Advanced Calculus: An Introduction to Linear Analysis

ADVANCED CALCULUS An Introduction to Linear Analysis Leonard F. Richardson ~WILEY ~INTERSCIENCE A JOHN WILEY & SONS,...

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ADVANCED CALCULUS An Introduction to Linear Analysis

Leonard F. Richardson

~WILEY ~INTERSCIENCE A JOHN WILEY & SONS, INC., PUBLICATION

ADVANCED CALCULUS

ADVANCED CALCULUS An Introduction to Linear Analysis

Leonard F. Richardson

~WILEY ~INTERSCIENCE A JOHN WILEY & SONS, INC., PUBLICATION

Copyright© 2008 by John Wiley & Sons, Inc. All rights reserved Published by John Wiley & Sons, Inc., Hoboken, New Jersey Published simultaneously in Canada No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section I 07 or I 08 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fcc to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., Ill River Street, Hoboken, NJ 07030, (20 I) 748-6011, fax (20 I) 748-6008, or online at http://www.wilcy.com/go/pcrmission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wilcy.com.

Library of Congress Cataloging-in-Publication Data: Richardson, Leonard F. Advanced calculus : an introduction to linear analysis I Leonard F. Richardson. p.cm. Includes bibliographical references and index. ISBN 978-0-470-23288-0 (cloth) I. Calculus. I. Title. QA303.2.R53 2008 515--dc22 2008007377 Printed in Mexico 10 9 8 7 6 5 4 3 2

To Joan, Daniel, and Joseph

CONTENTS

Preface

Xlll

Acknowledgments

XIX

Introduction

xxi

PART I

1

ADVANCED CALCULUS IN ONE VARIABLE

Real Numbers and Limits of Sequences

3

1.1

3 7

1.3

The Real Number System Exercises Limits of Sequences & Cauchy Sequences Exercises The Completeness Axiom and Some Consequences

1.4

Exercises Algebraic Combinations of Sequences

1.2

1.5

1.6

Exercises The Bolzano-Weierstrass Theorem Exercises The Nested Intervals Theorem

8 12 13

18 19 21 22 24 24 vii

viii

CONTENTS

1.7

Exercises The Heine-Borel Covering Theorem

1.8

Exercises Countability of the Rational Numbers Exercises

1.9

2

Test Yourself Exercises

Continuous Functions

39

2.1

Limits of Functions

2.2

Exercises Continuous Functions

2.3

Exercises Some Properties of Continuous Functions

2.4

Exercises Extreme Value Theorem and Its Consequences

2.5

The Banach Space C[a, b] Exercises

2.6


39 43 46 49 50 53 55 60 61 66 67 67

Riemann Integral

69

3.1

Definition and Basic Properties Exercises

3.2

The Darboux Integrability Criterion Exercises Integrals of Uniform Limits Exercises The Cauchy-Schwarz Inequality Exercises

69 74 76 81 83 87 90 93 95 95

Exercises

3

3.3 3.4 3.5

4

26 27 30 31 35 37 37


The Derivative

4.1 4.2

Derivatives and Differentials Exercises The Mean Value Theorem

99

99 103 105

CONTENTS

Exercises

4.3

The Fundamental Theorem of Calculus Exercises

4.4

Uniform Convergence and the Derivative Exercises

4.5

Cauchy's Generalized Mean Value Theorem Exercises Taylor's Theorem

4.6 4.7

5

Exercises Test Yourself Exercises

109 110 112 114 116 117 121 122 125 126 126

Infinite Series

127

5.1

127 132 134 137 138 146 148 153 154 157 158 161 162 167 169 173 174 174

5.2 5.3 5.4 5.5 5.6 5.7 5.8 5.9

Series of Constants Exercises Convergence Tests for Positive Term Series Exercises Absolute Convergence and Products of Series Exercises The Banach Space l 1 and Its Dual Space Exercises Series of Functions: The Weierstrass M-Test Exercises Power Series Exercises Real Analytic Functions and c= Functions Exercises Weierstrass Approximation Theorem Exercises Test Yourself Exercises PART II

6

ix

ADVANCED TOPICS IN ONE VARIABLE

Fourier Series

179

6.1

180 183 184 190

6.2

The Vibrating String and Trigonometric Series Exercises Euler's Formula and the Fourier Transform Exercises

X

CONTENTS

6.3

Bessel's Inequality and lz Exercises

6.4

Uniform Convergence & Riemann Localization Exercises

6.5

L 2 -Convergence & the Dual of l 2 Exercises

6.6


7

192 196 197 204 205 208 212 212

The Rlemann-Stieltjes Integral

215

7.1

216 220 223 227 228 230 231 239 241 241

Functions of Bounded Variation Exercises

7.2

Riemann-Stieltjes Sums and Integrals Exercises

7.3

Riemann-Stieltjes Integrability Theorems Exercises

7.4

The Riesz Representation Theorem

7.5

Test Yourself

Exercises Exercises

PART Ill ADVANCED CALCULUS IN SEVERAL VARIABLES 8

Euclidean Space

245

8.1

245 249 252 254 256 258 259 261 263 263

Euclidean Space as a Complete Norrned Vector Space Exercises

8.2

Open Sets and Closed Sets Exercises

8.3

Compact Sets Exercises

8.4

Connected Sets Exercises

8.5


9

Continuous Functions on Euclidean Space

265

9.1

265 268

Limits of Functions Exercises

CONTENTS

9.2

Continuous Functions Exercises

9.3

Continuous Image of a Compact Set Exercises Continuous Image of a Connected Set Exercises Test Yourself Exercises

9.4 9.5

10

270 272 274 276 278 279 280 280

The Derivative in Euclidean Space

283

10.1

283 286 289 295 298 300 301 303 305 309 311 317 322 327 328 328

10.2 10.3

10.4 10.5 10.6 10.7

11

xi

Linear Transformations and Norms Exercises Differentiable Functions Exercises The Chain Rule in Euclidean Space 10.3.1 The Mean Value Theorem 10.3.2 Taylor's Theorem Exercises Inverse Functions Exercises Implicit Functions Exercises Tangent Spaces and Lagrange Multipliers Exercises Test Yourself Exercises

Riemann Integration in Euclidean Space

331

11.1

331 336 338 341 342 344 346 349 351 355

11.2 11.3 11.4 11.5

Definition of the Integral Exercises Lebesgue Null Sets and Jordan Null Sets Exercises Lebesgue's Criterion for Riemann Integrability Exercises Fubini's Theorem Exercises Jacobian Theorem for Change of Variables Exercises

Xii

CONTENTS


357 357

Appendix A: Set Theory A. I Terminology and Symbols Exercises A.2 Paradoxes

359 359 363 363

Problem Solutions

365

References

379

Index

381

11.6

PREFACE

Why this Book was Written The course known as Advanced Calculus (or Introductory Analysis) stands at the summit of the requirements for senior mathematics majors. An important objective of this course is to prepare the student for a critical challenge that he or she will face in the first year of graduate study: the course called Analysis I, Lebesgue Measure and Integration, or Introductory Functional Analysis. We live in an era of rapid change on a global scale. And the author and his department have been testing ways to improve the preparation of mathematics majors for the challenges they will face. During the past quarter century the United States has emerged as the destination of choice for graduate study in mathematics. The influx of well-prepared, talented students from around the world brings considerable benefit to American graduate programs. The international students usually arrive better prepared for graduate study in mathematics-in particular better prepared in analysis-than their typical U.S. counterparts. There are many reasons for this, including (a) school systems abroad that are oriented toward teaching only the brightest students, and (b) the self-selection that is part of a student taking the step of travel abroad to study in a foreign culture. The presence of strongly prepared international students in the classroom raises the level at which courses are taught. Thus it is appropriate at the present time, in the early years of the new millennium, for college and university mathematics departments to xiii

XiV

PREFACE

reconsider their advanced calculus courses with an eye toward preparing graduates for the international environment in American graduate schools. This is a challenge, but it is also an opportunity for American students and international students to learn side-by-side with, and also about, one another. It is more important than ever to teach undergraduate advanced calculus or analysis in such a way as to prepare and reorient the student for graduate study as it is today in mathematics. Another recent change is that applied mathematics has emerged on a large scale as an important component of many mathematics departments. In applied and numerical mathematics, functional analysis at the graduate level plays a very important role. Yet another change that is emerging is that undergraduates planning careers in the secondary teaching of mathematics are being required to major in mathematics instead of education. These students must be prepared to teach the next generation of young people for the world in which they will live. Whether or not the mathematics major is planning an academic career, he or she will benefit from better preparation in advanced calculus for careers in the emerging world. The author has taught mathematics majors and graduate students for thirty-seven years. He has served as director of his department's graduate program for nearly two decades. All the changes described above are present today in the author's department. This book has been written in the hope of addressing the following needs. 1. Students of mathematics should acquire a sense of the unity of mathematics. Hence a course designed for senior mathematics majors should have an integrative effect. Such a course should draw upon at least two branches of mathematics to show how they may be combined with illuminating effect. 2. Students should learn the importance of rigorous proof and develop skill in coherent written exposition to counter the universal temptation to engage in wishful thinking. Students need practice composing and writing proofs of their own, and these must be checked and corrected. 3. The fundamental theorems of the introductory calculus courses need to beestablished rigorously, along with the traditional theorems of advanced calculus, which are required for this purpose. 4. The task of establishing the rigorous foundations of calculus should be enlivened by taking this opportunity to introduce the student to modern mathematical structures that were not presented in introductory calculus courses. 5. Students should learn the rigorous foundations of calculus in a manner that reorient 0 and for all n E N. Most of the current chapter, however, will deal with other consequences of completeness, that we will begin exploring right now. Definition 1.3.2 A number M is called an upper bound for a set A c JR. if and only iffor all a E A we have a :::; M. Similarly, a number m is called a lower bound for A if and only if for all a E A we have a ;::: m. A set A of real numbers is called bounded provided that it has both an upper bound and a lower bound. A least upper bound for a set A is an upper bound L for A with the property that no number L' < L is an upper bound of A. A least upper bound is denoted by lub( A).

Note that not every subset of JR. has an upper or a lower bound. For example, N has no upper bound, and Z has neither an upper nor a lower bound. It is important to bear in mind also that many bounded sets of real numbers have neither a largest nor a smallest element. For example, this is true for the set of numbers in the open interval (0, 1). The reader should prove this claim as an informal exercise. Theorem 1.3.1 bound£.

If a nonempty set S

has an upper bound, then S has a least upper

Remark 1.3.2 If S has an upper bound, then its least upper bound is denoted by lub(S). Iflub(S) exists, then it must have a unique value L. The reader should prove that no number greater or smaller than L could satisfy the definition of lub(S).

Proof: Since S f. 0, there exists s E S. Select any number a 1 < s so that a 1 is too small to be an upper bound for S. Let b1 be any upper bound of S. We will use a process known as interval halving, in which we will cut the interval [a1. b1 ] in half again and again without end. The midpoint between a 1 and b1 is a 1 1 •

;b

i. If~ is an upper bound for S, then let b2 ii. But if

a

1

;b

1

=

a1

;b

is not an upper bound for S, then let a2

1

and let a 2

=

= a1.

~ and let b2

= b1.

Thus we have chosen [a2, b2] to be one of the two half-intervals of [a 1 , b1 ], and we have done this in such a way that b2 is again an upper bound of S and a 2 is too small to be an upper bound for S. Now we cut [a 2 , b2] in half and select a half-interval of it to be [a3, b3] in the same way we did for [a 2, b2]. Note that

THE COMPLETENESS AXIOM AND SOME CONSEQUENCES

15

as N ----t oo. Thus if E > 0, there exists N E N, corresponding to E, such that IbN- aNI <E. But, if n ~ N, then an and bn E [aN, bN], son and m ~ N implies ian- ami < E and also Ibn- bml < E. Thus an and bn are Cauchy sequences. Hence an ----t a and bn ----t b, for some real numbers a, b. By Exercise 1.22, a and b are in [aN, bN], for all N. Thus 0 ~ Ia- bl < t:, for all t: > 0. Thus Ia- bl = 0 and a= b. We claim that the number L = a = b is the least upper bound of S. Note that for each k we have ak ~ L ~ bk, since for all j ~ k we have ai and bi E [ak, bk]. First, observe that if s E S, then s ~ L. In fact, if we did haves > L, then, since bk ----t L, for some big enough value of k we would have lbk - Ll < Is- Ll and so bk < s. But this is impossible, since bk is an upper bound of S. Thus s ~ Land L is an upper bound of S. Finally, we claim Lis the least upper bound of S. In fact, suppose L' < L. Then since ak ----t L, there exists k such that L' < ak. But ak is not an upper bound of S. Thus L' cannot be an upper bound of S. • Remark 1.3.3 The proof of Theorem 1.3 .I is the most difficult proof presented thus far in this book. It proceeds by the method of interval-halving. This method can be likened to the way that a first baseman and a second baseman in a baseball game will attempt to tag a base-runner out by throwing the ball back and forth between them, steadily reducing the distance between them until one baseman is close enough to tag the runner. Interval halving is a very useful method of calculating roots of equations with a computer, provided it is possible to tell from the endpoints of each half-interval which half would need to contain the root. The student should take careful note of how the method of interval-halving produces two natural Cauchy sequences, an and bn, corresponding to the left and right endpoints of the selected half-intervals. Corollary 1.3.1 If S is any nonempty set of real numbers that has a lower bound, then S has a greatest lower bound.

For the proof see Exercise 1.28 in this section. Remark 1.3.4 If S has a lower bound, then its greatest lower bound is denoted by glb(S).

Since not every subset S c R. has either an upper or a lower bound, least upper bounds and greatest lower bounds do not exist in every case. Thus we introduce the concepts ofthe supremum and the infimum of an arbitrary setS c R.. Definition 1.3.3 Let S be any nonempty subset of R Define the supremum of S, denoted sup(S), to be the least upper bound of S if Sis bounded above and define sup( B) = oo if S has no upper bound. Similarly, define the infimum of S, denoted inf(S), to be the greatest lower bound of S if S is bounded below, and define inf(S) = -oo if S has no lower bound.

Thus sup( B) = {

~b(S)

if S is bounded above, if S is not bounded above

16

REAL NUMBERS AND LIMITS OF SEQUENCES

and inf(S) = {

~~S)

if S is bounded below, if S is not bounded below .

• EXAMPLE 1.5

LetS= {xI 0 < x < 1} = (0, 1). Then sup(S) = 1 and inf(S) = 0. Proof:

Clearly, 1 is an upper bound of S. But if M < 1, then there exists

S n (M, 1). Thus M cannot be an upper bound of S. Hence 1 is the least • upper bound of S. The argument for inf(S) is similar. x

E

• EXAMPLE 1.6

Observe that sup(N) = oo and inf(N) = 1. This follows because N has no upper bound, but N does have a least element, namely 1.

Definition 1.3.4 We call a sequence Xn increasing provided n E N, and then we write this symbolically as

Xn

1.¥1 and IYn MI < l~l2. 1 1 Thus IYn MI is bounded by max { IY MI, ... , IYN~IMI, 1~12 }· 1

•

EXERCISES 1.47

Give examples of divergent sequences Xn and Yn such that Xn + Yn converges.

1.48 Let a E lR be arbitrary. Give examples of sequences Xn ___. oo and Yn ___. oo such that Xn - Yn ___. a. 1.49

Give examples of divergent sequences Xn and Yn such that XnYn converges.

1.50 Let the real number a ~ 0 be arbitrary. Give examples of sequences Xn ___. oo and Yn ___. oo such that :£n. ___. a. Yn

1.51 Prove or else give a counterexample: If Xn converges, then Xn converges and Yn converges. 1.52

+ Yn

converges and if Xn - Yn

Prove or else give a counterexample: If ad - be =f. 0 and if

axn

+ byn ___. L and CXn + dyn

___. M

as n ___. oo, then Xn converges and Yn converges.

1.53 Suppose for all n E N we have Yn =f. 0. Prove or else give a counterexample: If both XnYn and :£n. converge, then Xn converges and Yn converges. Yn 1.54

Prove or else give a counterexample: a) A bounded sequence times a convergent sequence must be convergent. b) A null sequence times a bounded sequence must be a null sequence.

1.55 a) Ifq(n) = bknk +bk_ 1 nk-l +· · ·+b 1 n+bo is a polynomial in the variable n E N with bk =f. 0, show that there exists N E N such that n ~ N implies q(n) =f. 0. b) Show that

provided that bk

=f. 0 and k is a positive integer.

22


1.56

t 5 Define the nth Cesaro mean of a sequence Xn by an

for all n EN. a) Suppose

Xn -+

ian- Ll =

L as

IE~=l

1

=-

n

n -+

Xk;;L

(x1

+ ... + Xn)

oo. Prove:

an -+

L as

n -+

oo. (Hint: Write

I·)

b) Give an example of a divergent sequence Xn for which an converges.

1.57

Let Xn and Yn be any two bounded sequences of real numbers. Prove that limsup(xn +

Yn):::;

limsupxn + limsup]fn·

Give an example in which strict inequality occurs.

1.58

Let Xn and Yn be any two bounded sequences of real numbers. Prove that lim inf(xn +

Yn)

:2: lim inf Xn +lim inf Yn·

Give an example in which strict inequality occurs.

1.5 THE BOLZANO-WEIERSTRASS THEOREM A subsequence of a sequence Xn is a sequence consisting of some (but not necessarily all) of the terms of the sequence Xn. The terms appear in the same order as they appeared in Xn, but with omissions. We formalize this concept in the following definition.

Definition 1.5.1 Let nk be any strictly increasing sequence of natural numbers, so that n1

< n2 < · · · < nk < · · ·.

Then we call Xnk a subsequence of Xn.

We remark that since n1 :2: 1, it follows that n 2 :2: 2, ... , and nk :2: k, for all k. An alternative way to think about and to notate subsequences is to write that nk = ¢(k), where¢ : N -+ N is a strictly increasing function, in the sense that j < k ~ ¢(j) < ¢(k). Then we could alternatively write Xnk as Xq,(k)· • EXAMPLE 1.7

Let Xn = n 2 , for all n E N. If nk squares of even natural numbers.

5 This

=

2k, then

Xnk

=

(2k ) 2 is the sequence of

exercise is used to develop the Fejer kernel for Fourier series in Exercise 6.47.

THE BOLZANO-WEIERSTRASS THEOREM

Theorem 1.5.1 If Xn converges to the limit L as Xnk ----> L as k ----> oo.

n ---->

23

oo, then every subsequence

Proof: Lett > 0. There exists N E N such that n 2: N implies Since nk 2: k for all k, it follows that k 2: N ===> lxnk - Ll < t.

lxn-

Ll
-1 as k ----> oo, We have learned previously that every convergent sequence is bounded. Although the student has seen several examples of bounded sequences that are not convergent, we do have the following very important theorem.

Theorem 1.5.2 (Balzano-Weierstrass) Let Xn be any bounded sequence ofreal numbers, so that there exists M E ~such that lxnl :::; M for all n. Then there exists a convergent subsequence Xnk of Xn· That is, there exists a subsequence Xnk that converges to some L E [-M, M]. Proof: We will use the method of interval-halving introduced previously to prove theexistenceofleastupperbounds. Leta1 = -Mandb1 = M. Soxn E [a1.b1],for all n E N. Let Xn 1 = x1. Now divide [a1, b1] in half using the midpoint a 1 %b 1 = 0. i. If there exist oo-many values of n such that Xn E [a1, OJ, then let a2 = a 1 and b2 = 0. n. But if there do not exist oo-many such terms in [a 1, OJ, then there exist oo-many such terms in [0, b1 ]. ln that case let a2 = 0 and b2 = b1. Now since there exist oo-many terms of Xn in [a2, b2], pick any n2 > n1 such that E [a2, b2]. Next divide [a2, b2] in half and pick one of the halves [a3, b3] having oo-many terms of Xn in it. Then pick n3 > n2 such that Xn 3 E [a3, b3]. Observe that

Xn 2

as k ____, oo. So if t > 0, there exists K such that k 2: K implies lbk - ak I < t. Thus if j and k 2: K, we have lxni - Xnk I < taswell. Hence Xnk is a Cauchy sequence and must converge. Since [- M, M] is a closed interval, we know from a previous • exercise that Xnk ----> L as k ----> oo for some L E [- M, M].

24


EXERCISES 1.59

Give an example of a bounded sequence that does not converge.

1.60 Use Corollary 1.5.1 to prove that the sequence converge.

Xn

= (-l)n +~does not

1.61 Suppose Xn --t oo. Prove that every subsequence Xnk --t oo as k --t oo as well. (Hint: The sequence Xn is divergent, so it is not enough to quote Theorem 1.5.1.) 1.62 Use the following steps to prove that the sequence Xn has no convergent subsequences if and only if lxnl --too as n --too. a) Suppose that the sequence Xn has no convergent subsequences. Let M > 0. Prove that there exist at most finitely many values of n such that Xn E [-M, M]. Explain why this implies lxnl --too as n --too. b) Suppose lxnl --t oo as n --t oo. Show that Xn has no convergent subsequence. (Hint: Exercise 1.61 may help.) 1.63

Give an example in which Yj

> 0 for all j and Yj --t 0 yet Yj is not monotone.

1.64 The following questions provide an easy, alternative proof of the BolzanoWeierstrass Theorem. a) Use the following steps to prove that every sequence Xn of real numbers has a monotone subsequence. Denote the nth tail of the sequence by

Tn={xjiJ2::n}. (i) Suppose the following special condition is satisfied: For each n E N, Tn has a smallest element. Prove that there exists an increasing subsequence Xn;. (ii) Suppose the condition above fails, so that there exists N E N such that TN has no smallest element. Prove that there exists a decreasing subsequence Xn;. b) Give an easy alternative proof of the Bolzano-Weierstrass Theorem.

1.65 Prove: A sequence Xn --t L E R if and only if every subsequence Xn, possesses a sub-subsequence Xn;. that converges to L as j --t oo. (Hint: To prove 3 the if part, suppose false and write out the logical negation of convergence of Xn to L.) 1.66 Prove or Give a Counterexample: A sequence Xn E R converges if and only if every subsequence Xn; possesses a sub-subsequence Xn;. that converges as j --t oo. 3

1.6 THE NESTED INTERVALS THEOREM Having used the method of interval-halving twice already, it is natural to consider the following theorem.

25

THE NESTED INTERVALS THEOREM

Theorem 1.6.1 (Nested Intervals Theorem) Suppose

is a decreasing nest of closed finite intervals. Suppose also that

Then there exists exactly one point L E n~ 1 [ak, bk]. Moreover, ak bk --+ L as k --+ oo.

--+

L and

Proof: Let E > 0. Then there exists K such that k ;::: K implies ibk - aki < E. But, for all k :2: K, ak E [aK, bK]· Thusj, k;::: K implies iaj- aki <E. Hence the sequence ak is a Cauchy sequence so there exists a point L such that ak --+ L. Since k :2: n implies for all n that ak E [an, bn]. it follows that L E [an, bn] for all nand that 00

Now, if

n 00

L'

E

[ak, bk]

k=l

also, then IL- L'i ::::; ibk- aki --+ 0, which implies L = L'. Hence the point Lis unique. Observe that ibk - Ll ::::; lbk - ak I --+ 0 so that bk --+ L as claimed. • The reader is aware that there are real numbers that are not rational. For example, we will prove that there is a square root of 2 in lR in Example 1.10. Yet we know that no rational number can be a square root of 2 as was shown in Exercise 1.11. Despite the fact that not every real number is rational, every finitely long decimal expansion represents a rational number, and common sense tells us that we may approximate any real number as closely as we wish by using a suitable but finitely long decimal expansion. This observation gives rise to the following definition of what it means for a subset S ~ lR to be dense in R Definition 1.6.1 A subset S ~ lR is called dense in lR if and only there exists a sequence Sk of elements of S such that Sk --+ x .

if for all x

E

• EXAMPLE 1.9 We will show that 1QJ is dense in R

=

Proof: Let x E R If x E IQJ, we could simply let Sk x so that Sk --+ x, being a constant sequence. So suppose x f/_ IQJ, so that x is irrational. Then there exists n E Z such that n < x < n + 1. Let a 1 =nand b1 = n + 1, both rational numbers. Then the midpoint is also a rational number, and x must lie in one half-interval but

R

26


not the other. Let [a2, b2] be the half-interval containing x. Now cut [a2, b2] in half again and select [a 3 , b3 ] containing x again. Note that 1 lbk- akl = 2k-l ___. 0.

Since x E [ak, bk] for all k, lak - xl ---> 0, so ak ---> x, and ak E Q for all k. Thus we have a sequence of rational numbers converging to x in this case as well. (Note that bk would have served just as well as ak.) •

Remark. Because lR is complete and because the set Q c lR is dense in IR, it follows that any set of numbers that contains limits for all its Cauchy sequences and that contains Q must also contain JR. For this reason lR is called the completion of Q.

• EXAMPLE 1.10 We will show that J2 exists in JR. Proof: Recall that in the first paragraph of Section 1.3 we constructed an increasing sequence Xk as follows: X1

1

X2

1.4

X3

1.41

X4

1.414

Here Xk is the largest k-digit decimal greater than 1 such that x% < 2. We could have constructed also a decreasing sequence Yk by letting Yk be the smallest > 2. Thus k-digit decimal such that

yz

1 IYk- Xkl ~ 1()k-l

-4 ()

ask---> oo We see that the intervals [xk, Yk] satisfy the hypotheses of Theorem 1.6.1. Thus there exists a unique

n 00

L E

[xk,Yk]

k=l

suchthatxk ___. Landyk ___. L. Hencexz ___. £ 2, sothat£ 2 ~ 2, andyz ___. £ 2, so that L 2 ~ 2. Thus L 2 = 2 and L = J2 exists in JR. •

EXERCISES

1.67

Give an example of a decreasing nest of nonempty open finite intervals

THE HEINE-BOREL COVERING THEOREM

such that

1.68

n%:1 (ak, bk) =

27

0, the empty set.

Give an example of a decreasing nest of open intervals

(a1, b1) ;:2 (a2, b2) ;:2 · · · such that bk- ak

---'>

0 yet

n%:1 (ak, bk) -=1- 0.

1.69 Give an example of a decreasing nest of infinite intervals with empty intersection. 1.70 Prove or give a counterexample: If an i, bn l, and (an, bn) is a decreasing nest of finite open intervals, then there exists L E lR such that

n 00

(an, bn) = {L}.

n=l

1.71 Show that every open interval (a, b) C IR, with 0 < b- a but no matter how small, must contain a rational number. (Hint: Apply Example 1.9.)

1.72 t Let I denote the set of all irrational numbers. The following steps will lead to the conclusion that I is dense in R (You may assume it is known that vl2 E I.) Let x E R We must show there exists a sequence 8k of elements of I converging to X.

a) Show that if ~ is any nonzero rational number then ~vl2 is irrational. (Hint: Suppose the claim is false, and deduce a contradiction.) b) Now suppose x is any real number. Explain why there exists a sequence tk of nonzero elements of Ql converging to ~- Define a sequence sk of elements of I converging to x.

1.73 Show that every open interval (a,b), with b- a > 0 but no matter how small, must contain an irrational number. (Hint: Use the result of Exercise 1. 72.) 1.74

Is the set { ;~ ImE Z, n EN} dense in IR? Prove your conclusion.

Let D -=1- 0 be a subset of the set of strictly positive real numbers, and let = {nd InEZ, dE D}. Prove: Sis dense in lR if and only if inf(D) = 0.

1.75

S

1.7 THE HEINE-BOREL COVERING THEOREM Although the study of continuous functions belongs to the next chapter, let us think in advance on an intuitive level about this concept. A function f : lR ---'> lR is said to be everywhere continuous provided that for each point p E IR, f (x) remains very close to f (p) provided that x is kept sufficiently close to p. For example, the set

S

= {x lif(x)- f(p)l < f}

should contain some sufficiently small open interval around p, although S may also include points far away from p.

28


Consider next an open interval (a, b) that is contained in the set of values achieved by f. Let 0 = {x I f (x) E (a, b)}. For each p E 0 there will be a corresponding small number E > 0 such that

(f(p)-

E,

f(p)

+e)~

(a, b).

Because f is continuous at p, there will be a small open interval around p that is contained in 0. This example motivates the concept of an open set, which generalizes the familiar notion of an open interval.

Definition 1.7.1 AsetO ~ !Riscalledanopen subsetof!Rprovidedforeachx there exists r x > 0 such that

E 0

Thus 0 is called open provided that each x E 0 has some (perhaps very small) open interval of radius rx > 0 around it that is entirely in 0 .

• EXAMPLE 1.11 We claim that every open interval (a, b) is an open set. In fact, if x E (a, b), then a < x < b and we can let

rx = min{lx- al, lx- bl}. Then (x- rx, x

+ rx)

~

(a, b).

~ lR is a union of (perhaps infinitely many) open intervals. Moreover; every union of open sets is an open set.

Theorem 1.7.1 Every open subset 0

Proof: Let 0 ~ lR be open. Then, using the notation of Definition I. 7 .I, you will show in Exercise 1.78 that 0= U(x-rx,x+rx)· xEO

To prove the second conclusion, let 0 = UaEA 0"' be any union6 of open sets. Let X E 0. We know there exists O:o E A such that X E oo• which is open. Thus there • exists rx > 0 such that (x- rx, x + rx) ~ Oa 0 ~ 0. Thus 0 is open.

Definition 1.7.2 An open cover of a setS 0 = {Oa 6 When

~

lR is a collection

I 0: E A}

denoting an arbitrary union of open sets, it is customary to use a so-called index set, such as the set A used here. One should think of A as being a set of labels, or names, used to tag, or identify the sets of which the union is being formed. One cannot always index sets by means of natural numbers, because there exist sets so large that they cannot be uniquely indexed by natural numbers. Even the infinite set N is too small. The reader will learn more about this in Theorem 1.15.

THE HEINE-BOREL COVERING THEOREM

29

of (perhaps infinitely many) open sets Oa., where a ranges over some index set A, such that S s;;; Ua.EA Oa.. In analysis, it is often necessary to try to control small-scale local variations of some structure defined on a domain D. Under suitable conditions, one can control variations by restricting ones view to a very small open set surrounding each given point of D. Then in the large we cover the whole domain D with a family 0 of these (possibly small) open sets whose union contains D. Usually 0 will have infinitely many open sets as members, or elements of itself. Within each one of the open sets that are elements of 0 the fine structure varies only slightly. We hope for the availability of a finite subcover, consisting of only finitely many of the open sets belonging to 0, so as to produce uniform controls on fine-scale variations for the entire large domain D. Below, we show an example of an open covering of a set for which there is no finite subcover. This will motivate the Heine-Borel Theorem which follows . • EXAMPLE 1.12 Consider the setS= (0, 2), a finite open interval. We claim that

In fact, for each x E (0, 2) there exists n E N such that x E (~, 2). (Make sure you see why this is so.) Thus 0 = { (~, 2) In EN} is an open cover of S. However, it is impossible to select any finite subset of 0 that covers S. The reason is that any finite subset of 0 would have a largest value no of n for which ~ would be the left hand endpoint of an interval belonging to the chosen finite subset of 0. Thus the finite subset would fail to cover any points to the left of...!.... no Remark 1.7.1 Note that the term finite interval means an interval of finite length. Any finite interval with strictly positive length has infinitely many distinct points within it. Thus the word finite infinite interval means the same thing as bounded. On the other hand, a finite set means a set with finitely many elements. In Example 1.12, a finite subset of a set of intervals means a collection of finitely many of those intervals. This does not mean that the intervals in question have finitely many points. The Heine-Borel theorem is one of the most important in advanced calculus. But it is the most abstract theorem presented thus far in this book, and the reader will need time and experience to absorb fully its significance. It is recommended to consider Exercise 1.80 below after reading the statement of the theorem. Theorem 1.7.2 (Heine-Borel) Suppose the closed finite interval

[a, b] s;;;

U Oa., o.EA

30


where 0 = { Oa I a E A} is an open cover of [a, b]. Then there exists a finite set F = {at, ... , an} ~ A such that n

[a, b] ~

U Oa = UOa;·

a.EF

i=t

The collection {Oa 1 , ••• , Oa.,.} ~ 0 is called a finite subcover of[a, b]. Proof: We suppose the theorem were false. We will deduce a logical selfcontradiction from that supposition. This will prove the theorem. So suppose the Heine-Bore) theorem were false: Thus we can assume the given cover does not admit a finite subcover of [a,b]. Let at = a and bt = b, and Jet c = ~. Then each of the intervals [a1, c] and [c, bt] is covered by Ua.EA Oa.. If both of these half-intervals had finite subcovers, then the whole interval [a,b] would have a finite subcover since the union of two finite families is still finite. Since we are supposing [a,b] has no finite subcover, pick a half-interval [a 2, b2] that has no finite subcover. Now cut [a2, b2] in half and reason the same way for [a 2, b2] as we did for [at, bt]. We obtain a decreasing nest of intervals

[at. bt] -.;2 • • • -.;2 [ak, bk] "-2 • • · such that each [ak, bk] is covered by Ua.EA Oa. but has no finite subcover. However,

as k

~

b-a

k-t ~o 2 oo. By the nested intervals theorem, there exists a unique

lbk-akl=

n 00

X

E

[ak, bk] ~ [a, bj.

k=t

Since x E [a, b], there exists a E A such that x E Oa. So there exists rx > 0 such that (x- rx, x + rx) ~ Oa.. Now pick k big enough so that bk- ak < rx. Thus

x E [ak, bk] C (x- rx, x

+ rx)

~

Oa.

and we have covered [ak, bk] with a single open set Oa from the original cover. This is a (very small) finite subcover. This contradicts the statement that [ak, bk] could not have a finite subcover. This contradiction proves the Heine-Bore) theorem. •

EXERCISES 1.76

Show that a closed finite interval [a,b] is not an open set.

1.77

Show that a half-closed finite interval (a, b) is not an open set.

1.78 Let 0 be any open subset of JR., and for each x E 0 Jet rx be defined as in the proof of Theorem 1. 7 .1. Complete the proof of that theorem by showing that 0 = UxEO(x- rx, X+ rx)·

COUNTABILITY OF THE RATIONAL NUMBERS

1.79

31

The empty set 0 satisfies the definition of being open. Explain.

1.80 Find an open cover of the interval ( -1, 1) that has no finite subcover. Justify your claims. 1.81 Find an open cover of the interval ( -oo, oo) that has no finite subcover. Justify your claims. 1.82 Let E ~ lR be any unbounded set. Find an open cover of E that has no finite subcover. Prove that you have chosen an open cover and that it has no finite subcover. 1.83 Let E = { ~ I n E N}. Find an open cover 0 = {On I n E N} of E that has no finite subcover, and prove that 0 is an open cover and that 0 has no finite subcover. 1.84 0 We call p a cluster point of E, provided that for all E such that 0 < le- PI< E.

> 0 there exists e

E E

(See Defintion 2.1.1.) Let E c lR be any set with the property that there is a cluster point p of E such that p ¢ E. Show that there exists an open cover of E that has no finite subcover. Justify your claims. (Note: Exercise 1.83 is an example of the claim of this exercise.) 1.85

True or False: Finitely many of the open sets in the collection

would suffice to cover [0, 1]. 1.86 Prove or give a counterexample: Every open cover of a finite subset of lR has a finite subcover. (Note: For the real line, the phrase finite subset does not mean the same thing as finite interval.)

1.8 COUNTABILITY OF THE RATIONAL NUMBERS Definition 1.8.1 A set S is called countable if it is an infinite set for which it is possible to arrange all the elements of S into a sequence. That is, S is countable if S = {s 1 , s 2 , .•• , Sk, ... } with each element of S listed exactly once in the sequence. Equivalently, we may say that Sis countable if and only if there exists a function s : N --> S that is both one-to-one, which is also called injective, and onto S. Onto maps are often called surjective. The terms., in the definition above would be s(n) in this notation.

32

•


EXAMPLE 1.13

Let E denote the set of all even natural numbers. Thus E ~ N. We claim that E is countable. In fact, the elements of E can be arranged into a sequence by means of a function s : n ~ 2n that is both an injection and a surjection of E onto N. That is, the sequence is given by sn = 2n. It may surprise the reader that the elements of an infinite set can be paired one-to-one with those of a proper subset. • EXAMPLE 1.14

We will prove the surprising and useful fact that the set Q of all rational numbers is countable. It is important to understand that if a sequence s n is to include all the rational numbers, then these numbers cannot be listed in size places. That is, if Sn < Bn+t• both in Q, then sn+;n+l lies between them and is again rational. Hence there is no next smallest rational number after s n· We can explain how to list the rational numbers in a sequence, disregarding the order relation, as follows. We are going to consider a table of numbers with infinitely many rows. The entry in the mth row and nth column will be the fraction fii. Here m E N and n E Z. Thus there will be a first row, in which each denominator is understood to be 1, but no last row. Each row will extend endlessly to left and to the right. We can draw only part of this table below.

-4 4 -2

-3 3 -2

4

3

-3 4

-4

-2 2 -2 2

-3

-3

3

-4

-4

2

-1 1

-2 1

-3 1

-4

0 0 2

1 1

2 3

3

3

1

0

4

3 3 2

2

1

0

3

2 2 2

4

3

2

3

4

4

4 4 2 4

3 4

4

We will describe a systematic expanding search pattern that reaches each term on the infinite table after some finite number of terms in the sequence described below. We will list side-by-side those terms fii for which

lml + lnl = k beginning with k = 1, k = 2, and so on. If parentheses are placed around a number, we are skipping that number because it was already listed previously. Here is the resulting list:

o-1 (Q = o) '

'

2

1 -2

' '

'

-~2' (Q3 =

o)

~

2 -3

'2' '

'

(-~2 = -1) ' -~3'

(~ =0) '~' (~ =1) ,3, .... It is clear that this expanding search pattern eventually reaches any rational number fii that one might choose, and each rational number is listed exactly

COUNTABILITY OF THE RATIONAL NUMBERS

33

once in the resulting endless sequence. The first several terms of the sequence sn, corresponding to k = 0, 1, 2, 3, 4, are

1 1 1 1 0, -1, 1, -2, -2, 2' 2, -3, -3, 3' 3, .... We will see several applications of the countability of Q in this book. However, for now we describe a startling example .

• EXAMPLE 1.15 We will describe a set 0 that is both open and dense in IR, yet which is quite small. Let E > 0, a small positive number. Consider the line segment [0, E] of length E. We will construct a sequence of intervals (ak, bk), each of length ~. That is, the first interval, ( a1, b1) will have length ~. This leaves half of [0, E) remaining. But for (a2, b2) we will use only half that remainder: namely, f. b3 - a 3 will be taken to be ~.or half of the remaining f from the original interval [0, E]. Let Q = {St, s2, ... , Bk, ... }, which can be arranged since Q is countable, as explained above. Let (a1,b1) be centered around St. (a2,b2) centered around s2, and in general (ak, bk) will be centered around the point Bk. For any finite subcollection of the intervals (ak, bk), k = 1, 2, 3, · · · , the sum of the lengths of each of the finitely many intervals chosen must be less than E. That is because the whole infinite sequence of intervals is chosen by cutting E in half again and again without end. Now consider that Q is dense in R But if we let 0 = 1 (ak, bk), then Q C 0 and so 0 is also dense in R Moreover, 0 is open by Theorem 1. 7.1. We claim that 0 is a smal1 set in the following sense. Let [a,b] be any closed finite interval of length~ E. We claim it is impossible for [a, b] ~ 0. In fact, if [a,b] were a subset of 0, then

U;::

00

[a, b] ~

U(ak, bk), k=l

an open cover of [a,b]. By the Heine-Bore! theorem, there must be a finite number of intervals from among the (ak, bk)'s that cover [a,b]. Yet the sum of the lengths of these finitely many intervals must be less than E ::; b - a. This is impossible. It is interesting to compare the preceding example with Exercise 1.91. The interested student can learn much more about surprising subsets of the line in the book by Gelbaum and Olmsted [7]. It is natural to wonder at this point whether or not perhaps every infinite set is countable. The answer is no, as is shown by the following surprising theorem.

Theorem 1.8.1 (Cantor) The set lR of real numbers is uncountable. (That is, it is impossible to include all the real numbers in a sequence.)

34


Proof: We begin by noting that the (possibly endless) decimal expansions of real numbers are not unique, because an infinite tail of 9's can always be replaced by an expansion ending in an infinite tail of O's. For example,

0.999 ...

= 1.000 ...

This is understood in the sense that if Xn = 0.999 ... 9 with n 9's then 1 lxn - 11 = -l()n ---+ 0 as n ---+ oo. But if we agree not to allow endless tails of9's, then decimal expansions of real numbers are unique. Moreover, every infinite decimal representation corresponds to a real number. The reason for this fact is as follows. Consider any infinite decimal expression. It could be written in terms of a whole number K in the form

K

+ O.d1d2d3 ... dn ... ,

where dn is the nth digit to the right of the decimal point. Then let

Xn = K

+ O.d1d2 ... dn.

It follows that if m and n are both greater than N, then

lxn-

Xml
0 small enough so that Ia- bl < 8 and lc- bl < 8 implies Ia- cl <E. 1.99 The sequence Xn begins as follows: 0, 1, ~, 2, ~, ~, 3, continues according to the same pattern. a) True or False: limn-.oo lxn - Xn+ll = 0. b) True or False: Xn is a Cauchy sequence.

i, 144 , ~, 4, ... and

1

1.100 Give an example of two sequences, Xn and Yn =/:. 0 such that XnYn converges, ~ converges, but neither Xn nor Yn converges. 1.101 Let Xn = (( -l)n + 1) limsupxn.

+ 2~

for all n E N. Find both liminf Xn and

1.102 Give an example of two sequences of real numbers Xn and Yn for which liminf(xn +Yn) = Obutliminfxn = -oo = liminfYn· 1.103 State True or Give a Counterexample: If Xn is an unbounded sequence, then Xn has no convergent subsequences. 1.104 Give an example of a decreasing nest of nonempty open intervals (an, bn) (an, bn) = 0. such that bn -an ---+ 0 but

n:1

1.105

True or False: The set S = {;:; I m E Z, n E N} is dense in R

1.106 Let E ={~In EN}. Find an open cover 0 ={On In EN} of E that has no finite subcover. 1.107

True or False: The set Q is closed in the real lineR

1.108 True or False: The setS = {O.d 1 d 2 ••• dn I n E N} of all .finitely long decimal expansions (with each di an integer between 0 and 9) is countable. 1.109

True or False: The setS= { ~V2~

~

E

Q} is uncountable.

7http://www-history.mcs.st-andrews.ac.uk/history/

38


1.110 Let Xn = 1 + implies lxn - 11 < f.

<Jn". Iff> 0 find aN EN sufficiently big so that n''?: N

1.111 True or Give a Counterexample: A bounded sequence times a convergent sequence must be a convergent sequence.

(

1.112

Find n~=I -oo, -n].

1.113

Give an example of an open cover 0 = {On

lin E N} of the set

such that S has no finite subcover from 0. 1.114

Let

E =

{~In E w} u {0}.

True or False: The set lR \ E, that is the complement of E, is an open subset of R

CHAPTER2

CONTINUOUS FUNCTIONS

2.1

LIMITS OF FUNCTIONS

During the 19th century, many mathematicians worked to identify those classes of functions to which useful techniques, such as differentiation, integration, and decomposition into infinite series could be applied correctly. Mathematicians and physical scientists had worked with many types of functions described by formulas. These included polynomials, quotients of polynomials, and combinations of these with roots and powers. Sines, cosines, tangents, exponential and logarithmic functions played a role too. But in order to discover a coherent body of theorems that would explain to which functions what techniques could be properly applied, it was necessary to identify the underlying properties that functions would need to have for certain theorems to work. Those properties could be shared by functions that might be described in terms of very different-looking formulas. The formal concept of a function includes all the familiar examples and many others that are not so easily described. A function f assigns a numerical value f (x) to each point x in the domain D f on which f is defined. The range off is the set RJ = {f(x)

Ix

E DJ}·

Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright© 2008 John Wiley & Sons, Inc.

39

40


Since derivatives, integrals, and infinite series are all defined in terms of limits, we need to define the concept of the limit of f (x) as x ~ a. In order to understand the subtleties of the definition that is required, we look ahead to one of the most important applications of this concept, which will be the definition of the derivative of a function. The student will recall that in elementary calculus the derivative of f at a point x is defined as the limit as h ~ 0 of the difference quotient

Q(h) = f(x

+ h~- f(x).

Thus f'(x) = Iimh .....o Q(h). It is important to note that we wish to define the limit of Q( h) as h ~ 0, although Q is not even defined at 0. Thus we are forced to define Iimx-+a f(x) in such a way that it will be irrelevant whether or not the function f is actually defined at a. On the other hand, for Iimx-+a f(x) to make sense, it will have to be possible for x to become as close as we like to a without x being a itself. Thus we formulate the following preliminary concept.

Definition 2.1.1 A point a is called a cluster point of the set D if and only if for all 8 > 0 there exists x E D such that 0 < lx - ai < 8. Thus a cluster point a of a set D has the property that it is always possible to find points x E D for which x -1- a and yet x is as close to a as we like .

• EXAMPLE 2.1 Let D = [0, 1), the interval closed at 0 but open at l. Then the set of cluster points of Dis the interval [0,1]. See Exercise 2.1.

• EXAMPLE 2.2 Let D = { ~ I n EN}. Then D has only one cluster point: namely, the point 0. Note that in this example 0 ~ D. See Exercise 2.2. We are ready to define the concept of the limit of a function, which the reader should compare with Exercise 2.18.

Definition 2.1.2 Let a be any cluster point of the domain D f of a function f. Then lim f(x) = L

x-+a

if and only if for each E > 0 there exists 8 > 0 such that x

E

DJ andO
lf(x)-

Ll < L

LIMITS OF FUNCTIONS

41

• EXAMPLE 2.3 We present four useful examples of limits. 1. We claim that for all m and b in JR. we have lim(mx +b)= ma +b.

x-+a

We understand implicitly that the domain of mx + b is the whole real line R Now, let E > 0. We need 8 > 0 such that 0 < lx- al < 8 implies

(ma + b) I
0, the inequality 0 < Ix - 0 I < 8 will be satisfied by points x for which f (x) can be either 1 or 0, and we cannot force both 1 and 0 to be simultaneously within arbitrarily small E > 0 of any one number L. In mathematics it is very important to have equivalent forms of definitions. For some purposes, one form of a definition is most convenient to use; for other purposes, another form is more suitable.

42


Theorem 2.1.1 Suppose a is a cluster point of the domain D 1 of f. following two statements are logically equivalent.

Then the

i. Iimx_,a f(x) = L. ii. For every sequence of points Xn E D 1 \ {a} such that Xn ---+ a, we have the sequence f(xn) ---+ L. (This is called the Sequential Criterion for Limits.)

Proof: Let us prove first that (i) implies (ii). So suppose (i) and consider any sequence of points Xn E D 1 \ {a} such that Xn ---+ a. We know for all f > 0 there exists 8 > 0 such that x E D 1 and 0 < lx - al < 8 implies lf(x) - Ll < f. But since Xn ---+ a, there exists N such that n 2 N implies 0 < lxn - al < 8. This implies lf(xn)- Ll < f, so that f(xn) ---+ L. Next we prove (ii) implies (i). So suppose (ii) is true. We need to show lirnx_,a f(x) = L. We will show that if this conclusion were false then a selfcontradiction would result. But if it were false that lirnx_,a f(x) = L, then there would exist f > 0 such that for all8 > 0 there exists x E D 1 such that 0 < Ix- a I < 8 and yet lf(x)- Ll 2 LIn particular, if we let 8n =~.then we get Xn such that 0

1 n

< lxn- al < - and lf(xn)- Ll 2

f.

Now, Xn E DJ \{a} and Xn---+ a yet f(xn) ~ L. This contradicts (ii) which was • assumed to be true. Remark 2.1.1 Since this is the first theorem of Chapter 2, we remind the reader of the importance of writing out a full analysis of each proof in the reader's own words and to his or her own satisfaction. This has been discussed in the Introduction on page xxiii. For the first implication, (i) implies (ii), the reasoning follows from a careful reading of the relevant definitions. The opposite implication is trickier and benefits much from the choice of an indirect method of proof. Without the indirect proof, it would be very difficult to establish that there is a suitable 8 corresponding to each f > 0, just from knowing that Xn ---+ a implies f ( Xn) ---+ f (a). If we have two functions f and g we can form the sum, difference, and product of these functions with the domain of this combination being DJ n D 9 • On the other hand, the domain D 1 of the quotient will be only those points of DJ n D 9 for which g

g(x) =/; 0. Corollary 2.1.1 Suppose a is a cluster point of the intersection D 1 domains off and g, and suppose further that lim f(x) = L and lim g(x) = M. x~a

Then

i. Iimx_,a(f ± g)(x) = L ± M

X-i-a

n D9

of the

EXERCISES

43

ii. limx-.a(fg)(x) = LM iii. limx-.a

f (x) = f:t provided that M -/=- 0 and that a is a cluster point of D

L. g

Proof: Consider first conclusion (i). Consider a sequence of points XnEDtnD 9 \{a} such that Xn

---t

a. Then

(! ± g)(xn) = f(xn) ± g(xn)

---t

L± M

by the corresponding theorem for limits of sequences. The proof of (ii) is very similar. The proof of (iii) is Exercise 2.3. • Remark 2.1.2 It is possible to make up many variations on the definition of limit of a function, for assorted specialized uses. Here are two examples, in which we adapt the sequential condition of Theorem 2.1.1 to generalize the concept of limit. The reader should compare the definitions below with the statements in Exercise 2.17. Definition 2.1.3 We define limits at infinity and also one-sided limits as follows.

i.

If the domain Dt is not bounded above, we say f(x) that for all sequences Xn E D f such that Xn

---t

---t Las x ---t oo provided oo we have f ( Xn) ---t L.

= L,providedthat a we have f(xn) ---t L.

ii. lfaisaclusterpointofDtn(a,oo), wesaylimx-.a+ f(x)

for all sequences Xn EDt n (a, oo) such that Xn In this case, we write f(a+) = limx-.a+ f(x).

---t

In both cases, L can be a real number, in which case we speak of convergence of f(x) to L. However, if Lis ±oo, then we speak of the divergence of f(x) to L.

EXERCISES 2.1 2.2

2.3

t Prove the claim of Example 2.1. t Prove the claim of Example 2.2. t Prove part (iii) of Corollary 2.1.1.

2.4 Find all the cluster points of the set [a, b] c IR, we denote C(D) by C[a, b]. There is a surprising degenerate case of continuity. If there exists 8 > 0 such that D 1 n (a- 8, a+ 8) = {a}, then the stipulation in the definition of continuity is true, in that there are no points in the designated intersection other than a itself, and of course lf(a)- f(a)l < t.

Definition 2.2.2 We call a point a E D an isolated point of D > 0 such that D n (a- 8, a+ 8) = {a}.

~

IR, provided that

there exists 8

In other words, if D 1 has any isolated points a, then f is automatically continuous at a. The more interesting case, however, is that of a nonisolated point a E D 1. Note that a E D 1 is a nonisolated point, provided that a E D 1 and that a is a cluster point of DJ.

Theorem 2.2.1 A function f is continuous at a cluster point a E D 1 if and only if limx-+a

f (x) exists and lim f(x)

x-+a

=

f(a).

Proof: First suppose f is continuous at a E DJ. Then for all E > 0 there exists 8 > 0 such that lx- al < 8 and x E DJ implies lf(x) - f(a)i < E. Hence for all x E DJ such that 0 < lx- al < 8 we have lf(x)- f(a)i < E, which implies limx-+a f(x) exists and limx-+a f(x) = f(a). Now suppose lirnx-+a f(x) exists and is f(a). Then for all E > 0 there exists 8 > 0 such that 0 < lx -a I < 8 and x E DJ implies lf(x)- f(a)l < E. This implies for all x E DJ n (a- 8, a+ 8) we have lf(x)- f(a)l < E so that f is continuous at

a.

•

Corollary 2.2.1 A function f : D ~ lR is continuous at a E D if and only if it satisfies the following Sequential Criterion for Continuity: For every sequence of points Xn ED such that Xn ~a we havef(xn) ~ f(a).


47

Proof: i. Suppose a is an isolated point of D. In this case there is little to prove, since f is automatically continuous at a and since Xn as in the Sequential Criterion must have the property that there exists N E N such that n ~ N ===} Xn = a and f(xn) = f(a), which implies that f(xn) -4 f(a). 11.

Suppose a is a cluster point of D. We will apply Theorem 2.1.1 and Theorem 2.2.1. Iff is continuous at a, then limx-+a f(x) exists and equals f(a). Hence the Sequential Criterion for limits implies that the sequential criterion for Continuity is satisfied. Conversely, if the Sequential Criterion for continuity is satisfied, then limx-+a f(x) exists and equals f(a).

• Theorem 2.2.2 Suppose both f and g are continuous at a E D f n D 9 • Then

i. f ± g is continuous at a. ii. f g is continuous at a. iii. ~ is continuous at a provided g(a)

-=/=

0.

iv. If, moreover, his continuous at f(a), then the composition h of is continuous at a, where h o f(x) = h(f(x)). Proof:

We will apply the Sequential Criterion for Continuity (Corollary 2.2.1) throughout. i. For the first case, bear in mind that DJ±g = DJ n D 9 . If Xn -4 a with all Xn E DJ n D 9 , then f(xn) -4 f(a) and g(xn) -4 g(a). We conclude that limx-+a(f ± g)(x) exists and equals

lim f(x) ± lim g(x) x-+a

x~a

= f(a) ± g(a).

ii. This part has a nearly identical proof. iii. This part is left to the reader in Exercise 2.19. iv. We note that Dhof = {x E DJ I f(x) E Dh}· Now let Xn E Dhof such that Xn -4 a. Then letting Yn = f(xn). we have

lim h(f(xn)) = lim h(yn) = h(f(a))

n---+CXJ

since Yn -4 f(a) as n respectively.

n-+CXJ

-4

oo since f and h are continuous at a and f(a),

•

48 •


EXAMPLE 2.4

Examples of Continuity and Discontinuity.

1. Let i(x) = x for all x E R Then i is continuous at every point a E R In fact, if Xn -+ a we have i(xn) = Xn -+a= i(a) for all a E R 2. Let f(x) = c, a constant. It is easy to check that f satisfies the Sequential Criterion for Continuity at every a E R so that f E C(IR.).

3. Let f(x) = x 2 , for all x E R By Theorem 2.2.2, f E C(R). This generalizes easily to show that g(x) = xn is in C(R) for all n E N. Now apply Theorem 2.2.2 to see that every polynomial p(x) = anxn + · · · a1x + ao is continuous on all ofR as well. Moreover, every rational function Q(x) = ~·where p and q are polynomials, is continuous wherever defined. 4. Let

f(x) = {

~

if X 2: 0, ifx

< 0.

Then f is discontinuous at 0 but continuous everywhere else. Why? 5. Let

f(x) = {

~

if X 2: 0, if X= -1.

Then f is continuous wherever it is defined. Note that -1 is an isolated point ofDJ.

6. Let

f(x) =

{

x2

ifx E Q,

-x2

if X¢_ Q.

We claim that f is continuous at x = p if and only if p = 0. Perhaps the easiest way to prove this is to utilize Corollary 2.2.1. Consider two sequences x n E Q and Xn ¢. Q such that Xn -+ p and Xn -+ p. Iff is continuous at p, then

f(xn) = x~ -+ p 2 = f(p) and f(xn) = -x~ -+ -p 2 = f(p). Iff is continuous at p, then p 2 = -p2 and p = 0. It remains to prove that f actually is continuous at 0. Suppose now that x n is any sequence converging to 0. We need to prove that f(x n) -+ f(O) = 0. Let E > 0. Note that the sequences x~ converges to 0, as does the sequence -x;;,. Moreover, there exists N E N, corresponding to E, such that n 2: N implies

EXERCISES

49

01

It follows also that n 2: N implies 1-x;_ < f. Now let n 2: N. If Xn E Q, then lf(xn) - Ol = x; 0, we seek 8 > 0 such that x 2: 0 and lx- al < 8 implies 2 IQ(x)- Q(a)l < E. Begin by showing that lvx- v'al lx- a!.)

s

2.22

Let

X!, X2

E

2.23

Let

f

Dj

be continuous at a, and

f

> 0. Show there exists 8 > 0 such that

n (a- 8, a+ 8) implies lf(xl)- f(x2)1

p+ f(x) = f(p+) and limx-->p- f(.r) = f(p-) both of which exist. Let

j(p) = f(p+)- f(p-) denote the height of the jump at p. Let

j(b) = f(b)- f(b-) and j(a) = f(a+)- f(a)

50


and let

En =

{X Ij (X) ~ ~ }

for all n EN. a) Prove: f is continuous at x E [a, b] {::} j(x) = 0. b) Show that the set E of points of discontinuity off is given by

C) If

a

=

0 there exists 8 > 0 such that for all x1 and x2 in D such that lx2 - x1l < 8 we have lf(x2) -!(xi) I < t:. That is, f is uniformly continuous on D provided there exists 8 > 0 that satisfies the requirement'> for continuity at x 1 uniformly, meaning at all x1 E D. Intuitively, uniform continuity of f on D imposes a restriction on how fast f can change its values on the entire domain D.

Theorem 2.3.2 Iff E C[a, b], then f is uniformly continuous on [a, b]. Proof: We suppose the theorem were false and we will deduce a contradiction. Thus we suppose f is not uniformly continuous on [a,b]. Hence there exists t: > 0 such that no 8 > 0 will suffice the meet the uniform continuity condition. Hence, if 8n = ~.there exists Xn, Yn E [a, b] such that lxn- Ynl < ~but

Since Xn is a bounded sequence, the Bolzano-Weierstrass Theorem guarantees the existence of a convergent subsequence Xnk -+ p E [a, b] as k -+ oo. Now,

Thus Ynk -+ p as well. Therefore, since f is continuous at p,

This means that

But lf(xnk)- f(Ynk)l 2:

t:

for all k and so we have a contradiction.

•

EXERCISES

53

EXERCISES 2.29 t Suppose f is continuous at p and f (p) > c. Prove: There exists 8 > 0 such that x E DJ n (p- 8,p + 8) implies f(x) >c. (Hint: Consider g(x) = f(x)- c.) 2.30 t Suppose f is continuous at p and f(p) < c. Prove: There exists 8 > 0 such that x E D f n (p- 8, p + 8) implies f(x) < c. (Hint: Consider g(x) = - f(x ).) 2.31 t Suppose f E C[a, b] and f(a) > k > f(b). Prove: There exists c E (a, b) such that f( c) = k. This will complete the proof of Theorem 2.3.1. (Hint: Consider

g(x) = - f(x).) 2.32 Letp(x) = x 3 + 3x 2 - 2x -1. Prove that the polynomial equationp(x) has a root somewhere in the interval (0, 1). Let p(x) = x 4 + x 3 - 2x 2 p(x) = Ohasarootin (-1,0). 2.33

+x+

=0

1. Prove that the polynomial equation

Let f(x) = cos x- x- sin x. Prove: There exists a solution to the equation f(x) = 0 on the interval [0, 1rj6]. (You may assume that sinx and cosx are

2.34

continuous on R) 2.35 Let p(x) = a2n+lx 2n+l + · · · + a1x + ao be any polynomial of odd degree. Prove: The equation p(x) = 0 has at least one realroot. (Hint: Prove p(x) --+ ±oo depending on whether x --+ ±oo and consider then that p must have both positive and negative values.) 2.36

A function

f : (-a, a)

--+

lR is called an odd function provided

f (- x) =

- f (x) for all x E (-a, a). A function f : (-a, a) --+ lR is called an even function provided f( -x) = f(x) for all x E (-a, a). Prove or give a counterexample for each of the following statements. a) Iff is an odd continuous function on (-a, a), then there exists a solution to the equation f(x) = 0. b) The composition oftwo odd functions must be odd. c) Every function of an even function is even. d) An even function of an odd function is even. e) An even function of any function is even. f) Every function f : (-a, a) --+ lR is the sum of an even function with an odd function. 2.37 t Prove the following fixed point theorem: Suppose f E C[O, 1] and suppose 0 :S f(x) :S 1 for all x E [0, 1]. Then there exists c E [0, 1] such that f(c) =c. (The pointe is thencalleda.fixedpointforthefunction f. Hint: Consider g(x) = f(x)-x.) 2.38 Provethefollowingtheorem: Suppose! E C[O, 1] andsupposeO :S f(x) :S 1 for all x E [0, 1]. Let n EN. Then there exists c E [0, 1] such that f(c) =en. (Hint: Compare this problem with Exercise 2.37.)

54


y

3.0

2.5

2.0

1.5

1.0

0.5

2

Figure 2.1

y = ~·

2.39 Suppose f and g are in C[a, b] and suppose [!(a)- g(a)][f(b)- g(b)] ~ 0. Prove that there exists c E [a, b] such that f(c) = g(c). (Hint: Compare this problem with Exercise 2.37.) 2.40 Suppose f is uniformly continuous on (a, m] and also on [m, b). Prove: uniformly continuous on (a, b).

f

is

2.41 Let f(x) = ~.for all x E (0, 1). Is f continuous on (0, 1)? Is f uniformly continuous on (0, 1)? Justify your conclusions. (See Fig. 2.1.) 2.42 Let f(x) = ~.for all x E (1,oo). Is f continuous on (1,oo)? Is f uniformly continuous on (1, oo )? Is f uniformly continuous on (0, oo )? Justify your conclusions. (See Fig. 2.1.) 2.43 Let f(x) = x 2 , for all x E R Is f E C(I~)? Is f uniformly continuous on ~? Justify your conclusions. 2.44 Suppose f is uniformly continuous on D and suppose E uniformly continuous on E as well. 2.45

c D.

f

is

= x 2 uniformly continuous on (0,1)? Why or why not? LetQ(x) = yx. Prove that the function Q is uniformly continuous on [0, oo ).

Is f(x)

2.46 (Hint: See Exercise 2.21.) 2.47

Prove:

Let .

f(x)

=

{

1

~m x

<x ~ ifx = 0.

ifO

1,

You may assume that the function sin x is continuous. (See Fig. 2.2.)

EXTREME VALUE THEOREM AND ITS CONSEQUENCES

Figure 2.2

y

55

= sin ~.

a) Show that f does have the intermediate value property (Definition 2.3.1)

on [0, 1], but f tj. C[O, 1]. C(O, 1] but f is not uniformly continuous on (0, 1].

b) Show f E

2.4


Definition 2.4.1 We say a function f is bounded on a domain D, provided that there exist m and M in lR such that

m:::;

f(x):::; M

for all xED.

Observe that f (x) = x is continuous on lR but f is not bounded on JR. And g( x) = ~ is continuous on (0, 1) but is not bounded, since it lacks an upper bound, though there is a lower bound, 0. We have, however, the following important theorem pertaining to continuous functions on closed, finite intervals. Theorem 2.4.1 Iff E

C[a, b], then f is bounded on [a, b].

Proof: By Theorem 2.3.2, f is uniformly continuous on [a, b]. Thus, for example, if we pick the positive number f = 1, there exists 6 > 0 such that for all x 1 , x E [a, b] such that lx 1 -xI < 6 we must have If (xi) - f (x) I < 1. Now consider the sequence of evenly spaced points defined as follows: XI

=

a, X2 = a + 6, X3 = a + 26, ... , Xk = a + (k - 1)6, ....

By the Archimedean Property of IR, there exists N E N such that a + N 6 > b, whereas a+ (N- 1)6 :::; b. In other words, although 6 may be a small positive number, if we take enough steps of size 6 we must eventually pass b. Now consider

56


the following inequalities.

x E [x1,x2)

=::}

x E [x2, x3)

=::}

x E [x3, x4)

=::}

< f(x) < j(x1) + 1 j(x2)- 1 < f(x) < j(x2) + 1 j(x3)- 1 < f(x) < j(x3) + 1

j(x1) -1

If we let m

= min{f(x1)- 1, j(x2)- 1, ... , f(xN)

- 1}

and if we let

M = max{f(x1) then we have for all x E [a, b], m

+ 1, j(x2) + 1, ... , f(xN) + 1}, < f(x) < M.

•

Observe that the function f (x) = x is bounded on D = (0, 1). However, f has no largest value and no smallest value, since for all x E (0, 1), there exists x' < x < x" withx',x" E (0, 1).

Definition 2.4.2 We say that f has a maximum value M on a domain D, provided that there exists x M E D such that f(x) S f(xM)

=M

for all x E D. Similarly, we say f has a minimum value m on D, provided that there exists Xm E D such that f(x) ~ f(xm) = m for all xED. The next theorem establishes that on a closed, finite interval [a, b], every continuous function must have both a maximum and a minimum value.

Theorem 2.4.2 (Extreme Value Theorem) Iff E C[a, b], then f has both a maximum and a minimum value on

[a, b].

Proof: By Theorem 2.4.1, we know that {f(x) I x E [a, b]} i= 0 is bounded both above and below. Let M = sup{f(x) I x E [a, b]}. Thus M is the least upper bound of the range of f. Hence for all k E N, M is too small to be an upper bound, so there exists Xk E [a, b] such that

k

M-

1

k < f(xk) S M.


57

Since Xk is a bounded sequence, the Bolzano-Weierstrass Theorem guarantees that there exists a convergent subsequence x nk. Thus there exists x M such that Xnk -> x M ask----> oo. Since [a, b] is closed, XM E [a, b]. But

and nk is an increasing sequence of natural numbers. Thus

-1

1

< f(xnk) :::; M, for all k E N. Since M- ---->Mask----> oo, and Hence M because f is continuous at XM, f(xM) = limk_,oo f(xnk) = M. For the minimum point, see Exercise 2.48. • •

EXAMPLE 2.6

Let p( x) = x 2 - x + 1. We claim that p must have a minimum value on JR.. The problem is that although p E C(IR), lR is not a finite closed interval so the Extreme Value Theorem does not apply directly. Consider

p(x)

=

x 2 ( 1 - -1 x

1) . +x2

The second factor approaches 1 as x ----> ±oo. Thus, there exists M such that lxl > M implies

>!. ( 1-~+2_) x x 2 2

Observe that p(O) = 1 and if lxl > J2 then x 2 > 2. Hence if we set a= max{ J2, M}, then lxl >a impliesp(x) > 1 = p(O). Now,p E C[-a, a] so the Extreme Value Theorem guarantees that there exists x m E [-a, a] such thatp(xm) :::; p(x) for all x E [-a, a]. But this implies that

p(xm) :::; p(O) = 1:::; p(x) for all valuesofx sincep(x) exceeds 1 outside [-a, a]. Hencep(xm):::; p(x), for all x E JR. Theorem 2.4.3 Let f E C[a, b]. Then the range off is a closed, finite interval.

Proof: Recall that the range off is f([a, b]) = {f(x) I x E [a, b]}. Since f achieves a maximum value M and a minimum value m, the Intermediate Value Theorem assures us that f achieves every value between M and m. Hence

f([a, b])

=

[m, M].

•

58


Definition 2.4.3 Iff is any function whatever (not necessarily continuous) on a domain of definition D, we define the sup-norm off to be llfllsup =sup {lf(x)ll XED}· Thus ll!llsup may be either positive real-valued or else oo. However, if f E E C[a, b] as well, since the absolute value function is continuous and compositions of continuous functions must be continuous. Thus, for all f E C[a, b], llfllsup < oo, since f is bounded. For any function f the reader will show in Exercise 2.50 that f is bounded if and only if llfllsup < oo.

C[a, b], then III

Theorem 2.4.4 Let f and g be any bounded functions on [a, b] and a

E

JR. Then

i. llfllsup = 0 if and only iff= 0. ii. llafllsup = lalllfllsup· iii. IIJ + gllsup ~ IIJIIsup + llgllsup• Item 3 is called the triangle inequality. Proof: Items (i) and (ii) are immediate from the definitions. For 3, consider that for all x E [a, b] we must have I(!+ g)(x)l ~ lf(x)l + lg(x)l ~ llfllsup + llgllsup· But II!+ gllsup is the least upper bound of the set of values I(!+ g)(x)l, and the • right side above is an upper bound of the same set of values. In Theorem 2.2.2 we learned that iff and g are in C[a, b], then f ± g E C[a, b] too, as is af for all a E R Thus continuous functions on [a,b] can be added and subtracted giving other continuous functions. The constant function 0 is an additive identity: f + 0 = f. And addition and subtraction of functions in C[a, b] as well as multiplication by constants enjoy all the usual properties of shared by any so-called vector space. We list the axioms of a general (real or complex) vector space in Table 2.1. (See [10] for the concept of a vector space.) In linear algebra, the student will have studied vectors in JR.n primarily, and also the concept of an abstract vector space. So in effect we have learned here that C[a, b] is a vector space. In a vector space, it is convenient to have defined a type of positive real-valued function on vectors called a norm. (In physics, the norm of a vector is called its magnitude.)

Definition 2.4.4 A normed vector space V, over a field F (which is either JR. or 0, then p has a minimum value on R 2.50 Prove: A function Definition 2.4.1.)

f

is bounded if and only if

11/llsup
n. Now apply the Bolzano-Weierstrass Theorem to Xn and deduce a contradiction of the fact that f E C[a, b].)

2.52

Prove that C[a, b] is a vector space.

2.53 Let B[a, b] denote the set of all bounded functions on [a, b]. Is B[a, b] a vector space? Prove your conclusion. 2.54 For each n E N, let Xn be the set of polynomials of degree equal to n, and let Pn = u~=O xk. Prove or give a counterexample: a) The set Xn is a vector space. b) The set Pn a vector space. Let P denote the set of all functions f on [0, 1] such that f(x) 2 0 for all x E [0, 1]. Is P a vector space? Justify your conclusion.

2.55

2.56 Let P denote the set of all functions f on [0, 1] such that there exists at least one point x E [0, 1], perhaps depending on f, for which f(x) > 0. Is P a vector space? Justify your conclusion. 2.57 t Let S = {1, x, x 2, x 3, ... , xn, ... }, a list of infinitely many continuous functions on [a,b]. Show that if n E N and if a 0 , 0:1, ••• , an are not all 0, then it is not possible for akxk = 0, the zero function. (Remark: This exercise shows that C[a, b] is an infinite-dimensional vector space, because it cannot have finite dimension n for any n E N.)

EZ=o

2.58

2.59

Given an example off E C(a, b) for which the range off is a) a finite open interval. b) an infinite open interval. Find

11/llsup if

61

THE BANACH SPACE C(a, bj

a) f(x)=xon(-1,!).

b) f(x) = {o_xz

if x E Q, ifx ~ Q.

2.5 THE BANACH SPACE C[a, b] We begin with a definition that is applicable whenever a sequence of functions f n is defined on a common domain, D. We do not require that the functions be continuous in the following definition.

Definition 2.5.1 Iff and f n are all defined on a domain D, we say that f n uniformly on D provided

--+

f

llfn- fllsup --+ 0 as n--+ oo. (Here the sup-norm is taken over the domain D.) A helpful way to visualize the concept of uniform convergence is to picture, for each given E > 0, the curves f(x) + E and f(x)- E. Then fn - t f uniformly on D if and only if for each E > 0 there exists N E N such that n ~ N implies that the graph of y = f n ( x) is sandwiched between the graph off ( x) + E and that off ( x) - E. The following theorem shows that uniform convergence behaves very well with respect to the continuity of functions.

Theorem 2.5.1 Suppose fn E C(D) for all n, and suppose also that fn uniformly on D. Then f E C(D).

--+

f

In words, this theorem says that a uniform limit of continuous functions must be continuous.

Proof: Let x E D be arbitrary. We must show f is continuous at x. Let E > 0. We need to show there exists 8 > 0 such that x' E D n (x- 8, x + 8) implies lf(x')- f(x)l < E. Since fn --+ f uniformly on D, there exists N EN such that k ~ N implies ll!k- fllsup < ~· In particular, this means E

llfN - fllsup < 3· Since fN is continuous at x, there exists 8 implies

> 0 such that x'

lfN(x')- fN(x)l
0. But if x = 1, x 11 = 111 ----> 1. Thus fn E C[O, 1] for all n EN, and fn----> f pointwise on D, where

f(x)={~

ifO:::; X< 1, ifx = 1.

So the continuous functions f n ----> f and yet f ¢ C[0, 1]. The function f could not have failed to be continuous on [0, I] if the convergence had been uniform . • EXAMPLE 2.8

Now consider fn(x)

=x

11

,

but on the domain x E [0,

llfn- Ollsup = fn

(~)

=

2~

H Here we see that

---->

0

H

so fn ----> 0 uniformly on D = [0, This illustrates how the question of uniform convergence is affected by the choice of domain as well as by the sequence of functions fn· Note that on [0, ~] the limit function 0 is indeed continuous.

Theorem 2.5.2 If fn

---->

f uniformly on D, then fn

---->

f pointwise on D, but not

conversely. Proof: Observe that for each x E D, we must have

lfn(x)- f(x)j :S llfn- Jllsup·

THE BANACH SPACE C[a, bj

63

Thus if llfn- fllsup----) 0, it follows that lfn(x)- f(x)l----) 0 for each xED. That the converse is false is shown by Example 2.7. • •

EXAMPLE 2.9

Let

fn(x)

=X

I+

I n.

It is not hard to see that fn(x) ----) f(x) = x pointwise for all x E [0, 1]. We claim that this convergence is actually uniform. For this purpose it can be convenient to make use of the derivative, even though we have not yet reached our chapter on this subject, but using what we recall from elementary calculus about derivatives. Let 9n(x) = f(x) - fn(x) for all x, so that 9n(O) = 0 = 9n(1) and 9n(x) 2: 0 for all x E [0, 1]. The reader can set g~(x) = 0 and check that 9n(x) achieves its maximum value at 1

and that

II!- fnllsup = 9n(xn) = (1 +1~r

(

1 ) 1- 1 + ~

----)

1 ~(11) = 0.

Thus f n ----) f uniformly on [0, 1]. The graph showing this sequence offunctions converging uniformly to f (x) = x is shown by the right-hand half of the graph in Fig. 4.5. This theorem assures us that the only possible uniform limit f for a sequence of functions f n would be the pointwise limit f. If the sequence fails to be pointwise convergent, then it cannot be uniformly convergent. However, a sequence offunctions can be pointwise convergent on a domain D without being uniformly convergent on D. Definition 2.5.3 In a vector space V equipped with a norm, we say that v n ----) v in the norm if and only if llv n -vii ----) 0 as n ----) oo. We call a sequence vn a Cauchy sequence if and only iffor all f > 0 there exists N E N such that n 2: m 2: N implies llvn- vmll < f. A normed vector space will be called complete if and only if every Cauchy sequence converges. A complete normed vector space is known also as a (real) Banach Space, after the discoverer, Stefan Banach. Theorem 2.5.3 In any normed vector space V, then it is a Cauchy sequence.

if a

sequence v n E V converges,

Proof: First, suppose Vn----) v, which we understand to mean that llvn- vii ----) 0. We need to show that the sequence v n is Cauchy. Let f > 0. Then there exists N E N, corresponding to f, such that n 2: N implies that llv n -vii < ~- Hence, if

64


nandm ;::-: N,

llvn- Vmll = ll[vn- v] + [v- VmJII ~ llvn- vii+ llv- Vmll f

f

0 there exists N EN such thatn ;::-: m ;::-: N implies llfn- fmllsup 0. Then there exists N E N such that n ;::-: m ;::-: N implies 11/n- fmllsup < ~- For each fixed x E [a, b],

Proof:

lfn(x) - fm(x)l ~ 11/n- fmllsup
0, so f n does not converge uniformly on [0, 1]. Although the question posed just above Example 2.11 is no, there is what may be called a partial converse to Theorem 2.5.1.

=

Theorem 2.5.5 (Dini) Suppose f n E C[a, b) ,for all n E N, and suppose

fn(x)

~

f(x)

at least pointwise on [a, b], where f E C[a, b] as well. If for each fixed x E [a, b] the sequence of numbers f n (x) is monotone decreasing, then f n ~ f uniformly on [a, b]. (There is also an increasing version of this theorem-see Exercise 2.70.) Proof: Let us suppose for all x E [a,b] the sequence of numbers fn(x) is a decreasing sequence converging to the limit f(x). We wish to show that llfn- fllsup ~ 0 as n ~ oo. For convenience, denote hn = fn - f, so hn E C[a, b] and for all x E [a, b], hn(x) is a decreasing sequence of positive numbers approaching 0. Let t > 0. For each fixed x E [a, b], there exists Nx E N such that n 2: Nx implies 0 ~ hn(x) < ~- Since hNx E C[a, b], there exists rx > 0 such that lx'- xl < rx and x' E [a, b] imply that lhNJx')- hNx(x)l

< ~-

Hence 0 ~ hNx(x') < t for all n 2: Nx and for all x' E (x- rx,x x' E [a, b]. Next observe that [a, b] C

+ rx)

with

U (x- rx, x + rx), xE[a,b]

an open cover of [a, b]. By the Reine-Borel Theorem, there exists a finite subcover: n

[a,b]

C

U(xk -rxk,x+rxk). k=l

Now let

N = max{Nxk

Ik

= 1, 2, ... ,n}.

If n 2: Nand x E [a, b], then there exists k E {1, ... , n} such that

X E (xk- rxk' Xk

+ rxk)

and N 2: Nxk as well. Thus 0 ~ hn(x) < t; and since hn E C[a, b], we have llhn- Ollsup 0 pointwise on lR but not uniformly on R However, prove the convergence is uniform on [0, 1].

=

x•~n. Prove: fn ---> 0 uniformly on R

2.61

Let fn(x)

2.62

Show that fn(x) = x~n converges uniformly on [0,1] but not on [0, oo ).

2.63

2.64

Let fn(x) = xe-nx for all x E [0, oo). a) Find llfnllsup for all n. b) Prove that .fn---> 0 uniformly on [0, oo). Let fn(x)

= nxe-nx for all x

E

[0, oo).

a) Find llfnllsup for all n. b) Determine whether or not fn converges uniformly on your claim.

[0, oo), and prove

2

2.65

Let fn(x) = xe-nx for all x E R a) Find llfn[[sup for all n. b) Determine whether or not fn converges uniformly on lR and prove your claim.

2.66

Let fn(x) = H:x2 for all X E R a) Find the pointwise limit of fn· b) Does fn converge uniformly on IR? Prove your conclusion.

2.67

Let fn(x) = sinn x and let~ > 0. a) Prove: fn converges uniformly on [0, ~ -~].but not uniformly on [0, ~]. b) In which sense does fn converge on [0, 1f /2)? Prove your claim.

2.68 Suppose f n (x) = 1 - xn. Decide whether or not f n converges uniformly on each interval, and prove your conclusion. a)

[0, 1]

c) [0, b], where 0 :S b < 1

2.69

Let

fn(x)

= {~

b) [0, ~] d) [0, 1)

ifO<x 0 such E.

= x sin ~is uniformly continuous on (0, 1).

Suppose f is a monotone increasing function on R True or Give a Counterexample: limx->O+ J(x) = f(O).

2.86

68


2.87 Let € > 0. Find a value of 8 > 0 for which whenever x and a are both in [o, oo) with lx - al < 8, this implies 1 val < €.

vx -

2.88 Let fn(x) = 1 + sinn 1r2x for all x E [0, 1]. True or False: The sequence fn is uniformly convergent on a) [o, !] b) [0, b] for each bE [0, 1). c)

[0, 1)

CHAPTER3

RIEMANN INTEGRAL

3.1

DEFINITION AND BASIC PROPERTIES

J:

The Riemann integral, denoted by f(x) dx, is an especially useful concept for both pure and applied mathematics. But it is much more difficult to define than the other, simpler limits studied earlier in this book. The integral is defined to meet the following objectives. Iff is a positive-valued function defined on an interval [a, b], then the integral should be the area of the region of the xy-plane above the interval [a, b] on the x-axis and below the graph of y = f(x). Iff is not strictly positive, then the integral is intended to be a signed area, by which we mean the area between the x-axis and the positive part of the graph off, minus the area between the x-axis and the negative part of the graph of y = f (x). To these ends, the integral is defined as a limit of so-called Riemann sums, which are determined by f, by a so-called partition P of [a, b], and by a choice of so-called evaluation points Xi. Now we define all these terms carefully. Definition 3.1.1 A partition P is an ordered list offinitely many points starting with a and ending with b >a. Thus P = {xo, x1, ... , xn}. where

a = xo
0 such that liP II < J implies

IP(f, {xi})- Ll

0

jb f(x) dx. a

But it must be understood that this limit is in the very intricate sense of Definition 3.1.2. The family of all so-called Riemann integrable functions f is denoted by R[a, b]. Thus, f E R[a, b] if and only if f(x) dx exists.

J:

It must be emphasized that the required inequality 3.1 must be satisfied independent of the choice of P and independent of the choice of {xi}, just so long as IIPII < J . • EXAMPLE 3.1 Let f(x) = c, a constant, for all x E [a, b]. Then f E R[a, b] and

1b

f(x) dx = c(b- a).

The proof is simple: Just observe that for all partitions P of [a, b], and for all {xi}, chosen evaluation points, we have n

n

P(f, {xi})= .L:cf:lxi = c Lf:lxi = c(b- a). i=l

i=l

71


•

EXAMPLE 3.2

Let

f(x)={

1 0

ifxEQn[0,1], if X E [0, 1] \ Q.

Then f ~ R[a, b]. This can be understood as follows. No matter how small we make IIPII, each subinterval determined by the partition must contain both rational and irrational numbers (see Section 1.6). We are free to pick all Xi E Q, in which case n

P(f, {xi}) =

L

lL~xi = 1- o = 1.

i=l

But we are free also to select all Xi ~ Q, in which case P(f, {Xi}) = 0. We cannot bring both values of the Riemann sums within E of the same number L if()< € :$ ~· Theorem 3.1.1 Iff E R[a, b], then f is bounded on [a, b]. Proof: We will suppose the theorem were false and deduce a contradiction. Denote by L the value of the f (x) dx. Then choose E = 1. There exists 8 > 0 such that IIPII < 8 implies IP(f, { xk})- Ll < 1, so that

J:

(3.2) Let us fix P of such mesh, so as to satisfy this inequality. But if f is unbounded on [a, b], there exists i E { 1, ... , n} such that f is also unbounded on [xi-1, xi]. However,

IP(f, {xk} )I=

It k=1

f(xk)fixkl2:: lf(xi)lfixi

-I L

f(xk)fixkl

ki-i

since Ia + ,61 2:: !lo:l-ltJI! (see Exercise 3.1 ). But we can hold Xk fixed for all k i= i, and vary Xi E [xi- 1 , xi] at will so as to make lf(xi)l as large as we wish, violating • the bound of inequality (3.2). Remark 3.1.1 We remind the reader again of the guidance for the proving of theorems in the Introduction on page xxiii. The reader should write out a full analysis of the proof of Theorem 3.1.1 as explained earlier, and just as should be done with regard to every theorem. Note how the indirect proof begins with the supposition that a Riemann integrable function exists that is unbounded. Then we appeal to the definition of the Riemann integral and show that there must be a partition with a subinterval on which the function is unbounded. This introduces an impossible variability in the contribution of that particular integral to the value of the Riemann sum, on account of the total freedom of choice of evaluation point in each interval of the partition.

72

RIEMANN INTEGRAL

• EXAMPLE 3.3

Let f(x) = {

Then f

tf R[O, 1], since f

~

ifO < x ~ 1, if X= 0.

is unbounded on [0, 1].

Theorem 3.1.2 Let f and g be in R[a, b] and c E R Then i. f

+ g E R[a,b] and

1b

(! + g)(x) dx =

ii. cf E R[a, b] and

1b

1b

cf(x) dx

f(x) dx

=c

1b

+

1b

g(x) dx.

f(x) dx.

Remark 3.1.2 Observe that Theorem 3.1.2 says R[a, b] is a vector space, since sums (and differences) of Riemann integrable functions are again Riemann integrable, and the same is true for constant multiples of Riemann integrable functions. Moreover, if we define a mapping T : R[a, b] ----. lR by T(f)

=

1b

f(x) dx

then Theorem 3.1.2 says that Tis a linear function, since T(cf +g)= cT(f) + T(g).

A linear function from a (real) vector space to lR is called a linear functional. Proof: i. Let E > 0, let

L

=

1b

f(x) dx and M

We know there exists 81 > 0 such that

=

1b

g(x) dx.

liP II < 81 implies

IP(f, {xi})- Ll
0 such that liP I < 82 implies IP(g, {xi})-

E

Ml < 2.


Now let

73

o=min{ 01, o2} > 0. Then liP II < oimplies

!P(f + g, {xi})- (L + M)l =I [P(f, {xi})- L]

+ [P(g, {xi})- MJI

:s; I [P(f, {xi})- LJI + I[P(g, {xi})- MJI < Thus f

f

f

2 + 2 =f.

+ g E 'R[a, b] and I:U + g)(x) dx =

L

+ M.

ii. This part is Exercise 3.2.

• Theorem 3.1.3 Suppose a

and

1c

:s; b :s; c. Iff E 'R[a, b] and f E 'R[b, c], then f E 'R[a, c]

f(x) dx =

1b

f(x) dx

I:

+

lc

f(x) dx.

I:

Proof: Let~: > 0, L = f(x) dx and M = f(x) dx. By hypothesis, there exist o1 > 0 and 02 > 0 such that if P1 is a partition of [a, b] with IIP1 11 < 1 and if P2 is a partition of [b, c] with IIP2II < 02, then we have

Let B denote Let

o

ll!llsup. where the sup-norm refers to the interval [a, c]

and is finite.

o=min { 01, o2 , 9~}.

Suppose now that Pis any partition of [a, c] with II Pll < o. Unfortunately, P needn't be the union of a partition of [a, b] with a partition of [b, c], since we might not have bE P. So we let P' = P U {b} = P1 U P2, so liP' II < too, and the same is true for P1 and P 2 , which are partitions of [a, b] and [b, c], respectively. Then

o

IP(f, {xi})- (L + M)l :s; !P(f, {xi})- P'(f, {xi} )I + IP'(f, {xi})- (L + M)l :s; IP(f,{xi}) -P'(f,{xi})l + IP1(!, {xi})- Ll + IP2(f,{xi})- Ml

00

3.12

b- a -I:! n n

a)

a +bk n

(

k=l

=

1b

f(x)dx.

a

Express

2k)

2 "cos n ( lim1+n L.J n

n-->oo

k=l

as an integral. (Be sure to include the lower and upper limits of integration.) 3.13 t Let f E C[a, b]. Show that there exists a sequence of step functions an (Exercise 3.8) such that an --+ f uniformly on [a, b]. (Hint: The function f is uniformly continuous on [a, b].) 3.14

0 Let 1 if x E { ~ In EN}, f(x)= { 0 ifxE[0,1]\{~InEN}.

Prove:

f

E R[O, 1] and

1 1

f(x) dx

= 0.

Hint: Let € > 0 and show there exists 8 > 0 such that

liP II < 8 implies

3.15 Let f be defined as in Exercise 3.14. If a is any step function (Exercise 3.8) on [0, 1], prove that II a- /II sup ~

!·

3.2 THE DARBOUX INTEGRABILITY CRITERION Here we will state and prove an alternative but equivalent criterion for the Riemann integrability of a function f : [a, b] --+ R The new criterion will be very useful for proving important theorems about the Riemann integral. In particular, the Riemann sum of a function f depends on the choice of a partition P and arbitrary evaluation points {Xi}. It is helpful to give a condition for Riemann integrability off that does not refer to evaluation points. Since every f E R[a, b] must be bounded on [a, b], we can define Upper Sums and Lower Sums for f as follows. Definition 3.2.1 Let f be any bounded function on [a, b] and let P be any partition of [a, b]. Let

THE DARBOUX INTEGRABILITY CRITERION

77

Note that Mi and mi are in JR. Define the upper sum and the lower sum by n

U(f, P) =

L Mitl.xi

n

and L(f, P) =

i=1

L mitl.xi. i=1

Observe that

L(f, P) ::; P(f, {xi}) ::; U(f, P) for all P and for all {xi}, since mi ::; f(xi) ::; Mi for all i. We can say more. Lemma 3.2.1

If P and P' are any two partitions of [a, b], then L(f, P) ::; U(f, P').

Proof: We begin with the special case in which P and P' differ by a single point. Let P = { xo, ... , Xn} be the first partition, and let z E ( x k-1 , x k), for one specific index k between 1 and n. Let P' = P U { z}. Thus P' differs from P only in that the kth interval of P has been subdivided into two smaller intervals by P', and those smaller intervals will be the kth and (k + 1)th determined by P'. Let m~. m~+ 1 , M£, and M£+1 denote the inf and sup off for the two smaller intervals. Now m~ ~ mk and m~+l ~ mk, whereas M£ ::; Mk and M£+1 ::; Mk. Thus

+ m~+ 1 (xk- z) ::; M£(z- Xk-1) + M£+ 1 (xk- z)::::; Mktl.xk.

mktl.xk ::; m~(z- Xk-d

Hence

L(f, P) ::; L(f, P') ::; U(f, P') ::; U(f, P). So far, we have shown that adding one point to P lowers the upper sum and raises the lower sum. If we add a finite number of points to P to form P', we use this argument repeated finitely often to show the same inequality. Now, suppose P and P' are any two partitions of [a, b]. Then consider their so-called mutual refinement P U P' = P". Then we have

L(f, P) ::; L(f, P") ::; U(f, P") ::; U(f, P').

• Definition 3.2.2 Define the upper integral and the lower integral off on [a, b] by

and

1b

f

1b

f = sup {L(f, P) I Pis a partition of [a, b]}

= inf{U(f, P) I Pis a partition of[a, b]},

78

RIEMANN INTEGRAL

respectively. Since every lower sum is less than or equal to every upper sum, it follows from Exercise 1.32 that

I: f ::; I: f.

Theorem 3.2.1 (Darboux Integrability Criterion) Let f be bounded on [a, b]. Then f E R[a, b] if and only if lim [U(f, P)- L(f, P)] = 0. IIPII-->O When this condition is satisfied,

I:

f (x) dx =

I: I: f =

f.

Proof: For the right-to-left implication, we suppose that

lim [U(f, P)- L(f, P)] IIPII--->0

= 0.

Since

for all P, we have

as

liP II

---4

0. It follows from the squeeze theorem that

which we denote by L. But then we know that if t > 0 there exists t5 > 0 such that for all P with mesh less than t5 we musthaveboth L andP(f, {xi}) between L(f, P) and U (!, P), and thus less than t apart, independent of the choice of evaluation points Xi. Thus letting liP II ---4 0, we see that

lim P(f, {xi})= L IIPII--->0

I:

so that f E R[a, b] and f(x) dx = L. For the left-to-right implication we suppose f E R[a, b], and write L = Lett

lb

f(x) dx.

> 0. There exists 8 > 0 such that liP II < t5 implies IP(f,{xi})-LI
0. Thus f E R[a, b]. Finally iff is decreasing, then- f is integrable by • the first part of this proof, and so f = -(-f) E R[a, b] as well. Here is a variant of the Darboux Integrability Criterion which is often useful. (See Exercises 3.26 and 3.27 for applications of this variant.)

Theorem 3.2.4 Let f be bounded on [a, b]. If for all E > 0 there exists a partition Po of[a, b] such that U(f, Po) - L(f, Po) < E, then f E R[a, b]. Proof: Let M =

ll!llsup E JR., and let E > 0. By hypothesis there exists Po = {xo, ... , x N}

such that U(f, Po) - L(j, Po)

0 such that f(x) 2: 8 for all x E [a, b], prove that

1

7

E'R[a,b].

Hint: On each interval [xk-1. xk] of any partition, compare 1 1 .l 1 MJ =sup f(x) and m£ = inf f(x)

with the corresponding numbers M{ and m{ Then use the Darboux criterion. b) Now suppose only that lf(x )I 2: 8 > 0 for all x E [a, b], for some fixed 8 > 0. Show again that

1

7 Hint: Apply (a) to 3.48

-j, and then use Theorem 3.4.1.

J0'i (1 + tanx)y'xsecxdx::::; ?T~.

Use the triangle inequality for

[1~ 3.50

R[a,b].

Use the Cauchy-Schwarz inequality to show: a) f011" v'x sin x dx ::::; 1r. b)

3.49

E

Prove:

(

II · ll2 to show 1

2

2

v'cosx + x) dx ]

::::;

1+

~-

Iff E R[O, 1], then 1 [ {1 Jo{1 xf(x) dx::::; v'3 Jo [f(x)f dx] !

94

RIEMANN INTEGRAL

3.51

Prove: Iff E 'R[O, 1], then

[1 (y'cosx+f(x)) dx] ! s; ~+ [1 (f(x)) dx] ! 1

1

2

2

3.52 tIff and g are in n[a, b], we say f is orthogonal tog, denoted by f and only if (!,g) = 0. Prove that

f

l_ g {:}

l_

g, if

II!+ gil~ = IIIII~ + llgll~·

(This is a modem analogue of the Pythagorean Theorem.)

3.53

Let f and g be in 'R[a, b], with llgll2 > 0. a) Find a constant c E IR such that (f - cg) l_

g. (Hint: Use the definition of orthogonality in Exercise 3.52.) b) Now let f(x) = x and g(x) = 1 on [0, 1]. Find the value of c for which (!- cg) l_ g. Using this value of c, would (!- cg) l_ g on the different interval [-1, 1]? 3.54 3.55

Let f(x) = sinx and g(x) = cosx on [0, n]. Prove f

l_

g.

Suppose llfll2llgll2 > 0, where f and g are in R.[a, b]. a) Prove: I(!, g) I = llfll2llgll2 if and only if there exists t E 1R such that

1b[g(x) + tf(xW dx = 0. (Hint: Let p(t) = (tf + g, tf +g) ;: : 0, as in the proof of the CauchySchwarz inequality.) b) The preceding part says that we get equality in the Cauchy-Schwarz inequality if and only if g is very close to being a constant multiple of f. What happens to the Cauchy-Schwarz inequality if llfll2 = 0 or if llgll2 = 0?

3.56

Let f(x) = sinx on [0, 1]. Give an example of g, with

l(f,g)l = ll!ll2ll.qll2· 3.57 Let J,g E n[a,b]. (!,g) = llfll2ll.qll2· 3.58

Show that

ll.qll2 > 0, for which

II!+ gll2 = 11!112 + llgll2

if and only if

Let g E R.[a, b], and define T 9 : C[a, b] ----. IR by

T 9 (!) = 1b f(x)g(x) dx for all f E C[a, b]. Prove: T9 is a bounded linear functional on the Banach space C[a, b], equipped with the sup-norm.

Remark 3.4.3 On a cover of the Notices of the American Mathematical Society [ 17] the reader can see a photograph of a clay tablet from the Yale Babylonian Collection.

TEST YOURSELF

95

The sketch on this tablet indicates that Babylonian mathematicians were aware of the geometrical reasoning that justifies the Pythagorean Theorem of plane geometry for the case of a right isosceles triangle. This was approximately one thousand years before the life of Pythagoras - about three thousand years before the present time. It is considered the earliest known record indicating a geometrical proof. We mention this because it is connected to an interesting circle of ideas that we have encountered. We have used the quadratic formula to prove the Cauchy-Schwarz inequality for the vector space R[a, b]. The reader could easily look ahead to Theorem 8.1.1 to see that essentially the same method of proof establishes the Cauchy-Schwarz inequality in every vector space equipped with a scalar product (x, y), which includes also the Euclidean plane and n-dimensional Euclidean space. The Cauchy-Schwarz inequality yields as a consequence the triangle inequality of plane geometry, but in the general context of a vector space with an inner product-a so-called inner product space. Although the concept of an inner product space appears at first abstract, the geometry that is employed hinges on the quadratic formula, which embodies the method of completing the square learned in high school algebra. This method has been used in school exercises for at least four thousand years. Again, the earliest records appear on clay tablets that were school-room exercise tablets from Babylon [15]. And in Exercise 3.52 above, the reader has proven a version of the Pythagorean Theorem for the vector space R[ a, b]. The same proof works in the Euclidean plane, in n-dimensional Euclidean space, or in any inner product space. It is a simple application of the inner product of two vectors. In other words, we have learned some elements of the geometry of the functionspace R[a, b] by applying the quadratic formula from four thousand years ago. Study of this abstract and modem subject from functional analysis has shed light on Euclidean Geometry. The tool on which this rests is four thousand years old, and preceded Euclidean Geometry by two millennia. So here is a question that is not part of Mathematics: Is this light shed upon Euclidean Geometry a new light, or an old one?

3.5 TEST YOURSELF

EXERCISES 3.59

Let f n ( x)

=

l+~x2 for all x E JR. Find:

= limn-+oo fn(x) for all x E JR. b) llfnllsup for each n E N. c) Is fn uniformly convergent on IR? (Yes or No) a) The pointwise limit f(x)

3.60

Does f n (x) = xn converge pointwise on a) (-1,0]?

b) [-1, 0]?

96

RIEMANN INTEGRAL

3.61

Let

f(x)

=

{12 ifif 01 ~ x < 2.1, ~X~

If E

> 0, find a 8 > 0 such that

independent of the choice of evaluation points Xi·

3.62

Evaluate

2

n ( lim- "cos

n-+oo

n L..J k=1

2k)

1+n

by expressing it as a suitable definite integral and evaluating this by means of the Fundamental theorem.

3.63

Let

f(x)

= {1 0

ifx = 1, ifxE[0,2]\{1}.

Find a partition P of [0, 2] for which U(f, P)- L(f, P)

3.64

f

E

3.65

True or Give a Counterexample: If

III

E

0

I~

.f, we

,

provided that this limit exists. (Note that in accordance with the concept of limit, h ----> 0 through values such that x 0 + h E D 1. It is also common to denote x = x 0 + h and to write the limit criterion as

!

'( x ) = 0

f(x)- f(xo) . 1nn

x->xo

x- Xo

Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright © 2008 John Wiley & Sons, Inc.

99

100

THE DERIVATIVE

if the limit exists.) This limit is often interpreted geometrically as representing the slope of a tangent line to the graph of y = f(x) at the point (x 0 , f(x 0 )). The difference quotient of which f' (x 0 ) is the limit is the slope of a chord joining ( x 0 , f (x 0 )) to the point (x 0 + h, f (x 0 + h)). In effect, if the limit defining f' (xo) exists, we take the tangent line to be defined by the equation y- f(x 0 ) = f'(x 0 )(x- x 0 ), which describes the unique line through the base point (x 0 , f(x 0 )) with the slope f'(xo) . • EXAMPLE 4.1

We present several examples of derivatives. i. Let f(x)

=

lxl- Then we claim f'(O) does not exist. In fact, lim

IO + hi - 101 = h

h--->0

lim h--->0

1M h

does not exist, since if hn --+ 0+ (i.e., from the right side ofO), then~ --+ 1, but if hn --+ 0- (i.e., from the left), then ~ --+ -1. Since these one-sided limits are different, the limit of the function does not exist. ii. Again we take f(x) = lxl, but we show that if xo > 0, f'(xo) = 1. In fact, we can use h such that Ihi < xo, and then lim lxo

iii. Now let f(x)

= xn, n

+ hl-lxol = h

h--->0

lim!!_ h

= 1.

h---tO

EN. Then

J'(x) = lim (x h--->0

+ h)n- xn

h nxn-lh + n(n-l)xn-2h2

=lim

h---tO

2

+ ... + hn

h

= nxn-1. iv. If f(x)

= c, a constant, then f'(x)

= 0. (See Exercise 4.3.)

There is another way to understand the concept of a tangent line that is also very useful, although the definition may seem a bit more technical at the beginning. Imagine now a straight line through ( x 0 , f (x 0 )) with slope m. Thus the rise of this line corresponding to an increment h in the x-variable is given by a linear function L(h) = mh. On the other hand, the rise of the graph of y = f(x) will be denoted by tJ.f = f(xo +h)- f(xo). The intuitive idea is that the graph y = f(x) has a tangent line at (x 0 , f(x 0 )), provided that there exists a linear function L(h) = mh that approximates tJ.f accurately for all sufficiently small values of h. Then the

DERIVATIVES AND DIFFERENTIALS

101

tangent line corresponds to the linear function L(h) that gives this approximation. We make the following formal definition based on this idea.

Definition 4.1.2 The function f is called differentiable at a cluster point x 0 E D f, provided that there exists a linear function L(h) = mh such that the function given by i(h) = D.f- L(h) --+ 0

h ash--+ 0.

The intuitive idea is that iff( h) than h--+ 0.

--+

0 as h

--+

0 then D.f - L( h)

--+

0 much faster

Theorem 4.1.1 A function f is differentiable at a clusterpointx0 E DJ if and only if f'(xo) exists. Iff is differentiable at xo, then L(h) = f'(xo)h is denoted by dfxo (h) and (4.1) D.f = f(xo +h)- f(xo) = dfxo(h) + t:(h)h, where f(h)--+ 0 ash--+ 0 (in such way that x 0 +hE DJ). Proof: For the implication from left to right, we suppose that f is differentiable at xo. Then there exists m E lR such that f

as h

--+

(h)

=

D.f- mh

h

--+

0

0. Thus D.f

=

f(xo +h)- f(xo)

h

= m + f(h)

-+ m

h

as h --+ 0, so f' ( xo) exists and equals m. For the opposite direction of implication, we suppose f'(xo) exists. Let m If we define L( h) = mh and let f

(

h)= D.f- L(h) = f(xo +h)- f(xo) _

h

then the hypothesis implies f(h)

h

--+

0 ash-+ 0.

=

f'(xo).

!'( )

Xo ,

•

Remark 4.1.1 We remind the student here to review the Introduction regarding Learning to Write Proofs on page xxiii. The reader should write out a careful analysis of what makes each proof in this course work. We will do this together for the first proof of the chapter. We are trying to prove that a function f is differentiable if and only if its increments D.f can be approximated well locally by the differential, df, in the technical sense described in the statement of the theorem. For the implication from left to right, we suppose f is differentiable, so that D.f can be approximated by df using a suitable error-function f. This enables us to express the difference quotient for the derivative

102

THE DERIVATIVE

in terms of df and t:, and the assumption that E __, 0 as h __, 0 implies that the limit that yields the derivative exists. For the opposite implication we assume f'(x) exists and we define E by means of Equation (4.1 ). All that remains is to prove that t: __, 0 as h __, 0. By writing E in terms of the difference quotient and the derivative, the existence of the derivative implies that f. __, 0 as h __, 0.

Corollary 4.1.1 Iff is differentiable at xo then f is continuous at xo. Proof: We need to prove limx->xo f(x) = f(x 0). This is equivalent to proving limx->xoU(x)- f(xo)) = 0. Denote h = .T- xo, and we have lim (f(xo +h) - f(xo)) = lim [dfxo (h)+ E(h)h]

h->0

h->0 =lim [f'(xo) h->0

+ E(h)]h = 0.

• Theorem 4.1.2 Suppose f'(x) and g'(x) both exist. Then i. (f ± g)'(x) exists and (f ± g)'(x) = f'(x) ± g'(x). ii. (fg)'(x) exists and (fg)'(x) = f'(x)g(x)

iii.

+ f(x)g'(x).

('&)' (x) exists and('&)' (x) = g(x)f'(~)(~{Jx)g'(x), ifx E Df is a cluster point of D L· g

Proof: Let us prove case (i) for f +g.

r

h~ .

(f

I un = h--->0

+ g)(x +h)- (f + g)(x) h

(f(x +h)- f(x) h

+

g(x +h)- g(x)) h

= f'(x) + g'(x) The proof for f

-

g is very similar. For (ii) and (iii) see Exercises 4.4 and 4.5.

•

Theorem 4.1.3 (The Chain Rule) Suppose g is differentiable at xo and f is differentiable at g(xo). Then (f o g)'(x0 ) exists and (f o g)'(xo) = f'(g(xo))g'(xo).

Proof: To show that the derivative of the composition f o g exists at x 0, we denote k = g( xo + h) - g( xo) __, 0 as h __, 0 since g is continuous at xo by Corollary 4.1.1.

EXERCISES

103

Also, k = g'(xo)h + F:(h)h, where f:(h) --+ 0 ash--+ 0. Next we form the following difference quotient noting that E( k) --+ 0 as k --+ 0:

f(g(xo +h))- f(g(xo)) h f'(g(xo))k + E(k)k h f'(g(xo))[g'(xo) + F:(h)]h + E(k)[g'(xo) h --+ J'(g(xo))g'(xo)

+ F:(h)]h

•

ash--+ 0. EXERCISES

< 0, showthatf'(xo) = -1. (SeeExample4.1(ii).) 4.2 Prove directly from Definition 4.1.1 that if f' (a) exists, then f must be 4.1

If f(x)

=

lxl andifxo

continuous at a. (Hint: Suppose Xn --+ a with each Xn E D f \ a. Prove that f(xn)- f(a) --+ 0 as Xn --+a.) 4.3

t Prove the conclusion of Example 4.1 (iv). t Prove conclusion (ii) of Theorem 4.1.2. (Hint:

4.4 of Corollary 4.1.1.) 4.5

You may use the conclusion

t a) Prove: if g'(x) exists and g(x)

=/=-

0, then

(i )' (x) exists and is equal to

~(~W. (Hint: You may use the conclusion of Corollary 4.1.1.) b) Now use part (a) above together with Theorem 4.1.2(ii) to prove conclusion (iii) of Theorem 4.1.2. 4.6

Let

ifx E (0,1], ifx =0, as in Fig. 4.1. a) Prove: f E C[O, 1]. (Be sure to consider x = 0. You can use the squeeze theorem for functions.) b) Prove: f'(x) exists for all x E (0, 1], but f'(O) does not exist. (We assume you know about the derivative of sin x from elementary calculus.) 4.7

t Let f (x)

See Fig. 7.1, p. 217.

= {

~2 sin ( ; )

ifx =/=- 0, ifx =0.

104

THE DERIVATIVE

Figure4.1

f(x) = xsin (~).with envelope u(x) = x, l(x) = -x

a) Prove: f'(x) exists for all x E JR. (Be sure to consider x b) Prove: f E C(JR). c) Prove: f' is not continuous at x = 0. 4.8

= 0.)

Let x2

f(x) =

{

0

ifx E Q, ifx tJ_ Q.

Prove that f is continuous at one and only one point x = a. Find the value of a and prove that f' (a) exists. Suppose f and g are differentiable and suppose that f(g(x)) x E D 9 is a cluster point of D 9 •

4.9

= x.

Suppose

a) Prove:

g'(x)

=

f'(gl(x)).

b) Let f be a restriction of the sine to a domain on which it is injective. Suppose that f and g = f- 1 are differentiable. Use the result of (a) to derive a familiar formula for g '. (This will be treated further in Theorem 10.4.3.)

THE MEAN VALUE THEOREM

105

4.2 THE MEAN VALUE THEOREM The theorems in this section play a vital role in such diverse applications as extreme value problems, proofs of inequalities, and as we shall see in Section 4.3, even the proof of the Fundamental Theorem of Calculus. Definition 4.2.1 The local maximum and local minimum points off, defined below, are called local extreme points. i. A function f is said to have a local maximum point at a E D 1 if and only there exists 8 > 0 such that x EDt n (a- 8, a+ 8) implies f(x) ~ f(a). ii. The point a E D 1 is called a local minimum point if and only 8 > 0 such that x EDt n (a- 8, a+ 8) implies f(x) ~ f(a).

if

if there exists

The idea behind the preceding definition is that a local extreme point need not be either the highest or lowest point on the entire graph of the function, but will be a local high point or a local low point-meaning just within its own vicinity. To study local extreme points with the derivative, we need the concept of an interior point of a domain. Definition 4.2.2 A point a is called an interior point of D 1 if and only if there exists 8 > 0 such that (a- 8, a+ 8) s:;; Dt· The set of all interior points of Dt is denoted by D/, the interior of D 1. In other words, an interior point a of a domain D is a point which has no points from the complement of D within some small specified radius 8 > 0 of a. Theorem 4.2.1 Iff is differentiable at a local extreme point J.L E

D/, then f' (J.L) =

0.

Proof: 1.

Consider first the case in which f has a local maximum point at J.L. Note that there exists 8 > 0 such that 0 < h < 8 implies f(J.L +h)- f(J.L)

h

~ 0,

whereas -8 < h < 0 implies

Thus lim f(J.L+h)-f(J.L) 0 h

-

'

0

106

THE DERIVATIVE

Because f' (Jl) exists, it follows that

f'(M) = lim f(Jl +h)- f(Jl) = 0 h->0 h since it is both nonnegative and nonpositive.

ii. If f has a local minimum at Jl, consider g at Jl, to reach the desired conclusion.

= - f, which has a local maximum

• Theorem 4.2.2 (Rolle's Theorem) Iff E C[a, b] is such that

f(a)

= 0 = f(b)

and such that f' (x) exists at least for all x E (a, b), then there exists Jl E (a, b) such that f' (Jl) = 0. Remark 4.2.1 Rolle's Theorem says that if the chord joining the two endpoints (a, f(a)) and (b, f(b) lies on the x-axis, then there exists a tangent line at some point of the graph that is parallel to the chord.

Proof: Iff : : : : 0 then the claim of the theorem is trivial. If there exists x 0 E [a, b] such that f(xo) > 0 then the maximum point off on [a, b] must occur at an interior point Jl E (a, b). By Theorem 4.2.1, f' (Jl) = 0. The remaining case, in which there • exists x 0 E [a, b] such that f(x 0 ) < 0, is similar. Theorem 4.2.3 (Mean Value Theorem) Suppose f E C[a, b] and differentiable at least on (a, b). Then there exists JL E (a, b) such that

f'(!-l)

= f(b)- f(a). b-a

Remark 4.2.2 The Mean Value Theorem states that under the given hypotheses there must be a tangent to the graph that is parallel to the chord joining the endpoints. (See Fig. 4.2.) Observe that this is a generalization of Rolle's Theorem.

Proof: Observe that the straight line that is the chord joining the endpoints of the graph off is described by the equation

y = f(a)

+

f(b)- f(a) (x- a). b-a

Now, for all x E [a, b], let

h(x)

= f(x)- y = f(x)-

f(a)-

f(b)- f(a) (x- a), b-a

THE MEAN VALUE THEOREM

107

[h) Figure 4.2

f (x) = v'l -

x 2 , chord with parallel tangent.

the difference in height between the graph off and the graph of the chord. Observe that hE C[a, b] and h(a) = 0 = h(b). Also, h' exists at least on (a, b). Thus Rolle's Theorem implies there exists J-t E (a, b) such that h' (J-t) = 0. This implies that

Corollary 4.2.1

Iff'

=0 on an interval I, then f is a constant function on I.

•

Proof: Fix any a E I and let x E I be arbitrary. Then the Mean Value Theorem implies there exists J-t E I such that

f(x)- f(a) = f'(J-t)(x- a)= O(x- a)= 0

=f(a), a constant function. • Corollary 4.2.2 If f' = g' on an interval I, then there exists c E IR such that f(x) =g(x) +c. Proof: Let h( x) = f (x) - g (x) and observe that h' =0 on I. Now use Corollary so that f(x)

4.2.1. Corollary 4.2.3 Suppose f E C[a, b] is differentiable at least on (a, b). Then

•

108

THE DERIVATIVE

i. Iff' (x) 2: 0 for all x E (a, b), then f is increasing (denoted by f /)on [a, b]. ii.

If!' (x) > 0 for all x

E

(a, b), then f is strictly increasing (denoted by f

i)

on [a,b]. iii. If f'(x) ::::; Oforall x E (a, b), then f is decreasing (denoted by f '\,)on [a, b]. iv. If f'(x) < Ofor all x E (a, b), then f is strictly decreasing (denoted by f on [a,b].

!)

Proof: i. Suppose that f'(x) 2: 0 for all x E (a, b). Then ifx1 < x2 are points of [a, b], the Mean Value Theorem implies there exists J.L E (a, b) such that

so that j(x2) 2: j(x1), and f is increasing on

[a, b].

For the remaining cases, see Exercises 4.10-4.12.

•

• EXAMPLE 4.2

If x

> 0, then sinx < x.

Proof: Let f(x) = x- sinx. Then f(O) = 0 and f'(x) = 1- cosx > 0 on (0, 27T). By Corollary 4.2.3, f is increasing strictly on [0, 27r], so that f(x) > 0 for all x E (0, 27T]. Hence sinx < x if 0 < x ::::; 27T. If x > 27T, however, it follows that sin x ::::; 1 < x as well. •

Definition 4.2.3 A function f : I

R

where I is an interval, is called monotone on I if and only iff is either increasing throughout I or else decreasing throughout I. The function f is called strictly monotone provided it is either strictly increasing or else strictly decreasing. --+

Definition 4.2.4 A function f : D --+ IR, where D 0 for all x E [a, b]. 4.14

Give an example off increasing strictly on [a, b] yet f

4.15

t Let n

E N. Prove:

(1-x

for all x E

4.16

¢ C[a, b].

2

t

[0, 1]. Hint: Let f(x) = (1-

Prove: log(1

4.17

Prove: 1 -

4.18

Prove:

~ 1-nx2 x2

t- (1- nx

2

).

+ x) < x for all x > 0. 2

x 2

< COS X for all X

=/:- 0.

X

- -2 < tan- 1 x < x 1+ x

for all x

4.19

> 0.

Let p be a polynomial of degree n and let

Prove iliat the number of elements in the set E, denoted by IEl, satisfies the inequality lEI ~ n + 1. (Hint: Apply Rolle's Theorem.)

4.20

Suppose If' (x) I ~ M E lR for all x E I, an interval. a) Prove: f is uniformly continuous on I. b) If I is a finite interval (a, b), prove that limx-.a+ f (x) exists. (Hint: It is enough for this problem to know that f is uniformly continuous, by the previous part. Consider first any sequence Xn ~a+ and prove that f(xn) is Cauchy.) c) Consider the function s ( x) = sin ~ on (0, 1). Does s satisfy ilie hypothesis of this exercise? Investigate statements (a) and (b) for the case of the functions.

4.21

4.22

Let q(x) = y'X for all x E [0, 1]. True or False: a) The function q is uniformly continuous on (0, 1). b) The function q' is bounded on (0, 1).

t Suppose f

E

C[a, b]. Suppose f is differentiable at a and at b, and also that J'(a)J'(b) < 0.

11 0

THE DERIVATIVE

Prove f is not injective on [a, b]. (Hint: Consider the extreme points of f.)

4.23 If D of D.

~

JR, then prove D 0 , the interior of D, is the union of all open subsets

4.3 THE FUNDAMENTAL THEOREM OF CALCULUS Theorem 4.3.1 (Fundamental Theorem, Version 1) Suppose f E R[a, b], and suppose also there exists F such that F'(x) = f(x) on [a, b]. Then

1b

f(x) dx = F(b)- F(a).

Remark 4.3.1 We will see by example below that it is possible for a derivative to exist throughout [a, b] and yet not be integrable, which implies that a derivative can exist without being either continuous or monotone, for example.

Proof: By hypothesis if E > 0 there exists 8

> 0 such that liP I < 8 implies

Let P = {xo, ... , Xn} be any partition of [a, b] such that liP II < 8. By the Mean Value Theorem in each sub-interval [xi- 1 , xi] we are free to pick Xi in such a way that F'(xi)Llxi = F(xi)- F(xi-d· Then n

P(f, {xi})=

L f(xi)Llxi i=1 n

= L:F'(xi)Llxi i=1 n

= L[F(xi)- F(xi-1)] i=1

= F(b)- F(a). Since this implies IF(b)- F(a)-

I:

I:

f(x) dxl

0, we must have •

• EXAMPLE 4.4 Let

F(x)

E

=

2

{x sin ()

;2

ifO < lxl:::; 1, if X= 0.

THE FUNDAMENTAL THEOREM OF CALCULUS

111

The student can show that F'(O) exists and equals 0, and F'(x) can be calculated using the Chain Rule at any x f:- 0. Thus F' exists throughout [-1, 1], yet F' fJ_ R[-1, 1] and also F' fJ_ C[-1, 1]. (See Exercise4.24.) Theorem 4.3.2 (Fundamental Theorem, Version 2) Suppose f E C[a, b], and define

F(x) = for all x

E

1x f(t) dt

[a, b]. Then F'(x) exists and F'(x)

Proof: The difference quotient F(x

+

hh-

F(x)

= f(x ).

=

*1x+h

=

h J(x)h

f(t) dt

1

= f(x) ____. f(x) as h ____. 0. Here x comes from the Mean Value Theorem for Integrals (Exercise 3.1 0), and f(x) ____. f(x) since x ____. x ash____. 0 and f E C[a, b]. • We have seen in Example 4.4 that a derivative can exist without being continuous. Nevertheless, all derivatives on intervals do share one property in common with continuous functions, as we see below. Theorem 4.3.3 (Intermediate Value Theorem for Derivatives) Iff' (x) exists for all x E I, an interval, then f' has the Intermediate Value Property on I: Namely, if a, bE I and ify lies strictly between f'(a) and f'(b), then there exists x between a

and b such that f'(x)

= y.

Proof: Suppose a < band f'(a) < y < f'(b). (For the opposite inequality, just consider g = -f.) Define 1>(x) = f(x) - yx so that¢ E C[a, b] and is also differentiable on [a, b]. It would suffice to show there exists x E (a, b) such that ¢'(x) = 0. By the Extreme Value Theorem, ¢must have both a maximum and a minimum point in [a, b]. If such an extreme point is also an interior point, we are done. So we will suppose the extreme points occur only at the endpoints a and b, and we will deduce a contradiction. We will present the argument in three cases. 1.

Suppose ¢(a) is a maximum point and ¢(b) is a minimum point. Then

A-'( )

.

1 '+'a=nn

h-->0+

¢(a+ h)- ¢(a) < 0 h _

so that f'(a):::; y. And ¢'(b)= lim ¢(b +h)- ¢(b) :::; 0 h-->0h

112

THE DERIVATIVE

so that f' (b) ::; y as well. This contradicts the hypothesis. ii. Suppose ¢(a) is a minimum and ¢(b) is a maximum. Then a similar contradiction can be deduced. (See Exercise 4.30.) iii. Finally, if one endpoint is both a maximum and a minimum point, then ¢ is a constant function and ¢' (x) 0.

=

•

EXERCISES 4.24 t Prove that F' (x) exists for all x E [-1, 1] in Example 4.4, yet F' is not bounded, and so F' tJ_ R[-1, 1] and also F' tJ_ C[-1, 1]. 4.25

Let

1

x2

F(x) =

2

e-t

for all x E R Find F'(x). (Hint: Let G(u) ThusF =Go U.)

4.26

2

= J0u e-t dt

where u

= U(x) =

x2•

Suppose f E R[a, b] and let

F(x) =

1x

f(t) dt

[a, b]. Prove: F E C[a, b]. (Hint: f must be bounded.)

for all x E

4.27

dt

Let

f(x) =

{~in~

Let

F(x) =

1x

if 0

<X

::;

1,

if X= 0.

f(t) dt

for all x E [0, 1]. Prove: FE C[O, 1]. (Hint: See Exercise 3.27.)

4.28 Suppose/ E R[a,b] andletF(x) = J: f(t)dtforallx E [a,b]. If f(x) 2 0 for all x E [a, b], prove F is increasing on [a, b]. 4.29

Suppose f' and g' exist and suppose f', g' E R[a, b]. Prove:

1b

f(x)g'(x) dx = f(b)g(b)- f(a)g(a)

(Hint: Consider J:(fg)'(x) dx.)

4.30

t Prove case (ii) of Theorem 4.3.3.

-1b

g(x)f'(x) dx.

EXERCISES

4.31

= 2xsin (~)-cos(~).

f(x)

Figure4.3

113

Suppose

f(x)

= {

~

x < 1,

ifO

~

if1

~X~

2.

a) Does there exist a function F on [0, 2] such that F'(x) = f(x)? b) Let F(x) = f(t) dt for all x E [0, 2]. Prove that FE C[O, 2] but that it is false that F'(x) = f(x) on [0, 2].

J;

4.32

Let 2 .

F(x) =

{

1

~ sm;

ifO < lxl ~ 1, ifx = 0

and let f(x) = F'(x). (See Fig. 4.3.) a) Prove that F' (x) exists for all x E [-1, 1]. b) Find f(x) for all x E [-1, 1], and prove f E R[-1, 1] \ C[-1, 1]. (Hint: Apply Exercise 3.26.) c) Find

4.33

Find

f~ 1 f(x) dx.

J; g(x) dx if g (X )

=

2x cos K X {0

+ 1r sin K X

ifx E (0, 1], ifx = 0.

114

THE DERIVATIVE

4.34 Suppose f : (-a, a) ---+ JR. is a differentiable function. Either prove the following statements or give counterexamples: a) Iff is an odd function, then f' is an even function. b) Iff is an even function, then f' is an odd function. c) 0 Iff' is odd, then f is even. (Hint: Use Theorem 4.5.1.) d) If f' is even, then f is odd. 4.4

UNIFORM CONVERGENCE AND THE DERIVATIVE

We have learned that if fn E C[a, b] and if llfn - !II sup ---+ 0, then f E C[a, b] as well. We have seen that if fn E 'R[a, b] and if llfn -!II sup ---+ 0, then f E 'R[a, b] and, moreover,

1b

fn(x) dx

---+

1b

f(x) dx.

It is understandable that the student now anticipates learning a virtually identical result for differentiation. The following examples show, however, that differentiation behaves in a more delicate manner with respect to uniform convergence (meaning sup-norm convergence) . • EXAMPLE 4.5

Let f n (x) = sinnnx for all x E JR., and let g( x)

llfn- Yllsup =sup {

lsinnxll n

=0. Then X

E lR

}

1 = ;;:

---+

0

so that fn ---+ g uniformly on R However, f~(x) = cosnx ~ g'(x) = 0 for some values of x. For example, if x = 0, cos nx cos 0 = 1 ~ 0, and this failure of f~ (x) to converge to g'(x) occurs at many other values of x as well. (See Fig. 4.4.)

=

y

Figure 4.4

f n ( x) =

~ sin nx, n

= 10, 20, 40.

UNIFORM CONVERGENCE AND THE DERIVATIVE

115

• EXAMPLE 4.6

Let

fn(x)

=

lxll+~

for all x E [-1, 1]. Then fn is increasing as n increases and fn(x)----> f(x) = lxl uniformly on [-1, 1] by Dini's Theorem. (An alternative approach to proving this uniform convergence is illustrated in Example 2.9.) If x > 0, f(x) x and J'(x) = 1. If x < 0, then f(x) -x and f' (x) = -1. Note that f' (0) does not exist, but

=

=

. . fn(x) - fn(O) x 1 +~ llm = 1lm x--->0+ X - 0 x--->0+ X = lim x~ = 0 1 X--->0+

and

( -x)l+~ . fn(x)- fn(O) . ltm = 1lm X - 0 x--->0X

X--->0-

=lim -(-x)~=O, x--->0-

so that

f~ (0)

= 0 ----> 0 and yet f' (0) does not exist.

(See Fig. 4.5.)

y

X

-1.0

Figure4.5

-0.5

0.5

1.0

lxl 1+!, lxl 1 +~, ... increases and approaches lxl.

Nevertheless, there is a useful theorem about uniform convergence and the derivative, as we see below. Theorem 4.4.1 Suppose f n is defined on a finite interval I and f~ E C (I). Suppose f~ converges uniformly on I. Suppose moreover that there exists at least one point a E I such that f n (a) is a convergent sequence of real numbers. Then there exists a differentiable function f such that f n ----> f uniformly on I, and

f'(x)

= lim f~(x) n->oo

116

THE DERIVATIVE

on!. Remark 4.4.1 Notice that the key unexpected hypothesis is that it is the sequence of

derivatives f~ which we must assume to be uniformly convergent. Proof: Let g denote temporarily the uniform limit off~ on I, so g E C[a, x] for each x such that [a, x] ~ I. By the Fundamental Theorem (Version 1) we can define f(x) for all x E I as follows: fn(x) = fn(a) --->

+

1x f~(t)

lim In( a)+

n-+oo

dt

r g(t) dt = f(x).

Ja

By the Fundamental Theorem (Version 2) f'(x) = g(x) for all x E I. Thus fn---> f at least pointwise. To show Ilin- /II sup ---> 0, observe that

lfn(x)- f(x)l = \Jn(a)

+

1x f~(t)

dt- f(a)

-1x

g(t) dti.

Denoting by L the finite length of I, we have 11/n- Jllsup :::; lfn(a) - J(a)l

+ L II/~- 9llsup ---> 0 + 0 =

0

•

as n---> oo.

EXERCISES 4.35

Let

fn(x) = .!sin(n 2 x). n Prove: f n converges uniformly on IR to a differentiable function, yet f~ (0) diverges.

4.36

In Example 4.5, find a value of x for which f~ (x) diverges as n ---> oo.

4.37

Give an example of a sequence fn for which f~ ---> 0 uniformly on IR, yet ER

fn(x) diverges for all x

4.38 Let fn(x) = sinn x for all x E [0, 1r]. Prove that f~ is not uniformly convergent on [0, 1r]. (Hint: Suppose false and apply Theorem 4.4.1 to deduce a contradiction.) 4.39 Denote by C1 [a, b] the space of functions having continuous derivatives on [a, b]. Define 11/11 = 11/llsup + IIJ'IIsup for each f E C1 [a, b]. a) Prove that II · II is a norm on the vector space C 1 [a, b]. b) a

= L,

f(x) · d . g(x) extsts an then l lmx-+a

lim f(x) = L. x-+ag(x)

Proof: Observe that under the hypotheses we must have f(a) Cauchy's Generalized MVT we have f(x) g(x) as x

---->

= f(x)- f(a) = f'(x) g(x)- g(a)

a because this forces x ----> a.

---->

=

0

= g(a).

By

L

g'(x)

•

Remark 4.5.2 We observe that a similar theorem could be proved with the limit as x----> band that g(x) can't be 0 in this theorem.

CAUCHY'S GENERALIZED MEAN VALUE THEOREM

119

• EXAMPLE4.7

We give some examples of limits that can be computed by means ofL'Hopital's rule. · l'1lllx--+1

I.

.. l'lmx--+0

11.

logx x~

-1·IIIlx--+1

1 -

1~cosx

1/x _ - 1- -

1.

-1'1mx--+0 2X sinx

~ -

cosx = 1'lm~;--+0 -2-

_ 1

2·

-

Next, we prove two of many possible variations on L'Hopital's Rule. (A similar theorem could be proven for x ~ -oo.) Theorem 4.5.3 Suppose f and g are both differentiable on ( b, oo) with g'( x) nowhere 0 and b > 0. Suppose that lim f(x)

=0=

X-+00

lim g(x) X--+00

and also that

= L.

lim f'(x) X--+00 g 1 (X)

Then

limx--+oo 1 f~')) _qx

Proof:

exists and is equal to L.

Define

F(u) = {

~ (~)

if()< u < if 'U = ()

f,,

if()< u
b that f and g are both differentiable with g'(x) nowhere 0. Suppose that lim f(x)

X--+00

= oo =

lim g(x)

X--+00

and that

f'(x) - L . l lm - - - . x---+oo

Then

g'(x)

.

f(x)

Inn - () x--->oo g X exists and is equal to L. Proof: If x

> x 0 > b, there exists x such that f'(x) g'(x)

f(x)- f(xo) g(x)- g(xo) f(x)

= g(x)

[1- ~] 1-

g(xo) g(x)

'

where the expression in brackets--> 1 as x --> oo and where can write this as

- [1

f(x) f'(x) g(x) = g'(x)

-

~]

x depends upon x.

We

g(x)

1 _ /(xo) f(x)

·

If E > 0, there exists xo such that x > xo implies E

f'(x)

L-- < - - < L 2 g'(x) Let 1

> TJ > 0. There exists B > x 0 such that x 1

since 1 - TJ

~

B implies

_ g(xo)

-____:;.,9 ~(x4-) 1 _ /(xo) f(x) Thus

E

+ -. 2

-

1 < TJ.

(1- TJ) < f(x) < (L + 2.) (1 + TJ) (L- 2.) 2 g(x) 2

> 0. By picking TJ > 0 sufficiently small we can guarantee that f(x)- Ll < E

l g(x)

for all x ~B. Thus ~ -->Las x--> oo.

•

EXERCISES

121

• EXAMPLE 4.8

Here are more examples utilizing l'Hopital's Rule.

i. lim

n----too

(1 + .!.) n

n

=

lim X-+00

(1 + .!.)

x

X

X-->00

= elimx--+oo~ l+x

.. rlffix-->oo ex xn - r nxn-1 - lffix-->00 ~

ll.

-- ... --

=e .

rlffix-->oo eXn! = 0 . Here n

E

N.

EXERCISES

4.40

Let f(t) = t 2 and g(t) = t 3 , for all t E [0, 1]. a) Find the value(s) ofl1 E [0, 1] such that

/(1)- f(O) =

f' (t1) (1- 0).

b) Find the value(s) ofl2 E [0, 1] such that

g(1)- g(O) = g' (£2) (1- 0). c) Find the value(s) off E [0, 1] such that

/(1)- f(O) g(1)- g(O)

!'(f) g' (f).

4.41 Let p be a polynomial of degree n and let E = {x I ex = p( x)}. Prove that the number of elements in the set E, denoted by lEI, satisfies the inequality lEI :::; n + 1. (Hint: Use Cauchy's Generalized Mean Value Theorem.) 4.42

Find the error in the following attempt to apply L'Hopital's Rule: 2x - 1 I" 2 . x 2 - x - 2 = I"lffi 1lffi - = lffi - = 1

x-->2 x2 - 2x

x-->2 2x - 2

2 -x-2 3 · f act 1"lffix-->2 xx2_ Show th at m 2x = 2·

4.43

Find

(Hint: See Example 4.8(i).)

x-->2 2

122

THE DERIVATIVE

> 0, find limx_,oo

4.44

If n E Nand if p

4.45

If P is any polynomial, find

r

h~t 4.46

P(x

(lo;:r'.

+ 3h) + P(x- 3h)- 2P(x) h2

•

Find

where kEN.

4.6 TAYLOR'S THEOREM We will see that Taylor's Theorem is another generalization of the Mean Value Theorem. Denote by cn(a- r, a+ r) the set of all functions f such that at least the first n derivatives off are continuous: Thus f(n) E C(a- r, a+ r). We would like to approximate f by means of a polynomial of degree n, expressed in powers of .r-a: f(x) ~Co+ c1(x- a)+···+ cn(x- a)n. If we actually had f(x) =eo+ c1(x- a)+···+ cn(x- a)n then it is easy to see by substitution that

f(a) =

C(),

f'(a) = CI, ... , j(n)(a) = n!Cn.

Thus we could in this case write

f(k)(a) Ck=--

k!

for all k = 0, ... , n, with the understandings that O! = 1 and f( 0)(a) = f(a), by definition. Definition 4.6.1 We define the nth Taylor Polynomial

Pn(x)

=

k L -J(k)(a) k-(x- a) 1 n

k=O

for all f E Cn(a- r, a+ r), where r > 0. The problem is to describe the difference f(x)- Pn(x), which we call Rn(x), the nth Taylor Remainder term. This remainder is zero only iff is actually equal to the polynomial Pn.

TAYLOR'S THEOREM

123

Theorem 4.6.1 (Taylor's Theorem) If f(n+l) (x) exists for all x in the interval (ar, a+ r), then we have f(x) = Pn(x) + Rn(x), where R (x) = n

f (n+l)( ) J.L (x- a)n+l (n+ 1)!

for some suitable value of J.L between a and x.

Remark 4.6.1 We can regard Taylor's Theorem as a generalization of the Mean Value Theorem in the following sense. The Mean Value Theorem says that if exists on (a- r, a+ r), then

f(x) = f(a)

f'

+ f'(J.L)(x- a)

for all x E (a- r, a+ r) and for some suitable J.L strictly between a and x. Note that f (a) = Po (x), a constant polynomial. And f' (J.L) (x - a) is in the correct form to be Ro(x). Thus the Mean Value Theorem implies the special case of Taylor's Theorem in which n = 0. Note that if f(n+ll(x) exists for all x E (a- r, a+ r), then f E Cn(a- r,a + r). Proof: Fix x for now. If x =a we see easily that f(a) On the other hand, if x -:f. a, then

= Pn(a), so thatRn(a) = 0.

(x- a)n+l (n+1)! -::j=.O, so there exists K E lR such that

R'71 (X ) -_

(

K ( n+ 1)!. X

a

-

)n+l

.

Our goal is to show there exists J.L between a and x such that K = f( n+ 1 l (J.L). The trick is to replace a by a variable o: and to define

If we set a

= a, we see that h(a) = f(x)- [Pn(x)

+

(n!

)! (x- a)n+l]

1

=0

by definition of K. On the other hand, if we set a = x, then we see that

h(x) = f(x)- [f(x) + 0 +···+OJ = 0.

124

THE DERIVATIVE

Moreover, as a function of a, his differentiable for all a between a and x. By Rolle's Theorem, there exists J.L between a and x such that h'(J.L) = 0. However,

=0-

h' (a)

{ (!' (a))

+ (- f' (a) + !" (a)( x -

+ ( -f"(a)(x-a)+ + ... + (n!

so that h'(J.L)

= 0 and K

=

(3)(

f

2

!a)(x-a) 2 )

f(n)(a) (x- a)n-l (n- 1)!

+ f(n+l)(a) (x- a)n ) =

a))

K

- -(x- a)n n!

}

f(n+l)(a) K (x- a)n- -(x- a)n, n! n!

•

f(n+l)(J.L).

• EXAMPLE 4.9 Let f(x) =ex and write f(x) = Pn(x) + Rn(x) in powers of x = x- 0. We claim that Rn (x) ----. 0 as n ----. oo for all x E JR, so that n 1 lim "'""' - xk n--+oo L...J k!

=

ex

k=O

forallx E R It suffices to show for all x that

R (x)

=

n

J(n+l)( ) J.L xn+l (n+1)!

=

l.t e xn+l ___. 0 (n+1)!

as n ----. oo. However, 0

< IR (x)l = -

n

(n

el.t

+ 1)!

lxln+l < elxl lxln+l . (n + 1)!

Thus it suffices to show that for all x E JR,

lxln+l

-,'----'-----.,..-, ___. 0 (n + 1)!

as n ----. oo. So fix x arbitrarily and observe that there exists N E N such that .J# < ~- Now, for all n 2": N, write n = N + k, and

lxln

lxiN

1

- 0, prove that

(Hint: Use ex= P1(x)

+ R1(x).)

4.49 Prove that e is irrational. (Hint: Suppose false, so that e = ~, where p, q E N. Write e = e 1 = Pn(1)+ Rn(1), multiply both sides by n!, and deduce a contradiction when n EN is sufficiently large.) 4.50 Expand the polynomial p( x) of (x- 1): That is, show that

= 3x 3 + 2x 2 - x + 1 as a polynomial in powers 3

p(x) = :~::::>k(x-

1)k

k=O

and find the values of the constants

eo, ... , c3 .

Let f(x) = sinx and find the nth Taylor Polynomial Pn(x) in powers of x = x- 0 (that is, use a= 0). Prove that Pn(x) ~ sinx as n ~ oo for all x E R (See Fig. 4.7.) Prove also that no polynomial P(x) = sinx on any interval of

4.51

positive length. y

Figure 4.7 function?

sin(x), P5 (x), and P7 (x) on [0, 1r]. Can you see which curve belongs to which

Let f(x) = cosx and find the nth Taylor Polynomial Pn(x) in powers of x = x- 0 (i.e., use a= 0). Prove that Pn(x) ~ cosx as n ~ oo for all x E R

4.52

126

THE DERIVATIVE

4.7 TEST YOURSELF

EXERCISES

+ x) < x for all x > 0.

4.53

True or False: ln(1

4.54

Give an example of a function F that is differentiable at every x E [0, 1] yet

F'(x) is not bounded on [0, 1]. 4.55

True or False: Iff E R[a, b] and F(x) for all x E [a, b].

=

J: f(t) dt for all x

F'(x)

= f(x)

4.56

True or False: If F'(x) exists and is Riemann integrable on

1b 4.57

E

[a, b], then

[a, b], then

F'(x) dx = F(b)- F(a).

Suppose that the derivative f' (x) exists and is bounded on the closed interval

[a, b] and that f' is continuous on the open interval (a, b) but is not continuous at a or at b. True or False: The function f' is Riemann integrable on [a, b]. 4.58 Give an example of an unbounded function f on [0, 1] that is equal everywhere to the derivative of another function F. 4.59

Define

IIIII = ll!llsup + ll!'llsup for each

f

E C1 [a,

b]. True or False: If T: C1 [a, b]

-->

lR is defined by

Tf = !' (a ; b) ' then T is continuous with respect to

I · II·

4.60 Let f(t) = t 2 and g(t) = t 3 , for all t E [0, 1]. Find the value(s) off E [0, 1] such that f(l)-f(O) = f'(B g(I)-g(O)

g'(t •

4.61

If Pis any polynomial, find limh-+O P(x+4h)+P~- 4 h)- 2 P(x).

4.62

Fmd hmx-+oo

4.63

Let T(f) =

.

.

(logox/'o x . ·

J01 f(x) sin~ dx for each f

E

R[O, 1]. Is T continuous?

CHAPTERS

INFINITE SERIES

An infinite series is a sum of infinitely many numbers. Infinite series appear throughout pure and applied mathematics, and they were important even long ago. For example, the student probably recalls that the infinite decimal expansion of the fraction ~ is 0.33.;l .... The underlined .;l connotes endless repetition of the digit 3. Such an endless decimal expansion can be understood as representing

3 10

3

3

+ 100 + 1000 + ... '

where again the 3 dots indicate that the additions continue without end. What does it mean to add infinitely many numbers? Can anyone actually perform an infinite number of additions? These are some of the questions we will address in the present chapter.

5.1

SERIES OF CONSTANTS

If xis a sequence of real numbers, we can think of x as being a junction defined on the natural numbers N as follows: for all k E N, x(k) = Xk E R Usually the kth term of the sequence xis denoted Xk rather than x(k), but it is useful to have the Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright© 2008 John Wiley & Sons, Inc.

127

128

INFINITE SERIES

symbol x for the sequence as a whole (viewed as a function, for example) so that we can write conveniently about various global properties of the sequence, as we will see in the next several sections. The first thing we wish to understand about a sequence x is whether or not it is possible to define in a useful way the concept of the sum of the infinitely many numbers Xk. What is clear is that for all n E N we can define the nth partial sum n Sn

= L X k = X ! + X2 + · · · + Xn k=l

by virtue of finitely many repetitions of the closure of lR under addition.

Definition 5.1.1 We say the infinite series 00

L

Xk

=

X!

+ X2 + · · · + Xn + · · ·

k=l

converges to the sum s provided the sequence of partial sums

as n ---t oo. If'L.~ 1 Xk converges to s, then we call the sequence x summable. A series that does not converge is called divergent

Theorem 5.1.1 (nth Term Test) If x is a summable sequence, then n ---t oo.

Xn ---t

0 as

Proof: Denote sn = EZ=l Xk. Since xis summable, there exists L E lR such that sn ---t Las n ---t oo. Hence the sequence Sn-l ---t Las well. Therefore Xn

=

Sn- Sn-1 ---t

L- L = 0.

• Remark 5.1.1 We remind the reader here, for one final time, of the importance of learning how to read mathematical proofs, and how this contributes to learning to prove theorems oneself. The reader should review the Introduction on page xxiii. It is very important for each reader to write out a careful analysis of each proof studied, taking careful note of exactly how it works and how each step is justified. The proof of the nth term test is very brief. Even so, it offers lessons. The reader Xn may diverge. will see in Example 5.1 that it is possible for Xn ---t 0, although The insight that enables us to prove this theorem is that we can express Xn in terms of Sn and Sn-1· Then one must recognize that if Sn ---t s, so also must Sn-1 ---t s. In fact, because sn ---t s we know that for each f > 0 there exists N E N such that

Er'

SERIES OF CONSTANTS

n ~ N implies that lsn- sl <E. Hence if n ~ N + 1 we must have lsn-1as well, showing that Sn-1 ----> s, lagging only one step behind sn itself.

129

sl < E

It is important to understand that the nth term test can be used to prove divergence of an infinite series, but it is a one-directional implication and does not prove convergence, as shown by the following important example .

• EXAMPLE 5.1 Let Xk = /c for all k E N. We claim that even though Xk = /c ----> 0 as k ----> oo, the so-called harmonic series L:;~ 1 Xk diverges. As usual, let sn denote the Nth partial sum. If s., were convergent, then sn would be bounded, and every subsequence, such as s2n, would be bounded as well. However,

n 2

>1+----->oo

as n ----> oo. This is a contradiction. Hence the harmonic series diverges. See Exercise 5.1.

• EXAMPLE 5.2 Since the harmonic series diverges, it is interesting to note that the so-called alternating harmonic series

converges. This is a consequence of the following theorem.

Theorem 5.1.2 (Alternating Series Test) Suppose the sequence Xk is a decreasing sequence converging to the limit 0. That is, suppose x1 ~ x2 ~ · · · and Xk ----> 0 as k----> oo. Then 00

:~:)-1)k+1Xk = .T 1

-

X2

+ X3- X4 + '''

k=1

converges. Moreover, if s denotes the sum of the infinite series, then we have the following estimate of the difference between the partial sum Sn and the sums:

for all n.

130

INFINITE SERIES

Proof: Observe that 81 = x1 2: 82 = x1 - x2. However, 83 = 82 + X3 :S 81 because the positive sequence Xk decreases. Extending this reasoning, we observe the following:

In other words, the subsequence 82n is increasing with every odd partial sum as an upper bound, and 82n-1 is decreasing with every even partial sum as a lower bound. Thus 82n / ' L, the least upper bound of the even partial sums. Hence L is a lower bound of the odd partial sums. And S2n-1 ""'

G 2: L,

where G is the greatest lower bound of the odd partial sums. And

Thus G- L = 0 and L = G. Moreover, since the successive terms of the sequence 8n are on opposite sides of L for all n

•

Thus sn ---) L = s and the theorem is proven.

One of the things we learn from the harmonic and the alternating harmonic series is that a series 1 Xk can converge even though 1 lxkl fails to converge. We see next why the opposite phenomenon cannot occur.

L%:

L%:

Definition 5.1.2 A series 00

:~:::>k k=1

is called absolutely convergent if the series 00

converges. In this case the sequence x is called absolutely summable. A series 00

that converges but fails to converge absolutely is called conditionally convergent, or conditionally summable.

Thus the alternating harmonic series is conditionally convergent.

SERIES OF CONSTANTS

131

Theorem 5.1.3 Every absolutely summable sequence x is summable. Proof: If n Sn

=

LXk, k=l

it will suffice to show that sn is a Cauchy sequence. What we know by hypothesis is that n

an=

L

lxkl

k=l

isCauchy. Thusift: > O,thereexistsNsuchthatn However, if n > rn ;:::: N, then n

n

k=m+l

k=m+l

> m;:::: Nimplieslan-aml <E.

• • EXAMPLE 5.3

We present the Geometric Series Test. We call the series 00

Lark

= a + ar· + ar 2 + · · · + arn + ...

k=O

a geometric series with common ratio r. Observe that if the constant a =1- 0 and lrl ;:::: 1, then arn f+ 0 as n _, oo, so E~o ark diverges by the nth term test. However, we claim that if lrl < 1 then the corresponding geometric series must be absolutely convergent, and thus convergent as well. To prove 00

converges provided lrl < 1, it suffices to prove n

Bn

=

Lark

= a + ar + ar 2 + · · · + arn

k=O

converges. But Bn - rsn = a - arn+l. Hence Sn

as n _, oo. That is

a 1-r

= - - - - _, - -

1-r

00

"""' ark = _a_' L....t 1- r k=O

(5.1)

132

INFINITE SERIES

provided that

lrl < 1, so 00

converges as well. For example, the series

L: -21)k 00

(

k=l

converges absolutely.

• EXAMPLE 5.4 The geometric series test enables us to resolve the famous problem known as Zeno's paradox. The ancient Greek philosopher Zeno proposed this scenario. The legendary warrior Achilles was challenged to a foot race against a tortoise. For fairness, the tortoise is given a head start of d yards. The claim is that Achilles can never catch the tortoise. Suppose the tortoise runs at r times the speed of Achilles, where 0 < r < 1. The idea is that if it takes Achilles t seconds to reach the point where the tortoise was at the beginning of the race, then when he reaches that point the tortoise will have advanced to a position dr yards ahead of Achilles new location. Then it will take tr seconds to reach the new location of the tortoise. However, during that time the tortoise will have advanced to a position dr 2 yards ahead of Achilles. In effect, each time Achilles reaches the previous position of the tortoise, the determined quadruped has moved a bit farther ahead. So Achilles can never reach the tortoise. The resolution is that the time required for Achilles to reach the tortoise is the sum of an infinite series:

T =

00

00

0

0

L trn = t L rn,

which is finite since lrl < 1, by the geometric series test. Of course, Zeno realized that the tortoise would lose the race to Achilles. Zeno's point was that in his era it was not yet understood how to deal with limits, such as the sum of an infinite series, with logical rigor.

EXERCISES 5.1 tIn Example 5.1, try regrouping s2n-1 in such a way as to show s2n-1 ::::; n, for all n. If BN 2": 100, how big must N be? Remark: The reader may be surprised at how many terms a computer would have to add to make the partial sums rise to a total of only 100, even though the harmonic series diverges to oo. How many years would your PC have to compute in order to add the required number of terms to make SN 2": 100? What would be the round-off error from so many additions? 5.2

Test for absolute convergence, conditional convergence, or divergence:

EXERCISES

133

a) 2::~ 1 (-1)k+l

b)

t

1

00

f;k(k+1) (Hint: Write 1 1 ---k+ 1 k

1 k(k

+ 1)

and calculate sn. This is an example of what is called a telescoping series.) c) 2::~ 1 ak, where if k is a perfect square, if k isn't a perfect square.

if k = 2j -1, if k

5.3

= 2j.

Using Exercise 5.2.b above as a model, find a formula for ak such that n Sn

= Lak = Vn k=1

for all n, so that sn diverges to oo although ~ ~ 0 as n ~ oo.

=

L~=t ak

=

log n, for all n.

5.4

Find a formula for ak so that sn

5.5

Give an example of Xk ~ 0 for which 2::~ 1 ( -1)k+lxk diverges.

5.6 If x andy are sequences and c E JR, we define (cx)k = cxk and (x Xk + Yk· Prove: If x andy are summable, then a) x + y is summable and 00

00

00

L(xk+Yk) k=1

+ Y)k =

= LXk+ LYk· k=1

k=1

b) ex is summable and 00

00

LCXk =cL:xk. k=1 k=t We note that this exercise shows that the family of summable sequences a is vector space.

134

INFINITE SERIES

5.7 A function f : lR ----> lR is called a contraction of lR if and only if there exists a constant r E [0, 1) such that for all x and x' in lR we have

lf(x)- f(x')l S rlx- x'l· Let

5.2

f

be a contraction of!R, with corresponding constant r. a) Prove that f is uniformly continuous on R b) Let x 0 E lR be arbitrary and define a sequence Xn by Xn = f(xn_I), for each n EN. Show that lxn+l - xnl S rnlx1- xol· c) Prove that the sequence Xn in 5.7.b is a Cauchy sequence. (Hint: Exercise 1.25 may be helpful.) d) Let p = limn-+oo Xn, and prove that f(p) = p. (A point q is called a fixed point off if and only if f(q) = q. You have just shown that every contraction of lR has a fixed point.) e) If p and q are both fixed points of J, prove that p = q. f) Let f : lR ----> lR be a differentiable function such that ll!'llsup = r < 1. Prove that f is a contraction of R

CONVERGENCE TESTS FOR POSITIVE TERM SERIES

Since every absolutely summable sequence is summable, it is desirable to have tests designed to determine whether or not a series of exclusively nonnegative terms converges. We will see that absolutely convergent series play a very important role in the applications of infinite series. Throughout the present section, all terms of infinite series will be nonnegative unless otherwise noted. We observe first that if Xk ~ 0 for all k EN, then

is a monotone increasing sequence: sn is increasing as n increases strictly. Thus we know that Sn----> sup{sm I mEN}. It follows that Sn is convergent if and only if sup{ Sm observation leads us to the following useful test.

Theorem 5.2.1 (Comparison Test) Suppose there exists K

Im

E N}

0 be arbitrary. There exists

0. But if m and n ~ N, we know that

Letting n

---->

oo, we see that

for all v E V. Thus

so T m

---->

T in norm. Moreover,

1\T\1 = 1\Tm- (Tm- T)\1 :::; 1\Tm\1

+ 1\Tm- T\1 < oo,

so T is bounded as claimed. Thus V' is a Banach space.

•

Now we will proceed to identify l~ as a Banach space. Definition 5.4.3 Denote by l 00 the set of all bounded sequences, and

1\Y\\oo =sup {\Yk\1 kEN} Vy E loo. The norm 1\ · 1\oo is also called a sup-nonn. Theorem 5.4.3 For ally E 100 andforall x E

h define

00

Ty(x)

=

2:::: YkXk. k=l

We claim that loo is a Banach space, the mapping T : y ----> Ty from loo is linear, injective (meaning one-to-one), and surjective (meaning onto) l~, and

which means that T preserves norms. Remark 5.4.3 Because of the properties described in Theorem 5.4.3, the mapping Tis called an isomorphism from the Banach space l00 to l~.

152

INFINITE SERIES

Proof: For all y E l00 00

Ty: X____, LYkXk

k=l is an absolutely convergent series (Exercise 5.10), and Ty : h Ty is bounded since

JTy(x)J =

----> JR. is clearly linear.

~~ YkXkl ~ ~ 1YkXk1 ~ JJyJJoollxJJl

for all x E h. Thus IITyJJ ~ IIYIIoo· The reader will prove that IITyJI = IIYIIoo (Exercise 5.34 ). The mapping T : y ----> Ty carries l00 linearly into l~. Since IITyll = IIYIIoo• Ty = 0 if and only if y = 0. That is, the kernel of the map T: y----> Ty is {0}, soT is one-to-one from l00 into l~. It remains to be proven that T is onto l~. So we let L E l~ be arbitrary, and we must show there exists y E l00 such that L = Ty. For each j E N, let e(j) E h be defined by if k

= j,

if k =I= j. Define a sequence y by letting Yi = L ( e(jl) for all j E N. Then

for allj. Hence y E l00 • We claim that for all x E Observe that

oo

L k=n+l

as n

h. L(x) = Ty(x), so that L = Ty. n

lxkl = llxll1- L lxkl----> 0 k=l

----> oo. Therefore

ao; n ----> oo, since every bounded linear functional L must be continuous. That is, L(x) = 1 YkXk = Ty(x). Finally, we explain briefly how the mapping y ----> Ty establishes that l00 must be a Banach space itself, and why this Banach space is said to be isomorphic to l~. We know already that the sup-norm is a norm, but this follows also from the properties ofT. For example,

E:

IIY + zlloo = IITy+zll = IITy + Tzll ~ IITyll + IITzll = IIYIIoo + llzlloo·

153

EXERCISES

The other properties of a norm can be established for the Z00 -norm similarly. To see that Zoo is complete, suppose y(n) is a sequence of bounded sequences that is Cauchy in the Z00 -norm. Thus Ty is a Cauchy sequence in Z~. Hence there exists y E Zoo such that Ty ---+ Ty. It follows that y(n) ---+ y since

The mapping T is called an isomorphism of Banach spaces because it preserves all the vector space operations, it preserves the norm, and it preserves convergence and completeness in that norm. •

EXERCISES 5.30

t Prove for all x E h and for all c E JR, llcxll1 = lclllxll1 < oo. t Show for all x E h that llxll1 ::;:: 0 and llxll1 = 0 if and only if x = 0, the

5.31 identically 0 sequence (ie, 5.32

Xk

= 0 for all k).

Let V be a normed linear space. a) tifT E V', prove IT(v)l :S: IITII·IIvll for all v E V. b) 1fT E V', prove T = 0, the zero functional, if and only if liT II

t

5.33

t In the proof of Theorem 5.4.2, we defined a function T

:V

---+

= 0.

lR by

T(v) = lim Tn(v), n->oo

where Tn E V'. Prove that Tis linear.

5.34 t If y E Z00 , x E h. and Ty(x) = L:~ 1 YkXk, prove IITyll = IIYIIoo· (Hint: The direction ":S:" was established in the proof of Theorem 5.4.3. Try applying Ty to e 0 as n----> oo on D, then we say the series '2.~ 1 fk converges uniformly to the sums on D.

•

EXAMPLE 5.7

Let fk(x) = xk, for all x E D = [0, 1). Since lxl 00

< 1,

00

~ fk(x) = ~ xk = ____::____ L...i L...i 1-x

k=l

k=l

pointwise convergent on D. However, we can see as follows that this series is not uniformly convergent on D. Suppose it were uniformly convergent. Then the sequence sn would be Cauchy in the sup-norm. Lett = 1, and there exists N such that n > m 2': N

155

SERIES OF FUNCTIONS: THE WEIERSTRASS M-TEST

implies llsn- smllsup < 1. That is, E~=m+ 1 xk < 1 for all n > m :2: Nand for all x E D. Hence n

lim

x---+1-

""' xk ~ 1. L...

k=m+1

However, limx---+1- E~=m+ 1 xk = n- m > 1 whenever n > m + 1. This is a contradiction. Hence the convergence is not uniform on [0, 1). Exercise 5.40 shows that the series in Example 5.7 does converge uniformly on a smaller domain. The next theorem shows that when we do have uniform convergence of an infinite series, this permits strong conclusions of interest in the calculus. Theorem 5.5.1 Let fk be defined on a domain D for all k E N. Then we have the following conclusions.

i. If !k E C(D) for all k and if E~ 1 fk converges uniformly to fonD, then f E C(D) as well. ii. If fk E R[a, b] for all k and if E~ 1 fk converges uniformly to f on [a, b], then f E R[a, b] and

1 b

00

f(x) dx

= {;

1 b

fk(x) dx.

iii. If fk E C1 [a, b] and if there exists c E [a, b] such that E~ 1 fk(c) converges and if E~ 1 f{ converges uniformly on [a, b], then E~ 1 fk converges uniformly to a differentiable function f on [a, b] and 00

J'(x)

=L

fk(x).

k=1

Proof: 1.

For all n E N, Sn = E~= 1 fk E C[a, b] and Thus f E C[a,b].

llsn- !II sup

-t

0 as n

-t

oo.

ii. This is Exercise 5.41. iii. By hypothesis, s~ = EZ=l f{ - t g uniformly on [a, b]. And sn(c) converges for some c E [a, b]. By the corresponding theorem on uniform limits and derivatives for sequences, we see that Sn converges uniformly to some differentiable function f and f'(x) g(x) = E~ 1 fk(x).

=

•

Because uniform convergence is so useful, it is helpful to have a convenient test that can identify many (though not all) uniformly convergent series. Theorem 5.5.2 (Weierstrass M-test) Let fk be defined on a domain D for all kEN. Let Mk = llfkllsup < 00

156

INFINITE SERIES

for all k. If"L-: 1 on D. Moreover,

Mk < oo then

E: !k converges both absolutely and uniformly 1

00

converges uniformly on D as well. Proof: Let E > 0. By hypothesis, an = E~=l exists N E N such that n > m ~ N implies ian

Mk is a Cauchy sequence, so there -ami < E. Denoting

we see that n

L

llsn- Smllsup =

/k

k=m+l

sup

n

k=m+l n

=

L

Mk = !an- ami < E.

k=m+l

E:

Thus Sn is Cauchy in the sup-norm, so 1 fk converges uniformly. The exact same argument as that just given still works if we replace !k by l!k I in every step. This implies that • 1 l!kl converges uniformly on D as well.

E:

• EXAMPLE 5.8 Let

1 f(x) = - 1 -x

for all x E ( -1, 1). We know from the Geometric Series formula that 00

provided that !xi

< 1. We wish to express

f

'(

1 x)=(1-x)2

as an infinite series in powers of x. Suppose we fix x 0 E ( -1, 1) and select a real number r for which !xol < r < 1. We apply the Weierstrass M-Test by

EXERCISES

letting Mk =

krk-l

157

to conclude that the series 00

Lkxk-1 k=l

of term-by-term derivatives converges uniformly on [-r, r] by applying the Ratio Test (Theorem 5.2.3) to L::;: 1 krk-l . It follows then from Theorem 5.5.1 that f'(x) exists on [-r, r] and, specifically, 00

J'(xo)

=

L kx~-

1

•

k=1

Note that this formula has been established for all values of x 0 E ( -1, 1). EXERCISES

5.39

Let

fk(x)

=

( 1)k+l -

k

xk.

a) Prove: L::;: 1 fk(x) converges uniformly on [0,1]. (Hint: Use the error estimate from Theorem 5.1.2.) b) Find ll!kllsup on [0,1]. Can the Weierstrass M-test be used to prove the uniform convergence of L::;: 1 fk(x) on [0,1]? Why or why not?

< 1, prove that L::;: 1 xk converges uniformly on [0, a].

5.40

If 0 ~

5.41

Prove part (ii) of Theorem 5.5.1.

a

5.42 If L::%"= 1 fk converges uniformly on D, prove: the converse true? Prove or give a counterexample.

llfnllsup __, 0 as n

__, oo. Is

5.43 Determine whether or not each of the following series converges uniformly on the indicated domain. a) L::;: 1 e-kx on [1, oo). b) d)

on lR

c) e)

·

L::;: 1 sink x on [0, 8], where

0

5.44

sin

kx uk=l ~

'"'00

0 such that

Thus for all x E

-lxii < -r:::; a:::; fJ:::; r < lxii· [a, ,8], we have lxl :::; r < Ixi I. so that lckxkl =

lckx~l·~~~k :S: lckx~l·l!_lk XI

XI

Since L~o ckxt converges, ckxt ____, 0, and so hxt I :::; B < oo for all k and for some B. Since 0:::; r·

< lx1l.

I;, I< 1, and

is a convergent geometric series. But

on [a, ,8], so L~o llckxk II sup < oo. Hence L~o ckxk converges both absolutely and uniformly on [a, fJ] by the Weierstrass M-test. Observe that every x E ( -lxii,

lxii) lies in some such interval

Thus we have also proven that L~o ckxk converges absolutely on the interval

(-lxii,Ixii). ii. Now let I be the set of all points at which L~o ckxk converges, and let

R = sup { lx1ll x1 If [a, fJ]

c (- R, R), then there exists xi

E

E

I} .

I such that

Hence the series converges uniformly and absolutely on [a, ,8]. And it follows that the series converges absolutely on (-R,R). On the other hand, if x (j. [-R,R], then lxl is too big to permit x E I. Hence x (j. [-R,R] implies x (j. I. That is, ( -R, R) .negative. (Hint: Use the boundary conditions in Equation (6.2).)

6.6

Explain why negative integers n are not needed in Equation (6.5).

6.7

Prove Equation (6.7).

6.8

Prove Equation (6.8).

6.2

EULER'S FORMULA AND THE FOURIER TRANSFORM

We will see that it is helpful both computationally and conceptually to recast the concept of trigonometric series so that it applies to complex-valued functions as well as to real-valued ones. First we remind the reader of some elementary properties of the set C of complex numbers. The properties are listed in Table 6.1. The first six axioms listed are identical to the field axioms for the real numbers. Axiom 7 applies to C but not to R Because of the fact that squares of complex numbers can be negative, there is no order relation for C and of course it follows that there is no Archimedean property for C either. Similarly, there is no concept of positivity or negativity for nonreal complex numbers, though these features are retained by the sub field of real numbers. We regard lR as being a subset of C consisting of all those complex numbers having the form X

Definition 6.2.1 If z = x

+ i0 = X + 0 =

X.

+ iy E C, then we define the conjugate z of z by Z = X - iy

z

and the modulus of to be the nonnegative real number

izl2 =

zz = x2 + y2.

izi given by

185


Table 6.1

Field of Complex Numbers

The set

C

= {x + iy I x

E

R, y E R}

of all complex numbers is a field with two operations, called addition and multiplication. These satisfy the following properties:

+ z E C and wz E C. Commutativity: If w and z are elements of C, w + z = z + w and wz = zw. Associativity: If v, w, and z are elements of C, v + (w + z) = (v + w) + z

i. Closure: If w and z are elements of C, then w

ii. iii.

v(wz)

and

= (vw)z.

+ z) = vw + 1JZ. + z = z and 1z = z, for all z E C.

iv. Distributivity: If v, w, and z are elements of C, v( w v. Identity: There exist elements 0 and 1 inC such 0 Moreover, 0 # 1.

vi. Inverses: If z E C, there exists -z E C such that -z + z there exists z- 1 = ~ E C such that z~ = 1. vii. The number i 2

= 0.

Also, for all z

#

0,

= -1, the additive inverse of the number 1.

We call x and y, respectively, the real and the imaginary parts of z these are denoted by

x =

~(z),

y=

=

x

+ iy and

~(z).

We remark that z z is a nonnegative real number for all z E C. The complex numbers can be modeled conveniently in a geometrical manner by identifying z = x + iy with the point (x, y) in the Cartesian plane. We think of the x-axis as the real axis and they-axis as the imaginary axis in this picture. The plane can be equipped with polar coordinates (r, 0) as well as with rectangular coordinates (x, y). The relationship between these two systems of coordinates is that x = r cos 0 andy = r sin 0. Thus each complex number z can be written in the polar form

z = r(cosO + isinO).

Definition 6.2.2 A sequence of complex numbers Zn is said to converge to z E C if and only iflzn- zl --+ 0 as n--+ oo. It is easily verified that Zn Exercise 6.1 0.)

--+

z if and only both ~Zn --+ ~z and ~zn --+ ~ z. (See

186

FOURIER SERIES

Definition 6.2.3 An infinite series E:=o Zn of complex numbers is said to converge to a complex numberS if and only if the sequence N

SN = LZn n=O

converges to S.

E:=o

We leave it to Exercise 6.11 for the reader to prove that Zn converges to ~Zn converges to ~Sand ~Zn a complex numberS if and only both converges to ~S.

E:=o

E:=o

Definition 6.2.4 For each complex number z E C we define n

oo

z' e = L..J t z

(6.9)

"""

n=O

n.

which is shown to converge in Theorem 6.2.1.

Theorem 6.2.1 For each complex number z, the series in Equation (6.9) converges. Moreover, for each real number x, we have Euler's formula:

L oo

eix =

n=O

Proof:

('

)n

~ n.

=

cosx + isinx.

(6.10)

E:=o 1.ftl

We note first that converges since the real exponential series converges absolutely for every real number. But it is immediate that I~ (zn) I ::; lznl and also that I~ (zn)l ::; lznl. Thus the sum of the real parts in Equation (6.9) converges absolutely, as does the sum of the imaginary parts. Now we apply Exercise 6.11 to the series appearing in Equation (6.9) in order to conclude that the series expansion that defines ez converges for every z E C. For Euler's formula, we observe that if n

= 2k

if n = 2k + 1. It follows that

= cosx + isinx.

•


187

Corollary 6.2.1 For each real number x, we have

eix

cosx

+ e-ix

(6.11)

2

eix _ e-ix sinx

(6.12)

2i

•

Proof: Apply Euler's formula.

We observe next that the Nth partial sum of the trigonometric series in Equation (6.6) can be rewritten using Corollary 6.2. I in the form N

L (an cos 21rnx + bn sin 21rnx)

(6. 13)

n=O N

L

Cne211"inx'

n=-N

where eo

= ao and

alnl - i sgn(n)blnl 2 if n # 0. (Here sgn denotes the signum function.) We would like to convert the formulae in Equation (6.7) for an and bn into direct formulae for the calculation of en. In order to do this, we will need a suitable definition for the integral of a complex-valued function of a real variable.

Cn

=

Definition 6.2.5 Suppose that f : [a, b] ~ 0_ Show that there is a step function a such that II!- all1 < ~· Next use a double application of the triangle inequality for the £ 1 -norm.)

6.40 Suppose f E C 1( (a, b), q for some (a, b) c [0, 1] and that f is a Riemann integrable function of period 1. Prove that the Fourier series S(f) converges uniformly to f on each proper closed subinterval [c, d] c (a, b). See Fig. 6.3. Note that in the figure a = -0.5 and b = 0_5 and [a, b) is also a valid domain for the study of functions of period 1. (Hint: Use Exercise 5_62 to create a smooth periodic function that agrees with f on (a, b)- Apply Theorem 6.4.3.)

6.5

L 2 -CONVERGENCE & THE DUAL OF l 2

We have shown in Equation (6.22) that iff E R([O, 1], q is a function of period 1, then the double sequence j( n) is in l 2 , and 00

L IJ(n)l

2

~ IIIII~

-oo

with equality holding if and only if IIBn(f)- fll2 ---> 0 as n ---> oo. The case of equality in Bessel's inequality is called the Plancherel identity, which we will establish with Theorem 6_5 _J_ The Plancherel identity can be interpreted as an infinitedimensional version of the Pythagorean Theorem for R[O, 1], which we interpret as an infinite-dimensional Hermitian inner product space, utilizing the equivalence relation f "'gin R[O, 1] if and only if lf(x)- g(x)l 2 dx = 0_ Some applications of this identity are given in Exercise 6.44_

J;

Theorem 6.5.1 Iff E R([O, 1], q is a function of period 1, then

as n ___, oo. Consequently we have the Plancherel identity:

L lf(n)l 00

2

=

11!11~-

(6.31)

-oo

Remark 6.5.1 A celebrated theorem of Lennart Carleson [5] established that the Fourier series of any square-integrable Lebesgue measurable function must converge to f (x) pointwise except on a set of Lebesgue measure zero, 23 and that theorem does apply to all the functions covered by our theorem above. However, a set of points 23 The reader will not need to know about Lebesgue measure here. The definition of a set of Lebesgue

measure zero-known also as a Lebesgue null set-is, however, provided in this book as Definition 11.2.1.

206

FOURIER SERIES

can have Lebesgue measure zero and still be an uncountably infinite set. There are examples known of continuous functions f for which Sn(f) is actually divergent for infinitely many values of x. And there is an example of a Lebesgue integrable function f for which the Fourier series diverges at each point x! The extraordinary pathologies of Fourier series in regard to pointwise convergence, even for continuous functions, make theorems like the one we are about to prove very interesting and useful. Proof: Suppose first that f is real-valued. By Exercise 3.29, there exist step functions a(x) :::; f(x) :::; a'(x) for all x E [a, b], so that

1b a(x) dx:::; 1b f(x) dx :::; 1b a'(x) dx such that I I: a'(x) dx- I: a(x) dxl

< E, implying also that

b

11f(x)- a(x)l dx < E and

1b lf(x)- a'(x)l dx <E. a

This follows immediately from the use of the upper sums and the lower sums from the Darboux integrability criterion. These sums are integrals of step functions with heights corresponding to the infimum and the supremum of f on the intervals of a partition. Consider next a step function having only two values, each on .a subinterval of strictly positive length. Let p be the point at which the single jump discontinuity occurs. By Exercise 5.62 we know that for each 8 > 0 there exists a function ¢> E coo [0, 1] such that ¢>(x) = a(x) except on an interval (p- 8, p + 8). In effect, we are connecting the two steps with a smooth (C 00 ) curve which departs from the lower step very near the point of jump discontinuity and joins the upper step smoothly only slightly to the opposite side of the jump discontinuity. This permits us to make 111>- alii as small as we like. For general step functions we can iterate the process just described at each of the finitely many jump discontinuities of a. Moreover, we can do this keeping llr/>llsup = llallsup :::; 11/llsup- (Iff is complex-valued, the same approximations can be produced by working separately with the real and imaginary parts, R(f) and SS(f).) In this manner we establish that there exists a sequence of functions rl>n E coo [0, 1] with period readily adjusted to be 1, and having the properties

II!- rl>nlll _, 0,

and

llrl>nllsup::::: 11/llsup =

M

< 00

for all n. It follows that

II/- r/>nll~ =

1If1

2

rPnl dx:::;

2MIIJ(x)- rPn(x)lh

_, 0

as n _, oo. By Theorem 6.4.2 we know that Sk(r/>n) _, rl>n uniformly on [0, 1] as k _, oo. By Exercise 6.41 it follows that Sk(rPn) _, rPn as k _, oo in the

£ 2 -CONVERGENCE & THE DUAL OF / 2

207

£ 2-norm. Thus for each E > 0 there exists a number K such that k 2: K implies that II!- Sk(¢)112 < E. By Exercise 6.29, the Fourier coefficients off provide optimal L 2 approximation to f. Thus

•

for all k 2: K. This implies the theorem.

Remark 6.5.2 Thanks to Theorem 6.5.1, we know that iff and g are in R([O, 1], oo. Now use the function l from Exercise 5.62, to define f E C00 [Xn, Xn-l] so that

We require that f'(x) ;:::: 0 for all x and that f' vanish at at all x in the interval

Xn,

at

Xn-1.

and

b) Now link together the segments of the graph off smoothly for all the intervals indexed by n let f(O) = 0. Prove that f' is unbounded although f' (x) exists for all x E [0, 1], including x = 0. (Hint: Use the Mean Value Theorem.) Suppose f'(x) exists for all x E [a, b], and suppose f' E R[a, b]. Use the Fundamental Theorem of Calculus to prove that f E BV[a, b] and

7.14

Vd'(f)::; 7.15

1b lf'(x)j

dx.

Let

g(x)

={sin(:;) ()

Let f(x) 3.27.)

0

= J; g(t) dt.

if()

< X :S 1,

if X= 0.

Use Theorem 7.1 to prove f E BV[O, 1] (Hint: Use Exercise

Let a[a, b] denote the family of step functions defined on [a, b] (Exercise 3.8), and let a[a, b] denote the set of all uniform limits of step functions. a) Prove: a[a, b] is a complete normed vector space equipped with the supnorm. b) Prove: BV[a, b] is not a complete normed vector space in the sup-norm. c) Prove: a[a, b] 2 C[a, b]. d) Prove: a[a, b] 2 BV[a, b]. e) Prove: a[a, b] 0. Since g' is uniformly continuous on [a, b], and since f is bounded, there exists 8 > 0 such that liP!! < 8 implies jxk - J.tkl < 8 which implies Since

I:

RIEMANN-5TIELTJES SUMS AND INTEGRALS

225

making the second sum in Equation (7.2) less than E. (Note that if b- a = 0 or if llfllsup = 0, the claim about the second sum is trivial.) •

Theorem 7.2.2 Suppose fiE RS([a, b],gi)fori andj in {1, 2}. Let c E R Then

i. I:(cfi

+ fz) dg1

exists and

1b ii. I: fid(cgl

(cfi

+ !z) dg1

= c

1b h

dg1

+

1b

1b h

dg1

+

1b !I

fz dg1.

+ gz) exists and

1b !I

d(cg1

+ gz) =

c

dgz.

I:

Remark 7.2.2 This theorem says that f dg is separately linear in each of its two variables f and g. It also establishes that RS([a, b], g) is always a vector space.

Proof: We will prove the first part here. (See Exercise 7.18 for the second part.)

IP(cj,

,; lciiP(j,

+

J,g,p)- [c l

g, p)

-l

j, dg,

---+

as

j, dg,

+

I+ IP(j,

l j,dg,ll

g, p)-

f.' j, I dg,

lciO + 0 = 0

IIPII _, o.

•

In some ways the Riemann-Stieltjes integral has surprisingly different properties from those of the Riemann integral. Consider the following example . • EXAMPLE 7.5 Let

f(x)={~

and let

g(x) =

if X E [0, 1], ifxE(1,2]

{ ()

if X E [0, 1),

1

ifxE[1,2].

Then we make the following observations. 1 i. I 0 f dg

[0, 1].

=

!(1) · 1

=

0, since

f

E C[O, 1] and g is a step function on

226 ii.

THE RIEMANN-8TIELTJES INTEGRAL

t f dg

= 0, since on [1, 2] we have

g'(x)

1 =1 2

2

f dg

iii.

= 0, so

f(x)g'(x) dx

= 0.

I02 fdg

does not exist (that is, f ~ RS([0,2],g)), in stark contrast to the properties of the Riemann integral. Let us prove this claim as follows. No matter how small we make IIPII. we can still have Xk-l < 1 < Xk, so that 6.gk = 1, and f(JLk) can be either 0 or 1 depending upon how we choose f.Lk· Thus P(f, g, JL) can be either 0 or 1, and cannot be forced to converge to a limit merely by requiring IIPII ---> 0. Note that f being Riemann-Stieltjes integrable on both [0, 1] and [1, 2] with respect tog fails to force f to be in RS([O, 2], g).

Theorem 7 .2.3 Let a < b < c. If

i"

I: f dg, Ibc f dg, and I: f dg all exist, then

J dg =

ib lc J dg +

J dg.

Let t > 0. There exists 81 > 0 such that if P 1 is a partition of [a, b] with < 8t. then

Proof: IIPtll

Pt(f,g,JL) And there exists 82

-ib

fdgl
0 such that if P2 is a partition of [b, c] with IIP2II < 82, then IP2(f,g,JL)

-lc

fdgl
0 such that if P3 is a partition of [a, c] with IIP311 < 83, then IP3(f, g, JL)- I~' f dgl < ~· So let 8 = min{81. 82, 83} > 0, and let P1 and P2 be any partitions of [a, b] and of [b, c], respectively, with IIPill < 8, i = 1, 2. And let P = P1 U P2, a partition of [a, c] with liP II < 8 also. Then we have

lie

Jdg-

(ib lc Jdg+

Jdg)l =

l(ic

Jdg-P(f,g,JL))

[ib lc -ib + -lc

+ (P(f,g,,L)-

: ; lie

J dg- P(f,g,JL)I

+ IPl(f,g,JL) t

J dgl

t

Jdg+

IP2(f,g, JL)

Jdgl

)I

J dgl

t

0 such that for all li > 0 there exists p/, x' with

c < 1i < x' < c + o, lg(x') - g(c)l ;:-:::

fg

and

lf(J-L')- f(c)l ;:-: : fJ· No matter how small we make liP II. the value of P(f, g, p.) can fluctuate by at least the fixed positive amount f Jfg by choosing P with x' in it and then choosing p.' as the evaluation point for f in the interval between c and x'. Thus The other cases are very similar.

I: f

dg fails to exist. •

EXERCISES 7.17

Suppose that f E

C[a, b], t

(a, b), and

E

g(x) =

Cl

if a 0.

1b

f dg

In fact,

(fg{- P(g, J, J.L) = f(b)g(b)- f(a)g(a) - { g(J.LI)[f(xi)- f(xo)]

+ g(J.L2)[f(x2) -

f(xi)]

+ · · · + g(J.Ln)[f(xn) - f(xn~I)]} = f(a)[g(J.LI)- g(a)] + f(xi)[g(J.L2)- g(J.LI)] + · · · + f(b)[g(b)- g(J.Ln)] ---t

as

1b f

dg

IIPII ---> 0 since {a, J.li. J.l2, ..• , J.ln, b} is a partition P' of [a, b] and IIP'II ~ 2IIPII 0. ---t

•

RIEMANN-8TIELTJES INTEGRABILITY THEOREMS

•

229

EXAMPLE 7.6

t

We evaluate I~ 1 xdlxl = 2121- (-1)1- 11- 1 lxl dx = 4 + 1- ~.where the right-hand integral can be read directly from a graph. Theorem 7.3.2 Iff E C[a, b] and 9 E BV[a, b], then f E RS([a, b], 9 ). Remark 7.3.2 By Theorem 7.3.1, this theorem implies also that 9 E RS([a,b],f).

Proof: Since 9 E BV[a, b], 9 = 91-92, where 91 is increasing and 9 2 is increasing on [a, b]. Thus it will suffice to prove the claim in the theorem for the case in which 9 is increasing on [a, b]. The proof will be very similar in concept to the proof of Riemann integrability of each f E C[a, b]. Let P be any partition of [a, b], let Mk = maxxE[xk_ 1 ,xk] f(x) and mk = minxE[xk_ 1 ,xk] f(x). Then n

U(f, 9, P) =

n

L Mki:!.9k and L(f, g, P) L mki:!.9k· =

k=l

k=l

Clearly,

L(f, 9, P) :::; P(f, 9, J-L) :::; U(f, 9, P) for all P and J-L. It is easy to show, just as we did for Riemann sums in Chapter 3, that

P' 2 P

===}

L(f, 9, P) :S L(f, 9, P') :::; U(f, 9, P') :::; U(f, 9, P).

Thus for all P and P' we have L(f, 9, P) :::; U(f, 9, P'). We define the upper integral

I:

I:

f d9 to be the infimum of all the upper sums, and the lower integral f d9 to be the supremum of all the lower sums, again just as for Riemann integration. Since every lower sum is less than or equal to every upper sum, we have

and we claim that these two are actually equal. Let f > 0. It would suffice to show that

1b

f d9 -

1b

f d9
0 such that liP II < 8 implies U(f, g, P) - L(f, g, P) < f. By uniform continuity off, there exists 8 > 0 such that lx- x'l < 8 implies

lf(x)- f(x')l
lR by T(f) = f(p), a so-called point evaluation. Prove that Tis a bounded linear functional on C[a, b] with liT II = 1.

7.26

THE RIESZ REPRESENTATION THEOREM

231

=

7.27 LetT be as in Exercise 7.26. Find a function g E BV[a, b] such that T_9 (!) T(f), where T 9 is defined in Theorem 7.3.3. Can you find gin this exercise in such a way that Vd'(g) = 1 = I!TI!? Explain. 7.28 Letg E BV[a,b] andsupposeh(x) = g(x) exceptatonepointx = p E (a, b). Show that hE BV[a, b] and Th T9 • Must Vd'(h) = Vd'(g)? If yes, prove it. If no, give a counterexample.

=

Let g E BV[O, 2] such that

7.29

ifO::S:x I uniformly on [0, 1]. c) Prove or give a counterexample: the uniform limit of a sequence of func-

tions of bounded variation must be of bounded variation. 7.32

Let {

O

if~ :':::X:'::: 1, ifO <Xoo for all f E B+[a, b]. If c > 0, clearly cfk / cf at each x too, so the set B+ [a, b] is closed under multiplication by positive scalars. Similarly, fr and h E B+ [a, b] implies !I + h E B+ [a, b] (Exercise 7.34). Now let

B[a,b]

= {f =!I- hI JI,h E B+[a,b]}.

Then B[a, b] is a vector space of functions (Exercise 7.6). Moreover, we claim that we can define

T(J) = T(JI) - T(h) for all f E B[a, b]. For this extension ofT to be well-defined, we need to know that if

f = !I - h = 91

- 92

are two representations off, where

then

T(JI)- T(h) = T(g1)- T(g2) and T so-defined is linear on B[a, b] (Exercise 7.36). iii. Next, we wish to show that T, as extended above, remains bounded on the vector space B[a, b]. We will show that

IT(J)I :::; Mllfllsup, whereM is still the normofTasgiveninitially onC[a.b]. So let j, g E n+ [a, b] and suppose fk, 9k E C[a, b] such that for all x E [a, b] we have fk(x) / f(x) and 9k(x) / g(x). Although fk- 9k --+ f(x)- g(x), it is not necessarily true that

[fk(x)- 9k(x)] / [f(x)- g(x)].

236

THE RIEMANN-STIELTJES INTEGRAL

What is more serious for our purposes is that we may have

llfk- 9kllsup > II/- 9llsup =

K.

But we can fix this last problem by a method known as truncation as follows. We define a new sequence of functions c/Jn E C[a, b] (see Exercise 7.37) by

c/Jn(x)

=

fn(x) 9n(x) + K { 9n(x)- K

if fn(x) - 9n(x)

S K, > K,

if fn(x)- 9n(x)

< -K.

if lfn(x)- 9n(x)l

(7.3)

Also,

llc/Jn- 9nllsup S K =

II/- 9llsup·

We need to know that c/Jn(x) / f(x) for all x E [a,b]. This can be seen from the geometrical meaning of the truncation defined above in Equation (7.3) as follows. Consider the band in the plane trapped between the graphs of 9n(x) + K and 9n(x) - K, over the interval [a, b]. The whole band moves upwards as n increases strictly since 9n (x) increases. If the graph of f n slips either over the top or under the bottom of the band, then we truncate the graph of fn with the upper or lower boundary curve, respectively. This produces c/Jn· It is clear that

c/Jn(x) S max{fn(x), 9n(x)} S f(x) for all n and for all x. To see that c/Jn (x) / , consider the fact that for each x and for each n, c/Jn (x) must be either the middle, the upper, or the lower value permitted by Equation (7.3). The only way it is conceivable that c/Jn(x) ~ c/Jn+I (x) is if c/Jn+I (x) is a lower value among the three possibilities than is c/Jn (X). For example, if c/Jn (X) = 9n (X) + K and c/Jn+ 1(X) = 9n+ 1(X) - K, then

c/Jn+I(x)

9n+I(x)- K > fn+I(x) ~ fn(x) > 9n(x) + K = c/Jn(x).

=

On the other hand, if c/Jn (x) = f n (x), we could have

c/Jn+I(x) = 9n+I(x)- K > fn+1(x) ~ fn(x) = c/Jn(x). In each case, c/Jn(x)

S c/Jn+I(x). Moreover, since

9n(x) / g(x), fn(x) / f(x) and lf(x)- g(x)l S K for all x, lim c/Jn(x) = lim fn(x) = f(x).

n-+oo

n-+oo

THE RIESZ REPRESENTATION THEOREM

237

Now we can reason as follows.

IT(!- g)l

=I

n-+CXJ

=

lim IT(¢n- 9n)l ~ MK

lim T(¢n)- lim T(gn)l n-+oo

n--->oo

Thus even on B[a, b] we have the extended T with

liT II =

M.

iv. Now we define a(t) = T(1[a,t)) for all t E [a, b), and a(b) = T(1[a,bj)· We observe that a(a) = T(1[a,a)) = T(O) = 0. The reason for the slight difference in the way a(b) is defined will become clear in part v of this proof. We claim that

vd'(a) ~

IITII =

M.

To prove this, we let P = { xo, x1, ... , xn} be any partition of [a, b], and we form n

P(a) =

2: la(xk)- a(Xk-I)I. k=l

Observe that for all k we can select a number Ek E { ±1} such that

la(xk)- a(Xk-dl

=

Ek[a(xk)- a(xk_I)]

=T

(Ek[1[a,xk)- 1[a,Xk-dl) ·

Thus we can write

P(a) = T

{tf.k

[1[a,xk) -1[a,xk_ 1 ) ] }

= T(f), where we denote by f the argument ofT. Note that lf(x) I ~ 1 for all x, so that IT(!) I :S MIIJIIsup = M. Hence

Vd'(a) :S

IITII·

If we can show that T = Tw then we will know from Theorem 7.3.3 that the opposite inequality holds as well, and then we will know that IITII = v;(a). v. We claim that T = Ta. To prove this, we let P = {xo,XI, ... ,xn} be any partition of [a, b]. Let f E C[a, b]. We need to show that T(f) = Ta(f). Consider

P(f, a, x)

=

n

n

k=l

k=l

2: f(xk)[a(xk)- a(Xk-d] = 2: f(xk)~ak

(7.4)

238

THE RIEMANN-STIELTJES INTEGRAL

This is a Riemann-Stieltjes sum for

I:

fda, with f.Lk selected to be Xk for fda as

all k = l, ... ,n. Thus the sum in Equation (7.4) converges to IIPII --; 0.

I:

On the other hand, P(f, a, x) = T( 0 such that liP I < 8 implies II!- oo. Prove that Tf3' = Tf3. 7.40 Let an be a Cauchy sequence in BVo [a, b], in the sense of the total variation norm: For all f > 0 there exists N E N such that n and m ;::: N implies

Prove that there exists a E BV0 [a, b] such that

as n----> oo. That is, prove that BV0 [a, b] is complete. (Hint: Consider the sequence

which is already known to be complete, and use the uniqueness of the correspondence between a and Ta in Remark 7.4.1.)

TEST YOURSELF

241

7.5 TEST YOURSELF

EXERCISES 7.41

Let

f(x)

f

True or False:

E

= {

~ 2 (sin(~) +sin(~))

if X =/= 0, if X= 0.

BV[O, 1]. Explain.

7.42 Give an example of an integrable function f (x) that is not of bounded variation on [0, 1].

I:

7.43 Find X d is no less than x.)

r l (Note: The ceiling function r l denotes the least integer that X

X

7.44

True or False: The Riemann-Stieltjes integral I~ 1 Lx Jd sgn x exists.

7.45

True or False: The Riemann-Stieltjes integral 1

1

7r

x 2 sin - d sgn x X

-1

exists. (We interpret the integrand function at x = 0 as having the value 0.)

= 1{1}·

7.46

Let g(x)

7.47 that

LetT : C[O, 2]

Find I 0 tan- 1 x dg.

---->

2

R by T(f)

1

Find a function g E BV[O, 2] such

2

T(f)

for all

7.48

= 2f(1). =

f

dg,

f E C[O, 2]. Suppose that f(x) =ex for all x. Let ifO ~ x < 1, if X= 1, if 1

Evaluate I 0

2

7.49

I: f

f

<X~

2.

dg.

True or Give a Counterexample: If dg exists and

I: f dg and Ibc f dg both exist,

then

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PART Ill

ADVANCED CALCULUS IN SEVERAL VARIABLES


CHAPTERS

EUCLIDEAN SPACE

8.1

EUCLIDEAN SPACE AS A COMPLETE NORMED VECTOR SPACE

Throughout pure and applied mathematics, it is necessary to consider functions of more than one variable. If a function depends on n real variables, x 1 , x 2 , ••• , Xn, it is possible to combine these n real variables into one vector variable

In this notation, x is not a real number, but rather an n-tuple of real numbers. In addition, it is often necessary to consider functions that have vector values instead of real values. Up to this point, our course has focused on real-valued functions of one real variable. In this chapter we will begin the rigorous study of vector-valued functions of vector variables. In the study of real-valued functions of a single real variable, we saw the advantages of considering normed vector spaces of more than one dimension. For example, the reader has seen that the vector space C[a, b] has infinitely many dimensions (Exercise 2.57). In the present chapter, we will focus on finite-dimensional vector spaces equipped with what is called the Euclidean norm. However, we begin with a more general context. Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright © 2008 John Wiley & Sons, Inc.

245

246

EUCLIDEAN SPACE

Definition 8.1.1 In any (real) vector space V (Tabel2.1, p. 59), we call a function (-,-):VxV-+lR. a scalar product if and only if it has the following three properties. i.

(ax+ y, z)

ii.

(x, y)

=

(y, x) for all x andy in V.

iii.

(x,x)

~

Oforallx E V and (x,x) = 0 {::}

=

a(x, z) + (y, z) for all a E JR. and for all x andy in V. x=

0 E V.

• EXAMPLE 8.1

In the vector space JR.n, define the Euclidean scalar product by n

(x, y) =

I:>i1/i·

(8.1)

i=l

The reader should verify that this product satisfies all three conditions to be called a scalar product.

Theorem 8.1.1 In any (real) vector space V equipped with a scalar product as defined in Definition 8.1 .I, we define

llxll = y'(x,x)

(8.2)

for all x E V. The function II · II as defined in Equation (8.2) is a norm, as in Definition 2.4.4, and the Cauchy-Schwarz Inequality is satisfied:

I(x, y) I :::; llxiiiiYII· Proof: To prove the Cauchy-Schwarz inequality, we fix x and y, and we proceed as follows. For all t E JR., define a polynomial

p(t) Observe that p( t)

~

p(t)

=

(tx + y, tx + y).

0 for all t. By linearity of (·, ·) in each variable we see that =

llxll 2 t 2 + 2(x, y)t + IIYII 2 = at2 + bt + c,

where a= llxll 2 , b = 2(x, y), and c = IIYII 2 • But the quadratic polynomialp(t) ~ 0 for all t E JR. if and only if b2 - 4ac :::; 0, which is equivalent to b2

:::;

4ac. Hence

EUCLIDEAN SPACE AS A COMPLETE NORMED VECTOR SPACE

The first two conditions of Definition 2.4.4 are easily verified for condition, the triangle inequality, is left for Exercise 8.1.

Definition 8.1.2 If a vector space V is equipped with a norm open ball of radius r ?: 0 about p E V by

Br(P)

II · II·

II · II,

247

The third •

we define the

= {v E V lllv- Pll < r} ·

Similarly, we define the closed ball of radius r ?: 0 about p E V by

Br-(P) = {v E V lllv- Pll

:S: r}

·

The bar above the symbol Br indicates that the set Br is the closed ball, meaning that it includes the spherical boundary surface .

• EXAMPLE 8.2 In the familiar Cartesian plane of Euclidean geometry, with the norm of a vector being its geometrical length, we have

(x, y)

= XIYI

+ X2Y2 = llxiiiiYII cosO,

where() is the angle between x andy. Thus the Cauchy-Schwarz inequality in the plane follows from the fact that Icos Bl ::; 1 for all 0. (TheCauchy-Schwarz inequality follows alternatively from the argument given in Theorem 8.1.1.) In the plane of Euclidean geometry, B 1 (0) is the region strictly inside the circle of radius 1 centered at the origin. The reader should check that B 0 (p) = 0, the empty set, for all points p.

Definition 8.1.3 lfx = (xi. x2, ... , Xn) andy = (YI. y2, ... , Yn) are in JR.n, the set of all n-tuples of real numbers, we define the Euclidean scalar product of x andy as in Equation (8.1) and the Euclidean norm ofx as in Equation (8.2). The Euclidean space lEn of n dimensions is defined to be the vector space JR.n equipped with the Euclidean scalar product and the Euclidean norm. The reader should have checked that the Euclidean scalar product as defined above does satisfy the three properties required to be a scalar product, and that the Euclidean norm is in fact a norm. We remark that it is very common when dealing with lEn to call it JR.n informally. This means that we should suppose that the Euclidean inner product and norm are in use unless stated explicitly to the contrary. (See Exercise 8.4.)

Theorem 8.1.2 The Euclidean normed vector space lEn has the following two properties.

i. The Euclidean space lEn is complete in the sense of Definition 2.5.3. ii. For a sequence of vectors x<Jl in lEn, the sequence X(j) ----> X E

!En

¢o?

X~)

----> Xk

248

EUCLIDEAN SPACE

as j----) oo,foreach k = 1, 2, ... , n.

Remark 8.1.1 The second part of the theorem tells us that convergence of a sequence in lEn is equivalent to convergence in each separate coordinate sequence. Proof: The reader will recall that the vectors x E !Rn, which are the vectors of lEn, comprise a vector space using the familiar operations under which

ex= (cx1. ... , cxn) and

X+ Y = (x1 + Yl, ... , Xn + Yn). We begin by proving claim (i). To see that every convergent sequence in any normed vector space is Cauchy in the sense of Definition 2.5.3, see Exercise 8.8. We will prove that every Cauchy sequence x(k) in V converges. By hypothesis, For each f > 0 there exists K such that j and k ;::: K implies llx(j) - x(k) < f. However, for each l = 1, ... , n, we have

II

x(j) - x(k) I

lI

I -
0 such that E ~ BR(O), which is, in tum, contained in the hypercube [-R, R] xn, ann-fold Cartesian product of the interval [- R, R]. It will be convenient in this proof to denote the hypercube by a pair of diagonally opposite comer-vectors a 1 = (- R,- R, ... ,- R) and b 1 = (R, R, ... , R). Thus we will denote the cube

Next we subdivide the cube [a1, b 1 ] into 2n congruent subcubes as follows. Simply bisect the interval [- R, R] on each of the n coordinate axes, and form all 2n possible Cartesian products of a half-interval from each of the axes. Each such subcube can be denoted [a, b], where the coordinates of a= (a 1 , a2 , ••• , an) denote the left-hand endpoints of the n chosen subintervals on the axes, and with a similar convention is employed forb using right-hand endpoints. Observe that for each such subcube [a, b] c [a 1 , b 1], the set En [a, b] is covered by 0. Among the 2n subcubes formed, there must exist at least one subcube [a2, b2] having the property that En [a2, b 2 ] has no finite subcover from 0. (Otherwise, there would exist a finite subcover for E itself.) Now we repeat the process by subdividing [a2, b 2 ] into 2n subcubes and we select a subcube [a3, b3] in the same manner as before. In this way, we obtain a decreasing nest [ai. b1] ::) [a2, b2] ::) ... ::) [ak, bk] ::) ... of subcubes having the property that for each k E N the set Ek = E covered by 0 but has no finite subcover.

n [ak, bk]

is

258

EUCLIDEAN SPACE

Select a point Pk E Ek for each k EN. If j and k ;::: N then Pj and Pk E EN. Hence

as N ----> oo. Thus the sequence Pk is a Cauchy sequence, and Pk ----> p E EN for each N E N. (Here we use the fact that EN is a closed set.) Since p E E, there exists a E A such that p E Oa. Since Oa is open, there exists r > 0 such that Br(P) ~ Oa. Now select N EN such that k;::: N implies

Ry'n 2k-2

< r,

so that p E EN C Br(P) ~ Oa. Thus we have covered EN with a single set Oa from 0, contradicting the claim that EN has no finite subcover from 0. •

EXERCISES 8.34 If Eisa compact subset of !En, prove (without using the Beine-Borel Theorem) that every closed subset of E is compact. 8.35 Use only the definition of compactness to show that every finite subset of lEn is compact. 8.36 a) LetS = { xUl Ij E N} c lEn be any convergent sequence. Show that Sis a compact set if and only if limj--->oo xUl E S. b) LetS= {xn=(1+ 2'!;, 2'!;) ln=0,1,2,3, ... } c!E 2. TrueorFalse: S is compact.

8.37 Let S = { x<j) Ij EN} C lEn be any sequence. Prove or give a counterexample: The set S is a compact set if and only if the sequence x(j) is convergent to an element of S.

t Prove part of the Heine-Bore! theorem by showing that if E ~ lEn is compact, then E is bounded. (Hint: Show how to cover E with suitable open balls

8.38

centered at 0.)

t Prove part of the Heine-Bore! theorem by showing that if E ~ lEn is compact, then E is closed. (Hint: Suppose that pis any point not in E. Show how to cover E with the complements of suitable closed balls centered at p. Conclude that Ec is open.)

8.39

8.40 t Let f be a real-valued function defined on a closed finite interval [a, b] in IE 1 . Define the graph Gf = { (x1, X2) E IE 2 j X2 = j(x1 ), X1 E [a, bl}. a) Prove: Iff E C[a, b], then G f is a compact subset of IE 2. (Hint: Use the Heine-Bore! Theorem.)

CONNECTED SETS

259

b) Let

g(xl) =

{

sin .K... 0 xl

if XI E (0, 1], if XI

= 0.

2

Is the graph G 9 a compact subset of IE ? Prove your conclusion.

8.41 Let EI 2 E2 2 ... 2 Ek 2 ... be a decreasing nest of nonempty closed subsets of lEn. a) Give an example to show it is possible for n:I Ek to be empty. b) If EI is compact, show that n:I Ek =/= 0. (Hint: Select a point xk E Ek for each k E N. Apply Exercise 8.ll.c. ) 8.42 For each of the following subsets of lEn, determine whether or not it is compact and justify your conclusion. a) Br(x), with r > 0. b) Br(x), with r > 0. c) sn-I = Br(x) \ Br(x), with r > o. Let K c lEn be compact. Suppose f and g are in C(K) and suppose that f(x) = 0 for each x E K such that g(x) = 0. Prove: Iff > 0, there exists M 0

8.43

z

such that

lf(x)l < Mlg(x)l

+f

for all x E K. (Hint: Use the Heine-Borel Theorem. For interesting applications of this exercise in approximation theory, see [13].)

8.4 CONNECTED SETS In order to generalize the Intermediate Value Theorem (Theorem 2.3.1) to functions defined on Euclidean space, it will be necessary to identify a class of subsets of lEn with properties sufficiently similar to those of intervals. Let us begin by writing a formal definition of the concept of interval. Definition 8.4.1 An interval is any subset I ofiR such that for all a and bin I with a -::; b the set {x E IR I a -::; x -::; b} ~ I.

This concept includes all open, closed, half-open and half-closed, finite or infinite intervals. (See Exercise 8.44.) The difficulty in generalizing this concept is that there is no natural linear ordering, or notion of inequality in lEn when n > 1. The concept we seek is that of being a connected set. In order to define this concept it is convenient to begin with its opposite: a set is called disconnected, in the sense that it can be separated into two parts, or components. Definition 8.4.2 We say that a set E ~ V, a normed vector space, is disconnected if there exists a pair of disjoint open subsets A and B of V for which E ~ A U B and such that both EI = E n A =/= 0 and E2 = En B =/= 0.

260

EUCLIDEAN SPACE

In this case, we say that A and B separate E. If E is not disconnected, then E is called connected .

• EXAMPLE 8.10 Let E = Q, the set of all rational numbers. We claim that Eisa disconnected subset of IE 1. In fact, let

A=

{x E IE1 Ix< J2}

and B =

{x E IE1 Ix> J2}.

Then A and Bare disjoint open sets that separate E. The key to this construction is that there exists E IE 1 \ Q. In Exercise 8.45 the reader will prove that a subset of IE 1 is connected if and only if it is an interval. The set Q, which is shown above to be disconnected, is not an interval. We show below that the concept of a set E being disconnected can be expressed without reference to open sets A and B as in the definition. This will give us a more visually intuitive concept of the meaning of connectivity.

J2

~ V, a normed vector space, is disconnected if and only if we can decompose EasE= E1 U E2, both sets nonempty, where E1 n E2 = 0 and where neither E 1 nor E2 contains any cluster points of the other set.

Theorem 8.4.1 A set E

Proof: First we suppose that E is separated. Thus A and B exist as in the definition. If e E E 1, then there exists r > 0 such that Br (e) ~ A. Since E 2 is disjoint from A, e is not a cluster point of E2. Similarly, E 2 has no cluster point of E 1. Next we suppose that E = E1 U E2, a disjoint union, where neither E1 nor E2 has any cluster point of the other, and each set is nonempty. Let e E E. Since e is in one of the two sets E1 or E2 without being a cluster point of the other, there exists re > 0 such that Bre (e) is disjoint from the set to which e does not belong. Let

A=

U BTf(e). eEE1

Similarly, we let

B=

U BTf(e). eEE2

It follows that A and B are open sets and that A n E = E 1 and B n E = E 2. We need prove only that A n B = 0. Suppose this claim were false. Then there exists pEA n B. Hence there exist e1 E E1 and e2 E E2 such that lle1- Pll < ~and lle2 - Pll < ~- By the triangle inequality for norms, it follows that

II e1- e2 II < This is a contradiction.

Te 1

+ Te 2

2

{

~max Te 1 ,Te2

}

•

•

EXERCISES

261

Theorem 8.4.2 Let f be any continuous real-valued function defined on an interval I. Let G f denote the graph off as defined in Exercise 8.40. We claim that G f is a connected subset ofiE 2 • Proof: We suppose the claim were false, and we will deduce a contradiction. If G f were disconnected, then G f = E1 U E2, a disjoint union of two nonempty sets, neither one containing any cluster point of the other. So there exist (xi, f(xi)) E Ei, fori = 1, 2. Suppose without loss of generality that x 1 < x 2 . (We know x 1 =1- x 2 , since G f is the graph of a function.) Define

c =sup { t

E

[xl. x2JI (q,f(q)) E E1. V'q E [x1. tl}.

Thus x1 :::; c:::; x2. If (c, f(c)) E E1 then c < x2. Hence for n EN there exists

Cn

E

(c,c+ ~) n [x1,x2]

for which (en, !(en)) E E2. This would imply that (cn.f(cn)) ----+ (c, f(c)), by continuity of f. Hence (c, f(c)) is a cluster point of E2, which is a contradiction. On the other hand, if (c, f(c)) E E2, then c > Xt. (c, f(c)) is a cluster point of E 1 . This yields a similar contradiction. •

Corollary 8.4.1 The Euclidean space lEn is connected, for each n

E

N.

Proof: Suppose the corollary were false. Then lEn = E1 U E2, a disjoint union of two sets, neither of which contains any of the other's cluster points. Let p E E 1 and q E E2. Let

¢(t) = p

+ t(q- p)

for each t E [0, 1]. Then ¢(0) E E1 and ¢(1) E E2. Define

c =sup { t E [0, 1JI (q, qy(q)) E Et, V'q E [0, tl}. Now complete the proof just as in Theorem 8.4.2.

•

Theorem 8.4.3 Suppose E and F are nondisjoint connected subsets of a normed vector space V. Then G = E U F is connected. Proof: Suppose false. Then there exist nonempty disjoint open sets A and B such that An G =1- 0 =1- B n G. Since E cannot be separated, E is contained entirely in one open set or the other. Without loss of generality, suppose E ~ A. Similarly, F is contained entirely in one of the open sets. Since B n G =1- 0, F ~ B. But this is • impossible since E n F =1- 0 and A n B = 0. EXERCISES

t Prove that if I ~ R. is an interval, as defined in Definition 8.4.1, then I must have one of the following forms: [a, b], (a, b), [a, b) or (a, b]. In this notation -oo :::; a :::; b :::; oo, but closed endpoints must be finite. (Hint: Consider sup( I) and inf(J).)

8.44

262 8.45

EUCLIDEAN SPACE

t

a) Prove that every connected subset S ~ JE1 is an interval. b) Prove that every interval I is a connected subset of lE 1. In particular, this includes the claim that the real line JE 1 is itself a connected set. (Hint: Consider Theorem 8.4.2 and the function f(x) = 0. Explain how connectedness in E 2 implies connectedness in E 1 .)

8.46

Prove that the union of the x 1-axis and the x 2-axis in lE 2 is a connected set.

8.47

Prove that 8 1

= {x E 1E2 IIIxll = 1} is a connected set.

8.48 Suppose E ~ JE2 such that for all x E E, x 2 = 0. Suppose E is a connected subset oflE1, which we identify with the x 1-axis oflE 2. Prove that E is also connected as a subset of lE 2. 8.49 Let E = { x E JE2 1 x 1x 2 > 0 }. Is E connected? Prove your conclusion. 8.50 Let E = { x E lE 2 1 x1x2 ::::: 0 }. Is E connected? Prove your conclusion. 8.51

Let E = { x

E lE2 Ixi

=

1}.

Is E connected? Prove your conclusion.

8.52 In E 2, let E = { x Ix1 = 0} U { x Ix1 x2 = 1, x1 Prove your conclusion. 8.53

> 0}. Is E connected?

Let .g () X

=

xsin.!!:. x {0

if X> 0, if X= 0.

Prove that the graph of g is a connected subset of lE 2.

8.54

Let if X> 0, if X= 0.

Prove that the graph off is a connected subset of lE 2. (Hint: Try writing G f ~ AU B with A and B open and disjoint. Prove that Gf must be contained entirely within one of the two open sets.)

8.55 Let E ~ lEn be any connected set and let p be a cluster point of E. Prove or disprove: E U {p} is connected. 8.56 Let E ~ lEn be any connected set and let disprove: E is connected.

E be the closure of E. Prove or

2

8.57 Let f (x) = e -x for all x E R Denote the graph off by Gf and the x-axis in lE 2 by Go. Prove or disprove: S = G f U G0 is a connected subset of JE2. (Hint: Apply Theorem 8.4.1.) 8.58 Prove or give a counterexample: Every open set 0 ~ lEn can be expressed as the union of a finite or countable family of disjoint open balls. (Hint: See Exercise 8.32.)

TEST YOURSELF

263

8.5 TEST YOURSELF

EXERCISES 8.59 Let x = (5, -2, 3) E IE3 . Give an example of a non-0 vector y E JE3 for 2 2 2 which llxll + IIYII = llx + Yll · 8.60

Let S

~

!En be a set of vectors for which each sequence

has a convergent subsequence. True or False: S must be bounded.

8.61 The taxicab norm in 1R 2 is defined by llxllt = lx1l + lx2l· Draw a sketch of the unit ball B 1 (0) in JR 2 using the taxicab norm. Label your axes. 8.62 D.

Give an example of a set D ~ IE2 and a point p which is an isolated point of

8.63 Give an example of a set S ~ IE 2 for which S is dense IE 2 but has empty interior. 8.64 True or False: The interior of the closure of an open set S S itself. 8.65

Let f(x) = sin i for all x

#

~

!En need not be

0. True or False: The graph G f is compact.

8.66 Let E 1 :2 E2 :2 ... :2 Ek :2 ... be a decreasing nest of closed subsets of !En. True or False: n~ 1 Ek must be non-0. 8.67 LetS = {Xn = (1 + ~, ~,) 0, 1, 2, 3, ... } C IE2. True or False: S is compact.

8.68

2 2

In=

Let

g(x)

=

x 2 cos.! x {0

# 0, ifx = 0. ifx

True or False: The graph G 9 is connected.

8.69 For each one of the following two sets, decide whether it is connected or not connected. a) S = {x E IE 2 jlx2l = lx1j}. b) T = {x E IE 2 jlx2l < lx1j}. 8.70 True or False: If S # 0 is open in !En, then S non empty disjoint open balls of finite radius.


CHAPTER9

CONTINUOUS FUNCTIONS ON EUCLIDEAN SPACE

9.1

LIMITS OF FUNCTIONS

We will consider vector-valued functions f : D ____, lEm, where the domain of definition off is a set D ~ V, a normed vector space. (Most often, V will be lEn.) Since for each x E D we have f(x) E lEm, we can write f(x)

= (!l(x), ... , fm(x)),

where fi(x) E R. for all x E D. In elementary courses in the calculus of several variables, Vis always lEn and it is common to write the real-valued function fi(x) in the form fi(xl, ... , Xn)· In order to define the concept limx-+a f(x), we need to restrict ourselves to the case in which a is a cluster point of D. Recall that a cluster point of D need not be an element of D. A cluster point a of a set D has the property that it is always possible to find points x E D for which x -=f. a and yet x is as close to a as we like. If a is not a cluster point, it will be impossible to define limx-+a f(x) because it is impossible for x to approach a. For example, if a were an isolated point of D, then for all sufficiently small t5 > 0 we obtain the set

{xI 0 < llx- all < t5,

xED}= 0,

Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright @ 2008 John Wiley & Sons, Inc.

265

266


the empty set. Every statement one might make about each element x of the empty set, such as the statement \\f(x) - L\\ < t, is vacuously true. Definition 9.1.1 Let a be any cluster point of the domain De of a function £. Then limx_,a f(x) = L E !Em if and only if for all t > 0 there exists 8 > 0 such that x E Dr and 0 < \\x- a\\ < 8 implies \\f(x) - L\\ < t. This definition is a natural adaptation of Definition 2.1.2 to vector-valued functions defined on Euclidean space. We have the following theorem that permits more such adaptations to be developed. Theorem 9.1.1 Suppose a is a cluster point ofthe domain De of£. Then the following two statements are equivalent. i. limx_,a f(x)

=L

ii. For every sequence of points X 71 E Dr \ {a} such that X 71 ~ a, we have the sequence f(xn) ~ L.

Proof: Let us prove first that (i) implies (ii). So suppose (i) and consider any sequence of points x 71 E Dr \ {a} such that X 71 ~ a. We know for each t > 0 there exists 8 > 0 such that x E DJ and 0 < \\x- a\\ < 8 implies \\f(x)- L\\ < t. But since x 71 ~ a, there exists N such that n 2: N implies 0 < \\xn - a\\ < 8. This implies \\f(x71 ) - L\\ < E, so that f(xn) ~ L. Next we prove (ii) implies (i). So suppose (ii) is true. We need to show limx->a f(x) = L. We will show that if this conclusion were false then a selfcontradiction would result. But if it were false that limx_,a f(x) = L, then there exists t > 0 such that for all 8 > 0 there exists x E Dr such that 0 < \\x - a\\ < 8 and yet \\f(x) - L\\ 2: f.. In particular, if we let 871 = ~ then we get x 71 such that 0 < l\x71 - a\\ < ~and yet \\f(x71 ) - L\\ 2: E. Now, Xn E Dr\ {a} and X 71 ~a, yet f(Xn) ft L. This contradicts (ii), which was assumed to be true. • The theorem above combined with Theorem 8.1.2 and Theorem 2.1.1 yields the following immediate corollary. Corollary 9.1.1 Let a be any cluster point of the domain De. Then limx_,a f(x) exists and equals L = (h, ... , lm) if and only iflimx->a fi(x) exists and equals li foralli = l, ... ,m. This corollary enables us to understand the concept of convergence of vectorvalued functions f in terms of convergence of each real-valued component function

k Remark 9.1.1 If two functions f and g are defined on subsets of the same space IE", with values in the same space !Em, then we can form the sum or difference of these functions with the domain of this combination being Den Dg. If, furthermore, f = f and g = g are real-valued functions of vector variables, then we can also multiply and divide the two functions. The domain D L of the quotient will be only 9

LIMITS OF FUNCTIONS

267

those points of Dt n D 9 for which g(x) =1- 0. (Multiplication and division are not defined, however, iff and g map lEn to !Em with m > 1.)

Corollary 9.1.2 Let f and g be defined on domains in V, a normed vector space, with values in !Em, as in the Remark above. Suppose a is a cluster point of the intersection DrnDg of the domains off and g, limx--+a f(x) = L andlimx--+a g(x) = M. Then

i. limx--+a(f ± g)(x) = L ± M. ii. Ifm

= 1, then limx_, (fg)(x)

= LM.

8

iii. If m = 1, then limx--+a L (x) g cluster point of D 1·

=

f:t,

provided that M =/:- 0 and that a is a

9

Proof: The proofs are very similar to those of Corollary 2.1.1. Here we will treat only conclusion (i). Consider a sequence of points Xn E Dr n Dg \ {a} such that Xn ---> a. Then (f ± g)(xn) = f(xn) ± g(xn) ---> L ± M, by the corresponding theorem for limits of sequences. The other proofs are similar. • This corollary suggests that limits of functions on !En are very similar in behavior to limits of real- valued functions of one real variable, subject only to certain essential hypotheses to insure the requisite operations are defined for the objects in question. However, the reader must be very careful: it is much harder for the limit of even a real-valued function of two or more variables to exist than it is for functions of only one variable. See Exercise 9.4.

Definition 9.1.2 Let x E IE" and let k integers. We define

= (k 1 , ... , kn) be any n-tuple of nonnegative

a real-valued monomial on lEn. A real-valued polynomial on IE" is any function of the form p(x) = LJ=l cixki, where each Cj E JR. A rational function is any ratio of two real-valued polynomials. A rational function is defined wherever the denominator is nonzero. In the exercises below, the reader will show that the behavior of polynomials and rational functions on !En with respect to limits is very similar to the behavior of such functions on JR.

Definition 9.1.3 Let f: D---> !Ern.

SupposeD~

IE", where n 2 2.

=L providedforallsequencesxn E Drsuchthatllxnll-> oowehavef(xn)---> L.

i. If the domain Dr is not a bounded subset of IE", we say limx--+oo f(x)

ii. /fa is a cluster pointoJDr we say limx--+a f(x) = oo provided for all sequences Xn E Dr such that Xn ---> a we have llf(xn) II ---> oo. Remark 9.1.2 In case (i) we say that f(x) converges to Las x diverges to infinity. In case (ii) we say that f(x) diverges to oo as x converges to a. The use of the word converges versus diverges hinges on whether or not the limit exists as a point of lEn.

268


Remark 9.1.3 The restriction ton ~ 2 in Definition 9.1.3 is employed in order to avoid conflict with the normal terminology in IE1 . For functions of a single real variable, it is customary to distinguish between x ----) oo and x ----) -oo and to distinguish both of these concepts from lxl ----) oo. The rationale for this is that in IE 1 there are only two directions in which to move away from 0 as far as one pleases. But in lEn for n ~ 2 there are infinitely many directions along which one might wander as far from 0 as one pleases. Thus distinguishing among directions in the concept of divergence to infinity becomes hopeless.

EXERCISES 9.1

Write a detailed proof for Corollary 9 .1.1.

9.2 Let p be any real-valued polynomial on lEn, as in Definition 9.1.2. In the parts below, prove that limx_,ap(x) exists and equals p(a). a) If m(x) =Xi, 1 :S i :S n, prove that limx---+a m(x) = m(a). b) If m(x) = xk, prove that limx_,a m(x) = m(a). c) Let p be any polynomial on lEn. Prove that limx_,a p( x) exists and equals p(a). That is, prove that a polynomial is continuous everywhere. (See Definition 9.2.1.) 9.3 Let Q(x) be any real-valued rational function on lEn, as in Definition 9.1.2. Prove that limx_,a Q(x) exists and equals Q(a), provided that a E DQ. That is, prove that a rational function is continuous wherever it is defined. (See Definition 9.2.1.) 9.4

t Let f : IE2 ----) lR by the formula ifx-#0, ifx = 0.

(See Figs. 9.1 and 9.2) a) Show that ifx----) 0 along either the x 1 - or the x2-coordinate axis, f(x) ----)

0. b) Show that if x ----) 0 along any straight line x2

=

kxt through the origin,

f(x)----) 0. c) Show that limx_,o f(x) does not exist. Hint: Show there exist points x arbitrarily close to 0 at which f(x) = ~· d) Does limx_, 00 f(x) exist? Prove your conclusion.

9.5

Let f

: IE2 ----) lR by the formula ifx -=J 0, ifx

= 0.

Prove that limx_,o f(x) = 0. (Hint: Let € > 0. Find 8 implies lf(x)l < E. Compare with Exercise 9.4.)

> 0 such that llxll < 8

EXERCISES

Figure9.1

f(xt,x2)

2 1

= xl~ +"'x22 2

269

inthefirstquadrant.

9.6

For x E E 2 , evaluate each of the following limits. a) limx_,o 11 11 . b) limx_,oo 11 11 .

9.7

Let f(x) = x sin~' f: E 1 a) limx--+0 f(x) = 0. b) limx--+0 = 00.

! !

\

{0}

--4

JR. Prove each statement True or False:

ix)

c) limx--->0 lflx)l =

00.

9.8 Let a be a cluster point of the domain Dr of a nonvanishing (that is nowhere-0) function f (x). Prove that

!~ f(x) = 0 {::} !~ 11£(~)11

= oo.

9.9 Suppose the domain D ~ En of a function f : D Prove: limx_, 00 f(x) = L E Em if, and only if, for each f such that x E D \ BM(O) implies llf(x)- Lll < f.

--4

Em is not bounded.

> 0 there exists M > 0

270


Figure 9.2

f(xt, x2)

2 1

= xl~ +"'x22 2

in all four quadrants.

9.10 Suppose that a is a cluster point of the domain D a f(x) = oo if, and only if, for each M > 0 there exists 8 > 0 such that x E (D \{a}) n Bo(a) implies llf(x)l! > M. 9.11 Prove the following Cauchy Criterion for the existence of the limit of a function. Let f : D ---+ lEm and let a be a cluster point of D 0 there exists 8 > 0 such that x and x' E D n (Bo(a) \ {a}) implies llf(x)- f(x')ll < E.

9.2 CONTINUOUS FUNCTIONS The following definition is analogous to Definition 2.2.1. Definition 9.2.1 Let V be any normed vector space. A function f : D ---+ lEm is called continuous at a point a E D 0 there exists a corresponding 8 > 0 such that x E D n B0 (a) implies

llf(x)- f(a)ll

<E.

Iffiscontinuousateverypointa ED, wesayf E C(D), thefamilyofallcontinuous functions on D with values in lEm. In this notation, the value ofm is fixed. If there is


271

more than one possible range-space lEm under discussion, we will denote the family of continuous functions instead as C (D, lEm ). If Dr has any isolated points a, then f is automatically continuous at a. The more interesting case, however, is that of a nonisolated point a E Dr.

Theorem 9.2.1 A function f is continuous at a cluster point a E D limx_,a f(x) exists and equals f(a).

if and only if

Proof: First suppose f is continuous at a E D. Then for each E > 0 there exists 8 > 0 such that llx- all < 8 and x E D implies llf(x) - f(a)ll < E. Hence for all x E D such that 0 < llx- all < 8 we have llf(x)- f(a)ll < Ewhich implies limx_,a f(x) exists and equals f(a). Now suppose limx--->a f(x) exists and is f(a). Then for all E > 0 there exists 8 > 0 such that 0 < llx- all < 8 and xED implies llf(x)- f(a)ll <E. This implies for all xED n Bo(a) we have llf(x)- f(a)ll < E so that f is continuous at a. •

This theorem enables us to prove some properties of combinations of continuous functions that are analogous to the results in Theorem 2.2.2 to the extent possible. Theorem 9.2.2 Let f and g be defined on domains in a normed vector space V with values in lEm. Suppose f and g are both continuous at a Then i. f ± g is continuous at a ii. If m = 1, then the function

iii.

/fm

= 1 and if g(a) =1- 0,

f g is continuous at a

then~ is continuous at a

iv. If we write f = (JI, ... , fm) in terms of its scalar-valued components, then f is continuous at a if and only if !I, ... , f m are all continuous at a Proof: This is a simple application of Corollary 9.1.2 and Corollary 9.1.1 and it is left to the Exercises. Note that the proof is trivial if a is not a cluster point of the • domain of the appropriate combination off and g.

The following definition is useful for the study of continuous functions in Euclidean space. Definition 9.2.2 A set E is said to be relatively open in D ~ V, a normed vector space, provided there exists an open set U ~ V such that E = U n D. Similarly, a set E is said to be relatively closed in D ~ V provided there exists a closed set K ~ V such that E = K n D .

•

EXAMPLE 9.1

In JE 1 , [0, 1) is relatively open in [0, 2], since [0, 1) = (-1, 1) n [0, 2]. Similarly, (0, 1] is relatively closed in (0, 2], since (0, 1J = [ -1, 1] n (0, 2].

272


Theorem 9.2.3 A subset E ~ Dis relatively open in D if and only iffor each x E E there exists 8x > 0 such that Box(x) n D ~E. Proof: First suppose that E is relatively open in D. Then there exists an open set U such that U n D = E. If x E E ~ U, it follows that there exists a corresponding 8x > 0 such that B 0Jx) ~ U. Thus B 0Jx) n D ~ U n D = E. This completes

the proof in one direction. For the other direction see Exercise 9.27. Corollary 9.2.1 A subset E ~ D is relatively open in D in D, D \ E, is relatively closed in D.

•

ifand only if its complement

The proof of this corollary is left to the Exercises. It is a remarkable and very useful fact that the continuity of a function f on a domain D is closely connected to the way in which open sets and closed sets are transformed by the inverse of f. In this terminology we make no supposition that f is invertible, as we see below.

Definition 9.2.3 Let f : D --+ !Em, where D ~ V, a normed vector space. We define the inverse image of E ~!Em by r- 1 (E) ={xED I f(x) E E}. Theorem 9.2.4 Let f be defined on a domain D ~ V, a normed vector space, where f: D--+ !Em. Then f E C(D) if and only if£- 1 (U) is relatively open in D for each open set U ~ !Em. Proof: First we suppose f E C(D). Let U ~ !Em be open. We need to show r- 1 (U) is relatively open in D. If r- 1 (U) = 0 then there is nothing to prove. Without loss of generality, let x E r- 1 (U), so that f(x) E U. Since U is open, there exists r > 0 such that Br(f(x)) ~ U. Because f is continuous at x, there exists 8x > 0 such that

f: Bo)x) n D--+ Br(f(x)).

Thus Box (x) n D ~ r- 1 (U). By Theorem 9.2.3 r- 1 (U) relatively open in D. Next we suppose r- 1 ( U) is relatively open in D for each open set U ~ !Em, and we must prove that for each x E D, f is continuous at x. Let E > 0. We need to show that there exists 8 > 0 such that f: B 0 (x) n D--+ Be(f(x)). Since r- 1 (Be(f(x))) is relatively open in D, there exists 8 > 0 such that B 0 (x)

nD

~

r- 1 (B.(f(x))).

(See Theorem 9.2.3.) Thus f is continuous at x.

•

EXERCISES 9.12

Define?Ti: lEn--+ !Rby?Ti(x) =

Xi,i

= l, ... ,n. Provethat?Tiiscontinuous.

9.13 t IfV is any normed vector space, define N: V--+ IR by N(x) = llxll· Prove that N is continuous. Hint: See Exercise 2.20. Show that illxll - IIYIII :S llx- Yll·

EXERCISES

9.14

273

Let f : IE 2 ----+ IE 2 by the formula ifx:f:O, ifx

= 0.

Determine whether or not f E C (1E 2 ) and prove your conclusion.

9.15

Let f : IE 2 ----+ lE2 by the formula f(x) = { (x1x2,sin (0,0)

(~))

ifx:f:O, ifx

= 0.

a) Determine whether or not f E C (1E 2 ) and prove your conclusion.

b) LetS = {xI f(x) conclusion.

= 0 }.

Is the complement of San open set? Prove your

9.16 Suppose f : lE 2 ----+ lR has the property that for each fixed value x 2 = b the function gb(xl) = f(xl, b) is a continuous function of x1. Suppose also that for each fixed value x 1 =a the function ha(x 2 ) = f(a, x 2 ) is a continuous function of x 2 • Does it follow that f E C (1E 2 , lR)? If yes, prove it. If no, give a counterexample and prove that it is a counterexample. (Remark: This question is often expressed verbally as follows. Does the continuity of f(x) in each variable x 1 , x 2 separately imply joint continuity, meaning continuity as a function of the vector variable x? See Problem 9.14.) 9.17

Prove all four parts of Theorem 9.2.2.

9.18

Prove Corollary 9.2.1.

9.19 Give an example of a continuous function f : JE 1 ----+ JE 2 that maps the open set ( -271', 211') onto the circle xi + x~ = 1, which is not an open set. ~

lEn is open then a relatively open subset E

9.20

Prove that if D

9.21

Prove that C(D, lEm) is a vector space.

9.22

Let D

= {x

~

D is open.

11 < jjxjj ::; 3} C lE Show that the set E = { x 11 < l!x!l ::; 2} 2

•

is relatively closed in D, and that the set F in D.

= {x

12 < l!xll ::; 3} is relatively open

9.23 Identify the relatively open subsets of the set Z contained in IE 1 . Identify the relatively closed subsets as well. 9.24 Give an example of a function such that f- 1 (U) is not open.

f

E

C([O, 1), IR) and an open subset U

~ lR

9.25 Let f be defined on a domain D ~ lEn, where f : D ----+ !Em. Prove that f E C(D) if and only if r- 1 (K) is relatively closed in D for each closed set K ~ lEm.

274 9.26


Is the set { x E JE4

1

x 1 X4 -

x2x3

= 1} open, closed, neither, or both?

9.27 t Complete the proof of Theorem 9.2.3 by showing that E ~ D is relatively open in D if for each x E E there exists 8x > 0 such that Box (x) n D ~ E. 9.28 Let f : D -+ lEm be continuous on its domain D ~ lEn, and suppose g : f(D) -+ JEP is continuous on f(D). Prove: The composition go f E C(D, JEP), where go f(x) = g(f(x)) E JEP, for each x E D. (Hint: Use Theorem 9.2.4.)

9.3 CONTINUOUS IMAGE OF A COMPACT SET In the Extreme Value Theorem (Theorem 2.4.2) we saw that a continuous realvalued function on a closed finite interval [a, b] must achieve both a maximum and a minimum value. The type of set analogous to [a, b] that enables us to prove an Extreme Value Theorem for real-valued functions on lEn is a compact set. (Iff were not real-valued but rather 1Em-valued with m > 1, then it would not make sense to speak of an extreme value for f since there is no natural order relation among the vectors of lEm .) We begin with a more general theorem that does not require real values, however. Theorem 9.3.1 Let D C lEn be compact, and suppose f : D -+ lEm is continuous. Then the set f(D) = { f(x) x E D} is a compact subset oflEm.

I

Proof: In words, this theorem states that the continuous image of a compact set is compact. We begin by letting 0 = { Oa I a E A} be an arbitrcli)' open cover of f(D), where A is an index set and each Oa is an open set in lEm. It will suffice to prove that there exists a finite subcover of f(D) selected from the given open cover 0. Since f is continuous, each set Va = r- 1 ( Oa) is relatively open in D. Thus there exists an open set Ua in lEn such that Va = Ua n D. Since f(D) is covered by 0, U = {Ua I a E A} is an open cover of D. Since Dis compact, there exists a finite subcover {Uj I j = 1,2, ... ,p} of D. Thus D = u~=1 ltj. and f(D) ~ U;=l Oj. This is a finite subcover of f(D), which must therefore be • compact. Now we can prove an Extreme Value Theorem for lE". Theorem 9.3.2 (Extreme Value Theorem for JEn) If D C lEn is compact and if f : D -+ JR. is continuous, then f achieves both a maximum value and a minimum

value on D.

Proof: Since f(D) is a compact subset of JR., it follows that f(D) is both closed and bounded. Let M =sup f(D) and m = inf f(D), both of which are necessarily real-valued (not infinite). It will suffice to prove that both M and mare elements of f(D). We will see that this follows from the fact that f(D) is closed. Since M is the least upper bound of f(D), M- fails to be an upper bound of f(D), and this

i

CONTINUOUS IMAGE OF A COMPACT SET

275

is true for every k E N. Thus, for each k E N we have a point M-

1

k
M, with none of the points of this sequence being M, so that M is a cluster point j(D). Hence M E f(D) as claimed. The similar argument for m is an exercise. • Here is a surprising theorem that can be quite useful. Theorem 9.3.3 Iff : D ----> !Em is a continuous, one-to-one function defined on a compact set D c !En, Then f- 1 is continuous.

Proof: Since f is one-to-one (also called injective), it makes sense to define f- 1 (y) = x if and only if f(x) = y. An invertible continuous function that has a continuous inverse is called a homeomorphism. Thus in words the theorem says that a continuous injective map of a compact set is a homeomorphism. It suffices to prove the continuity of f- 1 at any cluster pointy E f(D). If

we need to show that xj

= f- 1 (yj)

____. x

= f- 1 (y).

We claim that the entire sequence Xj must converge to x. If this were false, then there exists E > 0 and a subsequence Xj; for which llxj; - xll 2': E for all i. In that case, the bounded sequence Xj; must itself have a subsequence Xji 1 that converges to some vector x' -I- x, since llx' - xll 2': E. But f(x') = y = f(x). This is a contradiction since f is one-to-one. • We can extend the concept of the sup-norm to vector-valued functions as follows. Definition 9.3.1 Iff: D----> !Em where D

f uniformly on D provided that llfj - fllsup ----> 0 as j ----> oo.

The following theorem shows that uniform convergence behaves very well with respect to the continuity of functions. Theorem 9.3.4 Suppose fj E C(D, !Em) for all j, and suppose also that fj uniformly on D. Then f E C(D, !Em).

The proof is left to the Exercises.

---->

f

276


EXERCISES Prove or give a counterexample: If a continuous injective function f mapping (-1, 1) to JE 1 has closed range then its range must be bounded.

9.29

9.30 Prove or give a counterexample: If a continuous injective function f mapping ( -1, 1) to lE 1 has bounded range, then its range must be closed. 9.31 Give an example of a noncontinuous image of a compact set that is not compact because a) it is closed but not bounded. b) it is bounded but not closed.

The set E = (0, 1] is a bounded subsetofJE 1 and E is relatively closed in the set F = (0, 2]. Is E compact? Justify your answer. 9.32

In

9.33

Let D = { ~ EN} U {0} c JE 1 . Prove each of the following statements. a) D is compact. b) Iff E C(D, JEm), then function !If I must achieve both a maximum value and a minimum value on D. c) If g: D----> JEm, then g E C(D,JEm) if and only if g(~) ----> g(O) as n ----> oo. (Caution: Be sure to prove that if the latter condition is satisfied and if {xk IkE N} ~Dis such thatxk----> 0, then g(xk)----> g(O).)

9.34

Complete the proof of Theorem 9.3.2 by proving that m is a cluster point of

f(D). 9.35

Let¢: [0, 1]----> JE2 be defined by ¢(t) = (cos[27rt], sin[27rt]). Let f

= 0 and x2 > 0. See Fig. 9.3. (Hint: The idea is to reduce the problem to one on a compact domain. It may be helpful to consider the curves xix 2 = k for positive constants k, as well as the lines xi = a and x2 = b for constants a and b.)

9.4 CONTINUOUS IMAGE OF A CONNECTED SET We saw in Theorem 2.3.1 that every continuous function defined on an interval has the Intermediate Value Property, which is stated in Definition 2.3.1. In order to generalize the Intermediate Value Theorem to Euclidean Space, we will need to consider functions defined on a connected subset D ~ En, and we will need to restrict our attention to real-valued functions so that the concept of k being between f(a) and f(b) will make sense. But first we prove a more general theorem. Theorem 9.4.1 Let D ~ V, a normed vector space, be a connected set, and suppose f E C(D, Em). Then the range f(D) is a connected subset ofEm.

Proof: Suppose the theorem were false: We will deduce a contradiction. By Theorem 8.4.1, the range f(D) can be separated as f(D) = AUB, where A =f. 01- B, AnB = 0, and neither A nor B contains any cluster points of the other set. It follows thatD = r-I(A)uf-I(B), wherethetwonon-0setsA' = r-I(A)andB' = r-I(B) are disjoint. We claim that neither A' nor B' can have any cluster point of the other set. Suppose false: For example, suppose there exists a' E A' that is also a cluster point of B'. Then there exists a sequence bj E B' such that bj ---+ a'. By continuity, bi = f(bj) ---+a= f(a'). Hence A has a cluster point of B, which contradicts the hypothesis that D is connected. It follows that neither A' nor B' can have a cluster point of the other set, which contradicts the supposition that the theorem is false. Hence f(D) is connected. • Theorem 9.4.2 (Intermediate Value Theorem) Let D be a connected subset of V, a normed vector space, and suppose f E C(D, EI ), so that f is real-valued. /fa and bED and if f(a) < k < f(b), then there exists c E D such that f( c) = k.

Proof: By the preceding theorem, we see that f(D) is a connected subset of EI and is thus an interval I. (See Exercise 8.45.) Since f(a) and f(b) E I, it follows that k E I= f(D). Thus there exists c ED such that f(c) = k. •

EXERCISES

Figure 9.4

f(x1 ' x2)

= (x1

'

x2 -

279

1-).

'XIX2

EXERCISES

9.43 t A subset D 0, and the dots denote disjointness of the union. Prove: Each of the five sets in the union is connected. 9.45

Let f : D

---7

JE3 by

where D is the Euclidean plane without either of the two coordinate axes. Prove: f(D) can be written as the union of 4 mutually disjoint sets, each set connected and

280


each set maximal in the sense that it could not be enlarged with additional points from f(D) without losing connectivity. See Fig. 9.4. 9.46 Prove that each open ball Br (a) is a convex set in lEn. Do the same thing for each closed ball Br(a). 9.47

Consider the polynomial p E C

(1E2 , lE 1 )

p(x) = xix~ - 2xix2

given by

+ x1x2 -

3.

Show that there exists x E JE 2 such that p(x) = 0. 9.48 Consider the function f E C (1E 2 ,1E2 ) given by f(x) = ex 1 (cosx 2 ,sinx 2 ). Note that f(l, 0) = (e, 0) and f(l, 1r) = (-e, 0). Is there a point x E E 2 such that f(x) = (0, 0). Justify your claim. 9.49 Suppose D C lEn is compact and f E C (D, lEm) is one-to-one. Prove: If f(D) is connected, then Dis connected. 9.50

Show how to use Theorem 9.4.1 to give an alternative proof of Theorem 8.4.2.

9.51 Show how to use Theorem 9.4.1 to give an alternative proof that lEn is connected. (Hint: If false, show that the continuous image of an interval into lEn could fail to be connected, which is a contradiction.)

9.5 TEST YOURSELF

EXERCISES 9.52 True or Give a Counterexample: If a set S = AUB c lEn is the disjoint union of two non-0 sets, neither A nor B containing a cluster point of the other, then there is a positive distance 8 > 0 such that for all a E A and b E B we have II a- bll ~ 8. 9.53

Let if X> 0, if X= 0.

True or False: The graph of f is a connected set. 9.54

Give an example of a function

f : lE 2

__,

lim j(x1, kx1) X1-+0

lR for which

=0

for all k E JR, yet limx-+O f(x) does not exist. 9.55 State the Cauchy Criterion for the existence of limx-+a f(x) at a cluster point a of D1. 9.56

True or False: Iff : lEn __, lR is defined by f(x) =

"'n (

)i

L.t~=~ ~:112 xi ' 2

EXERCISES

281

then the setS= {xI f(x) = 0} is a closed set. 9.57

True or False: Iff E C (D, IEm) with D C lEn and if 0 is open in IEm, then

r- 1 (0) is an open set in lEn. 9.58

Let ifx

# 0,

ifx = 0. Either find limx---+O f(x) or else state that it does not exist.

In

9.59 Let D = {0} U { 2~ E N} C IE 1 • True or False: If g : D g E C(D, IEm) if and only if g ( 2~) --> g(O) as n--> oo.

-->

IEm, then

9.60 True or Give a Counterexample: If Sis a relatively closed subset of B 1 (0) c lEn, then S must be compact. 9.61

True or False: B 1 (0, 0) U B 1 (1, 1) is a connected subset of IE2 .

9.62 Let f E C(IEI, JE 2 ). Suppose f(O) = (1, 0) and f(1) = ( -1, 0). True or False: there exists a number x between 0 and 1 such that f(x) = (0, 0).

9.63

Let D = {0} U { ( Xt. then f(D) is connected.

sin;_) Ix 1 # 0 }·

True or False: Iff E C (D, IE 2 ),


CHAPTER 10

THE DERIVATIVE IN EUCLIDEAN SPACE

10.1

LINEAR TRANSFORMATIONS AND NORMS

We saw in Theorem 4.1.1 that a differentiable function f : D --+ JR. defined on a domain D

IR and that there is a

number ME IR such that j.§!;(x)j ::::; M for all x E U and for all j = 1, ... , n.

f E C(U, IR). (Hint: Let p E U and take r > 0 such that U. Prove that f is continuous at p. Adapt the technique from the proof of Theorem l 0.2.4.) b) Now suppose that f : U ----> !Em and that there is a number M E IR such that a) Prove that

Br(P)

~

~~(x)j ::::; M forallx E U and forallj Prove that f E C(U, !Em).

= 1, ... , nand all i =

1, ... , m.

10.42 Iff E C1 (lEn,lEm) and v and w E lEn, express Dtv+wf(x) in terms of Dvf(x), Dwf(x) and t, for each t E R Conclude that the mapping V :

298


Figure 10.3

C1 (JEn,lEm) ear map.

--7

2 1+"'~

xl

x2

in all four quadrants.

]Em defined by V(v) = Dvf(x) for fixed f and fixed xis a lin-

10.43 Let f and g E C1 (lE 2 , JE 2 ). Define¢ : JE 2 --7 JR. by ¢(x) = f(x) · g(x), a scalar product in JE 2 • Find the matrix [¢' (0)] given that

[f(O)] = (1, 1), [f'(O)] = ( !2

[g(O)]

r ).

=

(-1,-1),

[g'(O)] = ( _31 ~ ) .

10.3 THE CHAIN RULE IN EUCLIDEAN SPACE Here we generalize Theorem 4 .1. 3 (the Chain Rule) to lEn. Theorem 10.3.1 (The Chain Rule) SupposeD~ lE" and g: D --7 lEm is differentiable at x 0 E D. Suppose f: g(D) --7 JEP is differentiable at g(x0 ). Then (fog) is differentiable at xo and (fog)' (xo) = f' (g(xo))g' (xu), a composition of linear transformations. Proof: To show the composition f o g is differentiable at xo, we denote

k = g(xo +h) - g(xo)

--7

0

THE CHAIN RULE IN EUCLIDEAN SPACE

299

as h ___. 0 since g is continuous at x 0 by Theorem 10.2.2. Also, k = g'(xo)h + E"(h),

1

where 11f (~~11 ___. 0 ash___. 0. Next we observe that ~f

= f(g(xo +h))- f(g(x 0 )) = f'(g(x 0 ))k + E(k),

where ~ llkll ___. 0 as k ___. 0 . Therefore ~f

= f'(g(xo))g'(xo)h + f'(g(x 0 ))£(h) + E(g'(x0 )h + E"(h))

Denoting ~(h) = f'(g(x 0 ))£(h) ll~(h)ll llhll

+ E(g'(x0 )h + E"(h)) it suffices to show that

llf'(g(xo))E"(h)

+ E(g'(xo)h + E"(h))ll

llhll < llf'(g(xo))E"(h)ll + IIE(g'(xo)h + f.(h))ll llhll llf'(g(xo))f.(h)ll IIE(g'(xo)h + f.(h))ll 0 = llhll + llhll -----

as h ___. 0. Since llf'(g(xo))E"(h)ll < llf'(g(xo))IIIIE"(h)ll ___. 0 llhll llhll as h ___. 0, it suffices to show that IIE(g'(xo)h + E"(h))ll IIE(k)ll llhll =

lfhll ----- 0

ash ___. 0. Thus it suffices to show that if rJ > 0, then for all sufficiently smallllhll we have IIE(k)ll < 'fJIIhll· Since 1 > 0 there exists 81 > 0 such that llhll < 81 implies llf.(h)ll :S 1llhll, which implies in tum that llkll :S (1 + llg'(xo)ll)llhll· Also, if rJ

2

= 1 + llg'(xo)ll'

then there exists 82 > 0 such that llkll < 82 implies

300


Thus it suffices to pick h such that

since then

• 10.3.1

The Mean Value Theorem

In one-variable calculus, the Mean Value Theorem (Theorem 4.2.3) plays a very important role. In Exercise 10.49 the reader will see that a direct adaptation of that theorem to vector-valued functions is not possible. However, the following version of the theorem is true and is useful. Theorem 10.3.2 (Mean Value Theorem) Suppose f : D --t !Em is a differentiable function, where D is a convex subset of lEn, as defined in Exercise 9.43. Suppose M = supxED jjf'(x)jj < oo. Then,forall aandb in D,

llf(b)- f(a)ll ::; Mllb- all. Proof: Observe that the straight-line segment

S = {a+ t(b- a) I 0 ::; t ::; 1} ~ D because Dis convex. Define a differentiable function¢: [0, 1]

--t

JR. by

¢(t) = (f(b)- f(a)) · f(a + t(b- a)) for all t E [0, 1]. By the ordinary Mean Value Theorem from one-variable calculus together with Exercise 10.45.b, we obtain

llf(b)- f(a)l! 2 = ¢(1)- ¢(0) = ¢' (f)(1 - 0) = (f(b)- f(a)) · (f' (a+ f(b- a)) (b- a)) ::; llf(b)- f(a)l!l!f'(a + l(b- a))l!l!h- all ::; l!f((b)- f(a)IIMI!b- all, where the inequality comes from the Cauchy-Schwarz inequality combined with the Chain Rule and the property of the norm of a linear transformation. Therefore

l!f(b)- f(a)l! ::; Mllb- all

• The following corollary is an immediate consequence of the proof of the Mean Value Theorem.

THE CHAIN RULE IN EUCLIDEAN SPACE

301

Corollary 10.3.1 Suppose f : D ----> lEm is a differentiable function. Suppose x and y E D and suppose the straight-line segment L between x andy lies in D. Suppose M = supwEL llf'(w)ll < oo. Then

llf(x)- f(y)ll :::; Mllx- Yll· Remark 10.3.1 The corollary differs from the Mean Value Theorem in that D does not need to be convex. Instead, x andy are special points for which L lEm, with D ~has a local extreme point at x E D, provided that there is a number r > 0 such that f(x) is either the largest or the smallest value achieved by f in the open ball Br(x). Theorem 10.3.3 (Local Extreme Point) Suppose that D ~ has a local extreme point at X E D 0 , the interior of D. Iff is differentiable at X, then J'(x) = 0 E .C {lEn,lE1). Proof: Let v E lEn \ {0}. Define ¢(t) = f(x +tv). Thus ¢ : ~ ----> ~has a local extreme point at t = 0. By the first derivative test from one-variable calculus, combined with the multivariable chain rule, we see that ¢'(0) = f'(x)v = 0 for all nonzero vectors v. Thus f' = 0 E .C {lEn, lE 1 ). Note that the latter condition is • equivalent to the condition that ~ = 0 for all i = 1, ... , n.

10.3.2 Taylor's Theorem Next we will generalize Taylor's Theorem (Theorem 4.6.1). For this purpose we restrict our attention to functions f : D ----> ~. where D is an open, convex subset of lEn. Note that for such functions f'(x) is a 1 x n matrix, and its one row vector is the gradient of f, denoted by

We denote by CN + 1 ( D, ~) the set of all functions f : D ----> ~ such that f and each of its derivatives of order up to N has in tum a continuous derivative. (See Exercise 10.53 for elaboration of this concept, including its equivalence to the continuity of all partial derivatives of order up to and including N + 1.) We will let k denote an arbitrary n-tuple of nonnegative integers, and we will denote by lkl = E~=l ki. We will denote (1k1+··+kn J 0 ikiJ axk

= 8x~ 1

0

0

0

ax~n

302


and

Theorem 10.3.4 (Taylor's Theorem in lEn) Let f E CN+l(D,JR), where Dis an open, convex subset of !En. Let a and b E D. Then there exists a point J.l on the straight-line segment between a and b such that f(b)

=

(1 0.3)

where

~ (N: I)! (~(b;- a;) 8~;) N+> fix="·

RN(b)

Equivalently, we can express these formulas in the form

alkl f

f(b) =

(b- a)k

L 8xk (a) k1!k2! ... k ! + RN(b), lki:S:N

(10.4)

n

where

Proof: Define a function¢ : lR ~ lR by ¢(t) = f(a + t(b- a)) = (! o '1/J)(t), where '1/J(t) = a+ t(b- a) for all t E [0, 1]. Observe first that in matrix form

[¢'(t)] = [f'('!j;(t))]['l/J'(t)] =[/f;('l/J(t))

=

...

8 ((bl- al)-8 x1

[b1~a1]

Jt('l/J(t))J

bn- an 8 + ... + (bn- an)-8 ) x1

II . 1/J(t)

Repeating this argument k times for each of then summands in the last line above, we see that ¢{k)(t) =

(~(b·- a-)~) II 8x· ~

J=l

k

J

J

J

.p(t)

The rest of the proof of formula (10.3) is a matter of applying Taylor's Theorem (Theorem 4.6.1) for one variable to ¢(t). In order to prove formula (10.4), we expand the powers of the differential operator in formula (I 0.3), using the expansion given by the so-called multinomial theorem:

EXERCISES

303

where the coefficients

are the multinomial coefficients of degree K. (See [2].) However, in order to carry out this expansion, we need to know Clairaut's theorem. This theorem, which is proven as Exercise 11.46, assures under the hypothesis of continuity of the partial derivatives that there is independence of the order of composition of the differential operators. That is, continuity of the partial derivatives assures that

/J2f

/J2f

DxiDXj

DxjDxi ·

This permits the regrouping and collecting of the like partial derivatives in the expansions utilized above. • We remark that sometimes it is useful to know that the sum of the multinomial coefficients of order N is

This follows from the fact that the sum represents the number of ways to select N things from a set of n things. Equivalently, it is the number of elements in the set of functions from a set of N elements to a set of n elements. EXERCISES

10.44 Suppose f g(O) = Xo,

IE 2

-+

IE 2 and g : IE3

g'(o) = (

-+

IE2 are both differentiable. If

_!2 -,} ~ ) '

and

f'(xo) = (

~ ~1

) ,

find the matrix [(f o g)'(O)] using the standard bases. Suppose that T E .C(!Em, JEP) and that f : !En

10.45

-+

!Em is differentiable.

a) Prove that T o f is differentiable at each x E !En and that

(To f)'(x) =To f'(x). b)

10.46

t If a E !Em is a constant vector, prove that (a· f)'(x) exists and equals a· f'(x). Suppose f E C1(IE3, JE 2) and the matrix

[f'(O)] = (

1 g ~ ).

304


If (a, b) E JE2 , express the matrix [( (a, b) ·f)' (0)] in terms of a and b.

10.47 Suppose f: D----* lEm is differentiable at x E D and suppose there exists an inverse function g : f(D) ----* D that is differentiable at f(x). Find the relationship between f' and g'. 10.48 Suppose y = g(x) is differentiable in an open ball around x E lEn. Suppose z = f(y) is differentiable in an open ball around g(x), where g : lEn ----* lEm and f : lEm ----* JEP. Apply the chain rule to the differentiable function f o g : lEn ----* JEP to compute ~ for alll ::; i ::; p and all 1 ::; j ::; n. 3

10.49 t Letf: JE 1 ----* JE 2 bedefinedbyf(t) = (cost,sint). Showthattheredoes not exist a point l E [0, 211"] for which

f(27r)- f(O) = f'(l)(27r- 0). 10.50 Suppose that f'(x) exists and that llf'(x)ll is bounded on a convex set D ~ lEn, where f : D ----* JEm. Prove that f is uniformly continuous on D, as defined in Exercise 9.41. 10.51 Suppose f : E 1 ----* E2 by f(x) = (cosx, sin x). Prove or give a counterexample: For all real numbers a and b we have llf(b)- f(a)ll

:S: lb- ai.

10.52 Suppose f'(x) exists for all x in a nonempty open set D ~ lEn, where f: D----* Em. 0 E C(En, Em), prove that f is a constant a) SupposeD is convex. Iff'(x) function on D. b) Suppose next that Dis an open connected set in En, but not necessarily convex. If f'(x) := 0 E C(En,Em)

=

on D, prove that f is a constant function on D. (Hint: Show that if x E D, then there exists an r > 0 such that f remains constant on Br(x). Show then that the set r- 1 (c) is an open subset of En for each c E Em.)

10.53

t Let D be an open subset of En and f E C1 (D, Em). Thus f': D----* C(En, Em).

We can denote f'(x) =

~ ;:,; (x)Tij, t,J

where Tij is defined in Exercise 10.13. We say f E C 2 (D, Em) if and only iff' has a continuous derivative. Using {Tij} as a basis of C(En, lEm) to show that f E C 2 (D, Em) if and only if a::£~k exists and is continuous for all i, j, and k.

INVERSE FUNCTIONS

10.54

305

n

Let D = {X E lE 2 IIIxll > 1} {X E lE 2 1 X2 # 0 if Xi ~ 0}. a) Show that D is open, but not convex. b) Apply Exercise 5.62 to showthatthereexistsg E c=(JR) such thatg(O) := 1 if 0 ;::: g( 0) = -1 if 0 ~ and g' (0) is bounded on R c) Let f : D---+ JR. be defined by

i,

i,

g (tan-i

f(x) = {

(~))

if x E D with

xi

1

if x E D with xi

-1

if X E D with

> 0, ~

Xi ~

0, x 2 > 0, 0, X2 < 0.

Show that f E Ci(D) and that f' is bounded on D. < oo such that

d) Show that there does not exist M

lf(x)- f(x')l

~

Mllx- x'll

identically on D.

10.55 Let f : lE2 ---+ JR. be defined by f(x) = sin(xi + x 2 ). Selecting a = 0 in Taylor's Theorem, use either formula (10.3) or the expanded version to prove that RN(b)---+ 0 as N---+ oo. 10.56 The following exercise works in any vector space equipped with a scalar product, but the reader may let f E lEm and let {ek I k = 1, ... , n} be an orthonormal subset of lEm, where n ~ m. Note that (ej, ek) = 0 if j # k but equals 1 if j = k. Prove that

has an absolute minimum value on lEn, which is achieved by selecting ak = (f, ek) for all k = 1, ... , n. Hint: Apply Theorems 10.2.5 and 10.3.3 to

We remark that this exercise can be applied usefully in function spaces such as R[a, b], giving rise to the standard formulas for Fourier coefficients and also coefficients of other orthogonal-function expansions.

10.4 INVERSE FUNCTIONS Even for differentiable functions f : JR. ---+ JR., there is no need for f to be invertible. And if the function f has an inverse, there is no need for that inverse to be differentiable. For example, f(x) = x 2 is continuously differentiable on JR, yet it is not one-to-one and hence has no inverse. Another good example is this one: Let f(x) = x 3 . Now f is one-to-one on JR., but f-i(x) = ~.which has no derivative at the origin. On the other hand, suppose D is an open subset of JR. and suppose that

306


E C1 (D,R) and f'(a) =f. 0 at some a ED. By continuity off' we know that f' remains nonzero on some interval I = (a- 8, a+ 8) ~ D. By the Mean Value Theorem, the restriction f!I off to I is one-to-one and thus has an inverse. We will generalize this theorem to lEn and we will prove a theorem that shows under suitable hypotheses that the inverse exists and is also continuously differentiable on a suitably small open ball. We begin with the following theorem which provides an interesting contrast to the Mean Value Theorem (Theorem 10.3.2).

f

Theorem 10.4.1 (Magnification Theorem) Let f E C1 (D, lEn), where Dis an open subset of !En. Ifx E Dis such that dct f'(x) =/= 0, then there exists an open ball Br(x) and a constant a> 0 such that for ally and z in Br(x) we have

llf(y)- f(z)ll ~ ndiY- zll. Consequently, f is one-to-one and thus invertible on Br(x). Proof: Begin by denoting T = f' (x) E .C(lEn, lEn) c C1 (lEn, lEn). Observe that T is invertible and that both T and r- 1 have strictly positive norms. For ally and z in lEn we have IIY- zll

=

IIT-I(T(y)- T(z))ll:::; IIT-liiiiT(y)- T(z)ll.

Letting a = 21 1'j_ 111 , we have IIT(y)- T(z)ll ~ 2ally- zll. Suppose we could show that for sufficiently small r > 0 and for all y and z in Br(x) we have ll(f(y)- f(z))- (T(y)- T(z))ll ally- zll since if that were not true, then it would follow that IIT(y)- T(z)ll < 2ally- zll, which is false. Hence we will know that f is both one-to-one and invertible on Br(x) once we have proven Equation 10.5. However, for all y and z in Br (x) we have ll(f(y)- f(z))- (T(y)- T(z))ll

= ll(f- T)(y)- (f- T)(z)ll :::;MIIy-zll

where M =

sup wEBr(x)

ll(f- T)'(w)ll =

sup wEBr(x)

llf'(w)- f'(x)ll.

INVERSE FUNCTIONS

307

Here we are using the Mean Value Theorem 10.3.2 together with Exercises 10.30 and 10.32 for the basic properties of differentiation, as well as the convexity of a ball. Since f' is continuous, we can choose r > 0 so as to insure that M < a. • The next theorem will complete our preparation for the Inverse Function Theorem. Theorem 10.4.2 (Open Mapping Theorem) Let f E C1 (D, lEn), where Dis an open subset ofJEn. Suppose that for all xED we have det f'(x) =/:- 0. Then f(D) is an open subset oflEn.

Proof: Let x E Dandy = f(x). We seek an open ball Bp(y) 0 such that f is one-to-one even on a suitably chosen closed ball Br(x) · · · ,ym because of the block-triangular form of the matrix. Also, it is easy to see that f E C1 (D, JEn+m) and that the determinant of its derivative is a continuous function. By the Inverse Function Theorem together with the Magnification Theorem, there exists an open ball of the form

on which the Jacobian remains nonvanishing and on which f has a continuously differentiable inverse. Applying Exercise 10.83 we see that f(Br(x 0 ) x Br(Yo)) is open in JEn+ m. Thus there exists p > 0 such that

Thus for each x E Bp(Xo) there exists a unique y E Br(Yo) such that

(x, y) =

£- 1 (x, 0).

IMPLICIT FUNCTIONS

313

=

If we denote g(x) = y, we see that f(x, g(x)) 0. In the notation of the theorem, we can take U = Bp(Xo) and V = Br(Yo), the Cartesian product of these two sets • being in Br(Xo,Yo). Remark 10.5.1 If one is presented with an m x (n + m) matrix, it is difficult to see by inspection whether or not there exist m column vectors from this matrix that together would form a m x m matrix with nonzero determinant. Thus it is hard to judge by inspection whether or not the conditions of the implicit function theorem are satisfied. It is helpful to review the concept of rank from linear algebra. The rank(T) of a linear transformation T : V --+ W is dim T(V), the dimension of the image ofT. T can be represented as a matrix in many ways, depending on the chosen bases for V and for W, the domain and range vector spaces. But for any matrix, it is not hard to prove that the rank of the corresponding linear transformation is the dimension of the span of the set of all the column vectors of the matrix, which is called the column rank of the matrix. There is also a concept of row rank, which is the dimension of the span of the set of all the rows of the matrix. It is a fundamental theorem in linear algebra that rank(T), the column rank of [T], and the row rank of [T] must all be the same number. If the row rank or the column rank of [T]mx(n+m) is equal tom, then there must be m linearly independent columns and the student will then be able to look for those columns, whose determinant will yield the needed nonzero Jacobian as in the implicit function theorem. The row rank will be m if and only if the set of all m rows of the matrix is linearly independent. We remark also that the open sets U and V can be challenging to identify in any given example, unless it is a relatively simple example such as one in which it is easy to solve for the implicitly determined function explicitly. The delicacy of the problem is reflected in the difficulty of establishing the existence of U and V in the proof of the implicit function theorem.

Let us consider in detail an example in which we apply the implicit function theorem to a specific real-valued function of three real variables . • EXAMPLE 10.4

Let F : IE 3

--+

lR by

F(x)

.'!.

=

.'!.

.'!.

xf +xi +xj -1.

For which points on the surface defined by F(x) = 0 is it true that at least one variable is determined as a continuously differentiable function of the others, restricted to suitable open sets? Fig. I 0.5 shows just the upper half of this surface. The reader can see in the figure that the entire surface defined by F(x) = 0 has twelve sharp edges and six sharp points, or cusps. We will begin our analysis of the problem of this example by seeing what information we can get from the implicit function theorem. To apply this theorem, we need to have

314


~

£

Figure 10.5 Upper half of xt +xi

~

+ xg = 1, X3 2: 0.

F E C1. But with respect to the standard basis, the matrix of the derivative of Fis [F'(x)] = ~ [x~!,x;l,x~!], provided that x 1x2x3 =/=- 0. One sees that F' exists only in the open set D = {x I x1x2x3 =/=- 0}, which is the union of the eight open octants of three-dimensional space which exclude the three coordinate planes. Moreover, FE C1 (D, E 1 ). Since g~ =/=- 0 on D for any i = 1, 2, 3, each of the variables can be expressed as a C1 function of the other two, in suitably restricted open sets. For example, suppose F(xo) = 0, where x 0 = (a, b, c). If we solve for x 3 in terms of the first two, we can restrict (x1, x 2) to the open set U, which is that part of the open quadrant of the x1x2-plane to which (a, b) belongs and is strictly inside the curve 2

2

x[ +xi= 1. In order to insure the existence of a unique continuously differentiable solution for x3 over the domain U chosen as above, we restrict x 3 to the open set V = (0, 1) or else V = (-1, 0), choosing the interval to which c belongs. The reader sees that uniqueness of the solution is a delicate matter here, since if (xi. x2, X3) is on the full graph, then so is (xi. x2, -x3), which wrecks uniqueness unless x 3 is suitably restricted.

315

IMPLICIT FUNCTIONS

Finally, we mention that we have not shown yet that unique C1 solutions fail to exist if at least one of the three coordinates is zero. That information does not come from the implicit function theorem, but depends on elementary calculations, which we leave to Exercise 10.76. We can translate the Implicit Function Theorem into a statement about the partial derivatives of the component functions gi of the implicitly determined function g(x). Since f(x, g(x)) = 0 E lE"' for all x E U 0 there exists a partition P for which U(f, P)L(f,P) < t. The proof is left to Exercise 11.4.

Theorem 11.1.4 The set R[a, b] is a vector space, and the map T: R[a, b] defined by T(f)

= JB f(x) dx is linear.

-t

JR.

The proof is left to Exercise 11.5.

Theorem 11.1.5 Iff E C[a, b], then f E R[a, b]. Proof: Since f is continuous on the block B = [a, b], which is both closed and bounded, we know that f is uniformly continuous on B. Thus if f. > 0, there exists a 8 > 0 such that llx- x'll < 8 implies that

lf(x)- f(x')l < 2 p~B), where we assume p(B) > 0. Now if we partition B with a finite set of blocks P = {Bi I i = 1, ... ,p} with IIPII < 8, it follows that on each block Bi we have IMi -mil ::;

2p~B) ·

Hence

U(f, P)- L(f, P)::;

f.

"2 < t.

DEFINITION OF THE INTEGRAL

It follows from the Darboux Criterion that

f

E

R(B).

335

•

We tum our attention now to the definition of the Riemann integral for functions defined on any bounded subset Dt C lEn. Then the domain Dt is contained in some sufficiently large rectangular block [a, b]. We would like to make the following definition. Definition 11.1.6 Suppose that the domain D f of a function f is contained in some sufficiently large rectangular block [a, b]. We will extend the definition off so that f(x) = 0 at each point x E [a, b] \ D f· We define

J

f(x) dx

D1

= {

J[a,b]

f(x) dx,

provided that this integral exists.

=

Theorem 11.3.1 will show that for some sets D, even f(x) 1 is not integrable on D. Also, for Definition 11.1.6 to make sense, we must show that it is independent ofthe choice ofblock [a, b] :;2 Dt. For this purpose, it suffices to prove the following theorem. Theorem 11.1.6 Let f be any bounded real-valued function on a bounded domain Dt C lEn. Suppose Dt ~ [a, b] and Dt ~ [a', b']. Extend the definition off to be zero identically on the complement of D f in each block. Then

{

J[a,b]

f =

y- f J[a' ,b']

and {

J[a,b]

f = {

J[a' ,b']

f.

Proof: Let [A, B] be any closed rectangular block that contains both [a, b] and [a', b']. It will suffice to prove that

la,b/ = lA,B/

and

la,b/ = lA,B/

The two arguments are so similar that we will present a proof of only the first. Let E > 0. There exists a partition P of [A, B] for which

U(f,P)- f f
0. Then the set E ={xED I o(f, x) ~ 8} is a closed set. Proof: Suppose Xn E E for all n E N and suppose Xn -+ x as n -+ oo. We need to prove that x E E. For each t: > 0 there exists Xn E B,(x), and there exists r > 0

such that Br(xn)

~

B,(x). Also, sup

f(y) -

yEDnBr(Xn)

f(y)

inf

~

8.

yEDnBr(xn)

However, this implies that sup yEDnB,(x)

f(y) -

inf

f(y)

~

8,

yEDnB.(x)

which completes the proof since this is true for all t: > 0.

•

The following theorem is useful when studying the Jacobian Theorem for changes of variables in Riemann integrals on lEn.

EXERCISES

341

Theorem 11.2.4 Let 0 C lEn be an open set, B a closed block in 0 and let 4> be in Cl(O,lEn). (i) If S C B 0 is a Jordan null set, then 4>( S) is also a Jordan null set. (ii) If S C B 0 is a Lebesgue null set, then 4>(8) is also a Lebesgue null set.

Denote M = ll4>'llsup < oo and let E > 0. We begin by proving item (i). By hypothesis, there exists a finite sequence of closed blocks B1, ... , B N such that

Proof:

N

Be

U B'k k=l

N

and LJ1(Bk) < k=l

:

n·

2M n2

Note that the intersection of two rectangular blocks (with edges parallel to the axes) must again be a rectangular block. Thus without loss of generality we can assume that each block Bk ~B. It is easy to see that any closed rectangular block R is contained in a union of finitely many cubes for which the sum of the volumes of the cubes is less than twice 11(R). This is true even though the edges of R need not be commensurable, because we can slightly enlarge the edges of R to make all the (extended) edges commensurable. Thus, without loss of generality, we can assume that each block Bk used above is actually a cube. If we denote by Dk the diagonal measurement of Bk and sk its edge-length, then Dk = SkVn and

JL(Bk) =

Dn

!.

n2

On the other hand, by Theorem 10.3.2, each individual coordinate function of the image (Bk) is limited to at most an interval of length M Dk. Hence (Bk) is contained in a cubeB~ such that Jl(BD :::; Mnn ~ 11 (Bk)· By slightly enlarging the cubes B~ we can assure that 4> (Bk) C (B~t and that

Jl(BD:::; 2Mnn~J1(Bk)· Thus

N

4>(8)

~

N

U 4> (B'k) c U (B~t k=l

k=l

N

and Lll (BD

JR., D --->

~

lEn.

JR., D

~

!En, is the

U {xED Io(f, x) 2: ~}. nEN

11.28

Prove that every subset of a Lebesgue null set is a Lebesgue null set.

11.29 t Suppose that the set E of points of discontinuity off : D ---> JR., D ~ IE", is a Lebesgue null set, and suppose that D is a bounded subset of lEn. Let

Ek = {xED I o(f, x) 2:

~}.

Prove that Ek is a Jordan null set. (Hint: Use Theorem 11.2.3.)

11.3

LEBESGUE'S CRITERION FOR RIEMANN INTEGRABILITY

The main objective of this section is to prove the following theorem. Theorem 11.3.1 (Lebesgue's Criterion for Riemann Integrability) Let f : B ---> JR. be any bounded function defined on a closed, bounded rectangular block B c lEn. Let S = { x E B I f is discontinuous at x }. Then f E R(B) if and only if S is a Lebesgue null set.

Proof: First suppose that f E R(B). We will prove that Sis a Lebesgue null set. By Exercise 11.27 we know that s = ukEN sk. where

Sk

= {x

I

E B o(f, x)

2:

~} .

LEBESGUE'S CRITERION FOR RIEMANN INTEGRABILITY

343

By Exercise 11.16, we see that it suffices to prove for each k E N that Sk is a Lebesgue null set. By Exercise 11.22, it suffices to prove that Skis a Jordan null set. Let£ > 0. There exists a partition P = { B1, ... , Bp} of B for which

U(f,P)- L(f,P) < Denote

E

k'

p

P'(~t)

UBf i=l

and observe that B \ P' is a Jordan null set, being a union of degenerate blocks, and so is Sk \ P'. Thus it suffices to prove that

is a Jordan null set. If x E S~, then x lies in the interior of exactly one block Bi of the partition P, and on that block we have Mi-mi ~ o(j, x). Hence

U(f, P)- L(f, P)

~~

I:

JL(Bi)·

B;nS~-10

It follows that

I:

JL(Bi)
0. We need to show there exists a partition P of B such that

U(f, P) - L(f, P)
4JL(B). E

Without loss of generality, suppose llfllsup > 0. Since Sk is a Jordan null set, there exists a finite set of blocks B1, . .. , Bt such that Bf n B'j = 0 if i =I j

Hence l

I:(Mi- mi)JL(Bi) < i=l

~·

344

RIEMANN INTEGRATION IN EUCLIDEAN SPACE

Let

i=l

be a compact set. For each x E K we have

o(f,x)
JE 3 be the affine transformation defined by

T(x1, x2, x3) = (x1 +a, x2 + b, X3 + c + ax2) where a, b, and care real constants. Find the volume of T(B 1 (0)), where B 1 (0) is the unit ball around the origin. 11.53 Prove that the change of variables theorem remains valid even if the transformation g has the property that g' vanishes on a Jordan null set. (Hint: What can you say about g(S), if Sis a Jordan null set?)

356


11.54

Let B1 ( 0) denote the closed ball of radius 1 centered at 0 E lEn. Let

Z -11 1112 dx,

In =

lth(O)

X

which we define by

In= lim { r->O+ J{xEJEnlr:-;llxll:9}

~dx. llxll

For parts (a) and (b) below, use polar or spherical coordinates. a) Prove that In = oo if n = 1 or if n = 2. b) Prove that I3 < oo. c) Introduce spherical coordinates in lEn as follows. Let p =

n =cos

-1 Xn

-

p

l!xl!. Let

and Pn-1 = psinn

for 0 ~ n ~ 1r. Let

n-1 =cos for 0

~

n-l

~

-1 Xn-1

- - and Pn-2 = Pn-1 sinn-1

Pn-1

1r, ... , and let

A. X3 . A. '1'3 =cos -1 , an d P2 = p3sm'l'3 P3

for 0 ~ ¢3 ~ 1r. Let x1 = P2 cos(), x2 = P2 sin(), with 0 ~ () Prove that

< 21r.

o(x1, ... ,xn) o(p,0,¢3,····n)

is pn- 1 times a polynomial in the sines and cosines of(), ¢ 3, ... , n· d) Prove that if n > 3, then In < oo.

11.55 Let g E C1((a,b),IE 1) be nonsingular(meaning g' is nowhere zero) and let S ~ [c', d'] c g(a, b) be any bounded subset of g(a, b). Let f be any bounded real valued function on S. Prove:

rf(x) dx r- f(x)!g'(x)l dx.

ls

=

}y-l(S)

(Hint: Interpret f as being identically zero off S. Consider any partition P of [c', d'] = g[c, d], and compare upper sums over both domains.)

11.56 Fix any a E lEn and define T E £(!En) by T(x) = x +a. If B is any closed finite box in lEn and iff E R(B), then f E R(T- 1B) and

{ f(x) dx = {

jB

lT-lB

f(Tx) dx.

TEST YOURSELF

357

That is, the Riemann integral on !En is invariant under all translations.

11.57 Denote by O(n) the set of all n x n matrices A for which AAt = I, the identity matrix, where At denotes the transpose of the matrix A. Prove: a) 0( n) is a group under the operation of matrix multiplication. This is called the orthogonal group on !En. b) I det AI= 1 for all A E O(n). c) If B is any closed finite box in !En and iff E R(B), then f E R(A -l B) and

{ f(x) dx = {

jB

}A-lB

f(Ax) dx.

That is, the Riemann integral on !En is invariant under orthogonal transformations.

11.6 TEST YOURSELF

EXERCISES 11.58

Give an example of a function

f :B

---+ JR. for which

~~ =0 and~~= 1, where B = [0, 1]x 2 •

11.59

Give an example of functions

f

and g mapping B ---+ JR. such that

where B = [0, 1]x 2 •

11.60 Give an example of a function f : B ---+ JR. for which integrable but f- is Riemann integrable, where B = [0, 1] x 2 . 11.61

f

is not Riemann

Let the linear map T: R[a, b] ---+JR. be defined by

T(f) = {

f(x) dx.

J[a,b]

FindiiTII·

11.62 Let f = 1Qn[O,l]• the indicator function of the set of all rational numbers in the unit interval of IE 1 . Find the set of points of discontinuity off. 11.63

LetS= { x E IE 2 1 x1 ~ IQI, x 2 ~ IQI}. IsS Jordan measurable?

11.64

True or False: The set { ( -1)n +~In EN} is a Jordan null set in IE 1 .

358


11.65

Let

f(x) =

{sin~' 0,

X

X~

0,

=0.

Find the oscillation o(f, 0). 11.66 Find the Jordan measure of the set of points of discontinuity of the function shown in Fig. 11.1. 11.67

Suppose F(x 1 ,x 2 )

= f(x 1 )1Q(x2)·

Find all functions j(x1) for which

FE n[o, l]X 2 • 11.68

Let f(x) = ( -x, -y, xy- z) for all x E IE 3 . a) Find the matrix [f'(x)]. b) Find the volume of the image of a sphere of radius 2 under the function f in this exercise.

APPENDIX A SET THEORY

A.1

TERMINOLOGY AND SYMBOLS

When the author was young, it was easy for a professor to know whether or not students were familiar with set-theoretic symbols and language from high school. The answer was no. Thus professors took care to explain what these terms mean. Today it is less clear. High-schools teach some set theory, but maybe not very much. And since students seldom get to use this language in their high-school problem solving, they may not have developed facility with it. The same is true with introductory college courses. There is likely to be some use of set theory, but it is not clear what background is common to all students. The formal study of set theory refers particularly to the study of infinite sets (also called transfinite sets), and the reader is referred to [11] for a deep and serious study of that subject. Here we begin by defining the commonly used symbols that appear in this text and giving a few illustrative examples and theorems.

Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright © 2008 John Wiley & Sons, Inc.

359

360

SET THEORY

For reasons that we will sketch briefly in Example A.7, logic requires us to limit the frame of reference in any set-theoretic discussion to some universal set, which is chosen for convenience or suitability to a given purpose. In the study of sets in the abstract, this universal set is commonly designated by the letter X. It is common to designate a subset A of X, denoted by A C X by writing that A is the set of all elements of X that possess some property. • EXAMPLE A.l

Let A = {x E R I -1 :::; x :::; 1}. This is read A is the set of all those real numbers x that have the property that x lies between -1 and 1 inclusive. It is common to write the set A in interval notation for the real numbers as

A= [-1, 1]. Some authors use the symbol A c X to mean proper subset, which means that A is contained in X but A is not equal to X. The author of this book will usually write A s;; B for this however. If it is important to stress the possibility of a subset of X being equal to X itself, the author will often write A ~ X. Two very important operations between sets are union and intersection.

Definition A.l.l The union of A and B, denoted by A U B is the set of all x E X such that either x E A or x E B. This can be written as AU B = {x EX I x E A or x E B}. The intersection of A and B is

An B = {x EX I x E A and x E B}.

Note that in mathematics we use the word or in the inclusive sense, not in the exclusive sense. Thus x E AU B, provided that x E A, x E B, or both . • EXAMPLE A.2

Continuing Example A.l, in which A = [-1, 1] c R we can define B = { x E R I 0:::; x < 2} = [0, 2). Then we have AU B = [-1, 2) in interval notation, and An B = [0, 1].

Definition A.1.2 If A n B = 0, the empty set, we call A and B disjoint sets. Another important operation is the difference between two sets.

Definition A.1.3 The difference between two sets A and B, denoted by A\ B, is the set of all those elements of A that are not in B.

TERMINOLOGY AND SYMBOLS

•

361

EXAMPLE A.3

Continuing Example A.2, we have A \ B

= [-1, 0).

Definition A.1.4 Within the context of a universal set X, we denote X\ A = Ac, which is called the complement of A. Thus the complement of a set has meaning only within the context of a previously specified universal set. •

EXAMPLE A.4

Continuing Example A.2, the set Ac = ( -oo, -1) U ( 1, oo ), provided we are working with reference to the universal set R A different but sometimes useful concept of subtraction of two sets is the following.

Definition A.l.S The symmetric difference of two sets A and B is denoted and defined as follows: Atl.B =(A\ B) U (B \A) .

•

EXAMPLE A.5

In terms of Example A.2, Atl.B

= [-1, 0) U (1, 2).

Theorem A.l.l Atl.B =(AU B)\ (An B). Proof: This proof illustrates a frequently useful approach to proving that two sets, Atl.B and (AU B)\ (An B), are the same. The plan is to show that each of the latter two sets is a subset of the other. We will begin by proving that

A6B

~

(AU B)\ (An B).

By Definition A.l.S, if x E Atl.B, either x E A but x rJ. B, or vice versa. (Observe that in this case the union that defines Atl.B does not permit x to be in both sets.) If x E A butx rj. B, thenx E AUB butx rj. XnB. Hencex E (AUB) \(An B). On the other hand, ifx E B butx rJ. A, we can still concludethatx E (AUB) \(An B). Therefore, A6B ~ (Au B)\ (An B) as claimed. We leave it to the reader to verify by similar reasoning that

A6B

;:;:>

(AU B)\ (An B).

• In this book, we need to apply the concepts of union and intersection to infinite families of sets-that is, to infinite sets of sets. We will clarify this with an example.

362

SET THEORY

• EXAMPLE A.6 Suppose to each x E lR we associate an interval Ax = (x - 1, x + 1). If S c lR we can define the union over x E S of the sets Ax to be the set of all those real numbers that lie in at least one of the sets Ax with x E S. Thus in the example

U

IX

Ax= U{Ax

E

[0,3]} = (-1,4).

xE[0,3]

Similarly, the intersection of a set of sets is the set of all those elements that belong simultaneously to every one of the sets. Continuing the example of this paragraph, we have

n

Ax= (0.5, 1).

xE{0,1.5)

TheoremA.1.2 (DeMorgan'sLaws)Supposeforeachs E S, asetofindices, there is associated a set As ~ X. Then i.

(u

As)c =

sES

nA~

sES

ii.

Proof: To prove the first equality, we will prove that the set that is written as the left side is contained in the set that is the right side. Then we will prove the reverse inclusion, thereby establishing equality. So suppose that x E (UsES As This tells us that for all s E S we have x ~ A..,. This tells us in tum that x E A~ for each s E s. Consequently, X E nsES A~. which establishes that

r,

Toprovethereverseinclusion, supposethatx E nsES A~. This means thatx E A~ for each s E S. Hence x ~ As for any s E S. Thus x E (UsES Ast This tells us that

which completes the proof of equality of the left and right sides of the first law. We leave the second law to the reader to prove as Exercise A.l. •

EXERCISES

363

EXERCISES A.l

Prove that

A.2 naEA

If a set Ea is a closed subset of lR for each a E A, an index set, prove that Ea must be a closed set.

A.2

PARADOXES

We tum next to a brief introduction to some of the logical problems that can arise in set theory and in the logical use of language . • EXAMPLE A.7

We explain in this example why it is necessary to fix a frame of reference, called a universal set, at the beginning of any logical discussion involving set theory. At first glance it might seem reasonable to define a set S = {A I A is a set}. Thus we appear to have defined an extraordinarily big set S that has as its elements every set that exists. In other words, A E S ~· Such N exist because of the Archimedean property of the real number system.

PROBLEM SOLUTIONS

367

1.111 We give a counterexample. The constant sequence Xn = 1 is convergent, and the sequence Yn = ( -1 )n is bounded. But XnYn is divergent. 1.112

0.

1.113 For example, we can let 0 = { 1.114 True, since Ec

(~, 1) n EN}· J

= UnEN ( n~ 1 , ~) U ( -oo, 0) U ( (1, oo ), a union of open sets.

SOLUTIONS FOR CHAPTER 2 2.4 Each real number x E JR. is a cluster point of the set Q because the set Q is dense in the set R 2.13.b The limit is nan- 1 . 2.25 c = nan- 1 . The reader should prove this claim. 2.45 f(x)

= x2

is uniformly continuous on (0, 1). The reader should explain why.

2.58.b For example, let f(x) 2.59.b

llfllsup =

2.64.a

llfnllsup =

=

tanx for all

x E ( -~, ~).

00.

e- 1 .

2.68.d The sequence of functions does not converge uniformly. The student should take care to distinguish this case from the one that precedes it. 2.71 The set is countable. Write Q = { ri functions !a,b as

Ii

E N} and arrange the set of all possible

UUr,,rj I j

EN}.

iEN

Then invoke the fact that the union of countably many countable sets is countable. 2.72 True. Pick a suitable open interval for each rational number that lies in the given open set. 2.73 Counterexample:

f(x) =

{~in~

if 0 <X< 1, if X= 0.

The indicated limit fails to exist and thus is not equal to any number L. 2. 74 The set JR. is the set of all cluster points of the set Q.

368

PROBLEM SOLUTIONS

2.75 True. Apply the Intermediate Value theorem to g(x) = f(x)-

v'x on [0, 1].

2.76 For example, let fn(x) = logx- ~· 2.77 For example, f(x) =sin~ on (0, 1). This was a homework problem.

2.78.a This is not a vector space because the difference between two such polynomials can have non-odd degree.

2.78.b Yes, this is a vector space. 2.79

11/llsup = 1.

2.80.a True, since f n converges pointwise on lR because f n ( x)

-+

0 for each fixed x

as n-+ oo.

2.80.b False since fn

-+

0 pointwise but

11/n- Ollsup =

oo and does not approach

0 as n-+ oo.

2.81 You can use the derivative from elementary calculus for this exercise, to deterfor all n E N. mine that 11/nllsup =

;e

2.82.a True. 2.82.b False. 2.82.c False. It is very important to distinguish this case from the preceding one. 2.83 For example, let Xn = 4 n~l. 2.84 We can use 8 =

~·

2.85 True, because this function can be extended to be continuous on the closed, finite interval [0, 1].

2.86 We give a counterexample. Let

f(x) = {x x+1 2.87 We can use 8 = 2.88.a True. 2.88.b True. 2.88.c False.

2 E

because

Iv'x - val

if X~ 0, if X> 0.

~

..;rx=ar.

369

PROBLEM SOLUTIONS

SOLUTIONS FOR CHAPTER 3 3.12 limn->oo ~ E~=l cos ( 1 +

2 :)

= J13 cos x dx = sin- sin 1.

3.17 For example, we could choose P = {0, 0.95, 1.05, 2}. 3.33.a fn(x) converges to 1 if and only if x E Qn [0, 1]. Proofs are left to the reader. 3.33.b fn(x) 3.35 limn---.oo

----t

0 for all x E [0, 1]. Proofs are left to the reader.

J12 1 + (1+;2 )n dx ----> 1. Proofs are left to the reader.

3.45 For example, let f be the indicator function of a single point. 3.56 For example, let g(x) 3.59.a We find that f(x) 3.59.b llfnllsup

= 2 }n

3.59.c Thus f n

---->

= csin x for some constant c =1- 0.

= 0 for all x

----t

0 as n

----t

E

R

00.

0 uniformly on R

3.60.a For the first interval the answer is yes. In fact lfn(x) - 01 n ----t oo, provided that -1 < x :::; 1.

=

lxln

---->

0 as

3.60.b For the second interval, the answer is no since the sequence ( -1 )n diverges. 3.61 We can pick 8

=f.

Then

II PII < 8 implies that IP(f, {xi})- 31 0

= 2sin(l).

P(x+3h)+P(x-3h)-2P(x) _ h2

-

gp"( X ) •

4.53 True. Let f(x) = x- ln(l + x). We see that f(O) = 0 and f'(x) > 0 for all x > 0. Thus the Mean Value theorem tells us that f is increasing and f(x) > 0 for all x > 0. 4.54 For such an example we can use

F(x) = { ~2sin

(;b)

if 0 <X~ 1, if X= 0.

4.55 False. Use any step function with distinct values on two contiguous intervals to give a counterexample. 4.56 True. This is the first version of the Fundamental theorem of the calculus. 4.57 This is true for any function with the given properties-not only those functions which are derivatives.

PROBLEM SOLUTIONS

371

4.58 For example, we can use

f( X ) = 4.59 True:

{2

° 1 2 1 xsm ::::2-- cos ::::2

X

X

X

0

if 0 < X :S 1, if X= 0.

liT II :::; 1. The reader could show as an additional problem that II Til = 1.

4.60 t = ~4.61 The limit is 16P"(x). 4.62 The limit is 0. 4.63 Yes, because

IITII

:S 1.

SOLUTIONS FOR CHAPTER 5

5.3 Let a1 = 1 and let an = .jTi -

Jn -

1 for each n > 1.

5.9.a Divergent. 5.9.b Convergent. 5.12.a Divergent. 5.19.a Hint: Use the Ratio Test together with the definition of e as a limit. 5.19.b Convergent. 5.20

i·

5.43.a Converges uniformly by theM-test. 5.43.c Not uniformly convergent on [0, ~). 5.49 The sum is In 2. 5.50.e The interval of convergence is I = (-e, e). The reader should prove divergence at the endpoints by applying the nth term test and using the hint. 5.54 j< 100) (0)

= 0 and j< 101 ) (0) = 100!.

5.70.a Absolutely convergent. 5.70.b Conditionally convergent. (_J\k+l

5.71 Let Xk =

~·for example.

372

PROBLEM SOLUTIONS

5.72 True. 5.73 False. This set of series is not closed under subtraction, for example. 5.74.a Diverges. 5.74.b Converges (absolutely). 5.75 Let Xk =

p and Yk = f• for example.

5.76.a Diverges, by the ratio test. 5.76.b Converges, by the comparison test. 5.77 False, because the series is absolutely convergent. 5.78 For example, let Xk =

Yk =

f,..

Then

The student should be able to write another such example.

5.81.a

Mk

= e- 2 k, so the series L:~

5.8l.b

Mk

= 1, so the series L:~ Mk diverges.

Mk

converges.

SOLUTIONS FOR CHAPTER 6 6.15 Forexample,letf(x) = ixforallx E answers.

[a,b]. Thereareinfinitelymanypossible n E Z \ {0}, n=O.

6.19.c J(n)

=

{i;n,

n

=

~

{ 0,

2,..n

0,

n=O.

2,

6.19.d J( n)

=/=-

'

n

=/=-

0,

n=O.

PROBLEM SOLUTIONS

j( n) =

6.19.e

{

~ ~,. 2 n 2 ' 12,

373

n "I 0, n=O.

6.35 Hint: Write each trigonometric function using Euler's formula, and multiple using the multiplicative property of the exponential function.

6.51.a

Jl f(x) dx = 1.

6.51.c

J01 f(x)sin27rxdx =

6.52.b

J01 f (x) sin 61rx dx = i.

6.53.a

4-.

2

0.

6.53.b i. 6.53.c ~· 6.54 -~. 6.55 False. The product is not linear in the second variable. 6.56.a

f( n) is both even and real-valued.

6.56.b

f( n) is both odd and imaginary-valued.

6.57.a f(x) = ~ cos61rx + ~ cos27rx. 6.57.b

l

6.58 True, because lim Sn is continuous, hence Riemann integrable. 6.59.a 6.59.b

,.4

90 . 81!"4

729

.


7.19 1. 7.21 6. 7.24 2.

374

PROBLEM SOLUTIONS

7.27 For example, we can use g = 1(p,b] if a::; p < b, or g = 1p if p =b. 7.30 For example, let g = 1{1 }.

7.41 This is false as the reader should be able to deduce from the fact that BV[O, 1] is a vector space.

7.42 For example, let

f(x)=

sin 1!: x' { 0,

X =/:-

X=

0, 0.

7.43 6. 7.44 False.

7.45 True. 7.46 0. 7.47 We can choose g(x)

= 2 · 1[1, 2], twice the indicator function of the subinterval

[1, 2] c [0, 2]. 7.48

J02 f dg =

3e.

7.49 We can take f = 1(a,b) and g = 1(b,c)· Then on the full interval [a, c] both functions have a discontinuity at b. But on [a, b] or the interval [b, c], only one function or the other one has a discontinuity.

SOLUTIONS FOR CHAPTER 8 8.7.a The limit is (0, e). 8.7.b The limit does not exist. 8.17.a Open. 8.17.c Neither. 8.17.e Closed. 8.42.a Not compact. 8.42.c Compact. 8.59 Any y perpendicular to x will suffice. For example: let y = ( 1, 1, -1). 8.60 True. Ifnotthereexists a sequence for which llx(k) II ---+ oo, and no subsequence converges.

PROBLEM SOLUTIONS

375

8.61 This is a square box with vertices at (±1, 0) and (0, ±1). 8.62 For example, let D = {(m, n) an isolated point of D.

Im

E Z and

n E Z}. Then the point (0, 0) is

8.63 The setS= Qx 2 is an example. 8.64 True. For example, this is true for Example 1.15. 8.65 False. The graph is not closed, lacking any points on the y-axis between 1 and -1. 8.66 False. For example, let Ek 8.67 True, because Xn

---> XI

= lEn \ Bk(O).

as n

--->

oo.

8.68 True, because g is continuous. 8.69.a This means that x 2 = ±xi, the graph of which consists of two intersecting straight lines, each of which is a connected set. Thus Sis connected. 8.69.b The two open sets A = {x E IE 2 I XI > 0} and B = { x E IE 2 1 XI < 0} separate T, so that T is not connected. There is no point in T corresponding to XI= 0. 8.70 False. For example, this would make lEn disconnected! For the two open separating sets, just take any one of the open balls for A and the union of all the others for B. But lEn is connected.


9.6.a The limit does not exist. 9.19 For example, let f(x)

= (cosx, sinx).

9.26 Closed. 9.52 Here is a counterexample. Let A be the graph of x 2 = .!.. in IE 2 , and let B be XI the XI -axis. 9.53 True. This follows from the fact that any open set containing the point (0, 1) must intersect the graph of fl(o,oo) which is connected since the restriction off is continuous on (0, oo ). 9.54 Let ifx-j-0, ifx =0.

376

PROBLEM SOLUTIONS

9.55 limx--.a f(x) exists if and only if for each E > 0 there exists 8 > 0 such that x andy in B 0 (a) n (DJ \{a} implies

llf(x)- f(y)il 9.56 True since f E C(!En, JR) and S

<E.

= f- 1 ( {0} ), the pre-image of a closed set.

9.57 False, since the pre-image of an open set under a continuous function need only be relatively open in D. 9.58 This limit does not exist, since the limit along either axis is zero, but the limit along the line X2 = X1 is ~9.59 True. Necessity follows directly from continuity at 0. Sufficiency is nearly the same as Exercise 9.33.b.

9.60 Here is a counterexample. B 1 (0) is a relatively closed subset of itself, but it is not closed so it cannot be compact, although it is bounded. 9.61 True, since A and B are connected being convex, and their union is connected since An B f:. 0. 9.62 False. Letf(x) = (cos7rx,sin7rx). Then llf(x)ll

= 1.

9.63 True, because the continuous image of a connected set is connected.

SOLUTIONS FOR CHAPTER 10 10.5

II All = liB II = IIA + Bll

= 1.

10.25.a f'(x) exists for all x. 10.26.a det f' (x)

= x 1.

10.26.c detf'(x) = x~sinx 3 • 10.28 D(l,- 2Jf(1, 2) 10.44 (

-3 4

3 -4

= ( ~150 ).

6 ) -3 .

10.48 ~ = ~m_ ~£11.J... 8x; L..Jk-1 8yk 8x;

10.89

II All =

1=

liB II, and IIA + Bll

= 1.

PROBLEM SOLUTIONS

10.90 This is false, since

377

IIX +XII = II2XII = 2IIXII·

10.91 This is false since the function ci,I (A) = a 1,I is a continuous function of A and ci,l (S) is not a connected subset of IE 1. 10.92 Let T have the matrix (

~

8) in the standard basis.

10.93 This is false, since f'(x) is not linear on x E D. Rather f'(x) is a linear map from IE2 --> IE3 for each fixed x E IE2 • Moreover, D need not be a vector space, so that cx 1 + x 2 need not be in D.

-~ ) ( 2 ) _ e ( 2v'3- 1 )

~

1

-2

2+v'3

.

10.95 The matrix

10.96 This is true, because if we did have f(a) = f(b) for some a< b there would be an extreme point someplace in the open interval (a, b), because of the extreme value theorem and because f cannot be constant on [a, b]. For such an extreme point p and r > 0, f could not be injective on (p - r, p + r ). 10.97 By direct calculation, the Jacobian

8(Jr,h,h) 8(xi, X2, X3) =XI. Thus f has nonvanishing Jacobian off the plane XI = 0, and for each such x we can take a ballofradiusr = min(lxii,7r).

10.98 The Jacobian

Thus the equation

f(x)

=

f(xo)

has a local C1-solution for x 3 in terms of the other three variables in a neighborhood of Xo provided x4 =f. 0.

10.99 Let f(x) = (cos x, sinx) for all x E [0, 21r). 10.100 Denote g(xr) = (x1, x2 (xi), x3(xi)). Thus f o g(xr) [f'(xo)][g'(a)] = ( _\

B

~ ) ( ~~h~

)

= 0. Hence

= (

8)·

378

PROBLEM SOLUTIONS

This yields the equations

1 + x~(a) + 3x~(a) -1 + 2x~(a)

0, 0

and the solutions x~(a) = ~and x~(a) = -~.

SOLUTIONS FOR CHAPTER 11 11.6 For example, we can let

p

= {((0,0), (.95, 1)), ((.95,0), (1.05, 1)), ((1.05, 1), (2, 1))}.

11.48 g'(p, (), ¢) = p2 sin¢ and the volume is ~1ra 3 . 11.52 The volumes are identical. 11.58 For example, let f be the indicator function of the set of points for which both coordinates are rational. 11.59 Let f be the indicator of the points with both coordinates rational, and let g be the indicator of the points with both coordinates irrational. 11.60 For example, let coordinate rational.

11.62 The function

f

f

be the indicator function of the set of points having first

is not continuous anywhere on [0, 1].

11.63 No, Sis not Jordan measurable since its boundary is not a Lebesgue null set. 11.64 True, because it is a compact subset of a Lebesgue null set. 11.65 o(f, 0)

=2

11.66 The Jordan measure is zero. 11.67 We must have f(xi) = 0 except on a Lebesgue null set. This follows from Lebesgue's theorem. 11.68.a

[f'(x)] = ( 11.68.b The volume is 3 ~11".

-1

8 (/0

y )

_:1 .

REFERENCES

I. Tom M. Apostol, Mathematical Analysis: A Modem Approach to Advanced Calculus, Addison-Wesley, Reading, MA, 1957. 2. C. Berge, Principles ofCombinatorics, Academic Press, New York, 1971. 3. Carl B. Boyer, The History of the Calculus, Dover Publications, New York, 1949.

4. R. Creighton Buck, Advanced Calculus, McGraw-Hill, New York, 1956.

5. Lennart Carleson, On convergence and growth of partial sumas of Fourier series, Acta Mathematica, Vol. 116, 1966, pp. 135-157. 6. H. Dym and H. P. McKean, Fourier Series and Integrals, Academic Press, New York, 1972. 7. Bernard R. Gelbaum and John M. H. Olmsted, Counterexamples in Analysis, Holden-Day, San Francisco, 1964. 8. Casper Goffman and George Pedrick, First Course in Functional Analysis, Prentice-Hall, Englewood Cliffs, NJ, 1965. 9. Kenneth Hoffman, Analysis in Euclidean Space, Prentice-Hall, Englewood Cliffs, NJ, 1975. 10. Kenneth Hoffman and Ray Kunze, Linear Algebra, Prentice-Hall, Englewood Cliffs, NJ, 1971. II. E. Kamke, Theory of Sets, Dover Publications, New York, 1950. 12. E. Landau, Foundations of Analysis, Chelsea Publishing Co., Broomall, PA, 1960. Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright © 2008 John Wiley & Sons, Inc.

379

380

REFERENCES

13. H. E. Lomeli and C. L. Garcia, Variations on a theorem of Korovkin, American Math Monthly, Vol. 113, No. 8, 2006, pp. 744-750. 14. G. D. Mostow, J. H. Sampson and J.-P. Meyer, Fundamental Structures of Algebra, McGraw Hill, New York, 1963. 15. Otto Neugebauer, The Exact Sciences in Antiquity, Brown University Press, Providence, RI, 1957. 16. John M. H. Olmsted, The Real Number System, Appleton-Century-Crofts, New York, 1962. 17. J. Dauben, Review: The Universal History of Numbers, Notices of the American Mathematical Society, Vol. 49, No. I, January 2002, pp. 32-38. 18. Frigyes Riesz and Bela Sz.-Nagy, Functional Analysis, Frederick Ungar Publishing Co., New York, 1955. 19. Walter Rudin, Principles of Mathematical Analysis, McGraw-Hill, New York, 1964. 20. Michael Spivak, Calculus on Manifolds, W. A. Benjamin, New York, 1965.

INDEX

llfllsup, 275 11/112. 92, 193 11/111. 199 11/llsup. 58 llxll1. 148

Is, 83

AnB, 360 AUB, 360 A\ B, 360

Ac, 361 Br(p), 247

Z, 5

Dvf(x),

290 L(f,P), 77 Pn(x), 122 S(f), 188 253 8°, 105, 255 SN(/), 188 Txo(S), 323 U(f,P), 77 V,I'(f), 216 [a, b], 332 En, 247 'J'(z), 185 N, 3 IQI, 4 IQin, 256

Br(p), 247

s, 255

Xk· 187 {Jjk· 208

sc,

u, 279

0,

~.290 J

glb, 15 inf, 15 rxl, 241 LxJ, 183, 221 lim, 9 limx-->a±• 44 limx-->a-, 44 liminf, 18 limsup, 18 lub, 14 f'(x), 289

JR, 5 IRn, 247 R(z), 185

IITII,

27, 360

8Un±l•···.fntm) 312 8(yl, ... ,y,.,.) '

150

Advanced Calculus: An Introduction to Linear Analysis. By Leonard F. Richardson Copyright

© 2008

John Wiley & Sons, Inc.

381

382

INDEX

r- 1 {E), 212

Xnk,

C'[a, b], 232

IQI+, 4

BV[a,b], 216 C(D), 46, 271 C(D,Em), 271

xk, 267

22

(a, b), 332

C[a, b], 46, 58, 61 C1 {D,Em), 293 C1 {IR), 184 C1 [a, b], 116 C"", 163 :F, 195

Q.C(n, IR), 286 .C{En), 284 .C{En, Em), 284 0(2), 321 O(n), 357 RS([a, b], g), 223 R(B), 334 R{[a,b],IC), 187, 192 R[a,b], 70 S.C{2, IR), 321

J"sl, 334 1:1. 1s as,

255

_l, 209, 323

sgn, 187, 234 sup, 15 8 1, 334

f J:l, 78 lEI,

109 188 dlx 0 , 101

f,

l(a+),43 I( a-), 44

I L 108 I/, 10s I'\., 108

1 "'g,

193

I r, 108 1+, 81, 337 1-' 81, 337 h, 148 b, 195 loo, 151

m(B), 332 o(/,x), 340 !, 16 /,16 Xn '\., 16 Xn--> L, 9 Xn---> 00, II Xn i, 16 Xn Xn

X¢(k)•

22

affine subspace, 323 alternating series test, 129 analytic function, real, 163 approximate identity, 170 Archimedean ordered field, 5 Property, 5 Banach space, 63 isomorphism, 151 Bessel's inequality, 194 bijection, 239, 286 Bolzano-Weierstrass Theorem for En, 251 for IR, 23 bound greatest lower, 15 least upper, 14 boundary, 255 bounded function, 55 sequence, II sequence in En, 250 set, 14 set in En, 256 variation, 216 Cantor diagonalization, 36 paradox, 363 theorem, 33 Cauchy Mean Value Theorem, 117 product, 144 sequence, 10 Cauchy criterion for functions, 45, 270 for sequences, 13 Cauchy-Schwarz inequality b, 195 Euclidean space, 246 functions, 92 Hermitian scalar product, 193 ceiling function, 241 Cesaro means, 22 chain rule Euclidean space, 298 for IR, 102 change of variables, 352

INDEX

character, 187 Clairaut's theorem, 351 closed ball, 24 7 closed set, 35, 253 closure, 255 cluster point, 40, 253 column rank, 313 matrix, 313 compact set, 256 comparison test, 134 complement, 253 complete normed space, 63 Completeness Axiom for IR, 13 completion BV[a, b], 222 normed vector space, 222 of IQ, 26 complex number conjugate, 185 imaginary part, 185 modeled in plane, 185 modulus, 185 polar form, 185 real part, 185 connected, 260 connectivity of!En,261 continuity sequential criterion, 46 continuously differentiable, 293 contraction, fixed points, 134 convergence at infinity, 267 conditional, 130 function on !En, 267 pointwise, 62 uniform, 61 convergence test nth tern1, 128 nth-root test, 138 alternating series, 129 comparison, 134 geometric series, 132 integral, 135 limit comparison, 137 ratio, 136 convergent series, 128 convex set, 279 convolution, 170, 346 countability of IQ, 32 cover

finite sub-, 256 open, 29, 256 Darboux integrability criterion Euclidean space, 334 for R 78 DeMorgan 's Laws, 362 dense, 25, 256 derivative, I00 diffeomorphic, 309 differentiable, 10 I, 289 Differentiation of integral, 350 Dini's Theorem, 65 directional derivative, 290 Dirichlet kernel, 198 theorem, 139 disconnected, 259 discontinuity jump, 44, 220 divergence function on !En, 267 to infinity, 267 divergent sequence, 9 to oo, 12 series, 128 dot product, 90 double summations, 144 dual space, 150 Euclidean norm, 88, 247 Euclidean space, 247 standard basis, 284 Euler's formula, 186 evaluation point, 70 existence v'2. 26 p > 0, 51 expectation, 145 Extreme Value Theorem for !En, 274 for IR, 56

w.

Fejer kernel, 210 theorem, 209 field axioms Archimedean ordered, 5 complex numbers, 185 finite interval, 29 set, 29 subcover, 256 fixed point, 53

383

384

INDEX

floor function, 183 Fourier coefficients, 188 series, 188 transform, 188, 351 Fubini's Theorem, 347 function, 40 bijection, 239 bounded, 55 continuous, 46, 271 derivative, I 00 differentiable, 101, 289 domain, 40 even, 53 gradient vector, 323 graph of, 258 implicit, 311 increasing, 44 indicator, 83 injective, I 04, I 08 integrable, 334 inverse image, 272 Jacobian, 308 level surface, 322 limit, 40, 266 locally injective, 310 monomial, 267 monotone, 108 nonsingular, 308 odd, 53 of vector variable, 265 oscillation, 340 partial derivative, 290, 291 periodic, 183 polynomial in lEn, 267 range, 40 rational, 267 real analytic, 163 step, 75 strictly monotone, 108 surjection, 351 surjective, 151 vector-valued, 265 functional bounded, 86 continuous, 85 linear, 72, 85 functions integrable equivalent, 193 orthogonal, 94 fundamental theorem of calculus, 110

geometric series, 132 test, 132 graph, 258 greatest integer function, 227 lower bound, 15 glb, 15 group, 288 Hamel basis, 89 Heine-Bore! Theorem Euclidean space, 257 real line, 30 homeomorphism, 275 implicit differentiation, 311 function theorem, 312 functions, 311 index set, 28, 253 indicator function, 74, 83 infimum, 15 injection, 286 injective, 104, 108, 275 inner product, 90 Hermitian, 192 space, 95 integer, 5 integral differentiation of, 350 improper, 135 lower in lEn, 334 parts, 228 upper in lEn, 334 integral test, 135 integrator, 223 interior, 255 point, 105 intermediate value property, 51 theorem, 51 connected set, 278 derivatives, Ill interval, 259 finite, 29 Inverse Function Theorem, 307 inverse image, 272 invertible, 286 isolated point, 46, 253 isomorphism, Banach space, 151 Jacobian, 308, 312 Jordan

INDEX

content, 346 measurable, 346 null set, 339 jump discontinuity, 44 kernel Dirichlet, 198 of linear map, 288 Weierstrass approximation, 171 L'Hopital's Rule, 118 Lagrange multipliers, 324 lea~t upper bound, 14 Lebesgue criterion Riemann Integrability, 342 Henri, 339 null set, 339 level surface, 322 limit function, 40 sequential criterion, 42 inferior, 18 sequence, 9 superior, 18 limit comparison test, 137 linear functional, 72, 85 bounded, 86 continuous, 85 linear space, 59 linear transformation identity, 286 invertible, 286 Jacobian, 312 kernel, 288 magnification theorem, 306 open mapping theorem, 307 local extreme point, 30 I local-global duality, 202 locally injective, 310 lower bound, 14 integral, 78 in lEn, 334 sum on lEn, 333 on IR, 77 magnification theorem, 306 matrix invertible, 286 Mean Value Theorem Cauchy, 117 derivatives, I 06 Euclidean space, 300

integrals, 75 332 monomial, 267 monotone function, 108 increa~ing, 44 sequence, 16 multinomial coefficients, 303 mutual refinement, 77 mea~ure,

Nested Intervals theorem, 25 non singular map, 308 matrix, 317 norm, 59 £ 1 , 199 £ 2 , 193 taxicab, 249 Euclidean, 88, 247 of a functional, 150 of linear transformation, 285 sup, 58, 275 taxicab, 250 nth root test, 138 nth term test, 128 number algebraic, 36 even, 8 integer, 5 irrational

/2,8 natural, 3 odd, 8 rational, 4 real, 5, 13 transcendental, 36 open ball, 247 open cover, 29, 256 Open Mapping Theorem, 307 open set, 28, 252 orthogonal characters, 190 complement, 209, 323 functions, 94 group, 321, 357 vectors, 250 oscillation of function on lEn, 340 of function on IR, 50 paradox, 363 Cantor's, 363 Russell's, 364

385

386

INDEX

Zeno's, 132 Parallelogram Law, 251 Parseval's identity, 209 partial derivative, 290, 291 partition, 70 in lEn, 332 mesh, 70 mesh in lEn, 332 mutual refinement, 77 in lEn, 333 refinement in lEn, 332 period T, 183 Plancherel identity, 194, 205 point cluster, 40, 253 evaluation, 230 interior, I 05 isolated, 46, 253 local extreme, 105, 30 I pointwise convergence, 62 polynomial, 267 power series, 158 preservation of sign, 50 product of open sets, 322 proposition, 364 Pythagorean Theorem L 2 norm, 94 Euclidean space, 250 Plancherel identity, 205 radius of convergence, 158 random variable discrete, 145 expectation, 145 rank, 313 linear transformation, 313 ratio test, 136 rational function, 267 rectangular block closed, 332 open, 332 Riemann -Lebesgue lemma, 199 -Stieltjes integral, 223 integral, 70 circle, 183 complex valued, 187 Euclidean space, 334 general domain, 335 localization theorem, 201 sum, 70 Riesz -Fischer Theorem, 207

Representation Theorem, 232 Rolle's Theorl(m, 106 row rank, 313 matrix, 313 Russel's paradox, 364 sandwich theorem functions, 44 sequences, 12 scalar product, 90, 246 Euclidean, 247 Hermitian, 192 separated, 260 separation of variables, 181 sequence nth tail, 17 bounded, II Cauchy, 10 convergent, 9 decrea~ing, 16 divergent, 9 to oo, 12 increasing, 16 limit, 9 monotone, 16 null, 20 strictly decreasing, 16 strictly increasing, 16 sub, 22 sequential criterion continuity, 46 limits, 42 series uniformly convergent, 154 absolutely summable, 130 conditionally convergent, 130, 147 convergent, 128 divergent, 128 double, 144 Fourier, 188 harmonic, 129 power, 158 radius of convergence, 158 ratio test, 136 rearrangement, 139 summable, 128 telescoping, 133 set boundary, 255 bounded, 14 closed, 35, 253 closed ball, 247 closure, 255 complement, 361

INDEX

connected, 260 convex, 279 countable, 31 dense, 256 dense in IR, 25 disconnected, 259 empty, 360 finite, 29 index, 28, 253 infinite, 359 interior, 255 minus, 360 open, 28, 252 open ball, 247 orthonomml, 190 power, 363 relatively closed, 271 relatively open, 271 separated, 260 transfinite, 359 sets DeMorgan's Laws, 362 difference, 360 disjoint, 360 intersection, 360 product of open, 322 symmetric difference, 361 union, 360 signum, 234 space affine sub, 323 Banach, 63 complete normed, 63 dual, 150 self dual, 207, 252 Euclidean, 247 Hilbert, 207 inner product, 95 linear, 59 normed vector, 58 orthogonal complement, 323 vector, 59 spaces diffeomorphic, 309 spherical coordinates, En, 356 squeeze theorem functions, 44 sequences, 12 standard basis, 284 step function, 75 strictly monotone function, I 08 subset proper, 360

supremum, 15 surjection, 286, 351 surjective, 151 tail

nth, 17 tangent plane, 323 space, 323 vector, 323 Taylor polynomial, 122 theorem in En, 302 in IR, 123 triangle inequality complex space, 196 nom1s, 58, 59, 287 on IR, 6 Unconscious Statistician law of, 146 uncountability of IR, 33 uniform continuity in En, 277 in IR, 52 convergence in En, 275 in IR, 61 of series, 154 upper bound, 14 integral in En, 334 in IR, 78 sum on En, 333 on IR, 77 variation bounded, 216 total, 216 vector space, 59 vibrating string, 180 Weierstrass Approximation Theorem, 169, 210 M-test, 156 Zeno's Paradox, 132

387


Advanced Calculus: An Introduction to Linear Analysis

Advanced Calculus: An Introduction to Linear Analysis

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Advanced Calculus: An Introduction to Linear Analysis