A Complete Solution Guide to Real and Complex Analysis 9887415677, 9789887415671

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A Complete Solution Guide to Real and Complex Analysis

by Kit-Wing Yu, PhD [email protected]

c 2021 by Kit-Wing Yu. All rights reserved. No part of this publication may be Copyright reproduced, stored in a retrieval system, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the author. ISBN: 978-988-74156-6-4 (eBook) ISBN: 978-988-74156-7-1 (Paperback)

ii

About the author

Dr. Kit-Wing Yu received his B.Sc. (1st Hons), M.Phil. and Ph.D. degrees in Math. at the HKUST, PGDE (Mathematics) at the CUHK. After his graduation, he joined United Christian College (UCC) to serve as a mathematics teacher between 2000 and 2020. From 2002 to 2020, he also took the responsibility of the mathematics panel at UCC. Starting from Sept. 2020, Dr. Yu has been promoted to be the Vice Principal (Academic) at Kiangsu-Chekiang College (Kwai Chung). Furthermore, he was appointed as a part-time tutor (2002 – 2005) and then a part-time course coordinator (2006 – 2010) of the Department of Mathematics at the OUHK. Besides teaching, Dr. Yu has been appointed to be a marker of the HKAL Pure Mathematics and HKDSE Mathematics (Core Part) for over thirteen years. Between 2012 and 2014, Dr. Yu was invited to be a Judge Member by the World Olympic Mathematics Competition (China). In the research aspect, he has published research papers in international mathematical journals, including some well-known journals such as J. Reine Angew. Math., Proc. Roy. Soc. Edinburgh Sect. A and Kodai Math. J.. His research interests are inequalities, special functions and Nevanlinna’s value distribution theory. In the area of academic publication, he is the author of the following eight books: • A Complete Solution Guide to Complex Analysis • A Complete Solution Guide to Real and Complex Analysis I • A Complete Solution Guide to Real and Complex Analysis II • A Complete Solution Guide to Principles of Mathematical Analysis • Problems and Solutions for Undergraduate Real Analysis • Problems and Solutions for Undergraduate Real Analysis I • Problems and Solutions for Undergraduate Real Analysis II • Mock Tests for the ACT Mathematics

iii

iv

Preface

This present book A Complete Solution Guide to Real and Complex Analysis is the combined volume of author’s two books [125] and [126]. It provides solutions to all exercises of Rudin’s classical graduate book [100]. The primary aim of this book is to help every mathematics student and instructor to understand the ideas and applications of the theorems in Rudin’s book. To accomplish this goal, I have adopted the usual way I wrote all my published solution guides. In other words, I intend writing the solutions as comprehensive as I can so that you can understand every detailed part of a proof easily. Apart from this, I also keep reminding you what theorems or results I have applied by quoting them repeatedly in the proofs. By doing this, I believe that you will become fully aware of the meaning and applications of each theorem. Before you go into details of this book, I have two gentle reminders for you. Firstly, as a mathematics instructor at a college, I understand that the growth of a mathematics student depends largely on how hard he/she does exercises. When your instructor asks you to do some exercises from Rudin, you are not suggested to read my solutions unless you have tried your best to prove them seriously yourselves. Secondly, when I prepared this book, I found that some exercises require knowledge that Rudin did not cover in his book. To fill this gap, I suggest you to check the references provided by the solutions or read directly the reference list of this book. Of course, I will use my books [124], [127], [128], [125], [123] and [126] freely. The features of the book are as follows: • It covers all the 397 exercises from Chapters 1 to 20 with detailed and complete solutions. As a matter of fact, my solutions show every detail, every step and every theorem that I applied. • There are 40 illustrations for explaining the mathematical concepts or ideas used behind the questions or theorems. • Sections in each chapter are added so as to increase the readability of the exercises. • Different colors are used frequently in order to highlight or explain problems, lemmas, remarks, main points/formulas involved, or show the steps of manipulation in some complicated proofs. (ebook only) • Necessary lemmas with proofs are provided because some questions require additional mathematical concepts which are not covered by Rudin. • Many useful or relevant references are provided to some questions for your future research. v

vi Since the solutions are written solely by me, you may find typos or mistakes. If you really find such a mistake, please send your valuable comments or opinions to [email protected]. Then I will post the updated errata on my website https://sites.google.com/view/yukitwing/ irregularly.

Kit Wing Yu April 2021

List of Figures

2.1

The graph of gn on [−1, 1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

2.2

The graphs of gn,k on [0, 1]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

31

2.3

The pictures of V, E ′ , E ′ ∩ V and (E ′ ∩ V )c . . . . . . . . . . . . . . . . . . . . . .

57

(E ′

∩V

)c

2.4

The set

\ U. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

57

3.1

The distribution of x and ǫn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85

3.2

The geometric interpretation of a special case. . . . . . . . . . . . . . . . . . . . .

89

5.1

The unit circle in different p-norm. . . . . . . . . . . . . . . . . . . . . . . . . . . 144

5.2

The square K. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161

6.1

The graphs of gn,1 (x) and gn,2 (x).

7.1

The closed intervals En and En+1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

7.2

Construction of the sequence Ei (p). . . . . . . . . . . . . . . . . . . . . . . . . . 242

. . . . . . . . . . . . . . . . . . . . . . . . . . 194

10.1 The closed contour ΓA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 10.2 The contour ΓA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 324 10.3 The contours ΓA , Γ1 and Γ2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 325 10.4 The annulus A(r1 , r2 ) and the circles γ1 , γ2 . . . . . . . . . . . . . . . . . . . . . . 337 10.5 A non null-homotopic closed path Γ = γ1 − γ3 − γ2 + γ4 in Ω. . . . . . . . . . . . 346 11.1 The I divides ∂∆ into several triangles. . . . . . . . . . . . . . . . . . . . . . . . 362 12.1 The boundary ∂∆. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383 12.2 The sectors ∆1 , ∆2 and the ray Lα . . . . . . . . . . . . . . . . . . . . . . . . . . 390 13.1 The simply connected set Ω. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 398 13.2 The compact sets Dn and En . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 13.3 The compact sets An , Bn and Cn . . . . . . . . . . . . . . . . . . . . . . . . . . . 400 13.4 The disc ∆n , the arc Ln and its neighborhood Ωn . . . . . . . . . . . . . . . . . . 401 14.1 The region Ω bounded by C1 and C2 .

. . . . . . . . . . . . . . . . . . . . . . . . 433

14.2 The constructions of Ωn−1 , D(0; rn ) and αn . . . . . . . . . . . . . . . . . . . . . . 436 14.3 The construction of the symmetric point z ∗ of z. . . . . . . . . . . . . . . . . . . 448
14.4 The conformal mapping ψ(z) = φ ϕ(z) . . . . . . . . . . . . . . . . . . . . . . . 457 vii

viii

List of Figures π

π

14.5 The conformal mapping f : U → A(e− 2 , e 2 ). . . . . . . . . . . . . . . . . . . . . 458

14.6 The locus of f (z) for t ∈ (0, π). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 14.7 The locus of f (z) for t ∈ (π, 2π). . . . . . . . . . . . . . . . . . . . . . . . . . . . 459

14.8 The image of f (E) when r = 0.25. . . . . . . . . . . . . . . . . . . . . . . . . . . 460 15.1 The distribution of the zeros zk,n of exp(exp(z)). . . . . . . . . . . . . . . . . . . 465 15.2 The paths γ(z + h) and −γ(z). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 473 16.1 The paths βζ (I) and γζ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 496 16.2 The paths γz (I), γω (I) and γω,ζ (I). . . . . . . . . . . . . . . . . . . . . . . . . . . 496 16.3 The fundamental domain R of G. . . . . . . . . . . . . . . . . . . . . . . . . . . . 498 16.4 The regions Ωα and Ωβ if α < β. . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 16.5 The regions of convergence of the two series. . . . . . . . . . . . . . . . . . . . . . 513 19.1 The closed contour Γr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 569 20.1 The compact set X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 585

Contents

Preface

v

List of Figures

viii

1 Abstract Integration 1.1 Problems on σ-algebras and Measurable Functions . . . . . . . . . . . . . . . . . 1.2

1 1

Problems related to the Lebesgue’s MCT/DCT . . . . . . . . . . . . . . . . . . .

7

2 Positive Borel Measures 2.1 Properties of Semicontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

17 17

2.2

Problems on the Lebesgue Measure on R . . . . . . . . . . . . . . . . . . . . . . .

25

2.3 2.4

Integration of Sequences of Continuous Functions . . . . . . . . . . . . . . . . . . Problems on Borel Measures and Lebesgue Measures . . . . . . . . . . . . . . . .

30 36

2.5

Problems on Regularity of Borel Measures . . . . . . . . . . . . . . . . . . . . . .

43

2.6

Miscellaneous Problems on L1 and Other Properties . . . . . . . . . . . . . . . .

59

3 Lp -Spaces 3.1 Properties of Convex Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69 69

3.2

Relations among Lp -Spaces and some Consequences . . . . . . . . . . . . . . . .

71

3.3 3.4

Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12 . . . . . . . . . . . . . . . . . Hardy’s Inequality and Egoroff’s Theorem . . . . . . . . . . . . . . . . . . . . . .

87 91

3.5 3.6

Convergence in Measure and the Essential Range of f ∈ L∞ (µ) . . . . . . . . . . 106 A Converse of Jensen’s Inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . 111

3.7

The Completeness/Completion of a Metric Space . . . . . . . . . . . . . . . . . . 112

3.8

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 115

4 Elementary Hilbert Space Theory 123 4.1 Basic Properties of Hilbert Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . 123 4.2

Application of Theorem 4.14

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135

4.3

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 137

5 Examples of Banach Space Techniques 143 5.1 The Unit Ball in a Normed Linear Space . . . . . . . . . . . . . . . . . . . . . . . 143 5.2 5.3

Failure of Theorem 4.10 and Norm-preserving Extensions . . . . . . . . . . . . . 146 The Dual Space of X . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150 ix

x

Contents 5.4

Applications of Baire’s and other Theorems . . . . . . . . . . . . . . . . . . . . . 157

5.5

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

6 Complex Measures

183

6.1

Properties of Complex Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . 183

6.2

Dual Spaces of Lp (µ) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186

6.3

Fourier Coefficients of Complex Borel Measures . . . . . . . . . . . . . . . . . . . 191

6.4

Problems on Uniformly Integrable Sets . . . . . . . . . . . . . . . . . . . . . . . . 196

6.5

Dual Spaces of Lp (µ) Revisit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 204

7 Differentiation

207

7.1

Lebesgue Points and Metric Densities . . . . . . . . . . . . . . . . . . . . . . . . 207

7.2

Periods of Functions and Lebesgue Measurable Groups . . . . . . . . . . . . . . . 210

7.3

The Cantor Function and the Non-measurability of f ◦ T

7.4 7.5

. . . . . . . . . . . . . 216

Problems related to the AC of a Function . . . . . . . . . . . . . . . . . . . . . . 218 Miscellaneous Problems on Differentiation . . . . . . . . . . . . . . . . . . . . . . 232

8 Integration on Product Spaces

249

8.1

Monotone Classes and Ordinate Sets of Functions . . . . . . . . . . . . . . . . . . 249

8.2

Applications of the Fubini Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 252

8.3

The Product Measure Theorem and Sections of a Function . . . . . . . . . . . . 268

8.4

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 273

9 Fourier Transforms

281

9.1

Properties of The Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . 281

9.2

The Poisson Summation Formula and its Applications . . . . . . . . . . . . . . . 297

9.3

Fourier Transforms on Rk and its Applications . . . . . . . . . . . . . . . . . . . 301

9.4

Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 310

10 Elementary Properties of Holomorphic Functions 10.1 Basic Properties of Holomorphic Functions

315

. . . . . . . . . . . . . . . . . . . . . 315

10.2 Evaluation of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 320 10.3 Composition of Holomorphic Functions and Morera’s Theorem . . . . . . . . . . 327 10.4 Problems related to Zeros of Holomorphic Functions . . . . . . . . . . . . . . . . 331 10.5 Laurent Series and its Applications . . . . . . . . . . . . . . . . . . . . . . . . . . 336 10.6 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 344 11 Harmonic Functions

347

11.1 Basic Properties of Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . . 347 11.2 Harnack’s Inequalities and Positive Harmonic Functions . . . . . . . . . . . . . . 363 11.3 The Weak∗ Convergence and Radial Limits of Holomorphic Functions . . . . . . 371 11.4 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 374 12 The Maximum Modulus Principle

383

Contents

xi

12.1 Applications of the Maximum Modulus Principle . . . . . . . . . . . . . . . . . . 383 12.2 Asymptotic Values of Entire Functions . . . . . . . . . . . . . . . . . . . . . . . . 392 12.3 Further Applications of the Maximum Modulus Principle . . . . . . . . . . . . . 393 13 Approximations by Rational Functions

397

13.1 Meromorphic Functions on S 2 and Applications of Runge’s Theorem . . . . . . . 397 13.2 Holomorphic Functions in the Unit Disc without Radial Limits . . . . . . . . . . 401 13.3 Simply Connectedness and Miscellaneous Problems . . . . . . . . . . . . . . . . . 407 14 Conformal Mapping

413

14.1 Basic Properties of Conformal Mappings . . . . . . . . . . . . . . . . . . . . . . . 413 14.2 Problems on Normal Families and the Class S . . . . . . . . . . . . . . . . . . . 426 14.3 Proofs of Conformal Equivalence between Annuli . . . . . . . . . . . . . . . . . . 432 14.4 Constructive Proof of the Riemann Mapping Theorem . . . . . . . . . . . . . . . 436 15 Zeros of Holomorphic Functions

463

15.1 Infinite Products and the Order of Growth of an Entire Function . . . . . . . . . 463 15.2 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 472 15.3 Problems on Blaschke Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 478 15.4 Miscellaneous Problems and the M¨ untz-Szasz Theorem . . . . . . . . . . . . . . . 486 16 Analytic Continuation

493

16.1 Singular Points and Continuation along Curves . . . . . . . . . . . . . . . . . . . 493 16.2 Problems on the Modular Group and Removable Sets . . . . . . . . . . . . . . . 497 16.3 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 509 17 H p -Spaces

519

17.1 Problems on Subharmonicity and Harmonic Majoriants . . . . . . . . . . . . . . 519 17.2 Basic Properties of H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 522 17.3 Factorization of f ∈ H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 531

17.4 A Projection of Lp onto H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 538 17.5 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544 18 Elementary Theory of Banach Algebras

549

18.1 Examples of Banach Spaces and Spectrums . . . . . . . . . . . . . . . . . . . . . 549 18.2 Properties of Ideals and Homomorphisms . . . . . . . . . . . . . . . . . . . . . . 553 18.3 The Commutative Banach algebra H ∞ . . . . . . . . . . . . . . . . . . . . . . . . 563 19 Holomorphic Fourier Transforms

565

19.1 Problems on Entire Functions of Exponential Type . . . . . . . . . . . . . . . . . 565 19.2 Quasi-analytic Classes and Borel’s Theorem . . . . . . . . . . . . . . . . . . . . . 574 20 Uniform Approximation by Polynomials

585

Index

589

xii Bibliography

Contents 591

CHAPTER

1

Abstract Integration

1.1

Problems on σ-algebras and Measurable Functions

Problem 1.1 Rudin Chapter 1 Exercise 1.

Proof. Let X be a set. Assume that M was an infinite σ-algebra in X which has only countably many members. Then we have M = {∅, X, A1 , A2 , . . .}. Let x ∈ X be fixed, we define the set \ Sx = A, (1.1) x∈A

where x ∈ A ∈ M. (By this definition (1.1), it is clear that x ∈ Sx .) Since M is countable, the intersection in (1.1) is actually at most countable. Thus it follows from Comment 1.6(c) that Sx ∈ M. Next we define the set S = {Sx | x ∈ X} so that S ⊆ M. We need a lemma about the set S: Lemma 1.1 If Sx , Sy ∈ S and Sx ∩ Sy 6= ∅, then Sx = Sy .

Proof of Lemma 1.1. Let z ∈ Sx ∩ Sy . We want to show that Sz = Sx . On the one hand, since z ∈ Sx , we know from the definition (1.1) that z ∈ A for all A containing x. Now if w ∈ Sx , then w ∈ A for all A containing x. By the previous observation, such A must contain z, so we get w ∈ Sz and then Sx ⊆ Sz .

(1.2)

On the other hand, if w ∈ Sz , then w ∈ A for all A containing z. Thus the hypothesis z ∈ Sx ∩ Sy implies that w ∈ Sx ∩ Sy , i.e., Sz ⊆ Sx ∩ Sy . By this result and the definition (1.1), we obtain Sz ⊆ Sx ∩ Sy ⊆ Sx . (1.3) Combining the observations (1.2) and (1.3), we have the desired result that Sz = Sx .  Similarly, we can show that Sz = Sy . Hence we conclude that Sx = Sy . 1

2

Chapter 1. Abstract Integration

Now we return to the proof of the problem. By Lemma 1.1, we may assume that all elements in S are distinct. If B ∈ M, then the definition (1.1) certainly gives y ∈ Sy for every y ∈ B so that [ B⊆ Sy . (1.4) y∈B

Therefore, if the cardinality of S is N , then the cardinality of its power set 2S is 2N and so there are at most 2N elements in M by the inclusion (1.4), a contradiction. Thus the set S must be infinite. Recall that S ⊆ M and M is countable, S is also countable by [99, Theorem 2.8, p. 26]. Since M is a σ-algebra, it must contain the power set of S. However, it is well-known that the power set of an infinite countable set is uncountable [93, Problem 22, p. 16]. We conclude from these facts that M is uncountable, a contradiction. This completes the proof of the problem.  Problem 1.2 Rudin Chapter 1 Exercise 2.

Proof. Put f (x) = (f1 (x), . . . , fn (x)). Since f1 , . . . , fn : X → R, f maps the measurable space X into Rn . Let Y be a topological space and Φ : Rn → Y be continuous. Define
h(x) = Φ (f1 (x), . . . , fn (x)) .

Since h = Φ ◦ f , Theorem 1.7(b) shows that it suffices to prove the measurability of f .

To this end, we first consider f −1 (R) for an open rectangle R in Rn . By the definition, R = I1 × · · · × In , where Ii is an open interval in R for 1 ≤ i ≤ n. Now we know from [74, Exercise 3, p. 21] that f −1 (R) = {x ∈ X | f (x) ∈ R}

= {x ∈ X | fi (x) ∈ Ii for 1 ≤ i ≤ n}

= {x ∈ X | f1 (x) ∈ I1 } ∩ · · · ∩ {x ∈ X | fn (x) ∈ In } = f1−1 (I1 ) ∩ · · · ∩ fn−1 (In ).

Since each fi is measurable and each Ii is open in R, the set fi−1 (Ii ) is measurable in X by Definition 1.3(c). Thus this implies that f −1 (R) is measurable in X by Comment 1.6(c). Our proof will be complete if we can show that every open set V in Rn can be written as a countable union of open rectangles Rj in Rn . We need some topology. By [74, Exercise 8(a), p. 83], the countable collection B = {I = (a, b) | a < b and a, b ∈ Q} is a basis that generates the standard topology on R. Then we follow from [74, Theorem 15.1, p. 86] that the collection C = {I1 × · · · × In | Ii = (ai , bi ), ai , bi ∈ Q, 1 ≤ i ≤ n} Rn .a

(1.5)

is a basis for the (product) topology of Since elements in the collection (1.5) are open rectangles and it is countable by [99, Theorem 2.13, p. 29], every open set V in Rn is, in fact, a countable union of open rectangles Rj , i.e, f −1 (V ) = f −1

∞ [

j=1

a

For details, please read [74, §13 and §15].

∞  [ f −1 (Rj ). Rj = j=1

1.1. Problems on σ-algebras and Measurable Functions

3

Hence f −1 (V ) is also a measurable set in X by Definition 1.3(a)(iii). This completes the proof of the problem.  Problem 1.3 Rudin Chapter 1 Exercise 3.

Proof. Let f : X → R ⊂ [−∞, ∞] and M be the σ-algebra of X. Let α be real. By [99, Theorem 1.20(b), p. 9], we can find a sequence {rn } of rational numbers such that α < rn for all n ∈ N, rn → α as n → ∞. In other words, we have (α, ∞] =

∞ [

[rn , ∞]

n=1

which implies that f −1 ((α, ∞]) = f −1

∞ [

∞  [ f −1 ([rn , ∞]). [rn , ∞] = n=1

n=1

Recall that f −1 ([rn , ∞]) = {x | f (x) ≥ rn }

is assumed to be measurable for each n, so we have f −1 ([rn , ∞]) ∈ M and then we follow from Definition 1.3(a)(iii) that f −1 ((α, ∞]) ∈ M. Since α is arbitrary, we conclude from Theorem  1.12(c) that f is measurable. This finishes the proof of the problem. Problem 1.4 Rudin Chapter 1 Exercise 4.

Proof. (a) Let αk = sup{−ak , −ak+1 , . . .} for k = 1, 2, . . .. By the definition, we have αk ≥ −an for all n ≥ k and if α ≥ −an for all n ≥ k, then α ≥ αk . Note that this is equivalent to the fact that −αk ≤ an for all n ≥ k and if −α ≤ an for all n ≥ k, then −α ≤ −αk . In other words, we have −αk = inf{ak , ak+1 , . . . , } for k = 1, 2, . . .. Thus this implies that sup(−an ) = sup{−ak , −ak+1 , . . .} = − inf{ak , ak+1 , . . .} = − inf (an ). n≥k

n≥k

(1.6)

Similarly, we have inf (−cn ) = − sup(cn )

n≥k

(1.7)

n≥k

for a sequence {cn } in [−∞, ∞]. By applying the equality (1.6) and then the equality (1.7), we achieve that n o o o n n lim sup(−an ) = inf sup(−an ) = inf − inf (an ) = − sup inf (an ) = − lim inf (an ) n→∞ k≥1 k≥1 n≥k n≥k n→∞ k≥1 n≥k | {z } cn

which is our desired result.

4

Chapter 1. Abstract Integration (b) This part is proven in [124, Problem 3.5, pp. 32, 33]. (c) Since an ≤ bn for all n = 1, 2, . . ., we must have αk = inf (an ) ≤ inf (bn ) = βk n≥k

n≥k

(1.8)

for all k = 1, 2, . . .. Thus {αn } and {βn } are two sequences in [−∞, ∞] such that αn ≤ βn for all n = 1, 2, . . .. By a similar argument, we also have sup αk ≤ sup βk

k≥m

(1.9)

k≥m

for all m = 1, 2, . . .. Combining the inequalities (1.8) and (1.9), we have n o n o sup inf (an ) ≤ sup inf (bn ) k≥m

n≥k

k≥m

n≥k

for all m = 1, 2, . . .. By Definition 1.13, we have the desired result. For a counterexample to part (b), we consider an = (−1)n and bn = (−1)n+1 for all n = 1, 2, . . .. On the one hand, we have an + bn = 0 for all n = 1, 2, . . . so that lim sup(an + bn ) = 0.

(1.10)

n→∞

On the other hand, we have lim sup an = lim a2k = 1 and n→∞

k→∞

lim sup bn = lim b2k+1 = 1 n→∞

k→∞

which imply that lim sup an + lim sup bn = 2. n→∞

(1.11)

n→∞

Hence we obtain from the results (1.10) and (1.11) that the strict inequality can hold in part  (b). This completes the proof of the problem. Problem 1.5 Rudin Chapter 1 Exercise 5.

Proof. (a) Let MX be a σ-algebra of X. Further, suppose that Sf (±∞) = {x ∈ X | f (x) = ±∞}, and Sg (±∞) = {x ∈ X | g(x) = ±∞}. We see that Sf (∞) =

∞ \

{x ∈ X | f (x) > n}.

n=1

Since f is measurable and (n, ∞] is openb in [−∞, ∞], f −1 ((n, ∞]) = {x | f (x) > n} ∈ MX by Theorem 1.12(b) for each n ∈ N. Thus Sf (∞) ∈ MX by Comment 1.6(c). Similarly, all the other sets Sf (−∞), Sg (∞) and Sg (−∞) belong to MX too. Next, we let X ′ = X\(Sf (∞)∪Sf (−∞)∪Sg (∞)∪Sg (−∞)). Since Sf (∞), Sf (−∞), Sg (∞) and Sg (−∞) are measurable, X ′ ∈ M by Definition 1.3(a)(ii) and Comment 1.6(b). By b

See the proof of Theorem 1.12(c) in [100, p. 13]

1.1. Problems on σ-algebras and Measurable Functions

5

the comment following Proposition 1.24, we know that X ′ is itself a measure space and if we let MX ′ be a σ-algebra of X ′ , then MX ′ ⊆ MX .

(1.12)

Furthermore, the restricted mappings fX ′ : X ′ → R and gX ′ : X ′ → R are also measurable because of Definition 1.3(c) and the fact that any open set V in R is a countable union of segments of the type (α, β) so that V is also open in the extended number system [−∞, ∞].c In fact, we have −1 −1 fX (V ) ∈ MX ′ ′ (V ) = f

and

−1 −1 gX (V ) ∈ MX ′ . ′ (V ) = g

We need to prove one more thing: the mapping −g : X → [−∞, ∞] is measurable. Since g is measurable, we know from [99, Definition 11.13] that {x ∈ X | g(x) > −a} ∈ MX for every real a. It is obvious that {x ∈ X | − g(x) < a} = {x ∈ X | g(x) > −a} for every real a, so we deduce from [99, Theorem 11.15, p. 311] that −g is measurable. Now we are ready to prove the desired results. Notice that

{x ∈ X | f (x) = g(x)} = {x ∈ X | h(x) = 0} = h−1 (0), where h = f − g. Since f and −g are measurable, the new (real) function h = f − g is also measurable by Theorem 1.9(c). Since h−1 (0) =

∞ \

n=1

h−1



−

1 1  , n n

and h−1 ((− n1 , n1 )) ∈ MX ′ for every n ∈ N, we yield from this and the relation (1.12) that h−1 (0) ∈ MX ′ ⊆ MX . This shows the second assertion. For the first assertion, we note that {x ∈ X | f (x) < g(x)} = {x ∈ X | h(x) = f (x) − g(x) < 0} 
= h−1 (−∞, 0) ∪ [Sf (−∞) \ Sg (−∞)] ∪ [Sg (∞) \ Sf (∞)]  \ [Sf (∞) ∩ Sg (∞)] ∪ [Sf (−∞) ∩ Sg (−∞)] .

Recall that Sf (∞), Sf (−∞), Sg (∞), Sg (−∞) ∈ MX , so we have

Sf (−∞) \ Sg (−∞), Sg (∞) \ Sf (∞), Sf (∞) ∩ Sg (∞), Sf (−∞) ∩ Sg (−∞) ∈ MX . (1.13) By the measurability of h and the relation (1.12), we have h−1 ((−∞, 0)) ∈ MX .

(1.14)

Hence the facts (1.13) and (1.14) show that {x | f (x) < g(x)} ∈ MX . (b) This part is proven in [124, Problem 11.3, p. 339]. This completes the proof of the problem.



Problem 1.6 Rudin Chapter 1 Exercise 6. c

This can be seen, again, from the proof of Theorem 1.12(c) or from the comment following Proposition 1.24.

6

Chapter 1. Abstract Integration

Proof. We prove the assertions one by one. • M is a σ-algebra in X. We check Definition 1.3(a). Since X c = ∅, it is at most countable. Thus we have X ∈ M. Let A ∈ M. If Ac is at most countable, then Ac ∈ M. Similarly, if A is at most countable, then since (Ac )c = A, we have Ac ∈ M. Suppose that An ∈ M for n = 1, 2, . . .. Then we have either An or Acn is at most countable for n = 1, 2, . . .. If all An are at most countable, then we deduce from the corollary following [99, Theorem 2.12, p. 29] that the set A=

∞ [

An

(1.15)

n=1

is also at most countable so that A ∈ M. Otherwise, without loss of generality, we suppose that A1 is uncountable but Ac1 is at most countable. Then we consider Ac =

∞ [

An

n=1

c

=

∞ \

n=1

Acn ⊆ Ac1

which means that Ac is at most countable too. Hence Ac ∈ M and then M is a σ-algebra in X. • µ is a measure on M. We check Definition 1.18(a). Since ∅ is at most countable, we have µ(∅) = 0 so that µ is not identically ∞. In fact, it is clear that we have µ : M → {0, 1} ⊂ [0, ∞]. Let {An } be a disjoint countable collection of members of M. Case (i): All An are at most countable. Recall that the set A given by (1.15) is at most countable. By the definition of µ, we have µ(A) = µ(An ) = 0 for all n = 1, 2, . . .. Thus we have ∞ X µ(An ) (1.16) µ(A) = n=1

in this case.

Case (ii): There is at least one Ak is uncountable. Since Ak ∈ M, Ack must be at most countable. Since An ∩ Ak = ∅ for all n 6= k, we have An ⊆ Ack for all n 6= k. In other words, the measurable sets An are at most countable for all n 6= k. Therefore, we have µ(Ak ) = 1 and µ(An ) = 0 for all n 6= k. Since Ac =

∞ \

n=1

Acn ⊆ Ack ,

Ac

is at most countable and then µ(A) = 1. Hence the equality (1.16) also holds in this case.

This completes the proof that µ is a measure on M. • The determination of measurable functions and their integrals. Let f : X → R be a measurable function. Since X is uncountable, we have µ(X) = 1 which is the only thing that we know and start with. For every n ∈ Z, we know that   [n, n + 1) = R \ (−∞, n) ∪ [n + 1, ∞) .

d

By the fact thatd f −1 (A \ B) = f −1 (A) \ f −1 (B), we obtain


 f −1 [n, n + 1) = f −1 (R) \ f −1 (−∞, n) ∪ f −1 [n + 1, ∞) .

See [74, Exercise 2(d), p. 20].

(1.17)

1.2. Problems related to the Lebesgue’s MCT/DCT

7

By [99, Theorem 11.15, p. 311], each set on the right-hand side in (1.17) is measurable. This implies that
f −1 [n, n + 1) ∈ M.
Let En = f −1 [n, n + 1) , where n ∈ Z. By the definition of M and then the definition of µ, we have either µ(En ) = 0 or µ(En ) = 1. For every x ∈ X, we must have f (x) ∈ [n, n+1) for some n ∈ Z, i.e., x ∈ En for some n ∈ Z. Therefore, we have ∞ [

En = X.

n=−∞

It is clear that R =

∞ [

[n, n + 1), so {En } is a disjoint countable collection of members

n=−∞

of M. By Definition 1.18(a), we see that µ(X) = µ

∞  [

n=−∞

∞  X µ(En ). En =

(1.18)

n=−∞

The fact µ(X) = 1 and the equality (1.18) force that there exists an integer n0 such that µ(En0 ) = 1. Without loss of generality, we may assume that n0 = 0, i.e.,

µ(E0 ) = µ f −1 [0, 1) = 1.

(1.19)

1 1 Next if we write [0, 1) = [0,

2 ) ∪ [ 2 , 1), then the above

argument and the value (1.19) 1 −1 imply that either µ f [0, 2 ) = 1 or µ f −1 [ 21 , 1) = 1. This process can be done continuously so that a sequence of intervals {[an , bn )} is constructed such that



µ f −1 [an , bn ) = 1,

0 ≤ bn − a n ≤

1 2n

and

lim (bn − an ) = 0,

n→∞

where n = 1, 2, . . .. In other words, it means that
µ f −1 (a) = 1


and µ f −1 (b) = 0

for some a ∈ R and all other real numbers b 6= a, but this is equivalent to saying that f −1 (a) is uncountable and f −1 (b) is at most countable for all b 6= a. Now we have completely characterized every measurable function on X. Finally, it is clear that X \ f −1 (a) is a set of measure 0. Therefore, we have Z Z f dµ = a. f dµ = X

f −1 (a)

Hence, this completes the proof of the problem.

1.2

Problems related to the Lebesgue’s MCT/DCT

Problem 1.7 Rudin Chapter 1 Exercise 7.



8

Chapter 1. Abstract Integration


Proof. Let Ek = {x ∈ X | f1 (x) > k} = f1−1 (k, ∞] and E = {x ∈ X | f1 (x) = ∞}, where k = 1, 2, . . .. It is clear that each (k, ∞] is a Borel set in [0, ∞], so each Ek is measurable by Theorem 1.12(b). Since ∞ \ E= Ek , k=1

E is also measurable by Comment 1.6(c). If µ(E) > 0, then we know from the definition that Z f1 dµ = ∞. (1.20) E

By using the result (1.20), Proposition 1.24(b) and Theorem 1.33, we conclude that Z Z Z f1 dµ = ∞ f1 dµ ≥ |f1 | dµ ≥ E X X

which contradicts the hypothesis that f1 ∈ L1 (µ). In other words, we must have µ(E) = 0 and thus f1 ∈ L1 (µ) on X \ E.

Here we may assume that the form of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) is also valid for measurable functions defined a.e. on X.e Now we see that the measurable function f1 in the problem plays the role of g in the theorem and hence our desired result follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) immediately. For a counterexample, we consider X = R, In = (n, ∞), µ = m the Lebesgue measure (see [99, Definition 11.5, pp. 302, 303]) and define for each n = 1, 2, . . .,

fn (x) = χIn (x) =

  1, if x ∈ In ; 

0, if x ∈ / In .

It is clear that f1 ≥ f2 ≥ · · · ≥ 0 on X and Z Z Z fn dm = |fn | dm = R

R

In

dm = ∞

(1.21)

for every n = 1, 2, . . .. In particular, the integrals (1.21) show that f1 ∈ / L1 (m). Furthermore, we have f (x) = lim fn (x) = lim χIn (x) = 0 (1.22) n→∞

n→∞

for every x ∈ R. Thus we deduce from the expression (1.22) and Proposition 1.24(d) that Z f dm = 0. (1.23) R

Hence the inconsistence of the integrals (1.21) and (1.23) show that the condition “f1 ∈ L1 (µ)”  cannot be omitted. This completes the proof of the problem. Problem 1.8 Rudin Chapter 1 Exercise 8.

e Actually, Rudin [100, p. 29] assumed this fact in the proof of Theorem 1.38 or the reader may refer to the comment between Theorem 11.32 and its proof in [99, p. 321].

1.2. Problems related to the Lebesgue’s MCT/DCT

9

Proof. If x ∈ E, then for all k ∈ N, we have


lim f2k (x) = lim 1 − χE (x) = 0.

k→∞

k→∞

Similarly, if x ∈ / E, then for all k ∈ N, we have

lim f2k+1 (x) = lim χE (x) = 0.

k→∞

k→∞

Thus we have lim inf fn (x) = 0 n→∞

for all x ∈ X. However, we have Z

fn dµ =

X

=

 Z  χ dµ,    X E

if n is odd;

Z     (1 − χE ) dµ, if n is even.  X if n is odd;  µ(E), 

(1.24)

µ(X \ E), if n is even.

Therefore we obtain from the results (1.24) that Z fn dµ = min(µ(E), µ(X \ E)) 6= 0 lim inf n→∞

X

if we assume that µ(E) > 0 and µ(X \E) > 0. Hence the example here shows that the inequality in Fatou’s lemma can be strict, completing the proof of the problem.f  Problem 1.9 Rudin Chapter 1 Exercise 9.

Proof. If µ(X) = 0, then Proposition 1.24(e) implies that c =

Z

f dµ = 0, a contradiction. Let X

E = {x ∈ X | f (x) = ∞}. We claim that µ(E) = 0. Otherwise, Proposition 1.24(b) implies that Z Z f dµ = ∞ f dµ ≥ c= X

E

which is a contradiction. Therefore, in the following discussion, we may assume that x ∈ X \ E so that 0 ≤ f (x) < ∞. For each n = 1, 2, . . ., we defineg fn : X \ E → [0, ∞) by

h  f (x) α i . fn (x) = n log 1 + n

Since f is measurable on X \ E, g(x) = [1 + ( nx )α ] and h(x) = n log(1 + x) are continuous on [0, ∞), Theorem 1.7(b) implies that

f g

h  f (x) α i

fn (x) = n log 1 + = h g f (x) n

There is another example in [124, Problem 11.5, p. 340]. X \ E is itself a measure space by the remark following Proposition 1.24.

10

Chapter 1. Abstract Integration

is also measurable on X \ E.

Now we are going to show that when α ≥ 1, there exists a function g ∈ L1 (µ) such that |fn (x)| ≤ g(x)

holds for all n = 1, 2, . . . and all x ∈ X \ E. We first show the following lemma: Lemma 1.2 For each n ∈ N and α ≥ 1, we have h  x α i ≤ αx n log 1 + n

(1.25)

on [0, ∞).

Proof of Lemma 1.2. For x ∈ [0, ∞), we let

Then basic calculus gives

Since

h  x α i F (x) = αx − n log 1 + . n

 x α−1  α α−1 nxα−1  n  = α − αnx =α 1− α . F ′ (x) = α − x α α α n +x n + xα 1+ n

(1.26)

nxα−1 xα ≤ α 1. In this case, we apply L’Hospital’s rule [99, Theorem 5.13, p. 109] to conclude that h  f α i αf α y α−1 0 log(1 + f α y α ) lim n log 1 + = = lim = lim =0 (1.28) n→∞ n y 1+0 y→0+ 1 + f α y α y→0+ on X \ E. Thus it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that  Z  f dµ, if α = 1;   Z  X h  f α i n log 1 + dµ = lim Z n→∞ X  n    0 dµ, if α > 1, =

 X  c, if α = 1; 

0, if α > 1.

It remains the case that 0 < α < 1. In this case, Theorem 1.28 (Fatou’s Lemma) can be applied directly to get ) Z Z ( h  f α i n h  f α io n log 1 + dµ ≤ lim inf lim inf n log 1 + dµ. (1.29) n→∞ X n→∞ n n X Since 0 < α < 1, we have y α−1 =

1 y 1−α

so that the limit (1.28) becomes

h  f α i αf α = ∞. = lim 1−α lim n log 1 + n→∞ n (1 + f α y α ) y→0+ y

(1.30)

Since lim xn = ∞ if and only if lim sup xn = lim inf xn = ∞, the inequality (1.29) and the limit n→∞

n→∞

n→∞

(1.30) combine to imply that n h  f α io lim inf n log 1 + =∞ n→∞ n

and then it certainly gives

lim

n→∞

Z

h  f α i dµ = ∞. n log 1 + n X

This completes the proof of the problem. Problem 1.10 Rudin Chapter 1 Exercise 10.

h

We use the definition ex = lim

n→∞

 x n here. 1+ n



12

Chapter 1. Abstract Integration

Proof. Since fn → f uniformly on X, there exists a positive integer N such that n ≥ N implies |fn (x) − f (x)| ≤ 1 for all x ∈ X. By this, we have |fn (x)| ≤ |fn (x) − f (x)| + |f (x)| ≤ |f (x)| + 1

(1.31)

|f (x)| ≤ |f (x) − fN (x)| + |fN (x)| ≤ |fN (x)| + 1

(1.32)

and for all n ≥ N and x ∈ X. Combining the inequalities (1.31) and (1.32), we see that |fn (x)| ≤ |fN (x)| + 2

(1.33)

g(x) = max{|f1 (x)|, . . . , |fN −1 (x)|, |fN (x)| + 2}.

(1.34)

for all n ≥ N and x ∈ X. We define g : X → [0, ∞) by

Then it is easy to see from the inequality (1.33) and the definition (1.34) that |fn (x)| ≤ g(x) for x ∈ X and n = 1, 2, . . .. Next we want to show that g ∈ L1 (µ). By Theorem 1.9(b), |f1 (x)|, . . . , |fN −1 (x)|, |fN (x)| are measurable. Thus it follows from the corollaries following Theorem 1.14 that g is also measurable. Furthermore, we know from the hypothesis that |fn (x)| ≤ Mn on X for some constants Mn , where n = 1, 2, . . . , N . Therefore, this and the definition (1.34) certainly imply that g(x) = |g(x)| ≤ M on X for some constant M . Since it is obvious that g ≥ 0 on X, we get from Proposition 1.24(a) that Z Z M dµ = M µ(X) < ∞ |g| dµ ≤ 0≤ X

X

L1 (µ).

so that g ∈ In conclusion, our sequence of functions {fn } satisfies the hypotheses of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and hence the desired result follows immediately from this. For a counterexample, consider X = R and µ = m so that m(R) = ∞. For each n ∈ N, define fn : R → R by 1 fn (x) = . n Then it is easy to prove that fn → f ≡ 0 uniformly on R. Since Z Z m(R) =∞ fn dm = f dm = 0 and n R R for n = 1, 2, . . ., we conclude that lim

Z

n→∞ R

This finishes the proof of the problem.

fn dm 6=

Z

f dm.

R



1.2. Problems related to the Lebesgue’s MCT/DCT

13

Problem 1.11 Rudin Chapter 1 Exercise 11.

Proof. For each n ∈ N, define Bn =

∞ [

Ek .

k=n

Recall that A is the set of all x ∈ X which lie in infinitely many Ek . On the one hand, if x ∈ A, ∞ \ Bn . In other words, we have then x ∈ Bn for all n ∈ N so that x ∈ n=1

A⊆

On the other hand, if x ∈

∞ \

n=1

∞ \

Bn .

(1.35)

n=1

Bn , then x ∈ Bn for each positive integer n and this is equivalent

to the condition that x belongs to infinitely many Ek , i.e., ∞ \

n=1

Bn ⊆ A.

(1.36)

Hence the set relations (1.35) and (1.36) imply the desired result that A=

∞ \

Bn =

n=1

∞ [ ∞ \

Ek .

n=1 k=n

We have to show that µ(A) = 0. By the definition of Bn , we have B1 ⊇ B2 ⊇ B3 ⊇ · · · . Furthermore, our hypothesis and the subadditive property of a measure (see , for example, [105, Corollary 4.6, p. 26]) show that µ(B1 ) = µ

∞ [

k=1

∞  X µ(Ek ) < ∞. Ek ≤ k=1

Thus it follows from Theorem 1.19(e) and the subadditive property of µ again that 0 ≤ µ(A) = µ

∞ \

n=1

∞ ∞  [  X Bn = lim µ(Bn ) = lim µ Ek ≤ lim µ(Ek ) = 0. n→∞

n→∞

k=n

n→∞

Hence we must have µ(A) = 0, completing the proof of the problem. Problem 1.12 Rudin Chapter 1 Exercise 12.

k=n



14

Chapter 1. Abstract Integration

Proof. Let n be a positive integer. Define fn : X → [0, ∞] by
fn (x) = min |f (x)|, n

(1.37)

for every x ∈ X. This definition (1.37) clearly satisfies

0 ≤ fn (x) ≤ n

(1.38)

for every x ∈ X. Since f ∈ L1 (µ), it is obviously measurable. By the corollaries following Theorem 1.14, each fn is also measurable on X. Furthermore, if x0 ∈ X such that |f (x0 )| ≤ n, then the definition (1.37) implies that fn (x0 ) = |f (x0 )|

and fn+1 (x0 ) = |f (x0 )|;

(1.39)

if n < |f (x0 )| ≤ n + 1, then we know again from the definition (1.37) that fn (x0 ) = n

and fn+1 (x0 ) = |f (x0 )|;

(1.40)

and fn+1 (x0 ) = n + 1.

(1.41)

if n + 1 < |f (x0 )|, then we must have fn (x0 ) = n

Thus we can conclude from the computations (1.39), (1.40) and (1.41) that the inequality 0 ≤ fn (x) ≤ fn+1 (x)

(1.42)

holds for all n ∈ N and x ∈ X. By Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem), we have Z Z F dµ, fn dµ = lim n→∞ X

X

where F = lim fn . By the definition (1.37), we have F = |f | and so n→∞

lim

Z

n→∞ X

or equivalently, lim

fn dµ =

Z

n→∞ X

Z

X

|f | dµ

|fn − f | dµ = 0.

Thus for every ǫ > 0, there exists a positive integer N such that Z ǫ |fn − f | dµ < 2 X

(1.43)

for all n ≥ N . We fix this N . Since 0 ≤ |f | ≤ |f −fN |+|fN | by the triangle inequalityi , we apply Proposition 1.24(a) and then Theorem 1.27 and the property (1.38) to the inequality (1.43) to derive Z Z Z Z ǫ ǫ |fN | dµ < + |f − fN | dµ + N dµ = + N µ(E) |f | dµ ≤ (1.44) 2 2 E E E E ǫ for every E ∈ M. Therefore, if we take δ = 2N and µ(E) < δ, then our inequality (1.44) becomes Z |f | dµ < ǫ E

as desired. Hence we complete the proof of the problem. i

See [99, Definition 2.15, p. 30].



1.2. Problems related to the Lebesgue’s MCT/DCT

15

Problem 1.13 Rudin Chapter 1 Exercise 13.

Proof. Let f : X → [0, ∞] ⊂ [−∞, ∞] be a measurable function. For each n = 1, 2, . . ., we define fn : X → [0, ∞] ⊂ [−∞, ∞] by fn (x) = nf (x). It is clear that n α  α io i
fn−1 (α, ∞] = {x ∈ X | fn (x) ∈ (α, ∞]} = x ∈ X f (x) ∈ , ∞ = f −1 ,∞ n n
for all real α. Since f is measurable, we have f −1 ( αn , ∞] ∈ M for every real α, where M is a σalgebra in X. Thus we obtain from Theorem 1.12(c) that each fn is measurable for n = 1, 2, . . .. Furthermore, we have 0 ≤ f1 (x) ≤ f2 (x) ≤ · · · ≤ ∞ on X and fn (x) → ∞ · f (x) as n → ∞. By Proposition 1.24(c), we have Z Z Z f dµ nf dµ = n fn dµ = X

(1.45)

X

X

for n = 1, 2, . . .. Hence we conclude from the equality (1.45) and Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) that Z Z Z Z f dµ = ∞ · fn dµ = lim n f dµ. ∞ · f dµ = lim X

n→∞ X

n→∞

X

X

Thus Proposition 1.24(c) also holds when c = ∞ and so we complete the proof of the problem. 

16

Chapter 1. Abstract Integration

CHAPTER

2

Positive Borel Measures

2.1

Properties of Semicontinuity

Problem 2.1 Rudin Chapter 2 Exercise 1.

Proof. We have fn : R → [0, ∞) for all n ∈ N. In Proposition 2.5, we will prove a property which is equivalent to Definition 2.8 and our proof of Statement (a) below becomes simpler. However, we choose to apply Definition 2.8 to prove the statements here. • Statement (a): For any real α, β1 and β2 , let E(α) = {x ∈ R | f1 (x) + f2 (x) < α}

and

Fi (βi ) = {x ∈ R | fi (x) < βi },

(2.1)

where i = 1, 2. If α ≤ 0, then we see that E(α) = ∅ which is open in R. Similarly, Fi (βi ) = ∅ if βi ≤ 0. So in the following discussion, we assume that α > 0, βi > 0 and furthermore, E(α) 6= ∅ and Fi (βi ) 6= ∅. Now we claim that

[

E(α) =

β1 +β2 ≤α β1 ,β2 >0

To prove the claim, on the one hand, if [ x∈

[F1 (β1 ) ∩ F2 (β2 )].

(2.2)

[F1 (β1 ) ∩ F2 (β2 )],

β1 +β2 ≤α β1 ,β2 >0

then there exist real β1 and β2 with β1 + β2 ≤ α and β1 , β2 > 0 so that x ∈ F1 (β1 )∩ F2 (β2 ). By the definition (2.1), this x satisfies f1 (x) < β1 and f2 (x) < β2 and their sum implies f1 (x) + f2 (x) < β1 + β2 ≤ α. Thus we have x ∈ E(α), i.e.,

[

[F1 (β1 ) ∩ F2 (β2 )] ⊆ E(α).

β1 +β2 ≤α β1 ,β2 >0

17

18

Chapter 2. Positive Borel Measures On the other hand, if x ∈ E(α), then we let η = f1 (x) + f2 (x). Define the two numbers β1 and β2 by η+α η+α β1 = − f2 (x) and β2 = − f1 (x). 2 2 By direct computation, we know that βi >

f1 (x) + f2 (x) + f1 (x) + f2 (x) − fi (x) = f1 (x) + f2 (x) − fi (x) ≥ 0, 2

where i = 1, 2. Since η = f1 (x) + f2 (x), it is easy to check that f1 (x) + f2 (x) < α if and only if [f1 (x) + f2 (x)] + [f1 (x) + f2 (x)] < α + η if and only if f1 (x) + f2 (x) < η+α 2 if and only if f1 (x) < β1 . (2.3) Similarly, the above argument can be used to show that f1 (x) + f2 (x) < α if and only if f2 (x) < β2 .

(2.4)

Now we deduce from the inequalities (2.3) and (2.4) that x ∈ F1 (β1 )∩F2 (β2 ). Furthermore, since β1 + β2 = η + α − f1 (x) − f2 (x) = α, we have [ [F1 (β1 ) ∩ F2 (β2 )], x∈ β1 +β2 ≤α β1 ,β2 >0

i.e., E(α) ⊆

[

β1 +β2 ≤α β1 ,β2 >0

[F1 (β1 ) ∩ F2 (β2 )].

Hence the claim (2.2) holds. Since F1 (β1 ) and F2 (β2 ) are open in R, F1 (β1 ) ∩ F2 (β2 ) is also open in R. Since the union of any collection of open sets is open ([99, Theorem 2.24, p. 34]), we follow from this and the equality (2.2) that E(α) is open in R. By Definition 2.8, f1 + f2 is upper semicontinuous. • Statement (b): By a similar argument as in part (a), we can show that the sum of two lower semicontinuous functions is lower semicontinuous. • Statement (c): This is not true in general. We use Proposition A.7 to give a counterex1 , n1 ]. Then each χFn is upper semicontinuous because ample. For each n ∈ N, let Fn = [ n+1 Fn is closed in R. Consider the set ∞ X n 1o χFn (x) < E = x ∈ R . 2 n=1

If x ∈ Fn for some n ∈ N, then we have 1≤

∞ X

n=1

χFn (x) ≤ 2.

In other words, we have Fn * E for n = 1, 2, . . .. Since

(2.5) ∞ [

Fn = (0, 1],a we see that

n=1

(0, 1] * E. a

If x ∈ (0, 1], then there exists a positive integer k such that

1 k+1

(2.6) < x which implies that x ∈ F1 ∪ F2 ∪ · · · ∪ Fk .

2.1. Properties of Semicontinuity

19

However, if x ≤ 0 or x > 1, then x ∈ / Fn for every n ∈ N which means that χFn (x) = 0, i.e., (−∞, 0] ∪ (1, ∞) ⊆ E. (2.7) Hence, by combining the set relations (2.6) and (2.7), we conclude that E = (−∞, 0] ∪ (1, ∞) which is not open in R. By Definition 2.8,

∞ X

χFn is not upper semicontinuous.

n=1

• Statement (d): The set

Fn (α) = {x ∈ R | fn (x) > α}

is open for every real α. By applying part (b) repeatedly, we know that

N X

fn is lower

n=1

semicontinuous for every positive integer N . For every x ∈ [0, ∞), let f (x) = lim

N →∞

N X

fn (x) =

∞ X

fn (x),

n=1

n=1

E(α) = {x ∈ R | f (x) > α}, N X o n fn (x) > α FN (α) = x ∈ R

(2.8)

n=1

for real α and N ∈ N. We claim that

∞ [

E(α) =

FN (α)

(2.9)

N =1

for every real α. Similar to the proof of part (a), we suppose that α > 0, E(α) 6= R and FN (α) 6= R. Suppose that x ∈ E(α), i.e., f (x) > α. Then there exists a ǫ > 0 such that f (x) > α + ǫ. Since each fn is nonnegative,

N nX

n=1

o fn is an increasing sequence. By this and the definition ′

of f in (2.8), there exists a positive integer

N′

such that

N X

fn (x) > α + ǫ which implies

n=1

that x ∈ FN ′ (α), i.e., E(α) ⊆ To prove the other side, if x ∈ ′

N ′, that

i.e.,

N X

∞ [

N =1

∞ [

FN (α).

(2.10)

N =1

FN (α), then x ∈ FN ′ (α) for some positive integer

fn (x) > α. Again, the fact that

N nX

n=1

n=1

fn

o

′

f (x) ≥

N X

n=1

fn (x) > α,

is an increasing sequence implies

20

Chapter 2. Positive Borel Measures i.e., x ∈ E(α) and then

∞ [

N =1

FN (α) ⊆ E(α).

(2.11)

Hence the set relations (2.10) and (2.11) definitely imply the claim (2.9) is true and since each FN (α) is open in R by Definition 2.8, E(α) is also open in R. Since α is arbitrary, f is lower semicontinuous by Definition 2.8. In the proof of Statements (a) and (b) above, we don’t use the property that f1 and f2 are nonnegative. Therefore, they remain valid even if the word “nonnegative” is omitted. However, Statement (c) cannot hold anymore if the word “nonnegative” is omitted. In fact, we consider the sequence of real functions {fn } defined by f1 = χ[−1,1]

and

fn = −χ[ 1 ,

1 ] n n−1

(n = 2, 3, . . .).

1 Since [−1, 1 and [ n1 , n−1 ] are closed in R, we know that f1 , f2 , . . . are upper semicontinuous. It is clear that f1 (0) = −1, so f1 is not a nonnegative function on R. By definition, we have

f=

∞ X

n=1

fn = χ[−1,1] −

∞ X

n=2

χ[ 1 ,

1 ] n n−1

= χ[−1,0] + χ(0,1] −

∞ X

χ[

n=1

1 ,1] n+1 n

.

(2.12)

By the inequalities (2.5), we see that  −2, if x = 21 , 13 , . . .;     ∞  X χ[ 1 , 1 ] (x) = − −1, if x ∈ (0, 1] \ { 21 , 13 , . . .}; n+1 n   n=1    0, if x ≤ 0 or x > 1.

Thus we follow from the expressions (2.12) and (2.13) that  1 1   −1, if x = 2 , 3 , . . .;    f (x) = 0, if x ∈ (0, 1] \ { 12 , 13 , . . .} or x < −1 or x > 1;      1, if x ∈ [−1, 0].

(2.13)

(2.14)

Therefore, the expression (2.14) of f gives n 1o E = x ∈ R f (x) > = [−1, 0] 2

which is not open in R. By Definition 2.8, f is not upper semicontinuous. By Definition 2.8, a function f is lower semicontinuous if and only if −f is upper semicontinuous. This observation indicates that we can deduce a counterexample to Statement (d) from the counterexample (2.12).

Finally, the truths of the Statements (a), (b) and (d) depend only on the range of f and the fact that the union of any collection of open sets in a topological space X is open in X (see  the set equalities (2.2) and (2.9)). This completes the proof of the problem. Problem 2.2 Rudin Chapter 2 Exercise 2.

2.1. Properties of Semicontinuity

21

Proof. We formulate and prove the general statement: Let X be a topological space, U an open set in X containing the point x and f : X → C. Define ϕ(x, U ) = sup{|f (s) − f (t)| | s, t ∈ U } and ϕ(x) = inf{ϕ(x, U ) | U is open, x ∈ U }. (2.15) We claim that ϕ is upper semicontinuous, f is continuous at x ∈ X if and only if ϕ(x) = 0 and the set of points of continuity of an arbitrary complex function is a Gδ . • ϕ is upper semicontinuous. Let E = {x ∈ X | ϕ(x) < α}, where α ∈ R. We have to show that E is open in X. If E = ∅, then there is nothing to prove. Thus we suppose that E 6= ∅. In this case, we have p ∈ E so that ϕ(p) < α. This fact shows that there exists an open set U containing p andb ϕ(p, U ) < α. Pick q ∈ U \ {p}. Since U is open, there exists an open set V containing q such that q ∈ V ⊆ U . Since p, q ∈ U , we know from the definition (2.15) that ϕ(p, U ) = ϕ(q, U ).

(2.16)

Furthermore, we observe from the definition (2.15) that ϕ(x, U ′ ) ≤ ϕ(x, U ) for every open sets U, U ′ with U ′ ⊆ U .

(2.17)

Now these facts (2.16) and (2.17) imply that ϕ(q, V ) ≤ ϕ(q, U ) = ϕ(p, U ) < α and then ϕ(q) ≤ ϕ(q, V ) < α. In other words, q ∈ E. Since q ∈ U \ {p} is arbitrary, we have shown that p ∈ U ⊆ E, i.e., E is open for every α ∈ R and hence ϕ is upper semicontinuous by Definition 2.8. • f is continuous at x if and only if ϕ(x) = 0. Suppose that f is continuous at x. Recall from the definition of continuity ([74, Theorem 18.1, p. 104] or [100, p. 9]) that for every ǫ > 0, the neighborhood n ǫo (2.18) B(f (x), ǫ) = z ∈ C |f (x) − z| < 2

has a neighborhood Uǫ of x such that f (Uǫ ) ⊆ B(f (x), ǫ), i.e., f (y) ∈ B(f (x), ǫ) for all y ∈ Uǫ . Now we take this Uǫ in the definition (2.15): ϕ(x, Uǫ ) = sup{|f (s) − f (t)| | s, t ∈ Uǫ } and it follows from the definition (2.18) thatc ϕ(x, Uǫ ) ≤ ǫ.

(2.19)

By the definition (2.15), we have ϕ(x) ≤ ϕ(x, U ) for every open set U containing x. Therefore, we deduce from this and the inequality (2.19) that ϕ(x) ≤ ǫ. Since ǫ is arbitrary, we have the desired result that ϕ(x) = 0. b Otherwise, we have ϕ(p, U ) ≥ α for every open set U containing p and this implies that α ≤ ϕ(p), a contradiction. c Note that |f (s) − f (t)| ≤ |f (s) − f (x)| + |f (x) − f (t)| < ǫ for every s, t ∈ Uǫ .

22

Chapter 2. Positive Borel Measures Conversely, suppose that ϕ(x) = 0. Thus for every ǫ > 0, there exists a neighborhood Uǫ of x such that ϕ(x, Uǫ ) < ǫ. By the definition (2.15), this implies that |f (s) − f (t)| < ǫ

(2.20)

for every s, t ∈ Uǫ . In particular, if we take s = x to be fixed and t = y vary, then the inequality (2.20) can be rewritten as |f (x) − f (y)| < ǫ for every y ∈ Uǫ . By the definition ([74, Theorem 18.1, p. 104] or [100, p. 9]), f is continuous at x. • The set of points of continuity of an arbitrary complex function is a Gδ . By the previous analysis, we establish that G = {x ∈ X | f is continuous at x} = {x ∈ X | ϕ(x) = 0} = Since ϕ is upper semicontinuous, each set {x ∈ X | ϕ(x) < 1.11, we see that G is actually a Gδ .

1 n}

∞ n \

n=1

1o x ∈ X ϕ(x) < . n

is open in X. By Definition

This completes the proof of the problem.



Problem 2.3 Rudin Chapter 2 Exercise 3.

Proof. The first part is proven in [124, Problem 4.20, pp. 73, 74]. The function in the question can be used to prove Urysohn’s Lemma for any metric space X directly. Lemma 2.1 (Urysohn’s Lemma) Suppose that X is a metric space with metric ρ, A and B are disjoint nonempty closed subsets of X. Then there exists a continuous function f : X → [0, 1] such that 0 ≤ f (x) ≤ 1 for all x ∈ X, f (x) = 0 precisely on A and f (x) = 1 precisely on B.

Proof of Lemma 2.1. This lemma is proven in [124, Problem 4.22, pp. 75, 76]. Readers are recommended to read [74, Theorem 32.2, p. 202; Theorem 33.1, pp. 207 – 210].  This ends the proof of the problem.



Problem 2.4 Rudin Chapter 2 Exercise 4.

Proof. (a) Recall from [100, Eqn. (2), p. 42] that µ(E) = inf{µ(V ) | E ⊆ V and V is open}

(2.21)

which is defined for every subset E of X.d By the result [100, Eqn. (4), p. 42], we have µ(E1 ∪ E2 ) ≤ µ(E1 ) + µ(E2 ). d

Here we don’t require E to be a measurable set.

(2.22)

2.1. Properties of Semicontinuity

23

We need a lemma: Lemma 2.2 For disjoint open sets V1 and V2 , we have µ(V1 ∪ V2 ) = µ(V1 ) + µ(V2 ).

Proof of Lemma 2.2. The proof of the inequality µ(V1 ∪ V2 ) ≤ µ(V1 ) + µ(V2 ) can be found in [100, STEP I, p. 42], so we only prove the other side. Suppose that g ∈ Cc (X) and g ≺ V1 . Similarly, suppose that h ∈ Cc (X) and h ≺ V2 . By Definition 2.9, since Cc (X) is a vector space, f = g + h ∈ Cc (X). Furthermore, we know from Definition 2.9(a) that supp (f ) = supp (g + h) ⊆ supp (g) + supp (h) ⊆ V1 ∪ V2 .

(2.23)

By assumption, we have V1 ∩V2 = ∅ which means that g(x) = 0 on X \V1 and h(x) = 0 on X \ V2 . Thus we deduce from these facts that  g(x), if x ∈ V1 ;      f (x) = h(x), if x ∈ V2 ;      0, if x ∈ X \ (V1 ∪ V2 ).

In other words, we have

0 ≤ f (x) ≤ 1

(2.24)

on X. Therefore, we can conclude from the set relation (2.23) and the inequalities (2.24) that f ≺ V1 ∪ V2 . (2.25) Now, by Definition 2.1, the relation (2.25) and then [100, Eqn. (1), p.41], we obtain that Λ(g) + Λ(h) = Λ(g + h) = Λ(f ) ≤ µ(V1 ∪ V2 ). (2.26) We first fix the h in the inequality (2.26) and since it holds for every g ≺ V1 , the definition of supremum gives µ(V1 ) + Λ(h) ≤ µ(V1 ∪ V2 ).

(2.27)

Next, the inequality (2.27) holds for every h ≺ V2 , we establish from the definition of supremum that µ(V1 ) + µ(V2 ) ≤ µ(V1 ∪ V2 ). Hence we have µ(V1 ) + µ(V2 ) = µ(V1 ∪ V2 ).



Let’s go back to the original proof. We want to compute µ(E1 ∪ E2 ). By the definition (2.21), it needs to consider all open sets containing E1 ∪ E2 . Since V1 ⊆ V2 certainly implies µ(V1 ) ≤ µ(V2 ), we can restrict our attention to any open set W such that E1 ∪ E2 ⊆ W ⊆ V1 ∪ V2 . Since V1 ∩ V2 = ∅, we have W = (W ∩ V1 ) ∪ (W ∩ V2 ), where (W ∩ V1 ) ∩ (W ∩ V2 ) = ∅.

24

Chapter 2. Positive Borel Measures Furthermore, since W ∩ V1 and W ∩ V2 are open sets in X, it yields from Lemma 2.2 that µ(W ) = µ(W ∩ V1 ) + µ(W ∩ V2 ) ≥ µ(E1 ) + µ(E2 ).

(2.28)

Since W is arbitrary, µ(E1 ) + µ(E2 ) is a lower bound of the set in the definition (2.21). Hence we follow from this fact and the inequality (2.28) that µ(E1 ∪ E2 ) ≥ µ(E1 ) + µ(E2 ).

(2.29)

By the inequalities (2.22) and (2.29), we have the desired result that µ(E1 ∪ E2 ) = µ(E1 ) + µ(E2 ). (b) Define E1 = E. By [100, STEP V, p. 44], there exists a compact set K1 and an open set V1 such that 1 K1 ⊆ E1 ⊆ V1 and µ(V1 \ K1 ) < 2 . 2 By [100, STEP II, p. 43], we have K1 ∈ MF . Then we follow from [100, STEP VI, p. 44] that E1 \ K1 ∈ MF . Define E2 = E1 \ K1 so that E = E1 = E2 ∪ K1 , where µ(E2 ) = µ(E1 \ K1 ) ≤ µ(V1 \ K1 ) < 212 . By induction, we can show that there exists a sequence of compact sets {Kn } and a sequence of open sets {Vn } such that Kn ⊆ En ⊆ Vn , where En+1 = En \ Kn , En+1 ∈ MF and µ(En+1 ) < the following relations: E = En+1 ∪ K1 ∪ K2 ∪ · · · ∪ Kn

(2.30) 1 . 2n+1

Therefore, the set E satisfies

and µ(En+1 )
0. 2k+1 2

26

Chapter 2. Positive Borel Measures

For the second assertion, let v be lower semicontinuous and v ≤ χK . Assume that there was a p ∈ R such that v(p) > 0. Since v is lower semicontinuous, Definition 2.8 implies that the set A = {x ∈ R | v(x) > 0} is open in R. By the assumption, we have p ∈ A, so there exists a δ > 0 such that (p − δ, p + δ) ⊆ A. (2.34) Now we follow from the hypothesis v ≤ χK and the relation (2.34) that χK (x) > 0 on (p − δ, p + δ), therefore we have (p − δ, p + δ) ∈ K which contradicts to the fact that K contains no segment. Hence we have v ≤ 0 and this completes the proof of the problem.  Problem 2.7 Rudin Chapter 2 Exercise 7.

Proof. By the idea used in Problem 2.6, it is not hard to see that the construction of the compact ǫ sets Kn and K still work for removing the “middle 22n−1 th” segments. In this case, instead of the Lebesgue measure (2.33), we have m(Kn ) = 1 − and so

n X k=1

ǫ 22k−1

· 2k−1 = 1 − ǫ

n X 1 2k

(2.35)

k=1

m(K) = lim m(Kn ) = 1 − ǫ. n→∞

Since K is closed in [0, 1], the complement E = [0, 1] \ K

(2.36)

is definitely open in [0, 1]. Now it remains to show that E is dense in [0, 1]. We prove an equivalent definition of a dense set: Let A ⊆ B ⊆ R. We say A is dense in B means that for every x, y ∈ B with x < y, we can find z ∈ A such that x < z < y. In particular, suppose that x, y ∈ [0, 1] with x < y. If (x, y) ∩ E = ∅, then the definition (2.36) says that (x, y) ⊆ K which is impossible. Therefore, we have (x, y) ∩ E 6= ∅ and this means that there exists z ∈ E such that x < z < y. Hence E is dense in [0, 1],  completing the proof of the problem. Problem 2.8 Rudin Chapter 2 Exercise 8.

Proof. Let {rn } be the enumeration of all rationals in R. Let Fn be the segment given by  1 1  Fn = rn − 2n , rn + 2n , (2.37) 2 2 where n = 1, 2, . . .. Define the sets Gn = Fn \ (Fn+1 ∪ Fn+2 ∪ · · · ) = Fn \ We divide the proof into several steps.

∞ [

k=1

Fn+k .

(2.38)

2.2. Problems on the Lebesgue Measure on R

27

• Step 1: m(Gn ) > 0. By the definition (2.37), we know that m(Fn+1 ) = n = 1, 2, . . .. By the definition (2.38), for every n ∈ N, we have Fn = Gn ∪

∞ [

m(Fn ) 22

for every

Fn+k

k=1

so that the subadditive property of a measure (see the proof of Problem 1.11) implies that m(Fn ) ≤ m(Gn ) +

∞ X

m(Fn+k ) = m(Gn ) + m(Fn )

k=1

∞ X 1 m(Fn ) = m(Gn ) + . 22k 3 k=1

Therefore, we have m(Gn ) ≥

2m(Fn ) >0 3

(2.39)

for every positive integer n. n) • Step 2: Existence of a Borel subset An ⊂ Fn with m(An ) = m(F 2 . It is easy to 1 check from the definition (2.37) that m(Fn ) = 22n−1 . By Definition 2.19, we know thate

Fn′ =

 1 h 1 i Fn − rn − 2n = (0, 1). m(Fn ) 2

Thus, by the proof of Problem 2.7f , there exists a (totally disconnected) compact set A′n ⊂ Fn′ such that m(A′n ) = 21 . Then Theorem 2.20(c) implies that the set An given by  1  An = m(Fn )A′n + rn − 2n 2

is a (totally disconnected) compact subset of Fn with measure  
1  m(Fn ) m(An ) = m m(Fn )A′n + rn − 2n . (2.40) = m m(Fn )A′n = m(Fn )m(A′n ) = 2 2

Since An is compact, it is closed by the Heine–Borel theorem [99, Theorem 2.41, p. 40]. By Definition 1.11, it is also a Borel set, completing the proof of Step 2. Now we can construct a Borel set with the desired properties. The construction is as follows: For every positive integer n, we define En = Gn ∩ An

and E =

∞ [

En .

(2.41)

n=1

By the definition (2.38), since each Fn is Borel, each Gn is also Borel. Thus each En and their countable union E are also Borel sets. It remains to show that the E satisfies the requirements of the problem. • Step 3: 0 < m(E ∩ Fn ). We first show that m(En ∩ Fn ) > 0 for every positive integer n. To prove this claim, we note from the definitions (2.38) and (2.41) that En ⊆ Gn ⊆ Fn , so we have m(En ∩ Fn ) = m(En ) = m(Gn ∩ An ). Furthermore, since Gn ⊆ Fn and An ⊂ Fn (by Step 2), we have m(Gn ∪ An ) ≤ m(Fn ). e f

It is just a translation and an enlargement. Take ǫ = 21 in the measure (2.35).

(2.42)

28

Chapter 2. Positive Borel Measures Combining the inequality (2.39) and the expression (2.40), we can reduce the inequality (2.42) to m(Gn ∪ An ) < m(Gn ) + m(An ). (2.43) Now we apply a property of additive function [99, Eqn. (7), p. 302] to the inequality (2.43) to obtain m(Gn ∪ An ) < m(Gn ∪ An ) + m(Gn ∩ An ) which means that m(Gn ∩ An ) > 0 for every positive integer n. Therefore, we follow from this and the fact that En ∩ Fn ⊆ E ∩ Fn 0 < m(Gn ∩ An ) = m(En ∩ Fn ) ≤ m(E ∩ Fn ) for every positive integer n. • Step 4: m(E ∩ Fn ) < m(Fn ). Fix an integer n. For m < n, we apply the identity A \ B = B c ∩ A to the definition (2.38) to get c Gcm = [Fm \ (Fm+1 ∪ Fm+2 ∪ · · · )]c = (Fm+1 ∪ Fm+2 ∪ · · · ) ∪ Fm

which implies that Fn ⊆ Gcm for m = 1, 2, . . . , n − 1. Recall from the definition (2.41) that Enc = Gcn ∪ Acn , so we must have c Fn ⊆ Em for m = 1, 2, . . . , n − 1. In other words, it says that Fn ∩ Em = ∅ for m = 1, 2, . . . , n − 1. Therefore, this fact implies m(E ∩ Fn ) = m

∞  [

m=1

∞ ∞  [   [

 Am ∩ Fn . Em ∩ Fn ≤ m Em ∩ Fn = m

(2.44)

m=n

m=n

By the subadditive property of a measure again, we further deduce the inequality (2.44) to ∞ ∞ X X m(Am ). (2.45) m(Am ∩ Fn ) ≤ m(E ∩ Fn ) ≤ m=n

Recall the result (2.40) and the fact m(Fn+1 ) = (2.45) that m(E ∩ Fn ) ≤

m=n

m(Fn ) , 22

so we derive from the inequality

∞ ∞ X m(Fn ) X 1 2m(Fn ) m(Fn+m ) = < m(Fn ) = 2m 2 2 m=0 2 3 m=0

for every positive integer n. • Step 5: 0 < m(E ∩ I) < m(I) for every nonempty segment I. By Step 3 and Step 4, the inequalities 0 < m(E ∩ Fn ) < m(Fn ) (2.46) hold for every positive integer n, where Fn and E are given by (2.37) and (2.41) respectively. Lemma 2.3 Let I = (α, β) with α < β. Then there is a positive integer n0 such that Fn0 ⊆ I.

2.2. Problems on the Lebesgue Measure on R

29

γ+β Proof of Lemma 2.3. Let γ = α+β 2 and δ = 2 . By the density of Q in R, there exists a rational in (γ, δ). Let it be rn for some positive integer n. If Fn ⊆ I, then we are done. Otherwise, we may find another rational in (rn , δ), namely rn+1 . In fact, this process can be repeated m times, i.e., there is a rational rn+m in (rn+m−1 , δ). It is clear that 1 1 1 (2.47) rn+m + 2(n+m) ≤ δ + 2(n+m) < δ + 2m 2 2 2 and 1 1 1 rn+m − 2(n+m) > γ − 2(n+m) > γ − 2m . (2.48) 2 2 2 Now we may pick the m large enough such that

1

We return to the proof of the problem. By Lemma 2.3, we have E ∩ Fn0 ⊆ E ∩ I and the left-hand side of the inequalities (2.46) shows that 0 < m(E ∩ Fn0 ) ≤ m(E ∩ I).

(2.51)

Since I = Fn0 ∪ (I \ Fn0 ) and Fn0 ∩ (I \ Fn0 ) = ∅, we have E ∩ I = (E ∩ Fn0 ) ∪ [E ∩ (I \ Fn0 )]. Hence, by applying the right-hand side of the inequalities (2.46) and Theorem 1.19(b) twice, we obtain
m(E ∩ I) = m(E ∩ Fn0 ) + m E ∩ (I \ Fn0 ) < m(Fn0 ) + m(I \ Fn0 ) = m(I).

(2.52)

Now our desired result follows immediately if we combine the inequalities (2.51) and (2.52). • Step 6: m(E) < ∞. By the subadditive property of the measure m, the expression (2.40) and the definition (2.41), we get m(E) ≤

∞ X

n=1

m(En ) ≤

∞ X

m(An ) =

n=1

This completes the proof of the problem.g

∞

∞

n=1

n=0

1X m(F1 ) X 1 2m(F1 ) m(Fn ) = = < ∞. 2n 2 2 2 3 

g Instead of Borel sets of the real number line R, Rudin proved a similar result for measurable set A in [0, 1], see [101]. Furthermore, there are two interesting results ([25] and [57]) related to this problem and some of its applications can be found in [24], [35], [43], [67], [87] and [115].

30

Chapter 2. Positive Borel Measures

2.3

Integration of Sequences of Continuous Functions

Problem 2.9 Rudin Chapter 2 Exercise 9.

Proof. For every n ∈ N, we consider the function gn : [−1, 1] → [0, 1] defined by gn (x) =

  0, 

if x 6∈ [− n1 , n1 ];

(2.53)

1 − n|x|, if x ∈ [− n1 , n1 ].

It is clear that each gn is continuous on [−1, 1] and 0 ≤ gn ≤ 1 on [−1, 1]. The graph of gn is shown in Figure 2.1 below.

Figure 2.1: The graph of gn on [−1, 1]. Next, we define gn,k : [0, 1] → [0, 1] by  k  gn,k (x) = gn x − , 2n

(2.54)

k where x ∈ [0, 1] and k = 0, 1, . . . , 2n. It is easy to check that x − 2n ∈ [−1, 1] so that each gn,k is well-defined by (2.54). We claim that if we define the sequence {f1 , f2 , . . .} to be

{g1,0 , g1,1 , g1,2 , g2,0 , g2,1 , g2,2 , g2,3 , g2,4 , . . .}, then {fn } satisfies the hypotheses of the problem.

To this end, we first note that since each gn,k is continuous on [0, 1] and 0 ≤ gn,k ≤ 1 on [0, 1], each fn is continuous on [0, 1] and 0 ≤ fn ≤ 1. To see why lim

Z

1

n→∞ 0

we have to check the behaviour of

Z

0

fn (x) dx = 0,

1

gn,k (x) dx

(2.55)

2.3. Integration of Sequences of Continuous Functions

31

for every positive integer n and k = 0, 1, . . . , 2n. In fact, we know from the definitions (2.53) and (2.54) that  1 1  2 + nx, if x ∈ [0, 2n );    1    1 − n|x|, if x ∈ [0, n ); 3 1 3 (2.56) gn,0 (x) = and gn,1 (x) = 2 − nx, if x ∈ [ 2n , 2n ];    0, if x ∈ [ n1 , 1]    3 , 1]. 0, if x ∈ ( 2n

Similarly, we have

gn,2n−1 (x) =

and

   0,

  1 − n x −

gn,2n (x) =

  0, 



if x ∈ [0, 2n−3 2n );

2n−1 2n ,

(2.57)

if x ∈ [ 2n−3 2n , 1]

if x ∈ [0, n−1 n );

(2.58)

1 + n(x − 1), if x ∈ [ n−1 n , 1].

Finally, for k = 2, 3, . . . , 2n − 2, we have  0,       1 − n x − gn,k (x) =       0,



k 2n ,

if x ∈ [0, k−2 2n ); k+2 if x ∈ [ k−2 2n , 2n ];

(2.59)

if x ∈ ( k+2 2n , 1].

The graphs of these gn,k are shown as follows:

Figure 2.2: The graphs of gn,k on [0, 1]. In other words, it is clear from Figure 2.2 that {gn,0 , gn,1 . . . , gn,2n } is a family of tent functions 1 0 h , 2n , . . . , 2n centered at { 2n 2n }. Now we have the following cases. h We use the fact that each gn,k is a part or the whole of an isosceles triangle with height 1 and base less than or equal to n2 .

32

Chapter 2. Positive Borel Measures • When k = 0, we have Z

1

gn,0 (x) dx = 0

• When k = 1, we have Z

1

gn,1 (x) dx =

Z

1 2n

0

0

• When 2 ≤ k ≤ 2n − 2, we have Z 1

Z

1

gn (x) dx = 0

1

2



+ nx dx +

gn,k (x) dx =

• When k = 2n − 1, we have Z Z 1 gn,2n−1 (x) dx = • When k = 2n, we have Z Z 1 gn,2n (x) dx = 0

1 2n−3 2n



Z

(1 − nx) dx =

3 2n 1 2n

1 . 2n

(2.60)

 7 − nx dx = . 2 8n

3

1 2 1 × ×1= . 2 n n

gn (x − 1) dx =

Z

1 1 1− n

(2.61)

(2.62)

2n − 1  7 . 1 − n x − dx = 2n 8n

1 1 1− n

1 n

0

0

0

Z

[1 + n(x − 1)] dx =

(2.63)

1 . 2n

(2.64)

Combining the expressions (2.60) to (2.64), we may conclude that the limit (2.55) holds. Now it remains to show that {fn (x)} converges for no x ∈ [0, 1]. If x = 0, then we obtain from the definitions (2.56) and (2.57) that gn,0 (0) = 1 and

gn,2n−1 (0) = 0.

In other words, we can find subsequences {fnk (0)} and {fnl (0)} such that fnk (0) → 1 and fnl (0) → 0 as k → ∞ and l → ∞ respectively. By Definition 1.13, they imply that lim sup fn (0) = 1 and n→∞

lim inf fn (0) = 0.

(2.65)

n→∞

Similarly, if x = 1, then the definitions (2.56) and (2.58) show that gn,0 (1) = 0 and gn,2n (1) = 1 which then imply the limits (2.65). If x ∈ (0, 1) ∩ Q, then we have x = Thus it follows from the definitions (2.56) and (2.59) that p p = 0 and gnq,2np =1 gn,0 q q

p q

=

2p 2q ,

where 0 < p < q.

(2.66)

for n ≥ q. Therefore, the limits (2.65) also hold for x ∈ (0, 1) ∩ Q. For irrational x in (0, 1), the Archimedean Property shows that there exists N ∈ N such that x > N1 . Thus it implies that gn,0 (x) = 0 for all n ≥ N . By the density of rationals, there exists a sequence such that pn →x qn

(2.67) { pqnn }

of rationals in (0, 1)

2.3. Integration of Sequences of Continuous Functions

33

as n → ∞, where 0 < pn < qn for all n ∈ N. Since we always have pn h npn − 1 npn + 1 i , ∈ , qn nqn nqn we derive from the second expression (2.66) that p  n gnqn ,2npn =1 qn

(2.68)

for all n ∈ N. Combining the two results (2.67) and (2.68), we have shown that the limits (2.65) remain valid for irrationals x, completing the proof of the problem.  Problem 2.10 Rudin Chapter 2 Exercise 10.

Proof. Given that ǫ > 0. For each positive integer n, we define En = {x ∈ [0, 1] | fk (x) > ǫ for some k ≥ n}.

(2.69)

It is clear that each En is bounded and En ⊇ En+1 ⊇ · · · . If En0 = ∅ for some positive integer n0 , then we know from the definition (2.69) that fk (x) ≤ ǫ for every x ∈ [0, 1] and every positive integer k ≥ n0 . Therefore it implies that 0≤

Z

0

1

fk (x) dx ≤ ǫ

for every positive integer k ≥ n0 . Since ǫ is arbitrary, we obtain from this that Z 1 fn (x) dx = 0. lim n→∞ 0

Without loss of generality, we assume that En 6= ∅ for all n ∈ N. Let’s prove two properties of En first: • Property 1: En is open. Let p ∈ En . Then we have fk (p) > ǫ for some k ≥ n. Define ǫ′ = 12 (fk (p) − ǫ) > 0. Since fk is continuous on [0, 1], it is continuous at p. Thus for this particular ǫ′ , there is a δ > 0 such that |fk (x) − fk (p)| < ǫ′ for all points x ∈ [0, 1] with |x − p| < δ. Now the inequality (2.70) implies that fk (x) > fk (p) − ǫ′ =

fk (p) + ǫ > ǫ. 2

In other words, we have (p − δ, p + δ) ⊆ En as desired.

(2.70)

34

Chapter 2. Positive Borel Measures

• Property 2:

∞ \

n=1

En = ∅. Otherwise, there was a p ∈ [0, 1] such that p ∈ En for all

n ∈ N. Therefore, for each positive integer n, we have fk (p) > ǫ

(2.71)

for some k ≥ n. If n → ∞, then k → ∞ and the inequality (2.71) implies that 0 = lim fn (p) = lim fk (p) ≥ ǫ > 0, n→∞

k→∞

a contradiction. To finish our proof, we have to study the lengths of certain subsets of En . Let F be a finite union of bounded (open or closed) intervals of [0, 1]. Then we may write F =

m [

[ak , bk ] ∪

k=1

s [

(ar , br )

(2.72)

r=1

where 0 ≤ ak < bk ≤ 1 and 0 ≤ ar < br ≤ 1 for k = 1, 2, . . . , m and r = 1, 2, . . . , s. Define the length of F , denoted by ℓ(F ), to be ℓ(F ) =

m X k=1

m s s X
X
X (br − ar ). ℓ (ar , br ) = (bk − ak ) + ℓ [ak , bk ] + r=1

k=1

r=1

Given a nonempty finite union of bounded interval F expressed in the form (2.72) and assume ǫ that F has at least one open interval. In addition, we suppose that ℓ(F ) > ǫ. Let δ = 4s . Then the set m s [ [ [ar + δ, br − δ] G= [ak , bk ] ∪ r=1

k=1

is clearly a nonempty finite union of bounded and closed intervals of [0, 1]. Now it is easy to see that m s X X ǫ (br − ar − 2δ) = ℓ(F ) − 2sδ = ℓ(F ) − ℓ(G) = (bk − ak ) + 2 r=1 k=1

implying that ℓ(G) > ℓ(F ) − ǫ. Here we need a lemma about the sequence of bounded subsets (2.69): Lemma 2.4 Let Sn = {F | F is a finite union of bounded intervals of En } for every positive integer n and Ln = sup{ℓ(F ) | F ∈ Sn }. (2.73) Then we have lim Ln = 0.

n→∞

2.3. Integration of Sequences of Continuous Functions

35

Proof of Lemma 2.4. By Property 1, we know that each Sn is nonempty. It is obvious that {Ln } is a bounded decreasing sequence. By the Monotone Convergence Theorem [99, Theorem 3.14, p. 55], {Ln } converges in R. Assume that this limit was nonzero. • Step 1: The construction of a compact set Kn . There exists a δ > 0 such that Ln ≥ δ for every positive integer n. By the definition (2.73), Ln − 2δn is not an upper bound of {ℓ(F ) | F ∈ Sn }. In other words, there exists a Fn ∈ Sn such that ℓ(Fn ) > Ln − δ · 2−n (2.74) for every positive integer n. By the observation preceding Lemma 2.4, we may n \ Fk ⊆ Fn . Since each Fk assume further that each Fn is closed. Let Kn = k=1

is closed in R, Kn is closed in R too. By the Heine-Borel Theorem, the set Kn must be compact. Furthermore, we have Kn ⊇ Kn+1 for each n = 1, 2, . . ..

• Step 2: Each Kn is nonempty. There exists a F ∈ Sn such that ℓ(F ) ≥ δ

(2.75)

for each n ∈ N. Otherwise, there is a N ∈ N such that ℓ(F ) < δ for all F ∈ SN which means that δ is an upper bound of LN , but this is a contradiction. Now we are going to show that if Kn = ∅, then it is impossible to have the inequality (2.75). This contradiction bases on the following two facts: – Fact 1: Suppose that G is a finite union of bounded intervals of En \ Fn for every n ∈ N, where Fn are those sets considered in Step 1. Then we have G ∩ Fn = ∅ and G ∪ Fn ∈ Sn so that ℓ(G) + ℓ(Fn ) = ℓ(G ∪ Fn ) ≤ Ln . This and the inequality (2.74) give, for every n ∈ N, ℓ(G) < δ · 2−n .

(2.76)

– Fact 2: Suppose that F is a finite union of bounded intervals of En \ Kn for every n ∈ N. By De Morgan’s law (see [74, p. 11]), we have (F \ F1 ) ∪ · · · ∪ (F \ Fn ) = F \ (F1 ∩ · · · ∩ Fn ) = F \ Kn = F.

(2.77)

Note that F \ Fk is also a finite union of bounded intervals of Ek (and hence of Ek \ Fk ) for k = 1, 2, . . . , n, so it follows from the inequality (2.76) that ℓ(F \ Fk ) < δ · 2−k

(2.78)

for k = 1, 2, . . . , n. Hence it follows from the expression (2.77) and the inequality (2.78) that ℓ(F ) < δ. (2.79) If Km = ∅ for some m ∈ N, then the F considered in Fact 2 is a subset of Em , but the inequality (2.79) definitely contradicts the inequality (2.75). • Step 3: A contradiction to Property 2. By Step 1 and Step 2, we ∞ \ Kn 6= ∅. Since Fk ⊆ Ek for deduce from [99, Theorem 2.36, p. 38] that n=1

k = 1, 2, . . . , n, we must have Kn ⊆ En . However, these two facts contradict Property 2.

Hence the limit must be 0 which completes the proof of the lemma.



36

Chapter 2. Positive Borel Measures

We return to the proof of the problem. By Lemma 2.4, for ǫ > 0, there is a positive integer N such that for n ≥ N , we have ℓ(F ) ≤ Ln < ǫ, (2.80) where F is any finite union of bounded intervals of En . Now one may think that we can derive Z 1 fn (x) dx < 2ǫ 0

for n ≥ N directly from the uniform boundedness of fn , the definition (2.69) and the inequality (2.80). However, it fails because we have no way to estimate the integral Z fn (x) dx. En \F

To overcome this problem, we play the trick that since fn is Riemann integrable on [0, 1], the integral must be equal to its lower Riemann integral, see [4, §1.17, p. 74]. Thus we have to find nZ 1 o sup sn (x) dx 0 ≤ sn (x) ≤ fn (x) and sn is a step function , 0

where the sup is taken over all step functions sn below fn on [0, 1]. For every n ≥ N , we define

En = {x ∈ [0, 1] | sn (x) > ǫ} and Fn = [0, 1] \ En . Since sn is a step function, it is clear that En and Fn are finite unions of bounded intervals. Furthermore, since 0 ≤ sn (x) ≤ fn (x), we have En ⊆ En and En ∈ Sn . Then we deduce from the inequality (2.80) that ℓ(En ) < ǫ for all n ≥ N . Hence we obtain from this and repeated uses of [99, Theorem 6.12(c), p. 128] that for all n ≥ N , we have Z Z Z Z Z 1 ǫ dx ≤ ℓ(En ) + ǫ < 2ǫ. dx + sn (x) dx ≤ sn (x) dx + sn (x) dx = 0≤ 0

En

Fn

En

Fn

Since ǫ is arbitrary, we have shown that lim

Z

n→∞ 0

1

fn (x) dx = 0.

This completes the proof of the problem.



Remark 2.1 There are many mathematicians who have provided different proofs of Problem 2.10. See, for examples, [64], [66] and [106].

2.4

Problems on Borel Measures and Lebesgue Measures

Problem 2.11 Rudin Chapter 2 Exercise 11.

2.4. Problems on Borel Measures and Lebesgue Measures

37

Proof. We follow the given hint. Suppose that Ω is the family of all compact subsets Kα of X with µ(Kα ) = 1. Since X is compact and µ(X) = 1, Ω is not empty. Now we define \ K= Kα . (2.81) Kα ∈Ω

Since X is Hausdorff, each Kα is closed in X by the corollaries following Theorem 2.5. Thus K is closed in X and then K is Borel (see Definition 1.11), i.e., K ∈ M. Since K ⊆ X, Theorem 2.4 says that K is also compact. Suppose that K ⊆ V and V is open in X. Then V c is closed in X and so it is compact in X by Theorem 2.4. As we have mentioned in the previous paragraph that each Kα is closed in X, so each Kαc is open in X. By the definition (2.81) and the fact that K ⊆ V , we have [ V c ⊆ Kc = Kαc . Kα ∈Ω

In other words,

{Kαc }

forms an open cover of the compact set V c . Thus we have V c ⊆ Kαc 1 ∪ Kαc 2 ∪ · · · ∪ Kαc n

(2.82)

for some positive integer n. Since µ(Kαm ) = µ(X) = 1 for m = 1, 2, . . . , n, we have µ(Kαc m ) = 0 for m = 1, 2, . . . , n. Therefore, it follows from these and the set relation (2.82) that µ(V c ) ≤ i.e., µ(V that

c)

n X

µ(Kαc m ) = 0,

m=1

= 0 or equivalent µ(V ) = 1. Since µ is regular and K ∈ M, Definition 2.15 implies µ(K) = inf{µ(V ) | K ⊆ V and V is open} = 1

which proves the first assertion. For the second assertion, let H be a proper compact subset of K, i.e., H ⊂ K. Assume that µ(H) = 1. Then it means that H ∈ Ω and the definition (2.81) shows that K ⊆ H, a  contradiction. Hence we must have µ(H) < 1. This completes the proof of the problem. Problem 2.12 Rudin Chapter 2 Exercise 12.

Proof. Let B be the σ-algebra of all Borel sets in R and K be a nonempty compact subset of R. We have to show that there exists a measure µ on B such that K is the smallest closed subset with the property µ(K c ) = 0. Since every compact set K has a countable base (see [99, Exercise 25, Chap. 2, p. 45]), there exists a countable sequence F = {x1 , x2 , . . .} ⊆ K such that K = F. Next we define µ : F → [0, ∞] by

µ(xn ) =

1 2n

and for any E ∈ B, we define µ : B → [0, ∞] by µ(E) =

∞ X 1 χ (E), 2n {xn }

n=1

(2.83)

38

Chapter 2. Positive Borel Measures

where χ{xn } is the characteristic function of the set {xn }. Suppose that {Ei } is a disjoint countable collections of members of B and E=

∞ [

Ei .

i=1

If xn ∈ Ei for some i, then xn ∈ / Ej for all j 6= i. In other words, each element of F belongs to at most one element of {Ei }.i Thus it follows from the definition (2.83) that µ(E) =

X 1 χ (E), 2n {xn } n

(2.84)

where the summation runs through all n such that xn ∈ Ei for some i. Since the series ∞ X 1 2n n=1

converges absolutely, its rearrangement converges to the same limit (see [99, Theorem 3.55, p. 78]). Therefore, we can rewrite the expression (2.84) as µ(E) =

∞ X X 1 χ (Ei ), 2ik {xik }

(2.85)

i=1 ik

where the inner summation runs through all ik such that xik ∈ Ei . It may happen that the set F ∩ Ei is finite for some i = 1, 2, . . .. Since the inner summation in the expression (2.85) is exactly µ(Ei ), we deduce from the expression (2.85) that µ

∞ [

i=1

∞  X µ(Ei ). Ei = µ(E) = i=1

By Definitions 1.18(a) and 2.15, µ is a Borel measure on R. By the definition (2.83), it is easy to see that for every E ∈ B, we have µ(E) > 0 if and only if

K ∩ E 6= ∅

(2.86)

so that µ(K c ) = 0. Assume that there was a proper closed subset H ⊂ K in X such that µ(H c ) = 0. Since H ⊂ K, we have K c ⊂ H c . If H c ∩ K = ∅, then H c ⊆ K c , a contradiction. Thus H c ∩ K 6= ∅. Since H c is open in X, we have H c ∈ B. Therefore, we can conclude from the condition (2.86) that µ(H c ) > 0 which contradicts to our hypothesis. In other words, K is the smallest closed subset in R such that µ(K c ) = 0, or equivalently, K = supp (µ). This completes the analysis of the problem. Problem 2.13 Rudin Chapter 2 Exercise 13. i

However, Ei may contain more than one element of F .



2.4. Problems on Borel Measures and Lebesgue Measures

39

Proof. It is clear that the point set {0} is a compact subset of R. Assume f : R → R was a continuous function such that {0} = supp (f ). Let V = (−∞, 0) ∪ (0, ∞). We know from Definition 2.9 that supp (f ) = f −1 (V ), so {0} = f −1 (V )

(2.87)

so that f −1 (V ) 6= ∅. Since V is open in R and f is continuous on R, f −1 (V ) is open in R by Definition 1.2(c). Thus, if p ∈ f −1 (V ), then there exists a δ > 0 such that (p − δ, p + δ) ⊆ f −1 (V ).

(2.88)

Thus we deduce from the set relations (2.87) and (2.88) that [p − δ, p + δ] ⊆ f −1 (V ) = {0}, a contradiction. Hence there is no continuous function f such that {0} = supp (f ).

For the second assertion, we suppose that K is a compact subset of R. We claim that K is the support of a continuous function if and only if K is the closure of an open set V in X. If K is the support of a continuous function f : R → R, then Definition 2.9 shows that


K = {x ∈ R | f (x) 6= 0} = f −1 (−∞, 0) ∪ (0, ∞) = f −1 (−∞, 0) ∪ f −1 (0, ∞) .



Since f is continuous on R, f −1 (−∞, 0) and f −1 (0, ∞) are open in R. Thus this proves one direction. Conversely, if we have K=V (2.89) for some open set V , then we consider F = V c which is closed in R. Define ρF : R → R by ρF (x) = inf{|x − y| | y ∈ F }. By Problem 2.3, we know that ρF is uniformly continuous on R and ρF (x) = 0 if and only if x ∈ F . Therefore, we must have ρF (x) > 0

if and only if

x ∈ V.

(2.90)

Applying the relation (2.90) to the expression (2.89), we have K = {x ∈ R | ρF (x) 6= 0}, i.e., K is the support of the continuous function ρF . We note that the third assertion is not valid in other topological spaces. We consider the three-element set X = {a, b, c}. By [74, Example 1, p. 76], we see that {∅, {b}, X} is a topology of X. Let K = {a, b}. Then it is easy to check from the definition that K is compact. Now if K = V for some open set V , then K must be closed in X (see [74, p. 95]). However, since X \ K = {c} which is not open in X, K is not closed in X. This gives a counter-example to explain that the description “K is the closure of an open set V in X” cannot guarantee that K is the support of a continuous function in an arbitrary topological space and hence we complete the proof of the problem.  Problem 2.14 Rudin Chapter 2 Exercise 14.

40

Chapter 2. Positive Borel Measures

Proof. Let f : Rk → R. We follow the proof of Theorem 2.24 (Lusin’s Theorem). Suppose that 0 ≤ f < 1. Attach a sequence {sn } of simple measurable functions to f , as in the proof of Theorem 1.17 (The Simple Function Approximation Theorem). Put t1 = s1 and tn = sn − sn−1 for n = 2, 3, . . .. Then 2n tn is the characteristic function of a measurable set Tn ⊆ Rk and f (x) =

∞ X

tn (x) =

∞ X

2−n χTn (x)

(2.91)

n=1

n=1

on Rk . By Theorem 2.20(b), there exist An , Bn ∈ B in Rk such that An ⊆ Tn ⊆ Bn

and

m(Bn \ An ) = 0

(2.92)

for every positive integer n.j We define g : Rk → R and h : Rk → R by g(x) =

∞ X

2−n χAn (x)

and h(x) =

∞ X

2−n χBn (x).

(2.93)

n=1

n=1

We claim that g and h are Borel measurable functions on Rk . It is easy to check that  ∅, if α ≤ 0;      Acn , if 0 < α ≤ 1; {x ∈ Rk | χAn (x) < α} =      R, if α > 1.

Thus each χAn is a Borel measurable function by Definition 1.11. Similarly, each χBn is also a Borel measurable function. Since the partial sums gn and hn of g and h are just simple functions and An , Bn ∈ B, they are Borel measurable functions by the comment following Definition 1.16. Furthermore, for each n = 1, 2, . . ., we have |2−n χAn (x)| ≤ 2−n

and

|2−n χBn (x)| ≤ 2−n

∞ X 2−n converges, it follows from Weierstrass M -test (see [99, Theorem 7.10, p. on Rk . Since Xn=1 X 147]) that 2−n χAn and 2−n χBn converge uniformly on Rk to g and h respectively. By the Corollary (a) following Theorem 1.14, g and h are Borel measurable. This proves our claim.

By the measure (2.92), we know that g = h a.e. [m] on Rk . Obviously, if F ⊆ E, then we must have χF (x) ≤ χE (x). Therefore, we observe from the set relations (2.92) that χAn (x) ≤ χTn (x) ≤ χBn (x)

(2.94)

on Rk . Hence we apply the inequalities (2.94) to functions (2.91) and (2.93), we obtain g(x) ≤ f (x) ≤ h(x)

(2.95)

on Rk . Next, if 0 ≤ f < M for some positive constant M on Rk , then the above argument can f be applied to M so that the inequalities (2.95) hold with g and h replaced by M g and M h respectively. j

In fact, each An is an Fσ and each Bn is a Gδ , see Definition 1.11.

2.4. Problems on Borel Measures and Lebesgue Measures

41

In the general case, we consider the measurable sets EN = {x ∈ Rk | − N ≤ f (x) ≤ N }, where N ∈ N. Now we have χEN f → f as N → ∞ on Rk . In fact, the above argument shows that we can find Boreal functions gN and hN such that gN (x) = hN (x) a.e. [m] on Rk and gN (x) ≤ χEN f (x) ≤ hN (x) on Rk . By Theorem 1.14, we see that the functions g and h defined by g(x) = lim sup gN (x)

and

h(x) = lim sup hN (x)

N →∞

(2.96)

N →∞

are Borel measurable functions on Rk . Since gN (x) = hN (x) a.e. [m] on Rk , we have g(x) = h(x) a.e. [m] on Rk . Furthermore, we note that g(x) = lim sup gN (x) ≤ lim sup χEN f (x) ≤ lim sup hN (x) = h(x) N →∞

N →∞

(2.97)

N →∞

on Rk . Since lim χEN f (x) = lim sup χEN f (x) = f (x)

N →∞

N →∞

Rk ,

on the inequalities (2.95) follow immediately from the inequalities (2.96) and the fact (2.97). Hence our desired results also hold in this general case, completing the proof of the problem.  Problem 2.15 Rudin Chapter 2 Exercise 15.

Proof. For each positive integer n, we define fn : [0, ∞) → [0, ∞] by   x n x  e 2 , if x ∈ [0, n];  1− n fn (x) =   0, if x > n.

It is easy to see that each fn is continuous on [0, ∞) and so Definition 1.11 shows that it is Borel measurable. Fix a x ∈ R. Then the Archimedean Property ensures the existence of a positive integer N such that N ≥ x. By this fact and the fact that (1 + nx )n → ex as n → ∞, we have  x x x n x e 2 = e−x · e 2 = e− 2 . lim fn (x) = lim 1 − n→∞ n→∞ n x

Since e− 2 ∈ L1 (R) and

x

|fn (x)| ≤ e− 2

(2.98)

for all n = 1, 2, . . . and x ∈ [0, ∞), it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that Z ∞ Z ∞ x (2.99) lim fn dx = e− 2 dx = 2. n→∞ 0

0

By the inequality (2.98), we get Z n Z Z ∞ fn dx = fn dx − 0

0

∞

n

which implies that

lim

Z

n→∞ 0

∞

Z fn dx ≤

∞

n

fn dx = lim

Z

|fn | dx ≤

n→∞ 0

Z

∞

x

n

e− 2 dx = 2e− 2

n

n

fn dx.

(2.100)

42

Chapter 2. Positive Borel Measures

Combining the results (2.99) and (2.100), we obtain Z n fn dx = 2. lim n→∞ 0

For the second integral, we define gn : [0, ∞) → [0, ∞] by   x n −2x   1 + n e , if x ∈ [0, n];  gn (x) = x    2 , if x > n. e2x

Therefore each gn is continuous on [0, ∞). Furthermore, by using a similar argument as above, we have for x ∈ R,  x n −2x e = ex · e−2x = e−x . lim gn (x) = lim 1 + n→∞ n→∞ n

Since e−x ∈ L1 (R) and |gn (x)| ≤ e−x for all n = 1, 2, . . . and x ∈ [0, ∞), it follows from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) that Z ∞ Z ∞ e−x dx = 1. gn dx = lim n→∞ 0

0

Now we apply the same trick as obtaining the result (2.100), we are able to show that Z n gn dx = 1. lim n→∞ 0

We have completed the proof of the problem.



Problem 2.16 Rudin Chapter 2 Exercise 16.

Proof. Let T : Rk → Rk be linear, Y = T (Rk ) with dim Y = r < k. Let B = {b1 , . . . , br } be a basis of Y . Since B is a finite set of linearly independent vectors of Rk , it can be enlarged to a basis {b1 , . . . , br , br+1 , . . . , bk } for Rk .k Define the linear transformation T : Rk → Rk by T (α1 b1 + · · · + αk bk ) = (α1 , . . . , αk ). Then it is easy to see that T is one-to-one and so det T 6= 0. Furthermore, we have T (Y ) = {(α1 , . . . , αr , 0, . . . , 0) | α1 , . . . , αr ∈ R} so that T (Y ) =

∞ [

Wn ,

(2.101)

n=1

where Wn = {(ξ1 , . . . , ξr , 0, . . . , 0) | −n ≤ ξi ≤ n and i = 1, 2, . . . , r}. By the definition, we have Wn ⊆ Wn+1 k See, for example, [42, Theorem 30.19, p. 279]. Sometimes, this result is called the Basis Extension Theorem.

2.5. Problems on Regularity of Borel Measures

43

for every n ∈ N. It is trivial to see from Theorem 2.20(a) that m(Wn ) = vol (Wn ) = 0

(2.102)

for each n = 1, 2, . . .. In addition, since Rk is a Borel set, Y is also a Borel set. Hence we follow from this fact, the expressions (2.101), (2.102) and Theorem 1.19(d) that
m T (Y ) = lim m(Wn ) = lim vol (Wn ) = 0. (2.103) n→∞

n→∞

When we read the proof of Theorem 2.20(e) and §2.23 carefully, we see that the validity of the formula
m T (E) = | det T |m(E) (2.104)

when T is an one-to-one map of Rk onto Rk is independent of whether T (Rk ) is a subspace of lower dimension or not. As a consequence, we
may apply the formula (2.104) here. Hence we can conclude from m(Y ) = | det T |−1 m T (Y ) and the result (2.103) that m(Y ) = 0, completing  the proof of the problem.

2.5

Problems on Regularity of Borel Measures

Problem 2.17 Rudin Chapter 2 Exercise 17.

Proof. Let d be the mentioned distance in the problem, p = (x1 , y1 ) and q = (x2 , y2 ). We are going to prove the assertions one by one. • X is a metric space with metric d. We check the definition of a metric space, see [99, Definition 2.15, p. 30]: – If p = q, then d(p, q) = |y1 − y2 | = 0. Otherwise, we have  |y1 − y2 |, if x1 = x2 and y1 6= y2 ; d(p, q) = 1 + |y1 − y2 |, if x1 6= x2 , 6= 0.

– It is clear that d(p, q) = d(q, p) holds because |y1 − y2 | = |y2 − y1 |.

– Let r = (x3 , y3 ). Suppose that r 6= p and r 6= q. Otherwise, the triangle inequality is trivial. There are two cases: ∗ Case (i): x1 = x2 . Then we have d(p, q) = |y1 − y2 | and  |y1 − y3 |, if x3 = x1 and y3 6= y1 ; d(p, r) = 1 + |y1 − y3 |, if x3 6= x1 .

(2.105)

Similarly, we have d(q, r) =



|y2 − y3 |, if x3 = x2 and y3 6= y2 ; 1 + |y2 − y3 |, if x3 6= x2 .

(2.106)

Since |y1 − y2 | ≤ |y1 − y3 | + |y3 − y2 |, any combination of the distances (2.105) and (2.106) imply that d(p, q) ≤ d(p, r) + d(r, q).

(2.107)

44

Chapter 2. Positive Borel Measures ∗ Case (ii): x1 6= x2 . In this case, we have d(p, q) = 1 + |y1 − y2 |. If x3 = x1 , then x3 6= x2 so that d(p, r) = |y1 − y3 | and d(q, r) = 1 + |y2 − y3 |. The situation for x3 = x1 is similar. If x3 6= x1 and x3 6= x2 , then we see from the distances (2.105) and (2.106) again that d(p, r) = 1 + |y1 − y3 | and d(q, r) = 1 + |y2 − y3 |. Hence the triangle inequality (2.107) remains true. By the above analysis, we conclude that d is a metric. • X is locally compact. Let X = (R2 , d) and p = (x, y), q = (u, v) ∈ R2 . Fix p first. By the definition, d(p, q) < 1 implies that x = u and |y − v| < 1. Thus the neighborhood B(p, 1) of p is given by B(p, 1) = {q ∈ X | d(p, q) < 1}

= {q ∈ X | x = u and |y − v| < 1}

= {x} × (y − 1, y + 1).

(2.108)

Geometrically, B(p, 1) is the vertical line segment with half length less than 1 and (x, y) as its midpoint. In addition, this line segment is open in R with the usual metric. Furthermore, it is clear from the definition (2.108) that B(p, 1) = {x} × [y − 1, y + 1].

(2.109)

We want to show that B(p, 1) is compact with respect to the metric d. To this end, let {Vα } be an open cover of B(p, 1), i.e., [ B(p, 1) ⊆ Vα . α

Let a ∈ B(p, 1), where a = (s, t) with s = x and t ∈ [y − 1, y + 1]. Then (x, t) ∈ Vα for some α. Since Vα is open in X, there exists a δt ∈ (0, 1) such that B(a, δt ) = {q = (u, v) | u = s = x and |v − t| < δt } = {x} × (t − δt , t + δt ) ⊆ Vα (2.110) so that B(p, 1) ⊆

[

a∈B(p,1) a=(s,t)

B(a, δt ) ⊆

[

Vα .

(2.111)

α

Applying the expressions (2.109) and (2.110) to the set relation (2.111), we see that [ [ B(p, 1) = {x} × [y − 1, y + 1] ⊆ {x} × (t − δt , t + δt ) ⊆ Vα (2.112) t∈[y−1,y+1]

α

which means that {Ut }, where Ut = (t − δt , t + δt ) with t ∈ [y − 1, y + 1], is an open cover of [y − 1, y + 1]. Since [y − 1, y + 1] is compact with respect to the usual metric | · | by the Heine-Borel Theorem, there are finitely many indices t1 , . . . , tk such that [y − 1, y + 1] ⊆ (t1 − δt1 , t1 + δt1 ) ∪ (t2 − δt2 , t2 + δt2 ) ∪ · · · ∪ (tk − δtk , tk + δtk ). Suppose that (x, ti ) ∈ Vαi for i = 1, 2, . . . , k. Hence we follow immediately from the set relation (2.112) that B(p, 1) ⊆ Vt1 ∪ Vt2 ∪ · · · ∪ Vtk . In other words, the closure B(p, 1) is compact with respect to the metric d and since p is an arbitrary point in X, Definition 2.3(f) shows that X is locally compact.l l We remark that we cannot use the Heine-Borel Theorem directly to the closure (2.109) because we are working in the space X with metric d, not R2 with the usual metric.

2.5. Problems on Regularity of Borel Measures

45

• Construction of a positive linear functional. Let f ∈ Cc (X), i.e., f : (R2 , d) → C is a continuous function and supp (f ) is compact. First of all, we notice that the set E = {x ∈ R | f (x, y) 6= 0 for some y} is finite. To see this, let K = supp (f ) and 0 < δ < 1. Since [ K⊆ B(p, δ), p∈K

where B(p, δ) = {x} × (y − δ, y + δ). Now the compactness of K ensures that there exists some positive integer n such that K⊆

n [

B(pi , δ) =

n [

{q ∈ X | u = xi and |yi − v| < δ}

i=1

i=1

which forces E = {x1 , x2 , . . . , xn }, as claimed.

Next, we claim that the mapping Λ : Cc (X) → R defined by Λ(f ) =

n Z X j=1

∞

f (xj , y) dy

−∞

is a positive linear functional. To this end, let f, g ∈ Cc (X). Since Cc (X) is a vector space, we have αf + βg ∈ Cc (X) for any α, β ∈ C. Furthermore, let f (xi , y) 6= 0 and g(xj , y) 6= 0 for at least one y, where 1 ≤ i ≤ n and n + 1 ≤ j ≤ m, and f (zk , y)g(zk , y) 6= 0 for at least one y, where m + 1 ≤ k ≤ r. By the definition, we have f (xi , y) = g(xj , y) = 0, where n + 1 ≤ i ≤ m, 1 ≤ j ≤ n and all y, so we get Λ(αf + βg) =

r Z X s=1

"

=α

∞

n Z X s=1

+β

"

[αf (xs , y) + βg(xs , y)] dy

−∞

∞

f (xs , y) dy +

−∞

m Z X

∞

s=n+1 −∞

= αΛ(f ) + βΛ(g).

Z r X

∞

f (zs , y) dy

s=m+1 −∞ Z ∞ r X

g(xs , y) dy +

s=m+1 −∞

#

g(zs , y) dy

#

By Definition 2.1, Λ is a linear functional on Cc (X). If f ≥ 0, then Λ(f ) ≥ 0. Thus Λ is positive. This proves the claim. • The measures of the x-axis and its compact subsets. In order to apply Theorem 2.14 (The Riesz Representation Theorem), we have to show that X is Hausdorff. Given p, q ∈ X and p 6= q. Let p = (x, y) and q = (u, v). There are two cases: – Case (i): x = u but y 6= v. Then we define δ = |y − x| and it obtains from the definition (2.108) that  δ h    δ δ δ δ i h δ i ∩ B q, = {x} × y − , y + ∩ {x} × v − , v + = ∅. B p, 2 2 2 2 2 2

46

Chapter 2. Positive Borel Measures – Case (ii): x 6= u but y = v. Then the geometric property of the neighborhoods B(p, 1) and B(q, 1) ensures that their intersection must be empty. In conclusion, we have shown that X is Hausdorff. By Theorem 2.14 (The Riesz Representation Theorem), there exists a unique positive measure µ on M associated with Λ. Let E = {(x, 0) | x ∈ R}. Let δ ∈ (0, 1) and

Uδ =

[

[

B(p, δ) =

x∈R

p∈E

{x} × (−δ, δ).

Then Uδ is clearly an open set in X and E ⊆ Uδ . Take (x1 , 0), . . . , (xn , 0) ∈ E and construct


K = B (x1 , 0), δ ∪ B (x2 , 0), δ ∪ · · · ∪ B (xn , 0), δ


= {x1 } × [−δ, δ] ∪ {x2 } × [−δ, δ] ∪ · · · ∪ {xn } × [−δ, δ] .


Since each B (xi , 0), δ is compact, K is also compact and K ⊆ Uδ . By Theorem 2.12 (Urysohn’s Lemma), there exists an f ∈ Cc (X) such that K ≺ f ≺ Uδ . Since µ(Uδ ) = sup{Λ(f ) | f ≺ Uδ } (see [100, Eqn. (1), p. 41]), we have µ(Uδ ) ≥ Λ(f ) =

m Z X j=1

∞

−∞

f (x′j , y) dy,

(2.113)

where x′1 , . . . , x′m are those values of x for which f (x, y) 6= 0 for at least one y and for some positive integer m. Since K ≺ f , we have f (xi , y) = 1 for all i = 1, 2, . . . , n and y ∈ (−δ, δ). Thus we have {x1 , x2 , . . . , xn } ⊆ {x′1 , x′2 , . . . , x′m } and then the inequality (2.113) reduces to n Z δ X dy = nδ µ(Uδ ) ≥ i=1

−δ

so that µ(Uδ ) → ∞ as n → ∞. If V is an open set containing E, then V must contain the open set Uδ for some small δ and thus µ(V ) = ∞ for every open set V containing E. By Theorem 2.14(c), we have µ(E) = inf{µ(V ) | E ⊆ V , V is open} = ∞. Next, let K ⊆ E be compact. Choose δ ∈ (0, 1). Since [
K⊆W = B (a, 0), δ ,

(2.114)

(a,0)∈K

the compactness of K implies that K⊆

m [

i=1

B((ai , 0), δ) =

m [

({ai } × (−δ, δ))

i=1

for some positive integer m. In other words, we know from the definition (2.114) that K = {(a1 , 0), (a2 , 0), . . . , (am , 0)}.

2.5. Problems on Regularity of Borel Measures

47

Besides, since W is open in X, Theorem 2.12 (Urysohn’s Lemma) again ensures there is a g ∈ Cc (X) such that K ≺ g ≺ W. Recall from [100, Eqn. (7), p. 43] that µ(K) = inf{Λ(f ) | K ≺ f } ≤ Λ(g) =

m Z X i=1

∞

g(ai , y) dy = 2mδ.

(2.115)

−∞

Since m is fixed and δ is arbitrary, it follows from the inequality (2.115) that µ(K) = 0. This completes the proof of the problem.



Problem 2.18 Rudin Chapter 2 Exercise 18.

Proof. Recall that X has an order relation ‘ α1 > α2 > · · · .

(2.117)

in X, which is a contradiction to the hint. Therefore, we may assume that α1 ∈ / V0 m

See [74, p. 63].

and

(α1 , α0 ] ⊆ V0 .

(2.118)

48

Chapter 2. Positive Borel Measures Since V is a cover of [ω0 , ω1 ], there exists a V1 ∈ V containing α1 . By the above argument, we may find an α2 ∈ X such that α2 ∈ / V1

and

(α2 , α1 ] ⊆ V1 .

(2.119)

If α2 ∈ V0 , then since [α2 , α0 ] ⊆ V0 and α2 < α1 < α0 , we have α1 ∈ V0 which contradicts the result (2.118). Thus α2 ∈ / V0 and then we deduce from this fact, the results (2.118) and (2.119) that α2 ∈ / V0 ∪ V1 and (α2 , α0 ] ⊆ V0 ∪ V1 . Inductively, we can find αn+1 ∈ X such that αn+1 ∈ / V0 ∪ V1 ∪ · · · ∪ Vn

and

(αn+1 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ Vn .

If this process continues infinitely, then we can find an infinite decreasing sequence (2.117) again, a contradiction. Therefore, the sequence must stop eventually at the N th step and it must stop at αn+1 = ω0 for all n ≥ N , i.e., ω0 ∈ / V0 ∪ V1 ∪ · · · ∪ VN

and

(ω0 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ VN .

Since ω0 ∈ V for some V ∈ V, we conclude that [ω0 , ω1 ] = [ω0 , α0 ] ⊆ V0 ∪ V1 ∪ · · · ∪ VN ∪ V. In other words, X is compact. Next, we prove that X is Hausdorff and we need the help of the following lemma: Lemma 2.5 Define the mapping γ : [ω0 , ω1 ] → [ω0 , ω1 ] by   min(Sα ), if α ∈ [ω0 , ω1 ); γ(α) =  ω1 , if α = ω1 .

Then we have X \ Sα = Pγ(α) = [ω0 , α].

Proof of Lemma 2.5. For every α ∈ [ω0 , ω1 ), Sα 6= ∅ so that the mapping γ is welldefined. By the definition, we have Sα = [γ(α), ω1 ] with α < γ(α). Furthermore, there is no element between α and γ(α). Otherwise, γ(α) is not a smallest element of Sα anymore. Hence we deduce from this that X \ Sα = [ω0 , α] = [ω0 , γ(α)) = Pγ(α) . If α = ω1 , then Sω1 = ∅ and Pγ(ω1 ) = [ω0 , ω1 ] = X so that X \ Sω1 = Pγ(ω1 ) in this  case, completing the proof of the lemma. Let α, β ∈ X and we may assume that α < β. Then we see that α ∈ Pγ(α) = [ω0 , α] and β ∈ Sα . Thus Pγ(α) and Sα are neighborhoods of α and β respectively. By Lemma 2.5, they are disjoint and this shows that X is Hausdorff, as required.

2.5. Problems on Regularity of Borel Measures

49

(b) X \ {ω1 } is open but not σ-compact. Let X \ {ω1 } = [ω0 , ω1 ) and α ∈ [ω0 , ω1 ). As we have proven in part (a) that Pγ(α) = [ω0 , γ(α)) is a neighborhood of α. Hence X \ {ω1 } is open in X. Assume that X \ {ω1 } = [ω0 , ω1 ) was σ-compact. Let K be a compact subset of [ω0 , ω1 ). We know that the family {Px = [ω0 , x) | x < ω1 } is an open cover of K so that there are finitely many indices x1 , x2 , . . . , xk such that K ⊆ Px1 ∪ Px1 ∪ · · · ∪ Pxk . Without loss of generality, we may assume that x1 < x2 < · · · < xk . Thus we have K ⊆ [ω0 , xk ) ⊂ [ω0 , xk ]. Since xk < ω1 , our hypothesis shows that K is at most countable. Since X \ {ω1 } is σ-compact, it is a countable union of compact sets so that X \ {ω1 } is countable. However, we have
X = X \ {ω1 } ∪ {ω1 }

which means that X is countable. Evidently, this result contradicts our hypothesis. Hence X \ ω1 is not σ-compact. (c) Every f ∈ C(X) is constant on Sα for some α 6= ω1 . Suppose that x = f (ω1 ) and B(x, n1 ) = {z ∈ C | |z − x| < n1 }, where n
is a positive integer. Since f is continuous on X
and B(x, n1 ) is open in C, f −1 B(x, n1 ) is an open subset of X and ω1 ∈ f −1 B(x, n1 ) . By the topology (2.116), there exists an α1 ∈ X \ {ω1 } such that   1  . (2.120) (α1 , ω1 ] ⊆ f −1 B x, n By Lemma 2.5, we note that (α1 , ω1 ] is uncountable, so a sequence {αn } exists such that α1 < α2 < · · · < ω1 .

(2.121)

Now the results (2.120) and (2.121) together give   1  Sαn ⊆ Sαn+1 ⊆ f −1 B x, n

for all n ∈ N. Before we proceed further, we need the following result:n Lemma 2.6 Every well-ordered set A has the least upper bound property.

Proof of Lemma 2.6. Let B be a nonempty subset of A having an upper bound in A. Therefore, the set U of upper bounds of B is nonempty. Since U ⊆ A and A is well-ordered, U has a least element, completing the proof of Lemma 2.6.  By the relation (2.121), we see that the nonempty subset {αn } of X is bounded above by ω1 . Therefore, it follows from Lemma 2.6 that sup{αn } exists in X. Call it α. It is n∈N

clear that α ≤ ω1 and n

See [74, Exercise 1, p. 66]

Sα ⊆ Sαn

  1  ⊆ f −1 B x, n

(2.122)

50

Chapter 2. Positive Borel Measures for all n = 1, 2, . . .. In fact, it is impossible to have α = ω1 because the set Sω1 = {x ∈ Y | ω1 < x} is uncountable, where Y is the well-ordered set as described in the “Construction” in the question. As ω1 = α, we have Sω1 = Sα ⊆ Sαn so that Sαn is uncountable too, but it contradicts to the fact that αn is a predecessor of ω1 . Hence we obtain α < ω1 and we deduce from the set relations (2.122) that f (Sα ) ⊆ B(x, n1 ) for all n ∈ N and this means that ∞  1 \ B x, f (Sα ) ⊆ = {x}. n n=1

Now this is exactly our desired result.

(d) The intersection of {Kn } of uncountable compact subsets of X is also uncountable compact. We first prove a lemma which indicates a relationship between the cardinality of a compact set K and the topological property of ω1 in K. Lemma 2.7 Suppose that K is a nonempty compact subset of X. Then K is uncountable if and only if ω1 is a limit point of K.

Proof of Lemma 2.7. Let K be uncountable. By the proof of part (b), we know that K cannot lie inside [ω0 , α] for every predecessor α < ω1 . Otherwise, K will be at most countable, a contradiction. Hence this implies that K ∩ (α, ω1 ] 6= ∅ for every α < ω1 . By the definition (see [74, p. 97]), it follows that ω1 is a limit point of K. Conversely, suppose that ω1 is a limit point of K. By [74, Theorem 17.9, p. 99], K must be infinite. Assume that K was countable. Then K has the following representation K = {αn }, where α1 ≤ α2 ≤ · · · . Since K is compact and X is Hausdorff, K is closed in X by Corollary (a) following Theorem 2.5. Thus it must be ω1 ∈ K, i.e., every neighborhood Sα of ω1 satisfies
K ∩ Sα \ {ω1 } 6= ∅.
Let αN ∈ K ∩ Sα \ {ω1 } for some positive integer N , i.e., α < αN < ω1 . Recall that {αn } is increasing, so we must have αn ∈ Sα for all n ≥ N . By the definition (see [74, p. 98]), {αn } converges to ω1 . However, by an argument similar to the proof of part (c), this fact will imply the contradiction that Sαn is uncountable. Hence K is uncountable and we complete the proof of the  lemma. Let’s go back to the proof of part (d). Denote the intersection of {Kn } by K. Since X is Hausdorff, each Kn is closed in X. Since K ⊆ Kn , Theorem 2.4 implies that K is

2.5. Problems on Regularity of Borel Measures

51

compact. For each n ∈ N, let Kn′ = K1 ∩ K2 ∩ · · · ∩ Kn . It is easy to check that n \

Ki′ =

n \

(K1 ∩ · · · ∩ Ki ) =

i=1

i=1

n \

Ki ,

i=1

so we may assume that K1 ⊇ K2 ⊇ · · · .

(2.123)

Therefore, any finite subcollection of {Kn } is nonempty and Theorem 2.6 shows that K 6= ∅. By Lemma 2.7, ω1 is a limit point of every Kn . Then, using [74, Theorem 17.9, p. 99] again, it means that every neighborhood (α, ω1 ] of ω1 contains infinitely many points of every Kn . Combining this fact and the sequence (2.123), every neighborhood (α, ω1 ] of ω1 must contain infinitely many points of the compact set K. Hence ω1 is a limit point of K and we finally conclude from Lemma 2.7 that K is uncountable. (e) M is a σ-algebra containing all Borel sets in X. Suppose that M = {E ⊆ X | either E ∪ {ω1 } or E c ∪ {ω1 } contains an uncountable compact set}. – M is a σ-algebra. We check Definition 1.3(a). In fact, it is obvious that X ∈ M because X is itself compact uncountable. Let E ∈ M. Then either E ∪ {ω1 } or E c ∪ {ω1 } contains an uncountable compact set. Since (E c )c = E, it must be true that E c ∈ M. ∞ [ En . If there exists an To verify Definition 1.3(a)(iii), let En ∈ M and E = n=1

n0 ∈ N such that En0 ∪ {ω1 } contains an uncountable compact set K, then K ⊆ En0 ∪ {ω1 } ⊆

∞ [

n=1

 En ∪ {ω1 } = E ∪ {ω1 }

so that E ∈ M. Otherwise, all En ∪ {ω1 } contain no uncountable compact sets. This forces that every Enc ∪ {ω1 } contains an uncountable compact set Kn . We note that E c ∪ {ω1 } =

∞ \

n=1

∞  \
Enc ∪ {ω1 } , Enc ∪ {ω1 } = n=1

so it is true that K=

∞ \

n=1

Kn ⊆

∞ \

n=1


Enc ∪ {ω1 } = E c ∪ {ω1 }.

Now the result of part (d) illustrates that K is uncountable compact, thus E c ∈ M which implies that E ∈ M.

– M contains all Borel sets in X. If α < ω1 , then (α, ω1 ) 6= ∅. Let β ∈ (α, ω1 ). Since X = Pβ ∪ [β, ω1 ] and Pβ is countable, [β, ω1 ] must be uncountable. Since [β, ω1 ] = X \Pβ , [β, ω1 ] is closed in X and Theorem 2.4 verifies that [β, ω1 ] is compact. Furthermore, we note that [β, ω1 ] ⊆ Sα , (2.124) so these facts show that Sα ∈ M. Similarly, for every α > ω0 , we have X \ Pα = [α, ω1 ] and then X \ Pα ∈ M. Since M is a σ-algebra, M also contains Pα . Hence M contains the order topology τ of X which proves that B(X) ⊆ M as desired.

52

Chapter 2. Positive Borel Measures

(f) λ is a measure on M but not regular. Suppose that λ : M → [0, ∞] is defined by   1, if E ∪ {ω1 } contains an uncountable compact set; λ(E) = (2.125)  0, if E c ∪ {ω1 } contains an uncountable compact set.

If E ∈ M, then we know from the definition of M in part (e) that it is impossible for both E ∪ {ω1 } and E c ∪ {ω1 } containing uncountable compact sets, so this function is well-defined. – λ is a measure on M. Suppose that {Ei } is a disjoint countable collection of members of M. We claim that at most one Ei ∪ {ω1 } contains an uncountable compact set. To see this, assume that both E1 ∪ {ω1 } and E2 ∪ {ω1 } contained uncountable compact sets K1 and K2 respectively. Since E1 ∩ E2 = ∅, we have

K1 ∩ K2 ⊆ E1 ∪ {ω1 } ∩ E2 ∪ {ω1 }

= E1 ∩ {ω1 } ∪ E2 ∩ {ω1 } ∪ {ω1 } = {ω1 }.

(2.126)

However, part (d) tells us that K1 ∩ K2 is uncountable so that the result (2.126) is impossible. This proves the claim. By the previous analysis, if E1 ∪ {ω1 } is the only set containing an uncountable compact set K, then we get ∞ ∞  [
[ Ei ∪ {ω1 } = Ei ∪ {ω1 }, K⊆ i=1

i=1

so we follow from the definition (2.125) that ∞  [ λ Ei = 1. i=1

On the other hand, we know from the facts λ(E1 ) = 1 and λ(Ei ) = 0 for i = 2, 3, . . . that ∞ X λ(Ei ) = 1. i=1

Therefore, we have

∞ ∞  X [ λ(Ei ) λ Ei =

(2.127)

i=1

i=1

in this case. Similarly, if there is no Ei ∪ {ω1 } contains an uncountable compact set, then each Eic ∪ {ω1 } contains an uncountable compact set Ki so that λ(Ei ) = 0 for every i = 1, 2, . . .. Furthermore, we have ∞ ∞ ∞ ∞ c  [ \ \
\ Ei ∪ {ω1 } Eic ∪ {ω1 } = Eic ∪ {ω1 } = Ki ⊆ K= i=1

i=1

i=1

i=1

and part (d) ensures that K is uncountable compact. Therefore, we obtain ∞  [ Ei = 0 λ i=1

and then the formula (2.127) still holds in this case. Finally, since λ(X) = 1 < ∞. By Definition 1.18(a), λ is a positive measure.

2.5. Problems on Regularity of Borel Measures

53

– λ is not regular. For every α < ω1 , we know from the set relation (2.124) that Sα ∪ {ω1 } must contain an uncountable compact set, so the definition (2.125) gives that λ(Sα ) = 1. If E = {ω1 }, then E c ∪ {ω1 } = {ω1 }c ∪ {ω1 } = X so that {ω1 } ∈ M. Therefore, we derive from the definition (2.125) that λ({ω1 }) = 0. However, these facts give λ({ω1 }) 6= inf{λ(Sα ) | {ω1 } ⊆ Sα }. By Definition 2.15, λ is not regular. (g) The validity of the integral. Let f ∈ Cc (X). By part (c), there exists an α0 < ω1 such that f (x) = f (ω1 ) on Sα0 . Recall from part (e) that Sα0 ∈ M. By the set relation (2.124), every Sα ∪ {ω1 } contains an uncountable compact set and thus, in particular, λ(Sαc 0 ) = 0. By this, we gain Z

f dλ =

X

Z

f dλ + c Sα 0

Z

f dλ =

Sα0

Z

Sα0

f dλ = f (ω1 )λ(Sα0 ) = f (ω1 )

which is the required result. (h) The regular µ associates with the linear functional in part (f ). It is clear that Λ : Cc (X) → C defined by Λ(f ) = f (ω1 ) (2.128) is a linear functional on Cc (X). By Theorem 2.14 (The Riesz Representation Theorem) and Theorem 2.17(b), we have a unique regular positive Borel measure µ on X. By the result (2.128), we know that Z f dµ = f (ω1 ). X

By [100, Eqn. (1), p. 41], we have µ(V ) = sup{Λ(f ) | f ≺ V } = f (ω1 ) for every open set V in X. Thus this implies that µ(E) = inf{µ(V ) | E ⊆ V and V is open} = f (ω1 ) for every E ∈ M. We have completed the proof of the problem.



Remark 2.2 The measure considered in Problem 2.18 is called the Dieudonn´ e’s measure, see [19, Exampl 7.1.3, pp. 68, 69] and [31].

Problem 2.19 Rudin Chapter 2 Exercise 19.

54

Chapter 2. Positive Borel Measures

Proof. Suppose that X is a compact metric space with metric ρ and Λ is a positive linear functional on Cc (X), the space of all continuous complex functions on X with compact support. To begin with the construction of the class µ, we use Λ to define a set function µ∗ on every open set V in X by µ∗ (V ) = sup{Λ(f ) | f ≺ V }

= sup{Λ(f ) | f ∈ Cc (X), 0 ≤ f ≤ 1 on X and supp (f ) ⊆ V }

(2.129)

and for every subset E ⊆ X that µ∗ (E) = inf{µ∗ (V ) | V is open in X and E ⊆ V }.

(2.130)

Now we are going to present the proof by quoting several facts (some are with proofs and some are not). In fact, the idea of the following proof is stimulated by Feldman’s online article [38]. • Fact 1: The set function µ∗ is an outer measure. We check the definition [93, p. 346]. – Since supp (f ) ⊆ ∅ if and only if supp (f ) = ∅, we get from Definition 2.9 that f ≡ 0 on X which implies that µ∗ (∅) = sup{Λf | f ∈ Cc (X), 0 ≤ f ≤ 1 on X and supp (f ) ⊆ ∅} = Λ(0) = 0. – Suppose that E, F ⊆ X and E ⊆ F . Since any open set V containing F must also contain E, we obtain from the definition (2.130) that µ∗ (E) = inf{µ∗ (W ) | W is open in X and E ⊆ W } ≤ inf{µ∗ (V ) | V is open in X and F ⊆ V }

= µ∗ (F ). – The proof of the subadditivity µ∗

∞ [

i=1

∞  X µ∗ (Ei ) Ei ≤

(2.131)

i=1

follows exactly the same as the proof of Step I. Hence µ∗ is an outer measure on X. • Fact 2: µ∗ (K) < ∞ for every compact set K ⊆ X. Since X is open in X, we follow from the definition (2.129) that µ∗ (X) ≤ Λ(1). By Fact 1, we have µ∗ (K) ≤ µ∗ (X) ≤ Λ(1) for every compact set K ⊆ X. • Fact 3: A topological result in metric spaces. Now we need the following topological result about metric spaces which will be used in Fact 4: Lemma 2.8 Let X be a metric space with metric ρ and E a nonempty proper subset of X. Then the set V = {x ∈ X | ρE (x) < ǫ} is open in X, where ǫ > 0 and ρE (x) is defined in Problem 2.3.

2.5. Problems on Regularity of Borel Measures

55

Proof of Lemma 2.8. If V = ∅, then there is nothing to prove. Thus we assume that V 6= ∅ and pick x0 ∈ V so that ρE (x0 ) < ǫ. Then there exists a δ > 0 such that ǫ − δ > 0 and ρE (x0 ) < ǫ − δ. We consider the ball B(x0 , δ) = {x ∈ X | ρ(x, x0 ) < δ} and we claim that B(x0 , δ) ⊆ V . To see this, let x ∈ B(x0 , δ). For every y ∈ E, we have ρ(x, y) ≤ ρ(x, x0 ) + ρ(x0 , y) < δ + ρ(x0 , y) and therefore we obtain ρE (x) = inf{ρ(x, y) | y ∈ E}

≤ inf{ρ(x, x0 ) + ρ(x0 , y) | y ∈ E}

< δ + inf{ρ(x0 , y) | y ∈ E}

= δ + ρE (x0 ) < ǫ.

In other words, x ∈ V . Since x is arbitrary, B(x0 , δ) ⊆ V and then V is open in X,  proving Lemma 2.8. • Fact 4: If V is open in X, then V is measurable with respect to µ∗ . We recall from [93, p. 347] that V ⊆ X is measurable with respect to the outer measure µ∗ if for every subset E of X, we have µ∗ (E) = µ∗ (E ∩ V ) + µ∗ (E ∩ V c ).

(2.132)

It suffices to show the inequality µ∗ (E) ≥ µ∗ (E ∩ V ) + µ∗ (E ∩ V c ) − ǫ

(2.133)

holds for every ǫ > 0 because the other side follows directly from the subadditivity (2.131). By the definition (2.130), there is an open set E ′ such that E ⊆ E ′ and ǫ µ∗ (E) ≥ µ∗ (E ′ ) − . 2

(2.134)

By the property in Fact 1, we have µ∗ (E ∩ V ) ≤ µ∗ (E ′ ∩ V ) and µ∗ (E ∩ V c ) ≤ µ∗ (E ′ ∩ V c ).

(2.135)

Thus it is easy to see from the inequalities (2.134) and (2.135) that one can obtain the inequality (2.133) if we can show the following result holds ǫ µ∗ (E ′ ) ≥ µ∗ (E ′ ∩ V ) + µ∗ (E ′ ∩ V c ) − . 2

(2.136)

To this end, we first find bounds of µ∗ (E ′ ∩ V ) and µ∗ (E ′ ∩ V c ). Since E ′ ∩ V is open in X, the definition (2.129) implies the existence of a continuous function f1 : X → [0, 1] such that supp (f1 ) ⊆ E ′ ∩ V and ǫ µ∗ (E ′ ∩ V ) ≤ Λ(f1 ) + . 8

(2.137)


Let ǫ ∈ 0, 8Λ(1) and δ > 0 be a constant such that δ ≤ ǫ[8Λ(1) − ǫ]−1 . Then we have ǫ δ Λ(1) ≤ 1+δ 8

and F1 =

f1 . 1+δ

(2.138)

56

Chapter 2. Positive Borel Measures Recall the fact that the positivity of Λ implies the monotonicity of Λ. In particular, Λ(f ) ≤ Λ(1) for every f ≺ E ′ ∩ V . Thus we apply the results (2.138) and the monotonicity of Λ to the inequality (2.137) to get µ∗ (E ′ ∩ V ) ≤

1 δ ǫ δ ǫ ǫ Λ(f1 )+ Λ(f1 )+ ≤ Λ(F1 )+ Λ(1)+ = Λ(F1 )+ . (2.139) 1+δ 1+δ 8 1+δ 8 4

We replace E by the closed set (E ′ ∩V )c in Lemma 2.8 to get the open set U . Given δ > 0. Since f1 is uniformly continuous on X, there exists a η > 0 such that |f1 (x) − f1 (y)| ≤ δ for all x, y ∈ X with ρ(x, y) < η. Recall that f1 vanishes on the closed set (E ′ ∩ V )c , so if we take y ∈ (E ′ ∩ V )c , then we have f1 (x) ≤ δ for all x ∈ X with ρ(x, y) < η. Since ρ(E ′ ∩V )c (x) < η, Lemma 2.8 ensures that the set U = {x ∈ X | ρ(E ′ ∩V )c (x) < η} is open in X. In other words, we have established from Fact 3 that there exists an open set U such that (2.140) U ⊃ (E ′ ∩ V )c ⊇ V c and f1 (x) ≤ δ on U . Since E ′ ∩ U is open in X, the definition (2.129) again shows that we can find a continuous function f2 : X → [0, 1] such that supp (f2 ) ⊆ E ′ ∩ U and furthermore, by using similar argument as in the proof of the inequality (2.139), we obtain that µ∗ (E ′ ∩ V c ) ≤ µ∗ (E ′ ∩ U ) ≤ Λ(f2 ) + where F2 =

ǫ ǫ ≤ Λ(F2 ) + , 4 4

(2.141)

f2 1+δ .

Now we have found the bounds (2.139) and (2.141) of µ∗ (E ′ ∩ V ) and µ∗ (E ′ ∩ V c ) respectively. Next, we want to show that F3 ∈ Cc (X), where F3 = F1 + F2 . To this 1 end, we recall the facts that F1 , F2 : X → [0, 1+δ ] because 0 ≤ f1 ≤ 1 and 0 ≤ f2 ≤ 1. Furthermore, we note from the set relation (2.140) that V ∩ U 6= ∅. Therefore, we have the following facts: – Since supp (f1 ) ⊆ E ′ ∩ V , supp (F1 ) ⊆ E ′ ∩ V , i.e., F1 (x) = 0 on (E ′ ∩ V )c . Therefore, we conclude that 1 F3 = F2 ≤ ≤1 1+δ on (E ′ ∩ V )c . Similarly, since supp (F2 ) ⊆ E ′ ∩ U , we have F2 (x) = 0 on (E ′ ∩ U )c and then 1 ≤1 F3 = F1 ≤ 1+δ on (E ′ ∩ U )c .

– On (E ′ ∩ V ) ∩ U , we immediately follow from the inequality in (2.140) that f1 (x) ≤ δ on (E ′ ∩ V ) ∩ U ⊆ U . It is clear that f2 (x) ≤ 1 on (E ′ ∩ V ) ∩ U , so these facts imply that f2 1 δ f1 + ≤ + = 1. F3 = F1 + F2 = 1+δ 1+δ 1+δ 1+δ – On (E ′ ∩ V ) \ U , we have F1 ≤ 1 and F2 = 0 so that F3 ≤ 1.

2.5. Problems on Regularity of Borel Measures

57

– On U \ (E ′ ∩ V ), we have F1 = 0 and F2 ≤ 1 so that F3 ≤ 1. To have a better understanding of the above facts, we can draw some pictures. For examples, Figure 2.3(a) shows the sets V, E ′ and E ′ ∩ V , the shaded blue part in Figure 2.3(b) indicates the closed set (E ′ ∩ V )c and the part inside the circle in Figure 2.4 is the set (E ′ ∩ V )c \ U .

(a) The sets V, E ′ and E ′ ∩ V .

(b) The set (E ′ ∩ V )c .

Figure 2.3: The pictures of V, E ′ , E ′ ∩ V and (E ′ ∩ V )c .

Figure 2.4: The set (E ′ ∩ V )c \ U . Thus we have shown that 0 ≤ F3 (x) ≤ 1 on X. Next, since F1 and F2 are continuous on X, F3 is also continuous on X. By the definition, we know that x0 ∈ supp (F3 ) if and only if x0 ∈ supp (F1 ) ⊆ E ′ ∩ V or x0 ∈ supp (F2 ) ⊆ E ′ ∩ U . In addition, we note that (E ′ ∩ V ) ∪ (E ′ ∩ U ) ⊆ E ′ . On the other hand, if y ∈ E ′ but y ∈ / E ′ ∩ V , then we have ′ c y ∈ (E ∩ V ) and we follow from the set relation (2.140) that y ∈ U . Thus this implies that y ∈ E ′ ∩ U and so E ′ = (E ′ ∩ V ) ∪ (E ′ ∩ U ). In other words, it gives x0 ∈ supp (F3 ) if and only if x0 ∈ supp (F1 ) ∪ supp (F2 ) ⊆ E ′ , i.e., supp (F3 ) ⊆ E ′ .

Finally, it yields from the inequalities (2.139) and (2.141) as well as the definition (2.129)

58

Chapter 2. Positive Borel Measures that µ∗ (E ′ ∩ V ) + µ∗ (E ′ ∩ V c ) ≤ Λ(F1 ) + Λ(F2 ) +

ǫ ǫ ǫ = Λ(F3 ) + ≤ µ∗ (E ′ ) + 2 2 2

which is exactly the inequality (2.136). Hence we obtain the desired result that (2.132). • Fact 5: The set M∗ of all measurable sets (with respect to µ∗ ) is a σ-algebra and µ = µ∗ |B (X) is a (complete) measure. The first assertion follows directly from Carath´eodory’s Theorem [93, Theorem 8, p. 349]. By Fact 4, we have B(X) ⊆ M∗ which is exactly Step VII, where B(X) denotes the set of all Borel sets in X. By Carath´eodory’s Theorem again, it shows that µ is a (complete) measure on B(X) and this is the same as Step IX. Consequently, Steps III, IV, V, VI and VIII can all be skipped.o • Fact 6: Deduction of parts (b) and (d). Let K be a compact set of X. Since X is a metric space, K is also Hausdorff. By Corollary (a) following Theorem 2.5, we know that K is closed in X and so K ∈ B(X). Therefore, we deduce from Facts 2 and 5 that µ(K) = µ∗ (K) ≤ µ∗ (X) ≤ Λ(1) which is exactly part (b) of Theorem 2.14 (The Riesz Representation Theorem). If E is open in X, then E ∈ B(X). Thus we observe from the equation (2.132) and then the definition (2.130) that µ(E) = µ∗ (E) = µ∗ (X) − µ∗ (E c )

= µ∗ (X) − inf{µ∗ (V ) | V is open in X and E c ⊆ V }

= µ∗ (X) − inf{µ∗ (X) − µ∗ (V c ) | V c is closed in X and V c ⊆ E}

= sup{µ∗ (V c ) | V c is closed in X and V c ⊆ E} = sup{µ(V c ) | V c is closed in X and V c ⊆ E}.

(2.142)

Since X is compact, V c is compact by Theorem 2.4 so that the expression (2.142) can be expressed as µ(E) = sup{µ(K) | K compact and K ⊆ E } which is the same as part (d) of Theorem 2.14 (The Riesz Representation Theorem). • Fact 7: Deduction of part (a). As Rudin pointed out in Step X that it suffices to prove Z f dµ (2.143) Λ(f ) ≤ X

holds for f ∈ Cc (X) because the opposite inequality follows by changing the sign of f . In fact, the proof of our inequality (2.143) for compact metric X is essentially the same as that of Step X, except that Rudin applied Step II to show that X µ(K) ≤ Λ(hi ),

o We cannot say at this stage that Step II can be omitted because the proof of Step X still needs it, but we will see very soon in Fact 7 that it is eventually allowable to do so.

2.6. Miscellaneous Problems on L1 and Other Properties

59

P where hi ≺ Vi and hi = 1 on K.p However, this kind of argument can be replaced completely by using Fact 6 that µ(K) ≤ Λ(1) holds for every compact set K ⊆ X. Therefore, we finally arrive at the expected inequality (2.143) and furthermore, Step II can also be skipped. Hence we conclude that Steps II to IX can be replaced by a simpler argument (outer measure in the sense of Carath´eodory). In other words, only Steps I and X should be kept. Of course, the proof of the uniqueness of µ cannot be omitted. This completes the proof of the problem.  Remark 2.3 (a) We remark that a Radon measure is a measure µ defined on the σ-algebra M containing all Borel sets B(X) of the Hausdforff space X and it satisfies µ(K) < ∞ for all compact subsets K, outer regular on M and inner regular on open sets in X, see [40, p. 212] or [93, p. 455]. Hence the unique positive measure µ in Theorem 2.14 (The Riesz Representation Theorem) is in fact a Radon measure when X is Hausdforff. (b) You are also advised to read the paper [102] for a similar proof of Problem 2.19.

2.6

Miscellaneous Problems on L1 and Other Properties

Problem 2.20 Rudin Chapter 2 Exercise 20.

Proof. Let f = sup fn : [0, 1] → [0, ∞]. By Definition 1.30, f ∈ / L1 on [0, 1] is equivalent to n

Z

0

1

|f (x)| dx = ∞.

In fact, the inspiration of the construction of {fn } comes from the idea of Problem 2.9. For each positive integer n, we consider the continuous functions gn : [−1, 1] → [0, ∞) given by  if x ∈ / [− n12 , n12 ];  0, gn (x) =  n − n3 |x|, if x ∈ [− n12 , n12 ]. Next, we define fn : [0, 1] → [0, ∞) by  0, if x ∈ / [ n1 −  1  fn (x) = gn x − =  n n − n2 |nx − 1|, if x ∈ [ n1 −

1 1 , n2 n

+

1 ]; n2

1 1 , n2 n

+

1 ]. n2

(2.144)

Recall that fn (x) is the tent function centered at n1 and the graph of it is an isosceles triangle with height n and base n22 so that each fn is continuous on [0, 1] and its area is n1 which implies that Z 1 1 fn (x) dx = → 0 n 0 p

See Step X for the meanings of the notations used here.

60

Chapter 2. Positive Borel Measures

as n → ∞. Furthermore, if x = 0, then x ∈ / [ n1 −

1 1 , n2 n

+

1 ] n2

for all n ∈ N. and thus

fn (0) = 0 for all n ∈ N by the definition (2.144). If x ∈ (0, 1], then there is a positive integer N such that 1 1 / [ n1 − n12 , n1 + n12 ] for all n ≥ N . Therefore, n + n2 < x for all n ≥ N . In other words, we have x ∈ the definition (2.144) certainly shows that fn (x) → 0 as n → ∞ for all x ∈ (0, 1]. Now it remains to show that f ∈ / L1 on [0, 1]. To this end, we study the behaviour of f as x → 0. For every n ∈ N, we consider every x in [ n1 − n12 , n1 + n12 ]. If n is sufficiently large enough, we have 1 1 , (2.145) x= +o n n where o(x) is the little-o notation.q Therefore, it follows from the definition (2.144) and the relation (2.145) that 1 fn (x) = n + o(1) = + o(1) (2.146) x for large enough n and small positive x. By the definition and the expression (5.102), we have f (x) = sup fn (x) ≥ fn (x) = n

1 + o(1) x

(2.147)

/ L1 on [0, 1], so we conclude from the estimate that (2.147) for small positive x. Obviously, x1 ∈ 1 that f ∈ / L on [0, 1]. This completes the proof of the problem.  Problem 2.21 Rudin Chapter 2 Exercise 21.

Proof. Let α ∈ f (X) ⊆ R and


Eα = {x ∈ X | f (x) ≥ α} = f −1 [α, ∞) .

By the definition, it is clear that Eα is nonempty. Furthermore, since
Eα = X \ f −1 (−∞, α) ⊆ X,

the set Eα is closed in X. Now we consider the collection {Eα } of subsets of X. Suppose that α1 , α2 , . . . , αn ∈ f (X) and we define α = max(α1 , α2 , . . . , αn ). Then we have n \

k=1

Eαk =

n \

{x ∈ X | f (x) ≥ αk } = Eα 6= ∅.

k=1

Therefore, {Eα } has the finite intersection property and then since X is compact, we conclude thatr \ Eα 6= ∅. Thus suppose that x0 ∈

\ α

α

Eα which means that f (x0 ) ≥ α for all α ∈ f (X), i.e., f (x0 ) ≥ f (x)

for all x ∈ X. Hence this shows that f attains its maximum at some point of X, completing the proof of the problem.  q r

Recall that g(n) = o(f (n)) means fg(n) → 0 as n → ∞. (n) See, for instances, [74, Theorem 26.9, p. 169] or [127, Problem 4.23, p. 43].

2.6. Miscellaneous Problems on L1 and Other Properties

61

Remark 2.4 Problem 2.21 can be treated as the Extreme Value Theorem for upper semicontinuous functions. Similarly, we can show that if f : X → R is lower semicontinuous and X is compact, then f attains its minimum at some point of X.

Problem 2.22 Rudin Chapter 2 Exercise 22.

Proof. By the definition and the triangle inequality, for all p ∈ X, we have gn (x) ≤ f (p) + nd(x, p) ≤ [f (p) + nd(y, p)] + nd(x, y). In other words, gn (x) − nd(x, y) is a lower bound of {f (p) + nd(y, p) | p ∈ X}. Thus, by the definition of infimum, we must have gn (x) − nd(x, y) ≤ gn (y) or equivalently, gn (x) − gn (y) ≤ nd(x, y).

(2.148)

gn (y) − gn (x) ≤ nd(x, y).

(2.149)

Likewise, we have Therefore, we conclude from the inequalities (2.148) and (2.149) that |gn (x) − gn (y)| ≤ nd(x, y). Since f (x) ≥ 0 on X and d(x, y) ≥ 0 for every x, y ∈ X, we have gn (x) ≥ 0 for all positive integers n. Fix x ∈ X, we observe that f (p) + nd(x, p) ≥ f (p) + (n − 1)d(x, p) ≥ gn−1 (x) for every p ∈ X and n = 2, 3, . . .. In other words, this implies that gn (x) ≥ gn−1 (x) for every x ∈ X and n = 2, 3, . . .. Besides, it is trivial that gn (x) ≤ f (x) holds for all n ∈ N if f (x) = ∞. Without loss of generality, we may assume that f (x) < ∞. In this case, since f (p) ≤ f (p) + nd(x, p) for all p ∈ X, if we take p = x in the definition of gn , then we have gn (x) ≤ f (x) for every n = 1, 2, . . .. This proves property (ii). To prove property (iii), we need the following lemma: Lemma 2.9 A function f : X → R is lower semicontinuous on X if and only if given x ∈ X, for every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive integer N such that n ≥ N implies f (x) < f (xn ) + ǫ.

(2.150)

62

Chapter 2. Positive Borel Measures Proof of Lemma 2.9. For every α ∈ R, denote Eα = {x
∈ X | f (x) ≤ α}. Recall that
f −1 (α, ∞) is open in X if and only if X \ f −1 (α, ∞) = Eα is closed in X. In other words, f is lower semicontinuous on X if and only if Eα is closed in X. Let Eα be closed in X. Assume that the inequality (2.150) was invalid. Then it means that there exists a x0 ∈ X and a sequence {xn } in X converging to x0 such that for some ǫ > 0, the inequality f (x0 ) − ǫ ≥ f (xnk )

(2.151)

holds for infinitely many k. Take α ∈ (f (x0 ) − ǫ, f (x0 )). On the one hand, since f (x0 ) > α, x0 ∈ / Eα . On the other hand, we gain from the inequality (2.151) that f (xnk ) < α for infinitely many k and this implies that {xnk } ⊆ Eα . Since the set Eα is closed and xnk → x0 as k → ∞, we have x0 ∈ Eα , a contradiction. Hence the inequality (2.150) must hold for a lower semicontinuous function f . Conversely, we prove that the inequality (2.150) implies Eα is closed in X by showing that Eα contains all its limit points. To this end, pick α ∈ R such that Eα 6= ∅. Let {xn } ⊆ Eα \ {x} and xn → x as n → ∞. Given ǫ > 0. By the inequality (2.150), there exists a positive integer N such that f (x) − ǫ < f (xn )

(2.152)

for all n ≥ N . Since f (xn ) ≤ α, we deduce immediately from the inequality (2.152) that f (x) < α + ǫ. Since ǫ is arbitrary, we must have f (x) ≤ α or equivalently x ∈ Eα . This ends the  proof of the lemma. Remark 2.5 (a) Similarly, a function f : X → R is upper semicontinuous on X if and only if given x ∈ X, for every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive integer N such that n ≥ N implies f (x) > f (xn ) − ǫ. (b) Except the equivalent definition stated in Lemma 2.9, a lower semicontinuous function f on X can also be reformulated in terms of the lower limit of f : lim inf f (y) ≥ f (x) y→x

for every y → x in X \ {x}, see [122, pp. 69, 70] and [129, Definition 3.62, p. 64].

Now it is time to go back to the proof of the problem. On the one hand, since f is lower semicontinuous on X, Lemma 2.9 implies that for every ǫ > 0, there exists a δ > 0 such that f (y) > f (x) − ǫ

(2.153)

for all y ∈ B(x, δ). In this case, we always have f (y) + nd(x, y) > f (x) − ǫ

(2.154)

for every positive integer n. Since ǫ > 0 is arbitrary, the inequality (2.154) becomes f (y) + nd(x, y) ≥ f (x).

(2.155)

2.6. Miscellaneous Problems on L1 and Other Properties

63

On the other hand, we consider y ∈ / B(x, δ) so that d(x, y) ≥ δ. It is clear that nδ − ǫ > 0 holds for large enough n,s thus we get from the inequality (2.153) that f (y) + nd(x, y) ≥ f (y) + nδ > f (x) − ǫ + nδ > f (x) for large enough n. Therefore, we conclude that the inequality (2.155) always holds for all y ∈ X and for all large enough n. By the definition of limit inferior, we have lim inf gn (x) ≥ f (x).

(2.156)

lim sup gn (x) ≤ f (x)

(2.157)

n→∞

By property (ii), we see that n→∞

holds for every x ∈ X. Hence it deduces from the inequalities (2.156) and (2.157) that lim gn (x) = f (x)

n→∞

on X. By property (i), each gn is continuous on X. Next, property (ii) says that {gn } is increasing and bounded by f . Finally, property (iii) ensures that {gn } converges to f on X pointwisely.  This ends the proof of the problem. Remark 2.6 With the aid of Lemma 2.9, we can prove Problem 2.1(b) easily. In fact, given x ∈ X, for every {xn } ⊆ X \ {x} converging to x and for every ǫ > 0, there exists a positive integer N1 such that n ≥ N1 implies that ǫ (2.158) f1 (x) < f1 (xn ) + . 2 Similarly, there exists a positive integer N2 such that n ≥ N2 implies ǫ f2 (x) < f2 (xn ) + . 2

(2.159)

Take N = max(N1 , N2 ). If n ≥ N , then we derive from the inequalities (2.158) and (2.159) that f1 (x) + f2 (x) < f1 (xn ) + f2 (xn ) + ǫ. Hence it follows from Lemma 2.9 that f is lower semicontinuous on X.

Problem 2.23 Rudin Chapter 2 Exercise 23.

Proof. Let x ∈ Rk and f : Rk → R be defined by f (x) = µ(V + x). Since V + x is just a translation of V , it is also open in Rk , i.e., V + x ∈ B. s

Obviously, the n depends on δ and ǫ.

64

Chapter 2. Positive Borel Measures

We first construct a counter example which is not upper semicontinuous or continuous. By Example 1.20(b), we consider the unit mass concentrated at 0, i.e.,   1, if 0 ∈ E;

µ(E) =



0, if 0 ∈ / E.

If V = B(0, 1), then we have V + x = B(x, 1) and

µ(V + x) =

=

  1, if 0 ∈ B(x, 1);  0, if 0 ∈ / B(x, 1)   1, if |x| < 1; 

(2.160)

0, if |x| ≥ 1.

Thus it establishes from the result (2.160) that
f −1 (−∞, 1) = {x ∈ Rk | f (x) < 1} = {x ∈ Rk | µ(V + x) < 1} = Rk \ B(x, 1)

which is closed in Rk . By Definition 2.8, f is not upper semicontinuous or continuous.

However, we claim that f is always lower semicontinous. Let {xn } ⊆ Rk \ {x} be such that xn → x as n → ∞. Recall that both V + x and V + xn are open in Rk . We claim that y ∈ V + x implies that y ∈ V + xn for all but finitely many n. Let y = a + x for some a ∈ V . Since V is open in Rk , there exists a ǫ > 0 such that B(a, ǫ) ⊆ V. By the hypothesis, there is a positive integer N such that n ≥ N implies xn − x =

ǫ 2

or

ǫ xn − x = − . 2

In the first case, we have y =a+x=a−

ǫ ǫ ǫ + x + = a − + xn . 2 2 2

(2.161)

Since a − 2ǫ ∈ V , the expression (2.161) guarantees that y ∈ V + xn for all n ≥ N . The other case can be done similarly, so we omit the details here. This proves our claim. Now if χV +x (y) = 1, then we have χV +xn (y) = 1 for all but finitely many n. In other words, it means that lim inf χV +xn (y) ≥ χV +x (y) (2.162) n→∞

for every y ∈ Rk . By Proposition 1.9(d), each χV +xn : Rk → [0, ∞] is measurable, so we may apply Theorem 1.28 (Fatou’s Lemma) to conclude that  Z   Z (2.163) lim inf lim inf χV +xn dµ. χV +xn dµ ≥ n→∞

Rk

Rk

By Proposition 1.24(f), we have Z Z χV +xn dµ = Rk

V +xn

n→∞

dµ = µ(V + xn ).

(2.164)

2.6. Miscellaneous Problems on L1 and Other Properties

65

In addition, it follows from the inequality (2.162) and the application of Proposition 1.24(f) again that Z Z Z   dµ = µ(V + x). (2.165) χV +x (y) dµ = lim inf χV +xn dµ ≥ Rk

n→∞

Rk

V +x

Thus, by substituting the results (2.164) and (2.165) into the inequality (2.163), we achieve that lim inf µ(V + xn ) ≥ µ(V + x) n→∞

or equivalently, lim inf µ(V + y) ≥ µ(V + x) y→x

Rk \{x}.

for every y → x in Hence we conclude from Remark 2.5(b) that f is lower semicontinuos on Rk , completing the proof of the problem.  Problem 2.24 Rudin Chapter 2 Exercise 24.

Proof. Suppose that S1 and S2 are the sets of simple functions and step functions on R respectively. We divide the proof into several steps. • Step 1: S1 is dense in L1 (R). Let f = f + − f 1 . By the Corollary (b) following Theorem 1.14, both f + and f − are measurable. Since f + : R → [0, ∞], we follow from Theorem 1.17 (The Simple Function Approximation Theorem) that there is a sequence {sn } of nonnegative increasing simple functions such that for every x ∈ R, sn (x) → f + (x) as n → ∞. Thus Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) implies that Z ∞ Z ∞ f + dx (2.166) sn dx → −∞

−∞

|f + (x)

f + (x)

as n → ∞. It is clear that − sn (x)| = − sn (x) for every x ∈ R, so we havet Z ∞ Z ∞ [f + (x) − sn (x)] dx |f + (x) − sn (x)| dx = −∞ −∞ Z ∞ Z ∞ sn (x) dx. (2.167) f + (x) dx − = −∞

−∞

If we apply the limit (2.166) to the expression (2.167), then we gain Z ∞ |f + (x) − sn (x)| dx = 0. lim n→∞ −∞

Likewise, the same is true for f − , so we have Z ∞ |f − (x) − tn (x)| dx = 0 lim n→∞ −∞

where {tn } is a sequence of nonnegative increasing simple functions such that for every x ∈ R, tn (x) → f − (x) as n → ∞. Hence {gn = sn − tn } is a sequence of simple functions such that Z ∞ Z ∞ |f + − f − − sn + tn | dx |f − gn | dx = −∞

−∞

Since |f (x)| ≤ |f (x)| on R and f ∈ L (R), we have f + ∈ L1 (R). Thus this makes the second equality holds because the right-hand side is not in the form ∞ − ∞. t

+

1

66

Chapter 2. Positive Borel Measures

≤

Z

∞ −∞

|f + − sn | dx +

→0

Z

∞ −∞

|f − − tn | dx

as n → ∞. • Step 2: S2 is dense in S1 . To this end, it suffices to prove that for every measurable set E ⊂ R with m(E) < ∞ and every ǫ > 0, there exists a step function g on R such that Z ∞ |χE − g| dx < ǫ. (2.168) −∞

By Theorem 2.20(b), there exists an open set V in R such that E ⊆ V and m(V \ E) < 2ǫ . By [99, Exercise 29, Chap. 2, p. 45], we have V =

∞ [

(an , bn ),

(2.169)

n=1

where (ai , bi ) ∩ (aj , bj ) = ∅ if i 6= j. Assume that there was an unbounded segment in the expression (2.169). Then we have m(V ) = ∞. However, since m(V ) = m(V \ E) + m(E), the facts m(V ) = ∞ and m(V \ E) < 2ǫ will force that m(E) = ∞, a contradiction. Furthermore, the expression (2.169) implies that m(V ) =

∞ X

n=1


m (an , bn ) ,

so there exists a positive integer N such that m(V ) −

N X
ǫ m (an , bn ) . < 2

(2.170)

n=1

Define g : R → R by g(x) =

N X

χ(an ,bn ) (x)

n=1

and VN = (a1 , b1 ) ∪ · · · ∪ (aN , bN ). Let x ∈ (E \ VN ) ∪ (VN \ E). If x ∈ E \ VN , then we have χE (x) = 1 and g(x) = 0. Similarly, if x ∈ VN \ E, then we have χE (x) = 0 and g(x) = 1. Therefore, we have |g − χE | = 1 (2.171) on (E \ VN ) ∪ (VN \ E). On the other hand, we let

x ∈ (E \ VN )c ∩ (VN \ E)c = (VN ∪ E c ) ∩ (E ∪ VNc ) = (VN ∩ E) ∪ (VNc ∩ E c ). If x ∈ VN ∩ E, then we have χE (x) = g(x) = 1. Similarly, x ∈ VNc ∩ E c implies that χE (x) = g(x) = 0. Both cases give |g − χE | = 0

(2.172)

on (E \ VN )c ∩ (VN \ E)c . Now the facts E, VN ⊆ V and (2.170) definitely imply that ǫ ǫ m(VN \ E) ≤ m(V \ E) < and m(E \ VN ) ≤ m(V \ VN ) < . (2.173) 2 2 Hence we deduce from the results (2.171), (2.172) and the estimates (2.173) that Z Z Z ∞ dx = m(VN \ E) + m(E \ VN ) < ǫ. dx + |g − χE | dx = −∞

E\VN

This is the desired result (2.168).

VN \E

2.6. Miscellaneous Problems on L1 and Other Properties

67

• Step 3: S2 is dense in L1 (R). We need the following lemma: Lemma 2.10 Suppose that A, B and C are subsets of a metric space X. If A ⊆ B ⊆ C, A is dense in B and B is dense in C, then A is dense in C.

Proof of Lemma 2.10. Recall the definition that A is dense in S if A ⊆ S ⊆ A, where A denotes the closure of A in X. Then we have A ⊆ B ⊆ A and B ⊆ C ⊆ B which show immediately that A ⊆ C. Let p ∈ B. We claim that p ∈ A. To this end, we know from the assumption that B(p, δ) ∩ (B \ {p}) 6= ∅ for every δ > 0. Let q ∈ B(p, δ) ∩ (B \ {p}). Since B ⊆ A, A also contains q. If q ∈ A, then we have B(p, δ) ∩ (A \ {p}) 6= ∅ (2.174) for every δ > 0. Therefore, the definition implies that p ∈ A. Next, suppose that q ∈ A′ . Since B(p, δ) is open in X, there exists a ǫ > 0 such that B(q, ǫ) ⊆ B(p, δ). Since q is a limit point of A, the definition shows that B(q, ǫ) ∩ A 6= ∅ and thus the set relation (2.174) still holds in this case. Therefore, we have proven our claim that p ∈ A and this means that B ⊆ A. Hence we establish the set relations A⊆C⊆A which implies that A is dense in C, completing the proof of Lemma 2.10.  Let’s return to the original proof. By applying Lemma 2.10 to the results in Step 1 and Step 2, we conclude easily that S2 is dense in L1 (R) which is our desired result. Hence we have completed the proof of the problem.



Remark 2.7 By Definition 1.16, we remark that a simple function s is a finite linear combination of characteristic functions of an arbitrary set Ai and when all Ai are intervals, then s becomes a step function. In fact, the class of step functions is one of the building blocks for the theory of Riemann integration, see [99, Chap. 6].

Problem 2.25 Rudin Chapter 2 Exercise 25.

Proof. (a) We note that et > 0 and log(1 + et ) > 0 for every t > 0. Furthermore, we know that the functions ex and log x are increasing in their corresponding domains respectively. Thus

68

Chapter 2. Positive Borel Measures the inequality log(1 + et ) < c + t is equivalent to 1 + et < ec+t and finally, it is equivalent to 0 < 1 + e−t < ec . (2.175) Apply log to both sides of the inequality (2.175), we have log(1 + e−t ) < c

(2.176)

for every t > 0. Since the left-hand side of the inequality (2.176) is a decreasing function of t, the smallest value of c for the validity of the inequality (2.176) on (0, ∞) is found by lim log(1 + e−t ) = log 2.

t→0 t>0


(b) Let f ∈ L1 on [0, 1]. Define E = {x ∈ [0, 1] | f (x) > 0} = f −1 (0, ∞) . By Theorem 1.12(b), E is a measurable set in [0, 1]. Thus we follow from part (a) that, on E,

nf (x) = log enf (x) < log 1 + enf (x) < log 2 + nf (x) on (0, ∞) and this implies that Z Z Z log 2 1 nf f dµ. log(1 + e ) dµ < + f dµ < n E n E E

(2.177)

Next, we have 1 ≤ 1 + enf (x) ≤ 2 on [0, 1] \ E. Thus we have 0 ≤ log(1 + enf ) ≤ log 2

(2.178)

on [0, 1]\E. Since [0, 1]\E is also a measurable set in [0, 1], we deduce from the inequalities (2.178) that Z log 2 1 log(1 + enf ) dµ ≤ . 0≤ n [0,1]\E n Now the sum of the inequalities (2.177) and (2.178) give Z

f dµ < E

1 n

Z

1

log(1 + enf ) dµ
0 on (a, b). By [100, Theorem (c), p. 2], the function exp is a monotonically increasing positive function on R. Since (ex )′ = ex , it is convex on R by the paragraph following Definition 3.1. Hence, if log ϕ is convex on (a, b), then we conclude from the first assertion that ϕ = elog ϕ is also convex on (a, b). However, the converse of the second assertion is false. For instance, we know that ϕ(x) = x is convex on (0, ∞), but log x is not convex on (0, ∞) because (log x)′′ = −x−2 < 0 on (0, ∞), i.e., (log x)′ is not a monotonically increasing function on (0, ∞). We have completed the proof  of the problem. Problem 3.3 Rudin Chapter 3 Exercise 3.

Proof. This problem is proven in [124, Problem 4.24, pp. 79 – 81].



3.2. Relations among Lp -Spaces and some Consequences

71

Relations among Lp -Spaces and some Consequences

3.2

Problem 3.4 Rudin Chapter 3 Exercise 4.

Proof. (a) Since 0 < r < p < s, we can find λ ∈ (0, 1) such that p = λr + (1 − λ)s. By Theorem 3.5 (H¨older’s Inequality), we have Z |f |p dµ ϕ(p) = X Z |f |λr × |f |(1−λ)s dµ = X nZ o1−λ oλ n Z
1 1 ≤ |f |λr λ dµ × [|f |(1−λ)s ] 1−λ dµ X

X

= [ϕ(r)]λ × [ϕ(s)]1−λ .

(3.7)

Since r, s ∈ E, ϕ(r) and ϕ(s) are finite. Hence the inequality (3.7) ensures that ϕ(p) is also finite, i.e., p ∈ E. (b) We prove the assertion one by one. – Case (i): ln ϕ is convex in E ◦ . If E ◦ = ∅, then there is nothing to prove. Assume that E ◦ 6= ∅. By [99, Theorem 2.47, p. 42] and part (a), the set E is connected. Since E ⊆ (0, ∞), E is either an interval in one of the forms [a, b], [a, b), (a, b] or (a, b) for some positive a and b with a < b. In each of the cases, we always have E ◦ = (a, b). Let x, y ∈ (a, b) and λ ∈ [0, 1]. Since (a, b) is a convex set, λx + (1 − λ)y ∈ (a, b). Thus it follows from the inequality (3.7) that
ϕ λx + (1 − λ)y ≤ [ϕ(x)]λ × [ϕ(y)]1−λ . (3.8) If ϕ(p) = 0 for some p ∈ (0, ∞), then we have Z |f |p dµ = 0 X

so that |f (x)| = 0 almost everywhere on X. By the remark following Definition 3.7, it implies that kf k∞ = 0, a contradiction. Hence we have ϕ(p) > 0 and we are allowed to take the logarithm to both sides of the inequality (3.8) to get
 ln ϕ λx + (1 − λ)y ≤ ln [ϕ(x)]λ × [ϕ(y)]1−λ ≤ λ ln ϕ(x) + (1 − λ) ln ϕ(y). By Definition 3.1, ln ϕ is convex in (a, b).

– Case (ii): ϕ is continuous on E. By Theorem 3.2, ln ϕ is continuous on (a, b). Thus ϕ is also continuous on (a, b), so it remains to show that ϕ is continuous at the endpoints. Let a ∈ E. Then E is either [a, b) or [a, b], but no matter which case E is, a is a limit point of E so that we can find a decreasing sequence {pn } ⊆ (a, b) such that pn → a as n → ∞. By this, there exists a ǫ > 0 and a positive integer N such that pn ∈ (a, a + ǫ) ⊂ E for all n ≥ N . Fix this ǫ and then n ≥ N implies that   |f (x)|a+ǫ , if |f (x)| ≥ 1; pn |f (x)| ≤  |f (x)|a , if |f (x)| < 1.

Chapter 3. Lp -Spaces

72 Hence we obtain that |f (x)|pn ≤ |f (x)|a+ǫ + |f (x)|a

on X. We recall that a, a + ǫ ∈ E, so the definition implies that ϕ(a) and ϕ(a + ǫ) are finite. Since f is measurable, |f |a and |f |a+ǫ are measurable by Proposition 1.9(b) and Theorem 1.7(b). Thus these two facts show that |f |a , |f |a+ǫ ∈ L1 (µ). By Theorem 1.32, we have |f |a+ǫ + |f |a ∈ L1 (µ). For each x ∈ X, |f (x)| ≥ 0. Since an exponential function with nonnegative base is continuous on its domain, we have lim |f (x)|pn = |f (x)|a

n→∞

on X. In conclusion, we have shown that the sequence {|f |pn } satisfies the hypotheses of Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and then Z Z |f |a dµ = ϕ(a). |f |pn dµ = lim ϕ(pn ) = lim n→∞

n→∞ X

X

By definition, ϕ is continuous at a. Similarly, ϕ is continuous at b if b ∈ E. Consequently, ϕ is continuous on E.

(c) We claim that E can be any open or closed connected subset of (0, ∞). We consider X = (0, ∞), µ = m and 0 < a ≤ b. We need the following result (see [4, Examples 1 & 5, pp. 417, 419]): Lemma 3.1 Let x > 0 and b > 1. Then we have  1 − x1−α   Z 1  , if α 6= 1; 1−α t−α dt =  x   − ln x if α = 1 and

 1−β b −1   , if β = 6 1;  b 1 − β −β t dt =  1   ln b, if β = 1.

Z

Furthermore, the improper integrals Z 1 x−α dx and 0

Z

∞

x−β dx

1

are convergent if and only if α < 1 and β > 1 respectively.

– Case (i): E = (a, b). For x ∈ (0, ∞), let  1   x− b , if x ∈ (0, 1); f (x) =   x− a1 , if x ∈ [1, ∞). We have to find p ∈ (0, ∞) such that Z ϕ(p) =

0

∞

|f (x)|p dx < ∞.

3.2. Relations among Lp -Spaces and some Consequences We assume that the integral can be split as follows: Z 1 Z ∞ p − pb x dx + ϕ(p) = x− a dx. 0

73

(3.9)

1

Apply Lemma 3.1 to the two improper integrals in the expression (3.9), we know that ϕ(p) < ∞ if and only if pb < 1 and ap > 1 if and only if a < p < b. In other words, we have E = (a, b) and such split is allowable. – Case (ii): E = [a, b]. In this case, we need another lemma:a Lemma 3.2 The improper integral

Z

e−1

dx xα | ln x|β

0

converges if and only if (i) α < 1 or (ii) α = 1 and β > 1. The improper integral Z ∞ dx α (ln x)β x e converges if and only if (i) α > 1 or (ii) α = 1 and β > 1.

We consider  1  if x ∈ (0, e−1 );  1 1 ,   b x | ln x|1+ b       1 1  1 e− a − e b f (x) = b + (x − e−1 ), if x ∈ (e−1 , e); e  e − e−1        1   if x ∈ [e, ∞).  1 1 , x a (ln x)1+ a

(3.10)

It is clear that f is continuous on (0, ∞) and so it is measurable. Similar to Case (i), we assume that the integral can be split as follows: Z ∞ |f (x)|p dx ϕ(p) = =

Z

0

e−1

+

p(1+ 1b )

x | ln x| dx

0

Z

dx

p b

+

∞

p

e

1

x a (ln x)p(1+ a )

Z

e

e−1

h

1

eb +

1 1 ip e− a − e b −1 (x − e ) dx e − e−1

.

(3.11)

By Lemma 3.2, the first integral in the expression (3.11) converges if and only if (i) p p 1 b < 1 or (ii) b = 1 and p(1 + b ) > 1. The first condition gives p < b. If p = b, then 1 p(1 + b ) = 1 + b > 1 so that the second condition is actually p = b. In this case, we have p ≤ b. Similarly, we apply Lemma 3.2 to the third integral in the expression (3.11), so it converges if and only if (i) ap > 1 or (ii) ap = 1 and p(1 + a1 ) > 1. The first condition a The integrals in Lemma 3.2 are called Bertrand’s integrals, see the following webpage https://fr.wikipedia.org/wiki/Int%C3%A9grale_impropre.

Chapter 3. Lp -Spaces

74

shows p > a. If p = a, then p(1 + a1 ) = 1 + a > 1 so that the second condition implies that p = a. In this case, we have p ≥ a. Since the second integral in the expression (3.11) is finite, we combine these observations to conclude that ϕ(p) < ∞ if and only if p ∈ [a, b].

– Case (iii): E = {a}. If we take b = a in the definition (3.10), then we obtain p ≥ a and p ≤ a. Hence we establish E = {a}.

– Case (iv): E = ∅. Consider f ≡ 1 on (0, ∞). Since the integral Z ∞ |1|p dx 0

is obviously divergent for every p ∈ (0, ∞), we conclude that E = ∅ in this case. (d) Let r < p < s. By the proof of part (a), we see that p = λr + (1 − λ)s for some λ ∈ (0, 1) so that the inequality (3.7) holds. In other words, we have
λ
1−λ kf kpp ≤ kf krr × kf kss . (3.12) It is clear that

kf krr


λ

× kf kss


1−λ

≤

=


λ
1−λ  , if kf kr ≤ kf ks ;  kf krs × kf kss

  kf kr
λ × kf ks
1−λ , if kf k ≤ kf k s r r r  p  kf ks , if kf kr ≤ kf ks ; 

kf kpr , if kf ks ≤ kf kr .

(3.13)

Hence, by combining the inequalities (3.12) and (3.13), we get kf kp ≤ max(kf kr , kf ks ).

(3.14)

Let f ∈ Lr (µ) ∩ Ls (µ). By the definition, we have kf kr < ∞

and

kf ks < ∞.

Hence we follow from these and the inequality (3.14) that kf kp < ∞, i.e., f ∈ Lp (µ) and then Lr (µ) ∩ Ls (µ) ⊆ Lp (µ). (e) Recall that kf k∞ > 0. – Case (i): kf k∞ < ∞. Let Eα = {x ∈ X | |f (x)| ≥ α}, where α ∈ (0, kf k∞ ). It is clear that Eα is measurable because |f | is measurable. By Definition 3.7, kf k∞ is the smallest number such that µ({x ∈ X | |f (x)| > kf k∞ }) = 0. Thus we have µ(Eα ) > 0 for every α ∈ (0, kf k∞ ). Assume that µ(Eα ) = ∞. Then we have Z Z Z |f |r dµ ≥ αr µ(Eα ) = ∞ |f |r dµ + |f |r dµ = kf krr = X\Eα

Eα

X

which contradicts to our hypothesis. Thus we have 0 < µ(Eα ) < ∞ and then Proposition 1.24(b) implies that Z Z |f |p dµ ≥ αp µ(Eα ) |f |p dµ ≥ kf kpp = X

Eα

3.2. Relations among Lp -Spaces and some Consequences which shows that

1

kf kp ≥ α[µ(Eα )] p .

75

(3.15)

1 p

Since lim [µ(Eα )] = 1, we obtain from the inequality (3.15) that p→∞

lim inf kf kp ≥ α. p→∞

In particular, since α is arbitrary in (0, kf k∞ ), we have lim inf kf kp ≥ kf k∞ . p→∞

Next, if p > r, then we have Z Z |f |p dµ = kf kpp = |f |p−r × |f |r dµ.

(3.16)

(3.17)

X

X

By Definition 3.7, since |f (x)| ≤ kf k∞ for almost all x ∈ X, we get from the expression (3.17) and Proposition 1.24(b) that Z Z r p−r r |f |r dµ = kf kp−r kf kpp ≤ kf kp−r (3.18) ∞ × |f | dµ = kf k∞ × ∞ × kf kr , X\E

X\E

where E = {x ∈ X | |f (x)| > kf k∞ } is of measure 0. Rewrite the inequality (3.18) in the form r 1− r kf kp ≤ kf k∞ p kf krp which, by using the fact that kf kr < ∞, implies lim sup kf kp ≤ kf k∞ .

(3.19)

p→∞

Hence our desired result follows immediately from the inequalities (3.16) and (3.19). – Case (ii): kf k∞ = ∞. Then we deduce immediately from the inequality (3.16) that lim kf kp = lim inf kf kp = ∞ = kf k∞ .

p→∞

p→∞

Hence we have completed the proof of the problem.



Problem 3.5 Rudin Chapter 3 Exercise 5.

Proof. (a) Suppose that s = ∞. Since |f (x)| ≤ kf k∞ holds for almost all x ∈ X, we have µ(E) = µ({x ∈ X | |f (x)| > kf k∞ }) = 0 so that kf krr =

Z

X\E

|f |r dµ ≤

Z

X\E

kf kr∞ dµ = kf kr∞ · µ(X \ E) = kf kr∞ .

Hence we have kf kr ≤ kf k∞ in this case.

(3.20)

Chapter 3. Lp -Spaces

76

Next, we suppose that s < ∞. Since f is measurable, |f |r is also measurable on X by Proposition 1.9(b) and Theorem 1.7(b) for every r > 0. We apply Theorem 3.5 (H¨older’s Inequality) to the measurable functions |f |r and 1 to obtain that Z

X

|f |r dµ ≤ =

nZ Z


s

X

|f |r

X

|f |s dµ

= kf krs .

r

dµ

r

s

or

s

×

nZ

X

dµ

o1− r

s

(3.21)

Hence we conclude that kf kr ≤ kf ks . (b) Suppose that 0 < r < s < ∞. Since kf kr = kf ks , it means that the equality holds in the inequality (3.21). Recall that the inequality (3.21) is derived from Theorem 3.5 (H¨older’s Inequality), so there are constants α and β, not both 0, such that s

α(|f |r ) r = β

a.e. on X.

If α = 0, then it is trivial that β = 0, a contradiction. Thus α 6= 0 and then we have |f | =

β 1 s

α

a.e.

In other words, |f | is a constant a.e. on X. Next, suppose that s = ∞ and kf kr = kf k∞ < ∞, where 0 < r < ∞. Then the equality holds in the inequality (3.20) so that |f |r = kf kr∞ < ∞ a.e. on X, i.e., |f | = kf k∞

a.e. on X.

In other words, we achieve that |f | is a constant a.e. on X in this case. Consequently, we conclude that 0 < r < s ≤ ∞ and the conditions kf kr = kf ks < ∞ hold if and only if |f | is a constant a.e. on X (c) If f ∈ Ls (µ), then kf ks < ∞ and we know from part (a) that kf kr < ∞, i.e., f ∈ Lr (µ). Thus it is true that Ls (µ) ⊆ Lr (µ). Now the other inclusion Lr (µ) ⊆ Ls (µ) is a consequence of the following lemmab : Lemma 3.3 Suppose that (X, M, µ) is a measure space and M0 = {E ∈ M | µ(E) > 0}. Then the following conditions are equivalent: – Condition (1). inf{µ(E) | E ∈ M0 } > 0, – Condition (2). Lr (µ) ⊆ Ls (µ) for all 0 < r < s ≤ ∞.

b

This lemma is part of the statements in [120, Theorem 1] and our proof follows part of the argument there.

3.2. Relations among Lp -Spaces and some Consequences

77

Proof of Lemma 3.3. Suppose that Condition (1) is true. Suppose that f ∈ Lr (µ) and En = {x ∈ X | |f (x)| ≥ n} for every n ∈ N. We claim that there exists an N ∈ N such that µ(EN ) = 0. Otherwise, there exists ǫ > 0 such that µ(En ) ≥ ǫ for each ∞ \ En . By the construction of En , we have E1 ⊇ E2 ⊇ · · · . Since n ∈ N. Let E = n=1

µ(E1 ) ≤ µ(X) < ∞, it yields from Theorem 1.19(e) that

µ(E) = lim µ(En ) ≥ ǫ > 0.

(3.22)

n→∞

However, we deduce from the inequality (3.22) that 1 1  Z Z 1 r r |f |r dµ = ∞ · µ(E) r = ∞, |f |r dµ ≥ kf kr = E

X

a contradiction. Therefore, we have our claim that µ(EN ) = 0 for some N ∈ N. In other words, |f (x)| ≤ N holds for almost all x ∈ X and we obtain from Definition 3.7 that kf k∞ ≤ N , i.e., f ∈ L∞ (µ) and so f ∈ Ls (µ). Hence Condition (2) holds. Assume that Condition (1) was false. Thus there exists a sequence of {En } ⊆ M0 such that 1 (3.23) 0 < µ(En ) < n . 2 Put F1 = E1 and Fn = En \ (En−1 ∪ · · · ∪ E1 ), where n = 2, 3, . . .. Then it is easy to check that Fi ∩ Fj = ∅ for i 6= j and
1 0 < µ(Fn ) = µ En \ (En−1 ∪ · · · ∪ E1 ) ≤ µ(En ) < n 2

for every n = 1, 2, . . .. Therefore, we may assume that {En } is a sequence of disjoint measurable sets. ∞ [ En . Suppose first that Next, we define αn = µ(En ) for n = 1, 2, . . . and E = n=1

s < ∞. Then we define f : X → R by

∞ X

f (x) =

−1

αn s χEn (x).

(3.24)

n=1

Since r < s, we have 1 − rs < 1. Note that f (x) = 0 if x ∈ X \ E. Thus it reduces from the bounds (3.23) and Theorem 1.29 that kf krr =

Z

X\E

|f |r dµ +

which means f ∈

Lr (µ).

Z

E

|f |r dµ =

∞ Z X

−r

αn s dµ =

n=1 En

∞ X

1− rs

αn

r


2 2r for all n ∈ N, we conclude kf k∞ = ∞ and then 

Chapter 3. Lp -Spaces

78

(d) If 0 < p < q, then part (a) says that kf kp ≤ kf kq < ∞. Let E0 = {x ∈ X | |f (x)| = 0}. By Problem 1.5, E0 is measurable. – Case (i): µ(E0 ) > 0. Now the set E∞ = {x ∈ X | |f (x)| = ∞} is measurable by Problem 1.5. Pick p ∈ (0, r), so part (a) indicates that kf kp ≤ kf kr < ∞. If µ(E∞ ) > 0, then we have nZ o1 o1 nZ 1 p p p ≥ = [∞ · µ(E∞ )] p = ∞, |f |p dµ |f | dµ kf kp = E∞

X

a contradiction. Thus we have µ(E∞ ) = 0, i.e., |f (x)| < ∞ a.e. on X. Consequently, we have Z Z Z Z log |f | dµ (3.25) log |f | dµ + log |f | dµ = log |f | dµ = X\(E∞ ∪E0 )

X\E∞

X

E0

Since µ(X) = 1 and |f (x)| is nonzero finite on X \ (E∞ ∪ E0 ), the first integral on the right-hand side of the equation (3.25) is finite. However, the second integral on the right-hand side of the equation (3.25) is in the form −∞ so that the right-hand side of our desired result is in the form exp{−∞} which is defined to be 0. On the other hand, we notice that o1 nZ o1 n Z p p = |f |p χX\E0 dµ . (3.26) kf kp = |f |p dµ X

X

Apply Theorem 3.5 (H¨older’s Inequality) with u > 1 and v > 1 as fixed conjugate exponents to the right-hand side of the expression (3.26), we see that o1 o1 nZ o 1 nZ nZ p pu pv |f |p χX\E0 dµ (χX\E0 )v dµ (|f |p )u dµ ≤ X

X

X

1

= µ(X \ E0 ) pv kf kpu .

(3.27)

Combining the expression (3.26) and the inequality (3.27), we gain 1

kf kp ≤ µ(X \ E0 ) pv kf kpu . Since µ(X) = 1 and µ(E0 ) > 0, we have µ(X \ E0 ) < 1 and then 1

lim µ(X \ E0 ) pv = 0.

(3.28)

p→0

In addition, when p is chosen to be small enough such that pu < r, part (a) implies that kf kpu ≤ kf kr < ∞. Hence we observe from the limit (3.28) and this fact that 1

1

lim kf kp ≤ lim µ(X \ E0 ) pv kf kpu ≤ kf kr lim µ(X \ E0 ) pv = 0.

p→0

p→0

p→0

In other words, the desired result holds in this case. – Case (ii): µ(E0 ) = 0. Then |f (x)| > 0 a.e. on X. By the fact used in Case (i), we may assume that 0 < |f (x)| < ∞ a.e. on X. (3.29) This fact allows us to apply [100, Eqn. (7), p. 63] to the almost positive function |f |p to obtain o n1 Z o nZ o1 nZ p log |f | dµ ≥ exp log |f |p dµ = exp |f |p dµ kf kp = p X X X

3.2. Relations among Lp -Spaces and some Consequences for every p ∈ (0, r). In particular, lim kf kp ≥ exp

p→0+

nZ

X

o log |f | dµ .

79

(3.30)

For the reverse direction, we consider the function φ : (0, ∞) → R defined by φ(x) = x − 1 − log x.

Since φ′ (x) = 1 − x1 = 0 and φ′′ (x) = x12 , φ has the absolute minimum at x = 1, i.e., φ(x) ≥ φ(1) on (0, ∞) or equivalently log x ≤ x − 1

(3.31) kf kpp

on (0, ∞). Recall the fact (3.29), so we replace x by in the inequality (3.31) and then using the fact that µ(X) = 1 to obtain Z Z  Z |f |p − 1 kf kpp − 1 1 log kf kp ≤ dµ = |f |p dµ − = dµ. (3.32) p p X p X X Let E = {x ∈ X | |f (x)| = 1} and F = {x ∈ X | |f (x)| = 6 1}. Then we get Z Z Z p p |f | − 1 |f | − 1 |f |p − 1 dµ = dµ + dµ. p p p E X F

(3.33)

It is clear from Proposition 1.24(d) that Z |f |p − 1 dµ = 0, p E so it suffices to evaluate

Z

|f |p − 1 dµ. p→0+ F p To this end, we need the following lemma: lim

Lemma 3.4 Let |f | > 0 and kf kr < ∞ for some r > 0. Define ψ : R → R by ψ(p) = |f |p . Then we have |f |p − 1 ∈ L1 (µ) p for all p ∈ (0, r).

Proof of Lemma 3.4. Since ψ ′′ = |f |p (log |f |)2 > 0 on R, ψ is convex on R. Thus we see from [99, Exercise 23, p. 101] that, for 0 < p < r, ψ(p) − ψ(0) ψ(r) − ψ(0) ≤ p−0 r−0 which reduces to

and so

|f |p − 1 |f |r − 1 ≤ p r Z

X

|f |p − 1 dµ ≤ p

In other words, we have lemma.

|f |p −1 p

Z

X

|f |r − 1 1 1 dµ = kf krr − < ∞. r r r

∈ L1 (µ) for all p ∈ (0, r), completing the proof of the 

Chapter 3. Lp -Spaces

80

Let’s return to the proof of the problem. By Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem), we obtain that Z Z  |f |p − 1  |f |p − 1 lim dµ. (3.34) lim dµ = p→0+ F p p F p→0+ By L’ Hˆospital Rule, we derive |f |p − 1 = lim |f |p log |f | = log |f | p→0+ p→0+ p lim

(3.35)

on X. Therefore, when we combine the results (3.32), (3.33), (3.34) and (3.35), we establish that Z Z Z  |f |p − 1  lim log |f | dµ. (3.36) lim log kf kp ≤ log |f | dµ = dµ = p→0+ p F p→0+ X F Since log x is continuous for x > 0, we get from the inequality (3.36) that nZ o lim kf kp ≤ exp log |f | dµ . p→0+

(3.37)

X

Hence the required result follows immediately from the inequalities (3.30) and (3.37). This ends the proof of the problem.



Problem 3.6 Rudin Chapter 3 Exercise 6.

Proof. Let x > 0 and 0 ≤ c ≤ 1. We claim that the function Φ : [0, ∞) → R satisfies the relation cΦ(x) + (1 − c)Φ(1) = Φ(xc ).

(3.38)

Obviously, the relation (3.38) holds for c = 0 and c = 1. Without loss of generality, we may assume that 0 < c < 1 in the following discussion. Since f is bounded measurable and positive on [0, 1], there exists a positive constant M such that Z 1 |f | dx ≤ M < ∞. kf k1 = 0

Thus we get from Problem 3.5(d) that lim kf kp = exp

p→0

nZ

1

0

o log |f (t)| dt .

(3.39)

By putting the expression (3.39) into the relation in question, we obtain  nZ Φ exp

0

1

log |f (t)| dt

We define f : [0, 1] → R by f (t) =

o

=

Z

0

1


Φ f (t) dt.

  x, if x ∈ [0, c]; 

1, if x ∈ (c, 1].

(3.40)

(3.41)

3.2. Relations among Lp -Spaces and some Consequences

81

Since x > 0, the f is bounded measurable and positive on [0, 1]. Therefore, we substitute the function (3.41) into the relation (3.40) to yield
Φ(xc ) = Φ exp(c log x)  Z c  = Φ exp log x dt 0 Z 1 Z c Φ(1) dt Φ(x) dt + = 0

c

= cΦ(x) + (1 − c)Φ(1).

This proves the claim (3.38). Next, we prove a lemma which is useful of determining the explicit form of the function Φ: Lemma 3.5 The relation (3.38) holds for all x > 0 and c ≥ 0.

Proof of Lemma 3.5. The case 0 ≤ c ≤ 1 is clear by the above analysis. Let c > 1 so that 0 < 1c < 1. Then the relation (3.38) becomes

1

 1
1 1 Φ x c = Φ(x) + 1 − Φ(1). c c

(3.42)

Take y = x c . Since 1c > 0, y will take all positive real numbers. Therefore, it follows from the expression (3.42) that  1 1 Φ(1) Φ(y) = Φ(y c ) + 1 − c c

and after simplification, we have

Φ(y c ) = cΦ(y) + (1 − c)Φ(1). This proves the lemma.



We return to the proof of the problem. Define Ψ : [0, ∞) → R by Ψ(x) = Φ(x) − Φ(1), then the relation (3.38) can be rewritten as Ψ(xc ) = cΨ(x),

(3.43)

where x > 0 and c ≥ 0. Take x = e and c = ln y, where y ≥ 1. Then the expression (3.43) becomes Ψ(y) = Ψ(e) ln y, (3.44) where y ≥ 1. For 0 < y < 1, we put x = e−1 and c = ln(y −1 ) > 0 into the expression (3.43) so that −1
Ψ(y) = Ψ e− ln(y ) = −Ψ(e−1 ) ln y. (3.45) Combining the results (3.44) and (3.45), we establish that  if x ≥ 1;  Ψ(e) ln x, Ψ(x) =  −Ψ(e−1 ) ln x, if 0 < x < 1

Chapter 3. Lp -Spaces

82 which imply that Φ(x) =

  Φ(1) + [Φ(e) − Φ(1)] ln x, 

if x ≥ 1;

Φ(1) − [Φ(e−1 ) − Φ(1)] ln x, if 0 < x < 1,

completing the proof of the problem.



Problem 3.7 Rudin Chapter 3 Exercise 7.

Proof. In this problem, we assume that µ is a positive measure. We give examples one by one: Example 3.1. Let X = N with the counting measure µ. By Definition 3.6, we have ∞ o 1 n X p 0, where M0 = {E ∈ M | µ(E) > 0}. • Ls (µ) ⊆ Lr (µ) for 0 < r < s < ∞. Similar to the proof of Lemma 3.3, the following result is part of the statements in [120, Theorem 2]: Lemma 3.6 Suppose that (X, M, µ) is a measure space and M∞ = {E ∈ M | µ(E) is finite}. Then the following conditions are equivalent: – Condition (1). sup{µ(E) | E ∈ M∞ } < ∞, – Condition (2). Ls (µ) ⊆ Lr (µ) for all 0 < r < s < ∞.

Proof of Lemma 3.6. Suppose that Condition (1) holds. To begin with, let f ∈ Ls (µ) 1 ≤ |f (x)| < n1 } for each n = 1, 2, . . .. Then it is clear that En ∈ M. and En = {x ∈ X | n+1 Now for each fixed n, we see that Z Z Z µ(En ) 1 dµ = |f |s dµ ≥ |f |s dµ ≥ (n + 1)s En (n + 1)s En X which implies that µ(En ) ≤ (n + 1)s

Z

X

|f |s dµ = (n + 1)s kf kss < ∞.

Therefore, we conclude that En ∈ M∞ . Note that Ei ∩ Ej = ∅ if i 6= j and µ is a positive measure, if Fn = E1 ∪ E2 ∪ · · · ∪ En , then we deduce from Theorem 1.19(b) that µ(Fn ) = µ

n [

k=1

n  X µ(Ek ) < ∞. Ek = k=1

In other words, each Fn is an element of M∞ . Next, we know that F1 ⊆ F2 ⊆ · · · and n n [ [ Fk = Ek , so Theorem 1.19(d) implies that

k=1

k=1

lim µ(Fn ) = µ

n→∞

∞ [

n=1

∞ ∞  X  [ µ(En ). En = Fn = µ n=1

(3.47)

n=1

Given a positive integer n, it is clear that µ(Fn ) ≤ sup{µ(Fk ) | k ∈ N} ≤ sup{µ(E) | E ∈ M∞ }.

(3.48)

Furthermore, by combining Condition (1), the expression (3.47) and the inequality (3.48), it yields that ∞ X

n=1

µ(En ) = lim µ(Fn ) ≤ sup{µ(E) | E ∈ M∞ } < ∞. n→∞

Chapter 3. Lp -Spaces

84

On X \

∞ [

n=1

En , we claim that |f (x)| ≥ 1. Otherwise, there is a x0 ∈ X \

∞ [

En such that

n=1

|f (x0 )| < 1. By the definition, x0 ∈ EN for some N ∈ N, a contradiction. This observation implies that |f (x)|r ≤ |f (x)|s on X \ Z

∞ [

En . Finally, we have

n=1 r

X

|f | dµ =

Z

S X\ ∞ n=1 En

r

|f | dµ +

∞ Z X

r

n=1 En

|f | dµ ≤

Z

X

|f |s dµ +

∞ X µ(En )

n=1

nr

< ∞.

This proves that Condition (2) holds. Conversely, since Ls (µ) ⊆ Lr (µ) implies that Lαs (µ) ⊆ Lαr (µ) for every α ∈ (0, ∞), we may assume that r ≥ 1. It is known that if r, s ∈ [1, ∞] and Ls (µ) ⊆ Lr (µ), then the mapping T : Ls (µ) → Lr (µ) is continuous, see [120, Lemma 1]. This result guarantees the existence of a positive constant k such that kf kr ≤ kkf ks for all f ∈ Ls (µ), so rs

µ(E) ≤ k s−r for every E ∈ M∞ and this implies Condition (1).



We have completed the proof of the problem.



Problem 3.8 Rudin Chapter 3 Exercise 8.

Proof. For every n ∈ N, since g(x) → ∞ as x → 0, there exists a sequence {ǫn } of positive numbers such that g(x) ≥ n (3.49) for every x ∈ (0, ǫn ). Without loss of generality, we may assume that {ǫn } is decreasing and ǫn+1
Nx . Thus we have x ∈ (0, ǫn ) for all 1 ≤ n ≤ Nx and x ∈ / (0, ǫn ) for all n > Nx . See Figure 3.1 for an illustration of the distribution of x and ǫn .

3.2. Relations among Lp -Spaces and some Consequences

85

Figure 3.1: The distribution of x and ǫn .

These facts imply that

so that

  x  , if 1 ≤ n ≤ Nx ;  n 1− ǫn hn (x) =   0, if n > Nx

(3.51)

 x h(x) = sup hn (x) = sup{h1 (x), h2 (x), . . .} = max n 1 − < ∞, 1≤n≤Nx ǫn n

i.e., h is finite. Since each hn is convex on (0, 1), Problem 3.1 shows that the h is also convex on (0, 1). Next, we want to compare the magnitudes of hn (x) and g(x) at the fixed point x ∈ (0, 1). On the one hand, for 1 ≤ n ≤ Nx , we have x ∈ (0, ǫn ), so n(1 − ǫxn ) ≤ n and we deduce from the inequality (3.49) and the definition (3.51) that hn (x) ≤ g(x)

(3.52)

for 1 ≤ n ≤ Nx . On the other hand, if n > Nx , then x ∈ / (0, ǫn ) and we have hn (x) = 0, but the inequality (3.49) still holds for n = Nx (i.e., x ∈ (0, ǫNx )) so that the inequality (3.52) remains valid in this case. Hence we have established that h(x) ≤ g(x)

(3.53)

for the fixed point x ∈ (0, 1). Since x is arbitrary, the inequality (3.53) holds on (0, 1). Take n = Nx − 1 in the definition (3.51) to get

 hNx −1 (x) = (Nx − 1) 1 −

x ǫNx −1



.

(3.54)

By the hypothesis (3.50), we have 2ǫNx < ǫNx −1 . Then since x < ǫNx , we have x 1 x < . < ǫNx −1 2ǫNx 2

(3.55)

By substituting the estimate (3.55) into the expression (3.54), we obtain hNx −1 (x) >

Nx − 1 . 2

Since x → 0 if and only if Nx → ∞, we conclude that hNx (x) → ∞ as x → 0 and the definition implies that h(x) → ∞ as x → 0 as required. 1

The second assertion is false. To see this, consider g(x) = x 3 which satisfies g(x) → ∞ as x → ∞. Assume that there was a convex function h : (0, ∞) → R such that 1

h(x) ≤ x 3

(3.56)

Chapter 3. Lp -Spaces

86

on (0, ∞) and h(x) → ∞ as x → ∞. We know from [131, Exercise 6(c), p. 262]c that if h : (0, ∞) → R is convex on (0, ∞), then the ratio h(x) x tends to a finite limit or to infinity as x → ∞. If this limit is finite, since h(x) → ∞ as x → ∞, this limit must be positive. Thus it must be true that h(x) ≥ kx

(3.57)

as x → ∞ for some positive constant k. However, the inequality (3.57) definitely contradicts the inequality (3.56). This completes the proof of the problem.  Problem 3.9 Rudin Chapter 3 Exercise 9.

Proof. Let Φ : (0, ∞) → (0, ∞) be such that Φ(p) → ∞ as p → ∞. Suppose that E1 = (0, 21 ) and h1 1 1 1 1 1 h 1 1 + 2 + · · · + n−1 , + 2 + · · · + n = 1 − n−1 , 1 − n , En = 2 2 2 2 2 2 2 2 where n ≥ 2. Obviously, we have En ⊂ (0, 1) for every n ∈ N and Ei ∩ Ej = ∅ for i 6= j. Furthermore, each En is measurable and m(En ) =

1 > 0. 2n

Suppose that x ∈ (0, 1). If x < 21 , then x ∈ E1 . Otherwise, 12 ≤ x < 1 and there exists a δ > 0 such that x < δ < 1. Take N to be least positive integer such that N >− Then we have x < 1 −

1 2N

log(1 − δ) > 0. log 2

< 1 so that x ∈ E2 ∪ E3 ∪ · · · ∪ EN . In other words, we get (0, 1) =

∞ [

En .

n=1

Finally, we define fn , f : (0, 1) → (0, ∞) by fn (x) =

n X

kχEk (x)

and f (x) = lim fn (x) =

k=1

n→∞

∞ X

nχEn (x)

(3.58)

n=1

respectively. Since each En is measurable, each χEn is measurable by Proposition 1.9(d). Thus each function fn is measurable and then Theorem 1.14 implies that f is also measurable. Now it remains to show that the function f satisfies the required properties of the problem. • Property 1: kf kp → ∞. Assume that f ∈ L∞ ((0, 1)). Then there exists a positive constant M such that kf k∞ < M . By Definition 3.7, we see that |f (x)| < M c

a.e. on (0, 1).

See also [50, Theorem 126, p. 99] and [100, Eqn. (2), p. 62].

(3.59)

3.3. Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12

87

Since Ei ∩ Ej = ∅ for i 6= j, we have |f (x)| = f (x) ≥ fn (x) ≥ n

(3.60)

for every x ∈ En . Since n → ∞, there exists a positive integer N such that N ≥ M . Recall that m(EN ) = 21N > 0, so we obtain from the inequality (3.60) that |f (x)| ≥ M on EN , but this contradicts our assumption (3.59). Therefore, we must have f ∈ / L∞ ((0, 1)) and in fact kf k∞ = ∞. Furthermore, it follows from the Ratio Test (see [99, Theorem 3.34]) that Z 1 ∞ ∞ X X n n × m(En ) = f (x) dx = kf k1 = < ∞, n 2 0 n=1 n=1 so we have kf k1 < ∞. Hence we deduce from Problem 3.4(e) that kf kp → kf k∞ = ∞ as p → ∞. • Property 2: kf kp ≤ Φ(p) for sufficiently large p. Fix a positive integer n, we consider h n ip Φ(p)

for p > 0. Since Φ(p) → ∞ as p → ∞, when p is sufficiently large, we have that h n ip 1 × m(En ) ≤ n . Φ(p) 2

n Φ(p)

≤ 1 so (3.61)

Therefore, we see that

kf kpp =

Z

0

∞ 1hX

n=1

∞ ip X np × m(En ). nχEn (x) dx =

(3.62)

n=1

Consequently, we follow from the inequality (3.61) and the expression (3.62) that ∞ ∞ h ∞ X X X kf kpp n ip 1 1 p × m(E ) ≤ n × m(E ) = = =1 n n p p [Φ(p)] [Φ(p)] n=1 Φ(p) 2n n=1 n=1

for sufficiently large p, or equivalently, kf kp ≤ Φ(p) for sufficiently large p. We end the proof of the problem.

3.3

Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12

Problem 3.10 Rudin Chapter 3 Exercise 10.



Chapter 3. Lp -Spaces

88

Proof. First of all, we have to show that f ∈ Lp (µ). Recall that Lp (µ) is a metric space with metric k · k. By the triangle inequality kfn − fm kp ≤ kfn − f kp + kf − fm kp and the hypothesis kfn − f kp → 0 as n → ∞ together imply that {fn } is a Cauchy sequence in Lp (µ). Now the completeness of Lp (µ) ensures that {fn } converges to an element of Lp (µ). By the uniqueness of limits, this element must be f so that f ∈ Lp (µ). By Theorem 3.12, {fn } has a subsequence {fnk } such that fnk → f a.e. on X. Since fn → g a.e. on X as n → ∞, every subsequence of {fn } converges to g a.e. on X. In particular, we take this subsequence {fnk } and the uniqueness of limits again imply that f =g a.e. on X. We have completed the proof of the problem.



Problem 3.11 Rudin Chapter 3 Exercise 11.

Proof. Since f, g : Ω → [0, ∞], the hypothesis f g ≥ 1 implies that (H¨older’s Inequality), we see that Z Z p 2 Z f g dµ ≤ g dµ. f dµ × Ω

By Proposition 1.24(a), we have Z p Ω

Z

(3.63)

Ω

Ω

f g dµ ≥

√ f g ≥ 1. By Theorem 3.5

dµ = µ(Ω) = 1.

(3.64)

Ω

Now our desired result follows immediately by combining the inequalities (3.63) and (3.64). This completes the analysis of the problem.  Problem 3.12 Rudin Chapter 3 Exercise 12. √ Proof. Suppose that A = ∞. Since h ≤ 1 + h2 on Ω, we have Z p 1 + h2 dµ = ∞, Ω

so the √ equalities √ definitely hold in this case. Next, suppose that A = 0. By the fact that 1 ≤ 1 + h2 ≤ 1 + 2h + h2 = 1 + h, we achieve that Z Z Z p Z dµ. (3.65) 1 + h2 dµ ≤ (1 + h) dµ = dµ ≤ Ω

Ω

Ω

Since µ(Ω) = 1, we conclude from the inequalities (3.65) that Z p 1 + h2 dµ = 1. Ω

Thus the equalities also hold in this case.

Ω

3.3. Applications of Theorems 3.3, 3.5, 3.8, 3.9 and 3.12

89

Suppose that 0 < A < ∞ which means h(x) ∈ (0, ∞) a.e. on Ω. We consider the function ϕ : R → (0, ∞) defined by p ϕ(x) = 1 + x2 which implies that

ϕ′ (x) = √

x 1 + x2

and ϕ′′ (x) =

1 3

(1 + x2 ) 2

.

(3.66)

Since ϕ′′ (x) ≥ 0 on R, we know from [99, Exercise 14, p. 115] that ϕ is convex on R. In particular, ϕ is convex on (0, ∞). By Theorem 3.3 (Jensen’s Inequality), we obtain that p

1+

A2

=

s

1+

Z

Ω

h dµ ≤

Z p 1 + h2 dµ

(3.67)

Ω

which is exactly √ the left-hand side inequality. The right-hand side inequality is easy because we always have 1 + h2 ≤ 1 + h. Hence we have verified the validity of the inequalities.

For the second assertion, we suppose that µ = m on Ω = [0, 1], h = f ′ and h is continuous on Ω. Then we have Z 1 f ′ (x) dx = f (1) − f (0). A= 0

In addition, we notice that Z p Ω

1 + h2 dµ =

Z

0

1p

1 + [f ′ (x)]2 dx

which is the arc length of the curve from (0, f (0)) to (1, f (1)).d Thus the inequalities give bounds of such arc length. Geometrically, the upper bound consists of the length from (0, f (0)) to (1, f (0)) (which is 1) and the length from (1, f (0)) to (1, f (1)) (which is A). The lower bound is just the hypotenuse of the right triangle with vertices (0, f (0)), (0, f (1)) and (1, f (1)), see Figure 3.2 below:

Figure 3.2: The geometric interpretation of a special case. d

See, for examples, [4, p. 535] and [99, Theorem 6.27, p. 137].

Chapter 3. Lp -Spaces

90

Our first inequality (3.67) comes from the application of Theorem 3.3 (Jensen’s Inequality), so the equality of the first inequality (3.67) holds if and only if the equality in [100, Eqn. (2), p. 62] holds if and only if
ϕ(A) = ϕ h(x) . (3.68)

We see from the first derivative in (3.66) that ϕ′ (x) > 0 in (0, ∞) so that it is injective in (0, ∞). Recall that A ∈ (0, ∞) and h(x) ∈ (0, ∞) a.e. on Ω, so we deduce that the equality (3.68) holds if and only if h = A a.e. on Ω. √ Next we recall the second inequality is derived from the fact that 1 + h2 ≤ 1 + h and some simple computations show that the equality holds if and only if h≡0

a.e. on Ω.

Hence we have ended the proof of the problem.



Problem 3.13 Rudin Chapter 3 Exercise 13.

Proof. There are two cases. • Case (i): 1 < p < ∞. Then the inequality in Theorem 3.8 is just H¨older’s inequality. If kf kp = 0 or kgkp = 0, then we have f = 0 a.e. on X or g = 0 a.e. on X by Theorem 1.39(a). In either case, the equality holds trivially. Therefore, we may assume that both kf kp > 0 and kgkp > 0. Furthermore, by the remark following the proof of Theorem 3.5 (H¨older’s Inequality), we may further assume that 0 < kf kp < ∞ and

0 < kgkp < ∞.

In this case, we see that the equality in Theorem 3.8 holds if and only if there are positive constants α and β such that αf p = βgq a.e. on X. In Theorem 3.9, the inequality is in fact Minkowski’s inequality. By simple observation, it is clear that the equality holds if f = 0 a.e. on X or g = 0 a.e. on X. Therefore, we may assume that f 6= 0 and g 6= 0 on measurable sets E and F with µ(E) > 0 and µ(F ) > 0 respectively. By examining the proof of [100, Eqn. (2), Theorem 3.5, pp. 64, 65], we know that it applies H¨older’s inequality to the functions f and (f + g)p−1 as well as the functions g and (f + g)p−1 . Thus we deduce from the previous paragraph that the equality in Z o1 nZ o1 n Z p q (f + g)(p−1)q dµ f · (f + g)p−1 dµ ≤ f p dµ × X

X

X

holds if and only if there are positive constants α and β such that αf p = β(f + g)q

a.e. on X.

Similarly, the equality in Z o1 n Z o1 nZ p q gp dµ (f + g)(p−1)q dµ g · (f + g)p−1 dµ ≤ × X

X

X

(3.69)

3.4. Hardy’s Inequality and Egoroff ’s Theorem

91

holds if and only if there are positive constants λ and ν such that λg p = ν(f + g)q

a.e. on X.

(3.70)

By combining the equations (3.69) and (3.70), we conclude that fp =

β βλ p (f + g)q = g α αν

a.e.

which means that f = Kg a.e. for some positive constant K. • Case (ii): p = ∞. The inequality in Theorem 3.8 comes from the inequalitye |f (x)g(x)| ≤ kf k∞ · |g(x)| for almost all x. Therefore, it is easy to see that the equality in Theorem 3.8 holds if and only if |f (x)g(x)| = kf k∞ · |g(x)| for almost all x if and only if g(x) = 0 for almost x or f (x) = kf k∞ for almost x.f Now the triangle inequality

|f + g| ≤ |f | + |g| implies the inequality in Theorem 3.9. Thus the equality in Theorem 3.9 holds if and only if f and g are either nonnegative functions or nonpositive functions for almost all x. This completes the proof of the problem.

3.4



Hardy’s Inequality and Egoroff’s Theorem

Problem 3.14 Rudin Chapter 3 Exercise 14.

Proof. We follow the suggestions given by Rudin. (a) We divide the proof into two cases:
– Special Case: f ≥ 0 and f ∈ Cc (0, ∞) . By the First Fundamental Theorem of Calculus, the function F (x) is differentiable on (0, ∞) and xF ′ (x) = f (x) − F (x)

(3.71)

for every x ∈ (0, ∞). By Integration by Partsg and the expression (3.71), we have Z ∞ Z ∞ pxF p−1 (x)F ′ (x) dx. (3.72) F p (x) dx = [xF p (x)]∞ 0 − 0

0

Now we have to evaluate the limits: lim xF p (x) and

x→0 e

lim xF p (x).

x→∞

This is [100, Eqn. (2), Theorem 3.8, p. 66]. The latter case means that f is a nonnegative constant for almost all x. g Here we assume that the formula for integration by parts is also valid for improper integrals, see [5, p. 278]. f

Chapter 3. Lp -Spaces

92

Since supp (f ) = {x ∈ (0, ∞) | f (x) 6= 0} is compact, the Heine-Borel Theorem guarantees that supp (f ) = [a, b] ⊂ (0, ∞). By this and the continuity of f , we conclude that f is bounded on (0, ∞), i.e., |f (x)| ≤ M on (0, ∞) for some positive constant M . Thus we follow from this that Z 1 x |F (x)| ≤ |f (t)| dt ≤ M x 0 for every x ∈ (0, ∞), so we see that lim xF p (x) = 0.

(3.73)

x→0

Again, the boundedness of f implies that Z b Z ∞ Z x |f (t)| dt ≤ M (b − a) < ∞ |f (t)| dt = |f (t)| dt ≤ |xF (x)| ≤ 0

0

on (0, ∞). Since xF p = that

(xF )p xp−1

(3.74)

a

and p > 1, it reduces from these and the bound (3.74) lim xF p (x) = 0.

(3.75)

x→∞

Therefore, we deduce from the formula (3.72) and the two limits (3.73) and (3.75) that Z ∞ Z ∞ F p−1 (x)[f (x) − F (x)] dx F p (x) dx = −p 0 0 Z ∞ Z ∞ F p (x) dx F p−1 (x)f (x) dx + p = −p 0

0

so that (p − 1)

Z

∞

F p (x) dx = p

Z

∞

F p−1 (x)f (x) dx.

(3.76)

0

0

Note that 1 < p < ∞, let q be its conjugate exponent. Obviously, our hypotheses make sure that f and F p−1 have range [0, ∞], so we may apply Theorem 3.5 (H¨older’s Inequality) to them. In fact, it yields from the fact (p − 1)q = p and the expression (3.76) that Z ∞ F p (x) dx (p − 1)kF kpp = (p − 1) 0 o1 nZ ∞ o1n Z ∞ p q F (p−1)q (x) dx ≤p f p (x) dx 0

0

p q

= pkf kp · kF kp

which is exactly

p− pq

kF kp = kF kp

≤

p kf kp . p−1


– General Case: f ∈ Lp (0, ∞) . By Definition 3.6, we always have

kf kp = |f | p ,

(3.77)

so we may
assume that f ≥ 0 on
(0, ∞). Since 1 < p < ∞, Theorem 3.14 says that
Cc (0, ∞) is dense in Lp (0, ∞) . Thus there exists a sequence {fn } ⊆ Cc (0, ∞) such that kfn − f kp → 0 (3.78)

3.4. Hardy’s Inequality and Egoroff ’s Theorem

93

as n → ∞. Notice that fn may not be nonnegative, but since |fn | − |f | ≤ |fn − f | and f ≥ 0, we have |fn (x)| − f (x) ≤ |fn (x) − f (x)| (3.79)

for every x ∈ (0, ∞). Certainly, the inequality (3.79) implies that

|fn | − f ≤ kfn − f kp p

(3.80)

for every n = 1, 2, . . .. Now the result (3.78) and the inequality (3.80) ensure that

|fn | − f → 0 p

as n → ∞. In other words,
we may also assume that fn ≥ 0 for each n = 1, 2, . . .. Recall that fn ∈ Cc (0, ∞) , so we derive from the Special Case that kFn kp ≤

where Fn (x) =

p kfn kp , p−1

1 x

Z

1 x

Z

and n = 1, 2, . . .. Suppose that F (x) =

(3.81)

x

fn (t) dt

0

x

f (t) dt.

0

Since f ≥ 0 and fn ≥ 0, it is easily seen from the definition that F ≥ 0 and Fn ≥ 0. We claim that {Fn } converges pointwise to F on (0, ∞). To see this, fix x ∈ (0, ∞), then their definitions give Z 1 x |fn (t) − f (t)| dt. (3.82) |Fn (x) − F (x)| ≤ x 0 By the result (3.78), we know that for every ǫ > 0, there exists a positive integer N such that n ≥ N implies kfn − f kp < ǫ.

Apply Theorem 3.5 (H¨older’s Inequality) with 1p + 1q = 1 to the inequality (3.82), we get immediately that Z o1 n Z x o1 1n x p q |fn (t) − f (t)|p dt × 1q dt |Fn (x) − F (x)| ≤ x 0 0 1 1 ≤ kfn − f kp × x q x 1

= x− p kfn − f kp 1

< ǫx− p

(3.83)

for all n ≥ N . Since x is fixed and independent of ǫ, the estimate (3.83) verifies the truth of the claim. Next, recall that each fn is continuous on [a, x] for every a > 0, so the First Fundamental Theorem of Calculus shows that each Fn is continuous on [a, x]. As a continuous function, every Fn is measurable on [a, x].h Hence, it follows from the h

See §1.11 or [99, Example 11.14].

Chapter 3. Lp -Spaces

94

pointwise convergence of {Fn }, Theorem 1.28 (Fatou’s Lemma), the inequality (3.81) and then the result (3.78) that Z ∞ F p (x) dx kF kpp = 0 Z ∞ lim inf Fnp (x) dx = n→∞ 0 Z ∞ Fnp (x) dx ≤ lim inf n→∞ 0 Z ∞ Fnp (x) dx ≤ lim n→∞ 0

= lim kFn kpp n→∞  p p ≤ lim kfn kpp p − 1 n→∞  p p kf kpp ≤ p−1

which is equivalent to the desired result kF kp ≤

p kf kp . p−1



Hence the mapping T : Lp (0, ∞) → Lp (0, ∞) given by T (f ) = F is continuous.
(b) Suppose that f ∈ Lp (0, ∞) and kF kp =

p kf kp < ∞. p−1

(3.84)

Recall that kf kp = |f | p , so we may suppose further that f ≥ 0 on (0, ∞). By the
General Case of the proof in part (a), there is a nonnegative sequence {fn } ⊆ Cc (0, ∞) such that kfn −f kp → 0 as n → ∞. Now we replace F and f by Fn and fn in the expression (3.76) respectively, we gain Z ∞ Z ∞ p Fnp (x) dx = kFn kpp = Fnp−1 (x)fn (x) dx (3.85) p−1 0 0

for each n = 1, 2, . . .. Since fn ∈ Cc (0, ∞) , T (fn ) = Fn ∈ Lp (0, ∞) , where T is the continuous mapping considered in part
(a). By Theorem 3.5 (H¨older’s Inequality), it is easy to see that fn Fnp−1 ∈ Lp (0, ∞) for each n = 1, 2, . . .. Since the mapping T is continuous, it follows from the expression (3.85) that Z ∞ Z ∞ p F p−1 (x)f (x) dx. (3.86) F p (x) dx = p−1 0 0
In other words, the formula (3.76) holds for f ∈ Lp (0, ∞) . We apply Theorem 3.5 (H¨older’s Inequality) to the right-hand side of the inequality (3.86) and then using our hypothesis (3.84) to conclude that Z ∞ p kF kpp = F p−1 (x)f (x) dx p−1 0 Z o1 n Z ∞ o1 p n ∞ p p q f (x) dx × (3.87) F p (x) dx ≤ p−1 0 0

3.4. Hardy’s Inequality and Egoroff ’s Theorem

95

p p kf kp × kF kpq p−1 = kF kpp .

=

Hence the equality in the inequality (3.87) must hold. Consequently, there are constants α and β, not both 0, such that αf p = β(F p−1 )q = βF p a.e. on (0, ∞) and so 1

1

αpf = β pF

a.e. on (0, ∞).

(3.88)

Here we have several cases: – Case (i): β = 0. Then we have α 6= 0 and f = 0 a.e. on (0, ∞). Thus we are done.

– Case (ii): α = 0. Then we have β 6= 0 and F = 0 a.e. on (0, ∞). By the hypothesis (3.84), we see that kf kp = 0 which gives f = 0 a.e. on (0, ∞) and we are done again. – Case (iii): αβ 6= 0. By Theorem 3.8, we see that f ∈ L1 ([a, b]), where [a, b] is any bounded interval of (0, ∞). By the comment following the proof of Theorem 11.3 on [99, p. 324], we know that F is differentiable a.e. on [a, b] and
′ xF (x) = f (x) (3.89) a.e. on [a, b]. Since [a, b] is arbitrary, the formula (3.89) holds a.e. on (0, ∞). Combining the equations (3.88) and (3.89), we see immediately that xF ′ + xF = f which implies xf ′ = (c − 1)f a.e. on (0, ∞) (3.90) for some nonzero constant c. By solving the differential equation (3.90), we obtain that f (x) = γxc−1 a.e. on (0, ∞)

for some constant γ. Since γxc−1 ∈ Lp ((0, ∞)), it forces that γ = 0. Hence we have shown that the equality holds only if f = 0 a.e. on (0, ∞). 1

(c) Take f (x) = x− p χ[1,A] (x) for large A. Then we have kf kp =

nZ

0

∞

|x

− p1

χ[1,A] (x)|p dx

o1

p

=

nZ

A 1

x−1 dx

o1

p

1

= (log A) p .

Next, we know from the definition that Z 1 x − p1 t χ[1,A] (t) dt F (x) = x  0 0, if x ∈ (0, 1);       Z x   1  1 t− p dt, if x ∈ [1, A]; = x 1       1 Z A −1    t p dt, if x ∈ (A, ∞); x 1  if x ∈ (0, 1);   0,        p (x− p1 − x−1 ), if x ∈ [1, A]; p−1 =      1− 1  A p −1  p   · , if x ∈ (A, ∞). p−1 x

(3.91)

(3.92)

Chapter 3. Lp -Spaces

96 Note that A is large and 1 −

1 p

> 0, so we have 1

A1− p − 1 > 0.

(3.93)

Therefore, we deduce from the expressions (3.92) and the fact (3.93) that Z ∞ |F (x)|p dx kF kpp = 0 Z ∞ Z A p |F (x)|p dx |F (x)| dx + = 1

A

 p p Z ∞ A1− p1 − 1 p  p p Z A − p1 −1 p |x − x | dx + = dx p−1 p − 1 x 1 A Z  p p A
p 1 1 pp (x− p − x−1 )p dx + A1− p − 1 A1−p = p+1 p−1 (p − 1) 1  p p Z A 1 (x− p − x−1 )p dx. > p−1 1

Fix p first. Assume that the constant for some δ < 1, i.e.,

p p−1

kF kp ≤

(3.94)

could be replaced by a smaller number

δp kf kp . p−1

δp p−1

1

Take ǫ > 0 such that δ < ǫ < 1. Then it is clear that for large A, we have x p −1 < 1 − ǫ for x ≥ A2 or equivalently, x

for x ≥

A 2.

Z

− p1

− x−1 > ǫx

− p1

>0

(3.95)

By this estimate (3.95), we obtain

A

(x

− p1

1

−x

−1 p

) dx >

Z

A A 2

 A > ǫp log A. ǫp x−1 dx = ǫp log A − log 2

(3.96)

By substituting the inequality (3.96) into the inequality (3.94) and then using the fact (3.91), we can show that  ǫp p  δp p kF kpp > log A > kf kpp ≥ 0, p−1 p−1

but this implies

kF kp > a contradiction.

δp kf kp , p−1

(d) Since f (x) > 0 on (0, ∞), there exists a δ > 0 such that we can find an α ∈ (0, ∞) with Z α f (t) dt > δ > 0. 0

By the definition, we have F (x) > 0 on (0, ∞) so that Z ∞ Z ∞ Z ∞ Z ∞ Z x  δ 1 f (t) dt dx > dx = ∞. kF k1 = F (x) dx ≥ F (x) dx = 0 α x 0 α α x

Hence F ∈ / L1 . In other words, this shows that Hardy’s inequality does not hold when p = 1.

3.4. Hardy’s Inequality and Egoroff ’s Theorem

97

Hence we have completed the proof of the problem.



Problem 3.15 Rudin Chapter 3 Exercise 15.

Proof. We follow the hint. Suppose that {an } is decreasing. Let f : (0, ∞) → (0, ∞) be given by ∞ X an χ(n−1,n] (x). (3.97) f (x) = n=1

Firstly, we note that

Z

∞

f p (x) dx =

0

∞ Z X

n

n=1 n−1

apn dx =

∞ X

apn

(3.98)

n=1

∞ X
apn < ∞. This shows that the desired which implies that f ∈ Lp (0, ∞) if and only if n=1

inequality holds if f ∈ /

Lp

∞ X
apn < ∞ in (0, ∞) . Without loss of generality, we assume that n=1

the following discussion. Secondly, let N be a positive integer and x ∈ (N − 1, N ]. Then we deduce from the definition (3.97) that Z 1 x F (x) = f (t) dt x 0 Z x N −1 Z i 1h X n f (t) dt f (t) dt + = x N −1 n=1 n−1 Z Z N −1 i x 1h X n = aN dt an dt + x n=1 n−1 N −1 1 [a1 + a2 + · · · + aN −1 + aN (x − N + 1)]. x

=

(3.99)

It is clear from the expression (3.99) and the fact {an } is decreasing that N −1 N  i 1X 1h X an + aN (x − N + 1) = an − N aN + xaN x x n=1

n=1 N

 1X an − N aN + aN = x n=1

N  1 X ≥ an − N aN + aN N n=1

=

N 1 X an . N

(3.100)

n=1

Thus we obtain from the expression (3.99) and the inequality (3.100) that kF kp =

nZ

∞ 0

F p (x) dx

o1

p

=

∞ Z nX

N

N =1 N −1

F p (x) dx

o1

p

≥

∞  N nX 1 X p o 1p an . N n=1 N =1

(3.101)

Chapter 3. Lp -Spaces

98

Furthermore, by Problem 3.14(a) and the expression (3.98), we derive that kF kp ≤

p p n kf kp = p−1 p−1

Z

∞

f p (x) dx

0

o1

p

=

p n X p o p1 a . p − 1 n=1 n ∞

(3.102)

Hence our desired inequality follows immediately from the inequalities (3.101) and (3.102). ∞ X For the general case, recall the assumption that an > 0 for all n ∈ N and apn < ∞, so the n=1

sequence {apn } converges absolutely and every rearrangement converges to the same sum (see [99, Theorem 3.55, p. 78]). On the other hand, if we fix a positive integer N , then N  X k X 1 k=1

k

n=1

an

p

=

 a p

 a + a p  a + a + a p 1 2 1 2 3 + 1 2 3  a + a + · · · + a p 1 2 N , + ··· + N 1

+

so it follows from this form that the sum attains its maximum if and only if a1 ≥ a2 ≥ · · · ≥ aN . Therefore, we have ∞  ∞  ∞ ∞ N N X X 1 X p  p p X p  p p X p 1 X p αn = an , an ≤ αn ≤ N N p−1 p−1 n=1 n=1 n=1 n=1 N =1 N =1 {z } |

By the special case.

where {αn } is the decreasing rearrangement of {an }. This completes the proof of the problem.  Problem 3.16 Rudin Chapter 3 Exercise 16.

Proof. Let’s prove the assertions one by one. • A proof of Egoroff ’s Theorem. Put \ n 1o . S(n, k) = x ∈ X |fi (x) − fj (x)| < k

(3.103)

i,j>n

For each k ∈ N, we follow from the definition (3.103) that \ n \ n 1o 1o ∩ x ∈ X |fi (x) − fj (x)| < S(n, k) = x ∈ X |fi (x) − fj (x)| < k k i,j>n+1

\ n 1o ∩ x ∈ X |fi (x) − fj (x)| < k

i=n+1 j>n

j=n+1 i>n

which implies that S(n, k) ⊆ S(n + 1, k).

Next, for each fixed k ∈ N, we claim that X=

∞ [

n=1

S(n, k).

(3.104)

3.4. Hardy’s Inequality and Egoroff ’s Theorem

99

To this end, we see that the set inclusion ∞ [

S(n, k) ⊆ X

n=1

is evident. To prove the reverse direction, we note that for each k ∈ N and x ∈ X, we yield from our hypothesis fn (x) → f (x) that there exists a positive integer N such that i, j > N implies that 1 |fi (x) − fj (x)| < . k In other words, we must have x ∈ S(n, k), i.e., X⊆

∞ [

S(n, k).

n=1

Thus these prove the validity of our claim (3.104) and Theorem 1.19(d) gives
lim µ S(n, k) = µ(X) n→∞

for every k = 1, 2, . . ..

Given that ǫ > 0. For each k ∈ N, this fact allows us to choose nk such that
ǫ |µ S(nk , k − µ(X)| < k . 2

Define

E=

∞ \

S(nk , k).

(3.105)

k=1

Then it is clear that X \E = which implies that µ(X \ E) ≤ Pick a k with

1 k

∞ X k=1

∞ [

k=1


X \ S(nk , k)

∞ ∞
X
X ǫ µ(X) − µ S(nk , k) < µ X \ S(nk , k) ≤ = ǫ. 2k k=1

k=1

< ǫ. If x ∈ S(nk , k), then we have

|fi (x) − fj (x)| < ǫ

(3.106)

for all i, j > nk . By the definition (3.105), we know that E ⊆ S(nk , k) for every k ∈ N, so the inequality (3.106) holds on E. By the definition, fn → f uniformly on E, proving Egoroff’s Theorem. • A counterexample on a σ-finite space. Recall from Definition 2.16 that X is said to be a σ-finite space if X is a countable union of sets Ei with µ(Ei ) < ∞. It is clear that R=

∞ [

[n − 1, n) ∪

n=1

∞ [

(−n, 1 − n],

n=1

so R is σ-finite. Let fn (x) = nx on R. It is clear that the sequence converges pointwise to 0 at every point of R. Furthermore, we know that a measurable set E ⊆ R with m(R\E) < 1 must be unbounded. Otherwise, E ⊆ [−M, M ] for some M > 0 and this implies that
m(R \ E) ≥ m R \ [−M, M ] = ∞,

Chapter 3. Lp -Spaces

100

a contradiction. However, {fn } cannot converge uniformly to 0 on the unbounded set E. Thus this counterexample shows that Egoroff’s Theorem cannot be extended to σ-finite spaces. • An extension of Egoroff ’s Theorem. Suppose that {ft } is a family of complex measurable functions such that (i) lim ft (x) = f (x) and t→∞

(ii) Fix a x ∈ X. Then the function F : (0, ∞) → C defined by F (t) = ft (x) is continuous. For each n ∈ N, we consider the real function gn : X → R given by gn (x) = sup{|ft (x) − f (x)|}.

(3.107)

t≥n

Then, for every x ∈ X, gn (x) → 0 as n → ∞. Now it remains to show that each gn is measurable. To this end, let a ∈ R and Ea (n) = {x ∈ X | gn (x) < a}

= {x ∈ X | |ft (x) − f (x)| < a for all t ≥ n} \ = {x ∈ X | |fr (x) − f (x)| < a}.

(3.108)

r≥n r∈Q

Since fr is measurable, f is also measurable by Corollary (a) following Theorem 1.14. Thus |fr − f | is measurable by Proposition 1.9(b) and then the set {x ∈ X | |fr (x) − f (x)| < a} is measurable for every a ∈ R by Definition 1.3(c). By the expression (3.108), since there are countable such measurable sets, Comment 1.6(c) implies that Ea (n) is measurable for every real a. By Definition 1.3(c) again, each gn is measurable and thus Egoroff’s Theorem can be applied to conclude that for every ǫ > 0, there exists a measurable set E ⊆ X such that µ(X \ E) < ǫ and {gn } converges uniformly to 0 on E. By the definition (3.107), {ft } converges to f uniformly on E. This completes the proof of the problem.



Problem 3.17 Rudin Chapter 3 Exercise 17.

Proof. (a) Since the inequality is clearly holds if either α = 0 or β = 0, we assume, without loss of generality, that α and β cannot be both zero in the following discussion. Let 0 < p ≤ 1. We claim that the continuous function ϕ : [0, 1] → R defined by ϕ(x) = xp + (1 − x)p

3.4. Hardy’s Inequality and Egoroff ’s Theorem

101

satisfies ϕ(x) ≥ 1 on [0, 1]. To this end, we note that ϕ′ (x) = pxp−1 − p(1 − x)p−1 = 0 if and only if x = 21 . Since p − 1 ≤ 0, we know that ϕ′ (x) ≥ 0 and

ϕ′ (x) ≤ 0

on (0, 12 ) and ( 21 , 1) respectively. In other words, ϕ is increasing and decreasing on (0, 21 ) and ( 21 , 1) respectively. By the continuity of ϕ, these are also valid on [0, 12 ] and [ 21 , 1] respectively. Thus we observe ϕ(x) ≥ ϕ(0) = 1 on [0, 21 ] and

ϕ(x) ≥ ϕ(1) = 1 on [ 12 , 1]

which mean that xp + (1 − x)p ≥ 1

(3.109)

on [0, 1]. For arbitrary nonzero complex numbers α and β, we put x = inequality (3.109) to get |α|p |β|p + ≥1 p (|α| + |β|) (|α| + |β|)p

|α| |α|+|β|

into the

which implies (|α| + |β|)p ≤ |α|p + |β|p .

(3.110)

By the triangle inequality, it is true that |α − β| ≤ |α| + |β|, so this and the inequality (3.110) give |α − β|p ≤ |α|p + |β|p . (3.111) Next, let 1 < p < ∞. The function ψ : [0, 1] → R given by ψ(x) = xp has second derivative p(p − 1)xp−2 > 0 on (0, 1). By Definition 3.1, it is convex on (0, 1), i.e., [(1 − λ)x + λy]p ≤ (1 − λ)xp + λy p , (3.112) where x, y ∈ (0, 1) and λ ∈ [0, 1]. Put λ = 12 , x = (3.112), we obtain

|α| |α|+|β|

and y =

|β| |α|+|β|

into the inequality

 1 |α| + |β| p 1 |α|p |β|p 1 ≤ · · · + 2 |α| + |β| 2 (|α| + |β|)p 2 (|α| + |β|)p which reduces to (|α| + |β|)p ≤ 2p−1 (|α|p + |β|p ). Again, the triangle inequality and this show that |α − β|p ≤ 2p−1 (|α|p + |β|p ).

(3.113)

Hence the desired inequality follows from combining the inequalities (3.111) and (3.113). (b)

(i) If µ(X) = 0, then Proposition 1.24(e) implies that kf kp = 0 for every f ∈ Lp (µ) and there is nothing to prove. Thus, without loss of generality, we may assume that µ(X) > 0. We divide the proof into several steps:

Chapter 3. Lp -Spaces

102 ∗ Step 1: Lemma 3.7 and its application. Lemma 3.7

For every ǫ > 0, there exists a δ > 0 such that for every E ∈ M with µ(E) < δ, we have Z |f |p dµ ≤ ǫ. (3.114) E

Proof of Lemma 3.7. Note that |f | must be bounded a.e. on X. Otherwise, there exists an E ∈ M such that µ(E) > 0 and |f | = ∞ on E. Then Proposition 1.24(b) shows that Z Z |f |p dµ = ∞, |f |p dµ ≥ E

X

a contradiction. Thus there exists a positive constant M such that |f | ≤ M a.e. on X. Given ǫ > 0. Now for every E ∈ M with µ(E) < Mǫ p , we have Z |f |p dµ ≤ M p µ(E) ≤ ǫ a.e. on X E

 which is exactly what we want. By Lemma 3.7, there exists a δ > 0 such that the inequality (3.114) holds for every E ∈ M with µ(E) < δ. For this δ > 0, suppose that we can find a measurable set F such that fn → f a.e. on F and µ(F ) < ∞. (The existence of such a F will be clear at the end of Step 2 below.) Then Egoroff’s Theorem guarantees that there exists a measurable set B ⊆ F such that µ(F \ B) < δ and {fn } converges uniformly to f on B. Since F \ B ∈ M and µ(F \ B) < δ, we also have Z ǫ (3.115) |f |p dµ ≤ . 2 F \B Furthermore, since fn → f uniformly on B, there is a positive integer N such that n ≥ N implies that |fn (x) − f (x)| ≤

 ǫ 1 p µ(B)

for all x ∈ B and this means that Z

B

|fn − f |p dµ ≤ ǫ

(3.116)

for n ≥ N .

∗ Step 2: Lemma 3.8 and its application. Lemma 3.8 If f ∈ L1 (µ), then for every ǫ > 0, there exists an E ∈ M such that F = X \ E, µ(F ) < ∞ and Z |f | dµ ≤ ǫ. E

3.4. Hardy’s Inequality and Egoroff ’s Theorem

103

Proof of Lemma 3.8. Since |f | is measurable, Theorem 1.17 (The Simple Function Approximation Theorem) ensures that there exists a sequence of simple measurable functions {sk } on X such that 0 ≤ s1 ≤ s2 ≤ · · · ≤ |f |

(3.117)

and sk converges to |f | pointwisely. By Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem), we have Z Z |f | dµ. sk dµ = lim k→∞ X

X

Hence, for every ǫ > 0, there exists a positive integer N such that Z Z (|f | − sk ) dµ = (|f | − sk ) dµ ≤ ǫ.

(3.118)

X

X

By Definition 1.16, we have

sk =

nk X

αi χFi ,

i=1

where each αi is positive, Fi is measurable and Fi ∩ Fj = ∅ for all i 6= j. Recall that |f | ∈ L1 (µ), so the hypothesis (3.117) implies that sk ∈ L1 (µ) for every n ∈ N. Since each αi is positive, µ is positive on X (i.e., µ(E) ≥ 0 for every E ∈ M) and Z nk X sk dµ < ∞, αi µ(Fi ) = X

i=1

they force that µ(Fi ) < ∞ for every i = 1, 2, . . . , nk . We suppose that E = {x ∈ X | sk (x) = 0}

and F =

nk [

Fi .

i=1

Then the previous analysis shows that µ(F ) < ∞. Furthermore, it is clear that sk (x) > 0 if and only if x ∈ Fi for some i, so this fact implies that X \ E = F. Thus we deduce from the estimate (3.118) and the definition of E that Z Z Z Z Z sk dµ ≤ ǫ. (|f | − sk ) dµ + sk dµ ≤ |f | dµ = (|f | − sk ) dµ + E

E

E

X

E

This proves Lemma 3.8.  p p 1 Since f ∈ L (µ), we have |f | ∈ L (µ). Then Lemma 3.8 implies that there exists an E ∈ M such that F = X \ E, µ(F ) < ∞ and Z ǫ (3.119) |f |p dµ ≤ . 2 E ∗ Step 3: Constructions of A and B satisfying the hypotheses. Let E and F be defined as in Lemma 3.8 so that the estimate (3.119) holds. We remark that X = E ∪ F = E ∪ (F \ B) ∪ B = A ∪ B,

Chapter 3. Lp -Spaces

104

where B is the measurable set guaranteed by Egoroff’s Theorem in Step 1 and A = E ∪ (F \ B). Recall that B ⊆ F , so it is easy to see that µ(B) ≤ µ(F ) < ∞ and since E ∩ (F \ B) = ∅, we obtain from Theorem 1.29 and the two estimates (3.119) and (3.115) that Z Z Z |f |p dµ ≤ ǫ. (3.120) |f |p dµ + |f |p dµ = F \B

E

A

Since fn → f a.e. on X and particularly on B, this and Theorem 1.28 (Fatou’s Lemma) together show that Z Z Z |fn |p dµ (3.121) lim inf |fn |p dµ ≤ lim inf |f |p dµ = B n→∞

B

n→∞

B

Next, we know from the definition that A = (E ∪ F ) \ B = X \ B so that A ∩ B = ∅. Since kfn kp → kf kp as n → ∞, it deduces from this, Problem 1.4 and the inequalities (3.120) and (3.121) that Z Z   Z |fn |p dµ lim sup |fn |p dµ ≤ lim sup |fn |p dµ + lim sup − n→∞ n→∞ n→∞ B A Z ZX p |fn |p dµ = lim sup |fn | dµ − lim inf n→∞ B Zn→∞ X Z |f |p dµ |f |p dµ − ≤ B X Z |f |p dµ = A

≤ ǫ.

(3.122)

∗ Step 4: The establishment of kfn − f kp → 0 as n → ∞. By part (a), the estimates (3.116) and (3.122), we see that Z Z Z lim sup |fn − f |p dµ ≤ lim sup |fn − f |p dµ + lim sup |fn − f |p dµ n→∞ n→∞ n→∞ X AZ B ≤ γp lim sup (|f |p + |fn |p ) dµ + ǫ n→∞

A

≤ 2γp ǫ + ǫ.

Since ǫ is arbitrary, we conclude that lim kfn − f kp = lim

n→∞

n→∞

nZ

X

|fn − f |p dµ

o1

p

= 0.

(ii) Put hn = γp (|f |p + |fn |p ) − |f − fn |p . By part (a), we have hn ≥ 0 for all n ∈ N. Since f and fn are measurable, each hn is also measurable. By Theorem 1.28 (Fatou’s Lemma), we get Z Z   hn dµ. (3.123) lim inf hn dµ ≤ lim inf X

n→∞

f |p

n→∞

X

Since fn → f a.e. on X, |fn − → 0 and |fn |p → |f |p a.e. on X. Thus we must have lim hn = 2γp |f |p a.e. on X. (3.124) n→∞

3.4. Hardy’s Inequality and Egoroff ’s Theorem

105

Putting the limit (3.124) into the inequality (3.123), we obtain from Problem 1.4 that Z

X

2γp |f |p dµ =

Z 

ZX 

 lim hn dµ

n→∞

 lim inf hn dµ n→∞ X Z hn dµ = lim inf n→∞ X Z Z i h |fn − f |p dµ γp (|f |p + |fn |p ) dµ − ≤ lim inf n→∞ X X Z hZ i p p (|f | + |fn | ) dµ − lim sup |fn − f |p dµ ≤ γp lim inf n→∞ n→∞ X X Z
|fn − f |p dµ. (3.125) = γp lim inf kfn kpp + kf kpp − lim sup =

n→∞

n→∞

X

Since kfn kp → kf kp as n → ∞, we can further reduce the inequality (3.125) to 0 ≤ lim sup n→∞

Z

X

|fn − f |p dµ ≤ γp



 lim kfn kpp + kf kpp − 2γp kf kpp = 0.

n→∞

Hence this leads to the following lim kfn − f kp = lim sup kfn − f kp = 0

n→∞

n→∞

as desired. (c) Consider X = (0, 1) and µ = m the Lebesgue measure. For each n = 1, 2, . . ., we let En = (0, n1p ) and fn = nχEn : (0, 1) → R. If x ∈ (0, 1), then there exists a positive integer N such that x ∈ / En for all n ≥ N . In this case, we have fn (x) = 0 for all n ≥ N . In other words, we have fn (x) → f (x) ≡ 0
for every x ∈ (0, 1). It is clear that f ∈ Lp (0, 1) and kf kp = 0. Besides, we have kfn kp =

nZ

0

1

|fn (x)|p dx

o1

p

=

nZ

En

np dx

o1

p

=1

for every n ∈ N. Thus we know that kfn kp 9 kf kp as n → ∞. Finally, since kfn − f kp = kfn kp = 1, we see that kfn − f kp 9 0 as n → ∞, i.e., the conclusion of part (b) is false. We have completed the proof of the problem. Remark 3.1 We remark that there is a short and elementary proof of Problem 3.17(b) in [82].



Chapter 3. Lp -Spaces

106

Convergence in Measure and the Essential Range of f ∈ L∞ (µ)

3.5

Problem 3.18 Rudin Chapter 3 Exercise 18.

Proof. (a) Given small ǫ > 0. For each positive integer k, we let Ek = {x ∈ X | |fn (x) − f (x)| ≤ ǫ for all n ≥ k}

and E =

∞ [

Ek .

k=1

The definitions of Ek and E guarantee that X \E =

∞ \

(X \ Ek ) =

k=1

∞ \

k=1

{x ∈ X | |fn (x) − f (x)| > ǫ for some n ≥ k}.

(3.126)

Since fn (x) → f (x) a.e. on X, we deduce from the expression (3.126) that µ(X \ E) = 0. Furthermore, we have E1 ⊆ E2 ⊆ · · · , so Theorem 1.19 implies that µ(Ek ) → µ(E) as k → ∞. Thus it follows from this fact and the result µ(X \ E) = 0 that lim µ(Ek ) = µ(E) = µ(X) < ∞.

k→∞

This result ensures that there exists a N ∈ N such that µ(X \ EN ) < ǫ. {x ∈ X | |fn (x) − f (x)| > ǫ for all n > N } ⊆ X \ EN . Hence we must have µ({x ∈ X | |fn (x) − f (x)| > ǫ for all n > N }) < ǫ, i.e., fn → f in measure. (b) Suppose that 1 ≤ p < ∞. Since kfn − f kp → 0 as n → ∞, we must have fn − f ∈ Lp (µ) for every n ∈ N. Since fn ∈ Lp (µ), we also have f = (f − fn ) + fn ∈ Lp (µ) by Theorem 3.9. Given ǫ > 0, there exists a positive integer N such that n ≥ N implies that kfn − f kp < ǫp+1 .

(3.127)

Let Fn = {x ∈ X | |fn (x) − f (x)| > ǫ} for each n = 1, 2, . . .. Now it is easy to check thati Z Z ǫp dµ = ǫp µ(Fn ). (3.128) |fn (x) − f (x)|p dµ ≥ Fn

i

Fn

The inequality (3.128) is a consequence of the so-called Chebyshev’s Inequality: If f is a nonnegative, extended real-valued measurable function on X with measure µ, 0 < p < ∞ and ǫ > 0, then Z 1 µ({x ∈ X | f (x) ≥ ǫ}) ≤ p f p dµ. ǫ X See [40, Theorem 6.17, p. 193].

3.5. Convergence in Measure and the Essential Range of f ∈ L∞ (µ)

107

Hence we establish from the inequalities (3.127) and (3.128) that µ({x ∈ X | |fn (x) − f (x)| > ǫ}) = µ(Fn ) < ǫ for n ≥ N . By the definition, fn → f in measure.

Next, we suppose that p = ∞. Now kfn − f k∞ → 0 means the existence of a positive integer N such that n ≥ N implies |fn (x) − f (x)| < ǫ a.e. on X. By the definition, we have µ(Fn ) = 0 < ǫ for all n ≥ N , i.e., fn → f in measure in this case.

(c) Suppose that fn → f in measure, i.e., for every ǫ > 0, there exists a positive integer N such that n ≥ N implies that µ({x ∈ X | |fn (x) − f (x)| > ǫ}) < ǫ. In particular, for each positive integer k, there exists a positive integer Nk such that µ({x ∈ X | |fn (x) − f (x)| > 2−k }) < 2−k

(3.129)

for all n ≥ Nk . Now we may choose nk > Nk freely in the estimate (3.129) and consider the subsequence {fnk }. We define Ek = {x ∈ X | |fnk (x) − f (x)| > 2−k }

and E =

∞ [ ∞ \

Ek .

(3.130)

m=1 k≥m

Then it follows from the estimate (3.129) that µ(E) ≤ µ

∞  [

k≥m

∞ ∞  X X 1 1 Ek ≤ µ(Ek ) < = m−1 , k 2 2 k=m

k=m

where m = 1, 2, . . .. As a result, we must have µ(E) = 0. Take any p ∈ X \ E, the definitions (3.130) say that p∈

∞ \

(X \ Ek ) =

k≥m

for some m ∈ N and then

∞ \

{x ∈ X | |fnk (x) − f (x)| ≤ 2−k }

k≥m

|fnk (p) − f (p)| ≤ 2−k

for all k ≥ m. In other words, we have lim fnk (p) = f (p)

k→∞

which is our desired result. Here we propose two examples which say that the converses of parts (a) and (b) fail: • Failure of the converse of part (a). By Problem 2.9, there exists a sequence of continuous functions fn : [0, 1] → R such that 0 ≤ fn ≤ 1 and kfn − 0k1 → 0 as n → ∞, but there is no x ∈ [0, 1] such that {fn (x)} converges. In other words, fn (x) does not converge to 0 a.e. on
[0, 1]. However, since each fn is continuous [0, 1], it is R on [0, 1] so that fn ∈ L1 [0, 1] by [99, Theorem 11.33, p. 323]. Thus the sequence of functions {fn } satisfies the hypotheses of part (b) which shows that fn → 0 in measure. Hence this example implies the the converse of part (a) is false.

Chapter 3. Lp -Spaces

108

• Failure of the converse of part (b). Suppose that X = [0, 1], µ is the Lebesgue measure and fn = en χ[0, 1 ] n

for each n = 1, 2, . . .. Then fn → 0 a.e. on [0, 1] and part (a) implies that fn → 0 in measure. However, we note that n Z n1 o1 nZ o1 en p p p = = 1 →∞ enp dx kfn kp = |fn (x)| dµ 0 X np
p as n → ∞, so the sequence {fn } fails to converge in L [0, 1] .

Finally, by examining the proofs of parts (b) and (c), we see that they remain to be true in the case µ(X) = ∞. However, part (a) does not hold in this case. For instance, let X = [0, ∞) and En = [0, n] for each n ∈ N. Consider the functions fn = χEn . Then for every x ∈ [0, ∞), there exists a positive integer N such that x ∈ En for all n ≥ N , so it is true that fn (x) = 1 for all n ≥ N , i.e., fn (x) → f (x) ≡ 1 on [0, ∞). However, for every ǫ ∈ (0, 1), since fn (x) = 0 if and only if x ∈ (n, ∞), we see that {x ∈ [0, ∞) | |fn (x) − 1| > ǫ} = (n, ∞) for every n ∈ N. Hence it leads that m({x ∈ [0, ∞) | |fn (x) − 1| > ǫ}) = ∞ for every n ∈ N, i.e., fn does not converge in measure to f . Hence we end the proof of the problem.



Problem 3.19 Rudin Chapter 3 Exercise 19.

Proof. For every ǫ > 0, define Eǫ (w) = {x ∈ X | |f (x) − w| < ǫ} and \

Rf = {w ∈ C | µ Eǫ (w) > 0 for every ǫ > 0} = {w ∈ C | µ Eǫ (w) > 0}.

(3.131)

ǫ>0

We are going to prove the assertions one by one: • Rf is compact. Given {wn } ⊆ Rf and w ∈ C such that wn → w as n → ∞. For every ǫ > 0, there exists a positive integer N such that |wn − w| < 2ǫ . By the triangle inequality, we have ǫ |f (x) − w| ≤ |f (x) − wn | + |wn − w| < |f (x) − wn | + 2 which means that E 2ǫ (wn ) ⊆ Eǫ (w) or equivalently

(3.132) µ Eǫ (w) > µ E 2ǫ (wn ) > 0. Since ǫ is arbitrary, we know from the estimate (3.132) that w ∈ Rf . In other words, Rf is closed in C.

Recall that kf k∞ < ∞, so we pick a complex number w0 such that |w0 | > kf k∞ and then consider the positive number ǫ = |w0 | − kf k∞ . If x ∈ Eǫ (w0 ), then the triangle inequality shows that |w0 | − |f (x)| < |f (x) − w0 | < ǫ = |w0 | − kf k∞

3.5. Convergence in Measure and the Essential Range of f ∈ L∞ (µ)

109

which implies that |f (x)| > kf k∞ . Thus we have Eǫ (w0 ) ⊆ {x
∈ X | |f (x)| > kf k∞ }. By Definition 3.7, µ({x ∈ X | |f (x)| > kf k∞ }) = 0, so µ Eǫ (w0 ) = 0 too. By the definition (3.131), we have w0 ∈ / Rf and hence the set Rf is bounded by kf k∞ . By the Heine–Borel Theorem, we conclude that Rf is compact. • A relation between Rf and kf k∞ . By the previous analysis, it is obvious that the relation Rf ⊆ {w ∈ C | |w| ≤ kf k∞ } (3.133) holds, i.e., Rf lies in B(0, kf k∞ ). We claim that kf k∞ = max{|z| | z ∈ Rf }.

(3.134)

Given w ∈ C and ǫ > 0. Let B(w, ǫ) = {z ∈ C | |z − w| < ǫ}. Then we have
f −1 B(w, ǫ) = {x ∈ X | f (x) ∈ B(w, ǫ)} = {x ∈ X | |f (x) − w| < ǫ} = Eǫ (w).

By the definition, we see that \

Rf = {w ∈ C | µ f −1 B(w, ǫ) > 0} = C \ V,

(3.135)

ǫ>0



where V = {w ∈ C | µ f −1 B(w, ǫ) = 0 for some ǫ > 0}. Now we have the following two facts: – Fact 1: It is easy to show that V is the largest open subset of C such that
µ f −1 (V ) = µ({x ∈ X | f (x) ∈ V }) = 0,

i.e., V is the largest open subset of C such that f (x) ∈ / V for almost all x ∈ X. By this fact and the relation (3.135), we conclude that Rf is the smallest closed subset of C such that f (x) ∈ Rf for almost all x ∈ X.

– Fact 2: Recall from Definition 3.7 that the number kf k∞ is the minimum of the set {α ≥ 0} such that µ({x ∈ X | |f (x)| > α}) = 0, so we see from the relation (3.133) that it is the minimum radius of the closed disc centered at 0 containing the set Rf .

Hence we follow immediately from these two observations that the equality actually holds in the set relation (3.133). Since they are equal, our claim (3.134) is established. • Relations between Af and Rf . By the definition, we have o n 1 Z f dµ E ∈ M and µ(E) > 0 . Af = µ(E) E

(3.136)

We claim that Rf ⊆ Af . Given ǫ > 0 and w ∈ Rf . If w ∈ Af , then there is nothing to prove. Therefore, we may assume that w ∈ / Af . Consider Eǫ = {x ∈ X, | |f (x) − w| < ǫ}. By the definition of Rf , we have µ(Eǫ ) > 0. If µ(Eǫ ) = ∞ for every ǫ, then f (x) = w a.e. on X. In this case, we have Z 1 f dµ = w ∈ Af , µ(Eǫ ) Eǫ a contradiction. Therefore, we may assume that µ(Eǫ ) < ∞ for infinitely many ǫ > 0. In fact, we may take ǫ = n1 and let E(n) = E 1 . Then, since f ∈ L∞ (µ), we have f − w ∈ L1 (E(n)), where

Z
n L1 E(n) = f : X → C

n

E(n)

o |f | dµ < ∞ .

Chapter 3. Lp -Spaces

110

Consequently, it follows from Theorem 1.33 and the definition of E that Z Z 1 1 1

f dµ − w ≤ |f − w| dµ < . n µ E(n) E(n) µ E(n) E(n)

In other words, w is a limit point of Af and this means that w ∈ Af . Hence this proves the claim that Rf ⊆ Af . • Af is not always closed. For example, we consider
X = [0, 1], f (x) = x with µ = m the Lebesque measure. By Definition 3.7, f ∈ L∞ [0, 1] . In addition, for a measurable set E in [0, 1] with m(E) > 0, we have Z 1 f (x) dx. (3.137) w(E) = m(E) E We claim that w(E) can take any value in (0, 1). Indeed, if a, b ∈ (0, 1) and E = (a, b), then we have m(E) = b − a > 0 and it follows from the definition (3.137) that Z Z b 1 1 b2 − a 2 a+b 1 w(E) = f dx = × = x dx = m(E) E b−a a b−a 2 2 which implies the claim. Next, we claim that w(E) 6= 0. Assume that w(E0 ) = 0 for a measurable set E0 ⊆ [0, 1] with m(E0 ) > 0. Then we have Z f (x) dx = 0, E0

but Theorem 1.39(a) implies that f (x) = 0 a.e. on E0 which contradicts the fact that f (x) = 0 only at x = 0. Hence 0 is not a limit point of Af and Af is not closed in R. • A measure µ such that Af is convex for every f ∈ L∞ (µ). Consider X = {0} and µ the counting measure (see Example 1.20(a)). Then we have M = {∅, X} and µ(X) = µ({0}) = 1 > 0. By the definition, kf k∞ = |f (0)| < ∞ for every f ∈ L∞ (µ) so that f (0) is either kf k∞ or −kf k∞ . Now for every f ∈ L∞ (µ), we have Z 1 w(X) = f dµ = f (0). µ(X) X Therefore, the definition (3.136) gives either Af = {kf k∞ } or Af = {−kf k∞ }, but both cases are also convex sets. • A measure µ such that Af is not convex for some f ∈ L∞ (µ). We consider the set X = {0, 1}, µ the counting measure and f (x) = x. Then we have M = {∅, {0}, {1}, X}, µ({0}) = µ({1}) = 1 > 0 and µ(X) = 2 > 0. Now we have Z Z 1 1 w({0}) = f dµ = f (0) = 0, w({1}) = f dµ = f (1) = 1, µ({0}) {0} µ({1}) {1} Z 1 1 1 1 w(X) = f dµ = [f (0)µ({0}) + f (1)µ({1})] = [f (0) + f (1)] = . µ(X) X 2 2 2 Thus we have Af = {0, 1, 21 } is not convex. • The situations when L∞ (µ) is replaced by L1 (µ). We consider f : (0, ∞) → C by  1    √ , if x ∈ (0, 1]; x f (x) =    0, if x > 1

3.6. A Converse of Jensen’s Inequality

111


and take µ = m the Lebesgue measure. By Lemma 3.1, we have f ∈ L1 (0, ∞) . For any large positive integer N , if x ∈ (0, N12 ), then we have f (x) > N and

 1  1 0, 2 = 2 > 0. N N
= ∞, i.e., f ∈ / L∞ (0, ∞) . Given that ǫ > 0 and

m({x ∈ (0, ∞) | f (x) > N }) = m By Definition 3.7, we have kf k∞ sufficiently large N ∈ N.

– Rf is unbounded and not compact. If x ∈ have |f (x) − N | < ǫ so that m({x ∈ (0, ∞) | |f (x) − N | < ǫ}) = m



1 , 1 (N +ǫ)2 (N −ǫ)2




⊆ (0, 1), then we

 1 1 , > 0. (N + ǫ)2 (N − ǫ)2

Thus N ∈ Rf and then N ⊆ Rf , i.e., Rf is unbounded and not compact.

– Af is unbounded. Next, if E = (0, N42 ), then m(E) = w(E) =

1 m(E)

Z

f (x) dx = E

N2 4

Z

4 N2

1

x− 2 dx =

0

Therefore, N ⊆ Af , i.e., Af is unbounded.

4 N2

> 0 and

√ 42 N2 × 2 x N = N. 4 0

– Af is not always closed. In fact, the
previous example
that X = [0, 1], f (x) = x and µ = m satisfies both f ∈ L1 [0, 1] and f ∈ L∞ [0, 1] . Thus the same conclusion that 0 is not a limit point of Af is achieved. – A measure µ such that Af is convex for every f ∈ L1 (µ). The example that X = {0} and µ the counting measure also works for this case because if f ∈ L1 (µ), then |f (0)| = kf k1 < ∞. Thus we have either Af = {kf k1 } or Af = {−kf k1 }.

– A measure µ such that Af is not convex for some f ∈ L1 (µ). The example we considered for the case L∞ (µ) also works in this case.

We complete the proof of the problem.

3.6



A Converse of Jensen’s Inequality

Problem 3.20 Rudin Chapter 3 Exercise 20.

Proof. We check Definition 3.1. Let p, q ∈ R and λ ∈ [0, 1]. By the trick of Proposition 1.24(f), we see that Z Z χ[λ,1] (x) dx χ[0,λ] (x) dx + q λp + (1 − λ)q = p =

Z

[0,1] 1

0

[0,1]

[pχ[0,λ] (x) + qχ[λ,1] (x)] dx.

(3.138)

Suppose that f (x) = pχ[0,λ] (x) + qχ[λ,1] (x)

(3.139)

which is clearly a real and bounded function. Besides, since [0, λ] and [λ, 1] are Borel sets in [0, 1], χ[0,λ] and χ[λ,1] are measurable by Proposition 1.9(d). By the last paragraph in [100, §1.22,

Chapter 3. Lp -Spaces

112

p. 19], the function f , as the sum of two measurable functions, is also measurable. Hence, by substituting the expressions (3.138) and (3.139) into the inequality in question, we obtain Z

1


ϕ pχ[0,λ] (x) + qχ[λ,1] (x) dx 0 Z 1 Z λ ϕ(q) dx ϕ(p) dx + =


ϕ λp + (1 − λ)q ≤

0

λ

= λϕ(p) + (1 − λ)ϕ(q).

Hence ϕ is convex in R, completing the proof of the problem.



Remark 3.2 We note that Problem 3.20 is a converse of Theorem 3.3 (Jensen’s Inequality).

3.7

The Completeness/Completion of a Metric Space

Problem 3.21 Rudin Chapter 3 Exercise 21.

Proof. Let (X, d) and (X ∗ , ρ) be two metric spaces. An isometry of X into X ∗ is a mapping ϕ : X → Y such that
ρ ϕ(p), ϕ(q) = d(p, q) (3.140) for all p, q ∈ X. We notice immediately that an isometry ϕ is necessarily injective and continuous. If ϕ is surjective, then we call ϕ an isomorphism. Two metric spaces X and Y are called isomorphic if there is an isomorphism between them. Before stating the statement, we need: Lemma 3.9 Let (X, d) be a metric space with metric d and p ∈ X. Then the function f : X → R defined by f (x) = d(x, p) is continuous.

Proof of Lemma 3.9. Given ǫ > 0. Let x ∈ X. If y ∈ X satisfies d(x, y) < ǫ, then the triangle inequality implies that |f (x) − f (y)| = |d(x, p) − d(y, p)| ≤ d(x, y) < ǫ. Hence f is continuous, completing the proof of Lemma 3.9. Lemma 3.10 Suppose that (X, d) and (Y, ρ) are metric spaces and Y is complete. If E is dense in X and f : E → Y is an isometry, then there exists an isometry g : X → Y such that g|E = f .



3.7. The Completeness/Completion of a Metric Space

113

Proof of Lemma 3.10. If x ∈ E, we define g(x) = f (x). Let x ∈ X \ E. Since E is dense in X, we can choose a sequence {xn } ⊆ E such that xn → x. By [99, Theorem 3.11(a)], {xn } is Cauchy in X. Thus, given ǫ > 0, there exists a positive integer N such that n, m ≥ N imply that d(xm , xn ) < ǫ. Since f is an isometry, it follows from the expression (3.140) that
ρ f (xm ), f (xn ) < ǫ

for all n, m ≥ N . In other words, {f (xn )} is Cauchy in Y . Since Y is complete, {f (xn )} converges to a limit y in Y and we define g(x) = y in this case. Now this y is uniquely determined by x. Indeed, let {x′n } ⊆ E be another sequence converging to x and y ′ be the limit of {f (x′n )} in Y . For each m ∈ N, it follows from Lemma 3.9 and the expression (3.140) that

ρ y, f (x′m ) = lim ρ f (xn ), f (x′m ) = lim d(xn , x′m ) = d(x, x′m ). n→∞

n→∞

Then this implies that


ρ(y, y ′ ) = lim ρ y, f (x′m ) = lim d(x, x′m ) = 0. m→∞

m→∞

y′

Since ρ is a metric, we must have y = and so we obtain a mapping g : X → Y given by   if x ∈ E;  f (x), (3.141) g(x) =   lim f (xn ), if x ∈ X \ E, {xn } ⊆ X and xn → x. n→∞

Next, we show that g is an isometry and we consider the following situations:

• Case (i): x, y ∈ E. In this case, we know from the definition (3.141) that g(x) = f (x) and g(y) = f (y) and then the use of the expression (3.140) implies that

ρ g(x), g(y) = ρ f (x), f (y) = d(x, y). • Case (ii): x ∈ X \ E and y ∈ E. In this case, we have

ρ g(x), g(y) = ρ g(x), f (y) .

Choose a sequence {xn } in E such that xn → x. By Lemma 3.9 and the expression (3.140), we establish that

(3.142) ρ g(x), f (y) = lim ρ f (xn ), f (y) = lim d(xn , y) = d(x, y). n→∞

n→∞

• Case (iii): x, y ∈ X \ E. Let {xn }, {yn } ⊆ E be sequences converging to x and y respectively. Then we deduce from the definition (3.141) and the expression (3.142) that
ρ g(x), f (yn ) = d(x, yn ). (3.143) Apply Lemma 3.9 to (3.143), we see that

ρ g(x), g(y) = lim ρ g(x), f (yn ) = lim d(x, yn ) = d(x, y). n→∞

By the definition, g is an isometry and g|E = f .

n→∞



Chapter 3. Lp -Spaces

114 Lemma 3.11 The mapping g in Lemma 3.9 is unique up to isometry.

Proof of Lemma 3.11. Let g ′ : X → Y be an isometry such that g′ |E = f . If x ∈ E, then g(x) = f (x) = g ′ (x). If x ∈ X \ E, then there is a sequence {xn } ⊆ E such that xn → x as n → ∞. The triangle inequality, the expressions (3.140) and (3.142) indicate that



ρ g(x), g ′ (x) ≤ ρ g(x), g(xn ) + ρ g(xn ), g′ (xn ) + ρ g′ (xn ), g′ (x)


= ρ g(x), f (xn ) + ρ f (xn ), f ′ (xn ) + ρ f ′ (xn ), g′ (x)

= ρ g(x), f (xn ) + d(xn , xn ) + ρ f ′ (xn ), g′ (x) = d(x, xn ) + d(xn , x)

Hence when n → ∞, it is clear that ρ(g(x), g′ (x)) = 0, so g(x) = g′ (x) on X \ E and our result follows from this.  Let’s return to the proof of the problem. Let (Y1 , ρ1 ) and (Y2 , ρ2 ) be complete metric spaces, ϕ1 : X → Y1 and ϕ2 : X → Y2 be isometries such that ϕ1 (X) is dense in Y1 and ϕ2 (X) is dense in Y2 . We claim that (Y1 , ρ1 ) and (Y2 , ρ2 ) are isomorphic. Define f : ϕ1 (X) → Y2 by f = ϕ2 ◦ ϕ−1 1 .

(3.144)

If p, q ∈ X, x = ϕ1 (p) and y = ϕ1 (q), then we have



−1 ρ2 f (x), f (y) = ρ2 ϕ2 ϕ−1 1 (x) , ϕ2 ϕ1 (y)
= ρ2 ϕ2 (p), ϕ2 (q) = d(p, q)


= ρ1 ϕ1 (p), ϕ1 (q) = ρ1 (x, y).

Consequently, f is an isometry. Since ϕ1 (X) is dense in Y1 and Y2 is complete, Lemma 3.10 ensures that there exists an isometry g : Y1 → Y2 such that g|ϕ1 (X) = f. It remains to prove that g is surjective. To this end, let y ∈ Y2 . Since ϕ2 (X) = Y2 , we can select a sequence {yn } ⊆ ϕ2 (X) such that yn → y as n → ∞. Furthermore, there exists a sequence {xn } ⊆ X such that yn = ϕ2 (xn ) for all n ∈ N. Consider the corresponding sequence zn = ϕ1 (xn )

(3.145)

for n ∈ N. Since {yn } is Cauchy in ϕ2 (X) (and hence in Y2 ) and ϕ2 is an isometry, the expression (3.140) forces that {xn } is Cauchy in X. Then, since ϕ1 is an isometry, the expression (3.140) again forces that {zn } is Cauchy in ϕ1 (X) (and hence in Y1 ). Since Y1 is complete, we can find z ∈ Y1 such that zn → z as n → ∞. We claim that g(z) = y.

3.8. Miscellaneous Problems

115

Thus we deduce from Lemma 3.10, the definition (3.145) and the fact g|ϕ1 (X) = f that





(3.146) ρ2 g(z), y = lim ρ2 g(zn ), y = lim ρ2 g ϕ1 (xn ) , y = lim ρ2 f ϕ1 (xn ) , y . n→∞

n→∞

n→∞

Finally, the definition (3.144) reduces the limit (3.146) to



ρ2 g(z), y = lim ρ2 f ϕ1 (xn ) , y = lim ρ2 ϕ2 (xn ), y = lim ρ2 (yn , y) = ρ2 (y, y) = 0. n→∞

n→∞

n→∞

Therefore, g(z) = y and g is surjective. Hence g is an isomorphism and our claim follows, completing the proof of the problem.  Problem 3.22 Rudin Chapter 3 Exercise 22.

Proof. This problem is proven in [124, Problem 3.20, p. 50].

3.8



Miscellaneous Problems

Problem 3.23 Rudin Chapter 3 Exercise 23.

Proof. If αn0 = 0 for some n0 , then we must have f = 0 a.e. on X by Theorem 1.39(a). However, this implies that the measure of
f −1 (α, ∞] = {x ∈ X | f (x) > α}

is zero for every α > 0. By Definition 3.7, it means that kf k∞ = 0 which contradicts the hypothesis. Thus we have αn 6= 0 for all n ∈ N so that the limit in question is well-defined. By Definition 3.7, we see that |f (x)| ≤ kf k∞ holds for almost all x ∈ X. Therefore, we have αn+1 ≤ kf k∞ · αn

a.e. on X

(3.147)

for all n ∈ N. Since kf k∞ > 0, we can find a ǫ > 0 such that kf k∞ > ǫ > 0. By the definition, the measurable set F = {x ∈ X | |f (x)| ≥ kf k∞ − ǫ > 0} satisfies µ(F ) > 0. Apply Theorem 3.5 (H¨older’s Inequality) to the measurable functions f n and 1 with p = n+1 n , we obtain αn =

Z

X

|f |n dµ ≤

nZ

X

(|f |n )

n+1 n

o

n n+1

nZ

1n+1 dµ X

o

1 n+1

so that −1 1 αn+1 ≥ (αn+1 ) n+1 × µ(X) n+1 αn nZ o 1 −1 n+1 ≥ |f |n+1 dµ × µ(X) n+1

F

 1 −1 ≥ (kf k∞ − ǫ)n+1 µ(F ) n+1 × µ(X) n+1 

n

1

= (αn+1 ) n+1 × µ(X) n+1

Chapter 3. Lp -Spaces

116

=

 µ(F ) 

1 n+1

µ(X)

× (kf k∞ − ǫ).

(3.148)

Thus we follow from the inequalities (3.147) and (3.148) that  µ(F ) 

1 n+1

µ(X)

× (kf k∞ − ǫ) ≤

αn+1 ≤ kf k∞ αn

a.e. on X

(3.149)

1

µ(F ) n+1 ) tends to 1 as n → ∞. By for all n ∈ N. Since 0 < µ(F ) ≤ µ(X) < ∞, the number ( µ(X) taking limit to both sides of the inequality (3.149), we see immediately that

kf k∞ − ǫ ≤ lim inf n→∞

αn+1 αn+1 ≤ lim sup ≤ kf k∞ αn αn n→∞

a.e. on X.

Remember that ǫ is arbitrary, so this implies that lim

n→∞

αn+1 = kf k∞ αn

a.e. on X

and we complete the proof of the problem.



Problem 3.24 Rudin Chapter 3 Exercise 24.

Proof. Let x ≥ 0 and y ≥ 0. Put α − β = x and α = y into the equality (3.111), we gain |x|p − |y|p ≤ |x − y|p .

(3.150)

Next, we substitute α − β = y and α = x into the equality (3.111) to get −|x − y|p ≤ |x|p − |y|p . Obviously, the inequalities (3.150) and (3.151) together imply that |xp − y p | = |x|p − |y|p ≤ |x − y|p

(3.151)

(3.152)

if 0 < p < 1.

If x = y, then the inequality |xp − y p | ≤ p|x − y|(xp−1 + y p−1 ) certainly holds. Without loss of generality, we may assume that x < y. For p > 1, we consider the function ϕ(t) = tp on [x, y]. Since ϕ is differentiable in (x, y), the Mean Value Theorem implies that |y p − xp | ≤ |x − y||ϕ′ (ξ)| ≤ p|x − y|ξ p−1 (3.153)

for some ξ ∈ (x, y). Clearly, we have ξ p−1 ≤ xp−1 + y p−1 , so it reduces from the inequality (3.153) that |xp − y p | ≤ p|x − y|(xp−1 + y p−1 ). (3.154) Hence we establish from the inequalities (3.152) and (3.154) that  if 0 < p < 1;  |x − y|p , p p |x − y | ≤  p|x − y|(xp−1 + y p−1 ), if 1 ≤ p < ∞.

3.8. Miscellaneous Problems

117

(a) We prove the assertions one by one: – The truth of the inequality. Put x = |f | ≥ 0 and y = |g| ≥ 0 into the inequality (3.152) and then take integration, we get Z Z p |f | − |g|p dµ ≤ |f | − |g| p dµ. (3.155)

Since |a| − |b| ≤ |a − b| holds for every a, b ∈ R, we follow immediately from the inequality (3.155) that Z Z Z p |f | − |g|p dµ ≤ |f | − |g| p dµ ≤ |f − g|p dµ, (3.156)

where 0 < p < 1.

– ∆ is a metric. Define ∆ : Lp (µ) × Lp (µ) → C by Z ∆(f, g) = |f − g|p dµ. By Remark 3.10, Lp (µ) is a complex vector space, so f − g ∈ Lp (µ) if f, g ∈ Lp (µ). Thus we have Z 0 ≤ ∆(f, g) = |f − g|p dµ < ∞

for all f, g ∈ Lp (µ). Next, Theorem 1.39(a) ensures that ∆(f, g) = 0 if and only if |f − g|p = 0 a.e. on X, i.e., f = g a.e. on X. It is clear that Z Z ∆(f, g) = |f − g|p dµ = |g − f |p dµ = ∆(g, f ).

Finally, since |f |p − |g|p ≤ |f |p − |g|p , it is clear from the inequality (3.156) that Z Z Z p p p p |f | − |g| dµ ≤ |f − g|p dµ. (|f | − |g| ) dµ ≤ (3.157) By replacing f and g by f − g and h − g in the inequality (3.157), we deduce that Z ∆(f, g) − ∆(h, g) = (|f − g|p − |h − g|p ) dµ Z ≤ |(f − g) − (h − g)|p dµ Z = |f − h|p dµ = ∆(f, h).

After rearrangement, we have ∆(f, g) ≤ ∆(f, h) + ∆(h, g). Hence, by the definition, ∆ is in fact a metric. – (Lp (µ), ∆) is a complete metric space. Suppose that {fn } is a Cauchy sequence in Lp (µ) with respect to the metric ∆, i.e., for every ǫ > 0, there exists a positive integer N such that n, m ≥ N implies that ∆(fn , fm ) < ǫ.

(3.158)

Although ∆(f, g) = kf −gkpp here, we cannot apply Theorem 3.11 directly to conclude that Lp (µ) is complete with respect to ∆ because p ≥ 1 in Theorem 3.11.

Chapter 3. Lp -Spaces

118

We imitate the proof of Theorem 3.11. To start with, there is a subsequence {fni } such that ∆(fni+1 , fni ) < 2−i (3.159) for each i = 1, 2, . . .. We define gk : X → [0, ∞] and g : X → [0, ∞] by gk =

k X i=1

|fni+1 − fni | and g =

∞ X i=1

|fni+1 − fni |

respectively. It is clear that gk ≥ 0 and g ≥ 0 and furthermore, it follows from Theorem 3.5 (Minkowski’s Inequality) that ∆(gk , 0) = ∆

k X i=1

=

 |fni+1 − fni |, 0

Z X k i=1

≤ =

p |fni+1 − fni | dµ

k nZ hX i=1

k hX

|fni+1 − fni |p dµ 1

∆(fni+1 , fni ) p

i=1

ip

o 1 ip p

(3.160)

for each k = 1, 2, . . .. To proceed further, we need the following result: Lemma 3.12 Let p ∈ (0, 1). For nonnegative real numbers a1 , a2 , . . . , ak , we have (a1 + a2 + · · · + ak )p ≤ ap1 + ap2 + · · · + apk .

Proof of Lemma 3.12. Replace β by −β in the inequality (3.111), we get (α + β)p ≤ |α|p + |β|p for p ∈ (0, 1). In particular, if α ≥ 0 and β ≥ 0, then we have (α + β)p ≤ αp + β p

(3.161)

for p ∈ (0, 1). By applying the inequality (3.161) repeatedly to a1 , a2 , . . . , ak , we  derive the desired inequality. This completes the proof of Lemma 3.12. Apply Lemma 3.9 to the inequality (3.160) and then using the inequality (3.159) to get k k X X 2−i ≤ 1 (3.162) ∆(fni+1 , fni ) < ∆(gk , 0) ≤ i=1

for each k = 1, 2, . . ..

i=1

3.8. Miscellaneous Problems

119

Since gk (x) → g(x) as k → ∞ for every x ∈ X, we have gk (x)p → g(x)p as k → ∞ on X. We note that since each fni is measurable, each gkp is also measurable and we may apply Theorem 1.28 (Fatou’s Lemma) to {gkp } and the inequality (3.162) to get Z Z Z Z gp dµ = lim gkp dµ = lim inf gkp dµ ≤ lim inf gkp dµ = lim inf ∆(gk , 0) ≤ 1, k→∞

k→∞

k→∞

k→∞

implying that g ∈ Lp (µ). In particular, g < ∞ a.e. on X, so the series fn1 (x) +

∞ X [fni+1 (x) − fni (x)]

(3.163)

i=1

converges absolutely for almost every x ∈ X. Denote the sum (3.163) by f (x) for those x at which the sum (3.163) converges and put f (x) = 0 on the remaining set of measure zero. Since we have k−1 X [fni+1 (x) − fni (x)] = fnk (x), fn1 (x) + i=1

it implies that fni (x) → f (x) as i → ∞ a.e. on X. Now it remains to show that f ∈ Lp (µ) and ∆(fn , f ) → 0 as n → ∞. Given ǫ > 0. Since {fn } is Cauchy in Lp (µ) with respect to ∆, there exists a N ∈ N such that the estimate (3.158) holds for n, m ≥ N . Take m ≥ N , since fni → f as i → ∞ a.e. on X, Theorem 1.28 (Fatou’s Lemma) again implies that Z Z   lim inf |fni − fm |p dµ |f − fm |p dµ = i→∞ Z  ≤ lim inf |fni − fm |p dµ i→∞

= lim inf ∆(fni , fm ) < ǫ. i→∞

(3.164)

so we conclude that f − fm ∈ Lp (µ), hence that f ∈ Lp (µ). Finally, according to the estimate (3.164), it follows that ∆(f, fm ) < ǫ for all m ≥ N . Since ǫ is arbitrary, we conclude that lim ∆(f, fm ) = 0.

m→∞

This shows the completeness of the space Lp (µ) with respect to ∆. (b) Similar to part (a), a direct substitution of x = |f | and y = |g| into the inequality (3.154) and the use of the fact |x| − |y| ≤ |x − y| give Z Z p |f | − |g|p dµ ≤ p |f | − |g| × (|f |p−1 + |g|p−1 ) dµ Z (3.165) ≤ p |f − g|(|f |p−1 + |g|p−1 ) dµ. By Theorem 3.5 (H¨older’s Inequality) and the facts that kf kp ≤ R and kgkp ≤ R, we observe that Z Z Z |f − g|(|f |p−1 + |g|p−1 ) dµ ≤ |f − g| · |f |p−1 dµ + |f − g| · |g|p−1 dµ nZ o p−1 o1 n Z p p p (|f |p−1 ) p−1 dµ ≤ |f − g|p dµ

Chapter 3. Lp -Spaces

120

+

nZ

|f − g|p dµ

o1 n Z p

p

(|g|p−1 ) p−1 dµ

= kf − gkp · (kf kp−1 + kgkp−1 p p )

o p−1

≤ 2Rp−1 kf − gkp .

p

(3.166)

Combining the inequalities (3.165) and (3.166), we discover that Z p |f | − |g|p dµ ≤ 2pRp−1 kf − gkp

which is our expected result.

We have completed the proof of the problem.



Problem 3.25 Rudin Chapter 3 Exercise 25.

Proof. Suppose that E ⊆ X and 0 < µ(E) < ∞. Let ME be a σ-algebra in E. Then (E, ME , µ) is a measure space.j Define Z 1 dµ (F ∈ ME ). ϕ(F ) = F µ(E) Then we follow from Theorem 1.29 that ϕ is a measure on ME and Z Z 1 f dµ f dϕ = µ(E) E E

(3.167)

for every measurable f on E with range in [0, ∞]. It is trivial that ϕ(E) = 1 and Z Z 1 1 f dϕ ≤ f dµ = 0 on (0, ∞), we apply Theorem 3.3 (Jensen’s Inequality) with ψ and measure ϕ to conclude that Z op nZ f p dϕ. (3.169) f dϕ ≤ − − E

E

Next, we use the fact (3.167) to reduce the inequality (3.169) to Z n 1 Z op n 1 Z op 1 1 f p dµ ≤ f dµ ≤ f dµ = µ(E) E µ(E) E µ(E) X µ(E)p so that

Z

E

completing the proof of the problem.

f p dµ ≤ µ(E)1−p , 

Problem 3.26 Rudin Chapter 3 Exercise 26.

Proof. Let M be a σ-algebra in [0, 1]. Suppose that there exists an E ∈ M such that f (x) = ∞ and m(E) > 0, where m is the Lebesgue measure. Then Z

1

f (x) log f (x) dx =

0

Z

f (x) log f (x) dx +

E

Z

f (x) log f (x) dx.

(3.170)

[0,1]\E

On the one hand, since f > 0 on [0, 1] \ E, we know from Proposition 1.24(a) that the second integral on the right-hand side of the expression (3.170) is nonnegative. Since f (x) log f (x) = ∞ on E and m(E) > 0, we get from the expression (3.170) that Z

0

1

f (x) log f (x) dx = ∞.

(3.171)

On the other hand, by using similar argument, we can show that Z 1 Z 1 log f (t) dt = ∞. f (s) ds = ∞ and

(3.172)

0

0

Therefore, we conclude from the results (3.171) and (3.172) that Z

1

f (x) log f (x) dx = 0

Z

1

f (s) ds 0

Z

1

log f (t) dt

0

in this case. Next, we suppose that 0 < f (x) < ∞ a.e. on [0, 1]. Define ϕ : (0, ∞) → R and ψ : (0, ∞) → R by ϕ(x) = x log x and ψ(x) = − log x

respectively. Since ϕ′′ (x) = x1 > 0 and ψ ′′ (x) = x−2 > 0 on (0, ∞), they are convex on (0, ∞). Since 0 < f (x) < ∞ a.e. on [0, 1] and m([0, 1]) = 1, we apply Theorem 3.3 (Jensen’s Inequality) with ϕ to f to obtain o Z 1 nZ 1 on  Z 1 f (x) log f (x) dx (3.173) ≤ f (x) dx f (x) dx log 0

0

0

Chapter 3. Lp -Spaces

122

Similarly, we apply Theorem 3.3 (Jensen’s Inequality) with ψ to f to obtain −log

Z

1

Z  f (x) dx ≤ −

1

log f (x) dx.

(3.174)

0

0

Combining the inequalities (3.173) and (3.174), we see that nZ

1

f (x) dx 0

on Z

0

1

o Z log f (x) dx ≤

We have ended the proof of the problem.

1

f (x) log f (x) dx. 0



CHAPTER

4

Elementary Hilbert Space Theory

In the following problems, we use hx, yi to denote the inner product of the complex vectors x and y and k · k the norm with respect to the Hilbert space H. Besides, we use span (S) to denote the span of a set S.

4.1

Basic Properties of Hilbert Spaces

Problem 4.1 Rudin Chapter 4 Exercise 1.

Proof. Let x ∈ M . Recall that M ⊥ = {y ∈ H | y ⊥ x for all x ∈ M }, so x ⊥ y for every y ∈ M ⊥ which means that x ∈ (M ⊥ )⊥ , i.e., M ⊆ (M ⊥ )⊥ . Conversely, suppose that x ∈ (M ⊥ )⊥ . Since M is a closed subspace of H, it follows from Theorem 4.11(a) that x = y + z, where y ∈ M and z ∈ M ⊥ . On the one hand, since x ∈ (M ⊥ )⊥ and z ∈ M ⊥ , we have hx, zi = 0. On the other hand, we deduce from Definition 4.1 that hx, zi = hy + z, zi = hy, zi + hz, zi = hz, zi. Since hz, zi = 0, it must be z = 0 and then x ∈ M , i.e, (M ⊥ )⊥ ⊆ M . In conclusion, we obtain M = (M ⊥ )⊥ . Suppose that M is a subspace of H which may not be closed. We claim that M = (M ⊥ )⊥ . To see this, we recall from §4.8 that M is a closed subspace of H, so the first assertion implies that ⊥ M = (M )⊥ . (4.1) ⊥

Let x ∈ M . By §4.9, we see that x ⊥ M so that x ⊥ M particularly. In other words, we have M

⊥

⊆ M ⊥.

(4.2) ⊥

For the other direction, suppose that x ∈ M ⊥ , we want to show that x ∈ M , i.e., x ⊥ M . To this end, given y ∈ M , there exist a sequence {yn } ⊆ M such that yn → y as n → ∞. Since 123

124

Chapter 4. Elementary Hilbert Space Theory

x ∈ M ⊥ and yn ∈ M , we must have hyn , xi = 0. By Theorem 4.6, the mapping f : H → C defined by f (y) = hy, xi is continuous on H so that hy, xi = f (y) = lim f (yn ) = lim hyn , xi = 0. n→∞

In other words, x ⊥ M , i.e.,

n→∞

⊥

M⊥ ⊆ M .

(4.3)

Now the claim is derived by combining the set relations (4.2) and (4.3) and then using the result (4.1), completing the proof of the problem.  Problem 4.2 Rudin Chapter 4 Exercise 2.

Proof. We note that this is actually the Gram-Schmidt Process. Suppose that n = 1. Then the set {u1 } is clearly orthonormal and span (u1 ) = span (x1 ). Thus the statement is true for n = 1. Assume that the statement is true for n = k for some k ∈ N, i.e., {u1 , u2 , . . . , uk } is an orthonormal set and span (u1 , u2 , . . . , uk ) = span (x1 , x2 , . . . , xk ). (4.4) Let n = k + 1. Note that xk+1 ∈ / span (x1 , x2 , . . . , xk ) because {x1 , x2 , . . .} is linearly independent. By the assumption (4.4), it implies that xk+1 ∈ / span (u1 , u2 , . . . , uk ) so that vk+1 6= 0. By this fact and Definition 4.1, for j = 1, 2, . . . , k, we derive that   vk+1 , uj huk+1 , uj i = kvk+1 k + * k X 1 = hxk+1 , ui i ui , uj xk+1 − kvk+1 k i=1

=

1 kvk+1 k

hxk+1 , uj i −

1 kvk+1 k

k X i=1

hxk+1 , ui i hui , uj i

1 1 hxk+1 , uj i − hxk+1 , uj i huj , uj i kvk+1 k kvk+1 k = 0. =

Therefore, {u1 , u2 , . . . , uk+1 } is orthonormal. Next, by the assumption (4.4), we know that x ∈ span (x1 , x2 , . . . , xk , xk+1 ) if and only if x ∈ span (u1 , u2 , . . . , uk , xk+1 ). Since vk+1 = xk+1 −

k X i=1

hxk+1 , ui i ui

and uk+1 =

(4.5)

vk+1 , kvk+1 k

the xk+1 in the result (4.5) can be replaced by vk+1 and ultimately by uk+1 . Thus we have span (u1 , u2 , . . . , uk+1 ) = span (x1 , x2 , . . . , xk+1 ).

4.1. Basic Properties of Hilbert Spaces

125

Hence the statement is true for n = k + 1 if it is true for n = k. By induction, the construction yields an orthonormal set {u1 , u2 , . . .} with span (u1 , u2 , . . . , un ) = span (x1 , x2 , . . . , xn ) for all n ∈ N. This completes the proof of the problem.



Problem 4.3 Rudin Chapter 4 Exercise 3.

Proof. Recall the definition that a space is separable if it contains a countable dense subset. Suppose that 1 ≤ p < ∞. Since T is compact, the second paragraph following Definition 3.16 says that Cc (T ) = C0 (T ) = C(T ). Thus we deduce from this and Theorem 3.14 that C(T ) is dense in Lp (T ) in the norm k · kp . Let P be the set of all trigonometric polynomials on T . By Theorem 4.25 (The Weierstrass Approximation Theorem) and Definitions 4.23 (or Remark 3.15), P is dense in C(T ) in the norm k · k∞ . By Definition 3.7, we know that |f (x)| ≤ kf k∞ a.e. on T and then o1 n 1 Z π o1 n 1 Z π p p |f (t)|p dt kf kp∞ dt ≤ = kf k∞ . (4.6) kf kp = 2π −π 2π −π

In other words, the inequality (4.6) implies that P is also dense in C(T ) in the norm k · kp . By Lemma 2.10, we conclude that P is dense in Lp (T ) in the norm k · kp . Next we suppose that PQ is the set of all trigonometric polynomials with rational coefficients. Note thata n n o X PQ = P (t) = (ak + ibk )eikt a−n , . . . , an , b−n , . . . , bn ∈ Q and n ∈ N k=−n

=

∼

∞ n [

n=1 ∞ [

P (t) =

n X

o (ak + ibk )eikt a−n , . . . , an , b−n , . . . , bn ∈ Q

k=−n

Q2(2n+1) .

(4.7)

n=1

For each n ∈ N, since each Q2(2n+1) is countable, it follows from the set relation (4.7) that PQ is also countable. Given f ∈ P. Since the set {p + iq | p, q ∈ Q} is dense in C and f must be in the form n X ck eikt f (t) = k=−n

for some positive integer n, there exists a sequence {fnk } ⊆ PQ such that kfnk − f k∞ → 0 as nk → ∞. By the inequality (4.6), it is true that kfnk − f kp → 0

as nk → ∞. Consequently, PQ is dense in P in the norm k · kp and an application of Lemma 2.10 shows that PQ is dense in Lp (T ) in the norm k · kp . Since PQ is countable, Lp (T ) is separable. By Definition 4.23, every function defined on T
can be identified with a 2π-periodic function on R, so we may identity L∞ (T ) with L∞ [0, 2π] . Consider the set E = {fθ = χ[0,θ] | θ ∈ [0, 2π]}.

a

We write A ∼ B if there is a bijection between the two sets A and B, see [99, Definition 2.3, p. 25].

126

Chapter 4. Elementary Hilbert Space Theory



It is clear that χ[0,θ] ∈ L∞ [0, 2π] so that E ⊆ L∞ [0, 2π] . Since χ[0,θ1] 6= χ[0,θ2 ] if θ1 6= θ2 , E ∼ [0, 2π] and then it is uncountable. In fact, we have kχ[0,θ1 ] − χ[0,θ2 ] k∞ = 1

(4.8)

if θ1 < θ2 .
Assume
that L∞ [0, 2π] was separable. Let F = {fn } be a countable dense subset of ∞ L [0, 2π] . Then we must have [ 
1 E ⊆ L∞ [0, 2π] ⊆ B fn , , 2

(4.9)

fn ∈F

where B(fn , 12 ) = {f ∈ L∞ ([0, 2π]) | kf − fn k∞ < 12 }. If χ[0,θ1 ] , χ[0,θ2 ] ∈ B(fn , 21 ) with θ1 6= θ2 , then it follows from the result (4.8) that 1 = kχ[0,θ1 ] − χ[0,θ2 ] k∞ ≤ kχ[0,θ1 ] − fn k∞ + kfn − χ[0,θ2 ] k∞ < 1, a contradiction. Thus each B(fn , 12 ) contains at most one element of E. Since F is countable but E is uncountable, we have [  1 E* B fn , 2 fn ∈F


which definitely contradicts the set relation (4.9). Hence L∞ [0, π] is not separable and we have completed the proof of the problem.  Problem 4.4 Rudin Chapter 4 Exercise 4.

Proof. Suppose that H is separable. By the definition, it has a countable dense subset {un }. By Problem 4.2, we may assume that {un } is also an orthonormal set. Let {unk } be a maximal linearly independent subset of {un }. Note that {unk } is at most countable. Since we have span ({unk }) = {un }, span ({unk }) is also dense in H. By Theorem 4.18, the set {unk } is in fact an at most countable maximal orthonormal set in H. Conversely, we suppose that E = {un } is an at most countable maximal orthonormal set in H. By Theorem 4.18 again, span (E) is dense in H. Let span Q (E) be the span of the set E with rational coefficients. It is easy to see that span Q (E) =

n nX

o (ak + ibk )uk a1 , . . . , an , b1 , . . . , bn ∈ Q and n ∈ N

n=1 ∞ [

k=1

k=1

= ∼

∞ nX n [

Q2n .

o (ak + ibk )uk a1 , . . . , an , b1 , . . . , bn ∈ Q

n=1

As a result, span Q (E) must be countable. Next, since Q is dense in R, we conclude that span Q (E) is dense in span (E). By Lemma 2.10, span Q (E) is also dense in H and hence it is  separable. This completes the proof of the problem.

4.1. Basic Properties of Hilbert Spaces

127

Problem 4.5 Rudin Chapter 4 Exercise 5.

Proof. Suppose that M 6= H. If L = 0, then M = H so that L 6= 0. By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there is a unique z ∈ H such that L(x) = hx, zi for all x ∈ H. This z must be nonzero, otherwise we have L = 0 which contradicts our assumption. It is clear that M = {x ∈ H | hx, zi = 0} = L−1 (0),

(4.10)

so M is closed in H. Apart from this, if x, y ∈ M , then since L is a linear functional, Definition 2.1 shows that L(x + y) = L(x) + L(y) = 0 and L(αx) = αL(x) = 0 for a scalar α. Thus M is a closed subspace of H. Next, we note that span (z) = {αz | α ∈ C} and so


⊥ span (z) = {x ∈ H | hx, αzi = 0 for every α ∈ C}.

(4.11)

Since hx, αzi = α hx, zi, the two expressions (4.10) and (4.11) give
⊥ M = span (z)

and then

M⊥ =


⊥
⊥ span (z) .

(4.12)

We claim that span (z) is a closed subspace of H. In fact, it is a subspace of H follows easily from the definition of span (z). Suppose that {xn } is a sequence in span (z). Then we know that each xn has the form xn = αn z, where αn ∈ C. Let α = lim αn and x = αz. We want to show n→∞

that xn → x in H with respect to the norm k · k. Indeed, we have

kxn − xk2 = h(αn − α)z, (αn − α)zi = |αn − α2 | × kzk so that kxn − xk → 0 as n → ∞, i.e., lim xn = x

n→∞

as desired. In conclusion, span (z) contains all its limit points and therefore it is closed in H.
⊥
⊥ By Problem 4.1, we have span (z) = span (z). Combining this fact and the expression (4.12), we have M ⊥ = span (z)
and since {z} is a basis of span (z), we have dim span (z) = 1 and our required result follows.  We end the proof of the problem. Problem 4.6 Rudin Chapter 4 Exercise 6.

128

Chapter 4. Elementary Hilbert Space Theory

Proof. We prove the assertions one by one: • E = {un } is closed and bounded, but not compact. Since E is orthonormal, kun k ≤ 1 for every n ∈ N. Thus E is bounded by 1. If n 6= m, then it follows from Definition 4.1 that kun − um k2 = hun − um , un − um i = hun , un i + hum , um i = 2. (4.13) Assume that u was a limit point√ of E. Then there exists a positive integer N such that n ≥ N implies that kun − uk < 22 . If n, m ≥ N , then we have kun − um k ≤ kun − uk + ku − xm k
0, there is a N ′ ∈ N such that X ǫ δn2 < . (4.16) 8 ′ n>N

Then we obtain from the definition (4.14), the fact |ckn | ≤ δn for all k and the estimate (4.16) that ∞

2

X

[cknm (m) − cn ]un kxkm (m) − xk2 = n=1

=

∞ X

n=1

|cknm (m) − cn |2

′

=

N X

n=1

|cknm (m) − cn |2 +

∞ X

n=N ′ +1

′

≤ ≤

N X

n=1 N′ X

n=1 ′


0. n∈Ek

Now for n ∈ Ek , we define

1 an bn = n X o 1 × . k 2 a2n n∈Ek

c

See [99, Theorem 3.28, p. 62].

P

a2n = ∞. Therefore, for every (4.20)

4.1. Basic Properties of Hilbert Spaces

131

On the one hand, we have ∞ X

b2n

=

n=1

∞ X X

b2n

=

∞ X k=1

k=1 n∈Ek

1 X

a2n

×

X

n∈Ek

n∈Ek

a2n

1 × 2 k

!

=

∞ X 1 < ∞. k2 k=1

On the other hand, we obtain from the inequality (4.20) that ! ∞ n X ∞ X ∞ ∞ ∞ o1 1 X X X X X X 1 a2n 1 2 √ a2n an bn = a n bn = > n X o1 × k = k 2 k n=1 n∈Ek k=1 n∈Ek k=1 n∈Ek k=1 k=1 a2n n∈Ek

which is divergent, a contradiction. Hence we must have of the problem.

P

a2n < ∞ and it completes the proof 

Problem 4.8 Rudin Chapter 4 Exercise 8.

Proof. By Problem 4.2, H contains an orthonormal set A and then Theorem 4.22 (The Hausdorff Maximality Theorem) implies that A is contained in a maximal orthonormal set in H. Suppose that {uα | α ∈ A} and {vβ | β ∈ B} are maximal orthonormal sets in H1 and H2 respectively. Without loss of generality, we may assume that |A| ≤ |B|. Then we take a subset B ′ of B which is of the same cardinality as A. Now we consider H = span ({vβ | β ∈ B ′ }) ⊆ H2 . By Definition 4.13, span ({vβ | β ∈ B ′ }) is a subspace of H2 . Thus we follow from the last paragraph in §4.7 that H is a closed subspace of H2 . By the given hint, H is also Hilbert. Finally, we know from §4.19 that H1 and H are Hilbert space isomorphic to ℓ2 (A) and ℓ2 (B ′ ) respectively. Since A and B have the same cardinality, ℓ2 (A) is Hilbert space isomorphic to ℓ2 (B ′ ). In other words, H1 is Hilbert space isomorphic to H ⊆ H2 , completing the proof of the problem.  Problem 4.9 Rudin Chapter 4 Exercise 9.

Proof. Since A ⊆ [0, 2π] and A is measurable, we have χA ∈ L2 (T ). By Theorem 3.11, L2 (T ) is a complete metric space, so it is a Hilbert space by Definition 4.4. Since {un | n ∈ Z} is a maximal orthonormal set in the Hilbert space L2 (T ), Theorem 4.18(iii) implies that Z X hun , χA i 2 = kχA k2 =

n∈Z

2π 0

|χA (x)|2 dx = m(A) < ∞.

By [99, Theorem 3.23, p. 60], we know that lim hun , χA i = lim hun , χA i = 0. n→∞

n→−∞

(4.21)

132

Chapter 4. Elementary Hilbert Space Theory

By Definition 4.1, we have 

 un + u−n , χA = 2   un − u−n hsin nx, χA i = , χA = 2i

hcos nx, χA i =

1 (hun , χA i + hu−n , χA i), 2 1 (hun , χA i − hu−n , χA i). 2i

(4.22)

By applying the limits (4.21) to the inner products (4.22), we obtain that Z π Z (cos nt) · χA (t) dt = 2π lim hcos nx, χA i = 0 cos nt dt = lim lim n→∞ A

and lim

Z

n→∞ A

n→∞

n→∞ −π

sin nt dt = lim

Z

π

n→∞ −π

(sin nt) · χA (t) dt = 2π lim hsin nx, χA i = 0. n→∞

Thus we have completed the proof of the problem.



Problem 4.10 Rudin Chapter 4 Exercise 10.

Proof. This problem is proven in [124, Problem 11.16, p. 352].



Problem 4.11 Rudin Chapter 4 Exercise 11.

Proof. For each n = 1, 2, . . ., we define xn = of L2 (T ) because, by Definition 4.23,

n+1 n un

kxn k = kxn k2 =

and E = {xn } which is obviously a subset

n+1

= kxN +1 k, N N +1 this contradiction shows that E has no element of smallest norm. Hence we end the proof of the problem.  Remark 4.1 Problem 4.11 indicates that the condition “convexity” in Theorem 4.10 cannot be omitted.

Problem 4.12 Rudin Chapter 4 Exercise 12.

Proof. By the half-angle formula for cosine function, we have 1=

ck π

Z

 1 + cos t k

π

2

0

dt =

Z

2ck π

π 2

cos2k x dx.

(4.24)

0

A further substitution with z = sin x reduces the expression (4.24) to 2ck π

1=

Z

1 0

1

(1 − z 2 )k− 2 dz.

(4.25)

If we substitute y = z 2 and use the integral form of the beta function (see [99, Theorem 8.20, p. 193]), then the expression (4.25) can be written as 1=

ck π

Z

1 0

1

1

(1 − y)k− 2 y − 2 dy =

ck Γ(k + 21 )Γ( 12 ) ck Γ(k + 21 ) · =√ · . π Γ(k + 1) kΓ(k) π

(4.26)

Finally, we recall from [99, Exercise 30, p. 203] that lim

k→∞

Γ(k + c) =1 kc Γ(k)

(4.27)

for every real constant c. Take c = 12 in the limit (4.27) and then substitute the corresponding result into the expression (4.26), we get 1

lim k− 2 ck = lim

k→∞

This completes the proof of the problem. Problem 4.13 Rudin Chapter 4 Exercise 13.

k→∞

√

1

π·

√ k 2 Γ(k) = π. Γ(k + 12 ) 

134

Chapter 4. Elementary Hilbert Space Theory

Proof. We follow the hint. Consider the function f (t) = e2πikt , where k = 0, ±1, ±2, . . .. When k = 0, we have f ≡ 1 and then the formula holds trivially. Suppose that k 6= 0. Since α is irrational, kα ∈ / Z and then e2πikα 6= 1. By this, we have N N 1 X 2πiknα 1 X 2πikα n e2πikα e2πikN α − 1 e = (e ) = · 2πikα N n=1 N n=1 N e −1

so that

N 1 X 2 e2πiknα ≤ lim = 0. lim N →∞ N |e2πikα − 1| N →∞ N

(4.28)

n=1

On the other hand, we have

Z

1

e2πikt dt =

0

e2πikt 1 = 0. 2πik 0

(4.29)

Hence we deduce from the inequality (4.28) and the integral (4.29) that Z 1 N 1 X 2πiknα e = e2πikt dt N →∞ N 0 lim

(4.30)

n=1

for every k = 0, ±1, ±2, . . .. Therefore the formula (4.30) holds for every trigonometric polynomials in the form N X P (t) = ck e2πikt , (4.31) k=−N

where c−N , . . . , cN ∈ C.

Suppose that f : R → R is continuous and f (x + 1) = f (x) for every x ∈ R. By Theorem 4.25 (The Weierstrass Approximation Theorem), for every ǫ > 0, there exists a trigonometric polynomial P in the form (4.31) such that |f (t) − P (t)|
p, then we have nZ 1 o1 n Z t o1 1 2 2 2 kγ(t) − γ(p)k = |χ(p,t] (x)| dx = dx = (t − p) 2 . (4.54) 0

p

Similarly, if t < p, then we have kγ(t) − γ(p)k =

nZ

p

dx t

o1

2

1

= (p − t) 2 .

(4.55)

By combining the two expressions (4.54) and (4.55), we conclude that 1

kγ(t) − γ(p)k = |t − p| 2 and thus γ is continuous on [0, 1]. Let 0 ≤ a ≤ b ≤ c ≤ d ≤ 1. If a = b, then γ(a) − γ(b) = 0 and Z 1 0 × χ(c,d] (x) dx = 0. hγ(b) − γ(a), γ(d) − γ(c)i = h0, γ(d) − γ(c)i = 0

Similarly, if c = d, then hγ(b) − γ(a), γ(d) − γ(c)i = 0. Without loss of generality, we may assume that a < b and c < d. Then (a, b] ∩ (c, d] = ∅ and so we have Z 1

 hγ(b) − γ(a), γ(d) − γ(c)i = χ(a,b] , χ(c,d] = χ(a,b] (x)χ(c,d] (x) dx = 0. 0


Hence the desired result holds in the special case that H = L2 [0, 1] . To treat the general case, we first consider the function un (t) = e2πint for every n ∈ Z. It is clear that un ∈ H and  Z 1  1, if n = m; 2πi(n−m)t e dt = hun , um i =  0 0, if n 6= m.
Thus {un | n ∈ Z} is an orthonormal set in H. By Theorem 3.14, C [0, 1] is dense in H with respect to the norm induced by the inner product (4.53). By Theorem 4.25 (The Weierstrass Approximation Theorem) and Definition 4.23, we know that the set P of all trigonometric polynomials in the form N N X X cn e2πint cn un (t) = P (t) = n=−N

n=−N


is dense in C [0, 1] with respect to the norm k · k∞ . Note that P
= span ({un | n ∈ Z}). By Definition 3.7, it can be shown easily that P is also dense in C [0, 1] with respect to the norm induced by the inner product (4.53). Therefore, we derive from Lemma 2.10 that P is also dense in H in the norm induced by the inner product (4.53). Consequently, we conclude from Theorem 4.18(i) that {un | n ∈ Z} is in fact maximal and thene H∼ = ℓ2 (N).

(4.56)

Suppose that H ′ is a Hilbert space with an infinite maximal orthonormal set {vα | α ∈ A}, where A is countable or uncountable infinite. Then we have H′ ∼ = ℓ2 (A). e

We say A ∼ = B if A is Hilbert space isomorphic to B.

(4.57)

140

Chapter 4. Elementary Hilbert Space Theory

Applying Problem 4.8 to the Hilbert space isomorphisms (4.56) and (4.57), we conclude that H must be Hilbert space isomorphic to a subspace of H ′ and then the special case is applicable to this general case. Indeed, let Λ : H → K be a Hilbert space isomorphism, where K is a Hilbert subspace of H ′ . If we define λ : [0, 1] → H ′ by λ = Λ ◦ γ, then λ is injective and continuous. Furthermore, since hΛ(f ), Λ(g)i = hf, gi for all f, g ∈ H, we deduce from the special case above that





 hλ(b) − λ(a), λ(d) − λ(c)i = Λ γ(b) − Λ γ(a) , Λ γ(d) − Λ γ(c)







 = Λ γ(b) , Λ γ(d) − Λ γ(b) , Λ γ(c)







 − Λ γ(a) , Λ γ(d) + Λ γ(a) , Λ γ(c) = hγ(b), γ(d)i − hγ(b), γ(c)i − hγ(a), γ(d)i + hγ(a), γ(c)i

= hγ(b) − γ(a), γ(c) − γ(d)i

= 0.

This completes the proof of the problem.



Problem 4.18 Rudin Chapter 4 Exercise 18.

Proof. If r, s ∈ R and r 6= s, then we have Z A 2i sin[(r − s)A] 1 = 0. ei(r−s)t dt = lim hur , us i = lim A→∞ A→∞ 2A −A 2A(r − s) If r = s, then 1 A→∞ 2A

hus , us i = lim

Z

A

1 A→∞ 2A

eist e−ist dt = lim

−A

Z

(4.58)

A

dt = 1.

(4.59)

−A

Thus {us | s ∈ R} is an orthonormal set of H by Definition 4.13. Since f and g are finite combinations of us , we may suppose that f (t) =

n X

cp eisp t

and g(t) =

m X

dq eirq t .

(4.60)

q=1

p=1

Therefore, we obtain Z A X Z AX m n n X m   X 1 1 lim dq e−irq t dt = lim cp eisp t cp dq ei(sp −rq )t dt A→∞ 2A −A A→∞ 2A −A q=1

p=1

p=1 q=1

which is undoubtedly finite by the values (4.58) and (4.59). As a consequence, we have shown that hf, gi is well-defined for all f, g ∈ X. Next we are going to show that h · i satisfies Definition 4.1. Firstly, it is clear that 1 A→∞ 2A

hf, gi = lim

Z

A

−A

1 A→∞ 2A

f (t)g(t) dt = lim

Z

A −A

Secondly, for f, g, h ∈ X, we observe that Z A 1 [f (t) + g(t)]h(t) dt hf + g, hi = lim A→∞ 2A −A

g(t)f (t) dt = hg, f i.

4.3. Miscellaneous Problems

141

1 A→∞ 2A

= lim

Z

A

1 A→∞ 2A

f (t)h(t) dt + lim

−A

= hf, hi + hg, hi .

Z

A

g(t)h(t) dt −A

Thirdly, for any scalar α, we have Z A Z A 1 1 αf (t)h(t) dt = α lim f (t)h(t) dt = α hf, gi . hαf , gi = lim A→∞ 2A −A A→∞ 2A −A Fourthly, if f takes the representation (4.60), then we establish from the results (4.58) and (4.59) that Z A Z A n X 1 1 |cp |2 ≥ 0. (4.61) f (t)f (t) dt = lim |f (t)|2 dt = hf, f i = lim A→∞ 2A −A A→∞ 2A −A p=1

Finally, if hf, f i = 0, then the result (4.61) definitely shows that c1 = c2 = · · · = cp = 0. In other words, we have f ≡ 0 in this case. Hence this inner product certainly makes X into a unitary space. Let H be the completion of X, i.e., X is dense in H and H = X. Since span ({us | s ∈ R}) = X, Theorem 4.18(i) says that the set {us | s ∈ R} is in fact a maximal orthonormal in H. By the discussion in §4.19, we have H∼ (4.62) = ℓ2 (R). Assume that H was separable. By Problem 4.4, H has an at most countable maximal orthonormal system {vn | n ∈ N}. By the discussion in §4.19 again, we know that H∼ = ℓ2 (N).

(4.63)

Now the two isomorphic relations (4.62) and (4.63) imply that ℓ2 (R) ∼ = ℓ2 (N), but then R and N have the same cardinal number, a contradiction. Hence we complete the proof of the problem.  Problem 4.19 Rudin Chapter 4 Exercise 19.

Proof. If N = 1, then ω = e2πi = 1 and k can only take the value 0. Thus the orthogonality relation holds in this special case. Suppose that N > 1. If k = 0, then ω 0 = 1 so that N 1 X 0 ω = 1. N n=1

Let k = 1, 2, . . . , N − 1. We remark that z N − 1 = (z − 1)(z N −1 + · · · + z + 1). Putting z = ω k into the expression (4.64), we obtain  
(ω k )N − 1 = (ω k − 1) (ω k )N −1 + · · · + ω k + 1 = (ω k − 1) ω k(N −1) + · · · + ω k + 1 .

(4.64)

(4.65)

It is clear that (ω k )N = (ω N )k = 1. Furthermore, since 1 ≤ k ≤ N − 1, k is evidently not divisible by N which implies that ω k 6= 1. Therefore, it deduces from the expression (4.65) that ω k(N −1) + · · · + ω k + 1 = 0

142

Chapter 4. Elementary Hilbert Space Theory

or equivalently ω kN + ω k(N −1) + · · · + ω k = 0. In conclusion, we have

 N 1, if k = 0; 1 X nk  ω =  N n=1 0, if 1 ≤ k ≤ N − 1.

Let N ≥ 3. It is clear from the orthogonality relations and the properties in Definition 4.1 that N N 1 X 1 X kx + ω n yk2 ω n = hx + ω n y, x + ω n yi ω n N N n=1

n=1

N N N X X 1 hX hω n y, xi ω n hx, ω n yi ω n + hx, xi ω n + = N n=1

n=1

+

N X

n=1

=

hω n y, ω n yi ω n

X X X 1h hy, xi ω n × ω n hx, yi ω −n × ω n + ωn + hx, xi N N

n=1

+ hy, yi =

n=1

i

N X

1h hx, xi N

= hx, yi

ωn

n=1 N X

n=1

N

N

n=1

n=1

i

ω n + N hx, yi + hy, xi

N X

n=1

ω 2n + hy, yi

N X

ωn

n=1

i

(4.66)

as desired. Finally, Definition 4.1 gives D E kx + eiθ yk2 = x + eiθ y, x + eiθ y = hx, xi + e−iθ hx, yi + eiθ hy, xi + hy, yi so that

1 2π

Z

Z π   1 kx + e yk e dθ = kxk2 + kyk2 + e−iθ hx, yi + eiθ hy, xi eiθ dθ 2π −π Z−π Z π π 1 1 hx, yi dθ + e2iθ hy, xi dθ = 2π −π 2π −π π

iθ

2 iθ

= hx, yi .

This completes the proof of the problem.

(4.67) 

Remark 4.3 We note that the polarization identity (see [100, p. 86]) can be written in the form 3

hx, yi =

1 X −n i kx + in yk2 . 4 n=0

Thus the two identities (4.66) and (4.67) in Problem 4.19 are actually generalizations of this.

CHAPTER

5

Examples of Banach Space Techniques

5.1

The Unit Ball in a Normed Linear Space

Problem 5.1 Rudin Chapter 5 Exercise 1.

Proof. For 0 < p ≤ ∞, we suppose that B = {f ∈ Lp (µ) | kf kp ≤ 1}. There are two cases: • Case (i): 0 < p < ∞. In this case, we have o1 nZ p |f |p dµ kf kp = X


1 = |f (a)|p × µ({a}) + |f (b)|p × µ({b}) p 1 1 = 1 {|f (a)|p + |f (b)|p } p . 2p

Consequently, kf kp ≤ 1 if and only if |f (a)|p + |f (b)|p ≤ 2.

(5.1)

Therefore, it follows from the inequality (5.1) that the unit ball B is a circle if and only if p = 2. Besides, the unit ball B becomes a square if and only if p = 1. • Case (ii): p = ∞. By Definition 3.7, we know that kf k∞ = max(|f (a)|, |f (b)|)

(5.2)

max(|f (a)|, |f (b)|) ≤ 1

(5.3)

so that kf k∞ ≤ 1 if and only if

which is a square. See Figure 5.1 for an illustration. 143

144

Chapter 5. Examples of Banach Space Techniques

Figure 5.1: The unit circle in different p-norm. If µ({a}) 6= µ({b}), then the expression (5.1) is replaced by µ({a})|f (a)|p + µ({b})|f (b)|p ≤ 1 which cannot be a circle or a square for any p ∈ (0, ∞). However, we note that the expression (5.2) is still valid even in the case µ({a}) 6= µ({b}), so we have the inequality (5.3) and then B  is a square when p = ∞. This ends the analysis of the problem. Problem 5.2 Rudin Chapter 5 Exercise 2.

Proof. Let X be a normed linear space with norm k · k and B(0, 1) = {x ∈ X | kxk < 1} be the open unit ball. For every x, y ∈ B(0, 1) and t ∈ [0, 1], we follow from Definition 5.2 that k(1 − t)x + tyk ≤ (1 − t)kxk + tkyk < 1 so that (1 − t)x + ty ∈ B(0, 1). By §4.8, B(0, 1) is convex. Similarly, the convexity of the closed unit ball B(0, 1) = {x ∈ X | kxk ≤ 1} can be proven similarly. Hence we complete the proof of the problem.  Problem 5.3 Rudin Chapter 5 Exercise 3.

Proof. Suppose that kf kp = kgkp = 1 and f 6= g. By Theorem 3.9, we always have

1

1 1

khkp = (f + g) ≤ kf kp + kgkp = 1. 2 2 2 p

Assume that khkp = 1. Then it follows from Problem 3.13 that f = λg a.e. on X for some λ > 0. Since λkgkp = kf kp = kgkp , we have λ = 1 implying f = g a.e. on X, a contradiction. However, strictly convexity does not hold for L1 (µ), L∞ (µ) and C(X). Ignore trivialities, we suppose that X contains more than one point. In fact, we can pick E, F ∈ M \ {∅} such that E ∩ F = ∅ and µ(E), µ(F ) ∈ (0, ∞). Define f : X → C and g : X → C by f (x) =

1 χE (x) and µ(E)

g(x) =

1 χF (x) µ(F )

respectively. It is obvious that f 6= g. Furthermore, we have Z Z 1 |f (x)| dx = kf k1 = χE (x) dx = 1, µ(E) E X

(5.4)

5.1. The Unit Ball in a Normed Linear Space

145

Z 1 χF (x) dx = 1, µ(F ) F ZX Z Z

f + g 1 1

f (x) + g(x)

f (x) dx + g(x) dx = 1.



= dx = 2 2 2 E 2 F 1 X kgk1 =

Z

|g(x)| dx =

These show that the invalidity of strictly convexity in L1 (µ). For L∞ (µ), we replace the functions f and g in the definition (5.4) by f (x) = χE (x)

and g(x) = χE∪F (x)

respectively. Then it is also true that f 6= g. By Definition 3.7, we have

f + g

kf k∞ = kgk∞ =

= 1. 2 ∞ ∞ Therefore, the strictly convexity does not hold in L (µ). Finally, we consider the space C(X), where X is Hausdorff. By Definition 3.16, we see that X is assumed to be compact and so the norm k · k∞ of C(X) is given by kf k∞ = sup |f (x)|.

(5.5)

x∈X

Take f ∈ C(X) and suppose that f is nonconstant. Thus one can find a x0 ∈ X such that |f (x0 )| < kf k∞ .

(5.6)

Otherwise, we have |f (x)| ≥ kf k∞ on X, but this and the definition (5.5) will imply that |f (x)| = kf k∞ on X, i.e., f is a constant which is a contradiction. Define 1 3 a = kf k∞ + |f (x0 )| and 4 4

Consider the sets o n
K = x ∈ X |f (x)| ≥ a = |f |−1 [a, ∞)

and

1 3 b = kf k∞ + |f (x0 )|. 4 4

o n
E = x ∈ X |f (x)| ≤ b = |f |−1 (−∞, b] .

It is clear that x0 ∈ E, i.e., E 6= ∅. Since |f | : X → R is continuous, the Extreme Value Theorem ([74, Theorem 27.4, p. 174]) implies that kf k∞ = max |f (x)| = |f (p)| for some p ∈ X. x∈X

Consequently, we have

p ∈ K.

(5.7)

Furthermore, both K and E are closed in X by [74, Theorem 18.1, p. 104]. If K ∩ E 6= ∅, then we have 1 1 3 3 kf k∞ + |f (x0 )| ≤ kf k∞ + |f (x0 )| 4 4 4 4 but these imply that kf k∞ ≤ |f (x0 )| which contradicts the hypothesis (5.6). Hence we have K ∩ E = ∅.

What we have done in the previous paragraph is that V = E c is open in X and K ⊆ V . Since X is compact, Theorem 2.4 ensures that K is also compact. By Theorem 2.12 (Urysohn’s Lemma), there exists a h ∈ Cc (X) such that K ≺ h ≺ V , i.e., 0 ≤ h(x) ≤ 1 on X, h(x) = 1 on K and supp (h) ⊆ V (or equivalently h(x) = 0 on E). If we define g = h × f : X → C, then we must have g 6≡ f and furthermore, the fact (5.7) shows that kgk∞ = sup |h(x)f (x)| = |h(p)| · |f (p)| = |f (p)| = kf k∞ x∈X

and

f + g 1 1 1



= kf + hf k∞ = sup [1 + h(x)]f (x) = |1 + h(p)| · |f (p)| = |f (p)| = kf k∞ . 2 2 2 x∈X 2 ∞ Hence strictly convexity does not hold for C(X), completing the proof of the problem.



146

5.2

Chapter 5. Examples of Banach Space Techniques

Failure of Theorem 4.10 and Norm-preserving Extensions

Problem 5.4 Rudin Chapter 5 Exercise 4.

Proof. Define f : [0, 1] → C by

  −8x + 4, if x ∈ [0, 12 ];

f (x) =



if x ∈ ( 21 , 1].

0,

Then it is easy to see that f ∈ M , i.e., M is nonempty. Suppose that f, g ∈ M , t ∈ [0, 1] and h = (1 − t)f + tg. Then we have Z

1 2

0

h(x) dx −

Z

1 1 2

h(x) dx =

Z

1 2

0

[(1 − t)f (x) + tg(x)] dx −

hZ = (1 − t) +t =1

hZ

f (x) dx −

0

1

g(x) dx −

0

Z

1

Z

1

1

1 1 2

[(1 − t)f (x) + tg(x)] dx

f (x) dx

1 2

g(x) dx

1 2

Z

i

i

so that h ∈ M , i.e., M is convex.

Next, we note that C and M are metric spaces. By [99, Theorem 7.15, p. 151], we see that C is a complete metric space. Since M ⊆ C, M is also a complete metric space. Suppose that {fn } ⊆ M and kfn − f k = sup |fn (x) − f (x)| → 0 x∈[0,1]

as n → ∞. Then it follows froma [99, Theorem 7.9, p. 148] that fn → f uniformly on [0, 1]. Since each fn is continuous on [0, 1], f is also continuous on [0, 1]. Furthermore, we have lim

n→∞

Z

1 2

0



hZ

0

1 2

fn (t) dt −

 lim fn (t) dt −

n→∞

Z

1 2

0

Z

1 2

1

Z

1 1 2

i fn (t) dt = 1

 lim fn (t) dt = 1

n→∞

f (t) dt −

Z

1

1 2

f (t) dt = 1.

In other words, f ∈ M and M is closed in the space C.

Finally, we want to show that M contains no element of minimal norm. To this end, we notice that

a

Z 1=

1 2

0

f (t) dt −

Z

1 1 2

Z f (t) dt ≤

0

1 2

|f (t)| dt +

Or we can see the rephrased Theorem 7.9 on p. 151 in [99].

Z

1 1 2

|f (t)| dt = kf k1 ≤ kf k∞

(5.8)

5.2. Failure of Theorem 4.10 and Norm-preserving Extensions

147

for every f ∈ M . Assume that there was a f ∈ M such that kf k∞ = 1. Then we deduce from the inequality (5.8) that Z 1 Z 1 (1 − |f (t)|) dt = 0. (5.9) |f (t)| dt = 1 or 0

0

Since kf k∞ = 1, we have |f (x)| ∈ [0, 1] for all x ∈ [0, 1] so that the function g(x) = 1 − |f (x)| is continuous and nonnegative on [0, 1]. Applying Theorem 1.39(a) to the second integral in (5.9), we obtain g(x) = 0 a.e. on [0, 1]. Now the continuity of g forces that g(x) = 0 on [0, 1], i.e., |f (x)| = 1 on [0, 1]. In particular, we have
−1 ≤ Re f (x) ≤ 1 (5.10)

on [0, 1]. On the other hand, since f ∈ M , the definition gives 1 = Re

Z

1 2

f (t) dt −

0

or equivalently,

Z

1 2

0



Z

1 1 2

 Z f (t) dt =


 Re f (t) − 1 dt +

Z

1 2

0

1

1 2


Re f (t) dt −

Z

1 1 2


Re f (t) dt


 −1 − Re f (t) dt = 0.

(5.11)



Observe from the inequalities (5.10) that Re f (x) − 1 ≤ 0 and
−1 − Re f (x) ≤ 0 on
[0, 1]. Therefore, the equation (5.11) and the continuities of Re f (x) − 1 and −1 − Re f (x) imply that  if [0, 12 ];
 1, Re f (x) =  −1, if [ 21 , 1].
However, this says that Re f (x) is discontinuous at x = 21 , a contradiction. Hence no such f exists and this completes the proof of the problem.  Problem 5.5 Rudin Chapter 5 Exercise 5.
Proof. It is clear that 1 ∈ M , so M 6= ∅. Suppose that f, g ∈ M ⊆ L1 [0, 1] , t ∈ [0, 1] and h = (1 − t)f + tg. Then we have Z 1 Z 1 Z 1 |g(x)| dx < ∞ |f (x)| dx + t |h(x)| dx ≤ (1 − t) which means that h ∈ Z

1

h(x) dx =

L1

Z

0

0

0

0

0

1


[0, 1] . Besides, we also have

[(1 − t)f (x) + tg(x)] dx = (1 − t)

Z

0

1

f (x) dx + t

Z

1

g(x) dx = 1 0

so that h ∈ M , i.e., M is convex. Next, suppose that {fn } ⊆ M and
there exists a function f on [0, 1] such that kfn − f k1 → 0 as n → ∞. Thus fn − f ∈ L1 [0, 1] and it yields from Theorem 1.33 that Z 1 Z 1 Z 1 |fn (t) − f (t)| dt = kfn − f k1 f (t) dt ≤ fn (t) dt − 0

0

0

148

Chapter 5. Examples of Banach Space Techniques

which implies that lim

Z

n→∞ 0

1

fn (t) dt =

Z

1

f (t) dt.

0


In other words, we have f ∈ M and M is closed in L1 [0, 1] .
For every f ∈ M ⊆ L1 [0, 1] , we see from Theorem 1.33 that kf k1 ≥ 1. For each n = 1, 2, . . ., we define fn : [0, 1] → C by fn (x) = nxn−1 . By direct checking, it is clear that kfn k1 =

Z

1

ntn−1 dt = 1.

0

In addition, fn 6= fm if n 6= m. Hence M contains infinitely many elements of minimal norm, completing the proof of the problem.  Problem 5.6 Rudin Chapter 5 Exercise 6.

Proof. We note that M is not necessarily closed in H. Let f : M ⊆ H → C be a bounded linear functional. If we check the proof of Theorem 5.4 carefully, we see that the bounded linear transformation Λ : X → Y is in fact uniformly continuous on X. Thus f : M → C is uniformly continuous on M . Since H is a metric (in fact Hilbert) space, M is also a metric space. Since C is a complete metric space and M is dense in M , we follow from [99, Exercies 4 & 13, pp. 98, 99] that f can be uniquely extended to a continuous function fe : M → C.b Without loss of generality, we may assume that M is closed in H. As a closed subspace of a Hilbert space H, M is itself a Hilbert space (see the note in Problem 4.8). By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there corresponds a unique x0 ∈ M such that f (x) = hx, x0 i

(5.12)

for all x ∈ M . Recall from Definition 5.3 that kf k is the smallest number such that |f (x)| ≤ kf k · kxk for every x ∈ M . In particular, we must have kf (x0 )k ≤ kf k · kx0 k. On the other hand, the representation (5.12) gives |f (x0 )| = f (x0 ) = hx0 , x0 i = kx0 k2 = kx0 k · kx0 k, so we must have kx0 k = kf k.

(5.13)

By Theorem 4.11, every x ∈ H can be expressed uniquely as x = P x + Qx, where P x ∈ M and Qx ∈ M ⊥ . This means that H = M ⊕ M ⊥ .c Now if we define F : H → C by F (x) = hx, x0 i In fact, such fe is uniformly continuous on M . See also [74, Exercise 2, p. 270]. Let U and W be two vector subspaces of the vector space V . Then V is said to be the direct sum of U and W , denoted by V = U ⊕ W , if V = U + W and U ∩ W . See [7, pp. 95, 96]. b

c

5.2. Failure of Theorem 4.10 and Norm-preserving Extensions for every x ∈ H, then we obtain F (x) = hx, x0 i =

149

  f (x), if x ∈ M ; 

if x ∈ M ⊥ .

0,

Furthermore, by the same analysis in obtaining the representation (5.13), we get kF k = kx0 k and then using the representation (5.13) again, we conclude that kF k = kf k = kx0 k,

(5.14)

i.e., F is a norm-preserving extension on H of f which vanishes on M ⊥ . Suppose that F ′ : H → C is another norm-preserving extension of f that vanishes on M ⊥ . Then F ′ is bounded and thus continuous by Theorem 5.4. By Theorem 4.12 (The Riesz Representation Theorem for Hilbert Spaces), there corresponds a unique x1 ∈ H such that F ′ (x) = hx, x1 i for all x ∈ H and also Since F ′ (x) = f (x) on M , we have

kF ′ k = kf k = kx1 k.

(5.15)

hx, x1 i = hx, x0 i

on M . By Definition 4.1, hx, x1 − x0 i = 0 and then x1 − x0 ∈ M ⊥ . Since x1 = x0 + (x1 − x0 ), where x0 ∈ M and x1 − x0 ∈ M ⊥ , Theorem 4.11 shows that x1 and x1 − x0 are unique and kx0 k2 = kx1 k2 + kx0 − x1 k2 .

(5.16)

By putting the numbers (5.14) and (5.15) into the formula (5.16), we see that kf k2 = kf k2 + kx0 − x1 k2 and this reduces to kx0 − x1 k2 = 0. By Definition 4.1, we must have x0 = x1 , i.e., F ′ = F . This  proves the uniqueness part and so we have completed the proof of the problem. Problem 5.7 Rudin Chapter 5 Exercise 7.

Proof. Consider X = [−1, 1] with µ = m, the Lebesgue measure. Then the vector space
L1 [−1, 1] consists of all complex measurable function on [−1, 1] such that kf k1 = We define

Z

1

−1

|f (x)| dx < ∞.


M = {f ∈ L1 [−1, 1] | f is real-valued and f (x) = 0 on [−1, 0]}.

Now the function χ[0,1] is obviously an element
of M , so M 6= ∅ and also kχ[0,1] k1 = 1. It is also clear that M is a subspace of L1 [−1, 1] . Next, we define the functional Λ : M → C by Λ(f ) =

Z

1

−1

f (x) dx =

Z

1

f (x) dx 0

(5.17)

150

Chapter 5. Examples of Banach Space Techniques

which is linear. By Definition 5.3, the definition (5.17) and the fact that χ[0,1] ∈ M , we have kΛk = sup{|Λ(f )| | f ∈ M and kf k1 = 1} Z Z 1 nZ 1 = sup |f (x)| dx = f (x) dx f ∈ M and −1

0

= 1.

Thus Λ is bounded.
Let δ ∈ (0, 1). If we define Λδ : L1 [−1, 1] → C by Λδ (f ) =

Z

1

−1

Re [f (x)]χ(−δ,1) (x) dx =

Z

1 0

o f (x) dx = 1

1

Re [f (x)] dx, −δ


then it satisfies Definition 2.1 so that Λδ is linear on L1 [−1, 1] . It is also clear that Λδ |M = Λ. A direct computation also shows that Z 1 Z 1 Z 1 |f (x)| dx |Re [f (x)]| dx ≤ Re [f (x)] dx ≤ |Λδ (f )| = −δ

−δ

−1

which implies that


kΛδ k = sup{|Λδ (f )| | f ∈ L1 [−1, 1] and kf k1 = 1} = 1 = kΛk.
Thus Λδ is a norm-preserving extension on L1 [−1, 1] of Λ. Hence if α 6= β, then we have Λα 6= Λβ . This completes the proof of the problem. 

5.3

The Dual Space of X

Problem 5.8 Rudin Chapter 5 Exercise 8.

Proof. (a) We first show that X ∗ is a normed linear space by checking Definition 5.2. For all f, g ∈ X ∗ , since |f (x) + g(x)| ≤ |f (x)| + |g(x)| for all x ∈ X, it must be true that kf + gk ≤ kf k + kgk. Next, for a scalar α and f ∈ X ∗ , we have |αf (x)| = |α| · |f (x)| for all x ∈ X which implies that kαf k = |α| · kf k. Finally, suppose that f ∈ X ∗ is such that kf k = 0. By [100, Eqn. (3), p. 96], we have |f (x)| ≤ kf k · kxk = 0 holds for every x ∈ X. Thus f ≡ 0 on X. By Definition 5.2, X ∗ is a normed linear space.

It remains to show that X ∗ is complete. To this end, let {Λn } ⊆ X ∗ be Cauchy. Then given ǫ > 0, there corresponds a positive integer N such that n, m ≥ N imply that kΛn − Λm k < ǫ

5.3. The Dual Space of X

151

and for arbitrary but fixed x ∈ X, this and [100, Eqn. (3), p. 96] together imply that |Λn (x) − Λm (x)| = |(Λn − Λm )(x)| ≤ kΛn − Λm k · kxk < ǫkxk.

(5.18)

Thus {Λn (x)} ⊆ C is Cauchy. Since C is complete, {Λn (x)} converges to a point in C and it is reasonable to define the pointwise limit Λ : X → C by Λ(x) = lim Λn (x). n→∞

Now our target is to show that Λ ∈ X ∗ and {Λn } converges to Λ in X ∗ , i.e., kΛn −Λk → 0 as n → ∞. Let α, β ∈ C and x, y ∈ X. Since X is a normed linear space, αx + βy ∈ X and thus Λ(αx + βy) = lim Λn (αx + βy) n→∞

= lim [αΛn (x) + βΛn (y)] n→∞

= α lim Λn (x) + β lim Λn (y) n→∞

n→∞

= αΛ(x) + βΛ(y). By Definition 2.1, Λ is linear. By the hypothesis, {Λn } is a Cauchy sequence in X ∗ so that it is bounded, i.e., there exists a positive constant M such that kΛn k ≤ M

(5.19)

for every n = 1, 2, . . .. Therefore, it follows from the triangle inequality, then [100, Eqn. (3), p. 96] and finally the inequalities (5.19) that |Λ(x)| ≤ |Λn (x)| + |Λ(x) − Λn (x)|

≤ kΛn k · kxk + |Λ(x) − Λn (x)|

≤ M kxk + |Λ(x) − Λn (x)|

(5.20)

for all n > N and for all x ∈ X. Recall that if we take m → ∞ in the inequality (5.18), then we have |Λn (x) − Λ(x)| ≤ ǫkxk (5.21) for all n > N . Thus, by putting the inequality (5.19) into the inequality (5.20), we get |Λ(x)| ≤ (M + ǫ)kxk for every x ∈ X. Hence we obtain kΛk < ∞, i.e., Λ ∈ X ∗ . Since the inequality (5.21) is true for all x ∈ X, it must be true for those x with kxk ≤ 1 and so n > N implies that kΛn − Λk = sup{|Λn (x) − Λ(x)| | x ∈ X and kxk ≤ 1} ≤ ǫ. Since ǫ is arbitrary, we have {Λn } converges to Λ in X ∗ . In other words, X ∗ is complete and we conclude that X ∗ is a Banach space. (b) Fix the x ∈ X, define Λ∗x : X ∗ → C by Λ∗x (f ) = f (x). If x = 0, then it is clear that Λ∗0 (f ) = f (0) = 0 on X ∗ . Therefore, Λ∗0 must be linear and of norm 0. Without loss of generality, we may assume that x 6= 0 in the following discussion. For any f, g ∈ X ∗ and α, β ∈ C, we have Λ∗x (αf + βg) = (αf + βg)(x) = αf (x) + βg(x) = αΛ∗x (f ) + βΛ∗x (g).

152

Chapter 5. Examples of Banach Space Techniques By Definition 2.1, Λ∗x is a linear functional on X ∗ . On the one hand, by [100, Eqn. (3), p. 96], we know that |Λ∗x (f )| = |f (x)| ≤ kf k · kxk ≤ kxk (5.22) for all f ∈ X ∗ with kf k = 1. Thus it means that kΛ∗x k ≤ kxk. On the other hand, since X is a normed linear space, and x 6= 0, Theorem 5.20 ensures the existence of a g ∈ X ∗ such that kgk = 1 and g(x) = kxk, so we have kΛ∗x k ≥ |Λ∗x (g)| = |g(x)| ≥ kxk,

(5.23)

where kgk = 1. Combining the inequalities (5.22) and (5.23), we get the desired result that kΛ∗x k = kxk. (c) Suppose that {xn } ⊆ X and {f (xn )} is bounded for every f ∈ X ∗ . We consider the mapping Λ∗xn : X ∗ → C given by Λ∗xn (f ) = f (xn ). By part (a), X ∗ is a Banach space. By part (b), kΛ∗xn k = kxn k for every n ∈ N. By the hypothesis, for each f ∈ X ∗ , the set {Λ∗xn (f )} = {f (xn )} is bounded by a positive constant Mn , so you can’t find a f ∈ X ∗ such that sup |Λ∗xn (f )| = ∞. n∈N

Therefore, we deduce from Theorem 5.8 (The Banach-Steinhaus Theorem) that kΛ∗xn k ≤ M for all n ∈ N, where M is a positive constant. Hence the sequence {kxn k} is bounded.  This completes the proof of the problem. Problem 5.9 Rudin Chapter 5 Exercise 9.

Proof. (a) We are going to prove the assertions one by one. – kΛk = kyk1 . Fix y = {ηi } ∈ ℓ1 . Define Λ : c0 → C by Λ(x) =

∞ X

ξi ηi ,

(5.24)

i=1

for every x = {ξi } ∈ c0 ⊆ ℓ∞ . Since ξi → 0 as i → ∞, the sequence {ξi } is bounded by a positive constant M (see [99, Theorem 3.2(c), p. 48]). Since y ∈ ℓ1 , we have ∞ ∞ ∞ X X X |ηi | < ∞ |ξi ηi | ≤ M ξi ηi ≤ i=1

i=1

i=1

so that the series (5.24) converges (absolutely) and thus the map Λ is well-defined.

5.3. The Dual Space of X

153

Let z = {ωi } ∈ c0 and α, β ∈ C. Since c0 is a vector space, we have αx + βz ∈ c0 . By the definition, we have Λ(x + z) =

∞ X (αξi + βωi )ηi

(5.25)

i=1

By the analysis in the previous paragraph, the series (5.25) is allowed to split into two series so that Λ(x + z) = α

∞ X

ξi ηi + β

∞ X

ωi ηi = αΛ(x) + βΛ(z),

i=1

i=1

i.e., Λ is linear. By the definition (5.24), it is easy to see that |Λ(x)| ≤ kxk∞ · kyk1 for every x ∈ c0 . Thus it is true that kΛk ≤ kyk1 . For the reverse direction, since y ∈ ℓ1 , we know that |ηi | → 0 as i → ∞. Next, for each n ∈ N, we consider the sequence xn = {ξn,i } defined by  ηi   , if i = 1, 2, . . . , n and ηi 6= 0;  |ηi | (5.26) ξn,i =    0, if i > n or η = 0. i

Since | |ηηii | | ≤ 1 for all i ∈ N, we have xn ∈ ℓ∞ for each n ∈ N. In fact, we have kxn k∞ ≤ 1. Furthermore, for every fixed n ∈ N, one can find i ∈ N such that i > n, thus we deduce easily from the definition (5.26) that ξn,i → 0

as i → ∞, i.e., xn ∈ c0 . Now we derive from Definition 5.2 that kΛk = sup{|Λ(x)| | x ∈ c0 and kxk∞ ≤ 1} ≥ |Λ(xn )| =

n X i=1

ξn,i ηi =

n X i=1

|ηi |

for every n ∈ N, i.e., kΛk ≥ kyk1 . In conclusion, we have kΛk = kyk1 < ∞ and thus Λ ∈ (c0 )∗ . – (c0 )∗ is isometric isomorphic to ℓ1 . In the first assertion, we have shown that given a y ∈ ℓ1 , if Λ is defined in the form (5.24), then Λ ∈ (c0 )∗ . More precisely, the mapping Φ : ℓ1 → (c0 )∗ given by Φ(y) = Λy is well-defined, where Λy is in the form (5.24). ∗ Step 1: Φ is linear. For every y = {ηi } ∈ ℓ1 , z = {θi } ∈ ℓ1 and α, β ∈ C, we have αy + βz = {αηi + βθi } ∈ ℓ1 . If x = {ξi } ∈ c0 , then we deduce from the definition (5.24) that Λαy+βz (x) =

∞ X i=1

ξi (αηi + βθi ) = α

∞ X

ξi ηi + β

i=1

Consequently, Φ(αy + βz) = αΦ(y) + βΦ(z).

∞ X i=1

ξi θi = αΛy (x) + βΛz (x).

154

Chapter 5. Examples of Banach Space Techniques ∗ Step 2: Φ is surjective. Given Λ ∈ (c0 )∗ . We want to show that there exists a y ∈ ℓ1 such that Φ(y) = Λ. It is well-known that the sequence {ei }∞ i=1 of standard unit vectors of c0 is a P d In fact, we have basis for c0 and that {ξi } = ξi ei whenever {ξi }∞ i=1 ∈ c0 . ei = {δi,k }∞ , where δ is the Kronecker delta function. Then it is routine to i,k k=1 check that ei ∈ c0 and kei k∞ = 1. Next, we define the sequence y = {ηi }∞ i=1 by ηi = Λ(ei )

(5.27)

for every i = 1, 2, . . .. Our aim is to show that y ∈ ℓ1 . If the sequence xn = {ξn,i }ni=1 is given by the form (5.26), then we immediately know that xn ∈ c0 and kxn k∞ = 1. Since n ∞ X X ξn,i ei , ξn,i ei = xn = i=1

i=1

the linearity of Λ shows that Λ(xn ) =

n X

ξn,i Λ(ei ) =

n X

ξn,i ηi =

i=1

i=1

n X i=1

|ηi |.

(5.28)

Therefore, we apply [100, Eqn. (3), p. 96] to the expression (5.28) to obtain n X i=1

|ηi | = |Λ(xn )| ≤ kΛk · kxn k∞ = kΛk

holds for every n = 1, 2, . . .. In particular, we have kyk1 =

∞ X i=1

|ηi | ≤ kΛk < ∞,

i.e, y ∈ ℓ1 . Finally, for every x = {ξi } ∈ c0 , we have x=

∞ X

ξi ei

i=1

and the application of the expressions (5.27) and the linearity of Λ indicate that ∞ ∞ ∞  X X X ξi ηi ξi Λ(ei ) = ξi ei = Λ(x) = Λ i=1

i=1

i=1

which is exactly in the form (5.24). Hence the map Φ is surjective. ∗ Step 3: Φ is injective. Let y = {ηi } ∈ ℓ1 , z = {θi } ∈ ℓ1 . Suppose further that Φ(y) = Φ(z). Since Φ is linear, we have Λy−z = Φ(y − z) = 0. By this, for every j = 1, 2, . . ., we get Λy−z (ej ) =

∞ X i=1

δj,i (ηi − θi ) = ηj − θj = 0

so that ηj = θj . In other words, we conclude that y = z, as desired. d The sequence {ei } is a Schauder basis for c0 (and also for ℓ1 ). In fact, a sequence {xi }∞ i=1 in a Banach space X is called a Schauder basis if for each x ∈ X, there is a unique sequence {αi } of scalars such that ∞ n X X x= αi xi = lim αi xi . See, for instance, [70, Example 4.1.3, p. 351]. i=1

n→∞

i=1

5.3. The Dual Space of X

155

∗ Step 4: Φ is an isometry. This follows from the first assertion directly because kΦ(y)k = kΛy k = kyk1 . Hence we have (c0 )∗ = ℓ1 . (b) For each y = {ηi } ∈ ℓ∞ , we define Θ : ℓ1 → C by Θ(x) =

∞ X

ξi ηi

(5.29)

i=1

for every x = {ξi } ∈ ℓ1 . Similar to the proof of the first assertion in part (a), we can show that the series (5.29) converges (absolutely) and Θ is linear. It is clear from the definition that |Θ(x)| ≤ kyk∞ · kxk1 , so we have kΘk ≤ kyk∞ < ∞.

(5.30)

This means that Θ ∈ (ℓ1 )∗ .

On the other hand, suppose that Θ ∈ (ℓ1 )∗ , i.e., kΘk < ∞. If we consider the Schader basis {ei } for ℓ1 and the relationships (5.27) with Λ replaced by Θ, then we have |ηi | = |Θ(ei )| ≤ kΘk · kei k1 = kΘk < ∞

(5.31)

for every i = 1, 2, . . .. Put z = {ηi }. Then the inequality (5.31) immediately implies that z ∈ ℓ∞ . In fact, it shows further that kzk∞ ≤ kΘk. Now for every x = {ξi } ∈ ℓ1 , we have x=

∞ X

ξi ei

i=1

and then Θ(x) =

∞ X

ξi ηi

i=1

which is exactly in the form (5.29), so the inequality (5.30) holds for z. Therefore, we have kΘk = kzk∞ , i.e., Θ ∈ ℓ∞ . If we define the mapping Ψ : ℓ∞ → (ℓ1 )∗ by

Ψ(y) = Θy , where Θy is given by the form (5.29), then the previous paragraph actually shows that Ψ is surjective. By using similar argument as in the proofs of Step 1, Step 3 and Step 4 in part (a), it is easy to conclude that Ψ is an isometric vector space isomorphism. (c) Let y = {ηi } ∈ ℓ1 . Define the functional Λ : ℓ∞ → C by Λ(x) =

∞ X

ξi ηi ,

i=1

where x = {ξi } ∈ ℓ∞ . Since this series converges (absolutely), it is a linear functional. Since |Λ(x)| ≤ kxk∞ · kyk1 , we know from [100, Eqn. (2), p. 96] that kΛk ≤ kyk1 , i.e., Λ ∈ (ℓ∞ )∗ . ℓ∞

Assume that ℓ1 was isometric isomorphic to (ℓ∞ )∗ . It is clear that x = {1, 1, . . .} ∈ \ c0 . By [70, Example 1.2.13, p. 14], we know that c0 is a proper closed subspace of

156

Chapter 5. Examples of Banach Space Techniques ℓ∞ so that ℓ∞ \ c0 6= ∅. Take x0 ∈ ℓ∞ \ c0 . Then Theorem 5.19 implies that there is a f ∈ (ℓ∞ )∗ such that kf k = 1, f (x) = 0 on c0 and f (x0 ) 6= 0.

(5.32)

If f ∈ ℓ1 , then we have f ∈ (c0 )∗ by part (a). Since f (x) = 0 on c0 , f is in fact the trivial bounded functional on c0 . By the assumption and part (a), (c0 )∗ is isometric isomorphic to (ℓ∞ )∗ so that the trivial bounded functional on c0 corresponds to the trivial bounded functional on ℓ∞ . In other words, f (x) = 0 on ℓ∞ which contradicts the result (5.32). Hence we conclude that ℓ1 is not isometric isomorphic to (ℓ∞ )∗ . (d) We prove the assertions one by one. – c0 is separable. For each k ∈ N, we define Sk = {{ξ1 , . . . , ξk , 0, 0, . . .} | ξ1 , . . . , ξk ∈ Q} and S =

∞ [

k=1

Sk .

Since each Sk is countable, S is also countable. We claim that S is dense in c0 . Pick y = {ηi } ∈ c0 . Given ǫ > 0. Since ηi → 0 as i → ∞, there exists a positive integer N such that i ≥ N implies |ηi | < ǫ. (5.33) By the density of Q in R, we can take x = {ξ1 , . . . , ξN , 0, 0, . . .} ∈ SN such that |ξi − ηi | < ǫ

(5.34)

for all i = 1, 2, . . . , N . Thus we obtain from the definition of k · k∞ , the inequalities (5.33) and (5.34) that kx − yk∞ = sup |ξi − ηi | ≤ ǫ. i∈N

Hence this proves the claim and then c0 is separable. – ℓ1 is separable. Take y = {ηi } ∈ ℓ1 . Given that ǫ > 0. Since

∞ X i=1

|ηi | < ∞, the

Cauchy Criterion (see [99, Theorem 3.22, p. 59]) shows that one can find a positive integer N ′ such that i ≥ N ′ implies ∞ X

i=N ′

|ηi |
0 be so small such that 4n (x + h) ∈ [0, 1]. Then we obtain ϕn (x + h) − ϕn (x) h
1 ϕ 4n (x + h) − ϕ(4n x) = n· 2 h  n (x + h) − 2 · 4n x 2 · 4 1   n· , if 4n x, 4n (x + h) ∈ [0, 12 ];   h  2     1 1 − 2 · 4n h − 1 = · , if 4n x = 12 and 4n (x + h) = n  2 h      n n    1 · −2 · 4 (x + h) + 2 · 4 x , 4n x, 4n (x + h) ∈ [ 1 , 1], 2 n h  2 n n x, 4n (x + h) ∈ [0, 1 ]; 2 · 2 , if 4  2     −2 · 2n , if 4n x = 12 and 4n (x + h) = 21 + 4n h; =      −2 · 2n , 4n x, 4n (x + h) ∈ [ 12 , 1]

1 2

+ 4n h;

which implies that |ϕ′n (x+)| ≥ 2n . Similarly, |ϕ′n (x−)| ≥ 2n , completing the proof of  Lemma 5.1. • Step 3: Construction of a e g ∈ C with large slopes and ke g − f k < 2ǫ. We define ge = g + ϕk ,

where the k will be determined later. We claim that gb satisfies the required properties. To this end, let Mk = max k|f (xi+1 ) − f (xi )|. 0≤i≤k−1

Now for every x ∈ [0, 1], if we take h > 0 small enough such that x + h ∈ [xi , xi+1 ], then we see from the definition (5.65) and the expression (5.66) that g(x + h) − g(x) |λ x+h − λx | · |f (xi+1 ) − f (xi )| = k|f (xi+1 ) − f (xi )| ≤ Mk = h h

(5.68)

which implies that |g′ (x+)| ≤ Mk on [0, 1]. Similarly, we have |g′ (x−)| ≤ Mk for every x ∈ [0, 1].f f

If x = 1, then g ′ (1+) does not exist. Similarly, if x = 0, then g ′ (0−) does not exist.

5.4. Applications of Baire’s and other Theorems

167

Suppose that N ′ ∈ N. Since f is bounded on [0, 1], we must have Mk ≤ kM for some constant M > 0 and then k can be chosen so large that 2k ≥ kM ≥ Mk + N ′

and

ǫ 1 < . 2k 2

(5.69)

Since g and ϕk are continuous on [0, 1], ge is also continuous on [0, 1]. Next, for x ∈ [0, 1) and h > 0 so small such that x + h ∈ [0, 1], we derive from Lemma 5.1 and the inequality (5.68) that ge(x + h) − ge(x) g(x + h) − g(x) ϕ(x + h) − ϕ(x) + = h h h ϕ(x + h) − ϕ(x) g(x + h) − g(x) ≥ − h h ≥ 2k − Mk > N′

(5.70)

which yields |e g ′ (x+)| > N ′ for all x ∈ [0, 1). Similarly, we have |e g′ (x−)| > N ′ for all x(0, 1]. Since N ′ can be chosen arbitrary large, our e g is continuous function on [0, 1] with large slopes.

Finally, it remains to show that ke g − f k < ǫ. On [0, 1], we know from Steps 1 and 2 that |e g(x) − f (x)| ≤ |g(x) − f (x)| + |ϕk (x)| < ǫ + 2−k |ϕ(4k x)| < ǫ +

ǫ < 2ǫ. 2

Thus we conclude that ke g − f k < 2ǫ. By the definition of Xn and the lower bound (5.70), we note that ge ∈ / Xn if we pick N ′ ≥ n.

• Step 4: Xn is closed in C. Let {fk } ⊆ Xn and fk → f in C. Then one can find a sequence {tk } ⊆ [0, 1] such that |fk (s) − fk (tk )| ≤ n|s − tk |

(5.71)

for every s ∈ [0, 1] and k ∈ N. Since [0, 1] is compact, we may assume that tk → t ∈ [0, 1]. By Theorem 3.17, C is a metric space and then we recall from the rephrased Theorem 7.9 [99, p. 151] that fk → f uniformly on [0, 1]. Hence it follows from [99, Exercise 9, p. 166] that lim fk (tk ) = f (t) k→∞

and we deduce from the inequality (5.71) that |f (s) − f (t)| ≤ n|s − t|, for every s ∈ [0, 1]. In other words, we conclude that f ∈ Xn which means Xn is closed in C. • Step 5: Construction of B(e g , ǫ′ ) ⊆ C such that B(e g, ǫ′ ) ∩ Xn = ∅. By Step 3 and our choice of ǫ > 0, we know that e g ∈ B(f, 2ǫ) ⊆ V and ge ∈ / Xn . Since ge ∈ Xnc and Xnc is open in C by Step 4. There exists a δ > 0 such that B(e g, δ) ⊆ Xnc , i.e., B(e g, δ) ∩ Xn = ∅. g − f k, 2ǫ − ke g − f k) and h ∈ B(e g , ǫ′ ), then we have If we take ǫ′ = 21 min(ke kh − f k ≤ kh − gek + ke g − f k < ǫ′ + ke g − fk ≤

which means B(e g, ǫ′ ) ⊆ B(f, 2ǫ) ⊆ V .

1 (2ǫ − ke g − f k) + ke g − f k < 2ǫ 2

168

Chapter 5. Examples of Banach Space Techniques

Hence we have shown the first assertion. To prove the second assertion, we notice that Xn◦ = ∅; otherwise, since Xn◦ is open in C, the first assertion shows the existence of a non-empty open set G ⊆ Xn◦ ⊆ Xn such that G ∩ Xn = ∅, a contradiction. Hence each Xnc is open and dense in C. Since C is a complete metric space, we establish from Theorem 5.6 (Baire’s Theorem) that the set X=

∞ \

Xnc

n=1

is a dense Gδ in C. Suppose that f ∈ C and f is differentiable at p ∈ [0, 1). Then there exists a δ > 0 such that for all 0 < |h| < δ, we haveg f (p + h) − f (p) − f ′ (p) ≤ 1 h

which implies that

Obviously, if |h| ≥ δ, then

f (p + h) − f (p) ≤ 1 + |f ′ (p)|. h

f (p + h) − f (p) |f (p + h)| + |f (p) 2kf k ≤ . ≤ h h δ

(5.72)

(5.73)

Combining the inequalities (5.72) and (5.73), we obtain that |f (x) − f (p)| ≤ N |x − p|

if N = max(1 + |f ′ (p)|, 2δ−1 kf k). In the case that p = 1, we will replace f (p + h) by f (p − h) in the inequalities (5.72) and (5.73). c Consequently, f ∈ Xn for every n ≥ N . By this, we must have f ∈ / X. Otherwise, f ∈ XN or equivalently, f ∈ / XN which is a contradiction. Hence we have completed the proof of the problem. 

Problem 5.15 Rudin Chapter 5 Exercise 15.

Proof. Recall from Problem 5.9 that c0 is the subspace of ℓ∞ consisting of all x = {ξi } ∈ ℓ∞ for which ξi → 0 as i → ∞. By the definition, we have {σi } = A(x) whenever the series converges (but ξi not necessarily converging to 0). We are going to prove the results one by one: • Necessity part: Suppose that A transforms every convergent sequence {sj } to a sequence {σi } which converges to the same limit. To prove Condition (a), it suffices to prove the case when both {sj } and {σi } converge to 0 because we may replace sj and σi by sj − L and σi − L respectively, where L is the g

Of course, only one-sided derivative f ′ (0+) will be considered if p = 0.

5.4. Applications of Baire’s and other Theorems

169

common limit of {sj } and {σi }. In other words, we work on the space c0 . In fact, for each fixed i ∈ N, we define Λi : c0 → C by Λi (x) = σi =

∞ X

aij sj

j=0

for all x ∈ c0 so that A(x) = {Λi (x)}. Since {σi } is convergent, it is bounded so that |Λi (x)| ≤ M

(5.74)

for all x ∈ c0 and i = 1, 2, . . . for some positive constant M . For every k = 0, 1, 2, . . ., we take sj = δjk (the Kronecker delta function). Then we have sj → 0 as j → ∞ and we obtain ∞ X 0 = lim σi = lim aij sj = lim aik i→∞

i→∞

i→∞

j=0

which is Condition (a). If sj = 1 for all j ∈ N, then sj → 1 as j → ∞ and thus 1 = lim σi = lim i→∞

i→∞

∞ X

aij

j=0

which is Condition (c). To prove Condition (b), we first show that

∞ X j=0

|aij | < ∞ for every i = 1, 2, . . .. Assume

that it was not the case. Then one is able to find a strictly increasing sequence of integers {nk } such that nk+1 X |aij | > k. j=nk +1

Let k ∈ N and define

sj =

 0,  

if j = 1, 2, . . . , nk ;

  sgn (aij ) , if j = n + 1, n + 2, . . . , n . k k k+1 k

(5.75)

Obviously, j → ∞ if and only if k → ∞. Then we have sj → 0 as j → ∞. Thus x = {sj } ∈ c0 and so lim σi = 0. However, we notice from the construction (5.75) that i→∞

∞ nX ∞ k+1 X X |aij | aij sj = |Λi (x)| = |σi | = =∞ k j=0

k=1 j=nk +1

which contradicts the inequality (5.74). Thus we must have ∞ X j=0

|aij | < ∞.

By the inequality (5.74) again, we have Λi ∈ c∗0 and since c0 is Banach, we may apply Theorem 5.8 (The Banach-Steinhaus Theorem) to conclude that kΛi k = sup{|Λi (x)| | x ∈ c0 ⊆ ℓ∞ and kxk∞ = 1} ≤ M

(5.76)

170

Chapter 5. Examples of Banach Space Techniques

for all i ∈ N. Recall that

∞ X j=0

|aij | < ∞ for every i = 1, 2, . . ., so yi = {aij } ∈ ℓ1 for every

i = 1, 2, . . . and Problem 5.9(a) implies that kΛi k = kyi k1 =

∞ X j=0

|aij |

(5.77)

for every i = 1, 2, . . .. Hence Condition (b) follows from the inequality (5.76) and the equality (5.77). • Sufficiency part: Let x = {sj } and sj → L as j → ∞. By Condition (b), the series ∞ X aij sj converges absolutely for every i ∈ N. We write σi = j=0

σi =

∞ X j=0

aij (sj − L) + L ·

By Condition (c), we have lim L ·

i→∞

∞ X

∞ X

aij .

(5.78)

j=0

aij = L.

(5.79)

j=0

Let the supremum in Condition (b) be M . Given ǫ > 0, we choose a N ∈ N such that |sj − L| < Mǫ for all n ≥ N . Thus we see that N ∞ N ∞ X X X X |aij | · |sj − L| + ǫ. |aij | · |sj − L| ≤ |aij | · |sj − L| + aij (sj − L) ≤ j=0

n=N +1

j=0

j=0

By Condition (a), we know that

lim

i→∞

so that

N X j=0

|aij | · |sj − L| = 0

∞ X aij (sj − L) ≤ ǫ. lim

i→∞

j=0

Since ǫ is arbitrary, we conclude from the representation (5.78) and the limit (5.79) that lim σi = L.

i→∞

• Two examples. It is easy to check that the examples satisfy the conditions of the problem. To show the last assertion, we consider the sequence {sj } defined by  √ if j = 2k;  k, sj =  √ − k, if j = 2k − 1.

Then it is clear that {sj } is unbounded and direct computation shows that, for each i = 1, 2, . . ., σi =

∞ X j=0

aij sj

5.5. Miscellaneous Problems

171

=

i X j=0

=

=

1 sj i+1

 2k  1 X    sj , if i = 2k;     2k + 1 j=0

  2k−1  1 X    sj , if i = 2k − 1;   2k j=0   0, if i = 2k;   1    − √ , if i = 2k − 1. 2 k

Since i → ∞ if and only if k → ∞, we have σi → 0 as i → ∞, i.e., {σi } is convergent.

For the other example, pick δ to be a number such that ri < δ < 1 for every i ∈ N. If sj = (−δ)−j , then {sj } is divergent and we have σi =

∞ X j=0

aij sj = (1 − ri )

∞  X j=0

−

ri j 1 − ri = →0 δ 1 + δ−1 ri

as i → ∞. We complete the proof of the problem.



Remark 5.2 Classically, Problem 5.15 is called the Silverman-Toepliotz Theorem. For more information or other proofs about this theorem, please refer to [49, §3.2], [80, Chap. 4] and [133, §1.2].

5.5

Miscellaneous Problems

Problem 5.16 Rudin Chapter 5 Exercise 16.

Proof. We follow Rudin’s hint. Let X ⊕Y = {(x, y) | x ∈ X and y ∈ Y } with addition and scalar multiplication defined componentwise. We define the norm k · k on X ⊕ Y byh k(x, y)k = kxkX + kykY . We check Definition 5.2. For (x1 , y1 ), (x2 , y2 ) ∈ X ⊕ Y , we have k(x1 , y1 ) + (x2 , y2 )k = k(x1 + x2 , y1 + y2 )k

= kx1 + x2 kX + ky1 + y2 kY

≤ kx1 kX + kx2 kX + ky1 kY + ky2 kY = k(x1 , y1 )k + k(x2 , y2 )k.

h

Of course, k · kX and k · kY denote the norms in X and Y respectively.

(5.80)

172

Chapter 5. Examples of Banach Space Techniques

Next, if (x, y) ∈ X ⊕ Y and α is a scalar, then we have kα(x, y)k = k(αx, αy)k = kαxkX + kαykY = |α| · kxkX + |α| · kykY = |α| · k(x, y)k. Finally, if k(x, y)k = 0, then kxkX + kykY = 0. Since kxkX and kykY are nonnegative for every x ∈ X and y ∈ Y , kxkX + kykY = 0 implies that kxkX = kykY = 0 and thus (x, y) = (0, 0). Hence X ⊕ Y is a normed linear space.

Suppose that {(xn , yn )} ⊆ X ⊕ Y is Cauchy. Given ǫ > 0, Then there exists a positive integer N such that n, m ≥ N imply that k(xn , yn ) − (xm , ym )k < ǫ.

By the definition (5.80), we must have kxn − xm kX < ǫ and kyn − ym kY < ǫ for all n, m ≥ N . In other words, {xn } ⊆ X and {yn } ⊆ Y are also Cauchy. Since X and Y are Banach, there exist x ∈ X and y ∈ Y such that kxn − xkX → 0

and kyn − ykY → 0

as n → ∞. Now these limits and the definition (5.80) show that k(xn , yn ) − (x, y)k = k(xn − x, yn − y)k = kxn − xkX + kyn − ykY → 0 as n → ∞. By Definition 5.2, X ⊕ Y ia Banach.
Suppose
that G
= { x, Λ(x) | x ∈ X} ⊆ X ⊕ Y . Furthermore, we let α, β ∈ C and x, Λ(x) , y, Λ(y) ∈ G. Then the linearity of Λ says that




α x, Λ(x) + β y, Λ(y) = αx + βy, αΛ(x) + βΛ(y) = αx + βy, Λ αx + βy ∈ G.

Thus G is a linear subspace of X ⊕ Y
and it is also a metric
space. We claim that G is closed in X ⊕ Y . To see this, let { xn , Λ(xn ) } ⊆ G and k xn , Λ(xn ) − (x, y)k → 0 as n → ∞. By the definition (5.80), we have kxn − xkX → 0 and kΛ(xn ) − ykY → 0 as n → ∞, i.e., x = lim xn n→∞

and y = lim Λ(xn ). n→∞


By the hypothesis, we gain that y = Λ(x) and so (x, y) = x, Λ(x) ∈ G by the definition. Therefore, we have shown the claim that G is closed in X ⊕ Y . Since X ⊕ Y is complete, we deduce from [99, Theorem 3.11, p. 53] that G is also complete and thus Banach. Consider the mapping Φ : G → X defined by
Φ x, Λ(x) = x. Then it is clear that Φ is linear and bijective. Furthermore, for every x ∈ X, we see that

kΦ x, Λ(x) kX = kxkX ≤ kxkX + kΛ(x)kY = k x, Λ(x) k which implies that




kΦk = sup{kΦ x, Λ(x) kX | x, Λ(x) ∈ G and k x, Λ(x) k = 1} ≤ 1,

i.e., Φ is bounded. By Theorem 5.10, Φ−1 is also a bounded linear transformation of X onto G and it follows from Theorem 5.10’s proof that there exists a positive constant δ such that 1 kΦ−1 k ≤ . δ

5.5. Miscellaneous Problems

173

Besides, Theorem 5.4 shows that Φ−1 is continuous. Combining these two facts and [100, Eqn. (3), p. 96], we obtain
1 kxkX + kΛ(x)kY = k x, Λ(x) k = kΦ−1 (x)k ≤ kΦ−1 k · kxkX ≤ kxkX < ∞ (5.81) δ for every x ∈ X. Since kxkX and kΛ(x)kY are nonnegative, we get from the inequality (5.81) that δ < 1 and 1  0 ≤ kΛ(x)kY ≤ − 1 · kxkX (5.82) δ on X. Hence we conclude from the inequality (5.82) that 1 < ∞ − 1. δ Consequently, Λ is bounded and it is continuous on X by Theorem 5.4, completing the proof of the problem.  kΛk = sup{kΛ(x)kY | x ∈ X and kxkX = 1} ≤

Problem 5.17 Rudin Chapter 5 Exercise 17.

Proof. We prove the assertions one by one. • kMf k ≤ kf k∞ . Given f ∈ L∞ (µ). Define Mf : L2 (µ) → L2 (µ) by Mf (g) = f g. By Definition 3.7, we have |f g| ≤ kf k∞ · |g| for all f ∈ L2 (µ) and then Remark 3.10 shows that o1 nZ o1 n Z 2 2 2 ≤ kf k2∞ · |g|2 dµ = kf k∞ · kgk2 . (5.83) kMf (g)k2 = kf gk2 = |f g| dµ X

X

By the inequality (5.83) and Definition 5.3, we obtain immediately that

kMf k = sup{kMf (g)k2 | g ∈ L2 (µ) and kgk2 = 1} ≤ kf k∞ .

(5.84)

• Measures µ with kMf k = kf k∞ for all f ∈ L∞ (µ). We call the measure µ semifinite if for each E ∈ M with µ(E) = ∞ one can find a F ∈ M with F ⊂ E such that 0 < µ(F ) < ∞, see [40, p. 25]. Now we claim that kMf k = kf k∞ for all f ∈ L∞ (µ) if and only if the measure µ is semifinite.

– Suppose that µ is semifinite. Since f ∈ L∞ (µ), we have kf k∞ < ∞. Let α = kf k∞ . If α = 0, then the inequality (5.84) forces that kMf k = kf k∞ = 0 and we are done. If α > 0, then we see from Problem 3.19 that α = max{|z| | z ∈ Rf }, where Rf denotes the essential range of f . Thus there exists a z0 ∈ C such that α = |z0 |. Without loss of generality, we may assume that z0 = α and so µ{x ∈ X | |f (x) − α| < ǫ} > 0

(5.85)

for every ǫ > 0. Combining Definition 3.7 and the result (5.85), we establish that the measure of the set E = {x ∈ R | |f (x)| > α − ǫ} is nonzero.

174

Chapter 5. Examples of Banach Space Techniques ∗ Case (i): µ(E) < ∞. Now it is trivial that the function g=p

1 µ(E)

χE

satisfies the conditions that g ∈ L2 (µ) and kgk2 = 1. Furthermore, we derive from the definition that o 1 n (α − ǫ)2 Z o1 n 1 Z 2 2 |f χE |2 dµ dµ > = α − ǫ. (5.86) kMf (g)k2 = µ(E) E µ(E) E Since ǫ is arbitrary, the estimate (5.86) implies that kMf k = sup{kMf (g)k2 | g ∈ L2 (µ) and kgk2 = 1} ≥ α and hence kMf k = kf k∞ by the inequality (5.84). ∗ Case (ii): µ(E) = ∞. Then the hypothesis ensures that there exists a F ∈ M with F ⊂ E and 0 < µ(F ) < ∞. Now the function 1 g = √ χF F also satisfies the estimate (5.86) and thus kMf k = kf k∞ holds in this case.

– Suppose that µ is not semifinite. Then there is a E ∈ M such that µ(E) = ∞ and every F ⊂ E satisfies either µ(F ) = 0 or µ(F ) = ∞. Take g ∈ L2 (µ) so that Z Z |g(x)|2 dµ < ∞. |g(x)|2 dµ ≤ E

X

If there is a F ⊂ E such that |g(x)| > 0 on F and g(x) = 0 on E \ F , then µ(F ) 6= ∞ by Proposition 1.24(a). Recall that µ is not semifinite, we have µ(F ) = 0. In other words, it must be true that g = 0 a.e. on E. Let f = χE . Then we have f g = 0 a.e. on E for every g ∈ L2 (µ). By the definition, we obtain kMf k = 0. On the other hand, kf k∞ = 1 so that kMf k < kf k∞ . Hence we have proven the claim. • Functions f ∈ L∞ (µ) such that Mf is onto. Suppose that µ satisfies the condition that every measurable set E of positive measure contains a measurable subset F with 0 < µ(F ) < ∞.i We claim that the map Mf is onto if and only if f1 ∈ L∞ (µ). – If Mf is onto, then we let E = {x ∈ X | f (x) = 0}.

i

(5.87)

We claim that µ(E) = 0. Assume that µ(E) > 0. By the hypothesis, there exists a F ⊆ E with 0 < µ(F ) < ∞. Thus χF ∈ L2 (µ). Since f g = 0 on F for every g ∈ L2 (µ), it implies that
χF ∈ / Mf L2 (µ)

Some books take this as the definition of a semifinite measure µ. See, for example, [14, Exercise 25.9]

5.5. Miscellaneous Problems

175

which contradicts the surjective property of Mf . Thus we conclude that µ(E) = 0. If g ∈ L2 (µ) is such that Mf (g) = f g = 0, then it follows from the definition (5.87) that g = 0 a.e. on E c . Since µ(E) = 0, we obtain fact that g = 0 a.e. on X and this means that Mf is one-to-one. Since Mf is assumed to be onto, it is bijective. On the one hand, recall that L2 (µ) is Banach and kMf k ≤ kf k∞ < ∞ by the first assertion, Theorem 5.10 ensures that there corresponds a δ > 0 such that kMf (g)k2 ≥ δkgk2

(5.88)

for every g ∈ L2 (µ). On the other hand, we consider F = {x ∈ X | |f (x)| < 2δ }. If µ(F ) 6= 0, then our hypothesis tells us that there is a G ⊂ F such that 0 < µ(G) < ∞ and thus 1 kχG k2 = [µ(G)] 2 which verifies that kMf (χG )k2 = kf χG k2 =

nZ

2

G

|f | dµ

o1

2

n Z δ2 o 1 1 δ δ 2 < dµ = [µ(G)] 2 = kχG k2 , 4 2 2 G

but it contradicts the inequality (5.88). Hence we have µ(F ) = 0 if and only if 1 ≤ 2δ a.e. on X if and only if f1 ∈ L∞ (µ). |f (x)| ≥ 2δ > 0 a.e. on X if and only if |f (x)|

– Suppose that f1 ∈ L∞ (µ). Then it is clear that M 1 is the inverse operator of Mf so f that Mf is bijective. In particular, Mf is surjective. Hence we have completed the proof of the problem.



Problem 5.18 Rudin Chapter 5 Exercise 18.

Proof. Let x ∈ X and ǫ > 0. Since E is dense in the normed linear space X, we can find y ∈ E such that ǫ . (5.89) kx − yk < 2M Since {Λn (y)} converges in the Banach space Y , there exists a positive integer N such that n, m ≥ N imply that ǫ kΛn (y) − Λm (y)k < . (5.90) 2 Therefore, if n, m ≥ N , then we deduce immediately from the inequalities (5.89) and (5.90) that kΛn (x) − Λm (x)k ≤ kΛn (x) − Λn (y)k + kΛn (y) − Λm (y)k + kΛm (y) − Λm (x)k ǫ < kΛn (x − y)k + + kΛm (y − x)k 2 ǫ ≤ kΛn k · kx − yk + kΛm k · kx − yk + 2 ǫ ǫ < + 2 2 = ǫ. This means that {Λn (x)} is Cauchy in Y . Since Y is Banach, {Λn (x)} converges in Y and we complete the analysis of the problem. 

176

Chapter 5. Examples of Banach Space Techniques

Problem 5.19 Rudin Chapter 5 Exercise 19.

Proof. Given f ∈ C(T ), i.e., f is a continuous complex function on T . Recall that sn (f ; x) =

1 2π

Z

π

−π

f (t)Dn (x − t) dt

and Dn (t) =

n X

k=−n

eikt =

sin(n + 21 )t . sin 2t

(5.91)

By §5.11, C(T ) is Banach relative to the supremum norm kf k∞ .j Next, since f ∈ C(T ), we follow from the equations (5.91) that sn (f ; x) ∈ C(T ) for each n ∈ N. Let X = Y = C(T ) and for each n = 2, 3, . . ., we define Λn : C(T ) → C(T ) by Λn (f ) =

sn (f ; x) . log n

Since f ∈ C(T ), there exists a M > 0 such that |f (x)| ≤ M for all x ∈ T so that Z π Z 1 M π |sn (f ; x)| ≤ |f (t)| · |Dn (x − t)| dt ≤ |Dn (x − t)| dt. 2π −π 2π −π

(5.92)

Since Dn (x − t) = eikx Dn (t) and Dn (t) is an even function, the estimate (5.92) can be replaced by Z Z Z M π M π sin(n + 21 )t M π |Dn (t)| dt = |Dn (t)| dt = (5.93) |sn (f ; x)| ≤ dt. 2π −π π 0 π 0 sin 2t Next, we consider f (x) = sin x2 − x4 on [0, π]. Using differentiation, we can show that sin x2 ≥ x4 on [0, π]. Thus the estimate (5.93) can be further written as Z 4M π | sin(n + 12 )t| |sn (f ; x)| ≤ dt π 0 t 1 Z 4M (n+ 2 )π | sin t| dt = π 0 t Z n Z (k+1)π X 4M  π sin t | sin t|  (5.94) ≤ dt + dt . π t t 0 kπ k=1 1 t

On the interval [kπ, (k + 1)π], it is easy to see that n Z X k=1

(k+1)π

kπ

X 1  | sin t| dt ≤ t kπ n

k=1

Z

0

is bounded by Z

π 0

π 2

sin t dt = t

π 2

1 kπ ,

(k+1)π kπ 2 π

Besides, we get from [99, Problem 8.6, p. 197] that Z

≤

where k = 1, 2, . . . , n, so we have

n  2X 1 | sin t| dt = . π k

(5.95)

k=1


0 such that kΛn k ≤ M ′ for all n = 2, 3, . . .. That is, {Λn } satisfies the first hypothesis of Problem 5.18.

By Theorem 4.25 (The Weierstrass Approximation Theorem), the set of all trigonometric polynomials, namely P, is dense in C(T ). Let Pm (t) = eimt for some m ∈ Z. If n ≥ m, then we follow from the result [100, Eqn. (8), p. 89] that sn (Pm ; x) =

Therefore, if P (t) =

N X

1 2π

Z

π

e−imt

−π

n X

eikt dt =

Z π n X 1 ei(k−m)t dt = 1. 2π −π

(5.98)

k=−n

k=−n

cm Pm (t), then we get immediately from the result (5.98) that

m=−N

sn (P ; x) =

N X

cm sn (Pm ; x) =

N X

cm

m=−N

m=−N

for every n ≥ N and this implies that Λn (P ) =

N X 1 sn (P ; x) cm → 0 = · log n log n m=−N

as n → ∞. Thus {Λn } satisfies the second hypothesis of Problem 5.18. Hence we establish from Problem 5.18 that {Λn (f )} converges to 0 for each f ∈ C(T ) and as a consequence, we have lim

n→∞

ksn k∞ = lim kΛn k = 0. n→∞ log n

For the second assertion, we note that if we define Λn f =

sn (f ; 0) , λn

then we may apply an argument similar to that used in §5.11 to show that kΛn k =

kDn k1 |λn |

holds. Furthermore, we follow from the hypothesis

λn log n

→ 0 as n → ∞ thatk

4 X1 4 4  log n γ  kDn k1 →∞ ≥ 2 ≥ 2 (log n + γ) = 2 + |λn | π |λn | k π |λn | π |λn | |λn | n

k=1

k

We have applied the estimate of kDn k1 used in [100, p. 102] in the first inequality in (5.99).

(5.99)

178

Chapter 5. Examples of Banach Space Techniques

as n → ∞, where γ is the famous Euler constant. Hence Theorem 5.8 (the Banach-Steinhaus Theorem) ensures that the sequence n s (f ; 0) o n λn

is unbounded for every f in some dense Gδ set in C(T ), completing the proof of the problem.  Problem 5.20 Rudin Chapter 5 Exercise 20.

Proof. (a) Assume that such a sequence of continuous positive functions {fn } existed. Let x ∈ R and n, k ∈ N. Furthermore, let U = {x ∈ R | {fn (x)} is unbounded} and We claim that U=

Uk = {x ∈ R | fn (x) > k for some n ∈ N}. ∞ \

Uk .

(5.100)

k=1

On the one hand, if x ∈ Uk for all k ∈ N, then for every k ∈ N, there exists a n ∈ N such that fn (x) > k. In other words, we have x ∈ U . On the other hand, if x ∈ U , then since {fn (x)} is unbounded, for every k ∈ N, there exists a n ∈ N such that fn (x) > k which is equivalent to saying that x ∈ Uk for all k ∈ N. Thus this proves the claim. Next, it is easy to see that ∞ [ {x ∈ R | fn (x) > k}. Uk = n=1

Since each fn is continuous, the set {x ∈ R | fn (x) > k} is open in R and therefore, each Uk is open in R. By the definition (5.100), U is a Gδ set. Thus we follow from the assumption that Q=U is also a Gδ set. We deduce from the definition (5.100) that R = Q ∪ Qc = Q ∪

∞ [

k=1

Ukc =

∞ [

{qk } ∪

k=1

∞ [

Ukc

(5.101)

k=1

Clearly, each qk is a nowhere dense subset of R. Furthermore, it is trivial that each Ukc is closed in R. Assume that Ukc0 was not a nowhere dense subset of R for some k0 . Let Vk0 be a nonempty open subset of Ukc0 . Then there exists a p ∈ Vk0 and a δ > 0 such that (p − δ, p + δ) ⊆ Vk0 ⊆ Ukc0 ⊆ Qc , but this means that Q ∩ Qc 6= ∅, a contradiction. Thus every Ukc is also a nowhere dense subset of R and the representation (5.101) shows that R is a set of the first category which contradicts Theorem 5.6 (Baire’s Theorem) that no complete metric space is of the first category. (b) Suppose that Q = {q1 , q2 , . . .} and for each n ∈ N, we define fn : R → R by fn (x) = min(n|x − q1 | + 1, n|x − q2 | + 2, . . . , n|x − qn | + n).

(5.102)

5.5. Miscellaneous Problems

179

It is clear that fn is continuous on R and fn (x) ≥ 1 for all x ∈ R, i.e., {fn } is a sequence of continuous positive functions on R. Suppose that θ is irrational and N ∈ N. Consider the number α = min(|θ − q1 |, |θ − q2 |, . . . , |θ − qN |) > 0. Then the Archimedean Property ([99, Theorem 1.20(a)]) implies that there is a positive integer n such that nα > N and thus fn (θ) = min(n|θ − q1 | + 1, n|θ − q2 | + 2, . . . , n|θ − qn | + n) > N + 1. Since N is arbitrary, we have fn (θ) → ∞ as n → ∞ so that {fn (θ)} is unbounded.

To prove the other direction (i.e., {fn (x)} is unbounded implies x is irrational), we prove its contrapositive. Let qk be a rational number. If n ≥ k, then we have fn (qk ) = min(n|qk − q1 | + 1, . . . , n|qk − qk | + k, . . . , n|qk − qn | + n) ≤ k
so that fn (qk ) ≤ min f1 (qk ), . . . , fk−1 (qk ), k . In other words, the set {fn (qk )} is bounded.

(c) The sequence of the functions given by (5.102) shows that the assertion is true for irrationals. For the rational numbers, we first prove the case on [0, 1].l To begin with, suppose that Qn = {q1 , q2 , . . . , qn } ⊆ [0, 1] ∩ Q, where q1 = 0. Furthermore, we let δn =

1 min{|qi − qj | | 1 ≤ i < j ≤ n} > 0. 2n+1

Clearly, we have lim δn = 0 and

n→∞

(qi − δn , qi + δn ) ∩ (qj − δn , qj + δn ) = ∅

(5.103)

for all 1 ≤ i < j ≤ n. (If qi = 0 or qi = 1, then (qi − δn , qi + δn ) are replaced by [0, δn ) or (1 − δn , 1] respectively.) n [ (qi − δn , qi + δn ) and fn : [0, 1] → R is defined by Suppose that En = i=1

 n  (x − qi + δn ), if x ∈ (qi − δn , qi ] and 1 ≤ i ≤ n;   δn     n fn (x) = − (x − qi − δn ), if x ∈ [qi , qi + δn ) and 1 ≤ i ≤ n;  δ  n      0, if x ∈ [0, 1] \ En .

Thus fn is a continuous function on [0, 1], zig-zag on (qi − δn , qi + δn ) and fn (qi ) = n for each i = 1, 2, . . . , n.m Therefore, if x ∈ [0, 1] ∩ Q, then x = qk for some k ∈ N and thus we obtain lim fn (x) = ∞. n→∞

Next, suppose that x is irrational. Assume that there was an N ∈ N such that x ∈ En for all n ≥ N . By the definition, it means that x ∈ (qk − δN , qk + δN ) l m

The following argument is stimulated by the papers of Fabrykowski [37] and Myerson [77]. The graph of the fn looks like the graph shown in Figure 2.1.

(5.104)

180

Chapter 5. Examples of Banach Space Techniques for some k ∈ {1, 2, . . . , N }. Since x ∈ En for all n ≥ N , the set relation (5.104) shows that x ∈ (qk − δn , qk + δn ) for all n ≥ N . By the limit (5.103), we know that x = qk ∈ Q, a contradiction. Hence we must have x ∈ / En for infinitely many n, i.e., fn (x) = 0 for infinitely many n so that lim fn (x) = ∞ is impossible. n→∞

In conclusion, we have constructed the sequence of continuous functions fn : [0, 1] → R which satisfies fn (x) → ∞ as n → ∞ if and only if x ∈ Q. If we suppose that fn is a function of period 1, then the domain of fn can be extended to R and we obtain the desired result. Hence we have completed the proof of the problem.



Problem 5.21 Rudin Chapter 5 Exercise 21.

Proof. Since Q is a countable union of closed √ sets of R, we have Q ∈ B. It is well-known that m(Q) = 0. Obviously, the translate Q + 2 does not intersect Q. This gives an affirmative answer to the first assertion. Assume that there was a homeomorphismn h : R → R such that e = h(E) ∩ E 6= ∅ E

for every measurable E ⊆ R with m(E) = 0. Given Q = {qk } and ǫ > 0. We first construct a particular measurable set E with m(E) = 0. To this end, for all qk ∈ Q, we consider the neighborhoods (qk − 2−k ǫ, qk + 2−k ǫ) and their union E(ǫ) =

∞ [

(qk − 2−k ǫ, qk + 2−k ǫ).

k=1

It is easy to see that every
E(ǫ) is nonempty open in R. In addition, since Q ⊆ E(ǫ), each E(ǫ) is dense in R with m E(ǫ) ≤ 2ǫ. Hence it follows from Theorem 5.6 (Baire’s Theorem) that the set ∞ 1 \ (5.105) E E= n n=1

is a dense Gδ set in R and m(E) = 0.

Suppose that E is the set (5.105). Since h is a homeomorphism, it is an open map which implies that ∞    \ 1 h E h(E) = n n=1


e is also a dense Gδ set is also a dense Gδ set in R with m h(E) = 0. Thus the intersection E
1 e ⊆E of measure zero. Since E n is countable for all n ∈ N, E is also countable and thus E is a countable dense Gδ in R of measure zero. However, R is a complete metric space which e certainly contradicts Theorem 5.13. Hence no such has no isolated points, the existence of E  homeomorphism h exists and this completes the proof of the problem. n

By the definition, h is a continuous bijection and h−1 is continuous.

5.5. Miscellaneous Problems

181

Problem 5.22 Rudin Chapter 5 Exercise 22.

Proof. Suppose that f ∈ C(T ), f ∈ Lip α and f (0) = 0. We need to show that lim sn (f ; x) = f (x).

n→∞

(5.106)

To achieve the goal, we first quote the following stronger form of the Riemann-Lebesgue Lemma whose proof can be found in [5, Theorem 11.16, p. 313]. Lemma 5.2 (Riemann-Lebesgue Lemma) For every f ∈ L1 (T ), we define 1 fb(n) = 2π

Z

π

f (t)e−i(n+β)t dt,

−π

where β ∈ R. Then we have fb(n) → 0 as |n| → ∞. To begin with, since f ∈ C(T ), we have f ∈ L1 (T ). Furthermore, the hypothesis f ∈ Lip α implies that |f (s) − f (t)| ≤ Mf |s − t|α for all s, t ∈ [−π, π], where Mf is finite and α ∈ (0, 1]. Particularly, take s = 0 so that |f (t)| ≤ Mf |t|α for all t ∈ [−π, π] which implies Z π Z π f (t) i 2M π α hZ 0 f α−1 α−1 α−1 tα−1 dt = < ∞. |t| dt = Mf (−1) t dt + dt ≤ Mf t α 0 −π −π −π

Z

π

f (t) t

In other words, we have

∈ L1 (T ). Since we have

 i 1 1 h i(n+ 1 )t 1 2 t= e sin n + − e−i(n+ 2 )t , 2 2i it yields Z π Z π 1 Z π sin(n + 12 )t 1 f (t) i(n+ 1 )t f (t) −i(n+ 1 )t 1 2 2 dt = e e f (t) · dt − dt π −π t 2πi −π t 2πi −π t Z Z 1 π π f (t) i(n+ 1 )t 1 f (t) −i(n+ 1 )t 2 2 dt + dt . ≤ e e 2π −π t 2π −π t

(5.107)

By Lemma 5.2, each of the integrals on the right-hand side of the inequality (5.107) tends to 0 as |n| → ∞. Thus it is true that 1 π

Z

π

−π

f (t) ·

sin(n + 12 )t dt → 0 = f (0) t

as |n| → ∞.o

Next, we claim that

o

Integrals of the form

Z

0

sn (f ; 0) → f (0) b

g(t)

(5.108)

sin αt dt are called Dirichlet integrals, where α > 0 and g is defined on [0, b]. t

182

Chapter 5. Examples of Banach Space Techniques

as n → ∞. To this end, we consider

Z Z π Z sin(n + 12 )t 1 sin(n + 12 )t 1 π 1 π dt = dt f (t) · f (t)Dn (t) dt − f (t) sn (f ; 0) − π −π t 2π −π π −π t 1 Z π 1    1 1 = − t f (t) sin n + t dt t 2π −π sin 2 2 2 1 Z π  1  t dt , (5.109) F (t)f (t) sin n + = 2π −π 2

where F : [0, π] → R is defined by F (t) =

  

1 sin 2t

− 1t , if t ∈ [−π, π] \ {0};

  0,

2

if t = 0.

Clearly, F is continuous on [−π, π] and so F f ∈ L1 (T ). Using Lemma 5.2 to the right-hand side of the equation (5.109), we see that Z π  1 1 t dt = 0 F (t)f (t) sin n + lim n→∞ 2π −π 2 and this guarantees the validity of the claim (5.108). For the general case, we consider the function g : [−π, π] → R defined by g(t) = f (x + t) − f (x)

(5.110)

for every x ∈ R. Clearly, the real function g satisfies the conditions g ∈ C(T ), g ∈ Lip α and g(0) = 0. By the above argument, we obtain lim sn (g; 0) = 0.

n→∞

(5.111)

By the definition (5.110), we gain
sn (g; 0) = sn f (x + t) − f (x); 0 Z π Z π 1 1 = f (x + t)Dn (−t) dt − f (x)Dn (−t) dt 2π −π 2π −π Z π 1 f (x + t)Dn (−t) dt − f (x) = 2π −π Z π 1 = f (x − t)Dn (t) dt − f (x) 2π −π = sn (f ; x) − f (x).

(5.112)

By applying the limit (5.111) to the equation (5.112), we obtain our desired result (5.106),  completing the proof of the problem.

CHAPTER

6

Complex Measures

6.1

Properties of Complex Measures

Problem 6.1 Rudin Chapter 6 Exercise 1.

Proof. By the definition, we have λ(E) = sup

n nX k=1

k o [ Ej . |µ(Ei )| E1 , E2 , . . . , Ek are mutually disjoint and E = j=1

By Definition 6.1, we have |µ|(E) = sup

∞ nX k=1

∞ o [ Ej . |µ(Ei )| E1 , E2 , . . . are mutually disjoint and E = j=1

Obviously, we have λ(E) ≤ |µ|(E) for every E ∈ M.

For the other direction, we suppose that {Ei } is a partition of E ∈ M. Given ǫ > 0. Since µ is a complex measure, we get from Definition 6.1 that the series ∞ X

µ(Ei )

i=1

converges absolutely. Thus there exists a N ∈ N such that ∞ X

n=N

|µ(En )| < ǫ.

(6.1)

′ ′ Define E1′ = E1 , E2′ = E2 , . . . , EN −1 = EN −1 and EN = EN ∪ EN +1 ∪ · · · . By Definition 1.3(a), ′ we have EN ∈ M. Then we have ∞ [

i=1

′ ′ Ei = E1′ ∪ E2′ ∪ · · · ∪ EN −1 ∪ EN

which implies, with the aid of the estimate (6.1), that ∞ X i=1

|µ(Ei )| =

N −1 X i=1

|µ(Ei′ )| +

∞ X

i=N

|µ(Ei )|
n En Thus we must have λ(En ) ∈ [0, ∞) or equivalently, En is finite or En = ∅. Let E=

∞ [

n=1

En = {x ∈ (0, 1) | h(x) > 0}.

(6.5)

Therefore, the set E is countable. There are two cases for consideration: • Case (i): E 6= ∅. Let E = {x1 , x2 , . . .}. On the one hand, we know from [99, Remark 11.11(f), p. 309] that µ(F ) = 0 for every countable subset F ⊂ (0, 1), so we deduce from the representation (6.4) and the definition (6.5) that Z ∞ X h(xn ) 6= 0, h dλ = 0 = µ(E) = E

n=1

a contradiction. • Case (ii): E = ∅. In this case, we have h(x) = 0 a.e. [λ] on (0, 1) and the representation (6.4) again shows that Z
h dλ = 0, 1 = µ (0, 1) = (0,1)

a contradiction.

a

See Definition 1.35 for the meaning of the notation a.e. [λ].

6.1. Properties of Complex Measures

185

Hence no such h exists and this ends the proof of the problem.



Problem 6.3 Rudin Chapter 6 Exercise 3.

Proof. We first show that if µ and λ are complex regular Borel measures, then both µ + λ and αµ are complex regular Borel measures too. By §6.18, it is equivalent to show that both |µ + ν| and |αµ| are regular Borel measures. To this end, we follow from §6.18 again that both |µ| and |λ| are regular Borel measures. Let E ∈ B and ǫ > 0. By Definition 2.15, there exist open sets V1 , V2 ⊇ E such that |µ|(V1 ) < |µ|(E) +

ǫ 2

ǫ and |λ|(V2 ) < |λ|(E) + . 2

(6.6)

Similarly, there are compact sets K1 , K2 ⊆ E such that |µ|(E) < |µ|(K1 ) +

ǫ 2

ǫ and |λ|(E) < |λ|(K2 ) + . 2

(6.7)

The triangle inequality certainly implies |µ(E) + λ(E)| ≤ |µ(E)| + |λ(E)| and then Definition 6.1 gives ∞ ∞ X X   |µ(E)| + |λ(E)| , (6.8) |µ(E) + λ(E)| ≤ sup |µ + λ|(E) = sup i=1

i=1

where the supremum being taken over all partitions {Ei } of E. Since the series ∞ X i=1

∞ X i=1

|µ(E)| and

|λ(E)| converge, we deduce from the expression (6.8) that |µ + λ|(E) ≤ sup

∞ X i=1

|µ(E)| + sup

∞ X i=1

|λ(E)| = |µ|(E) + |λ|(E).

Thus we have |µ + λ| ≤ |µ| + |λ|.

(6.9)

Let V = V1 ∩ V2 and K = K1 ∪ K2 . Then K is compact and V is open in X. Now we follow from the estimates (6.6) and (6.7) and the inequality (6.9) that |µ + λ|(V ) = |µ + λ|(E) + |µ + λ|(V \ E)

≤ |µ + λ|(E) + |µ|(V \ E) + |λ|(V \ E)

≤ |µ + λ|(E) + |µ|(V1 \ E) + |λ|(V2 \ E)

< |µ + λ|(E) + ǫ.

Since ǫ and E are arbitrary, the measure |µ + λ|(E) is in fact outer regular. Similarly, we have |µ + λ|(K) = |µ + λ|(E) − |µ + λ|(E \ K)

≥ |µ + λ|(E) − |µ + λ|(E \ K1 ) − |µ + λ|(E \ K2 ) > |µ + λ|(E) − ǫ.

In other words, |µ + λ| is also inner regular. By Definition 2.15, µ + λ is a regular complex Borel measure, i.e., µ + λ ∈ M (X). Now the regularity of |αµ| is easy to prove, so we omit the details here.

186

Chapter 6. Complex Measures

Now it is time to prove the assertion in the question. Since X is a locally compact Hausdorff space, Theorem 6.19 (The Riesz Representation Theorem) ensures every Φ ∈ C0 (X)∗ is represented by a unique µΦ ∈ M (X) in the sense that Z f dµΦ (6.10) Φ(f ) = X

for every f ∈ C0 (X). By [100, Eqn. (3), p. 130], we see that Z Z Z Z f d(µΦ + µΨ ) f dµΨ = f dµΦ + f dµΦ+Ψ = (Φ + Ψ)(f ) = Φ(f ) + Ψ(f ) = X

X

X

and

Therefore, they imply that

Z

f dµαΦ = αΦ(f ) =

Z

X

f d(αµΦ ).

X

X

µΦ+Ψ = µΦ + µΨ

and

µαΦ = αµΦ .

(6.11)

By the previous analysis, we see that µΦ+Ψ , µαµ ∈ M (X), so we may define the mapping F : C0 (X)∗ → M (X) by F (Φ) = µΦ and it is easy to see from the results (6.11) that F (Φ + Ψ) = F (Φ) + F (Ψ) and

F (αΦ) = αF (Φ).

Furthermore, Theorem 6.19 (The Riesz Representation Theorem) also implies that F is a bijection and kΦk = |µΦ |(X) = kµΦ k = kF (Φ)k. Consequently, F is actually an isometric vector space isomorphism, i.e., C0 (X)∗ ∼ = M (X). Since C0 (X) is a Banach space with the supremum norm, Problem 5.8 guarantees that C0 (X)∗ is Banach. Hence M (X) is Banach and we have completed the proof of the problem. 

6.2

Dual Spaces of Lp (µ)

Problem 6.4 Rudin Chapter 6 Exercise 4.

Proof. Since µ is positive and σ-finite, we can write X=

∞ [

Xn ,

(6.12)

n=1

where {Xn } is an increasing sequence of measurable sets and µ(Xn ) < ∞ for all n ∈ N, see [105, Definition 4.22, p. 22]. Let A = {x ∈ X | |g(x)| = ∞} and An = {x ∈ Xn | |g(x)| = ∞}, where n ∈ N. Since g is measurable, every An and A are measurable by Problem 1.5. Besides, it is evident that A1 ⊆ A2 ⊆ · · · and if x ∈ A, then x ∈ XN0 for some N0 ∈ N so that x ∈ AN0 . As a result, we get ∞ [ An . A= n=1

Now we are going to divide the proof into several steps:

6.2. Dual Spaces of Lp (µ)

187

• Step 1: µ(A) = 0. Otherwise, it follows from the construction and Theorem 1.19(c) that lim µ(An ) = µ(A) > 0.

n→∞

This means that one can find a N1 ∈ N such that 0 < µ(AN1 ) ≤ µ(XN1 ) < ∞. We know that χAN1 ∈ Lp (µ), so our hypothesis implies that χAN1 g ∈ L1 (µ), but Z Z |g| dµ = µ(AN1 ) · ∞ = ∞ |χAN1 g| dµ = kχAN1 gk1 = AN1

X

which is a contradiction. Hence µ(A) = 0, i.e., g is finite a.e. on X. • Step 2: g ∈ Lq (µ) when 1 < p < ∞. In this case, we have q ∈ (1, ∞). Here we define En = {x ∈ Xn | |g(x)| ≤ n}. Obviously, {En } is also an increasing sequence of measurable sets. Furthermore, if x0 ∈ X \ A, then x0 ∈ XN2 for some N2 ∈ N by the definition (6.12) so that x0 ∈ Xn for all n ≥ N2 . Since g is finite a.e. on X, there exists a positive integer N3 such that |g(x)| ≤ N3 for almost all x ∈ X. Take N4 = max(N2 , N3 ), then x0 ∈ XN4 and |g(x0 )| ≤ N4 which imply that x0 ∈ EN4 . Consequently, we have shown that ∞ [

X=

En .

n=1

Let gn = χEn g : X → C, where n = 1, 2, . . .. Then each gn is measurable on X and Z Z q q |g|q dµ ≤ nq µ(En ) ≤ nq µ(Xn ) < ∞ (6.13) |gn | dµ = kgn kq = En

X

for each n = 1, 2, . . .. Thus we have gn ∈ Lq (µ) for all n ∈ N. Next, we define Λn : Lp (µ) → C by Z f gn dµ

Λn (f ) =

X

which is linear for n ∈ N. Since f gn = (f χEn )g and f χEn ∈ Lp (µ), we have f gn ∈ L1 (µ). By Theorem 1.33 and Theorem 3.8, we obtain Z Z |f gn | dµ = kf gn k1 ≤ kf kp × kgn kq , (6.14) f gn dµ ≤ X

X

where n = 1, 2, . . .. By Definition 5.3 and the inequality (6.14), we know that kΛn k = sup{|Λn (f ) | f ∈ Lp (µ) and kf kp = 1} ≤ kgn kq , where n = 1, 2, . . .. Recall that 1 < q < ∞, so we may consider −q

f0 = kgn kq p |gn |q−2 gn . (q−1)p so that Then we have |f0 |p = kgn k−q q |gn | Z Z Z (q−1)p −q |f0 |p dµ = kgn k−q |g | dµ = kg k |gn |q dµ = 1 kf0 kp = n n q q X

X

and Λn (f0 ) =

Z

X

X

−q

kgn kq p |gn |q−2 gn · gn dµ

(6.15)

188

Chapter 6. Complex Measures − pq

= kgn kq

− pq

= kgn kq

= kgn kq

Z

X

|gn |q dµ

× kgn kqq

(6.16)

for all n ∈ N. Combining the results (6.13), (6.15) and (6.16), we have established the fact that kΛn k = kgn kq < ∞, (6.17) where n ∈ N. Hence {Λn } is a family of bounded linear transformations of Lp (µ) into C. Since Lp (µ) is Banach (see Definition 5.2) and Z Z Z Z |f g| dµ = kf gk1 < ∞ |f g| dµ ≤ |f gn | dµ = f gn dµ ≤ |Λn (f )| = X

X

En

X

for every f ∈ Lp (µ) and n ∈ N, Theorem 5.8 (The Banach-Steinhaus Theorem) implies that there is a M > 0 such that kΛn k ≤ M (6.18) for all n ∈ N. Using the results (6.17) and (6.18), we conclude that kgn kq ≤ M

(6.19)

for all n ∈ N.

By Step 1 and the definition of gn , we have |gn (x)| ≤ |g(x)| < ∞ a.e. on X for all n ∈ N and |g1 (x)| ≤ |g2 (x)| ≤ · · · for every x ∈ X. Now the Monotone Convergence Theorem [99, Theorem 3.14, p. 55] implies that |gn (x)| → |g(x)| < ∞ a.e. on X as n → ∞. Hence we gain from Theorem 1.26 (Lebesgue’s Monotone Convergence Theorem) that Z Z q q |g|q dµ = kgkqq . (6.20) |gn | dµ = lim kgn kq = lim n→∞

n→∞ X

X

Hence it follows from the inequality (6.19) and the limit (6.20) that kgkq ≤ M < ∞, i.e. g ∈ Lq (µ). • Step 3: g ∈ L∞ (µ) when p = 1. In this case, q = ∞. We recall from Step 1 that g is finite a.e. on X, so there exists a M > 0 such that |g(x)| ≤ M a.e. on X. By Definition 3.7, we assert that kgk∞ ≤ M , i.e., g ∈ L∞ (µ). • Step 4: g ∈ L1 (µ) when p = ∞. Define f1 : X → C by

 g(x)    , if g(x) 6= 0; |g(x)| f1 (x) =    0, otherwise.

Since |f1 (x)| ≤ 1 for all x ∈ X, we get from Definition 3.7 that kf1 k∞ ≤ 1, i.e., f1 ∈ L∞ (µ). By the hypothesis, we have f1 g ∈ L1 (µ). Since f1 g = |g|, we conclude that g ∈ L1 (µ).

6.2. Dual Spaces of Lp (µ)

189

Now we have completed the proof of the problem.



Problem 6.5 Rudin Chapter 6 Exercise 5.

Proof. The answer is negative. On the one hand, if f1 , f2 : X → C are defined by f1 (a) = 0,

f1 (b) = 1 and f2 (a) = 1,

f2 (b) = 0,

(6.21)

then we have f1 , f2 ∈ L∞ (µ) and f1 6≡ f2 . For every f ∈ L∞ (µ), we have f (a) = A + Bi and f (b) = C + Di for some A, B, C, D ∈ R. Obviously, we get from the definition (6.21) the representation f = (C + Di)f1 + (A + Bi)f2 . As a result, L∞ (µ) is a two-dimensional space spanned by f1 and f2 , i.e., dim L∞ (µ) = 2. On the other hand, if f ∈ L1 (µ), then kf k1 < ∞ and we follow from Definition 3.6 that f (b) = 0 and kf k1 = |f (a)| = |f (a)|f2 (a). Therefore, L1 (µ) is an one-dimensional space spanned by f2 . As a vector space (see Remark 3.10), we have dim L1 (µ)∗ = dim L1 (µ) = 1. Hence, L∞ (µ) 6= L1 (µ)∗ , completing the proof of the problem.



Problem 6.6 Rudin Chapter 6 Exercise 6.

Proof. We want to show that Lp (µ)∗ ∼ = Lq (µ) for 1 < p < ∞. Equivalently, we have to show p ∗ that for each Λ ∈ L (µ) , there exists a unique g ∈ Lq (µ) such that Z f g dµ (6.22) Λf = X

for all f ∈ Lp (µ). The following proof follows mainly Follan’s argument ([40, p. 190]): • Step 1: µ is finite. Let s be a simple function on X. Since µ is finite, we have µ({x ∈ X | s(x) 6= 0}) ≤ µ(X) < ∞ and then it follows from Theorem 3.13 that s ∈ Lp (µ). Let Λ ∈ Lp (µ)∗ , E be measurable and λ(E) = Λ(χE ). For every partition {Ei } of E, if we let Fn = E1 ∪ E2 ∪ · · · ∪ En , then we obtain Z Z Z χE\Fn dµ = µ(E \ Fn ) (6.23) |χE\Fn |p dµ = |χE − χFn |p dµ = X

X

X

for all n ∈ N. In fact, the expression (6.23) can be rewritten as 1

kχE − χFn kp = µ(E \ Fn ) p .

(6.24)

Since E \ F1 ⊇ E \ F2 ⊇ · · · and µ(E \ F1 ) is finite, it establishes from Theorem 1.19(e) that ∞ \  lim µ(E \ Fn ) = µ (E \ Fn ) = µ(∅) = 0. (6.25) n→∞

n=1

190

Chapter 6. Complex Measures Combining the results (6.24) and (6.25) and using the fact p ∈ (1, ∞), we derive that lim kχFn − χE kp = 0.

n→∞

Since χE ∈ Lp (µ) and each χEi g is measurable, we deduce from the representation (6.22) and then Theorem 1.27 that Z X Z ∞ ∞ Z ∞ X X λ(χEi ). χEi g dµ = (χEi g) dµ = χE g dµ = λ(E) = Λ(χE ) = X i=1

X

i=1

X

i=1

By Definition 6.1, λ is a complex measure. If E ∈ M satisfies µ(E) = 0, then χE = 0 in Lp (µ) and so λ(E) = Λ(0) = 0. By Definition 6.7, we have λ ≪ µ. Thus we know from Theorem 6.10 (The Lebesgue-RadonNikodym Theorem) that there exists a unique h ∈ L1 (µ) such that Z Z χE g dµ g dµ = Λ(χE ) = λ(E) = E

for every E ∈ M. Recall that Λ ∈

X

Lp (µ)∗ ,

Λ is linear and bounded so that Z sg dµ Λ(s) = X

Lp (µ).

for every simple function s ∈ By [100, Eqn. (3), p. 96], we know that Z sg dµ = kΛ(s)k ≤ kΛk · kskp < ∞. X

By [40, Theorem 6.14, p. 189], we conclude that g ∈ Lq (µ). Given that f ∈ Lp (µ). By Theorem 3.8, we have f g ∈ L1 (µ). By Theorem 3.13, there exists a sequence of simple functions {sn } such that |sn | ≤ |f | for every n ∈ N and sn → f in Lp (µ) as n → ∞. Since |sn g| ≤ |f g| and f g ∈ L1 (µ), we obtain from Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) and Theorem 5.4 that Z Z f g dµ. sn g dµ = Λ(f ) = lim Λ(sn ) = lim n→∞

n→∞ X

X

• Step 2: µ is σ-finite. This is exactly the hypothesis of Theorem 6.16, so we have Lp (µ)∗ ∼ = Lq (µ) in this case. • Step 3: µ is any measure and p > 1. In this case, q ∈ (1, ∞). As in Step 2, for each σ-finite subset E of X, there corresponds a unique gE ∈ Lq (E) such that Z f gE dµ Λ(f ) = X

Lp (E),

Lp (E)

for all f ∈ where = {f ∈ Lp (µ) | f (x) = 0 for all x ∈ / E} and Lq (E) is defined similarly. By Theorem 6.16, we know that

kgE kq = Λ|Lp (E) ≤ kΛk. (6.26) If F is a σ-finite subset of X containing E, then the uniqueness of gE implies that gF = gE a.e. on E so that kgE kq ≤ kgF kq . (6.27) Now we define M = sup{kgE kq | E a σ-finite subset of X}.

6.3. Fourier Coefficients of Complex Borel Measures

191

Then it follows from the inequality (6.26) that M ≤ kΛk holds. By the definition, we may select a sequence {En } of σ-finite subsets of X such that ∞ [ En which is obviously σ-finite subset of X so kgEn kq → M as n → ∞. Define F = n=1

that kgF kq ≤ M . Furthermore, we deduce from the inequality (6.27) that kgEn kq ≤ kgF kq

for all n ∈ N. Therefore, the definition of M implies that M ≤ kgF kq and then M = kgF kq . Finally, if A is a σ-finite subset of X containing F , we get Z Z Z Z |gF |q dµ |gA |q dµ = kgA kqq ≤ M q = |gA\F |q dµ = |gF |q dµ + X

X

X

X

which means that gA\F = 0 a.e. on X or equivalently, gA = gF a.e. on X.

(6.28)

If f ∈ Lp (µ), then the set A = F ∪ {x ∈ X | f (x) 6= 0} is clearly σ-finite. Therefore, we observe from the result (6.28) that Z Z f gF dµ. f gA dµ = Λ(f ) = X

X

Hence we may pick g = gF and the above argument makes sure this g is unique. We complete the proof of the problem.

6.3



Fourier Coefficients of Complex Borel Measures

Problem 6.7 Rudin Chapter 6 Exercise 7.

Proof. If µ is a real measure, then we have Z Z b(n) µ b(−n) = eint dµ(t) = e−int dµ(t) = µ

for every n ∈ Z. Therefore, our result follows in this case. Since µ is a complex Borel measure, Theorem 6.12 implies the existence of a Borel measurable function h such that |h(t)| = 1 on [0, 2π) and dµ = h d|µ|. Thus we have d|µ| = 1 · d|µ| = h · h d|µ| = h dµ. Since |µ| is real, the previous paragraph and the expression (6.29) imply that Z Z Z µ b(−n) = eint h d|µ| = e−int h(t) d|µ|(t) = e−int [h(t)]2 dµ(t)

(6.29)

192

Chapter 6. Complex Measures

for every n ∈ Z. By Theorem 3.14, C(T ) is dense in L1 (|µ|). Using Theorem 4.25 (The Weierstrass Approximation Theorem) and then Lemma 2.10, we deduce that the set of all trigonometric polynomials P is dense in L1 (|µ|). We want to apply Problem 5.18. For any f ∈ L1 (|µ|), we define Λn : L1 (|µ|) → C by Z Λn (f ) = f (t)e−int dµ(t) for all n ∈ N. Then we have Z Z Z |Λn (f )| ≤ |f | · | dµ(t)| = |f | · |h| d|µ(t)| = |f | d|µ(t)| = kf k1 .

(6.30)

By Definition 5.3, we see from the inequality (6.30) that kΛn k = sup{|Λn (f )| | f ∈ L1 (|µ|) and kf k1 = 1} ≤ 1 for every n ∈ N. Next, if g(t) = eikt , then we have Z Z Λn (g) = g(t)e−int dµ(t) = e−i(n−k)t dµ(t). Now our hypothesis ensures that Λn (g) → 0 as n → ∞ so that Λn (f ) → 0

(6.31)

as n → ∞ for each f ∈ P. Therefore, Problem 5.18 says that the limit (6.31) also holds for 2 every f ∈ L1 (|µ|). Obviously, we have h ∈ L1 (|µ|) and so Z 2 µ b(−n) = e−int [h(t)]2 dµ(t) = Λn (h ). Hence we conclude from the result (6.31) that

2

lim µ b(−n) = lim Λn (h ) = 0.

n→∞

n→∞

This completes the proof of the problem.



Problem 6.8 Rudin Chapter 6 Exercise 8.

Proof. Suppose that X = [0, 2π), B is the collection of all Borel sets in X and µ b is periodic with period k ∈ Z. Denote dλ(t) = (e−ikt − 1) dµ(t). Clearly, we have Z Z b (6.32) µ b(n + k) − µ b(n) = e−int (e−ikt − 1) dµ(t) = e−int dλ(t) = λ(n)

for all n ∈ Z. Particularly, take n = 0 in the result (6.32) and use the periodicity of µ b and Theorem 6.12 to get Z Z h d|λ| = dλ = 0, (6.33)

where h is a measurable function such that |h(x)| = 1 on X. If |λ|(X) 6= 0, then Theorem 1.39(a) and the form (6.33) together imply that h = 0 a.e. on X, a contradiction. This means that |λ|(X) = 0 and we assert from the fact |λ(E)| ≤ |λ|(X) = 0 that λ(E) = 0

(6.34)

6.3. Fourier Coefficients of Complex Borel Measures

193

for every E ∈ B.

Using Theorem 6.12 to write dµ = H d|µ|, where H is a measurable function with |H(x)| = 1 on X. Next, we recall the meaning of the notation dλ(t) = (e−ikt − 1) dµ(t) and we observe from the result (6.34) that Z Z 0 = λ(E) = (e−ikt − 1) dµ = (e−ikt − 1)H d|µ| (6.35) E

E

for every E ∈ B. Let A1 = {t ∈ X | cos kt − 1 = 0}, A2 = {t ∈ X | sin kt = 0} and A = {t ∈ X | e−ikt − 1 = 0} = A1 ∩ A2 . Since f1 (t) = cos kt − 1, f2 (t) = sin kt and g(t) = 0 are continuous on X, they are Borel measurable.b By Problem 1.5(a), we have A1 , A2 ∈ B and also A ∈ B. We claim that µ is concentrated on A. Assume that it was not the case, i.e., there is E0 ∈ B such that a E0 ∩ A = ∅ but µ(E0 ) 6= 0. We recall from the result (6.35) that (e−ikt − 1)H = 0 a.e. on any E with |µ|(E) > 0. Since |H| = 1 on X, we have e−ikt = 1 a.e. on any E with |µ|(E) > 0. Particularly, since µ(E0 ) 6= 0, we have |µ|(E0 ) > 0 and then e−ikt = 1

(6.36)

a.e. on E0 . However, E0 ∩ A = 0 = ∅ means that e−ikt 6= 1 on E0 which contradicts the expression (6.36). Hence we have proven our claim that µ is concentrated on A. On the other hand, if µ is concentrated on A, then for every E ∈ B, we write E = (E \ A) ∪ A and then Z Z Z −ikt −ikt (6.37) (e − 1) dµ + (e−ikt − 1) dµ. (e − 1) dµ = E

E\A

A

e−ikt

Notice that − 1 = 0 on A and (E \ A) ∩ A = ∅ so that µ(E \ A) = 0. Thus we deduce from Proposition 1.24(d) and (e) that all the integrals in the expression (6.37) are zero. This certainly implies that λ(E) = 0 for every E ∈ B and then Z b λ(n) = e−int dλ = 0 (6.38) for every n ∈ Z. Combining the expression (6.32) and the result (6.38), we may conclude that µ b(n + k) = µ b(n)

for all n ∈ Z.

Hence we have shown that µ is a complex Borel measure with periodic Fourier coefficient µ b with period k if and only if µ is concentrated on o n 2nπ A = {t ∈ X | e−ikt − 1 = 0} = t = n ∈ N ∪ {0} , k

completing the proof of the problem. Problem 6.9 Rudin Chapter 6 Exercise 9.

b

See §1.11.



194

Chapter 6. Complex Measures

Proof. The assertion is false. Take µ = m the Lebesgue measure on I. Since m ≪ m, if m ⊥ m, then Proposition 6.8(g) implies that m = 0, a contradiction. Now it remains to construct a sequence of {gn } satisfying the required properties. 2 . Define gn,k : [0, 1] → R and Suppose that n, k ∈ N and k = 1, 2, . . . , n. Let δn = n(2n+3) gn : [0, 1] → R by  h k δn k δn i   n + 1, if x ∈ ; − , +   n+1 2 n+1 2        k  δn δn k δn    ; − − , −  linear, if x ∈ n+1 2 2n + 2 n + 1 2 gn,k (x) =    k  δn k δn δn     ; + , + + linear, if x ∈   n+1 2 n+1 2 2n + 2       0, otherwise and

gn (x) =

n X

gn,k (x)

k=1

respectively. By direct computation, it is easy to check that 0
0. Thus A−1 exists. Let y = A−1 (x). Then we follow from Theorem 7.26 (The Change-of-variables Theorem) (with X = Y = V = Rk , T = A−1 which is one-to-one and differentiable on Rk , T (Rk ) = A−1 (Rk ) = Rk and y = A−1 (x)) that g (z) = f[ b ◦ A(z) Z
f A(y) exp(−iy · z) dmk (y) = k {z } |R Z This is

=

Z

Rk

f dm.

Y

f (x) exp[−iA−1 (x) · z] × | det A−1 | dmk (x) {z } | This is (f ◦ T )(x) = (f ◦ A

−1

(9.70)

)(x).

for every z ∈ Rk . We need a result from linear algebra: Lemma 9.7

 If A is a k × k matrix, then for all x, y ∈ Ck , we have hA(x), yi = x, AT (y) , where AT is the transpose of A. Proof of Lemma 9.7. Using [71, p. 109; Example 5.3.1, p. 286], it can be seen easily that

 hA(x), yi = [A(x)]T y = xT AT (y) = x, AT (y) ,

completing the proof of Lemma 9.7.

Thus we apply Lemma 9.7 to the expression (9.70) to deduce that Z
f (x) exp[−ix · A−T (z)] dmk (x) = | det A| · fb A−T (z) . g(z) = | det A| · b Rk



9.3. Fourier Transforms on Rk and its Applications

307

Consequently, we conclude that f\ ◦ A = | det A| · (fb ◦ A−T ).

(9.71)

Particularly, let A be a rotation matrix of Rk . Then it preserves the distance of x from the origin, i.e., hA(x), A(x)i = hx, xi. By [7, Proposition 5.1.13, p. 134], A is orthogonal which means AT = A−1 . Furthermore, this implies that | det A| = 1, so the formula (9.71) can be simplified to f\ ◦ A = fb ◦ A.

(9.72)

In other words, if f is invariant under rotations, then f ◦A = f and the expression (9.72) implies fb ◦ A = fb,

i.e., fb is also invariant under rotations. This completes the proof of the problem.



Problem 9.16

Rudin Chapter 9 Exercise 16.

Proof. We make the following assumption: Suppose that f ∈ S(Rk ), where S(Rk ) is the class of all functions f : Rk → C such that f ∈ C ∞ and xβ ∂ α f is bounded for all multi-indices α ∂f and β.k Let 1 ≤ j ≤ k, x = (xj , xj ) and F (x) = , where x = (x1 , . . . , xj , . . . , xk ) and ∂xj xj = (x1 , . . . , xj−1 , xj+1 , . . . , xk ). We note that x · y = x1 y1 + x2 y2 + · · · + xk yk = xj · yj + xj yj . For every y ∈ Rk , we follow from this, the definition in Problem 9.14 and Lemma 9.4 that Z ∂f Fb(y) = (x) exp(−ix · y) dmk (x) Rk ∂xj Z i h Z ∂f (xj , xj ) exp(−ixj · yj ) dm(xj ) exp(−ixj · yj ) dmk−1 (xk ) = ∂xj Rk−1 | R {z } =

Z

Apply Lemma 9.4 to this.

Rk−1 Z

= iyj

i h Z f (x) exp(−ixj · yj ) dm(xj ) exp(−ixj · yj ) dmk−1 (xk ) iyj

Rk

R

f (x) exp(−ix · y) dmk (x)

= iyj fb(y)

which implies that

Therefore, we have

k

2f ∂d (y) = −yj2 fb(y). ∂x2j

g(x) = − b

k X j=1

x2j fb(x) = −|x|2 · fb(x)

In fact, S is the Schwartz space on Rk , see Remark 9.2.

(9.73)

308

Chapter 9. Fourier Transforms

for all x ∈ Rk .

Suppose that f has continuous second derivatives so that the Laplacian ∆f is well-defined. Suppose further that supp (f ) is compact. In this case, f ∈ L1 (Rk ) so that fb is well-defined on Rk . Furthermore, we suppose that A is a rotation of Rk about the origin 0 and gA = ∆(f ◦ A). Then we deduce from the formulas (9.72) and (9.73) that 2 \ 2 2 b b gc b(Ax) = g[ ◦ A(x). A (x) = −|x| · f ◦ A(x) = −|Ax| · (f ◦ A)(x) = −|Ax| · f (Ax) = g

\ \ By the definition, we have ∆(f ◦ A) = (∆f ) ◦ A. To go further, we need the analogue of Theorem 9.12 (The Uniqueness Theorem): Lemma 9.8 If f ∈ L1 (Rk ) and fb(t) = 0 for all t ∈ Rk , then f (x) = 0

(9.74)

a.e. on Rk . Furthermore, if f is continuous on Rk , then the expression (9.74) holds everywhere on Rk .

Proof of Lemma 9.8. Since fb = 0 on Rk , we have fb ∈ L1 (Rk ) and the result (9.74) follows from the Inversion Theorem for Rk (Problem 9.14). If f is continuous on Rk , the definition of continuity ensures that the relation (9.74) holds everywhere on Rk .  Now we let F = ∆(f ◦ A) − (∆f ) ◦ A. It is clear that f ◦ A ∈ S(Rk ) if f ∈ S(Rk ). Similar to the proof of Problem 9.7, we can show that ∂ α f ∈ S(Rk )

and ∂ α f ∈ L1 (Rk ).

Thus we have ∆(f ◦ A) ∈ L1 (Rk ) and (∆f ) ◦ A ∈ L1 (Rk ) so that F ∈ L1 (Rk ). Since Fb(t) = 0 for all t ∈ Rk and F is continuous on Rk , Lemma 9.8 implies that ∆(f ◦ A) = (∆f ) ◦ A

(9.75)

holds everywhere on Rk . It is well-known that any rotation about a point p is the composition of a rotation about the origin 0 and a translation. By this fact, the analysis preceding Lemma 9.8 and the hypothesis that ∆ commutes with translations, we conclude that the expression (9.75) is still valid for every rotation about every point p. It is time to consider the general situation. Fix a p ∈ Rk . There exists a r > 0 such that p ∈ Kr = B(0, r) ⊂ Vr = B(0, 2r). By [99, Exercise 6, p. 289], there exists functions ψ1 , . . . , ψs ∈ C ∞ (Rk ) such thatl • 0 ≤ ψi ≤ 1 for 1 ≤ i ≤ s; • supp (ψi ) ⊂ Vr ; • ψ1 (x) + ψ2 (x) + · · · + ψs (x) = 1 for every x ∈ Kr . If we define Ψ : Rk → R by l

Ψ = ψ1 + ψ2 + · · · + ψs ,

For a proof of this, please read [124, Problem 10.6, pp. 266, 267 ].

9.3. Fourier Transforms on Rk and its Applications

309

then the above conditions imply that Ψ ∈ Cc∞ (Rk ) and Ψ(x) = 1 on Kr . Thus we have f · Ψ = f on Kr . Particularly, we have ∂Ψ ∂2Ψ (p) = 0 (p) = ∂xj ∂x2j

(f · Ψ)(p) = f (p) and

for every 1 ≤ j ≤ k. As a consequence, these imply that ∂ 2 (f · Ψ) ∂Ψ ∂2Ψ ∂2f ∂f ∂2f (p) · (p) + f · (p) = Ψ(p) · (p) + 2 (p) = (p) 2 2 2 ∂xj ∂xj ∂xj ∂xj ∂xj ∂x2j

(9.76)

for 1 ≤ j ≤ k. By the definition of the Laplacian and the result (9.76), we yield k k X X
∂ 2 (f · Ψ) ∂2f ∆(f · Ψ) (p) = (p) = (p) = (∆f )(p). 2 ∂xj ∂x2j j=1 j=1

(9.77)

Note that f · Ψ is compactly supported in this case. If Ap = q, then q ∈ Kr . On the one hand, the expressions (9.75) (with f replaced by f · Ψ) and (9.77) give

 ∆ f · Ψ) ◦ A (p) = [(∆(f · Ψ)) ◦ A](p) = ∆(f · Ψ) (q) = (∆f )(q) = [(∆f ) ◦ A](p). (9.78) On the other hand, since

[(f · Ψ) ◦ A](p) = [(f ◦ A)(p)] · [(Ψ ◦ A)(p)] = (f · Ψ)(q) = f (q) = (f ◦ A)(p) on Kr , we have (f · Ψ) ◦ A = f ◦ A on Kr so that 
   ∆ (f · Ψ) ◦ A (p) = ∆(f ◦ A) (p)

(9.79)

Finally, by combining the expressions (9.78) and (9.79) and using the fact that p is arbitrary, we may conclude that our desired formula (9.75) also holds everywhere on Rk in this general  case. This completes the proof of the problem. Problem 9.17 Rudin Chapter 9 Exercise 17.

Proof. We prove the case for Rk directly. Suppose that ϕ : Rk → C is a Lebesgue measurable character. We define Φ : L1 (Rk ) → C by Z f (x)ϕ(x) dmk (x). (9.80) Φ(f ) = Rk

Now direct computation gives Z Z (f ∗ g)(x)ϕ(x) dmk (x) = Φ(f ∗ g) = Rk

Rk

Z

Rk

f (x − y)g(y)ϕ(x) dmk (y) dmk (x).

(9.81)

Since ϕ is a character, we have ϕ(x) = ϕ(x − y)ϕ(y). Substituting this into the expression (9.81), we see that Z Z f (x − y)g(y)ϕ(x − y)ϕ(y) dmk (y) dmk (x) Φ(f ∗ g) = k k ZR h RZ i f (x − y)ϕ(x − y) dmk (x) g(y)ϕ(y) dmk (y) = Rk

Rk

310

Chapter 9. Fourier Transforms

=

Z

Rk

Φ(f )g(y)ϕ(y) dmk (y)

= Φ(f ) · Φ(g). Therefore, Φ is a linear functional on L1 (Rk ). In other words, Φ is a complex homomorphism of the Banach algebra L1 (Rk ). Hence the analogue of Theorem 9.23 in Problem 9.14 shows that one may find a β ∈ L∞ (Rk ) and a unique y ∈ Rk such that Z f (x) exp(−iy · x) dmk (x) = fb(y). (9.82) β(x) = exp(−iy · x) and Φ(f ) = Rk

Next, we deduce from the formulas (9.80) and (9.82) that Z f (x)[ϕ(x) − exp(−iy · x)] dmk (x) = 0

(9.83)

Rk

holds for all f ∈ L1 (Rk ). Since f (x) = e−|x| is in L1 (Rk ) and f (x) 6= 0 on Rk , the result (9.83) implies that ϕ(x) = exp(−iy · x) (9.84)

a.e. on Rk . To proceed further, we need the following result which is a generalization of Problem 7.6 and its proof uses the Steinhaus Theorem in Rk (see Remark 7.2), but we don’t present it here. Lemma 9.9 Suppose that G is a subgroup of Rk (relative to addition), G 6= Rk , and G is Lebesgue measurable. Then mk (G) = 0.

To apply this lemma, we consider h(x) = ϕ(x) − exp(−iy · x) and G = {x ∈ Rk | h(x) = 0}. For a, b ∈ G, if a + b ∈ / G, then we have h(a + b) 6= 0 and ϕ(a)ϕ(b) = ϕ(a + b) 6= exp[−iy · (a + b)] = exp(−iy · a) exp(−iy · b) = ϕ(a)ϕ(b) which implies 1 6= 1, a contradiction. Thus a + b ∈ G which means G is a subgroup of Rk (relative to vector addition). Besides, since h is obviously Lebesgue measurable by Proposition 1.9(c) and G = h−1 (0), G is a Lebesgue measurable set. Since mk (G) > 0, Lemma 9.9 forces that G = Rk so that the equation (9.84) holds for all x ∈ Rk . Now the continuity of the exponential function implies that ϕ is continuous on Rk . This completes the proof of the problem. 

9.4

Miscellaneous Problems

Problem 9.18 Rudin Chapter 9 Exercise 18.

Proof. First of all, the equation f (x + y) = f (x) + f (y) for all x, y ∈ R is called the Cauchy functional equation.

(9.85)

9.4. Miscellaneous Problems

311

• Existence of real discontinuous functions f satisfying the equation (9.85). It is well-known that the Hausdorff Maximality Theorem, the Axiom of Choice and Zorn’s Lemma are equivalent to each other. To prove this part, we would like to apply Zorn’s Lemma which says that Lemma 9.10 (Zorn’s Lemma) If X is a partially ordered set and every totally ordered subset of X has an upper bound, then X has a maximal element.

As a consequence, Zorn’s Lemma can be used to prove that every vector space has a basis. In fact, if A is a linearly independent subset of a vector space V over a field K, then there is a basis B of V that contains A.m By this result, there exists a basis of R over Q with the usual addition and scalar multiplication.n Let x ∈ R and H be a basis of R over Q. Thus x has a unique representation x = r 1 b1 + r 2 b2 + · · · + r n bn , where b1 , b2 , . . . , bn ∈ H and r1 , r2 , . . . , rn ∈ Q. Define f : R → R by f (x) = r1 + r2 + · · · + rn . It is obvious that f satisfies the equation (9.85). Assume that f was continuous on R. Then f (R) must be connected (see [99, Theorem 4.22, p. 93]), but f (R) = Q which is a contradiction. • If f is Lebesgue measurable and satisfies the equation (9.85), then f is continuous. We use the following special form of Theorem 2.24 (Lusin’s Theorem):o Lemma 9.11 Suppose that f : [a, b] → R is Lebesgue measurable. For every ǫ > 0, there is a compact set K ⊆ [a, b] such that m([a, b] \ K) < ǫ and f |K is continuous. By Lemma 9.11, there exists a compact set K ⊆ [0, 1] such that m(K) >

2 3

(9.86)

and f |K is continuous. Since K is compact, f is actually uniformly continuous on K. Given ǫ > 0. There is a δ > 0 such that |f (x) − f (y)| < ǫ for every x, y ∈ K with |x − y| < δ. Without loss of generality, we may assume that δ < 13 . Let η ∈ (0, δ) and K − η = {x − η | x ∈ K}. Then K − η ⊆ [−η, 1 − η]. Assume that K ∩ (K − η) = ∅. Since K, K − η ⊆ [−η, 1], we have
m(K) + m(K − η) = m K ∪ (K − η) ≤ m([−η, 1]) = 1 + η. (9.87) By Theorem 2.20(c), m(K − η) = m(K), so we deduce from the inequalities (9.86) and (9.87) that 1 + η ≥ 2m(K) > 43 and thus η > 13 , a contradiction. Hence we have K ∩ (K − η) 6= ∅. m n o

For a hint of this proof, please refer to [74, Exercise 8, p. 72] This kind of basis is called a Hamel basis. Read [40, Exercise 44, p. 64] or [93, p. 66]

312

Chapter 9. Fourier Transforms Let p ∈ K ∩ (K − η). Then p, p + η ∈ K and |p + η − p| = η < δ so that |f (η) − f (0)| = |f (p) + f (η) − f (p)| = |f (p + η) − f (p)| < ǫ.

(9.88)

In other words, f is continuous at 0. Now for any x ∈ [0, 1], the inequality (9.88) implies clearly that |f (x + η) − f (x)| = |f (x) + f (η) − f (x)| = |f (η) − f (0)| < ǫ. Hence f is continuous on [0, 1]. Next, if z ∈ [1, 2], then we have z = x+y for some x, y ∈ [0, 1]. Since f (z) = f (x)+f (y), we have |f (z + η) − f (z)| = |f (x + η) − f (x)| < ǫ and the above paragraph shows that f is also continuous on [1, 2]. Now this kind of argument can be applied repeatedly and we conclude that f is continuous on R. • If graph (f ) is not dense in R2 and satisfies the equation (9.85), then f is continuous. Suppose that f is discontinuous. Then f cannot be of the form f (x) = cx for any constant c. We have
graph (f ) = { x, f (x) | x ∈ R}.

Choose a nonzero real number x1 . Then there exists another nonzero real number x2 such that f (x2 ) f (x1 ) 6= x1 x2 or equivalently, x1 f (x1 ) x2 f (x2 ) 6= 0.

In other words, the vectors u = x1 , f (x1 ) and v = x2 , f (x2 ) are linearly independent and thus span the space R2 . Let p ∈ R2 . Given ǫ > 0. Since Q2 is dense in R2 , one can find q1 , q2 ∈ Q such that |p − q1 u − q2 v| < ǫ. Clearly, we have Therefore, the set


q1 u + q2 v = q1 x1 + q2 x2 , f (q1 x1 + q2 x2 ) .
G = { x, f (x) ∈ R2 | x = q1 x1 + q2 x2 and q1 , q2 ∈ Q}

is dense in R2 . Since G ⊆ graph (f ), graph (f ) is dense in R2 .

• The form of all continuous functions satisfying the equation (9.85). In fact, by a similar argument as in the proof used in [124, Problem 8.6, p. 178], it can be shown that the function f satisfies f (r) = rf (1) for every r ∈ Q. Now for every x ∈ R, since Q is dense in R, we can find a sequence {rn } ⊆ Q such that rn → x as n → ∞. Since f is continuous, we have f (x) = lim f (rn ) = lim rn f (1) = xf (1). n→∞

n→∞

Hence every continuous solution of the equation (9.85) must be in the form f (x) = f (1) · x.

9.4. Miscellaneous Problems

313

We have completed the proof of the problem.



Problem 9.19 Rudin Chapter 9 Exercise 19.

Proof. Let f = χA and g = χB . Since f, g ∈ L1 , their convolution h = f ∗ g is well-defined by Theorem 8.14. Suppose that p and q are conjugate exponents. Direct computation gives Z ∞ Z m(A) 1 dx = √ 0

ϕ(x+) + ϕ(x−) 2

for every x ∈ R. Hence the formula (11.36) asserts that  ϕ(x),         0,     lim [f (x + iǫ) − f (x − iǫ)] = ϕ(a+) ǫ→0  ,   ǫ>0  2         ϕ(b−) , 2

(11.36)

if x ∈ (a, b); if x ∈ R \ I; if x = a; if x = b.

• The case when ϕ ∈ L1 . In this case, we apply Theorem 9.10 to the expression (11.35)

11.1. Basic Properties of Harmonic Functions

361

to get lim

Z

∞

ǫ→0 −∞ ǫ>0

|f (x + iǫ) − f (x − iǫ) − ϕ(x)| dm(x) = lim kϕ ∗ Pǫ − ϕk1 = 0. ǫ→0 ǫ>0

(11.37)

Denote f (x + iǫ) = fǫ+ (x) and f (x − iǫ) = fǫ− (x). Then the result (11.37) means that lim kfǫ+ − fǫ− k1 = kϕk1 .

ǫ→0 ǫ>0

• The case when ϕ(x+) and ϕ(x−) exist at x. This case has been settled already in the formula (11.36). We have completed the proof of the problem.  Remark 11.1 The integral considered in Problem 11.10 is an example of the so-called Cauchy type integrals. For more details of this subject, please refer to Muskhelishvili’s book [76].

Problem 11.11 Rudin Chapter 11 Exercise 11.

Proof. We note that the following proof uses only the techniques from Chapter 10, not from Chapter 11. Actually, the author admits that he is not able to apply the theory of harmonic functions to prove this problem. • By Theorem 10.17 (Morera’s Theorem), it suffices to prove that Z f (z) dz = 0

(11.38)

∂∆

for every closed triangle ∆ ⊆ Ω. There are three cases. – Case (i): ∆ ∩ I = ∅. Then ∆ ⊆ Ω \ I. Since Ind ∂∆ (x) = 0 for every x ∈ I, Theorem 10.35 (Cauchy’s Theorem) implies that the result (11.38) holds trivially. – Case (ii): ∂∆ has a side lying on I. Since ∆ is compact, C \ Ω is closed in C and ∆ ∩ (C \ Ω) = ∅, Problem 10.1 ensures that there exists a δ1 > 0 such that d(∆, C \ Ω) = δ1 > 0. Therefore, it is true that ∆n = ∆ +

i ⊆Ω n

(11.39)

for all n > δ11 . Thus we have ∆n ∩ I = ∅ so that the result (11.38) holds for every ∂∆n with n > δ11 . Suppose that ∂∆ = [a, b]+[b, c]+[c, a], where a, b and c are vertices of the triangle ∆. Then we have  i  i  i ∂∆n = [a, b] + + [b, c] + + [c, a] + . n n n Using [100, Eqn. (4), p. 202], we know that Z 1  Z  i f a + + (b − a)t dt f (z) dz = (b − a) n 0 [a,b]+ i n

362

Chapter 11. Harmonic Functions so that Z

[a,b]

f (z) dz −

Z

[a,b]+ ni

f (z) dz

Z 1 h  
i i f a + + (b − a)t − f a + (b − a)t dt = |b − a| · n 0 Z 1  
i ≤ |b − a| · f a + + (b − a)t − f a + (b − a)t dt. n 0

Since f is continuous on Ω, it is uniformly continuous on any compact subset of Ω. Given ǫ > 0, there exists a δ2 > 0 such that n > δ12 implies  
i ǫ f a + + (b − a)t − f a + (b − a)t < n 3|b − a|

and then

Z

[a,b]

f (z) dz −

Z

[a,b]+ ni

ǫ f (z) dz < . 3

(11.40)

Thus, if n > max( δ11 , δ12 ), then both the conditions (11.39) and (11.40) hold simultaneously. Since the inequality (11.40) also holds for [b, c] and [c, a] for large enough n, we obtain immediately that Z f (z) dz < ǫ ∂∆n

which means

Z

f (z) dz = lim ∂∆

Z

n→∞ ∂∆ n

f (z) dz = 0.

– Case (iii): ∆ intersects with I at only two points. Then I divides ∆ into a triangle and a quadrangle or another triangle. If it is a quadrangle, then it can be further divided into two triangles, see Figure 11.1 for an illustration. Since ∂∆ is a sum of two or three boundaries of triangles, it follows from Case (i) and Case (ii) that our result (11.38) remains true in this case.

(a) ∂∆ = ∂∆1 + ∂∆2 .

(b) ∂∆ = ∂∆1 + ∂∆2 + ∂∆3 .

Figure 11.1: The I divides ∂∆ into several triangles.

11.2. Harnack’s Inequalities and Positive Harmonic Functions

363

• Removable sets for holomorphic functions of class C (Ω). Let N ∈ N. It is clear that the above argument can be applied to the compact set in the form K = I1 ∪ I2 ∪ · · · ∪ IN and f ∈ H(Ω \ K), where each Ij = [aj , bj ] is a subset of Ω for 1 ≤ j ≤ N . We have completed the analysis of the problem.f



Remark 11.2 (a) Classically, Problem 11.11 is a topic of the so-called removable sets for holomorphic functions. To say it more precisely, let F be a class of functions from Ω to C. Then a compact set K ⊆ Ω is said to be removable for holomorphic functions of class F if every f ∈ F such that f ∈ H(Ω \ E) can be extended to a holomorphic function in Ω. Examples of F are L∞ (Ω), C (Ω) and Lip α (Ω), where 0 < α ≤ 1. (b) It is clear that Theorem 10.20 is a positive result for bounded and holomorphic functions. Because of this, Painlev´e was motivated and studied some more general problems: “Which subsets of C are removable? What geometric characterization(s) must these subsets satisfy?” In fact, Painlev´e proved a sufficient condition that if a compact set K ⊆ Ω is of one-dimensional Hausdorff measure [18, pp. 215, 216], then it is removable for bounded and holomorphic functions in Ω \ K. (c) For the class Lip α (Ω) with 0 < α < 1, Dolˇzenko [32] has shown in 1963 that a compact set K is removable for holomorphic functions of this class if and only if the (1 + α)dimensional Hausdorff measure is zero. For the remaining case α = 1, Uy [119] verified in 1979 that K is removable if and only if m(K) = 0. (d) Besides the approach of Hausdorff measure, Ahlfors [1] introduced the analytic capacity (a purely complex-analytic concept) of a compact set K to study removable compact sets for holomorphic functions of class L∞ . In fact, he proved that K is removable for bounded analytic functions if and only if its analytic capacity vanishes.

11.2

Harnack’s Inequalities and Positive Harmonic Functions

Problem 11.12 Rudin Chapter 11 Exercise 12.

Proof. • Proof of Harnack’s Inequalities. We first prove the special case that if u is harmonic in D(a; R) and u > 0 in D(a; R), then for every 0 ≤ r < R and z = a + reiθ ∈ D(a; R), we have R+r R−r u(a) ≤ u(z) ≤ u(a). (11.41) R+r R−r f

See also Problem 16.10.

364

Chapter 11. Harmonic Functions To see this, choose ρ > 0 such that r < ρ < R. Using the second set of inequalities on [100, p. 236], we know that ρ−r ρ+r u(a) ≤ u(z) = u(a + reiθ ) ≤ u(a) ρ+r ρ−r

(11.42)

for every θ ∈ R. Letting ρ → R in the inequalities (11.42), we obtain the desired results (11.41). Since Ω is a region and K ⊆ Ω, C \ Ω is closed in C and (C \ Ω) ∩ K = ∅. By Problem 10.1, there exists a δ > 0 (depending on K and Ω) such that d(C \ Ω, K) = 2δ > 0. Clearly, we have [ [ K⊆ D(z; δz ) ⊆ D(z; δ) ⊆ Ω, z∈K

z∈K

where 0 < δz < δ. Since K is compact, one can find a finite set {z1 , z2 , . . . , zm } with positive numbers δ1 , δ2 , . . . , δm such that K⊆

m [

k=1

D(zk ; δk ) ⊆

m [

k=1

D(zk ; δ) ⊆ Ω.

Now we apply the special case (11.41) to each disc D(zk ; δ) and consider only points z ∈ D(zk ; δk ) to get δ − δk δ−r δ+r δ + δk u(zk ) ≤ u(zk ) ≤ u(z) ≤ u(zk ) ≤ u(zk ). δ + δk δ+r δ−r δ − δk

Take positive numbers α and β such thatg α=

δ − δk 1 · min u(zk ) u(z0 ) 1≤k≤m δ + δk

and β =

δ + δk 1 · max u(zk ). u(z0 ) 1≤k≤m δ − δk

Then we establish

for every z ∈

m [

k=1

αu(z0 ) ≤ u(z) ≤ βu(z0 )

(11.43)

D(zk ; δk ). In particular, the inequalities (11.43) are true for all z ∈ K.

• The behavior of {un } in Ω \ {z0 } if un (z0 ) → 0. Let un (z) → u(z) for every z ∈ Ω and a ∈ Ω \{z0 }. Since Ω \{z0 } is open in C, there exists a R > 0 such that D(a; R) ⊆ Ω \{z0 }. Obviously, D(a; R) is a compact subset of Ω. By the inequalities (11.43), we have αun (z0 ) ≤ un (z) ≤ βun (z0 )

(11.44)

for every z ∈ D(a; R) and n = 1, 2, . . ., where α and β depend on z0 , K and Ω only. Take n → ∞ in the inequalities (11.44) and then use the hypothesis, we get u(z) = 0

(11.45)

for all z ∈ D(a; R). Particularly, u(a) = 0. Since a is arbitrary, we conclude that u ≡ 0 in Ω \ {z0 }. Finally, the continuity of u ensures that u(z0 ) = 0 and then u ≡ 0 in Ω. • The behavior of {un } in Ω \ {z0 } if un (z0 ) → ∞. Instead of the result (11.45), we obtain u(z) = ∞ for all z ∈ D(a; R). Hence, using similar argument as the previous assertion, we conclude that u(z) = ∞ in Ω. g

Since δ depends on K and Ω, α and β trivially depend on z0 , K and Ω.

11.2. Harnack’s Inequalities and Positive Harmonic Functions

365

• The positivity of {un } is essential. For each n ∈ N, we consider u(x, y) = nx − ny in C. Then it is easily checked that ∆un = 0 so that un is harmonic in C. As un (0, 1) = − n1 < 0 and un (1, 0) = n > 0, each un is neither positive nor negative. Furthermore, un (0, 1) → 0 but un (1, 0) → ∞ as n → ∞. Hence this counterexample shows that the positivity of {un } cannot be omitted for these results.  We have completed the analysis of the problem. Problem 11.13 Rudin Chapter 11 Exercise 13.

Proof. By the Poisson formula, we have Z π 1 1 − r2 u(reiθ ) = u(eit ) dt, 2π −π 1 − 2r cos(θ − t) + r 2 where 0 ≤ r < 1 and −π ≤ θ ≤ π. Pick r = u( 12 ) = Since

3 5+4

≤

3 5−4 cos t

≤

1 2π

Z

3 5−4

π

−π

1 2

(11.46)

and θ = 0 in the formula (11.46) to get

1 − 41 1 u(eit ) dt = 2π 1 − cos t + 14

Z

π

−π

3 u(eit ) dt. 5 − 4 cos t

(11.47)

on [−π, π], the integral (11.47) gives

1 1 · 3 2π

Z

π −π

u(eit ) dt ≤ u( 12 ) ≤ 3 ·

1 2π

Z

π

u(eit ) dt.

(11.48)

−π

Put r = 0 in the Poisson formula (11.46), we obtain Z π 1 u(eit ) dt, u(0) = 2π −π therefore, we conclude from the hypothesis u(0) = 1 and the inequalities (11.48) that 1 ≤ u( 21 ) ≤ 3. 3 This completes the proof of the problem.



Problem 11.14 Rudin Chapter 11 Exercise 14.

Proof. By translation and/or rotation, we may assume that L1 is the real axis (i.e., y = 0) and if L1 and L2 intersect, the intersection point is the origin. • Case (i): L1 and L2 are parallel. Suppose that the equation of L2 is y = A for some A ∈ R \ {0}. Then we consider the function πx

u(x, y) = e A sin

πy . A

It is obvious that u(x, 0) = u(x, A) = 0 for all x ∈ R. Furthermore, direct computation gives uxx + uyy = 0 in R2 so that u is a harmonic function in R2 .

366

Chapter 11. Harmonic Functions • Case (ii): L1 and L2 are perpendicular. Suppose that L2 : x = 0. Then it is easy to check that the function u(x, y) = xy satisfies all the requirements. • Case (iii): L1 and L2 are non-parallel and non-perpendicular. Suppose that the m angle between L1 and L2 is a rational multiple of π, say mπ n , where m, n ∈ N and n is 1 n not a multiple of 2 . Since z = un (x, y) + ivn (x, y) is entire, its imaginary part vn (x, y) is harmonic in R2 by Theorem 11.4. On L1 , we have xn = (x + i · 0)2 = un (x, 0) + ivn (x, 0) for all x ∈ R. This implies that vn (x, 0) = 0 on R. Similarly, since points on L2 are in the mπ form z = cos mπ n + i sin n , so we have (−1)m =



cos

  mπ mπ mπ n mπ  mπ  mπ + ivn cos = un cos + i sin , sin , sin n n n n n n

which implies that vn (x, y) = 0 on L2 .

Now suppose that the angle between them is an irrational multiple of π, say απ for some α ∈ R \ Q. Assume that u(x, y) was a harmonic function in R2 vanishing on L1 ∪ L2 . We need to develop a harmonic version of Theorem 11.14 (The Schwarz reflection principle). To this end, we recall the following concept: Let L be a straight line passing through the origin. We say that a pair of points are symmetric with respect to L if L is the perpendicular bisector of the line segment joining these points. For each z = (x, y) ∈ C, it is easy to see that there exists a unique zL = (xL , yL ) ∈ C such that z and zL are symmetric with respect to L. Next, the following result is taken from [9, Theorem 4.12, p. 68]: Lemma 11.3 Let z0 = (a, b) ∈ C and consider the line L = {(x, y) ∈ C | (x, y) · (a, b) = c} for some a, b, c ∈ R. Define L+ = {(x, y) ∈ C | (x, y) · (a, b) > c}. Suppose that Ω ⊆ C is a region symmetric with respect to L. If u is continuous on Ω∩L+, u is harmonic on Ω ∩ L+ and u = 0 on Ω ∩ L, then the function  if (x, y) ∈ Ω ∩ L+ ;  u(x, y), U (x, y) =  −u(xL , yL ), if (x, y) ∈ Ω ∩ L−

is harmonic in Ω.

Applying Lemma 11.3 to our u with L = L2 , we see immediately that u vanishes on the line L3 : y = (tan απ)x. In fact, repeated applications of Lemma 11.3 show that u vanishes on lines in the form y = (tan nαπ)x (11.49) for every n ∈ Z.

To finish the proof, we have to show that the collection of the straight lines (11.49), denoted by L, is dense in R2 . To see this, let α > 0. Given that 0 < θ < 12 , δ > 0 and ǫ > 0. By the Kronecker’s Approximation Theoremh , we find that there exist m, n ∈ N such that |nαπ − mπ − θπ| < δ. h

See, for example, [6, §7.4, pp. 148, 149].

11.2. Harnack’s Inequalities and Positive Harmonic Functions

367

Note that tan(nαπ − mπ) = (−1)m tan(nαπ). The continuity of tan x implies that |(−1)m tan(nαπ) − tan(θπ)| < ǫ.

(11.50)

If m is odd, then we can replace (−1)m tan(nαπ) by tan(−nαπ) in the estimation (11.50). If − 21 < θ < 0, then a similar argument gives the following estimation |(−1)m+1 tan(nαπ) − tan(θπ)| < ǫ. In this case, if m is even, then (−1)m+1 tan(nαπ) will be replaced by tan(−nαπ). In other words, for every θ ∈ (− 12 , 12 ) \ {0}, we always have | tan(nαπ) − tan(θπ)| < ǫ

(11.51)

for some n ∈ Z. If θ = 0, then we follow from the Dirichlet’s Approximation Theoremi π that for a positive integer N with N < δ, there exist m, n ∈ N with 0 < n ≤ N such that |nαπ − mπ|
0 and then the inequality (11.51) remains valid with the integer −n. Finally, since u is continuous on R2 and vanishes on L, we conclude that u ≡ 0 on R2 . This completes the proof of the problem.



Problem 11.15 Rudin Chapter 11 Exercise 15.

Proof. We first prove the following lemma: Lemma 11.4 Given ǫ ∈ (0, 1). There exists a constant M > 0 such that P1−ǫ (t) ≥ for all t ∈ [−ǫ, ǫ].

i

Read [127, Problem 2.4, p. 12].

M ǫ

(11.52)

368

Chapter 11. Harmonic Functions Proof of Lemma 11.4. Recall from the series expansion of the Poisson kernel that Pr (t) = Pr (−t). We know from [100, Eqn. (4), p. 233] that Pr (t) is decreasing on [0, π] and [−π, 0]. Therefore, it suffices to prove that the inequality (11.52) holds for t = ǫ. In fact, we have 1 − (1 − ǫ)2 = 2ǫ − ǫ2 ≥ ǫ and 1 − 2(1 − ǫ) cos ǫ + (1 − ǫ)2 = 2(1 − ǫ) − 2(1 − ǫ) cos ǫ + ǫ2

= 2(1 − ǫ)(1 − cos ǫ) + ǫ2  ǫ2 ǫ4  = 2(1 − ǫ) − + · · · + ǫ2 2! 4! ǫ2 ≤ M

for some constant M > 0. Thus they imply that ǫP1−ǫ (ǫ) = ǫ ·

Mǫ 1 − (1 − ǫ)2 ≥ǫ· 2 =M 1 − 2(1 − ǫ) cos ǫ + (1 − ǫ)2 ǫ

which gives the desired result.



Let’s return to the proof of the problem. When u is considered in D(0; r) for 0 < r < 1, [11, Theorem 16.5, p. 227] implies that u attains its maximum on C(0; r). This fact and the positivity of u imply that u(reiθ ) ≤ u(seiθ )

for every θ ∈ [−π, π] if 0 ≤ r ≤ s < 1. Using this result and the hypothesis, if θ 6= 0, then we obtain Z Z     1 1 sup kur k1 = sup u(reiθ ) dθ ≤ lim u(reiθ ) dθ = 0. 2π [−π,π]\{0} 0 0. Recall the definition [100, Eqn. (3), §11.19, p. 241], we conclude that (M µ)(1) ≤ π(Mrad u)(1) as desired. (c) In fact, the results of part (b) are valid for every point eiθ on T if we apply the special case to the rotated measure µθ (E) = µ(eiθ E). In particular, we deduce from the inequalities (11.63) that
µ(Iδ ) ≤ u (1 − δ)eiθ ≤ (Mrad u)(eiθ ), (11.64) πσ(Iδ ) where eiθ ∈ T and Iδ is the open arc with center eiθ and length 2δ. Next, we know from [100, Eqn. (2), §11.17 p. 240] that

kur k1 ≤ kµk = |µ|(T ) < ∞ for every 0 < r < 1. Therefore, we follow from Theorem 11.30(A) that µ is a Borel measure on T . By the hypothesis, µ is positive. Since µ ⊥ m, it follows from Theorem 7.15 that (Dµ)(eiθ ) = ∞

a.e. [µ].

(11.65)

Finally, we observe from the inequality (11.64), the result (11.65) and [100, Eqn. (4), p. 241] that
lim u (1 − δ)eiθ = ∞ a.e. [µ] δ→0

which is exactly the required result. This ends the proof of the problem.



11.3. The Weak∗ Convergence and Radial Limits of Holomorphic Functions

373

Problem 11.20 Rudin Chapter 11 Exercise 20.

Proof. Since m(E) = 0, for each n ∈ N, we know from [99, Remark 11.11(b), p. 309] that there exists an open set Vn ⊆ T containing E such that m(Vn ) ≤ 21n . Define χVn (eiθ ) = and ϕ : T → R by

  1, if eiθ ∈ Vn ; 

ϕ(eiθ ) =

0, otherwise

∞ X

χVn (eiθ ).

n=1

Obviously, we have

∞ Z X

χVn (eiθ ) dm =

n=1 T

∞ X

n=1

m(Vn ) ≤

∞ X 1 = 1, 2n n=1

so Theorem 1.38 implies that Z

T

|ϕ(eiθ )| dm =

∞ Z X

χVn (eiθ ) dm = 1.

n=1 T

In other words, ϕ ∈ L1 (T ) and kϕk1 = 1.

Define u : U → R by u = P [ϕ] which is harmonic in U by Theorem 11.7. Now, for each eiθ ∈ E, we have eiθ ∈ Vn for every n ∈ N, so it follows from the definition that 1 lim u(re ) = lim r→1 r→1 2π iθ

Z

π −π

P (reiθ , eit )ϕ(eit ) dt = ∞.

Since U is simply connected, u is the real part of a holomorphic function g in U , see [11, Theorem 16.3, p. 226]. Let f = e−g . Then we have f ∈ H(U ) and |f | = e−Re g = e−u so that lim f (reiθ ) = lim e−u = 0

r→1

r→1

for every eiθ ∈ E. By the definition of f , f (z) 6= 0 for every z ∈ U . Thus f is nonconstant. Since ϕ ≥ 0, we have u ≥ 0 and consequently, f ∈ H ∞ . If f (0) 6= 1, then we can replace f by e fe(z) = ff (z) (0) so that f (0) = 1. This completes the proof of the problem. 

Remark 11.4 For further information, the reader can refer to [89, p. 295] and [133, pp. 105, 276].

Problem 11.21 Rudin Chapter 11 Exercise 21.

374

Chapter 11. Harmonic Functions

t+is−1 Proof. We first show that g ∈ / H ∞ . In fact, fix s ∈ R, consider zt = t+is+1 for 0 < t < ∞. Then 1+zt it is easy to check that |zt | < 1 and 1−zt = t + is. Therefore, we have

g(zt ) =

2 exp(−et+is ) (t + 1) + is

which gives |g(zt )| = p

2 (t + 1)2 + s2

exp(−et cos s).

(11.66)

Now we put s = π in the expression (11.66) to get |g(zt )| → ∞ as t → ∞. Consequently, g∈ / H ∞. Next, if eiθ ∈ T and 0 ≤ r < 1, then we have 1 − r 2 + 2i sin θ 1 + reiθ = iθ 1 − re 1 − 2r cos θ + r 2

and

f (reiθ ) = exp

 1 − r 2 + 2i sin θ  . 1 − 2r cos θ + r 2

By the definition of g, we have h  1 − r 2 + 2i sin θ i g(reiθ ) = (1 − reiθ ) · exp − exp 1 − 2r cos θ + r 2 and so  h  i sin θ i  , if θ = 6 0;  (1 − eiθ ) exp − exp 1 − cos θ g∗ (eiθ ) = lim g(reiθ ) = r→1   0, if θ = 0.

(11.67)

Hence g∗ (eiθ ) exists for every eiθ ∈ T . By the representation (11.67), we know that g ∗ (eiθ ) is continuous at every θ ∈ (0, 2π). Thus it suffices to show that g∗ (eiθ ) is continuous at θ = 0. To see this, since cos θ − sin θ sin θ = lim = lim = 0, lim θ→0 sin θ θ→0 cos θ θ→0 1 − cos θ

the representation (11.67) gives

h  i sin θ i lim g∗ (eiθ ) = lim (1 − eiθ ) exp − exp θ→0 θ→0 1 − cos θ h  i sin θ i = lim (1 − eiθ ) lim exp − exp θ→0 θ→0 1 − cos θ = 0 · e−1 = 0.

As a consequence, we establish the fact that g∗ ∈ C(T ) and this completes the proof of the  problem.

11.4

Miscellaneous Problems

Problem 11.22 Rudin Chapter 11 Exercise 22.

11.4. Miscellaneous Problems

375

Proof. For each 0 ≤ r < 1, the function ur : T → C is continuous, so u−1 r (R) is measurable. − + − Suppose that ur = u+ r − ur , where ur and ur are the positive and negative parts of ur , see Definition 1.15. Since {ur } is uniformly integrable, there exists a δ > 0 such that Z ur dm < 1 (11.68) E

iθ + whenever 0 ≤ r < 1 and m(E) < 2δ. Let Er+ = {θ ∈ [0, 2π] | u+ r (e ) > 0}. Then Er is measurable and Z Z π ur dm. (11.69) u+ dm = r Er+

−π

(k−1)δi to l Let N be the least positive integer such that 2π N ≤ δ. Define Ik to be the arc from e kδi (k−1)δi kδi e including e but not e , where k = 1, 2, . . . , N − 1. Similarly, we define IN to be the arc from e(N −1)δi to e2πi = 1 including e(N −1)δi but not e2πi = 1. Clearly, the central angles of I1 , I2 , . . . , IN −1 are exactly δ and the central angle of IN is less than or equal to δ. Next, we denote + Er,k = Er+ ∩ Ik + + + } forms a disjoint measurable subsets of Er+ and so that {Er,1 , Er,2 , . . . , Er,N + ) ≤ m(Ik ) ≤ δ < 2δ m(Er,k

for k = 1, 2, . . . , N . Hence we follow from the inequality (11.68) and the result (11.69) that Z

π −π

N Z X u+ dm = r k=1

+ Er,k

N Z X ur dm ≤ k=1

for every 0 ≤ r < 1. Similarly, it can be shown that Z π u− r dm ≤ N

+ Er,k

ur dm ≤ N

−π

for every 0 ≤ r < 1. Thus they imply that Z π Z Z π  N 1 1  π + iθ iθ iθ kur k1 = u− ur (e ) dθ + |ur (e )| dθ ≤ r (e ) dθ ≤ 2π −π 2π −π π −π for every 0 ≤ r < 1. Hence we apply Theorem 11.30(a) to obtain a unique complex Borel measure µ such that u = P [ dµ]. Next, we know form Theorem 11.24 that there exists a f ∈ L1 (T ) such that lim u(zj ) = f (eiθ )

j→∞

for almost all points eiθ ∈ T , where zj → eiθ and zj ∈ eiθ Ωα for α < 1. Particularly, this implies that urj (eiθ ) = u(rj eiθ ) → f (eiθ ) as j → ∞ a.e. on T , where {rj } ⊆ [0, 1) and rj → 1 as j → ∞. Furthermore, since σ(T ) < ∞ and {ur } ⊆ L1 (T ) is uniformly integrable, Problem 6.10(d) ensures that Z |urj (eiθ ) − f (eiθ )| dσ(eiθ ) → 0 kurj − f k1 = T

l

It is obvious that N is independent of r.

376

Chapter 11. Harmonic Functions

as j → ∞ or equivalently, we have Z Z f (eiθ ) dσ(eiθ ). urj (eiθ ) dσ(eiθ ) = lim j→∞ T

(11.70)

T

Finally, we take g = 1 in [100, Eqn. (3), p. 247] to get Z Z dµ(eiθ ). urj (eiθ ) dσ(eiθ ) = lim j→∞ T

(11.71)

T

Combining the results (11.70) and (11.71), we establish dµ = f dσ and hence u = P [f ] for some f ∈ L1 (T ) which completes the proof of the problem.



Remark 11.5 We can also apply Theorem 17.13 (F. and M. Riesz Theorem) on [100, p. 341] to prove that µ ≪ σ. Problem 11.23 Rudin Chapter 11 Exercise 23. 2 Proof. Given z ∈ U , since |eiθ − z|2 ≥ |eiθ | − |z| = (1 − |z|)2 , we have Thus the series

1 − |z|2 1 + |z| 1 − |z|2 = < ∞. |P (z, eiθ )| = iθ ≤ 2 |e − z| (1 − |z|)2 1 − |z|

v(z) =

∞ X

n=1

n−2 P (z, eiθn ) and w(z) =

∞ X

n−2 P (z, e−iθn )

n=1

converge absolutely so that we may split the representation of u into the difference of v and w. For each n ∈ N, we define vn (z) =

n X k=1

k−2 P (z, eiθk ) and wn (z) =

n X

k−2 P (z, e−iθk ).

k=1

Using Problem 11.4, we know that P (z, eiθk ) and P (z, e−iθk ) are harmonic in U for k = 1, 2, . . . , n so that vn and wn are also harmonic in U . Fix 0 ≤ r < 1. If z ∈ D(0, r), then we have 1 − |z| > 1 − r and thus ∞ ∞ X P (z, eiθn ) 1 + r X 1 · ≤ < ∞. n2 1 − r n=1 n2 n=1

By the Weierstrass M -test, we see that {vn } converges uniformly to v in D(0; r). Let K be a compact subset of U . Then there exists a 0 < r < 1 such that K ⊆ D(0, r), so it follows from Theorem 11.11 (Harnack’s Theorem) that v is harmonic in U . By a similar argument, we are able to show that w is also harmonic in U .

11.4. Miscellaneous Problems

377

Since P (z, eit ) > 0 for every z ∈ U and eit ∈ T , we have v > 0 and w > 0. Now, for every 0 ≤ r < 1, we observe that Z π 1 kvr k1 = v(reiθ ) dθ 2π −π Z π hX ∞ i 1 n−2 P (reiθ , eiθn ) dθ = 2π −π n=1 ∞ h 1 Z π i X n−2 = P (reiθ , eiθn ) dθ 2π −π n=1 =

∞ X

n−2

n=1

< ∞.

Similarly, we have sup kwr k1 < ∞ and hence v and w satisfy the requirements of Theorem 0 0 is small. In this case, cos2 (11.73) becomes P (z, eiθ ) − P (z, e−iθ ) ≥

θ 2

> 12 , so the inequality

2 cos2 θ2 sin2 θ 1 1 1 · = > = . 2 θ sin θ sin θ sin θ ǫ 2 sin 2

(11.74)

If z = x + iy ∈ U with x > 0 and y > 0, then it is easy to see from [100, Eqn. (6), p. 233] that P (x + iy, eiθ ) − P (x + iy, e−iθ ) > 0

(11.75)

for every θ ∈ (0, π2 ). By the definition of u, if we put z = 1 − ǫ + iǫ, then we may apply the estimate (11.75) to get u(1 − ǫ + iǫ) > n−2 [P (1 − ǫ + iǫ, eiθn ) − P (1 − ǫ + iǫ, e−iθn )]

(11.76)

for every n ∈ N. Now we take ǫn = sin θn = sin 21n in the inequality (11.76) and then apply the estimate (11.74) as well as the fact that sin θn ≤ θn for every n ∈ N to obtain u(1 − ǫn + iǫn ) > for every n ∈ N. Since that

2n n2

1 1 2n = ≥ 1 n 2 ǫn n2 n2 sin 2n

(11.77)

→ ∞ as n → ∞, we conclude immediately from the inequality (11.77) lim u(1 − ǫn + iǫn ) = ∞,

n→∞

completing the proof of the problem.



Problem 11.24 Rudin Chapter 11 Exercise 24.

Proof. Most assertions of this problem come from [99, Exercise 15, p. 199] and their solutions are shown by the author in [124, Problem 8.15, pp. 187 – 190], so we just prove the necessary assertions here. • Proof of KN −1 (t) ≤ LN (t). Since sin 2θ =

1−cos θ , 2

we know that

1  sin N2t 2 · . N sin 2t

KN −1 (t) =

(11.78)

By [99, Exercise 8, p. 197], we always have | sin nθ| ≤ n| sin θ| for n = 0, 1, 2, . . . so that  sin N t 2 sin

for N2 |t| ≤ that

π 2.

2 t 2

=

| sin N2t |2 ≤ N2 | sin 2t |2

If 0 ≤ t ≤ π, then the power series of sin x implies that sin 2t ≥  sin N t 2 sin

2 t 2

≤

1 sin2

t 2

≤

π2 . t2

(11.79) t π

≥ 0 so (11.80)

11.4. Miscellaneous Problems

379

If −π ≤ t ≤ 0, then we let t = −s so that 0 ≤ s ≤ π. In this case, we have  sin N t 2 sin

2 t 2

=

 sin N s 2 sin

2 s 2

≤

1 sin2

s 2

≤

π2 π = 2. s2 t

(11.81)

Hence, by putting the inequalities (11.79), (11.80) and (11.81) into the expression (11.78), π , then we conclude that if |t| ≤ N KN −1 (t) ≤ if

π N

1 · N2 = N; N

(11.82)

≤ |t| ≤ π, then

π2 . N t2 By the definition of LN , we see that KN −1 (t) ≤ LN (t) for every t ∈ [−π, π]. R • Proof of T LN dσ ≤ 2. Using the estimates (11.82) and (11.83), we have Z Z π 1 LN dσ = Ln (t) dt 2π −π T Z Z π dt 1 N dt + = 2π |t|≤ π 2N π ≤|t|≤π t2 KN −1 (t) ≤

N

(11.83)

N

1 =2− N ≤2

for every N ∈ N. • Proof of Fej´ er’s Theorem. The result about the convergence of the arithmetic means is called Fej´ er’s Theorem.m Recall from [100, Eqn. (1), p. 101] that Z Z
f ei(θ−t) Dn (t) dσ, f (eit )Dn (θ − t) dσ = sn (f ; θ) = T

T

so it is easy to check that σN (f ; θ) =

Z

f ei(θ−t) T


h

Z N i
1 X f ei(θ−t) KN (t) dσ. Dn (t) dσ = N +1 T

(11.84)

n=0

Suppose that eiθ is a Lebesgue point of f ∈ L1 (T ). Imitating the proof of Theorem 11.23 (Fatou’s Theorem), we may assume without loss of generality that f (eiθ ) = 0. Thus it suffices to prove that lim σN (f ; θ) = 0. N →∞

It is easy to check that the KN satisfies KN (t) ≥ 0, KN (t) is even and Z KN (t) dσ = 1 T

m

for every N ∈ N. Thus it follows from the expression (11.84) that 1 Z π
f ei(θ−t) KN (t) dt |σN (f ; θ)| = 2π −π

See, for instances, [99, Exercises 15, 16, p. 199] or [133, Theorem 3.4, p. 89].

380

Chapter 11. Harmonic Functions Z Z 0

1 π f ei(θ−t) KN (t) dt + f ei(θ−t) KN (t) dt 2π 0 −π Z π i(θ−t)

1 f e + f ei(θ+t) · KN (t) dt. ≤ 2π 0

Put g(t) = f ei(θ−t) + f ei(θ+t) and Z x g(t) dt, G(x) = =

0

where 0 ≤ x ≤ π. We have 0< and since that

eiθ

G(x) 1 ≤ x x

Z

0

x


1 |f ei(θ−t) | dt + x

Z

0

x


|f ei(θ+t) | dt

is a Lebesgue point of f , we know from the definition [100, Eqn. (5), p. 241] lim

x→0

G(x) = 0. x

(11.85)

Given ǫ > 0. Firstly, the result (11.85) means that we may choose a δ > 0 such that G(x) < ǫx for all 0 < x < δ. If N > 1δ , then we deduce from (11.82) that Z 1 Z 1 N N N +1 ǫ < 2ǫ. (11.86) g(t)KN (t) dt ≤ (N + 1) g(t) dt < N 0 0 Secondly, the property (11.83) actually holds for all 0 < |t| ≤ π so that Z δ Z π 2 δ g(t) dt g(t)KN (t) dt < 1 N 1 t2 N N Z π 2 δ dG(t) = N 1 t2 N Z δ 1 G(t) i π 2 h G(δ) 2 + 2 − N G dt = 1 N δ2 N t3 N Z δ dt  π2  ǫ < + 2ǫ 1 t2 N δ N  π2  ǫ + 2ǫN < N δ < 3π 2 ǫ.

(11.87)

Thirdly, we note that Z π Z π π2 g(t) dt g(t)KN (t) dt ≤ N δ2 δ δ Z Z
i
π2 h i(θ−t) ≤ |f ei(θ+t) | dt |f e | dt + 2 Nδ T T π2 · 2kf k1 = N δ2 0. Using [100, Eqn. (3), §9.7, p. 183], we find that Z ∞ Z 1 1 ∞ λf (x − y) u(z) = u(x, λ) = (f ∗ hλ )(x) = √ dy. f (x − y)hλ (y) dy = π −∞ y 2 + λ2 2π −∞

(11.89)

We prove the problem by showing the following steps: • Step 1: ϕ(z, t) is harmonic in Π+ for every t ∈ R. By the change of variable t = x− y, the expression (11.89) can be written as Z Z 1 ∞ λf (t) 1 ∞ u(z) = u(x, λ) = dt = ϕ(x + iλ, t)f (t) dt, π −∞ (x − t)2 + λ2 π −∞ where ϕ : Π+ × R is given by

ϕ(x + iλ, t) =

  1 λ . = Im (x − t)2 + λ2 t − (x + iλ)

(11.90)

1 Since t−(x+iλ) is holomorphic in Π+ for every t ∈ R, Theorem 11.4 ensures that ϕ is + harmonic in Π .n Hence ϕ(z, t) is continuous on Π+ and then it satisfies the mean value property for every t ∈ R: Z π 1 ϕ(z, t) = ϕ(z + reiθ , t) dθ, 2π −π

where r is any positive number such that D(z, r) ⊆ Π+ . • Step 2: ϕ(z, t) ≤ such that

Mz 1+t2

for some Mz > 0. We claim that there exists a constant Mz > 0 0 < ϕ(z, t) = ϕ(x + iλ, t) ≤

Mz 1 + t2

(11.91)

for all t ∈ R. To see this, simple calculation shows that

holds for all t ∈ R if and only if

λ Mz ≤ 2 2 (x − t) + λ 1 + t2

(λ − Mz )t2 + 2Mz xt + (λ − Mz x2 − Mz λ2 ) ≤ 0

for all t ∈ R if and only if

(2Mz x)2 − 4(λ − Mz )(λ − Mz x2 − Mz λ2 ) ≤ 0 λMz2 − (x2 + λ2 + 1)Mz + λ ≥ 0

and the inequality (11.92) is true for large Mz > 0. This proves the claim. n

In fact, it can be shown directly from the definition (11.90) that, for each t ∈ R, we have ϕxx =

which give ∆ϕ = 0.

6λ(x − t)2 − 2λ3 [(x − t)2 + λ2 ]3

and

ϕλλ =

[(x − t)2 + λ2 ](−2λ) − 4λ[(x − t)2 − λ2 ] [(x − t)2 + λ2 ]3

(11.92)

382

Chapter 11. Harmonic Functions • Step 3: ϕ(z, t) ∈ Lq (R) for every 1 ≤ q ≤ ∞. By Step 2, if 1 ≤ q < ∞, then we have 1 ≤ 1 + t2 ≤ (1 + t2 )q so that Z ∞ Z ∞ Z ∞ Mzq 1 q ϕq (z, t) dt ≤ dt ≤ M · dt < ∞. (11.93) z 2 q 2 −∞ (1 + t ) −∞ −∞ 1 + t In other words, we have ϕ(z, t) ∈ Lq (R1 ). By the inequality (11.91), since ϕ(z, t) is obviously bounded by Mz , we know that ϕ(z, t) ∈ L∞ (R1 ). • Step 4: u satisfies the mean value property. Suppose that z ∈ Π+ and r is a positive number such that D(z; r) ⊆ Π+ . If q is the conjugate exponent of p, then we obtain from Theorem 3.8 and Step 3 that Z ∞ |ϕ(z + reiθ , t)f (t)| dt ≤ kf kp × kϕ(z + reiθ , ·)kq < ∞. kϕ(z + reiθ , ·)f (·)k1 = −∞

Next, we observe from the inequality (11.92) that we can pick Mζ in such a way that the set {Mζ | ζ = z + reiθ and −π ≤ θ ≤ π} is bounded. Therefore, we get Z π kϕ(z + reiθ , ·)f (·)k1 dθ < ∞. −π

Consequently, we apply Theorem 8.8 (The Fubini Theorem) and Step 1 to conclude that Z π Z ∞ Z π 1 1 u(z + reiθ ) dθ = 2 ϕ(z + reiθ , t)f (t) dt dθ 2π −π 2π −π −∞ Z Z π  1 ∞ 1 = ϕ(z + reiθ , t) dθ f (t) dt π −∞ 2π −π Z 1 ∞ ϕ(z, t)f (t) dt = π −∞ = u(z).

• Step 5: u is continuous on Π+ . Let {zn } be a sequence of Π+ converging to z0 ∈ Π+ . Then we must have ϕq (zn , t) → ϕq (z0 , t) as n → ∞. Now the inequality (11.93) guarantees that we may use Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) to conclude that Z 1 ∞ lim u(zn ) = lim ϕ(zn , t)f (t) dt n→∞ n→∞ π −∞ Z ∞h i 1 lim ϕ(zn , t) f (t) dt = π −∞ n→∞ Z 1 ∞ ϕ(z0 , t)f (t) dt = π −∞ = u(z0 ).

Thus u is continuous on Π+ . • Step 6: u is harmonic in Π+ . In fact, this follows immediately from Steps 4, 5 and Theorem 11.13. Hence we have completed the analysis of the proof.



CHAPTER

12

The Maximum Modulus Principle

12.1

Applications of the Maximum Modulus Principle

Problem 12.1 Rudin Chapter 12 Exercise 1.

Proof. Let f (z) = (z − a)(z − b)(z − c). Then f is nonconstant and entire. By Theorem 10.24 (The Maximum Modulus Theorem), we know that max |f (z)| = max |f (z)|. z∈∆

z∈∂∆

Let L = |a − b| = |b − c| = |c − a| and t = t(z) = |z − a| ∈ [0, L], see Figure 12.1 below.

Figure 12.1: The boundary ∂∆. If z ∈ [a, b], then we have |f (z)|2 = |z − a|2 · |z − b|2 · |z − c|2 √ h L 2  3L 2 i 2 2 + = t (L − t) · t − 2 2 383

384

Chapter 12. The Maximum Modulus Principle = t2 (L − t)2 (t2 − Lt + L2 ).

Define the function F : [0, L] → R by F (t) = t2 (L − t)2 (t2 − Lt + L2 ). Elementary differentiation shows that F ′ (t) = 6t5 − 15Lt4 + 16L2 t3 − 9L3 t2 + 2L4 t = t(2t − L)(3t3 − 6Lt2 + 5L2 t − 2L3 )

= t(2t − L)(t − L)(3t2 − 3Lt + 2L2 ).

Thus F ′ (t) = 0 if and only if t = 0, L2 , L. By the First Derivative Test, it is easily seen that F 3 6 L at t = L2 . Hence we conclude that attains its maximum 64 r   √ p L 3 3 max |f (z)| = max |f (z)| = max F (t) = F = L . z∈∆ z∈∂∆ 2 8 t∈[0,L] This completes the analysis of the problem.



Problem 12.2 Rudin Chapter 12 Exercise 2.

Proof. Suppose that f (i) = α. If |α| = 1, then Theorem 10.24 (The Maximum Modulus Theorem) forces that f (z) = α in Π+ . In this case, we have |f ′ (i)| = 0. Suppose that α ∈ U . By Definition 12.3, ϕα (α) = 0. Furthermore, we know from [11, Theorem 13.16, p. 183] that the mapping h : Π+ → U given by z − β  , (12.1) h(z) = eiθ z−β where Im β > 0 and θ ∈ R, is a bijection. Particularly, we take β = i in the definition (12.1). Clearly, we have h(i) = 0. Next, we consider the mapping F = ϕα ◦ f ◦ h−1 : U → U . Then we have F ∈ H ∞ , kF k∞ ≤ 1 and


F (α) = ϕα f h−1 (0) = ϕα f (i) = ϕα (α) = 0. Hence it follows from Theorem 12.2 (Schwarz’s Lemma) that |F ′ (0)| ≤ 1. Since h−1 (z) = i · that

z+eiθ eiθ −z

(12.2)

, we have (h−1 )′ (0) = 2ie−iθ . Consequently, we see from Theorem 12.4 F ′ (0) = ϕ′α (α) × f ′ (i) × (h−1 )′ (0) =

so the inequality (12.2) implies that |f ′ (i)| ≤

2ie−iθ f ′ (i) , 1 − |α|2

(12.3)

1 − |α|2 . 2

Thus |f ′ (i)| attains the maximum 12 when α = 0. In this case, we observe from the expression (12.3) that |F ′ (0)| = 1, so Theorem 12.2 (Schwarz’s Lemma) implies that
F (z) = λz for some constant λ with |λ| = 1. Since ϕ0 (z) = z, we conclude that f h−1 (z) = λz. Now if we put z = h(ζ), then it asserts that ζ − i . (12.4) f (ζ) = λh(ζ) = λeiθ ζ +i

Since λ = eiφ for some φ ∈ R, we may simply replace λeiθ by eiθ in the representation (12.4) which gives all extremal functions with |f ′ (i)| = 12 . This completes the proof of the problem. 

12.1. Applications of the Maximum Modulus Principle

385

Problem 12.3 Rudin Chapter 12 Exercise 3.

Proof. If f is constant, then there is nothing to prove. Thus, without loss of generality, we may assume that f is nonconstant. In this case, f has a local minimum in Ω if and only if f has a zero in Ω. Assume that f was a non-vanishing function in Ω. Then f1 ∈ H(Ω) and |f | has a local minimum at z0 ∈ Ω if and only if |f1 | has a local maximum at z0 . By Theorem 10.24 (The Maximum Modulus Theorem), f1 is forced to be constant which is impossible by our hypothesis.  Hence f has a zero in Ω, completing the proof of the problem. Problem 12.4 Rudin Chapter 12 Exercise 4.

Proof. (a) Assume that f was non-vanishing in D. By Theorem 10.24 (The Maximum Modulus Theorem), f 6= 0 on ∂D. Thus it is true that f 6= 0 in D. Denote M = |f (z)| on ∂D. On the one hand, Theorem 10.24 (The Maximum Modulus Theorem) again implies that |f (z)| ≤ M for all z ∈ D. On the other hand, since f1 ∈ H(D) and 10.24 (The Maximum Modulus Theorem) that

(12.5) 1 f

∈ C(D), we follow from Theorem

1 1 ≤ |f (z)| M

(12.6)

for all z ∈ D. Combining the inequalities (12.5) and (12.6), we conclude that |f (z)| = M in D, or equivalently, f (D) ∈ {−M, M } which contradicts the Open Mapping Theorem. Hence f has at least one zero in D. (b) This part is proven in [123, Problem 7.5, pp. 85 – 87]. We end the proof of the problem.



Problem 12.5 Rudin Chapter 12 Exercise 5.

Proof. Given that ǫ > 0. Since fn → f uniformly on ∂Ω, there exists an N ∈ N such that m, n ≥ N imply that |fn (z) − fm (z)| < ǫ (12.7) for every z ∈ ∂Ω. Since Ω is bounded, we may apply Theorem 10.24 (The Maximum Modulus Theorem) to assert that the inequality (12.7) holds for every z ∈ Ω, i.e., max |fn (z) − fm (z)| = max |fn (z) − fm (z)| < ǫ z∈Ω

z∈∂Ω

(12.8)

for m, n ≥ N . According to the Cauchy Criterion for Uniform Convergence [99, Theorem 7.8, p. 147], the inequality (12.8) ensures that {fn } converges uniformly on Ω which ends the proof  of the problem.

386

Chapter 12. The Maximum Modulus Principle

Problem 12.6 Rudin Chapter 12 Exercise 6.

Proof. By the definitions, Γ∗ is closed in C and then C \ Γ∗ is open in C. By Definition 10.1, C \ Γ∗ is a union of disjoint open connected sets, and hence components. Suppose that V is a ˙ · · · +γ ˙ n , where each γj component of C \ Γ∗ such that Ind Γ (z) 6= 0 for every z ∈ V . Let Γ = γ1 + is a closed path in Γ. Assume that V was unbounded. If z ∈ V , then it follows from Theorem 10.10 that Ind γj (z) = 0 for every 1 ≤ j ≤ n. Since Ind Γ (z) =

n X

Ind γj (z),

j=1

we have Ind Γ (z) = 0, a contradiction. Therefore, V must be bounded. Since Ind Γ (α) = 0 for all α ∈ / Ω, we get V ⊆ Ω. (12.9) Let ζ ∈ ∂V . Then ζ ∈ / V . If ζ lies in another component U , then there exists a δ > 0 such that D(ζ; δ) ∩ V 6= ∅ and D(ζ; δ) ∩ U 6= ∅, but this is a contradiction by [74, Theorem 25.1, p. 159]. Hence ζ ∈ / C \ Γ∗ , i.e., ζ ∈ Γ∗ and then ∂V ⊆ Γ∗ . Since Γ is a cycle in Ω, we have Γ∗ ⊆ Ω.

(12.10)

Combining the set relations (12.9) and (12.10), we conclude that V ⊆ Ω. Recall that V is bounded, V is compact. By the hypothesis, we have |f (ζ)| ≤ 1 for every ζ ∈ ∂V ⊆ Γ∗ . By Theorem 10.24 (The Maximum Modulus Theorem), we observe that |f (z)| ≤ 1 for every z ∈ V with Ind Γ (z) 6= 0. This completes the proof of the problem.



Problem 12.7 Rudin Chapter 12 Exercise 7.

Proof. Suppose that Ω = {x + iy | a < x < b, y ∈ R} and M (a) = 0. We have to show that M (x) = 0 for all x ∈ (a, b). Consider the function g(x + iy) = f (y + a + ix) which is defined in the horizontal strip Ω+ g = {x + iy | −∞ < x < ∞ and 0 < y < b − a}. By the definition, we know that |g(x)| = |f (a + ix)| = 0 for all x ∈ R. In particular, g is real on the segment L = (0, 1). By Theorem 11.14 (The − Schwarz Reflection Principle), there exists a function G holomorphic in Π = Ω+ g ∪ L ∪ Ωg such that G(x) = g(x) = 0 for every x ∈ L. By Theorem 10.18, we have G(z) = 0 in Π. Since G(z) = g(z) = 0 in Ω+ g , we obtain f (z) = 0 in Ω which implies the required result that M (x) = 0 for all x ∈ (a, b) and hence Theorem 12.8 (The Hadamard’s Three-Line Theorem) is also true if M (a) = 0. This completes the proof of  the problem.

12.1. Applications of the Maximum Modulus Principle

387

Problem 12.8 Rudin Chapter 12 Exercise 8.

Proof. Suppose that Ω = {z = x + iy | c < x < d and y ∈ R} for some −∞ < c < d < ∞. By the hypothesis, the exponential function ζ = ez maps Ω onto A(R1 , R2 ). We are given that R1 < a < r < b < R2 . Then there exist c < α < β < d such that ζ = ez maps the closed strip Ω′ = {z = x + iy | α ≤ x ≤ β and y ∈ R} onto the closed annulus A(a, b). Thus we have eα = a We define F : Ω′ → C by

and eβ = b.

(12.11)

F (z) = f (ez ) = f (ζ)

which is continuous on Ω′ and F ∈ H(Ω′ ). By Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem, there is a positive constant B > 0 such that |F (z)| < B for all z ∈ Ω′ . In other words, we may apply Theorem 12.8 (The Hadamard’s Three-Line Theorem) to our function F to get M (x)β−α ≤ M (α)β−x × M (β)x−α , (12.12) where the two expressions (12.11) give M (x) = sup{|F (x + iy)| | α ≤ x ≤ β and y ∈ R} = sup{|f (ex · eiy )| | α ≤ x ≤ β and y ∈ R}

= sup{|f (reiy )| | a ≤ r ≤ b and y ∈ R} = M (r).

Therefore, it deduces from the inequality (12.12) that (β − α) log M (r) ≤ (β − x) log M (a) + (x − α) log M (b) log M (r) ≤

log rb

log

b a

log M (a) +

log

log

r a b a

log M (b)

(12.13)

which is the required result. We claim that the equality (12.13) holds if and only if f (ζ) = Aζ λ for some A ∈ C and λ ∈ Z. Obviously, the equality holds if f is of this form. Conversely, the we can rewrite the equality (12.13) as log

b r b log M (r) = log log M (a) + log log M (b) a r a  b r r = log − log · [log M (a)] + log log M (b) a a a b r M (b) = log log M (a) + log log a a M (a) log ar M (b) . log M (r) = log M (a) + log M (a) log ab

If we denote λ = (log ab )−1 log

M (a) M (b)

(12.14)

∈ R, then the expression (12.14) can be further simplified to

log M (r) = log M (a) − λ log

r a

388

Chapter 12. The Maximum Modulus Principle

M (r) =

 a λ r

M (a).

(12.15)

If we combine the Extreme Value Theorem and the expression (12.15), then one can show that there exists an |ζ0 | = r ∈ (a, b) such that |f (ζ0 )| = M (r) = ( ar )λ M (a) which can be rewritten as |ζ0λ f (ζ0 )| = aλ M (a).

(12.16)

Notice that ζ λ f (ζ) may not be holomorphic because λ may not be an integer. However, remember that ζ = ez , so we can express the expression (12.16) as |eλz0 f (ez0 )| = aλ M (a) = eλα M (α), where ζ0 = ez0 ∈ Ω′ . Now eλz f (ez ) ∈ H(Ω′ ) and continuous on Ω′ , so we may apply Theorem 12.4 (The Maximum Modulus Theorem) to this function to conclude that eλz f (ez ) = c

(12.17)

in Ω′ , where c is some constant. Since Ω′ is a region, Theorem 10.18 asserts that the result (12.17) also holds in Ω. Using the substitution ζ = ez again, the result (12.17) becomes f (ζ) = cζ −λ
in A(R1 , R2 ). Recall that f ∈ H A(R1 , R2 ) , it forces that λ ∈ Z. This ends the analysis of the  problem. Problem 12.9 Rudin Chapter 12 Exercise 9.

Proof. We prove the assertions one by one. • |f (z)| ≤ 1 in the open right half plane Π. If α = 0, then |f (z)| ≤ Ae in Π. Since Π is unbounded and ∂Π is exactly the imaginary axis, we may apply Problem 12.11 to conclude that |f (z)| ≤ 1 in Π. Next, if α < 0, then there exists a R > 0 such that |z| ≥ R implies |f (z)| ≤ 1. (12.18) Since D(0; R) ∩ Π is compact, it is bounded. Therefore, Theorem 10.24 (The Maximum Modulus Theorem), the Extreme Value Theorem and the hypothesis |f (iy)| ≤ 1 for all y ∈ R ensure that the inequality (12.18) is also valid for all z ∈ Π. Thus we may assume that 0 < α < 1 in the following discussion. Denote Ω = {x + iy | x ∈ R, |y| < π2 }. Consider the mapping ϕ : Ω → Π defined by ϕ(z) = ez

(12.19)

which is clearly an isomorphism and ϕ ∈ H(Ω). Next, we define g : Ω → C by g = f ◦ ϕ. By Problem 10.14, we know that g ∈ H(Ω). By the definition of (12.19), we see that ϕ(∂Ω) = ∂Π. Since ϕ and f are continuous on Ω and Π respectively, g is continuous on Ω. Furthermore, we have   πi   πi  g x ± = f exp x ± = |f (±iex )| ≤ 1 2 2

12.1. Applications of the Maximum Modulus Principle

389

for all x ∈ R. Finally, if z ∈ Ω, then we have


|g(z)| = |f ϕ(z) | < A exp(|ϕ(z)|α ) < A exp(|ez |α ) = A exp(eαx ) ≤ A exp(eα|x| ). (12.20)

Choose B such that B − 1 ≥ log A. Note that eα|x| ≥ 1 for all x ∈ R. Then it is easy to see that log A ≤ (B − 1)eα|x|

log A + eα|x| ≤ Beα|x|

A exp(eα|x| ) ≤ exp(Beα|x| ).

(12.21)

Combining the inequalities (12.20) and (12.21), the inequality |g(z)| < exp(Beα|x| ) holds for all z = x + iy ∈ Ω. Hence Theorem 12.9 (The Phragmen-Lindel¨ of Theorem) asserts that |g(z)| ≤ 1 in Ω which means that the inequality (12.18) holds in Π. z

• The conclusion is false for α = 1. It is easy to check that the function f (z) = ee gives a counterexample to the result. • The modified result. Suppose that ∆ is an open sector between two rays from the origin with sectoral angle βπ < π for some β > 1. Suppose that f is continuous on ∆, f ∈ H(∆) and there are constants A < ∞ and α ∈ (0, β) such that |f (z)| < A exp(|z|α )

(12.22)

for all z ∈ ∆. Furthermore, if |f (z)| ≤ 1 on ∂∆, then we have |f (z)| ≤ 1 in ∆. To see this, let θ be the angle between the real axis and the ray nearest to it. Then we see that the mapping φ : ∆ → Π defined by φ(z) = −i(e−iθ z)β = −ie−iβθ exp(β log z) is clearly an isomorphism such that φ maps the boundary of ∆ onto the boundary of Π. Since 0 ∈ / ∆, we can define a branch for log z so that φ ∈ H(∆). By Theorem 10.33, we have φ−1 ∈ H(Π). Next, the map F : Π → C defined by F = f ◦ φ−1 1

1

1

is continuous on Π and F ∈ H(Π). Since φ−1 (ζ) = eiθ i β ζ β , we have |φ−1 (ζ)| = |ζ| β and thus the hypothesis (12.22) implies

where

α β


α |F (ζ)| = |f φ−1 (ζ) | < A exp(|φ−1 (ζ)|α ) = A exp(|ζ| β ),

< 1. If ζ = iy for some y ∈ R, then since φ−1 (iy) lies on ∂∆, we get
|F (iy)| = |f φ−1 (ζ) | ≤ 1.

Hence we establish from our first assertion that |F (ζ)| ≤ 1 in Π or equivalently, |f (z)| ≤ 1 in ∆. We have completed the analysis of the problem.



390

Chapter 12. The Maximum Modulus Principle

Problem 12.10 Rudin Chapter 12 Exercise 10.

Proof. For each n = 1, 2, 3, . . ., we define gn (z) = f (z)enz . Suppose that Lα = {z = reiα | r ≥ 0}, o n n πo π . and ∆2 = z ∈ Π α < arg z < ∆1 = z ∈ Π − < arg z < α 2 2

Certainly, ∆1 and ∆2 have the sectoral angles α + Figure 12.2:

π 2

< π and

π 2

− α < π respectively. Refer to

Figure 12.2: The sectors ∆1 , ∆2 and the ray Lα . In order to use Problem 12.9 as stated in the hint, it is necessary to assume that f is continuous on the closure of Π. Since f ∈ H(Π) and ∆1 , ∆2 are proper subsets of Π, each gn is 2π continuous on ∆j and gn ∈ H(∆j ) for j = 1, 2. Choose an γ ∈ (0, 2α+π ) and let Fn (x) = nx − xγ for x ≥ 0. By elementary calculus, the Fn attains its maximum value n 1 γ − 1 γ−1 Mn = n γ γ 1

at x = (nγ −1 ) γ−1 > 0. Therefore, if we pick An = 1 + eMn , then we have exp(n|z| − |z|γ ) = eFn (|z|) ≤ eMn < An

(12.23)

for all z ∈ ∆1 . Since |f (z)| < 1 for all z ∈ Π, for each fixed n ∈ N, we note from the inequality (12.23) that |gn (z)| < exp(n|z|) < An exp(|z|γ )

for all z ∈ ∆1 . Consequently, each gn satisfies the inequality (12.22). Furthermore, we observe from the definition of gn that log |f (reiα )| log |gn (reiα )| = + n cos α → −∞ r r as r → ∞, so |gn (z)| is bounded on Lα . Without loss of generality, we may assume that the bound is 1. Obviously, we know from the additional assumption that π
π
|gn re±i 2 | = |f re±i 2 | < 1 (12.24)

12.1. Applications of the Maximum Modulus Principle

391

for all r ≥ 0. In other words, gn is bounded by 1 on ∂∆1 . By the modified result of Problem 12.9, we conclude that each gn is bounded by 1 in ∆1 . Similarly, each gn is also bounded by 1 on ∂∆2 and then in ∆2 . Since Π = ∆1 ∪ Lα ∪ ∆2 , what we have shown is that each gn is bounded by 1 in Π, so for every z ∈ Π, this implies that |f (z)| < e−nr cos θ for all n ∈ N which means f (z) = 0 because cos θ > 0 for θ ∈ (− π2 , π2 ), i.e., f = 0 as required. This completes the proof of the problem.  Problem 12.11 Rudin Chapter 12 Exercise 11.

Proof. Since the result is trivial if f is constant, we assume that f is nonconstant in the following discussion. Besides, without loss of generality, we may assume that |f (z)| ≤ 1 on Γ = ∂Ω. It suffices to prove that |f (ω)| ≤ 1 (12.25) for every ω ∈ Ω. We choose a ∈ Ω and consider f (z) − f (a) . fe(z) = z−a

Using Theorem 10.16 and then Theorem 10.6, we see that fe ∈ H(Ω). Furthermore, the continuity of f on Ω ∪ Γ certainly implies that fb is also continuous on Ω ∪ Γ. Now the boundedness of f guarantees that fe(z) → 0 as z → ∞. In other words, there is a positive constant C such that |fe(z)| ≤ C

(12.26)

in Ω ∪ Γ. Next, let ΩR = D(0; R) ∩ Ω ⊆ Ω for R > 0 and Fe(z) = f N (z)fe(z) for some N ∈ N. Clearly, we have Fe ∈ H(Ω).

By the boundedness of f and the fact fe(z) → 0 as z → ∞, we can find R large enough such that ω ∈ ΩR and |Fe(z)| ≤ C for all z ∈ ∂ΩR = C(0; R) ∩ Γ ⊆ Γ from the bound (12.26). By Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem, we assert that |Fe (ω)| ≤ C. (12.27) e If f (ω) 6= 0, then we deduce from the inequality (12.27) that 1

CN |f (ω)| ≤ 1 . e |f (ω)| N

(12.28)

Taking N → ∞ in the inequality (12.28) which yields the required result (12.25) immediately. Consequently, the inequality |f (ω)| ≤ 1 holds for all ω ∈ Ω \ Zfe.

Assume that Zfe had a limit point in Ω. Now Theorem 10.18 ensures that fe ≡ 0 in Ω and then f (z) = f (a) for all z ∈ Ω, a contradiction to our hypothesis that f is nonconstant. Therefore, Zfe is discrete and hence the continuity of f forces definitely that the inequality (12.25) remains  valid in Ω which completes the proof of the problem.

392

12.2

Chapter 12. The Maximum Modulus Principle

Asymptotic Values of Entire Functions

Problem 12.12 Rudin Chapter 12 Exercise 12.

Proof. Let E1 = {z ∈ C | |f (z)| > 1}. Since f is nonconstant, E1 6= ∅. Let F1 be a component of E1 . By Definition 10.1, F1 is an open set. By the continuity of f , it is true that |f (z)| ≥ 1 on ∂F1 . Assume that |f (z0 )| > 1 for some z0 ∈ ∂F1 . Obviously, z0 ∈ E1 . Furthermore, we also have |f (ω)| > 1 for all ω ∈ D(z0 ; δ) for some δ > 0 by the Sign-preserving Property [127, Problem 7.15, p. 112]. Therefore, D(z0 ; δ) ⊆ E1 . By [74, Theorem 25.1, p. 159], we know that D(z0 ; δ) intersects only F1 which is impossible by the definition of a boundary point of a set [5, Definition 3.40, p. 64]. Hence we have |f (z)| = 1 on ∂F1 . Assume that F1 was bounded. Since F1 is compact, it follows from Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem that f attains its maximum on ∂F1 . Thus |f (z)| ≤ 1 on F1 , contradicting to the fact that F1 ⊆ E1 . Consequently, every component of E1 is unbounded. Now for every n = 2, 3, . . ., we define En = {z ∈ C | |f (z)| > n}. By similar argument, it can be shown easily that every component of En is unbounded. Since f is unbounded on F1 , En 6= ∅ for every n = 2, 3, . . .. Clearly, we have En+1 ⊆ En for every n ∈ N. Let Fn+1 be a component of En+1 . By similar argument of the previous paragraph, Fn+1 is open in C so that it is a region. By the definition, we have Fn+1 ⊆ En and then it must lie entirely in a component of En , namely Fn . Hence, we obtain a sequence of regions F1 ⊇ F2 ⊇ · · · .

(12.29)

For each k = 1, 2, . . ., pick zk ∈ Fk . Now the sequence (12.29) ensures that zn ∈ Fk for all n ≥ k. Using [11, Proposition 1.7, p. 14], one can find a continuous mapping γk : [k − 1, k] → Fk connecting zk and zk+1 . We note that the definition of Ek asserts that |f (z)| > k for all z ∈ Fk , so
particularly, it is also true on γk , i.e., |f γ(t) | > k for every t ∈ [k −1, k]. Define γ : [0, ∞) → F1 by ˙ γ2 + ˙ ··· . γ = γ1 +

t ), then γ e is a Then it is easy to see that f (γ(t)) → ∞ as t → ∞. Finally, if we set γ e(t) = γ( 1−t well-defined continuous function on [0, 1) and satisfies

lim f (e γ (t)) = ∞,

t→1

completing the proof of the problem.



Problem 12.13 Rudin Chapter 12 Exercise 13.

Proof. If z = x < ∞, then |ez | = e−x → 0 as |z| → ∞. By the definition, 0 is an asymptotic value of ez . Since ez is nonconstant entire, ∞ is also one of its asymptotic value by Problem 12.12. Hence it remains to show that if α is an asymptotic value of ez , then α is either 0 or ∞. Let γ : [0, 1) → C be a continuous curve such that γ(t) → ∞ and exp(γ(t)) → α as t → 1. Let γ(t) = a(t) + ib(t), where a and b are continuous real-valued functions. Then we note that a2 (t) + b2 (t) → ∞ as t → 1.

12.3. Further Applications of the Maximum Modulus Principle

393

If a(t) → +∞ as t → 1, then it is clear that | exp(γ(t))| = exp(a(t)) → ∞ as t → 1 so that α = ∞. Next, if a(t) → −∞ as t → 1, then we | exp(γ(t)) = exp(a(t)) → 0 as t → 1 and so α = 0 in this case. Finally, if a(t) → A as t → 1 for some finite A, then b(t) → ∞ as t → 1. Since b is continuous, there exist sequences {tn } and {t′n } in [0, 1) such that b(tn ) = 2nπ and b(t′n ) = (2n + 1)π respectively. Thus we have eγ(tn ) = ea(tn ) × eib(tn ) → A and

′

′

′

eγ(tn ) = ea(tn ) × eib(tn ) → −A

as n → ∞. These imply that α = A = −A and so α = A = 0. In conclusion, we have shown that exp has exactly two asymptotic values: 0 and ∞. For the entire functions sin z and cos z, we notice that sin z =

eiz − e−iz 2i

and

cos z =

eiz + e−iz . 2

Therefore, ∞ is the only asymptotic value of sin z and cos z, completing the proof of the problem.  Problem 12.14 Rudin Chapter 12 Exercise 14.

Proof. Suppose that f is nonconstant. Since f (z) 6= α for all z ∈ C, the function F (z) =

1 f (z) − α

is nonconstant entire. Now Problem 12.12 implies that F has ∞ as an asymptotic value which means that α is an asymptotic value of f . This completes the proof of the problem. 

12.3

Further Applications of the Maximum Modulus Principle

Problem 12.15 Rudin Chapter 12 Exercise 15.

Proof. The case is trivial if f is constant. So we may assume that f is nonconstant. Suppose first that Z(f ) = ∅. Then we have f1 ∈ H(U ). For every n ∈ N, we have 0 < 1 − n1 < 1. By combining Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem, we see that 1 1 ≤ max |f (0)| z∈C(0;1− n1 ) |f (z)| or equivalently, |f (0)| ≥

min

1 ) z∈C(0;1− n

|f (z)|.

We simply take zn = 1 − n1 which gives |f (zn )| ≤ |f (0)| for all n ∈ N. Obviously, |zn | → 1 as n → ∞, so our assertion is true in this case. Next, we suppose that Z(f ) 6= ∅. Then there are two cases.

394

Chapter 12. The Maximum Modulus Principle • Case (i): Z(f ) is infinite. Since Z(f ) ⊆ U which is bounded, the Bolzano-Weierstrass Theorem [127, Problem 5.25, pp. 68, 69] ensures that Z(f ) has a convergent subsequence {ζk }. Now Theorem 10.18 forces that ζk → ζ ∈ C(0, 1). Since f (ζk ) = 0 for all k ∈ N, the assertion remains true in this case. • Case (ii): Z(f ) is finite. Suppose that Z(f ) = {ζ1 , ζ2 , . . . , ζN } for some N ∈ N. Suppose further that mk is the order of zero of f at ζk , where 1 ≤ k ≤ N . Consider g(z) =

f (z) N Y

.

(12.30)

mk

(z − ζk )

k=1

Therefore, we know that g ∈ H(U ) and Z(g) = ∅. Thus the special case implies that there is a sequence {zn } ⊆ U and a positive constant M such that |zn | → 1 and |g(zn )| ≤ M for all n ∈ N. Hence it follows from the representation (12.30) that |f (zn )| ≤ M

N Y

k=1

|zn − ζk |mk ≤ M

N Y

(1 + |ζk |) < ∞

k=1

for all n ∈ N. Consequently, we have completed the proof of the problem.



Problem 12.16 Rudin Chapter 12 Exercise 16.

Proof. The result is always true if f is a constant function. Without loss of generality, we may assume that f is nonconstant. Let α = sup{|f (z)| | z ∈ Ω}. If |f (ζ)| = α for some ζ ∈ Ω, then f is constant by Theorem 10.24 (The Maximum Modulus Theorem), a contradiction. Thus we always have |f (z)| < α (12.31) for all z ∈ Ω. By the definition, there is a sequence {zn } ⊆ Ω such that |f (zn )| → α as n → ∞. Since Ω is bounded, the Bolzano-Weierstrass Theorem ensures that {zn } contains a convergent subsequence {znk }. Let znk → z0 . If z0 ∈ Ω, then α = |f (z0 )| which is impossible. Therefore, we must have z0 ∈ ∂Ω and our hypothesis gives α ≤ M . By the inequality (12.31), we conclude that |f (z)| < α ≤ M for all z ∈ Ω. This ends the proof of the analysis.



Problem 12.17 Rudin Chapter 12 Exercise 17.

Proof. By the definitions, we have Φ = {f ∈ H(U ) | 0 < |f (z)| < 1 for all z ∈ U } and where 0 < c < 1.

Φc = {f ∈ Φ | f (0) = c},

12.3. Further Applications of the Maximum Modulus Principle

395

• The value of M . Without loss of generality, we may assume that f (0) > 0. Otherwise, we can consider the function fb = eiθ f , where θ = − arg f (0). Then fb(0) = |f (0)| > 0 and fb ∈ Φ. Let Ω− = {z ∈ C | Re z < 0}. Since 0 < |f (z)| < 1 for z ∈ U , the mapping f1 = log f maps U into Ω− . Next, the mapping f2 (z) = −iz clearly maps Ω− onto the upper half plane Π+ . Finally, for Im α > 0, we know that the mapping f3 (z) =

z−α z−α

maps Π+ into U . Hence the mapping F = f3 ◦ f2 ◦ f1 maps U into U . Since f1 ∈ H(U ), f2 ∈ H(Ω− ) and f3 ∈ H(Π+ ), it is true that F ∈ H(U ). Clearly, the definition of F implies that F ∈ H ∞ , kF k∞ ≤ 1 and F (0) =

−i log f (0) − α . −i log f (0) − α

(12.32)

If we take α = −i log f (0), then Im α = − log f (0) > 0 because 0 < f (0) < 1. Thus the expression (12.32) gives F (0) = 0 in this case. By Theorem 12.2 (Schwarz’s Lemma), we have |F (z)| ≤ |z| for all z ∈ U and |F ′ (0)| ≤ 1.

Now the explicit formula of F is given by F (z) =

(z) log ff (0) −i log f (z) + i log f (0) = −i log f (z) − i log f (0) log[f (0)f (z)]

so that F ′ (z) =

f ′ (z) log f (0)2 · . f (z) {log[f (0)f (z)]}2

(12.33)

Since |F ′ (0)| ≤ 1, we see from the formula (12.33) that

|f ′ (0)| ≤ 2|f (0) log f (0)|.

(12.34)

Elementary calculus shows that the function g : (0, 1) → R defined by g(x) = x log x attains its absolute minimum −e−1 at x = e−1 . Therefore, we have M ≤ 2e−1 . Next, we claim that M = 2e−1 . To see this, we consider the function f (z) = e−2z−1 which is holomorphic in U . Since e−2 < |e2z | = e2r cos θ < e2 for all z = reiθ ∈ U , it is easy to see that 0 < e−3 < |f (z)| < e−1 < 1 in U so that f ∈ Φ. As f ′ (z) = −2e−2z−1 , we have |f ′ (0)| = 2e−1 = M as required.

• The value of M (c). Since f (0) = c, it follows from the inequality (12.34) that   2c| log c|, if c < e−1 ; M (c) =  −1 2e , if e−1 ≤ c < 1.

Hence we complete the analysis of the problem.



Remark 12.1 The first assertion of Problem 12.17 is called Rogosinski’s Theorem. See, for instances, [94] and [23, Exercise 6.36, pp. 213, 214] for a different proof.

396

Chapter 12. The Maximum Modulus Principle

CHAPTER

13

Approximations by Rational Functions

13.1

Meromorphic Functions on S 2 and Applications of Runge’s Theorem

Problem 13.1 Rudin Chapter 13 Exercise 1.

Proof. Suppose that f is meromorphic on S 2 and A ⊆ S 2 is the set of poles of f . If A is infinite, then the compactness of S 2 implies that A has a limit point in S 2 . However, this contradicts the note following Definition 10.41 and so A must be finite. Let A = {a1 , a2 , . . . , aN } for some N ∈ N and mk be the order of ak for 1 ≤ k ≤ N . If we define P (z) = f (z) ·

N Y

(z − ak )mk ,

k=1

then this expression implies that P (z) is entire and has at most a pole at ∞. If ∞ is not a pole of f , then P is a constant by Theorem 10.23 (Liouville’s Theorem). Otherwise, the function P ( z1 ) has a pole at 0. If P (z) = c0 + c1 z + c2 z 2 + · · · , then we have P ( 1z ) = c0 + cz1 + zc22 + · · · and the nature of its singularity at 0 implies that cp = cp+1 = · · · = 0 for some p ∈ N. In other words, P must be a polynomial. Since f (z) =

P (z) N Y

, mk

(z − ak )

k=1

f must be rational which completes the proof of the problem.  Problem 13.2 Rudin Chapter 13 Exercise 2.

Proof. (a) It is clear that Ω is simply connected, so S 2 \ Ω is connected by Theorem 13.11. Hence Theorem 13.9 (Runge’s Theorem) implies that there exists a sequence {Pn } of polynomials such that Pn → f uniformly on compact subsets of Ω. See Figure 13.1 for details. 397

398

Chapter 13. Approximations by Rational Functions

Figure 13.1: The simply connected set Ω. (b) The answer is negative. Consider the function f (z) =

1 z−

1 2

(13.1)

which belongs to H(Ω). Assume that there was a sequence {Pn } of polynomials such that Pn → f uniformly in Ω. Take γ = C(− 34 ; 14 ). Then we have γ ⊆ Ω, see Figure 13.1 again. On the one hand, we have Z 1 = 2πi. f (z) dz = 2πi · Ind γ 2 γ Furthermore, Theorem 10.12 gives

Z

Pn (z) dz = 0

γ

for every n ∈ N. On the other hand, the uniform convergence shows that Z Z Pn (z) dz = f (z) dz = 2πi, 0 = lim n→∞ γ

γ

a contradiction. Hence no such sequence exists. (c) The answer is still negative. The function (13.1) considered in part (b) is in fact holomorphic in C \ { 12 } which is open in C and it contains Ω. We complete the proof of the problem.  Problem 13.3 Rudin Chapter 13 Exercise 3.

Proof. For every n ∈ N, let Dn = {z ∈ D(0; n) | |Im z| ≥ n1 } and En = [ n1 , n] × {0}, see Figure 13.2. It is clear that both Dn and En are compact, so the set Kn = Dn ∪ En ∪ {0} is compact 1 . Then the sets too. Furthermore, we note that S 2 \ Kn is connected. Take 0 < δn < 2n o   n 1 1 and En′ = Dn′ = z ∈ D(0; n + δn ) |Im z| ≥ − δn − δn , n + δn × (−δn , δn ) n n

13.1. Meromorphic Functions on S 2 and Applications of Runge’s Theorem

399

are open sets containing Dn and En respectively. Obviously, the set Dn′ ∪ En′ is open in C and is disjoint from Un = (−δn , δn ) × (−δn , δn ). Define the function fn : Ωn = Dn′ ∪ En′ ∪ Un → C by   1, if z ∈ Un ; fn (z) =  0, if z ∈ Dn′ ∪ En′ .

Then we have fn ∈ H(Ωn ) and Ωn is an open set containing Kn . According to Theorem 13.7, one can find a polynomial Qn such that |Qn (z) − fn (z)| < n1 for all z ∈ Kn . In fact, we get   |Qn (z) − 1|, if z ∈ Un ; |Qn (z) − fn (z)| = (13.2)  |Qn (z)|, if z ∈ Dn′ ∪ En′ .

If we define Pn (z) = Qn (z) − Qn (0) + 1, then the definition (13.2) implies immediately that Pn (0) = Qn (0) − Qn (0) + 1 = 1 for n = 1, 2, . . .. Since we have C \ {0} =

∞ [

(Dn′ ∪ En′ ),

n=1

if z 6= 0, then there exists an N ∈ N such that z ∈ Dn′ ∪ En′ for all n ≥ N and thus the definition (13.2) implies that |Pn (z)| = |Qn (z) − Qn (0) + 1| ≤ |Qn (z)| + |Qn (0) − 1|
0 such that the modified tubes Ω′n of Ln with widths 2(2n+2) 4 + δn and Πn = D(0; 1 −

1 2n

+ δn ) also satisfy Ω′n ∩ Πn = ∅. Notice that Ω′n ⊆ Ωn .

Figure 13.4: The disc ∆n , the arc Ln and its neighborhood Ωn . We apply induction to construct the sequence of polynomials {Qn } and a holomorphic function f such that Qn → f uniformly on U : Consider Q0 ≡ 0 and   Q0 (z), if z ∈ Π◦1 ; f1 (z) =  1, if z ∈ Ω′1 . Obviously, we have f1 ∈ H(Π◦1 ∪ Ω′1 ). Since C \ (Π◦1 ∪ Ω′1 ) is connected, it follows from Theorem 13.9 (Runge’s Theorem) that one can find a polynomial Q1 such that |Q1 (z) − f1 (z)|
0.11

(13.11)

for all m = 3, 4, . . .. Next, we know that m−1 m−1  h  1 nk iθnk 1  nk i X 5k X 5k ≤ . 1 − − S 1 − 1 − e m−1 5m nm 5m nm

(13.12)

k=1

k=1

Using differentiation, we always have 1 − (1 − x)n ≤ nx for every x ∈ [0, 1] and n ∈ N. By the definition of nm , we see that nm > 2j+1 (m − 1)(m − 2) · · · (m − j − 1)nm−j−1 for j = 0, 1, . . . , m − 2. Applying these to the inequality (13.12) to obtain m−1 m−1  1 nk iθnk 1 X k nk X 5k 1 − − S e 5 · ≤ m−1 5m nm 5m nm k=1

≤

1 5m

k=1 m−1 X k=1

5k 2m−k (m − 1)(m − 2) · · · k

m−1 1 X 1 · 5k−m ≤ m−1 2m−k k=1  1  1 · 1 − m−1 ≤ 8(m − 1) 5 < 0.06

(13.13)

for every m = 3, 4, . . .. Clearly, we deduce from the upper bound (13.10) that ∞ ∞ nk 1 nk iθnk X 5k 1 X 5k 5 5k  nm e = · a 1 − + ≤ nk . n m+1 5m nm 5m k 5m n e nm k=m+1 k=m+2 e m k=m+1

∞ X

(13.14)

406

Chapter 13. Approximations by Rational Functions As m ≥ 3, we get

nm+1 nm

≥ 6 which implies that 5 e

≤

nm+1 nm

5 < 0.0124. e6

(13.15)

Since nk > 2(k − 1)nm for k = m + 2, m + 3, . . ., we have nk

e nm > e2(k−1) and then the inequality (13.14) becomes ∞ X

k=m+2

∞ 5k  e2 X  5 k 1 nk iθnk 25 e ≤ < 0.026 1 − = 2 5m nm 5m e2 (e − 5)e2m

(13.16)

k=m+2

for every m ≥ 3. Finally, by combining the bounds (13.11), (13.13), (13.15) and (13.16), we conclude immediately that h (1 − 1 )eiθ
m−1  1 nk iθnk X 5k nm ≥ 1 − − S e |S + a | − m−1 m−1 m 5m 5m nm k=1

∞ X −

k=m+1

1 nk iθnk 5k  1 − e 5m nm

≥ 0.11 − 0.06 − 0.0124 − 0.026

> 0.

Consequently, there exists a constant C > 0 such that |h(z)| > C · 5m holds for every z with |z| = 1 −

1 nm

and m ≥ 3.

• Proof that h has no finite radial limits. If zm = (1 − assertion shows that  1  iθ  e > C · 5m h 1 − nm

1 iθ nm )e ,

then the previous

for every m ≥ 3 and θ ∈ [0, 2π]. Therefore, it guarantees that  1  iθ  e = ∞. lim h 1 − m→∞ nm

In other words, h has no finite radial limits.

• The h has infinitely many zeros in U . Assume that h had only finitely many zeros α1 , α2 , . . . , αp in U . Then the function ϕ(z) = z m

p Y

k=1

ϕαk (z) = z m

p Y z − αk 1 − αk z

(13.17)

k=1

has exactly the same zeros as h counted with multiplicity, where m is the order of zero of h at the origin. If h has no zero in U , then we let ϕ = 1. Now the function f = ϕh satisfies f ∈ H(U ). By the proof of Theorem 12.4, we know that |ϕαk (z)| < 1 if |z| < 1, so the definition (13.17) implies that |f (z)| > C · 5m (13.18)

13.3. Simply Connectedness and Miscellaneous Problems

407

for every z with |z| = 1− n1m and m ≥ 3. Furthermore, f has no zero in U so that f1 ∈ H(U ). Combining this fact, the inequality (13.18) and Theorem 10.24 (The Maximum Modulus Theorem), we see immediately that 1 1 < f (z) C · 5m for all z ∈ D(0; 1 − n1m ) and m ≥ 3. Since nm → ∞ as m → ∞, we conclude from this 1 that f (z) = 0 in U which is impossible.

• The function h assumes every complex number α infinitely many times in U . Define b h(z) = f (z) − α. Then b h ∈ H(U ). For large enough m, 5m > C2 |α| so that C |b h(z)| = |h(z) − α| ≥ |h(z)| − |α| > C · 5m − |α| > · 5m 2

for |z| = 1 − n1m . Therefore, we can apply similar argument as above to obtain the desired result. This completes the analysis of the problem.



Remark 13.1 (a) A sequence {nk } of positive integers is said to be lacunary if there is a constant c > 1 such that nk+1 > cnk for all k ∈ N. A power series ∞ X

ak z n k

(13.19)

k=1

is called a lacunary power series or a power series with Hadamard gaps. Thus our h is an example of this type of power series with c = 2. See, for instance, [133, Chap. V]. (b) In [52, Problem 5.36 & Update 5.36, p. 113], it is pointed out that the best known result concerning the number of zeros of a lacunary power series inside U is due to Chang [26], who proved that if ∞ X k=0

|ak |2+ǫ = ∞

for some ǫ > 0, then the series (13.19) has infinitely many zeros in any sector. See also [44], [75] and [121]

13.3

Simply Connectedness and Miscellaneous Problems

Problem 13.7 Rudin Chapter 13 Exercise 7.

Proof. Suppose that A intersects each component of S 2 \ Ω. Choose a sequence of compact sets Kn in Ω with the properties specified in Theorem 13.3. Fix a positive integer n. Let V be a

408

Chapter 13. Approximations by Rational Functions

component of S 2 \ Kn . By the proof of Theorem 13.9 (Runge’s Theorem), it suffices to prove that V ∩ A 6= ∅. Since every component of S 2 \ Kn contains a component of S 2 \ Ω, we have U ⊆V for at least one component U of

S2

(13.20)

\ Ω. Since A intersects every component of A ∩ U 6= ∅.

S2

\ Ω, we have (13.21)

Now the set relations (13.20) and (13.21) together imply that V ∩ A 6= ∅. If V ∩ A 6= ∅, then we are done. Otherwise, p ∈ V ∩ A′ , where A′ is the set of limit points of A. By the openness of V and the definition of limit points, there exists a δ > 0 such that q ∈ A ∩ D ′ (p; δ) and D(p; δ) ⊆ V . This means that V ∩ A 6= ∅. Hence we have obtained the requirement and this completes the analysis of the problem.  Problem 13.8 Rudin Chapter 13 Exercise 8.

Proof. Let Ω = C. For every n = 1, 2, . . ., we denote Kn = D(0; n) which is compact. Put A1 = A ∩ K1 and An = A ∩ (Kn \ Kn−1 ) for n = 2, 3, . . .. Since An ⊆ Kn and A has no limit point in C (hence none in Kn ), every An is a finite set. Put X Pα (z), Qn (z) = α∈An

where n = 1, 2, . . .. Since An is finite, each Qn is a rational function and the poles of Qn lie in An for n ≥ 2. In particular, Qn is holomorphic in an open set V containing Kn−1 . By the known fact given in Definition 10.5, the power series of Qn at 0 converges uniformly to Qn in Kn−1 . This means that for each n = 2, 3, . . ., there exists a polynomial Rn such that |Rn (z) − Qn (z)|
0,a then the above argument shows that there exists a nonempty open subset D of D(z0 ; δ) such that f ∈ H(D). Therefore, D ⊆ V which implies the contradiction D(z0 ; δ) ∩ V 6= ∅. As a result, it means that  V is a dense open subset of Ω which ends the proof of the problem. Remark 13.2 The result of Problem 13.11 is sometimes called Osgood’s Theorem [83]. In fact, Problem 10.5 is the well-known Vitali Convergence Theorem, see [117, p. 168].

Problem 13.12 Rudin Chapter 13 Exercise 12.

Proof. We regard C as R2 and consider the two-dimensional Lebesgue measure m2 . Suppose that f : R2 → R2 is measurable. Let f = u + iv and RN = [−N, N ] × [−N, N ]. Consider the measurable function u|RN : RN → R, where N ∈ N. By [111, Theorem 4.3, p. 32], there exists a sequence {ψN,k } of step functions converging to u|RN for almost every z ∈ RN , i,e., there corresponds an pN ∈ N such that k ≥ pN implies u|R (z) − ψN,k (z) < 1 (13.27) N 2N +1 for almost every z ∈ RN . Note that each ψN,k has the form ψN,k (z) =

mk X

αN,j χRN,j (z),

j=1

a

Without loss of generality, we may assume further that D(z0 ; δ) ⊆ Ω.

13.3. Simply Connectedness and Miscellaneous Problems

411

where {RN,j } forms a set of disjoint open rectangles and mk m     m  [k [ [k RN,j \ RN m2 RN ∆ RN,j ∪ RN,j = m2 RN \ j=1

j=1

j=1

Suppose that ΩN =

m [k

!


0, c2 + d2

so i is mapped into Π+ which proves that the transformation f maps Π+ onto itself. The converse part can be found in [123, Problem 13.16, p. 181], completing the proof of the problem.  Problem 14.2 Rudin Chapter 14 Exercise 2.

Proof. Denote Π+ to be the upper half plane. Let z ∈ U . We have to make clear the meaning of the reflection, namely z ∗ , of z with respect to the arc L. In fact, by the discussion in [11, pp. 102, 103]a , we know that z ∗ lies on the same ray as z and |z ∗ | = |z|−1 . In other words, we have z∗ = a

1 . z

See also [30, pp. 50, 51]

413

414

Chapter 14. Conformal Mapping

(a) We have the following analogous reflection theorem for this part: Lemma 14.1 Suppose that Ω ⊆ Π+ , L = R and every point t ∈ L is the center of an open disc Dt such that Π+ ∩ Dt ⊆ Ω. Let Ω− be the reflection of Ω, i.e., Ω− = {z ∈ C | z ∈ Ω}. If f ∈ H(Ω) and |f (zn )| → 1 for every {zn } in Ω which converges to a point of L, then there exists a function F , holomorphic in Ω ∪ L ∪ Ω− , such that  f (z), if z ∈ Ω ∪ L;    F (z) = (14.2) 1   , if z ∈ Ω− .  f (z) Proof of Lemma 14.1. Fix a point t ∈ L. By the hypothesis |f (z)| → 1 as z → t ∈ L, it is legitimate to select a disc Dt so small that f (z) 6= 0 in Π+ ∩ Dt . Then the function g(z) = i log f (z) is well-defined and holomorphic in Π+ ∩ Dt . Furthermore, we know that Im g(zn ) = log |f (zn )| → 0 for every {zn } in Ω converging to a point of L. By the application of Theorem 11.14 (The Schwarz Reflection Principle), we see that one can find a function G, holomorphic in Ω ∪ L ∪ Ω− , such that G(z) = g(z) in Ω and satisfies G(z) = G(z)

(14.3)

for all z ∈ Ω ∪ L ∪ Ω− . Define F (z) = e−iG(z) . Since f (z) = e−ig(z) , if z ∈ Ω, then we have F (z) = e−iG(z) = e−ig(z) = f (z). Next, if z ∈ Ω− , then z ∈ Ω and we deduce from the equation (14.3) and the definition of F that 1 1 F (z) = e−iG(z) = e−iG(z) = eiG(z) = = F (z) f (z) which is exactly the equation (14.2).



(b) Recall from [100, Eqn. (1), p. 281] that ψ(z) =

z−i z+i

is a conformal one-to-one mapping of Π+ onto U and ψ(R) ⊆ T . So the inverse ψ −1 (ζ) =

i(1 + ζ) 1−ζ

is a conformal one-to-one mapping of U onto Π+ . For every θ ∈ [0, 2π], we know that ψ −1 (eiθ ) =

sin θ i(1 + eiθ ) ∈ R. =− 1 − eiθ 1 − cos θ

14.1. Basic Properties of Conformal Mappings

415

b = ψ −1 (L) and Ω b = ψ −1 (Ω) are a segment of R and a region in Π+ respectively. Thus L

Define the map

b ⊆ Π+ → C. fb = f ◦ ψΩb : Ω

(14.4)

b converging to a z0 ∈ L, b the points ζn = ψ(zn ) ∈ Ω ⊆ U converging Then for every {zn } ∈ Ω to ζ0 = ψ(z0 ) ∈ L ⊆ T , so the hypothesis guarantees that
|fb(zn )| = f ψ(zn ) = |f (ζn )| → 1

as n → ∞. Hence it follows from Lemma 14.1 that there exists a function Fb, holomorphic b ∪L b∪Ω b − , such that in Ω  b ∪ L; b fb(z), if z ∈ Ω    Fb(z) = 1  b −,  , if z ∈ Ω  b f (z)
 b ∪ L; b f ψ(z) , if z ∈ Ω    (14.5) = 1  b −. 
, if z ∈ Ω  f ψ(z) b − = {z ∈ C | z ∈ Ω}. b If z ∈ Ω b − , then z ∈ Ω b and so we note from the definition of ψ Here Ω that 1 . (14.6) ψ(z) = ψ(z)

Hence the formula (14.5) becomes

where Ω∗ = {ζ ∈ C | ζ

−1

  f (ζ), if ζ ∈ Ω ∪ L;    F (ζ) = 1  , if ζ ∈ Ω∗ ,   −1
 f ζ

(14.7)

∈ Ω}.

(c) We have the following analogous reflection theorem for U : Lemma 14.2 (The Schwarz Reflection Principle for U ) Every eit ∈ L ⊆ T is the center of an open disc Dt such that Dt ∩ U lies in Ω. Denote Ω∗ to be the reflection of Ω, i.e., o n 1 Ω∗ = z ∈ C z ∗ = ∈ Ω . z

Suppose that f ∈ H(Ω) and Im f (zn ) → 0 for every sequence {zn } in Ω which converges to a point of L. Then there exists a function F , holomorphic in the set Ω ∪ L ∪ Ω∗ , such that   f (z), if z ∈ Ω ∪ L;   F (z) = (14.8) 1   ∗  f , if z ∈ Ω . z

416

Chapter 14. Conformal Mapping Proof of Lemma 14.2. With the same function ψ : Π+ → U as in part (b), we see that b and z → L” b is equivalent to saying that “ψ(z) ∈ Ω and ψ(z) → L”. This “z ∈ Ω property implies that the function (14.4) satisfies
Im fb(z) = Im f ψ(z) → 0 b and z → L. According to Theorem 11.14 (The Schwarz Reflection Principle), as z ∈ Ω b ∪L b∪Ω b − , such that there is a function Fb , holomorphic in Ω 
b ∪ L; b   f ψ(z) , if z ∈ Ω (14.9) Fb(z) =   f ψ(z
), if z ∈ Ω b −. Using the formula (14.6), the function (14.9) can be expressed in the following form:  f (ζ), if ζ ∈ Ω ∪ L;    F (ζ) = 1    f , if ζ ∈ Ω∗ . ζ

This completes the proof of Lemma 14.2.



Now, since ( α1 )−1 = α, it is easy to conclude from the expression (14.7) that if f (α) = 0 for some α ∈ Ω, then α1 ∈ Ω∗ so that 1 1 F = ∞, = α f (α) i.e., F has a pole at α1 . By the expression (14.8), the analogue of part (c) is that F ( α1 ) = f (α) = 0. For part (a), if f (α) = 0 for some α ∈ Ω, then the expression (14.2) implies that F (α) =

1 f (α)

= ∞,

i.e., F has a pole at α. This finishes the proof of the problem.



Problem 14.3 Rudin Chapter 14 Exercise 3.

b Proof. If |z| = 1, then it is easy to see that z = z1 and the rational function R(z) = R(z) · R( z1 ) satisfies 1 b = R(z) · R(z) = |R(z)|2 = 1 R(z) = R(z) · R z b = P , where P and Q are polynomials, we have P (z) = Q(z) on the unit if |z| = 1. Since R Q circle. Now the Corollary following Theorem 10.18 guarantees that P (z) ≡ Q(z) in C and this implies that R(z) = 1 for every z ∈ C, i.e., 1 1 = R z R(z) for every z ∈ C. Therefore, ω is a zero of order m of R if and only if ω1 is a pole of order m of R. This fact shows that the zeros and poles of R inside U completely determines all zeros and poles of R in C.

14.1. Basic Properties of Conformal Mappings

417

Next, suppose that R has a zero at z = 0 of order m. Suppose, further, that {0, α1 , α2 , . . . , αk } are the distinct zeros and poles of R inside U . We consider the product B(z) = z m ·

k Y z − αn 1 − αn z n=1

which is a rational function having the same zeros and poles of the same order as R. Recall z−αn | = 1 for |z| = 1 and thus |B(z)| = 1 for from the proof of Theorem 12.4, we know that | 1−α nz |z| = 1. Consequently, the quotient R(z) f (z) = B(z) |R(z)| =1 is a rational function without zeros or poles in D(0; r) for some r > 1. Since |f (z)| = |B(z)| on |z| = 1, we get from the Corollary following Theorem 10.18 that f (z) = c for some constant c with |c| = 1 in D(0; r) and hence in C \ { α11 , α12 , . . . , α1n }, i.e.,

R(z) = cB(z) = cz m ·

k Y z − αn 1 − αn z n=1

as desired. This ends the proof of the problem.



Problem 14.4 Rudin Chapter 14 Exercise 4.

Proof. We have R(z) > 0 on |z| = 1. By the hint, R must have the same number of zeros as poles in U . Let α1 , α2 , . . . , αN and β1 , β2 , . . . , βN be the zeros and poles of R inside U , where N is a positive integer. Next, we consider the rational function f (α, β, z) =

(z − α)(1 − αz) , (z − β)(1 − βz)

(14.10)

where α, β ∈ U . Obviously, if |z| = 1, then z · z = |z|2 = 1 and 1 − αz 6= 0 which imply that f (α, β, z) =

(1 − αz)(1 − αz) |1 − αz|2 z (z − α)(1 − αz) · = = > 0. z (z − β)(1 − βz) |1 − βz|2 (1 − βz)(1 − βz)

Now the representation (14.10) indicates that α and β are the only zeros and poles of f (α, β, z) inside U respectively. Therefore, the rational function Q(z) =

N Y

f (αn , βn , z)

n=1

has the same numbers of zeros and poles as those of R inside U . Consequently, the quotient F (z) =

R(z) Q(z)

is a rational function without zeros or poles in U , i.e., F ∈ H(U ). Since f (αn , βn , z) > 0 on |z| = 1 for every n = 1, 2, . . . , N , we also have Q(z) > 0 on |z| = 1 and hence F (z) > 0 on |z| = 1. In other words, Im F ≡ 0 on |z| = 1. Recall that Im F is a continuous real-valued

418

Chapter 14. Conformal Mapping

function on U and is harmonic in U , so we follow from [9, Corollary 1.9, p. 7] that Im F ≡ 0 in U . Finally, using [11, Proposition 3.6, p. 39], there is a positive constant c such that F (z) = c

(14.11)

in U . By the Corollary following Theorem 10.18, we conclude that the result (14.11) holds in Ω = C \ {β1 , β2 , . . . , βN , β1 , β1 , . . . , β1 } which means that 1

2

N

R(z) = cQ(z) = c

N Y (z − αn )(1 − αn z) (z − βn )(1 − βn z) n=1

(14.12)

in Ω, completing the analysis of the problem.



Problem 14.5 Rudin Chapter 14 Exercise 5.

Proof. Suppose that g(ζ) =

n X

ak ζ n which is a rational function. Then g(eiθ ) = f (θ), so g is

k=−n

positive on T . By the representation (14.12), one can find a positive constant c such that g(ζ) = c

n Y (ζ − αj )(1 − αj ζ) (ζ − βj )(1 − βj ζ) j=1

(14.13)

in Ω = C \ {β1 , β2 , . . . , βn , β1 , β1 , . . . , β1 }. Here {α1 , α2 , . . . , αn } and {β1 , β2 , . . . , βn } are sets of n 1 2 zeros and poles of g inside U respectively. Since ζ n g(ζ) is a polynomial, the expression (14.13) implies that β1 = β2 = · · · = βn = 0. Hence we obtain f (θ) = g(eiθ ) n Y (eiθ − αj )(1 − αj eiθ ) =c eiθ =c =c =c

j=1 n Y

(eiθ − αj )(e−iθ − αj )

j=1 n Y

(eiθ − αj )(eiθ − αj )

j=1 n Y

j=1

|eiθ − αj |2

n √ Y 2 = c (eiθ − αj ) j=1

where P (z) =

√

= |P (eiθ )|2 ,

c(z − α1 )(z − α2 ) · · · (z − αn ). We have completed the proof of the problem. 

Problem 14.6 Rudin Chapter 14 Exercise 6.

14.1. Basic Properties of Conformal Mappings

419

Proof. If α = 0, then ϕ0 (z) = z, so the fixed points of ϕ0 are U . Next, we suppose that α 6= 0. By Definition 12.3, ϕα (z) = z if and only if z 2 = αα if and only if z=±

α . |α|

This gives our first assertion. For the second assertion, we note that since ϕα is a special case of the linear fractional transformation ϕ(z) = az+b cz+d , we consider the general case for ϕ. By a suitable rotation, we may assume that the straight line is the real axis. We claim that ϕ maps R ∪ {∞} into R ∪ {∞} if and only if a, b, c and d are real. It is easy to see that if a, b, c and d are real, then we have ϕ(R∪{∞}) ⊆ R∪{∞}. Conversely, suppose that ϕ(R∪{∞}) ⊆ R∪{∞}. Since ϕ(0) ∈ R∪{∞}, we have either d = 0 or db ∈ R. • Case (i): d = 0. Since ϕ(z) → ac as z → ∞ along R, we have ac ∈ R or c = 0. In the latter case, the transformation is ϕ(z) = ∞ for all z ∈ R ∪ {∞} and we can rewrite it as 1 . For the case ac ∈ R, since ϕ(1) = ac + bc ∈ R with c 6= 0, we have bc ∈ R. ϕ(z) = 0·z+0 Therefore, we obtain a z + bc az + b = c . ϕ(z) = cz 1·z+0 • Case (ii):

b d

∈ R. Note that d 6= 0. Let B = db . Then we have az + Bd . cz + d

ϕ(z) =

(14.14)

Using ϕ(z) → ac as z → ∞ along R again, we know that A = ac ∈ R or c = 0. In the latter case, we have ϕ(z) = ad z + db . Since ϕ(1) = ad + db ∈ R and db ∈ R, we have ad ∈ R so that ϕ(z) = For the case A =

a c

+ db . 0z + 1

a dz

∈ R, the representation (14.14) becomes ϕ(z) =

Acz + Bd . cz + d

(14.15)

If c + d = 0, then c = −d 6= 0 so that the representation (14.15) reduces to ϕ(z) =

−Adz + Bd −Az + B = . −dz + d −z + 1

d d Otherwise, ϕ(1) = Ac+Bd c+d = A + (B − A) · c+d ∈ R if and only if c+d ∈ R. Since d 6= 0, c d c+d ∈ R if and only if d ∈ R. Let c = Cd for some C ∈ R. Now the representation (14.15) reduces to ACdz + Bd ACz + B ϕ(z) = = . Cdz + d Cz + 1

This proves our claim. Return to our original problem. Suppose that ϕα maps a straight line L into itself. It is clear that eiθ L = R for some θ ∈ [0, 2π], so we assume that eiθ ϕα maps R into R. Write eiθ ϕα (z) =

eiθ z − eiθ α . (eiθ α)z + eiθ

Then the above claim says that eiθ , eiθ α and eiθ α are real, or equivalently, both eiθ and α are  real. Hence we end the analysis of the problem.

420

Chapter 14. Conformal Mapping

Problem 14.7 Rudin Chapter 14 Exercise 7.

Proof. Suppose that z, ω ∈ U . By the definition, fα (z) = fα (0) = 0 so that z = 0 and α is arbitrary in this case. Thus, without loss of generality, we may assume that z, ω 6= 0. Now fα (z) = fα (ω) if and only if (z − ω)(1 − αzω) = 0 if and only if z=ω

or

α=

1 . zω

(14.16)

1 Since |z| < 1 and |ω| < 1, we have |zω| > 1. Suppose that |α| ≤ 1. Then the result (14.16) leads to us that z = ω, i.e., fα is one-to-one in U . Let |α| ≤ 1. It is easy to check that fα (0) = 0 and fα′ (0) = 1. Besides, we have f ∈ H(U ). Combining these facts and the previous result, we know that fα ∈ L and

f (z) = z − αz 3 + α2 z 5 + · · · .

Therefore, it deduces from Theorem 14.14 that D(0; 14 ) ⊆ fα (U ). This ends the analysis of the  problem. Problem 14.8 Rudin Chapter 14 Exercise 8.

Proof. Let z = reiθ , where r > 0 and 0 ≤ θ < 2π. Then we have   e−iθ 1 1 f (reiθ ) = reiθ + cos θ + i r − sin θ. = r+ r r r Suppose that   1 1 cos θ and y = r − sin θ. (14.17) x= r+ r r
If r = 1, then x = 2 cos θ and y = 0 so that f C(0; 1) = [−2, 2]. Suppose, otherwise, that r 6= 1, so we obtain y2 x2 =1 1 2 + (r + r ) (r − 1r )2

which is an ellipse. Next, suppose that θ is a fixed number. Denote Lθ = {reiθ | 0 ≤ r < ∞}. If θ = 0, then cos θ = 1 and sin θ = 0 and thus it easily follows from the representations (14.17) that f (L0 ) = (0, ∞). Similarly, if θ = π, then cos θ = −1 and sin θ = 0 so that f (Lπ ) = (−∞, 0). If θ =

π 3π 2, 2 ,

then cos θ = 0 and sin θ = ±1. Simple computation verifies that

f L π2 = f L 3π = iR. 2

Finally, if θ ∈ [0, 2π) \ {0, representations (14.17) that

π 3π 2 , π, 2 },

then we have cos θ · sin θ 6= 0 and we deduce from the

y2 x2 − =4 cos2 θ sin2 θ which is trivially a hyperbola. This completes the analysis of the problem.



14.1. Basic Properties of Conformal Mappings

421

Problem 14.9 Rudin Chapter 14 Exercise 9.

Proof. Define Ω1 = {z ∈ C | 0 < Im z < π} and Π+ = {z ∈ C | Im z > 0}. πi ′ (a) Notice that the map f1 : Ω → Ω1 defined by f1 (z) = πi 2 (z + 1) satisfies f (z) = 2 6= 0 in Ω. By Theorem 14.2, it is a one-to-one conformal mapping of Ω onto Ω1 . Next, we know from [11, p. 176] that f2 (z) = ez is a one-to-one conformal mapping of Ω1 onto the z−i is a one-to-one upper half plane Π+ . Recall from [100, Eqn. (1), p. 281] that f3 (z) = z+i conformal mapping of Π+ onto U . Hence the mapping f = f3 ◦ f2 ◦ f1 : Ω → U is the required mapping. Explicitly, we have

f (z) =

exp( iπ 2 z) − 1

exp( iπ 2 z) + 1

.

(b) The inverse f −1 : U → Ω is given by f −1 (z) = −

1 + z  1 + z  2i 2i 1+z 2 − log log = arg . π 1−z π 1−z π 1−z

(14.18)

If f −1 = u + iv, then we have u(z) =

1 + z 1 + z  2 2 and v(z) = − log arg . π 1−z π 1−z

Since f −1 ∈ H(U ), u must be bounded and harmonic in U . Its harmonic conjugate v is unbounded in U because |v(z)| → ∞ as z → ±1. It remains to prove that u can be extended continuously to U . The definition of f shows that it can be extended to a continuous function of Ω onto U . Hence its inverse, which is u, can also be extended to a continuous function on U . (c) Since our f −1 and g satisfy the hypotheses of Problem 14.10, we establish immediately that

g D(0; r) ⊆ f −1 D(0; r) (14.19)

for all 0 < r < 1. Of course, we observe from the definition (14.18) that 1 + z π − Im f −1 (z) = log (14.20) 2 1−z ∞ X for all z ∈ U . Let f −1 (z) = cn z n , where z ∈ U . Since n!cn = (f −1 )(n) (0) for every n=0

n = 0, 1, 2, . . ., it is easy to see that f −1 has the form f −1 (z) = −

∞

4i X z 2n−1 π 2n − 1 n=1

for all z ∈ U . Thus for every z ∈ D(0; r), we deduce from the expression (14.20) that ∞ 1 + |z| 2 2 1+r 4 X |z|2n−1 = −Im f −1 (|z|) = log . |f −1 (z)| ≤ = log π n=1 2n − 1 π 1 − |z| π 1−r Finally, the set relation (14.19) asserts that |g(z)| ≤ for all z ∈ D(0; r).

2 1+r log π 1−r

422

Chapter 14. Conformal Mapping

(d) Now suppose that Ω = {x + iy | − π2 < y < π2 } and h : Ω → Ω is a conformal bijective mapping such that h(a + iβ) = 0. It is known from Problem 14.32 that the mapping ψ(z) = log

1+z 1−z

sends U conformally, one-to-one and onto the horizontal strip Ω with ψ(0) = 0. Thus its inverse ez − 1 ψ −1 (z) = z e +1 is a conformal one-to-one mapping of Ω onto U and ψ −1 (0) = 0. Denote ψ −1 (α + iβ) = A so that eα+iβ − 1 = A. (14.21) eα+iβ + 1 It is well-known [11, Theorem 13.15, p. 183] that the conformal mapping ϕ of U onto itself and ϕ(A) = 0 is represented by ϕ(z) = eiθ ·

z−A 1 − Az

for some θ ∈ R. Therefore, the composition h = ψ ◦ ϕ ◦ ψ −1 is a conformal one-to-one mapping from Ω onto itself and



h α + iβ = ψ ϕ ψ −1 (α + iβ) = ψ ϕ(A) = ψ(0) = 0.

Notice that

ψ ′ (z) =

2 , 1 − z2

(ψ −1 )′ (z) =

2ez (ez + 1)2

and ϕ′ (z) = eiθ ·

1 − |A|2 . (1 − Az)2

By the Chain Rule, we have


h′ (α + iβ) = ψ ′ ϕ ψ −1 (α + iβ) · ϕ′ ψ −1 (α + iβ) · (ψ −1 )′ (α + iβ) 2eα+iβ (eα+iβ + 1)2 iθ 2eα+iβ e × α+iβ . =2× 2 1 − |A| (e + 1)2

= ψ ′ (0) × ϕ′ (A) ×

(14.22)

According to the value (14.21) and the expression (14.22), we see that  |eα+iβ − 1|2 −1 2eα |h′ (α + iβ)| = 2 1 − α+iβ × |e + 1|2 |eα+iβ + 1|2 4eα = α+iβ |e + 1|2 − |eα+iβ − 1|2 1 = cos β which is the desired result. Consequently, we have completed the proof of the problem.



Remark 14.1 Problem 14.9 contributes to the theory of the Principle of Subordination, read [23, Chap. VI, §5, pp. 207 - 215 ], [34, §1.5, pp. 10 – 13] or [81, Chap V, §9, pp. 226 – 236].

14.1. Basic Properties of Conformal Mappings

423

Problem 14.10 Rudin Chapter 14 Exercise 10.

Proof. Since f is one-to-one,
f −1 : Ω → U exists. Let h = f −1 ◦ g. Then it is clear that h ∈ H(U ), h(U ) = f −1 g(U ) ⊆ U and h(0) = 0. Thus Theorem 12.2 (Schwarz’s Lemma) ensures that |h(z)| ≤ |z| (14.23) for all z ∈ U . For every 0 < r < 1, if z ∈ D(0; r), then the
inequality (14.23) implies that h(z) ∈ D(0; r). Equivalently, this means that g(z) ∈ f D(0; r) and henceb g(D(0; r)) ⊆ f (D(0; r)),

completing the proof of the problem.



Problem 14.11 Rudin Chapter 14 Exercise 11.

Proof. Now we have Ω = {z ∈ U | Im z > 0}. Using [11, Example 1, p. 180; Theorem 13.16, p. 183], if f : Ω → U is a conformal bijective mapping, then f has the representation f (z) = eiθ ·

(z − 1)2 + 4α(z + 1)2 , (z − 1)2 + 4α(z + 1)2

(14.24)

(z − 1)2 + i(z + 1)2 (z − 1)2 − i(z + 1)2

(14.25)

where θ ∈ [0, 2π] and Im α > 0. Putting f (−1) = −1, f (0) = −i and f (1) = 1 into the formula (14.24), we obtain that θ = π and α = 4i , so f (z) = −

is the desired conformal mapping.√By the representation (14.25), it is easily seen that if z ∈ Ω satisfies f (z) = 0, then z = (−1 + 2)i. Furthermore, simple algebra gives f ( 2i ) = 7i which ends  the proof of the problem. Problem 14.12 Rudin Chapter 14 Exercise 12.

Proof. For convenience, we let u(z) = Re f ′ (z) : C → R. We prove the assertions as follows: • f is one-to-one in Ω when u(z) > 0 for all z ∈ Ω. Choose a, b ∈ Ω and a 6= b. Since Ω is convex, the path γ(t) = a + (b − a)t for all t ∈ [0, 1] is in Ω. Then we know from the Fundamental Theorem of Calculus that Z 1 Z
f ′ a + (b − a)t dt = f (b) − f (a) f ′ (z) dz = (b − a) 0

γ

so that

Re

h f (b) − f (a) i

=

Z

1


u a + (b − a)t dt > 0.

b−a 0 Consequently, Re f (a) 6= Re f (b) which implies that f (a) 6= f (b). Since the pair of points {a, b} is arbitrary, we assert that f is one-to-one in Ω. b

Rudin used the notation “⊂” to mean “⊆”, see [99, Definition 1.3, p. 3].

424

Chapter 14. Conformal Mapping • f is either one-to-one or constant in Ω when u(z) ≥ 0 for all z ∈ Ω. Since f ∈ H(Ω), u has continuous derivative of all orders. Thus the set S = {z ∈ Ω | u(z) = 0} = u−1 (0)

(14.26)

is closed in C. Let p ∈ S. Since u is harmonic in Ω, it follows from the mean value property that Z 2π 1 u(p + Reiθ ) dθ (14.27) 0 = u(p) = 2π 0 for some R > 0 such that D(p; R) ⊆ Ω. If there exists a measurable set E ⊆ [0, 2π] such that m(E) > 0 and u(p + Reiθ ) > 0 for every θ ∈ E, then we have Z Z Z 2π u(p + Reiθ ) dm > 0 u(p + Reiθ ) dm + u(p + Reiθ ) dθ = 0

[0,2π]\E

E

which contradicts the result (14.27). Therefore, no such E exists and then u(p + Reiθ ) = 0 a.e. on [0, 2π]. Now the continuity of u on Ω forces that u(z) = 0 for every z ∈ D(p; R). In other words, D(p; R) ⊆ S which means that S is open in C.

Since S is both open and closed in C, we have either S = ∅ or S = C. Suppose that S = ∅, then Re f ′ (z) > 0 in Ω so that f is one-to-one in Ω by the previous assertion. Next, we suppose that S = C, then the definition (14.26) implies f ′ (Ω) is purely imaginary. Since f ′ ∈ H(Ω), the Open Mapping Theorem ensures that f ′ (Ω) = A for some constant A. Finally, if we consider g(z) = f (z) − Az, then g ∈ H(Ω) and g ′ ≡ 0 there. Thus it follows from [11, Exercise 5, p. 42] that g is a constant B. Consequently, we have f (z) = Az + B. If A 6= 0, then f is linear and so it is one-to-one. Otherwise, f ≡ B in Ω. • The condition “convex” cannot be replaced by “simply connected”. The following example can be found in [53].c Take β = π2 + δ for very small δ > 0. Define 1+ π Ω = {z ∈ C | − β < arg z < β} and f (z) = z 2β . Then it is clear that f ∈ H(Ω) and π π f ′ (z) = (1 + 2β )z 2β so that   π arg z  π π  2β Re f ′ (z) = 1 + |z| cos . (14.28) 2β 2β

z π Since −β < arg z < β, we have − π2 < π arg 2β < 2 and it follows from the expression (14.28) that Re f ′ (z) > 0 for every z ∈ Ω. Suppose that z1 = reiθ1 and z2 = reiθ2 are points of Ω, where θ1 6= θ2 . Clearly, 1+

π

1+

π

z1 2β = z2 2β .
π ) = 1 if and only if if and only if exp i(θ1 − θ2 )(1 + 2β θ1 − θ2 =

If − π2 − δ < θ2 < − π2 + θ1 = θ2 +

δ2 π+δ


0 such that z2 ∈ / D(z1 ; δ) and D(z1 ; δ) ⊆ Ω. Since every fn is one-to-one in D(z1 ; δ), none of the functions fn (z) − fn (z2 ) has a zero in D(z1 ; δ). By the hypotheses, fn (z) − fn (z2 ) → f (z) − f (z2 ) uniformly on every compact subset in D(z1 ; δ), so we deduces from Problem 10.20 (Hurwitz’s Theorem) that either f (z) − f (z2 ) 6= 0 for all z ∈ D(z1 ; δ) or f (z) − f (z2 ) = 0 in D(z1 ; δ). Since f (z1 ) = f (z2 ), we have f (z) = f (z2 ) in D(z1 ; δ) and this means that
D(z1 ; δ) ⊆ Z f − f (z2 ) .

By Theorem 10.18, we conclude immediately that f is constant in Ω. If fn (z) = n1 ez for every n ∈ N, then each fn is entire and one-to-one in C. Since fn (z) → 0 pointwise in C, we have f ≡ 0. On each compact set K of C, since ez is bounded on K, fn → f 1 uniformly on K. In this case, our limit function f is a constant. Next, we consider fn (z) = ez+ n in C, then each fn is entire and one-to-one in C. Besides, it is easily checked that fn → f = ez uniformly on every compact subset of C. In this case, we have f (z) = ez which is one-to-one in  C. This completes the proof of the problem. Problem 14.14 Rudin Chapter 14 Exercise 14.

Proof. Let’s answer the questions step by step. • f (x + iy) → 0 as x → ∞ for all y ∈ (−1, 1). Assume that one could find an y ′ ∈ (−1, 1) \ {0} such that the limit lim f (x + iy ′ ) is nonzero. Since |f | < 1, the Bolzanox→∞

Weierstrass Theorem ensures the existence of a sequence {xn } and a nonzero complex number L such that xn → ∞ as n → ∞ and lim f (xn + iy ′ ) = L.

n→∞

(14.31)

Consider the family F = {fn } ⊆ H(Ω), where fn (z) = f (z + xn ). Now the boundedness of f implies that F is uniformly bounded on each compact subset of Ω. By Theorem 14.6 (Montel’s Theorem), it is a normal family and then there exists a subsequence {nk } and an F ∈ H(Ω) such that fnk → F uniformly on every compact subset of Ω. It is clear that {x, iy ′ } is compact. On the one hand, the hypothesis gives F (x) = lim fnk (x) = lim f (x + xnk ) = 0. k→∞

k→∞

On the other hand, it follows from the limit (14.31) that F (iy ′ ) = lim fnk (iy ′ ) = lim f (xnk + iy ′ ) = L. k→∞

k→∞

Thus it is a contradiction and we obtain our desired result.

426

Chapter 14. Conformal Mapping • The passage to the limit is uniform if y is confined to [−α, α], where α < 1. Assume that the limit was not uniform in Kα = {x + iy | x ∈ R and y ∈ [−α, α]} for some α < 1. Then there exists some ǫ > 0 so that for all N ∈ N, one can find xN ≥ N and yN ∈ [−α, α] such that |fN (iyN )| = |f (xN + iyN )| > ǫ. (14.32) As we have shown above that {fN } has a subsequence {fNk } which converges uniformly on compact subsets of Ω to a holomorphic function g, and g 6≡ 0 in view of the inequality (14.32). By the previous assertion, we have
fNk (x + iy) = f xNk + (x + iy) → 0

as k → ∞ for every (x, y) ∈ [−α, α]2 ⊆ Kα ⊆ Ω, so this means that g(z) = 0 for all z ∈ [−α, α]2 and then the Corollary following Theorem 10.18 implies that g(z) = 0 in Ω, a contradiction. Hence the limit must be uniform in Kα . • Boundary behavior of a function g ∈ H ∞ with a radial limit.d Let g ∈ H ∞ . Without loss of generality, we may assume that |g(z)| < C1 for all z ∈ U and g(reiθ ) → C2 as r → 1 for some θ, where C1 and C2 are some constants. Using the mapping (14.124), we know that π e2z − 1 κ(z) = π z e2 + 1 is a conformal one-to-one mapping of Ω onto U . Since κ(x) → 1 as x → ∞, the composite
g κ(z)eiθ − C2 h(z) = C1 + C2 is a mapping from Ω into U satisfying h ∈ H(Ω), |h(z)| < 1 for all z ∈ Ω and
g κ(x)eiθ − C2 →0 h(x) = C1 + C2 as x → ∞. Hence the first assertion implies that lim h(x + iy) = 0 or

x→∞


lim g κ(x + iy)eiθ = C2

x→∞

(14.33)

for every y ∈ (−1, 1). We observe from the definition that |κ(x+iy)| < 1 and κ(x+iy) → 1 as x → ∞. By §11.21, the limit (14.33) means that g has non-tangential limit C2 at eiθ . The analogue of the second assertion can be stated similarly and we omit the details here. We complete the proof of the problem.

14.2



Problems on Normal Families and the Class S

Problem 14.15 Rudin Chapter 14 Exercise 15.

Proof. Let Π be the right half plane. Recall that the ϕ given by [100, Eqn. (6), p. 281] is a conformal one-to-one mapping of U onto Π. Thus ϕ−1 : Π → U is conformal and bijective. Since −1 ◦ f : U → U is holomorphic and ϕ−1 (z) = z−1 z+1 and f : U → Π is holomorphic, we have g = ϕ g(0) = ϕ−1 (f (0)) = ϕ−1 (1) = 0. d

Recall from Theorem 11.32 (Fatou’s Theorem) that our g has radial limits almost everywhere on T .

14.2. Problems on Normal Families and the Class S

427

According to Theorem 12.2 (Schwarz’s Lemma), we always have |g(z)| ≤ |z|

(14.34)

for all z ∈ U . Thus the auxiliary family G = {g = ϕ−1 ◦ f | f ∈ F } is uniformly bounded on each compact subset of U . Particularly, Theorem 14.6 (Montel’s Theorem) implies that G is a normal family. Let K be a compact subset of U . Then there exists a constant 0 < R < 1 such that K ⊆ D(0; R), so the inequality (14.34) gives |g(z)| ≤ R for
every g ∈ G and all z ∈ K. Since ϕ is conformal, it is continuous on U . Therefore, ϕ D(0; R) is bounded by a positive constant M and then

f (z) = ϕ g(z) ∈ ϕ D(0; R) ⊆ D(0; M ) (14.35)

for all z ∈ K. As g runs through G , f runs through F . Hence, it yields from the result (14.35) that F is uniformly bounded on K. Again, Theorem 14.6 (Montel’s Theorem) implies that F is normal. The condition “f (0) = 1” can be omitted or replaced by “|f (0)| ≤ 1”. In fact, we suppose that F ′ = {f ∈ H(U ) | Re f > 0} and the auxiliary family G ′ = {g = e−f | f ∈ F ′ }. It is evident that 1 |g(z)| = |e−f (z) | = Re f (z) ≤ 1 e

for all z ∈ U , the family G ′ is uniformly bounded on U . Using similar argument as the previous paragraph, it can be shown that the family F ′ is also normal. This completes the proof of the  problem. Problem 14.16 Rudin Chapter 14 Exercise 16.

Proof. Let p ∈ U . Then there exists a R > 0 such that D(p; 2R) ⊆ U . For every z ∈ D(p; R), we have D(p; r) ⊆ U for all 0 < r ≤ R. Since f ∈ H(U ), it is harmonic in U by Theorem 11.4. By the mean value property, we have f (z) =

1 2π

Z

2π

f (z + reit ) dt

0

which implies Z 2π 1 rf (z + reit ) dt 2π 0 Z R Z R Z 2π 1 rf (z) dr = rf (z + reit ) dt dr 2π 0 0 0 Z R Z 2π 1 R2 f (z) = rf (z + reit ) dt dr. 2 2π 0 0 rf (z) =

Applying Theorem 3.5 (H¨older’s Inequality) to the expression (14.36), we obtain Z Z 1 R 2π it rf (z + re ) dt dr πR2 0 0 Z Z o 1 n Z R Z 2π √ o1
2 1 n R 2π √ 2 2 2 · r) dr dt r|f (z + reit )| dr dt ( ≤ 2 πR 0 0 0 0

|f (z)| =

(14.36)

428

Chapter 14. Conformal Mapping

=

n 1 √ · πR · 2 πR

ZZ

|f (z)|2 dx dy

U

o1 2

1 ≤√ πR

(14.37)

for all z ∈ D(p; R). Let K ⊆ Ω be compact and {D(p; 2R)} be an open cover of K, where D(p; 2R) ⊆ U . Then there exist finitely many points p1 , p2 , . . . , pN and positive numbers R1 , R2 , . . . , RN such that K ⊆ D(p1 ; R1 ) ∪ D(p2 ; R2 ) ∪ · · · ∪ D(pN ; RN ). If R = min(R1 , R2 , . . . , RN ), then we conclude from the inequality (14.37) that F is uniformly 1 bounded by √πR on K. Hence Theorem 14.6 (Montel’s Theorem) shows that F is a normal family and we complete the proof of the problem.  Problem 14.17 Rudin Chapter 14 Exercise 17.

Proof. The conclusion is affirmative. To this end, we need the following version of Hurwitz’s Theorem [30, p. 152]: Lemma 14.3 (Hurwitz’s Theorem) Let Ω be a region and fn ∈ H(Ω) for n = 1, 2, . . .. If {fn } converges to f uniformly on compact subsets of Ω, f 6≡ 0, D(a; R) ⊆ Ω and f (z) 6= 0 on C(a; R), then there is an N ∈ N such that f and fn have the same number of zeros in D(a; R) for all n ≥ N. Fix an z0 ∈ Ω and let a ∈ Ω. Then the functions gn = fn − fn (z0 ) converge uniformly to g = f − f (z0 ) on compact subsets of Ω. Since f is one-to-one in Ω, g(z) 6= 0 on C(a; r) for every r > 0 such that D(a; r) ⊆ Ω. By Lemma 14.3 (Hurwitz’s Theorem), there corresponds an N (a, r) ∈ N such that gn and g have the same number of zeros in D(a; r) for all n ≥ N (a, r). In other words, we obtain fn (z) 6= fn (z0 ) (14.38) in D(a; r) \ {z0 } and for all n ≥ N (a, r). Suppose that K ⊆ Ω is compact, p ∈ K and {D(a; r)} is an open covering of K, where each D(a; r) is a subset of Ω. Then we have K ⊆ D(a1 ; r1 ) ∪ D(a2 ; r2 ) ∪ · · · ∪ D(am ; rm ) for some positive integer m. Let N (K) = max{N (a1 , r1 ), N (a2 , r2 ), . . . , N (am , rm )}. It follows from the result (14.38) that fn (z) 6= fn (p) (14.39) in K \ {p} for all n ≥ N (K). Since p is an arbitrary point of K, the result (14.39) means that fn is one-to-one in K for all n ≥ N (K), completing the proof of the problem.



Problem 14.18 Rudin Chapter 14 Exercise 18.

14.2. Problems on Normal Families and the Class S

429

Proof. Suppose that f, g : Ω → U and f (z0 ) = g(z0 ) = 0. Then F = g ◦ f −1 : U → U is a bijective conformal mapping such that
g′ (z0 ) . F (0) = g f −1 (0) = g(z0 ) = 0 and F ′ (0) = ′ f (z0 )

(14.40)

By [11, Theorem 13.15, p. 183], we know that

 z−α  1 − αz

F (z) = eiθ ·

(14.41)

for some |α| < 1 and θ ∈ [0, 2π]. By the conditions (14.40), we have α = 0 and eiθ = which imply that
g′ (z0 ) g′ (z0 ) f (z) = ′ ϕ0 f (z) g(z) = ′ f (z0 ) f (z0 )

g ′ (z0 ) f ′ (z0 )

for all z ∈ Ω. For the case that f (z0 ) = g(z0 ) = a, the conditions (14.40) are replaced by F (a) = a

and

F ′ (a) =

By the form (14.41) again, we can show that eiθ = g(z) =

g ′ (z0 ) f ′ (z0 )

g ′ (z0 ) . f ′ (z0 ) ·

(1−αa)2 1−|α|2

which gives


g′ (z0 ) (1 − αa)2 · ϕα f (z) f ′ (z0 ) 1 − |α|2

for all z ∈ Ω. This ends the proof of the problem.



Problem 14.19 Rudin Chapter 14 Exercise 19.

Proof. We claim that the mapping given by f (z) = z exp



i  1 − |z|

(14.42)

is a homeomorphism of U onto U . If we consider z = reiθ and represent the function (14.42) as  1  , (14.43) f (r, θ) = r, θ + 1−r

then the inverse function f −1 is given by

 f −1 (r, θ) = r, θ −

1  . 1−r

Now it is easy to see that both f and f −1 are continuous and bijective. In other words, f is a homeomorphism. However, this homeomorphism cannot be extended to U continuously. Otherwise, we assume that F
: U → U was a continuous extension to f , i.e., F = f on U . ′ Therefore, we have F C(0; 1) = C(0; 1). Take F (eiθ ) = (1, 0) for some θ ′ ∈ [0, 2π] and V any ′ neighborhood of (1, 0). Suppose that i ∈ / V . Then there exists a neighborhood W of eiθ such ′ iθ e that F (W ) ⊆ V . Indeed, we can find a sequence {rn e } in W such that rn < 1 for all n ∈ N, ′

rn → 1 and F (rn eiθ ) = f (rn , θ ′ ) → (1, 0) e

See, for example, [74, Theorem 18.1, p. 104].

430

Chapter 14. Conformal Mapping

as n → ∞. Using the formula (14.43), we have  f (rn , θ ′ ) = rn , θ ′ +

h 1  1 i = rn exp i θ ′ + . (14.44) 1 − rn 1 − rn
1 is continuous on [0, 1) and s [0, 1) = [θ ′ + 1, ∞), if we take Since the function s(r) = θ ′ + 1−r 1 the points θ ′ + 1−r = (2n + 1)π for all large n, then we deduce from the formula (14.44) that n h f (rn , θ ′ ) = i 1 −

i 1 ′ (2n + 1)π − θ

which means that f (rn , θ ′ ) cannot be contained in the neighborhood V of (1, 0) for all large n.  Hence no such continuous extension exists and we complete the proof of the problem. Problem 14.20 Rudin Chapter 14 Exercise 20.

Proof. Write f (z) = zϕ(z). Then ϕ ∈ H(U ), ϕ(0) = 1 and ϕ has no zero in U . By Problem 13.9, there exists an h ∈ H(U ) such that hn (z) = ϕ(z) and h(0) = 1. Put g(z) = zh(z n )

(14.45)

g n (z) = z n hn (z n ) = z n ϕ(z n ) = f (z n )

(14.46)

in U . Then we know that

for every z ∈ U . It is clear that g(0) = 0 and g ′ (0) = h(0) = 1.

To prove g ∈ L , it suffices to show that g is one-to-one in U . Suppose that z, ω ∈ U and g(z) = g(ω). Since f is one-to-one in U , the formula (14.46) ensures that z n = ω n or equivalently, z=e

2kπi n

ω,

where k = 0, 1, . . . , n − 1. Put this into the equation (14.45) to get g(z) = e

2kπi n

ωh(e2kπi ω n ) = e

2kπi n

  2kπi · ωh(ω n ) = e n g(ω).

(14.47)

Recall that g(z) = g(ω), so it follows from the equation (14.47) that we have g(z) = g(ω) = 0 2kπi or e n = 1. In the latter case, we have z = w Otherwise, since g(z) = 0 if and only if z = 0 in  U , we have z = w = 0, completing the proof of the problem. Problem 14.21 Rudin Chapter 14 Exercise 21.

Proof. By Definition 14.10, we have f (z) = z +

∞ X

n=2

an z n .

14.2. Problems on Normal Families and the Class S

431

(a) Since f ∈ S , it is one-to-one. By Theorems 10.33 and 14.2, f is conformal and its inverse f −1 : f (U ) → U exists and is conformal.f As U ⊆ f (U ), we consider g = f −1 |U : U → U . Clearly, we have g(0) = f −1 (0) = 0 and g ′ (0) =

1 f′


=

f −1 (0)

1 = 1. f ′ (0)

By Theorem 12.2 (Schwarz’s Lemma), we conclude that g(z) = z and consequently, f (z) = z in U . (b) We claim that there is no element f ∈ S with U ⊆ f (U ). Assume that f was such a function. By part (a), we know that f (z) = z so that f (U ) = U which contradicts our assumption that U ⊆ f (U ). (c) Consider the function F given in [100, Eqn. (1), p. 286]. If |α1 | = 1, then Theorem 14.13 (The Area Theorem) implies that αn = 0 for all n = 2, 3, . . .. In this case, we have F (z) =

1 + α0 + eiθ z. z

Now we know from the proof of Theorem 14.14 that |a2 | = 2 is equivalent to |α1 | = 1, so we have 1 1 = G(z) = + α0 + eiθ z. g(z) z By Theorem 14.12, we have f (z 2 ) = g 2 (z) which implies definitely that α0 = 0 and then f (z 2 ) =

z2 . (1 + eiθ z 2 )2

Consequently, f must be in the form f (z) =

z . (1 + eiθ z)2

We complete the proof of the problem.



Problem 14.22 Rudin Chapter 14 Exercise 22.

Proof. Let f : U → S be a one-to-one conformal mapping, where S is a square with center at 0. Now it is clear that both if : U → S and f −1 : S → U are conformal. We consider g = f −1 ◦ (if ) : U → U


which is also a one-to-one conformal mapping of U onto itself. Clearly, g(0) = f −1 if (0) = 0. By Remark 10.3, we have if ′ (z)
g′ (z) = ′ f if (z) which gives |g′ (0)| = |i| = 1. Hence Theorem 12.2 (Schwarz’s Lemma) implies that the formula if (z) = f (λz) f

Or it can be seen directly from [11, Theorem 13.8, p. 174].

(14.48)

432

Chapter 14. Conformal Mapping

holds in U for some constant λ with |λ| = 1. By the note following Theorem 14.2, we have f ′ (0) 6= 0. Now we observe from this fact and the expression (14.48) that if ′ (z) = λf ′ (λz) so that λ = i if we put z = 0 into this equation. Thus we get what we want if (z) = f (iz) for every z ∈ U . Suppose that f (z) =

P

∞ X

n=1

(14.49)

cn z n . By the expression (14.49), we have

icn (1 − in−1 )z n = 0

for all z ∈ U . If n − 1 is not a multiple of 4, then in−1 6= 1, so it yields from Theorem 10.18 that cn = 0 for such n. A generalization is as follows: Let S be a simply connected region with rotational symmetry of order N and center at 0, i.e., exp( 2πi N )S = S. Let f : U → S be a one-to-one conformal −1 : S → U are conformal. By mapping with f (0) = 0. Thus both exp( 2πi N )f : U → S and f )f ] : U → U which is clearly a one-to-one conformal mapping of considering g = f −1 ◦ [exp( 2πi N U onto itself. Obviously, we have   2πi   g(0) = f −1 exp f (0) = 0. N

We observe from Remark 10.3 that

g′ (z) =

f′

′ exp( 2πi N )f (z)
, 2πi exp( N )f (z)

so |g ′ (0)| = 1. Hence Theorem 12.2 (Schwarz’s Lemma) implies that e

2πi N

f (z) = f (λz)

(14.50)

holds for some constant λ with |λ| = 1. Since f ′ (0) 6= 0, we take differentiation to both sides of the formula (14.50) to conclude that λ = exp( 2πi N ) and hence e

2πi N

f (z) = f e

2πi N

z




for every z ∈ U . Next, if n − 1 is not a multiple of N and f (z) = (14.51) gives ∞ X  2π(n−1)i  2πi e N cn 1 − e N zn = 0

(14.51) P

cn z n , then the formula

n=1

2π(n−1)i N

6= 1 if n − 1 is not a multiple of N , Theorem 10.18 implies that for all z ∈ U . Since e cn = 0 for such n. This completes the analysis of the problem. 

14.3

Proofs of Conformal Equivalence between Annuli

Problem 14.23 Rudin Chapter 14 Exercise 23.

Proof. Suppose that ∂Ω = C1 ∪C2 , where C1 lies in the inside of C2 . By appropriate translation, rotation and homothety, we assume that C2 = T (the unit circle) and the center of C1 lies on the real axis with x-intercepts a and b, where |a| < b < 1. See Figure 14.1 below:

14.3. Proofs of Conformal Equivalence between Annuli

433

Figure 14.1: The region Ω bounded by C1 and C2 . By Theorem 12.4, ϕα carries T onto itself and U onto U , where |α| < 1. Since ϕα is a linear fractional transformation, §14.3 ensures that it will send C1 ⊂ U onto a circle or a line. Now the condition ϕα (U ) = U implies that ϕα (C1 ) must be a circle. Suppose that α ∈ R. Since C1 intersects the real axis at a and b perpendicularly and ϕα is a conformal map, ϕα (a) and ϕα (b) are the end-points of a diameter of ϕα (C1 ). Furthermore, since ϕα (R) = R, both ϕα (a) and ϕα (b) are real. Thus if ϕα (C1 ) is a circle centered at 0, then we must have ϕα (a) = −ϕα (b) which gives b−α a−α =− 1 − αa 1 − αb 2(1 + ab) 2 α − α + 1 = 0. a+b Solving this equation to get 1 + ab ± α± = a+b

r

1 + ab 2 − 1. a+b

(14.52)

Since |a| < b < 1, we always have a(1 − b) < 1 − b or equivalently, 1 + ab > a + b > 0. Combining this and the formulas (14.52), it follows that 1 + ab α+ = + a+b

r

1 + ab 2 − 1 > 1. a+b

Since α+ · α− = 1, we conclude that 0 < α− < 1. Take
this α− . Then the conformal map ϕα− carries U onto U , T onto T and C1 onto C 0; ϕα− (a) . In other words, it is a one-to-one
 conformal mapping of Ω onto A ϕα− (a), 1 . This completes the proof of the problem.

434

Chapter 14. Conformal Mapping

Remark 14.2 An example of this kind of conformal mappings can be found in [123, Problem 13.20, pp. 182, 183].

Problem 14.24 Rudin Chapter 14 Exercise 24.

Proof. Since 1 < R2 < R1 , we have A(1, R2 ) ⊆ A(1, R1 ). Assume that f : A(1, R1 ) → A(1, R2 ) was a bijective conformal mapping. By the first half of the proof of Theorem 14.22, we may assume without loss of generality that |f (z)| → 1 as |z| → 1 and |f (z)| → R2 as |z| → R1 . Consider the family of holomorphic functions F = {fn }, where f1 = f

and

fn+1 = f ◦ fn : A(1; R1 ) → A(1; R2 )

for all n ∈ N. Thus each fn is bijective. By the definition, we have


f A(1, R1 ) = A(1, R2 ) ⊆ A(1, R1 ) and


fn A(1, R1 ) ⊆ A(1, R1 )

(14.53)

for every n = 1, 2, . . ., so the family F is uniformly bounded on compact subsets of A(1, R1 ). Therefore, it follows from Theorem 14.6 (Montel’s Theorem) that F is normal and then there exists a subsequence {fnk } converging uniformly on compact subsets of A(1, R1 ) to a holomorphic function g : A(1, R1 ) → A(1, R2 ). Denote Ω = A(R2 , R1 ). We claim that fn (Ω) ∩ fm (Ω) = ∅,

(14.54)

where n, m ∈ N and n 6= m. Fix m. Assume that ω ∈ fm (Ω) ∩ fn (Ω). Then we can find p, q ∈ Ω such that fn (p) = ω and fm (q) = ω. If n < m, then we have
fn (p) = fm (q) = fn fm−n (q)


which gives fm−n (q) = p ∈ Ω, but it contradicts the fact that fm−n A(1, R1 ) ∩ Ω = ∅. If n > m, then we have
fm (q) = fn (p) = fm fn−m(p)

which gives fn−m (p) = q ∈ Ω, a contradiction again. Consequently, we prove the claim (14.54). Assume that the range of g contained a nonempty open set. This means that g is not constant and so we pick an p ∈ Ω ⊆ A(1, R1 ) such that g(p) = ω. By the Open Mapping

Theorem, g A(1, R1 ) is an open set so that there exists a δ > 0 such that 0 ∈ / g D(p; δ) . Suppose that hk (z) = fnk (z) − ω and h(z) = g(z) − ω.
We observe that hk , h ∈ H A(1, R1 ) , {hk } converges to h uniformly on compact subsets of A(1, R1 ) and h 6≡ 0. Furthermore, we can select δ if necessary so that h(z) 6= 0 on C(p; δ).g Hence it follows from Lemma 14.3 (Hurwitz’s Theorem) that there corresponds an N ∈ N such g

Otherwise, h(z) = 0 for all z ∈ A(1, R1 ) by Theorem 10.18 which means that g is constant.

14.3. Proofs of Conformal Equivalence between Annuli

435

that if k ≥ N , then hk and h have the same number of zeros in D(p; δ). Since h(p) = g(p)−ω = 0, there exists an zk ∈ D(p; δ) ⊆ Ω such that k ≥ N implies fnk (zk ) = ω but this contradicts the fact (14.54). Therefore, the range of g cannot contain any nonempty open set and the Open Mapping Theorem ensures that g is constant. √ On the other hand, g cannot be constant on the circle C(0; R1 ). Otherwise, Theorem√10.24 (The Maximum Modulus Theorem)√and its Corollary establish that g is constant in A(1, R1 ). Let K be a compact subset of A(1, R1 ). Since fnk → g uniformly on K, fnk is also constant in K for large enough k. However, this contradicts the fact that fn is injective for every n = 1, 2, . . .. √ Therefore, g is not constant on C(0; R1 ). Now the above two results are contrary, so they force that no such f exists and we have completed the proof of the problem.  Problem 14.25 Rudin Chapter 14 Exercise 25.

Proof. We have f : A(1, R1 ) → A(1, R2 ), where 1 < R2 < R1 . By the first half of the proof of Theorem 14.22, we may assume without loss of generality that lim |f (z)| = 1

|z|→1

and

lim |f (z)| = R2 .

|z|→R1

Applying Problem 14.2(b), the reflection across the inner circle (i.e., the unit circle) extends f to a conformal mapping f1 : A(R1−1 , R1 ) → A(R2−1 , R2 )

and f1 satisfies

lim

|z|→R−1 1

|f1 (z)| = R2−1 ,

lim |f1 (z)| = 1 and

|z|→1

lim |f1 (z)| = R2 .

|z|→R1

Next, by considering the function g1 (z) =

f1 (R1 z) , R2

we see that g1 is holomorphic in the region Ω = {z ∈ C | R1−2 < |z| < 1} such that |g1 (z)| → 1 as |z| → 1 and |g(z)| → R2−2 as |z| → R1−2 . Thus Problem 14.2(b) may be applied to extend g1 to a conformal mapping f2 : A(R1−2 , R12 ) → A(R2−2 , R22 ) and f2 satisfies lim

|z|→R−2 1

|f2 (z)| = R2−2 ,

lim |f2 (z)| = 1 and

|z|→1

lim |f2 (z)| = R22 .

|z|→R21

This process can be repeated infinitely many times and finally we obtain a conformal mapping F of the punctured plane C \ {0}. If F has a pole of order m at the origin, then z m F (z) is entire and |F (z)| = 1 whenever |z| = 1. Now Problem 12.4(b) asserts that F (z) = αz n for some |α| = 1 and some n ∈ Z \ {0}. Assume that n ≥ 2. Let a ∈ A(1, R1 ). Take ζ 6= 1 to be an n-root of unity. Then we have F (a) = αan = α(ζa)n = F (ζa), (14.55) but a 6= ζa. This contradicts the fact that F is one-to-one in A(1, R1 ). Assume that n ≤ −1. The equation (14.55) also holds in this case which in turn contradicts the injectivity of F again. In other words, F (z) = αz which implies that R1 = R2 , a contradiction. Next, if F has a removable singularity at the origin, then F is actually entire and we use the same argument as above to show that no such F exists. Hence no such mapping f exists which completes the proof  of the problem.

436

Chapter 14. Conformal Mapping

Remark 14.3 (a) Theorem 14.22 is sometimes called Schottky’s Theorem. Besides the proofs given in the text (Problems 14.24 and 25), you can also find a simple and elegant proof ofs this theorem in [8]. (b) Besides the analytical proofs provided in the text and the problems, one can find a pure algebraic proof in [95].

14.4

Constructive Proof of the Riemann Mapping Theorem

Problem 14.26 Rudin Chapter 14 Exercise 26.

Proof. (a) Suppose that the regions Ω0 , Ω1 , . . . , Ωn−1 and functions f1 , f2 , . . . , fn are constructed such that Ωk = fj (Ωk−1 ), where k = 1, 2, . . . , n. Define rn = inf{|z| | z ∈ C \ Ωn−1 }.

(14.56)

Then rn is the largest number such that D(0; rn ) ⊆ Ωn−1 and the definition shows that there is an αn ∈ ∂Ωn−1 with |αn | = rn .h See Figure 14.2 below.

Figure 14.2: The constructions of Ωn−1 , D(0; rn ) and αn . h

Geometrically, αn is a point on ∂Ωn−1 nearest the origin.

14.4. Constructive Proof of the Riemann Mapping Theorem

437

Choose βn2 = −αn and put Fn = ϕ−αn ◦ s ◦ ϕ−βn : U → U, where ϕα (z) = By the Chain Rule, we have

z−α 1 − αz

(14.57)

and s(ω) = ω 2 .




Fn′ (z) = ϕ′−αn s ϕ−βn (z) × s′ ϕ−βn (z) × ϕ′−βn (z)

= 2ϕ′−αn s ϕ−βn (z) × ϕ′−βn (z) × ϕ−βn (z).

By Theorems 10.33 and 12.4, ϕ′−αn (z) 6= 0 and ϕ′−βn (z) 6= 0 for every z ∈ Ωn−1 . Thus Fn′ (z) 6= 0 for all z ∈ Ωn−1 and it follows from Theorem 10.30(c) that Fn has a holomorphic inverse Gn in Ωn−1 . (b) By the definition and Theorem 12.4, we may write −1 −1 ◦ ϕαn : Ωn−1 → U, Gn = ϕ−1 ◦ ϕ−1 −αn = ϕβn ◦ s −βn ◦ s

where s−1 (z) =

(14.58)

√

z. Combining the Chain Rule and Theorem 12.4, we get


G′n (0) = ϕ′βn s−1 ϕαn (0) × (s−1 )′ ϕαn (0) × ϕ′αn (0)
1 = ϕ′βn s−1 (−αn ) × √ × (1 − |αn |2 ) 2 −αn 1 − rn2 = × ϕ′βn (βn ) 2βn 1 − rn2 . = 2(1 − |βn |2 )βn

Put fn = λn Gn : Ωn−1 → U , where λn = fn′ (0) = λn G′n (0) = |G′n (0)| =

|G′n (0)| G′n (0) .

(14.59)

Now the formula (14.59) implies that

1 − rn2 1 + rn 1 − rn2 = √ = √ . 2 2(1 − |βn | ) · |βn | 2(1 − rn ) rn 2 rn

Using the A.M. ≥ G.M., it is easy to see that fn′ (0) > 1. (c) We prove the assertions one by one. – Each ψn : Ω → Ωn is bijective. By the definition, we have ψn = fn ◦ fn−1 ◦ · · · ◦ f1 . Since Ω = Ω0 and fn (Ωn−1 ) = Ωn ⊆ U , we have ψn (Ω) = fn (fn−1 (· · · f1 (Ω0 ))) = fn (fn−1 (· · · f2 (Ω1 ))) = fn (Ωn−1 ) = Ωn ⊆ U. By the representation (14.58) and Theorem 12.4, each Gn and hence each fn is injective on Ωn−1 . Consequently, each ψn : Ω → Ωn ⊆ U is injective on Ω. – {ψn′ (0)} is bounded. Since ϕ−αn , ϕ−βn , s ∈ H(U ), the definition (14.57) gives Fn ∈ H(U ). Furthermore, it is clear that
Fn (0) = ϕ−αn ϕ2−βn (0) = ϕ−αn (βn2 ) = ϕ−αn (−αn ) = 0, (14.60)

438

Chapter 14. Conformal Mapping so we know from Theorem 12.2 (Schwarz’s Lemma) that |Fn (ω)| ≤ |ω| and |Fn′ (0)| ≤ 1

(14.61)

for all ω ∈ U . Put ω = Gn (z) into the inequality (14.61), we get that |G′n (0)| ≥ 1

|Gn (z)| ≥ |z| and

(14.62)

hold for all z ∈ Ωn−1 . Lemma 14.4 For every n = 1, 2, . . ., we have 0 < r1 ≤ r2 ≤ · · · ≤ 1.

Proof of Lemma 14.4. Notice that we have fn (Ωn−1 ) = Ωn , D(0; rn ) ⊆ Ωn−1 and D(0; rn+1 ) ⊆ Ωn . By the definition (14.56), there exists a boundary point α ∈ ∂Ωn such that |α| = rn+1 . Select {zn−1,k } ⊆ Ωn−1 such that fn (zn−1,k ) → α as k → ∞. Assume that {zn−1,k } had a limit point β in Ωn−1 . Since fn is obviously continuous on Ωn−1 , we have α = fn (β) which means that Ωn ∩ ∂Ωn 6= ∅, a contradiction. Therefore, the sequence {zn−1,k } cannot have a limit point in Ωn−1 . Since D(0; rn ) ⊆ Ωn−1 , we must have lim sup |zn−1,k | ≥ rn . k→∞

Since |fn (z)| = |Gn (z)|, we derive from the first inequality (14.62) that rn+1 = |α| = lim |fn (zn−1,k )| = lim |Gn (zn−1,k )| ≥ lim sup |zn−1,k | ≥ rn k→∞

k→∞

k→∞

as desired. This ends the proof of the lemma.

 z r1

Now Lemma 14.4 ensures that D(0; r1 ) lies in every Ωn . Since φ(z) = maps D(0; r1 ) one-to-one and onto U , the map Ψn = ψn ◦ φ−1 : U → U satisfies the hypotheses of Theorem 12.2 (Schwarz’s Lemma) so that |Ψ′n (0)| ≤ 1 which means that |ψn′ (0)| ≤

1 r1

(14.63)

for every n = 1, 2, . . .. In other words, {ψn′ (0)} is bounded. – A formula of ψn′ (0). By the aid of the Chain Rule and part (b), we establish easily that n Y 1 + rk ′ ψn′ (0) = fn′ (0) × fn−1 (0) × · · · × f1′ (0) = (14.64) √ . 2 rk k=1

– The sequence {rn } converges to 1. For m > n ≥ 1, we define ψm,n = fm ◦ fm−1 ◦ · · · ◦ fn+1

which is holomorphic in Ωn and hence on D(0; rn+1 ). In view of the value (14.60), we know that Gn (0) = 0 and then fn (0) = 0 for every n ∈ N. Recalling the fact ψn = fn ◦ fn−1 ◦ · · · ◦ f1 , we therefore have ψn (0) = 0 for every n ∈ N which implies ψm,n (0) = 0. Using similar attack as in proving the inequality (14.63),i we can obtain ′ |ψm,n (0)| ≤

1 . rn+1

i That is, D(0; rn+1 ) lies in every Ωm for all m ≥ n + 1, the map Ψm,n = ψm,n ◦ φ−1 : U → U satisfies the −1 z maps D(0; rn+1 ) one-to-one and onto U . hypotheses of Theorem 12.2 (Schwarz’s Lemma), where φ(z) = rn+1

14.4. Constructive Proof of the Riemann Mapping Theorem

439

Now the Chain Rule and part (b) assert that ′ |ψm,n (0)| =

so that 1
Since 0


2n X

k=n+1

√ √ 2n h X (1 − rk )2 (1 − rk )2 i > > 0. log 1 + √ 2 rk 3

(14.72)

k=n+1

By elementary calculus again, we know that log(1+x) is strictly decreasing for x > −1. Combining x this fact and Lemma 14.4, it is true that for all n ≥ 1, log rn+1 log[1 + (rn+1 − 1)] log[1 + (r1 − 1)] log r1 = ≤ = rn+1 − 1 rn+1 − 1 r1 − 1 r1 − 1 which implies

As (1 +

√

− log rn+1 ≤ (1 − rn+1 ) · rk )2 ≤ 4, we have (1 −

log r1 log r1−1 ≤ (1 − rn ) · . r1 − 1 1 − r1

√ 2 (1 − rk )2 (1 − rk )2 rk ) = . √ 2 ≥ (1 + rk ) 4

(14.73)

(14.74)

log r −1

Let A = 1−r11 . Now we observe by substituting the inequalities (14.73) and (14.74) into the inequality (14.72) that 0
0 such that |nβ − m − α| < ǫ. θ Take α = 0 and β = 2π . For every k ∈ N, we obtain from Lemma 14.5 that there exist integers nk and mk with nk > 0 such that

n θ 1 k − mk < 2π 2kπ

or equivalently

|nk θ − 2mk π|
0 z |z|

that z ∗ lies on the ray L = {tz | t ∈ R}. Geometrically, see Figure 14.3 for the construction of the point z ∗ .

Figure 14.3: The construction of the symmetric point z ∗ of z. (f) Let α, β, γ ∈ C. Then it follows from part (c) and the definition that [ϕ(z ∗ ), ϕ(α), ϕ(β), ϕ(γ)] = [z ∗ , α, β, γ] = [z, α, β, γ] = [ϕ(z), ϕ(α), ϕ(β), ϕ(γ)]. Hence ϕ(z ∗ ) and ϕ(z) are symmetric with respect to ϕ(C). This finishes the analysis of the problem.



Problem 14.31 Rudin Chapter 14 Exercise 31.

Proof. (a) Given ϕ, ψ, φ ∈ Λ by ϕ(z) =

az + b , cz + d

ψ(z) =

αz + β γz + δ

and φ(z) =

– Composition as group operation. It is easy to see that
(αa + βc)z + αb + βd ψ ϕ(z) = (γa + δc)z + γb + δd

Az + B . Cz + D

14.4. Constructive Proof of the Riemann Mapping Theorem

449

and its determinant is (αa + βc)(γb + δd) − (γa + δc)(αb + βd) = (ad − bc)(αδ − βγ) 6= 0 so that ψ ◦ ϕ ∈ Λ. – Associativity. Simple algebra verifies [ϕ(z) + φ(z)] + ψ(z) = ϕ(z) + [φ(z) + ψ(z)]. – The identity element. Now the usual identity map id : C → C is the identity element of Λ because its determinant is 1 and id ◦ ϕ = ϕ ◦ id = ϕ. – The inverse of ϕ. The equation ω = ϕ(z) has exactly one solution and indeed, it is z = ϕ−1 (ω) =

dω − b . cω − a

Since the determinant of ϕ−1 is −ad + bc 6= 0, we have ϕ−1 belongs to Λ. Clearly, we know that ϕ−1 ◦ ϕ = ϕ ◦ ϕ−1 = id. By the definition (see [42, Definition 4.1, pp. 37, 38]),
Λ is indeed a group.
If1 we take which ϕ(z) = z + 1 and ψ(z) = z1 , then we see that ϕ ψ(z) = z1 + 1 and ψ ϕ(z) = z+1 imply that Λ is not commutative. (b) Let ϕ ∈ Λ be given by

az + b cz + d and ϕ 6= id. Define ∆ = ad − bc 6= 0. Since we may write ϕ(z) =

ϕ(z) =

√a z ∆ √c z ∆

+ +

√b ∆ √d ∆

,

we may assume without loss of generality that ad − bc = 1. Now the equation z = ϕ(z) is equivalent to saying that cz 2 + (d − a)z − b = 0. (14.92) – Case (i): c = 0. Thus we have ad 6= 0 and ϕ(∞) = ∞ so that ∞ is a fixed point of ϕ. If a 6= d, then by solving the equation (14.92), we get one more (finite) fixed point which is b . z= d−a Otherwise, a = d implies that b ϕ(z) = z + (14.93) d whose fixed point is also ∞. Since ϕ 6= id, b 6= 0 so that ϕ has only a unique (infinite) fixed point in this case. – Case (ii): c 6= 0. Then the equation (14.92) has two roots p p a − d ± (d − a)2 + 4bc a − d ± (a + d)2 − 4 z= = . (14.94) 2c 2c Since ϕ(∞) = ac , ∞ is not transformed into itself. This means that ϕ has either one or two finite fixed points on S 2 depending on whether a + d = ±2 or not. In the case of the unique finite fixed point, it is given by z=

a−d . 2c

450

Chapter 14. Conformal Mapping In conclusion, ϕ has either one or two fixed points on S 2 .

(c) We consider two cases. – Case (i): ϕ has a unique fixed point. Given ϕ1 (z) = z + 1 which is obviously an element of Λ. If c = 0, then we follow from part (b) that ϕ has a unique (infinite) fixed point if and only if it takes the form (14.93). Define ψ(z) = db z. Recall that bd 6= 0, so ψ ∈ Λ and ψ −1 (z) = db z. Furthermore, it is clear that

d  b b ψ −1 ϕ ψ(z) = · = z + 1 = ϕ1 (z). z+ b d d

Hence we have shown that ϕ is conjugate to ϕ1 in this case. Next, if c 6= 0, then we observe from the roots (14.94) that ϕ has a unique fixed point z1 = a−d 2c if and only if a + d = ±2. Define the linear fractional transformation S(z) =

1 z − z1

which carries z1 to ∞. Therefore, the linear fractional transformation T = S ◦ ϕ ◦ S −1

(14.95)

has ∞ as its only fixed point because if p is a fixed point of T , then S −1 (p) will be a fixed point of ϕ so that p = S(z1 ) = ∞. Hence it follows from part (b) that T (z) = z + B for some B ∈ C \ {0}. If we take P (z) = Bz, then P −1 (z) = Bz and so ϕ1 = P −1 ◦ T ◦ P.

(14.96)

By combining the expressions (14.95) and (14.96), we conclude that ϕ1 = (P −1 ◦ S) ◦ ϕ ◦ (S −1 ◦ P ) = (S −1 ◦ P )−1 ◦ ϕ ◦ (S −1 ◦ P ). Hence we have ψ = S −1 ◦ P . Since P, S ∈ Λ, we have ψ ∈ Λ.

– Case (ii): ϕ has two distinct fixed points. Consider the linear fractional transformation φα (z) = αz, where α is a non-zero complex number which will be determined soon. By part (b), we have either “c = 0 and a 6= d” or “c 6= 0 and a + d 6= ±2”. ∗ Subcase (i): c = 0 and a 6= d. We notice that ϕ(z) =

az + b . d

b which is obviously a linear fractional transformation and Take ψ(z) = z + d−a b −1 ψ (z) = z − d−a . Direct computation gives


a b  b a b ϕ ψ(z) = z+ + = z+ d d−a d d d−a

14.4. Constructive Proof of the Riemann Mapping Theorem

451

and then

a b b a − = z = φ ad (z). ψ −1 ϕ ψ(z) = z + d d−a d−a d

Consequently, ϕ is conjugate to φα with

α=

a . d

(14.97)

Particularly, 0 is the finite fixed point if and only if b = 0, so we have ϕ(z) = ad z and ψ(z) = z. ∗ Subcase (ii): c 6= 0 and a + d 6= ±2. The two distinct finite fixed points are given by (14.94). Let z1 and z2 be the roots corresponding to the negative square root and the positive square root respectively. Now the linear fractional transformation z − z1 S(z) = z − z2

maps the ordered pair {z1 , z2 } into {0, ∞}. Then the linear fractional transformation T = S ◦ ϕ ◦ S −1 fixes 0 and ∞. By the particular case of Subcase (i), we know that φα = ψ −1 ◦ T ◦ ψ = T for some complex α. Since S −1 (z) =

z2 z−z1 z−1 ,

we obtain from the definition that


(az2 + b)z − (az1 + b) ϕ S −1 (z) = . (cz2 + d)z − (cz1 + d)

and T (z) =

(az − dz1 + 2b)z + cz12 + (d − a)z1 − b  22  . − cz2 + (d − a)z2 − b z − (az1 − dz2 + 2b)

(14.98)

Since z1 , z2 are roots of the equation (14.92), the formula (14.98) simplifies to T (z) = −

(az2 − dz1 + 2b) z. (az1 − dz2 + 2b)

(14.99)

Using the formula (14.94), the expression (14.99) can further reduce to p (a + d)2 − 4 + (a + d) z. T (z) = − p (a + d)2 − 4 − (a + d)

Hence we obtain the formula

p

α = −p

(a + d)2 − 4 + (a + d)

(a + d)2 − 4 − (a + d)

.

(14.100)

Finally, α is determined by either (14.97) or (14.100). (d) We prove the assertions one by one. – The existence of β. Since ϕ has only a unique finite fixed point, the analysis of part (b) leads us to the result that c 6= 0 and a + d = ±2. In this case, we have α = a−d 2c . Then we have d = a − 2αc. (14.101) Since α is a root of the equation (14.92), we have b − dα = cα2 − aα = α(cα − a).

(14.102)

452

Chapter 14. Conformal Mapping Obviously, cα − a 6= 0. Otherwise, put cα = a into the equation (14.101) will give a + d = 0 which is impossible. Now we note that 1 = ϕ(z) − α

1 az+b cz+d

−α

=

cz + d cz + d = az + b − cαz − dα (a − cα)z + (b − dα)

(14.103)

Substituting the values (14.101) and (14.102) into the expression (14.103) to get cz + a − 2αc c(z − α) + a − αc 1 c 1 = = = + (14.104) ϕ(z) − α (a − cα)z + α(cα − a) (z − α)(a − αc) z − α a − αc which means that β= With the aid of α =

a−d 2c

c ∈ C. a − αc

and a + d = ±2, we can further show that β = c.

(14.105)

– Gα is a subgroup of Λ. Let Gα = {ϕ ∈ Λ | ϕ(α) = α} ∪ {id} ⊆ Λ. Let ϕ and φ be elements of Gα with the corresponding constant βϕ and βφ respectively. Therefore, we see that 1 1 1
= + βϕ = + βϕ + βφ (14.106) φ(z) − α z−α ϕ φ(z) − α so that ϕ ◦ φ ∈ Gα . By the definition, we have id ∈ Gα . Furthermore, since ϕ has α as its only finite fixed point, so is ϕ−1 . Thus we have 1 1 = − β. ϕ−1 (z) − α z−α In other words, it means that ϕ−1 ∈ Gα . By [42, Theorem 5.14, p. 52], Gα is a subgroup of Λ. – Gα is isomorphic to (C, +). Define f : Gα → C by f (ϕ) = βϕ , where βϕ is the complex number satisfying the equation 1 1 = + βϕ . ϕ(z) − α z−α

(14.107)

As the expression (14.106) shows definitely that f (ϕ ◦ φ) = βϕ + βφ , 1 1 if and only = z−α so f is a homomorphism. Next, suppose that β = 0. Then ϕ(z)−α if ϕ = id. In other words, the kernel of f is {id}. Finally, given β ∈ C \ {0}. we consider a = 1 + αβ, b = −α2 β, c = β and d = 1 − αβ. Direct computation gives ad − bc = 1. Besides, the linear fractional transformation

ϕβ (z) =

(1 + αβ)z − α2 β βz + (1 − αβ)

(14.108)

fixes α only and satisfies the equation (14.107).m Consequently, we have ϕβ ∈ Gα and f (ϕβ ) = β, i.e., f is surjective. Hence f is in fact an isomorphismn . m

In fact, we establish from the value (14.105) that the representation (14.108) becomes ϕ(z) =

n

See, for instance, [42, p. 132].

(1 + cα)z − cα2 . cz + (1 − cα)

14.4. Constructive Proof of the Riemann Mapping Theorem

453

(e) Now we have Gα,β = {ϕ ∈ Λ | ϕ(α) = α and ϕ(β) = β}. This refers to the case c 6= 0 and a + d 6= ±2. – Every ϕ ∈ Gα,β satisfies the required equation. Since α and β are roots of the equation (14.92), the formula (14.102) also holds forpβ. Obviously, a − βc 6= 0. Otherwise, it implies the contradiction that a + d = ± (a + d)2 − 4. We observe that ϕ(z) − α = ϕ(z) − β

az+b cz+d az+b cz+d

−α −β

(a − αc)z + (b − dα) = (a − βc)z + (b − dβ) (a − αc)z + α(cα − a) = (a − βc)z + β(cβ − a) z−α , =γ· z−β

where γ=

a − αc ∈ C.o a − βc

(14.109)

– Gα,β is a subgroup of Λ. Since id fixes α and β, we have id ∈ Gα,β . For every ϕ, φ ∈ Gα,β , let γϕ and γφ be their corresponding complex numbers respectively. Since φ(z) − α z−α ϕ(φ(z)) − α = γϕ · = γϕ · γφ · , (14.110) ϕ(φ(z)) − β φ(z) − β z−β we have ϕ ◦ φ ∈ Gα,β . Assume that ϕ ∈ Gα,β was a constant map. Then it implies that α = β, a contradiction. In addition, γϕ 6= 0. Otherwise, ϕ(z) = α for all z ∈ S 2 which is impossible. Next, if ϕ ∈ Gα,β , then ϕ−1 also fixes α and β, and we have ϕ−1 (z) − α 1 z−α = · . ϕ−1 (z) − β γϕ z − β Consequently, these imply that ϕ−1 ∈ Gα,β . Hence Gα,β is a subgroup of Λ.

– Gα,β is isomorphic to (C \ {0}, ×). Define g : Gα,β → C \ {0} by g(ϕ) = γϕ , where γϕ is the complex number satisfying the equation z−α ϕ(z) − α = γϕ · . ϕ(z) − β z−β

(14.111)

The equation (14.110) implies that g(ϕ ◦ φ) = γϕ × γφ so that g is a homomorphism. If g(ϕ) = 1, then we have a − αc = a − βc so that c = 0 and ϕ takes the form ϕ(z) =

az + b . d

Put this into the equation (14.111) with γϕ = 1 and after simplification, we conclude that ϕ(z) = z, i.e., the kernel of g is {id}. Let γ ∈ C \ {0}. By changing the subject of the formula (14.109) to c and using the formula α + β = a−d c , we can show that d= o

αγ − β a. α − βγ

This number is called the multipler of the transformation ϕ, read [41, pp. 15, 16].

454

Chapter 14. Conformal Mapping Next, we apply the fact

√

1 γ+ √ =a+d γ

to represent a, c and d in terms of α, β and γ as follows: √ α − βγ 1  γ+√ , a= (1 + γ)(α − β) γ √ 1  αγ − β γ+√ , d= (1 + γ)(α − β) γ √ 1−γ 1  c= γ+√ . (1 + γ)(α − β) γ

(14.112)

Finally, we employ the formula αβ = − cb to obtain b=−

αβ(1 − γ) √ 1  γ+√ . (1 + γ)(α − β) γ

(14.113)

Now it is a routine task to check that the linear fractional transformation ϕ with the coefficients given by the formulas (14.112) and (14.113) has α and β as its fixed points, ad − bc = 1 and satisfies g(ϕ) = γ. In other words, the map g is surjective and hence, an isomorphism. (f) By the hypothesis, the fixed points are finite. The following proof is due to Drazin [33] az+b has invariant circles if who verified that the linear fractional transformation ϕ(z) = cz+d and only if its determinant ∆ 6= 0 and

(a+d)2 ∆

is real.

– Case (i): ϕ has a unique finite fixed point. Recall from the explicit form (14.108) that (1 + βα)z − βα2 , (14.114) ϕ(z) = βz + (1 − βα) where β = c 6= 0. Simple algebra gives

βϕ(z) = 1 + βα −

1 . βz + (1 − βα)

(14.115)

Define ϕ∗ = βϕ + (1 − βα)

and

ζ = βz + (1 − βα).

(14.116)

Then the equation (14.115) becomes 1 ϕ∗ (ζ) = 2 − . ζ

(14.117)

Consequently, this change of variables establishes a one-to-one correspondence between the invariant circles of the linear fractional transformations (14.114) and (14.117). Let C ∗ = {ζ ∈ C | |ζ − ρeiθ | = R} be an invariant circle of ϕ∗ , where ρ, θ, R are real and R > 0. Consider the two points (ρ − R)eiθ and (ρ + R)eiθ which are the
endpoints of a diameter
of C ∗ . Since ϕ∗ is conformal, the points P = ϕ∗ (ρ − R)eiθ and Q = ϕ∗ (ρ + R)eiθ are also endpoints of a diameter of C ∗ so that |P − Q| = 2R and 21 (P + Q) = ρeiθ . Notice that P =2−

1 (ρ − R)eiθ

and Q = 2 −

1 , (ρ + R)eiθ

(14.118)

14.4. Constructive Proof of the Riemann Mapping Theorem so we have ρ2 − R2 = ±1 and

2 = ρeiθ +

455

ρe−iθ . − R2

ρ2

If ρ2 − R2 = −1, then we have 1 = iρ sin θ which is impossible. Therefore, we must have R2 = ρ2 − 1 and 1 = ρ cos θ. In this case, C ∗ are circles with centers 1 ± iR. By the transformation (14.116), the invariant circles of ϕ satisfy the equations  R R  , = z − α ± i β |β|

where R > 0.

– Case (ii): ϕ has two finite fixed points. By the expressions (14.112) and (14.113), the explicit form of ϕ (after the cancellation of the common coefficient) is given by ϕ(z) =

(α − βγ)z − αβ(1 − γ) (1 − γ)z + (αγ − β)

(14.119)

and ∆ = (α − βγ)(αγ − β) + αβ(1 − γ)2 = γ(α − β)2 6= 0. Now we have

√ ∆ (α − βγ)[(1 − γ)z + (αγ − β)] − ∆ = (α − βγ) − , (1 − γ)ϕ(z) = (1 − γ)z + (αγ − β) ζ

where ζ=

(1 − γ)z + (αγ − β) √ . ∆

(14.120)

1

Define ϕ∗ = ∆− 2 [(1 − γ)ϕ + (αγ − β)]. Then we have

√ i h 1 1+γ 1 ∆ ϕ∗ (ζ) = ∆− 2 (α − β)(1 + γ) − = √ − . ζ γ ζ

(14.121)

Similar to Case (i), this change of variables establishes a one-to-one correspondence between the invariant circles of the linear fractional transformations (14.119) and (14.121). Instead of the expressions (14.118), we have 1 1+γ P = √ − γ (ρ − R)eiθ so that ρ2 − R2 = ±1 Denote χ =

1+γ √ 2 γ.

and

1+γ 1 and Q = √ − γ (ρ + R)eiθ 1+γ ρe−iθ . √ = ρeiθ + 2 γ ρ − R2

Thus we have either R2 = ρ2 + 1 and χ = iρ sin θ

(14.122)

R2 = ρ2 − 1 and χ = ρ cos θ.

(14.123)

or Since ρ and θ are real, the expressions involving χ in (14.122) and (14.123) show that it is either purely real or purely imaginary.

456

Chapter 14. Conformal Mapping √ |. By the equations (14.122), ∗ Subcase (i): χ2 < 0. Here χ = it, where t = | 21+γ γ the invariant circle of ϕ∗ has the form p
ζ − ± R2 − t2 − 1 + it = R,

where R2 ≥ 1 − χ2 = 1 + t2 . Transforming back to the original system (using (14.120)), we get (1 − γ)z + (αγ − β) p
√ − ± R2 − t2 − 1 + it = R ∆ √ √
p (β − αγ) + ∆ ± R2 − t2 − 1 + it R|α − β| |γ| z − = 1−γ |1 − γ|   q √ 1+γ 1+γ 2 2 (β − αγ) + ∆ ± R − | 2√γ | − 1 + i| 2√γ | R|α − β|p|γ| . z − = 1−γ |1 − γ|

∗ Subcase (ii): χ2 = 0. In this subcase, we know that γ = −1. Furthermore, it can be seen from the expressions involving χ in (14.122) and (14.123) that ρ = 0 and then R = 1. Consequently, we have |ζ| = 1 which gives α + β |α − β| . = z − 2 2

√ is real and we get from the expression ∗ Subcase (iii): χ2 > 0. Then χ = 21+γ γ (14.123) that p
ζ − χ ± i R2 + 1 − χ2 = R,

where R2 ≥ χ2 − 1. Hence, after transforming back to the original system, we assert that  q √  p √ ± i R2 + 1 − ( 1+γ √ )2 (β − αγ) + ∆ 21+γ γ 2 γ R|α − β| |γ| z − , = 1−γ |1 − γ|

where R2 ≥

(1+γ)2 4γ

−1 =

(1−γ)2 4γ .

Now we have completed the analysis of the problem.



Remark 14.5 (a) The expressions (14.107) and (14.111) are called the normal forms of the linear fractional transformation ϕ. (b) The number of finite fixed points can be used to classify the linear fractional transformations ϕ. In fact, we rewrite the multiplier (14.109) as ρeiθ . If ρ 6= 1 and θ = 2nπ, then ϕ is called a hyperbolic transformation. If ρ = 1 and θ 6= 2nπ, then it is called an elliptic transformation. If ρ > 0 but ρ 6= 1 and θ 6= 2nπ, then it is called a loxodromic transformation. The case for one finite fixed point is called a parabolic transformation and it can be thought as corresponding to ρ = 1 and θ = 2nπ. See [1, §3.5, pp. 84 – 89] and [41, pp. 15 – 23] for further details. Problem 14.32 Rudin Chapter 14 Exercise 32.

14.4. Constructive Proof of the Riemann Mapping Theorem

457

Proof. We notice from §14.3 that the linear fractional transformation ω = ϕ(z) =

1+z 1−z

is a conformal one-to-one mapping of U onto the open right half plane Π = {ω = X + iY ∈ C | X > 0}. The images of the upper semi-circle and the lower semi-circle under ϕ are the positive Y -axis and the negative Y -axis respectively. Next, the mapping ζ = φ(ω) = log ω = log maps Π conformally onto the horizontal strip

1+z 1−z

S = {ζ = u + iv ∈ C | u ∈ R and − π2 < v < π2 }. Furthermore, the positive Y -axis is mapped onto the line v = mapped onto the line v = − π2 . Consequently, the mapping

π 2

and the negative Y -axis is


1+z ψ(z) = φ ϕ(z) = log 1−z

(14.124)

sends U conformally onto the horizontal strip S, the images of the upper semi-circle and the lower semi-circle under ψ are the lines v = π2 and v = − π2 respectively. See Figure 14.4 for the illustration.


Figure 14.4: The conformal mapping ψ(z) = φ ϕ(z)

Finally, it is easy to see that the mapping ζ 7→ iζ carries the horizontal strip S conformally onto the vertical strip H = {ω = X + iY ∈ C | − π2 < X
1. Consequently, they converge absolutely and uniformly in U so that an interchange of integration and summation is legitimate. In terms of polar coordinates, this means that 1 π

Z

2 ′ ϕα dm = 1 − |α| π U

=

Z

∞ X

U n,k=0

∞ Z X

|α|2

1− π

(αz)n (αz)k dm (αz)n (αz)k dm

n,k=0 U ∞ Z 2π 2 |α| X

1− = π = If n 6= k, then

n,k=0 0 ∞ 2 X |α|

1− π

Thus the expression becomes 1 π

2π 0

1

n,k=0

Z

1

(αreiθ )n (αre−iθ )k r dr dθ

0

(αn αk )

·

Z

Z

Z

2π 0

Z

1

r n+k+1 ei(n−k)θ dr dθ.

(14.127)

0

r n+k+1 ei(n−k)θ dr dθ = 0.

0

Z 2π Z 1 ∞ 2 X ′ 2n ϕα dm = 1 − |α| · |α| r 2n+1 dr dθ π 0 0 U

Z

=

|α|2

1− π

·

n=0 ∞ X n=0

|α|2n

π n+1

∞
X |α|2n . = 1 − |α|2 n+1

(14.128)

n=0

Consider the power series expansion of log(1 + z) about z = 0, we know that log(1 + z) =

∞ X

(−1)n

n=0

z n+1 n+1

which has radius of convergence 1. Substituting this with z = −|α|2 into the right-hand side of the formula (14.128), we obtain finally that Z 2 ′ 1 1 ϕα dm = 1 − |α| log . 2 π U |α| 1 − |α|2

We have ended the proof of the problem.



Remark 14.7 Problem 14.33(a) is a special case of the classical result Lusin Area Integral, see [61, p. 150].

CHAPTER

15

Zeros of Holomorphic Functions

15.1

Infinite Products and the Order of Growth of an Entire Function

Problem 15.1 Rudin Chapter 15 Exercise 1.

Proof. Let S be the set in which the infinite product converges uniformly. Define un (z) = Note that bn − a n z − an =1+ = 1 + un (z). z − bn z − bn

Suppose that

bn −an z−bn .


δ = d S, {bn } = inf{|z − ω| | z ∈ S and ω ∈ {bn }} > 0.

Then it is easy to see that

1 |un (z)| ≤ |bn − an | < ∞ δ for every n ∈ N and z ∈ S. Furthermore, we know that ∞ X

n=1

|un (z)| =

∞ X |bn − an |

n=1

|z − bn |

≤

∞

1X |bn − an | < ∞ δ n=1

for every z ∈ S. By Theorem 15.4, the infinite product f (z) =

∞ Y z − an z − bn n=1

(15.1)

converges uniformly on S. Clearly, S ◦ is an open set in C, fn (z) = component of S ◦ . By the above paragraph, ∞ X

n=1

z−an z−bn

∈ H(S ◦ ) for n = 1, 2, . . . and fn 6≡ 0 in any

|1 − fn (z))| =

∞ X

n=1

|un (z)|

converges uniformly on every compact subset of S ◦ . By Theorem 15.6, the infinite product  (15.1) is holomorphic in S ◦ . This ends the proof of the problem. 463

464

Chapter 15. Zeros of Holomorphic Functions

Problem 15.2 Rudin Chapter 15 Exercise 2.

Proof. Denote λ to be the order of the entire function f . By the definition, we have λ = inf{ρ | |f (z)| < exp(|z|ρ ) holds for all large enough |z|}. Using the fact an =

f (n) (0) n!

and Theorem 10.26 (Cauchy’s Estimates), we have λ

|an | ≤

er rn

(15.2)

for large enough r. Let g(r) = r −n exp(r λ ), where r > 0. Applying elementary differentiation, we can show that g attains its minimum λn λ

n

1 λ

at r = n λ (15.2) that

− λ1

exp

n λ

. Note that r is large if and only if n is large. Thus it follows from the inequality |an | ≤

holds for all large enough n.

λn λ

n

exp

n λ

=

 eλ  n λ

n

Consider the entire functions f (z) = exp(z k ), where k = 1, 2, . . .. It is clear that λ = k. By k 1 the power series expansion of ez , we have ank = n! . By induction, we obtain |ank | =

 e n 1 < n! n

for every large enough n. Consequently, the above bound is not close to best possible. This completes the analysis of the proof.  Problem 15.3 Rudin Chapter 15 Exercise 3.

z

Proof. The part of finding solutions of ee = 1 has been solved in [123, Problem 3.19, pp. 44 – 45]. In fact, they are given by   π  , if k > 0 and n ∈ Z; ln(2kπ) + i 2nπ +   2 (15.3) z=      ln(−2kπ) + i 2nπ + 3π , if k < 0 and n ∈ Z. 2

Denote the zeros (15.3) by zk,n , where k, n ∈ Z and k 6= 0, see Figure 15.1.

Assume that f was an entire function of finite order having a zero at each (15.3). Suppose further that f 6≡ 0. Consider the disc D(0, RN ), where RN = N π and N is a sufficiently large positive integer. Then we have zk,n ∈ D(0, RN ), where exp(N π) exp(N π) −1≤k ≤ . 2π 2π

15.1. Infinite Products and the Order of Growth of an Entire Function

465

Figure 15.1: The distribution of the zeros zk,n of exp(exp(z)). Therefore, we gain

so that


exp(N π) n RN ≥ 4π


log n RN N π − log 4π →∞ ≥ log RN log N π

as N → ∞, but it means that f is of infinite order, a contradiction. Hence no such entire  function exists and we finish the proof of the problem. Problem 15.4 Rudin Chapter 15 Exercise 4.

Proof. We prove the assertions one by one.

466

Chapter 15. Zeros of Holomorphic Functions • Both functions have a simple pole with residue 1 at each integer. Since sin πz has a simple zero at every integer N , we have Res (π cot πz; N ) = Res

 π cos πz

 π cos πN ;N = =1 sin πz π cos πN

which shows that π cot πz has a simple pole with residue 1 at each integer. We claim that

X

n∈Z n6=0

z 1 X 1 = + z−n n n(z − n)

(15.4)

n∈Z n6=0

converges absolutely and uniformly on compact subsets of C \ Z. To see this, let |z| ≤ R with R > 0. Then we have X

|n|≥2R

X X X 1 |z| R R ≤ ≤ 1. On the one hand, we have cot πz = i · so that

On the other hand, we write

e−2πy + e−2πix e−2πy − e−2πix

|e−2πy | + 1 < ∞. | cot πz| ≤ −2πy |e | − 1

f (z) =

∞

X 2(x + iy) 1 + . x + iy x2 − y 2 − n2 + 2ixy n=1

If y > 1 and |x| ≤

1 2,

√ we have |x + iy| ≤ 2y and

|x2 − y 2 − n2 + 2ixy| = Thus they imply that

p

1 y 2 + n2 . [x2 − (y 2 + n2 )]2 + 4x2 y 2 ≥ (y 2 + n2 ) − ≥ 4 2

|f (z)| ≤ 1 +

∞ X

2|x + iy| 2 − y 2 − n2 + 2ixy| |x n=1

468

Chapter 15. Zeros of Holomorphic Functions ∞ √ X ≤1+4 2

y y 2 + n2

n=1 ∞

√ Z ≤1+4 2

y dx . y 2 + x2

0

(15.5)

By the change of variable x = yt, it is easily checked that the integral in the inequality (15.5) becomes Z ∞ Z ∞ dt y dx = 2 + x2 y 1 + t2 0 0

which implies that |f (z)| ≤ M for some M > 0 and for all |x| ≤ 12 and y > 1. Similarly, f (z) is also bounded for all |x| ≤ 21 and y < −1. Consequently, we have shown that ∆(z) is a bounded entire function and Theorem 10.23 (Liouville’s Theorem) says that it is in fact a constant. To find this constant, we note that lim f (iy) = lim

y→∞

y→∞

h1

iy

= −2i lim

y→∞

= −2i = −πi

Z

∞ 0

+

2iy i −y 2 − n2 n=1 ∞ X

∞ X

n=1

y y 2 + n2

dt 1 + t2

and lim π cot iπy = −πi.

y→∞

Thus ∆(z) ≡ 0 and then π cot πz =

∞

1 X 2z + . z z 2 − n2

(15.6)

n=1

• The product representation of

sin πz . If g(z) = sin πz, then it is clear that πz g′ (z) = π cot πz. g(z)

Consequently, we observe from the representation (15.6) that ∞

g′ (z) 1 X 2z = + . g(z) z z 2 − n2

(15.7)

n=1

Next, we consider the infinite product P (z) = πz

∞  Y z2  1− 2 . n n=1

2

(15.8)

Now each Pn (z) = 1 − nz 2 ∈ H(C \ Z) and Pn 6≡ 0 in C \ Z. In addition, if K is a compact subset of C \ Z, then the Weierstrass M -test [99, Theorem 7.10, p. 148] guarantees that ∞ X

n=1

|1 − Pn (z)| =

∞ X |z|2 n=1

n2

15.1. Infinite Products and the Order of Growth of an Entire Function

469

converges uniformly on K. By Theorem 15.6, the product (15.8) is holomorphic in C \ Z. Using [30, Exercise 10, p. 174], we have ∞ ∞ ∞ P ′ (z) 1 X Pn′ (z) 1 X − n2z2 1 X 2z = + = + + = 2 P (z) z n=1 Pn (z) z n=1 1 − z 2 z n=1 z 2 − n2

(15.9)

n

for every z ∈ C \ Z. Substituting the result (15.9) into the formula (15.7), we see that g′ (z) P ′ (z) = g(z) P (z) and then it gives

h P (z) i′ g(z)

= 0.

Therefore, we have P (z) = cg(z) for some constant c in C \ Z. Since we have c = 1 and eventually, sin πz = πz

P (z) z

→ 1 as z → 0,

∞  Y z2  1− 2 n

(15.10)

n=1

in C \ Z. Since sin πz and πz whole plane.

∞ Y

(1 −

n=1

z2 ) agree on Z, the formula (15.10) holds in the n2

Hence we have completed the analysis of the problem.



Remark 15.1 Another way to prove Problem 15.4 is to consider the square CN with vertices (N + 12 )(±1±i) for N ∈ N. According to Theorem 10.42 (The Residue Theorem), we have Z   cot πz X cot πz 1 dz = ; zk , (15.11) Res 2πi CN z − ζ z−ζ k

πz where ζ ∈ C and the zk denotes a pole of g(z) = cot / Z and ζ to be z−ζ inside CN . Take ζ ∈ a point inside CN . Then it is clear that the poles of the function g(z) occur at z = ζ and at z = n ∈ {−N, −N + 1, . . . , 0, 1, . . . , N }, and they are all simple. By the basic method of evaluating residue [11, p. 129], we know that

Res

 cot πζ ;ζ = = cot πζ z−ζ 1

 cot πz

and Res

 cot πz z−ζ

 ;n =

1 . π(n − ζ)

By putting these values into the equation (15.11), we get 1 2i

Z

CN

N X 1 cot πz dz = π cot πζ − . z−ζ ζ −n n=−N

It can be shown that lim

Z

N →∞ CN

cot πz dz = 0 z−ζ

which implies the desired result. For details, please refer to [90, Lemma 7.22, pp. 181, 182].

470

Chapter 15. Zeros of Holomorphic Functions

Problem 15.5 Rudin Chapter 15 Exercise 5.

Proof. By Theorem 15.9, our f is an entire function having a zero at each point zn . We claim that M (r) < exp(|z|k+1 ) for sufficiently large enough |z|. If |z| < 12 , then h zk i z2 + ··· + log |Ek (z)| = Re log(1 − z) + z + 2 k   1 1 k+1 k+2 = Re − z − z − ··· k+1 k+1  1  |z| |z|2 ≤ |z|k+1 + + + ··· k+1 k+2 k+3   1 1 ≤ |z|k+1 1 + + 2 + · · · 2 2 ≤ 2|z|k+1 .
Since |Ek (z)| ≤ 1 + |z| exp |z| +

which gives

|z|2 2

+ ··· +

|z|k
k ,

(15.12)

we have


|z|k |z|2 + ··· + log |Ek (z)| ≤ log 1 + |z| + |z| + 2 k lim

z→∞

log |Ek (z)| =0 |z|k+1

so that if M1 > 0, then there exists a R > 0 such that log |Ek (z)| < M1 |z|k+1

(15.13)

for all |z| > R. On the set S = {z ∈ C | 12 ≤ |z| ≤ R}, the function g(z) = |z|−(k+1) log |Ek (z)| is continuous except at z = 1, where g(z) → −∞ as z → 1. Thus there is a constant M2 > 0 such that log |Ek (z)| ≤ M2 |z|k+1 (15.14) for all z ∈ S. Let M = max{2, M1 , M2 }. Now we combine the inequalities (15.12), (15.13) and (15.14) to get log |Ek (z)| ≤ M |z|k+1 (15.15) for every z ∈ C. By the hypothesis, one can find an N ∈ N such that ∞ X

n=N +1

1 1 < 2M |zn |k+1

and

N X

1 ≤ N. |z |k+1 n n=1

Using the inequality (15.15), we obtain ∞ X

n=N +1

∞  z  z k+1 |z|k+1 X log Ek . < ≤M zn zn 2 n=N +1

(15.16)

15.1. Infinite Products and the Order of Growth of an Entire Function

471

Since M1 can be chosen arbitrary, we deduce from the inequality (15.13) that there exists a R1 > 0 such that |z|k+1 log |Ek (z)| ≤ 2N for all |z| > R1 . Let R2 = max{|z1 |R1 , |z2 |R1 , . . . , |zN |R1 }. Then it is obvious that N X

n=1

 z  |z|k+1 log Ek ≤ zn 2

(15.17)

for all |z| > R2 . Finally, the inequalities (15.16) and (15.17) give log |f (z)| =

∞ X

n=1

 z  log Ek ≤ |z|k+1 zn

for all |z| > R2 , and it is equivalent to saying that

|f (z)| < exp(|z|k+1 ) for all |z| > R2 . This proves our claim and thus f is of finite order. This completes the analysis  of the problem. Problem 15.6 Rudin Chapter 15 Exercise 6.

Proof. Given ǫ > 0. Notice that X

|zn |≥1

|zn |−p−ǫ =

∞  X k=0

X

2k ≤|zn | 0 and all sufficiently large enough r. Combining the inequalities (15.18) and (15.19), we see that X

|zn |≥1

|zn |−p−ǫ < C

Hence we have

∞ X k=0

2−k(p+ǫ) · 2(k+1)p = C · 2p

∞ X

n=1

Rudin Chapter 15 Exercise 7.

k=0

2ǫ

< ∞.

|zn |−p−ǫ < ∞

for every ǫ > 0, completing the proof of the problem. Problem 15.7

∞   X 1 k



472

Chapter 15. Zeros of Holomorphic Functions

Proof. Without loss of generality, we may assume that
f 6≡ 0. By the definition (see Problem 15.2), f is of finite order. In the disc
D 0; N + 21 for large enough positive integer N , the number of zeros of f inside D 0; N + 12 is at least N 2 , i.e., n(N + 21 ) ≥ N 2 . By the hypothesis, we know that M (r) < exp(|z|α ), so it follows from [100, Eqn. (4), p. 309] that 2 ≤ lim sup N →∞

log n(N + 21 ) ≤ α. log N

Therefore, if 0 < α < 2, then no entire function can satisfy the hypotheses of the problem. In other words, f (z) = 0 for all z ∈ C if 0 < α < 2, completing the proof of the problem. 

15.2

Some Examples

Problem 15.8 Rudin Chapter 15 Exercise 8.

Proof. Let A = {zn }. We are going to verify the results one by one. • f is independent of the choice of γ(z). Suppose that γ, η : [0, 1] → C are paths from 0 to z and they pass through none of the points zn . Then Γ = γ − η is a simple closed path. Let {z1 , z2 , . . . , zN } be the set of points which are surrounded by Γ for some N ∈ N. Now we follow from Theorem 10.42 (The Residue Theorem) that 1 2πi which gives

Z

Z

g(ζ) dζ = Γ(z)

N X

Res (g; zn ) =

γ(z)

N X

mn

n=1

n=1

g(ζ) dζ = 2πi

N X

mn +

n=1

Z

g(ζ) dζ. η(z)

Since the summation is a positive integer, it is true that exp

nZ

g(ζ) dζ γ(z)

o

Z N n X mn + = exp 2πi n=1

g(ζ) dζ η(z)

o

= exp

nZ

η(z)

g(ζ) dζ

o

and this means that f is independent of the choice of γ(z). • f ∈ H(C \ A). The definition of f shows that the holomorphicity of f depends on the holomorphicity of the integral. Let z ∈ C \ A and denote Z g(ζ) dζ. G(z) = γ(z)

Now there exists a disc D(0; R) containing z. Let h be so small that z + h ∈ D(0; R) and [z, z + h] ∩ A = ∅. By Definition 10.41, D(0; R) contains only finitely many points of A. Without loss of generality, we may assume that {z1 , z2 , . . . , zN } ⊆ D(0; R) for some positive integer N . Then both γ(z) and γ(z + h) must lie in Ω = D(0; R) \ {z1 , z2 , . . . , zN }. Suppose that γ(z) consists of only horizontal or vertical line segments in Ω and γ(z + h) shares the same path with γ(z) until the point z. Since [z, z +h]∩A = ∅, we can connect z and z + h by another set of horizontal or vertical line segments in a way that only triangles are produced. Figure 15.2 illustrates this setting.

15.2. Some Examples

473

Figure 15.2: The paths γ(z + h) and −γ(z). Now we consider the difference G(z + h) − G(z) = =

Z

γ(z+h) m Z X

g(ζ) dζ −

Z

g(ζ) dζ +

g(ζ) dζ

γ(z)

Z

g(ζ) dζ

(15.20)

[z,z+h]

k=1 ∂∆k

where ∆1 , ∆2 , . . . , ∆m are the triangles produced. Obviously, the compactness of the set ∆1 ∪ ∆2 ∪ · · · ∪ ∆m ensures that there corresponds an open set Ω′ in Ω such that ∆k ⊆ Ω′ for every k = 1, 2, . . . , m. Since g ∈ H(Ω), Theorem 1013 (The Cauchy’s Theorem for a Triangle) implies that each integral in the summation (15.20) is 0. As g is continuous at z, we can write g(ζ) = g(z) + ǫ(ζ), where ǫ(ζ) → 0 as ζ → z. Therefore, the expression (15.20) becomes Z Z Z ǫ(ζ) dζ. (15.21) ǫ(ζ) dζ = hg(z) + g(z) dζ + G(z + h) − G(z) = Since

[z,z+h]

[z,z+h]

[z,z+h]

Z

[z,z+h]

ǫ(ζ) dζ ≤

sup ζ∈[z,z+h]

|ǫ(ζ)| · |h|,

the last integral actually tends to 0 as h → 0. Consequently, we have proved that lim

h→∞

G(z + h) − G(z) = g(z), h

i.e., G is holomorphic at z. Since z is arbitrary, G ∈ H(C \ A) which implies the desired result that f ∈ H(C \ A). • f has a removable singularity at each zn . It suffices to verify that lim (z − zn )f (z) = 0.

z→zn

Since zn is a simple pole of g with residue mn , there exists a δn > 0 such that mn + h(z) g(z) = z − zn

(15.22)

(15.23)

474

Chapter 15. Zeros of Holomorphic Functions
for all |z − zn | < 2δn and h ∈ H D(zn ; 2δn ) .a In fact, we can choose δn small enough such that D(zn ; 2δn ) contains only the pole zn . Suppose that z lies on the line segment joining 0 and zn , z ∈ D(zn ; δn ) and θn = arg zn . We split the path γ(z) into two paths γ1 (z) and γ2 (z) as follows: The path γ1 (z) is the line segment from 0 to zn − δn eiθn . If it passes through a pole of g, then we can make a small circular arc around that pole. The path γ2 (z) is the line segment from zn − δn eiθn to z.

On γ1 (z), since δn is fixed and g is continuous on γ1 (z), there exists a positive constant M1 such that Z g(ζ) dζ ≤ M1 . (15.24) γ1 (z)

By the definition, we parameterize γ2 (z) : [0, 1] → C as

γ2 (z; t) = z + (1 − t)(zn − δn eiθn − z) so that γ2 (z; 0) = zn − δn eiθn and γ2 (z; 1) = z. Substitute γ2 (z; t) into the Laurent series (15.23) to getb 
 Z 1n Z o Z z − zn − δn eiθn dt 
 g(ζ) dζ = mn h(ζ) dζ + t z − zn − δn eiθn − δn eiθn 0 γ2 (z) γ2 (z) Z 1 h(ζ) dζ = mn ln{t[z − (zn − δn eiθn )] − δn eiθn } + 0 γ (z) Z 2   h(ζ) dζ. (15.25) = mn ln(z − zn ) − i(θn + π) − ln δn + γ2 (z)


Since h ∈ H D(zn ; 2δn ) , there exists a positive constant M2 such that Z h(ζ) dζ ≤ M2 γ2 (z)

which gives

nZ exp

γ2 (z)

o g(ζ) dζ ≤ |z − zn |mn eM2 × exp(−mn ln δn ) .

(15.26)

Combining the estimate (15.24) and the expression (15.26), we establish o o nZ nZ g(ζ) dζ × exp g(ζ) dζ |f (z)| = exp γ1 (z)

γ2 (z)

≤ eM1 +M2 · exp(−mn ln δn ) × |z − zn |mn

which implies the limit (15.22).

• The extension of f has a zero of order mn at zn . Since f has a removable singularity at each zn , it follows from Definition 10.19 that f can be extended to be holomorphic at zn . The analysis in the previous part further shows that nZ o nZ o f (z) = exp g(ζ) dζ × exp g(ζ) dζ γ1 (z) γ2 (z) nZ o nZ o = exp g(ζ) dζ × exp h(ζ) dζ × δn−mn e−imn (θn +π) (z − zn )mn . γ1 (z)

γ2 (z)

In other words, f has a zero of order mn at zn . a b

See, for example, [11, Corollary 9.11, p. 124]. If zn is real, then θn = 0 and z is also real. In this case, the integrated result (15.25) becomes ln |z −zn |−ln δn .

15.2. Some Examples

475

Thus we have completed the analysis of the problem.



Problem 15.9 Rudin Chapter 15 Exercise 9.

Proof. Suppose that z1 , z2 , . . . , zn are the zeros of f , listed according to their multiplicities, such that z1 , z2 , . . . , zn ∈ D(0; β). Define g(z) =

n Y z − zk . 1 − zk z

k=1

Then g is clearly holomorphic in a neighbourhood V containing U and |g(z)| = 1 on T by Theorem 12.4. Now the function f (z) h(z) = g(z) must be holomorphic in U . Since f (U ) ⊆ U , we deduce from Theorem 10.24 (The Maximum Modulus Theorem) that |h(z)| ≤ 1 for all z ∈ U . If z = 0, then we get α ≤1 |z1 | × |z2 | × · · · × |zn | which implies 0 < α ≤ β n . Consequently, we obtain n≤ (a) Put α = β =

1 2

log α . log β

(15.27)

into the inequality (15.27), we conclude that n ≤ 1.

(b) Similarly, by the inequality (15.27) again, we know that n ≤ 2. (c) In this case, we have n = 0. (d) In this case, we have n ≤ 3. This completes the analysis of the problem.



Problem 15.10 Rudin Chapter 15 Exercise 10.

Proof. Let I be the ideal generated by the set {gN | N ∈ N}. By Definition 15.14, every element of I is of the form fN1 gN1 + fN2 gN2 + · · · + fNk gNk (15.28) for some increasing sequence {Nk } of positive integers, where fN1 , fN2 , . . . , fNk are entire. By the definition of gN , if N < M , then there exists an entire function hM such that gN = hM gM . Consequently, the element (15.28) can be expressed as fN1 gN1 + fN2 hN2 gN1 + · · · + fNk hNk gN1 = f gN1 ,

476

Chapter 15. Zeros of Holomorphic Functions

where f is entire. Thus we have I = {f gN | f is entire and N ≥ 1}.

(15.29)

Assume that I was principal. One can find an entire function g ∈ I such that I = [g]. By the representation (15.29), we know that g = f gN for some entire f and some N ∈ N. Thus we may assume that there exists some positive integer N such that I = {f gN | f is entire}. Then we have gN +1 = f gN (15.30) for some entire f . However, since gN (N ) = 0 but gN +1 (N ) 6= 0, the equation (15.30) is a contradiction. Hence I is not principal and we have completed the proof of the problem.  Problem 15.11 Rudin Chapter 15 Exercise 11.

z−1 Proof. Recall from [100, Eqn. (6), p. 281] that ϕ−1 (z) = z+1 is a conformal one-to-one mapping of Π = {z ∈ C | Re z > 0} onto U . Therefore, we can reduce the problem to the existence of a bounded holomorphic function f in U which is not identically zero and its zeros are precisely at

αn =

iyn , 2 + iyn

where n = 1, 2, . . .. In this case, §15.22 shows that {αn } satisfies ∞ X

(1 − |αn |) < ∞.

n=1

(a) We have ∞ p ∞  i log n  X X 4 + (log n)2 − log n p 1− = 2 + i log n 4 + (log n)2 n=1 n=1

=

∞ X

n=1

For n ≥ 3, we have

p

p

4

+ (log n)2

·

4 + (log n)2 ≤ 2 log n so that

p

4

4 + (log n)2 + log n

∞ ∞  ∞ i log n  X X 2 2X1 1− . ≥ ≥ 2 + i log n 3(log n)2 3 n=3 n n=3 n=3

.

(15.31)

Hence the series (15.31) diverges and thus no such bounded holomorphic function in Π. (b) In this case, we know from the A.M. ≥ G.M. that √ ∞ √ ∞  i√n  X X 4+n− n √ = √ 1− 2+i n 4+n n=1 n=1 = ≤

∞ X

n=1 ∞ X n=1

√ √

4+n· 4+n·

4 √ √
4+n+ n

2 p 4

(4 + n)n

15.2. Some Examples

477

=2 0 2 for all z ∈ K. Thus the inequality (15.34) reduces to α + |α |z

2(1 + M ) n n . · 1 − |αn | ≤ 1 − |αn | · (1 − αn z)αn δ |1 − αn z| ≥

for all z ∈ K and all n ≥ N . Using the Blaschke condition and the Weierstrass M -test, we conclude immediately that the series (15.33) converges uniformly on K. Eventually, Theorem 10.28 ensures that B ∈ H(Ω), and we end the analysis of the problem. 

15.3

Problems on Blaschke Products

Problem 15.13 Rudin Chapter 15 Exercise 13.

Proof. Since αn are real, we have B(z) = z k

∞ Y αn − z 1 − αn z

n=1

for some k ∈ N. Suppose that αN −1 < r < αN . Since 0 < αn < 1 and αn < αn+1 for all positive integers n, we obtain r k < 1 and α −r αn − r n r−αn > 0 and 1−αn r > 1−αn for each n = 1, 2, . . . , N −1, the inequality (15.36) reduces to |B(r)| < Since

αN −αn 1−αn

=1−

n2 N2

N −1 Y n=1

αN − αn . 1 − αn

and log x ≤ x − 1 for x > 0, the inequality (15.37) becomes |B(r)|
0.

p→∞

(15.41)

480

Chapter 15. Zeros of Holomorphic Functions

Now here is the trick: For a positive integer p, we first replace αp+1 , αp+2 , . . . , α4p−1 by α4p . Then we have 1 > S2 (p) =

4p−1 Y

n=p+1

 α − x 3p−1  4 3p−1 α4p − x2p 4p 2p = ≥ 1 − 2p 1 − α4p x2p 1 − α4p x2p e +2

so that lim S2 (p) = 1.

(15.42)

p→∞

Next, we select a subsequence {pk } of positive integers such that 4pk − 1 < pk+1 + 1 for each k = 1, 2, . . .. This makes sure that {αpk +1 , αpk +2 , . . . , α4pk −1 } ∩ {αpk+1 +1 , αpk+1 +2 , . . . , α4pk+1 −1 } = ∅ for each k = 1, 2, . . .. Then the modified sequence {α′n } will be the one that replaces only the terms αpk +1 , αpk +2 , . . . , α4pk −1 by α4pk from the original sequence {αn } for k = 1, 2, . . .. Since ∞ X

(1 − |α′n |) ≤

n=1

∞ X

(1 − |αn |) < ∞,

n=1

Thus it follows from Theorem 15.21 that B(z) =

∞ Y α′k − z 1 − α′k z

(15.43)

k=1

is an element of H ∞ and has no zeros except at α′k . It is no doubt that |B(α′n )| → 0 as n → ∞, so lim inf |B(r)| = 0. (15.44) r→1

Besides, we apply the representation (15.39) to our Blaschke product (15.43) to obtain |B(x2pk )| =

pk Y x2pk − α′n × 1 − α′n x2pk

n=1

4pY k −1

n=pk +1

= T1 (pk ) · S2 (pk ) · T3 (pk ).

∞ α −x Y α′n − x2pk 4pk 2pk × 1 − α4pk x2pk 1 − α′n x2pk n=4pk

Combining the limits (15.40), (15.41) and (15.42), we conclude immediately that lim |B(x2pk )| = 1

k→∞

which means lim sup |B(r)| = 1.

(15.45)

r→1

Hence the two results (15.44) and (15.45) imply that the function (15.43) has no radial limit at  z = 1, completing the analysis of the problem. Problem 15.15 Rudin Chapter 15 Exercise 15.

Proof. As a linear fractional transformation, ϕ is one-to-one and ϕ ∈ H(U ). Since ϕ(U ) = U , it follows from Theorem 12.6 that z−α (15.46) ϕ(z) = λ · 1 − αz for some α ∈ U and |λ| = 1.

15.3. Problems on Blaschke Products

481

(a) If α = 0, then λ 6= 1 because ϕ is not the identity function. In this case, 0 is the unique fixed point and ϕ(z) = λz. However, it implies that ∞ X

n=1

1 1. In this case, for sufficiently large enough n, the expression (15.49) implies that (b − b1 )z + ( 1 − 1)b b 1 2 2 kn kn |ϕn (z)| = (1 − k1n )z + ( kbn2 − b1 ) (b − 1 )z + ( 1 − b )b 2 1 2 kn kn ≈ z − b1 1 ≈1− n |k| which means 1 − |ϕn (z)| ≈ |k|1n . Hence the ϕ also satisfies the Blaschke condition in this case. ∗ Subcase (iii): |k| = 1 and k is an N th root of unity for some N . Put z = 21 into the expression (15.49), we have  1  1 = ϕN 2 2 so that

∞ h X

n=1

∞ h  1  i X  1  i 1 − ϕpN 1 − ϕn ≥ = ∞. 2 2 p=1

Thus ϕ does not satisfy the Blaschke condition in this case. ∗ Subcase (iv): |k| = 1 and k is not an nth root of unity for all n. We claim that the set S = {kn | n ≥ 0} is dense on T . To see this, we write k = eiθ , where θ is an irrational multiple of 2π.d Therefore, we obtain S = {einθ | n ∈ N}. Given ǫ > 0. Let ℓ be an arc on T with angle ǫ. Choose N ∈ N such that 2π N < ǫ. By the hypothesis, the (N + 1) points 1, eiθ , e2iθ , . . . , eN iθ are all distinct. As a result, two of them must have a counterclockwise angle less than 2π N , i.e., (p − q)θ < d

Otherwise, we have θ =

2πp q

2π m · |1 − knm | > 2 for large enough m. This shows that ϕ satisfies the Blaschke condition.

1 m

(b) We claim that ϕ satisfies ϕn (z) = z

(15.50) H ∞,

for some n ∈ N. To this end, if f ◦ ϕ = f for some nonconstant f ∈ have



f ϕn (z) = f ϕ ϕn−1 (z) = f ϕn−1 (z) = · · · = f (z)

then we must

for every z ∈ U and every n = 1, 2, . . ..

– Case (i): ϕ has a unique fixed point in U . Recall from the equation (15.46) that ϕ has a unique fixed point in U if and only if α = 0. In this case, we have ϕ(z) = λz for some λ such that |λ| = 1. Obviously, we know that ϕn (z) = λn z

for every n ∈ N and z ∈ U . If λ is an N th root of unity, then the expected result (15.50) holds immediately for some N ∈ N. Otherwise, we fix z = z0 ∈ U \ {0} so that f (λn z0 ) = f (z0 ) for every n ≥ 0. By part (a) Subcase (iv) above, the set S ′ = {λn | n ≥ 0} is dense on T which implies that f (z) = f (z0 ) (15.51) on C(0; |z0 |). By Theorem 10.18, the result (15.51) shows that f (z) = f (z0 ) for every z ∈ U and this contradicts the Open Mapping Theorem. – Case (ii): ϕ has a unique fixed point b on T . In this case, it follows from the equation (15.48) that ϕn (z) → b as n → ∞ for every z ∈ U . This means that
f (z) = lim f ϕn (z) = f (b) n→∞

for every z ∈ U , but it also contradicts the Open Mapping Theorem. – Case (iii): ϕ has two distinct fixed points b1 and b2 on T . Then it yields from the equation (15.49) that for every z ∈ U , we have   b1 , if |k| < 1; lim ϕn (z) = n→∞  b2 , if |k| > 1.

By similar argument to Case (ii), we can show that it is impossible for these two cases so that |k| = 1. Suppose that k is an N th root of unity for some N ∈ N. Then we see from the equation (15.49) again that ϕN (z) = z. Otherwise, part (a) Subcase (iv) ensures that the set S = {k n | n ≥ 0} is dense on T . Using similar argument as Case (i), we conclude that this is impossible.

484

Chapter 15. Zeros of Holomorphic Functions Consequently, the linear fractional transformation ϕ must satisfy the claim (15.50).

Hence we have completed the analysis of the problem.



Problem 15.16 Rudin Chapter 15 Exercise 16.

Proof. Note that 1 − |αj | = so we have

Z

1

dr, |αj |

∞ Z ∞ X X (1 − |αj |) = j=1

j=1

1

dr.

(15.52)

|αj |

Define the characteristic function χj (r) = Then formula (15.52) becomes

  1, if r ≥ |αj |; 

0, otherwise.

∞ Z ∞ X X (1 − |αj |) = j=1

j=1

Observe that

∞ X

1

χj (r) dr.

(15.53)

0

χj (r) = n(r).

(15.54)

j=1

Since each χj : [0, 1] → [0, ∞] is measurable for every j = 1, 2, . . ., we apply Theorem 1.27 to interchange the summation and the integration in the equation (15.53) and then use the formula (15.54) to gain Z 1 Z 1 X ∞ ∞  X n(r) dr. χj (r) dr = (1 − |αj |) = 0

j=1

j=1

0

This ends the proof of the problem.



Problem 15.17 Rudin Chapter 15 Exercise 17.

Proof. Assume that B(z) =

∞ X k=0

ck z k was a Blaschke product with ck ≥ 0 for all k = 0, 1, 2, . . .

and B(α) = 0 for some α ∈ U \ {0}. Combining Theorem 15.24 and the facte that |f (x)| ≤ λ holds for almost all x if and only if kf kL∞ ≤ λ, we have kB ∗ kL∞ (T ) ≤ 1 or e

Refer to [100, p. 66]

B(eiθ ) ∈ L∞ (T ).

15.3. Problems on Blaschke Products

485

Furthermore, we also have An (eiθ ) = e−inθ B(eiθ ) ∈ L2 (T ) for every n = 1, 2, . . .. Clearly, we have L∞ (T ) ⊆ L2 (T ). Recall that L2 (T ) is an inner product space. On the one hand, we may apply [100, Eqn. (7) & (8), p. 89] to get Z π 1 hB, An i = B(eiθ ) · An (eiθ ) dθ 2π −π Z π X ∞ ∞  X  1 ck eikθ × ck ei(n−k)θ dθ = 2π −π k=0

=

∞ X

k=0

ck cn+k .

(15.55)

k=0

On the other hand, the fact |B(eiθ )| = 1 a.e. on T gives h 1 Z π o2 i2 n 1 Z hB, An i = einθ dθ , einθ |B(eiθ )|2 dθ = 2π −π 2π T \N

(15.56)

where |B(eiθ )| = 6 1 on N with m(N ) = 0. Recall from Remark 3.10 that L2 (T ) is a space whose elements are equivalence classes of functions, so we can express (15.56) by o2 n 1 Z einθ dθ = 0. (15.57) hB, An i = 2π T

Combining the two results (15.55) and (15.57), we conclude that ck = 0 for all k ∈ N, which is impossible. Hence no such Blaschke product exists and we complete the proof of the problem.  Problem 15.18 Rudin Chapter 15 Exercise 18.

Proof. Obviously, we have f ′ (z) = 2(z − 1)B(z) + (z − 1)2 B ′ (z). Thus it suffices to show that (z − 1)B ′ (z) is bounded in U . Suppose that {αn } is the sequence of zeros of B. Then we know that {αn } ⊆ (0, 1) and it satisfies the Blaschke condition ∞ X

(1 − α2n ) < ∞.

(15.58)

n=1

Without loss of generality, we may assume that 0 < α1 ≤ α2 ≤ · · · < 1.

(15.59)

Furthermore, suppose that B(z) =

∞ Y αn − z 1 − αn z

n=1

and

Bn (z) =

∞ Y αk − z . 1 − αk z

k=1 k6=n

We yield from [30, Exercise 10, p. 174] that B ′ (z) X  αn − z ′  αn − z −1 = B(z) 1 − αn z 1 − αn z n=1 ∞

486

Chapter 15. Zeros of Holomorphic Functions

B ′ (z) = −

∞ X

n=1

1 − α2n Bn (z). (1 − αn z)2

It is well-known that |Bn (z)| < 1 for every z ∈ U . Now we observe from the assumption (15.59) that δ = min{α1 , α2 , . . .} = α1 > 0. Geometrically, if z ∈ U and p > 1, then |p − z| > |1 − z|.

(15.60)

Put p = α1n into the estimate (15.60) to get |1 − αn z|2 ≥ δ2 |1 − z|2 for every z ∈ U . Hence, the condition (15.58) implies |(z − 1)2 B ′ (z)| ≤ |1 − z|2 ·

∞ ∞ X 1 X 1 − α2n (1 − α2n ) < ∞ · |B (z)| ≤ n 2 2 (1 − α z) δ n n=1 n=1

holds for all z ∈ U . Consequently, the modulus |(z − 1)2 B ′ (z)| is bounded in U which shows the  boundedness of f ′ in U , completing the proof of the problem. Remark 15.2 Blaschke products serve as an important subclass of H(U ). If you are interested in the literature of Blaschke products, you are suggested to read the book by Colwell [29].

15.4

Miscellaneous Problems and the M¨ untz-Szasz Theorem

Problem 15.19 Rudin Chapter 15 Exercise 19.

Proof. Let 0 < r < 1. On the one hand, we notice that  reiθ + 1  r2 − 1 = 2 . log |f (reiθ )| = log exp Re iθ re − 1 r − 2r cos θ + 1

Thus [100, Eqn, (3), p. 233] asserts that Z π Z π 1 r2 − 1 1 µr (f ) = dθ = − Pr (θ) dθ = −1. 2π −π r 2 − 2r cos θ + 1 2π −π

(15.61)

On the other hand, we have f ∗ (eiθ ) = lim f (reiθ ) = lim exp r→1

r→1

so that log |f ∗ (eiθ )| = Re Therefore, we obtain

 reiθ + 1  reiθ − 1

 eiθ + 1  eiθ − 1

µ∗ (f ) = 0.

= exp

 eiθ + 1  eiθ − 1

= 0.

(15.62)

15.4. Miscellaneous Problems and the M¨ untz-Szasz Theorem

487

Hence it follows from the results (15.61) and (15.62) that lim µr (f ) < µ∗ (f ),

r→1

completing the proof of the problem.



Problem 15.20 Rudin Chapter 15 Exercise 20.

Proof. If λN < 0 for some N ∈ N, then there exists a δ > 0 such that λN < δ < 0. By the hypothesis, we actually have λn < δ < 0 for all n ≥ N , but this contradicts another hypothesis that λn → 0 as n → ∞. Therefore, we have λ1 > λ2 > · · · > 0 which implies that 0
− 12 . Then the set of all finite linear combinations of the functions tλ0 , tλ1 , tλ2 , . . . is dense in L2 (I) if and only if

∞ X

2λn + 1 = ∞. 2 (2λ n + 1) + 1 n=0 To this end, let m ∈ N and m ∈ / {λn }. By [20, p. 173], we know that

Thus we see that

n−1 n−1

Y m − λk X 1

ak tλk = √ min tm − . ak ∈C 2 1 + 2m k=0 m + λk + 1 k=0

tm ∈ span {tλ0 , tλ1 , . . .}

(15.63)

488

Chapter 15. Zeros of Holomorphic Functions

if and only if lim sup n→∞

if and only if lim sup n→∞

n−1 Y

k=0 λk >m

1 −

n−1 Y k=0

m − λk =0 m + λk + 1

2m + 1 × m + λk + 1

n−1 Y

k=0 − 12 m

1 −

2m + 1 = 0 or m + λk + 1

lim sup n→∞

n−1 Y k=0

− 21 0. If λk > m, then we have λk >

− 12 ,

we have

2λk +1 m+λk +1 ∞ X

k=0 λk >m

1 −

2λk + 1 = 0. m + λk + 1

2m+1 m+λk +1

(15.64)

∈ (0, 1). Otherwise, since

∈ (0, 1). By Theorem 15.5, the results (15.64) hold if and only if

2m + 1 =∞ m + λk + 1

∞ X

or

k=0 − 12 m. Then we have λk +

1 2

2λk + 1 = ∞. m + λk + 1

(15.65)

< m + λk + 1 < 2λk + 1 and we get

1 1 2 < < 2λk + 1 m + λk + 1 2λk + 1 which means that the first summation (15.65) is equivalent to ∞ X

k=0 λk >m

1 = ∞. 2λk + 1

(15.66)

If − 12 < λk < m, then we have 1 < m + λk + 1 ≤ 2m + 1 so that the second summation (15.65) is equivalent to ∞ X (2λk + 1) = ∞. (15.67) k=0 − 21 m, then it is clear that 1 2λk + 1 < 2λk + 1. < 2(2λk + 1) (2λk + 1)2 + 1 Similarly, if − 12 < λk ≤ m, then it is easy to see that 2m + 1 2λk + 1 2λk + 1 2m + 1 ≤ < . ≤ (2m + 1)2 + 1 (2λk + 1)2 + 1 (2λk + 1)2 + 1 2λk + 1 Therefore, we establish that one of the summations (15.66) and (15.67) holds if and only if ∞ X k=0

2λk + 1 =∞ (2λk + 1)2 + 1

holds. Finally, our desired result is proven if we combine Theorem 3.14 and the Weierstrass Approximation Theorem [99, Theorem 7.26, p. 159]. This completes the proof of the problem. 

15.4. Miscellaneous Problems and the M¨ untz-Szasz Theorem

489

Remark 15.3 The proof of Problem 15.21 follows basically the proof of Theorem 2.2 in [21]. For the version of a complex sequence {λn }, please refer to [85, §12, pp. 32 – 36]. Problem 15.22 Rudin Chapter 15 Exercise 22.

Proof. Let M be the set of all finite linear combinations of the functions fn . It is clear that M is a subspace of L2 (0, ∞). Let g ∈ L2 (0, ∞) be orthogonal to each fn , i.e., Z ∞ fn (t)g(t) dt = 0 hfn , gi = 0

for each n = 1, 2, . . .. Define F (z) =

Z

∞

e−tz g(t) dt

0

on Π = {z ∈ C | Re z > 0}. For every fixed z ∈ Π, since Z ∞ Z ∞ 1 e−2tRe z dt = √ |e−tz |2 dt = < ∞, 2 Re z 0 0 we have e−tz ∈ L2 (0, ∞). Using Theorem 3.8, we see that |F (z)| = ke−tz gk1 ≤ ke−tz k2 · kgk2 < ∞. In other words, F is well-defined in Π. Next, we have to compute F ′ (z) in Π. To this end, we consider Z ∞ −tω e − e−tz F (ω) − F (z) = · g(t) dt. ω−z ω−z 0 Given ǫ > 0. Let t > 0 and ω = z − ζ,

(15.68)

where |ζ| < ǫ. Then we have etζ − 1 e−t(z−ζ) − e−tz = −e−tz · . z−ζ−z ζ Suppose that h(ζ) =

etζ −1 ζ .

The power series expansion of h is given by h(ζ) =

∞ n X t

n=1

n!

ζ n−1 ,

(15.69)


so we have h ∈ H D(0; ǫ) by Theorem 10.6. Since h is continuous on D(0; ǫ), we obtain from Theorem 10.24 (The Maximum Modulus Theorem) and the Extreme Value Theorem that the maximum of |h(ζ)| occurs on the boundary |ζ| = ǫ. By the expansion (15.69), we see that max |h(ζ)| ≤ |ζ|=ǫ

∞ n X t

n=1

n!

· |ζ|n−1 =

etǫ − 1 . ǫ

(15.70)

490

Chapter 15. Zeros of Holomorphic Functions

By the Mean Value Theorem for Derivatives, we know that etǫ − 1 = ǫtetξ for some ξ ∈ (0, ǫ), so we induce from the estimate (15.70) that etζ − 1 ≤ tetξ < tetǫ , ζ

where |ζ| < ǫ and t > 0. If z ∈ Π, then Re z > 2ǫ for some ǫ > 0. With this ǫ > 0, we may pick ω ∈ Π satisfying the condition (15.68). Therefore, we have e−tω − e−tz −tz etζ − 1 · = e < te−tRe z · etǫ = te−t(Re z−ǫ) ≤ te−tǫ ω−z ζ

for t > 0. Since te−tǫ , g ∈ L2 (0, ∞), Theorem 3.8 implies that te−tǫ ·g ∈ L1 (0, ∞). Consequently, Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) can be applied to show the limit and the integral can be interchanged in the following deduction: F (ω) − F (z) ω−z Z ∞ −ωt e − e−tz = lim · g(t) dt ω→z 0 ω−z Z ∞ e−ωt − e−tz · g(t) dt lim = ω→z ω−z Z0 ∞ −tetz g(t) dt. =

F ′ (z) = lim

ω→z

0

Since z is arbitrary, we have proven that F ∈ H(Π). In fact, this process can be repeated to get Z ∞ F (n) (z) = (−1)n tn e−tz · g(t) dt,

(15.71)

0

where n ∈ N and z ∈ Π. By the hypothesis and the formula (15.71), we have Z ∞ (n) n F (1) = (−1) (tn e−t )g(t) dt = (−1)n hfn , gi = 0 0

so that F ≡ 0. Particularly, if we denote G(t) = χ(0,∞) (t)e−t g(t), then we see that √ Z ∞ −t √ √ Z ∞ b G(t)e−ity dt = 2π e g(t)e−ity dt = 2πF (1 + iy) = 0, G(y) = 2π −∞

0

L1 (0, ∞),

where y ∈ R. As G ∈ we observe from Theorem 9.12 (The Uniqueness Theorem) that G(t) = 0 a.e. on R which implies that g(t) = 0 (15.72) a.e. on (0, ∞). Recall that L2 (0, ∞) is Hilbert and M is a subspace of L2 (0, ∞). If we have M 6= L2 (0, ∞), according to the Corollary of Theorem 4.11, there corresponds an g ∈ L2 (0, ∞) such that g 6= 0 and hfn , gi = 0 for every n = 1, 2, . . .. However, this definitely contradicts the above conclusion (15.72). Hence M = L2 (0, ∞), as desired. This completes the analysis of the  problem. Problem 15.23 Rudin Chapter 15 Exercise 23.

15.4. Miscellaneous Problems and the M¨ untz-Szasz Theorem

491

Proof. Since f (0) = 0, λn 6= 0 for all n = 1, 2, . . . , N . Suppose that B(z) =

Since

N X

N Y λn − z |λn | · . 1 − λn z λn n=1

(1 − |λn |) < ∞, Theorem 15.21 implies that B ∈ H ∞ and B has no zeros except at

n=1

B the points λn . Consider the function g = 1−f . Then we have g ∈ H(U ). By Theorem 12.4, g is continuous on U and λ − eiθ n =1 1 − λn eiθ

for every n = 1, 2, . . . , N . Therefore, we see that |g(eiθ )| =

1 1 |B(eiθ )| ≤ ≤ . |1 − f (eiθ )| |f (eiθ )| − 1 2

Consequently, it yields from Theorem 10.24 (The Maximum Modulus Theorem) that |λ1 λ2 · · · λN | = |g(0)| < We end the proof of the problem.

1 . 2 

492

Chapter 15. Zeros of Holomorphic Functions

CHAPTER

16

Analytic Continuation

16.1

Singular Points and Continuation along Curves

Problem 16.1 Rudin Chapter 16 Exercise 1.

Proof. By Theorem 16.2, f has a singularity at some point eiθ . If we consider the power series for f about the point 21 , then the representation f (z) =

∞ X k=0

∞  1 k 1 k X f (k) ( 21 )  z− = bk z − 2 k! 2

(16.1)

k=0

holds in D( 21 ; 21 ). Thus the radius of convergence of the power series (16.1) must be 12 . Otherwise, the power series would define a holomorphic extension of f beyond eiθ , a contradiction. Assume that f was regular at z = 1. Then the series ∞ X f (k) ( 21 )  1 k x− k! 2 k=0

converges at some x > 1. For every k ≥ 1, we derive from the representation f (z) = that ∞ 1 X n(n − 1) · · · (n − k + 1)an = . f (k) 2 2n−k

P

an z n

n=k

Now we deduce from the binomial theorem that ∞ X f (k) ( 1 )  2

k=0

k!

x−

∞ ∞ 1 k X h X n(n − 1) · · · (n − k + 1) an i  1 k = × n−k × x − 2 k! 2 2 k=0

=

n=k

∞ ∞ X X

k=0 n=k

Ckn ·

an 2n−k

 1 k . · x− 2

(16.2)

an Since an ≥ 0, Ckn · 2n−k ·(x− 12 )k ≥ 0. Consequently, the order of the summation in the expression (16.2) can be switched (see [99, Exercise 3, p. 196]) so that n ∞ ∞ ∞ n−k  X f (k) ( 12 )  1 k X h X n  1 1 k i X an xn an x− = Ck = −0 · x− k! 2 2 2 k=0

n=0

n=0

k=0

493

494

Chapter 16. Analytic Continuation

P which means that the radius of convergence of f (z) = an z n is greater than 1. This is a contradiction to our hypothesis and we have completed the analysis of the problem.  Problem 16.2 Rudin Chapter 16 Exercise 2.

Proof. Suppose that (f, D) and (g, D) can be analytically continued along γ to (fn , Dn ) and (gm , Dm ) respectively. According to Definition 16.9, there exist chains Cf = {D0 , D1 , . . . , Dn } ′ }, where D = D ′ = D. Furthermore, there are numbers and Cg = {D0′ , D1′ , . . . , Dm 0 0 0 = s0 < s1 < · · · < sn = 1 and

0 = t0 < t1 < · · · < tm = 1

′ , such that γ(0) is the center of D0 = D0′ , γ(1) is the center of Dn and Dm

γ [sj , sj+1 ] ⊆ Dj and γ [tk , tk+1 ] ⊆ Dk′

for j = 0, 1, . . . , n − 1 and k = 0, 1, . . . , m − 1. We notice that there are function elements ′ ) for j = 0, 1, . . . , n−1 and k = 0, 1, . . . , m−1, (fj , Dj ) ∼ (fj+1 , Dj+1 ) and (gk , Dk′ ) ∼ (gk+1 , Dk+1 where f0 = f and g0 = g.

Since P f0 (z), g0 (ζ) = 0 for all z, ζ ∈ D0 , we have P f1 (z), g0 (ζ) = 0 for all z ∈ D0 ∩ D1 and ζ ∈ D0 . For each fixed ζ ∈ D0 , P is a polynomial in z so that P ∈ H(D1 ). By the Corollary to Theorem 10.18, we have
P f1 (z), g0 (ζ) = 0 (16.3)

for all z ∈ D1 . Since ζ is arbitrary, the equation (16.3) is actually true for all ζ ∈ D0 . Repeat this process, we conclude that P (fn (z), g0 (ζ)) = 0 for all z ∈ Dn and ζ ∈ D0 . Next, we fix a z ∈ Dn and now P is a polynomial in ζ so that P ∈ H(D1′ ). Similar argument shows that
P fn (z), g1 (ζ) = 0 holds for all ζ ∈ D1′ and hence also for every z ∈ Dn . Repeat the process also implies that the equation
P fn (z), gm (ζ) = 0 (16.4)

holds in Dn and Dm . Finally, we can establish the required equation of the problem if we replace fn and gm by f1 and g1 in the equation (16.4). Obviously, this can be extended to n function elements (f1 , D), (f2 , D), . . . , (fn , D) provided that P (z1 , z2 , . . . , zn ) is a polynomial in n variables, f1 , f2 , . . . , fn can be analytically continued along a curve γ to g1 , g2 , . . . , gn and P (f1 , f2 , . . . , fn ) = 0 in D. In fact, our above proof only uses the holomorphicity of the polynomial P , so similar results can be established if we only require that P is a function of n variables such that P (. . . , zj , . . .) is holomorphic in each variable  zj for j = 1, 2, . . . , n. This completes the analysis of the problem. Problem 16.3 Rudin Chapter 16 Exercise 3.

Proof. By Theorem 11.2, the function f = ux − iuy is holomorphic in Ω. Since Ω is simply connected, Theorem 13.11 ensures that there corresponds an F ∈ H(Ω) such that F ′ = f . If F = A + iB, then we have F ′ (z) = Ax + iBx = Ax − iAy = ux − iuy

16.1. Singular Points and Continuation along Curves

495

so that A(x, y) = u(x, y) + C for some constant C. Hence u is the real part of F − C ∈ H(Ω).

Next, we suppose that Ω is a region, but not simply connected. Let f ∈ H(Ω) and f (z) 6= 0 for every z ∈ Ω. Then log |f | is harmonic in Ω by Problem 11.5. If it has harmonic conjugate, then there corresponds an F ∈ H(Ω) such that |eF (z) | = eRe F (z) = elog |f (z)| = |f (z)| which means that |f e−F | = 1

in Ω. According to [11, Proposition 3.7, p. 39], f (z)e−F (z) = eiθ for some constant θ in Ω. Now we can write f (z) = eg(z) for some g ∈ H(Ω). By Theorem 13.11, Ω is simply connected, a contradiction. Hence this shows that the statement of the problem fails in every region that is not simply connected, completing the proof of the problem.  Problem 16.4 Rudin Chapter 16 Exercise 4.

Proof. Denote I = [0, 1]. Let α : I → C \ {0} be an arbitrary path from 1 to f (0). By Definition 10.8, without loss of generality, we may assume further that α′ is continuous on I. Define Z 1 ′ α (t) dt . (16.5) g(0) = α(t) 0 We note that g(0) =

Z

1 0


d ln α(t) = log f (0)

and it means that f (0) = eg(0) . Let ζ ∈ X \ {0} and βζ : I → X be the line segment joining 0 and ζ. Next, we define γζ : I → C \ {0} by  if t ∈ [0, 21 ];  α(2t), γζ (t) = (16.6)
 f βζ (2t − 1) , otherwise.

Finally, we define

g(ζ) =

Z

0

1

γζ′ (t) dt γζ (t)

.

Clearly, we know that Z 1  
d log γζ (t) = log γζ (1) − log γζ (0) = log f βζ (1) − log α(0) = log f (ζ) g(ζ) =

(16.7)

0

so that f (ζ) = eg(ζ) . See Figure 16.1 for the paths βζ (I) and γζ . Now it suffices to prove that the function g : X → C defined by the equations (16.5) and (16.7) is continuous on X. To this end, let z, ω ∈ X and suppose that βω,z : I → X is the line segment from ω to z. We also define γω,z : I → C \ {0} by
γω,z (t) = f βω,z (t) .

496

Chapter 16. Analytic Continuation

Figure 16.1: The paths βζ (I) and γζ . If γ is a path in X from x0 to x1 , and if λ is a path in X from x1 to x2 , then we define the product γ ∗ λ of γ and λ to be the path η given by the equations  if t ∈ [0, 21 ];  γ(2t), η(t) =  λ(2t − 1), otherwise.

Thus both βz and βω ∗ βω,z are paths in X from 0 to z. Since X is simply connected, we follow from Theorem 16.14 or [74, p. 323] that βz and βω ∗ βω,z are (path) homotopic in
X, i.e., βz ≃p βω ∗ βω,z in X. Since f is continuous on X, we must have f (βz ) ≃p f βω ∗ βω,z in f (X). Recall that α is arbitrary in the definition (16.6), it establishes that γz ≃p γω ∗ γω,z in f (X) ∪ α(I), see Figure 16.2 for the paths γz (I), γω (I) and γω,z (I)

Figure 16.2: The paths γz (I), γω (I) and γω,ζ (I). Since X is compact and f (z) 6= 0 for all z ∈ X, there exists a constant m > 0 such that κ < ǫ. By the continuity of f , there m = min |f (z)|. Given ǫ > 0. Choose κ > 0 such that m z∈X

corresponds a δ > 0 such that for z, ω ∈ X and |z − ω| < δ, the length of γω,z is less than κ. Let

16.2. Problems on the Modular Group and Removable Sets ℓ(z, ω) be the length of γω,z . Hence, if |z − ω| < δ, then we obtain Z 1 γ ′ (t) dt Z 1 γ (t) dt ω z |g(z) − g(ω)| = − γz (t) γω (t) 0 0 Z Z 1 (γ ∗ γ )′ (t) dt 1 ′ γω (t) dt ω ω,z − = (γω ∗ γω,z )(t) γω (t) 0 0 Z Z dt dt = − t γω ∗γω,z γω t Z Z Z dt dt dt + − = γω t γω,z t γω t Z dt = . γω,z t

497

(16.8)

Applying the definition of m to the integral (16.8), we get

ℓ(z, ω) κ < < ǫ. m m In other words, g is continuous at ω so that it is actually continuous on X, as required. This  completes the proof of the problem. |g(z) − g(ω)| ≤

16.2

Problems on the Modular Group and Removable Sets

Problem 16.5 Rudin Chapter 16 Exercise 5.

Proof. Let τ (z) = z + 1, σ(z) = − z1 and ϕ(z) =

az + b ∈ G. cz + d

Then we have a, b, c, d ∈ Z and ad − bc = 1.a Furthermore, we notice that τ −1 (z) = z − 1 and σ −1 (z) = − z1 . • Case (i): a = 0. We have bc = −1 so that b = −c. Obviously, b = ±1 if and only if c = ∓1. If b = 1, then c = −1 and we have
1 1 1 =− = − −d = σ τ −d (z) , ϕ(z) = −z + d z−d τ (z) i.e., ϕ = σ ◦ τ −d . Similarly, we have ϕ = σ ◦ τ d if b = −1 and c = 1.

az+b • Case (ii): a = ±1. Since −az−b −cz−d = cz+d , we may only consider the case that a = 1. Thus z+b we have d − bc = 1 and ϕ(z) = cz+d . We note that  −cz − d  −cz − d
−cz − d −(c − 1)z − (d − b) σ ϕ(z) = = and τ +1= z+b z+b z+b z+b

which imply that




−(d − bc) 1 =− = σ τ b (z) . τ c σ ϕ(z) = z+b z+b

a

Hence we have ϕ = σ −1 ◦ τ −c ◦ σ ◦ τ b .

With the aid of Problem 16.7, we remark that G = SL2 (Z).

498

Chapter 16. Analytic Continuation
• Case (iii): |a| > 1. We may take |a| > |c|. Otherwise, we consider σ ϕ(z) = − cz+d az+b az+b . Now it is easy to see that one can find an N ∈ Z satisfying instead of ϕ(z) = cz+d 0 ≤ |a − N c| < |c| < |a|. Since

we establish that


az + b (a − c)z + (b − d) τ −1 ϕ(z) = −1= , cz + d cz + d



ϕ1 (z) = σ τ −N ϕ(z) =

−cz − d a1 z + b1 = . (a − N c)z + (b − N d) c1 z + d1

Simple algebra shows that ϕ1 ∈ G, |a1 | < |a| and 0 ≤ |c1 | < |c|. If |c1 | = 0, then a1 d1 = 1 so that a1 = ±1 which goes back to Case (ii). Otherwise, we can repeat the above process finitely many times, say m times, to get ϕm (z) =

am z + bm ∈ G, cm z + dm

where either am = 0 or am = ±1. Therefore, the ϕm , and hence ϕ, is generated by τ and σ. Consequently, this proves the first assertion that τ and σ generate the modular group G. For the second assertion, suppose that R1 = {z = x + iy | |x| < 12 , y > 0 and |z| > 1}, R2 = {z = x + iy | − 12 ≤ x ≤ 0 and |z| = 1}, n o 1 R3 = z = − + iy y > 0 and |z| ≥ 1 . 2

(16.9)

Then we have R = R1 ∪ R2 ∪ R3 , see Figure 16.3.

Figure 16.3: The fundamental domain R of G. We check Theorem 16.19(a) and (b). Based on Apostol’s description [6, p. 30], two points ω, ω ′ ∈ Π+ = {z ∈ C | Re z > 0} are said to be equivalent under G if ω ′ = ϕ(ω) for some ϕ ∈ G. With this terminology, property (a) means that no two distinct points of R are equivalent under G and property (b) implies that for every ω ∈ Π+ , there exists a z ∈ R such that z is equivalent to ω.

16.2. Problems on the Modular Group and Removable Sets

499

Lemma 16.1 Suppose that z1 , z2 ∈ R, z1 6= z2 and z2 = ϕ(z1 ) for some ϕ ∈ G. Then we have Re z1 = ± 21 and z2 = z1 ∓ 1 or

|z1 | = 1 and z2 = − z11 .

Proof of Lemma 16.1. Without loss of generality, we may assume that Im z2 ≥ Im z1 by symmetry. Let ϕ(z) = az+b cz+d . Combing the assumption and the relation Im ϕ(z) =

Im z |cz + d|2

(16.10)

to get the condition |cz1 + d|2 ≤ 1.

(16.11)

√

3 2 .

Thus |c| · Since z1 ∈ R, it is easy to see that Im z1 ≥ As c ∈ Z, this forces that either c = 0 or |c| = 1.

√

3 2

≤ |c|Im z1 ≤ |cz1 + d| ≤ 1.

• Case (i): c = 0. Then ad = 1 and since a, d ∈ Z, we have a = d = ±1. In this case, the relation (16.10) shows that Im z2 = Im z1 . Furthermore, ϕ(z) = z ± b so that Re z2 = Re z1 ± b. Since b is an integer, the definition (16.9) shows b = 1 which implies that Re z1 = ± 21 and hence z2 = z1 ∓ 1. • Case (ii): |c| = 1. Then the condition (16.11) becomes |z1 ± d|2 ≤ 1 or equivalently, (Re z1 ± d)2 + (Im z1 )2 ≤ 1. (16.12) Further reduction implies that (Re z1 ± d)2 ≤ 1 − (Im z1 )2 ≤ 1 − so that |Re z1 ± d| ≤

3 1 = , 4 4

1 . 2

(16.13)

Since − 21 ≤ Re z1 ≤ 12 , we have |d| ≤ 1 which means either d = 0 or |d| = 1. – Subcase (i): |d| = 1. Using the inequality (16.13), we have |Re z1 ± 1| = 21 and then Re z1 = ± 21 . Next, it follows from the inequality (16.12) that 0 ≤ Im z1 ≤√

z1 =

± 12

+

√

i 3 2

3 2 ,

so actually we have Im z1 =

√

3 2 .

Consequently, we have

and then both our results hold in this case.

– Subcase (ii): d = 0. Now the condition (16.11) implies that |z1 | ≤ 1. Since z1 ∈ R by the definition (16.9), we actually have |z1 | = 1. Simple calculation gives 1 z2 = ±a − , z1 where a ∈ Z. Let z1 = x + iy. Thus z2 = ±a − x + iy. If a 6= 0, then since z1 , z2 ∈ R and z1 6= z2 , we get z1 = − 21 + iy and z2 = 21 + iy. Otherwise, a = 0 so that z1 = x + iy and z2 = −x + iy for every 0 < x ≤ 12 . Obviously, this case satisfies |z1 | = 1 and z2 = − z11 . This completes the proof of the lemma.



500

Chapter 16. Analytic Continuation

Combining the definition (16.9) of R and Lemma 16.1, we see immediately that no two distinct points of R are equivalent under G which is property (a). For proving property (b), we need the following result whose proof can be found in [6, Lemma 1, pp. 31, 32]: Lemma 16.2 ω′

Given ω1′ , ω2′ ∈ C with ω2′ not real. Let Ω = {mω1′ + nω2′ | m, n ∈ Z}. Then there 1 exist ω1 , ω2 ∈ C such that ω2 = aω2′ + bω1′ and ω1 = cω2′ + dω1′ , where ad − bc = 1, |ω2 | ≥ |ω1 | and |ω1 ± ω2 | ≥ |ω2 |. Now we go back to the proof of our problem. If ω1′ = 1 and ω2′ = ω ∈ Π+ , then it is easy to ω′ see that ω2′ ∈ / R. By Lemma 16.2, there exist ω1 and ω2 with |ω2 | ≥ |ω1 | and |ω1 ± ω2 | ≥ |ω2 | 1 such that ω2 = aω + b and ω1 = cω + d. Let z =

ω2 ω1 .

These relations give z=

aω + b = ϕ(ω) cω + d

(16.14)

with ad − bc = 1, |z| ≥ 1 and |z ± 1| ≥ |z|. The relation (16.14) means that there exists a point z ∈ R equivalent to ω ∈ Π+ under G which is exactly property (b). Hence we obtain the result that R is a fundamental domain of G and we end the proof of the problem.  Problem 16.6 Rudin Chapter 16 Exercise 6.


Proof. Since ψ ϕ(z) = z + 1, it follows from Problem 16.5 that G is also generated by ϕ and ψ. It is easy to see that
ϕ2 (z) = ϕ ϕ(z) = z

and

ψ 2 (z) = −

1 z−1

and ψ 3 (z) = z.

Hence ϕ has period 2 and ψ has period 3. This completes the proof of the problem.



Problem 16.7 Rudin Chapter 16 Exercise 7.

Proof. For each linear fractional transformation ϕ(z) = Mϕ =



a b c d



az+b cz+d ,

we associate the 2 × 2 matrix

.

Here we identify each matrix with its negative because Mϕ and −Mϕ represent the same transformation. If Mϕ and Mψ are the matrices associated with the linear fractional transformations ϕ and ψ respectively, then it is easy to see that the matrix product Mϕ Mψ is associated with the function composition ϕ ◦ ψ.

16.2. Problems on the Modular Group and Removable Sets

501

• An algebraic proof of Theorem 16.19(c). Now the group Γ is generated by the matrices     1 0 1 2 A= and B = . 2 1 0 1 If M ∈ Γ, then we have M = An1 Bm1 An2 Bm2 · · · Anp Bmp ,

(16.15)

where the nk , mk are integers. Direct computation gives     1 0 1 −2 A−1 = and B−1 = −2 1 0 1 so that =



1 (−2)nk

0 1

Ank Bmk =



1 (−2)nk

(−2)mk (−2)mk +nk + 1

A

nk



and

mk



B

=



1 (−2)mk 0 1

=



1 2Nk 2Mk 2Lk + 1



and then 

,

where Nk , Mk and Lk are integers. Thus we obtain   1 + 2Nk,j 2Mk,j Ank Bmk Anj Bmj = , 2Pk,j 1 + 2Lk,j where Nk,j , Mk,j , Lk,j and Pk,j are integers. Hence we apply this to the expression (16.15) to conclude immediately that if   a b M= , c d then a and d are odd, b and c are even. • Proof of the first part of Problem 16.5. Note that the transformations z 7→ z + 1 and z 7→ − z1 correspond to the matrices T=



1 1 0 1



and S =



0 −1 1 0



respectively. We claim that if M ∈ G, then it has the form M = Tn1 STn2 S · · · Tnp S,

(16.16)

where the nk are integers. To this end, we first notice that S2 = I,b so this explains why only the S appears in the form (16.16). Next, it suffices to prove those matrices   a b M= c d with c ≥ 0. If c = 0, then ad = 1 or equivalently, a = d = ±1 so that     ±1 b 1 ±b M= = = T±b . 0 ±1 0 1 b

Remember that we have identified I = −I.

502

Chapter 16. Analytic Continuation Next, if c = 1, then ad − b = 1 so that b = ad − 1 and       1 d 0 −1 1 a a ad − 1 = Td STd . = M= 0 1 1 0 0 1 1 d Assume that the form (16.16) is true for all matrices with lower left-hand element less than c for some c ≥ 1. Since ad − bc = 1, c and d must be coprime so that d = cq + r for some q ∈ Z and 0 < r < c. Since      a b 1 −q a −aq + b −q MT = = , c d 0 1 c r we have MT−q S =



−aq + b −a r −c



.

(16.17)

By the hypothesis, the matrix (16.17) has the form (16.16) which implies that M can be expressed in the form (16.17). This completes the proof of the problem.



Problem 16.8 Rudin Chapter 16 Exercise 8.

Proof. Since E ⊆ R is compact, we can define A = max x x∈E

and B = min x. x∈E

Notice that Ω = C \ E. Denote R = max(|A|, |B|). (a) Let x, y ∈ R such that x < y < A. Then x 6= y and Z Z  dt 1  1 dt = (x − y) − . f (x) − f (y) = t−y E (t − x)(t − y) E t−x

(16.18)

Since x, y ∈ R and E ⊂ R, the integrals in the equation (16.18) are real integrals. Clearly, 1 1 1 1 1 1 t−x ≥ A−x > 0 and t−y ≥ A−y > 0 for all t ∈ E. Let δx = A−x and δy = A−y . Then it follows from the expression (16.18) that Z δx δy dt = (x − y)δx δy m(E) > 0, f (x) − f (y) ≥ (x − y) E

i.e., f (x) 6= f (y). Consequently, f is nonconstant. (b) The answer is negative. Assume that f could be extended to an entire function. Since 1 |t| ≤ R on E, we have | t−z | → 0 as |z| → ∞ for every t ∈ E. In other words, we see that f (z) → 0 as |z| → ∞ which means f is bounded in C. By Theorem 10.23 (Liouville’s Theorem), f is constant which contradicts part (a). Hence we conclude that f cannot be extended to an entire function. z on E. We claim that (c) Given ǫ > 0 and z ∈ C \ D(0; R + Rǫ ). Define g(z, t) = − z−t g(z, t) → −1 uniformly on E. In fact, we see that

|g(z, t) + 1| =

|t| |z − t|

16.2. Problems on the Modular Group and Removable Sets

503

on E. Since |z| > R, we have |z − t| ≥ |z| − |t| ≥ |z| − R > 0 so that |g(z, t) + 1| = on E. Since |z| > R +

R ǫ,

|t| R ≤ |z − t| |z| − R

(16.19)

the inequality (16.19) implies that |g(z, t) + 1| < ǫ

for all t ∈ E. This proves our claim which asserts that Z g(z, t) dt + m(E) |zf (z) + m(E)| = E Z = [g(z, t) + 1] dt E Z |g(z, t) + 1| dt ≤ E

< ǫm(E)

for every |z| ≥ R +

R ǫ.

(16.20)

Since ǫ is arbitrary, we conclude from the estimate (16.20) that lim zf (z) = −m(E).

z→∞

(d) The compactness of E implies that Ω is open in C. We have to show that Ω is connected. Since E ⊂ R, we have C \ R ⊆ Ω. Thus the upper half plane Π+ lies in a component of Ω. Similarly, the lower half plane Π− must lie in a component of Ω. Since E 6= R, one can have a real number a lying in Ω. Since Π+ is connected, it follows from [74, Theorem 23.4, p. 150] that Π+ ∪ {a} is also connected. Similarly, the set Π− ∪ {a} is also connected. According to [74, Theorem 23.3, p. 150], the union Π+ ∪ {a} ∪ Π− = (C \ R) ∪ {a} is connected. Finally, since (C \ R) ∪ {a} ⊆ Ω ⊆ C, the connectedness of Ω can be deduced again from [74, Theorem 23.4, p. 150]. Assume that f had a holomorphic square root in Ω. By the definition, Ω is a region. Furthermore, we observe from Theorem 13.11 that Ω is simply connected. However, the closed curve C(0; 2R) is not null-homotopic in Ω because E lies inside C(0; 2R). By the definition, Ω is not simply connected and hence f has no holomorphic square root in Ω. (e) Assume that Re f was bounded in Ω. We use part (f) in advance that f will be bounded in Ω. This implies that f can be extended to a bounded entire function because E is compact. Hence it contradicts part (a) and then Re f is unbounded in Ω. (f) Suppose that z = a + ib ∈ Ω. Then it is easy to see that Z Z Z (t − a) dt b dt dt = + i . f (z) = 2 2 2 2 E (t − a) + b E (t − a) + b E (t − a) − ib For every z ∈ Ω, we deduce from the expression (16.21) that Z b dt |Im f (z)| = 2 2 E (t − a) + b Z ∞ b dt ≤ 2 2 −∞ (t − a) + b

(16.21)

504

Chapter 16. Analytic Continuation  ∞  −1 t − a = tan b −∞ π  π = − − 2 2 = π.

(g) Suppose that γ is a positively oriented circle which has E in its interior. Since γ ∗ (the range of γ) is closed in C and γ ∗ ∩ E = ∅, we have δ = inf∗ |t − z| > 0 so that z∈γ t∈E

Z Z γ

E

Z Z ℓ(γ)m(E) dt  1 dz ≤ dt dz = < ∞, t−z δ γ E δ

where ℓ(γ) is the circumference of the circle γ. Hence Theorem 8.8 (The Fubini Theorem) and Theorem 10.11 together assert that Z Z Z Z Z Z  1 1 dt = 2πm(E)i. 2πiInd γ (t) dt = 2πi dt dz = dz dt = γ t−z E E E γ E t−z (h) By parts (e) and (f), we see that f (Ω) ⊆ {z = x + iy | x ∈ R and −π ≤ y ≤ π}. Define g(z) = eiz and ϕ = g ◦ f : Ω → C. Now part (a) ensures that ϕ is not constant. Furthermore, it is clear that
ϕ(Ω) = g f (Ω) ⊆ {reiθ | e−π ≤ r ≤ eπ and θ ∈ [0, 2π]}

so that ϕ is bounded on Ω. Thus it remains to show that ϕ ∈ H(Ω) and this follows from the result f ∈ H(Ω). To see this, we write Z ψ(z, t) dt, f (z) = E

1 where ψ(z, t) = t−z . For each fixed z ∈ Ω, ψ(z, t) is measurable. For each fixed t ∈ E, we have ψ(z, t) ∈ H(Ω). Furthermore, for each z0 ∈ Ω, we have inf |t − z0 | > 0. Let this t∈E

number be 2δ. Then we have

δ=

inf z∈D(z0 ;δ) t∈E

|t − z|

which implies that |t − z| ≥ δ for every z ∈ D(z0 ; δ) and t ∈ E. Therefore, we get Z Z dt 1 m(E) sup dt ≤ = < ∞. |t − z| δ δ E z∈D(z0 ;δ) E In other words,

Z

E

|ψ(z, t)| dt is locally bounded. Hence we conclude that f ∈ H(Ω).c

We end the proof of the problem. Problem 16.9 Rudin Chapter 16 Exercise 9. c

See the online paper http://www.nieuwarchief.nl/serie5/pdf/naw5-2001-02-1-032.pdf or [36].



16.2. Problems on the Modular Group and Removable Sets

505

Proof. (a) It is easy to see that f (−2) =

Z

1 −1

1 dt = log(t + 2) = log 3 t+2 −1

Thus f is not constant in Ω.

1 5 and f (−4) = log(t + 4) = log . 3 −1

(b) Similar to Problem 16.8(b), f cannot be extended to an entire function. (c) According to Problem 16.8(c), the value of the limit is −2 because m(E) = 2. (d) Similar to Problem 16.8(d), f has no holomorphic square root in Ω. (e) Similar to Problem 16.8(e), Re f is unbounded in Ω. (f) Since E = [−1, 1], we see that Im f =

Z

1 −1

1 − a  −1 − a   t − a  1 b − tan−1 . dt = tan−1 = tan−1 2 2 (t − a) + b b b b −1

(g) We have the exact result

Z Z γ

1 −1

1 dt dz = 2πm(E)i = 4πi. t−z

(h) By Problem 16.18(h), the nonconstant bounded holomorphic function ϕ in Ω is given by  Z ϕ(z) = exp i

1

−1

 z − 1 dt  = exp i log . t−z z+1

This completes the proof of the problem. Problem 16.10 Rudin Chapter 16 Exercise 10.

Proof. (a) We first need the following lemma: Lemma 16.3 Suppose that E is compact and has no interior, and K satisfies the following two conditions: – Condition (1): K ⊆ E is compact (K can possibly be empty) and – Condition (2): Each f ∈ H(C \ E) can be extended to an fK ∈ H(C \ K). Let E ′ be the intersection of all such compact subsets K of E. Then E ′ also satisfies the conditions.



506

Chapter 16. Analytic Continuation Proof of Lemma 16.3. Obviously, we have E ′ ⊆ K and E ′ is compact. This means that E ′ satisfies Condition (1). To show that E ′ also satisfies Condition (2), let f ∈ H(C \ E) and z ∈ C \ E ′ . Then z ∈ / K (or equivalently z ∈ C \ K) for some compact K ⊆ E satisfying Condition (2). Thus our f ∈ H(C \ E) can be extended to an fK ∈ H(C \ K). Since E has no interior, there exists a sequence {zn } ⊆ C \ E such that zn → z and fK (z) = lim f (zn ) n→∞

which implies that the value fK (z) is uniquely determined by the limit, i.e., all the values fK (z) must be equal for all compact K ⊆ E satisfying z 6∈ K and the conditions. Hence we may define fb : C \ E ′ → C by fb(z) = fK (z)

(16.22)

for any such compact K. Recall that fK ∈ H(C \ K) and E ′ ⊆ K, so the expression  (16.22) ensures that fb ∈ H(C \ E ′ ), completing the proof of Lemma 16.3

Suppose that E is countable compact, i.e., E = {z1 , z2 , . . .}. Since {zn } has no interior, it is nowhere dense and we observe from the Baire Category Theorem (see §5.7) that E has no interior. Let E ′ be the set in Lemma 16.3. Assume that E ′ 6= ∅. Notice that E′ =

∞ [

n=1


E ′ ∩ {zn } .

Since E ′ is compact, E ′ is closed in C so that it is a complete metric space by [99, Theorem 3.11, p. 53]. If each E ′ ∩ {zn } has no interior, then it is nowhere dense and the Baire Category Theorem shows that E ′ is of the first category, a contradiction. Thus {zN } = E ′ ∩ {zN } has a nonempty interior for some N ∈ N which is impossible. Hence we have E ′ = ∅ and we deduce from Lemma 16.3 that every f ∈ H(C \ E) can be extended to an entire function and we denote it by the same notation f . Particularly, if f is bounded, then the corresponding entire function f is also bounded and Theorem 10.23 (Liouville’s Theorem) forces that it is a constant. By the definition, E is removable. (b) Let E ⊆ R be compact and m(E) = 0. Let f ∈ H(C \ E) be bounded by a positive constant M . By the proof of Theorem 13.5, there exists a cycle Γ in C \ E such that Ind Γ (z) = 1 for every z ∈ E. Suppose that V is the union of the collection of those components of C \ Γ intersecting E. Thus we have E ⊆ V . Define g : V → C by Z f (ζ) 1 dζ. F (z) = 2πi Γ ζ − z for every z ∈ V . By an argument similar to part (c) below, we see that F ∈ H(V ).

Fix α ∈ V \ E. Since the set (C \ V ) ∪ {α} is closed
in C and [(C \ V ) ∪ {α}] ∩ E = ∅,
d E, (C\V )∪{α} > 0. Let 0 < ǫ < d E, (C\V )∪{α} . Since E is compact and m(E) = 0, E can be covered by a finite number of open intervals I1 = (a1 , b1 ), I2 = (a2 , b2 ), . . ., In = (an , bn ) whose total length is less than ǫ. Without loss of generality, we may assume that I1 , I2 , . . . , In are pairwise disjoint and intersect E. Let γk denote the counterclockwise circle having Ik as its diameter and Γǫ =

n [

k=1

γk .

16.2. Problems on the Modular Group and Removable Sets

507

We notice that the length of Γǫ is less than πǫ. By applying Theorem 10.35 (Cauchy’s Theorem) to the cycle Γ − Γǫ in C \ E, we obtain Z Z 1 f (ζ) f (ζ) 1 dζ = F (α) − dζ. (16.23) f (α) = 2πi Γ−Γǫ ζ − α 2πi Γǫ ζ − α It is easy to see that

1 Z f (ζ) 1 M πǫ dζ ≤ × . 2πi Γǫ ζ − α 2π d(α, E)

Since ǫ is arbitrary small, the second integral in the equation (16.23) is actually zero, we get f (α) = F (α) for every α ∈ V \ E. Since F ∈ H(V ), we conclude immediately that f ∈ H(C) which implies that it is a constant. By the definition, E is removable. (c) Suppose that
f : Ω \ E → C is bounded by M . We fix z0 ∈ Ω \ E. Let Γ1 be a cycle in Ω \ E ∪ {z0 } with winding number 1 around
E ∪ {z0 } and zero around C \ Ω. Similarly, suppose that Γ2 is a cycle in Ω \ E ∪ {z0 } with winding number 1 around E and zero around (C \ Ω) ∪ {z0 }. Since Ind Γ1 (α) = Ind Γ2 (α) = 0 for every α ∈ / Ω, Theorem 10.35 (Cauchy’s Theorem) asserts that our construction guarantees Z Z f (ζ) f (ζ) 1 1 dζ − dζ. f (z0 ) = 2πi Γ1 ζ − z0 2πi Γ2 ζ − z0 Define f1 , f2 : Ω \ E → C by Z f1 (z) =

Γ1

f (ζ) dζ ζ−z

and f2 (z) =

Z

Γ2

f (ζ) dζ. ζ −z

We claim that f1 ∈ H(Ω) and f2 ∈ H(C \ E). To this end, we first note from Theorem 10.35 (Cauchy’s Theorem) that f1 is independent of Γ1 . Next, we take z ∈ Ω and fix the cycle Γ1 as constructed above. Denote
the length of Γ1 to be ℓ(Γ1 ). Since E ∪ {z} lies entirely inside Γ1 ,
we have Γ∗1 ∩ E ∪ {z} = ∅. Recall that E is compact, so is E ∪ {z} and then d Γ∗1 , E ∪{z} > 0. Let this number be 2δ. If h is very small such that z +h ∈ D(z; δ), then we have Z f (ζ) f1 (z + h) − f1 (z) = dζ. (16.24) h (ζ − z)(ζ − z − h) Γ1 Clearly, for every ζ ∈ Γ1 , we have

f (ζ) M ≤ 2. (ζ − z)(ζ − z − h) 2δ

Using this and the fact that ℓ(Γ1 ) < ∞, we may apply Theorem 1.34 (Lebesgue’s Dominated Convergence Theorem) to the expression (16.24) to conclude that Z f (ζ) dζ. f1′ (z) = 2 Γ1 (ζ − z) Since z ∈ Ω is arbitrary, we get the desired result that f1 ∈ H(Ω). Using a similar argument, we can show that f2 ∈ H(C \ E) which proves the desired claim.

Therefore, we have f = f1 − f2 on Ω \ E. Now the boundedness of f certainly implies the boundedness of f2 . Since E is removable, f2 is a constant. Consequently, we obtain f ∈ H(Ω).

508

Chapter 16. Analytic Continuation

(d) Suppose that E ⊂ C is compact and m2 (E) = 0.d Then E is removable. Here we need the following lemma to prove this result. Lemma 16.4 E ⊆ C is removable if and only if every bounded holomorphic function f on C \ E satisfies f ′ (∞) = 0.

Proof of Lemma 16.4. By the definition, we know that E is removable if and only if every bounded holomorphic function f on C\E is constant. Obviously, if f ∈ H(C\E) is constant, then f ′ (∞) = 0. Conversely, let g : C \ E → C be nonconstant and bounded. Then there exists an z0 ∈ C \ E such that g(z0 ) 6= g(∞). Define f (z) =

g(z) − g(z0 ) z − z0

on C \ E. Obviously, f is also a bounded and nonconstant function on C \ E and f (∞) = lim f (z) = 0. z→∞

Consequently, we establish f ′ (∞) = lim z[f (z) − f (∞)] z→∞

z[g(z) − g(z0 )] z − z0 z z g(z) − g(z0 ) lim = lim z→∞ z − z0 z→∞ z − z0 = g(∞) − g(z0 ) 6= 0,

= lim

z→∞

completing the proof of Lemma 16.4.



We return to the proof of the problem. Let f be a bounded holomorphic function on C \E, i.e., |f (z)| ≤ M on C \ E for some positive constant M . Given ǫ > 0. Then E can be covered by open discs D1 , D2 , . . . , Dn of radii r1 , r2 , . . . , rn respectively such that n X

rk < ǫ.

k=1

Let Γ = ∂D1 ∪ ∂D2 ∪ · · · ∪ ∂Dn . Using [118, Eqn. (1.2), p. 16], we have n 1 Z X f (z) dz ≤ M rk < M ǫ. |f ′ (∞)| = 2πi Γ

(16.25)

k=1

Since ǫ is arbitrary, the inequality (16.25) guarantees that f ′ (∞) = 0. Now we conclude from Lemma 16.4 that E is in fact removable. (e) Suppose first that E ⊂ C is compact and removable. If F ⊆ E is a connected component of E containing more than one point, then it follows from Theorem 14.8 (The Riemann Mapping Theorem) that there exists a conformal mapping f : C \ F → U which is nonconstant. Thus f |C\E ∈ H(C \ F ) and f |C\E must be bounded. By the definition, E is d

Here m2 denotes the Lebesgue measure in two dimensional space C.

16.3. Miscellaneous Problems

509

non-removable, a contradiction. Hence connected components of E are one-point sets, i.e., E is totally disconnected, see [74, Exercise 5, p. 152]. Now if E ⊂ C is a connected subset with more than one point, then the above paragraph ensures that E must be non-removable.e We have completed the analysis of the proof of the problem.



Remark 16.1 Recall that we have studied the special case of removable sets in Problem 11.11. See also Remark 11.2.

16.3

Miscellaneous Problems

Problem 16.11 Rudin Chapter 16 Exercise 11.

Proof. By the definition, we have Ωα ⊂ Ωβ if α < β. In Figure 16.4, Ωβ is the union of Ωα and the region shaded by straight lines.

Figure 16.4: The regions Ωα and Ωβ if α < β. e

Recall from point-set topology [78, p. 3] that a nonempty compact connected metric space is a continuum.

510

Chapter 16. Analytic Continuation • fβ is an analytic continuation of fα if α < β. Let ζ ∈ Ωα . Then there exists a δ > 0 such that D(ζ; δ) ⊆ Ωα and D(ζ; δ) ∩ Γ∗α = ∅. Define ψα : D(ζ; δ) × Γ∗α → C by ψα (z, ω) =

exp(eω ) . ω−z

If ω = t + πi for t ∈ [α, ∞), then we have eω = −et so that | exp(eω )| = exp(−et ) ≤

1 0 and an ζ ∈ Ωα . Since Ωα is open in C, one can find a δζ > 0 such that D(ζ; δζ ) ⊆ Ωα . Let γ be a curve in C with parameter interval [0, 1] that starts at the center of D(ζ; δζ ). Note that this may happen that γ([0, 1]) * Ωα . However, the compactness of γ([0, 1]) ensures that there corresponds
an β > α such that γ([0, 1]) ⊆ Ωβ . Hence the first assertion guarantees that fα , D(ζ; δ) can be analytically continued along the curve γ in C. By Theorem 16.15 (The Monodromy Theorem), there exists an entire function f such that f (z) = fα (z) for all z ∈ D(ζ; δζ ). By the Corollary to Theorem 10.18, we have f = fα on Ωα . • f (reiθ ) → 0 as r → ∞ for every eiθ 6= 1. Suppose that r > 0 and θ is real. By the second assertion, we know that f (z) = f1 (z) on Ω1 . By the assumption, we have reiθ ∈ Ω1 for large enough r > 0. Write Γα = γα− + Lα + γα+ , where γα− = −t − πi for t ≤ −α, γα+ = t + πi for t ≥ α and Lα = α + πit α for t ∈ [−α, α]. Therefore, we see that Z exp(eω ) 1 dω |f (reiθ )| = 2π Γ1 ω − z Z Z Z 1 exp(eω ) exp(eω ) exp(eω ) 1 1 ≤ dω + dω + dω 2π γ1− ω − z 2π L1 ω − z 2π γ1+ ω − z Z Z 1 1 exp(e cos πt) · exp(ie sin πt) 1 −1 exp(−e−t ) dt + πi dt = iθ iθ 2π −∞ −t − πi − re 2π −1 1 + πit − re Z 1 ∞ exp(−et ) + dt (16.27) . 2π 1 t + πi − reiθ Since |ω − reiθ | ≥ r sin θ − 1 for large enough r > 0 and for every ω ∈ Γ∗1 , the inequality (16.27) reduces to Z ∞ Z 1 i h Z −1 1 exp(−et ) dt exp(e cos πt) dt + exp(−e−t ) dt + |f (reiθ )| ≤ 2π(r sin θ − 1) −∞ 1 −1

16.3. Miscellaneous Problems

≤

h 1 2 2π(r sin θ − 1)

511 Z

∞ 1

i exp(−et ) dt + 2ee .

(16.28)

Since et ≤ exp(et ) for every t ≥ 0, the inequality (16.28) further reduces to |f (reiθ )| ≤

1 (ee + e−1 ). π(r sin θ − 1)

Since eiθ 6= 1, sin θ 6= 0 which implies that lim f (reiθ ) = 0.

r→∞

• f is not constant. Fix r > 0. Let 0 < r < α < R and Γ = Γα ∪ LR , where LR = R + πit R for t ∈ [−R, R]. Assume that f was constant. Since f is entire, the third assertion forces that f (z) = 0 in C. In particular, we have 1 0 = f (r) = 2πi

Z

Γα

exp(eω ) dω ω−r

(16.29)

for every α > r. It is clear that Γ is closed and Ind Γ (r) = 0. Using Theorem 10.35 (Cauchy’s Theorem), we know that 1 2πi

Z

Γα

1 exp(eω ) dω + ω−r 2πi

which implies

Z

Z

LR

LR

exp(eω ) 1 dω = ω−r 2πi

Z

Γ

exp(eω ) dω = 0 ω−r

exp(eω ) dω = 0 ω−r

(16.30)

for every R > r. Write Z 1 exp(eω ) dω 2πi ΓR ω − r Z Z Z 1 exp(eω ) 1 1 exp(eω ) exp(eω ) = dω + dω + dω 2πi γR+ ω − r 2πi γR− ω − r 2πi LR ω − r Z ∞ Z ∞ Z exp(−et ) exp(−et ) exp(eω ) 1 1 1 dt − dt + dω = 2πi R t + πi − r 2πi R t − πi − r 2πi LR ω − r Z Z ∞ exp(eω ) 1 exp(−et ) dt + dω. =− 2 2 2πi LR ω − r R (t − r) + π

f (r) =

Using the results (16.29) and (16.30), we immediately see that Z

∞

R

− exp(−et ) dt = 0 (t − r)2 + π 2

for every R > r. Since exp(−et ) ≤ e−t and Z

∞ R

1 (t−r)2 +π 2

1 − exp(−et ) dt ≥ − (t − r)2 + π 2 (R − r)2 + π 2

Z

∞

≤

1 , (R−r)2 +π 2

e−t dt =

R

which contradicts the result (16.31). Hence f (r) 6= 0.

(16.31) we get

1 >0 eR [(R − r)2 + π 2 ]

512

Chapter 16. Analytic Continuation • g(reiθ ) → 0 as r → ∞ for every eiθ . If eiθ 6= 1, then the third assertion implies that lim g(reiθ ) = lim f (reiθ ) exp[−f (reiθ )] = 0 · 1 = 0.

r→∞

Next, suppose that

eiθ

r→∞

= 1. Since f (r) → ∞ as r → ∞, we see immediately that lim g(r) = lim

r→∞

r→∞

f (r) = 0. exp[f (r)]

This gives the fifth assertion. • Existence of an entire function h with the required properties. By the fourth and the fifth assertions, we know that g is a nonconstant entire function such that g(reiθ ) → 0 as r → ∞ for every eiθ . If g has a zero of order N at z = 0, then we write g(z) = z N G(z). Thus G is nonconstant entire, G(0) 6= 0 and lim G(reiθ ) = 0

r→∞

for every eiθ . Define h(z) =

G(z) G(0) .

(16.32)

Therefore, h is nonconstant entire and h(0) = 1.

Furthermore, if z 6= 0, then we write z = reiθ for some r > 0. Since h(nz) = h(nreiθ ) =

G(nreiθ ) , G(0)

we follow from the limit (16.32) that h(nz) → 0 as n → ∞. If g(0) 6= 0, then we consider the nonconstant entire function h(z) = g(z) g(0) which satisfies h(0) = 1 and h(nz) → 0 as n → ∞. In conclusion, there exists an entire function h such that   1, if z = 0; lim h(nz) = n→∞  0, if z 6= 0.

We have ended the analysis of the problem.



Problem 16.12 Rudin Chapter 16 Exercise 12.

Proof. Suppose that f is represented by the series ∞  X z − z 2 3k

2

k=1

.

2

(16.33)

Evidently, if |z − z 2 | < 2, then | z−z 2 | < 1 so that the series (16.33) converges by [99, Theorem 3.26, p. 61]. Furthermore, if |z − z 2 | > 2, then the series (16.33) diverges. The red shaded part in Figure 16.5 indicates the regionf of convergence of the power series (16.33). k

Next, suppose that Pk (z) = [z(1 − z)]3 , so f (z) = f

This is (x − x2 + y)2 + (y − 2xy)2 < 4.

∞ X 1 P (z). 3k k 2 k=1

16.3. Miscellaneous Problems

513

Figure 16.5: The regions of convergence of the two series. Note that the highest power and the lowest power of z in Pk (z) and in Pk+1 (z) are 2 · 3k and 3k+1 respectively. Since 3k+1 − 2 · 3k = 3k > 0 for every k ≥ 1, the polynomial Pk (z) contains no power of z that appear in any other Pj (z) for all j 6= k. If we replace every Pk (z) by its expansion in powers of z, then we get the power series f (z) =

∞ X

an z n

(16.34)

n=1

with the property that a1 = a2 = 0 and for each positive integer k, we have  3k (−1)n+1 Cn−3  k   , if n = 3k , 3k + 1, . . . , 2 · 3k ; k 3 2 an =    0, if 2 · 3k < n < 3k+1 .

(16.35)

If both n and r tend to infinity, then it follows from Stirling’s formula [99, Eq. (103), p. 194] that r nn n · r . Crn ∼ 2πr(n − r) r (n − r)n−r k

3 Recall that Cn−3 k takes its maximum value when nk =

3k 2π(nk −

3k )[3k

− (nk −

3k )]

= =

so it is true that

3k 2π(nk − 3k 2π ·

2 π 2 ∼ π

=

3k+1 +1 , 2

·

3k +1 2 3k

3k )(2 ·

32k − 1 1 · k 3

· 3k − nk )

3k −1 2

(16.36)

514

Chapter 16. Analytic Continuation

and k

(3k )3 k +1 2

3 k ( 3 2+1 )

k

k −1 2

3 k ( 3 2−1 )

∼ 23 .

(16.37)

1

1

Since g(x) = x x is decreasing for x > e and x x → 1 as x → ∞ and nk ∼ 1.5 × 3k for large k, it yields from the estimates (16.36) and (16.37) that 1 C 33k +1 ! 1.5×3 k k

lim |ank |

1 nk

k→∞

2

= lim

23k

k→∞

1

= lim

k→∞

= lim

k→∞

= 1.

× C 33k +1

2 3

·

2 1 2

k

2 3

2




2 1 3k+1

π

1 1.5×3k

·

1  1  k+1 2 3 · 23 k 3

In other words, the radius of convergence of the power series is 1. Check the blue shaded part in Figure 16.5. Let λ = 3. Let pk = 2 · 3k and qk = 3k+1 for k = 1, 2, . . .. Then they satisfy λqk > (λ + 1)pk and an = 0 for pk < n < qk for all positive integers k. Now the power series (16.33) and (16.34) assert that there exists a δ > 0 such that ∞  X z − z 2 3k

2

k=1

=

∞ X

an z n

n=1

for all z ∈ D(0; 1) ∩ D(1; δ). By Definition 16.1, it means that 1 is a regular point of f . Finally, it concludes from Theorem 16.5 that the sequence {spk (z)} converges in a neighborhood of 1, where sp (z) is the pth partial sum of the power series (16.34). By Figure 16.5 again, we know that all boundary points of T , except z = −1, are regular points of f . Observe from the representation (16.34) and the definition (16.35) that −f (−z) =

∞ X

bn z n ,

n=1

where bn ≥ 0 for every n ≥ 1. Thus Problem 16.1 ensures that −f (−z) has a singularity at −z = 1. Hence z = −1 is the singular point of f which is nearest to the origin, so we have  completed the proof of the problem. Problem 16.13 Rudin Chapter 16 Exercise 13.

Proof. For each positive integer n, we have Xn = {f ∈ H(Ω) | f = g(n) for some g ∈ H(Ω)}. (a) If f ∈ X1 , then f = g′ for some g ∈ H(Ω). Since γ is a closed path lying in Ω, Theorem 10.12 implies that Z Z (16.38) f (z) dz = g′ (z) dz = 0. γ

γ

16.3. Miscellaneous Problems

515

Conversely, suppose that the integral (16.38) holds. Since f ∈ H(Ω) and Ω is an annulus, Problem 10.25 shows that f admits the Laurent series f (z) = f1 (z) + f2 (z) =

−1 X

cn z n + f2 (z),

−∞



where f1 ∈ H C \ D(0; 21 ) and f2 ∈ H D(0; 2) . Since cn =

1 2πi

Z

γ

f (z) dz, z n+1

the integral (16.38) implies that c−1 = 0. Thus we obtain f1 (z) =

∞ X

n=2

c−n z −n .

If we let ω = z1 , then the function F1 (ω) = f1

1 ω

∞ X

=

n=2

c−n ω n

is holomorphic in {ω ∈ C | |ω| < 21 }. Therefore, it is true that lim sup

implies that the radius of convergence of the series G(ω) =

n→∞

1 √ n c which −n ≥ 2

∞ X c−n n−1 ω 1−n n=2

is at least 21 too. Next, it is clear from Theorem 10.6 that G′ (ω) = −ω −2 F1 (ω) in the disc{ω ∈ C | |ω| < 12 }. By transforming back to the variable z, we see that f1 (z) =

d  X c−n 1−n  z dz n=2 1 − n ∞

(16.39)

holds in C \ D(0; 21 ). Similarly, we can show that f2 (z) =

d  X cn n+1  z dz n=0 n + 1 ∞

(16.40)

holds in D(0; 2). Finally, if we define g(z) =

∞ ∞ X c−n 1−n X cn n+1 z + z 1−n n+1 n=0 n=2

for all z ∈ Ω, then the facts (16.39) and (16.40) combine to imply immediately that f (z) = g ′ (z) in Ω, i.e., f ∈ X1 . (b) Since f ∈ H(Ω) and Ω is an annulus, Problem 10.25 shows that f admits a representation f (z) =

∞ X

n=−∞

an z n .

(16.41)

516

Chapter 16. Analytic Continuation (m)

If f ∈ Xm , then f = gm i.e.,

for some gm ∈ H(Ω). Again, gm has the Laurent series in Ω, gm (z) =

∞ X

bn,m z n

n=−∞

which gives ∞ X

(m) an z n = gm (z)

n=−∞

= =

∞ X

n=−∞ ∞ X

n(n − 1) · · · (n − m + 1)bn,m z n−m

(n + m)(n + m − 1) · · · (n + 1)bn+m,m z n .

n=−∞

Therefore, it means that a−1 = a−2 = · · · = a−m = 0. As m runs through all positive integers, the Laurent series (16.41) reduces to f (z) =

∞ X

an z n

n=0


which implies that f ∈ H D(0; 2) .


Conversely, suppose that there exists an g ∈ H D(0; 2) such that f (z) = g(z) for all z ∈ Ω. By Theorem
13.11, the simply connectedness of D(0; 2) ensures that one can find an g1 ∈ H D(0; 2) such that g1′ = g. In fact, this argument can be repeated to achieve
(n) the existence of an gn ∈ H D(0; 2) with gn = g for each positive integer n. Hence we obtain f (z) = gn(n) (z)

for all z ∈ Ω and this means that f ∈ Xn for every positive integer n. We have completed the proof of the problem.



Problem 16.14 Rudin Chapter 16 Exercise 14.

Proof. Our proof here basically follows that in [104, §2.7, pp. 54 – 56]. Since normality is a local property, we may assume that Ω is the unit disc U . Suppose that F = {f ∈ H(U ) | |f (p)| ≤ R and 0, 1 ∈ / f (U )}. Recall from Theorem 16.20 that the modular function λ is invariant under Γ (i.e., λ ◦ ϕ = λ for all ϕ ∈ Γ) and maps Π+ onto C \ {0, 1}. Since f (U ) ⊆ C \ {0, 1}, the function λ−1 ◦ f has a local branch defined in a sufficiently small neighbourhood of f (0). Then this function element may be analytically continued in U , so we assert from Theorem 16.15 (The Monodromy Theorem) that there exists a holomorphic function fb : U → Π+ such that λ ◦ fb = f.

(16.42)

Let {fn } ⊆ F . Since |fn (p)| ≤ R for all n ∈ N, the Bolzano-Weierstrass Theorem [127, Problem 5.25, pp. 68, 69] ensures that there is a convergent subsequence {fnk (p)}. Let this limit be ℓ.

16.3. Miscellaneous Problems

517

• Case (i): ℓ 6= 0 and ℓ 6= 1. Then we can fix a branch of λ−1 in a neighborhood of ℓ and c use this to define the functions fc nk by the equation (16.42). Since Im fnk > 0, we have c c Re (−ifc ) = Im f > 0 so that the family {−i f | k ∈ N} is normal by Problem 14.15. nk nk nk Consequently, the family c = {fc F nk | k ∈ N} is also normal. For simplicity, we may assume that fc nk converges normally to g ∈ H(U ). Clearly, we have g(U ) ⊆ Π+ . By the equation (16.42) again, we conclude that
g(p) = lim fbnk (p) = lim λ−1 fnk (p) = λ−1 (ℓ). k→∞

k→∞

Π+ ,

Recall that the domain of λ is so the Open Mapping Theorem implies that g(U ) ⊆ Π+ and λ ◦ g : U → C is well-defined such that

lim fnk (z) = lim λ fc nk (z) = λ g(z) k→∞

k→∞

for all z ∈ U . Hence {fnk } is the required subsequence.

• Case (ii): ℓ = 1. Since fnk ∈ H(U ) and 0 ∈ / fnk (U ), we have fn1 ∈ H(U ). Since U is k simply connected, we deduce from Theorem 13.11 that each fnk has a holomorphic square root hk in U . We choose the branch such that lim hk (p) = −1.

(16.43)

k→∞

/ hk (U ) and |hk (p)| ≤ Since fnk = h2k , we have 0, 1 ∈

√

R. Consider the family

H = {hk | k ∈ N}. Now the limit (16.43) guarantees that we can apply Case (i) to H to obtain a convergent subsequence hkj in U . In conclusion, the limit lim fnkj (z)

j→∞

exists for all z ∈ U . • Case (iii): ℓ = 0. In this case, we may apply Case (ii) to the sequence {1 − fnk | k ∈ N}. Hence we have completed the proof of the problem.



Remark 16.2 Problem 16.14 is classically called the Fundamental Normality Test.

Problem 16.15 Rudin Chapter 16 Exercise 15.

Proof. Without loss of generality, we may assume that D is the unit disc and D ⊆ Ω = D(0; R) for some R > 1. Let (f, D) be analytically continued along every curve in Ω that starts at the origin 0. Since f ∈ H(D), f has a power series expansion at 0, i.e., f (z) =

∞ X

n=0

an z n .

518

Chapter 16. Analytic Continuation

Suppose that r is the radius of convergence of this power series. Clearly, we have 1 ≤ r. If r = 1, then we know from Theorem 16.2 that f has at least one singular point on the unit circle T . Let ω be a singular point of f on T and γ be a curve in Ω starting at 0 and passing through ω. Now the assumption ensures that (f, D) can be analytically continued along γ in Ω, so ω is a regular point of f , a contradiction. As a result, 1 < r ≤ R. Next, if r < R, then similar argument can be applied to conclude that no point on C(0; r) is a singular point, but this contradicts Theorem 16.2. Hence we must have r ≥ R and it means that Theorem 16.5 holds in this special case. This proves the first assertion. Let Ω be any simply connected region (other than the complex plane itself), D ⊆ Ω and (f, D) be analytically continued along every curve in Ω that starts at the center of D. Let z0 ∈ D. By Theorem 14.8 (The Riemann Mapping Theorem) or [11, §14.2, pp. 200 – 204], one can find a (unique) conformal mapping F : Ω → U such that F (z0 ) = 0 and F ′ (z0 ) > 0. Let S = F (D). Obviously, we have G = F −1 |S : S ⊂ U → D so that G(0) = z0 . We consider the mapping g = f ◦ G : S ⊂ U → C. (16.44) Since S is an open set containing the origin, we may assume that it is an open disc centered at 0 so that S and U are concentric. Since f ∈ H(D), we conclude that g ∈ H(S). Furthermore, since (f, D) can be analytically continued along every curve in Ω, the function element (g, S) can also be analytically continued along every curve in U . Therefore, the first assertion guarantees that there corresponds a h ∈ H(U ) such that h(z) = g(z) for all z ∈ S. Using the definition (16.44), we have h ◦ F ∈ H(Ω) and for all z ∈ D,



h F (z) = h G−1 (z) = h g −1 f (z) = f (z).

This proves the second assertion and we end the analysis of the problem.



CHAPTER

17

H p-Spaces

17.1

Problems on Subharmonicity and Harmonic Majoriants

Problem 17.1 Rudin Chapter 17 Exercise 1.

Proof. Let u : Ω → R be an upper semicontinuous subharmonic function. By Definition 2.8, for every real α, the set {z ∈ C | u(z) < α} is open in C. • Let K ⊂ Ω be compact and h : K → R be continuous such that h is harmonic in V = K ◦ and u(z) ≤ h(z) for all boundary points of K. Put u1 = u − h. Assume that u1 (ζ) > 0 for some ζ ∈ V . Since h is continuous on K, −h is upper semicontinuous on K. Thus u1 is also upper semicontinuous on K by Problem 2.1. Since K is compact, Problem 2.21 ensures that u1 attains its maximum m on K. Since u1 ≤ 0 on the boundary of K, the set E = {z ∈ K | u1 (z) = m} is a nonempty compact subset of V . Let z0 be a boundary point of E. Since E is compact, V is open in C and E ⊂ V , there exists an r > 0 such that D(z0 ; r) ⊂ V . Now some subarc of the boundary of D(z0 ; r) lies in V \ E. Hence we have Z π 1 u1 (z0 ) = m > u1 (z0 + reiθ ) dθ 2π −π which means that u1 is not subharmonic in V . However, since u and h are subharmonic and harmonic in V respectively, it follows from the mean value property that u1 is also subharmonic in V , a contradiction. This proves that no such ζ exists and then u(z) ≤ h(z) for all z ∈ K. • By Definition 17.1, it suffices to prove that Theorem 17.5 is true for subharmonic function in U . Let 0 ≤ r < 1. Then K = D(0; r) ⊂ U and u : K → R is subharmonic. In particular, u is an upper semicontinuous function. We need the following result:a Lemma 17.1 (Baire’s Theorem on Semicontinuous Functions) Let K ⊂ C be compact and −∞ ≤ u(z) < ∞ on K. If u is upper semicontinuous, then it is the limit of a monotone decreasing sequence of continuous functions {un } on K. a

Read https://encyclopediaofmath.org/wiki/Baire_theorem#Baire.27s_theorem_on_semi-continuous_functions .

519

Chapter 17. H p -Spaces

520

Let 0 ≤ r1 < r2 < 1. Using this lemma, we know that u(z) ≤ un (z) on C(0; r2 ). By Theorem 11.8, Hun ∈ C(K), Hun is harmonic in D(0; r2 ) and (Hun )|C(0;r2 ) = un . Clearly, we have u(z) ≤ un (z) = (Hun )(z) on C(0; r2 ), so the first assertion implies that u(z) ≤ (Hun )(z) for all z ∈ K. Furthermore, the mean value property gives Z π Z π 1 1 (Hun )(r2 eit ) dt = un (r2 eit ) dt. (Hun )(0) = 2π −π 2π −π Combining the inequality (17.1) and the formula (17.2) we obtain Z π Z π Z π 1 1 1 it it u(r1 e ) dt ≤ (Hun )(r1 e ) dt = (Hun )(0) = un (r2 eit ) dt. 2π −π 2π −π 2π −π

(17.1)

(17.2)

(17.3)

Finally, we apply Problem 1.7b to the inequality (17.3) to get Z π Z π Z π 1 1 1 u(r1 eit ) dt ≤ lim un (r2 eit ) dt = u(r2 eit ) dt. n→∞ 2π −π 2π −π 2π −π This completes the proof of the problem.



Problem 17.2 Rudin Chapter 17 Exercise 2.

Proof. Let u(z) = log 1 + |f (z)| = log 1 + elog |f (z)| . Define ϕ : R → R by ϕ(x) = log(1 + ex ). Then we can write
u(z) = ϕ log |f (z)| . (17.4)

Since f ∈ H(Ω), it follows from Theorem 17.3 that log |f (z)| is subharmonic in Ω. Evidently, ex ′ ′ ϕ′ (x) = 1+e x > 0 for all x ∈ R and ϕ (s) < ϕ (t) if s < t, so ϕ is a monotonically increasing convex function on R. By applying Theorem 17.2 to the function (17.4), we conclude that u is  subharmonic in Ω, as required. This completes the analysis of the problem. Problem 17.3 Rudin Chapter 17 Exercise 3.

Proof. It seems that the hypothesis 0 < p ≤ ∞ should be replaced by 0 < p < ∞ because if p = ∞, then the function f (z) ≡ 2 belongs to H ∞ . In this case, we have |f (z)|∞ = ∞ for any z ∈ U. • f ∈ H p if and only if |f (z)|p ≤ u(z) for some harmonic function u in U . Suppose that there exists a harmonic function u in U such that |f (z)|p ≤ u(z) for all z ∈ U . Combining this and the mean value property, we get o1 n Z nZ o1 1 p p |f (reiθ )|p dσ kfr kp = u(reiθ ) dσ ≤ = [u(0)] p < ∞ T

T

for every 0 ≤ r < 1. By Definition 17.7, we have f ∈ H p .

b

In fact, it is

17.1. Problems on Subharmonicity and Harmonic Majoriants

521

Conversely, suppose that f ∈ H p . Now it is easy to see that Z Z |fr |p dσ = kfr kpp |frp | dσ = kfrp k1 = T

T

which means kf p k1 = kf kpp < ∞. As (5), p. 344] directly to f p , we have

a consequence, we have f p ∈ H 1 . Applying [95, Eqn.

|f (z)|p ≤ |Qf p (z)| n 1 Z eit + z o p ∗ it = exp log |(f ) (e )| dt 2π T eit − z Z io n h 1 eit + z log |(f p )∗ (eit )| dt = exp Re it 2π T e − z o n 1 Z Pr (θ − t) log |(f p )∗ (eit )| dt = exp 2π T

(17.5)

for all z ∈ U . Using the same argument as in proving the inequality in the proof of Theorem 17.16(c), we obtain from the inequality (17.5) that Z 1 |f (z)|p ≤ Pr (θ − t)|(f p )∗ (eit )| dt = P [(f p )∗ ](z) 2π T in U .c By Theorem 17.11(b), (f p )∗ ∈ L1 (T ), so Theorem 11.7 ensures that P [(f p )∗ ] is harmonic in U . • The existence of a least harmonic majorant. Let uf = P [(f p )∗ ]. In fact, this is a least harmonic majorant. To see this, let u be a harmonic majorant in U , i.e., u is harmonic in U and |f (z)|p ≤ u(z) for every z ∈ U . For any ρ < 1, we have Z Z 1 1 Pr (θ − t)|f (ρeit )|p dt ≤ Pr (θ − t)u(ρeit ) dt = u(ρz), 2π T 2π T where 0 ≤ r < 1. As ρ → 1, we getd

Z 1 Pr (θ − t)|(f p )∗ (eit )| dt 2π T Z 1 Pr (θ − t)lim |f p (ρeit )| dt = ρ→1 2π T h 1 Z i Pr (θ − t)|f (ρeit )|p dt = lim ρ→1 2π T ≤ u(z)

uf (z) =

for every z ∈ U . 1

• kf kp = uf (0) p . Let 0 < p < ∞. By the previous assertion, we know that Z P (z, eit )|f (eit )|p dσ. uf (z) = T

1

When z = 0, we have P (0, eit ) = 1 so that uf (0) = kf kpp , i.e., kf kp = uf (0) p . c

Recall that P [f ] is the Poisson integral of f . We can interchange the limit and the integral because Theorem 10.24 (The Maximum Modulus Theorem) asserts that Fρ (t) = Pr (θ − t)|f (ρeit )|p is increasing with respect to ρ, so we may apply Theorem 1.26 (The Lebesgue’s Monotone Convergence Theorem). d

Chapter 17. H p -Spaces

522 We have completed the proof of the problem.



Problem 17.4 Rudin Chapter 17 Exercise 4.

Proof. On the one hand, if log+ |f | has a harmonic majorant in U , then there exists a harmonic function u in U such that 0 ≤ log+ |f (z)| = log+ |f (z)| ≤ u(z). Combining this and the mean value property, we obtain  Z Z u(reit ) dσ = u(0) < ∞ log+ |fr (eit )| dσ ≤ 0 ≤ kfr k0 = exp T

T

for all z = reit ∈ U and all 0 ≤ r < 1. By Definition 17.7, we get kf k0 < ∞ so that f ∈ N . On the other hand, suppose that f ∈ N . If f ≡ 0 so that log+ |f (z)| = 0 in U , then there is nothing to prove. With the aid of the result in §17.19, there correspond two functions b1 , b2 ∈ H ∞ such that b2 has no zero in U and f=

b1 . b2

Without loss of generality, we may assume that kb1 k∞ ≤ 1 and kb2 k∞ ≤ 1. Since U is simply connected and b12 ∈ H(U ), we deduce from Theorem 13.11 that there exists an g ∈ H(U ) such that b2 = eg . Since kb2 k∞ = eRe g ≤ 1, the function u(z) = Re g(z) is less than or equal to zero in U and this implies log |f (z)| = log |b1 (z)| − log |b2 (z)| ≤ − log eu(z) = −u(z)

(17.6)

for all z ∈ U . Since −u(z) ≥ 0, the inequality (17.6) and the definition in §15.22 yield log+ |f (z)| ≤ −u(z) in U . By Theorem 11.4 and g ∈ H(U ), −u is harmonic in U . Hence, log+ has a harmonic  majorant in U and this completes the proof of the problem.

17.2

Basic Properties of H p

Problem 17.5 Rudin Chapter 17 Exercise 5.

Proof. Since f ∈ H(U ), it is true that f ◦ ϕ ∈ H(U ). Since f ∈ H p , Problem 17.3 ensures that there is a harmonic function u in U such that |f (z)|p ≤ u(z) for all u ∈ U . Thus this shows that

f ϕ(z) p ≤ u ϕ(z)

for all z ∈ U . Applying Problem 11.7(b) with Φ = u and f = ϕ there, we see that ∆[u ◦ ϕ] = [(∆u) ◦ ϕ] × |ϕ′ |2 = 0.

By the definition, u ◦ ϕ is harmonic in U and then Problem 17.3 asserts that f ◦ ϕ ∈ H p .

17.2. Basic Properties of H p

523

The assertion is also true when we replace H p by N . To see this, Problem 17.4 ensures that there exists a harmonic function u in U such that log+ |f (z)| ≤ u(z) holds for all z ∈ U . Then we have

log+ f ϕ(z) ≤ u ϕ(z) (17.7)

in U . Applying Problem 11.7 to u ◦ ϕ and then using the fact that ∆u = 0 in U , we see immediately that ∆[u ◦ ϕ] = [(∆u) ◦ ϕ] · |ϕ′ |2 = 0.

In other words, u ◦ ϕ is harmonic in U . Combining the inequality (17.7) and Problem 17.4, we  conclude that f ◦ ϕ ∈ N , completing the proof of the problem. Problem 17.6 Rudin Chapter 17 Exercise 6.

Proof. Let α > 0 and consider the functione fα (z) =

1 . (1 − z)α

Put z = reit , we have Iα (r) =

Z

T

|fα (reit )| dt =

Z

π −π

1 dt. |1 − reit |α

We want to estimate Iα (r) as r → ∞. Of course, it depends on the value of α.

Since r → 1, we may assume that r > 21 . By considering the triangle formed by 1, r and reit . Clearly, this is an obtuse triangle. If t ∈ [−π, π], then we have |1 − reit | > max{1 − r, r|1 − eit |}
1 ≥ 1 − r + r|1 − eit | 2  1 1 > 1 − r + |1 − eit | 2 2 1 t  = 1 − r + sin . 2 2

(17.8)

Recall the fact [99, Exercise 7, p. 197] that | sin x| ≥ π2 |x| for every x ∈ [− π2 , π2 ], so the inequality (17.8) can further reduce to |1 − reit | ≥

|t|  1 1 1−r+ ≥ (1 − r + |t|). 2 π 2π

(17.9)

On the other hand, the triangle inequality gives

|1 − reit | ≤ |1 − eit | + |eit − reit | = |1 − eit | + 1 − r ≤ |t| + 1 − r.

(17.10)

Combining the inequalities (17.9) and (17.10), we obtain


Here we note that 1 − z 6= 0 in U , so we may take the branch such that (1 − z)α = exp α log(1 − z) , where π < arg(1 − z) < 2 .

e

− π2

1 1 (2π)α
α ≤
α ≤ it α |1 − re | |t| + 1 − r |t| + 1 − r

Chapter 17. H p -Spaces

524 Z

π −π

dt
α ≤ Iα (r) ≤ (2π)α |t| + 1 − r

Z

π −π

dt
α . |t| + 1 − r

(17.11)

Since |t| is an even function in t, we get Z π Z π dt dt
α = 2
α t+1−r −π |t| + 1 − r 0    if α = 1;  2 log(1 − r + π) − log(1 − r) , =   2  1−α − (1 − r)1−α , otherwise. 1−α (1 − r + π) 1−α

If α < 1, then the integral in the inequalities (17.11) tends to 2π1−α as r → 1 so that Iα (r) is bounded as r → 1. If α = 1, then we observe from the inequalities (17.11) that 0 < m1 ≤

Iα (r) ≤ M1 log(1 − r)

as r → 1 for some positive constants m1 and M1 . Similarly, if α > 1, then the inequalities (17.11) tells us that Iα (r) 0 < m2 ≤ ≤ M2 (1 − r)1−α as r → 1 for some positive constants m2 and M2 . We notice that n 1 Z o 1  I (r)  1 p p αp = , k(fα )r kp = |fα (reit )|p dt 2π T 2π

so the previous paragraph indicates that fα ∈ H p if and only if αp < 1. Thus, for 0 < r < s < ∞, if we take α ∈ ( 1s , 1r ), then it is easy to see that fα ∈ H r but fα ∈ / H s , i.e., H s ⊂ H r . The case for s = ∞ is obvious because we have fα ∈ / H ∞ for every α > 0. In particular, if we take 1 / H ∞ . This completes the analysis of the problem.  α = 2r , then we have f 1 ∈ H r but f 1 ∈ 2r

2r

Problem 17.7 Rudin Chapter 17 Exercise 7.

Proof. Now Problem 17.6 guarantees that H ∞ ⊂ H p for every 0 < p < ∞. By Definition 17.7, H p ⊆ N holds for every 0 < p < ∞, so we have H ∞ ⊂ N . Hence we conclude easily that \ \ H∞ ⊂ N ∩ Hp ⊆ H p. 0 0. m=n

Then we claim that

Fα (z) =

∞ eiθ X m(m − 1) · · · (m − n + 1)|α|m−n z m Cα,n m=n

are the extremal functions for α 6= 0. In fact, we know that kFα k22 =

1 |Cα,n |2

∞ X

m=n

|m(m − 1) · · · (m − n + 1)αm−n |2 =

1 × |Cα,n |2 = 1. |Cα,n |2

Furthermore, direct computation gives |Fα(n) (α)| =

1 Cα,n

×

∞ X

m=n

|m(m − 1) · · · (m − n + 1)αm−n |2 = Cα,n

and we prove the claim. If α = 0, then it is easily seen that F0 (z) = eiθ n!z n (n)

are the extremal functions in this case because |F0 (α)| = (n!)2 . Therefore, we have completed the proof of the problem.  Problem 17.12 Rudin Chapter 17 Exercise 12.

17.2. Basic Properties of H p

529

Proof. Since p ≥ 1, we know that f ∈ H 1 and Theorem 17.11 tells us that Z π 1 f (z) = P (z, eit )f ∗ (eit ) dt 2π −π for all z ∈ U . Now the hypothesis guarantees that f is real a.e. in U . By the Open Mapping Theorem, f must be constant. 1+z Consider f (z) = i 1−z in U . It is easy to see that f ∈ H p for every 0 < p < 1 and

f ∗ (eit ) = lim f (reit ) = − cot r→1

t 2

is real a.e. on T , but f is not constant. Hence we have completed the proof of the problem.  Problem 17.13 Rudin Chapter 17 Exercise 13.

Proof. Since |f (reit )| = |γr (t)| ≤ M for every 0 ≤ r < 1 and t ∈ [−π, π], f is bounded in U . Thus f ∈ H ∞ so that f ∈ H 1 . Furthermore, we also have the fact that f ∗ is bounded on T . Let f ∗ (eit ) = µ(t). Since f (reπi ) = f (re−πi ), it is easy to see that µ(π) = f ∗ (eπi ) = lim f (reπi ) = lim f (re−πi ) = f ∗ (e−πi ) = µ(−π). r→1

r→1

On the one hand, if f (z) =

∞ X

(17.17)

an z n ,

n=0

then we have an =

1 2πi

Z

C(0;r)

f (ζ) 1 dζ = ζ n+1 2π

Z

π

r −n e−int f (reit ) dt

(17.18)

−π

for every n ≥ 0 and 0 < r < 1. On the other hand, it follows from Theorem 17.11 that f ∗ ∈ L1 (T ), so the Fourier coefficients of f ∗ are given by Z π 1 e−int f ∗ (eit ) dt (17.19) fc∗ (n) = 2π −π for every n ∈ Z. By observing the coefficients (17.18) and (17.19), we know that Z Z |f (reit ) − f (eit )| dσ = kfr − f ∗ k1 → 0 |r n an − fc∗ (n)| = e−int [f (reit ) − f ∗ (eit )] dσ ≤ T

T

as r → 1 by Theorem 17.11. In other words, we get   an , if n ≥ 0; fc∗ (n) =  0, if n < ∞.

This implies that

for every n = 1, 2, . . ..

Z

π

int

e −π

µ(t) dt =

Z

π

−π

eint f ∗ (eit ) dt = 0

(17.20)

Chapter 17. H p -Spaces

530 Using the two facts (17.17) and (17.20), we find Z Z π Z π µ(t) d(eint ) = −in eint dµ(t) = [eint µ(t)]π−π − −π

−π

π

eint µ(t) dt = 0

−π

holds for every n = 1, 2, . . .. Next, Theorem 17.13 (The F. and M. Riesz Theorem) shows that µ is absolutely continuous with m, i.e., µ(E) = 0 if m(E) = 0. Recall that f ∗ (eit ) = µ(t), so f ∗ ∈ C(T ). In fact, f ∗ is AC on T because of Theorem 7.18 because f maps sets of measure 0 to sets of measure 0. Finally, according to Theorem 17.11, we have Z π Z π 1 1 f (z) = P (z, eit )µ(t) dt = P (z, eit )f ∗ (eit ) dt = P [f ∗ ](z) 2π −π 2π −π for all z ∈ U . Thus if we define the function Hf ∗ : U → C as [100, Eqn. (1), p. 234], then Theorem 11.8 shows that Hf ∗ ∈ C(U ) and the restriction of Hf ∗ to T is exactly f ∗ . This is a  required extension of f and we end the analysis of the problem. Problem 17.14 Rudin Chapter 17 Exercise 14.

Proof. Recall from Problem 2.11 that the support of a measure µ is the smallest closed set K ⊆ T such that µ(T \ K) = 0. Without loss of generality, we may assume that µ 6≡ 0. Given that K is a proper closed subset of T . Note that T \ K is nonempty open in T , so σ(T \ K) > 0. Our goal is to show that µ(T \ K) > 0. (17.21) Now we observe from the proof of Theorem 17.13 (The F. and M. Riesz Theorem) that dµ = f ∗ (eit ) dσ, i.e., Z 1 f ∗ (eit ) dt µ(E) = 2π E

holds for every measurable subset E of T , where f = P [f ∗ ] ∈ H 1 and f ∗ ∈ L1 (T ). Thusit suffices to show that f ∗ doesn’t vanish on T \ K. Let g(t) = f ∗ (eit ). Assume that g(T \ K) = 0. Then it means that log |g| = ∞ on T \ K, but it contradicts Theorem 17.17 (The Canonical Factorization Theorem) that log |g| = log |f ∗ | ∈ L1 (T ).

Hence f ∗ does not vanish on T \ K with σ(T \ K) > 0 which implies the result (17.21). This completes the analysis of the problem.  Problem 17.15 Rudin Chapter 17 Exercise 15.

Proof. Denote CK to be the set of all continuous functions on K.g Then the problem is equivalent to show that the set of polynomials PK on K is dense in CK with respect to the norm kf k∞ = sup{|f (z)| | z ∈ K}. This is well-defined because f is continuous on the compact set K. Assume that PK was not dense in CK . In other words, there exists an g ∈ CK such that g ∈ / PK . It is clear that PK g

See Definition 3.16, p. 70.

17.3. Factorization of f ∈ H p

531

is a linear subspace of the normed linear space CK . Then Theorem 5.19 implies that there is a bounded linear functional Φ on CK such that Φ(P ) = 0 for all P ∈ PK and Φ(g) 6= 0. According to Theorem 6.19, there exists a unique regular complex Borel measure ν on K such that Z f dν (17.22) Φ(f ) = K

for every f ∈ CK . This measure ν must be nonzero because of Φ(g) 6= 0.

Define µ(E) = ν(E ∩ K) for E ∈ MT . Then it is easily checked that µ is also a complex Borel measure on T . Combining this and the representation (17.22), we get Z Z Z P dµ = Φ(P ) = 0 P dν + P dµ = T

T \K

K

for every P ∈ PK . In particular, we have Z

e−int dµ = 0

T

for every n = 1, 2, . . .. Since µ 6≡ 0, Problem 17.14 asserts that the support of µ is exactly all of T so that µ(T \ K) > 0 because K is a proper compact (hence closed) subset of T . However, the definition of µ implies that µ(T \ K) = ν(∅) = 0, a contradiction. This completes the analysis of the problem. 

17.3

Factorization of f ∈ H p

Problem 17.16 Rudin Chapter 17 Exercise 16.

Proof. Suppose that 0 < p < 1. Recall from the first paragraph of the proof of Theorem 7.17 that we may assume that f has no zeros in U . Therefore, we deduce from Theorem 17.10 that 2 one can find a zero-free function h ∈ H 2 such that f = h p . Note that h = Mh Qh by Theorem 17.17, so we have 2

2

f = Mhp Qhp . By the definition of an inner function, we know that n 2 Z eit + z n Z eit + z o o 2 2 2 p exp − dµ (t) = c dµ (t) , Mhp (z) = c p exp − h f it p T eit − z T e −z

(17.23)

(17.24)

where µh is a finite positive Borel measure on T and µf = p2 µh . Clearly, µf is also a finite positive Borel measure on T , so it follows from Theorem 17.15 that the right-most function is in fact an inner function. Let it be Mf . Next, according to [100, Eqn. (1), p. 344], we see that n 1 Z eit + z 2 o 2 ∗ it Qhp (z) = exp · · log |h (e )| dt 2π T eit − z p n 1 Z eit + z o 2 log (h∗ ) p (eit ) dt = exp it 2π T e − z

Chapter 17. H p -Spaces

532

= exp

o n 1 Z eit + z log |f ∗ (eit )| dt . it 2π T e − z

(17.25)

Since log |h∗ | ∈ L1 (T ), we immediately have log |f ∗ | ∈ L1 (T ). By Definition 17.14, the function (17.25) is an outer function and we let it be Qf . Since Qh ∈ H 2 , Theorem 17.16 implies that |h∗ | ∈ L2 (T ) and thus |f ∗ | ∈ Lp (T ). Using Theorem 17.16 again, we conclude that Qf ∈ H p . By substituting the expressions (17.24) and (17.25) into the formula (17.23), we obtain f = Mf Qf . Finally, we note that the inequality log |h(0)| ≤

1 2π

Z

log |h∗ (eit )| dt

(17.26)

T

log |f (0)| ≤

1 2π

Z

log |f ∗ (eit )| dt.

(17.27)

T

is equivalent to the inequality

Hence equality holds in (17.27) if and only if equality holds in (17.26) if and only if Mh is constant if and only if Mf is constant too. Consequently, we have completed the proof of the  problem. Problem 17.17 Rudin Chapter 17 Exercise 17.

Proof. (a) Assume that ϕ1 ∈ H p for some p > 0. Since ϕ1 has no zero in U and nonconstant, Theorem 17.10 (The Riesz Factorization Theorem) implies that there is a zero-free function f ∈ H 2 such that 2 1 = f p. ϕ Therefore,

1 p ϕ2

= f is in H 2 . Let f (z) =

∞ X

an z n .

n=0

By the Parseval Theorem [100, Eqn. (6), p. 91] and also the proof of Theorem 17.12, we get Z Z Z ∞ X 1 1 |an |2 = lim |fr |2 dσ = lim dσ = dσ. ∗ |p r→1 T r→1 T |ϕ|p |ϕ T n=0

Since ϕ is an inner function in U , Definition 17.14 implies that |ϕ∗ | = 1 a.e. on T and then ∞ X |an |2 = 1. n=0

In particular, we have |f (0)| = |a0 | ≤ 1 or equivalently, |ϕ(0)| ≥ 1. By Theorem 17.15, every inner function M satisfies |M (z)| ≤ 1 in U . Combining this fact and Theorem 10.24 (The Maximum Modulus Theorem), we establish that ϕ is constant, a contradiction. / H p for all p > 0. Consequently, ϕ1 ∈

17.3. Factorization of f ∈ H p

533

(b) By the hypotheses, we have ϕ(z) = c exp

n

−

Z

T

o eit + z dµ(t) , it e −z

where |c| = 1, µ is a finite positive Borel measure on T , and µ ⊥ m. By Theorem 17.15 and Theorem 10.24 (The Maximum Modulus Theorem), we know that log |ϕ| is always negative, i.e., 0 < |ϕ(z)| < 1 for every z ∈ U . Since ϕ ∈ H(U ) and ϕ(z) 6= 0 for all z ∈ U , it follows from Problem 11.5 that log |ϕ| is harmonic in U . Recall from [100, Eqn. (2), §11.5, p. 233] that u(z) = − log |ϕ(z)| =

Z

Re T

 eit + z  eit − z

dµ(t) =

Z

P (z, eit ) dµ(t) T

which means u = P [ dµ]. Since µ ⊥ m, it follows from Problem 11.19 that u(reiθ ) → ∞ a.e. [µ]. Consequently, there exists an eiθ ∈ T such that lim ϕ(reiθ ) = 0.

r→1

We have completed the analysis of the problem.



Problem 17.18 Rudin Chapter 17 Exercise 18.

Proof. Suppose that ϕα (z) =

ϕ(z) − α 1 − αϕ(z)

for every z ∈ U . It is clear that ϕα has no zero in U because α ∈ / ϕ(U ). If we can show that ϕα is nonconstant and inner, then it follows directly from Problem 17.17(b) that there is at least one eiθ ∈ T such that ϕα (reiθ ) → 0 as r → 1. Equivalently, it means that lim ϕ(reiθ ) = α.

r→1

(17.28)

To this end, we directly apply the following result from [112, p. 323]: Lemma 17.2 If ψ and ϕ are inner functions in U , then ψ ◦ ϕ is also inner. z−α Since ψα (z) = 1−αz is clearly inner and ϕα = ψα ◦ ϕ, we follow from Lemma 17.1 that ϕα is also inner. Now ϕα is nonconstant because ϕ is also nonconstant. Therefore, we conclude that  the result (17.28) holds and we have completed the proof of the problem.

Problem 17.19 Rudin Chapter 17 Exercise 19.

Chapter 17. H p -Spaces

534

Proof. Let g = f1 . Since f, g ∈ H 1 and f, g are not identically 0, we deduce from Theorem 17.17 (The Canonical Factorization Theorem) that f = Mf Qf

and g = Mg Qg

which imply that 1 = (Mf Mg )(Qf Qg ). By Definition 17.14 and Theorem 17.15, finite products of inner functions and outer functions remain inner and outer respectively. Thus we can write 1 = MQ

(17.29)

for some inner and outer functions M and Q, where M has no zero in U . We claim that this factorization (17.29) is unique up to a constant of modulus 1. Suppose that we have 1 = M1 Q1 = M2 Q2 . By Theorem 17.15, M1 and M2 can be expressed in the form [95, Eqn. (1), p. 342] which gives |M1 (z)| = |M2 (z)| = 1 on T . Therefore, we also have |Q1 (z)| = |Q2 (z)| = 1 on T . Since Q1 M2 = Q2 M1 Q1 Q2 Q2 and Q1 Q1 (z) | = Since | Q 2 (z)

both

Theorem) that

and

Q2 M1 = , Q1 M2

are inner functions without zero in U . In other words, we have

Q2 (z) |Q | = 1 (z) Q1 (z) | Q2 (z) | ≤ 1

Q1 Q2 Q2 , Q1

∈ H(U ).

1 on T , it follows from Theorem 10.24 (The Maximum Modulus 2 (z) and | Q Q1 (z) | ≤ 1 in U which imply that

Q1 (z) = cQ2 (z) in U for some constant c with |c| = 1. Since Q is unique up to a constant of modulus 1, M is also unique up to a constant of modulus 1 in the factorization (17.29), as required. By the definition of g and Theorem 17.17 (The Canonical Factorization Theorem), we know that Qg = Q1f . Consequently, this fact and the above claim show immediately that Qf Qg = Mf Mg = 1. By Theorem 17.15 again, Mf Mg = 1 implies that Mf = 1 and hence f = Qf , completing the proof of the problem.  Problem 17.20 Rudin Chapter 17 Exercise 20.

Proof. Given ǫ > 0. Define fǫ (z) = f (z) + ǫ in U . Then Clearly, Problem 17.18 implies that fǫ = Qfǫ , i.e., fǫ (z) = c exp

1 fǫ

is bounded in U so that

n 1 Z π eit + z o (fǫ )∗ (eit ) dt log 2π −π eit − z

1 fǫ

∈ H 1.

(17.30)

for some constant c with |c| = 1. By the definition, the functions log |(fǫ )∗ | decrease to log |f ∗ | as ǫ → 0. Next, we know from Theorem 17.17 (The Canonical Factorization Theorem) that log |f ∗ | ∈ L1 (T ). Finally, we apply Problem 1.7 to the expression (17.30) to conclude that f (z) = c exp

o n 1 Z π eit + z log |f ∗ (eit )| dt = Qf (z) it 2π −π e − z

for all z ∈ U . This completes the proof of the problem.



17.3. Factorization of f ∈ H p

535

Problem 17.21 Rudin Chapter 17 Exercise 21.

Proof. Suppose first that f = hg , where g, h ∈ H ∞ . There is no loss of generality to assume that |g(z)| ≤ 1 and |h(z)| ≤ 1. By the definition of log+ , it is easy to see that Z π Z π Z π log |h(reit )| dt (17.31) log+ |f (reit )| dt ≤ log+ |fr | dt = −π

−π

−π

for 0 ≤ r < 1. Since h(z) 6= 0 in U , Theorem 15.18 (Jensen’s Formula) gives Z π 1 log |h(reit )| dt = log |h(0)|. 2π −π

(17.32)

Combining the inequality (17.31) and the result (17.32), we conclude that f ∈ N . Conversely, let f ∈ N and f 6≡ 0. According to Theorem 17.9, we may assume that f has no zero in U . By Problem 11.5, u = log |f | is harmonic in U . Thus the mean value property gives Z 1 u(0) = u(r) dt 2π T Z 1 log |fr (eit )| dt = 2π T Z Z 1 1 log+ |fr (eit )| dt − log− |fr (eit )| dt = 2π T 2π T Z 1 log− |fr (eit )| dt (17.33) ≤ log kf k0 − 2π T for all 0 ≤ r < 1. Since f ∈ N and the left-hand side of the equation (17.33) is independent of r, we know immediately that Z 1 sup log− |fr (eit )| dt < ∞ 0≤r k k nk ek

(17.55)

and the property (17.53) shows that k−1 k−1 X np z np −1 X np |z|np −1 ≤ p p=1 p=1

≤

≤ ≈

k−1 X np  p=1

p

k−1 X np h p=1

1 np −1 2np

1−

p

1−

1 2np −2 i 12 2np

k−1 X np √ ep p=1

(k − 1)nk−1 √ e nk . < 4ek ≤

(17.56)

If p ≥ k + 1, then np − 1 = exp(exp · · · exp nk ) − 1. Therefore, we obtain | {z } (p − k) iterations

np −1 ∞ ∞ X X 1 2nk i 2nk np h np z np −1 1− ≤ p p 2nk

p=k+1

≈

p=k+1 ∞ X

p=k+1

np

p exp(

np −1 . 2nk )

(17.57)

Besides the property (17.53), we require that the sequence {nk } satisfies exp

np − 1
4eknp > 2p−k × 2nk pnk

for every p ≥ k + 1. Then the estimate (17.57) becomes

∞ ∞ X np z np −1 nk nk X 1 = . < p−k p 4ek 2 4ek p=k+1

(17.58)

p=k+1

Substituting the estimates (17.55), (17.56) and (17.58) into the inequality (17.54), we get |f ′ (z)| > in 1 −

1 nk

< |z| < 1 −

nk nk nk nk nk − − = > ek 4ek 4ek 2ek 10k

1 2nk .

• The divergence of the integral. It is clear from the first assertion that Z

1 0

|f ′ (reiθ )| dr ≥

∞ Z X k=1

1− 2n1

k

1− n1

k

|f ′ (reiθ )| dr

Chapter 17. H p -Spaces

546

≥

∞ Z X k=1

1− 2n1

k

1− n1 k

nk dr 10k

∞ X nk  1 1  = − 10k nk 2nk k=1

∞ X 1 = 20k k=1

for every θ. Hence we have

Z

for every θ.

1 0

|f ′ (reiθ )| dr = ∞

(17.59)

• The convergence of the limit. Let 0 < R < 1. By the definition of f , we have Z RX Z R ∞ nk r nk −1 i(nk −1)θ f ′ (reiθ ) dr = e dr k 0 k=1 0 Z ∞ X nk  R nk −1  i(nk −1)θ r dr e = k 0 k=1

=

∞ X R nk k=1 −iθ

=e As

∞ X k=1

k

ei(nk −1)θ

f (Reiθ ).

(17.60)

1 < ∞, Theorem 17.12 implies that f ∈ H 2 and then we follow from Theorem k2

17.11 that f ∗ (eiθ ) exists a.e. on T . Thus we deduce from the expression (17.60) that Z R f ′ (reiθ ) dr = e−iθ f ∗ (eiθ ) lim R→1 0

exists for almost all θ. • The geometrical meaning of the integral (17.59). If f ∈ H(U ), we denote Z 1 |f ′ (reiθ )| dr V (f ; θ) = 0

which is the total variation of f on the radius of U terminating at the point eiθ . Hence the result (17.59) tells us that the length of the curve which is the image of the radius with angle θ under f is always infinite. We complete the proof of the problem.



Problem 17.29 Rudin Chapter 17 Exercise 29.

Proof. Let g ∈ Lp (T ). Suppose that g(eit ) = f ∗ (eit ) a.e. for some f ∈ H p . Let f (z) =

∞ X

n=0

an z n

17.5. Miscellaneous Problems

547

and fc∗ (n) be the Fourier coefficients of its boundary function f ∗ (eit ), i.e., Z π 1 fc∗ (n) = e−int f ∗ (eit ) dt 2π −π

for all n ∈ Z. The Taylor coefficients of f can be expressed in the form Z π Z π 1 1 f (reit ) an = dt = r −n e−int f (reit ) dt 2π −π r n eint 2π −π

(17.61)

(17.62)

for every 0 < r < 1. Combining the coefficients (17.61) and (17.62), we get Z π Z π n   1 f (reit ) − f ∗ (eit ) dt r an − fc∗ (n) = 1 e−int f (reit ) − f ∗ (eit ) dt ≤ 2π −π 2π −π for every n ∈ Z. Applying Theorem 17.11, we obtain n r an − fc∗ (n) ≤ kfr − f ∗ k1 → 0

as r → 1. Therefore, we have

fc∗ (n) =

  an , if n ≥ 0; 

0,

if n < 0.

By the hypothesis, we conclude that Z π Z π 1 1 e−int g(eit ) dt = e−int f ∗ (eit ) dt = 0 2π −π 2π −π

(17.63)

for all negative integers n. Conversely, we suppose that the formula (17.63) holds for all negative integers n. In other words, gb(n) = 0 for all negative integers n. Let Z π 1 Pr (θ − t)g(eit ) dt, (17.64) f (reiθ ) = 2π −π where 0 < r < 1. Since the Poisson kernel has the expansion Pr (t) = 1 +

∞ X

r n (eint + e−int ),

n=1

it follows from the formula (17.63) that Z π 1 f (reiθ ) = Pr (θ − t)g(eit ) dt 2π −π Z π Z π ∞ ∞ X X 1 1 r n ein(θ−t) g(eit ) dt + e−in(θ−t) g(eit ) dt = 2π 2π −π −π n=1 n=0 ∞ h 1 Z π i X (reiθ )n = e−int g(eit ) dt 2π −π =

n=0 ∞ X

n=0

which means

gb(n)(reiθ )n f (z) =

∞ X

n=0

g(n)z n b

Chapter 17. H p -Spaces

548

for every z ∈ U and then f ∈ H(U ). By the integral (17.64) and [100, Eqn. (3), p. 233], it is easy to see that Z π 1 kf k1 = |f (reiθ )| dθ 2π −π Z πh Z π i 1 1 Pr (θ − t)|g(eit )| dt dθ ≤ 2π −π 2π −π Z πh Z π i 1 1 Pr (θ − t) dθ · |g(eit )| dt = 2π −π 2π −π Z π 1 = |g(eit )| dt 2π −π so that f ∈ H 1 . By Theorem 17.11, we see that f ∗ ∈ L1 (T ). Now we consider Z π 1 Pr (θ − t)f ∗ (eit ) dt = P [f ∗ ](z). Φ(z) = 2π −π Let z ∈ U . For any fixed 0 < ρ < 1, since f ∈ H(U ), it follows from Theorem 11.4 and the mean value property that Z π 1 Pr (θ − t)f (ρeit ) dt. f (ρz) = 2π −π We observe from Theorem 17.11 that Z π lim |f (ρeit ) − f ∗ (eit )| dt → 0, ρ→1 −π

so we obtain f (z) = lim f (ρz) = Φ(z) ρ→1

for every z ∈ U , i.e.,

1 f (re ) = 2π iθ

Z

π −π

Pr (θ − t)f ∗ (eit ) dt.

(17.65)

Comparing the two integrals (17.64) and (17.65), we conclude immediately that g(z) = f ∗ (z) a.e. on T . Since g ∈ Lp (T ), we know that f ∗ ∈ Lp (T ) and it yields from Problem 17.8 that  f ∈ H p , as required. Now we have completed the proof of the problem.

CHAPTER

18

Elementary Theory of Banach Algebras

18.1

Examples of Banach Spaces and Spectrums

Problem 18.1 Rudin Chapter 18 Exercise 1.

Proof. Let X be a Banach space and B(X) = {A : X → X | A is linear and bounded with mentioned conditions}.

(18.1)

Denote k · kX to be the norm of the Banach space X. The hypotheses ensure that an associative and distributive multiplication is well-defined in B(X). For every α ∈ C, A1 , A2 ∈ B(X) and x ∈ X, we see that α(A1 A2 )(x) = αA1 (A2 x) = A1 (αA2 x) = (αA1 )(A2 x), i.e., α(A1 A2 ) = A1 (αA2 ) = (αA1 )A2 . Thus B(X) is a complex algebra. We check Definition 5.2. By the definition, kAk must be nonnegative. By the definition (18.1), there exists a positive constant M such that kAxkX ≤ M kxkX for all x ∈ X. Therefore, we have kAxkX ≤ M, kAk = sup kxkX i.e., kAk is a real number. Next, for all A1 , A2 ∈ B(X), we apply the fact that X is a Banach, so we have k(A1 + A2 )(x)kX kxkX kA1 x + A2 xkX = sup kxkX kA1 xkX + kA2 xkX ≤ sup kxkX = kA1 k + kA2 k.

kA1 + A2 k = sup

Furthermore, if α ∈ C and A ∈ B(X), then we obtain kαAk = sup

kAxkX k(αA)(x)kX = |α| sup = |α| · kAk. kxkX kxkX 549

550

Chapter 18. Elementary Theory of Banach Algebras

Let kAk = 0. This means that kAxkX = 0 for all x ∈ X so that Ax = 0 for all x ∈ X. Consequently, it must be the case A = 0 and then B(X) is a normed linear space. For A1 , A2 ∈ B(X), we see that k(A1 A2 )(x)kX kxkX kA2 xkX kA1 (A2 x)kX × = sup kA2 xkX kxkX kA1 (A2 x)kX kA2 xkX ≤ sup × sup kA2 xkX kxkX ≤ kA1 k · kA2 k.

kA1 A2 k = sup

Hence B(X) is also a normed complex algebra. We claim that B(X) is a complete metric space. Fix x ∈ X. Given ǫ > 0. Let {An } be Cauchy with respect to the norm k · k. Then it suffices to show that there exists an A ∈ B(X) such that kAn − Ak → 0

as n → ∞. Since there exists an N ∈ N such that kAm − An k < 1 for all n, m ≥ N . Using the triangle inequality of the norm k · k, we see that kAn k < 1 + kAN k for all n ≥ N . Denote M = max{kA1 k, kA2 k, . . . , kAN −1 k, 1 + kAN k}. Thus we get kAn k < 1 + kAN k for all n ∈ N. This means that

kAn xkX ≤ M kxkX

(18.2)

for all x ∈ X and n ∈ N. Now, for all x ∈ X, we can establish kAm x − An xkX = k(Am − An )xkX ≤ kAm − An k · kxkX → 0 as n, m → ∞. In other words, {An x} is a Cauchy sequence in X. Since X is Banach, the sequence converges to an element in X, namely Ax. Then we can define the operator A : X → X by this and the linearity of taking limits implies the linearity of A. Besides, we follow from the inequality (18.2) that kAxkX = lim kAn xkX ≤ M kxkX n→∞

for every x ∈ X. Consequently, it means that A ∈ B(X). Now it remains to verify that kAm − Ak → 0 as m → ∞. Recall that {An } is Cauchy, so given ǫ > 0, there exists an N ∈ N such that kAm − An k ≤ ǫ whenever n, m ≥ N . Therefore, for every x ∈ X, we have kAm x − An xkX ≤ kAm − An k · kxkX < ǫ · kxkX whenever n, m ≥ N . If n → ∞, then we see that kAm x − AxkX ≤ ǫ · kxkX for all m ≥ N and all x ∈ X. By the definition, we conclude that kAm − Ak ≤ ǫ for all m ≥ N . Hence this proves our claim that B(X) is complete and then it is a Banach algebra, completing  the proof of the problem.

18.1. Examples of Banach Spaces and Spectrums

551

Problem 18.2 Rudin Chapter 18 Exercise 2.

Proof. Suppose that we have X = {v = (z1 , z2 , . . . , zn ) | z1 , z2 , . . . , zn ∈ C} = Cn is equipped with the norm k · kCn and B(Cn ) is the algebra of all bounded linear operators on Cn . By Problem 18.1, B(Cn ) is also a Banach algebra with the norm k · k given by kAk = sup

kAvkCn . kvkCn

Our target is to find σ(A) = {λ ∈ C | A − λI is not invertible}. According to the explanation in [62, pp. 96, 97], every bounded linear operator A can be represented by a matrix with entries in C. We also denote this matrix by A. Therefore, A − λI is not invertible if and only if det(A − λI) = 0. (18.3) Since det(A−λI) = 0 is an equation in λ of order n, the Fundamental Theorem of Algebra ensures that it has at most n complex roots. Hence σ(A) consists of at most n complex numbers and they are exactly the solutions of the equation (18.3), completing the proof of the problem.  Problem 18.3 Rudin Chapter 18 Exercise 3.

Proof. Suppose that C is a positive constant such that |ϕ(x)| ≤ C a.e. on R. Define the mapping Mϕ : L2 → L2 by Mϕ (f ) = ϕ × f. Of course, it is true that ϕf ∈ L2 , so Mϕ is well-defined. Furthermore, the linearity of Mϕ is clear. Recall that L2 is a normed linear space with the norm kf k2 . Since we have Z |ϕ(x)f (x)|2 dx ≤ Ckf k2 , kϕf k2 = R

so we get  kMϕ k = sup kMϕ (f )k | f ∈ L2 and kf k2 = 1  = sup kϕf k2 | f ∈ L2 and kf k2 = 1 ≤ C.

By Definition 5.3, Mϕ is bounded. For the second assertion, recall from [100, Exercise 19, p. 74] that 
Rϕ = λ ∈ C m {x | |ϕ(x) − λ| < ǫ} > 0 for every ǫ > 0 .

(18.4)

Let I be the identity operator on L2 . We are required to prove that

σ(Mϕ ) = {λ ∈ C | Mϕ − λI is not invertible} = Rϕ .

(18.5)

552

Chapter 18. Elementary Theory of Banach Algebras

On the one hand, let λ ∈ / Rϕ . Then |ϕ(x) − λ| ≥ ǫ for some ǫ > 0 a.e. on R, so we have 1 ∞ (R) and this implies that the operator M is bounded. Furthermore, it is easy to ∈ L 1 ϕ−λ ϕ−λ see that
M 1 Mϕ (f ) − λf = M 1 (ϕf − λf ) = f. ϕ−λ

Thus we have M

1 ϕ−λ

ϕ−λ

is the inverse of Mϕ − λI which means λ ∈ / σ(Mϕ ).

On the other hand, let λ ∈ Rϕ . For any n ∈ N, we denote

Sn = {x | |ϕ(x) − λ| < 2−n }. The definition (18.4) reveals that m(Sn ) > 0. Suppose that there exists an N ≥ n such that 0 < m(SN ) < ∞. Otherwise, m(Sn ) = ∞ for all n ≥ 1 and this means that ϕ(x) = λ for almost all x ∈ R. In this case, we know that Rλ = {λ}. Clearly, Mλ f − λf = 0, so λ ∈ σ(Mλ ). If µ 6= λ, then Mλ (f ) − µf = (λ − µ)f so that
M 1 Mλ (f ) − µf = f. λ−µ

Consequently, we obtain σ(Mλ ) = {λ} and then the equality (18.5) holds. Let 0 < m(SN ) < ∞. Take φn = χSN . Then we have Z 2 2 |ϕ(x) − λ|2 · |φn (x)|2 dx ≤ 2−2n kφn k2 k(Mϕ − λI)(φn )k = kϕφn − λφn k = SN

so that the operator (Mϕ − λI)−1 is not bounded, i.e., Mϕ − λI is not invertible. Hence we conclude that λ ∈ σ(Mϕ ) and we have established the equality (18.5), completing the proof of the problem.  Problem 18.4 Rudin Chapter 18 Exercise 4.

Proof. Recall that ℓ2 = given by

n

∞ o o1 nX 2 < ∞ and S : ℓ2 → ℓ2 is |ξn |2 x = {ξ0 , ξ1 , ξ2 , . . .} kxk = n=0

Sx = {0, ξ0 , ξ1 , . . .}

which is a bounded linear operator on ℓ2 and kSk = 1.a We want to determine σ(S) = {λ ∈ C | S − λI is not invertible}. Since kSk = 1, Corollary 3 to Theorem 18.4 implies that σ(S) ⊆ U .

(18.6)

Take 0 < |λ| < 1. Assume that λ ∈ / σ(S). Then S − λI is invertible so that the equation (S − λI)x = y

(18.7)

has a unique solution x ∈ ℓ2 for every y ∈ ℓ2 . If y = (1, 0, 0, . . .), then the equation (18.7) takes the system −λξ0 = 1 and ξn − λξn+1 = 0 a

In fact, S is called the right-shift operator.

18.2. Properties of Ideals and Homomorphisms

553

for every n = 0, 1, 2, . . .. Solving it, we obtain ξn = −λ−(n+1) for every n ∈ N. Since |λ| < 1, we have |ξn | = |λ|−(n+1) > and thus x ∈ / ℓ2 which is a contradiction. Hence we must have λ ∈ σ(S), i.e., {λ ∈ C | 0 < |λ| < 1} ⊆ σ(S). (18.8) Finally, we observe from Theorem 18.6 that σ(S) is a closed set, so we conclude from the set relations (18.6) and (18.8) that σ(S) = U . We end the analysis of the problem.

18.2



Properties of Ideals and Homomorphisms

Problem 18.5 Rudin Chapter 18 Exercise 5.

Proof. Let M be an ideal of the commutative complex algebra A. Then M is a vector space and we note from §4.7 that M is also a vector space. Thus it remains to show that aM ⊆ M and M a ⊆ M for all a ∈ A. Fix a ∈ A and consider b ∈ M . Then there exists a sequence {bn } ⊆ M such that bn → b as n → ∞. By considering the sequences {abn } and {bn a}, since M is an ideal, it is true that abn ∈ M and bn a ∈ M for all n = 1, 2, . . .. Since a is fixed, the mapping x → 7 ax and x 7→ xa are both continuous on M . Therefore, abn → ab and bn a → ba as n → ∞. In other words, it is true that ab, ba ∈ M .  This completes the proof of the problem. Problem 18.6 Rudin Chapter 18 Exercise 6.

Proof. Let C(X) be the algebra of all continuous complex functions on X with pointwise addition multiplication and the supremum norm. The constant function 1 is the unit element. Let I be an ideal in C(X). We claim that either I = C(X) or there exists a p ∈ X such that I = {f ∈ C(X) | f (p) = 0}.

(18.9)

Assume that, for every p ∈ X, there exists a continuous function f ∈ I such that f (p) 6= 0. Since X is compact Hausdorff, the continuity of f implies that there exists an open set Vp containing p such that f (x) 6= 0 for all x ∈ Vp . Therefore, the collection {Vp } forms an open covering of X. Since X is compact, there must exist a finite subcover. Call this subcover V1 , V2 , . . . , VN and the corresponding functions f1 , f2 , . . . , fN for some N ∈ N. Define 2 F (x) = f12 (x) + f22 (x) + · · · + fN (x).

(18.10)

Since fk ∈ I and I is an ideal, it follows from Definition 18.12 that F ∈ I. For every p ∈ X, we have fk (p) 6= 0 for some k ∈ {1, 2, . . . , N } so that F (p) 6= 0. Since F is continuous on the compact set X, it must attain a minimum. By the form (18.10), it is trivial that F (x) > 0 for 1 is the inverse of F in C(X). However, we note all x ∈ X. This implies that F −1 (x) = F (x) from Definition 18.12 that no proper ideal contains an invertible element, so we have I = C(X). Consequently, we have obtained our claim.

554

Chapter 18. Elementary Theory of Banach Algebras

If I is maximal, then it has the form (18.9) for some p ∈ X. Assume that I ⊂ J for some ideal J in C(X), where I 6= J. Then there corresponds an f ∈ J such that f (p) 6= 0. Since f is continuous, one can find a neighborhood Vp of p such that f (x) 6= 0 for all x ∈ Vp . The point set {p} is compact by Theorem 2.4. According to Urysohn’s Lemma, there exists an g ∈ C(X) such that {p} ≺ g ≺ Vp , i.e., g(p) = 1 and g(x) = 0 for all x ∈ X \ Vp . Take h = 1 − g which is also an element of C(X) and it satisfies h(p) = 0 and h(x) = 1 for all x ∈ X \ Vp . As J has the unit, we have h ∈ J. Next, we define H(x) = f 2 (x) + h2 (x) which is an element of J. Obviously, it is easy to check that H(x) > 0 on X. Since X is compact, H attains its minimum in X and thus H is bounded from below by a positive number. Therefore, its inverse H1 belongs to C(X) which asserts that J = C(X) by the previous paragraph. Hence we have the expected conclusion that the ideal in the form (18.9) is maximal. This completes  the analysis of the problem. Problem 18.7 Rudin Chapter 18 Exercise 7.

Proof. Let e be the unit of A. Given λ ∈ / σ(x). Then x − λe is invertible so that (x − λe)−1 ∈ A. Since A is generated by a single element x, this means that there are polynomials Pn such that Pn (x) → (x − λe)−1

(18.11)

as n → ∞ in A. If z ∈ σ(x), then Theorem 18.17(b) ensures that h(x) = z for some h ∈ ∆. Since h is a complex homomorphism of A, we have h(xm ) = z m for every m ∈ Z. By this and Theorem 18.17(e), we establish that
|Pn (z) − (λ − z)−1 | = h Pn (x) − (λ − x)−1 ≤ kPn (x) − (λ − x)−1 k. (18.12)

Applying the result (18.11) to the inequality (18.12), we obtain the result that Pn (z) → (λ−z)−1 uniformly on σ(x).

Assume that C \ σ(x) was disconnected. Let Ω be a (non-empty) bounded component of it. Fix λ ∈ Ω, i.e., λ ∈ / σ(x). Choose {Pn } as above. For every n ∈ N and z ∈ σ(x), we have |(z − λ)Pn (z) − 1| = |z − λ| · Pn (z) − (z − λ)−1 ≤ ℓ · Ln , (18.13)

where

ℓ = sup |z − λ| and z∈σ(x)

Ln = sup |Pn (z) − (z − λ)−1 |. z∈σ(x)

The compactness of σ(x) by Theorem 18.6 asserts that both ℓ and Ln are finite. Furthermore, ∂Ω ⊆ σ(x). Next, we use Theorem 10.24 (The Maximum Modulus Theorem) to see that the inequality (18.13) also holds on Ω. In particular, we get |Pn (z) − (z − λ)−1 | ≤

ℓ · Ln |z − λ|

(18.14)

for all z ∈ Ω \ {λ}. Since Ω is a component, one can find a δ > 0 small enough such that the circle C(λ; δ) lies in Ω. Therefore, we conclude from the estimate (18.14) that Z Z   dz ℓLn Pn (z) − (z − λ)−1 dz ≤ ℓ · Ln 2πi = = · 2πδ = 2πℓLn . (18.15) |z − λ| δ C(λ;δ) C(λ;δ)

18.2. Properties of Ideals and Homomorphisms

555

Notice that Ln → 0 as n → ∞, so the inequality (18.15) implies a contradiction. Hence the set C \ σ(x) must be connected, as required. This completes the proof of the problem.  Problem 18.8 Rudin Chapter 18 Exercise 8.

Proof. Since

∞ X

n=0

that

|cn | < ∞, there exists a positive constant M such that |cn | ≤ M . This implies 1

lim sup |cn | n ≤ 1, n→∞

i.e., the radius of convergence R of the power series satisfies R ≥ 1. By Theorem 10.6, both f and then f1 are holomorphic in a region containing U . Define Cn = cn for all n ≥ 0 and Cn = 0 for all n < ∞. It is clear that f (eit ) =

∞ X

cn eint =

∞ X

Cn eint

∞ X

and

n=−∞

n=−∞

n=0

|Cn | =

∞ X

n=0

|cn | < ∞.

Now the hypothesis |f (z)| > 0 for every z ∈ U implies that f (eit ) 6= 0 for every real t, so Theorem 18.21 (Wiener’s Theorem) guarantees that f satisfies ∞ X 1 γn eint = f (eit ) n=−∞

Since

1 f

∞ X

and

n=∞

|γn | < ∞.

(18.16)

is holomorphic in a region containing U , we have γn = 0 for all n < 0 and ∞

∞

n=0

n=0

X X 1 γn eint an eint = = f (eit ) which implies immediately that an = γn for all n ≥ 0 by the Corollary following Theorem 10.18. Hence the second condition (18.16) gives the desired result that ∞ X

n=0

|an | < ∞.

This completes the proof of the problem.



Problem 18.9 Rudin Chapter 18 Exercise 9.

Proof. We note from Example 9.19(d) that we define the multiplication in L1 (R) by convolution. Let f, g ∈ L1 (R) and φ ∈ L∞ (R). Here we employ the proof of [98, pp. 157, 158]: If I is a translation invariant subspace of L1 (R), then we say that φ annihilates I if Z f (x − y)φ(y) dm(y) = 0 (f ∗ φ)(x) = R

for all f ∈ I and x ∈ R. On the one hand, we note that Z (f ∗ g ∗ φ)(0) = [(f ∗ g) ∗ φ](0) = (f ∗ g)(0 − y)φ(y) dm(y). R

(18.17)

556

Chapter 18. Elementary Theory of Banach Algebras

On the other hand, recall from [100, Example 9.19(d)] that f ∗ g = g ∗ f , so we have Z g(0 − y)(f ∗ φ)(y) dm(y). (f ∗ g ∗ φ)(0) = [g ∗ (f ∗ φ)](0) =

(18.18)

R

In other words, the two integrals (18.17) and (18.18) are equal, i.e., Z Z g(0 − y)(f ∗ φ)(y) dm(y). (f ∗ g)(0 − y)φ(y) dm(y) =

(18.19)

R

R

Let I be a closed translation invariant subspace of the Banach space L1 (R) and φ ∈ L1 (R) annihilate f ∈ I. Then f ∗ φ = 0 and we see from the right-hand side of the expression (18.19) that (f ∗ g) ∗ φ = 0 (18.20)

for every g ∈ L1 (R).b Recall the basic fact from Theorem 6.16 that L1 (R) is isometrically
∗ ∞ 1 ∼ isomorphic to the dual space of L (R), i.e., L (R) = L∞ (R). By Remark 5.21, every φ ∈ L∞ (R) is a bounded linear functional on L1 (R). Assume that f ∗ g ∈ / I. Since I = I, there corresponds an ϕ ∈ L∞ (R) such that f ∗ ϕ = 0 and (f ∗ g) ∗ ϕ 6= 0 by Theorem 5.19, but this contradicts the result (18.20). Hence we conclude that f ∗ g ∈ I so that I is an ideal. Conversely, let I be a closed ideal and f ∗ φ = 0 for all f ∈ I. Assume that F = fx0 ∈ /I=I for some x0 ∈ R.c By Theorem 5.19, there exists an ϕ ∈ L∞ (R) such that f ∗ ϕ = 0 for all f ∈ I but F ∗ ϕ 6= 0. (18.21)

Now the hypotheses show that we have f ∗ g ∈ I for every g ∈ L1 (R), so the left-hand side of the expression (18.19) gives f ∗ φ annihilates every g ∈ L1 (R). This implies that f ∗ φ = 0 or Z f (x − y)φ(y) dm(y) 0 = (f ∗ φ)(x) = R

for x ∈ R. In other words, φ annihilates every translate of f . Particularly, this implies that F ∗ ϕ = 0 which contradicts the result (18.21). Consequently, fx ∈ I for every x ∈ R which  means it is translation invariant. Hence we have completed the analysis of the problem. Remark 18.1 As [98, Theorem 7.1.2, p. 157] indicates, the result of Problem 18.9 is also valid if we replace R by any locally compact abelian group. In particular, Problem 18.9 remains true for the unit circle T . Problem 18.10 Rudin Chapter 18 Exercise 10.

Proof. We prove the assertions one by one. • L1 (T ) is a commutative Banach algebra. Suppose that f, g, h ∈ L1 (T ). It is clear that f ∗ (g + h) = f ∗ g + f ∗ h, (f + g) ∗ h = f ∗ h + g ∗ h and α(f ∗ g) = f ∗ (αg) = (αf ) ∗ g for every α ∈ C. It also satisfies the associative law by an application of Theorem 8.8 (The Fubini Theorem)d Thus L1 (T ) is a complex algebra. By Theorem 8.14, we know that f ∗ g ∈ L1 (R). Recall that fx0 (y) = f (y − x0 ), see Theorem 9.5. d See also [100, Example 9.19(d), p. 190]. b

c

18.2. Properties of Ideals and Homomorphisms

557

Recall from [100, p. 96] that L1 (T ) is a Banach space normed by kf k1 . Using Theorem 8.8 (The Fubini Theorem) again, we know that Z π 1 kf ∗ gk1 = |(f ∗ g)(t)| dt 2π −π Z Z π 1 1 π = f (t − s)g(s) ds dt 2π −π 2π −π Z π Z π 1 1 ≤ |f (t − s)| · |g(s)| ds dt 2π −π 2π −π Z π Z π  1 1 = |f (t − s)| dt · |g(s)| ds 2π −π 2π −π Z π 1 = kf k1 · |g(s)| ds 2π −π = kf k1 · kgk1 .

In other words, L1 (T ) is a Banach algebra by Definition 18.1. The definition implies immediately that f ∗ g = g ∗ f . Hence L1 (T ) is commutative. • L1 (T ) does not have a unit. Assume that e ∈ L1 (T ) was a unit. Then e∗f = f for every f ∈ L1 (T ). Using similar argument as in the proof of Theorem 9.2(c) (The Convolution Theorem), we can show that b h(n) = fb(n) · gb(n) (18.22) if f, g ∈ L1 (T ). Therefore, we have

eb(n) = 1

for every n ∈ Z, but it contradicts the Riemann-Lebesgue Lemma [100, §5.14, p. 103]. • Complex homomorphisms of L1 (T ). Denote ∆T to be the set of all complex homomorphisms of L1 (T ). Let ϕ ∈ ∆T and ϕ 6= 0. By Theorem 18.17(e), ϕ is bounded by 1, so it follows from Theorem 6.16 that there is a unique β ∈ L∞ (T ) such that Z π 1 f (t)β(t) dt. (18.23) ϕ(f ) = 2π −π On the one hand, we have Z π Z π Z π 1 1 f (t − s)g(s)β(t) ds dt. (f ∗ g)(t)β(t) dt = ϕ(f ∗ g) = 2π −π (2π)2 −π −π On the other hand, we obtain i h 1 Z π i h 1 Z π f (t)β(t) dt × g(s)β(s) ds 2π −π 2π −π Z πZ π 1 f (t − s)g(s)β(t − s)β(s) ds dt. = (2π)2 −π −π

ϕ(f )ϕ(g) =

The fact ϕ(f ∗ g) = ϕ(f )ϕ(g) asserts that β(t) = β(t − s)β(s) a.e. on T or equivalently, β(x + y) = β(x)β(y) a.e. on T . Employing similar analysis as in the proof of Theorem 9.23, since β is periodic with period 1, it has the form β(x) = e−iαx

558

Chapter 18. Elementary Theory of Banach Algebras for a unique α ∈ R.e Since we must have β(x + 2π) = β(x), α must be an integer. Put α = n. Substituting this back into the integral (18.23), we get Z π 1 ϕ(f ) = f (t)e−int dt = fb(n) 2π −π for a unique integer n.

• IE is a closed ideal in L1 (T ). Let E ⊆ Z and  IE = f ∈ L1 (T ) fb(n) = 0 for all n ∈ E .

(18.24)

For any f ∈ IE and g ∈ L1 (T ), if we write h = f ∗ g, then the formula (18.22) implies that b h(n) = fb(n) · gb(n) = 0

for every n ∈ E. Thus we have f ∗ g ∈ IE and IE is an ideal in L1 (T ). For every α ∈ R, we follow from the definition that fb(n − α) = fb(n)e−iαn = 0

for every n ∈ E, so f (x − α) ∈ IE . This means that IE contains every translate of f and Remark 18.1 ensures that IE is closed. • Every closed ideal I in L1 (T ) has the form (18.24). Let I be a closed ideal of L1 (T ). We have to prove that I = IE for some set E ⊆ Z. Suppose each n ∈ Z, there  −1that for 1 (T ). Then we have exists an fn ∈ I such that c fn (n) 6= 0. Put gn (t) = eint fc (n) ∈ L n (gn ∗ fn )(t) =

1 1 · c fn (n) 2π

Z

π

−π

fn (s)ein(t−s) ds =

eint 1 · 2π fc n (n)

Z

π

fn (s)e−ins ds = eint .

−π

Since I is an ideal, we have eint ∈ I for every n ∈ Z. Using Theorems 3.14, 4.25 and the fact that I is closed, we know that the set {eint | n ∈ Z} is dense in L1 (T ). Thus we conclude that I = L1 (T ). Without loss of generality, we may assume that I 6= L1 (T ). Then there exists an n ∈ Z such that fb(n) = 0 for all f ∈ I. Let the collection of such integers be E, i.e., E = {n ∈ Z | fb(n) = 0 for all f ∈ I}.

(18.25)

Thus it is easy to see that I ⊆ IE . If n ∈ / E, then there exists an g ∈ I such that b g(n) 6= 0. For simplicity, we may assume that this number is 1. Regarding eint as an element of L1 (T ), we see that (g ∗ ein )(t) = −eint . Since I is an ideal, we have g ∗ ein ∈ I and then eint ∈ I. Thus I contains every trigonometric polynomial of the form X an eint provided that an = 0 for all n ∈ E. Suppose that f ∈ IE and we consider the set  Z(f ) = n ∈ Z fb(n) = 0 .

Now the definitions (18.24) and (18.25) imply that E ⊆ Z(f ), so it follows from Theorem 9.2(c) that if P is a trigonometric polynomial on T and n ∈ E, then we have

e

See also [40, Theorem 8.19, p. 247].

(f[ ∗ P )(n) = fb(n) × Pb(n) = 0,

18.2. Properties of Ideals and Homomorphisms

559

i.e., E ⊂ Z(f ∗ g). Using [98, Theorem 2.6.6, p. 51], we see that kf − f ∗ P k1 can be made as small as we want. Therefore, the closeness of I asserts that f ∈ I which implies IE ⊆ I. Hence we conclude that I = IE as desired. We end the proof of the problem.



Remark 18.2 For other classes of complex homomorphisms of specific Banach algebras, please refer to [130, §9, pp. 39 – 43]. Problem 18.11 Rudin Chapter 18 Exercise 11.

Proof. Notice that λ, µ ∈ C \ σ(x). On the one hand, we have     (x − λe) R(λ, x) − R(µ, x) (x − µe) = (x − λe)(λe − x)−1 − (x − λe)(µe − x)−1 (x − µe) = −e(x − µe) − (x − λe)(µe − x)−1 (x − µe)

= −x + µe + x − λe

= (µ − λ)e.

(18.26)

On the other hand, we see that     (x − λe) (µ − λ)R(λ, x)R(µ, x) (x − µe) = (µ − λ) (x − λe)R(λ, x)R(µ, x)(x − µe) = (µ − λ)e.

(18.27)

It yields from the results (18.26) and (18.27) that R(λ, x) − R(µ, x) = (µ − λ)R(λ, x)R(µ, x)

(18.28)

holds for all λ, µ ∈ C \ σ(x).

We follow from the identity (18.28) that (x − µe)−1 − (x − λe)−1 R(λ, x) − R(µ, x) = = R(λ, x)R(µ, x) → R(λ, x)2 = (x − λe)−2 µ−λ µ−λ

as µ → λ. This is exactly [100, Eqn. (3), p. 359], so the argument in the proof of Theorem 18.5 can be applied directly. This completes the proof of the problem.  Remark 18.3 The result in Problem 18.11 is called Hilbert’s identity.

Problem 18.12 Rudin Chapter 18 Exercise 12.

560

Chapter 18. Elementary Theory of Banach Algebras

Proof. Denote M be the set of maximal ideals of A. Let M be a maximal ideal of A. By Theorem 18.17, we have M = ker h for some h ∈ ∆. Conversely, if h ∈ ∆, then it follows from the First Isomorphism Theorem [42, Theorem 16.2, p. 145] that A/ ker h ∼ = h(A) = C. Since C is a field, ker h is a maximal ideal of A. In other words, there exists an one-to-one correspondence between M and ∆. Let rad A be the radical of A. Suppose that x ∈ rad A, i.e., \ x∈ M. (18.29) M ∈M

Now the previous paragraph yields that the set relation (18.29) is equivalent to the condition \ x∈ h−1 (0) h∈∆

which means that h(x) = 0 for every h ∈ ∆. Consequently, statements (a) and (c) are equivalent. Next, Theorem 18.17(b) means that the spectrum σ(x) is exactly the set {h(x) | h ∈ ∆}. Since x ∈ rad A if and only if h(x) = 0 for every h ∈ ∆, this implies that x ∈ rad A if and 1 only if σ(x) = {0} if and only if kxn k n → 0 as n → ∞ by Theorem 18.9 (The Spectral Radius Formula). Hence we have shown the three statements are equivalent, completing the proof of  the problem. Problem 18.13 Rudin Chapter 18 Exercise 13.

Proof. Let X = C([0, 1]). Then X is a Hilbert space (and hence a Banach space). For each f ∈ X, we define Z t f (s) ds, T (f )(t) = 0

where t ∈ [0, 1]. Since f ∈ X, T (f ) ∈ X so that T ∈ B(X), the algebra of all bounded linear operations on X. By Problem 18.1, B(X) is a Banach space. Particularly, we take f (x) = 1. Then we observe Z t Z tZ s t2 s ds = . du ds = T 2 (f )(t) = 2 0 0 0 More generally, for n = 1, 2, . . ., we obtain T n (f )(t) =

tn n!

which implies that T n (f ) 6= 0 for all n > 0 and kT n k = sup

kT n (f )k∞ 1 = . kf k∞ n!

(18.30)

1

Since (n!) n → ∞ as n → ∞, we conclude immediately from the result (18.30) that 1

lim kT n k n = 0.

n→∞

This completes the analysis of the problem.



18.2. Properties of Ideals and Homomorphisms

561

Problem 18.14 Rudin Chapter 18 Exercise 14.

Proof. By the definition, the function x b : ∆ → C is given by x b(h) = h(x) and we have the set b = {b b by G(x) = x A x | x ∈ A}. Denote the surjective mapping G : A → A b.

• The mapping G is a homomorphism. Suppose that x, y ∈ A, α ∈ C and h ∈ ∆. Then we see that d (αx)(h) = h(αx) = αh(x) = (αb x)(h)

and

\ (x + y)(h) = h(x + y) = h(x) + h(y) = x b(h) + yb(h) = (b x + yb)(h) x cy(h) = h(xy) = h(x)h(y) = x b(h)b y (h) = (b xyb)(h).

Therefore, the map G is a homomorphism. Its kernel consists of those x ∈ A such that h(x) = 0 for every h ∈ ∆. By Problem 18.12, it is exactly the radical of A, i.e., ker x b= rad A.

Combining the First Isomorphism Theorem and the previous result, we see that b A/rad A = A/ ker x b∼ = A.

b and G becomes an isomorphism. Thus if rad A = {0},f then we get A ∼ =A

• ρ(x) = kb xk∞ = sup{|b x(h)| | h ∈ ∆}. By Theorem 18.17(e), we have ρ(x) ≥ |h(x)| = |b x(h)| for every h ∈ ∆ which means that ρ(x) ≥ kb xk∞ . For the other direction, λ belongs to the range of x b means that λ = x b(h) = h(x) for some h ∈ ∆ and it follows from Theorem 18.17(b) that this happens if and only if λ ∈ σ(x), so Definition 18.8 establishes ρ(x) = sup{|λ| | λ ∈ σ(x)} ≤ sup{|b x(h)| | h ∈ ∆} = kxk∞ . • The range of x b is σ(x). The analysis in the previous part also implies that the range of the function x b is exactly the spectrum σ(x).

We have completed the proof of the problem.



Problem 18.15 Rudin Chapter 18 Exercise 15.

Proof. Let A1 = {(x, λ) | x ∈ A and λ ∈ C} and k(x, λ)k = kxk + |λ|. For any (x, λ), (y, µ) ∈ A1 , we define the multiplication in A1 by (x, λ)(y, µ) = (xy + µx + λy, λµ). f

In this case, A is called semisimple.

(18.31)

562

Chapter 18. Elementary Theory of Banach Algebras • A1 is a commutative Banach algebra with unit. It is easily checked that this is associative and distributive. Thus A1 is a complex algebra. Furthermore, the element (0, 1) is a unit for this multiplication because (x, λ)(0, 1) = (x · 0 + 1 · x + λ · 0, λ · 1) = (x, λ) = (0, 1)(x, λ). For any (x, λ), (y, µ) ∈ A1 , we see that k(x, λ) + (y, µ)k = k(x + y, λ + µ)k

= kx + yk + |λ + µ|

≤ kxk + kyk + |λ| + |µ|

= k(x, λ)k + k(y, µ)k. If α ∈ C, then we have

kα(x, λ)k = k(αx, αλ)k = kαxk + |αλ| = |α| · kxk + |α| · |λ| = |α| · k(x, λ)k. As k(x, λ)k = 0 if and only if kxk + |λ| = 0 if and only if kxk = 0 if and only if x = 0 and λ = 0, A1 is a normed linear space by Definition 5.2. Since kxyk ≤ kxk · kyk, we see that k(x, λ)(y, µ)k = k(xy + µx + λy, λµ)k

= kxy + µx + λyk + |λµ|

≤ kxyk + kµxk + kλyk + |λ| · |µ|

≤ kxkkyk + |µ| · kxk + |λ| · kyk + |λ| · |µ|

= kxk + |λ| · kyk + |µ|

= k(x, λ)k · k(y, µ)k.

Since the spaces A and C are complete, A1 is obviously complete and then it is a Banach algebra with unit by Definition 18.1. It is commutative because A and C are commutative so that (y, µ)(x, λ) equals to the right-hand side of the expression (18.31). • The mapping x 7→ (x, 0) is an isometric isomorphism of A onto a maximal ideal of A1 . Let Φ be this mapping. It is trivial surjective. If Φ(x) = Φ(y), then (x, 0) = (y, 0) which means that (x − y, 0) = 0. Since k(x − y, 0)k = 0, we get x = y and thus Φ is injective. Is is easily checked that Φ satisfies Φ(x + y) = Φ(x) + Φ(y), so Φ is an isomorphism onto Φ(A). It is also isometric because kΦ(x) − Φ(y)k = kΦ(x − y)k = k(x − y, 0)k = kx − yk. Now we may identify A with Φ(A) ⊆ A1 . Since (x, λ)(y, 0) = (xy + λy, 0) ∈ Φ(A) for every (x, λ) ∈ A1 and (y, 0) ∈ Φ(A), Φ(A) is an ideal of A1 by Definition 18.12. Since A1 = A ⊕ C, we have A1 /Φ(A) ∼ = A1 /A ∼ = C which implies that Φ(A) is a maximal ideal of A1 as required. This completes the proof of the problem.



18.3. The Commutative Banach algebra H ∞

18.3

563

The Commutative Banach algebra H ∞

Problem 18.16 Rudin Chapter 18 Exercise 16.

Proof. It is clear that H ∞ is a commutative complex algebra. Recall from §11.31 that its norm is defined by kf k∞ = sup{|f (z)| | z ∈ U }.

This norm makes H ∞ satisfy Definition 5.2. Thus H ∞ is a normed linear space, so it is a normed complex algebra. The fact that H ∞ is complete has been shown in [100, Remark 17.8(c), p. 338], so H ∞ is a commutative Banach algebra by Definition 18.1. The element 1 ∈ H ∞ is easily seen to be its unit which gives the first assertion. Suppose that |α| < 1. Define Φα : H ∞ → C by Φα (f ) = f (α).

(18.32)

Then it satisfies Φα (f g) = f (α)g(α) = Φα (f )Φα (g). For every constant a, the function f (z) = a gives Φα (f ) = a so it is surjective. In other words, Φα ∈ ∆. To see that there are complex homomorphisms of H ∞ other than the point homomorphisms (18.32), we let I be the set of functions f ∈ H ∞ such that f (α) → 0 as α → 1 and α > 0. Then it is easy to see that I is a proper ideal of H ∞ . By Theorem 18.13, I is contained in a maximal ideal J of H ∞ which means that there exists a complex homomorphism of H ∞ , say ϕ ∈ ∆ such that ϕ(f ) = 0 for all f ∈ I by Theorem 18.17(a). However, ϕ 6= Φα for all α ∈ U because there is no α such that  Φα (f ) = f (α) = 0 for every f ∈ I. We have finished the proof of this problem. Problem 18.17 Rudin Chapter 18 Exercise 17.

Proof. By Problem 18.16, H ∞ is a commutative Banach algebra. Let I = {(z − 1)2 f | f ∈ H ∞ }. For any f, g ∈ H ∞ , we know that f g ∈ H ∞ . Thus if (z − 1)2 f ∈ I and g ∈ H ∞ , then we have g · (z − 1)2 f = (z − 1)2 f g ∈ I. By Definition 18.12, I is an ideal of H ∞ . Given ǫ > 0. The function fǫ (z) = (1 + ǫ − z)−1 belongs to H ∞ because |1 + ǫ − z| ≥ |1 + ǫ| − |z| > ǫ so that |fǫ (z)| < ǫ−1 for all z ∈ U . We observe that

(1 − z)2 (1 + ǫ − z)−1 − (1 − z) = ǫ(1 − z) < ǫ 1−z+ǫ

for all z ∈ U . This means that (z − 1)2 fǫ (z) converges uniformly to 1 − z in U . However, we know that 1− z ∈ / H ∞ which means that I is not closed. This ends the proof of the problem.  Problem 18.18 Rudin Chapter 18 Exercise 18.

564

Chapter 18. Elementary Theory of Banach Algebras

Proof. Denote I = {ϕf | f ∈ H ∞ }. Of course, we have ϕf ∈ H ∞ . Since f g ∈ H ∞ for any f, g ∈ H ∞ , the space I is an ideal of H ∞ by Definition 18.12. Let {fn } be a sequence in H ∞ such that kϕfn − gk∞ → 0

as n → ∞, where g ∈ H ∞ . Thus {ϕfn } is a Cauchy sequence by [99, Theorem 3.11(a), p. 53] and hence so is {fn }. Observing from Remark 17.8(c) that H ∞ is Banach, so it is complete. Then we have fn → f ∈ H ∞ as n → ∞ and this means that ϕf ∈ I. Hence we complete the proof of the problem. 

CHAPTER

19

Holomorphic Fourier Transforms

19.1

Problems on Entire Functions of Exponential Type

Problem 19.1 Rudin Chapter 19 Exercise 1.

Proof. By the hypothesis, we know that there exist some constants A and C such that |f (z)| ≤ CeA|z| for all z ∈ C. Suppose for simplicity that ϕ(0) < ∞, i.e., Z ∞ |f (x)|2 dx < ∞. −∞

By Theorem 19.3 (The Paley and Wiener Theorem), there exists an F ∈ L2 (−A, A) such that Z A f (z) = F (t)eitz dt −A

for all z ∈ C. Define F1 and F2 by   F (t), if 0 ≤ t < A; F1 (t) =  0, if t ≥ A

and F2 (t) =

Then we may express f as

f (z) = f1 (z) + f2 (z) =

Z

∞

  F (t), if −A < t ≤ 0; 

F1 (t)eitz dt +

0,

Z

0

if t ≤ −A.

F2 (t)eitz dt.

(19.1)

−∞

0

By the definition, it is clear that F1 ∈ L2 (0, ∞), so we know from [100, Eqn. (3), p. 372] that Z ∞ Z ∞ 1 |F1 (t)|2 dt < ∞ (19.2) |f1 (x + iy)|2 dx ≤ 2π −∞ 0 for every y > 0. For the second integral of the equation (19.1), we write Z ∞ Z 0 f2 (t)e−itz dt, F F2 (t)eitz dt = f2 (z) = 0

−∞

565

566

Chapter 19. Holomorphic Fourier Transforms

f2 (t) = F2 (−t). Since F2 ∈ L2 (−∞, 0), we have F f2 ∈ L2 (0, ∞). By similar argument as where F in [100, pp. 371, 372], we can show that f2 is holomorphic in the lower half plane Π− and if we write Z ∞   f2 (t)ety · e−itx dt, F f2 (x + iy) = 0

regard y as fixed, then Theorem 9.13 (The Plancherel Theorem) implies that Z ∞ Z ∞ Z ∞ 2 2 1 f f F2 (t) dt < ∞ F2 (t) e2ty dt ≤ |f (x + iy)|2 dx = 2π −∞ 0 0

(19.3)

for every y < 0. Now we substitute the estimates (19.2) and (19.3) into the expression (19.1), we see from Theorem 3.8 that Z ∞ 1 1 ϕ(y) = |f (x + iy)|2 dx 2π 2π −∞ Z ∞ Z 1 ∞ 1 |f1 (x + iy)|2 dx + |f1 (x + iy)| · |f2 (x + iy)| dx ≤ 2π −∞ π −∞ Z ∞ 1 |f2 (x + iy)|2 dx + 2π −∞ Z Z ∞ o1 n Z ∞ o1 1n ∞ 2 2 |f1 (x + iy)|2 dx · |F1 (t)|2 dt + ≤ |f2 (x + iy)|2 dx π −∞ 0 −∞ Z ∞ 2 f F2 (t) dt + 0



2 = kF1 k22 + 2kF1 k2 · f F2 2 + f F2 2


2 f2 = F1 k2 + kF 2 R. For |z| ≤ R, we let M = CeAR so that |f (z)| ≤ M exp(|z|α ) for all z ∈ ∆. Thus Problem 12.9 reveals that |f (z)| is bounded in ∆ which implies that f is a bounded entire function. Hence it follows from Theorem 10.23 (Liouville’s Theorem) that f is  constant, as desired. This ends the proof of the problem. Problem 19.4 Rudin Chapter 19 Exercise 4.

Proof. Now f is an entire function of exponential type. • The series converges if |w| > A. By the hypothesis, it is true that |f (z)| < exp(|z|λ ) for all large enough |z|, where λ > 1. According to Problem 15.2, f is of order 1. We note from [16, Eqns. (2.1.6) & (2.2.12), pp. 8, 12] that 1

lim sup |f (n) (0)| n = A. n→∞

a

Notice that f (eiθ z) is also an entire function satisfying the inequality (19.5).

568

Chapter 19. Holomorphic Fourier Transforms This means that

1

lim sup |n!an | n = A n→∞

and so the power series Φ converges if 1 1 1 < , 1 = |w| A lim sup |n!an | n n→∞

i.e., |w| > A.b • The function f can be expressed as an integral. By the power series of Φ, we see that Z Z X ∞ ∞ n!an   X z k wk  1 1 dw Φ(w)ewz dw = × 2πi Γ 2πi Γ n=0 wn+1 k! k=0 Z X ∞ am z m  1 dw = 2πi Γ w m=0 Z ∞ 1 X dw m = am z 2πi Γ w m=0

=

∞ X

am z m

m=0

= f (z), where the term-by-term integration being justified by the uniform convergence of the series on Γ.c • Φ is the function which occurred in the proof of Theorem 19.3. Recall from Remark 19.4 that the functions Φα are restrictions of a function holomorphic in the complement of the interval [−iA, iA]. Thus it suffices to prove that Φ is such function. Our first assertion and Theorem 10.6 ensure that the Borel transform Φ is holomorphic in the complement of [−iA, iA]. It remains to show that Φ|Πα = Φα for every real α.d If Re (weiα ) > 3A, then we have
| exp(−wseiα )| = exp − sRe (weiα ) < e−3As . Furthermore, if we define Mf (r) =

(19.6)

(19.7)

max |f (z)|, then we follow from Theorem 10.26

z∈D(0;r)

(Cauchy’s Estimates) that |an | ≤ and the remainder Rn (s) =

Mf (r) rn ∞ X

ak s k

k=n+1 b

The function Φ(w) in question is called the Borel transform of the function f (z), see [16, §5.3, p. 73] or [63, §20, p. 84]. c The integral is sometimes called the P´ olya representation of the function f . d Indeed, the half plane Πα is given by {w = x + iy | x cos α − y sin α > A}.

19.1. Problems on Entire Functions of Exponential Type

569

of the power series of f satisfies |Rn (s)| ≤ Mf (r)

∞   X Mf (r)  s n+1 s k = . · r 1 − rs r

k=n+1

By putting r = 2s, we get e2s 2n Therefore, we deduce from the inequalities (19.7) and (19.8) that the series |Rn (s)| ≤

∞ X

(19.8)

an e−wz z n

n=0

converges uniformly on the ray Γα = {seiα | s ≥ 0}. Consequently, for every w ∈ Πα , we obtain Z Z ∞ ∞ X X n!an an z n e−wz dz = e−wz f (z) dz = Φα (ω) = = Φ(w) n+1 w Γα Γα n=0

n=0

which is exactly the expression (19.6). We have completed the proof of the problem.



Problem 19.5 Rudin Chapter 19 Exercise 5.

Proof. Suppose that f ∈ H(Π+ ) and 1 sup 2π 0 0. Notice that f (ξ + iǫ) |f (ξ + iǫ)| ≤ ξ + iǫ − z |ξ − z|

for every ξ ∈ (−∞, ∞) and 0 < ǫ < y. Thus it follows from Theorem 3.8 that Z ∞ Z ∞ |f (ξ + iǫ)| f (ξ + iǫ) dξ dξ ≤ ξ + iǫ − z |ξ − x| −∞ −∞ Z o1 n Z ∞ n ∞ dξ o 12 2 |f (ξ + iǫ)|2 dξ ≤ × 2 −∞ −∞ (ξ − x) Z 1 n ∞ √ dξ o 2 ≤ 2Cπ · 2 −∞ (x − ξ) < ∞,

so Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) ensures that Z ∞ ∗ f (ξ) 1 dξ. f (z) = 2πi −∞ ξ − z We complete the proof of the problem.



Problem 19.6 Rudin Chapter 19 Exercise 6.

Proof. Here we follow mainly [85, Theorem XII, pp. 16 – 20]. Since 0 < ϕ < eϕ , we have log ϕ < ϕ. Combining the hypothesis and Theorem 3.5 (H¨older’s Inequality), we obtain Z ∞ Z ∞ nZ ∞ o1 n Z ∞ ϕ(x) dx o 21 log ϕ(x) 2 2 × < ∞. dx < dx ≤ |ϕ(x)| dx −∞ < 2 2 2 2 −∞ 1 + x −∞ (1 + x ) −∞ −∞ 1 + x

572

Chapter 19. Holomorphic Fourier Transforms

In other words, we have

Z

∞ −∞

| log ϕ(x)| dx < ∞. 1 + x2

We write z = x + iy with y > 0 and consider Z y 1 ∞ log ϕ(t) dt. u(z) = π −∞ (x − t)2 + y 2

(19.17)

Using the half-plane version of Fatou’s Theorem [91, Theorem 5.5, pp. 86, 87], we see that the function (19.17) is harmonic in Π+ and lim u(x + iy) = log ϕ(x)

y→0

or

lim |f (x + iy)| = ϕ(x)

y→0

(19.18)

holds for almost all x ∈ R.e Let v(z) be its harmonic conjugate and write f (z) = exp(u(z) + iv(z)). Because of [100, Eqn. (7), p. 63], we see that |f (x + iy)| = eu(z) ≤

1 π

Z

∞ −∞

ϕ(t)y dt, (x − t)2 + y 2

it follows from Theorem 3.5 (H¨older’s Inequality) that Z Z ∞ ϕ(t)y ϕ(s)y 1 ∞ × dt ds |f (x + iy)|2 ≤ 2 2 2 2 2 π −∞ (x − t) + y −∞ (x − s) + y Z Z o1n ∞ o1 1n ∞ |ϕ(t)|2 y y 2 2 ≤ 2 dt dt 2 + y2 2 + y2 π (x − t) (x − t) −∞ −∞ nZ ∞ o 1 nZ ∞ o1 |ϕ(s)|2 y y 2 2 ds ds × 2 2 2 2 −∞ (x − s) + y −∞ (x − s) + y Z |ϕ(t)|2 y 1 ∞ dt = π −∞ (x − t)2 + y 2 which implies Z

∞ −∞

Z Z 1 ∞ ∞ |ϕ(t)|2 y dt dx π −∞ −∞ (x − t)2 + y 2 Z ∞ Z ∞ y 1 dx |ϕ(t)|2 dt · = 2 2 π −∞ −∞ (x − t) + y Z ∞ |ϕ(t)|2 dt. =

|f (x + iy)|2 dx ≤

−∞

In other words, we have established sup y>0

Z

∞ −∞

|f (x + iy)|2 dx < ∞.

According to Theorem 19.2 (The Paley-Wiener Theorem), there exists an F ∈ L2 (−∞, ∞) vanishing on (−∞, 0) such that Z ∞ F (t)eitz dt f (z) = −∞

e

Here the function

y (x − t)2 + y 2

is the Poisson kernel in the upper half plane Π+ . See also [9, pp. 145 – 147; Theorem 7.28, pp. 160, 161].

19.1. Problems on Entire Functions of Exponential Type for all z ∈ Π+ . In particular, we have lim f (x + iy) = f (x) =

y→0

Z

∞

itx

F (t)e

dt =

Z

∞

−∞

0

F (t)eitx dt = Fb(−x)

573

(19.19)

for x ∈ R. If we denote G(x) = Fb (−x), then we combine the results (19.18) and (19.19) to get |G(x)| = ϕ(x).

b Since G ∈ we derive from Lemma 9.3 that G(x) = F (−x) which vanishes on [0, ∞). b Conversely, we suppose that there exists an f with f = ϕ such that f (x) = 0 for all x ≤ 0. Let us write Z ∞ 1 fb(x) = √ f (t)e−ixt dt (19.20) 2π −∞ and Z ∞ 1 f (t)e−izt dt, (19.21) ψ(z) = √ 2π −∞ L2 (−∞, ∞),

where z ∈ Π+ and the integral in (19.21) is taken along a horizontal line in the z-plane. Certainly, we have ψ ∈ H(Π+ ) by §19.1. Suppose that we map Π+ (conformally) onto U by z = i ζ+1 ζ−1 . Write  ζ + 1 k(ζ) = ψ(z) = ψ i and K(eiθ ) = fb(x), ζ −1 iθ

+1 . Then it is easily seen from Theorem 12.12 (The where ζ = reiθ and 0 ≤ r < 1 and x = i eeiθ −1 Hausdorff-Young Theorem) that Z ∞ Z π Z ∞ b 2

2 |f (x)| dx ≤ 2 |fb(x)|2 dx = 2 fb 2 ≤ 2kf k22 = 2kϕk22 < ∞. |K(eiθ )|2 dθ = 2 2 −∞ −π −∞ 1 + x

Therefore, we have K ∈ L2 (T ). On the other hand, if z = x + iy, then the integral (19.20) implies that Z π Z 1 y 1 ∞ fb(t) dt K(eiθ )Pr (θ − φ) dθ = 2π −π π −∞ (x − t)2 + y 2 Z Z ∞ y 1 1 ∞ √ × f (ξ)e−itξ dξ dt = π −∞ (x − t)2 + y 2 2π 0 Z ∞  1 Z ∞ e−itξ y 1 dt dξ f (ξ) × =√ π −∞ (x − t)2 + y 2 2π 0 Z ∞ 1 =√ f (ξ)e−ixξ+yξ dξ 2π 0 Z ∞ 1 f (ξ)e−izξ dξ =√ 2π 0 = ψ(z) = k(reiθ ).

By Definition 11.6, k is the Poisson integral of K, i.e., k = P [K]. Since K ∈ L2 (T ) and k = P [K], it follows from Theorem 11.16 that Z π Z π Z π |K(reiθ )|2 dθ. (19.22) |k(reiθ )|2 dθ ≤ log+ |k(reiθ )| dθ ≤ −π

−π

−π

If k(0) 6= 0, then we apply Theorem 15.18 (Jensen’s Formula) to obtain Z π 1 log |k(0)| ≤ log |k(reiθ )| dθ. 2π −π

(19.23)

574

Chapter 19. Holomorphic Fourier Transforms

Now the formula Z π Z π Z π 1 1 1 iθ + iθ log |k(re )| dθ = log |k(re )| dθ + log− |k(reiθ )| dθ 2π −π 2π −π 2π −π clearly implies Z π Z π Z π 1 1 log |k(reiθ )| dθ = 1 log+ |k(reiθ )| dθ − log− |k(reiθ )| dθ 2π −π 2π −π 2π −π Z Z π 1 1 π log+ |k(reiθ )| dθ − log |k(reiθ )| dθ. = π −π 2π −π Substituting the inequalities (19.22) and (19.23) into the formula (19.24), we obtain Z π Z π 1 log |k(reiθ )| dθ ≤ 1 |K(reiθ )|2 dθ − log |k(0)| 2π −π π −π

(19.24)

(19.25)

for all 0 ≤ r < 1. If k has zero at 0 with multiplicity m, then the inequality (19.25) becomes Z π Z π k(ζ) 1 log |k(reiθ )| dθ ≤ 1 − m log r. (19.26) |K(reiθ )|2 dθ − log m 2π −π π −π ζ ζ=0

for all 0 ≤ r < 1. By the definition, log |k(reiθ )| → log |K(eiθ )| as r → 1 almost everywhere, so we conclude from the inequality (19.26) that Z Z Z π 1 ∞ log |fb(x)| 1 1 ∞ | log ϕ(x)| log |K(eiθ )| dθ < ∞. dx = dx = 2 2 π −∞ 1 + x π −∞ 1 + x 2π −π

Consequently, this ensures that

Z

∞

−∞

log ϕ(x)

dx > −∞. 1 + x2

Hence we have completed the proof of the problem.

19.2



Quasi-analytic Classes and Borel’s Theorem

Problem 19.7 Rudin Chapter 19 Exercise 7.

Proof. It is trivial that Condition (a) implies Condition (b). Conversely, suppose that Condition (b) holds. For each α ∈ E, we fix a neighborhood Vα of α and put [ Ω= Vα . α∈E

It is clear that E ⊂ Ω and Ω is an open set. Now we define F : Ω → C as follows: Given z ∈ Ω. Then we have z ∈ Vα for some Vα and we define F (z) = Fα (z).

(19.27)

We claim that F ∈ H(Ω) and F (z) = f (z) for z ∈ E. If z ∈ Vβ for β 6= α, then we have Vα ∩ Vβ 6= ∅. Since Vα ∩ Vβ is an open set, there exists a δ > 0 such that z ∈ D(0; δ) ⊆ Vα ∩ Vβ , so Theorem 10.18 says that Fα ≡ Fβ in Vα ∩ Vβ and thus the formula (19.27) is well-defined.

19.2. Quasi-analytic Classes and Borel’s Theorem

575

Furthermore, it is clear that F is holomorphic at every point of Ω. Finally, if ζ ∈ E, then ζ ∈ Vζ . Since Fζ (z) = f (z) for all z ∈ Vζ ∩ E, we must have F (ζ) = Fζ (ζ) = f (ζ). This ends the proof of the problem.



Problem 19.8 Rudin Chapter 19 Exercise 8.

Proof. Since n! ≤ nn for every n ≥ 1, we have kD n f k∞ ≤ βf Bfn n! ≤ βf Bfn nn . By Definition 19.6, we have C{n!} ⊆ C{nn }. Recall the approximation to Stirling’s formula [99, Exercise 20, p. 200] that 1 nn ∼√ en n! 2πn for large n. Thus there exists a constant M > 0 such that nn ≤ M en n! for all n ≥ 1. If f ∈ C{nn }, then we have kD n f k∞ ≤ βf Bfn nn ≤ (M βf )(eBf )n n! which means f ∈ C{n!}. Consequently, we obtain the desired result that C{n!} = C{nn }, completing the proof of the problem.  Problem 19.9 Rudin Chapter 19 Exercise 9.

Proof. Let M0 = 1, M1 = 1, M2 = 2 and Mn = n!(log n)n for every n ≥ 3. Thus we always have Mn2 ≤ Mn−1 Mn+1 for all n = 1, 2, . . .. Using Theorem 19.11 (The Denjoy-Carleman Theorem), we know that C{Mn } is quasi-analytic. If f ∈ C{n!}, then there exist positive constants βf and Bf such that kD n f k∞ ≤ βf Bfn n! for all n ≥ 0. By the definition of {Mn }, we also have kD n f k∞ ≤ βf Bfn Mn for every n ≥ 0. Thus we have

C{n!} ⊆ C{Mn }.

The construction of an example f belonging to C{Mn }, but f ∈ / C{n!} is basically motivated for every n = 0, 1, 2, . . .. Since by [114, Theorem 1, p. 4]. Put mn = MMn+1 n mn − mn−1 =

Mn+1 Mn Mn+1 Mn−1 − Mn2 − = ≥0 Mn Mn−1 Mn Mn−1

for every n ≥ 0. Thus {mn } is a positive increasing sequence. It is clear that f ∈ C ∞ . For every n, k ∈ N, if k ≤ n, then we see that 1 1 1 1 × × ··· × = mn mn mn mnn−k

576

Chapter 19. Holomorphic Fourier Transforms 1 1 1 × × ··· × mn−1 mn−2 mk Mk Mn−1 Mn−2 × × ··· × = Mn Mn−1 Mk+1 Mk = . Mn ≤

If k > n, then we have 1 mnn−k

= mnk−n ≤ mn × mn+1 × · · · × mk−1 Mn+1 Mn+2 Mk = × × ··· × Mn Mn+1 Mk−1 Mk = . Mn

In other words, we obtain the estimate 1 mnn−k

≤

Mk . Mn

(19.28)

Define fn (x) =

Mn 2mn ix e (2mn )n

and

f (x) =

∞ X

fn (x)

n=0

for x ∈ R. We first show that f ∈ C{Mn }. For every k ≥ 1, we have fn(k) (x) =

ik Mn e2mn ix . (2mn )n−k

Combining this and the estimate (19.28), we get |f (k) (x)| ≤

∞ X

n=0

|fn(k) (x)| =

∞ X

n=0

∞

∞

n=0

n=0

X Mn Mk X 1 Mn ≤ · = Mk · ≤ 2 · 2k Mk (19.29) n−k n−k (2mn ) 2 Mn 2n−k

for every k ≥ 1 and x ∈ R. Furthermore, we also have |f (x)| ≤

∞ X

n=0

By induction, we can show that

|fn (x)| =

Mnn+1 n Mn+1

∞ X

∞

X 1 M n+1 Mn = · n . n n Mn (2m ) 2 n n+1 n=0 n=0

≤ 1 for each n ≥ 0. Therefore, the estimate (19.30) gives

|f (x)| ≤ 2 = 2 · 20 M0 for every x ∈ R. Now we conclude from the estimates (19.29) and (19.30) that kD k f k∞ ≤ 2 · 2k Mk holds for every k ≥ 0 so that f ∈ C{Mn } as required.

Next, we want to show that |f (k) (0)| ≥ Mk for every k ≥ 0. It is obvious that f (0) =

∞ X

(19.30)

Mn M1 = M0 + + · · · ≥ 1 = M0 . n (2m ) 2m n 1 n=0

19.2. Quasi-analytic Classes and Borel’s Theorem In addition, if k ≥ 1, then since every term

Mn (2mn )n−k

|f (k) (0)| = |ik |

∞ X

n=0

577

is positive for every n = 0, 1, 2, . . ., we have

Mn ≥ Mk . (2mn )n−k

(19.31)

Hence we have obtained what we want. If f ∈ C{n!}, then it must be true that |f (k) (0)| ≤ βf Bfk k! for some positive constants βf and Bf . However, if k is sufficiently large so that (log k)k ≥ βf Bfk , then this will certainly contradict the estimate (19.31). Hence we conclude that f ∈ / C{n!} and then we complete the proof of the problem.  Problem 19.10 Rudin Chapter 19 Exercise 10.

Proof. Suppose that λ =

∞ X

λn is positive finite. Recall from Definition 2.9 that Cc (R) is the

n=1

collection of all continuous complex functions on R whose support is compact. Now we let g0 to be the function modified from the (19.38) in such the way that g0 (x) = 1 for −λ ≤ x ≤ λ, g0 (x) = 0 for |x| ≥ 2λ and 0 ≤ g0 (x) ≤ 1 if x ∈ [−2λ, −λ] ∪ [λ, 2λ]. Then g0 ∈ Cc (R) and g0 is integrable in R. Write
gn (x) = g λ1 , λ2 , . . . , λn ; g0 (x) Z λ1 Z λn Z λ2 1 dt1 g0 (x − t1 − t2 − · · · − tn ) dtn . (19.32) dt2 · · · = n 2 λ1 λ2 · · · λn −λ1 −λn −λ2 Since |g0 (x)| ≤ 1 for every x ∈ R, the definition (19.32) ensures that |gn (x)| ≤ 1 for every n = 0, 1, 2, . . . and x ∈ R. In other words, the family {gn } is (uniform) bounded in R. Besides, if |x| ≥ 3λ, then |x − λ1 − λ2 − · · · − λn | ≥ 2λ so that gn (x) = 0 there. Obviously, we have


g λ1 , λ2 , . . . , λn ; g0 (x) = g λ1 , λ2 , . . . , λk ; g λk+1 , λk+2 , . . . , λn ; g0 (x) . (19.33)

By the definition (19.32) again, we see that g1 (x) =

1 2λ1

Z

λ1

−λ1

g0 (x − t1 ) dt1 =

1 2λ1

Z

x+λ1

g0 (t) dt,

x−λ1

so the Fundamental Theorem of Calculus yields that g1 (x) is differentiable in R and g1′ (x) =

1 [g0 (x + λ1 ) − g0 (x − λ1 )]. 2λ1

Next, we assume that the function gn−1 (x) is continuous for any n ≥
2 in R and if n ≥ 2, then

the function gn (x) = g λn ; g(λ1 , λ2 , . . . , λn−1 ; g0 (x) = g λn ; gn−1 (x) is differentiable in R and it follows from the formula (19.33) that gn′ (x) =


d g λ1 , λ2 , . . . , λn ; g0 (x) dx

578

Chapter 19. Holomorphic Fourier Transforms

d g λ1 ; g λ2 , λ3 , . . . , λn ; g0 (x) dx

 1  = g λ2 , λ3 , . . . , λn ; g0 (x + λ1 ) − g λ2 , λ3 , . . . , λn ; g0 (x − λ1 ) 2λ1
1 = g λ2 , λ3 , . . . , λn ; g0 (x + λ1 ) − g0 (x − λ1 ) . 2λ1 =

(19.34)

This also implies that gn (x) has continuous derivatives of order 0, 1, . . . , n−1 in R. Furthermore, if we combine the formula (19.34) and the Mean Value Theorem for Derivatives, we get, for every x ∈ R, that
′ |gn′ (x)| ≤ max |g′ λ2 , λ3 , . . . , λn ; g0 (x) | = max |gn−1 (x)| ≤ · · · ≤ max |g2′ (x)| < ∞. (19.35) x∈R

x∈R

x∈R

For every n ≥ 2 and any x, y ∈ R, the bound (19.35) asserts that

|gn (x) − gn (y)| = |x − y| · |gn′ (ξ)| ≤ |x − y| · max |g2′ (x)| x∈R

which shows that the family {gn } is equicontinuous on R. Recall that gn (x) = 0 outside [−3λ, 3λ], so {gn } is actually equicontinuous on [−3λ, 3λ]. The fact |gn (x)| ≤ 1 in R guarantees that {gn } converges pointwise on R. Hence it asserts from [99, Exercise 16, p. 168] that {gn } converges uniformly to a continuous function g on [−3λ, 3λ], i.e., g(x) = lim gn (x) n→∞

for every x ∈ [−3λ, 3λ]. Since the argument also applies to any compact interval of R, the function g is also continuous at the end points ±3λ. It is trivial that if x ∈ / [−3λ, 3λ], then g(x) = 0. Hence we must have g(±3λ) = 0.
Denote g(x) = g λ1 , λ2 , . . . ; g0 (x) . Then we know that
g(x) = lim g λ1 , λ2 , . . . , λn ; g0 (x) n→∞ Z λ1
1 g λ2 , λ3 , . . . , λn ; g0 (x − t) dt = lim n→∞ 2λ1 −λ 1 Z λ1
1 = g λ2 , λ3 , . . . ; g0 (x − t) dt 2λ1 −λ1 is true for all x ∈ [−3λ, 3λ] which means that



 1  g λ2 , λ3 , . . . ; g0 (x + λ1 ) − g λ2 , λ3 , . . . ; g0 (x − λ1 ) 2λ1  g0 (x + λ1 ) − g0 (x − λ1 )  = g λ2 , λ3 , . . . ; 2λ1

g′ (x) =

(19.36)

holds in [−3λ, 3λ]. Using similar reasoning as the previous paragraph, it can be shown that g is also differentiable at the end points ±3λ and the formula (19.36) holds in R. In conclusion, we have g ∈ C ∞ . Since |λ1 + λ2 + · · · + λn | < λ, we obtain g0 (−t1 − t2 − · · · − tn ) = 1 for all −λk ≤ tk ≤ λk , where 1 ≤ k ≤ n. Thus we note that Z λ2 Z λn Z λ1 1 dtn = 1, dt2 · · · dt1 g(0) = lim n n→∞ 2 λ1 λ2 · · · λn −λ −λn −λ2 1 so g is not identically zero in R.

19.2. Quasi-analytic Classes and Borel’s Theorem

579

Put G0 (x) = g0 (x) G0 (x + λ1 ) − G0 (x − λ1 ) G1 (x) = , 2λ1 .. . Gn−1 (x + λn ) − Gn−1 (x − λn ) . Gn (x) = 2λn Thus the formula (19.36) can be written as

In fact, it is true that


g′ (x) = g λ2 , λ3 , . . . ; G1 (x) .


g (n) (x) = g λn+1 , λn+2 , . . . ; Gn (x)

(19.37)

for every n = 0, 1, 2, . . . and x ∈ R. Recall that |g0 (x)| ≤ 1 on R, so we have |Gn (x)| ≤

1 = Mn . λ1 λ2 · · · λn

on R. Thus it follows from the formula (19.37) that

|g(n) (x)| ≤ Mn for all n = 0, 1, 2, . . . and x ∈ R. Consequently, g ∈ C{Mn } which completes the proof of the problem.  Remark 19.1 The construction in Problem 19.10 follows basically the unpublished work of H. E. Bray which was quited in Mandelbrojt’s article [68, pp. 79 – 84]. See also [55].

Problem 19.11 Rudin Chapter 19 Exercise 11.

Proof. An example of a function ϕ ∈ C ∞ with the required properties can be found in [124, Problem 10.6, pp. 266, 267]. In fact, we start with  −1  e x , if x > 0; g(x) =  0, if x ≤ 0. It is known that g ∈ C ∞ and g(m) (0) = 0 for all m = 1, 2, . . .. Define ϕ : R → R by ϕ(x) =

g(2 − |x|) . g(2 − |x|) + g(|x| − 1)

(19.38)

Now it is easy to see that ϕ ∈ C ∞ . Furthermore, we have ϕ(x) = 1 for −1 ≤ x ≤ 1, ϕ(x) = 0 for |x| ≥ 2 and 0 ≤ ϕ(x) ≤ 1 if x ∈ [−2, −1] ∪ [1, 2] so that supp ϕ ⊆ [−2, 2]. This completes the proof of the problem.



580

Chapter 19. Holomorphic Fourier Transforms

Problem 19.12 Rudin Chapter 19 Exercise 12.

Proof. Let ϕ be as in Problem 19.11. Set β = fn (x) =

αn n!

and gn (x) = βn xn ϕ(x). Take

gn (λn x) = βn xn ϕ(λn x), λnn

where λn is large enough. Fix the non-negative integer n, we notice that (D k fn )(x) = βn

k X

n! C k λk−m xn−m ϕ(m) (λn x), (n − m)! m n m=0

(19.39)

where k = 0, 1, 2, . . . , n − 1. Recall from the definition of ϕ in Problem 19.11 that supp ϕ(m) ⊆ supp ϕ ⊆ [−2, 2] holds for every m = 0, 1, . . . , k. If λn x ∈ / [−2, 2], then ϕ(m) (λn x) = 0 so that (D k f )(x) = 0. If 2 (m) is continuous on [−2, 2], there is a positive constant M λn x ∈ [−2, 2], then |x| ≤ λn . Since ϕ such that |ϕ(m) (λn x)| ≤ M . Thus we obtain |(D k fn )(x)| ≤ |βn |M ≤ |βn |M ≤

k X

n−m n! k k−m 2 Cm λn · n−m (n − m)! λn m=0 k X 2n (n!)2 λnn−k m=0

n2n (n!)2 |βn |M λn

(19.40)

for all x ∈ R and k = 0, 1, . . . , n − 1. Since λn can be chosen large enough, we observe from the estimate (19.40) that 1 (19.41) kD k fn k∞ < n 2 for all k = 0, 1, . . . , n − 1. Take f = f0 + f1 + · · · .

It is clear that f0 (0) + f1 (0) + · · · = α0 . Besides, the result (19.41) ensures that the series {f0′ + f1′ + · · · + fn′ } converges uniformly on R. Using [99, Theorem 7.17, p. 152], termwise differentiation is legitimate so that f ′ = f0′ + f1′ + · · · . Now this argument can be applied repeatedly to show that f ∈ C ∞ . Next, it follows from the expression (19.39) that (D k fn )(0) = 0 for k = 0, 1, . . . , n − 1. Since ϕ(x) = 1 on [−1, 1], ϕ(n) (0) = 0 for all n = 1, 2, . . . and this implies that (D n fm )(0) = 0 for m = 0, 1, . . . , n − 1. Hence we have   (D n f )(0) = (D n f0 )(0) + (D n f1 )(0) + · · · + (D n fn−1 )(0) + (D n fn )(0)  n  + (D fn+1 )(0) + (D n fn+2 )(0) + · · · = n!βn = αn for every n = 0, 1, 2, . . ., as required. This completes the proof of the problem.f f

Please also read [73] and [92].



19.2. Quasi-analytic Classes and Borel’s Theorem

581

Remark 19.2 Problem 19.12 is called Borel’s Theorem which says that every power series is the Taylor series of some smooth function, see, for examples, [79, Theorem 1.5.4, p. 30] and [86].

Problem 19.13 Rudin Chapter 19 Exercise 13.

Proof. It suffices to prove that lim sup n→∞

 |(D n f )(a)|  1

n

n!

= ∞.

(19.42)

We follow the suggestion. Let ck = λ1−k k , where the sequence {λk } satisfies ck λkk = λk > 2

k−1 X j=1

cj λkj = 2(λk1 + λ2k−1 + · · · + λ2k−1 ) and

λk > k2k > 1.

Fix the non-negative integer n, we have ∞ X k=1

ck λnk =

n+1 X

λkn+1−k +

k=1

∞ X

λkn+1−k
. 2 k=1 k6=n

k=1 k6=n

Combining this and Stirling’s formula, for large enough n, we get  |(D n f )(a)|  1

n

n!

>

 n2n  1

n

2n!

≈

 en n2n  1 en n √ = 2 2 2nn 2πn (8π) n · n n

which implies the result (19.42). Hence the power series ∞ X (D n f )(a)

n=0

n!

(x − a)n

has radius of convergence 0 for every a ∈ R, completing the proof of the problem.



582

Chapter 19. Holomorphic Fourier Transforms

Remark 19.3 Let S = {2n | n ∈ N}. Define

g(x) =

X

e−

√

k

cos(kx).

k∈S

Then it can be shown that g also satisfies the requirements of Problem 19.13.

Problem 19.14 Rudin Chapter 19 Exercise 14.

Proof. Suppose that f ∈ C{Mn } has infinitely many zeros {xn } in [0, 1]. Then {xn } has a convergent subsequence by the Bolzano-Weierstrass Theoremg . Without loss of generality, we may assume that {xn } is itself convergent, distinct, increasing and its limit is α. The continuity of f gives f ′ (α) = 0. By the Mean Value Theorem for Derivatives, we see that f ′ (ξn ) = 0 for some ξn ∈ (xn , xn+1 ) for all n = 1, 2, . . .. The fact xn → α as n → ∞ ensures that ξn → α as n → ∞. Since f ∈ C ∞ , the continuity of f ′ implies that f ′ (α) = lim f ′ (ξn ) = 0. n→∞

This argument can be repeated to show that f (n) (α) = 0 for all n = 0, 1, 2, . . .. Since C{Mn } is quasi-analytic, Definition 19.8 gives f (x) ≡ 0 for all x ∈ R, completing the proof of the  problem. Problem 19.15 Rudin Chapter 19 Exercise 15.

Proof. Suppose that X = {f ∈ H(C) | |f (z)| ≤ Ceπ|z| for some C > 0, f ∈ L2 ([−π, π])}. Recall from Definition 3.6 that the sequence space ℓ2 is given by ∞ X o n |f (n)|2 < ∞ . ℓ2 = {f (n)} n=−∞

Define the map Φ : X → ℓ2 by

Φ(f ) = {f (n)}.

(19.43)

For any α, β ∈ C and f, g ∈ X, it is clear that |f (z)| ≤ C1 eπ|z| and |g(z)| ≤ C2 eπ|z| for some positive constants C1 and C2 . Therefore, we have |αf (z) + βg(z)| ≤ (|α|C1 + |β|C2 )eπ|z| and kαf + βgk2 ≤ |α| · kf k2 + β · kgk2 < ∞ g

See [127, Problem 5.25, pp. 68, 69]

19.2. Quasi-analytic Classes and Borel’s Theorem

583

by Theorem 3.9. This means that αf + βg ∈ X. Besides, the fact ∞ X

n=−∞

|αf (n) + βg(n)|2 ≤ 2|α|2

∞ X

n=−∞

|f (n)|2 + 2|β|2

∞ X

n=−∞

|g(n)|2 < ∞

says that {αf (n) + βg(n)} ∈ ℓ2 . Thus the relation Φ(αf + βg) = {αf (n) + βg(n)} = α{f (n)} + β{g(n)} = αΦ(f ) + βΦ(g) holds, i.e, Φ is linear. For every f ∈ L2 (T ), we define F (z) =

1 2π

Z

π

f (t)e−izt dt.

−π

Using the analysis of §19.2, one can show that F is entire, |F (z)| ≤ Ceπ|z| for all z ∈ C and F ∈ L2 (−∞, ∞). In other words, this means that F ∈ X. Furthermore, we note that F = fb. Next, for each n ∈ Z, recall from Definition 4.23 that {un (t) = eint | n ∈ Z} forms an orthonormal set in L2 (T ). Furthermore, we have Z π ei(n−z)π − e−i(n−z)π sin[(z − n)π] 1 1 × = (19.44) ei(n−z)t dt = Un (z) = u bn (z) = 2π −π 2π i(n − z) (z − n)π holds for every n ∈ Z. Recall from [100, Example 4.5(b), p. 78] that Z π 1 f (t)g(t) dt hf, giT = 2π −π

defines an inner product in L2 (T ). Then it is easily checked that we can induce a norm k · k to X by defining p p (19.45) kF kX = hF, F iX = hf, f iT = kf kT , so it makes X to be Hilbert. Of course, it follows from the expression (19.45) that   1, if n = m; hUn , Um iX = hun , um iT =  0, otherwise.

In other words, {Un | n ∈ Z} forms an orthonormal set in X. Since {un | n ∈ Z} is maximal in L2 (T ), {Un | n ∈ Z} is also maximal in X. By Theorem 4.18 (The Riesz-Fischer Theorem), it is true that X F (z) = ck Uk (z), (19.46) k∈Z

where c1 , c2 , . . . are some scalars. Since F is entire, we have X F (n) = lim ck Uk (z) = cn Un (n) = cn z→n

k∈Z

for every n ∈ Z. Finally, as a Hilbert space X with a maximal orthonormal set {Un | n ∈ Z}, we conclude from §4.19 that the mapping F 7→ hF, Un i = cn = F (n) is a Hilbert space isomorphism of X onto ℓ2 (Z). This means that our map Φ is a bijection, completing the analysis of the problem. 

584

Chapter 19. Holomorphic Fourier Transforms

Problem 19.16 Rudin Chapter 19 Exercise 16.

Proof. Since |f (x)| ≤ e−|x| on R, f ∈ L2 (−∞, ∞). By the analysis of §19.1, its Fourier transform Z ∞ f (t)eitz dt fb(z) = −∞

is holomorphic in Π+ . In particular, we have Z b f (x) =

∞

f (t)eitx dt

−∞

for every x ∈ R. For every n ≥ 0, the hypothesis implies that differentiation under the integral sign is legitimateh so that Z ∞ Z ∞ e−|t| · |t|n dt = 2n!. f (t)(it)n eitx dt ≤ |fb(n) (x)| = −∞

−∞

Consequently, the power series of fb

∞ X

n=0

cn (z − a)n

has at least 1 as its radius of convergence for every a ∈ R which means that fb is also holomorphic on R. If fb has compact support, then fb vanishes on a set with a limit point. Hence Theorem 10.18 forces that f ≡ 0 a.e. on R. This completes the proof of the problem. 

h Of course, it follows from the Leibniz’s Rule by Problem 10.16, where ϕ(z, t) = f (t)eitz . See also [3, Theorem 24.5, pp. 193, 194].

CHAPTER

20

Uniform Approximation by Polynomials

Problem 20.1 Rudin Chapter 20 Exercise 1.

Proof. We want to prove that if ǫ > 0, S 2 \ K has finitely many components, f ∈ C(K) and f ∈ H(K ◦ ), then there exists a rational function R such that |f (z) − R(z)| < ǫ

(20.1)

for all z ∈ K.

Indeed, everything up to [100, p. 392] in the proof of Theorem 20.5 (Mergelyan’s Theorem) remains the same. Let S1 , S2 , . . . , Sm be the (connected) components of S 2 \ K, i.e., S 2 \ K = S1 ∪ S2 ∪ · · · ∪ Sm .

Pick δ > 0 very small. Recall also that X = {z ∈ supp Φ | dist(z, S 2 \ K) ≤ δ} is compact, so X contains no point which is “far within” K, see Figure 20.1 which shows that X is exactly the yellow part.

Figure 20.1: The compact set X. Now we can cover X by finitely many open discs D1 (p1 ; 2δ), D2 (p2 ; 2δ), . . . , Dn (pn ; 2δ), where p1 , p2 , . . . , pn are points in S 2 \ K. We may assume that n = m and pj ∈ Sj for j = 1, 2, . . . , n. 585

586

Chapter 20. Uniform Approximation by Polynomials

Since each Sj is connected, there must be a curve from pj to a point of ∂Dj (pj ; 2δ) that is not of K. In other words, one can find a set Ej ⊂ Dj (pj ; 2δ) such that Ej is a compact and connected subset of Sj , diam Ej ≥ 2δ, S 2 \ Ej is connected and K ∩ Ej = ∅, where j = 1, 2, . . . , n. We apply Lemma 20.2 with r = 2δ and follow the proof in [100, pp. 393, 394], we can obtain |F (z) − Φ(z)| < 6000ω(z)

and |f (z) − Φ(z)| < ω(δ)

(20.2)

for all z ∈ Ω, where Ω = S 2 \ (E1 ∪ E2 ∪ · · · ∪ En ) which is an open set containing K. By the definition, we have S 2 \ Ω = E1 ∪ E2 ∪ · · · ∪ En . Since Ej ⊂ Sj for j = 1, 2, . . . , n, one gets the set A = {p1 , p2 , . . . , pn }. Since F ∈ H(Ω) and K ⊆ Ω, Theorem 13.9 (Rung’s Theorem) implies that there exists a rational function R(z) with poles only in A such that |F (z) − R(z)| < ω(δ) (20.3) for all z ∈ K. Combining this and the inequalities (20.2) and (20.3), we have |f (z) − R(z)| ≤ |f (z) − Φ(z)| + |Φ(z) − F (z)| + |F (z) − R(z)| < 10000ω(δ) for all z ∈ K. Since ω(δ) → 0 as δ → 0, it yields the inequality (20.1) by choosing sufficiently  small δ. Hence we have completed the proof of the problem. Problem 20.2 Rudin Chapter 20 Exercise 2.

Proof. The set K is known as a Swiss cheese set.a The construction of such a sequence {Dn } in U with the specific properties can be found in [45, pp. 344, 345]. In fact, we can also assume that ∞ X rn2 < 1. (20.4) n=1

• L is a bounded linear functional on C(K). It is easy to see that L is a linear functional on C(K). According to [11, Theorem 4.10, p. 49], we have Z Z f (z) dz ≤ 2πrn · kf k∞ f (z) dz ≤ 2π · kf k∞ and Γ

γn

for every f ∈ C(K) and n ∈ N. Therefore, we get

∞   X rn · kf k∞ < ∞ |L(f )| ≤ 2π 1 + n=1

which shows that L is bounded. • L(R) = 0 for every rational function R whose poles are outside K. Let z0 be a pole of R. Since K = U \ V , we have either z0 lies outside U or z0 ∈ Dn for exactly one n. If z0 lies outside U , then since Γ([0, 2π]), γn ([0, 2π]) ⊆ U , the integrals in L(R) are both zero so that L(R) = 0 in this case. If z0 ∈ Dm , then we have z0 ∈ / Dn for every n 6= m. In this case, we know that Ind Γ (z0 ) = Ind γm (z0 ) = 1 and Ind γn (z0 ) = 0 for every n 6= m. Consequently, the integrals cancel for the principal part of R at the pole z0 which gives L(R) = 0 when we express R in its partial fraction decomposition. a This shows that Mergelyan’s Theorem does not hold anymore if the finiteness of the components of S 2 \ K is dropped.

587 • There exists an f ∈ C(K) for which L(f ) 6= 0. We take f (z) = z which belongs to C(K). Obviously, we have Z Z z dz = 1 and z dz = 2πirn2 Γ

γn

for every n ∈ N. Therefore, we obtain

∞   X rn2 6= 0 L(f ) = 2πi 1 − n=1

by the hypothesis (20.4). This completes the proof of the problem.



Problem 20.3 Rudin Chapter 20 Exercise 3.

Proof. Suppose that E ⊆ D(0; r) is compact and connected, where r > 0. Let diam E ≥ r and Ω = S 2 \ E be connected. Denote X = {f ∈ H(Ω) | zf (z) → 1 as z → ∞}. Now we recall the definitions of the conformal mappings F : U → Ω and g : U → D(0; |a|−1 ) that F (ω) =

∞

a X cn ω n + ω n=0

and g(z) =

1 −1 F (z), a

(20.5)

where ω ∈ U and z ∈ Ω. Without loss of generality, we may assume that a > 0. Assume that there was an f ∈ X such that kgk∞ > kf k∞ . (20.6)

Since F −1 is a conformal mapping of Ω onto U , we have kF −1 k∞ = 1 and thus the definition (20.5) gives kgk∞ = a−1 . (20.7)

These two facts (20.6) and (20.7) combine to give f (Ω) ⊆ D(0; a−1 ). Next, we define the mapping ϕ : U → U by
ϕ(ω) = af F (ω) .
Then it is easily checked that ϕ ∈ H ∞ and ϕ(0) = af F (0) = af (∞) = 0. Besides, we observe that
ϕ′ (ω) = af ′ F (ω) · F ′ (ω). By the definition (20.5), we have

F ′ (ω) = −

∞

X a ncn ω n−1 . + ω 2 n=1

Since zf (z) → 1 as z → ∞, f has the form f (z) =

∞

1 X a−n + z zn n=0

so that f ′ (z) =

∞

−1 X −na−n + . z2 z n+1 n=0

588

Chapter 20. Uniform Approximation by Polynomials

By the definition (20.5), we get
ϕ′ (0) = lim af ′ F (ω) · F ′ (ω) = 1. ω→0

Hence it follows from Theorem
12.2 (The Schwarz Lemma) that ϕ(ω) = λω for some constant λ with |λ| = 1 so that af F (ω) = λω. Substituting ω = F −1 (z) into this equation, we obtain f (z) = λ

F −1 (z) = λg(z) a

which implies that kf k∞ = kgk∞ , a contradiction to the inequality (20.6). Put ω = F −1 (z). Then the definitions (20.5) imply
z = F F −1 (z) =

a F −1 (z)

+ c0 +

∞ X

n=1

 n cn F −1 (z) =

Rewrite it as zg(z) = 1 + c0 g(z) +

∞ X

∞ X 1 cn an gn (z). + c0 + g(z) n=1

cn an g n+1 (z).

(20.8)

n=1

Since b=

1 2πi

Z

zg(z) dz,

Γ

where Γ is the positively oriented circle with center 0 and radius r, we may substitute the formula (20.8) into the integral to get b=

1 2πi

Z h

1 + c0 g(z) +

Γ

∞ X

n=1

i cn an gn+1 (z) dz.


Since g ∈ X, g has a simple zero at ∞ which shows that Res (g; ∞) = 1 and Res gn+1 (z); ∞ = 0 for all n ≥ 1. Hence we conclude from Theorem 10.42 (The Residue Theorem) that b=

c0 2πi

Z

Γ

g(z) dz +

∞ X

n=1

cn an ·

h 1 Z i gn+1 (z) dz = c0 2πi Γ

as desired. This proves the second assertion. To prove the third assertion, we notice that since F (0) = ∞, we observe that F maps CR = {ω | |ω| = R} into the disk D(0; r) for some R < 1 and sufficiently close to 1. Therefore, we obtain 1 Z F (ω) 1 2πRr dω < × =r |b| = 2πi CR ω 2π R as desired, completing the proof of the problem.



Index

A algebra, 262 analytic capacity, 363 annihilate, 555 approximate identity, 295 arc length, 89 atom, 199 atomicless measures, 199 Axiom of Choice, 311

Dirichlet integrals, 181, 274 Dirichlet’s Approximation Theorem, 367 discrete measure, 225 dyadic rationals, 218 E elliptic transformation, 456 equivalent under G, 498 error function, 301 Euler constant, 178 extreme point, 370 Extreme Value Theorem, 61

B Baire’s Theorem on Semicontinuous Functions, 519 basis, 2 Basis Extension Theorem, 42 Bertrand’s integrals, 73 Bolzano-Weierstrass Theorem, 394, 425, 516, 582 Borel transform, 568 Borel’s Theorem, 581 bump function, 217, 293

F Fej´er’s Theorem, 379 finite intersection property, 60 Fubini’s Theorem on Differentiation, 236 Fundamental Normality Test, 517 Fundamental Theorem of Algebra, 551 G Gaussian function, 289 Gram-Schmidt Process, 124

C Cauchy functional equation, 310 Cauchy type integrals, 361 Chebyshev’s Inequality, 106 connected sets, 25 continuum, 509 converge strongly, 371 converge weakly, 371 convex combinations, 162 Convolution Theorem, 557

H Haar measure, 263 Hadamard gaps, 407 Hadamard’s Three-Line Theorem, 386 Hamel basis, 311 harmonic conjugate, 495, 538 Harnack’s inequalities, 363 Hausdorff measure, 363 Heine-Borel Theorem, 35 Hilbert’s identity, 559 Hurwitz’s Theorem, 332, 425, 428 hyperbolic transformation, 456

D De Morgan’s law, 35 dense, 67 determinant of ϕ, 445 Dieudonn´e’s measure, 53 Dirac delta function at x, 371 direct sum, 148

I ideal, 262 Invariance of Domain, 217 isometric isomorphism, 157 589

590 isometry, 112 isomorphism, 112 K Koebe mapping, 442 Kolmogorov’s Theorem, 539 Kronecker delta function, 154 Kronecker’s Approximation Theorem, 212, 366, 445 L lacunary power series, 407 Lebesgue function associated with the Cantor-like set E, 219 Leibniz’s Rule, 584 limit point, 50 little-o notation, 60 locally compact abelian, 556 locally compact group, 263 locally integrable functions, 240 lower densities of E at 0, 210 lower Riemann integral, 36 loxodromic transformation, 456 Lusin Area Integral, 462 M Mean Value Theorem, 116 Minkowski’s Inequality, 256 mollifiers, 295 multipler of the transformation, 453 N normal form, 456 O order topology, 47 Osgood’s Theorem, 410 outer measure, 54 P P´ olya representation, 568 parabolic transformation, 456 Poisson kernel, 296, 572 Poisson kernel for the upper half-plane, 359 power set, 2 Principle of Subordination, 422 Product Measure Theorem, 270 push-forward measure, 264 R Radon measure, 59 Removable sets for holomorphic functions, 363

Index Riemann-Lebesgue Lemma, 285, 557 Riesz-Fischer Theorem, 236 right-shift operator, 552 Rogosinski’s Theorem, 395 S Schauder basis, 154 Schottky’s Theorem, 436 Schwartz space, 289 Schwarz Integral Formula, 355 Schwarz Reflection Principle for U , 415 second-countable, 273 semifinite, 173 semisimple, 561 sequentially compact, 128 Silverman-Toepliotz Theorem, 171 singular function, 218 standard topology, 2 Stone-Weierstrass Theorem, 284, 344 strong convergence, 371 subadditive, 13, 27 subalgebra, 262 Swiss cheese set, 586 T tent function, 59 The Basic Connectedness Lemma, 356 The Steinhaus Theorem, 215, 310 total variation of f , 546 totally disconnected, 509 triangle inequality, 14, 43 U upper densities of E at 0, 210 Urysohn’s Lemma, 22 usual metric, 44 V Vitali Convergence Theorem, 410 W weak convergence, 371 weak∗ convergence, 371 Weierstrass Approximation Theorem, 125, 139 Weyl’s Equidistribution Theorem, 135 Y Young’s Convolution Inequality, 256 Z Zorn’s Lemma, 311

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