A Complete Solution Guide to Real and Complex Analysis II 9789887415640, 9789887415657, 9887415650

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A Complete Solution Guide to Real and Complex Analysis II

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: 978-988-74156-4-0 (eBook) ISBN: 978-988-74156-5-7 (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 was 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 seven books: • A Complete Solution Guide to Complex Analysis • A Complete Solution Guide to Real and Complex Analysis I • 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 is the continuum of my book A Complete Solution Guide to Real and Complex Analysis I. It covers the “Complex Analysis” part of Rudin’s graduate book. In fact, we study all exercises of Chapters 10 to 20. Same as A Complete Solution Guide to Real and Complex Analysis I, 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 way I wrote the solution guides of Baby Rudin and the first part of Papa Rudin. 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 read 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 refer to some other analysis or topology books such as [2], [9], [15], [18], [23], [42] and [65]. Other useful references are [3], [27], [28], [37], [69], [79], [80], [82] and [83]. Of course, we will use the exercises in Baby Rudin and the first part of Papa Rudin freely and if you want to read proofs of them, you are strongly advised to read my books [77] and [78]. As you will expect, this book always keeps the main features of my previous books [77] and [78]. In fact, its features are as follows: • It covers all the 221 exercises from Chapters 10 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 29 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. v

vi • Many useful or relevant references are provided to some questions for your future research. 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

10.1 The closed contour ΓA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

7

10.2 The contour ΓA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

10

10.3 The contours ΓA , Γ1 and Γ2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11

10.4 The annulus A(r1 , r2 ) and the circles γ1 , γ2 . . . . . . . . . . . . . . . . . . . . . .

23

10.5 A non null-homotopic closed path Γ = γ1 − γ3 − γ2 + γ4 in Ω. . . . . . . . . . . .

32

11.1 The I divides ∂∆ into several triangles. . . . . . . . . . . . . . . . . . . . . . . .

48

12.1 The boundary ∂∆. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

69

12.2 The sectors ∆1 , ∆2 and the ray Lα . . . . . . . . . . . . . . . . . . . . . . . . . .

76

13.1 The simply connected set Ω. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

84

13.2 The compact sets Dn and En . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

85

13.3 The compact sets An , Bn and Cn . . . . . . . . . . . . . . . . . . . . . . . . . . .

86

13.4 The disc ∆n , the arc Ln and its neighborhood Ωn . . . . . . . . . . . . . . . . . .

87

14.1 The region Ω bounded by C1 and C2 .

. . . . . . . . . . . . . . . . . . . . . . . . 119

14.2 The constructions of Ωn−1 , D(0; rn ) and αn . . . . . . . . . . . . . . . . . . . . . . 122 14.3 The construction of the symmetric point z ∗ of z. . . . . . . . . . . . . . . . . . . 134
14.4 The conformal mapping ψ(z) = φ ϕ(z) . . . . . . . . . . . . . . . . . . . . . . . 143 π

π

14.5 The conformal mapping f : U → A(e− 2 , e 2 ). . . . . . . . . . . . . . . . . . . . . 144 14.6 The locus of f (z) for t ∈ (0, π). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

14.7 The locus of f (z) for t ∈ (π, 2π). . . . . . . . . . . . . . . . . . . . . . . . . . . . 145

14.8 The image of f (E) when r = 0.25. . . . . . . . . . . . . . . . . . . . . . . . . . . 146 15.1 The distribution of the zeros zk,n of exp(exp(z)). . . . . . . . . . . . . . . . . . . 151 15.2 The paths γ(z + h) and −γ(z). . . . . . . . . . . . . . . . . . . . . . . . . . . . . 159 16.1 The paths βζ (I) and γζ . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182 16.2 The paths γz (I), γω (I) and γω,ζ (I). . . . . . . . . . . . . . . . . . . . . . . . . . . 182 16.3 The fundamental domain R of G. . . . . . . . . . . . . . . . . . . . . . . . . . . . 184 16.4 The regions Ωα and Ωβ if α < β. . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 16.5 The regions of convergence of the two series. . . . . . . . . . . . . . . . . . . . . . 199 19.1 The closed contour Γr . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255 vii

viii

List of Figures 20.1 The compact set X. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 271

Contents

Preface

v

List of Figures

viii

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

1

. . . . . . . . . . . . . . . . . . . . .

1

10.2 Evaluation of Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

10.3 Composition of Holomorphic Functions and Morera’s Theorem . . . . . . . . . .

13

10.4 Problems related to Zeros of Holomorphic Functions . . . . . . . . . . . . . . . .

17

10.5 Laurent Series and its Applications . . . . . . . . . . . . . . . . . . . . . . . . . .

22

10.6 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

30

11 Harmonic Functions

33

11.1 Basic Properties of Harmonic Functions . . . . . . . . . . . . . . . . . . . . . . .

33

11.2 Harnack’s Inequalities and Positive Harmonic Functions . . . . . . . . . . . . . .

49

11.3 The Weak∗ Convergence and Radial Limits of Holomorphic Functions . . . . . .

57

11.4 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

60

12 The Maximum Modulus Principle

69

12.1 Applications of the Maximum Modulus Principle . . . . . . . . . . . . . . . . . .

69

12.2 Asymptotic Values of Entire Functions . . . . . . . . . . . . . . . . . . . . . . . .

78

12.3 Further Applications of the Maximum Modulus Principle . . . . . . . . . . . . .

79

13 Approximations by Rational Functions 13.1 Meromorphic Functions on

S2

83

and Applications of Runge’s Theorem . . . . . . .

83

13.2 Holomorphic Functions in the Unit Disc without Radial Limits . . . . . . . . . .

87

13.3 Simply Connectedness and Miscellaneous Problems . . . . . . . . . . . . . . . . .

93

14 Conformal Mapping

99

14.1 Basic Properties of Conformal Mappings . . . . . . . . . . . . . . . . . . . . . . .

99

14.2 Problems on Normal Families and the Class S . . . . . . . . . . . . . . . . . . . 112 14.3 Proofs of Conformal Equivalence between Annuli . . . . . . . . . . . . . . . . . . 118 14.4 Constructive Proof of the Riemann Mapping Theorem . . . . . . . . . . . . . . . 122 15 Zeros of Holomorphic Functions

149 ix

x

Contents 15.1 Infinite Products and the Order of Growth of an Entire Function . . . . . . . . . 149 15.2 Some Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 15.3 Problems on Blaschke Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . 164 15.4 Miscellaneous Problems and the M¨ untz-Szasz Theorem . . . . . . . . . . . . . . . 172

16 Analytic Continuation

179

16.1 Singular Points and Continuation along Curves . . . . . . . . . . . . . . . . . . . 179 16.2 Problems on the Modular Group and Removable Sets . . . . . . . . . . . . . . . 183 16.3 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195 17 H p -Spaces

205

17.1 Problems on Subharmonicity and Harmonic Majoriants . . . . . . . . . . . . . . 205 17.2 Basic Properties of H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208 17.3 Factorization of f ∈ H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 217 17.4 A Projection of Lp onto H p . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 224 17.5 Miscellaneous Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230 18 Elementary Theory of Banach Algebras

235

18.1 Examples of Banach Spaces and Spectrums . . . . . . . . . . . . . . . . . . . . . 235 18.2 Properties of Ideals and Homomorphisms . . . . . . . . . . . . . . . . . . . . . . 239 18.3 The Commutative Banach algebra H ∞ . . . . . . . . . . . . . . . . . . . . . . . . 249 19 Holomorphic Fourier Transforms

251

19.1 Problems on Entire Functions of Exponential Type . . . . . . . . . . . . . . . . . 251 19.2 Quasi-analytic Classes and Borel’s Theorem . . . . . . . . . . . . . . . . . . . . . 260 20 Uniform Approximation by Polynomials

271

Index

275

Bibliography

277

CHAPTER

10

Elementary Properties of Holomorphic Functions

10.1

Basic Properties of Holomorphic Functions

Problem 10.1 Rudin Chapter 10 Exercise 1.

Proof. In fact, this is [61, Exercise 21, p. 101]. For a solution of it, please refer to [77, p. 75]. This completes the proof of the problem.  Problem 10.2 Rudin Chapter 10 Exercise 2.

Proof. Let a ∈ C and f (z) =

∞ X

n=0

cn (z − a)n .

(10.1)

By [62, Eqn. (8), p. 199], we have n!cn = f (n) (a) for all n = 0, 1, 2, . . . and so the hypothesis implies that f (n) (a) = 0 for some n ∈ N ∪ {0}. For every n ∈ N ∪ {0}, let Zn = {z ∈ C | f (n) (z) = 0} ⊆ C. Now the previous paragraph implies that ∞ [ Zn . (10.2) C= n=0

f (n)

Since f is entire, every is also entire. If f (n) 6≡ 0 for every n ∈ N ∪ {0}, then Zn 6= C for every n ∈ N ∪ {0} by Theorem 10.18. Furthermore, each Zn is at most countable and it deduces from the set relation (10.2) that C is countable, a contradiction. Thus there exists an N ∈ N such that f (n) ≡ 0 for all n > N and so the representation (10.1) implies that f is a polynomial of degree at most N , completing the proof of the problem.  Problem 10.3 Rudin Chapter 10 Exercise 3.

1

2

Chapter 10. Elementary Properties of Holomorphic Functions

Proof. We claim that f (z) = cg(z) for some |c| ≤ 1. If g 6≡ 0 in C, then we follow from Theorem 10.18 that Z(g) has no limit point in C. Consider h : C \ Z(g) → C given by h(z) =

f (z) . g(z)

For each a ∈ Z(g), we have h ∈ H(D ′ (a; r)) and |h(z)| ≤ 1 in D ′ (a, r) for some r > 0. Now we follow from Theorem 10.20 that h has a removable singularity at a and then h can be defined at a so that it is holomorphic in D(a; r). Since it is true for every a ∈ Z(g), h is in fact entire and |h(z)| ≤ 1 in C. Therefore, Theorem 10.23 (Liouville’s Theorem) asserts that h(z) = c for some constant c such that |c| ≤ 1. Consequently, we obtain f (z) = cg(z) as required. This proves the  claim and we end the proof of the problem. Problem 10.4 Rudin Chapter 10 Exercise 4.

Proof. For every n = 0, 1, 2, . . ., we apply the hint given in Problem 10.2, Theorem 10.26 (The Cauchy’s Estimates) and the hypothesis to get f (n) (0) A + Rk |cn | = , ≤ n! Rn

(10.3)

where R > 0. If n > k, then we take R → ∞ to both sides of the inequality (10.3) to conclude cn = 0 for all n > k. In other words, f is a polynomial of degree at most k. This completes the proof of the problem.  Problem 10.5 Rudin Chapter 10 Exercise 5.

Proof. Since {fn } is uniformly bounded in an open subset Ω ⊆ C, there exists a M > 0 such that |fn (z)| ≤ M for all z ∈ Ω and all n ∈ N. Let a ∈ Ω and f (z) = lim fn (z). Since Ω is open n→∞

in C, one can find a δ > 0 such that D(a; 4δ) ⊆ Ω. Define γ : [0, 2π] → Ω by γ(t) = a + 2δeit

(10.4)

which is clearly a circle centered at a with radius 2δ. If z ∈ D(a; δ), then we have |γ(t) − z| > δ for all t ∈ [0, 2π]. Therefore, we get from this fact and the representation (10.4) that γ ′ (t) 2δ =2 < γ(t) − z δ

(10.5)

for all t ∈ [0, 2π]. Next, for every n ∈ N, we define Fn : [0, 2π] → C and F : [0, 2π] → C by Fn (t) = fn (γ(t))

and

F (t) = f (γ(t))

respectively. Clearly, the pointwise convergence of {fn } on Ω implies the pointwise convergence of {Fn } on [0, 2π]. Besides, since all fn and γ are continuous functions, every Fn is complex

10.1. Basic Properties of Holomorphic Functions

3

measurable on the measurable space [0, 2π]. Since 2M ∈ L1 (m) and |Fn (t)| ≤ M for all n ∈ N, we deduce from Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) that Z 2π Z 2π F (t) dt. (10.6) Fn (t) dt = lim n→∞ 0

Now the bound (10.5) ensures that certainly have

0

γ ′ (t) γ(t)−z

Fn (t) ·

is continuous on [0, 2π] so that for all z ∈ D(a; δ), we

γ ′ (t) γ ′ (t) → F (t) · γ(t) − z γ(t) − z ′

γ (t) pointwisely on [0, 2π] and each Fn (t)· γ(t)−z is complex measurable on [0, 2π]. Since 2M ∈ L1 (m) ′

γ (t) | ≤ 2M for all n ∈ N, further application of Theorem 1.34 (The Lebesgue’s and |Fn (t) · γ(t)−z Dominated Convergence Theorem) also gives Z 2π Z 2π γ ′ (t) γ ′ (t) F (t) · dt = dt. Fn (t) · lim (10.7) n→∞ 0 γ(t) − z γ(t) − z 0

Recall that z ∈ / γ ∗ , so Ind γ (z) = 1. Then we obtain from Theorem 10.15 (The Cauchy’s Formula in a Convex Set) and the limit (10.7) thata f (z) = lim fn (z) n→∞ Z 1 fn (ζ) = lim dζ n→∞ 2πi γ ζ − z Z 2π γ ′ (t) 1 dt fn (γ(t)) · = lim n→∞ 2πi 0 γ(t) − z Z 2π γ ′ (t) 1 Fn (t) · dt = lim n→∞ 2πi 0 γ(t) − z Z 2π γ ′ (t) 1 F (t) · dt. = 2πi 0 γ(t) − z

(10.8)

Given that ǫ > 0. The result (10.6) guarantees that there is an N ∈ N such that n ≥ N implies Z 2π |Fn (t) − F (t)| dt < ǫπ. 0

In this case, for every z ∈ D(a, δ), we obtain from the expression (10.8) that Z 2π 1 Z 2π 1 γ ′ (t) γ ′ (t) |fn (z) − f (z)| = dt − dt Fn (t) · F (t) · 2πi 0 γ(t) − z 2πi 0 γ(t) − z Z 2π ′ 1 γ (t) |Fn (t) − F (t)| · ≤ dt 2π 0 γ(t) − z 0} positively oriented, see Figure 10.1

10.2. Evaluation of Integrals

7

Figure 10.1: The closed contour ΓA . Clearly, if A is large enough, then ΓA will contain all zeros of Q(z) lying in the upper half plane. Hence it follows from Theorem 10.42 (The Residue Theorem) that Z X R(z) dz = 2πi Res (R; zk ), (10.19) ΓA

k

where {zk } is the set of all zeros of Q in the upper half plane. In fact, we can write the expression (10.19) in the form Z A Z X R(x) dx = 2πi R(z) dz + Res (R; zk ). −A

CA

k

Since CA is a semi-circle of radius A, its length is πA. Using this fact and deg Q − deg P ≥ 2, we obtain from the estimate [62, Eqn. (5), Definition 10.8, p. 202] that Z πM M (10.20) R(z) dz ≤ 2 · πA = A A CA for some positive constant M . Taking A → ∞ in the inequality (10.20), we get Z R(z) dz = 0. lim A→∞ CA

Finally, we combine the results (10.18), (10.19) and (10.21) to conclude that Z ∞ X R(x) dx = 2πi Res (R; zk ). −∞

(10.21)

(10.22)

k

For the analogous statement for the lower half plane, the formula (10.22) will be replaced by Z ∞ X R(x) dx = −2πi Res (R; zk ), −∞

k

where the set {zk } now consists of all zeros of Q in the lower half plane.c To compute the integral, we note from the formula (10.22) and some basic facts of calculating residuesd that Z ∞  z2 h  z2  πi   3πi i x2 + Res dx = 2πi Res ; exp ; exp 4 1 + z4 4 1 + z4 4 −∞ 1 + x c

Here we have the negative sign in the formula because the corresponding semi-circle in the lower half plane is negatively oriented. d See, for examples, [9, pp. 129, 130] or [65, pp. 75, 76].

8

Chapter 10. Elementary Properties of Holomorphic Functions  πi   3πi i πi h exp − + exp − 2 4 4 πi −2i × √ = 2 2 π =√ . 2 =

This completes the analysis of the problem.



Problem 10.9 Rudin Chapter 10 Exercise 9.

Proof. Let ΓA be the closed contour consisting of the real segment [−A, A] and the upper semicircle CA = {z ∈ C | |z| = A and Im z > 0}, see Figure 10.1. Furthermore, we let P and Q be polynomials such that deg Q − deg P ≥ 1, Q(x) 6= 0 (except perhaps at zeros of cos x or sin x) P (x) . By the discussion of Type II integrals in [9, pp. 144 – 146], we see that and R(x) = Q(x) Z

∞

Z

ix

R(x)e dx = lim

A→∞ ΓA

−∞

R(z)eiz dz = 2πi

X

Res (R(z)eiz ; zk ),

(10.23)

k

where the points zk are the poles of R(z) in the upper half plane. Suppose that t ≥ 0. By the substitution y = tx, we have Z ∞ Z ∞ eitx t dx = eiy dy. 2 2 + y2 1 + x t −∞ −∞ Set R(z) =

t t2 +z 2 .

Then it follows from the representation (10.23) that Z

∞ −∞

 t  eitx π iz dx = 2πiRes e ; ti = t. 2 2 2 1+x t +z e

Next, if t = −u for some u > 0, then we have Z ∞ −iux Z ∞ e eitx dx = dx 2 2 −∞ 1 + x −∞ 1 + x Z ∞ u eiy dy = 2 2 −∞ u + y  u  iz = 2πiRes 2 e ; ui u + z2 π = u. e

(10.24)

(10.25)

Combining the two expressions (10.24) and (10.25), we conclude that Z

∞

−∞

π eitx dx = |t| . 2 1+x e

Using the theory of Fourier transforms, we notice that if f (t) = Definition 9.1 that Z ∞ 1 fb(x) = √ f (t)e−ixt dt 2π −∞

(10.26) pπ

−|t| , 2e

then we know from

10.2. Evaluation of Integrals 1 = 2 =

Z

∞

9 e−|t| e−ixt dt

−∞ 0

Z 1h

e(1−ix)t dt +

Z

∞

e−(1+ix)t dt

i

2 −∞ 0 1 n exp[(1 − ix)t] 0 exp[−(1 + ix)t] ∞ o = · + 2 1 − ix −1 + ix −∞ 0 1  1 1 + = 2 1 − ix 1 + ix 1 = . 1 + x2 Now it is clear that f, fb ∈ L1 (R), so we follow from Theorem 9.11 (The Inversion Theorem) that g(t) = f (t), where Z ∞ Z ∞ 1 eixt 1 ixt b √ √ g(t) = dx. f (x)e dx = 2π −∞ 2π −∞ 1 + x2 Consequently, we obtain

Z

∞

−∞

√ eixt dx = 2π · 2 1+x

r

π π −|t| e = |t| 2 e

which is consistent with the result (10.26). Hence we have completed the proof of the problem.  Problem 10.10 Rudin Chapter 10 Exercise 10.

Proof. Let f (z) = (ez − e−z )z −4 . Then f is holomorphic in C \ {0}. By the power series expansion of ez (see [62, Eqn. (1), p. 1]), we have 1  ez − e−z 1 z = 2 + + + · · · z4 z 3 3!z 5!

so that f has a pole of order 3 at 0. By Theorem 10.21(b), the difference 1 1  f (z) − 2 3 + z 3!z has a removable singularity at 0. Thus there exists an entire function g such thate 1 1  f (z) − 2 3 + = g(z) z 3!z which gives Z Z  Z 1 1 1 1  1 dz + f (z) dz = + g(z) dz. 2πi γ πi γ z 3 3!z 2πi γ

(10.27)

Applying Theorems 10.10 and 10.12 to the right-hand side of the expression (10.27), we establish that Z z Z Z 1 e − e−z dz 1 1 1 dz = f (z) dz = = . 2πi γ z4 2πi γ 6πi γ z 3

This ends the proof of the problem. e



Obviously, we have g(z) = 2

∞ X

n=2

z 2n−3 . (2n + 1)!

10

Chapter 10. Elementary Properties of Holomorphic Functions Problem 10.11

Rudin Chapter 10 Exercise 11.

Proof. Since |α| = 6 1, we have either |α| < 1 or |α−1 | < 1. If z = eiθ , then cos θ = 21 (z + z1 ). Using [9, Eqn. (5), p. 150], we see that Z 2π Z dθ dz 1 · = 1 1 2 2 1 − 2α cos θ + α iz 0 |z|=1 1 − 2α · 2 (z + z ) + α Z dz =i 2 2 |z|=1 αz − (1 + α )z + α    1  −2πRes ; α , if |α| < 1;    αz 2 − (1 + α2 )z + α =     1  −1  −2πRes , if |α−1 | < 1 ; α αz 2 − (1 + α2 )z + α  2π     − α2 − 1 , if |α| < 1; =   2π   , if |α−1 | < 1. 2 α −1  Hence we complete the analysis of the problem. Problem 10.12 Rudin Chapter 10 Exercise 12.

Proof. Let ΓA be the path obtained by going from −A to −1 along the real axis, from −1 to 1 along the lower half of the unit circle C and from 1 to A along the real axis, see Figure 10.2 below.

Figure 10.2: The contour ΓA . We note that Z −1 Z 1 Z A Z A sin2 x itx sin2 x itx sin2 x itx sin2 x itx e dx = e dx + e dx + e dx. 2 2 x2 x2 −A −1 x 1 −A x

(10.28)

Since z −2 · sin2 z · eitz is entire for every t ∈ R, it follows from Theorem 10.14 (The Cauchy’s Theorem in a Convex Set) that Z −1 Z sin2 x itx sin2 z itz e dz + e dx = 0 2 x2 1 C z

10.2. Evaluation of Integrals or equivalently,

Z

1 −1

11

sin2 x itx e dx = x2

Z

C

sin2 z itz e dz. z2

(10.29)

Combing the integral relations (10.28) and (10.29), we see immediately that Z

A

−A

Next, we write 2i sin z = eiz − e−iz so that Z

ΓA

sin2 z itz e dz = z2

Z

ΓA

Now we define ϕA (s) =

Z

Z

sin2 z itz e dz. z2

(10.30)

e2iz − 2 + e−2iz itz e dz. −4z 2

(10.31)

sin2 x itx e dx = x2

ΓA

ΓA

eisz dz, z2

(10.32)

so the expression (10.31) becomes Z

ΓA

sin2 z itz 1 1 e dz = − [ϕA (t + 2) + ϕA (t − 2)] + ϕA (t). 2 z 4 2

(10.33)

If we combine (10.30) and (10.33), then we have Z

A

−A

sin2 x itx 1 1 e dx = ϕA (t) − [ϕA (t + 2) + ϕA (t − 2)]. 2 x 2 4

(10.34)

Complete ΓA to a closed path in two different ways: Firstly, we consider the semi-circle Γ1 from A to −Ai and then to −A; secondly, we consider the semi-circle Γ2 from A to Ai and then to −A, see Figure 10.3.

Figure 10.3: The contours ΓA , Γ1 and Γ2 .

12

Chapter 10. Elementary Properties of Holomorphic Functions

It is easily checked that the function eisz · z −2 has a pole of order 2 at 0, and so the residue is is. Thus in the first way, we have Z Z eisz eisz dz + dz = 0 2 2 ΓA z Γ1 z

so that if z = Aeiθ , where θ ∈ [−π, 0], then we deduce from the definition (10.32) that Z 0 Z exp(isAeiθ ) eisz dz = i dθ (10.35) ϕA (s) = 2 Aeiθ −π ΓA z and in the second way, Theorem 10.42 (The Residue Theorem) yields Z Z  eisz  eisz eisz dz + dz = 2πiRes ; 0 = −2πs 2 2 z2 ΓA z Γ2 z which implies that

ϕA (s) = Since

Z

ΓA

eisz dz = −2πs − i z2

Z

π 0

exp(isAeiθ ) dθ. Aeiθ

(10.36)

exp(isAeiθ ) exp(−As sin θ) →0 ≤ Aeiθ A as A → ∞ if s and sin θ have the same sign. Thus it follows from Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) that the integral (10.35) tends to 0 if s < 0 and the one in (10.36) tends to 0 if s > 0. In other words, we obtain   −2πs, if s > 0; lim ϕA (s) = (10.37)  A→∞ 0, if s < 0.

Finally, we apply the result (10.37) to the expression (10.34) to get  if |t| > 2;  Z ∞  0, 2 sin x itx (10.38) e dx = 2  −∞ x  π − πt , if −2 < t < 0 or 0 < t < 2. 2 When t = 0, we know from the integral (10.35) that ϕA (0) = − A2 , so we establish from the expression (10.34) that  Z ∞  π, if t = 0; 2 sin x itx e dx = (10.39) 2  −∞ x 0, if t = ±2. By combining the results (10.38) and (10.39), we  0,      Z ∞  sin2 x itx πt e dx = π− , 2  x 2 −∞      π,

achieve

if |t| ≥ 2; if −2 < t < 0 or 0 < t < 2; if t = 0.

We have completed the proof of the problem.



Problem 10.13 Rudin Chapter 10 Exercise 13.

Proof. This has been solved on [76, p. 143] which completes the proof of the problem.



10.3. Composition of Holomorphic Functions and Morera’s Theorem

10.3

13

Composition of Holomorphic Functions and Morera’s Theorem

Problem 10.14 Rudin Chapter 10 Exercise 14.

Proof. Both cases can be negative. Let Ω1 = C \ {0} and Ω2 = C. Define f (z) = z in Ω1 and   z, if z 6= 0; g(z) =  1, otherwise.

Then f (Ω1 ) ⊆ Ω2 and h(z) = g(f (z)) = g(z) = z in Ω1 . Hence f and h are holomorphic in Ω1 , but g is discontinuous in Ω2 . Next, we consider Ω1 = Ω2 = C. Define   −1, if z 6= 0; f (z) =  1, otherwise

and g(z) = z 2 . Obviously, we have f (Ω1 ) = {±1} ⊆ Ω2 and h(z) = g(f (z)) = 1 for all z ∈ Ω1 . It is clear that both g and h are holomorphic in Ω2 and Ω1 respectively, but f is not continuous in Ω1 . This ends the analysis of the proof.  Remark 10.1 A problem similar to Problem 10.14 but for (uniform) continuity has been discussed in [61, Exercise 26, p. 102].

Problem 10.15 Rudin Chapter 10 Exercise 15.

Proof. According to Theorem 10.18, we have f (z) = (z − ω0 )m h(z), where h ∈ H(ϕ(Ω)) and h(ω0 ) 6= 0. Then we have g(z) = f (ϕ(z)) = [ϕ(z) − ϕ(z0 )]m h(ϕ(z)).

(10.40)

Suppose that n ≥ 1 is the order of the zero of ϕ(z) − ϕ(z0 ). Now Theorem 10.18 implies that ϕ(z) − ϕ(z0 ) = (z − z0 )n φ(z), where φ ∈ H(Ω) and φ(z0 ) 6= 0. Assume that n ≥ 2. We follow from ϕ′ (z) = (z − z0 )n−1 [(z − z0 )φ′ (z) + φ(z)]

(10.41)

that ϕ′ (z0 ) = 0, a contradiction. Consequently, n = 1 and we can write the expression (10.40) as g(z) = (z − z0 )m · φm (z)h(ϕ(z)). (10.42)

Finally, since φm (z0 )h(ϕ(z0 )) = φm (z0 )h(ω0 ) 6= 0, the representation (10.42) ensures that g has a zero of order m at z0 .

14

Chapter 10. Elementary Properties of Holomorphic Functions

If ϕ′ has a zero of order k at z0 , then the expression (10.41) will imply that n = k + 1 so the representation (10.40) becomes g(z) = (z − z0 )m(k+1) · φm (z)h(ϕ(z)). In conclusion, g has a zero of order m(k + 1) at z0 . This completes the proof of the problem.  Problem 10.16 Rudin Chapter 10 Exercise 16.

Proof. Since ϕ is bounded on Ω × X, there exists a M > 0 such that |ϕ(z, t)| ≤ M for all (z, t) ∈ Ω × X. Let z0 ∈ Ω. Since Ω is open in C, there exists a ǫ > 0 such that D(z0 ; 3ǫ) ⊆ Ω. Then we have D(z0 ; 2ǫ) ⊆ Ω. We claim that for every pair z, ω ∈ D(z0 ; ǫ), z 6= ω and p ∈ X, we have ϕ(z, p) − ϕ(ω, p) 2M . ≤ z−ω ǫ

(10.43)

To this end, we consider the closed curve γ(t) = z0 + 2ǫeit , where t ∈ [0, 2π]. Obviously, since ϕ(z, t) ∈ H(D(z0 ; 3ǫ)) for each t ∈ X, we establish from Theorem 10.15 (The Cauchy’s Formula in a Convex Set) that if z ∈ D(z0 ; 2ǫ) ⊆ D(z0 ; 3ǫ) and p ∈ X, then z ∈ / γ ∗ and Z Z 2π 1 ϕ(ζ, p) ϕ(γ(t), p) ′ 1 ϕ(z; p) = dζ = · γ (t) dt. (10.44) 2πi γ ζ − z 2πi 0 γ(t) − z Note that z ∈ D(z0 ; ǫ) implies |γ(t) − z| > ǫ. Thus we follow from the formula (10.44) that Z 2π ϕ(z, p) − ϕ(ω, p) 1 1 1 · − ≤ · |ϕ(γ(t), p)| · |γ ′ (t)| dt z−ω 2π|z − ω| 0 γ(t) − z γ(t) − ω Z 2π z−ω 1

· 2M ǫ dt · ≤ 2π|z − ω| 0 γ(t) − z γ(t) − ω Z 2π Mǫ dt ≤ · π |γ(t) − z| · |γ(t) − ω| 0 Z 2π Mǫ dt ≤ 2 ǫ π 0 2M = ǫ which is exactly the inequality (10.43). Recall that µ is a complex measure, so Theorem 6.12 tells us that there is a measurable function h such that |h(x)| = 1 in X and dµ = h d|µ|. This fact ensures that Z 2M 2M |µ|(X) d|µ| = < ∞, ǫ X ǫ i.e.,

2M ǫ

∈ L1 (|µ|). Suppose that {zn } ⊆ D(z0 ; ǫ) \ {z0 } satisfies zn → z0 . Define gn (x) =

ϕ(zn , x) − ϕ(z0 , x) · h(x). zn − z0

By the hypotheses, we know that each gn is measurable of x and lim gn (x) = ϕ′ (z0 , x) · h(x).

n→∞

10.3. Composition of Holomorphic Functions and Morera’s Theorem

15

In other words, the sequence {gn } satisfies the conditions of Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem), so we conclude that ϕ′ (z0 , x) · h(x) ∈ L1 (|µ|) and furthermore, Z ϕ(zn , x) − ϕ(z0 , x) f (zn ) − f (z0 ) = lim · h(x) d|µ| lim n→∞ n→∞ zn − z0 zn − z0 ZX gn (x) d|µ| = lim n→∞ X Z ϕ′ (z0 , x) · h(x) d|µ|. = X

Consequently, f ′ (z0 ) exists. Since z0 is arbitrary, we establish that f ∈ H(Ω), completing the proof of the problem.  Problem 10.17 Rudin Chapter 10 Exercise 17.

Proof. We apply Problem 10.16 to the functions one by one. • The function f (z). We write f (z) =

Z

ϕ(z, t) dm,

X

were X = [0, 1] and ϕ(z, t) = (1 + tz)−1 . Notice that if z0 ∈ (−∞, −1], then ϕ(z0 , − z10 ) is unbounded. Thus it is reasonable to take Ω = C\(−∞, −1]. Now the function ϕ(z, t) satisfies all hypotheses of Problem 10.16 except the boundedness condition because ϕ(z, 1) → ∞ as z → −1 in Ω. Instead, it is really bounded locally. In fact, for every z ∈ Ω, there exists a δ > 0 such that D(z; δ) ⊂ D(z; 2δ) ⊆ Ω so that ϕ is bounded on D(z; δ) × X. Hence Problem 10.16 implies that f ∈ H(D(z; δ)). Since z is arbitrary, it yields that f ∈ H(Ω). • The function g(z). We have g(z) =

Z

ϕ(z, t) dm, X

where X = [0, ∞) and ϕ(z, t) = etz (1 + t2 )−1 . Since |etz | = etRe z , if Re (z0 ) > 0, then etRe (z0 ) = ∞. t→∞ 1 + t2

lim |ϕ(z0 , t)| = lim

t→∞

In other words, ϕ(z, t) is unbounded in any set containing a point of the right half plane. Thus we may take Ω to be the left half plane which is an open set in C. We want to apply Morera’s Theorem and Fubini’s Theorem.f Let z0 , zn ∈ Ω for all n ∈ N, where zn → z0 as n → ∞. Fix t ∈ [0, ∞), then it is easy to see that ϕ(z, t) is continuous at z0 . In addition, we know that |ϕ(z, t)| ≤ f

1 1 + t2

(10.45)

Problem 10.16 cannot be applied directly in this case because m(X) = ∞, i.e., m is not a complex measure on X.

16

Chapter 10. Elementary Properties of Holomorphic Functions 1 1 for all z ∈ Ω and 1+t 2 ∈ L (m). Then we deduce from Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) that Z Z ϕ(z0 , t) dm = g(z0 ), ϕ(zn , t) dm = lim g(zn ) = lim n→∞

n→∞ X

X

i.e., g is continuous at z0 . Since z0 is arbitrary, g is continuous in Ω. Next, we suppose that ∆ ⊆ Ω is a closed triangle and ∂∆ is parameterized by a piecewise continuously differentiable curve γ : [a, b] → Ω. We have Z

g(z) dz = ∂∆

Z

b

′

g(γ(x))γ (x) dx =

a

Z bZ a

∞

ϕ(γ(x), t)γ ′ (x) dt dx.

(10.46)

0

Define φ(x, t) = ϕ(γ(x), t) · γ ′ (x) on [a, b] × [0, ∞). Clearly, both [a, b] and [0, ∞) are σfinite measure spaces. Since γ is piecewise continuously differentiable on [a, b], there exists a M > 0 such that |γ ′ (x)| ≤ M on [a, b]. Furthermore, φ(x, t) is piecewise continuous on [a, b] × [0, ∞) so that φ is a measurable function on [a, b] × [0, ∞). Finally, for each x ∈ [a, b], we know from the inequality (10.45) that M |φ|x = |ϕ(γ(x), t) · γ ′ (x)| ≤ 1 + t2 which gives ∗

φ (x) =

Z

∞ 0

and

|φ|x dt ≤ M Z

b

a

Z

∞ 0

Mπ dt = 2 1+t 2

φ∗ dx < ∞.

Consequently, we may apply Theorem 8.8 (The Fubini Theorem) to change the order of integration in the integral (10.46) and get Z

g(z) dz = ∂∆

Z

0

∞hZ b a

Z i ϕ(γ(x), t)γ ′ (x) dx dt =

0

∞Z

ϕ(z, t) dz dt.

(10.47)

∂∆

Since ϕ(z, t) is holomorphic in Ω for every t ∈ [0, ∞), we conclude from Theorem 10.13 (The Cauchy’s Theorem for a Triangle) that Z ϕ(z, t) dz = 0, ∂∆

so the integral (10.47) reduces to Z

g(z) dz = 0.

∂∆

Finally, we apply Theorem 10.17 (Morera’s Theorem) to obtain the desired conclusion that g ∈ H(Ω). • The function h(z). We have h(z) =

Z

X

ϕ(z, t) dm,

10.4. Problems related to Zeros of Holomorphic Functions

17

where X = [−1, 1] and ϕ(z, t) = etz (1 + t2 )−1 . For every z ∈ C, we have z ∈ D(z; 1) ⊂ C. We know that |ϕ(z, t)| ≤ e|Re z| < ∞ in D(z; 1) × X. Thus the function ϕ satisfies all the requirements of Problem 10.16 in D(z; 1) × X, so h ∈ H(D(z; 1)). Since z is arbitrary, we conclude that h ∈ H(C), i.e., h is entire. We complete the proof of the problem.

10.4



Problems related to Zeros of Holomorphic Functions

Problem 10.18 Rudin Chapter 10 Exercise 18.

Proof. By [76, Problem 10.11, p. 131], we know that Z ′ X 1 f (z) p z dz = mk zkp , 2πi γ f (z) k

where the sum is taken over all the zeros of f inside γ and mk is the multiplicity of the zero zk of f . This is the answer of the first assertion. ′

(z) For the second assertion, let F (z) = ff (z) ϕ(z). Then F is a meromorphic function in Ω. Let A ⊆ Ω be the set of poles of F . Since f 6= 0 on γ ∗ , γ is a cycle in Ω \ A and Ind γ (α) = 0 for all α ∈ Ω. In other words, our F satisfies Theorem 10.42 (The Residue Theorem) which implies that Z ′ X f (z) 1 ϕ(z) dz = Res (F ; zk ), (10.48) 2πi γ f (z) k

where the zk denote the isolated singularities of F inside γ which are exactly the zeros of f inside γ. By Theorem 10.18, there exists a g ∈ H(Ω) and a unique positive integer mk such that f (z) = (z − zk )mk g(z) and g(zk ) 6= 0. Clearly, we have F (z) =

mk ϕ(z) g′ (z) f ′ (z) ϕ(z) = + ϕ(z). f (z) z − zk g(z)

If ϕ(zk ) = 0, then F is holomorphic at zk . Otherwise, F has a simple pole at zk and it yields from [9, Eqn. (1), p. 129] that Res (F ; zk ) = lim (z − zk )F (z) = mk ϕ(zk ). z→zk

(10.49)

Substituting the residues (10.49) into the formula (10.48), we get Z ′ X f (z) 1 ϕ(z) dz = mk ϕ(zk ), 2πi γ f (z) k

where the sum is taken over all the zeros of f which are not zeros of ϕ inside γ. This completes the proof of the problem.  Problem 10.19 Rudin Chapter 10 Exercise 19.

18

Chapter 10. Elementary Properties of Holomorphic Functions

Proof. We claim that f = cg for some nonzero constant c. To this end, we consider h = fg in U . Since f (z) 6= 0 on U , we know that h(z) 6= 0 on U . In addition, since f, g ∈ H(U ) and g(z) 6= 0 on U , we conclude that h ∈ H(U ). Direct differentiation gives f ′ (z)g(z) − f (z)g ′ (z) f (z) h f ′ (z) g′ (z) i h (z) = = · − g 2 (z) g(z) f (z) g(z) ′

so that h′ ( n1 ) = 0 on {1, 2, . . .}. By Theorem 10.18, we conclude that h′ (z) = 0 in U . Next, the Fundamental Theorem of Calculus [9, Proposition 4.12, p. 51] shows that Z h′ (ζ) dζ = h(0) 6= 0 h(z) = h(0) + [0,z]

for every z ∈ U , where [0, z] is a path connecting 0 and z in U . By the definition, we establish that f (z) = h(0)g(z) on U . This ends the proof of the problem.



Problem 10.20 Rudin Chapter 10 Exercise 20.

Proof. Suppose that f (z0 ) = 0 for some z0 ∈ Ω. Assume that f 6≡ 0. Then there exists a circle C(z0 ; R) for some R > 0 such that f (z) 6= 0 on C(z0 ; R). By Theorem 10.28, we know that fn′ → f ′ uniformly on compact subsets of Ω. Thus the convergence f′ fn′ → fn f is also uniform on C(z0 ; R). Next, Theorem 10.43(a) gives Z Z 1 f ′ (z) fn′ (z) 1 Nf = dz and Nfn = dz. 2πi C(z0 ;R) f (z) 2πi C(z0 ;R) fn (z) Our hypotheses give Nf = 1 but Nfn = 0 which is a contradiction, so we establish the result f ≡ 0 on Ω. For the second assertion, suppose that

∞ [

n=1

fn (Ω) ⊆ Ω′ .

Choose a point a ∈ C \ Ω′ . Define Fn : Ω → C by Fn = fn − a for every n = 1, 2, . . .. Then each Fn is holomorphic in Ω, Fn (z) 6= 0 for all z ∈ Ω and Fn → f − a uniformly on compact subsets of Ω. Thus the first assertion and the fact that f is nonconstant imply that f (z) 6= a in Ω. In other words, we must have f (Ω) ⊆ Ω′ which completes the proof of the problem.



Remark 10.2 We note that Problem 10.20 is classically called Hurwitz’s Theorem, see [9, Theorem 10.13, p. 139] or [18, p. 152].

10.4. Problems related to Zeros of Holomorphic Functions

19

Problem 10.21 Rudin Chapter 10 Exercise 21.

Proof. Let g(z) = f (z) − z and h(z) = −z on Ω. Since D(0; 1) ⊆ Ω and |f (z)| < 1 on |z| = 1, we have |g(z) − h(z)| = |f (z)| < 1 = |h(z)| on |z| = 1. By Theorem 10.43(b) (Rouch´e’s Theorem), we conclude that Ng = Nh = 1 in the  disc D(0; 1). This completes the analysis of the proof. Problem 10.22 Rudin Chapter 10 Exercise 22.

Proof. Assume that f (z) 6= 0 for all z ∈ Ω. By the Corollary to Theorem 10.24 (The Maximum Modulus Theorem), we see that 1 = |f (0)| ≥ min |f (reiθ )| > 2, θ

a contradiction. Hence f has at least one zero in the unit disc, completing the proof of the  problem. Problem 10.23 Rudin Chapter 10 Exercise 23.

Proof. Here we list some observations about the zeros of Pn and Qn . • By Theorem 10.25 (The Fundamental Theorem of Algebra), both Pn and Qn have precisely n zeros in C. • The definitions of Pn and Qn guarantee that Pn and Qn cannot have any common zeros. • Since Qn (0) = 0, Qn always has a zero at 0 for every n. • By [9, Exercise 12, p. 142], we know that for every R > 0, if n is large enough, Pn (z) has no zeros in |z| ≤ R, i.e., all zeros ζ of Pn satisfy |ζ| > R for large n. In fact, it has been shown further in [33] that every zero ζ of Pn lie in the annulus n < |ζ| < n e2 for large enough n. • Applying [13, Corollary 1.2.3, p. 13] to Qn , all the zeros z 6= 0 of Qn (z) = z + lie inside the annulus 2=

 zn z z n−1  z2 + ··· + = z 1 + + ··· + 2! n! 2! n!

min (k + 2) ≤ |z|
0 such that |ez − 1| > ǫk . (10.50) Since Pn (z) → ez uniformly in D(0; Rk ), there exists a Mk ∈ N such that n ≥ Mk implies |ez − 1 − Qn | = |ez − Pn | < ǫk .

(10.51)

Thus the inequalities (10.50) and (10.51) give |ez − 1 − Qn | < |ez − 1| for all n ≥ Mk and for all z ∈ C(0; Rk ). By Theorem 10.43(b) (Rouch´e’s Theorem), we conclude that NQn = Nez −1 = 2k + 1 inside C(0; Rk ) for all n ≥ Mk . This also means that NQn = n − 2k − 1

(10.52)

outside C(0; Rk ) for all n ≥ Mk .

Similarly, since 2(k + 1)π < Rk + 2π < 2(k + 2)π, Theorem 10.43(b) (Rouch´e’s Theorem) tells us that NQn = Nez −1 = 2k + 3

inside C(0; Rk + 2π) for all n ≥ Mk+1 , or equivalently, NQn = n − 2k − 3

(10.53)

outside C(0; Rk + 2π) for all n ≥ Mk+1 . Now if we combine the results (10.52) and (10.53), there exists a Mk′ ∈ N such that NQ n = 2 in the annulus A = {z ∈ C | Rk < |z| < Rk + 2π} for all large enough n ≥ Mk′ , where k = 1, 2, 3, . . .. We complete the proof of the problem.



Remark 10.3 There are some books concerning the location of the zeros of polynomials. For instances, [13, Chap. 1], [39] and [40, Chap. 3].

Problem 10.24 Rudin Chapter 10 Exercise 24.

Proof. We have Ω = K ◦ which is an open set in C. Put ϕ(z) = |f (z) − g(z)| − |f (z)| − |g(z)|, E = {z ∈ K | ϕ(z) = 0} and {zn } ⊆ E with zn → z0 as n → ∞. We divide the proof into several steps.

10.4. Problems related to Zeros of Holomorphic Functions

21

• Step 1: E ⊆ Ω. Our hypothesis implies that |f (z) − g(z)| < |f (z)| + |g(z)|

(10.54)

on ∂Ω = K \ Ω and this ensures that f and g cannot have any zero on ∂Ω. In addition, the continuity of f and g imply the continuity of ϕ and then z0 ∈ E. Thus E is a closed, hence compact, subset of K by Theorem 2.4. By the inequality (10.54), we know that E ∩ ∂Ω = ∅ which means that E ⊆ Ω and |f (z) − g(z)| < |f (z)| + |g(z)|

(10.55)

in K \ E. • Step 2: The numbers of zeros of f and g are finite. Suppose that Z(f ) = {a ∈ Ω | f (a) = 0}. Assume that Z(f ) was infinite. Then [79, Problem 5.25, p. 68] leads to us that Z(f ) has a convergent subsequence and Theorem 10.18 says that Z(f ) = Ω, but the continuity of f implies immediately that f ≡ 0 on K which is impossible. Consequently, Z(f ) is finite. Furthermore, since f (a) = 0 implies ϕ(a) = 0, all zeros of f lie in E. Similarly, Z(g) is also finite and all zeros of g belong to E. • Step 3: A lemma and its application. Here we need the following result whose proof can be found in [63, IX.8 & 9, pp. 115 – 118] or [67, Lemma 5.8, pp. 61, 62]: Lemma 10.1 Let G be an open subset of C and K a compact subset of G. Then there exists a cycle Γ in G \ K such that K ⊆ int Γ ⊆ G and Z f (ζ) 1 dζ (10.56) f (z) = 2πi Γ ζ − z for every f ∈ H(G) and z ∈ K. By Step 1, we see that E and Ω satisfy the roles of K and G of Lemma 10.1 respectively, so it ensures that there exists a cycle Γ in Ω \ E such that E ⊆ int Γ ⊆ Ω and the formula (10.56) holds for every f ∈ H(Ω) and every z ∈ E.

As an application, let Fe ∈ H(Ω), z ∈ E and F (ζ) = (ζ − z)Fe(ζ). Obviously, we have F ∈ H(Ω) and F (z) = 0, so the formula (10.56) gives Z

Γ

Fe(ζ) dζ = 0.

(10.57)

• Step 4: The calculation of |Z(f )| and |Z(g)|. Recall from the hypotheses and Step 2 that f ∈ H(Ω) and f has only finitely many zeros a1 , a2 , . . . , aN with multiplicities p1 , p2 , . . . , pN respectively in E. Then the function N

f ′ (z) X pk fe(z) = − f (z) z − ak k=1

22

Chapter 10. Elementary Properties of Holomorphic Functions can be shown to have a removable singularity at each ak by Theorem 10.18 and Definition 10.19. Thus fe is holomorphic in Ω. Hence it follows from the result (10.57) that Z

Γ

N

X f ′ (ζ) pk dζ = f (ζ) k=1

Z

Γ

dζ . ζ − ak

Next, we apply Lemma 10.1 with f ≡ 1 and z = ak to get Z dζ 2πi = . ζ − ak Γ

(10.58)

(10.59)

By combining the integrals (10.58) and (10.59), we conclude that 1 2πi

Z

N

Γ

X f ′ (ζ) dζ = pk = |Z(f )|. f (ζ)

(10.60)

k=1

Similarly, we obtain 1 2πi

Z

Γ

g′ (ζ) dζ = |Z(g)|. g(ζ)

(10.61)

By the strict inequality (10.55), it is impossible that f = −N ′ g for some N ′ ∈ N ∪ {0} on Ω \ E. Therefore, the function h : Ω \ E → C \ (−∞, 0] given by h(z) = log

f (z) g(z)

is well-defined and holomorphic in the open set Ω \ E. Taking differentiation, we have h′ =

f ′ g′ − f g

on Ω \ E. Since Z(f ), Z(g) * Ω \ E, h′ is continuous in Ω \ E. Since Γ is a cycle in Ω \ E, Theorem 10.12 guarantees that Z h′ (ζ) dζ = 0 Γ

or equivalently,

Z

Γ

f ′ (ζ) dζ = f (ζ)

Z

Γ

g′ (ζ) dζ. g(ζ)

(10.62)

Finally, by substituting the results (10.60) and (10.61) into the expression (10.62), we have established that |Z(f )| = |Z(g)|. Now we complete the proof of the problem.

10.5

Laurent Series and its Applications

Problem 10.25 Rudin Chapter 10 Exercise 25.



10.5. Laurent Series and its Applications

23

Proof. Define A(r1 , r2 ) = {z ∈ C | r1 < |z| < r2 }. Let ǫ > 0 be such that r1 + ǫ < r2 − ǫ. Furthermore, we define ρr : [0, 2π] → C by ρr (t) = reit , where r > 0. Particularly, we have γ1 = −ρr1 +ǫ and γ2 = ρr2 −ǫ . (a) Note that γ1 and γ2 are negatively oriented and positively oriented circles respectively. By Theorem 10.11, we have   −1, if z ∈ D(0; r1 + ǫ); Ind γ1 (z) =  0, if z ∈ / D(0; r1 + ǫ) and

Ind γ2 (z) =

  1, if z ∈ D(0; r2 − ǫ); 

0, if z ∈ / D(0; r2 − ǫ).

Let Γ = γ1 + γ2 which is the sum of two circles in A(r1 , r2 ). Let α ∈ / A(r1 , r2 ). If |α| ≤ r1 , then we have α ∈ D(0; r1 + ǫ) ⊆ D(0; r2 − ǫ). Thus we follow from [62, Eqn. (8), p. 218] that Ind Γ (α) = Ind γ1 (α) + Ind γ2 (α) = 0. (10.63) Similarly, if |α| ≥ r2 , then we still have the result (10.63). Next if z ∈ A(r1 + ǫ, r2 − ǫ), then z ∈ / D(0; r1 + ǫ) and z ∈ D(0; r2 − ǫ). Thus it is easy to check that Ind Γ (z) = Ind γ1 (z) + Ind γ2 (z) = 1. See Figure 10.4 for an illustration below.

Figure 10.4: The annulus A(r1 , r2 ) and the circles γ1 , γ2 . Hence, Theorem 10.35 (Cauchy’s Theorem) and [62, Eqn. (5), p. 217] assert that

24

Chapter 10. Elementary Properties of Holomorphic Functions

1 f (z) = 2πi

Z

for every z ∈ A(r1 + ǫ, r2 − ǫ).

Γ

f (ζ) 1  dζ = ζ −z 2πi

Z

Z  f (ζ) + dζ γ2 ζ − z γ1

(10.64)

(b) Let R1 = r1 + ǫ and R2 = r2 − ǫ so that r1 < R1 < |z| < R2 < r2 . Define Z Z 1 f (ζ) f (ζ) 1 f1 (z) = dζ and f2 (z) = dζ. 2πi −ρR ζ − z 2πi ρR ζ − z 1

(10.65)

2

Then the formula (10.64) simplifies to f = f1 + f2 . – Step 1: f1 and f2 are well-defined. To see this, we first recall from the hypothesis that f ∈ H(A(r1 , r2 )). Next, we fix z and take r1 < R1 < R1′ < |z| which means that z∈ / D(0; R1′ ). Let Γ1 = ρR1 − ρR′1 . By Theorem 10.11, we know that Ind Γ1 (α) = Ind ρR1 (α) − Ind ρR′ (α) = 0 1

for every α ∈ / A(r1 , r2 ). Thus it yields from Theorem 10.35 (Cauchy’s Theorem)g that Ind Γ1 (z) = 0 − 0 = 0 and then Z Z 1 f (ζ) 1 f (ζ) dζ = dζ. 2πi ρR ζ − z 2πi ρR′ ζ − z 1

1

In other words, f1 (z) is uniquely determined by z, not by R1 . Since ǫ is arbitrary, this means that f1 is actually well-defined in C \ D(0; r1 ). Similarly, we take |z| < R2 < R2′ < r2 so that z ∈ D(0; R2 ) ⊆ D(0; R2′ ). Let Γ2 = ρR2 − ρR′2 . By Theorem 10.11 again, we have Ind Γ2 (α) = Ind ρR2 (α) − Ind ρR′ (α) = 0 2

for every α ∈ / A(r1 , r2 ). Hence we have Ind Γ2 (z) = 1 − 1 = 0 and similar argument shows that f2 (z) is well-defined in D(0; r2 ). – Step 2: f2 ∈ H(D(0; r2 )). As suggested by the proof of Theorem 10.16, we may apply Theorem 10.7 to the integral representation (10.65) of f2 with X = [0, 2π], ϕ = ρR2 and dµ(t) = f (ρR2 (t))ρ′R2 (t) dt to establish the fact that f2 is representable by a power series in D(0; R2 ). Since ǫ is arbitrary, f2 is representable by a power series in D(0; r2 ) and hence Theorem 10.6 concludes that f2 ∈ H(D(0; r2 )).

– Step 3: f1 ∈ H(C \ D(0; r1 )). For f1 , we consider the function g(ζ) = ζ1 f ( 1ζ ) on
|ζ| = R11 . Tale X = [0, 2π], ϕ = ρ 1 and dµ(t) = g ρ 1 (t) ρ′ 1 (t) dt in Theorem 10.7. R1

Obviously, we have ϕ([0, 2π]) = ρ

and thus the function

R1

1 R1

R1

([0, 2π]) = C(0; R11 ), so ϕ([0, 2π]) ∩ D(0; R11 ) = ∅

1 g(z) = 2πi

Z

ρ

1 R1

dµ(ζ) ζ−z

is representable by power series in D(0; R11 ). Furthermore, for ρ have 1 2πi g

Z

ρ

1 R1

dµ(ζ) 1 = ζ −z 2πi

Z

2π

R1 e−it f (R1 e−it )

0

In fact, we have applied the formula [62, Eqn. (2), p. 219] here.

i it e · R1

1 R1

(t) =

dt −z

1 it R1 e

1 it R1 e ,

we

10.5. Laurent Series and its Applications 1 = 2πi

Z

2π 0

25

if (R1 e−it ) ·

R1 e−it dt ( R11 eit − z)R1 e−it

Z 2π 1 1 if (R1 e−it )R1 e−it dt · 1 −it z 2πi 0 z − R1 e Z f (ζ) 1 = dζ z −ρ 1 ζ − z1 R1  1 1 . = f1 z z =

Therefore, the function 1z f1 ( z1 ) is representable by power series in D(0; R11 ). Again, since ǫ is arbitrary, z1 f1 ( 1z ) can be represented by power series in D(0; r11 ) and then Theorem 10.6 ensures that z1 f1 ( 1z ) ∈ H(D(0; r11 )). Hence, this certainly shows that f1 ∈ H(C \ D(0; r1 )) as required.

– Step 4: Uniqueness of the decomposition. Suppose that f = g1 + g2 , where g1 ∈ H(C \ D(0; r1 )) and g2 ∈ H(D(0; r2 )). Then we have g1 − f1 = f2 − g2

(10.66)

in A(r1 , r2 ). Define h : C → C by   f2 (z) − g2 (z), if z ∈ D(0; r2 ); h(z) =  g1 (z) − f1 (z), if z ∈ C \ D(0; r1 ).

Now the equation (10.66) shows that the two definitions of h actually agree in A(r1 , r2 ), so h is well-defined in C and in fact it is entire. Thus it suffices to prove that h ≡ 0. Since f1 (z), g1 (z) → 0 as |z| → ∞, we have h(z) → 0 as |z| → ∞ and thus h is bounded. By Theorem 10.23 (Liouville’s Theorem), we conclude that h(z) = 0 for all z ∈ C and this means that f1 (z) = g1 (z) and f2 (z) = g2 (z) in C \ D(0; r1 ) and D(0; r2 ) respectively. This proves the uniqueness of the decomposition. (c)

– Existence of a Laurent series. On γ2 , we have |ζ| > |z| so that 1 z z2 1 1 = + 2 + 2 + ··· . z = ζ −z ζ(1 − ζ ) ζ ζ ζ

(10.67)

On −γ1 , since |ζ| < |z|, we have 1 ζ ζ2 1 1 =− = − − 2 − 3 − ··· . ζ ζ−z z z z z(1 − z )

(10.68)

Substituting the series (10.67) and (10.68) into the formula (10.64), we get 1 f (z) = 2πi

Z

∞ X f (ζ)z n

γ2 n=0

ζ n+1

1 dζ + 2πi

Z

−∞ X f (ζ)z n dζ. ζ n+1

(10.69)

−γ1 n=−1

Since the convergence of the series (10.67) and (10.68) are uniform, we can switch the order of integration and summation in the expression (10.69) to obtain f (z) =

∞ X −∞

cn z n ,

26

Chapter 10. Elementary Properties of Holomorphic Functions where 1 cn = 2πi

Z

γ

f (ζ) dζ ζ n+1

and γ =

  γ2 , 

if n = 0, 1, 2, . . .; (10.70)

−γ1 , if n = −1, −2, −3, . . ..

– Uniqueness of the Laurent series. We claim that if A, then an = cn for all n ∈ Z. To this end, since on γ, we have

Z

γ

∞

X f (z) dz = z k+1 −∞

Z

∞ X

∞ X

an z n converges to f in

−∞

an z n converges uniformly to f

−∞

an z n−k−1 dz = 2πiak , γ

where k ∈ Z. Hence it asserts from the definitions (10.70) that cn = an hold for all n ∈ Z. This proves the claim and then the uniqueness follows.

– Uniform convergence on compact subsets of A. Let K be a compact subset of A. It is easy to check that C \ A is closed in C and (C \ A) ∩ K = ∅. Thus we deduce from Problem 10.1 that there is a δ > 0 such that d(K, C \ A) = 2δ > 0. In addition, we have K ∩ C(0; r1 + δ) = ∅ and K ∩ C(0; r2 − δ) = ∅ which imply that K ⊆ A(r1 + δ, r2 − δ) ⊆ A(r1 , r2 ). For r1 + δ ≤ |z| ≤ r2 − δ, we know that ∞ X −∞

n

|cn z | ≤ ≤ ≤

−1 X

n=−∞ ∞ X n=1 ∞ X

n=1

< ∞.

n

|cn z | +

|cn | + |z|n

∞ X

n=0

∞ X

n=0

|cn z n |

|cn |(r2 − δ)n ∞

X |cn | |cn |(r2 − δ)n + (r1 + δ)n n=0

Hence it follows from the Weierstrass M -Test [9, Theorem 1.9, p. 15] or [61, Theorem 7.10, p. 148] that the series converges to f uniformly on K. (d) Let r1 < s1 < s2 < r2 . Firstly, since f2 ∈ H(D(0; r2 )) and D(0; s2 ) is a compact subset of D(0; r2 ), f2 is bounded in D(0; s2 ). Secondly, since f1 (z) → 0 as |z| → ∞, there exists a M > s1 such that |f1 (z)| < 1 (10.71)

on C \ D(0; M ). The set D(0; M ) \ D(0; s1 ) = {z ∈ C | s1 ≤ |z| ≤ M } is a closed and bounded subset in C, so it is compact. Since f1 ∈ H(C \ D(0; r1 )), f1 is bounded in D(0; M ) \ D(0; s1 ). This fact and the bound (10.71) combine to say that f1 is bounded in C \ D(0; s1 ). Now we write

f2 = f − f1 on D(0; r2 ) \ D(0; s2 ). Since D(0; r2 ) \ D(0; s2 ) ⊆ C \ D(0; s1 ), f1 is bounded there. Since f is bounded in A, f2 is bounded in D(0; r2 ) \ D(0; s2 ). Using the first result in the previous paragraph, we conclude that f2 is bounded in D(0; r2 ). Similarly, we write

10.5. Laurent Series and its Applications

27

f1 = f − f2 on D(0; s1 ) \ D(0; r1 ). Clearly, we know that D(0; s1 ) \ D(0; r1 ) ⊆ D(0; s2 ), so f2 is bounded there and thus f1 is also bounded there. Using the second assertion in the previous paragraph, we see that f1 is bounded in C \ D(0; r1 ). (e) All the foregoing parts remain valid if r1 = 0, or r2 = ∞ or both. In the case r2 = ∞, we note that f2 represents an entire function and thus it reduces to 0 by Theorem 10.23 (Liouville’s Theorem). (f) Suppose that 0 < r1 < r2 < r3 < · · · < rn−1 < rn < ∞. Each A(rk , rk+1 ) is an annulus, n−1 [ A(rk , rk+1 ) and f ∈ H(A). Then the foregoing where k = 1, 2, . . . , n − 1. Let A = k=1

results can be applied to f in each A(rk , rk+1 ) and obtain the corresponding Laurent series in each A(rk , rk+1 ).

Hence we have completed the analysis of the problem.



Problem 10.26 Rudin Chapter 10 Exercise 26.

Proof. Let f be the function in the question. The function f is holomorphic in C \ {−1, 1, 3}. By Problem 10.25, we have to consider the following three regions D(0; 1) = {z ∈ C | |z| < 1},

A = {z ∈ C | 1 < |z| < 3}

Note that if |z| < 1, then we have

and

B = {z ∈ C | |z| > 3}.

∞

X 1 z 2n . = 1 − z2

(10.72)

n=0

If |z| > 1, then | 1z | < 1 so that

∞

X 1 1 1 1 =− 2 · =− . 2 −2 1−z z 1−z z 2n

(10.73)

n=1

Similarly, if |z| < 3, then we have

| z3 |

< 1 so that

1 1 1 = · 3−z 3 1−

=

z 3

∞

1 X zn 3 n=0 3n

(10.74)

and if |z| > 3, then | 3z | < 1 which gives 1 1 1 =− · 3−z z 1−

3 z

=−

∞

1 X 3n . z zn n=0

Therefore, for every z ∈ D(0; 1), we follow from the series (10.72) and (10.74) that f (z) =

∞ X

n=0

Thus we have

z 2n +

1 X z n X h 1 + (−1)n 1 i n = + z . 3 3n 2 3n+1 ∞

∞

n=0

n=0

 1 1 + (−1)n   + n+1 , if n = 0, 1, 2, . . .; 2 3 cn =   0, otherwise.

(10.75)

28

Chapter 10. Elementary Properties of Holomorphic Functions

Next, if z ∈ A, then we deduce from the series (10.73) and (10.74) that ∞

∞ ∞ X X 1 1 X zn f (z) = − cn z n , + = 2n n z 3 3 n=−∞ n=1 n=0

where cn =

 1     3n+1 ,

if n ≥ 0;

 n+1    −1 + (−1) , otherwise. 2

Finally, if z ∈ B, then the series (10.73) and (10.75) imply that

∞ ∞ ∞ X X 1 X 3n 1 cn z n , − = f (z) = − z 2n z z n n=−∞ n=0

n=1

where cn =

 0,  

if n ≥ 0;

h i n   − 1 + (−1) + 3|n|−1 , otherwise. 2

Hence we have completed the analysis of the problem.



Problem 10.27 Rudin Chapter 10 Exercise 27.

Proof. Suppose that ζ = e2πiz

or

z=

1 i arg ζ − log |ζ|, 2π 2π

(10.76)

where 0 < arg ζ < 2π. Set F (ζ) = f (z). We claim that F (ζ) does not depend on the choice of arg ζ. In fact, it is easy to see from the second equation (10.76) that i arg ζ i arg ζ + 2π − log |ζ| = +1− log |ζ| = z + 1. 2π 2π 2π 2π Since f is a function of period 1, we have F (ζ) = f (z) = f (z + 1) = F (ζe2πi ) which proves our claim. Denote the annulus A = {ζ ∈ C | e−2πb < |ζ| < e−2πa }. Then the mapping G(z) = e2πiz clearly maps the horizontal strip Ω = {z ∈ C | a < Im z < b} onto A. Furthermore, G is holomorphic in Ω and f = F ◦ G. Here we need a positive result to Problem 10.14: Lemma 10.2 Suppose that Ω1 and Ω2 are two regions, f and g are nonconstant complex functions defined in Ω1 and Ω2 respectively. Put h = g ◦ f . If f and h are holomorphic in Ω1 and f (Ω1 ) = Ω2 , then g is also holomorphic in Ω2 .

10.5. Laurent Series and its Applications

29

Proof of Lemma 10.2. We consider S = {z ∈ Ω2 | every ω ∈ Ω1 with f (ω) = z implies f ′ (ω) = 0}.

(10.77)

Let β ∈ Ω2 \ S. Since f (Ω1 ) = Ω2 , there exists an α ∈ Ω1 such that f (α) = β. By the definition (10.77), f ′ (α) 6= 0. Using Theorem 10.30, there exist neighborhoods V ⊆ Ω1 and W ⊆ Ω2 of α and β respectively such that f : V → W is a bijection and a holomorphic inverse f −1 : W → V exists. Thus we have g = g ◦ (f ◦ f −1 ) = h ◦ f −1 on W which guarantees that g is holomorphic in W and thus g is holomorphic in Ω2 \ S. Next, we want to show that every point b of S is isolated. This means that one can find a neighborhood W ⊆ Ω2 of b such that W ∩ S = {b}. By the hypothesis, one can find an a ∈ Ω1 such that f (a) = b. By the definition (10.77), we have f ′ (a) = 0. Since f is nonconstant, f ′ is nonzero. Since f ′ ∈ H(Ω1 ), Theorem 10.18 implies that Z(f ′ ) has no limit point in Ω1 . In other words, Ω1 contains a neighborhood U of a such that f ′ (z) 6= 0

(10.78)

for all z ∈ U \ {a}. By the Open Mapping Theorem, f (U ) ⊆ Ω2 is our wanted neighborhood because if f (U ) contains a c ∈ S \ {b}, then we have f (ζ) = c for some ζ ∈ U \ {a} but the definition (10.77) shows that f ′ (ζ) = 0 which contradicts the result (10.78). Given ǫ > 0. Since h is continuous at a, there exists a δ > 0 such that ω ∈ D(a; δ) implies h(ω) ∈ D(h(a); ǫ). (10.79) Since f is holomorphic in Ω1 , f (D(a; δ)) is open in C by the Open Mapping Theorem. Since b ∈ f (D(a; δ)), there is a δ′ > 0 such that D(b; δ′ ) ⊆ f (D(a; δ)). Obviously, if z ∈ D(b; δ′ ), then we have f (ω) = z for some ω ∈ D(a; δ). Combining this fact with the property (10.79), we conclude that g(z) = g(f (ω)) = h(ω) ∈ D(h(a); ǫ) = D(g(f (a)); ǫ) = D(g(b); ǫ). Consequently, g is continuous on S. According to the Riemann’s Principle of Removable Singularities [9, Theorem 9.3, p. 118], every point of S is a removable singularity  of g. Hence g can be extended to a function holomorphic in Ω2 , as required. Since f and G are holomorphic in Ω and G(Ω) = A, Lemma 10.2 ensures that F is analytic in A. By Problem 10.25(c), F admits the Laurent series F (ζ) =

∞ X

cn ζ n

(10.80)

−∞

in A and this implies f (z) = F (ζ) =

∞ X

cn e2nπiz .

(10.81)

−∞

Finally, we denote Ω′ = {z ∈ C | a + ǫ ≤ Im z ≤ b − ǫ} and A′ = {ζ ∈ C | e−2π(b−ǫ) ≤ |ζ| ≤ e−2π(a+ǫ) }. Then we have G(Ω′ ) = A′ and A′ is a compact subset of A. By Problem 10.25(c), the series (10.80) converges uniformly in A′ and hence the corresponding series (10.81) converges uniformly in Ω′ . This completes the proof of the problem. 

30

Chapter 10. Elementary Properties of Holomorphic Functions

10.6

Miscellaneous Problems

Problem 10.28 Rudin Chapter 10 Exercise 28.

Proof. If we consider γ = Γ − α, then we observe immediately from [62, Eqn. (2), p. 203] that 1 Ind Γ (α) = 2πi

Z

2π 0

Γ′ (s) 1 ds = Γ(s) − α 2πi

Z

2π

0

γ ′ (s) ds = Ind γ (0). γ(s)

Thus, without loss of generality, we may assume that α = 0 in the following discussion. Now we prove the assertions one by one. • Ind Γn (0) = Ind Γm (0) if m and n are sufficiently large. Since Γ : [0, 2π] → C is a closed curve, it is a continuous function with period 2π. Thus the Stone-Weierstrass Theorem [61, Theorem 8.15, p. 190] asserts that there exists a sequence of trigonometric polynomials {Γn } converges to Γ uniformly in [0, 2π]. In other words, choose |Γ(θ)| > δ > 0, there exists an N ∈ N such that n ≥ N implies |Γ(θ) − Γn (θ)|
2ǫ

(10.82)

for all (s, t) ∈ I 2 and there exists a positive integer n such that |Γt (s) − Γt′ (s)| < ǫ

(10.83)

for all t, t′ ∈ I with |t − t′ | ≤ n1 and all s ∈ I. Let t0 = 0, tn = 1 and tk = tk−1 + n1 , where k = 1, 2, . . . , n. Then the set {t0 , t1 , . . . , tn } forms a partition of I and tk − tk−1 = n1 for each k = 1, 2, . . . , n. Therefore, it follows from the inequalities (10.82) and (10.83) that |Γtk (s) − Γtk−1 (s)| < ǫ < |Γtk−1 (s)| h

The author has proven it in [77, Problem 8.26, pp. 204 – 206].

10.6. Miscellaneous Problems

31

for every s ∈ I, where k = 1, 2, . . . , n. By the improved version of Lemma 10.39, we see that Ind Γtk (0) = Ind Γtk−1 (0), where k = 1, 2, . . . , n. Consequently, we have the desired result that Ind Γ0 (0) = Ind Γt0 (0) = Ind Γt1 (0) = · · · = Ind Γtn (0) = Ind Γ1 (0). This ends the proof of the problem.



Problem 10.29 Rudin Chapter 10 Exercise 29.

Proof. Suppose that z 6= 0, so we can write 1 f (z) = π 1 π

=

1 π

=

Z Z Z

0

0

0

1Z π

−π 1Z π −π 1Z π

Z

r dθ dr +z

reiθ

re−iθ dθ dr r + ze−iθ

r −iθ ze r −iθ −π z + e Z π 1 h

dθ dr

i −ieiθ r dθ dr r −iθ − (− ) −π e 0 z z Z π i 1 h r 1 −ieiθ dθ dr r −iθ − (− ) 0 z 2πi −π e z Z 1  −r  r dr = −2 · Ind γ z 0 z

1 =− πi Z = −2

(10.84)

where γ(θ) = e−iθ with −π ≤ θ ≤ π which is the negatively oriented circle with center at 0 and radius 1. By Theorem 10.11, we know that  r  −1, if |z| < 1;  −r   Ind γ =  z r  0, if |z| > 1. If |z| < 1, then the expression (10.84) reduces to hZ f (z) = −2 =2

Z

|z|

0

|z|

0

Z

 −r  r dr + · Ind γ z z

r dr z

2 |z| r dr z 0 |z|2 = z = z. =

Z

1

|z|

 −r  i r dr · Ind γ z z

32

Chapter 10. Elementary Properties of Holomorphic Functions

Next, if |z| ≥ 1, then we always have Ind γ (− zr ) = −1 so that 2 f (z) = z

Z

1

r dr = 0

1 . z

Finally, if z = 0, then it is easy to see that Z Z 1 1 π −iθ e dθ dr = 0. f (0) = π 0 −π We have completed the analysis of the problem.



Problem 10.30 Rudin Chapter 10 Exercise 30.

Proof. Without loss of generality, we may assume that Ω = C \ {−1, 1} which is clearly open in C. Consider the boundaries γ1 , γ2 , γ3 and γ4 of the discs D(−1; 1), D(1; 1), D(−2; 2) and D(2; 2) respectively. Each starts and ends at 0 with the orientation as shown in Figure 10.5. Then

Figure 10.5: A non null-homotopic closed path Γ = γ1 − γ3 − γ2 + γ4 in Ω. Γ = γ1 − γ3 − γ2 + γ4 is a closed curve in Ω and we observe easily that Ind Γ (±1) = 0. Assume that Γ was null-homotopic to the constant map 0 in Ω. By §10.38, there exists a continuous map Γt : [0, 1] → Ω connecting Γ and 0 such that Γ0 = 0 and Γ1 = Γ. However, the continuity of Γt forces that it must pass through one of the omitted points 1 and −1, a contradiction. This completes the proof of the problem. 

CHAPTER

11

Harmonic Functions

11.1

Basic Properties of Harmonic Functions

Problem 11.1 Rudin Chapter 11 Exercise 1.

Proof. Let u, v : Ω → R. We prove the assertions one by one. • uv is harmonic if and only if u + icv ∈ H(Ω) for some c ∈ R. Direct computation gives ∆(uv) = (uv)xx + (uv)yy = (uvxx + 2ux vx + uxx v) + (uvyy + 2uy vy + uyy v) = 2(ux vx + uy vy ) = 2(ux , uy ) · (vx , vy ). Thus uv is harmonic if and only if (ux , uy ) · (vx , vy ) = 0.

(11.1)

If u + icv ∈ H(Ω), then it yields from the Cauchy-Riemann equations that ux = cvy and uy = −cvx so that ux vx + uy vy = cvy vx − cvx vy = 0 in Ω. This means that ∆(uv) = 0. Conversely, suppose that uv is harmonic in Ω. If u is constant, then Theorem 11.2 ensures that ux = uy = 0 in Ω which gives the equation (11.1). The case for v being constant is similar. Therefore, without loss of generality, we may assume that both u and v are nonconstant. Clearly, the equation (11.1) implies that there exists a function c : R2 → R such that (ux , uy ) = c(x, y)(vy , −vx ) in Ω, or equivalently, ux = c(x, y)vy

and uy = −c(x, y)vx

(11.2)

in Ω. On the one hand, since ∆u = 0 in Ω, we obtain cx vy − cy vx = 0 in Ω. On the other hand, uxy − uyx = 0 and ∆v = 0 imply that cy vy + cvyy − (−cx vx − cvxx ) = 0 33

(11.3)

34

Chapter 11. Harmonic Functions and thus cy vy − cx vx = 0.

(11.4)

Eliminating vx and vy from the equations (11.3) and (11.4), we get (c2x + c2y )vx = 0 and

(c2x + c2y )vy = 0

(11.5)

in Ω. Since v is harmonic in Ω, the function f = vx − ivy is holomorphic in Ω by Theorem 11.2. If Z(f ) = Ω, then vx = vy = 0 in Ω so that v is constant in Ω, a contradiction. By Theorem 10.18, Z(f ) has no limit point in Ω. Pick z ∈ Ω \ Z(f ). Then there exists a δ > 0 such that D(z; δ) ∩ Z(f ) = ∅. By the equations (11.5), we have (c2x + c2y )f = 0 which implies that c2x + c2y = 0 in D(z; δ). Since z is arbitrary, we conclude that c2x + c2y = 0

(11.6)

in Ω \ Z(f ). Since f = vx − ivy ∈ H(Ω), f has continuous derivatives of all orders. Thus both vx and vy have continuous partial derivatives of all orders in Ω. Similarly, both ux and uy have continuous partial derivatives of all orders in Ω. Therefore, it follows from any one of the equations (11.2) that both cx and cy are continuous on Ω. Hence the equation (11.6) holds in Ω so that cx = cy = 0 in Ω. In conclusion, c ∈ R and hence u + icv satisfies the Cauchy-Riemann equations. • u2 cannot be harmonic in Ω unless u is constant. This part is shown in [76, Problem 16.3, pp. 199, 200]. • |f |2 is harmonic. Since f ∈ H(Ω), we know from Theorem 11.4 that both u and v are harmonic in Ω. Note that |f |2 is harmonic in Ω if and only if u2 + v 2 is harmonic in Ω. Clearly, we have ∆(u2 + v 2 ) = (u2 + v 2 )xx + (u2 + v 2 )yy = (2uux + 2vvx )x + (2uuy + 2vvy )y = (2uuxx + 2u2x + 2vvxx + 2vx2 ) + (2uuyy + 2u2y + 2vvyy + 2vy2 ) = 2(u2x + vx2 + u2y + vy2 ). Therefore, ∆(u2 + v 2 ) = 0 in Ω if and only if ux = uy = vx = vy = 0 in Ω if and only if both u(x, y) and v(x, y) are constant in Ω. In other words, |f |2 is harmonic in Ω if and only if f is constant in Ω. This completes the proof of the problem.



Problem 11.2 Rudin Chapter 11 Exercise 2.

Proof. Let f = u + iv. The result is obvious if f is constant, so without loss of generality, we may assume that f is nonconstant. Since f is harmonic in Ω, both u and v are harmonic in Ω. Similarly, since f 2 = u2 − v 2 + 2iuv is harmonic in Ω, both u2 − v 2 and uv are harmonic in Ω. Thus ∆(u2 − v 2 ) = 0 and ∆(uv) = 0 imply that u2x + u2y = vx2 + vy2

and ux vx + uy vy = 0

in Ω respectively. Consequently, they show that (ux + ivx )2 + (uy + ivy )2 = 0

11.1. Basic Properties of Harmonic Functions

35

(ux + ivx )2 − [i(uy + ivy )]2 = 0

[ux + ivx + i(uy + ivy )][ux + ivx − i(uy + ivy )] = 0 (ux − vy + ivx + iuy )(ux + vy + ivx − iuy ) = 0

holds in Ω. Next, the equation (ux − ivx )2 + (uy − ivy )2 = 0 gives (ux − vy − ivx − iuy )(ux + vy − ivx + iuy ) = 0

(11.7)

g = ux − vy − i(vx + uy ) and h = ux + vy + i(uy − vx ).

(11.8)

in Ω. We denote

It is obvious that (ux − vy )x = uxx − vyx = −uyy − vxy = −(uy + vx )y and (ux − vy )y = uxy − vyy = uyx + vxx = −[−(uy + vx )]x hold in Ω. In other words, g ∈ H(Ω). Similarly, we have h ∈ H(Ω). Assume that g 6≡ 0 and h 6≡ 0 in Ω. Then Theorem 10.18 ensures that Z(g) and Z(h) are at most countable. Therefore, Z(gh) = Z(g)Z(h) is also at most countable. Note that zeros of h and h are identical. Therefore, the equation (11.7) says that Z(gh) = Ω, a contradiction. Hence we have g ≡ 0 or h ≡ 0 in Ω. If g ≡ h ≡ 0 in Ω, then we deduce from the definition (11.8) that ux = uy = 0 in Ω, i.e., u is constant in Ω. By this and the fact f ∈ H(Ω), we obtain immediately from Theorem 11.2 that v is also constant in Ω. Hence f is also constant in H(Ω), a contradiction. In other words, we have either g ≡ 0 or h ≡ 0 in Ω. • Case (i): g ≡ 0 in Ω. By the definition, g ≡ 0 implies that ux − vy = i(vx + uy ) in Ω. Since ux , uy , vx and vy are real functions, we get ux = vy and uy = −vx in Ω. By Theorem 11.2, we conclude that f ∈ H(Ω). • Case (ii): h ≡ 0 in Ω. Similarly, this shows that ux = −vy and uy = vx in Ω and these mean that f = u − iv ∈ H(Ω). We have completed the proof of the problem.



Problem 11.3 Rudin Chapter 11 Exercise 3.

Proof. Let u be a real function. Suppose that V = {z ∈ Ω | grad u = 0} = {z ∈ Ω | ux (z) = uy (z) = 0}

and f = ux − iuy .

Since ux and uy are continuous in Ω, V is closed in Ω. By Theorem 11.2, f ∈ H(Ω). It is trivial that V = Z(f ). By Theorem 10.18, either V = Ω or V has no limit point in Ω.a This ends the  analysis of the problem. a

Note that the case V = Ω implies that u is constant in Ω.

36

Chapter 11. Harmonic Functions Problem 11.4

Rudin Chapter 11 Exercise 4.

Proof. We prove the assertions one by one. • Every partial derivative of a harmonic is harmonic. See [76, Problem 16.2, p. 199]. • Pr (θ − t) is a harmonic function of reiθ for a fixed t. It can be shown easily that the Laplacian equation in polar form is given by 1 1 ∆u = urr + ur + 2 uθθ . r r

(11.9)

Fix t. Write P = Pr (θ − t) and A = A(r, θ) = 1 − 2r cos(θ − t) + r 2 for convenience. Then [62, Eqn. (2), §11.5, p. 233] can be written as P A = 1 − r 2 . Direct differentiation gives Pr 2[r − cos(θ − t)] 2 =− P− , r rA A 4[r − cos(θ − t)] 2 2 Prr = − Pr − P − , A A A 8 sin2 (θ − t) 2 cos(θ − t) Pθθ = P− P. r2 A2 rA

(11.10)

Put these equations (11.10) into the Laplacian equation (11.9) to get 4[r − cos(θ − t)] 2 2 2[r − cos(θ − t)] 2 Pr − P − − P− A A A rA A 2 cos(θ − t) 8 sin2 (θ − t) P− P + A2 rA 8r[r − cos(θ − t)] 4 4 8 sin2 (θ − t) 8[r − cos(θ − t)]2 P + − P − + P = A2 A2 A A A2 8r[r − cos(θ − t)] − 4A 8[r − cos(θ − t)]2 − 4A + 8 sin2 (θ − t) P+ = 2 A A2 2 2 2 8[r − 2r cos(θ − t) + 1] − 4A 4[r + r − 2r cos(θ − t)] − 4A = P+ 2 A A2 2 4(r + A − 1) − 4A 4P + = A A2 2 4(1 − r ) 4(r 2 − 1) = + A2 A2 = 0.

∆P = −

Hence Pr (θ − t) is a harmonic function of reiθ for a fixed t. • P [ dµ] is harmonic in U . Suppose that µ is a finite Borel measure on T and u = P [ dµ]. It suffices to show that Z Z Z ∂ ∂ Pr (θ − t) dµ(eit ) = Pr (θ − t) dµ(eit ) = [Pr (θ − t)]r dµ(eit ) (11.11) ur = ∂r T T ∂r T and ∂ uθ = ∂θ

Z

T

it

Pr (θ − t) dµ(e ) =

Z

T

∂ Pr (θ − t) dµ(eit ) = ∂θ

Z

T

[Pr (θ − t)]θ dµ(eit ) (11.12)

11.1. Basic Properties of Harmonic Functions

37

because they certainly yield that Z Z it urr = [Pr (θ − t)]rr dµ(e ) and uθθ = [Pr (θ − t)]θθ dµ(eit ) T

T

so that ∆u =

Z

T

∆Pr (θ − t) dµ(eit ) = 0.

In other words, u = P [ dµ] is harmonic in U . Fix r ∈ [0, 1). For very small h > 0 such that 1 − r − 2h > 0, we note that Z u(r + h, θ) − u(r, θ) Pr+h (θ − t) − Pr (θ − t) = dµ(eit ) h h T Z Pr+h (θ − t) − Pr (θ − t) ur = lim dµ(eit ). h→0 T h We observe that lim

h→0

(11.13)

∂ Pr+h (θ − t) − Pr (θ − t) = Pr (θ − t) h ∂r

and Pr+h (θ − t) − Pr (θ − t) h i 1 − r2 1 − (r + h)2 1 h − = · h 1 − 2(r + h) cos(θ − t) + (r + h)2 1 − 2r cos(θ − t) + r 2 [1 − (r + h)2 ][1 − 2r cos(θ − t) + r 2 ] − (1 − r 2 )[1 − 2(r + h) cos(θ − t) + (r + h)2 ] = h[1 − 2(r + h) cos(θ − t) + (r + h)2 ] · [1 − 2r cos(θ − t) + r 2 ] 2h cos(θ − t) + 2r 2 − 2(r + h)2 + 2hr(r + h) cos(θ − t) = h[1 − 2(r + h) cos(θ − t) + (r + h)2 ][1 − 2r cos(θ − t) + r 2 ] 2h cos(θ − t) − 2h(2r + h) + 2hr(r + h) cos(θ − t) = h[1 − 2(r + h) cos(θ − t) + (r + h)2 ] · [1 − 2r cos(θ − t) + r 2 ] 2 cos(θ − t) − 2(2r + h) + 2r(r + h) cos(θ − t) = . (11.14) [1 − 2(r + h) cos(θ − t) + (r + h)2 ] · [1 − 2r cos(θ − t) + r 2 ] Since 1 − r − 2h > 0 implies 1 − (r + h) >

1−r 2 ,

this and the expression (11.14) give

P (θ − t) − P (θ − t) 2 + 2(2 + h) + 2(1 + h) 4(8 + 4h) 40 r+h r ≤ ≤ ≤ h (1 − r)2 [1 − (r + h)]2 (1 − r)4 (1 − r)4

for every eit ∈ T . Since µ is a finite Borel measure on T , we have Z 40 40 dµ(eit ) = · µ(T ) < ∞ 4 (1 − r) (1 − r)4 T

40 1 Hence Theorem 1.34 (The Lebesgue’s Dominated which means that (1−r) 4 ∈ L (T ). Convergence Theorem) ensures that the order of integration and the limit in (11.13) can be changed and this action shows that the formula (11.11) holds. Similarly, it is easily to check that

P (θ + h − t) − P (θ − t) −4r(1 − r 2 ) sin(θ − t + h2 ) sin h2 r r = 2 2 h h[1 − 2r cos(θ + h − t) + r ][1 − 2r cos(θ − t) + r ] 2r(1 − r 2 ) sin h2 · h ≤ (1 − r)4 2

38

Chapter 11. Harmonic Functions 3 4 · (1 − r)3 2 6 . = (1 − r)3 ≤

6 1 Again, the finiteness of µ implies that (1−r) 3 ∈ L (T ) and we can apply Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) to conclude that the formula (11.12) holds.

This ends the proof of the problem.



Problem 11.5 Rudin Chapter 11 Exercise 5.

Proof. Let f = u + iv. Since f ∈ H(Ω), both u and v are harmonic in Ω. Since |f | = 6 0 in Ω, 1 2 2 2 log |f | is a well-defined real function and in fact log |f | = log(u + v ) . Using ∆u = ∆v = 0 and the Cauchy-Riemann equations, we get 1

1

∆(log |f |) = [log(u2 + v 2 ) 2 ]xx + [log(u2 + v 2 ) 2 ]yy

 1 = log(u2 + v 2 ) xx + log(u2 + v 2 ) yy 2  uu + vv   uu + vv  y y x x = + u2 + v 2 x u2 + v 2 y (u2 + v 2 )(u∆u + u2x + u2y + v∆v + vx2 + vy2 ) − 2(uux + vvx )2 − 2(uuy + vvy )2 = (u2 + v 2 )2 2 2 2 2 2 2 (u + v )(ux + uy + vx + vy ) − 2[u2 (u2x + u2y ) + 2uv(ux vx + uy vy ) + v 2 (vx2 + vy2 )] = (u2 + v 2 )2 2 2 2 2 2 2 2 2(u + v )(vx + vy ) − 2(u + v )(vx + vy2 ) = (u2 + v 2 )2 =0 in Ω. Hence log |f | is harmonic in Ω.

If f is holomorphic and non-vanishing in Ω, then (log f )′

f′ f ,

1 ′ f,f

∈ H(Ω). Thus we obtain

f′ f

∈ H(Ω).

we conclude log f ∈ H(Ω) so that its real part, which is log |f |, is harmonic Since = in Ω, completing the proof of the problem.  Problem 11.6 Rudin Chapter 11 Exercise 6.

Proof. Let A(Ω) be the area of the region Ω = f (U ). Referring to [2, §2.4, pp. 75, 76],b we know that ZZ |f ′ (z)|2 dx dy, A(Ω) = U

b

See also the discussion on [59, Eqn. (6), p. 11].

11.1. Basic Properties of Harmonic Functions

39

where f (z) = u(x, y) + iv(x, y). Put z = reiθ , where 0 ≤ r < 1 and 0 ≤ θ ≤ 2π. Then the formula of A(Ω) becomes Z

A(Ω) =

Z

2π 0

1

0

|f ′ (r cos θ, r sin θ)|2 r dr dθ.

(11.15)

Since |f ′ (z)|2 = f ′ (z) · f ′ (z), we see that |f ′ (r cos θ, r sin θ)|2 = =

∞   X mcm r m−1 e−i(m−1)θ ncn r n−1 ei(n−1)θ ×

∞ X

m=1

n=1

∞ X

nmcn cm r n+m−2 ei(n−m)θ .

m,n=1

Recall from Definition 4.23 that {einθ } forms an orthonormal set, the integral (11.15) becomes A(Ω) =

Z

2π

0

′

Since f (z) =

∞ X

n=1

Z

0

∞ 1X

n=1

2

2 2n−1

n |cn | r



dr dθ.

(11.16)

ncn z n−1 ∈ H(U ), we have lim sup n→∞

which implies that lim sup n→∞

p n n|cn | = 1

p n

n2 |cn |2 = 1.

Therefore, the radius of convergence of the series

∞ X

n=1

n2 |cn |2 r 2n−1 is 1 and then the order of

integration in the integral (11.16) can be switched so that A(Ω) =

Z

0

∞ Z 1 2π X n=1

0

∞ ∞  X X 1 n2 |cn |2 × n2 |cn |2 r 2n−1 dr dθ = 2π n|cn |2 , =π 2n n=1

n=1

as required. This completes the proof of the problem.



Problem 11.7 Rudin Chapter 11 Exercise 7.

Proof. Suppose that f = u + iv. (a) Let ψ be a twice differentiable function on (0, ∞). Then f f = u2 + v 2 = |f |2 . Using the formulas [62, Eqn. (3), p. 231], we see that ∂  ∂ ∂  1 ∂ ψ(|f |2 ) −i +i 4 ∂x ∂y ∂x ∂y  1 ∂ ∂  ′ = ψ (|f |2 ) · (uux + vvx ) + iψ ′ (|f |2 ) · (uuy + vvy ) −i 2 ∂x ∂y 1 n ′′ 2ψ (|f |2 ) · (uux + vvx )2 + ψ ′ (|f |2 ) · (u2x + uuxx + vx2 + vvxx ) = 2

∂∂[ψ ◦ (f f )] =

40

Chapter 11. Harmonic Functions + iψ ′′ (|f |2 ) · (uux + vvx )(uuy + vvy )

+ iψ ′ (|f |2 ) · (ux uy + uuxy + vx vy + vvxy )  + −iψ ′′ (|f |2 ) · (uuy + vvy )(uux + vvx )



−iψ ′ (|f |2 )(uy ux + uuxy + vx vy + vvxy )

o + 2ψ ′′ (|f |2 ) · (uuy + vvy )2 + ψ ′ (|f |2 ) · (u2y + uuyy + vy2 + vvyy ) . (11.17)

Since f ∈ H(Ω), both u and v are harmonic in Ω. Furthermore, we note from Theorem 11.2 that |f ′ |2 = u2x + vx2 = u2y + vy2 , so the expression (11.17) becomes ∂∂[ψ ◦ (f f )] =

  1 n ′′ 2ψ (|f |2 ) · u2 (u2x + u2y ) + v 2 (vx2 + vy2 ) 2  o + ψ ′ (|f |2 ) · u2x + u2y + vx2 + vy2 + u(uxx + uyy ) + v(vxx + vyy )

 1  ′′ 2ψ (|f |2 ) · |f |2 · |f ′ |2 + 2ψ ′ (|f |2 ) · |f ′ |2 2 = [|f |2 ψ ′′ (|f |2 ) + ψ ′ (|f |2 )] · |f ′ |2

=

= (ϕ ◦ |f |2 ) · |f ′ |2

(11.18)

as required. α

Now the function ψ(t) = t 2 is clearly twice differentiable on (0, ∞) with ϕ(t) =

 α αα α α α2 α −1 − 1 t 2 −1 + t 2 −1 = t2 , 2 2 2 4

so we combine the formula (11.18) and [62, Eqn. (3), p. 232] to get ∆(|f |α ) = 4∂∂(|f |α ) = α2 |f |α−2 |f ′ |2 . (b) Suppose that Φ : f (Ω) → C is defined by Φ(f ) = Φ(u + iv) = Φ(u(x, y), v(x, y)). Since uxx + uyy = vxx + vyy = 0, we have ∆[Φ ◦ f ] = ∆Φ(f )

∂2 ∂2 Φ(u, v) + Φ(u, v) ∂x2 ∂y 2 ∂ ∂ (Φu · ux + Φv · vx ) + (Φu · uy + Φv · vy ) = ∂x ∂y ∂ ∂ = ux Φu + Φu · uxx + vx Φv + Φv · vxx ∂x ∂x ∂ ∂ + uy Φu + Φu · uyy + vy Φv + Φv · vyy ∂y ∂y = ux (Φuu · ux + Φuv · vx ) + Φu · uxx + vx (Φuv · ux + Φvv · vx ) + Φv · vxx =

+ uy (Φuu · uy + Φuv · vy ) + Φu · uyy + vy (Φuv · uy + Φvv · vy ) + Φv · vyy

= Φuu · (u2x + u2y ) + Φvv · (vx2 + vy2 ) + 2Φuv · ux vx + 2Φuv · uy vy .

(11.19)

According to the Cauchy-Riemann equations, we can further reduce the expression (11.19) to ∆[Φ ◦ f ] = Φuu · (u2x + vx2 ) + Φvv · (vx2 + u2x ) = (Φuu + Φvv ) · |f ′ |2

= [(∆Φ) ◦ f ] · |f ′ |2

(11.20)

11.1. Basic Properties of Harmonic Functions

41

as desired. Finally, we take Φ(w) = |w| so that Φ(w) = Φ(|w|). Suppose that f = u + iv and α w = |f |α = (u2 + v 2 ) 2 . Direct differentiation implies that α

α

α

Φuu + Φvv = 2α(u2 + v 2 ) 2 −1 + α(α − 2)u2 (u2 + v 2 ) 2 −2 + α(α − 2)v 2 (u2 + v 2 ) 2 −2 α

= α2 (u2 + v 2 ) 2 −1

and thus (∆Φ) ◦ f = α2 |f |α−2 . Substituting this into the formula (11.20), we have established that ∆(|f |α ) = α2 |f |α−2 · |f ′ |2 . This ends the analysis of the problem.



Problem 11.8 Rudin Chapter 11 Exercise 8.

Proof. The proof of this problem will be divided into two steps as follows: • Step 1: {fn (z)} converges at every point of Ω. Suppose that S1 = {z ∈ Ω | {fn (z)} converges} Let a ∈ Ω and R > 0 be such that D(a; 2R) ⊆ Ω.

(11.21)

We consider the functions u cn (z) = un (z + a) and c fn (z) = fn (z + a) which are defined in D(0; 2R) and in D(0; 2R) respectively. By [9, Theorem 16.9, p. 233]c , we have Z π 2Reit + z  1 c c u cn (2Reit ) dt fn (z) = iIm fn (0) + 2π −π 2Reit − z and thus

fn (z + a) = iIm fn (a) +

1 2π

Z

 2Reit + z  un (a + 2Reit ) dt, it − z 2Re −π π

where z ∈ D(0; 2R). If we take z = Reiθ , then it is easy to see that

and this implies that

c

2Reit + z 2Reit + Reiθ ≤3 = it it iθ 2Re − z 2Re − Re

|fn (z + a) − fm (z + a)| ≤ Im [fn (a) − fm (a)] 1 Z π  2Reit + z  it it [u (a + 2Re ) − u (a + 2Re )] dt + n m 2π −π 2Reit − z Z π 3 un (a + 2Reit ) − um (a + 2Reit ) dt ≤ |fn (a) − fm (a)| + 2π −π

In fact, this is a generalization of Theorem 11.9 and it is called the Schwarz Integral Formula. See, for example, [37, p. 408].

42

Chapter 11. Harmonic Functions ≤ |fn (a) − fm (a)| + 3 · max un (a + 2Reit ) − um (a + 2Reit ) . −π≤t≤π

Consequently, we conclude that sup z∈D(0;R)

|fn (z + a) − fm (z + a)| ≤ 3 · max un (a + 2Reit ) − um (a + 2Reit ) −π≤t≤π

+ |fn (a) − fm (a)|

or equivalently, sup z∈D(a;R)

|fn (z) − fm (z)| ≤ 3 ·

max

ζ∈C(a;2R)

|un (ζ) − um (ζ)| + |fn (a) − fm (a)|.

(11.22)

Given ǫ > 0. Now our hypothesis asserts that there is an N1 ∈ N such that n, m ≥ N1 imply ǫ |un (ζ) − um (ζ)| < 6 for all ζ ∈ D(a; 2R). Thus it follows from the inequality (11.22) that |fn (z) − fm (z)| ≤ |fn (a) − fm (a)| +

ǫ 2

(11.23)

for all z ∈ D(a; R).

Particularly, if a ∈ S1 , then there exists an N2 ∈ N such that n, m ≥ N2 imply |fn (a) − fm (a)|
0 such that for every a ∈ A and D(a; 2R) ⊆ Ω for some R > 0, we have D(a; 2θR) ⊆ A, then A is relatively open and closed in Ω. Furthermore, if Ω is a region, then we have A = Ω.

Put A = S1 , the above set relations (11.21) and (11.25) mean that we can take θ = 21 in Lemma 11.1 (The Basic Connectedness Lemma). Hence we conclude immediately that S1 = Ω, as desired. • Step 2: {fn } converges uniformly on compact subsets of Ω. Let K be a compact subset of Ω. Clearly, [ K⊆ D(a; Ra ), a∈K

11.1. Basic Properties of Harmonic Functions

43

where Ra > 0 is a number satisfying the set relation (11.21). Therefore, it follows from Step 1 that the inequality (11.24) holds in D(a; Ra ) and all n, m ≥ N (a, Ra ) (of course, the positive integer N (a, Ra ) depends on a and Ra ) for every a ∈ K. Since K is compact, there exists a finite set {a1 , a2 , . . . , ap } ⊆ K such that K⊆

p [

j=1

D(aj ; Rj ) ⊆ Ω,

where Rj ∈ {Ra | a ∈ K} for 1 ≤ j ≤ p. In particular, if we take N = max N (aj , Rj ), 1≤j≤p

then it is true that n, m ≥ N imply

|fn (z) − fm (z)| < ǫ for all z ∈ K. Using [61, Theorem 7.8, p. 147] again, the sequence {fn } converges uniformly on K. This completes the analysis of the problem.



Problem 11.9 Rudin Chapter 11 Exercise 9.

Proof. Let D(a; r) ⊆ Ω. Since u is locally in L1 , the double integral considered in the question is well-defined. If u is harmonic in Ω, then Definition 11.12 says that it satisfies Z π 1 u(a) = u(a + reiθ ) dθ (11.26) 2π −π for every r > 0 with D(a; r) ⊆ Ω. Multiplying both sides of the expression (11.26) by ρ and integrating from 0 to r, we obtain Z r r2 u(a)ρ dρ u(a) = 2 Z0 r  Z π  1 u(a + ρeiθ ) dθ ρ dρ = 2π 0 Z r Z π−π 1 = u(a + ρeiθ )ρ dθ dρ 2π 0 −π which gives exactly 1 u(a) = 2 πr

Z

0

r

Z

π

1 u(a + ρe )ρ dθ dρ = 2 πr −π iθ

ZZ

u(x, y) dx dy.

(11.27)

D(a;r)

Conversely, suppose that the formula (11.27) holds for any D(a; r) ⊆ Ω. On the one hand, applying the polar coordinates, we can write Z rZ π ZZ u(x, y) dx dy = u(a + ρeiθ )ρ dθ dρ. (11.28) 0

D(a;r)

On the other hand, we have 2

r u(a) = 2

−π

Z

r

ρu(a) dρ. 0

(11.29)

44

Chapter 11. Harmonic Functions

Substituting the expressions (11.28) and (11.29) into the formula (11.27), we get Z r Z r ρu(a) dρ ρ dρ = u(a) 0 0 Z rZ π 1 = u(a + ρeiθ )ρ dθ dρ 2π 0 −π Z r 1 = U (a, ρ, θ)ρ dρ, 2π 0 where U (a, ρ, θ) =

Z

π

(11.30)

(11.31)

u(a + ρeiθ ) dθ.

−π

As the integral on the left-hand side in the formula (11.30) is differentiable with respect to r, so is the integral (11.31). Consequently, we derive from the First Fundamental Theorem of Calculus [79, p. 161] that Z π r 1 · U (a, r, θ)r = u(a + reiθ ) dθ. (11.32) ru(a) = 2π 2π −π After cancelling the r in the expression (11.32), we find that Z π 1 u(a) = u(a + reiθ ) dθ 2π −π whenever D(a; r) ⊆ Ω. To finish the proof, we have to show that u is continuous in Ω. To this end, fix z ∈ Ω. Given that ǫ > 0. Since u ∈ L1loc (Ω),d we must have u ∈ L1 (D(z; 2r ′ )) for some r ′ > 0 such that D(z; 2r ′ ) ⊆ Ω. We fix this r ′ . By Problem 1.12, there exists a δz > 0 such that Z |u| dm < π(r ′ )2 ǫ (11.33) E

whenever m(E) < δz and E ⊆ D(z; 2r ′ ). Clearly, we may assume that δz < r ′ . For every ω ∈ D(z; r ′ ), it is always true that D(ω; r ′ ) ⊆ D(z; 2r ′ ) and

m D(ω; r ′ ) \ D(z; r ′ ) = m D(z; r ′ ) \ D(ω; r ′ ) < δz

if ω is very close to z. Therefore, we follow from the formula (11.27) and the inequality (11.33) that ZZ ZZ 1 u(x, y) dx dy u(x, y) dx dy − |u(z) − u(ω)| = π(r ′ )2 D(z;r ′ )

=

= ≤ < d

D(ω;r ′ )

Z Z 1 u dm u dm − π(r ′ )2 D(z;r′ ) ′ D(ω;r ) Z Z 1 u dm u dm − 2π(r ′ )2 D(z;r′ )\D(ω;r′ ) D(ω;r ′ )\D(z;r ′ ) # "Z Z 1 |u| dm |u| dm + 2π(r ′ )2 D(z;r′ )\D(ω;r′ ) D(ω;r ′ )\D(z;r ′ )  1  ′ 2 π(r ) ǫ + π(r ′ )2 ǫ ′ 2 2π(r )

The notation L1loc (Ω) is the set of all locally integrable functions on Ω.

11.1. Basic Properties of Harmonic Functions

45

= ǫ. By the definition, u is continuous at z. Since z is arbitrary, u is actually continuous in Ω. Finally, Theorem 11.13 asserts that u is harmonic in Ω, completing the proof of the problem. 

Problem 11.10 Rudin Chapter 11 Exercise 10.

Proof. • By the definition, we have Z b   1 1 1 f (x + iǫ) − f (x − iǫ) = dt − ϕ(t) 2πi a t − x − iǫ t − x + iǫ Z b ǫ ϕ(t) = · dt π (x − t)2 + ǫ2 a Z b Pǫ (x − t)ϕ(t) dt = a Z ∞ ϕ(x − t)Pǫ (t) dt, =

(11.34)

−∞

where Pǫ (t) =

ǫ 1 · 2 π t + ǫ2

relates to the formula [62, Eqn. (3), §9.7, p. 183] (in fact, it is the Poisson kernel for the upper half-plane, see [66, p. 149]) and ϕ(t) = 0 if t ∈ R \ I. Using the convolution notation introduced in Theorem 8.14, the expression (11.34) becomes f (x + iǫ) − f (x − iǫ) = (ϕ ∗ Pǫ )(x). Since |ϕ(x)| ≤ max |ϕ(t)| for every x ∈ R, we get ϕ ∈ L∞ (R). t∈I

To proceed, we need to modify Theorem 9.9 and givese Lemma 11.2 If g ∈ L∞ , then we have lim (g ◦ hλ )(x) =

λ→0

g(x+) + g(x−) , 2

where hλ is the formula [62, Eqn. (3), §9.7, p. 183].

e

If g is continuous at x, then g(x+) = g(x−) and it is exactly Theorem 9.9.

(11.35)

46

Chapter 11. Harmonic Functions Proof of Lemma 11.2. Following the proof of Theorem 9.9, we have Z 0 h g(x+) + g(x−) i g(x+) + g(x−) g(x − λs) − (g ∗ hλ )(x) − h1 (s) dm(s) = 2 2 −∞ Z ∞h g(x+) + g(x−) i g(x − λs) − + h1 (s) dm(s). 2 0 Since the integrands are dominated by 2kgk∞ h1 (s) and the integrals converge pointwise for every s as λ → 0, we deduce from Theorem 1.34 (The Lebesgue’s Dominated Convergence Theorem) that h g(x+) + g(x−) i lim (g ∗ hλ )(x) − λ→0 2 Z 0 h g(x+) + g(x−) i h1 (s) dm(s) g(x − λs) − = lim λ→0 −∞ 2 Z ∞h g(x+) + g(x−) i g(x − λs) − + lim h1 (s) dm(s) λ→0 0 2 Z 0 h g(x+) + g(x−) i lim g(x − λs) − = h1 (s) dm(s) 2 −∞ λ→0 Z ∞ h g(x+) + g(x−) i h1 (s) dm(s) lim g(x − λs) − + 2 0 λ→0 Z 0 h g(x+) + g(x−) i g(x+) − = h1 (s) dm(s) 2 −∞ Z ∞h g(x+) + g(x−) i g(x−) − + h1 (s) dm(s) 2 0 Z ∞ Z 0 g(x−) − g(x+) g(x+) − g(x−) h1 (s) dm(s) + h1 (s) dm(s) = 2 2 0 −∞ Z Z g(x+) − g(x−) ∞ g(x−) − g(x+) ∞ = h1 (s) dm(s) + h1 (s) dm(s) 2 2 0 0 =0 which implies the desired result. We complete the proof of Lemma 11.2.



By Lemma 11.2, we conclude at once that lim [f (x + iǫ) − f (x − iǫ)] =

ǫ→0 ǫ>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

47

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 [44].

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  + [b, c] + + [c, a] + . ∂∆n = [a, b] + n n n Using [62, Eqn. (4), p. 202], we know that Z Z f (z) dz = (b − a) [a,b]+ ni

1 0

  i f a + + (b − a)t dt n

48

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

49

• 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 [12, 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 [19] 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 [74] 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.

50

Chapter 11. Harmonic Functions To see this, choose ρ > 0 such that r < ρ < R. Using the second set of inequalities on [62, 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 =

1 δ + δk · 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

51

• 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 it 1 u(e ) dt = 2π 1 − cos t + 4

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

it

u(e ) dt ≤

u( 12 )

1 ≤3· 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 .

52

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 = un cos + ivn 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 [7, 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, [5, §7.4, pp. 148, 149].

11.2. Harnack’s Inequalities and Positive Harmonic Functions

53

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 [79, Problem 2.4, p. 12].

M ǫ

(11.52)

54

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 [62, 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! 2 ǫ ≤ M

for some constant M > 0. Thus they imply that ǫP1−ǫ (ǫ) = ǫ ·

Mǫ 1 − (1 − ǫ)2 ≥ǫ· 2 =M 2 1 − 2(1 − ǫ) cos ǫ + (1 − ǫ) 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, [9, 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 iθ sup kur k1 = sup u(re ) dθ ≤ lim u(reiθ ) dθ = 0. 2π [−π,π]\{0} 0 0. Recall the definition [62, 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 [62, 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 [62, 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

59

Problem 11.20 Rudin Chapter 11 Exercise 20.

Proof. Since m(E) = 0, for each n ∈ N, we know from [61, 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

iθ

χVn (e ) dm =

n=1 T

∞ X

n=1

∞ X 1 = 1, m(Vn ) ≤ 2n n=1

so Theorem 1.38 implies that Z

iθ

T

|ϕ(e )| dm =

∞ Z X

χVn (eiθ ) dm = 1.

n=1 T

In other words, ϕ ∈ L1 (T ) and kϕk1 = 1. eiθ

Define u : U → R by u = P [ϕ] which is harmonic in U by Theorem 11.7. Now, for each ∈ 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 [9, 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 [53, p. 295] and [84, pp. 105, 276].

Problem 11.21 Rudin Chapter 11 Exercise 21.

60

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 t it is easy to check that |zt | < 1 and 1+z = t + is. Therefore, we have 1−zt

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 = 1 − reiθ 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 −1 =0·e = 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

61

Proof. For each 0 ≤ r < 1, the function ur : T → C is continuous, so u−1 r (R) is measurable. − , where u+ and u− are the positive and negative parts of u , see Suppose that ur = u+ − u r r r r r Definition 1.15. Since {ur } is uniformly integrable, there exists a δ > 0 such that Z (11.68) ur dm < 1 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) ur dm = 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 ekδi including e(k−1)δi but not ekδi , 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− dm ≤N r

+ 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− (e ) dθ ≤ ur (e ) dθ + |ur (e )| dθ ≤ r 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.

62

Chapter 11. Harmonic Functions

as j → ∞ or equivalently, we have Z Z iθ iθ f (eiθ ) dσ(eiθ ). urj (e ) dσ(e ) = lim j→∞ T

(11.70)

T

Finally, we take g = 1 in [62, Eqn. (3), p. 247] to get Z Z iθ iθ dµ(eiθ ). urj (e ) dσ(e ) = 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 [62, 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

1 − |z|2 1 + |z| 1 − |z|2 = < ∞. ≤ |P (z, e )| = iθ 2 2 |e − z| (1 − |z|) 1 − |z| iθ

Thus the series

v(z) =

∞ X

n

−2

iθn

P (z, e

) and w(z) =

∞ X

n−2 P (z, e−iθn )

n=1

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

iθk

P (z, e

) 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

63

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 1 1 sin2 θ 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 [62, 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 = ≥ n 2 ǫn n2 n2 sin 21n

(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 [61, Exercise 15, p. 199] and their solutions are shown by the author in [77, 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 [61, 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

65

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 [62, Eqn. (1), p. 101] that Z Z
it f ei(θ−t) Dn (t) dσ, f (e )Dn (θ − t) dσ = sn (f ; θ) = T

T

so it is easy to check that σN (f ; θ) =

Z

i(θ−t)

f e 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, [61, Exercises 15, 16, p. 199] or [84, Theorem 3.4, p. 89].

66

Chapter 11. Harmonic Functions Z 0 Z

1 π i(θ−t) f ei(θ−t) KN (t) dt f e KN (t) dt + = 2π 0 −π Z π

1 f ei(θ−t) + 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 [62, Eqn. (5), p. 241] G(x) = 0. x→0 x lim

(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 h 1 2 G(t) i π 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) i(θ−t) ≤ |f e | dt |f e | dt + N δ2 T T π2 · 2kf k1 = N δ2 0. Using [62, 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)

68

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 q ϕ (z, t) dt ≤ dt ≤ Mz · dt < ∞. (11.93) 2 )q (1 + t 1 + t2 −∞ −∞ −∞ 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 ∞ iθ |ϕ(z + reiθ , t)f (t)| dt ≤ kf kp × kϕ(z + reiθ , ·)kq < ∞. kϕ(z + re , ·)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 iθ u(z + re ) 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 69

70

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 [9, 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 −1 constant λ with |λ| = 1. Since ϕ0 (z) = z, we conclude that f h (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

71

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 [76, 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 [61, Theorem 7.8, p. 147], the inequality (12.8) ensures that {fn } converges uniformly on Ω which ends the proof of the problem. 

72

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 [42, 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

73

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 ab

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  r b r · [log M (a)] + log log M (b) = log − log 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 b M (a) log a

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

74

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

75

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 −1 have φ ∈ 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.



76

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 |gn (reiα )| log |f (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
(12.24) |gn re±i 2 | = |f re±i 2 | < 1

12.1. Applications of the Maximum Modulus Principle

77

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) If fe(ω) 6= 0, then we deduce from the inequality (12.27) that 1

CN |f (ω)| ≤ 1 . |fe(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.

78

Chapter 12. The Maximum Modulus Principle

12.2

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 [79, Problem 7.15, p. 112]. Therefore, D(z0 ; δ) ⊆ E1 . By [42, 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 [4, 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 [9, 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

79

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 1 |f (0)| z∈C(0;1− n ) |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.

80

Chapter 12. The Maximum Modulus Principle • Case (i): Z(f ) is infinite. Since Z(f ) ⊆ U which is bounded, the Bolzano-Weierstrass Theorem [79, 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)

(z − ζk )mk

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

mk

|zn − ζk |

≤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

81

• 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

(z) log ff (0) −i log f (z) + i log f (0) = F (z) = −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, [57] and [15, Exercise 6.36, pp. 213, 214] for a different proof.

82

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

,

(z − ak )mk

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. 83

84

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

85

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− − Sm−1 ≤ 1− 1− e 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− − Sm−1 ≤ m e 5 · 5m nm 5 nm k=1

k=1 m−1 X

≤

1 5m

≤

1 m−1

5k 2m−k (m − 1)(m − 2) · · · k

k=1 m−1 X k=1

1 2m−k

· 5k−m

 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 ≤ nm+1 + m 1− = nk . 5m nm 5m k 5 n e nm k=m+1 k=m+2 e m k=m+1

∞ X

(13.14)

92

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 −

5k

k=m+1

5m



1−

1 nk iθnk e 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

93

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, [84, Chap. V]. (b) In [29, 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 [16], 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 [26], [43] and [75]

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

94

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

(13.20)

for at least one component U of S 2 \ Ω. Since A intersects every component of S 2 \ Ω, we have A ∩ U 6= ∅.

(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 [48]. In fact, Problem 10.5 is the well-known Vitali Convergence Theorem, see [72, 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 [67, 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

97

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

!

1


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 [76, 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 [9, 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

See also [18, pp. 50, 51]

99

1 . z

100

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 [62, 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. =− iθ 1−e 1 − cos θ

14.1. Basic Properties of Conformal Mappings

101

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

F (ζ) =

where Ω∗ = {ζ ∈ C | ζ

−1

∈ Ω}.

  f (ζ),       

if ζ ∈ Ω ∪ L;

1

f ζ

, if ζ ∈ Ω∗ , −1


(14.7)

(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

102

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 b (14.9) F (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

103

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 k Y z − αn B(z) = z · 1 − αn z n=1 m

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

104

Chapter 14. Conformal Mapping

function on U and is harmonic in U , so we follow from [7, Corollary 1.9, p. 7] that Im F ≡ 0 in U . Finally, using [9, 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

N Y (z − αn )(1 − αn z) R(z) = cQ(z) = c (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

105

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. 

106

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 (r + 1r )2 (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 = π2 , 3π 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

107

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 [9, p. 176] that f2 (z) = ez is a one-to-one conformal mapping of Ω1 onto the upper half plane Π+ . Recall from [62, Eqn. (1), p. 281] that f3 (z) = z−i z+i is a one-to-one 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 . iπ exp( 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 2 1−z for all z ∈ U . Let f −1 (z) =

∞ X

n=0

(14.20)

cn z n , where z ∈ U . Since n!cn = (f −1 )(n) (0) for every

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 2n − 1 π 1 − |z| π 1−r n=1

Finally, the set relation (14.19) asserts that |g(z)| ≤

for all z ∈ D(0; r).

2 1+r log π 1−r

108

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 [9, 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 2eα+iβ eiθ × . =2× 1 − |A|2 (eα+iβ + 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 [15, Chap. VI, §5, pp. 207 - 215 ], [21, §1.5, pp. 10 – 13] or [47, Chap V, §9, pp. 226 – 236].

14.1. Basic Properties of Conformal Mappings

109

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 [9, 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 [61, Definition 1.3, p. 3].

110

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 0 = u(p) = u(p + Reiθ ) dθ (14.27) 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 iθ iθ u(p + Reiθ ) dm > 0 u(p + Re ) dm + u(p + Re ) 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 [9, 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 [30].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.

112

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 [62, 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

113

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 1 f (z) = 2π

Z

2π

f (z + reit ) dt

0

which implies Z 2π 1 rf (z + reit ) dt rf (z) = 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 Applying Theorem 3.5 (H¨older’s Inequality) to the expression (14.36), we obtain Z Z 1 R 2π it |f (z)| = rf (z + re ) dt dr 2 πR 0 0 Z Z o 1 n Z R Z 2π √ o1
2 1 n R 2π √ 2 2 2 it r|f (z + re )| dr dt ( r) dr dt · ≤ πR2 0 0 0 0

(14.36)

114

Chapter 14. Conformal Mapping n 1 √ = πR · · πR2

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 [18, 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

115

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 [9, 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) ′ 2 f (z0 ) 1 − |α|

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 ′ that F (W ) ⊆ V .e Indeed, we can find a sequence {rn eiθ } 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, [42, Theorem 18.1, p. 104].

116

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

117

(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 1
= ′ = 1. f (0) f ′ f −1 (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 [62, 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 [9, Theorem 13.8, p. 174].

(14.48)

118

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 considering g = f −1 ◦ [exp( 2πi N )f ] : U → U which is clearly a one-to-one conformal mapping of 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) =

′ exp( 2πi N )f (z)


, f ′ exp( 2πi 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

119

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) α2 − 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.

120

Chapter 14. Conformal Mapping

Remark 14.2 An example of this kind of conformal mappings can be found in [76, 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

121

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.

122

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 [6]. (b) Besides the analytical proofs provided in the text and the problems, one can find a pure algebraic proof in [58].

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

123

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 + rn 1 − rn2 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)

124

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

.

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 hypotheses of Theorem 12.2 (Schwarz’s Lemma), where φ(z) = rn+1 z maps D(0; rn+1 ) one-to-one and onto U . i

14.4. Constructive Proof of the Riemann Mapping Theorem

125

Now the Chain Rule and part (b) assert that ′ |ψm,n (0)| =

so that

m Y

k=n+1

|fk′ (0)| =

m Y 1 + rk √ 2 rk

k=n+1

m Y 1 + rk 1 1< ≤ √ 2 rk rn+1

(14.65)

k=n+1

which implies the convergence of

∞ Y 1 + rn √ 2 rn

n=1

by taking m → ∞. By a basic fact about the convergence theory of infinite products [9, Note 1, p. 242], we see that lim

n→∞

1 + rn = 1. √ 2 rn

Combining this fact and the identity √ (1 − rn )2 1 + rn , √ −1= √ 2 rn 2 rn we conclude immediately that rn → 1 as n → ∞. (d) We verify the parts one by one. – |hn | ≤ |hn+1
| for each n ∈ N. By the definition of ψn , we have ψn+1 (z) = fn+1 ψn (z) so that
(14.66) zhn+1 (z) = fn+1 zhn (z) . Applying the first inequality (14.62) to the right-hand side of the expression (14.66), we assert that |zhn+1 (z)| ≥ |zhn (z)|

which implies our expected result immediately. – ψn → ψ converges uniformly on compact subsets of Ω. Recall from part (c) that each ψn is injective on Ω, so every hn is zero-free in Ω and we can define gn = log hn as a holomorphic function in Ω. Define un = Re gn = log |hn | which is harmonic in Ω. Since each ψn (Ω) is a subset of U , |hn (z0 )| < ∞ for some z0 ∈ Ω \ {0} and all n ∈ N. Now the inequality |hn | ≤ |hn+1 | guarantees that u1 ≤ u2 ≤ · · · . According to Theorem 11.11 (Harnack’s Theorem), we see immediately that {un } converges uniformly on compact subsets of Ω. Since ψn′ (z) = hn (z) + zh′n (z), the second fact in part (c) assures us that {hn (0)} is bounded and therefore {hn (0)} converges. Consequently, {gn (0)} converges and we conclude from Problem 11.8 that {gn } converges uniformly on compact subsets of Ω to g. By the definition, we know that ψn (z) = zegn (z) , so {ψn } also converges uniformly on compact subsets of Ω to ψ. In view of Theorem 10.28, it is true that ψ ∈ H(Ω).

126

Chapter 14. Conformal Mapping – The map ψ : Ω → U is surjective. Let ω ∈ U . By Lemma 14.4 and the fact rn → 1 as n → ∞, we may select an N ∈ N such that |ω| < rn for all n ≥ N . Therefore, for p ≥ 1, we have ω ∈ D(0, rN +p ) ⊆ ΩN +p−1

= fN +p−1 (ΩN +p−2 ) = ···

= fN +p−1 ◦ fN +p−2 ◦ · · · ◦ fN +1 ◦ fN (ΩN −1 ) = fN +p−1 ◦ fN +p−2 ◦ · · · ◦ fN +1 ◦ ψN (Ω)

which means that there is a zp ∈ Ω such that ω = fN +p−1 ◦ fN +p−2 ◦ · · · ◦ fN +1 ◦ ψN (zp ) = ψN +p−1 (zp ).

(14.67) (14.68)

Applying the inequality (14.62) repeatedly to the expression (14.67), we see that |ω| ≥ |ψN (zp )|

(14.69)

lim ψN +pj −1 (zpj ) = ψ(z).

(14.70)


−1 for all p ≥ 1. Clearly, D(0; |ω|) is a compact subset of ΩN , so ψN D(0; |ω|) is also a compact subset of Ω. Denote this set by K. By the result (14.69), we establish that {zp } ⊆ K. Then the Bolzano-Weierstrass Theorem [79, Problem 5.25, p. 68] suggests that there corresponds a subsequence {zpj } such that zpj → z ∈ K. Since ψn → ψ uniformly on K, it observes that j→∞

Combining the expression (14.68) and the limit (14.70), we conclude at once that ω = ψ(z) which means that ψ is surjective. – The map ψ : Ω → U is injective. Since ψ is surjective, it is not constant. Recall that each ψn is injective, so Problem 14.13 indicates easily that ψ is also injective. Hence we have completed the analysis of the problem.



Problem 14.27 Rudin Chapter 14 Exercise 27.

Proof. Taking logarithms in the inequality (14.64) with m = 2n and then using the hint to get 2n X

k=n+1

√ 2n h X (1 − rk )2 i 1 + rk 1 = log 1 + log √ ≤ log = − log rn+1 . √ 2 rk 2 rk rn+1 k=n+1

(14.71)

14.4. Constructive Proof of the Riemann Mapping Theorem Using the Mean Value Theorem for Derivatives, one can show that log(1 + x) > Since 0


2n X

k=n+1

√ √ 2n X (1 − rk )2 (1 − rk )2 i > > 0. log 1 + √ 2 rk 3 h

(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 − log rn+1 ≤ (1 − rn+1 ) · As (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 √ 2n 2n X (1 − rk )2 1 X  1 − rk 2 ≤ < − log rn+1 < A(1 − rn ) 0< 3 2 3 k=n+1

k=n+1

or equivalently,

2n  X 1 − rk 2 1 − rn < , 2B B

0
0 and an gn ∈ H D(0; rn ) such that for all z ∈ D(0; rn ), we have fn (z) = z + cN z N gn (z)

and gn (0) = n.

(14.83)

The expression (14.82) is just the case n = 1, so we assume that the result (14.83) holds for some positive integer n. Since fn (0) = 0, we can pick rn+1 ∈ (0, rn ) such
that fn D(0; rn+1 ) ⊆ D(0; rn ). Then it follows from the expressions (14.82) and (14.83) that if z ∈ D(0; rn+1 ), then
fn+1 (z) = f fn (z)
= fn (z) + cN [fn (z)]N g fn (z)  N
= z + cN z N gn (z) + cN z + cN z N gn (z) g fn (z)
  N = z + cN z N gn (z) + 1 + cN z N −1 gn (z) g fn (z) .

 N
Suppose that gn+1 (z) = gn (z) + 1 + cN z N −1 gn (z) g fn (z) . Recall that N − 1 ≥ 1, so we obtain
gn+1 (0) = gn (0) + g fn (0) = n + 1.

By induction, our claim follows. Next, we differentiate the expression (14.83) N times and put z = 0, we have fn(N ) (0) = N !cN gn (0) = nN !cN (N ) for every n = 1, 2, . . .. Since cN 6= 0, we have fn (0) → ∞ as n → ∞ which contradicts the fact (14.81). Consequently, this means that f (k) (0) = 0 for all k ≥ 2 which implies f (z) = z in D(0; r). By the Corollary following Theorem 10.18, it is actually true in Ω. (c) By the hypothesis, we know that f ′ (0) = eiθ for some θ ∈ [0, 2π]. If eiθ is an N -root of unity, then the integer nk = kN satisfies [f ′ (0)]nk = (eiN θ )k = 1.

14.4. Constructive Proof of the Riemann Mapping Theorem

131

Otherwise, we need the following form of the Kronecker’s Approximation Theorem [5, Theorem 7.8, p. 149]: Lemma 14.5 (The Kronecker’s Approximation Theorem) Given any real α, any irrational β and any ǫ > 0, there exist integers m and n with n > 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

135

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 [25, 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 (14.93) ϕ(z) = z + 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

136

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

137

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)

138

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 ϕ−1 (z)

−α

=

1 − β. z−α

In other words, it means that ϕ−1 ∈ Gα . By [25, 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, [25, p. 132].

(1 + cα)z − cα2 . cz + (1 − cα)

14.4. Constructive Proof of the Riemann Mapping Theorem

139

(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 = · . −1 ϕ (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 [24, pp. 15, 16].

140

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  αγ − β 1  √ d= γ+√ , (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 [20] 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 +

141

ρe−iθ . ρ2 − R2

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 ∗

ϕ (ζ) = ∆

− 12

h

√ i 1+γ 1 ∆ = √ − . (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.

142

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 √ |2 − 1 + i| 1+γ √ R|α − β|p|γ| (β − αγ) + ∆ ± R2 − | 21+γ γ 2 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 [24, pp. 15 – 23] for further details. Problem 14.32 Rudin Chapter 14 Exercise 32.

14.4. Constructive Proof of the Riemann Mapping Theorem

143

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 expression (14.126) are |α| 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

(αz)n (αz)k dm

U n,k=0 ∞ Z |α|2 X

1− π

(αz)n (αz)k dm

n,k=0 U ∞ Z 2π |α|2 X

1− = π = If n 6= k, then

n,k=0 0 ∞ |α|2 X

1− π

·

Z

Thus the expression becomes 1 π

Z

2π 0

Z

1

(αn αk )

n,k=0

Z

1

(αreiθ )n (αre−iθ )k r dr dθ

0

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 U 0 0 n=0 ∞ 1 − |α|2 X 2n π |α| = · π n+1 n=0

∞
X |α|2n 2 . = 1 − |α| 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 [34, 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. 149

150

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

er |an | ≤ n r

(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

1

exp

n λ

at r = n λ λ− λ . Note that r is large if and only if n is large. Thus it follows from the inequality (15.2) that  n   eλ  n λn λ λ = exp |an | ≤ n λ n holds for all large enough 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 [76, 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=    3π   ln(−2kπ) + i 2nπ + , 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

151

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.

152

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 + . 2 x + iy x − y 2 − n2 + 2ixy n=1

√ If y > 1 and |x| ≤ 21 , 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

n=1

2|x + iy| |x2 − y 2 − n2 + 2ixy|

154

Chapter 15. Zeros of Holomorphic Functions ∞ √ X ≤1+4 2

y2

n=1 ∞

√ Z ≤1+4 2

y + n2

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 2 y +x 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

+

∞ X

n=1 ∞ X

n=1

y2

2iy i −y 2 − n2 y + n2

dt 1 + t2

and lim π cot iπy = −πi.

y→∞

Thus ∆(z) ≡ 0 and then

∞

1 X 2z π cot πz = + . 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 [61, 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

155

converges uniformly on K. By Theorem 15.6, the product (15.8) is holomorphic in C \ Z. Using [18, Exercise 10, p. 174], we have ∞ ∞ ∞ P ′ (z) 1 X Pn′ (z) 1 X − n2z2 1 X 2z = + = + + = P (z) z n=1 Pn (z) z n=1 1 − z 22 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,

P (z) z

→ 1 as z → 0,

∞  Y z2  1− 2 sin πz = πz n n=1

in C \ Z. Since sin πz and πz whole plane.

∞ Y

(1 −

n=1

(15.10)

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 [9, 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 cot πz 1 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 [54, Lemma 7.22, pp. 181, 182].

156

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 = Re − z k+1 − z k+2 − · · · 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 < k+1 |zn | 2M

and

N X

n=1

1 ≤ N. |zn |k+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

157

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 ≤|z

n

| 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



158

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 [62, 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

n

= exp 2πi

N X

mn +

n=1

Z

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

159

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 XZ

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)

160

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) 1 Z iθn iθn h(ζ) dζ = mn ln{t[z − (zn − δn e )] − δn e } + 0 γ2 (z) Z   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 |f (z)| = exp g(ζ) dζ × exp g(ζ) dζ γ1 (z)

M1 +M2

≤e

which implies the limit (15.22).

γ2 (z)

· exp(−mn ln δn ) × |z − zn |mn

• 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, [9, 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

161

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 n Y z − zk . g(z) = 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 ,

162

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 [62, 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 2 + i log n 3(log n) 3 n n=3 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 √ √
4+n· 4+n+ n 4+n·

2 p 4

(4 + n)n

15.2. Some Examples

163

=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)

166

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

n=1

− |α′n |)

≤

∞ 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 so

H∞

and has no zeros except at α′k . It is no doubt that |B(α′n )| → 0 as n → ∞, 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

167

(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 (15.48) implies that (b − b1 )z + ( 1 − 1)b b 1 2 2 kn kn |ϕn (z)| = b2 1 (1 − kn )z + ( kn − 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 (15.48), we have

so that

∞ h X

n=1

 1  1 = ϕN 2 2

∞ h  1  i X  1  i 1 − ϕn 1 − ϕpN ≥ = ∞. 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., e(p−q)iθ
2 for large enough m. This shows that ϕ does not satisfy the Blaschke condition. (b) We claim that ϕ satisfies ϕn (z) = z

(15.49)

for some n ∈ N. To this end, if f ◦ ϕ = f for some nonconstant f ∈ H ∞ , then we must have f (ϕn (z)) = f (ϕ(ϕn−1 (z))) = f (ϕn−1 (z)) = · · · = f (z) 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 (15.49) holds immediately for N . Otherwise, we fix z = z0 ∈ U \ {0} so that f (λn z0 ) = f (z0 ) for every n ≥ 0. By Subcase (iv), the set S ′ = {λn | n ≥ 0} is dense on T which implies that f (z) = f (z0 ) (15.50) on C(0; |z0 |). By Theorem 10.18, the result (15.50) 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 (15.48) 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 (15.48) again that ϕN (z) = z. Otherwise, Subcase (iv) ensures that the set S = {kn | n ≥ 0} is dense on T . Using similar argument as Case (i), we conclude that this is impossible.

170

Chapter 15. Zeros of Holomorphic Functions Consequently, the linear fractional transformation ϕ must satisfy the claim (15.49).

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.

(15.51)

|αj |

Define the characteristic function χj (r) = Then formula (15.51) becomes

  1, if r ≥ |αj |; 

0, otherwise.

∞ Z ∞ X X (1 − |αj |) = j=1

j=1

Observe that

∞ X

1

χj (r) dr.

(15.52)

0

χj (r) = n(r).

(15.53)

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 (15.52) and then use the formula (15.53) 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 [62, p. 66]

B(eiθ ) ∈ L∞ (T ).

15.3. Problems on Blaschke Products

171

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 [62, 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 ikθ × ck e ck ei(n−k)θ dθ = 2π −π k=0

=

∞ X

k=0

ck cn+k .

(15.54)

k=0

On the other hand, the fact |B(eiθ )| = 1 a.e. on T gives h 1 Z π 2 i2  1 Z inθ iθ 2 hB, An i = einθ dθ , e |B(e )| dθ = 2π −π 2π T \N

(15.55)

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 (15.55) by 2  1 Z einθ dθ = 0. (15.56) hB, An i = 2π T Combining the two results (15.54) and (15.56), 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 ) < ∞.

(15.57)

n=1

Without loss of generality, we may assume that 0 < α1 ≤ α2 ≤ · · · < 1.

(15.58)

Furthermore, suppose that ∞ Y αn − z B(z) = 1 − αn z n=1

and

∞ Y αk − z Bn (z) = . 1 − αk z k=1 k6=n

We yield from [18, Exercise 10, p. 174] that B ′ (z) X  αn − z ′  αn − z −1 = B(z) 1 − αn z 1 − αn z ∞

n=1

172

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 (15.58) that δ = min(α1 , α2 , . . .) > 0. Geometrically, if z ∈ U and p > 1, then |p − z| > |1 − z|.

(15.59)

Put p = α1n into the estimate (15.59) to get |1 − αn z|2 ≥ δ2 |1 − z|2 for every z ∈ U . Hence, the condition (15.57) implies |B ′ (z)| ≤

∞ ∞ X X 1 1 − α2n (1 − α2n ) < ∞ · |B (z)| ≤ n 2 2 |1 − z|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 [17].

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 [62, Eqn, (3), §11.5, 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

(15.60)

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.

(15.61)

15.4. Miscellaneous Problems and the M¨ untz-Szasz Theorem

173

Hence it follows from the results (15.60) and (15.61) 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 , . . . if dense in L2 (I) if and only if ∞ X

2λn + 1 = ∞. 2+1 (2λ + 1) n n=0 To this end, let m ∈ N and m ∈ / {λn }. By [13, p. 173], we know that

Thus we see that

n−1 n−1

Y m − λk 1

m X λk √ min t − ak t = . ak ∈C 2 1 + 2m k=0 m + λk + 1 k=0

tm ∈ span {tλ0 , tλ1 , . . .}

(15.62)

174

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→∞

Since m + λk + 1 > 0 and λk > − 12 , we have results (15.63) hold if and only if ∞ X

k=0 λk >m

Since λk +

1 2

2m + 1 =∞ m + λk + 1

or

n−1 Y

k=0 − 21 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

(15.67)

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)

176

Chapter 15. Zeros of Holomorphic Functions

By the Mean Value Theorem, 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 ω−z Z ∞ −ωt e − e−tz · g(t) dt = lim ω→z 0 ω−z Z ∞ e−ωt − e−tz lim = · g(t) dt ω→z ω−z Z0 ∞ −tetz g(t) dt. =

F ′ (z) = lim

0

Since z is arbitrary, we have proven that F ∈ H(Π). In fact, this process can be repeated to get Z ∞ (n) n F (z) = (−1) 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 b G(y) =

√

2π

Z

∞

−ity

G(t)e

−∞

√ Z dt = 2π

∞

e−t g(t)e−ity dt =

√

2πF (1 + iy) = 0,

0

where y ∈ R. As G ∈ L1 (0, ∞), 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 a 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

177

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 iθ 1 − λn e

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 

178

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

bk

k=0



∞ 1 k 1 k X f (k) ( 21 )  z− = 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  1 k . · x − 2n−k 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 [61, 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

179

180

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

181

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 [9, 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) .

182

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 [42, 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)| = − γ (t) γ (t) z ω 0 0 Z 1 (γ ∗ γ )′ (t) dt Z 1 γ ′ (t) dt ω ω,z ω − = (γ ∗ γ )(t) γ (t) ω ω,z ω 0 0 Z Z dt dt = − γω ∗γω,z t γω t Z Z dt Z dt dt + − = γω,z t γω t γω t Z dt = . γω,z t

183

(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 note that we have d − bc = 1 and ϕ(z) = cz+d  −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 τ c σ ϕ(z) = =− = σ τ b (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).

184

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 [5, 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

185

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)

√

Since z1 ∈ R, it is easy to see that Im z1 ≥ 23 . Thus |c| · 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.



186

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 [5, 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

187

• 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 −1 −1 A = and B = −2 1 0 1 so that A

nk

=



1 (−2)nk

0 1



=



1 (−2)nk

(−2)mk (−2)mk +nk + 1

and

mk

B



=



1 (−2)mk 0 1

=



1 2Nk 2Mk 2Lk + 1

and then nk

A B

mk





,

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.

188

Chapter 16. Analytic Continuation Next, if c = 1, then ad − b = 1 so that b = ad − 1 and       a ad − 1 1 a 0 −1 1 d M= = = Td STd . 1 d 0 1 1 0 0 1 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 MT−q = = , 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 1 dt 1  f (x) − f (y) = dt = (x − y) − . t − x t − y (t − x)(t − y) E E

(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

189

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 Z E |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 [42, Theorem 23.4, p. 150] that Π+ ∪ {a} is also connected. Similarly, the set Π− ∪ {a} is also connected. According to [42, 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 [42, 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 + b2 (t − a) E Z ∞ b dt ≤ 2 + b2 (t − a) −∞

(16.21)

190

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 δ γ E

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 [22].



16.2. Problems on the Modular Group and Removable Sets

191

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    b −1 1 − a −1 −1 − a −1 t − a = tan − tan . dt = tan (t − a)2 + b2 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.



192

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 [61, 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

193

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 1 f (ζ) f (ζ) 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 Γ1 (ζ − z)(ζ − z − h) 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).

194

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 [73, Eqn. (1.2), p. 16], we have n 1 Z X rk < M ǫ. f (z) dz ≤ M |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

195

non-removable, a contradiction. Hence connected components of E are one-point sets, i.e., E is totally disconnected, see [42, 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 [45, p. 3] that a nonempty compact connected metric space is a continuum.

196

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 1 exp(eω ) |f (reiθ )| = dω 2π Γ1 ω − z Z Z Z exp(eω ) exp(eω ) exp(eω ) 1 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) . iθ 2π 1 t + πi − re 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)

197 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 ) f (r) = 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 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 ]

198

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→∞

r→∞

Next, suppose that eiθ = 1. Since f (r) → ∞ as r → ∞, we see immediately that f (r) = 0. r→∞ exp[f (r)]

lim g(r) = lim

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 k=1

2

.

2

(16.33)

Evidently, if |z − z 2 | < 2, then | z−z 2 | < 1 so that the series (16.33) converges by [61, 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 ∞ X 1 f (z) = k Pk (z). 23 k=1 f

This is (x − x2 + y)2 + (y − 2xy)2 < 4.

16.3. Miscellaneous Problems

199

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 [61, 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+1 +1 , 2

so it is true that

3k 3k = 2π(nk − 3k )[3k − (nk − 3k )] 2π(nk − 3k )(2 · 3k − nk ) 3k = k k 2π · 3 2+1 · 3 2−1 2 π 2 ∼ π

=

3k 32k − 1 1 · k 3 ·

(16.36)

200

Chapter 16. Analytic Continuation

and k

(3k )3 (

3k +1 2

)

3k +1 2

k

3k −1

(

2

1

)

3k −1 2

∼ 23 .

(16.37)

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

201

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 1 cn = 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

n→∞

1 √ n c which −n ≥ 2

∞ X c−n n−1 G(ω) = 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 d  X c−n 1−n  z f1 (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)

202

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 [64, §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 [79, Problem 5.25, pp. 68, 69] ensures that there is a convergent subsequence {fnk (p)}. Let this limit be ℓ.

16.3. Miscellaneous Problems

203

• 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 nk ) = Im fnk > 0 so that the family {−ifnk | k ∈ N} is normal by Problem 14.15. 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→∞

Recall that the domain of λ is Π+ , 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 .

204

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 [9, §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 [78, Problem 2.1, pp. 17, 18]. Since K is compact, [78, Problem 2.21, p. 60] 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 + reiθ ) dθ u1 (z0 ) = m > 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 .

205

Chapter 17. H p -Spaces

206

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 u(r1 eit ) dt ≤ (Hun )(r1 eit ) 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 nZ o1 n Z o1 1 p p iθ p iθ kfr kp = |f (re )| dσ ≤ u(re ) 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

207

Conversely, suppose that f ∈ H p . Now it is easy to see that Z Z p p |fr |p dσ = kfr kpp |fr | dσ = kfr k1 = T

T

which means kf p k1 = kf kpp < ∞. As a consequence, we have f p ∈ H 1 . Applying [58, Eqn. (5), p. 344] directly to f p , we have |f (z)|p ≤ |Qf p (z)| n 1 Z eit + z o p ∗ it = exp log |(f ) (e )| dt it 2π T e − z io n h 1 Z eit + z p ∗ it log |(f ) (e )| dt = exp Re 2π T eit − z n 1 Z o = exp Pr (θ − t) log |(f p )∗ (eit )| dt 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 Pr (θ − t)|(f p )∗ (eit )| dt = P [(f p )∗ ](z) |f (z)|p ≤ 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 it p Pr (θ − t)|f (ρe )| 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 uf (z) = 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)

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

208 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 in U , then there exists a harmonic + majorant + 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 + it u(reit ) dσ = u(0) < ∞ 0 ≤ kfr k0 = exp log |fr (e )| dσ ≤ 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

209

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

(17.7) log+ f ϕ(z) ≤ u ϕ(z)

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

it

|fα (re )| 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 [61, 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 < arg(1 − z) < π2 .

e

− π2

1 1 (2π)α
α ≤
α ≤ |1 − reit |α |t| + 1 − r |t| + 1 − r

Chapter 17. H p -Spaces

210 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 r ∞ / H . This completes the analysis of the problem.  α = 2r , then we have f 1 ∈ H 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

∞ eiθ X m(m − 1) · · · (m − n + 1)|α|m−n z m Fα (z) = 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

× |Cα,n |2 = 1.

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

215

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 −int it ∗ it n ∗ c |f (reit ) − f (eit )| dσ = kfr − f ∗ k1 → 0 [f (re ) − f (e )] dσ ≤ |r an − f (n)| = e T

T

as r → 1 by Theorem 17.11. In other words, we get   an , if n ≥ 0; ∗ c f (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

216 Using the two facts (17.17) and (17.20), we find Z Z π Z π int int int π µ(t) d(e ) = −in e dµ(t) = [e µ(t)]−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 it f (z) = P (z, e )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 [62, 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

217

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 p p exp − dµ (t) = c dµ (t) , Mh (z) = c 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 [62, Eqn. (1), p. 344], n 1 Z 2 p Qh (z) = exp 2π T n 1 Z = exp 2π T

we see that eit + z eit − z eit + z eit − z

o 2 · log |h∗ (eit )| dt p o 2 log (h∗ ) p (eit ) dt ·

Chapter 17. H p -Spaces

218 n 1 Z eit + z o ∗ it = exp log |f (e )| dt . 2π T eit − 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 1 log |h(0)| ≤ 2π

Z

(17.26)

T

log |h∗ (eit )| dt

1 2π

Z

(17.27)

T

log |f ∗ (eit )| dt.

is equivalent to the inequality log |f (0)| ≤

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 [62, Eqn. (6), p. 91] and also the proof of Theorem 17.12, we get Z Z Z ∞ X 1 1 2 2 |an | = lim |fr | dσ = lim dσ = dσ. p ∗ p r→1 T r→1 T |ϕ| T |ϕ | 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. Consequently, ϕ1 ∈ / H p for all p > 0.

17.3. Factorization of f ∈ H p

219

(b) By the hypotheses, we have ϕ(z) = c exp

n

−

Z

T

o eit + z dµ(t) , eit − 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 [62, 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 [68, 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

220

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 [58, 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.,

1 fǫ

is bounded in U so that

n 1 Z π eit + z o (fǫ )∗ (eit ) dt log fǫ (z) = c exp 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 o n 1 Z π eit + z ∗ it log |f (e )| dt = Qf (z) f (z) = c exp 2π −π eit − z

for all z ∈ U . This completes the proof of the problem.



17.3. Factorization of f ∈ H p

221

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 + it + log |h(reit )| dt (17.31) log |f (re )| 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(r) dt u(0) = 2π T Z 1 log |fr (eit )| dt = 2π T Z Z 1 1 + it log |fr (e )| 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 < ∞ 2π 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 n √p ≈ 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 4ek 2p−k 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

232

≥

∞ 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 ′ iθ f (re ) 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

k

ei(nk −1)θ

f (Reiθ ).

(17.60)

∞ X 1 < ∞, Theorem 17.12 implies that f ∈ H 2 and then we follow from Theorem k2 k=1

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 V (f ; θ) = |f ′ (reiθ )| dr 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

233

and fc∗ (n) be the Fourier coefficients of its boundary function f ∗ (eit ), i.e., Z π 1 ∗ c f (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 f (reit ) 1 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 −int it ∗ it r an − fc∗ (n) = 1 f (reit ) − f ∗ (eit ) dt e f (re ) − f (e ) 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 −int it e g(e ) 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 iθ Pr (θ − t)g(eit ) dt, (17.64) f (re ) = 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 n in(θ−t) it r e g(e ) dt + e−in(θ−t) g(eit ) dt = 2π 2π −π −π n=1 n=0 Z ∞ h 1 i π 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

234

for every z ∈ U and then f ∈ H(U ). By the integral (17.64) and [62, 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 Φ(z) = Pr (θ − t)f ∗ (eit ) dt = P [f ∗ ](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.,

f (reiθ ) =

1 2π

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 235

236

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

237

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 [35, 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 [62, 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)

238

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 |ϕ(x) − λ|2 · |φn (x)|2 dx ≤ 2−2n kφn k2 k(Mϕ − λI)(φn )k2 = kϕφn − λφn k2 = 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

239

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.

240

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 −1 Pn (z) − (z − λ) dz ≤ ℓ · Ln 2πi = = · 2πδ = 2πℓLn . (18.15) δ C(λ;δ) |z − λ| C(λ;δ)

18.2. Properties of Ideals and Homomorphisms

241

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 int γn eint a e = = n 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 [60, 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)

242

Chapter 18. Elementary Theory of Banach Algebras

On the other hand, recall from [62, 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
∗ isomorphic to the dual space of L∞ (R), i.e., L1 (R) ∼ = L∞ (R). By Remark 5.21, every ∞ 1 φ ∈ L (R) is a bounded linear functional on L (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 [60, 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 [62, Example 9.19(d), p. 190]. b

c

18.2. Properties of Ideals and Homomorphisms

243

Recall from [62, 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 [62, §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 ∗ g)(t)β(t) dt = f (t − s)g(s)β(t) ds 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

244

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 int 1 c c exists an fn ∈ I such that fn (n) 6= 0. Put gn (t) = e fn (n) ∈ L (T ). Then we have 1 1 · (gn ∗ fn )(t) = c fn (n) 2π

Z

π

−π

in(t−s)

fn (s)e

eint 1 ds = · c fn (n) 2π

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 [23, Theorem 8.19, p. 247].

(f[ ∗ P )(n) = fb(n) × Pb(n) = 0,

18.2. Properties of Ideals and Homomorphisms

245

i.e., E ⊂ Z(f ∗ g). Using [60, 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 [81, §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 [62, 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.

246

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 [25, 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 2 s ds = . du ds = T (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

247

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)

248

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

249

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 [62, 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.

250

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 [61, 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

∞

itz

F1 (t)e

  F (t), if −A < t ≤ 0; 

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 [62, 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

−∞

251

252

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 [62, 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, f2 (x + iy) = F 0

regard y as fixed, then Theorem 9.13 (The Plancherel Theorem) implies that Z ∞ Z ∞ Z ∞ 2 2 2ty 1 2 f f F2 (t) dt < ∞ F2 (t) e dt ≤ |f (x + iy)| 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 2 2 |f1 (x + iy)| dx · |F1 (t)| 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 [11, 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).

254

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 wn+1 k! n=0 k=0 Z X ∞  m 1 am z = 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 [11, §5.3, p. 73] or [36, §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

255

of the power series of f satisfies ∞   X Mf (r)  s n+1 s k |Rn (s)| ≤ Mf (r) = . · 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 −wz an z n e−wz dz = e f (z) dz = Φα (ω) = = Φ(w) wn+1 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 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| −∞ −∞ nZ ∞ o1 n Z ∞ dξ o 12 2 ≤ × |f (ξ + iǫ)|2 dξ 2 −∞ −∞ (ξ − x) Z n ∞ √ dξ o 21 ≤ 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 [50, 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

258

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 [55, 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 [62, 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 ∞ 2 |f (x + iy)| ≤ 2 dt × ds 2 2 2 2 π −∞ (x − t) + y −∞ (x − s) + y Z o 1 nZ ∞ 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 + y2 2 + y2 (x − s) (x − s) −∞ −∞ 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 ∞ 2 dx |ϕ(t)| 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 [7, 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)

259

(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 ∈ L2 (−∞, ∞), we derive from [78, Lemma 9.3, p. 286] that G(x) = F (−x) which vanishes on [0, ∞). Conversely, we suppose that there exists an f with fb = 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

1 ψ(z) = √ 2π

Z

∞

f (t)e−izt dt,

(19.21)

−∞

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 and K(eiθ ) = fb(x), k(ζ) = ψ(z) = ψ i ζ −1 iθ

+1 where ζ = reiθ and 0 ≤ r < 1 and x = i eeiθ −1 . Then it is easily seen from Theorem 12.12 (The Hausdorff-Young Theorem) that

Z

π

−π

|K(eiθ )|2 dθ = 2

Z

∞ −∞

|fb(x)|2 dx ≤ 2 1 + x2

Z

∞ −∞

2 |fb(x)|2 dx = 2 fb 2 ≤ 2kf k22 = 2kϕk22 < ∞.

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

260

Chapter 19. Holomorphic Fourier Transforms

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)

Now the formula Z π Z π Z π 1 1 1 log |k(reiθ )| dθ = log+ |k(reiθ )| 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 + iθ log |k(re )| 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 |K(reiθ )|2 dθ − log m − m log r. (19.26) 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)

19.2. Quasi-analytic Classes and Borel’s Theorem

261

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. 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 [61, 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 [70, Theorem 1, p. 4]. Put mn = MMn+1 n mn − mn−1 =

Mn Mn+1 Mn−1 − Mn2 Mn+1 − = ≥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

mnn−k

=

1 1 1 × × ··· × mn mn mn

262

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 ≤ · = M · ≤ 2 · 2k Mk (19.29) k (2mn )n−k 2n−k 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

n=0

(19.30)

Mn M1 = M0 + + · · · ≥ 1 = M0 . (2mn )n 2m1

19.2. Quasi-analytic Classes and Borel’s Theorem In addition, if k ≥ 1, then since every term |f

(k)

Mn (2mn )n−k

k

(0)| = |i |

∞ X

n=0

263

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

264

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 [61, 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 λn Z λ2 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

265

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 [38, pp. 79 – 84]. See also [32].

Problem 19.11 Rudin Chapter 19 Exercise 11.

Proof. An example of a function ϕ ∈ C ∞ with the required properties can be found in [77, 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.



266

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! k k−m n−m (m) Cm λn x ϕ (λn x), (n − m)! 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 (m) such that |ϕ (λn x)| ≤ M . Thus we obtain |(D k fn )(x)| ≤ |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 ≤ |βn |M 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 [61, 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)   + (D n 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 [41] and [56].



19.2. Quasi-analytic Classes and Borel’s Theorem

267

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, [46, Theorem 1.5.4, p. 30] and [51].

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

k=n+2


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 n 2n 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.



268

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)}.

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
π|z| positive constants C1 and C2 . Therefore, we have |αf (z) + βg(z)| ≤ |α|C1 + |β|C2 e and kαf + βgk2 ≤ |α| · kf k2 + β · kgk2 < ∞

g

See [79, Problem 5.25, pp. 68, 69]

19.2. Quasi-analytic Classes and Borel’s Theorem

269

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 < ∞

implies 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 in §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 × = 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 [62, Example 4.5(b), p. 78] that Z π 1 hf, giT = f (t)g(t) dt 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 (19.43) kF kX = hF, F iX = hf, f iT = kf kT , so it makes X Hilbert. Of course, it follows from the expression (19.43) 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), 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. 

270

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 in §19.1, its Fourier transform Z ∞ b f (t)eitz dt f (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 ∞ n itx (n) b e−|t| · |t|n dt = 2n!. f (t)(it) e dt ≤ |(f ) (x)| = −∞

−∞

Consequently, the power series

∞ 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. 

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]. h

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 [62, 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. 271

272

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 [62, 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 [27, 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 [9, 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 

|L(f )| ≤ 2π 1 +

∞ X

n=1



rn · kf k∞ < ∞

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. This shows that Mergelyan’s Theorem does not hold anymore if the finiteness of the components of S 2 \ K is dropped. a

273 • 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

∞

X a ncn ω n−1 . F (ω) = − 2 + ω n=1 ′

Since zf (z) → 1 as z → ∞, f has the form ∞

1 X a−n f (z) = + z zn n=0

so that

∞

−1 X −na−n . f (z) = 2 + z z n+1 ′

n=0

274

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

1 b= 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 1 b= 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 c0 b= 2πi

Z

Γ

g(z) dz +

∞ X

n=1

h 1 Z i cn a · gn+1 (z) dz = c0 2πi Γ n

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 analytic capacity, 49 annihilate, 241

Hurwitz’s Theorem, 18, 111, 114 hyperbolic transformation, 142 K Koebe mapping, 128 Kolmogorov’s Theorem, 225 Kronecker’s Approximation Theorem, 52, 131

B Baire’s Theorem on Semicontinuous Functions, 205 Bolzano-Weierstrass Theorem, 80, 111, 202, 268 Borel transform, 254 Borel’s Theorem, 267

L lacunary power series, 93 Leibniz’s Rule, 270 locally compact abelian, 242 loxodromic transformation, 142 Lusin Area Integral, 148

C Cauchy type integrals, 47 continuum, 195 converge strongly, 57 converge weakly, 57 Convolution Theorem, 243

M multipler of the transformation, 139 N normal form, 142

D determinant of ϕ, 131 Dirac delta function at x, 57 Dirichlet’s Approximation Theorem, 53

O Osgood’s Theorem, 96

E elliptic transformation, 142 equivalent under G, 184 extreme point, 56

P P´ olya representation, 254 parabolic transformation, 142 Poisson kernel, 258 Poisson kernel for the upper half-plane, 45 Principle of Subordination, 108

F Fej´er’s Theorem, 65 Fundamental Normality Test, 203 Fundamental Theorem of Algebra, 237

R Removable sets for holomorphic functions, 49 Riemann-Lebesgue Lemma, 243 right-shift operator, 238 Rogosinski’s Theorem, 81

H Hadamard gaps, 93 Hadamard’s Three-Line Theorem, 72 harmonic conjugate, 181, 224 Harnack’s inequalities, 49 Hausdorff measure, 49 Hilbert’s identity, 245

S Schottky’s Theorem, 122 Schwarz Integral Formula, 41 275

276 Schwarz Reflection Principle for U , 101 semisimple, 247 Stone-Weierstrass Theorem, 30 strong convergence, 57 Swiss cheese set, 272 T The Basic Connectedness Lemma, 42

Index total variation of f , 232 totally disconnected, 195 V Vitali Convergence Theorem, 96 W weak convergence, 57 weak∗ convergence, 57

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