A First Course in Sobolev Spaces (Graduate Studies in Mathematics)

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A First Course in Sobolev Spaces

Giovanni Leoni

Graduate Studies in Mathematics Volume IOS

American Mathematical Society

A First Course in Sobolev Spaces

A First Course in Sobolev Spaces Giovanni Leoni

Graduate Studies in Mathematics Volume 105

American Mathematical Society Providence, Rhode Island

Editorial Board David Cox (Chair) Steven G. Krantz Rafe Mazzeo Martin Scharlemann 2000 Mathematics Subject Classification. Primary 46E35; Secondary 26A24, 26A27, 26A30, 26A42, 26A45, 26A46, 26A48, 26B30.

For additional information and updates on this book, visit

www.ams.org/bookpages/gsm-105

Library of Congress Cataloging-in-Publication Data Leoni, Giovanni, 1967A first course in Sobolev spaces / Giovanni Leoni.

p. cm. - (Graduate studies in mathematics ; v. 105) Includes bibliographical references and index. ISBN 978-0-8218-4768-8 (alk. paper) 1. Sobolev spaces. I. Title.

QA323.L46 2009 515'.782-dc22

2009007620

Copying and reprinting. Individual readers of this publication, and nonprofit libraries acting for them, are permitted to make fair use of the material, such as to copy a chapter for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for such permission should be addressed to the Acquisitions Department, American Mathematical Society, 201 Charles Street, Providence, Rhode Island 02904-2294, USA. Requests can also be made by e-mail to reprint-peraiasionQams. org. © 2009 by the American Mathematical Society. All rights reserved. The American Mathematical Society retains all rights except those granted to the United States Government. Printed in the United States of America.

® The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability.

Visit the AMS home page at http://vw.ams.org/ 10987654321 14131211 1009

Contents

Preface

ix

Acknowledgments

xv

Part 1. Functions of One Variable Chapter 1. Monotone Functions §1.1. Continuity §1.2. Differentiability

3 3

8

Chapter 2. Functions of Bounded Pointwise Variation §2.1. Pointwise Variation §2.2. Composition in BPV (I) §2.3. The Space BPV (I) §2.4. Banach Indicatrix

39

Chapter 3. Absolutely Continuous Functions

73

§3.1.

AC(I) Versus BPV(I)

Chain Rule and Change of Variables §3.3. Singular Functions §3.2.

Chapter 4. Curves §4.1. Rectifiable Curves and Arclength §4.2. Frechet Curves §4.3. Curves and Hausdorff Measure §4.4. Jordan's Curve Theorem

39 55 59 66

73

94 107 115 115 130 134 146

v

Contents

vi

Chapter 5. Lebesgue-Stieltjes Measures §5.1. Radon Measures Versus Increasing Functions §5.2. Signed Borel Measures Versus BPV (1) §5.3. Decomposition of Measures §5.4. Integration by Parts and Change of Variables

155

Chapter 6. Decreasing Rearrangement §6.1. Definition and First Properties §6.2. Absolute Continuity of u*

187

§6.3.

Derivative of u*

Chapter 7. Functions of Bounded Variation and Sobolev Functions §7.1. BV (St) Versus BPV (1) §7.2. Sobolev Functions Versus Absolutely Continuous Functions

155 161

166 181

187 202

209 215 215

222

Part 2. Functions of Several Variables Chapter 8. §8.1. §8.2. §8.3.

Absolutely Continuous Functions and Change of

Variables The Euclidean Space RN Absolutely Continuous Functions of Several Variables Change of Variables for Multiple Integrals

231

231

234 242

255 Chapter 9. Distributions 255 §9.1. The Spaces DK (fl), D (1k), and D' (St) 264 §9.2. Order of a Distribution §9.3. Derivatives of Distributions and Distributions as Derivatives 266 §9.4.

Convolutions

Chapter 10. Sobolev Spaces §10.1. Definition and Main Properties §10.2. Density of Smooth Functions § 10.3. Absolute Continuity on Lines §10.4. Duels and Weak Convergence §10.5. A Characterization of W1,P (St) Chapter 11. Sobolev Spaces: Embeddings §11.1. Embeddings: 1 N

275 279

279 283 293 298

305 311

312 328 335

Contents

§11.4.

vii

Lipschitz Functions

Chapter 12. Sobolev Spaces: Further Properties §12.1. Extension Domains §12.2. Poincare Inequalities Chapter 13. Functions of Bounded Variation §13.1. Definition and Main Properties §13.2. Approximation by Smooth Functions § 13.3. Bounded Pointwise Variation on Lines §13.4. Coarea Formula for BV Functions §13.5. Embeddings and Isoperimetric Inequalities §13.6. Density of Smooth Sets §13.7. A Characterization of BV (fl) Chapter 14. Besov Spaces §14.1. Besov Spaces B,P,e, 0 < s < 1 §14.2.

Dependence of B8,P,O on s

§14.3.

The Limit of Bs,P.e as s - 0+ and s - 1-

§14.4.

Dependence of Bs.n,e on 0 Dependence of Bs.P.e on s and p Embedding of Bs,P'e into LQ Embedding of W1,P into B1.4 Besov Spaces and Fractional Sobolev Spaces

§14.5. § 14.6.

§14.7. §14.8.

Chapter 15. Sobolev Spaces: Traces §15.1. Traces of Functions in W',' (fl) §15.2. Traces of Functions in BV (St) §15.3. Traces of Functions in W1.P (1k), p > 1 §15.4. A Characterization of W' 1' (fl) in Terms of Traces Chapter 16. Sobolev Spaces: Symmetrization §16.1. Symmetrization in LP Spaces §16.2. Symmetrization of Lipschitz Functions §16.3. Symmetrization of Piecewise Affine Functions §16.4. Symmetrization in W1,P and BV Appendix A. Functional Analysis §A.1. Metric Spaces

341

349 349 359

viii

§A.2. §A.3. §A.4.

§A.5. §A.6.

Contents

Topological Spaces Topological Vector Spaces Normed Spaces Weak Topologies Hilbert Spaces

Appendix B. Measures §B.1. Outer Measures and Measures §B.2. Measurable and Integrable Functions §B.3. Integrals Depending on a Parameter §B.4. Product Spaces §B.5. Radon-Nikodym's and Lebesgue's Decomposition Theorems §B.6. Signed Measures §B.7.

LP Spaces §B.8. Modes of Convergence §B.9. Radon Measures §B.10. Covering Theorems in RN

Appendix C. The Lebesgue and Hausdorff Measures §C.1. The Lebesgue Measure §C.2. The Brunn-Minkowski Inequality and Its Applications

494

497 501

503 506

507 507 511

519

520 522 523

526

534 536

538 543 543 545

§C.3.

Conv olutions

550

§C.4.

Molli fiers

552

Diffe rentiable Functions on Arbitrary Sets §C.6. Maxi mal Functions §C.7. Anis otropic L Spaces §C.8. Haus dorff Measures §C.5.

Appendix D.

Notes

560 564

568 572 581

Appendix E. Notation and List of Symbols

587

Bibliography

593

Index

603

Preface The Author List, I. giving credit where credit is due. The first author: Senior grad student in the project. Made the figures. -Jorge Chaut, www.phdcomics.com

There are two ways to introduce Sobolev spaces: The first is through the elegant (and abstract) theory of distributions developed by Laurent Schwartz in the late 1940s; the second is to look at them as the natural development and unfolding of monotone, absolutely continuous, and BV functions' of one variable.

To my knowledge, this is one of the first books to follow the second approach. I was more or less forced into it: This book is based on a series of lecture notes that I wrote for the graduate course "Sobolev Spaces", which

I taught in the fall of 2006 and then again in the fall of 2008 at Carnegie Mellon University. In 2006, during the first lecture, I found out that half of the students were beginning graduate students with no background in functional analysis (which was offered only in the spring) and very little in measure theory (which, luckily, was offered in the fall). At that point I had two choices: continue with a classical course on Sobolev spaces and thus loose half the class after the second lecture or toss my notes and rethink the entire operation, which is what I ended up doing. I decided to begin with monotone functions and with the Lebesgue differentiation theorem. To my surprise, none of the students taking the class had actually seen its proof. I then continued with functions of bounded pointwise variation and absolutely continuous functions. While these are included in most books on real analysis/measure theory, here the perspective and focus are rather different, in view of their applications to Sobolev functions. Just to give an example, 1BV functions are functions of bounded variation. ix

x

Preface

most books study these functions when the domain is either the closed interval [a, b] or R. I needed, of course, open intervals (possibly unbounded). This changed things quite a bit. A lot of the simple characterizations that hold in [a, b) fall apart when working with arbitrary unbounded intervals. After the first three chapters, in the course I actually jumped to Chapter 7, which relates absolutely continuous functions with Sobolev functions of one variable, and then started with Sobolev functions of several variables. In the book I included three more chapters: Chapter 4 studies curves and arclength. I think it is useful for students to see the relation between rectifiable curves and functions with bounded pointwise variation. Some classical results on curves that most students in analysis have heard of, but whose proof they have not seen, are included, among them Peano's filling curve and the Jordan curve theorem. Section 4.3 is more advanced. It relates rectifiable curves with the flI Hausdorff measure. Besides Hausdorff measures, it also makes use of the Vitali-Besicovitch covering theorem. All these results are listed in Appendices B and C. Chapter 5 introduces Lebesgue-Stieltjes measures. The reading of this chapter requires some notions and results from abstract measure theory. Again it departs slightly from modern books on measure theory, which introduce Lebesgue-Stieltjes measures only for right continuous (or left) functions. I needed them for an arbitrary function, increasing or with bounded pointwise variation. Here, I used the monograph of Saks 11451. I am not completely satisfied with this chapter: I have the impression that some of the proofs could have been simplified more using the results in the previous chapters. Readers' comments will be welcome. Chapter 6 introduces the notion of decreasing rearrangement. I used some of these results in the second part of the book (for Sobolev and Besov functions). But I also thought that this chapter would be appropriate for the first part. The basic question is how to modify a function that is not monotone into one that is, keeping most of the good properties of the original function. While the first part of the chapter is standard, the results in the last two sections are not covered in detail in classical books on the subject. As a final comment, the first part of the book could be used for an advanced undergraduate course or beginning graduate course on real analysis or functions of one variable. The second part of the book starts with one chapter on absolutely continuous transformations from domains of RN into RN. I did not cover this chapter in class, but I do think it is important in the book in view of its ties with the previous chapters and their applications to the change of variables

Preface

xi

formula for multiple integrals and of the renewed interest in the subject in recent years. I only scratched the surface here. Chapter 9 introduces briefly the theory of distributions. The book is structured in such a way that an instructor could actually skip it in case the students do not have the necessary background in functional analysis (as was true in my case). However, if the students do have the proper background, then I would definitely recommend including the chapter in a course. It is really important. Chapter 10 starts (at long last) with Sobolev functions of several variables. Here, I would like to warn the reader about two quite common misconceptions. Believe it or not, if you ask a student what a Sobolev function is, often the answer is "A Sobolev function is a function in L" whose derivative is in Lv" This makes the Cantor function a Sobolev function :( I hope that the first part of the book will help students to avoid this danger. The other common misconception is, in a sense, quite the opposite, namely to think of weak derivatives in a very abstract way not related to the classical derivatives. One of the main points of this book is that weak derivatives of a Sobolev function (but not of a function in BV!) are simply (classical) derivatives of a good representative. Again, I hope that the first part of the volume will help here. Chapters 10, 11, and 12 cover most of the classical theorems (density, absolute continuity on lines, embeddings, chain rule, change of variables, extensions, duals). This part of the book is more classical, although it contains a few results published in recent years. Chapter 13 deals with functions of bounded variation of several variables.

I covered here only those parts that did not require too much background in measure theory and geometric measure theory. This means that the fundamental results of De Giorgi, Federer, and many others are not included here. Again, I only scratched the surface of functions of bounded variation. My hope is that this volume will help students to build a solid background, which will allow them to read more advanced texts on the subject. Chapter 14 is dedicated to the theory of Besov spaces. There are essentially three ways to look at these spaces. One of the most successful is to see them as an example/by-product of interpolation theory (see [7], [166], and [167]). Interpolation is very elegant, and its abstract framework can be used to treat quite general situations well beyond Sobolev and Besov spaces. There are two reasons for why I decided not to use it: First, it would depart from the spirit of the book, which leans more towards measure theory and real analysis and less towards functional analysis. The second reason

xii

Preface

is that in recent years in calculus of variations there has been an increased interest in nonlocal functionals. I thought it could be useful to present some techniques and tricks for fractional integrals. The second approach is to use tempered distributions and Fourier theory to introduce Besov spaces. This approach has been particularly successful for its applications to harmonic analysis. Again it is not consistent with the remainder of the book. This left me with the approach of the Russian school, which relies mostly on the inequalities of Hardy, Holder, and Young, together with some integral identities. The main references for this chapter are the books of Besov, Win, and Nikol'skii [18J, [191.

I spent an entire summer working on this chapter, but I am still not happy with it. In particular, I kept thinking that there should be easier and more elegant proofs of some of the results (e.g., Theorem 14.32, or Theorem 14.29), but I could not find one. In Chapter 15 I discuss traces of Sobolev and BV functions. Although

in this book I only treat first-order Sobolev spaces, the reason I decided to use Besov spaces over fractional Sobolev spaces (note that in the range of exponents treated in this book these spaces coincide, since their norms are equivalent) is that the traces of functions in Wk,1(1) live in the Besov space $k-1,1(8 ) (see [28] and [120]), and thus a unified theory of traces for Sobolev spaces can only be done in the framework of Besov spaces. Finally, Chapter 16 is devoted to the theory of symmetrization in Sobolev

and BV spaces. This part of the theory of Sobolev spaces, which is often missing in classical textbooks, has important applications in sharp embedding constants, in the embedding N = p, as well as in partial differential equations. In Appendices A, B, and C I included essentially all the results from functional analysis and measure theory that I used in the text. I only proved those results that cannot be found in classical textbooks.

What is missing in this book: For didactical purposes, when I started to write this book, I decided to focus on first-order Sobolev spaces. In my original plan I actually meant to write a few chapters on higher-order Sobolev and Besov spaces to be put at the end of the book. Eventually I gave up: It would have taken too much time to do a good job, and the book was already too long. As a consequence, interpolation inequalities between intermediate derivatives are missing. They are treated extensively in [7]. Another important theorem that I considered adding and then abandoned for lack of time was Jones's extension theorem [92].

Preface

xiii

Chapter 13, the chapter on BV functions of several variables, is quite minimal. As I wrote there, I only touched the tip of the iceberg. Good reference books of all the fundamental results that are not included here are (10], (54], and [182].

References: The rule of thumb here is simple: I only quoted papers and books that I actually read at some point (well, there are a few papers in German, and although I do have a copy of them, I only "read" them in a weak sense, since I do not know the language). I believe that misquoting a paper is somewhat worse than not quoting it. Hence, if an important and relevant paper is not listed in the references, very likely it is because I either

forgot to add it or was not aware of it. While most authors write books because they are experts in a particular field, I write them because I want to learn a particular topic. I claim no expertise on Sobolev spaces.

Web page for mistakes, comments, and exercises: In a book of this length and with an author a bit absent-minded, typos and errors are almost inevitable. I will be very grateful to those readers who write to [email protected] indicating those errors that they have found. The AMS is hosting a webpage for this book at http://www.ams.org/bookpages/gsm-105/ where updates, corrections, and other material may be found. The book contains more than 200 exercises, but they are not equally distributed. There are several on the parts of the book that I actually taught, while other chapters do not have as many. If you have any interesting exercises, I will be happy to post them on the web page. Giovanni Leoni

Acknowledgments The Author List, II. The second author: Grad student in the lab that has nothing to do with this project, but was included because he/she hung around the group meetings (usually for the food). The third author: First year student who actually did the experiments, performed the analysis and wrote the whole paper. Thinks being third author is "fair". -Jorge Cham, www.phdcomies.com

I am profoundly indebted to Pietro Siorpaes for his careful and critical reading of the manuscript and for catching 2a° mistakes in previous drafts. All remaining errors are, of course, mine. Several iterations of the manuscript benefited from the input, suggestions, and encouragement of many colleagues and students, in particular, Filippo Cagnetti, Irene Fonseca, Nicola Fusco, Bill Hrusa, Bernd Kawohl, Francesco Maggi, Jan Maly, Massimiliano Morini, Roy Nicolaides, Ernest

Schimmerling, and all the students who took the Ph.D. courses "Sobolev spaces" (fall 2006 and fall 2008) and "Measure and Integration" (fall 2007 and fall 2008) taught at Carnegie Mellon University. A special thanks to Eva Eggeling who translated an entire paper from German for me (and only after I realized I did not need it; sorry, Eva!). The picture on the back cover of the book was taken by Monica Montagnani with the assistance of Alessandrini Alessandra (always trust your high school friends for a good laugh... at your expense). I am really grateful to Edward Dunne and Cristin Zannella for their constant help and technical support during the preparation of this book. I would also like to thank Arlene O'Sean for editing the manuscript, Lori Nero for drawing the pictures, and all the other staff at the AMS I interacted with.

xv

xvi

Acknowledgments

I would like to thank three anonymous referees for useful suggestions that led me to change and add several parts of the manuscript. Many thanks must

go to all the people who work at the interlibrary loan of Carnegie Mellon University for always finding in a timely fashion all the articles I needed. I would like to acknowledge the Center for Nonlinear Analysis (NSF Grant Nos. DMS-9803791 and DMS-0405343) for its support during the preparation of this book. This research was partially supported by the National Science Foundation under Grant No. DMS-0708039. Finally, I would like to thank Jorge Cham for giving me permission to use some of the quotes from www.phdcomics.com. They are really funny.

Part 1

Functions of One Variable

Chapter 1

Monotone Functions Undergradese, I. "Is it going to be an open book exam?" Translation: "I don't have to actually memorize anything, do I?" -Jorge Cham, www.phdcomics.coui

In this chapter we study continuity and differentiability properties of monotone functions. The central result of this chapter is the Lebesgue differentiation theorem.

1.1. Continuity In this section we study regularity properties of monotone functions.

Definition 1.1. Let E C R. A function u : E -p R is called (i) increasing if u (x) u (y) for all x, y E E with x < y, (iv) strictly decreasing if u (x) > u (y) for all x, y E E with x < y, (v) monotone if any of the above holds. A monotone function is not continuous in general, so a natural question is how discontinuous it can be. The answer is given by the following theorem.

In what follows an interval I C R is a set of R such that if x, y E I and x < y, then [x, y] C I.

Theorem 1.2. Let I C R be an interval and let u : I - R be a monotone function. Then u has at most countably many discontinuity points. Conversely, given a countable set E C R, there exists a monotone function u : R - R whose set of discontinuity points is exactly E. 3

1. Monotone Functions

Proof. Step 1: Assume that I = (a, b] and, without loss of generality, that u is increasing. For every x E (a, b) there exist lim u (y) =: u+ (T),

lim u (y) _: u_ (x) .

vex

Let S (x) := u+ (x) -u- (x) > 0 be the jump of u at x. Then u is continuous at x if and only if S (x) = 0. For each n E N define

En:=

S(x) >

! }, 1

JJ1

let J C E. be any finite subset, and write

J = {xi,...) xk}, where x1 xn.

1.1. Continuity

5

Note that un is discontinuous only at the point xn. Set 00

u(x):_Euf,(x), xER. n=1

n for all x E R, the series of functions is uniformly convergent, and so it is continuous at every point at which all the functions u, are continuous. In particular, u is continuous in R \ E. We now prove that u is discontinuous at every point of E. Indeed, for every k E N write Since Iu,, (x)I

u=uk+>un. n5&k

Then E un is continuous at xk while uk is not. Hence, u is discontinuous n#k

at each x E E. To conclude, observe that u is increasing, since it is the pointwise limit of a sequence of increasing functions.

By taking E :_ Q, we obtain the following result.

Corollary 1.3. There exists an increasing function u : K --> R that is continuous at all irrational points and discontinuous at all rational points.

Definition 1.4. Given an interval I C R, an increasing function u : I - 1[8 is called an increasing saltus or jump function if it can be written in the form

u (x) _

un (x) ,

x E I,

nEE

where E C N,

U. (x) =

0 8n

ifx < an, ifx = an.,

sn+t,, ifx>an,

for some sequences {an}nEE C I, {sn}nEE C [0, oo), {tl}nEE C [0, oo), with

>0 for allnEE. Exercise 1.5 (The jump function). Let I C P. be an interval and let u : I-t R be increasing. For each x E I define

uj(x) = F, (u+(p)-u-(V))+u(x)-u-(x), vr=I, y<s

where u_ (inf I) := u (inf I) if inf I E I and u+ (sup I) := u (sup I) if sup I E I. Prove that uJ is a jump function, that uj and u - uj are increasing, and that u - uJ is continuous. The function uj is called the jump function of u.


Exercise 1.6. Let I C R be an interval and let xo E I. Prove that every increasing function u : I - R can be written uniquely as the sum of an increasing jump function v : I -* R and of an increasing continuous function

w:I->Rwith w(xo)=0. We conclude this section by studying the continuity of the inverse of an increasing function.

Theorem 1.7 (Inverse of an increasing function). Let I C R be an interval bounded from below, let u : I -, R be an increasing function, let J C R be the smallest interval that contains u (I), and let v : J -+ R be the function defined by (see Figure 1)

v(y):=inf{zEI: u(z)>y}, yEJ.

(1.1) Then

(i) v is increasing and left continuous, (ii) v jumps at some point yo E J \ {sup,, u} if and only if u (x) - yo for all x in some interval (xi, x2) C I, with vi < x2,

(iii) v (u (x)) < x for every x E I, with the strict inequality holding if and only if u is constant on some interval [z, x] C I, with z < x, (iv) v (y) - xo for ally in some interval (yl, y2) C J, with yi < y2, and for some xo E I° if and only if u jumps at xo and (yl, y2) C (u- (v0) , u+ (xo)). In particular, if the function u is strictly increasing, then v is a left inverse of u and is continuous. Proof. (i) If y1, y2 E J, with yi < y2, then

{zEI: u(z) 2y2}C{zEI: u(z) - V1 I, and so

v(yi)=inf{zEI: u(z)>-yi}-y2}=v(y2), which shows that v is increasing. To prove that v is left continuous, fix yo E J, with yo > inf J = inf r u. By the definition of v (yo) we have that u (z) < yo for all z E I with z < v (yo). Let e > 0 and fix zo E Ifl [v (yo) - e, v (yo)). Then for every y E (u (zo) , yo), we have that

v(y)=inf{zEI: u(z)>y}>zo>v(yo)-e, which shows that v is left continuous at yo.

(ii) Assume that v (yo) < v+ (yo) for some yo E J, with yo < sup J = sups u. By the definition of v (yo) and the fact that u is increasing, we have

1.1. Continuity

7

y d

c

I b+

x=v(y)

a}z

d

Figure 1. The graphs of u and its generalized inverse v.

that u (z) >_ yo for all z > v (yo). On the other hand, if v (yo) < x < v+ (yo),

then, since v is increasing, v (y) > x for all y > yo, and so u (x) < y for all y > yo, which implies that u (x) < yo. Thus, u (z) - yo for all z E (v (yo) , v+ (yo)).

Conversely, assume that u (x) - to in some interval (xl, x2) C I, with xl < x2 and yo _ x by the definition of v (y). Letting x -r x2 , we get that v (y) > x2 for ally E (yo, sup J). In particular, v+ (yo) > x2. Thus, we have shown that v (yo) < X1 < X2:5 v+ (yo) .

(iii) Taking y = u (x) in the definition of v (y), yields v (u (x)) < x. If v (u (x)) < x for some x E I, then there exists z E I, with z < x, such that u (z) >_ u (x), but since u is increasing, it follows that u - u (x) in (z, x]. Conversely, if u -const. in some interval [z, x] c I, then v (u (x)) < z < x.


(iv) Assume that v = xo E I° in some interval (yi i 3(2) C J, with y1 < y2.

If z E I with z > xo, then u (z) > y for all y E (yl, y2) by the definition of v (y). Letting y y2 , we get that u(z) > y2, and so u+ (xo) > 112. On the other hand, if z E I with z < xo, then u (z) < y for all y E (y1, y2) by the definition of v (y). Letting y - yi , we get that u (z) < y1, and so u_ (xo) < Y1 -

Conversely, let y E (u_ (xo) , u+ (xo)). If z < xo, then u (z) < y, and so v (y) > xo. On the other hand, since u and v are increasing and by part (iii), for every x E I with x > xo,

xo < v (y) < v (u+ (xo)) < v (u (x)) < X.

Letting x -+ xo , we get that v (y) = xo for all y E (u_ (xo) , u+ (xo)). Finally, assume that u is strictly increasing. Then by parts (i) and (ii) the function v is continuous. By part (iii), v (u (x)) = x for every x E I, which proves that v is a left inverse of u. This concludes the proof.

Remark 1.8. Note that the fact that I is bounded from below is used to guarantee that the function v does not take the value -oo.

Exercise 1.9. Let I = [a, b], let u : [a, b] - R be increasing and left continuous, and let v [u (a), u (b)] -> I8 be defined as in the previous theorem. Prove that u (x) = inf {y E [u (a), u (b)] : v (y) > x), x E [a, b) . Remark 1.10. The previous theorem continues to hold if u is decreasing, with the only changes that the function v : J -* R is now defined by v (y) := inf {z E I : u (z) < Y1,

that in part (i) one should replace increasing and left continuous with decreasing and right continuous, that in part (ii) one should take J \ (infj u) in place of J \ {sup, u}, and that in part (iv) the interval (u_ (xo) , u+ (xo)) should be replaced by (u+ (xo) , u_ (xo)).

1.2. Differentiability We now study the differentiability of monotone functions.

Definition 1.11. Let E C R and let xo E E be an accumulation point of E. Given a function u : E - R, if there exists in [-oo, oo] the limit lim

u(x-u(xo) W

xEE,x- o x - xp then the limit is called the derivative of u at xo and is denoted u' (xo) or du (xo). The function u is difemntsable at xo if ui (xo) exists in R.

1.2. Differentiability

9

From now on we will write lima o in place of liznxEE, x_,xo . The central theorem of this section is Lebesgue's theorem, which shows that a monotone function is differentiable everywhere except possibly on a set of Lebesgue measure zero. Note that this result relies strongly on monotonicity. Indeed, most (continuous) functions are nowhere differentiable. A classical example is due to Weierstrass.

Theorem 1.12 (Weierstrass). Let 0 < a < 1 and let b E N be an odd integer such that ab > 1 + 7r. Then the function

i

00

u (x) := E an cos (bn7rx)

,

x E 18,

n=o is continuous, but nowhere differentiable.

Proof. Since Ia" cos (b"irx) I C a' and the geometric series E00 0 a'L converges (0 < a < 1), we have that the series of functions converges uniformly, and so u is continuous. To prove that u is nowhere differentiable, fix x E R. Since u (x) = u (-x), we can assume, without loss of generality, that x > 0. For every h E R\ {0} and m E N we write 1a (x + h) - u (x) =

00

1 E an [cos (bn7r (x + h)) - cos (bn7ra:)] n=o

M-1

= h1

a" [cos (bnlr (x + h)) - cos (bmirx)] n--o 00

+h

a' [cos (bnir (x + h)) - cos (b'Lirx)] n=m

Since Icos xl - cosx2I < Ixl - x21, we have that m-1 (ab)

1Im1 < 7r

1L

= (ab)et - 1

(ab)t

ab-1 < gab -1'

-o

1L

where we have used the fact that ab > 1. To estimate write bmx = km + rm, where km E No andz- < rm < .

Define

hm :=

1

bra m

0+

m -r

as m - oo. We will estimate Ilm for h = h,n. For every n > m we have bn (x + hm) = bn-m (bmx + b'nh,n) = bn-m (k", + 1)

,

and so, since b = 2p + 1 is odd, it follows that cos

(bnir (x + hm)) = cos ((2p +

1)'°-m

Tr (km + 1)) _ (-1)k"'+1,


10

while (7rbn-m (bmx))

cos (bnirx) = cos

= cos (7rbn-'' (km +

cos (7rbn-mk,n) cos (7rr-mrm) = (-1)k- cos (?rbn-mrm)

,

where we have used the fact that sin (7rb"-'k,,,) = 0. Hence, taking h = hm in II,,, we have

_1km+l 00 IIm =

a" [1 + Cos (7rbn-mrm) J

(

,

and, in turn, 00

III,nI =

hm n=m

Since -2 < r,,,

am [1 + cos (7rrm)]

an [1 + cos (7rbn-mrm)]

0, and so III1I > -hin am. It

2, we have that cos(7rrm)

follows that

Iu(x+hm) - u(x)I

.

IIImI - IjmI >

hm

am _7r

1

h,,,

(ab)m

ab-1

Recalling that -1 co and using

Since ab > 1 + 27r, we have that s the fact that ab > 1, we have that

m o lu(x+hm)-u(x)I-0O. h

0

This concludes the proof.

Remark 1.13. More generally one can show that the same result holds if 0 < a < 1, b > 1 (not necessarily an integer) and ab > 1 (see (82)). Exercise 1.14. Modify the proof of the Weierstrass theorem to show that for every x E R,

u (y) - u (x) lim inf u (_) - U (x) < lim sup Y--+x

y-x

y"z

y-x

so that u' (x) does not exist in (-oo, oo] for every x E R.


11

Remark 1.15. It is actually possible to construct a continuous function u : (-1,1) -> R such that u (x + h) - u (x) u (x + h) u (x) _ hm inf

< lim sup

h

h

h-+0

= o0

for all x E (-1,1) (see 1128]). Next we show that the Weierstrass function is actually Holder continuous

of exponent a for every a E (0, 1). More generally, we will show that u belongs to the Zyginund space Al (R).

Definition 1.16. Given an interval I C R, we say that a function u : I - R belongs to the Zygmund space Al (1) if IItIIAI(I) =_ up lu (x)I + IUTAI(I) < 00, XEI

where IuI A, (I) :=

sup

Iu (x + h) -2u(x) + u (x - h)I

hAo,zElh

Ihl

and Ih:= {xE I: x+h,x-hE I}. Remark 1.17. Note that if u : I -r R is a Lipschitz function, then I1IA1(I) < oo. Indeed, for every h # 0 and x E Ih, we have that

Iu(x+h)-2u(x)+u(x-h)I 0 n=1

n=m+1

as m-4ooby (1.8). Remark 1.32. Note that (1.9) also shows that from every cover J as in the statement we may extract a countable family {Jn}M 1 of pairwise disjoint closed intervals such that M

E C UJn, n=1

where in is the closed interval with the same center of J. and five times its length.

Remark 1.33. The same proof continues to work if we replace Cl with an outer measure µ* with the property that 1z* (5E) < cµ* (E) for all E C R and for some constant c > 1. This is called a doubling property. For more general covering theorems we refer to 1651.

We are now ready to give the second proof of Lebesgue's theorem.


23

Second proof of Lebesgue's theorem. Step 1: Assume that I = [a, b] and, without loss of generality, that u is increasing. For each S > 0 let J76 :_ {J is a closed interval, J C [a, b] , 0 < diam J < d}

,

and for x E [a, b] define

u (max J) - u (min J) diam J b>a JEJ6:xEJ u (max J) - u (lain J) D-u (x) := sup inf a>0 JEJ6. rEJ diem J Note that if 0 < b1 < b2, then J6, C j j,, and so u (max J) - u (min J) u (max J) - u (min J) sup diam J diam J < JEJ : EJ JEJ6 P EJ

D+u(x)

inf

sup

Hence,

u (max J) - u (min J) diam J 6--'0+ JEJJ: xEJ

D4-u(x) = lim

sup

D-u (x) = limn

inf

(1.10)

and, similarly, (1.11)

u (max J) - u (min J)

diamJ

6-,0+ JE9a:xEJ

In particular, 0 < D-u (x) < D+u (x) < oo for all x E (a, b). Observe that if D-u (x) = D+u (x) < oo, then u' (x) exists and is finite. Hence, to prove the result, it is enough to shows that the set

F :_ {x E (a, b) : either D-u (x) # D+u (x) or D+u (x) = oo} has Lebesgue outer measure zero. Let

P. := {xE(a,b): D+u(x)=oo}, and for every 0 < r < R, with r, R E Q, define

Fr,R := {x E (a, b) : D-u (x) < r < R < D+u (x)} Since

FCFQQU

U

Fr,R,

0 R diam J > 0}.

If x E Fr,R fl u, then D+u (x) > R, and so by (1.10) for every d > 0 there

exists J E .F, with diam J < S, such that x E J. By Vitali's theorem,


25

for every e > 0 there exists a countable family of pairwise disjoint closed intervals {Ik} C.F such that (1.14)

0=£ (F,.,Rf1U)\Ulk =Go Fr,R\UIk k

k

where we have used (1.12) and (1.13). Then by (1.12), (1.13), and (1.14),

r1-. uns (x) - Un, (x0) X - xo x - xo :=1 It follows that lim sup zyzo

z=i

x - anf - (xo - an,) = 1. x - xo

u (x) - u (xo) > x - XO

for every 1 E N. Hence, lim sup zPzo

u (x) - u (xo) = oa,

x - xo

which implies that u is not differentiable at xo.

0

Remark 1.36. Note that the set of points at which the increasing function u (defined in (1.17)) is not differentiable contains the set E, but in general it may be larger. To the author's knowledge an exact characterization of


27

the class of sets that are sets of nondifferentiability for increasing functions is still an open problem (see Exercise 1.41 below for some properties of this set). We refer to [27] and [171] for more information on this subject.

As a simple consequence of Lebesgue's theorem we obtain that the derivative of a monotone function is always locally Lebesgue integrable_

Corollary 1.37. Let I C R be an interval and let u : I - lR be a monotone function. Then u' is a Lebesgue measurable function and for every [a, b] C I, b

(1.18)

Ja

lu' (x) I dx < ]u (b) - u (a)] .

Moreover, if u is bounded, then u' is Lebesgue integrable and l u' (x) l dx
b.

{x E [a, b] : v is not differentiable at x}. By Lebesgue's theorem,

41 (E) = 0. Moreover,

v' (x) = u' (x)

(1.21)

for all x E (a, b) \ E.

Since v is increasing, we have that v' (x) > 0 for all x E [a, b] \ E and v(x)_ ]in-,m

v(x+n)-v(x) 1

00

n

By Theorem 1.2 the nonnegative functions

y(x+n)1 -v(x)' Wn(x)

xE[a,b],

W

are continuous except for a countable set, and so they are Borel measurable functions. Hence, V: [a, b] \ E -> [0, oo) is a Lebesgue measurable function. Since v is increasing, by (1.20) for every h > 0 we have fb

[v (x + h) - v (x)] dx =

(1.22) h

< h

fJ

b+h v

(x) dx -

ja+h v (x) dx }

h{(u(b)-u(a))h}=u(b)-u(a).


28

Taking h:= 1, it follows from Fatou's Lemma, (1.21), and (1.22) that

0
R be a monotone function, and let h > 0. Then for every [a, b] C I, with b - a > h, b- h

(1.23)

1Iu(x+h) -u(x)[ dx h, and let v be defined as in (1.20). Then by (1.22),

hj

b-h [u

p

(x + h) - u (x)] dx < J b [v (x + h) - v (x)] dx a

0 the set

U"a:={xEE: there isyEIsuch that 0 0} =

n

U

U.,,6,

where

e>O, eEQ S>0, SEQ

Du (x)

lim sup

u (y) - u (x)

y-x

Y--+X

(v) {x. E E : Du (x) > a}, {x E E : Du (x) < a) are Borel sets for every a E R, where Du (x) := lim inf u (y)

y-s

- u (x)

y-x

(vi) the set F is a Borel set, (vii) the function v : I -> ]R, defined by uf x) ifxEF, v (x) :=

otherwise,

0

is Borel measurable.

Exercise 1.42 (Expansion in base b). Let b E N be such that b > 2 and let

(i) Prove that there exists a sequence {an} of nonnegative integers such that 0 < a < b for all n E N and 00

an bL

n=1

Prove that the sequence {an} is uniquely determined by x, unless x is of the form x = vA for some k, m E N, in which case there are exactly two such sequences. (ii) Conversely, given a sequence {an} of nonnegative integers such that 0 < an < b for all n E N, prove that the series EO0_1 converges to a number x E [0,1]. The next example shows that the inequality in (1.18) may be strict, and so for continuous monotone functions the fundamental theorem of calculus for Lebesgue integration fails.

Example 1.43 (The Cantor function). Divide [0,1) into three equal subintervals, and remove the middle interval Ii,l (N, Divide each of the two remaining closed intervals [0, 1] and [2,1] into three equal subintervals, and remove the middle intervals 11,2 := (3 , 3 ) and 12,2 (a , a ). Continuing in this fashion, at each step n remove 2n-1 middle intervals Il,n, .. , 12n-',,, each of length a . The Cantor set D is defined as 9).

00 2`1

D := 10,1] \ U U Ik,n. n=1 k=1


31

M III II,

3

I

4

r ---i

II

1

PlII

I

I

II

1

I

II

-9

9

3

9

2:

1

Figure 3. The Cantor function.

The set D is closed (since its complement is given by a family of open intervals), and 00 2"-1

00 2*-1

G1(1D)=1-LE diem(Ik,n)=1- 1: 1: n=1 k=1

n=1 k=1

1

00 28-1 gn=1-E 3n n=1

=0.

Thus, ® has Lebesgue measure zero. To prove that IID is uncountable, observe

that x E D if and only if 00

x=L,3

(1.25)

n=1

where each c,, E {0, 2} (see the previous exercise). For every x E D of the form (1.25) define 00

oft 8=1

where 1

a'`

ifc,,=2,

0

The function u : IlD -p [0, 1] is well-defined (why?), increasing, continuous, and has the same values at the endpoints of each removed interval Ik,n and thus u extends to a continuous function on 10, 11. This function is called the Cantor function (see Figure 3). Since u (D) = [0, 11, it follows that D is uncountable. Note also that ul (x) = 0 for all x E Ikon, k, n E N, so that u' = 0 except on a set of Lebesgue measure zero. Since to (0) = 0 and u (1) = 1, it follows that the inequality in (1.18) is strict, and so the fundamental theorem of calculus for Lebesgue integration fails.


32

Exercise 1.44. Consider the Banach space X :_ {u : [0,1] - R : u is Continuous, u (0) = 0, and a (1) = 1) , where we take the supremum norm

IIuII, := max Iu(x)I Consider the operator T : X -i X, defined by T (u) (x)

-ju(3x)

if OCxn +

00

2Rn

2Rn

n-o

(1.30)

1

27n

1

+

2akm+j '

j=1

-

1

=

tan +

n-

1

00

1

j=1

2m+j'

By (1.28) and the fact that ak,,, +1 > m + 1, we get that akm +j > m + j for all j N, and so 00

xm < = MM + 2a En- xm +k = xm + urn 00

1

n=k,n+1

zm.

k=m+1

, we have that (why?)

Since xm and zm differ only by

rn

00

(1.31)

1

1

u(z,,)-u(x»i)=

rkn+l 1+r)n+m-km

n=k,,,+1

(1+r}'n.

Using the fact that km < m + 1, it follows that u (z,n) - u (xm) -, 0 as m -* oo, and since u is increasing and xm < x < z,n, this shows that u is continuous at x.

To prove continuity at x = 0, it is enough to observe that as x 0+, where x is given by (1.27), the corresponding coefficient ao = ao (x) approaches infinity, and so U (x)

1

_ (1+r)

00

n --0 00

n by (1.28).


35

Finally, we shall prove that if u is differentiable at some x E (0, 1], then necessarily u' (x) = 0. Indeed, assume by contradiction that u' (x) = e # 0. Then by (1.29) and (1.31),

fl (z,,,) - u (x,) _

zf1-xm

m

2

\l+r

rk,»+1

On the other hand, by the previous exercise, u (zm) - t (Xm) - Q zm. - xm

as m -i oo, and so

(0. This implies that r k_+ m

_1 -11

1

+r 2

Since r # 1 and k,,, is a nonnegative integer, we have that km - km-1 s r= No, so for all m > mo we have that km = k,,,-, + a. Note that a = 0 or s = 1 leads to 1 = 1-' are both impossible. If s > 2. 2 and r =, which 2 we have that km

m mo + (m - mo) as m -> oo, but since km C m + 1, we again have a contradiction.

Exercise 1.48. Let C c R be a closed set and let u (x) := dist (x, C) = inf {fix - tI : y E C},

x E R.

Prove that (i) u is 1-Lipschitz, that is,

Iu (x) - u (01 5 Ix

- uI

for all x)yER, (ii) there exists an increasing function v : R -> R such that v' is continuous and

C={xER:v'(x)=0}, (iii) if C contains no intervals, then every such v is strictly increasing. Another important consequence of the Lebesgue differentiation theorem is the following result.

Theorem 1.49 (Fubini). Let I C R be an interval and let be a sequence of increasing functions, u : I - it Assume that the series of


36

functions E u, converges pointwise in I. Then E u, converges uniformly n=1

n=1

on compact sets of I, the sum of the series

u(x) := E u,, (x) , x E I, n=1

is differentiable C 1-a. e. in I, and 00

u' (x) _

for G1-a. e. x E I.

un (x) n=1

Proof. Since the result is local, without loss of generality, we may assume that I = [a, b]. Step 1: Assume that un (a) = 0 for all n E N so that uTZ > 0. For every

IENset 81 :_

Ztn.

n=1

Then for x E [a, b], 00

00

0 Rd, then we can define its variation exactly as in Definition 2.1, with the only difference that the absolute value is now replaced by the norm in Rd, which we still denote I . I. The space of all functions u : I -> Rd of bounded pointwise variation (respectively, locally bounded pointwise variation) is denoted by BPV (I; Rd) (respectively, BPVIOC (I; Rd)).

The following facts are left as an exercise. related) function space. This book studies both spaces, so we really had to change the notation for one of them.

2.1. Pointwise Variation

41

Exercise 2.3. (i) If u : (a, b] -* R is (everywhere) differentiable with bounded derivative, then u E BPV ([a, b]).

(ii) If u : (a, bJ -* R is (everywhere) differentiable and u' is Riemann integrable, then u e BPV ([a, b]) and rb Var u = lu' (x) I dx.

Ja

Compare this with the Katznelson-Stromberg theorem below.

Exercise 2.4. Let u : (0,1] -, R be defined by it

(X)

xasiny if0<x 0. Assume that I does not contain its right endpoint (the other case is similar) and fix 0 < t < Var u. By the definition of Var u we may find a partition


43

P:= {x0, ... , xn} of I such that n / t <EIu(xi) -u(xi-1)I i=1

For any x E (x,,, sup I), we have that P is a partition of I fl (-oo, x], and so by part (ii),

t
J be monotone. Prove that f o u belongs to BPYOC (I).

Exercise 2.22. Let

unx)=

1 1

2

76

z

s

,

xEIB.

(i) Calculate Var u,s. (ii) Let oo

u(x):=Eun(x), xER. ft=1

Prove that u E BPV (R). (iii) Prove that F,-, u;, (x) does not converge uniformly in [-1, 1]. (iv) Find a formula for u'. (v) What is the relevance of this exercise? The following result is a consequence of Theorem 2.18.

Corollary 2.23. Let I C R be an interval and let u E BPV°C (I). Then for every x E I the limits lim u (y) =: u+ (x) , lim u (y) =: u_ (x) 2With the obvious changes if x is an endpoint.

2. Functions of Bounded Pointwise Variation

48

exist in R, u has at most countably many discontinuity points and is differentiable Ll -a. e. in I, and for every [a, b] C I, /b J' (x) I dx < Var1Q,bi u. /

(2.8)

a

If, in addition, u E BPV (I), then u is bounded, the limits lim

(2.9)

u (y) ,

lim

y-(aup')exist in R, u' is Lebesgue integrable, and y-4inf I)+

dx < f IV'I dx

(2.10)

I

u (y)

sup V - inf V I

Var u.

Proof. Step 1: Since by Theorem 2.18 every function in BPVOC (I) is the difference of two increasing functions, from Theorem 1.2 and Lebesgue's theorem we obtain that u+ (x) and u_ (x) exist in R for every x E I, that u has at most countably many discontinuity points, and that it is differentiable L -a.e. in I. To prove (2.8), let [a, b] C I and consider the increasing function V defined in (2.2). Fix X E (a, b) such that both it and V are differentiable at x. Then from (2.3) it follows that for y > x, lu (y) - u (x)I < V (y) - V (x)

y-x

y-x

and letting y - x+, we obtain Iu' (x)I < V' (X).

(2.11)

By Corollary 1.37 applied to V and (2.3), we obtain

f IV'I dx < V (b) b

(2.12)

f b Iu'I dx < a

- V (a) = Var1a,bj u.

a

This proves the first part of the corollary. Step 2: If it E BPV (I), then again by Theorem 2.18, it is the difference of two bounded increasing functions. Hence u is bounded and the limits (2.9) exist in R. To prove (2.10), we proceed as in the last part of the proof of Proposition 2.10 to conclude from (2.12) that

dxf IV'I dxslpV-infV=Varu, JI

I

where in the last equality we have used Exercise 2.14. This concludes the

0

proof.

Remark 2.24. Note that in the second part of the proof, one could have used the function V. defined in (2.6). Indeed, as in Step 1 we have that for L1-a.e. X E I, (2.13)

In'(x)I

1--


53

by (2.27), so that (2.23)1 holds. To check (2.24), note that by (2.26) and

(2.28) in

(a3) + is + 1 -

Utz = 'fin-1 +'On
j,

/ F' (.3j) _

00

00

OA: (03)

.1'

='n-1 (J3,) + F, '`Yk (0i) k=n

k=1 00

0. Next we claim that f is Lipschitz in (-r, r]. Indeed, assume by contradiction that this is not the case. Then we may find two sequences {sn}, {tn} C [-r, r] such that s,a # tn. and (2.31)

If (8,,,) - f (tJ > 2 (n2 + n) Isn.-tnl

for all n E N. Since {s,,} is bounded, we may extract a subsequence (not relabeled) such that sn -+ soo. Take a further subsequence (not relabeled) such that (2.31) continues to hold and

Is -acc I
- If (8n) - f (tn)I > 2 (n2 + n) Isn - tnI

Hence, (2.34)

0 < d0 :=

Isn - tn1 (b - a)

21i4

b.

For every n E N, let

b-a 2M,. - 8n(n2+n) Isn-tnl(n2+n) >2 an by (2.33) and set m = max {j E No : j < Q. Note that Mn is the number of times that the function u takes the value sn in the interval In. Since (2.36)

Pn:=

diam In

En > 2, we have

f2

(2.37)

< Mn < Qn.

Consider the partition P. of In = [a +

1, a + --n-] given by

Pn.={a+n+1+21bn: i=0,...,2m,, o

,...

2,n

Then by (2.33)-(2.37), 2

Var7 (fOu)>EIf(u(xr))-1 (U(-T'

an-tn4M,., and so for every n E N by Remark 2.7 we obtain that n

Var1Q.,b] (f ou)Varjk(fou)>4Mn->oo k=1

asn -oo. On the other hand, since u is a step function on In, by (2.32)-(2.37) we have that (why?) Var jn u < 2mn I8n - to l + I8n+1 - Sn I + I Sn+1 - tnl

R is said to be of f-bounded pointwise variation if 9-1 ([a, b]) E 7t for all [a, b] C R.

(i) Prove that if g : 10,1] -> R is of I -bounded pointwise variation and

if Ig(x)I<Mfor allxE[0,11,then Varg E Varg,,. n=1

2.3. The Space BPV (1)

59

(v) Prove that if g : [0,1] -+ 10,11 is such that u o g is of bounded pointwise variation for all u : 10,11 -- R of bounded pointwise variation, then g : 10, 1] - R is of £-bounded pointwise variation for some e E N. Hint: If not, then for each n E N find an interval In C [0,1] such that g 0 3Jn and take 00

u:=Eun for an appropriate choice of u,a.

2.3. The Space BPV (I) Note that by (2.7), u H Varu has most of the properties for being a norm in the space of BPV (I). What is missing is the fact that Varu = 0 does not imply that u = 0, but only that u is a constant. However, we have the following.

Corollary 2.34. Let I C R be an interval. Then for any fixed c E I, the application u

Hull := Iu (c) I + Varu

is a norm in the space BPV (I). Proof. By (2.7), Iltull = Iti lull and 11u + ull no there must exist m distinct intervals Jl("), ... , Jmn) E .F, such that x; E J(n) for all i = 1, ... , m and n > n6. Thus, for all n > n6, m

m=

X,,(J(n)) (F!) N. (y; I) in the previous inequality to conclude that

N. (y; I) < lnm inf E Xt(J) (y). -400 JEFn

On the other hand, by the definition of N. (y; I), for every integer n E N, we have that (2.46)

E Xu(J) (y) JR be continuous. For every 5 > 0 let P6 be a partition of [a, b], with

xo,s := a < xl,g < ... < xna,a b, such that xi,a - xj-j,,5:5 6 for all i = 1, ... , no. Then n6

Var u = Jim E [u (xs,d) - u (xs_l,a) I a-,o+

s=i

.

2.4. Banach Indicatrix

69

Proof. Without loss of generality we may assume that Var is > 0, since otherwise is is constant and there is nothing to prove. Fix any 0 < t < Varu of [a, b], with xo = a and x,n = b (see and find a partition P = {xo, ... , Remark 2.2), such that m

S :_ E Iu (xj) - u (xj_1)1 > t.

(2.49)

j=1

Since is is uniformly continuous, there exists t > 0 such that (2.50)

Iu (x) - is (x') I < 2m t for all x, x' E [a, b] with Ix - x'l < q. Let 1

-min { ri, min (xj - xj-1) 2 l 3=1,...,m If 0 < 6 C 60, then for each j = 1, ... , nt - 1 there exists a unique index k, then ij E { 1, ... , n6} such that xj E (Ci,-1,5, xi?,d Moreover, if j 0 < So

<Setio:=0andi,,,:=n6.

ij74 ix,andi1 Since

- xj

I

:5 xi,,6 - xi,-1+6

n,

by (2.50) we have

-

l u (xj) -u (xj-1)I :5 1'U (xj) u (xi3,6) I + Is (xif,6) - 46 (xis-1,6) I+ Iu (xi,-1,6) - u (xj-1)I +

S-t rn

It(xis,6) -u(xi,

Summing over all j = 1, ... , m and using (2.49), we get S=

E lu (x3) - u (x,-1}1 < S - t + E I's (Xij,6) - u (xz,-1,6) I

j=1

j=1

n6

<S-t+

lu (xi,6) - 2i (xi-1,6)I , i=1

and so

n6

u(x:,6) -u(xi-1,6)I 1. Given a function u : [a, b] -> R, define n

Varrp°) u := lira sup b-,0+

11r

lU (.Ti) - u (Xi-1)Ip i=1

where for each fixed 6 > 0 the supremum is taken over all partitions P {x0, ... , x,,} of I such that Ixi - xi-1 I < b, i = 1, ... , n, n E N. (i) Prove that Varp°) u < oo if and only if Vari, u < oo.

(ii) Prove that there exists a function u for which Varp°} u < Var , u < oo.

(iii) Prove that if u is continuous and Varo u < oo, then Var( °)u = 0 for q > p. Hint: Use uniform continuity.

Proof of Banach's theorem. Step 1: Assume that I = [a, bJ . For every n E N let F be the partition of [a, b] given by

I1, := [a, a+

b

2na1

Ik,n:= La+(k-)b2na,a+kb2"aJ

fork=2,...,2".

Since u is continuous, by the Weierstrass theorem and the intermediate value theorem for every interval [a, Q] C [a, b] we may find t1, t2 E [a,131 such that

u([a,/3]) = [u(t1),u(t2)]. Hence, by (2.3), Iu (0) - u MI < £' (u ([a, 01)) = Iu (t1) - u (t2)I oo and using the previous lemma, we obtain

VarlQ,bl u = l!ino E Iu (sup J) - u (inf J)I JEF.

< lim

C1(u(J)) < Var1a,bl u, JEFn

and so, by Federer's theorem with p given by the Lebesgue measure,

jN(v; [a, b]) dy = lm

,V 1(u (J)) = Var[n,b] u.

JEF Step 2: If I is an arbitrary interval, construct an increasing sequence of intervals [ak, bk] such that

ak \ inf I, bk / sup I, and

00

I= U [ak, bk] k=1

By the definition of N,, (y; -) for every y /E l[i; we have that N. (y; [ak, bk]) C N. (y; [ak+1, bk+1])

and N. (y; [ak, bk]) - N. (y; I). Hence, by Proposition 2.6, Exercise 2.8, the previous step, and the Lebesgue monotone convergence theorem, we get

Varl u =

>n Var(Q,,,bkl u = k m f N(y; [ak, bk]) dy = f N(y;1) dy

This concludes the proof.

11

Corollary 2.51. Let I C R be an interval and let u E BPV (I) be continuous. Then the set o f values y E ]R f o r which u 1({y}) is infinite has Lebesgue measure zero.

Proof. Since by the previous theorem the function Nu is integrable, it follows that it must be finite for £1-a.e. Y E R.

The previous corollary expresses the fact that a function in BPV (I) cannot oscillate too much.

Exercise 2.52. Consider the Cantor set 00 P-1

®:_ [0,1] \ U U Ik,nn=1 k=1

For every interval Ik,, = (ak,n, bk,n), n r= N, k = 1, ... , 2n-1, define 9A;, (x)

(x - 4k,.) -2'1 sin 27r , bk,n - ak,n

x E [ak,nr bk,n]

2. Functions of Bounded Pointwise Variation

72

(i) Prove that the function g : [0, 11 -R, defined by gk,n (x) if x E Ik,n, n E N, k = 1, ... , 2n-1, g (x)

0

ifxED,

is continuous.

(ii) Let f := g+u, where u is the Cantor function. Prove that for every n E N,

[0,1] = f

U Ik,n k 0 there exists b > 0 such that (3.1)

Iu(bk) - u(ak)I < E k=1

for every finite number of nonoverlapping intervals (ak, bk), k with [ak, bkl C I and

(bk-ak) R is denoted by

AC(I). 73

3. Absolutely Continuous Functions

74

Remark 3.2. Note that since a is arbitrary, we can also take e = oa, namely, replace finite sums by series.

A function u : I -+ R is locally absolutely continuous if it is absolutely continuous in [a, b] for every interval [a, b] C I. The space of all locally absolutely continuous functions u : I -, R is denoted by ACID,, (I). Note

that ACID.. ([a,b]) = AC([a,b]).

If u : I --+ Rd, then we can define the notion of absolute continuity exactly as in Definition 3.1, with the only difference that the absolute value is now replaced by the norm in Rd. The space of all absolutely continuous functions u : I -> Rd (respectively, locally absolutely continuous) is denoted by AC (I; Rd) (respectively, ACID, (I; Rd)). If 0 C R is an open set, then we define the notion of absolute continuity for a function u : SZ -* R as in Definition 3.1, with the only change that we now require the intervals [ak, bkj to be contained in [Z in place of I. The space of all absolutely continuous functions u : 11-> R is denoted by AC (Q).

Exercise 3.3. Let I C R be an interval and let u : I - R. (i) Prove that u belongs to AC (I) if and only if for every s > 0 there

exists b > 0 such that e

k=1

for every finite number of nonoverlapping intervals (ak, bk), k = with [ak, bk] C I and

>(bk-ak) <S. k=1

(ii) Assume that for every > 0 there exists 6 > 0 such that

k=1

for every finite number of intervals (ak, bk), k = 1,...J, with [ak, bk] Cl and

>(bk-ak) R be differentiable with bounded derivative. Prove that u belongs to AC (I). Exercise 3.6. Let u, v E AC ([a, b]). Prove the following.

(i) u±vEAC([a,b]). (ii) uv E; AC ([a, b]).

(iii) If v (x) > 0 for all x E [a, b], then v E AC ([a, b]). (iv) What happens if the interval [a, b] is replaced by an arbitrary interval I C R (possibly unbounded)? We now turn to the relation between absolutely continuous functions and functions of bounded pointwise variation. In Corollary 2.23 we have proved

that if u : I - llt has bounded pointwise variation, then u is bounded and u' is Lebesgue integrable. However, the function u (x) := x, x E R, is absolutely continuous, but it is unbounded and u' (x) = 1, which is not Lebesgue integrable. Also the function u (x) := sin x, x E R, is absolutely continuous, bounded, but ui is not Lebesgue integrable. These simple examples show

that an absolutely continuous function may not have bounded pointwise variation. Proposition 3.8 below will show that this can happen only on unbounded intervals.

Exercise 3.7. Let I C R be an interval and let u : I -i R be uniformly continuous.

(1) Prove that u may be extended uniquely to I in such a way that the extended function is still uniformly continuous.

(ii) Prove that if u belongs to AC (I), then its extension belongs to AC (I). (iii) Prove that there exist A, B > 0 such that for all x E I, Iu(x)l O such that u is differentiable for all x E E, with

lu'(x)I<M forallxEE. Then

Go(u(E)) :5 M41(E). Proof. Without loss of generality we may assume that E C I°. Fix s > 0 and for each n E N let E. be the set of points x E E such that (3.2)

Go (u (J)) R be a Lebesgue integrable function. Fix xo E 1 and let x

U (X) :=

Jz:V(t) dt,

Then the function u is absolutely continuous in I and u' (x) = v (x) for G1-a.e. x E I. Proof. The facts that u is absolutely continuous and differentiable for G1a.e. x E I follow from Exercise 3.11 and Proposition 3.8, respectively. In the remainder of the proof we show that u' (x) = v (x) for Ll-a.e. x E I. Step 1: Assume first that v = xE for some Lebesgue measurable set E C R. Fix a bounded open interval J C I containing xa in its closure. We claim

that

u'(x)=1 forLl-a.e. xEEf1J. By the definition of Lebesgue outer measure we may find a decreasing sequence {U,,} of open sets such that Un ) E fl J and (3.6)

,C1(UU\(EflJ))=0, whereUU:= flUn. n=1

By replacing Un with Un fl j, we may assume that U. C J. Define

f Xu (t) dt, a

un (x)

x E J.


86

Since U1 C J, which is bounded, we are in a position to apply Lebesgue's dominated convergence theorem and (3.6) to conclude that for all x e J,

urn nn (x) n m Jxu(t) dt = Jxu. y =

(t) dt

o

o

f XEnJ (t) dt = J x

px

x0

xo

XE (t) dt = u (x)

where in the fourth equality we have used the fact that the open interval of endpoints x and xo is contained in J. Hence, in the interval J we may write u in terms of the telescopic series 00

2a = 261 + E (un+1 - un) . n=1

Note that since U. D U,a+1, (un - un+1) (x) =

(t) dt,

x E J,

xo

and since 0, the function un - tan+1 is monotone in the intervals (-oo, xo)nJ and [x0, oo, )nJ. By F ubini's theorem (applied in each interval)

we get that for 41-a.e. x E J, 00

4L' (x) = ti (x) + E lun+l (x) -'un (x)) n=1

= n noun

(x) .

On the other hand, if x E Un, then u;, (x) = 1 (why?), and so, if x E U"., then un (x) = 1 for all n E N. Hence, we have proved that u' (x) = 1 for ,C1-a.e. X E Uo and so, in particular, for £1-a.e. T E E fl J. Next we show that u' (x) = 0 for G1-a.e. x E J \ E. Since for all x E J, a

x

XE (t) dt =

U (x) = ITO

L.

(1

- XJ\E) (t) dt,

by applying what we just proved to X.\E, we conclude that u' (x) = 1-1 = 0 for G1-a.e. x E J \ E. Thus, 26' (x) = XE (x) for G1-a.e. x. E J. By letting

J / I, we obtain the same result in I. Step 2: By the linearity of the derivatives and Step 1 we conclude that if v is a simple function, then u' (x) = v (x) for C'-a.e. x E I. If v is a nonnegative Lebesgue measurable function, then we may construct an

'

increasing sequence {sn} of nonnegative simple functions such that s,, (x) v (x) for Cl-a.e. x r= I. Then by Lebesgue's dominated convergence theorem for all x E 1, lim

n-+°0

f8(t) dt = Jv(t) dt = u (x) , x o

o

87

3.1. AC (I) Versus BPV (1)

and so we may proceed as in the first step (using telescopic series) to show that u' (x) = v (x) for L1-a.e. x E 1. In the general case, it suffices to write v = v+ - v-. We are now ready to prove Theorem 3.30.

Proof of Theorem 3.30. Assume that u E ACID, (I). In view of Theorem 3.12, it remains to prove (iii). Let [a, b] c I be so large that xo E [a, b] and define

w (x) = u (x) - (u(a)

+ 1.00 U' (t)

dt)

,

x E [a, b] .

By Lemma 3.31 and Theorem 3.12, there exists a Lebesgue measurable set

E C [a, b], with ,C1 (E) = 0, such that for all x E [a, b] \ E the function w is differentiable at x and w' (x) = 0. By Corollary 3.14 we have that ,C1 (w ([a, b] \ E)) = 0. On the other hand, since to is absolutely continuous in [a, b] (see Exercise 3.6 and Lemma 3.31), by Theorem 3.12 it sends sets of Lebesgue measure zero into sets of Lebesgue measure zero, and so G1 (w (E)) = 0. Thus, we have shown that ,C' (w ([a, b])) = 0. But since to is a continuous function, by the intermediate value theorem w ([a, bJ) is either a point or a proper interval. Thus, it has to be a point. In conclusion, we have proved that to (x) = const. Since to (xo) = 0, it follows that to = 0, and so (iii) holds for all x E [a, b]. Given the arbitrariness of [a, b], we have that (iii) holds for all x E I. Conversely, assume that (i)-(iii) are satisfied. Then again by Lemma 3.31, u belongs to ACID, (I) and the proof is complete. The next corollary follows from the previous theorem and Corollary 3.22.

Corollary 3.32. Let I C R be an interval and let u : I -- R be everywhere differentiable. I f u ' E L10C (I), then for all x, xo E I, U (X) = U (X0) + j u' (t) dt. :ox

Using the previous corollary, we are in a position to complete the proof of the Katznelson-Stromberg theorem. We begin with some well-known results on Riemann integration.

Exercise 3.33 (Riemann integration, I). Let u : [a, b] -p R be a bounded function and for x E [a, b] define

w(x)

.

allies sup{Iu(xl) -u(x2)I : xlox2 E [a,b], IXl - xI < 6, Ix2 - XI 0 such that the set

E1:={xEE: w(x)>a} has positive Lebesgue outer measure.

(iii) Let t = C1(E1) > 0 and prove that for every partition P the intervals containing points of El in their interior have total length greater than or equal to t. (iv) Deduce that if u is Riemann integrable, then E has Lebesgue measure zero.

Exercise 3.34 (Riemann integration, II). Let u : [a, b] -> R be a bounded function.

(i) Prove that if g : [a, b] -* R is Lipschitz and 9'(x) = 0 for £1-a.e. x E [a, b], then g is constant. (ii) Let x a

u t dt-

t dt a

a<x IV (bk) - V (ak)I, k


92

Conversely, if u E AC ([a, bJ), then by Theorem 3.30 for every partition of [a,b],

E Iu (xi) - u (xi-1)I = i=1

i=1

f

xi

px;

4s1 dx 0 and let 6 > 0 be as in Definition 3.1. Consider a Lebesgue measurable set E C 1, with L1 (E) < '6. By the outer regularity of the Lebesgue measure we may find an open set A D E such that Ll (A) < J. Decompose A into a countable family {Jk} of pairwise disjoint intervals. By replacing each JA.' with Jk fl I, we may

assume that Jk C I. Let Wk = [ak, bk]. Using the fact that C1 (A) < 6, we have that 1: Ibk - ak] < d. k

Consider a partition Pk = xOk), ... , x '.k } of [ak, bk] . Since x(k) k

i=1

- x`kll

=>Ibk-ak] J be such that f, u, and f o u are differentiable G1-a.e. in their respective domains. If f maps sets of Lebesgue measure zero into sets of Lebesgue measure zero, then for G1-a.e. x E I,

(f o u)' (x) = f (u (x)) u' (x) , where f' (u (x)) u' (x) is interpreted to be zero whenever n' (x) = 0 (even if f is not differentiable at u (x)). (3.14)

3.2. Chain Rule and Change of Variables

95

In the proof we will show that (f o u)' (x) = 0 and u' (x) = 0 for £1-a.e. x E I such that f is not differentiable at u (x). To prove Theorem 3.44, we need an auxiliary result, which is a converse of Corollary 3.14.

Lemma 3.45. Let I C l[t be an interval and let u : I -p R. Assume that u has derivative (finite or infinite) on a set E C I (not necessarily measurable), with £1(u (E)) = 0. Then u' (x) = 0 for C1-a.e. X E E.

Proof. Let E* := {x E E : Iu' (x) I > 0}. We claim that Ll (E') = 0. For every integer k E N let

Ek:= {xEE*: Iu(x)-u(y)I? Ixkvl for all yE (x-k,x+ I)nI}. Noting that 00

E*=UEl, k=1

we fix k and we let F:= J fl Ek, where J is an interval of length less than To prove that £1(E") = 0, it suffices to show that ,C' (F) = 0. Since

C1(u (E)) = 0 and F C E, for every e > 0 we may find a sequence of intervals { Jn} such that 00

00

u(F) C U Jn, E V (.fin) < n=1

n=1

Let E. := u-1(Jn) n F. Since {En} covers F, we have 00

00

Go (F) E 40 (En) C E sup Ix - T!I n=1 x,yEE.

n=1 00

k sup Iu(x) -u(y)I =:I, n=1

x,yEE

where we have used the fact that En c J n E. Since u (En) C Jn, we have sup Iu (x) - u (y) I R is a measurable, bounded function and that u : [a, b] - [c, ci] is absolutely continuous. Then (g o u) u' is integrable and the change of variables formula (3.16) holds.

Exercise 3.60. Prove that under the hypotheses of the previous corollary the function f o u is absolutely continuous and then prove the corollary.

Corollary 3.61. Assume that g : [c, d] - R is an integrable function, that u : [a, b] - [c, d] is absolutely continuous, and that (g o u) u' is integrable. Then the change of variables formula (3.16) holds.

Proof. Let

n gn (z) :=

g (z)

I -n

if g (z) > n,

if - n < g (z) < n, if g (z) < -n.

Applying the Lebesgue dominated convergence theorem and the previous corollary, we obtain rn fUW) g (z) dz = lim gn (z) dz ' '°° Ju(Q)

(")

= slim

f

a

r

gn (u (x)) u' (x) dx

= Ja

(u (x)) u.' (x) dx.

0 To extend the previous results to arbitrary intervals, we consider two functions g : J R and u : I J and we assume that there exist in J the limits Urn

x-(inf I)+

lim

U (X) = P,

u (x) = L.

X--+ (sup 1)

In this case, the analog of (3.16) becomes (3.18)

dt

L g (z) dz =

Indeed, we have the following result.

l

f g (u (x)) u' (x) dx.


100

Theorem 3.62. Let I, J C R be two intervals, let g : J -> R be an integrable function, and let is : I - J be differentiable G1-a.e. in I and such that there exist in R the limits (3.19)

lim u (x) = Q, x-(inf I)+

lim u (x) = L. x-'(sup I)-

Then (g o u) u' is integrable on I and the changes of variables (8.16) and (3.18) hold for all [a, /3] C I if and only if the function f o u belongs to AC (I) fl BPV (I), where

rf

f(z):= Jmz J g(t)dt,

zEJ.

Proof. Assume that f o is E AC (I) fl BPV (I). Then, we can proceed as in the proof of Theorem 3.54 to show that (3.16) holds for all [a, 0] C I. Since f o is E BPV (I), by Corollary 2.23 and (3.17) its derivative (g o u) u' is integrable on I. Hence, by taking limits as a and 6 approach the endpoints of I and using (3.19) and the Lebesgue dominated convergence theorem, we conclude that (3.18) holds. Conversely, if (g o u) u' is integrable and (3.16) and (3.18) hold for all [a, $] C I, then

(f o u) (/j) - (f o u) W=

J

g (u (x)) u' (x) dx

for all a,# E I. As in Theorem 3.54 we deduce that fou E ACI0c {I). In turn, by Corollaries 3.41 and 3.42, it follows that f o is r: AC (I) fl BPV (1). 0

Closely related to the change of variables formula is the area formula, which will be discussed next.

Definition 3.63. Let X be a nonempty set and let 0 : X - [0, oo] be a function. We define the infinite sum of i!i over X as

/ (t) := sup SEX

4' (t) : Y C X finite

.

tEY

Exercise 3.64. Let X be a nonempty set and let 4' : X -' [0, oo] be a function. Prove that if EtEX 0 (t) < oo, then the set

{tEX:4'(t)>0} is countable.

Theorem 3.65 (Area formula). Let I C R be an interval, let 0 : I - [0, oo] be a Borel function, and let u : I - R be differentiable LI-a.e. in 1 and


101

such that u maps sets of Lebesgue measure zero into sets of Lebesgue measure

zero. Then

E 1) (t) dy =

(3.20) dddR

tEu-1({y})

fI

0 (x) lu' (x)I dx.

Proof. Step 1: Assume first that I = (a, b) and that u E C' (I). Consider the open set A :_ {x E I : u' (x) 01 and let {(ak,bk)} be the countable family of connected components of A. If u' > 0 in (ak, bk) and 0 E Loo ((ak, bk)), then by Corollary 3.59 (applied in (ak, bk) to the functions u and g o v, where v (ttI(Okbk))') we get u(bk)

bk

1

j tb (x) u' (x) dx = fu(ak) + ((1LI(O,b))

(3.21)

ll (y))

dy.

k

O n the other hand, since ul(ak

is strictly decreasing and continuous, for bk) every y E (u (ak) , u (bk)) there is one and only one t E (ak, bk) such that u (t) = y, so that (ak, bk) n u-' ({y}) = {t}, while, if y E ]R \ (u (ak) , u (bk)), then (ak, bk) n u-1 ({y}) _ 0. This shows that

f

1

u(bk)

(t) dy = f

(t) dy

u(a,,) tE(ak,bk)nu-1({y})

tE(ak,bk)n*a-1({y})

'++(bk)

=f

( ak )

i ((u) (y)) 1

dy.

Combining this equality with (3.21) gives fk

0 (x) ta' (x) dx =

(3.22)

k tE(ak,bk)nu 1{{y))

0 (t) dy.

To remove the additional assumption that 10 E LO0 ((ak, bk)), it suffices to apply (3.22) to 0,, := min {0, n} and to use the Lebesgue monotone convergence theorem.

A similar argument shows that if u' < 0 in (ak, bk), then bk

ftb(x)Iu'(x)Idx=J k

1:

tP (t) dy.

tE(ak,bk)nu-1({3l})

Adding over k, we obtain fA

i/i (x) I u' (x) I dx =

f

(t) dy tC-Anu-1 Q

)


102

Since u' (x) = 0 in I \ A, by Corollary 3.14, G1 (u (I \ A)) = 0. Hence, the previous equality can be rewritten as

J iP (x) Iu' (x) I dx = J 1u(I\A) tEAlu-1({y})

(t)

dy.

On the other hand, if y E R \ u (I \ A), then

u-1({y}) _ (An u -I ({y})) u ((I \ A) flu 1({y})) = A fl u 1({y}), and so

L\u(l\A) tEAflu -1((y})

i (t) dy

(t) dy = fflt\u(1\A) tEu -1({y))

tEu (W)

(t) dy,

1

where in the last equality we have used the fact that G1 (u (I \ A)) = 0 once more.

This shows that (3.20) holds. Step 2: Assume next that I = (a, b), that there exists a compact set K C I such that i{' = 0 on I \ K, that u is differentiable for all x in K, and that inn

u (y)

yEK,y-'x

-u l/-x

(x) = u' (x)

uniformly for x E K.

Then by Exercise 3.66 below there exists a function v E Q1 (I) such that

v = u and v' = u' on K. Applying the previous step to v and withi replaced by XK',, we obtain

fKh1 dx = f XK (x) V) (x)

I v' (x) I dx

E XK (t)(t) fftEKrW-I([yj)

dy

tEv-1((y))

, (t)

1:

ti (t) dy.

R tEKnu-1({y})

0 on I \ K, we have that (3.20) holds. Step 3: Assume that I = (a, b). Since u is differentiable G1-a.e. in I, the Since

sequence of functions un (x)

sup

I u (y) - u {x)

Y-X

- u' (x)I , x E I,


103

converges to 0 for G1-a.e. x E I. It follows by Egoroff's theorem that there exists an increasing sequence of compact sets {Kj} C I such that ,C1

(o)

=0

and such that {u'} converges to zero uniformly in Kj for every j E W. In particular, u (g) - u (x) = u' (x) uniformly as x E Kj lim &oEKj,y--w

y-x

for all j E N. By the previous step with h 0 replaced by 'XK;,

ii (x) lu' (x) I dx = J

-0 (t) dy. tEKjnu '({y})

Letting j -> oo, it follows by the Lebesgue monotone convergence theorem

that

I

U,

f

ik(x) lu'(x)I dx = 1

Ki

E

ip (f) dy.

tEU,°;°_, Kjnu- I ({y})

Since G1 (I \ U;°_1 Kj) = 0, by hypothesis we have that 00 ,C1

(.u (r\

Kj

I

= 0,

j=1

and so we obtain (3.20).

Step 4: If I is an arbitrary interval, let (a.,, b,) C I be such that a,= (inf I)+, b. -, (sup 1) -. By the previous step, (3.20) holds in each (an, ba). Formula (3.20) now follows in I0 from the Lebesgue monotone convergence

theorem. If one or both endpoints of I belong to I, we can proceed as in last part of Step I to show that (3.20) holds in 1. Choosing tfi (x) := g (u (x)) in (3.20), where g : R -> 10, ooj is a Borel function, yields JRg(y)N.(y;1)

dy=

J(u(x))It/(x)I dx,

where, we recall, N. I) is the Banach indicatrix of u. In particular, for g = 1, we get the analog of Banach's theorem (see (2.48)), that is,

JR N (y; I) dy J u'(x)I dx.


104

Exercise 3.66. Let K C (a, b) be a compact set and let u : K - R be such that u is differentiable on K and lim

yEK,y-yz

u (x) = u' (x) y-x

u (p)

uniformly for x E K.

Prove that there exists a function v : (a, b) -t R, with v E C1 ((a, b)), such that v = u and v' = u' on K. Hint: On each connected component (at., bk) of (a, b) \ K define v to be a suitable third-order polynomial. We conclude this section by discussing the analog of Theorem 2.31. The following exercise (see Exercise 2.30) shows that the composition of absolutely continuous functions is not absolutely continuous (see however Exercise 3.51).

Exercise 3.67. Let f : R --+ R be defined by

ifz 1,

and let u : [-1, 1] - R be the function U (X)

.-_

r x2 sin i

0

if x # 0, if x = 0.

Prove that f and u are absolutely continuous but their composition f o u is not.

The next result gives necessary and sufficient conditions on f : R - R for f o u to be absolutely continuous for all absolutely continuous functions

u : [a, b] - R.

Theorem 3.68 (Superposition). Let I C R be an interval and let f : R R. Then f o u E ACIoc (I) for all functions u E ACI,v (I) if and only if f is locally Lipschitz. In particular, if f is locally Lipschitz and u E ACI0, (I), then the chain rule (3.14) holds.

Proof. Step 1: Assume that f is locally Lipschitz and let u E ACtoc (I). Fix an interval [a, b]. In particular, Iul is bounded in [a, b] by some constant e, and so there exists L > 0 such that (3.23)

If (zi) - f (z2)1:5 L Izl - z2I

for all zl, z2 G [-I, UJ. We claim that f o t is absolutely continuous in [a, b]. Indeed, since u E AC ([a, b] ), for every e > 0 there exists b > 0 such that

> Iu (bk) - u (ak) I < k


105

for every finite number of nonoverlapping intervals (ak, bk) C [a, b], with

E(bk - ak) < S. k

Hence, by (3.23),

1: I(fou)(bk)-(f ou)(ak)I 0. We claim that f is bounded in [-r, r]. Indeed, for every zo E [-r, r] consider the function

I zo+x-e ifxE[a,b], u (x) :=

zo - 2n

if x < a,

zo+2

ifx>b.

Since u E ACID, (I), by hypothesis (f o u) E AC ([a, b]). In particular, it is bounded in (a, b]. Thus, there exists a constant M,, = M,,, (a, b) > 0 such

that

If( zo+x-a2b)ICMto for all x E [a, b], which implies that if (z)I < MV6

for all z E (zo - j, zo + 2a) . A compactness argument shows that f is as

bounded in [-r, r] by some constant M,. > 0.

Next we claim that f is Lipschitz in (-r, r]. Indeed, assume by contradiction that this is not the case. Then we may find two sequences {a,:} , {tn} C [-r, r] such that s # t,, and If (sn) - f N) I > 2(n 2 + n) Isn - tnI for all n E N. Since {sn} is bounded, we may extract a subsequence (not relabeled) such that sn -+ s.. Take a further subsequence (not relabeled) such that (3.24) continues to hold and (3.24)

(3.25)

Ian - soo I
- If (sn) - f (tn) I > 2 (n2 + n) Is. - tnI.

(3.26)

Hence,

02

2AI,.

Isn - tnI

(n2

+ n)

and set m, := max I j E No : j < en}. Since en > 2, we have

In<mn
2m. If (sn) - f (t, )I > 2en (n2 + n) Isn - tnI = 41vl',-,

and so for every n E N by Remark 2.7 we obtain that n

Val k (f o u)

Var[a,bJ (f o u) >

4M,.n - 00

k=1

as n -p oo. Hence, we have obtained a contradiction.

0

Remark 3.69. Note that in the necessity part of the theorem we have actually proved a much stronger result, namely that if f : R - R is such that f o u E BPYOC (I) for all functions u E AC (I) (z ACIOC (I), then f is locally Lipschitz.

Remark 3.70. Note that the previous proof continues to hold if f : Rd + R; namely f o u belongs to ACi0 (I) for all functions u E ACIOC (I; Rd) if and only if f is locally Lipschitz.

3.3. Singular Functions In this section we prove that every function of bounded pointwise variation may be decomposed into the sum of an absolutely continuous function and a singular function.

Definition 3.71. Let I C R be an interval. A nonconstant function u : I -> R is said to be singular if it is differentiable at,C1-a.e. x E I with u' (x) = 0 for C1-a.e. x E I. The jump function uj of a function u E BPVIOC (I) is an example of a singular function. Another example is the Cantor function or the function given in Theorem 1.47.


108

The following theorem provides a characterization of singular functions.

Theorem 3.72 (Singular functions). Let I C R be an interval and let u : I - R be a nonconstant function such that u' (x) exists (possibly infinite) for,C' -a. e. x E I. Then u is a singular function if and only if there exists a Lebesgue measurable set E C I such that C' (I \ E) = 0 and Ll (u (E)) = 0.

Proof. Assume that u is singular and let E := {x E I : u' (x) = 0}. Then

G'(I\ E) = 0. By Corollary 3.14 we have that G1 (u (E)) = 0. Conversely, assume that there exists a Lebesgue measurable set E C I such that

I' (I \ E) = 0 and £' (u (E)) = 0. Then by Lemma 3.45, u' (x) = 0 for £1-a.e. X E E. Since G1 (I, E) = 0, we have that u' (x) = 0 for ,C1-a.e. As an application of Lemma 3.31 we obtain the standard decomposition of a monotone function into an absolutely continuous monotone function and a singular monotone function.

Theorem 3.73. Let I C ]R be an interval and let u : I -p R be an increasing function. Then u may be decomposed as the sum of three increasing functions

u= uAC+uc+uJ,

(3.30)

where uAC E ACioc (I), uc is continuous and singular, and uj is the jump function of u.

Proof. Define v := u - uj. By Exercises 1.5 and 1.50 we have that v is increasing, continuous, and v' (x) = u' (x) for L1-a.e. x E I. Fix x0 E I and for every x E I define (3.31) uAC (x) :=

x so

v' (t) dt = J x u' (t) dt, uc (x) := v (x) - uAC (x). ..11 xo

Then the decomposition (3.30) holds. Moreover by Lemma 3.31 we have (x) = 0 for 1.'-a.e. X E I. It remains to show that uC is increasing. Let x, y E I, with x < y. By Corollary 1.37, uAC (Y) - uAC (x) = I v v' (t) dt < v (y) - v (x),

and so uc (y) > uc (x) by (3.31)2.

O

The function uC is called the Cantor part of u. Since every function with bounded pointwise variation may be written as a difference of two increasing functions, an analogous result holds for functions of bounded pointwise variation.

3.3. Singular Functions

109

Corollary 3.74. Let I C ][8 be an interval and let u E BPVoc (I). Then u may be decomposed as the sum of three functions in BPVI,, (I), i.e.,

u=uAC+uC+uJ,

(3.32)

where uAC E ACID, (I), uc is continuous and singular, and (3.33)

uJ (x) := E (u+ (y) - u_ (y)) + u (x) - u_ (x) . gE 1, g<s

Moreover for every interval [a, b] C I, (3.34)

Varla,bl u = Varla,bl uAC + Varla,bl uc + Var1a,b) uJ,

where b

(

3.3)

Iu' (x)

Varies bl uAC = Ia

(3.36)

(Iu+ (x) - u (x)I + Iu (x) - u_ (x)I)

Varlatbl uJ = xe(a,b)

+Iu(a)-u+(a)I+Iu(b) -u_(b)I. If, in addition, u E BPV (I), then Var it = Var tAC + Var tic + Var uJ

=J Jul(x)I dx+Varuc+F(Iu+(x)-u(x)I+Iu(x)-u-(x)I), aEI

where u- (inf I) := u (inf I) if inf I E I and u+ (sup I) := u (sup I) if sup I E I.

Proof. The decomposition (3.32) follows either by modifying the proof of the previous theorem or by writing u as a difference of two increasing functions (see Theorem 2.18) and applying the previous theorem to each increasing function. We leave the details as an exercise. By (2.7) for every interval J C I we have Vary u < Vary uAC + Vari uc + Varj uJ.

Thus, to prove (3.34), it remains to show (3.37)

Varlafbl u > Varla,bl UAC + Varla,bl uc + Varla,b) uJ.

We divide the proof of (3.37) into five steps.

Step 1: Assume first that u is continuous, so that uJ = 0. We claim that for every interval [a,,61 C I, (3.38)

Var1Qr,sl it > Varl,,,,,,l UAC + Iuc (Q)

- tic (a)I

To see this, let

E := {xE(a,,3): uc(x)=0}.


110

Fix e > 0 and let d > 0 be as in Definition 3.1 for the absolutely continuous function uAC. Using the definition of differentiability, for every x E E we may find an interval (ax, br) C [a,,31 such that ax and bx are rational numbers, and if ax < xi < x < x2 < bzj then

IuC(x2)-uC(xi)I i=1

cF

i=1

_ i=1

J

Va [,,,,b j uAC-

I (uAC)' I dx = i=1

Using (2.7), we obtain that Var[b,-,,,,] u > Varjb,_,,ail uc - Var]b,_,,,,, UAC

>- Iuc (ai) - uc (b;-i)I - Vary_,,,,) uAC, which, together with (3.41) and (3.42), yields n

(3.43)

Var1«,A1

n+l

Varla;tbsj uAC + E Iuc (ai) - uc (bi-1)

U> i=1

-

icl

n+1

Varlb{_, a4j ''AC-

i-1

I

3.3. Singular Functions

111

By (3.40) we have n+l

a-

(ai-bi-1) =

n (bi - ai) < S. i=1

i=1

Hence (see the proof of Corollary 3.41), n+1

(3.44)

Var[b;-i,0,j UAC < E i=1

On the other hand, by (3.39), n+1

n

Iuc (bi) - uc (ai)I

Iuc (ai) - uc (bi-1) I ? Iuc (3) - uc (a) I i=1

i=1

n

? Iuc (i3) - uc (a)I -

(3.45)

E

3

(bi

- ai)

> Iuc (3) - uc (a)I - e. Combining (3.43), (3.44), and (3.45) and using Remark 2.7 for UAC, we obtain u > Var[Q,31 uAC + Iuc (3) - uc (a)I - 3e.

By letting e -' 0+, we obtain (3.38). Step 2: Fix an interval [a, b] C I and consider a partition P of [a, b], with

a=YO R2 such that u ([0,11) = [0,1]2. Proof. For every n r= N divide the interval [0, 1] into 4'h closed intervals Ik,n, k = 1, ... , 4n, of length aL and the square [0, 1] 2 into 4n closed squares Qk,,,, k = 1, ... , 4n, of side length 1. Construct a bijective correspondence between the 4'z intervals IA-,,n and the 41 squares Qk,,, in such a way that (see Figure 1)

(i) to two adjacent intervals there correspond adjacent squares,

4.1. Rectifiable Curves and Arclength

117

Figure 1. Peano's curve.

(ii) to the four intervals of length a contained in some interval IA,,,, there correspond the four squares of side length 2 -TT contained in the square corresponding to Ik,,,. By relabeling the squares, if necessary, we will assume that the square Qk,n corresponds to the interval Ik,n. Let .F

{Ik,,, : n E N, k = 1, ... , 4n} ,

9: = Qk,n: nEN,k=1,. .,4"}. By the axiom of continuity of the reals, if {Jj} C F is any infinite sequence of intervals such that Jj+1 C Jj for all j E N and {R1} C 9 is the corresponding sequence of squares, then there exist unique t E [0, 1] and (x, y) E [0,1]2 such

that

00

00

{t} = n Jj, {(x, y)} = n Rj. j=1

j=1

We set the point t and the point (x, y) in correspondence. We claim that this correspondence defines a continuous function u R2 with all the desired properties. Indeed, a point t E 10, 1J that is [0, 1] not an endpoint of any interval determines uniquely a sequence {Jj} C F to which it belongs and hence a point (x, y) belonging to the corresponding

sequence {Rj } C 9. The same is true for t = 0 and t = 1. A point t that is common to two different intervals Ik,,,n and Ik,,,,, for some m E N is also common to t w o different intervals Ik,n, k = 1, ... , 41, for all n > m. Hence, it belongs to two different sequences {J1}, {J?} C F. Since the squares Rj and R'j, corresponding to Jj and Jj, respectively, are adjacent by property (i), it follows that 00

00

nRj=n ki. j=1

j=1

Thus, to every t E [0, 1J there corresponds a unique (x, y) E [0,1]2 that we denote u (t).

4. Curves

118

Since every (x, y) E [0,1]2 belongs to one, two, three, or four sequences {R, } C 9, there exists one, two, three, or four t E [0, 1J such that u (t) _ (x, y). Hence, u ([0,1]) = [0,1]2.

To prove that u is continuous, write u (t) = (x (t) , y (t)), t E [0,1]. By conditions (i) and (ii) we have that Ix (ti)

- x 001
Rd be a continuous function. Prove that if to E [a, 0], then I f p g (t) dt >_ J l

R

Ig (t) l dt - 2

J'8

Ig (t) - g (to) I dt.

(iii) Using part (ii), prove that if u is of class Cl ([a, b] ; Rd), then 'y is rectifiable and b

L (y) = f Iu' (t) I dt. a

(iv) Prove that if each component of u is absolutely continuous, then fb

L (7) =

(u' (t) I dt. a

Remark 4.10. If f : [a, b] - R is absolutely continuous, then by the previous exercise we have that the curve y with parametric representation u (t) :_ (t, f (t)), t E [a, 6], is rectifiable and

jIi b

L('Y)=

+I f'(t)Iadt.

4. Curves

120

Exercise 4.11. Given a continuous convex function f : [a, b] - R, prove that the curve -y of parametric representation a (t) :_ (t, f (t)), t E [a, b], is rectifiable and that

L(y) < f (a) -min f + f (b) -mi f + b - a. la,

We prove that the length of a curve is lower semicontinuous.

Proposition 4.12. Let ry, yk, k E N, be curves with parametric representations u, uk : I -* R", k E N, respectively, where I C ][8 is an interval. If u (t) = klim Uk (t)

(4.2)

for all t E I, then L (y) < 1 m f L (yk) .

(4.3)

Proof. The proof is the same as that of Proposition 2.38.

0

Remark 4.13. As in Remark 2.39, if u is continuous, then it is enough to assume that (4.2) holds for all t in a dense set E of I. The next exercise shows that the equality sign does not necessarily hold in (4.3).

Exercise 4.14. For t E [0,1], let u (t) :_ (t, 0) and uk (t) := (t, "sin 27rk't), k E N, r > 0. Let y and -yk be the corresponding curves. (i) Estimate L (yk) for all k E N and r > 0. (ii) Prove that if r = 2, then L (y) _ ?lim L (y23) -000

.

(iii) Prove that for r = 1 or r = 2 we have strict inequality in (4.3). In the special case of graphs of absolutely continuous functions, equality in (4.3) implies convergence in L' of {u4}. Indeed, we have the following result.

Theorem 4.15. Let u,'uk

:

[a, b] -> R, k E N, be absolutely continuous

functions such that

u (t) = lim uk (t)

(4.4)

for all t E [a, b] and (4.5)

lim /

k->oo

Q

b

a


Then

b

121

1+IukI2-

kl'n°1°J a

and lim

Zb Iuk-u'I

k-roo

dt=0. 2

1 + luk (t) I2 - 2

1 + lu' (t)12 +

fk (t)

1 + 2 (u' (t) + uk (t)) I

.

1 + y2 is strictly convex, we have that fk (t) > 0 Since the function y unless u' (t) = uk (t). Consider the sequence

u(t)+uk(t),

Vk(t)

2

tE [a,b].

+u'n

By hypothesis, vi. (t) -+ u (t) for every t E [a, b], and so, by Exercise 4.9 and the previous proposition applied to the sequence {vk},

1m fJ1bt/1+

2

(u'+uk)2dt>_

fb 1

V

which, together with (4.5), implies that fb

jb

1+I

limsupJ fkdt= lim k-,oo k->oo

I2

dt

n

-2lim

fb

2

I21

uk))

f

dt 0, we have that fk - 0 in L' ([a, b]). Extract a subsequence { fk1 } of { fk} such that fkj (t) -> 0 for Gl-a.e. t E [a, b]. Let E be the set of t E [a, b] for which A. (t) -> 0.

We claim that ukj (t) -> u' (t) for all t E E. Indeed, assume by contradiction that this is not true. Then there exist to E E and a subsequence

(not relabeled) such that either ukj (to) -' e E R, with e # u' (to), or It kj (to)

I

- oo. In the first case, we have that 2 fkj(to)-

1+Iu'(to)I2+

1+e2-2 1+I2(u'(to)+t)I >0,

which contradicts the fact that fkj (to) -> 0. Thus, assume that Hki (to) oo. Fix y, z e R, and consider the function g (s) :=

1

1Iy + szI2,

s E (0, oo) .

I

4. Curves

122

Since g is convex and differentiable, the function s E (0, oo) H g (s) - 2g ( ) is increasing. To see this, it suffices to differentiate and use the fact that g' is an increasing function. Using this fact with y = u' (to) and z = u' (to) u' (to), we have that for all 0 < s < 1, f k;

(to) =

1 + lu' (to) I2 + 9 (1) - 2g

(2

1+Iul(to)I2+g(s) -2g( .)

>

1+lu'(tp)l2+ -2

1+

I

u'(to)+8(ukf(to)-u,'(t0))12

- u' (to)) 2 R(to)

1 + Jul i

Taking s := Ini (to) - u' (to) I

f

12

.

E (0, 1) for all j sufficiently large, we have

that 1 + lie (to)12 +

fk, (to) ? where

j .-

V+tj ukf(to)_'

(to)

1 V+ Iu' (ta) + k,12 - 2 V1+ I U, (to) +

Since k

relabeled) such that (kJ inequality yields

I

I

= 1, we may find a subsequence (not

with

o=3 m fk,(to)?

2

1. Letting j -r oo in the previous

1+Iu'(to)I2+

-2

2

V1+Ju'(to)+ 1

>0,

where the last inequality follows from the fact that # 0. Thus, we have reached a contradiction even in this case, and so the claim is proved. Next we show that

lb lim

I V- -+

j +00

l+lu'12

at = 0.

Using the fact that 1181 - 18 - rll c Irl for all s, r E R, we have

1+Iuk7I2-

RI

1+luk?I2-

1+lu'l2 < V1+I4,


123

and thus, by the Lebesgue dominated convergence theorem and the fact that uki (t) --# u' (t) for ,C1-a.e. t E [a, b], hm

f

2

1+ u'kp

2

1+lu'I2

+ luk1) =

+

dt dt,

Ia

which, together with (4.5), implies (4.6). Finally, we prove that slim

J

dt =

b Iuki I

b

J

lull dt.

By Fatou's lemma,

1b11 1im i

b

dt >

J

lull dt.

Thus, if (4.7) fails, then there is a subsequence (not relabeled) and n > 0 such that

fiu"i for all j E N. Since fa that (4.9)

dt + 3rl (b - a) < fb

dt I

1 + lu'I2 dt < oo, we may find E = E (rl) > 0 such

f/1+Iui2dt jE. In turn, by (4.8) for all j > jj,

flull dt+3rd(b-a) < Ja

lukil

rb lull

(Ee f

dt=J

lukjl dt+ J

Vl

dt+rl(b-a)+

e

L4 I dt 2

dt.

4. Curves

124

It follows that

2r7(b-a)< E.

F

2

li4Jl dt

l+lukJl2- l+1u'I2 dt

< JEc

oo, then there exist c > 0 and

S>0such that g(s)>cI al forall]a]>S. Prove that if u, uk : [a, b] --+ R, k E N, are absolutely continuous functions such that u (t) = slim uk (t) for all t E [a, b] and

lim j g (uk (t)) dt =

k--.oo a

Ia g (u' (t)) dt < oo,

then

lim

/'b J

Ig (uk (t))

- g (u (t)) I dt = 0.

(You may use the fact that by (4.4),

lim inf z bg (uk (t)) dt >

Jo

(u' (t)) dt;

see [8] and see also the proof of (13.24) in Chapter 13). Prove that if in part (ii) we also assume that g (s) oo as ]s[ -i oo, then b

lim

k-'oo

fa

u'k-u'Idt=0.


125

Next we introduce the notion of arclength of a rectifiable curve. Given

a rectifiable curve y, let u : I - Rd be a parametric representation, where I C R is an interval. For every t E I, let s (t) denote the length of the curve of parametric representation ulin(..,tl With a slight abuse of notation, we will denote this curve by We observe that the function s : I -* [0, L (-y)] is increasing, although it may not be strictly increasing, in general. We call s a length function for y.

Remark 4.17. Given a rectifiable curve y, let u : I -. Rd be a parametric representation, where I C R is an interval. For every interval [ti, t2] C I, the difference s (t2) - s (tl) is the length of the curve lilt,,t2l and by (4.1) we have d

(4.11)

V1 (t2)

- Vi (t1) < s (t2) - s (t1) < E (Vj (t2) - Y j (ti)) ,

j-1 1ui(t2) -uY(t1)] < Iu(t2) -u(tl)] C S(t2) -s(ti), for all i = 1,...,d, where, Vi (t) := Varin(-oo,t] ui, t E I. (4.12)

In particular, if u is continuous at some to E I, then by Exercise 2.14 and (4.11), the length function a is continuous at to f=- I. Conversely, in view of (4.12), ifs is continuous at to E I, then so is u. Thus, u and s have the same set of discontinuity points. Moreover, by (4.11), s E AC (I) if and only if Vi E AC (I) f o r all i = 1 , . ,dd,, and hence, by Remark 3.40, if and only if

ui E AC (I) for all i= 1,...,d. The next result is the vectorial version of Theorem 3.39.

Theorem 4.18 (Tonelli). Given a rectifiable curve y with parametric representation u : I Rd, where I C R is an interval, then s' (t) = lu' (t) I

(4.13)

for C1-a.e. t E I and

r (4.14)

J

lu' (t) I dt < L (-y)

.

The equality holds in (4.14) if and only if ui E AC (I) for all i = 1, ... , d.

Proof. By (4.1), ui E BPV (I) for all i = 1, ... , d, and so by Corollary 2.23 the derivative u' (t) exists for ,C1-a.e. t E I and is integrable. Similarly, since

s : I - [0, L (y)] is increasing, we have that s' (t) exists for £1-a.e. t r= I and is integrable. Since for all 4, t2 E I, with t1 < t2,

(4.15)

8 (t2)-8 (tl)> Iu(t2)-u(tl)i,

4. Curves

126

it follows that s' (t) > Jul (t) I whenever all the derivatives involved exist at t, i.e., for L1-a.e. t E I. Let

E := It E I : s' (t) and Iu1(t)I exist and s' (t) > Iu te(tI. For every k E N, let Ek be the set of all points t E E such that s (t2) - s (ti) > Iu (t2) - u (t1) I + 1 (4.16)

t2 - tl

t2

tl

k

for all intervals [tl, t2] such that t E [tl, t2] and 0 < t2 -11
0 and consider a partition P :_ {to, ... , t,a} of I such that (4.17)

Iu (ti)

- u (ti-i)I > L (y) - E.

i=1

By adding more points to the partition, without loss of generality, we may assume that I ti - ti-1 I < for all i = 1, ... , n. Fix i E (1,.. . , n-}. We distinguish two cases. If the interval [ti_ 1, ti] contains points of Ek, then multiplying (4.16) by ti - ti_1 yields (ti) - (ti-1) (ti) - u + ti - ti-1 a 8 Iu (4-01 k while if [ti-1, ti] does not intersect Ek, then

8(t1) -a(t4-1)

Iu(ti) -u(ti-1)I

by (4.15). Summing these two inequalities and using (4.17) gives n

L(7) _

n

1

Iu(ti) -u(ti-1)I+ it,(E,)

(a (ti) -a(ti-1)) i=1

i=1

L(y)-E+Lo(Ek), ke. Given the arbitrariness of e > 0, we which implies that Lo (Ek) conclude that L1 (Ek) = 0. This proves (4.13). It now follows from (4.13) and Corollary 1.37 applied to a that

Iu'(t)I dt= f 1(t) dt <sups - infs= L (y).

1, Assume next that

p

J I u' (t) I dt = L (-y) I


127

Then

J s' (t) di = s jp 8 - inf s, I and so by Theorem 3.39, s belongs to AC (I). In turn, by Remark 4.17, u, E AC (I) for all i = 1, ... , d. Conversely, assume that us E AC (I) for all i = 1, ... , d. Then by Theorem 3.30, for every partition P = {t0,... , tn} of I we have

t

n

n

Iu (ta) - u (t _i)I = j=1

9=1

f

Rd, as in Corollary 3.74, we may decompose each component ui, i = 1,... , d, into an absolutely continuous part, a Cantor part, and a jump part. Thus, we obtain u = UAC + ttc + uJ, where for a fixed to E I and for all t E I, t

UAC (t) :=

f u' (8) ds, to

(u+ (g) - u_ (y)) + u (t) - u_ (t) ,

u, (t) := y Rd, where I C R is an interval, then

L(7) = Lj(7AC) + L (7c) + L(3)

= J lu'(t)I tEl

4. Curves

128

where u_ (inf I) := u (inf I) if inf I E I and u+ (sup I) := u (sup I) if sup I E 1.

Proof. The proof follows exactly as in the proof of Corollary 3.74.

Remark 4.20. If f E BPV ([a, b] ), then its graph is given by the curve y with parametric representation

u:[a,b]->R2, t --p (t, f (t)) Thus, if we decompose f = fAC + f c + f j as in Corollary 3.74, we have that UAC (t) _ (t, AC (t)) and tic (t) _ (0, fc (t)) for all t E [a, b]. Thus,

jb1

L( Grf)=L(y)+If '(t)I2dt+Varfc (If+ (t) - f MI + If (t) - f_ (t) I) .

+ tEfa,b]

In particular, if f is continuous and increasing and if f' (x) = 0 for G1-a.e. x E [a, b], then f = fc, and so by Proposition 2.10, rb

L(Grf) = J 1 dt + Var fc = b - a + f (b) - f (a). a

Hence, if we consider the Cantor function f : 10, 1] i R defined in Example 1.43, then L (Gr f) = 2.

Definition 4.21. Given a rectifiable curve y, we say that y is parametrized

by arclength if it admits a parametrization v : J -> Rd, where J is an interval of endpoints 0 and L (-y), such that v is Lipschitz continuous, for every r E J the length of v : J fl [0,,r] -> Rd is exactly r, and Iv' (T) I = 1 for

G'-a.e. r E J. Note that, if y is parametrized by arclength, then by Tonelli's theorem, Iv'(T)I

Given a rectifiable curve y with parametric representation u : I - Rd, where I C R is an interval, by Remark 4.17 the length function s : I -+ [0, L (y)] corresponding to u is increasing. Moreover, if y (or equivalently u) is continuous, then so is a (again by Remark 4.17). It follows from the intermediate value theorem that s (l) is an interval. However, s may not be injective. Indeed, s is constant on any interval [tl, t2] if and only if u is constant in the same interval. Hence, if we also assume that every point of the curve is simple, so that y is a continuous rectifiable simple arc, then


129

s : I -, s (I) becomes a continuous bijection (see Theorem 1.7). In turn, the function v : s (I) - Rd, defined by v (T) := u (s 1 (T)) ,

(4.18)

T E s (I) ,

is equivalent to u. In this case we say that y is parametrized by arclength.

Theorem 4.22 (Arclength, I). If y is a continuous rectifiable simple arc, then y can be parametrized by arclength.

Proof. It suffices to consider the case L (y) > 0. Let u : I -' Rd, where I C R is an interval, be any continuous parametric representation of y. By the discussion preceding the theorem, the function v defined in (4.18) is a parametric representation of y. Set J := s (I). Then, for all T, T-' E J, with ,r 0, then there exists an interval (am, bra) such that a,, = s (to) and b,n = s+ (to), and we define v(T)

'u+(to)-u(to) (T-a.)+u(to), T E (ara,bra) bra - ara

Thus, we have extended v to (0, L (-y)] in such a way that v is Lipschitz continuous with Lipschitz constant less than or equal to 1 and (v'(T)( = 1 for 41-a.e. T E (0, L (y)]. The curve y with parametric representation v : [0, L (y)] - Rd has length L (y), and its range contains the range of y.

4.2. Frechet Curves To extend the previous results to rectifiable curves that are not simple arcs, we need to modify the definition of a curve by giving a more general concept of equivalence.

4.2. Frechet Curves

131

Definition 4.26. Given two intervals I, J C R and two functions u : I - Rd and v : J -> Rd, we say that they are &F chet equivalent if for every e > 0 there exists a continuous, bijective function ¢ : I -* J such that

Iu(t) - v(0(t))I < e for all t E I. We write u N V.

Exercise 4.27. Prove that N is an equivalence relation. F

A Frdchet curve y is an equivalence class (with respect to ) of parametric representations. We say that y is continuous (respectively, rectifiable), if it has a representative u that is continuous (respectively, of bounded pointwise variation).

Exercise 4.28. Prove that if a Frechet curve y has a continuous representative, then all its representative are continuous.

The next exercise shows that two functions u : I - Rd and v : J -+ R' may be Frechet equivalent but not (Lebesgue) equivalent. Exercise 4.29. For t, T E 10, 1] let u (t) :_ (t, 0) and /(37-,0) if 0 < T < I, if. CTCg V(T):= 12,0) (1+ (T-1),0) if 'S Rd such that u is continuous, u (a) = z1, u (b) = z2, and u ([a, b]) C E. We are interested in finding a curve of minimal length, that is, a solution of the problem

inf {L (y) : y E TI. Any solution (if it exists) of the previous minimization problem is called a (4.25)

geodesic joining zl, z2.

4. Curves

134

Exercise 4.36. Prove that the value of the infimum in (4.25) does not depend on the choice of the interval [a, b].

Theorem 4.37 (Existence of geodesics). Let C C Rd be a closed set, let zl, z2 E C, and let F be the family of &echet curves y that admit a parametric representation u : 10, 1] -p Rd such that u is continuous, u (0) = zl, u (1) = z2, and u ([0,1]) C C. If the family F is nonempty, then problem (4.25) admits a solution.

Proof. Let L := inf {L (y) : -y E F).

If L = oo, then any -y E F will do, while if L = 0, then z1 = z2 and the problem becomes trivial. Thus, assume that 0 < L < oo and let {'yn} CT be a family of Frechet curves such that

lim L (yn) = L. n-co By Theorem 4.31, for each n E N there is a representative vn : [0, L (yn)] Rd of yn such that vn is Lipschitz continuous and Iv,,,(T)l = 1 for L1-a.e. T E [0,L(yn)]. Define wn (s)

vn (sL (yn))

,

s E [0,1] .

Then w : 10,1] -, Rd is another representative of yn with Iw; (s)I = L for Ll-a.e. s E [0,1]. In particular, w,, is still Lipschitz continuous. Since wn (0) = zl and Var wn = L (yn) C L + 1 for all n sufficiently large, we can apply the Helly selection theorem (one component at a time) to find a function w : [0,1] --+ Rd and a subsequence (not relabeled) such that wn (a) w (a) as n -+ oo for all a E [0, 1]. Using the facts that

- W. (82)1:5 L (yn) I 81 -821 for all 81, 82 E [0, 1] and that L (yn) - L, letting n - cc, we obtain that w is Lipschitz continuous with Lipschitz constant at most L. Moreover, I wn (81)

w (0) = z1, w (1) = z2i and w ([0,1]) C C, since C is closed. Thus, the curve

y parametrized by w belongs to F, and so L (-y) > L. On the other hand, by Proposition 4.12,

L (y) < lim inf L (yn) = lim L (yn) = L, n-poo

noo

and so L (y) = L and the proof is concluded.

4.3. Curves and Hausdorff Measure In this section we study the relation between the length of a curve and the Hausdorff measure N1 and then we characterize continua.

4.3. Curves and Hausdorff Measure

135

In what follows, given two points z1, z2 E Rd, we denote by z zl segment joining them, that is,

the

z1

== {tz1 + (1 - t) Z2 : t E [0, 1]} .

Theorem 4.38. If y is a continuous simple are -y with range F C Rd, then

W' (r) = L ('y) . In particular, y is rectifiable if and only if its range has finite 1-l1 measure.

Lemma 4.39. Let u : [a, b] -> Rd be a continuous function. Then Iu (b) - u (a)I < ii' (u ([a, b])) < Var(Q,bl u.

(4.26)

Proof. Let proj : Rd -.. Rd be the orthogonal projection from line through u (a) and u (b). Then

R.d

onto the

(proj z1 - proj z2I

Rd

be as in Theorem 4.22. Prove that if E is a Lebesgue

measurable set of J, then v (E) is an fl1-measurable subset of 11d and

N,(v(E)) _'c1(E). Deduce that a simple arc admits a tangent line at 7d1-a.e. point of the range of y. If the continuous curve y is not a simple are, then the second part of the previous proof continues to hold so that

fl1(I') < L (y)

(4.27)

,

but the opposite inequality is no longer true. Indeed, it is enough to consider the curve y given in Example 4.4. The natural extension in this case is the following generalization of Banach's theorem (see Theorem 2.47), due to Federer [57]. Theorem 4.42 (Federer). Given a continuous curare -y with parYametric rep-

resentation u : I - Rd, where I C R is an interval, then Nu function and

I) is a Borel

Nu (y; I) d7-l' (y) = Var u = L (y)

(4.28)

In particular, y is rectifiable if and only if N,,

I) is Lebesgue integr'able

with respect to 7d1.

Proof. The proof is very similar to the one of Theorem 2.47 and we only indicate the main changes. In Step 1 define .F, as before. By Theorem 2.46, with µ given by 7{1, we have that

J N. (y; [a,b]) d7l1(y) = "lim°° d

it 1 (U (j)) JEFr,

By (4.26),

Iu (sup J) - u (inf J) 1 < E 7{1 (u (J)) G E Var j u = Var[a,bj u, JEF.,

JEF.,

JE.F..


137

where in the last equality we have used Remark 2.7. Letting n oo and using Lemma 2.48 (which continues to hold with no changes for continuous functions u : [a, b] -> Rd) gives the desired result in the case I = [a, b]. Step 2 of the proof is the same as that for Theorem 2.47 and we omit it. Since for a simple arc -f we have that N., (y; I) = 1 for all y in the range of it, Theorem 4.38 is a special case of the previous theorem. Next we study compact connected sets with finite V measure.

Definition 4.43. A set E C Rd is (i) a continuum if it is compact and connected, (ii) pathwise connected if for all zi, z2 E E there exists a continuous curve joining zl and z2 and with range contained in E.

Exercise 4.44. Let E C Rd. (i) Prove that if E is pathwise connected, then E is connected. (ii) Prove that the set E = El U E2 of R2, where

El := {(0,x2) : -1 < x2 < 11,

E2 := { X1, sin x ) : xl > 0

\

1///

JJJ

is connected but not pathwise connected. (iii) Assume that E is open and connected, fix zo r= E, and consider the sets

A :_ {z E E : there exists a continuous curve joining z and zo and with range contained in E}

and B := E \ A. Prove that A and B are open and conclude that E is pathwise connected.

Exercise 4.45. Let K C Rd be a continuum and let z, w E K. Prove that for every e > 0 there exists a chain of points (yo,.. ., C K, with yo := z and y,,:= w, such that Iy, - yz_1I < E for all i = 1,... , n.

Theorem 4.46. Let K C Rd be a continuum with J(1 (K) < oo. Then K is pathwise connected.

Lemma 4.47. Let K C Rd be a continuum and let zo, wo E K. Then W1 (K n B (zo, Izo

- woI)) ? Izo - wol .

Proof. Let r := Izo - wo I. Consider the function f : Rd -> (0, oo) defined by f (y) := Izo - yI, y E Rd. Then f is Lipschitz continuous with Lipschitz constant 1. We claim that the set f (K fl B (zo, r)) contains the interval

4. Curves

138

Indeed, if not, then there would exist 0 < p < r such that p 0 f (K fl B (zo, r)) . But then we could write K as the union of the two [0, r].

nonempty disjoint closed sets K fl B (zo, p) and K \ B (zo, p), which would contradict the fact that K is connected. Hence the claim holds. Since f is Lipschitz with Lip f = 1, by Proposition C.40 in Appendix C and what we just proved,

l1 (KflB(zo,r)) 2: 7 1 (f (KriB(zo,r))) >?J1([O,rJ)=r. 0 Proof of Theorem 4.46. Fix two distinct points z, w E K. By the previous exercise for every 0 < e < [z - wI there exists a chain of points {yo, ... , C K with yo := z and y := w such that [yi - Vi-11 < s for all i = 1, ... , n. By deleting some of the points, if necessary, we may assume that Iii - y1I > e for all Ii - jI > 2. This implies that no point of Rd lies in more than two balls B (yi, ), i = 1'...,n, and so, by the previous lemma

2f1(K)>E7dKf1B

yip2

>n2.

i=1

Let y£ be the polygonal curve obtained by joining yo,..., yy.. Using the fact that Iyi - y3I > s for all [i - j[ > 2, we have that yE is a simple arc, with n

L(76) = 1: Iyi - yi-I[ < nE Rd of -In with Iw' (s)I = L (y,) for f-1-a.e. s E [0, 11 and a function w : [0,1] -. Rd such that, up to a subsequence, wn (8) -> w (a) as n -> oo for all s E [0, 1] and [w (81) - w (82)1:5 L 181 - 82l

for all 81, 82 E (0, L). Moreover, w (0) = z, w (1) = w. Since L (yn) L, it follows that {wn} converges to w uniformly. To conclude the proof, it remains to show that w ([0,1]) C K. Fix 6 > 0 and let y E w (10, 11). Then

there exists a E [0,1) such that w (s) = y. Since wn (s) - w (s) = y as n -, oo, there exists naa r= N such that Iwn (s) - y[ < 2 for all n > ns,a. By taking n > na,a sufficiently large, we have that Fn < . From the construction of -yE we may find an endpoint yi E K of one of the


139

segments of -yr. such that Iw7z (s) - yi, I < en < .2, and so IYi,.

- yl

di, where

d, := max disc zEK

(zSJr2). .9--O

n

do+1 :- maxdist z, U ri zEK

i=0

n

K=Uri i=o

and the proof is completed. If do+1 > 0, let z,,+l E K and E Lli!--O ri be such that Izn+1 - w.+1I = do+,. Using the previous theorem once more, we may find a continuous rectifiable curve i'n+1 joining z,+1 and wn+l, with Rd be a continuous range f n+1 contained in K. Let un+1 : [0,

4. Curves

140

parametric representation of ')'n+1 with u,,+1 (0) = zn+1 W,+1. Let bn+1 := min t E [0, bn+l]

:

and un+1 (+) _

iin+1(t) E U r,

.

==o

Let un+1 be the restriction of tis,,+1 to the interval (0, brs+1J, let yn+1 be the

corresponding curve, and let r,+, be its range. Then the set (4.30)

rn+1 n

Or3)

=o

consists of a single point. Note that by Lemma 4.47,

xl (rn+1) 2 d,1. Thus, we have constructed by induction a countable family of curves

C

K. It remains to show that if 4 > 0 for all n E N, then

nl (K\ Urn

= 0.

Step 2: We claim that 00

K=

(4.31)

Urn.

n=o

Indeed, let z E K and assume by contradiction that d := diet

(zOrn) > 0. n=o

Since 00

00

E dn r. n=*n+1


141

To see this, note that since z E K, by (4.31) there exist a subsequence nk and a sequence of points zk E rnk such that zk -> z. Let k1 E N be so large that jzk - zI < r for all k > k1. Then by (4.32) it follows that nk. > m + 1 for all k > k1. Define nA;

Kk:= Urn. n=O

It follows by (4.30) that Kk is a continuum. Fix 0 < r < r'. Since zk -, z, if k > k1 is sufficiently large, say k > k2, we have that B (z, r) D B (zk, r'), and 00

o0

xl B (y, r) n U r

(4.34)

=

W1

B (z, r) n U I'n

.=m+1

n=0

>711(B(zk,r')nKk) where in the first equality we have used (4.32). Fix k > k2. Then B (z, r) J B (zk, r'), and so by (4.32), the set Kk contains points outside B (zk, r'). Since Kk is a continuum, it follows by Lemma 4.47 that

xl

(K&. n B

(z&., r')) > r',

which, together with (4.34), implies that 00

B(z,r)n U rn >r'. n=n,+1

Letting r' / r proves (4.33). Step 4: Fix m E N. We want to construct a fine cover of closed balls centered at K \ U 0 I'f,. Fix e > 0. By the definition of '1 there exists 6 > 0 such that

'HI (K\Ur) < n , (K\Ura) + e.

(4. 35)

n=0

n=0

Let Tn

Y,

z E K 1 Urn, {zir: B n=0

0 0. Since w is Lipschitz, 1t1 (w([-n, nj)) < oo. Let ' be the continuous rectifiable F rechet curve parametrized by w : [-n, n] -* Rd. In view of Theorem 4.31 there exists a representative v : [0, L (y)] - Rd such that v is Lipschitz continuous, for every r E J the length of v : 10, 'T] - Rd is exactly T, and Iv' (r) I = 1 for G1-a.e. r E [0, L (y)].

By property (ii) we have that v' (T) = 0 for 41-a.e. T E v-1(E). Since Iv' (7-)I = 1 for G1-a.e. T E [0, L (y)], it follows that the set

0:= Jr E v-1(E) : v is differentiable at r and v'(7-) = 01 has Lebesgue measure zero. Hence, 41 (v-1(E)) = 0. Since v is Lipschitz, it follows from Proposition C.40 that

f'(E n w([-n, n])) = xl (v (v-1(E))) 2 it is easy to construct a set of 44-measure zero that contains the range of an absolutely continuous curve (see Example 3.53). Thus, to recover (3.15), stronger restrictions on the set Ef are needed. The following theorem is due to Marcus and Mizel [117.

Theorem 4.54 (Chain rule). Let f : Rd -r R be a locally Lipschitz continuous function and let I C R be an interval. Then f o u E ACloc (I) for every u e ACir (I;Rd) . Moreover, if the set Ef is purely 11-unrectifiable, then d

(4.38)

(f 0 u)' (x)

forG1-a.e. x E I, where (x) = 0.

(u (x)) u; (x)

(u (x)) ui (x) is interpreted to be zero whenever

Proof. The fact that f o U. E ACto, (I) follows as in Step 1 of the proof of Theorem 3.68. By Theorem 3.12, we have that u and f o u are differentiable for G1-a.e. x E I. Thus, it suffices to prove (4.38) for G1-a.e. x E I such that f o u and u are differentiable at x. If at any of these points x E I, u (x) E Rd \ Ef, then f is differentiable at u (x), and so (4.38) follows by the classical chain rule.

Thus, it remains to consider those x E I such that x E u 1(Ef) and f o u and u are differentiable at x. By Theorem 4.52, we have that

u' (x) = 0 for G1-a.e. x e u 1(Ef ).

Fix X E I° such that x E u-1(Ef ), u' (x) = 0, and f o u and u are differentiable at x. To conclude the proof, we need to show that (f o u)' (x) = 0. Since u is bounded in [x - 6, x + S] C I, let LL be the Lipschitz constant of

4. Curves

146

f in the ball containing u ([x - 5, x + 91). Then, for all 0 < Ih1 < 6,

I (f ou)(x+h)-(f ou)(x)I yo}, meets r, with the exception that if one of these points is a vertex vi (and hence only one by our assumption that the x-coordinates of all vertices vi are different), then it is not counted if vi-1 and vi+1 lie to the same side of the vertical through zo. We say that zo is even (respectively, odd) if m (zo) is even (respectively, odd) (see Figure 2). Consider the function

ir :R2\r->{0,1}, defined by (z)

1 0 if m (z) is even, I if m (z) is odd.

We claim that it is continuous. To see this, it suffices to show that for every zo = (xo, yo) E R2 \ r there exists 6 > 0 such that it (z) _ 7r (zo) for all z E R2 \ r with I z - zo I < 6. There are two cases. If xo xi for all i = 1,... n, then by continuity the same property will hold for all z E R2\ r sufficiently close to zo, and so we actually have that m (z) = m (zo) for all z E It21r sufficiently close to zo. If, say, xo = x1, then we need to distinguish a few cases. If v1 lies below zo (so that the upward vertical ray from zo does

4. Curves

148

not meet v1) or (x - xl) (x2 - xl) < 0 (so that v2 and v,, lie on opposite sides of the vertical through zo), then by continuity the same property will hold for all z E R2 \ r sufficiently close to zo, and so we actually have that m (z) = m (zo) for all z r= 1R2 \ P sufficiently close to zo.

Finally, if v1 is not below zo (so that the upward vertical ray from zo meets v1) and (x,t - x1) (x2 - xi) > 0 (so that v2 and v a lie on the same side of the vertical through zo, both to the left of z), then for all z = (x, y) E R2\P sufficiently close to zo one has m (z) = m (zo) if x > x1 and m (z) = m (z8)+2 if x < x1. This proves the claim.

Since it is continuous, it follows that it is constant on each connected component of R2\P. In particular, no continuous curve with range contained in R2 \ r connects an odd point to an even point. Step 2: Next we show that R2 \ r has at least two connected components. Consider an edge, say E1, and choose a point zo = (xo,1lo) E El such that xo # x1, x.,,. This can be done, since E1 is not vertical. Let

r:=diet (zoCiE1). i=2

Note that r > 0, since zo is not one of the vertices of El. Consider the vertical line x = xo- Since E1 is not vertical, we can choose two points zi = (xo, yi)

and z2 = (xo, y2) in (R2 \ r) n B (zo, r), but on opposite sides of El. By construction In (z1) - m (z2)I = 1, and so we have that it (zr) # it (z2). This proves that R2 \ r has at least two connected components. Step 3: To conclude the proof, it remains to show that two even (respectively, odd) points in R2 \ r can be joined by a continuous curve with range contained in R2 \ P. Thus, let zi, x2 E R2 \ P be such that yr (zi) = lr (z2). If the segment joining zi and z2 does not intersect r, then we are done. Thus, assume that the segment Ti-z2' intersects r and let wl and w2 be the first

and last intersections of this segment with r, respectively. By relabeling the edges, if necessary, we may assume that wl E El and that 2x12 E E. for

some 1<m 2, the (possibly empty) set ui 1(Si) is contained in u-1(Si), and thus, again by (4.41), it has diameter less than xF3. In particular, if u-11 (S2) is nonempty, then it is contained in a unique minimal arc A2 shorter than s . We now replace u2 in A2 with a segment joining U1 (a2) and U1 02), where a2, b2 E S' are the endpoints of the arc A2. Let y2 be the new curve and u2 : S1 - R2 its parametric representation. If ui 1 (S2) is empty, we set y2 := yl and u2 Ul. Continuing in this fashion, we construct a continuous closed simple polygonal curve y,z with parametric representation v := u,z : S1 -+ R2.

To conclude the proof, it remains to show (4.39). Let T E S1 be such that u (T) # v (T). Then we may find i = 1, ... , n such that v (T) = ui (T) # UL-1 (T) (uo := u). By construction r belongs to the are Ai whose endpoints are ai, bi E S1. Since, again by construction, ui (ai) = u (ai) and ui (b,) _ u (bi), we have (4.42)

l u (T) - v (T) I = Iu ('r) - u (ai) + ui (ai) - ui ('r) l < I'u (T) - u (ai) I + E2 :5- J u (T) - u (ai) I +

2$

where we have used the fact that ui (ai) and ui (T) belong to the square Si whose diameter is E2. Since Iu (bi) - u (ai)I

E2 < E1,

it follows from (4.41) that

Ir - aiI < Ibi - aiI < S1, and so by (4.40), In (T) - u (ai)I < E, which, together with (4.42), completes the proof.

Lemma 4.59. Given a closed simple continuous polygonal curve y in R2 with parnmetric representation u : S1 -, R2 and with range r, the interior component of R2 \ I' contains an open ball, on whose boundary there are two points u (T1) and u ('r2), rl, T2 E S1, such that ITl - T2I ? V3-.

Proof. Construct an open ball B contained in the interior component of R2 \r touching r in at least two points a (ri) and u (T2), Tl, n E S1, and with the property that the distance ITl - r2I is maximal. Assume by contradiction that In - T2 I < f. Then ri, r2 are the endpoints of an are A C S1 of length

4.4. Jordan's Curve Theorem

151

greater than air. Note that by the maximality of ITi -raI, the boundary of B cannot meet u (A) \ {u (rl) , u (T2) }, since (4.43)

max {ITl -71, I7'2 - TI} > Iri - 721

for every T E A\ {Ti, r2}. Let u u (a,,,) be the vertices of'' in u (A), as met moving along the curve y when the parameter T goes from Ti to T2.

If al _ Tl and a,,,

T2, then a circle touching the segments u (ri) u (ai) and u (o,,, very close to u (Ti) and u (T2), in points u (Ti) and u (ri), will meet r only at those points. Since In' r2I > Irl - r21, we obtain a contradiction.

-

If, say, al 36 Ti and a,n = T2, then a circle touching the segments u (Ti) u (al) very close to u (rl) and passing through u (T2) will give a contradiction by (4.43). Finally, if ai = ri and a. = r2, consider a circle through u (ri) and u (T2) whose center c moves from the center of B into the domain D bounded by the radii through u (Ti) and u (T2) together with u (A) (note that this domain D exist in view of Lemma 4.57). Eventually the ball either meets u (A) at points other than u (71) and u (r2) or becomes tangent to one of the segments u (Ti) u (a,) and u (o,n) u (r2). In both cases we have reached a contradiction by (4.43) and the proof is complete. O Consider a continuous closed simple polygonal curve ry in R2 with range

r and two points zi and z2 belonging to the same connected component D of R2 \ r. For every chord C, contained in D except for its endpoints, D \ C consists of two connected components (this follows from the proof of Lemma

4.57). Let the distance between r and {zi, z2} be at least 1 and assume that for every chord C of length less than 2 and contained in D except for its endpoints, zi and z2 are in the same connected component of D \ C. Then we have

Lemma 4.60. Under the previous assumptions there exists a continuous curve 71 from zi to z2 with range ri such that dist (r, rl) > 1. Proof. We first observe that if zi is any point of R2, connected to zl by a continuous curve with range r, and such that dist (r, 1") > 1, then the pair zi and z2 satisfy the same conditions of zi and z2. Indeed, for every chord C of length less than 2 and contained in D except for its endpoints, since C does not meet r, (since dist (r, v) > 1), we have that the points zi and z1 are in the same connected component of D \ C, and so are zi and z2 by hypothesis.

4. Curves

152

Hence, by first moving zl and then z2, without loss of generality, we may assume that dist (zl, r) = list (z2, r) = 1. Let rl, r2 E S1 be such that lzl - u (Tl)I = 1z2 - u (T2)I = l and consider an open ball of radius 1 and moving center c initially placed in zl. The desired curve yl will be the curve described by the center c as the boundary of the ball B (c, 1) (confined to D U r) rolls along r starting at u (TI) until c falls in z2.

We first claim that B (c, 1) arrives at least at u (T2). If not, then for some c, the ball B (c, 1) and r will have a common chord C = u (al) u (0,2) of length less than 2. Then D \ C consists of two connected components D1 and D2, with, say, the center of the ball c in D1 and u (T2) on the boundary of D2, somewhere on r, strictly between u (ol) and u (02). Since by construction the center of the ball is in the same connected component of zl, by hypothesis zl and z2 are both in D1. Moreover z2 and c are on the same side of the line through the chord C. Consider the ball B (z2,1) and its radius toward u (T2). This radius starts in Dl and then goes to D2 until it meets r at u (r2). Hence, it crosses the chord C. Since B (z2, 1) does not contain either u (al) or u (a2) in its interior (recall that disc (z2, r) = 1), B (z2, 1) must intersect the chord C at two points. But since c and z2 are on the same side of C and the balls B (c, 1) and B (z2, 1) have both radius 1, we get a contradiction. Thus the claim is proved, and so there is c in D such that B (c, 1) touches u (T2)-

Next we claim that c reaches z2. If u (1-2) is not a vertex, then since disc (z2, r) = 1, necessarily c = z2.

If u (T2) is a vertex, then we claim that even in this case c reaches z2. Indeed, the only problem is if B (c, 1) and r have a common chord C of length less than 2, with u (T2) as one endpoint, and pointing into the angle cu (T2) z2. In this case, however, c and z2 are in different connected components of D \ C, contrary to the hypothesis (c and z1 are in the same connected component of D \ C). This completes the proof. We now turn to the proof of the Jordan curve theorem.

Proof of Jordan's curve theorem. Step 1: We claim R2 \r has at least two connected components. Since r is compact, its complement has an unbounded connected component. To prove the existence of a bounded connected component of R2 \ r, consider a ball B (0, ra) that contains r. Let u : S1 - R2 be a parametric representation of -y and apply Lemma 4.58 to construct a sequence {yn} of continuous closed simple polygonal curves

4.4. Jordan's Curve Theorem

153

with parametric representations vn : S1 -+ R2 such that (4.44)

Iu (T) - v,L (T)I
0, we have that the range I`,, of -y,,, is contained in B (0, ro) for all n sufficiently large. By Lemma 4.58 for every n E N the

bounded connected component of R2 \ r,, contains an open ball B (zn, rn) on whose boundary there are two points un (T,3) and u (a,,.), T,,, a E S such that I rn - an I > v f3-. Passing to a subsequence, we may assume that z,, -* z E B(0,ro). Since u is uniformly continuous and injective, we may finds > 0 such

that if

Iu(TI)-u(T2)I <E, for some Tl, n E S1, then ITl - T2I < -/3-. Hence Iu (Tn) - u (an) I > s, and so Iun (T,.) - u (a,,.) I > a for all n sufficiently large by (4.44). In turn, 2rn > z and dist (zn, Pn) > . It follows that for all n large the points z and zn are in the same connected component of R2 \ r,, (and also of R2 \ r).

Assume by contradiction that z is in the unbounded connected component of R2 \ IF. Then there exists a continuous curve with parametric representation v : 10,11 -' Rd such that its range is contained in R2 \ r, v (O) = z, and v(1) E 11:2 \ B (0, ro). Let d := dirt (v ([0,1]) , r) > 0. Then for all n sufficiently large dist (v ([0,1]) , r,,) > 7, which shows that for all n large the points z and z,= are in the unbounded connected component of R2 \ .rn. This contradiction proves the claim.

Step 2: To prove that R2 \ r has at most two connected components, assume by contradiction that there exist more than two, and let zl, z2, and z3 be points from three distinct connected components of R2 \ r. Define co := dist ({zl, z2, z3 } , I') > 0. Construct {yn} as in the previous step. Then dist ({zl, z2, z3 1, r) > f" for all n sufficiently large and by Lemma 4.57 two of the three points have to be in the same connected component Dn of R2 \ r,,. Passing to a subsequence, we may assume that zj and z2 are in D for all n E N. There are now two cases. Assume first that there exist 6 E (0, eo) and infinitely many n such that zl is connected to z2 by a continuous curve with range r,, such that dist (rn, r,) > 6. Then dist (I ;L, r) > for all n sufficiently large, which would imply that zl and z2 are in the same2connected

component of R2 \ r, which is a contradiction. If no such 6 exists, then the hypotheses of Lemma 4.60 (with 1 replaced by 6) must be violated, and so there exist a sequence of chords {Ck } and an increasing sequence {nk} such that z1 and z2 are in different connected components of Ck, u., (Tk) - u,,, (ak) -> 0 as k - oo, where u, (Tk)

154

4. Curves

and

(ak) are the endpoints of Ck. It follows from (4.44) that u (Tk) -

u (ak) - 0 as k - oo, which implies that 1k - ok -> 0 as k - oo, since u is uniformly continuous and injective. Then for infinitely many k, say, zi belongs to the connected components of D,ak \ Ch bounded by Ck and by u,,, (Ak), where Ak is the small are on S' with endpoints Tk and ak. Since Tk - ak -* 0, again by (4.44), diem un.; (Ak) - 0, so that diam ufzk (Ak) < co for all k sufficiently large. In particular, we have - u (ak)I < CO for all k sufficiently large, which contradicts the definition of EO and completes the proof. IZi

Chapter 5

Lebesgue-Stieltjes Measures Prospective Grad Students, I: "Will my teachers take personal interest in my learning, or will I be spending all-nighters working on problem sets made by frustrated faculty who would rather be doing research instead?"

-Jorge Cham, www.phdcomics.com

In the previous chapter we have extended the notion of pointwise variation to vector-valued functions u : I -r Rd, where I C R. The next natural question

is the possibility of extending this notion to functions u : R -p Rd, where now R C R is a rectangle of RN. In the literature there have been several attempts in this direction, but they have never been quite satisfactory. As we will see in the next chapters, a more successful approach is to give a different characterization of functions of bounded pointwise variation, namely as the

class of functions whose "derivative" is a finite measure. We will make this concept more precise in the Chapter 7. In this chapter we describe the correspondence between functions of bounded pointwise variation and Radon measures. We will start with increasing functions. We will show that there is a one-to-one correspondence between increasing, right continuous functions and (positive) Radon measures. In this chapter

I C R is an open interval. We recall that a Radon measure p : B (I) -+ 10, oo] is a Borel measure finite on compact sets contained in I.

5.1. Radon Measures Versus Increasing Functions The next result shows that every (positive) Radon measure gives rise to a right (or left) continuous increasing function. 155

5. Lebesgue-Stieltjes Measures

156

Theorem 5.1. Let I C R be an open interval and let p : ,B (I) -> [0, ooj be a Radon measure. Then for every a E I and y E R, the function uµ : I -> R, defined by (5.1)

uj, (x) := y + j

((a, XD -pA ((x, a])

if X >- a a,

if x
R be incr'+easing, and let pu : B (I) - [0, oo] be the Lebesgue-Stieltjes measure generated by u. Then for all a, b E I, with a < b, µu ([a, b]) = u+ (b) - u_ (a). (5.7) Moreover, if u is bounded, then iau is finite with

µu (I) = sup u - inf u. I I

Proof. To prove (5.7), fix a, b E I, with a < b. Since I is open, there exists an integer no E N such that (a - ) b + 1) C I. Hence, the interval (a

-

,b

is an admissible cover of [a, b] for n

+ m)

p.([a,b])

no. By (5.5) we have

0 such that [a - e, b + e] C Un 1(an, bn). We claim that the set 00

U [u (an) , u {bn)] n=1

covers (u (a - e) , u (b + e)). Indeed, if t E (u (a - e) , u (b + e)), then there exists x E [a - E, b + E] such that t E [u_ (x), u+ (a)]. Since [a - e, b + e] C U'_1 (an, bn), there exists n E N such that x E (an, bn). Hence, u (an) < u_ (x) < t < u+ (x) G1 U [u (an) , u (bn)] n=1

G' ((u(a-e),u(b+e)))

=u(b+e)-u(a-e)_u+(b)-u_(a). Taking the supremum over all admissible sequences {(an, bn)} and using (5.5), we obtain that i ([a, b]) > u+ (b) - u_ (a). Thus, (5.7) holds.

5.1. Radon Measures Versus Increasing Functions

159

Finally, if u is bounded, then letting a , inf I and b / sup I in (5.7) and using Proposition B.9(i) gives (5.8).

Remark 5.6. Note that under the hypotheses of the previous lemma, taking a = b E I in (5.7) gives (5.9)

A. ({a}) = u+ (a) - u- (a) . Hence, also by (5.7), if a, b E I, with a < b, then yu ((a, bJ) = FLu. ([a, b)) - i`, ({a}) = u+ (b) - u+ (a) , Itu ((a, b)) = pu. ((a, b)) - µu ({b}) = u- (b) - u+ (a).

We are now ready to prove Theorem 5.3.

Proof of Theorem 5.3. Let Pu be the restriction to B (I) of the outer measure u* defined in (5.5). By the preceding discussions, p is a Radon measure and by Remark 5.6, (5.3) holds. To prove (5.4), it is enough to let a \ inf I and b / sup I in (5.3) and to use Proposition B.9(i). Uniqueness follows from Corollary B.16. This concludes the proof.

Remark 5.7. (i) If I c R is an open interval and u, v : I -* R are two increasing functions, then by the uniqueness established in the previous theorem, Au+v = 11u + Pt,

(ii) If the interval I is not open, then it contains one of its endpoints, say a := inf I. In this case, if u : I -> R is increasing, then we can extend u to (-co, a) by setting u (x) := u (a) for all x < a. Thus, we can construct the Lebesgue-Stieltjes measure of the extension of u defined on the Borel a-algebra of some open interval containing

I and then consider its restriction to B (I). (iii) If it : B (I) --+ [0, oo] is a Radon measure and uµ : I -y l is the function defined by (5.1), then by the uniqueness proved in the previous theorem, it follows that p is the Lebesgue-Stieltjes measure generated by u1, that is, Au,, = p,. Exercise 5.8. Construct the Lebesgue-Stieltjes outer measures corresponding to ul (x)

- ` (.xJ l0

forx > 0, forx < 0,

where LxJ is the integer part of x. The following is an important property of icu.

forx>0, for x < 0,


160

Proposition 5.9. Let I C R be an open interval and let u : I --+ R be an increasing function. Then .Co (u (E)) < A* (E)

for every set E C I. Moreover, if E C I is such that u is continuous at all points of E, then Co (u (E)) = µu (E)

(5.12)

Proof. Step 1: Fix e > 0 and let an, bn E I, a,,. < bn, be such that 00

E C U (a1,f b7,) and 00

(u (bn)

- u (an)) < p, (E) + e.

n=1

Then the set U°=1 [u (an.) , u (bn)J covers u (E), and so 00

00

.C (u (E)) < ,C1 U [u (an) , u (bn)] < E C' ([u (an) , u (bn)J) (n=1

n=1

00

(u (b,,,) -u(an)) a,

-(i-v)((x,a]) ifx L for all x E E (respectively, u' (x) < L). Then au (E) > LIC1(E) (respectively, au (E) < LL1 (E)). In particular, if there exists u' on a Borel set E C I such that either GL (E) = 0 or u' (x) = 0 for all x E E, then 1,\.l (E) = 0. Proof. The proof is very similar to the one of Lemma 3.13 and we indicate only the main changes.

Step 1: Assume that E C (a,,(3) CC I. Fix e > 0 and for each n E N let En be the set of points x E E such that (5.32)

u (b) - u (a) > (L - E) (b - a)

for all intervals (a, b) C I such that x E (a, b) and 0 < b - a < n. Note that E. C En+1. Reasoning as in the proof of Lemma 3.13, we have that 00

E = U E.n. n=1

Using (5.5) for the function V defined in (2.2), we may find an open set U. such that Un D En, (5.33)

1a`v (Un) < t4, (En) + E.

By the outer regularity of Go, taking U,, smaller if necessary, we may also assume that (5.34)

C' (Un) - (L - e) Go (U') .

(5.36)

On the other hand, by (5.35), (5.33), and Proposition B.105, we get

pv (U.) - LG1(E). Step 2: To remove the additional assumption that E C (a"13) CC I, we apply the previous result to E f1 (an, b,J, where a,,, b,b E I, an < b,,, and

then let a,, \ inf I, bn / sup I and use Proposition B.9(i) and Theorem B.72.

Step 3: To prove the final part of the statement, assume that u is differentiable on a Borel set E C I with C1 (E) = 0. Let F C E be measurable. Since

F= U Fk, Fk:= {xEF: ko) Urdx+A,,(Io)- J

In(u' R2 such that IW' (r)I = 1 for L'-a.e. T E J. In particular, the set H of points T E J such that either one of the derivatives X' (T) or Y' (T) does not exist in R or both exist and vanish has Lebesgue measure zero. Since X is strictly increasing, if T E a (I) \ H, then for all h :/ 0 sufficiently small we may write

u(X(T+h))-u(X(T))X(T+h)-X(T)

Y(T+h)-Y(T)

X (T + h) -X(T)

h

h

or, equivalently,

Y(T+h)-Y(T)

h

u(X(T+h))-u(X(T))

h

X (T + h) - X (T)

X (T + h) - X (T)

If u is continuous at X (T), then, since X' (T) and Y' (T) exist, at most one of them does vanish, u is continuous at X (T), and X is strictly increasing (and continuous at r), letting h -+ 0, it follows that at the point x = X (T) the derivative u' (x) exists and Y' (T) /X' (T) if X' (T)

u'(x)

0,

00

if X'(T)=0and Y'(T)>0,

-co

if X' (T) = 0 and Y' (T) < 0.

Therefore, Inoder c X (s (I) fl H), or, equivalently, 8 (Inodr) C s (I) fl H, and hence C1 (8 (Inoder)) = 0.

It follows by (4.11) with i = 2 that for all a, b E I, with a < b,

V (a) - V (b) < a (b) - a (a) , and so, by (5.5),

AV 0 we may find an open set A such that u (E) C A and JA

N.(y;I) dy<e.

Let {A,} be the family of all connected components of the open set I f1 U-1 (A). Then

1: Nu (y; A.) -5 N. (y; I) n

and

L1 V (UAn))

>G'(V(An))=E JNii(;An)dy =

f

R n

N,, (y; An) dy < f N. (y; I) dy < e. A

Since E C Un An, it follows that

'C' (V (E)) < e. Given the arbitrariness of e, we conclude that £' (V (E)) = 0. In turn, by Proposition 5.9, we have that uv (E) = 0.

Lemma 5.35. Let I C R be an interval and let u : I -> R. Assume that u' (x) exists in (0, oo] (respectively, in [-oo, 0)) for all x in a set E C I (not necessarily measurable), with £1(u (E)) = 0. Then £1 (E) = 0.

Proof. By Lemma 3.45, we have that u' (x) = 0 for C'-a.e. x E E. Hence,

L,(E)=0. We are now ready for the proof of Theorem 5.33.

Proof of Theorem 5.33. Step 1: Assume that there exist two Borel sets F. and F such that (5.44) holds. By Lemma 5.34, (5.46)

p vu (Fu) = p v.

0-

For simplicity of notation, we abbreviate sets of the type

{x E I : there exists u` (x) E (0,00]}


178

by {u' > 0}. We note that by (5.44),

({u'>o}\ F,)n({v' 0} \ Fu) U ({v' > IAu+uI (({u < 0} \ Fu) U ({v' < 0} \ F.))

=.Z+ZZ. By Remark 5.14 and (5.47), we may rewrite.T as

Z= au+yI(({u >0,v'>0}\F,,)U(lie >0,v'>0}1F.,)) +IAu+Az,I({u'>0,v' 0}\F,) +IA.,, I1 + 12 + 13

By Remark 5.14 we have that

=Au

A.({u'>0,v'?0})+xVQt >-0,v'>0}}, where we have used (5.46) and Theorem 5.13. On the other hand, since {u' > 0, v' < 0} \ Fu C F,, by (5.44), by (5.46)2 we have that Au (F) = 0 for every Borel set F contained in {u' > 0, v' < 0} \ F.u, and so 12

IauI({u'>0, /Xu({u'>0,v' -au ({u' < o}) - A. ({v' < o}) ,

5.3. Decomposition of Measures

179

and so we have proved that

Var(u+v)>Au({u'>0})-Au({u'o})-A,,({v' 0} n {v' < 0} n I-. By Lemma 5.24, for every Borel set E C FO we have that Au (E) > 0, and so we can define the (positive) measure p : B (I) -+ [0, oo) as

FE(E) := Au(EnFo),

E E 13(I).

Since u is continuous, it follows from (5.9) that p ({x}) = 0. By Theorem 5.1 there exists an increasing continuous function to : I R such that p, = p. By Lemma 5.24, for every E E 13 (I) we have that

=0 ifECFo, >0 ifEC{u'>0}, :50 ifEC{u'0},

where for the second inequality we have used the fact that if E C {v < 0}, then

(A,,,+p,)(E) = (Av+Au)(EnFo)+Aq,(E\Fo) = Au+, (EnFo)+A4,(E\Fo) 0}\Fo)

-(Au -A.) ({u' < 0}) =AU

(FO)

({u'>0}\F0)-Au

Var (u + v) - 2pu, (Fo),

where we have used (5.49), Theorem 5.30, and the fact that p,o is concentrated in Fo C Jul > 0} n {v' < 0}. This implies that ,a,, (Fo) = Au (Fo) = 0. Given the arbitrariness of the Borel set Fo in {u' > 01 n {v' < 0} n r-, it follows that

IA,I (Jul > o} n {v' < 0} n I-) = 0.

(5.50)

Similarly, we can prove that

IAv1({u'>0}nIV 0} n {v' < 0} n I-) U Jul does not exist),

52:= ({u'>0}n{v' 0 for 41-a.e. X E I.

5.4. Integration by Parts and Change of Variables Next we prove a formula of integration by parts for Lebesgue-Stieltjes integrals.

Proposition 5.39. Let p : B ((a, b)) -+ (0, oo) and v : B ((a, b)) -> [0, oo) be two finite Bored measures, and for x E (a, b) define

f (x) :=

1

1

(p ((a, x]) + p ((a, x)))

,

g (x)

(v ((a, x]) + v ((a, x))) . 2

Then

km f dv+kgdp= FA((a,b))v((a,b)). Proof. Since the functions f and g are increasing, nonnegative, and bounded, they are Borel measurable and Lebesgue integrable. Let E := { (x, y) E (a, b) x (a, b) : x >_ y), and for every x, y E (a, b) let

E.

{y E (a, b) : (x, y) E E}

,

Ey{xE(a,b): (x,y)EE}. By Tonelli's theorem,

f ,a(Ey) dv(y) = a,b)

v(Ex) dp(x), (a,b)

or, equivalently, (5.52)

J(a,b)

A([y, b)) dv (y) =

J(a,b)

v ((a, xJ) dA(x) .


182

Since (a, b) = (a, y) U [y, b) for y E (a, b), the left-hand side of the previous identity can be written as

f

{a,b}

f

t ([y, b)) dv (y) =

{a,b}

p ((a, b)) dv (y) -

= !L ((a, b)) v ((a, b)) -

f

(a,b)

f

p ((a, y)) dv (y)

IL ((a, y)) dv (y)

(a,b)

.

Combining this identity with (5.52), we have 1(a,b)

f

y ((a, y)) dv (y) +

ab) v ((a, x]) dp (x) = p ((a, b)) v ((a, b)) .

Interchanging the roles of µ and v, we can also write

f

a,b)

A ((a, y]) dr. (y) +

J(a,b}

v ((a, x)) 4 (x) = p ((a, b)) v ((a, b))

To complete the proof, it suffices to add the last two identities and divide by 2.

Corollary 5.40 (Integration by parts). Let I C R be an open interval and let u, v E BPV0, (I). Then for every [a, b] C I,

J

a,b)

u+

+2

u_

da +

v+

k'

b)

+ 2

v

dAu = u_ (b) v_ (b) - v+ (a) u+ (a).

In particular, if there are no points in (a, b) at which both u and v are discontinuous, then

f(a,b) udA,+ f(a,b) vdau=u- (b)v-(b)-v+ (a) u+(a).

P roof. Since u and v are differences of increasing functions, by (5.19) it suffices to assume that u and v are increasing. Then by Proposition 5.39 and (5.3), 1 (u+ (x) - u+ (a) + u- (x) - u+ (a))

f(a,b) 2

+

dµ (x)

f 1 (v+ (x) - v+ (a) + v_ (x) - v+ (a)) dpu (a,b) 2

= (u_ (b) - u+ (a)) (v- (b) - v+ (a)) , or, equivalently,

f

a,b)

1(u++u-) d,t + J 2

(a,b)

12 (v++v-) dµu

= u_ (b) v_ (b) - v+ (a) u+ (a).

Assume next that there are no points in (a, b) at which both u and v are discontinuous. By (5.9), we have that E.,6 ({x}) > 0 if and only if v is

5.4. Integration by Parts and Change of Variables

183

discontinuous at x. By hypothesis, if x E (a, b) is such that u+ (x) > u- (x),

then µ ({x}) = 0, and so using the fact that the set S of discontinuity points of u in (a, b) is countable, we have that p., (S) = 0. Hence,

k

1 (u+ + u-)

(a ,b) 2

1 (u+ + u-) dpv

km\s 2

=f

J (a,b)\S

and, similarly,

11(v+ + v-) dµu = (a,b) 2

udµv=J

udl4'v, (a,b)

J (a,b)

v dµ.,

which completes the proof.

Corollary 5.41. Let I C R be an open interval and let U E BPVo, (I). Then for every cp E Cc, (I),

I

ucp' dx = - J cp dAu.

(5.53)

Proof. Let [a, bI C I be such that supp cp C (a, b). Since cp_ (b) = cp+ (a) _ 0, applying the previous corollary, with v = cp, we get

f ud) +J,b) WdAu.=0. (a

a b)

Using the fact that (see Exercise 5.21)

AS(E)=IE cp'dx for all Borel sets E C I, by a standard argument in measure theory (first

taking u to be a characteristic function, then a simple function, etc.) we conclude that r u dA9, _ ( u dx.

4,b)

J(a,b)

Since cp = cp' = 0 outside (a, b), the proof is concluded.

0

The following change of variables formula transforms a Lebesgue-Stieltjes integral into an integral with respect to the Lebesgue measure.

Theorem 5.42. Let I C II8 be an open interval bounded from below, let u : I -p R be a right continuous increasing function, and let f : I ---). [0, oo) be a Borel function. Then for every a, b r= I, with a < b, (5.54)

p J J(a,bJ

/

u(b)

f (x) dµu (x) =

u(a)

f (v (1!)) dy,

where pu is the Lebesgue-Stieltjes measure generated by u and v : J -> R is the inverse of u defined as in Theorem 1.7.


184

Proof. Step 1: We claim that if a, /3 E I, with a < /3, then a < v (g) < /3 if and only if u (a) < y : u (/3). Indeed, if a < v (y), then by the definition of v (see (1.1)) we have that u (a) < y. On the other hand, if v (y) < 0, let s > 0 be so small that /3 + e E I. Again by the definition of v and the fact that u is increasing we have that a (/3 + s) >- y. Lettings -+ 0+ and using the fact that u is right continuous gives u (/3) > y. Thus, if a < v (y) < (3, then u (a) < y a such that u (x) < y. Hence, v (y) > x > a by the definition of v. Finally, if a (/3) > y, then v (y) < /3, again by the definition of v. This shows that if a (a) < y [0, oo] be a Lebesgue measurable function. The distribution function of u is the function Pu : [0, oo) -p [0,. ' (E)], defined by Qu (s) := 'C' QT E E : u (x) > s}),

s > 0.

The function u is said to vanish at infinity if it is Lebesgue measurable and pu (s) < oo for every s > 0. Note that for a function u vanishing at infinity the value o,,(0) could be infinite. For example, if E has infinite measure 187

6. Decreasing Rearrangement

188

and u is everywhere positive, then Pu (0) = oo. Moreover, (6.1)

Lou (s) = 0

for all s >_ esssup u, E

Pu (s) = G1(E)

(6.2)

for all s < essinf u.

Note that if E has infinite measure and u : E - [0, oo] vanishes at infinity, then essinfE U = 0. Some important properties of Lou are summarized in the next proposition.

Proposition 6.1. Let E C R be a Lebesgue measurable set and let u, v, un : E -+ 10, oo], n E N, be Lebesgue measurable functions. Then the following properties hold: (i) The function Pu : [0, oo) -> [0,,Cl (E)] is decreasing and right continuous. (ii) If it vanishes at infinity, then lim Pu (s) = 0

e-poo

and Pu is continuous at s > 0 if and only if

£'({xE E: u(x) = s})=0. (iii) If u (x) < v (x) for ,C1-a.e. x E E, then p,,, :5 at,. In particular, if u (x) = v (x) for G1-a.e. x E E, then Lou = P,,.

(iv) If un (x) / u (x) for G1-a.e. x E E, then Pu / Lou. Proof. In what follows, for s >_ 0 we set

E. := {x E E : u (x) > s} .

(6.3)

(i) If 0 s}=E n=1

and so by Proposition B.9(i) we have that 00

lien Pu (Sn) = liIn

n-+o0

G1

(E,n) = 'Cl

Ean

Since Pu is decreasing,

(Pu)+ (s0) = n Pu which shows that Vu is right continuous.

Pu (s) ,

= Pu (S) -

6.1. Definition and First Properties

189

(ii) Assume that u vanishes at infinity. Since Qu is decreasing, there exists

s

info (s) ,

Q" (s) = 8>0

and thus, to evaluate it, it suffices to consider the sequence s = n Then En ) E,,,+1 and

oo.

00

n1E =O. Since 41 (En) < 00, it follows by Proposition B.9(ii) that

litn Qu(sn)= lim C1(En)=C1((b)=0.

n-,00

Similarly, if s > 0, consider an increasing sequence {sn} such that sn -> S--

Then E8, D E..+, and 00

u(x)>s}. n=1

Since L1

(6.4)

oo, we conclude as before that

(ou)_ (so) _ m p.

,C 1 ({x E E : u (x) >_ 8})

= Lou (s) + L1({xEE

: u (x) = s}).

Hence, Lou is continuous at s if and only if C1 ({x E E : u (x) = s}) = 0. (iii) For every s >_ 0,

{xEE: u(x)>s}C{xEE: v(x)>s}UF, where L1 (F) = 0, and so Qu (s) < ov (s). (iv) In view of part (iii), by modifying each un on a set of measure zero, we can assume that un (x) / u (x) for all x r= E. For every s > 0 and n E N

set En := {x E E : un (x) > s}. Since un < uri+1, we have that E C En+1 for all n E N, and since un (x) - u (x) for all x E E, it follows that 00

UEn={xEE: u(x)>s}. n=1

Hence, again by Proposition B.9(i), (6.5)

Limo ou,, (e) = lim

(En) = C1 ({x E E : u (x) > 8}) = Lou (a).

Exercise 6.2. Give an example of a function u not vanishing at infinity for which o, is not left continuous.


190

Let E C R be a Lebesgue measurable set and let u : E - (0, oo) be a Lebesgue measurable function. Since by the previous proposition the distribution function e. : [0, oo) -> [0, oo] is decreasing, it admits a left inverse u* : 10, oo) - (0, oo) (see Theorem 1.7 and Remark 1.10), called the decreasing rearrangement of u. Precisely, for t > 0 we set u* (t) := inf {s E [0, 00) : gu (a) < t} .

(6.6)

Note that if u vanishes at infinity, then 8u (s) -, 0 as s -> oo by the previous proposition, and so for t > 0 the set {s E [0, oo) : LOU (a) < t}

is nonempty. Thus, if (t) < oo for all t > 0, while u* (0) could be infinite. Observe that if G1 (E) < oo, then it follows from the definition of u*

that

u*(t)=0 for all t >V (E).

(6.7)

In what follows, for every Lebesgue measurable set F C R we define

F* := [0, L1(F)) .

(6.8)

Note that if G1 (F) = oo, then F* = (0, oo). In view of Proposition 6.1 we have the following result.

Proposition 6.3. Let E C R be a Lebesgue measurable set and let u : E (0, ooJ be a Lebesgue measurable function. Then the following properties hold:

(i) The function u* : (0, oo) -> (0, oc] is decreasing and right continuous,

u' (0) = esssup u, E

and

u'(t)>essiinfu forallt 0 if and only if

G1({xEE: u(x)=so})>0 and (tl, t2) C (Au (so) , (Au)- (so)) .

(iii) For all a, t > 0, u* (t) > s if and only if pu (s) > t.


191

(iv) The functions u and u* are equi-measurable; that is, for all s > 0,

£'({tEE*: u*(t)>s})=,£1({xEE: u(x)>s}). In particular, if u vanishes at infinity, then so does u*. (v) cl ({t E E* : u* (t) = 0}) < V ({x E E : u (x) = 0}) with equality holding if and only if either

£' ({x E E : u (x) > 0}) < oo or

G1({xEE: u(x)>0})=0o and £'({xEE: u(x)=0})=0. If£1({xEE: u(x)>0})=oo, then {tEE*: u*(t)=0} is empty.

Proof. (i) If 0 < tl < t2, then 18E[0,oo): Au(s) 0. E

If esssupE u = oo, then there is nothing to prove. Thus, assume that esssupE u < oo. Then pu (esssupE u) = 0 by (6.1). Hence, (6.10) holds by (6.6).

It follows from (6.10) that if esssupE u = 0, then u* = 0, and so (6.6) follows. Thus, assume that esssupE u > 0 and fix any 0 < so < esssupE u.


192

By the definition of essential supremum there exists a Lebesgue measurable

subset F C E, with to :_ ,C1 (F) > 0, such that 'u (x) > so for all x E F. Hence pu (so) > to. In turn, by (6.6), for all 0 < t < to we have that u* (t) > so, which shows that so < (u`)+ (0) < U* (0) < esssup U. E

Letting 80 -, esssupE u, we conclude that (6.9) holds.

On the other hand, if essinfE u > 0, then pu (s) = Gl (E) for all s < essinfE u by (6.1), and so if t < £' (E), then u* (t) > essinfE u by (6.6). (ii) The first two statements follow from Remark 1.10 and the last from Remark 1.10 and Proposition 6.1. (iii) Assume that u* (t) > s for some s, t > 0. Then pu (s) > t by (6.6). Conversely, assume that pu (s) > t. Since Lou is right continuous, we have that ou (r) > t for all r E is, s + d] for some 6 > 0. Hence u* (t) > s by (6.6). (iv) By part (iii) for a >_ 0, we have that u* (t) > s for t > 0 if and only if pu (s) > t, and so

{t > 0 : u* (t) > s} = It > 0 : pu (s) > t} = [0, pu (s)) . Hence,

G1({t>0: u*(t)>s})=Lou(s)_,C1({xEE: u(x)>s}) by the definition of ou. The result now follows from (6.7) and (6.8). (v) If ,C1({x E E : u (x) > 0)) < oo, then by part (iv) and (6.8),

Gl ({x E E : u (x) = 0}) = G1(E) -V ({x E E : u (x) > 0}) = C1(E*) - C' ({s E E* : u* (s) > 0})

=G1({sEE*: u*(a)=0}). On the other hand, if G1 ({x E E : u (x) > 0}) = oo, then by part (iv),

us (t)>0})

C1({xEE: u(x)>0})=oo.

Since u* is decreasing, it follows that u"` (t) > 0 for all t > 0. Hence, the set 0 it E E* : u* (t) = 0} is empty.

Remark 6.4. Note that parts (iv) and (v) of the previous proposition imply,

in particular, that (6.7) can be strengthened as follows. Let E C R be a Lebesgue measurable set and let u : E -+ [0, co] be a Lebesgue measurable function. Recall that Lou (0) = G1({x E E : u (x) > 0}) .

If pu (0) = oo, then u* (t) > 0 for all t > 0, while if Lou (0) < oo, then u* (t) > 0 for all t E [0, ou (0)) and u* (t) = 0 for all t > Lou (0).


193

Exercise 6.5. What happens to part (ii) of the previous proposition if we remove the assumption that u vanishes at infinity? Corollary 6.6. Let E C R be a Lebesgue measurable set and let u E LOO (E) be nonnegative. Then u* belongs to LOO (E*) and IIuIIL-(E)

IIt

Proof. By Proposition 6.3 we have that IIu*IIL-(E*) = u* (0) = IIuIIL-(E) < °O

Exercise 6.7. Let E C R be a Lebesgue measurable set, let u : E -* (0, oo) be a function vanishing at infinity, and let increasing function. Prove that (W a u)* = TI o u*.

1

:

10, oo) -+ [0, oo) be an

Exercise 6.8. Let a > 0 and b E R and consider the function u : R defined by it (x)

(b- ++a) 0

R,

ifx 0. Suppose that u* (t) > 0 for some t > 0 and fix a E 10, u* (t)). By

Proposition 6.3(iii), gu (s) > t. Hence, by Proposition 6.1(iv) there exists k E N such that put (s) > t. Again by Proposition 6.3(iii) we have that u* (t) > s and since u,*, < un+l < u* for all n E N, it follows that urn un (t) > a.

n-+oo

Letting s / u* (t), we conclude that lim u,,*, (t) > u* (t), which, together n--too with (6.11), implies that lim u, (t) = u* (t)

n-oo

for all t > 0.

Remark 6.11. It follows from part (1) of Proposition 6.10 that modifying a nonnegative Lebesgue measurable function on a set of measure zero does not change its decreasing rearrangement. In what follows, we will use this fact without further mention. For the same reason, we can replace the set E with a measurable subset F C E such that G1 (E \ F) = 0. The next exercise shows that the decreasing rearrangement of a simple function is still a simple function.

Exercise 6.12. Let E C R be a Lebesgue measurable set and let u . E [0, oo) be a simple function vanishing at infinity (see Figure 1); that is, k

u = E cjXE;, i=o

> ck > 0, Ei C E are pairwise disjoint Lebesgue measurable sets with 0 < V (Ei) < oo, i = 1, ... , k. For every i = 1, ... , k set where co >

i Fi = U E?. a=o


195

CO Cl

C2

x

E2

Co C1

C2 r2

-t

Figure 1. The graphs of u and u*.

(i) Let ri

£' (F1), i = 1, ... , k. Prove that k

i

i=0

j=0

OU _

C1 (Ei)

X[Cj+I1cf)l

where ck+1 := 0, and that L.

U* =

CiX[ri-,,r& i=o

where r_1 := 0. (ii) Let {po, ... , pk}, {qo,... , qk} C ][t and prove that k

k

pigi = i=0

f

PPQi>

i=0 i

where Pi := pi -Pi+1 for i = 0, ... , k -1, Pk := pk, and Qi = > q1

fori=0,...,k.

2=0


196

(iii) Prove that (6.12)

U=

aiXFf, i=0

where ai := ci - ci+1, i = I,-, k, and that

'u _

aiXFi .

Using the previous exercise, it is possible to give a simple proof of a classical result of Hardy and Littlewood.

Theorem 6.13 (Hardy-Littlewood's inequality). Let E C R be a Lebesgue measurable set and let u, v : E - 10, oo) be Lebesgue measurable functions. Then

L Proof. Step 1: Assume first that u, v : E i [0, oo) are simple functions vanishing at infinity and, using the notation (6.12), write k

I

U = E aiXF,, V = E QjXGj, j=0

i=0

so that by part (iii) of the previous exercise,

k k

us =

I

aiXF' '

V* = E I3jXG f . j=0

i=0

Then k

f

JE

I

f

uv dx =

cri$j J XF{XG, (fix i=0 j=0 k

I

E k

I

_ 1: 1: ai$jLI ( Fi n Gj) > aiQj min {L1(Fi) , L' (0j) i=0 j=0

i= j=0


197

while rk

I

fu*v*dt=aif3,JET XF, XGs dt i--O j=0 k

_ E ai$jLI (F' n G.') i=0 j=0 k

_

ai$jmin{L1 i--O j=0 k

E E ai,3j min {L1 (Fi), L1 (GA) , i=0 j=0

where we have used (6.8) and the fact that min{L1(F,1*),L1 (G3)},

L1

since both sets are half-open intervals of the form [0, a) for some a > 0. Step 2: Now let u, v : E -* [0, oo) be two Lebesgue measurable functions and construct two increasing sequences fu j and {v..,} of nonnegative simple functions vanishing at infinity, below u and v, and converging pointwise to u and v, respectively (see Corollary B.37). By Proposition 6.10 we have that unv; < for all n E N and that {un*vn} converges pointwise to u*v*. By Step 1, for each n E N we have

u vtt dt

L

and the result follows by the Lebesgue monotone convergence theorem. 0

Exercise 6.14. Let E C a be a Lebesgue measurable set and let u, v : E [0, oo) be Lebesgue measurable functions. Prove that for every r > 0,

L u (z') Xtv jE

u* (t) X{vimo}

where f (0),C' ({x re E : u (x) = 0}) is understood to be zero if f (0) = 0, independently of the value of L1 ({x E E : u(x) = 0}). Similarly, (6.17)

j f (u* (t)) dt E f (u* (t)) dt + f (0),C' (It E E* : u* (t) = 0)).

Hence, it follows from the previous step, (6.16), (6.17), and Proposition 6.3(v) that (6.13) holds. Moreover, if f (0) = 0 or f (0) > 0 and

L, (It E E' : u* (t) = 0}) = 'C' ({x E E : u (x) = 0)), then r

f f (u (x)) dx = fE. f (u* (t)) dt.

(6.18)

Thus, the result now follows by ProposProposition 6.3(v) once more.

Remark 6.16. It follows from the previous proof that equality holds in (6.13) if and only if one of the following conditions holds:

(i) f{.>0} f (u (x)) dx = oo. (ii) f{n>o} f (u (x)) dx < oo and f (0) = 0.

f

(iii) 00.

f (0) > 0, and £' ({x E E : u (x) > 0})
o} f (u (x)) dx < oo, f (0) > 0, and (6.14) holds.

Exercise 6.17. Let u(x) = e-'x1o,.) (x) and let f = X{o} Find u* and prove directly that

0 = j f (u* (t)) dt
r2 ? 0 and s1 > g2 > 0, then

IF (I ri-$i1)+T(1r2-821) SW(Ir1-821) +T([r2-811) >s, 2:0, (ii) Prove that ifri > r2 >- ... >-r,, >0, s1 > 82 > nEN, and if

f :{1,...,n}-+{1,...,n}, g:{1,...,n}-+{1,...,n} are two bijections, then n

n

(Iri - sil)
[0, oo) vanishing at infinity.

6.2. Absolute Continuity of u* In this section we prove that if u is absolutely continuous (respectively, singular), then so is u*. We first show that if u has the Lusin (N) property, then so does u*. Indeed, we prove that there exists a measure-preserving transformation connecting u with u*. Definition 6.20. Let E C RN and F C [-oo, oo] be two Lebesgue measurable sets. A Lebesgue measurable function V : E -* F is called measurepreserving if LN

(W-1 (G)) = G1 (G)

for every Lebesgue measurable set G C F.

Exercise 6.21. Let E C R be a Lebesgue measurable set with L' (E) < 00 and let a > 0. Prove that the function yo : E -p [a, a + C1(E)] , defined by W(x) := a + C1(E n (-oo, x])

,

x E E,

is measure-preserving.

Exercise 6.22. Given a sequence of measure-preserving functions cp,, E7, -* Fn, where {En}, {Fn} C R are two sequences of pairwise disjoint Lebesgue measurable sets, prove that the function 00

00

W : U En _4 U Fn n=1

n=1

defined by W (x) := V. (X)

if x E E,,,

is measure-preserving. Prove that the result continues to hold if the sets Fn are only assumed to be pairwise disjoint up to sets of Lebesgue measure zero.

Theorem 6.23. Let E C R be a Lebesgue measurable set and let u : E (0, oo) be a Lebesgue measurable function such that (6.25)

L' ({x c- E : u (x) > 0}) < oo.

Then there exists a measure-preserving function W : E -' [0, oo) n [0, C1(E)]

such that for C1-a.e. x E E, (6.26)

u* (gyp (x))

= u (x) .

6.2. Absolute Continuity of u*

203

Proof. Step 1: Assume first that £' (E) < oo. In view of Remark 6.11 and Corollary B.124, without loss of generality we may assume that every point of E has density one. Let cp : E -* [0, t' (E)] be defined by (6.27)

V (x) := LO. (u (x)) +.C1 ({yEE : y < x, u (y) = u (x)})

.

We claim that cp is Lebesgue measurable. Since P,,, is decreasing, it is Borel measurable, and so p o u is Lebesgue measurable. It remains to prove that the second function on the right-hand side of (6.27) is Lebesgue measurable. Since 41 (E) < oo, by Proposition B.12 we have that G1({yEE E. U (Y) = s}) = 0

for all s > 0 except at most a countable number. Let {sk}k C [0, co) be the set of all s > 0 such that the set {y E E : u (y) = s} has positive measure and define

Ek:={yEE: u(y)=sh}.

(6.28)

Then the second function on the right-hand side of (6.27) may be written as

E XEk (x) k

JEn(-oo,z] XE,, (Y) dy,

which is a Lebesgue measurable function.

Step 2: Let 0 < or < ,C1 (E). We claim that there exist two extended real numbers xo E [-oo, oo[ and so E 10, oo) such that (6.29)

R.(so)+-C1({yEE: ycxo,u(,)=ao})=a

and (6.30)

p,s(s)+G1({yEE: y<x,u(y)=s})>a

if either x E E, with x > xo, and s = so, or x E E is arbitrary and 0 < s < so. To see this, let so := u` (a) = inf Is > 0 : p,s (s) < al. (6.31) Note that if s < so, then p,, (a) > a by the definition of so. Since Lou is right continuous, we have that (6.32)

pu (so) = (Lu)+ (so) < a < (Au)- (so) . There are now two cases. If pu (so) = a, then we take (6.33)

xo:=sup{xEE: C1({yEE: y<x,u(y)=so})=0}

and the claim is proved. If Au (so) < a, then by (6.4) and (6.32),

u(y)?so}), and so C1({y E E : u(y) = so)) > 0. In this case, we take

xo:=sup{xEE: C'({yEE: y<x,u(y)=so})so}) so or u (x) = so and x so or u (x) = so and x so, then since Au (so) < a by (6.32) and P,, is decreasing by Proposition 6.1, it follows that Pu (s) < a for all s > so, and thus by (6.4), G' ({y E E : u (y) > u (x)}) =

8lim

La. (s) < a,

cp (x) < a. On the other hand, if u (x) = so and x < xo, then cp (x) < a by (6.29). This concludes the proof of (6.35). Step 4: We are ready to prove that cp is measure-preserving. Consider the measure p : B ([0, . ' (E)]) - 10, oo], defined by )u(s)-which implies that

p (B) :_ .C' (co 1(B))

,

B E 13([0,,C1(E)])

.

We claim that for every 0 < a < ,C1 (E) we have that (6.36)

p ([0, a]) _ £1 ((p-' (10, a])) = V ({x E E : cp (x.) < a}) = a.

6.2. Absolute Continuity of u*

205

If a > 0 or a = 0 and esssupE u < oo, then the claim follows by (6.29) and (6.35). If a = 0 and esssupE u = oo, then we claim that the set {x E E : W (x) = 0} is empty. Indeed, if esssupE u = oo, which contradicts the fact that u is real-valued. Thus, the set {x E E : cp (x) = 0} is empty, and so even in this case

p({0}) = L1({xE E: cp(x)=0}) = 0. Thus, we have shown that p and L1 coincide on all intervals [0, a] contained

in [0,,C' (E)]. Note that the family of these intervals generates the Borel o-algebra ([O,L1(E)]).

B

Let {au} C (0, oo) be an increasing sequence such that a. / G1(E). By Proposition B.9(i),

p([0,L1(E)]) =slim it([0,an]). Since p ([0, an]) = an < oo by (6.36), we have that p is a-finite. Thus, we are in a position to apply Corollary B.16 and Remark B.17 to conclude that p and G1 coincide on B ([0, L1 (E)] ). Step 5: We prove that (6.26) holds. Assume first that u (x) = 0. Note that by (6.25), 8u (0) < oo. By Remark 6.4 we have that u* (Pu (0)) = 0 and since u* is decreasing and cp (x) > Qu (0), it follows that 0 < u* (cp (x)) < u* (9, (0)) = 0. Thus, it remains to show that (6.26) holds for L1-a.e. x E E such that u (x) > 0. We first prove that (6.26) holds L1-a.e. in the sets Ek defined in (6.28). If x E Ek and sk > 0, write P (x) = Ate. (8k) + L1 ({y E E : y < x, u (y) = 8k}) .

Since L1 (Ek) > 0, it follows from Proposition 6.3 and the fact that u* is right continuous that u* (t) = ak for all /t E [Nu (sk), (pu)_ (8k)). Since (Qu)- (sk) = Au (8k) + L1 ({yEE : u (y) = 8k}) by (6.4) and

L1({yEE: y<x,u(y)=sk}) 0, and u (x) = bn for some n, then by construction of Ek, we must have that C1({y E E : u(y) = bn}) = 0. Together with (6.37), this implies that the set of x E E \ Uk Ek such that cp (x) > 0 and u* (Lou (u (x))) esssupE u. Since

C1( y E E: u(y)>esssupu1)= E l

0,

we c an assume that u (x) = esssupE u. On the other hand, since x E E , Uk Ek, we have that esssupE u is not one of the values sk, and so

41QVEE: u(y)=esssupu} }

=

0.

Hence, the set of x E E \ Uk E, such that go (x) = 0 has r1-measure zero. Step 6: Assume next that C1 (E) = oo. Define

Eo:={sEE: u(x)=0}, Fo:={tEE*: u*(x)=0}. By Proposition 6.3(v) and (6.25) we have that C' (Eo) =V (Fo). Partition Eo into a sequence {G..} of disjoint sets of finite measure,

Gn:={xEEo: n-1 0 depending only on p such that (7.6)

fiu (x) -:s(o,d) p g (x) dx < CPd& J d

for all u E W 1-P ((0, d)), where u(o'd)

fog (x) dx

f

a

I u' (x) Ip g (x) dx

u (x) g (x) dx.

Proof. If g = 0 on some interval [0, b] (respectively, [b, d]), then all the integrals involved reduce to integrals over [b, d] (respectively, (0, b)). Thus, we

can assume that g is strictly positive in (0, d). Also, by a scaling argument, it is enough to prove the result for d = 1. Finally, by dividing the inequality (7.6) by fag (x) dx, we may assume that I

A By Corollaries 7.14 and 7.15, without loss of generality we may assume that u is absolutely continuous. By' Theorem 3.30 and (7.7),

Lf

u (x) - u(o,l) = u (x) - J 1u (t) g (t) dt = fo 1 [u (x) - u (t)] g (t) dt

f Jf= 0

u' (a) g (t) dsdt

=

JJu'(8)9(t) d8dt - J

= fo u' (s)1 g (t) dtds o

f

x

f

d8 dt

x

u' (s) J g (t) dtds, e

7.2. Sobolev Functions Versus Absolutely Continuous Functions

227

where in the last equality we have used F ubini's theorem. Hence,

Iu (x) - u(o.l) I
g (x), and so for c < x < 1 we

7. Functions of Bounded Variation and Sobolev Functions

228

have that

(LX Iu' (s) I f9(t)

_

J c I u' (s) I

dtds)

r g (t) dtds + J y I u' (8) I

1. The proof in the case p = 1 is simpler, since there is no need to use Holder's inequality.

Part 2

Functions of Several Variables

Chapter 8

Absolutely Continuous Functions and Change of Variables Prospective Grad Students, IV. "Can you really live comfortably in this major metropolitan area with that stipend, or will I find myself living out of a closet working part time as a shoe sales-

man?" - Jorge Cham, www.phdcomics.com

In this chapter we extend some of the concepts and results of Chapter 3 to functions of several variables, including the notion of absolute continuity, the Lusin (N) property, and some change of variables formulas.

8.1. The Euclidean Space RI In the remainder of this book the implicit space is the Euclidean space RN. The elements of RN are N-tuples of real numbers x = (Si, ... , xN).1 Given

x = (x1, ... , XN) E RN and y = (yl, ... , YN) E RN, the Euclidean inner product of x and y is the real number defined by N

x'y

xiq

ili7

i=1

iWhen there is no possibility of confusion, we will also use xl,z2, etc., to denote different points of RN. Thus, depending on the context, z; is either a point of RN or the ith coordinate of the point z E RN. 231

S. Absolutely Continuous Functions and Change of Variables

232

while the Euclidean norm of x is the number N

txz.

IxI:=

When working with Euclidean spaces of different dimension, we will sometime use the notation IxI N for Ixl. We will often use the Cauchy's inequality: Ix - V1 and IT (x) -'f` (v)I < k 12., - VI

for all y E fl such that [x - yI < n. Since %F is differentiable on E, it follows

that

00

E = U En,k, n,k=1

and so it suffices to show that £N ('P (En,k)) = 0. To see this, fix n, k E N and e > 0. Since .CN (En,k) = 0, there exists a sequence {Q (xi, ri)} of cubes with center xi and side length ri < 2 1Nn such that

En,k C UQ(Xi,ri) i and

rv 0 we use the notation Z. :_ {x E RN : dist (x, F) < e}.

Lemma 8.7. Let L : RN RN be a linear transformation and let F L (Q(xo,r)), where xo E RN and r > 0. Then for every e > 0, LN (Fir) 0. We claim that for every e > 0, LN(Es) !5

2N(r+e)N-1e.

In view of the previous remark, by applying a rigid motion, we may assume

that yo = 0 and that H = {xN = 0}. It follows that Ee is contained in the rectangular parallelepiped R (-e - r, E + r)N-1 x (-e, e), and so LN (Ea) C LN (E) = 2Ne (e +r) N-1

This proves the claim.

Step 2: To prove the lemma, assume first that det L = 0. In this case F is contained in a hyperplane H. Taking yo := L (xo), for all x E Q (xo, r) we have

IL (x) -yoI = IL(x - xo)I S IILII Ix - xol
0 and choose an open subset A C f such that E C A and (8.6)

Vv (A) < Lo (E) + e.

We claim that for every xo E E there exists r 0 > 0 such that Q (xo, r) C A and (8.7)

Lo (41 (Q(xo,r))) < (M + e) LN (Q(xor))

8.2. Absolutely Continuous Functions of Several Variables

239

for all 0 < r < r,,,. Since 'I' is differentiable at xo, for every 8 > 0 there exists rya > 0 such that for all 0 < r < rx0 we have that Q (so, rxo) C A and

I'p(x)-'p(xo)-V'I'(xo)(x-xo)I 0 so small that C (N) (I VT (xo)I +

6)N-1

b < s,

we have proved the claim. Let F be the family of all closed cubes contained in A that are centered

at x E E and satisfy (8.7). By the Vitali-Besicovitch covering theorem (see Theorem B.118) there exists a countable family {Q,,} C F of pairwise disjoint cubes such that

Lrv E\UQ,,

0.

Let F:=E\U,LQ,,. Then ,Co (10 (E)) L


240

By Proposition 8.2 we have that ,rCN ('W (F)) = 0, while by Lemma 8.7, the fact that the cubes are disjoint and contained in A, and (8.6),

£o

(u)) C E Qn

n

CN

(w

.

(Q ))

: 5 (M + E)

E'LN

(en)

n

N, has an absolutely continuous representative (respectively, locally absolutely continuous if fl C RN is an arbitrary open set). Another important class of absolutely continuous functions is given by Lipschitz con-

tinuous functions' : 1-> RN.

8.3. Change of Variables for Multiple Integrals In this section we prove a change of variables formula for multiple integrals. The proof will make use of Brouwer's fixed point theorem [261.

Theorem 8.15 (Brouwer's fixed point theorem). Let K C RN be a nonempty compact convex set and let IT : K -> K be a continuous transformation. Then there exists x E K such that IQ (x) = x. The proof that we present here is due to Lax (1041 and makes use of an interesting change of variables formula (see Theorem 8.17 below).

8.3. Change of Variables for Multiple Integrals

243

Exercise 8.16. Let W : RN RN be of class C2 and let Ml,..., MN be the cofactors of the first row of the Jacobian matrix J9.2

(i) Prove that E

can be written symbolically as

i=1

V V",2 det

OWN (ii) Let N = 2. Using part (i), prove that 0M1 8x1

0.412

+ ax2 = o

(iii) Let N > 2 and prove that

det

V'P2

N

=

V'P2

E det k=2

VWN

OWN

where the subscript k means that the differential operator V in the first row acts only on the kth row. (iv) Prove that V 0'Q2 = 0

det

WN k for all k = 2, ... , N and conclude that N 8n'it o.

Theorem 8.17 (Lax). Let W E C2 (RN; RN) be such that W is the identity outside some sphere, say, the unit sphere:

+l; (x) = x for Jxj > 1. 2If A = (a;f)j

9=j... N

is an N x N matrix, the cofactor of the entry a{j is defined by Mr1:_(-1):+?detCj,

i,7=1,...,N,

where Cj is the (N - 1) x (N - 1) matrix obtained from A by removing the ith row and the jtb column.

244

8. Absolutely Continuous Functions and Change of Variables

Then the following change of variables formula holds for every function u E C'I (EN): (8.9)

jen u (y) dy = LN u (W (x)) J'P (x) dx.

Proof. We use notation (8.2). Define (8.10)

v (y) = v (y'i, yi)

y E RN.

J-700 u (yi, t) dt,

Then v is differentiable and by the fundamental theorem of calculus, We claim that

Z

= U.

V(VOW) (8.11)

VW2

(u o T) JT = det

V'N By the chain rule, for all x E RIV, IV

v (v o 1Y) (x)

= i-1 F,

Ov i (W (x)) VWi (x)

.

Thus, the vector V (v o 41) can be written as a linear combination of the vectors v W 1, ... , V'F jv. The last N - 1 of these vectors are the last N - 1 rows of the matrix on the right-hand side of (8.11), and therefore these can be subtracted from V (v o') without altering the value of the determinant (8.11). This leaves us with

2 (V0W)VW;

b

let

v'P2

= let

I

O W)

\\

0q'2

VW1

VWN

= (i-oW) J'k

1 (

noI)JW,

where in the second identity we have factored out the scalar 7g o W. Hence, (8.11) holds. Since u has compact support, there exists r > 0 such that u - 0 outside

the cube Q (0, r). Without loss of generality, we may take r > 2, so that is the identity outside the unit Q (0, r) contains the unit ball. Since sphere, it suffices to restrict the integrals in (8.9) to Q (0, r). By expanding


245

the determinant on the right-hand side of (8.11) according to the first row, we get

(uo')J = Mla(v°T) + axl

(8.12)

+MNa(yo11),

axN where All, ... , MN are the cofactors of the first row of the Jacobian matrix

JW. By Fubini's theorem and a one-dimensional integration by parts, for all i = 2, ... , N, 8 {v o

Q(OMi

,r)

(x)

}

(x) dx =

axi

J( -a,2) rr

N-`

j r lbli (x)

a(voty) axe

,

{ e} dxidx;

2

(8.13)

(v o') (x) aMi (x) dx (9xi

Q(O,r)

+

rr

(

ri (xi, xi) (v o ) (xi

N_1

dx4

2'2)

Since W is the identity on 8Q (0, r) and u - 0 outside the cube Q (0, r), for

all i=2,...,Nwe have (see 8.10)

v(W(x,± )) -v\xi'±2/ =0 N-1.

for all x; E (- 2 ,

Hence, (8.13) reduces to

2)

(8.14)

(x)

4%

a(u 0 T) 0-Ti

Q(o,r) )

(x) dx = -

(v o W) (x) JQ(O,r)

aM 8xi

(x) dx

for alli=2,...,N. On the other hand, for i = 1 we have

v \ \xl' 2I11I = v r\x21} rr

f-00 u (xi, t) dt,

while

r

1l1l

v(11 (x''-2/I =v-2/ _J0o u(xi,t) dt=0, (XI,

where we have used the fact that u 0 outside Q (0, r). Since Ml (x) = 1 when IF (x) = x, again by Fubini's theorem we have that l

r r

\-a 2 )

r%

r

(

(X' X1) (v O `Y) (xl>xl) I "2 ,

22

J oo u (xi, t) dt dx1 = JQ(O,r) U (X) dx,


246

and so (8.13) for i = 1 becomes

I

1bi18(yoT)

dx =

8x1

Q(O,r)

j - f rQ(O,r) (voW) 8Ml dx+ J 8x1

udx.

Q(O,r)

Summing this identity together with the N - 1 ones in (8.14) and using the previous exercise and (8.12) yields N

Mya(y0T)dx

i=1

Q(O,r)

8xi

(v o T) E

J

(O,r)

i=1

Ldx + J 8xi

4

Q(O,r)

u dx.

This proves the result.

Exercise 8.18. Using mollifiers (see Appendix C) prove that the previous theorem continues to hold if E C1 (RN; RN) is the identity outside some sphere and u E CC (RN).

Although the transformation T in the previous theorem (and more generally in the previous exercise) is not assumed to be one-to-one or onto, it actually turns out that 12 is onto. Corollary 8.19. Let 11 E C' (]RN; RN) be the identity outside the unit ball. Then 1F is onto.

Proof. Assume that there exists a point yo E RN \ %p (RN). Since 19 is the identity outside B (0, 1), it follows that yo E B (0, 1). By the continuity of 'P we have that the set 91 (B (0,1)) is closed, and thus we may find a ball B (yo, ro) C B (0, 1) that does not intersect IF (B (0, 1)) (and, in turn, T (RN) either). Let 0 < r < ro and define U(Y)

to

if y E B (yo, r) , otherwise.

Note that supp u C B (yo, r) , and so it does not intersect 'P (RN). Thus u o 91 = 0. On the other hand, fRN a (y) dy > 0 and this contradicts the previous theorem.

We now turn to the proof of Brouwer's fixed point theorem.

Proof of Theorem 8.15. Step 1: Assume that K = B (0, 1) and let IF E C (B (0,1); RN) be the identity on the unit sphere. We claim that


247

T (.(o 1)) D B (0, 1). Extend W to be the identity outside B (0, 1). Using standard mollifiers, we may construct a sequence of transformations and {'P.} C C1 (RN; RN) such that Pn is the identity outside B (0, 1 + Wn -4 'P uniformly on compact sets. By the previous corollary (which continues to hold when the unit ball is replaced by any ball) we have that each 'Pn is onto, and so B (0,1) C'Pn (B (0,1 + n)} . Hence, for each y E B (0,1)

(x,) = y. Let {xnk} be a subsethere is x E B (0, 1 + such that quence of {x} such that xnk - x E B (0, 1). By uniform convergence in B (0, 2) we have that v = T.k (xTk } 1 'f` (x) as k --> oo. It follows that 19 (x) = y. Step 2: Let W : B (0, 1) -t B (0,1) be a continuous transformation. Assume

x for all x E B (0, 1). Define 4P : B (0,1) -> SN-1 as follows. For each x E B (0,1) let 41> (x) be the intersection with the sphere SN-1 of the ray from ' (x) to

by contradiction that T has no fixed point, that is, that ' (x) x, precisely3,

-b (x):=x+F(x)(x-*(x)), where

x

(x) -

(x) - I X12 +

Ix1212 -I- Ix -'P (x)12

IT -'P(x)I Then fi E C (B (0,1); RN), 0 and let 'P : B (0, R) -, B (0, R) be a continuous transformation. To obtain a fixed point, it suffices to apply the previous step to the resealed function

TR (x) := R9 (Rx) ,

x E B (0, 1).

Step 4: Let K C RN be a nonempty compact convex set and let' : K -> K be a continuous transformation. Find R > 0 such that K C B (0, R) and 3To obtain F, we consider the line t1Y (x) + (1 - t) z,

t E R,

through the distinct points II' (x) and x and then find t E R such that

+(1-f)n12 = tt Ix - T (x)l2 + 2t (z %Y (x) - 1x12) + 1x12 It suffices to solve for t.

.

248


for each x E B (0, R) consider the continuous transformation

(x) :_ ' (II (x)) x E B (0, R), where I1 : RN -+ K is the projection onto the convex set K. Note that -P (K) C K C B (0, R), and so by the continuity of II we have that I : B (0, R) -i B (0, R) is continuous. By the previous step there exists x E B (0, R) such that

x=c(x)=W(II (x)).

On the other hand, since 11) (K) C K, we have that x E K, and so II (x) = x. Thus, the previous identity reduces to x = %P (x) and the proof is completed. 17

As a corollary of Brouwer's fixed point theorem we have the following result, which will be needed in the proof of the change of variables formula below.

Corollary 8.20. Let T E C (B (0,1); RN) be such that 14, (x) - xI < e

for all x E SN-1, where 0 < e < 1. Then IF (B (0,1)) J B (0,1- e). Proof. Assume by contradiction that there exists a point yo E B (0, 1 - e) 1 IF (B (0,1)). By hypothesis IW (x) I > 1 - E if x E SN-1. Thus, yo T (SN-1), and therefore W (x) # yo for all x E B (0, 1). Define : B (0, 1) SN-1 by

yo-%F(x) Iyo - W (x)I' Then W is continuous. To reach a contradiction in view of Brouwer's fixed point theorem, it remains to show that has no fixed points. Since

(B

(0,1)) C

sN-1,

the only possible fixed points lie on the unit sphere. On the other hand, if SN-1, then xE x 4 (x) < 0. Hence, c cannot have fixed points in which is a contradiction and completes the proof.

SN-1,

We are now ready to prove the change of variables formula for multiple integrals.

Theorem 8.21 (Change of variables for multiple integrals). Let n C 1R' be an open set and let W : n -> RN be continuous. Assume that there exist two Lebesgue measurable sets F, G C n on which W is differentiable and one-to-one, respectively. If


(i) GN (SZ

(ii)

£N

249

F) = 0,

\ (W (IZ 1 F)) = 0,

(iii) LN (,p (S \ G)) = 0, then the change of variables formula

I

(E)

u (y) dy = fE u (I (x)) I J (x) I dx

holds for every Lebesgue measurable set E C f2 and for every Lebesgue measurable function u : ID (E) - [-oo, oo], which is either Lebesgue integrable or has a sign.

Proof. Step 1: We begin by showing that IV maps Lebesgue measurable subsets of F into Lebesgue measurable sets. Let E C S l be a Lebesgue measurable set. In view of property (ii) and the completeness of the Lebesgue

measure, it suffices to show that E n F is Lebesgue measurable. But this follows from Theorem 8.10.

Step 2: For every Lebesgue measurable set E C RN define

µ(E):=CN(IF (EnFnG)). Since CN is countably additive and IV is one-to-one on G, it follows that µ is a Borel measure. Moreover, since IQ is continuous, it maps compact sets into compact sets, and so µ is finite on compact sets. Thus, is is a Radon measure. Moreover, p. is absolutely continuous with respect to the Lebesgue measure. Indeed, if E C RN is such that CN (E) = 0, then by Proposition 8.2 the set T (E n F n G) has Lebesgue measure zero. Thus, by the Radon-Nilcodym theorem (see Theorem B.65) there exists a nonnegative locally integrable function

:

dGN

A (E)

RN

10, ooj such that

= fE dLN

(x) dx

for all Lebesgue measurable sets E C IR'. In particular, if E C SZ is a Lebesgue measurable set, then

T (E) =' (EnFnG) U IF (E\ F)u'P (E\G). By (ii) and (iii) we get that (8.15)

CN(,p (E))=,CN('(EnFnG)) =

(E) = JE dCN (x)

for every Lebesgue measurable set E C S1.

dx


250

Let now H be a Borel set in 19 (fl) with finite measure. Since ' is continuous, the set E := W-1 (H) is Lebesgue measurable, and so by (8.15), (8.16)

(q, (E))

401) XH (H) dy = GN (H) = GN

= Jb) dx = fy XH ('I' (x))

(x) dx,

where we have used the fact that XE (x) = XH (I' (x)) for all x E 1. On the other hand, if H is only a Lebesgue measurable set in W (12) with finite measure, then '-1 (H) may not be Lebesgue measurable (see Exercise 3.56). In this case find two Borel subsets B1 and B2 of (St) such that B1 C H C B2 and GN (Bi) = GN (H) = GN (B2) < oo (see Proposition C.3). Then by (8.16),

dx = j

XB2

OF (x)) fly (x) dx < 001

and since

XB, ('IP (x)) N (x) N, and that are one-to-one except at

most on a set of Lebesgue measure zero. Another extremely useful class of transformations is given by (locally) Lipschitz continuous functions T : S2 - RN, which are again one-to-one except at most on a set of Lebesgue measure zero.

Perhaps one of the most important applications of Theorem 8.21 and Remark 8.22 is given by spherical coordinates in RN. We proceed by induc-

tion. For N = 2 we consider the standard polar coordinate system. Given a point x = (x1, ... , xy.) E RN, let r Ixl, let 01 be the angle from the positive x1-axis to x, precisely, 81 := cos 1 (21 }

r

,

0 0 for all nonnegative functions 0 E V (fl), then there exists a unique Radon measure µ : 13 (f2) -> [0, oo] such that

r

T (0) = J 0dµ for all 0 E D (1l) . s2

(ii) If T has order zero, then there exist two Radon measure 14,111 B (1) - [0, ool such that

T(0) =

Jif

0dµ1 -

r

Jsa

0 dµ2 for all q E D (Q)

.

Proof. (i) We claim that T has order zero. Fix a compact set K C f1 and find an open set 01 such that

KC01CC11. Construct a smooth cut-off function c E C°O (f) such that co - 1 on K, supp cp C SZl, and 0 < Sp < 1 (see Exercise C.22). In particular, Sp E D (SZ). Since 9 1 on K and rp > 0, for every 0 E DK (Il) we have that 10 W1 >_0,

T(0+II0IIK,oSp)

?0;

that is, by the linearity of T, IT(0)I 0 such that (9.9)

IT (0)1 5 CGII0IIKi,o

for all 0 E DK; (0). If 0 E C, (Sts), then dist (supp 0, c( 1) > 0, and thus if we consider 0n := gyp, *0, where cp are standard mollifiers (with a := 1) and n

n

9. Distributions

266

n < dist (supp gyp, 8Sti), we have that {4,,a} C DK, (f2) and

uniformly

on Ki. It follows by (9.9) that IT(4,n.-41)1 !5 Cz110. -01IIK,,a--+ o

as 1, n -* oo. Hence, IT (4,,a)} is a Cauchy sequence, and therefore it converges to a limit that we denote by T, (0). Moreover, if 4, > 0, then 0, > 0 also, and so Ti (0) > 0. Note that, again by (9.9), Ti (0) is independent of the choice of the approximating sequence {4,,a}.

By the linearity of T it follows that Ti : CC (a,) - R is linear and positive, while by (9.9) we have that IT (0)I 5 Ci II4I1c,A)

for all 0 E C, (St,). Since Il, C St,+i, it follows that T,+i (4') = T, (¢)

for all 4' E CC (Sti)

.

Thus {T,} defines a unique linear positive extension of T to the union of all CC (SZi), which coincides with CC (0). The result now follows from the Riesz representation theorem in CC (C') for positive linear functionals (see Theorem B.115). (ii) The second part of the proof of (i) continues to hold in this case. By (9.9) and the fact that {(l,} covers St, we have that the extended functional is locally bounded, and so the result now follows from the Riesz representation theorem in Cc (12) for locally bounded linear functionals (see Theorem B.115). 0

9.3. Derivatives of Distributions and Distributions as Derivatives We now define the notion of a derivative of a distribution.

Definition 9.14. Let St C RN be an open set and let T E D' (S2). Given a multi-index a, we define the ath derivative of T as a 8z:a

(4,) = (-1)IaIT

0 E D(1Z).

For j E N the symbol D'T stands for the collection of all ath distributional derivatives of T with I a I = j.

Remark 9.15. It can be verified that is still a distribution. Indeed, let K C 12 be a compact set. By Theorem 9.10 there exist an integer j E No and a constant CK > 0 such that IT (0)10 for all 0 E V (Cl) with 0 > 0. By Theorem 9.13 there exists a unique (positive) Radon measure µ : B (Cl) - [0, oo] such that

TDu (0) = in uA4 dx =

Jn

0 dµ for all

E D (S2)

Thus, with an abuse of notation, we may write Au = ia. Similarly, for U E Lic (Cl; RN) one can define the divergence of u in the sense of distributions.

Definition 9.17. Let Cl C R v be an open set, let u E Li (Cl), and let a be a multi-index. If there exists a function va E L je (Cl) such that a

a u(0)

for all 0ED(Cl),

then vQ is called the ath weak, or distributional, derivative of Tu. We write 8

.= Va.

9. Distributions

268

Thus, a function va E Li°C (Q) is the nth weak derivative of u r= Ll (12)

if F

(9.10)

in

Ovcdx = (-1)I"I In uLO- dx

for all 0 E C'° (SZ).

Exercise 9.18. Let u : R -i R be defined as follows:

if -7r <x 0 such that IT (0)15 CK II0IIKj

for all 0 E DK (Il). On the other hand, if 0 E D (Il), then z'¢ E DK A, and so, since OT = T, IT (0)1 = IT(OO)I < CK IIt4)IIK,J

By the Leibnitz formula (see Exercise 9.11), II7P0II K j 5 C.'O II4II KJ) and so IT (4))I 2 T ( n=1

00

(-1)1°I

n'1')

J fan)8xa dx

n=1 a

n=1

(.

E

LoJ 00 a 8xa 8xo a a :a=1 This shows (ii). Finally, if T has finite order $, then -O.T has finite order R. < Q, and so it suffices to consider only multi-indices a < ,l3 := (f + 2, ... , e + 2). Exercise 9.26. Let f2 C RN be an open set. Prove that every continuous linear functional on C00 (1l) is of the form f H T (f), where T is a distribution with compact support.

Exercise 9.27. Let 1 = (a1, bl) x ... x (aN, bN). (i) Prove that 0 E V (0) is such that

1 O (x) dx = 0 if and only if N

80s

Eax; i=1 for some 01, ... , ON E V (SZ). Hint: Use induction on N and look at Step 1 of the proof of Lemma 7.3.

(ii) Prove that if T E D' (12) is such that = 0 for all i = 1, ... , N, then there exists a constant cin R such that T (O) = c

0 ( x) dx

for all 0 E V (f2), i.e., T is constant.

9.4. Convolutions

275

(iii) Prove that if Q C RN is an open connected set and T E D' (fI) is such that = 0 for all i = 1, ... , N, then T is constant.

9.4. Convolutions In this section we work mostly with c = RN. We recall that given two functions 0, cp E C°° (RN), one of which with compact support, their convolution is the function 0 * cp defined by

(sp * 0) (x) = j

(9.14)

11R N

co (x - y) 0 (y) dy,

x E RN.

Motivated by this formula, given T E D' (RN) and ¢ E V (RN), we define the convolution of T and 0 by (T * 0) (x) := T (9.15) x E RN,

(x - y), y E RN. Note that T * ¢ is a function. In the special case in which T = Tp for some rp E CO° (RN), the two formulas (9.14) and (9.15) agree. where 4,X (y)

Exercise 9.28. Let 0, cp E V (RN). For h > 0 define uh (x) ._ $N

tp (x - hy) c (hy) ,

x E RN.

YIEZN

Prove that {uh} converges uniformly to cp * . Hint: Use Riemann sums.

Theorem 9.29. Let T E D' (RN) and -0, ip E V (RN). Then (i) T * 0 E Coo (RN), (ii) supp (T * 0) C supp T + supp 4), (iii) for every multi-index a,

a (T*4))=T* (iv)

Q10) = L

/

Q *4)

(T*gyp)*4)=T*((p*O).

Proof. If xf,, --+ x in RN, then for every y E RN, 4)

(y) = 4(xn.-y) - 4'(x-y) = 9x(y)

and conditions (i) and (ii) of Theorem 9.8 are satisfied. Hence, {pxn} converges to 9 with respect to r, and so by Theorem 9.10,

(9)

(T * 0) (xn) = T -* T (9) = (T * 4') (x) which proves that T * 0 is a continuous function. To prove (ii), note that if x E RN is such that supp 4'X fl supp T = 01

9. Distributions

276

then (T * 4)) (x) = 0. Thus, supp (T * 4i) C (x E RN supp 4x fl supp T # 0) = supp T + supp 0.

Next we prove (iii). Let e; be an element of the canonical basis of RN and for every x E RN and h # 0 consider the function Oa,h,i (y) :_

4) (x + he; - y) - 0 (x - y),

yE

h

1[$N.

# (x - y) for all y E RN and conditions (i) and (ii) of Theorem 9.8 are satisfied (why?). Hence, {4) converges to (A. ) x with respect to r as h -p 0, and so, by the linearity of T and As h -* 0, we have that 0x.h,i (y)

Theorem 9.10, (T * 4)) (x + he;) - (T * 4)) (x) h

T

T

(8" \ Ox:

as h -p 0, which proves that z = T *. Moreover, since for all x, y E RN,

(ax )z (y) = for all x

8x (x - y) =

-a (x - y)

av (y),

RN we have a

* 4)/ (x) =

a

(4)x)

=TI _(

_ -T

&Yi

)= (T*-)

N-i which, together with an inducti\on ar\gument, gi ves (iii).

Finally, to prove (iv), we approximate the Riemann sum

(

x),

function cp * 0 with the

uh (x) := hN E Sp (x - hy) 0 (hy) ,

x E RN,

YEZN

where h > 0. Note that for every multi-index a, a

8xh (x) = hN 1/E$N axa Since {

} converges uniformly to

8xa

(x. - hy)

(hy) ,

x E IlBN.

* 0 by the previous exercise and

*0_ &60*0) 0x°i

9.4. Convolutions

277

(why?), by Theorem 9.8 we have that {uh) converges to V*4) with respect to r as h -> 0. It follows that for every x E RN, {(nh)'} converges to (cp * Of with respect to r as h --+ 0. By the linearity of T and by Theorem 9.10 we have that

(T * (tP * )) (x) = T ((co * Of) = lim T ((uhf) = lim (T * Uh) (x) h-,0

= lim hN h-,0

h"O

(T * gyp) (x - hy) 4) (hy) YEZN

_ ((T * 9) * 0) (x), where we have used the previous exercise. This completes the proof.

0

As a consequence of the previous theorem, we can approximate convolutions with C°° functions.

Theorem 9.30. Let T E V1(RN) and let {rpE} f, e > 0, be a family of standard mollifiers2. Then {T * cp} converges to T in the sense of distributions as E -> 0+; that is,

sh 0 for every 0 E D (RN).

I

N

(T * coe) (x) 0 (x) dx = T (0)

Proof. By Theorems C.19(i) and C.20 and Theorem 9.8 we have that for every 0 r =D (RN) the sequence {ape * 4i} converges to 0 with respect to r as

For every 0 E D (RN) define fi (x)

4) (-x) ,

xE

RAr.

Then by Theorem 9.10, T (0) = (T * ) (0) = T

Jim T ((APE * )c)

= Urn (T * (cpE * )) (0) = Eli m ((T * we) * ) (0)

= lim f

e-.O+ f1,N

(T * cpe) (y)

(0 - y) dy

= lim f (TN * (pe) (y) 0 (y) dy, e 4O+

where we have used Theorem 9.29(iv).

Exercise 9.31. Let Il C RN be an open set and let T E D` (fl). 2See Appendix C.

O

278

9. Distributions

(i) Prove that there exists a sequence {T,,} C V (ft) such that each T, has support compactly contained in f and {T.} converges to T in the sense of distributions. (ii) Prove that CLOD (S2) is dense in D' (S2) with respect to the weak star

topology of V (ci).

Chapter 10

Sobolev Spaces Newton's first law of gradrwtion: A grad student in procrastination tends to stay in procrastination unless an external force is applied to it. -Jorge Cham, www.phdcomics.com

In this chapter we define Sobolev functions on domains of RN, N > 1, and begin to study their properties.

10.1. Definition and Main Properties Definition 10.1. Let ft C RN be an open set and let 1 < p < oo. The Sobolev space W" (S2) is the space of all functions u E LP (Q) whose dis-

tributional first-order partial derivatives belong to V (0); that is, for all i = 1, ... , N there exists a function gi E 1/ (S2) such that (10.1)

u-Lo-

in exi

dx = - in gi4dx

for all 0 E CC° (S2). The function gi is called the weak, or distributional, partial derivative of u with respect to xi and is denoted O. In terms of distributions the previous definition means that for all i = 1, . . . , N (see Definition 9.17).

= Ti

Remark 10.2 (Important). Following the literature, we use the same notation to indicate the weak and the classical (in the sense of (8.4)) partial derivatives of a function. Unfortunately, this results in endless confusion for students. In the remainder of this book, when u E W4(92), unless otherwise specified, is the weak partial derivative of u. 279

10. Sobolev Spaces

280

For u E W1P (f2) we set

Vu :=

au

au.

X_I'

XN)

As usual, we define W 1,' (n; Rd) as the space of all functions u = (ui,... , ud) such that u; E W 1°P (ff) for all i = 1, ... , d. Also, 1 (SZ) :_ {u E L, a (11): u E Wl,P (l?) for all open sets S)' CC fl).

Exercise 10.3. Let Il = B (0,1). Show that the function u : 0

R,

defined by

u (x) = u (XI, ... , xrv):=

1

ifxN>0,

0

if xN < 0,

does not belong to W1,P(1Z) for any 1
0 and construct N-simplexes 1, ... , At with pairwise disjoint interiors such that

t

KCUAiCB(0,R) i=1 and

diem Ai > 4'i

for all i = 1, ... , t, where ri is the radius of the inscribed ball in Ai. Let v be the continuous piecewise affine function that coincides with u in all the vertices of the simplexes. Then by Taylor's formula sup Iu (x) - v (x) I C Ch2 sup I V 2u (x) I < Ch2 sup I V2u (x) I , xEA,

WEA;

xERN

sup IVu(x) - Vv(x)I < Ch sup IV2u(x)I 0}

(RN). Prove that the function

t

v (x) .

u(x)

ifxN>0,

IL W, -XN) if xN < 0

belongs to W 1,P (RN) and that for all i = 1, ... , N and for CN-a.e.

xEIItN

ifxN>0,

8 (x)

8v (x) axt

(-1)61N 8u

8xi (x', -XN)

ifxN fi (x) 9i (x) i=1

for all g E LP (S2; RN+1) and N

IILII(wl.P(n)}+ = IIL1II(Ln(.;eN+'))!

_

j7 Ilfzll

'(n}

It follows that

L (u) =

r

Jsz

N

(foX)X)+f(X)X)) dx i=1

0

for all u E W1'P (S2).

Remark 10.42. Note that the previous theorem does not imply that the dual of W 1,P (0) is LP' (S2; Indeed, if f = Yo,..., fN) belongs to RN+1).

LP' (c'; RN+1), then we have shown that the functional L defined in (10.23) belongs to (W' P (9))' and that (10.24) holds. On the other hand, by Theorem 10.41 there exists h = (ho, ... , hN) in LP' (S2; RN+1) (possibly different

from f) such that T

N

IIhiII .,

IILII(wl.n(n)}'

cz

i-0

Thus, we have that N

IILII(wl.P(ca)' = min E IIhiiiLq-(

h E L" (s2; RN+1)

i=0 such that (10.25) holds

To explore the possible lack of uniqueness, assume that h = (h4, ... , hN) E iY (S2; RN+1)

.

10. Sobolev Spaces

302

is such that N

au (fou+fi.)ax

dx

J

" au hou+Fh,a

ax

for all u E W1' (Sl). In particular, we obtain that

J (fo+fi:) dx = J

rtx

i=1

for all we get

i=1

E V (]ESN). If we rewrite this identity in the sense of distributions, N

Tfoxn (0) -

a`y (10) = Thoxn ( i=1

)i=1

8T

8

;

n

(ib)

for all ¢ E D (RN), or, equivalently,

foxn - div (FXn) = hoXa - div (Hxn)

for all 0 E D (RN), where F := (fl,..., fN) and H := (hl,..., hN). Hence, we have shown that given f = (fo,..., fN) E LP' (fl; RN+'-), any solution h = (ho,..., hN) E Z' (Q; RN+1) (in the sense of distributions) of the partial differential equation div (HXn) - hoX& = div (Fxn) - foxsa in RN will give rise to a different representation in (10.25).

In view of Theorem 10.41, we can characterize weak convergence in

W''()).

Exercise 10.43. Let Il C RN be an open set, let 1 < p < oo, and let {u, } C W 1,P (1k). Prove that u,a -s u in W 1.P (f) if and only if u, -s u in LP (S2) and Vu,, - Vu in LP (fl; RN). If SZ has additional properties, in the previous exercise one can replace

un - u in LP (1) with u, - u in LP (1) (see Exercise 11.14). Theorem 10.44 (Compactness). Let 11 C RN be an open set and let 1 < p < oo. Assume that {u,s} C W1"(fl) is bounded. Then there exist a subsequence {ufk } of {u,,} and u E W 1.P (Cl) such that u, - u in W 1,P (11).

Proof. Since NO and {Vun} are bounded in the reflexive Banach spaces L" (Cl) and LP (Cl; RN), respectively, we may select the subsequence {u., }

such that u,,,, -i u in LP (fl) and

s vi in LP (1Z) for all t = 1, ... , N

10-4. Duals and Weak Convergence

303

and for some functions u, v1, ... , vN E LP (il). It remains to show that u e W 1,P (SZ). For every ¢ E (0), i = 1, ... , N, and k E N we have

Jin unx

a

20dx =-

J a k 0 dx. i

Letting k -> oo in the previous equality yields

Ju±-dx = ax; which shows that

n

vidx,

0

= vi. Hence, U E W1,P (a).

Remark 10.45. The previous result fails for p = 1. Indeed, in this case, L' (SZ) is not reflexive (so we do not have weak sequential compactness of bounded sequences) and it is not the dual of a separable space (so we do not have weak star sequential compactness of bounded sequences). To recover some compactness, we use the embedding L1(SZ) -> Mb (1l)

U -' A. (E) where

Au (E) := JE (x) dx,

E E B (il)

.

Note that (why?)

r L Jul dx = IIAUIIMb(n) Thus, given a bounded sequence {u.} C L1 (Q), one can only conclude that A in there exist a subsequence {i } and A E Mb (f) such that W1,1 (11), Mb (SZ). In particular, given a bounded sequence one will C only recover some compactness in the space BV (a). See Theorem 13.35.

Exercise 10.46. Let SZ = B (0,1). Construct a bounded sequence {un} C W1,1 (il) converging strongly in L1 (1Z) to the function given in Exercise 10.3.

Every functional L E (W 1'P (0))' is the extension of a distribution T E More precisely, if L has the form (10.25), then its restriction to D (it) is given by D' (11).

(10.28)

T (0) := Tfo (0)

for all 0 E D (SZ) . e=1

Conversely, a distribution T E D' (SZ) of the form (10.28), where fo,... , fN E

LP' (fl), can be extended to an element of (W1,P (1k))', but this extension may not be unique, unless W 1,p (SZ) = Wo'1' (SZ).

Definition 10.47. For 1 < p:5 oo let p' be its Holder conjugate exponent. The dual space of TV0'r (SZ) is denoted by W-1,p (n).

10. Sobolev Spaces

304

Exercise 10.48. Let fZ C RN be an open set, let 1 < p < oo, and let T E D' (fl) be of the form (10.28), where fo, ... , fN E L/ (1). Prove that T may be uniquely extended to a functional L E W-1,p (a). In view of the previous exercise we have the following characterization of the dual of W3"P (S2).

Corollary 10.49 (Riesz's representation theorem in Let fl C RN be an open set and let 1 < p < oo. Then W-1,P(fl) can be identified with the subspace of distributions T of the form N

T=Tfo_1: fir`, i=1

where fo, ... , fN E LP' (fl).

Next we study the case p = oo. We endow W1,1 (1k) with the equivalent norm IIuIIW1,00(n) := max

...

IIUIIL-(sz)' II ax,

II aXN

By Theorem B.96 the dual of L°O (i]) can be identified with the space of all bounded finitely additive signed measures that are absolutely continuous with respect to the Lebesgue measure restricted to fl. Thus, if Ao, ... , AN are any such measures, then the functional N

r

udAo+> J

L(u):=,

dAi,

uEW1'OO(fl),

belongs to (W1" (S2))'. Indeed, since each Ao E (L°° (Il))', we have that N

IL (u)j< IIAOII(L-(n))' IIUIIL-(n) + Max 0 0, let

flh,i:={xEfl: x+teiEfl forall0 1 and r > 0, then there exists a constant C = C (E, p) > 0 such that

(a+b)p 0. Hint: Use the convexity of the function g (t) = ItJP,

tGR. Theorem 10.55. Let 1 C

RN be an open set and let u E W ',P (Il). 1 0, Iu (x + hei) - u (x)I P dx < fah'i

Iu (x + hei) - u (x)IP Jszh ,

18u (x)IP dx ax,

J

hP

"

dx

=

hP

(J

18u (x) Ip dx) si

axi

A

< o0

Conversely, if u re L" (0), 1 < p < oo, is such that (10.32)

lim inf

Iu (x + he - u (x)IP

(foh'i

dx

for every i = 1, ... , N, then u E W '.P (cl).

Proof. Step 1: Assume that u E c, (11) n W 1'p (Q) and fix a compact set KCSZ. Let d := min {dirt (K, 00) ,1 } . By Taylor's formula there exists a constant CK > 0 (depending on u) such 2

that

u (x + hei) - u (x) - a (x) hI < CKh2 for all x E K and 0 < h < d. Hence,

su

(x)Ih 0 we have

au (x) hP < (1 + 6) Iu (x + hei) - u (x)IP + CK,rh2P axi for all x E K and 0 < h < d. Dividing by hP and integrating over K yields I

I

axi

(x) I P dx < (1 + 8) fK

0 and for GN-1-a.e. x E RN-1 for which f2x{ is nonempty, by the fundamental theorem of calculus we have 1

Iu (x;, xi + h) - u (xi, xi) I = I f

th)) dtl dt (u (x:, x1 + "U

_
0 define the resealed function tar (x) := u (rx) ,

x E RN.

If (11.1) holds for ur, we get Q

(LN

Iu(rx)IQ dx J

= (.LN CC

Iur (x)IQ

dx) Q

U. Ivur (x) 1P dx

=C rP

f.."

Ivu(rx)Ipdx

,

d) 0, (11.2)

GN({xEE: Iu(x)I>t})n1.

lu (x) I - n if

vn (x) :=

By the chain rule (see Exercise 10.37(1) and (vi)) for LN-a.e. T E RN, Ivv,z(x)I =

pvu(x)I

therwlsex)I < n,

and so Vvn E LP (RN; RN), while

I

Iv Idx = fi2i>} Ivldx { N fr -}} 0 such that 116 (u) II W 1.P(RN)

< C IIuII W 1.P(n)

for all u E W 1,P (S2). On the other hand, by Theorem 11.2, for all p < q < p* we have that IIuIIL4(n) = IIE (u)IILQ(n) _< IIe (U)IILQ(RN)

C1 IIe (u)IIW1.P(RN)

< C1C IIuIIw1.P(n) ,

where C1 = C1 (N, p) and we have used the fact that S (u) (x) = u (x) for CN-a.e. x E SZ.

Next we show that if p < q < p* and Il is an extension domain for W1,P (S2) with finite measure, then the embedding W1,P (cZ) -' Lq (12)

u'-* u is actually compact.

Theorem 11.10 (Rellich-Kondrachov). Let 1 < p < N and let S2 C RN be an extension domain for W" (Il) with finite measure. Let {un} C W1,P (SZ) be a bounded sequence. Then there exist a subsequence {un,, } of {un} and a function u E LP' (St) such that un,r - u in Lq (S2) for all 1 < q < p*.

The proof makes use of the following auxiliary results.

11.1. Embeddings: 1 < p < N

321

Lemma 11.11. Let 1 < p < oo and let u E W1.p (RN). Then for all

hERN\{0},

f

v

Iu (x + h) - ti (x)I" dx < IhIpf IVu (x) I'' dx. N

Proof. Assume that u E W4 (RN) n C- (RN). For x E RN and h E RN \ {0} by the fundamental theorem of calculus we have that Iu (x + h) - u (x) l =

I

d (u (x + th)) dtl dt fo 1

i

u in LP (0). For simplicity, for every v E LP (RN) we Set v(k) := u * Wk. By the previous lemma and the fact that {ua} is bounded in W1.P (RN), we get slip J

nEN RN

dx < C (N, p) sup f

(nn)(k) -

dx
L4 (tI)

uHu is compact for all 1 < q < p.

Exercise 11.18. Let fl C RN be an open set and let 1 < p < oo. Prove that the embedding W "P (l1) - LP (SI)

UI-'u is compact if and only if sup

fn\nn Jul" dx :

n and IxI < n}.

11.1. Embeddings: 1 < p < N

325

The Rellich-Kondrachov theorem and its variations hold for special domains with finite measure. The next two exercises show that if we restrict our attention to the class of radial functions, then we have compactness in the entire space RN for all N < q < p", but not for q = N or q = p".

Exercise 11.10 (Radial functions, I). Let N > 2, let f E C1(10, oo)), and let

u(x)

(11.10)

f (IzI), -TERN . (i) Let 1 0. Prove that for r > 0, (r2a f2 (r)),

< [(raf

(r))12

+ (1,f

(r))2

=r2a [(fi (r))2 + f2 (r)] + a (r2a-1 f2 (r))' - a (a - 1)

r2a-2 f2 (r).

(iii) Prove that for every N E N with N > 3 and for all r > N - 1, rN-1 f2 (r) < 2

J0

r tN-1 [(f' (t))2 + (f (t))2] dt.

(iv) Prove that for all r > 1, 00

r f 2 (r) < 2

Jr

t [(f' (t))2 + (f (t))2] dt.

(v) Prove that if the function u defined in (11.10) belongs to W1,2 (RN),

N > 2, then C

It(x)I: iN) I

IIUIIw1.2(RN)

a

forallxERN,with IxI>N-1. C W 1,2 (RN) n C' (RN)

Exercise 11.20 (Radial functions, II). Let be a sequence of radial functions with sup IItIIw'.2(itN) < 00n

(i) Prove that 21

urn sup

R-oo a

dx = 0.

RN\B(o,R)

(ii) Prove that there exist a subsequence {u,yk} of and a function u E W1.2 (RN) such that u in Lq (RN) for all 2 < q < N-2' 2N

11. Sobolev Spaces: Embeddings

326

(iii) Let V E C' (R) be such that supp cp C 10, 11, cp # 0, and define u,3 (x) = an.cp (IxI - n),

x E RN.

Find an in such a way that {u.} is bounded in W1"2 (RN) but does not converge in L2 (RN). (iv) Let cp be as in part (iii) and define

um (x)=a W(2'LIxI-1),

xERN.

Find an in such a way that {un} is bounded in W1.2 (RN) but does not converge in L2* (RN).

We conclude this section by showing that the Rellich-Kondrachov theorem continues to hold for bounded domains with boundary of class C.

Theorem 11.21. Let 1 < p < oo and let n C RN be a bounded domain whose boundary is of class C. Then the embedding W 1,P (0) -- LP (1)

is compact.

Proof. For every xo E M there exist a neighborhood Ax0 of xo, local coordinates y = (y', yN) E RN-1 x R, with y = 0 at x = xo, and a function f E C (QN_1 (0, r)), r > 0, such that n fl Ayo = { (y', YN) E f n A.. : y' E QN-1 (0, r) , 1N > f (y') )

By taking r > 0 smaller, if necessary, and t > 0 sufficiently small, we have that the open set U (xo, r, t) defined in (10.10) is contained in A.,a, so that, by eventually replacing A,,, with U (xo, r, t), without loss of generality, we may assume that

(11.11) nn Axa

= { (Y1', Y!N) : 1/ E QN-1 (0, r) , f (Y1') < YIN 0 l(x)l' dx < 2PM45 tl\f,a

f

as S -> 0, and so the result follows from Exercise 11.18.

0

11.2. Embeddings: p = N The heuristic argument at the beginning of the chapter shows that when p > N, we cannot expect an inequality of the form IIuIILs(RN) Lq (RN)

that is, inequalities of the type IIUIILQ(IlCN) 5

CIIUIIwl.p(gN).

We now show that this is the case when p = N. We begin by observing

that when p / N, then p* / oo, and so one would be tempted to say W1,N (RN), then u E LOO (RN). For N = 1 this is true since if that if u E u E W1,1 (R), then a representative u is absolutely continuous in R so that

z

-C(x) = U (O) +

jii(s) ds,

and since u = u' E Ll (R), we have that I is bounded and continuous. For N > 1 this is not the case, as the next exercise shows.

Exercise 11.22. Let n = B (0,1) C RN, N > 1. Prove that the function U (X) := log (log (1 + ICI )} '

x E B (O,1) \ {0} ,

belongs to W1,N (B (0, 1)) but not to L00 (B (0, 1)). However, we have the following result.

Theorem 11.23. The space W1,N (RN) is continuously embedded in the space Lq (RN) for all N < q < oo.

11.2. Embeddings: p = N

329

Proof. Let u E W1,N (RN). Define v := lult, where t > 1 will be determined so that v E W1,1 (RN) By Theorem 11.2 with p = 1 and Exercise 10.37(i), N-1

(

fRN

July dx

N-1

_1 !

N

N

r dx

IvI'

N, the function u must decays faster than algebraically at infinity. Indeed, we will show that it must have exponential decay. For every m E N consider the function m-1

00

(11.14)

expm (s) :=

c 1n.s" = exp (s) - E 1n.sn,

n=m

1

s E R,

1

n=0

and let (11.15)

7N

NQN

,

where, we recall, QN is the surface area of the unit sphere, that is, ON

HN-1 (SN-1) =

N

2w 2

r(2)

11.2. Embeddings: p = N

331

The next theorem, which is due to Adachi and Tanaka [4], gives an embedding of the space W1,N (RN) into the Orlicz space generated by the function expN_1 (sN/(N-1)) We introduce the notion of Orkcz space.

Definition 11.27. Let E C RN be a Lebesgue measurable set and let [0, oo) -i [0, oo] be a convex, lower semicontinuous function such that 4) (0) = 0 and 4) is not identically zero or infinity. The Orlicz space L" (E) generated by the Orlicz function 4) is the space of all Lebesgue measurable 4)

:

functions u : E - R such that u (x)I l dx < oo \ s /l Ì

1

for some s > 0 (depending on u).

Exercise 11.28. Let E and 4) be as in the previous definition. (i) Prove that Lt (E) is(ra normed space with the norm

11u04,:=inf{s>0:

(11.16)

f

dx ro by (11.21), we have that or,

ZZ=QN n=N-1 W!

" Jr.

00

< 13N

n=N-1

yL.

v'

= ON (expN-1 y)

J°° o

f

oo

o

wN (r)

rN-1

dr

wN (r) rN-1 dr.

11. Sobolev Spaces: Embeddings

334

Combining this estimate with (11.22) and (11.24), we get (11.25)

JRN eXpN-1 (7vN' (x)) dx 0. By (11.21) and the fact that w is decreasing, we have that w (r) > w (ro) = 1 if and only if 0 < r < ro. Hence,

{xERN: v'(x)> 1} ={xERN: w(Ixl)> 1}=B(0,ro). By Proposition 16.6, weI have that i1 aNrp = GN ({x E RN : v* (x) > 1})

=RCN({xER": v(x)>1}) <J

vN(x)dx,

{v>1}

and so we have proved (11.19) and, in turn, (11.17). By (11.17), the number s = IIVUIILN(RN;RNxN) is admissible in the defi0 nition of IIufl*, (see (11.16)), and so (11.18) follows.

The previous theorem is complemented by the following exercise, which shows that (11.17) fails for all 7 > 7N.

Exercise 11.31. Suppose N > 2 and let r > 0. (i) Construct a sequence C W1,N (RN) of nonnegative radial functions such that supp u, C B (0, r), IlovnllLN(RN;RNXN) = 1, and

fRN expN-1 7N< (x) dx

-00.

IIvnIILN(RN)

(ii) Prove that for every open set 1 C RN and for every C > 0, the inequality

eXpN-1 7N

Iu (x) I

N'

IIVt IILN(Q;RN)

IIUIILN(n)

dx < C (IIIILN(fl;RN))

fails for some u E Wo'N (St) \ {0}.

N

11.3. Embeddings: p > N

335

Remark 11.32. The analog of this result for bounded domain can be found in Exercise 12.20. See also Remark 12.21 for more references on this topics.

11.3. Embeddings: p > N We recall that, given an open set SZ C RN, a function u : SZ -* R is Holder continuous with exponent a > 0 if there exists a constant C > 0 such that

lu(r)-u(y)I 0. (i) Prove that if a > 1 and if fZ is connected, then any function that is Holder continuous with exponent a is constant.

(ii) Prove that the space C°'a R, 0 < a < 1, is a Banach space with the norm IIUIICO.a(

}

sup Iu (x) I + su z, ysup#7J

lu ( I

-

yl(01

Note that if SZ is bounded, then every function u : SZ -+ R that is Holder

continuous with exponent a > 0 is uniformly continuous and thus it can be uniquely extended to a bounded continuous function on RN. Thus, in the definition of CO,' (SZ) one can drop the requirement that the functions are bounded. The next theorem shows that if p > N, a function u E Wln (RN) has a representative in the space C0,1 P (RN).

Theorem 11.34 (Morrey). Let N < p < oo. Then the space W1,p (RN) is continuously embedded in C°"- a (RN). Moreover, if u E 1v'.P (RN) and u is its representative in C°'1 p (RN), then lim u (x) = 0. jxI-'oo

Proof. Let u E W4 (RN) n C°° (RN) and let Qr be any cube with sides of length r parallel to the axes. Fix x, y E Q,, and let

g(t):=u(tx+(1-t)y), 0 N

337

consider a cube Q1 containing x and of side length one. By (11.26) we get (11.28)

lu (x)I s ItQI I + Iu (x) - tQ' 1:5 L1' (x) dx I + C (N, p) L

S IItIIL,(Q,) + C (N,p) IIVuIIl (RN;RN) s C (N,p) IIuIIW,.p(RN) ,

where we have used Holder's inequality. Next we remove the extra hypothesis that u E C' (RN). Given any u r= W 1'P (R'V ), let x, y E RN be two Lebesgue points of u and let u£ := u * rp,, where V£ is a standard mollifier. By (11.27) we have that

Iu£(x)-uE(y)I N

339

Exercise 11.37. Let fZ C RN be an open set and let p > N. Prove that if IQ is an extension domain for W 1'' (0), then the space W l,-' (f') can be N

continuously embedded in C0,1- P (S2) and that if u E W i.p (1) and u is its N representative in CO,1- v (), then ii is differentiable at GN-a.e. X E fl and, if Sl is unbounded, u (x) = 0. lim 00

ZEU, IsH

Exercise 11.38. Let St C RN be an open set and let p > N. Prove that if 11 is a bounded extension domain for W 1p (Ii), then the RellichKondrachov theorem (Theorem 11.10) continues to hold for p > N, that is, for all 0 < a < 1 - n the embedding

W' (0) -' Co,a P) is compact.

Exercise 11.39. Let Ii C RN be an open set and let p > N. Prove that if Il is an extension domain for W 1,p (Ii) with finite measure, then the embedding

WI.P (a) -+ II' (ci) is compact.

The following exercise shows that if u E W1,N (f2), where i C RN is an open set, then one cannot have differentiability C,v-a.e. in i (if u E W 1P (S2), with p > N, then this is true by Corollary 11.36).

Exercise 11.40 (Indian mystic's bed of nails). Let Q :_ (0,1)2 be the unit cube in R2, subdivide it into 4" subcubes each having side length 2-k, and {x(1),... , x(4k) } be th e set of centers of the subcubes. Define let Ck e-2k

gk

(t)

in

t+ 1+e-2k ,

e-2 k I

n

+

e-2k

if

Ix

- t 0 such that for every Lebesgue measurable set E C RN the set T (E) is Lebesgue measurable and GN

(E)) < C (GN (E))- A

(E;RNxN).

In particular, i has the (N) property; that is, it maps sets of LN-measure zero into sets of £N-measure zero.

Proof. In the proof of Theorem 11.34, we have seen that for every cube Q with sides parallel to the axes and for every x, y E Q, NP J

(x) - Ì`4 (y) 1 < C (N,p) (diamQ)1

IIV'&=IILP(Q,;RN)

foralli=l,...,N. Hence, (11.31)

CN (1 (Q)) < 2N (diam 1 < C (N, p)

(Q))N

(diamQ)(1- p)N

IIoTIIL

(Q;RNxN)

Next, if E is an arbitrary Lebesgue measurable set, for every e > 0 find a countable family {Q,,} of pairwise disjoint cubes such that

ECUQ,, =:E0 n and

GN (Qn) R ie differentiable at £ N-a.e. x in the (possibly empty) set Eu

x E RN . lim_sup Iu (ly)

I(x) < oo}

_

-

Proof. Assume that E,, is nonempty. Let {B (x., rn)}n be the family of all balls with xn E QN and r1 > 0, r. E Q, such that u : B (x, , rn) -* R is bounded. It follows from the definition of Eu that C E,y

00

U B (xn., rn)

.

n=1

Set Bn := B (xn, N). For every x E Bn define vn (x) := sup {v (x) : v : B,, -> R is Lipschitz,

Lip v < n, and v < uIB. }

11.4. Lipschitz Fbnctions

345

and

W. (x) := inf {w (x) : w : Bn -t R is Lipschitz, Lip w < n, and to > ulBn } .

Then (exercise) Lip v < n, Lip wf, < n, and w,, < ul B. < vn. By Exercise 11.44 and Rademacher's theorem, v,, and wn are differentiable at GN-a.e. x E B. Define Fn :_ {x E B,z : vn and wn are differentiable at x)

.

Then LN (Bn \ Fn) = 0. By Exercise 10.37(iv) applied to v - wn, we have that Vvn (x) = Vwn (x) for LN-a.e. x E Bn such that v,, (x) = wn (x). Set C. := {x E F. : vn (x) = wn (x) and Vvn (x) 96 Vwn (x)} . Then RCN (Gn) = 0. Hence, the set w

E:= U n=m

has Lebesgue measure zero.

We claim that u is differentiable at all x e R. \ E. To see this, fix x E Et, \ E. By definition of Eu, there exist r > 0 and L > 0 such that ju (y) - u (x)I < L ly - xI

(11.33)

for all y E B (x, r). Let n > L be such that B C B (x, r) and x E B,,. Since x 0 E, it follows that x E F. By definition of vn and wn and by (11.33) we have that

u(x) - nly -xj
1. We now prove that if 8 is sufficiently regular,

then it is possible to construct an extension operator C that works for all Sobolev spaces W1,P (1). We begin with the important special case in which 11 := `(x', XN) E

e-1 x R : x.Nr > f (x') } ,

where f : RN-1 -> R is a Lipschitz function.

Theorem 12.3. Let f : RN-1 - R be a Lipschitz function and let (12.1) 0:= l(x', xN) E RN-1 x R : xN > f (xr)} . Then for all 1 W 1P (RN) such that for all u E W 1'' (a), E (u) (x) = u (x) and (12.2) (12.3)

for EN-a.e. x r= St

IIC(u)IIr,.(RN) = 2IlullLVfnl /

IIVC(u)Ilp'(mN;OtNxN) S (2+Lip

f)IIVUlIr'(O;RNxN).

12.1. Extension Domains

351

Proof. The idea of the proof is to first flatten the boundary to reduce to the case in which SZ = RN and then use a reflection argument (see Exercise 10.37). We only prove the case 1 < p < 00 and leave the easier case p = 00 as an exercise. Consider the transformation

IF :RN,RN (z', ZN) '-' (z', zN + f (z')) Note that W is invertible, with inverse given by

.

tI,-1:RNRN (X', xN)

(x', xN

- f (x')) .

Moreover, for all y, z E RN,

(J)

(z)

(V,YN + f (Y))

(z',zN + f (z'))

= I (F!' - z', f (J') - f (z') - YIN + Z.) I
0, M E N, and a locally finite countable open cover {52n} of 8St such that

(i) if x E 852, then B (x, e) C Stn for some n E N, (ii) no point of RN is contained in more than M of the Stn's,

(iii) for each n there exist local coordinates y = (y',yN) E RN-1 X R and a Lipschitz function f : RN-1 - R (both depending on n), with Lip f < L, such that ERN-1 xR: yN> f(y')}. nn nfl =52nn{(y',yN)

12.1. Extension Domains

355

Exercise 12.11. Let 12 C RN be an open set such that 812 is bounded. Prove that 8S2 is uniformly Lipschitz if and only if it is Lipschitz.

Exercise 12.12. Let 11 C R be such that

Q = U In' n

where the In are open intervals such that length I,, > S for all n E N and

dist(In,Ik)>Sfor all n,kENwith n

k and for some 6 > 0.

(i) Prove that 812 is uniformly Lipschitz. (ii) Prove that the condition length In > S for all n E N is necessary to have an extension operator from W'-' (12) to W 'm (R). (iii) Prove that the condition dist (In, Ik) > S for all n, k e N is necessary to have an extension operator from W1'O° (12) to W1'O° (R).

In the next exercise, by a finite cone having vertex v E RN, length e, and vertex angle 2a E (0, ir) we mean a set of the form Ki,,R := v + RK, where R is an orthogonal N x N matrix and

K:_ {(x',XN)

ERN-1

x R: xN>x1cota}flB(0,Q).

Exercise 12.13. Let S2 C RN be an open set whose boundary 812 is uniformly Lipschitz.

(i) Prove that for every x E 811 there exists a finite cone KK,R, with vertex x, length e, and vertex angle depending on L, such that KX,R \ {x} C S2.

(ii) Prove that if 12 is unbounded, then GN (ft) = oo. The next theorem shows that domains with uniformly Lipschitz boundary are extension domains for W1.P (S2) for all 1 < p < oo. In the proof we will use the following exercise.

Exercise 12.14. Let wn : RN -' [0, oa] be a sequence of Lebesgue measurable functions and let

w(x):=Ewn(x), xERN. n=1

Assume that there exists an integer M E N such that for every x E most M terms wn (x) are nonzero. Prove that for every 1 0 such that for every bounded open set N1 (12.20)

fn exp

lu (x)I

dx

C"CN (a)

IIVuIILN(n;ItN)

for all uE Wo'N(St) \{0}. (ii) Prove that the previous inequality fails if 7 > yN. Remark 12.21. The inequality (12.20) was first established by Trudinger in [168]. In (130], Moser proved that for bounded domains the inequality (12.20) also holds for y = 7N. The proof is not immediate and we refer to [130] for more details. See also the recent paper of Li and Ruf [107] and the bibliography contained therein for some recent results on unbounded domains.

12.2. Poincare Inequalities

361

Exercise 12.22. Assume that the open set fZ C RN has finite width and let 1 v in LP (fl) for some function v E Lp (S2) with IIVIILP(f) = 1,

uE = 0.

12. Sobolev Spaces: Further Properties

362

f

Moreover, f o r every

f

Jn

v

axi

E C 1 (fZ) and i = 1, ... , N, by Holder's inequality ,0

a dx. = lim

dx = lim

loco fn

Oxi

li o Un

IVv,,kI

r

sunk dx 8x{

k-400

\ Jn jiff dxA

dx J p y fl)

0,

and so v r= W 1,' (f2) with Vv = 0. Since f2 is connected, this implies that v is constant, but since vE = 0, then, necessarily, v = 0. This contradicts the fact that IIyIILp(,) = 1 and completes the proof.

Exercise 12.24. Let 1 0 such that for all u E W 1,P (l),

(fIux) - uEI"" dxl

n

0 such that for all u E W l,p (ft),

(L1u(x) -uEIq dx ° 0, the set f2i is convex and

dist( ,8Sl)=S>0. Consider a sequence of standard mollifiers {cps}E>O. For 0 < e < d define us := u * cps in fl.. Then us E C00 (17a) by Theorem C.20, and so, if (12.23) holds for the pair u£, SZa, we get

f,l,u,(x)-(,u,).,I'"dx ELN (ft,,.),

fd

fUJ (0,YN)9(VN) 1

dyNI

d

(

9(yN) dyN0

-1+1a

do

u (0, YN) 9 (YN) dyN -

< sLN (nn)

Jst., u (y) dy \ Jo l

g (yN) d11N 1

1

//

`+P

g (yN) dyN 1

(10

(SZn))

/I

while if f d" g (yN) dyN < C,CN (Stn), then

fd- u(0,yN)9(yN) dyN

ZZ

p

\Jo

J0"9(1!N) dyN

<MUI

/

9(yN) dyNI

1

< M (ELN (Stn)) P,

9 (YN) dyN)

` 0

where

M :=

a

Jul+maxlouI+1 0. Then there exists a constant C = C (N) > 0 such that for all u E W1"p (S2),

I

R ) N-1

Iu (x) - of I" dx < CRP (

r R

in lVu (x) Ip dx.

Proof. By the Meyers-Serrin theorem (Theorem 10.15), without loss of generality, we may assume that u E COO (52) n W IM (Sl) and that xo = 0. Hence, we may write yESN-1,OSp 0: pyE 11}.

12.2. Poincare Inequalities

371

Note that for all y E SN-1 we have that f (y) y E Otl and 2r < f (y) < R by (12.39).

Step 1: Assume that u vanishes in B (0, r). Since u (ry) = 0 for all y E SN-1, using polar coordinates and the fundamental theorem of calculus, for

all r < p < f (y) we have that P OP

U (P?I) = u (Py) - u (ry) =

(ty) dt,

J

and so by Holder's inequality, Iu (Py) r=

fu r

(Jr,!±(y) I d t ) < p

0+ (see Theorem C.19), by Exercise 13.3, IDuI (cl') < limiinf IDuEI

(13.12)

Since 12E D 1' for all 0 < e < dist (Q', 8Q), it follows that IDuI (SZ') < lim inf IDuEI (f2E)

.

Letting 11' / SZ and using Proposition B.9 gives (13.13)

IDuI (SZ) < lim inf IDuEI (QE)

,

E--,O+

which, together with (13.11), gives (13.4). Step 2: Fix an open set fl' C fl, with dist (fZ', On) > 0 and IDuI (8St') = 0. In view of (13.12), it remains to show that lim sup I Du, I (ii') < IDuI o+

We proceed as in the previous step, with the only change that we now consider E C,° (SZ'; RN) such that II-,D IICo(nr;mN) 5 1. If 0 < e < dist (SZ', 8t ), then by (13.9), supp,pF C (0')E := {x E SZ : dirt (x, St') < E } C SZ,

and so (13.10) should be replaced by IDuEI (f') u in Li° (RN).

13.2. Approximation by Smooth Functions

383

We observe that if u = XE, then (13.14) gives lim

a_.0+

IV (XE)EI dx = P (E).

N

This identity can be used to define the perimeter of a set. This approach was taken by De Giorgi [46], [45], who considered a somewhat different family of mollifiers.

Exercise 13.12. Let f1 C RN be an open set, let u E BV (0), and let {u,,} C BV (11) be a sequence of functions converging in L L (Ii) to u and such that IDul (fl)

n I Dun.I (s1) .

Prove that D2u j Diu in Mb (0) for all i = 1, ... , N and that IDul (K') = n-oo lien IDu,,l (n') for every open set 11' C Il, with I DuI (as1' fl n) = 0.

Exercise 13.13. Let S1 C RN be an open set and let u E BV (fl). Fix

1 0 and ID:uI (an') = 0, lin+ o fn'

I axe I

dx = l

(iii) Prove that if u E Lt1'0 (RN) and its distributional gradient Du belongs to Mb (RN; RN), then aue

lim RN f axi e-0+ I

I

dx = IDsuI (RN).

13. Functions of Bounded Variation

384

Exercise 13.14. Let Q C RN be an open set. Prove that if u E BV (f2)

and¢ECO°(f),then ¢uEBV(f)and ID (Ou) I (n) R, defined by w

(x)

U (x) { V (x)

if xpr > 0, if xN < 0,

belongs to BV (RN).

Exercise 13.24. Let 0, 0' C RN be open sets, let' :1Z' --+ 0 be invertible, with fi and (D-1 Lipschitz functions, and let u E BV (1k). Prove that uofi E BV (V). 1The key point here is that given two sequences of real numbers {a_} and {b, ), in general one can only conclude that liim sup (a. + bn) < lim sup nn + Jim sup b. ,

n-oo n-00 ft-00 with the strict inequality possible. However, if one of the two sequences converges, say {an}, then lim sup (a., + 84,) = lim an + lint sup b,,. n-eo n-*o n-cc

13.4. Coarea Formula for BV Functions

397

13.4. Coarea Formula for BV Functions In this section we prove the coarea formula for a function u E BV (a). This formula relates the total variation measure IDul with the perimeter of its superlevel sets {x E 12 : u (x) > t}.

Theorem 13.25 (Coarea formula). Let fI C RN be an open set and let u E L10C (Sl). Then

V(u,1l)=J P({xE 1 : u(x)>t}, 12) dt. R

In particular, if u E BV (Q), then the set {x E 12 : u (x) > t} has finite perimeter in 12 for G1-a.e. t E R and (13.40)

IDul (1l) = J P ({x E 12 : u (x) > t}, 11) dt. R

We begin by proving the theorem for linear functions.

Lemma 13.26. Let u : RN - R be an affine function of the form

xERN, where a E R and b E ]RN. Then for every Lebesgue measurable set E C RN the function

t E R - RN-1 (E n u 1({t})) is Lebesgue measurable and (13.41)

IbI GN (E) =

J

RN-1 (E n u-1({t})) dt.

Proof. If b = 0, then there is nothing to prove, since both sides of (13.41) are zero. Thus, assume that b 54 0. Hence, u-1 ({t}) = {x E RN : a + b . x = t} is the translate of a hyperplane in RN, and so the measure xN-1 restricted to u-1 ({t}) coincides with the measure RCN-1. It follows by Tonelli's theorem

that the function

t E R f-, ?LN-1 (E n ul ({t})) is Lebesgue measurable. Since GN and 7-1N-1 are both rotation invariant, without loss of generality we may assume that n = e1. By the change of


398

variables t = JbI a + a we obtain

(En u-1 ({t})) dt = IbI = IbI

futxN-1 (En u-1({jbI s + a})) ds

J %N-1 ({x E E : xl = s}) ds R

= IbI J LN-1({x E E : xl = s}) de R

_ JbI LN(E), where in the last inequality we have used Tonelli's theorem.

Remark 13.27. Note that Vu - b. Exercise 13.28. Let fl C RN be an open set and let u E LloC (Sl). Prove that the function

u(x)>t},Q) is Lebesgue measurable.

We are now ready to prove Theorem 13.25. In what follows, for t E R we set

SZt:={xEft:u(x)>t}. Proof of Theorem 13.25. Step 1: We claim that if {u,,} C LbC (St) converges to u in Li C (f2), then there exists a subsequence (u,,,} of {u,,} such that P (flt, fZ) : lim inf P (S2t k, n) k-*oo

f o r G1-a.e. t E R and JR

P ((Zt,11) dt t}. Indeed, for every x E SZ and u E N,

f

IXn, (x) - Xn, (x) I dt

= Jmin{u,.(s),u(x)}

dt = Iun (x) -,u (x) I .

Hence, by Fubini's theorem, for every ft' CC f2,

jf

,

Ixn, (x) - xn, (--)I dxdt = fol IUn (x) - u (x)I dx -r 0.

By considering an increasing sequence fj' / ft, with fZj CC fl and using a diagonal argument, we may find a subsequence of {u,, } such that Xni k

Xns in

L'10i

(11) for

1-a.e. t E R. It follows by Exercise 13.3 that

p (ftt, fZ} < lim inf p (SZi k, ft) k-+oo

13.4. Coarea Formula for BV Functions

399

for G1-a.e. t E R. In turn, by Fatou's lemma, P (nt, Cl) dt < lim inf k-1oo

JR

f

P (f k, Z) dt.

R

Step 2: We prove that if u is piecewise affine in the sense of Definition 10.32, then

I. ID (xn,)I (f) dt = f Ivul dx. To see this, let

be N-simplexes with pairwise disjoint interiors e

such that the restriction of u to each Ai is affine and u = 0 outside U a;, 1=1

O;, where asERand bi ERN. Then by the previous lemma, applied to the affine function ui (x) = ai + bi x, x E RN,

f

IDul dx =

Ibil £N (CZ n o;) _

1-IN-1

(fZn, 1nu-1 Qt})) dt.

a_1 fIR

i_1

Since by Exercise 13.5 (note that in Cl n a; the function u coincides with the smooth function v (x) := a; + bi x, x E RN) NN-1 (a n Ai n u-1 ({t})) , ID (xct)I (l n Di) = we have

E jnN_1 () nt1nu 1({t})) dt=

jlD(x)l(ni) dt i=1

= L1D(x)1(c1) dt. Step 3: We prove that if u E coo (CZ) n W,.',' (CZ), then

fit

P A, fZ) dt < f lVul dx.

It suffices to assume that fn Ioul dx < oo. Fix an open set Sl' CC Q. By Remark 10.34 we may find a sequence C W1,1 (CZ') of piecewise affine functions such that u,, -> u in W 1,1 (CZ'). By the previous two steps (with f? replaced by CZ') we may find a subsequence (not relabeled) of {u,,} such

that JR D 7rsa,) I (CZ') dt < lim oof

j ID (% (n% } I {St') dt

_ lim. f IVu,,l dx = Sz,

f

IVul dx
0. Then for CN-a.e. x E 0, 00

Xsa, (x) dt,

U (X) = 0

and so, by F ubini's theorem 00

u (x) div

(x)

dx =

J°xa= (x) div r 00

=J

f

(x)

Xc, (x) div 415 (x) dxdt

f fn 00

=

div ' (x) dxdt.

o

Similarly, if u < 0, then for £ "-a.e. x E Sl, U (X) =

f

to 00

dtdx

(xn, (x) -1) dt,

13.5. Embeddings and Isoperimetric Inequalities

401

and so

Ju(x)div4'(x) dx =

f f°(x(x) -1) div 4' (x) dt dx Z

l J

=

(X., (x) -1) div 4 (x) dxdt 0

=J

div 4 (x) dxdt, 0o

Izt

where we have used the fact that f. div 4' dx = 0 by the divergence theorem

and the fact that - E q° (fi; RN). In the general case, write u = u+ - u-. Note that for t > 0,

Sit={xEf2: u(x)>t}={xE0: u+ (x)>t}, while for t < 0,

fit={xEfi: u(x)>t}= {xEfl: -u (x)>t}. Hence,

If,

u (x) div 4' (x) dx = / (u+ (x) - u (x)) div 4' (x) dx r

= JR fl

div 4) (x) dxdt,

which proves the claim.

Step 6: We show that for u E Lj (fi), V (u, fi)
2, N 1* = N-1'


402

Theorem 13.30. Let u E LloC (RN), N > 2, be a function vanishing at 1

infinity such that its distributional gradient Du belongs to Mb (RN; RN) Then them exists a constant C = C (N) > 0 such that

(LNIU(x)I' dx}

0 (independent of R and xo) such that for all u E BV (S1),

(flux) - u&iI'* dx l 3

< C IDul (0).

As a corollary of Theorem 13.30 we can prove the following isoperimetric inequality (see also Theorem C.13).

Theorem 13.40 (Isoperimetric inequality). Lei E C RN, N > 2, be a set of finite perimeter. Then either E or RN \ E has finite Lebesgue measure and (13.42)

min{LN(E),LN (Rv \E)}

for some constant C = C (N) > 0.

l (CN (BR n E)) 1Z

= 2 min{LN(BRnE),LN(BR \E)}T". The other case is analogous. By applying Poincare's inequality for balls (see the previous exercise), we get that the left-hand side of the previous inequality is bounded from above by C (N) ID (XE)I (Bp), and so

2min{LN(BRnE),LN(BR\E)}1 0 such that 1W

(LNIU(x)11 ,

dx)

For every t > 0, consider the function ut := min {u, t} and let

f (t)

UN ( ut

(x))1* dx

1

.

Note that f is finite, since

J (u(x))dx < tl.-1 N

f

u (x) dx + t1GN (A) < oo.

{u 0, by Minkowski's inequality,

0< f(t+h)-f(t)r2>...>rp. Put it := 1 and discard all the balls that intersect B (xl, rl). Let i2 be the first integer, if it exists, such that B (xi2, riz) does not intersect B (xi, ri). If i2 does not exist, then set I :_ {i1}, while if i2 exists, discard all the balls B (xi, ri), with i > i2, that intersect B (xiz, ri2). Let i3 > i2 be the first integer, if it exists, such that B (xi3, ri3) does not intersect B (xiz, riz). If i3 does not exist, then set I :_ {i1, i2}; if i3 exists, continue the process. Since there are only a finite number of balls, the process stops after a finite number of steps and we obtain the desired set I C {1, ... , e}. By construction, the balls B (xi, ri) with i E I are pairwise disjoint. To prove the last statement, let x E E. Then there exists i = 1, ... , e such that x E B (xi, ri). If i E 1, then there is nothing to prove. If i V Z, then B (xi, ri) is one of the balls that has been discarded and thus there exists j E Z such

13.6. Density of Smooth Sets

409

that rj >- ri and B (x3, r3) n B (xi, ri) # Ql. Let z E B (x3, rj) fl B (xi, ri). Then

Ix - xjI P(At) n-00


412

for Gl-a.e. t E R, where for t E R,

At:={xERN: XE(x)>t}, At:={xERN: ufn(x)>t}. Since

At _

E ift 1,

it follows that

lim inf P (At) > P (E) -oo

(13.48)

for £'-a.e. t E [0,1]. On the other hand, by the coarea formula, Fatou's lemma, and the fact that 0 < uEn < 1,

P (E) = lim

I VuEn I dx = lim

n-oo J fo Together with (13.48), this implies that

1

P (At) dt >

r

J

lim inf P (At) dt.

lim inf P (A') = P (E) n-oo for Gl-a.e. t E [0, 1]. Since uEn E C°° (RN), it follows by Sard's theorem that the sets

{xERN: U",(x)=t} are C°O-manifolds for Gi-a.e. t E [0, 1J. Hence, we may find t E (0,1) such

that lim inf P (At) = P (E) and the sets {x E RN : uEn (x) = t} are C°O-manifolds for all n E N. Define En := Ant. It remains to show that GN (EnLE) - 0. To see this,

note that

UEn(x)-XE(x)>t for X E En \ E, while

XE(x)-uen(x)>1-t for x EE \ En. Hence, f/ N

IUEn - XEI dx >

JEn\E

IUEn - XEI

dx+ fE\En

IuEn - XEI dX

2: tGN(En\E)+(1-t)GN(E\En), which implies that CN (EnOE) -> 0, since 0 < t < 1. This concludes the proof.

0

Exercise 13.47. Prove that in general we cannot approximate E with smooth sets contained outside E (or inside). Hint: Consider the set A given in Exercise 13.8.

13.7. A Characterization of BV (12)

413

13.7. A Characterization of BV (f2) In this section we give a characterization of BV ((2) in terms of difference quotients. This is the extension of Corollaries 2.17 and 2.43 in higher dimension. A similar characterization has been given in Section 10.5 for the Sobolev space W1" (S2), 1 0, let

f1h,i:={xEf2: x+heiEf2}. Theorem 13.48. Let Sl C RN be an open set and let u E BV (12). Then for every i = N and h > 0, lu (x + hei) - u (x)I dx < h IDiuI (R)

J and

(13.49)

u (x + h h) - u (x)I dx = ID:uI

hli lim

(11).

nh,t

Conversely, if u E L' (S2) is such that

r

(13.50)

liminf h-Of Jflh i

Iu(x+hei)-u(x)I h

dx < 00

for every i = 1, ... , N, then u E BV (S2). Proof. The proof is very similar to the one of Theorem 10.55 and we only indicate the main changes. Step 1 of that proof remains unchanged. In Step 2 we assume that 0' satisfies the additional hypothesis IDuI (Oil') = 0 and we proceed as before to obtain (10.33), which now reads (13.51)

r 8u£ 8xi

(x) dx < limo f

Iu (x + hei) - u (x)

dx.

Inh's

We are now in a position to apply Lemma 13.10 to obtain (13.52)

I Diul (St')
v' (x, h) for LN-a.e. x e RN and L1-a.e. h > 0, i = 1, ... , N. It follows that for ,CN-a.e. x r= W' and LI-a.e. h > 0 the function v' (x, h) coincides with the function w1. (x, h) :=

0, u (x)

1

.

hB+s

Hence, u E Bs,P,e (RN) and IIt

- ullBe.n.e(RN) -+ 0.

Exercise 14.4. Prove that Bs"°° (RN) is a Banach space. Next we discuss the density of smooth functions in

Bs,r,e (RN).

Proposition 14.5. Let 0 < s < 1, 1 < p < oo, and 1 < 0 < oo. For any u E Ll (RN), let uE :_ SPa * u, where (pr is a standard mollifier. Then (14.3)

IueIB8.D.e(RN)

0 and (14.4)

olio IuEIBe.v..(RN) = IuIBs,D.e(RN)

Moreover, if p < oo, 0 < oo, and u E gs,n,e (RN), then (14.5)

lim o+IuE - ulB,.P.e(RN) = 0.

In particular, if p < oo and 0 < oo, then Coo (R1') fl Bs,P,O (RN) is dense in Bs,P,e (RN).

Proof. Since 4z uE _ E * / u, by Theorem C.19 we have that (14.6)

IIOs

IILP(RIN)

Oande>Oand (14.7)

ue IIL-V(R.V)

11A- u 11LV(RN)

as a -> 0+ for all h > 0. It follows from (14.6) that (14.3) holds. In turn, limSUP IuEIBs.v.B(RN) 6--,Q+

IulBs.v.9(RN)

14. Besov Spaces

418

To prove the opposite inequality, assume first that 0 < oo. For h > 0 and e > 0 we define fE (h)

II2nuEIILP(RN)'

h1+490

f (h)

h1+s8 FA'uIILP(RN)

Since fe (h) -> f (h) for every h > 0 by (14.7), by Fatou's lemma we have

that 00

00

10

f (h) dh = fo

00

lim fe (h) dh < lim inf

t -«0+

fe (h) dh. a

Thus, (14.4) holds.

for all h > 0 by

If 0 = oo, then since (14.7), we have that h$

III

eEm h8

LP(RN}

lim inf sup s Ilk cell N 4F-p0+ h>0 h LP(R )

for all h > 0. It follows that sup

a

a

< lim inf sup LP(RN) - e--'0+ h>0 h

I

h>o h

11'&

IIL-I(RN)

and so (14.4) is satisfied.

Assume next that p < oo, 0 < oo, and u E B',P,e (RN). By Theorem C.19(iv), (14.8)

IIA4 ne - At uII

LP(RN)

-' 0.

For h > 0 and e > 0 define the functions 1

ge (h) := h1+ 0 lohue

- a,

g (h) == h1+ae

uIIe

L P(RN)

'

I.

By hypothesis g E L' ((0, oo)), and by (14.6), Minkowski's inequality, and the convexity of the function Iyle we have that ge (h) 5 leg (h)

for all h>0. SincegE(h)->0forallh>0by(14.8),weareinapositionto apply the Lebesgue dominated convergence theorem to conclude that (14.5) holds.

14.2. Dependence of B',P,e on s

419

(RN) Exercise 14.6. Let f : R - R be a Lipschitz function and let u r= Ll 10

be such that IUIB,,p,a(N) < oc for some 0 < 8 < 1, 1 < p < oo, and 1 < 0 < oo. Prove that If o UIB5.P.e(RN) :5 (Lip f) I t6I B..p,o(RN)

14.2. Dependence of B',P,e on s In this section we prove that for 0 < t < s < 1, W1,P (RN) C Ba,P,O (RN) C Bt,P,e (1N)

Theorem 14.7. Let 0 < t < 8
0 such that IUIBa,P,B(RN) + C IIUIIL.(RN)

IUIBt.P.e(RN)

for all u E B',P,B (RN). In particular, Bs,P,e (RN) C Bt,P,e (RN).

Proof. Assume first that 1 < 0 < oo. Then by Exercise 10.54, the fact that hs < ht for 0 < h < 1, and Remark 14.2, we have r poo

hdh \ a

r`J

x h

(14.9) I

l+te 1

dh uJJO

LP(RN)

h1+ts) +

(/

h

oo

00

6

hl+ )

< (J0

dh h1+te}

g

+ 2 IIIIILP(RN)

U

1

h1+te

dh)

This proves the desired inequality when 1 < 0 < oo. When 0 = cc, it is enough to replace the integrals by suprema, precisely,

o 0 such that IkIB+.P.a(RN) !5 C IIUIIW1.P(RN)

for all u E W 1,P (RN). In particular, W 1,P (RN) C B',P,e (RN).

1 4. Besov Spaces

420

Proof. If 1 < p < oo, then by Theorem 10.55, h

(14.10)

I1Ai UIILP(RN) < h

11

axi LP(RN) .

The same inequality holds for p = oo, since in this case u has a representative u that is Lipschitz continuous by Exercise 11.46.

Assume first that 1 < 0 < oo. Reasoning as in (14.9), by (14.10) and Remark 14.2, we have (14.11) dh 0

II

a

1

IILN(RN)

00

(8

(L

+ 2 11U16(RN)

1)9)

hl

\ Ji

g

h1+8° dh1

1

+2

"U = ((113) 0) 1 II ax

( TO1

IIUIILP(RN)

IILP(RN)

while if 0 = oo, sup h8 h>0

( RN )

 fa. Letting f, / f gives the desired result.

0

Exercise 14.12. Prove (14.13)2 and (14.14). Exercise 14.13. Let g : [0, oo) - [0, oo) be a Lebesgue measurable function such that IIgII,,,,. < oc for some so E (0, 1).

(i) Prove that if g is decreasing, then there exists eli

o IIgLI,,L = him g (h).

What happens if we remove the hypothesis that g is decreasing?

14.3. The Limit of Ba.P,e as s - 0+ and s - 1-

423

(ii) Prove that if EfL is increasing, then there exists lim II9Ila,oo

lim g (h) = h-o+ h

What happens if we remove the hypothesis that 2V is increasing?

Lemma 14.14. If u E LP (RN), 1 0 we may find a function v E CC (RN) such that

(LN lu (x ) - v (x)Idx)

(14.16)

Hence by Minkowski's inequality, IQhu(x)I P \P
0 such that for every u E LP (RN), 1 0, h

11'&lUIL

(RN)

h

f Ila,UlI

P(RN)

Proof. We only prove the case 1 < p < oo. By Fubini's theorem we have that for RCN-1-a.e. x' E RN'' the function u (xz, -) belongs to LP (I8). Fix any such x= E RN-i. Applying Lemma 14.18 to u(xi, ) gives h

C Ilahu (4, ) IILa(R) < h

p

4

II' &7U (x, -)IILp(R) d77.

14. Besov Spaces

428

Taking the norm in LP (RN-1) on both sides in the variable x; and using Corollary B.83 and Tonelli's theorem yields 1Jp

C II

II

LP

C W (JRN-i

c

dldxi

Il'&i u (xi, .) IILP(R)

p

P

0

Jh

W

IIazUIILP(RN) dal.

Exercise 14.20. Prove the case p = oo. Hint: Use the definition of essential supremum and Fubini's theorem. Exercise 14.21. Let 1 < 91 < 82 < oo.

of nonnegative numbers, prove that

(i) Given a sequence 00

oo

Ean n=1

Eany (?z=I

(ii) Given an increasing function g : [0, r) 10, oo), r > 0, and q > 1, prove that there exists a constant C > 0 independent of g, q, r, 81, and 02 such that

(fr

62

\hq}

7

dh

0 define the function

U(x,h) :=

rh

Ja

u(x+y)co (h)

dy.

Since

ds=1,

hJoh'Plhl by the continuity of u we obtain

liin U (x, h) =

h

fh m

u (x + y) cP

hdy = u x) .

By the fundamental theorem of calculus we get Ph

u (x) = U (x, h) -

(14.24)

We now compute au x'{

.

Since 9

have

\/

au (X, )

dC.

= cp (1) = 0, by Theorem B.53 we

f u(x+y) w'\ } dy

aUa

_ - 3 J u(x+y) [e,() +yco' (f)J dy _-g

f

u (x + y)

ay (4Yc

\

)) dy.

14.5. Dependence of B',P,O on s and p

431

Integrating by parts and using Fubini's theorem and again the fact that cp (1) = 0, we obtain OU(( 2

f f u'(x+Y)mo ( ) dy

f'u,(X+Y)w (

) Jo

dr dy

where

T:= {(y,77) ER2: 00. Proof. By Lemma 14.23, for all x E R and h > 0 we have

Iu (x)I < -

jh

Iu (x + y)I dy

+Cf

( 14.26)

0

j j jWu 0

(x + y) I drjdydc

0

f (x) + g (x)

.

To estimate the norm of the function f in LQ, for all x E R we write C

f (x) =

fI x[o,h[ (y) Iu (x + y) I dy =

fI X(o,h) (z - x) Iu (z) I dz,

where we have used the change of variableskz = x + y. Applying the general form of Young's inequality (see Theorem C.16) with 1

r

:= 1

-fl

p

q -1)

(note that the roles of q and r are exchanged here), we obtain (14.27)

h

Ilf IILQ(m}

f

m

lu (x)Ip dx) p

hpC qf1R

Iu (x)I

fR IX[°'h) (x)

I.

dx )

n dx) p

Similarly, by Tonelli's theorem and the change of variables z = x + y, h

g (x) = C f

ff £

3

X[°,EJ (z - x) Ia"14 (z) I

dzdrjd.

14.5. Dependence of B*,P,e on s and p

433

Using Corollary B.83 twice and reasoning as in (14.27), we get

ff

f1

h

II9IIL4(R) 0, where q' is the Holder conjugate exponent of q. Define

a:= +--p 1

(14.34)

1

1

9

q

and

IF (y, 0 :=

g (x)

T-00

(y-x+t)

2

dx,

y E R, e > 0.

By Tonelli's theorem, the change of variables y = x + z (so that dy = dz), and Holder's inequality applied twice, we have that

f

00

f (x) 9 (x) dx = 1 and 9' > p'. Hence, if 0 > 1, by Corollary B.83 (where the exponent p there is replaced here with 0' > 1) we have that P

Z=

dS

[r° UR Iba41 (y,c)IP' dy

f

ff

0. Thus, in both cases we obtain

Z 1, we are in a position to apply Proposition C.31 (with N = 1 and where the numbers a, p in the proposition are replaced here with p - 4 and q') to conclude that the right-hand side of the previous inequality is bounded by C II9II Lv' (R)

Step 3: If p = 1, then p' = oo, and so for every fixed t; > 0, IIÌ'

esssup yER

f

g (x) by (x) dx,

R

where 1

by (x)

(x)

(y - x + b)2'

x E R.

14. Besov Spaces

440

Let g* : [0, oo) - (0, oo) and h; : [0, oo) -> (0, oo) be the decreasing arrangement of g and hy, respectively. Then

h; (z) = (z

re-

z > 0,

and so, by the Hardy-Littlewood inequality (see Theorem 6.13),

f

00

g (x) by (x) dx
0,

([0

T

e'

9* (z)

JOO

= C II9IIL¢(R) , where we have used (14.34), Proposition C.31 in the second inequality, and Theorem 6.15 in the last equality. If 8 = 1, then by (114.36), C II9RII1,4'((0,00))

2 < esssup l 2- 9 £>o

l

9# (z) )2 dff

o

(z +

] dzJJ/

1

< (>O

II9'IIq((0,Q)) (L°° (z

(

C II9'IILQ'((0,00)) = C II9IILQ'(R)

)

dz Y

where we have used Holder's inequality and Theorem 6.15.

Step 4: Finally, if p > 1 and 8' < p', then 00

0. We claim that

1

1P 1q q

P

1

r:

Indeed, by (14.39) and the facts that N > 2 and p < q we have that (14.49)

1-t-(P

(N-1)(p-9)>0,

q)

and so 1-1

_

t-1

1-1

P

q

q

(14.50)

1-t

0. First, observe that 1-1- + 1 > 0 by (14.49) and the fact that 1p< t. Next, by (14.39) and the fact that p p1-N+N=t, q p p q which implies that 1-1 q,

P

and so the claim holds.

t-1 I

P

q

>I-1 +-, p 1

14. Besov Spaces

446

To prove (14.46), it now suffices to choose a number r > r. > q so close

to r. that 1- l - p + ,1-, > O and such that 11-1

q

_

P 0.

t-tq+Z+p

q-p

q

///

Since the vectors q := (q... .. , q) and r and the real numbers i and t satisfy

the relations 1 0}. If U E Wi,P (iE) with p > N, using a reflection argument (see Exercise 10.37(iii)), we can extend u to W1,P (RN) and then apply Morrey's theorem

to conclude that u has a Holder continuous representative u. Thus, the value of u on the boundary of ft, namely, on the hyperplane xN = 0, is well-defined.

The situation is quite different when 1 < p : N (unless N = 1). Note that in dimension N = 2, by Theorem 10.35 we know that there is a representative a such that u x2) is absolutely continuous for C1-a.e. X2 E R, but we do not know apriori if x2 = 0 is an admissible value, so that in general the pointwise value of u 0) may not make sense. In what follows, we need to distinguish the cases p = 1 and p > 1. We begin with p = 1.

15.1. Traces of Functions in W 1,1 (Q) Theorem 15.1. Let N > 2 and let X be the family of all functions u E L1,1 (iE) vanishing at infinity. Then there exists a linear operator Tr:X-,L1(WN-1)

451

15. Sobolev Spaces: Traces

452

such that

(i) Tr (u) (x') = u (x', 0) for all x' E RN-1 and for all u E xnc (R+}, (ii) for all u E X,

(x')I dx' < jN

(15 1)

la8u

(x) dx,

RN-1

(iii) for all V; E CC (RN), u E X, and i = 1, ... , N, (15-2)

uOx dx = Je+ where v = -eN .

%b_

dx + f

ip Tr(u)vidx',

RN-1

Proof. Step 1: Assume first that u E

LA11 (RN) n C' (RN) with Vu E

L1 (R)V;RN). Reasoning as in the first step of the proof of Theorem 11.2, for every x' E we have that R-1v-1

Iu (x', 0) I 0, we may extend Tr uniquely as a linear operator TV : X -r L' (RN-1) satisfying properties

The function Tr (u) is called the trace of it on xN = 0.

0

15.1. 'Daces of Functions in IV',' (l)

453

Remark 15.2. In particular, it follows from the previous theorem that the linear operator

Tr : W1'1 (R+) - L'

(RN-1)

is continuous and satisfies (i)-(iii).

Exercise 15.3. Adapt the previous proof to conclude that there exists a linear operator a : Wlo1-1c (R+r) --, Lloc (RN-1) such that (i)-(iii) hold (with the obvious modifications).

Exercise 15.4. Let u, v E

W1,1 (RN). Prove that for

all i= 1'...,N

ua!dx=_ f va- dx+ fTr(u)Tr(v)v:dx', f +8xi U(

Ox{

RN-1

where v = -eN. Exercise 15.5. Let u E (W1,1 (RN) and v E W1,1 (RN), where RN

{ (x', XN) E

RN-1 X R : xN < 01.

(i) Prove that the function w : RN - R, defined by W (x) :_

!

if x E RN,

v (x)

belongs to BV (RN).

(ii) Prove that the function w belongs to W1,1 (RN) if and only if Tr (u) = Tr (v).

Next we prove that the operator Tr is onto.

Theorem 15.6 (Gagliardo). Let g E L1 (RN-1), N > 2. Then for every 0 < e < 1 there exists a function u E W 1'1 (R+) such that Tr (u) = g and

f+ Iu (x)I dx < e

f

IVu (x) I

RN-1 I g (x+) I dx',

dz < (1 + e)

&.,. RN-1 Ig (x') I

Proof. If g = 0, it suffices to take u = 0. Thus, assume that g ,-6 0. By Theorem C.23 there exists a sequence {gn} C C:° (RN-1) such that gf1 -p g in V (RN-1). For each k E N there exists N1. E N such that for all n >_ NA,, C

11 9- - 9 IIL1 (RN-1) 5

2I

II9IIL1(RN-1)

.


454

and ho := 0. The

Let nk := max{Nk,Nk_1 + 1} and define hk := sequence {hk} satisfies the inequalities (15.3)

IIhk+1-

hkllL1(RN-1)

IIhkIIL1(RN-1)

< 2k IIgIILI(RN_1)

< (1 + E)

for all k E N,

II9IILI(RN-1)

for all kEN0.

Construct a strictly decreasing sequence {ti.} C (0, 1), k E No, such that

tk->0and (15.4)

Itk+1 - tkI
to, by (15.3) and (15.4) we have that X00

tk

r JR+ Iul dx Ik0

fJJff

+l

JN-1 Iul dx+dN

00

C

Itk+1 - t,I (lIhk+1IlLtN_1) + IIhkIILI(RN-1)) k=0 00

0

11.

k=O 00

jN-1

E Itk+1 - tkI k=0

<E

ax=

dx'dXN ahk

ahk+1

(x') D dzr 8xi (x1) + axt

LAr-1

II9IIL1(RN-1)

by (15.4) and (15.5). Since u is locally absolutely continuous on GN-1-a.e. lines of R that are parallel to the coordinate axes (why?), by Theorem 10.35 we have that u E W1"1 (R+). It remains to show that Tr (u) = g. Reasoning as in the proof of Theorem 15.1, we have that J,aN1

Iu (x', ZN) - Tr (u) (x') I dx'
f (x')}

.

There exists a continuous linear operator flN-1)

TY : W1,1 (n) -> L1 (an, such that

(i) Tr (u) = u on an for all u E W 1,1 (n) n c (ii) for all u E W1,1 (n),

dN-1 < i + (Lip f I2 (x) l dx, IT (u) Jill JS2 OXN (iii) for all hii E C'I (RN), u E W1,1 (n), and i = 1, ... , N, I

fu2±dx=_f!.dx+f

rJ

where v is the outward unit normal to an, that is, for RCN-1-a.e. x' E RN-1 (15.8)

v (x" f (x')) =

VIIf (x)

(VII-+

-1

(x'F V11-+ Ivx-f (x')I2

15.1. paces of Functions in W1,1 (n)

457

Proof. Set w(z) := u($(z))= u(z',zN+f (z')),

z r= R+,

where 'Y : RN -> RN is defined by

FY (z) := (z', zN + f (z'))

z E RN.

,

As in the proof of Theorem 12.3 we have that given by 'P

is invertible, with inverse

-1: RN SRN (x', XN) - (x', xN - f (x+) )

and that P and '-1 are Lipschitz, W (RN) = 0, and det V (z', zN) = 1 for GN-1-a.e. z' E RN-1 and for all ZN E R. Hence, by Exercise 10.37(iv), we have that w r= W1,1 (RN), with (15.9)

(z)

T

ax{ N(z)=

a

''N + .f (z')) + 8xN (2, zN + f (z')) 8f (z') ,

(z',ZN+f (z'))

It follows by Theorem 15.1 and Exercise 15.4 that Tr (w) E L1 (RN-1) (15.10)

LN....l Tr (w) (2)1 dz'
2, be an open set whose boundary O1 is uniformly Lipschitz. There exists a continuous linear operator

Tr : W1'1(f) -, L'

(an,?-1N-1)

such that (i) Tr (u) = u on c7S2 for all u E W 1"1(1) fl C (Sa), (ii) for all ip r= C1 (RN), u E W 1-I (12), and i = 1,... , N,

r 8x1

dx _-

au Ip

dX

+f 0Tr(U)v'dhN-1, ,

8x1 n where v is the outward unit normal to OIL

Proof. Let e, L > 0, M E N, and

be given as in Definition 12.10.

We proceed exactly as in the proof of Theorem 12.15 and define 4)7L1 Ilo, Qt,

4)o, 4f, i/if as in (12.8), (12.11), (12.12), (12.15), respectively. Then for all x E Cl we may write (see (12.17) ) (15.14)

'4)+ (x) 4)n (x)

u (x) _

c 4)k (X)

u (x) + t/i_ (x) u (x)

u,z (x) + u_ (x) . m

By (12.15) the support of 0_ is contained in ft Hence, we define the function u_ satisfies (ii), that is

Tr (u_) ;= 0. We claim (15.15)

8x1

If,

E CI (RI"). To see this, construct a cut-off function 4) E C' (Il) E CC (RN) we have that 0 E C' (Cl), and so, by the definition of weak derivative, for all i = 1, ... , N for all

such that 0 = 1 on supp L. Then for every we have ia u_

OW) dx = ax{

n

¢

dx. 8x1

Since 4) = 1 on supp t/i_, we have that = 0 on the support of u_, and so the previous equality reduces to (15.15). This proves the claim.

15.1. 'Daces of Functions in IV1,1 (fl)

461

Next we study the functions un. Fix n. By property (iii) of Definition 12.10 there exist local coordinates y = (y', yN) E RN-1 x R and a Lipschitz function f : RN-1 -> R (both depending on n), with Lip f < L, such that

fto n ft = fin rl An, where

A.:= {(Y',yN) E

RN-1 X

R: YN > f (%()I Since by (12.9) the support of un is contained in fln, we may extend un to be zero in An \f2n. Thus, we are in a position to apply the previous theorem to obtain a function Tr (un) E L1 (&A., 9`(N-1) such that (i)-(iii) of Theorem 15.8 hold (with An and un in place of ft and u). Since -

k+ In

U.

02

it follows by Exercise 15.9 that Tr (u.n) =

,,2 u

01I8An 11

k

Since the support of On is contained in fLn C An, it follows that b,IOA,, = on OA n fl f2. Hence, the same holds for Tr (un). This shows that

f

(15.16)

n

dhN-1
2 8xi

= - >2 n

J

_

u 8xiL dx + fin u_ ± dx 8xi

Jin

92!1 d x axi

8xi

d2,

+l

ast

j (

,11

sut

1I axi

(fix + >2 1"'n&A:.. n

,h Tr l(u)1 vi

'1

Tr (Yin) vz

dnN-1

dl.{N-1

for all' E C' (RN) and i = 1, ... , N, where we have used the fact that the support of 0 intersects only finitely many Stn, since {Stn} is locally finite. On the other hand, from (15.16),

JITr(u)I dnN1 n

n

2

1

n

Jana, IVunl dx.

To estimate the right-hand side of the previous inequality, we use the fact that since {Stn} is locally finite, any bounded neighborhood of every point x E RN intersects only finitely many Stn's. Hence, in 0 fl1 On, by (15.14), 4 VYSn =

} (0 VV;+ + 2vy+0.Von)

-1;+.0n

s

'

(;:5)

u

2

+ '0+0n Vu. E 02 k

k

Since

1, using (12.10), (12.13), and (12.19), we get that in On Sin,,

Iounl
0 and n, m E N we have that

JNi

ITr (un - um) (x) I dx'
roo, we obtain ITr (un - urn) (x') I dx' < 2 IDul lim sup n,rn-oo RN-1

J

(RN-1 x (0,.

15.3. Traces of Functions in W 1,P (fl), p > 1

465

Letting a -> 0+ in the previous inequality, we have that {Tr (u a)} is a Cauchy sequence in L1 (RN-1) and thus it converges to a function Tr (u) in

L' (RN-1). Moreover, by (15.1) and (15.2) we have that for all n E N,

f

v_1

IT (U-) (x') I dx' < J

N

l

a

dx,

N (x)

and for all .0 E CC (RN) and it = 1, ... , N,

8xi J un.a0dxb+

R+

Oxi

f dx+ J RN-1 ipTY(un)v4dxl.

Letting n -+ oo gives (ii) and (iii). Since W1"' (RN) C BV (R+), it follows from Theorem 15.6 that

Tr : BV (R+) - Ll (RN-') is onto.

The analog of Theorem 15.10 is given by the following theorem. The proof is very similar to the one of Theorem 15.10 and is left as an exercise.

Theorem 15.16. Let Q C RN, N > 2, be an open set whose boundary 8f1 is uniformly Lipschitz. There exists a continuous linear operator Tr : BV (fZ) - L1 (8ft, xN-1) such that

(i) Tr (u) = u on t7I for all u E BV (C2) n c (St), (ii) for all V; E C' (RN), u E BV (0), and i = 1, ... , N,

-J ODiudx+

f

bTr

(u)VidfN-1

where v is the outward unit normal to O.Q.

15.3. Traces of Functions in

Wl,P (0),

p>1

In this section we study the trace of functions in bV'4r' (ft) for 1 < p < N. We will see that the situation is quite different from the case p = 1.

Theorem 15.17. Let 1 0 such that 1 (i) Tr (u) (x') = u (x', 0) for all u E X. fl C

R

),

466


(ii) for all u E Xp,

I(x')

N-y

I"('N

//p

-T,(.) d.1//f AIN-1'
1 we have that ju

(x', 0) I'

< rf

Iu (x', XN)

00

O

Ir-'

OXN

(x , xN) dxN.

Integrate both sides with respect to x' and use Tonelli's theorem to conclude

that Iu (x', 0) Ir dx' Lq

(RN-1)

such that (i)-(iii) of the previous theorem hold (with the obvious modifications).

Unlike the case p = 1, when 1 < p < oo, the trace operator Tr : W1,P (R+) Lp (RN-1) is not onto. We now show that if u E W1P (RN), 1 2. Then there exists a constant C = C (p, N) > 0 such that for all u E L1.' (I8+), (15.20)

IVu (x)Ip dx

Imo' (uIBh1(RN_t) 0. Proof. Let cp E C' ° (RN-1) be such that supp V C BN_ 1(0,1) and cp (x') dx' = 1.

LN-I For x' G RN-1 and xN > 0 define

x ,_ xNpr

V (2, ZN)

1

xN-1 N

9 (p) dp'-

iP

JIRN_1

By Theorem C.19 (with xN in place of E) we have that for all xN > 0,

f

(15.22)

N-1

IV (x',xN)Ip do;' < J

Ig (x') I" dx'. N-1

With a slight abuse of notation, for every i = 1, ... , N - 1 we write

x' = (x'i, xi) E

RN-2

X It

By Theorem C.20 (where xN plays the role of E), for any i = 1, ... , N - 1 we have that

I

RN8cp

_, xai

X

x'

dpr

-

g (VI)

-IN

aIP('N11l

19

1

.

xN JRN-i

L (N) [g (x? - y') - g (xi' - y!', xi) ] dy',

where in the second equality we used the fact that

a R axi

x' xN

g

(Vill,

xi) dyi = g (t!"$ xi)

8cp x - 3 dz/i = 0 fJR axi xN I 1\

Since supp cp C BN_ 1 (0,1), we have that (15.23)

av axi (x) I

:5

c

N

xN

I g (xi' BN-1(o,xN)

Vi", x;

- fi) - 9 (Z - y"o xi) I

dgf'.

15.3. Traces of Functions in W1,P (n), p > 1

471

Raising both sides to the power p, integrating in x over R+, and using Holder's inequality, we get

f

P

r

(x)I dx

JR+ C7xi

C

g

Np

R+ xN

B

(x_ y') - g (xi' _

XN

J

LN_I(O,XN) I g

xP

f

dx

(x'

- y') - g (x- y', xi) IP dy'dx

kg (x' - ii) - (x- y'xi)Idydydx

,-1

f

P+N 'r'N

)xN

< C RN R+ C2,

P

I dy'

I

(N-1)(p-1)

1

473

and so, by (15.24), we obtain that the estimate (15.24) also holds for -. On the other hand, ou

8

2A

(x) = e

axN (x) -

-e 'v (x),

and so, again by (15.22) and (15.25), IP

t

I axN

(1001

dX)

1.

Exercise 15.28. Let Q':= (0,1)N-1, Q = (0,1)N, and 1 0, (a) V(X) if xN < 0. 0 By Theorems 15.1 and 15.17 and the fact that Tr (u) = 0, for all E Q' (I8N) and i = 1, ... , N we have that It N v Ox

=

JR

f+

u ax dx

lox

)

which shows that v E W1,P (RN), with

8v Ox,

(x) =

r

(x) if xN > 0,

t0

if xN < 0,

i = I,-, N. We now translate v upwards. More precisely, for t > 0 and x E RN define

Vt(x):=V(x,2N-t) Note that the support of vt is a compact set contained in RN-1 x (2, oo). For each 71 > 0, by Lemma 10.30 we may find t so small that IIu - VIII 4 l.°(RN) o of vt. Find

0 < e < a so small that II (vt)E - VIII 41.P(RN) s}), s>0.

Similarly, we define (16.2)

c (s) := RCN ({x E E : u (x) > s}) ,

s > 0.

We recall that u : E - [0, oo] vanishes at infinity if u is Lebesgue measurable and g,, (s) coo for s > 0 (gu (0) can be infinite). The proof of the following proposition is very similar to that of Proposition 6.1 and is left as an exercise. 477

16. Sobolev Spaces: Symmetrization

478

Proposition 16.1. Let E C RN be a Lebesgue measurable set and let u, v, u : E -, 10, oo], n G N, be Lebesgue measurable functions. Then the following properties hold:

(i) The function gu : [0, oo) -> [0, RCN (E)] is decreasing and right continuous, while cu : [0, oo) - [0,GN (E)] is decreasing. (ii) If u vanishes at infinity, then cu is left continuous, lim Su (s) = 0, Iim Qu (s) = S-»oo

a+oo

and Vu and c are continuous at s > 0 if and only if

GN({xEE: u(x)=s})=0. In particular, Su (s) = Lou (a) for all s > 0 except for at most a countable number.

(iii) If u (x) < v (x) for LN-a.e. x E E, then Lou < p,,. In particular, if u (x) = v (x) for GN-a.e. x E E, then pu = Q,,.

(iv) If u,b (x) / u (x) for LN-a.e. xEE, then p / Lou, Let E C RN be a Lebesgue measurable set and let u : E - [0, oo] be a Lebesgue measurable function. As in Chapter 6, the function u* : [0, oo) -[0, ool, defined by

u* (t) := inf {s E [0, oo) : Lou (8):5 t},

(16.3)

t > 0,

is the decreasing rearrangement of u.

Proposition 16.2. Let E C RN be a Lebesgue measurable set and let u : E -, [0, ool be a measurable function. Then the following properties hold:

(i) The function u* is decreasing and right continuous and (16.4)

u* (0) = esssup U. E

(ii) For all s, t > 0, u* (t) > s if and only if ou (s) > t. (iii) If u vanishes at infinity, then for all 0 < t < pu (0), t < su (u* (t)). (iv) If u vanishes at infinity, then for all t > 0, pu (u* (t)) < t, and if Lou (u* (ti)) < ti for some tl > 0, then u* is constant on [to, t1J, where (16.5)

to

pu (u* (tl))

Proof. Properties (i) and (ii) follow as in the proof of Proposition 6.3. (iii) If pu (0) > t > 0, then u* (t) > 0 by property (ii). Fix 0 < s < u* (t). Then pu (s) > t by part (ii), and since pu cu, we have S. (s) > t. Using the fact that Su is left continuous, letting s / u* (t), we get -;. (u* (t)) > t.

16.1. Symmetrization in LP Spaces

479

(iv) If u* (t) < s, then by part (ii), ou (s) < t. Taking s = u* (t) gives Lou (u* (t)) < t. Assume next that Lou (u* (t1)) < tl for some tl > 0 and let to be defined as in (16.5). For to < t < tl we have Lou (u* (t1)) = to < t and hence u* (t1)

is an admissible s in the definition of u* (t), and so u* (t) < u* (t1). But since u* is decreasing, this implies that u* (t) = u* (t1) in [to, ti].

0

Remark 16.3. If in the previous proposition we assume that u is also bounded, then by (16.4), u* (0) < oo. Hence, reasoning as in part (iv), we have that ou (u* (0)) = 0.

Exercise 16.4. State and prove the analogous statement of part (iv) of Proposition 16.2 for the function cu.

Definition 16.5. Let E C RN be a Lebesgue measurable set and let u : E [0, oo] be a Lebesgue measurable function. The function u# : RN -> [0, oo], defined by (16.6)

u# (y) := u* (aN I yI N) = inf Is E [0, 00) : ou (s) [0, oo] be a measurable function. Then the following properties hold:

(i) For all s > 0,

{yERhh1:u(Y)>8}={xEE:u(x)>s}I.


480

(ii) For all s > 0, GN ({y E R'V : u# (y) > s}) = ,CN ({x E E : u(x) > s})

(16.8)

.

In particular, if u vanishes at infinity, then so does uo. Moreover,

LN ({y E E1t : ul (y) = 0}) < LN ({xEE : u (x) = 0}) with equality holding if and only if either

£N({xEE: u(x)>0})0})=oo and ,CN({xEE: u(x)=0})=0. (iii) The function ub is lower semicontinuous and (16.9)

up (0) = esssup u. E

(iv) If v : E -r [0, oo] is another Lebesgue measurable function with u < v ,CN-a.e. on E, then u# < v1 on RN. (v) If {un} is an increasing sequence of nonnegative Lebesgue measur-

able functions u : E -r [0, oo] such that u moo U,,, (x) = u (x)

for all x E E, then lim t4 (y) = u# (y)

n-+oo

for all y E RN.

Proof. (i) For s > 0,

IV R

: up (y) > s} =

{y

RN : u* (ON jY1N) > s}

By Proposition 16.2 we have that u* (ON I yI N) > a if and only if OU (s) > ON IYIN, and so

N:

p

YER N yERN : IyI
s}))N ON

={xEE: u(x)>s}p, where in the last equality we have used (16.7).

(ii) By part (i) the equality (16.8) holds for 0 a > 0. The second part follows as in the proof of Proposition 6.3(v).

16.1. Symmetrization in L" Spaces

481

(iii) To prove that ui is lower semicontinuous, it suffices to show that the set {y E RN : ul (y) > s} is open for all s E R. Ifs > 0, then by part (i) the set {y E RN : ua (y) > s} is an open ball or empty or RN, while if ss} = RN since uO > 0. Properties (iv) and (v) follow as in the proof of Proposition 6.10(1) and (ii), respectively.

Corollary 16.7. Let E C RN be a Lebesgue measurable set and let u E LOO (E) be nonnegative. Then ul belongs to L°° (En) and IIUIILOO(E)

Proof. By (16.6) and Propositions 16.2 and 16.6 we have that up (0) = if (0) = IIUIIL-(E) < oo. IIu'IIL-(EI) = 11

Exercise 16.8 (Uniqueness of Schwarz symmetrization). (i) Let fl, f2

:

(0, oo) - [0, oo] be two decreasing right continuous functions such

that

£N({xERN: fl (IxI)>s}) ='CN({xERN: f2 (IxI)>s}) for all s > 0. Prove that f, = f2. (ii) Let u : RN -- [0, oo] be a function vanishing at infinity such that

u (x) = f (IxI)

for LN-a.e. x E RN for some decreasing function f : (0, 00) -+ [0, 00] [0, oo). Prove that there exists a unique function v : RN vanishing at infinity such that (a) v (x) = f (IxI) for all x E RN \ {0} for some decreasing right continuous function f : (0, 00) - [0, 00), (b) v (0) = clim f (t), (c) u (x) = v (x) for CN-a.e. x E RN. Prove also that the function v is lower semicontinuous. [0, oo] be as in part (ii). Prove that u (x) = ul (x) (iii) Let u : RN for GN-a.e. x E RN. The proofs of the following results are very similar to the analogous ones in Chapter 6 and are left as an exercise.


482

Theorem 16.9 (Hardy-Littlewood's inequality). Let E C RN be a Lebesgue measurable set and let u, v : E -> [0, oo) be two Lebesgue measurable functions. Then IE u (x) v (x) dx < f u# (v) v# (y) dy.

Theorem 16.10. Let E C RN be a Lebesgue measurable set, let u : E -> (0, oo) be a function vanishing at infinity, and let f : [0, oo) - [0, oo) be a Borel function. Then

ff

(16.10)

(uU

(y)) dy < fE f (u (x)) dx,

with equality holding if f (0) = 0 or LP' ({x E E : u (x) > 0}) < oo or

CN({xEE: u(x)>0})=oo andCN({xEE: u(x)=0})=0. In particular, for any p > 0,

L (UI (y)) P dy = L (u (x))" dx. Theorem 16.11. Let Q : R -> [0, oo) be a convex function such that 41 (0) _

0, let E C RN be a Lebesgue measurable set, and let u, v : E - [0, oo) be two functions vanishing at infinity. Then

f In particular,

T

f

(ul (l!) - vI (y)) dy < fE' (u (x) - v (x)) dx.

(y)I'

u# (y) - v n

dy
1, then by Remark 16.13 and Exercise 5.21 we have that

A. (B) = fB ( u*)' (t) dt for every Borel set B C 10, oo), and so by Theorem 5.42 and Remark 5.43, N-1

IVu# (y) lp dy = pN J

f (s)

IN-1

Lou S

N

ds

N

where (16.14)

f(S):=

P-1) N1 I(u')'(9u(8))I'

1 (-0,1(3))' C(N

sE(yi-lryi)


486

Note that f is a Borel function by Exercise 1.41 and Proposition 16.1(i) and (ii). By Exercise 16.15 and Proposition 16.1 (ii), for s E (7s_1, -ti),

iN-1 (-1({}))

(Qu(8)'\l

(16.15)

aN and so by Exercise 16.16,

f

Uts

ON

fy

IVu# (y) I' dy < ON 1

1

7i_1

-ON 1

fui

?jN-1 (u 1

f (a)

l ds ({s}))

f (u (.T)) IVu(x)I dx-

We now use Holder's inequality and Theorem 16.10 to conclude that

J©u() Idy < f3N 1 (fU, IVu (x)Id)

{

\Ju{ If (u (x))l'

l

'

P

= aN 1 UU, Iou(x)IY

dx) ())P'

(fu: If

n

dy

We claim that 1

I"("

f (u# (y)) < #N

(16.16)

for GN-a.e. y E U;. By (16.6) and Proposition 16.2(iv) we have that Lou

(Ul (y)) = Au. (u* (aN IyIN)) u in L' (RN) and DuI (RN)

lim fRMN IVun (x)I dx fl-'

By Theorem 10.33 for every n E N there exists a piecewise affine function v E W111 (RN) with compact support such that 1

IIV, - urallwl.l(RN) < n

Hence vn -), u in L1 (RN) and lim

n-4oo fRX

lava (x)I dx = IDul (18N) .

We now proceed as in the previous theorem. By Theorem 16.14, vn E w1,1 (RN),

hN I() I dy =

f

N

Ivn (x)I dx,

and

II'ILlNNxN)
92})

x-i

1

x + y

and

R xX - X, (t, x) H tx

are continuous with respect to r.

Remark A.M. (i) In a topological vector space a set U is open if and only if x + U is open for all x E X. Hence, to give a base, it is enough to give a local base at the origin.

(ii) Using the continuity of addition and scalar multiplication, it is possible to show that each neighborhood U of the origin is absorbing

and it contains a neighborhood W of zero such that W + W C U and W C U, as well as a balanced neighborhood of zero. As a corollary of Theorem A.14 we have the following:

A. Functional Analysis

498

Corollary A.17. A topological vector space X is metrixable if and only if (i) singletons are closed, (ii) X has a countable base.

Definition A.18. Let X be a topological vector space. A set E C X is said to be topologically bounded if for each neighborhood U of 0 there exists

t > 0 such that E C W. Note that when the topology r is generated by a metric d, sets bounded in the topological sense and in the metric sense may be different. To see this, it suffices to observe that the metric dl := i generates the same topology

as d, but since dl < 1, every set in X is bounded with respect to dl. We now define Cauchy sequences in a topological vector space.

Definition A.19. Let X be a topological vector space. A sequence {x,, } C X is called a Cauchy sequence if for every neighborhood U of the origin there exists an integer W E N such that

xn - x,n E U for all n, m > W. The space X is complete if every Cauchy sequence is convergent.

Note that Cauchy (and hence convergent) sequences are bounded in the topological sense.

Proposition A.20. Let X be a topological vector space and let {xn} C X be a Cauchy sequence. Then the set {xn : n E N} is topologically bounded.

Topologically bounded sets play an important role in the normability of locally convex topological vector spaces (see Theorem A.36 below).

Definition A.21. A topological vector space X is locally convex if every point x E X has a neighborhood that is convex. Proposition A.22. A locally convex topological vector space admits a local base at the origin consisting of balanced convex neighborhoods of zero.

Let X be a vector space over R and let E C X. The function pE : X - R defined by

pE (x) := inf {t > 0 : x E tE},

x E X,

is called the gauge or Minkowaki functional of E.

Definition A.23. Let X be a vector space over R. A function p : X -' R is called a seminorm if p (x + y) 0 such that

forallxEX.

CIIxII1 0.

Proposition B.19. Let X be a metric space and let µ* : P (X) -* [0, oo] be a metric outer measure. Then every Borel set is µ*-measurable.

B.2. Measurable and Integrable Functions In this section we introduce the notions of measurable and integrable functions.

Definition B.20. Let X and Y be nonempty sets and let fit and 91 be algebras on X and Y, respectively. A function u : X -' Y is said to be measurable if u-' (F) E fit for every set F E 9!.

If X and Y are topological spaces, fit := B (X) and M := B (Y), then a measurable function u : X --> Y will be called a Borel function, or Borel measurable. If in a topological space (usually, RN, R, or [-oo, oo]) the aalgebra is not specified, it is understood that we take the Borel a-algebra B (X).

Proposition B.21. If fit is a a-algebra on a set X and 9! is the smallest aalgebra that contains a given family G of subsets of a set Y, then u : X - Y is measurable if and only if ul (F) E 9)! for every set F E G.

B. Measures

512

Remark B.22. In particular, if in the previous proposition Y is a topological space and 91= B (Y), then it suffices to verify that u-1 (A) E fit for every open set A C Y. Moreover, if Y = R (respectively, Y = [-oo, 001), then it suffices to check that u1 ((a, oo)) E fit (respectively, u-1 ((a, oo]) E fit) for every a E R.

Remark B.23. If X and Y are topological spaces, then, in view of the previous remark, every continuous function u : X --> Y is a Borel function.

Proposition B.24. Let (X, M), (Y, 91), (Z, Q) be measurable spaces and let u : X -> Y and v : Y --+ Z be two measurable functions. Then the function v o u = X -> Z is measurable.

Corollary B.25. Let (X, fit) be a measurable space and let u : X -> R (respectively, u : X - [-oo, oo]) be a measurable function. Then uz, lul,

u+, u , cu, where c E R, are measurable. Remark B.26. If c = 0 and u : X

[-oo, oo], the function cu is defined

to be identically equal to zero. Given two measurable spaces (X, MI) and (Y, 92), we denote by MI 091 C

P (X x Y) the smallest a-algebra that contains all sets of the form E x F, where E E fit, F E 7t. Then fit 091 is called the product a-algebra of fit and 91.

Proposition

B.27. Let (X, M?), (Y1, 911), ... , (Yn, 91..) be measurable spaces and consider (Yi x ... x Yt, 9T1 0 ... (9 9Tn) .

Then the vector-valued function u : X -+ Y1 x - x Yn is measurable if and only if its components uu : X -> Yi are measurable functions for all

i= 1,...,n. Corollary B.28. Let (X, fit) be a measurable space and let u : X -+ R and v : X -). R be two measurable functions. Then u + v, uv, min {u, v}, max {u, v} are measurable.

Remark B.29. The previous corollary continues to hold if R is replaced by [-oo, oo], provided u + v are well-defined, i.e., (u (x), v (x)) {± (oo, -oo)} for all x E X. Concerning uv, we define (uv) (x) := 0 whenever u (x) or v (x) is zero.

Proposition B.30. Let (X, WI) be a measurable space and let un : X [-oo, oo], n E N, be measurable functions. Then the functions sup un, inf un, lim inf u,t, lim sup u,i n

are measurable.

n

n-yoo

B.2. Measurable and Integrable Functions

513

Remark B.31. The previous proposition uses in a crucial way the fact that fit is a v-algebra. Let (X, fit, µ) be a measure space. We will see later on that the Lebesgue integration does not "see" sets of measure zero. Also, several properties (existence of pointwise limit of a sequence of functions, convergence of a series of functions) do not hold at every point x E X. Thus, it is important to work with functions u that are defined only on X \ E with µ (E) = 0. For this reason, we extend Definition B.20 to read as follows.

Definition B.32. Let (X, fit) and (Y, 91) be two measurable spaces and let

p : fit - [0, oo] be a measure. Given a function u : X \ E - Y where p (E) = 0, u is said to be measurable over X if u-1 (F) E fit for every set

FE91. In general, measurability of u on X \ E does not entail the measurability of an arbitrary extension of u to X unless p is complete. However, if we define W (x)

U(X) ifxEX\E,

if x E E, yi where yl E Y, then w is measurable. Hence, in general there are extensions

of u that are measurable and others that are not. The next result shows that if the measure p is complete, then this cannot happen. In what follows, if p is a measure, we write that a property holds µ-a.e.

on a measurable set E if there exists a measurable set F C E such that p (F) = 0 and the property holds everywhere on the set E \ F. Proposition B.33. Let (X, fit) and (Y, 91) be two measurable spaces and let u : X -> Y be a measurable function. Let p : fit -> [0, oo] be a complete measure.

If v : X - Y is a function such that u (x) = v (x) for p-a.e.

x E X, then v is measurable.

Going back to the setting in which u : X \ E -+ Y with p (E) = 0, since Lebesgue integration does not take into account sets of measure zero, we will see that integration of u depends mostly on its measurability on X \ E.

Corollary B.34. Let (X, fit) be a measurable space and let u : X [-oo, oo], n E N, be measurable functions. Let p : fit -> [0, oo] be a complete measure. If there exists limn_oo u (x) for p-a.e. x E X, then limn_oo u,, is measurable.

We are now in a position to introduce the notion of integral.

Definition B.35. Let X be a nonempty set and let fit be an algebra on X. A simple function is a measurable function s : X -- R whose range consists of finitely many points.

B. Measures

514

If cl, ... , ee are the distinct values of s, then we write

t

s=EenXE"' n=1

where 71 CE" is the characteristic function of the set

En:={xEX:8(x)=cn}, if x E En, otherwise. If µ is a finitely additive (positive) measure on X and s > 0, then for every measurable set E E X12 we define the Lebesgue integral of s over E as I 1

XE " (x)

0

ad/L>cnp(EnnE),

(B.2)

n=1

where if c = 0 and it (En n E) = oo, then we use the convention c.A (E,, n E) := 0.

Theorem B.36. Let X be a nonempty set, let WI be an algebra on X, and let u : X - [0, oo] be a measurable function. Then there exists a sequence {sn} of simple functions such that

081(x) <s2(x) < C s,,(a')-*u(x) for every x E X. The convergence is uniform on every set on which u is bounded from above.

Corollary B.37. Let (X, 0, p) be a measure space and let u : X - [0, oo] be a measurable function. If the set {x E X : u (x) > 01 has u-finite measure and u is finite µ-a.e., then there exists a sequence of simple functions, each of them bounded and vanishing outside a set of finite measure (depending on n), such that 0 'C 81 (x) < 82 (x) 'C ... < sn (x) -' u (x) for every x C= X.

In the remainder of this section, WI is a c-algebra and U a (countably additive) measure. In view of the previous theorem, if u : X -. [0, oo] is a

IJJJ

measurable function, then we define its (Lebesgue) integral over a measurable

set E as

IJE udu:=sup{ fE sdp: ssimple,0<s 0 for all n, k E N, we have 00

00

00

00

E E ank = > E ank'

n=1k=1

k=1n=1

B. Measures

516

To see this, it suffices to consider X = N with the counting measurer and to define un : N - [0, oo] by un (k) := ank. Then

JX

00 ank, un dµ = E

k=1

and the result now follows from the previous corollary.

Lemma B.44 (Fatou's lemma). Let (X, fit, µ) be a measure space. (i) If un X - [[0, oo] is a sequence of measurable functions, then lim inf un dµ < lim inf

JX n- oo

n-too

X

u,1 dµ.

(ii) If un X - [-oo, oo] is a sequence of measurable functions such that

un lim sup un dµ. n-oo JX Jx n-oo

Corollary B.45. Let (X, fit, it) be a measure space and let u : X -' [0, oo] be a measurable function. Then

Ix udtt=0 if and only if u (x) = 0 for µ-a. e. X E X X.

In order to extend the notion of integral to functions of arbitrary sign,

consider u : X - [-oo, oo]. Note that u = u+ - u- and Jul = u+ + u- and that u is measurable if and only if u+ and u are measurable. Also, if u is bounded, then so are u+ and u-, and in view of Theorem B.36, u is then the uniform limit of a sequence of simple functions.

Definition B.46. Let (X, fit, Et) be a measure space and let u : X [-oo, oo] be a measurable function. Given a measurable set E E fit, if at least one of the two integrals fE u+ dp and fE u- dµ is finite, then we define the (Lebesgue) integral of u over the measurable set E by

JE

:=

fEU+dit

-

fEu

dµ.

'Given a set X, the counting measure u : 7, (X) - 10, ool is defined by

A(E):= for every E C X.

the number of elements of E 00

if E is a finite set, otherwise

B.2. Measurable and Integrable Functions

517

If both fE u+ dµ and fE u- dµ are finite, then u is said to be (Lebesgue) integrable over the measurable set E.

A measurable function u : X -+ [-oo, oo] is Lebesgue integrable over the measurable set E if and only if

1 Jul dµ [-oo, oo] is Lebesgue integrable, then the set {x E X : In (x) I = oo} has measure zero, while the set {x E X : In (x) I > 0} is a-finite. If (X, 911, µ) is a measure space, with X a topological space, and if fit contains B (X), then u : X - [-oo, oo] is said to be locally integrable if it is Lebesgue integrable over every compact set.

Proposition B.48. Let (X, 9A,µ) be a measure space and let u, v : X [-oo, oo] be two integrable functions.

(i) If a,,8 E R, then au +,Qv is integrable andr

J(+

dµ = a

Jx udµ + f3 xJ v d.

(ii) If u (x) = v (x) for µ--a.e. x E X, then

fud/L=JvdIL(iii)

(B.3)

I fx udµl [0, oo) such that If (x,y)1 [0, oo) such that

of (x,y)I < h(x) for 14-a. e. x E X and for ally E Y. Then the function F : Y - R, defined by

F (y) = Ix f

(x,

y) dA (x) ,

yEY,

is well-defined and differentiable, with F' (y)

= Ix d'+J

(x,

y) dµ (x) .

The next example shows that, without the integrability of g, Theorem B.52 fails.

520

B. Measures

Exercise B.54. Consider the function IILY4

f (x, y) _

0

if IVI < IxI , if IyI ? IxI

Prove that the function

F (y) =

ff(xu) dx,

Y E R,

is well-defined and is not continuous at y = 0.

B.4. Product Spaces We recall that, given two measurable spaces (X, 911) and (Y, 91), we denote by 931 0 9Z C P (X x Y) the smallest a-algebra that contains all sets of the

form E x F, where E E 931, F E 97. Then 931 0 91 is called the product or-algebra of 9R and 91.

Exercise B.55. Let X and Y be topological spaces and let B (X) and B (Y) be their respective Borel a-algebras. Prove that

B(X)0B(Y) CB(X xY). Show also that if X and Y are separable metric spaces, then

B(X)®B(Y)=B(X xY). In particular, B (RN) = B (R) 0... ®B (R). Let (X, 931, jc) and (Y, 91, v) be two measure spaces. For every E E X x Y define 00

(B.5)

(p x v)* (E) := inf

p (F.) v (Gn) : {Fn} C 931, {G,,} C 91, n=1 00

ECU (Fn X Gn)

.

n=1

By Proposition B.3, (p x v)` : P (X) - [0, oo] is an outer measure, called the product outer measure of ,a and v. By Caratheodory's theorem, the restriction of (p x v)* to the a-algebra 911 x 91 of (p x v)*-measurable sets is a complete measure, denoted by p x v and called the product measure of It and v. Note that 911 x 91 is, in general, larger than the product v-algebra 9'11®9.

Theorem B.56. Let (X, fit, p) and (Y, 97, v) be two measure spaces.

(i) If F E M and G E 91, then F x G is (,u x v)`-measurable and (B.6)

(pxv)(FxG)=p(F)v(G).

B.4. Product Spaces

521

(ii) If u and v are complete and E has a-finite (p x v)-measure, then for u-a.e. x r: X the section

E,:_ {y e Y : (x, y) E E} belongs to the cr-algebra 9 and for v-a.e. y E Y the section

Ey := {x E X : (x, y) E E} belongs to the o-algebra fit. Moreover, the functions y u-r us (Ev) v (E.,) are measurable and and x

(p x v) (E) =

f(E) dv (y) _ Jv(Ex) dµ (x)

.

The previous result is a particular case of Tonelli's theorem in the case

that u = XE Theorem B.57 (Tonelli). Let (X, fit, u) and (Y, 97, v) be two measure spaces. Assume that p. and v are complete and or-finite and let u : X x Y [0, ooj be an (9A x 91) -measurable function. Then for p-a. e. x E X the function it (x, ) is measurable and the function fj, u y) dv (y) is measurable. Similarly, for v-a.e. y E Y the function it y) is measurable and the function fX u (x, ) dA (x) is measurable. Moreover,

IXXY

u (x, y) d (u x v) (x, y)

fX =

\JY u (x, y) dv (y))

f (f' ) dµ

du (x)

(x)) dv (y).

Throughout this book, for simplicity, we write

XY

u (x7'll) dv (y) du (x)

fX fy u

(x, y) dv (y)dp (x)

Exercise B.58. Prove that in the case that it : X x Y -+ [0, oo] is (91 0 T)measurable, then Tonelli's theorem still holds even if the measures p and v are not complete, and the statements are satisfied for every x E X and

y E Y (as opposed to for p-a.e. xEXand for va.e. yEY). Theorem B.59 (Fubini). Let (X, Wt, A) and (Y, 01, v) be two measure spa-

ces. Assume that u and v are complete and let u : X x Y -* [-oo, oo] be (p x v)-integrable. Then for p-a.e. x E X the function is vintegrable and the function fy u

y) dv (y) is ,a-integrable.

522

B. Measures

Similarly, for v-a.e. y E Y the function u y) is µ-integrable and the function fX u (x, ) dµ (x) is v-integrable. Moreover,

J xY u (x, y) d (µ x v) (x, y) = J X JY u (x,1!) dv (y) dA (x) J u (x, y) dp (x) dv (y)

f

Exercise B.60. Prove that in the case that u : X x Y -' [-oo, oo] is (9R (9 9t)-measurable, then Fubini's theorem still holds even if the measures µ and v are not complete. The following result is a simple consequence of Tonelli's theorem.

Theorem B.61. Let (X, 9Jt, µ) be a measure space and let u : X -' [0, oo] be a measurable function. Then 00

fud=j({xEX:u(x)>t})dt.

(B.7)

B.5. Radon-Nikodym's and Lebesgue's Decomposition Theorems Definition B.62. Let (X, sJJ) be a measurable space and let µ, v : 97t [0, oo] be two measures. The measure v is said to be absolutely continuous with respect to µ, and we write v 0 there exists 6 > 0 such that v (E) < e

(B.8)

for every measurable set E C X with µ (E) < 5.

Proposition B.64. Let (X, 9R) be a measurable space and let µ, v : Wi [0, oo] be two measures. For every E E 91t define (B.9)

v(

(E) := sup { J u dµ : u : X -' [0, oo] is measurable,

I

E

fE' udµ < v(E) for allE'CE,E'E9Jt}. 1JJ

Then vac is a measure, with vas [-oo, oo] be a signed measure. Then for every E E 9)1, CO

IAI(E)=sup EIA(E.)I n=1

B.6. Signed Measures

525

where the supremum is taken over all partitions { E } C 911 of E.

Definition B.74. Let (X, 9931) be a measurable space. A set function A = (Al, ... , Am) : fit

RI is a vectorial measure if each component A, : 9912 - R

is a signed measure, i = 1, . . . , m. Proposition B.75. Let (X, 9931) be a measurable space and let A : fit -> R'" be a signed measure. Then the function IAA :9N -p [0, oo], defined by W

11\1(E):=sup E IA (En)d

E E'931,

1n=1

where the supremum is taken over all partitions {En} C 9971 of E, is a measure.

Definition B.76. Let (X, 9912) be a measurable space, let p : fit - [0, oo] be a measure, and let A :9931- [-oo, oo] be a signed measure. (i) A is said to be absolutely continuous with respect to p, and we write

A«p,if A(E)=0whenever EE991Iand a(E)=0. (ii) A and p are said to be mutually singular, and we write A 1 it, if there exist two disjoint sets X, , Xa E 9932 such that X = X. U Xa and for every E E 9932 we have

p(E)=p(EnX,,), A(E)=A(EnXA). Note that if A

1

Iwvkl dp = 1

( (Ek))1

1

(1A (El.))

1

J Jul du k

= n (p (Ek))n -, 00

as k - oo. This contradiction shows that the set {x E X : u (x) # 0} has u-finite measure.

Step 2: We prove (B.19) in the case 1 < p < 00. In view of the previous proposition, it remains to show that if tc 0 LP (X), then the right-hand side

B. Measures

530

of (B.19) is infinite. Thus, assume by contradiction that (B.20) holds. Let

{ u (x) if Iu (x)I < n and x E Xn,

vn (x)

0

otherwise,

where the Xn are the sets defined in (B.21). Then

Ix IunI" dA < n"p (Xn) < 00, and so by the previous proposition we may find vn E LP' (X) such that IIVnIILP' IIunIILP(X) -1. Since

JX IuvnI dit >

fx IunvnI dµ ? IIunIILP(X) - 1

and IIUnIILP(X) - IIUIILP(X) = oo as n -'

oo, we obtain that C = 00. This contradiction completes the proof in the case 1 I Unvn dx >- IIUnIILQQ(X) -1- 00

asn -oo. As a corollary of Proposition B.81 we have the following result.

Corollary B.83 (Minkowski's inequality for integrals). Let (X, 9R,µ) and (Y, IN, v) be two measure spaces. Assume that p and v are complete and a-finite. Let u : X x Y - [0, oo] be an (9R x 9't)-measurable function and let I < p < oo. Then

III Iu (x, )I X

[-oo, oo] is said to belong to L10C (X) if u E LA (K) for every compact set K C X. A sequence {un} C Lo,, (X) is said to converge to u in L a (X) if u,, - u in LP (K) for every compact set K C X. Theorem B.91 (R.iesz's representation theorem in LP). Let (X,9)1, p) be a measure space, let 1 < p < oo, and let p' be its Holder conjugate exponent. Then every bounded linear functional L : LP (X) R is represented by a unique v E L'1 (X) in the sense that (B.22)

L (u) = f uv dp for every u E Lp (X) . x

Moreover, the norm of L coincides with IIVIIL,y. Conversely, every functional of the form (B.22), where v E V" (X), is a bounded linear functional on Lp (X). Thus, the dual of LP (X) may be identified with &(X). In particular, LP (X) is reflexive.

Definition B.92. Let (X, fit, p) be a measure space and let {EQ}QE J be a family of measurable sets of X. A measurable set E,, is called the essential supremum of the family {Ea}a.E J if the following hold.

(i) E. D Ea (up to a set of p-measure zero) for every a E J. (ii) If E E 811 is such that E D E. (up to a set of p-measure zero) for every a E J, then E,,,,, D E. (up to a set of p-measure zero). Definition B.93. Let (X, M1, p) be a measure space. The measure p is said to be localizable if every family of measurable sets admits an essential union. Proposition B.94. Let (X, s9)1, p) be a measure space. If p is u-finite, then it is localizable.

B.7. LA Spaces

533

Theorem B.95 (Riesz's representation theorem in L1). Let (X, fit, µ) be a measure space. Then the dual of L' (X) may be identified with L°O (X) if and only if the measure it is localizable and has the finite subset property. In particular, if p is a-finite, then the dual of L1 (X) may be identified with LOD (X).

The next result shows that L1 (X) is not reflexive, since in general its bidual is larger than L' (X). Let (X, WI, µ) be a measure space. The dual of LO0 (X) may be identified with the space ba (X, 9)1, µ) of all bounded finitely additive signed measures

absolutely continuous with respect to µ, that is, all maps A : 9l? - R such that (i) A is a finitely additive signed measure, (ii) A is bounded, that is, its total variation norm l

IAI (X) := Sup E IA (En)I 1n=1

where the supremum is taken over all finite partitions {En}1 C fit of X, I E N, is finite, (iii) A (E) = 0 whenever E E 9R and µ (E) = 0. Given a measure space (X, 9)t,µ), A E ba (X, fii, µ), and u E L°O (X), by Theorem B.36 we may find a sequence {sk} C L' (X) of simple functions that converges uniformly to u. For every E E fit, we define fE sk dA as in (B.2) with A and sk in place of It and s, respectively. Then

I fE (sk - Sm) dal

0 there exists E C X with E E D1 such that µ (E) < 0o and

IunIpdp<e

fx\E for all n.

Remark B.102. Note that condition (iii) is automatically satisfied when X has finite measure.

In view of Vitali's theorem it becomes important to understand equiintegrability.

Theorem B.103. Let (X, VA, µ) be a measure space and let.F be a family of integrable functions u : X - [-oo, oo]. Consider the following conditions:

(i) F is equi-integrable. (ii)

lim sup f

(B.25)

Jul dA = 0.

t'OO uE.7 .!{xEX: jug>t}

(iii) (De la Vail& Poussin) There exists an increasing function y : [0, oo) -> [0, ooI with lim y (t) = 00

(B.26)

t-4oo

such that (B.27)

sup

f

ue7 JX

t

y (I ul) dji < oo.

B. Measures

536

Then (ii) and (iii) are equivalent and either one implies (i). If in addition we assume that sup J dp < oo, uE.F IX X

then (i) implies (ii) (and so all three conditions are equivalent in this case).

B.9. Radon Measures Definition B.104. An outer measure µ* : P (X) -> [0, oo] is said to be regular if for every set E C X there exists a µ*-measurable set F C X such that E C F and µ* (E) = µ* (F). An important property of regular outer measures is the fact that Proposition B.9(i) continues to hold.

Proposition B.105. Let µ* : P (X)

[0, oo] be a regular outer measure. If {En} is an increasing sequence of subsets of X, then 00

* U En = lim µ* (En) n-oo n=1

Definition B.106. Let X be a topological space and let µ* : P (X) -+ [0, oo] be an outer measure.

(i) A set E C X is said to be inner regular if µ* (E) = sup {µ* (K) : K C E, K is compact}, and it is outer regular if µ* (E) = inf {µ* (A) : A D E, A is open}.

(ii) A set E C X is said to be regular if it is both inner and outer regular.

Definition B.107. Let X be a topological space and let µ* : P (X) --> [0, oo] be an outer measure.

(i) µ* is said to be a Borel outer measure if every Borel set is µ*measurable. (ii) µ* is said to be a Borel regular outer measure if µ* is a Borel outer

measure and for every set E C X there exists a Borel set F C X such that E C F and µ* (E) = µ* (F). Definition B.108. Let X be a topological space and let µ* : P (X) - [0, oo] be an outer measure. Then µ* is said to be a Radon outer measure if is a Borel outer measure, (ii) µ* (K) < oo for every compact set K C X, (i)

537

B.9. Radon pleasures

(iii) every open set A C X is inner regular, (iv) every set E C X is outer regular.

Remark B.109. Note that a Radon outer measure is always Borel regular.

We investigate the relation between Radon outer measures and Borel regular measures.

Proposition B.110. Let X be a locally compact Hausdorff space such that every open set is a-compact. Let p* : P (X) -> [0, ool be a Borel outer measure such that p* (K) < oo for every compact set K C X. Then every Borel set is inner regular and outer regular. If, in addition, p* is a Borel regular outer measure, then it is a Radon outer measure. We now introduce analogous regularity properties for measures.

Definition B.111. Let (X, 931, p) be a measure space. If X is a topological space, then the following hold.

(i) p is a Borel regular measure if it is a Borel measure and if for every set E E 9)1 there exists a Borel set F such that E C F and p (E) = it (F). (ii) A Borel measure p : fit [0, oo] is a Radon measure if (a) p (K) < oo for every compact set K C X, (b) every open set A C X is inner regular, (c) every set E E 9)1 is outer regular.

Hausdorff measures if, s > 0, represent an important class of regular Borel measures that are not Radon measures. Proposition B.112. Let X be a locally compact Hausdorff space such that every open set is a-compact. Let it : B (X) -> [0, oo] be a measure finite on compact sets. Then p is a Radon measure and every Borel set E is inner regular.

Definition B.113. Let (X, 9R) be a measurable space and let A : 9)1 [-oo, oc] be a signed measure. If X is a topological space, then A is a signed [0, oo] is a Radon measure. Radon measure if IAI : 932

If X is a topological space, then Mb (X; R) is the space of all signed finite Radon measures A : B (X) - R endowed with the total variation norm. It can be verified that Mb (X; R) is a Banach space with the norm II'IIMb(X;R) := IAI (X)

B. Measures

538

Similarly, if (X, fit) is a measurable space and X is a topological space, then A : fit - R1 is a vectorial Radon measure if each component Ai : fit -, R is a signed Radon measure. The space Mb (X; Rm) of all vectorial Radon measures A : B (X) - R n is a Banach space with the norm IIAIIM4(x;Rm) = IAI (X)

Theorem B.114 (Riesz's representation theorem in CO). Let X be a locally compact Hausdorff space. Then for every bounded linear functional L Co (X; Rm) --+ R there exists a unique A E Mb (X; R'n) such that

L (u) = J u dA for every u E CO (X; R).

(B.28)

Ix

Moreover, the norm of L coincides with IAI (X). Conversely, every functional of the form (B.28), where A E Mb (X; R), is a bounded linear functional on CO (X; Rfb).

Theorem B.115 (Riesz's representation theorem in Ce). Let X be a locally

compact Hausdorf space and let L : Cc (X) - ]R be a linear functional. Then

(i) if L is positive, that is, L (v) > 0 for all v E CC (X), then there exists a unique (positive) Radon measure µ : B (X) - [0, oo] such that

L (u) =

u dµ for every u E Cc (X) , fX (ii) if L is locally bounded, that is, for every compact set K C X there exists a constant CK > 0 such that IL (u)1:5 CK IlvIlcc(x)

for all v E CC (X) with supp v C K, then there exist two (positive) Radon measures Al, µ2 B (X) - [0, oo] such that ::

L (u) = J u dµ1 - Ix u dµ2 for every u E CC (X) x Note that since both µl and µ2 could have infinite measure, their difference is not defined in general, although on every compact set it is a well-defined finite signed measure.

B.10. Covering Theorems in R" Theorem B.116 (Besicovitch's covering theorem). There exists a constant f, depending only on the dimension N of RN, such that for every collection 7 of (nondegenerate) closed balls with (B.29)

sup { diam B : B E F} < 00

B.10. Covering Theorems in Riv

539

there exist F1,. .. , .gyp C F such that each .7;,, n family of disjoint balls in F and

is a countable

I

EC UU n=1 BEF'J

where E is the set of centers of balls in F.

Definition B.117. Given a set E C RN, a family.F of nonempty subsets of RN is said to be (i) a cover for E if

ECUF, FEF

(ii) a fine cover for E if for every x E E there exists a subfamily .Fy C F of sets containing x such that

inf {diem F : F E .Fx} = 0.

(B.30)

The following result is an important consequence of the Besicovitch covering theorem.

Theorem B.118 (Vitali-Besicovitch's covering theorem). Let E C RN and let .F be a family of closed balls such that each point of E is the center of arbitrarily small balls, that is,

inf {r : B (x, r) E F} = 0 for every x E E. Let A* : P (RN) - (0, ooj be a Radon outer measure. Then there exists a countable family {B (xn, r,,)} C F of disjoint closed balls such that

µ* E 1 U B (xn, rn) = 0. n

Theorem B.119 (Besicovitch's derivation theorem). Let µ, v : 8 (RN) [0, oo] be two Radon measures. Then there exists a Borel set M C RN, with tc (M) = 0, such that for every x e RN \ M, (B.31)

v (B (x, r)}

dv

°C (x) = lim

IA(r(B -(T, r))

and ve 11m

r-r0+

(.(z,r)) (B(x,r))

- 0,

E Ilt

B. Measures

540

where v = vac + U3,

(B.32)

Vac [0, oo] be a Radon measure, let u : RN -+ [-oo, oo] be a locally integrable function. Then there exists a Borel set M C RN, with µ (M) = 0, such that RN \ M C {x E R" : u (x) E R} and for every x E RN \ M, lim

r+

1

(

,

r))\

J

u(y)dpc(y)=u(x). (z,r)

By enlarging the "bad" set M, we can strengthen the conclusion of the previous theorem.

Corollary B.122. Let it : B (RN) - [0, oo] be a Radon measure and let u : RN - [-oo, oo] be a locally integrable function. Then there exists a Borel set M C RN, with tt (M) = 0, such that

RN\MC {xERN: u(x) ER} and for every x E RN \ M, (B.33)

lim 1 r0+ J Iu (y) - u (x) I dµ (y) = 0. µ (ii (x, r)B(x,r)

A point x E RN for which (B.33) holds is called a Lebesgue point of u.

Corollary B.123. Let p : B (RN) - [0, oo] be a Radon measure, let 1 < p < oo, and let u E L' (RN). Then there exists a Bored set M C RN, with p (M) = 0, such that RN \ M C {x E RN : u (x) E R}, and for every x E RN \ M, (B.34)

rlim+

1

p (B (x, r))

f

B(x,r)

j u (y)

- u (x) I P dp (y) = 0.

A point x E RN for which (B.34) holds is called a p-Lebesgue point of u. By applying Theorem B.121 to XE, we obtain the following result.

B.10. Covering Theorems in RN

541

Corollary B.124. Let it : B (RN) - [0, oo] be a Radon measure and let E C RN be a Borel set. Then there exists a Bored set M C RN, with p (M) = 0, such that for every x E RN \ M,

lim p (B (x, r) f E) = XE (x) p (B (x, r))

r-0+

A point x E E for which the previous limit is one is called a point of density one for E. More generally, for any t E [0, 1] a point x E RN such that

(x, r) n E) _ t

lim p (B r-.0+ p (B (x, r))

is called a point of density t for E.

Appendix C

The Lebesgue and Hausdorff Measures The Analysis of Value must be used carefully to avoid the following two types of errors. Type I: You incorrectly believe your research is not Dull. Type H. No conclusions can be made. Good luck graduating. -Jorge Cham, www.phdcomics.com

C.1. The Lebesgue Measure In the Euclidean space RN consider the family of elementary sets

9:= {Q (x, r) : x E RN, 0 < r < oo} U {0} and define p (Q (x, r)) := rN and p (0) := 0, where

Q (r, r) := x +

rrN

(-ii)

For each set E C RN define

0 ,C (E) := inf

(rn.)N : {Q (xn, rm)} C G, E C 11t=1

UQ n=1

By Proposition B.3, Lo is an outer measure, called the N-dimensional Lebesgue outer measure. Using Remark B.4, it can be shown that

(C.i)

Go (Q (x, r)) = p (Q (x, r)) = rN

and that Go is translation-invariant, i.e., Co (x + E) = Lo (E) for all x E RN and all E C RN. The class of all Co -measurable subsets of RN is called the a-algebra of Lebesgue measurable sets, and by Caratheodory's theorem, Lp restricted to this a-algebra is a complete measure, called the N-dimensional Lebesgue measure and denoted by LN. Since LN (RN) > 543

C. The Lebesgue and Hausdorff Measures

544

oo, we conclude that GN is not a finite measure. However, it is o--finite, since

Go (Q (x, r)) = rN, by sending r

00

RN

= U Q (0, n) n=1

and GN (Q (0, n)) _ nN < oo.

Exercise C.1. Prove that for every b > 0 and for each set E C RN, 00

00

Go (E) = inf E (rn.)N : {Q (x,,, rn)} C 9, rn 10, oo] is not a complete measure. Using the axiom of choice, it is also possible to construct sets that are not Lebesgue measurable. Exercise C.2 (A non-Lebesgue measurable set). On the real line consider

the equivalence relation x - y if z - y E Q. By the axiom of choice we may construct a set E C (0,1) that contains exactly one element from each equivalence class. Let

F:=

U

(r + E) C (-1, 2).

re(-1,1)f1Q

(1) Prove that F D (0, 1). (ii) Prove that if r, q E Q, with r # q, then (r + E) fl (q + E) _ 0. (iii) Prove that E is not Lebesgue measurable. We observe that the Lebesgue outer measure is a Radon outer measure. Indeed, outer regularity of arbitrary sets follows from (C.1) and the definition of 'C' N' while inner regularity of open sets is an immediate consequence of the fact that each open set can be written as a countable union of closed cubes with pairwise disjoint interiors.

Proposition C.3. Go is a Radon outer measure. Moreover, every Lebesgue measurable set E is the union of a Borel set and a set of Lebesgue measure zero.

Using the notation introduced in Section B.4, we have

Proposition C.4. Let N = m+ k, where N, n, m E N. Then (,Cl x G"')"` _ GN. 0

C.2. The Brunn-Minkowski Inequality and Its Applications

545

If E C RN is a Lebesgue measurable set, fit is the a-algebra of all Lebesgue measurable subsets of E, Y is a nonempty set, and 91 is an algebra on Y, then a measurable function u : E --+ Y is called Lebesgue measurable.

Proposition C.5. Let E C RN be a Lebesgue measurable set and let u : E -* R be a Lebesgue measurable function. Then there exists a Borel func-

tion v : E - R such that v (x) = u (x) for iN-a.e. x E E.

If E C RN is a Lebesgue measurable set and u : E -' J-oo, ooJ is a Lebesgue measurable function, then, whenever it is well-defined, we denote fE u simply by u dx. JE

If N = 1 and I is an interval of endpoints a and b, we also write fa u dx for fr u dz.

C.2. The Brunn-Minkowski Inequality and Its Applications We leave the following preliminary result as an exercise.

Exercise C.6. Let E, F C RN be two compact sets. (i) Construct a decreasing sequence of sets {En} and {Fn} such that 00

00

E= n E,,, En, F= n1'., n=1

n=1

and all E. and F,, consist of finite unions of rectangles with sides parallel to the axes. (ii) Prove that

LN(E" +Fn) LN(E+F) as

Theorem C.7 (Brunn-Minkowski's inequality). Let E, F C RN be two Lebesgue measurable sets such that

E+F:={x+y: xEE,yEF} is also Lebesgue measurable. Then

(f-'v (E))"' + (LN(F))"' < (.CN(E+F))N

.


546

Proof. Step 1: Assume that E and F are rectangles whose sides are parallel to the axes and let x; and yi, i = 1, ... , N, be their respective side lengths. Then N

GN (E)

N

N

= fl xi, GN (F) = II yi, £ N (E + F) = fi (xi + yi) i=1

i=1

i-1

By the arithmetic-geometric mean inequality

(n) N

N . 1 xi + yi +

i_1 xi + tfi

+

1N

N

NN

xi + 3Ji

Ti + Vi

lv

which gives the Brunn-Minkowski inequality for rectangles.

Step 2: We now suppose that E and F are finite unions of rectangles as in Step 1. The proof is by induction on the sum of the numbers of rectangles in E and in F. By interchanging E with F, if necessary, we may assume that E has at least two rectangles. By translating E if necessary, we may also assume that a coordinate hyperplane, say {XN = 01, separates

two rectangles in E. Let E+ and E- denote the union of the rectangles formed by intersecting the rectangles in E with the half spaces {XN > 01 and {XN < 01, respectively. Translate F so that

GN (Et) _ GN (Ft) GN (E)

GN (F)

where F+ and F- are defined analogously to E+ and E-. Then E+ + F+ C

{xN > 0} and E- + F- C {xN < 0}, and the numbers of rectangles in E+ U F+ and in E- U F- are both smaller than the number of rectangles in E U F. By induction and Step 1 we obtain GN (E +.F) > GN (E+ + F+) + GN (E- + F-) >

((GN

(E+))

N

(CN (F+)) x +

+ ((CN (E-)) x +

(V"" (F-))

N

)

N

= GN (E+)

1+

(F)) 1 N

= GN (E)

1+ (

GN (E))A

1+

=

N N

GN (E)

(GN (E)) 'W

(F)) A

+

(GN (E)) N

((N(E))+(N(F))*)N

C.2. The Brunn-Minkowski Inequality and Its Applications

547

Step 3: Assume next that E and F are compact sets. Let {E} and {Fn} be as in the previous exercise. Then for all n E N (,CN (E,)) *

(GN

*

(En + Fn)) + (,CN (F'n)) < It now suffices to let n - oo. Finally, if E and F are Lebesgue measurable sets, with E + F measurable, fix two compact sets K1 C E, K2 C F. Then K1 + K2 C E + F, and so 77

(,CN (,CN (KI + K2)) ' < (GN (E + F)) a N' (Ki)) W + (K2)) < Using the inner regularity of the Lebesgue measure, we obtain the desired (,CN

result.

Remark C.8. (i) Note that the hypothesis that E+F is measurable cannot be omitted. Indeed, Sierpinsky [154] constructed an example of two Lebesgue measurable sets whose sum is not measurable. However, if E and F are Borel sets, then E + F is an analytic set, and so it is measurable. We refer to [65] for more information on this topics.

(ii) Fix 0 E (0,1). By replacing E with OE and F with (1 - 0) F and using the N-homogeneity of the Lebesgue measure, we obtain that

0(GN(E))77 +(1-0) (GN(F))* -L

-L

= (GN(0E))N + (GN((1-0)F))N

(G^'(0E+(1-0)F))p. Thus, the function f (t) := (GN (tE + (1 - t) F)) is concave in [0,1]. In particular, if C C RN+1 is a convex set and E and F are the intersections of C with the hyperplanes xl = 1 and xl = 0 (we write x = (x1i x2, ... , xN+1) E RN+1), then tE + (1 - t) F is 717

the intersection of the convex hull of E and F with the hyperplane xl = t and is therefore contained in the intersection of C with the hyperplane xl = t. Hence, the function giving the nth root of the volumes of parallel hyperplane sections of an (n + 1)-dimensional convex set is concave (see [711). We used this fact in the proof of Poincare's inequality in convex domains (see Theorem 12.30).

Exercise C.S. Let

C={xER3:

VX2

ll JJJJ

Prove that C is convex but the function 9 (xl)

fR2 XE (x1, x2, x3) dx1dx2

is not concave (although its square root is by the previous remark).


548

Using the Minkowski inequality, we can prove the isodiametric inequality.

Theorem C.10 (Isodiametric inequality). Let E C RN be a Lebesgue measurable set. Then

(diarnE)N

LN (E)

0, we have that LN (AE) = ANON (E) and (diam (AE))N = AN diam E, so without loss of generality we may assume that diam E = 1. Let

F:={-x:xEE} By the previous remark the function f (t) := (LN (tE + (1 - t) F)) N is concave in [0,1], and so (LN

(LN

(E)) N +

(F))'v < ( LN (2E + 2 F)) N

.

\\

2 2 But LN (F) = LN (E), and so the previous inequality becomes

(E+F).

LN(E)=GN(F) 0 we define

(X),

X E P.N.

The functions W. are called mollifiers. Note that supp We C B (0, e). Hence, given an open set fl C Rv and a function u E Liar (Il), we may define (C.6)

De (x) = (2s * VC) (x) = fa we (x - Y) U (Y) dy

C.4. Mollifiers

553

for x E flet where the open set fl, is given by SfE :_ {x E fl : dist (x, Ofl) > e} .

(C.7)

The function uE : Sle - R is called a mollification of u. Note that if x E Sl, then ue (x) is well-defined for all 0 < e < dist (x, 090).

Thus, it makes sense to talk about lime.0+ uE (x). We will use this fact without further mention. Remark C.17. Note that if n = RN, then S1E = RN; thus uE is defined in the entire space RN. Remark C.18. In the applications we will consider two important types of mollifiers:

(i) cp is the (renormalized) characteristic function of the unit ball, that is, 1

w (x) := aN XB(o,1) (x) , x E RN, (ii) cp is the function of class C°° defined by (C.8)

c exp (xI_1) if IxI < 1,

P (x)

if IxI > 1, where we choose c > 0 so that (C.5) is satisfied. In this case, the functions cpe are called standard mollifiers. 0

The following theorem is the first main result of this subsection.

Theorem C.19. Let S2 C RN be an open set, let W E Li (RN) be a nonnegative bounded function satisfying (C.5), and let u E L (S).

I

(i) If u E C (S2), then ue - u as e -- 0+ uniformly on compact subsets of Q.

(ii) For every Lebesgue point x E 12 (and so for LN-a.e. x E fl), as

(iii) If 1 0 and IItIILP(n) as a - 0+

(C.10) IIUEIILP(ne)

(iv) If u E L" (Sl), 1 0, ue u in L" (fl').


554

Proof. (i) Let K C RN be a compact set. For any fixed

0 < rj < dist (K, M), let

K,i :_ {x E RN : dist (x, K) < ,i} so that K1 C 92. Note that fore > 0 sufficiently small we have that K,r C Q . Since K., is compact and u is uniformly continuous on K,1, for every p > 0 there exists 5 = 5 (,, K, p) > 0 such that P

1u(X) - u(y)I R is a measurable function, then (C.28)

IIIILP = sup I

!RN

IuvI dx : v E &(R"), IIt'IILp < 1}

Proof. If u = 0, then both sides of (C.28) are zero, and so there is nothing to prove. Hence, without loss of generality, we may suppose that IIuIILP > 0. By Holder's inequality, fR

V

Iuvl dx :5 IIVIILP, IIUIILP :5 IIuIILP

for all v E LI' (RN) with I I V I I LP' < 1, and so

M := sup {LN IuvI d : v E L" (RN) , IIILP, !5

I

IIuIILP

To prove the reverse inequality, it suffices to consider the case that M < 00. Let 0 < m < N be the number of indexes pi that are infinite. Assume by contradiction that IIuIILP > M. Then there exists s > 0 so small that IIuIILP > M (1 + e)rn .

FornENdefine u,, : R-"' --,R by

u (x)

u(x) if Iu(x)I M (1 + 07z .


570

Set R = (-n, n)N. For every 1 < i < N write R'-1 xRx x = (&/j, xi, zi') E Pi = (PI)...,Pi-1),

RN-',

(pN-ii.. ,pN),

P4' _

and

R=R;x(-n,n) xki'CR[ -'

xRxRN-i

Note that p = (p;, pi, The cases i = 1 and i = N are simpler. If 1 < pi < oO, define vi : R x RN-i _, R as 11U. (', xi, z')IILP:(Ri-1)

vi (xi, Zi") :=

m -1

,

RN-i

(xi, 4) E R X

,

11

if I1 u

# 0, and vi := 0 otherwise. For i = 1 we would

z,') II L(;,i"") (R`)

take instead

Iu'n(x1,zi)I

vi (xl, z)

P1-1 ,

II

(xl, zl) E R X

RN-1,

IILP1(R))

If pi = oo, let Ei,E :=

1

(xi, 41) E R X

PtN-i

II''nz,')II

(1+e) IMu

and consider the section {xi E R : (xi, z,') E Ei,,t }

(E:,e)z Define vi : R x RN-i

l[2 as 1

vi (xi, z,') :=

G1

XE,,, (xi, zz')

otherwise.

0

One should replace I U"' definition of the set E1,E. Finally, set

if L1 ((Ei,E),1) , 0,

with I un (x 1, zi ) I for i = 1 in the

xi, xs') II Lp:

N

V:_Evi. i=1

We leave it as an exercise to check that v is zero outside a compact set, v r= Li' (RN) n L00 (RN), IIVIILp, = 1, and

fgN IId

(1+C) "+'IIfhILP>M,

C.7. Anisotropic LP Spaces

571

where in the last inequality we have used (C.29). On the other hand,

M>

j

IuvIdx> Jilt"

and so we have reached the desired contradiction.

0

Theorem C.3T (Minkowski's inequality). Let p = (PI, ... ,PN), 1 < pi < 00, i = 1, ... , N, and let u, v : RN - R be measurable functions. Then, IIU + uIILP C II'IILP + IIvIILP

In particular, if u, v E LP (W'), then u + v E LP (RN). Proof. The result now follows by successive applications of Minkowski's inequality with respect to each variable x;, starting from i = 1. 0 By identifying functions with their equivalence classes [u], it follows from Minkowski's inequality that is a norm on LP (RN).

Theorem C.38. Let p = (pi, ... , pN), 1 C p= < oo, i = 1, ... , N. Then LP (RN) is a Banach space.

Proof. Let {u,,} c LP (RN) be a Cauchy sequence; that is, (C.30)

hm m,n->00 IIU

-4

IILp

0.

By Corollary C.36 for every 1, in, n E N, IIUm - U,IILP =

sup 0 and find n E N such that Ilum - U-IILp < 6


572

for all m, n E N with m, n > n. Let v E LP' (RN), with IIVIILP, < 1, and let m, k E N be such that m, uk > rt. Then by Corollary C.36, NI(U,, - unk)vl dx 7( 6(E) + 1-C6 (F)

.

To conclude, it suffices to let S - 0+ and to use (C.32).

O

Remark C.43. We remark that unlike W,",, 7-f6" is not a metric outer measure for 0 < a < N. As an example, consider the outer measure 7H6, given by 00

1 : E C U En, diam En < S

(E) := inf E"'00

E C RN.

n=1

If we take E1 = {x} and E2 = {y}, where 0 < Ix - yi < b, then to cover E1UE2 it is enough to consider one set, so Na (El U E2) = 1, while HO, (El) _ 1, NO (E2) = 1.

6(ElUE2) = 1 0. Hence, fl (E) = 0.

This implies that 71N is absolutely continuous with respect to GN.

0

We turn to the proof of the theorem.

Proof. (i) Since ao = 1, we have 00

7{0(E) := inf E 1: E C U En, diamEn < b

.

n=1

EnOO

If y := card E is finite and S is sufficiently small, then we need at least

y sets En to cover E. Thus 71a (E) = card E. On the other hand, if card E is infinite, let En be a subset of E with exactly n elements. Then

(ii) Fix 5 > 0 and consider any sequence {En} C RN such that E C UO_1 En and diam En < 8 for all n E N. Since diam En = diam En, without

loss of generality, we may assume that the sets En are closed and thus Lebesgue measurable.

Using the monotonicity and subadditivity of Go together with the isodiametric inequality, we have 00

G0 (E) < L. U En < n=1

00

00

aN

GN (En) < n=1

n=1

( diam2 En ) N J \

Taking the infimum over all admissible sequences {En} gives (C.33)

Go (E) < 7{0 (E).


576

To prove the other inequality, fix e > 0 and 6 > 0 and by Exercise C.1 find a sequence of cubes {Q (x., rn)} with diameter less than S such that 00

EGN(Q(xn,rn)) e}. PA: a particular family of continuous piecewise affine functions. p* := N P: Sobolev critical exponent. Ll (E): an Orlicz space; CO," (ii): the space of all bounded Holder continuous functions with exponent a, E : W 1,P (S2) -- W 1,P (RN) : an extension operator.

dreg: the regularized distance. Given a function u and a set E, UE is the integral average of u over

a set E. BV (12): the space of functions of bounded variation. Given u E BV (11), Diu is the weak (or distributional) partial derivative of u with respect to xi, Du is the weak (or distributional) gradient of u, IDul is the total variation measure of the measure Du. V (u, f2): the variation of a function u in Q. P (E, fl): the perimeter of a set E in fl. O'u (x) := u (x + he-j) - u (x). Bs,P,e (RN): a Besov space; W8'P (RN): a fractional Sobolev space. Tr (u): the trace of a Sobolev function. uO: the spherically symmetric rearrangement (or Schwarz symmetric rearrangement) of a function u. M (u): maximal function of u.

Bibliography In the end, we will remember not the words of our enemies, but the silence of our friends.

-Martin Luther King Jr. [1] E. Acerbi, V. Chiado Piat, G. Dal Maso, and D. Percivale, An extension theorem from connected sets, and homogenization in general periodic domains, Nonlinear Anal. 18 (1992), no. 5, 481-496. [2] G. Acosta and RG. Dur£n, An optimal Poincart inequality in L' for convex domains, Proc. Amer. Math. Soc. 132 (2004), 195-202. [3] P. Acquistapace, Appunti di Analisi convessa, 2005. [4] S. Adachi and K. Tanaka, Trudinger type inequalities in R' and their best exponents, Proc. Amer. Math. Soc. 128 (2000), no. 7, 2051-2057. [5] C.R. Adams, The space of functions of bounded variation and certain general spaces, Trans. Amer. Math. Soc. 40 (1936), no. 3, 421-438. [6] R.A. Adams, Sobolev spaces, Pure and Applied Mathematics, 65. A Series of Monographs and Textbooks, New York-San Francisco-London: Academic Press, Inc., a subsidiary of Harcourt Brace Jovanovich, Publishers. XVIII, 1975. [7] R.A. Adams and J.J.F. Fburnier, Sobolev spaces, Second edition, Academic Press (Elsevier), 2003.

[8] F.J. Almgren and E.H. Lieb, Symmetric decreasing rearrangement is sometimes continuous, J. Amer. Math. Soc. 2 (1989), no. 4, 683-773. [9] L. Ambrosio and G. Dal Maso, A general chain rule for distributional derivatives, Proc. Amer. Math. Soc. 108 (1990), no. 3, 691-702. [10] L. Ambrosio, N. Fusco, and D. Pallara, Functions of bounded variation and free discontinuity problems, Oxford Mathematical Monographs, The Clarendon Press, Oxford University Press, New York, 2000.

[11] L. Ambrosio and P. Tilli, Topics on analysis in metric spaces, Oxford Lecture Series in Mathematics and its Applications, 25, Oxford University Press, Oxford, 2004.

[12] C.J. Amick, Decomposition theorems for solenoidal vector fields, J. London Math. Soc. (2) 15 (1977), no. 2, 288-296. [13] T. Aubin, Problimes isopesrimetriquee et espaces de Sobolev, J. Differential Geometry 11 (1976), no. 4, 573-598. 593

Bibliography

594

[141 S. Banach, Sur lea lignes rectifiables et lea surfaces dart faire est fine, F undamenta Mathematicae 7 (1925), 225-236. 1151 R.G. Bartle, The elements of real analysis, Second edition, John Wiley & Sons, New York-London-Sydney, 1976.

[161 M. Bebendorf, A note on the Poincare inequality for convex domains, Z. Anal. Anwend. 22 (2003), no. 4, 751-756. [171 J.J. Benedetto, Real variable and integration. With historical notes, Mathematische Leitfaden. B. G. Teubner, Stuttgart, 1976. 1181 O.V. Besov, V.P. I1'in, and S.M. Nikol'skil, Integral representations of functions and imbedding theorems, Vol. I. Translated from the Russian, Scripta Series in Mathematics. Edited by Mitchell H. Taibleson, V. H. Winston & Sons, Washington, D.C.; Halsted Press [John Wiley & Sons], New York-Toronto, Ontario-London, 1978.

[191 O.V. Besov, V.P. Il'in, and S.M. Nikol'skii, Integral representations of functions and imbedding theorems, Vol. II. Translated from the Russian, Scripts Series in Mathematics. Edited by Mitchell H. Taibleson. V. H. Winston & Sons, Washington, D.C.; Halsted Press [John Wiley & Sons], New York-Toronto, Ontario-London, 1979.

[201 L. Boccardo and F. Murat, Remarques sur l'homogdndisation de certain problemes quasi-lineairaes, Portugal. Math. 41 (1982), no. 1-4, 535-562 (1984). [211 M.W. Botsko, An elementary proof of Lebesgue's differentiation theorem, Amer. Math. Monthly 110 (2003), no. 9, 834-838. 1221 M.W. Botsko, An elementary proof that a bounded a, e. continuous function is Riemann integr able, Amer. Math. Monthly 95 (1988), no. 3, 249-252. [231 J. Bourgain, H. Brezis, and P. Mironescu, Another look at Sobolev spaces, J.L.

Menaldi, E. Rofman et A. Sulem, eds. Optimal control and partial differential equations. In honour of Professor Alain Bensoussan's 60th birthday. Proceedings of the conference, Paris, France, December 4, 2000. Amsterdam: IOS Press; Tokyo: Ohmsha. 439-455 (2001). [241 H. Brezis, Analyse fonctionnelle. Thdorie et applications, Collection Math4matiques Appliqu4es pour la Maitrise, Masson, Paris, 1983. [251 H. Brezis, How to recognize constant functions. A connection with Sobolev spaces.

Uspekhi Mat. Nauk 57 (2002), no. 4(346), 59-74; translation in Russian Math. Surveys 57 (2002), no. 4, 693-708. [261 L.E.J. Brouwer, Uber Abbildung von Mannigfaltigkeiten, Math. Ann. 71 (1911), no. 1, 97-115.

[271 A. Bruckner, Differentiation of real functions, second edition, CRM Monograph Series, 5, American Mathematical Society, Providence, RI, 1994. [28] V.I. Burenkov, Sobolev spaces on domains, Teubner-Texte zur Mathematik, 137. B. G. Teubner Verlagsgesellachaft mbH, Stuttgart, 1998. [291 F.S. Cater, When total variation is additive, Proc. Amer. Math. Soc. 84 (1982), no. 4, 504-508. [301 L. Cesari, Rectifiable curves and the Weierstrass integral, Amer. Math. Monthly 85 (1958), no. 7, 485-500. 1311 G. Chiti, Rearrangements of functions and convergence in Orlicz spaces, Applicable Anal, 9 (1979), no. 1, 23-27. [321 M. Chlebfk, A. Cianchi, and N. Fusco, The perimeter inequality under Steiner symmetrization: Cases of equality, Ann. of Math. (2) 162 (2005), no. 1, 525-555.

Bibliography

595

[33] S.K. Chua and R.I. Wheeden, Sharp conditions for weighted 1-dimensional Poincare inequalities, Indiana Univ. Math. J. 49 (2000), no. 1, 143-175.

[34] S.K. Chua and R.I. Wheeden, Estimates of best constants for weighted Poincard inequalities on convex domains, Proc. London Math. Soc. (3) 93 (2006), no. 1, 197226.

135J A. Cianchi, Second-order derivatives and rearrangements, Duke Math. J. 106 (2000), no. 3, 355-385. [361 A. Cianchi and A. Ferone, A strengthened version of the Hardy-Littlewood inequality,

J. Lord. Math. Soc. (2) 77 (2008), no. 3, 581-592. [371 A. Cianchi and N. Fusco, Functions of bounded variation and rearrangements, Arch. Ration. Mech. Anal. 165 (2002), no. 1, 1-40. [381 A. Cianchi, N. Fusco, F. Maggi, and A. Pratelli, The sharp Sobolev inequality in quantitative form, to appear in the Journal of the European Mathematical Society. [391 J. Ciemnoczolowski and W. Orlicz, Composing functions of bounded cp-variation, Proc. Amer. Math. Soc. 98 (1986), no. 3, 431-436. [401 D. Cioranescu and J. Saint Jean Paulin, Homogenization in open sets with holes, J. Math. Anal. Appl. 71 (1979), no. 2, 590-607. 1411 M.G. Crandall and L. Tartar, Some relations between nonexpansive and order preserving mappings, Proc. Amer. Math. Soc. 78 (1980), no. 3, 385-390. [421 M. Csornyei, Absolutely continuous functions of Rado, Reichelderfer, and Matt, J. Math. Anal. Appl. 252 (2000), no. 1, 147-166. 1431 B.E.J. Dahlberg, Total curvature and rearrangements. Posthumous paper prepared for publication by Vilhelm Adolfsson and Peter Kumlin. Ark. Mat. 43 (2005), no. 2, 323-345. [441 G. Dal Maso, BV functions, SISSA. [451 E. De Giorgi, Definizione ed espressione analitica del perimetro di un insieme, Atti Accad. Naz. Lincei. Rend. CL Sci. Fis. Mat. Nat. (8) 14 (1953), 390-393. [461 E. De Giorgi, Selected papers, edited by Luigi Ambrosio, Gianni Dal Maso, Marco Forti, Mario Miranda, and Sergio Spagnolo. Springer-Verlag, Berlin, 2006. 1471 M.C. Delfour and J.P.

Shapes and geometries. Analysis, differential calculus,

and optimization, Advances in Design and Control, 4. Society for Industrial and Applied Mathematics (SIAM), Philadelphia, PA, 2001. [481 E. DiBenedetto, Real analysis, Birkhaueer Advanced Texts: Basler Lehrbucher, Birkhauser Boston, Inc., Boston, MA, 2002. [491 0. Dovgoshey, 0. Martio, V. Ryazanov, and M. Vuorinen, The Cantor function, Expo. Math. 24 (2006), no. 1, 1-37. [501 G.F.D. Duff, Differences, derivatives, and decreasing rearrangements, Canad. J. Math. 19 (1967), 1153-1178. [511 Y. Ebihara and T.P. Schonbek, On the (non)compactness of the radial Sobolev spaces, Hiroshima Math. J. 16 (1986), 665-669. (52J R.E. Edwards, Functional analysis. Theory and applications, corrected reprint of the 1965 original, Dover Publications, Inc., New York, 1995.

[531 L.C. Evans, Partial differential equations, Graduate Studies in Mathematics, 19, American Mathematical Society, Providence, RI, 1998. [541 L.C. Evans and R.F. Gariepy, Measure theory and fine properties of functions, Studies in Advanced Mathematics, CRC Press, Boca Raton, FL, 1992.

596

Bibliography

[551 K.J. Falconer, The geometry of fractal sets, Cambridge Tracts in Mathematics, 85. Cambridge University Press, Cambridge, 1986. 1561 C.A. Faure, The Lebesgue differentiation theorem via the rising sun lemma, Real Anal. Exchange 29 (2003/04), no. 2, 947-951. [571 H. Federer, Surface area. 1, Trans. Amer. Math. Soc. 55 (1944), 420-437. [581 H. Federer, Geometric measure theory, Die Grundlehren der mathematischen Wissenschaften, Band 153, Springer-Verlag New York Inc., New York, 1969. [591 C. Fefferman, Whitney's extension problem for C', Ann. of Math. (2) 164 (2006), no. 1, 313-359. [601 A. Ferone and R. Volpicelli, Polar factorization and pseudo-rearrangements: Applications to P6lya-Szeg6 type inequalities, Nonlinear Anal. 53 (2003), no. 7-8, 929-949. 1611 W.E. Fleming and R. Rishel, An integral formula for total gradient variation, Arch. Math. 11 (1960), 218-222. [621 W.E. Fleming, Functions of several variables, second edition. Undergraduate Texts in Mathematics. Springer-Verlag, New York-Heidelberg, 1977. [631 T.M. Flett, On transformations in R" and a theorem of Said, Amer. Math. Monthly 71 (1964), 623-629. 1641 G.B. Folland, Real analysis. Modern techniques and their applications, second edition, Pure and Applied Mathematics, A Wiley-Interscience Series of Texts, Monographs, and Tracts, New York, 1999. 1651 1. Fonseca and G. Leoni, Modern methods in the calculus of variations: La spaces, Springer Monographs in Mathematics, Springer, New York, 2007. 1661 L.E. Fraenkel, On regularity of the boundary in the theory of Sobolev spaces, Proc. London Math. Soc. (3) 39 (1979), no. 3, 385-427. [671 G. Preilich, Increasing continuous singular functions, Amer. Math. Monthly 80 (1973), 918-919. [681 N. Fusco, F. Maggi, and A. Pratelli, The sharp quantitative Sobolev inequality for functions of bounded variation, J. Funct. Anal. 244 (2007), no. 1, 315-341. [691 E. Gagliardo, Caratterizzazioni delle tracce sulfa frontiera relative ad alcune ctasai di funzioni in n variabei, Rend. Sem. Mat. Univ. Padova 27 (1957), 284-305. [701 E. Gagliardo, Propriet4 di alcune classi di funzioni in pirl variabili, Ricerche Mat. 7 (1958), 102-137. [711 R.J. Gardner, The Brunn-Minkowski inequality. Bull. Amer. Math. Soc. (N.S.) 39 (2002), no. 3, 355-405. [721 F.W. Gehring, A study of a-variation, Trans. Amer. Math. Soc. 76 (1954), 420-443. 1731 D. Gilbarg and N.S. Trudinger, Elliptic partial differential equations of second order, reprint of the 1998 edition, Classics in Mathematics, Springier-Verlag, Berlin, 2001. [741 E. Giusti, Minimal surfaces and functions of bounded variation, Monographs in Mathematics, 80. Birkh5user Verlag, Basel, 1984. [751 C. Goffinan, On functions with summable derivative, Amer. Math. Monthly 78 (1971), 874-875. 1761 G.S. Goodman, Integration by substitution, Proc. Amer. Math. Soc. 70 (1978), no. 1,89-91. [77] L.M. Graves, The theory of functions of real variables, first edition, McGraw-Hill Book Company, Inc., New York and London, 1946.

Bibliography

597

[781 H. Hajaiej and C.A. Stuart, Symmetrization inequalities for composition operators of Caratheodory type, Proc. London Math. Soc. (3) 87 (2003), no. 2, 396-418. [791 H. Hajaiej, Cases of equality and strict inequality in the extended Hardy-Littlewood inequalities, Proc. Roy. Soc. Edinburgh Sect. A 136 (2005), no. 3, 643-661. 1801 P. Hajlasz, Change of variables formula under minimal assumptions, Colloq. Math. 64 (1993), no. 1, 93-101. 1811 P. Hajlasz and P. Koekela, Sobolev ,net Poincard, Mem. Amer. Math. Soc. 145 (2000), no. 688, [821 G.H. Hardy, Weierstrass's non-differentiable function, Trans. Amer. Math. Soc. 17 (1916), no. 3, 301-325. 1831 L.I. Hedberg, On certain convolution inequalities, Proc. Amer. Math. Soc. 35 (1972), 505-510. 1841 J. Heinonen, Nonsmooth calculus, Bull. Amer. Math. Soc. (N.S.) 44 (2007), no. 2, 163-232.

1851 G. Helmberg, An absolutely continuous function in L' (R) \ (R), Amer. Math. Monthly 114 (2007), no. 4, 356-357. [861 S. Hencl and J. Malt, Absolutely continuous functions of several variables and difjeomorphisms, Cent. Eur. J. Math. 1 (2003), no. 4, 690-705. 1871 G.A. Heuer, The derivative of the total variation function, Amer. Math. Monthly 78 (1971), 1110-1112.

[881 E. Hewitt, Integration by parts for StieUje8 integrals, Amer. Math. Monthly 67 (1960), 419-423. [891 E. Hewitt and K. Stromberg, Beat and abstract analysis. A modern treatment of the theory of functions of a real variable, Springer-Verlag, New York, 1965. [901 D. Hilbert, Ueber die stetige Abbildung einer Line auf ein Flachenstiick, Math. Ann. 38 (1891), no. 3, 459-460. [911 K. Hildbn, Symmetrization of functions in Sobolev spaces and the isoperimetric inequality, Manuscripts Math. 18 (1976), no. 3, 2155-235.

[921 P.W. Jones, Quasiconformal mappings and extendabiiity of functions in Sobolev spaces, Acta Math. 147 (1981), no. 1-2, 71-88. [931 M. Josephy, Composing functions of bounded variation, Proc. Amer. Math. Soc. 83 (1981), no. 2, 354-356. [941 W.J. Kaczor and M.T. Nowak, Problems in mathematical analysis. II. Continuity and differentiation, translated from the 1998 Polish original, revised and augmented by the authors. Student Mathematical Library, 12, American Mathematical Society, Providence, RI, 2001. [951 G.E. Karadzhov, M. Milman, and J. Xiao, Limits of higher-order Besov spaces and sharp reiteration theorems, J. Funct. Anal. 221 (2005), no. 2, 323-339. [961 Y. Katznelson and K. Stromberg, Everywhere differentiable, nowhere monotone, functions, Amer. Math. Monthly 81 (1974), 349-354. [971 B. Kawohl, Rearrangements and convexity of level sets in PDE, Lecture Notes in Mathematics, 1150, Springer-Verlag, Berlin, 1985. [981 S. Kesavan, Symmetrization & applications, Series in Analysis, 3, World Scientific Publishing Co. Pte. Ltd., Hackensack, NJ, 2006. [991 H. Kober, On singular functions of bounded variation, J. London Math. Soc. 23 (1948),222-229.

Bibliography

598

[100] T. Kolsrud, Approximation by smooth functions in Sobolev spaces, a counterexample, Bull. London Math. Soc. 13 (1981), no. 2, 167-169.

[101] S.G. Krantz, Lipschitz spaces, smoothness of functions, and approximation theory, Exposition. Math. 1 (1983), no. 3, 193-260. [102] K. Krzyiewsld, On change of variable in the Denjoy-Perron integral, 1, Colloq. Math. 9 (1962), 99-104. [103] U. Lang, Introduction to geometric measure theory, 2005. [104] P.D. Lax, Change of variables in multiple integrals, Amer. Math. Monthly 106 (1999), no. 6, 497-501. [105] P.D. Lax, Change of variables in multiple integrals. II, Amer. Math. Monthly 108 (2001), no. 2, 115-119. [106] G. Leoni and M. Morini, Necessary and sufficient conditions for the chain rule in

IV] (RN;Rd) and BWo, (RN ;Rd), J. Eur. Math. Soc. (JEMS) 9 (2007), no. 2, 219-252.

[107] Y. Li and B. Ruf, A sharp Trudinger-Moser type inequality for unbounded domains in R", Indiana Univ. Math. J. 57 (2008), no. 1, 451-480. [108] E.H. Lieb and M. Loss, Analysis, second edition, Graduate Studies in Mathematics, 14. American Mathematical Society, Providence, RI, 2001.

[109] S. Lindner, Additional properties of the measure vf, Tatra Mt. Math. Publ. 28 (2004), part II, 199-205. [110] G. Lu and R.I. Wheeden, Poincare inequalities, isoperimetric estimates, and representation formulas on product spaces, Indiana Univ. Math. J. 47 (1998), no. 1, 123-151.

[111] R.. Maehara, The Jordan curve theorem via the Bmuwer fixed point theorem, Amer. Math. Monthly 91 (1984), no. 10, 641-643. [112] F. Maggi and C. Villani, Balls have the worst best Sobolev inequalities, J. Geom. Anal. 15 (2005), no. 1, 83-121. [113] J. Malt, Absolutely continuous functions of several variables, J. Math. Anal. Appl.

231 (1999), no. 2, 492508. [114] J. Malt, A simple proof of the Stepanov theorem on differentiability almost everywhere, Exposition. Math. 17 (1999), no. 1, 59-61. [115] J. Malt and 0. Martio, Lusin's condition (N) and mappings of the class Wi'", J. Reine Angew. Math. 458 (1995), 19-36. [116] M. Marcus and V.J. Mizel, Tiansformations by functions in Sobolev spaces and lower aemicontinuity for parametric variational problems, Bull. Amer. Math. Soc. 79 (1973), 790-795. (117] M. Marcus and V.J. Mizel, Absolute continuity on tracks and mappings of Sobolev spaces, Arch. Rational Mech. Ana L 45 (1972), 294-320. [118] M. Marcus and V.J. Mizel, Complete characterization of functions which act, via superposition, on Sobolev spaces, Trans. Amer. Math. Soc. 251 (1979), 187-218. (119] J. Martin, M. Milman, and E. Pustylnik, Sobolev inequalities: Symmetrization and self-improvement via truncation, J. Funct. Anal. 252 (2007), no. 2, 677-695. [120] V.G. Maz`ja, Sobolev spaces, translated from the Russian by T. 0. Shaposhnilowa, Springer Series in Soviet Mathematics, Springer-Verlag, Berlin, 1985. [121] V.G. Maz'ja and T. Shaposhnikova, On the Brezis and Mironescu conjecture concerning a Gagliardo-Nirenberg inequality for fractional Sobolev norms, J. Math.

Bibliography

599

Pures Appl. (9) 81 (2002), no. 9, 877-884. Erratum J. Funct. Anal. 201 (2003), no. 1, 298-300. (122) N.G. Meyers and J. Serrin, H = W, Proc. Nat. Acad. Sci. U.S.A. 51 (1964), 10551056.

[123] N. Merentes, On the composition operator in AC[a,b], Collect. Math. 42 (1991), no. 3, 237-243 (1992). [124] J.W. Milnor, Topology from the differentiable viewpoint, based on notes by David W. Weaver. Revised reprint of the 1965 original. Princeton Landmarks in Mathematics. Princeton University Press, Princeton, NJ, 1997.

(125] E.H. Moore, On certain crinkly curves, Trans. Amer. Math. Soc. 1 (1900), no. 1, 72-90. Errata, Trans. Amer. Math. Soc. 1 (1900), no. 4, 507. [126] M. Morini, A note on the chain rule in Sobolev spaces and the differentiability of Lipschitz functions, preprint. [127] A.P. Morse, Convergence in variation and related topics, Trans. Amer. Math. Soc. 41 (1937), no. 1, 48-83. Errata, Trans. Amer. Math. Soc. 41 (1937), no. 3, 482. (128] A.P. Morse, A continuous function with no unilateral derivatives, Trans. Amer. Math. Soc. 44 (1938), no. 3, 496-507. (129] A.P. Morse, The behavior of a function on its critical set, Ann. of Math. (2) 40 (1939), no. 1, 62-70. [130] J. Moser, A sharp form of an inequality by N. Trudinger, Indiana Univ. Math. J. 20 (1970/71), 1077-1092. [131] B. Muckenhoupt, Weighted norm inequalities for the Hardy maximal function, Trans. Amer. Math. Soc. 165 (1972), 207-226.

(132] I.P. Natanson, Theory of functions of a real variable, translated by Leo F. Boron with the collaboration of Edwin Hewitt, Frederick Ungar Publishing Co., New York, 1955.

(133] J. Nbcas, Les mr thodes directes en theorie des equations elliptiques, Masson et Cie, Editeurs, Paris, Academia, Editeurs, Prague, 1967.

[134] E. Novak, Two remarks on the decreasing rearrangement of a function, J. Math. Anal. Appl. 122 (1987), no. 2, 485-486. [135] L.E. Payne and H.F. Weinberger, An optimal Poincare inequality for convex domains, Arch. Rational Mech. Anal. 5 (1960), 286-292. [136] G. Peano, Sur une courbe, qui remplit toute une aire plane, Math. Ann. 36 (1890), no. 1, 157-160. [137] J. Peetre, A counterexample connected with Gagliardo's trace theorem, special issue dedicated to Wladyslaw Orlicz on the occasion of his seventy-fifth birthday, Comment. Math. Special Issue 2 (1979), 277-282.

(138] A. Pinkus, Weierstrass and approximation theory, J. Approx. Theory 107 (2000), no. 1, 1-66. [139] A.C. Ponce, A new approach to Sobolev spaces and connections to r-convergence, Calc. Var. Partial Differential Equations 19 (2004), no. 3, 229-255. [140] H. Rademacher, Uber partielle and totale differenzierbarkeit von 1%nktionen mehrerer Variabeln and caber die Transformation der Doppelintegrale, Math. Ann. 79 (1919), no. 4, 340-359. (141] F. Riesz, Sur l'existence de la derivr a des functions monotones et sur quelques problt mes qui s'y rattachent, Acta Sci. Math. 5 (1930-1932), 208-221.

Bibliography

600

[142] J.V. Ryff, Measure preserving transformations and rearrangements, J. Math. Anal. Appl. 31 (1970), 449-458. [143] W. Rudin, Real and complex analysis, third edition, McGraw-Hill Book Co., New York, 1987.

[144] W. Rudin, Functional analysis, second edition, International Series in Pure and Applied Mathematics, McGraw-Hill, Inc., New York, 1991. [145] S. Saks, Theory of the integral, second revised edition, English translation by L. C. Young, with two additional notes by Stefan Banach, Dover Publications, Inc., New York 1964.

[146] R. Salem, On some singular monotonic functions which are strictly increasing, Trans. Amer. Math. Soc. 53 (1943), 427-439. [147] B. Sandberg and R.A. Christianson, Problems and solutions: Solutions of advanced problems: 6007, Amer. Math. Monthly 83 (1976), no. 8, 663-664. [148] A. Sard, The measure of the critical values of differentiable maps, Bull. Amer. Math. Soc. 48 (1942), 883-890. [149] A. Sard, Images of critical sets, Ann. of Math. (2) 68 (1958), 247-259. [150] J. Schwartz, The formula for change in variables in a multiple integral, Amer. Math. Monthly 81 (1954), 81-85. [151] J. Serrin, On the differentiability of functions of several variables, Arch. Rational Mech. Anal. 7 (1961), 359-372. [152] J. Serrin, Strong convergence in a product space, Proc. Amer. Math. Soc. 13 (1962), 651-655.

[153] J. Serrin and D.E. Varberg, A general chain rule for derivatives and the change of variables formula for the Lebesgue integral, Amer. Math. Monthly 76 (1969), 514-520.

[154] W. Sierpinski, Sur la question de la mesurabilite de la base de M. Hamel, Fund. Math. 1 (1920), 105-111. [155] C.G. Simader, Sobolev's original definition of his spaces revisited and a comparison with nowadays definition, Boundary value problems for elliptic and parabolic operators (Catania, 1998). Matematiche (Catania) 54 (1999), suppl., 149-178. [156] W. Smith and D.A. Stegenga, Holder domains and Poincare domains, Trans. Amer. Math. Soc. 319 (1990), no. 1, 67-100. [157] V.A. Solonnikov, A priori estimates for second-order parabolic equations, Amer. Math. Soc., Ttansl., II. Ser. 65, 1967, 51-137. [158] V.A. Solonnikov, Inequalities for functions of the classes Wn (R"), J. Soy. Math. 3 (1975), 549-564. [159] E.J. Sperner, Our Symmetrisierung von Funktionen auf Spharen, Math. Z. 134 (1973), 317-327. [160] E.M. Stein, Singular integrals and differentiability properties of functions, Princeton Mathematical Series, no. 30, Princeton University Press, Princeton, N.J, 1970. [161] E.M. Stein and A. Zygmund, On the differentiability of functions, Studia Math. 23 (1963/1964),247-283. [162] W. Stepanoff, Uber totale Differenxierbarkeit, Math. Ann. 90 (1923), no. 3-4, 318320.

[163] W.A. Strauss, Existence of solitary waves in higher dimensions, Commun. Math. Phys. 55 (1977), 149-162.

Bibliography

601

[1641 L. Takkcs, An increasing continuous singular function, Amer. Math. Monthly 85 (1978), no. .35-37. [1651 G. Talenti, Best constant in Sobokv inequality, Ann. Mat. Pura Appl. (4) 110 (1976), 353-372. [1661 L. Tartar, An introduction to Sobolev spaces and interpolation spaces, Lecture Notes of the Unione Matematica Italiana, 3. Springer, Berlin; UMI, Bologna, 2007. [167] H. Triebel, Interpolation theory, function spaces, differential operators, second edition. Johann Ambrosius Barth, Heidelberg, 1995. [1681 N.S. Trudinger, On imbeddings into Orlicz spaces and some applications, J. Math. Mech. 17 (1967), 47:3-483.

[1691 H. Tverberg, A proof of the Jordan curve theorem, Bull. London Math. Soc. 12 (1980), no. 34-38. [1701 S.V. Uspenskil, Imbedding theorems for weighted classes, Amer. Math. Soc., Transl., .

II. Ser. 87, 1970, 121-145; translation from Trudy Mat. Inst. Steklov 60 (1961), 282-303. [1711 A.C.M. van Rooij and W.H. Schikhof, A second course on meal functions, Cambridge University Press, Cambridge-New York, 1982. [1721 F.S. Van Vleck, A remark concerning absolutely continuous functions, Amer. Math. Monthly 80 (1973), 286-287. [1731 D.E. Varberg, On absolutely continuous functions, Amer. Math. Monthly 72 (1965), 831-841.

[1741 D.E. Varberg, On differentiable transformations in R", Amer. Math. Monthly 73 (1966), no. 4, part II, 111-114. [1751 D.E. Varberg, Change of variables in multiple integrals, Amer. Math. Monthly 78 (1971), 42-45. [1761 H. Whitney, Analytic extensions of differentiable functions defined in closed sets, Trans. Amer. Math. Soc. 36 (1934), no. , 63-89.

[1771 H. Whitney, A function not constant on a connected set of critical points, Duke Math. J. (1935), no. 4, 514-517. [1781 H. Whitney, Differentiable functions defined in arbitrary subsets of Euclidean space, Trans. Amer. Math. Soc. 40 (1936), no. , 309-317. [1791 N. Wiener, The quadratic variation of a function and its Fourier coefficients, Mass. J. of Math. 3 (1924), 72-94. [1801 B.B. Winter, 7iunsformations of Lebesgue-Stieltjes integrals, J. Math. Anal. Appl. 205 (1997), no. 471-484. [1811 K. Yosida, Functional analysis, reprint of the sixth (1980) edition, Classics in Mathematics, Springer-Verlag, Berlin, 1995. [1821 W.P. Ziemer, Weakly differentiable functions. Sobolev spaces and functions of bounded variation, Graduate Texts in Mathematics, 120. Springer-Verlag, New York, 1989.

[1831 P.R. Zingano and S.L. Steinberg, On the Hardy-Littlewood theorem for functions of bounded variation, SIAM J. Math. Anal. 33 (2002), no. 5, 1199-1210.

Index absolute continuity of u", 208 of a function, 73, 241 of a measure, 522 of a signed measure, 525 absorbing set, 497 accumulation point, 494 algebra, 508 arclength of a curve, 128, 132 area formula, 100 atom, 510 background coordinates, 232 balanced set, 256, 497 Banach indicatrix, 66 Banach space, 501 Banach-Alaoglu's theorem, 504 base for a topology, 495 Besicovitch's covering theorem, 538 Besicovitch's derivation theorem, 539 bidual space, 499 Borel function, 511 boundary Lipschitz, 354 locally Lipschitz, 354 of class C, 287 uniformly Lipschitz, 354 Brouwer's theorem, 242 Brunn-Minkowski's inequality, 545

Cantor diagonal argument, 60 Cantor function, 31 Cantor part of a function, 108 Cantor set, 30 Carathdodory's theorem, 510 Cauchy sequence, 494, 498

Cauchy's inequality, 232 chain rule, 94, 145 change of variables, 98, 183, 346 for multiple integrals, 248 characteristic function, 514 closed curve, 116 simple, 116 closed set, 494 closure of a set, 494 coarea formula, 397 cofactor, 243 compact embedding, 320 compact set, 495 complete space, 494, 498 connected component, 14 exterior, 146 interior, 146 connected set, 14 continuous function, 495 continuum, 137 convergence almost everywhere, 534 almost uniform, 534 in measure, 534 in the sense of distributions, 264 strong, 494 weak, 503 weak star, 504 convergent sequence, 494 convolution, 275, 550 of a distribution, 275 counting function, 66 cover, 539 curve, 116 continuous, 116 parameter change, 115 603

Index

604

parametric representation, 115 cut-off function, 496, 559

finite width, 359 first axiom of countability, 495 FYdchet curve, 131

De Is Vallee Poussin's theorem, 173, 535 decreasing function, 3 decreasing rearrangement, 190, 478 delta Dirac, 264 dense set, 494 derivative, 8

of a distribution, 266 differentiability, 8 differentiable transformation, 233 differential, 233 Dini's derivatives, 20 directional derivative, 233 disconnected set, 14 diet-once, 493

distribution, 264 order infinite, 264 distribution function, 187, 477 distributional derivative, 215, 222, 267 distributional partial derivative, 279, 377 doubling property, 22 dual space, 499 dual spaces D' (ft), 284 Mb (X;R), 537 of W - (ft), 299

(l), 303 duality pairing, 499 Eberlein-Smulian's theorem, 505 edge of a polygonal curve, 146 Egoroff's theorem, 534 embedding, 502 compact, 503 equi-integrability, 535 equi-integrable function, 76 equivalent curves Frdchet, 131 Lebesgue, 115 equivalent function, 526 equivalent norms, 502 essential supremum, 526, 532

essential variation, 219 Euclidean inner product, 231 Euclidean norm, 232 extension domain for BV (fl), 402 for W '-P (fd), 320 extension operator, 320

Fo set, 29 Fatou's lemma, 516 fine cover, 539 finite cone, 355

Fubini's theorem, 35, 521 function of bounded pointwise variation, 39 in the sense of Cesari, 389 function of bounded variation, 377 function spaces ACv ((a, b]), 94

AC(1), 73 Agac (1), 74 AC (1; Rd), 74

B',P,e (R-), 415 B'-p-s (On), 474 BV P (I; Rd), 40 BV P (1), 39 BVP1OC (1), 40

BV ((2), 215, 377

BVI, (fl), 220 Co.- (f2), 335 C(X;Y), 495 Co(X),501 C, (X), 501 C°O (ft), 255

C'° (il), 255 Cm (E), 561 C'" (f2), 256 C," (ft), 255 CC (X), 496 D (cl), 259

DK (ft), 255 L1,P (fl), 282

LP (RN), 568 L°O (X), 526 LP (X), 526

LP,, 632 L'V (E), 331

PA, 292 Al (1), 11 W - (fl), 222 ylrs,P (RN), 44$ Wt,P (0), 279

W," (ci), 282 function vanishing at infinity, 187, 312, 477 functional locally bounded, 538 positive, 538 fundamental theorem of calculus, 85 Gd set, 29

Gagliardo's theorem, 453 Gamma function, 572 gauge, 498 geodesic curve, 133 gradient, 233

Index

7{k-rectifiable set, 143 Hahn-Banach's theorem, analytic form, 500 Hahn-Banach's theorem, first geometric form, 500 Hahn-Banach's theorem, second geometric form, 501 Hamel basis, 12 Hardy-Littlewood's inequality, 196, 482 Hausdorff dimension, 578 Hausdorff measure, 574 Hausdorff outer measure, 573 Hausdorff space, 494 Helly's selection theorem, 59 Hilbert space, 506 Hilbert's theorem, 118 Holder's conjugate exponent, 527, 568 Holder continuous function, 335 Holder's inequality, 527, 588 immersion, 502 increasing function, 3 indefinite pointwise variation, 44 infinite sum, 100 inner product, 506 inner regular set, 536 integrals depending on a parameter, 519 integration by parts, 89, 181 interior of a set, 494 interval, 3 inverse of a monotone function, 6 isodiametric inequality, 548 isoperimetric inequality, 405, 549

Jacobian, 233 Jensen's inequality, 518 Jordan's curve theorem, 146 Jordan's decomposition theorem, 524 Josephy's theorem, 55 jump function, 5

Kakutani'a theorem, 505 Katznelson-Stromberg's theorem, 50 Laplacian, 267 Lax's theorem, 243 Lebesgue integrable function, 517 Lebeegue integral of a nonnegative function, 514 of a simple function, 514 of a real-valued function, 516 Lebesgue measurable function, 545 Lebeegue measurable eat, 543 Lebesgue measure, 543 Lebeegue outer measure, 543 Lebesgue point, 540 Lebesgue's decomposition theorem, 523, 525

605

Lebesgue's dominated convergence theorem, 518 Lebesgue's monotone convergence theorem, 515

Lebesgue's theorem, 13 Lebesgue-Stieltjes measure, 157 Lebesgue-Stieltjes outer measure, 157 Leibnitz formula, 264 length function, 125 length of a curve, 118 o-finite, 118 Lipschitz continuous function, 342 local absolute continuity of a function, 74 local base for a topology, 495 local coordinates, 232 locally bounded pointwise variation, 40 locally compact space, 496 locally convex space, 498 locally finite, 496 locally integrable function, 517 locally rectifiable curve, 118 lower variation of a measure, 524 Lusin (N) property, 77, 208, 234, 340

It'-measurable set, 508 maximal function, 564 measurable function, 511, 513 measurable space, 509 measure, 509 v-finite, 509 absolutely continuous part, 526 Bore], 509 Borel regular, 537 complete, 509 counting, 516 finite, 509 finitely additive, 509 localizable, 532 nonatomic, 510 product, 520 Radon, 537 semifinite, 510 signed Radon, 537

singular part, 526 with the finite subset property, 510 measure space, 509 measure-preserving function, 202 measures mutually singular, 523, 525 metric, 493 metric space, 493 metrizable space, 497 Meyers-Serrin's theorem, 283 Minkowski content lower, 549 upper, 549 Minkowski functional, 498

Index

606

Minkowski's inequality, 531, 571 for integrals, 530 mollification, 553 mollifier, 552 standard, 553 monotone function, 3 Morrey's theorem, 335, 437 Muckenhoupt's theorem, 373 multi-index, 255

point of density one, 541 of density t, 541 pointwise variation, 39 polygonal curve, 146 positive pointwise variation, 45 precompact set, 496 principal value integral, 268 purely ?lk-unrectifiable set, 143

multiplicity of a point, 116

N-simplex, 291 negative pointwise variation, 45 neighborhood, 494 norm, 501 normable space, 501 normal space, 495 normed space, 501

open ball, 232, 493 open cube, 232 open set, 494 operator bounded,500 compact, 502 linear, 499 order of a distribution, 264 orthonormal basis, 232 outer measure, 507 Borel, 536 Borel regular, 536 metric, 511 product, 520 Radon, 536 regular, 536 outer regular set, 536 p-equi-integrability, 535 p-Lebeague point, 540 p-variation, 54 parallelogram law, 506 parameter of a curve, 115 partial derivative, 233 partition of an interval, 39 partition of unity, 496 locally finite, 497 smooth, 557 subordinated to a cover, 497 pat hwise connected set, 137 Peano's theorem, 116 perimeter of a set, 379 Poincar4's inequality, 225, 361, 405 for continuous domains, 363 for convex sets, 364 for rectangles, 363 for star-shaped sets, 3T0 in W.'D, 359

Rademacher's theorem, 343 radial function of a star-shaped domain, 370

Radon measure, 155 Radon-Nikodym's derivative, 523 Radon-Nikodym's theorem, 523 range of a curve, 116 rectifiable curve, 118 reflexive space, 505 regular set, 536 regularized distance, 358 relatively compact set, 496 Rellich-Kondrachov'a theorem, 320, 402 for continuous domains, 326 Riemann integration, 87 Riesz's representation theorem in Cc, 538 in Co, 538

in Lt, 533 in L°0,533 in V', 532 in W , in W in W

, 304

in Wo'OO, 305

Riesz's rising sun lemma, 14 rigid motion, 232 o-algebra, 508 Borel, 509 product, 512, 520 a-compact set, 496 a-locally finite, 496 ealtus function, 5 Sard's theorem, 408 Schwarz symmetric rearrangement, 479 second axiom of countability, 495 section, 521 segment property, 286 seminorm, 498 separable space, 494 sequentially weakly compact set, 505 Serrin's theorem, 389 set of finite perimeter, 379 sherically symmetric rearrangement, 479 signed Lebesgue-Stielties measure, 162 signed measure, 524

Index

bounded, 524 finitely additive, 523 simple arc, 116 simple function, 513 simple point of a curve, 116 singular function, 107, 212 Sobolev critical exponent, 312 Sobolev function, 222 Sobolev-Gagliardo-Nirenberg's embedding theorem, 312 spherical coordinates, 253 spherically symmetric rearrangement of a set, 479 star-shaped set, 370 Stepanoff's theorem, 344 strictly decreasing function, 3 strictly increasing function, 3 subharmonic function, 267 superposition, 104 support of a distribution, 271 surface integral, 578

tangent line, 119 tangent vector, 119 testing function, 259 Tonelli's theorem, 91, 125, 521 topological space, 494 topological vector space, 497 topologically bounded set, 498 topology, 494 total variation measure, 378 total variation norm, 533 total variation of a measure, 524 trace of a function, 452 upper variation of a measure, 524 Urysohn's theorem, 495 vanishing at infinity, 312 Varberg's theorem, 240 variation, 378 vectorial measure, 525 Radon, 538 vertex of a polygonal curve, 146 vertex of a symplex, 291 Vitali's convergence theorem, 535 Vitali's covering theorem, 20, 408 Vitali-Besicovitch's covering theorem, 539 weak derivative, 215, 222, 267 weak partial derivative, 279, 377 weak star topology, 503 weak topology, 503 Weierstrass's theorem, 9 weighted Poincar6's inequality, 226 Whitney's decomposition, 564 Whitney's theorem, 561

607

Young's inequality, 527, 551 Young's inequality, general form, 551

Sobolev spaces are a fundamental tool in the modern study of partial differential equations. In this book, Leoni takes a novel

approach to the theory by looking at Sobolev spaces as the natural development of monotone, absolutely continuous, and BV functions of one variable. In this way, the majority of the text wit hout t eh prerequisite o fa course in nctrona

can b a rea d analysis.

117.

e

I

The first part of this text is devoted to studying functions of one

variable. Several of the topics treated occur in courses on real analysis or measure theory. Here, the perspective emphasizes their applications to Sobolev functions, giving a very different flavor to the treatment. This elementary start to the book

makes it suitable for advanced undergraduates or beginning graduate students. Moreover, the one-variable part of the book helps to develop a solid background that facilitates the reading and understanding of Sobolev functions of several variables.

The second part of the book is more classical, although it also contains some recent

results. Besides the standard results on Sobolev functions, this part of the book includes chapters on BV functions, symmetric rearrangement, and Besov spaces. The book contains over 200 exercises.

ISBN 978-0-8218-4768-8

I

For additional information and updates on this book, visit

www.ams.org/bookpages/gsm- I OS 9

78082111847688 GSM/ 105

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