Univalent Functions: A Primer 9783110560961, 9783110560091

Table of contents :
Preface
Contents
List of Symbols
1. Univalent Functions – the Elementary Theory
2. Definitions of Major Subclasses
3. Fundamental Lemmas
4. Starlike and Convex Functions
5. Starlike and Convex Functions of Order α
6. Strongly Starlike and Convex Functions
7. Alpha-Convex Functions
8. Gamma-Starlike Functions
9. Close-to-Convex Functions
10. Bazilevič Functions
11. B1(α) Bazilevič Functions
12. The Class U(λ)
13. Convolutions
14. Meromorphic Univalent Functions
15. Loewner Theory
16. Other Topics
17. Open Problems
Concluding Remarks
Bibliography
Index

Citation preview

Derek K. Thomas, Nikola Tuneski, Allu Vasudevarao Univalent Functions

De Gruyter Studies in Mathematics

| Editor in Chief Carsten Carstensen, Berlin, Germany Gavril Farkas, Berlin, Germany Nicola Fusco, Napoli, Italy Fritz Gesztesy, Waco, Texas, USA Niels Jacob, Swansea, United Kingdom Zenghu Li, Beijing, China Karl-Hermann Neeb, Erlangen, Germany

Volume 69

Derek K. Thomas, Nikola Tuneski, Allu Vasudevarao

Univalent Functions | A Primer

Mathematics Subject Classification 2010 30C45, 30C55, 30C80 Authors Derek K. Thomas Swansea University Department of Mathematics Singleton Park Swansea SA2 8PP UK [email protected]

Allu Vasudevarao Indian Institute of Technology Kharagpur Department of Mathematics Kharagpur – 721302 India [email protected]

Nikola Tuneski Ss. Cyril and Methodius University in Skopje Department of Mathematics and Informatics Faculty of Mechanical Engineering 1000 Skopje Republic of Macedonia [email protected]

ISBN 978-3-11-056009-1 e-ISBN (PDF) 978-3-11-056096-1 e-ISBN (EPUB) 978-3-11-056012-1 ISSN 0179-0986 Library of Congress Control Number: 2018935017 Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2018 Walter de Gruyter GmbH, Berlin/Boston Typesetting: le-tex publishing services GmbH, Leipzig Printing and binding: CPI books GmbH, Leck ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

Preface The study of univalent functions dates back to the early years of the twentieth century and is one of the most popular areas of research in complex analysis. Initiated by the work of Bieberbach and his contemporaries, the famous conjecture of 1916 became one of the most celebrated problems in mathematics. The eventual solution of the Bieberbach conjecture by de Branges in 1984 employed nonelementary methods from several branches of analysis, the understanding of which probably presents a challenge to postgraduate students and young researchers. There are several excellent books on the theory of univalent functions, notably those of Hayman [70], Duren [45], and Pommerenke [198]. All these give accounts of the elementary material, and thereafter present more advanced and deeper ideas, many of which could again be difficult for postgraduate students to assimilate. The books by Goodman [58, 59] provided a summary of known results to date for a very wide range of subclasses of univalent functions. An extensive number of subclasses were considered, and many of the important results were given, which therefore restricted the number of proofs that could be included. Naturally, since its publication in 1983 many more results have been discovered, and so an update seems timely. Since the field of univalent functions is large and the aim of this book is to included as many proofs as possible, its content must necessarily be highly selective. The book concentrates on what the authors consider to be the important subclasses of univalent functions and is primarily directed toward researchers who are new to the field of univalent functions. The aim, therefore, is to describe the important ideas and techniques used in solving problems, while at the same time updating as far as possible much of what is known for a variety of subclasses of univalent functions. Since the material included in this book is chosen to be accessible to new researchers, results requiring deeper methods are omitted. However in order to introduce the reader to one of the most significant advanced methods, we include an introduction to the basic ideas behind Loewner Theory, which was used by de Branges in his proof of the Bleberbach conjecture. The set of starlike functions is perhaps the most natural subclass of univalent functions, and over the last century many of the fundamental and significant properties of these functions have been settled, resulting in a range of rich and beautiful theorems. Other important subclasses of univalent functions have also been developed and explored, and many other interesting results obtained, but at the same time leaving a great number of unsolved problems. Since many of the solutions to problems rely on the use of a relatively small number of ideas and techniques, the authors believe that a primer on univalent functions illustrating these ideas is timely, and would appeal to the ever growing number of researchers in this field. Changes in the means of publication and the spread in popularity of the study of univalent functions over recent years has led to an ever increasing number of pubhttps://doi.org/10.1515/9783110560961-201

VI | Preface

lished papers, many of which could be considered simple extensions to existing theorems, without the introduction of any new significant ideas or results. We believe that providing a resource describing the methods commonly found in proofs may help to slow down this hyper-production of papers. To this end, the final chapter of this book includes a number of unsolved problems. Finding the solution to any of these would represent a significant addition to the subject.

February 2018

Derek K. Thomas, Swansea Nikola Tuneski, Skopje Allu Vasudevarao, Kharagpur

Contents Preface | V List of Symbols | XI 1 1.1 1.2 1.3

Univalent Functions – the Elementary Theory | 1 Definitions and Basic Properties | 1 Bieberbach’s Conjecture and Related Topics | 3 Growth and Distortion Theorems | 8

2 2.1 2.2 2.3 2.4 2.5

Definitions of Major Subclasses | 13 Convex and Starlike Functions | 13 Close-to-Convex Functions | 16 Bazilevič Functions | 18 The Class U | 20 Rotational Invariance | 20

3 3.1 3.2 3.3

Fundamental Lemmas | 22 Functions with Positive Real Part | 22 Subordination | 31 The Clunie–Jack Lemma | 35

4 4.1 4.1.1 4.1.2 4.1.3

Starlike and Convex Functions | 37 Starlike Functions | 37 Coefficient Theorems | 37 Refined Growth Theorems | 42 Theorems Concerning lim arg f(re iθ ) | 45 r→1

4.1.4

The Radial Limit lim f(re iθ ) | 47

4.1.5 4.1.6 4.2 4.2.1 4.2.2

Length and Integral Mean Problems | 50 Some Subclasses of Starlike Functions | 54 Convex Functions | 58 Growth and Distortion Theorems | 59 Coefficient Inequalities | 60

5 5.1 5.2 5.3

r→1

Starlike and Convex Functions of Order α | 65 Definitions and Growth and Distortion Theorems | 65 Inclusion Relationships | 67 Coefficient Theorems | 75

VIII | Contents

5.4 5.4.1 5.4.2

Sufficient Conditions | 78 Sufficient Conditions on |f 󸀠󸀠 (z)| | 78 On a Class Defined by Silverman | 81

6 6.1 6.2 6.3 6.4 6.5

Strongly Starlike and Convex Functions | 83 Definitions | 83 Strongly Starlike Functions | 83 Coefficient Theorems | 87 Strongly Convex Functions | 93 Inclusion Relationships | 95

7 7.1 7.2 7.3

Alpha-Convex Functions | 99 Definition and Integral Representation | 99 Distortion and Growth Theorems | 101 Coefficient Problems | 106

8 8.1 8.2 8.2.1 8.2.2 8.2.3

Gamma-Starlike Functions | 112 Definition and Basic Properties | 112 Coefficient Inequalities | 113 Logarithmic Coefficients | 116 Inverse Coefficients | 118 The Second Hankel Determinant | 119

9 9.1 9.2 9.3 9.3.1 9.3.2 9.3.3 9.3.4 9.3.5 9.3.6 9.4 9.5 9.5.1 9.5.2

Close-to-Convex Functions | 121 Definitions and Basic Properties | 121 Distortion Theorems | 124 Coefficient Problems | 125 The Fekete–Szegő Problem | 125 The Zalcman Conjecture | 127 Difference of Coefficients | 132 Robertson’s Conjecture | 134 Logarithmic Coefficients | 136 The Second Hankel Determinant | 143 Growth Estimates | 143 Ozaki Close-to-Convex functions | 149 Growth and Area Estimates | 151 Strongly Ozaki Close-to-Convex Functions | 151

10 10.1

Bazilevič Functions | 153 Definition and Basic Properties | 153

Contents | IX

10.2 10.2.1 10.2.2 10.3 10.4

Growth Theorems | 154 Coefficients of Powers of Functions in B(α) | 159 Logarithmic Coefficients | 162 The Fekete–Szegő Problem | 162 Sufficient Conditions for f ∈ B(α) | 163

11 11.1 11.2 11.3 11.4 11.5

B1 (α) Bazilevič Functions | 165 Definition and Basic Properties | 165 Distortion Theorems | 165 Growth Estimates | 169 Coefficients | 171 Other Inequalities | 174

12 12.1 12.2 12.3

The Class U(λ) | 179 Definition and Geometrical Properties | 179 Sufficient Conditions and Univalence | 182 Coefficients | 188

13 13.1 13.2

Convolutions | 195 Definition and the Pólya–Schoenberg Conjecture | 195 Subordination and Convolution | 201

14 Meromorphic Univalent Functions | 205 14.1 The Class Σ | 205 14.1.1 Coefficients and the Clunie Constant | 205 14.1.2 Coefficients of the Inverse Function | 209 14.1.3 Distortion Theorems | 209 14.2 Subclasses of Σ | 210 14.2.1 Meromorphic Starlike Functions | 210 14.2.2 Meromorphic Close-to-Convex Functions | 211 14.2.3 Meromorphic Bazilevič Functions | 213 15 15.1 15.2

Loewner Theory | 215 The Loewner Equation | 215 Applications | 216

16 16.1 16.2

Other Topics | 224 Harmonic Univalent Functions | 224 Bi-univalent Functions | 225

X | Contents

16.3 16.4 16.5 16.5.1 16.5.2 17

Functions of Bounded Boundary Rotation | 226 Differential Subordinations | 229 Operators | 230 The Sălăgean Operator | 230 The Libera Operator and Generalizations | 232 Open Problems | 235

Concluding Remarks | 238 Bibliography | 239 Index | 250

List of Symbols ≪ – majorization 75 ≺ – subordination 31 2 F 1 – Gaussian hypergeometric function 94 A – class of normalized analytic functions 1 A(r) – area enclosed by C(r) 39 A n – n-th coefficient of the inverse function 7 B(α) – class of Bazilevič functions when β = 0 18 B(α, β) – class of Bazilevič functions of type (α, β) 18 B1 (α) – subclass of B(α) when g(z) = z 19 C – class of normalized convex functions 13 ℂ – complex plane 1 C(α) – class of normalized convex functions of order α 65 C(A, B) – class of convex functions introduced by Janowski 65 C(r) – image of the circle |z| = r under a given function 3 C∗∗ (α) – class of strongly convex functions of order α 83 Cg (h) – class of generalized convex functions defined with convolution 202 co U – closed convex hull of U 128 co U – convex hull of U 128 CR+ – subclass of K0 for g(z) = k(z) 20 CR+ := K0 (k) – class of close-to-convex functions with respect to the Koebe function with argument 0 140 𝔻 – unit disk 1 Dn [f] – Sălăgean operator 230 D∗n [f] – Ruscheweyh operator 231 D n – Toeplitz determinant 26 D−n 61 D+n 61 ∆ – set of points z such that |z| > 1 3 Er – disk with radius r, Er = {z : |z| < r} 201 F(λ) – class of Ozaki close-to-convex functions 149 F(λ, α) – class of strongly Ozaki close-to-convex functions 151 f ∗ g – convolution of f and g 195 F1 – class of close-to-convex functions with respect to odd starlike functions 141 F2 – subclass of the class of close-to-convex functions 142 F3 – subclass of the class of close-to-convex functions 142 Gλ – class defined by Silverman 81 Gλ,γ – class defined by Singh and Tuneski 82 γ n – n-th logarithmic coefficient 6 g ⊀ f – g is not subordinate to f 201 https://doi.org/10.1515/9783110560961-202

XII | List of Symbols

H 2 (2) – second Hankel determinant 41 H 3 (1) – third Hankel determinant 42 H q (n) – qth Hankel determinant 8 HK – closed convex hull of K 128 In [f] – integral operator related to the Sălăgean operator 231 Jα [f](z) – operator that defines the class of α-convex functions 99 K – class of normalized close-to-convex functions 17 ̃ K(α) – class of strongly close-to-convex functions of order α 121 K(α, z) – α-convex Koebe function 101 k(z) – Koebe function 1 K(z; ϵ; α) – general α-convex Koebe function 109 Kβ – class of normalized close-to-convex functions with argument β 121 Kβ (k) – class of close-to-convex functions with respect to the Koebe function with argument β 140 K0 – class of normalized close-to-convex functions with argument 0 17 Kg (h) – class of generalized close-to-convex functions defined with convolution 202 L(r) – length of the curve C(r) 51 L[f] – Libera integral operator 75 Lc [f] – Bernardi operator 232 Λ λ (f) – Fekete–Szegő functional 131 Li2 – dilogarithm function 192 M(r) – modulus maximum of a function on the circle |z| = r 36 Mγ – class of gamma-starlike functions 112 Mα – class of α-convex functions 99 M p (r, f) – integral mean 32 ℕ – the positive integers (natural numbers) 39 N δ (f) – δ-neighborhood 203 ℕ0 = {0} ∪ ℕ 230 P – class of normalized functions with positive real part 13 P(α) – class of normalized functions with real part > α 22 P[f](z) = f(z)/f 󸀠 (z) 187 Φ λ (f) – functional Φ λ (f) = λa2n − a2n−1 131 R – class of normalized functions whose derivative has positive real part 121 S – class of normalized univalent functions 1 S∗ – class of normalized starlike functions 13 S∗ (α) – class of normalized starlike functions of order α 65 S∗ (A, B) – class of starlike functions introduced by Janowski 65 S∗∗ (α) – class of strongly starlike functions of order α 83 S∗g (h) – class of generalized starlike functions defined by convolution 202 S∗L – subclass of the class of starlike functions 55 S∗n (α) – generalization of the class S∗ using the Sălăgean operator 231

List of Symbols

̃ – class of bi-univalent functions 225 Σ Σ – class of meromorphic functions 3 Σ󸀠 – subclass of Σ with b 0 = 0 3 Σ∗ – class of meromorphic starlike functions 210 Σ B(α) – class of meromorphic Bazilevič functions 213 ΣK0 – class of meromorphic close-to-convex functions 211 𝕋 – unit circle, 𝕋 = {z : |z| = 1} 199 U – the class U := U(1) 20 U(λ) – the class U(λ) 20 Vk – class of functions of bounded boundary rotation 226 (x)k – factorial polynomial, (x)k = x(x − 1) . . . (x − k + 1), (k = 1, 2, . . . ; (x)0 = 1) 108 ⟨x⟩k – factorial polynomial, ⟨x⟩k = x(x + 1) . . . (x + k − 1), (k = 1, 2, . . . ; ⟨x⟩0 = 1) 107

|

XIII

1 Univalent Functions – the Elementary Theory 1.1 Definitions and Basic Properties A domain is an open connected subset of the complex plane ℂ. A complex-valued function f of a complex variable is called univalent in a domain D if it does not take the same value twice, so that for z1 , z2 ∈ D, f(z1 ) ≠ f(z2 )

for

z 1 ≠ z 2 .

(1.1.1)

Let D be a simply connected domain, which is a proper subset of the complex plane and z be a given point in D. Then the Riemann mapping theorem guarantees that there is a unique function f which maps D conformally onto the unit disk, with the properties f(z) = 0 and f 󸀠 (z) > 0. A necessary condition for a function f analytic in D to be univalent in D is that its derivative does not vanish on D, i.e., f is univalent in D

⇒

f 󸀠 (z) ≠ 0 for all z ∈ D.

A proof of this can be obtained using Rouche’s theorem and is given in Titchmarsh [249, p. 198]. The opposite is not true, as the function f(z) = z2 defined on D = ℂ \ {0} shows. Thus, if f is analytic and univalent in D, then without loss of generality we can assume that D is the unit disk 𝔻 = {z : |z| < 1}, with Taylor expansion ∞

f(z) = z + ∑ a n z n .

(1.1.2)

n=2

The normalization f(0) = f 󸀠 (0) − 1 = 0 is obtained by the transformation f(z) − f(0) . f 󸀠 (0) We denote by A, the class of functions f analytic in 𝔻 and normalized by f(0) = − 1 = 0. When f is univalent, we denote this a subset of A by S (the symbol S reflects the original German word schlicht used for univalent). The theory of univalent functions is an old, but still very active field, dating back over a century. Much of its history is related in one form or other to the famous Bieberbach conjecture that |a n | ⩽ n for n ⩾ 2, inspired by the so-called Koebe function k ∈ S, defined for z ∈ 𝔻 by ∞ z = z + ∑ nz n , k(z) = 2 (1 − z) n=2 f 󸀠 (0)

which maps 𝔻 onto ℂ, slit along the negative real axis from −1/4 to −∞, i.e., k(𝔻) = ℂ \ (−∞, −1/4]. To see this, note that k(z) =

1 1+z 2 1 z = ( ) − , 2 4 1−z 4 (1 − z)

https://doi.org/10.1515/9783110560961-001

2 | 1 Univalent Functions – the Elementary Theory and observe that (1 + z)/(1 − z) maps the unit disk onto the right half of the complex plane. We also note that using (1.1.1), it is an elementary exercise to show that k is univalent, and so k ∈ S. Apart from the Koebe function, other examples of functions in S are (i) f(z) = z, the identity map; z , with range f(𝔻) = {w : Re w > −1/2}; (ii) f(z) = 1−z z (iii) f(z) = , with range f(𝔻) = ℂ \ {(−∞, −1/2] ∪ [1/2, +∞)}; 1 − z2 1 1+z (iv) f(z) = log , with range f(𝔻) = {w : | Im w| < π/4}; 2 1−z 1 (v) f(z) = z − z2 , with range f(𝔻), the interior of a cardioid. 2 Univalence is preserved under a variety of transformations, i.e., if f ∈ S, then direct verification of (1.1.1) shows that g ∈ S, where n (i) g(z) = f(z) = z + ∑∞ n=2 a n z (conjugation) (ii) g(z) = e−iθ f(e iθ z) (rotation) 1 (iii) g(z) = f(rz), with 0 < r < 1 (dilation) r z+α f( ) − f(α) 1 + αz (iv) g(z) = , with |α| < 1 (disk automorphism) (1 − |α|2 )f 󸀠 (α) (v) g = ψ∘f where ψ is analytic and univalent in the range of f , and ψ(0) = ψ󸀠 (0)−1 = 0 (range transformation) ωf(z) (vi) g(z) = , with ω ∉ f(𝔻) (omitted-value transformation) ω − f(z) (vii)g(z) = z√ f(z2 )/z2 (square root transformation). Apart from (vii), all of the above are easily directly verified using (1.1.1). We explain the square root transformation as follows. Since f(z) = 0 only when z = 0, we choose a single-valued branch of the square root so that for z ∈ 𝔻, g(z) = z√f(z2 )/z2 = z ⋅ √1 + a2 z2 + a3 z4 + . . . = z + c3 z3 + c5 z5 + . . . , i.e., g(−z) = −g(z). Thus for z1 , z2 ∈ 𝔻 with 0 ≠ z1 , z2 , g(z1 ) = g(z2 )

⇒

f (z21 ) = f (z22 )

⇒

z21 = z22

⇒

z1 = ±z2 .

If z1 = −z2 , then g(z1 ) = g(z2 ) = g(−z1 ) = −g(z1 ), so that g(z1 ) = 0, and z1 = 0. Thus, g(z1 ) = g(z2 ) only when z1 = z2 , and so g is univalent. On the other hand, the sum (mean) of two univalent functions may also be univalent, but not always. For example, it is trivial to show that the sum of the two univalent

1.2 Bieberbach’s Conjecture and Related Topics | 3

functions z and z − z2 /2 is also univalent. On the other hand, the sum of the two univalent functions z/(1 − z) and z/(1 + iz) is not univalent, since the derivative of their sum is 1/(1 − z)2 + 1/(1 + iz)2 , which vanishes when z = (1 + i)/2 ∈ 𝔻. Closely related to S is the class Σ of meromorphic functions of the form ∞

g(z) = z + b 0 + ∑ b n z−n ,

(1.1.3)

n=1

which are analytic and univalent in the exterior of 𝔻, i.e., in ∆ = {z : |z| > 1}, except for a simple pole at infinity with residue 1. The subclass Σ󸀠 of Σ consists of functions g with the additional property that g(z) ≠ 0 for all z ∈ ∆. We note at this point that the so-called inverse transformation 1/f(1/z), given by 1 = z − a2 + (a22 − a3 )z−1 + . . . , f(1/z)

(1.1.4)

establishes a one-to-one correspondence between S and Σ󸀠 , (the univalence is obvious, and the inverse transformation maps a function f ∈ S onto an analytic function of the form (1.1.3), which omits zero, and vice versa).

1.2 Bieberbach’s Conjecture and Related Topics Before discussing Bieberbach’s famous conjecture and its eventual solution, we first establish the so-called area theorem proved by Gronwall [62] in 1914. Theorem 1.2.1 (Area theorem). If g ∈ Σ, then ∞

∑ n|b n |2 ⩽ 1. n=1

Equality holds if, and only if, the set E = ℂ \ g(∆) of values omitted by g has a twodimensional Lebesgue measure of zero. Proof. Let C(r) = g(|z| = r), r > 1 be the image of the circle |z| = r. Since g is univalent, it follows that C(r) is a simple closed curve that encloses a domain E(r) which contains the set E = ℂ \ g(∆) of values omitted by g. With z = re iθ , using Green’s theorem we have 1 1 ∫ w dw = ∫ g(z)g󸀠 (z)dz area of E(r) = 2i 2i |z|=r

C(r) 2π

∞ 1 = ∫ [re−iθ + ∑ b n r−n e inθ ] 2 n=0 0

∞

⋅ [1 − ∑ kb k r−k−1 e−i(k+1)θ ] re iθ dθ k=1

4 | 1 Univalent Functions – the Elementary Theory 2π

=

∞ 1 ∫ [r2 + ∑ b n r1−n e iθ(1+n)] 2 n=0 0

∞

⋅ [1 − ∑ kb k r−k−1 e−i(k+1)θ ] dθ. k=1

Since

2π

∫ e i(n−k)θ dθ = 0 when n ≠ k, 0

and 2π

∫ b n r1−n e iθ(1+n) ⋅ kb k r−k−1 e−i(k+1)θ dθ = 2πn|b n |2 r−2n

when n = k,

0

we obtain

∞

area of E(r) = π [r2 − ∑ n|b n |2 r−2n ] . n=1

Letting r → 1 we obtain

∞

m(E) = π [1 − ∑ n|b n |2 ] , n=1

and since m(E) ⩾ 0, we have

∑∞ n=1

n|b n

|2

⩽ 1.

An immediate corollary is Corollary 1.2.1. If g ∈ Σ, then |b 1 | ⩽ 1. Equality holds if, and only if, g(z) = z + b 0 + b 1 /z with |b 1 | = 1. We now show how, using the area theorem, Bieberbach in 1916 [16] proved that if f ∈ S, then |a2 | ⩽ 2, and we then briefly describe the history of his famous conjecture that for f ∈ S, |a n | ⩽ n for n ⩾ 2. Theorem 1.2.2 (Bieberbach’s theorem). If f ∈ S, then |a2 | ⩽ 2. Equality holds if, and only if, f is a rotation of the Koebe function. Proof. Let f ∈ S and g(z) = z/√z2 f(1/z2 ). It is easy to see that g ∈ Σ, and so |b 1 | ⩽ 1, with equality if, and only if, g(z) = z + b 0 + b 1 /z with |b 1 | = 1. However, since b 0 = 0, and 2b 1 = −a2 , the inequality |a2 | ⩽ 2 follows at once. It is easy to check that the rotation of the Koebe function k(z) = z/(1 + b 1 z)2 is such that g(z) = z/√z2 k(1/z2 ). Bieberbach’s conjecture was the catalyst from which the study of univalent functions grew, and during the last century it developed into the extensive field of mathematical research we can see today. n Bieberbach’s conjecture. If f ∈ S, and is given by f(z) = z + ∑∞ n=2 a n z , then |a n | ⩽ n for any integer n ⩾ 2. Equality holds if, and only if, f is a rotation of the Koebe function.

1.2 Bieberbach’s Conjecture and Related Topics |

5

This conjecture became one of the most celebrated mathematical problems of the twentieth century. Progress was slow, and by 1968 the conjecture had been established only for n = 2, 3, 4, 5, and 6. For n ⩾ 3, the methods of proof required the deployment of nonelementary methods. The case n = 3 was settled in 1923, by Loewner [125], who developed an ingenious method of considering a class of functions defined via a differential equation, which are dense in S. This method not only solved the Bieberbach conjecture in the case n = 3, but was also used to find the sharp bounds for the coefficients of the inverse function of f ∈ S (see, for example, Hayman [70]). Even more importantly, Loewner’s theory was eventually used by de Branges in his celebrated proof of the Bieberbach conjecture [40]. During the twentieth century, others tried to find the smallest constant C, such that |a n | < Cn, and by 1976 the best known bound for C was approximately 1.0691, which remained the best known bound for all n ⩾ 2 until de Branges found a proof of the Bieberbach conjecture in 1984. The so-called Koebe 1/4 theorem, follows easily from the inequality |a2 | ⩽ 2. Theorem 1.2.3 (The Koebe 1/4 theorem). Let f ∈ S, and f(z) ≠ w for z ∈ 𝔻. Then |w| ⩾ 1/4, with equality only possible when f is a rotation of the Koebe function, which omits a value of modulus 1/4. Proof. Let f ∈ S, and f omit the value ω ∈ ℂ, i.e., f(z) ≠ ω for z ∈ 𝔻. Then by the omitted-value transformation, the function g(z) = ωf(z)/(ω − f(z)) is also in S. Since the second coefficient of g(z) equals a2 + 1/w, it follows that 󵄨󵄨 1 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨a2 + 󵄨󵄨󵄨 ⩽ 2, 󵄨󵄨 ω 󵄨󵄨 and so

󵄨󵄨 1 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 ⩽ 󵄨󵄨 + a2 󵄨󵄨󵄨 + |a2 | ⩽ 4. 󵄨󵄨 󵄨󵄨 ω 󵄨󵄨 󵄨󵄨 ω

Returning to (1.1.4), we notice at once the sharp inequality |a3 −a22 | ⩽ 1 for f ∈ S, which no doubt inspired the following deeper theorem of Fekete and Szegő ([48] or [45, Theorem 3.8]), the proof of which again required the use of the Loewner theory, together with an elegant lemma of Valiron and Landau ([103] or [45, p. 104]). Theorem 1.2.4 (The Fekete–Szegő theorem). Let f ∈ S and be given by f(z) = z + n ∑∞ n=2 a n z . Then for 0 < μ < 1, |a3 − μa22 | ⩽ 1 + 2e−2μ/(1−μ) . The inequality is sharp for each μ. Since the Koebe function is not the extremal function in the above Fekete–Szegő theorem, the theory of univalent functions is more than just showing that k is the extremal function for questions in S.

6 | 1 Univalent Functions – the Elementary Theory

Establishing Fekete–Szegő theorems for subclasses of univalent functions has become a popular area of research in recent years, and we will include some of these in this book. As was noted above, de Branges’s proof of the Bieberbach conjecture was far from simple. The proof not only requires the use of the Loewner method, but is also based on a consideration of the coefficients of the function log( f(z)/z), together with some powerful inequalities of Lebedev and Milin [104, 142] concerning general power series. Subsequently, refinements of de Branges’s proof have been found, a good account of which can be found in Hayman [70]. Since we will use the third Lebedev–Milin inequality in a subsequent chapter, we state it below. Also, since we will consider the logarithmic function log( f(z)/z), and the inverse function f −1 of a function f ∈ S at various points throughout this book, we define these now in terms of power series. Theorem 1.2.5 (Third Lebedev–Milin inequality). Let ∞

ϕ(z) = ∑ α k z k k=1

be an arbitrary power series with a positive radius of convergence, and be such that ϕ(0) = 0. Also let ∞

ψ(z) = e ϕ(z) = ∑ β k z k . k=0

Then for n ⩾ 1,

n

|β n |2 ⩽ exp { ∑ (k|α k |2 − k=1

Equality occurs if, and only if, α k = |γ| = 1.

γ k /k

1 )} . k

for k = 1, 2, . . . , n, for some constant γ, with

The Logarithmic Function Definition 1.2.1. Let f ∈ S. Then the logarithmic coefficients γ n of f are defined for z ∈ 𝔻 by ∞ f(z) (1.2.1) = 2 ∑ γn zn . log z n=1 Relatively little exact information is known about the coefficients of log( f(z)/z) when f ∈ S. The Koebe function leads to the natural conjecture that for n ⩾ 1, |γ n | ⩽ 1/n. However, this is false, even in order of magnitude (see Duren [45, Section 8.1]), with most results obtained being of an averaging nature, Milin [140, 142]. It is important to point out at this stage that Milin [140, 142] conjectured the following (which implies Bieberbach’s conjecture), and that this was established by de Branges in his proof of the Bieberbach’s conjecture.

1.2 Bieberbach’s Conjecture and Related Topics |

7

Milin’s conjecture. For each f ∈ S, n

m

∑ ∑ (k|γ k |2 − m=1 k=1

1 ) ⩽ 0, k

for any positive integer n. For subclasses, the situation is not a great deal better. However, we will include some exact results in this book, which in most cases will involve only the initial coefficients of log( f(z)/z). Since most of these results concern the first three coefficients, we calculate these in terms of the coefficients of f(z) by equating coefficients in (1.2.1) as follows: 1 γ 1 = a2 2 1 1 γ2 = (a3 − a22 ) 2 2 1 1 (1.2.2) γ3 = (a4 − a2 a3 + a32 ) . 2 3

Inverse Functions We first note that because f ∈ S is univalent, f −1 exists in some disk |ω| < r0 (f). Let ∞

f −1 (ω) = ω + ∑ A n ω n .

(1.2.3)

n=2

The problem of determining the sharp bounds for the coefficients |A n | when f ∈ S was solved by Loewner [125] (see also [70, p. 222]) who proved the following: Theorem 1.2.6. Let f ∈ S, and f −1 be defined by (1.2.3). Then for n ⩾ 2, |A n | ⩽

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅ ⋅ (2n − 1) ⋅ 2n . (n + 1)!

(1.2.4)

Equality holds when f −1 is the inverse of the Koebe function. It is clear that if the Koebe function k is contained in any subclass of S, then (1.2.4) provides the solution to the inverse coefficient problem. On the other hand, if k is not a member of a particular subclass, then the problem is not the same, resulting in a different solution. We will consider inverse coefficient problems of this nature in this book, most of which only have solutions for the initial coefficients of f −1 . By equating coefficients in f(f −1 (ω)) = ω, we give expressions for the initial coefficients of f −1 in terms of the coefficients of f as follows: A2 = −a2 A3 = 2a22 − a3 A4 = −5a32 + 5a2 a3 − a4 A5 = 14a42 − 21a22 a3 + 3a23 + 6a2 a4 − a5 .

(1.2.5)

8 | 1 Univalent Functions – the Elementary Theory

Hankel Determinants We first give the definition of the Hankel determinant, whose elements are the coefficients of a function f ∈ S. Definition 1.2.2. Let f ∈ S. Then the q-th Hankel determinant of f is defined for q ⩾ 1, and n ⩾ 0 by 󵄨󵄨 a a n+1 . . . a n+q−1 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨 n 󵄨 󵄨󵄨 a a n+2 . . . a n+q 󵄨󵄨󵄨󵄨 󵄨󵄨 n+1 󵄨 󵄨 H q (n) = 󵄨󵄨󵄨 . .. .. 󵄨󵄨󵄨 . 󵄨󵄨 .. . . 󵄨󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨󵄨a n+q−1 a n+q . . . a n+2q−2 󵄨󵄨󵄨 󵄨 󵄨 Although Hankel determinants do not play a major role in the class S, in recent years a great deal of attention has been given to Hankel determinants whose elements are the coefficients of functions in subclasses of S. For S, finding sharp results for H q (n) is clearly a very difficult task, and little is known in this direction. Perhaps the most significant result is due to Hayman [69], who showed that H2 (n) ⩽ An1/2 , where A is an absolute constant, and that this rate of growth is the best possible. Finding the correct rate of growth of H q (n) remains an open problem. Although the exact upper bound for H2 (2) is unknown for S, the problem for subclasses has proved more accessible, and we will include examples of sharp results in this book.

1.3 Growth and Distortion Theorems An important consequence of Bieberbach’s theorem is that it provides sharp lower and upper bounds of |f(z)| and |f 󸀠 (z)|, usually referred to as growth, and distortion theorems respectively. The description “growth” for the estimates of |f(z)| comes naturally, and the word “distortion” for the estimates of |f 󸀠 (z)| follows from the fact that |f 󸀠 (z)| and |f 󸀠 (z)|2 are, respectively, infinitesimal magnification factors of the arc length and the area under the mapping f . We first prove the following. Theorem 1.3.1. For f ∈ S, and z = re iθ ∈ 𝔻, 󵄨󵄨 󸀠󸀠 󵄨 2r2 󵄨󵄨󵄨 4r 󵄨󵄨 zf (z) 󵄨󵄨 󸀠 󵄨󵄨 ⩽ − 2 󵄨󵄨 f (z) 󵄨󵄨 1 − r2 . 1 − r 󵄨 󵄨 Proof. Let f ∈ S, and 0 < |α| = r < 1. Then the disk automorphism g defined by g(z) =

z+α ) − f(α) 1 + αz (1 − |α|2 )f 󸀠 (α)

f(

(1.3.1)

1.3 Growth and Distortion Theorems | 9

is also univalent. Direct calculations show that the second coefficient of g(z) is f 󸀠󸀠 (α) 1 [(1 − |α|2 ) 󸀠 − 2α] . 2 f (α) Multiplying the above by 2α/(1 − |α|2 ), and applying Bieberbach’s theorem, we obtain 󵄨󵄨 󸀠󸀠 󵄨 2|α|2 󵄨󵄨󵄨 4|α| 󵄨󵄨 αf (α) 󵄨󵄨 󸀠 󵄨󵄨 ⩽ . − 1 − |α|2 󵄨󵄨󵄨 1 − |α|2 󵄨󵄨󵄨 f (α) Replacing α with z, and |α| by r completes the proof of the theorem. We are now able to prove the sharp distortion theorem for S. Theorem 1.3.2 (Distortion theorem). For f ∈ S, and z = re iθ ∈ 𝔻, 1+r 1−r ⩽ |f 󸀠 (z)| ⩽ . (1 + r)3 (1 − r)3

(1.3.2)

Equality occurs if, and only if, f is a rotation of the Koebe function. Proof. Let f ∈ S. From (1.3.1) we have − i.e.,

4r zf 󸀠󸀠 (z) 4r 2r2 ⩽ Re , − [ ]⩽ f 󸀠 (z) 1 − r2 1 − r2 1 − r2 2r2 − 4r zf 󸀠󸀠 (z) 2r2 + 4r ⩽ Re [ 󸀠 . ]⩽ 2 f (z) 1−r 1 − r2

(1.3.3)

Since f(0) = 0 and f 󸀠 (0) = 1, we are able to choose a single-valued branch of log f 󸀠 (z), which vanishes at the origin. Also, for z = re iθ and fixed θ, we have r and r

∂ ∂ log f 󸀠 (z) = r [log |f 󸀠 (z)| + i arg f 󸀠 (z)] , ∂r ∂r

d ∂z ∂ f 󸀠󸀠 (z) iθ zf 󸀠󸀠 (z) log f 󸀠 (z) = r [log f 󸀠 (z)] =r 󸀠 e = 󸀠 . ∂r dz ∂r f (z) f (z)

Equating the real parts of the right hand sides of the two last expressions, we obtain Re [

∂ zf 󸀠󸀠 (z) ] = r Re [log f 󸀠 (z)] . f 󸀠 (z) ∂r

(1.3.4)

Combining this with (1.3.3), we have 2r − 4 ∂ 󵄨 2r + 4 󵄨 ⩽ , log 󵄨󵄨󵄨󵄨f 󸀠 (re iθ )󵄨󵄨󵄨󵄨 ⩽ ∂r 1 − r2 1 − r2 which, after integrating with respect to r from 0 to R, gives

(1.3.5)

10 | 1 Univalent Functions – the Elementary Theory

log

1−R 1+R 󵄨 󵄨 ⩽ log 󵄨󵄨󵄨󵄨f 󸀠 (Re iθ )󵄨󵄨󵄨󵄨 ⩽ log . (1 + R)3 (1 − R)3

Exponentiating and replacing R with r, gives the desired result. Since k 󸀠 (z) = (1 + z)/(1 − z)3 , a direct calculation shows that equality holds in both inequalities when f is the Koebe function (or a rotation). To see that rotations of the Koebe function are the only functions where equality holds, we proceed as follows. For a function f ∈ S where |f 󸀠 (z)| is equal to the lower (upper respectively) estimate for some z = Re iθ , equality occurs in the lower (upper respectively) estimate in (1.3.5) for all r ∈ [0, R]. Thus choosing z = r = 0 in (1.3.5), and combining it with (1.3.4), we obtain e iθ f 󸀠󸀠 (0) e iθ f 󸀠󸀠 (0) Re [ 󸀠 ] = −4, or Re [ 󸀠 ] = 4, f (0) f (0) which implies |a2 | = 2, and so f is a rotation of the Koebe function. We now use the above distortion theorem to obtain the growth estimates. Theorem 1.3.3 (Growth theorem). For f ∈ S, and z = re iθ ∈ 𝔻, r r ⩽ |f(z)| ⩽ . (1 + r)2 (1 − r)2

(1.3.6)

Equality occurs if, and only if, f is a suitable rotation of the Koebe function. Proof. We follow the proof given by Goodman [58]. We begin with the right hand inequality. Using Theorem 1.3.2, and the fact that f(0) = 0, integration gives 󵄨󵄨 z 󵄨󵄨 󵄨󵄨 r 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 󵄨 󵄨 |f(z)| = 󵄨󵄨󵄨∫ f 󸀠 (s) ds󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ f 󸀠 (ρe iθ )e iθ dρ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨0 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 r

r

0

0

1+ρ r 󵄨 󵄨 dρ = . ⩽ ∫ 󵄨󵄨󵄨󵄨f 󸀠 (ρe iθ )󵄨󵄨󵄨󵄨 dρ ⩽ ∫ (1 − ρ)3 (1 − r)2 The left hand inequality is more delicate. Let C(r) = f(|z| = r) and let w1 be a point of C(r) that is closest to the origin. By employing a suitable rotation of f , we can assume that w1 > 0. Further, let Λ be the line segment 0 ⩽ w ⩽ w1 , and z1 and L be the pre-images of w1 and Λ under f , respectively. Integrating along the curve L with respect to z from 0 to z1 , we have f 󸀠 (z) dz = dw > 0, and w1

z1

z1

w1 = ∫ dw = ∫ f 󸀠 (z) dz = ∫ |f 󸀠 (z)| ⋅ |dz| 0 r

0

0 r

⩾ ∫ |f 󸀠 (ρe iθ )| dρ ⩾ ∫ 0

0

1−ρ r dρ = . (1 + ρ)3 (1 + r)2

1.3 Growth and Distortion Theorems |

11

If for some f ∈ S, equality occurs in the lower or upper bound of the growth estimate (1.3.6), then equality also occurs in the corresponding inequality of the distortion estimate (1.3.2). Thus from the distortion theorem it follows that f must be the Koebe function or a suitable rotation. The following result combines the growth theorem, and the distortion theorem. Theorem 1.3.4. For f ∈ S, and z = re iθ ∈ 𝔻, 󵄨 󵄨 1 − r 󵄨󵄨󵄨 zf 󸀠 (z) 󵄨󵄨󵄨 1 + r 󵄨󵄨 ⩽ ⩽ 󵄨󵄨󵄨 . 1 + r 󵄨󵄨 f(z) 󵄨󵄨󵄨 1 − r

(1.3.7)

Equality occurs if, and only if, f is a suitable rotation of the Koebe function. Proof. As in the proof of Theorem 1.3.1, let g be the function obtained by the disk automorphism transformation, so that for some α, with |α| < 1, g(z) =

z+α ) − f(α) 1 + αz . 2 (1 − |α| )f 󸀠 (α)

f(

(1.3.8)

Then g ∈ S, and the growth theorem, with z = −α, implies that 󵄨󵄨 󵄨󵄨 |α| −f(α) |α| 󵄨 󵄨󵄨 ⩽ |g(−α)| = 󵄨󵄨󵄨 󸀠 , 󵄨󵄨 ⩽ 2 2 (1 + |α|) 󵄨󵄨 f (α)(1 − |α| ) 󵄨󵄨 (1 − |α|)2 i.e.,

1 − |α| 󵄨󵄨󵄨󵄨 f(α) 󵄨󵄨󵄨󵄨 1 + |α| ⩽󵄨 . 󵄨⩽ 1 + |α| 󵄨󵄨󵄨 αf 󸀠 (α) 󵄨󵄨󵄨 1 − |α| Replacing α by z, completes the proof of (1.3.7). Once more we now show that equality occurs only for the Koebe function or a suitable rotation. If the upper bound in (1.3.7) is attained for some function f ∈ S and some z = α ∈ 𝔻, then equality is obtained for g ∈ S when |g(−α)| =

|α| , (1 − |α|)2

which by the growth theorem implies that g is suitable rotation of the Koebe function, that is, z g(z) = , (1.3.9) (1 + e iθ z)2 where θ = − arg α. Equating the right hand sides of (1.3.8) and (1.3.9), and noting that w=

z+α 1 + αz

⇔

z=

w−α , 1 − αw

it follows that f(w) is a constant multiple of (1 + |α|)2 w . ⋅ 1 − |α| (1 + e−iθ w)2 Thus, f is a rotation of the Koebe function.

12 | 1 Univalent Functions – the Elementary Theory

A similar argument applies to the lower bound of (1.3.7), the details of which we omit. We end this section by noting that taking imaginary instead of real parts in the proof of Theorem 1.3.2 leads to the inequality −

∂ 4 4 ⩽ , arg f 󸀠 (re iθ ) ⩽ 1 − r2 ∂r 1 − r2

which, on integrating along a radius, gives the following “rotation” theorem, that for functions in S, 1+r 󵄨󵄨 󵄨 󵄨󵄨arg f 󸀠 (re iθ )󵄨󵄨󵄨 ⩽ 2 log . 󵄨 󵄨 1−r However, this inequality is not sharp, and it is again necessary to employ the Loewner theory to obtain the following curious, but sharp result (see Duren [45]), which once more is outside the scope of this book. Theorem 1.3.5. For f ∈ S, and z = re iθ ∈ 𝔻, { 4 arcsin r, 󵄨 { 󵄨󵄨 󵄨󵄨arg f 󸀠 (z)󵄨󵄨󵄨 ⩽ { r2 󵄨 { 󵄨 π + log , { 1 − r2

r ⩽ 1/√2 r ⩾ 1/√2.

2 Definitions of Major Subclasses In this chapter we introduce what we consider to be the important subclasses of S. Although this choice is subjective, many of the significant ideas and methods of proof that have been developed over the years in the study of subclass of S were discovered while studying these classes. We discuss the classes of convex, starlike, close-toconvex, and Bazilevič functions in the context of the wider class S, together with an interesting subclass, which is known simply as the class U. The connection between these classes and the class P of functions with positive real part in 𝔻 is also introduced.

2.1 Convex and Starlike Functions We begin by defining the notions of a convex and a starlike set. Definition 2.1.1. (i) A set G in the complex plane is called convex if the line segment joining any two points in G lies in G, i.e., tz1 + (1 − t)z2 , where z1 , z2 ∈ G and t ∈ [0, 1]. (ii) A set G in the complex plane is called starlike with respect to the origin if the straight line joining any point in G to the origin lies in G, i.e., tz ∈ G, when z ∈ G and t ∈ [0, 1]. We now give the definitions of starlike and convex functions. Definition 2.1.2. (i) A function f ∈ A is called convex in 𝔻 if f(𝔻) is a convex set. (ii) A function f ∈ A is called starlike in 𝔻 if f(𝔻) is a set that is starlike with respect to the origin. We denote these sets of functions by C and S∗ , respectively. Without doubt, the two central and most studied subclasses of univalent functions are C and S∗ , and in the following we shall see that both C and S∗ are subsets of S and as such provide examples of natural subclasses of S. Both subclasses have a long history and date back to the early years of the twentieth century. The first reference to convex functions appears to be in paper of Study [236] in 1913, and starlike functions were first introduced by Nevanlinna in 1921 [159]. Starlike functions are particularly interesting, and since the Koebe function is a member of S∗ , starlike functions can grow as fast as those in S. This, and the geometry imposed on the image domain of starlike functions, has resulted in a rich and beautiful set of properties for functions in S∗ , some of which are true for the wider class S, while other are either false or pose open problems in S. The classes C and S∗ are very closely related to the class P of functions with positive real part, which will be discussed in Chapter 3. https://doi.org/10.1515/9783110560961-002

14 | 2 Definitions of Major Subclasses

We begin by giving the well-known analytic description of starlike functions in terms of functions with positive real part. In the following we have in part chosen to follow the excellent exposition given by Duren [45]. Theorem 2.1.1. Let f ∈ A. Then f ∈ S∗ if, and only if, zf 󸀠 (z)/f(z) ∈ P, i.e., Re [

zf 󸀠 (z) ] > 0 (z ∈ 𝔻). f(z)

(2.1.1)

Proof. Suppose first that f ∈ S∗ . We first show that f maps each disk |z| < ρ < 1 onto a starlike set. Thus, we need to show that g(z) = f(ρz) is starlike in 𝔻, i.e., that for each fixed t ∈ [0, 1] and each z ∈ 𝔻, the point tg(z) lies in the range of g. At this point we need to use the well-known Schwarz lemma, viz: Schwarz lemma. Let ω be analytic in 𝔻, with ω(0) = 0 and |ω(z)| < 1 in 𝔻, i.e., ω is a Schwarz function. Then, |ω󸀠 (0)| ⩽ 1 and |ω(z)| ⩽ |z| in 𝔻. Strict inequality holds in both the estimates unless ω(z) = e iθ z. Clearly there exists a Schwarz function ω, such that tf(z) = f(ω(z)), and so from the Schwarz lemma we have |ω(z)| ⩽ |z|. Hence, tg(z) = tf(ρz) = f(ω(ρz)) = g(ω1 (z)), where ω1 (z) = ω(ρz)/ρ and |ω1 (z)| ⩽ |z|. Thus, we have shown that g(z) = f(ρz) is starlike in 𝔻, i.e., f maps each circle |z| = ρ < 1 onto a curve C(ρ), which bounds a starlike set. We can now obtain an analytic representation of functions in S∗ in terms of P as follows. From the above it follows that as z moves around the circle |z| = ρ in the positive direction, arg f(z) increases. Thus ∂ arg f(ρe iθ ) ⩾ 0. ∂θ Since for z = ρe iθ , ∂ ∂ arg f(ρe iθ ) = Im [ log f(ρe iθ )] ∂θ ∂θ izf 󸀠 (z) zf 󸀠 (z) = Im [ ] = Re [ ], f(z) f(z) it follows from the maximum principle for harmonic functions that zf 󸀠 (z)/f(z) ∈ P. To prove the converse, suppose that f ∈ A, where zf 󸀠 (z)/f(z) ∈ P. Then f has a simple zero at z = 0 and no other zeros in 𝔻. Reversing the above argument shows that for each ρ < 1, ∂ arg f(ρe iθ ) > 0, ∂θ when 0 ⩽ θ ⩽ 2π.

2.1 Convex and Starlike Functions

| 15

Thus, as z travels around the circle |z| = ρ in an anticlockwise direction, the point f(z) moves around a closed curve C(ρ) with increasing argument. Since f has only one zero inside |z| = ρ, the argument principle tells us that C(ρ) traverses the origin once. Hence, C(ρ) cannot intersect itself, and so is a simple closed curve bounding a starlike domain D(ρ), which is the interior of C(ρ), and f assumes each value in D(ρ) exactly once in the disk |z| < ρ. Since this is true for every ρ < 1, it follows that f is univalent and starlike in 𝔻. We note that the above shows that if f ∈ A satisfies Re[zf 󸀠 (z)/f(z)] > 0 for z ∈ 𝔻, then f is univalent in 𝔻, i.e., S∗ ⊂ S. Thus, in this sense, S∗ is a natural subset of S. A similar analytic expression holds for the convex functions, the proof of which is similar, and again follows that given by Duren [45]. Theorem 2.1.2. Let f ∈ A. Then, f ∈ C if, and only if, 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) ∈ P, i.e., Re [1 +

zf 󸀠󸀠 (z) ] > 0 (z ∈ 𝔻). f 󸀠 (z)

(2.1.2)

Proof. Suppose that f ∈ C. Then as in Theorem 2.1.1, we first show that each subdisk |z| < r < 1 is mapped by f onto a convex set as follows. Suppose z1 and z2 are points in 𝔻 such that |z1 | < |z2 | < r. Let w1 = f(z1 ) and w2 = f(z2 ), and w0 = tw1 + (1 − t)w2 , where 0 < t < 1. Then, since f maps 𝔻 on a convex set, there is a point z0 ∈ 𝔻, where f(z0 ) = w0 . Thus we need to show that |z0 | < r. As before, we appeal to the Schwarz lemma as follows. First note that the function g defined by g(z) = tf(z1 z/z2 ) + (1 − t)f(z) is analytic in 𝔻, with g(0) = 0 and g(z2 ) = w0 . Also, since f ∈ C, it is univalent (since every convex set is starlike with respect to any interior point) and so possesses an inverse function h, where h(z) = f −1 (g(z)). Since h(0) = 0 and |h(z)| < 1, z ∈ 𝔻, we can apply the Schwarz lemma to deduce that |h(z)| ⩽ |z|, z ∈ 𝔻. Thus |z0 | = |h(z2 )| ⩽ |z2 | < r. We have, therefore, shown that f maps each circle |z| = r < 1 onto a curve C(r), which bounds a convex set; this implies that the slope of the tangent vector to C(r) is nondecreasing as the curve C(r) is traversed in the positive direction. Analytically this can be expressed as 0⩽

∂ ∂ ∂ [arg ( f(re iθ ))] = (arg[ire iθ f 󸀠 (re iθ )]) ∂θ ∂θ ∂θ =

∂ {Im[log(ire iθ f 󸀠 (re iθ ))]} ∂θ

= Im {

∂ [log(ire iθ f 󸀠 (re iθ ))]} ∂θ

16 | 2 Definitions of Major Subclasses

= Im {

ire iθ [if 󸀠 (re iθ ) + ire iθ f 󸀠󸀠 (re iθ )] } ire iθ f 󸀠 (re iθ )

= Re [1 +

zf 󸀠󸀠 (z) ], f 󸀠 (z)

where z = re iθ . Thus, using the maximum principle for harmonic functions we have shown that 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) ∈ P. To prove the converse, suppose that f is normalized as before and 1+zf 󸀠󸀠 (z)/f 󸀠 (z) ∈ P. From the above, the slope of the tangent to C(r) increases, but as a point makes a complete revolution around C(r), the argument of the tangent vector satisfies, with z = re iθ , 2π

∫ 0

2π

∂ zf 󸀠󸀠 (z) ∂ [arg ( f(re iθ ))] dθ = ∫ Re [1 + 󸀠 ] dθ ∂θ ∂θ f (z) 0

{ } { zf 󸀠󸀠 (z) dz } ] } = Re { ∫ [1 + 󸀠 { f (z) iz } } {|z|=r = 2π. This shows that for any r < 1, C(r) is a simple closed curve surrounding a convex set, and f is a univalent function, whose range is a convex set. We also note that C ⊂ S∗ ⊂ S. The following, known as Alexander’s theorem, follows immediately from (2.1.1) and (2.1.2). Theorem 2.1.3. Let f ∈ A. Then f ∈ C if, and only if, zf 󸀠 ∈ S∗ .

2.2 Close-to-Convex Functions In 1952, Kaplan [84] generalized convex and starlike functions and introduced the idea of close-to-convex functions as follows. Definition 2.2.1. A function f ∈ A is called close-to-convex, if there exists a convex function ϕ such that f 󸀠 (z) Re [ 󸀠 ] > 0 (z ∈ 𝔻). (2.2.1) ϕ (z) Kaplan’s definition does not require that the function ϕ is normalized, but since the majority of results obtained for close-to-convex functions assume this, we will suppose that ϕ is normalized, so that ϕ(0) = 0 and ϕ󸀠 (0) = 1.

2.2 Close-to-Convex Functions

| 17

It follows at once from Theorem 2.1.3 that an equivalent formulation of Kaplan’s definition is that there exists g ∈ S∗ , such that Re [

zf 󸀠 (z) ] > 0 (z ∈ 𝔻). g(z)

We shall denote this class of close-to-convex functions by K0 and note that later in the book, K will denote a wider class of close-to-convex functions. From the above definition, choosing g(z) = f(z), it is clear that S∗ ⊂ K0 , and so C ⊂ S∗ ⊂ K0 . We note that the definition of close-to-convex functions does not require a priori that f is univalent. However, as in the case of C and S∗ , we shall now show that if (2.2.1) holds, then f is necessarily univalent, and so K0 (and also K) are natural subclasses of S. In order to do this we use the well-known, much used theorem of Noshiro [164] and Warschawski [260], a simple proof of which we include below. Theorem 2.2.1. Let f be analytic in a convex domain D, satisfying Re f 󸀠 (z) > 0 for z ∈ D. Then f is univalent in D. Proof. Suppose that z1 and z2 are distinct points in D. Then f is defined on the line segment joining z1 and z2 . Thus, z2

f(z1 ) − f(z2 ) = ∫ f 󸀠 (z)dz z1 1

= (z2 − z1 ) ∫ f 󸀠 (tz2 + (1 − t)z1 )dt ≠ 0, 0

since Re f 󸀠 (z) > 0, z ∈ D. Hence f(z1 ) ≠ f(z2 ), and so f is univalent in D. Using Theorem 2.2.1 it is an easy step to show that every close-to-convex function is univalent as follows. Since f is close-to-convex, there exists some convex function ϕ such that (2.2.1) holds. Let D be the range of ϕ, and consider the function h(w) = f(ϕ−1 (w)) Then h󸀠 (w) =

(w ∈ D).

f 󸀠 (z) f 󸀠 (ϕ−1 (w)) = 󸀠 , 󸀠 −1 ϕ (ϕ (w)) ϕ (z)

and so Re h󸀠 (w) > 0 for w ∈ D. Thus by Theorem 2.2.1, h is univalent and, so, f is univalent. As in the case for convex and starlike functions, close-to-convex functions have an interesting geometrical property as follows.

18 | 2 Definitions of Major Subclasses Suppose that f is analytic for z ∈ 𝔻, and let C(r) be the image of 0 < |z| = r < 1 under f . Then the condition that f is close-to-convex means that as θ increases, the curve C(r) does not turn back upon itself by more than π, or in other words, C(r) cannot make a reverse hairpin turn. Clearly if C(r) did turn back more than π, it could then intersect with itself, which would mean that f would then not be univalent. (We note that in the case of convex functions, the curve C(r) does not turn back on itself at all as θ increases.) Analytically this means that as θ increases, the tangent direction arg(∂f(re iθ )/∂θ) cannot decrease by as much as π from any previous value. Using the fact that for z = re iθ , zf 󸀠󸀠 (z) ∂ ∂ [arg f(re iθ )] = Re [1 + 󸀠 ], ∂θ ∂θ f (z) Kaplan [84] proved the following necessary and sufficient condition for f ∈ K. Theorem 2.2.2. Let f be analytic and locally univalent in 𝔻. Then f is close-to-convex if, and only if, for z = re iθ ∈ 𝔻, θ2

∫ Re [1 + θ1

zf 󸀠󸀠 (z) ] dθ > −π, f 󸀠 (z)

for each r ∈ (0, 1), and for each pair of real numbers θ1 and θ2 such that θ1 < θ2 . We omit the proof, an excellent account of which can also be found in Duren [45].

2.3 Bazilevič Functions A significant extension to the close-to-convex functions was made by Bazilevič [13], who in 1955, introduced the following functions, which have become known as the Bazilevič functions. Definition 2.3.1. The class B(α, β) of Bazilevič functions of type (α, β) consists of functions f ∈ A such that z

{ } f(z) = {(α + iβ) ∫ g α (t)p(t)t iβ−1 dt} 0 { }

1/(α+iβ)

(z ∈ 𝔻),

(2.3.1)

where g is starlike, p ∈ P, and where α and β are real. When β = 0 and α ⩾ 0, (2.3.1) reduces to the following subclass B(α) of Bazilevič functions, which has been widely studied.

2.3 Bazilevič Functions

Definition 2.3.2. Let f ∈ A. Then, f ∈ B(α) if, and only if, zf 󸀠 (z) ] > 0 (z ∈ 𝔻), Re [ 1−α f (z)g α (z) where g is starlike and α ⩾ 0.

| 19

(2.3.2)

Clearly, B(1) = K0 and B(0) = S∗ . Bazilevič showed that functions defined by (2.3.1), and therefore (2.3.2), are univalent, thus providing another natural subclass of S. A great many results have been obtained for functions defined by (2.3.2), and these are the Bazilevič functions we will be most concerned with in this book. A geometrical interpretation of the image domain of functions in B(α) is not simple, and is outside the scope of this book. We refer the reader to Prokhorov [204], where a geometrical characterization of functions in B(α) in terms of the complement of f(𝔻) is given. We also note that, analogous to Kaplan’s characterization of close-to-convex functions given in Theorem 2.2.2, Sheil-Small [226] proved the following characterization of Bazilevič functions. Theorem 2.3.1. Let f be Bazilevič of type (α, β). Then with z = re iθ and each 0 < r < 1, θ2

∫ {1 + Re [ θ1

−β Im [

re iθ f 󸀠 (re iθ ) re iθ f 󸀠󸀠 (re iθ ) + (α − 1) Re ] [ ] f 󸀠 (re iθ ) f(re iθ )

re iθ f 󸀠 (re iθ ) ]} dθ > −π, f(re iθ )

(2.3.3)

whenever θ2 > θ1 . Conversely, if f is analytic in 𝔻, with f(0) = 0, f(z) ≠ 0 (0 < |z| < 1), f 󸀠 (z) ≠ 0 when z ∈ 𝔻, and f(z) satisfies the condition (2.3.3) for 0 < r < 1, where α ⩾ 0 and β is real, then f is univalent in 𝔻 and is Bazilevič of type (α, β) in the case α > 0. We will see in Chapters 9 and 10 that the presence of the independent starlike function g in the definitions of close-to-convex and Bazilevič functions often creates difficulties when we try to extend known results from S∗ to K and B(α). However, taking g(z) ≡ z in (2.3.2) provides an interesting subclass of B(α), which has received a great deal of attention in recent years. We denote this class of functions by B1 (α), noting that α = 0 gives S∗ and α = 1, functions whose derivative has positive real part in 𝔻. Functions in B1 (α) are also referred to as Bazilevič functions with logarithmic growth, in view of the fact that the maximum modulus M(r) of functions in B1 (α) grows as a power of the logarithmic function log(1/(1 − r)) as r → 1. We therefore define the class B1 (α) below, and we will exhibit many of the interesting properties of functions in B1 (α) in Chapter 11. Definition 2.3.3. Let f ∈ A. Then for α ⩾ 0, f ∈ B1 (α) if, and only if, Re {f 󸀠 (z) [

f(z) α−1 ] } > 0 (z ∈ 𝔻). z

20 | 2 Definitions of Major Subclasses

2.4 The Class U Another interesting and somewhat curious subclass of S which has received attention in recent years is the so-called class U. Although not strictly related to either C or S∗ , its definition resembles that of B1 (α) when α = −1. We give the definition below in a more general setting. Definition 2.4.1. Let f ∈ A and 0 < λ ⩽ 1. Then f ∈ U(λ) if, and only if, 󵄨󵄨 󵄨 󵄨󵄨U f (z)󵄨󵄨󵄨 < λ

(z ∈ 𝔻),

where U f (z) = [

z 2 󸀠 ] f (z) − 1. f(z)

We write U := U(1). We will see in Chapter 12 that U(λ) also forms a natural subclass of S and that U contains the Koebe function.

2.5 Rotational Invariance We end this chapter by explaining the notion of rotational invariance, which is very useful, particularly when considering coefficient problems. We will use this idea at various points in this book. For f ∈ A, its rotation is defined by f θ (z) := e−iθ f(e iθ z), for θ ∈ ℝ. A family F ⊂ A is said to be rotationally invariant if for every f ∈ F, the function f θ ∈ F for every θ ∈ ℝ. We give some examples. Let f ∈ S∗ and f θ (z) = e−iθ f(e iθ z), for θ ∈ ℝ. Then, f θ󸀠 (z) = f 󸀠 (e iθ z), and so Re [

zf θ󸀠 (z) e iθ zf 󸀠 (e iθ z) ]>0 ] = Re [ f θ (z) f(e iθ z)

z ∈ 𝔻.

Thus the class S∗ is rotationally invariant. Similarly, one can show that C, K, B(α), B1 (α), and U(λ) are all rotationally invariant. An example of a noninvariant class is the following. Let CR+ denote the subclass of class of close-to-convex functions K0 with g(z) = k(z), i.e., CR+ = {f ∈ A : Re [(1 − z)2 f 󸀠 (z)] > 0, z ∈ 𝔻} . Then functions in CR+ are convex in the positive direction of the real axis. It is easy to see that the class CR+ is not rotationally invariant.

2.5 Rotational Invariance | 21

We note that even if a particular class of functions is rotationally invariant, it does not necessarily imply that all functionals are rotationally invariant. For example, if f ∈ S and f θ (z) = e−iθ f(e iθ z) for θ ∈ ℝ as above, with f θ (z) = z + α 2 z2 + . . . , the functional |a2 a4 − a23 | is rotationally invariant, since |α 2 α 4 − α 23 | = |a2 e iθ a4 e3iθ − (a3 e2iθ )2 | = |a2 a4 − a23 |. On the other hand, the functional |a23 −a22 | is not rotationally invariant, since |α 23 −α 22 | = |(a3 e2iθ )2 − (a2 e iθ )2 | ≠ |a23 − a22 | for all θ ∈ ℝ.

3 Fundamental Lemmas Before embarking on the study of the various classes of univalent functions, we present some fundamental results which will be used in this book. Not all of the material is utilized but it is included so as to provide a useful resource. Most of this material can be found in the literature, and we include proofs where we think appropriate. References to omitted proofs are provided. Classical theorems in complex analysis such as the maximum modulus principle are not included and can be found in standard textbooks. When dealing with subclasses of univalent functions, the most used class of functions are those whose real part is positive in 𝔻. These functions are the main concern of this chapter.

3.1 Functions with Positive Real Part Definition 3.1.1. Let p be analytic in 𝔻, with p(0) = 1. For 0 ⩽ α < 1, denote by Pα the class of functions p with Taylor series expansion ∞

p(z) = 1 + ∑ p n z n

(3.1.1)

n=1

satisfying Re p(z) > α

(z ∈ 𝔻).

Functions in P0 , which we denote by P, are referred as functions with positive real part, or Carathéodory functions. An obvious example of a function p ∈ Pα is p(z) =

∞ 1 + (1 − 2α)z 1 + z = (1 − α) + α = 1 + 2(1 − α) ∑ z n , 1−z 1−z n=1

which maps 𝔻 onto the half-plane {w : Re w > α}. Another is p(z) = 1 − (1 − α)z n , n ∈ ℕ, which belongs to Pα , but for n ⩾ 2 is not univalent. A more general function is p(z, A, B) =

∞ 1 + Az = 1 + (A − B) ∑ (−1)n−1 B n−1 z n , 1 + Bz n=1

with − 1 ⩽ B < A ⩽ 1,

which maps 𝔻 onto a convex domain, symmetric with respect to the real axis, with diametric end points (1 − A)/(1 − B) and (1 + A)/(1 + B).

https://doi.org/10.1515/9783110560961-003

3.1 Functions with Positive Real Part

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23

Specifying values of A and B provides more examples of functions with positive real part, for example, 1 + (1 − 2α)z , (A = 1 − 2α and B = −1), discussed above; (i) p(z) = 1−z 󵄨󵄨 1 − p(z) 󵄨󵄨 1 + Az 󵄨 󵄨󵄨 (ii) p(z) = , (0 < A = −B ⩽ 1), such that 󵄨󵄨󵄨 󵄨 < A for all z ∈ 𝔻; 󵄨󵄨 1 + p(z) 󵄨󵄨󵄨 1 − Az (iii) p(z) =

1+z , (A = −B = 1), which maps the unit disk onto the right half-plane; 1−z

(iv) p(z) = 1 + Az, (0 < A ⩽ 1 and B = 0), so that p(𝔻) is the open disk with center 1, and radius A; 1 , (A = 0 and −1 ⩽ B < 0), so that p(𝔻) is the open disk with center 1 + Bz 2 1/(1 − B ) and radius −B/(1 − B2 ).

(v) p(z) =

Unlike the class of univalent functions, the mean of two functions in P belongs to P (in fact, a linear combination of functions in P, when the coefficients are nonnegative with sum 1, lies in P). This, however, does not hold for the product of two functions in P, as the functions 1 + z and (1 + z)2 show. The following transformations are preserved in the class P. Thus, if p, p1 , and p2 belong to P, then the functions g given below are also in P. (i) g(z) = p(e iα z) with α real; (ii) g(z) = p(rz), with −1 ⩽ r ⩽ 1; (iii) g(z) = 1/p(z); (iv) g(z) = [p1 (z)]r1 ⋅ [p2 (z)]r2 , with r1 , r2 ⩾ 0 and r1 + r2 ⩽ 1; z+α 1 [p ( ) − bi], with |α| < 1 and p(α) = a + bi; a 1 + αz p(z) + ib (vi) g(z) = , with b real. 1 + ibp(z) (v) g(z) =

We proceed by giving sharp growth and distortion estimates for functions in P, which can be proved in several ways. Theorem 3.1.1 (Growth and distortion theorems for the class P). Let p ∈ P with z = re iθ ∈ 𝔻. Then, 1−r 1+r ⩽ Re p(z) ⩽ |p(z)| ⩽ , 1+r 1−r and 2 󵄨󵄨 󸀠 󵄨󵄨 Re p(z), 󵄨󵄨󵄨p (z)󵄨󵄨󵄨 ⩽ 1 − r2 so that 2 . (3.1.2) |p󸀠 (z)| ⩽ (1 − r)2 The inequalities are sharp.

24 | 3 Fundamental Lemmas Proof. Let p ∈ P, with z = re iθ ∈ 𝔻. Define ω(z) =

1 − p(z) . 1 + p(z)

It is clear that ω is analytic in 𝔻, with ω(0) = 0, and that |ω(z)| < 1 for all z ∈ 𝔻. Hence, by the Schwarz lemma, |ω(z)| ⩽ |z| = r. Since 1 − ω(z) p(z) = , 1 + ω(z) and since |ω(z)| ⩽ |z| = r implies 1 − |ω(z)|2 ⩾ 1 − r2 and 1 − r ⩽ 1 − |ω(z)| ⩽ |1 ± ω(z)| ⩽ 1 + |ω(z)| ⩽ 1 + r, it follows that Re p(z) = and

1 − |ω(z)|2 1 − r ⩾ |1 + ω(z)|2 1 + r

󵄨󵄨 1 − ω(z) 󵄨󵄨 1 + r 󵄨󵄨 󵄨 . |p(z)| = 󵄨󵄨󵄨 󵄨⩽ 󵄨󵄨 1 + ω(z) 󵄨󵄨󵄨 1 − r

It also follows from the Schwarz–Pick lemma, see e.g., [122] (a sharpened form of the Schwarz lemma) that for |z| = r < 1, |ω󸀠 (z)| ⩽

1 − |ω(z)|2 , 1 − |z|2

which, since p(z) =

1 − ω(z) , 1 + ω(z)

gives |p󸀠 (z)| =

2|ω󸀠 (z)| 2 1 − |ω(z)|2 2 ⩽ ⋅ = Re p(z). 2 2 2 |1 + ω(z)| 1 − |z| |1 + ω(z)| 1 − r2

Combining this with the right hand side of the growth estimate for |p(z)| above gives (3.1.2). In 1907, Carathéodory [28] proved the following sharp estimate for the coefficients of functions in P, which is fundamental in the study of functions in the subclasses of S.

3.1 Functions with Positive Real Part

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25

Theorem 3.1.2. Let p ∈ P, with p(z) given by (3.1.1). Then, |p n | ⩽ 2 (n ∈ ℕ). The inequality is sharp. Proof. Let n be a positive integer. Without loss of generality we can, via a suitable rotation, assume that p n ⩾ 0. Since Re p(z) > 0 for z = re iθ , it is clear that the integral 2π

I(r) = ∫ [1 − cos(nθ)]p(re iθ )dθ 0

satisfies Re I(r) > 0. Further, the orthogonality of the trigonometric functions implies that 2π

∞

I(r) = ∫ [1 − cos(nθ)] [1 + ∑ p k r k [cos(kθ) + i sin(kθ)]] dθ = 2π − πr n p n . 0

k=1

Thus, 2π − πr n p n = Re I(r) > 0, and so r n p n < 2. Letting r → 1 we obtain p n ⩽ 2, which proves the theorem. Another widely used inequality concerning the coefficients of functions in P is the following. Theorem 3.1.3. Let p ∈ P, with p(z) given by (3.1.1). Then, 󵄨󵄨 1 󵄨󵄨󵄨 1 󵄨󵄨 󵄨󵄨p2 − p21 󵄨󵄨󵄨 ⩽ 2 − |p1 |2 . 󵄨󵄨 2 󵄨󵄨 2 The inequality is sharp. Proof. Since p ∈ P, we can write p(z) =

1 + ω(z) , 1 − ω(z)

(3.1.3)

where ω is a Schwarz function, i.e., |ω(z)| < 1 for z ∈ 𝔻, and where we can write ω(z) = ω1 z + ω2 z2 + . . . .

(3.1.4)

Using the Schwarz lemma, it follows that |ω2 | ⩽ 1 − |ω1 |2 (see Ahlfors [3, p. 136, Exercise 1]), and equating coefficients in (3.1.3) and (3.1.4) gives 2ω1 = p1 , and 2ω2 = p2 − p21 /2. The result now follows.

26 | 3 Fundamental Lemmas The following necessary and sufficient condition for a function to belong to P is due to Carathéodory and Toeplitz, a proof of which can be found in Grenander and Szegő [60]. Apart from the intrinsic interest of the result, two useful, and much used identities expressing the coefficients p2 and p3 in terms of p1 follow. Theorem 3.1.4. Let p ∈ P and be given by (3.1.1). Then the power series for p(z) converges in 𝔻 to a function in P if, and only if, for n ∈ ℕ, the Toeplitz determinants 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 󵄨󵄨󵄨 p1 D n = 󵄨󵄨󵄨󵄨 . 󵄨󵄨 .. 󵄨󵄨 󵄨󵄨 󵄨󵄨p n

p1 2 p n−1

p2 p1

... ...

p n−2

...

p n 󵄨󵄨󵄨󵄨 󵄨 p n−1 󵄨󵄨󵄨󵄨 󵄨 .. 󵄨󵄨󵄨 . 󵄨󵄨󵄨 󵄨󵄨 2 󵄨󵄨󵄨

(3.1.5)

it k are all nonnegative. They are strictly positive except for p(z) = ∑m k=1 ρ k p o (e z), ρ k > 0, t k real and t k ≠ t j for k ≠ j, where p0 (z) = (1 + z)/(1 − z); in this case, D n > 0 for n < m − 1, and D n = 0 for n ⩾ m.

We now consider three important consequences. Rewriting (3.1.5) for the case n = 2 gives 󵄨󵄨 󵄨󵄨 2 󵄨󵄨 D2 = 󵄨󵄨󵄨p1 󵄨󵄨 󵄨󵄨p2 󵄨

p1 2 p1

󵄨 p2 󵄨󵄨󵄨 󵄨󵄨 p1 󵄨󵄨󵄨 = 8 + 2p21 Re p2 − 2|p2 |2 − 4p21 ⩾ 0, 󵄨󵄨 2 󵄨󵄨󵄨

which is equivalent to |2p2 − p21 | ⩽ 4 − p21 , i.e., to 2p2 = p21 + x(4 − p21 ),

(3.1.6)

for some x, with |x| ⩽ 1. When n = 3, we obtain 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 󵄨󵄨p1 D3 = 󵄨󵄨󵄨 󵄨󵄨󵄨p2 󵄨󵄨 󵄨󵄨p3

p1 2 p1 p2

p2 p1 2 p1

p3 󵄨󵄨󵄨󵄨 󵄨 p2 󵄨󵄨󵄨󵄨 󵄨 ⩾ 0, p1 󵄨󵄨󵄨󵄨 󵄨 2 󵄨󵄨󵄨

which is equivalent to 󵄨 󵄨󵄨 󵄨󵄨(4p3 − 4p1 p2 + p3 ) (4 − p2 ) + p1 (2p2 − p2 )2 󵄨󵄨󵄨 ⩽ 2(4 − p2 )2 − 2 󵄨󵄨󵄨󵄨2p2 − p2 󵄨󵄨󵄨󵄨2 , (3.1.7) 1 1 󵄨󵄨 1 1󵄨 1 󵄨󵄨 󵄨 which, together with (3.1.6), gives the relation 4p3 = p31 + 2 (4 − p21 ) p1 x − p1 (4 − p21 ) x2 + 2 (4 − p21 ) (1 − |x|2 ) ζ,

(3.1.8)

for some value of ζ with |ζ| ⩽ 1, and some value of x with |x| ⩽ 1. To verify this, we write X = 4 − p21 and use (3.1.6) to eliminate p2 from (3.1.7) to obtain 󵄨 󵄨󵄨 󵄨󵄨(4p3 − p31 − 2xXp1 ) X + x2 p1 X 2 󵄨󵄨󵄨 ⩽ 2X 2 (1 − |x|2 ) . 󵄨 󵄨 Dividing by X, and noting that p1 = 2 implies that p2 = p3 = 2, now establishes (3.1.8).

3.1 Functions with Positive Real Part |

27

We note at this point that (3.1.6) and (3.1.8) give expressions for p2 and p3 in terms of p1 , which will be used repeatedly in subsequent chapters. Using (3.1.6) and (3.1.8) enables a limited number of problems to be solved, and the key to further advances is to express the coefficients p n for n ⩾ 4 in terms of p1 . This presents a much more difficult problem. However a recent advance in this direction has been made by Kwon et al. [102], who found the following expression for p4 in terms of p1 , 8p4 = p41 + (4 − p21 )ζ [p21 (ζ 2 − 3ζ + 3) + 4ζ ] − 4(4 − p21 )(1 − |ζ|2 ) [p1 (ζ − 1)x + ζ x2 − (1 − |x|2 )ξ ] ,

(3.1.9)

for some ζ , x, ξ , with |ζ| ⩽ 1, |x| ⩽ 1, |ξ| ⩽ 1. Applying (3.1.9) necessarily involves increasingly complicated arguments, and we will state some applications later in this book. Another significant necessary and sufficient condition for a function p to belong to P is given by Leverenz [108]. Theorem 3.1.5. A function p given by (3.1.1) belongs to P if, and only if, 󵄨󵄨2 󵄨󵄨 ∞ 󵄨󵄨2 ∞ {󵄨󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨󵄨 } 󵄨 ∑ {󵄨󵄨󵄨2z j + ∑ p k z k+j 󵄨󵄨󵄨 − 󵄨󵄨󵄨 ∑ p k+1 z k+j 󵄨󵄨󵄨 } ⩾ 0, 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 j=0 {󵄨 k=1 󵄨k=0 󵄨 󵄨 } for every sequence {z k } of complex numbers satisfying lim sup |z k |1/k < 1. k→∞

Suitably choosing the sequence {z k } in Theorem 3.1.5 provides many useful inequalities concerning the coefficients of functions in P. In the following theorem we give a selection, some of which will be used in this book, noting that other choices of {z k } would generate more. Theorem 3.1.6. Let p ∈ P and be given by (3.1.1). Then 󵄨󵄨 󵄨 󵄨󵄨p2 − p21 󵄨󵄨󵄨 ⩽ 2, 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨p3 − 2p1 p2 + p31 󵄨󵄨󵄨 ⩽ 2, 󵄨 󵄨

󵄨 󵄨󵄨 󵄨󵄨p4 + 3p21 p2 − 2p1 p3 − p22 − p41 󵄨󵄨󵄨 ⩽ 2, 󵄨 󵄨

󵄨󵄨 󵄨 󵄨󵄨p5 − 2p2 p3 − 2p1 p4 − 4p31 p2 + 3p21 p3 + 3p1 p22 + p51 󵄨󵄨󵄨 ⩽ 2, 󵄨 󵄨 󵄨󵄨 󵄨󵄨p6 + 6p1 p2 p3 + 3p21 p4 + 5p41 p2 + p32 − p23 󵄨 󵄨 −2p2 p4 − 2p1 p5 − 4p31 p3 − 6p21 p22 − p61 󵄨󵄨󵄨󵄨 ⩽ 2.

(3.1.10) (3.1.11) (3.1.12) (3.1.13)

(3.1.14)

All the inequalities are sharp. Proof. We first remark that the inequalities in Theorem 3.1.6 can also be proved on noting that the reciprocal of a function in P also belongs to P, and then calculating the coefficients of the reciprocal.

28 | 3 Fundamental Lemmas We choose a suitable sequence {z k } in Theorem 3.1.5 as follows. For (3.1.10), choose z0 = −p1 , z1 = 1, and z k = 0 for k ⩾ 2. For (3.1.11), choose z0 = p21 − p2 , z1 = −p1 , z2 = 1, and z k = 0 for k ⩾ 3. For (3.1.12), choose z0 = p3 − 2p1 p2 + p31 , z1 = −p21 + p2 , z2 = p1 , z3 = −1, and z k = 0 for k ⩾ 4. For (3.1.13), choose z0 = p41 + p22 + 2p1 p3 − 3p21 p2 − p4 , z1 = −p3 + 2p1 p2 − p31 , z2 = −p2 + p21 , z3 = −p1 , z4 = 1, and z k = 0 for k ⩾ 5. For (3.1.14), choose z0 = p5 + p51 + 3p1 p22 + 3p21 p3 − 4p31 p2 − 2p1 p4 − 2p2 p3 , z1 = −p41 − p22 − 2p1 p3 + 3p21 p2 + p4 , z2 = p3 − 2p1 p2 + p31 , z3 = −p21 + p2 , z4 = p1 , z5 = −1, and z k = 0 for k ⩾ 6. Coefficient inequalities that do not arise from calculating the coefficients of the reciprocal of a function in P can also be obtained using Theorem 3.1.5. Choosing z n−1 = 1 and z k = 0 for k ≠ n − 1, is a simple example and provides another proof of the inequality |p n | ⩽ 2 for n ⩾ 1 (Theorem 3.1.2). Theorem 3.1.5 can also be used to prove the following well-known and widely used inequality, which can be proved in other ways. Theorem 3.1.7. Let p ∈ P and be given by (3.1.1). Then for real μ, 󵄨󵄨 󵄨 󵄨󵄨p − μ p2 󵄨󵄨󵄨 ⩽ max{2, 2|μ − 1|} = { 2, 󵄨󵄨 2 2 1 󵄨󵄨 { 󵄨 󵄨 2|μ − 1|, { The inequalities are sharp.

0 ⩽ μ ⩽ 2, elsewhere.

Proof. In Theorem 3.1.5, choose z0 = −μp1 /2, z1 = 1, and z k = 0 for k ⩾ 2. Then 󵄨2 󵄨󵄨 󵄨󵄨p − μ p2 󵄨󵄨󵄨 ⩽ |p |2 ⋅ (|μ − 1|2 − 1) + 4, (3.1.15) 1 󵄨󵄨 2 1 󵄨󵄨 2 󵄨 󵄨 and the result follows on noting that |μ −1| ⩽ 1, when 0 ⩽ μ ⩽ 2, and that |p1 | ⩽ 2. Using Theorem 3.1.5, the following very useful inequalities were obtained by Ali in [8]. (We note that inequalities (3.1.17) and (3.1.18) are sharp, but that (3.1.16) is not). Theorem 3.1.8. Let p ∈ P and be given by (3.1.1). If 0 ⩽ B ⩽ 1 and B(2B − 1) ⩽ D ⩽ B, then 󵄨󵄨 󵄨 󵄨󵄨p3 − 2Bp1 p2 + Dp31 󵄨󵄨󵄨 ⩽ 2 (3.1.16) 󵄨 󵄨 and

󵄨 󵄨󵄨 󵄨󵄨p3 − 2Bp1 p2 + Bp31 󵄨󵄨󵄨 ⩽ 2. 󵄨 󵄨

(3.1.17)

Also, for all real μ, 󵄨 󵄨󵄨 󵄨󵄨p3 − (1 + μ)p1 p2 + μp31 󵄨󵄨󵄨 ⩽ max{2, 2|2μ − 1|} 󵄨 󵄨 { 2, ={ 2|2μ − 1|, {

0 ⩽ μ ⩽ 1, elsewhere.

(3.1.18)

3.1 Functions with Positive Real Part

| 29

Proof. First note that choosing z0 = αp21 − βp2 , z1 = −γp1 , z2 = 1, and z k = 0 for k ⩾ 3, Theorem 3.1.5 gives 󵄨󵄨󵄨p − (β + γ)p p + αp3 󵄨󵄨󵄨2 1 2 󵄨󵄨 3 1 󵄨󵄨 󵄨2 󵄨 󵄨2 󵄨 ⩽ 4 + 4γ(γ − 1)|p1 |2 + 󵄨󵄨󵄨󵄨(2α − γ)p21 − (2β − 1)p2 󵄨󵄨󵄨󵄨 − 󵄨󵄨󵄨󵄨p2 − γp21 󵄨󵄨󵄨󵄨 󵄨󵄨 ν 󵄨󵄨2 (α − βγ)2 = 4 + 4γ(γ − 1)|p1 |2 + 4β(β − 1) 󵄨󵄨󵄨󵄨 p2 − p21 󵄨󵄨󵄨󵄨 − (3.1.19) |p1 |4 , 2 󵄨 β(β − 1) 󵄨 where ν=

α(β − 1) + β(α − β) . β(β − 1)

Next, note that if B = 0, then D = 0, and so there is nothing to prove in (3.1.16). Also if B = 1, then D = 1, in which case, (3.1.16) becomes (3.1.11). Thus we can assume that 0 < B < 1, so that B(B − 1) < 0. With γ = β = B and α = D, from (3.1.19) we obtain 󵄨󵄨 󵄨2 󵄨󵄨p3 − 2Bp1 p2 + Dp31 󵄨󵄨󵄨 󵄨 󵄨

󵄨󵄨 ν 󵄨󵄨2 (D − B2 )2 ⩽ 4 + 4B(B − 1)|p1 |2 + 4B(B − 1) 󵄨󵄨󵄨󵄨 p2 − p21 󵄨󵄨󵄨󵄨 − |p1 |4 2 󵄨 B(B − 1) 󵄨 ⩽ 4 + bx + ax2 := ϕ(x),

where x = |p1 |2 ∈ [0, 4], a = −(D − B2 )2 /(B(B − 1)) and b = 4B(B − 1). Since a ⩾ 0, it follows that ϕ(x) ⩽ ϕ(0), provided ϕ(0) ⩾ ϕ(4), i.e., b + 4a ⩽ 0. This is equivalent to |D − B2 | ⩽ B(B − 1), which completes the proof of (3.1.16). Next, note that (3.1.17) follows at once on putting D = B in (3.1.16). For (3.1.18), we first note that when 0 ⩽ μ ⩽ 1, (3.1.18) follows from (3.1.16) with D = μ and 2B = μ + 1. Outside this interval, i.e., μ ∉ [0, 1], take α = β = μ and γ = 1 in (3.1.19). Since μ(μ − 1) > 0, from (3.1.15) and (3.1.19) we obtain 󵄨󵄨 󵄨2 󵄨2 󵄨 󵄨󵄨p3 − (μ + 1)p1 p2 + μp31 󵄨󵄨󵄨 ⩽ 4 + 4μ(μ − 1) 󵄨󵄨󵄨 p2 − p21 󵄨󵄨󵄨 ⩽ 4(2μ − 1)2 . 󵄨 󵄨 󵄨 󵄨 We next give an important result of MacGregor [134], which was used by Leung [107] to solve the difference of coefficient problems for starlike functions (see Theorem 4.1.4 in Chapter 4). Theorem 3.1.9. Let p ∈ P, be given by (3.1.1), and λ n ⩾ 0, for n ∈ ℕ. If q(z) = n ∑∞ n=1 λ n p n z is analytic in 𝔻, and Re q(z) ⩽ M for some M > 0 and all z ∈ 𝔻, then ∞

∑ λ n |p n |2 ⩽ 2M. n=1

Proof. Write p n = c n + id n , ∞

u(r, θ) = Re p(re iθ ) = 1 + ∑ (c n cos nθ − d n sin nθ)r n n=1

30 | 3 Fundamental Lemmas

and

∞

𝑣(r, θ) = Re q(re iθ ) = ∑ λ n (c n cos nθ − d n sin nθ)r n . n=1

Using the orthogonality of the trigonometric functions and the fact that 2π

2π

∫ cos2 nθ dθ = ∫ sin2 nθ dθ = π, 0

0

together with the absolute and uniform convergence of the two series on the interval 0 ⩽ θ ⩽ 2π, we obtain 2π

∞

∞

n=1

n=1

∫ u(r, θ)𝑣(r, θ)dθ = π ∑ λ n (c2n + d2n )r2n = π ∑ λ n |p n |2 r2n . 0

Since u ⩾ 0 and 𝑣 ⩽ M, 2π

2π

∫ u(r, θ)𝑣(r, θ)dθ ⩽ M ∫ u(r, θ)dθ = 2Mπ, 0

0

2 2n ⩽ 2M, which on letting r → 1 proves the from which we deduce that ∑∞ n=1 λ n |p n | r theorem.

Corollary 3.1.1. For every p ∈ P and every positive integer n, there exists ζ , with |ζ| = 1, such that n n 1 1 ∑ |p k − ζ k |2 ⩽ ∑ . k k k=1 k=1 Proof. Applying Theorem 3.1.9 with n

1 pk zk , k k=1

q(z) = ∑ gives n

n n 1 1 1 |p k − ζ k |2 = ∑ |p k |2 − 2 Re [q(ζ )] + ∑ k k k k=1 k=1 k=1

∑

n

1 . k k=1

⩽ 2M − 2 Re [q(ζ )] + ∑

Choosing ζ with |ζ| = 1, so that Re q(ζ ) = M = max Re [q(ζ)] gives the result. |ζ|=1

We end this section by giving the representation of a function in P as a Poisson– Stieltjes integral. This follows from the Herglotz representation theorem for positive harmonic functions (see Duren [45]).

3.2 Subordination

| 31

Theorem 3.1.10 (Representation theorem for the class P). Every p ∈ P can be represented as a Poisson–Stieltjes integral 2π

p(z) = ∫ 0

1 + e−it z dν(t), 1 − e−it z

(3.1.20)

2π

where dν(t) ⩾ 0 and ∫0 dν(t) = 1. This representation theorem has important implications in the study of subclasses of univalent functions and has been used to great effect, particularly in the case of starlike functions. Since all functions in P can be represented by (3.1.20) in terms of the probability measure ν, properties for functions in P can also be obtained from this representation. As an example, expressing the integrand in (3.1.20) as an infinite series, gives the inequality |p n | ⩽ 2, for n ⩾ 1 at once. Below are two other useful inequalities for functions in P, which can easily be proved using the above representation theorem. Proofs are given by Pommerenke in [196]. Theorem 3.1.11. Let p ∈ P for z = re iθ ∈ 𝔻. Then 2π

1 + 3r2 1 ∫ |p(z)|2 dθ ⩽ , 2π 1 − r2 0

and

2π

1 2 ∫ |p󸀠 (z)|dθ ⩽ . 2π 1 − r2 0

3.2 Subordination The concept of subordination between two functions of a complex variable is a natural generalization of the idea of an inequality between functions of a real variable. The idea of subordination can be traced back to Lindelöf [120], but it was Littlewood [121, 122] and Rogosinski [214, 215] who introduced the term and first provided the basic properties. Definition 3.2.1. Let f and g be analytic in the unit disk 𝔻. A function f is said to be subordinate to g, written as f ≺ g or f(z) ≺ g(z), if there exists a function ω, analytic in 𝔻, with ω(0) = 0, |ω(z)| < 1, and f(z) = g(ω(z)) for z ∈ 𝔻. If the superordinate function g is univalent, then the concept of subordination implies that the image under g of each disk |z| < r < 1 contains the image under f of the same disk. The following result is basic.

32 | 3 Fundamental Lemmas Theorem 3.2.1. Let f be analytic in 𝔻 and g be univalent in 𝔻. Then f ≺ g if, and only if, f(0) = g(0), and f(𝔻) ⊆ g(𝔻). Proof. Let f ≺ g. Then from Definition 3.2.1, there exists a function ω such that ω(𝔻) ⊆ 𝔻, ω(0) = 0, and f(z) = g(ω(z)) for z ∈ 𝔻, which implies that f(0) = g(0) and f(𝔻) ⊆ g(𝔻). For the converse, consider the function ω(z) = g−1 (f(z)). Then ω is well defined since f(𝔻) ⊆ g(𝔻), and g is univalent in 𝔻. Moreover, the inverse function g−1 is analytic in f(𝔻), so that ω is analytic in 𝔻. Finally, we note that ω(0) = g−1 (f(0)) = g−1 (g(0)) = 0 and |ω(z)| < 1 for z ∈ 𝔻, since ω(𝔻) = g−1 (f(𝔻)) ⊆ g−1 (g(𝔻)) = 𝔻. This completes the proof of the theorem. If f ≺ g, then the Schwarz lemma tells us that ω satisfies |ω󸀠 (0)| ⩽ 1 and |ω(z)| ⩽ |z| for z ∈ 𝔻. It follows that |f 󸀠 (0)| ⩽ |g󸀠 (0)| and f(|z| ⩽ r) ⊆ g(|z| ⩽ r)

(0 ⩽ r < 1).

This is known as the Lindelöf subordination principle, which implies that max |f(z)| ⩽ max |g(z)| |z|=r

|z|=r

(0 ⩽ r < 1).

This important result was generalized by Littlewood [121] into an inequality concerning integral means defined by 2π

{ 1 󵄨 󵄨p } M p (r, f) = { ∫ 󵄨󵄨󵄨󵄨f(re iθ )󵄨󵄨󵄨󵄨 dθ} 2π 0 { }

1/p

(0 < p < ∞),

a proof of which can be found in [45, p. 191]. Theorem 3.2.2 (Littlewood’s subordination theorem). If f ≺ g, and p is a positive real number, then M p (r, f) ⩽ M p (r, g)

(0 ⩽ r < 1).

Equality holds if, and only if, r = 0, or if r ≠ 0 but g is constant, or ω(z) = e iα z. ∞ n n It is natural to ask that if f ≺ g (with f(z) = ∑∞ n=1 a n z and g(z) = ∑ n=1 b n z ), then does this imply that |a n | ⩽ |b n | for n ⩾ 1? However, the simple example z2 ≺ z shows that this is not the case. In spite of this, the following remarkable domination result was proved by Rogosinski [214, 215]. ∞ n n Theorem 3.2.3 (Rogosinski’s theorem). If f(z) = ∑∞ n=1 a n z , g(z) = ∑ n=1 b n z , and f ≺ g, then for any integer n ⩾ 1, n

n

∑ |a k |2 ⩽ ∑ |b k |2 . k=1

k=1

3.2 Subordination

| 33

k Proof. Setting t n (z) = ∑nk=1 b k z k , we have g(z) = t n (z) + r n (z), r n (z) = ∑∞ k=n+1 b k z and

f(z) = g(ω(z)) = t n (ω(z)) + r n (ω(z)). Further, with s n (z) = ∑nk=1 a k z k , since ω(0) = 0, it follows that there exists a sequence {c n } such that ∞

t n (ω(z)) = s n (z) + ∑ c k z k , k=n+1

and so by Parseval’s theorem, 2π

n ∞ 1 ∫ |t n (ω(re iθ ))|2 dθ = ∑ |a k |2 r2k + ∑ |c k |2 r2k . 2π k=1 k=n+1 0

On the other hand, it is clear that t n ∘ ω ≺ t n , and so Littlewood’s theorem gives 2π

2π

0

0

n 1 1 ∫ |t n (ω(re iθ ))|2 dθ ⩽ ∫ |t n (re iθ )|2 dθ = ∑ |b k |2 r2k . 2π 2π k=1

Thus, ∑nk=1 |a k |2 r2k ⩽ ∑nk=1 |b k |2 r2k , and letting r → 1 gives the result. Rogosinski’s theorem holds only for the exponent 2 (see Duren [45, Exercises 5 and 6, p. 212]), however Goluzin [56], using Rogosinski’s theorem, obtained the following more general result. ∞ n n Theorem 3.2.4. Let f(z) = ∑∞ n=1 a n z , g(z) = ∑ n=1 b n z , f ≺ g, and λ1 ⩾ λ2 ⩾ . . . ⩾ 0. Then, ∞

∞

k=1

k=1

∑ λ k |a k |2 ⩽ ∑ λ k |b k |2 . If λ1 > λ2 , b 1 ≠ 0, and the series on the right is finite, then equality occurs only when f(z) = g(αz), for |α| = 1. If the superordinate function is convex, starlike, or close-to-convex, then the modulus of the coefficients of the subordinate function (which necessarily needs only to be analytic), have the same upper bound as that for the corresponding class of functions. For convex and starlike functions, this was proved by Rogosinski [215], while the result for close-to-convex functions is due to Robertson [208]. Thus we have the following. ∞ n n Theorem 3.2.5. Let f(z) = ∑∞ n=1 a n z , g(z) = ∑ n=1 b n z and f ≺ g. Then (i) g ∈ C implies |a n | ⩽ 1, n = 1, 2, . . . ; (ii) g ∈ S∗ implies |a n | ⩽ n, n = 1, 2, . . . ; (iii) g ∈ K0 implies |a n | ⩽ n, n = 1, 2, . . . .

This motivated Rogosinski to conjecture that if f ≺ g with g ∈ S, then |a n | ⩽ n for all integers n ⩾ 1. This became known as the Rogosinski or generalized Bieberbach conjecture.

34 | 3 Fundamental Lemmas

The Rogosinski conjecture clearly implies the Bieberbach conjecture and follows from the Robertson conjecture (1936) for odd univalent functions, which again follows from the Milin conjecture (1971) for the logarithmic coefficients of functions in S. In his famous paper of 1985, de Branges [40] proved the Milin conjecture, thus proving all these conjectures at once. We next give two very useful, and much-used theorems. The first concerns subordination and is due to Mocanu [148, Lemma 2.2d, p. 24]; the second is due to Nunokawa [166]. Other variations and extensions of Mocanu’s theorem can also be found in the literature. Theorem 3.2.6. Let f be analytic in 𝔻 and g be analytic and univalent in 𝔻, with f(0) = g(0). If f ⊀ g, there exists z0 ∈ 𝔻 and ζ0 ∈ ∂𝔻, such that f(|z| < |z0 |) ⊂ g(𝔻), f(z0 ) = g(ζ0 ), and z0 f 󸀠 (z0 ) = mζ0 g󸀠 (ζ0 ), for m ⩾ 1. Theorem 3.2.7. Let p be analytic in 𝔻, with p(z) ≠ 0 and p(0) = 1. If there exists z0 ∈ 𝔻, such that | arg p(z)| < απ/2 for |z| < |z0 | and | arg p(z0 )| = απ/2 for some α > 0, then z0 p󸀠 (z0 ) = ikα, p(z0 ) where

and

απ , 2

1 1 (a + ) 2 a

when

arg p(z0 ) =

1 1 k ⩽ − (a + ) 2 a

when

arg p(z0 ) = −

k⩾

απ , 2

and where p1/α (z0 ) = ±ia for a > 0. We omit the proofs, both of which depend on the application of the Schwarz lemma. Nunokawa’s lemma can be used to provide a simple proof of the following useful theorem. Theorem 3.2.8. Let p be analytic in 𝔻, p(0) = 1, and p(z) ≠ 0 for z ∈ 𝔻. Then for α > 0, Re [p(z) + α

zp󸀠 (z) ] > 0 (z ∈ 𝔻), p(z)

(3.2.1)

implies Re p(z) > 0, for z ∈ 𝔻. Proof. Suppose there exists a point z0 ∈ 𝔻, such that | arg p(z)| < π/2, for |z| < |z0 |, and | arg p(z0 )| = π/2. Then by Theorem 3.2.7, z0 p󸀠 (z0 ) = ik, p(z0 ) where

1 (a + 2 1 k ⩽ − (a + 2 and p(z0 ) = ±ia for a > 0. k⩾

1 ), a 1 ), a

when when

π , 2 π arg p(z0 ) = − , 2 arg p(z0 ) =

3.3 The Clunie–Jack Lemma | 35

Thus, if arg p(z0 ) = π/2, then arg {p(z0 ) + α

z0 p󸀠 (z0 ) z0 p󸀠 (z0 ) )} } = arg {p(z0 ) (1 + α 2 p(z0 ) p (z0 ) ik π π = + arg (1 + α ) = , 2 ia 2

where p(z0 ) = ia (a > 0) and k ⩾ (1 + 1/a)/2, which contradicts (3.2.1). Next if arg p(z0 ) = −π/2, then arg {p(z0 ) + α

z0 p󸀠 (z0 ) z0 p󸀠 (z0 ) )} } = arg {p(z0 ) (1 + α 2 p(z0 ) p (z0 ) ik π π = − + arg (1 − α ) = − , 2 ia 2

where p(z0 ) = −ia (a > 0) and k ⩽ −(a + 1/a)/2, which also contradicts (3.2.1). Thus, there is no point z0 ∈ 𝔻, such that | arg p(z0 )| < π/2, for |z| < |z0 |, and | arg p(z0 )| = π/2, which establishes the result. We end with a useful and sharp result due to Rogosinski [215, p. 70], which is similar to Theorem 3.2.5 (i). ∞ n n Theorem 3.2.9. Let f ≺ g, with f(z) = ∑∞ n=1 a n z and g(z) = ∑ n=1 b n z . If g is univalent in 𝔻 and g(𝔻) is convex, then

|a n | ⩽ |b 1 | (n = 1, 2, . . . ). Proof. First note that when n = 1, Theorem 3.2.3 gives |a1 | ⩽ |b 1 |. Now put η = e2πi/n . Since g(𝔻) is convex we have h(z n ) =

f(ηz) + f(η2 z) + . . . + f(η n z) = a n z n + a2n z2n + . . . ≺ g(z), n

and so h(z) = a n z + a2n z2 + . . . ≺ g(z). Thus, h ≺ g, and it follows that |a n | ⩽ |b 1 |.

3.3 The Clunie–Jack Lemma The following powerful and useful result is very widely used. It was first published by Jack in 1971 [77] and is known as Jack’s lemma or the Clunie–Jack lemma. Suffridge independently discovered this result in [238]. The special case (when z0 = 1) appeared in the Pólya–Szegő collection of theorems and problems from 1925, [190, Problem 291, p. 162]. A more general form is given in [148, Lemma 2.2a(i), p. 19]. Lemma 3.3.1 (Clunie–Jack lemma). Let ω be a nonconstant and analytic function in 𝔻 with ω(0) = 0. If |ω| attains its maximum value on the circle |z| = r at z0 , then z0 ω󸀠 (z0 ) = kω(z0 ), for some k ⩾ 1.

36 | 3 Fundamental Lemmas Proof. Let M(r) be the maximum value of |ω(z)| on the circle |z| = r. Valiron in [257, Theorem 9, p. 20] gives a result of Hadamard that log M(r) is a continuous, convex, and strictly increasing function of log r. Writing z = re iθ and |ω(z)| = R, for the points where |ω(z)| = M(r), we have ∂R = 0, ∂θ i.e., for R ≠ 0, 0=

∂ ∂ re iθ ω󸀠 (re iθ ) 1 ∂R ]. = log R = Re [ log ω(z)] = − Im [ R ∂θ ∂θ ∂θ ω(re iθ )

So, if z0 is a point on the circle |z| = r at which ω(z) attains its maximum value, then z0 ω󸀠 (z0 )/ω(z0 ) is real and does not depend on θ, i.e., z0 ω󸀠 (z0 ) = k(|z0 |), ω(z0 ) where k is real function. Note next that k(0) = n, where n is the first term with nonzero coefficient in the Taylor expansion of ω(z) and is greater than or equal to 1, since ω(0) = 0, i.e., the 0-th coefficient is 0. Thus, in order to complete the proof, it is enough to show that k(r) is an increasing function of r on [0, +∞). Since log M(r) is a convex and strictly increasing function of log r, its inverse exists, and d(log M(r))/dr rM 󸀠 (r) = d(log r)/dr M(r) is an increasing function of log r, and so of r at those points for which d log M(r) d log r exists. Again, from Valiron [257, p. 21] we know that at those points for which this derivative does not exist, the left and right derivatives exist, and the left derivative does not exceed the right derivative. Therefore, rM 󸀠 (r)/M(r) is an increasing function of r (not necessarily continuous). Thus k(r) = Re [

󵄨󵄨 󵄨󵄨 ∂ r ∂R 󵄨󵄨󵄨󵄨 z0 ω󸀠 (z0 ) ∂ 󵄨 󵄨 ] = r (log R)󵄨󵄨󵄨 = ] = r Re [ log ω󵄨󵄨󵄨 󵄨 󵄨󵄨z=z0 󵄨󵄨 z=z0 ω(z0 ) ∂r ∂r R ∂r 󵄨󵄨󵄨z=z0

󵄨 is an increasing function of r since R󵄨󵄨󵄨z=z0 = M(r), and the proof is complete.

4 Starlike and Convex Functions 4.1 Starlike Functions As a result of the simple expression zf 󸀠 (z)/f(z) = p(z) of starlike functions in terms of a function p ∈ P given in Chapter 2 and the representation of functions in P in terms of a probability measure given in Chapter 3, it is possible to obtain a great many complete and elegant results for functions f in S∗ .

4.1.1 Coefficient Theorems We begin by giving a simple sufficient condition concerning the coefficients of f(z) for a function f to be starlike, due to Clunie and Keogh [32]. ∞ n Theorem 4.1.1. Let f ∈ A and f(z) = z + ∑∞ n=2 a n z , then ∑ n=2 n|a n | ⩽ 1 implies that ∗ f ∈S .

Proof. Write 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ∞ ∞ 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 |zf 󸀠 (z) − f(z)| − |f(z)| = 󵄨󵄨󵄨 ∑ (n − 1)a n z n 󵄨󵄨󵄨 − 󵄨󵄨󵄨z + ∑ a n z n 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨n=2 n=2 󵄨 󵄨 󵄨 󵄨 ∞

∞

n=2 ∞

n=2

< ∑ (n − 1)|a n | − (1 − ∑ |a n |) = ∑ n|a n | − 1 ⩽ 0. n=2

Hence,

󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨 zf (z) 󵄨󵄨 󵄨󵄨 󵄨󵄨 < 1 − 1 󵄨󵄨 f(z) 󵄨󵄨 󵄨 󵄨

(z ∈ 𝔻),

which means that zf 󸀠 (z)/f(z) lies inside a circle center (1, 0), with radius 1, for z ∈ 𝔻, and so f ∈ S∗ . We next show that the Bieberbach conjecture has a very simple proof when f ∈ S∗ , which, using an ingenious and much copied idea of Clunie [30], can be extended into a more general and elegant inequality concerning the coefficients of f(z). We prove the following. Theorem 4.1.2. Let f ∈ S∗ and be given by (1.1.2). Then for n ⩾ 2, |a n | ⩽ n, and further n

(n + 1)2 |a n |2 ⩽ 4 (1 + ∑ k|a k |2 ) . k=2

Equality holds if, and only if, f is a rotation of the Koebe function. https://doi.org/10.1515/9783110560961-004

(4.1.1)

38 | 4 Starlike and Convex Functions Proof. Since f ∈ S∗ , write zf 󸀠 (z) = f(z)p(z) for p ∈ P, by equating coefficients we have n−1

na n = a n + ∑ p n−k a k

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

k=1

where a1 = 1. We now use mathematical induction and suppose |a k | ⩽ k for k = 1, 2, . . . , n − 1, where n ⩾ 2. Then |p n−k | ⩽ 2 and n−1

n−1

(n − 1)|a n | ⩽ ∑ |p n−k ‖a k | ⩽ 2 ∑ k = n(n − 1). k=1

Thus, |a n | ⩽ n by induction. Next we write

k=1

zf 󸀠 (z) 1 + ω(z) = , f(z) 1 − ω(z)

k where ω is a Schwarz function, with ω(z) = ∑∞ k=1 ω k z . Then,

[zf 󸀠 (z) + f(z)]ω(z) = zf 󸀠 (z) − f(z), i.e.,

∞

∞

∞

k=2

k=1

k=2

[2z + ∑ (k + 1)a k z k ] ∑ ω k z k = ∑ (k − 1)a k z k .

(4.1.2)

Equating coefficients, for n ⩾ 2 we obtain 2ω n−1 + 3a2 ω n−2 + . . . + na n−1 ω1 = (n − 1)a n . This means that the coefficient a n on the right hand side of (4.1.2) depends only on a2 , a3 , . . . , a n−1 on the left hand side. So, we can rewrite this expression as n−1

∞

n

[2z + ∑ (k + 1)a k z k ] ω(z) = ∑ (k − 1)a k z k + ∑ b k z k , k=2

k=2

k=n+1

for some b k . Squaring the moduli of both sides, integrating around the circle |z| = r, using Parseval’s theorem, and the fact that |ω(z)| < 1 gives n

∞

n−1

k=n+1

k=2

∑ (k − 1)2 |a k |2 r2k + ∑ |b k |2 r2k < 4 + ∑ (k + 1)2 |a k |2 . k=2

Now let r → 1 to obtain n

n−1

∑ (k − 1)2 |a k |2 ⩽ 4 + ∑ (k + 1)2 |a k |2 , k=2

k=2

i.e., n−1

(n − 1)2 |a n |2 ⩽ 4 (1 + ∑ k|a k |2 ) . k=2

Now add 4n|a n

|2

to both sides to obtain (4.1.1).

4.1 Starlike Functions

|

39

We note that a trivial induction argument again shows that |a n | ⩽ n for n ⩾ 2 follows from (4.1.1). We remark that using a different method, Holland and Thomas [74], generalized (4.1.1) and showed that for any μ ∈ ℂ and n ∈ ℕ, n

|(n + 1)a n − μna n−1 |2 ⩽ 4 Re ∑ (a k − μa k−1 )(ka k − μ(k − 1)a k−1 ), k=1

where a0 = 0 and a1 = 1. The elegant inequality (4.1.1) provides information about the coefficients of bounded starlike functions. To see this, we need the following analogue of the area theorem for functions f ∈ S. We omit the proof, which is very similar to that of Theorem 1.2.1 and can be found in Nehari [156, p. 218]. Theorem 4.1.3. Let f ∈ S. For z = re iθ , 0 ⩽ r < 1, write C(r) = f(|z| = r), and let A(r) be the area enclosed by C(r). Then r 2π

∞

A(r) = π (r2 − ∑ n|a n |2 r2n ) = ∫ ∫ |f 󸀠 (ρe iθ )|2 ρ dθ dρ. n=2

0 0

Suppose now that f ∈ S∗ and |f(z)| < 1, then since A(r) ⩽ π, (4.1.1) shows that |a n | ⩽

2 n+1

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

(4.1.3)

The estimate (4.1.3) is the best possible in two senses. Clunie and Keogh [32] constructed an example of a bounded starlike function satisfying a n > 1/n for infinitely many n, thus showing that the rate of growth 1/n is best possible, and Pommerenke [191] gave an example of a function in S∗ satisfying |f(z)| < 1, where lim supn→∞ (n + 1)|a n | > 2 − ϵ, showing that 2 is the best possible constant. Finding the best upper bound for the difference of coefficients ‖a n+1 | − |a n ‖ when f ∈ S is unsolved. Hayman [68] showed that ‖a n+1 | − |a n ‖ ⩽ A, for n ⩾ 2, where A is an absolute constant. Milin [141], using a different method, showed that A < 9, Ilina [76] improved this to A < 4.26 . . . , and Grinšpan [61] to A < 3.61 . . . . On the other hand, when n = 2, Goluzin [54, 55], Jenkins [81] and Duren [45, p. 114] showed, in a theorem reminiscent of the sharp bound for the Fekete–Szegő theorem, that for f ∈ S, the following sharp inequalities hold: −1 ⩽ |a3 | − |a2 | ⩽

3 + e−λ 0 (2e−λ 0 − 1) = 1.029 . . . , 4

where λ0 is the unique value of λ satisfying 0 < λ < 1, and 4λ = e λ . Thus when f ∈ S, the constant A cannot be reduced to 1. However, for starlike functions, Leung [107], employing the third Lebedev–Milin inequality, showed that the best possible bound for A is 1, and we give his proof in the following theorem.

40 | 4 Starlike and Convex Functions Theorem 4.1.4. Let f ∈ S∗ and be given by (1.1.2). Then, ‖a n+1 | − |a n ‖ ⩽ 1, for n ⩾ 2. Equality holds for fixed n, when f(z) = f0 (z), where f0 (z) =

z (1 − γz)(1 − ζz)

for some γ and ζ , with |γ| = |ζ| = 1. Proof. Since f ∈ S∗ , we can write zf 󸀠 (z) = p(z), f(z) n with p(z) = 1 + ∑∞ n=1 p n z , for some p ∈ P. Then, integration gives z

∞ f(z) p(t) − 1 1 log =∫ dt = ∑ p k z k . z t k k=1 0

Also for |ζ| = 1, log {(1 − ζz)

∞ f(z) } = ∑ αk zk , z k=1

where α k = 1k (p k − ζ k ), and (1 − ζz)

∞ f(z) = ∑ βk zk , z k=0

where β k = a k+1 − ζa k . Thus ∞

∞

k=0

k=1

∑ β k z k = exp { ∑ α k z k } . We now apply the third Lebedev–Milin inequality (see Theorem 1.2.5) to obtain n

n

1 }, k k=1

|β n |2 ⩽ exp { ∑ k|α k |2 − ∑ k=1

and so

n

n 1 1 |p k − ζ k |2 − ∑ } . k k k=1 k=1

|a n+1 − ζa n |2 ⩽ exp { ∑

(4.1.4)

By Corollary 3.1.1 we can choose ζ with |ζ| = 1 so as to make the exponent in (4.1.4) nonpositive, to obtain |a n+1 − ζa n |2 ⩽ 1. Finally, we note that since ‖a n+1 | − |a n ‖ ⩽ |a n+1 − ζa n | for all |ζ| = 1, the proof is complete. A detailed discussion concerning the sharpness of ‖a n+1 | − |a n ‖ ⩽ 1 can be found in Leung [107]. We next show that the Fekete–Szegő problem has a simple solution for functions in S∗ .

4.1 Starlike Functions

| 41

Theorem 4.1.5 (The Fekete–Szegő theorem). Let f ∈ S∗ with coefficients given by (1.1.2). Then for μ ∈ ℂ, |a3 − μa22 | ⩽ max{1, |4μ − 3|}. The inequality is sharp. Proof. Equating coefficients in the expression zf 󸀠 (z) = f(z)p(z), where p(z) = 1+ p1 z + p2 z2 + ⋅ ⋅ ⋅ ∈ P, we obtain a2 = p1 1 a3 = (p2 + p21 ) 2 1 1 1 a4 = p31 + p1 p2 + p3 6 2 3

(4.1.5)

(although we do not need the expression for a4 at this point; we will use it in subsequent chapters, so we include it here). Thus using Theorem 3.1.3, 󵄨󵄨 󵄨 1󵄨 󵄨 󵄨󵄨a3 − μa22 󵄨󵄨󵄨 = 󵄨󵄨󵄨p2 + (1 − 2μ)p21 󵄨󵄨󵄨 󵄨 󵄨 2󵄨 󵄨 󵄨 p21 󵄨󵄨󵄨󵄨 1 1 󵄨󵄨󵄨 󵄨󵄨 + |p1 |2 |4μ − 3|) ⩽ (󵄨󵄨󵄨p2 − 2 󵄨󵄨 2 󵄨󵄨󵄨 2 |p1 |2 1 1 + |p1 |2 |4μ − 3|) . ⩽ (2 − 2 2 2

(4.1.6)

Maximizing the above expression in |p1 | over [0, 2] gives the result. We finally note that when p1 = p2 = 2 in (4.1.6), we have |a3 − μa22 | = |4μ − 3|, and when p1 = 0 and p2 = 2, then |a3 − μa22 | = 1, and so the inequality is sharp. We next give the sharp bound for the second Hankel determinant H2 (2) when f ∈ S∗ . We follow the method given by Janteng et al. [80]. Theorem 4.1.6. Let f ∈ S∗ . Then, H2 (2) ⩽ 1. The inequality is sharp. Proof. Since H2 (2) = |a2 a4 − a23 |, (4.1.5) gives 󵄨󵄨 󵄨 󵄨󵄨 p1 p3 p22 p41 󵄨󵄨󵄨 󵄨󵄨 . − − H2 (2) = 󵄨󵄨󵄨 󵄨󵄨 3 4 12 󵄨󵄨󵄨󵄨 󵄨 Since H2 (2) is rotationally invariant, we can assume that p1 ⩾ 0. So writing p1 = p, we use (3.1.6) and (3.1.8) from Theorem 3.1.4 to express the coefficients p2 and p3 in

42 | 4 Starlike and Convex Functions terms of p to obtain with |x| ⩽ 1, and |ζ| ⩽ 1, 󵄨󵄨 4 xp2 (4 − p2 ) p2 x2 (4 − p2 ) x2 (4 − p2 )2 󵄨 p H2 (2) = 󵄨󵄨󵄨󵄨− + − − 24 12 16 󵄨󵄨 16 󵄨󵄨 2 p(4 − p ) 󵄨 (1 − |x|2 )ζ 󵄨󵄨󵄨󵄨 + 6 󵄨󵄨

p4 |x|p2 (4 − p2 ) p2 |x|2 (4 − p2 ) |x|2 (4 − p2 )2 + + + 16 24 12 16 p(4 − p2 ) 2 (1 − |x| ) := ϕ(|x|) ⩽ ϕ(1) = 1, + 6

⩽

since ϕ󸀠 (|x|) = (4 − p2 )(p2 + 12|x| − 8p|x| + p2 |x|)/24 ⩾ 0, when 0 ⩽ p ⩽ 2 and 0 ⩽ |x| ⩽ 1. Obtaining best possible bounds for the third Hankel determinant has proved more difficult, and many authors have given nonsharp estimates for a variety of subclasses of S. However, as was pointed out in Section 3.1, finding an expression for the fourth coefficient of functions in P in terms of p1 (Kwon et al. [102]) should help in finding the best possible bounds for the third Hankel determinant. Using this expression, Kowalczyk et al. [97] showed the following. Theorem 4.1.7. If f ∈ S∗ , then H3 (1) ⩽ 8/9, where 󵄨󵄨 󵄨󵄨a1 󵄨󵄨 H3 (1) = 󵄨󵄨󵄨a2 󵄨󵄨 󵄨󵄨a3 󵄨

a2 a3 a4

󵄨 a3 󵄨󵄨󵄨 󵄨󵄨 a4 󵄨󵄨󵄨 , 󵄨󵄨 a5 󵄨󵄨󵄨

with a1 = 1. We note, however, that although this is the best estimate to date for H3 (1) when f ∈ S∗ , it is not sharp. Since the Koebe function belongs to S∗ , the problem of estimating the coefficients of the inverse function for f ∈ S∗ has the same solution as that in S. For f ∈ S∗ , the sharp bound |γ n | ⩽ 1/n for n ⩾ 1 for the coefficients of log(f(z)/z) is obtained at once on differentiating and noting that |p n | ⩽ 2 for n ⩾ 1.

4.1.2 Refined Growth Theorems Since the Koebe function belongs to S∗ , the standard sharp growth and distortion theorems for f ∈ S given in Chapter 1 hold for f ∈ S∗ . However, when f ∈ S∗ these can be refined (see Singh [230] and Twomey [254]), and we give some of these important results in the following two theorems.

4.1 Starlike Functions

| 43

Theorem 4.1.8. Let f ∈ S∗ . Then for z = re iθ ∈ 𝔻, (1 + r)2 |f(z)|] 󵄨󵄨 󸀠 󵄨󵄨 r log [ r 󵄨󵄨 zf (z) 󵄨󵄨 󵄨󵄨 󵄨󵄨 ⩽ + 1. 󵄨󵄨 f(z) 󵄨󵄨 1+r 󵄨 󵄨 (1 − r) log 1−r The inequality is sharp for the Koebe function and z = r. Proof. We first show that for |z| = r < 1, 1+r 2r log 󵄨󵄨 1 + z 󵄨󵄨 |1 − z| 󵄨󵄨 󵄨󵄨 + 1. 󵄨󵄨 ⩽ 󵄨󵄨 󵄨󵄨 1 − z 󵄨󵄨 (1 − r) log 1 + r 1−r

(4.1.7)

To see this we note that when |1 + z| ⩽ |1 − z|, the inequality is trivial. Next, since (x − 1)/ log x increases for x > 1, it follows that with u=

󵄨󵄨 1 + z 󵄨󵄨 󵄨 󵄨󵄨 and 𝑣 = 󵄨󵄨󵄨 󵄨, 󵄨󵄨 1 − z 󵄨󵄨󵄨

1+r , 1−r

[(u − 1)/ log u] log 𝑣+ 1 ⩾ [(𝑣− 1)/ log 𝑣] log 𝑣+ 1 = 𝑣 when u ⩾ 𝑣 > 1, which establishes (4.1.7) when |1 + z| > |1 − z|. With p(z) = zf 󸀠 (z)/f(z) in Theorem 3.1.10, integration gives 2π

f(z) 1 dν(t), log = 2 ∫ log z 1 − e−it z 0

and so taking real parts, and adding log(1 + r)2 to both sides gives 2π

(1 + r)2 1+r log { |f(z)|} = 2 ∫ log dν(t). r |1 − e−it z|

(4.1.8)

0

Also from Theorem 3.1.10 we obtain 󵄨 󵄨󵄨 󸀠 󵄨󵄨 2π 󵄨󵄨 󵄨 1 + e−it z 󵄨󵄨󵄨 󵄨󵄨 zf (z) 󵄨󵄨 󵄨 dν(t), 󵄨󵄨 ⩽ ∫ 󵄨󵄨󵄨 󵄨󵄨 −it 󵄨󵄨 󵄨 󵄨󵄨 f(z) 󵄨󵄨 󵄨 0 󵄨󵄨 1 − e z 󵄨󵄨 󵄨 2π

and so using (4.1.8) and (4.1.7) (with z replaced by e−it z), and noting that ∫0 dν(t) = 1, the result follows. Other inequalities similar to that in Theorem 4.1.8 can be obtained using the same kind of arguments. The following theorem contains examples due to Singh [230], the proofs of which we omit.

44 | 4 Starlike and Convex Functions Theorem 4.1.9. Let f ∈ S∗ . Then for z = re iθ ∈ 𝔻, 1 − r2 zf 󸀠 (z) |f(z)| ⩽ Re [ ] ⩽ 2ρ(r) − 1, r f(z) 󵄨 󵄨󵄨 zf 󸀠 (z) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 ⩽ 2ρ(r), 󵄨󵄨1 + 󵄨󵄨 f(z) 󵄨󵄨󵄨 󵄨 and

󵄨󵄨 󸀠 󵄨󵄨2 󵄨󵄨 zf (z) 󵄨󵄨 2 2 2 󵄨 󵄨󵄨 󵄨󵄨 f(z) 󵄨󵄨󵄨 ⩽ 2(1 + r ) ρ (r) − 1, 󵄨 󵄨

where

(1 + r)2 |f(z)|] r 1 ρ(r) = . + 1+r 1+r (1 − r2 ) log 1−r All the inequalities are sharp for the Koebe function. r log [

We recall now that for f ∈ S the bound |a2 | ⩽ 2 for the second coefficient ultimately determined the growth and distortion theorems. We show next how, for starlike functions (and later for convex functions), this relationship can be refined. In order to do this we need an improvement of the Schwarz lemma given by Finkelstein [49]. We include the proof for completeness. Lemma 4.1.1. Suppose that ω is a Schwarz function with ω(z) = ω1 z + . . . and |z| = r. Then, r(r + |ω1 |) . |ω(z)| ⩽ 1 + |ω1 |r The inequality is sharp when ω(z) =

e it z(z + ω1 e it ) . 1 + ω1 ze it

Proof. Let g(z) = ω(z)/z = ω1 + ω2 z + . . . . Then the maximum principle shows that |g(z)| < 1 for z ∈ 𝔻. Now let h(z) = (g(z) − ω1 )/(1 − ω1 g(z)). Then since |g(z)| < 1, z ∈ 𝔻, it follows that |h(z)| < 1, for z ∈ 𝔻, and so by the Schwarz lemma, |h(z)| ⩽ r, for z ∈ 𝔻. The result now follows on noting that 󵄨󵄨 h(z) + ω 󵄨󵄨 1 󵄨󵄨 󵄨 |ω(z)| = r 󵄨󵄨󵄨 󵄨. 󵄨󵄨 1 + ω1 h(z) 󵄨󵄨󵄨 We now give the Finkelstein theorem [49]. Theorem 4.1.10. Let f ∈ S∗ , with f(z) = z + a2 z2 + . . . . Then for z = re iθ ∈ 𝔻, |f(z)| ⩽

1+r ρ r ⋅( ) , 2 1−r 1−r

where 2ρ = |a2 |, so that 0 ⩽ ρ ⩽ 1. The inequality is sharp.

4.1 Starlike Functions

|

45

Proof. Without loss in generality we can, via a rotation, assume that a2 ⩾ 0. Write ψ(r) = (1 − r)1+ρ (1 + r)1−ρ and let 󵄨󵄨 f(z) 󵄨󵄨 󵄨 󵄨󵄨 ϕ(z) = log 󵄨󵄨󵄨 󵄨 + log ψ(r). 󵄨󵄨 z 󵄨󵄨󵄨 Note that ϕ(0) = 0, and that ϕ is a continuous function of z. Then, 󵄨󵄨 f(z) 󵄨󵄨 1 + ρ 1 − ρ ∂ϕ ∂ 󵄨󵄨 󵄨 = log 󵄨󵄨󵄨 + . 󵄨− 󵄨󵄨 z 󵄨󵄨󵄨 1 − r 1 + r ∂r ∂r However, r

󵄨󵄨 f(z) 󵄨󵄨 zf 󸀠 (z) ∂ 󵄨 󵄨󵄨 log 󵄨󵄨󵄨 ] − 1. 󵄨󵄨 = Re [ 󵄨󵄨 z 󵄨󵄨 ∂r f(z)

Since f ∈ S∗ , we have Re[zf 󸀠 (z)/f(z)] > 0 for z ∈ 𝔻, and so we can write 2ω(z) zf 󸀠 (z) −1= f(z) 1 − ω(z) for some Schwarz function ω. Taking the modulus, using Lemma 4.1.1, and noting that ω1 = ρ, we obtain 󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨 zf (z) 󵄨󵄨 2r(r + ρ) 󵄨󵄨 󵄨󵄨 ⩽ , − 1 󵄨󵄨 f(z) 󵄨󵄨 1 − r2 󵄨 󵄨 so that r

∂ϕ 2r(r + ρ) 1+ρ 1−ρ −r ⩽ +r = 0, ∂r 1−r 1+r 1 − r2

which completes the proof. Equality is attained for the starlike function f(z) =

1+z ρ z ( ) . 1 − z2 1 − z

We note that a corresponding sharp lower bound for |f(z)|, together with bounds for |f 󸀠 (z)| can also be obtained in a similar fashion (see Finkelstein [49]).

4.1.3 Theorems Concerning lim arg f(re iθ ) r→1

The Herglotz representation theorem gives a necessary and sufficient condition for a function f to belong to S∗ in terms of the measure ν. Thus a detailed examination of ν should provide information about functions in S∗ . By analyzing the finite number of jumps in ν, in an important paper of Pommerenke [191] showed that 𝑣(θ) = lim arg f(re iθ ) r→1

(4.1.9)

46 | 4 Starlike and Convex Functions exists for all real θ, and if απ is the largest jump of ν, so that 0 ⩽ α ⩽ 2. Then log M(r) , log(1/(1 − r))

α = lim

r→1

(4.1.10)

where M(r) = max |f(z)|. Clearly, α = 0 if, and only if, ν is continuous, and α = 2 if, |z|=r

and only if, f is (a rotation of) the Koebe function. A stronger version of (4.1.10) is the following, which is again due to Pommerenke [196], who showed that M 󸀠 (r) α = lim(1 − r) , (4.1.11) r→1 M(r) where M 󸀠 (r) is the left derivative. We omit the proofs of (4.1.10) and (4.1.11), which require detailed analysis of the measure ν. It is easy to see that if α > 0, then f(𝔻) contains at least one sector of opening πα, and no sector of larger opening. Thus, if α > 0, the area of f(𝔻) is infinite. However, α = 0 does not necessarily imply that the area of f(𝔻) is finite. For z = re iθ , write C(r) = f(|z| = r) and let A(r) be the area enclosed by C(r). Then Pommerenke’s results (4.1.10) and (4.1.11) were extended by Holland and Thomas in [74] and London and Thomas [128] to show that 2α = lim

r→1

log A(r) , log(1/(1 − r))

and the stronger version 2α = lim(1 − r) r→1

A󸀠 (r) . A(r)

We again omit the proofs, which involve a detailed analysis of the measure ν. Other interesting results of this nature have also been obtained, the proofs of which are nontrivial. However, we end this section by proving a connection between α and the coefficients of f(z) (see Holland and Thomas [75]), the proof of which is relatively straightforward. Theorem 4.1.11. Let f ∈ S∗ , with coefficients given by (1.1.2). Then lim sup n→∞

log+ n|a n | = α. log n

Proof. The inequality n|a n | ⩽ n2 implies that for n ⩾ 2, 0⩽

log+ n|a n | ⩽ 2. log n

Write τ = lim sup n→∞

log+ n|a n | , log n

4.1 Starlike Functions

| 47

and let ϵ > 0 be given. Then there exists N such that n|a n | < n τ+ϵ for n > N. Thus ∞

N

M(r) ⩽ ∑ |a n |r n + ∑ |a n |r n n=1

n=N+1

N

∞

n=1

n=N+1

< ∑ |a n |r n + ∑ n τ+ϵ−1 r n = O(1 − r)−τ−ϵ , as r → 1. Thus since ϵ > 0 is arbitrary, we obtain lim sup r→1

log M(r) ⩽ τ, log(1/(1 − r))

and so from (4.1.10) α ⩽ τ. On the other hand, it also follows that there exists an increasing sequence of integers {n k } such that for k ⩾ 1, n k |a n k | > n τ−ϵ k . Next note that a slight modification of the proof of Theorem 4.1.2 shows that for 0 ⩽ r < 1, (n + 1)|a n |r n ⩽ 2M(r), and so with r k = 1 − 1/n k , 2M(r k ) ⩾ (1 − 1/n k )n k (1 − r k )−τ+ϵ , which again since ϵ > 0 is arbitrary, gives lim sup r→1

log M(r) ⩾ τ, log(1/(1 − r))

and so α ⩾ τ. Thus α = τ.

4.1.4 The Radial Limit lim f(re iθ ) r→1

We saw in Section 4.1.3 that lim arg f(re iθ ) exists and is finite for every real θ. Since r→1 with z = re iθ ∈ 𝔻, zf 󸀠 (z) ∂ ] > 0, r log |f(re iθ )| = Re [ ∂r f(z) |f(re iθ )| increases with r for every fixed θ, lim |f(re iθ )| exists for every fixed θ. Thus, r→1

the radial limit ̂f , defined as follows also exists for every θ, ̂f (θ) = lim f(re iθ ). r→1

Recalling from Chapter 3 that every f ∈ S∗ can be represented via the Herglotz representation theorem (Theorem 3.1.10) in terms of the measure ν, we give an example of the connection between the radial limit ̂f (θ), the measure ν, and the coefficients of f(z) in the following (see Holland and Thomas [74]).

48 | 4 Starlike and Convex Functions Theorem 4.1.12. Let f ∈ S∗ and be given by (1.1.2). Then with a1 = 1, 2π

∞

∫ |̂f (θ)|2 dν(t) = ∑ n|a n |2 ,

(4.1.12)

n=1

0

in the sense that if either side of (4.1.12) exists, then so does the other. Proof. For f ∈ S∗ with p(z) = zf 󸀠 (z)/f(z), treating (|f(z)|2 )/2 as a sectorial element in the f(z) plane, we have 2π

1 A(r) = ∫ |f(z)|2 d[arg f(z)] 2 0 2π

=

1 ∫ |f(z)|2 Re p(z)d[arg z], 2

(4.1.13)

0

where as before A(r) is the area inside f(|z| = r). Next define the sequence {s n (t)} for n ∈ ℕ and t ∈ [0, 2π], by n

s n (t) = ∑ a k e ikt ,

(4.1.14)

k=1

where a1 = 1. Taking real parts in the Herglotz representation theorem we obtain 2π

1 − r2 zf 󸀠 (z) dν(t) Re [ ]= ∫ f(z) |1 − ze−it |2

(z ∈ 𝔻).

(4.1.15)

0

Also ∞ ∞ n f(z) = ∑ z n ∑ a k e−i(n−k)t = ∑ e−int s n (t)z n . −it 1 − ze n=1 n=1 k=1

(4.1.16)

Using (4.1.14), (4.1.15), (4.1.16), and Parseval’s theorem, with z = re iθ we obtain ∞

n

n=1

k=1

∞

2π ∑ r2n ∑ k|a k |2 = 2π(1 − r2 )−1 ∑ n|a n |2 r2n n=1 2π

= (1 − r2 )−1 ∫ |f(re iθ )|2 Re [

zf 󸀠 (z) ] dθ f(z)

0 2π 2π 󵄨

󵄨󵄨 f(re iθ ) 󵄨󵄨󵄨2 󵄨󵄨 dθdν(t) = ∫ ∫ 󵄨󵄨󵄨󵄨 󵄨 󵄨󵄨 1 − re i(θ−t) 󵄨󵄨󵄨 0 0 2π ∞

= 2π ∫ ∑ r2n |s n (t)|2 dν(t) 0 n=1 ∞

= 2π ∑ r n=1

2π 2n

∫ |s n (t)|2 dν(t), 0

(4.1.17)

4.1 Starlike Functions |

49

and so equating coefficients of r2n in (4.1.17) we obtain 2π

n

∫ |s n (t)|2 dν(t) = ∑ k|a k |2 .

(4.1.18)

k=1

0

2 We next assume that A(r) is finite, so that ∑∞ k=1 k|a k | < ∞. Then the sequence {δ n } defined for n ⩾ 1 by nδ n = ∑nk=1 k|a k |, and the sequence {ϵ n } for n ⩾ 0 by ϵ2n = 2 ∑∞ k=n+1 k|a k | ; both tend to 0 as n → ∞. Now, ∞

n

s n (t) − f(re it ) = ∑ a k (1 − r k )e ikt − ∑ a k r k e ikt , k=1

k=n+1

and so with t ∈ [0, 2π], using the Cauchy–Schwarz inequality, we have ∞

n

max |s n (t) − f(re it )| ⩽ (1 − r) ∑ k|a k | + ∑ |a k |r k

t∈[0,2π]

k=1

k=n+1 1/2

∞

⩽ n(1 − r)δ n + ( ∑ k|a k |2 ) k=n+1

⩽ n(1 − r)δ n +

ϵn √n(1 − r)

∞

1/2

r2k ) k k=n+1

( ∑

.

Choosing r = r n = 1 − 1/n, we deduce that max |s n (t) − f(re it )| → 0 as

t∈[0,2π]

n → ∞.

We are now able to complete the proof of Theorem 4.1.12 as follows. 2 First suppose that ∑∞ k=1 k|a k | < ∞. Then by the Lebesgue monotone convergence theorem and (4.1.18), 2π

n

lim ∑ k|a k |2 = lim ∫ |s n (t)|2 dν(t)

n→∞

n→∞

k=1

0 2π

= lim ∫ |f(r n e it )|2 dν(t) n→∞

0 2π

= ∫ |̂f (θ)|2 dν(t), 0

where r n = 1 − 1/n. Thus, (4.1.12) is proved in this case. 2π In order to complete the proof, we need to show that if ∫0 |̂f (θ)|2 dν(t) is finite, 2 then ∑∞ k=1 k|a k | is also finite. Suppose then that the integral is finite, so that for 0 ⩽ r < 1, 2π

2π

∫ |f(re )| dν(t) ⩽ ∫ |̂f (θ)|2 dν(t). it 2

0

0

(4.1.19)

50 | 4 Starlike and Convex Functions n it Next write zf 󸀠 (z) = f(z)p(z), with p(z) = 1 + ∑∞ n=1 p n z . Then if z = re and 0 ⩽ ρ < r < 1, 2π

2π

∞ zf 󸀠 (z) zf 󸀠 (z) ∫ |f(ρe )| Re [ ] dt = ∑ a n a m ρ n ρ m ∫ e−i(m−n)t Re [ ] dt f(z) f(z) n,m=1 it 2

0

0

∞

= π ∑ a n a m ρ n+m p m−n r m−n n,m=1 2π

∞

= 2π ∑

a nn (ρ/r)a m m (ρr)

n,m=1

∫ e−i(m−n)t dν(t) 0

2π

= 2π ∫ f(ρe it /r)f(rρe it )dν(t). 0

Fix r with 0 < r < 1. Then for all t, lim f(ρe it /r)f(rρe it ) = ̂f (t)f(r2 e it ), and so by (4.1.19), ρ→r

|̂f (t)f(r2 e it )| is integrable with respect to ν. Hence, by the Lebesgue dominated convergence theorem, it follows that 2π

2π

lim ∫ |f(ρe it )|2 Re [ ρ→r

zf 󸀠 (z) ] dt = 2π ∫ ̂f (t)f(r2 e it )dν(t), f(z)

0

0

i.e., 2π

2π

zf 󸀠 (z) ∫ |f(re )| Re [ ] dt = 2π ∫ ̂f (t)f(r2 e it )dν(t). f(z) it 2

0

0

The result now follows by letting r → 1, noting (4.1.13), and using Lebesgue’s theorem once more.

4.1.5 Length and Integral Mean Problems For f ∈ S, and p > 0, let 2π

I p (r, f ) =

1 ∫ |f(re iθ )|p dθ. 2π 0

Baernstein [11] showed that the Koebe function k maximizes I p (r, f ) for all p > 0, i.e., I p (r, f ) ⩽ I p (r, k). On the other hand, the corresponding problem of maximizing I p (r, f 󸀠 ) remains open, and it is known that I p (r, f 󸀠 ) ⩽ I p (r, k 󸀠 ) is false for p < 1/3 (see Lohwater et al. [126]). For starlike functions, I p (r, f 󸀠 ) ⩽ I p (r, k 󸀠 ) is valid for p > 0, the case p = 1 being solved by Marx in [139], and the remaining values of p by Clunie and Pommerenke [33], where the proofs are valid for the wider class of close-to-convex functions.

4.1 Starlike Functions |

51

The case I1 (r, f 󸀠 ), which we will consider next, is of particular interest. For f ∈ S, denote by L(r) the length of the curve C(r) = f(|z| = r), and A(r) the area inside C(r). Then for z = re iθ , 2π

L(r) = I1 (r) = ∫ |zf 󸀠 (z)|dθ.

(4.1.20)

0

In this section we find bounds for L(r) in terms of M(r) = max |f(z)| and A(r). We |z|=r consider first the bound in terms of M(r) (see Keogh [87]). Theorem 4.1.13. Let f ∈ S∗ . Then with z = re iθ ∈ 𝔻, L(r) ⩽ M(r) (2π + 4 log

1+r ). 1−r

Proof. Write zf 󸀠 (z) = f(z)p(z), so that 2π

2π

L(r) = ∫ |f(z)p(z)|dθ ⩽ M(r) ∫ |p(z)|dθ. 0

0

Since p(z) ≺

1+z , 1−z

it follows from Theorem 3.2.2 that 2π

󵄨󵄨 1 + z 󵄨󵄨 󵄨󵄨 󵄨 L(r) ⩽ M(r) ∫ 󵄨󵄨󵄨 󵄨 dθ 󵄨󵄨 1 − z 󵄨󵄨󵄨 0

2π

󵄨󵄨 1 + z 󵄨󵄨 󵄨󵄨 1 + z 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ⩽ M(r) ∫ (󵄨󵄨󵄨Re 󵄨 + 󵄨Im 󵄨) dθ 󵄨󵄨 1 − z 󵄨󵄨󵄨 󵄨󵄨󵄨 1 − z 󵄨󵄨󵄨 0

2π

π

= M(r) ( ∫ P(r, θ)dθ + 2 ∫ Q(r, θ)dθ) , 0

0

where P(r, θ) =

1 − r2 1 − 2r cos θ + r2

Q(r, θ) =

2r sin θ 1 − 2r cos θ + r2

is positive and

is an odd function on (0 < θ < π). Since 2π

∫ P(r, θ) = 2π 0

the theorem is proved.

π

and

∫ Q(r, θ) = 2 log 0

1+r , 1−r

52 | 4 Starlike and Convex Functions

Theorem 4.1.13 is best possible in two senses. First, it was shown by Hayman [67] that there exists a bounded function in S∗ such that L(r) >0 lim sup log(1/(1 − r)) r→1 which, in particular shows that the O(1) cannot, in general, be replaced by o(1). The problem of replacing lim sup by lim inf was subsequently settled by Lewis [112], but finding an example of a function in S∗ such that lim sup r→1

L(r) > 0, M(r) log(1/(1 − r))

or

L(r) >0 M(r) log(1/(1 − r))

lim inf r→1

remains open. Suppose next that f is a bounded function in S∗ , with |f(z)| < 1. Then Theorem 4.1.13 can be written as lim sup r→1

L(r) ⩽ 4. 1+r log 1−r

The question of whether the constant 4 is best possible was answered by Crane and Markose [36], where a function in S∗ satisfying |f(z)| < 1 was constructed, such that for any ϵ > 0, L(r) ⩾ 4 − ϵ. lim inf 1+r r→1 log 1−r In [241], Thomas showed that Theorem 4.1.13 can be improved as follows. Theorem 4.1.14. Let f ∈ S∗ . Then with z = re iθ , 0 ⩽ r < 1, 2√πA(r) ⩽ L(r) ⩽ 2√πA(r) (1 + log

1+r ). 1−r

In particular, L(r) ∼ 2√πA(r) as r → 0. Proof. Again, write zf 󸀠 (z) = f(z)p(z) and then use the Cauchy–Schwarz inequality to obtain 2π

1/2

2π 2

L(r) = ∫ |f(z)‖p(z)|dθ ⩽ ( ∫ |f(z)| |p(z)|dθ) 0

2π

( ∫ |p(z)|dθ)

0

1/2

.

(4.1.21)

0

From the proof of Theorem 4.1.13 we have 2π

∫ |p(z)|dθ ⩽ 2π + 4 log 0

1+r , 1−r

and so it remains to estimate 2π

I(r) := ∫ |f 2 (z)p(z)|dθ. 0

(4.1.22)

4.1 Starlike Functions

| 53

With z = ρe iθ , r 2π

r 2π 󸀠

I(r) ⩽ 2 ∫ ∫ |f (z)f(z)p(z)|dθdρ + ∫ ∫ |f 2 (z)p󸀠 (z)|dθdρ 0 0

0 0

r 2π

r 2π

= 2 ∫ ∫ |f 󸀠 (z)|2 ρdθdρ + ∫ ∫ |f 2 (z)p󸀠 (z)|dθdρ, 0 0

0 0

on using zf 󸀠 (z) = f(z)p(z). Hence, since r 2π

A(r) = ∫ ∫ |f 󸀠 (z)|2 ρdθdρ, 0 0

we have

r 2π

I(r) ⩽ 2A(r) + ∫ ∫ |f 2 (z)p󸀠 (z)|dθdρ.

(4.1.23)

0 0

Thus we need to estimate r 2π

J(r) := ∫ ∫ |f 2 (z)p󸀠 (z)|dθdρ. 0 0

To do this, we use Theorem 3.1.1 to obtain r 2π

J(r) ⩽ 2 ∫ ∫ |f(z)|2 [Re p(z)]dθ 0 0

dρ . 1 − ρ2

Now write f(z) = Re iϕ . Then, since f ∈ S∗ , Re p(z) = Re [

zf 󸀠 (z) ∂ϕ , ]= f(z) ∂θ

and so r 2π

J(r) ⩽ 2 ∫ ∫ R2 0 0 r

dρ ∂ϕ dθ ∂θ 1 − ρ2

= 2 ∫ ∫ R2 dϕ 0 |z|=ρ

r

dρ A(ρ) = 4∫ dρ, 1 − ρ2 1 − ρ2 0

since 2A(ρ) = ∫|z|=ρ R2 dϕ. Since A(ρ) increase with ρ, we obtain J(r) ⩽ 2A(r) log

1+r . 1−r

(4.1.24)

54 | 4 Starlike and Convex Functions

Thus from (4.1.21), (4.1.22), (4.1.23), and (4.1.24) we deduce that L(r) ⩽ 2√πA(r) (1 + log

1+r ). 1−r

We finally note that the left hand inequality in Theorem 4.1.14 follows at once from the isoperimetric inequality. The proof of the next theorem by Twomey [255], which we omit, uses a simple variation of that used in Theorem 4.1.8. It not only provides an estimate for I2 (r, f 󸀠 ) in terms of the maximum modulus, but also gives an improved bound for the order of growth when f is bounded, reminiscent of Theorem 4.1.8. Theorem 4.1.15. Let f ∈ S∗ . Then for 0 ⩽ r < 1, r2 I2 (r, f 󸀠 ) ⩽ CA(r)λ(r, f ), where C > 0 is an absolute constant, and λ(r, f ) =

r log((4/r)M(r)) . (1 − r) log(1/(1 − r))

We remark that when f is bounded, Theorem 4.1.15 shows that r2 I2 (r, f 󸀠 ) = O[(1 − r) log(1/(1 − r))]−1 , which is an improvement to the expected bound r2 I2 (r, f 󸀠 ) = O(1 − r)−1 obtained by 2π using ∫0 |p(re iθ )|2 dθ = O(1 − r)−1 .

4.1.6 Some Subclasses of Starlike Functions By suitably choosing subsets of the right hand half-plane Re p(z) > 0 in the definition of starlike functions, it is easy to construct subclasses of S∗ . We give two examples, both of which give subclasses consisting of bounded functions. Theorem 4.1.16. Let f ∈ A and satisfy 󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨 zf (z) 󵄨󵄨 󵄨󵄨 󵄨󵄨 < 1 − 1 󵄨󵄨 f(z) 󵄨󵄨 󵄨 󵄨

(z ∈ 𝔻).

(4.1.25)

Then f ∈ S∗ and with z = re iθ ∈ 𝔻, re−r ⩽ |f(z)| ⩽ re r

(4.1.26)

(1 − r)e−r ⩽ |f 󸀠 (z)| ⩽ (1 + r)e r ,

(4.1.27)

and |a n | ⩽ 1/(n − 1), for n ⩾ 2. All the inequalities are sharp.

4.1 Starlike Functions

|

55

Proof. We have seen in the proof of Theorem 4.1.1 that such functions belong to S∗ , with points in the half-plane restricted to lie inside a circle center (1, 0) with radius 1. To see that these functions are bounded, write zf 󸀠 (z) − 1 = zω(z), f(z)

(4.1.28)

so that |ω(z)| < 1 for z ∈ 𝔻. Integrating (4.1.28) gives z

f(z) = z exp (∫ ω(t)dt) , 0

and so with z = re iθ ,

re−r ⩽ |f(z)| ⩽ re r .

Differentiating the above expression for f(z) also gives (1 − r)e−r ⩽ |f 󸀠 (z)| ⩽ (1 + r)e r . Both (4.1.26) and (4.1.27) are sharp when f(z) = ze z . Using the method of Clunie described in Theorem 4.1.2, it is a relatively simple exercise to show that the coefficients of functions defined by (4.1.25) satisfy the sharp inequality |a n | ⩽ 1/(n − 1), for n ⩾ 2. We now give a more interesting example (Sokol and Thomas [234]) where the region in the half-plane Re p(z) > 0 lies inside the right hand half of the lemniscate of Bernoulli. Some coefficient estimates are given, while other results are included without proofs. Theorem 4.1.17. Let f ∈ A and denote by S∗L the class of functions f satisfying the condition zf 󸀠 (z) √ ≺ 1 + z =: q(z), f(z) where the branch of the square root is chosen to be q(0) = 1. Then f ∈ S∗ , and the sharp inequality |a n | ⩽ 1/(2(n − 1)) holds for n = 2, 3, 4, 5. Also for n ⩾ 2, n−1

(n − 1)2 |a n |2 ⩽ ∑ |a k |2 {(kδ − 1)2 − (k − 1)2 } ,

(4.1.29)

k=1

where δ = √2 − 1. Proof. First we note that the set q(𝔻) lies in the region bounded by the right loop of the lemniscate of Bernoulli given by (x2 + y2 )2 − 2(x2 − y2 ) = 0. That f ∈ S∗ is obvious from the definition of S∗L . For a proof of |a n | ⩽ 1/(2(n − 1)) when n = 2, 3, 4, 5, see Ravichandran and Shelly Verma [205] and Sokol [233], noting that the inequalities are sharp for the function

56 | 4 Starlike and Convex Functions f n ∈ S∗L defined by

zf n󸀠 (z) = √1 + z n f n (z)

(n = 1, 2, . . . ).

We once more use the method of Clunie to prove the inequality (4.1.29), which for large values of n gives estimates near the conjecture |a n | ⩽ 1/(2(n − 1)) for n ⩾ 2 made by Sokol in [233]. Since for z = re iθ , zf 󸀠 (z) 1+z ≺ √1 + z ≺ , f(z) 1 + δz we have

1 + ω(z) zf 󸀠 (z) = , f(z) 1 + δω(z)

where ω(0) = 0 and |ω(z)| < 1 for z ∈ 𝔻, and ∞

ω(z) = ∑ ω k z k . k=1

Thus, we obtain zf 󸀠 (z) − f(z) = ω(z) [f(z) − δzf 󸀠 (z)] , so that with a1 = 1, ∞

∞

k=1

k=1

∑ (k − 1)a k z k = ω(z) ∑ (1 − kδ)a k z k . We now write ∞

n

∑ (k − 1)a k z k + ∑ (k − 1)a k z k k=1

k=n+1 ∞

n−1

= ω(z) { ∑ (1 − kδ)a k z k + ∑ (1 − kδ)a k z k } , k=1

k=n

which can be written as n

∞

∞

k=n+1

k=n

∑ (k − 1)a k z k + ∑ (k − 1)a k z k − ω(z) ∑ (1 − kδ)a k z k k=1 n−1

= ω(z) ∑ (1 − kδ)a k z k . k=1

Applying Clunie’s method, we now write n

∞

n−1

k=n+1

k=1

∑ (k − 1)a k z k + ∑ b k z k = ω(z) ∑ (1 − kδ)a k z k , k=1

for some b k , n + 1 ⩽ k < ∞.

4.1 Starlike Functions

| 57

This gives 󵄨󵄨2 󵄨󵄨 󵄨󵄨2 󵄨󵄨 n ∞ n−1 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨 ∑ (k − 1)a k z k + ∑ b k z k 󵄨󵄨󵄨 = 󵄨󵄨󵄨ω(z) ∑ (1 − kδ)a k z k 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨k=1 k=n+1 k=1 󵄨󵄨2 󵄨󵄨n−1 󵄨󵄨 󵄨󵄨 ⩽ 󵄨󵄨󵄨 ∑ (1 − kδ)a k z k 󵄨󵄨󵄨 , 󵄨󵄨󵄨 󵄨󵄨󵄨 k=1 where

∞

∞

k=n+1

k=1

n

∑ (k − 1)a k z k + ∑ b k z k := ∑ d k z k k=1

is analytic in 𝔻. Parseval’s theorem now gives 2π 󵄨 ∞ 󵄨󵄨

󵄨󵄨2 ∞ 󵄨󵄨 󵄨 ∫ 󵄨󵄨󵄨 ∑ d k (re iθ )k 󵄨󵄨󵄨 dθ = 2π ∑ |d k |2 r2k , 󵄨󵄨 󵄨󵄨 k=1 󵄨 0 󵄨 k=1 and so integrating with respect to θ from 0 to 2π, we obtain n

∞

n−1

k=n+1

k=1

∑ (k − 1)2 |a k |2 r2k + ∑ |b k |2 r2k ⩽ ∑ (1 − kδ)2 |a k |2 r2k . k=1

Therefore, n

n−1

∑ (k − 1)2 |a k |2 r2k ⩽ ∑ (1 − kδ)2 |a k |2 r2k . k=1

k=1

Letting r → 1 gives n

n−1

∑ (k − 1)2 |a k |2 ⩽ ∑ (kδ − 1)2 |a k |2 , k=1

k=1

which gives (4.1.29). Corollary 4.1.1. If f ∈ S∗L , then for n ⩾ 2, |a n | ⩽

2 − √2 . n−1

Proof. From (4.1.29) and Theorem 4.1.17, we have n−1

(n − 1)2 |a n |2 ⩽ ∑ |a k |2 {(kδ − 1)2 − (k − 1)2 } k=1 n−1

= (δ − 1)2 − ∑ |a k |2 {(k − 1)2 − (kδ − 1)2 } k=2

⩽ (δ − 1)

2

= (√2 − 2)2 .

(4.1.30)

58 | 4 Starlike and Convex Functions For n = 6, (4.1.30) becomes 2 − √2 = 0.117 . . . , 5 which compares well with the conjectured bound 1/10 for the sixth coefficient, and that for n ⩾ 10 the bound (2 − √2)/(n − 1) in (4.1.30) does not exceed the conjectured bound 1/(2(n − 1)) by more than 1/100. In the next theorem, we state other results for functions in S∗L , the proofs of which can be found in Ma and Minda [132] and Sokol and Thomas [234]. |a6 | ⩽

Theorem 4.1.18. If f ∈ S∗L . Then, the second Hankel determinant H2 (2) = |a2 a4 − a23 | ⩽

1 . 16

The inequality is sharp when f(z) = f2 (z). Also, 1 λ { − , { { { 16 4 { { { 1 |a3 − λa22 | ⩽ { , { 4 { { { { { λ− 1 , { 4 16 The inequalities are sharp when f(z) = f1 (z). Finally, for z = re iθ ∈ 𝔻,

λ⩽− −

3 4

3 5 ⩽λ⩽ 4 4 5 λ⩾ . 4

−f1 (−r) ⩽ |f(z)| ⩽ f1 (r), and f1󸀠 (−r) ⩽ |f 󸀠 (z)| ⩽ f1󸀠 (r). Moreover,

󵄨󵄨 f1 (z) f(z) 󵄨󵄨󵄨󵄨 󵄨󵄨 }󵄨󵄨 ⩽ max arg { }. 󵄨󵄨arg { 󵄨 󵄨󵄨 z 󵄨 |z|=r z In each of the above inequalities, equality holds for some z = re iθ ∈ 𝔻 if, and only if, f(z) is a rotation of f1 (z).

4.2 Convex Functions Since Alexander’s theorem tells us that f ∈ C if, and only if, zf 󸀠 ∈ S∗ , we can immediately deduce some properties for functions in C. First we note that integrating the Herglotz representation theorem for f ∈ S∗ shows that any f ∈ C can be expressed parametrically as z

2π

f(z) = ∫ exp [−2 ∫ log(1 − e−it ζ)dν(t)] dζ, 0 [ 0 ] 2π

where dν(t) ⩾ 0, and ∫0 dν(t) = 1.

4.2 Convex Functions |

59

4.2.1 Growth and Distortion Theorems Let f ∈ C for z ∈ 𝔻, with

∞

f(z) = z + ∑ a n z n .

(4.2.1)

n=2

Since the Bieberbach conjecture is true when f ∈ S∗ , we deduce at once that |a n | ⩽ 1 for n ⩾ 2, when f ∈ C, and that this inequality is sharp when f(z) = z/(1 − z). Analogously to starlike functions, the following growth and distortion inequalities hold for f ∈ C. Theorem 4.2.1. Let f ∈ C, with z = re iθ ∈ 𝔻. Then, r r ⩽ |f(z)| ⩽ 1+r 1−r 1 1 ⩽ |f 󸀠 (z)| ⩽ (1 + r)2 (1 − r)2 󵄨󵄨 󸀠 󵄨󵄨 1 1 󵄨 zf (z) 󵄨󵄨 󵄨󵄨 ⩽ ⩽ 󵄨󵄨󵄨󵄨 . 󵄨 1 + r 󵄨󵄨 f(z) 󵄨󵄨 1 − r

(4.2.2)

All the inequalities are sharp when f(z) = z/(1 − z), or a rotation. Proof. The inequalities for |f 󸀠 (z)| follow at once from Alexander’s theorem, and the right hand side of the first inequality is trivial because |a n | ⩽ 1. For the left hand side of the first inequality, take a point on f(|z| = r), closest to the origin, and consider the line segment between this point and the origin. Let L be the pre-image of this line segment. Then, |z| 󸀠

|f(z)| = ∫ |f (z)‖dz| ⩾ ∫ 0

L

1 r dr = . 1+r (1 + r)2

We omit the proof of the third pair of inequalities in (4.2.2), which is exactly analogous to the method used in the proof of Theorem 1.3.4 in Chapter 1. At this point we note that for f ∈ S, with z = re iθ ∈ 𝔻, (1.3.3) gives Re [1 +

zf 󸀠󸀠 (z) 1 − 4r + r2 . ]⩾ 󸀠 f (z) 1 − r2

Since 1 − 4r + r2 > 0 when r < 2 − √3, this shows that when f ∈ S, f maps the disk |z| < 2 − √3 onto a convex set. The number 2 − √3 is called the radius of convexity for the class S. When f is the Koebe function 1+

zk 󸀠󸀠 (z) 1 + 4z + z2 , = k 󸀠 (z) 1 − z2

which shows that the bound 2 − √3 is sharp.

60 | 4 Starlike and Convex Functions

The corresponding result for starlike functions is surprisingly much more difficult to obtain and needs the power of the Loewner theory, which as we have mentioned, is outside the scope of this book. A proof that the radius of starlikeness for S is tanh π/4 can be found in Duren [45]. The above illustrates the idea of a radius problem, and there are many such results in the literature. The idea is to find the largest radius |z| = r < 1 for which one class is included in another. We will not be concerned with these problems in this book and refer the reader to Goodman [58, 59] for some interesting examples. In the case of starlike functions, we saw how the size of the second Taylor coefficient influences the growth and distortion theorems. A similar situation holds for convex functions (see Finkelstein [49]), which follows at once from Theorem 4.1.10 by using Alexander’s theorem. An illustration of this is given the following, Theorem 4.2.2. Let f ∈ C, with f(z) = z + a2 z2 + . . . and z = re iθ ∈ 𝔻. Then |f 󸀠 (z)| ⩽

1 1+r ρ ( ) , 2 1−r 1−r

where ρ = |a2 |, so that 0 ⩽ ρ ⩽ 1. The inequality is sharp. We next give the analogue of the Koebe 1/4 theorem for functions in C. Theorem 4.2.3. Let f ∈ C and f(z) ≠ w for z ∈ 𝔻. Then, |w| ⩾ 1/2, with equality only possible when f(z) = z/(1 − z) or a rotation, which omits a value of modulus 1/2. Proof. Let f ∈ C, with f(z) ≠ ω and g(z) = [ f(z) − ω]2 . Then, g(z1 ) = g(z2 ) implies that either f(z1 ) = f(z2 ), or [ f(z1 ) + f(z2 )] = 2ω. Since f(z) omits the value ω, the latter is impossible, and so h ∈ S, where ω2 − g(z) . 2ω Since g(z) ≠ 0, it follows that h(z) ≠ ω/2, and so from the Koebe 1/4 theorem |ω|/2 ⩾ 1/4, i.e., |ω| ⩾ 1/2. h(z) =

4.2.2 Coefficient Inequalities In addition to the inequality |a n | ⩽ 1, for n ⩾ 2 mentioned above, Theorem 4.1.1 and 2 Alexander’s theorem at once give the sufficient condition ∑∞ n=2 n |a n | ⩽ 1 for a function f to belong to C. We can also deduce an immediate analogue to Theorem 4.1.2 as follows. Theorem 4.2.4. Let f ∈ C, and be given by (4.2.1). Then for n ⩾ 2, n

n2 (n + 1)2 |a n |2 ⩽ 4 (1 + ∑ k 3 |a k |2 ) . k=2

The inequality is sharp when f(z) = z/(1 − z).

4.2 Convex Functions

|

61

The complete solution to finding the best bounds for ‖a n+1 | − |a n ‖ when f ∈ C does not follow immediately from Alexander’s theorem. We give the following theorem due to Li and Sugawa [116]. Theorem 4.2.5. Let D+n = sup (|a n+1 | − |a n |) , f ∈C

and D−n = sup (|a n | − |a n+1 |) . f ∈C

Then D+n = 1/(n + 1)

(n ⩾ 2),

(4.2.3)

D−2 = 1/2, and D−3 = 1/3. All the inequalities are sharp. Also 1/n < D−n < 2/(n + 1), for n ⩾ 4. The above inequality for D−n shows that the expected result D−n = 1/n is false when n ⩾ 4, leaving open a significant problem for functions in C. Proof. We prove (4.2.3) as follows. Alexander’s theorem and the inequality for the difference of coefficients for starlike functions (Theorem 4.1.4) give −1 ⩽ (n + 1)|a n+1 | − n|a n | ⩽ 1, and so |a n+1 | − |a n | ⩽

1 , n+1

which proves (4.2.3). To show that the inequalities for D+n are sharp, define l ϕ (z) =

∞ sin nϕ n 1 1 − e−iϕ z log ∑ z . = n sin ϕ e iϕ − e−iϕ 1 − e iϕ z n=1

(4.2.4)

Then equality is attained in (4.2.3) when f(z) = l π/n (z), in which case a n = 0 and a n+1 = −1/(n+1), and in the inequalities for D−2 and D−3 for the functions f(z) = l π/3 (z), and f(z) = l π/4 (z), respectively [116]. The inequalities concerning D−n for n ⩾ 4 require further analysis, which can be found in [116]. Related to the above problem is finding the sharp bound for |a n+1 − a n |. Observing that this difference is 2 when f(z) = z/(1 + z) and 0 when f(z) = z/(1 − z), the following theorem by Robertson [210] is of interest. Theorem 4.2.6. Let f ∈ C, with coefficients given by (1.1.2). Then for n ⩾ 2, |a n+1 − a n | ⩽

2n + 1 ⋅ |1 − a2 |. 3

The factor (2n + 1)/3 cannot be replaced by a smaller one independent of f .

62 | 4 Starlike and Convex Functions Proof. We first note that if p ∈ P, then the function q defined by q(z) =

1 − z2 (1 − z)2 − p(z) z z

is also in P. To see this, let ρ ∈ (0, 1) and zq ρ (z) = (1 − z2 ) − (1 − z)2 p(ρz). Then, q ρ (z) → q(z) as ρ → 1, and q ρ (z) is analytic for z ∈ 𝔻. Take z = e iθ , so that Re q ρ (z) = 2(1 − cos θ) Re p(ρz) ⩾ 0. Since the minimum of a harmonic function occurs on the boundary, Re q ρ (z) ⩾ 0 for z ∈ 𝔻, and so letting ρ → 1, the result follows. n Now let s(z) = zf 󸀠 (z), so s ∈ S∗ , and write s(z) = z + ∑∞ n=2 b n z , and t(z) = ∞ 2 n (s(z)/z)(1 − z) = 1 + ∑n=1 t n z . Then, taking p(z) = zs󸀠 (z)/s(z) and q(z) = −zt󸀠 (z)/s(z), we have − Re[zt󸀠 (z)/ s(z)] > 0 for z ∈ 𝔻, unless t󸀠 (z) ≡ 0, in which case t(z) ≡ 1 and s(z) = z/(1 − z)2 . Since s(z)/z = t(z)/(1 − z)2 , equating coefficients we obtain, with t1 = b 2 − 2, b n+1 = n + 1 + nt1 + (n − 1)t2 + ⋅ ⋅ ⋅ + 2t n−1 + t n . Thus, 1 b n+1 b n − = [t1 + 2t2 + ⋅ ⋅ ⋅ + (n − 1)t n−1 + nt n ], n+1 n n(n + 1) and so 󵄨󵄨 b 󵄨 2n + 1 |t1 | 󵄨󵄨 n+1 b n 󵄨󵄨󵄨 − [12 + 22 + ⋅ ⋅ ⋅ + n2 ] = 󵄨󵄨 󵄨⩽ 󵄨󵄨 n + 1 n 󵄨󵄨󵄨 n(n + 1) 3

󵄨󵄨 b 2 󵄨󵄨󵄨󵄨 󵄨 ⋅ 󵄨󵄨󵄨1 − 󵄨. 󵄨󵄨 2 󵄨󵄨󵄨

Since na n = b n , the result follows. To see that the inequality is the best possible, we note that for the function l ϕ ∈ C, defined by (4.2.4), 󵄨󵄨 a 󵄨 󵄨 n+1 − a n 󵄨󵄨󵄨 2n + 1 lim 󵄨󵄨󵄨 󵄨󵄨 = 3 . ϕ→0 󵄨󵄨 1 − a2 󵄨󵄨 Recalling that for f ∈ C, we can write 1+

zf 󸀠󸀠 (z) = p(z); f 󸀠 (z)

for p ∈ P, equating coefficients gives a2 =

p1 , 2

a3 =

p2 + p21 , 6

a4 =

p3 p1 p2 p31 + + . 12 8 24

(4.2.5)

We next state the Fekete–Szegő theorem for convex functions. The proof, which we omit, uses the above expressions for a2 and a3 , and is similar to that for starlike functions.

4.2 Convex Functions

| 63

Theorem 4.2.7 (The Fekete–Szegő theorem for convex functions). Let f ∈ C. Then for μ ∈ ℂ, 1 |a3 − μa22 | ⩽ max{1, 3|1 − μ|}. 3 The inequality is sharp. Finding the sharp bound for the second Hankel determinant H2 (2) when f ∈ C uses the same technique as in the case of starlike functions. Although the proof is similar, we include it for completeness. Theorem 4.2.8. Let f ∈ C, then H2 (2) = |a2 a4 − a23 | ⩽

1 . 8

The inequality is sharp. Proof. From (4.2.5) we obtain 󵄨󵄨 󵄨 󵄨󵄨 p1 p3 p21 p2 p22 p4 󵄨󵄨󵄨 + − − 1 󵄨󵄨󵄨 . H2 (2) = 󵄨󵄨󵄨 󵄨󵄨 24 144 36 144 󵄨󵄨󵄨 󵄨

(4.2.6)

Without loss of generality we can assume that p1 ⩾ 0. So, writing p1 = p, we use (3.1.6) and (3.1.8) from Theorem 3.1.4 to express the coefficients p2 and p3 in terms of p, and as in the proof of Theorem 4.1.6, use the triangle inequality to obtain H2 (2) ⩽

p p2 p2 (4 − p2 )(1 − |x|2 ) + (4 − p2 )|x| + (4 − p2 )|x|2 48 96 96 1 + (4 − p2 )2 |x|2 := ϕ(|x|). 144

Since ϕ󸀠 (|x|) ⩾ 0, when 0 ⩽ p ⩽ 2 and 0 ⩽ |x| ⩽ 1, it follows that H2 (2) ⩽ ϕ(1) =

1 p2 p4 + − , 9 36 72

which has a maximum value 1/8 when p = 1. To see that the bound 1/8 is sharp, choose p1 = 1, p2 =−1, and p3 = −2 in (4.2.6). As was noted in Section 4.1.1, using (3.1.7), (3.1.8), and (3.1.9) gives a more refined method applicable to finding sharp estimates for the third Hankel determinant H3 (1). Using these expressions in the case when f ∈ C, Kowalczyk et al. [97] recently proved the following. Theorem 4.2.9. If f ∈ C, then H3 (1) ⩽ 4/135. The result is sharp. We saw in Theorem 4.2.5 that for f ∈ C, coefficient differences exhibited irregular behavior. A similar situation arises for the coefficients of the inverse function, as was shown by Libera and Zlotkiewicz [119].

64 | 4 Starlike and Convex Functions Theorem 4.2.10. Let f ∈ C and f −1 be the inverse function of f , with Taylor expansion given by ∞

f −1 (ω) = ω + ∑ A n ω n , n=2

defined on some disk |ω| < r0 ( f ). Then, |A n | ⩽ 1, for 2 ⩽ n ⩽ 7, and |A n | > 1, when n ⩾ 10. The inequalities for 2 ⩽ n ⩽ 7 are sharp. Proof. The complete proof of this theorem is highly technical, and so in order to demonstrate the methods used, we give the solutions for the cases n = 2, 3 and 4. Recall from Chapter 1 that |A2 | = |a2 |,

|A3 | = |a3 − 2a22 |,

|A4 | = |a4 − 5a2 a3 + 5a32 |.

Since |a2 | ⩽ 1, the inequality |A2 | ⩽ 1 is trivial. Next, |A3 | = |p2 − 2p21 |/6 ⩽ (|p2 − p21 | + |p1 |2 )/6 ⩽ 1, on using the inequalities |p1 | ⩽ 2, and (3.1.10). Similarly, we obtain 24|A4 | = |6p31 − 7p1 p2 + 2p3 | ⩽ 2|p31 − 2p1 p2 + p3 | + |p1 |3 + 3|p1 ||p21 − p2 |, and the result now follows on using (3.1.10), (3.1.11), and |p1 | ⩽ 2. The inequalities for n = 5, 6 and 7 follow in a similar fashion, using the inequalities in Theorem 3.1.6. The cases n = 8 and 9 appear to be open. However, using the fact that the reciprocal of a function in P is also in P, it is possible to find suitable inequalities involving the coefficients of p(z) up to and including p7 and p8 , similar to those in Theorem 3.1.6. Further analysis would then be needed to show that |A8 | ⩽ 1 and |A9 | ⩽ 1. For n ⩾ 10, an explicit example of a function f ∈ C was constructed by Kirwan and Schober [91], where |A n | > 1.248 . . . , showing that the inequality |A n | ⩽ 1 is false when n ⩾ 10.

5 Starlike and Convex Functions of Order α As was pointed out in the last chapter, one of the simplest ways of extending the idea of starlike and convex functions is to restrict the image of the function p to certain subsets of the right half-plane. In [79] Janowski introduced the following idea that gives generalizations of starlike and convex functions. For f ∈ A, let zf 󸀠 (z) 1 + Az ≺ , f(z) 1 + Bz and 1+

zf 󸀠󸀠 (z) 1 + Az ≺ , f 󸀠 (z) 1 + Bz

with −1 ⩽ B < A ⩽ 1, noting that (1 + Az)/(1 + Bz) maps the unit disk univalently onto a convex region in the right hand half-plane that is symmetric with respect to the real axis and has the diameter end points (1 − A)/(1 − B) and (1 + A)/(1 + B). These classes are normally denoted by S∗ (A, B) and C(A, B), respectively. Choosing A = 1 and B = −1 gives the classes of starlike and convex functions, while A = 1 − 2α and B = −1, with 0 ⩽ α < 1, gives the classes of starlike and convex functions of order α, denoted by S∗ (α) and C(α), respectively, which are the subject of this chapter.

5.1 Definitions and Growth and Distortion Theorems We are thus led to the following definitions. Definition 5.1.1. Let f ∈ A. Then, f ∈ S∗ (α) if, and only if, for 0 ⩽ α < 1, Re [

zf 󸀠 (z) ]>α f(z)

(z ∈ 𝔻).

Definition 5.1.2. Let f ∈ A. Then f ∈ C(α) if, and only if, for 0 ⩽ α < 1, Re [1 +

zf 󸀠󸀠 (z) ]>α f 󸀠 (z)

(z ∈ 𝔻).

The following extension to Alexander’s theorem is obvious. Theorem 5.1.1. f ∈ C(α) if, and only if, zf 󸀠 ∈ S∗ (α). Upper and lower bounds for |f(z)| and |f 󸀠 (z)| when f ∈ C(α) and f ∈ S∗ (α) are easily obtained (see Robertson [211]) as follows. https://doi.org/10.1515/9783110560961-005

66 | 5 Starlike and Convex Functions of Order α Theorem 5.1.2 (Growth and distortion theorems for C(α) and S∗ (α)). If f ∈ C(α), then for 0 ⩽ α < 1 and 0 < |z| = r < 1, 1 1 ⩽ |f 󸀠 (z)| ⩽ . (5.1.1) 2(1−α) (1 + r) (1 − r)2(1−α) If α ≠ 1/2, then (1 + r)2α−1 − 1 1 − (1 − r)2α−1 ⩽ |f(z)| ⩽ , (5.1.2) 2α − 1 2α − 1 and if α = 1/2, then log(1 + r) ⩽ |f(z)| ⩽ − log(1 − r). (5.1.3) The above inequalities are sharp when f(z) = f α (z), where 1 − (1 − z) { { , α ≠ 1/2, 2α − 1 f α (z) = { { α = 1/2. { − log(1 − z), Similar results hold when f ∈ S∗ (α), for example, r r ⩽ |f(re iθ )| ⩽ . (1 + r)2(1−α) (1 − r)2(1−α) 2α−1

(5.1.4)

(5.1.5)

These inequalities are sharp for the function zf α󸀠 (z). Proof. We note that if f ∈ C(α), then we can write zf 󸀠󸀠 (z) 1 [1 + 󸀠 − α] = p(z), 1−α f (z) where p ∈ P. From the growth and distortion estimates for the class P (Theorem 3.1.1), for 0 < |z| = r < 1, we have 1+r 1−r ⩽ Re p(z) ⩽ , 1+r 1−r so that −2r(1 − α) zf 󸀠󸀠 (z) 2r(1 − α) ⩽ Re [ 󸀠 . ]⩽ 1+r f (z) 1−r Since ∂ zf 󸀠󸀠 (z) ∂ Re 󸀠 = r Re[log f 󸀠 (z)] = r log |f 󸀠 (z)|, f (z) ∂r ∂r we obtain 2(1 − α) ∂ −2(1 − α) ⩽ log |f 󸀠 (z)| ⩽ , 1+r ∂r 1−r and by integrating from 0 to R and replacing R with r, (5.1.1) follows. Using similar techniques as in the proof of the growth theorems for univalent functions (Theorem 1.3.3) we easily obtain (5.1.2) and (5.1.3). For f ∈ S∗ (α), consider the function g defined by f(z) = zg󸀠 (z). Theorem 5.1.1 tells us that g is convex of order α, and so from (5.1.1) we obtain 1 1 ⩽ |g󸀠 (z)| ⩽ . (1 + r)2(1−α) (1 − r)2(1−α) Multiplying this expression by |z| = r gives (5.1.5).

5.2 Inclusion Relationships | 67

5.2 Inclusion Relationships We first give the celebrated sharp result of Marx [139] and Strohhäcker [235], which states that every convex function is starlike of order 1/2, together with an additional property of starlike functions of order 1/2. We give proofs different to the originals, illustrating the power of the Clunie–Jack lemma. Theorem 5.2.1. (i) If f ∈ C, then f ∈ S∗ (1/2). (ii) If f ∈ S∗ (1/2), then Re [f(z)/z] > 1/2 for z ∈ 𝔻. The function f(z) = z/(1 − z) shows that both implications are sharp, i.e., the constant 1/2 cannot be replaced by a bigger one in (i) or by a smaller number in (ii), so that the result remains true. Proof. Let f ∈ C and put p(z) = Then Re [1 +

zf 󸀠 (z) 1 = . f(z) 1 − ω(z)

zf 󸀠󸀠 (z) zp󸀠 (z) 1 + zω󸀠 (z) = Re + p(z)] = Re ] [ [ ]. f 󸀠 (z) p(z) 1 − ω(z)

(5.2.1)

We need to show that |ω(z)| < 1 for z ∈ 𝔻. If not, then by the Clunie–Jack lemma, there exists z0 such that |ω(z0 )| = 1 and z0 ω󸀠 (z0 ) = kω(z0 ) for k ⩾ 1. For such z0 , from (5.2.1) we have Re [

z0 p󸀠 (z0 ) 1 + z0 ω󸀠 (z0 ) 1 + kω(z0 ) 1−k + p(z0 )] = Re [ ⩽ 0, ] = Re [ ]= p(z0 ) 1 − ω(z0 ) 1 − ω(z0 ) 2

which is a contradiction to the assumption that f ∈ C, and so (i) is established. The same approach with p(z) = f(z)/z proves (ii). It was noted at the end of Section 4.1.1 that when f ∈ S∗ , the logarithmic coefficients γ n of f(z) satisfy the sharp inequality |γ n | ⩽ 1/n for n ⩾ 1. The corresponding inequality for functions in C is an easy consequence of Theorem 5.2.1 (i) as follows. We omit the simple proof. Corollary 5.2.1. Let f ∈ C, then the logarithmic coefficients γ n of f(z), satisfy the sharp inequality |γ n | ⩽ 1/(2n) for n ⩾ 1. In [77], Jack posed the more general problem of finding the order of starlikeness of convex functions of given order, i.e., finding the largest number β = β(α) so that C(α) ⊂ S∗ (β(α)). Bearing in mind that the function f α defined in (5.1.4) is extremal for the growth and distortion estimates for the classes C(α) and S∗ (α) (Theorem 5.1.2), Jack conjectured that it will also be extremal in this case. MacGregor [135] determined the exact value of β(α) as the infimum over the unit disk 𝔻 of the real part of a specific analytic function, and conjectured that this infimum is attained at z = −1. More precisely, he proved the following.

68 | 5 Starlike and Convex Functions of Order α Theorem 5.2.2. If f ∈ C(α), 0 ⩽ α < 1, then zf 󸀠 (z) zf α󸀠 (z) ≺ ≡ G α (z), f(z) f α (z) where (2α − 1)z , { { { (1 − z)2(1−α) [1 − (1 − z)2α−1 ] G α (z) = { { −z { , { (1 − z) log(1 − z)

1 , 2 1 α= , 2 α ≠

(5.2.2)

and β(α) = inf Re G α (z). z∈𝔻

In advance of the proof, MacGregor obtained the following three lemmas. Lemma 5.2.1. Let g and h be analytic functions in 𝔻, with g(0) = h(0) = 0, so that h maps 𝔻 onto a (possibly many-sheeted) region which is starlike with respect to the origin. Then, (i) If Re[g󸀠 (z)/h󸀠 (z)] > γ for z ∈ 𝔻, then Re[g(z)/h(z)] > γ for z ∈ 𝔻. (ii) If Re[g󸀠 (z)/h󸀠 (z)] < γ for z ∈ 𝔻, then Re[g(z)/h(z)] < γ for z ∈ 𝔻. Proof. Part (i) of the lemma in the case γ = 0 was proved by Libera [117]. He used essentially the same idea as Sakaguchi (in the case when h is univalent) [223] and Robinson [213, Lemma, p. 30]. For the general case when γ ≠ 0, first note that g󸀠 (z)/h󸀠 (z) is subordinate to a suitably chosen Möbius transform which maps the unit disk onto the half-plane {w : Re w > γ} and maps the disk |z| < r, 0 < r < 1, onto an open disk with center a(r) and radius b(r). Thus, by the Lindelöf subordination principle, 󵄨󵄨󵄨 󵄨󵄨󵄨 g󸀠 (z) 󵄨󵄨 − a(r)󵄨󵄨󵄨󵄨 < b(r), 󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨 h (z) for all |z| < r. We now use Libera’s approach and put A(z) =

g󸀠 (z) − a(r) h󸀠 (z)

so that |A(z)| < b(r) when |z| < r. Further, for fixed z0 ∈ 𝔻, let L be the segment joining 0 and h(z0 ), which lies in one of the sheets of the starlike region h(𝔻). Let L−1 be the pre-image of L under h and r = max{|z| : z ∈ L−1 }. Then, 󵄨󵄨󵄨 z0 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󸀠 󸀠 󵄨 |g(z0 ) − a(r)h(z0 )| = 󵄨󵄨∫ [g (t) − a(r)h (t)] dt󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨 = 󵄨󵄨󵄨󵄨 ∫ h󸀠 (t)A(t)dt󵄨󵄨󵄨󵄨 ⩽ b(r) ∫ |dh(t)| 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 L−1 L = b(r)|h(z0 )|,

5.2 Inclusion Relationships

| 69

which proves part (i) of the lemma. Part (ii) follows from (i), by replacing g with −g. Lemma 5.2.2. If 0 ⩽ α < 1, then G α defined by (5.2.2) is univalent in 𝔻. Proof. When α ≠ 1/2, we can put G α (z) =

2α − 1 , ̂ G α (z) + 1

where

(1 − z)q − 1 ̂ , G α (z) = z and q = 2 − 2α, with 0 < q ⩽ 2 and q ≠ 1. Noting that z

q ̂ G α (z) = −q + ∫[1 − (1 − t)q−1 ]dt, z 0

it follows that ̂ G α is convex (and univalent), and so using a result of Libera [117, Theorem 2, p. 757] that states that z

φ∈C

⇒

2 ∫ φ(t)dt ∈ C, z

(5.2.3)

0

with φ(z) =

1 [1 − (1 − z)q−1 ], 1−q

it follows that G α is univalent in 𝔻. The univalence of G1/2 follows as a limiting case from the univalence of G α , when α ≠ 1/2. The next result is the last needed for the proof of Theorem 5.2.2 and is concerned with the univalence of a certain linear combination of zf α󸀠 (z)/f α (z) and 1 + zf α󸀠󸀠 (z)/f α󸀠 (z). In the interest of brevity, we omit the rather complicated proof, which can be found in MacGregor [135, Lemma 4, p. 533]. ̂ k be defined for z ∈ 𝔻 by Lemma 5.2.3. Let the function H ̂ k (z) = kH α (z) + (1 − k)G α (z), H where G α is given by (5.2.2), and H α (z) = 1 +

zf α󸀠󸀠 (z) 1 + (1 − 2α)z = , 1−z f α󸀠 (z)

(5.2.4)

̂ k is univalent in 𝔻, and where f α (z) is defined by (5.1.4). If k ⩾ 1 and 0 ⩽ α < 1, then H ̂ H α is subordinate to H k .

70 | 5 Starlike and Convex Functions of Order α

We can now give the proof of Theorem 5.2.2. Proof of Theorem 5.2.2. Let f ∈ C(α), with f(z) = z+a z z2 +. . . and f α (z) = z+A2 z2 +. . . . Then, |a2 | ⩽ A2 = 1 − α (this follows when n = 2 in Theorem 5.3.1 in the next section), with equality only when f(z) = f α (xz)/x and |x| = 1. Writing g(z) = and G α (z) =

zf 󸀠 (z) = 1 + b1 z + . . . f(z) zf α󸀠 (z) = 1 + B1 z + . . . , f α (z)

we obtain |b 1 | = |a2 | ⩽ A2 = B1 , with equality only when g(z) = G α (xz), and |x| = 1. Since for |x| = 1, G(xz) ≺ G(z), we can assume without loss of generality that |b 1 | < B1 . Now let 𝔻r = {z : |z| < r} be the open disk with radius r. Then |b 1 | < B1 implies g(𝔻r ) ⊂ G α (𝔻r ) for all sufficiently small values of r (this follows because G α is univalent in 𝔻 implies that g is subordinate to G α in 𝔻r ). Thus, there exists a function ω, analytic on 𝔻r , with ω(0) = 0 and |ω(z)| < r for all z ∈ 𝔻r such that ω(z) = G−1 α (g(z)).

(5.2.5)

Since this holds for all sufficiently small values of r, there exists ρ = sup {r ∈ (0, 1) : ω is analytic on 𝔻r , |ω(z)| < r on 𝔻r } . In order to complete the proof, it is enough to show that ρ = 1, so we assume, on the contrary, that ρ < 1. First note that ω is analytic for |z| < ρ and, indeed, for |z| ⩽ ρ. Thus, knowing that g is subordinate to G α in 𝔻ρ , it follows that g(𝔻 ρ ) ⊂ G α (𝔻 ρ ), which together with G α (𝔻ρ ) ⊂ G α (𝔻) implies that g(𝔻ρ+ϵ ) ⊂ G α (𝔻) for all sufficiently small values of ϵ > 0. Next, note that the univalence of G α on 𝔻 implies that ω defined by (5.2.5), is analytic for |z| ⩽ ρ. Thus from the definition of ρ, there exists z1 such that |z1 | = ρ and |ω(z1 )| = ρ, i.e., |ω(z1 )| = max |ω(z)|, and so from the Clunie–Jack lemma (Lemma 3.3.1) there exists a |z|=ρ

real number k ⩾ 1 such that z1 ω󸀠 (z1 ) = kω(z1 ). Further, f being convex of order α means that h(z) = 1 +

zf 󸀠󸀠 (z) f 󸀠 (z)

5.2 Inclusion Relationships | 71

is subordinate to H α (see (5.2.4)), i.e., h(z) = H α (ϕ(z)) for some function ϕ analytic in 𝔻, with ϕ(0) = 0 and |ϕ(z)| < 1 for z ∈ 𝔻 with h(z) = g(z) +

zg󸀠 (z) = H α (ϕ(z)). g(z)

Now from (5.2.5) we have g(z) = G α (ω(z)) and g󸀠 (z) = G󸀠α (ω(z)) ω󸀠 (z). Thus (5.2.5) becomes kω(z1 )G󸀠α (ω(z1 )) G α (ω(z1 )) + = H α (ϕ(z1 )), G α (ω(z1 )) which, since H α (z) = G α (z) +

zG󸀠α (z) , G α (z)

is equivalent to ̂ k (ω(z1 )) = H α (ϕ(z1 )), H

(5.2.6)

̂ k is defined in Lemma 5.2.3. and where H ̂ −1 (H α (ϕ(z))) is analytic in 𝔻, Lemma 5.2.3 shows that the function ψ(z) = H k ψ(0) = 0 and |ψ(z)| < 1 for z ∈ 𝔻. From (5.2.6), with ψ instead of ϕ, we therefore ̂ k (ψ(z1 )), which shows that ω(z1 ) = ψ(z1 ), since H ̂ k is univalent ̂ k (ω(z1 )) = H have H in 𝔻 and both ω(z1 ) and ψ(z1 ) are in 𝔻. Thus |ω(z1 )| = |ψ(z1 )| = ρ = |z1 |. On the other hand, since z1 ≠ 0, we deduce from the Schwarz lemma (see Section 2.1) that equality in |ω(z1 )| ⩽ |z1 | = |ψ(z1 )| occurs only when ω(z) = xz, with |x| = 1. Further, ̂ k (ψ) and H α (ϕ), if ϕ(z) = c1 z + c2 z2 + . . . , then equating the coefficients of z in H gives (1 − α)(k + 1)x = 2(1 − α)c1 and k = 2c1 /x − 1, which is greater than, or equal to 1, i.e., c1 ⩾ x. This, together with |x| = 1 and |c1 | = |ϕ󸀠 (0)| ⩽ 1, (see the Schwarz lemma with ϕ in place of ω) implies c1 = x. The Schwarz lemma also implies that equality occurs in |c1 | ⩽ 1 only when ϕ(z) = c1 z, i.e., when ϕ(z) = xz, and |x| = 1. Thus we have shown that h(z) = H α (xz) and g(z) = G α (xz), which contradicts our assumption that g does not have this form. Thus ρ = 1, and the proof is complete. We end this section by showing how Wilken and Feng in [263] proved MacGregor’s conjecture, thus solving the problem posed by Jack in 1971. Theorem 5.2.3. β(α) = inf Re G α (z) = G α (−1) when 0 ⩽ α < 1. z∈𝔻

In order to prove this theorem, Wilken and Feng used the following three lemmas, also given in [263]. Lemma 5.2.4. Let g be analytic and Re g(z) > 0 for z = re iθ ∈ 𝔻, and let g(−r) be real. Then the following two statements are equivalent: 1 1 (i) ⩾ , when |z| ⩽ r. Re g(z) g(−r) (ii) |g(z) − g(−r)/2| ⩽ g(−r)/2, when |z| ⩽ r.

72 | 5 Starlike and Convex Functions of Order α

Proof. The conclusion follows since 󵄨󵄨 g(−r) 󵄨󵄨󵄨󵄨2 g2 (−r) 󵄨󵄨 󵄨 ⩽ 󵄨󵄨g(z) − 󵄨󵄨 2 󵄨󵄨󵄨 4

⇔

|g(z)|2 − g(−r) Re g(z) ⩽ 0

⇔

1 1 ⩾ . Re g(z) g(−r)

The next lemma makes use of the following assumptions: ∙ μ is a positive measure on [0, 1]; ∙

g(t, z) is analytic in 𝔻 for each t ∈ [0, 1], and integrable with respect to t for each z ∈ 𝔻, and almost all t ∈ [0, 1];

∙

Re g(t, z) > 0 for z ∈ 𝔻, and t ∈ [0, 1];

∙

Re g(t, −r) > 0 is real, and Re[1/g(t, z)] ⩾ 1/g(t, −r) for all |z| ⩽ r, and t ∈ [0, 1].

Lemma 5.2.5. Let g(z) = ∫ g(t, z)dμ(t). Then, Re

1 1 ⩾ g(z) g(−r)

when |z| ⩽ r. Proof. It is obvious that Re g(z) > 0 when z ∈ 𝔻, and that g(−r) is real. So, by Lemma 5.2.4, 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 1 1 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨g(z) − g(−r)󵄨󵄨󵄨 = 󵄨󵄨󵄨∫ g(t, z)dμ(t) − ∫ g(t, −r)dμ(t)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 2 󵄨 󵄨󵄨 󵄨 1 󵄨 󵄨 ⩽ ∫ 󵄨󵄨󵄨g(t, z) − g(t, −r)󵄨󵄨󵄨 dμ(t) 2 󵄨󵄨 󵄨󵄨 1 1 ⩽ ∫ g(t, −r)dμ(t) = g(−r). 2 2 The lemma follows by applying Lemma 5.2.4 again. Now let γ = 2α − 1, 0 < α < 1, so that −1 < γ < 1, and G(z, γ) be given by G(z, γ) =

z γ [ ] 1 − z (1 − z)−γ − 1

(γ ≠ 0),

so that

γz = G(z, −γ) (|z| < 1, 0 < γ < 1). 1−z With this notation, Theorem 5.2.3 can, therefore, be rewritten in the following equivalent form. G(z, γ) +

Theorem 5.2.4. inf Re G(z, γ) = G(−1, γ), when −1 ⩽ γ < 1. z∈𝔻

The theorem holds for γ = 0 since lim G(z, γ) = G(z, 0) uniformly in |z| ⩽ r, 0 < r < 1. γ→0

The next lemma shows that it is enough to prove that the theorem holds only when 0 < γ < 1.

5.2 Inclusion Relationships | 73

Lemma 5.2.6. If Theorem 5.2.4 holds for 0 < γ < 1, then it also holds for −1 < γ < 1/2. Proof. If 0 < γ < 1, then Re G(z, −γ) = Re G(z, γ) + γ Re

z 1−z

1 ⩾ G(−1, γ) + γ (− ) = G(−1, −γ). 2 We can now give the proof of Theorem 5.2.4. Proof of Theorem 5.2.4 in the case when 0 < γ < 1. The result of Marx and Strochacker ([139] and [235]) shows that the theorem holds for α = 0, i.e., when γ = −1. Let 0 < γ < 1. Then since 1

1 1 Γ(1) = ∫ t γ−1 (1 − t)−γ dt, (1 − z)γ Γ(γ)Γ(1 − γ) 1 − tz 0

we have G(z, γ) =

1 , 1−z ∫ dμ(t) 1 − tz

[0,1]

where dμ(t) =

Γ(1) 1 t γ (1 − t)−γ dt. γ Γ(γ)Γ(1 − γ)

For the functions g(t, z) =

1−z 1 − tz

and

g(z) = ∫ g(t, z)dμ(t), [0,1]

we have that Re g(t, z) > 0, g(t, −r) is real, and also 1 1 − tz 1 + tr 1 = Re ⩾ = , Re g(t, z) 1−z 1+r g(t, −r) for |z| ⩽ r, 0 < r < 1 and 0 ⩽ t ⩽ 1. So from Lemma 5.2.5 we obtain Re G(z, γ) ⩾ G(−r, γ) for |z| ⩽ r and 0 < r < 1, and that Re G(z, γ) ⩾ G(−1, γ) for |z| < 1. Combining the results of MacGregor and Wilken and Feng given in this section we now obtain the answer to the question posed by Jack in 1971, formulated in the following theorem. Theorem 5.2.5. If f ∈ C(α), then f ∈ S∗ (β), where 1 − 2α { { { 22(1−α) − 2 , β = β(α) ≡ { 1 { { , 2 log 2 { The result is sharp for the function f α given in (5.1.4).

1 , 2 1 α= . 2

α ≠

74 | 5 Starlike and Convex Functions of Order α

In 1973 Brickman et al. generalized the second part of the Marx–Strohhäcker result (Theorem 5.2.1 (ii)). We state the result below without proof, which can be found in Brickman et al. [23, Theorem 10, p. 425, Corollary, p. 426]. Theorem 5.2.6. If f ∈ C(α), then for all z = re iθ ∈ 𝔻, { { { f(z) Re [ ]⩾{ { z { {

1 − (1 + r)2α−1 , (1 − 2α)r

α ≠

1 , 2

log(1 + r) , r

α=

1 . 2

The inequalities are sharp. Halim and Thomas [64], using a quite elementary method, generalized the above result as follows. Theorem 5.2.7. Let f ∈ C(α) and z = re iθ ∈ 𝔻. For n = 1, 2, . . . , define z

I n (z) =

1 ∫ I n−1 (t)dt, z 0

where I0 = f(z)/z. Then for n ⩾ 0, Re I n (z) ⩾ γ n (r), where

∞ (−r)k−1 A k (α) 1 < 1, ⩽ γ n (r) = ∑ 2 kn k=1

and An =

1 n ⋅ ∏(k − 2α), n! k=2

A1 = 1.

The result is sharp. Proof. From Theorem 5.2.6, for 0 ⩽ α < 1, we have ∞

Re I0 (z) ⩾ ∑ (−r)k−1 A k (α) = γ0 (r). k=1

Next, with t = ρe iθ , using induction we have Re I n (z) = Re

z

r

0

0

1 1 ∞ (−ρ)k−1 A k (α) dρ ∫ I n−1 (t)dt ⩾ ∫ ∑ z r k=1 k n−1

∞

(−r)k−1 A k (α) = ∑ = γ n (r). kn k=1 For n ⩾ 0 and 0 ⩽ α < 1, γ n (r) is absolutely convergent for 0 ⩽ r < 1, and so rearranging the terms appropriately shows that 1/2 < γ n (r) < 1.

5.3 Coefficient Theorems | 75

Finally it is worth pointing out that (5.2.3) states an important result, namely that the Libera integral operator, L : A → A, defined by z

2 L[f](z) = ∫ f(t)dt, z 0

preserves the class of convex functions, i.e., L[C] ⊂ C. The same holds for the classes of starlike and close-to-convex functions, also proved by Libera [117]. We also note that Mocanu et al. [155], and Miller and Mocanu in [148, Theorem 3.3g, p. 116] showed that the order of starlikeness of L[S∗ (α)], when −1/2 ⩽ α < 1, is α(2α − 1) −1 2(2−2α + α − 1) 1 − 1 = 0.629 . . . 2(1 − log 2) 1 − 1 = 0.294 . . . 2(2 log 2 − 1)

if

α ≠

if

1 and α ≠ 0, 2 1 α= , 2

if

α = 0.

5.3 Coefficient Theorems Sharp estimates for the coefficients of starlike and convex functions of order α were given by Robertson [211] in 1936, who used the idea of majorization defined as follows. ∞ n n Definition 5.3.1. For z ∈ 𝔻, let ϕ(z) = ∑∞ n=0 ϕ n z and ψ(z) = ∑ n=0 ψ n z , where ψ n ⩾ 0 for all n ⩾ 0. Then ϕ(z) is majorized by ψ(z), written as ϕ(z) ≪ ψ(z), if for n ⩾ 0, |ϕ n | ⩽ ψ n . n ∗ Theorem 5.3.1. Let f(z) be given by f(z) = z + ∑∞ n=2 a n z . If f ∈ S (α), then for n ⩾ 2,

|a n | ⩽

n 1 ⋅ ∏(k − 2α), (n − 1)! k=2

and if f ∈ C(α), |a n | ⩽

1 n ⋅ ∏(k − 2α). n! k=2

Both inequalities are sharp. Proof. Let f ∈ S∗ (α) with

∞

f(z) = z + ∑ a n z n , n=2

and g(z) = and let p(z) =

∞ zf 󸀠 (z) = 1 + ∑ bn zn , f(z) n=1

∞ 1 [g(z) − α] = 1 + ∑ p n z n . 1−α n=1

76 | 5 Starlike and Convex Functions of Order α Then b n = p n (1 − α), Re g(z) > α

and

Re p(z) > 0

(z ∈ 𝔻),

and |p n | ⩽ 2 for n ⩾ 1 (see Theorem 3.1.2), and so for n ⩾ 1, |b n | ⩽ 2(1 − α). Further, since ∞ n zf 󸀠 (z) z + ∑∞ n=2 na n z = 1 + ∑ b n z n = g(z), = n f(z) z + ∑∞ n=2 a n z n=1

it follows that for each integer n, a n is a polynomial in b i , with positive multiples of the coefficients b i . Thus, |a n | is less than or equal to the expression obtained by replacing |b i | with 2(1 − α) in that polynomial. Therefore, 1 + (1 − 2α)z zf 󸀠 (z) ≪ f(z) 1−z and f(z) ≪ φ(z) := Thus, |a n | ⩽ c n =

∞ z = z + ∑ cn zn . (1 − z)2(1−α) n=2

(2 − 2α)(3 − 2α) ⋅ ⋅ ⋅ (n − 2α) . (n − 1)!

The function φ(z) above shows that the result is sharp. The result for f ∈ C(α) follows at once from Alexander’s theorem, and is sharp when f(z) = ψ(z), given by 1 − (1 − z)2α−1 { , { { φ(t) 2α − 1 ψ(z) = ∫ dt = { { t { 0 − log(1 − z), { z

1 , 2 1 α= . 2

α ≠

Sharp estimates for the Fekete–Szegő problem for starlike and convex functions of given order, were provided by Keogh and Merkes in 1969 [88] as follows: n Theorem 5.3.2. Let f(z) = z + ∑∞ n=2 a n z , 0 ⩽ α < 1 and μ be any complex number. (i) If f ∈ S∗ (α), then |a3 − μa22 | ⩽ (1 − α) max{1, |2(1 − α)(2μ − 1) − 1|}. 1 (ii) If f ∈ C(α), then |a3 − μa22 | ⩽ (1 − α) max{1, |3(1 − μ) + α(3μ − 2)|}. 3 Both inequalities are sharp.

We omit the proofs, which are similar to that used in Theorem 4.1.5, noting that (i) is sharp when f(z) = z/(1 − z)2(1−α) or f(z) = z/(1 − z2 )1−α , obtained when p(z) = (1 + z)/ (1 − z) and p(z) = (1 + z2 )/(1 − z2 ) in [zf 󸀠 (z)/f(z) − α]/(1 − α) = p(z), respectively. Inequality (ii) is sharp when f(z) is given by either zf 󸀠 (z) = z/(1 − z)2(1−α) , or 󸀠 zf (z) = z/(1 − z2 )1−α , both of which are convex of order α (see (5.2.3)).

5.3 Coefficient Theorems | 77

As mentioned in Chapter 4, since S∗ ⊂ S and k ∈ S∗ , the problem of finding sharp bounds for the inverse coefficients of f ∈ S∗ is solved. For S∗ (α) and C(α), the situation is different and presents a highly technical problem, and we refer the reader to the work of Kapoor and Mishra [85], and Ali and Vasudevarao [6] for the class S∗ (α). Since sharp bounds have only been obtained for the initial inverse coefficients when f ∈ C, the problem for C(α) is clearly more difficult. We next consider bounds for the second Hankel determinant H2 (2) = |a2 a4 − a23 | for starlike and convex functions of order α. For f ∈ S∗ (α), Krishna and Ramreddy [258] dealt with the case 0 ⩽ α ⩽ 1/2. We give the sharp bound for α ⩾ 0 [232]. For f ∈ C(α), we state the estimate also found in [258], which is probably not sharp. n Theorem 5.3.3. Let f(z) = z + ∑∞ n=2 a n z and α ⩾ 0. 1 (i) If f ∈ S∗ (α), then H2 (2) ⩽ (1 − α)2 |(3 − 2α)(2α − 1)|. 3 (1 − α)2 17α 2 − 36α + 36 (ii) If f ∈ C(α), then H2 (2) ⩽ . ⋅ 144 α 2 − 2α + 2 Inequality (i) is sharp.

Proof. Let f ∈ S∗ (α). Then from the proof of Theorem 5.3.1 it is a simple matter to find expressions for a2 , a3 and a4 in terms of the coefficients of p(z) to obtain 󵄨 (1 − α)2 󵄨󵄨 󵄨󵄨a2 a4 − a23 󵄨󵄨󵄨 = 󵄨 󵄨 12 Now use (3.1.6) and (3.1.8), so that

󵄨 󵄨 ⋅ 󵄨󵄨󵄨󵄨4p1 p3 − 3p22 − (1 − α)2 p41 󵄨󵄨󵄨󵄨 .

󵄨 (1 − α)2 󵄨󵄨 󵄨󵄨a2 a4 − a23 󵄨󵄨󵄨 = 󵄨 󵄨 12 󵄨󵄨 1 󵄨 󵄨 ⋅ 󵄨󵄨4p1 ⋅ [p31 + 2(4 − p21 )p1 x − p1 (4 − p21 )x2 + 2(4 − p21 )(1 − |x|2 )ζ ] 󵄨󵄨 4 󵄨󵄨 1 2 󵄨 − 3 ⋅ [p21 + x(4 − p21 )] − (1 − α)2 p41 󵄨󵄨󵄨 , 󵄨󵄨 4 for some complex-valued x with |x| ⩽ 1 and some complex-valued ζ with |ζ| ⩽ 1. A simple rotation allows us to assume that, without loss in generality, we can write p1 = p, with 0 ⩽ p ⩽ 2, and so applying the triangle inequality we obtain 󵄨󵄨 󵄨 (1 − α)2 󵄨󵄨󵄨 󵄨󵄨a2 a4 − a23 󵄨󵄨󵄨 = ⋅ 󵄨󵄨󵄨 − (3 − 8α + 4α 2 )p4 + 8p(4 − p2 ) 󵄨 󵄨 48 󵄨 󵄨󵄨 2 + 2p (4 − p2 )|x| − (2 − p)(6 − p)(4 − p2 )|x|2 󵄨󵄨󵄨󵄨 󵄨 (1 − α)2 ⩽ ⋅ [|3 − 8α + 4α 2 | ⋅ p4 + 8p(4 − p2 ) 48 + 2p2 (4 − p2 )|x| + (2 − p)(6 − p)(4 − p2 )|x|2 ] := F(p, |x|). Since F(p, |x|) is an increasing function of |x| on [0, 1], it follows that (1 − α)2 󵄨 󵄨󵄨 󵄨󵄨a2 a4 − a23 󵄨󵄨󵄨 ⩽ F(p, 1) = ⋅ [(|3 − 8α + 4α 2 | − 3)p4 + 48] . 󵄨 󵄨 48

78 | 5 Starlike and Convex Functions of Order α Finally for α ⩾ 0, we deduce that 󵄨 1 󵄨󵄨 󵄨󵄨a2 a4 − a23 󵄨󵄨󵄨 ⩽ (1 − α)2 |(3 − 2α)(2α − 1)|. 󵄨 3 󵄨 The inequality (i) is sharp when f(z) = f(z) = z/(1 − z)2(1−α) , which is starlike of order α. Inequality (ii) follows by repeating the above argument using p(z) =

∞ 1 zf 󸀠󸀠 (z) [1 + 󸀠 − α] = 1 + ∑ p n z n . 1−α f (z) n=1

It is not clear whether the inequality (ii) is sharp. A common approach in the study of analytic functions is to restrict the elements of the Taylor expansion of f(z) so that, for example, they are all of the same sign or certain coefficients are missing (lacunary or gap series). In this direction, we give an example of a result similar to Theorem 4.1.1, which considers normalized analytic functions, whose Taylor expansion has only negative coefficients, Chatterjea [29]. We omit the proof. k Theorem 5.3.4. Let n be a positive integer and f(z) = z − ∑∞ k=n+1 a k z be analytic in 𝔻 with a k ⩾ 0 for all integers k > n. Then (i) f ∈ S∗ (α) for 0 ⩽ α < 1 if, and only if, ∞

∑ (k − α)a k ⩽ 1 − α. k=n+1

(ii) f ∈ C(α) for 0 ⩽ α < 1 if, and only if, ∞

∑ k(k − α)a k ⩽ 1 − α. k=n+1

5.4 Sufficient Conditions for Starlikeness and Convexity of Certain Order Many authors have given sufficient conditions for functions to belong to S∗ (α) or C(α). In this section, we give a selection of what we consider are some of the more interesting examples.

5.4.1 Sufficient Conditions on |f 󸀠󸀠 (z)| The problem of finding λ > 0 such that the condition |f 󸀠󸀠 (z)| ⩽ λ, for z ∈ 𝔻, implies f ∈ S∗ was first considered by Mocanu [151], who showed that the implication was

5.4 Sufficient Conditions

| 79

valid when λ = 2/3. Later, Ponnusamy and Singh improved this to λ = 2/√5, and finally Obradović [173] showed that λ = 1 could be taken, and that this constant is the best possible. Tuneski [252], using similar techniques to Obradović [173], considered the problem of finding λ such that the condition |f 󸀠󸀠 (z)| ⩽ λ, for z ∈ 𝔻, implies f ∈ S∗ (α) or f ∈ C(α). Tuneski’s results below uses the following subordination properties, which are of interest in themselves. Lemma 5.4.1 (Hallenbeck and Ruscheweyh [66]). Let G be convex and univalent in 𝔻, with G(0) = 1, and F be analytic in 𝔻, with F(0) = 1. Then if F(z) ≺ G(z) in 𝔻 for all n ⩾ 0, z

z

(n + 1)z−n−1 ∫ t n F(t)dt ≺ (n + 1)z−n−1 ∫ t n G(t)dt. 0

0

Lemma 5.4.2 (Miller et al. [150]). Let F and G be functions analytic in the unit disk 𝔻 with F(0) = G(0). If H(z) = zG󸀠 (z) is starlike in 𝔻 and zF 󸀠 (z) ≺ zG󸀠 (z), then z

F(z) ≺ G(z) = G(0) + ∫

H(t) dt. t

0

Theorem 5.4.1. If f ∈ A, 0 ⩽ α < 1, and |f 󸀠󸀠 (z)| ⩽

2(1 − α) := k 2−α

(z ∈ 𝔻),

then f ∈ S∗ (α). The result is sharp. Proof. Noting that the condition of the theorem is equivalent to zf 󸀠󸀠 (z) ≺ kz and choosing F(z) = zf 󸀠󸀠 (z) + 1, G(z) = kz + 1, and n = 0 in Lemma 5.4.1, we obtain f 󸀠 (z) −

f(z) kz ≺ , z 2

which is equivalent to kz 󸀠 f(z) 󸀠 ] ≺ z (1 + ) , z 2

(5.4.1)

f(z) zf 󸀠 (z) k [ − 1] ≺ z. z f(z) 2

(5.4.2)

z[ and to

From (5.4.1) and Lemma 5.4.2, taking F(z) = f(z)/z and G(z) = 1 + kz/2, we obtain f(z)/z ≺ 1 + kz/2, which implies 1 − k/2 < |f(z)/z| < 1 + k/2 for z ∈ 𝔻. From this relation and (5.4.2), we conclude that (1 −

󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 k 󵄨󵄨󵄨 zf 󸀠 (z) 󵄨 󵄨 f(z) 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨 zf 󸀠 (z) 󵄨 k ) 󵄨󵄨󵄨 − 1󵄨󵄨󵄨󵄨 < 󵄨󵄨󵄨 − 1󵄨󵄨󵄨󵄨 < 󵄨󵄨 󵄨󵄨󵄨 󵄨 󵄨 2 󵄨󵄨 f(z) 󵄨󵄨 󵄨 z 󵄨 󵄨󵄨 f(z) 󵄨󵄨 2

(z ∈ 𝔻),

80 | 5 Starlike and Convex Functions of Order α

i.e.,

󵄨󵄨 󸀠 󵄨󵄨 k 󵄨󵄨 zf (z) 󵄨󵄨 󵄨󵄨 󵄨󵄨 < − 1 󵄨󵄨 f(z) 󵄨󵄨 2 − k (z ∈ 𝔻). 󵄨 󵄨 󸀠 Thus Re{zf (z)/f(z)} > 1 − k/(2 − k) = α, for z ∈ 𝔻, i.e., f ∈ S∗ (α). Further, the function f(z) = z + (k + ϵ)z2 /2, 0 < k ⩽ 1, 0 < ϵ < 1, shows that the result is sharp, i.e., k as defined in the statement of the theorem is the largest value possible for a given α, since |f 󸀠󸀠 (z)| = k + ϵ > k, and zf 󸀠 (z) 2[1 + (k + ϵ)z] = f(z) 2 + (k + ϵ)z is smaller than α when z is real and close to −1, in which case f ∉ S∗ (α). Theorem 5.4.2. If f ∈ A, 0 ⩽ α < 1 and |f 󸀠󸀠 (z)| ⩽

1−α := k 2−α

(z ∈ 𝔻),

then f ∈ C(α). The result is sharp. Proof. The condition |f 󸀠󸀠 (z)| ⩽ k, for z ∈ 𝔻, is equivalent to zf 󸀠󸀠 (z) ≺ kz

(5.4.3)

for z ∈ 𝔻, and again using Lemma 5.4.2 with F(z) = f 󸀠 (z) and G(z) = 1 + kz, we deduce that (5.4.4) f 󸀠 (z) ≺ 1 + kz. From (5.4.3), with p(z) = 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) we conclude that [p(z) − 1] f 󸀠 (z) ≺ kz.

(5.4.5)

Now suppose that there exists z0 ∈ 𝔻 such that p(z0 ) = α + ix. From (5.4.4) and (5.4.5) it follows that 1 − k < |f 󸀠 (z0 )| < 1 + k (5.4.6) and

󵄨󵄨 󵄨 󵄨󵄨(p(z0 ) − 1) f 󸀠 (z0 )󵄨󵄨󵄨 < k. 󵄨 󵄨

(5.4.7)

Further, using (5.4.6) we obtain 󵄨2 󵄨󵄨 󵄨󵄨(p(z0 ) − 1) f 󸀠 (z0 )󵄨󵄨󵄨 = |α − 1 + ix|2 |f 󸀠 (z0 )|2 󵄨 󵄨 > [(α − 1)2 + x2 ] (1 − k)2 = (α − 1)2 (1 − k)2 + x2 (1 − k)2 ⩾ (α − 1)2 (1 − k)2 = k 2 , for α = (1 − 2k)/(1 − k) (i.e., k = (1 − α)/(2 − α)), which contradicts (5.4.7). Therefore, we have proved that under the conditions of the theorem Re[1 + zf 󸀠󸀠 (z)/f 󸀠 (z)] > α is true for any z ∈ 𝔻, i.e., f ∈ C(α).

5.4 Sufficient Conditions |

81

To show that the result is sharp, let f(z) = z + (k + ϵ)z2 /2, 0 < k ⩽ 1/2, and ϵ > 0, so that |f 󸀠󸀠 (z)| = k + ϵ > k, and Re [1 +

1 + 2z(k + ϵ) zf 󸀠󸀠 (z) ]= f 󸀠 (z) 1 + z(k + ϵ)

is smaller than α when z is real and close to −1, i.e., f ∉ C(α). We note that other expressions such as f 󸀠 (z) − (1 − γ)f(z)/z and zf 󸀠󸀠 (z) − γf 󸀠 (z), which imply f ∈ S∗ (α) and f ∈ C(α) are considered by Tuneski in [253].

5.4.2 On a Class Defined by Silverman Silverman [227] investigated the properties of functions defined in terms of the quotient of the analytical representations of convexity and starlikeness, i.e., 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) . zf 󸀠 (z)/f(z) Precisely, he considered 󵄨󵄨 󵄨󵄨󵄨 󵄨 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) 󵄨󵄨 < b, z ∈ 𝔻} , Gb = {f ∈ A : 󵄨󵄨󵄨󵄨 − 1 󵄨󵄨 󵄨󵄨 zf 󸀠 (z)/f(z) 󵄨󵄨 and claimed that G b ⊂ S∗ (

(5.4.8)

2 ) √ 1 + 1 + 8b

is sharp for all 0 < b ⩽ 1. Using the Clunie–Jack lemma Obradović and Tuneski [178] corrected this result by proving the following. Theorem 5.4.3. If 0 < b ⩽ 1, then G b ⊂ S∗ (

1 ). 1+b

The result is sharp. Proof. We prove the inclusion by showing that zf 󸀠 (z) 1 ≺ . f(z) 1 + bz Put p(z) = zf 󸀠 (z)/f(z) = 1/(1 + bω(z)). Then 󵄨󵄨 󵄨󵄨 󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) 󵄨󵄨 󵄨󵄨 zp (z) 󵄨󵄨 󸀠 󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨 − 1 󵄨󵄨 zf 󸀠 (z)/f(z) 󵄨󵄨 󵄨󵄨 p2 (z) 󵄨󵄨󵄨 = | − bzω (z)|. 󵄨 󵄨 󵄨 󵄨

(5.4.9)

We need to show that |ω(z)| < 1, for z ∈ 𝔻. Supposing not, then by the Clunie–Jack lemma, there exists z0 such that |ω(z0 )| = 1 and z0 ω󸀠 (z0 ) = kω(z0 ), for k ⩾ 1. For

82 | 5 Starlike and Convex Functions of Order α such z0 it follows from (5.4.9) that |z0 p󸀠 (z0 )/p2 (z0 )| = | − bkω(z0 )| = bk ⩾ b, which contradicts the assumption that f ∈ Gb . To see that the result is the best possible, let f(z) = z/(1 + bz), so that (5.4.9) gives |zp󸀠 (z)/p2 (z)| = | − bz| < b, i.e., f ∈ Gb and zf 󸀠 (z)/f(z) = 1/(1 + bz). A more general class was introduced by Singh and Tuneski in [231], who considered 󵄨󵄨 󵄨󵄨 󵄨 1 − γ + zf 󸀠󸀠 (z)/f 󸀠 (z) 󵄨󵄨 󵄨󵄨 < λ, z ∈ 𝔻} , Gλ,γ = {f ∈ A : 󵄨󵄨󵄨󵄨 − (1 − γ) 󸀠 󵄨󵄨 zf (z)/f(z) 󵄨󵄨 󵄨

(5.4.10)

for γ ⩾ 0 and λ > 0. They proved that Gλ,γ ⊆ S∗ (α)

when

λ < 1 + γ and α = (1 + γ − 2λ)/(1 + γ − λ),

(5.4.11)

and Gλ,γ ⊆ C(γ)

when

0 ⩽ γ < 1 and λ = (1 − γ2 )/√2(1 + γ2 ).

When γ = 1 in (5.4.11) we obtain the following criteria concerning the expression f(z)f 󸀠󸀠 (z)/f 󸀠2 (z) due to Robertson [212, Theorem 2, p. 139] and Miller and Mocanu [148, Theorem 5.3a, p. 284]. Theorem 5.4.4. Let f ∈ A, with f(z)/z ≠ 0. If there exists 0 < k ⩽ 2, such that 󵄨󵄨 󸀠󸀠 󵄨󵄨 󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨 zf (z) 󵄨󵄨 󵄨 zf (z) 󵄨󵄨 󵄨󵄨 󸀠 󵄨󵄨 ⩽ k 󵄨󵄨󵄨 󵄨 󵄨󵄨 f (z) 󵄨󵄨 󵄨󵄨 f(z) 󵄨󵄨󵄨 󵄨 󵄨 󵄨 󵄨 then

(z ∈ 𝔻),

2 zf 󸀠 (z) ≺ , f(z) 2 + kz

and so f ∈ S∗ (α), with α = 2/(2 + k).

6 Strongly Starlike and Convex Functions 6.1 Definitions In the last chapter we saw how results with functions f ∈ S∗ and C could be extended when the half-plane Re p(z) > 0 was generalized so that Re p(z) > α, where 0 ⩽ α < 1. In this chapter, we consider another natural extension to the definition of S∗ and C, by restricting the values taken by Re p(z) > 0, so that they lie in the sector | arg p(z)| < πβ/2, where 0 < β ⩽ 1. Functions satisfying this condition are called strongly starlike (convex), which we define as follows. Definition 6.1.1. Let f ∈ A. Then f is strongly starlike of order β if, and only if, for 0 < β ⩽ 1, 󵄨󵄨 󵄨 zf 󸀠 (z) 󵄨󵄨󵄨 πβ 󵄨󵄨 󵄨󵄨arg 󵄨󵄨 < (z ∈ 𝔻). 󵄨󵄨 f(z) 󵄨󵄨󵄨 2 󵄨 We denote these functions by S∗∗ (β). Similarly Definition 6.1.2. A function f is strongly convex of order β if, and only if, for 0 < β ⩽ 1, 󵄨󵄨 󵄨 zf 󸀠󸀠 (z) 󵄨󵄨󵄨 πβ 󵄨󵄨 󵄨󵄨arg (1 + 󸀠 )󵄨󵄨󵄨 < (z ∈ 𝔻). 󵄨󵄨 f (z) 󵄨󵄨 2 󵄨 We denote these functions by C∗∗ (β). As before, the following extension to Alexander’s theorem is obvious. Theorem 6.1.1. f ∈ C∗∗ (β) if, and only if, zf 󸀠 ∈ S∗∗ (β). We first note that from the above definition of S∗∗ (β), we can express functions f ∈ S∗∗ (β) via zf 󸀠 (z) (6.1.1) = p β (z), f(z) where p ∈ P. Here and throughout the chapter, powers are taken as principal values. We are, therefore, led at once to consider the problem of raising an infinite series to a power, which clearly creates difficulties. Thus, properties of functions in S∗∗ (β) and C∗∗ (β) should be more difficult to obtain than those for functions in S∗ (α) and C(α). In this chapter, we include what we consider the (relatively few) significant known results for functions in S∗∗ (β) and C∗∗ (β).

6.2 Strongly Starlike Functions A complete set of sharp upper and lower bounds for |f(z)|, |f 󸀠 (z)| and |zf 󸀠 (z)|/|f(z)| when f belongs to S∗∗ (β) and C∗∗ (β) are more difficult to obtain. Noting that S∗∗ (1) = https://doi.org/10.1515/9783110560961-006

84 | 6 Strongly Starlike and Convex Functions S∗ , we first give the following sharp upper bounds for |f(z)|, |f 󸀠 (z)|, and |zf 󸀠 (z)|/|f(z)| when f ∈ S∗∗ (β) (see Brannan and Kirwan [21]), which, in particular, shows that if 0 < β < 1, then functions in S∗∗ (β) are bounded. Theorem 6.2.1. Let f ∈ S∗∗ (β) for 0 < β < 1. Then for z = re iθ ∈ 𝔻, ∞

1 } := rM(β). (2k + 1)(2k + 1 − β) k=0

(6.2.1)

1+r β ) , 1−r

(6.2.2)

|f(z)| < r exp {2β ∑ Also

|zf 󸀠 (z)| < rM(β) ( and

󵄨󵄨 󸀠 󵄨󵄨 1+r β 󵄨󵄨 zf (z) 󵄨󵄨 󵄨󵄨 < ( 󵄨󵄨 (6.2.3) ) . 󵄨󵄨 f(z) 󵄨󵄨 1−r 󵄨 󵄨 The inequalities are sharp in the sense that there exists a function in S∗∗ (β) for which the supremum over the open unit disk of the left hand sides of the inequalities equals their right hand sides. Proof. First note that from (6.1.1), z

log

f(z) ds = ∫[p β (s) − 1] , z s 0

and so taking real parts, and using the fact that p(z) ≺ (1 + z)/(1 − z), we obtain (with z = re iθ ) 󵄨󵄨 f(z) 󵄨󵄨 r dt 󵄨 󵄨󵄨 log 󵄨󵄨󵄨 󵄨 = ∫ Re{p β (te iθ ) − 1} 󵄨󵄨 z 󵄨󵄨󵄨 t 0

r

⩽ ∫ [( 0 1

< ∫ [( 0

1+t β dt ) − 1] 1−t t 1+t β dt ) − 1] 1−t t

1

∞ ∞ β β dt = ∫ { ∑ ( )t k − ∑ ( )(−1)k t k } (1 − t)−β k k t k=0 k=0 0

∞

=2∑( k=0 ∞

=2∑( k=0

1

β ) ∫ t2k (1 − t)−β dt 2k + 1 0

β )B(2k + 1, 1 − β), 2k + 1

(6.2.4)

6.2 Strongly Starlike Functions

|

85

where B(p, q) denotes the classical beta function. Using elementary properties of the beta function, it follows that (

β β(β − 1) . . . (β − 2k) Γ(2k + 1)Γ(1 − β) ⋅ , )B(2k + 1, 1 − β) = (2k + 1)! Γ(2k + 2 − β) 2k + 1 β . = (2k + 1)(2k + 1 − β)

Substituting into (6.2.4) and exponentiating, gives the first inequality of the theorem. The second inequality follows at once from (6.1.1), and the third inequality follows on using the subordination principle. The equation 1+z β zf 󸀠 (z) =( ) f(z) 1−z defines a function f ∈ S∗∗ (β) for which the supremum over 𝔻 of the moduli in the left hand sides of (6.2.1), (6.2.2), and (6.2.3), equals the corresponding right hand sides, respectively.

Noting that functions in S∗∗ (β) when 0 < β < 1 are bounded for z = re iθ ∈ 𝔻, we now find a bound for the length L(1) of the image of |z| = 1 under f (see Brannan and Kirwan [21]). Theorem 6.2.2. Let f ∈ S∗∗ (β) and 0 < β < 1. Then 2π

L(1) = ∫ |f 󸀠 (e iθ )|dθ ⩽ 2πM(β) sec (

πβ ), 2

0

with M(β) as in Theorem 6.2.1. Proof. In view of Theorem 6.2.1, we have 2π

2π

L(1) = ∫ |f 󸀠 (e iθ )|dθ ⩽ M(β) ∫ |p(e iθ )|β dθ 0

0 2π 󵄨

󵄨󵄨 1 + e iθ 󵄨󵄨󵄨β 󵄨󵄨 dθ ⩽ M(β) ∫ 󵄨󵄨󵄨󵄨 󵄨 󵄨󵄨 1 − e iθ 󵄨󵄨󵄨 0 2π

= M(β) ∫ | cot(θ/2)|β dθ 0

86 | 6 Strongly Starlike and Convex Functions π/2

= 4M(β) ∫ (cot ϕ)β dϕ 0

1+β 1−β , ) 2 2 1+β 1−β = 2M(β) Γ ( ) Γ( ) 2 2 πβ = 2πM(β) sec ( ) . 2 = 2M(β) B (

We next give the sharp inequality for |arg(f(z)/z)| when f ∈ S∗∗ (β) (see Ma and Minda [130]). Theorem 6.2.3. Let f ∈ S∗∗ (β). Then for z = re iθ ∈ 𝔻, 󵄨󵄨 f0 (z) f(z) 󵄨󵄨󵄨󵄨 󵄨󵄨 ), 󵄨 ⩽ max (arg 󵄨󵄨arg z 󵄨󵄨󵄨 |z|=r z 󵄨󵄨

where f0 ∈ S∗∗ (β) is defined by

zf0󸀠 (z) 1+z β =( ) . f0 (z) 1−z Proof. As in the proof of Theorem 6.2.1 we can write z

{ f(z) ds } = exp {∫ [p β (s) − 1] , z s } {0 } for p ∈ P, and so

z

{ ds } f(z) ] = Im {∫ [p β (s) − 1] . arg [ z s } {0 } Next, note that since p β (z) − 1 is subordinate to (

1+z β ) − 1, 1−z

it follows that z

z

ds 1+s β ds ∫ [p (s) − 1] ≺ ∫ [( ) − 1] := g(z). s 1−s s β

0

0

zg󸀠

is convex in 𝔻, so is g, whose power series has real coefficients. Thus by Since the subordination principle, it follows that 󵄨󵄨 󵄨󵄨 z 󵄨󵄨 f(z) 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 { ds }󵄨󵄨󵄨󵄨 󵄨󵄨 β 󵄨󵄨󵄨arg z 󵄨󵄨󵄨 = 󵄨󵄨󵄨Im {∫ [p (s) − 1] s }󵄨󵄨󵄨 󵄨 󵄨 󵄨󵄨 󵄨 󵄨 {0 }󵄨󵄨 ⩽ max Im g(z) |z|=r

= max [arg |z|=r

f0 (z) ]. z

6.3 Coefficient Theorems |

87

We note that in using the subordination principle, inequality becomes equality only when p(z) = (1 + e iϕ z)/(1 − e iϕ z) for some real ϕ, and so equality can occur only for some rotation of f0 . On the other hand, inequality becomes equality for some |z| = r, provided that f is a rotation of f0 .

6.3 Coefficient Theorems In Chapter 5, it was shown that sharp inequalities can be obtained for the coefficients of the Taylor series when f ∈ S∗ (α). However, as was mentioned above, since the definition of functions in S∗∗ (β) involves raising a power series to a power, obtaining exact inequalities for the coefficients of the Taylor series when f ∈ S∗∗ (β) is more difficult. Indeed, sharp estimates are only known for the first four coefficients (Ali and Singh [10], Brannan et al. [20], and Lecko and Sim [105]). We note at this point that there was an error in the final part of the proof of the estimate for |a4 | in [10], which was corrected by Lecko and Sim [105]. In the following, we give the correction. n Theorem 6.3.1. Let f ∈ S∗∗ (β) and be given by f(z) = z + ∑∞ n=2 a n z . Then

|a2 | ⩽ 2β,

{ { β, |a3 | ⩽ { { 3β 2 , {

2β { , { { 3 |a4 | ⩽ { { { 2β (1 + 17β 2 ), { 9

0< β⩽

1 ⩽ β ⩽ 1, 3 0< β⩽√

√

1 , 3

2 , 17

2 ⩽ β ⩽ 1. 17

All the inequalities are sharp. n Proof. Let p(z) = 1 + ∑∞ n=1 p n z . Then equating coefficients in (6.1.1) gives

a2 = βp1 β 1 − 3β 2 (p2 − p1 ) 2 2 β 5β − 2 17β 2 − 15β + 4 3 a4 = (p3 + p1 p2 + p1 ) . 3 2 12 a3 =

(6.3.1)

Since |p1 | ⩽ 2, the inequality |a2 | ⩽ 2β is trivial, and the inequality for |a3 | follows easily from Theorem 3.1.7. For the sharp inequality of |a4 |, we give a modification of the proof given by Ali and Singh [10]. We first note from (6.3.1) that the coefficients of p3 , p1 p2 and p31 in the expression for a4 are all positive provided 2/5 ⩽ β ⩽ 1, and so using the inequality |p n | ⩽ 2 for

88 | 6 Strongly Starlike and Convex Functions n = 1, 2 and 3, in this case we obtain 󵄨 󵄨 β 󵄨󵄨 5β − 2 17β 2 − 15β + 4 3 󵄨󵄨󵄨 2β p1 p2 + p1 󵄨󵄨󵄨 ⩽ (1 + 17β 2 ). |a4 | = 󵄨󵄨󵄨󵄨p3 + 3 󵄨󵄨 2 12 9 󵄨󵄨 We use Theorem 3.1.8 and write B = (2 − 5β)/4, so that 0 ⩽ B ⩽ 1 is satisfied when 0 < β ⩽ 2/5. Then from Theorem 3.1.8, and the inequality |p1 | ⩽ 2, provided that β ⩾ √2/17, we obtain, 󵄨 󵄨 β 󵄨󵄨 5β − 2 17β 2 − 15β + 4 3 󵄨󵄨󵄨 p1 p2 + p1 󵄨󵄨󵄨 |a4 | = 󵄨󵄨󵄨󵄨p3 + 3 󵄨󵄨 2 12 󵄨󵄨 󵄨󵄨 󵄨󵄨 2 17β − 2 3 󵄨󵄨 β󵄨 = 󵄨󵄨󵄨󵄨p3 − 2Bp1 p2 + Bp31 + p1 󵄨󵄨󵄨 3 󵄨󵄨 12 󵄨󵄨 2β 2 β (1 + 17β 2 ), ⩽ [2 + (17β 2 − 2)] = 3 3 9 which establishes the inequality for |a4 | on the interval √2/17 ⩽ β ⩽ 2/5. Finally, with B = (2 − 5β)/4 and D = (17β 2 − 15β + 4)/12 in (6.3.1), we have a4 =

β (p3 − 2Bp1 p2 + Dp31 ) . 3

Since 0 ⩽ B ⩽ 1 is satisfied when 0 < β ⩽ 2/5, and B(2B − 1) ⩽ D ⩽ B is satisfied when 0 < β ⩽ √2/17, applying Theorem 3.1.8 now gives the inequality for |a4 | on the interval 0 < β ⩽ √2/17, which completes the proof. To see that the above inequalities are sharp, we choose f(z) as follows. The inequality |a2 | ⩽ 2β is sharp if, and only if, β

zf 󸀠 (z) 1 + e iϕ z =( ) . f(z) 1 − e iϕ z When 0 < β < 1/3, the inequality |a3 | ⩽ β is sharp if, and only if, β

zf 󸀠 (z) 1 + e iϕ z2 =( ) . f(z) 1 − e iϕ z2 When 1/3 < β ⩽ 1, the inequality |a3 | ⩽ 3β 2 is sharp if, and only if, β

zf 󸀠 (z) 1 + e iϕ z =( ) . f(z) 1 − e iϕ z When β = 1/3, the inequality |a3 | ⩽ 1/3 is sharp if, and only if, zf 󸀠 (z) 1 + e2iϕ z2 1 + e iϕ z )] ) + (1 − λ) ( = [λ ( f(z) 1 − e iϕ z 1 − e2iϕ z2 where 0 ⩽ λ ⩽ 1.

1/3

,

6.3 Coefficient Theorems | 89

The first inequality for |a4 | is sharp if, and only if, β

1 + e iϕ z3 zf 󸀠 (z) =( ) , f(z) 1 − e iϕ z3 and the second inequality for |a4 | is sharp if, and only if, β

zf 󸀠 (z) 1 + e iϕ z ) . =( f(z) 1 − e iϕ z The method used in extending the Fekete–Szegő theorem from S∗ to S∗∗ (β) (Ma and Minda [130]) differs from that given by Keogh and Merkes [88], and so we include the complete proof in the following. n Theorem 6.3.2. Let f ∈ S∗∗ (β) and be given by f(z) = z + ∑∞ n=2 a n z . Then

{ β 2 (3 − 4μ), { { { { { { |a3 − μa22 | ⩽ { β, { { { { { 2 { β (4μ − 3), {

3 − 1/β , 4 3 + 1/β 3 − 1/β ⩽μ⩽ , 4 4 3 + 1/β ⩽ μ. 4 μ⩽

All the inequalities are sharp. Proof. Since functions in S∗∗ (β) are invariant under rotation, it is sufficient to prove the theorem for Re(a3 − μa22 ). From (6.3.1) we have a3 − μa22 =

β 1 [p2 + (3β − 4μβ − 1)p21 ] . 2 2 2π

Equating coefficients in (3.1.20) we obtain p n = 2 ∫0 e−ins dν(s) for n ⩾ 1, and so Re{a3 −

μa22 }

2π 2π { { [ = β { ∫ cos(2s)dν(s) + (3β − 4μβ − 1) [( ∫ cos(s)dν(s)) { {0 [ 0

2

2

} ]} − ( ∫ sin(s)dν(s)) ]} . } 0 ]} If μ ⩽ (3 − 1/β)/4, then 3β − 4μβ − 1 ⩾ 0, and so 2π

Re{a3 −

μa22 }

2π 2π { { ⩽ β { ∫ cos(2s)dν(s) + (3β − 4μβ − 1) ( ∫ cos(s)dν(s)) { 0 {0 2π

2π

{ } ⩽ β { ∫ cos(2s)dν(s) + (3β − 4μβ − 1) ∫ cos2 (s)dν(s)} 0 {0 } ⩽ β 2 (3 − 4μ).

2

} } } } }

90 | 6 Strongly Starlike and Convex Functions Next, when μ ⩾ (3 − 1/β)/4, then 4μβ + 1 − 3β ⩾ 0, and so Re{a3 −

μa22 }

2π 2π { { ⩽ β {− ∫ cos(2s)dν(s) + (4μβ + 1 − 3β) ( ∫ cos(s)dν(s)) { 0 { 0 2π

2

} } } } }

2π

{ } ⩽ β {− ∫ cos(2s)dν(s) + (4μβ + 1 − 3β) ∫ cos2 (s)dν(s)} 0 { 0 } 2π

⩽ β ∫ [1 + (4μβ − 1 − 3β) cos2 (s)]dν(s), 0

which is less than or equal to β if μ ⩽ (3 + 1/β)/4, and less than or equal to β 2 (4μ − 3) if μ ⩾ (3 + 1/β)/4. To see that the inequalities are sharp, we first note that if equality holds in one of the inequalities for some function f , then there is a rotation of f that is extremal for the functional Re(a3 − μa22 ). First consider the interval μ ⩽ (3 − 1/β)/4. If Re(a3 − μa22 ) = β 2 (3 − 4μ), then 2π

2

2π

∫ cos(2s)dν(s) = ( ∫ cos(s)dν(s)) = 1, 0

0

2π

which implies that ∫0 cos2 (s)dν(s) = 1. Thus the probability measure ν can be written as ν = λν0 + (1 − λ)ν π , where 0 ⩽ λ ⩽ 1 and ν0 and ν π are point masses at 0 and π, respectively. Thus 2π

2

1 = ( ∫ cos(s)dν(s)) = (2λ − 1)2 , 0

which implies that λ = 0, or 1, and so ν = ν0 , or ν = ν π , in which case f(z) = f0 (z) or f(z) = −f0 (−z). Similar arguments by Ma and Minda [130] show that the inequalities are sharp on the remaining intervals. Using similar methods, other coefficient results for f ∈ S∗∗ (β) were proved by Thomas in [246], of which the following are of interest. Theorem 6.3.3. Let f ∈ S∗∗ (β), then the second Hankel determinant H2 (2) = |a2 a4 − a23 | ⩽ β 2 . The inequality is sharp. Proof. We omit the proof, which follows a similar argument to that in Theorem 4.1.6.

6.3 Coefficient Theorems | 91

Theorem 6.3.4. Let f ∈ S∗∗ (β), then the logarithmic coefficients γ n , defined by (1.2.1), satisfy the sharp inequality n|γ n | ⩽ β, for n ⩾ 1. Proof. We include the simple proof as follows. Write 1 + ω(z) β zf 󸀠 (z) =( ) , f(z) 1 − ω(z) where ω is a Schwarz function. This can be written as z (log or z (log

f(z) 󸀠 1 + ω(z) β ) =( ) −1 z 1 − ω(z)

f(z) 󸀠 1+z β ) ≺( ) − 1 = 2βz + 2β 2 z2 + . . . . z 1−z

Applying Theorem 3.2.9 gives 2n|γ n | ⩽ 2β. The inequality is sharp when p n = 2 for n ⩾ 1.

Coefficients of the Inverse Function We next consider the problem of finding bounds for the coefficients of the inverse function when f ∈ S∗∗ (β). Recall that any f ∈ S has an inverse function f −1 defined inside some disk |ω| < r0 (f). Let ∞

f −1 (ω) = ω + ∑ A n ω n .

(6.3.2)

n=2

It was pointed out in Chapter 1 that when f ∈ S, (and therefore in S∗ ), the problem of finding sharp bounds for |A n | for all n ⩾ 2 was solved by Loewner (see Theorem 1.1.6, Chapter 1 from [125]). For f ∈ S∗∗ (β), the problem appears far from simple, and to date is solved only when n = 2, 3 and 4, which we present in the following theorem by Ali [8]. Theorem 6.3.5. Let f ∈ S∗∗ (β) with inverse coefficients given by (6.3.2), then |A2 | ⩽ 2β,

{ { β, |A3 | ⩽ { {5β 2 , {

1 0 0 and Re(c − a) > 0. Proof of Theorem 6.4.1. Since f ∈ C∗∗ (β) if, and only if, zf 󸀠 ∈ S∗∗ (β), we deduce from Theorem 6.2.1 that when 0 < β < 1, r

|f(z)| ⩽ ∫ |f 󸀠 (te iθ )|dt 0 r

< M(β) ∫ ( 0 1

< M(β) ∫ ( 0

1+t β ) dt 1−t 1+t β ) dt 1−t

M(β) = 2 F 1 (1, −β, 2 − β; −1). 1−β The second inequality follows at once from Theorem 6.2.1. We next give more information concerning functions in C∗∗ (β), which is not immediately obvious from the relationship f ∈ C∗∗ (β) if, and only if, zf 󸀠 ∈ S∗∗ (β). Since the proofs follow the same kind of methods as those used for functions in S∗∗ (β), they are omitted. We recall from Chapter 4 that when f ∈ C, the bounds for the modulus of the coefficients A n of the inverse function f −1 are the same (i.e., unity) as those for f , provided that 2 ⩽ n ⩽ 7. In the spirit of this result, we list below some invariance properties concerning the coefficients of functions in C∗∗ (β) (see Thomas and Verma [248]) .

6.5 Inclusion Relationships |

95

Theorem 6.4.2. Let f ∈ C∗∗ (β) with a n and A n the coefficients of f(z) and f −1 (ω), respectively, then β 1 { { , 0 1. Proof. Let p(z) = zf 󸀠 (z)/f(z). Then differentiating logarithmically, we obtain zp󸀠 (z) zf 󸀠󸀠 (z) zf 󸀠 (z) =1+ 󸀠 − . p(z) f (z) f(z) Thus, Jα [f](z) = (1 − α)p(z) + α [

zp󸀠 (z) zp󸀠 (z) + p(z)] = p(z) + α . p(z) p(z)

Therefore, from (7.1.1), with z = re iθ , we have Re [p(z) − iα

∂ log p(z)] > 0 ∂θ

(z ∈ 𝔻).

(7.1.2)

Suppose now that there exists a point z0 = r0 e iθ0 in 𝔻 such that Re p(z) ⩾ 0 for |z| ⩽ r0 , and Re p(z0 ) = 0. Then, arg p(r0 e iθ ) is either maximum or minimum for θ = θ0 , and so ∂ arg p(z0 )/∂θ = 0. https://doi.org/10.1515/9783110560961-007

100 | 7 Alpha-Convex Functions Combining this last remark with Re p(z0 ) = 0, we conclude that the left hand side of (7.1.2), and hence of (7.1.1) must vanish for z = z0 . However, this contradicts (7.1.1) (and (7.1.2)). Since p(0) = 1, Re p(z) does not vanish in 𝔻, and so p(z) = zf 󸀠 (z)/f(z) has a positive real part in 𝔻. Hence, f is univalent and starlike in 𝔻 for all real α. Now set z = 1/ξ and g(ξ) = 1/f(1/ξ) in (7.1.1). Then with |ξ| > 1, ξg󸀠 (ξ) ξg󸀠 (ξ) ξg󸀠󸀠 (ξ) + α (2 −1− 󸀠 )] g(ξ) g(ξ) g (ξ) ξg󸀠 (ξ) ξg󸀠󸀠 (ξ) = Re [(1 + α) − α (1 + 󸀠 )] > 0. g(ξ) g (ξ)

Re Jα [f](z) = Re [(1 − α)

(7.1.3)

Since f ∈ S∗ , so is g(ξ). Hence, from (7.1.3) we obtain the inequality Re [1 + ξ

g󸀠󸀠 (ξ) ]>0 g󸀠 (ξ)

(|ζ| > 1),

provided that 1 + α ⩽ 0. Hence, g(ξ) = 1/f(1/ξ) is convex for |ξ| > 1 if α ⩽ −1. If α ⩾ 1, then the first term on the left hand side of (7.1.1) is nonpositive. From this it follows that f is convex. If α = ±∞, then (7.1.1), with an application of the maximum principle for harmonic functions, implies that f(z) = z. This completes the proof. In view of Theorem 7.1.1 we make the following remark. Remark 7.1.1. (i) Mα ⊆ S∗ for any real α. (ii) Mα ⊆ Mβ ⊆ S∗ for 0 ⩽ β/α ⩽ 1. Todorov [251] obtained the following integral representation for α-convex functions. Theorem 7.1.2. f ∈ Mα for α > 0 if, and only, if there exists a starlike function F such that α z 1 1/α −1 ] [ ∫ F (ξ)ξ dξ (z ∈ 𝔻), (7.1.4) f(z) = α ] [ 0 where powers are meant as principal values. Proof. If f is α-convex, then there exists a Carathéodory function p ∈ P such that Jα [f](z) = p(z). Let the function F ∈ A be defined by zF 󸀠 (z)/F(z) = p(z). Then F is starlike and F(z) = zP(z), where z

P(z) = exp (∫

p(t) − 1 dt) , t

(7.1.5)

0

and so

α

f(z) zf 󸀠 (z) [ ] = P(z). z f(z)

(7.1.6)

7.2 Distortion and Growth Theorems | 101

By using the identity

d f(z) zf 󸀠 (z) = 1+z log , f(z) dz z

and making the substitution f(z) = u α (z) z

(7.1.7)

u P1/α (z) du =− + . dz αz αz

(7.1.8)

in (7.1.6), we obtain for α ≠ 0,

From (7.1.7) and (7.1.8) we, therefore, obtain for α > 0, α

z

α

z

1 1 f(z) = [ ∫ P1/α (ξ)ξ 1/α−1 dξ ] = [ ∫ F 1/α (ξ)ξ −1 dξ ] . α α ] [ 0 ] [ 0

(7.1.9)

Conversely, let F be a starlike function and f be given by (7.1.4). Then a simple computation shows that zF 󸀠 (z) Jα [f](z) = , F(z) which shows that f is α-convex. This completes the proof. In looking for the solutions of extremal problems in the class Mα , it is natural to examine the function that corresponds to the Koebe function k(z) = z/(1 − z)2 . Using the Koebe function in (7.1.4), we obtain z

α

1 ξ 1/α K(α, z) := [ ∫ dξ ] . α ξ(1 − ξ)2/α [ 0 ]

(7.1.10)

This function is called the α-convex Koebe function.

7.2 Distortion and Growth Theorems For α > 0 it follows from (7.1.10) that r

1 K(α, r) = [ ∫ ρ 1/α−1 (1 − ρ)−2/α dρ ] α [ 0 ] α 1 2 1 = r [2 F1 ( , , + 1; r)] , α α α

α

(7.2.1)

where 2 F1 is the Gaussian hypergeometric function. In 1973, Miller [143] obtained the following distortion theorems in terms of K(α, r) for α-convex functions.

102 | 7 Alpha-Convex Functions Theorem 7.2.1. Let f ∈ Mα for α > 0. Then with z = re iθ ∈ 𝔻, − K(α, −r) ⩽ |f(z)| ⩽ K(α, r),

(7.2.2)

where K(α, r) is defined by (7.2.1). Equality holds in both cases for the α-convex function f θ (α, z), where α

z

f θ (α, z) := e

−iθ

1 K(α, e z) = [ ∫ ξ 1/α−1 (1 − ξe iθ )−2/α dξ ] . α [ 0 ] iθ

Proof. It suffices to prove the theorem for z = r, since the general case can be reduced to this by considering the function f(ηz)/η and by suitably choosing η such that |η| = 1. Since f ∈ Mα , Theorem 7.1.2 shows that there exists a starlike function F such that z

α

1 F 1/α (ξ) ] f(z) = [ ∫ dξ , α ξ [ 0 ] and so taking z = r and integrating along the positive real axis, we obtain r

α

1 F 1/α (ρ) ] f(r) = [ ∫ dρ . α ρ [ 0 ] Since F is starlike, in view of (1.3.6), ρ ρ ⩽ |F(ρ)| ⩽ , 2 (1 + ρ) (1 − ρ)2 and so

(7.2.3)

r

|f(r)|

1/α

1 ⩽ ∫ ρ 1/α−1 (1 − ρ)−2/α dρ = [K(α, r)]1/α . α 0

To prove the left hand inequality of (7.2.2), we consider the straight line Λ joining 0 to f(z) = Re iϕ . Since f is starlike, Λ is the image of a Jordon arc γ in 𝔻 connecting 0 and z = re iθ . The image of γ under the mapping f 1/α (z) will, in general, consist of many line segments emanating from the origin, each of length R1/α = |f(z)|1/α = ∫γ |df 1/α (ξ)/dξ‖dξ|. Since f ∈ Mα , from the integral representation we know there exists a starlike function F such that df 1/α (ξ)/dξ = (1/α)F 1/α (ξ)/ξ . Thus, if ρ = |ξ|, we deduce from (7.2.3) that 󵄨 󵄨 1 󵄨󵄨 F 1/α (ξ) 󵄨󵄨󵄨 1 󵄨󵄨 |dξ| ⩾ ∫ ρ 1/α−1 (1 + ρ)−2/α |dξ| R1/α = ∫ 󵄨󵄨󵄨󵄨 α 󵄨󵄨 ξ 󵄨󵄨󵄨 α γ γ r

1 ⩾ ∫ ρ 1/α−1 (1 + ρ)−2/α dρ, α 0

and by substituting ρ = ru we obtain |f(z)| ⩾ −K(α, −r).

7.2 Distortion and Growth Theorems |

103

That the right hand inequality in (7.2.2) is sharp can be seen by considering the function f θ (α, z) with θ = 0 and z = r. For the left hand inequality, we consider f θ (α, z) with θ = π and z = r. Remark 7.2.1. (i) If α = 1 then (7.2.2) reduces to r/(1 + r) ⩽ |f(z)| ⩽ r/(1 − r), the well-known result for convex functions. (ii) If α = 2 then (7.2.2) reduces to 2

1 1 + √r (arctan √r)2 ⩽ |f(z)| ⩽ [ log ( )] . 2 1 − √r (iii) If α > 2, then by (7.2.2), we have |f(z)|

1/α

r

1

0

0

1 1 ⩽ ∫ ρ 1/α−1 (1 − ρ)−2/α dρ ⩽ ∫ ρ 1/α−1 (1 − ρ)−2/α dρ α α 1 Γ(1/α)Γ(1 − 2/α) = , α Γ(1 − 1/α)

i.e.,

1 Γ(1/α)Γ(1 − 2/α) α (7.2.4) ] , α Γ(1 − 1/α) and so if f ∈ Mα with α > 2, then f is bounded. The bound in (7.2.4) is sharp for f θ (α, z) with θ = 0 and z = r when r → 1. |f(z)| ⩽ [

Theorem 7.2.2. Let f ∈ Mα for α ⩾ 1. Then for z = re iθ ∈ 𝔻, α−1

r

[ 1 ∫ ρ 1/α−1 (1 + ρ)−2/α dρ ] α ] [ 0 r1−1/α (1 + r)2/α

=

∂ ∂ K(α, −r) ⩽ |f 󸀠 (z)| ⩽ K(α, r) ∂r ∂r α−1

r

[ 1 ∫ ρ 1/α−1 (1 − ρ)−2/α dρ ] α [ 0 ] = r1−1/α (1 − r)2/α

.

Both inequalities are sharp. Proof. From the integral representation we have |f 󸀠 (z)| =

|F(z)|1/α |f(z)|1−1/α , |z|

where F is starlike. By (7.2.3) and Theorem 7.2.1, we obtain for |z| = r (0 < r < 1), |f 󸀠 (z)| ⩽

1/α 1 r ] [K(α, r)]1−1/α [ 2 r (1 − r) r

1 [ 1 ∫ ρ 1/α−1 (1 − ρ)−2/α dρ ] = 1−1/α r (1 − r)2/α α [ 0 ]

α−1

=

∂ K(α, r), ∂r

104 | 7 Alpha-Convex Functions

which proves the right hand inequality. The left hand inequality is proved by using the corresponding inequalities in (7.2.3) and Theorem 7.2.1. The functions f0 (α, z) and f π (α, z) at z = r show that the inequalities are sharp. Remark 7.2.2. (i) When α = 1, we obtain the well-known distortion theorem for f ∈ C, 1 1 ⩽ |f 󸀠 (z)| ⩽ . (1 + r)2 (1 − r)2 (ii) When α = 2, we obtain log ((1 + r1/2 )/(1 − r1/2 )) arctan r1/2 󸀠 (z)| ⩽ ⩽ |f . r1/2 (1 + r) 2r1/2 (1 − r) We next apply Theorem 7.2.1 to give the following analogue of the 1/4-theorem for α-convex functions. Theorem 7.2.3. If f ∈ Mα for α > 0, then the image of the unit disk 𝔻 under the mapping w = f(z) always contains the disk |w| < d(α), where 1 { , { { 4 { d(α) = { α { Γ2 (1/α) { { [ ] , { 2αΓ(2/α)

α = 0, α > 0.

The result is the best possible in the sense that d(α) cannot be made any larger. Proof. The case α = 0 holds for all f ∈ S, and is best possible. Let α > 0 and let w0 be a point on the boundary of f(𝔻), which is nearest to the origin. Let Λ1 denote the straight line from 0 to w0 , and Λ be its pre-image in 𝔻. Then we have |w0 | > |f(z)| for z ∈ Λ ∩ 𝔻. Since the circle |z| = r, for 0 ⩽ r < 1, intersects Λ at least once, by Theorem 7.2.1 we have |w0 | > −K(α, −r) for all 0 ⩽ r < 1, i.e., α

r

α 1 2 1 1 |w0 | > r [2 F1 ( , , + 1; −r)] = [ ∫ ρ 1/α−1 (1 + ρ)−2/α dρ ] , α α α α [ 0 ]

for 0 ⩽ r < 1. Since last expression is an increasing function of r, we obtain α

1

α 1 1 |w0 | ⩾ [ ∫ ρ 1/α−1 (1 + ρ)−2/α dρ ] := [ H(α)] . α α [ 0 ]

Substituting ρ = 1/u in H(α), we obtain ∞

1

H(α) = ∫ ρ 1/α−1 (1 + ρ)−2/α dρ = ∫ u 1/α−1 (1 + u)−2/α du 0

1 ∞

=

Γ2 (1/α) 1 ∫ u 1/α−1 (1 + u)−2/α du = . 2 2Γ(2/α) 0

(7.2.5)

7.2 Distortion and Growth Theorems | 105

Hence from (7.2.5) we obtain α

|w0 | ⩾ [

Γ2 (1/α) ] = d(α). 2αΓ(2/α)

That d(α) cannot be made any larger can be seen by considering the function f θ (α, z) with θ = π. Another application of Theorem 7.2.1 gives the following growth results due to Miller [143]. Theorem 7.2.4. Let f ∈ Mα for α > 0, and M(r) = max |f(z)|. Then as r → 1, |z|=r

2−α { { O(1/(1 − r)) , M(r) = { { O(log(1/(1 − r)))2 , {

If α > 2, then M(r) ⩽ [

0 ⩽ α < 2, α = 2.

1 Γ(1/α)Γ(1 − 2/α) α ] . α Γ(1 − 1/α)

Proof. From (7.2.3) we have M(r) ⩽ r/(1 − r)2 , which proves the theorem for the case α = 0. If α > 0, then from Theorem 7.2.1 we have the sharp inequality M(r) ⩽ K(α, r) = r[2 F1 (1/α, 2/α, 1/α + 1; r)]α .

(7.2.6)

Now assume that 0 < α < 2; then from a result in Whittaker and Watson [261, p. 299], 2 F 1 (1/α, 2/α, 1/α r→1 (1 − r)1−2/α

lim

+ 1; r)

=

1 . 2−α

Combining this with (7.2.6) we obtain M(r) = O(r/(1 − r)2−α ) = O(1/(1 − r))2−α ,

as

r → 1.

If α = 2, then we again have from Whittaker and Watson [261, p. 299], lim

r→1

(1/2, 1, 3/2; r) 1 = , log(1/(1 − r)) 2

2 F1

and combining this once more with (7.2.6), we obtain M(r) = O(r(log(1/(1 − r)))2 ) = O(log(1/(1 − r)))2 as r → 1. Finally, when α > 2, (7.2.4) gives M(r) ⩽ [

1 Γ(1/α)Γ(1 − 2/α) α ] , α Γ(1 − 1/α)

which completes the proof of the theorem.

106 | 7 Alpha-Convex Functions

7.3 Coefficient Problems With the application of Theorem 7.2.1, Miller [143] obtained the following sharp upper bound of the second coefficient in the Taylor expansion for f ∈ Mα . n Theorem 7.3.1. Let f ∈ Mα for α > 0 and be given by f(z) = z + ∑∞ n=2 a n (α)z . Then, |a2 (α)| ⩽ 2/(1 + α). The inequality is sharp.

Proof. Since the class Mα is rotationally invariant, i.e., e−iθ f(e iθ z) ∈ Mα when f(z) ∈ Mα and θ ∈ ℝ, we can assume that a2 (α) is real. A simple calculation in (7.2.1) shows that α 1 2 1 K(α, r) = r [2 F1 ( , , + 1; r)] = r + (2/(1 + α))r2 + O(r3 ). α α α 2 Since f(r) = r + a2 r + O(r3 ), from Theorem 7.2.1 we can deduce that r + a2 r2 + O(r3 ) ⩽ r + (2/(1 + α))r2 + O(r3 ), and so a2 (α) ⩽ 2/(1 + α). The sharpness of the inequality follows by considering the function f0 (α, z), and noting that f0 (α, r) = K(α, r). We now discuss (without complete proofs), the important theorems concerning estimates for the general coefficients a n (α). Taking the principal value of the powers in (7.1.10), we can write K(α, z) = zH(z), where α ∞

H(z) = [1 + ∑ c n z n ] , n=1

with cn =

n−1 1 ∏ (2 + kα). n! α n (1 + nα) k=0

(7.3.1)

n Thus if f ∈ Mα and is given by f(z) = z + ∑∞ n=2 a n (α)z , it is easy to see that for n ⩾ 1, |a n+1 (α)| ⩽ H n (0)/n!. In fact, the following result holds. n Theorem 7.3.2. Let f ∈ Mα and be given by f(z) = z + ∑∞ n=2 a n (α)z . Also, let S(n) be the set of all n-tuples (x1 , x2 , . . . , x n ) of nonnegative integers in which ∑ni=1 ix i = n, and for each such n-tuple, define q by ∑ni=1 x i = q. If 𝛶(α, q) = α(α − 1)(α − 2) ⋅ ⋅ ⋅ (α − q) with 𝛶(α, 0) = α, then for n = 1, 2, . . . x

|a n+1 (α)| ⩽ ∑

x

x

𝛶(α, q − 1)c11 c22 ⋅ ⋅ ⋅ c nn , x1 !x2 ! ⋅ ⋅ ⋅ x n !

where the summation is taken over all n-tuples in S(n), and c n is given by (7.3.1). For the proof of Theorem 7.3.2, see Kulshrestha [101]. For f ∈ Mα with α ≠ 0, the function P defined by (7.1.6) is analytic near the origin. Let ∞

[P(z)]1/α = 1 + ∑ P n (α)z n . n=1

(7.3.2)

7.3 Coefficient Problems |

107

From (7.1.9), it is easy to see that in order to find expressions for the coefficients of the function (7.1.4), we must first find the coefficients of (7.3.2) in terms of the coefficients of p(z) in (3.1.1). The following explicit form of the coefficients P n (α) and its estimate was obtained by Todorov [251]. Theorem 7.3.3. The coefficients P n (α), n = 1, 2, . . . (α ≠ 0) of the function (7.3.2) have the explicit representation n

P n (α) = ∑ k=1

p1 p n−k+1 1 C nk ( , . . . , ), 1 n−k+1 αk

(7.3.3)

where C nk (

n−k+1 1 ps νs p1 p n−k+1 ,..., )=∑ ∏ ( ) , 1 n−k+1 νs ! s s=1

(7.3.4)

and where the sum is taken over all solutions in nonnegative integers ν1 , . . . , ν n−k+1 of the system ν1 + ν2 + ⋅ ⋅ ⋅ + ν n−k+1 = k, (7.3.5) ν1 + 2ν2 + ⋅ ⋅ ⋅ + (n − k + 1)ν n−k+1 = n. Proof. It follows from (3.1.1) and (7.1.5) that the function (7.3.3) can be represented by the composite function ∞ pn n P1/α (z) = e u ∘ ∑ z , αn n=1 where ∘ denotes the substitution ∞

u= ∑ n=1

pn n z . αn

Using a more precise version of the Faà di Bruno formula for the derivatives of composite functions, Todorov [250, Theorem 1] applied this to the n-th derivative of the composite function above at the point z = 0, to obtain the representation n

n−k+1

D nz=0 [P(z)]1/α = n! ∑ ∑ ∏ k=1

s=1

1 ps νs ( ) , ν s ! αs

(7.3.6)

where the interior sum is taken over all solutions in nonnegative integers ν1 , . . . , ν n−k+1 of the system (7.3.5). From (7.3.6) we obtain the representations (7.3.3) and (7.3.4) for P n (α). This completes the proof. For an arbitrary x, let ⟨x⟩k denote the factorial polynomial ⟨x⟩k = x(x + 1) ⋅ ⋅ ⋅ (x + k − 1)

(k = 1, 2, . . . ; ⟨x⟩0 = 1).

By making use of Theorem 7.3.3, Todorov [251] obtained the following sharp upper bound for |P n (α)|, when n ⩾ 1, for functions given by (7.3.2).

108 | 7 Alpha-Convex Functions Theorem 7.3.4. For α > 0, the coefficients P n (α), for n ⩾ 1 of the function (7.3.2) satisfy the sharp inequalities ⟨2/α⟩n , (7.3.7) |P n (α)| ⩽ n! where equality holds only for the function P∗ given by ∞

⟨2/α⟩n n n ϵ z , n! n=1

P∗ (z) = (1 − ϵz)−2/α = 1 + ∑ 1/α

(|ϵ| = 1).

(7.3.8)

Proof. From (7.3.3) and (7.3.4) we obtain n

|P n (α)| ⩽ ∑ k=1

1 |p1 | |p n−k+1 | C nk ( ,..., ). 1 n−k+1 αk

(7.3.9)

Since p ∈ P, using the inequality |p s | ⩽ 2 for s = 1, 2, . . . , n, in (7.3.9) gives n

|P n (α)| ⩽ ∑ k=1

1 2 2 C nk ( , . . . , ), 1 n−k+1 αk

(7.3.10)

where by (7.3.3) – (7.3.5), equality holds in (7.3.10) if, and only if, p s = 2ϵ s (|ϵ| = 1), s = 1, 2, . . . , n. However, if p1 = 2ϵ holds for the Carathéodory function (3.1.1), then p s = 2ϵ s

for

s = 1, 2, . . . .

Therefore, equality holds in (7.3.10) only if the function (3.1.1) has the form p∗ (z) =

∞ 1 + ϵz = 1 + 2 ∑ ϵn zn 1 − ϵz n=1

(|ϵ| = 1),

or, by (7.1.5), only if the function (7.3.2) has the form (7.3.8). Using n

1 2 2 ⟨2/α⟩n C ( ,..., )= , k nk 1 n − k + 1 n! α k=1 ∑

(n = 1, 2, . . . ; α > 0),

the inequality (7.3.10) becomes (7.3.7). This completes the proof. For an arbitrary number x, let (x)k denote the factorial polynomial (x)k = x(x − 1) ⋅ ⋅ ⋅ (x − k + 1),

(k = 1, 2, . . . ; (x)0 = 1),

and as above, let f be given for z ∈ 𝔻 by ∞

f(z) = z + ∑ a n (α)z n

(α > 0).

(7.3.11)

n=2

. By making use of Theorems 7.3.3 and 7.3.4, we obtain the following explicit form of the coefficients in (7.3.11) in terms of the coefficients of the function given in (7.3.2).

7.3 Coefficient Problems

| 109

Theorem 7.3.5. The Taylor coefficients a n+1 (α), n = 1, 2, . . . of the function (7.1.4) for an arbitrary α > 0 have the explicit representation n

a n+1 (α) = ∑ (α)k C nk ( k=1

P1 (α) P n−k+1 (α) ,..., ), α+1 (n − k + 1)α + 1

(7.3.12)

where C nk (

n−k+1 1 P1 (α) P n−k+1 (α) P s (α) ν s ,..., )=∑ ∏ ( ) , α+1 (n − k + 1)α + 1 ν s ! sα + 1 s=1

(7.3.13)

and where the sum is taken over all solutions in nonnegative integers ν1 , . . . , ν n−k+1 of the system (7.3.5). If α is a positive integer, we also have the explicit representation a n+1 (α) = α! C n+α,α (P0 (α),

P1 (α) P n (α) ,..., ), α+1 nα + 1

P0 (α) = 1,

(7.3.14)

for n = 1, 2, . . . , where C n+α,α (P0 (α),

n P1 (α) P n (α) P s (α) ν s+1 1 , ,..., ) = ∑∏ ( ) α+1 nα + 1 ν ! sα + 1 s=0 s+1

(7.3.15)

and where the sum is taken over all nonnegative integers ν1 , . . . , ν n+1 satisfying ν1 + ν2 + ⋅ ⋅ ⋅ + ν n+1 = α, ν1 + 2ν2 + ⋅ ⋅ ⋅ + (n + 1)ν n+1 = n + α.

(7.3.16)

For the proof of Theorem 7.3.5, see Todorov [251]. Next note that from Theorem 7.3.5 and (7.3.8), we can obtain the coefficient of the most general α-convex Koebe function α

z

1 K(z; ϵ; α) = [ ∫ ξ 1/α−1 (1 − ϵξ)−2/α dξ ] α [ 0 ]

(|ϵ| = 1; α > 0),

(7.3.17)

for arbitrary α > 0 as follows. For z ∈ 𝔻, write ∞

K(z; ϵ; α) = z[ϕ(z; ϵ; α)]α = z + ∑ K n+1 (ϵ; α)z n+1 ,

(7.3.18)

n=1

where

∞

ϕ(z; ϵ; α) = 1 + ∑ ϵ n c n (α)z n ,

(7.3.19)

n=1

with c n (α) =

⟨2/α⟩n , n!(nα + 1)

for

n = 1, 2, . . . .

(7.3.20)

110 | 7 Alpha-Convex Functions Note now that the substitution ξ = zt in (7.3.17) yields the identity α 2 1 1 K(z; ϵ; α) = z [2 F1 ( , , 1 + ; ϵz)] , α α α

where

(|ϵ| = 1; α > 0),

(7.3.21)

1

2 1 1 1 ∫ t1/α−1 (1 − ϵzt)−2/α dt, 2 F 1 ( , , 1 + ; ϵz) = α α α α 0

which is the analytic continuation of the series (7.3.19) into the z-plane cut along the ray z = ρe−i arg ϵ , ρ ⩾ 1. In particular, if α is a positive integer, then from (7.3.18) we obtain K(z; ϵ; α) = z1−α [zϕ(z; ϵ; α)]α

(7.3.22)

∞

= z + ∑ K n+1 (ϵ; α)z n+1

(z ∈ 𝔻).

n=1

As a corollary to Theorem 7.3.5, we have the following result due to Todorov [251] concerning the coefficients of the function (7.3.17) (or (7.3.21)) and the function (7.3.22), respectively. Theorem 7.3.6. The coefficients K n+1 (ϵ; α) of the general α-convex Koebe functions (7.3.17) for an arbitrary α > 0 have the explicit representation K n+1 (ϵ; α) = ϵ n K n+1 (α)

(n = 1, 2, . . . ),

(7.3.23)

n

K n+1 (α) = ∑ (α)k C nk (c1 (α), . . . c n−k+1 (α)),

(7.3.24)

k=1

where n−k+1

C nk (c1 (α), . . . c n−k+1 (α)) = ∑ ∏ s=1

(c s (α))ν s , νs !

and where c1 (α), . . . c n−k+1 (α) are defined in (7.3.20), and the sum is taken over all solutions in nonnegative integers ν1 , . . . , ν n−k+1 of the system (7.3.5). If α is a positive integer, then we also have the representation (7.3.23), but with the simpler expression K n+1 (α) = α!C n+α,α (c0 (α), c1 (α), . . . , c n (α)),

c0 (α) = 1

(7.3.25)

for n ⩾ 1, where n

(c s (α))ν s+1 , ν s+1 ! s=0

C n+α,α (c0 (α), c1 (α), . . . , c n (α)) = ∑ ∏

and the sum is taken over all solutions in nonnegative integers ν1 , . . . , ν n+1 of the system (7.3.16).

7.3 Coefficient Problems

| 111

We note that when α is a positive integer, (7.3.24) reduces to (7.3.25). Using the coefficients of the function (7.3.17) in Theorem 7.3.6 leads to the following conjecture of Todorov [251]. Conjecture 7.3.1. The coefficients a n+1 (α) of the α-convex functions given by (7.3.11) satisfy the sharp inequalities |a n+1 (α)| ⩽ K n+1 (α),

(n = 1, 2, . . . )

(7.3.26)

where the K n+1 (α) are given by (7.3.24) for arbitrary α > 0 and by (7.3.25) for a positive integer α, respectively. Equality holds only for the α-convex Koebe function (7.3.17). Our final theorem shows that Theorems 7.3.4, 7.3.5, and 7.3.6 verifies Conjecture 7.3.1 in some special cases. Theorem 7.3.7. The coefficients a n+1 (α) of the α-convex functions given by (7.3.11) satisfy the sharp inequalities (7.3.26) for n = 1, 2, . . . , [α] + 1, if α > 0 and is not an integer ([α] denotes the greatest integer less than α), and for all n ⩾ 1 if α is a positive integer, where the K n+1 (α) are given by (7.3.24) and (7.3.25), respectively. Equality holds only for the α-convex Koebe functions given by (7.3.17). Proof. If α > 0 is not a positive integer, then from (7.3.12) and (7.3.13) we obtain, for n = 1, 2, . . . , [α] + 1, the estimates n

|a n+1 (α)| ⩽ ∑ (α)k C nk ( k=1

|P1 (α)| |P n−k+1 (α)| ,..., ). α+1 (n − k + 1)α + 1

(7.3.27)

We can now use the inequalities (7.3.7) in (7.3.27) to obtain n

|a n+1 (α)| ⩽ ∑ (α)k C nk (c1 (α), . . . , c n−k+1 (α)) = K n+1 (α).

(7.3.28)

k=1

Equality holds in (7.3.28), according to Theorems 7.3.4 and 7.3.5, only if the function (7.1.9) has the form (7.3.17). If α is a positive integer, the above proof is valid for all n ⩾ 1. Indeed, (α)k = 0 for k > α, and hence the summation in (7.3.27) and (7.3.28) is taken over k = 1, 2, . . . , n, and if 1 ⩽ n ⩽ α, over k = 1, 2, . . . , α if n > α. If α is a positive integer, an alternative proof can be obtained from (7.3.14) and (7.3.15) as follows. From (7.3.14) and (7.3.15) for n ⩾ 1, we obtain |a n+1 (α)| ⩽ α!C n+α,α (|P0 (α)|,

|P1 (α)| |P n (α)| ,..., ). α+1 nα + 1

(7.3.29)

Again using (7.3.7), (7.3.20), (7.3.25), and (7.3.29), we obtain |a n+1 (α)| ⩽ α!C n+α,α (c0 (α), c1 (α), . . . , c n (α)) = K n+1 (α), where equality holds according to Theorems 7.3.4 and 7.3.5 only if the function (7.1.9) is of the form (7.3.17). This completes the proof of the theorem.

8 Gamma-Starlike Functions 8.1 Definition and Basic Properties In the last chapter, we saw how combining the expressions for starlike and convex functions in terms of a linear combination of zf 󸀠 (z)/f(z) and 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) not only produced a generalization of S∗ and C, but also resulted in a significant subclass Mα of S∗ . In this chapter we use a similar idea to generate an analogous extension of S∗ and C, but this time using a combination of powers of zf 󸀠 (z)/f(z) and 1 + zf 󸀠󸀠 (z)/f 󸀠 (z). This also produces a significant subclass of S∗ , whose elements have become known as gamma-starlike functions. We note that some authors have called these functions alpha-logarithmically convex. Definition 8.1.1. Let f ∈ A, with f(z) ≠ 0, f 󸀠 (z) ≠ 0, and 1 + zf 󸀠󸀠 (z)/f 󸀠 (z) ≠ 0 for 0 < |z| < 1. Then, if for γ ∈ ℝ, f satisfies γ

Re [(1 +

zf 󸀠 (z) zf 󸀠󸀠 (z) ) ( ) 󸀠 f (z) f(z)

1−γ

]>0

(z ∈ 𝔻),

we say that f is gamma-starlike and denote this class of functions by Mγ . As we saw when we considered strongly starlike functions in Chapter 6, the presence of powers in the definition creates difficulties. This problem also applies to Mγ , resulting in relatively few complete theorems having been obtained to date. Clearly, M0 = S∗ and M1 = C, but more importantly and fundamentally, it was first shown by Lewandowski et al. [110] that Mγ ⊂ S∗ for all real γ. Subsequently, various authors have given other proofs of this, all of which are essentially the same. As an illustration, we give the following proof, which uses the lemma of Nunokawa (Theorem 3.2.7). Theorem 8.1.1. Mγ ⊂ S∗ for all real γ. Proof. We show that if f ∈ Mγ , then 󵄨󵄨 γ 1−γ 󵄨󵄨 󸀠󸀠 󸀠 󵄨 󵄨󵄨 󵄨󵄨arg {(1 + zf (z) ) ( zf (z) ) }󵄨󵄨󵄨 < π 󵄨󵄨 2 󵄨󵄨 󸀠 f (z) f(z) 󵄨󵄨 󵄨󵄨 implies

󵄨󵄨 󵄨 zf 󸀠 (z) 󵄨󵄨󵄨 π 󵄨󵄨 󵄨󵄨arg 󵄨󵄨 < 󵄨󵄨 f(z) 󵄨󵄨󵄨 2 󵄨

Let p(z) =

(z ∈ 𝔻).

zf 󸀠 (z) , f(z)

where p is analytic in 𝔻, p(0) = 1, and p(z) ≠ 0 for z ∈ 𝔻. https://doi.org/10.1515/9783110560961-008

(z ∈ 𝔻),

(8.1.1)

8.2 Coefficient Inequalities | 113

Suppose there exists a point z0 ∈ 𝔻, such that | arg p(z)| < π/2 for |z| < |z0 | and | arg p(z0 )| = π/2. Then by Nunokawa’s lemma, z0 p󸀠 (z0 ) = ik, p(z0 ) where

1 (a + 2 1 k ⩽ − (a + 2 and p(z0 ) = ±ia for a > 0. Thus, if arg p(z0 ) = π/2, then k⩾

arg {(

1 ), a 1 ), a

z0 f 󸀠 (z0 ) ) f(z0 )

π , 2 π when arg p(z0 ) = − , 2 when arg p(z0 ) =

1−γ

γ

(1 +

z0 f 󸀠󸀠 (z0 ) ) } f 󸀠 (z0 )

= (1 − γ) arg p(z0 ) + γ arg [p(z0 ) + = (1 − γ)

z0 p󸀠 (z0 ) ] p(z0 )

π π + γ arg (ia + ik) = , 2 2

where p(z0 ) = ia, and k ⩾ (a + 1/a)/2. Next if arg p(z0 ) = −π/2, then arg {(

z0 f 󸀠 (z0 ) ) f(z0 )

1−γ

γ

(1 +

z0 f 󸀠󸀠 (z0 ) ) } f 󸀠 (z0 )

z0 p󸀠 (z0 ) ] p(z0 ) π π = −(1 − γ) + γ arg (−ia + ik) = − , 2 2 = (1 − γ) arg p(z0 ) + γ arg [p(z0 ) +

where p(z0 ) = −ia, and k ⩽ −(a + 1/a)/2. So, we have a contradiction. Thus, there is no point in z0 ∈ 𝔻, such that | arg p(z0 )| < π/2 for |z| < |z0 | and | arg p(z0 )| = π/2, which establishes (8.1.1). Apart from the above, relatively little seems to be known about the basic properties of functions in Mγ . For example, distortion theorems and simple bounds for |f(z)| appear not to have been determined. Similarly, there are few known meaningful sufficient conditions for a function f ∈ A to belong to Mγ .

8.2 Coefficient Inequalities As a result of the nature of the definition of functions in Mγ containing powers, finding sharp bounds for all coefficients of the Taylor series expansion of f(z) is far from simple. However, when γ ⩾ 0, the following sharp inequalities for the initial coefficients can be determined.

114 | 8 Gamma-Starlike Functions n Theorem 8.2.1. Let f ∈ Mγ for γ ⩾ 0 and be given by f(z) = z + ∑∞ n=2 a n z . Also let γ 0 2 3 4 be the unique positive root of the equation 15 + 98γ + 265γ − 14γ − 4γ = 0. Then,

2 |a2 | ⩽ , 1+γ

3(1 + 3γ) , { { { (1 + γ)2 (1 + 2γ) |a3 | ⩽ { { 1 { , { 1 + 2γ

γ2 − 7γ − 2 ⩽ 0, γ2 − 7γ − 2 ⩾ 0,

2(18 + 113γ + 292γ2 + 7γ3 + 2γ4 ) { { , { { 9(1 + γ)3 (1 + 2γ)(1 + 3γ) |a4 | ⩽ { { 2 { { , 3(1 + 3γ) {

γ ⩽ γ0 , γ ⩾ γ0 .

All the inequalities are sharp. Proof. Write γ

[1 +

1−γ

zf 󸀠 (z) zf 󸀠󸀠 (z) ] [ ] 󸀠 f (z) f(z)

= p(z)

n where p ∈ P, and p(z) = 1 + ∑∞ n=1 p n z . Then, equating coefficients gives

a2 =

p1 1+γ

a3 =

(2 + 7γ − γ2 )p21 p2 + 4(1 + γ)2 (1 + 2γ) 2(1 + 2γ)

a4 =

(6 + 23γ + 154γ2 − 47γ3 + 8γ4 )p31 36(1 + γ)3 (1 + 2γ)(1 + 3γ) p3 (3 + 19γ − 4γ2 )p1 p2 + . + 6(1 + γ)(1 + 2γ)(1 + 3γ) 3(1 + 3γ)

(8.2.1)

The inequality for |a2 | is trivial. The first inequality for |a3 | is obvious on noting that the coefficient of p21 is positive when γ2 − 7γ − 2 ⩽ 0, and applying the inequalities |p1 | ⩽ 2 and |p2 | ⩽ 2. The second inequality follows by a simple application of Theorem 3.1.7. For |a4 |, write a4 =

1 (4γ2 − 19γ − 3)p1 p2 (6 + 23γ + 154γ2 − 47γ3 + 8γ4 )p31 + ]. [p3 − 3(1 + 3γ) 2(1 + γ)(1 + 2γ) 12(1 + γ)3 (1 + 2γ)

Then, since the coefficients of p1 p2 and p31 are positive when γ ⩽ (19 + √409)/8, the first inequality for |a4 | is valid on this interval on using the inequalities |p n | ⩽ 2 for n = 1, 2 and 3. We now use Theorem 3.1.8 with B=

4γ2 − 19γ − 3 . 4(1 + γ)(1 + 2γ)

8.2 Coefficient Inequalities |

115

First note that 0 ⩽ B ⩽ 1, when γ ⩾ (19 + √409)/8, and so writing a4 =

1 [p3 − 2Bp1 p2 + Bp31 + (D − B)p31 ] , 3(1 + 3γ)

with D=

6 + 23γ + 154γ2 − 47γ3 + 8γ4 , 12(1 + γ)3 (1 + 2γ)

we see that D − B ⩾ 0 when (19 + √409)/8 ⩽ γ ⩽ γ0 . Applying Theorem 3.1.8 and again using the inequalities |p n | ⩽ 2 for n = 1, 2 and 3 gives the first inequality for |a4 | on the interval (19 + √409)/8 ⩽ γ ⩽ γ0 . Thus, it remains to establish the inequality for |a4 | on the interval γ ⩾ γ0 . We again use Theorem 3.1.8. It is easy to see that 0 ⩽ B ⩽ 1 and B(2B − 1) ⩽ D ⩽ B hold when γ ⩾ γ0 , and so applying Theorem 3.1.8 gives the second inequality for |a4 | at once. To see that the above inequalities are sharp, we note that equality is attained in the inequality for |a2 |, and the first inequalities for |a3 | and |a4 | when f(z) is given by γ

[1 +

1−γ

zf 󸀠󸀠 (z) zf 󸀠 (z) [ ] ] f 󸀠 (z) f(z)

=

1+z . 1−z

=

1 + z2 , 1 − z2

=

1 + z3 . 1 − z3

The second inequality for |a3 | is sharp when γ

[1 +

1−γ

zf 󸀠󸀠 (z) zf 󸀠 (z) ] [ ] 󸀠 f (z) f(z)

and the second inequality for |a4 | is sharp when γ

[1 +

1−γ

zf 󸀠󸀠 (z) zf 󸀠 (z) ] [ ] 󸀠 f (z) f(z)

Finding a solution to the Fekete–Szegő problem involves only the coefficients a2 and a3 , and a complete solution is relatively easy to obtain. We therefore omit the proof of the following theorem in Darus and Thomas [38]. n Theorem 8.2.2. Let f ∈ Mγ , and be given by f(z) = z + ∑∞ n=2 a n z . Then for γ, μ ∈ ℝ,

3(1 + 3γ) − 4μ(1 + 2γ) { , { { { (1 + 2γ)(1 + γ)2 { { { { { 1 |a3 − μa22 | ⩽ { , { 1 + 2γ { { { { { { { 4μ(1 + 2γ) − 3(1 + 3γ) , (1 + 2γ)(1 + γ)2 { All the inequalities are sharp.

μ⩽

2 + 7γ − γ2 , 4(1 + 2γ)

2 + 7γ − γ2 4 + 11γ + γ2 ⩽μ⩽ , 4(1 + 2γ) 4(1 + 2γ) μ⩾

4 + 11γ + γ2 . 4(1 + 2γ)

116 | 8 Gamma-Starlike Functions

8.2.1 Logarithmic Coefficients We saw in Chapter 4 that when γ = 0 or 1, obtaining a complete solution to the problem of finding sharp estimates for the modulus of the coefficients of log( f(z)/z) is a simple matter. For other values of γ differentiating log( f(z)/z) does not give a convenient expression, and the problem is much more difficult. We show next that it is possible to obtain sharp bounds for the modulus of the initial coefficients of log( f(z)/z) when f ∈ Mγ , for γ ⩾ 0. To avoid confusion with (1.2.1), we will here write log( f(z)/z) = n 2 ∑∞ n=1 δ n z . Theorem 8.2.3. Let f ∈ Mγ for γ ⩾ 0 and the coefficients of log( f(z)/z) be given by (1.2.2). Then,

|δ1 | ⩽

1 , 1+γ

1 + 5γ , { { { 2(1 + γ)2 (1 + 2γ) |δ2 | ⩽ { { 1 { , { 2(1 + 2γ)

3 + 11γ + 121γ2 + 7γ3 + 2γ4 { { , { { 9(1 + γ)3 (1 + 2γ)(1 + 3γ) |δ3 | ⩽ { { 1 { { , 3(1 + 3γ) {

0 ⩽ γ ⩽ 3, γ ⩾ 3, γ ⩽ γ1 , γ ⩾ γ1 ,

where γ1 = 3.3751 . . . is the unique positive root of the equation 2−47γ+7γ2 +2γ3 = 0. All the inequalities are sharp. Proof. The inequality for |δ1 | is trivial. For |δ2 |, we note as before that δ2 =

1 1 (a3 − a22 ) , 2 2

and so the inequalities for |δ2 | follow at once from the Fekete–Szegő theorem (Theorem 8.2.2). We note that the inequalities for |δ2 | can also be easily obtained by applying Theorem 3.1.7. For |δ3 |, recall from (1.2.2) that δ3 =

1 1 (a4 − a2 a3 + a32 ) , 2 3

which on using (8.2.1), gives δ3 =

γ(γ − 1)(17 − 6γ + 4γ2 )p31 γ(5 − 2γ)p1 p2 p3 + + . 36(1 + γ)3 (1 + 2γ)(1 + 3γ) 6(1 + γ)(1 + 2γ)(1 + 3γ) 6(1 + 3γ)

As in Theorem 8.2.1, we first note that the coefficients of p31 , p1 p2 , and p3 are all positive on 1 ⩽ γ ⩽ 5/2, and so using the inequality |p n | ⩽ 2 for n = 1, 2, 3, the first inequality for |δ3 | follows when 1 ⩽ γ ⩽ 5/2.

8.2 Coefficient Inequalities | 117

Next, write the above expression for δ3 as δ3 = where B=

1 (p3 − 2Bp1 p2 + Dp31 ), 6(1 + 3γ)

γ(2γ − 5) 2(1 + γ)(1 + 2γ)

and

D=

γ(−17 + 23γ − 10γ2 + 4γ3 ) . 6(1 + γ)3 (1 + 2γ)

We again use Theorem 3.1.8, so that 0 ⩽ B ⩽ 1, when γ ⩾ 5/2, and B(2B − 1) ⩽ D ⩽ B, when γ ⩾ γ1 , and so Theorem 3.1.8 gives the second bound for |δ3 | in Theorem 8.2.3 when γ ⩾ γ1 . Next write δ3 =

1 (p3 − 2Bp1 p2 + Bp31 + (D − B)p31 ), 6(1 + 3γ)

and note that D − B ⩾ 0 when 0.428 . . . ⩽ γ ⩽ γ1 . We again use Theorem 3.1.8 with B = D, and recalling that since 0 ⩽ B ⩽ 1, we also require that γ ⩾ 5/2, to obtain the first inequality for |δ3 | on the interval 5/2 ⩽ γ ⩽ γ1 . Thus we are left to prove the first inequality for |δ3 | on the interval 0 ⩽ γ ⩽ 1. We now use Theorem 3.1.4 to express the coefficients p2 and p3 in terms of p1 . After simplification, and since Mγ is rotationally invariant, normalizing the coefficient p1 so that p1 = p where 0 ⩽ p ⩽ 2, and finally using the triangle inequality, we have |δ3 | ⩽

(1 + 8γ)p(4 − p2 )|y| (3 + 11γ + 121γ2 + 7γ3 + 2γ4 )p3 + 12(1 + γ)(1 + 2γ)(1 + 3γ) 72(1 + γ)3 (1 + 2γ)(1 + 3γ) +

p(4 − p2 )|y|2 (4 − p2 )(1 − |y|2 ) + := ϕ(p, |y|). 24(1 + 3γ) 12(1 + 3γ)

where |y| ⩽ 1. We now use elementary calculus to find the maximum of the above expression. It is easily verified that differentiating ϕ(p, |y|) with respect to p and then |y|, and equating to zero shows that the only admissible turning points when |γ| ⩽ 1 are when p = |y| = 0, and when p = 2 and |y| =

(1 + γ + 103γ2 − 7γ3 − 2γ4 ) , 4(1 + γ)2 (1 + 8γ)

which correspond to a maximum point and a saddle point, respectively. Thus, when p = |y| = 0 we are led to the second required inequality for |δ3 |, and when p = 2 and (1 + γ + 103γ2 − 7γ3 − 2γ4 ) |y| = 4(1 + γ)2 (1 + 8γ) to the first inequality. Finally, we consider the end points of [0, 2] × [0, 1].

118 | 8 Gamma-Starlike Functions

First note that for any value of γ, ϕ(0, |y|) = and ϕ(2, |y|) =

1 1 − |y|2 ⩽ , 3(1 + 3γ) 3(1 + 3γ)

(3 + 11γ + 121γ2 + 7γ3 + 2γ4 ) . 9(1 + γ)3 (1 + 2γ)(1 + 3γ)

Next, ϕ(p, 0) =

(3 + 11γ + 121γ2 + 7γ3 + 2γ4 )p3 (4 − p2 ) + , 3 12(1 + 3γ) 72(1 + γ) (1 + 2γ)(1 + 3γ)

and ϕ󸀠 (p, 0) ⩾ 0 on 0 ⩽ p ⩽ 2 if 0 ⩽ γ ⩽ 1, again giving the first inequality for |δ3 |. Finally, ϕ(p, 1) =

(3 + 11γ + 121γ2 + 7γ3 + 2γ4 )p3 p(4 − p2 ) + 24(1 + 3γ) 72(1 + γ)3 (1 + 2γ)(1 + 3γ) (1 + 8γ)p(4 − p2 ) . 12(1 + γ)(1 + 2γ)(1 + 3γ)

+

The only critical point of this expression when 0 ⩽ γ ⩽ 1 is when p = 0 and checking the values at the end points gives the first inequality for |δ3 | once more. The first inequality is sharp when p1 = p2 = p3 = 2, and the second is sharp when p1 = 0 and p3 = 2.

8.2.2 Inverse Coefficients It was seen in Theorem 8.2.1 that finding sharp bounds for the coefficients of f(z) when f ∈ Mγ is not simple. In the next theorem we see that the same problem applies to the coefficients of the inverse function f −1 . The following gives sharp bounds for the initial coefficients, which completes estimates found for these coefficients in Darus and Thomas [39]. Theorem 8.2.4. Let f ∈ Mγ for γ ⩾ 0, and f −1 be the inverse function of f , defined on some disk |ω| < r0 (f), and be given by (1.2.3). Then, |A2 | ⩽

2 , 1+γ

5 + 7γ , { { { (1 + γ)2 (1 + 2γ) |A3 | ⩽ { { 1 { , { 1 + 2γ

2(63 + 77γ + 3γ2 + γ3 ) { { , { { 9(1 + γ)3 (1 + 3γ) |A4 | ⩽ { { 2 { { , 3(1 + 3γ) { All the inequalities are sharp.

1 (5 + √41), 2 1 γ ⩾ (5 + √41), 2

0⩽ γ⩽

0 ⩽ γ ⩽ 5, γ ⩾ 5.

8.2 Coefficient Inequalities |

119

Proof. We use the expressions for A2 , A3 , and A4 given in (1.2.5). The inequality for |A2 | is trivial. Using (8.2.1) we obtain 󵄨󵄨 (6 + 9γ + γ2 )p21 󵄨󵄨󵄨󵄨 p2 󵄨 󵄨󵄨 |A3 | = 󵄨󵄨󵄨󵄨 − 󵄨󵄨 2(1 + 2γ) 4(1 + γ)2 (1 + 2γ) 󵄨󵄨󵄨 󵄨󵄨 (6 + 9γ + γ2 )p21 󵄨󵄨󵄨󵄨 1 󵄨󵄨 󵄨󵄨 . 󵄨󵄨p2 − = 󵄨󵄨 2(1 + 2γ) 󵄨󵄨󵄨 2(1 + γ)2 󵄨 A simple application of Theorem 3.1.7 with μ = (6 + 9γ + γ2 )/(1 + γ)2 gives the inequalities for |A3 |. Using (8.2.1) we can write the expression for A4 as A4 = where B=

−1 (p3 − 2Bp1 p2 + Dp31 ) , 3(1 + 3γ)

6+γ 2(1 + γ)

and

D=

48 + 73γ + 21γ2 + 2γ3 . 6(1 + γ)3

First note that 0 ⩽ B ⩽ 1, when γ ⩾ 4, and B(2B − 1) ⩽ D ⩽ B, when γ ⩾ 5, and so applying Theorem 3.1.8 gives the second inequality for |A4 |. Next write A4 =

−1 (p3 − 2Bp1 p2 + Bp31 + (D − B)p31 ) , 3(1 + 3γ)

then since D−B ⩾ 0, when 0 ⩽ γ ⩽ 5, Theorem 3.1.8 once more gives the first inequality for |A4 | on the interval 4 ⩽ γ ⩽ 5. For the remaining interval 0 ⩽ γ ⩽ 4, we use the last part of Theorem 3.1.8. Write A4 =

−1 (18 + 13γ − 9γ2 + 2γ3 ) 3 p1 ] , [p3 − (1 + μ)p1 p2 + μp31 + 3(1 + 3γ) 6(1 + γ)3

with μ = 5/(1 + γ). Since μ lies outside [0, 1], when 0 ⩽ γ ⩽ 4, and noting that 18+13γ−9γ2 +2γ3 ⩾ 0, when γ ⩾ 0, applying Theorem 3.1.8 gives the first inequality for |A4 | on this interval, which completes the proof of the theorem. The inequality for |A2 | and the first inequalities for |A3 | and |A4 | are sharp when p1 = p2 = p3 = 2; the second inequalities for |A3 | and |A4 | are sharp when p1 = 0 and p2 = p3 = 2.

8.2.3 The Second Hankel Determinant The problem of finding sharp bounds for the second Hankel determinant H2 (2) = |a2 a4 − a23 | for subclasses of univalent functions has received much attention in recent

120 | 8 Gamma-Starlike Functions

years, and we gave examples in previous chapters. Most authors have employed the technique developed in Janteng et al. [80], which was used to find the sharp bounds for functions in S∗ and C. The same techniques give the sharp bounds for H2 (2) when f ∈ Mγ , provided that 0 ⩽ γ ⩽ 1, noting that γ = 0 and γ = 1 correspond to the results for S∗ and C above. We state the result without proof. n Theorem 8.2.5. Let f ∈ Mγ for 0 ⩽ γ ⩽ 1, and be given by f(z) = z + ∑∞ n=2 a n z . Then

{ { { H2 (2) ⩽ { { { {

(1 − γ)(9 + 142γ + 257γ2 + 80γ3 + 16γ4 ) , 9(1 + γ)4 (1 + 2γ)2 (1 + 3γ) 1 , 8

The inequality is sharp.

γ ≠ 1, γ = 1.

9 Close-to-Convex Functions In Chapter 2, we gave the basic definition of the class of close-to-convex functions, noting that they are an extension of the class S∗ of starlike functions and showed that they form a subclass of S. In this chapter, we first give a more detailed definition of close-to-convex functions and subsequently consider some of the known results for S∗ that extend to close-to-convex functions and some that do not.

9.1 Definitions and Basic Properties Definition 9.1.1. A function f ∈ A is said to be close-to-convex with argument β, if there exists a starlike function g ∈ S∗ of the form ∞

g(z) = z + ∑ b n z n ,

(9.1.1)

n=2

such that Re [e iβ

zf 󸀠 (z) ]>0 g(z)

(z ∈ 𝔻).

(9.1.2)

Let Kβ denote the set of all close-to-convex functions with argument β. The condition (9.1.2) implies that |β| < π/2, and we write K = ⋃ Kβ . |β| 0 for z ∈ 𝔻, which we denote by R. A great deal of attention has been given to this relatively simple class of functions, and many straightforward results have been obtained. We, therefore, omit a discussion of the class R, apart from noting the significant paper by London [127]. Clearly, since K ⊂ S and k ∈ K, the basic distortion theorems are valid for f ∈ K and are the best possible. Also, as in Chapter 5, it is natural to extend the definition of K to consider close-to-convex functions of, say, order α. However, relatively little significant new information is obtained by doing this. Instead, as in Chapter 6, we define the class of strongly close-to-convex functions as follows, for which interesting properties are known. Definition 9.1.2. A function f ∈ A is said to be strongly close-to-convex of order α > 0, if there exist a real number β ∈ (−π/2, π/2) and a starlike function g ∈ S∗ , such that 󵄨󵄨 󵄨 zf 󸀠 (z) 󵄨󵄨󵄨 απ 󵄨󵄨 󵄨󵄨arg [e iβ 󵄨󵄨 < ] (z ∈ 𝔻). (9.1.3) 󵄨󵄨 g(z) 󵄨󵄨󵄨 2 󵄨 https://doi.org/10.1515/9783110560961-009

122 | 9 Close-to-Convex Functions ̃ ̃ ̃ We denote this class by K(α) and observe that K(1) = K, and if α < 1, then K(α) is properly contained in K. We can permit α > 1, but in this case f is not necessarily univalent. In 1956, using the same method employed by Kaplan [84], Reade [206] established the following necessary and sufficient condition for a function f to belong ̃ to K(α). Theorem 9.1.1. Let f ∈ A and f 󸀠 (z) ≠ 0 for z = re iθ ∈ 𝔻. Then, a necessary and ̃ sufficient condition that f ∈ K(α) is that for every r ∈ (0, 1), and every pair of θ1 , θ2 with 0 ⩽ θ2 − θ1 ⩽ 2π, θ2

∫ Re [1 + re iθ θ1

f 󸀠󸀠 (re iθ ) ] dθ > −απ. f 󸀠 (re iθ )

Following Kaplan’s original paper [84], in 1958 Lewandowski [109] gave another geometric interpretation of close-to-convexity. However, in 1936, Biernacki [17] had introduced the notion of linearly accessible functions (see below), and in effect showed that these functions were the same as close-to-convex functions. Definition 9.1.3. A domain Ω is said to be linearly accessible, if its complement is the union of rays that are disjoint except that the initial point of one ray may lie on another such ray. A univalent function in the unit disk 𝔻 is said to be linearly accessible, if its range f(𝔻) is a linearly accessible domain. The concepts of a linearly accessible domain and linearly accessible functions have since been generalized by Koepf [95] as follows. A domain Ω is called (angularly) accessible of order α, α ∈ [0, 1], if it is the complement of the union of rays that are pairwise disjoint, except that the origin of one ray may lie on another one of the rays, and such that every ray is the bisector of a sector of angle (1 − α)π, which wholly lies in the complement of Ω. For α = 1, Ω is called (strictly) linearly accessible, if it is the complement of the union of rays that are pairwise disjoint, except that the origin of one ray may lie on another one of the rays. A function f ∈ A is said to be (angularly) accessible of order α, α ∈ [0, 1] if f(𝔻) is accessible of order α. In 1989, Koepf [95] proved that a function f ∈ A is close-to-convex of order α, 0 < α ⩽ 1, if, and only if, f(𝔻) is accessible of order α. ̃ ̃ We first show that the integral of a function in K(α) also belongs to K(α), for which we will need the following lemma. Lemma 9.1.1. Let f be analytic and h be convex (not necessarily normalized) in 𝔻. If for some 0 < α ⩽ 1, 󵄨 󵄨󵄨 f 󸀠 (z) 󵄨󵄨󵄨 απ 󵄨󵄨 (z ∈ 𝔻), 󵄨󵄨󵄨arg 󸀠 󵄨󵄨󵄨 < h (z) 󵄨󵄨 2 󵄨󵄨 then, for any z1 , z2 ∈ 𝔻,

󵄨󵄨 f(z2 ) − f(z1 ) 󵄨󵄨󵄨󵄨 απ 󵄨󵄨 . 󵄨󵄨arg 󵄨< 󵄨󵄨 h(z2 ) − h(z1 ) 󵄨󵄨󵄨 2

9.1 Definitions and Basic Properties

| 123

Proof. Let H be the inverse function of w = h(z). Then, H is an analytic function in the convex region h(𝔻). Let g(z) = f(H(w)). With w j = h(z j ) for j = 1, 2, we can write 1

f(z2 ) − f(z1 ) g(w2 ) − g(w1 ) = ∫ g󸀠 (w1 + (w2 − w1 )t) dt. = h(z2 ) − h(z1 ) w2 − w1 0

g󸀠 (w)

f 󸀠 (H(w))/h󸀠 (H(w))

Since s = = lies in the convex sector {s : | arg s| < απ/2}, the same is true of its integral mean value, which proves the lemma. By a repeated application of Lemma 9.1.1, Pommerenke [193] proved the following interesting result. ̃ Theorem 9.1.2. Let f ∈ K(α), 0 < α ⩽ 1 and z1 , z2 ∈ 𝔻. Then z

∫ 0

f(z2 ξ) − f(z1 ξ) ̃ dξ ∈ K(α). z2 ξ − z1 ξ

(9.1.4)

̃ Proof. If f ∈ K(α) then (9.1.3) holds. Since h1 (z) := e−iβ h(z) is a convex function, it follows that (zh󸀠 (z))󸀠 ]>0 (z ∈ 𝔻). Re [ 󸀠1 h1 (z) Applying Lemma 9.1.1 with f(z) = zh󸀠1 (z), α = 1 and z j z instead of z j (j = 1, 2), we obtain z2 zh󸀠1 (z2 z) − z1 zh󸀠1 (z1 z) Re [ (z ∈ 𝔻). (9.1.5) ]>0 h1 (z2 z) − h1 (z1 z) Let

z

h∗ (z) = ∫ 0

h1 (z2 ξ) − h1 (z1 ξ) dξ. z2 ξ − z1 ξ

Then,

z[zh󸀠∗ (z)]󸀠 z2 zh󸀠1 (z2 z) − z1 zh󸀠1 (z1 z) = , h1 (z2 z) − h1 (z1 z) zh󸀠∗ (z) and so h∗ is convex by (9.1.5). If f∗ (z) denotes the function defined by (9.1.4), then Lemma 9.1.1 shows that 󵄨 󸀠 󵄨󵄨 󵄨 󵄨󵄨󵄨 󵄨󵄨arg f∗ (z) 󵄨󵄨󵄨 = 󵄨󵄨󵄨󵄨arg f(z2 z) − f(z1 z) 󵄨󵄨󵄨󵄨 < απ (z ∈ 𝔻). 󵄨󵄨 󵄨󵄨 󵄨󵄨 󸀠 h1 (z2 z) − h1 (z1 z) 󵄨󵄨󵄨 2 󵄨󵄨 h∗ (z) 󵄨󵄨 󵄨 ̃ Thus, f∗ ∈ K(α). For the particular choice z1 = 0 and z2 = 1, Theorem 9.1.2 leads to the following corollary due to Sakaguchi [223]. ̃ Corollary 9.1.1. If f ∈ K(α), 0 < α ⩽ 1, then z

∫ 0

f(ξ) ̃ dξ ∈ K(α). ξ

124 | 9 Close-to-Convex Functions

9.2 Distortion Theorems As was noted above, if f ∈ K, the standard distortion theorems for f ∈ S given in ̃ Theorems 1.3.2, 1.3.3, and 1.3.4 hold. An extension for f ∈ K(α) can be obtained as follows. ̃ Theorem 9.2.1. If f ∈ K(α), then for z = re iθ ∈ 𝔻, (1 + r)α (1 − r)α ⩽ |f 󸀠 (z)| ⩽ , α+2 (1 + r) (1 − r)α+2 and | arg f 󸀠 (z)| ⩽ 2(α + 1) arcsin r. The results are sharp for each α > 0 and each r ∈ [0, 1), and equality occurs for the function k α given by 1 1 + z α+1 k α (z) = − 1] . [( ) 2(α + 1) 1−z ̃ Proof. Writing g(z) = zh󸀠 (z) in Definition 9.1.2, if f ∈ K(α), there exist a real number β ∈ (−π/2, π/2) and a convex function h ∈ C such that e iβ

f 󸀠 (z) = p α (z), h󸀠 (z)

(9.2.1)

where p is an analytic function with p α (0) = e iβ , and Re p(z) > 0 for z = re iθ ∈ 𝔻. It is easy to verify that α

e−iβ p α (z) ≺ (

1 + e−2iβ z ) . 1−z

Consequently, (

1+r α 1−r α ) ⩽ |e−iβ p α (z)| ⩽ ( ) 1+r 1−r

(z ∈ 𝔻),

and 󵄨󵄨 󵄨 󵄨󵄨arg [e−iβ p α (z)]󵄨󵄨󵄨 ⩽ α| arg(1 + e−2iβ z) − arg(1 − z)| ⩽ 2α arcsin r 󵄨 󵄨 Since h ∈ C, from Chapter 4 we have 1 1 ⩽ |h󸀠 (z)| ⩽ (1 + r)2 (1 − r)2 and so | arg h󸀠 (z)| ⩽ 2 arcsin r. The desired results now follow from (9.2.1).

(z ∈ 𝔻),

(z ∈ 𝔻).

9.3 Coefficient Problems

| 125

9.3 Coefficient Problems Before considering coefficient estimates for functions in K, we note that when f ∈ ̃ K(α) for α ≠ 1, little is known about the coefficients of the Taylor series for f(z). Although Pommerenke [193] showed that if λ ⩾ 0, then M(r) < (1 − r)−λ ⇒ n|a n | ⩽ C(α)n λ , finding sharp bounds for the coefficients seems difficult. It is an easy exercise to obtain sharp bounds for |a2 | and |a3 |, but finding sharp bounds for |a n | when n ⩾ 4 appears to be more difficult, and remains an open problem. This is in contrast to the class S∗ (α) where, as we saw in Chapter 5, sharp bounds for the initial coefficients |a2 |, |a3 |, and |a4 | have been found, together with exact bounds for the corresponding inverse coefficients and other functionals involving the fourth coefficient. We first note that since K ⊂ S, the Bieberbach conjecture holds for functions in K. The proof follows using a simple induction argument and the result for functions in S∗ . We omit the details, but state the theorem formally. Theorem 9.3.1. Let f ∈ K and be given by ∞

f(z) = z + ∑ a n z n .

(9.3.1)

n=2

Then, |a n | ⩽ n for n ⩾ 2, with equality for the Koebe function k(z) = z/(1 − z)2 . Using the same techniques, it is easy to see that the following extension holds for functions in Kβ . Theorem 9.3.2. Let f ∈ Kβ and be given by (9.3.1). Then, |a n | ⩽ 1 + (n − 1) cos β for n ⩾ 2. The bound is sharp for n ⩾ 2 and for all β in (−π/2, π/2).

9.3.1 The Fekete–Szegő Problem The presence of the independent starlike function g in the definition of close-toconvex functions can sometimes create significant problems in establishing sharp coefficient estimates for functions in K, but can also lead to results that differ from the case of starlike functions. One such case is the solution of the Fekete–Szegő problem. Let ω be a Schwarz function with ∞

ω(z) = ∑ α n z n . n=1

Then, for any complex number s, we have |α 2 − sα 21 | ⩽ |α 2 | + |s‖α 21 | ⩽ 1 + (|s| − 1)|α 21 | ⩽ max(1, |s|).

(9.3.2)

126 | 9 Close-to-Convex Functions Moreover, the functions ω(z) = z and ω(z) = z2 , respectively, show that the result is sharp for |s| ⩾ 1 and for |s| < 1. In 1969, Keogh and Merkes [88] partially solved the Fekete–Szegő problem for close-to-convex functions, as we state below. Theorem 9.3.3. Let f ∈ K and be of the form (1.1.2) and μ be a real number. Then, |a3 − μa22 | ⩽ max{1, 3|μ − 1|, |4μ − 3|}. If μ is outside the interval (0, 2/3), there is a close-to-convex function for which equality holds. In the same paper, Keogh and Merkes [88] proved the Fekete–Szegő inequality for the class K0 when μ is real as follows. Theorem 9.3.4. Let f ∈ K0 and be of the form (1.1.2), and μ be any real number. Then, { { { { { { { { { 2 |a3 − μa2 | ⩽ { { { { { { { { { {

3 − 4μ, 4 1 + , 3 9μ 1, 4μ − 3,

1 , 3 1 2 ⩽μ⩽ , 3 3 2 ⩽ μ ⩽ 1, 3 μ ⩾ 1. μ⩽

(9.3.3)

For each μ, there is a function in K0 such that equality holds. Proof. If f ∈ K0 , then there exists a starlike function g ∈ S∗ of the form (9.1.1) such that the function ∞ f 󸀠 (z) − g(z)/z ω(z) = 󸀠 = ∑ αn zn f (z) + g(z)/z n=1 is a Schwarz function. Comparing coefficients in the various power series expansions for the functions in this identity shows that 2a2 = b 2 + 2α 1

and 3a3 = b 3 + 2(α 1 b 2 + α 2 + α 21 ).

Thus, we have a3 − μa22 =

1 3 2 3 (b 3 − μb 22 ) + [α 2 + (1 − μ) α 21 ] 3 4 3 2 2 + (μ − ) α 1 b 2 . 3

We now use (9.3.2) and (4.1.6) as in the proof of Theorem 4.1.5 for the starlike case, to obtain |a3 − μa22 | ⩽

1 2 1 1 [1 + (|3μ − 3| − 1)|b 2 |2 ] + [1 + (|3μ − 2| − 2)|α 21 |] 3 4 3 2 1 + |3μ − 2‖α 1 ‖b 2 |. 3

9.3 Coefficient Problems | 127

If 1/3 ⩽ μ ⩽ 2/3, this becomes |a3 − μa22 | ⩽ 1 + ⩽1+

1 {(2 − 3μ)|b 22 | + 4(2 − 3μ)|α 1 ||b 2 | − 12μ|α 21 |} 12 2 − 3μ 2 1 4 |b 2 | ⩽ + , 18μ 3 9μ

since |b 2 | ⩽ 2. The result is sharp, since there is a starlike function with b 2 = 2 and b 3 = 3 and a Schwarz function with α 1 = (2 − 3μ)/3μ and α 2 = 1 − α 21 , provided that 1/3 ⩽ μ ⩽ 2/3. For 0 ⩽ μ ⩽ 1/3, we have 󵄨󵄨 a2 󵄨󵄨󵄨 󵄨 |a3 − μa22 | ⩽ 3μ 󵄨󵄨󵄨󵄨a3 − 2 󵄨󵄨󵄨󵄨 + (1 − 3μ)|a3 | ⩽ 3 − 4μ. 3 󵄨󵄨 󵄨󵄨 For the remaining choice of μ, (9.3.3) is a consequence of Theorem 9.3.3. The sharpness for μ not in the interval (1/3, 2/3) follows from Theorem 4.1.5, since S∗⊂ K0 . Subsequently, Koepf [93] proved that Theorem 9.3.4 is, indeed, true for the whole class K of close-to-convex functions, the proof of which we omit.

9.3.2 The Zalcman Conjecture In 1960, L. Zalcman posed the conjecture that if f ∈ S and is given by (9.3.1), then |a2n − a2n−1 | ⩽ (n − 1)2 ,

for n ⩾ 2,

(9.3.4)

with equality only for the Koebe function or its rotations. That the Zalcman conjecture implies the Bieberbach conjecture was first pointed out by Brown and Tsao [25]. To see this, let ϵ > 0 be such that |a n | < n(1 + ϵ) for all n ∈ ℕ (one may choose ϵ = e − 1). If the inequality (9.3.4) holds for all n, then, |a n |2 ⩽ |a2n − a2n−1 | + |a2n−1 | < (n − 1)2 + (2n − 1)(1 + ϵ) < n2 (1 +

ϵ 2 ) , 2

and iterating this procedure gives |a n | ⩽ n. A well-known consequence of the area theorem (see Theorem 1.2.1) shows that (9.3.4) holds for n = 2. For f ∈ S, Krushkal [98, 99] proved the conjecture for n = 3, and later for n = 4, 5, and 6. For a simple and elegant proof of the conjecture for n = 3, we refer to [99]. The conjecture for f ∈ S remains open for n ⩾ 7. For subclasses of S, the situation is much better. In 1986, Brown and Tsao [25] proved the conjecture for both starlike and typically real functions, and in 1988, Ma [131] proved that the conjecture is also true for close-to-convex functions. In fact,

128 | 9 Close-to-Convex Functions

Ma’s result established the conjecture for a wider class of functions, as we shall now demonstrate. Before we proceed we need some definitions. Suppose that X is a linear topological space and U ⊂ X. The set U is called convex if tx + (1 − t)y ∈ U whenever x, y ∈ U and 0 ⩽ t ⩽ 1. The convex hull of U, denoted by co U, is defined as the intersection of all convex sets containing U, and the closed convex hull of U, denoted by co U, is defined as the intersection of all closed convex sets containing U. It is easy to verify that the convex hull of U consists of all elements of the form ∑nk=1 t k x k , where x k ∈ U, t k ⩾ 0, and ∑nk=1 t k = 1 with varying n. Moreover, if F is a family of analytic functions in 𝔻, then co F consists of all functions that are limits of functions of the form described above. For U ⊂ V ⊂ X, we say that U is an extremal subset of V if u = tx + (1 − t)y, where u ∈ U, x, y ∈ V, and 0 < t < 1, then x and y both belong to U. An extremal subset of V consisting of just one point is called an extreme point of V. We denote the set of extreme points of U by E(U). Thus u ∈ E(U) if u ∈ U, and u = tx + (1 − t)y where x, y ∈ U, 0 < t < 1 implies x = y. Let HK denote the closed convex hull of the class K of close-to-convex functions. Then Brickman et al. [24] showed that HK consists of functions represented by f(z) = ∫ [z −

1 is (e + e it )z2 ] (1 − e it z)−2 dμ(t, s), 2

(9.3.5)

R

where μ varies over the set of probability measures on R = [0, 2π] × [0, 2π]. It was for this larger class HK that Ma [131] proved the Zalcman conjecture. Theorem 9.3.5. Let f ∈ HK and be of the form (9.3.1). Then, |a2n − a2n−1 | ⩽ (n − 1)2

for n ⩾ 4.

(9.3.6)

The inequality (9.3.6) is sharp. The proof of the Theorem 9.3.5 relies on the following lemma. Lemma 9.3.1. For −1 ⩽ x, y ⩽ 1 and n ⩾ 4, let F n (x, y) = (n2 − 6n + 1)x2 − 2(n − 1)2 xy + 4(n − 1)√1 − x2 √1 − y2 + (n − 1)2 y2 .

(9.3.7)

Then, F n (x, y) ⩽ 4(n − 1)2 − 4n. Equality holds if, and only if, (x, y) = (1, −1) or (x, y) = (−1, 1). Proof. By applying the inequality 2ab ⩽ a2 + b 2 , we have F n (x, y) ⩽ 4(n − 1) + G n (x, y), where G n (x, y) = (n2 − 8n + 3)x2 − 2(n − 1)2 xy + (n2 − 4n + 3)y2 .

(9.3.8)

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| 129

If n ⩾ 8, it is clear that G n (x, y) ⩽ 4n2 − 16n + 8,

(9.3.9)

which implies (9.3.8). Equality holds in (9.3.9) if, and only if, (x, y) = (1, −1) or (x, y) = (−1, 1), and in (9.3.8), since 1 − x2 = 1 − y2 at (1, −1) and (−1, 1). Using elementary calculus, it is not difficult to show that the functions G5 (x, y) = −12x2 − 32xy + 8y2 ,

G6 (x, y) = −9x2 − 50xy + 15y2

and G7 (x, y) = −4x2 − 72xy + 24y2 are maximum only at (1, −1) and (−1, 1). This gives (9.3.8) for n = 5, 6 and 7. From (9.3.7), we note that for −1 < x < 1 and −1 < y < 1, ∂F4 (x, y) = −14x − 18y − 12x√(1 − y2 )/(1 − x2 ), ∂x ∂F4 (x, y) = −18x + 18y − 12y√(1 − x2 )/(1 − y2 ). ∂y Both of the above vanish only if (x + y)(7x − 9y) = 0. For x = −y and −1 < y < 1, G4 (x, y) = −13x2 − 18xy + 3y2 = 8y2 < 8. For x = 9y/7 and −7/9 < y < 7/9, G4 (x, y) = −2040y2 /49 < 8. Thus F4 (x, y) ⩽ 12 + G4 (x, y) < 20, if −1 < x < 1 and −1 < y < 1. Finally, it is easy to check that F4 (x, y) maximizes on the boundary of {(x, y) : − 1 ⩽ x, y ⩽ 1} only at (1, −1) and (−1, 1), which completes the proof of the lemma. Proof of Theorem 9.3.5. Since the functional |a2n − a2n−1 | is invariant under rotations, it suffices to prove that Re(a2n − a2n−1 ) ⩽ (n − 1)2 . If f ∈ HK, and is of the form (9.3.1), then from (9.3.5), we have (writing μ = μ(t, s)) a n = ∫ [ne i(n−1)t −

1 is (e + e it )(n − 1)e i(n−2)t ] dμ 2

R

=

1 ∫ [(n + 1)e i(n−1)t − (n − 1)e i(s+(n−2)t)] dμ, 2 R

130 | 9 Close-to-Convex Functions

and so Re(a2n − a2n−1 ) } 1{ = {∫ [(n + 1) cos((n − 1)t) − (n − 1) cos(s + (n − 2)t)] dμ } 4 {R }

2

} 1{ − {∫ [(n + 1) sin((n − 1)t) − (n − 1) sin(s + (n − 2)t)] dμ } 4 } {R

2

− ∫ [n cos(2(n − 1)t) − (n − 1) cos(s + (2n − 3)t)] dμ.

(9.3.10)

R

Applying the Cauchy–Schwarz inequality to the first integral in (9.3.10), we obtain Re(a2n − a2n−1 ) ⩽

1 ∫ [(n + 1)2 cos2 ((n − 1)t) + (n − 1)2 cos2 (s + (n − 2)t) 4 R

− 2(n + 1)(n − 1) cos((n − 1)t) cos(s + (n − 2)t)] dμ − ∫ [n cos(2(n − 1)t) − (n − 1) cos(s + (2n − 3)t)] dμ R

=n+

1 ∫ [(n2 − 6n + 1) cos2 ((n − 1)t) + (n − 1)2 cos2 (s + (n − 2)t) 4 R

− 2(n − 1)2 cos((n − 1)t) cos(s + (n − 2)t) − 4(n − 1) sin((n − 1)t) sin(s + (n − 2)t)] dμ ⩽n+

1 ∫ F n (cos((n − 1)t), cos(s + (n − 2)t)) dμ, 4 R

where F n is defined by (9.3.7). It therefore follows from Lemma 9.3.1 that Re(a2n − a2n−1 ) ⩽ n +

1 ∫[4(n − 1)2 − 4n] dμ = (n − 1)2 . 4 R

If equality holds, we must have cos((n − 1)t) = 1

and

cos(s + (n − 2)t) = −1,

or cos((n − 1)t) = −1

and

cos(s + (n − 2)t) = 1.

(9.3.11)

9.3 Coefficient Problems

| 131

Moreover, the support of μ is contained in E1 or E2 , where E1 = {(t, s) : t =

2kπ 2(n − 2)kπ , s = (2m − 1)π − , k = 0, 1, . . . , n − 1, n−1 n−1

and m is an integer in [ E2 = {(t, s) : t =

1 (n − 2)k 3 (n − 2)k + , + ]} , 2 n−1 2 n−1

(2k + 1)π (n − 2)(2k + 1)π , s = 2mπ − , k = 0, 1, . . . , n − 2, n−1 n−1

and m is an integer in [

(n − 2)(2k + 1) (n − 2)(2k + 1) ,1+ ]} . 2(n − 1) 2(n − 1)

Since sin((n − 1)t) and sin(s + (n − 2)t) vanish on E1 and E2 , it follows that (9.3.11) is sharp in HK, and so is (9.3.6). This completes the proof. If f ∈ K is of the form (9.3.5), and μ has its support in E i (i = 1, 2), then the univalence of f implies that f has only one pole of order 2 on |z| = 1, from which we can show that f(z) =

z − (e is + e it )z2 /2 , (1 − e it z)2

(9.3.12)

where (t, s) is in E i . Therefore, noting that K ⊂ HK, we have the following corollary. Corollary 9.3.1. If f ∈ K and is of the form (9.3.1), then |a2n − a2n−1 | ⩽ (n − 1)2

for

n ⩾ 4,

with equality if, and only if, f(z) is given by (9.3.12). Also since S∗ ⊂ K, we have the following result for starlike functions. Corollary 9.3.2. If f ∈ S∗ and is of the form (9.3.1), then |a2n − a2n−1 | ⩽ (n − 1)2

for

n ⩾ 4,

with equality when f is the Koebe function. More general versions of the Zalcman conjecture have also been considered by Brown and Tsao [25], and Ma [133], where the authors considered the functional Φ λ (f) = λa2n − a2n−1 for a real parameter λ. As pointed out by Pfluger [186], the quantity λa2n − a2n−1 arises quite naturally when considering coefficient problems. Indeed, if f ∈ S, then the coefficients of (f(z2 ))1/2 and 1/f(1/z) contain expressions of the form Φ λ (f). Note that when n = 2, the functional Φ λ (f) reduces to the classical Fekete–Szegő functional Λ λ (f) = λa22 − a3 which, as we have seen, has a long and rich history. Below we state a result of Li and Ponnusamy [114], which extends Theorem 9.3.5 to the functional Φ λ (f).

132 | 9 Close-to-Convex Functions Theorem 9.3.6. Let f ∈ K, and be of the form (9.3.1) and n ⩾ 3. 3n + √5n2 − 4n , then (i) If λ ⩾ n2 + n |λa2n − a2n−1 | ⩽ λn2 − (2n − 1), with equality when f is the Koebe function. 3n + √5n2 − 4n 2n 5 for some function in S, showing that Robertson’s conjecture is not true for all functions in S. In order to prove Robertson’s conjecture for functions in K, Leung made use of the following three lemmas. The first two are needed to prove the third, which is then used in the proof of the theorem. ∞ k k Lemma 9.3.2. If ∑∞ k=0 D k z = exp(∑ k=1 A k z ), and both functions are analytic for z ∈ 𝔻, then, n−1 n k 2 |A k |2 1 ∑ |D k |2 ⩽ n exp { ∑ (k|A k |2 − − ) + 1} , n k k=0 k=1

with equality if, and only if, A k = c k /k, |c| = 1, k = 1, 2, . . . , n − 1. Lemma 9.3.3. With the same notation as in Lemma 9.3.2, n−1

∑ |D k |2 + k=0

⩽

|D n |2 2

n n 1 k2 2n + 1 k2 1 exp { ∑ ( + k − ) |A k |2 − ∑ − + 1} , 2 n(2n + 1) n k 2n +1 k=1 k=1

with equality if, and only if, A k = c k /k, |c| = 1, k = 1, 2, . . . , n. n Lemma 9.3.4. If g ∈ S∗ with g(z) = z + ∑∞ n=2 b n z , then for each positive integer n, there exists ζ with |ζ| = 1 such that n−1

1 + ∑ |b k+1 − ζb k |2 + k=1

|b n+1 − ζb n |2 1 ⩽n+ . 2 2

In Chapter 1, we referred to the Lebedev–Milin inequalities. Lemma 9.3.2 is a variant of one of these, which was proved in Aharonov’s lecture notes [2, Theorem 1.2]. Alternatively, a similar proof can be found in Pommerenke’s book [198]. For the proof of Lemma 9.3.3 and Lemma 9.3.4 we refer to Leung [106].

9.3 Coefficient Problems |

135

Leung’s theorem establishing Robertson’s conjecture for the class K is the following. k Theorem 9.3.8. If f ∈ K, with f(z) = z + ∑∞ k=2 a k z , then,

|n|a n | − m|a m ‖ ⩽ |n2 − m2 | for all positive n and m. Strict inequality holds for all n and m, with n ≠ m, unless f is a rotation of the Koebe function. Proof. We use the same notation as Leung. Since f ∈ K, we can write e iα zf 󸀠 (z) = r(z)g(z), where Re r(z) > 0, z ∈ 𝔻, and ∞

r(z) = e iα + ∑ r k z k . k=1

Thus, for |ζ| = 1, we can write e iα (1 − ζz)f 󸀠 (z) = (1 − ζz)

g(z) r(z), z

or ∞

∞

∞

k=1

k=1

e iα (1 + ∑ [(n + 1)a n+1 − ζna n ]z n ) = (1 + ∑ (b k+1 − ζb k )z k ) (e iα + ∑ r k z k ) . n=1

Comparing the coefficients of

zn

on both sides, we obtain

e iα [(n + 1)a n+1 − ζna n ] = r n + r n−1 (b 2 − ζb 1 ) + ⋅ ⋅ ⋅ + e iα (b n+1 − ζb n ) n

= ∑ r n−k D k , k=0

where r0 = e iα and D k = b k+1 − ζb k . To the first n terms of the right side we apply the inequality |ab| ⩽ |a/2|2 + |b|2 and to the last term the inequality |b| ⩽ 1/2 + |b|2 /2. This gives 󵄨󵄨 r n 󵄨󵄨2 󵄨󵄨 r n−1 󵄨󵄨2 󵄨󵄨 |(n + 1)a n+1 − ζna n | ⩽ 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 + 1 + 󵄨󵄨󵄨󵄨 󵄨 󵄨2󵄨 󵄨 2 󵄨󵄨

1 1 + |b n+1 − ζb n |2 2 2 n−1 1 1 ⩽ n + + 1 + ∑ |b k+1 − b k |2 + |b n+1 − ζb n |2 , 2 2 k=1 + |b 2 − b 1 |2 + ⋅ ⋅ ⋅ +

where we have applied the inequality |r k | ⩽ 2, for k = 1, 2, . . . . By Lemma 9.3.4 |(n + 1)a n+1 − ζna n | ⩽ n +

1 1 + n + = 2n + 1, 2 2

for some ζ with |ζ| = 1, but by the triangle inequality |(n + 1)|a n+1 | − n|a n ‖ ⩽ |(n + 1)a n+1 − ζna n |,

136 | 9 Close-to-Convex Functions

and so |(n + 1)|a n+1 | − n|a n ‖ ⩽ 2n + 1. The general case of the inequality now follows from this by induction. For equality, a close inspection of the proof shows that r(z) has the form (1 + tz)/(1 − tz), with |t| = 1. Thus, zf 󸀠 (z) = ((1 + tz)/(1 − tz))g(z). Multiplying both sides by (1 − tz), we obtain ∞

∞

n=1

n=1

(1 − tz) ∑ na n z n = (1 + tz) ∑ b n z n . Comparing the coefficient of z n+1 in the above gives (n + 1)a n+1 − tna n = b n+1 + tb n . Hence, 2n + 1 = |(n + 1)|a n+1 | − n|a n ‖ ⩽ |b n+1 + tb n |. Thus, the starlike function g must be a rotation of the Koebe function with the form z/(1 − tz)2 , and so f(z) = z/(1 − tz)2 also.

9.3.5 Logarithmic Coefficients For a function f ∈ S, we recall that the logarithmic coefficients γ n for n ⩾ 1 are defined by ∞ f(z) log (z ∈ 𝔻), = 2 ∑ γn zn z n=1 and play a central role in the theory of univalent functions. Finding exact bounds for |γ n | has proved difficult, and more attention has been given to results of an average nature (see, e.g., Duren and Leung [43], Duren [45]). For the Koebe function k(z) = z/(1 − z)2 , γ n = 1/n, and since the Koebe function plays the role of the extremal function for many extremal problems in the class S, it could be expected that |γ n | ⩽ 1/n holds when n ⩾ 1 for functions in S. This is not true, even in order of magnitude. Indeed, there exists a bounded function f ∈ S with logarithmic coefficients γ n ≠ O(n−0.83 ) (see Duren [45, Theorem 8.4]), and the correct order of growth of γ n as n → ∞ is unknown. Using the Fekete–Szegő inequality for f ∈ S in (1.2.1), Duren [45, Theorem 3.8], we can obtain the sharp estimate |γ2 | ⩽

1 (1 + 2e−2 ) = 0.635 . . . . 2

For n ⩾ 3, the problem seems much harder, and no significant upper bound for |γ n | when f ∈ S appears to be known.

9.3 Coefficient Problems

| 137

As noted in Chapter 4, when f ∈ S∗ , the inequality |γ n | ⩽ 1/n, for n ⩾ 1 is a trivial consequence of the representation of f in terms of functions of positive real part. It was claimed by Elhosh [47] that this inequality extends to functions in K0 , but this was later shown not to be the case by Girela [51]. We note that if f ∈ K, then the sharp inequality |γ1 | ⩽ 1 follows at once from (1.2.2), with equality for the Koebe function. Again, using the Fekete–Szegő inequality for functions in K in (1.2.2) (see Koepf [93]), it follows at once that 1 󵄨󵄨󵄨 1 󵄨󵄨󵄨 11 |γ2 | ⩽ 󵄨󵄨󵄨a3 − a22 󵄨󵄨󵄨 ⩽ = 0.6111 . . . 2 󵄨󵄨 2 󵄨󵄨 18 and the inequality is sharp for the function f0 defined by f0󸀠 (z) =

1 1 1 + z 2 1 + z2 ( ). + 2 3 1 − z 3 1 − z2 (1 − z)

(9.3.19)

At this point we note that this result demonstrates the irregular behavior of the logarithmic coefficients of functions in K compared to those in S∗ referred to above. It was shown by Thomas [247] that if f ∈ K0 , then |γ3 | ⩽ 7/12, with the added restriction that the second coefficient of the associated starlike function g is real. However, the following theorem by Ali and Vasudevarao [5] shows that the bound 7/12 can be improved, and without the added restriction, with the bound obtained close to the one conjectured later in this section. Theorem 9.3.9. If f ∈ K0 , then |γ3 | ⩽ (3 + 4√2)/18 = 0.4809 . . . . Proof. Let f ∈ K0 be of the form (9.3.1). Then there exists a starlike function g of the form (9.1.1) and a Carathéodory function p ∈ P of the form ∞

p(z) = 1 + ∑ c n z n

(9.3.20)

n=1

such that zf 󸀠 (z) = g(z)p(z).

(9.3.21)

Comparing the coefficients on the both sides of (9.3.21) gives 1 (b 2 + c1 ) 2 1 a3 = (b 3 + b 2 c1 + c2 ) 3 1 a4 = (b 4 + b 3 c1 + b 2 c2 + c3 ). 4

a2 =

From (1.2.2) 2γ3 = a4 − a2 a3 + =

1 3 a 3 2

1 1 [(6b 4 − 4b 2 b 3 + b 32 ) + 2c1 (b 3 − b 22 ) + b 2 (2c2 − c21 ) + c31 − 4c1 c2 + 6c3 ] . 24 2

138 | 9 Close-to-Convex Functions

Using (3.1.6) and (3.1.8) to write c2 and c3 in terms of c1 , we obtain 48γ3 = (6b 4 − 4b 2 b 3 + b 32 ) + 2c1 (b 3 − +

1 2 b ) + b 2 x(4 − c21 ) 2 2

3 1 3 c + c1 x(4 − c21 ) − c1 x2 (4 − c21 ) + 3(4 − c21 )(1 − |x|2 )t, 2 1 2

(9.3.22)

where |x| ⩽ 1 and |t| ⩽ 1. If γ3 (g) denotes the third logarithmic coefficient of g(z), then, as was pointed out above, 1 󵄨󵄨󵄨 1 󵄨󵄨󵄨 1 |γ3 (g)| = 󵄨󵄨󵄨b 4 − b 2 b 3 + b 32 󵄨󵄨󵄨 ⩽ . 2 󵄨󵄨 3 󵄨󵄨 3 Since g ∈ S∗ , using this inequality and Theorem 4.1.5 we obtain 󵄨󵄨 1 1 󵄨󵄨󵄨 󵄨 |6b 4 − 4b 2 b 3 + b 32 | ⩽ 6|b 4 − b 2 b 3 + b 32 | + 2|b 2 | 󵄨󵄨󵄨b 3 − b 22 󵄨󵄨󵄨 ⩽ 8. 󵄨󵄨 3 2 󵄨󵄨

(9.3.23)

Since the class K0 and the functional |γ3 | are invariant under rotation, without loss of generality we can assume that c1 = c, where 0 ⩽ c ⩽ 2. Taking the modulus on both the sides of (9.3.22), applying the triangle inequality, and then using the inequality (9.3.23) and Theorem 4.1.5, it follows that 󵄨󵄨 1 󵄨󵄨 3 󵄨 󵄨 48|γ3 | ⩽ 8 + 2c + 2|x|(4 − c2 ) + 󵄨󵄨󵄨 c3 + cx(4 − c2 ) − cx2 (4 − c2 )󵄨󵄨󵄨 + 3(4 − c2 )(1 − |x|2 ), 󵄨󵄨 2 󵄨󵄨 2 where we have also used the fact that |t| ⩽ 1. Now let x = re iθ , where 0 ⩽ r ⩽ 1 and 0 ⩽ θ ⩽ 2π. For simplicity, we write cos θ = p, to obtain 󵄨 󵄨 48|γ3 | ⩽ ψ(c, r) + 󵄨󵄨󵄨ϕ(c, r, p)󵄨󵄨󵄨 =: F(c, r, p),

(9.3.24)

where ψ(c, r) = 8 + 2c + 2r(4 − c2 ) + 3(4 − c2 )(1 − r2 ) and 9 1 ϕ(c, r, p) = [ c6 + c2 r2 (4 − c2 )2 + c2 r4 (4 − c2 )2 + c4 (4 − c2 )rp 4 4 1/2 3 − c4 r2 (4 − c2 )(2p2 − 1) − 3c2 (4 − c2 )r3 p] . 2 Thus, we need to find the maximum value of F(c, r, p) over the rectangular cube R := [0, 2] × [0, 1] × [−1, 1]. We first find the maximum value of F(c, r, p) on the boundary of R, i.e., on the six faces of R. Let R1 = [0, 1] × [−1, 1], R2 = [0, 2] × [−1, 1] and R3 = [0, 2] × [0, 1]. By using elementary calculus one can verify that max F(0, r, p) = max ψ(0, r) = ψ (0,

(r,p)∈R 1

0⩽r⩽1

1 64 )= = 21.33, 3 3

max F(2, r, p) = 16,

(r,p)∈R 1

2 16 (9 + √6) = 20.3546, max F(c, 0, p) = F ( (3 − √6), 0, p) = 3 9 2 8 max F(c, 1, p) = F ( (2√7 − 1), 1, −1) = (403 + 112√7) = 23.023. 9 243 (c,p)∈R 2 (c,p)∈R 2

9.3 Coefficient Problems | 139

On the face p = −1, we have { { ψ(c, r) + η1 (c, r) F(c, r, −1) = { { ψ(c, r) − η1 (c, r) {

for

η1 (c, r) ⩾ 0,

for

η1 (c, r) < 0,

where η1 (c, r) = c3 (3r2 + 2r + 1) − 4cr(3r + 2), and (c, r) ∈ R3 := [0, 2] × [0, 1]. A routine calculation now shows that 1 8 max F(c, r, −1) = F (2(√2 − 1), (1 + √2), −1) = (3 + 4√2) = 23.0849. 3 3

(c,r)∈R 3

In a similar way, one can verify that 1 1 16 (28 + 19√19) = 21.89. max F(c, r, 1) = F ( (10 − 2√19), , 1) = 3 3 81

(c,r)∈R 3

Let S󸀠 = {(c, r, p) ∈ R : ϕ(c, r, p) = 0}. Then, max F(c, r, p) ⩽ max ψ(c, r) = ψ ( (c,r)∈R 3

(c,r,p)∈S󸀠

3 1 649 , )= = 21.6333. 10 3 30

We now show that F(c, r, p) has no maximum at any interior point of R \ S󸀠 . Suppose that F(c, r, p) has a maximum at an interior point of R\S󸀠 . Then at such a point ∂F/∂c = 0, ∂F/∂r = 0, and ∂F/∂p = 0. Note that ∂F/∂c, ∂F/∂r, and ∂F/∂p may not exist at points in S󸀠 . From ∂F/∂p = 0 (for points in the interior of R \ S󸀠 ), a straightforward, but laborious calculation gives p=

3c2 r2 + c2 − 12r2 . 6c2 r

(9.3.25)

Substituting the value of p in (9.3.25) into ∂F/∂c = 0 and ∂F/∂r = 0 and simplifying (again, a long and laborious calculation), we obtain 3√6c3 (1 − 3r2 ) + 12[c(3r2 − 2r − 3) + 1]√c2 + 2 + 4√6c = 0, 6√c2 + 2

(9.3.26)

and (4 − c2 ) {[√6(c2 + 2) − 6] r + 2} = 0.

(9.3.27)

Since 0 < c < 2, solving the equation (9.3.27) for r, gives r=

2 . √ 6 − 6(c2 + 2)

Substituting the value of r in (9.3.28) into (9.3.26), and simplifying gives 3c3 + 6c − (6c − 2)√6 (c2 + 2) = 0, which on squaring gives 3(c2 + 2)(3c4 − 66c2 + 48c − 8) = 0.

(9.3.28)

140 | 9 Close-to-Convex Functions Clearly, c2 + 2 ≠ 0 in 0 < c < 2. On the other hand, the polynomial q(c) = 3c4 − 66c2 + 48c − 8 has exactly two roots in (0, 2), one in (0, 1/3), and the another in (1/3, 1/2). This can be seen using the well-known Sturm theorem for isolating real roots (for the sake of brevity, we omit the details). By solving the equation q(c) = 0 numerically, we obtain two approximate roots as 0.2577 and 0.4795 in (0, 2). However, the corresponding values of p obtained from (9.3.28) and (9.3.25) are −23.6862 and −6.80595, which do not belong to (−1, 1). This proves that F(c, r, p) has no maximum in the interior of R \ S󸀠 . Thus combining all the above cases we find that max F(c, r, p) = F (2(√2 − 1),

(c,r,p)∈R

1 8 (1 + √2), −1) = (3 + 4√2) = 23.0849, 3 3

and, hence, from (9.3.24) we obtain |γ3 | ⩽

1 (3 + 4√2) = 0.4809, 18

which completes the proof. By fixing the starlike function g, the inequality Re [e iβ

zf 󸀠 (z) ]>0 g(z)

(z ∈ 𝔻)

(9.3.29)

gives specific subclasses of close-to-convex functions. One such subclass is the class of close-to-convex functions with respect to the Koebe function, where we choose g(z) = k(z) = z/(1 − z)2 , which we denote by Kβ (k). In this case, the inequality (9.3.29) becomes (z ∈ 𝔻). (9.3.30) Re [e iβ (1 − z)2 f 󸀠 (z)] > 0 Several authors (e.g., Hengartner and Schober [73], Kowalczyk and Lecko [96], and Marjono and Thomas [138]) have studied the class Kβ (k), satisfying the condition (9.3.30), which is naturally easier to deal with. Geometrically, (9.3.30) says that the function h := e iβ f has the boundary normalization lim h−1 (h(z) + t) = 1, t→∞

and h(𝔻) is a domain such that {w + t : t ⩾ 0} ⊂ h(𝔻) for every w ∈ h(𝔻). Clearly, the image domain h(𝔻) is convex in the positive direction of the real axis. Let CR+ := K0 (k) be the class of close-to-convex functions with respect to the Koebe function with argument 0 (see Section 2.5), that is, CR+ = {f ∈ A : Re(1 − z)2 f 󸀠 (z) > 0, for z ∈ 𝔻} . Then, functions in CR+ are convex in the positive direction of the real axis, but we note that the class CR+ is not rotationally invariant.

9.3 Coefficient Problems |

141

Now let S∗2 denote the class of odd starlike functions in S, and F1 denote the class of close-to-convex functions with respect to odd starlike functions, with argument 0. That is, F1 = {f ∈ A : Re [

zf 󸀠 (z) ] > 0, for z ∈ 𝔻, and for some g ∈ S∗2 } . g(z)

We note that the class F1 is rotationally invariant. Ali and Vasudevarao [5, 7] have found sharp upper bounds for |γ n |, when n = 1, 2, 3, for functions in the classes CR+ and F1 . We note that since CR+ ⊂ K and both the Koebe function and the function f defined by (9.3.19) belong to the class CR+ , the sharp inequalities |γ1 | ⩽ 1 and |γ2 | ⩽ 11/18 also hold for functions in CR+ . The result for γ3 in CR+ is the following. Theorem 9.3.10. Let f ∈ CR+ and be of the form (9.3.1) with 1 ⩽ a2 ⩽ 2. Then, |γ3 | ⩽

1 (28 + 19√19) = 0.4560. 243

(9.3.31)

The inequality is sharp. Proof. We sketch the proof, which is similar to that used in Theorem 9.3.9. If f ∈ CR+ , then there exists a Carathéodory function p ∈ P of the form (9.3.20) such that zf 󸀠 (z) = g(z)p(z), where g(z) = z/(1 − z)2 . Following the same method as used in the proof of Theorem 9.3.9, and noting that g(z) = k(z) = z+2z2 +3z3 +4z4 +. . . , a simple computation in (9.3.22) shows that 48γ3 = 8 + 2c1 +

1 3 3 c + (4 − c21 ) (2x + c1 x − c1 x2 ) + 3(4 − c21 )(1 − |x|2 )t, (9.3.32) 2 1 2

where |x| ⩽ 1 and |t| ⩽ 1. Since 1 ⩽ a2 ⩽ 2 and 2a2 = 2 + c1 , then 0 ⩽ c1 ⩽ 2. Taking the modulus on both the sides of (9.3.32), applying the triangle inequality, and writing c1 = c, it follows that 󵄨󵄨 󵄨󵄨 1 3 󵄨 󵄨 48|γ3 | ⩽ 󵄨󵄨󵄨8 + 2c + c3 + (4 − c2 ) (2x + cx − cx2 )󵄨󵄨󵄨 + 3(4 − c2 )(1 − |x|2 ), 󵄨󵄨 󵄨󵄨 2 2 where we have also used the fact that |t| ⩽ 1. Let x = re iθ where 0 ⩽ r ⩽ 1 and 0 ⩽ θ ⩽ 2π, and again for simplicity, write cos θ = p, so that 󵄨 󵄨 48|γ3 | ⩽ ψ(c, r) + 󵄨󵄨󵄨ϕ(c, r, p)󵄨󵄨󵄨 =: F(c, r, p),

(9.3.33)

where ψ(c, r) = 3(4 − c2 )(1 − r2 ) and ϕ(c, r, p) = [(8 + 2c +

9 1 3 2 2 2 c ) + r (4 − c2 ) (4 + c2 + c2 r2 + 4c − 6crp − 3c2 rp) 2 4

+2(4 − c2 ) (8 + 2c +

1/2 1 3 3 c ) (2rp + crp − cr2 (2p2 − 1))] . 2 2

142 | 9 Close-to-Convex Functions

Thus, we need to find the maximum value of F(c, r, p) over the rectangular cube R = [0, 2] × [0, 1] × [−1, 1]. By proceeding similarly as in the proof of Theorem 9.3.9, we can prove that F(c, r, p) has no maximum at an interior point of R, and 1 1 16 max F(c, r, p) = F ( (10 − 2√19), , 1) = (28 + 19√19) = 21.8902 . . . , 3 3 81 (c,r,p)∈R and hence from (9.3.33) we obtain 1 |γ3 | ⩽ (28 + 19√19) = 0.4560 . . . . 243 An examination of the proof shows that equality holds in (9.3.31) if we choose c1 = c = (10−2√19)/3, x = 1/3 and t = 1 in (9.3.32). For such values of c1 , x and t, the relations (3.1.6) and (3.1.8) give c2 = 2(97−20√19)/27 and c3 = (2050−362√19)/243. A function p ∈ P with the first three coefficients c1 , c2 and c3 above is given by 1+z 1 + uz 1 + uz +λ +λ 1−z 1 − uz 1 − uz 2 1 = 1 + (10 − 2√19)z + (97 − 20√19)z2 3 27 1 + (9.3.34) (2050 − 362√19)z3 + . . . , 243 where λ = (−13 + 4√19)/18 and u = α + i√1 − α 2 , with α = −(1 + √19)/9. Hence, the inequality (9.3.31) is sharp for a function f defined by (1 − z)2 f 󸀠 (z) = p(z), where p(z) is given by (9.3.34). This completes the proof. p(z) = (1 − 2λ)

In view of the above two theorems, Ali and Vasudevarao [5] made the following conjecture. Conjecture 9.3.1. The sharp bounds for |γ3 | when f ∈ K0 and f ∈ CR+ are the same, i.e., 1 (28 + 19√19) = 0.4560. |γ3 | ⩽ 243 Using the same method employed in Theorem 9.3.9 and Theorem 9.3.10, Ali and Vasudevarao [7] obtained the following sharp inequalities for the first three logarithmic coefficients when f ∈ F1 . We state the results below. Theorem 9.3.11. Let f ∈ F1 and be of the form (9.3.1). Then, |γ1 | ⩽ 1/2, |γ2 | ⩽ 1/2 and 1 |γ3 | ⩽ (95 + 23√46). 972 The inequalities are sharp. Similar results have been found by Pranav Kumar and Vasudevarao [202] for the following two subclasses of close-to-convex functions: F2 := {f ∈ A : Re[(1 − z)f 󸀠 (z)] > 0, for z ∈ 𝔻} F3 := {f ∈ A : Re[(1 − z + z2 )f 󸀠 (z)] > 0, for z ∈ 𝔻} .

9.4 Growth Estimates

| 143

9.3.6 The Second Hankel Determinant We saw in Chapter 4 that when f ∈ S∗ , the second Hankel determinant |H2 (2)| = |a2 a4 − a23 | ⩽ 1, which is of course sharp when f is the Koebe function. It is a relatively simple matter to obtain a poor upper bound for |H2 (2)| for the whole class of closeto-convex functions, but finding the sharp upper bound of |H2 (2)| remains an open problem. Again, the question arises as to whether this upper bound is 1. In a recent paper by Rˇaducanu and Zaprawa [217], it was shown that when f ∈ K0 , |H2 (2)| < 1.242 . . . , which is the best bound known to date. We also note that when f ∈ CR+ , Marjono and Thomas [138] proved the following. Theorem 9.3.12. Let f ∈ CR+ be of the form (9.3.1), with a2 real. Then, |H2 (2)| = |a2 a4 − a23 | ⩽ 1. The inequality is sharp.

9.4 Growth Estimates As previously, denote by C(r) the closed curve that is the image of the circle |z| = r < 1 under the mapping w = f(z) and let L(r) be the length of C(r). Let M(r) = max |f(z)|. |z|=r

Noting that α ≠ 1, we begin by giving the following result of Pommerenke [193], ̃ which shows that if f ∈ K(α), then L(r) and M(r) have essentially the same growth rate as r → 1. ̃ Theorem 9.4.1. Let f ∈ K(α), with 0 < α < 1. If 0 ⩽ r < 1, then 2π

󵄨 󵄨 L(r) = ∫ 󵄨󵄨󵄨󵄨rf 󸀠 (re iθ )󵄨󵄨󵄨󵄨 dθ ⩽ A(α)M(r),

(9.4.1)

0

where A(α) depends only on α. Since trivially L(r) ⩾ 2√πM(r), inequality (9.4.1) shows that L(r) and M(r) have essentially the same growth. ̃ Proof. If f ∈ K(α), then (9.2.1) holds. Let 𝑣(r, θ) = arg(e i(θ−β) h󸀠 (re iθ )). Then from (9.2.1) we have 󵄨󵄨 󵄨󵄨 απ iθ 󸀠 iθ (9.4.2) 󵄨󵄨󵄨arg[e f (re )] − 𝑣(r, θ)󵄨󵄨󵄨 < 2 . Let m = 1 + 8/(1 − α). For fixed r < 1, define θ k by 𝑣(r, θ k ) =

2kπ , m

k = 0, 1, . . . , m − 1,

and θ m = θ0 +2π. Since 𝑣(r, θ) increases with θ, we see that θ0 < θ1 < ⋅ ⋅ ⋅ < θ m = θ0 +2π and that 0 ⩽ 𝑣(r, θ) − 𝑣(r, θ k ) ⩽

2π π < (1 − α) , m 4

for

θ k ⩽ θ ⩽ θ k+1 .

144 | 9 Close-to-Convex Functions

Together with (9.4.2), this gives | arg[e iθ f 󸀠 (re iθ )] − 𝑣(r, θ k )| < (1 + α)

π π < , 4 2

for

θ k ⩽ θ ⩽ θ k+1 .

Hence, θ k+1

π 󵄨 󵄨 cos ((1 + α) ) ∫ 󵄨󵄨󵄨󵄨rf 󸀠 (re iθ )󵄨󵄨󵄨󵄨 dθ 4 θk θ k+1

󵄨 󵄨 ⩽ ∫ 󵄨󵄨󵄨󵄨rf 󸀠 (re iθ )󵄨󵄨󵄨󵄨 cos (arg(e iθ f 󸀠 (re iθ )) − 𝑣(r, θ k )) dθ θk θ k+1

[ ] = Re [e−i𝑣(r,θ k) ∫ re iθ f 󸀠 (re iθ ) dθ] θk [ ] −i𝑣(r,θ k)−iπ/2 (f(re iθ k+1 ) − f(re iθ k ))] = Re [e

⩽ 2M(r), for k = 0, 1, . . . , m − 1. By summation we obtain θ 0 +2π

∫ |rf 󸀠 (re iθ )| dθ ⩽ θ0

2mM(r) 18M(r) ⩽ , cos ((1 + α)π/4) (1 − α) sin ((1 − α)π/4)

which completes the proof. When α = 1, i.e., for functions in K, the equivalence of L(r) and M(r) no longer holds, and we need the following result, the proof of which can be found in Clunie and Pommerenke [33, Theorem 1], to prove our next theorem. Lemma 9.4.1. Let f ∈ K and g ∈ S∗ be the associated starlike function such that zf 󸀠 (z) = e−iβ g(z)p(z)

(z ∈ 𝔻),

(9.4.3)

where Re p(z) > 0 in 𝔻 and β ∈ (−π/2, π/2). Then 2π

∫ |g(z)| Re p(z) dθ ⩽ 2πM(r). 0

Making use of this lemma, Thomas [240] proved the following, thus extending the known result for functions in S∗ given in Chapter 4. We note that his proof is for functions in K0 , but a simple alteration enables us to prove the theorem for functions in K as follows. Theorem 9.4.2. If f ∈ K, and is of the form (9.3.1), then L(r) = O (M(r) log

1 ) 1−r

as

r → 1.

9.4 Growth Estimates |

145

The logarithmic factor cannot be removed. In particular, if f is bounded for z ∈ 𝔻, then

L(r) = O (log

1 ) 1−r

r → 1.

as

Proof. Since f ∈ K, from (9.4.3) we have 2π

2π 󸀠

L(r) = ∫ |zf (z)| dθ = ∫ |g(z)p(z)| dθ 0

0

2π r

⩽ ∫ ∫ (|g󸀠 (z)p(z)| + |g(z)p󸀠 (z)|) dρdθ 0 0

say, where now z = ρe iθ .

= J 1 (r) + J 2 (r)

Since g ∈ S∗ , we may write zg󸀠 (z) = g(z)ϕ(z), where ϕ ∈ P with ϕ(z) = 1 + ϕ n z n . Then,

∑∞ n=1

2π r

2π r 󸀠

J 1 (r) = ∫ ∫ |g (z)p(z)| dρdθ = ∫ ∫ |f 󸀠 (z)ϕ(z)| dρdθ 0 0 r

0 0 1/2

2π 󸀠

2

⩽ ∫ ( ∫ |f (z)| dθ) 0

0 r

2

( ∫ |ϕ(z)| dθ)

dρ

0 1/2

∞

= 2π ∫ ( ∑ n2 |a n |2 ρ 2n−2 ) 0

1/2

2π

1/2

∞

( ∑ |ϕ n |2 ρ 2n )

n=1

dρ,

n=0

where ϕ0 = 1. Since ϕ ∈ P, we have |ϕ n | ⩽ 2 for n ⩾ 1, and so r

1/2

∞

J 1 (r) ⩽ 4π ∫ ( ∑ n2 |a n |2 ρ 2n−2 ) 0 r

n=1 ∞

1/2 2

2 2n−2

2

2 2n−2

)

(

n=1 r ∞

1/2 1 ) dρ 1 − ρ2

1/2

⩽ 4π (∫ ∑ n |a n | ρ 0 n=1

dρ

n=0

= 4π ∫ ( ∑ n |a n | ρ 0

1/2

∞

( ∑ ρ 2n )

dρ)

r

1/2

1 (∫ dρ) 1 − ρ2 0

146 | 9 Close-to-Convex Functions 1/2

∞

n2 |a n |2 r2n−1 ) 2n − 1 n=1

= 2√2π ( ∑

1/2

∞

⩽ 2√2π ( ∑ n|a n |2 r2n−1 )

(log

n=1

⩽ 2√2π (

(log

1 + r 1/2 ) 1−r

1 + r 1/2 ) 1−r

A(r) 1/2 1 + r 1/2 ) (log ) , πr 1−r

2 2n is the area of the image of |z| < r under w = f(z). where A(r) = π ∑∞ n=1 n|a n | r 2 Since A(r) ⩽ πM(r) , we have

1 1 + r 1/2 ) J 1 (r) ⩽ 2√2πM(r) ( log r 1−r 1 1/2 = O (M(r) (log ) ) as 1−r

r → 1.

Also, since Re p(z) > 0 for z ∈ 𝔻 and p(0) = e iβ , the Herglotz integral representation theorem gives 2π

1 + ze−is 1 p(z) − i sin β ∫ dμ(s), = cos β 2π 1 − ze−is

̃ (z) := p

0

for some probability measure μ on the unit circle. Thus, 2π

̃ (z)]󸀠 = [p

1 1 e−is dμ(s), p󸀠 (z) = ∫ cos β π (1 − ze−is )2 0

and so 2π r

r 2π

J 2 (r) = ∫ ∫ |g(z)p󸀠 (z)| dρdθ = ∫ ∫ |g(z)p󸀠 (z)| dθ dρ 0 0

0 0

r 2π 2π

⩽

cos β|g(z)| 1 ∫∫∫ dμ(s) dθ dρ. π |1 − ze−is |2 0 0 0

̃ ∈ P, it follows from Heins [72, p. 68] that Since p 2π

1 − ρ2 1 ̃ (z) = dμ(s), Re p ∫ 2π |1 − ze−is |2 0

and, therefore, r 2π

J 2 (r) ⩽ 2 ∫ ∫ |g(z)| Re p(z) dθ 0 0

dρ . 1 − ρ2

9.4 Growth Estimates | 147

Hence, by Lemma 9.4.1 we have r

J 2 (r) ⩽ 4π ∫ 0

r

M(ρ) dρ 1+r dρ ⩽ 4πM(r) ∫ = 2πM(r) log . 1−r 1 − ρ2 1 − ρ2

(9.4.4)

0

From the above estimates for J 1 (r) and J 2 (r), we therefore obtain L(r) = O (M(r) log

1 ) 1−r

r → 1.

as

The same function as discussed in Theorem 4.1.13 shows that the logarithmic factor cannot be removed. Using a similar argument, Thomas [240] extended a known result of Pommerenke [191] for functions in S∗ to K. We include the proof so as to illustrate the slightly different method. Theorem 9.4.3. Let f ∈ K and be of the form (9.3.1), and suppose that M(r) = max |f(z)| ⩽ (1 − r)−λ ,

0 < λ ⩽ 2.

|z|=r

Then, L(r) ⩽ A1 (λ)(1 − r)−λ , where A1 (λ) depends only on λ. Proof. Throughout the proof, C will denote an absolute constant, not necessarily the same each time. As in the proof of Theorem 9.4.2, we have r 2π

L(r) ⩽ ∫ ∫ (|f 󸀠 (z)ϕ(z)| + |g(z)p󸀠 (z)|) dθdρ = J 1 (r) + J 2 (r)

again.

0 0

Now

r 2π

J 1 (r) = ∫ ∫ |f 󸀠 (z)ϕ(z)| dθdρ 0 0 r 2π

⩽ ∫ ∫ |f 󸀠 (z)ϕ(z)| dθdρ + C, δ 0

where δ is fixed and 0 < δ < 1. Proceeding as in Theorem 9.4.2 we obtain r

∞

1/2 2

2 2n−2

J 1 (r) ⩽ 2π ∫ ( ∑ n |a n | ρ δ r

n=1

)

1/2

∞

2 2n

( ∑ |ϕ n | ρ ) 1/2

∞

⩽ 2π ∫ ( ∑ n|a n |2 ρ n . max nρ n−2 ) δ

n=1

dρ + C

n=0

n

∞

1/2

( ∑ |ϕ n |2 ρ 2n ) n=0

dρ + C.

148 | 9 Close-to-Convex Functions Since log nρ n has a maximum at a point n0 = 1/ log(1/ρ) ⩽ 1/(1 − ρ), it follows that nρ n−2 ⩽

eρ 2

1 1 1 ⩽ . 2 1 −ρ log(1/ρ) eρ

(9.4.5)

This, together with |ϕ n | ⩽ 2, n ⩾ 1, implies r

J 1 (r) ⩽ C ∫ δ

√A1 (√ρ) ρ(1 − ρ)

r

dρ + C ⩽ C ∫ δ

M(√ρ) dρ + C. ρ(1 − ρ)

Since M(√ρ) ⩽ (4/√ρ)M(ρ) (see Pommerenke [194, Lemma 2]), we have r

J 1 (r) ⩽ C ∫ δ r

⩽ C∫ δ r

⩽ C∫ δ

M(ρ) dρ + C − ρ)

ρ 3/2 (1

M(ρ) dρ + C, 1−ρ dρ + C, (1 − ρ)λ+1

since 0 < δ < ρ

by hypothesis

⩽ A1 (λ)(1 − r)−λ .

(9.4.6)

In estimating J 2 (r) we obtain (9.4.4), then since M(ρ) ⩽ (1 − ρ)−λ , we obtain r

J 2 (r) ⩽ 4π ∫ 0

dρ 4π ⩽ (1 − r)−λ . λ (1 − ρ)λ+1

(9.4.7)

From (9.4.6) and (9.4.7), it follows that L(r) ⩽ A1 (λ)(1− r)−λ , which completes the proof of the theorem. Remark 9.4.1. In Theorem 4.1.14 we saw that when f ∈ S∗ , it is possible to replace M(r) by √A(r) in the growth estimate L(r) = O(M(r) log(1/(1 − r))) as r → 1, which raises the question as to whether this also holds for f ∈ K. Using the same kind of techniques as above, it is a relatively simple exercise to show that if f ∈ K, then L(r) = O(√A(r)(log(1/(1 − r)))3/2 ) as r → 1. The question of whether this order of growth is the best possible remains open. Remark 9.4.2. Employing the same techniques used in Theorems 9.4.2 and 9.4.3, it is possible to show that when f ∈ K, na n = O(M((n + 1)/n)) as n → ∞, which also leads us to ask the natural question of whether we can show that na n = O(√ A((n + 1)/n)) as n → ∞. This, again, is an open problem.

9.5 Ozaki Close-to-Convex functions |

149

9.5 Ozaki Close-to-Convex functions Although the class K was first formally introduced by Kaplan in 1952 [84], in 1941 Ozaki [180] considered functions in A satisfying the condition Re [1 +

zf 󸀠󸀠 (z) 1 ]>− f 󸀠 (z) 2

(z ∈ 𝔻).

(9.5.1)

It follows from the original definition of Kaplan [84] (Definition 2.2.1) that functions satisfying (9.5.1) are close-to-convex, and, therefore, members of S. Kargar and Ebadian [86] considered the following generalization to (9.5.1). Definition 9.5.1. Let f ∈ A and be locally univalent for z ∈ 𝔻 and −1/2 < λ ⩽ 1. Then, f ∈ F(λ) if, and only if, Re [1 +

zf 󸀠󸀠 (z) 1 ]> −λ f 󸀠 (z) 2

(z ∈ 𝔻).

(9.5.2)

Clearly, when −1/2 < λ ⩽ 1/2 functions defined by (9.5.2) provide a subset of C, with F(1/2) = C, and since 1/2 − λ ⩾ −1/2 when λ ⩽ 1, functions in F(λ) are close-toconvex when 1/2 ⩽ λ ⩽ 1. On the other hand, it was pointed out by Ponnusamy et al., [200] that functions in F(1) are not necessarily starlike. We shall call members of f ∈ F(λ) when λ ∈ [1/2, 1], Ozaki close-to-convex functions. We note that in the paper by Ponnusamy et al., [200], sharp coefficient bounds and distortion theorems for f ∈ F(1) were given, and it was also shown that every partial sum (or section) s n (z) = z + ∑nk=2 a k z k of a function f ∈ F(1) given by (1.1.2) belongs to C in the disk |z| < 1/6, and that this radius is the best possible. We therefore first find sharp bounds for the coefficients of functions in F(λ). Theorem 9.5.1. Let f ∈ F(λ) for 1/2 ⩽ λ ⩽ 1 and be given by (1.1.2). Then for n ⩾ 2, |a n | ⩽

1 n ∏(k + 2λ − 1). n! k=2

The inequality is sharp when f(z) = f λ (z) = Proof. Write 1+ and let p(z) =

1 1 − 1] . [ 2λ (1 − z)2λ

∞ zf 󸀠󸀠 (z) ∑ c n z n := g(z), = 1 + f 󸀠 (z) n=1

∞ 2 1 [g(z) − + λ] = 1 + ∑ p n z n . 1 + 2λ 2 n=1

(9.5.3)

150 | 9 Close-to-Convex Functions Then, Re p(z) > 0 and Re g(z) > 1/2 − λ for z ∈ 𝔻, and |p n | ⩽ 2 for n ⩾ 1, and since c n = (1/2 + λ)p n , we have |c n | ⩽ 1 + 2λ for n ⩾ 1. Since for each integer n, the coefficients a n are polynomials with positive coefficients in c n , it follows that |a n | will be less than or equal to the result by replacing |c n | by 1 + 2λ. Thus, by the principle of majorization (see, e.g., [89]), we have 1+ and f(z) ≪

1 + 2λz zf 󸀠󸀠 (z) ≪ , f 󸀠 (z) 1−z

∞ 1 1 − 1] := z + ∑ dn zn . [ 2λ (1 − z)2λ n=2

Therefore, |a n | ⩽ d n =

1 n ∏(k + 2λ − 1), n! k=2

which is (9.5.3). We next give distortion results for functions in F(λ). Theorem 9.5.2. Let f ∈ F(λ) for 1/2 ⩽ λ ⩽ 1. Then for z = re iθ ∈ 𝔻, 󵄨󵄨 󸀠󸀠 󵄨󵄨 󵄨󵄨 zf (z) 󵄨󵄨 (1 + 2λ)r 󵄨󵄨 󸀠 󵄨󵄨 ⩽ 1−r 󵄨󵄨󵄨 f (z) 󵄨󵄨󵄨 and

1 1 ⩽ |f 󸀠 (z)| ⩽ . 1+2λ (1 + r) (1 − r)1+2λ The inequalities are sharp when f(z) = f λ (z). Proof. From (9.5.2) we can write 1+

zf 󸀠󸀠 (z) 1 1 = ( + λ) p(z) + − λ, f 󸀠 (z) 2 2

so that 1+ and so

zf 󸀠󸀠 (z) 1 + 2λz ≺ , f 󸀠 (z) 1−z

zf 󸀠󸀠 (z) (1 + 2λ)z ≺ . f 󸀠 (z) 1−z

Thus, we can write

(9.5.4)

zf 󸀠󸀠 (z) (1 + 2λ)ω(z) = , f 󸀠 (z) 1 − ω(z) where |ω(z)| ⩽ |z|. The first inequality in Theorem 9.5.2 now follows. To prove the second inequality, we use Lemma 5.4.2 and integrate (9.5.4) to obtain f 󸀠 (z) ≺

1 . (1 − z)1+2λ

The inequalities for |f 󸀠 (z)| now follow in the same way.

9.5 Ozaki Close-to-Convex functions | 151

9.5.1 Growth and Area Estimates For f ∈ S, and z = re iθ ∈ 𝔻, let M(r) = max |f(z)|, C(r) be the curve f(|z| = r) and A(r) |z|=r

be the area enclosed by C(r). As was pointed out in Section 9.4, a long-standing open problem for functions in K is whether M(r) can be replaced by √A(r) in the growth estimate L(r) = O(M(r) log(1/(1 − r))) as r → 1, which is a result that is already known for functions in S∗ . Similarly, replacing M(r) by √A(r) in the known estimate na n = O(M((n + 1)/n)) as n → ∞ for functions in K remains an open question [240, 241]. Since the definition of Ozaki close-to-convex functions does not include an independent starlike function, it is relatively easy to show that both these growth estimates can be improved when f ∈ F(λ). We omit the proofs, which use the techniques used in Theorem 9.4.2. Theorem 9.5.3. Let f ∈ F(λ) for 1/2 ⩽ λ ⩽ 1 and be given by (1.1.2), with M(r), L(r) and A(r) defined as above. Then, L(r) = O (√A(r) log

1 ) 1−r

as

r → 1,

and na n = O (√A((n + 1)/n))

as

n → ∞.

9.5.2 Strongly Ozaki Close-to-Convex Functions We have seen in Chapter 6 that some sharp coefficient estimates can be obtained for the classes of strongly starlike and convex functions. However, sharp bounds for the coefficients of functions in the class K(α) of strongly close-to-convex functions seem more difficult. It is a relatively simple exercise to obtain sharp bounds for the coefficients |a2 | and |a3 | when f ∈ K(α), but finding sharp bounds for |a4 | appears to be a more difficult problem. In this section, we show how it is possible to obtain sharp bounds for the coefficients |a2 |, |a3 |, and |a4 |, and the corresponding inverse coefficients for strongly Ozaki close-to-convex functions, thus providing sharp inequalities for the fourth coefficient of a class of strongly close-to-convex functions. We therefore make the following definition. Definition 9.5.2. Let f ∈ A with 0 < α ⩽ 1 and 1/2 ⩽ λ ⩽ 1. Then, f is called strongly Ozaki close-to-convex if, and only if, 󵄨󵄨 󵄨 2λ − 1 2 zf 󸀠󸀠 (z) 󵄨󵄨󵄨 απ 󵄨󵄨 󵄨󵄨arg [ 󵄨󵄨 < + + (z ∈ 𝔻). (1 )] 󵄨󵄨 󵄨󵄨 2λ + 1 2λ + 1 f 󸀠 (z) 2 󵄨 󵄨 We denote this class of functions by F(λ, α). Note that when α = 1, the above definition reduces to (9.5.2).

152 | 9 Close-to-Convex Functions

Making use of Theorems 3.1.7 and 3.1.8 it is possible to obtain sharp bounds for the coefficients a2 , a3 , and a4 , and the corresponding inverse coefficients. We omit the proofs of the following theorems. Theorem 9.5.4. Let f ∈ F(λ, α) for 0 < α ⩽ 1 and 1/2 ⩽ λ ⩽ 1 be given by (1.1.2). Then, α |a2 | ⩽ (1 + 2λ), 2

α { (1 + 2λ), { { { 6 |a3 | ⩽ { 2 { α { { (1 + λ)(1 + 2λ), { 3

α { { (1 + 2λ), { { 12 |a4 | ⩽ { { { { α (1 + 2λ)(1 + 8α 2 + 15α 2 λ + 6α 2 λ2 ), { 36

0< α⩽

1 , 2(1 + λ)

1 ⩽ α ⩽ 1, 2(1 + λ)

0 0 (z ∈ 𝔻). f 1−α (z)g α (z)

We recall from Chapter 2, that B(α) ⊂ S, and note again that B(1) = K0 and B(0) = S∗ . Since the Bieberbach conjecture is settled for f ∈ S, and the Koebe function k ∈ B(α), the inequality |a n | ⩽ n for n ⩾ 2 follows at once, and the distortion theorems for f ∈ S given in Chapter 1 also hold for functions in B(α), and are all the best possible. Although the definition of the B(α) Bazilevič functions involves the notion of powers, and is therefore more complicated than that of close-to-convex functions, we shall see in the following that some of the methods used in the case of close-to-convex functions extend, to give corresponding results. We begin by giving an example of a function f ∈ B(α), which is not close-toconvex, Keogh and Miller [89]. Let ϕ be an odd close-to-convex function that maps 𝔻 onto the w plane slit along the half-lines Re w ⩾ 0, Im w = c, and Re w ⩽ 0, Im w = −c, for some c > 0. If ϕ(z) = z√ f(z2 )/z2 , then f ∈ B(1/2), but the transformation ζ = ξ +iη maps 𝔻 onto the ζ -plane slit along the portion of the parabola ξ = (η/2c)2 − c2 for η ⩾ 0, which clearly cannot be expressed as the union of half-lines, and so f is not close-to-convex.

https://doi.org/10.1515/9783110560961-010

154 | 10 Bazilevič Functions

10.2 Growth Theorems Using the methods developed for close-to-convex functions by Clunie and Pommerenke [33], Thomas [242] showed that the known result na n = O(1) as n → ∞ for bounded functions in S∗ and K0 extends to the class of B(α). These methods have been developed to great effect, as we shall see in the following theorems. It is well known (see, e.g., Theorem 3.3 in Hayman [70]) that if f ∈ S, then M(r) = max |f(z)| = O((1 − r)−λ ) implies na n = O(n λ ) as n → ∞ for 1/2 ⩽ λ ⩽ 2, and is false |z|=r

when 0 < λ < 1/2. Pommerenke [195] improved this in the case of functions in S∗ by showing that the implication is valid when 0 ⩽ λ ⩽ 2. Clunie and Pommerenke [33] subsequently showed that this could be improved even further for functions in K0 (and, therefore, also in S∗ ), by showing that na n = O(M(n/(n + 1))) as n → ∞. For f ∈ B(α), it was similarly shown by Thomas [242] that M(r) = O((1 − r)−λ ) implies na n = O(n λ ) as n → ∞ for 0 ⩽ λ ⩽ 2, and that when f is bounded in 𝔻, na n = O(1) as n → ∞. The first theorem in this section (Noor [163]), uses elements of the proofs contained in [242] to show that na n = O(M(n/(n + 1))) as n → ∞ when f ∈ B(α) and 0 ⩽ α ⩽ 1, thereby also providing a proof for f ∈ K0 . Theorem 10.2.1. Let f ∈ B(α), for 0 ⩽ α ⩽ 1 and for z = re iθ ∈ 𝔻, be given by f(z) = n z + ∑∞ n=2 a n z . Then, na n = O(M(n/(n + 1))),

as

n → ∞.

Proof. The last part of the proof requires that α > 0, but when α = 0 this reduces to the result for functions in S∗ mentioned above. Write zf 󸀠 (z) = f 1−α (z)g α (z)p(z), (10.2.1) where g ∈ S∗ and p ∈ P. Then, [zf 󸀠 (z)]󸀠 = (1 − α)f −α (z)f 󸀠 (z)g α (z)p(z) + αf 1−α (z)g α−1 (z)g󸀠 (z)p(z) + f 1−α (z)g α (z)p󸀠 (z). Since

2π

1 n an = ∫ z(zf 󸀠 (z))󸀠 e−inθ dθ, 2πr n 2

0

it follows that 2π

(1 − α) 󵄨󵄨 󸀠 󵄨 ∫ 󵄨󵄨󵄨zf (z)f −α (z)g α (z)p(z)󵄨󵄨󵄨󵄨 dθ n |a n | ⩽ 2πr n 2

0 2π

+

α 󵄨 󵄨 ∫ 󵄨󵄨zg󸀠 (z)f 1−α (z)g α−1 (z)p(z)󵄨󵄨󵄨󵄨 dθ 2πr n 󵄨󵄨 0 2π

+

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨󵄨f(z)1−α g(z)α zp󸀠 (z)󵄨󵄨󵄨󵄨 dθ, n 2πr 0

= I1 (r) + I2 (r) + I3 (r) say.

(10.2.2)

10.2 Growth Theorems | 155

Using (10.2.1) once more, we have 2π

I1 (r) =

(1 − α) 󵄨󵄨 󸀠 󵄨 ∫ 󵄨󵄨󵄨zf (z)f −α (z)g α (z)p(z)󵄨󵄨󵄨󵄨 dθ 2πr n 0 2π

=

(1 − α) ∫ |zf 󸀠 (z)|2 |f(z)|−1 dθ. 2πr n 0

To estimate the last integral we note from Lemma 3.1 in Hayman [70], that when f ∈ S and 1/2 < r < 1, there exists ρ such that 2r − 1 ⩽ ρ ⩽ r, and 2π

4M(r) 1 󵄨 󵄨2 󵄨 󵄨−1 ∫ 󵄨󵄨f 󸀠 (ρe iθ )󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨f(ρe iθ )󵄨󵄨󵄨󵄨 dθ ⩽ . 2π 󵄨󵄨 1−r 0

Since r ⩽ (1 + ρ)/2 and M(r) increases with r, we have M(r) ⩽ M((1 + ρ)/2), and 1/(1 − r) ⩽ 2/(1 − ρ). Thus, choosing ρ = (n − 1)/(n + 1), we obtain for n ⩾ 2, I1 (r) ⩽ 8(n + 1)(1 − α) (1 +

1 n ) M(n/(1 + n)) < 8en(1 − α)M(n/(n + 1)). n

Next, it follows from (10.2.1) that since g ∈ S∗ , we can write zg󸀠 (z) = g(z)ϕ(z) for ϕ ∈ P, so that 2π

α I2 (r) = ∫ |zf 󸀠 (z)ϕ(z)|dθ 2πr n 0 1/2

2π

1 α ⩽ n( ∫ |f 󸀠 (z)|2 dθ) r 2π 0

2π

1 ( ∫ |ϕ(z)|2 dθ) 2π

1/2

,

(10.2.3)

0

on using the Cauchy–Schwarz inequality. For the first of the above integrals, using (9.4.5), we have 2π

∞ ∞ 1 ∫ |f 󸀠 (z)|2 dθ = ∑ n2 |a n |2 r2n−2 ⩽ ∑ n|a n |2 r n ⋅ (max nr n−2 ) n 2π n=1 n=1 0

∞

⩽ ∑ n|a n |2 r n n=1

1 er2 (1 − r)

(see (9.4.5)). Next note that for f ∈ S, the area theorem states that A(√r) ⩽ πM 2 (√r) and also M(r) ⩽ 4M(r2 )/r. The latter inequality, which is often useful, can easily be verified by observing that r

f 󸀠 (ρe iθ ) ] 󵄨 󵄨 󵄨 󵄨 dρ , log 󵄨󵄨󵄨󵄨f(re iθ )󵄨󵄨󵄨󵄨 = log 󵄨󵄨󵄨󵄨f(r2 e iθ )󵄨󵄨󵄨󵄨 + Re [∫ e iθ f(ρe iθ ) 2 [r ]

156 | 10 Bazilevič Functions and so using the distortion Theorem 1.3.4 for f ∈ S in Chapter 1, we have r󵄨 󵄨󵄨 f 󸀠 (ρe iθ ) 󵄨󵄨󵄨 󵄨󵄨 dρ log M(r) ⩽ log M(r2 ) + ∫ 󵄨󵄨󵄨󵄨 󵄨 󵄨󵄨 f(ρe iθ ) 󵄨󵄨󵄨 2 r r 2

⩽ log M(r ) + ∫ r2

= log (

M(r2 ) r

1+ρ dρ ρ(1 − ρ)

(1 + r)2 ) ,

from which the inequality follows. Thus, we have 2π

M 2 (r) 1 . ∫ |f 󸀠 (z)|2 dθ ⩽ 2 2π er (1 − r) 0

For the second of the integrals in (10.2.3) we appeal to Theorem 3.1.11, so that 2π

1 4 ∫ |ϕ(z)|2 dθ ⩽ . 2π 1 − r2 0

Thus, 2α M(r) . r n−1 √e 1 − r Choosing r = n/(n + 1) and arguing as above, we obtain I2 (r) ⩽

I2 (r) ⩽ 2α √e(n + 1)M(n/(n + 1)) < 4α √enM(n/(n + 1)). Next we use (10.2.1) and the inequality |p󸀠 (z)| ⩽ 2 Re p(z)/(1 − r2 ) from Theorem 3.1.1 so that 2π

I3 (r) =

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨f 1−α (z)g α (z)zp󸀠 (z)󵄨󵄨󵄨󵄨 dθ 2πr n 󵄨 0 2π

M 1−α (r) 󵄨󵄨 α 󵄨 ∫ 󵄨󵄨󵄨g (z)zp󸀠 (z)󵄨󵄨󵄨󵄨 dθ ⩽ n 2πr 0 2π

M 1−α (r) 󵄨 󵄨 ⩽ n−1 ∫ 󵄨󵄨g α (z)󵄨󵄨󵄨 Re p(z)dθ πr (1 − r2 ) 󵄨 0

2π

=

M 1−α (r) Re ∫ zf 󸀠 (z)f α−1 (z)e−iα arg g(z) dθ n−1 πr (1 − r2 ) 0

=

M 1−α (r) πr n−1 (1 − r2 )

2M(r) , ⩽ n−1 αr (1 − r)

2π

Re ∫ f α (z)e−iα arg g(z) d θ {arg g(z)} 0

10.2 Growth Theorems |

157

provided that α > 0, and where we have used integration by parts and the fact that 2π since g ∈ S∗ , ∫0 d θ {arg g(z)} = 2π. Choosing r = n/(n + 1) again, we obtain I3 (r) ⩽

2(n + 1) 2(n + 1)e 1 n (1 + ) M(n/(1 + n)) < M(n/(n + 1)). α n α

Using the above estimates for I1 (r), I2 (r) and I3 (r) in (10.2.2) completes the proof of the theorem. The following is now obvious and extends the similar result for functions in K0 discussed in Chapter 9. Corollary 10.2.1. Let f ∈ B(α) and be bounded in 𝔻, Then, na n = O(1) as n → ∞. Remark 10.2.1. We note that when α = 1 (i.e., when f ∈ K), Corollary 10.2.1 shows that na n = O(1) as n → ∞, mentioned above. An examination of the above analysis shows that for f ∈ B(α) if the area of f(𝔻) is finite, then na n = O(log n)1/2 as n → ∞, and the question arises as to whether the logarithmic factor can be removed. We also remark that Campbell and Pearce [27] observed that the methods used in [242] can be employed to prove a more general theorem as follows. 2π

Theorem 10.2.2. Let f, g ∈ A. If α ∈ ℝ, sup ∫0 |d θ {arg g(z)}| < ∞, f is bounded and Re [

zf 󸀠 (z) ]>0 f(z)1−α g(z)α

(z ∈ 𝔻),

then na n = O(1) as n → ∞. Note that this theorem does not require that g ∈ S∗ and that α can be negative. Thus, the result na n = O(1) as n → ∞ holds for a wider class of functions, which, since g is not required to belong to S∗ , are not necessarily univalent. We now consider the problem of estimating the length L(r) of f(|z| = r), recalling the results for f ∈ S∗ and K0 in Chapter 4 and Chapter 9, respectively. We will show that it is possible to extend the results for f ∈ S∗ and K0 in terms of M(r), but as in the case of close-to-convex functions, finding a best possible growth estimate for L(r) in terms of the area A(r) of f(|z| = r) remains an open problem. Theorem 10.2.3. Let f ∈ B(α), for 0 ⩽ α ⩽ 1, and for z = re iθ , denote by L(r) the length of the curve C(r) = f(|z| = r). Then as r → 1, L(r) = O (M(r) log

1 ). 1−r

In particular, L(r) = O(log(1/(1 − r))) as r → 1 when f is bounded.

158 | 10 Bazilevič Functions

Proof. The proof, which we sketch below, is a variation of that used in Theorem 10.2.1. Using (10.2.1) we have, with z = re iθ , and again writing zg󸀠 (z) = g(z)ϕ(z) for ϕ ∈ P, 2π

L(r) = ∫ |zf 󸀠 (z)|dθ 0 2π r

󵄨 󵄨 ⩽ (1 − α) ∫ ∫ ρ 󵄨󵄨󵄨󵄨f 󸀠2 (ρe iθ )f −1 (ρe iθ )󵄨󵄨󵄨󵄨 dθdρ 0 0 2π r

󵄨 󵄨 + α ∫ ∫ 󵄨󵄨󵄨󵄨f 󸀠 (ρe iθ )ϕ(ρe iθ )󵄨󵄨󵄨󵄨 dθdρ 0 0 2π r

󵄨 󵄨 + ∫ ∫ 󵄨󵄨󵄨󵄨f 1−α (ρe iθ )g α (ρe iθ )p󸀠 (ρe iθ )󵄨󵄨󵄨󵄨 dθdρ 0 0

= I4 (r) + I5 (r) + I6 (r), say, where now z = ρe iθ . For I4 (r), we apply the method used to estimate I1 (r) in Theorem 10.2.1, use the fact that M(r) increases with r, and then integrate the resulting expression with respect to ρ to obtain the required result. The expression for I5 (r) is treated in the same way as I2 (r) above, followed again with an integration with respect to ρ. Similarly, I6 (r) is estimated using the method in I3 (r), followed once more with an integration with respect to ρ. An examination of the above proof, easily establishes the following. Corollary 10.2.2. Let f ∈ B(α), for 0 ⩽ α ⩽ 1, and for z = re iθ , denote by L(r) the length of the curve C(r) = f(|z| = r), and let A(r) be the area inside C(r). Then as r → 1, L(r) = O (√A(r) (log

1 3/2 ) ). 1−r

In particular, the above shows that when the area of f(𝔻) is finite, then as r → 1, L(r) = O (log

1 3/2 ) . 1−r

As in the corresponding theorem in Chapter 9 for close-to-convex functions, the question arises as to whether the exponent 3/2 in the above corollary can be reduced, ultimately to 1.

10.2 Growth Theorems | 159

10.2.1 Coefficients of Powers of Functions in B(α) For f ∈ S, and λ > 0, suppose that for z ∈ 𝔻, [

∞ f(z) λ ] = 1 + ∑ a n (λ)z n , z n=1

[

∞ k(z) λ ] = 1 + ∑ k n (λ)z n , z n=1

and

(10.2.4)

where k is the Koebe function, and k n (λ) =

Γ(n + 2λ) Γ(n + 1)Γ(2λ)

Hayman and Hummel [71] posed the question of whether |a n (λ)| ⩽ k n (λ) for n ⩾ 1. De Branges’s proof of the Bieberbach conjecture [40] easily extends to show that |a n (λ)| ⩽ k n (λ) is true when λ ⩾ 1. Hayman and Hummel [71], however, showed that this inequality is false for λ < 1, when n = 2. For subclasses of S, it is known (see Klein [92]) that for starlike functions, |a n (λ)| ⩽ k n (λ) is true for n ⩾ 1, and λ > 0 and false for close-to-convex functions when n = 2, Jahangiri [78], and hence for functions in B(α) also. Although strictly concerning close-to-convex functions, we give Jahangiri’s result here for completeness, the proof of which is instructive. Theorem 10.2.4. Let f ∈ K0 . Then, { 2λ2 + λ, { { |a2 (λ)| ⩽ { λ(11 − 3λ) { { , { 9(1 − λ)

λ ⩾ 1/3, λ ⩽ 1/3.

The inequalities are sharp. Proof. We first note that adding (μ + 1/2)|p1 |2 to both sides of the inequality in Theon rem 3.1.3 shows that if p ∈ P with p(z) = 1 + ∑∞ n=1 p n z , then for μ ⩾ −1/2, |p2 + μp21 | ⩽ 2 + μ|p1 |2 .

(10.2.5)

From the definition of functions in K0 , we can write zf 󸀠 (z) = g(z)p(z), with g ∈ S∗ , n and so writing zg󸀠 (z) = g(z)q(z), with q ∈ P and q(z) = 1 + ∑∞ n=1 q n z , and equating coefficients gives a1 (λ) = λ(p1 + q1 )/2 and a2 (λ) = λ [

q2 p2 (1 + 3λ) 2 λ − 1 2 1 + 3λ + + q1 + p1 + p1 q1 ] . 6 3 24 8 12

Since |p1 | ⩽ 2 and |q1 | ⩽ 2 we have |a1 (λ)| ⩽ 2λ, which is obviously sharp.

160 | 10 Bazilevič Functions For a2 (λ), |a2 (λ)| ⩽

λ 6

⩽

󵄨󵄨 (1 + 3λ) 2 󵄨󵄨󵄨󵄨 λ 󵄨󵄨󵄨󵄨 3(λ − 1) 2 󵄨󵄨󵄨󵄨 λ(1 + 3λ) 󵄨󵄨 q1 󵄨󵄨 + 󵄨󵄨p2 + p1 󵄨󵄨 + |p1 q1 | 󵄨󵄨q2 + 󵄨󵄨 󵄨󵄨 3 󵄨󵄨 󵄨󵄨 4 8 12

λ(1 + 3λ) λ(7 + 3λ) λ(λ − 1) |p1 |2 + |p1 | + := Φ(|p1 |), 8 6 6

where we have used (10.2.5) and the inequalities |q1 | ⩽ 2 and |q2 | ⩽ 2. Elementary calculus now shows that the quadratic expression Φ(|p1 |) is maximum at |p1 | = 2(1 + 3λ)/(3(1 − λ)) provided that λ ⩽ 1/3 and at |p1 | = 2 when λ ⩾ 1/3, from which the result follows. To show that the inequalities are sharp, take p1 = p2 = q1 = q2 = 2 when λ ⩾ 1/3. For λ ⩽ 1/3, let p1 = 2(1 + 3λ)/(3(1 − λ)), and p2 = q1 = q2 = 2. Corollary 10.2.3. The inequality |a n (λ)| ⩽ k n (λ) is false for f ∈ K0 when n = 2, and λ < 1/3. Although the inequality |a n (λ)| ⩽ k n (λ) is false in general for f ∈ B(α), Keogh and Miller [89] showed that it is true when λ = α. Following Keogh and Miller [89], we shall use the notion of majorization (see Definition 5.3.1). Theorem 10.2.5. Let f ∈ B(α) with [

∞ f(z) α ] = 1 + ∑ a n (α)z n , z n=1

[

∞ k(z) α ] = 1 + ∑ k n (α)z n , z n=1

and

where k is the Koebe function and k n (α) =

Γ(n + 2α) . Γ(n + 1)Γ(2α)

Then, for n ⩾ 1, |a n (α)| ⩽ k n (α). Proof. Write F(z) = [f(z)/z]α , so that differentiating and using (10.2.1), we obtain zF 󸀠 (z) + αF(z) = α [

g(z) α ] p(z). z

It follows from a result of Klein [92] that when g ∈ S∗ , [

1 g(z) α . ] ≪ z (1 − z)2α

Also, since p ∈ P, p(z) ≪

1+z , 1−z

10.2 Growth Theorems | 161

and since multiplication is preserved by majorization, we obtain zF 󸀠 (z) + αF(z) ≪ α ⋅

1+z . (1 − z)2α+1

Comparing coefficients we have 2α + n 2α + n − 1 )+( )] , (n + α)|a n (α)| ⩽ α [( n n−1 and so

2α + n − 1 ), |a n (α)| ⩽ ( n

which is equivalent to [

f(z) α 1 ] ≪ z (1 − z)2α

as required. We remark that in [89], Keogh and Miller used the same ideas to give a more general result concerning m-fold symmetric functions in B(α). For f ∈ S, the question of finding the correct order of growth of a n (λ) given by (10.2.4) as n → ∞ is not difficult to resolve. Using Prawitze’s theorem (see, e.g., Duren [45]), Cauchy’s integral formula gives |a n (λ)| ⩽

2π

λ

λ

2πr n+λ−1

∫ |f(z)| dθ ⩽ 0

λ2 r n+λ−1

r

∫

M λ (ρ) dρ. ρ

0

Since M(ρ) ⩽ ρ/(1 − ρ)2 , integrating the above expression and taking r = 1 − 1/n, gives a n (λ) = O(n2λ−1 ) as n → ∞, provided λ ≠ 1/2. The case λ = 1/2 is equivalent to showing that the coefficients of odd functions in S are bounded, which was proved in 1932 by Littlewood and Paley [123]. Thus, if f ∈ S, then a n (λ) = O(n2λ−1 ) as n → ∞, for λ > 0, which is clearly the best possible when f is the Koebe function. For completeness, we state this formally. Theorem 10.2.6. Let f ∈ S, and for λ > 0 define [

∞ f(z) λ ] = 1 + ∑ a n (λ)z n . z n=1

Then, a n (λ) = O(n2λ−1 ) as n → ∞. The index 2λ − 1 is the best possible. We remark at this point that for f ∈ B(α), Deng [41] claimed the weaker result that if 0 < λ ⩽ 1, then a n (λ) = O(n2λ−1 (log n)3/2 ) as n → ∞, the proof of which is not clear.

162 | 10 Bazilevič Functions

10.2.2 Logarithmic Coefficients As has been mentioned at various points in this book, apart from the case of starlike functions, finding information about the logarithmic coefficients of functions in S is difficult. Few exact results have been found, and even obtaining the correct rate of growth has resulted in limited success. In this section, we state the best known estimate for the rate of growth of the logarithmic coefficients for f ∈ B(α), noting that the result is valid for the wider class B(α, β) (see Deng [42]), and that the same result for f ∈ K0 has been proved by Ye [264]. For f ∈ S, we recall once more from Definition 1.2.1 that the logarithmic coefficients γ n of f are defined by log

∞ f(z) = 2 ∑ γn zn . z n=1

(10.2.6)

The following result of Deng [42], which uses the method of Ye [264], appears to be the best to date. Theorem 10.2.7. Let f ∈ B(α) and the logarithmic coefficients of f be given by (10.2.6). Then, as n → ∞, γ n = O(n−1 log n). The exponent −1 is best possible.

10.3 The Fekete–Szegő Problem Establishing a Fekete–Szegő theorem for functions in B(α), although complicated, follows the use of similar techniques as those employed in previous chapters. We state the result without proof, due to Eenigenburg and Silvia [46], observing that when α = 1, the result corresponds to that given in Keogh and Merkes [88]. n Theorem 10.3.1. Let f ∈ B(α) and for z ∈ 𝔻 be given by f(z) = z + ∑∞ n=2 a n z . Then,

3 − 4μ, { { { { { { { 8α 2 (μ 0 − μ)2 4α 2 (μ 0 − μ) { { 1+ + , 2 2 |a3 − μa2 | ⩽ { (1 + α) (1 + α)2 (2μ + α − 1) { { { { 1, { { { { 4μ − 3, { where μ0 = (3 + α)/(2(2 + α)). All the inequalities are sharp.

μ⩽

1 , 2+α

1 ⩽ μ ⩽ μ0 , 2+α μ 0 ⩽ μ ⩽ 1, μ ⩾ 1,

10.4 Sufficient Conditions for f ∈ B(α) | 163

10.4 Sufficient Conditions for f ∈ B(α) We end this chapter by giving a sufficient condition due to Kim and Sugawa [90] for a function f to belong to B(α), noting that the given condition is a special case of a more general result proved for the larger class B(α, β). Define the slit domain U = ℂ \ {iy : y ⩽ −1 or y ⩾ 1}, noting that U is starlike with respect to the origin. For γ ∈ ℂ and f ∈ A, define P[γ, f](z) = 1 +

zf 󸀠󸀠 (z) zf 󸀠 (z) + (γ − 1) . f 󸀠 (z) f(z)

We need the following result of Miller and Mocanu [148], which follows from Theorem 3.4h, p. 132 in [148]. Lemma 10.4.1. Let ψ be analytic in 𝔻 and h be univalent in 𝔻 with ψ(z) ≠ 0 and h(z) ≠ 0 for z ∈ 𝔻. Also, let ψ(0) = h(0) and zψ󸀠 (z)/ψ(z) be starlike. If zh󸀠 (z) zψ󸀠 (z) ≺ , h(z) ψ(z) then h ≺ ψ, and ψ is the best dominant. Lemma 10.4.2. If ψ(z) = (1 + z)/(1 − z), then ψ maps 𝔻 onto the right half-plane and Q(z) = zψ󸀠 (z)/ψ(z) maps 𝔻 onto the domain U. Proof. The first part is obvious. Next, since Q(z) = it follows that

2z , (1 + z)(1 − z)

zQ󸀠 (z) z z =1+ − . Q(z) 1−z 1+z

Since Re[z/(1 − z)] > −1/2 for z ∈ 𝔻, then Re[zQ󸀠 (z)/Q(z)] > 0 i.e., Q ∈ S∗ . Finally, when z = e iθ , i , Q(z) = sin θ and since −1 ⩽ sin θ ⩽ 1, the boundary values of Q(z) form the set {iy : y ⩾ 1} ∪ {∞} ∪ {iy : y ⩽ −1}, and so Q maps 𝔻 onto U as required. We are now able to state and prove our theorem. Theorem 10.4.1. Let f ∈ A and α > 0. If for some p ∈ P, P[α, f](z)− αp(z) ∈ U for z ∈ 𝔻, then f ∈ B(α). In particular, if f ∈ A and α > 0, then f ∈ B(α) provided Re P[α, f](z) > 0, for z ∈ 𝔻.

164 | 10 Bazilevič Functions We note that the condition Re P[α, f](z) > 0, z ∈ 𝔻, means that f is 1/α convex and so every 1/α-convex function belongs to B(α). Proof. Suppose that f ∈ A and for some p ∈ P, P[α, f](z) − αp(z) ∈ U for z ∈ 𝔻. Take g ∈ S∗ , so that zg󸀠 (z)/g(z) = p(z), and let h(z) = [ then

zf 󸀠 (z) f(z) α ]⋅[ ] , f(z) g(z)

zh󸀠 (z) = P[α, f](z) − αp(z). h(z)

Thus by the assumption and Lemma 10.4.2, we therefore have zψ󸀠 (z) zh󸀠 (z) ≺ Q(z) = . h(z) ψ(z) Applying Lemma 10.4.1 we have h ≺ ψ, i.e., h ∈ P, and so f ∈ B(α).

11 B1(α) Bazilevič Functions 11.1 Definition and Basic Properties In the last two chapters we saw how the independence of the starlike function g created difficulties in fully extending known results for starlike functions to close-toconvex and Bazilevič functions. In order to overcome this difficulty an obvious way forward is to specify g. As was pointed out, assuming g(z) ≡ z in the definition of K leads to the rather simple class R of functions satisfying Re f 󸀠 (z) > 0 for z ∈ 𝔻. However, the same choice of g in the case of the Bazilevič functions B(α) provides a subclass B1 (α) of B(α), which is not only easier to deal with, but contains a rich source of results and unsolved problems. We, therefore, make the following definition. Definition 11.1.1. Let f ∈ A and α ⩾ 0. Then, f ∈ B1 (α) if, and only if, Re {f 󸀠 (z) [

f(z) α−1 ] }>0 z

(z ∈ 𝔻).

(11.1.1)

The restriction α ⩾ 0 is included in the above definition; however, much of what follows requires that 0 ⩽ α ⩽ 1. We note that B1 (0) = S∗ and that B1 (1) = R. So B1 (α), for α ∈ [0, 1], makes a transition between the classes S∗ and R. Also, although functions in R are univalent, a counter example by Krzyz [100] shows that neither S∗ ⊆ R nor R ⊆ S∗ . A great deal has been written about the class B1 (α) in recent years, and many significant results have been obtained. However, the presence of powers in Definition 11.1.1 again creates difficulties, especially when finding sharp coefficient bounds. In this chapter, we list some of the more interesting properties of functions in B1 (α), and give proofs where appropriate. We begin by giving some distortion theorems.

11.2 Distortion Theorems To find sharp upper and lower bounds for |f(z)| and |f 󸀠 (z)| is relatively simple, and the proofs follow in the usual way, using the distortion theorems for functions with positive real part given in Chapter 3. We omit the proofs. n iθ Theorem 11.2.1 (Thomas [244]). Let f ∈ B1 (α), with f(z) = z + ∑∞ n=2 a n z for z = re ∈ 𝔻. Then, if α > 0, [Q2 (r)]1/α ⩽ |f(z)| ⩽ [Q1 (r)]1/α .

If 0 < α ⩽ 1, then r α−1 (

1−r 1+r ) [Q2 (r)](1−α)/α ⩽ |f 󸀠 (z)| ⩽ r α−1 ( ) [Q1 (r)](1−α)/α , 1+r 1−r

https://doi.org/10.1515/9783110560961-011

166 | 11 B1 (α) Bazilevič Functions and if α ⩾ 1, then r α−1 (

1−r 1+r ) [Q1 (r)](1−α)/α ⩽ |f 󸀠 (z)| ⩽ r α−1 ( ) [Q2 (r)](1−α)/α , 1+r 1−r

where

r

Q1 (r) = α ∫ ρ α−1 ( 0

and

r

Q2 (r) = α ∫ ρ α−1 ( 0

1+ρ ) dρ, 1−ρ 1−ρ ) dρ. 1+ρ

Equality holds in all cases for f0 ∈ B1 (α), given by 1

f0 (z) = z (α ∫ t 0

1/α

+ tz dt) 1 − tz

α−1 1

.

(r)]1/α

→ r/(1 − r)2 and [Q2 (r)]1/α → r/(1 + r)2 , and that We note that as α → 0, [Q1 functions in B1 (α) have order of growth O(log 1/(1−r))1/α as r → 1. Also the following corollary, which can be proved using familiar techniques used in previous chapters, is an easy consequence of Theorem 11.2.1. Corollary 11.2.1. Let f ∈ B1 (α) and f(z) ≠ ω for z ∈ 𝔻. Then |ω| ⩾ [Q2 (1)]1/α ∼ 1/4, as α → 0. Finding sharp bounds for |zf 󸀠 (z)/f(z)| requires deeper analysis (see London and Thomas [129]), and uses ideas so far not contained in this book. We, therefore, include the proof of the following theorem. Theorem 11.2.2. Let f ∈ B1 (α) for α > 0. Then for z = re iθ ∈ 𝔻, 1

1−r 1 − tr (∫ t α−1 dt) α(1 + r) 1 + tr

−1

−1

1 󵄨󵄨 󸀠 󵄨󵄨 1 + tr 1+r 󵄨 zf (z) 󵄨󵄨 󵄨󵄨 ⩽ ⩽ 󵄨󵄨󵄨󵄨 ∫ t α−1 ( dt) 1 − tr 󵄨󵄨 f(z) 󵄨󵄨󵄨 α(1 − r) 0

0

.

The left and right hand inequalities are sharp at z = −r and z = r, respectively, when f(z) = f0 (z) is defined above. We shall need the following lemma. −1

z

Lemma 11.2.1. Let F(z) = 1 − z α (α ∫0 ζ α−1 /(1 − ζ)dζ ) and G(z) = [1 − F(z)]/(1 − z). Then, F and G have nonnegative Taylor coefficients about z = 0, and in particular for z = re iθ ∈ 𝔻, |F(z)| ⩽ F(r) < limF(t) = 1, (11.2.1) 󸀠

t→1 󸀠

|F (z)| ⩽ F (r),

(11.2.2)

|G(z)| ⩽ G(r).

(11.2.3)

and

11.2 Distortion Theorems | 167

Proof. Let z

ζ α−1 dζ ) H(z) = F(z) − 1 = −z (α ∫ 1−ζ α

−1

.

0

Then, (1 − z)[zH 󸀠 (z) − αH(z)] = αH 2 (z). With H(z) =

∑∞ k=0

ck

z k , (11.2.4)

(11.2.4)

implies that k

(k − α)c k = (k − 1 − α)c k−1 + α ∑ c j c k−j , j=0

where c−1 = 0. Thus, c0 = −1,

c1 =

α , α+1

and for k ⩾ 3, (k + α)c k = (k +

c2 =

α , (2 + α)(α + 1)2

α 2 − 2α − 1 ) c k−1 + b k , α+1

(11.2.5)

(11.2.6)

where b 3 = 0, and for k ⩾ 4, k−2

b k = α ∑ c j c k−j . j=2

Since 3 + (α 2 − 2α − 1)/(α + 1) > 0, a simple induction argument using (11.2.5) and (11.2.6) shows that c k > 0 when k ⩾ 1. Thus, the coefficients of F(z) are nonnegative, and so (11.2.1) and (11.2.2) follow. k Finally, writing G(z) = ∑∞ k=0 d k z , it follows that k

k

d k = 1 − ∑ c j = 1 − lim ∑ c j t j ⩾ 1 − limF(t) = 0, j=1

t→1

j=1

t→1

which establishes (11.2.3). We are now able to prove Theorem 11.2.2. Proof of Theorem 11.2.2. From (11.1.1) we have, with p ∈ P and z ∈ 𝔻, zf 󸀠 (z) p(z) p(z) . = = z α−1 1 −α f(z) αz ∫0 ζ p(ζ)dζ α ∫0 t α−1 p(tz)dt

(11.2.7)

It follows from the duality principle, Ruscheweyh [221, Theorem 1.1, Corollary 1.1 and Theorem 1.6] that any value assumed by the right hand side of (11.2.7) for z ∈ 𝔻 is assumed for this value of z when p(z) is of the form (1 + xz)/(1 + yz), where |x| = 1 and |y| = 1. Clearly, in seeking lower and upper bounds for |zf 󸀠 (z)/f(z)|, we may take p(z) = for |x| = 1.

1 + xz , 1−z

(11.2.8)

168 | 11 B1 (α) Bazilevič Functions

We first obtain the lower bound in Theorem 11.2.2. Use (11.2.7) and (11.2.8) to write z

1

0

0

1 + xζ 1−z α 1−z 1 + txz f(z) ∫ ζ α−1 = α dζ = α ∫ t α−1 ( )( ) dt. 󸀠 zf (z) z 1 + xz 1−ζ 1 + xz 1 − tz Clearly, when 0 ⩽ t ⩽ 1 and |z| < 1, 1 + t|z| 󵄨󵄨󵄨󵄨 1 + tz 󵄨󵄨󵄨󵄨 1 − t|z| ⩽󵄨 . 󵄨⩽ 1 + |z| 󵄨󵄨󵄨 1 + z 󵄨󵄨󵄨 1 − |z| Hence

󵄨󵄨 1 + txz 1 − z 󵄨󵄨 1 − t|z| 1 + |z| 󵄨󵄨 󵄨󵄨 , 󵄨⩽ 󵄨󵄨 󵄨󵄨 1 + xz 1 − tz 󵄨󵄨󵄨 1 − |z| 1 + t|z|

and so 1 󵄨󵄨 f(z) 󵄨󵄨 1 − tr 1+r 󵄨 󵄨󵄨 ∫ t α−1 dt, 󵄨󵄨 󸀠 󵄨󵄨󵄨 ⩽ α 󵄨󵄨 zf (z) 󵄨󵄨 1−r 1 + tr 0

which is the required lower bound. For the upper bound we use (11.2.8), together with F defined in Lemma 11.2.1 to write z

z

α ∫ ζ α−1 p(ζ)dζ = α ∫ ζ α−1 (−x + 0

0

1 + xF(z) x+1 ) dζ = z α . 1−ζ 1 − F(z)

Thus (11.2.7) and (11.2.8) give 1 + xz zf 󸀠 (z) = G(z) , f(z) 1 + xF(z) where G(z) = (1 − F(z))/(1 − z). Since (1 + az)/(1 + bz) maps the closed unit disk onto a closed disk with center (1 − ab)/(1 − |b|2 ) and radius |a − b|/(1 − |b|2 ), provided that |b| < 1, we deduce that 󵄨󵄨 󸀠 󵄨󵄨 |z − F(z)| + |1 − F(z)z|̄ 󵄨󵄨 zf (z) 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨 f(z) 󵄨󵄨󵄨 ⩽ |G(z)| 1 − |F(z)|2 󵄨 󵄨 󵄨󵄨󵄨 |G(z)| F(z) 󵄨󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 F(z) 󵄨󵄨󵄨󵄨 = [r 󵄨󵄨󵄨1 − )󵄨 ] 󵄨󵄨 + 󵄨󵄨1 − r2 + r2 (1 − 2 󵄨 󵄨 󵄨 z 󵄨 󵄨 z 󵄨󵄨󵄨 1 − |F(z)| 󵄨 󵄨󵄨 F(z) 󵄨󵄨󵄨󵄨 |G(z)| 󵄨󵄨 [r(1 + r) 1 − ⩽ 󵄨 󵄨 + (1 − r2 )] 󵄨 󵄨󵄨 z 󵄨󵄨󵄨 1 − |F(z)|2 =

r 1+r [ |F 󸀠 (z)| + (1 − r)|G(z)|] , 1 − |F(z)|2 α

where we have used F 󸀠 (z) = αG(z)(1 − F(z)/z).

11.3 Growth Estimates |

169

It now follows from Lemma 11.2.1 that the last expression takes its maximum value when z = r, and so 󵄨󵄨 󸀠 󵄨󵄨 󵄨󵄨 zf (z) 󵄨󵄨 (1 + r)G(r) 1 + r 1 − F(r) 󵄨󵄨 ⩽ 󵄨󵄨 = 1 + F(r) 1 − r 1 + F(r) 󵄨󵄨󵄨 f(z) 󵄨󵄨󵄨 r

ζ α−1 1+r (−1 + 2αr−α ∫ dζ ) = 1−r 1−ζ

−1

0

1

= (1 + r) (α(1 − r) ∫ t 0

α−1 1 +

tr dt) 1 − tr

−1

,

which completes the proof of Theorem 11.2.2. Before leaving this topic we note that when α = 1, i.e., when f belongs to the class R discussed above, the upper bound in Theorem 11.2.2 reduces to 1 󵄨󵄨 󸀠 󵄨󵄨 1 + tr 󵄨󵄨 zf (z) 󵄨󵄨 1 + r 󵄨󵄨 󵄨󵄨 ⩽ ∫ ( dt) 󵄨󵄨 f(z) 󵄨󵄨 1 − r 1 − tr 󵄨 󵄨 0

−1

=−

1+r 2 [1 + log(1 − r)] . 1−r r

However, the following stronger version of the above inequality was obtained by London [127], who showed that for f ∈ R, 1+r 2 |f 󸀠 (z)| ⩽− [1 + log(1 − r)] , Re[f(z)/z] 1−r r and it is interesting to speculate whether a corresponding improved inequality holds for f ∈ B1 (α).

11.3 Growth Estimates It was seen in previous chapters that the classical growth result of Littlewood and Paley (see, e.g., Theorem 3.3 by Hayman [70]), estimating the growth of the coefficients of functions in S in terms of the maximum modulus could be improved when f ∈ S∗ , K0 , or B(α). Since functions in B1 (α) exhibit a slower rate of growth, better results are to be expected in this case. The following shows that this is, indeed, the case. Theorem 11.3.1. Let f ∈ B1 (α) for 0 < α ⩽ 1, and for z = re iθ ∈ 𝔻 be given by f(z) = n z + ∑∞ n=2 a n z . Then, for n ⩾ 2, n|a n | ⩽ C(α)[M(1 − 1/n)]1−α ,

(11.3.1)

where C(α) is a constant depending on α, and M(r) = max |f(z)|. Also, |z|=r

α

1−α

L(r) ⩽ r [M(r)]

{2π + 4 log[(1 + r)/(1 − r)]},

where L(r) is the length of the curve f(|z| = r).

(11.3.2)

170 | 11 B1 (α) Bazilevič Functions

Proof. Following the method used in Chapter 10, from (11.1.1) we can write z(zf 󸀠 (z))󸀠 = z(z α f(z)1−α p(z))󸀠 , for some p ∈ P. Since

2π

1 n an = ∫ z(zf 󸀠 (z))󸀠 e−inθ dθ, 2πr n 2

0

it follows that 2π

n2 |a n | ⩽

2π

α 1 − α 󵄨󵄨 1+α 󵄨 󵄨 󵄨 ∫ 󵄨󵄨󵄨z α [f(z)]1−α p(z)󵄨󵄨󵄨󵄨 dθ + ∫ 󵄨󵄨z [f(z)]−α f 󸀠 (z)p(z)󵄨󵄨󵄨󵄨 dθ 2πr n 󵄨 2πr n 󵄨 0

0 2π

+

1 󵄨 󵄨 ∫ 󵄨󵄨󵄨󵄨z1+α [f(z)]1−α p󸀠 (z)󵄨󵄨󵄨󵄨 dθ n 2πr 0

= I1 (r) + I2 (r) + I3 (r) say. Clearly, 2π

α α 1+r [M(r)]1−α ∫ |p(z)|dθ ⩽ [M(r)]1−α (2π + 4 log I1 (r) ⩽ ), 2πr n−α 2πr n−α 1−r 0

where we have used an inequality established in Theorem 4.1.13. For I2 (r), we use Theorem 4.1.13 to obtain 2π

I2 (r) =

1 − α 󵄨󵄨 2α 󵄨 ∫ 󵄨z [f(z)]1−2α p2 (z)󵄨󵄨󵄨󵄨 dθ 2πr n 󵄨󵄨 0 2π

⩽

1−α 󵄨 󵄨 ∫ 󵄨󵄨󵄨z α [f(z)]1−α p2 (z)󵄨󵄨󵄨󵄨 dθ 2πr n Q2 (1) 󵄨 0

2π

(1 − α)[M(r)]1−α ∫ |p(z)|2 dθ ⩽ 2πr n−α Q2 (1) 0

⩽

(1 − + 3r2 ) , n−α r Q2 (1)(1 − r2 ) α)[M(r)]1−α (1 2π

where we have used the inequality (2π)−1 ∫0 |p(z)|2 dθ ⩽ (1 + 3r2 )/(1 − r2 ) from Theorem 3.1.11. Finally 2π

[M(r)]1−α 2[M(r)]1−α 󸀠 , ∫ |p (z)|dθ ⩽ I3 (r) ⩽ 2πr n−α−1 r n−α−1 (1 − r2 ) 0

where we have again used Theorem 3.1.11.

11.4 Coefficients | 171

Choosing r = 1 − 1/n in the above estimates for I1 (r), I2 (r), and I3 (r) completes the proof of (11.3.1). The proof of (11.3.2) is simple, viz: 2π

2π 󸀠

L(r) = ∫ |zf (z)|dθ ⩽ r [M(r)] 0

α

1−α

∫ |p(z)|dθ, 0

1+r ⩽ r α [M(r)]1−α (2π + 4 log ), 1−r as before. Corollary 11.3.1. Let f ∈ B1 (α), then na n = O(log n)(1−α)/α as n → ∞, and the order of growth is best possible. Proof. This is an immediate consequence of Theorems 11.2.1 and 11.3.1, and an example in Littlewood [122, p. 44] shows that the coefficients of f0 (z) defined in Theorem 11.2.1 have an order of growth n−1 (log n)(1−α)/α as n → ∞.

11.4 Coefficients In what was probably the first paper to consider functions in B1 (α), Singh [229] established the following sharp inequalities for the first four coefficients of the Taylor series for f(z). n Theorem 11.4.1. Let f ∈ B1 (α), and for z ∈ 𝔻 be given by f(z) = z + ∑∞ n=2 a n z . Then, for α ⩾ 0,

2 |a2 | ⩽ , 1+α

2(3 + α) { , { { (2 + α)(1 + α)2 |a3 | ⩽ { { 2 { , { 2+α

2 4(1 − α)(5 + 3α + α 2 ) { { { 3 + α + 3(2 + α)(1 + α)3 , |a4 | ⩽ { { 2 { , { 3+α

0 ⩽ α ⩽ 1, α ⩾ 1, 0 ⩽ α ⩽ 1, α ⩾ 1.

All the inequalities are sharp. For 0 ⩽ α ⩽ 1, the proofs of the inequalities in Theorem 11.4.1 follow relatively easily using the results in Theorems 3.1.7 and 3.1.8, and we therefore omit the proofs. When α ⩾ 1 further analysis is needed, which can be found in Singh’s paper [229]. Finding exact bounds for |a n | when n ⩾ 5 has proved to be more difficult. However, the exact bound for |a5 | when 0 ⩽ α ⩽ 1/2 was recently obtained by Marjono et al. [137].

172 | 11 B1 (α) Bazilevič Functions n Theorem 11.4.2. Let f ∈ B1 (α), and for z ∈ 𝔻 be given by f(z) = z + ∑∞ n=2 a n z . Then, for 0 ⩽ α ⩽ 1/2,

|a5 | ⩽

2 2(1 − α)(3α 5 + 22α 4 + 68α 3 + 113α 2 + 121α + 81) . + 4+α 3(1 + α)4 (2 + α)2 (3 + α)

The inequality is sharp. Since the method of proof involves a significant variation of those used in Theorem 11.4.1; we give the complete proof. Proof. As before, we can write [

∞ f(z) α−1 󸀠 f (z) = p(z) = 1 + ∑ p n z n , ] z n=1

(11.4.1)

for p ∈ P, and so equating coefficients in (11.4.1) gives a5 =

p22 p4 p1 p3 + (1 − α) [ + 4+α (1 + α)(3 + α) 2(2 + α)2 +

(3α − 1)(2α − 1)p41 (1 − 2α)p21 p2 + ]. 2(1 + α)2 (2 + α) 24(1 + α)4

(11.4.2)

Suppose first that 0 ⩽ α ⩽ 1/3. Then, using |p n | ⩽ 2, we obtain 2 2 1 + 2(1 − α) [ + 4+α (1 + α)(3 + α) (2 + α)2 (3α − 1)(2α − 1) 2(1 − 2α) + + ] (1 + α)2 (2 + α) 3(1 + α)4 2 2(1 − α)(3α 5 + 22α 4 + 68α 3 + 113α 2 + 121α + 81) , = + 4+α 3(1 + α)4 (2 + α)2 (3 + α)

|a5 | ⩽

which is the first inequality for |a5 | when 0 ⩽ α ⩽ 1/3. Next suppose that 1/3 ⩽ α ⩽ 1/2. Then, from (11.4.2), we have 2 2 1 ] + 2(1 − α) [ + 4+α (1 + α)(3 + α) (2 + α)2 󵄨 (3α − 1)p21 󵄨󵄨󵄨󵄨 (1 − α)(1 − 2α)|p1 |2 󵄨󵄨󵄨 p2 󵄨󵄨 󵄨󵄨 . + − 2(1 + α)2 󵄨󵄨󵄨 2 + α 12(1 + α)2 󵄨󵄨󵄨 We now use Theorem 3.1.3 to obtain |a5 | ⩽

|p1 |2 2 2(1 − α)(11 + 12α + 3α 2 ) (1 − α)(1 − 2α)|p1 |2 + (2 − + ) 2 2 4+α 2 (1 + α)(2 + α) (3 + α) 2(1 + α) (2 + α) 󵄨 󵄨 (1 − α)(1 − 2α)|p1 |2 󵄨󵄨󵄨 p21 (3α − 1)(2 + α) 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 + − ) (1 2 2 󵄨󵄨 2(1 + α) 6(1 + α) 󵄨󵄨󵄨 2(2 + α) 󵄨 2 2(1 − α)(11 + 12α + 3α ) (1 − α)(1 − 2α)|p1 |2 2 + + = 4+α (1 + α)(2 + α)2 (3 + α) (1 + α)2 (2 + α) (1 − α)(1 − 2α)(2 − 5α − 3α 2 )|p1 |4 := ϕ(|p1 |). + 24(1 + α)4 (2 + α)

|a5 | ⩽

11.4 Coefficients | 173

Elementary calculus shows that ϕ󸀠 (|p1 |) ⩾ 0 on [0, 2] for 1/3 ⩽ α ⩽ 1/2, so that |a5 | ⩽ ϕ(2), which gives the required inequality. It is easy to see that the inequality for |a5 | is sharp when p(z) = 1 + 2z + 2z2 + . . .. For 1/2 ⩽ α ⩽ 1 the problem becomes more difficult; however it was also shown in [137] that Theorem 11.4.2 holds on the interval 0 ⩽ α ⩽ α 1 , where α 1 = 0.96942 . . . is the smallest positive root of the equation 12(1 + α)4 − 24(1 + α)2 (2α − 1)(2 + α) + 5(3α − 1)(2α − 1)(2 + α)2 = 0. We remark that using similar techniques to those employed in the proof of Theorem 11.4.2, it is possible to prove the following sharp bound for |a6 | (see, again, [137]). n Theorem 11.4.3. Let f ∈ B1 (α), and for z ∈ 𝔻 be given by f(z) = z + ∑∞ n=2 a n z . Then, for 0 ⩽ α ⩽ 1/2,

2 + (1 − α) (32A + 16B + 8C + 8D + 4E + 4F) 5+α 2 + 4(1 − α) = 5+α 1008 + 1839α + 2573α 2 + 2264α 3 + 1207α 4 + 391α 5 + 72α 6 + 6α 7 ⋅( ), 15(1 + α)5 (2 + α)2 (3 + α)(4 + α)

|a6 | ⩽

where A=

(1 − 2α)(1 − 3α)(1 − 4α) , 120(1 + α)5

B=

(1 − 2α)(1 − 3α) , 6(1 + α)3 (2 + α)

C=

1 − 2α , 2(1 + α)(2 + α)2

D=

1 − 2α , 2(1 + α)2 (3 + α)

E=

1 , (2 + α)(3 + α)

F=

1 . (1 + α)(4 + α)

The inequality is sharp. The proof follows the same arguments as those used in Theorem 11.4.2, by considering α on each of the intervals [0, 1/4], [1/4, 1/3], and [1/3, 1/2], and using the expressions for A, B, C, D, E, and F above. Also as before, the inequality for |a6 | is sharp when p(z) = 1 + 2z + 2z2 + . . . . The above results suggest the following obvious conjectures. Conjecture 11.4.1. Let 0 ⩽ α ⩽ 1. Suppose that g ∈ S and is given by g(z) = z+b 2 z2 +. . . and ∞ g(z) α−1 = 1 + 2 ∑ zn . g󸀠 (z) [ ] z n=1 n If f ∈ B1 (α) with f(z) = z + ∑∞ n=2 a n z , then for n ⩾ 2,

|a n | ⩽ b n .

174 | 11 B1 (α) Bazilevič Functions n Conjecture 11.4.2. Let α ⩾ 1 and f ∈ B1 (α), with f(z) = z + ∑∞ n=2 a n z . Then, for n ⩾ 2,

|a n | ⩽

2 . n−1+α

Finally, we note that Singh [229] gave a complete solution to the Fekete–Szegő problem for both real and complex μ. We state the complex case below. n Theorem 11.4.4. Let f ∈ B1 (α), for α ⩾ 0, and for z ∈ 𝔻 be given by f(z) = z+∑∞ n=2 a n z . Then, for any μ ∈ ℂ,

|a3 − μa22 | ⩽

2 |3 + α − 2μ(2 + α)| } ⋅ max {1, 2+α (1 + α)2

2 = 2+α

1, { { ⋅ { |3 + α − 2μ(2 + α)| { , { (1 + α)2

|3 + α − 2μ(2 + α)| ⩽ (1 + α)2 , |3 + α − 2μ(2 + α)| ⩾ (1 + α)2 .

The inequalities are sharp.

Other coefficient inequalities Finding sharp bounds for the coefficients of the inverse function when f ∈ B1 (α) turns out to be a little easier. When 0 ⩽ α ⩽ 1, sharp inequalities for the first five inverse coefficients can again be found in Marjono et al. [137]. Also using the methods of previous chapters, the sharp bound for the second Hankel determinant, together with the initial coefficients of log( f(z)/z) for f ∈ B1 (α) follow relatively easily (see Thomas [245]).

11.5 Other Inequalities Singh, in his early paper [229], proved various properties of functions in B1 (α), and in particular that if f ∈ B1 (α) and α is a positive integer, then Re[(f(z)/z)α ] > 0 for z ∈ 𝔻. This simple result was subsequently improved by Obradović in 1986 [167] to Re[(f(z)/z)α ] > 1/(1 + 2α) (z ∈ 𝔻), for α > 0, then finally sharpened and generalized by Halim and Thomas [64]. We first give a useful lemma, due to Sakaguchi [223], which is essentially the same as Lemma 5.2.1 (i) and has been used repeatedly in the investigation of subclasses of functions in S. Lemma 11.5.1. Let F and G be analytic in 𝔻, with F(0) = G(0), and suppose that G maps 𝔻 onto a (many-sheeted) region, which is starlike with respect to the origin. Then, Re [

F 󸀠 (z) F(z) ] > 0 (z ∈ 𝔻) implies Re [ ] > 0 (z ∈ 𝔻). G󸀠 (z) G(z)

11.5 Other Inequalities | 175

We give Singh’s observation below to illustrate a use of Lemma 11.5.1, noting that the lemma is only applicable when α is a positive integer. Theorem 11.5.1. Let f ∈ B1 (α), then Re[(f(z)/z)α ] > 0 for z ∈ 𝔻. Proof. Simply note that f 󸀠 (z) [

f(z) α−1 (f α (z))󸀠 = , ] z (z α )󸀠

and apply Lemma 11.5.1. The sharpened and generalized version referred to above [64] is the following. Theorem 11.5.2. Let f ∈ B1 (α) and for z ∈ 𝔻 and n ⩾ 1, define z

1 I n (z) = ∫ I n−1 (t)dt, z 0

and I0 (z) = (f(z)/z)α . Then, Re I n (z) ⩾ γ n (r) > γ n (1), where

∞

0 < γ n (r) = 2α ∑ k=1

(−1)k+1 r k−1 − 1 < 1. k n (k − 1 + α)

Equality occurs when 1/α

1

1 + tz f0 (z) = z (α ∫ t α−1 dt) 1 − tz

.

0

We note that when n = 0, we obtain r

∞ (−1)k+1 r k−1 α 1−ρ f(z) α ] ⩾ α ∫ ρ α−1 ( ) dρ = 2α ∑ − 1, Re [ z r 1+ρ k−1+α k=1 0

which in the case α = 1, i.e., when f ∈ R reduces to (2/r) log(1 + r) − 1, showing that for f ∈ R, 2 f(z) ] ⩾ log(1 + r) − 1 > 2 log 2 − 1, Re [ z r which is a result proved by Hallenbeck [65]. Proof. First note that for p ∈ P we can write z

[

f(z) α α ] = α ∫ t α−1 p(t)dt, z z 0

so that with t = Re [

ρe iθ ,

we have r

r

0

0

f(z) α α α 1−ρ ] = α ∫ ρ α−1 Re p(ρe iθ )dρ ⩾ α ∫ ρ α−1 ( ) dρ, z r r 1+ρ

on using Theorem 3.1.1.

176 | 11 B1 (α) Bazilevič Functions

Next note that r

Re I0 (z) = Re [

f(z) α α 1−ρ ] ⩾ α ∫ ρ α−1 ( ) dρ, z r 1+ρ 0

and that z

r

0

0

1 1 Re I n+1 (z) = Re ∫ I n (t)dt = ∫ Re I n (ρe iθ )dρ z r r

⩾

∞ (−1)k+1 ρ k−1 1 − 1) dρ = γ n+1 (r), ∫ (2α ∑ n r k (k − 1 + α) k=1 0

where the inequality follows by an induction argument. Now set ∞ (−1)k+1 r k−1 ϕ n (r) = α ∑ n . k (k − 1 + α) k=1 This series is absolutely convergent for n ⩾ 0 and 0 < r < 1, and so suitably rearranging the pairs of terms shows that 1/2 < ϕ n (r) < 1. Thus 0 < γ n (r) < 1. We finally note that since for n ⩾ 1, r

rϕ n (r) = ∫ ϕ n−1 (ρ)dρ, 0

induction shows that ϕ󸀠n (r) < 0, and so γ n (r) decreases with r as r → 1 for fixed n and increases to 1 as n → ∞ for fixed r. We end this chapter with a subordination result due to Marjono [136], noting that the result holds for all f ∈ A. Theorem 11.5.3. Let f ∈ A. Then, for α > 0, γ > 0 and z ∈ 𝔻, f 󸀠 (z) [

f(z) α−1 1 + z β(γ) ≺( ] ) z 1−z

implies [

1+z γ f(z) α ] ≺( ) , z 1−z

where β(γ) = γ +

2 γ arctan ( ) . π α

Proof. Write p(z) = [

f(z) α ] , z

(11.5.1)

(11.5.2)

11.5 Other Inequalities | 177

so that p is analytic in 𝔻, p(0) = 1, and p(z) +

f(z) α−1 󸀠 zp󸀠 (z) f (z). =[ ] α z

Thus, we need to show that p(z) +

1+z β zp󸀠 (z) ≺( ) α 1−z

implies p(z) ≺ (

1+z γ ) , 1−z

when β := β(γ) is given by (11.5.2). For z ∈ 𝔻, let h(z) = [(1 + z)/(1 − z)]β(γ) and q(z) = [(1 + z)/(1 − z)]γ , so that | arg h(z)| < β(γ)π/2, and | arg q(z)| < γπ/2. Suppose that p ⊀ q. Then from Theorem 3.2.6, there exists z0 ∈ 𝔻 and ζ0 ∈ ∂𝔻, such that p(z0 ) = q(ζ0 ), p(|z| < |z0 |) ⊂ q(𝔻) and z0 p󸀠 (z0 ) = kζ0 q󸀠 (ζ0 ), for k ⩾ 1. Since p(z0 ) = q(ζ0 ) ≠ 0, it follows that ζ0 ≠ ±1. Thus, we can write ri = (1 + ζ0 )/ (1 − ζ0 ) for r ≠ 0. Hence, p(z0 ) +

z0 p󸀠 (z0 ) kζ0 q󸀠 (ζ0 ) = q(ζ0 ) + α α

for k ⩾ 1. Differentiating q(z) we obtain p(z0 ) +

z0 p󸀠 (z0 ) kγ(1 + r2 ) = (ri − ) (ri)γ−1 α 2α = (ri −

kγ(1 + r2 ) γ−1 (γ − 1)π (γ − 1)π + i sin ) r (cos ). 2α 2 2

Since cos((γ − 1)π/2) = sin(γπ/2) and sin((γ − 1)π/2) = − cos(γπ/2), we have p(z0 ) +

z0 p󸀠 (z0 ) γπ kγ(1 + r2 ) γ−1 γπ = (ri − ) r (sin − i cos ) α 2α 2 2 =

γπ γπ kγ r γ−1 cos (1 + r2 ) tan [2r − 2 2 α 2 γπ kγ + (1 + r2 ))] , + i (2r tan 2 α

and so taking arguments in (11.5.3), we obtain arg (p(z0 ) +

z0 p󸀠 (z0 ) 2r tan(γπ/2) + kγ(1 + r2 )/α ) = arctan α 2r − kγ(1 + r2 ) tan(γπ/2)/α ⩾ arctan

2r tan(γπ/2) + γ(1 + r2 )/α := Φ(r). 2r − γ(1 + r2 ) tan(γπ/2)/α

(11.5.3)

178 | 11 B1 (α) Bazilevič Functions Now write Φ(r) = arctan(U(r)/V(r)), so that U(r) = 2r tan(γπ/2) + γ(1 + r2 )/α, and V(r) = 2r − γ(1 + r2 ) tan(γπ/2)/α. Next, note that since Φ󸀠 (r)[U 2 (r) + V 2 (r)] = V(r)U 󸀠 (r) − U(r)V 󸀠 (r), it follows that V(r)U 󸀠 (r) − U(r)V 󸀠 (r) = 0 when r = 1. Thus, Φ(r) attains its minimum when r = 1, and so γπ +γ β(γ)π 2 Φ(r) ⩾ Φ(1) = arctan ( γπ ) = 2 , α − γ tan 2 α tan

which contradicts the fact that |h(z)| < β(γ)π/2, provided that (11.5.1) holds. A discussion of whether (11.5.2) defines the best dominant, can be found in Marjono [136].

12 The Class U(λ) In this chapter, we consider a class of univalent functions, whose definition resembles that of the B1 (α) functions considered in Chapter 11. These functions have received considerable attention in recent years, mainly through the work of Obradović and Ponnusamy.

12.1 Definition and Geometrical Properties Definition 12.1.1. Let f ∈ A and 0 < λ ⩽ 1. Then f ∈ U(λ) if, and only if, 󵄨 󵄨󵄨 (z ∈ 𝔻), 󵄨󵄨U f (z)󵄨󵄨󵄨 < λ where U f (z) = [

(12.1.1)

z 2 󸀠 ] f (z) − 1. f(z)

We write U := U(1). Since f 󸀠 (z)[z/f(z)]2 ≠ 0 in 𝔻, it follows that every f ∈ U(λ) is nonvanishing in 𝔻 \ {0}. It is also clear that every f ∈ U(λ) is locally univalent. From Definition 2.3.3 we see that functions in U have a similar definition to the B1 (α) Bazilevič functions and are sometimes referred to as non-Bazilevič functions. Obradović et al. [175] studied properties of functions in U(λ) and proved the following. Theorem 12.1.1. The class U(λ) is preserved under rotation, conjugation, dilation, and omitted-value transformations, but is not preserved under the n-th root transformation for any n ⩾ 2. Proof. Let f ∈ U(λ), and define g(z) = e−iθ f(ze iθ ), h(z) = f(z) and ψ(z) = r−1 f(rz). Then, g󸀠 (z) = f 󸀠 (ze iθ ), h󸀠 (z) = f 󸀠 (z), ψ󸀠 (z) = f 󸀠 (rz), and 2

[

ze iθ z 2 󸀠 ] f 󸀠 (ze iθ ) − 1 ] g (z) − 1 = [ g(z) f(ze iθ )

[

z 2 󸀠 z 2 󸀠 ] h (z) − 1 = [ ] f (z) − 1 h(z) f(z)

[

rz 2 󸀠 z 2 󸀠 ] ψ (z) − 1 = [ ] f (rz) − 1. ψ(z) f(rz)

Since f ∈ U(λ), the above three relations show that each of the functions g, h and ψ belong to U(λ). Next, letting f ∈ U(λ) and f(z) ≠ c for some c ≠ 0, it is not difficult to show that the omitted-value transformation F defined by F(z) = https://doi.org/10.1515/9783110560961-012

cf(z) c − f(z)

180 | 12 The Class U(λ) belongs to S. Thus, z/F(z) is nonvanishing in 𝔻. A simple computation gives U F (z) = U f (z), and since f ∈ U(λ), it follows that F ∈ U(λ). To see that the class U is not preserved under the n-th root transformation, consider the function f n defined for n ⩾ 2 by z (z ∈ 𝔻). f n (z) = 1 + (1/n)z + (−1)n (1/(n + 1))z n+1 Clearly, z/f n (z) ≠ 0 in 𝔻, and U f n (z) = (−1)n+1 (n/(n + 1))z n+1 . Therefore |U f n (z)| < 1, and so f n ∈ U. Now let the n-th root transformation of f n be given by g1 , where n n f n (z n ) = z √ g1 (z) = √

f n (z n ) . zn

Thus, 2 z U g1 (z) = [ ] g󸀠1 (z) − 1 = g1 (z)

n n2 +n z n+1 − 1. 1 n2 +n (n−1)/n 1 [1 + z n + (−1)n+1 ] z n 1+n 1 + (−1)n

It is not difficult to see that g1 does not belong to U for each n ⩾ 2. In particular, we demonstrate this in the simple case when n = 2. Here, U g1 (z) → (5√6 − 3)/3 > 1 as z → i, which shows that U g1 (𝔻) is not a subset of 𝔻, and so g1 does not belong to U when n = 2. Aksentév [4] and Ozaki and Nunokawa [181] showed that functions in U(λ) are univalent, i.e., U(λ) ⊂ S for all 0 < λ ⩽ 1. However, surprisingly, U is not a subset of S∗ , which might be expected from the similarity in their analytic representations and the fact that functions in the set Sℤ given by Sℤ = {z,

z z z z , , , }, (1 ± z)2 1 ± z 1 ± z2 1 ± z + z2

all belong to both S∗ and U. (For a discussion of functions in S with integral coefficients, see Friedman [50].) To see that U 󳠬 S∗ , let the function f1 be given by z f1 (z) = ∈U 1 + z/2 + z3 /2 and then for z0 = (−1 + i)/√2, |z0 | = 1; it is easy to see that z0 f1󸀠 (z0 )/f1 (z0 ) has negative real part, so that f1 ∉ S∗ . On the other hand, we saw in the proof of Theorem 12.1.1 that the square root transform of f2 is not in U, while f2 is starlike since zf2󸀠 (z) 6 − 4z3 , = f2 (z) 6 + 3z + 2z3 and so is its square root transform. It is a relatively easy exercise to prove the following characterization of the class U given by Obradović and Ponnusamy [170].

12.1 Definition and Geometrical Properties | 181

Theorem 12.1.2 (Characterization of the class U). Every function f ∈ U has the representation z

ω(t) z = 1 − a2 z − z ∫ 2 dt, f(z) t 0

where a2 = 1 for z ∈ 𝔻.

f 󸀠󸀠 (0)/2, and the function

ω is analytic in 𝔻, ω(0) = ω󸀠 (0) = 0, and |ω(z)|
1/2 for z ∈ 𝔻. In [176], Obradović et al. generalized this result to the class U n defined by U n := {f ∈ U : f(z) = z + a n+1 z n+1 + ⋅ ⋅ ⋅ } , showing that when f ∈ U n , Re [(

f(z) n/2 1 ) ]> z 2

(z ∈ 𝔻).

12.2 Sufficient Conditions and Univalence As was shown in the previous section, there are functions that are starlike but do not belong to U. The next result, due to Obradović and Ponnusamy [170], gives a sufficient condition for a function to belong to U, which is identical to that given in Theorem 4.1.1 for starlike functions. ∞ n Theorem 12.2.1. Let f ∈ A and be given by f(z) = z+∑∞ n=2 a n z for z ∈ 𝔻. If ∑ n=2 n|a n | ⩽ 1, then f ∈ U. The result is sharp, i.e., the constant 1 cannot be replaced by a larger one so that the conclusion holds. n Proof. Since f(z) = z + ∑∞ n=2 a n z , we have 2 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 ∞ ∞ 󵄨󵄨 f(z) 2 󵄨󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨 󸀠 󵄨󵄨f (z) − [ 󵄨󵄨 = 󵄨󵄨1 + ∑ na n z n−1 − (1 + ∑ a n z n−1 ) 󵄨󵄨󵄨 ] 󵄨󵄨 󵄨 󵄨󵄨 󵄨 z 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 n=2 n=2 󵄨 󵄨󵄨 ∞ 2 󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 = 󵄨󵄨󵄨󵄨 ∑ (n − 2)a n z n−1 − ( ∑ a n z n−1 ) 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨n=2 n=2 󵄨 󵄨 󵄨󵄨 ∞ 2 󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 = |z|2 󵄨󵄨󵄨󵄨 ∑ (n − 2)a n z n−3 − ( ∑ a n z n−2 ) 󵄨󵄨󵄨󵄨 . 󵄨󵄨 󵄨󵄨󵄨n=3 n=2 󵄨

Applying ∑∞ n=2 n|a n | ⩽ 1 in (12.2.1) we obtain 2 󵄨 󵄨󵄨 ∞ f(z) 2 󵄨󵄨󵄨 ∞ 󵄨󵄨 󸀠 󵄨 󵄨󵄨f (z) − [ ] 󵄨󵄨 < ∑ (n − 2)|a n | + ( ∑ |a n |) 󵄨󵄨 z 󵄨󵄨 n=2 󵄨 n=2 ∞

∞

2

⩽ 1 − 2 ∑ |a n | + ( ∑ |a n |) n=2 ∞

n=2 2

⩽ (1 − ∑ |a n |) n=2

󵄨󵄨 f(z) 󵄨󵄨2 󵄨󵄨 󵄨 ⩽ 󵄨󵄨󵄨 󵄨 . 󵄨󵄨 z 󵄨󵄨󵄨 This shows that |U f (z)| < 1, and hence f ∈ U.

(12.2.1)

12.2 Sufficient Conditions and Univalence

| 183

To see that the constant bound 1 cannot be replaced by 1+ϵ for any ϵ > 0, consider the function 1+ϵ n (n ⩾ 2), z f(z) = z + n and observe that since ϵ > 0, f 󸀠 (z) = 1 + (1 + ϵ)z n−1 has a zero in 𝔻. This shows that the result is sharp. The next result, again due to Obradović et al. [171], gives a sufficient condition for functions f ∈ A of the form z f(z) = , (12.2.2) ϕ(z) n where ϕ(z) = 1 + ∑∞ n=1 b n z has no zeros in 𝔻, to belong to U(λ). The case λ = 1 was considered in Obradović [168], while its sharpness was proved by Ali et al. [9]. Theorem 12.2.2. Let f ∈ A for 0 < λ ⩽ 1 and be of the form (12.2.2). If ∞

∑ (n − 1)|b n | ⩽ λ,

(12.2.3)

n=2

then, f ∈ U(λ). For λ = 1, the result is the best possible in the sense that the constant 1 cannot be replaced by a larger one so that the conclusion holds. Proof. We give a simple proof when λ = 1 (the general case is a straightforward extension). From (12.2.2) and the inequality (12.2.3), we obtain 󵄨󵄨 󵄨󵄨 z z 󸀠 󵄨 󵄨 |U f (z)| = 󵄨󵄨󵄨󵄨−z [ ] + − 1󵄨󵄨󵄨󵄨 f(z) f(z) 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ∞ 󵄨 󵄨󵄨 󵄨 = 󵄨󵄨󵄨− ∑ (n − 1)b n z n 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 n=1 󵄨 󵄨 ∞

⩽ ∑ (n − 1)|b n | ⩽ 1, n=1

which shows that f ∈ U. In order to prove the second part, it suffices to show that there exist ε > 0 and f of the form (12.2.2), such that ∞

∑ (n − 1)|b n | = 1 + ε, n=2

with f ∈ ̸ S. Let f(z) = z − az2 , where a = (√1 + ε)/(1 + √1 + ε) with ε > 0. Then, a ∈ (1/2, 1) and ∞ 1 z = = 1 + ∑ bn zn , f(z) 1 − az n=1 where b n = a n . Thus, ∞

∞

n=2

n=2

∑ (n − 1)b n = ∑ (n − 1)a n =

a2 = 1 + ε. (1 − a)2

Since f 󸀠 (z) = 1 − 2az, it follows that f 󸀠 (x0 ) = 0 at x0 = 1/(2a) ∈ (1/2, 1). This shows that f is not univalent in 𝔻. Thus, the constant 1 in (12.2.3) is the best possible.

184 | 12 The Class U(λ) The above result is closely related to one obtained by Reade et al. [207], with 1 − |b 1 | instead of 1 on the right hand side of (12.2.3) implying that f ∈ S∗ (as well as f ∈ U). Further, the condition (12.2.3) is sufficient but not necessary for f to belong to U. To see this, consider the function f defined by √5 3 1 4 z 1 = 1 + z2 + iz + z . f(z) 3 6 9 Then, z/f(z) is nonvanishing in 𝔻, since 󵄨 󵄨󵄨 z 󵄨󵄨 󵄨󵄨󵄨 √5 1 √5 1 1 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 iz + z2 󵄨󵄨󵄨󵄨 ⩾ 1 − (1 + + ) > 0. 󵄨󵄨 󵄨󵄨 ⩾ 1 − |z|2 󵄨󵄨󵄨󵄨1 + 󵄨󵄨 f(z) 󵄨󵄨 3 2 3 󵄨󵄨 3 2 3 󵄨󵄨 Also, |U f (z)| < 1 for z ∈ 𝔻, since |U f (z)| =

1 󵄨󵄨 2 󵄨 1 󵄨󵄨−z (1 + √5iz + z2 )󵄨󵄨󵄨 < max |2 cos θ + √5i| = 1. 󵄨 3 θ∈[0,2π) 3󵄨

Therefore, f ∈ U, but

∞

∑ (n − 1)|b n | = n=2

1 √5 1 + + > 1. 3 3 3

The condition (12.2.3) is not only sufficient, but also necessary for f to belong to U, provided that the coefficients b n are all nonnegative. More precisely, the following much stronger result, which is due to Obradović and Ponnusamy, holds; the proof can be found in [169, Theorem 6]. Theorem 12.2.3. Let f ∈ A and be of the form (12.2.2), with b n ⩾ 0. Then we have the following equivalence. (i) f ∈ S. f(z)f 󸀠 (z) ≠ 0 for all z ∈ 𝔻. z ∞ (iii) ∑n=2 (n − 1)b n ⩽ 1.

(ii)

(iv) f ∈ U. As a consequence of the area theorem (Theorem 1.2.1), it follows that for functions 2 f ∈ S of the form (12.2.2), ∑∞ n=2 (n − 1)|b n | ⩽ 1. In [9], Ali et al. proved that the opposite is true only in a disk |z| < 1/√2, with 1/√2 being the largest possible radius. The corresponding results for the class U were obtained by Ali et al. [9] and are given in the following theorem. Theorem 12.2.4. Let f ∈ A and be of the form (12.2.2). Then, 2 2 (i) If f ∈ U, ∑∞ n=2 (n − 1) |b n | ⩽ 1. 2 2 −1 (ii) If ∑∞ n=2 (n − 1) |b n | ⩽ 1, the function g, given by g(z) = r f(rz) belongs to U, for

0 < r ⩽ r0 = √(√5 − 1)/2 = 0.786 . . . . The result is best possible, i.e., r0 is the largest constant with this property.

12.2 Sufficient Conditions and Univalence

| 185

Proof. (i) From the power series representation of f ∈ U given by (12.2.2), we have 󵄨󵄨 ∞ 󵄨󵄨 󵄨󵄨 󵄨󵄨 (z ∈ 𝔻). |U f (z)| = 󵄨󵄨󵄨 ∑ (n − 1)b n z n 󵄨󵄨󵄨 ⩽ 1 󵄨󵄨n=2 󵄨󵄨 󵄨 󵄨 With z = re iθ ∈ 𝔻, the last inequality gives ∞

∑ (n − 1)2 |b n |2 r2n = n=2

2π 󵄨 ∞ 2π 󵄨󵄨2 󵄨󵄨 󵄨󵄨 1 1 󵄨 ∫ 󵄨󵄨󵄨 ∑ (n − 1)b n z n 󵄨󵄨󵄨 dθ ⩽ ∫ dθ = 1. 󵄨󵄨 2π 󵄨󵄨󵄨n=2 2π 󵄨 0 0

The desired inequality now follows by letting r → 1. (ii) Since g(z) = r−1 f(rz) has the form ∞ z rz = = 1 + ∑ bn rn zn , g(z) f(rz) n=1

it follows that ∞

∞

n=2

n=2

1/2

∑ (n − 1)|b n |r n ⩽ [ ∑ (n − 1)2 |b n |2 ] ∞

⩽ [ ∑ r2n ] n=2

∞

1/2

[ ∑ r2n ] n=2

1/2

=

r2 √1 − r 2

⩽ 1,

if r4 + r2 − 1 ⩽ 0, i.e., if 0 < r ⩽ r0 = √(√5 − 1)/2, which establishes (ii). To prove the sharpness, we consider the function f0 defined for z ∈ 𝔻 by ∞ rn z = 1 + ∑ 0 z n = 1 − r0 z log(1 − r0 z). f0 (z) n−1 n=2

It is easy to see that Re(z/f0 (z)) > 0 for z ∈ 𝔻, showing that f0 (z) ≠ 0 for 0 < |z| < 1. Also, ∞ ∞ r2n r40 0 ∑ (n − 1)2 |b n |2 = ∑ (n − 1)2 = = 1. (n − 1)2 1 − r20 n=2 n=2 On the other hand, for |z| < r0 we see that 󵄨 󵄨󵄨 󵄨󵄨󵄨 ∞ 󵄨󵄨 r0n n 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 r20 z2 󵄨󵄨󵄨󵄨 r40 󵄨󵄨 󵄨󵄨 󵄨󵄨 z 2 󸀠 󵄨󵄨 = 󵄨󵄨 󵄨 󵄨󵄨 < 󵄨󵄨[ f (z) − 1 = ∑ (n − 1) = 1, z ] 󵄨 󵄨 0 󵄨 󵄨 n − 1 󵄨󵄨󵄨 󵄨󵄨󵄨 1 − r0 z 󵄨󵄨󵄨 1 − r20 󵄨󵄨󵄨 󵄨󵄨󵄨n=2 󵄨󵄨󵄨 f0 (z) while for r0 < |z| = r < 1, 󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨 r2 r2 󵄨󵄨 󵄨󵄨 󵄨󵄨 z 2 󸀠 󵄨󵄨 󵄨󵄨 = 0 󵄨󵄨[ f (z) − 1 ] > 1. 0 󵄨󵄨 󵄨󵄨 󵄨󵄨 f0 (z) 󵄨 󵄨󵄨z=r 1 − r0 r 󵄨 Thus, g0 defined by g0 (z) = r−1 f0 (rz) belongs to U, and so |U f (z)| ⩽ 1 holds in the disk |z| < r0 , but not in any larger one. Since f0󸀠 (z) =

1 − r0 z − r20 z2 , (1 − r0 z)(1 − r0 z log(1 − r0 z))2

and f0󸀠 (r0 ) = 0, then f0 is not univalent in any disk larger than |z| < r0 .

186 | 12 The Class U(λ) The following simple criteria for starlikeness of functions in U(λ) in terms of their second coefficient was obtained by Obradović et al. in [171]. Theorem 12.2.5. If f ∈ U(λ), a := |a2 | = |f 󸀠󸀠 (0)|/2 ⩽ 1, and 0 < λ ⩽ (√2 − a2 − a)/2, then f ∈ S∗ . Proof. Suppose that f ∈ U(λ) for 0 < λ ⩽ 1. Then from the definition of U(λ), we have 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 z 2 󸀠 󵄨󵄨 < λ, 󵄨󵄨[ f (z) − 1 (z ∈ 𝔻). (12.2.4) ] 󵄨󵄨 󵄨󵄨 f(z) 󵄨 󵄨 Using a similar method to that used in the proof of Theorem 12.1.2, f ∈ U(λ) has the following representation 1

z ω(tz) = 1 − a2 z − λ ∫ 2 dt. f(z) t 0

Since ω is a Schwarz function, with ω󸀠 (0) = 0 for z ∈ 𝔻, we have 󵄨󵄨 󵄨 1 󵄨󵄨 z 󵄨󵄨 󵄨󵄨󵄨 ω(tz) 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 − 1󵄨󵄨 = 󵄨󵄨−a2 z − λ ∫ 2 dt󵄨󵄨 󵄨󵄨 󵄨󵄨 f(z) 󵄨󵄨 󵄨󵄨 󵄨󵄨 t 󵄨󵄨 󵄨󵄨 0 ⩽ |z‖a2 | + λ|z|2 = |z|(|a2 | + λ|z|)
1 + λ and for any such ω, there exists z0 ∈ 𝔻, such that z0

1 − a2 z0 + λz0 ∫ ω(t) dt = 0. 0

Let |a2 | =

1+λ , r

r ∈ (0, 1).

(12.3.3)

We prove that F defined by z

a2 F(z) = 1 + λz ∫ ω(t) dt, 0

is a contraction map of 𝔻r into 𝔻r , where 𝔻 r = {z : |z| ⩽ r}. For z ∈ 𝔻r , we have r |F(z)| = 1+λ

󵄨󵄨 󵄨󵄨 z 󵄨󵄨 r(1 + λ|z|2 ) 󵄨󵄨 󵄨 󵄨󵄨 < r, 󵄨󵄨1 + λz ∫ ω(t) dt󵄨󵄨󵄨 ⩽ 󵄨󵄨 󵄨󵄨 1+λ 󵄨󵄨 󵄨󵄨 0

and for z1 , z2 ∈ 𝔻r , 󵄨󵄨 󵄨󵄨 z1 z2 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 󵄨󵄨z1 ∫ ω(t) dt + (−z1 + z1 − z2 ) ∫ ω(t) dt󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 0 0 󵄨󵄨 z1 󵄨󵄨 󵄨󵄨 z2 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 λr 󵄨 󵄨 󵄨 󵄨 (|z1 | 󵄨󵄨󵄨∫ ω(t) dt󵄨󵄨󵄨 + |z1 − z2 | 󵄨󵄨󵄨∫ ω(t) dt󵄨󵄨󵄨 ) ⩽ 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 1+λ 󵄨󵄨z2 󵄨󵄨 󵄨󵄨 0 󵄨󵄨 λr (|z1 | + |z2 |)|z1 − z2 | ⩽ 1+λ ⩽ r2 |z1 − z2 |,

λr |F(z1 ) − F(z2 )| = 1+λ

and so F is a contraction map of 𝔻r into 𝔻r . From the Banach fixed point theorem, there exists z0 ∈ 𝔻r such that F(z0 ) = z0 , which contradicts (12.3.2) at the point z0 ∈ 𝔻, and so (12.3.3) is not true for any r ∈ (0, 1). Thus, |a2 | ⩽ 1 + λ for f ∈ U(λ). If |a2 | = 1 + λ, then we must show that ω in (12.3.2) takes the form ω(z) = e iθ for some θ ∈ [0, 2π] and all z ∈ 𝔻. Assume on the contrary that ω(0) = a ∈ 𝔻, and f is as in (12.3.2). Then, from the Schwarz–Pick lemma (see, for example, [122]) applied to ω, we have 󵄨󵄨 a − ω(z) 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 ⩽ |z| 󵄨󵄨 󵄨󵄨 1 − aω(z) 󵄨󵄨󵄨

(z ∈ 𝔻),

from which we immediately obtain |ω(z)| ⩽

|a| + |z| 1 + |az|

(z ∈ 𝔻).

190 | 12 The Class U(λ)

Thus,

󵄨󵄨 z 󵄨󵄨󵄨 |z| 󵄨󵄨 2 󵄨󵄨 󵄨󵄨 󵄨󵄨 ⩽ ∫ |a| + s ds = |z| − 1 − |a| log(1 + |az|) ∫ ω(t) dt 󵄨󵄨 󵄨󵄨󵄨 2 1 + |a|s |a| |a| 󵄨󵄨 󵄨󵄨 󵄨 0 󵄨0 ⩽

1 − |a|2 1 log(1 + |a|) =: 𝑣(|a|) < 1. − |a| |a|2

Now let

z

F(z) = and define For z ∈ 𝔻r , we have

1 + λz ∫0 ω(t) dt a2

,

1 + λ𝑣(|a|) =: r < 1. 1+λ

1 + λr𝑣(|a|) < r, 1+λ and for z1 , z2 ∈ 𝔻r , we obtain (as above) 󵄨󵄨 z1 󵄨󵄨 z2 󵄨󵄨 λ 󵄨󵄨󵄨󵄨 󵄨 |F(z1 ) − F(z2 )| = 󵄨󵄨z1 ∫ ω(t) dt + (z1 − z2 ) ∫ ω(t) dt󵄨󵄨󵄨 󵄨󵄨 1 + λ 󵄨󵄨󵄨 󵄨󵄨 󵄨 z2 0 1 ⩽ (|z1 | + |z2 |)|z1 − z2 | ⩽ r|z1 − z2 |. 2 Hence F has a fixed point in 𝔻r , which contradicts the fact that f ∈ U(λ). Finally, we consider for fixed ϕ, ψ ∈ [0, 2π], |F(z)| ⩽

z = 1 − (1 + λ)e iϕ z + λe iψ z2 =: p(ϕ, ψ, z), f(z) and show that p(ϕ, ψ, z) is nonvanishing in 𝔻 if, and only if, ψ = 2ϕ. Without loss of generality, we may assume ϕ = 0 and show that among the functions p(0, ψ, z), the only nonvanishing one in 𝔻 is p(0, 0, z). Consider q ψ (z) := (1 + λ)z − λe iψ z2 . Since for z = re iτ , r ∈ [0, 1) and τ ∈ [0, 2π], we have Re q󸀠ψ (z) = 1 + λ − 2λr cos(ψ + τ) > 0, the function q ψ is univalent in 𝔻 (i.e., it belongs to the class R). In this case, q ψ (∂𝔻) is a Jordan curve, and q ψ (𝔻) is the simply connected domain bounded by this curve. Also, |q ψ (e iτ )| ⩾ 1 + λ − λ = 1, τ ∈ [0, 2π], and the minimum modulus is attained if, and only if, e iτ = e i(ψ+2τ), i.e., τ = −ψ. Hence, 1 ∈ ̸ q ψ (𝔻) if, and only if, Re q ψ (e−iψ ) = (1 + λ) cos ψ − λ cos ψ = cos ψ = 1, which is satisfied if, and only if, ψ = 0. Thus, f must be of the form (12.3.1).

12.3 Coefficients |

191

k Recently Obradović et al. [177] conjectured that for functions in U(λ), |a n | ⩽ ∑n−1 k=0 λ , and verified this conjecture when n = 2, 3, and 4 as follows. n Theorem 12.3.2. Let f ∈ U(λ) for 0 < λ ⩽ 1, and be given by f(z) = z + ∑∞ n=2 a n z . Then if 0 < λ < 1, 1 − λn , for n = 2, 3, 4. (12.3.4) |a n | ⩽ 1−λ

If λ = 1, then |a n | ⩽ n for n ⩾ 2. All the inequalities are the best possible. Proof. From (12.1.2) with λ = 1, we have ∞ f(z) 1 = ∑ nz n−1 , ≺ z (1 − z)2 n=1

and so by Theorem 3.2.5 (ii), it follows that |a n | ⩽ n for n ⩾ 2. Next let 0 < λ < 1. (Note that the case n = 2 has already been proved in Theorem 12.3.1.) Again from (12.1.2), we have ∞ f(z) 1 − λ n+1 n 1 ≺ =1+ ∑ z , z (1 − z)(1 − λz) 1−λ n=1

and f(z) 1 = , z (1 − zω(z))(1 − λzω(z)) where ω is analytic in 𝔻 and |ω(z)| ⩽ 1 for z ∈ 𝔻. Now set ω(z) = c1 + c2 z + . . . and rewrite the last relation as ∞

∞

n=1

n=1

∑ (1 − λ)a n+1 z n = ∑ (1 − λ n+1 )(c1 + c2 z + . . . )n z n .

(12.3.5)

Comparing coefficients of z n for n = 1, 2, 3 in (12.3.5), we obtain (1 − λ)a2 = (1 − λ2 )c1 (1 − λ)a3 = (1 − λ2 )c2 + (1 − λ3 )c21 (1 − λ)a4 = (1 − λ2 )(c3 + μc1 c2 + νc31 ), where μ=2

1 − λ3 , 1 − λ2

and

ν=

1 − λ4 . 1 − λ2

From the first equation in (12.3.6), and the fact that |c1 | ⩽ 1, we obtain (1 − λ)|a2 | = (1 − λ2 )|c1 | ⩽ 1 − λ2 , which gives another proof of the inequality |a2 | ⩽ 1 + λ.

(12.3.6)

192 | 12 The Class U(λ) Using the second equation in (12.3.6), |c1 | ⩽ 1, and the inequality |c2 | ⩽ 1 − |c1 |2 , we obtain (1 − λ)|a3 | ⩽ (1 − λ2 )|c2 | + (1 − λ3 )|c1 |2 ⩽ (1 − λ2 )(1 − |c1 |2 ) + (1 − λ3 )|c1 |2 = 1 − λ2 + (λ2 − λ3 )|c1 |2 ⩽ 1 − λ3 , which implies |a3 | ⩽ 1 + λ + λ2 . Finally, we prove (12.3.4) when n = 4. To do this, we use the sharp upper bounds for the functional |c3 + μc1 c2 + νc31 | when μ and ν are real, as proved by Prokhorov and Szynal [203]. It was shown in [203] that if 2 ⩽ |μ| ⩽ 4 and ν ⩾ (μ2 + 8)/12, then |c3 + μc1 c2 + νc31 | ⩽ |ν|, but the third equation in (12.3.6) ensures that these conditions are satisfied, and so (1 − λ)|a4 | = (1 − λ2 )|c3 + μc1 c2 + νc31 | ⩽ (1 − λ2 ) (

1 − λ4 ) = 1 − λ4 , 1 − λ2

which proves the inequality |a4 | ⩽ 1 + λ + λ2 + λ3 . In the same paper [177], Obradović et al. obtained the following sharp result concerning the logarithmic coefficients of functions in U(λ), together with other interesting properties concerning these and related functions. Theorem 12.3.3. For 0 < λ ⩽ 1, the logarithmic coefficients γ n (f) of f ∈ U(λ) satisfy the inequality ∞ 1 π2 ∑ |γ n (f)|2 ⩽ [ (12.3.7) + 2Li2 (λ) + Li2 (λ2 )] , 4 6 n=1 where Li2 denotes the dilogarithm function given by ∞

1

zn log(1/t) Li2 (z) = ∑ 2 = z ∫ dt. 1 − tz n n=1 0

The inequality (12.3.7) is sharp. Further, there exists a function in U such that |γ n (f)| > (1 + λ n )/(2n), for some n. Proof. Let f ∈ U(λ). Then from (12.1.2), we have z ≺ (1 − z)(1 − λz), f(z) which clearly gives ∞

∑ 2γ n (f)z n = log n=1

∞ 1 f(z) ≺ − log(1 − z) − log(1 − λz) = ∑ (1 + λ n )z n . z n n=1

12.3 Coefficients | 193

By Rogosinski’s theorem (Theorem 3.2.3), we obtain ∞

∞

∞ ∞ 2n 1 λn λ 1 ∞ 1 n 2 ∑ ∑ ∑ (1 + λ ) = + 2 + ), ( 2 2 2 4 4n n n n2 n=1 n=1 n=1 n=1

∑ |γ n (f)|2 ⩽ ∑ n=1

from which (12.3.7) follows. For the function g λ given by g λ (z) =

z , (1 − z)(1 − λz)

we find that γ n (g λ ) = (1 + λ n )/(2n) for n ⩾ 1, and so we have equality in (12.3.7). We now construct a function f ∈ U such that |γ n (f)| > (1 + λ n )/(2n), for some n. Let the function f λ be defined for z ∈ 𝔻 by z f λ (z) = . (1 − z)(1 − λz)(1 + (λ/(1 + λ))z) Then λ − (1 + λ)2 λ2 3 z =1+ z+ z , f λ (z) 1+λ 1+λ and z 2 󸀠 2λ2 3 (1 + 2λ)(1 − λ) 3 [ ] f λ (z) − 1 = − z = − [1 − ]z , f λ (z) 1+λ 1+λ which shows that f λ ∈ U(λ). Moreover, for this function, we have log

∞ f λ (z) λ = − log(1 − z) − log(1 − λz) − log (1 + z) = 2 ∑ γ n (f λ )z n , z 1+λ n=1

where

λn 1 1 + λn ]. [ + (−1)n 2 n n(1 + λ)n This contradicts the inequality |γ n (f)| ⩽ (1 + λ n )/(2n), at least for even integer values of n ⩾ 2. Also for n ⩾ 1, we have γ n (f λ ) =

∞

∑ |γ n (f λ )|2 n=1 n 2n 1 ∞ (1 + λ n )2 (−1)n λ2 λ 1 λ ∑[ + 2 (( + ( ) + ( ], ) ) ) 4 n=1 1+λ 1+λ n2 n2 n2 1 + λ n

=

from which it follows that for 0 < λ ⩽ 1, ∞

∑ |γ n (f λ )|2 = n=1

1 π2 ( + 2Li2 (λ) + Li2 (λ2 )) 4 6 +

−λ2 −λ λ2 1 1 ) [Li2 ( ) + Li2 ( )] + Li2 ( 2 1+λ 1+λ 4 (1 + λ)2

=

1 1 π2 + 2Li2 (λ) + Li2 (λ2 )) + A(λ) ( 4 6 4


0 for t ∈ [0, 1), so that A(1) < 0. On the other hand, since N(λ, t) is a decreasing function of t for t ∈ [0, 1], it follows that N(λ, t) ⩾ N(λ, 1) = (1 + λ)3 − (3 − λ)(1 + λ) − 4λ3 = 1 − λ3 + λ2 (1 − λ) > 0, for 0 < λ < 1. Consequently, B(λ, t) > 0 for t ∈ [0, 1], and 0 < λ < 1. This shows that A(λ) < 0 for 0 < λ ⩽ 1. Thus the proof is complete. We end by noting that when λ = 1, we obtain the following. Corollary 12.3.1. The logarithmic coefficients γ n (f) of f ∈ U satisfy ∞

∞

n=1

n=1

∑ |γ n (f)|2 ⩽ ∑

π2 1 = . 2 6 n

The inequality is sharp, with equality for the Koebe function. Also, there exists a function f ∈ U such that |γ n (f)| > 1/n for some n.

13 Convolutions The theory of univalent functions and geometric function theory covers a wide variety of concepts. One such concept is the idea of the convolution of analytic functions, which is the subject of this chapter. We will present what we consider to be the most important convolution results relating to univalent functions, and refer the reader to Ruscheweyh’s excellent monograph [221] for more information.

13.1 Definition and the Pólya–Schoenberg Conjecture The classical result of Hadamard says that if the functions f and g are analytic in the ∞ n n unit disk 𝔻, with expansions f(z) = ∑∞ n=1 a n z and g(z) = ∑ n=1 b n z , respectively, for z ∈ 𝔻, then ∞

∑ an bn zn = n=1

1 dζ ∫ f(z/ζ)g(ζ) , 2πi ζ

(|z| < r < 1).

|ζ|=r

This gives rise to the following definition. n Definition 13.1.1. Let f and g be analytic in 𝔻 and be given by f(z) = ∑∞ n=1 a n z and ∞ n g(z) = ∑n=1 b n z , respectively, for z ∈ 𝔻. Then, their convolution or Hadamard product, denoted by f ∗ g, is defined for z ∈ 𝔻 by ∞

(f ∗ g)(z) = ∑ a n b n z n . n=1

It is obvious that convolution has the algebraic properties of ordinary multiplication, n with identity element s(z) = z/(1 − z) = ∑∞ n=1 z , so that s ∗ f = f ∗ s = f . Perhaps the most important problem involving the convolution of analytic functions is the idea of convolution invariance, i.e., finding classes of functions V such that the convolution of any two functions in V, belongs to V, i.e., f, g ∈ V

⇒

f ∗ g ∈ V.

(13.1.1)

In 1958, Pólya and Schoenberg [189] conjectured that if f, g ∈ C, where C denotes the class of convex functions, then f ∗ g ∈ C. A significant contribution to solving the problem was made by Suffridge [239] in 1966, who showed that the convolution of any two convex functions is close-to-convex, and hence univalent. The Pólya–Schoenberg conjecture was finally proved in 1973 by Ruscheweyh and Sheil-Small, [218], who also showed in the same paper that the convolution of a convex and a close-to-convex function is close-to-convex, i.e., f ∈ C, g ∈ K https://doi.org/10.1515/9783110560961-013

⇒

f ∗ g ∈ K.

(13.1.2)

196 | 13 Convolutions

In [221], Ruscheveyh gave other results similar to (13.1.1) and (13.1.2) for a variety of classes V using an approach similar to that used in [218], but with different proofs, based on the “duality principle.” In the following, we present the results of Ruscheweyh and Sheil-Small [218], the first of which is a proof of the Pólya–Schoenberg conjecture. Theorem 13.1.1. (i) If f ∈ C and g ∈ C, then f ∗ g ∈ C. (ii) If f ∈ C and g ∈ K, then f ∗ g ∈ K. (iii) If f ∈ C and g ∈ S∗ , then f ∗ g ∈ S∗ . Following the methods used in [218], we first prove that (i) and (iii) are equivalent, and after extensive preparation, present the proof of (ii) and (iii). Proof that Theorem 13.1.1 (i) and Theorem 13.1.1 (iii) are equivalent. Suppose that (i) is true. Let f ∈ C and g ∈ S∗ . Then, h defined by z

h(z) = ∫ 0

∞ g(t) bn n dt = ∑ z t n n=1

belongs to C, and so f ∗ h is also in C. Therefore, ∞

∞

n=1

n=1

(f ∗ g)(z) = ∑ a n b n z n = z ( ∑

󸀠

an bn n z ) = z(f ∗ h)󸀠 (z) n

belongs to S∗ . Similarly assume that (iii) is true, and that f, g ∈ C. Then, zg󸀠 ∈ S∗ , and ∞

f(z) ∗ (zg󸀠 (z)) = ∑ na n b n z n = z(f ∗ g)󸀠 (z) n=1

belongs to S∗ , i.e., f ∗ g ∈ C. Now follows the preparatory work needed for the proofs of (ii) and (iii). We begin with the following lemma, which is a modification of a lemma in Ruscheweyh [219]. Lemma 13.1.1. Let φ and g be analytic in 𝔻, with φ(0) = g(0) = 0, φ󸀠 (0) ≠ 0 and g󸀠 (0) ≠ 0. If for each σ and α on the unit circle φ(z) ∗

1 + ασz g(z) ≠ 0 1 − σz

(0 < |z| < 1),

(13.1.3)

then for each function F, analytic and with positive real part on 𝔻, we have Re

(φ ∗ Fg)(z) >0 (φ ∗ g)(z)

(z ∈ 𝔻).

(13.1.4)

13.1 Definition and the Pólya–Schoenberg Conjecture |

197

Proof. For α = −1 in (13.1.3), we have (φ ∗ g)(z) ≠ 0 when 0 < |z| < 1, so we consider the quotient 1 + σz g(z) φ(z) ∗ 1 − σz , G(z) ≡ φ(z) ∗ g(z) and show that this has positive real part on 𝔻. Indeed, if |α| = 1 and α ≠ −1, then φ(z) ∗

1 + ασz 1 1 + σz 1 g(z) = (1 + α)φ(z) ∗ g(z) + (1 − α)φ(z) ∗ g(z), 1 − σz 2 1 − σz 2

and from (13.1.3), G(z) ≠ −(1 − α)/(1 + α) for z ∈ 𝔻. Thus, G(z) takes no value on the imaginary axis, which together with G(0) = 1, implies Re G(z) > 0 for z ∈ 𝔻. Now, without loss in generality, we can assume that |F(0)| = 1 and apply the Herglotz representation formula (Theorem 3.1.10) to obtain e−iγ F(z) = ∫ 𝕋

1 + βσz dμ(σ), 1 − σz

where 𝕋 is the unit circle, μ is a probability measure, β and γ are unique constants such that |β| = 1, β ≠ −1, cos γ > 0, and e iγ (1 + βz)/(1 − z) has positive real part on 𝔻. Then, e−iγ (φ ∗ Fg)(z) = ∫ φ(z) ∗ g(z) 𝕋

=

1 + βσz dμ(σ) 1 − σz

1 + σz 1 1 (1 + β) ∫ φ(z) ∗ g(z) dμ(σ) + (1 − β)(φ ∗ g)(z) 2 1 − σz 2 𝕋

1 1 = [φ(z) ∗ g(z)] [ (1 + β) ∫ H σ (z)dμ(σ) + (1 − β)] , 2 2 𝕋 ] [ where H σ (0) = 1. Since Re G(z) > 0, for z ∈ 𝔻, it follows that Re H σ (z) > 0, for z ∈ 𝔻, i.e., (φ ∗ Fg)(z) 1 1 e−iγ = (1 + β)K(z) + (1 − β), (φ ∗ g)(z) 2 2 where K(0) = 1 and Re K(z) > 0, for z ∈ 𝔻, and so (13.1.4) now follows. We next give some properties of convex functions, which first appeared in [218] (see above), but follow the style in Duren [45, Section 8.3]. Lemma 13.1.2. If f ∈ C, then Re [

ζ 1 z ζ − w f(z) − f(w) − ]> z − ζ z − w f(ζ) − f(w) z − ζ 2

(z, ζ, w ∈ 𝔻).

Proof. Consider the function F(z, ζ, w) =

2ζ 2z ζ − w f(z) − f(w) ⋅ ⋅ − − 1, z − ζ z − w f(ζ) − f(w) z − ζ

198 | 13 Convolutions which is holomorphic in the polydisk 𝔻3 = {(z, ζ, w) : z, ζ, w ∈ 𝔻}. Further, if a, b ∈ (0, 2π), α = e ia and β = e ib , then Re F(αw, βw, w) = −

sin(b/2) f(αw) − f(w) ⋅ Im [ ]. sin((a − b)/2) ⋅ sin(a/2) f(βw) − f(w)

Since f is convex, so is f(𝔻ρ ), where 𝔻ρ = {w : |w| ⩽ ρ < 1}, and so arg [

{ { (0, π), f(αw) − f(w) ]∈{ { f(βw) − f(w) (−π, 0), {

thus sin

a−b 2

and

Im

0 < a < b < 2π, 0 < b < a < 2π,

f(αw) − f(w) f(βw) − f(w)

have opposite signs. Therefore, since the real part of F(z, ζ, w) attains its minimum on the boundary 𝕋3 = {(z, ζ, w) : |z| = |ζ| = |w| = 1} of the polydisk 𝔻3 , it follows that Re F(z, ζ, w) > 0, for (z, ζ, w) ∈ 𝔻3 . Lemma 13.1.2 leads to three useful corollaries. Choosing ζ = 0 and replacing w with ζ , we obtain the following. Corollary 13.1.1. If f ∈ C, then f(z) − f(ζ) z 1 Re [ ⋅ ]> , (z, ζ ∈ 𝔻). z−ζ f(z) 2 Next, note that since 1 1 f(z) ∗ s(αz) = f(αz), α α where s(z) = z/(1 − z), and 1 1 f(αz) − f(βz) 1 ( − )= , f(z) ∗ α − β 1 − αz 1 − βz α−β Corollary 13.1.1 can be rewritten in the following equivalent form. Corollary 13.1.2. If f ∈ C, then Re [

f(z) ∗ [z(1 − αz)−1 (1 − βz)−1 ] 1 ]> 2 f(z) ∗ [z(1 − αz)−1 ]

(|α| ⩽ 1, |β| ⩽ 1).

If in Lemma 13.1.2 we replace z, ζ , and w with αz, βz, and γz, respectively, and reduce the resulting expression to β α f(z) ∗ [z(1 − αz)−1 (1 − γz)−1 ] − ⋅ α − β f(z) ∗ [z(1 − βz)−1 (1 − γz)−1 ] α − β =

1 f(z) ∗ [αz(1 − αz)−1 (1 − γz)−1 − βz(1 − βz)−1 (1 − γz)−1 ] ⋅ α−β f(z) ∗ [z(1 − βz)−1 (1 − γz)−1 ]

=

f(z) ∗ [z(1 − αz)−1 (1 − βz)−1 (1 − γz)−1 ] ; f(z) ∗ [z(1 − βz)−1 (1 − γz)−1 ]

we can rewrite the lemma in the following equivalent form.

13.1 Definition and the Pólya–Schoenberg Conjecture

| 199

Corollary 13.1.3. If f ∈ C, then Re [

1 f(z) ∗ [z(1 − αz)−1 (1 − βz)−1 (1 − γz)−1 ] ]> −1 −1 2 f(z) ∗ [z(1 − αz) (1 − βz) ]

(|α| ⩽ 1, |β| ⩽ 1, |γ| ⩽ 1).

The properties of convex functions established above will be used in the proof of the next result, which is slightly different to the form appearing in Ruscheweyh [219]. Lemma 13.1.3. Let h be analytic in the unit disk 𝔻, with h(0) = 0. If there exist constants α and β on the unit circle such that Re [(1 − αz)(1 − βz)

h(z) ]>0 z

(z ∈ 𝔻),

then for every convex function φ, φ(z) ∗ h(z) ≠ 0

(0 < |z| < 1).

Proof. Again using the Herglotz representation formula (Theorem 3.1.10), we can see that there exists γ on the unit circle 𝕋 such that h(z) = h󸀠 (0) ∫ 𝕋

z(1 + γσz) dμ(σ) (1 − αz)(1 − βz)(1 − σz)

(z ∈ 𝔻),

where μ is a probability measure on 𝕋. Therefore, 1 z(1 + γσz) (φ(z) ∗ h(z)) = ∫ φ(z) ∗ dμ(σ) h󸀠 (0) (1 − αz)(1 − βz)(1 − σz) 𝕋

= ∫ φ(z) ∗ [(1 + γ)z(1 − σz)−1 (1 − αz)−1 (1 − βz)−1 𝕋

− γz(1 − αz)−1 (1 − βz)−1 ] dμ(σ) = ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ [φ(z) ∗ z(1 − αz)−1 (1 − βz)−1 ] I1

φ(z) ∗ z(1 − σz)−1 (1 − αz)−1 (1 − βz)−1 ⋅ [(1 + γ) ∫ dμ(σ) − γ] . φ(z) ∗ z(1 − αz)−1 (1 − βz)−1 𝕋 [ ] ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ I2

Applying Corollary 13.1.2 to I1 and Corollary 13.1.3 to I2 , the lemma follows from the fact that I1 and I2 do not vanish on the punctured disk 𝔻 \ {0}. We now give a final lemma needed for the proof of a theorem, which will immediately imply parts (ii) and (iii) of Theorem 13.1.1. Lemma 13.1.4. Let g ∈ S∗ and α be on the unit circle. Then there exist β and γ on the unit circle such that Re [γ(1 − αz)(1 − βz)

g(z) ]>0 z

(z ∈ 𝔻).

200 | 13 Convolutions Proof. A compactness argument shows that since g ∈ S∗ , we can assume without loss in generality that it is continuous on the closed unit disk. Also, since g ∈ S∗ , g(0) = 0, and arg g(re iθ ) is an increasing function of θ. Thus, after a suitable rotation, it follows that Re g(e it ) > 0 when a < t < b and Re g(e it ) < 0 when b < t < a + 2π. Setting u(e it ) = Re g(e it ), ω = e it , α = e ia , and β = e ib , we obtain 2π

2π

0

0

ω+z α+z 1 ω+z 1 ∫ ∫( u(ω)dt = − ) u(ω)dt, g(z) = 2π ω−z 2π ω−z α−z i.e., 2π

2π

0

0

g(z) 1 α−ω 1 π(α − z) =∫ u(ω)dt = ∫ (α − ω) ( − ) u(ω)dt. z ω−z ω−z β−z So, with h(z) = π(α − z)(β − z)g(z)/z, we have 2π

h(z) = ∫ 0

and

(α − ω)(β − ω) u(ω)dt, ω−z

a+2π

e−(a+b)i/2 h(z) = 4 ∫ sin ( a

t−a b−t u(e it ) dt. ) sin ( ) 2 2 1 − ze−it

The lemma now follows since 1/(1 − ζ) has positive real part on the unit disk, and so also the integrand in the above expression, when a < t < b and b < t < a + 2π. All the preparatory work is now complete in order to prove the following theorem, which leads to a proof of the Pólya–Schoenberg conjecture. Theorem 13.1.2. Let φ be convex, and g be starlike. If F is analytic with positive real part for z ∈ 𝔻, then (φ ∗ Fg)(z) Re [ ]>0 (z ∈ 𝔻). (φ ∗ g)(z) Proof. In view of Lemma 13.1.1, it is sufficient to show that for every α and σ on the unit circle, 1 + ασz g(z) ≠ 0 (z ∈ 𝔻). φ(z) ∗ 1 − σz By Lemma 13.1.3, this follows from the existence of constants a and b on the unit circle such that 1 + ασz 1 ]>0 (z ∈ 𝔻). Re [(1 − az)(1 − bz) 1 − σz z (The existence of the above constants follows from Lemma 13.1.4.)

13.2 Subordination and Convolution

| 201

We end this section with a simple proof that the previous theorem implies parts (ii) and (iii) of Theorem 13.1.1, thus proving the Pólya–Schoenberg conjecture. Proof of (ii) and (iii) in Theorem 13.1.1. Let f be a close-to-convex function. Then there exists a starlike function g, such that zf 󸀠 (z) = g(z)F(z), where F has a positive real part. Thus, Theorem 13.1.2 gives Re [

(φ ∗ Fg)(z) z(φ ∗ f)󸀠 (z) ] = Re [ ]>0 (φ ∗ g)(z) (φ ∗ g)(z)

(z ∈ 𝔻),

where φ is convex. Hence, in order to prove (ii), it is enough to show that φ ∗ g is starlike. This follows from Theorem 13.1.2, using the starlikeness of g, taking F(z) = zg󸀠 (z)/g(z), and the fact that φ(z) ∗ [F(z)g(z)] = φ(z) ∗ [zg󸀠 (z)] = z(φ ∗ g)󸀠 (z). This proves (ii) and by showing that φ ∗ g is starlike, we have also proved (iii).

13.2 Subordination and Convolution We consider next the connection between subordination and convolution. Suppose that f, g ∈ C and h is an analytic function defined on 𝔻. Wilf [262] proposed that if h ≺ f , then g ∗ h ≺ g ∗ f , and this conjecture was proved by Ruscheweyh and SheilSmall in 1973 [218]. To prove the conjecture, we will need the following useful result concerning subordination. Theorem 13.2.1 (Ruscheweyh and Sheil-Small [218]). Let f ∈ C. Then, g ≺ f if, and only if, zf 󸀠 (z) Re [ ]>0 (z ∈ 𝔻, |x| < |z|). (13.2.1) f(z) − g(x) Proof. Let f ∈ C and g ≺ f . Then, f(𝔻) is a convex domain and, so, starlike with respect to each of its points (including g(x)), which is analytically expressed by (13.2.1). Next let f ∈ C and (13.2.1) hold. Also, assume that g is not subordinate to f . Then there exists x0 ∈ 𝔻 such that g(x0 ) ∉ f(Er ), where Er = {z : |z| < r} for certain r with |x0 | < r < 1. Since the function h(z) =

zf 󸀠 (z) f(z) − g(x0 )

is analytic in Er with positive real part on |z| = r, i.e., on |z| ⩽ r, this contradicts the fact that h(0) = 0, and so the assumption g ⊀ f is false. We now apply Theorem 13.2.1 to prove Wilf’s conjecture.

202 | 13 Convolutions Theorem 13.2.2. If f, g ∈ C and h ≺ f , then g ∗ h ≺ g ∗ f . Proof. Set

(g ∗ f)(z) − (g ∗ h)(x) . z(g ∗ f)󸀠 (z) Theorem 13.2.1 means that it is enough to show that Re ψ(z) > 0, for z ∈ 𝔻 and |x| < |z|. The same theorem also shows that for zf 󸀠 (z) , φ(z) = f(z) − h(x) ψ(z) =

Re φ(z) > 0, for z ∈ 𝔻 and |x| < |z|. Finally, Theorem 13.1.2, together with zf 󸀠 (z) ] φ(z) , ψ(z) = g(z) ∗ [zf 󸀠 (z)] g(z) ∗ [

implies Re ψ(z) > 0. Ruscheweyh and Stankiewicz [222] generalized Wilf’s conjecture by showing that for any two convex univalent functions f and g, and any two functions h1 and h2 analytic in the unit disk, h1 ≺ f and h2 ≺ g ⇒ h1 ∗ h2 ≺ f ∗ g. Shanmugam [225] used subordination and convolution to obtain generalizations of Wilf’s conjecture to starlikeness, convexity, and close-to-convexity. The motivation comes from the fact that f ∈ C, if, and only if, zf 󸀠 (z) = f(z) ∗ z/(1 − z)2 ∈ S∗ , as well as f(z) = f(z) ∗ z/(1 − z). For a fixed function g ∈ A, and an analytic function h, with h(0) = 1, the following classes were considered [225]. ∙ S∗g (h) ⊂ A, the class of functions f such that ((g ∗ f)(z))/z ≠ 0 for z ∈ 𝔻, and z(g ∗ f)󸀠 (z) ≺ h(z). (g ∗ f)(z) ∙

Cg (h) ⊂ A, the class of functions f such that (g ∗ f)󸀠 (z) ≠ 0 for z ∈ 𝔻, and 1+

∙

z(g ∗ f)󸀠󸀠 (z) ≺ h(z). (g ∗ f)󸀠 (z)

Kg (h) ⊂ A, the class of functions f such that ((g ∗ f)(z))/z ≠ 0 for z ∈ 𝔻, and z(g ∗ f)󸀠 (z) ≺ h(z), (g ∗ ψ)(z) for some ψ ∈ S∗g (h).

Clearly, choosing g(z) = z/(1 − z) and h(z) = (1 + z)/(1 − z) leads to the classes S∗ , C, and K, respectively. The case when g(z) = z/(1 − z)a was studied by Padmanabhan and Parvatham [185], and Padmanabhan and Manjini [184]. The case with the additional condition h(z) = (1 + z)/(1 − z) was previously studied by Owa et al. [179]. The generalized classes defined above [225] mirror the same inclusion properties found in classes of univalent functions, i.e.,

13.2 Subordination and Convolution |

∙ ∙ ∙

203

Cg (h) ⊆ S∗g (h) f ∈ S∗g (h) if, and only if, zf 󸀠 (z) ∈ Cg (h) Cg (h) ⊆ Kg (h)

As well as the convolution invariance properties ∙ f ∈ S∗g (h) and ψ ∈ C ⇒ ψ ∗ f ∈ S∗g (h) ∙ f, ψ ∈ Cg (h) ⇒ ψ ∗ f ∈ Cg (h) ∙ f ∈ Kg (h) and ψ ∈ C ⇒ ψ ∗ f ∈ Kg (h) Another application of convolution in the theory of univalent functions arises from a result of Silverman et al. [228], which says that f is strongly starlike of order α, 0 ⩽ α < 1, if, and only if, x + 2α − 1 z 1+ f(z) 2(1 − α) ≠ 0 (z ∈ 𝔻, |x| = 1). (13.2.2) ∗ z (1 − z)2 An application of this result gives interesting information about the δ-neighborhood of convex functions defined as follows. ∞ n n Definition 13.2.1. For f, g ∈ A given by f(z) = z + ∑∞ n=2 a n z and g(z) = z + ∑ n=2 b n z , respectively, we say that the function g is in the δ-neighborhood N δ (f) of f if ∞

∑ n|a n − b n | < δ. n=2

Theorem 13.2.3. If f ∈ C, then N1/4 (f) ⊂ S∗ . Proof. Consider the convex function h x defined for z ∈ 𝔻 by x−1 ∞ 1+ z 2 = z + ∑ c n (x)z n , h x (z) = z (1 − z)2 n=2 such that |c n (x)| ⩽ n

(n ⩾ 2, |x| = 1),

and f ∗ h x ∈ C ⊂ S. Then by the Koebe 1/4 theorem, 󵄨󵄨 (h ∗ f)(z) 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 x (z ∈ 𝔻). 󵄨󵄨 > 󵄨󵄨 z 󵄨󵄨 4 󵄨󵄨 Further, for g ∈ N1/4 (f), using (13.2.3) and (13.2.4), we have 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 1 󵄨󵄨 󵄨󵄨 1 󵄨 󵄨 󵄨 󵄨 󵄨 󵄨󵄨 󵄨󵄨 (g ∗ h x )(z)󵄨󵄨󵄨 ⩾ 󵄨󵄨󵄨 (f ∗ h x )(z)󵄨󵄨󵄨 − 󵄨󵄨󵄨 [(g − f) ∗ h x ](z)󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 z 󵄨󵄨 󵄨󵄨 z 󵄨󵄨 󵄨󵄨 z ∞ 1 > − ∑ |c n (x)| ⋅ |b n − a n | ⋅ |z|n−1 4 n=2 >

1 ∞ − ∑ n|b n − a n |, 4 n=2

and so applying (13.2.2), it follows that g ∈ S∗ .

(13.2.3)

(13.2.4)

204 | 13 Convolutions We end this section with a convolution result for the class U due to Barnard [12, Theorem 3.9]. Theorem 13.2.4. Let f, g ∈ S. If F(z) =

z z ∗ ≠ 0 f(z) g(z)

(z ∈ 𝔻),

then z/F(z) ∈ U, and F is univalent in 𝔻. Proof. Let

∞ z = 1 + ∑ bn zn f(z) n=1

and

∞ z = 1 + ∑ cn zn . g(z) n=1

Then, from an equivalent form of the area theorem (see Goodman [59, Theorem 11 on p. 193]), we have ∞

∑ (n − 1)|b n |2 ⩽ 1,

∞

and

n=2

∑ (n − 1)|c n |2 ⩽ 1. n=2

Further, using the Cauchy–Schwarz inequality, we obtain ∞

∞

n=2

n=2

∑ (n − 1)|b n c n | ⩽ [ ∑ (n − 1)|b n |2 ]

1/2

∞

⋅ [ ∑ (n − 1)|c n |2 ]

1/2

⩽ 1.

n=2

The last inequality, together with Theorem 12.2.2 and the analyticity of z/F(z) on 𝔻, shows that z/F(z) ∈ U.

14 Meromorphic Univalent Functions The main aim of this chapter is to summarize some of the significant results for the class Σ of meromorphic univalent functions introduced in Chapter 1. We will then present a selection of some of the important properties of the main subclasses of Σ to include starlike, close-to-convex, and Bazilevič meromorphic functions.

14.1 The Class Σ We recall from Chapter 1, that the class Σ of functions g, which are analytic and univalent in ∆ = {z : |z| > 1}, with a simple pole at infinity and given by ∞

g(z) = z + b 0 + ∑ b n z−n ,

(14.1.1)

n=1

played an important role in determining fundamental results for the class S. We also note that if f ∈ S for z ∈ 𝔻, then g given by g(z) = f(z−1 )−1 belongs to Σ. Conversely if g ∈ Σ and f is given by f(z) = (g(z−1 ) − c)−1 , for c ∈ ℂ \ g(∆), then f ∈ S. Thus, there is a close relationship between Σ and S.

14.1.1 Coefficients and the Clunie Constant In Chapter 1, The Area Theorem 1.2.1 was used to obtain fundamental results concern2 ing the class S, and we note that the convergence of ∑∞ n=1 n|b n | implies that |b n | ⩽ −1/2 −1/2 n and b n = O(1)n as n → ∞. Although the Bieberbach conjecture is now solved for functions in S, relatively little exact information is known about the coefficients b n of functions in Σ. It is clear that |b 1 | ⩽ 1, with equality when g(z) = [k(z−2 )]−1/2 = z − z−1 , where k is the Koebe function. Deeper methods are needed to establish the inequality |b 2 | ⩽ 2/3 (Goluzin [53], Schiffer [224]), which is sharp when g(z) = [k(z−3 )]−1/3 = z − (2z−2 )/3. This suggests that the inequality |b n | ⩽ 2/(n + 1) for n ⩾ 1 could be valid, with extreme function given by g(z) = [k(z−n−1 )]−1/(n+1) = z −

2 −n z +.... n+1

This, however, is false, since a relatively simple argument using the Fekete–Szegő theorem for functions in S, establishes the sharp inequality |b 3 | ⩽ 1/2 + e−6 (see Theorem 14.1.2 below). It was noted by Goodman [59] that this result for b 3 was generalized by Bazilevič [14], who showed that for k-symmetric functions in Σ, given by ∞

c nk−1 , nk−1 n=1 z

f(z) = z + ∑ https://doi.org/10.1515/9783110560961-014

206 | 14 Meromorphic Univalent Functions then for k ⩾ 2, |c nk−1 | ⩽

1 2 + 2(k+1)/(k−1) , k ke

and that this inequality is sharp for each k ⩾ 2. Even though the above conjecture is false in general, an early result of Goluzin [53] showed that the following weaker theorem is true. We also note that Duren [44], using the Grunsky inequalities, gave another proof of this (see Duren [45]). Theorem 14.1.1. Let g ∈ Σ and be given by (14.1.1) and suppose that b 1 = b 2 = . . . = b m−1 = 0 for some m ⩾ 1. Then, 2 |b n | ⩽ , n+1 for n = m, m + 1, . . . , 2m. On the other hand, even the correct rate of growth for b n remains an open question. Surprisingly, the growth estimate |b n | ⩽ n−1/2 , proved in 1916, remained all that was known until 1966, when Clunie, using a clever major–minor arc argument, showed 1/4 that b n = O(n−1/2 ) could be improved to b n = O(n−1/2−log n ). A developed version of the same argument appeared in 1967, where Clunie and Pommerenke [34] showed that b n = O(n−1/2−1/300 ). Earlier, Clunie [31] constructed a function in Σ where b n ≠ O(n−0.98 ), and later Pommerenke [197] improved this to b n ≠ O(n−0.83 ). Other slight improvements to these estimates have since been given, but finding the correct constant γ such that b n = O(n−γ ), known as the Clunie constant, remains one of the most significant unsolved problems in the theory of univalent functions, and probably compares in difficulty to that of solving the Bieberbach conjecture. We give Clunie and Pommerenke’s proof that b n = O(n−1/2−1/300 ) below, which illustrates the use of the major–minor arc technique, a commonly used tool in complex analysis. We begin with the case n = 3. Theorem 14.1.2. Let g ∈ Σ and be given by (14.1.1), then |b 3 | ⩽ 1/2 + e−6 . The equality is sharp. Proof. Write g(z) = z[z2 f(z−2 )]−1/2 = z + b 1 z−1 + b 3 z−3 + . . . , so that f ∈ S. It follows that b 3 = −(a3 − 3a22 /4)/2, and the Fekete–Szegő theorem with μ = 3/4 for functions in S (see Chapter 1, Theorem 1.2.4) now gives the result. Equality follows from the sharpness of the Fekete–Szegő theorem. We next prove the result of Clunie and Pommerenke. Theorem 14.1.3. Let g ∈ Σ and be given by (14.1.1), then |b n | ⩽ A n−1/2−1/300 for n ⩾ 1, where A is an absolute constant.

14.1 The Class Σ

| 207

Proof. For convenience, write z = r−1 e iθ with r = 1/ρ, so that 0 < r < 1. Let 0 < δ < 1/4, and A1 , A2 , . . . be absolute constants. Then the Cauchy–Schwarz inequality gives 2π

1/2

2π 󸀠

∫ |g (z)|

1+δ

󸀠

dθ ⩽ ( ∫ |g (z)| dθ)

0

1/2

2π 󸀠

( ∫ |g (z)| dθ)

2

0

2δ

.

(14.1.2)

0

We first note that the area theorem gives 2π

∞ A1 1 ∫ |g󸀠 (z)|2 dθ = 1 + ∑ k 2 |b k |2 r2k+2 ⩽ . 2π 1 −r k=1

(14.1.3)

0

To estimate the last integral in (14.1.2), write ∞

[g󸀠 (z)]δ = ∑ c k z−k , k=0

where c0 = 1, and |z| > 1. Then, 2π

ψ(r) :=

∞ 1 ∫ |g󸀠 (z)|2δ dθ = ∑ |c k |2 r2k , 2π k=0

(14.1.4)

0

where 0 < r < 1. Since

ψ󸀠 (r)

is positive, differentiation gives ∞

rψ󸀠󸀠 (r) ⩽ rψ󸀠󸀠 (r) + ψ󸀠 (r) = 4 ∑ k 2 |c k |2 r2k−1 . k=1

From a result of Goluzin [57, p. 139, expresson (29)], it follows that if g ∈ Σ, then for some absolute constant r0 , with 1 < |z| ⩽ 1/r0 , 󵄨󵄨 󸀠󸀠 󵄨󵄨 3 󵄨󵄨 zg (z) 󵄨󵄨 8|z|2 − 2 󵄨󵄨 󸀠 󵄨 (14.1.5) 󵄨󵄨 g (z) 󵄨󵄨󵄨 ⩽ |z|2 − 1 ⩽ |z| − 1 + 8. 󵄨 󵄨 Then, using (14.1.5), we have ∞

2

2 2k+2

∑ k |c k | r k=1

2π 2π 󵄨 󵄨2 󵄨󵄨 d 󵄨2 1 δ2 󵄨󵄨󵄨 g󸀠󸀠 (z) 󵄨󵄨󵄨 󸀠 2δ 󵄨󵄨 󸀠 δ 󵄨󵄨󵄨 = ∫ 󵄨󵄨 [g (z)] 󵄨󵄨 dθ = ∫ 󵄨󵄨󵄨 󸀠 󵄨󵄨󵄨 |g (z)| dθ 󵄨󵄨 2π 󵄨󵄨 dz 2π 󵄨󵄨 g (z) 󵄨󵄨 0 0 2π

2 3r r2 δ 2 ( + 8) ∫ |g󸀠 (z)|2δ dθ. 2π 1 − r

⩽

0

Thus, we obtain r2 ψ󸀠󸀠 (r) ⩽ 4δ2 (

2 3r + 8) ψ(r). 1−r

Write 2π

2π

0

0

1 1 ∫ |g󸀠 (z)|2δ dθ = ( ∫ |g󸀠 (z)‖g󸀠 (z)|2δ−1 dθ) . ψ(r) = 2π 2π

(14.1.6)

208 | 14 Meromorphic Univalent Functions Noting that 0 < δ ⩽ 1/2, we now apply the Cauchy–Schwarz inequality to (14.1.4), use (14.1.3), and the second inequality in Theorem 14.1.5 below to obtain ψ(r) ⩽ A2 (1 − r)−1/2 . Hence, (14.1.6) gives ψ󸀠󸀠 (r) ⩽

36δ2 A3 ψ(r) + , (1 − r)2 (1 − r)3/2

for r0 ⩽ r < 1. Integration by parts now gives r 󸀠

󸀠

ψ (r) ⩽ ψ (r0 ) + ∫ r0

36δ2 A4 ψ(t) dt + (1 − t)2 (1 − r)1/2 r

⩽

36δ2 󸀠 A4 36δ2 + ψ(r) − ∫ ψ (t) dt. 1/2 1−r 1−t (1 − r) r0

Since the last term is negative, and ψ(r) ⩾ |c0 ψ󸀠 (r) ψ(r)

⩽

|2

= 1, it follows that

36δ2 A4 + , (1 − r)1/2 1 − r

and so for 0 ⩽ r < 1, integration gives ψ(r) ⩽ A5 (1 − r)−36δ . 2

(14.1.7)

Thus, combining the estimates in (14.1.3) and (14.1.7) into (14.1.2) we have 2π

∫ |g󸀠 (z)|1+δ dθ ⩽ A6 (1 − r)−1/2−18δ . 2

(14.1.8)

0

We now use the major–minor arc technique as follows. Divide the interval [0, 2π] into two sets so that [0, 2π] = E1 (r) ∪ E2 (r), where E1 (r) = {θ : |g󸀠 (r−1 e iθ )| ⩽ (1 − r)−β }, E2 (r) = {θ : |g󸀠 (r−1 e iθ )| ⩾ (1 − r)−β }, and β is to be chosen. Then using (14.1.8), we have 2π

∫ |g󸀠 (z)|dθ = ∫ |g󸀠 (z)|dθ + ∫ |g󸀠 (z)|dθ E 1 (r)

0

E 2 (r)

2π + (1 − r)βδ ∫ |g󸀠 (z)|1+δ dθ ⩽ (1 − r)β E 2 (r)

−β

⩽ 2π(1 − r)

2

+ A6 (1 − r)1/2−βδ+18δ .

Now choose β = 1/2 − 1/300 and δ = 1/72, so that −1/2 + βδ − 18δ2 < β, which shows that for 1 < ρ < ∞ 2π

∫ |g󸀠 (ρe iθ )|dθ ⩽ A (1 −

1 −1/2+1/300 , ) ρ

0

and so setting ρ = 1 + 1/n in Cauchy’s integral formula we obtain the result.

14.1 The Class Σ

|

209

We remark that an analogous proof for bounded functions in S can be found in Duren [45].

14.1.2 Coefficients of the Inverse Function We saw in Chapter 1 that the problem of finding sharp bounds for the coefficients of the inverse function f −1 of f ∈ S was settled by Loewner [125] in 1923. For f ∈ Σ, the corresponding problem of finding sharp bounds for the coefficients of the inverse of functions in Σ was similarly resolved relatively early in the study of univalent functions. In this section we summarize this result. First note that since functions g ∈ Σ are univalent, they possess an inverse function g−1 , such that g−1 (g(z)) = z for z ∈ ∆, and g(g−1 ω) = ω is defined on some set {ω : M < |ω| < ∞ and M > 0}. Thus g−1 can be expressed as ∞

̃ n ω−n , g−1 (ω) = ω + ∑ b n=0

for M < |ω| < ∞. ̃ n | in the case of meromorphic starOn the basis of known sharp inequalities for |b like functions (see later in this chapter), in 1963 Royster [216] conjectured that when g ∈ Σ, the coefficients of the inverse function g−1 satisfy the sharp inequality ̃n| ⩽ |b

1 2n ( ), n n+1

(14.1.9)

for n ⩾ 1. This was subsequently proved by Netanyahu [157] in 1965, with equality in (14.1.9) obtained for the inverse of the function g(z) = z(1 + λz−1 )2 , where |λ| = 1. At this point, we note that the inequality (14.1.9) is identical to that obtained by Loewner for the inverse of functions in S. This rather unexpected result follows from the following observation. ̃ Theorem 14.1.4. Let F be the inverse function of f ∈ S. Then, F(1/ω)−1 := F(ω) is the −1 inverse function of f(1/z) := ̃f (z). ̃ maps ̃ Proof. The function F maps f(𝔻) := D onto 𝔻, and so F D := {ω : 1/ω ∈ ̃ ̃f (z)) = [F(1/̃f (z))]−1 = D, and F( D} onto ∆. Also, ̃f (z) = f(1/z)−1 maps ∆ onto ̃ [F(f(1/z))]−1 = (1/z)−1 = z. Thus, F̃ is the inverse of ̃f .

14.1.3 Distortion Theorems Before leaving the class Σ, we list some classical distortion theorems for functions g ∈ Σ, dating back to the early days in the study of univalent functions. These can be found in Goluzin [57], and with more accessible proofs, together with a discussion about the sharpness of the inequalities, in Goodman [59].

210 | 14 Meromorphic Univalent Functions Theorem 14.1.5. Let g ∈ Σ and be given by (14.1.1), then for |z| = r > 1, 󵄨󵄨 g(z) 󵄨󵄨 󵄨 󵄨󵄨 (1 − r−1 )2 ⩽ 󵄨󵄨󵄨 󵄨 ⩽ (1 + r−1 )2 , 󵄨󵄨 z 󵄨󵄨󵄨 1 − 1/r2 ⩽ |g󸀠 (z)| ⩽ (1 − 1/r2 )−1 .

(14.1.10) (14.1.11)

All the inequalities are sharp. We now give a brief summary of the important coefficient results for the major subclasses of Σ.

14.2 Subclasses of Σ 14.2.1 Meromorphic Starlike Functions We first give the definition of meromorphic starlike functions, which is analogous to that for functions in S∗ . Definition 14.2.1. Let g be analytic in ∆ = {z : 1 < |z| < ∞}, be given by (14.1.1), and satisfy zg󸀠 (z) Re [ (z ∈ ∆). ]>0 g(z) Then g is said to be a meromorphic starlike function. We denote the class of meromorphic starlike functions by Σ∗ . Analogous to the class S∗ of starlike functions, it is easy to see that functions in Σ∗ map ∆ onto a region whose (compact) complement is starlike with respect to the origin, and because of the reciprocal relationship between functions in Σ and S, functions in Σ∗ are necessarily univalent. We first note that the extreme functions in Theorem 14.1.5 show that (14.1.10) and (14.1.11) remain the best possible when g ∈ Σ∗ . As was seen in the case of functions g ∈ Σ, a subject of major interest in the theory of meromorphic univalent functions is to find information about the coefficients. We recall the false conjecture that if g ∈ Σ and is given by (14.1.1), then |b n | ⩽ 2/(n + 1). Clunie [30], in the first demonstration of his seminal method, showed that this conjecture is true for functions in Σ∗ , by proving the following. Theorem 14.2.1. Let g ∈ Σ∗ and be given by (14.1.1), then for z ∈ ∆ and n ⩾ 1, |b n | ⩽

2 , n+1

with equality when g(z) = z(1 + z−n−1 )2/(n+1) = z +

2 −n z +.... n+1

14.2 Subclasses of Σ

|

211

We omit the proof, the method being more or less the same as that given in Theorem 4.1.2. An important paper by Pommerenke [192], extended the above result by considering functions in Σ∗ (α) defined by the relationship Re [

zg󸀠 (z) ] > α, g(z)

(0 ⩽ α < 1),

and in particular established the sharp inequality |b n | ⩽ 2α/(n + 1) for n ⩾ 1. Other significant results are contained in this paper. A similar extension of Theorem 14.2.1 to the class of strongly meromorphic starlike functions, defined by | arg(zg󸀠 (z)/g(z))| < απ/2 was obtained by Brannan et al. in [20], yielding the same inequality |b n | ⩽ 2α/(n + 1) for n ⩾ 1.

14.2.2 Meromorphic Close-to-Convex Functions We first give the definition of meromorphic close-to-convex functions, which is again analogous to the class K0 of analytic functions. We note, however, that unlike the class Σ∗ , these functions are not necessarily univalent in ∆. Definition 14.2.2. Let g be analytic in ∆ = {z : 1 < |z| < ∞} and be given by (14.1.1). Then, g is said to be a meromorphic close-to-convex function, if there exists a function h ∈ Σ∗ such that zg󸀠 (z) ]>0 (z ∈ ∆). Re [ h(z) We denote this class by ΣK0 . We note that the function g0 defined by g󸀠0 (z) = (1−z−2 )/(1+z−2 ) belongs to ΣK0 , since in this case, h(z) = z and g0 (z) = z + 2z−1 + . . . . Since the coefficient of z−1 is greater than 1, it follows that g0 is not univalent. Thus members of ΣK0 are not necessarily univalent. With the added condition that functions g and h are univalent, but that h is not necessarily starlike, Libera and Robertson [118], and Pommerenke [195], using an extension of Clunie’s method, independently proved the following. Theorem 14.2.2. Let g and h be analytic and univalent in ∆. Let g be given by (14.1.1), and zg󸀠 (z) ]>0 (z ∈ 𝔻). Re [ h(z) Then, for n ⩾ 1, |b n | ⩽

2√2 . n

To date, this constant appears to be the best known, and the question of whether 2√2 can be improved (possibly to 2) remains an open problem.

212 | 14 Meromorphic Univalent Functions

In both papers, Libera and Robertson [118] and Pommerenke [195], employed Clunie’s method, obtaining a slight generalization in terms of the coefficients of g and h. Both authors showed the following, which does not require that functions in ΣK0 are univalent, or that h ∈ Σ∗ . Theorem 14.2.3. Let g and h be analytic in ∆, and for z ∈ ∆ be given by g(z) = z + −n and h(z) = z + ∑∞ h z−n , respectively. Further, suppose that ∑∞ n=0 b n z n=0 n Re [

zg󸀠 (z) ]>0 h(z)

(z ∈ ∆).

Then, for n ⩾ 0, n−1

|nb n + h n |2 ⩽ 4 (1 − ∑ k Re[b k h k ]) .

(14.2.1)

k=1

We omit the proof, which uses the quantitative elements of Clunie’s method, rather than the nature of the functions g and h. We now show that the proof of Theorem 14.2.2 can be easily deduced from Theorem 14.2.3 as follows. Proof of Theorem 14.2.2. From (14.2.1) we have n−1

n2 |b n |2 + 2n Re[b n h n ] + |h n |2 ⩽ 4 (1 − ∑ k Re[b k h k ]) , k=1

and so n−1

n2 |b n |2 ⩽ 4 + 4 ∑ k|b k ‖h k | + 2|b n ‖h n |n k=1 n−1

n

k=1

k=1

= 4 + 2 ∑ √k|b k |√k|h k | + 2 ∑ √k|b k |√k|h k |.

(14.2.2)

Since Theorem 14.2.2 assumes that both g and h are univalent, it follows from the area ∞ 2 2 principle that ∑∞ k=1 k|b k | ⩽ 1 and ∑ k=1 k|h k | ⩽ 1. Thus, using the Cauchy–Schwarz inequality, it follows from (14.2.2) that n−1

1/2

n2 |b n |2 ⩽ 4 + 2 ( ∑ k|b k |2 )

1/2

n

+ 2 ( ∑ k|b k |2 )

k=1

k=1

⩽ 4 + 2(1 − n|b n |2 )1/2 + 2 ⩽ 6 + (2 − n|b n |2 ). Thus, |b n |2 ⩽

8 , n(n + 1)

so that

|b n | ⩽

2√2 √n(n + 1)


0 h(z)

then for n ⩾ 1, |b n | ⩽

(z ∈ ∆),

21 . n

Further detailed properties concerning meromorphic close-to-convex functions can be found in Pommerenke [195].

14.2.3 Meromorphic Bazilevič Functions Relatively little attention has been given to extending the above results to meromorphic functions analogous to those discussed in Chapter 10. We give the following definition. Definition 14.2.3. Let g be analytic in ∆ and be given by (14.1.1). Also, let h be analytic −n in ∆ and be given by h(z) = z + ∑∞ n=0 h n z . Then, if for 0 ⩽ α ⩽ 1, there exists a ∗ function h ∈ Σ such that Re [

zg󸀠 (z) g1−α (z)h α (z)

]>0

(z ∈ ∆),

g is said to be a meromorphic Bazilevič function of type α. We denote this class by Σ B(α) . The restriction 0 ⩽ α ⩽ 1 in Definition 14.2.3 is not necessary but is included since significant results have only been found in this case. Thus, Σ B(0) = Σ∗ and Σ B(1) = ΣK0 , and so functions in Σ B(α) are therefore not necessarily univalent. Finding numerical bounds for the coefficients of functions in Σ B(α) has proved more difficult; the following is probably the best result to date, Thomas [243]. Theorem 14.2.5. Let g ∈ Σ and be given by (14.1.1) for z ∈ ∆. Also, let h ∈ Σ and be given −n for z ∈ ∆. Suppose in addition that g(z) ≠ 0 and h(z) ≠ 0 for by h(z) = z + ∑∞ n=0 h n z z ∈ ∆. If for 0 ⩽ α ⩽ 1, Re [ then nb n = O(1) as n → ∞.

zg󸀠 (z) g1−α (z)h α (z)

]>0

(z ∈ ∆),

214 | 14 Meromorphic Univalent Functions

As in the case of Theorem 14.2.2, both functions g and h are required to be univalent, and h does not necessarily need to belong to Σ∗ . Thus, in particular, Theorem 14.2.5 is valid for functions in Σ B(α) . In order to prove Theorem 14.2.5, we require the following theorem of Thomas [243], which is of interest in itself. We omit the proof, which uses an area principle theorem due to Grunsky [63]. Theorem 14.2.6. Let g, h ∈ Σ and suppose that g(z) ≠ 0, h(z) ≠ 0, and 0 ⩽ α ⩽ 1. For sufficiently large z, define G for z ∈ ∆ by ∞

G(z) = g(z)1−α h(z)α = z + ∑ c n z−n . n=0

Then,

∑∞ n=1

n|c n |2 < ∞.

Note that in the above definition, since g and h have no zeros in ∆, the function G is analytic in ∆. The proof of Theorem 14.2.5 can now easily be obtained using the method in Theorem 14.2.3 as follows. In this case, (14.2.2) becomes n−1

n

k=1

k=1

n2 |b n |2 ⩽ 4 + 2 ∑ √k|b k |√k|c k | + 2 ∑ √k|b k |√k|c k |, ∞ 2 2 and since ∑∞ n=1 n|b n | ⩽ 1 and ∑ n=1 n|c n | < ∞, applying the Cauchy–Schwarz inequality gives nb n = O(1) as n → ∞.

15 Loewner Theory 15.1 The Loewner Equation In previous chapters we have endeavoured to use only elementary ideas of complex analysis to establish many of the basic properties in the theory of univalent functions. However, as was mentioned in the Preface, significant advances have been made using more advanced ideas, one of which, the so-called Loewner theory, plays a central role in the subject, and in particular for functions in S. As has been mentioned, these ideas were used to solve several important problems, including eventually providing a basis for the proof the Bieberbach conjecture. A great deal of detailed analysis is required to fully understand the fundamental ideas behind Loewner theory, and good accounts can be found in the books of Duren [45], Hayman [70], and Pommerenke [198]. However, as we shall see, if one assumes these deeper ideas, then the details of the proofs of many of the applications of Loewner theory, although technical in nature, are not difficult to follow. We will, therefore, give a summary of the basic idea involved, and show how this can be applied to solve some fundamental results for functions in S. The basic idea of Loewner theory is simple. Find a dense subset S󸀠 of S such that 󸀠 every function f ∈ S can be approximated by a sequence {f n }∞ n=1 of functions in S , so that f n (z) → f(z) as n → ∞ uniformly on every compact subset of 𝔻. It would then follow that the derivatives of f n (z) at any point z ∈ 𝔻 would tend to those of f(z), and also for example, so would the coefficients of f n (z) tend to the those of f(z) as n → ∞. In this way, bounds true for functionals in S󸀠 , will also be true in S. Loewner’s clever idea was to consider a dense subset S󸀠 of S consisting of all single slit-mappings defined by S󸀠 := {f ∈ S : f(𝔻) = ℂ \ Γ}, where Γ is a Jordan curve extending to ∞. Loewner’s remarkable theorem is as follows. Theorem 15.1.1. Let f ∈ S be a single-slit mapping, and f(z, t) be defined by ∞

f(z, t) = e−t (z + ∑ a n (t)z n ) . n=2

Then f(z, t) satisfies the differential equation ∂ 1 + κ(t)z f(z, t) = − f(z, t), ∂t 1 − κ(t)z

f(z, 0) = z,

(15.1.1)

where κ(t) is a continuous complex-valued function with |κ(t)| = 1, for 0 ⩽ t < ∞.

https://doi.org/10.1515/9783110560961-015

216 | 15 Loewner Theory

Furthermore, lim e t f(z, t) = f(z)

t→∞

(z ∈ 𝔻),

where the convergence is uniform on each compact subset of 𝔻. As mentioned above, a detailed proof of this theorem requires a significant amount of preparation, which can be found in any of the books by Duren [45], Hayman [70], and Pommerenke [198]. We will give some applications of Theorem 15.1.1, which illustrate the power of the method, noting that the most important application was that given by de Branges in his proof of the Bieberbach conjecture. We will mostly follow the excellent version given in Duren [45].

15.2 Applications We present some examples of how to apply Loewner’s equation (15.1.1) to obtain results for functions in S that are not accessible using elementary methods. The applications that we illustrate mostly rely on the following ingenious result of Valiron and Landau [103]. Lemma 15.2.1 (Valiron–Landau lemma). Let ϕ(t) be real valued and continuous for t ⩾ 0, |ϕ(t)| ⩽ e−t , and for 0 ⩽ λ < ∞, satisfy ∞

∫[ϕ(t)]2 dt = (λ +

1 −2λ )e . 2

(15.2.1)

0

Then,

󵄨󵄨 ∞ 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ⩽ (λ + 1)e−λ . ∫ ϕ(t)dt 󵄨󵄨 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨 󵄨0

Equality holds if, and only if, ϕ(t) = ±ψ(t), where ψ(t) = e−λ when 0 ⩽ t ⩽ λ, and ψ(t) = e−t when λ < t < ∞. Proof. First note that |ϕ(t)| ⩽ e−t implies that ∞

∞

∫ [ϕ(t)] dt ⩽ ∫ e−2t dt = 2

0

1 . 2

0

Since (λ + 1/2)e−2λ decreases from 1/2 to 0 as λ increases from 0 to ∞, it follows that λ is uniquely determined by (15.2.1).

15.2 Applications | 217

Next, note that for all t ⩾ 0, the function F given by F(t) = [ψ(t) − ϕ(t)] ⋅ [2e−λ − ψ(t) − |ϕ(t)|] is non-negative, and so ∞

0 ⩽ ∫ F(t)dt 0 ∞

= 2e

−λ

∞

∞

{ } { ∫ ψ(t)dt − ∫ |ϕ(t)|dt} 0 {0 } ∞

− ∫ [ψ(t)]2 dt + ∫ [ϕ(t)]2 dt 0

0 ∞

{ } = 2e−λ {(λ + 1)e−λ − ∫ |ϕ(t)|dt} , 0 { } which proves Lemma 15.2.1. As has been seen at various points in this book, a great deal of attention has been given to solving Fekete–Szegő problems for subclasses of S, all of which originated from the initial 1933 paper of Fekete–Szegő [48], stated in Theorem 1.2.4. Our first application is to give a proof of this rather curious, but sharp result. We restate the theorem, and follow the proof in Duren [45]. n Theorem 15.2.1. Let f ∈ S and be given by f(z) = z + ∑∞ n=2 a n z for z ∈ 𝔻. Then for 0 < μ < 1, 󵄨 󵄨󵄨 󵄨󵄨a3 − μa22 󵄨󵄨󵄨 ⩽ 1 + 2e−2μ/(1−μ) . 󵄨 󵄨

The inequality is sharp for each μ. Proof. Since S is rotationally invariant, it is enough to consider Re(a3 − μa22 ). We now rewrite (15.1.1) as ∂ (15.2.2) (1 − κz) f(z, t) = − (1 + κz) f(z, t). ∂t With u(t) = e−t κ(t), comparing coefficients in (15.2.2) gives a󸀠2 (t) = −2u(t), a󸀠3 (t) = −4u(t)a2 (t) + u(t)a󸀠2 (t). Integrating gives ∞

a2 = −2 ∫ e−t κ(t)dt 0

(15.2.3)

218 | 15 Loewner Theory

and ∞

a3 = −2 ∫ e

2

∞

−2t

0

{ } [κ(t)] dt + 4 { ∫ e−t κ(t)dt} , {0 } 2

(15.2.4)

where κ(t) is a piecewise continuous complex-valued function, with |κ(t)| = 1 for all t. Now, write κ(t) = e iθ(t), so that 2

Re(a3 −

μa22 )

∞ ∞ { { −t = 4(1 − μ) {( ∫ e cos θ(t)dt) − ( ∫ e−t sin θ(t)dt) { 0 { 0 ∞

2

} } } } }

− 4 ∫ e−2t cos2 θ(t)dt + 1 0 2

∞

∞

} { ⩽ 4(1 − μ) { ∫ ϕ(t)dt} − 4 ∫ [ϕ(t)]2 dt + 1, 0 } {0 where ϕ(t) = e−t cos θ(t). Thus, with ∞

∫ [ϕ(t)]2 dt = (λ +

1 −2λ )e , 2

0

it follows from Lemma 15.2.1 that Re(a3 − μa22 ) ⩽ 4e−2λ [(1 − μ)(λ + 1)2 − (λ +

1 )] + 1. 2

Elementary calculus now shows that the above has a maximum value of 1 + 2e−2μ/(1−μ) at λ = μ/(1 − μ), which completes the proof. An account of the sharpness of the inequality in Theorem 15.2.1 can be found in any of the books [45, 70, 198]. We note that the inequality |a2 | ⩽ 2 follows at once from (15.2.3), and the case μ = 0 in Theorem 15.2.1 would imply that |a3 | ⩽ 3. However, this can be proved using (15.2.4) in a similar way as follows. Again, note that it is sufficient to consider Re a3 , so that from (15.2.4) we have ∞

Re a3 ⩽ 2 ∫ e 0

∞

−2t

2

{ } [1 − 2 cos θ(t)] dt + 4 { ∫ e−t cos θ(t)dt} . {0 } 2

15.2 Applications | 219

We now use the Cauchy–Schwarz inequality to obtain ∞

Re a3 ⩽ 1 − 4 ∫ e−2t cos2 θ(t)dt 0 ∞

∞

}{ } { + 4 { ∫ e−t dt} { ∫ e−t cos2 θ(t)dt} } {0 } {0 ∞

⩽ 1 + 4 ∫ (e−t − e−2t ) cos2 θ(t)dt 0 ∞

⩽ 1 + 4 ∫ (e−t − e−2t )dt = 3. 0

We have seen in this book that finding exact bounds for the difference of coefficients for functions in S, or even subclasses of S has proved difficult, and only when f ∈ S∗ has a complete solution been found. It is, therefore, not surprising that the problem for functions in S is far from easy. We next give another example of how the use of the Valiron–Landau lemma can be employed in conjunction with the Loewner equation to obtain sharp upper and lower bounds for |a3 | − |a2 | when f ∈ S. n Theorem 15.2.2. Let f ∈ S and be given by f(z) = z + ∑∞ n=2 a n z for z ∈ 𝔻. Then,

−1 ⩽ |a3 | − |a2 | ⩽

3 + e−λ 0 (2e−λ 0 − 1) = 1.029 . . . , 4

where λ0 is the unique value of λ satisfying 0 < λ < 1 and 4λ = e λ . Both inequalities are sharp. Proof. The lower bound is simple, since if |a2 | ⩽ 1, then clearly |a2 | − |a3 | ⩽ 1. On the other hand, if |a2 | ⩾ 1, then the inequality |a3 − a22 | ⩽ 1 (see Chapter 1) implies that |a2 | − |a3 | = |a2 |2 − |a3 | + |a2 |(1 − |a2 |) ⩽ 1. Equality occurs when, for example, f(z) = z/(1 + z + z2 ). For the upper bound we need to use the Loewner equation, and once more appeal to the Valiron–Landau lemma. Rotational invariance allow us to assume that a3 ⩾ 0; so, again letting κ(t) = e iθ(t), we can write ∞

a2 = −2 ∫ e−t κ(t)dt = −2(u + i𝑣) 0

and

∞

a3 = −2 ∫ e 0

∞

−2t

2

{ } [κ(t)] dt + 4 { ∫ e−t κ(t)dt} , {0 } 2

220 | 15 Loewner Theory

so that |a3 | − |a2 | = Re a3 − |a2 | ∞

= −4 ∫ e−2t cos2 θ(t)dt + 1 + 4(u 2 − 𝑣2 ) − 2(u 2 + 𝑣2 )1/2 0 ∞

⩽ 1 + 4u 2 − 2|u| − 4 ∫ e−2t cos2 θ(t)dt. 0

We now note that if the last integral has the value (λ + 1/2)e−2λ for 0 ⩽ λ < ∞, then the Valiron–Landau lemma shows that |u| ⩽ (λ + 1)e−λ . Thus, we consider two cases. Case (i): (λ + 1)e−λ ⩽ 1/2, in which case |u| ⩽ 1/2 and 4u 2 − 2|u| ⩽ 0. Thus, |a3 | − |a2 | ⩽ 1. Case (ii): (λ + 1)e−λ > 1/2. Then, since 4u 2 − 2|u| is positive and increasing on the interval 1/2 < |u| < ∞, it follows that |a3 | − |a2 | ⩽ 1 + 4(λ + 1)2 e−2λ − 2(λ + 1)e−λ − 4(λ + 1/2)e−2λ = 1 + 2(2λ2 + 2λ + 1)e−2λ − 2(λ + 1)e−λ := ν(t). Thus, we need to find the maximum of ν(t) when (λ+1)e−λ > 1/2. Elementary calculus shows that ν has critical points at λ = 0, and at λ0 and λ1 , where 4λ = e λ and 0 < λ0 < 1 and λ1 > 2. Since (λ1 +1)e−λ 1 < 1/2, we can disregard the point λ1 . It is an easy exercise to show that λ0 is a maximum point, which on substituting back gives |a3 | − |a2 | ⩽

3 + e−λ 0 (2e−λ 0 − 1) = 1.029 . . . , 4

which completes the proof of the theorem. We omit the details showing that the upper bound for |a3 | − |a2 | is sharp, which can be found in [45, Theorem 3.11]. We have seen that using a combination of the Valiron–Landau lemma and the Loewner equation, the proof of the inequality |a3 | ⩽ 3 for f ∈ S follows in a relatively simple way. It was, therefore, natural to assume that the Bieberbach conjecture might follow in a similar way. Although the inequality when n = 3 was proved by Loewner as early as 1923 [125], it was only in 1985 that de Branges finally proved the conjecture. Also, the corresponding problem of finding sharp upper bounds for the coefficients of the inverse function f −1 of functions in S, which was stated in Chapter 1, was also solved in 1923 by Loewner [125]. There is no doubt that this supported the view that the key to solving the Bieberbach conjecture lay somewhere in Loewner theory. It was, therefore, no surprise to many that de Branges’s solution relied heavily of the use of the Loewner equation. (A proof of the inverse coefficient problem, which does not rely on the use of the Valiron–Landau lemma, can be found in Hayman [70].)

15.2 Applications | 221

In the process of proving the Bieberbach conjecture, de Branges was able to prove a long-standing conjecture of Robertson (see, e.g., [45]) concerning the coefficients of odd functions in S, i.e., functions in S whose Taylor series contains only odd coefficients. Robertson’s conjecture, which implies the Bieberbach conjecture, states that if 2n−1 for z ∈ 𝔻, then for n ⩾ 1, g ∈ S and is given by g(z) = z + ∑∞ n=1 c 2n−1 z n

∑ |c2k−1 |2 ⩽ n, k=1

where c1 = 1. To see that Robertson’s conjecture implies the Bieberbach conjecture, we simply write, with c1 = 1, ∞

g(z) = √f(z2 ) = ∑ c2n−1 z2n−1 , n=1

where f ∈ S is given by f(z) =

z+∑∞ n=2

an

z n . Then, equating coefficients gives for n

⩾ 1,

n

a n = ∑ c2k−1 c2(n+1−k)−1 . k=1

Now we apply the Cauchy–Schwarz inequality to obtain for n ⩾ 1 n

|a n | ⩽ ∑ |c2k−1 |2 . k=1

Thus, if n

∑ |c2k−1 |2 ⩽ n, k=1

then the Bieberbach conjecture is true. As another illustration of the use of the Valiron–Landau lemma, we now show how the Loewner equation can be used to prove Robertson’s conjecture when n = 3. Theorem 15.2.3. Let g ∈ S and be an odd function with Taylor series g(z) = z + 2n−1 for z ∈ 𝔻. Then, ∑∞ n=1 c 2n−1 z |c1 |2 + |c3 |2 + |c5 |2 ⩽ 3, where c1 = 1. Proof. At a first glance, the involvement of the coefficient c5 should make the analysis more complicated. However, since c3 = a2 /2, and c5 = (4a3 − a22 )/8, using (15.2.3) and (15.2.4), we can write ∞

c3 = − ∫ e−t κ(t)dt 0

222 | 15 Loewner Theory

and

2

∞

∞

} 3{ c5 = { ∫ e−t κ(t)dt} − ∫ e−2t [κ(t)]2 dt. 2 0 } {0 Since we are allowed one rotation, we now assume that c5 ⩾ 0, and since c1 = 1, we need to show that |c3 |2 + [Re c5 ]2 ⩽ 2. The remainder of the proof now follows the pattern of Theorem 15.2.2. As before, put κ(t) = e iθ(t), and write ∞

∫ e−t κ(t)dt = u + i𝑣, 0

so that |c3 |2 = u 2 + 𝑣2 and ∞

3 2 1 (u − 𝑣2 ) − 2 ∫ e−2t cos2 θ(t)dt + . 2 2

Re c5 =

0

Now, for λ ⩾ 0 let

∞

∫ e−2t cos2 θ(t)dt = (λ +

1 −2t )e , 2

0

and apply the Valiron–Landau lemma to obtain |u| ⩽ (λ+1)e−λ . Thus, since u 2 +𝑣2 ⩽ 1, we have |c3 |2 + |c5 |2 ⩽ B(λ), where B(λ) = min {1 − 𝑣2 , (λ + 1)2 e−2λ } + 𝑣2 +

1 2 [β(λ) + 1 − 3𝑣2 ] , 4

(15.2.5)

with β(λ) = (3λ2 + 2λ + 1)e−2λ , and we therefore need to show that B(λ) ⩽ 2. It is easily seen that β(λ) takes its maximum value when λ = 1/3, and so β(λ) ⩽ 2e−2/3 = 1.026 . . . . We now consider the two minimum values in (15.2.5) and distinguish two cases. Case (i): 0 ⩽ 1 − 𝑣2 ⩽ (λ + 1)2 e−2λ . Here, B(λ) = 1 +

1 {β(λ) + 1 − 3𝑣2 }2 . 4

Using β(λ) = (3λ2 + 2λ + 1)e−2λ , and considering the above expression as a quadratic expression in 𝑣2 , it is easily seen that B(λ) ⩽ 2. Case (ii): (λ + 1)2 e−2λ ⩽ 1 − 𝑣2 ⩽ 1. Here, B(λ) = (λ + 1)2 e−2λ + 𝑣2 +

1 2 [β(λ) + 1 − 3𝑣2 ] . 4

15.2 Applications |

223

Again, considering the above expression as a quadratic in 𝑣2 , it is easy to see that its maximum value occurs when either 𝑣2 = 0 or 𝑣2 = 1 − (λ + 1)2 e−2λ , and in the latter case, we have already seen that B(λ) ⩽ 2. Thus, we are left with the case 𝑣2 = 0. Here, B(λ) becomes B(λ) = λ(1 − 3λ)(3λ2 + 2λ + 1)e−4λ − λ(5λ + 1)e−2λ . When 1/3 ⩽ λ < ∞, B󸀠 (λ) < 0, and when 0 < λ < 1/3 (using the fact that e−4λ < e−2λ ), it follows that B󸀠 (λ) < −3λ2 (3λ2 + 2λ + 2)e−2λ e−2λ < 0. Thus, B(λ) ⩽ B(0) = 2 in this case also, since 𝑣2 = 0. We end by noting that the logarithmic coefficients of functions in S and its subclasses defined in Chapter 1 play an important role in the theory of univalent functions, and it was considering these coefficients that eventually led to the solution of the Bieberbach conjecture. On the other hand, finding sharp bounds for the modulus of individual logarithmic coefficients γ n when f ∈ S remains a difficult problem. Since γ1 = a2 /2, then |γ1 | ⩽ 1 and is sharp. Also, since 4γ2 = 2a3 − a22 , the Fekete–Szegő theorem with μ = 1/2 gives the sharp inequality |γ2 | ⩽

1 (1 + 2e−2 ). 2

Although no other sharp bounds for |γ n | when f ∈ S are known for n ⩾ 3, the following interesting sharp results of Upadrashta and Vasudevarao [256] provide some information concerning the initial coefficients γ n in an averaging sense. We note that the highly technical proofs are derived directly from the Loewner equation and do not use the Valiron–Landau lemma. Theorem 15.2.4. Let f ∈ S and γ n be given by (1.2.1). Then, 󵄨 1 󵄨󵄨 󵄨󵄨γ3 − 2γ1 γ2 + γ31 󵄨󵄨󵄨 ⩽ , 󵄨 3 󵄨 󵄨󵄨 4 󵄨󵄨󵄨 1 󵄨󵄨 󵄨󵄨γ4 − 2γ22 − 2γ1 γ3 + 4γ21 γ2 − γ41 󵄨󵄨󵄨 ⩽ , 󵄨󵄨 3 󵄨󵄨 4 󵄨󵄨 23 23 5 󵄨󵄨󵄨󵄨 1 󵄨󵄨 γ31 γ2 + γ 󵄨⩽ . 󵄨󵄨γ5 − 2γ1 γ4 − 4γ2 γ3 + 4γ21 γ3 + 7γ22 γ1 − 󵄨󵄨 3 12 1 󵄨󵄨󵄨 5 All the inequalities are sharp.

16 Other Topics This book has addressed what we consider to be the important problems concerning the major subclasses of S. Naturally, there are other important topics which have not been included, and we end by listing five such topics where significant unsolved problems remain. We merely give definitions and references to some important results, problems, and papers.

16.1 Harmonic Univalent Functions A topic that has attracted considerable attention in recent years is the idea of harmonic univalent functions introduced by Clunie and Shiel-Small in 1984 [35]. Their definition is as follows. Definition 16.1.1. A continuous function f = u + i𝑣, defined in a subset D of the complex plane is harmonic in D if both u and 𝑣 are real and harmonic in D. In any simply connected subdomain of D, we can write f = h + g, where both f and g are analytic, and where g is the complex conjugate of g. When f = h + g, we call h the analytic part of f , and g the co-analytic part of f . The Jacobian of f is then given by J f (z) = |h󸀠 (z)|2 − |g󸀠 (z)|2 , and f is locally univalent if J f (z) ≠ 0 in D. A result of Lewy [113] shows that the converse is true for harmonic mappings, and so f is locally univalent and sense preserving in D, if, and only if, |g󸀠 (z)| < |h󸀠 (z)|

(z ∈ D).

We call such mappings locally univalent and say that f is univalent in D if it is univalent and is sense preserving in D. We denote the class of all sense preserving maps on 𝔻 with f(0) = g(0) = 0 and f 󸀠 (0) = 1 by SH and by S0H functions in SH with g󸀠 (0) = 0. Thus, S = {f = h + g ∈ SH : g(z) ≡ 0} and S ⊂ S0H ⊂ SH . For f ∈ S0H , let ∞

∞

n=2

n=2

f(z) = h(z) + g(z) = z + ∑ a n z n + ∑ b n z n .

(16.1.1)

The question of finding sharp upper bounds for the coefficients |a n | and |b n | raised in the original paper of Clunie and Shiel-Small remains an open problem. Indeed, even the case n = 2 is unsolved. By considering subclasses of S0H such as starlike harmonic functions, i.e., when f(𝔻) is a starlike domain, Clunie and Shiel-Small were led to the following conjecture. https://doi.org/10.1515/9783110560961-016

16.2 Bi-univalent Functions

|

225

The Clunie–Shiel-Small Conjecture For f ∈ S0H , with coefficients given by (16.1.1). Then for n ⩾ 2, |a n | ⩽

(n + 1)(2n + 1) , 6

|b n | ⩽

(n − 1)(2n − 1) , 6

and ‖a n | − |b n ‖ ⩽ n.

We note that when g(z) ≡ 0, this conjecture reduces to the Bieberbach conjecture proved by de Branges, and so the Clunie–Shiel-Small conjecture is the harmonic equivalent of the Bieberbach conjecture. The Clunie–Shiel-Small conjecture has been verified for a number of subclasses of f ∈ S0H , and we refer the reader to the paper of Ponnusamy and Sairam Kaliraj [199] for a summary of the history and progress of the problem, and to Ponnusamy et al. [201] for some recent numerical estimates for the coefficients |a n | and |b n | when n ⩾ 3. Other important contributions to harmonic univalent functions can be found in the references in each of these papers. A further set of interesting problems and conjectures concerning harmonic functions can be found in Bshouty and Lyzzaik [26], most of which remain open.

16.2 Bi-univalent Functions Another area of interest is that where both f and its inverse f −1 are univalent, and where again few significant results have been found. The basic definition is as follows. n Definition 16.2.1. Let f ∈ A with f(z) = z + ∑∞ n=2 a n z . Then f is bi-univalent in 𝔻 if −1 both f and f are univalent in 𝔻.

̃ we note that examples of functions Denoting the class of bi-univalent functions by Σ, ̃ are in Σ 1 1+z z , − log(1 − z), and log , 1−z 2 1−z ̃ as well as simple functions in S such but that the Koebe function does not belong to Σ, as z2 z z− . and 2 1 − z2 The first reference to bi-univalent functions appears to be in an oral communication by Nehari to Lewin (see [111]), who raised the natural question as to what are the best upper bounds for |a n | in the Taylor expansion for f . This turned out to be a significantly difficult problem, and relatively little is known about it even half a century later. Lewin showed that |a2 | < 1.51 . . . and it was conjectured by Brannan and Clunie that |a2 | ⩽ √2. On the other hand, Netanyahu [158] showed that the exact upper bound for ̃ consisting of all functions that are bi-univalent where |a2 | = 4/3, for the subclass of Σ their ranges contain the unit disk 𝔻. Little significant progress appears to have been made on these early results.

226 | 16 Other Topics ̃ As was to be expected, attention was then given to considering subclasses of Σ, such as the class of strongly bi-starlike functions defined as follows by Brannan and Taha [22]. n Definition 16.2.2. Let f ∈ A with f(z) = z + ∑∞ n=2 a n z . Then, f is strongly bi-starlike of order α in 𝔻 if for 0 < α ⩽ 1, each of the following are satisfied 󵄨󵄨 󵄨󵄨 󸀠 ̃ and 󵄨󵄨󵄨󵄨arg [ zf (z) ]󵄨󵄨󵄨󵄨 < απ (z ∈ 𝔻), f ∈Σ f(z) 󵄨󵄨󵄨 2 󵄨󵄨󵄨

and

󵄨 󵄨󵄨 ωg󸀠 (ω) 󵄨󵄨󵄨 απ 󵄨󵄨 󵄨󵄨 < 󵄨󵄨arg [ ] 󵄨󵄨 g(ω) 󵄨󵄨󵄨 2 󵄨 where g is the inverse of f .

(ω ∈ 𝔻),

Brannan and Taha [22] found nonsharp estimates for |a2 | and |a3 |, but finding sharp ̃ such as this, remains a diffibounds for the coefficients |a n |, even for subclasses of Σ cult problem.

16.3 Functions of Bounded Boundary Rotation The idea of a function of bounded boundary rotation originated with Loewner in 1917 [124], but it was Paatero who in 1931 and 1933 [182, 183] subsequently developed the concept. The basic idea is to consider locally univalent functions f in 𝔻, which map 𝔻 onto a domain with bounded boundary rotation, i.e., where the total variation in the direction of the angle of the tangent to the boundary curve is bounded under a complete circuit. More precisely, Paatero gave the following definition. n 󸀠 Definition 16.3.1. Let f ∈ A with f(z) = z + ∑∞ n=2 a n z and f (z) ≠ 0 for z ∈ 𝔻. For k ⩾ 2, suppose that the image domain of f has boundary rotation at most kπ. Then, f ∈ Vk if, and only if, for z = re iθ ∈ 𝔻, 2π 󵄨

󵄨󵄨 (zf 󸀠 (z))󸀠 󵄨󵄨󵄨 󵄨󵄨 dθ ⩽ kπ. ∫ 󵄨󵄨󵄨󵄨Re 󵄨 f 󸀠 (z) 󵄨󵄨󵄨 󵄨󵄨 0 Alternatively, f ∈ Vk if, and only if, for z ∈ 𝔻, 2π

} {1 f(z) = exp { ∫ log(1 − ze−t )dμ(t)} , π } { 0 where μ(t) is a real-valued function of bounded variation satisfying 2π

∫ |dμ(t)| ⩽ kπ 0

2π

and

∫ dμ(t) = 2π. 0

(16.3.1)

16.3 Functions of Bounded Boundary Rotation

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227

Clearly, if k 1 < k 2 , then Vk1 ⊂ Vk2 and V2 = C, the class of convex functions. It was shown by Pinchuck [188] that when 2 ⩽ k ⩽ 4, functions in Vk belong to K the class of close-to-convex functions and are, therefore, univalent in 𝔻. Moreover, Brannan [18] showed that, again, when 2 ⩽ k ⩽ 4, Vk forms a subclass of K(α) the class of close-toconvex functions of order α with α = k/2 − 1. (Note that here, the class K(α) denotes ̃ the half-plane analogue of K(α), defined in Chapter 9.) It is easy to see that from the above definition, a further alternative representation for functions in Vk is the following, which is a very useful tool in dealing with functions in Vk . Theorem 16.3.1. f ∈ Vk if, and only if, there exists functions s1 , s2 ∈ S∗ such that f 󸀠 (z) =

[s1 (z)/z]k/2+1/2 . k/2−1/2 [s2 (z)/z]

Paatero [183] gave sharp distortion theorems for f ∈ Vk , showing that F k ∈ Vk defined below serves as the extreme function F k (z) =

∞ 1 1 + z k/2 [( ) − 1] = z + ∑ B n (k)z n . k 1−z n=2

Finding sharp bounds for the coefficients |a n | proved more difficult, but the problem was eventually solved by Aharonov and Friedland [1] and Brannan [19], who showed that |a n | ⩽ B n (k) for all k ⩾ 2 and n ⩾ 2. Although much is known about functions in Vk , there remain many interesting and significant problems. For example, although the correct rate of growth of the difference of coefficients is known (see Noonan and Thomas [160]), there are no known precise bounds for ‖a n+1 | − |a n ‖ for any value of k or n. This appears to be a difficult problem in view of the fact that even when k = 2, i.e., when f ∈ C, this problem remains partially unsolved (see Theorem 4.2.5). Since the result concerning the growth estimate for ||a n+1 | − |a n || involves an important and much used inequality of Golusin [54, 57], we give the proof below, first stating Golusin’s inequality as a lemma. Lemma 16.3.1. Let f ∈ S, and z = re iθ ∈ 𝔻. Then there exists ζ ∈ 𝔻, such that for 0 0, z ∈ 𝔻. Theorem 16.4.3. Let Ω ⊂ ℂ and suppose that the function ψ : ℂ2 × 𝔻 → ℂ satisfies ψ(Me iθ , Ke iθ ; z) ∉ Ω for all K ⩾ M, θ ∈ ℝ, and z ∈ 𝔻. If p is analytic in 𝔻, p(0) = 1 and ψ(p(z), zp󸀠 (z); z) ∈ Ω for all z ∈ 𝔻, then |p(z)| < M, z ∈ 𝔻.

16.5 Operators Given a function f ∈ A, the most natural operators acting on f are perhaps the first and second derivatives f 󸀠 and f 󸀠󸀠 , and a discussion of the properties of these derivatives form a fundamental element of any class of functions in A. Similarly, the integral of f(z) with respect to z plays an important role. Generalizing these basic operators when f is univalent so as to give meaningful and significant areas of research has been discussed extensively in the literature. In this chapter, we will define some of the more important ones, noting that in recent years, a great many related operators have been defined, which are more or less simple extensions of these ideas, but in some cases do lead to interesting connections with the special functions of complex analysis.

16.5.1 The Sălăgean Operator We begin with a most natural extension to the differential operator due to Sălăgean [237]. k Definition 16.5.1. Let f ∈ A, with f(z) = z + ∑∞ k=2 a k z , then for n ⩾ 1, the Sălăgean n differential operators D are defined inductively by

Dn [f](z) = D(D n−1 [f](z)),

16.5 Operators

| 231

n k where D1 [f](z) = zf 󸀠 (z) and D0 [f](z) = f(z), so that Dn [f](z) = z + ∑∞ k=2 k a k z for n ∈ ℕ0 = {0} ∪ ℕ.

With k(z) as the Koebe function z/(1 − z)2 , these operators can be defined via convolution as ⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ ∗ k ∗ ⋅ ⋅ ⋅ ∗ k ∗f)(z). Dn [f](z) = (k n times

We note that D[f](z) = k∗f(z) is special case of the operator defined by Ruscheweyh in 1975 [220], D∗n [f](z) =

∞ z n+k−1 ∗ f(z) = z + ∑ ( )a k z k , n (1 − z)n+1 k=2

n ∈ ℕ0 .

The Sălăgean operator is closely related to the integral operator given as follows. Definition 16.5.2. Let f ∈ A. We define inductively the integral operators In , n ∈ ℕ0 , by In [f](z) = I(I n−1 [f](z)), z

where I0 [f](z) = f(z) and I1 [f](z) = I[f](z) = ∫0 f(t)/t dt. k If f(z) = z + ∑∞ k=2 a k z , then when l(z) = − log(1 − z), ∞

1 a zk . n k k k=2

∗ l ∗ ⋅ ⋅ ⋅ ∗ l ∗f)(z) = z + ∑ In [f](z) = (l⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟⏟ n times

Moreover we have Dn In [f](z) = In Dn [f](z) = f(z). An obvious generalization of the class S∗ using the Sălăgean operator is the following. Definition 16.5.3. A function f ∈ A belongs to S∗n (α) if for n ∈ ℕ0 , 0 ⩽ α < 1, Re [

Dn+1 [f](z) ]>α Dn [f](z)

(z ∈ 𝔻),

and so S∗0 (α) = S∗ (α) and S∗1 (α) = C(α). Using a result from the theory of differential subordination, Sălăgean proved the following. Theorem 16.5.1 ([237]). Let h be analytic and convex univalent in 𝔻, with h(0) = 1 and Re h(z) > 0, for z ∈ 𝔻, and let f ∈ A and n ∈ ℕ0 . (i) If Dn+2 [f](z) ≺ h(z), Dn+1 [f](z) then

Dn+1 [f](z) ≺ h(z). Dn [f](z)

232 | 16 Other Topics

(ii) If the differential equation q(z) +

zq󸀠 (z) = h(z), q(z)

q(0) = 1,

has a univalent solution q(z), then Dn+2 [f](z) Dn+1 [f](z)

≺ h(z)

Dn+1 [f](z) ≺ q(z) ≺ h(z), Dn [f](z)

⇒

and q(z) is the best dominant. Using part (i) of the above theorem it is easy to show that S∗n+1 (α) ⊂ S∗n (α) ⊂ C(α) = S∗1 (α) ⊂ S∗0 (α) = S∗ (α)

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

16.5.2 The Libera Operator and Generalizations In 1965 Libera [117] proved that the integral operator L[f], defined as follows preserves various classes of univalent functions. Definition 16.5.4. Let f ∈ A, then for z ∈ 𝔻, the Libera operator L[f] is defined by z

L[f] =

2 ∫ f(t) dt. z 0

The following shows that L[f] preserves the important subclasses of univalent functions. Theorem 16.5.2 ([117]). If f ∈ A, then (i) If f ∈ S∗ ⇒ L[f] ⊂ S∗ , (ii) If f ∈ K ⇒ L[f] ⊂ K, (iii) If f ∈ C ⇒ L[f] ⊂ C. In 1969, Bernardi [15] generalized the Libera integral operator by showing that for f ∈ A, integrals of the form z

Lc [f] =

c+1 ∫ f(t)t c−1 dt, zc 0

with c = 1, 2, 3, . . . , also satisfy the conclusions of Theorem 16.5.2. Even more, (Lc [f])󸀠 ∈ P. In 1978, Miller et al. [149] proved a more general result, which again we only state.

16.5 Operators

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Theorem 16.5.3. Let ψ(z) = 1 + . . . and let ϕ(z) = 1 + . . . be analytic functions defined in 𝔻, with the property ψ(z)ϕ(z) ≠ 0 there. Let α, β, γ and δ be real constants such that α ⩾ 0,

β > 0,

δ ⩾ 0,

α + δ > 0,

and α + δ = β + γ. If there exists a nonnegative constant J satisfying J ⩾ γ + Re [ δ + Re [ where λ(J) ≡

zψ󸀠 (z) ], ψ(z)

β + γ > J,

zϕ󸀠 (z) ] ⩾ max{0, J − λ(J)}, ϕ(z)

1 β+γ−J J max { , }, 2 J β+γ−J

λ(0) = 0,

and if f ∈ S∗ , then there exists a function F ∈ S∗ , which satisfies the identity z

β+γ F(z) ≡ [ γ ∫ f α (t)ϕ(t)t δ−1 dt] z ψ(z) 0 ] [

1/β

= z+....

(16.5.1)

In (16.5.1) all powers are principal ones. A weaker form of the above theorem, which is easier to apply, is the following corollary, which follows directly from Theorem 16.5.3. Corollary 16.5.1. Let ψ(z) = 1 + . . . and let ϕ(z) = 1 + . . . be analytic functions defined in 𝔻, with ψ(z) and ϕ(z) ≠ 0 there. Let α, β, γ, δ, and J be real constants satisfying the conditions α > 0, (α + δ) = (β + γ), and δ + Re [

zϕ󸀠 (z) zψ󸀠 (z) ] ⩾ J ⩾ γ + Re [ ] ϕ(z) ψ(z)

(z ∈ 𝔻).

(16.5.2)

If f ∈ S∗ , then F(z) given by (16.5.1), is also in S∗ . It is also possible to deduce the following. Theorem 16.5.4. If β > 0, γ ⩾ 0 and f ∈ S∗ , then F(z) defined by z

β+γ F(z) ≡ [ γ ∫ f β (t)t γ−1 dt] z 0 ] [

1/β

= z+...

is again an element of S∗ . Proof. If we choose α = β, γ = δ, ϕ(t) ≡ ψ(z) ≡ 1, and J = γ, then (16.5.2) is satisfied, and hence, by Corollary 16.5.1, F ∈ S∗ .

234 | 16 Other Topics

We note that Theorem 16.5.4 was first proved by Singh [229] for β and γ positive integers. We end this section by noting the following, which is a simple consequence of Corollary 16.5.1 in the case α + δ = β + γ = 1, ϕ(t) ≡ ψ(z) ≡ 1, γ ⩽ J ⩽ δ, where J = 1 − α ⩾ 0. Theorem 16.5.5. If 0 ⩽ α ⩽ 1, α ⩽ β, and if f ∈ S∗ , then the function z

F(z) ≡ [z [ is again an element of S∗ .

β−1

f(t) α ] ∫( ) dt t 0 ]

1/β

= z+...

17 Open Problems In this final section we list some open problems that arose in the previous chapters. The list is by no means exhaustive, and we have tried to identify those open problems that we consider to be significant. There are, of course, many other interesting outstanding problems in the theory of univalent functions, which are outside the scope of this book; many of these can be found in the books of Duren [45], Goodman [58, 59], and Pommerenke [198].

Chapter 3 1.

Use Theorem 3.1.4 to find an expression for p n in terms of p1 , when n ⩾ 5.

Chapter 4 2. 3. 4. 5. 6.

Find the sharp bound for H3 (1) when f ∈ S∗ . In Theorem 4.1.17, find sharp bounds for |a n |, when n ⩾ 6. Also improve and/or sharpen the estimate (4.1.30). In Theorem 4.2.5, find sharp expressions for D+n and D−n when n ⩾ 4. In Theorem 4.2.10, find sharp bounds for |A n | when n ⩾ 8. Solve the Zalcman conjecture for functions in the classes S∗ (A, B) and C(A, B).

Chapter 5 7.

In Theorem 5.3.3, find the sharp bound for |a2 a4 − a23 | when f ∈ C(α).

Chapter 6 8. In Theorem 6.2.1, find corresponding sharp lower bounds for the inequalities (6.2.1), (6.2.2) and (6.2.3). 9. In Theorem 6.2.2, is the inequality for L(1) sharp? 10. In Theorem 6.3.1, find sharp bounds for |a n |, when n ⩾ 5. 11. In Theorem 6.3.5, find sharp bounds for |A n |, when n ⩾ 5. 12. In Theorem 6.4.1, are the inequalities for |f(z)| and |f 󸀠 (z)| sharp? 13. In Theorem 6.4.1, find corresponding sharp lower bounds for |f(z)| and |f 󸀠 (z)|. Also, find sharp upper and lower bounds for |zf 󸀠 (z)/f(z)|. 14. In Theorem 6.4.2, find sharp inequalities for |a n | and |A n |, when n ⩾ 5. 15. In Theorem 6.5.2, find the sharp value of λ. https://doi.org/10.1515/9783110560961-017

236 | 17 Open Problems

Chapter 7 16. Solve the Zalcman conjecture for the class Mα . 17. Find the best upper bound for |a n+1 − a n | when f ∈ Mα for α > 0. 18. If f ∈ Mα for α > 0, find the best upper bound for |n|a n | − m|a m ‖ for all positive n and m. 19. Find the best possible bounds for the logarithmic coefficients |γ n | for n ⩾ 4, when f ∈ Mα and α > 0.

Chapter 8 20. In Theorem 8.2.1, find sharp inequalities for |a n |, when n ⩾ 5. 21. In Theorem 8.2.3, find sharp inequalities for |δ n |, when n ⩾ 4. 22. In Theorem 8.2.4, find sharp inequalities for |A n |, when n ⩾ 5.

Chapter 9 23. Prove Conjecture 9.3.1. ̃ 24. For f ∈ K(α), show that |a4 | ⩽ (1 + α)(3 + 2α + α 2 )/3. (Note that sharp bounds for |a2 | and |a3 | are easily obtained using Theorem 3.1.7.) 25. Are the estimates (ii) and (iii) in Theorem 9.3.6 the best possible? If not, find them. 26. In Theorem 9.3.9, find the best upper bound for |γ n | for n ⩾ 4. 27. Find the best upper bound for |γ n | for n ⩾ 3 for the whole class of close-to-convex functions. 28. In Theorem 9.3.10, find the best upper bound for |γ n | for n ⩾ 4. 29. In Theorem 9.3.11, find the best upper bound for |γ n | for n ⩾ 4. 30. Find the best upper bounds for |γ n | for n ⩾ 4 for functions in the classes F2 and F3 , respectively. 31. In Remark 9.4.1, can we improve the index 3/2 (possibly to 1) in the estimate for L(r)? 32. In Remark 9.4.2, show that na n = O(√(A((n + 1)/n)) ) as n → ∞. 33. In Theorem 9.5.4, find sharp inequalities for |a n |, when n ⩾ 5. 34. In Theorem 9.5.5, find sharp inequalities for |A n |, when n ⩾ 5.

Chapter 10 35. When f ∈ B(α) and the area of f(𝔻) is finite, improve the index 1/2 in the estimate na n = O(log n)1/2 as n → ∞ in Remark 10.2.1, possibly to 0.

17 Open Problems

| 237

36. Is the index 3/2 in Corollary 10.2.2 the best possible? 37. In Theorem 10.2.4, find sharp bounds for |a n (λ)|, when n ⩾ 3.

Chapter 11 38. In Theorem 11.4.2, find sharp bounds for |a5 |, when 1/2 ⩽ α ⩽ 1. 39. In Theorem 11.4.3, find sharp bounds for |a6 |, when 1/2 ⩽ α ⩽ 1. 40. Prove Conjectures 11.4.1 and 11.4.2.

Chapter 12 41. Is Theorem 12.2.2 sharp when λ ≠ 1? If not, find the sharp version. 42. Extend Theorem 12.3.2 to prove that (12.3.4) holds for n ⩾ 5. 43. Find the sharp upper bound for |γ n | for n ⩾ 1 when f ∈ U(λ).

Chapter 13 44. Let f be convex univalent and g be starlike univalent function such that f ∗ g is convex. Also, let h1 and h2 be analytic in the unit disk such that h1 ≺ f and h2 ≺ g. Show that h1 ∗ h2 ≺ f ∗ g. Conjectured in [187]. 45. Is Theorem 13.2.3 sharp, i.e., is 1/4 the largest constant so that the inclusion holds? If not, find the sharp version.

Chapter 14 46. In Theorem 14.1.3, find the Clunie constant, or improve on the known upper and lower bounds. 47. In Theorem 14.2.2, is the constant 2√2 the best possible, or can this be reduced to 2? 48. In Theorem 14.2.4, improve on the constant 21. 2 49. In Theorem 14.2.6, find a constant C(α) such that ∑∞ n=1 n|c n | ⩽ C(α).

Chapter 16 50. Prove the Clunie–Shiel-Small conjecture.

Concluding Remarks In this book, we have attempted to provide an introduction to some of the fundamental ideas and basic results in the theory of univalent functions, paying particular attention to major subclasses such as C, S∗ , and K. In an area with over a century of research activity, it has naturally not been possible to include a comprehensive account of the literally hundreds, if not thousands of papers that have appeared in the literature. As was mentioned in the Introduction, the book is directed at those starting their research careers, while at the same time, providing an up-to-date summary of a many of the basic properties and results for the important subclasses of univalent functions. Over the years, there have been many examples of authors considering rather simple extensions to the more important results for the subclasses considered in this book. These have consisted mostly of either considering subclasses, or simple generalization of classes such as C, S∗ , or K. These papers, although producing new results, have often been of a rather trivial nature, merely copying known methods of proof, and resulting in no real significant advance. On the other hand, some of these results, in particular when considering operators, have found interesting and intriguing connections between univalent functions, and, for example, special functions such as the hypergeometric functions. In general, finding new and novel methods of proof in mathematics usually produces major advances, and there have, of course, been a multitude of such ideas developed, many of which are outside the scope of this book. It has often been said that analysis can be compared with “a bag of tricks,” a prime example being Clunie’s method dealing with coefficient problems, and another the use of the maximum– minimum arc technique. Finding new tricks is, of course, difficult, but usually reaps great rewards. However, using known methods in a creative way can often lead to worthwhile results, and we have given many examples in this book. Thus, we have attempted to show how basic questions can be addressed by using important known techniques, in the hope that new researchers will concentrate on finding solutions to significant outstanding problems, rather than being concerned with less interesting ones.

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Index Alexander’s theorem 16, 65, 83 alpha (α) – convex functions 99 – coefficient representation 109 – distortion theorem 103 – general coefficient estimate 106 – growth result 105 – growth theorem 101 – inclusion properties 100 – integral representation 100 – second coefficient estimate 106 alpha (α) – convex Koebe function 101 Area theorem – analogue of 39 area theorem 3 Bazilevič function of type (α, β) 18, 153 – characterization 19 Bazilevič functions, class B(α) 19, 153 – coefficients of powers of functions 159 – estimate of L(r) 157 – estimate of na n 154 – Fekete–Szegő theorem 162 – logarithmic coefficients 162 – sufficient condition 163 Bazilevič functions, class B1 (α) 19, 165 – estimate of |a2 |, |a3 |, |a4 | 171 – estimate of |a5 | 172 – estimate of |a6 | 173 – estimate of L(r) 169 – estimate of n|a n | 169 – Fekete–Szegő theorem 174 – growth and distortion theorems 165 Bernardi operator 232 Bieberbach conjecture 4 Bieberbach theorem 4 bi-univalent functions 225 class defined by Silverman 81, 97 class defined by Singh and Tuneski 97 class R of normalized functions whose derivative has positive real part 165 class S∗L 55 – coefficient estimates 55 – second Hankel determinant 58 – subordination 55 https://doi.org/10.1515/9783110560961-020

class U 20, 179 – characterization 180 – equivalency result 184 – Koebe transform 187 – relation with S∗ 187 – sufficient condition 182 class U(λ) 20, 179 – criteria for starlikeness 186 – estimate of |a2 | 188 – estimate of |a n | 191 – estimates of logarithmic coefficients 192 – geometrical properties 179 – subordination 181 – sufficient condition 183 closed convex hull 128 closed convex hull of K 128 close-to-convex function with argument 0 – coefficients of powers of functions 159 – Fekete–Szegő theorem 126 close-to-convex functions 16, 121 – Bieberbach conjecture 33, 125 – difference of coefficients 132 – estimates of L(r) 144, 147 – Fekete–Szegő theorem 126 – logarithmic coefficients 136 – Robertson’s conjecture 135 – with respect to odd starlike functions 141 – logarithmic coefficients 142 – with respect to the Koebe function with argument β 140 – with respect to the Koebe function with argument 0 140 – logarithmic coefficients 140 – second Hankel determinant 143 close-to-convex functions with argument β 121 Clunie–Jack lemma 35 Clunie–Shiel-Small conjecture 225 convex function 13 – analytic description 15 convex functions 58 – Bieberbach conjecture 33 – difference of coefficients 61 – Fekete–Szegő theorem 63 – growth and distortion theorems 59 – inverse coefficients 63

Index |

– second Hankel determinant 63 – third Hankel determinant 63 convex functions introduced by Janowski 65 convex functions of order α 65 – Fekete–Szegő problem 76 – growth and distortion theorems 66 – second Hankel determinant 77 – sufficient condition 80 convex hull 128 convolution 195 delta (δ) – neighborhood 203 difference of coefficients 39 differential subordination (first order) 229 – best dominant of 229 – dominant of 229 – solution of 229 extremal subset 128 extreme point 128 factorial polynomial 107, 108 Fekete–Szegő theorem 5 functions of bounded boundary rotation 226 functions with positive real part 22 – coefficient estimates 25 – coefficient inequalities 27, 28 – coefficients inequalities 25, 28 – distortion theorem 23 – growth theorem 23 – Herglotz representation theorem 31 – necessary and sufficient condition 27 functions with real part > α 22 gamma-starlike functions 112 – Fekete–Szegő theorem 115 – first, second and third logarithmic coefficient estimates 116 – inclusion property 112 – second Hankel determinant 119 – second, third, and fourth coefficient estimates 113 – second, third, and fourth inverse coefficient estimates 119 Gaussian hypergeometric function 94 general α-convex Koebe function 109 – coefficients representation 110 Goluzin’s theorem 33

Hadamard product 195 Hankel determinant (qth) 8 harmonic function 224 integral mean 32 inverse function 7 Jack’s conjecture 67 – solution 73 Koebe 1/4 theorem 5 – analogue for convex functions 60 – analogue for α-convex functions 104 Libera integral operator 75 Lindelöf subordination principle 32 linearly accessible domain / function 122 Littlewood’s subordination theorem 32 locally univalent 224 Loewner equation 215 Loewner theory 215 logarithmic coefficients 6 majorization 75 Marx–Strohhäcker theorem 67 meromorphic Bazilevič functions 213 – coefficient estimates 213, 214 meromorphic close-to-convex functions 211 – coefficient estimates 211–213 meromorphic functions 3, 205 – estimate of |b 3 | 206 – estimates of coefficients 206 – growth and distortion theorems 209 – inverse coefficients 209 meromorphic starlike functions 210 – coefficient estimates 210 Milin’s conjecture 7 Mocanu’s theorem 34 Nunokawa’s lemma 34 operators 230 Ozaki close-to-convex functions 149 – coefficient estimates 149 – distortion theorem 150 – growth and area estimates 151 Pólya–Schoenberg conjecture 195

251

252 | Index

radial limit of f(re iθ ) 45, 47 radius of convexity of univalent functions 59 Riemann mapping theorem 1 right hand half of the lemniscate of Bernoulli 55 Robertson’s conjecture 134 Rogosinski’s theorem 32 rotational invariance 20 Ruscheweyh operator 231 Sălăgean operator 230 Schwarz lemma 14 – improvement 44 Schwarz–Pick lemma 24, 189 sense preserving 224 starlike function 13 – analytic description 14 starlike functions – Bieberbach conjecture 33 – coefficient theorems 37 – difference of coefficients 40 – estimates of L(r) 51, 52 – Fekete–Szegő theorem 41 – refined growth theorems 42 – second Hankel determinant 41 – third Hankel determinant 42 starlike functions introduced by Janowski 65 starlike functions of order α 65 – coefficient theorem 75 – Fekete–Szegő problem 76 – second Hankel determinant 77 – sufficient condition 79 strongly bi-starlike functions 225 strongly close-to-convex functions of order α 121 – coefficient estimates 125 – distortion theorem 124 – estimates for L(r) 143 – necessary and sufficient condition 122 strongly convex functions 83 – Fekete–Szegő theorem 95 – growth and distortion theorems 93 – second Hankel determinant 95 – second, third, and fourth coefficient estimates 94 – second, third, and fourth inverse coefficient estimates 94

strongly Ozaki close-to-convex functions 151 – second, third, and fourth coefficient estimates 152 – second, third, and fourth inverse coefficient estimates 152 strongly starlike functions 83 – Fekete–Szegő theorem 89 – growth and distortion theorems 83 – logarithmic coefficients 91 – second Hankel determinant 90 – second, third, and fourth coefficient estimate 87 – second, third, and fourth inverse coefficient estimate 91 – the length L(1) 85 subclasses of starlike functions 54 – growth and distortion theorems 54 subordination 31 third Lebedev–Milin inequality 6 Todorov conjecture 111 Toeplitz determinant 26 – important consequences 26 transformations of univalent functions 2 – conjugation 2 – dilation 2 – disk automorphism 2 – inverse transformation 3 – omitted-value transformation 2 – range transformation 2 – rotation 2 – square root transformation 2 univalent functions 1 – distortion theorem 9 – growth theorem 10 – inverse coefficient estimates 7 Valiron–Landau lemma 216 Wilf’s conjecture 201 Zalcman Conjecture 127

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