

n−1 X

Φ (tk ) (W (t ∧ tk+1 ) − W (t ∧ tk )) .

k=0

It is also denoted by Φ · W (t) . The above definition is the same as saying that for t ∈ (tm , tm+1 ], Z

t

Φ (s) dW (s) = 0

m−1 X

Φ (tk ) (W (tk+1 ) − W (tk ))

k=0

+Φ (tm ) (W (t) − W (tm )) .

(53.1.3)

The following lemma will be useful. Lemma 53.1.3 Let f, g ∈ L2 (Ω; H) and suppose g is G measurable and f is F measurable where F ⊇ G. Then E ((f, g)H |G) = (E (f |G) , g)H a.e. Similarly if Φ is G measurable as a map into L (U, H) with Z 2 ||Φ|| dP < ∞ Ω

and f is F measurable as a map into H such that f ∈ L2 (Ω; H) E (Φf |G) = ΦE (f |G) . Proof: Let A ∈ G. Let {gn } be a sequence of simple functions, measurable with respect to G, mn X gn (ω) ≡ ank XEkn (ω) k=1 2

which converges in L (Ω; H) and pointwise to g.Then Z Z (E (f |G) , g)H dP = lim (E (f |G) , gn )H dP A

n→∞

A

1624 = =


lim

n→∞

lim

Z X mn ¡ A k=1 Z X mn

n→∞

A k=1

E (f |G) , ank XEkn

¢

n→∞

lim

n→∞

A

A k=1

E ((f, ank )H |G) XEkn dP

ÃÃ m ! ! Z n ³¡ ´ X ¢ n n E f, ak XEkn H |G dP = lim E f, ak XEkn |G dP n→∞

Z =

Z X mn

dP = lim H

Z

E ((f, gn )H |G) dP = lim

n→∞

A

A

k=1

Z

(f, gn )H dP =

A

H

(f, g)H dP

which shows (E (f |G) , g)H = E ((f, g)H |G) as claimed. Consider the other claim. Let Φn (ω) =

mn X

Φnk XEkn (ω) , Ekn ∈ G

k=1

where

Φnk

∈ L (U, H) be such that Φn converges to Φ pointwise in L (U, H) and also Z 2 ||Φn − Φ|| dP → 0. Ω

Then letting A ∈ G and using Corollary 23.2.6 as needed, Z Z Z X mn ΦE (f |G) dP = lim Φn E (f |G) dP = lim Φnk E (f |G) XEkn dP n→∞

A

= =

lim

n→∞

lim

n→∞

k=1 mn X k=1

Z =

lim

n→∞

n→∞

A

mn X

Z Φnk

E (f |G) X

Ekn

A

A k=1

dP = lim

Z Φnk

A

XEkn f dP = lim

n→∞

Z n→∞

A k=1

k=1

Z Φnk

A

¡ ¢ E XEkn f |G dP

Φnk XEkn f dP

Z

Φn f dP = lim A

n→∞

Z X mn

mn X

Φf dP ≡ A

E (Φf |G) dP A

Since A ∈ G is arbitrary, this proves the lemma. With this lemma, here is a major theorem. Theorem 53.1.4 Let Φ (t) be an elementary process as defined in Definition 53.1.2 and let W (t) be a Q Wiener process. Then Φ·W (t) is a continuous square integrable H valued martingale with respect to the σ algebras of 53.1.2 on [0, T ] and Ã∞ ! ³ ´ Z t X 2 2 E |Φ · W (t)|H = E λi |Φ (s) ei |H ds 0

where Q=

∞ X i=1

i=1

λi e i ⊗ e i .


1625

Proof: Let Φ (t) be an elementary process as described in Definition 53.1.2 so the stochastic integral Φ · W (t) is of the form n−1 X

Φ (tk ) (W (t ∧ tk+1 ) − W (t ∧ tk ))

k=0

Why is Φ · W (t) a continuous martingale? It is clearly continuous because it is the sum of continuous functions. It is also adapted to Ft from the definition because t ∧ tk , t ∧ tk+1 are both no larger than t. It remains to verify it is a martingale. Suppose then that t ∈ (tm , tm+1 ] and consider s < t. By Lemma 52.4.2 W (t) is a martingale with respect to the σ algebras of 53.1.2. This follows from the estimate ³ ´ 2 E ||W (t) − W (s)||U

= =

∞ X k=1 ∞ X

³ ´ 2 E (W (t) − W (s) , ei ) (t − s) (Qei , ei ) ≤ (t − s)

k=1

∞ X

λk .

k=1

By adding in both s and t into {tk } and keeping Φ the same at the possibly new points, Φ · W (t) and Φ · W (s) are both unchanged. Therefore, there is no loss of generality in assuming s = tl , t = tm where m > l. Thus E (Φ · W (t) |Fs ) = E (Φ · W (tm ) |Ftl ) Ãm−1 ! X =E Φ (tk ) (W (tk+1 ) − W (tk )) |Ftl

=E

Ã l−1 X k=0

k=0

Φ (tk ) (W (tk+1 ) − W (tk )) +

m−1 X

! Φ (tk ) (W (tk+1 ) − W (tk )) |Ftl

k=l

Consider a typical term of the second sum. Since W (t) is a martingale, E (Φ (tk ) (W (tk+1 ) − W (tk )) |Ftl ) = E (E (Φ (tk ) (W (tk+1 ) − W (tk )) |Ftk ) |Ftl ) = E (Φ (tk ) E ((W (tk+1 ) − W (tk )) |Ftk ) |Ftl ) = E (Φ (tk ) · 0|Ftl ) = 0. Thus this reduces to Ã l−1 ! X E Φ (tk ) (W (tk+1 ) − W (tk )) |Ftl = k=0

l−1 X

Φ (tk ) (W (tk+1 ) − W (tk ))

k=0

= Φ · W (tl ) = Φ · W (s) This proves the claim about Φ · W (t) being a martingale. Now consider the estimate. To save space, let ∆k W (t) ≡ W (t ∧ tk+1 ) − W (t ∧ tk )

1626


Then

³ ´ 2 E |Φ · W (t)|H = 



E

(Φ (tk ) (∆k W (t)) , Φ (tj ) (∆j W (t)))

n−1 X n−1 X k=0 j=0

Consider the expectation of a mixed term for which j < k. E ((Φ (tk ) ∆k W (t) , Φ (tj ) ∆j W (t)))

(53.1.4)

Recall that from the definition of elementary functions, each of Φ (tj ) and Φ (tk ) take only finitely many values in L (U, H) so there is no question that (Φ (tk ) ∆k W (t) , Φ (tj ) ∆j W (t)) ∈ L1 (Ω) . Thus 53.1.4 above equals ¡ ¡ ¢¢ E E (Φ (tk ) (∆k W (t)) , Φ (tj ) (∆j W (t))) |Ftj+1 By Lemma 53.1.3 this equals ¡ ¡ ¢ ¢ E E Φ (tk ) ∆k W (t) |Ftj+1 , Φ (tj ) ∆j W (t) Now consider

¡ ¢ E Φ (tk ) ∆k W (t) |Ftj+1

By properties of conditional expectation this is E (E (Φ (tk ) ∆k W (t) |Ftk ) |Fj+1 ) = E (Φ (tk ) E (∆k W (t) |Ftk ) |Fj+1 ) = E (Φ (tk ) 0|Fj+1 ) = 0 because W (t) is a martingale by Lemma 52.4.2. Therefore the mixed term in 53.1.4 vanishes. It follows Ãn−1 ! ³ ´ X 2 2 |Φ (tk ) (∆k W (t))|H (53.1.5) E |Φ · W (t)|H = E k=0

Let {fk } be a complete orthonormal basis for H and let {ek } be the complete orthonormal basis for U such that Q=

∞ X

λk ek ⊗ ek

(53.1.6)

k=1

Then 53.1.5 reduces to





E

2 (Φ (tk ) (∆k W (t)) , fj ) 

n−1 ∞ XX k=0 j=1


=

n−1 ∞ XX

E

³¡

1627 ∗

(∆k W (t)) , Φ (tk ) fj

¢2 ´

(53.1.7)

k=0 j=1

Claim: ³¡ ¢2 ´ ¡¡ ¢¢ ∗ ∗ ∗ E (∆k W (t)) , Φ (tk ) fj = (t ∧ tk+1 − t ∧ tk ) E QΦ (tk ) fj , Φ (tk ) fj Proof of claim: First note that there is nothing to prove if t < tk since both sides are 0 in this case. This would be a lot easier if Φ (tk ) were a constant, but unfortunately it is not. Consider the simpler formula in which Φ (tk ) is replaced with ΦXA where A ∈ Ft∧tk , Φ ∈ L (U, H). Then ³ ´ ³ ´ 2 2 E (∆k W (t) , Φ∗ XA fj )U = E XA (∆k W (t) , Φ∗ fj )U (53.1.8) Now A ∈ ∩r>t∧tk σ (W (s) − W (r) : 0 ≤ r < s ≤ r) and so by Lemma 52.4.2, 53.1.8 equals Z 2 P (A) (∆k W (t) , Φ∗ fj )U dP Ω

= P (A) (tk+1 ∧ t − tk ∧ t) (QΦ∗ fj , Φ∗ fj )U By assumption Φ (tk ) is a finite sum of functions of this form. Say Φ (tk ) (ω) =

mn X

Φj XAnj (ω)

j=1

Then the above computation implies Z 2 (Φ (tk ) ∆k W (t) , fj )H dP Ω

= =

mn Z X j=1 mn X j=1

=

Ω

P

¡ ¢2 XAnj (ω) ∆k W (t) , Φ∗j fj U dP

¡

Anj

¢

Z Ω

¢2 ¡ ∆k W (t) , Φ∗j fj U dP

(tk+1 ∧ t − tk ∧ t)

mn X

¡ ¢¡ ¢ P Anj QΦ∗j fj , Φ∗j fj U

j=1

Z =

(tk+1 ∧ t − tk ∧ t) Ω

This proves the claim.

¡ ¢ ∗ ∗ QΦ (tk ) fj , Φ (tk ) fj U dP

1628


Now returning to 53.1.7 and 53.1.5 and using the claim, ´ ³ 2 E |Φ · W (t)|H =

n−1 ∞ XX

³¡

E

∗

(∆k W (t)) , Φ (tk ) fj

¢2 ´

k=0 j=1

=

n−1 ∞ XX

(tk+1 ∧ t − tk ∧ t) E

¡¡

∗

∗

QΦ (tk ) fj , Φ (tk ) fj

¢ ¢ U

k=0 j=1

=

n−1 ∞ XX

³³ (tk+1 ∧ t − tk ∧ t) E

∗

∗

Q1/2 Φ (tk ) fj , Q1/2 Φ (tk ) fj

´ ´

k=0 j=1 n−1 ∞ XX

=

µ¯ ¯2 ¶ ¯ 1/2 ¯ ∗ (tk+1 ∧ t − tk ∧ t) E ¯Q Φ (tk ) fj ¯ U

k=0 j=1 n−1 ∞ XX

=

Ã (tk+1 ∧ t − tk ∧ t) E

k=0 j=1 n−1 ∞ XX

=

Ã (tk+1 ∧ t − tk ∧ t) E

k=0 j=1

 = E

=

n−1 ∞ XX

E

k=0

Z

t

=

E 0

∞ ³ X

Q

1/2

∗

Φ (tk ) fj , ei

i=1 ∞ ³ X

∗

Φ (tk ) fj , Q

i=1

k=0 i=1 n−1 X

U

Ã∞ X

i=1 Ã∞ X

λi

∞ X

1/2

ei

´2

!

U

´2

!

U

 2

(fj , Φ (tk ) ei )H (tk+1 ∧ t − tk ∧ t)

j=1

!

2 λi |Φ (tk ) ei |H

(tk+1 ∧ t − tk ∧ t)

! 2 λi |Φ (s) ei |H

ds

i=1

This proves the theorem. Consider the expression ∞ X

2

λi |Φ (s) ei |H =

i=1

∞ ¯ ¯2 X p ¯ ¯ ¯Φ (s) λi ei ¯ i=1

H

(53.1.9)

which occurs in the conclusion of the above theorem. It is convenient to define a new Hilbert space called U0 ⊆ U such that this expression equals 2

||Φ (s)||L2 (U0 ,H) From 53.1.9 this will be so if and only if np o λi ei : λi 6= 0


1629

is an orthonormal basis for U0 . This is the motivation for the following definition. Definition 53.1.5 Let U be a separable Hilbert space and let Q be a positive nuclear operator. That is ∞ X Q= λk ek ⊗ ek k=1

P∞

∞

where each λk ≥ 0, k=1 λk < ∞, and {ek }k=1 is an orthonormal basis for U . Assume without loss of generality the λk form a decreasing sequence. (See Theorem 23.3.9.) Let L = sup {k : λk > 0} Thus L ≤ ∞. Define 2

ρ (x) ≡

¶2 L µ X ek x, √ λk U k=1

and let W0 consist of all x ∈ U such that ρ (x) < ∞ Then U0 will be defined as ( U0 ≡

x ∈ W0 : x =

L X

) (x, ek )U ek

, ||x||U0 ≡ ρ (x)

k=1

Lemma 53.1.6 In the above definition, U0 is a Hilbert space properly contained in U and C ||x||U0 ≥ ||x||U The inner product in U0 is (x, y)U0

¶ µ ¶ L µ X ek ek = y, √ x, √ λk U λk U k=1

and an orthonormal basis for U0 is if and only if L = ∞.

©√

λi e i

ªL i=1

(53.1.10)

L

≡ {gi }i=1 . Also U0 is dense in U

Proof: It is clear U0 is contained in U from the definition. Why is ρ a norm on U0 ? It is clear it comes from the bilinear ¶ µ ¶ L µ X ek ek x, √ y, √ (x, y) ≡ λk U λk k=1 Therefore, the triangle inequality holds as well as ||αx||U0 = |α| ||x||U0 . If ρ (x) ≡ ||x||U0 = 0, why does x = 0? Since x ∈ U0 , x=

L X k=1

(x, ek )U ek =

¶ p L µ X ek x, √ λk ek λk U k=1

1630


and so ||x||U ≤

¶ ¯p L ¯µ X ¯ ¯ ¯ x, √ek ¯ λk ¯ ¯ λ k

k=1

U

Ã ≤

=

!1/2 ¶2 !1/2 ÃX L µ L X ek x, √ λk λk U k=1 k=1 Ã L !1/2 Ã L !1/2 X X ρ (x) λk =0 λk =0 k=1

k=1

Thus U0 is indeed an inner product space. Why is it complete? Let {xn } be a Cauchy sequence in U0 . From the definition of ρ, ρ (xn − xm )

≥ =

1 √ λ1

Ã

L X

!1/2 (xn −

2 xm , ek )U

k=1

1 √ ||xn − xm ||U λ1

(53.1.11)

which shows {xn } is Cauchy in U also. Let xn → x in U . Let ε > 0 be given. Then there exists nε such that if m, n > nε , then ρ (xn − xm ) < ε. Now pick m ≥ nε . By Fatou’s lemma ¶2 ¶2 L µ L µ X X ek ek ≤ lim inf ≤ ε2 x − xm , √ xn − xm , √ n→∞ λ λ k k U U k=1 k=1 Thus Ã

Ã

L µ X k=1

ek x, √ λk

Ã

¶2 !1/2 ≤ U

¶2 L µ X ek + xm , √ λk U k=1

!1/2
L and then for all x ∈ U0 , 2 ||em − x||U ≥ 1 If L = ∞, then if ε > 0 is given ¯¯ ¯¯ m ¯¯ ¯¯ X ¯¯ ¯¯ (x, ek ) ek ¯¯ < ε ¯¯x − ¯¯ ¯¯ k=1

U

whenever m is large enough. However, each ek is in U0 and so U0 is indeed dense in U . This proves the lemma. Note this last part about density adapts easily if U is a finite dimensional Hilbert space. In this case, just replace ∞ with n where n is the dimension. Now with the above lemma and the description of the new Hilbert space U0 , Theorem 53.1.4 can be restated in the following form. Corollary 53.1.7 Let Φ (t) be an elementary process as defined in Definition 53.1.2 Then Φ · W is a continuous square integrable H valued martingale with respect to the σ algebras of 53.1.2 on [0, T ] and ³ ´ Z t ³ ´ 2 2 E |Φ · W (t)|H = E ||Φ (s)||L2 (U0 ,H) ds

(53.1.12)

0

where Q=

∞ X

λi e i ⊗ e i

i=1

is the covariance of W ©√(t) . Here ª∞ U0 is a Hilbert space contained in U for which the nonzero elements of λi ei i=1 is an orthonormal basis for U0 . This Hilbert space U0 equals the Hilbert space Q1/2 U where this Hilbert space is discussed in Theorem 15.2.2 for L = Q1/2 .

1632


Proof: It only remains to verify the claim about U0 = Q1/2 U. Suppose then that x ∈ U0 so that ¶ p L µ L X X ek x, √ (x, ek )U ek = x= λk ek λk U k=1 k=1 Is x = Q1/2 y for some y ∈ U ? Q1/2 y =

L p X

λk ek ⊗ ek (y) =

¡√

(ek , y)

p

λk e k

k=1

k=1

and so if y is such that

L X

¢ λk ek , y = (x, ek ) , then Q1/2 y = x as desired. Let y=

L X 1 √ (x, ek )U ek λk k=1

This is in U because by definition, since x ∈ U0 , ¶2 L µ X ek < ∞. x, √ λk U k=1 Next suppose x ∈ Q1/2 U so for some y ∈ U, x=

L p X

L X

k=1

k=1

λk ek ⊗ ek (y) =

and so it follows x is of the form x=

L X

PL k=1

(ek , y)U

p

λk e k

ak ek . Since {ek } is an orthonormal basis,

(x, ek )U ek , (x, ek )U = (ek , y)U

p

λk

k=1

Also 2

ρ (x) ≡

L µ X k=1

ek x, √ λk

¶2 = U

L X

2

(ek , y)U < ∞

k=1

so x ∈ U0 . What about the norm? Is ||x||U0 = ||x||Q1/2 U ? By definition and Theorem 15.2.2 2 ||x||U0

≡

L µ X k=1

and 2

||x||Q1/2 U

ek x, √ λk

¶2 = U

L X

2

λ−1 k (x, ek )U

k=1

³

´ Q−1/2 x, Q−1/2 x U Ã L ! L X X −1/2 −1/2 = λk ek (x, ek ) , λk ek (x, ek ) =

k=1

k=1

U


=

L X

1633

2

λ−1 k (x, ek )U

k=1

which is the same thing. This proves the corollary. This shows you can think of U0 as the space Q1/2 U where this is a Hilbert space defined in Definition 15.2.1. Now this is an interesting observation because one can define such a space without assuming Q is trace class. See Theorem 15.10.4 on Page 462. The formula 53.1.12 is called the Ito isometry. Note ∞ p X p Q1/2 = λi ei ⊗ ei , Q1/2 ek = λk ek . i=1

©√

For Ψ ∈ L2 (U0 , H) , and using that {gk } = U0 , 2

||Ψ||L2 (U0 ,H)

≡

∞ X ∞ X

2

(Ψgk , fj )H =

k=1 j=1

=

λk e k

ª

is an orthonormal basis for

∞ X ∞ ³ ´2 X ΨQ1/2 ek , fj

H

k=1 j=1

¯¯ ¯¯2 ¯¯ ¯¯ ¯¯ΨQ1/2 ¯¯

L2 (U,H)

¯¯³ ´∗ ¯¯2 ¯¯ ¯¯ = ¯¯ Q1/2 Ψ ¯¯

(53.1.13)

L2 (H,U )

Thus also, ¯¯³ ´ ¯¯ ¯¯ 1/2 ∗ ¯¯2 ¯¯ Q Ψ ¯¯

L2 (H,U )

= =

∞ X ∞ ³ X

Q1/2 Ψ∗ fj , ek

´2 U

=

j=1 k=1 ∞ X

∞ ³ X

j=1

j=1

(ΨQΨ∗ fj , fj )H =

∞ ³ X

Q1/2 Ψ∗ fj , Q1/2 Ψ∗ fj

j=1

³ ´∗ ´ ΨQ1/2 ΨQ1/2 fj , fj

´ U

H

and this last expression is defined to be trace (ΨQΨ∗ ) by analogy with the case of matrices. Definition 53.1.8 Define L (U, H)0 to be the restrictions of Φ ∈ L (U, H) to U0 . Lemma 53.1.9 L (U, H)0 is dense in L2 (U0 , H) , the space of Hilbert Schmidt operators mapping U0 to H. Also, letting ρ0 (Φ) ≡ sup {||Φ (x)||H : x ∈ U0 , ||x||U ≤ 1} Then if ρ0 (Φn ) → 0, it follows ||Φn ||L0 → 0. 2

Proof: First why is L (U, H)0 contained©√ in L2 (U ª0 , H)? Let Φ ∈ L (U, H)0 . From the above, an orthonormal basis for U0 is λk ek for λk > 0. Then L ¯ p L ¯2 X X ¯ ¯ 2 λk ||Φ|| < ∞ ¯Φ λk e k ¯ ≤ k=1

H

k=1

1634


and so this shows Φ ∈ L2 (U0 , H) . Why is L (U, H)P (U0 , H)? Consider Φ ∈ L2 (U0 , H) . Then by The0 dense in L2p orem 15.7.12 Φ = λ e where {fj } is an orthonormal basis in H ij fi ⊗ i,j a ©√ ª j j and as explained above, λi ei is an orthonormal basis in U0 , the sum converging in L2 (U0 , H) . Then replacing the infinite sum with a suitable finite sum, Φ is approximated in L2 (U0 , H) by something in L (U, H)0 , X

aij fi ⊗

p λj ej

i,j 0 and define Gs0 ≡ {S ∈ P∞ : Ss0 ∈ Fs0 } where Ss0 ≡ {ω ∈ Ω : (s0 , ω) ∈ S} .

s0

It is clear Gs0

Ω S s0 is a σ algebra. The next step is to show Gs0 contains the sets (s, t] × F, F ∈ Fs

(53.2.14)

{0} × F, F ∈ F0 .

(53.2.15)

and It is clear {0} × F is contained in Gs0 because ({0} × F )s0 = ∅ ∈ Fs0 . Similarly, if s ≥ s0 or if s, t < s0 then ((s, t] × F )s0 = ∅ ∈ Fs0 . The only case left is for s < s0 and t ≥ s0 . In this case, letting As ∈ Fs , ((s, t] × As )s0 = As ∈ Fs ⊆ Fs0 . Therefore, Gs0 contains all the sets of the form given in 53.2.14 and 53.2.15 and so since P∞ is the smallest σ algebra containing these sets, it follows P∞ = Gs0 . The case where s0 = 0 is entirely similar but shorter. Therefore, if X is predictable, letting A ∈ B (E) , X −1 (A) ∈ P∞ or PT and so ¡ −1 ¢ −1 X (A) s ≡ {ω ∈ Ω : X (s, ω) ∈ A} = X (s) (A) ∈ Fs showing X (t) is Ft adapted. This proves the proposition. In the situation of the above definition, a predictable stochastic process will be called E predictable. Here is an interesting lemma, also like the sort of thing which occurs with product measure.

1636


Lemma 53.2.3 Let E be a separable Banach space and let Φ : [0, ∞) × Ω → E be E predictable. Then the following iterated integrals make sense and are equal. Z ∞Z Z Z ∞ φ (Φ) dP dt = φ (Φ) dtdP (53.2.16) 0

Ω

Ω

0

where φ is any nonnegative continuous or Borel measurable function defined on E. Proof: Let B ∈ P∞ be such that Z ∞Z Z Z XB dP dt = 0

Ω

Ω

∞

XB dtdP

(53.2.17)

0

and both integrals make sense. Also let K denote sets of the form (s, t] × As where As ∈ Fs . Thus K is a π system because (s, t] × As ∩ (s1 , t1 ] × As1 = (s ∨ s1 , t ∧ t1 ] × As ∩ As1 and so As ∩ As1 ∈ Fs∨s1 . The iterated integrals make sense for sets in K and are also equal. Also if G denotes those sets for which 53.2.17 holds, then G ⊇ K and it is clear that G is closed under countable disjoint unions and also for complements. Therefore, by the Lemma on π systems, Lemma 9.11.3 on Page 275 it follows G ⊇ σ (K) = P∞ . If Φ is E predictable then φ (Φ) is R predictable. Hence, approximating with simple functions, the iterated integrals in 53.2.16 make sense and are equal. This proves the lemma. A similar lemma holds if you replace [0, ∞) with [0, T ] in the above. Now here is an important observation. Lemma 53.2.4 Let f : [0, T ] × Ω → [0, ∞) be real predictable. Then letting Z F (t, ω) ≡

t

f (s, ω) ds, 0

it follows F is also real predictable. Proof: First suppose f (t, ω) = X(p,q]×Ap (t, ω) = X(p,q] (t) XAp (ω) , Ap ∈ Fp Is the assertion true in this case? Here (p, q] × Ap is one of the sets which generates PT . Doing the integral yields that in this case F (t, ω) = l (t) XAp (ω) where l is piecewise linear and continuous, equal to 0 till p and then increasing to q − p and remaining constant for t > q. Now let α ∈ [0, q − p) and consider [F > α] This is of the form (r, q] × Ap

(53.2.18)

53.2. INTEGRALS OF MORE GENERAL PROCESSES

1637

for some r ∈ (p, q) . This is a set which generates PT . If α ≥ q − p, 53.2.18 is ∅ also in PT . Let K consist of the sets of this sort which generate PT . These sets form a π system. Let G denote those sets E of PT for which Z t (t, ω) → XE (s, ω) ds (53.2.19) 0

makes sense and is real predictable. Then it was shown above that K ⊆ G. It is also the case that G is closed under countable disjoint unions because if {Ei } is a countable sequence of disjoint sets of G then letting E = ∪∞ i=1 Ei XE (s, ω) =

∞ X

XEi (s, ω)

i=1

and so by the monotone convergence theorem, Z t ∞ Z t X XE (s, ω) ds = XEi (s, ω) ds 0

i=1

0

and so the integral makes sense and the function of 53.2.19 is predictable. It is also easy to see that G is closed under complements. By the lemma on π systems, Lemma 9.11.3,it follows G = PT . In the general case, let fn be an increasing sequence of R predictable simple functions converging pointwise to f. Thus from the above Z t (t, ω) → fn (s, ω) ds 0

makes sense and is predictable. By the monotone convergence theorem, the same must be true of Z t (t, ω) → f (s, ω) ds 0

This proves the lemma. Lemma 53.2.5 The elementary functions are L2 (U0 , H) predictable. Proof: First of all, the elementary functions have values in L (U, H) and by Lemma 53.1.9 this means the restriction to U0 is in L2 (U0 , H) . Consider a typical term in an elementary function X(s,t] Φ where Φ has values in L (U, H) and is measurable with respect to Fs and has finitely many values. Say X(s,t] Φ = X(s,t]

m X

Φk XAk (ω)

k=1

where each Ak ∈ Fs . Then taking the restrictions of each Φk to U0 , it follows from Lemma 53.1.9 the restricted simple function is of the form m X k=1

Φk X(s,t] XAk (ω) =

m X k=1

Φk X(s,t]×Ak (t, ω)

1638


which is clearly predictable because the sets (s, t] × Ak are among those which generate PT . This proves the lemma. The following proposition is like the theorem which says the simple functions are dense in Lp except that here it will be elementary functions and these are like step functions. It is this proposition which shows the need for Φ to be predictable in order to define the stochastic integral in terms of limits of integrals of elementary functions. Proposition 53.2.6 Let Ft be the filtration determined for a Q Wiener process as above. If Φ is a L2 (U0 , H) ≡ L02 predictable process such that for some p ≥ 1, Z TZ p ||Φ (t)||L0 dP dt < ∞ 0

Ω

2

then there exists a sequence {Φn } of elementary processes such that Z TZ p lim ||Φ (t) − Φn (t)||L0 dP dt = 0. n→∞

0

Ω

2

Proof: Let Φ be L2 (U0 , H) predictable. Then there exists a sequence of predictable simple functions, {Φn } converging to Φ pointwise such that ||Φn (t, ω)||L2 (U0 ,H) < ||Φ (t, ω)||L2 (U0 ,H) Then adjusting the finitely many values of Φn and using Lemma 53.1.9, the density of L (U, H)0 in L2 (U0 , H) , it can also be assumed that the values of these simple functions lie in L (U, H)0 . Unfortunately, the simple functions are not elementary. The sets on which they are constant are not products of an interval, (s, t] and a set A ∈ Fs . I will argue that if B ∈ PT , then it differs from a finite disjoint union of sets of the form (s, t] × As where As ∈ Fs on a set of arbitrarily small measure. Let K denote the set of all finite disjoint unions of sets of the form (s, t] × As as just described. If (s, t] × As and (s1 , t1 ] × As1 are two of these sets having nonempty intersection, the intersection being (s ∨ s1 , t ∧ t1 ] × As ∩ As1 , then As ∩ As1 ∈ Fs∨s1 and so such an intersection is one of the sets whose finite disjoint unions constitute K. Thus it is routine to show that K is a π system, but it is also an algebra. To see it is an algebra, let s < s1 < t < t1 so there is possibly a nonempty intersection of two sets of the sort whose finite disjoint unions yield K, (s, t] × As , (s1 , t1 ] × As1 then (s1 , t1 ] × As1 \ (s, t] × As =

((s1 , t] × (As1 \ As )) ∪ ((t, t1 ] × As1 ) ∈ K.

53.2. INTEGRALS OF MORE GENERAL PROCESSES

1639

By Lemma 9.9.2 on Page 264, K is an algebra. Now letting Γ denote such a set of K, let G denote those sets, B ∈ PT such that for any ε > 0 there exists some Γ, such that (m1 × P ) ((B \ Γ) ∪ (Γ \ B)) < ε It is clear K ⊆ G because if B ∈ K then you could simply pick Γ = B. Now suppose {Bk } is a disjoint union of sets of G and pick Γk such that (m1 × P ) (Bk \ Γk ) + (m1 × P ) (Γk \ Bk ) < ε/2k+1 Then

n so that τ m ≥ τ n ≥ t. By Lemma 53.4.4, X[0,τ n ] Φ · W (t) = = =

X[0,τ n ] X[0,τ m ] Φ · W (t) X[0,τ m ] Φ · W (t ∧ τ n ) X[0,τ m ] Φ · W (t) .

Φ · W (t) is Ft adapted because for U an open set in H, −1

Φ · W (t)

(U ) = ∪∞ n=1

³¡ ´ ¢−1 X[0,τ n ] Φ · W (t) (U ) ∩ [τ n ≥ t] ∈ Ft

This proves the lemma. Lemma 53.5.4 Let Φ be predictable or progressively measurable and satisfy Ã"Z #! T

P 0

2

||Φ (s)||L2 (U0 ,H) ds < ∞

=1

Then if σ is a stopping time, Z

t∧σ

ΦdW (s)

≡ Φ · W σ (t)

0

Z = X[0,σ] Φ · W (t) =

t 0

X[0,σ] ΦdW (s)

Proof: Let {τ n } be the localizing sequence described above for which, when the local martingale is stopped, it results in a martingale, and suppose ω is such that τ n (ω) > t. Then by definition, for that ω, Φ · W (t) = X[0,τ n ] Φ · W (t)

1654


and so from Lemma 53.4.4, for that ω Z t∧σ ΦdW (s) ≡ Φ · W σ (t) ≡ X[0,τ n ] Φ · W σ (t) 0

= = ≡

X[0,σ] X[0,τ n ] Φ · W (t) Z t X[0,τ n ] X[0,σ] ΦdW (s) 0 Z t X[0,σ] ΦdW (s) 0

This proves the lemma. Rt The lemma says that even when 0 Φ (s) dW (s) is only a local martingale because Φ is only stochastically integrable, it is still the case that Z t∧σ Z t ΦdW (s) = X[0,σ] ΦdW (s) . 0

0

An important corollary of ¢Lemma 53.4.4 concerns the quadratic variation of ¡ 2 Φ · W (t) for Φ ∈ NW 0, T ; L02 . ¢ ¡ 2 Corollary 53.5.5 Let Φ ∈ NW 0, T ; L02 , Φ being L02 predictable or progressively measurable. Then the quadratic variation, [Φ · W ] is given by the formula Z t 2 [Φ · W ] (t) = ||Φ (s)||L2 (U0 ,H) ds 0

If Φ · W (t) is only a local martingale, because Φ is predictable or progressively measurable but only satisfies Ã"Z #! T

P 0

2

||Φ (s)||L2 (U0 ,H) ds < ∞

=1

the same conclusion holds. Proof: Let σ be a stopping time having values in [0, T ]. Then using Lemma 53.4.4 and the Ito isometry, Theorem 53.2.8,   ¯¯ ¯¯2 Z ¯¯Z T ∧σ ¯¯ T ∧σ ¯¯ ¯¯ 2 ||Φ (s)||L2 (U0 ,H) ds E  ¯¯ Φ (s) dW (s)¯¯ − ¯¯ 0 ¯¯ 0 ¯¯  ¯¯2 Z ¯¯Z T ¯¯ T ¯ ¯ ¯ ¯ 2 = E ¯¯ X[0,σ] Φ (s) dW (s)¯¯ − X[0,σ] ||Φ (s)||L2 (U0 ,H) ds ¯¯ 0 ¯¯ 0 Z

T

Z

= 0

Ω

¯¯ ¯¯ ¯¯X[0,σ] Φ (s)¯¯2 0 dP ds − L2

Z Z Ω

T 0

2

X[0,σ] ||Φ (s)||L2 (U0 ,H) dsdP

53.6. TAKING OUT A LINEAR TRANSFORMATION Z

T

Z

= 0

Ω

¯¯ ¯¯ ¯¯X[0,σ] Φ (s)¯¯2 0 dP ds −

Z Z

L2

Ω

0

T

1655

¯¯ ¯¯ ¯¯X[0,σ] Φ (s)¯¯2

L2 (U0 ,H)

dsdP

and this equals 0 because the integrands are assumed product measurable which implies the iterated integrals are equal. Therefore, by Lemma 51.1.1, the lemma about recognizing a martingale when you see one, the stochastic process ¯¯Z t ¯¯2 Z t ¯¯ ¯¯ 2 ¯¯ Φ (s) dW (s)¯¯¯¯ − ||Φ (s)||L2 (U0 ,H) ds ¯¯ 0

0

is a martingale because you could also take σ = t, the constant and the result would still hold. The process Z t

0

2

||Φ (s)||L2 (U0 ,H) ds

is adapted because the integrand is predictable. It is also increasing. Therefore by uniqueness of the quadratic variation, Theorem 51.4.4, it follows this process must be the quadratic variation. Finally, consider the case where Φ·W (t) is only a local martingale. Then letting {τ n } be a localizing sequence, the above result along with Lemma 53.5.4 applies to the stopped process. Thus by Theorem 51.4.4 and letting {τ n } be a localizing sequence, then from the above and the localization result, ¡ ¢ Φ · W τ n (t) = X[0,τ n ] Φ · W (t) , where τ n (ω) > t, it follows ³³ martingale

=

||Φ · W ||

2

´ − [Φ · W ]

´τ n

(t)

2

||Φ · W τ n || (t) − [Φ · W τ n ] (t) Z t ³ ´τ n ¯¯ ¯¯ 2 ¯¯X[0,τ ] (s) Φ (s)¯¯2 ds ||Φ · W || (t) − n 0 Z t∧τ n ³ ´τ n 2 2 ||Φ · W || (t) − ||Φ (s)|| ds

= = =

0 2

Therefore, ||Φ · W || (t) − part of this theorem

Rt 0

2

||Φ (s)|| ds is a local martingale and by the uniqueness Z

t

[Φ · W ] (t) =

2

||Φ (s)|| ds 0


53.6

Taking Out A Linear Transformation

When is

Z L

Z

T

T

ΦdW (s) = 0

LΦdW (s)? 0

1656


Here I am using the notation Z

T

ΦdW (s) ≡ Φ · W (T ) 0

and it is assumed L ∈ L (H, H1 ) where H1 is another separable real Hilbert space. First of all, here is a lemma which shows LΦ · W (T ) at least makes sense. 2 2 (0, T ; L2 (U0 , H1 )) . (0, T ; L2 (U0 , H)) . Then LΦ ∈ NW Proposition 53.6.1 Let Φ ∈ NW Furthermore, for each t ∈ [0, T ] Z t Z t LΦdW (s) = L ΦdW (s) 0

0

Proof: First note that if Φ ∈ L2 (U0 , H) , then LΦ ∈ L2 (U0 , H1 ). This follows because if {gk } is an orthonormal basis for U0 , then X X 2 2 2 ||LΦgk ||H1 ≤ ||L|| ||Φgk ||H < ∞ k

k

because Φ ∈ L2 (U0 , H) . Also this shows the map Φ → LΦ is continuous. It follows LΦ is L2 (U0 , H1 ) predictable. All that remains is to check the appropriate integral. Z Z T Z Z T 2 2 2 ||LΦ||L2 (U0 ,H1 ) dtdP ≤ ||L|| ||Φ||L2 (U0 ,H) dtdP < ∞ Ω

0

Ω

0

2 NW

and so this proves LΦ ∈ (0, T ; L2 (U0 , H1 )). 2 If Φ ∈ NW (0, T ; L2 (U0 , H)) it follows one can consider Z

T

LΦdW (s) . 0

Letting {Φn } be an approximating sequence of elementary functions satisfying ÃZ ! T

E 0

it is also the case that

ÃZ

2

||Φn − Φ||L2 (U0 ,H) dt

!

T

E

→ 0,

||LΦn − 0

2 LΦ||L2 (U0 ,H1 )

dt

→0

By Theorem 53.2.8, for each t Z t Z t Z t LΦdW (s) = lim LΦn dW (s) = lim L Φn dW (s) n→∞ 0 n→∞ 0 0 Z t Z t = L lim Φn dW (s) = L ΦdW (s) n→∞

0

0

53.7. THE CASE WHERE Q IS NOT TRACE CLASS

1657

The second equality is obvious for elementary functions. Now consider the case where Φ ∈ NW (0, T ; L2 (U0 , H)) so that all is known is that #! Ã"Z T

P 0

2

||Φ||L2 (U0 ,H) dt < ∞

= 1.

Then define τ n as above ½ Z t ¾ 2 τ n ≡ inf t : ||Φ||L2 (U0 ,H) dt ≥ n 0

This sequence of stopping times works for LΦ also. Recall there were two conditions the sequence of stopping times needed to satisfy. The first is obvious. Here is why the second holds. Z T Z T ¯¯ ¯¯ ¯¯ ¯¯ 2 ¯¯X[0,τ ] LΦ¯¯2 ¯¯X[0,τ ] Φ¯¯2 dt ≤ ||L|| dt n n L2 (U0 ,H1 ) L2 (U0 ,H) 0 Z0 τ n 2 2 2 = ||L|| ||Φ||L2 (U0 ,H) dt ≤ ||L|| n 0

Then let t be given and pick n such that τ n ≥ t. Then from the first part Z t Z t L ΦdW (s) ≡ L X[0,τ n ] ΦdW (s) 0 0 Z t = LX[0,τ n ] ΦdW (s) 0 Z t Z t = X[0,τ n ] LΦdW (s) ≡ LΦdW (s) 0

0

This proves the proposition.

53.7

The Case Where Q Is Not Trace Class

A positive self adjoint continuous linear operator has a unique square root according to Theorem 15.10.4. Thus using Theorem 15.2.2, one can still consider the Hilbert space U0 ≡ Q1/2 U. Is there some way to define a stochastic integral even though the covariance Q is not of trace class? This involves the concept of a cylindrical Wiener process. It is a cheap trick which causes great confusion. One generalizes the definition by using another space which does have a Q Wiener process for Q of trace class and one defines the integral in terms of the Wiener process on this space. Also note that this is an issue which is never a problem in finite dimensions. Let U0 ≡ Q1/2 U using the construction of Theorem 15.2.2. Then let J : U0 → U1 be a one to one Hilbert Schmidt operator where U1 is a separable real Hilbert space.

1658


Lemma 53.7.1 Let Q be a positive self adjoint bounded linear transformation defined on U a separable real Hilbert space. Let U0 = Q1/2 U as in Theorem 15.2.2. Then there exists a one to one Hilbert Schmidt operator J : U0 → U1 where U1 is a separable real Hilbert space. In fact you can take U1 = U . P∞ Proof: Let αk > 0 and k=1 α2k < ∞. Then let {gk } be an orthonormal basis for U0 ≡ Q1/2 U, the inner product and norm given in Theorem 15.2.2, and let ∞ X

Jx ≡

(x, gk )U0 αk gk

k=1

Then it is clear that J ∈ L (U0 , U ) . This is because by the norm estimate of Theorem 15.2.2 ||Jx||U

≤

∞ X ¯ ¯ ¯(x, gk ) ¯ αk ||gk || U0 U k=1 =1

≤

∞ X ¯ ¯ z }| { ¯(x, gk ) ¯ αk ||gk || C U0 U0 k=1

Ã

≤

C

!1/2 Ã

∞ X

U0

Ã =

∞ X ¯ ¯ ¯(x, gk ) ¯2

C

k=1 ∞ X

k=1

!1/2 α2k

!1/2 α2k

||x||U0

k=1

Also, by the norm estimate of Theorem 15.2.2, ∞ X

2

||Jgj ||U =

j=1

∞ X

2

α2j ||gj ||U ≤ C

j=1

∞ X

α2j < ∞

j=1

and so J is also a Hilbert Schmidt operator which maps U0 to U . It is clear that J is one to one because each αk > 0. If U0 is finite dimensional, just make the above sum finite. (When you run out of gk , quit summing.) This proves the lemma. Now consider the operator Q1 ≡ JJ ∗ which is defined on U1 . This is a positive self adjoint operator and furthermore, this operator is trace class. To see this, let {fm } be an orthonormal basis for U1 . Then letting {ek } be a complete orthonormal basis for U0 , ∞ X

∗

(JJ fk , fk )U1 =

k=1

=

∞ X ∞ X

∞ X

∗

∗

(J fk , J fk )U =

k=1 2

(J ∗ fk , ej ) =

k=1 j=1

because J is Hilbert Schmidt.

∞ X ∞ X j=1 k=1

∞ X

2

||J ∗ fk ||U

k=1 2

(fk , Jej ) =

∞ X j=1

2

||Jej || < ∞


1659

1/2

Lemma 53.7.2 J : Q1/2 U = U0 → Q1 U1 = JU0 is an isometry, the norms on these spaces given by Definition 15.2.1. U1 ∪ U

Q1/2

→

J

Q1/2 U = U0

→

JU0

1/2

= Q1 U1

1/2

Q1

←

U1

Proof: There are two things which need to be shown. One is that JU0 = 1/2 Q1 U1 ⊆ U1 and the other is that J is an isometry. These things follow from Theorem 15.2.4. For x ∈ U1 , 2

||J ∗ x||U0

= (J ∗ x, J ∗ x)U0 = (JJ ∗ x, x)U1 ¯¯ ¯¯ ¯¯³ ´ ¯¯ ¯¯ 1/2 ¯¯2 ¯¯ 1/2 ∗ ¯¯2 = (Q1 x, x)U1 = ¯¯Q1 x¯¯ = ¯¯ Q1 x¯¯ U1

U1

1/2

and so it follows from Theorem 15.2.4 that Q1 (U1 ) = J (U0 ) . In applying this theorem, the H there is the same as U1 here and the two Ui are U0 and U1 . Also from this theorem ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ −1/2 ¯¯ ||x||JU0 ≡ ¯¯J −1 x¯¯U0 = ¯¯Q1 x¯¯ ≡ ||x||Q1/2 U1 U1

1

1/2

for all x ∈ Q1 U1 = JU0 . The notation is in the sense of Definition 15.2.1 but in this case since J is one to one, J −1 has the usual meaning. Now let y ∈ U0 so 1/2 Jy ∈ Q1 U1 = JU0 . Then by Definition 15.2.1 ¯¯ ¯¯ ||Jy|| 1/2 = ||Jy|| ≡ ¯¯J −1 Jy ¯¯ = ||y|| Q1

JU0

U1

U0

U0

This proves the lemma. L L Now let {gk }k=1 be any orthonormal basis for U0 and consider {Jgk }k=1 . Since L J is an isometry this implies {Jgk }k=1 is also an orthonormal basis for JU0 . This is because ´ 1³ 2 2 (Jgk , Jgl )JU0 = ||Jgk + Jgl || − ||Jgk − Jgl || 4 ´ 1³ 2 2 = ||gk + gl || − ||gk − gl || = (gk , gl )U0 4 Consider W (t) ≡

∞ X

ψ k (t) Jgk

k=1

It follows from Theorem 52.5.2 this is a Q1 Wiener process in U1 . However, it is interesting to review why this is. First of all, why does the series converge? Since the {Jgk } are orthonormal, ¯2 Z ¯¯ X Z X n n ¯ ¯ ¯ 2 2 (ψ k (t) − ψ k (s)) Jgk ¯ dP = (ψ k (t) − ψ k (s)) ||Jgk ||U1 ¯ ¯ ¯ Ω Ω k=m

U1

k=m

1660


= (t − s)

n X

2

||Jgk ||U1

k=m

and as m, n → ∞ this converges to 0 because of the assumption that J is a Hilbert Schmidt operator. Why is its covariance Q1 ?

=

¡ ¢ E (a, W (t − s))U1 (b, W (t − s))U1 Ã  ! n n X X¡ ¢ lim E  a, (ψ k (t) − ψ k (s)) Jgk b, ψ j (t) − ψ j (s) Jgj  n→∞

 =

=

lim E 

n→∞

n X

(a, Jgk ) (ψ k (t) − ψ k (s))

n X

 ¢ (b, Jgj ) ψ j (t) − ψ j (s)  ¡

j=1

k=1

lim (t − s)

n→∞

j=1

k=1

n X

(a, Jgk ) (b, Jgk ) = (t − s)

k=1

∞ X

(J ∗ a, gk ) (J ∗ b, gk )

k=1

= (t − s) (J ∗ a, J ∗ b) = (t − s) (JJ ∗ a, b) = (t − s) (Q1 a, b) This shows W (t) is indeed a Q1 Wiener process. To get an idea how to define an integral here is a lemma. Lemma 53.7.3

³ ´ Φ ∈ L2 Q1/2 U, H = L2 (U0 , H)

if and only if

´ ³ 1/2 Φ ◦ J −1 ∈ L2 Q1 U1 , H 1/2

where the norms of Q1 U1 and Q1/2 U are in the sense of Theorem 15.2.2. Furthermore, ¯¯ ¯¯ ¯¯Φ ◦ J −1 ¯¯2

³ ´ 1/2 L2 Q1 U1 ,H

2

2

= ||Φ||L2 (Q1/2 U,H ) = ||Φ||L2 (U0 ,H)

L

Proof: Let {gk }k=1 be an orthonormal basis for U0 ≡ Q1/2 U. What about 1/2 {Jgk }? Is it an orthonormal basis in JU0 = Q1 U1 ? (Jgk , Jgl )JU0 ≡ (gk , gl )U0 1/2

from the definition of the inner product in JU0 = Q1 U1 . Thus {Jgk } is orthonormal. Is it complete? Yes because by Lemma 53.7.2 J is an isometry of U0 = Q1/2 U

53.7. THE CASE WHERE Q IS NOT TRACE CLASS 1/2

1661 1/2

and Q1 U1 . It follows that if Jx is an arbitrary thing in Q1 ! Ã L L X X Jx = J (x, gk )U0 gk = (x, gk )U0 Jgk k=1

=

k=1

L L X X ¡ −1 ¢ J Jx, J −1 Jgk U0 Jgk ≡ (Jx, Jgk )JU0 Jgk k=1

=

(U1 ) ,

L X

k=1

(Jx, Jgk )Q1/2 U1 Jgk 1

k=1

Now ¯¯ ¯¯ ¯¯Φ ◦ J −1 ¯¯2

³ ´ 1/2 L2 Q1 U1 ,H

L X ¯¯ ¯¯ ¯¯Φ ◦ J −1 (Jgk )¯¯2

=

H

k=1 L X

=

2

2

||Φgk ||H = ||Φ||L2 (Q1/2 U,H )

k=1

This proves the lemma. With this lemma the stochastic integral can be defined in the case that Q is not of trace class. Definition 53.7.4 Let U , H be separable real Hilbert spaces and suppose Q is a nonnegative continuous self adjoint operator defined on U . Let U0 = Q1/2 U as in Definition 15.2.1 on Page 427. Also let Φ be L2 (U0 , H) predictable and satisfy Z TZ 2 ||Φ||L2 (U0 ,H) dP dt < ∞. 0

Ω

Let J : U0 → U1 be a Hilbert Schmidt operator where U1 is another separable real Hilbert space. Then define W (t) ≡

L X

ψ k (t) Jgk

(53.7.27)

k=1

where gk is an orthonormal basis for U0 and the ψ k are the independent real Wiener processes of Lemma 52.3.1 so that W (t) is a Wiener process with values in U1 . Then define the stochastic process Φ · W (t) , having values in H as follows Z t Z t Φ · W (t) ≡ ΦdW (s) ≡ Φ ◦ J −1 dW (s) ≡ Φ ◦ J −1 · W (t) 0

0

where the last definition is just the stochastic integral defined earlier in terms of the Wiener process W (t) . In the case that Φ is L2 (U0 , H) predictable and it is only assumed Ã"Z #! T

P 0

2

||Φ||L2 (U0 ,H) dt < ∞

=1

1662


³ ´ 1/2 then by Corollary 53.1.7 U0 = Q1/2 U and by Lemma 53.7.3 Φ◦J −1 is L2 Q1 U1 , H predictable and in fact Ã"Z #! T ¯¯ ¯¯ −1 ¯¯2 ³ ´ ¯ ¯ P Φ◦J dt < ∞ =1 1/2 0

L2 Q1

U1 ,H

so one can still define Z t Z t Φ · W (t) ≡ ΦdW (s) ≡ Φ ◦ J −1 dW (s) ≡ Φ ◦ J −1 · W (t) . 0

0

This definition is independent of J and U1 . Here is why. Let Φ be an elementary function having values in L (U, H). Then Φ (tk ) ◦ J −1 is also an elementary function with values in L (U1 , H) . From the definition given above and 53.7.27, it follows from Theorem 52.5.2, letting N denote the subsequence there giving uniform convergence of the sum off a set of measure zero that for a.e. ω   n−1 ∞ X X ¡ ¢ Φ · W (t) = Φ (tk ) ◦ J −1  ψ j (t ∧ tk+1 ) − ψ j (t ∧ tk ) Jgj  j=1

k=0

=

n−1 X

Φ (tk ) ◦ J −1

k=0

=

=

  N X ¡ ¢ lim  ψ j (t ∧ tk+1 ) − ψ j (t ∧ tk ) Jgj 

N →∞

j=1

  n−1 N X X ¡ ¢ lim Φ (tk ) ◦ J −1  ψ j (t ∧ tk+1 ) − ψ j (t ∧ tk ) Jgj 

N →∞

j=1

k=0

  n−1 N X X ¡ ¢ lim Φ (tk )  ψ j (t ∧ tk+1 ) − ψ j (t ∧ tk ) gj 

N →∞

k=0

j=1

which does not depend on J. It seems the reason for this generalization is to allow the definition of a stochastic integral having values in H in the case where Q is not trace class, including the case where Q = I.

The Ito Formula 54.1

The Case Of A Q Wiener Process

First recall 52.5.22 where it is shown that for every α α

E (|W (t) − W (s)| ) ≤ C |t − s|

α/2

,

ˇ and so by Kolmogorov Centsov continuity theorem |W (t) − W (s)| ≤ Cγ |t − s|

γ

(54.1.1)

for every γ < 1/2. Recall it was important that W be a Q Wiener process where Q was nuclear. That is X Q= λk ek ⊗ ek where

P k

k

λk < ∞. The Ito formula is the following.

Theorem 54.1.1 Let Φ be an L02 valued predictable or progressively measurable process which is stochastically integrable in [0, T ] because Ã"Z #! T

P

2

||Φ|| dt < ∞

=1

0

and let φ : [0, T ] × Ω → H be predictable or progressively measurable process which is Bochner integrable on [0, T ] and let X (0) be F0 measurable and H valued. Let Z t X (t) ≡ X (0) + φ (s) ds + Φ · W (t) . 0

Let F : [0, T ] × H × Ω → R1 have continuous partial derivatives Ft , Fx , Fxx which are uniformly continuous on bounded subsets of [0, T ] × H independent of ω ∈ Ω and let it be progressively measurable. Then the following formula holds for a.e. ω. F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ · W (t) + 1663

1664

THE ITO FORMULA

Z

t

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) + 0

+

¡ ¢ª 1 ∗ trace Φ (s) Fxx (s, X (s)) Φ (s) Q ds 2

The dependence of F on ω is suppressed. To begin with, here is a lemma which has most of the important ideas. Lemma 54.1.2 Suppose the simpler case that X (t) = X (0) + φ0 t + Φ0 · W (t) where Φ0 is constant in t and takes finitely many values in L (U, H). (Recall from Lemma 53.1.9 the restrictions of such operators to U0 are dense in L2 (U0 , H) .) Then the theorem holds. Proof: Let {t0 , · · · , tm } be a partition of [0, t] with |ti − ti−1 | = t/m. Then F (t, X (t)) − F (0, X (0))

m−1 X

=

(F (tk+1 , X (tk+1 )) − F (tk , X (tk+1 )))

k=0

+

m−1 X

(F (tk , X (tk+1 )) − F (tk , X (tk )))

k=0

which by Taylor’s formula and mean value theorem equals m−1 X

Ft (tk , X (tk+1 )) ∆tk +

k=0

+

m−1 X

¡ ¡ ¢ ¢ Ft tek , X (tk+1 ) − Ft (tk , X (tk+1 )) ∆tk +

k=0

m−1 X k=0

1 Fx (tk , X (tk )) ∆Xk + Fxx (tk , X (tk )) ∆Xk (∆Xk ) + 2

m−1 Xµ k=0

¶ ³ ´ 1 1 f Fxx tk , Xk − Fxx (tk , X (tk )) ∆Xk (∆Xk ) 2 2

(54.1.2)

(54.1.3)

(54.1.4)

fk is between X (tk ) and X (tk+1 ), similarly for tek . By assumptions on F, Here X the expressions in 54.1.4 and the last term in 54.1.2 both converge to 0 as m → ∞. It remains to consider the other terms. The first term of 54.1.2 converges to Z t Ft (s, X (s)) ds (54.1.5) 0

thanks to the continuity of X. First consider the first term of 54.1.3. ∆Xk

=

φ0 (tk+1 − tk ) + Φ0 · W (tk+1 ) − Φ0 · W (tk )

=

φ0 (tk+1 − tk ) + Φ0 W (tk+1 ) − Φ0 W (tk )

(54.1.6)

54.1. THE CASE OF A Q WIENER PROCESS m−1 X

Fx (tk , X (tk )) ∆Xk =

k=0

+

m−1 X

1665

Fx (tk , X (tk )) φ0 (tk+1 − tk )

k=0

m−1 X

Fx (tk , X (tk )) Φ0 W (tk+1 ) − Fx (tk , X (tk )) Φ0 W (tk )

k=0

By continuity of X and Fx , this converges to Z

t

Fx (s, X (s)) φ (s) ds + Fx (·, X) Φ0 · W (t)

(54.1.7)

0

where in this case φ (s) = φ0 , the constant. It remains to consider the remaining term in 54.1.3. From 54.1.6 this term is m−1 X k=0

m−1 X k=0

+

1 Fxx (tk , X (tk )) (φ0 (tk+1 − tk )) (φ0 (tk+1 − tk )) 2

m−1 X k=0

+

m−1 X k=0

+

1 Fxx (tk , X (tk )) ∆Xk (∆Xk ) = 2

m−1 X K=0

1 Fxx (tk , X (tk )) (Φ0 ∆W (tk )) (Φ0 ∆W (tk )) 2

(54.1.8) (54.1.9)

1 Fxx (tk , X (tk )) (φ0 ) (Φ0 ∆W (tk )) (tk+1 − tk ) 2

(54.1.10)

1 Fxx (tk , X (tk )) (Φ0 ∆W (tk )) (φ0 ) (tk+1 − tk ) 2

(54.1.11)

Of course, by continuity of Fxx the two last terms are the same and they each 2 converge to 0 in L2 (Ω) thanks to 54.1.1. Because of the factor (tk+1 − tk ) in 54.1.8, this term also converges to 0 as m → ∞. It remains to consider the expression in 54.1.9. Fxx (tk , X (tk )) ∈ L (H, L (H, R)) so Fxx (tk , X (tk )) (Φ0 ∆W (tk )) ∈ L (H, R) and so recalling Φ∗0 maps H to U,

=

Fxx (tk , X (tk )) (Φ0 ∆W (tk )) (Φ0 ∆W (tk )) (Φ∗0 (Fxx (tk , X (tk )) (Φ0 ∆W (tk ))) , ∆W (tk ))U

=

(Φ∗0 Fxx (tk , X (tk )) Φ0 ∆W (tk ) , ∆W (tk ))U

1666

THE ITO FORMULA

Thus this expression equals m−1 X k=0

1 ∗ (Φ Fxx (tk , X (tk )) Φ0 ∆W (tk ) , ∆W (tk ))U 2 0

To save on notation let ζ k ≡ Φ∗0 Fxx (tk , X (tk )) Φ0 . Thus ζ k is a self adjoint operator defined on U and the above equals m−1 X k=0

1 (ζ ∆W (tk ) , ∆W (tk ))U 2 k

(54.1.12)

I need to find what this converges to. Let {ej } be the orthonormal basis such that Qei = λi ei . Let A ∈ Fk . Z Z (E ((ζ k ∆W (tk ) , ∆W (tk ))U |Fk )) dP = (ζ k ∆W (tk ) , ∆W (tk ))U dP A

A

 z ∞ X

∆W (tk )

}|

 {

 Z     = (∆W (tk ) , ej ) ej , ∆W (tk ) ζ k  dP A  j=1  

Z



= A

=

∞ Z X j=1

Ω

∞ X

 (∆W (tk ) , ej ) (ζ k ej , ∆W (tk )) dP

j=1

XA ((∆W (tk ) , ej ) (ζ k ej , ∆W (tk ))) dP

Now by independence of XA and ∆W (tk ) the above equals ∞ Z X = P (A) ((∆W (tk ) , ej ) (ζ k ej , ∆W (tk ))) dP j=1

Ω

If ζ k were not a random variable but was just a constant operator mapping U to U, this would reduce to P (A)

∞ X

∆tk (ζ k ej , Qej ) =

j=1

P (A) ∆tk

∞ X

(ej , ζ k Qej )

j=1

=

P (A) ∆tk trace (ζ k Q) .

I claim this is still the case even though ζ k is a random variable. The reason for this is you can get a sequence of simple functions, {sn } converging uniformly to ζ k which are of the form pn X cni XAni (ω) i=1

54.1. THE CASE OF A Q WIENER PROCESS

1667

where Ani is measurable with respect to Fk . Thus by independence of the increments in the Wiener process, the functions XAni and ∆W (tk ) are independent. Therefore, P (A)

= lim P (A) n→∞

Ω

∞ Z X

Ã

Ãp n X

∞ Z X j=1

= lim P (A)

(∆W (tk ) , ej )

Ω

Ãp Z n X

j=1

i=1

pn ∞ X X

(tk )

i=1

Ω

dP !

XAni (∆W (tk ) , ej ) (cni ej , ∆W (tk )) dP ! XAni (∆W

(tk ) , ej ) (cni ej , ∆W

(tk )) dP

Z P (Ani )

j=1 i=1

= lim P (A) ∆tk n→∞

=

!! cni XAni ej , ∆W

i=1

∞ X

= lim P (A)

Ãp n X

Ω

j=1

n→∞

n→∞

((∆W (tk ) , ej ) (ζ k ej , ∆W (tk ))) dP

j=1

= lim P (A)

n→∞

∞ Z X

Ω

(∆W (tk ) , ej ) (cni ej , ∆W (tk )) dP

pn ∞ X X

P (Ani ) (Qcni ej , ej )

j=1 i=1

lim P (A) ∆tk

n→∞

∞ Z X

(Qsn ej , ej ) dP

Ω

j=1

∞ Z X ¡ ¢ = ∆tk P (A) Qζ j ej , ej dP j=1

= P (A) ∆tk

∞ X

Ω

(ej , ζ k Qej )

j=1

≡ P (A) ∆tk trace (ζ k Q) =

Z A

∆tk trace (ζ k Q) dP

Now return to 54.1.12, m−1 X k=0

1 (ζ ∆W (tk ) , ∆W (tk ))U 2 k

I have just shown ¶ µ 1 1 E (ζ ∆W (tk ) , ∆W (tk ))U |Ftk = ∆tk trace (ζ k Q) . 2 k 2 Now here is an interesting lemma.

(54.1.13)

1668

THE ITO FORMULA

¡ ¢ Lemma 54.1.3 Suppose η j are random variables E η 2j < ∞, such that η j is measurable with respect to Gk for all k > j where {Gk } is increasing. Then " #2  m−1 X E η k − E (η k |Gk ) 

(54.1.14)

k=0

=E

Ãm−1 X

! 2

η 2k − E (η k |Gk )

k=0

Proof of the lemma: First consider a mixed term i < k. E ((η i − E (η i |Gi )) (η k − E (η k |Gk ))) This equals E (η i η k ) − E (η i E (η k |Gk )) − E (η k E (η i |Gi )) + E (η i |Gi ) E (η k |Gk ) = E (η i η k ) − E (E (η i η k |Gk )) − E (η k E (η i |Gi )) + E (η k E (η i |Gi ) |Gk ) = E (η i η k ) − E (E (η i η k |Gk )) − E (η k E (η i |Gi )) + E (η k E (η i |Gi )) = 0. Thus 54.1.14 equals m−1 X

³ ´ 2 E (η k − E (η k |Gk ))

k=0

which equals m−1 X

³ ´ ¡ ¢ 2 E η 2k − 2E (η k E (η k |Gk )) + E E (η k |Gk )

k=0

=

m−1 X

³ ´ ¡ ¢ 2 E η 2k − 2E (E (η k E (η k |Gk )) |Gk ) + E E (η k |Gk )

k=0

=

m−1 X

³ ´ ¡ ¢ 2 E η 2k − 2E (E (η k |Gk ) E (η k |Gk )) + E E (η k |Gk )

k=0

=

m−1 X

³ ´ ³ ´ ¡ ¢ 2 2 E η 2k − 2E E (η k |Gk ) + E E (η k |Gk )

k=0

=

m−1 X

³ ´ ¡ ¢ 2 E η 2k − E E (η k |Gk )

k=0

and this proves the lemma.


1669

Now from this lemma and 54.1.13, this implies Ã !2  m−1 X 1 1 E (ζ ∆W (tk ) , ∆W (tk ))U − ∆tk trace (ζ k Q)  2 k 2 k=0 Ãm−1 ! X 1 2 2 = E (ζ k ∆W (tk ) , ∆W (tk ))U − (∆tk trace (ζ k Q)) 4 k=0

≤ C

m−1 X

³ ´ 4 E |∆W (tk )| + ∆tk2

k=0

which converges to 0 as m → ∞ by 54.1.1. Thus m−1 X

lim

m→∞

=

1 2

Z

t

k=0

1 Fxx (tk , X (tk )) (Φ0 ∆W (tk )) (Φ0 ∆W (tk )) 2

¡ ¢ ∗ trace Φ (s) Fxx (s, X (s)) Φ (s) Q ds,

0

the convergence taking place in L2 (Ω). This proves the lemma. Now from this lemma, the following is immediate. Lemma 54.1.4 Let Φ, φ be elementary having values in L (U, H) and H respectively. Then the conclusion of Theorem 54.1.1 holds. Proof: You apply what was just shown on each subinterval. Lemma 54.1.5 Now assume Φ is L02 predictable or progressively measurable and φ is H predictable or progressively measurable, and Z T Z T 2 |φ (s)|H ds < C, ||Φ||L0 dt < C a.e. ω (54.1.15) 0

2

0

Also letting

Z X (t) ≡ X (0) +

t

φ (s) ds + Φ · W (t) ,

(54.1.16)

0

assume that for a.e. ω |X (t, ω)|H < C

(54.1.17)

for all t. Then the conclusion of Theorem 54.1.1 holds. Proof: By Proposition 53.2.6 there exists a sequence of predictable elementary functions {Φn } having values in L (U, H) and a sequence of predictable elementary functions {φn } having values in H such that Z 0

T

Z

Z Ω

|φn − φ| dP ds,

T 0

Z 2

Ω

||Φn − Φ||L0 dP ds 2

1670

THE ITO FORMULA

both converge to 0. One can also assume 54.1.15 holds for φn , Φn in place of φ and Φ. By taking a further subsequence if needed, φn (s, ω) → φ (s, ω) , Φn (s, ω) → Φ (s, ω) pointwise a.e. Also let Xn (t) satisfy 54.1.16 with φ and Φ replaced with φn and Φn respectively. I need to verify there exists a subsequence still denoted by {Xn (t)} such that for a.e. ω this sequence converges uniformly to X (t) in H on [0, T ]. Recall from Theorem 53.2.8, both {Φn · W (t)} and {Φ · W (t)} are martingales. Therefore, from the maximal estimate of Theorem 50.3.5, and the Ito isometry of Theorem 53.2.8 Ã" #! 1 P sup |Φn · W (t) − Φ · W (t)| > α ≤ E (|(Φn − Φ) · W (T )|) α t∈[0,T ] Z Z ´ 1 ³ 1 T 2 2 E |(Φn − Φ) · W (T )| = ||Φn − Φ||L0 dP dt. 2 α α 0 Ω It follows there exists a subsequence, still denoted by n such that #! Ã" ≤

sup |Φn · W (t) − Φ · W (t)| > 2−n

P

(54.1.18)

≤ 2−n .

t∈[0,T ]

(For α = 2−k in 54.1.18, you let nk be so large that the right side is less than 2−k for every n ≥ nk . Then denote nk by k to save notation.) By the Borel Cantelli lemma, Lemma 47.1.2, the collection of ω which are in infinitely many of the sets # " sup |Φn · W (t) − Φ · W (t)| > 2−n t∈[0,T ]

has measure 0. Therefore, {Φn · W (t)} is uniformly Cauchy with values in H for all ω not in a set of measure 0 and it converges to Φ · W (t) for these ω. Therefore, for these ω it can be assumed |Φn · W (t)| < D

(54.1.19)

for some constant D. This is because by 54.1.15 and 54.1.17, there is a set of measure zero such that for ω not in this set, |Φ · W (t)| < D and now the uniform convergence implies for all n large enough, 54.1.19. Similarly, Ã" #! ¯Z t ¯ ¯ ¯ P sup ¯¯ (φ (s) − φn (s)) ds¯¯ > α t∈[0,T ] 0 H Ã"Z #! T

≤ P

|φ (s) − φn (s)|H ds > α

0

2−k

< 2−k

showing that off a possibly larger set of measure 0 it follows that ¾ ½Z t φnk (s) ds

(54.1.20)

0

¯ Rt RT ¯ converges uniformly on [0, T ] to 0 φ (s) ds and that 0 ¯φnk − φ¯H ds also converges to 0. Thus there exists a set of measure zero such that for ω not in this set, there is a constant D such that for all k large enough ¯Z t ¯ ¯ ¯ ¯ φnk (s) ds¯¯ < D ¯ 0

for all t. Similarly, one can obtain a set of measure 0 and a further subsequence such that off this set of measure zero, Z T 2 ||Φnk − Φ||L0 ds (54.1.21) 2

0

converges to 0 and

Z 0

T

2

||Φ||L0 ds < ∞. 2

Letting

Z

t

Xn (t) ≡ X (0) + 0

φn (s) ds + Φn · W (t) ,

it follows from the above observations, {Xnk (t, ω)} converges uniformly to X (t, ω) on [0, T ] for ω off a set of measure 0 and in fact there is some constant such that for ω not in this exceptional set, |Xn (t, ω)| < D

(54.1.22)

for all t ∈ T . For simplicity, refer to this sequence as {Xn (t)} . From Lemma 54.1.4, the Ito formula holds for each Xn . That is F (t, Xn (t)) = F (0, X (0)) + Fx (·, Xn (·)) Φn · W (t) + Z

t

0

+

{Ft (s, Xn (s)) + Fx (s, Xn (s)) (φn (s)) +

¡ ¢ª 1 ∗ trace Φn (s) Fxx (s, Xn (s)) Φn (s) Q ds 2

(54.1.23)

1672

THE ITO FORMULA

The uniform convergence of Xn to X on [0, T ] and the dominated convergence theorem along with the fact Fx is uniformly continuous on bounded subsets of [0, T ] × H independent of ω ∈ Ω implies with 54.1.22 that Z

Z

T

2

lim

n→∞

0

Ω

||Fx (·, Xn (·)) Φn − Fx (·, X (·)) Φ||L0 dP dt = 0 2

Therefore, the Ito isometry implies Fx (·, Xn (·)) Φn · W (t) → Fx (·, X (·)) Φ · W (t) in L2 (Ω). It follows then from the usual use of the maximal inequality and the fact these Ito integrals are martingales there exists a set of measure zero such that for ω not in this set, the above convergence is actually uniform in t ∈ [0, T ]. It remains to consider the other terms in 54.1.23. The limit Z t Ft (s, Xn (s)) + Fx (s, Xn (s)) (φn (s)) ds 0

Z

t

→

Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) ds a.e. ω 0

is also routine from the uniform convergence of the various functions above and the boundedness of the Xn (t, ω) for ω not in a set of measure zero. It only remains to consider the term involving Fxx . Z t ¡ ¢ 1 ∗ trace Φn (s) Fxx (s, Xn (s)) Φn (s) Q ds (54.1.24) 0 2 and show it converges to Z t ¡ ¢ 1 ∗ trace Φ (s) Fxx (s, X (s)) Φ (s) Q ds. 2 0 From the definition of what is meant by trace, ¡ ¢ ∗ trace Φn (s) Fxx (s, Xn (s)) Φn (s) Q =

X

(Fxx (s, Xn (s)) Φn (s) Qek , Φn (s) ek )

k

=

X

λk (Fxx (s, Xn (s)) Φn (s) ek , Φn (s) ek )

k

Similarly, ¡ ¢ ∗ trace Φ (s) Fxx (s, X (s)) Φ (s) Q X λk (Fxx (s, X (s)) Φ (s) ek , Φ (s) ek ) = k

(54.1.25)


1673

Since Xn is bounded there exists a constant M from Fxx such that ¯ ¡ ¢ ¡ ¢¯ ¯trace Φn (s)∗ Fxx (s, Xn (s)) Φn (s) Q − trace Φ (s)∗ Fxx (s, X (s)) Φ (s) Q ¯ is dominated by an expression of the form X

=

k

+

λk M ||Φn − Φ|| ||Φn ||

X

λk ||Fxx (s, Xn (s)) − Fxx (s, X (s))|| ||Φn || ||Φ||

k

It follows the norm of the difference of the integrals of these terms is dominated by an expression of the form X

Z

||Φn − Φ|| ||Φn || ds 0

k

+

Z

X

t

λk M

t

λk

||Fxx (s, Xn (s)) − Fxx (s, X (s))|| ||Φn || ||Φ|| ds 0

k

From Holder’s inequality the first term is dominated by an expression of the form Ã X

! µZ

¶1/2 µZ

t

λk M

2

||Φn − Φ|| dt 0

k

t

¶1/2 ||Φn || dt 2

0

and by 54.1.21 this converges to 0. Consider the second term. Z

t

||Fxx (s, Xn (s)) − Fxx (s, X (s))|| ||Φn || ||Φ|| ds 0

µZ ≤

t

¶1/2 µZ t ¶1/2 2 2 (||Fxx (s, Xn (s)) − Fxx (s, X (s))|| ||Φ||) ds · ||Φn || ds

0

0

The first term in the product converges to 0 by the uniform convergence of Xn (s) to X (s) which implies ||Fxx (s, Xn (s)) − Fxx (s, X (s))|| → 0 uniformly. Then this term converges to zero by the dominated convergence theorem and the boundedness of Fxx (s, Xn (s)) which follows from the boundedness of Xn . The second term in the product is bounded by 54.1.21 which says that Z

T

lim

n→∞

2

||Φn − Φ|| ds = 0. 0

Therefore, one can pass to the limit in 54.1.23 and obtain the Ito formula, F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ · W (t) +

1674

THE ITO FORMULA

Z

t

+

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) 0

+


(54.1.26)

This proves the lemma. Now with these lemmas, it is easy to prove Theorem 54.1.1, the Ito formula. By now, it is likely one has forgotten what this theorem said. Therefore, here it is stated again. Theorem 54.1.6 Let Φ be an L02 valued predictable or progressively measurable process which is stochastically integrable in [0, T ] because Ã"Z #! T

P

2

||Φ|| dt < ∞

=1

0

and let φ : [0, T ] × Ω → H be predictable or progressively measurable process which is Bochner integrable on [0, T ] and let X (0) be F0 measurable and H valued. Let Z

t

X (t) ≡ X (0) +

φ (s) ds + Φ · W (t) . 0

Let F : [0, T ] × H × Ω → R1 have continuous partial derivatives Ft , Fx , Fxx which are uniformly continuous on bounded subsets of [0, T ] × H independent of ω ∈ Ω and let it be progressively measurable. Then the following formula holds for a.e. ω. F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ · W (t) + Z

t

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) + 0

+


The dependence of F on ω is suppressed. Proof: Define a stopping time, ½ ¾ Z t Z t 2 τ (ω) = inf t ≥ 0 : |X (t, ω)|H > C or ||Φ|| ds > C or |φ| ds > C 0

0

Then by the localization lemma, Lemma 53.4.4, Z t∧τ X (t ∧ τ ) = X (0) + φ (s) ds + Φ · W (t ∧ τ ) 0 Z t = X (0) + X[0,τ ] (s) φ (s) ds + X[0,τ ] Φ · W (t) . 0

54.2. THE CASE WHERE Q IS NOT TRACE CLASS Thus

Z Z Ω

and

T

0

Z Z Ω

0

T

¯ ¯ ¯X[0,τ ] φ (s)¯ dsdP = ¯¯ ¯¯ ¯¯X[0,τ ] Φ¯¯2 0 dsdP = L 2

Z Z

1675

τ

|φ (s)| dsdP ≤ CP (Ω) = C Ω

0

Z Z Ω

τ

0

2

||Φ||L0 dsdP ≤ CP (Ω) = C 2

In addition, |X (t ∧ τ )|H ≤ C and so, letting XC (t) ≡ ΦC (t) =

X (t ∧ τ ) , φC (t) ≡ X[0,τ ] φ (t) , X[0,τ ] Φ (t) ,

These functions satisfy the conditions of Lemma 54.1.5 and so since X (0) = XC (0) , F (t, XC (t)) = F (0, X (0)) + Fx (·, XC (·)) ΦC · W (t) + Z

t 0

+

{Ft (s, XC (s)) + Fx (s, XC (s)) (φC (s)) +

¡ ¢ª 1 ∗ trace ΦC (s) Fxx (s, XC (s)) ΦC (s) Q ds 2

(54.1.27)

Now by assumption, there is a set of measure zero such that for ω not in this set, Z

T

Z

T

2

||Φ|| ds < ∞ and

|φ| ds < ∞

0

0

Thus for fixed ω not in this set of measure zero, it follows that for large enough C, Z

T

Z

T

2

||Φ|| ds < C and 0

|φ| ds < C. 0

Also for these ω, t → X (t, ω) is continuous and so eventually, for large enough C, |X (t, ω)|H < C for all t ∈ [0, T ] . Thus letting C → ∞, the expression in 54.1.27 implies the desired formula. This proves the theorem.

54.2

The Case Where Q Is Not Trace Class

In this case Q ∈ L (U, U ) and Q is self adjoint and nonnegative. Then let U0 ≡ Q1/2 U and J : U0 → U1 is a Hilbert Schmidt operator. Then W (t) is a Q1 Wiener process converging in U1 where Q1 ≡ JJ ∗ and Q1 is trace class. Suppose as above Z t Z t X (t) = X (0) + φ (s) ds + Φ (s) dW (s) 0

0

1676

THE ITO FORMULA

where.

Ã"Z

T

P 0

#! 2 ||Φ||L2 (U0 ,H)

dt < ∞

=1

The definition of the last integral is Z t Φ (s) ◦ J −1 dW (s) 0

What does the Ito formula say in this setting? According to Theorem 54.1.1 under the conditions stated there for F, F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ ◦ J −1 · W (t) + Z

t

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) + 0

+

³¡ ó ¢∗ 1 trace Φ (s) ◦ J −1 Fxx (s, X (s)) Φ (s) ◦ J −1 Q1 ds 2

By definition of the integral, this is the same as F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ · W (t) + Z

t

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) + 0

+

³¡ ó ¢∗ 1 trace Φ (s) ◦ J −1 Fxx (s, X (s)) Φ (s) ◦ J −1 Q1 ds 2

Consider that messy term at the end. Letting Q1 fk = λk fk for {fk } orthonormal in U1 , it equals ∞

¢ 1X ¡ λk Fxx (s, X (s)) Φ (s) ◦ J −1 fk , Φ (s) ◦ J −1 fk 2 = =

1 2 1 2

k=1 ∞ ³ X

Fxx (s, X (s)) Φ (s) ◦ J −1

k=1 ∞ ³ X

p

λk fk , Φ (s) ◦ J −1

1/2

p

´ λk f k

1/2

Fxx (s, X (s)) Φ (s) ◦ J −1 Q1 fk , Φ (s) ◦ J −1 Q1 fk

´

k=1

Consider Fxx . This is in L (H, L (H, R)) . Suppose it is just multiplication by a scalar function of s called f (s). Then the above reduces right away to ´ f (s) X ³ 1/2 1/2 Φ (s) ◦ J −1 Q1 fk , Φ (s) ◦ J −1 Q1 fk 2 ∞

k=1

=

¯¯2 f (s) ¯¯¯¯ f (s) 2 ||Φ (s)||L2 (U0 ,H) Φ (s) ◦ J −1 ¯¯L2 ³Q1/2 U1 ,H ´ = 1 2 2


1677

by Lemma 53.7.3. In particular if one considers the case where Q = I, then this term reduces to f (s) 2 ||Φ (s)||L2 (U,H) 2 which is relatively pleasant to behold since it does not involve U0 or any other noxious and intricate Banach space. This is summarized in the following theorem. Theorem 54.2.1 Let F : [0, T ] × H × Ω → R be progresively measurable and have continuous partial derivatives Ft , Fx , Fxx which are uniformly continuous on bounded subsets of [0, T ]×H independent of ω ∈ Ω. Also suppose Φ is an L2 (U0 , H) predictable or progressively measurable process which is stochastically integrable in [0, T ] , Ã"Z #!2 T P ||Φ|| dt < ∞ dt = 1, 0

where U0 ≡ Q1/2 U for Q ∈ L (U, U ) a continuous self adjoint operator (Qx, x)U ≥ 0. Also let φ be an H valued predictable or progressively measurable process which is Bochner integrable on [0, T ] and let X (0) be F0 measurable and H valued. Let Z t Z t X (t) ≡ X (0) + φ (s) ds + Φ (s) dW (s) 0

0

where W (t) is a Q1 Wiener process defined on U1 where J : U0 → U1 is Hilbert Schmidt and Q1 ≡ JJ ∗ is trace class. Then one has the following Ito formula. Z t F (t, X (t)) = F (0, X (0)) + Fx (s, X (s)) Φ (s) dW (s) + 0

Z

t

[Ft (s, X (s)) + Fx (s, X (s)) φ (s) + 0

³¡ ´¸ ¢∗ 1 trace Φ (s) ◦ J −1 Fxx (s, X (s)) Φ (s) ◦ J −1 Q1 ds 2 The stochastic integral on the right is defined as Z t Fx (s, X (s)) Φ (s) ◦ J −1 dW (s) . 0

(The most important observation is that it is a Martingale or submartingale.) In addition to this, in the case where Fxx (s, x) reduces to multiplication by a scalar function of s, the above formula simplifies to Z t F (t, X (t)) = F (0, X (0)) + Fx (s, X (s)) Φ (s) dW (s) + 0

¸ Z t· f (s) 2 ||Φ (s)||L2 (U0 ,H) ds Ft (s, X (s)) + Fx (s, X (s)) φ (s) + 2 0

1678

THE ITO FORMULA

In the special case where Q = I the identity, it simplifies even more to an expression of the form Z t F (t, X (t)) = F (0, X (0)) + Fx (s, X (s)) Φ (s) dW (s) + 0

Z t· Ft (s, X (s)) + Fx (s, X (s)) φ (s) + 0

¸ f (s) 2 ||Φ (s)||L2 (U,H) ds. 2

Stochastic Differential Equations 55.1

Theorems Based On Contraction Mapping

55.1.1

Predictable And Stochastic Continuity

First of all, here is an important observation. Proposition 55.1.1 Let X (t) be a stochastic process having values in E a complete metric space and let it be Ft adapted and left continuous where Ft is a normal filtration. Then it is predictable. If t → X (t, ω) is continuous for all ω ∈ / N, P (N ) = 0, then (t, ω) → X (t, ω) XN C (ω) is predictable. Also, if X (t) is stochastically continuous and adapted on [0, T ] , then it has a predictable version. If X ∈ C ([0, T ] ; Lp (Ω; F )) , p ≥ 1 for F a Banach space, then X is stochastically continuous. Proof: First suppose X is continuous for all ω ∈ Ω. Define Im,k ≡ ((k − 1) 2−m T, k2−m T ] if k ≥ 1 and Im,0 = {0} if k = 1. Then define m

Xm (t) ≡

2 X

¡ ¢ X T (k − 1) 2−m X((k−1)2−m T,k2−m T ] (t)

k=1

+X (0) X[0,0] (t) Here the sum means that Xm (t) has value X (T (k − 1) 2−m ) on the interval ((k − 1) 2−m T, k2−m T ]. Thus Xm is predictable because each term in the formal sum is. Thus ¡ ¢ ¢−1 m ¡ −1 Xm (U ) = ∪2k=1 X T (k − 1) 2−m X((k−1)2−m T,k2−m T ] (U ) 1679

1680

STOCHASTIC DIFFERENTIAL EQUATIONS

¡ ¡ ¢¢−1 m = ∪2k=1 ((k − 1) 2−m T, k2−m T ] × X T (k − 1) 2−m (U ) , a finite union of predictable sets. Since X is left continuous, X (t, ω) = lim Xm (t, ω) m→∞

and so X is predictable. Now suppose that for ω ∈ / N , P (N ) = 0, t → X (t, ω) is continuous. Then applying the above argument to X (t) XN C it follows X (t) XN C is predictable by completeness of Ft , X (t) XN C is Ft measurable. Next consider the other claim. Since X is stochastically continuous on [0, T ] it is uniformly stochastically continuous on this interval by Lemma 50.1.1. Therefore, there exists a sequence of partitions of [0, T ] , the mth being 0 = tm,0 < tm,1 < · · · < tm,n(m) = T such that for Xm defined as above, then for each t ¡£ ¤¢ P d (Xm (t) , X (t)) ≥ 2−m ≤ 2−m

(55.1.1)

Then as above, Xm is predictable. Let A denote those points of PT at which Xm (t, ω) converges. Thus A is a predictable set because it is just the set where Xm (t, ω) is a Cauchy sequence. Now define the predictable function Y ½ limm→∞ Xm (t, ω) if (t, ω) ∈ A Y (t, ω) ≡ 0 if (t, ω) ∈ /A From 55.1.1 it follows from the Borel Cantelli lemma that for fixed t, the set of ω which are in infinitely many of the sets, £ ¤ d (Xm (t) , X (t)) ≥ 2−m has measure zero. Therefore, for each t, there exists a set of measure zero, N (t) such that for ω ∈ / N (t) and all m large enough £ ¤ d (Xm (t, ω) , X (t, ω)) < 2−m Hence for ω ∈ / N (t) , (t, ω) ∈ A and so Xm (t, ω) → Y (t, ω) which shows d (Y (t, ω) , X (t, ω)) = 0 if ω ∈ / N (t) . The predictable version of X (t) is Y (t). Finally consider the claim about the specific example where X ∈ C ([0, T ] ; Lp (Ω; F )) . Z p P ([||X (t) − X (s)||F ≥ ε]) εp ≤ ||X (t) − X (s)||F dP ≤ εp δ Ω

provided |s − t| sufficiently small. Thus P ([||X (t) − X (s)||F ≥ ε]) < δ when |s − t| is small enough. This proves the proposition.

55.1. THEOREMS BASED ON CONTRACTION MAPPING

1681

Definition 55.1.2 Let U, H be real separable Hilbert spaces and let Q be a self adjoint nonnegative operator and U0 ≡ Q1/2 U Let σ : [0, T ] × H × Ω → L2 (U0 , H)

(55.1.2)

||σ (t, x2 , ω) − σ (t, x1 , ω)||L2 (U0 ,H) ≤ K |x1 − x2 |H

(55.1.3)

(s, ω) → ||σ (s, 0, ω)||L2 (U0 ,H) ∈ L2 ([0, T ] × Ω)

(55.1.4)

σ is progressively measurable

(55.1.5)

satisfy

Recall from Proposition 53.3.6 this means that the restriction of σ to [0, t] × H × Ω is B ([0, t]) × B (H) × Ft measurable. Let Ft be a normal filtration corresponding to a Wiener process W ¡ (t) defined on U ¢ 1 where J : U0 → U1 is a Hilbert Schmidt operator. Also let C [0, T ] ; L2 (Ω, U ) denote the space of continuous functions defined on in L2 (Ω, H) . Then let VF denote those functions ¡ [0, T ] 2having values ¢ X in C [0, T ] ; L (Ω, H) with the property that X (t) is Ft measurable. (X (t) is adapted.) I will often suppress the dependence of σ on ω in what follows. Lemma 55.1.3 Let σ be given in the above definition and let X ∈ VF . Then (t, ω) → σ (t, X1 (t, ω) , ω) is progressively measurable where X1 is a predictable version of X and Z

T

Z 2

0

Ω

||σ (t, X1 (t, ω) , ω)||L2 (U0 ,H) dP dt < ∞

Proof: Since X ∈ VF it is continuous and Ft adapted, it has a predictable version X1 . Therefore, by Proposition 53.3.6 the first claim follows right away. Next I need to verify the claim about the integral. However, Z 0

Z

T

Z ³

≤2 0

Ω

T

Z 2

Ω

||σ (t, X1 (t, ω))||L2 (U0 ,H) dP dt

´ 2 2 ||σ (t, X1 (t, ω)) − σ (t, 0)||L2 (U0 ,H) + ||σ (t, 0)||L2 (U0 ,H) dP dt

Z ≤ 2K 2 0

T

Z

Z

T

2

Ω

||X1 (t, ω)||H dP dt + 2


0

2

||σ (t, 0)||L2 (U0 ,H) dt < ∞.

1682


It follows from this lemma that for σ, VF given in Definition 55.1.2 and X ∈ VF , one can define the stochastic integral Z t σ (s, X1 (s)) dW (s) 0

The following will be assumed on b (t, x, ω) . |b (t, x2 , ω) − b (t, x1 , ω)|H ≤ K |x1 − x2 |H

(55.1.6)

(s, ω) → |b (s, 0, ω)|H ∈ L2 ([0, T ] × Ω)

(55.1.7)

b : [0, T ] × H × Ω → H is progressively measurable.

(55.1.8)

Now here is a useful function space described in the following lemma. ¡ ¢ Lemma 55.1.4 Let C [0, T ] ; L2 (Ω; H) denote the space of continuous functions ¡ ¢ 2 2 with values in L (Ω; H) and for each λ ≥ 0, denote for X ∈ C [0, T ] ; L (Ω; H) n o ||X||λ ≡ max e−λt ||X (t)||L2 (Ω;H) : t ∈ [0, T ] ¡ ¡ ¢ ¢ 2 Then all these norms are equivalent and C ¡ [0, T ] ;2L (Ω; H) ¢ , ||·||λ is a Banach space. If VF denotes those functions in C [0, T ] ; L (Ω; H) which are adapted to Ft , a normal filtration, then VF is a closed subspace of ¡ ¢ C [0, T ] ; L2 (Ω; H) . Proof: The assertion about the norms and the Banach space are all obvious. ∞ It only remains to consider the last claim about VF . Suppose {Xn }n=1 ⊆ VF and ||Xn − Y ||λ → 0. I need to verify Y is Ft adapted. However, taking a subsequence, still called n and letting λ = 0 for convenience, it can be assumed ||Xn − Y ||0 < 2−n and so it follows for each t ∈ [0, T ] , Xn (t) (ω) → Y (t) (ω) a.e. ω. It follows since (Ω, Ft , P ) is complete, Y (t) is Ft measurable for each t. This proves the lemma. Lemma 55.1.5 Let X1 be a predictable version of X ∈ VF as above and let b be as above. Then there exists a set of measure 0 N such that for ω ∈ / N, Z t t→ b (s, X1 (s, ω) , ω) ds 0

is continuous with values in H and adapted. The above stochastic process Z t F (t) (ω) ≡ b (s, X1 (s, ω)) ds 0

is Ft adapted and Thus F ∈ VF .

¡ ¢ F ∈ C [0, T ] ; L2 (Ω; H) .

55.1. THEOREMS BASED ON CONTRACTION MAPPING

1683

Proof: First note that (s, ω) → b (s, X1 (s, ω) , ω) is progressively measurable because X1 is and b is progressively measurable which implies this follows from Proposition 53.3.6. That F (t) given above is adapted follows easily from Proposition ?? and Pettis theorem applied ¡ to positive and ¢negative parts of (b (s, X1 (s, ω)) , h)H . It only remains to verify F ∈ C [0, T ] ; L2 (Ω; H) . Let s < t. ¯2 Z ¯Z t ¯ ¯ ¯ ¯ dP b (r, X (r, ω) , ω) dr 1 ¯ ¯

2

||F (t) − F (s)||L2 (Ω;H) = Z Z

Ω

s

H

t

≤ (t − s)

2

|b (r, X1 (r, ω) , ω)| drdP Ω s T Z

Z

2

≤ (t − s)

|b (r, X1 (r, ω) , ω)| dP dr 0

(55.1.9)

Ω

and from the aboe assumptions on b, Z

T

Z

Z 2

|b (r, X1 (r, ω) , ω)| dP dr 0

Ω

Z

T

2

a n→∞

n→∞

which is a contradiction. If lim inf n→∞ φ (xn ) = ∞, then ∞ > a = lim inf an = ∞ n→∞

another contradiction. Therefore, epi (φ) is closed and so φ is lower semicontinuous as claimed. Therefore, ¡ ¢ −1 X = Y \ ∩∞ ((n, ∞)) n=1 φ and since φ is lower semicontinuous, each φ−1 ((n, ∞)) is open. Hence X is a Borel subset of Y . This proves the proposition.

55.3

An Ito Formula

This Ito formula seems to be the fundamental idea which allows one to obtain solutions to stochastic partial differential equations using a variational point of view.

1690


I am following the treatment found in [55]. The following lemma is fundamental to the presentation. It approximates a function with a sequence of two step functions Xkr , Xkl where Xkr has the value of X at the right end of each interval and Xkl gives the value X at the left end of the interval. The lemma is very interesting for its own sake. You can obviously do this sort of thing for a continuous function but here the function is not continuous and in addition, it is a stochastic process depending on ω also. Lemma 55.3.1 Let X : [0, T ] × Ω → V, be B ([0, T ]) × F measurable and suppose X ∈ K ≡ Lp ([0, T ] × Ω; V ) . Then there exists a sequence of nested partitions, Pk ⊆ Pk+1 , © ª Pk ≡ tk0 , · · · , tkmk such that the step functions given by Xkr

(t) ≡

mk X

¡ ¢ X tkj X[tkj−1 ,tkj ) (t)

j=1

Xkl (t) ≡

mk X

¡ ¢ X tkj−1 X[tkj−1 ,tkj ) (t)

j=1

both converge to X in K as k → ∞ and ¯ ©¯ ª lim max ¯tkj − tkj+1 ¯ : j ∈ {0, · · · , mk } = 0 k→∞

Proof: For t ∈ R let γ n (t) ≡ k/2n and let δ n (t) ≡ (k + 1) /2n where t ∈ C [k/2n , (k + 1) /2n ). Also suppose X is defined to equal 0 on [0, T ] × Ω. There exists a set of measure zero N such that for ω ∈ / N, t → X (t, ω) is in Lp (R). Therefore by continuity of translation, as n → ∞ it follows that for ω ∈ / N, and t ∈ [0, T ] , Z p

R

||X (γ n (t) + s) − X (t + s)||V ds → 0

The above is dominated by Z p p 2p−1 (||X (s)|| + ||X (s)|| ) X[−2T,2T ] (s) ds < ∞ R

Consider

Z Z Ω

0

T

µZ R

||X (γ n (t) + s) − X (t +

p s)||V

¶ ds dtdP

By the dominated convergence theorem, this converges to 0 as n → ∞. Now Fubini. This yields ¶ Z Z T µZ p ||X (γ n (t) + s) − X (t + s)||V ds dtdP Ω

0

R

55.3. AN ITO FORMULA

1691

Next change the variable in the integrand. ¶ Z Z T µZ p = ||X (γ n (t − s) + s) − X (t)||V ds dtdP Ω

0

R

which converges to 0 as n → ∞. Therefore, Z Z ÃZ T

Ω

T

0

||X (γ n (t − s) + s) − X

0

! p (t)||V

ds dtdP → 0

also. Now interchange the order of integration again to obtain ! Z ÃZ Z T

T

0

Ω

p

||X (γ n (t − s) + s) − X (t)||V dtdP

0

ds → 0

At this point note that t → X (γ n (t − s) + s) is a step function of the form Xnl (t) for each s ∈ [0, T ] and t → X (δ n (t − s) + s) is a step function of the form Xnr (t). This is because γ n (t − s) + s ≤ t and for fixed s, γ n (t − s) + s is constant on small intervals of the form [a, b) where b − a = 2−n while δ n (t − s) + s > t and is similarly constant on small intervals. Also γ n (t − s) + s ≥ −2−n for every t ∈ [0, T ] . It follows the function Z Z

T

s→ Ω

p

||X (γ n (t − s) + s) − X (t)||V dtdP

0

(55.3.11)

converges to 0 in L1 ([0, T ]) . Similarly, the function Z Z

T

s→ Ω

0

p

||X (δ n (t − s) + s) − X (t)||V dtdP

converges to 0 in L1 ([0, T ]) . The problem here is that γ n (t − s) + s could be negative. Therefore, replace it + with (γ n (t − s) + s) . This will not affect the convergence of the integral to 0 in 55.3.11 as n → ∞ because the modification takes place at most on an interval of length 2−n provided X (0) ∈ Lp (Ω; V ). If this is not the case, simply redefine it at 0 to equal 0. Therefore, there exists a subsequence nk such that Z Z Ω

T 0

¯¯p ¯¯ ³ ¡ ¢+ ´ ¯¯ ¯¯ p − X (t)¯¯ + ||X (δ nk (t − s) + s) − X (t)||V dtdP ¯¯X γ nk (t − s) + s V

converges to 0 for a.e. s ∈ [0, T ] where X (0) is redefined to equal 0 if it happens that X (0) ∈ / Lp (Ω; V ) . Pick such an s. Let ³¡ ¢+ ´ Xkl (t) ≡ X γ nk (t − s) + s , Xkr (t) ≡ X (δ nk (t − s) + s) . This proves the lemma.

1692


Remark 55.3.2 The mesh points other than 0 of each of the k th partitions can be assumed to not be in some given set of measure zero N . The variable s must avoid a set of measure 0 already. If you have another set of measure zero simply adjust s. When you slide s to the right or to the left, you are just moving the mesh points, except for the first one which is always 0. For each n there exists a closed interval [an , bn ] such that for s in this interval, the mesh points for Pn avoid the set of measure zero n. You can also have these intervals be nested. Therefore, there exists a value of s such that the partition points of the step functions determined as above avoid the set of measure zero for all n. The interesting case is the set of measure zero where X is a particular measurable representative of X. The set to avoid would be those values of t > 0 where X (t) is not equal to X (t) a.e. In the case where it is also the case that X ∈ L2 ([0, T ] × Ω; H) it follows that s can be chosen such that these step functions also converge in this space in addition to K. Whenever convenient, this will be assumed. Now consider the following situation. Situation 55.3.3 Let X satisfy the following. Z t Z t X (t) = X0 + Y (s) ds + Z (s) dW (s) , 0

(55.3.12)

0

X0 ∈ L2 (Ω; H) and is F0 measurable, where Z is L2 (U, H) predictable and Z TZ 2 ||Z (s)||L2 (U,H) dP dt < ∞ 0

Ω

so that the stochastic integral can be defined. Also X has a measurable representative X (For a.e. t, X (t) = X (t) for P a.e. ω) such that X ∈ L2 ([0, T ] × Ω, B ([0, T ]) × F, H) ∩ Lp ([0, T ] × Ω, B ([0, T ]) × F, V ) so for each t, X (t) ∈ Lp (Ω; V ) ∩ L2 (Ω; H) . Also W (t) is a Wiener process on 0 another Hilbert space U1 and Q = I. Suppose Y ∈ K 0 = Lp ([0, T ] × Ω; V 0 ) where 0 0 1/p + 1/p = 1 and Y is V predictable. Thus X is continuous into V 0 although it has values in V and it is also V 0 predictable thanks to Propositon 55.1.1. The stochastic integral is defined as Z t Z (s) ◦ J −1 dW (s) 0

where J ∈ L2 (U, U1 ) and W (t) is a Wiener process with values in U1 . Also Q : U → U is just the identity. Recall also that by Lemma 53.7.3 ¯¯ ¯¯ ||Φ||L2 (U,H) = ¯¯Φ ◦ J −1 ¯¯L ³Q1/2 U ,H ´ 2

1

1

where Q1 = JJ ∗ and the last norm is the one which occurs in the usual Ito isometry. The goal is to prove the following Ito formula. Z t³ ´ ® 2 2 2 |X (t)| = |X0 | + 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0


1693

Z

t

+2

R

³¡ ´ ¢∗ Z (s) ◦ J −1 X (s) dW (s)

(55.3.13)

0

where R is the Riesz map which takes U1 to U10 . The main thing is that the last term above be a local martingale. l

Lemma 55.3.4 Let X be as in Situation 55.3.3 and let X k be as in Lemma 55.3.1 l corresponding to the measurable representative X above. Then X k is predictable. l In determining X k , the mesh points always are chosen such that for a > 0 one of them X (a) = X (a) a.e. Note a > 0 Proof: This is a step function and a typical term is of the form X (a) X[a,b) (t) . I will try and show this is predictable and then it will follow the given step function is predictable. Let an be an increasing sequence converging to a and let bn be an increasing sequence converging to b. Then for a.e. ω, X (an ) X(an ,bn ] (t) → X (a) X[a,b) (t) in V 0 due to the fact that t → X (t) is continuous into V 0 for a.e. ω. Therefore, letting v ∈ V be given it follows that for a.e. ω ® ® X (an ) X(an ,bn ] (t) , v → X (a) X[a,b) (t) , v and since the filtration is a normal filtration in which (P, Ft ) is complete, this shows ® (t, ω) → X (a) X[a,b) (t) , v is real predictable because it is the pointwise limit of real predictable functions, those in the sequence being real predictable because of the continuity of X (t) into V 0 and Propositon 55.1.1. Now since H ⊆ V 0 it follows that for all v ∈ V, ¡ ¢ (t, ω) → X (a) X[a,b) (t) , v is real predictable. Now this holds for h ∈ H replacing v in the above because V is dense in H. By the Pettis theorem, this proves the lemma. Lemma R55.3.5 In Situation 55.3.3 the following formula holds for 0 < s < t where t M (t) ≡ 0 Z (u) dW (u). In this formula it is assumed s, t are such that each of X (s) = X (s) , X (t) = X (t) so these are each in Lp (Ω; V ) ∩ L2 (Ω; H) . |·| denotes the norm in H and h·, ·i denotes the duality pairing between V, V 0 . Z t 2 2 |X (t)| = |X (s)| + 2 hY (u) , X (t)i du + 2 (X (s) , M (t) − M (s)) s 2

+ |M (t) − M (s)| − |X (t) − X (s) − (M (t) − M (s))|

2

(55.3.14)

Also for t > 0 Z 2

2

|X (t)| = |X0 | + 2

t

hY (u) , X (t)i du + 2 (X0 , M (t)) 0 2

2

+ |M (t)| − |X (t) − X0 − M (t)|

(55.3.15)

1694


Proof: For a.e. t, X (t) = X (t) a.e. Assume s, t are not in the exceptional set just described. The formula is a straight forward computation which holds a.e. ω. 2

2

|M (t) − M (s)| − |X (t) − X (s) − (M (t) − M (s))| + 2 (X (s) , M (t) − M (s)) 2

2

2

= |M (t) − M (s)| − |X (t) − X (s)| − |M (t) − M (s)| +2 (X (t) − X (s) , M (t) − M (s)) + 2 (X (s) , M (t) − M (s)) 2

= − |X (t) − X (s)| + 2 (X (t) , M (t) − M (s)) ¿Z t À 2 = − |X (t) − X (s)| + 2 (X (t) , X (t) − X (s)) − 2 Y (u) du, X (t) s

=

2

2

2

− |X (t)| − |X (s)| + 2 (X (t) , X (s)) + 2 |X (t)| − 2 (X (t) , X (s)) Z t −2 hY (u) , X (t)i du s

Z 2

2

t

= |X (t)| − |X (s)| − 2

hY (u) , X (t)i du s

Now it is time to prove the other assertion. 2

2

|M (t)| − |X (t) − X0 − M (t)| + 2 (X0 , M (t)) 2

= − |X (t) − X0 | + 2 (X (t) − X0 , M (t)) + 2 (X0 , M (t)) 2

= − |X (t) − X0 | + 2 (X (t) , M (t)) ¿Z = − |X (t) − X0 | + 2 (X (t) , X (t) − X0 ) − 2 ¿Z t À 2 2 = |X (t)| − |X0 | − 2 hY (s) , X (t)i ds

À

t

2

Y (s) ds, X (t) 0

0

This proves the lemma. Noting that X (0) = X0 the first formula works in both cases. Lemma 55.3.6 In the Situation 55.3.3, Ã E

sup |X (t)|H

t∈[0,T ]

Thus each X (t) ∈ L2 (Ω; H).

! < ∞.


1695

Proof: Consider the formula in Lemma 55.3.5. Z t 2 2 |X (t)| = |X (s)| + 2 hY (u) , X (t)i du + 2 (X (s) , M (t) − M (s)) s 2

+ |M (t) − M (s)| − |X (t) − X (s) − (M (t) − M (s))|

2

(55.3.16)

Now let tj denote a point of Pk from Lemma 55.3.1. It follows from the construction in this lemma, it can be assumed that Pk is such that for all t ∈ Pk , X (t) = X (t) provided t > 0. When t = 0, X (0) ≡ X0 . At the positive partition points X = X. Also X (0) = X0 ∈ L2 (Ω; H) . 2

2

|X (tm )| − |X0 | =

m−1 X

2

|X (tj+1 )| − |X (tj )|

2

j=0

Using the formula in Lemma 55.3.5 for t = tm this yields 2

2

|X (tm )| − |X0 | = 2

m−1 X Z tj+1 tj

j=0 m−1 X

+2

ÃZ

−

!

tj+1

Z (u) dW (u) , X (tj ) tj

j=0

m−1 X

hY (u) , Xkr (u)i du+

+

m−1 X

H

|M (tj+1 ) − M (tj )|

2

j=0 2

|X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj ))|

(55.3.17)

j=0

This can be written as Z 2

2

|X (tm )| − |X0 | ≤ 2 0

Z

tm

+2

R 0

tm

hY (u) , Xkr (u)i du

m−1 ³¡ ´ X ¢∗ l 2 bk (u) dW (u) + Z (u) ◦ J −1 X |M (tj+1 ) − M (tj )| j=0

−

m−1 X

2

|X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj ))|

(55.3.18)

j=0

where R is the Riesz map from U1 to U10 which satisfies Rx (y) ≡ (y, x)U1 . b l = X l at positive mesh points but at b l is a step function for which X and X k k k b l (0) ≡ X0 . That stochastic integral makes sense because the integrand is 0 X k predictable. Here is why. It follows from Lemma 55.3.5 that Xkl is predictable and b l is predictable. That the second term on since X0 is F0 measurable, it follows X k

1696


the right in 55.3.13 does reduce to what is claimed follows from noting that it is a sum of expressions of the form µZ t ¶ Z (u) ◦ J −1 dW (u) , X (s) s

H

If Z (u) were an elementary function, this is a sum of things like ³ ´ ¡ ¢∗ ∆W (u) , Z (u) ◦ J −1 X (s) U1

=R

³¡

´ ¢∗ Z (u) ◦ J −1 X (s) ∆W (u)

where R : U1 → U10 is the Riesz map, and so, approximating Z (u) by elementary functions which have terms like the above and passing to the limit, one obtains this term in the case where the appropriate ³¡ integrability¢ condition ´ holds. What −1 ∗ X (s) is appropriately about the question whether the function R Z (u) ◦ J integrable? This is not clearly the case because X (s) might not be uniformly bounded. However, this term is a local martingale. You could define stopping times τ k (ω) ≡ inf {t : |X (t, ω)| > k} and this would be a localizing sequence. Thus the second term on the right is a local martingale of the form claimed above. Now it follows on discarding the negative terms, Z T 2 2 sup |X (tj )| ≤ |X0 | + 2 |hY (u) , Xkr (u)i| du tj ∈Pk

0

¯Z t ³ ¯ mk −1 ¯¯Z ´ X ¯ ¯ ¯ ¡ ¢ −1 ∗ b l ¯ +2 sup ¯ R Z (u) ◦ J Xk (u) dW (u)¯¯ + ¯ ¯ t∈[0,T ]

0

tj+1

tj

j=0

where there are mk + 1 points in Pk . R Do Ω to both sides. Using the Ito isometry, this yields ! Z Ã mX k −1 Z ³ ´ 2 2 r sup |X (tj )| dP ≤ E |X0 | +2 ||Y ||K 0 ||Xk ||K + Ω

tj ∈Pk

j=0

Z Ã +2 Ω

t∈[0,T ]

Ω

tj

Z 2

||Z (u)|| dP du Ω

0

Z

+2

tj+1

¯Z t ³ ¯! ´ ¯ ¯ ¡ ¢∗ l −1 b (u) dW (u)¯ dP sup ¯¯ R Z (u) ◦ J X k ¯ T

Z 2

≤C+ Z Ã

¯2 ¯ ¯ Z (u) dW (u)¯ ¯

||Z (u)|| dP du+ 0

Ω

¯Z t ³ ¯! ´ ¯ ¯ ¡ ¢∗ l −1 bk (u) dW (u)¯ dP sup ¯¯ R Z (u) ◦ J X ¯

t∈[0,T ]

0

55.3. AN ITO FORMULA Z Ã ≤C +2

1697

¯! ¯Z t ³ ´ ¯ ¯ ¡ ¢∗ l −1 bk (u) dW (u)¯ dP X sup ¯¯ R Z (u) ◦ J ¯

t∈[0,T ]

Ω

0

where the result of Lemma 55.3.1 that Xkr converges to X in K shows the term 2 ||Y ||K 0 ||Xkr ||K is bounded. The term involving the stochastic integral is next. Applying the Burkholder Davis Gundy inequality, Theorem 51.4.7 for F (r) = r along with the description of the quadratic variation of the Ito integral found in Corollary 53.5.5 !1/2 Z Z Ã Z T ¯¯ ³ ´¯¯2 ¢ ¡ ¯¯ ¯¯ 2 −1 ∗ b l dP sup |X (tk )| dP ≤ C + C Xk (u) ¯¯ du ¯¯R Z (u) ◦ J Ω tj ∈Pk

Ω

Z ÃZ

0

T

≤C +C Ω

0

¯ ¯2 ¯ bl ¯ ||Z (u)|| ¯X (u) ¯ du k

!1/2

2

dP

b l (u) and they are X (tj ) Now for each ω, there are only finitely many values of X k for tj ∈ Pk . Therefore, the above is dominated by !1/2 ÃZ !1/2 Z Ã sup |X (tj )|

Ω

≤C+ and so

1 2

T

2

C +C

dP

0

Z

Z Z

T

2

sup |X (tj )| + C

Ω tj ∈Pk

1 2

2

||Z (u)|| du

tj ∈Pk

Ω

0

2

||Z (u)||L2 (U,H) dudP

Z 2

sup |X (tk )| dP ≤ C

Ω tj ∈Pk

R RT 2 for some constant C independent of Pk dependent on Ω 0 ||Z (u)||L2 (U,H) dudP . Let D denote the union of all the Pk except for 0. Thus D is a dense subset of [0, T ] and it has just been shown that µ ¶ 2 E sup |X (t)| ≤ C. t∈D

Let {ej } an orthonormal basis for H which is also contained in V . I claim that for y ∈ V 0 n X 2 2 |y| = sup |hy, ej i| n

j=1

This is certainly true if y ∈ H because in this case hy, ej i = (y, ej ) If y ∈ / H, then the series must diverge. If not, you could consider the infinite sum y≡

∞ X j=1

hy, ej i ej ∈ H

1698


and it would satisfy 2

|y| =

∞ X

2

|hy, ej i|

j=1

It follows 2

|X (t)| = sup n

n X

2

|hX (t) , ej i|

j=1

and for a.e. ω this is just the sup of continuous functions of t. Therefore, for given ω, 2 t → |X (t)| is lower semicontinuous. Hence letting t ∈ [0, T ] and tj → t where tj ∈ D, 2

|X (t)| ≤ lim inf |X (tj )|

2

n→∞

so it follows for a.e. ω 2

2

2

sup |X (t)| ≤ sup |X (t)| ≤ sup |X (t)| t∈D

t∈[0,T ]

Hence

t∈[0,T ]

Ã

! sup |X (t)|

E

2

≤ C.

t∈[0,T ]

This proves the lemma. Observation 55.3.7 Note the above shows that for a.e. ω, supt∈[0,T ] |X (t)|H < ∞ so that for such ω, X (t) has values in H. Say |X (t, ω)| ≤ C (ω) Hence if v ∈ V, then for a.e. ω lim (X (t) , v) = lim hX (t) , vi = hX (s) , vi = (X (s) , v)

t→s

t→s

Therefore, since for such ω, |X (t, ω)| is bounded, the above holds for all h ∈ H as well. Therefore, for a.e. ω, t → X (t, ω) is weakly continuous with values in H. Eventually, it is shown that in fact it is continuous with values in H. I am working toward the Ito formula 55.3.13. In order to get this, there is a technical result which will be needed. b l (s) ≡ ∆k (s) . Then the following limit occurs. Lemma 55.3.8 Let X (s) − X k Ã" #! ¯Z t ³ ¯ ´ ¯ ¯ ¡ ¢∗ −1 lim P sup ¯¯ R Z (s) ◦ J ∆k (s) dW (s)¯¯ ≥ ε = 0. (55.3.19) k→∞ t∈[0,T ]

0

¯R ¯ ³¡ ´´ ¢∗ ³ ¯ t b l (s) dW (s)¯¯ converges to 0 That is, supt∈[0,T ] ¯ 0 R Z (s) ◦ J −1 X (s) − X k in probability. Also the stochastic integral makes sense because X is H predictable.


1699

Proof: First note that from Lemma 55.3.6, for a.e. ω, X (t) has values in H for t ∈ [0, T ] and so it makes sense to consider it in the stochastic integral provided it is H predictable. However, as noted in Situation 55.3.3, this function is automatically V 0 predictable. Therefore, hX (t) , vi = (X (t) , v) is real predictable for every v ∈ V. Now if h ∈ H, let vn → h in H and so for each ω, (X (t, ω) , vn ) → (X (t, ω) , h) By the Pettis theorem, X is H predictable. Let {ek } be an orthonormal basis for H such that each ek is in V. Let Pn be the orthogonal projection onto span (e1 , · · · , en ). First note that for each n ÃZ ! T ¯ ³ ´¯2 ¯ ¯ l b lim E ¯Pn X (s) − Xk (s) ¯ d [M ] (s) = k→∞

Z

T

lim

k→∞

0

0

µ¯ ³ ´¯2 ¶ ¯ bkl (s) ¯¯ d [M ] (s) = 0 E ¯Pn X (s) − X

(55.3.20)

This follows because the ek are in V and s → X (s) is continuous in V 0 for a.e. b l (s) equals X evaluated at ω and so the integrand converges to 0. After all, X k the mesh³ point of Pk no larger than s which is closest to s. Also, it was shown ´ 2 above E supt∈[0,T ] |X (t)| is bounded and so the dominated convergence theorem bl implies the ³ above converges ´ to 0. Therefore, in 55.3.19 if X (s)−Xk (s) were replaced b l (s) , the desired conclusion would follow because convergence with Pn X (s) − X k in L1 (Ω) implies convergence in probability. Here is why. By the Burkholder Davis Gundy theorem, Theorem 51.4.7 and Corollary 53.5.5 which describes the quadratic variation of the stochastic integral ¯Z t ³ ¯! Z Ã ´ ¯ ¯ ¡ ¢∗ −1 sup ¯¯ R Z (s) ◦ J Pn ∆k (s) dW (s)¯¯ dP Ω

t∈[0,T ]

0

Z ÃZ ≤

T

C Ω

0

Ã Z

=

T

CE  0

ÃZ ≤

T

CE 0

!1/2 ¯ ³ ´¯2 ¯ ¯ 2 l b dP ¯Pn X (s) − Xk (s) ¯ ||Z (s)|| ds !1/2  ¯ ³ ´¯2 ¯ bkl (s) ¯¯ d [M ] (s)  ¯Pn X (s) − X

¯ ³ ´¯2 ¯ bkl (s) ¯¯ d [M ] (s) ¯Pn X (s) − X

!1/2

1700


Therefore, in order to obtain the desired result, it suffices to verify the following two expressions are small provided n is large. Ã" #! ¯Z t ³ ¯ ´ ¯ ¯ ¡ ¢∗ −1 P sup ¯¯ R Z (s) ◦ J (1 − Pn ) X (s) dW (s)¯¯ ≥ ε (55.3.21) t∈[0,T ]

Ã" P

0

#! ¯ ¯Z t ³ ´ ¯ ¯ ¢ ¡ ∗ b l (s) dW (s)¯ ≥ ε sup ¯¯ R Z (s) ◦ J −1 (1 − Pn ) X k ¯

t∈[0,T ]

(55.3.22)

0

By Corollary 51.4.8 which depends on the Burkholder Davis Gundy inequality and Corollary 53.5.5 which describes the quadratic variation of the stochastic integral, 55.3.21 is dominated by Ã  !1/2 Z T C  2 2 E ||Z (s)|| |(1 − Pn ) X (s)| ds ∧ δ ε 0 Ã  !1/2 Z T 2 2 +P  ||Z (s)|| |(1 − Pn ) X (s)| ds > δ  0

Ã

Cδ ≤ + P  ε

Z



!1/2

T

2

2

||Z (s)|| |(1 − Pn ) X (s)| ds

> δ 

(55.3.23)

0

Let η > 0 be given. Then let δ be small enough that the first term is less than η. Fix such a δ. Consider the second of the above terms. By Corollary 53.5.5 the integral inside it equals Z T 2 |(1 − Pn ) X (s)| d [M ] (s) 0

For a.e. ω,

Z

T

lim

n→∞

2

|(1 − Pn ) X (s)| d [M ] (s) = 0 0

because for a.e. ω, t → X (t, ω) is in H and so the integrand converges to 0 and is dominated by sup |X (t, ω)| < C (ω) . t∈[0,T ]

by Lemma 55.3.6. It follows that as n → ∞, the second term in 55.3.23 converges to 0. Let n be large enough that this term is less than η also. Similarly the expression in 55.3.22 is small for large enough n. This proves the lemma. Now at long last, here is the first version of the Ito formula. Lemma 55.3.9 In Situation 55.3.3, let D be as above, the union of all the positive mesh points for all the Pk . Also assume X0 ∈ L2 (Ω; H)∩Lp (Ω; V ) . Then for every t ∈ D, Z t³ ´ ® 2 2 2 |X (t)| = |X0 | + 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0


1701

Z

t

+2

R

³¡ ´ ¢∗ Z (s) ◦ J −1 X (s) dW (s)

(55.3.24)

0

Proof: Let t ∈ D. Then t ∈ Pk for all k large enough. Consider 55.3.18, Z t 2 2 hY (u) , Xkr (u)i du |X (t)| − |X0 | = 2 0

Z

t

+2

R

³¡

Z (u) ◦ J

¢ −1 ∗

0

qX k −1 ´ 2 l b Xk (u) dW (u) + |M (tj+1 ) − M (tj )| j=0

−

qX k −1

2

|X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj ))|

(55.3.25)

j=0

where tqk = t. By Lemma 55.3.8 the second term on the right, the stochastic integral, converges to Z t ³ ´ ¡ ¢∗ 2 R Z (u) ◦ J −1 X (u) dW (u) 0

in probability. The first term on the right converges to Z t ® 2 Y (u) , X (u) du 0

in L1 (Ω) because Xkr → X in K. Therefore, this also happens in probability. Consider the next term. ¯   ¯2  qX qX k −1 k −1 ¯Z tj+1 ¯ ¯ ¯ 2 E ¯ E |M (tj+1 ) − M (tj )|  = Z (s) dW (s)¯  ¯ tj ¯ j=0

j=0

=

qX k −1 Z tj+1 j=0

Z t ³ ³ ´ ´ 2 2 E ||Z (s)|| ds = E ||Z (s)|| ds

tj

0

Rt

2

and so this term converges in L1 (Ω) to 0 ||Z (s)|| ds which implies this term also Rt 2 converges in probability to 0 ||Z (s)|| ds. In fact, for all k large enough, it equals this. Thus all the terms in 55.3.25 converge in probability except for the last term which also must converge in probability because it equals the sum of terms which do. It remains to find what this last term converges to. Thus Z t ® 2 2 |X (t)| − |X0 | = 2 Y (u) , X (u) du 0

Z

t

+2

R 0

³¡

Z t ´ ¢∗ 2 ||Z (s)|| ds − a Z (u) ◦ J −1 X (u) dW (u) + 0

1702


where a is the limit in probability of the term qX k −1

|X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj ))|

2

j=0

Let Pn be the projection onto span (e1 , · · · , en ) as before where {ek } is an orthonormal basis for H with each ek ∈ V . Then using Z tj+1 X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj )) = Y (s) ds tj

the troublesome term above is of the form qX k −1 Z tj+1 j=0

−

qX k −1

hY (s) , X (tj+1 ) − X (tj ) − Pn (M (tj+1 ) − M (tj ))i ds

(55.3.26)

tj

(X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj )) , (1 − Pn ) (M (tj+1 ) − M (tj )))

j=0

(55.3.27) It is here that X0 has values in V is used. When j = 0, X (tj ) = X (0) = X0 and this needs to have values in V to do Y (s) to it in 55.3.26. The sum in 55.3.27 is dominated by 

qX k −1



1/2 2 |X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj ))| 

·

j=1

 

qX k −1

1/2 2 |(1 − Pn ) (M (tj+1 ) − M (tj ))| 

(55.3.28)

j=0

Pqk −1 2 Now it is known that j=1 |X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj ))| converges in probability to a. If you take the expectation of the other factor it is  ¯ ¯2  Z tj+1 qX k −1 ¯ ¯ ¯ ¯ E Z (s) dW (s)¯  ¯(1 − Pn ) ¯ ¯ tj j=1



qX k −1

=E

∞ X

j=1 m=n+1

 =E

qX k −1

ÃZ

tj+1

!2  Z (s) ◦ J −1 dW (s) , em 

tj

¯Z ¯2  ∞ ¯ tj+1 ³¡ ¯ ´ X ¢ ∗ ¯ ¯ R Z (s) ◦ J −1 em dW (s)¯  ¯ ¯ tj ¯

j=1 m=n+1


=

qX k −1

∞ X

j=1 m=n+1

1703

¯ ¯2  ¯Z tj+1 ³¡ ¯ ´ ¢ ∗ ¯ ¯ E ¯ R Z (s) ◦ J −1 em dW (s)¯  ¯ tj ¯ Z ∞ X

≤

T

m=n+1 0 Z T X ∞

=

0

µ¯ ¯2 ¶ ¢ ¯¡ ¯ −1 ∗ E ¯ Z (s) ◦ J em ¯ ds µ¯ ¶ ¢∗ ¯¯2 ¯¡ E ¯ Z (s) ◦ J −1 em ¯ ds

m=n+1

As n → ∞ the sum converges to 0 as n → ∞ independent of k. In fact the sum is just the tail of the series which defines ¯¯¡ ¯¯ ¯¯2 ¢∗ ¯¯¯¯2 ¯¯ 2 ´ ´ = ¯¯Z (s) ◦ J −1 ¯¯ ³ 1/2 = ||Z (s)||L2 (U,H) , ¯¯ Z (s) ◦ J −1 ¯¯ ³ 1/2 L2 Q U1 ,H L2 H,Q1

U1

1

and this is given to be integrable so by the dominated convergence theorem the term converges to 0. Thus the expression in 55.3.28 is of the form fk gnk where fk converges in probability to a as k → ∞ and gnk converges in probability to 0 as n → ∞ independently of k. Now this implies fk gnk converges in probability to 0. Here is why. P ([|fk gnk | > ε]) ≤ P (2δ |fk | > ε) + P (2Cδ |gnk | > ε) ≤ P (2δ |fk − a| + 2δ |a| > ε) + P (2Cδ |gnk | > ε) where δ |fk | + Cδ |gkn | > |fk gnk | and limδ→0 Cδ = ∞. Pick δ small enough that ε − 2δ |a| > ε/2. Then this is dominated by ≤ P (2δ |fk − a| > ε/2) + P (2Cδ |gnk | > ε) Fix n large enough that the second term is less than η. Now taking k large enough, the above is less than η. It follows the expression in 55.3.28 and consequently in 55.3.27 converges to 0 in probability. Now consider the other term, 55.3.26 using the n just determined. This term is of the form qX k −1 Z tj+1 D ¡ ¢E bkl (s) − Pn Mkr (s) − Mkl (s) ds Y (s) , Xkr (s) − X j=0

Z tD = 0

tj

¡ ¢E bkl (s) − Pn M r (s) − M l (s) ds Y (s) , Xkr (s) − X k k

The term

Z 0 1

t

¡ ¢® Y (s) , Pn Mkr (s) − Mkl (s) ds

converges to 0 in L (Ω) as k → ∞. This is because the integrand converges to 0 thanks to the continuity of M (t) and also since this is a projection onto a finite

1704


dimensional subspace of V, Pn (M (s)) is uniformly bounded in V so there is no problem getting a dominating function for the dominated convergence theorem. Now consider Z tD E bkl (s) ds Y (s) , Xkr (s) − X 0

1

This converges to 0 in L (Ω) from the construction in Lemma 55.3.1. This is because X0 ∈ Lp (Ω; V ) . Thus it equals Z 0

t

® Y (s) , Xkr (s) − Xkl (s) ds +

Z

t1k

® Y (s) , X (0) − X0 ds

0

and the expectation of the absolute value of the second term converges to 0 as k → ∞. While the expectation of the absolute value of the first term converges to 0 by the construction of the two step functions. Recall they both converge in K to X. Therefore, the expression qX k −1

|X (tj+1 ) − X (tj ) − (M (tj+1 ) − M (tj ))|

2

j=0

converges to 0 in probability. This proves the lemma. In fact, the formula 55.3.24 is valid for all t ∈ [0, T ] Theorem 55.3.10 In Situation 55.3.3, also assume X0 ∈ Lp (Ω; V ) ∩ L2 (Ω; H) . Then for every t ∈ [0, T ] , 2

2

|X (t)| = |X0 | +

Z t³ ´ ® 2 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0

Z

t

+2

R

³¡ ´ ¢∗ Z (s) ◦ J −1 X (s) dW (s)

(55.3.29)

0

Furthermore, for t ∈ [0, T ] , t → X (t) is continuous as a map into H. In the above X is the representative of X which is in Lp ([0, T ] × Ω, B ([0, T ]) × F, V ) ∩ L2 ([0, T ] × Ω, B ([0, T ]) × F, H) . In addition to this, µZ t ³ ³ ´ ³ ´ ´ ¶ ® 2 2 2 E |X (t)| = E |X0 | + E 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0

Proof: Let t ∈ / D. For t > 0, let t (k) denote the largest point of Pk which is less than t. Suppose t (m) < t (k). Then Z X (t (m)) = X0 +

Z

t(m)

t(m)

Y (s) ds + 0

Z (s) dW (s) , 0


1705

a similar formula holding for X (t (k)) . Thus for t > t (m) , Z t Z t X (t) − X (t (m)) = Y (s) ds + Z (s) dW (s) t(m)

t(m)

which is the same sort of thing studied so far except that it starts at t (m) rather than at 0 and X0 = 0. Therefore, from Lemma 55.3.9 it follows Z 2

t(k)

|X (t (k)) − X (t (m))| =

³ ´ ® 2 2 Y (s) , X (s) − X (t (m)) + ||Z (s)|| ds

t(m)

Z

´ ³¡ ¢∗ Z (s) ◦ J −1 (X (s) − X (t (m))) dW (s)

t(k)

+2

R

(55.3.30)

t(m)

Consider that last term. It equals Z

t(k)

2

R t(m)

³¡ ¢´ ¢∗ ¡ l X (s) − Xm (s) dW (s) Z (s) ◦ J −1

(55.3.31)

This is dominated by ¯Z ¯ t(k) ³¡ ´´ ¢∗ ³ ¯ l bm 2¯ R Z (s) ◦ J −1 X (s) − X (s) dW (s) ¯ 0 ¯ Z t(m) ³ ¯ ´´ ¡ ¢ ³ −1 ∗ l bm (s) dW (s)¯¯ −2 R Z (s) ◦ J X (s) − X ¯ 0

≤

¯Z ¯ ¯ t(k) ³¡ ¯ ³ ´´ ¢ ∗ ¯ ¯ l bm 2¯ R Z (s) ◦ J −1 X (s) − X (s) dW (s)¯ ¯ 0 ¯ ¯Z ¯ ¯ t(m) ³¡ ¯ ´´ ¢ ³ ¯ −1 ∗ l bm (s) dW (s)¯¯ +2 ¯ R Z (s) ◦ J X (s) − X ¯ 0 ¯

¯Z t ³ ¯ ´´ ¯ ¯ ¡ ¢ ³ −1 ∗ l b ¯ ≤ 4 sup ¯ R Z (s) ◦ J X (s) − Xm (s) dW (s)¯¯ t∈[0,T ]

0

In Lemma 55.3.8 the above expression was shown to converge to 0 in probability. Therefore, by the usual appeal to the Borel Canteli lemma, there is a subsequence still referred to as {m} , such that it converges to 0 pointwise in ω for all ω off some set of measure 0 as m → ∞. It follows there is a set of measure 0 such that for ω not in that set, 55.3.31 converges to 0 in H. Note that t > 0 is arbitrary. Similar reasoning shows the first term in the non stochastic integral of 55.3.30 is dominated by an expression of the form Z T ¯ ®¯ l ¯ Y (s) , X (s) − Xm 4 (s) ¯ ds 0

1706


l which clearly converges to 0 for ω not in some set of measure zero because Xm converges in K to X. Finally, it is obvious that Z t(k) 2 ||Z (s)|| ds = 0 for a.e. ω lim m→∞

t(m)

due to the assumptions on Z. This shows that for ω off a set of measure 0 2

lim |X (t (k)) − X (t (m))| = 0

m,k→∞ ∞

and so {X (t (k))}k=1 is a convergent sequence in H. Does it converge to X (t)? Let ξ (t) ∈ H be what it converges to. Let v ∈ V then (ξ (t) , v) = lim (X (t (k)) , v) = lim hX (t (k)) , vi = hX (t) , vi = (X (t) , v) k→∞

k→∞

and now, since V is dense in H, this implies ξ (t) = X (t). Now for every t ∈ D, Z t³ ´ ® 2 2 2 |X (t)| = |X0 | + 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0

Z +2

t

R

³¡ ´ ¢∗ Z (s) ◦ J −1 X (s) dW (s)

0

and so, using what was just shown along with the obvious continuity of the functions of t on the right of the equal sign, it follows the above holds for all t ∈ [0, T ]. It only remains to verify t → X (t) is continuous with values in H. However, 2 the above shows t → |X (t)| is continuous and it was shown in Lemma 55.3.6 that t → X (t) is weakly continuous into H. Therefore, from the uniform convexity of the norm in H it follows t → X (t) is continuous.This is very easy to see in Hilbert space. Say an * a and |an | → |a| . From the parallelogram identity. 2

2

2

2

|an − a| + |an + a| = 2 |an | + 2 |a| so

³ ´ 2 2 2 2 2 |an − a| = 2 |an | + 2 |a| − |an | + 2 (an , a) + |a|

Then taking lim sup both sides, ³ ´ 2 2 2 2 2 0 ≤ lim sup |an − a| ≤ 2 |a| + 2 |a| − |a| + 2 (a, a) + |a| = 0. n→∞

Of course this fact also holds in any uniformly convex Banach space. Now consider the last claim. If the last term in 55.3.29 were a martingale, then there would be nothing to prove. This is because if M (t) is a martingale which equals 0 when t = 0, then E (M (t)) = E (E (M (t) |F0 )) = E (M (0)) = 0.

55.4. FINITE TO INFINITE DIMENSIONS

1707

However, that last term is unfortunately only a local martingale. One can obtain a localizing sequence as follows. τ n (ω) ≡ inf {t : |X (t, ω)| > n} where as usual inf (∅) ≡ T . Then stopping both processes on the two sides of 55.3.29 with τ n , Z t∧τ n ³ ´ ® 2 2 2 |X (t ∧ τ n )| = |X0 | + 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0

Z

t∧τ n

+2

R

³¡

Z (s) ◦ J −1

¢∗

´ X (s) dW (s)

0

Now from Lemma 53.5.4, Z 2

t

2

|X (t ∧ τ n )| = |X0 | + 0

Z +2 0

³ ´ ® 2 X[0,τ n ] (s) 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds

t

X[0,τ n ] (s) R

³¡ ´ ¢∗ Z (s) ◦ J −1 X (s) dW (s)

That last term is now a martingale and so you can take the expectation of both sides. This gives ³ ´ ³ ´ 2 2 E |X (t ∧ τ n )| = E |X0 | µZ t ³ ´ ¶ ® 2 +E X[0,τ n ] (s) 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0

Letting n → ∞ and using the dominated convergence theorem yields the desired result. This proves the theorem. Notation 55.3.11 The stochastic integrals are unpleasant to look at. Z t ³ ´ ¡ ¢∗ R Z (s) ◦ J −1 X (s) dW (s) 0 Z t ≡ (X (s) , Z (s) dW (s)) . 0

55.4

Finite To Infinite Dimensions

55.4.1

Some Nonlinear Operators.

Definition 55.4.1 For V a real Banach space, A : V → V 0 is a pseudomonotone map if whenever un * u (55.4.32) and lim sup hAun , un − ui ≤ 0 n→∞

(55.4.33)

1708


it follows that for all v ∈ V, lim inf hAun , un − vi ≥ hAu, u − vi. n→∞

(55.4.34)

The half arrows denote weak convergence. Definition 55.4.2 A : V → V 0 is monotone if for all v, u ∈ V, hAu − Av, u − vi ≥ 0, and A is Hemicontinuous if for all v, u ∈ V, lim hA (u + t (v − u)) , u − vi = hAu, u − vi.

t→0+

Theorem 55.4.3 Let V be a Banach space and let A : V → V 0 be monotone and hemicontinuous. Then A is pseudomonotone. Proof: Let A be monotone and Hemicontinuous. First here is a claim. Claim: If 55.4.32 and 55.4.33 hold, then limn→∞ hAun , un − ui = 0. Proof of the claim: Since A is monotone, hAun − Au, un − ui ≥ 0 so hAun , un − ui ≥ hAu, un − ui. Therefore, 0 = lim inf hAu, un − ui ≤ lim inf hAun , un − ui ≤ lim sup hAun , un − ui ≤ 0. n→∞

n→∞

n→∞

Now using that A is monotone again, then letting t > 0, hAun − A (u + t (v − u)) , un − u + t (u − v)i ≥ 0 and so hAun , un − u + t (u − v)i ≥ hA (u + t (v − u)) , un − u + t (u − v)i. Taking the lim inf on both sides and using the claim and t > 0, t lim inf hAun , u − vi ≥ thA (u + t (v − u)) , (u − v)i. n→∞

Next divide by t and use the Hemicontinuity of A to conclude that lim inf hAun , u − vi ≥ hAu, u − vi. n→∞

From the claim, lim inf hAun , u − vi = lim inf (hAun , un − vi + hAun , u − un i) n→∞

n→∞

= lim inf hAun , un − vi ≥ hAu, u − vi. n→∞

This proves the theorem. Monotonicity is very important in the above proof. The next example shows that even if the operator is linear and bounded, it is not necessarily pseudomonotone.


1709

Example 55.4.4 Let H be any Hilbert space and let A : H → H 0 be given by hAx, yi ≡ (−x, y)H . Then A fails to be pseudomonotone. ∞

Proof: Let {xn }n=1 be an orthonormal set of vectors in H. Then Parsevall’s inequality implies ∞ X 2 2 ||x|| ≥ |(xn , x)| n=1

and so for any x ∈ H, limn→∞ (xn , x) = 0. Thus xn * 0 ≡ x. Also lim sup hAxn , xn − xi = n→∞

³ lim sup hAxn , xn − 0i = lim sup n→∞

n→∞

− ||xn ||

2

´ = −1 ≤ 0.

If A were pseudomonotone, we would need to be able to conclude that for all y ∈ H, lim inf hAxn , xn − yi ≥ hAx, x − yi = 0. n→∞

However, lim inf hAxn , xn − 0i = −1 < 0 = hA0, 0 − 0i. n→∞

Now the following proposition is useful. Proposition 55.4.5 Suppose A : V → V 0 is pseudomonotone and bounded where V is separable. Then it must be demicontinuous. This means that if un → u, then Aun * Au. Proof: Since un → u is strong convergence and since Aun is bounded, it follows lim sup hAun , un − ui = lim hAun , un − ui = 0. n→∞

n→∞

Suppose this is not so that Aun converges weakly to Au. Since A is bounded, there exists a subsequence, still denoted by n such that Aun * ξ. I need to verify ξ = Au. From the above, it follows that for all v ∈ V hAu, u − vi ≤ lim inf hAun , un − vi n→∞

= lim inf hAun , u − vi = hξ, u − vi n→∞

Hence ξ = Au. This proves the proposition.

1710

55.4.2


The Situation

First recall that in the notion of the cylindrical Wiener process, Z t Z t Φ (s) dW (s) ≡ Φ (s) ◦ J −1 dW (s) 0

0

where W (t) is a Wiener process having values in U1 , a separable Hilbert space such that J : U → U1 is a Hilbert Schmidt operator and W (t) is a Q1 Wiener process where Q1 ≡ JJ ∗ is defined on U1 . Also recall that W (t) ≡

∞ X

ψ k (t) Jgk

k=1

where {gk } is an orthonormal basis for U. Here U = U0 = Q1/2 U because it is assumed Q = I. Let n X Wn (t) = ψ k (t) gk k=1 0

Let V ⊆ H ⊆ V be a Gelfand triple. Let A : K → K 0 where K = Lp ([0, T ] × Ω; V ) 0 and so K 0 = Lp ([0, T ] × Ω; V 0 ). Also let σ : H →L2 ([0, T ] × Ω; L2 (U, H)) where H ≡ L2 ([0, T ] × Ω; H) . Let A : [0, T ] × V × Ω → V 0 , σ : [0, T ] × H × Ω → L2 (U ; H) be progressively measurable as in Proposition 53.3.6 so that by this proposition, if X is H progressively measurable and X is V progressively measurable, then ¡ ¢ (t, ω) → A t, X (t, ω) , ω , (t, ω) → σ (t, X (t, ω) , ω) (55.4.35) will be V 0 and L2 (U, H) progressively measurable. Also assume X → σ (t, X, ω)

(55.4.36)

is continuous from H ¡ to L2 (U ¢; H) . As¡ in [55] I will¢ suppress the dependence on ω. Thus I will write A t, X (t, ω) for A t, X (t, ω) , ω . The interest is in the stochastic differential equation dX = A (t, X) dt + σ (t, X) dW (t) , X (0) = X0 ∈ H. Other assumptions will be on A and σ. Assume the following. [55] 1. (Hemicontinuity) For a.e. ω λ → hA (t, u + λv, ω) , zi is continuous. 2. (Weak monotonicity) There exists c ∈ R such that for all u, v ∈ V, 2

2 hA (·, u) − A (·, v) , u − vi + |σ (·, u) − σ (·, v)|L2 (U,H) ≤

2

c |u − v|H on [0, T ] × Ω


1711

3. (Coercivity) There exist p ∈ (1, ∞), c1 ∈ R, c2 ∈ (0, ∞) , and f ∈ L1 ([0, T ] × Ω) which is B ([0, T ]) × F measurable and is also Ft adapted such that for all v ∈ V, t ∈ [0, T ] , 2

p

2

2 hA (t, v) , vi + ||σ (t, v)||L2 (U,H) ≤ c1 |v|H − c2 ||v||V + f (t) 0

4. (Boundedness) There exist c3 ∈ [0, ∞) and an Ft adapted process g ∈ Lp ([0, T ] × Ω) which is B ([0, T ]) × F measurable such that for all v ∈ V, t ∈ [0, T ] p−1

||A (t, v)||V 0 ≤ g (t) + c3 ||v||V

.

Here 1/p + 1/p0 = 1. 5. (Measurability) Let A : [0, T ] × V × Ω → V 0 , σ : [0, T ] × H × Ω → L2 (U ; H) be progressively measurable. Observation 55.4.6 4 and 3 imply the following estimate holds. 2

||σ (t, v)||L2 (U,H)

2

2

≤ c1 |v|H − c2 ||v||V + f (t) − 2 hA (t, v) , vi ³ ´ 2 2 p−1 ≤ c1 |v|H − c2 ||v||V + f (t) + 2 g (t) + c3 ||v||V ||v||V 2

p

≤ c1 |v|H + f (t) + 2g (t) ||v||V + 2c3 ||v||V

(55.4.37)

Another important observation is the following which holds for a.e. ω. Proposition 55.4.7 The above hypotheses imply that u → A (t, u) is demicontinuous. This means that if un → u in V then A (t, un ) * A (t, u) where the arrow denotes weak convergence. Proof: To place this in a more familiar context and write more simply, let B1 (u) ≡ −A (t, u) . Then from 2 hB1 u − B1 v, u − vi ≥ −c |u − v|

2

Therefore, if B = cI + B1 it follows B is monotone and hemicontinuous as well as bounded. That B is bounded is obvious from 4. Therefore, the conclusion follows from Proposition 55.4.5. Note that as usual, the above holds for a.e. ω.

1712

55.4.3


The Finite Dimensional Problem, Estimates

In this section I will assume there exists a solution to a finite dimensional approximate problem and consider how to pass to a limit and obtain existence for the problem which is really of interest. Let {ek } be an orthonormal basis for H such that each ek ∈ V. Let Hn = span (e1 , · · · , en ) . Let Pn be the orthogonal projection onto Hn . Thus n X Pn y ≡ hy, ek i ek k=1

and so for v ∈ Hn and w ∈ V , Ã (Pn A (t, w) , v)H

=

n X

! hA (t, w) , ek i ek , v

k=1

* =

A (t, w) ,

n X

+H ek (ek , v)

k=1

=

hA (t, w) , vi

(55.4.38)

What does Pn σ mean for σ ∈ L2 (U, H)? It will mean that for u ∈ U, Pn σ (u) ≡

n X

(σu, ek ) ek

k=1

Thus Pn σ ∈ L2 (U, Hn ) and letting {gk } be the orthonormal basis for U 2

||Pn σ||L2 (U,Hn ) ≡

∞ X

2

|Pn σ (gj )|Hn

j=1

¯ ¯2 ∞ ¯X n ¯ X ¯ ¯ ≡ (σgj , ek ) ek ¯ ¯ ¯ ¯ j=1 k=1

≤

Hn

∞ X

2

=

∞ X n X

2

(σgj , ek )

j=1 k=1 2

|σgk |H = ||σ||L2 (U,H)

(55.4.39)

k=1

Assumption 55.4.8 Assume there exists a solution Xn to dXn Xn (0)

= Pn A (t, Xn (t)) + Pn σ (t, Xn (t)) dWn (t) = Pn X0

where X0 ∈ V . Since Xn is continuous for a.e. ω, it must be predictable by Proposition 55.1.1. Now estimates are obtained. First note that from 55.4.38, for v ∈ Hn , hPn A (t, v) , vi = hA (t, v) , vi


1713

By the Ito formula derived earlier, Z 2

t

2

2

|Xn (t)| = |Pn X0 | +

2 hA (s, Xn (s)) , Xn (s)i + ||Pn σ (s, Xn (s))||L2 (U,H) ds

0

Z

t

(Xn (s) , Pn σ (s, Xn (s)) dWn (s)) ,

+2 0

the last term being a local martingale as discussed earlier. Now let τ k be a stopping time τ k (ω) ≡ inf {t : |Xn (t, ω)| > k} By the localization theorems or versions of them, Z |Xn (t ∧ τ k )|

2

t

2

= |Pn X0 | + 0

X[0,τ k ] (s) 2 hA (s, Xn (s)) , Xn (s)i 2

+X[0,τ k ] (s) ||Pn σ (s, Xn (s))||L2 (U,H) ds Z

t

¡

+2 0

X[0,τ k ] (s) Xn (s) , Pn σ (s, Xn (s)) dWn (s)

¢

Then take expectation of both sides. That last term is a martingale so its expectation is 0. Thus ³ ´ ³ ´ 2 2 E |Xn (t ∧ τ k )| = E |Pn X0 | + Z

t 0

³ h i´ 2 E X[0,τ k ] (s) 2 hA (s, Xn (s)) , Xn (s)i + ||Pn σ (s, Xn (s))||L2 (U,H) ds

Then

³ ´ ³ ´ 2 2 E e−c1 t |Xn (t ∧ τ k )| − E |Pn X0 | = Z

´´ d ³ ³ −c1 s 2 E e |Xn (s ∧ τ k )| ds = 0 ds Z t ³ ´ 2 −c1 E e−c1 s |Xn (s ∧ τ k )| ds+ t

0

Z 0

t

³

h i´ 2 e−c1 s E X[0,τ k ] (s) 2 hA (s, Xn (s)) , Xn (s)i + ||Pn σ (s, Xn (s))||L2 (U,H) ds

Now apply 3 along with 55.4.38 and 55.4.39 to obtain ³ ´ ³ ´ Z t ³ ´ 2 2 2 E e−c1 t |Xn (t ∧ τ k )| − E |Pn X0 | ≤ −c1 E e−c1 s |Xn (s ∧ τ k )| ds+ 0

Z

t 0

³ ³ ´´ 2 p e−c1 s E X[0,τ k ] (s) c1 |Xn (s ∧ τ k )| − c2 ||Xn (s ∧ τ k )||V + f (s) ds

1714


Two of these terms cancel when X[0,τ k ] (s) = 1. Since e−c1 s ≤ 1, Z t ³ ´ 2 p −c1 t E e |Xn (t ∧ τ k )| + c2 ||Xn (s ∧ τ k )||V ds ³ ≤

2

E |Pn X0 |

0

Z

´

T

+

|f (t)| dt 0

Let k → ∞ and use Fatou’s lemma to conclude µZ t ¶ ³ ´ 2 p E e−c1 t |Xn (t)| + c2 E ||Xn (s)||V ds ³ ≤ E |Pn X0 |

2

³ ´ 2 ≤ E |X0 | +

0

Z

´

T

+ Z

|f (t)| dt 0 T

|f (t)| dt 0

From 4 it follows A (·, Xn ) is bounded in K 0 . This has proved the following lemma. Lemma 55.4.9 There exists a constant C independent of n such that ³ ´ 2 sup E |Xn (t)| + ||Xn ||K + ||A (·, Xn )||K 0 ≤ C. t∈[0,T ]

55.4.4

Passing To The Limit

With this estimate, it follows there exists a subsequence which will still be denoted by n such that Xn * X weakly in L2 ([0, T ] × Ω; P; H) , (55.4.40) Xn * X weakly in K ≡ Lp ([0, T ] × Ω; P; V ) , 0

Pn A (·, Xn ) * Y weakly in K 0 ≡ Lp ([0, T ] × Ω; P; V 0 )

(55.4.41) (55.4.42)

Thus the limit functions are all progressively measurable, P denoting the σ algebra of progressively measurable sets. Letting Pn0 be the projection in U onto span (g1 , · · · , gn ), 55.4.37 implies there is a further subsequence, still denoted by n such that for some Z Pn σ (·, Xn ) Pn0 ◦ J −1 * Z ◦ J −1 weakly in L2 ([0, T ] × Ω; P; L2 (U1 , H)) Now recall Z Xn (t) = Pn X0 +

Z

t

t

Pn A (s, Xn (s)) ds + 0

Pn σ (s, Xn (s)) dWn (s) . 0

Consider the last term. It equals Z t Pn σ (s, Xn (s)) Pn0 ◦ J −1 dW (s) 0

(55.4.43)


1715

Thus 55.4.43 is of the form Z t Z t Xn (t) = Pn X0 + Pn A (s, Xn (s)) ds + Pn σ (s, Xn (s)) Pn0 ◦ J −1 dW (s) 0

Let

0

(55.4.44)

¡ ¢ I : L2 ([0, T ] × Ω; P; L2 (U1 , H)) → C [0, T ] ; L2 (Ω; H)

given by

Z

t

Iσ (t) ≡

σ (s) dW (s) . 0

Then by the Ito isometry this is a linear continuous map and so I is continuous from¡ the weak topolgy¢ on L2 ([0, T ] × Ω; P, L2 (U1 , H)) to the weak ∗ topology on L∞¡ 0, T ; P, L2 (Ω, H)¢ since this is a dual space of the separable Banach space L1 0, T ; P, L2 (Ω, H) . See Corollary 13.4.6 and Theorem 23.8.6. Therefore, Z

t

Pn σ (s, Xn (s)) dWn (s) = 0

Z

(·) 0

Z Pn σ (s, Xn (s)) Pn0

◦J

−1

(·)

dW (s) *

Z (s) ◦ J −1 dW (s)

0

¡ ¢ weak ∗ in L∞ 0, T ; P; L2 (Ω, H) . 0 the ¡ Next consider ¢ deterministic integral in 55.4.43. The mapping J : K → 2 0 C [0, T ] ; L (Ω; V ) given by Z JY (t) ≡

t

Y (s) ds 0

is linear and continuous. Therefore, as in the ¡ ¢ stochastic integral, this is weak ∗ continuous as a map into L∞ 0,¡T ; L2 (Ω; V 0 ) and ¢ so the deterministic integral in 55.4.43 converges weak ∗ in L∞ 0, T ; L2 (Ω; V 0 ) to Z

(·)

Y (s) ds. 0

¡ ¢ It follows the right side of 55.4.43 converges weak ∗ in L∞ 0, T ; L2 (Ω; V 0 ) to Z

Z

(·)

(·)

Z (s) dW (s) .

Y (s) ds +

X0 +

0

0

Now define X (t) by Z X (t) ≡ X0 +

Y (s) ds + 0

Thus X ∈ C ([0, T ] ; V 0 ) .

Z

t

t

Z (s) dW (s) 0

1716


Is it the case that X ∈ K above is a measurable representative of X? Since the ¡ ¢ right side of 55.4.43 converges weak ∗ in L∞ 0, T ; L2 (Ω; V 0 ) , it¡must be the case ¢ that the left side of this equation also converges ¡weak ∗ in L∞ ¢0, T ; L2 (Ω; V 0 ) . Thus it is also the case it converges weakly in L2 0, T ; L2 (Ω; V 0 ) Let η ∈ L2 ([0, T ] × Ω; V ) . Then

Z

T

0

Z

® X (t) , η (t) L2 (Ω;V ) dt =

Z

T

X (t) , η (t)

0

Z

Z

T

¡

T

n→∞

Ω

Z n→∞

0

Ω

Z (Xn (t) , η (t))L2 (Ω;H) dt = lim

n→∞

Z = 0

dt =

Xn (t, ω) η (t, ω) dP dt 0

T

= lim

L2 (Ω;H)

Z

X (t, ω) η (t, ω) dP dt = lim 0

¢

T

hXn (t) , η (t)i dt 0

T

hX (t) , η (t)iL2 (Ω;V ) dt

Therefore, X (t) = X (t) a.e. t. Thus X is a measurable representative of X as hoped. Therefore, the situation of the Ito formula is obtained. It remains to identify Y (s) and Z (s) . First, from the Ito formula, Theorem ??, it follows 2

2

|X (t)| = |X0 | +

Z t³ ´ ® 2 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds + M (t) 0

where M (t) is a local martingale. Also from this theorem, µZ t ³ ´ ¶ ³ ´ ³ ´ ® 2 2 2 E |X (t)| = E |X0 | + E 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 0

and so

³ ´ ³ ´ Z t³ ³ ´´0 2 2 2 E e−ct |X (t)| − E |X0 | = e−cs E |X (s)| ds 0

Z

t 0

³ ³ ³ ´ ´´ ® 2 2 −ce−cs E |X (s)| + e−cs E 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds

which implies

µZ =E

t

e 0

³ ´ ³ ´ 2 2 E e−ct |X (t)| − E |X0 | −cs

³

´ ¶ ® 2 −c |X (s)| + 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) ds 2

(55.4.45)


1717

Now consider 55.4.44. Using 55.4.38, the same reasoning applied to this equation leads to ³ ´ ³ ´ 2 2 E e−ct |Xn (t)| − E |Pn X0 | µZ t ³ 2 e−cs −c |Xn (s)| + 2 hA (s, Xn (s)) , Xn (s)i = E 0 ´ ´ 2 + ||Pn σ (s, Xn (s)) Pn0 ||L2 (U,H) ds Deleting those projection operators only makes the norm larger and so ³ ´ ³ ´ 2 2 E e−ct |Xn (t)| − E |Pn X0 | µZ t ³ 2 ≤ E e−cs −c |Xn (s)| + 2 hA (s, Xn (s)) , Xn (s)i 0 ´´ 2 + ||σ (s, Xn (s))||L2 (U,H) ds At this point, it is desired to exploit the monotonicity assumption on A. Let φ (s) ∈ K ∩ L2 ([0, T ] × Ω; H) where φ is progressively measurable. Then inserting this and patching things up by adding other terms, ³ ´ ³ ´ 2 2 E e−ct |Xn (t)| − E |Pn X0 | ≤ µZ E

t

³ 2 e−cs −c |Xn (s) − φ (s)| + 2 hA (s, Xn (s)) − A (s, φ (s)) , Xn (s) − φ (s)i

0

´ ´ 2 + ||σ (s, Xn (s)) − σ (s, φ (s))||L2 (U,H) ds µZ

t

+E

(55.4.46)

³ 2 e−cs c |φ (s)| − 2c (Xn (s) , φ (s)) + 2 hA (s, φ (s)) , Xn (s) − φ (s)i

0

+2 hA (s, Xn (s)) , φ (s)i + 2 (σ (s, Xn (s)) , σ (s, φ (s)))L2 (U,H) ´ ´ 2 − ||σ (s, φ (s))|| ds dP Now from 2 the expression in 55.4.46 is nonpositive and so the whole mess above reduces to ³ ´ ³ ´ 2 2 E e−ct |Xn (t)| − E |Pn X0 | ≤ Z +E( 0

t

³ 2 e−cs c |φ (s)| − 2c (Xn (s) , φ (s)) + 2 hA (s, φ (s)) , Xn (s) − φ (s)i

1718


+2 hA (s, Xn (s)) , φ (s)i + 2 (σ (s, Xn (s)) , σ (s, φ (s)))L2 (U,H) ´ ´ 2 − ||σ (s, φ (s))|| ds dP Now let ψ (t) ∈ L∞ (0, T ), just a real valued function. Both sides of the above without the E are progressively measurable because this is true of Xn , φ and 55.4.35. RT Then pass to the limit in the above, ∗ do 0 ψ (t) ∗ dt Then use Fubini’s theorem to obtain ÃZ ! T ³ ´ 2 2 −ct ψ (t) e E |X (t)| − |X0 | dt 0

ÃZ ≤

Z

T

E

t

ψ (t) 0

³ ¡ ¢ 2 e−cs c |φ (s)| − 2c X (s) , φ (s)

0

® +2 A (s, φ (s)) , X (s) − φ (s) +2 hY (s) , φ (s)i + 2 (Z (s) , σ (s, φ (s)))L2 (U,H) ´ ´ 2 − ||σ (s, φ (s))|| dsdt Now from 55.4.45, the left side of the above inequality can be replaced with the right side of 55.4.45 to obtain ÃZ ! Z t T ´ ³ ® 2 2 −cs E ψ (t) e −c |X (s)| + 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) dsdt 0

0

ÃZ ≤

Z

T

E

t

ψ (t) 0

³ ¢ ¡ 2 e−cs c |φ (s)| − 2c X (s) , φ (s)

0

® +2 A (s, φ (s)) , X (s) − φ (s) +2 hY (s) , φ (s)i + 2 (Z (s) , σ (s, φ (s)))L2 (U,H) ´ ´ 2 − ||σ (s, φ (s))|| dsdt

(55.4.47)

Recall that φ was arbitrary. Let it equal X. Then the above reduces to ÃZ ! Z t T ³ ´ ® 2 2 E ψ (t) e−cs −c |X (s)| + 2 Y (s) , X (s) + ||Z (s)||L2 (U,H) dsdt 0

0

ÃZ ≤E

Z

T

ψ (t) 0

0

t

³ ¯ ¯2 e−cs −c ¯X (s)¯


1719

® ¡ ¡ ¢¢ +2 Y (s) , X (s) + 2 Z (s) , σ s, X (s) L (U,H) 2 ´ ¯¯ ¡ ¢¯¯2 ´ ¯ ¯ ¯ ¯ − σ s, X (s) dsdt Recall X (t) = X (t) a.e. and so the above simplifies to ÃZ ! Z t T ³ ´ 2 −cs ψ (t) e E ||Z (s)||L2 (U,H) dsdt ≤ 0

ÃZ

Z

T

E

t

ψ (t)

e

0

−cs

0

0

³ ¡ ¡ ¢¢ 2 Z (s) , σ s, X (s) L

2 (U,H)

−e

−cs

which simplifies further to ÃZ Z t T ³ ¯¯ ¡ ¢¯¯2 E ψ (t) e−cs ¯¯Z (s) − σ s, X (s) ¯¯L 0

0

! ¯¯ ¡ ¢¯¯2 ´ ¯¯σ s, X (s) ¯¯ ds dt

!

´ 2 (U ;H)

dsdt

≤0

¡ ¢ and so for a.e. ω, Z (s) = σ s, X (s) a.e. s and so for a.e. ω, Z (s) = σ (s, X (s)) a.e. s. Now 55.4.47 simplifies to ÃZ ! Z t T ³ ®´ 2 −cs E ψ (t) e −c |X (s)| + 2 Y (s) , X (s) dsdt 0

0

ÃZ ≤

Z

T

E

t

ψ (t) 0

³ ¢ ¡ 2 e−cs c |φ (s)| − 2c X (s) , φ (s)

0

® ¢´ + 2 A (s, φ (s)) , X (s) − φ (s) + 2 hY (s) , φ (s)i

(55.4.48)

Next let φ = X − εηv where η ∈ L∞ ([0, T ] × Ω) and v ∈ V in 55.4.48. ÃZ ! Z t T ³ ´ ® 2 E ψ (t) e−cs −c |X (s)| + 2 Y (s) , X (s) dsdt 0

ÃZ ≤

E

0

Z

T

t

ψ (t) 0

³ ¯ ¯2 ¡ ¢ e−cs c ¯X (s) − εη (s) v ¯ − 2c X (s) , X (s) − εη (s) v

0

¡ ¢ ® ®¢´ + 2 A s, X (s) − εη (s) v , εη (s) v + 2 Y (s) , X (s) − εη (s) v Simplifying this yields ÃZ 0

≤

E

Z

T

ψ (t) 0

0

t

³ ´ 2 e−cs cε2 |η (s) v| − 2ε hY (s) , η (s) vi

¡ ¢ ®¢ ´ + 2 A s, X (s) − εη (s) v , εη (s) v ds

1720


First divide by ε and get ÃZ

Z

T

0

≤ E

ψ (t) 0

t

³ ´ 2 e−cs cε |η (s) v| − 2 hY (s) , η (s) vi

0

¡ ¢ ®¢ ´ + 2 A s, X (s) − εη (s) v , η (s) v ds Next let ε → 0 and use Hemicontinuity, 1 and the dominated convergence theorem to conclude ÃZ ! Z t T ¡ ¡ ¢ ® ¢ −cs ψ (t) e 0 ≤ 2E A s, X (s) , v − hY (s) , vi η (s) dsdt 0

0

Since this is true for all ψ ∈ L∞ (0, T ) , it follows for a.e. ω, Z

t

e−cs

¡ ¡ ¢ ® ¢ A s, X (s) , v − hY (s) , vi η (s) ds = 0

0

for all t. In particular, this is true if t = T. Since η is arbitrary, this shows that for any v ∈ V, ¡ ¢ ® A s, X (s) , v − hY (s) , vi = 0 a.e. s Since V is separable, there exists a set of measure zero such that for s not in this set, the above equals 0 for all v ∈ V . This has shown that for a.e. ω, ¡ ¢ A s, X (s) = Y (s) a.e. s. This proves the existence part of the following theorem. Theorem 55.4.10 Suppose 1 - 5 on the operators A and σ and suppose also there exists a solution to the finite dimensional stochastic equation given in 55.4.8. Then there exists a unique solution to the stochastic differential equation dX = X (0) =

A (t, X (t)) dt + σ (t, X (t)) dW, X0 ∈ Lp (Ω; V ) ∩ L2 (Ω; H)

More precisely, Z X (t)

t

= X0 +

¡ ¢ A s, X (s) ds +

0

X0

∈

Z

t

σ (s, X (s)) dW (s) , 0

Lp (Ω; V ) ∩ L2 (Ω; H) , X0 is F0 measurable

where X is a measurable representative of X which is in Lp ([0, T ] × Ω; V ) ∩ L2 ([0, T ] × Ω; H) and is V and H progressively measurable.


1721

Proof: It only remains to verify uniqueness. Suppose then that Xi , i = 1, 2 are solutions corresponding to the initial condition X0i . Then by the Ito formula and Theorem ?? ´ ³ ´ ³ 2 2 E |X1 (t) − X2 (t)| = E |X01 − X02 | + µZ

t

¡ ¢ ¡ ¢ ® 2 A s, X 1 (s) − A s, X 2 (s) , X 1 (s) − X 2 (s) + 0 ¶ 2 ||σ (s, X1 (s)) − σ (s, X2 (s))|| ds

E

Now from 2 and Fubini’s theorem, Z t ³ ³ ´ ³ ´ ¯ ¯2 ´ 2 2 E |X1 (t) − X2 (t)| ≤ E |X01 − X02 | + c E ¯X 1 (s) − X 2 (s)¯ ds ³ = E |X01 − X02 |

2

Z

´

0 t

+c

³ E |X1 (s) − X2 (s)|

2

´ ds

0

When X01 = X02 uniqueness is obtained from Gronwall’s inequality. That is, for each t, X1 (t) = X2 (t) in L2 (Ω; H) . This proves the theorem. Of course it should also be possible to relax the assumption that X0 ∈ V. One should only have to assume X0 ∈ L2 (Ω; H) and measurable in F0 . That a theorem should hold in this more general case follows right away from the above formula for uniqueness. Let X0n → X0 ∈ L2 (Ω; H) the convergence being in L2 (Ω; H). Then an adaption of the argument given to prove existence and in particular, passing to the limit shows existence. Uniqueness is then immediate from a repeat of the uniqueness argument given in the proof of Theorem 55.4.10 above in the simpler case where X01 = X02 . This proves the following corollary. Corollary 55.4.11 In the situation of Theorem 55.4.10 it is only necessary to assume X0 ∈ L2 (Ω; H). Suppose in addition to the above, that ∗

u → σ (t, u) h

(55.4.49)

is Lipschitz continuous for all h ∈ H. Suppose also that in addition to hemicontinuity, 1 v → hA (t, v) , wi (55.4.50) is Lipschitz continuous for all w ∈ V. Then you can show the finite dimensional problem has a unique solution because it is in the situation of Theorem 55.1.8. Therefore, there exists a solution to the stochastic differential equation of Theorem 55.4.10. Here is a simple example.

1722


Example 55.4.12 Let H = L2 (W ) where W is a bounded open subset of Rn and let V be a closed subspace of H 1 (W ) which contains Cc∞ (W ). Let A : V → V 0 satisfy Z hAu, vi ≡ −

∇u · ∇vdx W

Let σ ∈ L2 (U, H) be a constant and let W (t) be a Wiener process defined on U1 where J : U → U1 is a Hilbert Schmidt operator. Then the assumptions 1 - 5 all hold if K = L2 ([0, T ] × Ω; V ). The only one of these assumptions which is not immediately obvious is 3, the one labeled coercivity. However, using the definition of the norm in V, 2 2 − hAu, ui + |u|H = ||u||V and so

2

2

2

2

2 hAu, ui + ||σ||L2 (U,H) = − ||u||V + |u|H + ||σ||L2 (U,H) which is sufficient. The conditions 55.4.49 and 55.4.50 both hold and this is sufficient to use Theorem 55.1.8 to get existence of a solution to the finite dimensional problem. Therefore, by Corollary 55.4.11, there exists a unique solution to the equation dX = AXdt + σdW (t) , X (0) = X0 ∈ H The unique solution to this is called the Ornstein Uhlenbeck process. Maybe you might right this as dX = ∆Xdt + σdW (t) , X (0) = X0 ∈ H.

Of course Corollary 55.4.11 and Theorem 55.1.8 include many more examples than this, including time dependent operators.

55.5

Finite Dimensional Problems

55.5.1

The Ito Formula

Before starting, it is a good idea to review the Ito formula, Theorem 54.1.1. Theorem 55.5.1 Let Φ be an L2 (U0 , H) valued predictable or progressively measurable process which is stochastically integrable in [0, T ] because Ã"Z #! T

P

2

||Φ|| dt < ∞

=1

0

and let φ : [0, T ] × Ω → H be predictable or progressively measurable process which is Bochner integrable on [0, T ] and let X (0) be F0 measurable and H valued. Let Z t X (t) ≡ X (0) + φ (s) ds + Φ · W (t) . 0

55.5. FINITE DIMENSIONAL PROBLEMS

1723

Let F : [0, T ] × H × Ω → R1 have continuous partial derivatives Ft , Fx , Fxx which are uniformly continuous on bounded subsets of [0, T ] × H independent of ω ∈ Ω and let it be progressively measurable. Then the following formula holds for a.e. ω. F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ · W (t) + Z

t

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) + 0


+

The dependence of F on ω is suppressed. In the case of interest here H = Rd so it is a finite dimensional problem and Q ≡ I which is trace class because of the finite dimensions. Thus in the case of interest here, the formula is F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ · W (t) + Z

t

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) + 0

+

¡ ¢ª 1 ∗ trace Φ (s) Fxx (s, X (s)) Φ (s) ds 2

and also U0 = U = H = Rd and the space of Hilbert Schmidt operators in the statement of the theorem reduces to the space of d × d1 matrices with the Frobenius norm.

55.5.2

An Unbelievably Technical Lemma

This is all from [55], see also [43]. Here W (t) is a Wiener process on Rd1 with respect to the usual normal filtration Ft . Thus U ≡ Rd1 and Q ≡ I. Let H ≡ Rd . (t, x, ω) → σ (t, x, ω) is progressively measurable and continuous in x into the space of d × d1 matrices with the usual Frobenius norm. Similarly (t, x, ω) → b (t, x, ω) is progressively measurable and continuous in x into Rd . Assume also the following integrability condition. Z

T

sup 0

|x|≤R

n o 2 ||σ (t, x)|| + |b (t, x)| dt < ∞ on Ω

(55.5.51)

1724


The dependence on ω is suppressed. In addition to this, some other assumptions are made. For |x| , |y| ≤ R, 2

2 (x − y, b (t, x) − b (t, y)) + ||σ (t, x) − σ (t, y)|| ≤ Kt (R) |x − y| ³ ´ 2 2 2 (x, b (t, x)) + ||σ (t, x)|| ≤ Kt (1) 1 + |x|

2

(55.5.52) (55.5.53)

In the above, Kt (R) is an [0, ∞) valued Ft adapted process for each R ≥ 0. It is assumed to satisfy Z T Kt (R) dt < ∞ a.e. ω (55.5.54) 0

Also define for each R ≥ 0 and t ∈ [0, T ] Z t Ks (R) ds. αt (R) ≡

(55.5.55)

0

To begin with, here is a useful lemma which is not unbelievably technical. Lemma 55.5.2 Let (Ω, F, P ) be a probability space and let {fn } be a sequence of measurable functions. Then fn → 0 in probability and only if for every subsen if o quence, {fnk } there exists a further subsequence fnkl which converges pointwise a.e. to 0. Also, if fn → 0 in probability and if g is any real valued measurable function then gfn also converges in probability to 0. Proof: First of all, if {fn } converges in probability, then every subsequence converges in probability also. Now if {fn } does converge in probability, then there exists a subsequence fnk such that ¡£ ¤¢ P |fnk | ≥ 2−k ≤ 2−k £ ¤ Then by the Borel Cantelli lemma, letting Ek denote the set |fnk | ≥ 2−k , it follows the set of ω which is in infinitely many of these sets has measure zero and so for a.e. ω, |fnk (ω)| < 2−k for all k and so this subsequence {fnk } converges to 0 a.e. Now suppose every subsequence has an a.e. convergent subsequence which converges to 0. Why does {fn } converge to 0 in probability? Suppose it doesn’t. Then there is an ε > 0 and a δ > 0 and a subsequence {fnk } such that P ([|fnk | ≥ ε]) ≥ δ. But now, by hypothesis, there is a subsequence fnkl which converges to 0 a.e. and ¯ ¯ ¯ ¯ so for a.e. ω, ¯fnkl (ω)¯ < ε for all l large enough. Consider ¯ ©¯ ª gl (ω) ≡ sup ¯fnkr (ω)¯ : r ≥ l Then {gl } is a decreasing sequence of measurable functions and it converges to 0 a.e. Furthermore, P ([gl ≥ ε]) ≥ δ


1725

and the sets [gl ≥ ε] decrease to a set of measure zero. Hence the above is a contradiction. The last claim follows from this. If fn → 0 in probability, this means every subsequence has an a.e. convergent subsequence converging to 0. This last condition is unchanged if fn is replaced by gfn . This proves the lemma. Note that limn→∞ P ([|gfn | ≥ ε]) = 0 and limn→∞ P ([|fn | ≥ ε]) = 0 could be replaced in the last claim with lim supn→∞ with no change in the conclusion. Some sort of convergence in probability to 0 is preserved when you multiply by a nonzero measurable function. Here is a lemma which will be needed. This is not the unbelievably technical lemma either. Lemma 55.5.3 Let Y (t) ∈ [0, ∞), t ∈ [0, ∞) be a continuous Ft adapted stochastic process and let γ be a stopping time. Also let ε > 0. Let β ε ≡ inf {t : Y (t) ≥ ε} The usual conventinon that inf ∅ = ∞ is followed. Then Ã" #! 1 P sup Y (t) ≥ ε ≤ E (Y (β ε ∧ γ)) . ε t∈[0,γ] Proof: From the continuity of Y (t) and definition of β ε , " # sup Y (t) ≥ ε = [β ε ≤ γ]

t∈[0,γ]

To see this show instead that "

# sup Y (t) < ε = [β ε > γ]

t∈[0,γ]

Why is this so? If ω ∈ left side, then for some δ > 0, Y (t) < ε − δ for all t ∈ [0, γ] . By continuity, if Y (t) ≥ ε, it follows t > γ + η for some η > 0. Hence taking the inf of such t gives β ε ≥ γ + η and so the left side is contained in the right side. Now suppose ω is in the right side. Then β ε > γ + η for some η > 0. Therefore, if Y (t) ≥ ε, it follows t ≥ γ + η and so if t ≤ γ you can’t have Y (t) ≥ ε so for such t, Y (t) < ε. Therefore, the right side is contained in the left side. Now it follows " # sup Y (t) ≥ ε = [β ε ≤ γ] ⊆ [Y (β ε ∧ γ) ≥ ε]

t∈[0,γ]

because if β ε ≤ γ, then Y (β ε ∧ γ) = Y (β ε ) and by continuity, Y (β ε ) ≥ ε. Now if Y (β ε ∧ γ) ≥ ε, then you can’t have γ < β ε and so in fact the inclusion in the above

1726


is an equality. Now the desired result follows. Ã" #! εP

sup Y (t) ≥ ε

=

εP ([Y (β ε ∧ γ) ≥ ε])

≤

E (Y (β ε ∧ γ))

t∈[0,γ]

and this proves the lemma. Note the conclusion of the lemma is valid if t → Y (t) is only continuous off a set of measure zero. Next is a localization lemma. This lemma seems like it has very elaborate hypotheses but the conclusion is very attractive. In this lemma, the pn (t) is going to be of the form µ ¶ 1 Xn t − − Xn (t) n because the lemma is a technical result related to obtaining existence to a finite dimensional stochastic process by using the Euler method. The litany of technical assumptions are things which end up holding for the proof of existence using the Euler method. The conclusion of this lemma is that sup |Xn (t) − Xm (t)| → 0 as m, n → ∞ t∈[0,T ]

in probability. This is very nice indeed. The proof of the lemma proceeds by first showing that sup |Xn (t) − Xm (t)| → 0 as m, n → ∞ t∈[0,γ n,m ] in probability where γ n,m (R) is a suitable stopping time. Lemma 55.5.4 Let n ∈ N and Xn (t) be a continuous Rd valued Ft adapted process such that X (0) = X0 an F0 measurable function and dXn (t) = b (t, Xn (t) + pn (t)) dt + σ (t, Xn (t) + pn (t)) dW (t) for t ∈ [0, T ] for pn (t) a progressively measurable process. For n ∈ N and R ≥ 0, let τ n (R) be stopping times such that |Xn (t)| + |pn (t)| ≤ R if t ∈ (0, τ n (R)] a.e. ω ÃZ ! T ∧τ n (R) lim E |pn (s)| ds = 0 n→∞

(55.5.56) (55.5.57)

0

There exists a function r : [0, ∞) → [0, ∞) with limR→∞ r (R) = ∞ and Ã " #! lim lim sup P

R→∞

n→∞

[τ n (R) ≤ T ] ∩

sup

|Xn (s)| ≤ r (R)

=0

(55.5.58)

s∈[0,τ n (R)]

for all t ∈ [0, T ] . Then sup |Xn (t) − Xm (t)| → 0 t∈[0,T ]

in probability as n, m → ∞.

(55.5.59)


1727

Proof: First note that from 55.5.51 you can replace Kt (R) with ³ ´ 2 Kt (R) ∨ sup ||σ (t, x)|| + |b (t, x)| |x|≤R

Since σ and b are continuous as functions of x, you could replace the above sup with the supremum over rational numbers less than or equal to R, a countable set and so the above sup delivers a Ft measurable function. Therefore, with no loss of generality, assume sup |b (t, x)| ≤ Kt (R) . (55.5.60) |x|≤R

You can also replace Kt (R) with Kt (R) ∨ Kt (1) if needed and without loss of generality assume it is also true that for R ≥ 1, Kt (1) ≤ Kt (R) .

(55.5.61)

Next define stopping times τ (R, u) ≡ inf {t ≥ 0 : αt (R) > u} Then from 55.5.54 it follows that for a.e. ω, the integral in 55.5.54 is less than some u and so τ (R, u) = ∞ for such u. Thus lim τ (R, u) = ∞.

u→∞

Let u (R) denote the smallest positive integer such that P ([τ (R, u (R)) ≤ R]) ≤

1 R

Thus letting τ (R) ≡ τ (R, u (R)) , it follows the set of ω such that αt (R) exceeds u (R) for some t ≤ R has measure no more than R−1 . Thus P ([τ (R) > R]) > 1 −

1 R

(55.5.62)

and so τ (R) → ∞ in probability as R → ∞. Also αt∧τ (R) (R) ≤ u (R)

(55.5.63)

for all t ∈ [0, T ]. This is because τ (R) is the first time αt equals u (R) so in the above, if t < τ (R) , then strict inequality takes place and if t ≥ τ (R) then equality takes place. Now it is desired to show that the litany of technical conditions holds for the stopping times τ n (R) ∧ τ (R) . This is obvious for 55.5.56 and 55.5.57 because by making the stopping time smaller, it only makes these two conditions hold better. Consider the condition 55.5.58. For convenience, this condition is Ã " #! lim lim sup P

R→∞

n→∞

[τ n (R) ≤ T ] ∩

sup s∈[0,τ n (R)]

|Xn (s)| ≤ r (R)

=0

1728


I have to show it still holds if τ n (R) is replaced with τ n (R) ∧ τ (R) . However, Ã P

"

#!

[τ n (R) ∧ τ (R) ≤ T ] ∩

sup

|Xn (s)| ≤ r (R)

s∈[0,τ n (R)∧τ (R)]

  smaller one is τ n (R) z }| { " #      sup |Xn (s)| ≤ r (R) ∩ [τ n (R) ≤ τ (R)] ≤ P [τ n (R) ≤ T ] ∩  s∈[0,τ n (R)]  



 τ n (R)>τ (R) z }| {   +P [τ n (R) > τ (R)] ∩ [τ (R) ≤ T ]

This last converges to 0 as R → ∞ independently of n because as shown above in 55.5.62 τ (R) → ∞ in probability and it is smaller than P ([τ (R) ≤ T ]) . Then it follows Ã lim lim sup P

R→∞

n→∞

" [τ n (R) ∧ τ (R) ≤ T ] ∩

sup

|Xn (s)| ≤ r (R)

s∈[0,τ n (R)∧τ (R)]

Ã ≤ lim lim sup P R→∞

#!

n→∞

"

[τ n (R) ≤ T ] ∩

#! sup

|Xn (s)| ≤ r (R)

= 0.

s∈[0,τ n (R)]

This shows τ n (R) can be replaced with τ n (R) ∧ τ (R). Hence making this change, assume without loss of generality αt∧τ n (R) (R) ≤ u (R) , u (R) → ∞. Claim:

ÃZ lim E

n→∞

(55.5.64)

!

T ∧τ n (R)

|pn (s)| Ks (R) ds

= 0.

0

Proof of claim: First recall that from the technical assumption 55.5.56, |pn (t)| ≤ R


1729

if t ≤ τ n (R). Therefore, for m a positive integer, ÃZ

!

T ∧τ n (R)

E

|pn (s)| Ks (R) ds 0

z ≤

ÃZ

Ks (R)<m

}|

!{

T ∧τ n (R)

mE

|pn (s)| ds

(55.5.65)

0

Ã Z +E R

!

T ∧τ n (R)

Ks (R) X[m,∞) (Ks (R)) ds

0

(55.5.66)

Consider the second term on the right in the above. The expression inside the expectation is dominated by ¡ ¢ R αT ∧τ n (R) (R) ≤ u (R) R by 55.5.64. Now from 55.5.54 which says the integral of Kt (R) is finite, the dominated convergence theorem implies Z

T ∧τ n (R)

lim R

m→∞

Ks (R) X[m,∞) (Ks (R)) ds = 0

0

and now by the dominated convergence theorem again, Ã Z

!

T ∧τ n (R)

lim E

R

m→∞

0

Ks (R) X[m,∞) (Ks (R)) ds

=0

and so the last term of 55.5.66 is less than ε for large enough m. Pick such an m. Then by technical assumption 55.5.57, ÃZ ! T ∧τ n (R)

lim sup E n→∞

≤

|pn (s)| Ks (R) ds 0

ÃZ

lim sup mE n→∞

!

T ∧τ n (R)

|pn (s)| ds

+ε=ε

0

Since ε is arbitrary, this proves the claim. Now it is assumed Z t Z t Xn (t) = X0 + b (s, Xn (s) + pn (s)) ds + σ (s, Xn (s) + pn (s)) dW (s) 0

0

and so Z

t

Xn (t) − Xm (t) =

(b (s, Xn (s) + pn (s)) − b (s, Xm (s) + pm (s))) ds 0

1730


Z

t

+

(σ (s, Xn (s) + pn (s)) − σ (s, Xm (s) + pm (s))) dW (s)

(55.5.67)

0 2

Let ψ t (R) ≡ exp (−2αt (R) − |X0 |) and let F (t, X) ≡ ψ t (R) |X| . Thus d (ψ (R)) = exp (−2αt (R) − |X0 |) (−2Kt (R)) dt t and Fxx = 2ψ t (R) id The Ito formula is F (t, X (t)) = F (0, X (0)) + Fx (·, X (·)) Φ · W (t) + Z

t

{Ft (s, X (s)) + Fx (s, X (s)) (φ (s)) + 0

+

¡ ¢ª 1 ∗ trace Φ (s) Fxx (s, X (s)) Φ (s) ds 2

Then using the Ito formula on 55.5.67 this F, 2

ψ t (R) |Xn (t) − Xm (t)| = Z th 0

2

ψ s (R) (−2Ks (R)) |Xn (s) − Xm (s)| + 2ψ s (R) (Xn (s) − Xm (s) ,

b (s, Xn (s) + pn (s)) − b (s, Xm (s) + pm (s)))Rd ] ds Z + 0

t

(55.5.68) 2

ψ s (R) ||σ (s, Xn (s) + pn (s)) − σ (s, Xm (s) + pm (s))|| ds + M (t) (55.5.69)

where M (t) is a continuous local martingale with M (0) = 0. It isn’t a martingale because the right sort of integrability does not necessarily hold on the term Fx (·, Xn (·) − Xm (·)) (σ (·, Xn (·)) − σ (·, Xm (·))) . However, it can be localized by stopping times of the form τ r ≡ inf {t : |Xn (t) − Xm (t)| + |Xn (t)| + |Xm (t)| ≥ r} . This is where you use the monotonicity technical assumption 55.5.52. Xn (s) − Xm (s)

= Xn (s) + pn (s) − (Xm (s) + pm (s)) +pm (s) − pn (s)

Then substituting this in to 55.5.68, 55.5.69 yields 2

ψ t (R) |Xn (t) − Xm (t)| =

55.5. FINITE DIMENSIONAL PROBLEMS Z

t 0

Z

1731

{ψ s (R) (−2Ks (R)) |Xn (s) − Xm (s)|

2

t

ψ s (R) (pm (s) − pn (s) , b (s, Xn (s) + pn (s)) − b (s, Xm (s) + pm (s))) ds

+ 0

+2ψ s (R) (Xn (s) + pn (s) − (Xm (s) + pm (s)) , b (s, Xn (s) + pn (s)) − b (s, Xm (s) + pm (s)))Rd ds} Z

t

2

+

ψ s (R) ||σ (s, Xn (s) + pn (s)) − σ (s, Xm (s) + pm (s))|| ds + M (t) (55.5.70)

0

Of course M (t) depends on m, n but I am going to take expectation and it will end up being 0 so I will suppress this dependence. Hence from 55.5.52, 2

ψ t (R) |Xn (t) − Xm (t)| ≤ Z

(55.5.71)

t

2ψ s (R) (pm (s) − pn (s) , b (s, Xn (s) + pn (s)) − b (s, Xm (s) + pm (s))) ds

0

Z + Z + 0

2

ψ s (R) (−2Ks (R)) |Xn (s) − Xm (s)| ds

0 t

(55.5.72)

t

³ ´ 2 ψ s (R) Ks (R) |Xn (s) − Xm (s) + (pn (s) − pm (s))| + M (t)

(55.5.73)

At this point, use 55.5.60 which says sup |b (t, x)| ≤ Kt (R) . |x|≤R

If t ≤ τ n (R) ∧ τ m (R) , then from 55.5.56, |Xn (s)| + |pn (s)| ≤ R, |Xm (s)| + |pm (s)| ≤ R and so the integral in 55.5.72 is dominated by Z t 4ψ s (R) Ks (R) |pm (s) − pn (s)| ds 0

It follows the whole mess from 55.5.71 - 55.5.73 implies for t ≤ τ n (R) ∧ τ m (R) Z t 2 ψ t (R) |Xn (t) − Xm (t)| ≤ 4ψ s (R) Ks (R) |pm (s) − pn (s)| ds 0

Z + 0

t

2

ψ s (R) (−2Ks (R)) |Xn (s) − Xm (s)| ds Z

+ 0

t

2

ψ s (R) Ks (R) |Xn (s) − Xm (s)| +

1732


Z

t

0

2ψ s (R) Ks (R) (Xn (s) − Xm (s) , pn (s) − pm (s)) ds Z

t

+ 0

(55.5.74)

2

ψ s (R) Ks (R) |pn (s) − pm (s)| ds + M (t)

Now 55.5.74 is dominated by Z t Z t 2 2 ψ s (R) Ks (R) |Xn (s) − Xm (s)| ds + ψ s (R) Ks (R) |pn (s) − pm (s)| ds 0

0

Replace 55.5.74 with this and observe several things cancel and this implies Z t 2 ψ t (R) |Xn (t) − Xm (t)| ≤ 4ψ s (R) Ks (R) |pm (s) − pn (s)| ds+ 0

Z

t

0

2

2ψ s (R) Ks (R) |pn (s) − pm (s)| ds + M (t)

(55.5.75)

Now note that |pk (s)| ≤ R if k = m, n because t ≤ τ n (R) ∧ τ m (R) and so 2

|pn (s) − pm (s)| ≤ 2R (|pn (s)| + |pm (s)|) It follows from 55.5.75 and that ψ s (R) ≤ 1 2

ψ t (R) |Xn (t) − Xm (t)| Z t ≤ (4 + 2R) Ks (R) (|pn (s)| + |pm (s)|) ds + M (t) 0

Now let δ k be a stopping time like δ k ≡ inf {t : |Xn (t) − Xm (t)| + |Xn (t)| + |Xm (t)| ≥ k} for k ∈ N. Then stopping with δ k and using the localization lemma, Lemma 53.5.4, it follows 2

ψ t∧δk (R) |Xn (t ∧ δ k ) − Xm (t ∧ δ k )| Z t ≤ (4 + 2R) X[0,δk ] (s) Ks (R) (|pn (s)| + |pm (s)|) ds + M (t ∧ δ k ) 0

where the last term is now a martingale. Letting τ be any stopping time, τ ≤ τ n (R) ∧ τ m (R) ∧ t, Then you can replace t with τ in the above using localizaton on the last term and then taking the expectation of both sides ´ ³ 2 E ψ τ ∧δk (R) |Xn (τ ∧ δ k ) − Xm (τ ∧ δ k )| µZ ≤ (4 + 2R) E 0

τ

¶ X[0,δk ] (s) Ks (R) (|pn (s)| + |pm (s)|) ds


1733

See Theorem 50.5.8. Letting k → ∞ and using the dominated convergence theorem (Recall the pk (s) are even bounded.) and Fatou’s lemma on the left, it follows µZ τ ¶ ³ ´ 2 E ψ τ (R) |Xn (τ ) − Xm (τ )| ≤ (4 + 2R) E Ks (R) (|pn (s)| + |pm (s)|) ds 0

(55.5.76) and it was shown in the above claim that as m, n → ∞, the right side of this expression converges to 0. Let τ n (R) ∧ τ m (R) ∧ T ≡ γ n,m . Here is where Lemma 55.5.3 will be used.   P 

sup

s∈[0,γ m,m ]

2 ψ s (R) |Xn (s) − Xm (s)| ≥ ε ≤

¯ ¡ ¢ ¡ ¢¯2 ´ 1 ³ E ψ γ n,m ∧β ε (R) ¯Xn γ n,m ∧ β ε − Xm γ n,m ∧ β ε ¯ ε n o 2 β ε ≡ inf s : ψ s (R) |Xn (s) − Xm (s)| ≥ ε .

where

For fixed ε it follows from 55.5.76 that the term on the right of the inequality in the above converges to 0 as m, n → ∞. Of course for t ∈ [0, T ] , the definition of ψ t (R) implies it is bounded below by a positive number a (ω) > 0 and so this shows from Lemma 55.5.2 that 2 sup |Xn (s) − Xm (s)| (55.5.77) s∈[0,γ m,n ] converges in probability to 0 as m, n → ∞. You can multiply by sups∈[0,γ

m,m

in this lemma. Note also γ n,m = γ n,m (R) . Now consider the following. " # 



2

sup |Xn (t) − Xm (t)| ≥ ε ⊆ 

2

sup

t∈[0,γ n,m (R)]

t∈[0,T ]



] (ψ s (R))

|Xn (t) − Xm (t)| ≥ ε



∪

sup

t∈[γ n,m (R),T ]

 ⊆

³ ´ 2 |Xn (t) − Xm (t)| ≥ ε ∩ [τ n (R) ≤ T ] ∪ [τ m (R) ≤ T ] 

sup

t∈[0,γ n,m (R)]

³ ´ 2 |Xn (t) − Xm (t)| ≥ ε ∪ [τ n (R) ≤ T ] ∪ [τ m (R) ≤ T ]

The reason for the inclusion £ of the union ¤ of the two terms on the end is that if both τ n (R) , τ m (R) > T, then 0, γ n,m (R) = ∅ and so   

sup

t∈[0,γ n,m (R)]

2

|Xn (t) − Xm (t)| ≥ ε = ∅.

−1

1734


Thus Ã" P



#! 2

sup |Xn (t) − Xm (t)| ≥ ε



≤ P 

sup

t∈[0,γ n,m (R)]

t∈[0,T ]

2

|Xn (t) − Xm (t)| ≥ ε

+ P ([τ n (R) ≤ T ]) + P ([τ m (R) ≤ T ])

(55.5.78)

Suppose it is shown that lim lim sup P ([τ n (R) ≤ T ]) = 0.

R→∞

(55.5.79)

n→∞

Then there exists R0 such that if R > R0 , lim sup P ([τ n (R) ≤ T ]) < η n→∞

Then fixing such an R, there exists N such that for n > N, P ([τ n (R) ≤ T ]) < η. Then for that R and for m, n > N, the sum of the two terms in 55.5.64 is less than 2η. This with 55.5.77 will imply the desired conclusion of the lemma in 55.5.59. Therefore, it suffices to show 55.5.79. This is done by using the coercivity assumption 55.5.53 instead of the monotonicity assumption. As pointed out ealier, it can be assumed Kt (R) ≥ Kt (1) for all R ≥ 1. Recall the equation solved by Xn (t) , Z Xn (t) = X0 +

Z

t

b (s, Xn (s) + pn (s)) ds + 0

t

σ (s, Xn (s) + pn (s)) dW (s) 0 2

Also recall ψ t (1) ≡ exp (−2αt (1) − |X0 |) , and this time let F (t, X) ≡ ψ t (1) |X| . Then using the Ito formula for F (t, Xn (t)) Z 2

2

2

ψ t (1) |Xn (t)| = |X0 | e−|X0 | +

0

t

h 2 ψ s (1) (−2Ks (R)) |Xn (s)| + 2

2ψ s (1) (Xn (s) , b (s, Xn (s) + pn (s))) +ψ s (1) ||σ (s, Xn (s) + pn (s))||

i ds

+M (t) where as before, M (t) is a local martingale, M (0) = 0. It depends on R and n but as before, this is suppressed because I am just going to take expectation and it is going to disappear. Using 55.5.53 the above implies Z t ³ 2 2 −|X0 |2 2 ψ t (1) |Xn (t)| ≤ |X0 | e + ψ s (1) (−2Ks (1)) |Xn (s)| + ³ 0 ´´ 2 ψ s (1) Ks (1) 1 + |Xn (s) + pn (s)|

55.5. FINITE DIMENSIONAL PROBLEMS Z

1735

t

+ 0

2ψ s (1) (−pn (s) , b (s, Xn (s) + pn (s))) ds + M (t)

(55.5.80)

Now for t ≤ T ∧τ n (R) ,this implies |Xn (t)|+|pn (t)| ≤ R and so |b (s, Xn (s) + pn (s))| ≤ Ks (R) . Recall Ks (R) ≥ Ks (1). Now it is elementary to show 2

2

|Xn (s) + pn (s)| ≤ 2 |Xn (s)| + 2 |pn (s)|

2

and using these observations along with ψ s (1) ≤ 1 in 55.5.80, this implies Z t 2 2 2 2 ψ t (1) |Xn (t)| ≤ |X0 | e−|X0 | + 2Ks (1) |pn (s)| + ψ s (1) Ks (1) ds 0

Z

t

+

2Ks (1) |pn (s)| ds + M (t) 0

Z 2

2

≤ |X0 | e−|X0 | +

t

2Ks (R) |pn (s)| (1 + |pn (s)|) + e−2αs (1) Ks (1) ds + M (t)

0

The reason for that term e−2αs (1) is that ψ s (1) ≤ e−2αs (1) Now this is convenient because Z t Z e−2αs (1) Ks (1) ds ≤ 0

∞ 0

e−2αs (1) Ks (1) ds =

1 2

Therefore, the above inequality implies Z t 1 2 2 −|X0 |2 ψ t (1) |Xn (t)| ≤ |X0 | e + + 2Ks (R) |pn (s)| (1 + |pn (s)|) ds + M (t) 2 0 As before, let τ be an arbitrary stopping time with τ ≤ T ∧ τ n (R) . Also let δ k be a localizing sequence such that M δk (t) is a martingale. Then first locallizing with δ k and then with τ , it follows from Theorem 50.5.8 that ¡ ¢ ¡ ¡ ¢¢ ¡ ¢ E M δk (τ ) = E E M δk (τ ) |F0 = E M δk (0) = 0 and so letting k → ∞, and using Fatou’s lemma and dominated convergence theorem as before, µZ τ ¶ ³ ´ ³ ´ 1 2 2 Ks (R) |pn (s)| ds E ψ τ (1) |Xn (τ )| ≤ E |X0 | + + 2 (1 + R) E 2 0 (55.5.81) By the claim, the limit as n → ∞ of that last term is 0. Therefore, from Lemma 55.5.3, Ã" #! ´ 1 ³ 2 2 P sup |Xn (t)| ψ t (1) ≥ c ≤ E ψ τ (1) |Xn (τ )| c t∈[0,T ∧τ n (R)]

1736


where here τ ≡ inf {t : Y (t) ≥ c} ∧ T ∧ τ n (R) . From 55.5.81 and claim 1 which says the last term of 55.5.81 converges to 0, Ã" #! µ ³ ´ 1¶ 1 2 2 lim sup P sup |Xn (t)| ψ t (1) ≥ c ≤ E |X0 | + . c 2 n→∞ t∈[0,T ∧τ n (R)] Therefore, Ã"

#!

lim sup P n→∞

2

sup

|Xn (t)|

t∈[0,T ∧τ n (R)]

inf ψ t (1) ≥ c

≤

t∈[0,T ]

1 c

µ ³ ´ 1¶ 2 E |X0 | + 2 (55.5.82)

claim 2: Ã"

#!

lim lim sup P

R→∞

sup

n→∞

2

|Xn (t)| ≥ r (R)

=0

(55.5.83)

t∈[0,T ∧τ n (R)]

Proof of claim: Let g (ω) ≡ inf t∈[0,T ] ψ t (1) . This is a positive function, g (ω) ∈ (0, 1). Let 2 fn (ω) ≡ sup |Xn (t)| t∈[0,T ∧τ n (R)]

Then 55.5.82 is of the form lim sup P ([fn g ≥ c]) ≤ n→∞

1 c

µ ³ ´ 1¶ 2 E |X0 | + 2

Recall r (R) → ∞ so the above implies lim lim sup P ([fn g ≥ r (R)]) = 0

R→∞

n→∞

(55.5.84)

Is the same true for fn g replaced with fn alone? This is the content of the claim. If not, there exists ε > 0 and an increasing sequence Rk → ∞ such that lim sup P ([fn ≥ r (Rk )]) ≥ 2ε n→∞

This requires there a subsequence nk such that P ([fnk ≥ r (Rk )]) > ε Also from 55.5.84, lim P ([fnk g ≥ r (Rk )]) = 0

k→∞

Thus there exists a further subsequence, nkl such that ³h i´ P fnkl g ≥ r (Rkl ) < 2−l


1737

It follows from the h Borel Cantellii lemma, the set N of ω which are in infinitely many of the sets fnkl g ≥ r (Rkl ) has measure 0. Thus for ω ∈ / N it follows that for all l large enough, fnkl (ω) g (ω) < r (Rkl ) Since g (ω) > 0, it follows that for all l large enough fnkl (ω) < r (Rkl ) h i which shows the set of ω contained in infinitely many of the sets fnkl ≥ r (Rkl ) is contained in N . Thus £ ¤ ∞ ∩∞ l=1 ∪m=l fnkm ≥ r (Rkm ) ⊆ N but this is impossible because ¡ £ ¤¢ ∞ P ∩∞ = l=1 ∪m=l fnkm ≥ r (Rkm )

¡ £ ¤¢ lim P ∪∞ m=l fnkm ≥ r (Rkm ) l→∞ ³h i´ lim inf P fnkl ≥ r (Rkl ) ≥ ε

≥

l→∞

and N has measure zero. This proves the claim. Now from Ã" lim lim sup P

R→∞

sup

n→∞

#!

2

|Xn (t)| ≥ r (R)

= 0,

(55.5.85)

t∈[0,T ∧τ n (R)]

it follows lim lim sup P ([τ n (R) ≤ T ]) ≤

R→∞

Ã lim lim sup P

R→∞

n→∞

n→∞

"

[τ n (R) ≤ T ] ∩

R→∞

n→∞

sup

|Xn (t)| ≤ r (R)

t∈[0,T ∧τ n (R)]

Ã + lim lim sup P

#! 2

" [τ n (R) ≤ T ] ∩

#! sup

2

|Xn (t)| > r (R)

t∈[0,T ∧τ n (R)]

and the second of these on the right equals 0 by 55.5.85 and the first equals 0 from 55.5.58. Therefore, 55.5.79 is obtained and this proves the Lemma.

55.5.3

An Existence Theorem

Here an existence theorem is presented for finite dimensional stochastic differential equations. To review, W (t) is a Wiener process on Rd1 with respect to the usual normal filtration Ft . Thus U ≡ Rd1 and Q ≡ I. Let H ≡ Rd . (t, x, ω) → σ (t, x, ω)

1738


is progressively measurable and continuous in x into the space of d × d1 matrices with the usual Frobenius norm. Similarly (t, x, ω) → b (t, x, ω) is progressively measurable and continuous in x into Rd . Assume also the following integrability condition. Z

T

sup 0

n o 2 ||σ (t, x)|| + |b (t, x)| dt < ∞ on Ω

(55.5.86)

|x|≤R

The dependence on ω is suppressed. In addition to this, some other assumptions are made. For |x| , |y| ≤ R, 2

2 (x − y, b (t, x) − b (t, y)) + ||σ (t, x) − σ (t, y)|| ≤ Kt (R) |x − y| ³ ´ 2 2 2 (x, b (t, x)) + ||σ (t, x)|| ≤ Kt (1) 1 + |x|

2

(55.5.87) (55.5.88)

In the above, Kt (R) is an [0, ∞) valued Ft adapted process for each R ≥ 0. It is assumed to satisfy Z T Kt (R) dt < ∞ a.e. ω (55.5.89) 0

Also define for each R ≥ 0 and t ∈ [0, T ] Z αt (R) ≡

t

Ks (R) ds.

(55.5.90)

0

Here is the main theorem. Theorem 55.5.5 Under the above conditions there exists a solution to the stochastic differential equation dX = A (t, X) dt + σ (t, X) dW, X (0) = X0 where X0 is F0 measurable and in L2 (Ω). If the monotonicity condition 55.5.52 is modified to have Z T Kt (R) dt ≡ ln C < ∞ (55.5.91) sup sup R≥0 ω∈Ω

0

then the solution is unique. In fact if X0k → X0 in probability, then letting Xk denote the above solution corresponding to initial condition X0k , it follows sup |Xk (t) − X (t)| → 0 t∈[0,T ]

in probability.

(55.5.92)


1739

Proof: Consider Xn the solution to the following approximate problem ¶¶ ¶¶ µ µ Z t µ Z t µ 1 1 Xn (t) = X0 + b s, Xn s − ds + σ s, Xn s − dW (s) n n 0 0 The explanation for this is that X (−s) ≡ X0 for all s ≥ 0. This last term is only a local martingale. One can use a localizing sequence of the form ) ( Z ¯¯ µ µ ¶¶¯¯2 t ¯¯ ¯¯ 1 ¯ ¯ ¯ ¯ τ m = inf t : ¯¯σ s, Xn s − n ¯¯ ds > m 0 Let

µ pn (t) ≡ Xn

and define

1 t− n

¶ − Xn (t)

½ ¾ R τ n (R) ≡ inf t ≥ 0 : |Xn (t)| > 3

Then if t ∈ (0, τ n (R)],

¯ ¶¯ µ ¯ 1 ¯¯ 2R |pn (t)| ≤ ¯¯Xn t − + |Xn (t)| ≤ ¯ n 3

and so τ n (R) is such that for t ≤ τ n (R) |Xn (t)| + |pn (t)| ≤ R These stopping times then satisfy the condition 55.5.56 of Lemma 55.5.4. Let R r (R) ≡ . 4 That horribly technical condition 55.5.58 is satisfied trivially. It involved the limit as R → ∞ of the lim sup of the following is equal to 0. Ã " #! P

[τ n (R) ≤ T ] ∩

sup

|Xn (s)| ≤ r (R)

s∈[0,τ n (R)]

However, the set inside is ∅. This is because if τ n (R) ≤ T, then |Xn (s)| > R/4 = r(R) whenever s is close enough to τ n (R) and so the intersection of the two sets above is empty because if ω is in the first of them, then it is not in the second. Next it is necessary to consider 55.5.57. Recall this condition is ÃZ ! T ∧τ n (R) lim E |pn (s)| ds = 0. n→∞

0

First note that |pn (s)| is bounded by R for all s ≤ τ n (R). The idea is to show something about convergence in probability. By definition, µ µ ¶¶ µ µ ¶¶ Z t Z t 1 1 −pn (t) = b s, Xn s − ds + σ s, Xn s − dW (s) n n t−1/n t−1/n

1740


Therefore, P

¡£ ¤¢ |pn (t)| X[0,τ n (R)] (t) ≥ 2ε = P ([|pn (t)| ≥ 2ε] ∩ [t ≤ τ n (R)])

Since only t ≤ τ n (R) are in the above set, |Xn (t)| , |Xn (t − 1/n)| ≤ R #! Ã"Z t

sup |b (s, x)| ds ≥ ε

≤P

+

t−1/n |x|≤R

Ã" P

¯Z ¯ #! ¯ r∧τ n (R) ¯ ¯ ¯ sup ¯ X[t−1/n,T ] (s) σ (s, Xn (s − 1/n)) dW (s)¯ ≥ ε ¯ r∈[0,t] ¯ 0 Ã"Z #! t

=P

sup |b (s, x)| ds ≥ ε

+

t−1/n |x|≤R

Ã" P

¯Z ¯ sup ¯¯

r∈[0,t]

r 0

#! ¯ ¯ X[0,τ n (R)] (s) X[t−1/n,T ] (s) σ (s, Xn (s − 1/n)) dW (s)¯¯ ≥ ε

Now this last term can be estimated by the exotic Corollary 51.4.8 which comes from the amazing Burkolder Davis Gundy inequality of Theorem 51.4.7. It is dominated by ´ ¤1/2 C ³£ E X[0,τ n (R)] X[t−1/n,T ] σ (s, Xn (· − 1/n)) (t) ∧ δ ε ³ ´ £ ¤1/2 +P X[0,τ n (R)] X[t−1/n,T ] σ (·, Xn (· − 1/n)) (t) > δ Then using the characterization of the quadratic variation found in Corollary 53.5.5 this is dominated by an expression of the form Ã  !1/2 Z t C 2 δ + P  sup ||σ (s, x)|| ds > δ  ε t−1/n |x|≤R Thus ¡£ ¤¢ P |pn (t)| X[0,τ n (R)] (t) ≥ 2ε ≤ P

Ã"Z

#!

t

sup |b (s, x)| ds ≥ ε t−1/n |x|≤R

 Ã !1/2 Z t C 2 + δ + P  sup ||σ (s, x)|| ds > δ  ε t−1/n |x|≤R Now first let n → ∞. Then by 55.5.86 lim sup P n→∞

¡£

¤¢ C |pn (t)| X[0,τ n (R)] (t) ≥ 2ε ≤ δ ε


1741

Next since δ is arbitrary, it follows ¡£ ¤¢ lim sup P |pn (t)| X[0,τ n (R)] (t) ≥ 2ε = 0 n→∞

Thus pn (t) X[0,τ n (R)] (t) → 0 in probability. This is enough to verify 55.5.57. By definition of τ n (R) , |pn (t)| X[0,τ n (R)] (t) ≤ R Then using the distribution function, Theorem 9.7.4, ÃZ ! Z Z T ∧τ n (R) T E |pn (s)| ds = X[0,τ n (R)] (t) |pn (t)| dP dt 0

0

Z

T

Z

R

=

P 0

0

Ω

¤¢ ¡£ X[0,τ n (R)] (t) |pn (t)| > λ dλdt

and now by the convergence in probability and the above bound along with the dominated convergence theorem, this converges to 0. This verifies 55.5.57. It follows the conditions of Lemma 55.5.4 are satisfied and so sup |Xn (t) − Xm (t)| → 0 t∈[0,T ]

in probability. Let nk be such that if m, n ≥ nk , then #! Ã" sup |Xn (t) − Xm (t)| ≥ 2−k

P

≤ 2−k

t∈[0,T ] ∞

It follows there exists a set of measure zero N such that if ω ∈ / N then {Xnk (t)}k=1 is uniformly Cauchy. Thus there exists a continuous t → X (t) such that Xnk converges uniformly to X. Now recall Z Xnk (t) = X0 +

0

Z

t

b (s, Xnk (s − 1/n)) ds +

t 0

σ (s, Xnk (s − 1/nk )) dW (s)

(55.5.93) and it is necessary to pass to the limit. Consider the deterministic integral first. For ω ∈ /N ¯Z t ¯ Z t ¯ ¯ ¯ b (s, Xnk (s − 1/nk )) ds − b (s, X (s)) ds¯¯ ¯ 0

≤

0

¯Z t ¯ Z t ¯ ¯ ¯ b (s, Xnk (s − 1/nk )) ds − b (s, X (s − 1/nk )) ds¯¯ ¯ 0 0 ¯Z t ¯ Z t ¯ ¯ ¯ +¯ b (s, X (s − 1/nk )) ds − b (s, X (s)) ds¯¯ (55.5.94) 0

0

1742


In order to pass to a limit, define S (t) ≡ sup |Xnk (t − 1/nk )| k∈N

Now by uniform convergence, |Xnk (t − 1/nk )| ≤ |Xnk (t − 1/nk ) − X (t − 1/nk )| + |X (t − 1/nk )| Hence S (t) is bounded independent of t ∈ [0, T ] for each ω ∈ / N. Now define τ (R) ≡ inf {t ∈ [0, T ] : S (t) > R} Then it follows from the integrability condition 55.5.86 and the continuity of b in x and 55.5.94 that Z t Z t lim X[0,τ (R)] b (s, Xnk (s − 1/nk )) ds = X[0,τ (R)] b (s, X (s)) ds k→∞

0

0

Next consider the stochastic term. Let another stopping time be defined. ( ) Z t 2 τ N (R) ≡ inf t ∈ [0, T ] : sup ||σ (s, x)|| ds > N ∧ τ (R) 0 |x|≤R

Then by 55.5.86 again and dominated convergence theorem ÃZ τ N (R)

lim E

k→∞

0

! 2

||σ (s, Xnk (s − 1/nk )) − σ (s, X (s))|| ds

=0

Note how 55.5.86 provides the dominating function and then the convergence of the integrand comes from continuity of σ in the second argument. By the Ito isometry, it follows as k → ∞, Z t Z t X[0,τ N (R)] σ (s, Xnk (s − 1/nk )) dW (s) → X[0,τ N (R)] σ (s, X (s)) dW (s) 0

0

in L2 (Ω) . Hence this convergence also takes place in probability. Thus for each N ∈ N there exists a set of measure zero, SN and a further subsequence denoted by {j} such that for ω ∈ / SN Z 0

Z

t

X[0,τ N (R)] σ (s, Xj (s − 1/j)) dW (s) →

0

t

X[0,τ N (R)] σ (s, X (s)) dW (s)

pointwise. It is eventually the case that for N large enough, τ N (R) = τ (R) . This follows from the integrability condition again 55.5.86, ( ) Z t

inf

t ∈ [0, T ] :

2

sup ||σ (s, x)|| ds > N 0 |x|≤R

=∞


1743

if N is large enough. Then if S ≡ ∪N ∈N SN it is a set of measure zero and using a Cantor diagonalization procedure if necessary, there exists a subsequence {i} such that if ω ∈ / S, then Z 0

Z

t

X[0,τ N (R)] σ (s, Xi (s − 1/i)) dW (s) →

t

X[0,τ N (R)] σ (s, X (s)) dW (s)

0

pointwise for each N . Using this subsequence in 55.5.93, it follows that for a.e. ω, Z

Z

t

X (t) = X0 + 0

X[0,τ (R)] b (s, X (s)) ds +

t

X[0,τ (R)] σ (s, X (s)) dW (s)

0

where τ (R) ≡ inf {t ∈ [0, T ] : S (t) > R} . Since sup S (t) < ∞ t∈[0,T ]

for a.e. ω, it follows that for any t ∈ [0, T ] , eventually for R large enough, τ (R) = ∞ and so the above equation implies Z

Z

t

X (t) = X0 +

t

b (s, X (s)) ds + 0

σ (s, X (s)) dW (s) 0

This proves the existence part of the theorem. Now consider the uniqueness part, 55.5.92. From assumption Z Xk (t) = X (t) =

Z

t

t

X0k + b (s, Xk (s)) ds + σ (s, Xk (s)) dW (s) (55.5.95) 0 0 Z t Z t X0 + b (s, X (s)) ds + σ (s, X (s)) dW (s) (55.5.96) 0

0

Since X0k → X0 in probability, it follows sup |X0k | < ∞ k

for a.e. ω. This is because X0 has values in Rd . Consider then the positive function 2

2

F (t, x) ≡ e−αt (R)−(supk |X0k |+|X0 |) |x| ≡ φt (R) |x| where

Z αt (R) ≡

t

Ks (R) ds. 0

Note

µ µ ¶¶ 1 φτ (R) ≥ φT (R) ≥ exp − sup |X0k | + |X0 | C k

1744


which is positive. As before, Ft (t, x) Fxx (t, x)

= =

2

(−Kt (R)) φt (R) |x| , Fx (t, x) = 2φt (R) x, 2φt (R) id

Then by the Ito formula 54.1.1 applied to the difference of 55.5.95 and 55.5.96, φt (R) |Xk (t) − X (t)| =

2 2

e−(supk |X0k |+|X0 |) |X0k − X0 | Z t + 2φs (R) (b (s, Xk (s)) − b (s, X (s)) , Xk (s) − X (s)) ds 0

Z + 0

t

2

φs (R) ||σ (s, Xk (s)) − σ (s, X (s))|| ds + M (t)

where M (t) is a local martingale equal to 0 when t = 0 which equals a suitable Ito integral. Define the stopping time τ k (R) τ k (R) ≡ inf {t ∈ [0, T ] : |X (t)| + |Xk (t)| > R} Then by 55.5.52 it follows that if t ≤ τ k (R) , then the estimate shown there holds. Thus φt (R) |Xk (t) − X (t)|

2

2

≤ e−(supk |X0k |+|X0 |) |X0k − X0 | Z t 2 + Ks (R) φs (R) |Xk (s) − X (s)| ds + M (t) 0

Now by Gronwall’s inequality, ³ ´ 2 2 φt (R) |Xk (t) − X (t)| ≤ C e−(supk |X0k |+|X0 |) |X0k − X0 | + M (t) where C is given as in 55.5.91. Thus CM (t) is also a local martingale and to save notation denote this product by M (t). Localizing with τ a stopping time ≤ τ k (R) and then taking the expectation using Theorem 50.5.8, it follows E (M (τ ) |F0 ) = M (0) = 0 and so E (M (τ )) = E (E (M (τ ) |F0 )) = E (M (0)) = 0 so the above implies ³ ´ ³ ´ 2 2 E φτ (R) |Xk (τ ) − X (τ )| ≤ E Ce−(supk |X0k |+|X0 |) |X0k − X0 | It follows from Lemma 55.5.3 that Ã" P

sup t∈[0,T ∧τ k (R)]

#! 2

φt (R) |Xk (t) − X (t)| ≥ ε

=


1745

Ã"

#! 2

P

sup φt∧τ k (R) (R) |Xk (t ∧ τ k (R)) − X (t ∧ τ k (R))| ≥ ε

t∈[0,T ]

´ 1 ³ ´ 1 ³ 2 2 E φτ (R) |Xk (τ ) − X (τ )| ≤ E Ce−(supk |X0k |+|X0 |) |X0k − X0 | ε ε (55.5.97) where n o 2 τ ≡ inf t ∈ [0, T ] : |Xk (t) − X (t)| ≥ ε ∧ τ k (R) ≤

2

Now Ce−(supk |X0k |+|X0 |) |X0k − X0 | converges to 0 in probability as k → ∞. Also there is some constant K independent of k such that ³ ´ 2 2 2 e−(supk |X0k |+|X0 |) |X0k − X0 | ≤ 2e−(supk |X0k |+|X0 |) |X0k | + |X0 | ≤ K Thus from the distribution function, Theorem 9.7.4 ³ ´ 2 E Ce−(supk |X0k |+|X0 |) |X0k − X0 | Z K ³h i´ 2 = C P e−(supk |X0k |+|X0 |) |X0k − X0 | > x dx 0

and so by the dominated convergence theorem and the assumption X0k → X0 in probability, this converges to 0. Thus from 55.5.97 and the observation that φt (R) ≥ φT (R) is positive, it follows Ã" #! 2

sup |Xk (t ∧ τ k (R)) − X (t ∧ τ k (R))| ≥ ε

lim sup P k→∞

= 0.

t∈[0,T ]

It only remains to get rid of the τ k (R) in the above. P ([τ k (R) < T ]) ≤ Ã"

#!

P

sup |Xk (t ∧ τ k (R))| + |X (t ∧ τ k (R))| ≥ R t∈[0,T ]

Now |Xk (t ∧ τ k (R))| + |X (t ∧ τ k (R))| ≤ |Xk (t ∧ τ k (R)) − X (t ∧ τ k (R))| + 2 |X (t ∧ τ k (R))| and so P ([τ k (R) < T ]) ≤

Ã" P

#!

sup |Xk (t ∧ τ k (R)) − X (t ∧ τ k (R))| ≥ 1 t∈[0,T ]

Ã" +P

#! 2 sup |X (t)| ≥ R − 1 t∈[0,T ]

(55.5.98)

1746


By 55.5.98 lim supk→∞ of the first of these on the right is 0 and since X is continuous, limR→∞ of the second is also 0. Therefore, lim lim sup P ([τ k (R) < T ]) = 0.

R→∞

k→∞

Now with this, the desired result follows. Ã" #! P

sup |Xk (t) − X (t)| ≥ ε

≤ P ([τ k (R) < T ]) +

t∈[0,T ]

Ã"

#

P

!

sup |Xk (t ∧ τ k (R)) − X (t ∧ τ k (R))| ≥ ε ∩ [τ k (R) ≥ T ] t∈[0,T ]

Ã" ≤P

#! sup |Xk (t ∧ τ k (R)) − X (t ∧ τ k (R))| ≥ ε

+ P ([τ k (R) < T ])

t∈[0,T ]

Now take lim supk→∞ and use 55.5.98 to conclude Ã" #! lim sup P k→∞

sup |Xk (t) − X (t)| ≥ ε

≤ lim sup P ([τ k (R) < T ])

t∈[0,T ]

Then take limR→∞ of both sides to conclude Ã" lim sup P k→∞

k→∞

#!

sup |Xk (t) − X (t)| ≥ ε

=0

t∈[0,T ]

and this proves the uniqueness part of the theorem.

55.6

Infinite Dimensional Problems

First Gelfand triples were considered. Then a general Ito formula was presented. Next assumptions were placed on operators which would allow one to pass from finite to infinite dimentional problems assuming a solution to appropriate finite dimensional problems was known. Finally, a general existence and uniqueness theorem was presented for a finite dimensional problem. It only remains to verify that the conditions on the operators which allowed passage from finite dimensional problems to the infinite dimensional problems are sufficient to give a unique solution to the corresponding finite dimensional problem. Recall the conditions which were used to go from finite to infinite dimensional problems. These were as follows. First you have a Gelfand triple V, H, V 0 . Next there is a Q1 ≡ JJ ∗ Wiener process defined on U1 where J : U → U1 is a Hilbert Schmidt operator for U a separable real Hilbert space. W (t) ≡

∞ X k=1

ψ k (t) Jgk

55.6. INFINITE DIMENSIONAL PROBLEMS Wn (t) ≡

n X

1747

ψ k (t) gk

k=1

Next there are operators A, σ as follows. A : [0, T ] × V × Ω → V 0 , σ : [0, T ] × H × Ω → L2 (U, H) 1. (Hemicontinuity) For a.e. ω λ → hA (t, u + λv, ω) , zi is continuous. 2. (Weak monotonicity) There exists c ∈ R such that for all u, v ∈ V, 2

2 hA (·, u) − A (·, v) , u − vi + |σ (·, u) − σ (·, v)|L2 (U,H) ≤

2

c |u − v|H on [0, T ] × Ω

3. (Coercivity) There exist p ∈ (1, ∞), c1 ∈ R, c2 ∈ (0, ∞) , and f ∈ L1 ([0, T ] × Ω) which is B ([0, T ]) × F measurable and is also Ft adapted such that for all v ∈ V, t ∈ [0, T ] , 2

2

p

2 hA (t, v) , vi + ||σ (t, v)||L2 (U,H) ≤ c1 |v|H − c2 ||v||V + f (t) 0

4. (Boundedness) There exist c3 ∈ [0, ∞) and an Ft adapted process g ∈ Lp ([0, T ] × Ω) which is B ([0, T ]) × F measurable such that for all v ∈ V, t ∈ [0, T ] p−1

||A (t, v)||V 0 ≤ g (t) + c3 ||v||V

.

Here 1/p + 1/p0 = 1. 5. (Measurability) Let A : [0, T ] × V × Ω → V 0 , σ : [0, T ] × H × Ω → L2 (U ; H) be progressively measurable. Next recall the finite dimensional problems. First there is the orthogonal projection Pn which maps V 0 onto Hn , a finite dimensional subspace of V which consists ∞ of the span of the first n vectors in {ek }k=1 , an orthonormal basis for H with the property that each ek ∈ V . This is given by Pn y ≡

n X

hy, ek i ek .

k=1

For σ ∈ L2 (U, H) , Pn σ (u) ≡

n X k=1

(σu, ek ) ek

1748


Then the finite dimensional problem is of the form dXn Xn (0)

= Pn A (t, Xn (t)) + Pn σ (t, Xn (t)) dWn (t) = Pn X0

where to begin with X0 ∈ V. For each Xn ∈ Hn , Pn σ (t, Xn (t)) ∈ L2 (U, Hn ). In particular, it can be considered in L2 (Un , Hn ) by taking its restriction to span (g1 , · · · , gn ) where {gk } is an orthonormal basis for U . Thus identifying Hn with Rn and span (g1 , · · · , gn ) with Rn also, one can regard Pn σ (t, Xn (t)) as an n × n matrix or linear transformation from Rn to Rn . It only remains to verify the conditions for existence for finite dimensional problems. That Pn A (t, x) is a continuous as a function of x ∈ Hn follows from the fact proved in Proposition 55.4.5 that X → A (t, X) is demicontinuous and the observation that Hn is finite dimensional. That Pn σ (t, x) is continuous as a function of x ∈ Hn follows from 55.4.36. What about the monotonicity assumptions necessary to prove the above finite dimensional problem? For |x|Hn , |y|Hn ≤ R, 2

2 (x − y, Pn A (t, x) − Pn A (t, y))Hn + ||Pn σ (t, x) − Pn σ (t, y)|| 2

2

≤ 2 hA (t, x) − A (t, y) , x − yi + ||σ (t, x) − σ (t, y)|| ≤ c |x − y|Hn by the property assumed on A in 2. This certainly implies the monotonicity condition used in proving both existence and uniqueness in the finite dimensional problem 55.5.87, 55.5.91. In fact this gives a stronger condition because there was no provision for fussing with the size of |x| , |y| . Next consider whether the coercivity condition 55.5.88, 2

2 (x,Pn A (t, x)) + ||Pn σ (t, x)|| 2

≤ 2 hA (t, x) , xi + ||σ (t, x)||

and by the coercivity condition 3, this is dominated by 2

p

2

≤ c1 |x|Hn − c2 ||x||V + f (t) ≤ c1 |x|Hn + f (t)

(55.6.99)

³ ´ ³ ´ 2 2 ≤ f (t) 1 + |x|Hn + c1 1 + |x|Hn ³ ´ 2 = (c1 + f (t)) 1 + |x|Hn and so condition 55.5.53 is verified with Kt (1) = (c1 + f (t)) . The integrability condition 55.5.51 is satisfied because of 4 and 55.4.37. This completes the proof of the following theorem. Theorem 55.6.1 Suppose 1 - 5 on the operators A and σ. Then there exists a unique solution to the stochastic differential equation dX

=

X (0) =

A (t, X (t)) dt + σ (t, X (t)) dW, X0 ∈ Lp (Ω; V ) ∩ L2 (Ω; H)

55.6. INFINITE DIMENSIONAL PROBLEMS

1749

More precisely, Z X (t)

t

= X0 +

¡

¢ A s, X (s) ds +

0

X0

∈

Z

t

σ (s, X (s)) dW (s) , 0

Lp (Ω; V ) ∩ L2 (Ω; H) , X0 is F0 measurable

where X is a measurable representative of X which is in Lp ([0, T ] × Ω; V ) ∩ L2 ([0, T ] × Ω; H) and is V and H progressively measurable. Proof: Under the conditions 1 - 5 there exists a unique solution to the finite dimensional problem 55.4.8. Therefore, the conclusion of this theorem follows from Theorem 55.4.10.

1750


Another Approach, Semigroups 56.1

Stochastic Fubini Theorem

Let (t, ω, x) → Φ (t, ω, x) be PT × B (E) measurable. This is product measure where PT is the predictable σ algebra and E is a real separable Banach space. Let µ be a finite measure defined on B (E). Suppose also that Z ÃZ Z E

Ω

T 0

!1/2 2 ||Φ||L0 2

dtdP

dµ < ∞

(56.1.1)

R RT 2 It follows that for µ a.e. x, Ω 0 ||Φ|| dtdP < ∞ and so for these values of x it makes perfect sense to write the stochastic integral Φ (·, ·, x) · W (T ) . Does it also make sense to write Z Φ (·, ·, x) · W (T ) dµ?

(56.1.2)

E

You would need to have a set of µ measure 0 such that for E0 the complement of this set, x → XE0 (x) Φ (·, ·, x) · W (T ) (ω) is Borel measurable and also Z |XE0 (x) Φ (·, ·, x) · W (T )|H dµ < ∞ E

at least for a.e. ω. 1751

(56.1.3)

1752

ANOTHER APPROACH, SEMIGROUPS

From the Ito Isometry and 56.1.1, Z ÃZ Z ∞

T

> E

Ω

0

Z µZ ³ = E

Ω

!1/2 2 ||Φ||L0 2

dtdP

dµ

¶1/2 ´ 2 |Φ (·, ·, x) · W (T )|H dP (ω) dµ (x)

Z µZ ³ = ZE Z ≥ E

|XE0 (x) Φ (·, ·, x) · W

Ω

Ω

2 (T )|H

´

¶1/2 dP (ω) dµ (x)

|XE0 (x) Φ (·, ·, x) · W (T )|H dP (ω) dµ (x)

If it is possible to interchange the integrals, it would follow 56.1.3 would hold for a.e. ω. Thus it suffices to show for some set E0 , the complement of a set of µ measure 0, (ω, x) → XE0 (x) Φ (·, ·, x) · W (T ) (ω) is FT × B (E) measurable because this will establish both 56.1.2 and 56.1.3. Here is an approximation lemma. Lemma 56.1.1 Suppose (t, ω, x) → Φ (t, ω, x) is PT × B (E) measurable and Z ÃZ Z E

Ω

T 0

!1/2 2 ||Φ||L0 2

dtdP

dµ < ∞

(56.1.4)

Then there exists a sequence {Φn } of mappings which are of the form Φn (t, ω, x) ≡

Jn X Kn X Ln X

n (ω) αnkl Hj φnjk (x) X(snkl ,tnkl ] (t) XFkl

j=1 k=1 l=1

where the intervals (snkl , tnkl ] are contained in [0, T ] , αnkl are constants, Hj are in n L2 (U, H) , φnjk (x) are bounded Borel measureable functions, and Fkl ∈ Fsnkl such that also !1/2 Z ÃZ Z T

lim

n→∞

E

Ω

0

2

||Φ (t, ω, x) − Φn (t, ω, x)||L0 dtdP 2

dµ = 0.

(56.1.5)

Now assuming the approximation lemma, it is obvious from the formula that (ω, x) → Φn (·, ·, x) · W (T ) (ω) is FT × B (E) measurable and in any other way desired. From 56.1.5 Z ÃZ Z lim

m→∞

!1/2

T

||Φm (t, ω, x) − E

Ω

0

2 Φ (t, ω, x)||L0 2

dtdP

dµ = 0

56.1. STOCHASTIC FUBINI THEOREM

1753

and so by the Ito isometry, Z µZ lim

m→∞

|Φm (·, ·, x) · W (T ) − Φ (·, ·, x) · W E

Ω

2 (T )|H

¶1/2 dP

and there exists a subsequence, mk such that Ã"µZ µ Ω

|Φmk (·, ·, x) · W (T ) − Φ (·, ·, x) · W

2 (T )|H

dµ = 0

#!

¶1/2 dP

(56.1.6)

≥2

−k

≤ 2−k

and so for x off a set of measure zero, Z 2 lim |Φmk (·, ·, x) · W (T ) − Φ (·, ·, x) · W (T )|H dP = 0 k→∞

Ω

which implies Z lim

k→∞

Ω

|Φmk (·, ·, x) · W (T ) − Φ (·, ·, x) · W (T )|H dP = 0.

Letting E0 denote the complement of this exceptional set, the following holds for all x. Z lim |XE0 (x) Φmk (·, ·, x) · W (T ) − XE0 (x) Φ (·, ·, x) · W (T )|H dP = 0 k→∞

Ω

showing that (ω, x) → XE0 (x) Φ (·, ·, x) · W (T ) (ω) is FT × B (E) measurable. It follows it makes sense to write Z Φ (·, ·, x) · W (T ) dµ. E

Now it is reasonable to ask whether this equals Z Φ (·, ·, x) dµ · W (T ) .

(56.1.7)

E

First consider whether this expression makes sense. It clearly does for Φn in place of Φ because by construction, Z Φn (·, ·, x) dµ E

is L02 predictable and satisfies ¯¯Z ¯¯2 ¯¯ ¯¯ ¯¯ Φn (t, ω, x) dµ¯¯ dtdP ¯¯ ¯¯ 0 Ω 0 E L2 µZ ¶ Z Z T 2 C ||Φn (t, ω, x)||L0 dµ dtdP < ∞

Z Z

≤

Ω

T

0

E

2

1754


Also from 56.1.5 Z ÃZ Z

!1/2

T

lim

||Φ (t, ω, x) −

n→∞

E

Ω

0

2 Φn (t, ω, x)||L0 2

dtdP

dµ = 0

and so by Minkowski’s inequality, (Remember Φ was PT × B (E) measurable.) 

Z Z

T

lim 

n→∞

Ω

0

1/2 Ã¯¯Z ¯¯ ! 2 ¯¯ ¯¯ ¯¯ (Φ (t, ω, x) − Φn (t, ω, x)) dµ¯¯ dtdP  =0 ¯¯ ¯¯ 0 E

which shows

L2

Z

Z Φn (·, ·, x) dµ → E

Φ (·, ·, x) dµ E

in L2 (Ω × [0, T ] ; PT ) . It follows 56.1.7 at least makes sense and also, from the Ito isometry, Z Z Φn (·, ·, x) dµ · W (T ) → Φ (·, ·, x) dµ · W (T ) E

E

2

in L (Ω × [0, T ] ; PT ). From 56.1.6 and the measurability result above, Minkowski’s inequality yields Z µZ 0 = lim

m→∞

|Φm (·, ·, x) · W (T ) − Φ (·, ·, x) · W E

Ω

2 (T )|H

¶1/2 dP

dµ

ÃZ ¯Z !1/2 ¯2 ¯ ¯ ¯ ¯ ≥ lim ¯ (Φm (·, ·, x) · W (T ) − Φ (·, ·, x) · W (T )) dµ¯ dP m→∞ Ω

so that

E

H

Z

Z Φm (·, ·, x) · W (T ) dµ (x) →

E

Φ (·, ·, x) · W (T ) dµ (x) E

in L2 ([0, T ] × Ω, PT ) . Thus Z Φn (·, ·, x) dµ · W (T ) ZE =

Φn (·, ·, x) · W (T ) dµ ZE

→

Φ (·, ·, x) · W (T ) dµ (x) E

Hence

Z

Z Φ (·, ·, x) · W (T ) dµ (x) = E

Now it is time to prove Lemma 56.1.1.

Φ (·, ·, x) dµ · W (T ) . E

56.1. STOCHASTIC FUBINI THEOREM

1755

Lemma 56.1.2 Suppose (t, ω, x) → Φ (t, ω, x) is PT × B (E) measurable and Z ÃZ Z E

Ω

T 0

!1/2 2 ||Φ||L0 2

dtdP

dµ < ∞

(56.1.8)

Then there exists a sequence {Φn } of mappings which are of the form Φn (t, ω, x) ≡

Jn X Kn X Ln X

n (ω) αnkl Hj φnjk (x) X(snkl ,tnkl ] (t) XFkl

(56.1.9)

j=1 k=1 l=1

where the intervals (snkl , tnkl ] are contained in [0, T ] , αnkl are constants, Hj are in n L2 (U, H) , φnjk (x) are bounded Borel measureable functions, and Fkl ∈ Fsnkl such that also Z ÃZ Z lim

n→∞

!1/2

T

||Φ (t, ω, x) − E

Ω

0

2 Φn (t, ω, x)||L0 2

dtdP

dµ = 0.

(56.1.10)

Proof: First without loss of generality, it can be assumed Φ (t, ω, x) is bounded. Next, there exists a complete orthonormal set {Hj } for L2 (U0 , H) such that each Hj ∈ L2 (U, H). The reason for this is that L2 (U, H) is dense in L2 (U0 , H) . Letting {Gk } be an orthonormal basis for L2 (U, H) , you can use the Gram Schmidt process with the inner product in L2 (U0 , H) to obtain an orthonormal basis for L2 (U0 , H), {Hk } in which each Hk happens to be in L2 (U, H). Then one can obtain an approximation Jn X Hj φj (t, ω, x) j=1

where φj (t, ω, x) ≡ (Φ (t, ω, x) , Hj )L0 2

and Jn is such that the sum up to Jn is closer to Φ in the norm of 56.1.8 than 1/n. Now since (t, ω) → Φ (t, ω, x) is L02 predictable, (t, ω) → φj (t, ω, x) is in L2 ([0, T ] × Ω, PT ). Also from the assumption that Φ is product measurable with respect to PT × B (E) , the mapping x → φj (t, ω, x) is B (E) measurable. Now let {εk (t, ω)} be an orthonormal basis in L2 ([0, T ] × Ω, PT ) . Then there exists Kn such that Jn X Kn X j=1 k=1

¡ ¢ Hj φj (·, ·, x) , εk (·, ·) L2 ([0,T ]×Ω) εk (t, ω)

1756


is closer to Φ in the norm of 56.1.8 than 1/n. Denote by φnj,k (x) the inner product in the above expression to have Jn X Kn X

Hj φnj,k (x) εk (t, ω)

j=1 k=1

is closer to Φ in the norm of 56.1.8 than 1/n and x → φnj,k (x) is a bounded Borel measurable function having values in R due to the assumption that Φ was bounded as well as Borel measurable in x. Finally use Proposition 53.2.6 to approximate the predictable functions εk (t, ω) with elementary functions. Doing this, there exists Ln such that Jn X Kn Ln X X n Hj φnj,k (x) αnk,l X(snk,l ,tnk,l ] (t) XFk,l j=1 k=1

l=1

is closer to Φ in the norm of 56.1.8 than 1/n. This proves the lemma. This completes the proof of the stochastic Fubini theorem. Theorem 56.1.3 Let Φ be PT × B (E) measurable for E a separable Banach space and PT the predictable σ algebra. Also assume !1/2 Z ÃZ Z T

E

Ω

0

2

||Φ||L0 dtdP 2

dµ < ∞

where µ is a finite measure on B (E). Let {W (t)} be a Q Wiener process. Then the following equation holds and both sides make sense. Z Z Φ (·, ·, x) · W (T ) dµ (x) = Φ (·, ·, x) dµ (x) · W (T ) . E

56.2

E

Correlation Operators

Recall that for a Q Wiener process E ((W (t) − W (s) , h) (W (t) − W (s) , k)) = (t − s) (Qh, k) where Q is a trace class operator. Also the Ito integral was discussed above. The correlation operators ¡are a little ¢ like this but involving two different Ito integrals. 2 Recall that Φ ∈ NW 0, T ; L02 meant (t, ω) → Φ (t, ω) is L02 ≡ L2 (U0 , H) (where U0 = Q1/2 U ) predictable and Z Z T 2 ||Φ (t, ω)||L0 dtdP < ∞. Ω

2

0

Then for such Φ the Ito ¡integral, ¢Φ · W (t) was defined. It was also defined using locallization for Φ ∈ NW 0, T ; L02 which meant that Ã"Z #! T 2 P ||Φ (t, ω)||L0 dt < ∞ = 1. 0

2

56.2. CORRELATION OPERATORS

1757

¡ ¢ 2 Definition 56.2.1 For Φi , i = 1, 2 in NW 0, T ; L02 , the correlation operators are defined by Z Z t∧s ³ ´³ ´∗ V (t, s) ≡ Φ2 (r) Q1/2 Φ1 (r) Q1/2 drdP ∈ L (H, H) (56.2.11) Ω

0

Lemma 56.2.2 The above definition makes sense. Proof: It is necessary to show Z t∧s ³ ´³ ´∗ ω→ Φ2 (r) Q1/2 Φ1 (r) Q1/2 dr

(56.2.12)

0

is measurable and that the above function of ω makes sense. To do this, it suffices to show Z t∧s ¯¯³ ´³ ´ ∗ ¯¯ ¯¯ ¯¯ dr < ∞ ¯¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 ¯¯ L(H,H)

0

because then 56.2.12 will exist, there being no measurability problems due to the assumption that each Φi is predictable. Let f, h ∈ H ¯³³ ´³ ´∗ ´ ¯ ¯ ¯ ¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 f, h ¯ H

¯³ ´ ∗ ¯ ¯³ ´∗ ¯ ¯ ¯ ¯ ¯ ¯ Φ2 (r) Q1/2 h¯ ¯ Φ1 (r) Q1/2 f ¯ U U ¯¯³ ¯¯³ ´ ∗ ¯¯ ´∗ ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ 1/2 1/2 ¯¯ Φ2 (r) Q ¯¯ ¯¯ Φ1 (r) Q ¯¯

≤ ≤

L2 (H,U )

which shows

L2 (H,U )

|f |H |h|H

¯¯³ ´³ ´ ∗ ¯¯ ¯¯ ¯¯ ¯¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 ¯¯ L(H,H) ¯¯³ ¯¯³ ´∗ ¯¯ ´ ∗ ¯¯ ¯¯ ¯ ¯ ¯ ¯ ¯¯ ¯¯ Φ2 (r) Q1/2 ¯¯ ¯¯ Φ1 (r) Q1/2 ¯¯ L2 (H,U ) L2 (H,U ) ¯¯ ¯¯ ¯¯ ¯¯ ¯¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯¯Φ2 (r) Q1/2 ¯¯ ¯¯Φ1 (r) Q1/2 ¯¯

≤ =

L2 (U,H)

Thus

L2 (U,H)

Z ¯¯³ ´³ ´∗ ¯¯ ¯¯ ¯¯ ¯¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 ¯¯ Ω

L(H,H)

dP

µZ ¯¯ ¶1/2 µZ ¯¯ ¶1/2 ¯¯2 ¯¯2 ¯¯ ¯¯ 1/2 ¯¯ 1/2 ¯¯ ≤ ¯¯Φ1 (r) Q ¯¯ dP ¯¯Φ2 (r) Q ¯¯ dP L2

Ω

µZ 2

= Ω

since

||Φ1 (r)||L0 dP 2

L2

Ω

¶1/2 µZ

¶1/2

2

Ω

||Φ2 (r)||L0 dP

(56.2.13)

2

¯¯ ¯¯2 ¯2 X ¯¯ ¯¯ ¯¯ ¯ ¯¯Φ (r) Q1/2 ¯¯ ≡ ¯Φ (r) Q1/2 ek ¯ L2

k

H

1758


=

X³

Φ (r) Q1/2 ek , fj

k,j

X

´2

=

X³

Φ (r)

p

λk e k , f j

´2

k,j 2

2

(Φ (r) gk , fj ) = ||Φ (r)||L0 2

k,j

where C is the largest of the λk . From 56.2.13 Z T Z ¯¯³ ´³ ´ ∗ ¯¯ ¯¯ ¯¯ dP dr ¯¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 ¯¯ L(H,H) 0 Ω Z Z T ¯¯³ ´³ ´ ∗ ¯¯ ¯¯ ¯¯ = drdP < ∞ ¯¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 ¯¯ Ω

L(H,H)

0

Thus for a.e. ω, Z

T

¯¯³ ´³ ´ ∗ ¯¯ ¯¯ ¯¯ ¯¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 ¯¯

0

and so for those ω

Z

T

³

Φ2 (r) Q1/2

´³

L(H,H)

Φ1 (r) Q1/2

´∗

dr < ∞

dr

0

makes sense. Also this makes sense for T replaced with t ∧ s. Also this shows Z Z t∧s ³ ´³ ´∗ Φ2 (r) Q1/2 Φ1 (r) Q1/2 drdP Ω

0

makes sense because Z ¯¯Z t∧s ³ ´³ ´∗ ¯¯¯¯ ¯¯ 1/2 1/2 ¯¯ Φ (r) Q Φ (r) Q dr¯¯¯¯ dP 2 1 ¯¯ Ω 0 Z Z t∧s ¯¯³ ´³ ´∗ ¯¯ ¯¯ ¯¯ ≤ ¯¯ Φ2 (r) Q1/2 Φ1 (r) Q1/2 ¯¯ drdP < ∞ Ω

0

This proves the lemma. With this lemma, here is the main result. ¡ ¢ 2 Theorem 56.2.3 Let Φi , i = 1, 2 be in NW 0, T ; L02 . Then E ((Φ1 · W (t) , a)H (Φ2 · W (s) , b)H ) = (V (t, s) a, b)H

(56.2.14)

where V (t, s) is given in Definition 56.2.1. Proof: First suppose Φi is elementary. Then by adding in points on [0, T ] it can be assumed Φ1 (r, ω) = Φ2 (r, ω) =

n X i=0 n X i=0

X(ti ,ti+1 ] (r) Φi (ω) X(ti ,ti+1 ] (r) Ψi (ω)

56.2. CORRELATION OPERATORS

1759

where Φi Ψi are Fti measurable. Then Φ1 · W (t) ≡ Φ2 · W (t) ≡

n X i=0 n X

Φi (ω) (W (t ∧ ti+1 ) − W (t ∧ ti )) Ψi (ω) (W (s ∧ ti+1 ) − W (s ∧ ti ))

i=0

The left side of 56.2.14 equals ÃÃ n !Ã n !! X X E Φi (W (t ∧ ti+1 ) − W (t ∧ ti )) , a Ψi (W (s ∧ ti+1 ) − W (s ∧ ti )) , b i=0

=

n X

i=0

E ((Φi (W (t ∧ ti+1 ) − W (t ∧ ti )) , a) (Ψi (W (s ∧ ti+1 ) − W (s ∧ ti )) , b))

i=0

because of independence of the increments for the Wiener process. If Φi and Ψi were constants, this would reduce to n X

((t ∧ s ∧ ti+1 ) − (t ∧ s ∧ ti )) (QΦ∗i a, Ψ∗i b)

i=0

=

n X

((t ∧ s ∧ ti+1 ) − (t ∧ s ∧ ti )) E ((QΦ∗i a, Ψ∗i b))

i=0 t∧s

Z

(QΦ∗1 a, Φ∗2 b) dP dr

= 0

= =

µZ

µZ

Z Ω t∧s

E 0 µ µZ E

t∧s

=E 0

¶ (QΦ∗1 a, Φ∗2 b) dr

¶ µZ t∧s ³³ ´³ ´∗ ´ ¶ Q1/2 Φ∗1 a, b dr = E Φ2 Q1/2 Φ1 Q1/2 a, b dr 0 ¶ t∧s ³ ´³ ´∗ ¶ Φ2 Q1/2 Φ1 Q1/2 a, b ≡ (V (t, s) a, b) ³³

Q1/2 Φ∗2

´∗

´

0

but unfortunately, these are not constants. However, the expression still reduces to n X

((t ∧ s ∧ ti+1 ) − (t ∧ s ∧ ti )) E ((QΦ∗i a, Ψ∗i b))

i=0

=

(V (t, s) a, b)

and this is shown now. There exists two sequences of simple functions measurable in Fti , © such that

φik

ª∞ k=1

© ª∞ , ψ ik k=1

¯¯ ¯¯ ¯¯ ¯¯ lim ¯¯φik − Φi ¯¯L2 (Ω) = lim ¯¯ψ ik − Ψi ¯¯L2 (Ω) = 0.

k→∞

k→∞

1760


Consider ¢¢ ψ ik (W (s ∧ ti+1 ) − W (s ∧ ti )) , b ¡¡ ¢¡ ¢¢ E (W (t ∧ ti+1 ) − W (t ∧ ti )) , φi∗ (W (s ∧ ti+1 ) − W (s ∧ ti )) , ψ i∗ k a k b E

=

¡¡

φik (W (t ∧ ti+1 ) − W (t ∧ ti )) , a

¢¡

Pm Pm i∗ Say φi∗ k a = j=1 aj XAj , ψ k b = j=1 bj XBj where Aj , Bj ∈ Fti . Then the above reduces to an expression of the form X ¡ ¢ E ((W (t ∧ ti+1 ) − W (t ∧ ti )) , aj ) ((W (s ∧ ti+1 ) − W (s ∧ ti )) , bl ) XAj XBl j,l

Now by independence, this is X E (((W (t ∧ ti+1 ) − W (t ∧ ti )) , aj ) ((W (s ∧ ti+1 ) − W (s ∧ ti )) , bl )) P (Aj ∩ Bl ) j,l

= ((t ∧ s ∧ ti+1 ) − (t ∧ s ∧ ti ))

X

(Qaj , bl ) P (Aj ∩ Bl )

j,l

= ((t ∧ s ∧ ti+1 ) − (t ∧ s ∧ ti )) E

¡¡

i∗ Qφi∗ k a, ψ k b

= ((t ∧ s ∧ ti+1 ) − (t ∧ s ∧ ti )) E

¢¢

³³³ ´³ ´∗ ´´ ψ ik Q1/2 φik Q1/2 a, b

Now consider 56.2.13. Applying this estimate the differences of φik Q1/2 and Φi Q1/2 and ψ ik Q1/2 and Ψi Q1/2 , it follows lim E

k→∞

³³³ = lim E k→∞

³³³ ´³ ´∗ ´´ ψ ik Q1/2 φik Q1/2 a, b − E ((QΦ∗i a, Ψ∗i b))

ψ ik Q1/2

´³

φik Q1/2

´∗

´´ a, b

−E

³³³ ´³ ´∗ ´´ Ψi Q1/2 Φi Q1/2 a, b = 0.

Therefore for Φ1 , Φ2 elementary functions, E ((Φ1 · W (t) , a)H (Φ2 · W (s) , a)H ) = (V (t, s) a, b)H where

µZ

t∧s

³

V (t, s) ≡ E

Φ2 (r) Q

1/2

´³ Φ1 (r) Q

1/2

´∗

¶ dr .

0

Now by Proposition 53.2.6 there exists sequences of elementary functions {Φn1 } and {Φn2 } satisfying Z TZ 2 lim ||Φni − Φi ||L0 dP dt = 0 n→∞

0

Ω

2

and now by the estimate 56.2.13 applied again to the differences, Φni − Φi , it follows µ µZ (Vn (t, s) a, b) ≡ E

t∧s 0

¶ ´∗ ¶ ´³ ³ n 1/2 n 1/2 dr a, b Φ1 (r) Q Φ2 (r) Q

56.3. WEAK SOLUTIONS TO STOCHASTIC DIFFERENTIAL EQUATIONS

1761

converges to (V (t, s) a, b) ≡

µ µZ E

t∧s

³

Φ2 (r) Q1/2

´³

Φ1 (r) Q1/2

´∗

¶ ¶ dr a, b

0

and also by the Ito isometry, E ((Φn1 · W (t) , a) (Φn2 · W (s) , b)) = (Vn (t, s) a, b) converges to E ((Φ1 · W (t) , a) (Φ2 · W (s) , b)) which shows (V (t, s) a, b) ≡

µ µZ E

t∧s

³ Φ2 (r) Q

1/2

´³ Φ1 (r) Q

1/2

´∗

¶

¶

dr a, b

0

=

E ((Φ1 · W (t) , a) (Φ2 · W (s) , b))


56.3

Weak Solutions To Stochastic Differential Equations

For A the generator of S a strongly continuous semigroup, the stochastic differential equation dx (t) = [Ax (t) + f (t)] dt + BdW (t) , x (0) = ξ is said to have a strong solution for t ∈ [0, T ] if x (t) has values in D (A) a.e. and Z

T

|Ax (s)| ds < ∞ a.e. 0

Z

t

x (t) = ξ +

[Ax (s) + f (s)] ds + BW (t) a.e. 0

It is always assumed that f is a predictable process (H predictable) and t → f (t, ω) is Bochner integrable on finite intervals. Also ξ is F0 measurable. Recall the filtration was derived from the Wiener process. That is Ft = σ (W (b) − W (a) : 0 ≤ a ≤ b ≤ t) . Also B ∈ L (U, H).

1762

56.3.1


Convolution With A Semigroup

Recall the definitions of U and U0 ≡ Q1/2 U where {ej } was an orthonormal basis 1/2 for U such that Qek = λk ek and {gk } given by gk = λk ek was an orthonormal basis for U0 . Also let {fj } be an orthonormal basis for H. Letting S ∈ L (H, H) and B ∈ L (U, H) X X 2 2 2 ||SB||L0 ≡ |SBgk |H = (SBgk , fj ) 2

=

X³

SB

k

p

kj

λk e k , f j

´2

=

X ³p

λk ek , B ∗ S ∗ fj

kj

=

X³

´2

kj

Q1/2 ek , B ∗ S ∗ fj

´2

=

XX³ j

kj

ek , Q1/2 B ∗ S ∗ fj

´2

k

¯¯2 X ³ ´ X ¯¯¯¯ ¯¯ = Q1/2 B ∗ S ∗ fj , Q1/2 B ∗ S ∗ fj ¯¯Q1/2 B ∗ S ∗ fj ¯¯ = j

j

=

X

(fj , SBQB ∗ S ∗ fj ) = trace (SBQB ∗ S ∗ )

(56.3.15)

j

In fact, the following lemma holds. Lemma 56.3.1 The function t → S (t) B is continuous with values in L2 (U0 , H). Proof: Letting {ek }√be the orthonormal basis for U and {gk } the orthonormal basis for U0 with gk = λk ek as described earlier and {fj } an orthonormal basis for H, 2

||S (t) B − S (s) B||L0 2

≡

X³

(S (t) B − S (s) B)

´2 p λk e k , f j

kj

=

XX k

=

X

2

λk ((S (t) B − S (s) B) ek , fj )

j 2

λk |(S (t) − S (s)) Bek |H

k

and P as s → t, this converges to 0 from the dominated convergence theorem. Recall k λk < ∞. This proves the lemma. For S a strongly continuous semigroup as described above and B ∈ L (U, H) , Z WA (t) ≡ S (t − (·)) B · W (t) ≡

t

S (t − s) BdW (s) 0

That is, you fix t and for s ∈ [0, t] you consider the constant process Φ (s) ≡ S (t − s) B and WA (t) is just the Ito integral Φ · W (t) . Because of Lemma 56.3.1


1763

and the definition of the Ito integral, WA (t) is approximated in L2 (Ω) by sums of the form m X S (t − tk ) B (W (t ∧ tk+1 ) − W (t ∧ tk )) k=0

because the elementary functions corresponding to these sums converge in the norm Z TZ 2 2 ||Φ|| ≡ ||Φ||L0 dP dt. 0

Also

Z

Ω

2

b

S (c − r) BdW (r) a

means

S (c − (·)) B · W (b) where a takes the place of 0 in the definition of the Ito integral and sums of the above form can be used to approximate this random variable also. Theorem 56.3.2 Suppose A generates a strongly continuous semigroup as described above. Suppose also that Z t Z t 2 ||S (r) B||L0 ds = trace (S (r) BQB ∗ S ∗ (r)) dr < ∞ (56.3.16) 2

0

0

Then the process {WA (t)} is Gaussian, continuous in mean square and has a predictable version. Recall Gaussian means the random variables (WA (t) , a) are normally distributed. Continuous in mean square means ³ ´ 2 lim E |WA (t) − WA (s)|H = 0 s→t

and predictable means that WA−1 (U ) ∈ PT , the smallest σ algebra containing the sets of the form (s, r] × As where As ∈ Fs and {0} × A0 , where A0 ∈ F0 . Also µµZ t ¶ ¶ ∗ ∗ E ((WA (t) , a) (WA (t) , b)) = S (r) BQB S (r) dr a, b (56.3.17) 0

Thus the covariance is the linear operator Z t S (r) BQB ∗ S ∗ (r) dr ∈ L (H, H) . 0

In addition to this, for a.e. ω, t → WA (t, ω) is in L2 (0, T ; H) with covariance operators Q defined by Z

T

Z

T

= 0

0

µµZ

(Qφ, ψ)H ≡ E ((WA , φ)H (WA , ψ)H ) ¶ ¶ s∧r ∗ ∗ S (r − t) BQB S (s − t) dt φ (s) , ψ (r) dsdr.

0

1764


Proof: I want to show WA (t) is mean square integrable. For s ≤ t Z

Z

t

WA (t) − WA (s) =

s

S (t − r) BdW (r) +

(S (t − r) − S (s − r)) BdW (r)

s

0

Now the random variables defined by the two integrals are independent. This is because of the independence of the increments in Wiener process and the definition of the Ito integral as limits of things which clearly are independent. Thus Ã ¯Z ¯2 ! ³ ´ ¯ ¯ t 2 S (t − r) BdW (r)¯¯ E |WA (t) − WA (s)| = E ¯¯ s

Ã ¯Z ¯ + E ¯¯

s

0

¯2 ! ¯ (S (t − r) − S (s − r)) BdW (r)¯¯

(56.3.18)

Consider the first expression on the right in the above. By the Ito isometry this equals Z Z t 2 ||S (t − r) B||L0 drdP Ω

s

2

and by 56.3.15 this equals Z Z Ω

t

trace (S (t − r) BQB ∗ S ∗ (t − r)) drdP

s

Z Z

t−s

= Z

Ω 0 t−s

=

trace (S (r) BQB ∗ S ∗ (r)) drdP

trace (S (r) BQB ∗ S ∗ (r)) dr

0

which converges to 0 as s → t due to the assumption 56.3.16 which says r → trace (S (r) BQB ∗ S ∗ (r)) is L1 ([0, T ]). Next consider the second term in 56.3.18. Ã ¯Z ¯2 ! ¯ s ¯ E ¯¯ (S (t − r) − S (s − r)) BdW (r)¯¯ 0

Z Z

s

= ZΩs = 0

Z =

0

0

2

||(S (t − r) − S (s − r)) B||L0 drdP 2

2

||(S (t − r) − S (s − r)) B||L0 dr 2

s

2

||(S (t − s + r) − S (r)) B||L0 dr 2

(56.3.19)


1765

Which converges to 0 by Lemma 56.3.1 as s → t. This shows the claim about mean square continuity. Why is WA (t) Gaussian? That is, why is (WA (t) , a) normally distributed? Observe that WA (t) is just an Ito integral. Thus it is a limit in L2 (Ω; H) of expressions of the form m X

Φi (W (t ∧ ti+1 ) − W (t ∧ ti ))

i=0

and Ã

m X

! Φi (W (t ∧ ti+1 ) − W (t ∧ ti )) , a

=

i=0

m X

((W (t ∧ ti+1 ) − W (t ∧ ti )) , Φ∗i a)

i=0

Now by independence of the increments this is a finite sum of normally distributed independent random variables and so it is Gaussian with zero mean. Now in general if X is the limit of a sequence of H valued random variables in L2 (Ω; H) and each Xn is Gaussian with zero mean, then passing to a subsequence still denoted by n such that the convergence is also pointwise, the dominated convergence theorem implies ³ ´ ³ ´ lim E eit(a,Xn ) = E eit(a,X) n→∞

so the characteristic functions must converge to the characteristic function of X. However, ³ ´ 1 2 2 E eit(a,Xn ) = e− 2 t σn ¡ ¢ 1 2 2 If a subsequence σ nk → ∞, e− 2 t σn → 0 and it is not possible that 0 = E eit(a,X) which must equal 1 when t = 0. Therefore, there must be some upper bound to the σ 2n and so a further subsequence has the property that σ 2n → σ 2 . Therefore, ³ ´ ³ ´ 1 2 2 1 2 2 e− 2 t σ = lim e− 2 t σn = lim E eit(a,Xn ) = E eit(a,X) n→∞

n→∞

and so X is also Gaussian with mean 0. Thus WA (t) is Gaussian. Consider the claim about a predictable version. This follows from Proposition 50.2.6 which is stated here. Proposition 56.3.3 Let X be Ft adapted and left continuous. Then it is predictable. Also, if X is stochastically continuous and adapted on [0, T ] , then it has a predictable version. Recall what it means to be stochastically continuous. Definition 56.3.4 X is stochastically continuous at t0 ∈ I means: for all ε > 0 and δ > 0 there exists ρ > 0 such that P ([||X (t) − X (t0 )|| ≥ ε]) ≤ δ whenever |t − t0 | < ρ, t ∈ I. X is stochastically continuous if it is stochastically continuous at every t ∈ I.

1766


{WA (t)} is stochastically continuous thanks to the assertion proved above that it is mean square continuous, and adapted. To see this, let ε, δ > 0 be given and pick t0 . Then if |t − t0 | is small enough, µZ 1/2

εδ

¶1/2

2

≥

|WA (t) − WA (t0 )| dP Ω

ÃZ

!1/2 2

≥

|WA (t) − WA (t0 )| dP [|WA (t)−WA (t0 )|≥ε]

≥

1/2

εP ([|WA (t) − WA (t0 )| ≥ ε])

which shows that WA is stochastically continuous. Therefore, from that proposition stated above it has a predictable version. Next consider the claim about the correlation. I need to consider E ((WA (t) , a) (WA (t) , b)) . From Theorem 56.2.3 there was a formula like the following. µ µZ E

(V (t, s) a, b) ≡

t∧s

³

Φ2 (r) Q1/2

´³

Φ1 (r) Q1/2

´∗

¶ ¶ dr a, b

0

=

E ((Φ1 · W (t) , a) (Φ2 · W (s) , b))

In the situation of interest here, t = s and µZ t ³ V (t, t) =

E

S (t − r) BQ

Z t³ = 0

Z t³ =

0

S (r) BQ1/2

´³

1/2

´³ S (t − r) BQ

S (r) BQ1/2

´∗

1/2

´∗

¶ dr

dr

´ S (r) BQ1/2 Q1/2 B ∗ S ∗ (r) dr

0

Z

t

=

S (r) BQB ∗ S ∗ (r) dr

0

This proves the claim about the covariance. It only remains to prove the last claim. It was shown above there is a measurable predictable version of WA . By Fubini’s theorem applied to this predictable version, ÃZ

T

E

! 2

|WA (t)| dt 0

Z = 0

T

³ ´ 2 E |WA (t)| dt

(56.3.20)


1767

Then from what was shown above about the covariance, ! Z T ÃX 2 (WA (t) · fk ) dt = E 0

Z

k

X

T

XZ

= 0

Z

k

= 0

Z

(S (r) BQB ∗ S ∗ (r) fk , fk ) drdt

Z tX

0

0 T

t

0

k

T

= Z

³ ´ 2 E (WA (t) · fk ) dt

T

Z

= 0

(S (r) BQB ∗ S ∗ (r) fk , fk ) drdt

k t

trace (S (r) BQB ∗ S ∗ (r)) drdt

0

which by 56.3.15 equals Z

T 0

Z

t

0

2

||S (r) B||L0 drdt < ∞. 2

It follows from 56.3.20 that for a.e. ω, Z T 2 |WA (t)| dt < ∞. 0

Thus for a.e. ω,

Z

T

WA (t) dt 0

makes sense. Modifying this on a set of measure zero, it follows the above integral is a random variable having values in H. Similarly, if g ∈ L2 (0, T ; H) ≡ H Z T (WA (t) , g (t)) dt 0

is measurable and so by the Pettis theorem, WA is strongly measurable having values in H. I want to show that WA is actually Gaussian. That is, if g ∈ H then (WA , g)H has a normal distribution. To do this, use the definition of the Ito integral and Lemma 56.3.1 to approximate WA (t) as n X

S (t − ti ) B (W (t ∧ ti+1 ) − W (t ∧ ti ))

i=0

where {t0 , t1 , · · · , tn } is a partition of [0, T ] having each |ti+1 − ti | = T /n. Of course S is only a semigroup so define S (t − ti ) ≡ I whenever t < ti . Then letting g be in C ([0, T ] ; H) , a continuous function, an approximation to Z T (WA (t) , g (t)) dt 0

1768


is

Z

T

Ã n X

0

! S (t − ti ) B (W (t ∧ ti+1 ) − W (t ∧ ti )) , g (t) dt

i=0

I will show later that as n → ∞, the limit of the second expression converges to the m first. For now, consider the second expression. For fixed n let {τ k }k=0 be a uniform partition of [0, T ] and approximate the above integral by m X n X

(S (τ k − ti ) B (W (τ k ∧ ti+1 ) − W (τ k ∧ ti )) , g (τ k )) (τ k − τ k−1 )

k=1 i=0

Thus from continuity of S and g, the above double sum converges for a.e. ω as m → ∞ to ! Z T ÃX n S (t − ti ) B (W (t ∧ ti+1 ) − W (t ∧ ti )) , g (t) dt (56.3.21) 0

i=0

and so it follows that if for all m the above double sum is symmetric (mean =0) and normally distributed, the same must be true of the above integral. This follows from looking at the characteristic functions as was done earlier in the proof. The double sum is obviously normally distributed because it is of the form m X n X ¡ ¢ ∗ (W (τ k ∧ ti+1 ) − W (τ k ∧ ti )) , B ∗ S (τ k − ti ) (τ k − τ k−1 ) g (τ k ) , k=1 i=0

the second term in every inner product being independent of ω. By modifying the sum by adding in points as required, and rearranging, one can express the above double sum as a sum of the form p X

((W (bl ) − W (al )) , gl )H

l=1

where the intervals (al , bl ) are disjoint. Then it is the sum of independent normally distributed random variables having mean 0 and is therefore normally distributed with mean 0. Thus the integral in 56.3.21, ! Z T ÃX n S (t − ti ) B (W (t ∧ ti+1 ) − W (t ∧ ti )) , g (t) dt, (56.3.22) 0

i=0

is normally distributed and has mean 0 also. I need to pass to the limit and argue that Z T (WA (t) , g (t)) dt 0

is normally distributed and has mean 0. To save space, let Ã n ! X Σn (t) ≡ S (t − ti ) B (W (t ∧ ti+1 ) − W (t ∧ ti )) i=0


1769

It follows from the definition of the Ito integral that when n → ∞ the integrand in 56.3.22 converges in L1 (Ω) to (WA (t) , g (t))H for each t. Next, ¯ Z T Z ¯¯Z T ¯ ¯ ¯ (WA (t) , g (t)) dt − (Σn (t) , g (t)) dt¯ dP ¯ ¯ 0 Ω¯ 0 Z

Z

T

|(WA (t) , g (t)) − (Σn (t) , g (t))| dP dt

≤ 0

Ω

ÃZ

!1/2

Z

T

2

≤ CT

|(WA (t) − Σn (t) , g (t))| dP dt 0

Ω

Then the Ito isometry can be used on the inside integral to obtain ÃZ Z !1/2 T

≤

2

CT ÃZ

=

2

|WA (t) − Σn (t)| |g (t)| dP dt 0

Ω

CT

2

|g (t)| 0

Z Z

T

= CT

|g (t)| 0 n X i=0

|WA (t) − Ω

"Z

−

!1/2

Z

T

2

2 Σn (t)|H

dP dt

t

||S (t − s) B Ω

0

1/2 ¯¯2 ¯¯ ¯¯ S (t − ti ) BX(ti ,ti+1 ] (s)¯¯ dsdP dt ¯¯ 0 L2

Then by Lemma 56.3.1 about continuity of t → S (t) B with values in L02 and consequent uniform continuity on bounded intervals, the integrand converges uniformly to 0 as n → ∞ and so the above expression converges to 0. It follows that for a suitable subsequence Z T Z T (WA (t) , g (t)) dt (Σn (t) , g (t)) dt → 0

0

a.e. and so it follows the random variable Z T (WA (t) , g (t)) dt 0

is normally distributed having mean 0. I need to verify this is true for all g ∈ H, not just for continuous g. However, C ([0, T ] ; H) is dense in H due to regularity of

1770


Lebesgue measure and so if g is arbitrary there exists a sequence gn converging to g in H such that gn is continuous. Then ¯ Z ¯¯Z T Z T ¯ ¯ ¯ (WA (t) , g (t)) dt − (WA (t) , gn (t)) dt¯ dP ¯ ¯ ¯ Ω 0 0 Z Z T ≤ |(WA (t) , g (t) − gn (t))| dtdP Ω

ÃZ Z

0

!1/2 ÃZ Z

T

≤

|WA (t)| dtdP Ω

Ω

RT

2

|g (t) − gn (t)| dtdP

0

and so the random variables

!1/2

T

2

0

(WA (t) , gn (t)) dt converges in L1 (Ω) to

0

Z

T

(WA (t) , g (t)) dt 0

so this random variable is normal with zero mean. It only remains to consider the covariance operators. Let φ, ψ ∈ H ÃÃZ

(Qφ, ψ) ≡ E ((WA , φ)H (WA , ψ)H ) = ! ÃZ

T

T

E

(WA (s) , φ (s)) ds

(WA (r) , ψ (r)) dr

0

Z

T

Z

0 T

=

E ((WA (s) , φ (s)) (WA (r) , ψ (r))) dsdr 0

!!

(56.3.23)

0

Suppose s > r. Then

Z

s

WA (s) =

S (s − t) BdW (t) 0

Z

r

=

Z

s

S (s − t) BdW (t) + 0

S (s − t) BdW (t) r

and the second of these integrals is independent of WA (r) and by reasoning similar to the above is symmetric and Gaussian so µ·µZ =E

r

E ((WA (s) , φ (s)) (WA (r) , ψ (r))) ¶¸ ¶ Z s S (s − t) BdW (t) + S (s − t) BdW (t) , φ (s) (WA (r) , ψ (r))

0

r

µµZ

=

¶ ¶ S (s − t) BdW (t) , φ (s) (WA (r) , ψ (r)) µµ 0 ¶ ¶ Z r E S (s − r) S (r − t) BdW (t) , φ (s) (WA (r) , ψ (r))

=

E ((S (s − r) WA (r) , φ (s)) (WA (r) , ψ (r)))

=

E ((WA (r) , S ∗ (s − r) φ (s)) (WA (r) , ψ (r)))

=

r

E

0

(56.3.24)


1771

Now recall from 56.3.17 shown above, µµZ

r

E ((WA (r) , a) (WA (r) , b)) =

¶ ¶ S (t) BQB ∗ S ∗ (t) dt a, b

0

Thus 56.3.24 equals µµZ

¶ ¶ S (t) BQB ∗ S ∗ (t) dt S ∗ (s − r) φ (s) , ψ (r) µµZ0 r ¶ ¶ = S (r − t) BQB ∗ S ∗ (r − t) dt S ∗ (s − r) φ (s) , ψ (r) µµZ0 r ¶ ¶ = S (r − t) BQB ∗ S ∗ (s − t) dt φ (s) , ψ (r) (56.3.25) r

0

which follows from changing the variables in the integral. If s < r similar considerations would yield the following for the integrand of 56.3.23. µµZ

s

¶ ¶ S (s − t) BQB ∗ S ∗ (r − t) dt ψ (r) , φ (s)

0

which equals µZ

= = =

¶ S (s − t) BQB ∗ S ∗ (r − t) ψ (r) , φ (s) dt µZ0 s ¶ ψ (r) , S (r − t) BQB ∗ S ∗ (s − t) φ (s) dt µ 0 µZ s ¶ ¶ ψ (r) , S (r − t) BQB ∗ S ∗ (s − t) dt φ (s) 0 µµZ s ¶ ¶ ∗ ∗ S (r − t) BQB S (s − t) dt φ (s) , ψ (r) s

0

It follows the integrand of 56.3.23 is always of the form µµZ

s∧r

¶ ¶ S (r − t) BQB ∗ S ∗ (s − t) dt φ (s) , ψ (r) .

0

It follows (Qφ, ψ)H ≡ E ((WA , φ)H (WA , ψ)H ) Z

T

Z

T

µµZ

s∧r

= 0

0

0


¶ ¶ S (r − t) BQB ∗ S ∗ (s − t) dt φ (s) , ψ (r) dsdr.

1772

56.3.2


Continuity Of WA (t)

The convolution with a continuous semigroup is mean square continuous as shown above. However, it is also the case that for a.e. ω, t → WA (t) is continuous with values in H provided B = I. The following lemma is an essential part of the argument. Lemma 56.3.5 Let X be a random variable having values in H a real separable Hilbert space which is Gaussian and symmetric with trace class covariance operator Q, X Q= λk ek ⊗ ek k

where {ek } is orthonormal and Qek = λk ek , λk > 0, (h, X (ω)) is normally distributed having mean 0 and

P k

λk < ∞. Thus ω →

E ((X, a) (X, b)) = (Qa, b) Then for each m ∈ N, the positive integers, ³ ´ µ 2m! ¶ ³ ³ ´´m 2m 2 E |X| ≤ E |X| 2m m!

(56.3.26)

Proof: First consider the case where X is normally distributed with mean 0 and variance σ 2 so X has values in R. Then 1 2

E (exp (itX)) = e− 2 t

σ2

=

∞ X

k

(−1)

k=0

t2k σ 2k 2k k!

It follows upon taking derivatives and setting t = 0 to compute moments that ¡ ¢ ¡ ¢ (2m)! (2m)! ¡ ¢m E X 2m+1 = 0, E X 2m = m σ 2m = m E X 2 . 2 m! 2 m! Next consider the general case. Letting {ek } be as above, ³ ´ 2 E (X, ek ) ≡ (Qek , ek ) = λk . Therefore, from what was just shown ³ ´ (2m)! ³ ´m (2m)! 2m 2 E (X, ek ) = m E (X, ek ) = m λm 2 m! 2 m! k Now using Minkowski’s inequality and the above, ³ E |X|

2m

´1/m

=E

ÃÃ X k

!m !1/m 2

(X, ek )


1773

³ ´1/m X µ (2m)! ¶1/m 2m E (X, ek ) = λm 2m m! k k k ¶1/m X µ ¶1/m ³ µ ´ (2m)! (2m)! 2 λ = = E |X| k 2m m! 2m m! X

≤

k

since

³ ´ X ³ ´ X 2 2 E |X| = E (X, ek ) = λk k

k

and this proves the lemma. Next I will present a simple formula which will be used to factor the integral defining WA (t). Lemma 56.3.6 The following holds for α ∈ (0, 1) . Z

t

(t − s)

α−1

−α

(s − σ)

ds =

σ

π sin (πα) −1

Proof: First change variables to get rid of the σ. Let y = (t − σ) Then the integral becomes Z 1 α−1 −α (t − [(t − σ) y + σ]) (t − σ) y −α (t − σ) dy Z

0 1

=

α−1

((t − σ) (1 − y)) Z

(s − σ) .

(t − σ)

−α

y −α (t − σ) dy

0 1

=

α−1

(1 − y)

y −α dy

0

Next let y = x2 . The integral is Z 1 ¡ ¢α−1 1−2α 2 1 − x2 x dx 0

Next let x = sin θ Z 12 π Z 2α−1 2 (cos (θ)) sin(1−2α) (θ) dθ = 2 0

0

1 2π

µ

cos (θ) sin (θ)

¶2α−1 dθ

Now change the variable again. Let u = cot (θ) . Then this yields Z ∞ 2α−1 u 2 du 1 + u2 0 This is fairly easy to evaluate using contour integrals. Consider the following contour called ΓR for large R. As R → ∞, the integral over the little circle converges to 0 and so does the integral over the big circle. There is one singularity at i.

1774


−R

−R−1

R−1

R

Thus Z

e(ln|z|+i arg(z))(1−2α) dz = 1 + z2

lim

R→∞

ΓR

Z

∞

u2α−1 du + i sin ((1 − 2α) π) 1 + u2

= (1 + cos (1 − 2α) π) 0

Z

∞

0

u2α−1 du 1 + u2

´ ³π ´´ ³ ³π (1 − 2α) + i sin (1 − 2α) = π cos 2 2 Then equating the imaginary parts yields Z sin ((1 − 2α) π) 0

∞

³π ´ u2α−1 du = π sin (1 − 2α) 1 + u2 2

and so using the trig identities for the sum of two angles, Z 0

∞

u2α−1 du 1 + u2

= =

¡ ¡ ¢¢ π sin π2 (1 − 2α) ¡ ¢ ¡ ¢ 2 sin π2 (1 − 2α) cos π2 (1 − 2α) π π ¢= ¡ 2 sin (πα) 2 cos π2 (1 − 2α)

This proves the lemma. Lemma 56.3.7 Let W (t) be a Q Wiener process on H. Recall this means Q is a positive trace class operator defined on H and H0 ≡ Q1/2 H. In terms of the usual context, U = H. As before, let L02 denote L2 (H0 , H) , the Hilbert Schmidt operators mapping H0 to H. Also let S (t) be a stongly continuous semigroup on H. Then for any α ∈ (0, 1) , Z T 2 t−α ||S (t)||L0 dt < ∞ 0

2


1775

Proof: 2

XX

≡

||S (t)||L0 2

k

l

X

=

λl

X

l

l

≤

C

(S (t) el , ek )

2

k

X

=

2

(S (t) gl , ek )

2

λl |S (t) el |H

X

λl eβt

l

because ||S (t)|| has exponential growth thanks to Theorem 15.12.3 on Page 468. This proves the lemma. Theorem 56.3.8 Let W (t) be a Q Wiener process on H. Recall this means Q is a positive trace class operator defined on H and H0 ≡ Q1/2 H. In terms of the usual context, U = H. As before, let L02 denote L2 (H0 , H) , the Hilbert Schmidt operators mapping H0 to H. Let S (t) be a strongly continuous semigroup whose generator is A, a closed densely defined operator mapping D (A) ⊆ H to H. Then Z

t

t → WA (t) ≡

S (t − s) dW (s) 0

has the property that for a.e. ω, the function is continuous having values in H.1 Proof: In the argument, m will be a positive integer such that m>

1 2α

where α ∈ (0, 1/2) . Using Lemma 56.3.6, Z

t

WA (t) = WA (t) ≡

1 sin (πα) π

Z tZ

S (t − σ) dW (σ) 0

t

α−1

S (t − σ) (t − s) 0

(s − σ)

−α

dsdW (σ) .

σ

Now use the Stochastic Fubini theorem, Theorem 56.1.3 to write this as Z tZ s 1 α−1 −α sin (πα) S (t − σ) (t − s) (s − σ) dW (σ) ds π 0 0 = 1 Recall

1 sin (πα) π

Z

zZ

t

S (t − s) (t − s) 0

α−1

Y (s) s

}|

{ −α

S (s − σ) (s − σ)

dW (σ)ds (56.3.27)

0

it was shown earlier this function was mean square continuous. This is something more.

1776


³ ´ 2m Now consider the random variable Y (s) I want to show E |Y (s)|H < C for some C independent of s. From this it will follow that for a.e. ω Z

T 0

2m

|Y (s)|H ds < ∞

Then when this has been shown it will follow that for these ω, t → WA (t) will be continuous having values in H. To show this I need to find the covariance of Y (s) . This random variable will end up being Gaussian as in the proof of Theorem 56.3.2. First I will show this. The integral equal to Y (s) is the limit in L2 (Ω, H) of sums of the form p X −α S (s − σ j ) (W (σ j+1 ) − W (σ j )) (s − σ j ) j=0

and each of these is Gaussian. The fact this limit exists follows because −α

σ → S (s − σ) (s − σ)

¢ ¡ is in L2 0, T ; L02 by Lemma 56.3.7. Therefore, using the trick involving characteristic functions as in Theorem 56.3.2, it follows Y (s) must also be Gaussian. I need to find and estimate the trace of the covariance. Consider for a, b ∈ H, E ((Y (s) , a) (Y (s) , b)) µZ ¶ µZ s ¶ Z s −α −α ≡ S (s − σ) (s − σ) dW (σ) , a S (s − τ ) (s − τ ) dW (τ ) , b dP Ω

0

0

(56.3.28)

Consider one of these integrals on the inside. Z s −α S (s − σ) (s − σ) dW (σ) 0

is the limit in L2 (Ω, H) of sums like p X

S (s − σ j ) (W (σ j+1 ) − W (σ j )) (s − σ j )

−α

j=0

and so

³R

s 0

−α

S (s − σ) (s − σ) p ³ X

´ dW (σ) , a is a limit in L2 (Ω) of sums like −α

S (s − σ j ) (W (σ j+1 ) − W (σ j )) (s − σ j )

´ ,a

j=0

=

p X j=0

−α

((W (σ j+1 ) − W (σ j )) , S ∗ (s − σ j ) a) (s − σ j )


1777

Thus the expression in 56.3.28 is the limit as p → ∞ of  p X −α  E ((W (σ j+1 ) − W (σ j )) , S ∗ (s − σ j ) a) (s − σ j ) · j=0 p X

! −α

∗

((W (σ i+1 ) − W (σ i )) , S (s − σ i ) b) (s − σ i )

i=0

By independence of the increments this is of the form p X

(s − σ j )

−2α

E [((W (σ j+1 ) − W (σ j )) , S ∗ (s − σ j ) a)

j=0

((W (σ j+1 ) − W (σ j )) , S ∗ (s − σ j ) b)] =

p X

−2α

(s − σ j )

(σ j+1 − σ j ) (QS ∗ (s − σ j ) a, S ∗ (s − σ j ) b)

j=0

and this converges to Z

s

−2α

(s − σ) 0

Z = =

(QS ∗ (s − σ) a, S ∗ (s − σ) b) dσ

s

u−2α (QS ∗ (u) a, S ∗ (u) b) du 0 µµZ ¶ ¶ s u−2α S (u) QS ∗ (u) du a, b 0

which shows the covariance of Y (s) is the operator µZ s ¶ −2α ∗ u S (u) QS (u) du .

(56.3.29)

0

Does it even make sense? For each u, the integrand is in L (H0 , H). Also XX ¯¯ −2α ¯¯2 2 ¯¯u S (u) QS ∗ (u)¯¯L0 ≡ u−4α (ek , S (u) QS ∗ (u) gl ) 2

k

= u−4α

X l

= u−4α

X

l

λl

´2 X³ Q1/2 S ∗ (u) ek , Q1/2 S ∗ (u) el k

¯ ¯2 ¯ ¯ λl ¯Q1/2 S ∗ (u) el ¯

H

l

¯¯2 X ¯¯¯¯ ¯¯ ≤ u−4α λl ¯¯S (u) Q1/2 ¯¯ = u−4α C l

and so

¯¯ −2α ¯¯ ¯¯u S (u) QS ∗ (u)¯¯

L02

≤ Cu−2α

1778


which is an integrable function since α < 1/2. It follows 56.3.29 at least makes sense and is in L2 (H0 , H) . Furthermore, its norm in L2 (H0 , H) is bounded independent of s. What about the trace of this covariance operator? Let {en } be the orthonormal basis for H used above. Then if Q (s) denotes this covariance operator, X

(Q (s) ek , ek )H

XZ

=

k

k

XZ

=

Z

s

0

Z

X k

s

=

X

0

Z

s

X

0

=

X³

u−2α

u

u−2α λl

H

Q1/2 S ∗ (u) ek , el ek , S (u) Q1/2 el

´2

´2

X

du

du

2

(ek , S (u) el ) du

k

−2α

0

¯ ¯2 ¯ ¯ u−2α ¯Q1/2 S ∗ (u) ek ¯ du

k

l s

s

l

l

= Z

X³

u−2α

u−2α (S (u) QS ∗ (u) ek , ek ) du

0

0

k

=

s

X

Z

λl |S

2 (u) el |H

du ≤ C

l

T

u−2α du < ∞

0

because of the assumption that α < 1/2. Therefore, Q ³(s) , the ćovariance of Y (s) 2 has trace which is bounded independent of s. Thus E |Y (s)| which equals the trace of Q (s) is bounded independent of s. Now recall Lemma 56.3.5. ³ ´ 2 Since there exists a bound for trace (Q (s)) = E |Y (s)| which is independent of s, it follows there exists a constant C independent of s ∈ [0, T ] such that ³ ´ 2m E |Y (s)| ≤C

(56.3.30)

With 56.3.30 it becomes possible to finish the proof. From 56.3.27 WA (t) =

1 sin (πα) π

Z

t

S (t − s) (t − s)

α−1

Y (s) ds

0

where thanks to 56.3.30, there is a set of measure 0 such that off this set, Z 0

T

2m

|Y (s)|H ds < ∞.

Assume ω is such that the above holds. First note that if s → Y (s) is continuous, then t → WA (t) must also be contin-


1779

uous. Next suppose Y ∈ L2m (0, T, H) . By Holder’s inequality µZ t ¶2m/(2m−1) 2m/(2m−1) (α−1)2m/(2m−1) |WA (t)| ≤ C ||S (t − s)|| (t − s) ds · 0

ÃZ

!1/2m

T

|Y 0

ÃZ

T

≤C

2m (s)|H

ds !2m/(2m−1) ÃZ

2m/(2m−1)

||S (s)||

(α−1)2m/(2m−1)

(s)

!1/2m

T

ds

|Y

0

0

2m (s)|H

ds

the first integral is finite because ||S (t)|| has exponential growth due to its being a continuous semigroup and the assumption that m > 1/ (2α) implies (α − 1) 2m ((1/2m) − 1) 2m > = −1. 2m − 1 2m − 1 Thus

ÃZ sup |WA (t)| ≤ C t∈[0,T ]

!1/2m

T

|Y 0

2m (s)|H

.

ds

Letting Wn (t) be defined as WA (t) but with Y (s) replaced with Yn (s) where {Yn (t)} is a sequence of continuous H valued functions converging in L2m (0, T ; H) to Y, it follows sup |WA (t) − Wn (t)| ≤ C ||Yn − Y ||L2m (0,T ;H)

t∈[0,T ]

which shows that since uniform convergence occurs, it must be the case that t → WA (t) is continuous. This proves the theorem. Note α ∈ (0, 1/2) was arbitrary and it was shown that if m > 1/2α, then ³ ´ 2m sup E |WA (t)| 2. Note that 2m must be larger than 2. Corollary 56.3.9 In the situation of the above theorem, let {An } denote a sequence of Yosida approximations such that for all x ∈ H, eAn t x → S (t) x Then for all p > 2

Ã lim E

n→∞

! sup |WA (t) − WAn (t)|

t∈[0,T ]

p

= 0.

1780


Proof: It is no loss of generality to assume at the outset that S (t) is a bounded semigroup since if not, you could consider e−γt S (t) for a suitable γ which would be a bounded semigroup due to the exponential growth of any continuous semigroup. Then you could do everything for this bounded semigroup and note that the desired result would only involve a factor of eγt . More specifically, approximating by sums you can assert that for each t ≤ T and Sn (t) the semigroup generated by An ,

≥

¯ ¯Z t ¯ ¯ −γ(t−s) ¯ ¯ e (S (t − s) − S (t − s)) dW (s) n ¯ ¯ 0 ¯Z t ¯ ¯ ¯ e−γT ¯¯ (S (t − s) − Sn (t − s)) dW (s)¯¯ 0

and so, if you have shown the result for the bounded semigroups, e−γt S (t) , it will follow also for the unbounded semigroup. Thus from the proof of the Hille Yosida theorem, ¯¯ ¯¯ ||S (t)|| , ¯¯eAn t ¯¯ ≤ M. This is convenient. Also let m be a large positive integer larger than p corresponding to α such that m > 1/2α. Now as in the proof of the above theorem, 1 WA (t) = sin (πα) π where

Z

Z

t

S (t − s) (t − s)

Y (s) ds

0

s

Y (s) ≡

α−1

−α

S (s − σ) (s − σ)

dW (σ)

0

and 1 WAn (t) = sin (πα) π where

Z Yn (s) ≡

Z

t

α−1

Sn (t − s) (t − s)

Yn (s) ds

0

s

−α

Sn (s − σ) (s − σ)

dW (σ)

0

for Sn (t) ≡ eAn t . Then WA (t) − WAn (t) =

Z sin (πα) t α−1 (S (t − s) − Sn (t − s)) (t − s) Y (s) ds π 0 Z sin (πα) t α−1 Sn (t − s) (t − s) (Yn (s) − Y (s)) ds + π 0 ≡ In (t) + Jn (t)


1781

First consider In (t) . For fixed t, µZ |In (t)| ≤

t

C

|(S (t − s) − Sn (t − s)) Y (s)| 0

µZ

¶1/2m ds ·

¶(2m−1)/2m

t

(t − s)

(α−1)(2m)/(2m−1)

0

ÃZ

2m

ds

!(2m−1)/2m

T

≤ C

s

(α−1)(2m)/(2m−1)

ds

·

0

µZ

t

2m

|(S (t − s) − Sn (t − s)) Y (s)|

¶1/2m ds

0

ÃZ

!1/2m

T

= Cm

|(S (t − s) − Sn (t − s)) Y (s)|

2m

ds

0

where define both S (r) and Sn (r) to equal I if r < 0. By the choice of α, (α − 1) (2m) / (2m − 1) > −1. Thus for every t, Z T 2m 2m |In (t)| ≤ Cm 2M |Y (s)| ds < ∞ 0

and it follows sup |In (t)|

2m

< ∞.

t∈[0,T ]

Now choose tn such that from the above, sup |In (t)|

2m

n 2 t∈[0,T ] p

#! ≤

1 2n

Therefore, by the Borel Cantelli lemma, there is a set of measure 0 such that for ω not in this set, ω is in only finitely many of the Bn and so for a.e. ω, WAn (t) converges uniformly to WA (t) on [0, T ] . Now taking a countable union of sets of measure 0 corresponding to T = 1, 2, 3, · · · , it follows there is a set of measure 0 such that WAn (t) converges uniformly to WA (t) on [0, ∞). Now recall the definition of the Yosida approximations. An = Aλn (λn I − A)

−1

≡ AJn

and |Jn x − x| converges to 0 for each x ∈ D (A) and ||Jn || is bounded because by Proposition 15.12.5, ¯¯ ¯¯ M ¯¯ −1 ¯¯ ¯¯(λn I − A) ¯¯ ≤ |λn − α|

1786


for some constants, α, M. Therefore, |Jn x − x| converges to 0 for each x ∈ H, not just for those in D (A) . By 56.3.34 Z WAn (t)

t

= W (t) + An

Sn (t − s) W (s) ds 0

Z =

t

Sn (t − s) W (s) ds

W (t) + AJn 0

and it was shown above that WAn (t) → WA (t) uniformly and so it follows from the above that Z t AJn Sn (t − s) W (s) ds → WA (t) − W (t) 0

Also ¯ Z t ¯ Z t ¯ ¯ ¯J n ¯ S (t − s) W (s) ds − S (t − s) W (s) ds n ¯ ¯ 0 0 ¯ Z t ¯ Z t ¯ ¯ ¯ ≤ ¯J n Sn (t − s) W (s) ds − Jn S (t − s) W (s) ds¯¯ 0 0 ¯ Z t ¯ Z t ¯ ¯ ¯ + ¯Jn S (t − s) W (s) ds − S (t − s) W (s) ds¯¯ 0

≤

0

¯Z t ¯ Z t ¯ ¯ ¯ M¯ Sn (t − s) W (s) ds − S (t − s) W (s) ds¯¯ 0 0 ¯ Z t ¯ Z t ¯ ¯ + ¯¯Jn S (t − s) W (s) ds − S (t − s) W (s) ds¯¯ 0

0

which converges to 0. Thus Z t Z t Jn Sn (t − s) W (s) ds → S (t − s) W (s) ds, 0 0 Z t Jn Sn (t − s) W (s) ds ∈ D (A) 0

and

Z

t

Sn (t − s) W (s) ds → WA (t) − W (t) .

AJn 0

Since A is closed,

Z

t

S (t − s) W (s) ds ∈ D (A) 0

and

Z A

t

S (t − s) W (s) ds = WA (t) − W (t) . 0

56.3. WEAK SOLUTIONS TO STOCHASTIC DIFFERENTIAL EQUATIONS Hence WA (t)

1787

Z t = W (t) + A S (t − s) W (s) ds 0 µZ t ¶ d = S (t − s) W (s) ds . dt 0

the reason for this last step is as follows. Taking a difference quotient for Z t S (t − s) W (s) ds 0

yields 1 h =

1 h

Z

t+h

S (h) S (t − s) W (s) ds − Z

0 t+h t

1 h

Z

t

S (t − s) W (s) ds 0

1 S (t + h − s) W (s) ds + (S (h) − I) h

Z

t

S (t − s) W (s) ds 0

and taking a limit, µZ t ¶ d S (t − s) W (s) ds dt Ã0 Z ! Z t 1 t+h 1 = lim S (t + h − s) W (s) ds + (S (h) − I) S (t − s) W (s) ds h→0 h t h 0 Z t = W (t) + A S (t − s) W (s) ds = WA (t) 0

Rt It follows from this formula that t → 0 S (t − s) W (s) ds is in C 1 ([0, ∞), H) for a.e. Rt ω because t → WA (t) is continuous. Also the formula shows t → 0 S (t − s) W (s) ds is continuous with values in D (A) . This proves the corollary.

56.3.3

Temporal Regularity Analytic Semigroups

Here the semigroups will be analytic semigroups, not just continuous semigroups. This will make possible results on regularity in time and later, some other results involving fractional powers of operators which amount to spacial regularity. First the regularity in time will be considered. This presentation will make use of Corollary 30.2.7 on Page 894 which is the basic result on analytic semigroups. The theorem is very interesting because it says the convolution integral has the same sort of temporal regularity as the Wiener process. Theorem 56.3.11 Let W (t) be a Q Wiener process where Q is of trace class and has all positive¡ eigenvalues as described above and let A be a sectorial operator. For ¢ arbitrary α ∈ 0, 21 , there is a set of measure zero such that for ω not in this set, the convolution integral, WA (t) has α Holder continuous trajectories on any finite interval, [0, T ] .

1788


Proof: Fix T > 0 and let T ≥ t > s > 0. Then ³ ´ 2 E |WA (t) − WA (s)| Ã¯Z ¯ = E ¯¯

Z

s

S (t − r) dW (r) + 0

Z

t

s

S (t − r) dW (r) − s

0

¯2 ! ¯ S (s − r) dW (r)¯¯

By the usual argument involving sums approximating the integrals along with the independence of the increments, (Formally, the dW (r) for r > s are independent of the dW (r) for r < s.) the above expression equals Ã ¯Z Ã ¯Z ¯2 ! ¯2 ! ¯ s ¯ ¯ t ¯ ¯ ¯ ¯ E ¯ (56.3.36) (S (t − r) − S (s − r)) dW (r)¯ + E ¯ S (t − r) dW (r)¯¯ 0

s

Approximating by sums in the usual way, the following formal argument is justified. Ã¯Z ¯2 ! µZ t ¶ ¯ t ¯ E ¯¯ S (t − r) dW (r)¯¯ =E (S (t − r) dW (r) , S (t − r) dW (r))H s

Z

Ã

t

=

s

E

X

s

=

! 2 (r) , ek )H

(S (t − r) dW

Z =

s

E

X¡

s

k

Z tX

Ã

t

¡ ¢2 ∗ E dW (r) , S (t − r) ek H =

dW (r) , S (t − r) ek

¢2

!

H

k

Z tX ¡ s

k

∗

¢ ∗ ∗ QS (t − r) ek , S (t − r) ek dr

k

Z t X¯ Z t XX³ ¯2 ´2 ¯ 1/2 ¯ ∗ ∗ = Q1/2 S (t − s) ek , ei dr ¯Q S (t − s) ek ¯ dr = s

=

s

k

Z t XX³ s

i

ek , S (t − s) Q1/2 ei

´2

i

k

dr =

Z tX s

k

=

Z tX ∞

λi

X

i

2

(ek , S (t − s) ei ) dr

k 2

λi |S (t − s) ei | dr.

(56.3.37)

s i=1

With exactly similar reasoning, the other term in 56.3.36 is of the form Z 0

∞ sX

2

λi |(S (t − r) − S (s − r)) ei | dr.

(56.3.38)

i=1

The idea is to estimate these terms. First consider the one in 56.3.37. From Corollary 30.2.7 on Page 894, ||S (t − s)|| is bounded with a bound which depends on T and so Z tX ∞ 2 λi |S (t − s) ei | dr ≤ MT |t − s| trace (Q) . s i=1


1789

Now consider the term in 56.3.38. This one will lead to an estimate which, while usable, is not so good. Using Corollary 30.2.7 again, ¯2 Z sX ∞ X Z s ¯¯Z t−r ¯ 2 ¯ ¯ dr λi |(S (t − r) − S (s − r)) ei | dr = λi AS (σ) e dσ i ¯ ¯ 0

i=1

i

0

0

i

X Z s ¯¯Z ¯ λi ¯

¯2 t−r X Z ¯ |AS (σ) ei | dσ ¯¯ dr ≤ NT2 λi

s−r

i

≤ ≤

t−r

s−r

µ

1 + aMT σ

¶

¯2 ¯ dσ ¯¯ dr

¯Z t−r ¯2 ¯ ¯ 1 ¯ trace (Q) + aMT dσ ¯¯ dr ¯ σ s−r 0 ¯2 Z s ¯Z t−r ¯ ¯ 1 2 ¯ NT trace (Q) dσ + aMT |t − s|¯¯ dr ¯ s−r σ 0 ¯2 Z s ¯Z t−r ¯ 1 ¯¯ ¯ 2NT2 trace (Q) dσ ¯ ¯ dr 0 s−r σ Z

=

0

s−r

s ¯¯Z ¯ ¯

NT2

s

2

+2T (aMT ) NT2 trace (Q) |t − s|

2

Now pick γ ∈ (0, 1/2) . Z =

2NT2

trace (Q) 0

Z ≤

2NT2 trace (Q) 2NT2 trace (Q)

s

0

Z ≤

s

0

s

¯2 ¯ 1 dσ ¯¯ dr γ 1−γ s−r σ σ ¯Z t−r ¯2 ¯ ¯ 1 γ−1 ¯ σ dσ ¯¯ dr 2γ ¯ (s − r) s−r ¯Z t−r ¯2 ¯ ¯ 1 γ−1 ¯ σ dσ ¯¯ dr 2γ ¯ (s − r) s−r

¯Z ¯ ¯ ¯

t−r

Now s − r ≥ 0 and the exponent γ − 1 < 0. Therefore, ¯Z t−s ¯2 Z s ¯ ¯ 1 γ−1 ¯ ¯ dr ≤ 2NT2 trace (Q) σ dσ 2γ ¯ ¯ 0 (s − r) 0 Z s 1 1 2γ = 2NT2 trace (Q) 2 (t − s) 2γ dr γ 0 (s − r) Z T 1 1 2γ ≤ 2NT2 trace (Q) 2 (t − s) 2γ dr γ (r) 0 1 T 1−2γ 2γ (t − s) = 2NT2 trace (Q) 2 γ 1 − 2γ Thus ³ ´ 1 T 1−2γ 2 2γ E |WA (t) − WA (s)| ≤ 2NT2 trace (Q) 2 |t − s| γ 1 − 2γ 2

MT |t − s| trace (Q) + 2T (aMT ) NT2 trace (Q) |t − s|

2

(56.3.39)

1790


and it was shown earlier that WA (t) − WA (s) is a symmetric Gaussian random variable. In general, if X is a symmetric Gaussian random variable having trace class covariance Q, then letting {ek } be the eigenvectors for Q, Ã ! ³ ´ ´ X X ³ 2 2 2 E |X| = E (X, ek ) = E (X, ek ) X

≡

k

k

(Qek , ek ) = trace (Q) .

k

From Lemma 56.3.5, if m is a positive integer, it follows from 56.3.39 that for t, s ∈ [0, T ] , since 2γ < 1, ³ ´ 2m E |WA (t) − WA (s)| µ 1 T 1−2γ 2γ ≤ 2NT2 trace (Q) 2 |t − s| + γ 1 − 2γ 2

2

MT |t − s| trace (Q) + 2T (aMT ) NT2 trace (Q) |t − s| ≤ Cm,γ,T,Q |t − s|

´m

2γm

ˇ Now recall the Kolmogorov Centsov continuity theorem, Theorem 50.1.4 on Page 1472. This theorem implies that for a.e. ω, t → WA (t) is Holder continuous with exponent 2γm − 1 1 =γ− 2m m ¡ 1¢ Thus if α < 1/2, you can choose γ ∈ α, 2 and then choose m so large that γ − 1/m > α also. This proves the theorem on [0, T ]. To get Holder continuous trajectories off a set of measure zero on any finite interval, [0, T ], let the set of measure zero be the union of the sets of measure zero corresponding to T = 1, 2, 3, · · · . This proves the theorem.

56.3.4

Spacial Regularity

Next a different sort of regularity will be established. It will be shown that WA (t) α is in D ((−A) ) for fractionial powers of (−A) . The necessary theory of analytic semigroups is found starting on Page 902. To begin with, here is a proposition. γ

Proposition 56.3.12 Let γ ∈ (0, 1/2) . Then WA (t) ∈ D ((−A) ) a.e. Also γ (−A) WA (t) is a Gaussian random variable and its covariance can be computed. Proof: First note

Z

t

γ

(−A) S (t − s) dW (s) 0


1791

γ

is well defined. This follows because s → ||(−A) S (t − s)|| is in L2 ([0, t]) due to the estimate, M γ ||(−A) S (t − s)|| ≤ γ |t − s| γ

from Theorem 30.2.18 on Page 906. This estimate shows s → (−A) S (t − s) is in L2 ([0, T ] ; L (H, H)) and so the stochastic integral makes sense. Rt Approximating WA (t) ≡ 0 S (t − s) dW (s) with sums, m X

S (t − sk ) (W (sk+1 ) − W (sk )) ≡ Sm (t) ,

k=0

it follows since S (t) is an analytic semigroup, γ

Sm (t) ∈ D ((−A) ) and γ

(−A) Sm (t) =

m X

γ

(−A) S (t − sk ) (W (sk+1 ) − W (sk ))

k=0 2

which converges in L (Ω; H) to Z t

γ

(−A) S (t − s) dW (s) 0

while Sm (t) converges in L2 (Ω; H) to Z t WA (t) ≡ S (t − s) dW (s) 0 γ

γ

Since (−A) is closed, it follows that for a.e. ω, WA (t) ∈ D ((−A) ) and Z t γ γ (−A) WA (t) = (−A) S (t − s) dW (s) . 0 γ

Next consider the claim that (−A) WA (t) is Gaussian. This follows from noting γ that (−A) Sm (t) is Gaussian. This follows because γ

=

((−A) S (t − sk ) (W (sk+1 ) − W (sk )) , a) ¡ ∗ ¢ γ W (sk+1 ) − W (sk ) , ((−A) S (t − sk )) a γ

and this is a normal random variable having mean 0. Then (−A) Sm (t) is a finite sum of such random variables and by the independence of the increments, it must also be a normal random variable having mean 0. Then taking a subsequence, γ γ (−A) Sm (t) converges pointwise a.e. to (−A) WA (t) and so using the characteristic functions it follows from the dominated convergence theorem that γ

γ

E (exp (iλ ((−A) Sm (t) , a)H )) → E (exp (iλ ((−A) WA (t) , a)H ))

1792


Now

1

γ

E (exp (iλ ((−A) Sm (t) , a)H )) = e− 2 λ

2

σ 2m

Then σ 2m must be bounded since otherwise, there would exist a subsequence for γ which σ 2m → ∞ and 0 = E (exp (iλ ((−A) WA (t) , a)H )) which is impossible because this last must equal 1 when λ = 0. Therefore, there is a subsequence, still called σ 2m which converges to σ 2 and so 1

2

e− 2 λ

σ2

= =

1

2

lim e− 2 λ

m→∞

E

σ 2m

γ

= lim E (exp (iλ ((−A) Sm (t) , a)H ))

m→∞ γ (exp (iλ ((−A) WA (t) , a)H )) . γ

Since a ∈ H is arbitrary, it follows (−A) WA (t) is a Gaussian random variable. Now it is time to compute its covariance. Let a, b ∈ H. I will compute something a little more general, γ

γ

γ

γ

E (((−A) WA (t) − (−A) WA (s) , a) ((−A) WA (t) − (−A) WA (s) , b)) , where s ≤ t. µµZ t ¶ Z s γ γ E (−A) S (t − r) dW (r) − (−A) S (s − r) dW (r) , a 0 0 µµZ t ¶¶¶ Z s γ γ · (−A) S (t − r) dW (r) − (−A) S (s − r) dW (r) , b 0

0

Now using the independence of the increments and justifying all the manipulations by considering sums and passing to limits, the above reduces to µµZ s ¶ γ E (−A) (S (t − r) − S (s − r)) dW (r) , a µµZ 0s ¶¶¶ γ · (−A) (S (t − r) − S (s − r)) dW (r) , b 0

µµZ

t

+E

¶ µZ γ

(−A) S (t − r) dW (r) , a s

t

¶¶ γ

(−A) S (t − r) dW (r) , b s

Consider the top first. Approximating by sums and taking the meaning suggested by the following symbols, the following formal argument yields the following computation. Z s γ γ E (((−A) (S (t − r) − S (s − r)) dW (r) , a) ((−A) (S (t − r) − S (s − r)) dW (r) , b)) 0

Z

s

=

E Z0 s

= 0

¡¡

∗ ¢¡ ∗ ¢¢ γ γ dW (r) , ((−A) (S (t − r) − S (s − r))) a dW (r) , ((−A) (S (t − r) − S (s − r))) b

¡ ∗ ∗ ¢ γ γ Q ((−A) (S (t − r) − S (s − r))) a, ((−A) (S (t − r) − S (s − r))) b dr

56.3. WEAK SOLUTIONS TO STOCHASTIC DIFFERENTIAL EQUATIONS µ·Z

¸

s

=

1793

γ

¶

∗

γ

(−A) (S (t − r) − S (s − r)) Q ((−A) (S (t − r) − S (s − r))) dr a, b 0

Next consider the bottom. Using a similar formal computation, ¶ µZ t ¶¶ µµZ t γ γ (−A) S (t − r) dW (r) , a (−A) S (t − r) dW (r) , b E s

Z

s

t

=

γ

γ

E (((−A) S (t − r) dW (r) , a)) ((−A) S (t − r) dW (r) , b) Z

s t

=

E Z

s

Z

s

t

=

¡

¡¡

∗

γ

dW (r) , ((−A) S (t − r)) a

¢¢ ¡

γ

∗

dW (r) , ((−A) S (t − r)) b

¢

∗ ∗ ¢ γ γ Q ((−A) S (t − r)) a, ((−A) S (t − r)) b dr

t−s

¡

∗ ∗ ¢ γ γ Q ((−A) S (σ)) a, ((−A) S (σ)) b dσ 0 µ·Z t−s ¸ ¶ ∗ γ γ = (−A) S (σ) Q ((−A) S (σ)) dσ a, b

=

0

Thus the covariance is the operator taking H to H given by ·Z t−s ¸ ∗ γ γ (−A) S (σ) Q ((−A) S (σ)) dσ 0 ·Z ¸ s ∗ γ γ + (−A) (S (t − r) − S (s − r)) Q ((−A) (S (t − r) − S (s − r))) dr 0

This proves the proposition. γ γ To reiterate, WA (t) ∈ D ((−A) ) whenever γ < 1/2 and you can take (−A) inside the stochastic convolution integral. For A a sectorial operator for the sector S−a,φ where −a < 0, so that the fracγ tional powers are defined, you can define a scale of Banach spaces, Xγ ≡ D ((−A) ) with the norm γ ||x||γ ≡ |(−A) x| γ

These are Banach spaces because (−A) is a closed operator. Recall that for α ≤ β, ³ ´ α β D ((−A) ) ⊆ D (−A) . Earlier it was shown that t → WA (t) was Holder continuous having exponent γ < 1/2 with values in H. The next theorem shows that t → WA (t) is also Holder continuous with exponent α < 1/2 − γ with values in Xγ . Theorem 56.3.13 Let A be as just described. Then for all γ ∈ (0, 1/2) and α ∈ (0, 1/2 − γ) t → WA (t) is α Holder continuous with values in Xγ on any finite interval.

1794


ˇ Proof: The proof will depend on the Kolmogorov Centsov continuity theorem, Theorem 50.1.4. Let [0, T ] be a finite interval. It is necessary to estimate for 0 < s < t ≤ T, Ã¯Z ¯2 ! Z s ¯ t ¯ γ γ E ¯¯ (−A) S (t − r) dW (r) − (−A) S (s − r) dW (r)¯¯ . 0

0

Using the independence of the increments and approximating the integrals with sums and then taking a limit, the above equals Ã¯Z ¯2 ! ¯ t ¯ γ E ¯¯ (−A) S (t − r) dW (r)¯¯ + (56.3.40) s

Ã¯Z ¯ E ¯¯

s

0

¯2 ! ¯ (−A) (S (t − r) − S (s − r)) dW (r)¯¯ γ

(56.3.41)

First consider 56.3.40. Approximating by sums, the following formal argument can be made precise. The expression equals Z t ³ ´ 2 γ E |(−A) S (t − r) dW (r)| s

The integrand of this term is Ã ! X¡ ¢2 ∗ γ dW (r) , ((−A) S (t − r)) ek E =

X k

=

X k

=

X

k

E

³¡ ¢2 ´ ∗ γ dW (r) , ((−A) S (t − r)) ek

¡ ¢ ∗ ∗ γ γ dr Q ((−A) S (t − r)) ek , ((−A) S (t − r)) ek ¯ ¯2 ∗ ¯ ¯ γ dr ¯Q1/2 ((−A) S (t − r)) ek ¯

k

= dr

XX³ j

k

= dr

XX³ k

= dr

X

j

λj

j

= dr

∗

γ

Q1/2 ((−A) S (t − r)) ek , ej

X

∗

γ

((−A) S (t − r)) ek , Q1/2 ej

X

´2

´2

2

γ

(ek , (−A) S (t − r) ej )

k γ

λj |(−A) S (t − r) ej |

2

j

≤ dr

X

M2 |t − r|

2γ j

λj = trace (Q)

M2 |t − r|

2γ

dr


1795

where the estimate found in Theorem 30.2.18 is being used. Thus 56.3.40 is dominated by Z trace (Q) M

t

2

2γ

(t − r)

s

Z

1

dr

t

1 dr 2γ r s Z t−s 1 ≤ trace (Q) M 2 dr 2γ r 0 1 1−2γ |t − s| = trace (Q) M 2 1 − 2γ = trace (Q) M 2

Next consider 56.3.41. Formally, it equals Z s ³ ´ 2 γ E |(−A) (S (t − r) − S (s − r)) dW (r)| 0

Using exactly the same computations, the integrand equals X ¡ ¢2 ∗ γ E dW (r) , ((−A) (S (t − r) − S (s − r))) ek k

X¡

∗

γ

∗

γ

=

dr

Q ((−A) (S (t − r) − S (s − r))) ek , ((−A) (S (t − r) − S (s − r))) ek

=

¯2 X ¯¯ ∗ ¯ γ dr ¯Q1/2 ((−A) (S (t − r) − S (s − r))) ek ¯

k

k

= dr

XX³ j

k

= dr

X

λj

j

= dr

X

∗

γ

((−A) (S (t − r) − S (s − r))) ek , Q1/2 ej

X

2

γ

(ek , (−A) (S (t − r) − S (s − r)) ej )

k γ

λj |(−A) (S (t − r) − S (s − r)) ej |

2

j

Thus 56.3.41 is dominated by X Z s 2 γ λj |(−A) (S (t − r) − S (s − r)) ej | dr 0

j

=

X j

Z

s

λj 0

¯¯Z ¯¯ ¯¯ ¯¯

s−r

Z ≤

t−r

trace (Q) 0

s

¯¯Z ¯¯ ¯¯ ¯¯

¯¯2 ¯¯ (−A) AS (σ) dσ ¯¯¯¯ dr

t−r s−r

γ

¯¯2 ¯¯ (−A) AS (σ) dσ ¯¯¯¯ dr γ

´2

¢

1796


Using Theorem 30.2.18 again, this is dominated by Z

s

µZ

t−r

≤ trace (Q) s−r

0

M dσ σ 1+γ

¶2 dr

Now this requires some estimating. Let ξ ∈ (γ, 1/2) and write it as Z ≤

0 s

trace (Q) Z trace (Q)

2ξ

s−r

µZ

1 2ξ

(s − r)

2

M trace (Q) M 2 trace (Q)

1 dσ σξ

σ 1+γ−ξ µZ t−r

(s − r)

Ã

=

M

1

s 0

=

t−r

s−r

0

≤

µZ

trace (Q) Z

≤

s

ξ−γ

(t − s) ξ−γ T

dr ¶2

M

σ

dσ 1+γ−ξ

t−s

0

¶2

M σ 1+γ−ξ

!2 Z

s

0

1−2ξ

dr ¶2

dσ

dr

1 dr r2ξ 2ξ−2γ

(1 − 2ξ) (ξ − γ)

2

(t − s)

It follows Ã ¯Z Z ¯ t γ ¯ E ¯ (−A) S (t − r) dW (r) − 0

≤ trace (Q) M 2

0

s

¯2 ! ¯ (−A) S (s − r) dW (r)¯¯ γ

T 1−2ξ 1 2ξ−2γ 1−2γ |t − s| + M 2 trace (Q) 2 (t − s) 1 − 2γ (1 − 2ξ) (ξ − γ)

Now to simplify things note 1 − 2γ ≥ 2γ − 2ξ and so, adjusting the constant, there exists a constant CT such that ³ ´ 2 2ξ−2γ E ||WA (t) − WA (s)||Xγ ≤ CT |t − s| Letting m be a positive integer, it follows from Lemma 56.3.5 that ³ ´ µ 2m! ¶ 2m 2mξ−2mγ E ||WA (t) − WA (s)||Xγ ≤ CTm |t − s| 2m m! ˇ Let m be large enough that 2mξ − 2mγ > 1. It follows from Kolmogorov Centsov continuity theorem that t → WA (t) is Holder continuous with exponent α for any α such that 2mξ − 2mγ − 1 1 0

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