Counterexamples on Uniform Convergence: Sequences, Series, Functions, and Integrals [1 ed.] 1119303389, 9781119303381, 166-133-133-5

Table of contents :
Content: Preface ix List of Examples xi List of Figures xxix About the Companion Website xxxiii Introduction xxxv I.1 Comments xxxv I.1.1 On the Structure of This Book xxxv I.1.2 On Mathematical Language and Notation xxxvii I.2 Background (Elements of Theory) xxxviii I.2.1 Sequences of Functions xxxviii I.2.2 Series of Functions xli I.2.3 Families of Functions xliv 1 Conditions of Uniform Convergence 1 1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset 1 1.2 Uniform Convergence of Sequences and Series of Squares and Products 15 1.3 Dirichlet s and Abel s Theorems 31 Exercises 39 Further Reading 42 2 Properties of the Limit Function: Boundedness, Limits, Continuity 45 2.1 Convergence and Boundedness 45 2.2 Limits and Continuity of Limit Functions 51 2.3 Conditions of Uniform Convergence. Dini s Theorem 68 2.4 Convergence and Uniform Continuity 79 Exercises 88 Further Reading 93 3 Properties of the Limit Function: Differentiability and Integrability 95 3.1 Differentiability of the Limit Function 95 3.2 Integrability of the Limit Function 117 Exercises 128 Further Reading 131 4 Integrals Depending on a Parameter 133 4.1 Existence of the Limit and Continuity 133 4.2 Differentiability 144 4.3 Integrability 154 Exercises 162 Further Reading 166 5 Improper Integrals Depending on a Parameter 167 5.1 Pointwise, Absolute, and Uniform Convergence 167 5.2 Convergence of the Sum and Product 176 5.3 Dirichlet s and Abel s Theorems 185 5.4 Existence of the Limit and Continuity 192 5.5 Differentiability 198 5.6 Integrability 202 Exercises 210 Further Reading 214 Bibliography 215 Index 217

Citation preview

Counterexamples on Uniform Convergence

Counterexamples on Uniform Convergence Sequences, Series, Functions, and Integrals

Andrei Bourchtein Ludmila Bourchtein

Copyright © 2017 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 107 or 108 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 fee 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., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permissions. 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.wiley.com. Library of Congress Cataloging-in-Publication Data Names: Bourchtein, Andrei. | Bourchtein, Ludmila. Title: Counterexamples on uniform convergence : sequences, series, functions, and integrals / Andrei Bourchtein, Ludmila Bourchtein. Description: Hoboken, New Jersey : John Wiley & Sons, Inc., c2017. | Includes bibliographical references and index. Identifiers: LCCN 2016038687 | ISBN 9781119303381 (cloth) | ISBN 9781119303428 (epub) | ISBN 9781119303404 (epdf ) Subjects: LCSH: Mathematical analysis–Problems, exercises, etc. | Calculus–Problems, exercises, etc. | Sequences (Mathematics) | Functions. | Integrals. Classification: LCC QA301 .B68 2017 | DDC 515–dc23 LC record available at https://lccn.loc.gov/2016038687 Cover Image credit: Abscent84/gettyimages Set in 10/12pt Warnock by SPi Global, Pondicherry, India Printed in the United States of America 10

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To Haim and Maria with the warmest memories; To Maxim with love and inspiration; To Valentina for always yummy breakfast; To Victoria for amazing rainbow flower

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Contents Preface ix List of Examples xi List of Figures xxix About the Companion Website Introduction xxxv

I.1 I.1.1 I.1.2 I.2 I.2.1 I.2.2 I.2.3

xxxiii

Comments xxxv On the Structure of This Book xxxv On Mathematical Language and Notation xxxvii Background (Elements of Theory) xxxviii Sequences of Functions xxxviii Series of Functions xli Families of Functions xliv

1

Conditions of Uniform Convergence 1

1.1

Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset 1 Uniform Convergence of Sequences and Series of Squares and Products 15 Dirichlet’s and Abel’s Theorems 31 Exercises 39 Further Reading 42

1.2 1.3

2

Properties of the Limit Function: Boundedness, Limits, Continuity 45

2.1 2.2 2.3 2.4

Convergence and Boundedness 45 Limits and Continuity of Limit Functions 51 Conditions of Uniform Convergence. Dini’s Theorem 68 Convergence and Uniform Continuity 79 Exercises 88 Further Reading 93

viii

Contents

3

Properties of the Limit Function: Differentiability and Integrability 95

3.1 3.2

Differentiability of the Limit Function 95 Integrability of the Limit Function 117 Exercises 128 Further Reading 131

4

Integrals Depending on a Parameter

4.1 4.2 4.3

Existence of the Limit and Continuity Differentiability 144 Integrability 154 Exercises 162 Further Reading 166

5

Improper Integrals Depending on a Parameter

5.1 5.2 5.3 5.4 5.5 5.6

Pointwise, Absolute, and Uniform Convergence Convergence of the Sum and Product 176 Dirichlet’s and Abel’s Theorems 185 Existence of the Limit and Continuity 192 Differentiability 198 Integrability 202 Exercises 210 Further Reading 214 Bibliography Index 217

215

133 133

167 167

ix

Preface Looking for counterexamples is one of important things a mathematician does. Once a conjecture is proposed, if no proof can be found, the next step is to look for a counterexample. If no counterexample can be found, the next step is to try to find a proof again, and so on. On a more routine level, counterexamples play an important role for students as they learn new mathematical concepts. To best understand a theorem, it can be useful to see why each of the hypotheses of the theorem is necessary by finding counterexample when the hypotheses fails. In this book, we present counterexamples related to different concepts and results on the uniform convergence usually studied in advanced calculus and real analysis courses. It includes the convergence of sequences, series and families of functions, and also proper and improper integrals depending on a parameter. The corresponding false statements are not formulated explicitly, but instead are invoked implicitly by the form of counterexamples. The text is divided into six parts: the introductory chapter and five chapters of counterexamples. The first part contains some introductory material such as comments on notations, presentation form, and background theory. Chapter 1 considers conditions of uniform convergence. Chapter 2 deals with such properties of the limit functions as boundedness, existence of the limit and continuity. Chapter 3 analyzes the conditions of differentiability and integrability of the limit functions. Chapters 4 and 5 consider the properties of integrals (proper and improper) depending on a parameter. The goal of the book is threefold. First, it provides a brief survey and discussion of principal results of the theory of uniform convergence in real analysis. Second, it supplies a material for a deeper study of the concepts and theorems on uniform convergence using counterexamples as a main technique. Finally, the text shows to the reader how such important mathematical tool as counterexamples can be used in different situations. We restricted our exposition to the main definitions and theorems in order to explore different versions (wrong and correct) of the fundamental concepts. Hence, many interesting (but more

x

Preface

specific and applied) problems not related directly to the main notions and results are left out of the scope of this manuscript. The selection and exposition of the material are directed, in the first place, to those advanced calculus and analysis students who are interested in a deeper understanding and broader knowledge of the topics of uniform convergence. We think the presented material may also be used by instructors that wish to go through the examples (or their variations) in class or assign them as homework or extracurricular projects. To this end, the main text is accompanied by the Instructor’s Solutions Manual containing the detailed solutions to all the exercises proposed at the end of each chapter. It is assumed that a reader has knowledge of a traditional university course of calculus. In order to make the majority of the examples and solutions accessible to calculus and analysis students, we tried to keep the level of reasoning as simple as possible. As in the majority of the mathematics books, the logical sequence of the material just follows the chapter sequence, that is, the content of the next chapter may depend on the previous text, but not vice-versa. The book is not appropriate as the main textbook for a course, but rather, it can be used as a supplement that can help students to master important concepts and theorems. So we think the best way to use the book is to read its parts while taking a respective calculus/analysis course. On the other hand, the students already familiarized with the subjects of university calculus can find here deeper interpretation of the results and finer relation between concepts than in standard presentations. Also, more experienced students will better understand provided examples and ideas behind their construction. To facilitate the reading of the main text (containing counterexamples) and make the text self-contained, and also to fix terminology, notation, and concepts, we gather the relevant definitions and results in the introductory chapter. For many examples, we make explicit references to the concepts/theorems to which they are related. A short (but representative) list of bibliography can be found at the end of the book, including both collections of problems and textbooks in calculus/analysis. On the one hand, these references are the sources of some examples collected here, although it was out of our scope to trace all the original sources. On the other hand, they may be used for finding further information (examples and theory) on various topics. Some of these references are classic collections of the problems, such as that by Demidovich [6] and by Gelbaum and Olmsted [8]. Our preparation of the text was inspired, in the first place, by the latter book. We tried to extend its approach to the specialized topics of the uniform convergence, which frequently are sources of misunderstanding and confusion for fresh mathematics students. We hope that both students and professionals will find our book useful and (at least partly) challenging.

xi

List of Examples Chapter 1. Conditions of Uniform Convergence Example 1. A function f (x, y) defined on X × Y converges pointwise on X as y approaches y0 , but this convergence is nonuniform on X. . . . . . . . . . . . . . . . . 1 A sequence of functions converges (pointwise) on a set, but this convergence is nonuniform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1 A series of functions converges (pointwise) on a set, but this convergence is nonuniform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2 Example 2. A series of functions converges on X and a general term of the series converges to zero uniformly on X, but the series converges nonuniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Example 3. A sequence of functions converges on X and there exists its subsequence that converges uniformly on X, but the original sequence does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 Example 4. A function f (x, y) defined on (a, b) × Y converges on (a, b) as y approaches y0 and this convergence is uniform on any interval [c, d] ⊂ (a, b), but the convergence is nonuniform on (a, b). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5 A sequence of functions defined on (a, b) converges uniformly on any interval [c, d] ⊂ (a, b), but the convergence is nonuniform on (a, b) . . . . . . . . . . . . . . . . 6 A series of functions converges uniformly on any interval [c, d] ⊂ (a, b), but the convergence is nonuniform on (a, b). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 Example 5. A sequence converges on X, but this convergence is nonuniform on a closed interval [a, b] ⊂ X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6 A series converges on X, but this series does not converge uniformly on a closed subinterval [a, b] ⊂ X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

xii

List of Examples

A function f (x, y) defined on X × Y has a limit lim f (x, y) = 𝜑(x) for ∀x ∈ X, y→y0

but f (x, y) converges nonuniformly on a subinterval [a, b] ⊂ X. . . . . . . . . . . . . 8 Example 6. A sequence converges on a set X, but it does not converge uniformly on any subinterval of X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 A series converges on X, but it does not converge uniformly on any subinterval of X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9 Example 7. A series converges uniformly on an interval, but it does not converge absolutely on the same interval. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 Example 8. A series converges absolutely on an interval, but it does not converge uniformly on the same interval. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10 ∑ Example 9. A series un (x) converges absolutely and uniformly on [a, b], ∑ but the series |un (x)| does not converge uniformly on [a, b]. . . . . . . . . . . . 11 ∑ Example 10. A series un (x) converges absolutely and uniformly on X, but there is no bound of the general term un (x) on X in the form |un (x)| ≤ an , ∀n ∑ such that the series an converges. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 Example 11. A sequence fn (x) converges uniformly on X to a function f (x), but fn2 (x) does not converge uniformly on X to f 2 (x). . . . . . . . . . . . . . . . . . . . . . 15 Sequences fn (x) and gn (x) converge uniformly on X, but fn (x)gn (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16 Example 12. Sequences fn (x) and gn (x) converge nonuniformly on X to f (x) and g(x), respectively, but fn (x) ⋅ gn (x) converges to f (x) ⋅ g(x) uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 Example 13. A sequence fn2 (x) converges uniformly on X, but fn (x) diverges on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 A sequence |fn (x)| converges uniformly on X, but fn (x) diverges on X. . . . 19 A sequence fn2 (x) converges uniformly on X and fn (x) converges on X, but the convergence of fn (x) is nonuniform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19 Example 14. A sequence fn (x) ⋅ gn (x) converges uniformly on X to 0, but neither fn (x) nor gn (x) converges to 0 on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20 Example 15. A sequence fn (x) converges uniformly on X to a function f (x), fn (x) ≠ 0, f (x) ≠ 0, ∀x ∈ X, but f 1(x) does not converge uniformly on X to

1 . f (x)

n

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

List of Examples

Example 16. A sequence fn (x) is bounded uniformly on ℝ and converges uniformly on [−a, a], ∀a > 0, to a function f (x), but the numerical sequence sup fn (x) does not converge to sup f (x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21 x∈ℝ

x∈ℝ

Example 17. Suppose each function fn (x) maps X on Y and function g(y) is continuous on Y ; the sequence fn (x) converges uniformly on X, but the sequence gn (x) = g(fn (x)) does not converge uniformly on X. . . . . . . . . . . . . . 22 Suppose functions fn (x) map X on Y and function g(y) is continuous on Y ; the sequence fn (x) converges nonuniformly on X, but the sequence gn (x) = g(fn (x)) converges uniformly on X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 ∑ 2 Example 18. A series un (x) converges uniformly on X, but the series ∑ un (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23 ∑ 2 Example 19. A series un (x) converges uniformly on X, but the series ∑ un (x) does not converge (even pointwise) on X. . . . . . . . . . . . . . . . . . . . . . . . 23 ∑ Example 20. A series un (x)vn (x) converges uniformly on X, but at least one ∑ ∑ of the series un (x) or vn (x) does not converge uniformly on X. . . . . . . . 24 ∑ Example 21. A series un (x)vn (x) converges uniformly on X, but neither ∑ ∑ un (x) nor vn (x) converges (even pointwise) on X. . . . . . . . . . . . . . . . . . . . 25 ∑ ∑ Example 22. Series un (x) and vn (x) converge nonuniformly on X, but ∑ un (x)vn (x) converges uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 ∑ ∑ Example 23. A series un (x) converges uniformly on X, but u2n (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26 ∑ ∑ Both series un (x) and vn (x) converge uniformly on X, but the series ∑ un (x)vn (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . 28 ∑ ∑ Example 24. A series un (x) converges uniformly on X, but u2n (x) does not converge (even pointwise) on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 ∑ ∑ Both series un (x) and vn (x) converge uniformly on X, but the series ∑ un (x)vn (x) does not converge (even pointwise) on X. . . . . . . . . . . . . . . . . . . 30 ∑ ∑ Example 25. Both series un (x) and vn (x) are nonnegative for ∀x ∈ X, u (x) lim n = 1 and one of these series converges uniformly on X, but another n→∞ vn (x) series does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30 ∑ Example 26. The partial sums of un (x) are bounded for ∀x ∈ X, and the sequence vn (x) is monotone in n for each fixed x ∈ X and converges uniformly ∑ on X to 0, but the series un (x)vn (x) does not converge uniformly on X. . . 32

xiii

xiv

List of Examples

∑ Example 27. The partial sums of un (x) are uniformly bounded on X and ∑ the sequence vn (x) converges uniformly on X to 0, but the series un (x)vn (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32 ∑ Example 28. The partial sums of un (x) are uniformly bounded on X, and the sequence vn (x) is monotone in n for each fixed x ∈ X and converges on X ∑ to 0, but the series un (x)vn (x) does not converge uniformly on X. . . . . . . . 33 ∑ Example 29. The partial sums of un (x) are not uniformly bounded on X, and the sequence vn (x) is not monotone in n and does not converge uniformly ∑ on X to 0, but still the series un (x)vn (x) converges uniformly on X. . . . . . 34 ∑ A series un (x) diverges at each point of X, and a sequence vn (x) is not mono∑ tone in n and diverges on X, but still the series un (x)vn (x) converges uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35 ∑ Example 30. A series un (x) converges on X, and a sequence vn (x) is monotone in n for each fixed x ∈ X and uniformly bounded on X, but the series ∑ un (x)vn (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . 36 ∑ Example 31. A series un (x) converges uniformly on X, and a sequence vn (x) ∑ is uniformly bounded on X, but the series un (x)vn (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 ∑ Example 32. A series un (x) converges uniformly on X, and a sequence vn (x) ∑ is monotone in n for each fixed x ∈ X, but the series un (x)vn (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 ∑ A series un (x) converges uniformly on X, and a sequence vn (x) is monotone ∑ and bounded in n for each fixed x ∈ X, but the series un (x)vn (x) does not converge uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37 ∑ Example 33. A series un (x) does not converge uniformly on X, and a sequence vn (x) is not monotone in n and is not uniformly bounded on X, but ∑ still the series un (x)vn (x) converges uniformly on X. . . . . . . . . . . . . . . . . . . . 38 ∑ A series un (x) diverges at each point of X, and a sequence vn (x) is not mono∑ tone in n and is not bounded at each point of X, but still the series un (x)vn (x) converges uniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

Chapter 2. Properties of the Limit Function: Boundedness, Limits, Continuity Example 1. A sequence of bounded on X functions converges on X to a function, which is unbounded on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

List of Examples

Example 2. A sequence of functions is not uniformly bounded on X, but it converges on X to a function, which is bounded on X. . . . . . . . . . . . . . . . . . . . 46 Example 3. A sequence of unbounded and discontinuous on X functions converges on X to a function, which is bounded and continuous on X. . . . . 47 Example 4. A sequence of unbounded and discontinuous on X functions converges on X to an unbounded and discontinuous function, but the convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 A sequence of unbounded and continuous on X functions converges on X to an unbounded and continuous function, but the convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Example 5. A sequence of unbounded and discontinuous on X functions converges on X to an unbounded and discontinuous function, but this convergence is uniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49 Example 6. A sequence of uniformly bounded on X functions converges on X, but the convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50 ∑ Example 7. A series un (x) converges on X and lim un (x) exists for each n, x→x0 ∑ ∑ but lim un (x) ≠ lim un (x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51 x→x0

x→x0

A sequence fn (x) converges on X and lim fn (x) exists for each n, but x→x0

lim lim fn (x) ≠ lim lim fn (x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

x→x0 n→∞

n→∞ x→x0

A function f (x, y) is defined on X × Y , converges on X to a limit function 𝜑(x), as y → y0 , and lim f (x, y) exists for each y ∈ Y , but x→x0

lim lim f (x, y) ≠ lim lim f (x, y). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

x→x0 y→y0

y→y0 x→x0

x→x0

x→x0

∑ Example 8. A series un (x) converges on X and lim un (x) exists for each n, x→x0 ∑ ∑ but lim un (x) ≠ lim un (x) since the left-hand side limit is infinite. . . . . 53 A sequence fn (x) converges on X and lim fn (x) exists for each n, but x→x0

lim lim fn (x) ≠ lim lim fn (x) since the left-hand side limit is infinite. . . . . . . . 54

x→x0 n→∞

n→∞ x→x0

A function f (x, y) is defined on X × Y , converges on X to a limit function 𝜑(x), as y → y0 , and lim f (x, y) exists for each y ∈ Y , but lim lim f (x, y) ≠ x→x0

x→x0 y→y0

lim lim f (x, y) since the left-hand side limit is infinite. . . . . . . . . . . . . . . . . . . . 55

y→y0 x→x0

Example 9. A series

∑

un (x) converges on X, also lim un (x) exists for each x→x0 ∑ n and one of the following two conditions is satisfied: either lim un (x) x→x0

xv

xvi

List of Examples

converges or lim

x→x0

∑

un (x) exists, but nevertheless the remaining condition

does not hold. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55 Suppose fn (x) converges on X and lim fn (x) exists for each n. Even if one x→x0

of the limits— lim lim fn (x) or lim lim fn (x)—exists, another one may not n→∞ x→x0

x→x0 n→∞

exist. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 56 Assume a function f (x, y) is defined on X × Y , converges on X to a limit function 𝜑(x), as y → y0 , and lim f (x, y) exists for each y ∈ Y . Even though one of x→x0

the two iterated limits—lim lim f (x, y) or lim lim f (x, y)—exists, another one y→y0 x→x0

x→x0 y→y0

may not exist. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57 Example 10. A series converges nonuniformly on X, but still a limit of this series can be calculated term by term. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 58 A sequence fn (x) converges nonuniformly on X, but still lim lim fn (x) = x→x0 n→∞

lim lim fn (x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 59

n→∞ x→x0

A function f (x, y) defined on X × Y converges nonuniformly on X to a limit function 𝜑(x), as y → y0 , but lim lim f (x, y) = lim lim f (x, y). . . . . . . . . . . . . . 59 x→x0 y→y0

y→y0 x→x0

Example 11. A sequence of discontinuous functions converges uniformly on X, but the limit function is continuous on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60 All the terms of a uniformly convergent on X series are discontinuous at some point x0 ∈ X, but the series sum is continuous at x0 . . . . . . . . . . . . . . . . . . . . . . 60 A function f (x, y) defined on X × Y has a discontinuity at x0 ∈ X for any y ∈ Y and converges uniformly on X to a limit function 𝜑(x), as y → y0 , but 𝜑(x) is continuous at x0 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61 Example 12. A sequence of discontinuous functions converges uniformly on X, and the limit function is also discontinuous on X. . . . . . . . . . . . . . . . . . . . . . 62 A series of discontinuous functions converges uniformly on X, and the sum of the series is also discontinuous on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62 A discontinuous in x function f (x, y) converges uniformly on X to a limit function 𝜑(x), as y → y0 , which is also discontinuous on X. . . . . . . . . . . . . . . . 62 Example 13. A sequence fn (x) converges uniformly on X to a continuous function, but the functions fn (x) have infinitely many points of discontinuity on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63 A series converges uniformly on X to a continuous function, but the terms of the series possess infinitely many discontinuities on X. . . . . . . . . . . . . . . . . . . 63

List of Examples

A function f (x, y) defined on X × Y converges uniformly on X to a continuous function 𝜑(x), as y → y0 , but f (x, y) possesses infinitely many points of discontinuity on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 Example 14. A function f (x, y), defined on X × Y , is continuous with respect to x on X for any fixed y ∈ Y , and it has a limit function 𝜑(x) on X as y approaches y0 , but 𝜑(x) is discontinuous on X. . . . . . . . . . . . . . . . . . . . . . . . . . . 64 A sequence of continuous functions converges on X, but the limit function is discontinuous on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64 A series of continuous functions converges on X, but the sum of this series is a discontinuous function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65 Example 15. A function f (x, y), defined on X × Y , is continuous with respect to x on X for any fixed y ∈ Y , and it has a continuous limit function 𝜑(x) on X as y approaches y0 , but the convergence is nonuniform on X. . . . . . . . . . . . . . 66 A sequence of continuous functions converges on X to a continuous function, but the convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66 A series of continuous functions converges on X to a continuous function, but the convergence is nonuniform on X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67 Example 16. A sequence of functions fn (x), which are monotone in n for any fixed x ∈ X, converges on a compact set X to a continuous function, but the convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 Example 17. A sequence of continuous functions fn (x), which are monotone in n for any fixed x ∈ X, converges on a compact set X, but this convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 69 Example 18. A sequence of continuous functions converges on a compact set X to a continuous function, but this convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Example 19. A sequence of continuous functions fn (x), which are monotone in n for any fixed x ∈ X, converges on a set X to a continuous function, but this convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70 Example 20. A sequence of functions violates all the four conditions of Dini’s theorem on a set X, but nevertheless converges uniformly on this set. . . . . . 71 Example 21. A series of continuous nonnegative functions converges on a compact set X, but this convergence is nonuniform on X. . . . . . . . . . . . . . . . . 71 Example 22. A series of nonnegative functions converges on a compact set X to a continuous function, but this convergence is nonuniform on X. . . . . 72

xvii

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List of Examples

Example 23. A series of continuous functions converges on a compact set X to a continuous function, but this convergence is nonuniform on X. . . . . . . 73 Example 24. A series of continuous nonnegative functions converges on X to a continuous function, but this convergence is nonuniform on X. . . . . . . 74 Example 25. Some conditions of Dini’s theorem are violated, but a series still converges uniformly. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 A series violates all the four conditions of Dini’s theorem on a set X, but nevertheless converges uniformly on this set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75 Example 26. A sequence of continuous functions converges on X to a function f (x), which has infinitely many points of discontinuity. . . . . . . . . . . . . . . 76 Example 27. A sequence of uniformly continuous on X functions converges on X, but the limit function is discontinuous on X. . . . . . . . . . . . . . . . . . . . . . . 79 A series of uniformly continuous on X functions converges on X, but the sum of the series is a discontinuous on X function. . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 Example 28. A sequence of uniformly continuous on X functions converges on X to a continuous function f (x), but the limit function is not uniformly continuous on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80 A series of uniformly continuous on X functions converges on X to a continuous function f (x), but f (x) is not uniformly continuous on X. . . . . . . . . . . . 81 Example 29. A sequence of uniformly continuous on X functions converges on X to an uniformly continuous function, but this convergence is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81 A series of uniformly continuous on X functions converges on X to an uniformly continuous function, but this convergence is nonuniform. . . . . . . . . . 82 Example 30. A sequence of nonuniformly continuous on X functions converges on X, but the limit function is uniformly continuous on X. . . . . . . . . 82 A series of of nonuniformly continuous on X functions converges on X, but the sum of series is uniformly continuous on X. . . . . . . . . . . . . . . . . . . . . . . . . . 83 A sequence of nonuniformly continuous on X functions converges on X to a nonuniformly continuous function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84 A series of nonuniformly continuous on X functions converges on X to a nonuniformly continuous function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85 Example 31. A sequence of nonuniformly continuous on X functions converges uniformly on X, but nevertheless the limit function is uniformly continuous on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

List of Examples

A series of nonuniformly continuous on X functions converges uniformly on X, but nevertheless the sum of the series is uniformly continuous on X. . . . 86 A sequence of nonuniformly continuous on X functions converges uniformly on X to a nonuniformly continuous function. . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 A series of nonuniformly continuous on X functions converges uniformly on X to a nonuniformly continuous function. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87

Chapter 3. Properties of the Limit Function: Differentiability and Integrability Example 1. A sequence fn (x) of differentiable on X functions converges on X to f (x), but the limit function is not differentiable on X or f ′ (x) ≠ lim fn′ (x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 n→∞

A function f (x, y) defined on X × Y is differentiable in x ∈ X at any fixed y ∈ Y , and there exists lim f (x, y) = 𝜑(x), but 𝜑(x) is not differentiable on X y→y0

or 𝜑′ (x) ≠ lim fx (x, y). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95 y→y0

A series of differentiable functions converges on X, but this series cannot be differentiated term by term on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97 Example 2. A sequence fn (x) of differentiable on X functions converges uniformly on X to f (x), but the limit function is not differentiable on X or f ′ (x) ≠ lim fn′ (x). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98 n→∞

A function f (x, y) defined on X × Y is differentiable in x ∈ X at any fixed y ∈ Y , and f (x, y) converges uniformly on X to a function 𝜑(x) as y approaches y0 , but 𝜑′ (x) ≠ lim fx (x, y), ∀x ∈ X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 y→y0

A series of differentiable functions converges uniformly on X, but this series cannot be differentiated term by term on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . 100 Example 3. A sequence fn (x) of differentiable on X functions converges on X to a function f (x) and f ′ (x) = lim fn′ (x), ∀x ∈ X, however, the convergence of n→∞ fn′ (x) to f ′ (x) is nonuniform on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 A function f (x, y) defined on X × Y is differentiable in x ∈ X at any fixed y ∈ Y , f (x, y) converges on X to a function 𝜑(x), as y approaches y0 , and 𝜑′ (x) = lim fx (x, y), ∀x ∈ X, however, the convergence of fx (x, y) is not y→y0

uniform. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103

xix

xx

List of Examples

∑ A series un (x) of differentiable functions converges on X and can be dif∑ ferentiated term by term on X, however, the series u′n (x) converges nonuniformly on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 104 Example 4. A sequence fn (x) of differentiable functions converges nonuniformly on X to a function f (x), but nevertheless f ′ (x) = lim fn′ (x) n→∞ on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 A function f (x, y) defined on X × Y is differentiable in x ∈ X at any fixed parameter y ∈ Y , f (x, y) converges nonuniformly on X to a function 𝜑(x) as y approaches y0 , but nevertheless 𝜑′ (x) = lim fx (x, y) on X. . . . . . . . . . . . . . . 107 y→y0

A series of differentiable functions converges nonuniformly on X, but nevertheless this series can be differentiated term by term on X. . . . . . . . . . . . . . . 107 A series of differentiable functions converges nonuniformly on X and both the partial sums and the sum of the series cannot be expressed through elementary functions, but nevertheless this series can be differentiated term by term on X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 109 Example 5. A sequence fn (x) of infinitely differentiable functions converges uniformly on X, but the sequence fn′ (x) diverges at each point of X. . . . . . . 111 Example 6. A sequence fn (x) of infinitely differentiable functions converges uniformly on X, but the sequence of the derivatives fn′ (x) is convergent and divergent at infinitely many points of X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 Example 7. A series of continuous functions converges uniformly on X, but the sum of this series is not differentiable at an infinite number of points in X. (A function is continuous on an interval, but it can be nondifferentiable at infinitely many points of this interval.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113 ∑ Example 8. A series un (x) of continuous functions converges uniformly on X, but nevertheless the sum of this series is nondifferentiable at any point of X. (There exist continuous on an interval functions f (x) that are nondifferentiable at any point of this interval.) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114 Example 9. A sequence of Riemann integrable on [a, b] functions converges on [a, b], but the limit function is not Riemann integrable on [a, b]. . . . . . . 117 Example 10. A sequence fn (x) converges on [a, b] to f (x), and each of the functions fn (x) is not Riemann integrable on [a, b], but f (x) is Riemann integrable on [a, b]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 Example 11. A sequence of continuous functions fn (x) converges on [a, b] to b b f (x), but lim ∫a fn (x)dx ≠ ∫a f (x)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 118 n→∞

List of Examples

A function f (x, y) defined on X × Y = [a, b] × Y is continuous in x ∈ X b at any fixed y ∈ Y , and there exists lim f (x, y) = 𝜑(x), but ∫a 𝜑(x)dx ≠ y→y0

b lim ∫ y→y0 a

f (x, y)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

A series of continuous functions converges on [a, b], but it cannot be integrated term by term on [a, b]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120 Example 12. A sequence of continuous functions fn (x) converges on [a, b] b b to f (x) and lim ∫a fn (x)dx = ∫a f (x)dx, but the convergence is nonuniform on n→∞ [a, b]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 A function f (x, y) defined on X × Y = [a, b] × Y is continuous in x ∈ X at any b b fixed y ∈ Y , there exists lim f (x, y) = 𝜑(x), and ∫a 𝜑(x)dx = lim ∫a f (x, y)dx, y→y0

y→y0

but f (x, y) converges nonuniformly on X to 𝜑(x). . . . . . . . . . . . . . . . . . . . . . . . 123 A series of continuous functions converges on [a, b] and it can be integrated term by term, but this convergence is nonuniform on [a, b]. . . . . . . . . . . . . . 124 Example 13. The elements of a convergent on X sequence and its limit function have infinitely many discontinuity points on X, but the formula of term-by-term integration holds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127 The terms of a convergent on X series and its sum have infinitely many discontinuity points on X, but the formula of term-by-term integration holds. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127

Chapter 4. Integrals Depending on a Parameter Example 1. A function f (x, y) defined on [a, b] × Y is continuous in x ∈ [a, b] at any fixed y ∈ Y , and there exists lim f (x, y) = 𝜑(x), but 𝜑(x) is not integrable y→y0

on [a, b] or

b ∫a

𝜑(x)dx ≠

b lim ∫ y→y0 a

f (x, y)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 133

Example 2. A function f (x, y) defined on [a, b] × Y is continuous in x for any fixed y, and the function 𝜑(x) = lim f (x, y) is continuous on [a, b], but y→y0

b

b

lim ∫a f (x, y)dx ≠ ∫a lim f (x, y)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

y→y0

y→y0

Example 3. A function f (x, y) defined on [a, b] × Y is continuous in b x ∈ [a, b] at any fixed y ∈ Y , there exists lim f (x, y) = 𝜑(x) and ∫a 𝜑(x)dx = y→y0

b

lim ∫a f (x, y)dx, but f (x, y) converges nonuniformly on [a, b]. . . . . . . . . . . . 135

y→y0

xxi

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List of Examples

Example 4. A function f (x, y) is continuous on ℝ2 except at only one point, b but F(y) = ∫a f (x, y)dx is not continuous. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137 Example 5. A function f (x, y) is continuous in y for any fixed x and also in x b for any fixed y, but F(y) = ∫a f (x, y)dx is not continuous. . . . . . . . . . . . . . . . . 137 Example 6. A function f (x, y) is discontinuous at infinitely many points of b [a, b] × Y , but F(y) = ∫a f (x, y)dx is continuous on Y . . . . . . . . . . . . . . . . . . . 139 A function f (x, y) is discontinuous and unbounded at infinitely many points b of [a, b] × Y , but still F(y) = ∫a f (x, y)dx is continuous on Y . . . . . . . . . . . . . 140 A function f (x, y) is discontinuous at each point of [a, b] × Y , but nevertheb less F(y) = ∫a f (x, y)dx is continuous on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143 Example 7. Both f (x, y) and fy (x, y) are continuous in y for any fixed x and also in x for any fixed y, but

d dy

b

b

∫a f (x, y)dx ≠ ∫a fy (x, y)dx. . . . . . . . . . . . . . . 144

Example 8. Both f (x, y) and fy (x, y) are continuous in ℝ2 except at one point, but

d dy

b

b

∫a f (x, y)dx ≠ ∫a fy (x, y)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 146

Example 9. A function f (x, y) is continuous on ℝ2 and fy (x, 0) exists for b b d ∫a f (x, y)dx|y=0 ≠ ∫a fy (x, 0)dx. . . . . . . . . . . . . . . . . . . . . . . . . . 147 ∀x ∈ ℝ, but dy Example 10. A function f (x, y) has infinitely many points of discontinuity on b b d [a, b] × Y , but dy ∫a f (x, y)dx = ∫a fy (x, y)dx, ∀y ∈ Y . . . . . . . . . . . . . . . . . . . . 151 Example 11. d dy

b

Although

fy (x, y)

is

discontinuous

at

(x0 , y0 ),

but

b

∫a f (x, y)dx|y=y0 = ∫a fy (x, y0 )dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151

Both the function and its partial derivative are discontinuous at some points, but still the integration and partial differentiation can be interchanged. . . 153 Example 12. A function f (x, y), defined on [a, b] × Y , converges on [a, b] to a function 𝜑(x) as y approaches y0 , and f (x, y) is not Riemann integrable on [a, b] for each fixed y ∈ Y , but 𝜑(x) is Riemann integrable on [a, b]. . . . . . . . . . . . 154 Example 13. A function f (x, y) is defined on [a, b] × [c, d] and one of the d b b d iterated integrals—∫c dy ∫a f (x, y)dx or ∫a dx ∫c f (x, y)dy—exists, but another does not exist. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 A function f (x, y) is discontinuous at only one point of [a, b] × [c, d], but d b b d both iterated integrals—∫c dy ∫a f (x, y)dx and ∫a dx ∫c f (x, y)dy—do not exist. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156

List of Examples

Example 14. A function f (x, y) is defined on [a, b] × [c, d] and both iterated d b b d integrals ∫c dy ∫a f (x, y)dx and ∫a dx ∫c f (x, y)dy exist, but they assume different values. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 156 Example 15. A function f (x, y) is defined on [a, b] × [c, d] and b b d ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy, but f (x, y) is discontinuous on [a, b] × [c, d]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 d

A function f (x, y) is defined on [a, b] × [c, d] and both iterated integrals exist and are equal, but nevertheless the double integral does not exist. . . . . . . . 159 Example 16. A function f (x, y) has infinitely many discontinuity points in d b b d [a, b] × [c, d], but ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy. . . . . . . . . . . . . . . . 160 A function f (x, y) is discontinuous and unbounded at infinitely many d b ∫c dy ∫a f (x, y)dx = points in [a, b] × [c, d], but nevertheless b d ∫a dx ∫c f (x, y)dy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

Chapter 5. Improper Integrals Depending on a Parameter +∞

Example 1. An integral ∫a f (x, y)dx converges on a set Y , but it does not converge uniformly on this set. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 +∞

Example 2. An integral ∫a f (x, y)dx converges on a (finite or infinite) interval (c, d) and converges uniformly on any interval [c1 , d1 ] ⊂ (c, d), but it does not converge uniformly on (c, d). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 +∞

Example 3. An integral ∫a f (x, y)dx converges on a (finite or infinite) interval (c, d), but it does not converge uniformly on an interval [c1 , d1 ] ⊂ (c, d). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 167 +∞

Example 4. An integral ∫a f (x, y)dx converges absolutely on an interval (c, d), but it does not converge uniformly on this interval. . . . . . . . . . . . . . . . 170 +∞

Example 5. An integral ∫a f (x, y)dx converges uniformly on an interval Y , but it does not converge absolutely on this interval. . . . . . . . . . . . . . . . . . . . . 171 +∞

Example 6. An integral ∫a f (x, y)dx converges uniformly and absolutely +∞ on an interval Y , but ∫a |f (x, y)|dx does not converge uniformly on this interval. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172 +∞

Example 7. An integral ∫a f (x, y)dx converges uniformly and absolutely on Y , but there is no bound of f (x, y) such that |f (x, y)| ≤ 𝜑(x), ∀y ∈ Y , where the +∞ improper integral ∫a 𝜑(x)dx converges. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173

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xxiv

List of Examples

Example 8. Suppose functions f (x, y) and g(x, y) are positive and continuous +∞ f (x,y) on [a, +∞) × Y , and lim g(x,y) = 1, ∀y ∈ Y . One of the integrals ∫a f (x, y)dx x→+∞

+∞

or ∫a g(x, y)dx converges uniformly on Y , but another converges nonuniformly. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174 +∞

+∞

Example 9. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx con+∞ verge nonuniformly on Y , but the integral ∫a f (x, y) + g(x, y)dx converges uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 176 +∞

+∞

Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge nonuniformly +∞ on Y , and the integral ∫a f (x, y) + g(x, y)dx also converges nonuniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177 +∞

+∞

Example 10. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge +∞ nonuniformly on Y , but the integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 178 +∞

An integral ∫a f (x, y)dx converges nonuniformly on Y , but the integral converges uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179

+∞ ∫a f 2 (x, y)dx

+∞

+∞

Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge nonuniformly +∞ on Y , and the integral ∫a f (x, y)g(x, y)dx also converges nonuniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 +∞

+∞

Example 11. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge +∞ uniformly on Y , but the integral ∫a f (x, y)g(x, y)dx converges nonuniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 179 +∞

An integral ∫a f (x, y)dx converges uniformly on Y , but the integral does not converge uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . 180

+∞ ∫a f 2 (x, y)dx

+∞

+∞

Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge uniformly on +∞ Y , and ∫a f (x, y)g(x, y)dx also converges uniformly on Y . . . . . . . . . . . . . . . 180 +∞

Example 12. An improper integral ∫a f (x, y)dx converges uniformly +∞ and ∫a g(x, y)dx converges nonuniformly on Y , but the improper integral +∞ ∫a f (x, y)g(x, y)dx converges uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . 181 +∞

+∞

An improper integral ∫a f (x, y)dx converges uniformly and ∫a g(x, y)dx +∞ converges nonuniformly on Y , but the improper integral ∫a f (x, y)g(x, y)dx converges nonuniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 +∞

+∞

Example 13. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx diverge +∞ on Y , but nevertheless the integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182

List of Examples +∞

+∞

Example 14. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx con+∞ verge uniformly on Y , but the improper integral ∫a f (x, y)g(x, y)dx diverges on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183 A

Example 15. For each fixed y ∈ Y , integral ∫a f (x, y)dx is bounded and function g(x, y) is monotone in x ∈ [a, +∞); additionally, g(x, y) converges +∞ uniformly to 0 on Y as x → +∞, but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185 A

Example 16. The integral ∫a f (x, y)dx is uniformly bounded on Y and g(x, y) converges uniformly to 0 on Y as x → +∞, but the improper integral +∞ ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . . . . . . . . . . . . . . . . . . 186 A

Example 17. The integral ∫a f (x, y)dx is uniformly bounded on Y , and the function g(x, y) is monotone in x and converges to 0 for ∀y ∈ Y as x → +∞, +∞ but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 The violation of one of the conditions of Dirichlet’s theorem can lead +∞ not only to nonuniform convergence of ∫a f (x, y)g(x, y)dx but even to divergence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 187 Example 18. All the conditions of Dirichlet’s theorem are violated, but +∞ nevertheless the improper integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 188 +∞

Example 19. For each fixed y ∈ Y , integral ∫a f (x, y)dx is convergent and function g(x, y) is monotone in x ∈ [a, +∞); additionally, g(x, y) is uniformly +∞ bounded on [a, +∞) × Y , but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 189 +∞

Example 20. The integral ∫a f (x, y)dx is uniformly convergent and g(x, y) +∞ is uniformly bounded on Y , but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 +∞

Example 21. The integral ∫a f (x, y)dx converges uniformly on Y , and the function g(x, y) is monotone in x and bounded for each fixed y ∈ Y , +∞ but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 190 The violation of only one condition in Abel’s theorem results in the divergence +∞ of ∫a f (x, y)g(x, y)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 191 Example 22. All the conditions of Abel’s theorem are violated, but nev+∞ ertheless the improper integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192

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List of Examples

Example 23. A function f (x, y) is continuous in x on [a, +∞) for any fixed y ∈ Y and converges uniformly on [a, +∞) to a function 𝜑(x), as y approaches +∞ +∞ y0 ; also both improper integrals ∫a f (x, y)dx and ∫a 𝜑(x)dx are convergent, +∞ +∞ but lim ∫a f (x, y)dx ≠ ∫a lim f (x, y)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192 y→y0

y→y0

Example 24. A function f (x, y) is continuous in x on [a, +∞) for any fixed y ∈ Y , converges uniformly on [a, +∞) to 𝜑(x), as y approaches y0 , +∞ +∞ +∞ and lim ∫a f (x, y)dx = ∫a lim f (x, y)dx, but ∫a f (x, y)dx converges y→y0

y→y0

nonuniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 Example 25. A function f (x, y) is continuous in two variables on X × Y = +∞ [a, +∞) × [c, d] and the improper integral F(y) = ∫a f (x, y)dx converges on Y , but F(y) is discontinuous on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 197 Example 26. A function f (x, y) is continuous in two variables on [a, +∞) × Y , +∞ where Y is an interval, and F(y) = ∫a f (x, y)dx is continuous on Y , but the +∞ improper integral ∫a f (x, y)dx does not converge uniformly on Y . . . . . . 197 Example 27. Both f (x, y) and fy (x, y) are continuous on [a, +∞) × Y , +∞ where Y is an interval, and ∫a f (x, y)dx converges uniformly on Y , but +∞ +∞ (∫a f (x, y)dx)y ≠ ∫a fy (x, y)dx on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 198 Example 28. Both f (x, y) and fy (x, y) are continuous on [a, +∞) × Y , where +∞ +∞ Y is an interval, and both integrals ∫a f (x, y)dx and ∫a fy (x, y)dx converge +∞ +∞ on Y , but (∫a f (x, y)dx)y ≠ ∫a fy (x, y)dx on Y . . . . . . . . . . . . . . . . . . . . . . . . 199 Example 29. Both f (x, y) and fy (x, y) are continuous on [a, +∞) × Y , where +∞ +∞ Y is an interval, and the integrals ∫a f (x, y)dx and ∫a fy (x, y)dx converge +∞ +∞ on Y and ∫a fy (x, y)dx = (∫a f (x, y)dx)y , ∀y ∈ Y , but the improper integral +∞ ∫a fy (x, y)dx converges nonuniformly on Y . . . . . . . . . . . . . . . . . . . . . . . . . . . 200 +∞

+∞

Both integrals ∫a f (x, y)dx and ∫a fy (x, y)dx converge nonuniformly on Y , +∞ +∞ but ∫a fy (x, y)dx = (∫a f (x, y)dx)y for ∀y ∈ Y . . . . . . . . . . . . . . . . . . . . . . . . 200 Example 30. A function f (x, y) is continuous on [a, +∞) × [c, d] and +∞ d +∞ the integral ∫a f (x, y)dx converges on [c, d], but ∫c dy ∫a f (x, y)dx ≠ +∞ d ∫a dx ∫c f (x, y)dy. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202 Example 31. A function f (x, y) is continuous on [a, +∞) × [c, d] and +∞ +∞ d ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy, but the improper integral +∞ ∫a f (x, y)dx converges nonuniformly on [c, d]. . . . . . . . . . . . . . . . . . . . . . . . . 203 d

List of Examples

Example 32. A function f (x, y) is continuous on [a, +∞) × [c, +∞), and the +∞ +∞ +∞ +∞ improper integrals ∫c dy ∫a f (x, y)dx and ∫a dx ∫c f (x, y)dy converge, but they assume different values. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204 Example 33. A function f (x, y) is continuous and nonnegative on [a, +∞) × +∞ +∞ +∞ +∞ [c, +∞) and ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy, but at least one of +∞ +∞ the functions F(y) = ∫a f (x, y)dx and G(x) = ∫c f (x, y)dy is discontinuous on [c, +∞) or [a, +∞), respectively. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 207 Example 34. A function f (x, y) is continuous on [a, +∞) × [c, +∞), and both +∞ +∞ improper integrals F(y) = ∫a f (x, y)dx and G(x) = ∫c f (x, y)dy do not con+∞ +∞ +∞ +∞ verge uniformly, but ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy. . . . . . . 207

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xxix

List of Figures

Figure 1.1 Examples 1, 4, and 28, sequence fn (x) = xn . . . . . . . . . . . . . . . . . . . . 2 ∑∞ Figure 1.2 Examples 1 and 4, series n=0 xn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 ∑∞ n Figure 1.3 Examples 2, 26, 27, and 30, series n=1 xn . . . . . . . . . . . . . . . . . . . . 4 2 ∑∞ Figure 1.4 Example 7, series n=1 (−1)n x n+n . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2 Figure 1.5 Examples 9 and 10 (second counterexample), series ∑∞ n n n=0 (−1) (1 − x)x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13 ∑∞ Figure 1.6 Example 10, series n=1 un (x), { 0, x ∈([0, 2−n−1)] ∪ [2−n 1] ( , −n−1 ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14 un (x) = 1 2 n+1 sin 2 𝜋x , x ∈ 2 , 2−n n nx Figure 1.7 Examples 11 and 17, sequence fn (x) = ln n+1 . . . . . . . . . . . . . . . . . 17

Figure 1.8 Example 11, sequence of squares fn2 (x) = ln2 2

nx . n+1

. . . . . . . . . . . . 17

Figure 1.9 Example 16, sequence fn (x) = e−(x−n) . . . . . . . . . . . . . . . . . . . . . . . . ∑∞ 1 Figure 1.10 Examples 19, 20, and 21, series n=1 x+n ................... ∑∞ 1 Figure 1.11 Example 19, series of squares n=1 (x+n) ................... 2 ∑∞ Figure 1.12 Examples 22, 5, and 18, series n=1 sinnnx . . . . . . . . . . . . . . . . . . . ∑∞ Figure 1.13 Examples 22 and 24, series n=1 sin√nx . . . . . . . . . . . . . . . . . . . . . . n

23 24 25 27 27

2 ∑∞ Figure 1.14 Example 22, series of products n=1 sinn3∕2nx . . . . . . . . . . . . . . . . . . 28 n ∑∞ Figure 1.15 Example 23, series n=1 (−1)n √3x . . . . . . . . . . . . . . . . . . . . . . . . . . 29

n

Figure 1.16 Example 23, series of squares

∑∞

x2n n=1 n2∕3 . .

. . . . . . . . . . . . . . . . . . . 29

xxx

List of Figures

Figure 1.17 Example 24, series of squares

∑∞

sin2 nx .. n

. . . . . . . . . . . . . . . . . 31 ∑∞ ∑∞ Figure 1.18 Examples 26, 29, and 33, series n=1 un (x) = n=1 xn . . . . . . . 33 ∑∞ Figure 1.19 Examples 26, 27, 30, 31, and 32, series n=1 un (x)vn (x) = n ∑∞ x n=1 n . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34 n=1

Figure 1.20 Examples 29 and 33, sequence vn (x) =

(−1)n . xn2

. . . . . . . . . . . . . . . 35 ∑∞ Figure 1.21 Examples 29 and 33, series n=1 un (x)vn (x) = n−1 ∑∞ nx n=1 (−1) n2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36 { ( ) min n, 1x , 0 < x ≤ 1 Figure 2.1 Example 1, sequence fn (x) = . . . . . . 46 0, x = 0 [ ] ⎧ n2 x + n − n3 , x ∈ n − 1 , n [ n ] ⎪ Figure 2.2 Example 2, sequence fn (x) = ⎨ −n2 x + n + n3 , x ∈ n, n + 1 . 47 n ⎪ ⎩ 0, otherwise { 1 , 00

0, x = 0

. . . . . . . . 194

+∞

Figure 5.12 Example 23, integral F(y) = ∫0

f (x, y)dx. . . . . . . . . . . . . . . . . 195 ( ) y Figure 5.13 Examples 28 and 29, function f (x, y) = x + x12 e−xy . . . . . . . 201 +∞

Figure 5.14 Examples 28 and 29, integrals F(y) = ∫1 f (x, y)dx, G(y) = +∞ ∫1 fy (x, y)dx. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 202

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About the Companion Website This book is accompanied by a companion website: www.wiley.com/go/bourchtein/counterexamples_on_uniform_convergence The website includes: • Solution Manual

xxxv

Introduction I.1

Comments

I.1.1 On the Structure of This Book

This book consists of the introductory chapter and five chapters of counterexamples. The Introduction fixes terminology and notations, and it contains some comments on presentation form and a brief overview of the background theory. The content of each of the five main chapters is as follows: 1) In Chapter 1, different conditions involving the uniform convergence of sequences, series, and functions depending on a parameter are considered—relations between different types of convergence, between convergence on a given set and its subsets, implications involving the sequences/series of the squares and products of the original terms, analysis of the conditions of Dirichlet’s theorem, and Abel’s theorem on uniform convergence of series. 2) In Chapter 2, the properties of the boundedness and continuity of the limit functions are investigated—relations between boundedness of the terms of a convergent sequences/series/functions depending on a parameter and that of the limit function, the conditions that do and do not guarantee the continuity of the limit function, and analysis of the conditions of Dini’s theorem on uniform convergence. 3) Chapter 3 concerns the differentiability and integrability of the limit functions; first, the relations between differentiablity of the terms of a sequence/series/function depending on a parameter and that of the limit function are studied under the conditions of uniform/nonuniform convergence, including the possibility of term-by-term differentiation; then, similar results are presented for the Riemann integral. 4) In Chapter 4, the properties of the Riemann integrals depending on a parameter are considered; it includes the conditions on a function of two variables that do and do not guarantee the existence of a limit of such integral, its continuity, differentiability, and integrability with respect to

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Introduction

parameter, and the possibility to interchange the order of operations; some examples involve improper integrals on a finite interval, but an analysis of the uniform/nonuniform convergence of such integrals is postponed to the last part. 5) Chapter 5 deals with the properties of improper integrals depending on a parameter, many of them similar to those in preceding chapters—the nature of the convergence of improper integrals, relations between pointwise, absolute, and uniform convergence, conditions of existence/nonexistence of the limit, and the properties of continuity, differentiability, and integrability; the exposition of the material is focused on the improper integrals of the first kind (on an infinite interval) since the improper integrals of the second kind (of unbounded functions) can be reduced to the former type by a simple change of a variable. Each chapter is divided into sections corresponding to the (conditional) division of the material in main topics. At the end of each chapter, supplementary exercises of different levels of complexity are provided, the most difficult of them with a hint to the solution. All the solutions to the proposed exercises can be found in the Instructor’s Solutions Manual. The logical sequence of the material just follows the chapter sequence, which means that the reading of a specific chapter/section does not require any knowledge of the subjects of next chapters/sections. At the same time, different concepts, results, and examples considered and analyzed in earlier chapters/sections are frequently used in the subsequent sections. Since the use of counterexamples in the book is aimed to analyze important concepts and theorems in calculus and analysis, the selection of the material is restricted by those examples that have direct connection with main concepts. Therefore, more specific and applied problems are left out of the scope of the manuscript. The interested reader can find such exercises in different collections of problems on real analysis. Some well-known books of this type, containing problems of different levels of difficulty, are indicated in the bibliography: [2], [6], [10], [13], [14], and [15]. The following structure is chosen for the presentation of each counterexample: 1) Example statement that implicitly invokes the false general statement aimed to be disproved by this counterexample. 2) Solution (counterexample(s)), which provides a complete analytic solution to the posed problem. 3) Remark(s) (optional), which offer additional explanations, extensions of the counterexamples, and comparisons with other similar situations, and which also indicate links to a general theory and make comparisons with the correct statements.

Introduction

4) Figure(s) (optional, for some examples involving more geometric and complex constructions) give geometric illustrations of the analytic solutions and clarify the analytic arguments in some complex cases. Each example statement is formulated in the way that resembles the corresponding general false statement. Due to this choice, some formulations seem to be more complicated than they could be, but we decided to keep this form in order to make a clear association with implicitly invoked false statements. For instance, the example statement “a sequence of discontinuous functions converges uniformly on X, but the limit function is continuous on X” can be reformulated in a simpler form as “a sequence of discontinuous functions converges uniformly on X to a continuous function,” but we chose the first version since it is directly associated with the corresponding false statement “if a sequence of discontinuous functions converges uniformly on X, then the limit function is discontinuous on X.” For the same reasons of the logic and clarity of exposition, the well-known example of “an everywhere continuous and nowhere ∑ differentiable function” is initially formulated in the form: “a series un (x) of continuous functions converges uniformly on X, but nevertheless the sum of this series is not differentiable at any point of X.” The latter form allows us to stress the connection with the analyzed concept of the uniform convergence and to recall (implicitly) the corresponding general false statement: “if a series ∑ un (x) of continuous functions converges uniformly on X, then the sum of this series is differentiable at least at one point of X.” (Of course, without a reference to the series convergence, this false statement has a simpler form: “if a function is continuous everywhere, then it should be differentiable at least at one point.”) In the choice of figures for illustration of the analytic solutions, we don’t try to provide as many pictures as possible or to keep an equal distribution of pictures among chapters and sections. Instead, we follow the principle of choosing the figures that can be most helpful for understanding the presented counterexamples (at least from our point of view). In the first place, it is related to the nature of each counterexample (if it is more analytic or geometric) and to the complexity and singularity of the functions employed in the solution. I.1.2 On Mathematical Language and Notation

The language in the manuscript, just as in the majority of mathematical books, is a mixture of a natural language with mathematical terminology and symbolism used traditionally for a concise expression of numerous concepts, results, and logical reasoning. Depending on the topic, the use of mathematical symbols and terms can be more or less concentrated. In order to facilitate access to these symbols/terms inside the text, the list of notations and subject index are provided.

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Throughout this work, we try to follow the standard terminology and notation used in calculus/analysis books. To avoid any misunderstanding and ambiguity, the main concepts and results, along with the corresponding terminology and notation, are set out in Section I.2 containing background material. Besides, a preliminary list of symbols (those used most frequently, somewhat ambiguous or nontraditional) is presented below: 1) ∀—any, every, each, for all; 2) ℕ, ℤ, ℚ, 𝕀, ℝ—set of natural, integer, rational, irrational, and real numbers, respectively; 3) fn (x)—sequence of functions; ∑ 4) un (x)—series of functions; 5) f (x, y)—function of two variables or{function depending on a parameter y; 1, x ∈ ℚ 6) D(x)—Dirichlet’s function, D(x) = ; 0, x ∈ 𝕀 {1 , x = mn ∈ ℚ n 7) R(x)—Riemann function, R(x) = , where m is integer, n is 0, x ∈ 𝕀 m natural, and n is in lowest terms; 8) fx —(first-order) partial or ordinary derivative in x; 9) const—constant value, for instance, f (x) = const means a constant function.

I.2 Background (Elements of Theory) The background material contains some basic concepts and results from the theory of (infinite) sequences and series of functions and also functions and integrals depending on a parameter. Some parts of this material can be found in different analysis books, and more complete treatments can be found in monographs on sequences and series of functions. In particular, one can consult the analysis textbooks indicated in the bibliography list, which treat the subjects with different levels of strictness, abstraction, and generalization: [1], [3], [7], [9], [12], [16], and [17]. The classical monographs on the subject include [4], [5], and [11]. I.2.1 Sequences of Functions I.2.1.1 Basic Definitions

Remark. In this text, we consider only the real-valued functions. Definition. A sequence of functions. If each n ∈ ℕ is associated with a function fn (x) defined on a set X ⊂ ℝ, then fn (x) is called a sequence of functions defined on X. More formally, a sequence of functions is a mapping from ℕ into a set of functions Φ that assigns to each n ∈ ℕ exactly one function in Φ.

Introduction

Remark. The definition is usually extended to domains that include additional integers or exclude some naturals. The main point in these extensions is to keep the domain with the properties of ordering of ℕ: all the elements of a sequence should be indexed (with an integer index), the first element with an initial index must exist, and each following element has the index equal to the index of the preceding element plus one. Definition. Convergence of a sequence of functions at a point. Let a sequence of functions fn (x) be defined on a set X. The sequence fn (x) is said to be convergent/divergent at a point x0 ∈ X if the numerical sequence fn (x0 ) is convergent/divergent. Accordingly, x0 is called a point of convergence/divergence. Definition. Convergence of a sequence of functions on a set. Let a sequence of functions fn (x) and a function f (x) be defined on a set X. The sequence fn (x) is said to be convergent pointwise (or simply convergent) on X to f (x) if for each x ∈ X the numerical sequence fn (x) converges to f (x). Restated in more detailed 𝜖 − N form, the (pointwise) convergence of the sequence fn (x) on X means that for every fixed x ∈ X and any 𝜖 > 0, there exists a number N ∈ ℕ (dependent on 𝜖 and, perhaps, on x) such that the condition n > N implies |fn (x) − f (x)| < 𝜖. The function f (x) is called the (pointwise) limit function of the sequence fn (x) on X. Definition. Uniform convergence of a sequence of functions. A sequence fn (x) converges uniformly on a set X to a limit function f (x) if, for any 𝜖 > 0, there exists a number N ∈ ℕ (dependent on 𝜖, but independent of x) such that the condition n > N implies |fn (x) − f (x)| < 𝜖 simultaneously for all x ∈ X. The function f (x) is called the uniform limit of fn (x) on X. Evidently, the uniform convergence implies the pointwise convergence. Definition. Nonuniform convergence of a sequence of functions. A sequence fn (x) converges pointwise but nonuniformly on a set X to a limit function f (x), if there exists 𝜖 > 0 such that for any N there are nN > N and xN ∈ X for which |fnN (xN ) − f (xN )| ≥ 𝜖. Remark. In some cases, the following more simple condition may be applied to show that the convergence is nonuniform: there exists 𝜖 > 0 such that for any n there exists xn ∈ X for which |fn (xn ) − f (xn )| ≥ 𝜖. Evidently, this condition implies that given in the definition of nonuniform convergence.

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I.2.1.2 Conditions of the Uniform Convergence

Criterion of uniform convergence of a sequence. A sequence fn (x) converges uniformly on X to a function f (x) if and only if sup|fn (x) − f (x)| → 0 as n → ∞, x∈X

that is, the numerical sequence of the maximum deviations of fn (x) from f (x) on X converges to zero. Cauchy criterion for uniform convergence of a sequence. A sequence of functions fn (x) converges uniformly on X if and only if for every 𝜖 > 0 there exists N (independent of x) such that for all n > N and all p > 0 the inequality |fn+p (x) − fn (x)| < 𝜖 holds for all x ∈ X simultaneously. Dini’s theorem on uniform convergence. Let fn (x) be a sequence of continuous functions on a compact set X. If fn (x) converges to a continuous function f (x) on X, and if this convergence is monotone with respect to n for every fixed x ∈ X, then fn (x) converges uniformly on X. I.2.1.3 Properties of the Uniformly Convergent Sequences

Arithmetic properties of uniform convergence. 1) If fn (x) and gn (x) converge uniformly on X, then fn (x) + gn (x) also converges uniformly on X. 2) If fn (x) converges uniformly on X and g(x) is a bounded function on X, then g(x)fn (x) converges uniformly on the same set. Theorem. Passage to limit term by term. If a sequence of functions fn (x) converges uniformly on X to f (x) and for every n there exists a finite limit lim fn (x) = an , then the limit function f (x) also has a limit at x0 , the numerical x→x0

sequence an is also convergent and lim lim fn (x) = lim lim fn (x). x→x0 n→∞

n→∞ x→x0

Theorem. Continuity and uniform convergence. If a sequence fn (x) converges uniformly on X to f (x), and if each fn (x) is continuous at x0 ∈ X, then f (x) is also continuous at x0 . If a sequence of continuous on X functions fn (x) converges uniformly on X to f (x), then the limit function f (x) is also continuous on X. Theorem. Uniform continuity and uniform convergence. If a sequence of uniformly continuous on X functions fn (x) converges uniformly on X to f (x), then the limit function f (x) is also uniformly continuous on X. Theorem. Interchange of the limit and integration signs. If a sequence of integrable (by Riemann) on [a, b] functions fn (x) converges uniformly on [a, b] to a function f (x), then f (x) is integrable (by Riemann) on [a, b] and b

∫a

b

lim fn (x)dx = lim

n→∞

n→∞ ∫a

fn (x)dx.

Introduction

That is, one can interchange the order of integration and limit calculation. This formula is also called term-by-term integration for sequences. Theorem. Interchange of the limit and differentiation signs. Let fn (x) be a sequence of differentiable on a finite interval I functions, and assume fn′ (x) converges uniformly on I to a function g(x). If there exists a point x0 ∈ I at which fn (x0 ) is convergent, then fn (x) converges uniformly on I. Moreover, the limit function f (x) = lim fn (x) is differentiable and satisfies f ′ (x) = g(x), that is, (

n→∞

)′ lim fn (x) = lim fn′ (x), ∀x ∈ I.

n→∞

n→∞

That is, one can pass to the derivative under the limit sign. This formula is also called term-by-term differentiation for sequences. I.2.2 Series of Functions

Remark. The definitions for a series of functions and statements on its convergence follow the usual scheme, which reduces the series behavior to that of the sequences of its partial sums. In this way, almost all definitions and results stated for the convergence of sequences of functions can be reformulated for a series of functions. I.2.2.1 Basic Definitions

Definition. A series of functions. The sum of all the elements of a sequence of functions un (x) is called series of functions. The standard notation is ∑+∞ ∑+∞ ∑ un (x). The n=1 un (x) or n=i un (x), where i is the initial index, or simply function un (x) is called the general term of a series. Definition. Partial sums. As in the case of numerical series, the sum of the ∑n first n terms of a series is called the nth partial sum: fn (x) = k=1 uk (x) or fn (x) = ∑n+i−1 uk (x), n ≥ 1. k=i Remark. Even when a series starts from the index i, the nth partial sum can ∑n be defined as fn (x) = k=i uk (x), n ≥ i. The choice of one of the two options for nominating the partial sums has no influence on the main properties of series, including convergence/divergence properties. Definition. Convergence of a series of functions at a point. Let a sequence ∑ of functions un (x) be defined on a set X. The series un (x) is convergent/divergent at a point x0 ∈ X if the sequence of its partial sums fn (x) is convergent/divergent at x0 , that is, if the numerical sequence fn (x0 ) is convergent/divergent.

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Definition. Convergence of a series of functions on a set. Let a sequence of ∑ functions un (x) be defined on a set X. The series un (x) is (pointwise) convergent on X if the sequence of its partial sums fn (x) is (pointwise) convergent on X. In 𝜖 − N terms, it can be formulated as follows: for each x ∈ X and for every 𝜖 > 0, there exists a number N (perhaps, dependent on x) such that ∑∞ |rn (x)| = |f (x) − fn (x)| = || k=n+1 uk (x)|| < 𝜖 whenever n > N. If a series converges, the limit of the partial sums f (x) is called the sum of the ∑ series: un (x) = f (x). The difference between f (x) and fn (x) is called the nth residual: rn (x) = f (x) − fn (x). Remark. For the sake of brevity, in the text, the sum of the series is also called the limit function (as for a sequence), and the difference f (x) − fn (x) between the limit function of a convergent sequence and its general term is called the residual (as for a series). Definition. Uniform convergence of a series of functions. The series ∑ un (x) is uniformly convergent on X if the sequence of the partial sums fn (x) is uniformly convergent on X to its sum f (x). It means that for every 𝜖 > 0, there exists a number N (independent of x) such that ∑∞ |rn (x)| = |f (x) − fn (x)| = || k=n+1 uk (x)|| < 𝜖 for all x ∈ X simultaneously whenever n > N. Evidently, the uniform convergence implies the pointwise convergence. Definition. Nonuniform convergence of a series of functions. A series ∑ un (x) converges pointwise but nonuniformly on a set X to its sum f (x) if there exists 𝜖 > 0 such that for any N there are nN > N and xN ∈ X for which |rnN (xN )| = |f (xN ) − fnN (xN )| ≥ 𝜖. Remark. In some cases, the following more simple condition may be applied to show that the convergence is nonuniform: there exists 𝜖 > 0 such that for any n there exists xn ∈ X for which |rn (xn )| = |f (xn ) − fn (xn )| ≥ 𝜖. Evidently, this condition implies the condition given in the last Definition. I.2.2.2 Conditions of the Uniform Convergence

∑ Criterion of uniform convergence of a series. A series un (x) is uniformly convergent on X to its sum f (x) if and only if sup|f (x) − fn (x)| = x∈X ∑∞ sup || k=n+1 uk (x)|| → 0 as n → ∞. x∈X Cauchy criterion for uniform convergence of a series. A series of functions ∑ un (x) is uniformly convergent on X if and only if for every 𝜖 > 0 there exists N (independent of x) such that for all n > N and all p > 0 the inequality |fn+p (x) − |∑n+p | fn (x)| = | k=n+1 uk (x)| < 𝜖 is fulfilled simultaneously for all x ∈ X. | |

Introduction

Necessary condition for uniform convergence of a series. If a series of ∑ functions un (x) is uniformly convergent on X, then the sequence un (x) converges uniformly to 0 on X. ∑ Definition. Majorant (dominant) series. A numerical series Mn with nonnegative terms is said to be a majorant (or dominant) series for a series ∑ of functions un (x) on X if the inequalities sup|un (x)| ≤ Mn hold for all x∈X

sufficiently large indices n ∈ ℕ. ∑ Weierstrass M-test. If for a series of functions un (x) defined on X there ∑ ∑ exists a convergent majorant series Mn , then the series un (x) converges uniformly on X. Definition. Uniform boundedness. A sequence un (x) defined on X is uniformly bounded on X if there exists a constant C (independent of x) such that |un (x)| ≤ C for ∀n and ∀x ∈ X. ∑ Dirichlet’s theorem. Let a series un (x)vn (x) be defined on X. If the partial ∑ sums of the series un (x) are uniformly bounded on X, and the sequence vn (x) is monotone with respect to n (for each fixed x ∈ X) and converges uniformly ∑ to 0 as n → +∞, then un (x)vn (x) converges uniformly on X. ∑ Abel’s theorem. Let a series un (x)vn (x) be defined on X. If the series ∑ un (x) converges uniformly on X, and the sequence vn (x) is uniformly bounded on X and monotone with respect to n (for each fixed x ∈ X), then ∑ un (x)vn (x) converges uniformly on X. ∑ Dini’s theorem on uniform convergence. Let un (x) be a series of continuous functions on a compact set X. If un (x) ≥ 0 on X for ∀n, and if the series ∑ converges to a continuous function f (x) on X, then un (x) converges uniformly on X. I.2.2.3 Properties of the Uniformly Convergent Series

Arithmetic properties of uniform convergence. ∑ ∑ ∑ 1) If un (x) and vn (x) converge uniformly on X, then (un (x) + vn (x)) also converges uniformly on X. ∑ 2) If un (x) converges uniformly on X and v(x) is a bounded function on X, ∑ then v(x)un (x) converges uniformly on the same set. ∑ Theorem. Passage to limit term by term in series. If a series un (x) converges uniformly on X and for every n there exists a finite limit lim un (x) = an , x→x0 ∑ ∑ then the sum of series f (x) = un (x) also has limit at x0 , the series an is ∑ ∑ also convergent, and lim un (x) = lim un (x). It means that in a uniformly x→x0

x→x0

convergent series it is allowable to use term-by-term passage to the limit.

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Theorem. Continuity and uniform convergence. If all the terms of a series ∑ un (x) are continuous functions on X and the series is uniformly convergent on X, then its sum is also continuous on X. Theorem. Uniform continuity and uniform convergence. If a series ∑ un (x) of uniformly continuous on X functions un (x) converges uniformly on X, then its sum is also uniformly continuous on X. Theorem. Term-by-term integration of series. Let un (x) be integrable (by ∑ Riemann) on [a, b] functions. If a series un (x) is uniformly convergent on [a, b] to a function f (x), then f (x) is (Riemann) integrable on [a, b] and b

∫a

∑

un (x)dx =

∑

b

∫a

un (x)dx,

that is, the uniformly convergent series can be integrated term by term. Theorem. Term-by-term differentiation of series. Let un (x) be differen∑ tiable on a finite interval I functions, and assume u′n (x) converges uniformly ∑ to g(x) on I. If there exists a point x0 ∈ I where un (x0 ) converges, then ∑ the series un (x) converges uniformly on I to a differentiable function f (x) ∑ ∑ satisfying f ′ (x) = g(x), that is, ( un (x))′ = u′n (x), ∀x ∈ I. In other words, the series can be differentiated term by term. I.2.3 Families of Functions I.2.3.1 Basic Definitions

Definition. A family of functions. A function f (x, y) defined on X × Y ⊂ ℝ2 is called a family of functions depending on a parameter y if one of the variables, say y, is set apart for some reason. Remark. Frequently, a family of functions depending on a parameter is also called a function depending on a parameter. Definition. Convergence of a family of functions at a point. A family of functions f (x, y) is said to be convergent/divergent at a point x0 ∈ X as y approaches y0 if the function f (x0 , y) converges/diverges as y approaches y0 . Definition. Convergence of a family of functions on a set. A family of functions f (x, y) is said to be convergent pointwise (or simply convergent) on X to 𝜑(x) if for each fixed x ∈ X the function f (x, y) converges to 𝜑(x) as y approaches y0 . Restated in more detailed 𝜖 − 𝛿 form, it means that for each given x ∈ X and for any 𝜖 > 0 there exists a number 𝛿 > 0 (dependent on 𝜖 and, perhaps, on x) such that |f (x, y) − 𝜑(x)| < 𝜖 whenever 0 < |y − y0 | < 𝛿.

Introduction

The function 𝜑(x) is called the limit function of the family f (x, y) on X. Definition. Uniform convergence of a family of functions. A family of functions f (x, y) is said to be uniformly convergent on X to 𝜑(x) if for any 𝜖 > 0, there exists a number 𝛿 > 0 (dependent on 𝜖, but independent of x) such that if 0 < |y − y0 | < 𝛿 then |f (x, y) − 𝜑(x)| < 𝜖 for all x ∈ X simultaneously. The function 𝜑(x) is called the uniform limit of f (x, y) on X. Evidently, the uniform convergence implies the pointwise convergence. Definition. Nonuniform convergence of a family of functions. A family f (x, y) converges pointwise but nonuniformly on a set X to a limit function 𝜑(x) if there exists 𝜖 > 0 such that for any 𝛿 > 0 there are x𝛿 ∈ X and y𝛿 ∈ Y , 0 < |y𝛿 − y0 | < 𝛿 such that |f (x𝛿 , y𝛿 ) − 𝜑(x𝛿 )| ≥ 𝜖. Remark. In some cases, the following more simple condition may be applied to show that the convergence is nonuniform: there exists 𝜖 > 0 such that for ∀y ∈ Y there exists xy ∈ X for which |f (xy , y) − 𝜑(xy )| ≥ 𝜖. Evidently, this condition implies the condition given in the definition of nonuniform convergence. I.2.3.2 Conditions of the Uniform Convergence

Criterion of uniform convergence of a family. A family f (x, y) converges uniformly on X to a function 𝜑(x) as y approaches y0 if and only if lim sup|f (x, y) − y→y0 x∈X

𝜑(x)| = 0. Cauchy criterion for uniform convergence of a family. A family f (x, y) converges uniformly on X as y approaches y0 if and only if for every 𝜖 > 0, there exists 𝛿 > 0 (independent of x) such that ∀y1 , y2 ∈ Y the conditions 0 < |y1 − y0 | < 𝛿 and 0 < |y2 − y0 | < 𝛿 imply |f (x, y1 ) − f (x, y2 )| < 𝜖 for all x ∈ X simultaneously. Dini’s Theorem on uniform convergence. Let f (x, y) be continuous on a compact set X for each fixed y ∈ Y . If f (x, y) converges on X to a continuous function 𝜑(x) as y → y0 , and if this convergence is monotone with respect to y for every fixed x ∈ X, then f (x, y) converges uniformly to 𝜑(x) on X. I.2.3.3 Properties of the Uniformly Convergent Families

Arithmetic properties of uniform convergence. 1) If f (x, y) and g(x, y) converge uniformly on X, then f (x, y) + g(x, y) also converges uniformly on X. 2) If f (x, y) converges uniformly on X and g(x) is a bounded function on X, then g(x)f (x, y) converges uniformly on the same set. Theorem. Passage to limit. Let a family of functions f (x, y) be defined on X × Y and x0 , y0 be the limit points of X and Y , respectively. If f (x, y) converges uniformly on X to 𝜑(x) as y approaches y0 and for each y ∈ Y there exists

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the limit lim f (x, y) = 𝜓(y), then the limits lim 𝜑(x) and lim 𝜓(y) exist and are x→x0

x→x0

y→y0

equal, that is, lim lim f (x, y) = lim lim f (x, y). x→x0 y→y0

y→y0 x→x0

Theorem. Continuity and uniform convergence. If f (x, y) converges uniformly on X to 𝜑(x) as y approaches y0 and if each f (x, y) is continuous in x on X for any fixed y ∈ Y , then the limit function 𝜑(x) is continuous on X. Theorem. Interchange of the limit and differentiation signs. Let f (x, y) be a family of functions defined on I × Y , where I is a finite interval, and differentiable in x ∈ I at any fixed y ∈ Y . If fx (x, y) converges uniformly on I to 𝜓(x) and f (x, y) converges at least at one point x0 ∈ I, as y approaches y0 , then f (x, y) converges uniformly on I to a differentiable function 𝜑(x) and 𝜑′ (x) = 𝜓(x), ∀x ∈ I, d that is, dx lim f (x, y) = lim fx (x, y). y→y0

y→y0

I.2.3.4 Integrals Depending on a Parameter

Definition. Integral depending on a parameter. Let f (x, y) be a family of functions defined on [a, b] × Y ⊂ ℝ2 . Integral depending on a parameter has b the form ∫a f (x, y)dx. Theorem. Interchange of the limit and integration signs. Let f (x, y) be a family of integrable on x ∈ [a, b] functions at any fixed y ∈ Y , and assume f (x, y) converges uniformly on [a, b] to 𝜑(x) as y approaches y0 . Then the limit function 𝜑(x) is also integrable on [a, b] and b

b

lim f (x, y)dx =

∫a

y→y0

b

𝜑(x)dx = lim

∫a

y→y0

f (x, y)dx.

∫a

Corollary. Let f (x, y) be a family of functions defined on [a, b] × Y . Suppose for any fixed y ∈ Y the function f (x, y) is continuous in x ∈ [a, b], and for any fixed x ∈ [a, b] the function f (x, y) is monotone in y ∈ Y . Assume also f (x, y) converges pointwise on [a, b] to a continuous function 𝜑(x) as y approaches y0 . Then b

∫a

b

lim f (x, y)dx =

y→y0

∫a

b

𝜑(x)dx = lim

y→y0

∫a

f (x, y)dx.

Theorem. On continuity of integral. If f (x, y) is continuous on [a, b] × [c, d], b then F(y) = ∫a f (x, y)dx is continuous on [c, d]. Theorem. On differentiability under the integral sign. If f (x, y) is continuous in x ∈ [a, b] for any fixed y ∈ [c, d], and if fy (x, y) is continuous b

on [a, b] × [c, d], then F(y) = ∫a f (x, y)dx is differentiable on [c, d] and b F ′ (y) = ∫a fy (x, y)dx.

Introduction

Theorem. On integrability of the integral function. If f (x, y) is continb uous on [a, b] × [c, d], then F(y) = ∫a f (x, y)dx is integrable on [c, d] and d b b d ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy. I.2.3.5 Improper Integrals Depending on a Parameter

Definition. Improper integral depending on a parameter. Let f (x, y) be a family of functions defined on [a, +∞) × Y ⊂ ℝ2 . Improper integral depending +∞ on a parameter has the form F(y) = ∫a f (x, y)dx. We assume that the improper integral F(y) is convergent for each y ∈ Y . Definition. Uniformly convergent improper integral. Consider an auxilA iary function G(y, A) = ∫a f (x, y)dx for ∀A > a. The improper integral F(y) is uniformly convergent on Y if the function G(y, A) converges uniformly on Y as A → +∞, that is, for ∀𝜖 > 0 there exists A𝜖 > a (independent of y) such that |G(y, A) − F(y)| < 𝜖 for any A > A𝜖 and for all y ∈ Y simultaneously. Using the residual of the improper integral R(y, A) = F(y) − G(y, A) = +∞ ∫A f (x, y)dx for each y ∈ Y , the last definition can be restated in the form: the improper integral F(y) is uniformly convergent on Y if the function R(y, A) converges uniformly on Y to 0 as A → +∞, that is, for ∀𝜖 > 0 there exists +∞ A𝜖 > a (independent of y) such that |R(y, A| = | ∫A f (x, y)dx| < 𝜖 for any A > A𝜖 and for all y ∈ Y simultaneously. I.2.3.6 Conditions of the Uniform Convergence of Improper Integral

+∞

Cauchy criterion for uniform convergence. The integral F(y) = ∫a f (x, y)dx converges uniformly on Y if and only if for ∀𝜖 > 0, there exists A𝜖 > a (indeA pendent of y) such that | ∫A 2 f (x, y)dx| < 𝜖 for any A1 , A2 > A𝜖 and for all y ∈ Y 1 simultaneously. Definition. Majorant (dominant) function. A nonnegative function 𝜑(x) defined on [a, +∞) is called a majorant (or dominant) function on Y for a family of functions f (x, y) if |f (x, y)| ≤ 𝜑(x), ∀x ∈ [a, +∞) and ∀y ∈ Y . Weierstrass test. If for a family of functions f (x, y) there exists a majorant +∞ function with convergent improper integral ∫a 𝜑(x)dx, then the integral +∞ F(y) = ∫a f (x, y)dx converges uniformly on Y . Definition. Uniform boundedness. A family of functions f (x, y) is uniformly bounded on X × Y if there exists M > 0 such that |f (x, y)| ≤ M for ∀x ∈ X and ∀y ∈ Y . A An integral G(y, A) = ∫a f (x, y)dx is uniformly bounded on Y if there exists A M > 0 such that |G(y, A)| = | ∫a f (x, y)dx| ≤ M for ∀A > a and ∀y ∈ Y . A Dirichlet’s theorem. Let an integral G(y, A) = ∫a f (x, y)dx be uniformly bounded on Y . If a family of functions g(x, y) is monotone in x ∈ [a, +∞) for

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each fixed y ∈ Y , and g(x, y) converges uniformly to 0 on Y as x → +∞, then +∞ the improper integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . +∞ Abel’s theorem. Let an integral (F(y))= ∫a f (x, y)dx be uniformly convergent on Y . If a family of functions g x, y is uniformly bounded on [a, +∞) × Y and monotone in x ∈ [a, +∞) for each fixed y ∈ Y , then the improper integral +∞ ∫a f (x, y)g(x, y)dx converges uniformly on Y . I.2.3.7 Properties of the Uniformly Convergent Improper Integrals

Arithmetic properties of uniform convergence. +∞

+∞

1) If F(y) = ∫a f (x, y)dx and G(y) = ∫a g(x, y)dx converge uniformly on Y , +∞ then ∫a f (x, y) + g(x, y)dx also converges uniformly on Y . +∞ 2) If F(y) = ∫a f (x, y)dx converges uniformly on Y and g(y) is a bounded func+∞ tion on Y , then ∫a g(y)f (x, y)dx converges uniformly on the same set. Theorem. Passage to limit. Let f (x, y) be continuous in x ∈ [a, +∞) for each fixed y ∈ Y , and assume f (x, y) converges uniformly on [a, A], ∀A > a to a func+∞ tion 𝜑(x) as y → y0 . If F(y) = ∫a f (x, y)dx converges uniformly on Y , then the +∞ +∞ +∞ integral ∫a 𝜑(x)dx converges and lim ∫a f (x, y)dx = ∫a lim f (x, y)dx. y→y0

y→y0

Theorem. Continuity with respect to parameter. Let f (x, y) be continuous +∞ on [a, +∞) × [c, d]. If F(y) = ∫a f (x, y)dx converges uniformly on [c, d], then the function F(y) is continuous on [c, d]. Remark. Instead of [c, d], one can use any type of interval. Theorem. Differentiation with respect to parameter. Let f (x, y) and fy (x, y) +∞ be continuous functions on [a, +∞) × [c, d]. If F(y) = ∫a f (x, y)dx converges +∞ (pointwise) on [c, d] and G(y) = ∫a fy (x, y)dx converges uniformly on [c, d], then F(y) is differentiable on [c, d] and F ′ (y) = G(y), ∀y ∈ [c, d], that is, +∞ +∞ d ∫ f (x, y)dx = ∫a fy (x, y)dx. dy a Remark. Instead of [c, d], one can use any type of interval. Theorem. Integration with respect to parameter on a finite interval. Let +∞ f (x, y) be continuous function on [a, +∞) × [c, d]. If F(y) = ∫a f (x, y)dx converges uniformly on [c, d], then F(y) is continuous on [c, d] and d +∞ +∞ d ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy. Corollary. Let f (x, y) be a continuous and sign-preserving function on +∞ [a, +∞) × [c, d]. If F(y) = ∫a f (x, y)dx is continuous function on [c, d], then d +∞ +∞ d ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy.

Introduction

Theorem 1. Integration with respect to parameter on an infinite interval. Let f (x, y) be continuous and nonnegative on [a, +∞) × [c, +∞), +∞ and assume that F(y) = ∫a f (x, y)dx is continuous on [c, +∞) and +∞ G(x) = ∫c f (x, y)dy is continuous on [a, +∞). If, additionally, one of the +∞ +∞ integrals ∫c F(y)dy or ∫a G(x)dx converges, then another integral also converges and these two integrals have the same value. Theorem 2. Integration with respect to parameter on an infinite interval. Let f (x, y) be a continuous function on [a, +∞) × [c, +∞), and +∞ assume that F(y) = ∫a f (x, y)dx converges uniformly on [c, d], ∀d > c and +∞ G(x) = ∫c f (x, y)dy converges uniformly on [a, b], ∀b > a. If, additionally, +∞ +∞ +∞ +∞ at least one of the integrals ∫c dy ∫a |f (x, y)|dx or ∫a dx ∫c |f (x, y)|dy +∞ +∞ +∞ +∞ converges, then both integrals ∫c dy ∫a f (x, y)dx and ∫a dx ∫c f (x, y)dy are convergent and equal to each other.

xlix

1

CHAPTER 1 Conditions of Uniform Convergence 1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset Example 1. A function f (x, y), defined on X × Y , has a limit for any fixed x ∈ X as y approaches y0 , that is, f (x, y) converges pointwise to a limit function 𝜑(x) as y approaches y0 , but the convergence of f (x, y) to 𝜑(x) is nonuniform on X. Solution xy Let us consider f (x, y) = x2 +y2 defined on [0, 1] × (0, 1] and choose y0 = 0. If x = 0, then f (0, y) = 0 and consequently lim f (0, y) = lim 0 = 0. If x ≠ 0, then y→0

xy

y→0

lim f (x, y) = lim x2 +y2 = 0. Therefore, the limit function is defined for any x ∈ y→0

y→0

[0, 1] and it is zero: 𝜑(x) = lim f (x, y) = 0. However, the convergence to 𝜑(x) y→0

is not uniform on X = [0, 1]. Indeed for ∀y ∈ Y = (0, 1], there exists xy = y ∈ (0, 1] such that |f (xy , y) − 𝜑(xy )| = that is, for 𝜀0 =

1 2

y2 1 = ↛ 0, 2y2 2 y→0

whatever radius 𝛿 is chosen, there exists the point xy = y ∈

(0, 1] such that |f (xy , y) − 𝜑(xy )| = convergence is not uniform.

y2 2y2

≥ 𝜀0 although |y| < 𝛿. It means that the

Remark 1. In the case of Y = ℕ, a similar example can be formulated as follows: a sequence of functions fn (x) converges (pointwise) on a set X, but this convergence is nonuniform. One of the counterexamples is fn (x) = xn , X = (−1, 1). Since |x| < 1, one gets lim fn (x) = lim xn = 0 = f (x), ∀x ∈ X. To show that this n→∞ n→∞ ( ) convergence is nonuniform, let us pick up xn = 1 − n1 ∈ X, for ∀n ∈ ℕ, n ≥ 2; and for these points, we obtain ( ) 1 n |fn (xn ) − f (xn )| = 1 − → e−1 ≠ 0. n n→∞ Counterexamples on Uniform Convergence: Sequences, Series, Functions, and Integrals, First Edition. Andrei Bourchtein and Ludmila Bourchtein. © 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. Companion website: www.wiley.com/go/bourchtein/counterexamples_on_uniform_convergence

2

Chapter 1 Conditions of Uniform Convergence

1 Limit function f(x) = 0 0.8

Functions fn(x) = xn Pn = ( xn, fn( xn)) = (1 – 1n , (1 –

1 )n ) n

0.6 f1

f2

f3

0.4 P3

P2

0.2

0.2

0.4

0.6

P4

0.8

f10 P10

1

Figure 1.1 Examples 1, 4, and 28, sequence fn (x) = x n .

In other words, for 𝜀0 = 14 whatever natural number N is chosen, for some (in this case, actually, for any) n ∈ ℕ, n ≥ 2, there exists the point xn = 1 − n1 ( )n such that |fn (xn ) − f (xn )| = 1 − n1 ≥ 14 , that is, the convergence is not uni( )n form. (In the last inequality, we have used the fact that the sequence 1 − n1 is increasing.) Remark 2. A similar formulation can be made in the case of series: a series of functions converges (pointwise) on a set, but this convergence is nonuniform. ∑∞ The respective counterexample can be given with the series n=0 xn , x ∈ X = (−1, 1). It is well known that the geometric series is convergent for |x| < 1 and ∑∞ n 1 n=0 x = 1−x = f (x). To analyze the character of this convergence, first let us n+1 ∑n find the partial sums fn (x) = k=0 xk = 1−x and the corresponding remain1−x n+1

1 ders rn (x) = f (x) − fn (x) = x1−x . Choosing now xn = 1 − n+1 , ∀n ∈ ℕ, we obtain ( )n+1 1 1 − n+1 ( )n+1 1 rn (xn ) = = (n + 1) 1 − → ∞. 1 n→∞ n+1 1 − 1 + n+1

Therefore, the convergence is nonuniform.

1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset

5.5 5 4.5 4 3.5

Exact sum f(x) = 1 1– x fn(x), partial sums rn(x), residuals

P10

Pn = (xn, rn(xn))

3 2.5 2

f3 f2

1.5

f1

1 r1

P1

0.5 0.2

0.4

Figure 1.2 Examples 1 and 4, series

P4 P3

P2 r2 r3

0.6

∑∞ n=0

0.8

1

xn .

Example 2. A series of functions converges on X and a general term of the series converges to zero uniformly on X, but the series converges nonuniformly on X.

Solution ∑∞ n Let us consider the series n=1 xn on X = [0, 1). This series converges for ∀x ∈ n ∑∞ X, because 0 ≤ xn ≤ xn , ∀n, and the geometric series n=1 xn is convergent for |x| < 1. We can even find the sum of the series if we recall that the nfunction ∑∞ ln (1 + x) has expansion in Taylor’s series ln (1 + x) = n=1 (−1)n−1 xn conver∑∞ n gent on (−1, 1]. Then, replacing x by −x, we obtain ln (1 − x) = − n=1 xn with convergence on [−1, 1) and, in particular, on X = [0, 1). Further, the general n term un (x) = xn converges to 0 uniformly on X = [0, 1), because lim n1 = 0 and n→∞

evaluation |un (x)| = |x|n < n1 holds for ∀x ∈ X. Hence, the conditions of the statement are satisfied. However, the series is not convergent uniformly on X that can be shown by verifying the Cauchy criterion of the uniform convergence. In fact, for ∀x ∈ X and for ∀n, p ∈ ℕ we have the following evaluation: n

n+p |∑ xk || xn+1 xn+p xn+p | +···+ >p . | |= | | n+p n+p |k=n+1 k | n + 1

3

4

Chapter 1 Conditions of Uniform Convergence

3 Exact sum f(x) = –In(1 – x) fn(x), partial sums

2.5

rn(x), residuals 2

Pn = ( xn, rn( xn))

1.5 f3 f2

1

f1

0.5 r1 0.2

0.4

Figure 1.3 Examples 2, 26, 27, and 30, series

P1 r3

r2 0.6

∑∞

0.8 n

x n=1 n

Now, for ∀n ∈ ℕ, choosing pn = n and xn =

P2 P3

P10

1

. 1 √ n

∈ X, we get

2

( √ )2n n 1∕ 2 n k| | n+p ∑ x 1 | n| = ↛ 0, | |>n | | k 2n 8 n→∞ |k=n+1 | which means that the Cauchy criterion is not satisfied and, therefore, the series does not converge uniformly on X. Remark 1. The series

∑∞

xn √ n

considered on X = (−1, 1) provides a similar | n| counterexample. First, it converges for ∀x ∈ X, because 0 ≤ || √x || ≤ |x|n , ∀n, | n| ∑∞ and the geometric series n=1 |x|n is convergent for |x| < 1. Second, the | n| inequality |un (x)| = || √x || < √1 holds for ∀x ∈ X; since lim √1 = 0, it implies n n→∞ n | n| n the uniform convergence of √x to 0 on X. Hence, the statement conditions n hold. To analyze the nature of the convergence of the series, let us evaluate the k ∑n+p sum k=n+1 √x for ∀n ∈ ℕ, pn = n, and xn = √n1 ∈ X: k

n=1

3

( √ )2n n | n+p | 1∕ 3 n k | ∑ xn | 1 √ | | = √ n → ∞. √ |>n √ | n→∞ |k=n+1 k | 2n 9 2 | |

1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset

This means that the Cauchy criterion is not satisfied and, therefore, the series does not converge uniformly on X. ∑ Remark 2. The converse general statement is true: if a series un (x) converges uniformly on X, then its general term converges to zero uniformly on X. Example 3. A sequence of functions converges on X and there exists its subsequence that converges uniformly on X, but the original sequence does not converge uniformly on X. {

Solution Let us consider the sequence fn (x) =

x , n = 2k − n 1 , n = 2k n

1

, ∀k ∈ ℕ, X = ℝ.

For any fixed x ∈ ℝ, we have two partial limits: if n = 2k − 1, then x lim f2k−1 (x) = lim 2k−1 = 0; and if n = 2k, then lim f2k (x) = lim 2k1 = 0. k→∞

k→∞

k→∞

k→∞

Therefore, this sequence converges to 0 on ℝ: lim fn (x) = 0 = f (x). Note also n→∞ that the subsequence f2k (x) converges uniformly on ℝ, since the same evaluation |f2k (x) − f (x)| < 𝜀 holds [ ]simultaneously for all x ∈ ℝ. Or equivalently, for ∀𝜀 > 0 there exists K𝜀 = 2𝜀1 such that for ∀k > K𝜀 and simultaneously for all x ∈ ℝ it follows that |f2k (x) − f (x)| < 𝜀. That is, the definition of the uniform convergence is satisfied for f2k (x). Nevertheless, the sequence fn (x) does not converge uniformly on ℝ. Indeed, whatever large index N we choose, there exists the index nN = 2N − 1 > N and the real point xN = 2N − 1 such that |f2N−1 (xN ) − f (xN )| =

2N − 1 = 1 ↛ 0. N→∞ 2N − 1

Hence, the convergence of fn (x) is not uniform on ℝ. Example 4. A function f (x, y) defined on (a, b) × Y converges to a limit function 𝜑(x) as y approaches y0 , and this convergence is uniform on any interval [c, d] ⊂ (a, b), but the convergence is nonuniform on (a, b). Solution We can employ here the same functions used in Example 1. First, we consider xy f (x, y) = x2 +y2 defined on (0, 1) × (0, 1) and choose y0 = 0. As in Example 1, the limit function is zero: 𝜑(x) = lim f (x, y) = 0, ∀x ∈ (0, 1). However, the convery→0

gence to 𝜑(x) is not uniform on (0, 1), because for ∀y ∈ (0, 1) there exists xy = y such that |f (xy , y) − 𝜑(xy )| =

y2 1 = ↛ 0. 2y2 2 y→0

5

6

Chapter 1 Conditions of Uniform Convergence

On the other hand, for any interval [c, d] ⊂ (0, 1), the convergence is uniform. Indeed, for all x ∈ [c, d] and for any y > 0, it follows that |f (x, y) − 𝜑(x)| =

x2

xy xy 1 ≤ 2 ≤ y. 2 +y x c

Therefore, for any 𝜀 > 0, there exists 𝛿𝜀 = c𝜀 > 0 (which is the same for all points in [c, d]) such that if 0 < y < 𝛿, then |f (x, y) − 𝜑(x)| ≤ 1c y < 𝜀 for all x ∈ [c, d] simultaneously. It means that the convergence is uniform on [c, d]. Remark 1. The sequence fn (x) = xn on x ∈ (−1, 1) from Example 1 provides the following counterexample: a sequence of functions fn (x), defined on (a, b), converges uniformly on any interval [c, d] ⊂ (a, b), but the convergence is nonuniform on (a, b). The fact that the convergence to the limit function f (x) = 0 is not uniform on (−1, 1) was already proved in Example 1. Let us show that the convergence is uniform on any [c, d] ⊂ (−1, 1). Since we can always construct the interval [−q, q], where q = max{|c|, |d|}, such that [c, d] ⊂ [−q, q] ⊂ (−1, 1), it is sufficient to prove the uniform convergence on [−q, q]. For this interval, we get |fn (x) − f (x)| = |xn | ≤ qn , for all x ∈ [−q, q] at the same time. Since lim qn = 0, that is, for any 𝜀 > 0 (it is sufficient to consider 𝜀 < 1), there exists n→∞ [ ] ln 𝜀 N𝜀 = ln such that qn < 𝜀 if n > N𝜀 , we can conclude that for any 𝜀 > 0 there q [ ] ln 𝜀 exists exactly the same N𝜀 = ln such that when n > N𝜀 , then |xn | ≤ qn < 𝜀 q for all x ∈ [−q, q] simultaneously. The last sentence is the definition of the uniform convergence on [−q, q], and consequently, on [c, d]. ∑∞ Remark 2. Finally, the series of Example 1 n=0 xn , x ∈ (−1, 1) is an example of the situation when a series of functions converges uniformly on any interval [c, d] ⊂ (a, b), but the convergence is nonuniform on (a, b). It was already shown in Example 1 that the convergence of the given series is not uniform on (−1, 1). Let us consider an interval [−q, q] ⊂ (−1, 1), q > 0 and show that the convergence is uniform on such an interval (this will imply the uniform convergence on any interval [c, d] ⊂ (−1, 1)). Since |xn | ≤ qn for any x ∈ [−q, q] and ∑∞ the numerical series n=0 qn is convergent (the geometrical series with |q| < 1), ∑∞ according to the Weierstrass test the series n=0 xn converges uniformly on [−q, q]. Remark 3. The nearly converse situation also takes place, as it is shown in Example 5. Example 5. A sequence fn (x) converges on X, but this convergence is nonuniform on a closed interval [a, b] ⊂ X.

1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset

Solution 2 2 One of the counterexamples is fn (x) = nxe−n x on X = ℝ. It is easy to show that this sequence approaches f (x) ≡ 0 on ℝ. In fact, for x = 0 one has fn (0) = 0, ∀n and, consequently, lim fn (0) = 0. For x ≠ 0, one can use the change of variable n→∞ t = nx and apply l’Hospital’s rule: lim fn (x) = lim

n→∞

t→±∞

t 1 = lim = 0. et 2 t→±∞ 2tet 2

Consider now [a, b] ⊂ ℝ such that a ≤ 0 < b. Choosing N > obtains the following evaluation for ∀n > N:

1 b

and xn = n1 , one

2 2

|fn (xn ) − f (xn )| = n|xn |e−n xn = e−1 ↛ 0, k→∞

which means that the convergence of fn (x) to 0 is nonuniform on such a closed interval. ∑ Remark 1. A similar example for a series goes as follows: a series un (x) converges on a set X, but this series does not converge uniformly on a closed subin∑∞ terval [a, b] ⊂ X. The series n=1 sinnnx provides an example. First, we show that ∑∞ sin nx ∑∞ it is convergent on ℝ. If xk = k𝜋, ∀k ∈ ℤ, then n=1 n k = n=1 0 = 0. For x ≠ k𝜋, we can apply Dirichlet’s theorem. For the partial sums Bn (x) = ∑n k=1 sin kx, the following evaluation holds: n | 1 ∑ x || | |Bn (x)| = | 2 sin kx sin | x | 2 sin 2 || | 2 k=1 n ( ) ( ) )| | 1 ∑( 1 1 | | =| cos k − x − cos k + x | x | 2 sin | 2 2 | | 2 k=1 ( ) | | 1 x 1 |cos − cos n + = x|| | x | || 2 2 | 2 |sin 2 | | | | | 1 |2 sin n + 1 x ⋅ sin n x| ≤ 1 = | | | 2 2 || ||sin x || 2 |sin x2 | | | | | 2|

(note that the division by sin 2x is possible, since x ≠ k𝜋). Therefore, the sums Bn (x) are bounded for any fixed x ≠ k𝜋. Besides, the numerical sequence cn = n1 is decreasing and approaches 0 as n → ∞. Hence, all the conditions of Dirichlet’s theorem are satisfied, and therefore the series is convergent on ℝ. Let us show that this convergence is nonuniform on (0, 2𝜋) (and consequently on [0, 2𝜋] or any other interval containing (0, 2𝜋)). To this end, we evaluate ∑n+p ∑n+p sin kx the sum k=n+1 uk (xn ) = k=n+1 k n in the Cauchy criterion of uniform con𝜋 vergence. Choosing in this sum pn = n and xn = 6n , noting that 𝜋6 < kxn ≤ 𝜋3

7

8

Chapter 1 Conditions of Uniform Convergence

for any k such that n < ( k ≤)n + pn = 2n, and recalling that sin t is positive and strictly increasing on 0, 𝜋2 , we obtain | 2n | n+p | || ∑ sin kxn || | ∑n | uk (xn )| = || | | |k=n+1 | | | | |k=n+1 k || ( ) ( ) 𝜋 sin 𝜋6 + 6n sin 𝜋6 + 2𝜋 sin 𝜋3 6n = + +···+ n+1 n+2 2n sin 𝜋6 sin 𝜋6 sin 𝜋6 > + +···+ n+1 n+2 2n ( ) 1 1 1 1 n 1 = +···+ > = . 2 n+1 2n 2 2n 4 𝜋 Hence, there exists 𝜀0 = 14 such that for ∀n there are pn = n and xn = 6n ∈ ∑ n+p | | n (0, 2𝜋) such that | k=n+1 uk (xn )| > 𝜀0 . This means that the Cauchy criterion | | is not satisfied on (0, 2𝜋) and, consequently, the series does not converge uniformly on this interval. At the same time, the application of Dirichlet’s theorem of uniform convergence reveals that the series converges uniformly on the interval [a, 2𝜋 − a] for any a ∈ (0, 𝜋). Indeed, since sin 2x > 0, ∀x ∈ [a, 2𝜋 − a], we can apply the same ∑n evaluations as above for the partial sums Bn (x) = k=1 sin kx to obtain

|Bn (x)| ≤

1 1 1 = ≤ , |sin x∕2| sin x∕2 sin a∕2

∀x ∈ [a, 2𝜋 − a],

that is, the sums Bn (x) are uniformly bounded on [a, 2𝜋 − a]. Since cn = n1 → 0 n→∞ and cn is strictly decreasing, all the conditions of Dirichlet’s theorem of uniform convergence are satisfied. Hence, the series converges uniformly on any interval [a, 2𝜋 − a], a ∈ (0, 𝜋). Remark 2. For functions depending on a parameter, the corresponding formulation is as follows: a function f (x, y) defined on X × Y has a limit lim f (x, y) = y→y0

𝜑(x) for ∀x ∈ X, but f (x, y) converges to 𝜑(x) nonuniformly on a subinterval x2 y2 [a, b] ⊂ X. The function f (x, y) = x4 +y4 considered on ℝ × (0, +∞) with the limit point y0 = 0 provides the counterexample. This function converges to 𝜑(x) ≡ 0 on ℝ as y → 0: for x = 0, one has f (0, y) = 0, ∀y ∈ (0, +∞) which implies lim f (0, y) = 0; and for x ≠ 0, one obtains by the arithmetic rules of the limy→0

x2 y2

its lim x4 +y4 = y→0

0 x4

= 0. Choose now [a, b] ⊂ ℝ such that a ≤ 0 < b and evaluate

the difference |f (x, y) − 𝜑(x)| for ∀y ∈ (0, b) and xy = y ∈ [a, b]: |f (xy , y) − 𝜑(xy )| =

y4 1 = ↛ 0. 2y4 2 y→0

1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset

This result shows that the convergence is nonuniform on a chosen closed interval. Remark 3. A strengthened versions of these statements are presented in Example 6. Example 6. A sequence fn (x) converges on a set X, but it does not converge uniformly on any subinterval of X. Solution To construct a counterexample, let us place all the rational numbers of the interval [0, 1] in a specific order of a numerical sequence rn , n = 1, 2, · · · (this can be done, since the set of all the rational numbers of any interval is { countable). Define now the functions fn (x) on [0, 1] as follows: 1, x = r1 , r2 , · · · , rn fn (x) = . This sequence is monotone in n for any fixed 0, otherwise x ∈ [0, 1] (since fn (rn+1 ) = 0 < 1 = fn+1 (rn+1 ) and fn (x) = fn+1 (x), ∀x ≠ rn+1 ) and bounded (since 0 ≤ fn (x) ≤ 1, ∀n ∈ ℕ, ∀x ∈ [0, 1]). Therefore, { this sequence is 1, x ∈ ℚ convergent at any fixed x ∈ [0, 1] and f (x) = lim fn (x) = = D(x). 0, x ∈ 𝕀 n→∞ The convergence is nonuniform on [0, 1] since for ∀n there exist xn = rn+1 such that |fn (xn ) − f (xn )| = |fn (rn+1 ) − f (rn+1 )| = 1 ↛ 0. n→∞

Let us show that the convergence is also nonuniform on any interval [a, b] ⊂ [0, 1], which will imply that the convergence is nonuniform on any interval in [0, 1]. In fact, since any interval contains infinitely many rational points, in [a, b] there are infinitely many points of the sequence r1 , r2 , r3 , · · ·, which form a subsequence rn1 , rn2 , rn3 , · · · , rnk ∈ [a, b], ∀k ∈ ℕ. Then for any k ∈ ℕ, there exist nk > k and xk = rnk+1 ∈ [a, b] such that |fnk (xk ) − f (xk )| = |fnk (rnk+1 ) − f (rnk+1 )| = 1 ↛ 0, k→∞

which means that the convergence is nonuniform on [a, b]. Remark 1. The corresponding example for a series that converges on X, but does not converge uniformly on any subinterval of X, can be easily constructed using the sequence of the given counterexample as partial { sums of the series. ∑ 1, x = rn For instance, the series un (x) with the terms un (x) = defined on 0, x ≠ rn [0, 1] has the partial sums fn (x) of the above counterexample and, consequently, this series converges on [0, 1], but does not converge uniformly on any subinterval of [0, 1].

9

10

Chapter 1 Conditions of Uniform Convergence

Remark 2. Another example of this type, albeit for a sequence of continuous on X functions fn (x), is given in Example 26 of Chapter 2. ∑ Example 7. A series un (x) converges uniformly on an interval, but it does not converge absolutely on the same interval. Solution 2 ∑∞ The series n=1 (−1)n x n+n converges uniformly on the interval [−a, a], ∀a > 0. 2 In fact, for any fixed x ∈ ℝ, this is an alternating series, which converges by 2 2 Leibniz’s test: lim x n+n = 0 and x n+n is strictly decreasing in n for any fixed x ∈ 2 2 n→∞ ℝ. For alternating series, the remainder can be evaluated through its first term: 2 +n+1 |rn (x)| ≤ |un+1 (x)| = x(n+1) . Therefore, for all x ∈ [−a, a], we get 2 |rn (x)| ≤

x2 + n + 1 a2 1 ≤ + → 0, 2 (n + 1) (n + 1)2 n + 1 n→∞

which implies the uniform convergence of the series on [−a, a]. However, ∑∞ ∑∞ 2 the series of the absolute values n=1 |un (x)| = n=1 x n+n diverges for any x: 2 ∑∞ 1 for x = 0, the series n=1 n is harmonic and divergent; for ∀x ≠ 0, the series ∑∞ ( x2 1 ) ∑∞ 2 ∑∞ is the sum of the two series— n=1 nx2 and n=1 n1 —where the n=1 n2 + n former is convergent (p-series with p = 2) and the latter is divergent (the harmonic series), which implies the divergence of the sum. Hence, the given series is convergent on ℝ, uniformly convergent on [−a, a], ∀a > 0, but it does not possess absolute convergence at any point. Remark. The converse situation is considered in Example 8. ∑ Example 8. A series un (x) converges absolutely on an interval, but it does not converge uniformly on the same interval. Solution ∑∞ The series n=0 (−1)n xn converges absolutely on the interval X = [0, 1), because ∑∞ ∑∞ for ∀x ∈ [0, 1) the series n=0 |un (x)| = n=0 xn is the geometric series with the nonnegative ratio less than 1. However, the convergence is not uniform on 1 [0, 1), because for ∀n we can choose xn = 1 − n+1 ∈ [0, 1) that gives the following evaluation: ( )n+1 1 1 − n+1 | (−1)n+1 xn+1 | 1 | n | |rn (xn )| = | → e−1 ≠ 0. |= 1 | 1 + xn | n→∞ 2 1 + 1 − n+1 | |

1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset

3

f3,abs

2.5

f2,abs

2

f1,abs

1.5 1

n

fn,abs = ∑ k= 1

–0.8

–0.6

–0.4

r1 r3

rn = f – f n

0.5

–1

x2 + k k2

–0.2 –0.5

0.2

0.4

0.6

0.8

1 r2

–1 n fn = ∑ k = 1 (–1)k

–1.5 –2

Figure 1.4 Example 7, series

∑∞ n=1

f=

f2 x2 + k k2

∞ n x2 + n ∑n= 1 (–1) n2

f3 f1

2

(−1)n x n+n . 2

∑ A series un (x) converges absolutely and uniformly on [a, b], but Example 9. ∑ the series |un (x)| does not converge uniformly on [a, b]. Solution ∑∞ ∑∞ Let us consider the series n=0 un (x) = n=0 (−1)n (1 − x)xn on [0, 1]. If x = 1, then un (1) = 0 and the series converges at this point. If x ≠ 1, then for each fixed x we have the geometric series with the ratio q = −x, and since |q| = ∑∞ |x| < 1, the series is convergent. The series of the absolute values n=0 |un (x)| = ∑∞ n n=0 (1 − x)x is convergent on [0, 1] for the very same reasons. Let us analyze the uniform convergence of the original series. Since this series is alternating, we have the following evaluation for its remainder rn (x) for any n: |rn (x)| ≤ |un+1 (x)| = (1 − x)xn+1 , ∀x ∈ [0, 1]. Note that the continuous function h(x) = (1 − x)xn+1 is positive on (0, 1) and at the end points h(0) = h(1) = 0. Therefore, h(x) achieves its global maximum in some interior point of [0, 1]. Solving the critical point equation h′ (x) = (n + 1)xn − (n + 2)xn+1 = xn [(n + 1) − (n + 2)x] = 0,

11

12

Chapter 1 Conditions of Uniform Convergence

we find the only critical point xn = n+1 on (0, 1), which is the global maximum n+2 point of h(x) on [0, 1]. Thus, for ∀x ∈ [0, 1], |rn (x)| ≤ (1 − x)xn+1 ≤ (1 − xn )xn+1 n ( )n+1 1 1 = 1− → 0 ⋅ e−1 = 0, n→∞ n+2 n+2 that is, the convergences is uniform on [0, 1]. ∑∞ ∑∞ Finally, let us show that the series n=0 |un (x)| = n=0 (1 − x)xn does not converge uniformly on [0, 1]. It is sufficient to show that the convergence is nonuniform on [0, 1), so let us evaluate the remainder r̃n (x) for x ∈ [0, 1): r̃n (x) =

∞ ∑

|uk (x)| =

k=n+1

(1 − x)xn+1 = xn+1 . 1−x

1 Choosing now the points xn = n+1√ ∈ [0, 1) for each n, we obtain r̃n (xn ) = 2 ( )n+1 1 √ = 12 ↛ 0, which shows that the convergence is nonuniform on n+1 2 n→∞ [0, 1) and, consequently, on [0, 1].

∑∞ Remark. The converse general statement is true: if a series n=0 |un (x)| con∑∞ verges uniformly on [a, b], then the series n=0 un (x) converges absolutely and uniformly on [a, b]. ∑ Example 10. A series un (x) converges absolutely and uniformly on X, but there is no bound of the general term un (x) on X in the form |un (x)| ≤ an , ∀n ∑ such that the series an converges. Solution ∑∞ One of { the counterexamples is the series n=1 un (x) with the general term 0, x ∈ [0, 2−n−1 ] ∪ [2−n , 1] un (x) = defined on X = [0, 1]. Note that 1 sin2 (2n+1 𝜋x), x ∈ (2−n−1 , 2−n ) n un (x) ≥ 0 for ∀x ∈ [0, 1], so the convergence and absolute convergence is −n the same[ thing ] for this series. At the points x = 0, x = 2 , ∀n ∈ ℕ, and for ∀x ∈ 12 , 1 , we get un (x) = 0, ∀n ∈ ℕ and the series converges to zero. ( ) If x ∈ 0, 12 , x ≠ 2−n , n ≥ 2, then each of such points lies in only one of the intervals (2−n−1 , 2−n ), because these intervals have no common points: (2−n−1 , 2−n ) ∩ (2−n−2 , 2−n−1 ) = ∅, ∀n ∈ ℕ. Therefore, there is only one nx such that x ∈ (2−nx −1 , 2−nx ). Then unx (x) = n1 sin2 (2nx +1 𝜋x) and un (x) = 0, x ∑∞ ∀n ≠ nx and, consequently, n=1 un (x) = unx (x), which shows the (absolute) convergence of this series on [0, 1].

1.1 Pointwise, Absolute, and Uniform Convergence. Convergence on a Set and Subset

1 Exact sum f = 11 +– xx

0.9

Exact sum fa = 1

0.8

Residuals rn = f – fn Residuals rn,a = fa – fn,a

0.7

Pn = (xn , rn (xn )) Pn,a = (xn , rn,a(xn ))

0.6 0.5 0.4

r1,a P1,a P2,aP3,a r2,a r3,a

0.3 0.2 r1

0.1 0.1

0.2

0.3

0.4

r3 r2

P1 0.6

P3 P2 0.8

Figure 1.5 Examples 9 and 10 (second counterexample), series

0.9

∑∞ n=0

1

(−1)n (1 − x)x n .

Applying the Cauchy criterion and employing similar reasoning, we can also prove that the convergence is uniform. Indeed, since for any fixed x ∈ [0, 1] at most only one term in the entire series is nonzero and this | | term satisfies the inequality |unx (x)| = | n1 sin2 (2nx +1 𝜋x)| ≤ n1 , we obtain the | x | x |∑n+p | 1 following evaluation | k=n+1 uk (x)| ≤ n+1 < n1 , which holds for ∀n, p ∈ ℕ | | [ ] and simultaneously for ∀x ∈ [0, 1]. Hence, for ∀𝜀 > 0, there exists N𝜀 = 1𝜀 such that for ∀n > N𝜀 , ∀p ∈ ℕ and simultaneously for all x ∈ [0, 1], it follows |∑n+p | that | k=n+1 uk (x)| < n1 < 𝜀, that is, the series converges uniformly on [0, 1] | | according to the Cauchy criterion of the uniform convergence. Nevertheless, the functions un (x) do not admit majoration on [0, 1] by the ∑∞ constants an such that the series n=1 an converges. Indeed, for ∀n ∈ ℕ, the inequality |un (x)| ≤ n1 is exact (in the sense that n1 is the lowest upper bound for |un (x)|) on [0, 1], because there exists the point xn = 3 ⋅ 2−n−2 ∈ (2−n−1 , 2−n ) ( ) ∑∞ 1 3 such that un (xn ) = n sin2 2 𝜋 = n1 , and the series n=1 n1 diverges. ∑∞ Another interesting counterexample is the series of Example 9: n=0 un (x) = ∑∞ n n n=0 (−1) (1 − x)x on [0, 1]. It was shown in Example 9 that this series converges absolutely and uniformly on [0, 1]. For each fixed n, the function |un (x)| > 0, ∀x ∈ (0, 1), and |un (0)| = |un (1)| = 0. Therefore, the continuous function |un (x)| achieves its global maximum in an interior point of [0, 1],

13

14

Chapter 1 Conditions of Uniform Convergence

P1

1 0.9 0.8

Exact sum Partial sums

0.7

Pn = (xn , un (xn ))

0.6 P2

0.5 0.4

u1

P3

u2

0.3 0.2

u3

0.1 0.1

0.2

0.3

Figure 1.6 Example 10, series

0.4

0.5

0.6

0.7

{

∑∞

n=1 un (x), un (x) =

0.8

0.9

1

0, x ∈ [0, 2−n−1 ] ∪ [2−n , 1] . ∈ (2−n−1 , 2−n )

1 sin2 (2n+1 𝜋x), x n

which can be found by solving the critical point equation: |un (x)|′ = (xn − xn+1 )′ = nxn−1 − (n + 1)xn ( ) n = (n + 1)xn−1 − x = 0. n+1 The unique solution on (0, 1) is xn =

n n+1

|un (x)| ≤ max|un (x)| = |un (xn )| = [0,1]

( Since lim 1 −

)n

and, consequently, ( )n 1 1 1− . n+1 n+1

∑∞ 1 = e−1 and the series n=0 n+1 diverges, according to the n→∞ )n ∑∞ 1 ( 1 comparison test, the series n=0 n+1 1 − n+1 also diverges. Note that for ( )n 1 1 each n, the majorant term an = n+1 1 − n+1 is exact (i.e., the minimum possible) for |un (x)| on [0, 1]. Therefore, there is no convergent majorant series ∑∞ n=0 an such that |un (x)| ≤ an . 1 n+1

Remark. The converse general statement is true and represents the famous Weierstrass M-test.

1.2 Uniform Convergence of Sequences and Series of Squares and Products

1.2 Uniform Convergence of Sequences and Series of Squares and Products Example 11. A sequence fn (x) converges uniformly on X to a function f (x), but fn2 (x) does not converge uniformly on X to f 2 (x). Solution nx The sequence fn (x) = ln n+1 converges on X = (0, +∞) to f (x) = ln x, because ( ) nx n lim ln n+1 = ln lim n+1 x = ln x. The following evaluation shows that this n→∞ n→∞ convergence is uniform: | | | nx n || |fn (x) − f (x)| = ||ln − ln x|| = ||ln | < 𝜀. | n+1 | | n + 1| [ ] So for ∀𝜀 > 0, there exists N𝜀 = e𝜀1−1 , which depends only on 𝜀, such that for ∀n > N𝜀 and simultaneously for all x ∈ X we have |fn (x) − f (x)| < 𝜀. Due to arithmetic properties of the limits, the sequence fn2 (x) also converges nx = to f 2 (x) = ln2 x for any fixed x ∈ X (it can also be shown directly: lim ln2 n+1 n→∞ ( ( ) )2 n ln lim n+1 x = ln2 x). However, this convergence is not uniform. In fact, n→∞ for each x ∈ X, we get | | nx |fn2 (x) − f 2 (x)| = ||ln2 − ln2 x|| n + 1 | | | | | | nx nx | | = |ln − ln x| ⋅ ||ln + ln x|| | n+1 | | n+1 | n + 1 || nx2 || = ln ⋅ |ln |. n | n + 1| Choosing now xn =

1 nen

∈ X, we obtain n+1 n n+1 = ln n n+1 = ln n

|fn2 (xn ) − f 2 (xn )| = ln

| | n | ⋅ ||ln | 2 2n | (n + 1)n e | ⋅ ln(n(n + 1)e2n ) ⋅ ln n(n + 1) + 2n ln

n+1 >1 n

for sufficiently large n (since the first term positive and the limit of the second ( is ) n

equals two: lim 2n ln n+1 = lim 2 ln 1 + n1 = 2 ln e = 2, that is, 2n ln n+1 > n n n→∞ n→∞ 1 for large n). Therefore, the convergence is not uniform: for 𝜀0 = 1 whatever N is chosen, it can be found that ñ > N and corresponding xñ ∈ X such that |fñ2 (xñ ) − f 2 (xñ )| > 𝜀0 = 1.

15

16

Chapter 1 Conditions of Uniform Convergence

Remark 1. Naturally, the following example can also be constructed: sequences fn (x) and gn (x) converge uniformly on X to f (x) and g(x), respectively, but fn (x)gn (x) does not converge uniformly on X to f (x)g(x). In the case fn (x) = gn (x), we have the original example with the square of function. nx For different sequences, we can use the same fn (x) = ln n+1 and slightly nx different gn (x) = ln 2n+5 . The sequence gn (x) converges to g(x) = ln 2x , and this convergence is uniform on X = (0, +∞) due to the evaluation | nx x| |gn (x) − g(x)| = ||ln − ln || 2n + 5 2 | | ( ) | 2n || 5 | = |ln = ln 1 + → 0. | 2n n→∞ | 2n + 5 | Consequently, fn (x)gn (x) converges to f (x)g(x) = ln x ln 2x for each fixed x ∈ X due to arithmetic rules of the limits. However, this convergence is not uniform on X, as it is shown below: for each x ∈ X, we have | nx nx x| |fn (x)gn (x) − f (x)g(x)| = ||ln ln − ln x ln || 2| | n + 1 2n + 5 | nx nx nx x = ||ln ln − ln ln n+1 2 | n + 1 2n + 5 nx x x| + ln ln − ln x ln || n+1 2 2| | nx 2n x n || = ||ln ln + ln ln , 2 n + 1 || | n + 1 2n + 5 and for the special choice of the points xn =

1 nen

∈ X, we obtain

| 1 2n 1 n || |fn (xn )gn (xn ) − f (xn )g(xn )| = ||ln ln + ln ln n n 2n + 5 2ne n + 1 || | (n + 1)e ( ) ( ) | 5 1 || = ||(n + ln(n + 1)) ln 1 + + (n + ln 2n) ln 1 + | 2n n | | ( ) ln(n + 1) ( ) 5 2n 5 2n 5 = ⋅ ln 1 + + ln 1 + 2 5 2n 2n∕5 5 2n ( ) ( ) 1 ln 2n 1 5 + n ln 1 + + n ln 1 + → ⋅1 n n n n→∞ 2 7 +0⋅1+1+0⋅1= . 2 In the evaluation of the last limit, we have used the following auxiliary limits: ( ) ( ) n 𝛼 𝛼 n∕𝛼 ln 1 + = lim ln 1 + = ln e = 1, n→∞ 𝛼 n→∞ n n lim

∀𝛼 ≠ 0

1.2 Uniform Convergence of Sequences and Series of Squares and Products

1 r1 0.5

r2 r3 0.2

0.4

0.6

0.8

1

–0.5 –1 –1.5

f3 f2

–2

Limit function f(x) = ln x

f1

Functions fn(x)

–2.5

Residuals rn(x) = f(x) – fn(x) –3

Figure 1.7 Examples 11 and 17, sequence fn (x) = ln

nx . n+1

Limit function f 2(x) = ln2 x

3

Functions f 2n(x) Residuals rn(2)(x) = f 2(x) – f 2n(x)

2

f 21 f 22 f 23

1

0.2

0.4

0.6

(2) r3 (2) r2

–1

P1

–2

P3

P2

(2)

r1

Pn = (xn, r(2) n (xn))

–3 nx Figure 1.8 Example 11, sequence of squares fn2 (x) = ln2 n+1 .

0.8

1

17

18

Chapter 1 Conditions of Uniform Convergence

according to the second remarkable limit, and lim

t→∞

𝛼∕(𝛼t + 𝛽) ln(𝛼t + 𝛽) = lim = 0, t→∞ t 1

∀𝛼 > 0, ∀𝛽

due to l’Hospital’s rule. Therefore, for 𝜀0 = 1 whatever N is chosen, there is ñ > N and corresponding xñ ∈ X such that |fñ (xñ )gñ (xñ ) − f (xñ )g(xñ )|> 𝜀0 = 1, that is, the convergence is nonuniform. Remark 2. The following general statement is true for the sum and difference: if fn (x) and gn (x) converge uniformly on X to f (x) and g(x), respectively, then fn (x) ± gn (x) converges uniformly on X to f (x) ± g(x). Remark 3. The following general statement is true for the product: if fn (x) and gn (x) converge uniformly on X to f (x) and g(x), respectively, and f (x) and g(x) are bounded on X, then fn (x) ⋅ gn (x) converges uniformly on X to f (x) ⋅ g(x). (Note the requirement of boundedness of the limit functions in this formulation.)

Example 12. Sequences fn (x) and gn (x) converge nonuniformly on X to f (x) and g(x), respectively, but fn (x) ⋅ gn (x) converges to f (x) ⋅ g(x) uniformly on X. Solution 1 Consider the sequences fn (x) = √ and gn (x) = nxe−nx on X = (0, +∞). Both n x sequences converge to 0 for any fixed x ∈ X: 1 lim fn (x) = lim √ = 0 = f (x); n→∞ n x t 1 lim gn (x) = lim nxe−nx = lim t = lim t = 0 = g(x). n→∞ n→∞ t→+∞ e t→+∞ e n→∞

Therefore, lim fn (x) ⋅ gn (x) = 0. n→∞ Let us investigate the nature of the convergence of these sequences. For fn (x), choosing xn = n12 ∈ X, we obtain 1 n |fn (xn ) − f (xn )| = √ = = 1 ↛ 0, n→∞ n n xn that is, this convergence is nonuniform on X. Similarly, choosing xn = we can show the nonuniform convergence of gn (x) on X: |gn (xn ) − g(xn )| = nxn e−nxn = 1 ⋅ e−1 ↛ 0. n→∞

1 n

∈ X,

1.2 Uniform Convergence of Sequences and Series of Squares and Products

√ Finally, for fn (x) ⋅ gn (x), we have |fn (x)gn (x) − f (x)g(x)| = xe−nx . The derivative of the right-hand side is ( ) ( ) √ √ 1 n −nx 1 −nx −nx ( xe )x = −x . √ −n x e = √ e 2n 2 x x 1 Therefore, the point xn = 2n ∈ X is the only local (and global) maximum of √ −nx xe on X. Consequently, √ √ 1 |fn (x)gn (x) − f (x)g(x)| ≤ max xe−nx = xn e−nxn = √ e−1∕2 → 0, n→∞ (0,+∞) 2n

that is, fn (x) ⋅ gn (x) converges uniformly on X to 0. Example 13. A sequence fn2 (x) converges uniformly on X, but fn (x) diverges on X. Solution Consider the sequence fn (x) = (−1)n n+1 x on X = (0, 1]. The sequence of squares n fn2 (x) =

(n+1)2 2 x n2

converges uniformly on (0, 1] to x2 , because

| (n + 1)2 2 | 2n + 1 2 1 |fn2 (x) − x2 | = || x − x2 || = x2 ≤ + 2 → 0. 2 2 n→∞ n n n n | | However, there is no limit of fn (x) for any fixed x ∈ (0, 1], since two partial limits give different results: f2n (x) = 2n+1 x → x and f2n+1 (x) = − 2n+2 x → −x. 2n 2n+1 n→∞

n→∞

Remark 1. The same sequence can be used to exemplify the following situation: a sequence |fn (x)| converges uniformly on X, but fn (x) diverges on X. Indeed, although fn (x) = (−1)n n+1 x does not converge on X = (0, 1], the n n+1 sequence |fn (x)| = n x converges uniformly to x on X = (0, 1]: |n + 1 | 1 1 ||fn (x)| − x| = || x − x|| = |x| ≤ → 0. n n→∞ | n | n Remark 2. Note that the inequality ||fn (x)| − |f (x)|| ≤ |fn (x) − f (x)| ensures the validity of the converse general statement: if fn (x) converges uniformly on X to f (x), then |fn (x)| converges uniformly on X to |f (x)|. Remark 3. The following example also takes place: a sequence fn2 (x) converges uniformly on X and fn (x) converges on X, but the convergence of fn (x) is nonuniform. Consider{the sequence fn (x) on [0, 1] similar to that 1, x = r1 , r2 , · · · , rn analyzed in Example 6: fn (x) = , where rn is the sequence −1, otherwise

19

20

Chapter 1 Conditions of Uniform Convergence

of all the rational points in [0, 1] ordered in some way. This sequence is monotone in n for any fixed x ∈ [0, 1] (since fn (rn+1 ) = −1 < 1 = fn+1 (rn+1 ) and fn (x) = fn+1 (x), ∀x ≠ rn+1 ) and bounded (since −1 ≤ fn (x) ≤ 1, ∀n ∈ ℕ, ∀x ∈ [0, 1]). Therefore,{this sequence is convergent at any fixed x ∈ [0, 1] 1, x ∈ ℚ and f (x) = lim fn (x) = . The sequence of the squares consists of −1, x ∈ 𝕀 n→∞ the same constant function fn2 (x) = 1, ∀n ∈ ℕ and, therefore, it converges uniformly on [0, 1] to f 2 (x) = 1. At the same time, using the same reasoning as in Example 6, one can show that the convergence of fn (x) is nonuniform on [0, 1] and on any subinterval of [0, 1]. Example 14. A sequence fn (x) ⋅ gn (x) converges uniformly on X to 0, but neither fn (x) nor gn (x) converges to 0 on X. Solution The sequences fn (x) = nx + (−1)n nx and gn (x) = nx − (−1)n nx are divergent for every fixed x ∈ X = (0, +∞): f2n (x) = 4nx → +∞, f2n+1 (x) = 0 → 0; n→∞

n→∞

g2n (x) = 0 → 0, g2n+1 (x) = (4n + 2)x → +∞. n→∞

n→∞

However, fn (x) ⋅ gn (x) = n x − n x = 0 converges uniformly to 0 on X. interesting counterexample includes the sequences fn (x) = {Another { x nx , x ∈ ℚ ∩ X , x ∈ ℚ∩X n+1 n and gn (x) = defined on X = (0, 1]. nx x , x ∈ 𝕀∩X , x ∈ 𝕀 ∩ X n+1 n Both sequences converge on X to nonzero functions f (x) = lim fn (x) = n→∞ { { x, x ∈ ℚ ∩ X 0, x ∈ ℚ ∩ X and g(x) = lim gn (x) = , respectively. At the 0, x ∈ 𝕀 ∩ X x, x ∈ 𝕀 ∩ X n→∞ 2 x same time, lim fn (x)gn (x) = lim n+1 = 0, for ∀x ∈ (0, 1] and this convergence n→∞ n→∞ is uniform on X, since the estimate 2 2

2 2

x2 1 ≤ → 0 n + 1 n + 1 n→∞ holds simultaneously for all x ∈ (0, 1]. |fn (x)gn (x) − 0| =

Example 15. A sequence fn (x) converges uniformly on X to a function f (x), 1 fn (x) ≠ 0, f (x) ≠ 0, ∀x ∈ X, but f 1(x) does not converge uniformly on X to f (x) . n

Solution The sequence fn (x) =

nx n+2

converges uniformly on X = (0, 1) to f (x) = x:

| nx | 2 2 |fn (x) − f (x)| = || − x|| = |x| < 0 and simultaneously for all x ∈ X if we choose ∀n > N𝜀 = 2𝜀 . On the other hand, | 1 1 || || n + 2 1 || 2 1 | | f (x) − f (x) | = | nx − x | = n |x| , |n | | | | 1 | 1 and for xn = n ∈ X, it follows that | f (x ) − f (x1 ) | = n n2 = 2, which means that the |n n n | convergence is nonuniform. Example 16. A sequence fn (x) is bounded uniformly on ℝ and converges uniformly on [−a, a], ∀a > 0, to a function f (x), but the numerical sequence sup fn (x) does not converge to sup f (x). x∈ℝ

x∈ℝ

Solution 2 Consider the sequence fn (x) = e−(x−n) , which is defined and uniformly 2 bounded on ℝ: 0 < e−(x−n) ≤ 1, ∀n, ∀x ∈ ℝ. This sequence converges to zero on ℝ, since for any fixed x ∈ ℝ one has (x − n)2 → +∞ and, consen→∞

quently, lim e−(x−n) = lim e−t = 0. Hence, f (x) ≡ 0 on ℝ and, consequently, n→∞ t→+∞ sup f (x) = 0. On the other hand, sup fn (x) = 1, ∀n ∈ ℕ, since fn (x) ≤ 1 and 2

x∈ℝ

x∈ℝ

fn (n) = 1. This means that sup fn (x) = 1 does not converge to sup f (x) = 0. It x∈ℝ

x∈ℝ

just remains to prove the uniform convergence of fn (x) on [−a, a], ∀a > 0. For any fixed a > 0, there exists the natural number Na > a. Then for ∀n > Na , one gets (x − n)2 ≥ (a − n)2 for each x ∈ [−a, a]. Therefore, 2

|fn (x) − f (x)| = e−(x−n) ≤ e−(a−n)

2

for all x ∈ [−a, a]. Since exp (−(a − n)2 ) → 0, the last inequality guarantees n→∞ the uniformity of the convergence on [−a, a], where a > 0 is arbitrary. Note, however, that the convergence of fn (x) is not uniform on ℝ, which is evident if one chooses xn = n leading to 2

|fn (xn ) − f (xn )| = e−(n−n) = 1. Remark 1. Two other interesting counterexamples are fn (x) = arctan nx and fn (x) = n2nx . For instance, for the first function the reasoning can be as follows. 2 +x2 | | First, note that the sequence is uniformly bounded on ℝ (|arctan nx | < 𝜋2 , | | ∀n ∈ ℕ and ∀x ∈ ℝ). Second, it converges to zero for any fixed x ∈ ℝ x ( lim arctan n = 0). Further, this convergence is uniform on [−a, a], ∀a > 0 due n→∞ to the evaluation | |x| x| a |fn (x) − f (x)| = ||arctan || = arctan ≤ arctan , n| n n |

21

22

Chapter 1 Conditions of Uniform Convergence

that holds for all x ∈ [−a, a] (here we used the properties that arctan t is an odd and a strictly increasing function). Since lim arctan an = 0, the last evaluation n→∞ implies the uniform convergence. Hence, all the conditions of the example are satisfied, but still the sequence sup fn (x) does not converge to sup f (x), because sup arctan nx = x∈ℝ

𝜋 2

x∈ℝ

x∈ℝ

for any n, while sup f (x) = sup 0 = 0. Note, that just like in the x∈ℝ

x∈ℝ

first counterexample, the convergence of fn (x) is not uniform on ℝ: for any n one can choose xn = n to obtain n 𝜋 |fn (xn ) − f (xn )| = arctan = arctan 1 = ≠ 0. n 4 Remark 2. The following general statement is true: if a sequence fn (x) is bounded uniformly on ℝ and converges uniformly on ℝ to a function f (x), then the numerical sequence sup fn (x) converges to sup f (x). (Note x∈ℝ

x∈ℝ

the requirement of uniform convergence on ℝ to the limit function in this formulation.) Remark 3. The condition of uniform convergence of a sequence fn (x) to a function f (x) on X is equivalent to the condition lim sup|fn (x) − f (x)| = 0. n→∞ x∈X

Example 17. Suppose each function fn (x) maps X on Y and function g(y) is continuous on Y ; the sequence fn (x) converges uniformly on X, but the sequence gn (x) = g(fn (x)) does not converge uniformly on X. Solution nx Let us consider the sequence fn (x) = ln n+1 on X = (0, +∞) and function g(y) = y e . Each of the functions fn (x) maps (0, +∞) on the entire real line and the function g(y) is continuous on ℝ. In Example 11, it was shown that the sequence fn (x) converges uniformly on X to the function f (x) = ln x. The corresponding nx sequence gn (x) = g(fn (x)) = n+1 converges for any fixed x ∈ (0, +∞): nx = x = g(f (x)), n+1 but this convergence is not uniform. In fact, | nx | 1 |gn (x) − g(f (x))| = || − x|| = x |n + 1 | n+1 and choosing xn = (n + 1) ∈ X, one gets xn |gn (xn ) − g(f (xn ))| = = 1. n+1 lim gn (x) = lim

n→∞

n→∞

1.2 Uniform Convergence of Sequences and Series of Squares and Products

1.2

Limit function f(x) = 0 Functions fn(x)

Pn = (xn, fn(xn)) = (n, 1) P1

1

P2

P3

0.8

0.6

f1

f2

f3

0.4

0.2

1

2

3

4

5

6

2

Figure 1.9 Example 16, sequence fn (x) = e−(x−n) .

Remark. The following example also takes place: suppose functions fn (x) map X on Y and function g(y) is continuous on Y ; the sequence fn (x) converges nonuniformly on X, but the sequence gn (x) = g(fn (x)) converges uniformly on X. In the trivial case, one can use an arbitrary nonuniformly convergent sequence fn (x) and the constant function g(y) ≡ 1. For a nonconstant function nx g(y), one can use the above sequence fn (x) = n+1 defined on X = (0, +∞) and the function g(y) = ln y. ∑ 2 Example 18. A series un (x) converges uniformly on X, but the series ∑ un (x) does not converge uniformly on X. Solution ∑∞ In Remark 1 to Example 5, it was shown that the series n=1 sinnnx converges 2 ∑∞ nonuniformly on ℝ. However, the series n=1 sinn2nx converges uniformly on ℝ according to the Weierstrass test: ∑∞ 1 n=1 n2 converges.

sin2 nx n2

≤

1 , n2

∀x ∈ ℝ and the majorant series

∑ 2 Example 19. A series un (x) converges uniformly on X, but the series ∑ un (x) does not converge (even pointwise) on X.

23

24

Chapter 1 Conditions of Uniform Convergence

4 3.5 3 2.5 Partial sum f100

2

n

Partial sums fn = ∑ k= 1

1 x+k

1.5 1

f3

0.5

f1

f2

1

2

3

4

5

Figure 1.10 Examples 19, 20, and 21, series

6

7

8

9

10

∑∞

1 n=1 x+n .

Solution ∑∞ Consider the series n=1

1 x+n 1∕n lim n→∞ 1∕(x+n)

on X = [0, +∞). This series is divergent at each ∑∞ point x ∈ X, since = 1 and the harmonic series n=1 n1 is divergent. ∑∞ 1 At the same time, the series n=1 (x+n) converges uniformly on X = [0, +∞) 2 ∑∞ 1 1 due to the Weierstrass test: (x+n)2 ≤ n2 , ∀x ∈ X and the majorant series n=1 n12 converges. ∑ ∑∞ Remark. If un (x) diverges, it may happen that n=1 u2n (x) also diverges. For ∑∞ 1 instance, the series n=1 √ √ diverges at each point x ∈ X = [0, 1] according x+ n √ ∑∞ 1∕ n to the comparison test: lim √ √ = 1 and the series n=1 √1 is divergent. n n→∞ 1∕( x+ n) Due to the very same arguments, the series of squares also diverges on X = ∑∞ [0, 1]: lim √1∕n√ 2 = 1 and the series n=1 n1 is divergent. n→∞ 1∕( x+ n)

∑ Example 20. A series un (x)vn (x) converges uniformly on X, but at least one ∑ ∑ of the series un (x) or vn (x) does not converge uniformly on X. Solution 1 1 Consider un (x) = x+n and vn (x) = x2 +n on X = [0, +∞). The series 2 ∑∞ u (x)v (x) converges uniformly on X due to the Weierstrass test: n n=1 n

1.2 Uniform Convergence of Sequences and Series of Squares and Products

1.8

1.5

∞ ∑n = 1 n12 =

1.2

π2 6

Exact sum Partial sums gn = ∑ nk = 1

0.9

1 (x + k)2

0.6 g3 g2

0.3

g1

1

2

3

4

5

Figure 1.11 Example 19, series of squares

|un (x)vn (x)| =

6

7

8

9

10

∑∞

1 n=1 (x+n)2 .

1 1 1 ≤ 3, 2 2 x+nx +n n

∀x ∈ [0, +∞)

∑ and n13 is a convergent series. The same reasoning shows the uniform con∑∞ vergence of the series n=1 vn (x) on X: |vn (x)| = and

∑

x2

1 1 ≤ 2, 2 +n n

∀x ∈ [0, +∞)

1 n2

is a convergent series. However, the series ∑ 1∕n since lim 1∕(x+n) = 1 and the series n1 is divergent.

∑∞ n=1

un (x) diverges on X,

n→∞

∑ Example 21. A series un (x)vn (x) converges uniformly on X, but neither ∑ ∑ un (x) nor vn (x) converges (even pointwise) on X. Solution 1 Consider un (x) = x+n and vn (x) = √ 1 √ on X = [0, +∞). The series x+ n ∑∞ u (x)v (x) converges uniformly on X due to the Weierstrass test: n n=1 n |un (x)vn (x)| =

1 1 1 , √ √ ≤ x + n x + n n3∕2

∀x ∈ [0, +∞)

25

26

Chapter 1 Conditions of Uniform Convergence

and

∑

1 n3∕2

is a convergent series. However, both

∑∞

∑∞

n=1 un (x) and 1∕n diverge on X according to the comparison test: lim 1∕(x+n) = 1 and n→∞ √ ∑1 ∑ 1 1∕ n diverges; lim √ √ = 1 and the series √ diverges. n n n→∞ 1∕( x+ n)

n=1 vn (x)

the series

∑ ∑ Example 22. Series un (x) and vn (x) converge nonuniformly on X, but ∑ un (x)vn (x) converges uniformly on X. Solution ∑∞ The series n=1 sinnnx converges nonuniformly on ℝ (see Remark 1 to Example ∑∞ 5), and so does the series n=1 sin√nx (applying for the latter series the same n reasoning as for the former in Remark 1 to Example 5). However, the series ∑∞ sin2 nx sin2 nx n=1 n3∕2 converges uniformly on ℝ according to the Weierstrass test: n3∕2 ≤ ∑∞ 1 1 , ∀x ∈ ℝ and the majorant series n=1 n3∕2 converges. n3∕2 ∑ ∑ Example 23. A series un (x) converges uniformly on X, but u2n (x) does not converge uniformly on X. Solution ∑∞ ∑∞ xn n√ The uniform convergence of the series n=1 un (x) = n=1 (−1) 3 n on X = (0, 1) can be proved by applying Abel’s theorem. In fact, the series ∑ (−1)n √31 converges by Leibniz’s test of alternating series (and this conn vergence is uniform on X, since the series does not depend on x), and the sequence xn is monotone in n for each x ∈ (0, 1) and is uniformly bounded, since xn < 1, ∀x ∈ (0, 1), ∀n ∈ ℕ. ∑∞ ∑∞ 2n On the other hand, the convergence of the series n=1 u2n (x) = n=1 nx2∕3 is nonuniform on X = (0, 1). In fact, this series converges on X = (0, 1), since 2n ∑ u2n (x) = nx2∕3 ≤ x2n , ∀n ∈ ℕ, and the series x2n is convergent by being a geometric series with the ratio in (0, 1). At the same time, applying the ( ) Cauchy 1 criterion of the uniform convergence with pn = n and xn = 1 − 4n ∈ X, one obtains n ( ) | n+p | n+p ∑n x2k x4n n1∕3 1 4n |∑ 2 | n n uk (xn )| = > n = 1 − → +∞, | 2∕3 | | n→∞ 4n (2n)2∕3 41∕3 |k=n+1 | k=n+1 k ∑ that is, the series u2n (x) converges nonuniformly on X = (0, 1). Remark 1. If, in the given counterexample, one changes the set X to (0, 1], n ∑∞ ∑∞ then n=1 un (x) = n=1 (−1)n √3x converges uniformly on X = (0, 1] (due to the n

1.2 Uniform Convergence of Sequences and Series of Squares and Products

Partial sums fn = ∑kn= 1 sinkx k

2

Partial sum f100 Partial sum f200

1.5

f3 1 f2 0.5

f1

0.5

1

1.5

2

Figure 1.12 Examples 22, 5, and 18, series

∑∞ n=1

2.5

3

sin nx . n

4 n

Partial sums gn = ∑ k = 1 sinkx √

3.5

k

Partial sum g100

3

Partial sum g200

2.5 2 1.5 g3 1 g2 0.5

g1 0.5

1

1.5

Figure 1.13 Examples 22 and 24, series

2

∑∞ n=1

sin nx √ . n

2.5

3

27

28

Chapter 1 Conditions of Uniform Convergence

Partial sums hn = ∑ k = 1 sin3/kx 2 2

n

2

k

Partial sum h100 Partial sum h200

1.5

1 h3 0.5

h2 h1

0.5

1

1.5

Figure 1.14 Example 22, series of products

2

∑∞ n=1

2.5

3

sin2 nx . n3∕2

∑∞ ∑∞ same reasoning as before), but n=1 u2n (x) = n=1 ∑∞ 2 ∑∞ 1 n=1 un (1) = n=1 n2∕3 is a divergent p-series.

x2n n2∕3

diverges at x = 1 since

Remark 2. Evidently, the following more general example can also be con∑ ∑ structed: both series un (x) and vn (x) converge uniformly on X, but ∑ the series un (x)vn (x) does not converge uniformly on X. In the particular case un (x) = vn (x), the counterexample is already provided above. Let us consider the case when un (x) ≠ vn (x). For instance, using the same argun ∑∞ ∑∞ ments as before, one can prove that both n=1 un (x) = n=1 (−1)n √3x and n ∑∞ ∑∞ xn n√ v (x) = (−1) converge uniformly on X = (0, 1), but the series n=1 n n=1 n ∑∞ ∑∞ x2n n=1 un (x)vn (x) = n=1 n5∕6 converges nonuniformly on X = (0, 1). For the last 2n

series, its convergence follows from the evaluation nx5∕6 ≤ x2n , ∀n ∈ ℕ and the ∑∞ convergence of the geometric series n=1 x2n for ∀x ∈ (0, 1), while the nonuniformity ( can be ) shown by the Cauchy criterion, choosing as above pn = n and 1 xn = 1 − 4n ∈ X: 2n n ( ) | n+p | ∑ x2k x4n n1∕6 1 4n |∑ | n uk (xn )vk (xn )| = > n n 5∕6 = 5∕6 1 − → ∞. | 5∕6 | | n→+∞ 4n (2n) 2 |k=n+1 | k=n+1 k

1.2 Uniform Convergence of Sequences and Series of Squares and Products

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

–0.1 f2

–0.2 –0.3 –0.4 –0.5 –0.6 –0.7

Partial sum f200

–0.8

Partial sum f100

–0.9

Partial sums fn = ∑ k = 1 (–1)k 3x

n

k

f3

√k

f1

–1

Figure 1.15 Example 23, series

∑∞ n=1

n

(−1)n √3x . n

14

P200

Partial sum g200 12

Partial sum g100 n

10

Partial sums gn =∑ k = 1

P100

x2k k2 / 3

1 Pn = (xn, gn(xn)), xn = 1– 4n

8 6 4 2

P1 0.1

0.2

0.3

0.4

0.5

Figure 1.16 Example 23, series of squares

0.6

∑∞

x 2n n=1 n2∕3 .

0.7

0.8

P2

g3 P 3 g2 g1

0.9

1

29

30

Chapter 1 Conditions of Uniform Convergence

∑ ∑ Example 24. A series un (x) converges uniformly on X, but u2n (x) does not converge (even pointwise) on X. Solution ∑∞ ∑∞ The series n=1 un (x) = n=1 sin√nx is uniformly convergent on X = [a, 𝜋 − a], n ( ) 𝜋 ∀a ∈ 0, 2 , which can be shown using the same considerations as ∑∞ for n=1 sinnnx in Remark 1 to Example 5. Let us prove that the series ∑∞ 2 ∑∞ sin2 nx ∑∞ is divergent on X. Note that the series n=1 cosn2nx is n=1 un (x) = n=1 n uniformly convergent on X just like the series in Remark 1 to Example 5. The ∑∞ ∑∞ 2nx series of squares can be rewritten in the form n=1 u2n (x) = n=1 1−cos , that 2n 2 is, the general term un (x) is the difference of the general term of the divergent harmonic series and uniformly convergent series. This implies the divergence 2 ∑∞ of the series n=1 sinnnx at each point of X. Remark. Naturally, the following more general situation also takes place: both ∑ ∑ ∑ series un (x) and vn (x) converge uniformly on X, but the series un (x)vn (x) does not converge (even pointwise) on X. In the particular case un (x) = vn (x), the counterexample is given above. Let us consider the case when un (x) ≠ vn (x). For instance, using the same arguments as before, one can prove that ( both ) ∑∞ cos nx ∑∞ cos nx √ √ and converge uniformly on X = [a, 𝜋 − a], ∀a ∈ 0, 𝜋2 , n=1 3 n n=1 4 n 2 ∑∞ ∑∞ nx but the series n=1 un (x)vn (x) = n=1 cos diverges at each point of X, because n7∕12 2

nx 2nx its general term can be represented in the form cos = 2n17∕12 + cos , where the n7∕12 2n7∕12 first summand is a general term of the divergent p-series, while the second is a general term of the uniformly convergent series.

∑ ∑ Example 25. Both series un (x) and vn (x) are nonnegative for ∀x ∈ X, u (x) lim n = 1 and one of these series converges uniformly on X, but another n→∞ vn (x) series does not converge uniformly on X. Solution 2 ∑∞ ∑∞ ∑∞ Consider the two nonnegative series n=1 un (x) = n=1 n4x+x4 and n=1 vn (x) = ∑∞ x2 u (x) lim vn(x) = n=1 n4 +x2 on X = ℝ. The limit of the general terms equals 1: n→∞ n 4 2 ∑∞ x2 lim nn4 +x = 1 for ∀x ∈ ℝ. Also, the series converges uniformly on 4 4 4 n=1 +x n +x n→∞

2 2

X = ℝ by the Weierstrass test, since un (x) = 2n1 2 n2n4 +xx 4 ≤ 2n1 2 , ∀x ∈ ℝ and the ∑ 1 series converges. Therefore, all the statement conditions hold. At the 2n2 2 2 2 ∑∞ same time, the second series n=1 n4x+x2 converges on X = ℝ since n4x+x2 ≤ nx4 , ∑ 1 ∀x ∈ ℝ and the series n4 converges. However, the convergence of the second

1.3 Dirichlet’s and Abel’s Theorems

3

2.5

n

Partial sums gn =∑ k = 1 sinkkx 2

Partial sum g100

2

Partial sum g200 1.5

1

0.5

g3 g2 g1 0.5

1

1.5

Figure 1.17 Example 24, series of squares

2

∑∞

2.5

3

2

n=1

sin nx . n

series is nonuniform since its general term does not converge to 0 uniformly: x2n for xn = n2 , one gets vn (xn ) = n4 +x = 12 ↛ 0. 2 n

n→∞

Remark. For numerical series and for the pointwise convergence of series of functions, the corresponding general statement is true and represents a partic∑ ular case of the Comparison test for nonnegative series: if both series un (x) ∑ u (x) and vn (x) are nonnegative for ∀x ∈ X, lim v n(x) = const > 0 and one of these n→∞ n series converges on X, then another series also converges on X.

1.3 Dirichlet’s and Abel’s Theorems Remark to Examples 26–29. In the following four examples, the conditions of Dirichlet’s theorem, which provides sufficient conditions for the uniform con∑ vergence of the series un (x)vn (x), are analyzed. It is shown that none of the three conditions stated in the theorem can be dropped. At the same time, these conditions are not necessary: all of them can be violated for an uniformly convergent series.

31

32

Chapter 1 Conditions of Uniform Convergence

∑ Example 26. The partial sums of un (x) are bounded for ∀x ∈ X, and the sequence vn (x) is monotone in n for each fixed x ∈ X and converges uniformly ∑ on X to 0, but the series un (x)vn (x) does not converge uniformly on X. Solution ∑∞ Let un (x) = xn and vn (x) = n1 be defined on X = (0, 1). The series n=1 un (x) = ∑∞ n n=1 x converges on X, since this is a geometric series with the ratio in (0, 1). Therefore, the partial sums of this series are bounded for each fixed x ∈ X. However, the boundedness is not uniform as is seen from the evaluation of the partial sums at the points xn = 1 − n1 , n > 1 lying in X: ( n ( ) ) ∑ xn 1 − 1∕n 1 n xkn = (1 − xnn ) = 1− 1− 1 − xn 1∕n n k=1 ( ( ) ) 1 n = (n − 1) 1 − 1 − → +∞, n→∞ n that is, the first condition in Dirichlet’s theorem is weakened. The remaining two conditions hold: vn (x) = n1 is monotone and vn (x) = n1 → 0 (and the last n→∞ convergence is uniform, because vn does not depend on x). ∑∞ ∑∞ xn The series of the products n=1 un (x)vn (x) = n=1 n converges on (0, 1), n since 0 < xn ≤ xn , ∀n ∈ ℕ and the geometric series converges for ∀x ∈ (0, 1). However, this convergence is nonuniform, which can be shown by the Cauchy 1 criterion: choosing pn = n and x̃ n = 1 − 2n ∈ X, ∀n ∈ ℕ, one obtains: 2n n ( ) | n+p | ∑ x̃ kn x̃ 2n 1 1 2n 1 |∑ | uk (̃xn )vk (̃xn )| = >n n = 1− → e−1 ≠ 0. | | | n→+∞ 2 k 2n 2 2n |k=n+1 | k=n+1

∑ Example 27. The partial sums of un (x) are uniformly bounded on X, and ∑ the sequence vn (x) converges uniformly on X to 0, but the series un (x)vn (x) does not converge uniformly on X. Solution n n √ Let un (x) = (−1) and vn (x) = (−1)n √x be defined on X = (0, 1). The series n n ∑∞ ∑∞ (−1)n √ u (x) = converges by Leibniz’s test and, consequently, its n=1 n n=1 n partial sums are bounded (and this boundedness is uniform on X, because n the general term does not depend on x). The sequence vn (x) = (−1)n √x conn | n | x n verges uniformly on X to 0 due to the following evaluation: ||(−1) √ || ≤ √1 , n n | | 1 ∀x ∈ (0, 1) and √ → 0. Thus, both conditions of the statement hold, but n n→+∞ ∑∞ ∑∞ n the series of the products n=1 un (x)vn (x) = n=1 xn converges nonuniformly n on (0, 1) as was shown in Example 26. Note that the sequence vn (x) = (−1)n √x n

1.3 Dirichlet’s and Abel’s Theorems

is not monotone in n, that is, the condition of the monotonicity of vn (x) in Dirihlet’s theorem is violated. ∑ Example 28. The partial sums of un (x) are uniformly bounded on X, and the sequence vn (x) is monotone in n for each fixed x ∈ X and converges on X ∑ to 0, but the series un (x)vn (x) does not converge uniformly on X. Solution Consider un (x) = (−1)n and vn (x) = xn on X = (0, 1). The partial sums ∑n k| |∑n | |∑n k=1 uk (x) are uniformly bounded on X: | k=1 uk (x)| = | k=1 (−1) | ≤ 1, n for ∀n ∈ ℕ and ∀x ∈ (0, 1). The sequence vn (x) = x is decreasing in n and vn (x) = xn → 0 for each fixed x ∈ (0, 1). Thus, the conditions of the statement n→+∞ are satisfied. However, the series of the products converges nonuniformly. ∑∞ ∑∞ In fact, n=1 un (x)vn (x) = n=1 (−1)n xn is a convergent geometric series on (0, 1) (the ratio −x ∈ (−1, 0)), but the evaluation of its residual shows that this 1 convergence is nonuniform: choosing xn = 1 − n+1 ∈ (0, 1), one obtains ∞ |∑ | | (−1)n+1 xn+1 | | | | n | (−1)k xkn | = | | | | | | 1 + xn | |k=n+1 | | | ( )n+1 1 1 1 = 1 − → e−1 ≠ 0. n→+∞ 2 n+1 2− 1 n+1

3.5

Exact sum ∑∞n = 1xn =

Partial sums fn = ∑nk = 1

3 2.5

x 1– x xk

Pn = (xn, fn(xn)), xn = 1–

f3

1 n

P4

2

f2 1.5

P3

1

f1

P2 0.5

0.1

0.2

0.3

0.4

0.5

Figure 1.18 Examples 26, 29, and 33, series

0.6

∑∞ n=1

0.7

un (x) =

0.8

∑∞ n=1

xn .

0.9

1

33

34

Chapter 1 Conditions of Uniform Convergence

Note that the third condition of Dirichlet’s theorem (the uniform convergence of vn (x)) is weakened in the above statement, and the chosen sequence vn (x) = xn converges nonuniformly to 0 on X = (0, 1), since for xn = 1 − n1 ∈ (0, 1), ∀n ∈ ( )n ℕ it follows |vn (xn )| = 1 − n1 → e−1 ≠ 0. n→+∞

∑ Example 29. The partial sums of un (x) are not uniformly bounded on X, and the sequence vn (x) is not monotone in n and does not converge uniformly ∑ on X to 0, but still the series un (x)vn (x) converges uniformly on X. Solution n Consider un (x) = xn and vn (x) = (−1) on X = (0, 1). Let us check the condixn2 ∑∞ ∑∞ tions of the statement. First, the partial sums of n=1 un (x) = n=1 xn are not uniformly bounded on X (see for details Example 26). Second, the sequence vn (x) is not monotone in n (it is alternating for each fixed x ∈ (0, 1)). Finally, n vn (x) = (−1) converges to 0 for each fixed x ∈ (0, 1), but this convergence is xn2 | n| not uniform, because choosing xn = n12 , ∀n ∈ ℕ one obtains | (−1) | = 1 ↛ 0. | x n n2 | n→+∞ In this way, all the conditions in Dirichlet’s theorem are violated. Neverthen−1 ∑∞ ∑∞ less, the series n=1 un (x)vn (x) = n=1 (−1)n xn2 converges uniformly on (0, 1)

n

x Exact sum ∑∞ = – ln(1 – x) n=1 n

2.5

Partial sums gn = ∑nk = 1 uk vk =∑nk = 1

xk k

2 ~ )), x~ = 1– 1 Pn = (x~n, gn(x n n

g3

2n

1.5

P3

g2

P2

1

g1

P1

0.5

0.1

0.2

0.3

0.4

0.5

0.6

Figure 1.19 Examples 26, 27, 30, 31, and 32, series

0.7

∑∞ n=1

0.8

0.9

un (x)vn (x) =

1

∑∞

xn n=1 n .

1.3 Dirichlet’s and Abel’s Theorems n−1 | | according to the Weierstrass test: |(−1)n xn2 | ≤ | | ∑ and the majorant series n12 converges.

1 , n2

for ∀n ∈ ℕ and ∀x ∈ (0, 1),

Remark. The functions un (x) = nx and vn (x) = (−1)n x considered on X = (0, 10] ∑∞ exhibit even “wilder” behavior. In fact, the partial sums of the series n=1 nx are not bounded at any point x ∈ (0, 10] since this series is positive and divergent at each x ∈ (0, 10]. The sequence (−1)n x is not monotone in n and diverges at each 2 ∑∞ ∑∞ x ∈ (0, 10]. Nevertheless, the series n=1 un (x)vn (x) = n=1 (−1)n xn converges uniformly on (0, 10] since the following evaluation of the residual (resulting from Leibniz’s test for alternating series) ∞ |∑ x2 || | x2 || 100 | |rn (x)| = | (−1)k | ≤ ||(−1)n+1 → 0 | ≤ n + 1 n→+∞ | | | k n + 1 | |k=n+1 |

is true for all x ∈ (0, 10] simultaneously.

Remark to Examples 30–33. In the next four examples, we analyze the sufficient conditions of Abel’s theorem for the uniform convergence of the ∑ series un (x)vn (x). The situation here is quite similar to that for Dirichlet’s 3

Limit function v(x) = 0

2.5

Functions vn(x) =

Q2

2

P4

xn2

Pn = (xn, vn(xn)) = ( n12 , (–1)n)

1.5 1

(–1)n

P2

~ , v (x ~ 1 n Qn = (x n n n)) = ( 3 , (–1) n) n

0.5

–0.5

v2 0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

P3

–1

v1

–1.5 –2 –2.5 –3

Q3

Figure 1.20 Examples 29 and 33, sequence vn (x) =

0.9 v3 1

(−1)n xn2

.

35

36

Chapter 1 Conditions of Uniform Convergence

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

–0.2 Partial sum g100 –0.4

k–1

Partial sums gn =∑nk = 1 uk vk =∑nk = 1(–1)k x 2 k

k

–0.6

Pn = (1, ∑nk = 1 (–1)2 ) k

g2 P 2 –0.8 g3 –1

g1

Figure 1.21 Examples 29 and 33, series

∑∞ n=1

un (x)vn (x) =

∑∞ n=1

P3 P1

n−1

(−1)n xn2 .

theorem: none of the three conditions can be dropped, but, at the same time, all of them can be violated for an uniformly convergent series. ∑ Example 30. A series un (x) converges on X, and a sequence vn (x) is monotone in n for each fixed x ∈ X and uniformly bounded on X, but the series ∑ un (x)vn (x) does not converge uniformly on X. Solution ∑∞ ∑∞ For un (x) = xn−1 and vn (x) = nx on X = (0, 1), the series n=1 un (x) = n=1 xn−1 is convergent at each point x ∈ (0, 1), since this is a geometric series with the ratio in (0, 1), and the sequence vn (x) = nx is monotone in n and uniformly | | bounded on X: | nx | ≤ n1 ≤ 1, ∀n ∈ ℕ, ∀x ∈ X. Thus, all the statement condi| | ∑∞ ∑∞ n tions are satisfied. However, the series n=1 un (x)vn (x) = n=1 xn converges nonuniformly on (0, 1) (see Example 26 for details). Note, that in the statement conditions, the first condition of Abel’s theorem (the uniform convergence of ∑ ∑∞ ∑∞ un (x)) is weakened, and the chosen series n=1 un (x) = n=1 xn−1 converges nonuniformly on X = (0, 1), which can be seen from the evaluation of the series residual for xn = 1 − n1 ∈ X, ∀n > 1: ∞ ( ) |∑ | xnn 1 n | | xk−1 =n⋅ 1− → ∞. | |= n | | 1 − xn n n→+∞ |k=n+1 |

1.3 Dirichlet’s and Abel’s Theorems

∑ Example 31. A series un (x) converges uniformly on X, and a sequence vn (x) ∑ is uniformly bounded on X, but the series un (x)vn (x) does not converge uniformly on X. Solution n Consider un (x) = (−1) and vn (x) = (−1)n xn on X = (0, 1). The series n n ∑∞ ∑∞ (−1) converges by Leibniz’s test, and this convergence n=1 un (x) = n=1 n is uniform since un (x) does not depend on x. The uniform boundedness of vn (x) = (−1)n xn is also easily verified: |(−1)n xn | ≤ 1, for ∀n ∈ ℕ, ∀x ∈ (0, 1). Thus, all the statement conditions hold.n However, as was shown in Example ∑∞ ∑∞ 26, the series n=1 un (x)vn (x) = n=1 xn converges nonuniformly on (0, 1). Note that the condition of monotonicity of vn (x) in Abel’s theorem is omitted in this example and, consequently, the choice of the nonmonotone sequence vn (x) = (−1)n xn resulted in nonuniform convergence of the series of the products. ∑ Example 32. A series un (x) converges uniformly on X, and a sequence vn (x) ∑ is monotone in n for each fixed x ∈ X, but the series un (x)vn (x) does not converge uniformly on X. Solution n−1 For un (x) = xn2 and vn (x) = nx on X = (0, 1), all the statement conditions are satisfied. In fact, the sequence vn (x) = nx is monotone in n for each fixed ∑∞ ∑∞ n−1 x ∈ (0, 1), and the series n=1 un (x) = n=1 xn2 converges uniformly on (0, 1) ∑ | n−1 | according to the Weierstrass test: | xn2 | ≤ n12 , ∀x ∈ (0, 1) and the series n12 is a | | ∑∞ n ∑∞ convergent p-series. However, the series n=1 un (x)vn (x) = n=1 xn converges nonuniformly on (0, 1) (see details in Example 26). Note that the condition of uniform boundedness of vn (x) in Abel’s theorem is omitted in the statement. The chosen sequence vn (x) = nx is not bounded for any x ∈ (0, 1), and this led ∑ to nonuniform convergence of the series un (x)vn (x). Remark. The following strengthened version of this example can also be ∑ constructed: a series un (x) converges uniformly on X, and a sequence vn (x) is monotone and bounded in n for each fixed x ∈ X, but the series ∑ un (x)vn (x) does not converge uniformly on X. The counterexample can 2 x2 be provided by un (x) = (1+x) and vn (x) = (3n2n+2)x on X = (0, +∞). The series n ∑∞ ∑∞ x2 n=0 un (x) = n=0 (1+x)n converges on X as a geometric series with the ratio 1 1+x

∈ (0, 1), ∀x ∈ X. To show the uniformity of this convergence on X, let us consider the residual ∞ ∑

x2

(1+x)n+1 x2 x rn (x) = = = 1 k (1 + x)n (1 + x) 1 − 1+x k=n+1

37

38

Chapter 1 Conditions of Uniform Convergence

and solve the critical point equation for each n > 1 fixed: 1 − (n − 1)x rn′ (x) = =0 (1 + x)n+1 1 that gives xn = n−1 . Since the derivative rn′ (x) is positive at the left of xn and negative at the right, the critical point xn is the maximum on X and, consequently, for each fixed n one gets the following evaluation of the residual: ( )−n 1 1 rn (x) ≤ max |rn (x)| = rn (xn ) = 1+ → 0 ⋅ e−1 = 0, n→∞ (0,+∞) n−1 n−1 ∑∞ that is, the series n=0 un (x) converges uniformly on X. 2 As for the sequence vn (x) = (3n2n+2)x , for each fixed x ∈ X, its terms are mono2

1 tonic (vn+1 (x) > vn (x)) and bounded (|vn (x)| = (3n2n+2)x < 3x ). Thus, all the conditions of the statement are satisfied. ∑∞ ∑∞ n2 x Nevertheless, the series converges n=0 un (x)vn (x) = n=0 3n2 +2 (1+x)n nonuniformly on X. Indeed, the convergence on X follows from the inequality 2 ∑ x x x 0 < 3nn2 +2 (1+x) < 13 (1+x) and the convergence of the geometric series n n (1+x)n for each fixed x ∈ X. Applying now the Cauchy criterion with pn = n and xn = n1 , one obtains

2n n | n+p | ∑ xn xn k2 n |∑ | u (x )v (x ) = > | | k n k n 2 + 2 (1 + x )k | | 3k 4 (1 + xn )2n n |k=n+1 | k=n+1 ( )−2n n1 1 1 = 1+ → e−2 ≠ 0, n→∞ 4n n 4 which means that the convergence is nonuniform on X = (0, +∞). Note that although vn (x) is bounded for each fixed x ∈ X, it is not uniformly bounded on X, since for xn = 3n21+2 ∈ X one gets vn (xn ) = n2 → +∞. n→∞

∑

Example 33. A series un (x) does not converge uniformly on X, and a sequence vn (x) is not monotone in n and is not uniformly bounded on X, but ∑ still the series un (x)vn (x) converges uniformly on X. Solution n Consider un (x) = xn and vn (x) = (−1) on X = (0, 1). Let us check the conditions xn2 of the statement. First, using the same reasoning as in Example 30, one can ∑∞ prove that the series n=1 xn converges nonuniformly on (0, 1). Then, the sequence vn (x) is not monotone in n (it is alternating for each fixed x ∈ (0, 1)). n Finally, vn (x) = (−1) converges to 0 for each fixed x ∈ (0, 1), but this sequence xn2 does not bounded uniformly on (0, 1), since for xn = n13 ∈ (0, 1) one has |vn (xn )| = n → ∞. Thus, all the conditions of Abel’s theorem are violated. n→+∞ n−1 ∑∞ ∑∞ Nevertheless, the series n=1 un (x)vn (x) = n=1 (−1)n xn2 converges uniformly on (0, 1) according to the Weierstrass test as was shown in Example 29.

Exercises

Remark. The conditions of Abel’s theorem√are violated even stronger for the functions un (x) = nx and vn (x) = (−1)n nx considered on X = (0, 2]. ∑∞ x In fact, the series n=1 n diverges at each x ∈ (0, 2]. The sequence √ n (−1) nx is not monotone in n and is unbounded at each x ∈ (0, 2] because √ √ |vn (x)| = |(−1)n nx| = nx → +∞, ∀x ∈ (0, 2]. Nevertheless, the series n→+∞ ∑∞ ∑∞ x2 n√ converges uniformly on (0, 2] as is seen from n=1 un (x)vn (x) = n=1 (−1) n the evaluation of the residual (following from Leibniz’s test for alternating series) |∑ | | | 2 | | ∞ | x2 || 4 k x | n+1 | | |rn (x)| = | (−1) √ | ≤ |(−1) √ → 0, |≤ √ |k=n+1 n + 1 || n + 1 n→+∞ k || || | which is satisfied for all x ∈ (0, 2] simultaneously.

Exercises 1 Show that Example 1 can be illustrated by the sequence fn (x) = X = [0, 1].

nx n2 x2 +1

on

2 Use the points xn = √n1 , ∀n ∈ ℕ to prove that the sequence fn (x) = xn of 2 Example 1 converges nonuniformly on X = (−1, 1). ∑∞ 3 Show that the series n=1 2n sin 51n x converges on X = (0, ∞), but the convergence is not uniform. 4 Check if the series

∑∞ n=1

5 Use the sequence fn (x) = Example 3.

xn √ 3 n

on X = (−1, 1) can be used for Example 2.

x+n+(−1)n x n2

on X = ℝ to illustrate the statement in

6 Construct a counterexample to the following false statement: “if f (x, y) defined on [a, b] × Y converges to a limit function 𝜑(x) as y approaches y0 , and this convergence is uniform on any interval [c, b], ∀c ∈ (a, b), then the convergence is also uniform on [a, b].” Compare with the statement in Example 4. Formulate similar false statements for sequences and series and disprove them by counterexamples. (Hint: for the functions 2xy2 depending on a parameter, try f (x, y) = x2 +y4 on [0, 1] × (0, 1] with the limit point y0 = 0; for the sequences—fn (x) = n2 xnx2 +1 on [0, 1]; and for the n ∑ ∑ series— un (x) = (1−x) on (0, 1].) n

39

40

Chapter 1 Conditions of Uniform Convergence

7 Verify that 2 a) the function f (x, y) = xy2 e−x∕y on X × Y = [0, +∞) × (0, 1] with the limit point y0 = 0 {x y sin x , x ≠ 0 y b) the function f (x, y) = on X × Y = ℝ × (0, 1] with the 1, x = 0 limit point y0 = 0 2n2 x c) the sequence fn (x) = 1+n on X = ℝ 4 2 ∑∞ ∑x ∞ sin nx d) the series n=1 un (x) = n=1 √ on X = ℝ n provide counterexamples for Example 5. 2 2 ∑∞ 8 Verify that the series n=1 (−1)n x n+n on X = [−1, 1] is one more coun3 terexample to the statement of Example 7.

9 Use the series

∑∞

n

n=1

(−1)n−1 xn , X = (−1, 1) for Example 8.

10

Show the feasibility of Example 9 by using counterexamples with ∑∞ x2 a) the series n=1 (−1)n−1 (1+x ,X =ℝ 2 )n ∑∞ n b) the series n=0 (−1) x(1 − x)n , X = [0, 1].

11

Use the series { a) un (x) =

∑∞

un (x) with −n 0, x (∈ [0, 3−n−1 ) ] ∪ [3 , 1]

n=1

1 3n+1 √ cos2 𝜋x 2 n 2 x (−1)n−1 (1+x on X 2 )n

, x ∈ (3−n−1 , 3−n )

on X = [0, 1]

b) un (x) = =ℝ to show the feasibility of Example 10. x2 n2 2n2 +5

12

Verify that the sequence fn (x) =

13

Check the statement of Example 16 for the sequence fn (x) = X = ℝ.

14

Verify whether the sequences fn (x), gn (x) and fn (x) ⋅ gn (x) are convergent

on X = (0, 1] specifies Example 15. 2nx n2 +x2

on

or divergent on X. In the case of the convergence, analyze its character: 2 a) fn (x) = ln (n n+1)x , gn (x) = ln 3n+2 x2 on X = (0, +∞) 2 n b) c) d) e)

2 2

x fn (x) = gn (x) = ln nn2 +1 on X = (0, +∞) x sin nx fn (x) = n , gn (x) = nx on X = (0, +∞) fn (x) = nx , gn (x) = ln (n+1)x on X = (0, +∞) n fn (x) = nx , gn (x) = sinnxnx on X = (0, 1] 2

+2 f ) fn (x) = gn (x) = (−1)n 5n x on X = (0, 1]. 3n2 +1

Exercises

Formulate false statements for which these sequences represent counterexamples. { 15 Show { 1

that

the

x2 , x ∈ ℚ n2 n2 +1 2 x ,x ∈ 𝕀 n2

sequences

fn (x) =

n2 +1 2 x ,x ∈ n2 2 x ,x ∈ 𝕀 n2

ℚ

and

gn (x) =

defined on X = [0, 1] provide a counterexample to

Example 14.

∑ ∑ ∑ 16 Verify whether the series un (x), vn (x) and un (x) ⋅ vn (x) are convergent or divergent on X. In the case of the convergence, analyze its character: a) un (x) = vn (x) = sin√nx on X = ℝ n

nx b) un (x) = vn (x) = sin on X = ℝ n2∕3 1 c) un (x) = vn (x) = x2∕3 +n2∕3 on X = [0, +∞) 1 1 d) un (x) = x+n , vn (x) = x+ln on X = (0, +∞) 2 n

( ) on X = [a, 𝜋 − a], ∀a ∈ 0, 𝜋2 ( ) f ) un (x) = sin√nx , vn (x) = sin√4 nx on X = [a, 𝜋 − a], ∀a ∈ 0, 𝜋2

e) un (x) = vn (x) = n

cos nx √ n

n

n

g) un (x) = vn (x) = (−1)n √x on X = (0, 1) n

n

n

h) un (x) = (−1)n lnx n , vn (x) = (−1)n √x , n ≥ 2 on X = (0, 1) n

i) j) k) l)

nx un (x) = sin , vn (x) = sinnnx on X = ℝ n2∕3 n un (x) = vn (x) = xn on X = [0, 1) n n un (x) = xn , vn (x) = nx1∕3 on X = [0, 1) n un (x) = vn (x) = (−1)n √3x on X = [0, 1) n

m) un (x) = vn (x) = (−1)n √x on X = (0, 1). n Formulate false statements for which counterexamples. 3 ∑∞ ∑∞ 17 Show that the series n=1 n62x+x6 and n=1 the statement in Example 25.

2x3 n6 +x3

these

series

represent

on X = (0, +∞) exemplify

18 For given un (x) and vn (x) on the specified set X, verify the conditions of Dirichlet’s theorem and investigate the character of the convergence of ∑ the series un (x) ⋅ vn (x): a) un (x) = xn−1 , vn (x) = nx on X = (−1, 1) b) un (x) = xn−1 , vn (x) = (−1)n nx on X = [0, 1) 1 c) un (x) = xn+1 , vn (x) = (−1)n n3∕2 on X = (0, 1) x

41

42

Chapter 1 Conditions of Uniform Convergence

d) un (x) = (−1)n xn−1 , vn (x) = (−1)n nx on X = [0, 1) [ ] 𝜋 19𝜋 e) un (x) = √1 , vn (x) = sin nx on X = 10 , 10 n ( 2 )n x f ) un (x) = (−1)n , vn (x) = 1+x on X = ℝ 2 (−1)n on X = (0, +∞) (1+x2 )n [ ] x sin nx 𝜋 11𝜋 , v (x) = on X = , n n x 6 6 (−1)n x2n , vn (x) = (1+x on X = (−1, 1). 2 )n

g) un (x) = x2 , vn (x) = h) un (x) =

i) un (x) = Formulate false statements for which these functions and series represent counterexamples. 19

For given un (x) and vn (x) on the specified set X, verify the conditions of Abel’s theorem and investigate the character of the convergence of the ∑ series un (x) ⋅ vn (x): ( )

a) un (x) = (−1)n , vn (x) =

x2 1+x2

n

on X = ℝ

, vn (x) = nx on X = (−1, 1) vn (x) = sinnxnx on X = (0, +∞) 2 [

n−1

b) un (x) = x c) un (x) = d) un (x) = e) un (x) = f ) un (x) = g) un (x) = h) un (x) = i) un (x) =

x2 , n ] √ 1 𝜋 19𝜋 , v (x) = nx sin nx on X = , n nx 10 10 (−1)n n , v (x) = (−1) sin nx on X = ℝ n n 2 2n+1 n x (−1) (1+x2 )n , vn (x) = (n+1)x2 on X = (0, +∞) sin nx 1 √ , vn (x) = √ on X = ℝ n n sin nx 3∕2 , v (x) = n on X = ℝ n n2 √ xn √ , vn (x) = x n on X = (0, 1) n n ( ) √ sin nx 𝜋 n , v (x) = (−1) n on X = 0, . n n 2

j) un (x) = Formulate false statements for which these functions and series represent counterexamples.

Further Reading S. Abbott. Understanding Analysis, Springer, New York, 2002. D. Bressoud, A Radical Approach to Real Analysis, MAA, Washington, DC, 2007. T.J.I. Bromwich, An Introduction to the Theory of Infinite Series, AMS, Providence, RI, 2005. B.M. Budak and S.V. Fomin, Multiple Integrals, Field Theory and Series, Mir Publisher, Moscow, 1978.

Further Reading

G.M. Fichtengolz, Differential- und Integralrechnung, Vol.1–3, V.E.B. Deutscher Verlag Wiss., Berlin, 1968. V.A. Ilyin and E.G. Poznyak, Fundamentals of Mathematical Analysis, Vol.1,2, Mir Publisher, Moscow, 1982. K. Knopp, Theory and Applications of Infinite Series, Dover Publication, Mineola, NY, 1990. C.H.C. Little, K.L. Teo and B. Brunt, Real Analysis via Sequences and Series, Springer, New York, 2015. W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, New York, 1976. V.A. Zorich, Mathematical Analysis I, II, Springer, Berlin, 2004.

43

45

CHAPTER 2 Properties of the Limit Function: Boundedness, Limits, Continuity 2.1 Convergence and Boundedness Example 1. A sequence of bounded on X functions converges on X to a function, which is unbounded on X. {

Solution Consider the sequence of functions fn (x) =

( ) min n, 1x , 0 < x ≤ 1

, each 0, x = 0 of which is bounded on [0, 1], because 0 ≤ fn (x) ≤ n, ∀x ∈ [0, 1]. Since for any fixed x ∈ (0, 1] there exists n0 ∈ ℕ such that n0 > 1x , for all n > n0 we have fn (x) = 1x and, consequently, lim fn (x) = 1x , ∀x ∈ (0, 1]. For x = 0, we get n→∞ {1 ,0 < x ≤ 1 x lim fn (0) = lim 0 = 0. Hence, the limit function is f (x) = , n→∞ n→∞ 0, x = 0 which is unbounded on [0, 1]. Note, that this convergence is not uniform, 1 since for any n ∈ ℕ one can choose the point xn = 2n ∈ [0, 1], which gives |fn (xn ) − f (xn )| = |n − 2n| = n → ∞. n→∞

Remark 1. Note that the functions in the counterexample are not uniformly bounded on [0, 1]. In fact, for any M > 0 there exist n > M and xn = n1 ∈ (0, 1] such that fn (xn ) = n > M. This is the cause of the unboundedness of the limit function. Remark 2. The corresponding general statement will be true if the condition of the uniform boundedness of fn (x) on X would be added: a sequence of uniformly bounded on X functions converges on X to a bounded function f (x). Remark 3. For bounded functions, the uniform convergence on X implies the uniform boundedness. Therefore, the additional condition of the uniform convergence of fn (x) on X is sufficient for the general statement becomes true: if a Counterexamples on Uniform Convergence: Sequences, Series, Functions, and Integrals, First Edition. Andrei Bourchtein and Ludmila Bourchtein. © 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. Companion website: www.wiley.com/go/bourchtein/counterexamples_on_uniform_convergence

46

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

10 9

Limit function f(x) =

,x≠0

8

Functions fn(x)

7

Residuals rn(x) = f (x) – fn(x)

1 x

1 Pn = (xn, rn(xn)) = ( 2n , n)

6 5 4 3

f3

2

f2

1

f1

P3 P2 P1 r3 0.2

r2 0.4

Figure 2.1 Example 1, sequence fn (x) =

r1 0.6

0.8

1

{ ( ) min n, 1x , 0 < x ≤ 1 . 0, x = 0

sequence of bounded on X functions converges uniformly on X, then the limit function f (x) is bounded on X. Remark 4. The general statement with the opposite conclusion—“if a sequence of nonuniformly bounded functions converges on X, then the limit function is not bounded on X”—is also false as shown in Example 2.

Example 2. A sequence of functions is not uniformly bounded on X, but it converges on X to a function, which is bounded on X. [ ] ⎧ n2 x + n − n3 , x ∈ n − 1 , n [ n ] ⎪ Consider the sequence of functions fn (x) = ⎨ −n2 x + n + n3 , x ∈ n, n + 1 n ⎪ ⎩ 0, otherwise defined on X = [0, +∞). Like in Example 1, each of the functions is bounded, because 0 ≤ fn (x) ≤ n, ∀x ∈ [0, +∞) for any fixed n, but the sequence is not uniformly bounded on [0, +∞), since for any M > 0 there exist n > M and xn = n ∈ [0, +∞) such that fn (xn ) = n > M. Let us find the limit function. For any fixed x ∈ [0, +∞) there exists n0 ∈ ℕ such that n0 > x + 1, so for all n > n0 Solution

2.1 Convergence and Boundedness

10 Limit function f(x) = 0 Functions fn(x)

9 8 7

Pn = (xn , fn (xn ) – f (xn )) = (n, n)

6 P5

5 P4

4 P3

3 P2

2 1

P1

f2

f4

f5

4

5

f3

f1 1

2

3

6

7

8

9

10

[ ] ⎧ n2 x + n − n3 , x ∈ n − 1 , n n [ ] ⎪ Figure 2.2 Example 2, sequence fn (x) = ⎨−n2 x + n + n3 , x ∈ n, n + 1 . n ⎪ ⎩ 0, otherwise

we have n − n1 > n − 1 > x, which implies that lim fn (x) = 0, ∀x ∈ [0, +∞). n→∞ Therefore, the limit function is bounded on [0, +∞). Note, that the convergence of this sequence is not uniform (just like in Example 1). Indeed, choosing for any n ∈ ℕ the point xn = n ∈ [0, +∞), we have |fn (xn ) − f (xn )| = n → ∞. n→∞

Remark to Examples 1 and 2. In these two examples, although the sequences of the functions are not bounded uniformly, each of the functions is bounded on X. Analogous examples can be constructed for sequences of unbounded on X functions. Example 3. A sequence of unbounded and discontinuous on X functions converges on X to a function, which is bounded and continuous on X. Solution

{

1 nx

,0 < x ≤ 1 considered on X = [0, 1] 0, x = 0 gives a counterexample. Each of these functions is unbounded on [0, 1], since The sequence of functions fn (x) =

47

48

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity 1 for any M > 0 there exists xn = n(M+1) ∈ [0, 1] such that fn (xn ) = M + 1 > M. Also, each of the functions has an essential discontinuity at x = 0, since 1 lim fn (x) = lim nx = +∞. However, this sequence converges to the bounded x→0+

x→0+

and continuous function f (x) ≡ 0. Indeed, for x = 0, we get lim fn (0) = n→∞

1 lim 0 = 0, and for any fixed x ∈ (0, 1], we obtain lim fn (x) = lim nx = 0. Note, n→∞ n→∞ n→∞ that the convergence is not uniform, since for any n ∈ ℕ there is the point xn = n1 ∈ [0, 1] such that |fn (xn ) − f (xn )| = 1 ↛ 0. n→∞

Remark 1. If the condition of discontinuity is dropped in the statement, then the same sequence considered on X = (0, 1] provides a counterexample. Remark 2. The strengthened version of this example with the added condition of the uniform convergence also takes place. The corresponding counterexample, albeit for the case of bounded functions, is given in Example 11. Remark 3. The general statement—“if a sequence of unbounded and discontinuous functions converges to an unbounded and discontinuous function, then the convergence is uniform”—is also false as shown in Example 4. 10 9 Limit function f(x) = 0 Functions fn(x)

8

1 Pn = (xn,fn(xn)) = ( –, n 1)

7 6 5 f1

4 3

f2

2

f3

1 0.2

P2

P3

0.4

0.6

{ Figure 2.3 Example 3, sequence fn (x) =

P1

1 ,0 nx

0, n n n n it follows that |fn (x) − f (x)| < n1 for ∀x ∈ [0, +∞), and since that fn (x) converges uniformly to f (x) on [0, +∞).

1 → 0, this means n n→∞

Remark. The condition of discontinuity in the statement can be dropped. The respective counterexample can be given with the same sequence considered on X = (0, +∞) or X = (0, 1]. Example 6. A sequence of uniformly bounded on X functions converges on X, but the convergence is nonuniform on X. {

Solution The sequence fn (x) =

1 nx sin nx ,x ≠ 0 is uniformly bounded on X = [0, 1] 1, x = 0

| sin 1 | since |fn (x)| = || 1 nx || ≤ 1, ∀n ∈ ℕ and ∀x ∈ [0, 1]. Further, for x = 0 one has | nx |

1.5 f1 1

f2 f3

0.5

0.5

1

1.5

–0.5

2

r3 r2 r1

–1

Limit function f(x) = lnx, x ≠ 0 Functions fn(x)

–1.5

Residuals rn(x) = f (x) – fn(x)

{ Figure 2.4 Example 5, sequence fn (x) =

ln

n+1 x n

, x ∈ (0, +∞) . 0, x = 0

2.5

3

2.2 Limits and Continuity of Limit Functions

lim fn (0) = lim 1 = 1 and for x ≠ 0, using the first remarkable limit, one n→∞ obtains n→∞

lim fn (x) = lim

1 sin nx 1 nx

n→∞

n→∞

= lim

t→0+

sin t = 1. t

Therefore, the limit of fn (x) exists and f (x) = lim fn (x) = 1, ∀x ∈ [0, 1]. Thus, the n→∞ statement conditions are satisfied, but the convergence on [0, 1] is nonuniform: choosing xn = n1 ∈ [0, 1] for ∀n ∈ ℕ, one gets | 1 | |fn (xn ) − f (xn )| = ||n ⋅ sin 1 − 1|| = 1 − sin 1 ↛ 0. n→∞ n | |

2.2 Limits and Continuity of Limit Functions Example 7. A series

∑

un (x) converges on X and lim un (x) exists for each n, x→x0

but the corresponding limit of the series cannot be calculated term by term, that ∑ is, one cannot interchange the order of the limit and infinite sum: lim un (x) ≠ x→x0 ∑ lim un (x) . x→x0

Solution ∑∞ Let us consider the series n=1 (xn − xn+1 ) on X = (−1, 1) with the limit point x0 = 1. The partial sums are easily found: fn (x) =

n ∑

uk (x) = (x − x2 ) + (x2 − x3 ) + · · · + (xn − xn+1 ) = x − xn+1 .

k=1

Since |x| < 1, it follows that lim xn = 0, and therefore, the series is convern→∞ gent on (−1, 1), with the sum of the series being f (x) = lim fn (x) = lim (x − n→∞

n→∞

xn+1 ) = x. All the required limits exist: lim un (x) = lim (xn − xn+1 ) = 0, ∀n ∈ ℕ. x→1−

However, ∞ ∑ n=1

lim un (x) =

x→1−

∞ ∑ n=1

lim (xn − xn+1 ) =

x→1−

x→1−

∞ ∑

0=0

n=1

≠ 1 = lim x = lim f (x) = lim x→1−

x→1−

x→1−

∞ ∑

un (x).

n=1

Let us verify the type of convergence of this series. Evaluating the remainder ∞ |∑ | | | |rn (x)| = | uk (x)| = |fn (x) − f (x)| = |x − xn+1 − x| = |x|n+1 | | |k=n+1 |

51

52

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

and choosing for ∀n ∈ ℕ the corresponding xn = |rn (xn )| = (−1, 1).

1 ↛ 2 n→∞

1 √

n+1

2

∈ (−1, 1), we obtain

0, which means that the convergence is nonuniform on

Remark 1. A similar example can be given for a sequence of functions: a sequence fn (x) converges on X and lim fn (x) exists for each n, but x→x0

lim lim fn (x) ≠ lim lim fn (x). It can be shown by the counterexample

x→x0 n→∞

n→∞ x→x0

2 2

with the sequence fn (x) = nn2 xx2 −1 defined on X = (0, 1) with the limit point +1 x0 = 0. This sequence converges to the limit function n2 x2 − 1 = 1, ∀x ∈ (0, 1) n→∞ n2 x2 + 1

f (x) = lim fn (x) = lim n→∞

and lim f (x) = lim 1 = 1. Also, each function has the limit at 0: lim fn (x) = x→0+ 2 2 lim n x −1 x→0+ n2 x2 +1

x→0+

x→0+

= −1, ∀n ∈ ℕ. Therefore, lim lim fn (x) = 1 ≠ −1 = lim lim fn (x). x→0+ n→∞

n→∞ x→0+

To see that the convergence of fn (x) to f (x) is not uniform, we can choose xn = n1 ∈ (0, 1) for any n ∈ ℕ in order to obtain | n2 ⋅ | |fn (xn ) − f (xn )| = || | n2 ⋅ |

1 n2 1 n2

| | − 1|| = 1 ↛ 0 . n→∞ | +1 | −1

Remark 2. A similar example for a function depending on a continuous parameter is formulated as follows: a function f (x, y) is defined on X × Y , converges on X to a limit function 𝜑(x), as y → y0 , and lim f (x, y) exists for each y ∈ Y , but x→x0

lim lim f (x, y) ≠ lim lim f (x, y). For a counterexample, one can use the func-

x→x0 y→y0

y→y0 x→x0

y2

tion f (x, y) = x2 +y2 defined on X × Y = (0, 1] × (0, 1] with the limit points x0 = 0 for X and y0 = 0 for Y . The limit function is 𝜑(x) = lim f (x, y) = lim y→0+

y→0+

y2 = 0, ∀x ∈ (0, 1], x2 + y2

and therefore, lim 𝜑(x) = lim 0 = 0. However, another order of limits gives a x→0+

x→0+

different result: lim lim f (x, y) = lim lim

y→0+ x→0+

y→0+ x→0+

y2 = lim 1 = 1. x2 + y2 y→0+

2.2 Limits and Continuity of Limit Functions

1

P∑lim = (1,∑lim un(x))

0.8

Plim∑ = (1,lim∑ un(x))

Plim∑

0.6

–1

r1

0.4

r3

0.2

P1 PP 23

P∑lim

–0.5

0.5

1

r2

–0.2

f2

–0.4

Partial sums fn(x) = x – x n+1

–0.6

Residuals rn(x) = f (x) – fn(x)

f3

–0.8

f1

Exact sum f(x) = x

Pn = (xn, rn(xn)) = (

–1

Figure 2.5 Example 7, series

∑∞

n=1 (x

n

1

n+1√

2

,1 ) 2

− x n+1 ).

Analyzing the convergence, we see that for ∀y ∈ (0, 1] there exists xy = y ∈ y2 (0, 1] that leads to |f (xy , y) − 𝜑(xy )| = y2 +y2 = 12 ↛ 0. This means that the cony→0+

vergence is nonuniform. Another interesting counterexample of a similar type is f (x, y) = arctan xy defined on X × Y = (0, +∞) × (0, 1) with the limit points x0 = 0 for X and y0 = 0 for Y .

Remark 3. In this example, both limits are finite, but assume different values. Example 8 shows that one of the limits can be infinite. ∑ Example 8. A series un (x) converges on X and lim un (x) exists for each n, x→x0 ∑ ∑ but lim un (x) ≠ lim un (x) since the left-hand side limit is infinite. x→x0

x→x0

Solution 2 2 2 2 ∑∞ x +2 For such a case, consider the series n=1 un (x) where un (x) = (n−1) − nn2 xx3 +2 (n−1)2 x3 +1 +1 are defined on X = (0, ∞). The partial sums of this telescoping series are found 2 2 ∑n ∑ ∞ in the form fn (x) = k=1 uk (x) = 2 − nn2 xx3 +2 , and its sum is f (x) = n=1 un (x) = +1

53

54

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

1

P3

r1

P2

P1

Plimx limn

r2

0.5

r3

0.2

f3

0.4

f1 Plimn limx

0.8

1

Limit function f(x) = 1 Functions fn(x)

f2

–0.5

–1

0.6

Residuals rn(x) = f (x) – fn(x) Pn = (xn, rn(xn)) = ( 1 –, 1) n

Plimx limn = (0, limxlimn fn(x)), Plimn limx = (0, limnlimx fn(x)),

Figure 2.6 Example 7, sequence fn (x) =

n2 x 2 −1 . n2 x 2 +1

lim fn (x) = 2 − 1x , ∀x ∈ X. Therefore, lim

n→∞

x→0+

∑∞ n=1

( ) un (x) = lim 2 − 1x = −∞. x→0+

Nevertheless, at (the same) limit point x = 0, the limit of each un (x) exists: 2 lim u1 (x) = lim 2 − xx3 +2 = 0, lim un (x) = 2 − 2 = 0, ∀n ≠ 1, and the +1 x→0+ x→0+ x→0+ ∑∞ ∑∞ right-hand side limit in the statement is finite: n=1 lim un (x) = n=1 0 = 0. x→0+

Note that the above series converges nonuniformly on (0, ∞), since for ∀n ∈ ℕ choosing xn = n1 ∈ (0, ∞), one gets | | | | n2 x2n + 2 || || 3 1 | | → ∞. |rn (xn )| = |f (xn ) − fn (xn )| = |2 − −2+ = − n | | | n→∞ 3 1 2 | | x n x + 1 | | + 1 n n | | |n | Remark 1. Analogous example with infinite limit can be given for a sequence: a sequence fn (x) converges on X and lim fn (x) exists for each n, x→x0

but lim lim fn (x) ≠ lim lim fn (x) since the left-hand side limit is infinite. If x→x0 n→∞

n→∞ x→x0

2 2

one considers the functions fn (x) = nn2 xx3 +2 on X = (0, ∞) (similar to the partial +1 sums in the first counterexample), then lim fn (x) = 1x and lim lim fn (x) = n→∞

1 x→0+ x

lim

x→0+ n→∞

= ∞, while at the same limit point x = 0, the limit of each fn (x) exists:

2.2 Limits and Continuity of Limit Functions

lim fn (x) = 2, ∀n ∈ ℕ, and the right-hand side limit in the example statement

x→0+

is finite: lim lim fn (x) = lim 2 = 2. Note that the convergence of fn (x) to n→∞ x→0+

n→∞

f (x) is nonuniform on X as shown by the following evaluation: choosing xn = n1 , ∀n ∈ ℕ, one gets | n2 ⋅ | |fn (xn ) − f (xn )| = || | n2 ⋅ |

1 n2 1 n3

| | − n|| → ∞. | n→∞ +1 | +2

Remark 2. For a function depending on a parameter one gets: a function f (x, y) is defined on X × Y , converges on X to a limit function 𝜑(x), as y → y0 , and lim f (x, y) exists for each y ∈ Y , but lim lim f (x, y) ≠ lim lim f (x, y) since the x→x0

x→x0 y→y0

y→y0 x→x0

left-hand side limit is infinite. For a counterexample, one can consider f (x, y) = x4 −3y4 on X × Y = (0, 1] × (0, 1] using the limit points x0 = 0 and y0 = 0. On the x5 +y4 one hand, the left-hand side limit in the statement is infinite, since 𝜑(x) = lim f (x, y) = lim y→0+

y→0+

x4 − 3y4 1 = , ∀x ∈ (0, 1] 5 4 x x +y

and lim 𝜑(x) = lim

x→0+

x→0+

1 = +∞. x

On the other hand, the right-hand side limit is finite: lim lim f (x, y) = lim lim

y→0+ x→0+

y→0+ x→0+

x4 − 3y4 = lim (−3) = −3. y→0+ x5 + y4

Thus, the limits not only cannot be interchanged, but one of them is infinite while another is finite. It is easy to show that the convergence of f (x, y) to 𝜑(x) is nonuniform on (0, 1]: choosing xy = y ∈ (0, 1] for any y ∈ (0, 1], one obtains | y4 − 3y4 1 | 1 2 | | |f (xy , y) − 𝜑(xy )| = | 5 − |= + → +∞. | y + y4 | y 1 + y y→0+ y | | Remark 3. Under the conditions of the statement, the limit in the left-hand side may even not exist (neither finite nor infinite), as shown in Example 9. Example 9. A series

∑

un (x) converges on X, also lim un (x) exists for each x→x0 ∑ n and one of the two conditions is satisfied: either lim un (x) converges or x→x0 ∑ lim un (x) exists, but nevertheless the remaining condition does not hold. x→x0

55

56

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Solution ∑∞ Let us consider the series { n=1 un (x) with functions un (x) defined on X = xn − xn+1 , x ∈ ℚ ∩ X (−1, 1) as follows: un (x) = . At the limit point x0 = 1, 0, x ∈ 𝕀 ∩ X one gets lim un (x) = lim (xn − xn+1 ) = 0 and lim un (x) = lim 0 = x→1− ,x∈ℚ x→1− ,x∈ℚ x→1− ,x∈𝕀 ∑∞x→1− ,x∈𝕀 ∑ ∞ 0. Therefore, lim un (x) = 0, ∀n ∈ ℕ and lim u (x) = n n=1 n=1 0 = 0. x→1−

However, f (x) =

∞ ∑

x→1−

{ un (x) =

(1−x)x 1−x

= x, x ∈ ℚ ∩ X , 0, x ∈ 𝕀 ∩ X

n=1

that gives

lim

x→1− ,x∈ℚ

f (x) =

lim

x→1− ,x∈ℚ

x = 1 and

lim f (x) =

x→1− ,x∈𝕀

lim 0 = 0, which

x→1− ,x∈𝕀

means that lim f (x) does not exist. Note that choosing xn = 1 − x→1−

1 n+1

∈

ℚ ∩ X, ∀n ∈ ℕ, one obtains ∞ |∑ | (1 − x )xn+1 | | n n |rn (xn )| = | xkn (1 − xn )| = | | 1 − xn |k=n+1 | ( )n+1 1 = 1− → e−1 ≠ 0, n→∞ n+1

which means that the convergence is nonuniform on (−1, 1). ∑∞ Another example for series is n=0 un (x) with un (x) = (−1)n xn on X = (0, 1). ∑∞ 1 In this case, lim n=0 (−1)n xn = lim 1+x = 12 . However, lim un (x) = (−1)n , x→1− x→1− x→1− ∑∞ ∑∞ ∀n ∈ ℕ and the series n=0 lim un (x) = n=0 (−1)n diverges. This happens x→1−

since the convergence of the series is nonuniform on X: if for an evaluation ∑n+p of k=n+1 uk (x), ∀n ∈ ℕ in the Cauchy criterion one chooses pn = 1 and 1 xn = n+1√ ∈ X, then 2

n+p |∑ | 1 | | uk (xn )| = |(−1)n+1 xn n+1 | = ↛ 0. | | | 2 n→∞ |k=n+1 |

Remark 1. Under similar conditions, any of the two iterated limits may not exist in the statement on sequences. Suppose fn (x) converges on X and lim fn (x) exists for each n. Even if one of the limits— lim lim fn (x) or x→x0

n→∞ x→x0

lim lim fn (x)—exists, another one may not exist. x→x0 n→∞ { 1, x ∈ ℚ ∩ [0, 1) In fact, consider the sequence fn (x) = on X = [0, 1). xn , x ∈ 𝕀 ∩ [0, 1) n Since lim fn (x) = lim 1 = 1 and lim fn (x) = lim x = 1, it follows that x→1− ,x∈ℚ

x→1−

x→1− ,x∈𝕀

x→1−

2.2 Limits and Continuity of Limit Functions

lim fn (x) = 1, ∀x ∈ [0, 1) and, consequently, lim lim fn (x) = 1. On the other

x→1−

n→∞ x→1−

lim fn (x) = lim 1 = 1 and lim fn (x) = lim xn = 0, which implies n→∞ n→∞ n→∞,x∈ℚ n→∞,x∈𝕀 { 1, x ∈ ℚ lim fn (x) = = D(x), ∀x ∈ [0, 1). Therefore, the second iterated limit 0, x ∈ 𝕀 n→∞ lim lim fn (x) = lim D(x) does not exist. It happens because the convergence hand,

x→1− n→∞

x→1−

of the sequence is not uniform on X = [0, 1): ∀n > 1 choosing xn = one gets |fn (xn ) − f (xn )| = xnn =

1 √ n 2

∈ 𝕀 ∩ X,

1 ↛ 0. 2 n→∞

Another counterexample is the sequence fn (x) = (−1)n sinnxnx defined on X = (0, ∞) with the limit point x0 = 0. For any fixed x > 0 the sequence converges to zero: f (x) = lim fn (x) = lim (−1)n sinnxnx = 0, and therefore, the n→∞ n→∞ first iterated limit exists and equals zero: lim lim fn (x) = lim f (x) = 0. Also, x→0+ n→∞

x→0+

lim fn (x) = (−1)n for each n ∈ ℕ, but lim lim fn (x) = lim (−1)n does not exist.

x→0+

n→∞ x→0+

n→∞

Such a behavior is related to nonuniform convergence of the sequence on X: for ∀n, there exists xn = n1 ∈ X such that | sin 1 || |fn (xn ) − f (xn )| = ||(−1)n = sin 1 ↛ 0. n→∞ 1 || | Remark 2. Similar situation may occur for functions depending on a parameter. Assume a function f (x, y) is defined on X × Y , converges on X to a limit function 𝜑(x), as y → y0 , and lim f (x, y) exists for each y ∈ Y . Even though one x→x0

of the two iterated limits—lim lim f (x, y) or lim lim f (x, y)—exists, another one y→y0 x→x0

may not exist.

x→x0 y→y0

{

y2 ,x x2 +y2

∈ℚ on X × Y = (0, 1] × 1, x ∈ 𝕀 (0, 1]. At the limit point (0, 0), the first iterated limit exists: First, consider the function f (x, y) =

lim

x→0+ ,x∈ℚ

f (x, y) =

lim

x→0+ ,x∈ℚ x2

y2 = 1, lim f (x, y) = lim 1 = 1, x→0+ ,x∈𝕀 x→0+ ,x∈𝕀 + y2

and, consequently, lim lim f (x, y) = lim 1 = 1. However, the second iterated y→0+ x→0+

y→0+

limit does not exist, since lim

y→0+ ,x∈ℚ

f (x, y) =

y2 = 0, lim f (x, y) = lim 1 = 1, y→0+ ,x∈ℚ x2 + y2 y→0+ ,x∈𝕀 y→0+ ,x∈𝕀 lim

57

58

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

̃ and the modified Dirichlet function D(x) = 1 − D(x) = lim f (x, y) = y→0+ { 0, x ∈ ℚ ∩ (0, 1] has no limit. Note that such a situation occurs because the 1, x ∈ 𝕀 ∩ (0, 1] ̃ convergence of f (x, y) to D(x) is nonuniform on X = (0, 1]: for ∀y ∈ ℚ ∩ (0, 1], one can choose xy = y to obtain | | |1 2 | 1 | | ̃ y )| = || y |f (xy , y) − D(x − 0 ↛ 0. | = | − 0|| = | x2y + y2 | |2 | 2 y→0+ | | yD(y) For a second counterexample, consider f (x, y) = x arctan xy on X × Y = (0, +∞) × (0, +∞) with the limit point (0, 0). Since D(y) and arctan xy are y

bounded functions, one obtains 𝜑(x) = lim f (x, y) = lim x D(y) arctan xy = 0 y→0+

y→0+

for any x ∈ (0, +∞). Therefore, one of the iterated limits exists: lim lim f (x, y) = x→0+ y→0+

lim 𝜑(x) = 0. Nevertheless, the second limit does not exist, since lim f (x, y) =

x→0+

lim D(y)

x→0+

arctan x y

x y

x→0+

= D(y), ∀y ∈ (0, +∞) and the limit of D(y) does not exist at any

point. Note that the convergence of f (x, y) to 𝜑(x) is nonuniform on X = (0, +∞): for ∀y ∈ ℚ ∩ (0, +∞), one can choose xy = y ∈ (0, +∞) such that 𝜋 |f (xy , y) − 𝜑(xy )| = arctan 1 = ↛ 0. 4 y→0+ Remark to Examples 7–9. In all these counterexamples, the convergence is nonuniform on X. If the convergence would be uniform, the corresponding general statements will be true, that is, the calculation of the limit with respect to x can be interchanged with summation (for a series), or with the calculation of the limit with respect to discrete or continuous variable (for a sequence or a function depending on a parameter, respectively). Nevertheless, the following example shows that the uniform convergence is sufficient but not necessary condition for commutative properties of the operators. Example 10. A series converges nonuniformly on X, but still a limit of this series can be calculated term by term. Solution ∑∞ Consider the series n=1 (nxe−nx − (n − 1)xe−(n−1)x ) on X = [0, 1] with the limit point x0 = 0. First, let us find out if this series is convergent. If x = 0, then ∑∞ un (x) = 0, ∀n ∈ ℕ, and f (0) = n=1 0 = 0. If x ∈ (0, 1], then fn (x) =

n ∑

uk (x)

k=1 −x

= xe

+ 2xe−2x − xe−x + · · · + nxe−nx − (n − 1)xe−(n−1)x = nxe−nx ,

2.2 Limits and Continuity of Limit Functions

and consequently, t 1 = lim t = 0 t t→+∞ e e (to calculate the last limit, we use the change of variables nx = t, which implies t → +∞ for any x > 0, and then apply l’Hospital’s rule). Hence, n→∞ the series is convergent to f (x) = 0, ∀x ∈ [0, 1] . To show that this conver∑∞ gence is not uniform, for the series remainder |rn (x)| = || k=n+1 uk (x)|| = |fn (x) − f (x)| = nxe−nx , we can choose xn = n1 ∈ [0, 1] for any given n ∈ ℕ and f (x) = lim fn (x) = lim nxe−nx = lim n→∞

n→∞

t→+∞

1

obtain |rn (xn )| = n ⋅ n1 e−n⋅ n = e−1 ↛ 0. However, the limit, as x approaches 0, n→∞ can be calculated term by term: lim un (x) = lim (nxe−nx − (n − 1)xe−(n−1)x ) = 0,

x→0+

x→0+

and therefore, ∞ ∑ n=1

lim un (x) =

x→0+

∞ ∑

0 = 0 = lim 0 = lim f (x) = lim

n=1

x→0+

x→0+

x→0+

∞ ∑

un (x).

n=1

Remark 1. The corresponding example for a sequence goes as follows: a sequence fn (x) converges nonuniformly on X, but still lim lim fn (x) = x→x0 n→∞

lim lim fn (x). It can be exemplified by the sequence fn (x) = xn − x2n

n→∞ x→x0

defined on X = [0, 1] with the limit point x0 = 1. For the limit function, we have: f (x) = lim fn (x) = lim (xn − x2n ) = 0 if x ∈ [0, 1), and n→∞ n→∞ f (1) = lim fn (1) = lim (1 − 1) = 0 if x = 1. Thus, the sequence conn→∞ n→∞ verges to f (x) ≡ 0 on [0, 1]. Let us calculate the required limits at 1: lim fn (x) = lim (xn − x2n ) = 1 − 1 = 0, ∀n ∈ ℕ, and lim f (x) = lim 0 = 0. x→1−

x→1−

x→1−

x→1−

Therefore, lim lim fn (x) = lim f (x) = 0 = lim 0 = lim lim fn (x).

x→1− n→∞

x→1−

n→∞

n→∞ x→1−

At the same time, the convergence is not uniform on [0, 1] due to the following argument: for any n ∈ ℕ, we can choose xn = √n1 ∈ [0, 1], which gets 2

)n ( )2n | |( | 1 | |1 1| 1 1 |=| − |= |fn (xn ) − f (xn )| = || √ − √ ↛ 0. | | 2 4 | 4 n→∞ n n | | | | 2 2 | | Remark 2. A similar example for a function depending on a continuous parameter can be formulated as follows: a function f (x, y) is defined on X × Y and converges nonuniformly on X to a limit function 𝜑(x), as y → y0 , but lim lim f (x, y) = lim lim f (x, y). A counterexample is provided by the function x→x0 y→y0

y→y0 x→x0

59

60

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

f (x, y) =

xy x2 +y2

defined on X × Y = (0, 1] × (0, 1] with the limit points x0 = 0 for xy y→0+ x2 +y2

X and y0 = 0 for Y . The limit function is 𝜑(x) = lim f (x, y) = lim ∀x ∈ (0, 1] and also lim f (x, y) = x→0+

y→0+

xy lim x→0+ x2 +y2

= 0,

= 0, ∀y ∈ (0, 1]. Therefore, the

limits can be interchanged: lim lim f (x, y) = lim 0 = 0 = lim 𝜑(x) = lim lim f (x, y).

y→0+ x→0+

y→0+

x→0+

x→0+ y→0+

Nevertheless, the convergence is not uniform on X = (0, 1], since for ∀y ∈ (0, 1] there exists xy = y ∈ (0, 1] such that |f (xy , y) − 𝜑(xy )| =

y2 1 = ↛ 0. y2 + y2 2 y→0+

Example 11. A sequence of discontinuous functions converges uniformly on X, but the limit function is continuous on X. {

Solution

< x < n1 defined on X = [0, 1]. It 0, x = 0, n1 ≤ x ≤ 1 converges to f (x) ≡ 0 on [0, 1] because Consider the sequence fn (x) =

1 ,0 n

f (0) = lim fn (0) = lim 0 = 0, f (1) = lim fn (1) = lim 0 = 0, n→∞

n→∞

n→∞

n→∞

and for any fixed x ∈ (0, 1) there is nx ∈ ℕ such that nx > 1x , which implies n1 < 1 < x for ∀n > nx and, consequently, f (x) = lim fn (x) = 0. Moreover, this conn n→∞

x

vergence is uniform, because |fn (x) − f (x)| ≤ n1 → 0 simultaneously for all x ∈ n→∞ [0, 1]. Nevertheless, the limit function f (x) ≡ 0 is continuous on [0, 1], while each of the functions fn (x) has discontinuities at the points x = 0 and x = n1 : lim fn (x) =

x→0+

( ) 1 1 1 ≠ 0 = fn (0), lim1 fn (x) = ≠ 0 = fn , ∀n ∈ ℕ. n n n x→ n −

Remark 1. For a series of functions, the respective example is as follows: all the ∑ terms of a uniformly convergent on X series un (x) are discontinuous at some point x0 ∈ X, but the series sum is continuous at x0 . One can use the counterexsgn x ample with the functions u0 (x) = −sgn x and un (x) = n(n+1) , ∀n ∈ ℕ defined on X = ℝ. All these functions have the discontinuity point x0 = 0, but the sum of this telescoping series is zero function continuous at any point. Indeed, the partial sums can be found as follows:

2.2 Limits and Continuity of Limit Functions

1

Q1,dis

P1,dis

f1 Limit function f(x) = 0 Functions fn(x)

0.8

Pn,dis = (xn,limxn fn(x))=(–1n , –1n ) – discontinuity points 1 Qn,dis = (0,lim0 fn(x)) = (0, – ) – discontinuity points n

0.6 Q2,dis 0.4

P2,dis

f2

Q3,dis

f3

P3,dis

0.2

0.2

0.4

0.6

{ Figure 2.7 Example 11, sequence fn (x) =

fn (x) =

n ∑

0.8

1

1 ,0 n

< x < 1n . 0, x = 0, 1n ≤ x ≤ 1

uk (x)

k=0

( ) sgn x 1 1 1 1 1 = −sgn x + sgn x 1 − + − + · · · + − =− , 2 2 3 n n+1 n+1 sgn x

and therefore, the sum f (x) = − lim n+1 = 0, ∀x ∈ ℝ. Note that the series conn→∞ verges uniformly on ℝ by the Weierstrass test, since for ∀x ∈ ℝ the following ∑ 1 | sgn x | 1 evaluation holds |un (x)| = | n(n+1) | ≤ n(n+1) < n12 and the majorant series n2 | | converges. Remark 2. A similar example for a function depending on a continuous parameter can be formulated as follows: a function f (x, y) defined on X × Y has a discontinuity at x0 ∈ X for any y ∈ Y and converges uniformly on X to a limit function 𝜑(x), as y → y0 , but 𝜑(x) is continuous at x0 . For a counterexample, consider the function f (x, y) = y sgn x defined on X × Y = ℝ × (0, +∞) with y0 = 0, which has a discontinuity at x0 = 0 ∈ X for ∀y ∈ Y . At the same time, the limit function is zero: lim f (x, y) = lim y sgn x = 0 and, y→0+

y→0+

consequently, has no discontinuity. Checking the character of the convergence |f (x, y) − 0| = |y sgn x| ≤ |y| → 0, we see that the convergence is uniform on ℝ.

y→0+

61

62

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Remark 3. The general statement with the opposite conclusion—“if a sequence of discontinuous functions converges uniformly, then the limit function is continuous”—is also false as shown in the next example. Example 12. A sequence of discontinuous functions converges uniformly on X, and the limit function is also discontinuous on X. Solution An elementary example is fn (x) = sgn x defined on X = ℝ. Evidently, this sequence converges uniformly to the same function f (x) = sgn x that has a jump discontinuity at the point x = 0. An example of the functions dependent on n is the sequence fn (x) = sgn x + n1 considered on X = ℝ. Each of the functions fn (x) has a discontinuity (a jump) at the point x = 0: lim fn (x) = 1 +

x→0+

1 1 ≠ = fn (0), n n

lim fn (x) = −1 +

x→0−

1 1 ≠ = fn (0). n n

The limit function f (x) = lim fn (x) = sgn x also has a jump discontinuity at 0. n→∞ At the same time, the given sequence converges uniformly on ℝ, since |fn (x) − f (x)| = n1 → 0 simultaneously for all x ∈ ℝ. n→∞

Remark 1. One can formulate analogous example for a series of functions: a series of discontinuous functions converges uniformly on X, and the sum of the series is also discontinuous on X. A simple counterexample is the series ∑∞ sgn x n=1 n(n+1) considered on X = ℝ. All the terms of the series have a discontinuity (a jump) at x = 0, and the uniform convergence of the series on ℝ follows from the evaluation shown in Remark 1 to Example 11 for a similar series. Thus, all the statement conditions hold, but the sum of the series is sgn x that has the jump discontinuity at 0: fn (x) =

n ∑

uk (x)

k=1

( ) ( ) 1 1 1 1 1 1 = sgn x 1 − + − + · · · + − = 1− sgn x, 2 2 3 n n+1 n+1 ( ) 1 and therefore, the sum f (x) = lim 1 − n+1 sgn x = sgn x. n→∞

Remark 2. For a function depending on a continuous parameter, the corresponding example is as follows: a discontinuous in x function f (x, y) converges uniformly on X to a limit function 𝜑(x), as y → y0 , which is also discontinuous on X. The function f (x, y) = cos y ⋅ sgn x + y2 cos x defined on X × Y = ℝ × (0, 𝜋2 ) with y0 = 0 exemplifies this statement. In fact, f (x, y) has a discontinuity (a jump) at x0 = 0 for each y ∈ Y . The limit function is sgn x that also has

2.2 Limits and Continuity of Limit Functions

a jump at 0: 𝜑(x) = lim f (x, y) = lim (cos y ⋅ sgn x + y2 cos x) = sgn x. Finally, y→0+

y→0+

the convergence is uniform on ℝ since |f (x, y) − 𝜑(x)| = | cos y ⋅ sgn x + y2 cos x − sgn x| = |sgn x ⋅ (cos y − 1) + y2 cos x| ≤ (1 − cos y) ⋅ |sgn x| + y2 ⋅ | cos x| ( y )2 y ≤ 2sin2 + y2 ≤ 2 + y2 2 2 3 = y2 → 0. 2 y→0+ Example 13. A sequence of functions fn (x) converges uniformly on X to a continuous function, but the functions fn (x) have infinitely many points of discontinuity on X. {

Solution

1 ,x n

∈ℚ is discontinuous at each point of 0, x ∈ 𝕀 ℝ (it is just a slight modification of the famous Dirichlet’s function D(x), fn (x) = n1 D(x), which does not have a limit at any point of ℝ). At the same time, the sequence of fn (x) converges to the continuous on ℝ function f (x) ≡ 0: for ∀x ∈ ℚ, it follows f (x) = lim fn (x) = lim n1 = 0, and for ∀x ∈ 𝕀, one has n→∞ n→∞ f (x) = lim fn (x) = lim 0 = 0. And moreover, this convergence is uniform, Each of the functions fn (x) =

n→∞

n→∞

because |fn (x) − f (x)| ≤

1 → 0 n n→∞

simultaneously for all x ∈ ℝ.

Remark 1. An analogous example for a series of functions is as follows: a series ∑ un (x) converges uniformly on X to a continuous function, but the terms of the series possess infinitely many discontinuities on X. For a counterexample, ∑∞ D(x) consider the series n=0 un (x), u0 (x) = −D(x), un (x) = n(n+1) , ∀n ∈ ℕ defined on X = ℝ. According to the Weierstrass test, this series converges uniformly on ℝ: ∑ |D(x)| |un (x)| = n(n+1) < n12 , ∀x ∈ ℝ and the series n12 is convergent. To find the sum of the series, note that the series is telescoping: fn (x) =

n ∑

uk (x)

k=0

( ) D(x) 1 1 1 1 1 = −D(x) + D(x) 1 − + − + · · · + − =− , 2 2 3 n n+1 n+1 and therefore, f (x) = lim − D(x) = 0, that is, the sum is a continuous function on n+1 n→∞ ℝ. Nevertheless, all the terms of the series have discontinuity at each point of ℝ.

63

64

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Remark 2. A similar example can be constructed for functions depending on a continuous parameter: f (x, y) defined on X × Y converges uniformly on X to a continuous function 𝜑(x), as y → y0 , but f (x, y) possesses infinitely many points of discontinuity on X. For a counterexample, consider the function f (x, y) = { y, x ∈ ℚ yD(x) = on X × Y = ℝ × (0, 1] with y0 = 0. The limit function is 0, x ∈ 𝕀 continuous on ℝ: 𝜑(x) = lim f (x, y) = lim yD(x) = 0. The convergence is uniform on ℝ since

y→0+

y→0+

|f (x, y) − 𝜑(x)| = |yD(x)| ≤ |y| → 0. y→0+

However, for any fixed y ∈ Y , the function f (x, y) is discontinuous at each real point. Example 14. A function f (x, y), defined on X × Y , is continuous with respect to x on X for any fixed y ∈ Y , and it has a limit function 𝜑(x) on X as y approaches y0 , but 𝜑(x) is discontinuous on X. Solution The function f (x, y) = ex∕y considered on X × Y = [−10, 0] × (0, 1) is continuous in x on [−10, 0] for any fixed y ∈ (0, 1). For the limit point y0 = 0, the limit function is obtained as follows: if x = 0, then 𝜑(0) = lim f (0, y) = lim 1 = 1; y→0+

y→0+

if x ∈ [−10, 0), then 𝜑(x) = lim f (x, y) = lim ex∕y = 0 y→0+

y→0+

(since the last power is negative). Therefore, the limit function has a jump discontinuity at 0. Let us show that the convergence is not uniform on [−10, 0]. Indeed, choosing for ∀y ∈ (0, 1) the point xy = −y ∈ (−1, 0) ⊂ [−10, 0], we obtain |f (xy , y) − 𝜑(xy )| = |e−y∕y − 0| = e−1 ↛ 0. y→0+

Two other counterexamples are the function f (x, y) = arctan xy defined on X × Y = [0, +∞) × (0, 1] with the limit point y0 = 0, and f (x, y) = ered on X × Y = [0, 1] × (0, 1] with the limit point y0 = 0.

y2 x2 +y2

consid-

Remark 1. A similar example can be given for a sequence: a sequence of continuous functions fn (x) converges on X, but the limit function is discontinuous

2.2 Limits and Continuity of Limit Functions

on X. It can be exemplified by the sequence fn (x) = xn considered on X = [0, 1]. In fact, since lim fn (x) = lim xn = 0 for x ∈ [0, 1) and lim fn (1) = lim 1 = 1, n→∞ n→∞ n→∞ { n→∞ 0, x ∈ [0, 1) the sequence converges on [0, 1] to the function f (x) = . 1, x = 1 Each function fn (x) = xn is continuous on [0, 1], but the limit function { 0, x ∈ [0, 1) f (x) = is discontinuous at x = 1. Note that the convergence is 1, x = 1 not uniform on [0, 1] due to the following argument: for any n ∈ ℕ there exists xn = √n1 ∈ [0, 1) ⊂ [0, 1] such that 2

( |fn (xn ) − f (xn )| =

)n 1 √ n 2

−0=

1 ↛ 0. 2 n→∞

Remark 2. In the case of series, the corresponding example assumes the ∑ following form: a series un (x) of continuous functions converges on X, but the sum of this series is a discontinuous function. The series of Example 7 ∑∞ n n=1 (1 − x)x considered on X = [0, 1] provides the required counterexample. All the considerations follow those used in Example 7, and we reproduce them here in a brief form just for the sake of completeness. First, the series is ∑∞ convergent, since for x = 1 one has n=1 0 = 0, and for x ∈ [0, 1) one obtains ∑∞ the convergent geometric series n=1 (1 − x)xn = (1−x)x = x. Hence, the sum 1−x { x, x ∈ [0, 1) f (x) = is discontinuous at x = 1, although all the functions 0, x = 1 n un (x) = (1 − x)x are continuous on [0, 1]. Finally, note that the convergence is not uniform on [0, 1], because for x ∈ [0, 1) the remainder can be written in the form |rn (x)| =

∞ ∑

(1 − x)xk =

k=n+1

and choosing xn =

1 √

n+1

2

(1 − x)xn+1 = xn+1 , 1−x

∈ [0, 1) for any given n ∈ ℕ we get |rn (xn )| =

1 ↛ 2 n→∞

0.

Remark 3. In all the above counterexamples, the convergence was not uniform on X, which results in discontinuity of the limit function and the series sum. If the convergence would be uniform on X, then all the corresponding general statements will be true. For instance, in the case of the sequences, the correct general statement goes as follows: if a sequence of continuous functions fn (x) converges uniformly on X, then the limit function is continuous on X. At the same time, the uniform convergence of continuous functions is just a sufficient condition. The next example shows that the limit function still can be continuous if this condition is not satisfied.

65

66

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

1

Pdis = (1,0) – discontinuity point Pn = (xn,rn(xn)) = ( n 1 , –1 ) √2 2

0.8

Limit function f(x) = 0, x ≠ 1 Functions fn(x)

0.6 P1

P2

P3

f1

0.4

0.2

f2 f3 0.2

0.4

Pdis 0.6

0.8

1

Figure 2.8 Examples 14, 17, 19, 27, and 29, sequence fn (x) = x n .

Example 15. A function f (x, y), defined on X × Y , is continuous with respect to x on X for any fixed y ∈ Y , and it has a continuous limit function 𝜑(x) on X as y approaches y0 , but the convergence is nonuniform on X. Solution xy The function f (x, y) = x2 +y2 defined on X × Y = [0, 1] × (0, 1] is continuous in x on [0, 1] for any fixed y ∈ (0, 1]. The limit function, as y approaches y0 = 0, is xy 𝜑(x) = lim f (x, y) = lim 2 = 0, ∀x ∈ [0, 1], y→0+ y→0+ x + y2 so 𝜑(x) is also continuous on [0, 1]. However, the convergence is not uniform on X = [0, 1], since for ∀y ∈ (0, 1] there exists xy = y ∈ (0, 1] such that |f (xy , y) − 𝜑(xy )| =

y2 1 = ↛ 0. y2 + y2 2 y→0+

Remark 1. A similar example for a sequence is as follows: a sequence of continuous functions fn (x) converges on X to a continuous function, but the convergence is nonuniform on X. The sequence of the continuous functions fn (x) = nxe−nx , x ∈ X = [0, 1] provides a counterexample. Indeed, for the limit function we have: f (0) = lim fn (0) = lim 0 = 0 if x = 0, and n→∞

n→∞

2.2 Limits and Continuity of Limit Functions

1

Pdis = (1,1) – discontinuity point

Pdis

Pn = (xn,rn(xn)) = ( n+11 , –1 ) √ 2

0.8

2

Exact sum f ( x ) = x , x ≠ 1 Partial sums f n ( x ) = (1 – x n)x

0.6

Residuals rn(x) = f(x) – fn(x)

P11

P22 P33

f33

0.4

f22

r11

0.2

f11

r21 r31 0.2

0.4

0.6

Figure 2.9 Examples 14, 21, 24, 27, and 29, series

0.8

∑∞

n=1 (1

1

− x)x n .

nx t 1 = lim = lim =0 enx t→+∞ et t→+∞ et if x ∈ (0, 1] (in the last limit, we apply the change of variable t = nx and l’Hospital’s rule). Thus, the sequence converges on [0, 1] to the continuous function f (x) ≡ 0, but the convergence is not uniform on [0, 1]: for any n ∈ ℕ we can choose xn = n1 ∈ [0, 1], which results in |fn (xn ) − f (xn )| = e−1 ↛ 0. f (x) = lim fn (x) = lim n→∞

n→∞

n→∞

Remark 2. For a series, the example can be formulated as follows: a series ∑ un (x) of continuous functions converges on X to a continuous function, but the convergence is nonuniform on X. As a counterexample, we can use the ∑∞ series of continuous functions n=1 xn−1 on X = (−1, 1). The sum of this con1 vergent geometric series f (x) = 1−x is also continuous on (−1, 1). However, the convergence is not uniform on (−1, 1), because the remainder has the form ∑∞ xn rn (x) = k=n+1 uk (x) = 1−x and choosing xn = 1 − n1 ∈ (−1, 1) for any given n ∈ ℕ, we obtain ( )n 1 − n1 ( ) 1 n |rn (xn )| = = n 1 − → +∞. n n→∞ 1−1+ 1 n

67

68

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

2.3 Conditions of Uniform Convergence. Dini’s Theorem Remark. In this section, we analyze the conditions of Dini’s theorem on uniform convergence first for sequences (Examples 16–20) and then for series of functions (Examples 21–25). Example 26 clarifies how many discontinuity points the limit function of a sequence of continuous functions may have. Example 16. A sequence of functions fn (x), which are monotone in n for any fixed x ∈ X, converges on a compact set X to a continuous function, but the convergence is nonuniform on X. Solution

( ) ⎧ 1 ⎪ 1, x ∈ 0, n [ ] on the compact set Let us consider the sequence fn (x) = ⎨ 1 ⎪ 0, x = 0, x ∈ n , 1 ⎩ X = [0, 1]. For any fixed x ∈ X, the sequence is monotone in n: fn+1 (x) ≤ fn (x). It converges on [0, 1] to the continuous function f (x) ≡ 0: f (0) = lim fn (0) = lim 0 = 0, n→∞

n→∞

f (1) = lim fn (1) = lim 0 = 0, n→∞

n→∞

Limit function f(x) = 0 0.4

Pn = (xn,fn(xn)) = (

Functions fn(x) P3

1 1 ,– ) n e

P2

f1

0.3

f2

0.2

f3

P1

0.1

0.2

0.4

0.6

Figure 2.10 Examples 15 and 18, sequence fn (x) = nxe−nx .

0.8

1

2.3 Conditions of Uniform Convergence. Dini’s Theorem

Exact sum f ( x ) = 1 1–x

5

n Partial sums fn ( x ) = 1–x 1–x

4

Residuals rn ( x ) = f ( x ) – fn( x )

f4

3 Pn = (xn, rn(xn)) = ( 1 –

1 n

1 n

, n(1 –

f3

) n)

2

f2 P4

r1 1

r P2 2 r3 r4

P1 0.2

Figure 2.11 Example 15, series

0.4

∑∞ n=1

0.6

P3

f1

0.8

1

x n−1 .

and for any given x ∈ (0, 1) there exists Nx ∈ ℕ such that Nx > 1x , and then for ∀n ≥ Nx we have x > N1 ≥ n1 , and therefore, fn (x) = 0 for ∀n ≥ Nx , which x implies that f (x) = lim fn (x) = 0. To see that the convergence is nonuniform n→∞ ( ) 1 on [0, 1], we can choose xn = 2n ∈ 0, n1 ⊂ [0, 1] for any given n ∈ ℕ, which results in |fn (xn ) − f (xn )| = 1 ↛ 0. n→∞

Remark. Note that the functions of the above sequence have a jump discontinuity at the points x = 0 and x = n1 : lim fn (x) = 1 ≠ 0 = fn (0),

x→0+

lim1 fn (x) = 1 ≠ 0 = fn

x→ n

−

( ) 1 , ∀n. n

Thus, this example shows that a weakened version of Dini’s theorem with the omitted condition of the continuity of fn (x) on X is not true. Example 17. A sequence of continuous functions fn (x), which are monotone in n for any fixed x ∈ X, converges on a compact set X, but this convergence is nonuniform on X.

69

70

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Solution The sequence fn (x) = xn used in Remark 1 to Example 14 consists of the continuous functions on the compact set X = [0, 1]. For any fixed x ∈ X, the corresponding numerical sequence is monotone in n: fn+1 (x) = xn+1 ≤ xn = fn (x). As shown in Remark { 1 to Example 14, this sequence converges on [0, 1] to the 0, x ∈ [0, 1) function f (x) = , but the convergence is not uniform on [0, 1]. 1, x = 1 Remark. Note that the limit function f (x) in this counterexample has a jump discontinuity at the point x = 1. Thus, it is seen that one cannot omit the condition of the continuity of the limit function f (x) on X in the conditions of Dini’s theorem. Example 18. A sequence of continuous functions fn (x) converges on a compact set X to a continuous function, but this convergence is nonuniform on X. Solution The sequence of the continuous functions fn (x) = nxe−nx , defined on the compact set X = [0, 2], converges to the continuous function f (x) ≡ 0, but the convergence is not uniform on [0, 2] (see Remark 1 to Example 15 for details, where similar results were derived for [0, 1]). Remark 1. The convergence is not uniform because the condition of monotonicity in Dini’s theorem is not satisfied: if n > 1, then for xn = n1 we have fn−1 (xn ) < e−1 = fn (xn ), while fn (xn ) = e−1 > fn+1 (xn ) (take into account that the function g(t) = te−t is strictly increasing for t < 1 and decreasing for t > 1). Remark 2. The general converse statement is true: if a sequence of continuous functions fn (x) converges uniformly on a set X, then the limit function is continuous on X. Example 19. A sequence of continuous functions fn (x), which are monotone in n for any fixed x ∈ X, converges on a set X to a continuous function, but this convergence is nonuniform on X. Solution The sequence fn (x) = xn consists of the continuous functions on the set X = [0, 1). The sequence is monotone in n for any fixed x ∈ X, and it converges to the continuous function f (x) ≡ 0 on [0, 1). However, the convergence is not uniform on [0, 1). (See details in Example 17 and in Remark 1 to Example 14, where the same sequence was considered on [0, 1].)

2.3 Conditions of Uniform Convergence. Dini’s Theorem

Remark. The convergence is not uniform because the set X = [0, 1) is not compact (this is one of the conditions of Dini’s theorem). Remark to Examples 16–19. These examples show that if at least one of the four conditions in Dini’s theorem is not satisfied, then the convergence may be nonuniform. At the same time, these conditions are sufficient, but not necessary ones for the uniform convergence. It may happen that one, two, or even all the conditions are violated (except for the condition of the convergence on X), and even so the convergence is uniform. For instance, in Example 11, each of { 1 , 0 < x < n1 n the functions fn (x) = is discontinuous at the points x = 0 0, x = 0, n1 ≤ x ≤ 1

and x = n1 , but the convergence to f (x) ≡ 0 is uniform on [0, 1]. In Example 12, three conditions of Dini’s theorem are violated: the functions fn (x) = sgn x + n1 and the limit function f (x) = sgn x are discontinuous at 0, and the set X = ℝ is not compact. However, the convergence is still uniform on ℝ. Slightly modifying the sequence in Example 12, it is possible to construct the example where all four conditions of Dini’s theorem do not hold, but the convergence is uniform, as shown in the next example. Example 20. A sequence of functions that violates all the four conditions of Dini’s theorem on a set X, but nevertheless converges uniformly on this set. Solution n Consider the sequence gn (x) = sgn x + (−1) defined on X = ℝ. The n limit function of this sequence is g(x) = sgn x. All the functions gn (x) and the limit function g(x) are discontinuous at 0, the set X = ℝ is not compact, and the sequence gn (x) is not monotone for any fixed x: 1 1 1 g2n (x) = sgn x + 2n > sgn x − 2n+1 = g2n+1 (x), while g2n+1 (x) = sgn x − 2n+1 < 1 sgn x + 2n+2 = g2n+2 (x), ∀n ∈ ℕ. Thus, all the four conditions of Dini’s theorem are violated. Nevertheless, the convergence is uniform on ℝ, since |gn (x) − g(x)| = n1 → 0 simultaneously for all x ∈ ℝ. n→∞

Example 21. A series of continuous nonnegative functions converges on a compact set X, but this convergence is nonuniform on X. Solution ∑∞ Consider the series n=1 xn (1 − x) on X = [0, 1] used in Example 7 and Remark 2 to Example 14. Each function un (x) = xn (1 − x) is nonnegative on [0, 1] and continuous on ℝ (and, in particular, on [0, 1]). { As shown in Remark 2 to x, 0 ≤ x < 1 Example 14, this series converges to f (x) = . Nevertheless, 0, x = 1

71

72

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

2 g2

1.5 r1

–0.8

r3

Pdis –0.6

r2

g3

0.5

r3 –1

r1

1

–0.4

g2

–0.2

g1

0.4

0.6

0.8

1

r2

–0.5

Limit function g(x) = sgn x

–1 g3

g1 0.2

Functions gn(x) Residuals rn(x) = g(x) – gn(x)

–1.5

Pdis = (0,0) – discontinuity point

–2

Figure 2.12 Example 20, sequence gn (x) = sgn x +

(−1)n . n

the convergence on [0, 1] is nonuniform, since for ∀n ∈ ℕ we can choose 1 xn = 1 − n+1 ∈ X such that ∞ |∑ | xn+1 (1 − x ) | | n n |rn (xn )| = | xkn (1 − xn )| = | | 1 − x n |k=n+1 | ( )n+1 1 = 1− → e−1 ≠ 0. n→∞ n+1 Note that one of the conditions of Dini’s theorem is violated—the sum of the series has a discontinuity at x = 1—that leads to nonuniformity of the convergence.

Example 22. A series of nonnegative functions converges on a compact set X to a continuous function, but this convergence is nonuniform on X. Solution ∑∞ Consider the series n=0 un (x) with the nonnegative on X = [0, 1] terms u0 (x) = ( ] ⎧ { 1 1, x ∈ n+1 , n1 ⎪ 1, x = 0 [ ] ( ] , ∀n ∈ ℕ. At the point x = 0, , u (x) = ⎨ 1 1 0, x ∈ (0, 1] n ⎪ 0, x ∈ 0, n+1 ∪ n , 1 ⎩ the sum is 1 since u0 (0) = 1 and ( un (0) = ] 0, ∀n ∈ ∞ℕ.(For any]fixed x ∈ (0, 1], there 1 1 1 exists nx ∈ ℕ such that x ∈ n +1 , n (since ∪ n+1 , n1 = (0, 1]). Moreover, x

x

n=1

2.3 Conditions of Uniform Convergence. Dini’s Theorem

( ] ( ] 1 1 such nx is unique because n+1 , n1 ∩ k+1 , 1k = ∅ for ∀n, k such that n ≠ k. Then unx (x) = 1 and uk (x) = 0, ∀k ≠ nx , that is, exactly one term is equal to 1 ∑∞ and all other terms are 0 in the series n=0 un (x). Since it is true for any fixed ∑∞ x ∈ (0, 1], we conclude that f (x) = n=0 un (x) = 1 on (0, 1] and, consequently, on [0, 1]. Thus, the sum of the series is continuous (constant) function on X. However, the convergence is nonuniform on [0, 1], which can be seen applying 1 the Cauchy criterion with pn = 1 and xn = n+1 ∈ X for each n ∈ ℕ: n ( ) | n+p | 1 |∑ | uk (xn )| = un+1 = 1 ↛ 0. | | | n→∞ n+1 |k=n+1 |

In this example the nonuniformity of convergence is caused by discontinuity of the series terms: u0 (x) has a jump at 0, u1 (x) at 12 and all other functions un (x) 1 have two jump points— n+1 and n1 . Note that the continuity of the series terms is one of the conditions of Dini’s theorem. Example 23. A series of continuous functions converges on a compact set X to a continuous function, but this convergence is nonuniform on X. Solution ∑∞ n−1 The series n(1 − x)xn is composed of the functions un (x) = n=1 (−1) n−1 n (−1) n(1 − x)x continuous on ℝ and, in particular, on X = [0, 1]. To find the sum of this series, we use the binomial expansion: (1 + x)𝛼 =

∞ ∑ 𝛼(𝛼 − 1) · · · (𝛼 − n + 1)

n!

n=0

xn , x ∈ (−1, 1).

Setting 𝛼 = −2, we get (1 + x)−2 =

∞ ∑

(−1)n (n + 1)xn =

n=0

∞ ∑

(−1)k−1 kxk−1 .

k=1

Comparing this expression with our series, we have ∞ ∑ n=1

(−1)n−1 n(1 − x)xn = x(1 − x)

∞ ∑ n=1

(−1)n−1 nxn−1 =

x(1 − x) , ∀x ∈ (−1, 1). (1 + x)2

At the point x = 1, all the terms are zero and the sum of series is also zero. Thus, ∑∞ for ∀x ∈ [0, 1], the sum of the series is f (x) = n=1 (−1)n−1 n(1 − x)xn = x(1−x) , (1+x)2 which is a continuous function on [0, 1].

73

74

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

x(1 – x) (1 + x)2

Exact sum f(x) = 0.4

Partial sums fn(x) Residuals rn(x) = f(x) – fn(x)

0.3

Pn = (xn, ∣un+1(xn)∣) = (1–

1 , n+1

f1

(1– n 1+ 1)n + 1) P2

P5 P3 P4

P1

0.2 f3 f5 f4

0.1

f2

0.2

0.4

r2 r4

r5

0.6

0.8

1

r3

–0.1

r1

–0.2

Figure 2.13 Example 23, series

∑∞ n=1

(−1)n−1 n(1 − x)x n .

Now let us check the character of the convergence on [0, 1]. Applying the 1 Cauchy criterion with pn = 1 and xn = 1 − n+1 ∈ X, for each n ∈ ℕ, we obtain n | n+p | |∑ | uk (xn )| = |(−1)n (n + 1)(1 − xn )xn+1 | n | | | |k=n+1 | ( )n+1 1 1 = (n + 1) ⋅ 1− → e−1 ≠ 0, n→∞ n+1 n+1 which means that the convergence is nonuniform. Comparing the conditions of the statement with the conditions of Dini’s theorem, one can see that the former does not satisfy the condition of the nonnegativity of the series terms.

Example 24. A series of continuous nonnegative functions converges on X to a continuous function, but this convergence is nonuniform on X. Solution ∑∞ Consider the series n=1 xn (1 − x) on X = [0, 1). Each term un (x) = xn (1 − x) of this series is a continuous nonnegative function on [0, 1). The sum of this series, ∑∞ found in Example 21, is the continuous on [0, 1) function f (x) = n=1 xn (1 − x) = x. However, in the same way as shown in Example 21, one can prove that the convergence to f (x) is nonuniform on [0, 1). Comparing to Dini’s theorem, one sees that the example is missing the condition of compactness of X.

2.3 Conditions of Uniform Convergence. Dini’s Theorem

Remark to Examples 21–24. In each of the last four examples, only one condition of Dini’s theorem was omitted, but it causes the nonuniformity of the convergence. Thus, only one condition missing invalidates the result of the Dini’s theorem. At the same time, the set of conditions of Dini’s theorem is sufficient and not necessary, meaning that a violation of one or a few of these conditions does not imply the nonuniformity of the convergence. This situation is illustrated in Example 25.

Example 25. Some conditions of Dini’s theorem are violated, but a series still converges uniformly. Solution First, let us consider an example when the condition of the nonnegativity of the n ∑∞ terms of series in Dini’s theorem is not satisfied. The series n=1 (−1)n−1 xn is a convergent expansion of ln(1 + x) on (−1, 1] and, in particular, on X = [0, 1]. n The terms (−1)n−1 xn and the sum ln(1 + x) of the series are continuous functions on the compact set [0, 1], but infinitely many terms are negative on (0, 1]. The type of the convergence can be find out by Leibniz’s test for alternating series: | xn+1 || 1 |rn (x)| ≤ |un+1 (x)| = ||(−1)n ≤ → 0, | n + 1 | n + 1 n→∞ | that is, the series converges uniformly, despite the fact that the condition of the nonnegativity of the terms of series in Dini’s theorem does not hold. Another example involves violation of { all the conditions of Dini’s theorem. ∑∞ 1, x ∈ ℚ 1 ̃ ̃ All the terms un (x) = n(n+1) D(x), D(x) = of the series n=1 un (x) −1, x ∈ 𝕀 are discontinuous functions at each point x ∈ X = ℝ. The partial sums of this series are fn (x) =

n ∑

uk (x)

k=1

( ) ( ) 1 1 1 1 1 1 ̃ ̃ = D(x) 1− + − +···+ − = D(x) 1− , 2 2 3 n n+1 n+1 and consequently, the sum is ( ̃ f (x) = lim fn (x) = lim D(x) 1− n→∞

n→∞

1 n+1

)

̃ = D(x),

that is, the sum is everywhere discontinuous function too. Further, the functions un (x) do not keep the sign: un (x) > 0 for ∀x ∈ ℚ, while un (x) < 0 for ∀x ∈ 𝕀. Finally, the set X = ℝ is not compact. Thus, all the conditions of Dini’s

75

76

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

f

1

f3 f2 f1

0.5 ˜ (x) Exact sum f(x) = D ˜ (x)(1– 1 ) Partial sums fn(x) = D n+1

–3

–2

–1

f1

1

2

3

–0.5

f2 f3 f

Figure 2.14 Example 25, series

–1

∑∞

n=1 un (x), un (x) =

1 ̃ D(x), n(n+1)

̃ D(x) =

{

1, x ∈ ℚ . −1, x ∈ 𝕀

theorem are violated. Nevertheless, the series converges uniformly on X = ℝ according to the Weierstrass test: | D(x) | 1 1 | ̃ | |un (x)| = | < , ∀x ∈ ℝ, |= | n(n + 1) | n(n + 1) n2 | | ∑ 1 and the series n2 is convergent. Example 26. A sequence of continuous functions converges on X to a function f (x), which has infinitely many points of discontinuity. Solution Suppose that any rational number in [0, 1] is represented in the form x = pq , p where p, q ∈ ℕ and the fraction q is in lowest terms for x ∈ (0, 1), and additionally, q = 1 for x = 0 and x = 1. In this case, the numbers p and q are uniquely p determined in the representation x = q . For each n ≥ 2, n ∈ ℕ, the construction of the function fn (x) defined on [0, 1] is as follows: for each fixed n, one considers the values q = 1, … , n − 1, each of which forms a set of the special p points q , p = 0, … , q, employed to define the nth function of the sequence

2.3 Conditions of Uniform Convergence. Dini’s Theorem

1 Functions fn(x)

0.9 0.8 0.7 0.6

f2

0.5 0.4

f3

0.3

f4 f5 f6

0.2 0.1

f1

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

( ) [ ( ) ] ⎧ 1 + 2n2 x − p , x ∈ p − 1 2 1 − 1 , p q q) q 2n q( n q ( [ )] ⎪ Figure 2.15 Example 26, sequence fn (x) = ⎨ 1 − 2n2 x − p , x ∈ p , p + 1 1 − 1 . 2 q q q 2n q n ⎪q 1 ⎩ , otherwise n

( ) [ ( ) ] ⎧ 1 + 2n2 x − p , x ∈ p − 1 2 1 − 1 , p q) q ( [ q 2n q ( n )] ⎪q fn (x) = ⎨ 1 − 2n2 x − p , x ∈ p , p + 1 1 − 1 . q q q 2n2 q n ⎪q 1 ⎩ , otherwise n p

By construction, at each special point q , q < n, the function fn (x) has a local maximum equal to

1 q

> n1 , it is increasing in the left-hand and decreasing in the p

right-hand neighborhood of q . The values of fn (x) are larger than n1 only in a ( ) p neighborhood of each q , q < n, and the diameter n12 1q − n1 of such a neighborhood decreases for larger values of n, approaching 0 as n approaches infinity. Note also that fn+1 (x) is smaller than fn (x) at all points except for the special p points q , q < n + 1, at which the two functions coincide. Since 0 < fn+1 (x) ≤ fn (x) for ∀n ≥ 2 and ∀x ∈ [0, 1], the sequence is convergent on [0, 1]: lim fn (x) = f (x) (because at each point we have the decreasing n→∞ numerical sequence bounded below). Note that fn (0) = fn (1) = 1, ∀n ≥ 2 and, consequently, f (0) = f (1) = 1. Let us consider a rational point x0 = pq in (0, 1)

77

78

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

1 Limit function f(x) (Riemann function)

0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0.1

0.2

0.3

0.4

0.5

0.6

0.7

{ Figure 2.16 Example 26, limit function f (x) = R(x) =

0.8

0.9

1

1 ,x q

= pq ∈ ℚ . 0, x ∈ 𝕀

p q

in lowest terms. For each n > q, the point x0 is used for ( ) p construction of the function fn (x), and therefore, fn (x0 ) = fn q = 1q , ∀n > q, ( ) ( ) that is, lim fn pq = 1q = f pq . n→∞ Now let us choose an irrational x0 ∈ (0, 1) and, by using the method of mathematical induction, show that there exists nk ∈ ℕ such that fn (x0 ) ≤ 1k for ∀n ≥ nk . For k = 2, denote the distance between x0 and the endpoints 0 and 1 by d0 and d1 , respectively (both distances are positive because x0 is irrational and the endpoints are rational). There exists n2 ∈ ℕ such that 2n1 2 < min(d0 , d1 ), which means that x0 lies outside of the intervals [ (2 )] [ ( ) ] 0, 2n1 2 1 − n1 and 1 − 2n1 2 1 − n1 , 1 , implying that fn (x0 ) ≤ 12 for with the fraction

2

2

2

2

∀n ≥ n2 . For k = 3, besides the endpoints, there appears the point 12 used in the construction of the functions fn (x), n ≥ 3. Denote the distance between x0 and 12 by d2 and choose n3 ∈ ℕ such that 2n1 2 < d2 and n3 ≥ n2 . Then x0 lies 3 [ ( ) ( )] outside the interval 12 − 2n1 2 12 − n1 , 12 + 2n1 2 12 − n1 , as well as outside 3[ 3 [ ( )] (3 )3 ] the intervals 0, 2n1 2 1 − n1 and 1 − 2n1 2 1 − n1 , 1 , and consequently, 3

3

3

3

2.4 Convergence and Uniform Continuity 1 fn (x0 ) ≤ 13 for ∀n ≥ n3 . Let us suppose now that at (k − 1)th step fn (x0 ) ≤ k−1 for ∀n ≥ nk−1 and prove that a similar inequality holds for k. p On kth step, a few (at most k − 2) new points k−1 , 1 ≤ p ≤ k − 2, associated with the denominator q = k − 1, appear in the construction of the functions fn (x), n ≥ k. Denote the smallest distance between x0 and each of these points by dk and choose nk ∈ ℕ such that 2n1 2 < dk and nk ≥ nk−1 . Therefore, k

p

x0 lies outside of each of the intervals around the points q , q < k, (the intervals used in the first two sentences of the definition of fn (x)) and consequently, fn (x0 ) ≤ 1k for ∀n ≥ nk . Thus, the last property is proved. Passing to the limit, as k approaches infinity, in the inequality fn (x0 ) ≤ 1k , we obtain lim fn (x0 ) = 0 = f (x0 ) at all irrational points in (0, 1). Thus, the limit function n→∞ { p 1 ,x = q ∈ ℚ q has the following form: f (x) = . This is the Riemann function, 0, x ∈ 𝕀 which is discontinuous at all the rational points (and continuous at all the irrational points) in [0, 1]. Therefore, the limit function has infinitely many discontinuity points on [0, 1], although each of the functions fn (x) is continuous on [0, 1]. Remark 1. A similar example can be given for a functions defined on ℝ, if we extend the definition of fn (x) periodically (with the period 1) on ℝ. Remark 2. Of course, the convergence of the considered sequence is not uniform on [0, 1] (or on any subinterval in [0, 1]), since otherwise the limit function would be continuous (see the corresponding theorem on the continuity of the limit function of a uniformly convergent sequence). The fact that the convergence is nonuniform ( ( on [0, ))1] is also easy to check by choosing the irrational 1 1 points xn ∈ 0, 2n3 1 − n ∀n ∈ ℕ, and finding the limit ( )) 1 1 1 − 3 ) n ( ) (2n 1 1 1 1 2 = 1 − 2n 1 − = 1 − 1 − → 1 ≠ 0. 2n3 n n n n→∞

|fn (xn ) − f (xn )| = fn (xn ) ≥ fn

(

2.4 Convergence and Uniform Continuity Example 27. Each of functions fn (x) is uniformly continuous on X and the sequence fn (x) converges on X to f (x), but the limit function is discontinuous on X.

79

80

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Solution Each function of the sequence fn (x) = xn is continuous on X = [0, 1] and, according to the Cantor theorem, is uniformly{ continuous on [0, 1]. However, 0, x ∈ [0, 1) n the limit function f (x) = lim fn (x) = lim x = is discontinuous 1, x = 1 n→∞ n→∞ at x =( 1. Note ) that this sequence converges nonuniformly on [0, 1]: choosing 1 xn = 1 − n ∈ [0, 1] for each n ∈ ℕ, one has ( ) 1 n |fn (xn ) − f (xn )| = xnn = 1 − → e−1 ≠ 0. n n→∞ Remark. For a series, we have an analogous example: a series of uniformly continuous functions converges on X, but the sum of the series is a discon∑∞ tinuous on X function. In fact, consider n=1 (1 − x)xn on X = [0, 1]. Each n term un (x) = (1 − x)x is a continuous function on the compact set X = [0, 1] and, consequently, is uniformly continuous on [0, 1]. Let us find the sum of this series f (x). At the end points, one has un (0) = un (1) = 0, ∀n ∈ ℕ and, ∑∞ therefore, f (0) = f (1) = n=1 0 = 0. For any x ∈ (0, 1), the series is a conver∑∞ gent geometric series with the sum f (x) = n=1 (1 − x)xn = (1−x)x = x. Thus, 1−x { x, x ∈ [0, 1) f (x) = , and this function has a discontinuity at x = 1. Checking 0, x = 1 the nature of the convergence, one can evaluate the residual at the points 1 xn = 1 − n+1 , ∀n ∈ ℕ ∞ ) n+1 |∑ | (1 − x )xn+1 ( 1 | | n n |rn (xn )| = | (1 − xn )xkn | = = 1− → e−1 ≠ 0 | | n→∞ 1 − x n + 1 n |k=n+1 | and find out that the convergence is not uniform.

Example 28. A sequence of uniformly continuous on X functions fn (x) converges on X to a continuous function f (x), but the limit function is not uniformly continuous on X. Solution n Consider the sequence fn (x) = nx+1 on X = (0, 1]. The functions fn (x) are continuous on a compact X = [0, 1] for ∀n ∈ ℕ and, according to the Cantor theorem, are uniformly continuous on [0, 1] and, consequently, on (0, 1]. The limit function n 1 f (x) = lim fn (x) = lim = , ∀x ∈ (0, 1] n→∞ n→∞ nx + 1 x is continuous on (0, 1], but this continuity is not uniform. In fact, for the pair 1 of points xk = k1 and xk+1 = k+1 , both in (0, 1], ∀k ∈ ℕ, one has |xk − xk+1 | =

2.4 Convergence and Uniform Continuity

|1 1 | | k − k+1 | → 0, but |f (xk ) − f (xk+1 )| = |k − (k + 1)| = 1 ↛ 0. Let us verify the | | k→∞ k→∞ nature of the convergence of fn (x) on (0, 1]. If one picks up xn = n1 ∈ (0, 1] for each n ∈ ℕ, then | n | n 1 | |n |fn (xn ) − f (xn )| = || − || = || − n|| = → ∞, | nxn + 1 xn | | 2 | 2 n→∞ which means that the convergence is not uniform on (0, 1]. Remark. An analogous example for a series is as follows: a series of uniformly continuous functions converges on X to a continuous function f (x), but f (x) is not uniformly continuous on X. For a counterexample, ∑∞ n consider on X = [0, 1). Each function un (x) = xn is continuous n=0 x on the compact set X = [0, 1] and, consequently, is uniformly continuous on [0, 1], which implies the uniform continuity on [0, 1). The series converges (as a geometric series with the ratio in [0, 1)) to the function ∑∞ 1 f (x) = n=0 xn = 1−x , which is continuous on [0, 1). However, this continuity 1 is not uniform: for the pair of points xk = 1 − 1k and xk+1 = 1 − k+1 , both | |1 1 1 | 1 | in [0, 1), ∀k ∈ ℕ, one has |xk − xk+1 | = |1 − k − 1 + k+1 | = | k − k+1 | → 0, | | | | k→∞ | 1 1 || | but |f (xk ) − f (xk+1 )| = | 1−x − 1−x | = |k + 1 − k| = 1 ↛ 0. Note that the k→∞ k+1 | | k series does not converge uniformly on [0, 1), because the general term does not converge uniformly to 0: choosing xn = √n1 ∈ [0, 1), ∀n ∈ ℕ, one gets 2 ( )n 1 1 |un (xn )| = √n = 2 ↛ 0. 2

n→∞

Remark to Examples 27 and 28. It is well known that the limit function/sum of a series is uniformly continuous on a set X if the terms of the sequence/series are uniformly continuous on X and the convergence is uniform on X. In the two above examples, the violation of the condition of the uniform convergence causes an absence of uniform continuity of the limit function/sum of the series (and even discontinuity in Example 27). At the same time, the condition of the uniform convergence is sufficient and not necessary as shown in the next example. Example 29. A sequence of uniformly continuous on X functions fn (x) converges on X to an uniformly continuous function f (x), but this convergence is nonuniform on X. Solution Consider the sequence fn (x) = xn on X = [0, 1). It was shown in Example 27 that each fn (x) is uniformly continuous on [0, 1] that implies the uniform continuity

81

82

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Q3

3

2.5

2

1.5

1

0.5

Limit function f(x) =

1 x

Functions fn(x)

f4

Residuals rn(x) = f(x) – fn(x)

f3

P4

f2

Q2 P3

P2

f1

Q1 r1

Pn = (xn, rn(xn)) = ( 1n , n2 )

r2 r3 r4

Qk = (xk, f(xk)) = ( 1k , k)

0.2

0.4

Figure 2.17 Example 28, sequence fn (x) =

0.6

0.8

P1

1

n . nx+1

on [0, 1). The limit function f (x) = lim fn (x) = 0, ∀x ∈ [0, 1) is uniformly conn→∞ tinuous on [0, 1). However, the convergence is nonuniform on [0, 1) (just apply the same reasoning as in Example 27). Remark. The corresponding formulation for a series: a series of uniformly continuous functions converges on X to an uniformly continuous function f (x), but ∑∞ this convergence is nonuniform. For a counterexample, use the series n=1 (1 − x)xn on X = [0, 1). It was shown in Remark to Example 27 that the functions un (x) = (1 − x)xn are uniformly continuous on [0, 1], that implies the uniform continuity on [0, 1), and also that the sum of the series is f (x) = x, ∀x ∈ [0, 1), which is a uniform continuous function on [0, 1). However, using the same reasoning as in the mentioned Remark, one can prove that the convergence is nonuniform on [0, 1). Example 30. A sequence of nonuniformly continuous on X functions converges on X, but the limit function is uniformly continuous on X. Solution n+1 Consider the sequence fn (x) = x n on X = [1, +∞). Each function fn (x) is continuous, but this continuity is nonuniform on [1, +∞). Indeed, the distance

2.4 Convergence and Uniform Continuity n

n

between the two points xk = k n+1 and xk+1 = (k + 1) n+1 converges to 0 as k approaches ∞: ( ) n n lim |xk+1 − xk | = lim (t + 1) n+1 − t n+1 t→+∞ k→∞ (( ) )n n 1 n+1 n+1 = lim t 1+ −1 t→+∞ t n ( ) ( ) n −1 1 n+1 n 1 n+1 1 1+ t −1 1+ t n+1 t2 = lim = lim n n − n −1 t→+∞ t→+∞ t − n+1 t n+1 n+1 1 ( )− n+1 1 1 = lim 1 + t − n+1 = 1 ⋅ 0 = 0. t→+∞ t (Here, we apply the change of the discrete variable k by the continuous variable t, then transform an indeterminate form ∞ − ∞ to ∞ ⋅ 0 and to 00 , consecutively, and finally apply l’Hospital’s rule to solve the last indeterminate form.) At the same time, the distance between the respective ordinates does not approaches 0: |fn (xk+1 ) − fn (xk )| = |k + 1 − k| = 1 ↛ 0, k→∞

which shows that the continuity is nonuniform on [1, +∞). On the other hand, the limit function f (x) = lim fn (x) = lim x n→∞

n→∞

n+1 n

= x, ∀x ∈ [1, +∞)

is uniformly continuous on [1, +∞). As for the character of the convergence, choosing xn = 2n ∈ [1, +∞) for each n ∈ ℕ, one obtains |fn (xn ) − f (xn )| = |2n+1 − 2n | = 2n → ∞, n→∞

which means that the convergence is not uniform on [1, +∞). Remark 1. An analogous example for a series: a series of functions un (x), each of which is nonuniformly continuous on X, converges on X, but the sum of series is uniformly continuous on X. For a counterexample, let us construct ∑∞ the series n=1 un (x) with the partial sums equal on X = [1, +∞) to fn (x) in the given counterexample. It is easy to show that the terms of such a series can be n+1 n chosen as u1 (x) = x2 , un (x) = x n − x n−1 , ∀n ≥ 2. To prove that the continuity of each function un (x), n ≥ 2 is nonuniform on [1, +∞), let us use the same points n n xk = k n+1 and xk+1 = (k + 1) n+1 as for the sequence. The distance between xk and

83

84

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

xk+1 approaches 0, but the respective distance between un (xk+1 ) and un (xk ) does not. In fact, | n2 n2 | lim |un (xk+1 ) − un (xk )| = lim ||k + 1 − (k + 1) n2 −1 − k + k n2 −1 || k→∞ k→∞ | | and, applying the same techniques as in the evaluation of lim |xk+1 − xk |, one can calculate the principal part of the limit as follows: ( ) ( ) n2 n2 n2 n2 lim (k + 1) n2 −1 − k n2 −1 = lim (t + 1) n2 −1 − t n2 −1 k→∞

t→+∞

( 1+

= lim

t→+∞

( = lim

t→+∞

1 t

)

n2 n2 −1

k→∞

−1

2 − n2n−1

t ) 1 1 n2 −1 n21−1 1+ t = ∞, t

that is, lim |un (xk+1 ) − un (xk )| = ∞, which means a nonuniform continuity of k→∞

un (x) on [1, +∞). (Of course, the proof of a nonuniform continuity of u1 (x) is √ √ trivial, for example, by using the points x̃ k = k and x̃ k+1 = k + 1.) On the other hand, it was shown in the above counterexample for the sequence that the series converges nonuniformly on [1, +∞) to the function f (x) = x, which is uniformly continuous on [1, +∞). Remark 2. It can also happen that a sequence of nonuniformly continuous on X functions converges on X to a nonuniformly continuous function. For 2n+1 such an example, consider the sequence fn (x) = x n on X = [1, +∞). Let us show that each function fn (x) is nonuniformly continuous on [1, +∞). The disn n tance between the two points xk = k 2n+1 and xk+1 = (k + 1) 2n+1 approaches 0 as k approaches ∞: ( ) n n lim |xk+1 − xk | = lim (t + 1) 2n+1 − t 2n+1 t→+∞ k→∞ ( ) n 2n+1 1 + 1t −1 = lim n t→+∞ t − 2n+1 ( ) n −1 n 1 2n+1 1 1 + 2n+1 t t2 = lim n n t→+∞ t − 2n+1 −1 2n+1 ( ) n+1 n+1 1 − 2n+1 − 2n+1 = lim 1 + t = 1 ⋅ 0 = 0. t→+∞ t

2.4 Convergence and Uniform Continuity

(Here, we apply the same techniques as for the evaluation of a similar limit in the first counterexample.) At the same time, |fn (xk+1 ) − fn (xk )| = |k + 1 − k| = 1 ↛ 0, k→∞

that is, the continuity is nonuniform on [1, +∞). The limit function f (x) = lim fn (x) = lim x n→∞

n→∞

2n+1 n

= x2 , ∀x ∈ [1, +∞)

is also continuous nonuniformly on [1, +∞) (as for u1 (x), one can use the √ √ points x̃ k = k and x̃ k+1 = k + 1). Finally, the convergence is not uniform on [1, +∞), since for the points xn = 2n ∈ [1, +∞), ∀n ∈ ℕ, one has |fn (xn ) − f (xn )| = |22n+1 − 22n | = 22n → ∞. n→∞

∑+∞ A similar counterexample for a series can be constructed using n=1 un (x) 2n+1 on X = [1, +∞) for which the functions fn (x) = x n represent the partial sums: 2n+1 2n−1 u1 (x) = f1 (x) = x3 , un (x) = x n − x n−1 , ∀n ≥ 2. As shown above, the sum of this series f (x) = lim fn (x) = x2 is nonuniformly continuous on [1, +∞) function n→∞ and the convergence to f (x) is nonuniform on [1, +∞). It remains to see that the terms of the series are nonuniformly continuous on [1, +∞). To this end, n n choosing the same points xk = k 2n+1 and xk+1 = (k + 1) 2n+1 , ∀n > 1, one obtains n(2n−1)

n(2n−1)

|un (xk+1 ) − un (xk )| = |k + 1 − (k + 1) (n−1)(2n+1) − k + k (n−1)(2n+1) | → +∞, k→∞

since (

n(2n−1)

n(2n−1)

lim (k + 1) (n−1)(2n+1) − k (n−1)(2n+1)

k→∞

(

)

1+ = lim

t→+∞

= lim

t

( t→+∞

)

n(2n−1) (n−1)(2n+1)

−1

n(2n−1) − (n−1)(2n+1)

2n2 −n 2n2 −n−1

t→+∞

= lim

1 t

( 1+

2n2 −n 2n2 −n−1

1 1+ t

1 t

)

2n2 −n −1 2n2 −n−1

1 t2

2n2 −n

t − 2n2 −n−1 −1

1 ) 2n2 −n−1

1

t 2n2 −n−1 = +∞.

Example 31. A sequence of nonuniformly continuous on X functions converges uniformly on X, but nevertheless the limit function is uniformly continuous on X.

85

86

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Solution Each function fn (x) = n1 sin 1x , ∀n ∈ ℕ, is continuous on X = (0, 1], but the con2 tinuity is not uniform since the distance between the two points xk = 𝜋(4k+1) and x′k =

2 , 𝜋(4k−1)

both in (0, 1], ∀k ∈ ℕ, approaches 0 as k approaches ∞, but

|fn (xk ) − fn (x′k )| =

| | 1| 1 1| 1 2 2 − sin ′ | = |1 − (−1)| = → ≠ 0. |sin n || xk n k→∞ n xk || n

The limit function is zero: f (x) = lim n1 sin 1x = 0, ∀x ∈ (0, 1], and the convern→∞ gence is uniform since for all x ∈ (0, 1] simultaneously we have |fn (x) − f (x)| = 1 | 1| 1 |sin x | ≤ n → 0. However, differently from fn (x), the limit function is unin | | n→∞ formly continuous on (0, 1]. Remark 1. An analogous example for a series: a series of nonuniformly continuous on X functions converges uniformly on X, but nevertheless the sum of the series is uniformly continuous on X. A counterexample can be constructed ∑∞ using the series n=0 un (x) with the following terms: u0 (x) = − sin 1x , un (x) = 1 sin 1x , ∀n ∈ ℕ, defined on X = (0, 1]. Each function un (x) is continuous on n(n+1) (0, 1], but the continuity is nonuniform (it can be shown in the same way as for fn (x)). The uniform convergence on (0, 1] follows from the Weierstrass test: ∑ 1 |un (x)| ≤ n(n+1) < n12 , ∀x ∈ (0, 1], and the p-series n12 is convergent. Therefore, the sum of this series is continuous due to the theorem on uniform convergent series of continuous functions. However, to evaluate if this continuity is uniform, we should find an explicit form of the sum of the series. The partial sums can be easily calculated using the telescoping form of this series: n n ( ) ∑ 1 ∑ 1 1 1 uk (x) = − sin + − sin x k k + 1 x k=0 k=1 ( ) 1 1 1 1 1 1 = sin −1 + 1 − + − + · · · + − x 2 2 3 n n+1 1 1 =− sin , n+1 x and, consequently, the sum is 1 1 sin = 0, ∀x ∈ (0, 1]. n+1 x Therefore, the sum is uniformly continuous function on (0, 1]. f (x) = lim fn (x) = − lim n→∞

n→∞

Remark 2. A sequence of nonuniformly continuous on X functions can also converge uniformly on X to a nonuniformly continuous function. For such an example, consider the sequence fn (x) = 1x + n1 on X = (0, +∞). Each function fn (x) is continuous, but nonuniformly on (0, +∞) since the distance between

2.4 Convergence and Uniform Continuity

1

P12 P11

f1

0.5

P22 P21

f2

P32 P31

f3

0.2

–0.5

0.4

0.6

0.8

P′32

P′31

Limit function f(x) = 0

P′22

P′21

Functions fn(x)

1

Pnk = (xk, fn(xk)) = ( π(4k2+ 1), 1n ) –1

P′12

P′nk = (x′k, fn(x′k)) = ( π(4k2 – 1), – 1n )

P′11

Figure 2.18 Example 31, sequence fn (x) =

the pair of the points xk = while

1 k

sin 1x .

and xk+1 =

|fn (xk+1 ) − fn (xk )| = |k + 1 + The limit function

1 n

1 k+1

approaches 0 as k approaches ∞,

1 1 − k − | = 1 ↛ 0. k→∞ n n

(

) 1 1 1 + = , ∀x ∈ (0, +∞) n→∞ n→∞ x n x is also continuous nonuniformly on (0, +∞), which can be proved using the same points xk and xk+1 . At the same time, the convergence is uniform on (0, +∞), since for all x ∈ (0, +∞) simultaneously one has |1 1 1| 1 |fn (x) − f (x)| = || + − || = → 0. | x n x | n n→∞ ∑+∞ 1 As for a series, one can use the series n=1 un (x) with un (x) = n(n+1) sin 1x defined on X = (0, 1] for a similar counterexample. f (x) = lim fn (x) = lim

Remark to Examples 30 and 31. These two examples show that if the terms of a sequence/series are nonuniformly continuous functions and the limit function/sum of the series is continuous, then the nature of the convergence has no effect on the nature of the continuity of the limit function/sum of the series.

87

88

Chapter 2 Properties of the Limit Function: Boundedness, Limits, Continuity

Exercises 1 Show that the sequence of the functions fn (x) = min(n, x) on X = [0, +∞) provides a counterexample to Example 1. Investigate the type of the convergence of fn (x) on X. 2 Verify that{ each of the following sequences n sin nx, x ∈ [0, 𝜋n ] a) fn (x) = on X = [0, +∞) 0, x ∈ ( 𝜋n , +∞) ( ) { n2 , x ∈ 0, n1 b) fn (x) = on X = [0, 1] 0, x = 0, x ∈ [ n1 , 1] { n3 x arctan n12 x , x ∈ (0, n1 ) c) fn (x) = on X = [0, +∞) 0, x = 0, x ∈ [ n1 , +∞) gives a counterexample to Example 2. {

1 ,x n2 x2

≠0 on X = [0, +∞) to illustrate 0, x = 0 Example 3. Analyze the form of convergence of fn (x) on X.

3 Use the sequence fn (x) =

{

n2 +1 ,0 n2 x2

1, then lim x1n = 0, and therefore, n→∞ (1∕x)n+1 lim fn′ (x) = lim (1∕x) = 0. Thus, the sequence of the derivatives converges on 2n +1 n→∞ n→∞

0; if |x| = 1, then fn′ (1) =

1 2

and fn′ (−1) =

{

1∕2, x = 1 . However, the 0, x ≠ ±1 convergence is not uniform on any interval containing the point x = 1, since for ∀n ∈ ℕ one can choose xn = √n1 , which gives ℝ ⧵ { −1} to the limit function g(x) = lim

n→∞

fn′ (x)

=

2

n−1

|fn′ (xn ) − g(xn )| =

(1∕2) n 2 → ≠ 0. 1 + 1∕4 n→∞ 5

Another sequence with similar properties is fn (x) = 1+nx2 x2 , x ∈ ℝ. It converges for any fixed x to f (x) = 0, and the convergence is uniform, because simultaneously for all x ∈ ℝ it holds that | | x | = 1 2n|x| ≤ 1 → 0. |fn (x) − f (x)| = || 2 x2 | 1 + n 2n n→∞ | | 2n 1 + n2 x2 2 2

1−n x At the same time, the derivatives fn′ (x) = (1+n at the point x = 0 give 2 x2 )2 the numerical sequence fn′ (0) = 1, ∀n ∈ ℕ, which does not converge to f ′ (0) = 0. Note, that { again the sequence of the derivatives converges on ℝ: 1, x = 0 g(x) = lim fn′ (x) = , but this convergence is not uniform on any 0, x ≠ 0 n→∞ 1 interval containing the point x = 0: for ∀n ∈ ℕ, one can choose xn = 2n to obtain

|fn′ (xn ) − g(xn )| = (

1 − n2 4n1 2 12 ↛ 0. )2 = 25 n→∞ 1 2 1 + n 4n2

99

100

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

Remark 1. For a function depending on a parameter, the analogous example is as follows: a function f (x, y) defined on X × Y is differentiable in x ∈ X at any fixed y ∈ Y , and f (x, y) converges uniformly on X to a function 𝜑(x) as y approaches y0 , but 𝜑′ (x) ≠ lim fx (x, y), ∀x ∈ X. y→y0

The function f (x, y) = xe−x ∕y considered on X × Y = ℝ × (0, 1] with y0 = 0 provides the required counterexample. The partial of f (x, y) with ( derivative ) 2x2 −x2 ∕y2 respect to x exists for any y ∈ (0, 1]: fx (x, y) = 1 − y2 e . Further, for any fixed x ∈ ℝ, this function converges to 0, as y approaches 0, so 𝜑(x) ≡ 0. Moreover, the convergence is uniform on ℝ as the following evaluation shows. 2 2 For an arbitrary fixed y ∈ Y , the critical points of f (x, y) − 𝜑(x) = xe−x ∕y are found from the equation ( ) 2 2 2x2 (f (x, y) − 𝜑(x))x = 1 − 2 e−x ∕y = 0, y 2

which implies x2 = |y| √ , 2

y2 , or xy 2

2

y

= ± √ . Since 1 − 2

2x2 y2

> 0 for |x|
0). Since |f (x, y) − 𝜑(x)| assumes the same value at xy+ and xy− , we obtain sup|f (x, y) − 𝜑(x)| = max|f (x, y) − 𝜑(x)| x∈ℝ

x∈ℝ

y y 2 2 2 2 = |xy+ e−xy+ ∕y | = √ e−y ∕2y = √ e−1∕2 → 0, y→0 2 2 which proves the uniform convergence on ℝ. Let us turn to the convergence of the partial derivatives. Since lim

x2

y→0 y2 x≠0

2

2

e−x ∕y = lim

t→+∞

t 1 = lim =0 et t→+∞ et 2

(apply the substitution t = xy2 and l’Hospital’s rule), we conclude that g(x) = lim fx (x, y) = 0 for ∀x ≠ 0. Additionally, at the point 0 we have y→0

g(0) = lim fx (0, y) = 1 ⋅ e0 = 1. At the same time, (lim f (x, y))x (x = 0) = 𝜑′ (0) = y→0

y→0

0 ≠ 1 = lim fx (0, y). Note, that the convergence of the derivatives is nonuniform y→0

on any interval with the point 0 included, because for ∀y ∈ (0, 1] one can choose xy = y ∈ X to obtain |fx (xy , y) − g(xy )| = e−1 ↛ 0. y→0

Remark 2. For a series of functions, the example can be formulated as follows: ∑ a series un (x) of differentiable functions converges uniformly on X, but this series cannot be differentiated term by term on X.

3.1 Differentiability of the Limit Function n−1 n ∑∞ It is substantiated by the counterexample with n=1 (−1)n x considered on X = [0, 1]. This series converges uniformly on [0, 1] according to the Abel n−1 ∑∞ test: the numerical alternating series n=1 (−1)n converges by Leibniz’s test and the sequence xn is (nonstrictly) decreasing and uniformly bounded on [0, 1]. Recalling the expansion of the logarithmic function in the Taylor series, n−1 n ∑∞ we immediately conclude that n=1 (−1)n x = ln(1 + x) = f (x) on [0, 1]. All n−1 n

the functions un (x) = (−1)n x and f (x) = ln(1 + x) are differentiable on [0, 1]: 1 u′n (x) = (−1)n−1 xn−1 and f ′ (x) = 1+x , but the series of derivatives and f ′ (x) give ∑∞ ′ ∑∞ different results at x = 1. Indeed, n=1 un (1) = n=1 (−1)n−1 is divergent, since the general term (−1)n−1 does not approach 0; at the same time, f ′ (1) = 12 . Note ∑∞ that the series n=1 u′n (x) converges on [0, 1), but nonuniformly: for ∀n ∈ ℕ, 1 there is xn = √n ∈ [0, 1) such that 2

∞ ∞ |∑ | |∑ | | | | | |rn (xn )| = | u′k (xn )| = | (−1)k−1 xk−1 | n | | | | |k=n+1 | |k=n+1 | n xn 1∕2 1 = = > ↛ 0. √ n 1 + xn 1 + 1∕ 2 4 n→∞

Remark 3. In all the above examples, the convergence of the original sequence (function with a parameter or series) is uniform on the considered domain, but the convergence of the corresponding sequence (function with a parameter or series) of the derivatives is not uniform on the same domain. This is the cause of the impossibility of differentiation under the sign of the limit (or differentiating the series term by term). Recall that in Example 1 we have considered such sequences (functions with a parameter or series) that do not converge uniformly on a set X. This implies that the convergence of derivatives (even when it takes place) cannot be uniform on X. The condition of the uniform convergence of a sequence (function with a parameter or series) of derivatives guarantees the validity of differentiation under the sign of the limit or infinite sum (an additional minor condition of the convergence of original sequence, function, or series in some point should also be fulfilled). At the same time, this condition, albeit very important, is sufficient but not necessary as shown in Example 3. Example 3. A sequence fn (x) of differentiable on X functions converges on X to a function f (x) and f ′ (x) = lim fn′ (x), ∀x ∈ X, however, the convergence of n→∞ fn′ (x) to f ′ (x) is nonuniform on X. Solution 1 Consider the sequence fn (x) = 2n ln(1 + n2 x2 ), X = [0, 1]. If x = 0, then fn (0) = 0, ∀n ∈ ℕ and lim fn (0) = 0. If x ≠ 0, then n→∞

101

102

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

Limit function f(x) = 0

0.7

Functions fn(x)

0.6

Pn = (xn, fn′(xn)) = ( 1n , 12 )

Derivatives fn′(x) P3

0.5

P2

P1

f3′

0.4

f2′

0.3 0.2

f3 f2

f1′

f1

0.1

f100 0.2

0.4

0.6

Figure 3.3 Example 3, sequence fn (x) =

1 2n

0.8

1

ln(1 + n2 x 2 ).

ln(1 + n2 x2 ) ln(1 + t 2 x2 ) = lim n→∞ t→+∞ 2n 2t 2tx2 = lim =0 t→+∞ 2(1 + t 2 x2 )

lim fn (x) = lim

n→∞

(the discrete variable n was substituted by the continuous variable t and then l’Hospital’s rule was applied). Thus, lim fn (x) = 0 = f (x), ∀x ∈ [0, 1]. For the n→∞

nx derivatives, we have: fn′ (x) = 1+nnx2 x2 , f ′ (x) = 0, and lim fn′ (x) = lim 1+n =0= 2 x2 n→∞ n→∞ ′ f (x), ∀x ∈ [0, 1]. Therefore, the conditions of the statement are satisfied, however, the convergence of the derivatives is nonuniform: for any n ∈ ℕ, there exists xn = n1 ∈ [0, 1] such that

|fn′ (xn )

′

− f (xn )| =

n⋅ 1+

n2

1 n

⋅

1 n2

=

1 ↛ 0. 2 n→∞

It can be shown that in this counterexample, the convergence of fn (x) to f (x) is uniform on X: for all x ∈ [0, 1], it is verified that 1 1 |fn (x) − f (x)| = ln(1 + n2 x2 ) ≤ ln(1 + n2 ) → 0. n→∞ 2n 2n However, it is easy to construct another counterexample where the convergence of the original sequence is nonuniform. For instance, if one considers the same sequence of functions on X = ℝ, then all the above properties are

3.1 Differentiability of the Limit Function

true, in particular, lim fn′ (x) = lim 1+nnx2 x2 = 0 = f ′ (x), ∀x ∈ ℝ, but the original n→∞ n→∞ sequence converges nonuniformly on X: choosing xn = en for each n ∈ ℕ, we obtain 1 1 1 |fn (xn ) − f (xn )| = ln(1 + n2 en ) > ln en = ↛ 0. 2n 2n 2 n→∞ Remark 1. The corresponding example for functions depending on a parameter goes as follows: a function f (x, y) defined on X × Y is differentiable in x ∈ X at any fixed y ∈ Y , f (x, y) converges on X to a function 𝜑(x), as y approaches y0 , and 𝜑′ (x) = lim fx (x, y), ∀x ∈ X, however, the convergence of fx (x, y) is not y→y0

uniform. 2 2 It can be illustrated by the function f (x, y) = y e−x ∕y considered on X × Y = 2 2 ℝ × (0, 1] with y0 = 0. Indeed, lim y e−x ∕y = 0 = 𝜑(x) for any fixed x ∈ ℝ. For y→0

the partial derivative fx (x, y) = − 2xy e−x ∕y , we have lim fx (0, y) = lim 0 = 0 and y→0 y→0 ( ) 2x −x2 ∕y2 −2t −2 lim fx (x, y) = lim − e = lim t 2 = lim =0 y→0+ y→0+ t→∞ e t→∞ 2tet 2 y 2

x≠0

2

x≠0

(apply the change of variable t =

x y

and l’Hospital’s rule). Thus,

lim fx (x, y) = 0 = g(x), ∀x ∈ ℝ,

y→0+

and, therefore 𝜑 (x) = ′

(

) lim f (x, y) = 0 = lim fx (x, y) = g(x), ∀x ∈ ℝ.

y→0+

x

y→0+

Let us check the type of convergence of f (x, y) and fx (x, y) on X = ℝ. Since the 2 2 evaluation |f (x, y) − 𝜑(x)| = |y|e−x ∕y ≤ |y| is true simultaneously for all x ∈ ℝ and |y| → 0, it follows that the convergence of f (x, y) is uniform on ℝ. On the y→0

other hand, for the derivatives one can choose xy = y ∈ X for ∀y ∈ (0, 1] to obtain 2y 2 2 |fx (xy , y) − g(xy )| = e−y ∕y = 2e−1 ↛ 0, y→0 y which means that the convergence of fx (x, y) is not uniform on ℝ (and on any interval containing 0). Another interesting ( ) counterexample for the same statement is the function x2 f (x, y) = y ln 1 + y2 , X × Y = [−1, 1] × (0, 1], y0 = 0. The limit function is 0: ( ) x2 lim f (x, y) = lim y ln 1 + 2 = 0 = 𝜑(x), ∀x ∈ ℝ y→0 y→0 y

103

104

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

(for x = 0 the result is evident and for x ≠ 0 one can apply l’Hospital’s rule for ln(1+x2 ∕y2 ) 2xy the ratio ). For the partial derivative fx (x, y) = x2 +y2 the limit function 1∕y is also 0: lim fx (x, y) = lim y→0

2xy = 0 = g(x) + y2

y→0 x2

(for x = 0 the result is evident and for x ≠ 0 just apply the arithmetic rules of the limits). Therefore, ( lim f (x, y))x = 0 = lim fx (x, y) for ∀x ∈ [−1, 1]. Thus, all y→0+

y→0+

the conditions of the statement hold. Additionally, the convergence of f (x, y) is uniform on X = [−1, 1]. In fact, for any fixed y, the function f (x, y) is strictly decreasing on X = [−1, 0] and strictly increasing on [0, 1], which ensures the following evaluation simultaneously for all x ∈ [−1, 1]: ( ) ( ) x2 1 |f (x, y) − 𝜑(x)| = y ln 1 + 2 ≤ y ln 1 + 2 . y y Since the right-hand side of the last inequality approaches 0: ( ) ln(1 + t 2 ) 1 2t lim y ln 1 + 2 = lim = lim =0 y→0+ t→+∞ t→+∞ 1 + t 2 y t (apply the substitution t = 1y and l’Hospital’s rule), it guarantees that the convergence of f (x, y) is uniform on [−1, 1]. Nevertheless, the convergence of the derivatives is not uniform on [−1, 1]: for ∀y ∈ (0, 1], there exists xy = y ∈ X such that |fx (xy , y) − g(xy )| =

2y ⋅ y = 1 ↛ 0. y→0 y2 + y2

( ) 2 If we consider a slightly modified counterexample—f (x, y) = y ln 1 + xy2 , X × Y = ℝ × (0, 1], y0 = 0 (only the set X was changed)—then all the above properties hold (in particular, ( lim f (x, y))x = 0 = lim fx (x, y) for ∀x ∈ X), y→0+

y→0+

but the convergence of f (x, y) to 𝜑(x) ≡ 0 is not uniform on X = ℝ. Indeed, 1 choosing xy = ye 2y for ∀y ∈ (0, 1], we obtain ( ) 1 |f (xy , y) − 𝜑(xy )| = y ln 1 + e y > 1 ↛ 0. y→0

Remark 2. For a series of functions, a similar example has the following for∑ mulation: a series un (x) of differentiable functions converges on X and can ∑ be differentiated term by term on X, however, the series u′n (x) converges nonuniformly on X.

3.1 Differentiability of the Limit Function

∑∞ ( Let us consider the series n=1 n1 xn − sums of this telescoping series, we have n ( ) ∑ 1 k 1 k+1 fn (x) = x − x k k+1 k=1

1 xn+1 n+1

) on X = [0, 1). For the partial

1 1 1 1 1 n+1 1 n+1 = x − x2 + x2 − x3 + · · · + xn − x =x− x . 2 2 3 n n+1 n+1 Then, f (0) = lim fn (0) = 0 and n→∞ ( ) 1 n+1 f (x) = lim fn (x) = lim x − x =x n→∞ n→∞ n+1 for x ∈ (0, 1). Thus, the sum of the series is f (x) = x for x ∈ [0, 1). Note that simultaneously for all x ∈ [0, 1), the following evaluation for the series remainder holds: 1 n+1 1 |rn (x)| = |fn (x) − f (x)| = x < → 0, n+1 n + 1 n→∞ which means the uniform convergence of the series. ∑∞ ∑∞ The series of the derivatives n=1 u′n (x) = n=1 (xn−1 − xn ) is also telescoping with the partial sums fn′ (x) =

n ∑

u′k (x) = 1 − x + x − x2 + · · · + xn−1 − xn = 1 − xn .

k=1

For x ∈ [0, 1), it follows that lim fn′ (x) = lim (1 − xn ) = 1. Thus, n→∞ n→∞ (∞ )′ ( ∞ ) ( ) ′ ∑ ∑ 1 1 n+1 n un (x) = x − x =1 n n+1 n=1 n=1 =

∞ ∞ ∑ ∑ (xn−1 − xn ) = u′n (x), ∀x ∈ [0, 1). n=1

n=1

At the same time, the series of derivatives does not converge uniformly on [0, 1), since for ∀n ∈ ℕ there exists xn = √n1 ∈ [0, 1) such that 2 ( )n 1 1 ′ ′ n |fn (xn ) − f (xn )| = |1 − xn − 1| = √ = ↛ 0. n 2 n→∞ 2 Remark 3. In the given examples, it is shown that the condition of the uniform convergence of a sequence (function with a parameter or series) of derivatives is sufficient, but not necessary for the differentiation under the sign of the limit (or differentiating the series term by term). It happens that the condition of the uniform convergence of the original sequence (function with a parameter or series) is not also necessary as shown in Example 4.

105

106

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

Derivative f′(x) = 1

Exact sum f(x) = x Partial sums fn(x)

Derivatives f′n(x)

f′(x)

1 f′3

f(x)

f′2

0.8

f3 f2

0.6

f′1

P1

P2

P3

f1

0.4

0.2 Pn = ( xn, f′( xn) – f′n(xn) ) = ( n 1 , 1 ) √2

0.2

Figure 3.4 Example 3, series

0.4

0.6

∑∞ ( 1 n=1

n

xn −

1 x n+1 n+1

2

0.8

1

) .

Example 4. A sequence fn (x) of differentiable functions converges nonuniformly on X to a function f (x), but nevertheless f ′ (x) = lim fn′ (x) n→∞ on X. Solution nx2 Consider the sequence fn (x) = 1+n on X = ℝ. If x = 0, then fn (0) = 0, ∀n ∈ ℕ 2 x4 and lim fn (0) = 0. If x ≠ 0, then n→∞

nx2 lim fn (x) = lim = lim n→∞ n→∞ 1 + n2 x4 n→∞ Thus,

x2 n 1 n2

+ x4

lim fn (x) = 0 = f (x), ∀x ∈ ℝ, and therefore, f ′ (x) = 0, ∀x ∈ ℝ.

n→∞

For the derivatives, we have: fn′ (x) = lim 0 = 0, and for x ≠ 0 we obtain n→∞

= 0.

2nx(1−n2 x4 ) , (1+n2 x4 )2

and again lim fn′ (0) = n→∞

( ) x 1 4 2 − x 2 n n 2nx(1 − n x ) lim fn′ (x) = lim = lim ( )2 = 0. n→∞ n→∞ (1 + n2 x4 )2 n→∞ 1 4 +x n2 2 4

Thus, lim fn′ (x) = 0 = f ′ (x), ∀x ∈ ℝ. Let us show that both sequences fn (x) and n→∞ fn′ (x) converge nonuniformly on X = ℝ (and on any interval containing 0).

3.1 Differentiability of the Limit Function

Indeed, for the first sequence, by choosing xn = |fn (xn ) − f (xn )| =

n⋅ 1n 1+n2 n12

for any n ∈ ℕ to get

=

1 ↛ 0, 2 n→∞

1 √ n

for any n ∈ ℕ, we obtain

and for the second we can choose x̃ n =

1 √ 2 n

( ) 2n ⋅ √1 1 − n2 4n1 2 2 n 12 √ |fn′ (̃xn ) − f ′ (̃xn )| = = n ↛ 0. ( )2 n→∞ 25 1 + n2 ⋅ 4n1 2

Remark 1. For a function depending on a parameter, the example can be formulated as follows: a function f (x, y) defined on X × Y is differentiable in x ∈ X at any fixed parameter y ∈ Y , f (x, y) converges nonuniformly on X to a function 𝜑(x) as y approaches y0 , but nevertheless 𝜑′ (x) = lim fx (x, y) on X. It can be illustrated by the function f (x, y) = ℝ × (0, 1] with y0 = 0. Indeed,

x2 y2 lim x4 +y4 y→0

y→y0 x2 y2 x4 +y4

= 0 = 𝜑(x) for any fixed x ∈ ℝ. For the 2xy2 (y4 −x4 ) y→0 (x4 +y4 )2

partial derivative, we have lim fx (x, y) = lim y→0

considered on X × Y =

= 0, ∀x ∈ ℝ. Thus, 𝜑′ (x) =

(lim f (x, y))x = 0 = lim fx (x, y) for ∀x ∈ ℝ. However, the convergence of f (x, y) y→0

y→0

and fx (x, y) is not uniform on X = ℝ (and on any interval containing 0). Indeed, for the first function, one can use xy = y ∈ X for ∀y ∈ (0, 1] to obtain |f (xy , y) − y4 y 𝜑(xy )| = 2y4 = 12 ↛ 0, and, for the derivatives, one can choose x̃ y = 2 ∈ X for y→0

∀y ∈ (0, 1] to get

( ) y y4 2 2 y2 y4 − 16 240 1 |fx (̃xy , y) − 𝜑′ (̃xy )| = ( ↛ 0. )2 = 289 y y→0 4 y 4 + y 16

Remark 2. For a series of functions, the corresponding formulation is as fol∑ lows: a series un (x) of differentiable functions converges nonuniformly on X, but nevertheless this series can be differentiated term by term on X. Let us consider the telescoping series ∞ ∑

un (x) =

n=1

∞ ∑ 2 4 2 4 (nx2 e−n x − (n − 1)x2 e−(n−1) x ) n=1

on X = ℝ. The partial sums are easily found as follows: n ∑ 2 4 2 4 fn (x) = (kx2 e−k x − (k − 1)x2 e−(k−1) x ) k=1 4

2 4

4

2 4

2 4

= x2 e−x + 2x2 e−2 x − x2 e−x + · · · + nx2 e−n x − (n − 1)x2 e−(n−1) x 2 4

= nx2 e−n x .

107

108

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

Then, f (0) = lim fn (0) = 0 and n→∞

nx2 t 1 = lim t2 = lim =0 n→∞ en2 x4 t→+∞ e t→+∞ 2tet 2

f (x) = lim fn (x) = lim n→∞

for x ≠ 0 (the change of the variable t = nx2 and l’Hospital’s rule have been applied). Thus, the sum of the series is f (x) = 0, ∀x ∈ ℝ. Let us consider the series of the derivatives: ∞ ∑

u′n (x)

=

n=1

∞ ∑

2 4

2 4

2 4

(2nxe−n x − 4n3 x5 e−n x − 2(n − 1)xe−(n−1) x

n=1 2 4

+ 4(n − 1)3 x5 e−(n−1) x ). This is again a telescoping series, and its partial sums are fn′ (x) =

n ∑

u′k (x)

k=1 4

4

2 4

2 4

= 2xe−x − 4x5 e−x + 2 ⋅ 2xe−2 x − 4 ⋅ 23 x5 e−2 x 4

4

2 4

− 2xe−x + 4x5 e−x · · · + 2nxe−n x 2 4

2 4

2 4

− 4n3 x5 e−n x − 2(n − 1)xe−(n−1) x + 4(n − 1)3 x5 e−(n−1) x 2 4

2 4

= 2nxe−n x − 4n3 x5 e−n x . Further, lim fn′ (0) = lim 0 = 0, and for x ≠ 0 we have n→∞

n→∞

2nx(1 − 2n2 x4 ) 2 t − 2t 3 = lim 2 4 n→∞ t→+∞ x en x et 2 ( ) 2 2 1 − 6t 1 1 6t = lim = lim − lim 2 2 2 t→+∞ et x t→+∞ 2tet x t→+∞ tet ( ) 1 1 6 = lim − lim = 0. x t→+∞ tet 2 t→+∞ 2tet 2

lim fn′ (x) = lim

n→∞

Therefore, ∞ ∑ n=1

( u′n (x)

′

= 0 = f (x) =

∞ ∑

)′ , ∀x ∈ ℝ.

un (x)

n=1

∑∞ ∑∞ Let us check the type of convergence of the series n=1 un (x) and n=1 u′n (x). 1 For the first series, by choosing xn = √ for ∀n ∈ ℕ, we obtain n

2 4

|rn (xn )| = |fn (xn ) − f (xn )| = nx2n e−n xn = e−1 ↛ 0, n→∞

3.1 Differentiability of the Limit Function

that is, this series converges nonuniformly on ℝ (or on any interval containing 0). Similarly, using the same xn = √1 for the second series, we get n

|∑ | | | u′k (xk )| = |fn′ (xn ) − f ′ (xn )| | | | |k=n+1 | √ 2 4 = |2nxn (1 − 2n2 x4n )e−n xn | = 2 ne−1 ↛ 0, ∞

n→∞

which shows that the convergence for the series of derivatives is also nonuniform. Remark 3. For a series of functions, the example can be strengthened to the ∑ following form: a series un (x) of differentiable functions converges nonuniformly on X and both the partial sums and the sum of the series cannot be expressed through elementary functions, but nevertheless this series can be differentiated term by term on X. ∑∞ For a counterexample in this case, let us consider the series n=1 un (x) = ∑∞ 1 n=1 nx on X = (1, +∞), which represents the real-valued Riemann zeta function (a restriction of the famous complex-valued Riemann zeta function to the real axis). Note that this series converges on X, since this is a p-series with p = x > 1. However, the convergence is not uniform on X, as is seen from the ∑n+p Cauchy criterion: substituting pn = n and xn = 1 + n1 for each n in k=n+1 uk (x), we obtain n | n+p | pn 1 1 |∑ | uk (xn )| = +···+ > | x | | (n + 1)xn n (n + pn ) (n + pn )xn |k=n+1 | n 1 1 = = ⋅ 2−1∕n n−1∕n → ≠0 n→∞ 2 2 (2n)1+1∕n

(notice that lim 2−1∕n = 1 and lim n−1∕n = lim t −1∕t = lim e− ln t∕t = n→∞ t→+∞ t→+∞ ( )n→∞ ( ) exp − lim lnt t = exp − lim 1t = e0 = 1). t→+∞ t→+∞ On the other hand, for any a > 1 the same series converges uniformly on Xa = [a, +∞) by the Weierstrass test: |un (x)| = n1x ≤ n1a , ∀x ∈ Xa and ∀n ∈ ℕ, ∑∞ and the numerical series n=1 n1a is convergent (p-series with p = a > 1). Note that the functions un (x) = n1x are infinitely differentiable and the m

m ln n mth-order derivative has the form u(m) . Let us consider the n (x) = (−1) nx ∑∞ (m) ∑ ∞ lnm n m series of the mth derivatives n=1 un (x) = (−1) n=1 nx and prove that for any m ≥ 1 this series converges nonuniformly on X = (1, +∞), and uniformly on Xa = [a, +∞), ∀a > 1. Let us start with convergence of the series of the derivatives on X = (1, +∞). Consider an arbitrary fixed x ∈ (1, +∞). Due to the density of the set of real numbers, there exists c such that 1 < c < x. The

109

110

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

series of the derivatives can be rewritten as follows: ∞ ∞ ∑ ∑ 1 lnm n m u(m) (x) = (−1) ⋅ . n nc nx−c n=1 n=1 m

Note that for ∀𝛼 = x − c > 0 and ∀m ∈ ℕ, the limit lim lnt 𝛼 t can be calculated t→+∞ just applying m times l’Hospital’s rule: lnm t mlnm−1 t = lim t→+∞ t 𝛼 t→+∞ 𝛼t 𝛼 m(m − 1)lnm−2 t m! = lim = · · · = lim m 𝛼 = 0. t→+∞ t→+∞ 𝛼 t 𝛼2 t𝛼 lim

m

m

Therefore, there exists N such that lnnx−cn < 1, ∀n > N, that is, n1c ⋅ lnnx−cn < n1c , ∀n > ∑∞ N. Since the series n=1 n1c converges (p-series with p = c > 1), by the com∑∞ m parison test of positive series it follows that the series n=1 lnnx n also converges ∑∞ for any fixed x ∈ (1, +∞), which implies the convergence of n=1 u(m) n (x) on X = (1, +∞). To show that this convergence is not uniform on X, we can use the Cauchy criterion with the same parameters pn = n and xn = 1 + n1 for each n as for the original series. Then lnm n > 1 for ∀n ≥ 3, and we obtain n | n+p | lnm (n + 1) lnm (n + pn ) | ∑ (m) | uk (xn )| = + · · · + | | | (n + 1)xn (n + pn )xn |k=n+1 | 1 1 > +···+ x n (n + 1) (n + pn )xn pn n > = (n + pn )xn (2n)1+1∕n 1 1 = ⋅ 2−1∕n n−1∕n → ≠ 0. n→∞ 2 2 This means that the convergence is not uniform on X. At the same time, this series converges uniformly on Xa = [a, +∞) for any a > 1. Indeed, choosing d between 1 and a (1 < d < a), evaluating the derivatives for sufficiently large n in the form lnm n lnm n 1 lnm n 1 |u(m) ≤ a = d ⋅ a−d ≤ d n (x)| = x n n n n n m

(again use the fact that lim lnn𝛼 n = 0, ∀𝛼 > 0, ∀m ∈ ℕ), and noting that the series n→∞ ∑∞ 1 n=1 nd is convergent (p-series with p = d > 1), we can conclude that by the ∑∞ Weierstrass test the series n=1 u(m) n (x) converges uniformly on Xa = [a, +∞), ∀a > 1. Finally, using the fact that the differentiability is a local property, we can show that the original series can be differentiated term by term on X = (1, +∞) as

3.1 Differentiability of the Limit Function

many times as we want. For any x ∈ X, there is a such that 1 < a < x. It was already proved that the original series and the series of mth derivatives, ∀m ∈ ℕ, converge uniformly on [a, +∞). Therefore, the original series is infinitely differentiable on [a, +∞) and its mth derivative can be found by differentiating term by term m times. Since x ∈ [a, +∞), the same is true at the point x. Recalling that x is an arbitrary point in X, it follows that the original series can be differentiated term by term m times on X = (1, +∞). Example 5. A sequence fn (x) of infinitely differentiable functions converges uniformly on X to a function f (x), but the sequence fn′ (x) diverges at each point of X. Solution Each of the functions fn (x) = sin√nx is infinitely differentiable on X = ℝ and n the limit function of this sequence is zero: lim fn (x) = 0 = f (x), ∀x ∈ ℝ. n→∞ Moreover, the convergence is uniform on ℝ, because simultaneously for all nx| √ x ∈ ℝ it holds that |fn (x) − f (x)| = | sin ≤ √1 → 0. Let us consider now the n n→∞ √n sequence of the derivatives fn′ (x) = n cos nx and show that it is divergent at √ ′ each x. If x = 0, then fn (0) = n → ∞, that is, the sequence diverges at 0. n→∞ If x ≠ 0, then the sequence of the derivatives is√ unbounded: if n ∈ ℕ is such √ n that | cos nx| ≥ 12 , then |fn′ (x)| = | n cos nx| ≥ 2 ; otherwise (if n ∈ ℕ is such 1 that | cos nx| < 2 ), we have | cos 2nx| = |2cos2 nx − 1| = 1 − 2cos2 nx > 12 , that √ √ 2n ′ is, |f2n (x)| = | 2n cos 2nx| > 2 . Since the sequence is unbounded, it is divergent.

Example 6. A sequence fn (x) of infinitely differentiable functions converges uniformly on X, but the sequence of the derivatives fn′ (x) is convergent and divergent at infinitely many points of X. Solution The sequence of infinitely differentiable functions fn (x) = formly on ℝ to zero:

cos nx n

converges uni-

| cos nx| 1 ≤ → 0, ∀x ∈ ℝ. n n n→∞ However, there are two infinite sets of points, the first of which contains the convergence points of the sequence fn′ (x) = − sin nx, and the second one—the divergence points. Evidently, the first set can be chosen as xk = k𝜋, ∀k ∈ ℤ, since fn′ (xk ) = 0, ∀n ∈ ℕ, and therefore lim fn′ (xk ) = 0, ∀xk = k𝜋, ∀k ∈ ℤ. lim fn (x) = 0;

n→∞

|fn (x) − f (x)| =

n→∞

111

112

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

Function derivatives f′n(x)

Limit function f(x) = 0 Functions fn(x)

2

–1.5 f 2 f1

f′3

1.5

f′2

1

f′1

P1 P2

0.5

f3 –2

n Pn = ( 12 , √ncos – ), Qn = ( 32 , √ncos 3n ) 2 2 Q4

–1

f100

–0.5

P3 0.5

–0.5

Q1 1

1.5 Q3

2

P4

–1

Q2

–1.5 –2

Figure 3.5 Example 5, sequence fn (x) =

sin nx √ . n

To construct the second set, let us prove that for any x ∈ ℝ such that sin x ≠ 0, that is, x ≠ k𝜋, ∀k ∈ ℤ, the limit lim sin nx does not exist. Suppose, by absurd, n→∞ that ∃x ∈ ℝ, sin x ≠ 0 such that lim sin nx exists and denote this limit by a. n→∞ √ Then the following two limits also exist: lim cos nx = lim ± 1 − sin2 nx = n→∞ n→∞ √ √ ± 1 − a2 and lim sin 2nx = lim 2 sin nx ⋅ cos nx = 2a ⋅ (± 1 − a2 ) (note that n→∞ n→∞ if a ≠ 0, then the square root should be chosen with the plus sign). On the other hand, lim sin 2nx is a partial limit of lim sin nx and, therefore, it equals n→∞ n→∞ √ a. From the equality a = 2a ⋅ (± 1 − a2 ), one gets a1 = 0 or 1 − a2 = 14 , that √ 3

is, a2,3 = ± 2 . In the first case, let us consider the additional trigonometric equality sin(n + 1)x = sin nx ⋅ cos x + cos nx ⋅ sin x and pass to√limit in both

sides as n → ∞. Since lim sin nx = 0 and lim cos nx = ± 1 − a21 = ±1, n→∞ n→∞ one obtains 0 = 0 ⋅ cos x ± 1 ⋅ sin x. It follows then that √sin x = 0 that con3 tradicts the choice of the point x. In the case a2,3 = ± 2 , let us consider the limit of the additional equality cos 2nx = cos2 nx − sin2 nx, which gives ( )2 ( √ )2 √ 3 1 1 1 = − ± = − (since lim cos nx = 1 − a22,3 = 12 ). There2 2 2 2 n→∞ fore, again one arrives to a contradiction. Thus, the supposition that

3.1 Differentiability of the Limit Function

lim sin nx exists is wrong, that is, the sequence sin nx diverges for ∀x ≠ k𝜋, ∀k ∈ ℤ. n→∞

∑ Example 7. A series un (x) of continuous functions converges uniformly on X, but the sum of this series is not differentiable at an infinite number of points in X.

Solution Consider all the rational points of the interval [0, 1] and order them in the form of a numerical sequence rn , n = 1, 2, … (this can be done, since the set of all the rational numbers of any interval is countable). Define the function f (x) as ∑∞ ∑∞ |x−r | follows: f (x) = n=1 un (x) = n=1 3n n , x ∈ [0, 1]. Since x, rn ∈ [0, 1], ∀n ∈ ℕ, ∑∞ |x−r | we have |x − rn | ≤ 1, that is, |un (x)| = 3n n ≤ 31n . The numerical series n=1 31n converges, and consequently, the series for f (x) converges uniformly on [0, 1] |x−r | by the Weierstrass test. Additionally, the continuity of the functions 3n n on [0, 1] for ∀n ∈ ℕ ensures the continuity of f (x) on [0, 1]. { −1, x < rn |x−r | Each of un (x) = 3n n is differentiable at any x ≠ rn , u′n (x) = 31n , 1, x > rn and has one-sided derivatives at the point x = rn : u′n (rn− ) = − 31n , u′n (rn+ ) = 31n . ∑∞ Since |u′n (x− )| = |u′n (x+ )| = 31n for any x and the series n=1 31n is convergent, ∑∞ ∑∞ the series n=1 u′n (x− ) and n=1 u′n (x+ ) converge uniformly (and absolutely) on ℝ (and, in particular, on [0, 1]). Let us prove that f (x) has one-sided derivatives at each point x0 in [0, 1]. For an arbitrary fixed x ∈ [0, 1), we choose h > 0 such that x + h ∈ [0, 1] and consider the series f (x + h) − f (x) ∑ un (x + h) − un (x) = h h n=1 ∞

=

∞ ∑ |x + h − rn | − |x − rn | n=1

h ⋅ 3n

=

∞ ∑

vn (h).

n=1

For any h ∈ (0, 1 − x), the following evaluation holds: ||x + h − rn | − |x − rn || |x + h − rn − (x − rn )| 1 ≤ = n, h ⋅ 3n h ⋅ 3n 3 ∑∞ and since the series n=1 31n converges, by the Weierstrass test it implies the ∑∞ uniform convergence of the series n=1 vn (h) on (0, 1 − x). Also, for any fixed ∑∞ u (x+h)−u (x) n we have lim vn (h) = lim n h n = u′n (x+ ), and the series n=1 u′n (x+ ) |vn (h)| =

h→0+

h→0+

113

114

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

converges. Therefore, the limit of the series exists and can be calculated term by term: f ′ (x+ ) = lim

h→0+

=

∞ ∑ n=1

′

∞ ∑ f (x + h) − f (x) = lim vn (h) h→0+ h n=1

lim vn (h) =

h→0+

∞ ∑

u′n (x+ ).

n=1

∑∞

′ n=1 un (x− )

Similarly, f (x− ) = for any x ∈ (0, 1]. Let us show now that f (x) is differentiable at each irrational point x in [0, 1]: |x−r | since x ≠ rn , every un (x) = 3n n is differentiable at x, which means that u′n (x+ ) = ∑∞ ∑∞ u′n (x− ), and consequently, f ′ (x+ ) = n=1 u′n (x+ ) = n=1 u′n (x− ) = f ′ (x− ), that is, f (x) is differentiable. Finally, let us show that f (x) is not differentiable at any rational point x in [0, 1]. For a rational point x0 , there exists k such that x0 = rk . The function |x−r | uk (x) = 3k k is not differentiable at x0 = rk , because the one-sided derivatives are different: u′k (rk− ) = − 31k ≠ 31k = u′k (rk+ ). Thus, at any rational point, one of the terms of the series is nondifferentiable, and all others are differentiable: f (x) =

∞ ∑ n=1

un (x) =

∞ ∑

un (x) + uk (x) ≡ gk (x) + uk (x).

n=1,n≠k

Using the same reasoning as above for an irrational point, we can deduce that the function gk (x) is differentiable at x0 = rk . Therefore, f (x) is not differentiable at x0 = rk as the sum of differentiable and nondifferentiable functions. Remark. This example can be also formulated in the form: a function f (x) is continuous on an interval, but it can be nondifferentiable at infinitely many points of this interval. ∑ Example 8. A series un (x) of continuous functions converges uniformly on X, but nevertheless the sum of this series is not differentiable at any point of X. Solution Let d(x) be the distance from x to the nearest integer, means that [ which ] 1 d(x) is a continuous 1-periodic function with the range 0, 2 defined on the [ ] ⎧ 1 ⎪ x, x ∈ 0, 2 [ ]. fundamental interval [0, 1] in the following form: d(x) = ⎨ 1 1 − x, x ∈ , 1 ⎪ 2 ⎩ 1 1 Accordingly, un (x) = 10n d(10n x) is a continuous 10n -periodic function with [ ] [ ] 1 the range 0, 2⋅10 defined on the fundamental interval 0, 101n in the form: n

3.1 Differentiability of the Limit Function

0.8

Partial sum f11(x)

Series terms un(x)

Partial sums fn(x) 0.6 f3 f2

0.4

0.2

u0

f1

u1

f0

u2 u3 0.2

0.4

Figure 3.6 Example 8, partial sums of series instead of un (x) =

1 d(10n x) 10n

0.6

∑∞ n=0

0.8

ũ n (x) with ũ n (x) =

1

1 d(2n x) 2n

for better visualization.

[ ] ⎧ 1 x, x ∈ 0, 2⋅10 ⎪ ∑ n [ ] . Let us consider the series ∞ un (x) = ⎨ 1 n=0 un (x) on 1 1 ⎪ 10n − x, x ∈ 2⋅10n , 10n ⎩ X = [0, 1]. Since |un (x)| ≤ 101n for all x ∈ [0, 1] and the numerical series ∑∞ 1 ∑∞ n=0 10n converges, by the Weierstrass test the series n=0 un (x) converges uniformly on [0, 1] to its sum f (x), and the continuity of un (x) implies the continuity of f (x). Now let us show that f (x) is differentiable nowhere in [0, 1]. Let us consider any fixed a in [0, 1) with the decimal expansion a = 0. a1 a2 · · · an an+1 · · ·, and choose a particular sequence hm approaching 0, as m approaches infinity, in the { −10−m , am = 4 or 9 following form: hm = . Then 10−m , otherwise f (a + hm ) − f (a) ∑ 1 d(10n (a + hm )) − d(10n a) = ⋅ hm 10n ±10−m n=0 ∞

=

∞ ∑

±10m−n ⋅ [d(10n (a + hm )) − d(10n a)].

n=0

Since 10n hm is an integer for n ≥ m, we get d(10n (a + hm )) − d(10n a) = 0. Therefore, the last series is actually a finite sum with m terms. For these terms

115

116

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

f11 f7

0.65

f3

0.6

0.55

0.5 0.2

0.25

0.3

0.35

0.4

0.45

0.5

0.55

0.6

0.65

Figure 3.7 Example 8, amplified view of partial sums of series

∑∞ n=0

0.7

0.75

0.8

ũ n (x).

(when n < m) we can represent 10n a = a1 · · · an + 0. an+1 an+2 · · · am · · · and 10n (a + hm ) = a1 · · · an + 0. an+1 an+2 · · · (am ± 1) · · ·, where a1 · · · an is the integer part of both numbers. (Note that except for mth decimal digit, other digits are not changed, because we have chosen hm = −10−m if am = 9). If it happens that 0. an+1 an+2 · · · am · · · < 12 , then also 0. an+1 an+2 · · · (am ± 1) · · · < 12 (in the special case n = m − 1, the last inequality is true because we have chosen hm = −10−m when am = 4). Therefore, d(10n (a + hm )) − d(10n a) = d(0.an+1 · · · (am ± 1) · · ·) − d(0.an+1 · · · am · · ·) = 0.an+1 · · · (am ± 1) · · · − 0.an+1 · · · am · · · = ±10n−m . Similarly, if 0. an+1 an+2 · · · am · · · ≥ 12 , then 0. an+1 an+2 · · · (am ± 1) · · · ≥ 12 (here, between two equivalent representations of the rational numbers 0.an+1 · · · ak 00 · · · = 0.an+1 · · · (ak − 1)99 · · · we use the former), and consequently d(10n (a + hm )) − d(10n a) = d(0.an+1 · · · (am ± 1) · · ·) − d(0.an+1 · · · am · · ·) = (1 − 0.an+1 · · · (am ± 1) · · ·) − (1 − 0.an+1 · · · am · · ·) = ∓10n−m .

3.2 Integrability of the Limit Function

Thus, for n < m we have 10m−n ⋅ [d(10n (a + hm )) − d(10n a)] = ±1, and, therefore m−1 m−1 ∑ f (a + hm ) − f (a) ∑ = ±10m−n ⋅ [d(10n (a + hm )) − d(10n a)] = ±1. hm n=0 n=0

The last sum is an even integer when m is even, and an odd integer when m ∑m−1 f (a+hm )−f (a) is odd. Therefore, the sequence = n=0 ±1 does not converge as m hm approaches infinity, because this is a sequence of integers that are alternately odd and even. Hence, we have constructed a sequence hm = ±10−m such that f (a+hm )−f (a) the limit of does not exists, which means that f (x) is not differenhm tiable at a. Remark 1. This is a strengthened version of Example 7. Remark 2. Of course, all these considerations and conclusions are also true on ℝ, because all the functions un (x) = 101n d(10n x), n = 0, 1, 2, … have the common period 1, and so does f (x). Therefore, their properties on [0, 1] are automatically extended to ℝ. Remark 3. This example can also be formulated in the following form: there exist continuous on an interval functions f (x) that are nondifferentiable at any point of this interval. The provided counterexample was first presented by Takagi in 1903 and later (in 1930) rediscovered by van der Waerden. The first (and more intricate) example of an everywhere continuous and nowhere differentiable function was constructed by Weierstrass as early as 1861.

3.2 Integrability of the Limit Function Example 9. A sequence fn (x) of Riemann integrable on [a, b] functions converges on [a, b] to a function f (x), but the limit function is not Riemann integrable on [a, b]. Solution Consider the set of the rational points in [0, 1] and order it to form a sequence { rn , n = 1, 2, …. Define the functions fn (x) on [0, 1] as follows: 1, x = r1 , r2 , … , rn fn (x) = . Each of fn (x) is bounded on [0, 1] and has the 0, otherwise finite number of the discontinuities (at the points r1 , … , rn ), which implies that fn (x) is Riemann integrable on [0, 1]. Since fn+1 (x) ≥ fn (x), ∀x ∈ [0, 1]

117

118

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

(fn+1 (rn+1 ) = 1 > 0 = fn (rn+1 ) and fn+1 (x) = fn (x), ∀x ≠ rn+1 ), the sequence fn (x) is increasing and bounded with respect to n, and therefore it is convergent for any fixed x ∈ [0, 1]. From the form of fn (x), it follows { that the limit function 1, x ∈ ℚ is Dirichlet’s function considered on [0, 1]: D(x) = , which is not 0, x ∈ 𝕀 Riemann integrable on any interval. Note, that the convergence of fn (x) to D(x) is not uniform on [0, 1]: for any n ∈ ℕ, there exists xn = rn+1 ∈ ℚ ∩ [0, 1] such that |fn (rn+1 ) − D(rn+1 )| = |0 − 1| = 1 > 12 = 𝜀0 . Example 10. A sequence fn (x) converges on [a, b] to a function f (x), and each of the functions fn (x) is not Riemann integrable on [a, b], but f (x) is Riemann integrable on [a, b]. {

Solution

1∕n, x ∈ ℚ considered on [0, 1] are scaled Dirichlet’s 0, x ∈ 𝕀 functions: fn (x) = n1 D(x) and, therefore, are not Riemann integrable on [0, 1]. At the same time, the sequence fn (x) converges to f (x) ≡ 0, ∀x ∈ [0, 1], which is a Riemann integrable function. Moreover, this convergence is uniform due to the following evaluation held simultaneously for all ∀x ∈ [0, 1]: |fn (x) − f (x)| ≤ 1 → 0. n The functions fn (x) =

n→∞

Example 11. A sequence of continuous functions fn (x) converges on [a, b] to b b a function f (x), but lim ∫a fn (x)dx ≠ ∫a f (x)dx. n→∞

Solution 2 Let us consider the sequence of continuous functions fn (x) = nxe−nx on [0, 1]. This sequence converges to f (x) ≡ 0, ∀x ∈ [0, 1], because for x = 0 we have fn (0) = 0, ∀n, and for x ≠ 0 we can apply l’Hospital’s rule to obtain nx tx x f (x) = lim fn (x) = lim nx2 = lim tx2 = lim 2 tx2 = 0. n→∞ n→∞ e t→+∞ e t→+∞ x e 1

Therefore, ∫0 f (x)dx = 0. Each of fn (x) is also integrable (since it is continuous) 1 2| 1 and ∫0 fn (x)dx = − 12 e−nx | = 12 − 12 e−n . However, |0 1

lim

n→∞ ∫0

1

1 1 (1 − e−n ) = ≠ 0 = f (x)dx. n→∞ 2 ∫0 2

fn (x)dx = lim

Let us check the type of convergence. Choosing xn = √1 ∈ [0, 1] for any n ∈ ℕ, n we get √ 1 |fn (xn ) − f (xn )| = n ⋅ √ e−n⋅1∕n = n ⋅ e−1 → +∞, n→∞ n which means that the convergence is nonuniform on [0, 1].

3.2 Integrability of the Limit Function

Limit function f(x) = 0 Functions fn(x)

1

An =

0.8

1 2

1

e–n– area ∫0 fn(x)dx 1

f4

P4

0.7

P3

f3

0.6

–

Alim = 12 – limit area limn ∫0 f n(x)dx 1 A0 = 0 – area limit function ∫0 f(x)dx

1 , √ne–1 Pn = (xn, fn(xn)) = ( √n )

0.9

1 2

P2

0.5

Alim A3 A2 P1

f2

0.4

A1

0.3 f1

0.2 0.1

A0 0.2

0.4

0.6

0.8

1

2

Figure 3.8 Example 11, sequence fn (x) = nxe−nx . 2nx Another counterexample can be given using the sequence fn (x) = 1+n on 2 x4 [0, 1]. These functions are continuous and the sequence converges to f (x) ≡ 0, 1 ∀x ∈ [0, 1], which implies that ∫0 f (x)dx = 0. The integral of each fn (x) is easy 1 to calculate ∫0 fn (x)dx = arctan nx2 |10 = arctan n, but the corresponding limit is different from 0: 1

lim

n→∞ ∫0

1

fn (x)dx = lim arctan n = n→∞

𝜋 ≠0= f (x)dx. ∫0 2

Note that the convergence of fn (x) is not uniform on [0, 1]: for any n ∈ ℕ, one can choose xn = √1 ∈ [0, 1] to obtain n

|fn (xn ) − f (xn )| =

2n ⋅

1 √ n

1 + n2 ⋅

1 n2

=

√ n → +∞. n→∞

Remark 1. Analogous example can be formulated for a function depending on a parameter: a function f (x, y) defined on X × Y = [a, b] × Y is continuous in b x ∈ X at any fixed y ∈ Y , and there exists lim f (x, y) = 𝜑(x), but ∫a 𝜑(x)dx ≠ y→y0

b

lim ∫a f (x, y)dx.

y→y0

The corresponding counterexample is f (x, y) = yx2 e−x ∕y on X × Y = [0, 1] × (0, 1] and y0 = 0. Since f (0, y) = 0 and for any fixed x ∈ (0, 1] the application of 2

2

119

120

Chapter 3 Properties of the Limit Function: Differentiability and Integrability x −x2 ∕y2 e y→0+ y2

l’Hospital’s rule gives lim f (x, y) = lim y→0+

t t→+∞ xet

= lim

= 0, one concludes 1

that lim f (x, y) = 0 = 𝜑(x), ∀x ∈ [0, 1] and, consequently, ∫0 𝜑(x)dx = 0. Fury→0+

ther, f (x, y) is continuous in x ∈ X, and therefore, Riemann integrable in x, for any fixed y ∈ Y : 1

∫0

1 2 1 2 2| 1 1 f (x, y)dx = − e−x ∕y || = − e−1∕y . 2 |0 2 2

However, 1

lim

y→0+

∫0

1

f (x, y)dx = lim

y→0+

2 1 1 (1 − e−1∕y ) = ≠ 0 = lim f (x, y)dx. ∫0 y→0+ 2 2

Note that the convergence is not uniform on [0, 1]: choosing xy = y ∈ [0, 1] for ∀y ∈ (0, 1], we obtain |f (xy , y) − 𝜑(xy )| =

y −y2 ∕y2 1 −1 e = e → +∞. y→0+ y2 y

Remark 2. The corresponding example for a series is as follows: a series of continuous functions converges on [a, b], but it cannot be integrated term by term on [a, b]. For a counterexample, one can use the series ∞ ∑

un (x) =

n=1

∞ ∑

2 2

2 2

2x[n2 e−n x − (n − 1)2 e−(n−1) x ]

n=1

on [0, 1]. The partial sums of this telescoping series can be easily found as follows: fn (x) = 2x

n ∑ 2 2 2 2 [k 2 e−k x − (k − 1)2 e−(k−1) x ] k=1 2

2 2

2

2 2

2 2

= 2x[e−x + 22 e−2 x − e−x + · · · + n2 e−n x − (n − 1)2 e−(n−1) x ] 2 2

= 2xn2 e−n x . Therefore, the sum of the series is f (x) = lim fn (x) = 0 for any x ∈ [0, 1], and n→∞

1

consequently, ∫0 f (x)dx = 0. At the same time, 1 2 2

∫0

2 2

and the numerical series ∞ ∑ n=1

2

2

un (x)dx = (−e−n x + e−(n−1) x )|10 = −e−n + e−(n−1)

1

∫0

un (x)dx =

∞ ∑ 2 2 (−e−n + e−(n−1) ) n=1

3.2 Integrability of the Limit Function

is telescoping with the partial sums sn =

n ∑ 2 2 (−e−k + e−(k−1) ) k=1 2

2

2

2

= −e−1 + 1 − e−2 + e−1 + · · · − e−n + e−(n−1) = 1 − e−n . Therefore, ∞ ∑ ∫ n=1 0

1

un (x)dx =

∞ ∑

2

2

(−e−n + e−(n−1) )

n=1

= lim sn = 1 ≠ 0 = n→∞

∞ 1∑

∫0

un (x)dx.

n=1

Note that the convergence of the series is nonuniform on [0, 1]: for ∀n ∈ ℕ, there is xn = n1 such that ∞ |∑ | | | |rn (xn )| = | uk (xn )| = |fn (xn ) − f (xn )| | | |k=n+1 | 1 2 −n2 ⋅1∕n2 =2⋅ ⋅n e = 2ne−1 → +∞. n→∞ n Another interesting counterexample is the telescoping series ∞ ∑

∞ ∑ un (x) = [nx(1 − x2 )n − (n − 1)x(1 − x2 )n−1 ]

n=1

n=1

considered on [0, 1]. Its partial sums are fn (x) = x(1 − x2 ) + 2x(1 − x2 )2 − x(1 − x2 ) · · · + nx(1 − x2 )n − (n − 1)x(1 − x2 )n−1 = nx(1 − x2 )n . For x = 0 and 1, we have fn (x) = 0, and for x ∈ (0, 1) we can solve the indeterminate form in the following way: lim fn (x) = lim x

n→∞

t→+∞

t 1 = x lim =0 2 −t 2 −t t→+∞ (1 − x ) −(1 − x ) ln(1 − x2 )

(here the discrete variable n was replaced by the continuous variable t and then l’Hospital’s rule was applied). Therefore, the sum of the series is f (x) = 1 lim fn (x) = 0 for any x ∈ [0, 1] and, consequently, ∫0 f (x)dx = 0. n→∞ At the same time, 1 ( )|1 1 n 1n−1 un (x)dx = − (1 − x2 )n+1 + (1 − x2 )n || ∫0 2n+1 2 n |0 1 n 1n−1 = − 2n+1 2 n

121

122

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

1 0.9

n – areas partial sums ∑n ∫ 1u (x)dx k=1 0 k 2 n+1 1 1 ∞ A lim = 2 – sum areas ∑n = 1 ∫0 u n(x)dx

An = 1

Sum of series f(x) = 0

Partial sums fn(x) Pn = (xn, rn(xn)) = ( 1n , n (1– 1n )n) A = 0 – area sum of series∫ 1 ∑ ∞ u (x)dx 0 n=1 n 0

0.8

f4

0.7

P4

f3

0.6

P3

0.5 f2

0.4

Alim A4 A3 A2

P2

0.3

A1

f1

0.2 0.1

P1A0 0.4

0.2

Figure 3.9 Example 11, series

0.6

∑∞

n=1 [nx(1

0.8

1

− x 2 )n − (n − 1)x(1 − x 2 )n−1 ].

and the resulting numerical series ∞ ∞ 1 ( ) ∑ ∑ 1 n n−1 un (x)dx = − ∫ 2 n+1 n n=1 0 n=1 is telescoping. Its partial sums are ( ) n ∑ 1 k k−1 1 n sn = − = , 2 k + 1 k 2n+1 k=1 and, therefore, ∞ ∞ 1 ( ) ∑ ∑ 1 n n−1 un (x)dx = − = lim sn n→∞ ∫ 2 n+1 n n=1 0 n=1 =

∞ 1∑ 1 ≠0= u (x)dx. ∫0 n=1 n 2

Note that the convergence of the series is nonuniform on [0, 1]: for ∀n ∈ ℕ, there is xn = √1 such that n ) √ ( 1 n |rn (xn )| = |fn (xn ) − f (xn )| = n ⋅ 1 − → +∞. n n→∞ Remark 3. In all the counterexamples, the impossibility to interchange the operations of the integration and the limit or infinite sum is caused by nonuniform

3.2 Integrability of the Limit Function

convergence of the sequence, function, or series. Although we see that the condition of the uniform convergence is important, it is not necessary for the integration under the sign of limit or infinite sum as shown in the next example. Example 12. A sequence of continuous functions fn (x) converges on [a, b] to b b a function f (x) and lim ∫a fn (x)dx = ∫a f (x)dx, but the convergence is nonunin→∞ form on [a, b]. Solution Let us consider the sequence of continuous functions fn (x) = nx(1 − x)n on [0, 1]. This sequence converges to f (x) ≡ 0, ∀x ∈ [0, 1]. Indeed, for x = 0 and x = 1 we have fn (0) = fn (1) = 0, ∀n, and for x ∈ (0, 1) we can apply l’Hospital’s rule to obtain t 1 f (x) = lim fn (x) = lim x = x lim = 0. n→∞ t→+∞ (1 − x)−t t→+∞ −(1 − x)−t ln(1 − x) 1

Therefore, ∫0 f (x)dx = 0. On the other hand, each fn (x) is also integrable (since it is continuous) and 1 ( )|1 n n fn (x)dx = − (1 − x)n+1 + (1 − x)n+2 || ∫0 n+1 n+2 |0 n n = − . n+1 n+2 Therefore, 1

lim

n→∞ ∫0

fn (x)dx = lim

n→∞

n =0 (n + 1)(n + 2)

1

=

∫0

1

lim fn (x)dx =

n→∞

∫0

f (x)dx.

Thus, all the statement conditions are satisfied. Nevertheless, the convergence of fn (x) is not uniform on [0, 1]: choosing xn = n1 ∈ [0, 1] for any n ∈ ℕ, we get ( ) 1 1 n |fn (xn ) − f (xn )| = n ⋅ 1− → e−1 ≠ 0. n n n→∞

Remark 1. Analogous example can be formulated for a function depending on a parameter: a function f (x, y) defined on X × Y = [a, b] × Y is continuous in x ∈ X at any fixed y ∈ Y , there exists lim f (x, y) = 𝜑(x), and y→y0

b

b

∫a 𝜑(x)dx = lim ∫a f (x, y)dx, but f (x, y) converges nonuniformly on X to 𝜑(x). y→y0

The corresponding counterexample is f (x, y) =

xy x2 +y2

on X × Y = [0, 1] × (0, 1]

123

124

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

Limit function f(x) = 0 Functions fn(x) 0.4

0.3

An =

– areas ∫0 f n(x)dx 1

1

Alim = 0 – limit area limn ∫0 f n(x)dx

Pn = (xn, fn(xn)) = ( 1n , (1– 1n )n) P4 P3 f4 f3

n n n + 1– n + 2

A0 = 0 – area limit function ∫0 f(x)dx 1

P2

f2

0.2

A2 A3 A4

f1

0.1

P1 A0 = Alim = 0 0.2

0.4

0.6

0.8

1

Figure 3.10 Example 12, sequence fn (x) = nx(1 − x)n .

and y0 = 0. For any fixed x ∈ [0, 1], we have lim f (x, y) = 0 = 𝜑(x) and, y→0+

1 ∫0

therefore, 𝜑(x)dx = 0. Further, f (x, y) is continuous in x ∈ X, and consequently Riemann integrable in x, for any fixed y ∈ Y : 1

∫0

f (x, y)dx =

Consequently, 1

lim

y→0+

∫0

|1 1 1 y ln(x2 + y2 )|| = y(ln(1 + y2 ) − ln y2 ). 2 |0 2

( ) 1 1 f (x, y)dx = lim y ln 1 + 2 y→0+ 2 y ln(1 + t 2 ) = lim =0 t→+∞ 2t 1

=

∫0

1

lim f (x, y)dx =

y→0+

∫0

𝜑(x)dx.

However, the convergence is not uniform on [0, 1]: choosing xy = y ∈ [0, 1] for y2 ∀y ∈ (0, 1], we obtain |f (xy , y) − 𝜑(xy )| = 2y2 = 12 ↛ 0. y→0+

Remark 2. The corresponding example for a series goes as follows: a series of continuous functions converges on [a, b], and it can be integrated term by term, but this convergence is nonuniform on [a, b].

3.2 Integrability of the Limit Function 1 ∑∞ 2n+1 − For a counterexample, consider the series n=1 un (x), un = x 1 x 2n−1 ( on [0, 1]. This ) is a1 telescoping series with the partial sums fn (x) = 1 1 ∑n 2k+1 2k−1 −x = x 2n+1 − x. For x = 0, all these sums are 0, and conk=1 x sequently, f (0) = lim fn (0) = 0; for x ∈ (0, 1], the total sum of the series is

n→∞

1

found in the form f (x) = lim x 2n+1 − x = 1 − x. Thus, the series sum is f (x) = n→∞ { ( ) |1 2 0, x = 0 1 1 , and therefore, ∫0 f (x)dx = ∫0 (1 − x)dx = x − x2 || = 12 1 − x, x ∈ (0, 1] |0 (the value of a function in a single point does not affect the value of the definite integral). On the other hand, (

) |1 | = 2n + 1 − 2n − 1 | ∫0 2n |0 2n + 2 ) ∑∞ ∑∞ ( 1 2n−1 and the numerical series n=1 ∫0 un (x)dx = n=1 2n+1 − is telescoping 2n+2 2n 1

un (x)dx =

2n 2n + 1 2n+2 2n − 1 2n−1 x 2n+1 − x 2n + 2 2n

with the partial sums sn = ∞ ∑ n=1

2n+1 2n+2

− 12 . Therefore, (

1

∫0

un (x)dx = lim sn = lim n→∞

n→∞

2n + 1 1 − 2n + 2 2

)

∞ 1∑ 1 = = u (x)dx, 2 ∫0 n=1 n

that is, the original series can be integrated term by term. Nevertheless, the convergence of the series is nonuniform on [0, 1]: for ∀n ∈ ℕ, there exists xn = 2−(2n+1) ∈ [0, 1] such that | 1 | |rn (xn )| = |fn (xn ) − f (xn )| = ||xn2n+1 − xn − 1 + xn || | | 1 1 −(2n+1) 2n+1 = 1 − (2 ) = ↛ 0. 2 n→∞ Another ∑∞ ( 1

∑∞ counterexample is the telescoping series n=1 un (x) = ) 1 − 1+(n+1)x considered on [0, 1]. Its partial sums are fn (x) =

n=1 1+nx 1 1 − 1+(n+1)x . 1+x

For x = 0, we have fn (0) = 0, ∀n, and for x ∈ (0, 1] we get { 0, x = 0 1 lim fn (x) = 1+x . Therefore, the sum of the series is f (x) = , and 1 , x ∈ (0, 1] n→∞ 1+x consequently, 1

∫0

1

f (x)dx =

1 dx = ln(1 + x)|10 = ln 2 ∫0 1 + x

(again we use the property that the value of the integral does not depend on the value of a function in a single point).

125

126

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

1 0.9

Sum of series f(x) = 1 1+ x , x≠ 0 Partial sums fn(x) 1 Alim = ln2 – sum areas ∑∞ ∫ un(x)dx n=1 0 1 Pn = (xn, rn(xn)) = ( n +1 1 , 12 ) A0 = ln2 – area sum of series∫0 ∑∞ u (x)dx n=1 n

An = ln2 – n 1+ 1 ln(2 + n) – areas partial sums ∑ nk = 1∫0 uk(x)dx 1

0.8

A0 = Alim = ln2

0.7 0.6 P4 P3

0.5 0.4

P2

P1

f4 f3

0.3 0.2 0.1

f2

A4 A3 A2

f1

A1

0.2

0.4

Figure 3.11 Example 12, series

On the other hand,

0.6

∑∞ ( n=1

1 1+nx

−

0.8 1 1+(n+1)x

1

) .

)|1 1 1 ln(1 + nx) − ln(1 + (n + 1)x) || ∫0 n n+1 |0 1 1 = ln(1 + n) − ln(1 + (n + 1)). n n+1 The resulting telescoping series ∞ ∞ ( 1 ) ∑ ∑ 1 1 un (x)dx = ln(1 + n) − ln(2 + n) ∫ n n+1 n=1 0 n=1 1

(

un (x)dx =

has the partial sums n ( ) ∑ 1 1 1 sn = ln(1 + k) − ln(2 + k) = ln 2 − ln(2 + n), k k+1 n+1 k=1 and therefore, ∞ 1 ∑ n=1

∫0

( un (x)dx = lim sn = lim ln 2 − n→∞

= ln 2 =

n→∞

∞ 1∑

∫0

n=1

) 1 ln(2 + n) n+1 1

un (x)dx =

∫0

f (x)dx.

3.2 Integrability of the Limit Function

Thus, the original series can be integrated term by term, but it does not con1 verge uniformly on [0, 1]: for ∀n ∈ ℕ, there is xn = n+1 such that |rn (xn )| = |fn (xn ) − f (xn )| | 1 | 1 1 | = || − + | 1 + x 1 + x 1 + (n + 1)x | n n n| 1 = ↛ 0. 2 n→∞ Example 13. The elements of a convergent on X sequence and its limit function have infinitely many discontinuity points on X, but the formula of term-by-term integration holds. Solution A simple example is fn (x) = R(x) on X = [0, 1], where R(x) is the Riemann function. Evidently, this sequence converges uniformly on [0, 1] to the limit function R(x). It is known that R(x) is discontinuous at all rational points and continuous at all irrational points. It follows from this fact that R(x) is Riemann integrable on [a, b] according to the Lebesgue criterion that asserts that a bounded function is Riemann integrable if the set of the points of its discontinuity is of the Lebesgue measure zero, which is the case of the Riemann function that has countably many discontinuities on any interval [a, b]. Further, we can find the value of the integral by noting that for an arbitrary partition a = x0 < x1 < · · · < xn = b of an interval [a, b] the lower Riemann sums are equal to 0: n ∑ i=1

inf

x∈[xi−1 ,xi ]

R(x) ⋅ (xi − xi−1 ) =

n ∑

0 ⋅ (xi − xi−1 ) = 0

i=1

(since any subinterval [xi−1 , xi ] contains irrational points). Therefore, since the Riemann integral exists, its value is equal to the lower Riemann integral, that is, b ∫a R(x)dx = 0. (Of course, anyone familiar with integrability properties of the Riemann function can appeal directly to the last formula.) Since each function fn (x) is Riemann integrable on [0, 1] and the convergence is uniform, the Theorem on term-by-term integration of sequences guarantees 1 1 1 1 that ∫0 R(x)dx = ∫0 lim fn (x)dx = lim ∫0 fn (x)dx = lim ∫0 R(x)dx = 0. n→∞ n→∞ n→∞ Based on the given counterexample, it is also easy to construct an example of a sequence with the terms dependent on n, which maintains the same properties: n fn (x) = n+1 R(x) on X = [0, 1]. Remark 1. The corresponding example for series is as follows: the terms of a convergent on X series and its sum have infinitely many discontinuity points on X, but the formula of term-by-term integration holds. A simple counterexam∑∞ ∑∞ ples involves again the Riemann function: n=1 un (x) = n=1 n12 R(x) considered

127

128

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

on [0, 1]. This series is convergent uniformly on [0, 1] by the Weierstrass test: ∑ 1 |1 | 1 | n2 R(x)| ≤ n2 (since |R(x)| ≤ 1) and the series n2 converges. The sum of this | | 2 series is f (x) = 𝜋6 R(x). Therefore, ∞ 1∑

∫0

un (x)dx =

n=1

=

∞ 1∑

∫0 ∞ ∑

n=1

0=

n=1

1

1 𝜋2 R(x)dx = R(x)dx = 0 n2 6 ∫0 ∞ ∑ n=1

1

∫0

∑ 1 R(x)dx = un (x)dx, ∫ n2 n=1 0 ∞

1

that is also guaranteed by the Theorem on term-by-term integration of series. Remark 2. Although these counterexamples are in full agreement with the theorems on term-by-term integration of sequences and series, they show that these results are also applied to some “irregular” functions that can be a bit surprising.

Exercises 1 Let fn (x) be a sequence of differentiable on X functions convergent on X to f (x). Verify whether f (x) is differentiable on X and, if so, whether the relation f ′ (x) = lim fn′ (x) holds. Analyze the character of the convergence n→∞ of fn (x) and fn′ (x) on X. Do so for the following sequences: n2 x a) fn (x) = 1+n ,X =ℝ 4 x2 b) fn (x) =

n2 x ,X =ℝ 1+n4 x4 n2 x 2 ,X =ℝ 1+n4 x4 1 ,X =ℝ 1+nx2 (−1)n 1+nx4 x4 , X = 1 −n4 x4 e ,X =ℝ n −n2 x2

c) fn (x) = d) fn (x) = e) fn (x) = ℝ f ) fn (x) = g) fn (x) = nxe ,X =ℝ h) fn (x) = x arctan nx, X = ℝ i) fn (x) = n sin nx , X = ℝ. Formulate the false statements for which these sequences represent counterexamples. Compare with Examples 1–4 for sequences. 2 Let f (x, y) be defined on X × Y and differentiable in x on X for each fixed y ∈ Y . Find the limit lim f (x, y) = 𝜑(x), where y0 is a limit point of Y . Very→y0

ify if 𝜑(x) is differentiable on X and, if so, whether the relation 𝜑′ (x) = lim fx′ (x, y) holds. Analyze the character of the convergence of f (x, y) and y→y0

fx (x, y) on X as y approaches y0 . Do so for the following functions: y a) f (x, y) = y+x2 , X = ℝ, Y = (0, +∞), y0 = 0

Exercises

b) f (x, y) = x arctan xy, X = ℝ, Y = ℝ, y0 = 0 c) f (x, y) = y arctan xy , X = ℝ, Y = (0, 1], y0 = 0 d) f (x, y) = e) f (x, y) = f ) f (x, y) = g) f (x, y) =

x3 y2 , x4 +y4

X = ℝ, Y = (0, 1], y0 = 0 y y cos x , X = (0, +∞), Y = (0, +∞), y0 = 0 y sin xy , X = ℝ, Y = (0, 1], y0 = 0 1 sin xy, X = ℝ, Y = (0, +∞), y0 = 0 y −x ye y , X = [0, +∞), Y = (0, 1], y0 = 0 −x e y , X = [0, +∞), Y = (0, 1], y0 = 0 −x xe y , X = [0, +∞), Y = (0, 1], y0 = 0 arctan xy , X = (0, +∞), Y = (0, 1], y0 = 0.

h) f (x, y) = i) f (x, y) = j) f (x, y) = k) f (x, y) = Formulate the false statements for which these functions represent counterexamples. Compare with Examples 1–4 for functions depending on a parameter. 3 Let functions un (x) be differentiable on X for ∀n. Investigate if the series ∑ ∑ un (x) and u′n (x) are convergent, the character of the convergence on ∑ X and the possibility to differentiate the series un (x) term by term on X. Do so for the following series: ∑∞ 2 a) n=0 (−1)n xe−nx , X = ℝ ∑∞ xn b) n=1 n , X = [0, 1) ) ∑∞ ( 1 1 1 c) n=1 n1 cos nx − n+1 cos (n+1)x , X = (0, +∞) ∑∞ d) n=1 (arctan nx − arctan(n − 1)x), X = (0, +∞) ∑∞ e) n=1 (arctan nx − arctan(n − 1)x), X = [0, +∞). Formulate the false statements for which these series represent counterexamples. Compare with Examples 1–4 for series. 4 Show that the sequence fn (x) = ample for Example 5.

√ sin nx √ , 4 n

X = ℝ gives one more counterex-

5 Verify whether the statement “if a series of infinitely differentiable functions converges uniformly on X, then the series of the derivatives converges at least in one point of X” is false or true. (Hint: compare this ∑∞ statement with that in Example 5. Use n=1 sinnnx , X = [a, 2𝜋 − a], ∀a ∈ (0, 𝜋).) 6 Prove that the statement in Example 6 can be exemplified by the sequence nx √ fn (x) = cos considered on ℝ. n

129

130

Chapter 3 Properties of the Limit Function: Differentiability and Integrability

7 Analyze if the following statement is true or false: “if a series of infinitely differentiable functions converges uniformly on X, then the series of the derivatives cannot converge at infinitely many points in X and, at the same time, diverge at infinitely many points in X.” (Hint: compare with the ∑∞ statement of Example 6. Use the series n=1 sinn2nx on X = ℝ.) 8 Show that the statement “if all the terms un (x) of a convergent on [a, b] ∑ series un (x) are Riemann integrable on [a, b], then the sum of this series is also Riemann integrable on [a, b]” is false by constructing a counterexample. (Hint: construct the series whose partial sums are equal to the functions fn (x) in Example 9.) 9 Show that the statement “if all the terms un (x) of a convergent on [a, b] ∑ series un (x) are not Riemann integrable on [a, b], then the sum of this series is not Riemann integrable on [a, b] also” is false by constructing ∑∞ a counterexample. (Hint: use the series n=0 un (x), where u0 (x) = −D(x), 1 un (x) = n(n+1) D(x), n ≥ 1 and D(x) is Dirichlet’s function on [a, b] = [0, 1]. Compare with Example 10.) 10

Find the limit function of a given sequence fn (x) and analyze the character of the convergence on X = [a, b]. Verify the possibility to interchange the order of the integration on [a, b] and the passage to the limit. Do so for the following sequences: a) fn (x) = arctan nx, X = [0, 1] b) fn (x) = 3n2∕3 x2 (1 − x3 )n , X = [0, 1] c) fn (x) = 3nx2 (1 − x3 )n , X = [0, 1] 2 4 d) fn (x) = n3∕2 x3 e−n x , X = [0, 1] 2 4 e) fn (x) = 4n2 x3 e−n x , X = [0, 1] 4n2 x f ) fn (x) = 1+n4 x4 , X = [0, 1]. Formulate the false statements for which these sequences represent counterexamples. Compare with Examples 11 and 12 for sequences.

11

Find the limit function of a function f (x, y) depending on a parameter y ∈ Y as y approaches y0 . Investigate the character of the convergence on X = [a, b]. Verify the possibility to interchange the order of the integration on [a, b] and the passage to the limit. Do so for the following functions: xy2 a) f (x, y) = x4 +y4 , X = [0, 1], Y = (0, 1], y0 = 0 b) f (x, y) = c) f (x, y) =

2xy3 , x4 +y4 x3 y , x4 +y4

X = [0, 1], Y = (0, 1], y0 = 0 X = [0, 1], Y = (0, 1], y0 = 0.

Further Reading

Formulate the false statements for which these functions represent counterexamples. Compare with Examples 11 and 12 for functions depending on a parameter. 12 Find the sum of a given series and analyze the character of its convergence on X = [a, b]. Verify the possibility of term-by-term integration of the series on [a, b]. Do so for the following series: ∑∞ a) n=1 x(1 − x)n , X = [0, 1] ∑∞ 2 b) n=1 (−1)n−1 2nxe−nx , X = [0, 1] ∑∞ c) n=1 (arctan nx − arctan(n − 1)x), X = [0, 1] ∑∞ x d) n=0 (1+x) , X = [0, 1] n ∑∞ 2 2 e) n=1 (2nxe−nx − 2(n − 1)xe−(n−1)x ), X = [0, 1]. Formulate the false statements for which these series represent counterexamples. Compare with Examples 11 and 12 for series.

Further Reading S. Abbott. Understanding Analysis, Springer, New York, 2002. D. Bressoud, A Radical Approach to Real Analysis, MAA, Washington, DC, 2007. T.J.I. Bromwich, An Introduction to the Theory of Infinite Series, AMS, Providence, RI, 2005. B.M. Budak and S.V. Fomin, Multiple Integrals, Field Theory and Series, Mir Publisher, Moscow, 1978. G.M. Fichtengolz, Differential- und Integralrechnung, Vol.1–3, V.E.B. Deutscher Verlag Wiss., Berlin, 1968. V.A. Ilyin and E.G. Poznyak, Fundamentals of Mathematical Analysis, Vol.1,2, Mir Publisher, Moscow, 1982. K. Knopp, Theory and Applications of Infinite Series, Dover Publication, Mineola, NY, 1990. C.H.C. Little, K.L. Teo and B. Brunt, Real Analysis via Sequences and Series, Springer, New York, 2015. W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, New York, 1976. V.A. Zorich, Mathematical Analysis I, II, Springer, Berlin, 2004.

131

133

CHAPTER 4 Integrals Depending on a Parameter

4.1 Existence of the Limit and Continuity Example 1. A function f (x, y) defined on X × Y = [a, b] × Y is continuous in x ∈ X at any fixed y ∈ Y , and there exists lim f (x, y) = 𝜑(x), but 𝜑(x) is not intey→y0

grable on X or

b ∫a

𝜑(x)dx ≠

b lim ∫ y→y0 a

f (x, y)dx.

(See analogous example in Remark 1 to Example 11 in Chapter 3, but with another counterexample.) Solution The corresponding counterexample is f (x, y) = xy (1 − x2 )1∕y on X × Y = [0, 1] × (0, 1] and y0 = 0. For any fixed x ∈ [0, 1], we have lim f (x, y) = 0 = 𝜑(x): y→0+

f (0, y) = f (1, y) = 0 and for x ∈ (0, 1) the application of l’Hospital’s rule gives (1 − x2 )t x (1 − x2 )1∕y = lim x y→0+ y t→+∞ 1∕t t 1 = x lim = x lim = 0. t→+∞ (1 − x2 )−t t→+∞ −(1 − x2 )−t ln(1 − x2 )

lim f (x, y) = lim

y→0+

1

Therefore, ∫0 𝜑(x)dx = 0. Further, f (x, y) is continuous in x ∈ X and, consequently, Riemann integrable in x for any fixed y ∈ Y : 1

∫0

f (x, y)dx = −

|1 y 1 1 ⋅ (1 − x2 )1+1∕y || = . 2y y + 1 |0 2(y + 1)

However, 1

lim y→0

∫0

1

f (x, y)dx = lim y→0

1 1 = ≠0= lim f (x, y)dx. ∫0 y→0 2(y + 1) 2

Counterexamples on Uniform Convergence: Sequences, Series, Functions, and Integrals, First Edition. Andrei Bourchtein and Ludmila Bourchtein. © 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. Companion website: www.wiley.com/go/bourchtein/counterexamples_on_uniform_convergence

134

Chapter 4 Integrals Depending on a Parameter

Note that the convergence is not uniform on [0, 1]: choosing xy = for ∀y ∈ (0, 1], we obtain √ y |f (xy , y) − 𝜑(xy )| = (1 − y)1∕y → +∞. y→0 y

√ y ∈ [0, 1]

The second counterexample illustrates the situation when 𝜑(x) is not intex2 −2xy grable on X. Consider the function f (x, y) = (x+y)4 continuous on the domain X × Y = [0, 1] × (0, +∞). Choosing the limit point y0 = 0, we can easily find x2 −2xy the limit function: 𝜑(x) = lim f (x, y) = lim (x+y)4 = x12 for ∀x ∈ (0, 1] and, addiy→0

y→0

tionally, 𝜑(0) = lim f (0, y) = 0. Therefore, the limit function 𝜑(x) is not intey→0

grable on [0, 1] (it is unbounded in a neighborhood of 0 and the corresponding improper integral diverges). 1 At the same time, the integral F(y) = ∫0 f (x, y)dx exists for each fixed y ∈ (0, +∞): 1

x2 − 2xy dx ∫0 (x + y)4 1 2 1 1 x + 2xy + y2 xy + y2 y2 = dx − 4 dx + 3 dx ∫0 ∫0 (x + y)4 ∫0 (x + y)4 (x + y)4 ( ) |1 2y y2 1 1 | = − + − , | =− x + y (x + y)2 (x + y)3 ||0 (1 + y)3

F(y) =

and, therefore lim F(y) = − lim y→0

y→0

1 = −1. (1 + y)3

Note that the convergence is not uniform on [0, 1]: choosing xy = y ∈ (0, 1] for ∀y ∈ (0, 1], we obtain | y2 − 2y2 1 || 17 | |f (xy , y) − 𝜑(xy )| = | − → +∞. |= 2 | (2y)4 | y | 16y2 y→0 | Example 2. A function f (x, y) defined on [a, b] × Y is continuous in x for any fixed y, and the function 𝜑(x) = lim f (x, y) is continuous on [a, b], but y→y0

b

b

lim ∫a f (x, y)dx ≠ ∫a lim f (x, y)dx.

y→y0

y→y0

Solution

{

x −x2 ∕y2 e ,y y2

≠0 on [0, 1] × ℝ and 0, y = 0 2 2 choose y0 = 0. For any fixed y ≠ 0, the continuity of f (x, y) = yx2 e−x ∕y follows Let us consider the function f (x, y) =

4.1 Existence of the Limit and Continuity

from the arithmetic and composition rules, and for y = 0, the continuity of f (x, 0) = 0 is evident. Calculating 𝜑(x) = lim f (x, y), we obtain for x ≠ 0: y→0

𝜑(x) = lim y→0

x −x2 ∕y2 1 t 1 1 e = lim t = lim t = 0 y2 x t→+∞ e x t→+∞ e 2

(the change of the variable t = xy2 was used with the subsequent application of l’Hospital’s rule); and for x = 0: 𝜑(0) = lim f (0, y) = lim 0 = 0. Therefore, y→0

y→0

𝜑(x) ≡ 0 is continuous on ℝ and, in particular, on [0, 1], and consequently 1 ∫0 𝜑(x)dx = 0. However, for any y ≠ 0 we have 1

∫0

1

f (x, y)dx =

∫0

2 x −x2 ∕y2 1 2 2| 1 1 e dx = − e−x ∕y || = − e−1∕y + , 2 y 2 2 2 |0

1

and, consequently (

1

lim y→0

∫0

f (x, y)dx = lim y→0

1 ) 1 1 −1∕y2 1 − e = ≠0= lim f (x, y)dx. ∫0 y→0 2 2 2

Let us show that f (x, y) does not satisfy all the conditions of the theorem about the passing to the limit under the sign of integral or the conditions of the corollary to this theorem. In fact, choosing xy = y ∈ [0, 1] for ∀y ∈ (0, 1], we obtain y 2 2 1 |f (xy , y) − 𝜑(xy )| = 2 e−y ∕y = e−1 → +∞, y→0+ y y that is, in the mentioned theorem the condition of the uniform convergence on [0, 1] is not satisfied. On the other hand, ( ) 2 2 2 2 2x 2x3 2x fy (x, y) = − 3 + 5 e−x ∕y = 5 (x2 − y2 )e−x ∕y , y y y which implies that for any fixed x the derivative fy (x, y) changes its sign in the points y = 0 and y = ±x. In particular, for ∀x ∈ (0, 1) it follows that fy (x, y) > 0 if 0 < y < x and fy (x, y) < 0 if x < y < 1, which means that f (x, y) is not monotone in y, that is, the monotonicity condition in the corollary to the theorem is not satisfied. Remark to Examples 1 and 2. The condition of the uniform convergence or the condition of monotonicity is important to guarantee passing to the limit under the sign of integral, but these conditions are sufficient and not necessary, as shown in Example 3. Example 3. A function f (x, y) defined on X × Y = [a, b] × Y is continuous in x ∈ X at any fixed y ∈ Y , there exists lim f (x, y) = 𝜑(x) and y→y0

135

136

Chapter 4 Integrals Depending on a Parameter

φ(x) = lim f y→0

1 0 1 P1 = ∫ [lim f ]dx 0 y→0 1 P2 = lim[∫ f dx ] y→0 0

F(y) = ∫ fdx

3 2.5 2

P2

f 1.5

P1

F(y)

1 0.5

φ(x)

0 0

0 0.2

0.4

0.6

y

0.6 0.8

{x y2

Figure 4.1 Example 2, function f (x, y) =

b

0.8

1 1

e−x

2

∕y 2

0, y = 0

,y ≠ 0

0.4

0.2

x

.

b

∫a 𝜑(x)dx = lim ∫a f (x, y)dx, but f (x, y) converges nonuniformly on X to y→y0

𝜑(x). (See analogous example in Remark 1 to Example 12 in Chapter 3, but with another counterexample.) Solution 2 2 Consider the function f (x, y) = xy e−x ∕y on X × Y = [0, 1] × (0, 1] and y0 = 0. If x = 0, then 𝜑(0) = lim f (0, y) = lim 0 = 0, and if x ≠ 0, then by using the change of the variable t =

y→0 x and y

y→0

l’Hospital’s rule, one gets

x 2 2 t 1 𝜑(x) = lim e−x ∕y = lim t2 = lim = 0. y→0 y t→+∞ e t→+∞ 2tet 2 Thus,

lim f (x, y) = 0 = 𝜑(x) for any fixed x ∈ [0, 1], and therefore,

y→0+

1

∫0 𝜑(x)dx = 0. On the other hand, f (x, y) is continuous in x ∈ X and, consequently, Riemann integrable in x, for any fixed y ∈ Y : 1

∫0

2 1 −x2 ∕y2 || 1 ye = y(1 − e−1∕y ). | 2 |0 2

1

f (x, y)dx = −

4.1 Existence of the Limit and Continuity

Therefore, we can pass to the limit under the sign of integral: 1

lim

y→0+

∫0

2 1 y(1 − e−1∕y ) = 0 2

f (x, y)dx = lim

y→0+ 1

=

∫0

1

lim f (x, y)dx =

y→0+

∫0

𝜑(x)dx.

Nevertheless, just like in Example 2, the convergence is not uniform on [0, 1], neither the function is monotone in y. In fact, choosing xy = y ∈ [0, 1] for 2 2 y ∀y ∈ (0, 1], we obtain |f (xy , y) − 𝜑(xy )| = y e−y ∕y = e−1 ↛ 0, which shows y→0+

that the convergence is nonuniform. Also, ( ) 2 2 2 2 x 2x3 x fy (x, y) = − 2 + 4 e−x ∕y = 4 (2x2 − y2 )e−x ∕y , y y y ( ) from which it follows that for any fixed x ∈ 0, √1 the derivative fy (x, y) > 0 2 √ √ when 0 < y < 2x and fy (x, y) < 0 when 2x < y < 1, which means that f (x, y) is not monotone in y. Example 4. A function f (x, y) is continuous on ℝ2 except at only one point, b but F(y) = ∫a f (x, y)dx is not continuous. Solution y(x2 +1) The function f (x, y) = x2 +y2 is continuous on ℝ2 ⧵ (0, 0) due to the arithmetic rules of continuous functions, and it is easy to show that f (x, y) is unbounded and does not have a limit at the origin: along the x-axis we have lim f (x, 0) = y

x→0

lim 0 = 0, while along the y-axis we get lim f (0, y) = lim y2 = ∞. The function x→0

y→0

y→0

1

1

F(y) = ∫0 f (x, y)dx is found as follows: for y = 0, we have F(0) = ∫0 0dx = 0, and for y ≠ 0 we obtain ( 1 2 ) 1 1 y(x2 + 1) x + y2 1 2 F(y) = dx = y dx + (1 − y ) dx ∫0 x2 + y2 ∫0 x2 + y2 ∫0 x2 + y2 1 1 x| 1 = yx|10 + y(1 − y2 ) arctan || = y + (1 − y2 ) arctan . y y |0 y Since

( lim F(y) = lim

y→0+

y→0+

y + (1 − y2 ) arctan

1 y

) =

𝜋 ≠ 0 = F(0), 2

F(y) is not continuous at zero.

Example 5. A function f (x, y) is continuous in y for any fixed x and also in x b for any fixed y, but the function F(y) = ∫a f (x, y)dx is not continuous.

137

138

Chapter 4 Integrals Depending on a Parameter

1

F(y) = ∫ 0 fdx 1 P1= ∫ 0 f(x,0)dx

4

1

P2 = lim[∫ 0 f(x,y)dx]

P2

y→0

3

f(0,y) F(y)

f 2

P1

1 f(x,0) 0 0

0.2

0.2

0

0.4

0.4

0.6

y

0.8

1

Figure 4.2 Example 4, function f (x, y) =

0.8

1

0.6 x

y(x 2 +1) . x 2 +y 2

Solution

{

y2 −y2 ∕x2 e ,x x3

≠0 on ℝ2 . For any fixed 0, x = 0 x0 ≠ 0, this function is continuous in y due to the arithmetic and composition rules of continuous functions. For x0 = 0, the definition gives f (0, y) = 0 for any y, so again the function is continuous. Analogously, for y = 0, the function f (x, 0) = 0 is continuous. Finally for y0 ≠ 0, the function f (x, y0 ) is continuous at each x ≠ 0 according to the arithmetic and composition rules, and at x = 0 one gets Let us consider the function f (x, y) =

lim f (x, y0 ) = lim x→0

x→0

=

y20 x

2

2

e−y0 ∕x = lim 3

t→∞

y20 t 3 2 2

ey0 t

= lim

t→∞

3y20 t 2 2 2

2ty20 ey0 t

3 t 3 1 lim 2 2 = lim = 0 = f (0, y0 ). 2 2 2 y t t→∞ t→∞ 2 2 e0 2ty0 ey0 t

(In order to calculate this limit, make the change of the variable x = 1t and then apply l’Hospital’s rule.) Hence, the conditions of the statement hold. Note, however, that f (x, y) is discontinuous at the origin as a function of two variables: approaching zero along the path y = x we have lim f (x, x) = 2

1

x→0

lim xx3 e−x ∕x = ∞. This causes a discontinuity of F(y) = ∫0 f (x, y)dx at zero. x→0

2

2

1

1

Indeed, for y = 0 we have F(0) = ∫0 f (x, 0)dx = ∫0 0 dx = 0, while for y ≠ 0,

4.1 Existence of the Limit and Continuity

we obtain 1 y2 −y2 ∕x2 1 −y2 ∕x2 || 1 2 e dx = e = e−y , | ∫0 x3 2 |0 2 and the limit at zero shows discontinuity: 1 2 1 lim F(y) = lim e−y = ≠ 0 = F(0). y→0 y→0 2 2 1

F(y) =

Remark to Examples 4 and 5. Both these examples show that the condition of continuity of f (x, y) as a function of two variables is important to guarantee b continuity of the function F(y) = ∫a f (x, y)dx. At the same time, this condition is only sufficient and F(y) still can be continuous when f (x, y) is discontinuous, as it is shown in the next example. Example 6. A function f (x, y) is discontinuous at infinitely many points of b [a, b] × Y , but F(y) = ∫a f (x, y)dx is continuous on Y . Solution

⎧ −1, x − y < 0 ⎪ Let us consider the function f (x, y) = sgn(x − y) = ⎨ 0, x = y on S = ⎪ 1, x − y > 0 ⎩ [0, 1] × ℝ. This function is discontinuous at each point of the line y = x,

f (x,y) = sgn(x – y)

F=1 1

1 0

F(y) = ∫ fdx

f = 1, x > y

2 –2

G(x) = ∫ fdy

0.5 f

F = 1 – 2y

0 –0.5 –1 –1

f = 0, x=y –0.5

0

f = –1, x < y F = –1

0.5 y

1

1.5

2

2

1.5

0.5

1 x

Figure 4.3 Examples 6 and 16, function f (x, y) = sgn(x − y).

0

–0.5

–1

G = 2x

139

140

Chapter 4 Integrals Depending on a Parameter 1

but F(y) = ∫0 f (x, y)dx is continuous on ℝ. In fact, for y < 0 it follows that 1 1 F(y) = ∫0 1dx = 1, for y > 1 one gets F(y) = ∫0 −1dx = −1, and for 0 ≤ y ≤ 1 the integral can be separated in two parts: 1

y

F(y) =

f dx +

∫0

∫y

1

y

f dx =

∫0

−1dx +

∫y

1dx

= −y + (1 − y) = 1 − 2y. ⎧ 1, y < 0 ⎪ Summarizing, F(y) = ⎨ 1 − 2y, 0 ≤ y ≤ 1 , which is a continuous function on ℝ. ⎪ −1, y > 1 ⎩ Remark 1. In the given counterexample, the function f (x, y) is discontinuous and bounded on S. One can also construct { 7∕3 a counterexample with unbounded xy , x2 + y2 ≠ 0 x4 +y4 function. For instance, f (x, y) = is discontinuous and 0, x2 + y2 = 0 unbounded on S = [0, 1] × ℝ, since lim f (x, x) = lim x→0

x→0

x10∕3 1 = lim 2∕3 = +∞ x→0 2x 2x4 1

(albeit f (x, y) is continuous in ℝ2 ⧵ (0, 0)). At the same time, F(0) = ∫0 0dx = 0 and for y ≠ 0 one gets |1 y1∕3 xy7∕3 y7∕3 x2 | 1 dx = arctan arctan 2 . | = ∫0 x4 + y4 2y2 y2 ||0 2 y Evidently, F(y) is continuous at every y ≠ 0, and additionally 1

F(y) =

lim F(y) = lim y→0

y→0

y1∕3 1 arctan 2 = 0 = F(0). 2 y

Therefore, F(y) is continuous on ℝ.

Remark 2. The original statement can be strengthened to the following form: a function f (x, y) is discontinuous and unbounded at infinitely many points b of [a, b] × Y , but still F(y) = ∫a f (x, y)dx is continuous on Y . For a counterex⎧ √y , y < x2 , x ≠ 0 ⎪ x ample, one can use the function f (x, y) = ⎨ 2xy2 , y ≥ x2 considered on ⎪ 0, x = 0 ⎩ [0, 1] × ℝ. This function is continuous on the sets D1 = {x ∈ (0, 1], y < x2 } and D2 = {x ∈ [0, 1], y > x2 }. Let us show that f (x, y) is discontinuous at all the

4.1 Existence of the Limit and Continuity

1

2 xy

–1

–0.5

1

Figure 4.4 Examples 6 and 16, domain of function f (x, y) and function F(y) = ∫0 f (x, y)dx.

points of the parabola P = {y = x2 , x ∈ (0, 1]} except for x1 = 2−2∕7 . Indeed, approaching the points of the parabola P from within D1 , one gets y lim f (x, y) = lim 2 √ = x3∕2 , (x,y)→(x,x ) (x,y)→(x,x2 ) x (x,y)∈D 1

while approaching P from within D2 , one has lim

(x,y)→(x,x2 ) (x,y)∈D2

f (x, y) =

lim

(x,y)→(x,x2 )

2xy2 = 2x5 = f (x, x2 ).

Then the continuity condition x3∕2 = 2x5 is satisfied for the only point x1 = 2−2∕7 of P. Besides, f (x, y) is discontinuous and unbounded at all the points of the negative part of the y-axis: y lim f (x, y) = lim √ = −∞. (x,y)→(0,y) (x,y)→(0,y) x y x2 and, consequently, 1

F(y) =

∫0

2xy2 dx = x2 y2 |10 = y2 ;

finally, for 0 ≤ y ≤ 1 one obtains √

F(y) =

∫0

1

y

2xy2 dx +

∫√y

y √ dx x

√ √ y = x2 y2 |0 + 2y x|1√y = y3 + 2y − 2y5∕4 .

⎧ 2y, y < 0 ⎪ The function F(y) = ⎨ y3 + 2y − 2y5∕4 , 0 ≤ y ≤ 1 is continuous in each interval ⎪ y2 , y > 1 ⎩ (−∞, 0), (0, 1), and (1, +∞) according to the rules of continuous functions. It is easy to check that it is also continuous at the points y = 0 and y = 1:

4.1 Existence of the Limit and Continuity

2

2

y ≥ x2

F (y) = y2, y > 1

1.5

1

1

y = x2

0.5

0.5

–1 –0.5

F (y) = y3+ 2y – 2y5/4, 0 ≤ y ≤ 1

0.5

1

1.5

2

–0.5 F (y) = 2y, y < 0

y < x2 –1

–1

Figure 4.6 Remark 2 to Example 6, Remark to Example 16, domain of function 1 f (x, y) and function F(y) = ∫0 f (x, y)dx.

lim F(y) = lim 2y = 0 = F(0) = lim (y3 + 2y − 2y5∕4 ) = lim F(y);

y→0−

y→0−

y→0+

3

lim F(y) = lim (y + 2y − 2y

y→1−

5∕4

y→1−

y→0+

2

) = 1 = F(1) = lim y = lim F(y). y→1+

y→1+

Thus, F(y) is continuous on Y = ℝ. Remark 3. Another direction for strengthening the original statement is to maximize the number of discontinuity points: a function f (x, y) is disb continuous at each point of [a, b] × Y , but nevertheless F(y) = ∫a f (x, y)dx is continuous on Y . The illustration of this situation can be given with the function f (x, y) = sgnx ⋅ D(y) defined on S̃ = [−1, 1] × ℝ (recall that D(y) is Dirichlet’s function). At each point (x0 , y0 ), x0 ≠ 0, this function is discontinuous just because f (x0 , y) = ±D(y) is discontinuous in y at each point. At the points (0, y0 ), the function f (x, y) is discontinuous because in any neighborhood of (0, y0 ) there are the points with x > 0, y ∈ ℚ, which lead to the partial limit equal to 1, different from the value of function ̃ Neverf (0, y0 ) = 0. Therefore, f (x, y) is discontinuous at each point of S. 1 1 theless, F(y) = ∫−1 f (x, y)dx = D(y) ∫−1 sgnx dx = 0, ∀y ∈ ℝ, that is, F(y) is a continuous function on ℝ.

143

144

Chapter 4 Integrals Depending on a Parameter

4.2 Differentiability Example 7. Both f (x, y) and fy (x, y) are continuous in y for any fixed x and also in x for any fixed y, but Fy (y) =

d dy

b

b

∫a f (x, y)dx ≠ ∫a fy (x, y)dx.

Solution

{ y3 −y2 ∕x2 e ,x ≠ 0 x3 Let us show that the function f (x, y) = and its partial 0, x = 0 {( 2 ) 2 2 3y 2y4 − x5 e−y ∕x , x ≠ 0 3 x derivative fy (x, y) = are continuous separately in 0, x = 0 each variable on ℝ2 . For x ≠ 0, the continuity of f (x, y) separately in x and in y follows directly from the arithmetic and composition rules. For x = 0, the continuity in y is evident (f (0, y) = 0), and the continuity in x can be checked by the definition: for y0 = 0, one has lim f (x, 0) = lim 0 = 0 = f (0, 0), and for x→0 x→0 y0 ≠ 0 one gets lim f (x, y0 ) = lim x→0

x→0

=

y30 x

2

2

e−y0 ∕x = lim 3

t→∞

y30 t 3 2 2

ey0 t

= lim

t→∞

3y30 t 2 2 2

2ty20 ey0 t

yt y0 3 3 lim 0 = lim = 0 = f (0, y0 ) 2 t→∞ ey20 t 2 2 t→∞ 2ty20 ey20 t2

(the change of the variable x = 1t and l’Hospital’s rule were applied). Similar considerations show the continuity of fy (x, y) separately in x and in y. Hence, the conditions of the statement hold. At the same time, note that f (x, y) and fy (x, y) are discontinuous at the origin as functions of two variables, since along the path y = x we have x3 −x2 ∕x2 e = e−1 ≠ 0 = f (0, 0) x3

lim f (x, x) = lim x→0

x→0

and

(

) 2 2 3x2 2x4 1 − e−x ∕x = lim e−1 = ∞. x→0 x→0 x→0 x x3 x5 This leads to the impossibility to change the order of integration and differentiation at zero. Indeed, 1 3 y −y2 ∕x2 y −y2 ∕x2 ||1 y −y2 F(y) = e dx = e | = 2e ∫0 x3 2 |0 and, consequently, 2 1 2 Fy (y) = e−y − y2 e−y , 2 which gives Fy (0) = 12 . On the other hand, fy (x, 0) = 0 for any x, so lim fy (x, x) = lim

1

∫0

1

fy (x, 0)dx =

∫0

0 dx = 0 ≠ Fy (0).

4.2 Differentiability

Remark. Note that under the assumptions of Example 7, the derivative Fy (y) b

may even not exist at some point y0 ∈ Y , although ∫a fy (x, y)dx exists for { y2 −y2 ∕x2 e ,x ≠ 0 x3 ∀y ∈ Y . To this effect, consider the function f (x, y) = on ℝ2 . 0, x = 0 {( ) 2 2 2y 2y3 − e−y ∕x , x ≠ 0 3 5 x x Its partial derivative is fy (x, y) = . The proof of the 0, x = 0 continuity of f (x, y) and fy (x, y) separately in each variable on ℝ2 can be done in the same way as in Example 7. At the same time, note that both functions are discontinuous and unbounded (as functions of two variables) at (0, 0): 1 lim f (x, x) = lim e−1 = ∞ x→0 x→0 x and 1 lim fy (x, 2x) = lim 2 x→0 x→0 x

(

4x 16x3 − 3 x x

) e−4 = lim x→0

−12 −4 e = −∞. x2

Let us calculate the required integrals on [0, 1]: 1

F(y) =

1

f (x, y)dx =

∫0

y2 −y2 ∕x2 e dx x3

∫0 1 | 1 2 2 1 2 = e−y ∕x || = e−y , ∀y ≠ 0, 2 |0 2 1

F(0) = 1

1

f (x, 0)dx =

∫0

∫0

0 dx = 0,

1

1 2y −y2 ∕x2 2y3 −y2 ∕x2 e dx − e dx ∫0 x3 ∫0 x5 1 1 2y −y2 ∕x2 y −y2 ∕x2 ||1 2y −y2 ∕x2 = e dx − e − e dx | 3 ∫0 x3 ∫ x2 |0 0 x y 2 2 2 2 = −y e−y + lim 2 e−y ∕x = −y e−y , ∀y ≠ 0, x→0 x

fy (x, y)dx =

∫0

1

1

fy (x, 0)dx =

∫0

∫0

0 dx = 0.

Therefore, 1 2

∫0

fy (x, y)dx = −y e−y , ∀y ∈ ℝ,

while F(y) is discontinuous at y = 0 (lim F(y) = y→0

quently, is not differentiable at 0.

1 2

≠ 0 = F(0)) and, conse-

145

146

Chapter 4 Integrals Depending on a Parameter

Example 8. Both f (x, y) and fy (x, y) are continuous in ℝ2 except at one point, but Fy (y) =

d dy

b

b

∫a f (x, y)dx ≠ ∫a fy (x, y)dx. {

Solution The function f (x, y) =

y3 , x2 x2 +y4 2

+ y2 ≠ 0 has partial derivative fy (x, y) = 0, x + y2 = 0

3x2 y2 −y6 (x2 +y4 )2

at each point (x, y) ∈ ℝ2 ⧵ (0, 0), and both functions f (x, y) and fy (x, y) are continuous on ℝ2 ⧵ (0, 0) due to the arithmetic properties of continuous functions. At the origin, f (x, y) is discontinuous and unbounded: 1 lim f (x, y) = lim 0 = 0, lim f (x, y) = lim = ∞. x→0,y=0 x→0 y→0,x=0 y→0 y The derivative fy (x, y) does not exist at (0, 0) and is unbounded in its neighbor−y6 y→0 y8

hood: lim fy (x, y) = lim y→0,x=0

= ∞. Thus, the example conditions are satisfied.

Since f (x, y) is continuous on ℝ with respect to x for any fixed y, it is Riemann integrable in x on an arbitrary interval [a, b]. In particular, for any fixed y ≠ 0 one has 1 1 1 y3 1 x || 1 F(y) = f (x, y)dx = y3 dx = arctan = y arctan 2 , | ∫0 ∫0 x2 + y4 y2 y2 ||0 y 1

1

and at y = 0 it is just F(0) = ∫0 f (x, 0)dx = ∫0 0 dx = 0. Therefore, F ′ (y) = arctan

2y2 1 − , y2 1 + y4

y≠0

and F(y) − F(0) 1 𝜋 = lim arctan 2 = . y→0 y y 2 At the same time, fy (x, 0) = 0, ∀x ≠ 0 and, consequently, is Riemann integrable 1 on any [a, b]. However, the value of integral ∫0 fy (x, 0)dx = 0 is different from Fy (0) = 𝜋2 . F ′ (0) = lim y→0

Remark. It may happen that under the example conditions, Fy (y) even does not { 2 y , x2 + y2 ≠ 0 x2 +y2 exist at some points. In fact, consider the function f (x, y) = 0, x2 + y2 = 0 2x2 y

that has partial derivative fy (x, y) = (x2 +y2 )2 at each point (x, y) ∈ ℝ2 ⧵ (0, 0), and, according to the arithmetic rules of continuous functions, both f (x, y) and fy (x, y) are continuous on ℝ2 ⧵ (0, 0). At the origin, both functions are discontinuous: 2x3 lim f (x, y) = 0 ≠ 1 = lim f (x, y), lim fy (x, y) = lim 4 = ∞ x→0,y=0 y→0,x=0 x→0,y=x x→0 4x (besides the fact that fy (x, y) does not exist at the origin).

4.2 Differentiability 1

For ∀y ≠ 0, the integral ∫0 fy (x, y)dx can be calculated using the integration by parts: 1 2x2 y xy ||1 y dx = − + dx | ∫0 (x2 + y2 )2 x2 + y2 |0 ∫0 x2 + y2 1 y y x || 1 =− + arctan =− + arctan , | 1 + y2 y |0 1 + y2 y

1

∫0

1

fy (x, y)dx =

1

1

and for y = 0 the result is immediate ∫0 fy (x, 0)dx = ∫0 0dx = 0. Thus, 1 ∫0 fy (x, y)dx exists for ∀y ∈ ℝ. 1 The integral ∫0 f (x, y)dx also exists for any fixed y: 1

F(y) =

∫0

1 x| 1 f (x, y)dx = y arctan || = y arctan , ∀y ≠ 0 y |0 y

and 1

F(0) =

∫0

1

f (x, 0)dx =

∫0

0 dx = 0.

However, F ′ (0) does not exist at y = 0: lim

y→0+

F(y) − F(0) F(y) − F(0) 1 𝜋 𝜋 = lim arctan = ≠ − = lim . y→0+ y y 2 2 y→0− y

Remark to Examples 7 and 8. In both examples, the functions f (x, y) and fy (x, y) are discontinuous at the origin, which made impossible the interchange of the operations of integration and partial derivative. However, it may happen that the interchange of the two operations is impossible even when f (x, y) is continuous on ℝ2 , as shown in Example 9. Example 9. A function f (x, y) is continuous on ℝ2 and fy (x, 0) exists for ∀x ∈ b b | d ℝ, but Fy (0) = dy ∫a f (x, y)dx| ≠ ∫a fy (x, 0)dx. |y=0 Solution Consider the function defined on ℝ2 as follows: if y ≥ 0, then f (x, y) = √ ⎧ x, x ∈ [0, y] √ √ √ ⎪ ⎨ −x + 2 y, x ∈ [ y, 2 y] , and if y < 0 then f (x, y) = −f (x, |y|). To show that ⎪ 0, otherwise ⎩

147

148

Chapter 4 Integrals Depending on a Parameter

0.25 0.2 0.15

D1

D2

0.1 0.05

–1

–0.8

D5 –0.6

D6 –0.4

–0.2 –0.05

0.2

0.4

0.6

0.8

1

–0.1 –0.15

D3

D4

–0.2 –0.25

Figure 4.7 Example 9, parts of the domain of f (x, y).

f (x, y) is continuous on ℝ2 , we divide all the plane in six domains, which have no common interior point: √ D1 = {(x, y) ∶ y ≥ 0, 0 ≤ x ≤ y}, √ √ D2 = {(x, y) ∶ y ≥ 0, y ≤ x ≤ 2 y}, √ D3 = {(x, y) ∶ y ≤ 0, 0 ≤ x ≤ |y|}, √ √ D4 = {(x, y) ∶ y ≤ 0, |y| ≤ x ≤ 2 |y|}, √ D5 = {(x, y) ∶ x ≤ 0}, D6 = {(x, y) ∶ x ≥ 2 |y|}. Inside each of Di the function is represented by a simple formula, which guarantees the continuity. Therefore, we should check the continuity on the boundaries of these domains. √ For y > 0, there are three boundary curves—x = 0, x = y, and x = √ 2 y—which we analyze separately as follows: 1) Any point (0, y0 ) ∈ D1 ∩ D5 can be approached by the points lying in D1 or D5 ; for the points of D1 , we have lim

(x,y)→(0,y0 )

f (x, y) =

lim

x = 0 = f (0, y0 ),

lim

0 = 0 = f (0, y0 ),

(x,y)→(0,y0 )

and for those in D5 , lim

(x,y)→(0,y0 )

f (x, y) =

(x,y)→(0,y0 )

which means that f (x, y) is continuous on x = 0;

4.2 Differentiability

fn(x,y) = f(x,y), (x,y) ∈ Dn

f5 = 0

0.3

f2 = – x + 2 y

0.2

f3 = –x

0.1

f

f6 = 0

0 –0.1

f1 = x

–1

f4 = x – 2 y

–0.5

–0.2 0

–0.3 0.5 –0.25

–0.2 –0.15

–0.1 –0.05

0

0.05

0.1 0.15

x

1

0.2 0.25

y

Figure 4.8 Example 9, function f (x, y).

2) Any point (x0 , x20 ) ∈ D1 ∩ D2 can be approached by the points lying in D1 or D2 ; for the points of D1 , we have lim

(x,y)→(x0 ,x20 )

f (x, y) =

lim

x = x0 = f (x0 , x20 ),

lim

√ (−x + 2 y) = −x0 + 2x0 = x0 = f (x0 , x20 ),

(x,y)→(x0 ,x20 )

and for those in D2 , lim

(x,y)→(x0 ,x20 )

f (x, y) =

(x,y)→(x0 ,x20 )

√ that is, f (x, y) is continuous on x = y; ( ) x2 3) Any point x0 , 40 ∈ D2 ∩ D6 can be approached by the points lying in D2 or D6 ; for D2 , we have √ lim lim ( ) f (x, y) = ( )(−x + 2 y) 2 2 (x,y)→ x0 ,

x

(x,y)→ x0 ,

0 4

x

0 4

x = −x0 + 2 0 = 0 = f 2

(

and for D6 we get lim ( ) f (x, y) x2 (x,y)→ x0 , 40

x0 ,

x20

lim ( )0 x2 (x,y)→ x0 , 40

=0=f

x0 ,

√ which proves the continuity of f (x, y) on x = 2 y.

,

4

( =

)

x20 4

) ,

149

150

Chapter 4 Integrals Depending on a Parameter

Due to the symmetry with respect to x-axis, the function is also continuous for y < 0 on the boundaries of the domains D3 , D4 , D5 , and D6 . It remains to show the continuity at the origin. Approaching by the points of D1 or D3 , we have lim

(x,y)→(0,0)

f (x, y) =

lim (±x) = 0 = f (0, 0).

(x,y)→(0,0)

For the points of D2 or D4 , we obtain lim

(x,y)→(0,0)

f (x, y) =

lim

√ ±(−x + 2 |y|) = 0 = f (0, 0).

lim

0 = 0 = f (0, 0).

(x,y)→(0,0)

For D5 or D6 , we have lim

(x,y)→(0,0)

Hence,

lim

f (x, y) =

(x,y)→(0,0)

(x,y)→(0,0)

f (x, y) = 0 = f (0, 0) for any path passing through (0, 0), that is,

f (x, y) is continuous at the origin. Now let us find fy (x, 0). Any point (x0 , 0), x0 ≤ 0 lies in D5 and, consequently, all the corresponding points (x0 , y) also lie in D5 . For any point (x0 , 0), x0 > 0, located in D6 , there exists a small neighborhood such that all the corresponding points (x0 , y) also lie in D6 . In either case, the values of f (x, y) at (x0 , 0) and (x0 , y) f (x ,y)−f (x ,0) are zero, and therefore, fy (x0 , 0) = lim 0 y−0 0 = 0 for any x0 ∈ ℝ. y→0 ( ) 1 The next step is the calculation of F(y) = ∫−1 f (x, y)dx for ∀y ∈ − 14 , 14 . First, ( ) if y ∈ 0, 14 , then 1

F(y) =

∫−1

f (x, y)dx √ y

0

=

∫−1

0 dx +

∫0

√ 2 y

xdx +

√ (−x + 2 y)dx +

∫√y √ ( 2 ) 2√y y √ || x2 || x = + − + 2x y | = y. |√ 2 ||0 2 | y ( ) Similarly, for y ∈ − 14 , 0 , we get

1

∫2√y

0 dx

1

F(y) =

∫−1

f (x, y)dx √ |y|

0

=

∫−1

0 dx +

√ 2 |y|

(−x)dx +

√ (x − 2 |y|)dx +

∫0 ∫√|y| √ √ ( 2 )|2 |y| |y| √ x2 || x | =− | + − 2x |y| | = −|y| = y. |√ 2 |0 2 | |y|

1

∫2√|y|

0dx

4.2 Differentiability

Additionally, 1

F(0) =

∫−1

1

f (x, 0)dx =

∫−1

0 dx = 0.

( ) 1 Thus, F(y) = ∫−1 f (x, y)dx = y and, consequently, Fy (y) = 1 for ∀y ∈ − 14 , 14 . In 1 1 1 | d ∫−1 f (x, y)dx| = 1, but ∫−1 fy (x, 0)dx = ∫−1 0 dx = 0. particular, Fy (0) = dy |y=0 Remark to Examples 7–9. The important condition to ensure the validity of differentiation under the sign of integral is the continuity of fy (x, y) in the considered domain, which was not satisfied in all these three examples. However, this condition is sufficient and not necessary as shown in Examples 10 and 11. Example 10. Although f (x, y) has infinitely many points of discontinuity on b b d [a, b] × Y , but dy ∫a f (x, y)dx = ∫a fy (x, y)dx, ∀y ∈ Y . Solution Consider the function f (x, y) = y + R(x) S = [0, 1] × ℝ (recall that R(x) is the { on 1 , x = mn ∈ ℚ n Riemann function defined as R(x) = , where m is integer, n is 0, x ∈ 𝕀 m natural, and n is in lowest terms). It is known that R(x) is discontinuous at all rational points and continuous at all irrational points. It is also known that b R(x) is Riemann integrable on an arbitrary interval [a, b] and ∫a R(x)dx = 0 (see explanations on integrability of R(x) in Example 13 of Chapter 3). Using the last formula on [0, 1], one obtains 1

F(y) =

1

f (x, y)dx =

∫0 1

=

∫0

∫0

(y + R(x))dx

1

y dx +

∫0

R(x)dx = y, ∀y ∈ ℝ, 1

1

that is, Fy (y) = 1, ∀y ∈ ℝ. On the other hand, ∫0 fy (x, y)dx = ∫0 1 dx = 1. Thus, 1 1 d ∫ f (x, y)dx = 1 = ∫0 fy (x, y)dx, ∀y ∈ ℝ. dy 0 Example 11. Although fy (x, y) is b b | d ∫ f (x, y)dx| = ∫a fy (x, y0 )dx. dy a |y=y0 Solution

{

y3

discontinuous

at

(x0 , y0 ),

but

, x2 + y2 ≠ 0 x2 +y2 The function f (x, y) = possesses the partial derivative 0, x2 + y2 = 0 { 22 4 3x y +y , x2 + y2 ≠ 0 (x2 +y2 )2 fy (x, y) = . The function is continuous on ℝ2 : for any 2 1, x + y2 = 0

151

152

Chapter 4 Integrals Depending on a Parameter

(x, y) ≠ (0, 0), it follows straightforward from the arithmetic properties, and for (x, y) = (0, 0) the continuity follows from the fact that f (0, 0) = 0 and |y3 | evaluation |f (x, y)| = x2 +y2 ≤ |y| → 0. The derivative fy (x, y) is continuous y→0

on ℝ2 ⧵ (0, 0) due to the arithmetic properties of continuous functions, and discontinuous at the origin, because lim fy (x, 0) = 0 ≠ 1 = lim fy (0, y). x→0

y→0

Let us find the required integrals on [0, 1]. First, 1

F(y) =

∫0

1 x| 1 f (x, y)dx = y2 arctan || = y2 arctan , ∀y ≠ 0 y |0 y

and 1

F(0) =

∫0

1

f (x, 0)dx =

∫0

0 dx = 0.

Therefore, Fy (y) = 2y arctan

y2 1 − , ∀y ≠ 0 y 1 + y2

and Fy (0) = lim y→0

F(y) − F(0) 1 = lim y arctan = 0. y→0 y y

On the other hand, 1

1

1 2x2 y2 1 2 dx + y dx ∫0 (x2 + y2 )2 ∫0 x2 + y2 1 |1 xy2 || 2y2 x| =− 2 arctan | | + x + y2 ||0 y y ||0 y2 1 =− + 2y arctan , ∀y ≠ 0, 1 + y2 y

fy (x, y)dx =

∫0

and 1

∫0

1

fy (x, 0)dx =

∫0

0dx = 0.

Thus, 1

1

d f (x, y)dx = f (x, y)dx, ∀y ∈ ℝ, ∫0 y dy ∫0 in particular, Fy (0) =

1 1 | d | f (x, y)dx| = 0 = f (x, 0)dx. | ∫0 y dy ∫0 |y=0

4.2 Differentiability

{

y4 −y2 ∕x e ,x x2

>0 defined on 0, x = 0 X × Y = [0, +∞) × ℝ shows that both the function and its partial derivative can be discontinuous at a chosen point, but still the integration and partial differentiation can be interchanged. Indeed, the given function is discontinuous at (0, 0): choosing the path x = y2 , we get The counterexample with the function f (x, y) =

lim f (y2 , y) = lim y→0

y→0

y4 −y2 ∕y2 e = e−1 ≠ 0 = f (0, 0). y4

The partial derivative in y is defined in the considered domain: {( 3 ) 2 4y 2y5 − e−y ∕x , x > 0 2 3 x x fy (x, y) = , 0, x=0 but it is also discontinuous at (0, 0): using the same path x = y2 , we obtain ( 3 ) 4y 2y5 2 2 2 2 lim fy (y , y) = lim − 6 e−y ∕y = lim e−1 = ∞. 4 y→0 y→0 y→0 y y y Let us calculate the integrals: 1

F(y) =

∫0

y4 −y2 ∕x 2 2 e dx = y2 e−y ∕x |10 = y2 e−y 2 x

and, consequently 2

Fy (y) = (2y − 2y3 )e−y , ∀y ∈ ℝ. At the same time, 1

1

fy (x, 0)dx =

∫0

0dx = 0,

∫0

and for ∀y ≠ 0 1

1

1 4y3 −y2 ∕x 2y5 −y2 ∕x e dx − e dx ∫0 ∫0 x2 ∫0 x3 1 1 1 4y3 −y2 ∕x 2y3 −y2 ∕x || 2y3 −y2 ∕x = e dx − e − e dx | 2 | ∫0 x ∫0 x2 x |0 1 1 2y3 −y2 ∕x || 2 −y2 ∕x | = 2ye e | − | = (2y − 2y3 )e−y . |0 | x |0 To calculate the last integral, we have applied the integration by parts with 2 y2 u = 1x and dv = x2 e−y ∕x dx, and also l’Hospital’s rule for the limit

fy (x, y)dx =

lim

x→0+

2y3 −y2 ∕x 2y3 t 2y3 e = lim y2 t = lim 2 y2 t = 0. t→+∞ e t→+∞ y e x

153

154

Chapter 4 Integrals Depending on a Parameter

Hence, 1

1

d f (x, y)dx = f (x, y)dx, ∀y ∈ ℝ, ∫0 y dy ∫0 in particular, Fy (0) =

1 1 | d | f (x, y)dx| = 0 = f (x, 0)dx, | ∫0 y dy ∫0 |y=0

although both f (x, y) and fy (x, y) are discontinuous at (0, 0).

4.3 Integrability Example 12. A function f (x, y), defined on [a, b] × Y , converges on [a, b] to a function 𝜑(x) as y approaches y0 , and f (x, y) is not Riemann integrable on [a, b] for each fixed y ∈ Y , but 𝜑(x) is Riemann integrable on [a, b]. (This is an analogue to Example 10 for a sequence in Chapter 3.) {

Solution

y, x ∈ ℚ considered on [0, 1] × (0, 1] with y0 = 0 con0, x ∈ 𝕀 verges to 0: 𝜑(x) = lim f (x, y) = 0, ∀x ∈ [0, 1]. Therefore, the limit function is The function f (x, y) =

y→0

1

Riemann integrable on [0, 1]: ∫0 𝜑(x)dx = 0. Moreover, the convergence is uniform on [0, 1], because |f (x, y) − 𝜑(x)| ≤ y → 0. At the same time, the original y→0

function can be represented in the form f (x, y) = y ⋅ D(x), where D(x) is Dirichlet's function, which is not Riemann integrable on any interval. Therefore, f (x, y) is not Riemann integrable on [0, 1] for ∀y ∈ (0, 1]. Example 13. A function f (x, y) is defined on [a, b] × [c, d] and one of the d b b d iterated integrals—∫c dy ∫a f (x, y)dx or ∫a dx ∫c f (x, y)dy—exists, but another does not exist. Solution

{

2y3 −6x2 y , x2 + y2 (x2 +y2 )3 2 2

≠0 on [0, 1] × [0, 1]. Recall 0, x + y = 0 that the integral value does not depend on the specification of function at a singular point; therefore, we can calculate each integral regardless of the second part in the function definition. The calculation of the first iterated integral is as follows. For each fixed y ∈ [0, 1], we get Consider the function f (x, y) =

4.3 Integrability 1

2y3 − 6x2 y dx ∫0 (x2 + y2 )3 1 1 2y(y2 + x2 ) 8yx2 = dx − dx ∫0 (x2 + y2 )3 ∫0 (x2 + y2 )3 1 1 2y 2xy ||1 2y = dx + − dx | 2 2 2 2 2 2 ∫0 (x2 + y2 )2 ∫ (x + y ) |0 0 (x + y ) 2y = (1 + y2 )2

1

∫0 f (x, y)dx =

x (the integration by parts with u = x and dv = (x2 +y dx was applied). This result 2 )3 is valid for y = 0, in which case the function f (x, 0) = 0 and the corresponding integral is also zero. Then 1

1

dy

∫0

∫0

1

f (x, y)dx =

∫0

1 2y 1 || 1 dy = − = . | (1 + y2 )2 1 + y 2 |0 2

For the second iterated integral, we have at each fixed x ∈ (0, 1]: 1

1

2y3 − 6x2 y dy ∫0 ∫0 (x2 + y2 )3 1 1 2y 8yx2 = dy − dy ∫0 (x2 + y2 )2 ∫0 (x2 + y2 )3 1 |1 1 || 1 2 | =− 2 + 2x x + y2 ||0 (x2 + y2 )2 ||0 1 2x2 1 =− 2 + 2 − . x + 1 (x + 1)2 x2 { 2 ,y ≠ 0 y3 For x = 0, the integral of f (0, y) = diverges 0, y = 0 G(x) =

f (x, y)dy =

1

∫0

1

f (0, y)dy =

∫0

1 2 1 || dy = − = +∞, y3 y2 ||0

so, formally, the last result is included in the integral formula for x ∈ (0, 1], because the last term x12 approaches infinity as x approaches 0, whereas the first two terms have finite limits. Again, the mere fact that the function is not defined at some point in the interval of integration does not affect the existence or the value of the integral. However, since lim G(x) = ∞, the function G(x) is x→0

1

not Riemann integrable on [0, 1]. Moreover, the integral ∫0 G(x)dx does not exist even in the sense of the improper integral, because the functions x21+1 and 2x2 (x2 +1)2

1 1 dx x2

are Riemann integrable on [0, 1], whereas the improper integral ∫0 is divergent.

155

156

Chapter 4 Integrals Depending on a Parameter

Note that f (x, y) is continuous on ℝ2 ⧵ (0, 0). At the origin, the function is discontinuous, because lim f (0, y) = lim y23 = ∞. y→0

y→0

Remark. Of course, one can construct a function discontinuous at only one point of the considered{domain for which both iterated integrals do not exist. 2xy , x2 + y2 ≠ 0 (x2 +y2 )2 The function f (x, y) = considered on [0, 1] × [0, 1] is one 0, x2 + y2 = 0 of such examples. This function is continuous on ℝ2 except at the origin, where 2 lim f (x, x) = lim 2x = ∞. The first iterated integral is calculated as follows. For x→0 x→0 4x4 each fixed y ∈ (0, 1], we get 1

1

f (x, y)dx =

∫0

2xy y 1 dx = − + ∫0 (x2 + y2 )2 1 + y2 y

and at y = 0 1

f (x, 0)dx = 0.

∫0 Therefore, 1

1

1 y 1 dy + dy. 2 ∫0 ∫0 ∫0 ∫ 1+y 0 y The first integral in the right-hand side exists

dy

1

∫0

−

1

f (x, y)dx =

−

|1 y 1 1 2 | dy = − ln(1 + y ) | = − 2 ln 2, 1 + y2 2 |0 1

but the second diverges ∫0 1y dy = ln y|10 = +∞, which implies the divergence of the first iterated integral. Due to the symmetry of the function, the same is true for the second iterated integral. Example 14. A function f (x, y) is defined on [a, b] × [c, d] and both iterated d b b d integrals ∫c dy ∫a f (x, y)dx and ∫a dx ∫c f (x, y)dy exist, but they assume different values. Solution

{

x2 −y2 , x2 + y2 (x2 +y2 )2 2 2

≠0 on [0, 1] × [0, 1]. The first 0, x + y = 0 iterated integral can be calculated as follows: 1 1 1 1 x2 − y2 x2 + y2 2x2 f (x, y)dx = dx = dx − dx ∫0 ∫0 (x2 + y2 )2 ∫0 (x2 + y2 )2 ∫0 (x2 + y2 )2 1 1 1 x || dx dx 1 =− 2 + − =− | x + y2 |0 ∫0 x2 + y2 ∫0 x2 + y2 1 + y2 Consider the function f (x, y) =

4.3 Integrability

F(y) = ∫ 1 fdx 0 1 0

G(x) = ∫ fdy

P2

3 2

P1 = ∫ 1 dy∫ 1 fdx 0 1 0

0 1 0

P2 = ∫ dx∫ fdy

P1

1

f

0

F(y)

G(x)

–1 –2 –3 0

0 0.2

0.2

0.4

0.4

0.6

0.6

0.8

0.8

1

y

{ Figure 4.9 Example 14, function f (x, y) =

x

1

x 2 −y 2 , x2 + (x 2 +y 2 )2 2 2

y2 ≠ 0 . 0, x + y = 0

2x (we applied the integration by parts with u = x and dv = (x2 +y dx) and, 2 )2 consequently, 1 1 1 dy 𝜋 dy f (x, y)dx = − = − arctan y|10 = − . ∫0 ∫0 ∫0 1 + y2 4

For the second iterated integral, we have 1 1 x2 − y2 y ||1 1 f (x, y)dy = dy = = 2 , | 2 2 ∫0 ∫0 (x2 + y2 )2 x + y |0 x + 1 and 1

∫0

1

dx

∫0

1

f (x, y)dy =

dx 𝜋 = arctan x|10 = . ∫0 1 + x2 4

Here, in the evaluation of both iterated integrals, we used again the fact that the result of calculation of integral does not depend on the definition of a function at a singular point. Thus, the iterated integrals exist, but have different values. Note that like in Example 13 the function f (x, y) is continuous on −y2 2 ℝ ⧵ (0, 0) and has the essential discontinuity at the origin: lim f (0, y) = lim y4 y→0 y→0 = −∞.

157

158

Chapter 4 Integrals Depending on a Parameter

Remark to Examples 13 and 14. These examples show that the discontinuity of f (x, y) at only one point in the considered domain is sufficient to cause inequality of the iterated integrals or nonexistence of one of them (or even both). At the same time, the continuity of f (x, y) on the considered domain is not necessary condition for the existence and equality of the iterated integrals as shown in the next example. Example 15. A function f (x, y) is defined on [a, b] × [c, d] and d b b d ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy, but f (x, y) is discontinuous on [a, b] × [c, d]. {

Solution

xy2 , x2 + y2 (x2 +y2 )2 2 2

≠0 is continuous on ℝ2 ⧵ (0, 0) due 0, x + y = 0 to the arithmetic properties of continuous functions, and it has the essential 3 discontinuity at the origin, because lim f (x, x) = lim 4xx 4 = ∞. Let us find the x→0 x→0 iterated integrals, taking again into account that the function values at isolated points do not affect the value of the Riemann integral. For the first iterated integral, we get The function f (x, y) =

1

1

f (x, y)dx =

∫0

∫0

=−

1 xy2 y2 1 || dx = − | (x2 + y2 )2 2 x2 + y2 ||0

y2 1 1 + = 2(1 + y2 ) 2 2(1 + y2 )

and, therefore, 1

1

dy

∫0

∫0

1

f (x, y)dx =

dy 1 1 𝜋 = arctan y|10 = . 2 ∫0 1 + y2 2 8

For the second iterated integral, we have 1

|1 x 1 dy xy2 xy | + dy = − 2 2 2 2 ∫0 +y ) 2(x + y ) ||0 2 ∫0 x2 + y2 y |1 x 1 =− 2 + arctan || 2(x + 1) 2 x |0 x 1 1 =− 2 + arctan , ∀x ≠ 0 2(x + 1) 2 x 1

f (x, y)dy =

∫0

(x2

(the integration by parts with u = y and dv = 1

∫0

1

f (0, y)dy =

∫0

0 dy = 0.

y dy (x2 +y2 )2

was applied), and

4.3 Integrability

Therefore, 1

∫0

1

dx

∫0

1

1

1 xdx 1 1 + arctan dx 2 ∫0 1 + x2 2 ∫0 x 1 1 1 1 xdx x 1 || 1 xdx =− + arctan + | 2 ∫ ∫ 2 0 1+x 2 x |0 2 0 1 + x2 1 𝜋 = arctan 1 = 2 8

f (x, y)dy = −

(again the integration by parts was applied, at this time with u = arctan 1x and dv = dx). Thus, both iterated integrals exist and are equal, despite the fact that the function is not continuous on [0, 1] × [0, 1]. Remark. It can be shown a stronger result that the existence and equality of the iterated integrals are not sufficient to guarantee the existence of the double integral: a function f (x, y) is defined on [a, b] × [c, d] and both iterated integrals exist and are equal, but nevertheless the double integral does not exist. For a counterexample, consider the function f (x, y) defined on the unit square { 1, (x, y) ∈ T S = [0, 1] × [0, 1] as follows: f (x, y) = , where T consists of the 0, (x, y) ∉ T ( ) points of S such that (x, y) = mp , np , m, n, p ∈ ℕ, where both fractions are in lowest terms (note that the denominator is the same). First, let us show that 1 1 the iterated integral ∫0 dy ∫0 f (x, y)dx exists. In fact, if we choose any irra1 tional y0 , then f (x, y0 ) = 0 and ∫0 f (x, y0 )dx = 0. On the other hand, for a rational y0 = np (with the fraction in lowest terms), f (x, y0 ) is different from zero only in the finite number of the points, because for a given natural p there are at most p numbers of the form x = mp , m ∈ ℕ, in [0, 1]. Therefore, again 1

∫0 f (x, y0 )dx = 0. Since the inner integral is zero for any y0 , the iterated inte1 1 gral exists and ∫0 dy ∫0 f (x, y)dx = 0. Due to the symmetry of the function and domain with respect to interchange of variables, the same result is true for the 1 1 second iterated integral: ∫0 dx ∫0 f (x, y)dy = 0. Now let us prove that the double integral ∫ ∫S f (x, y)dA does not exist. To this end, let us consider the special uniform partitions obtained by division of each side of S in p equal parts, where p is a prime number. Note that the diameter of such partitions approaches zero as p approaches infinity. Each elementary square Rij = [xi−1 , xi ] × [yj−1 , yj ], i, j = 1, · · · , p contains the points (ci , dj ) ∉ T and also the points (ci , dj ) ∈ T. The possibility of the former choice is evident, since in any interval [xi−1 , xi ] (or [yj−1 , yj ]) there

159

160

Chapter 4 Integrals Depending on a Parameter

are irrational points. The latter choice can be made by using the points of the upper right corner in each Rij : ( ) i j (ci , dj ) = (xi , yj ) = , , i, j = 1, · · · , p − 1, p p except for the elementary squares of the upper row (j = p) and the right column (i = p), for which we choose ( ) i p−1 (ci , dp ) = , , i = 1, · · · , p − 1, p p ( ) p−1 j (cp , dj ) = , , j = 1, · · · , p − 1, p p and

( (cp , dp ) =

p−1 p−1 , p p

) .

The corresponding double Riemann sums S(f ; P) =

p p ∑ ∑

f (ci , dj ) ⋅ Δxi Δyj

j=1 i=1

are equal 0 in the case (ci , dj ) ∉ T and their partial limit is 0, but these sums are equal 1 if (ci , dj ) ∈ T and the corresponding partial limit is 1. Since two partial limits of the Riemann sums are different, there is no limit of the double Riemann sums as the partition diameter approaches 0, which means that the double integral does not exist. Example 16. A function f (x, y) has infinitely many discontinuity points in d b b d [a, b] × [c, d], but ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy. Solution Consider the function f (x, y) = sgn(x − y) on [0, 1] × [−2, 2]. It was shown in ⎧ 1, y < 0 ⎪ 1 Example 6 that F(y) = ∫0 f (x, y)dx = ⎨ 1 − 2y, 0 ≤ y ≤ 1 for ∀y ∈ ℝ. Therefore, ⎪ −1, y > 1 ⎩ 2

∫−2

1

dy

∫0

2

f (x, y)dx =

∫−2

F(y)dy

0

=

∫−2

1

1 dy +

∫0

2

(1 − 2y)dy +

= y|0−2 + (y − y2 )|10 − y|21 = 1.

∫1

(−1)dy

4.3 Integrability

On the other hand, 2

G(x) =

∫−2

2

x

f (x, y)dy =

∫−2

1 dy +

(−1)dy = y|x−2 − y|2x = 2x

∫x

and, consequently, 1

2

dx

∫0

1

f (x, y)dy =

∫−2

1

G(x)dx =

∫0

∫0

2x dx = x2 |10 = 1.

Thus, the two iterated integrals are equal. Nevertheless, f (x, y) is discontinuous at each point of the line y = x, x ∈ [0, 1]. Remark. One can strengthen the above example to the following form: a function f (x, y) is discontinuous and unbounded at infinitely many points d b b d in [a, b] × [c, d], but nevertheless ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy. ⎧ √y , y < x2 , x ≠ 0 ⎪ x A counterexample is provided by the function f (x, y) = ⎨ 2xy2 , y ≥ x2 on ⎪ 0, x = 0 ⎩ [0, 1] × [−2, 2]. In Remark 2 to Example 6, it was shown that ⎧ 2y, y1 ⎩

1

F(y) =

∫0

Therefore, 2

∫−2

1

dy

2

f (x, y)dx =

∫0

∫−2

F(y)dy

0

1

2y dy +

=

∫−2

=

y2 |0−2

( +

∫0

2

(y3 + 2y − 2y5∕4 )dy +

y4 8 + y2 − y9∕4 4 9

∫1

y2 dy

) |1 2 y3 || 47 | | + | =− . | | 3 |1 36 |0

For another iterated integral, one gets x2

2

G(x) =

∫−2

f (x, y)dy = 2 |x

=

∫−2

2 y 2xy2 dy √ dy + ∫ 2 x x

2 y2 | 2 3| 1 7∕2 16 2 −1∕2 + x − x7 , ∀x ≠ 0 √ || + xy || = x − 2x 3 2 2 3 3 |x 2 x ||−2

161

162

Chapter 4 Integrals Depending on a Parameter

and 2

G(0) =

∫−2

2

f (0, y)dy =

∫−2

0 dy = 0.

Then 1

∫0

2

dx

∫−2

1

f (x, y)dy =

G(x)dx

∫0

) 1 7∕2 16 2 x − 2x−1∕2 + x − x7 dx ∫0 2 3 3 ( )|1 1 9∕2 8 1 47 = x − 4x1∕2 + x2 − x8 || = − . 9 3 12 36 |0 Thus, the two iterated integrals are equal. However, as shown in Remark 2 to Example 6, f (x, y) is discontinuous and unbounded at each point of the segment x = 0, y ∈ [−2, 0]. 1

(

=

Exercises 3

1 Show that the function f (x, y) = 4xy2 (1 − x4 )1∕y defined on X × Y = [0, 1] × (0, 1] with y0 = 0 fulfills all the conditions of the statement in 1 1 Example 1, but ∫0 lim f (x, y)dx ≠ lim ∫0 f (x, y)dx. Check the character y→y0

2

y→y0

of the convergence of f (x, y) on X = [0, 1] as y approaches y0 . Do the 2xy2 same for f (x, y) = (x2 +y2 )2 defined on [0, 1] × (0, 1] with y0 = 0. 2xy

2 Verify whether the function f (x, y) = x4 +y2 satisfies the conditions of Examples 1 and 2 on X × Y = [0, 1] × (0, 1] with y0 = 0. What about 1 1 the relation between the integrals ∫0 lim f (x, y)dx and lim ∫0 f (x, y)dx? y→y0

y→y0

Analyze the nature of the convergence of f (x, y) on X = [0, 1] as y approaches y0 and monotonicity of f (x, y) in y on Y = (0, 1]. 3 Show that the functions 2xy a) f (x, y) = x4 +y2 , X × Y = [−1, 1] × (0, 1], y0 = 0, b) f (x, y) = arctan xy , X × Y = [0, 1] × (0, +∞), y0 = 0, 2xy3

c) f (x, y) = (x2 +y2 )2 , X × Y = [0, 1] × (0, 1], y0 = 0 provide counterexamples to Example 3. Verify which conditions are violated in the results (the theorem and corollary) on passing to the limit under the integral sign. 4 Prove that the following statement is false: “if f (x, y) is continuous on b b [a, b] × Y , then ∫a lim f (x, y)dx = lim ∫a f (x, y)dx.” y→y0

y→y0

Exercises

(Hint: use the counterexamples with one of the following functions: 2 a) f (x, y) = y12 e−x∕y , X × Y = [0, 1] × (0, 1], y0 = 0, [ ] x −x∕y2 𝜋 b) f (x, y) = cos e , X × Y = 0, × (0, 1], y0 = 0. y2 2 Check the character of the convergence of f (x, y) on X = [a, b] as y approaches y0 .) 5 Verify whether the following statement is false: “if f (x, y) is continuous b b on [a, b] × Y , and ∫a lim f (x, y)dx = lim ∫a f (x, y)dx, then there exists a y→y0

y→y0

finite limit lim f (x, y)dx for any fixed x ∈ [a, b].” y→y0

(Hint: construct a counterexample using the function f (x, y) = 1y e−x∕y , X × Y = [0, 1] × (0, 1], y0 = 0. Investigate the character of the convergence of f (x, y) on X = [0, 1] as y approaches y0 . Another counterexample: 2 try f (x, y) = sgnye−x∕y , X × Y = [0, 1] × [−1, 0) ∪ (0, 1], y0 = 0.) 2

6 Verify whether the following statement is false: “if f (x, y) is discontinuous b b at each point of [a, b] × Y , then ∫a lim f (x, y)dx ≠ lim ∫a f (x, y)dx.” y→y0

y→y0

(Hint: construct a counterexample using the function f (x, y) = D(y)e−x∕y , X × Y = [0, 1] × (0, 1], y0 = 0, where D(y) is Dirichlet’s function.) 7 Verify whether the following statement is false: “if f (x, y) is continuous on [a, b] × Y , there exists 𝜑(x) = lim f (x, y)dx for ∀x ∈ [a, b] y→y0

b

b

and ∫a lim f (x, y)dx = lim ∫a f (x, y)dx, then f (x, y) converges to 𝜑(x) y→y0

y→y0

uniformly on [a, b].” (Hint: consider one of the following functions: 2 a) f (x, y) = yx2 e−x∕y , X × Y = [0, 1] × (0, 1], y0 = 0, [ ] b) f (x, y) = siny x e−x∕y , X × Y = 0, 𝜋2 × (0, 1], y0 = 0, to construct a counterexample.) 8 Show that each following functions { of the 2xy(x2 +2) , x2 + y2 ≠ 0 x4 +y2 a) f (x, y) = 0, x2 + y2 = 0 { 2x −x2 ∕y2 e ,y ≠ 0 y2 b) f (x, y) = 0, y = 0 { 2xy2 , x2 + y2 ≠ 0 (x2 +y2 )2 c) f (x, y) = 2 0, x + y2 = 0 provides a counterexample to Examples 4 and 5. Consider [a, b] = [0, 1].

163

164

Chapter 4 Integrals Depending on a Parameter

{

2x, y < x2 , find such a set X × Y that can 2xy2 , y ≥ x2 be used to provide a counterexample to Example 6.

9 For the function f (x, y) =

10

Show that the statement 1 to Example 6 can be exemplified by { in Remark 2xy8∕3 2 , x + y2 ≠ 0 (x2 +y2 )2 the function f (x, y) = considered on [0, 1] × ℝ, and 2 2 0, x + y = 0 ⎧ √2y , y < x, x ≠ 0 ⎪ 3x the statement in Remark 2—by the function f (x, y) = ⎨ 3x2 y, y ≥ x ⎪ 0, x = 0 ⎩ considered on [0, 1] × ℝ.

11

Use the following functions to construct counterexamples to Examples 7 and 8: { 5 y −y4 ∕x2 e ,x ≠ 0 x3 a) f (x, y) = on X × Y = ℝ2 , 0, x = 0 { √ y y −y∕x2 e , x ≠ 0 on X × Y = ℝ × [0, +∞), x3 b) f (x, y) = 0, x = 0 { 2x −x2 ∕y2 e ,y ≠ 0 y c) f (x, y) = on X × Y = ℝ2 , 0, y = 0 { 2xy3 , x2 + y2 ≠ 0 4 +y4 x d) f (x, y) = on X × Y = ℝ2 , 0, x2 + y2 = 0 { 2xy2 , x2 + y2 ≠ 0 x4 +y4 e) f (x, y) = on X × Y = ℝ2 , 0, x2 + y2 = 0 { y3 −y2 ∕x e ,x > 0 x2 f ) f (x, y) = on X × Y = [0, +∞) × ℝ, 0, x = 0 { y2 −y2 ∕x e ,x > 0 x2 g) f (x, y) = on X × Y = [0, +∞) × ℝ, 0, x = 0 { y2 −y2 ∕x2 e ,x ≠ 0 x3 h) f (x, y) = on X × Y = ℝ2 . 0, x = 0 In all cases, consider [a, b] = [0, 1].

12

Show that the functions {following 2xy4 2 , x + y2 ≠ 0 x4 +y4 a) f (x, y) = on X × Y = ℝ2 , 0, x2 + y2 = 0 { 6 y −y4 ∕x2 e ,x ≠ 0 x3 b) f (x, y) = on X × Y = ℝ2 0, x = 0 are counterexamples to Example 11. Consider [a, b] = [0, 1].

Exercises

13 Use the functions { 2 a) f (x, y) = { b) f (x, y) = { c) f (x, y) = { d) f (x, y) = {

3xy −x3 , x2 + y2 (x2 +y2 )3 2 2

≠0 on X × Y = [0, 1]2 , 0, x + y = 0

x3 −xy2 , x2 + y2 (x2 +y2 )3 2 2

≠0 on X × Y = [0, 1]2 , 0, x + y = 0

2x−y ,x (x+y)4

+y>0 on X × Y = [0, 1]2 , 0, (x, y) = (0, 0) x2 y−y3 , x2 + y2 (x2 +y2 )3 2 2

≠0 on X × Y = [0, 1] × [0, 1], 0, x + y = 0

x−2y ,x (x+y)4

+y>0 on X × Y = [0, 1] × [0, 1] 0, (x, y) = (0, 0) to construct counterexamples to Example 13.

e) f (x, y) =

14 Prove that the following statement is false: “if f (x, y) is continuous on [a, b] × [c, d], except at one point, then there exists at least one of the d b b d integrals ∫c dy ∫a f (x, y)dx and ∫a dx ∫c f (x, y)dy.” (Hint: construct a counterexample with the function f (x, y) = { 4x−y ,x+ y > 0 (x+y)4 on X × Y = [0, 1] × [0, 1]. Another function is 0, (x, y) = (0, 0) { x−y ,x + y > 0 (x+y)4 f (x, y) = on X × Y = [0, 1] × [0, 1].) 0, (x, y) = (0, 0) {

y2 −2xy ,x (x+y)4

+y>0 on X × Y = [0, 1]2 is 0, (x, y) = (0, 0) a{counterexample to Example 14. Do the same for the function f (x, y) = x2 +2xy−2y2 ,x + y > 0 (x+y)4 on X × Y = [0, 1]2 . 0, (x, y) = (0, 0)

15 Show that the function f (x, y) =

16 Use the following functions to provide counterexamples to Example 15: { 2xy , x2 + y2 ≠ 0 x4 +y2 a) f (x, y) = on X × Y = [0, 1]2 , 0, x2 + y2 = 0 { 3 2 2 2x −x ∕y e ,y ≠ 0 y3 b) f (x, y) = on X × Y = [0, 1]2 , 0, y = 0 { 2 x −4xy+y2 ,x+ y > 0 (x+y)4 c) f (x, y) = on X × Y = [0, 1]2 , 0, x2 + y2 = 0

165

166

Chapter 4 Integrals Depending on a Parameter

{ d) f (x, y) = { e) f (x, y) = { f ) f (x, y) =

2x−y ,x (x+y)2 2

+y>0 on X × Y = [0, 1]2 , 0, x + y2 = 0

2xy3 , x2 + y2 (x2 +y2 )2 2 2

≠0 on X × Y = [0, 1]2 , 0, x + y = 0

x−3y ,x (x+y)2

+y>0 on X × Y = [0, 1]2 . 0, (x, y) = (0, 0) {

2x, y < x2 on [0, 1] × [−2, 2] 2xy2 , y ≥ x2 provides one more counterexample to Example 16.

17

Show that the function f (x, y) =

18

⎧ √2y , y < x, x ≠ 0 ⎪ 3x Show that the function f (x, y) = ⎨ 3x2 y, y ≥ x on [0, 1] × [−2, 2] is ⎪ 0, x = 0 ⎩ one more counterexample to Remark in Example 16.

Further Reading B.M. Budak and S.V. Fomin, Multiple Integrals, Field Theory and Series, Mir Publisher, Moscow, 1978. G.M. Fichtengolz, Differential- und Integralrechnung, Vol.1–3, V.E.B. Deutscher Verlag Wiss., Berlin, 1968. V.A. Ilyin and E.G. Poznyak, Fundamentals of Mathematical Analysis, Vol.1,2, Mir Publisher, Moscow, 1982. V.A. Zorich, Mathematical Analysis I, II, Springer, Berlin, 2004.

167

CHAPTER 5 Improper Integrals Depending on a Parameter 5.1 Pointwise, Absolute, and Uniform Convergence +∞

Example 1. An integral ∫a f (x, y)dx converges on a set Y , but it does not converge uniformly on this set. Solution +∞ Consider F(y) = ∫0 ye−xy dx on Y = (0, +∞). This integral converges on +∞ (0, +∞): F(y) = ∫0 ye−xy dx = −e−xy |+∞ = 1. To show that the convergence 0 is not uniform, let us evaluate the remainder of the integral for ∀A > 0 and yA = A1 : +∞

∫

yA e−xyA dx = −e−xyA |+∞ = e−A⋅1∕A = e−1 ↛ 0. A A→+∞

A +∞

Example 2. An integral ∫a f (x, y)dx converges on a (finite or infinite) interval Y = (c, d) and converges uniformly on any interval [c1 , d1 ] ⊂ (c, d), but it does not converge uniformly on (c, d). Solution +∞ Consider the integral of Example 1: F(y) = ∫0 ye−xy dx on Y = (0, +∞). It was shown that it converges on Y = (0, +∞), but nonuniformly. At the same time, the convergence is uniform on an interval [a, +∞) for any a > 0. Indeed, for A > 0 and simultaneously for all y ∈ [a, +∞), we get +∞

∫

ye−xy dx = −e−xy |+∞ = e−Ay ≤ e−Aa → 0. A A→+∞

A +∞

Example 3. An integral ∫a f (x, y)dx converges on a (finite or infinite) interval Y = (c, d), but it does not converge uniformly on an interval [c1 , d1 ] ⊂ (c, d). Counterexamples on Uniform Convergence: Sequences, Series, Functions, and Integrals, First Edition. Andrei Bourchtein and Ludmila Bourchtein. © 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. Companion website: www.wiley.com/go/bourchtein/counterexamples_on_uniform_convergence

168

Chapter 5 Improper Integrals Depending on a Parameter

f(x,y) = ye–xy

2

+∞ fdx 0

F(y) = ∫ 1.5

An fdx, 0

Fn(y) = ∫ f

1

An = 2n

Pn = Fn(yn) = 1 – e –1, yn =

1 An

0.5 F(y) 0 0 F3 P3 P2 F2 F1

1 x

2 3 4

0

P1

0.5

2

1.5

1

2.5

3

3.5

4

y

Figure 5.1 Examples 1, 2, 9, 10, 25, 28, 31, and 33, function f (x, y) = ye−xy .

Solution +∞ Consider F(y) = ∫0

sin xy dx x

on Y = ℝ. Note that the point x = 0 is not sinsin xy , because x sin xy lim xy ⋅ y = y, x→0

gular for the function f (x, y) =

for y = 0 we have f (x, 0) = 0, and

for y ≠ 0 we get lim f (x, y) = that is, at x = 0 the discontinuity x→0 is removable for any y ∈ ℝ. Let us show that the integral converges on ℝ. For +∞ y = 0, we have F(0) = ∫0 0dx = 0; for y > 0, changing the variable by the formula xy = t, we obtain +∞

+∞

sin xy sin t 𝜋 F(y) = dx = dt = ∫ ∫ x t 2 0

0

(the last integral is a known result); and for y < 0, we take y = −|y| and make a similar substitution x|y| = t to get +∞

+∞

sin x|y| sin t 𝜋 F(y) = − dx = − dt = − . ∫ ∫ x t 2 0

0

⎧ −𝜋∕2, y < 0⎫ ⎪ ⎪ Hence, F(y) = ⎨ 0, y = 0 ⎬ = 𝜋2 sgn y, ∀y ∈ ℝ. ⎪ 𝜋∕2, y > 0 ⎪ ⎩ ⎭

5.1 Pointwise, Absolute, and Uniform Convergence

f(x,y) =

sin xy x

+∞ fdx 0

F(y) = ∫ 2 1.5 1 0.5 f

0

F(y) = 2π , y > 0

–0.5 –1 –1.5 F(y) = – 2π , y < 0

–2 –4

0 2

–2

4

0 y

6

2 4

x

8

Figure 5.2 Examples 3, 5, 25, and 27, function f (x, y) =

sin xy . x

Now we consider an arbitrary interval Y1 = [c1 , d1 ] such that c1 < 0 < d1 and prove that the convergence on Y1 is not uniform. To assess the integral remainder, for any A > √1 we choose yA = A12 ∈ [c1 , d1 ] and obtain d1

+∞

+∞

sin xyA sin t dx = dt ∫ ∫ x t A

AyA +∞

+∞

sin t sin t 𝜋 = dt → dt = ≠ 0, ∫ A→+∞ ∫ t t 2 1∕A

0

that is, the convergence is not uniform on [c1 , d1 ]. Note that the convergence is uniform on any interval [c, +∞), c > 0 according A to Dirichlet’s theorem: the integral ∫0 sin xydx is uniformly bounded, since |A | |( ) |A | | | | 1 2 2 | | | | ≤ , | sin xydx| = || − cos xy | || ≤ |∫ | | | | |y| y c |0 | | |0 | | | for ∀A > 0 and ∀y ∈ [c, +∞), and the function monotonically and independently of y.

1 x

approaches zero, as x → +∞,

169

170

Chapter 5 Improper Integrals Depending on a Parameter

2 F(y) = ∫0+∞fdx,

f=

1.5

Fn(y) = ∫ An fdx, An = 2n–1 0

Pn = ( yn, ∫A+∞ sinx xyn dx), n

yn =

Qn = ( ynmax, Fnmax)

–4

–3

–2

Q3

sin xy x

1 A2n

P3 P2

1

F1

Q1 F(y) = 2π

P1

0.5

–1

Q2

F(0) = 0 1

2

3

4

–0.5 F2 –1 F3 –1.5

F(y) = –

π 2

–2 +∞ sin xy dx. x

Figure 5.3 Examples 3, 5, 25, and 27, integral F(y) = ∫0 +∞

Example 4. An integral F(y) = ∫a f (x, y)dx converges absolutely on a (finite or infinite) interval Y = (c, d), but it does not converge uniformly on this interval. Solution +∞ Consider F(y) = ∫0 e−xy sin xdx on Y = (0, +∞). Since |f (x, y)| = |e−xy sin x| ≤ +∞ |+∞ e−xy , ∀x ∈ [0, +∞), and the integral ∫0 e−xy dx = − 1y e−xy | = 1y converges |0 on (0, +∞), it follows that the original integral converges absolutely on Y = (0, +∞). However, we will see that the convergence is not uniform. The antiderivative of f (x, y) can be found by integrating by parts twice: ∫

e−xy sin xdx = −e−xy cos x − y

∫

e−xy cos xdx

= −e−xy cos x − ye−xy sin x − y2

∫

e−xy sin xdx

(u = e−xy , dv = sin xdx is used the first time and u = e−xy , dv = cos xdx - the second). Therefore, −e−xy cos x − ye−xy sin x e−xy sin xdx = . ∫ 1 + y2 Then the evaluation of the remainder goes as follows. For any A > 0 and corresponding yA = A1 ∈ (0, +∞), we obtain

5.1 Pointwise, Absolute, and Uniform Convergence

1 +∞ fdx = 1 2 , 0 1+y

F(y) = ∫ 0.8

Fn(y) = ∫ An fdx = 0

0.6

f(x,y) = e–xy sin x

1 (1– e–Any), 1 + y2

An = 2nπ

f 0.4

F3 F2

0.2

F1

0 0 1 x

2

F(y) 3 4

0.5

0

1

1.5

2

2.5

3

4

3.5

y

Figure 5.4 Examples 4, 26 (second counterexample), and 34, function f (x, y) = e−xy sin x.

+∞

∫

e

−xyA

sin xdx = −e

A

−xyA

+∞ cos x + yA sin x || = e−1 | | 1 + y2A |A

(

1

sin A cos A + A 1 1 + A2 1 + A12

)

cos x+y sin x lim e−xyA 1+yA2 = 0). Note now x→+∞ A cos A 0, whereas lim 1+ does not exist (for the sequence 1 A→+∞ A2 cos An we have lim 1+ 1 = 1, but for Ak = (2k + 1)𝜋 → +∞ n→+∞ k→∞ 2

(the second term disappears since that lim

1 A

sin A

1 A→+∞ 1+ A2

=

An = 2n𝜋 → +∞, n→∞

cos Ak 1 k→+∞ 1+ A2

the result is different lim

An

= −1). Concluding, there is no limit of

k

+∞

∫A e−xyA sin xdx as A approaches infinity, which means that the integral +∞ ∫0 e−xy sin xdx does not converge uniformly on Y = (0, +∞).

+∞

Example 5. An integral ∫a f (x, y)dx converges uniformly on an interval Y , but it does not converge absolutely on this interval. Solution +∞ sin xy In Example 3, it was shown that ∫0 dx converges uniformly on any interx val Y = [c, +∞), c > 0. This implies the uniform convergence of the integral +∞ sin xy F(y) = ∫𝜋∕2 x dx on Y = [1, +∞).

171

172

Chapter 5 Improper Integrals Depending on a Parameter

1.5

F(y) = ∫ +∞fdx, f = e–xysin x 0

Fk(y)

An fdx, 0

Fn(y) = ∫

Ak fdx 0

Fk(y) = ∫

An = 2nπ, Ak = (2k + 1)π

1

rn (y) = F(y) – Fn(y), rk(y) = F(y) – Fk(y) Fn(y)

Pn = (yn, rn(yn)) = (

Pn

Pk = (yk, rk(yk)) = ( A1 , –

0.5

1 An k

rn(y)

0.5 rk(y)

,

)

e–1 1+1/An2

e–1 1+1/Ak2

) F(y)

1

1.5

2

2.5

3

3.5

Pk –0.5

Figure 5.5 Examples 4, 26 (second counterexample), and 34, integral +∞ F(y) = ∫0 e−xy sin xdx.

+∞ | sin xy|

+∞ cos 2xy

Let us analyze the behavior of ∫𝜋∕2 x dx. Note that ∫𝜋∕2 x dx converges (even uniformly) on Y = [1, +∞) by Dirichlet’s theorem: |A | | | | 1 |A || | | | | cos 2xydx| = | sin 2xy|| | |∫ | | 2y |𝜋∕2 || |𝜋∕2 | | | | 1 𝜋 ≤ | sin 2Ay − sin 𝜋y| ≤ 1, ∀A > , ∀y ∈ Y , 2 2 1 and x approaches 0 monotonically as x approaches infinity. Together with the inequality | sin xy| ≥ sin2 xy = 12 (1 − cos 2xy) and diver+∞ +∞ | sin xy| gence of ∫𝜋∕2 2x1 dx, it proves the divergence of ∫𝜋∕2 x dx, that is, the original integral converges uniformly but not absolutely. +∞

Example 6. An integral ∫a f (x, y)dx converges uniformly and absolutely on +∞ an interval Y , but ∫a | f (x, y)|dx does not converge uniformly on this interval. Solution +∞ x Consider F(y) = ∫1 cos dx on Y = (1, +∞). Since | cos x| ≤ 1 and the integral xy +∞ 1−y +∞ 1 x | 1 ∫1 xy dx = 1−y | = y−1 converges for ∀y ∈ Y , it follows that the integral |1

5.1 Pointwise, Absolute, and Uniform Convergence x| ∫1 | cos dx also converges, that is, the original integral converges absolutely xy for ∀y ∈ Y . The uniformity of the convergence follows from Dirichlet’s theorem: first, |A | | | | | | cos xdx| = |sin x|A1 | = |sin A − sin 1| ≤ 2, ∀A ∈ [1, +∞), ∀y ∈ (1, +∞) |∫ | | 1 | | | +∞

and, second, the function x1y is continuous and monotone in x on [1, +∞) for any fixed y ∈ Y and also approaches 0, as x approaches infinity, uniformly on Y , because 0 < x1y < 1x → 0 simultaneously for all y ∈ Y . x→+∞ +∞

x| Now let us show that ∫1 | cos dx does not converge uniformly on Y . In xy effect, it is sufficient to note that | cos x| ≥ cos2 x = 12 (1 + cos 2x) and the inte+∞ gral ∫1 cosxy2x dx converges uniformly on Y (it can be shown in the same way +∞ as for the original integral), whereas the integral ∫1 x1y dx converges on Y , but not uniformly. The last can be seen by evaluating the remainder with ∀A > 1 and yA = 1 + A1 ∈ Y : +∞

+∞ 1 x1−yA || A1−yA dx = = = A1−1∕A → +∞. | A→+∞ ∫ x yA 1 − yA |A yA − 1 A

+∞

Example 7. An integral ∫a f (x, y)dx converges uniformly and absolutely on Y , but there is no bound of f (x, y) such that |f (x, y)| ≤ 𝜑(x), ∀y ∈ Y , where the +∞ improper integral ∫a 𝜑(x)dx converges. {

Solution

1 ,x x

∈ [y, y + 1] defined on [1, +∞) × [1, +∞) is 0, otherwise nonnegative, so the convergence and absolute convergence of the integral +∞ ∫1 f (x, y)dx is the same thing for this function. Direct calculations The function f (x, y) =

y+1

+∞

∫

f (x, y)dx =

1

y+1 1 y+1 dx = ln x|y = ln ∫ x y y

show that the integral converges for each y ∈ [1, +∞). Noting that for each fixed y it holds f (x, y) ≤ 1x and f (x, y) is 0 outside the interval [y, y + 1] of the unit length, one can prove the uniformity of the convergence on Y = [1, +∞) applying the Cauchy criterion: for ∀𝜖 > 0, one can choose A𝜖 = 1𝜖 and arbitrary A2 > A

A1 > A𝜖 to obtain ∫A 2 f (x, y)dx < A1 = 𝜖 simultaneously for all y ∈ [1, +∞). 1 𝜖 At the same time, the function f (x, y) does not admit majoration on Y = [1, +∞) in the form |f (x, y)| ≤ 𝜑(x) such that the improper integral

173

174

Chapter 5 Improper Integrals Depending on a Parameter

3 y=1

x=y

2.5

2 x=y+1 f(x,y) = 1x

f(x,y) = 0

1.5

f(x,y) = 0

x=1

1 1

1.5

2

2.5

3

Figure 5.6 Example 7, domain of the function f (x, y).

+∞

∫a

𝜑(x)dx converges. In fact, the function definition can be rewritten in the {1 , y ∈ [x − 1, x] x form f (x, y) = . Therefore, for each fixed x, the lowest upper 0, otherwise bound of |f (x, y)| is the value 1x that f (x, y) assumes on [x − 1, x]. Thus, the smallest function 𝜑(x) for which |f (x, y)| ≤ 𝜑(x), ∀y ∈ Y is 𝜑(x) = 1x . However, +∞ the improper integral ∫1 1x dx diverges. Remark. The converse general statement is true and represents the Weierstrass test for improper integrals. Example 8. Suppose functions f (x, y) and g(x, y) are positive and conf (x,y) tinuous on [a, +∞) × Y , and lim g(x,y) = 1, ∀y ∈ Y . One of the integrals +∞

x→+∞

+∞

∫a f (x, y)dx or ∫a g(x, y)dx converges uniformly on Y , but another converges nonuniformly. Solution Consider f (x, y) =

2y2 x4 +y4

and g(x, y) =

2y2 x4 +y2

on [1, +∞) × ℝ. Both functions are f (x,y) x4 +y2 = lim x4 +y4 = 1, g(x,y) x→+∞ x→+∞ +∞ ℝ, and ∫1 x12 dx converges, it

positive and continuous on the chosen domain and lim ∀y ∈ ℝ. Since f (x, y) =

2y2 x4 +y4

=

1 2x2 y2 x2 x4 +y4

≤

1 , ∀y x2

∈

5.1 Pointwise, Absolute, and Uniform Convergence

+∞ y+1 fdx = ln y 1

F(y) = ∫ 1 f = 1x ,

0.8

y≤x≤y+1

0.6 f

0.4 F(y)

0.2 0 1

f = 0, x > y + 1 1.5

2 y

2.5

f = 0, x < y

3 3.5

4

3.5

4

{1 Figure 5.7 Example 7, function f (x, y) =

x

2.5 x

3

2

1.5

1

, x ∈ [y, y + 1] . 0, otherwise

+∞

+∞

2y2

follows that the integral ∫1 f (x, y)dx = ∫1 x4 +y4 dx converges uniformly on ℝ. The second integral converges pointwise on ℝ, because for any fixed y ∈ ℝ +∞ 2y2 2y2 it holds |g(x, y)| = x4 +y2 ≤ x4 and the integral ∫1 x14 dx converges. However, this convergence is nonuniform on ℝ according to the Cauchy criterion: for ∀A > 1, one can choose B = A + 1 and yA = (A + 1)2 to obtain |A+1 | A+1 | | 2y2A 2y2A | | dx > (A + 1 − A) | g(x, yA )dx| = |∫ | ∫ x4 + y2 (A + 1)4 + y2A |A | A A | | 2(A + 1)4 = = 1 ↛ 0. A→+∞ 2(A + 1)4

Remark. The following comparison theorem for the pointwise convergence of improper integrals is true: if for any fixed y ∈ Y , the functions f (x, y) and g(x, y) f (x,y) are positive on [a, +∞) and lim g(x,y) = C = const ≠ 0 (in particular, C = 1), +∞

x→+∞

+∞

then both integrals ∫a f (x, y)dx and ∫a g(x, y)dx are pointwise convergent or divergent. Example 8 shows that this property cannot be extended to uniform convergence.

175

176

Chapter 5 Improper Integrals Depending on a Parameter

5.2 Convergence of the Sum and Product +∞

+∞

Example 9. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge +∞ nonuniformly on Y , but the integral ∫a f (x, y) + g(x, y)dx converges uniformly on Y . Solution The idea of a trivial counterexample is simple. One can choose f (x, y), which gives rise to a nonuniformly convergent integral and then use g(x, y) = −f (x, y). Just to not stick with the zero function f (x, y) + g(x, y), one can choose any uniformly convergent integral and use the function h(x, y) in such an integral to define g(x, y) = h(x, y) − f (x, y). One of such examples with nonzero function h(x, y) is provided below. Consider the functions f (x, y) = ye−xy and g(x, y) = (y2 − y)e−xy on [0, +∞) × (0, +∞). The first improper integral converges +∞

∫

ye−xy dx = −e−xy |+∞ = 1, ∀y ∈ (0, +∞), 0

0

but this convergence is not uniform on (0, +∞), since for yA =

1 A

one has

+∞

∫

yA e−xyA dx = −e−xyA |+∞ = e−1 ↛ 0. A A→+∞

A

Similarly, +∞

∫

(y2 − y)e−xy dx = −(y − 1)e−xy |+∞ = y − 1, ∀y ∈ (0, +∞), 0

0

but for yA = A1 , +∞

∫

(y2A − yA )e−xyA dx = −(yA − 1)e−xyA |+∞ A

A

(

) 1 − 1 e−1 → −e−1 ≠ 0, A→+∞ A that is, the second improper integral converges nonuniformly on (0, +∞). The integral of f (x, y) + g(x, y) also converges on (0, +∞) (due to the arithmetic properties of improper integrals or direct calculations): =

+∞

∫ 0

y2 e−xy dx = −ye−xy |+∞ = y, ∀y ∈ (0, +∞). 0

5.2 Convergence of the Sum and Product

However, contrary to the first two integrals, this convergence is uniform on +∞ (0, +∞). This can be shown by evaluating the residual ∫A y2 e−xy dx = ye−Ay . Note that the critical point equation for the last function (ye−Ay )y = (1 − Ay)e−Ay = 0 gives the only critical point yA = A1 , which is also the maximum point in y. Since the function ye−Ay is positive for ∀y ∈ (0, +∞), ∀A > 0, the following evaluation holds +∞

∫

y2 e−xy dx = ye−Ay ≤ sup ye−Ay = (0,+∞)

1 −1 e → 0, A→+∞ A

A

which means the uniform convergence.

+∞

Remark 1. An example of nonuniformly convergent integral ∫a f (x, y) + +∞ +∞ g(x, y)dx when both improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converges nonuniformly on Y can also be provided. As a trivial counterexample, one can choose f (x, y) = g(x, y). In a bit more elaborated example, one can pick up any function h(x, y), which generates uniformly convergent integral and define g(x, y) = f (x, y) + h(x, y). For instance, f (x, y) = ye−xy and g(x, y) = ye−xy + y2 e−xy is one of such counterexamples.

f(x,y) = y2e–xy

2

0

An fdx 0

Fn(y) = ∫ f

F(y)

F(y) = ∫ +∞fdx = y

1.5

1

= y(1 – e – Any),

An = 2n–1

0.5 0 0 F3 F2 F1

1 x

2 3 4

0

0.5

1

1.5

2 y

Figure 5.8 Examples 9 and 10, function f (x, y) = y2 e−xy .

2.5

3

3.5

4

177

178

Chapter 5 Improper Integrals Depending on a Parameter

Remark 2. The same results are true for the difference of nonuniformly convergent integrals.

+∞

+∞

Example 10. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge +∞ nonuniformly on Y , but the integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . Solution +∞ It was shown in Example 9 that the improper integral ∫0 ye−xy dx converges nonuniformly on (0, +∞) to 1. In a similar way, it can be shown that the integral +∞ √ −xy ∫0 ye dx is also nonuniformly convergent on (0, +∞): |+∞ √ −xy 1 −xy || 1 ye dx = − √ e | = √ , ∀y ∈ (0, +∞), ∫ | y y |0 0 +∞

and for yA =

1 A

+∞

√ −xy e−AyA √ yA e A dx = √ = Ae−1 → +∞. A→+∞ ∫ yA A

1 +∞ fdx, f = y2e–xy 0 A Fn(y) = ∫ n fdx, An = 2n–1 0 rn(y) = ∫ +∞ fdx An

F(y) = ∫

F3

0.8

F2

Pn = (yn, rn(yn)) = ( A1 , A1 e–1)

0.6

n

n

F1 P1

0.4

r1

P2

0.2 P3

r2 r3 0.5

1

1.5

2 +∞

Figure 5.9 Examples 9 and 10, integral F(y) = ∫0

y 2 e−xy dx.

2.5

3

5.2 Convergence of the Sum and Product

Nevertheless, the integral +∞

∫

+∞

f (x, y)g(x, y)dx =

0

∫

y3∕2 e−2xy dx

0

converges uniformly on (0, +∞). The convergence can be verified by the direct calculation: +∞

∫

y3∕2 e−2xy dx = −

0

+∞ 1 √ −2xy || 1√ ye | = y, ∀y ∈ (0, +∞). 2 2 |0

1 To evaluate the character of the convergence, note that yA = 4A is the only √ −2Ay maximum point in y of the function ye , which is seen from the criti( √ −2Ay √ ) −2Ay 1 cal point equation ( ye )y = 2 √y − 2A y e = 0. Since the function √ −2Ay ye is positive for ∀y ∈ (0, +∞), ∀A > 0, the following evaluation for the integral residual holds: +∞

∫ A

y3∕2 e−2xy dx =

1 √ −2Ay 1 √ −2AyA 1 1 −1∕2 ye ≤ y e = √ e → 0, A→+∞ 2 2 A 2 4A

which means the uniform convergence. +∞

Remark 1. The following example also takes place: an integral ∫a f (x, y)dx +∞ converges nonuniformly on Y , but the integral ∫a f 2 (x, y)dx converges +∞ uniformly on Y . For instance, it was shown in Example 9 that ∫0 ye−xy dx +∞ 2 −2xy converges nonuniformly on (0, +∞). However, the integral ∫0 y e dx converges uniformly on the same interval that can be shown in the same way +∞ as for ∫0 y2 e−xy dx in Example 9. +∞

+∞

Remark 2. It is also not true that if ∫a f (x, y)dx and ∫a g(x, y)dx converge +∞ nonuniformly on Y , then the integral ∫a f (x, y)g(x, y)dx converges uniformly √ on Y . For a counterexample, consider the functions f (x, y) = g(x, y) = ye−xy . √ +∞ It is shown in the last example, that ∫0 ye−xy dx converges nonuniformly on +∞ (0, +∞), and the nonuniform convergence of the integral ∫0 f (x, y)g(x, y)dx = +∞ ∫0 ye−2xy dx can be shown in the same way as in Example 9. +∞

+∞

Example 11. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge +∞ uniformly on Y , but the integral ∫a f (x, y)g(x, y)dx converges nonuniformly on Y .

179

180

Chapter 5 Improper Integrals Depending on a Parameter

Solution Consider f (x, y) = g(x, y) =

sin x −xy √ e x

on [1, +∞) × (0, +∞). The improper inte-

+∞ x √ e−xy dx converges uniformly on (0, +∞) by Abel’s theorem, since gral ∫1 sin x +∞ x √ dx converges (uniformly, because it does not depend on y) the integral ∫1 sin x −xy

and the function e is monotone in x ∈ [1, +∞) for any fixed y ∈ (0, +∞) and is also uniformly bounded: |e−xy | ≤ 1, ∀x ∈ [1, +∞), ∀y ∈ (0, +∞). However, the integral +∞

∫ 1

+∞

f (x, y)g(x, y)dx =

sin2 x −2xy e dx ∫ x 1 +∞

+∞

1 1 −2xy 1 cos 2x −2xy = e dx − e dx 2∫ x 2∫ x 1

1

converges nonuniformly on (0, +∞), since the first integral in the right-hand side converges nonuniformly and the second - uniformly on (0, +∞). In fact, the uniform convergence of the second integral follows from Abel’s +∞ theorem: the integral ∫1 cosx2x dx converges uniformly on (0, +∞) and e−2xy is monotone in x ∈ [1, +∞) for any fixed y ∈ (0, +∞) and is also uniformly bounded (|e−2xy | ≤ 1 on [1, +∞) × (0, +∞)). The first integral converges since +∞ 0 < 1x e−2xy ≤ e−2xy , ∀x ∈ [1, +∞) and the integral ∫1 e−2xy dx converges. To prove that the convergence is not uniform, for any A > 1 and corresponding 1 B = 2A and yA = 2A , we perform the following evaluations: 2A |B | 2A | | 1 −2xyA 1 | 1 −2xy | e dx| = e dx ≥ e−2xyA dx | |∫ x | ∫ x ∫ 2A |A | A A | | 1 1 −2AyA −4AyA = (e −e ) = (e−1 − e−2 ) ↛ 0. A→+∞ 4AyA 2

According to the Cauchy criterion, it means that the convergence is nonuniform on (0, +∞). Remark 1. The given counterexample can also be used for the following +∞ example: an integral ∫a f (x, y)dx converges uniformly on Y , but the integral +∞ ∫a f 2 (x, y)dx does not converge uniformly on Y . Remark 2. One can also show that the following statement is false: if +∞ +∞ ∫a f (x, y)dx and ∫a g(x, y)dx converge uniformly on Y , then the integral +∞ ∫a f (x, y)g(x, y)dx converges nonuniformly on Y . For a counterexample, consider the functions f (x, y) = g(x, y) = sinx x e−xy on [1, +∞) × (0, +∞). The

5.2 Convergence of the Sum and Product +∞

integral ∫1 sinx x e−xy dx converges uniformly on (0, +∞) by Abel’s theorem: +∞ the integral ∫1 sinx x dx converges (uniformly, because it does not depend on y) and the function e−xy is monotone in x ∈ [1, +∞) for any fixed y ∈ (0, +∞) and is also uniformly bounded—|e−xy | ≤ 1 on [1, +∞) × (0, +∞). The integral +∞

∫

+∞

f 2 (x, y)dx =

sin2 x −2xy e dx ∫ x2 1

a

2

also converges uniformly by the Weierstrass test: the evaluation 0 ≤ sinx2 x e−2xy ≤ +∞ 1 holds for ∀x ∈ [1, +∞), ∀y ∈ (0, +∞) and the majorant integral ∫1 x12 dx x2 converges. Remark 3. The following general statement for the sum and difference is true: +∞ +∞ if improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge uniformly on Y , +∞ then the integral ∫a f (x, y) ± g(x, y)dx also converges uniformly on Y . +∞

Example 12. An improper integral ∫a f (x, y)dx converges uniformly and +∞ an integral ∫a g(x, y)dx converges nonuniformly on Y , but the integral +∞ ∫a f (x, y)g(x, y)dx converges uniformly on Y . Solution Consider f (x, y) = sinx x and g(x, y) = e−xy on [1, +∞) × (0, +∞). The improper +∞ integral ∫1 sinx x dx converges uniformly: it does not depend on y and its con| A | vergence is guaranteed by Dirichlet’s theorem, since |∫1 sin xdx| ≤ 2, ∀A > 1 | | and 1x converges monotonically to 0 as x approaches +∞. On the other hand, the +∞ integral ∫1 e−xy dx converges nonuniformly on (0, +∞), which can be proved in the same way as for similar integrals in Example 9: the direct calculation +∞

|+∞ e−y 1 e−xy dx = − e−xy || = , ∀y ∈ (0, +∞) ∫ y y |1 1

shows the convergence, and the residual evaluation with yA = +∞

∫ A

e−xyA dx =

e−AyA = Ae−1 → +∞ A→+∞ yA

1 A

181

182

Chapter 5 Improper Integrals Depending on a Parameter

reveals that this convergence is not uniform on (0, +∞). At the same time, the integral +∞

∫

+∞

f (x, y)g(x, y)dx =

1

∫

sin x −xy e dx x

1

converges uniformly on (0, +∞) (see Remark 2 to Example 11 for details). Remark 1. An example with the opposite conclusion also takes place: +∞ an improper integral ∫a f (x, y)dx converges uniformly and an integral +∞ +∞ ∫a g(x, y)dx converges nonuniformly on Y , but the integral ∫a f (x, y)g(x, y)dx converges nonuniformly on Y . It can be illustrated using the functions f (x, y) = sinx x and g(x, y) = xe−xy on [1, +∞) × (0, +∞). In fact, as shown above, +∞ ∫1 sinx x dx converges uniformly. The convergence of the second integral can be proved by integrating by parts: +∞

∫

+∞

xe

1

−xy

|+∞ 1 x 1 1 dx = − e−xy || + e−xy dx = e−y + 2 e−y , ∀y ∈ (0, +∞), ∫ y y y y |1 1

but the evaluation of the integral residual with yA = (

+∞

∫ A

xe

−xyA

dx =

1 A

) A 1 + yA y2A

e−AyA = 2A2 e−1 → +∞ A→+∞

reveals that this convergence is not uniform on (0, +∞). At the same time, the integral +∞

∫ 1

+∞

f (x, y)g(x, y)dx =

∫

sin xe−xy dx

1

converges nonuniformly on (0, +∞), as was shown in Example 4. Remark 2. The following general statement is true for the sum and difference: if +∞ +∞ an improper integral ∫a f (x, y)dx converges uniformly and ∫a g(x, y)dx con+∞ verges nonuniformly on Y , then the integral ∫a f (x, y) ± g(x, y)dx converges nonuniformly on Y . +∞

+∞

Example 13. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx diverge +∞ on Y , but nevertheless the integral ∫a f (x, y)g(x, y)dx converges uniformly on Y .

5.2 Convergence of the Sum and Product

Solution Consider f (x, y) = y sin xy and g(x, y) = improper integral, we have

1 xy

on [1, +∞) × [1, +∞). For the first

+∞

y sin xydx = − cos xy|+∞ = cos y − lim cos xy. 1

∫

x→+∞

1

The last limit does not exist for any y0 ∈ [1, +∞): choosing the sequence xn = 2n𝜋 → +∞, we get lim cos xn y0 = 1, while another sequence y n→+∞

0

n→+∞

xm = (2m+1)𝜋 → +∞ gives different result lim cos xm y0 = −1. Therey0 m→+∞ m→+∞ fore, the first integral diverges for ∀y ∈ [1, +∞). The second integral diverges as well: +∞

|+∞ 1 1 dx = ln x|| → +∞, ∀y ∈ [1, +∞). ∫ xy y |1 x→+∞ 1

Nevertheless, the integral +∞

∫

+∞

f (x, y)g(x, y)dx =

sin xy dx x

∫

1

1

converges uniformly on [1, +∞) by Dirichlet’s theorem: first, the integral of sin xy is uniformly bounded |A | | | | 1 |A || 2 | | | | sin xydx| = | − cos xy|| | ≤ ≤ 2, ∀A > 1, ∀y ∈ [1, +∞), |∫ | | y |1 || y |1 | | | | and, second, the function [1, +∞)) to 0 as x → +∞.

1 x

converges monotonically (and uniformly on

+∞

+∞

Example 14. Improper integrals ∫a f (x, y)dx and ∫a g(x, y)dx converge +∞ uniformly on Y , but the integral ∫a f (x, y)g(x, y)dx diverges on Y . Solution Consider f (x, y) = g(x, y) =

sin xy √ x

on [1, +∞) × [1, +∞). The uniform con-

+∞ sin xy √ dx on [1, +∞) can be shown by x | A | applying Dirichlet’s theorem: the integral |∫1 sin xydx| is uniformly bounded | | on A > 1 and y ∈ [1, +∞) (see details in Example 13), and the function √1 x

vergence of the improper integral ∫1

converges monotonically (and uniformly on [1, +∞)) to 0 as x → +∞.

183

184

Chapter 5 Improper Integrals Depending on a Parameter

1.6

P2 P1 P3 P4

1.4 1.2

Zero level error ∣rn(y)∣= ∣F(y) – Fn(y)∣+1

Pn = (ynmax, ∣rn∣max)

1 0.8

sin xy √x

F(y) = ∫ +∞fdx, f = 1

0.6

Fn(y) = ∫ An fdx, An = 2n 1

0.4 0.2 1

2

–0.2 –0.4

3 F3 F2 F4 F1

4

5

6

7

8

9

+∞ sin xy √ dx. x

Figure 5.10 Example 14, integral F(y) = ∫1

However, the integral +∞

+∞

+∞

+∞

1

1

1

1

sin2 xy cos 2xy 1 1 1 f (x, y)g(x, y)dx = dx = dx − dx ∫ ∫ x 2∫ x 2∫ x diverges for ∀y ∈ [1, +∞). In fact, the second integral in the right-hand side converges uniformly on [1, +∞) according to Dirichlet’s theorem: first, the integral of cos 2xy is uniformly bounded |A | | | | 1 |A || | | | | cos 2xydx| = | sin 2xy|| | |∫ | | 2y |1 || | 1 | | | | 1 ≤ |sin 2Ay − sin 2y| ≤ 1, ∀A > 1, ∀y ∈ [1, +∞), 2y and, second, the function 1x converges monotonically (and uniformly on [1, +∞)) to 0 as x → +∞. At the same time, the first integral in the right-hand +∞ sin2 xy side diverges for ∀y ∈ [1, +∞). It implies the divergence of ∫1 dx for x ∀y ∈ [1, +∞). Remark to Examples 10–14. Examples 10–14 show that the information +∞ +∞ on the nature of convergence of integrals ∫a f (x, y)dx and ∫a g(x, y)dx

5.3 Dirichlet’s and Abel’s Theorems

does not allow us to make a conclusion about the behavior of the integral +∞ ∫a f (x, y)g(x, y)dx.

5.3 Dirichlet’s and Abel’s Theorems Remark to Examples 15–17. In the following three examples, we analyze the conditions of Dirichlet’s theorem on uniform convergence of improper integrals. It is shown that an omission of any of the conditions of the theorem can +∞ lead to nonuniform convergence of ∫a f (x, y)g(x, y)dx. A

Example 15. For each fixed y ∈ Y , integral ∫a f (x, y)dx is bounded and function g(x, y) is monotone in x ∈ [a, +∞); additionally, g(x, y) converges +∞ uniformly to 0 on Y as x → +∞, but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . Solution Consider the functions f (x, y) = sin xy and g(x, y) = (0, +∞). The estimate

1 x

on X × Y = [1, +∞) ×

|A | |A | |( ) |A | | | | | | 1 2 | | | | | | 1 | f (x, y)dx| = | sin xydx| = || − cos xy | || = | cos y − cos Ay| ≤ |∫ | |∫ | | | | y y y | 1 | | 1 | | |1 | | | | | A

shows that the integral ∫1 f (x, y)dx is bounded for each y ∈ Y . The second function g(x, y) = 1x is monotone in x and converges (uniformly) to 0 as x +∞ sin xy dx x

approaches ∞. In Example 3, it was proved that ∫0 +∞ sin xy ∫1 dx x

converges on

ℝ, which implies the convergence of on (0, +∞). Let us show that this convergence is nonuniform on (0, +∞). Using the same reasoning as in Example 3, we evaluate the integral remainder, for any A > 1 and yA = A12 ∈ (0, +∞): +∞

+∞

sin xyA sin t dx = dt ∫ ∫ x t A

AyA +∞

+∞

sin t sin t 𝜋 = dt → dt = ≠ 0, A→+∞ ∫ ∫ t t 2 1∕A

0

that is, the convergence is not uniform on (0, +∞).

185

186

Chapter 5 Improper Integrals Depending on a Parameter

The convergence fails to be uniform due to the violation of the condition of A uniform boundedness of the integrals ∫a f (x, y)dx on Y : for ∀A > 1, choosing 𝜋 yA = 2A > 0 one gets |A | |A | | | | | 1 | | | | | f (x, yA )dx| = | sin xyA dx| = |cos yA − cos AyA | |∫ | |∫ | yA | 1 | |1 | | | | | 2A || 𝜋 || = cos → +∞, 𝜋 || 2A || A→+∞ A

that is, ∫1 sin xydx is not uniformly bounded on (0, +∞). A

Example 16. The integral ∫a f (x, y)dx is uniformly bounded on Y , and g(x, y) converges uniformly to 0 on Y as x → +∞, but the improper integral +∞ ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . Solution x x √ √ e−xy on X × Y = Consider the functions f (x, y) = sin and g(x, y) = sin x x | A | [1, +∞) × (0, +∞). Since |∫1 sin xdx| = ||− cos x|A1 || ≤ 2, ∀A > 1 and √1 conx | | verges monotonically to 0 as x → +∞, it follows from Dirichlet’s theorem for +∞ x √ dx converges, and this improper integrals (without a parameter) that ∫1 sin x convergence is uniform on ℝ (and, in particular, on (0, +∞)), because f (x, y) A x √ dx is uniformly bounded does not depend on y. Therefore, the integral ∫1 sin x on (0, +∞). The function g(x, y) converges uniformly on (0, +∞) to 0, since ∀y ∈ (0, +∞) one has |g(x, y)| =

| sin x| −xy 1 → 0. √ e ≤ √ x→+∞ x x +∞

It remains to prove that ∫1 f (x, y)g(x, y)dx does not converge uniformly on 2x (0, +∞). Due to the equality sin2 x = 1−cos , the last integral can be written as 2 follows: +∞

+∞

+∞

1 1 −xy 1 cos 2x −xy f (x, y)g(x, y)dx = e dx − e dx. ∫ 2∫ x 2∫ x 1

1

1

The second integral in the right-hand side converges uniformly on (0, +∞) A according to Dirichlet’s theorem. Indeed, the integral ∫1 cos 2xdx is uni| A | | |A | formly bounded on (0, +∞): |∫1 cos 2xdx| = || 12 sin 2x| || ≤ 1, and the | | | |1 | e−xy function x is monotone in x and converges uniformly on (0, +∞) to 0: | e−xy | 1 | x | ≤ x → 0. The first integral in the right-hand side converges for each | | x→+∞

5.3 Dirichlet’s and Abel’s Theorems −xy

fixed y ∈ (0, +∞) due to the comparison theorem: 0 < e x ≤ e−xy , ∀x ≥ 1 −y +∞ |+∞ and ∫1 e−xy dx = − 1y e−xy | = ey converges for ∀y ∈ (0, +∞). However, |1 the convergence of the first integral is nonuniform: according to the Cauchy 1 criterion, for any A > 1 one can choose B = 2A and yA = 2A ∈ (0, +∞) such that 2A

e−xyA 1 −2AyA 1 dx > e (2A − A) = e−1 ↛ 0. ∫ A→+∞ x 2A 2 A

Since the first integral converges nonuniformly and the second—uniformly, +∞ it follows that ∫1 f (x, y)g(x, y)dx converges nonuniformly on (0, +∞). Note that this result is caused by the fact that g(x, y) is not monotone in x, that is, the corresponding condition of monotonicity in Dirichlet’s theorem is violated. A

Example 17. The integral ∫a f (x, y)dx is uniformly bounded on Y , and the function g(x, y) is monotone in x and converges to 0 for ∀y ∈ Y as x → +∞, but +∞ the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . Solution Consider the functions f (x, y) = y sin xy and g(x, y) = xy1 on X × Y = [1, +∞) × (0, +∞). The following evaluation holds simultaneously for | A | ∀A > 1 and ∀y ∈ Y : |∫1 y sin xydx| = | −cosxy|A1 | ≤ 2, which means the | | A uniform boundedness of the integral ∫1 f (x, y)dx on (0, +∞). For each fixed y ∈ (0, +∞), the function g(x, y) = xy1 is decreasing in x on [1, +∞) and lim 1 x→+∞ xy

= 0, but the last convergence is not uniform on (0, +∞) since for

∀x > 1 there exists yx = 1x such that g(x, yx ) = xy1 = 1 ↛ 0. Thus, the condix→+∞ x tion of the uniform convergence of g(x, y) in Dirichlet’s theorem is not satisfied. +∞ +∞ sin xy This leads to nonuniform convergence of ∫1 f (x, y)g(x, y)dx = ∫1 dx, x which was already shown in Example 15. Remark 1 to Examples 15–17. The violation of one of the conditions of Dirichlet’s theorem can lead not only to nonuniform convergence of +∞ ∫a f (x, y)g(x, y)dx, but even to divergence. For instance, for the functions sin xy f (x, y) = y sin xy and g(x, y) = x considered on X × Y = [1, +∞) × (0, +∞), the only condition violated is the monotonicity of g(x, y) in x. Indeed, it was A A shown in Example 17 that the integral ∫1 f (x, y)dx = ∫1 y sin xydx is uniformly bounded on (0, +∞), and the uniform convergence of g(x, y) to 0 follows | sin xy| immediately from the estimate |g(x, y)| = x ≤ 1x → 0. Nevertheless, the x→+∞

187

188

Chapter 5 Improper Integrals Depending on a Parameter +∞

+∞ y

integral ∫1 f (x, y)g(x, y)dx = ∫1 x sin2 xydx is divergent that can be proved as follows. Represent this integral in the following form: +∞

+∞

+∞

y 2 y y cos 2xy 1 1 sin xydx = dx − dx. ∫ x 2∫ x 2∫ x 1

1

1

Note that the second integral in the right-hand side converges (even uniformly A on (0, +∞)) by Dirichlet’s theorem: the integral ∫1 y cos 2xydx is uniformly A | | A | | | bounded since |∫1 y cos 2xydx| = | 12 sin 2xy| || ≤ 1 simultaneously for ∀A > 1, | | | |1 | ∀y ∈ (0, +∞), and the function 1x converges monotonically and uniformly on +∞ (0, +∞) to 0. At the same time, the first integral in the right-hand side ∫1 y 1x dx +∞ y diverges for each fixed y ∈ (0, +∞). Therefore, the integral ∫1 x sin2 xydx diverges. Remark 2 to Examples 15–17. As shown in Examples 15–17 and Remark 1, the violation of one of the conditions of Dirichlet’s theorem can lead to +∞ nonuniform convergence and divergence of the integral ∫a f (x, y)g(x, y)dx. At the same time, these conditions are sufficient, but not necessary: the integral +∞ ∫a f (x, y)g(x, y)dx still can be uniformly convergent if one of the conditions or even all the conditions are not satisfied. This situation is illustrated in the next example. Example 18. All the conditions of Dirichlet’s theorem are violated, but never+∞ theless the improper integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . Solution y cos xy For the functions f (x, y) = x12 y and g(x, y) = x defined on X × Y = [1, +∞) × (0, +∞), all the conditions of Dirichlet’s theorem are violated. In fact, according to the evaluation |A | |A | ) | | | | | 1 |A | 1 ( 1 1 1 | | | | | | || = dx = − 1 − ≤ | f (x, y)dx| = | | | |∫ | |∫ x2 y | | xy ||1 | y A y | 1 | | 1 | | | | | | | A

the integrals ∫1 f (x, y)dx are bounded for each fixed y ∈ (0, +∞), but the boundedness is not uniform on (0, +∞), since for ∀A > 1 there exists yA = A1 ∈ (0, +∞) such that |A | ( ) | | 1 1 | | 1− = A − 1 → +∞. | f (x, yA )dx| = |∫ | yA A→+∞ A | 1 | | |

5.3 Dirichlet’s and Abel’s Theorems

Further, the functions g(x, y) are not monotone in x and, although g(x, y) converges to 0 as x approaches infinity for each fixed y, this convergence is not √ uniform, since for the sequence of the points (xk , yk ), xk = yk = 2k𝜋, ∀k ∈ ℕ x cos 2k𝜋 one has xk → +∞, while g(xk , yk ) = k x = 1 ↛ 0. Even so, the integral k→+∞

+∞

∫1

k→+∞ k +∞ cos xy dx converges uniformly 3 x +∞ 1 and ∫1 x13 dx converges. 3 x

f (x, y)g(x, y)dx = ∫1 | cos xy | Weierstrass test: | x3 | ≤ | |

on (0, +∞) due to the

Remark to Examples 19–22. In the next four examples, we analyze in a similar way the conditions of Abel’s theorem on uniform convergence of improper integrals. Examples 19–21 analyze what can happen if one of the conditions is violated, and Example 22 shows that all the conditions are sufficient but not +∞ necessary for the uniform convergence of ∫a f (x, y)g(x, y)dx. +∞

Example 19. For each fixed y ∈ Y , integral ∫a f (x, y)dx is convergent and function g(x, y) is monotone in x ∈ [a, +∞); additionally, g(x, y) is uniformly +∞ bounded on [a, +∞) × Y , but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . Solution Consider the functions f (x, y) = ye−xy and g(x, y) = e−xy on X × Y = [0, +∞) × (0, +∞). Since +∞

∫

+∞

f (x, y)dx =

0

∫

ye−xy dx = −e−xy |+∞ = 1, 0

0 +∞

the integral ∫0 f (x, y)dx converges for each y ∈ (0, +∞). However, this convergence is not uniform on (0, +∞), because for ∀A > 0 choosing yA = A1 ∈ (0, +∞), one gets +∞

∫

f (x, yA )dx = −e−xyA |+∞ = e−1 ↛ 0. A A→+∞

A

The conditions for g(x, y) = e−xy are satisfied: it is decreasing in x for each fixed y ∈ (0, +∞) and is uniformly bounded |g(x, y)| = e−xy ≤ 1, ∀(x, y) ∈ [0, +∞) × (0, +∞). The convergence of the integral of the product can be proved by direct calculation: +∞

∫ 0

+∞

f (x, y)g(x, y)dx =

|+∞ 1 1 ye−2xy dx = − e−2xy || = , ∫ 2 2 |0 0

189

190

Chapter 5 Improper Integrals Depending on a Parameter

but this convergence is not uniform, which is seen by choosing yA = (0, +∞) for ∀A > 0 and arriving to

1 2A

∈

+∞

|+∞ 1 1 f (x, yA )g(x, yA )dx = − e−2xyA || = e−1 ↛ 0. ∫ A→+∞ 2 2 |A A

+∞

The last result is caused by nonuniform convergence of ∫0 violates one of the conditions of Abel’s theorem.

f (x, y)dx, which

+∞

Example 20. The integral ∫a f (x, y)dx is uniformly convergent and g(x, y) is +∞ uniformly bounded on Y , but the improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . Solution The functions f (x, y) = sinx x and g(x, y) = e−xy sin x defined on X × Y = [1, +∞) × (0, +∞) satisfy the conditions of this example. In fact, the con+∞ +∞ vergence of ∫1 f (x, y)dx = ∫1 sinx x dx can be proved in the same way +∞ x √ dx in Example 16, and this convergence is uniform as that of ∫1 sin x since f (x, y) does not depend on y. Also, the function g(x, y) is uniformly bounded on [1, +∞) × (0, +∞) since |g(x, y)| = |e−xy sin x| ≤ 1 for ∀(x, y) ∈ 2 +∞ +∞ [1, +∞) × (0, +∞). However, the integral ∫1 f (x, y)g(x, y)dx = ∫1 sinx x e−xy dx converges nonuniformly as was shown in Example 16. The last integral fails to converge uniformly because of the violation of the monotonicity condition for g(x, y) in Abel’s theorem. +∞

Example 21. The integral ∫a f (x, y)dx converges uniformly on Y , and the function g(x, y) is monotone in x and bounded for each fixed y ∈ Y , but the +∞ improper integral ∫a f (x, y)g(x, y)dx does not converge uniformly on Y . Solution Consider the functions f (x, y) = y2 e−xy and g(x, y) = 1y e−xy on X × Y = [0, +∞) × (0, +∞). Straightforward calculation +∞

∫ 0

+∞

f (x, y)dx =

∫

y2 e−xy dx = −ye−xy |+∞ =y 0

0 +∞

shows the convergence of the integral ∫0 f (x, y)dx. Note that the function h(y) = ye−Ay for any fixed A > 0 has the critical point equation hy (y) = (1 − yA)e−Ay = 0 with the only critical point yA = A1 , and this point is the maximum

5.3 Dirichlet’s and Abel’s Theorems

one, since hy (y) > 0 for y < yA and hy (y) < 0 for y > yA . Therefore, the following evaluation of the residual |+∞ | | | | | = ye−Ay ≤ max ye−Ay | f (x, y)dx| = −ye−xy |+∞ A |∫ | y∈(0,+∞) |A | | | 1 = yA e−AyA = e−1 → 0 A→+∞ A +∞

proves the uniform convergence of ∫0 f (x, y)dx on (0, +∞). Further, for each fixed y ∈ (0, +∞), the function g(x, y) = 1y e−xy is monotone (decreasing) and bounded: |g(x, y)| = 1y e−xy ≤ 1y . However, the boundedness is not uniform on Y = (0, +∞) since choosing yx = 1x ∈ (0, +∞) for ∀x ∈ (0, +∞), one gets g(x, yx ) = xe−1 → +∞. x→+∞ Let us turn to the integral of the product: +∞

∫ 0

+∞

f (x, y)g(x, y)dx =

|+∞ 1 1 ye−2xy dx = − e−2xy || = , ∫ 2 2 |0 0

that is, the integral is convergent for each y ∈ (0, +∞). However, this conver1 gence is nonuniform, since for ∀A > 0 and yA = 2A , it follows the following evaluation of the residual: |+∞ | +∞ | | | | y e−2xyA dx | f (x, yA )g(x, yA )dx| = |∫ | ∫ A |A | A | | |+∞ 1 1 = − e−2xyA || = e−1 ↛ 0. A→+∞ 2 2 |A +∞

In this example, the nonuniformity of the convergence of ∫0 f (x, y)g(x, y)dx was caused by weakening one of the conditions in Abel’s theorem: the function g(x, y) is bounded for each y ∈ Y , but it is not bounded uniformly on Y . Remark 1 to Examples 19–21. In the same way as for Dirichlet’s theorem, it may happen that the violation of only one condition in Abel’s theorem will +∞ result in the divergence of ∫a f (x, y)g(x, y)dx. For instance, the functions y sin xy f (x, y) = x and g(x, y) = sin xy defined on X × Y = [1, +∞) × (0, +∞) illus+∞

trate this situation. First, the integral ∫1

+∞ y sin xy

f (x, y)dx = ∫1 dx converges x | A | uniformly on (0, +∞) by Dirichlet’s theorem: |∫1 y sin xydx| = |− cos xy|A1 | ≤ 2, | | A that is, the integral ∫1 f (x, y)dx is uniformly bounded on (0, +∞), the function 1x decreases on [1, +∞) and converges uniformly on (0, +∞) to 0 as x approaches +∞. Further, the function g(x, y) is uniformly bounded on [1, +∞) × (0, +∞): |g(x, y)| = | sin xy| ≤ 1, ∀(x, y) ∈ [1, +∞) × (0, +∞). Thus,

191

192

Chapter 5 Improper Integrals Depending on a Parameter

the only condition of Abel’s theorem, which is not satisfied, is the monotonicity of g(x, y). Nevertheless, as shown in Remark 1 to Examples 15–17, the integral +∞ +∞ y ∫1 f (x, y)g(x, y)dx = ∫1 x sin2 xydx is divergent at each y ∈ (0, +∞). Remark 2 to Examples 19–21. It was shown above that the violation of one of the conditions of Abel’s theorem can lead to nonuniform convergence and +∞ even divergence of the integral ∫a f (x, y)g(x, y)dx. At the same time, the conditions of Abel’s theorem are sufficient, but not necessary, as it is illustrated in Example 22. Example 22. All the conditions of Abel’s theorem are violated, but neverthe+∞ less the improper integral ∫a f (x, y)g(x, y)dx converges uniformly on Y . Solution For the functions f (x, y) = x12 y and g(x, y) = y cos xy defined on X × Y = [1, +∞) × (0, +∞), all the conditions of Abel’s theorem are violated. In fact, the direct calculation +∞

∫ 1

+∞

f (x, y)dx =

+∞ 1 1 || 1 dx = − = | ∫ x2 y xy |1 y 1

+∞

shows that the integral ∫1 f (x, y)dx converges, but this convergence is nonuniform on (0, +∞), since for ∀A > 1 and the corresponding yA = A1 ∈ (0, +∞) one gets +∞

∫ A

+∞

f (x, yA )dx =

1 1 dx = = 1 ↛ 0. A→+∞ ∫ x2 yA AyA A

Further, the function g(x, y) = y cos xy is not monotone in x. Finally, g(x, y) is bounded for each fixed y ∈ (0, +∞): |g(x, y)| = |y cos xy| ≤ y, but this boundedness is not uniform √ on (0, +∞) since for the sequence of the points (xk , yk ), xk = yk = 2k𝜋, ∀k ∈ ℕ one has xk → +∞, k→+∞ √ while g(xk , yk ) = 2k𝜋 cos 2k𝜋 → +∞. Nevertheless, the integral +∞

∫1

+∞ cos xy

k→+∞

f (x, y)g(x, y)dx = ∫1 dx converges uniformly on (0, +∞) by x2 +∞ | cos xy | 1 the Weierstrass test: | x2 | ≤ x2 , ∀y ∈ (0, +∞) and ∫1 x12 dx converges. | |

5.4 Existence of the Limit and Continuity Example 23. A function f (x, y) is continuous in x on [a, +∞) for any fixed y ∈ Y and converges uniformly on [a, +∞) to a function 𝜑(x), as y approaches y0 ,

5.4 Existence of the Limit and Continuity +∞

+∞

and also both improper integrals ∫a f (x, y)dx and ∫a +∞ +∞ but lim ∫a f (x, y)dx ≠ ∫a lim f (x, y)dx. y→y0

𝜑(x)dx are convergent,

y→y0

{

Solution

1

1 − 2x2 y e ,x x3 y

>0 on X × Y = [0, +∞) × (0, 1] 0, x = 0 and y0 = 0. This function is continuous in x on X for any fixed y ∈ Y . For x > 0, it follows from the arithmetic properties and composition rule of continuous functions, and for x = 0 it can be shown as follows: Consider the function f (x, y) =

1 − 2x12 y 1 t3 e = lim 3 x→0 x y t→+∞ y et 2 ∕2y 1 3t 2 1 = lim 1 = 3 lim 1 2 = 0 t→+∞ t→+∞ tet ∕2y t 2 ∕2y y 2te 2y y

lim f (x, y) = lim x→0

(the change of the variable t = 1x was applied with subsequent application of l’Hospital’s rule). Next, let us show that f (x, y) converges to 𝜑(x) ≡ 0 as y approaches 0: for x = 0, the result is immediate—f (0, y) = 0, ∀y ∈ Y —and for other x, it can be seen from the evaluation 1 − 1 2 t 21 lim f (x, y) = lim 3 e 2x2 y = lim = lim =0 y→0 y→0 x y t→+∞ x et t→+∞ x et (here, the change of the variable t = 2x12 y was followed by l’Hospital’s rule). Moreover, f (x, y) converges to 𝜑(x) ≡ 0 uniformly on X due to the following argument. Since f (x, y) > 0 for x > 0, f (0, y) = 0, and lim f (x, y) = 0, the x→+∞ maximum max |f (x, y) − 𝜑(x)| = max f (x, y) is attained at an interior x∈[0,+∞)

x∈[0,+∞)

point of [0, +∞) for each fixed y ∈ Y . The corresponding partial derivative 1 1−3x2 y − fx (x, y) = x6 y2 e 2x2 y vanishes at the points where 1 − 3x2 y = 0, or x = ± √1 . 3y

The only critical point in [0, +∞) is xy = point. Then, for all x ∈ [0, +∞), we have

1 √ , 3y

which therefore is the maximum

|f (x, y) − 𝜑(x)| ≤ max f (x, y) = f (xy , y) x∈[0,+∞) √ √ 3y 3y −3y∕2y = e = 3 3ye−3∕2 → 0, y→0 y which means the uniform convergence on X = [0, +∞). Finally, note that both improper integrals exist: +∞

∫ 0

+∞

lim f (x, y)dx = y→0

∫ 0

0dx = 0

193

194

Chapter 5 Improper Integrals Depending on a Parameter

P2 φ(x) = lim f y→0

0.8 f

F(y)

+∞ fdx 0 +∞ P1 = ∫ [lim f ]dx 0 y→0 +∞ P2 = lim[∫ fdx ] y→0 0

F(y) = ∫

1

0.6

P1

0.4 0.2 0

φ(x)

0

0.2

0.4

0.6

0.8

1 4

y

{ Figure 5.11 Example 23, function f (x, y) =

3.5

3

2.5

2

1.5

1

0.5

0

x

1 1 − 2x2 y e x3 y

,x > 0 . 0, x = 0

and +∞

∫

+∞

f (x, y)dx =

0

+∞ 1 − 2x12 y − 2x12 y | e dx = e = 1, | |0 ∫ x3 y 0

but +∞

lim y→0

∫

+∞

f (x, y)dx = 1 ≠ 0 =

0

∫

lim f (x, y)dx. y→0

0

Remark. In the given counterexample, the interchange of the limit and integral +∞ is impossible because the convergence of the integral ∫0 f (x, y)dx is not uniform on Y = (0, 1]. The last fact can be verified through the evaluation of the remainder: for ∀A > 1 one can choose yA = 2A1 2 ∈ Y to obtain +∞

∫ A

f (x, yA )dx = e

− 2x21y

A

1 |+∞ | = 1 − e− 2A2 yA = 1 − e−1 ↛ 0. | A→+∞ |A

Thus, the example shows that the condition of the uniform convergence of +∞ ∫a f (x, y)dx on Y is important for the possibility of the calculation of the limit

5.4 Existence of the Limit and Continuity

under the sign of an improper integral. However, this condition is sufficient but not necessary as shown in the next example. Example 24. A function f (x, y) is continuous in x on [a, +∞) for any fixed y ∈ Y and converges uniformly on [a, +∞) to 𝜑(x), as y approaches y0 , +∞ +∞ +∞ and lim ∫a f (x, y)dx = ∫a lim f (x, y)dx, but ∫a f (x, y)dx converges y→y0

y→y0

nonuniformly on Y . Solution ( ) Consider the function f (x, y) = yx2 − 1y e−x∕y on X × Y = [1, +∞) × [1, +∞) and y0 = +∞. This function is continuous in x on X for any fixed y ∈ Y (just apply the arithmetic and composition rules ( for )continuous functions). Also f (x, y) converges on X to 𝜑(x) ≡ 0: lim yx2 − 1y e−x∕y = 0, ∀x ∈ X. To show y→+∞

that this convergence is uniform, let us find the extrema with respect to x of f (x, y) − 𝜑(x) = f (x, y). The critical points are found from the equation 2y−x fx (x, y) = y3 e−x∕y = 0, which results in xy = 2y. If xy < 2y then fx (x, y) > 0, and if xy > 2y then fx (x, y) < 0, that is, xy = 2y is the maximum point. Note also that f (x, y) is strictly increasing on [1, 2y) and has negative values when x < y and positive when x > y. Therefore, max |f (x, y) − 𝜑(x)| = max{|f (1, y)|, |f (xy , y)|}.

x∈[1,+∞)

F(y)

1

F3 F2

0.8

0.6

P3 P

2

2 +∞ fdx, f = 13 e–1/2x y 0 x y An n–1 Fn(y) = ∫ fdx, An = 2 0 rn(y) = ∫ +∞fdx An

F(y) = ∫

P1 F1

0.4 r1

Pn = (yn, rn(yn)) = (

1 –1 , 1– e ) 2An2

0.2 r2 r3

0.5

1

1.5

2 +∞

Figure 5.12 Example 23, integral F(y) = ∫0

f (x, y)dx.

2.5

3

195

196

Chapter 5 Improper Integrals Depending on a Parameter

For the last two functions of y, we have ( ) 1 1 lim |f (1, y)| = lim − 2 e−1∕y = 0 y→+∞ y→+∞ y y and

( lim |f (xy , y)| = lim

y→+∞

y→+∞

2y 1 − y2 y

) e−2y∕y = lim

y→+∞

1 −2 e = 0. y

Thus, simultaneously for all x ∈ X, we get |f (x, y) − 𝜑(x)| ≤ max |f (x, y)| = max{|f (1, y)|, |f (xy , y)|} → 0, y→+∞

x∈[1,+∞)

which implies the uniform convergence of f (x, y) on X. Finally, let us check the condition of interchanging the integral and limit. The first integral is trivial: +∞

∫

+∞

lim f (x, y)dx =

y→+∞

1

∫

0dx = 0.

1

For the second, we have +∞

|+∞ 1 x f (x, y)dx = − e−x∕y || = e−1∕y , ∫ y y |1 1

where the indeterminate form at x → +∞ was evaluated using l’Hospital’s rule: x∕y x 1 lim − e−x∕y = lim − x∕y = lim − x∕y = 0. x→+∞ e x→+∞ e y

x→+∞

Therefore, +∞

+∞

1 lim f (x, y)dx = lim e−1∕y = 0 = y→+∞ ∫ y→+∞ y ∫ 1

lim f (x, y)dx.

y→+∞

1

Hence, all the conditions of the statement are satisfied. +∞ Nevertheless, the improper integral ∫1 f (x, y)dx converges nonuniformly on Y : for ∀A > 1, choosing yA = A ∈ Y we obtain +∞

∫ A

f (x, yA )dx = −

x −x∕yA +∞ A e |A = e−A∕yA = e−1 ↛ 0. A→+∞ yA yA

5.4 Existence of the Limit and Continuity

Example 25. A function f (x, y) is continuous in two variables on X × Y = +∞ [a, +∞) × [c, d] and the improper integral F(y) = ∫a f (x, y)dx converges on Y , but F(y) is discontinuous on Y . Solution The function f (x, y) = ye−xy is continuous on X × Y = [0, +∞) × [0, +∞) and +∞ the improper integral converges on Y : F(0) = ∫0 0dx = 0 and for y ≠ 0 +∞ +∞ we have F(y) = ∫0 ye−xy dx = −e−xy |0 = 1. However, F(y) is discontinuous at y = 0. +∞ Note that the improper integral ∫0 f (x, y)dx converges nonuniformly on Y : for ∀A > 0, there exists yA = A1 ∈ Y such that +∞

∫

yA e−xyA dx = −e−xyA |+∞ = e−AyA = e−1 ↛ 0. A A→+∞

A

{

sin xy ,x x

≠0 considered on X × Y = y, x = 0 [0, +∞) × ℝ. This function is continuous at every (x, y), x ≠ 0 due to the arithsin xy metic rules and at every (0, y) because lim f (x, y) = lim xy ⋅ y = y = f (0, y). x→0 x→0 Besides, the improper integral converges (albeit nonuniformly) on Y : +∞ sin xy F(y) = ∫0 dx = 𝜋2 sgny (see calculations for a similar function in x Example 3). However, F(y) is discontinuous at y = 0. Another counterexample is f (x, y) =

Remark. The condition of the uniform convergence is relevant for the continuity of an improper integral depending on a parameter, but it is not necessary as shown in Example 26. Example 26. A function f (x, y) is continuous in two variables on X × Y , where +∞ X = [a, +∞) and Y is an interval, and F(y) = ∫a f (x, y)dx is continuous on Y , +∞ but the improper integral ∫a f (x, y)dx does not converge uniformly on Y . Solution The function f (x, y) = function +∞

F(y) =

1 xy

is continuous on X × Y = [1, +∞) × (1, +∞). Also, the

+∞ 1 x1−y || 1 dx = = | y ∫ x 1 − y |1 y−1 1

197

198

Chapter 5 Improper Integrals Depending on a Parameter +∞

is continuous on Y = (1, +∞). However, the improper integral ∫1 f (x, y)dx converges nonuniformly on Y : for ∀A > 1, one can choose yA = 1 + A1 ∈ Y to obtain +∞

+∞ dx x1−yA || = = A1−1∕A → +∞. | A→+∞ ∫ xyA 1 − y A |A A

Another counterexample employs the function of Example 4: f (x, y) = e−xy sin x considered on X × Y = [0, +∞) × (0, +∞). Evidently, this function is continuous on ℝ2 , and in particular on [0, +∞) × (0, +∞). Using the calculations made in Example 4 for y > 0, we obtain +∞

F(y) =

∫

e−xy sin xdx = −e−xy

0

cos x + y sin x ||+∞ 1 | = 1 + y2 . 1 + y2 |0

This function is continuous on (0, +∞), but it was shown in Example 4 that the +∞ integral ∫0 e−xy sin xdx does not converge uniformly on Y = (0, +∞). Remark. Note that the continuity of F(y) on Y does not guarantee the validity of the interchange between the limit and the improper integral. 1 In fact, in the last counterexample the function F(y) = 1+y , and in partic2 +∞ −xy

ular, lim F(y) = lim ∫0 +∞

∫0

y→0+

y→0+

e

+∞

lim e−xy sin xdx = ∫0

y→0+

sin xdx = 1. However, the improper integral

sin xdx is divergent.

5.5 Differentiability Example 27. Both f (x, y) and fy (x, y) are continuous on X × Y , where +∞ X =([a, +∞) and Y)is an interval, and ∫a f (x, y)dx converges uniformly on Y , +∞

but ∫a

Solution

f (x, y)dx

+∞

y

≠ ∫a

fy (x, y)dx on Y . {

sin xy

,x ≠ 0 x In Examples 3 and 25, it was shown that f (x, y) = is cony, x = 0 { cos xy, x ≠ 0 tinuous on ℝ2 . Its partial derivative fy (x, y) = is also 1, x = 0 continuous on ℝ2 . In particular, both f (x, y) and fy (x, y) are continuous on X × Y = [0, +∞) × [1, +∞). Besides, in Example 3 it was proved that +∞ +∞ sin xy F(y) = ∫a f (x, y)dx = ∫0 dx converges uniformly on any interval x [c, +∞), c > 0, in particular, on Y = [1, +∞). Since F(y) = 𝜋2 for ∀y ∈ [1, +∞)

5.5 Differentiability

(see calculations in Example 3), it follows that Fy (y) = 0. However, it is not possible to calculate the derivative inside the integral, because +∞

+∞

+∞

1 f (x, y)dx = cos xydx = cos tdt ∫ y ∫ y∫ 0

0

0

diverges. Example 28. Both f (x, y) and fy (x, y) are continuous on X × Y , where +∞ X = [a, +∞) and Y is an interval, ( and both )integrals ∫a f (x, y)dx and +∞ +∞ +∞ ∫a fy (x, y)dx converge on Y , but ∫a f (x, y)dx ≠ ∫a fy (x, y)dx on Y . y

Solution ( ) y Let us consider f (x, y) = x + x12 e−xy on X × Y = [1, +∞) × [0, +∞). Both f (x, y) and fy (x, y) = −ye−xy are continuous on X × Y . Also, both improper integrals converge on Y = [0, +∞), which can be checked by straightforward calculations: +∞

F(y) =

∫

+∞

f (x, y)dx =

(y

∫

x

1

1

+∞

+∞

+

) 1 e−xy dx x2

|+∞ 1 = − e−xy || = e−y , ∀y ∈ (0, +∞), x |1 F(0) =

∫

f (x, 0)dx =

1 +∞

G(y) = = G(0) =

∫

1 +∞

fy (x, y)dx =

1 e−xy |+∞ 1 +∞

∫ 1

+∞ 1 1 || dx = − = 1, | ∫ x2 x |1

∫

−ye−xy dx

1

= −e , ∀y ∈ (0, +∞), −y

fy (x, 0)dx =

+∞

∫

0dx = 0.

1

Nevertheless, Fy (y) = −e−y and, consequently, Fy (0) = −1 ≠ 0 = G(0). +∞ Note that the improper integral ∫1 fy (x, y)dx converges nonuniformly on Y = [0, +∞) because for ∀A > 1 there exists yA = A1 ∈ Y such that the remainder does not approach 0: +∞

∫ A

fy (x, yA )dx = e−xyA |+∞ = −e−AyA = −e−1 ↛ 0. A A→+∞

199

200

Chapter 5 Improper Integrals Depending on a Parameter +∞

At the same time, the second improper integral ∫1 f (x, y)dx converges uniformly on Y = [0, +∞). Indeed, for y ∈ (0, +∞) one gets +∞

|+∞ 1 1 f (x, y)dx = − e−xy || = e−Ay , ∫ x A |A A

and for y = 0 +∞

∫

+∞

f (x, 0)dx =

A

+∞ 1 1 || 1 dx = − = . | ∫ x2 x |A A A

| +∞ | Therefore, the evaluation |∫A f (x, y)dx| ≤ | | ∀y ∈ [0, +∞).

1 → 0 A A→∞

holds simultaneously for

+∞

Remark. This example shows that the uniform convergence of ∫a fy (x, y)dx is an important condition to interchange the calculation of partial derivative and improper integral. At the same time, this condition is not necessary as shown in the following example. Example 29. Both f (x, y) and fy (x, y) are continuous on X × Y , where +∞ +∞ X = [a, +∞) and Y is an interval, the integrals ∫a f (x, y)dx and ∫a fy (x, y)dx ( ) +∞ +∞ converge on Y and ∫a fy (x, y)dx = ∫a f (x, y)dx , ∀y ∈ Y , but the +∞

improper integral ∫a

y

fy (x, y)dx converges nonuniformly on Y .

Solution ( ) y Consider the function of Example 28: f (x, y) = x + x12 e−xy but on the different domain X × Y = [1, +∞) × (0, +∞). It was shown in Example 28 +∞ that ∫1 fy (x, y)dx does not converge uniformly on [0, +∞), and for exactly the same reason it does not converge uniformly on (0, +∞). However, all the statement conditions hold: f (x, y) and fy (x, y) = −ye−xy are continuous on X × Y ; both improper integrals converge: +∞

F(y) =

∫ a

+∞

f (x, y)dx = e , G(y) = −y

∫

fy (x, y)dx = −e−y , ∀y ∈ (0, +∞);

a

and finally Fy (y) = −e−y = G(y), ∀y ∈ (0, +∞). +∞

Remark. ( In )the presented counterexample, the integral ∫a f (x, y)dx = +∞ y ∫1 + x12 e−xy dx converges uniformly on Y (as was shown in Example 28). x ( ) +∞ +∞ However, it may happen that the equality ∫a fy (x, y)dx = ∫a f (x, y)dx y

5.5 Differentiability

f(x,y) = ( xy +

1 ) e–xy x2 F(y) = ∫ +∞fdx = e–y 1 Fy(y) = – e–y, y ≥ 0 G(y) = ∫ +∞fydx = – e–y, 1

1

f

0.8 0.6 0.4 0.2 0 –0.2 –0.4 –0.6 –0.8 –1 1

y>0

F(y) G(0)= 0 G(y)= Fy(y), y > 0

2 x

3 4

Fy(0) = –1 0.5 0

1

( Figure 5.13 Examples 28 and 29, function f (x, y) =

2.5

2

1.5

y x

3

4

3.5

y

+

1 x2

)

e−xy .

+∞

+∞

holds for ∀y ∈ Y even when both integrals ∫a f (x, y)dx and ∫a fy (x, y)dx converge nonuniformly on Y . For such a case, let us consider the function f (x, y) = ye−xy on X × Y = [1, +∞) × (0, +∞), which is continuous together with its derivative fy (x, y) = (1 − xy)e−xy on ℝ2 and, in particular, on X × Y . The +∞ +∞ convergence of both integrals F(y) = ∫1 f (x, y)dx and G(y) = ∫1 fy (x, y)dx on (0, +∞) can be verified by the direct calculations: +∞

F(y) =

∫

ye−xy dx = −e−xy |+∞ = e−y , ∀y ∈ (0, +∞), 1

1 +∞

G(y) =

∫

(1 − xy)e−xy dx = xe−xy |+∞ = −e−y , ∀y ∈ (0, +∞). 1

1

Therefore, Fy (y) = −e−y = G(y), ∀y ∈ (0, +∞). However, for ∀A > 1 and yA = one gets |+∞ | | | | | | = e−AyA = e−1 ↛ 0 | f (x, yA )dx| = |−e−xyA |+∞ A |∫ | A→+∞ |A | | |

1 A

201

202

Chapter 5 Improper Integrals Depending on a Parameter

F(y) = ∫1+∞fdx, G(y) = ∫1+∞fydx, f = ( xy + 1x2 )e–xy

1 0.8 0.6 P1 0.4 P2 0.2 P3

–0.2 –0.4 –0.6

Fn(y) = ∫1An fdx, Gn(y) = ∫1An fydx, An = 2n

F3 F2

rn(y) = ∫A+∞ fdx, qn(y) = ∫A+∞ fydx n n

Pn = (

F1

0, A1 n

), Qn = (

)

1 ,– e–1 An

r1 r2 r3 q3 0.5

F(y) 1

1.5

2

2.5

G(y)

3

q2 G1 Q3 Q2 Q1 q1

G2 G3

–0.8 –1 +∞

Figure 5.14 Examples 28 and 29, integrals F(y) = ∫1

+∞

f (x, y)dx, G(y) = ∫1

fy (x, y)dx.

and | +∞ | | | | | | = Ae−AyA = Ae−1 → +∞, | fy (x, yA )dx| = ||xe−xyA |+∞ A | |∫ | A→+∞ |A | | | which means that both integrals converge nonuniformly on (0, +∞).

5.6 Integrability Example 30. A function f (x, y) is continuous on [a, +∞) × [c, d] and +∞ d +∞ the integral ∫a f (x, y)dx converges on [c, d], but ∫c dy ∫a f (x, y)dx ≠ +∞ d ∫a dx ∫c f (x, y)dy. Solution 2 2 The function f (x, y) = (2xy − 2x3 y3 )e−x y is continuous on ℝ2 , and in particular, on X × Y = [0, +∞) × [0, 1]. The improper integral converges on [0, 1] according to the direct calculations: +∞

F(y) =

2 2

∫ 0

2 2

(2xy − 2x3 y3 )e−x y dx = x2 ye−x y |+∞ = 0. 0

5.6 Integrability

However, the integrals assume different values: 1

∫

1

+∞

dy

f (x, y)dx =

∫

∫

0

0

0

+∞

1

+∞

1

F(y)dy =

∫

0dy = 0,

0

while ∫

dx

f (x, y)dy =

∫

0

0

2 2

∫

xy2 e−x y |10 dx

0 +∞

=

∫

2 1 2 1 xe−x dx = − e−x |+∞ = . 0 2 2

0 +∞

Note that the convergence of F(y) = ∫0 (2xy − 2x3 y3 )e−x y dx is not uniform on [0, 1]. Indeed, for ∀A > 1 and yA = A1 ∈ [0, 1] the remainder evaluation gives 2 2

|+∞ | | | 2 2 | | | | 3 3 −x2 y2A dx| = |x2 yA e−x yA |+∞ | (2xyA − 2x yA )e A || |∫ | | |A | | | 2 2 = A2 yA e−A yA = Ae−1 → +∞. A→+∞

Remark. The required equality will be satisfied if the convergence of F(y) = +∞ ∫a f (x, y)dx would be uniform on [c, d]. However, Example 31 shows that the last condition is not necessary. Example 31. A function f (x, y) is continuous on an infinite strip [a, +∞) × d +∞ +∞ d [c, d] and ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy, but the improper integral +∞ ∫a f (x, y)dx converges nonuniformly on [c, d]. Solution The function f (x, y) = ye−xy is continuous on ℝ2 , and consequently, on X × Y = [1, +∞) × [0, 1]. The improper integral converges on [0, 1]: +∞

F(y) =

∫

ye−xy dx = −e−xy |+∞ = e−y , ∀y ∈ (0, 1] 1

1 +∞ ∫1

and F(0) = 0dx = 0. However, this convergence is not uniform: for ∀A > 1 and yA = A1 ∈ [0, 1], we get +∞

∫ A

yA e−xyA dx = −e−xyA |+∞ = e−1 ↛ 0. A A→+∞

203

204

Chapter 5 Improper Integrals Depending on a Parameter

Let us check if it is possible to change the order of integration. On the one hand, 1

∫

1

+∞

dy

f (x, y)dx =

∫

0

1

1

F(y)dy =

∫

∫

0

e−y dy = 1 − e−1 .

0

On the other hand, 1

+∞

∫

dx

f (x, y)dy =

∫

1

1

+∞

0

dx

∫

∫

1 +∞

ye−xy dy

0

1 ⎡ y ⎤ |1 1 −xy ⎥ −xy ⎢− e | + = e dy dx | ∫ ⎢ x ∫ x ⎥ | 0 ⎦ 1 ⎣ 0 +∞ [ ] |1 1 1 = − e−x − 2 e−xy || dx ∫ x x |0 1 +∞

+∞

+∞

1 e−x 1 = − e−x dx − dx + dx ∫ ∫ ∫ x2 x x2 1

=

1

−x |+∞

e | x ||1

+

1

+∞

+∞

1

1

+∞ e−x e−x 1 || dx − dx − ∫ x2 ∫ x2 x ||1

= 1 − e−1 (the integration by parts was applied for integration in y with u = y, dv = e−xy dy and for integration in x with u = 1x , dv = e−x dx). Therefore, both integrals give the same result. Example 32. A function f (x, y) is continuous on [a, +∞) × [c, +∞), and the +∞ +∞ +∞ +∞ improper integrals ∫c dy ∫a f (x, y)dx and ∫a dx ∫c f (x, y)dy converge, but they assume different values. Solution x−y The function f (x, y) = (x+y)3 is continuous on X × Y = [1, +∞) × [2, +∞). Let us evaluate the required improper integrals. For the first integral, we have +∞

+∞

+∞

x−y x+y y F(y) = dx = dx − 2 dx 3 3 ∫ (x + y) ∫ (x + y) ∫ (x + y)3 1

( =

1

y 1 − + x + y (x + y)2

1

)|+∞ +∞ x || 1 | = , | =− | 2 | (x + y) (1 + y)2 | 1 |1

5.6 Integrability

and therefore, +∞

+∞

F(y)dy =

∫ 2

+∞ 1 1 || 1 dy = − = . | 2 ∫ (1 + y) 1 + y |2 3 2

The second integral gives +∞

+∞

+∞

x−y x+y 2x G(x) = dy = dy − dy ∫ (x + y)3 ∫ (x + y)3 ∫ (x + y)3 2

( =

2

2

)|+∞ y ||+∞ x 1 2 | − + = =− , | (x + y)2 x + y ||2 (x + y)2 ||2 (x + 2)2

and then +∞

+∞

G(x)dx =

∫ 1

+∞ −2 2 || 2 dx = =− . ∫ (x + 2)2 x + 2 ||1 3 1

Thus, both integrals are convergent, but different. Comparing the conditions of this example with the conditions of Theorem 1 (on integration with respect to parameter on infinite interval), one can see that the former does not require that f (x, y) keeps the same sign over the considered domain. It is used in the counterexample and leads to different results for the integrals. On the other hand, comparing the conditions of this example with the conditions of Theorem 2 (on integration with respect to parameter on infinite interval), one can note that the former does not require the convergence of the integrals of the absolute values. Applying the Weierstrass test, one can show that in the used counterexample the integral F(y) converges uniformly on Y = +∞ |x−y| [2, +∞): |f (x, y)| = (x+y)3 < x12 , ∀(x, y) ∈ [1, +∞) × [2, +∞), and ∫1 x12 dx converges. Using the same reasoning, one can prove that the integral G(x) con+∞ +∞ |x−y| verges uniformly on X = [1, +∞). However, both integrals ∫2 dy ∫1 (x+y)3 dx +∞

and ∫1

+∞ |x−y| dy (x+y)3

dx ∫2

diverge. Indeed,

+∞

y

+∞

1

1

y

|x − y| y−x x−y dx = dx + dx ∫ (x + y)3 ∫ (x + y)3 ∫ (x + y)3 y +∞ x || x || 1 1 = − = − , | | 2 2 (x + y) |1 (x + y) |y 2y (1 + y)2

205

206

Chapter 5 Improper Integrals Depending on a Parameter

and then +∞

+∞

+∞

+∞

2

1

2

2

|x − y| 1 1 dy dx = dy − dy, ∫ ∫ (x + y)3 ∫ 2y ∫ (1 + y)2 where the second integral converges, but the first diverges, implying in the +∞ +∞ divergence of the integral ∫2 dy ∫1 |f (x, y)|dx. Similar evaluations can be made for the second integral: +∞

∫

+∞

dx

|x − y| dy ∫ (x + y)3

1

2 2

=

∫

+∞

dx

1

y−x dy ∫ (x + y)3 2

+∞ ⎡ x ⎤ x − y y−x ⎥ + dx ⎢ dy + dy ∫ ∫ (x + y)3 ⎥ ⎢∫ (x + y)3 ⎣2 ⎦ 2 x 2 ( )|+∞ 1 x | = − + | dx ∫ x + y (x + y)2 ||2 1 +∞

+∞

+

(

∫

1 x − x + y (x + y)2

2 2

=

(

∫

1 x − x + 2 (x + 2)2

) |x ( )|+∞ 1 x | | + | + − | dx 2 | | x + y (x + y) |2 |x

) dx

1 +∞ (

+

∫

2 2 1 x − − + 2x 4x x + 2 (x + 2)2

) dx

2 +∞

2 +∞ 2 || 2 || 1 =− + + dx, ∫ 2x x + 2 ||1 x + 2 ||2 2

where the last integral diverges, resulting in the divergence of the integral +∞ +∞ ∫1 dx ∫2 |f (x, y)|dy. x2 −y2 Another counterexample can be provided with the function f (x, y) = (x2 +y2 )2 considered on X × Y = [1, +∞) × [1, +∞). Remark. In these counterexamples, the improper integrals have different values because neither the conditions of Theorem 1 nor those of Theorem 2 are

5.6 Integrability

satisfied. At the same time, the conditions in Theorems 1 and 2 are sufficient, but not necessary for equality of the improper integrals. Example 33. A function f (x, y) is continuous and nonnegative on [a, +∞) × +∞ +∞ +∞ +∞ [c, +∞) and ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy, but at least one of +∞ +∞ the functions F(y) = ∫a f (x, y)dx and G(x) = ∫c f (x, y)dy is discontinuous on [c, +∞) or [a, +∞), respectively. Solution The function f (x, y) = ye−xy is continuous and nonnegative on X × Y = [1, +∞) × [0, +∞). For F(y), we obtain +∞

F(y) =

ye−xy dx = −e−xy |+∞ = e−y , ∀y ∈ (0, +∞) 1

∫ 1

+∞ ∫1

and F(0) = 0dx = 0, that is, F(y) is discontinuous at y = 0. Therefore, one of the conditions of Theorem 1 is not satisfied (this also implies that the integral F(y) does not converge uniformly on [0, d], ∀d > 0, that is, one of the conditions of Theorem 2 is not satisfied either). Nevertheless, both iterated improper integrals exist and coincide: for the first integral, we have +∞

∫

+∞

dy

f (x, y)dx =

∫

0

+∞

1

∫

+∞

F(y)dy =

0

e−y dy = 1,

∫ 0

and for the second, we get +∞

y 1 1 ye−xy dy = − e−xy |+∞ − 2 e−xy |+∞ = 2 0 0 ∫ x x x

G(x) =

0

and +∞

∫ 1

+∞

dx

∫ 0

+∞

f (x, y)dy =

∫ 1

+∞

G(x)dx =

+∞ 1 1 || dx = − = 1. ∫ x2 x ||1 1

Example 34. A function f (x, y) is continuous on [a, +∞) × [c, +∞), and both +∞ +∞ improper integrals F(y) = ∫a f (x, y)dx and G(x) = ∫c f (x, y)dy do not con+∞ +∞ +∞ +∞ verge uniformly, but ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy. Solution The function f (x, y) = e−xy sin x is continuous on X × Y = [0, +∞) × [0, +∞), but does not maintain the sign there. For this reason, we check the

207

208

Chapter 5 Improper Integrals Depending on a Parameter

conditions of Theorem 2 only. Using the result found in Example 4: cos x+y sin x ∫ e−xy sin xdx = − 1+y2 e−xy , we immediately obtain +∞

F(y) =

∫

e−xy sin xdx = −

0

cos x + y sin x −xy ||+∞ 1 e | = , ∀y ∈ (0, +∞). 1 + y2 1 + y2 |0

+∞

+∞

Since the integral F(0) = ∫0 f (x, 0)dx = ∫0 sin xdx diverges, let us consider an auxiliary interval Y1 = [c, +∞), ∀c > 0 on which the integral F(y) converges 1 to 1+y . Then 2 +∞

∫

+∞

F(y)dy =

c

1 𝜋 dy = arctan y|+∞ = − arctan c, ∀c > 0. c ∫ 1 + y2 2 c

Finally, passing to the limit, we obtain the following result for the first iterated integral: +∞

∫

+∞

dy

0

+∞

f (x, y)dx =

∫

F(y)dy

∫

0

0 +∞

= lim

c→0+

( F(y)dy = lim

∫

c→0+

) 𝜋 𝜋 − arctan c = . 2 2

c

For the second integral, we have +∞

G(x) =

e−xy sin xdy = −

∫

sin x −xy +∞ sin x e |0 = , ∀x ∈ (0, +∞) x x

0

and +∞

G(0) =

+∞

f (0, y)dy =

∫

0dy = 0.

∫

0

0

The function G(x) has a discontinuity (removable) at x = 0, but it is continuous | | for ∀x > 0 and bounded on [0, +∞) (since | sinx x | ≤ 1, ∀x ∈ (0, +∞)). Therefore, | | G(x) is integrable and +∞

∫ 0

+∞

dx

∫ 0

+∞

f (x, y)dy =

∫ 0

+∞

G(x)dx =

∫

sin x 𝜋 dx = . x 2

0

Thus, two iterated integrals coincide. Nevertheless, the conditions of Theorem 2 are not satisfied. Indeed, in +∞ Example 4, it was shown that F(y) = ∫0 e−xy sin xdx does not converge

5.6 Integrability

uniformly on Y = (0, +∞). By the same reasoning, it does not converge +∞ uniformly on Y1 = [0, 1] either. Further, G(x) = ∫0 e−xy sin xdy also does not converge uniformly on X1 = [0, 1], since for any A > 1 and corresponding xA = A1 ∈ [0, 1] we obtain |+∞ | +∞ | | | | sin x | | | | | sin 1∕A || A −xA y || −xA y sin xA dy| = | − e | | = ||e−1 → e−1 ≠ 0. | e |∫ | | x 1∕A || A→+∞ |A || | A |A | | | | +∞ +∞ +∞ +∞ Besides, both integrals ∫0 dy ∫0 |f (x, y)|dx and ∫0 dx ∫0 |f (x, y)|dy are divergent. For the first integral, the evaluation goes as follows. Since | sin x| ≥ sin2 x = 12 (1 − cos 2x), it follows that +∞

∫

+∞

dy

∫

0

+∞ −xy

e

+∞

1 |sin x|dx ≥ dy e−xy dx ∫ 2∫

0

0

0 +∞

+∞

1 − dy e−xy cos 2xdx. ∫ 2∫ 0

0

The second integral in the right-hand side converges (one can carry out sim+∞ +∞ ilar calculation as for ∫0 dy ∫0 e−xy sin xdx), but the first integral diverges, because for ∀y > 0 one gets +∞

∫

+∞

dy

∫

0

+∞

−xy

e

0

+∞

|+∞ 1 1 dx = − e−xy || dy = dy. ∫ ∫ y y |0 0 +∞

0

+∞

Therefore, the integral ∫0 dy ∫0 e−xy |sin x|dx diverges. Using the analogous approach for the second iterated integral, we obtain +∞

∫

+∞

e

−xy

| sin x| 1 | sin x|dy = ≥ (1 − cos 2x), ∀x > 0, f (0, y)dy = 0, ∫ x 2x

0

0

and for ∀x > 0 +∞

∫ 0

+∞

dx

∫ 0

+∞ −xy

e

|sin x|dy =

∫ 0 1

| sin x| dx x +∞

| sin x| | sin x| = dx + dx ∫ ∫ x x 0 1

1 +∞

+∞

1

1

| sin x| 1 dx 1 cos 2x ≥ dx + − dx. ∫ x 2∫ x 2∫ x 0

209

210

Chapter 5 Improper Integrals Depending on a Parameter

The first and third integrals in the right-hand side are convergent, whereas the +∞ +∞ second is divergent. Thus, the integral ∫0 dx ∫0 e−xy | sin x|dy also diverges.

Exercises 1 Show that the following integrals converge on the given intervals and verify the character of the convergence: +∞ a) ∫0 e−xy cos xdx, Y = (0, +∞) +∞ cos xy b) ∫−∞ x2 +1 dx, Y = ℝ +∞

dx ,Y =ℝ (x−y)2 +1 +∞ cos x −xy ∫1 e dx, p > 0, Y = [0, +∞) xp +∞ sin x −xy √ ∫0 e dx, Y = [0, +∞) 4 x +∞ √ −x2 y ∫0 ye dx, Y = [0, +∞) +∞ sin x ∫1 dx, Y = (0, +∞) xy +∞ −xy ∫0 e sin xydx, Y = (0, +∞).

c) ∫0

d) e) f) g) h)

2 Show that the integrals a), c), g), and h) in Exercise 1 are counterexamples to Example 2. +∞ 2 −xy2

3 Use ∫0

ye

dx, Y = ℝ as a counterexample to Example 3.

4 Verify whether the integrals a) and h) of Exercise 1 are counterexamples to Example 4. +∞

5 Verify whether the integral ∫1 sinxy x dx considered on Y = [10−1 , 1] provides a counterexample to Example 5. +∞

6 Show that the integral ∫1 sinxy x dx converges absolutely and uniformly on +∞ x| Y = (1, +∞), but the integral ∫1 | sin dx does not converge uniformly xy on Y = (1, +∞). It provides one more counterexample to Example 6. 2xy

2xy

7 Show that the functions f (x, y) = x4 +y4 and g(x, y) = x4 +y2 defined on X × Y = [1, +∞) × [1, +∞) provide one more counterexample to Example 8. +∞

8 Check if the following statement is true: “if ∫a f (x, y) + g(x, y)dx con+∞ +∞ verges uniformly on Y , then the integrals ∫a f (x, y)dx and ∫a g(x, y)dx 2 2 also converge uniformly on Y .” (Hint: use the functions f (x, y) = 2xy2 e−x y 3 2 −x2 y2 and g(x, y) = (2xy − 2xy )e on X = [1, +∞), Y = ℝ. Compare with Example 9.)

Exercises +∞

9 Check if the following statement is true: “if ∫a f (x, y) + g(x, y)dx +∞ converges nonuniformly on Y , then the integrals ∫a f (x, y)dx and +∞ ∫a g(x, y)dx also converge nonuniformly on Y .” (Hint: use the functions 2 2 f (x, y) = y3 e−xy and g(x, y) = y2 e−xy on X = [0, +∞), Y = ℝ. Compare with Example 9.) 10 Verify if the integrals ∫X f (x, y)dx, ∫X g(x, y)dx and ∫X f (x, y)g(x, y)dx converge on Y , and the nature of the convergence if it takes place. Use the following functions and sets: 2 2 2 2 a) f (x, y) = 2xy2 e−x y , g(x, y) = (y − 2x2 y3 )e−x y , X = [1, +∞), Y = ℝ 2 b) f (x, y) = g(x, y) = ye−xy , X = [1, +∞), Y = ℝ 2 2 2 2 c) f (x, y) = 2xy2 e−x y , g(x, y) = (2x2 y2 − 1)e−x y , X = [1, +∞), Y = (0, +∞) 2 d) f (x, y) = g(x, y) = y2 e−xy , X = [1, +∞), Y = ℝ x −xy x −xy √ √ e) f (x, y) = cos e , g(x, y) = cos , X = [1, +∞), Y = (0, +∞) 3 2 e 3 x

x

f ) f (x, y) = g(x, y) = (y − xy2 )e−xy , X = [0, +∞), Y = [0, +∞) x −xy √ g) f (x, y) = g(x, y) = cos e , X = [1, +∞), Y = (0, +∞) 3 x

h) f (x, y) = g(x, y) = cosx x e−xy , X = [1, +∞), Y = (0, +∞) i) f (x, y) = cosx x e−xy , g(x, y) = sinx x e−xy , X = [1, +∞), Y = [0, +∞) x −xy x −xy √ √ j) f (x, y) = cos e , g(x, y) = sin e , X = [1, +∞), Y = [0, +∞) 3 3 x

k) l) m) n)

x

f (x, y) = g(x, y) = y2 e−xy , X = [0, +∞), Y = [0, +∞) 2 f (x, y) = cosx x , g(x, y) = e−xy , X = [1, +∞), Y = (0, +∞) f (x, y) = (xy3 − y2 )e−xy , g(x, y) = e−xy , X = [0, +∞), Y = (0, +∞) f (x, y) = cosx x , g(x, y) = xe−xy , X = [1, +∞), Y = (0, +∞) sin xy

o) f (x, y) = x2 , g(x, y) = x2 e−xy , X = [1, +∞), Y = (0, +∞) p) f (x, y) = y cos xy, g(x, y) = xy1 , X = [1, +∞), Y = [1, +∞) cos xy

cos xy

q) f (x, y) = √3 , g(x, y) = √3 2 , X = [1, +∞), Y = [1, +∞). x x Formulate false statements for which the given functions represent counterexamples. 11 Verify what conditions of Dirichlet’s theorem are violated for functions f (x, y) and g(x, y) on X × Y . Analyze the nature of the convergence of the integral ∫X f (x, y)g(x, y)dx if it takes place. Do this for the following functions and sets: a) f (x, y) = y2 e−xy , g(x, y) = y12 e−xy , X = [0, +∞), Y = (0, 1] b) c) d) e)

f (x, y) = x1y , g(x, y) = cos x, X = [1, +∞), Y = (1, +∞) f (x, y) = ye−xy , g(x, y) = e−xy , X = [0, +∞), Y = (0, +∞) f (x, y) = yx , g(x, y) = 1x , X = [1, +∞), Y = (0, 1) f (x, y) = e−xy , g(x, y) = sin x, X = [0, +∞), Y = [1, +∞)

211

212

Chapter 5 Improper Integrals Depending on a Parameter cos xy

f ) f (x, y) = y cos xy, g(x, y) = x , X = [1, +∞), Y = (0, +∞) g) f (x, y) = cos x, g(x, y) = cosx x e−xy , X = [1, +∞), Y = (0, +∞) h) f (x, y) = sin x, g(x, y) = e−xy , X = [0, +∞), Y = (0, 1). Compare with Examples 15–18. 12

Verify what conditions of Abel’s theorem are violated for functions f (x, y) and g(x, y) on X × Y . Analyze the nature of the convergence of the integral ∫X f (x, y)g(x, y)dx if it takes place. Do this for the following functions and sets: x a) f (x, y) = ye−xy , g(x, y) = sin , X = [1, +∞), Y = (0, +∞) x2 y −x2 y

b) f (x, y) = 2xy2 e−x y , g(x, y) = e y , X = [1, +∞), Y = (0, +∞) c) f (x, y) = x−y , g(x, y) = y cos xy, X = [1, +∞), Y = (1, +∞) y d) f (x, y) = √ cos xy, g(x, y) = cos xy, X = [1, +∞), Y = (0, +∞) 2

x

e) f (x, y) = cosx x , g(x, y) = e−xy cos x, X = [1, +∞), Y = (0, +∞) f ) f (x, y) = yx , g(x, y) = 1x , X = [1, +∞), Y = (0, 1) g) f (x, y) = e−xy sin x, g(x, y) = e−xy , X = [0, +∞), Y = (0, 1). Compare with Examples 19–22. 13

Show that the integrals +∞ √ −x2 y a) ∫0 ye dx, Y = [0, +∞), y0 = 0 2 +∞ b) ∫0 y2 e−xy dx, Y = ℝ, y0 = 0 are counterexamples to Example 23.

14

Prove that the integral ∫1 (y − 2x2 y3 )e−x y dx, Y = ℝ, y0 = 0 is a counterexample to Example 24.

15

Use the √ integrals 2 +∞ a) ∫0 ye−x y dx, Y = [0, +∞), y0 = 0 2 +∞ b) ∫1 y2 e−xy dx, Y = ℝ, y0 = 0 +∞ −xy c) ∫0 e sin xydx, Y = [0, +∞), y0 = 0 to give counterexamples to Example 25.

16

Use the integrals +∞ a) ∫0 e−xy cos xdx, Y = (0, +∞) 2 +∞ b) ∫1 y2 e−xy dx, Y = (0, +∞) +∞ c) ∫0 e−xy sin xydx, Y = (0, +∞) 2 2 +∞ d) ∫1 (y − 2x2 y3 )e−x y dx, Y = ℝ to illustrate Example 26.

+∞

2 2

Exercises

17 Prove that the { functions y sin xy ,x ≠ 0 x a) f (x, y) = , X = [0, +∞), Y = ℝ y2 , x = 0 2 b) f (x, y) = y3 e−xy , X = [1, +∞), Y = ℝ provide counterexamples to Example 27. Find the conditions in the Theorem on differentiation under the integral sign that are violated by the given functions. 18 Prove that the functions 2 2 a) f (x, y) = (y − 2x2 y3 )e−x y , X = [1, +∞), Y = ℝ 2 b) f (x, y) = y2 e−xy , X = [1, +∞), Y = ℝ 2 c) f (x, y) = y3 e−xy , X = [1, +∞), Y = ℝ provide counterexamples to Example 28. Investigate the nature of the con+∞ +∞ vergence of the integrals ∫1 f (x, y)dx and ∫1 fy (x, y)dx on Y . 19 Show that the functions 2 2 a) f (x, y) = (y − 2x2 y3 )e−x y , X = [1, +∞), Y = (0, +∞) 2 b) f (x, y) = y2 e−xy , X = [1, +∞), Y = (0, +∞) 2 c) f (x, y) = y3 e−xy , X = [0, +∞), Y = (0, +∞) 2 3 4 −x2 y2 d) f (x, y) = (2xy − 2x , X = [1, +∞), Y = ℝ ( ) y )e 1 1 −xy e) f (x, y) = x + x2 y e , X = [1, +∞), Y = (0, +∞) satisfy all the conditions of Example 29. Investigate the character of the +∞ +∞ convergence of the integrals ∫a f (x, y)dx and ∫a fy (x, y)dx on Y . 20 Show that the function f (x, y) = (2xy − x2 y2 )e−xy , X = [1, +∞), Y = [0, 2] provides a counterexample to Example 30. Investigate the character of the +∞ convergence of the integral ∫1 f (x, y)dx on Y . 21 Verify whether the functions a) f (x, y) = (y − xy2 )e−xy , X = [1, +∞), Y = [0, 1] b) f (x, y) = xy2 e−xy , X = [1, +∞), Y = [0, 1] satisfy all the conditions of Example 31. Investigate the character of the +∞ convergence of the integral ∫a f (x, y)dx. 22 Prove that the functions x2 −y2 a) f (x, y) = (x2 +y2 )2 , X = [1, +∞), Y = [0, +∞) b) f (x, y) = (2xy − x2 y2 )e−xy , X = [1, +∞), Y = [0, +∞) x2 +2xy−2y2 c) f (x, y) = (x+y)4 , X = [1, +∞), Y = [2, +∞) d) f (x, y) =

x2 −2xy , (x+y)4

X = [1, +∞), Y = [1, +∞)

213

214

Chapter 5 Improper Integrals Depending on a Parameter

provide counterexamples to Example 32. Find the conditions in both Theorems on the interchange of integration that are violated by these functions. 23

Show that the function f (x, y) = xy2 e−xy , X = [1, +∞), Y = [0, +∞) provides a counterexample to Example 33.

24

Analyze the following statement: “if f (x, y) is continuous on [a, +∞) × +∞ +∞ [c, +∞), the functions F(y) = ∫a f (x, y)dx and G(x) = ∫c f (x, y)dy are +∞ +∞ continuous on [c, +∞) and [a, +∞), respectively, ∫c dy ∫a f (x, y)dx = +∞ +∞ ∫a dx ∫c f (x, y)dy and one of the integrals F(y) or G(x) converges uniformly, then another one also converges uniformly.” (Hint: use the function f (x, y) = (y − xy2 )e−xy on X × Y = [1, +∞) × [0, +∞) to construct a counterexample.)

25

Analyze the following statement: “if f (x, y) is continuous on [a, +∞) × +∞ +∞ [c, +∞), the integrals F(y) = ∫a f (x, y)dx and G(x) = ∫c f (x, y)dy converge uniformly on [c, +∞) and [a, +∞), respectively, and +∞ +∞ +∞ +∞ ∫c dy ∫a f (x, y)dx = ∫a dx ∫c f (x, y)dy, then f (x, y) keeps the x2 −4xy+y2 sign on [a, +∞) × [c, +∞).” (Hint: use the function f (x, y) = (x+y)4 on X × Y = [1, +∞) × [1, +∞) to construct a counterexample. Check the conditions of Theorem 2 on the interchange of integration.)

26

Show that the function f (x, y) = xye−xy cos xy, X = [0, +∞), Y = [0, +∞) provides a counterexample to Example 34. Check the conditions of both Theorems on the interchange of integration.

Further Reading B.M. Budak and S.V. Fomin, Multiple Integrals, Field Theory and Series, Mir Publisher, Moscow, 1978. G.M. Fichtengolz, Differential- und Integralrechnung, Vol. 1–3, V.E.B. Deutscher Verlag Wiss., Berlin, 1968. V.A. Ilyin and E.G. Poznyak, Fundamentals of Mathematical Analysis, Vol. 1,2, Mir Publisher, Moscow, 1982. V.A. Zorich, Mathematical Analysis I, II, Springer, Berlin, 2004.

215

Bibliography 1 S. Abbott. Understanding Analysis. Springer, New York, 2002. 2 P. Biler and A. Witkowski. Problems in Mathematical Analysis. Marcel

Dekker, New York, 1990. 3 D. Bressoud. A Radical Approach to Real Analysis. MAA, Washington, DC,

2007. 4 T.J.I. Bromwich. An Introduction to the Theory of Infinite Series. AMS,

Providence, RI, 2005. 5 B.M. Budak and S.V. Fomin. Multiple Integrals, Field Theory and Series. Mir 6 7 8 9 10 11 12 13 14

Publisher, Moscow, 1978. B. Demidovich. Problems in Mathematical Analysis. Mir Publisher, Moscow, 1989. G.M. Fichtengolz. Differential- und Integralrechnung, Vol.1–3. V.E.B. Deutscher Verlag Wiss., Berlin, 1968. B.R. Gelbaum and J.M.H. Olmsted. Counterexamples in Analysis. Dover Publication, Mineola, NY, 2003. V.A. Ilyin and E.G. Poznyak. Fundamentals of Mathematical Analysis, Vol.1,2. Mir Publisher, Moscow, 1982. W.J. Kaczor and M.T. Nowak. Problems in Mathematical Analysis, Vol.1–3. AMS, Providence, RI, 2001. K. Knopp. Theory and Applications of Infinite Series. Dover Publication, Mineola, NY, 1990. C.H.C. Little, K.L. Teo, and B. Brunt. Real Analysis via Sequences and Series. Springer, New York, 2015. B.M. Makarov, M.G. Goluzina, A.A. Lodkin, and A.N. Podkorytov. Selected Problems in Real Analysis. AMS, Providence, RI, 1992. G. Polya and G. Szego. Problems and Theorems in Analysis, Vol.1–2. Springer, Berlin, 1998.

Counterexamples on Uniform Convergence: Sequences, Series, Functions, and Integrals, First Edition. Andrei Bourchtein and Ludmila Bourchtein. © 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. Companion website: www.wiley.com/go/bourchtein/counterexamples_on_uniform_convergence

216

Bibliography

15 T.L.T. Radulesku, V.D. Radulesku, and T. Andreescu. Problems in Real

Analysis: Advanced Calculus on the Real Axis. Springer, New York, 2009. 16 W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, New York, 1976. 17 V.A. Zorich. Mathematical Analysis I, II. Springer, Berlin, 2004.

217

Index a Abel’s theorem on uniform convergence for an improper integral xlviii, 189–192 for a series of functions xliii, 35–39 Absolute convergence of an improper integral 170–174 of a sequence of functions 19 of a series of functions 10–14

c Cauchy criterion of uniform convergence for a family of functions xlv for an improper integral xlvii for a sequence of functions xl for a series of functions xlii Convergence of a family of functions arithmetic properties xlv continuity xlvi, 61–64, 66 differentiation xlvi, 95–96, 100, 103–104, 107 integration xlvi, 119–120, 123–124 limit function xlv nonuniform xlv nonuniform, condition xlv passage to limit xlv, 52–53, 55, 57–59 at a point xliv pointwise xliv, 1, 5, 8

uniform xlv uniform limit xlv uniform, Cauchy criterion xlv uniform, criterion xlv uniform, Dini’s theorem xlv Convergence of an improper integral depending on a parameter absolute 170–174 arithmetic properties xlviii, 174–185 continuity xlviii, 197–198 differentiation xlviii, 198–202 integration on a finite interval xlviii, 202–204 integration on an infinite interval xlix, 204–210 passage to limit xlviii, 192–196 pointwise xlvii, 167–169 residual xlvii uniform xlvii uniform, Abel’s theorem xlviii, 189–192 uniform, Cauchy criterion xlvii uniform, Dirichlet’s theorem xlvii, 185–189 uniform, Weierstrass test xlvii Convergence of a sequence of functions absolute 19 arithmetic properties xl, 15–21 continuity xl, 60, 62–64, 66–67, 76–79

Counterexamples on Uniform Convergence: Sequences, Series, Functions, and Integrals, First Edition. Andrei Bourchtein and Ludmila Bourchtein. © 2017 John Wiley & Sons, Inc. Published 2017 by John Wiley & Sons, Inc. Companion website: www.wiley.com/go/bourchtein/counterexamples_on_uniform_convergence

218

Index

Convergence of a sequence of functions (cont’d ) differentiation xli, 95, 98–99, 101–103, 106–107, 111–113 integration xl, 117–119, 123, 127 limit function xxxix nonuniform xxxix nonuniform, condition xxxix passage to limit xl, 52, 54–57, 59 at a point xxxix pointwise xxxix, 1–2, 5–7, 9 residual xlii uniform xxxix uniform, Cauchy criterion xl uniform continuity xl, 79–87 uniform, criterion xl uniform, Dini’s theorem xl, 68–71 Convergence of a series of functions absolute 10–14 arithmetic properties xliii, 23–31 continuity xliv, 60–63, 65, 67 differentiation xliv, 97–98, 100–101, 104–105, 107–111, 113–117 integration xliv, 120–122, 124–128 nonuniform xlii nonuniform, condition xlii partial sums xli passage to limit xliii, 51–56, 58–59 at a point xli pointwise xlii, 2–7, 9 residual xlii sum xlii uniform xlii uniform, Abel’s theorem xliii, 35–39 uniform, Cauchy criterion xlii uniform continuity xliv, 80–87 uniform, criterion xlii uniform, Dini’s theorem xliii, 71–76 uniform, Dirichlet’s theorem xliii, 31–35

uniform, necessary condition xliii uniform, Weierstrass M-test xliii

d Dini’s theorem on uniform convergence for a family of functions xlv for a sequence of functions xl, 68–71 for a series of functions xliii, 71–76 Dirichlet’s theorem on uniform convergence for an improper integral xlvii, 185–189 for a series of functions xliii, 31–35 Divergence of a sequence of functions xxxix of a series of functions xli

f Family of functions xliv arithmetic properties xlv Cauchy criterion of uniform convergence xlv condition of nonuniform convergence xlv continuity xlvi, 61–64, 66 criterion of uniform convergence xlv differentiation xlvi, 95–96, 100, 103–104, 107 Dini’s theorem on uniform convergence xlv integration xlvi, 119–120, 123–124 limit function xlv nonuniform convergence xlv passage to limit xlv, 52–53, 55, 57–59 pointwise convergence xliv, 1, 5, 8 uniform convergence xlv uniform limit xlv

Index

i Improper integral depending on a parameter xlvii Abel’s theorem of uniform convergence xlviii, 189–192 absolute convergence 170–174 arithmeric properties xlviii, 174–185 boundedness 173–174 Cauchy criterion of uniform convergence xlvii continuity xlviii, 197–198 differentiation xlviii, 198–202 Dirichlet’s theorem of uniform convergence xlvii, 185–189 integration on a finite interval xlviii, 202–204 integration on an infinite interval xlix, 204–210 passage to limit xlviii, 192–196 pointwise convergence xlvii, 167–169 residual xlvii uniform convergence xlvii uniformly bounded xlvii Weierstrass test xlvii Integral depending on a parameter xlvi continuity xlvi, 137–141 differentiation xlvi, 142–154 integration xlvii, 154–162 passage to limit xlvi, 133–137

s Sequence of functions xxxviii absolute convergence 19 arithmetic properties xl, 15–21 boundedness 21–22, 45–51 Cauchy criterion of uniform convergence xl condition of nonuniform convergence xxxix

continuity xl, 60, 62–64, 66–67, 76–79 convergent at a point xxxix criterion of uniform convergence xl differentiation xli, 95, 98–99, 101–103, 106–107, 111–113 Dini’s theorem of uniform convergence xl, 68–71 divergent xxxix integration xl, 117–119, 123, 127 limit function xxxix nonuniform convergence xxxix passage to limit xl, 52, 54–57, 59 pointwise convergence xxxix, 1–2, 5–7, 9 residual xlii uniform continuity xl, 79–87 uniform convergence xxxix uniform limit xxxix Series of functions xli Abel’s theorem of uniform convergence xliii, 35–39 absolute convergence 10–14 arithmetic properties xliii, 23–31 boundedness 12–14 Cauchy criterion of uniform convergence xlii condition of nonuniform convergence xlii continuity xliv, 60–63, 65, 67 convergent at a point xli criterion of uniform convergence xlii differentiation xliv, 97–98, 100–101, 104–105, 107–111, 113–117 Dini’s theorem of uniform convergence xliii, 71–76 Dirichlet’s theorem of uniform convergence xliii, 31–35 divergent xli general term xli integration xliv, 120–122, 124–128

219

220

Index

Series of functions (cont’d ) limit function xlii majorant (dominant) series xliii nonuniform convergence xlii partial sums xli passage to limit xliii, 51–56, 58–59 pointwise convergence xlii, 2–7, 9 residual xlii sum xlii uniform continuity xliv, 80–87 uniform convergence xlii uniform convergence, necessary condition xliii uniform limit xlii Weierstrass M-test xliii

t Term-by-term differentiation for sequences xli, 95, 98–99, 101–103, 106–107, 111–113

differentiation for series xliv, 97–98, 100–101, 104–105, 107–111, 113–117 integration for sequences xl, 117–119, 123, 127 integration for series xliv, 120–122, 124–128 passage to limit for sequences xl, 52, 54–57, 59 passage to limit for series xliii, 51–56, 58–59

w Weierstrass test for an improper integral xlvii for a series of functions xliii

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