differential subordination and coefficients problems of certain

DIFFERENTIAL SUBORDINATION AND

COEFFICIENTS PROBLEMS OF CERTAIN ANALYTIC

FUNCTIONS

SHAMANI A/P SUPRAMANIAM

UNIVERSITI SAINS MALAYSIA

2014

DIFFERENTIAL SUBORDINATION AND

COEFFICIENTS PROBLEMS OF CERTAIN ANALYTIC

FUNCTIONS

by

SHAMANI A/P SUPRAMANIAM

Thesis Submitted in fulfilment of the requirements for the Degree ofDoctor of Philosophy in Mathematics

September 2014

ACKNOWLEDGEMENT

First and foremost, I am very grateful to God, for this thesis would not have

been possible without His grace and blessings.

I am most indebted to my supervisor, Dr. Lee See Keong, for his continuous

support and encouragement. Without his guidance on all aspects of my research,

I could not have completed my dissertation.

I would like to express my special thanks to my co-supervisor and the head

of the Research Group in Complex Function Theory USM, Prof. Dato’ Rosihan

M. Ali, for his valuable suggestions, encouragement and guidance throughout my

studies. I also deeply appreciate his financial assistance for this research through

generous grants, which supported overseas conference as well as my living expenses

during the term of my study.

I express my sincere gratitude to my field-supervisor, Prof. V. Ravichandran,

Professor at the Department of Mathematics, University of Delhi, for his constant

guidance and support to complete the writing of this thesis as well as the challeng-

ing research that lies behind it. I am also thankful to Prof. K. G. Subramaniam

and to other members of the Research Group in Complex Function Theory at USM

for their help and support especially my friends Abeer, Chandrashekar, Mahnaz,

Maisarah and Najla.

Also, my sincere appreciation to Prof. Ahmad Izani Md. Ismail, the Dean of

the School of Mathematical Sciences USM, and the entire staff of the school and

the authorities of USM for providing excellent facilities to me.

My research is supported by MyBrain (MyPhD) programme of the Ministry

i

of Higher Education, Malaysia and it is gratefully acknowledged.

Last but not least, I express my love and gratitude to my beloved family and

husband, for their understanding, encouragement and endless love. Not forgetting

Shakthie and Sarvhesh for always cheering up my dreary days.

ii

TABLE OF CONTENTS

Page

ACKNOWLEDGMENTS i

SYMBOLS iv

ABSTRAK vii

ABSTRACT viii

CHAPTER

1 INTRODUCTION 11.1 Univalent function 11.2 Subclasses of univalent functions 41.3 Differential subordination 81.4 Scope of thesis 10

2 COEFFICIENTS FOR BI-UNIVALENT FUNCTIONS 122.1 Introduction and preliminaries 122.2 Kedzierawski type results 132.3 Second and third coefficients of functions f when f ∈ ST (α, ϕ) and

g ∈ ST (β, ψ), or g ∈M(β, ψ), or g ∈ L(β, ψ) 142.4 Second and third coefficients of functions f when f ∈M(α, ϕ) and

g ∈M(β, ψ), or g ∈ L(β, ψ) 252.5 Second and third coefficients of functions f when f ∈ L(α, ϕ) and

g ∈ L(β, ψ) 312.6 Second and third coefficients of functions f when f ∈ HSB(ϕ) 34

3 BOUNDS FOR THE SECOND HANKEL DETERMINANTOF UNIVALENT FUNCTIONS 363.1 Introduction and preliminaries 363.2 Second Hankel determinant of Ma-Minda starlike/convex functions 383.3 Further results on the second Hankel determinant 49

4 APPLICATIONS OF DIFFERENTIAL SUBORDINATIONFOR FUNCTIONS WITH FIXED SECOND COEFFICIENT 604.1 Introduction and preliminaries 604.2 Subordinations for starlikeness 624.3 Subordinations for univalence 76

iii

5 CONVEXITY OF FUNCTIONS SATISFYING CERTAIN DIF-FERENTIAL INEQUALITIES AND INTEGRAL OPERATORS 795.1 Introduction and preliminaries 795.2 Convexity of functions satisfying second-order differential inequalities 815.3 Convexity of functions satisfying third-order differential inequalities 91

6 CLOSE-TO-CONVEXITY AND STARLIKENESS OFANALYTIC FUNCTIONS 996.1 Introduction and preliminaries 996.2 Close-to-convexity and Starlikeness 100

CONCLUSION 111

REFERENCES 113

PUBLICATIONS 127

iv

SYMBOLS

Symbol Description Page

Ap,n Class of all analytic functions f of the form 99

f(z) = zp + an+pzn+p + an+p+1z

n+p+1 + . . . (z ∈ D)

Ap Class of all p-valent analytic functions f of the form 2

f(z) = zp +∑∞

k=p+1 akzk (z ∈ D)

An,b Class of all functions f(z) = z + bzn+1 + an+2zn+2 + · · · 60

where b is a fixed non-negative real number.

A Class of analytic functions f of the form 1

f(z) = z +∑∞

k=2 akzk (z ∈ D)

C Complex plane 1

CV Class of convex functions in A 4

CV(α) Class of convex functions of order α in A 5

CV(ϕ)

f ∈ A : 1 +zf ′′(z)f ′(z) ≺ ϕ(z)

8

CCV Class of close-to-convex functions in A 6

D Open unit disk z ∈ C : |z| < 1 1

D∗ Open punctured unit disk z ∈ C : 0 < |z| < 1 6

∂D Boundary of unit disk D 9

Gα(ϕ)

f ∈ A : (1− α)f ′(z) + α(1 +

zf ′′(z)f ′(z)

)≺ ϕ(z)

54

H(D) Class of all analytic functions in D 1

H[a, n] = Hn(a) Class of all analytic functions f in D of the form 1

f(z) = a + anzn + an+1zn+1 + · · ·

v

H := H[1, 1] Class of analytic functions f in D of the form 1

f(z) = 1 + a1z + a2z2 + · · ·

Hµ,n Class of analytic functions p on D of the form 60

p(z) = 1 + µzn + pn+1zn+1 + · · ·

HSB(ϕ)f ∈ SB : f ′(z) ≺ ϕ(z) and 34

g′(w) ≺ ϕ(w), g(w) := f−1(w)

Hq(n) Hankel determinants of functions f ∈ A 36

L(α, ϕ)

f ∈ S :

(zf ′(z)f(z)

)α (1 +

zf ′′(z)f ′(z)

)1−α≺ ϕ(z)

13

M(α, ϕ)

f ∈ S : (1− α)zf ′(z)f(z) + α

(1 +

zf ′′(z)f ′(z)

)≺ ϕ(z)

13

P p ∈ H : with Re p(z) > 0, z ∈ D 6

P(α) p ∈ H : with Re p(z) > α, z ∈ D 6

R Set of all real numbers 2

Re Real part of a complex number 4

Rτγ(ϕ)

f ∈ A : 1 + 1

τ (f ′(z) + γzf ′′(z)− 1) ≺ ϕ(z)

49

S Class of all normalized univalent functions f in A 2

ST Class of starlike functions in A 5

ST (α) Class of starlike functions of order α in A 5

ST (ϕ)

f ∈ A :zf ′(z)f(z) ≺ ϕ(z)

8

ST (α, ϕ)

f ∈ S :

zf ′(z)f(z) + α

z2f ′′(z)f(z) ≺ ϕ(z)

13

Σ Class of all meromorphic functions f of the form 6

f(z) = f(z) = 1z +

∑∞n=0 anzn

Σn,b Class of all meromorphic functions f of the form 73

vi

f(z) = 1z + bzn + an+1z

n+1 + · · · (b ≤ 0)

≺ Subordinate to 6

SB Class of bi-univalent functions 12

Ψn[Ω, q] Class of admissible functions for differential subordination 9

Ψµ,n[Ω] Class of admissible functions for fixed second coefficient 60

vii

SUBORDINASI PEMBEZA DAN MASALAH PEKALI UNTUKFUNGSI-FUNGSI ANALISIS

ABSTRAK

Lambangkan A sebagai kelas fungsi analisis ternormal pada cakera unit D

berbentuk f(z) = z +∑∞

n=2 anzn. Fungsi f dalam A adalah univalen jika fungsi

tersebut ialah pemetaan satu ke satu. Tesis ini mengkaji lima masalah penye-

lidikan.

Fungsi f ∈ A dikatakan dwi univalen dalam D jika kedua-dua fungsi f dan

songsangannya f−1 adalah univalen dalam D. Anggaran pekali awal, |a2| dan |a3|,fungsi dwi univalen akan dikaji untuk f dan f−1 yang masing-masing terkandung

di dalam subkelas fungsi univalen tertentu. Seterusnya, batas penentu Hankel

kedua H2(2) = a2a4 − a23 untuk fungsi analisis f dengan zf ′(z)/f(z) dan 1 +

zf ′′(z)/f ′(z) subordinat kepada suatu fungsi analisis tertentu diperoleh.

Bermotivasikan kerja terdahulu dalam subordinasi pembeza peringkat kedua

untuk fungsi analisis dengan pekali awal tetap, syarat cukup bak-bintang dan uni-

valen untuk suatu subkelas fungsi berpekali kedua tetap ditentukan. Kemudian,

syarat cukup cembung untuk fungsi yang pekali keduanya tidak ditetapkan dan

yang memenuhi ketaksamaan pembeza peringkat kedua dan ketiga tertentu diper-

oleh.

Akhir sekali, subkelas fungsi multivalen yang memenuhi syarat bak-bintang

dan hampir cembung dikaji.

Beberapa aspek permasalahan dalam teori fungsi univalen dibincangkan dalam

tesis ini dan hasil-hasil menarik diperoleh.

viii

DIFFERENTIAL SUBORDINATION AND COEFFICIENTSPROBLEMS OF CERTAIN ANALYTIC FUNCTIONS

ABSTRACT

Let A be the class of normalized analytic functions f on the unit disk D, in

the form f(z) = z+∑∞

n=2 anzn. A function f in A is univalent if it is a one-to-one

mapping. This thesis discussed five research problems.

A function f ∈ A is said to be bi-univalent in D if both f and its inverse

f−1 are univalent in D. Estimates on the initial coefficients, |a2| and |a3|, of

bi-univalent functions f are investigated when f and f−1 respectively belong to

some subclasses of univalent functions. Next, the bounds for the second Hankel

determinant H2(2) = a2a4 − a23 of analytic function f for which zf ′(z)/f(z) and

1 + zf ′′(z)/f ′(z) is subordinate to certain analytic function are obtained.

Motivated by the earlier work on second order differential subordination for

analytic functions with fixed initial coefficient, the sufficient conditions for star-

likeness and univalence for a subclass of functions with fixed second coefficient

are obtained. Then, without fixing the second coefficient, the sufficient condition

for convexity of these functions satisfying certain second order and third order

differential inequalities are determined.

Lastly, the close-to-convexity and starlikeness of a subclass of multivalent func-

tions are investigated.

A few aspects of problems in univalent function theory is discussed in this

thesis and some interesting results are obtained.

ix

CHAPTER 1

INTRODUCTION

1.1 Univalent function

Let C be the complex plane and D := z ∈ C : |z| < 1 be the open unit

disk in C. A function f is analytic at a point z0 ∈ D if it is differentiable in

some neighborhood of z0 and it is analytic in a domain D if it is analytic at all

points in domain D. An analytic function f is said to be univalent in a domain

if it provides a one-to-one mapping onto its image: f(z1) = f(z2) ⇒ z1 = z2.

Geometrically, this means that different points in the domain will be mapped into

different points on the image domain. An analytic function f is locally univalent

at a point z0 ∈ D if it is univalent in some neighborhood of z0. The well known

Riemann Mapping Theorem states that every simply connected domain (which is

not the whole complex plane C), can be mapped conformally onto the unit disk D.

Theorem 1.1 (Riemann Mapping Theorem) [29, p. 11] Let D be a simply con-

nected domain which is a proper subset of the complex plane. Let ζ be a given

point in D. Then there is a unique univalent analytic function f which maps D

onto the unit disk D satisfying f(ζ) = 0 and f ′(ζ) > 0.

In view of this theorem, the study of analytic univalent functions on a simply

connected domain can be restricted to the open unit disk D.

Let H(D) be the class of analytic functions defined on D. Let H[a, n] be the

subclass of H(D) consisting of functions of the form

f(z) = a + anzn + an+1zn+1 + · · ·

with H ≡ H[1, 1].

Also let A denote the class of all functions f analytic in the open unit disk

D, and normalized by f(0) = 0, and f ′(0) = 1. A function f ∈ A has the Taylor

1

series expansion of the form

f(z) = z +∞∑

n=2anzn (z ∈ D).

For a fixed p ∈ N := 1, 2, . . . , let Ap be the class of all analytic functions of the

form

f(z) = zp +∞∑

k=1

ak+pzk+p,

that are p-valent (multivalent) in the open unit disk, with A := A1.

The subclass of A consisting of univalent functions is denoted by S. The

function k given by

k(z) =z

(1− z)2=

∞∑

n=1nzn (z ∈ D)

is called the Koebe function, which maps D onto the complex plane except for a

slit along the half-line (−∞,−1/4], and is univalent. It plays a very important

role in the study of the class S. The Koebe function and its rotations e−iβk(eiβz),

for β ∈ R, are the only extremal functions for various problem in the class S. In

1916, Bieberbach [19] conjectured that for f ∈ S, |an| ≤ n, (n ≥ 2). He proved

only for the case when n = 2.

Theorem 1.2 (Bieberbach’s Conjecture) [19] If f ∈ S, then |an| ≤ 2 (n ≥ 2)

with equality if and only if f is the rotation of the Koebe function k.

For the cases n = 3, and n = 4 the conjecture was proved by Lowner [58] and

Garabedian and Schiffer [34], respectively. Later, Pederson and Schiffer [98] proved

the conjecture for n = 5, and for n = 6, it was proved by Pederson [97] and

Ozawa [95], independently. In 1985, Louis de Branges [20], proved the Bieberbach’s

conjecture for all the coefficients n.

2

Theorem 1.3 (de Branges Theorem or Bieberbach’s Theorem) [20] If f ∈ S,

then

|an| ≤ n (n ≥ 2),

with equality if and only if f is the Koebe function k or one of its rotations.

Bieberbach’s theorem has many important properties in univalent functions. These

include the well known covering theorem: If f ∈ S, then the image of D under f

contains a disk of radius 1/4.

Theorem 1.4 (Koebe One-Quarter Theorem) [29, p. 31] The range of every func-

tion f ∈ S contains the disk w ∈ C : |w| < 1/4.

The Distortion theorem, being another consequence of the Bieberbach theorem

gives sharp upper and lower bounds for |f ′(z)|.

Theorem 1.5 (Distortion Theorem) [29, p. 32] For each f ∈ S,

1− r

(1 + r)3≤ |f ′(z)| ≤ 1 + r

(1− r)3(|z| = r < 1).

The distortion theorem can be used to obtain sharp upper and lower bounds for

|f(z)| which is known as the Growth theorem.

Theorem 1.6 (Growth Theorem) [29, p. 33] For each f ∈ S,

r

(1 + r)2≤ |f(z)| ≤ r

(1− r)2(|z| = r < 1).

Another consequence of the Bieberbach theorem is the Rotation theorem.

Theorem 1.7 (Rotation Theorem) [29, p. 99] For each f ∈ S,

| arg f ′(z)| ≤

4sin−1r, r ≤ 1√2,

π + log r2

1−r2 , r ≥ 1√2,

3

where |z| = r < 1. The bound is sharp.

The Fekete-Szego coefficient functional also arises in the investigation of univalency

of analytic functions.

Theorem 1.8 (Fekete-Szego Theorem) [29, p. 104] For each f ∈ S,

|a3 − αa22| ≤ 1 + 2e−2α/(1−α), (0 < α < 1).

1.2 Subclasses of univalent functions

The long gap between the Bieberbach’s conjecture in 1916 and its proof by de

Branges in 1985 motivated researchers to consider classes defined by geometric

conditions. Notable among them are the classes of convex functions, starlike func-

tions and close-to-convex functions.

A set D in the complex plane is called convex if for every pair of points w1

and w2 lying in the interior of D, the line segment joining w1 and w2 also lies in

the interior of D, i.e.

tw1 + (1− t)w2 ∈ D for 0 ≤ t ≤ 1.

If a function f ∈ A maps D onto a convex domain, then f is a convex function.

The class of all convex functions in A is denoted by CV . An analytic description

of the class CV is given by

CV :=

f ∈ A : Re

(1 +

zf ′′(z)

f ′(z)

)> 0

.

Let w0 be an interior point of D. A set D in the complex plane is called starlike

with respect to w0 if the line segment joining w0 to every other point w ∈ D lies

4

in the interior of D, i.e.

(1− t)w + tw0 ∈ D for 0 ≤ t ≤ 1.

If a function f ∈ A maps D onto a starlike domain, then f is a starlike function.

The class of starlike functions with respect to origin is denoted by ST . Analyti-

cally,

ST :=

f ∈ A : Re

(zf ′(z)

f(z)

)> 0

.

In 1936, Robertson [105] introduced the concepts of convex functions of order

α and starlike functions of order α for 0 ≤ α < 1. A function f ∈ A is said to be

convex of order α if

Re

(1 +

zf ′′(z)

f ′(z)

)> α (z ∈ D),

and starlike of order α if

Re

(zf ′(z)

f(z)

)> α (z ∈ D).

These classes are respectively denoted by CV(α) and ST (α).

Note that CV(0) = CV and ST (0) = ST . An important relationship between

convex and starlike functions was first observed by Alexander [1] in 1915 and

known later as Alexander’s theorem.

Theorem 1.9 (Alexander’s Theorem) [29, p. 43] Let f ∈ A. Then f ∈ CV if and

only if zf ′ ∈ ST .

From this, it is easily proven that f ∈ CV(α) if and only if zf ′ ∈ ST (α).

Another subclass of S that has an important role in the study of univalent

functions is the class of close-to-convex functions introduced by Kaplan [45] in

1952. A function f ∈ A is close-to-convex in D if there is a convex function g and

5

a real number θ, −π/2 < θ < π/2, such that

Re

(eiθ f ′(z)

g′(z)

)> 0 (z ∈ D).

The class of all such functions is denoted by CCV . The subclasses of S, namely

convex, starlike and close-to-convex functions are related as follows:

CV ⊂ ST ⊂ CCV ⊂ S.

The well known Noshiro-Warschawski theorem states that a function f ∈ A with

positive derivative in D is univalent.

Theorem 1.10 [82,131] For some real α, if a function f is analytic in a convex

domain D and

Re(eiαf ′(z)

)> 0,

then f is univalent in D.

Kaplan [45] applied Noshiro-Warschawski theorem to prove that every close-to-

convex function is univalent.

The class of meromorphic functions is yet another subclass of univalent func-

tions. Let Σ denote the class of normalized meromorphic functions f of the form

f(z) =1

z+

∞∑

n=0anzn,

that are analytic in the punctured unit disk D∗ := z : 0 < |z| < 1 except for a

simple pole at 0.

A function f is said to be subordinate to F in D, written f(z) ≺ F (z), if

there exists a Schwarz function w, analytic in D with w(0) = 0, and |w(z)| < 1,

such that f(z) = F (w(z)). If the function F is univalent in D, then f ≺ F if

f(0) = F (0) and f(U) ⊆ F (U).

6

Let P be the class of all analytic functions p of the form

p(z) = 1 + p1z + p2z2 + · · · = 1 +

∞∑

n=1pnzn

with

Re p(z) > 0 (z ∈ D). (1.1)

Any function in P is called a function with positive real part, also known as

Caratheodory function. The following lemma is known for functions in P .

Lemma 1.1 [29] If the function p ∈ P is given by the series

p(z) = 1 + p1z + p2z2 + p3z

3 + · · · ,

then the following sharp estimate holds:

|pn| ≤ 2 (n = 1, 2, . . . ).

The above fact will be used often in the thesis especially in Chapters 2 and 3.

More generally, for 0 ≤ α < 1, we denote by P(α) the class of analytic functions

p ∈ P with

Re p(z) > α (z ∈ D).

In terms of subordination, the analytic condition (1.1) can be written as

p(z) ≺ 1 + z

1− z(z ∈ D).

This follows since the mapping q(z) = (1 + z)/(1− z) maps D onto the right-half

plane.

Ma and Minda [59] have given a unified treatment of various subclasses con-

sisting of starlike and convex functions by replacing the superordinate function

7

q(z) = (1 + z)/(1 − z) by a more general analytic function. For this purpose,

they considered an analytic function ϕ with positive real part on D with ϕ(0) = 1,

ϕ′(0) > 0 and ϕ maps the unit disk D onto a region starlike with respect to 1,

symmetric with respect to the real axis. The class of Ma-Minda starlike functions

denoted by ST (ϕ) consists of functions f ∈ A satisfying

zf ′(z)

f(z)≺ ϕ(z)

and similarly the class of Ma-Minda convex functions denoted by CV(ϕ) consists

of functions f ∈ A satisfying the subordination

1 +zf ′′(z)

f ′(z)≺ ϕ(z), (z ∈ D).

respectively.

1.3 Differential subordination

Recall that a function f is said to be subordinate to F in D, written f(z) ≺ F (z),

if there exists a Schwarz function w, analytic in D with w(0) = 0, and |w(z)| < 1,

such that f(z) = F (w(z)). If the function F is univalent in D, then f ≺ F if

f(0) = F (0) and f(U) ⊆ F (U).

The basic definitions and theorems in the theory of subordination and certain

applications of differential subordinations are stated in this section. The theory of

differential subordination were developed by Miller and Mocanu [61].

Let ψ(r, s, t; z) : C3×D→ C and let h be univalent in D. If p is analytic in D

and satisfies the second order differential subordination

ψ(p(z), zp′(z), z2p′′(z); z

)≺ h(z), (1.2)

8

then p is called a solution of the differential subordination. The univalent function

q is called a dominant of the solution of the differential subordination or more

simply dominant, if p ≺ q for all p satisfying (1.2). A dominant q1 satisfying

q1 ≺ q for all dominants q of (1.2) is said to be the best dominant of (1.2). The

best dominant is unique up to a rotation of D.

If p ∈ H[a, n], then p will be called an (a, n)-solution, q an (a, n)-dominant,

and q1 the best (a, n)-dominant. Let Ω ⊂ C and let (1.2) be replaced by

ψ(p(z), zp′(z), z2p′′(z); z

)∈ Ω, for all z ∈ D, (1.3)

where Ω is a simply connected domain containing h(D). Even though this is a

differential inclusion and ψ(p(z), zp′(z), z2p′′(z); z

)may not be analytic in D, the

condition in (1.3) shall also be referred as a second order differential subordination,

and the same definition of solution, dominant and best dominant as given above

can be extended to this generalization. The monograph [61] by Milller and Mocanu

provides more detailed information on the theory of differential subordination.

Denote by Q the set of functions q that are analytic and injective on D\E(q),

where

E(q) = ζ ∈ ∂D : limz→ζ

q(z) = ∞

and q′(ζ) 6= 0 for ζ ∈ ∂D\E(q).

The subordination methodology is applied to an appropriate class of admissible

functions. The following class of admissible functions was given by Miller and

Mocanu [61].

Definition 1.1 [61, Definition 2.3a, p. 27] Let Ω be a set in C, q ∈ Q and m be

a positive integer. The class of admissible functions Ψm[Ω, q] consists of functions

ψ : C3 × D → C satisfying the admissibility condition ψ(r, s, t; z) 6∈ Ω whenever

9

r = q(ζ), s = kζq′(ζ) and

Re

(t

s+ 1

)≥ k Re

(ζq′′(ζ)

q′(ζ)+ 1

),

z ∈ D, ζ ∈ ∂D\E(q) and k ≥ m. In particular, Ψ[Ω, q] := Ψ1[Ω, q].

The next theorem is the foundation result in the theory of first and second-order

differential subordinations.

Theorem 1.11 [61, Theorem 2.3b, p. 28] Let ψ ∈ Ψm[Ω, q] with q(0) = a. If

p ∈ H[a, n] satisfies

ψ(p(z), zp′(z), z2p′′(z); z

) ∈ Ω,

then p ≺ q.

1.4 Scope of thesis

This thesis will discuss five research problems. In Chapter 2, estimates on the

initial coefficients for bi-univalent functions f in the open unit disk with f and its

inverse g = f−1 satisfying the conditions that zf ′(z)/f(z) and zg′(z)/g(z) are both

subordinate to a univalent function whose range is symmetric with respect to the

real axis. Several related classes of functions are also considered, and connections

to earlier known results are made.

In Chapter 3, the bounds for the second Hankel determinant a2a4 − a23 of

analytic function f(z) = z + a2z2 + a3z

3 + · · · for which either zf ′(z)/f(z) or

1 + zf ′′(z)/f ′(z) is subordinate to certain analytic function are investigated. The

problem is also investigated for two other related classes defined by subordina-

tion. The classes introduced by subordination naturally include several well known

classes of univalent functions and the results for some of these special classes are

indicated. In particular, the estimates for the Hankel determinant of strongly

starlike, parabolic starlike, lemniscate starlike functions are obtained.

10

In Chapter 4, several well known results for subclasses of univalent functions

was extended to functions with fixed initial coefficient by using the theory of dif-

ferential subordination. Further applications of this subordination theory is given.

In particular, several sufficient conditions related to starlikeness, meromorphic

starlikeness and univalence of normalized analytic functions are derived.

In Chapter 5, the convexity conditions for analytic functions defined in the

open unit disk satisfying certain second-order and third-order differential inequal-

ities are obtained. As a consequence, conditions are also determined for convexity

of functions defined by following integral operators

f(z) =

∫ 1

0

∫ 1

0W (r, s, z)drds, and f(z) =

∫ 1

0

∫ 1

0

∫ 1

0W (r, s, t, z)drdsdt.

In the final chapter, several sufficient conditions for close-to-convexity and

starlikeness of a subclass of multivalent functions are investigated. Relevant con-

nections with previously known results are indicated.

11

CHAPTER 2

COEFFICIENTS FOR BI-UNIVALENT FUNCTIONS

2.1 Introduction and preliminaries

For functions f ∈ S, let f−1 be its inverse function. The Koebe one-quarter theo-

rem (Theorem 1.4) ensures the existence of f−1, that is, every univalent function

f has an inverse f−1 satisfying f−1(f(z)) = z, (z ∈ D) and

f(f−1(w)) = w, (|w| < r0(f), r0(f) ≥ 1/4) .

A function f ∈ A is said to be bi-univalent in D if both f and f−1 are univalent

in D. Let SB denote the class of bi-univalent functions defined in D. Examples of

functions in the class SB are z/(1− z) and − log(1− z).

In 1967, Lewin [51] introduced this class SB and proved that the bound for

the second coefficients of every f ∈ SB satisfies the inequality |a2| ≤ 1.51. He

also investigated SB1 ⊂ SB, the class of all functions f = φ ψ−1, where φ and

ψ map D onto domains containing D and φ′(0) = ψ′(0). For an example that

shows SB 6= SB1, see [23]. In 1969, Suffridge [122] showed that a function in

SB1 satisfies a2 = 4/3 and thus conjectured that |a2| ≤ 4/3 for all functions in

SB. Netanyahu [69], in the same year, proved this conjecture for a subclass of

SB1. In 1981, Styer and Wright [121] showed that a2 > 4/3 for some function

in SB, thus disproved the conjecture of Suffridge. For bi-univalent polynomial

f(z) = z+a2z2+a3z

3 with real coefficients, Smith [114] showed that |a2| ≤ 2/√

27

and |a3| ≤ 4/27 and the latter inequality being the best possible. He also showed

that if z + anzn is bi-univalent, then |an| ≤ (n − 1)n−1/nn with equality best

possible for n = 2, 3. Kedzierawski and Waniurski [47] proved the conjecture

of Smith [114] for n = 3, 4 in the case of bi-univalent polynomial of degree n.

Extending the results of Srivasta et al. [118], Frasin and Aouf [33] obtained estimate

12

of |a2| and |a3| for bi-univalent function f for which

(1− λ)f(z)

z+ λf ′(z) and (1− λ)

g(w)

w+ λg′(w) (g = f−1)

belongs to a sector in the half plane. Tan [125] improved Lewin’s result to |a2| ≤1.485. For 0 ≤ α < 1, a function f ∈ SB is bi-starlike of order α or bi-convex of

order α if both f and f−1 are respectively starlike or convex of order α. These

classes were introduced by Brannan and Taha [22]. They obtained estimates on

the initial coefficients for functions in these classes. For some open problems and

survey, see [35,115]. Bounds for the initial coefficients of several classes of functions

were also investigated in [7,8,24–26,33,39,48,60,64,67,108,117–120,126,133,134].

2.2 Kedzierawski type results

In 1985, Kedzierawski [46] considered functions f belonging to certain subclasses

of univalent functions while its inverse f−1 belongs to some other subclasses of

univalent functions. Among other results, he obtained the following.

Theorem 2.1 [46] Let f ∈ SB with Taylor series f(z) = z + a2z2 + · · · and

g = f−1. Then

|a2| ≤

1.5894 if f ∈ S, g ∈ S,

√2 if f ∈ ST , g ∈ ST ,

1.507 if f ∈ ST , g ∈ S,

1.224 if f ∈ CV, g ∈ S.

Consider the following classes investigated in [7, 8, 14].

Definition 2.1 Let ϕ : D→ C be analytic and ϕ(z) = 1 + B1z + B2z2 + · · · with

B1 > 0. For α ≥ 0, let

M(α, ϕ) :=

f ∈ S : (1− α)

zf ′(z)

f(z)+ α

(1 +

zf ′′(z)

f ′(z)

)≺ ϕ(z)

,

13

L(α, ϕ) :=

f ∈ S :

(zf ′(z)

f(z)

)α (1 +

zf ′′(z)

f ′(z)

)1−α

≺ ϕ(z)

,

ST (α, ϕ) :=

f ∈ S :

zf ′(z)

f(z)+ α

z2f ′′(z)

f(z)≺ ϕ(z)

.

Suppose that f is given by

f(z) = z +∞∑

n=2anzn, (2.1)

then it is known that g = f−1 has the expression

g(w) = f−1(w) = w − a2w2 + (2a2

2 − a3)w3 + · · · .

Motivated by Theorem 2.1, we will consider the following cases and then will

obtain the estimates for the second and third coefficients of functions f :

1. f ∈ ST (α, ϕ) and g ∈ ST (β, ψ), or g ∈M(β, ψ), or g ∈ L(β, ψ),

2. f ∈M(α, ϕ) and g ∈M(β, ψ), or g ∈ L(β, ψ),

3. f ∈ L(α, ϕ) and g ∈ L(β, ψ),

where ϕ and ψ are analytic functions of the form

ϕ(z) = 1 + B1z + B2z2 + B3z

3 + · · · , (B1 > 0) (2.2)

and

ψ(z) = 1 + D1z + D2z2 + D3z

3 + · · · , (D1 > 0). (2.3)

2.3 Second and third coefficients of functions f when f ∈ ST (α, ϕ) and

g ∈ ST (β, ψ), or g ∈M(β, ψ), or g ∈ L(β, ψ)

We begin with the cases for f ∈ ST (α, ϕ).

14

Theorem 2.2 Let f ∈ SB and g = f−1. If f ∈ ST (α, ϕ) and g ∈ ST (β, ψ),

then

|a2| ≤ B1D1

√B1(1 + 3β) + D1(1 + 3α)√

|ρB21D2

1 − (1 + 2α)2(1 + 3β)(B2 −B1)D21 − (1 + 2β)2(1 + 3α)(D2 −D1)B2

1 |(2.4)

and

2ρ|a3| ≤ B1(3 + 10β) + D1(1 + 2α) + (3 + 10β)|B2−B1|+(1 + 2β)2B2

1 |D2 −D1|D2

1(1 + 2α)(2.5)

where ρ := 2 + 7α + 7β + 24αβ.

Proof. Since f ∈ ST (α, ϕ) and g ∈ ST (β, ψ), there exist analytic functions u, v :

D→ D, with u(0) = v(0) = 0, satisfying

zf ′(z)

f(z)+

αz2f ′′(z)

f(z)= ϕ(u(z)) and

wg′(w)

g(w)+

βw2g′′(w)

g(w)= ψ(v(w)). (2.6)

Define the functions p1 and p2 by

p1(z) :=1 + u(z)

1− u(z)= 1+c1z+c2z

2+· · · and p2(z) :=1 + v(z)

1− v(z)= 1+b1z+b2z

2+· · · ,

or, equivalently,

u(z) =p1(z)− 1

p1(z) + 1=

1

2

(c1z +

(c2 −

c212

)z2 + · · ·

)(2.7)

and

v(z) =p2(z)− 1

p2(z) + 1=

1

2

(b1z +

(b2 −

b212

)z2 + · · ·

). (2.8)

Then p1 and p2 are analytic in D with p1(0) = 1 = p2(0). Since u, v : D→ D, the

functions p1 and p2 have positive real part in D, and thus |bi| ≤ 2 and |ci| ≤ 2

15

(Lemma 1.1). In view of (2.6), (2.7) and (2.8), it is clear that

zf ′(z)

f(z)+

αz2f ′′(z)

f(z)= ϕ

(p1(z)− 1

p1(z) + 1

)and

wg′(w)

g(w)+

βw2g′′(w)

g(w)= ψ

(p2(w)− 1

p2(w) + 1

).

(2.9)

Using (2.7) and (2.8) together with (2.2) and (2.3), it is evident that

ϕ

(p1(z)− 1

p1(z) + 1

)= 1 +

1

2B1c1z +

(1

2B1

(c2 −

c212

)+

1

4B2c

21

)z2 + · · · (2.10)

and

ψ

(p2(w)− 1

p2(w) + 1

)= 1 +

1

2D1b1w +

(1

2D1

(b2 −

b212

)+

1

4D2b

21

)w2 + · · · . (2.11)

Since

zf ′(z)

f(z)+

αz2f ′′(z)

f(z)= 1 + a2(1 + 2α)z +

(2(1 + 3α)a3 − (1 + 2α)a2

2

)z2 + · · ·

and

wg′(w)

g(w)+

βw2g′′(w)

g(w)= 1− (1 + 2β)a2w +

((3 + 10β)a2

2 − 2(1 + 3β)a3

)w2 + · · · ,

it follows from (2.9), (2.10) and (2.11) that

a2(1 + 2α) =1

2B1c1, (2.12)

2(1 + 3α)a3 − (1 + 2α)a22 =

1

2B1

(c2 −

c212

)+

1

4B2c

21, (2.13)

−(1 + 2β)a2 =1

2D1b1 (2.14)

16

and

(3 + 10β)a22 − 2(1 + 3β)a3 =

1

2D1

(b2 −

b212

)+

1

4D2b

21. (2.15)

It follows from (2.12) and (2.14) that

b1 = −B1(1 + 2β)

D1(1 + 2α)c1. (2.16)

Multiplying (2.13) with (1 + 3β) and (2.15) with (1 + 3α), and adding the results

give

a22((1 + 3α)(3 + 10β)− (1 + 3β)(1 + 2α)) =

1

2B1(1 + 3β)c2 +

1

2D1(1 + 3α)b2

+1

4c21(1 + 3β)(B2 −B1) +

1

4b21(1 + 3α)(D2 −D1).

Substituting c1 from (2.12) and b1 from (2.16) in the above equation give

a22((1 + 3α)(3 + 10β)− (1 + 3β)(1 + 2α))

− a22

((1 + 3β)(1 + 2α)2(B2 −B1)

B21

+(1 + 2β)2(1 + 3α)(D2 −D1)

D21

)

=1

2B1(1 + 3β)c2 +

1

2D1(1 + 3α)b2

which lead to

a22 =

B21D2

1[B1(1 + 3β)c2 + D1(1 + 3α)b2]

2[ρB21D2

1 − (1 + 2α)2(1 + 3β)(B2 −B1)D21 − (1 + 2β)2(1 + 3α)(D2 −D1)B

21 ]

,

where ρ := 2 + 7α + 7β + 24αβ, which, in view of |b2| ≤ 2 and |c2| ≤ 2, gives us

the desired estimate on |a2| as asserted in (2.4).

Multiplying (2.13) with (3 + 10β) and (2.15) with (1 + 2α), and adding the

17

results give

2((1 + 3α)(3 + 10β)− (1 + 3β)(1 + 2α))a3 =1

2B1(3 + 10β)c2 +

1

2D1(1 + 2α)b2

+c214

(3 + 10β)(B2 −B1) +b214

(D2 −D1)(1 + 2α).

Substituting b1 from (2.16) in the above equation lead to

2ρa3 =1

2[B1(3 + 10β)c2 + D1(1 + 2α)b2]

+c214

[(3 + 10β)(B2 −B1) +

(1 + 2β)2B21(D2 −D1)

D21(1 + 2α)

],

and this yields the estimate given in (2.5).

Remark 2.1 When α = β = 0 and B1 = B2 = 2, D1 = D2 = 2, inequality (2.4)

reduces to the second result in Theorem 2.1.

In the case when β = α and ψ = ϕ, Theorem 2.2 reduces to the following

corollary.

Corollary 2.1 Let f given by (2.1) and g = f−1. If f, g ∈ ST (α, ϕ), then

|a2| ≤B1√

B1√|B2

1(1 + 4α) + (B1 −B2)(1 + 2α)2|, (2.17)

and

|a3| ≤B1 + |B2 −B1|

(1 + 4α). (2.18)

For ϕ given by

ϕ(z) =

(1 + z

1− z

)γ

= 1 + 2γz + 2γ2z2 + · · · (0 < γ ≤ 1),

we have B1 = 2γ and B2 = 2γ2. Hence, when α = 0, the inequality (2.17) reduces

to the following result.

18

Corollary 2.2 [22, Theorem 2.1] Let f given by (2.1) be in the class of strongly

bi-starlike functions of order γ, 0 < γ ≤ 1. Then

|a2| ≤2γ√1 + γ

.

On the other hand, when α = 0 and

ϕ(z) =1 + (1− 2β)z

1− z= 1 + 2(1− β)z + 2(1− β)z2 + · · ·

so that B1 = B2 = 2(1 − β), the inequalities in (2.17) and (2.18) reduce to the

following result.

Corollary 2.3 [22, Theorem 3.1] Let f given by (2.1) be in the class of bi-starlike

functions of order β, 0 < β ≤ 1. Then

|a2| ≤√

2(1− β) and |a3| ≤ 2(1− β).

Theorem 2.3 Let f ∈ SB and g = f−1. If f ∈ ST (α, ϕ) and g ∈M(β, ψ), then

|a2| ≤ B1D1

√B1(1 + 2β) + D1(1 + 3α)√

|ρB21D2

1 − (1 + 2α)2(1 + 2β)(B2 −B1)D21 − (1 + β)2(1 + 3α)(D2 −D1)B2

1 |(2.19)

and

2ρ|a3| ≤ B1(3+5β)+D1(1+2α)+(3+5β)|B2−B1|+(1 + β)2B2

1 |D2 −D1|D2

1(1 + 2α)(2.20)

where ρ := 2 + 7α + 3β + 11αβ.

Proof. Let f ∈ ST (α, ϕ) and g ∈ M(β, ψ), g = f−1. Then there exist analytic

19

functions u, v : D→ D, with u(0) = v(0) = 0, such that

zf ′(z)

f(z)+

αz2f ′′(z)

f(z)= ϕ(u(z)) and (1−β)

wg′(w)

g(w)+β

(1 +

wg′′(w)

g′(w)

)= ψ(v(w)),

(2.21)

Since

zf ′(z)

f(z)+

αz2f ′′(z)

f(z)= 1 + a2(1 + 2α)z + (2(1 + 3α)a3 − (1 + 2α)a2

2)z2 + · · ·

and

(1−β)wg′(w)

g(w)+β

(1 +

wg′′(w)

g′(w)

)= 1−(1+β)a2w+((3+5β)a2

2−2(1+2β)a3)w2+· · · ,

equations (2.10), (2.11) and (2.21) yield

a2(1 + 2α) =1

2B1c1, (2.22)

2(1 + 3α)a3 − (1 + 2α)a22 =

1

2B1

(c2 −

c212

)+

1

4B2c

21, (2.23)

−(1 + β)a2 =1

2D1b1 (2.24)

and

(3 + 5β)a22 − 2(1 + 2β)a3 =

1

2D1

(b2 −

b212

)+

1

4D2b

21. (2.25)


b1 = − B1(1 + β)

D1(1 + 2α)c1. (2.26)

Multiplying (2.23) with (1 + 2α) and (2.25) with (1 + 3α), and adding the results

20

give

a22(2 + 7α + 3β + 11αβ) =

B12

(1 + 2β)c2 +D12

(1 + 3α)b2

+c214

(1 + 2β)(B2 −B1) +b214

(1 + 3α)(D2 −D1)


a22(2 + 7α + 3β + 11αβ)

− a22(1 + 2α)2

B21

((1 + 2β)(B2 −B1) +

(1 + 3α)(D2 −D1)(1 + β)2B21

(1 + 2α)2D21

)

=B12

(1 + 2β)c2 +D12

(1 + 3α)b2

which lead to

a22 =

B21D2

1[B1(1 + 2β)c2 + D1(1 + 3α)b2]

2[ρB21D2

1 − (1 + 2α)2(1 + 2β)(B2 −B1)D21 − (1 + β)2(1 + 3α)(D2 −D1)B

21 ]

,

which gives us the desired estimate on |a2| as asserted in (2.19) when |b2| ≤ 2 and

|c2| ≤ 2.


results give

2a3(2 + 7α + 3β + 11αβ) =B12

(3 + 5β)c2 +D12

(1 + 2α)b2

+c214

(3 + 5β)(B2 −B1) +b214

(1 + 2α)(D2 −D1)

Substituting b1 from (2.26) in the above equation give

2a3(2 + 7α + 3β + 11αβ) =B12

(3 + 5β)c2 +D12

(1 + 2α)b2

+c214

((3 + 5β)(B2 −B1) +

(1 + β)2(D2 −D1)B21

D21(1 + 2α)

)

21

which lead to

2ρa3 =1

2[B1(3 + 5β)c2 + D1(1 + 2α)b2]

+c214

[(3 + 5β)(B2 −B1) +

(1 + β)2B21(D2 −D1)

D21(1 + 2α)

],

where ρ = 2 + 7α + 3β + 11αβ and this yields the estimate given in (2.20).

Theorem 2.4 Let f ∈ SB and g = f−1. If f ∈ ST (α, ϕ) and g ∈ L(β, ψ), then

|a2| ≤ B1D1

√2[B1(3− 2β) + D1(1 + 3α)]√

|ρB21D2

1 − 2(1 + 2α)2(3− 2β)(B2 −B1)D21 − 2(2− β)2(1 + 3α)(D2 −D1)B2

1 |(2.27)

and

|ρa3| ≤1

2B1(β

2 − 11β + 16) + D1(1 + 2α) +1

2(β2 − 11β + 16)|B2 −B1|

+(2− β)2B2

1 |D2 −D1|D2

1(1 + 2α)(2.28)

where ρ := 10 + 36α− 7β − 25αβ + β2 + 3αβ2.

Proof. Let f ∈ ST (α, ϕ) and g ∈ L(β, ψ). Then there are analytic functions

u, v : D→ D, with u(0) = v(0) = 0, satisfying

zf ′(z)

f(z)+

αz2f ′′(z)

f(z)= ϕ(u(z)) and

(wg′(w)

g(w)

)β (1 +

wg′′(w)

g′(w)

)1−β

= ψ(v(w)),

(2.29)

Using

zf ′(z)

f(z)+

αz2f ′′(z)

f(z)= 1 + a2(1 + 2α)z + (2(1 + 3α)a3 − (1 + 2α)a2

2)z2 + · · · ,

(wg′(w)

g(w)

)β (1 +

wg′′(w)

g′(w)

)1−β

22

= 1− (2− β)a2w +((8(1− β) +

1

2β(β + 5))a2

2 − 2(3− 2β)a3

)w2 + · · · ,

and equations (2.10), (2.11) and (2.29) will yield

a2(1 + 2α) =1

2B1c1, (2.30)

2(1 + 3α)a3 − (1 + 2α)a22 =

1

2B1

(c2 −

c212

)+

1

4B2c

21, (2.31)

−(2− β)a2 =1

2D1b1 (2.32)

and

[8(1− β) +β

2(β + 5)]a2

2 − 2(3− 2β)a3 =1

2D1

(b2 −

b212

)+

1

4D2b

21. (2.33)


b1 = − B1(2− β)

D1(1 + 2α)c1. (2.34)

Multiplying (2.31) with (3− 2β) and (2.33) with (1 + 3α), and adding the results

give

a22

(5− 7β

2+ 18α− 25αβ

2+

β2

2+

3αβ2

2

)=

B12

(3− 2β)c2 +D12

(1 + 3α)b2

+c214

(3− 2β)(B2 −B1) +b214

(1 + 3α)(D2 −D1)


a22

(5− 7β

2+ 18α− 25αβ

2+

β2

2+

3αβ2

2

)

− a2(1 + 2α)2

B21

((3− 2β)(B2 −B1) +

B21(2− β)2

D21(1 + 2α)2

(1 + 3α)(D2 −D1)

)

23

=B12

(3− 2β)c2 +D12

(1 + 3α)b2

which lead to

a22 =

B21D2

1[B1(3− 2β)c2 + D1(1 + 3α)b2]

ρB21D2

1 − 2(1 + 2α)2(3− 2β)(B2 −B1)D21 − 2(2− β)2(1 + 3α)(D2 −D1)B

21,

which again in view of |b2| ≤ 2 and |c2| ≤ 2 gives the desired estimate on |a2| as

asserted in (2.27). Multiplying (2.31) with [8(1 − β) + β2 (β + 5)] and (2.33) with

(1 + 2α), and adding the results give

a3(10 + 36α− 7β − 25αβ + β2 + 3αβ2) =B14

(β2 − 11β + 16)c2 +D12

(1 + 2α)b2

+c214

[8(1− β) +

β

2(β + 5)

](B2 −B1) +

b214

(1 + 2α)(D2 −D1)


a3(10 + 36α− 7β − 25αβ + β2 + 3αβ2) =B14

(β2 − 11β + 16)c2 +D12

(1 + 2α)b2

+c214

[8(1− β) +

β

2(β + 5)

](B2 −B1) +

c21(2− β)2B21(D2 −D1)

4D21(1 + 2α)

which lead to

ρa3 =B14

(β2 − 11β + 16)c2 +D12

(1 + 2α)b2

+c214

[1

2(β2 − 11β + 16)(B2 −B1) +

(2− β)2B21(D2 −D1)

D21(1 + 2α)

]

where ρ := 10 + 36α− 7β − 25αβ + β2 + 3αβ2 and this yields the estimate given

in (2.28).

24

2.4 Second and third coefficients of functions f when f ∈ M(α, ϕ) and

g ∈M(β, ψ), or g ∈ L(β, ψ)

Now we consider the case where f ∈M(α, ϕ) and its inverse g is either in M(β, ψ)

or L(β, ψ). By using the similar technique, we determine the bound of second and

third coefficients of f .

Theorem 2.5 Let f ∈ SB and g = f−1. If f ∈M(α, ϕ), g ∈M(β, ψ), then

|a2| ≤B1D1

√B1(1 + 2β) + D1(1 + 2α)√

|ρB21D2

1 − (1 + α)2(1 + 2β)(B2 −B1)D21 − (1 + β)2(1 + 2α)(D2 −D1)B

21 |

(2.35)

and

2ρ|a3| ≤ B1(3+5β)+D1(1+3α)+(3+5β)|B2−B1|+(1 + β)2(1 + 3α)B2

1 |D2 −D1|D2

1(1 + α)2

(2.36)

where ρ := 2 + 3α + 3β + 4αβ.

Proof. For f ∈ M(α, ϕ) and g ∈ M(β, ψ), there exist analytic functions u, v :


(1− α)zf ′(z)f(z) + α

(1 +

zf ′′(z)f ′(z)

)= ϕ(u(z)),

(1− β)wg′(w)g(w) + β

(1 +

wg′′(w)g′(w)

)= ψ(v(w)).

(2.37)

Since

(1−α)zf ′(z)

f(z)+α

(1 +

zf ′′(z)

f ′(z)

)= 1+(1+α)a2z+(2(1+2α)a3−(1+3α)a2

2)z2+· · ·

and

(1−β)wg′(w)

g(w)+β

(1 +

wg′′(w)

g′(w)

)= 1−(1+β)a2w+((3+5β)a2

2−2(1+2β)a3)w2+· · · ,

25

then (2.10), (2.11) and (2.37) yield

a2(1 + α) =1

2B1c1, (2.38)

2(1 + 2α)a3 − (1 + 3α)a22 =

1

2B1

(c2 −

c212

)+

1

4B2c

21, (2.39)

−(1 + β)a2 =1

2D1b1 (2.40)

and

(3 + 5β)a22 − 2(1 + 2β)a3 =

1

2D1

(b2 −

b212

)+

1

4D2b

21. (2.41)

b1 = −B1(1 + β)

D1(1 + α)c1. (2.42)

Multiplying (2.39) with (1 + 2β) and (2.41) with (1 + 2α), and adding the results

give

a22(2 + 3α + 3β + 4αβ) =

B12

(1 + 2β)c2 +D12

(1 + 2α)b2

+c214

(1 + 2β)(B2 −B1) +b214

(1 + 2α)(D2 −D1)


a22(2 + 3α + 3β + 4αβ) =

B12

(1 + 2β)c2 +D12

(1 + 2α)b2

+(1 + α)2a2

2B2

1

((1 + 2β)(B2 −B1) +

(1 + β)2(1 + 2α)(D2 −D1)B21

(1 + α)2D21

)

which lead to

a22 =

B21D2

1[B1(1 + 2β)c2 + D1(1 + 2α)b2]

2ρB21D2

1 − 2(1 + α)2(1 + 2β)(B2 −B1)D21 − 2(1 + β)2(1 + 2α)(D2 −D1)B

21,

which, in view of |b2| ≤ 2 and |c2| ≤ 2 gives the desired estimate on |a2| as asserted

26

in (2.35).


results give

2ρa3 =B12

(3 + 5β)c2 +D12

(1 + 3α)b2 +c214

(3 + 5β)(B2 −B1)

+b214

(1 + 3α)(D2 −D1)


2ρa3 =B12

(3 + 5β)c2 +D12

(1 + 3α)b2 +c214

(3 + 5β)(B2 −B1)

+(1 + β)2(1 + 3α)B2

1(D2 −D1)

D21(1 + α)2

]

where ρ := 2 + 3α + 3β + 4αβ and the estimate in (2.36) is obtained.

When β = α and ψ = ϕ, Theorem 2.5 reduces to the following corollary.

Corollary 2.4 Let f be given by (2.1) and g = f−1. If f, g ∈M(α, ϕ), then

|a2| ≤B1√

B1√(1 + α)|B2

1 + (1 + α)(B1 −B2)|(2.43)

and

|a3| ≤B1 + |B2 −B1|

1 + α. (2.44)

For α = 0, Corollary 2.4 gives the coefficient estimates for Ma-Minda bi-starlike

functions, while for α = 1, it gives the following estimates for Ma-Minda bi-convex

functions.

Corollary 2.5 Let f given by (2.1) be in the class of Ma-Minda bi-convex func-

27

tions. Then

|a2| ≤B1√

B1√2|B2

1 + 2B1 − 2B2|and |a3| ≤

1

2(B1 + |B2 −B1|).

For ϕ given by

ϕ(z) =1 + (1− 2β)z

1− z= 1 + 2(1− β)z + 2(1− β)z2 + · · · ,

evidently B1 = B2 = 2(1 − β), and thus when α = 1 (bi-convex functions),

Corollary 2.4 reduces to the following result.

Corollary 2.6 [22, Theorem 4.1] Let f given by (2.1) be in the class of bi-convex

functions of order β, 0 < β ≤ 1. Then

|a2| ≤√

1− β and |a3| ≤ 1− β.

Theorem 2.6 Let f ∈ SB and g = f−1. If f ∈M(α, ϕ) and g ∈ L(β, ψ), then

|a2| ≤ B1D1

√2[B1(3− 2β) + D1(1 + 2α)]√

|ρB21D2

1 − 2(1 + α)2(3− 2β)(B2 −B1)D21 − 2(2− β)2(1 + 2α)(D2 −D1)B2

1 |(2.45)

and

|ρa3| ≤B12

(β2 − 11β + 16) + D1(1 + 3α) +1

2(β2 − 11β + 16)|B2 −B1|

+(2− β)2(1 + 3α)B2

1 |D2 −D1|D2

1(1 + α)2(2.46)

where ρ := 10 + 14α− 7β + β2 + 2αβ2 − 10αβ.

Proof. Let f ∈ M(α, ϕ) and g ∈ L(β, ψ). Then there are analytic functions

28

u, v : D→ D, with u(0) = v(0) = 0, satisfying

(1− α)zf ′(z)f(z) + α

(1 +

zf ′′(z)f ′(z)

)= ϕ(u(z)),

(wg′(w)g(w)

)β (1 +

wg′′(w)g′(w)

)1−β= ψ(v(w)).

(2.47)

Since

(1−α)zf ′(z)

f(z)+α

(1 +

zf ′′(z)

f ′(z)

)= 1+(1+α)a2z+(2(1+2α)a3−(1+3α)a2

2)z2+· · ·

and

(wg′(w)

g(w)

)β (1 +

wg′′(w)

g′(w)

)1−β

= 1− (2− β)a2w +((8(1− β) +

1

2β(β + 5))a2

2 − 2(3− 2β)a3

)w2 + · · · ,


a2(1 + α) =1

2B1c1, (2.48)

2(1 + 2α)a3 − (1 + 3α)a22 =

1

2B1

(c2 −

c212

)+

1

4B2c

21, (2.49)

−(2− β)a2 =1

2D1b1 (2.50)

and

[8(1− β) +β

2(β + 5)]a2

2 − 2(3− 2β)a3 =1

2D1

(b2 −

b212

)+

1

4D2b

21. (2.51)


b1 = −B1(2− β)

D1(1 + α)c1. (2.52)

29

Multiplying (2.49) with (3− 2β) and (2.51) with (1 + 2α), and adding the results

give

a222

(10 + 14α− 7β + β2 + 2αβ2 − 10αβ) =B12

(3− 2β)c2 +D12

(1 + 2α)b2

+c214

(3− 2β)(B2 −B1) +b214

(1 + 2α)(D2 −D1)


a22

(1

2(10 + 14α− 7β + β2 + 2αβ2 − 10αβ)− (1 + α)2(3− 2β)(B2 −B1)

B21

− (2− β)2(1 + 2α)(D2 −D1)

D21

)=

B12

(3− 2β)c2 +D12

(1 + 2α)b2

which lead to

a22 =

B21D2

1[B1(3− 2β)c2 + D1(1 + 2α)b2]

ρB21D2

1 − 2(1 + α)2(3− 2β)(B2 −B1)D21 − 2(2− β)2(1 + 2α)(D2 −D1)B

21.

Since |b2| ≤ 2 and |c2| ≤ 2 the desired estimate on |a2| as asserted in (2.45) is

obtained.

Multiplying (2.49) with [8(1 − β) + β2 (β + 5)] and (2.51) with (1 + 3α), and

adding the results give

ρa3 =B14

(β2 − 11β + 16)c2 +D12

(1 + 3α)b2 +c214

(β2 − 11β + 16)(B2 −B1)

+b214

(1 + 3α)(D2 −D1)


ρa3 =B14

(β2 − 11β + 16)c2 +D12

(1 + 3α)b2 +c214

[(β2 − 11β + 16)(B2 −B1)

+(2− β)2(1 + 3α)B2

1(D2 −D1)

D21(1 + α)2

]

30

where ρ := 10 + 14α− 7β + β2 + 2αβ2 − 10αβ and this yields the estimate given

in (2.46).

2.5 Second and third coefficients of functions f when f ∈ L(α, ϕ) and

g ∈ L(β, ψ)

Theorem 2.7 Let f ∈ SB and g = f−1. If f ∈ L(α, ϕ) and g ∈ L(β, ψ), then

|a2| ≤ B1D1

√2[B1(3− 2β) + D1(3− 2α)]√

|ρB21D2

1 − 2(2− α)2(3− 2β)(B2 −B1)D21 − 2(2− β)2(3− 2α)(D2 −D1)B2

1 |(2.53)

and

2|ρa3| ≤ B1(β2 − 11β + 16) + D1(8− 5α− α2) + (β2 − 11β + 16)|B2 −B1|

+(2− β)2(α2 + 5α− 8)B2

1 |D2 −D1|D2

1(2− α)2(2.54)

where ρ := 24 + 3α2 + 3β2 − 17α− 17β − 2βα2 − 2αβ2 + 12αβ.

Proof. Let f ∈ L(α, ϕ) and g ∈ L(β, ψ). Then there are analytic functions u, v :


(zf ′(z)f(z)

)α (1 +

zf ′′(z)f ′(z)

)1−α= ϕ(u(z)),

(wg′(w)g(w)

)β (1 +

wg′′(w)g′(w)

)1−β= ψ(v(w)).

(2.55)

Since

(zf ′(z)

f(z)

)α (1 +

zf ′′(z)

f ′(z)

)1−α

= 1 + (2− α)a2z +

(2(3− 2α)a3 +

(α− 2)2 − 3(4− 3α)

2a22

)z2 + · · ·

31

and

(wg′(w)

g(w)

)β (1 +

wg′′(w)

g′(w)

)1−β

= 1− (2− β)a2w +((8(1− β) +

1

2β(β + 5))a2

2 − 2(3− 2β)a3

)w2 + · · · ,


a2(2− α) =1

2B1c1, (2.56)

2(3− 2α)a3 +1

2[(α− 2)2 − 3(4− 3α)]a2

2 =1

2B1

(c2 −

c212

)+

1

4B2c

21, (2.57)

−(2− β)a2 =1

2D1b1 (2.58)

and

[8(1− β) +β

2(β + 5)]a2

2 − 2(3− 2β)a3 =1

2D1

(b2 −

b212

)+

1

4D2b

21. (2.59)


b1 = −B1(2− β)

D1(2− α)c1. (2.60)

Multiplying (2.57) with (3− 2β) and (2.59) with (3− 2α), and adding the results

give

1

2a22(24 + 3α2 + 3β2 − 17α− 17β − 2βα2 − 2αβ2 + 12αβ)

=B12

(3− 2β)c2 +D12

(3− 2α)b2 +c214

(3− 2β)(B2 −B1) +b214

(3− 2α)(D2 −D1)

32


1

2a22(24 + 3α2 + 3β2 − 17α− 17β − 2βα2 − 2αβ2 + 12αβ)

=B12

(3− 2β)c2 +D12

(3− 2α)b2 +(2− α)2a2

2B2

1

((3− 2β)(B2 −B1)

+(2− β)2B2

1(3− 2α)(D2 −D1)

(2− α)2D21

)

and so

a22 =

B21D2

1[B1(3− 2β)c2 + D1(3− 2α)b2]

ρB21D2

1 − 2(2− α)2(3− 2β)(B2 −B1)D21 − 2(2− β)2(3− 2α)(D2 −D1)B

21,

which, again by using |b2| ≤ 2 and |c2| ≤ 2 gives the estimate on |a2| as asserted

in (2.53).

Multiplying (2.57) with (β2 − 11β + 16) and (2.59) with (α2 + 5α − 8), and

adding the results give

2ρa3 =B12

(β2 − 11β + 16)c2 +D12

(8− 5α− α2)b2 +c214

(β2 − 11β + 16)(B2 −B1)

+b214

(α2 + 5α− 8)(D2 −D1)


2ρa3 =B12

(β2 − 11β + 16)c2 +D12

(8− 5α− α2)b2 +c214

[(β2 − 11β + 16)(B2 −B1)

+(2− β)2(α2 + 5α− 8)B2

1(D2 −D1)

D21(2− α)2

]

where ρ := 24 + 3α2 + 3β2− 17α− 17β − 2βα2− 2αβ2 + 12αβ and this yields the

estimate given in (2.54).

When β = α and ψ = ϕ, Theorem 2.7 reduces to the following corollary.

33

Corollary 2.7 Let f be given by (2.1) and g = f−1. If f, g ∈ L(α, ϕ) then

|a2| ≤2B1

√B1√

|2(α2 − 3α + 4)B21 + 4(α− 2)2(B1 −B2)|

(2.61)

and

|a3| ≤2(3− 2α)

(B1 + |B1 −B2|

)

|(3− 2α)(α2 − 3α + 4)| . (2.62)

2.6 Second and third coefficients of functions f when f ∈ HSB(ϕ)

A function f ∈ A with Re(f ′(z)) > 0 is known to be univalent (Theorem 1.10).

This motivates the following class of functions studied in [118].

Definition 2.2 A function f ∈ SB is said to be in the class HSB(ϕ) if the fol-

lowing subordinations hold:

f ′(z) ≺ ϕ(z) and g′(w) ≺ ϕ(w),

where ϕ is as given in Definition 2.1

For functions in the class HSB(ϕ), we have the following coefficient bounds.

Theorem 2.8 If f ∈ HSB(ϕ) is given by (2.1), then

|a2| ≤B1√

B1√|3B2

1 − 4B2 + 4B1|and |a3| ≤

(1

3+

B14

)B1. (2.63)

Proof. Let f ∈ HSB(ϕ) and g = f−1. Similar computations as in earlier theorems

give

2a2 =1

2B1c1, (2.64)

3a3 =1

2B1

(c2 −

c212

)+

1

4B2c

21, (2.65)

34

−2a2 =1

2B1b1 (2.66)

and

3(2a22 − a3) =

1

2B1

(b2 −

b212

)+

1

4B2b

21. (2.67)

From (2.64) and (2.66), it follows that

c1 = −b1. (2.68)

Now (2.65), (2.66), (2.67) and (2.68) yield

a22 =

B31(b2 + c2)

4(3B21 − 4B2 + 4B1)

,

which, in view of |b2| ≤ 2 and |c2| ≤ 2 gives the desired estimate on |a2| in (2.63).

By subtracting (2.67) from (2.65), further computation using (2.64) and (2.68)

leads to

a3 =1

12B1(c2 − b2) +

1

16B2

1c21,

and this yields the estimate given in (2.63).

For ϕ given by

ϕ(z) =

(1 + z

1− z

)α

, 0 < α ≤ 1,

the inequalities in (2.63) reduce to the following result.

Corollary 2.8 [118, Theorem 1, p. 3] Let f ∈ HαΣ := HSB((1+z

1−z )α). Then

|a2| ≤ α

√2

α + 2and |a3| ≤

α(3α + 2)

3.

In the case when

ϕ(z) =1 + (1− 2β)z

1− z, 0 ≤ β < 1,

35

the inequalities in (2.63) reduce to the following result.

Corollary 2.9 [118, Theorem 2, p.4] Let f ∈ HΣ(β) := HSB(1+(1−2β)z

1−z ). Then

|a2| ≤√

2(1− β)

3and |a3| ≤

(1− β)(5− 3β)

3.

36

CHAPTER 3

BOUNDS FOR THE SECOND HANKEL DETERMINANT

OF UNIVALENT FUNCTIONS


Recall that A denote the class of all analytic functions defined on the open unit

disk D := z ∈ C : |z| < 1 of the form

f(z) = z + a2z2 + a3z

3 + · · · . (3.1)

The Hankel determinants Hq(n), (n = 1, 2, . . . , q = 1, 2, . . . ) of the function f

are defined by

Hq(n) :=

∣∣∣∣∣∣∣∣∣∣∣∣∣

an an+1 · · · an+q−1

an+1 an+2 · · · an+q

......

...

an+q−1 an+q · · · an+2q−2

∣∣∣∣∣∣∣∣∣∣∣∣∣

, (a1 = 1).

In general, Hankel determinants are useful in the study of the singularities and

in the study of power series with integral coefficients. For example, in showing

that a function of bounded characteristic in D, i.e., a function which is a ratio of

two bounded analytic functions, with its Laurent series around the origin having

integral coefficients, is rational [27]. For the use of Hankel determinant in the study

of meromorphic functions, see [132]. Various properties of these determinants can

be found in [129, Chapter 4]. In 1966, Pommerenke [99] investigated the Hankel

determinant of areally mean p-valent functions, univalent functions as well as for

starlike functions. In [100], he proved that the Hankel determinants of univalent

37

functions satisfy

|Hq(n)| < Kn−(12+β)q+3

2 (n = 1, 2, . . . , q = 2, 3, . . . ),

where β > 1/4000 and K depends only on q. Later, Hayman [43] proved that

|H2(n)| < An1/2, (n = 1, 2, . . . ; A an absolute constant) for areally mean univa-

lent functions. In [71–73], the estimates for Hankel determinant for areally mean

p-valent functions were investigated. ElHosh obtained bounds for Hankel deter-

minants of univalent functions with positive Hayman index α [30] and of k-fold

symmetric and close-to-convex functions [31]. For bounds on the Hankel determi-

nants of close-to-convex functions, see [78, 79, 81]. Noor and Al-Bany studied the

Hankel determinant of Bazilevic functions in [77] and of functions with bounded

boundary rotation in [74–76,80]. The Hankel determinant H2(1) = a3 − a22 is the

well known Fekete-Szego functional. For results related to this functional, see [7,8].

The second Hankel determinant H2(2) is given by H2(2) = a2a4 − a23. In the

recent years, several authors have investigated bounds for the second Hankel de-

terminant of functions belonging to various subclasses of univalent and multivalent

functions [16, 40–42,44,63,65,66].

In this chapter, the bounds for H2(2) are obtained for functions belonging to

subclasses of Ma-Minda starlike and convex functions in Section 2. In Section 3,

the problem is investigated for two other related classes defined by subordination.

In proving our results, we do not assume the univalence or starlikeness of ϕ as

they were required only in obtaining the distortion, growth estimates and the

convolution theorems. The classes introduced by subordination naturally include

several well known classes of univalent functions and the results for some of these

special classes are indicated as corollaries.

Recall the class P of functions with positive real part consisting of all analytic

functions p : D→ C satisfying p(0) = 1 and Re p(z) > 0.

38

Lemma 3.1 [37] If the function p ∈ P is given by the series p(z) = 1 + c1z +

c2z2 + c3z

3 + · · · , then

2c2 = c21 + x(4− c21), (3.2)

4c3 = c31 + 2(4− c21)c1x− c1(4− c21)x2 + 2(4− c21)(1− |x|2)z, (3.3)

for some x, z with |x| ≤ 1 and |z| ≤ 1.

Another result that will be required is the optimal value of a quadratic expres-

sion. Standard computations show that

max0≤t≤4

(Pt2 +Qt+R) =

R, Q ≤ 0, P ≤ −Q4 ;

16P + 4Q + R, Q ≥ 0, P ≥ −Q8 or Q ≤ 0, P ≥ −Q

4 ;

4PR−Q2

4P , Q > 0, P ≤ −Q8 .

(3.4)

3.2 Second Hankel determinant of Ma-Minda starlike/convex func-

tions

Various subclasses of starlike functions are characterized by the quantity zf ′(z)/f(z)

lying in some domain in the right half-plane. For example, f is strongly starlike

of order β if zf ′(z)/f(z) lies in a sector | arg w| < βπ/2 while it is starlike of order

α if zf ′(z)/f(z) lies in the half-plane Re w > α. The various subclasses of starlike

functions were unified by subordination in [59]. The following definition of the

class of Ma-Minda starlike functions is the same as the one in [59] except for the

omission of starlikeness assumption of ϕ.

Definition 3.1 Let ϕ : D→ C be analytic and the Maclaurin series of ϕ is given

by

ϕ(z) = 1 + B1z + B2z2 + B3z

3 + · · · , (B1 > 0). (3.5)

39

The class ST (ϕ) of Ma-Minda starlike functions with respect to ϕ consists of

functions f ∈ A satisfying the subordination

zf ′(z)

f(z)≺ ϕ(z).

Several choices of ϕ would reduce the class ST (ϕ) to some known subclasses.

For the function ϕ given by ϕα(z) := (1+(1−2α)z)/(1− z) , 0 < α ≤ 1, the class

ST (α) := ST (ϕα) is the well-known class of starlike functions of order α. Let

ϕ(z) = ϕPAR(z) := 1 +2

π2

(log

1 +√

z

1−√z

)2.

Then the class

ST P := ST (ϕPAR) =

f ∈ A : Re

(zf ′(z)

f(z)

)>

∣∣∣∣zf ′(z)

f(z)− 1

∣∣∣∣

is the parabolic starlike functions introduced by Rønning [106]. For a survey of

parabolic starlike functions and the related class of uniformly convex functions,

see [11]. For 0 < β ≤ 1, the class

STβ := ST((1 + z

1− z

)β)

=

f ∈ A :

∣∣∣∣arg

(zf ′(z)

f(z)

)∣∣∣∣ <βπ

2

is the familiar class of strongly starlike functions of order β. The class

ST L := ST (√

1 + z) =

f ∈ A :

∣∣∣∣∣(

zf ′(z)

f(z)

)2− 1

∣∣∣∣∣ < 1

is the class of lemniscate starlike functions studied in [116].

Now we determine the bound for the second Hankel determinant for the class

ST (ϕ).

Theorem 3.1 Let the function f ∈ ST (ϕ) be given by (3.1).

40

1. If B1, B2 and B3 satisfy the conditions

|B2| ≤ B1, |4B1B3 −B41 − 3B2

2 | − 3B21 ≤ 0,

then the second Hankel determinant of f satisfies

|a2a4 − a23| ≤

B21

4.


|B2| ≥ B1, |4B1B3 −B41 − 3B2

2 | − B1|B2| − 2B21 ≥ 0,

or the conditions

|B2| ≤ B1, |4B1B3 −B41 − 3B2

2 | − 3B21 ≥ 0,


|a2a4 − a23| ≤

1

12|4B1B3 −B4

1 − 3B22 |.


|B2| > B1, |4B1B3 −B41 − 3B2

2 | −B1|B2| − 2B21 ≤ 0,


|a2a4 − a23| ≤

B21

12

(3|4B1B3 −B4

1 − 3B22 | − 4B1|B2| − 4B2

1 −B22

|4B1B3 −B41 − 3B2

2 | − 2B1|B2| −B21

).

Proof. Since f ∈ ST (ϕ), there exists an analytic function w with w(0) = 0 and

41

|w(z)| < 1 in D such that

zf ′(z)

f(z)= ϕ(w(z)). (3.6)

Define the functions p1 by

p1(z) :=1 + w(z)

1− w(z)= 1 + c1z + c2z

2 + · · ·

or equivalently,

w(z) =p1(z)− 1

p1(z) + 1=

1

2

(c1z +

(c2 −

c212

)z2 + · · ·

). (3.7)

Then p1 is analytic in D with p1(0) = 1 and has positive real part in D. By using

(3.7) together with (3.5), it is evident that

ϕ

(p1(z)− 1

p1(z) + 1

)= 1 +

1

2B1c1z +

(1

2B1

(c2 −

c212

)+

1

4B2c

21

)z2 + · · · . (3.8)

Since

zf ′(z)

f(z)= 1 + a2z + (−a2

2 + 2a3)z2 + (3a4 − 3a2a3 + a3

2)z3 + · · · , (3.9)

it follows by (3.6), (3.8) and (3.9) that

a2 =B1c1

2,

a3 =1

8

[(B2

1 −B1 + B2)c21 + 2B1c2

],

a4 =1

48[(−4B2 + 2B1 + B3

1 − 3B21 + 3B1B2 + 2B3)c

31

+ 2(3B21 − 4B1 + 4B2)c1c2 + 8B1c3].

42

Hence

a2a4 − a23 =

B196

[c41

(−B3

12

+B12−B2 + 2B3 −

3B22

2B1

)

+2c2c21(B2 −B1) + 8B1c1c3 − 6B1c

22

].

Let

d1 = 8B1, d2 = 2(B2 −B1),

d3 = −6B1, d4 = −B31

2 + B12 −B2 + 2B3 − 3B2

22B1

,

T = B196 .

(3.10)

Then

|a2a4 − a23| = T |d1c1c3 + d2c

21c2 + d3c

22 + d4c

41|. (3.11)

Since the function p(eiθz) (θ ∈ R) is in the class P for any p ∈ P , there is no loss

of generality in assuming c1 > 0. Write c1 = c, c ∈ [0, 2]. Substituting the values

of c2 and c3 respectively from (3.2) and (3.3) into (3.11), gives

|a2a4 − a23| =

T

4

∣∣∣c4(d1 + 2d2 + d3 + 4d4) + 2xc2(4− c2)(d1 + d2 + d3)

+(4− c2)x2(−d1c2 + d3(4− c2)) + 2d1c(4− c2)(1− |x|2)z

∣∣∣ .

Replacing |x| by µ and substituting the values of d1, d2, d3 and d4 from (3.10),

yield

|a2a4 − a23| ≤

T

4

[c4

∣∣∣∣∣−2B31 + 8B3 − 6

B22

B1

∣∣∣∣∣ + 4|B2|µc2(4− c2)

+ µ2(4− c2)(2B1c2 + 24B1) + 16B1c(4− c2)(1− µ2)

]

= T[c4

4

∣∣∣∣∣−2B31 + 8B3 − 6

B22

B1

∣∣∣∣∣ + 4B1c(4− c2) + |B2|(4− c2)µc2

+B12

µ2(4− c2)(c− 6)(c− 2)]

(3.12)

43

≡ F (c, µ).

Note that for (c, µ) ∈ [0, 2] × [0, 1], differentiating F (c, µ) in (3.12) partially with

respect to µ yields

∂F

∂µ= T

[|B2|(4− c2)c2 + B1µ(4− c2)(c− 2)(c− 6)

]. (3.13)

Then for 0 < µ < 1 and for any fixed c with 0 < c < 2, it is clear from (3.13) that

∂F∂µ > 0, that is, F (c, µ) is an increasing function of µ. Hence for fixed c ∈ [0, 2],

the maximum of F (c, µ) occurs at µ = 1, and

max F (c, µ) = F (c, 1) ≡ G(c),

where

G(c) =B196

[c4

4

(∣∣∣∣∣−2B31 + 8B3 − 6

B22

B1

∣∣∣∣∣− 4|B2| − 2B1

)+ 4c2(|B2| −B1) + 24B1

].

Let

P =1

4

(∣∣∣∣∣−2B31 + 8B3 − 6

B22

B1

∣∣∣∣∣− 4|B2| − 2B1

),

Q = 4(|B2| −B1), (3.14)

R = 24B1.

44

Then G(c) = B196 (Pt2 + Qt + R), where t = c2. Using (3.4) we have

|a2a4 − a23| ≤

B196

R, Q ≤ 0, P ≤ −Q4 ;

16P + 4Q + R, Q ≥ 0, P ≥ −Q8 or Q ≤ 0, P ≥ −Q

4 ;

4PR−Q2

4P , Q > 0, P ≤ −Q8

where P, Q,R are given by (3.14).

Hence for |B2| ≤ B1, |4B1B3 −B41 − 3B2

2 | − 3B21 ≤ 0, then

|a2a4 − a23| ≤

B21

4.

For |B2| ≥ B1, |4B1B3 − B41 − 3B2

2 | − B1|B2| − 2B21 ≥ 0, or for |B2| ≤ B1,

|4B1B3 −B41 − 3B2

2 | − 3B21 ≥ 0, then

|a2a4 − a23| ≤

1

12|4B1B3 −B4

1 − 3B22 |.

And for |B2| > B1, |4B1B3 −B41 − 3B2

2 | −B1|B2| − 2B21 ≤ 0, then

|a2a4 − a23| ≤

B21

12

(3|4B1B3 −B4

1 − 3B22 | − 4B1|B2|+ 4B2

1 −B22

|4B1B3 −B41 − 3B2

2 | − 2B1|B2| −B21

).

This completes the proof.

In the case when B1 = B2 = B3 = 2, Theorem 3.1 reduces to the following

corollary.

Corollary 3.1 [44, Theorem 3.1] Let f ∈ ST . Then |a2a4 − a23| ≤ 1.

Corollary 3.2

1. If f ∈ ST (α), then |a2a4 − a23| ≤ (1− α)2.

45

2. If f ∈ ST L, then |a2a4 − a23| ≤ 1/16 = 0.0625.

3. If f ∈ ST P , then |a2a4 − a23| ≤ 16/π4 ≈ 0.164255.

4. If f ∈ STβ, then |a2a4 − a23| ≤ β2.

Proof. 1. If f ∈ ST (α), then ϕ(z) = (1 + (1 − 2α)z)/(1 − z), and so B1 = B2 =

B3 = 2(1− α). Hence |a2a4 − a23| ≤ (1− α)2.

2. If f ∈ ST L, then ϕ(z) =√

1 + z, and so B1 = 1/2, B2 = −1/8 and

B3 = 1/16. Hence |a2a4 − a23| ≤ 1/16.

3. If f ∈ ST P , then ϕ(z) = 1 + 2π2

(log

1+√

z1−√z

)2, and so B1 = 8/π2, B2 =

16/(3π2) and B3 = 184/(45π2). Hence |a2a4 − a23| ≤ 16/π4.

4. If f ∈ STβ , then ϕ(z) =(

1+z1−z

)β, and so B1 = 2β, B2 = 2β2 and

B3 = 2β(1 + 2β2)/3. Hence |a2a4 − a23| ≤ β2.

After considering the Ma-Minda starlike functions, now we consider the Ma-

Minda convex functions with respect to ϕ and its bound for the second Hankel

determinant can be determined similarly.

Definition 3.2 Let ϕ : D → C be analytic and is given as in (3.5). The class

CV(ϕ) of Ma-Minda convex functions with respect to ϕ consists of functions f

satisfying the subordination

1 +zf ′′(z)

f ′(z)≺ ϕ(z).

Theorem 3.2 Let the function f ∈ CV(ϕ) be given by (3.1).


B21 + 4|B2| − 2B1 ≤ 0, |6B1B3 + B2

1B2 −B41 − 4B2

2 | − 4B21 ≤ 0,

46


|a2a4 − a23| ≤

B21

36.


B21+4|B2|−2B1 ≥ 0, 2|6B1B3+B2

1B2−B41−4B2

2 |−B31−4B1|B2|−6B2

1 ≥ 0,

or the conditions

B21 + 4|B2| − 2B1 ≤ 0, |6B1B3 + B2

1B2 −B41 − 4B2

2 | − 4B21 ≥ 0,


|a2a4 − a23| ≤

1

144|6B1B3 + B2

1B2 −B41 − 4B2

2 |.


B21+4|B2|−2B1 > 0, 2|6B1B3+B2

1B2−B41−4B2

2 |−B31−4B1|B2|−6B2

1 ≤ 0,


|a2a4 − a23|

≤ B21

576

16|6B1B3 + B21B2 −B4

1 − 4B22 | − 12B3

1 − 48B1|B2| − 36B21

−B41 − 8B2

1 |B2| − 16B22

|6B1B3 + B21B2 −B4

1 − 4B22 | −B3

1 − 4B1|B2| − 2B21

.

Proof. Since f ∈ CV(ϕ), there exists an analytic function w with w(0) = 0 and

47


1 +zf ′′(z)

f ′(z)= ϕ(w(z)). (3.15)

Since

1 +zf ′′(z)

f ′(z)= 1 + 2a2z + (−4a2

2 + 6a3)z2 + (8a3

2− 18a2a3 + 12a4)z3 + · · · , (3.16)


a2 =B1c1

4,

a3 =1

24

[(B2

1 −B1 + B2)c21 + 2B1c2

],

a4 =1

192[(−4B2 + 2B1 + B3

1 − 3B21 + 3B1B2 + 2B3)c

31

+ 2(3B21 − 4B1 + 4B2)c1c2 + 8B1c3].

Therefore

a2a4 − a23 =

B1768

[c41

(−4

3B2 +

2

3B1 −

1

3B3

1 −1

3B2

1 +1

3B1B2 + 2B3 −

4

3

B22

B1

)

+2

3c2c

21(B

21 − 4B1 + 4B2) + 8B1c1c3 −

16

3B1c

22

].

By writing

d1 = 8B1, d2 = 23(B2

1 − 4B1 + 4B2),

d3 = −163 B1, d4 = −4

3B2 + 23B1 − 1

3B31 − 1

3B21 + 1

3B1B2 + 2B3 − 43

B22

B1,

T = B1768 ,

(3.17)

we have

|a2a4 − a23| = T |d1c1c3 + d2c

21c2 + d3c

22 + d4c

41|.

48

Similar as in Theorems 3.1, it follows from (3.2) and (3.3) that

|a2a4 − a23| =

T

4

∣∣∣c4(d1 + 2d2 + d3 + 4d4) + 2xc2(4− c2)(d1 + d2 + d3)

+(4− c2)x2(−d1c2 + d3(4− c2)) + 2d1c(4− c2)(1− |x|2)z

∣∣∣ .

Replacing |x| by µ and then substituting the values of d1, d2, d3 and d4 from (3.17)

yield

|a2a4 − a23|

≤ T

4

[c4

∣∣∣∣∣−4

3B3

1 +4

3B1B2 + 8B3 −

16

3

B22

B1

∣∣∣∣∣ + 2µc2(4− c2)

(2

3B2

1 +8

3|B2|

)

+ µ2(4− c2)

(8

3B1c

2 +64

3B1

)+ 16B1c(4− c2)(1− µ2)

]

= T[c4

3

∣∣∣∣∣−B31 + B1B2 + 6B3 − 4

B22

B1

∣∣∣∣∣ + 4B1c(4− c2) +1

3µc2(4− c2)(B2

1 + 4|B2|)

+2B13

µ2(4− c2)(c− 4)(c− 2)]

(3.18)

≡ F (c, µ).

Again, differentiating F (c, µ) in (3.18) partially with respect to µ yield

∂F

∂µ= T

[c2

3(4− c2)(B2

1 + 4|B2|) +4B13

µ(4− c2)(c− 4)(c− 2)

]. (3.19)

It is clear from (3.19) that ∂F∂µ > 0. Thus F (c, µ) is an increasing function of µ for

0 < µ < 1 and for any fixed c with 0 < c < 2. So the maximum of F (c, µ) occurs

at µ = 1 and

max F (c, µ) = F (c, 1) ≡ G(c).

49

Note that

G(c) = T

[c4

3

(∣∣∣∣∣−B31 + B1B2 + 6B3 − 4

B22

B1

∣∣∣∣∣−B21 − 4|B2| − 2B1

)

+4

3c2(B2

1 + 4|B2| − 2B1) +64

3B1

].

Let

P =1

3

(∣∣∣∣∣−B31 + B1B2 + 6B3 − 4

B22

B1

∣∣∣∣∣−B21 − 4|B2| − 2B1

),

Q =4

3(B2

1 + 4|B2| − 2B1), (3.20)

R =64

3B1,

By using (3.4), we have

|a2a4 − a23| ≤

B1768

R, Q ≤ 0, P ≤ −Q4 ;

16P + 4Q + R, Q ≥ 0, P ≥ −Q8 or Q ≤ 0, P ≥ −Q

4 ;

4PR−Q2

4P , Q > 0, P ≤ −Q8

where P, Q,R are given in (3.20).

For the choice of ϕ(z) = (1 + z)/(1− z), Theorem 3.2 reduces to the following

corollary.

Corollary 3.3 [44, Theorem 3.2] Let f ∈ CV. Then |a2a4 − a23| ≤ 1/8.

3.3 Further results on the second Hankel determinant

In this section, we determine the bound for the second Hankel determinant of

functions in two more classes related to subordination.

A function f is in the class Rτγ(A,B), −1 ≤ B < A ≤ 1 if it satisfies 1 +

50

1τ (f ′(z) + γzf ′′(z) − 1) ≺ 1+Az

1+Bz . This class Rτγ(A,B), is essentially motivated

by Swaminathan [124] and introduced by Bansal [17]. Bansal [18] also obtained

upper bound of second Hankel determinant for functions belonging to this class.

Now we define the following class.

Definition 3.3 Let ϕ : D→ C be analytic and be given as in (3.5). Let 0 ≤ γ ≤ 1

and τ ∈ C\0. A function f ∈ A is in the class Rτγ(ϕ) if it satisfies the following

subordination:

1 +1

τ(f ′(z) + γzf ′′(z)− 1) ≺ ϕ(z).

The following theorem provides a bound for the second Hankel determinant of

the functions in the class Rτγ(ϕ).

Theorem 3.3 Let 0 ≤ γ ≤ 1, τ ∈ C \ 0 and suppose the function f as in (3.1)

is in the class Rτγ(ϕ). Also, let

p =8

9

(1 + γ)(1 + 3γ)

(1 + 2γ)2.


2|B2|(1− p) + B1(1− 2p) ≤ 0, |B1B3 − pB22 | − pB2

1 ≤ 0,


|a2a4 − a23| ≤

|τ |2B21

9(1 + 2γ)2.


2|B2|(1− p) + B1(1− 2p) ≥ 0, 2|B1B3− pB22 | − 2(1− p)B1|B2| −B1 ≥ 0,

51

or the conditions

2|B2|(1− p) + B1(1− 2p) ≤ 0, |B1B3 − pB22 | −B2

1 ≥ 0,


|a2a4 − a23| ≤

|τ |28(1 + γ)(1 + 3γ)

|B3B1 − pB22 |.


2|B2|(1− p) + B1(1− 2p) > 0, 2|B1B3− pB22 | − 2(1− p)B1|B2| −B2

1 ≤ 0,


|a2a4 − a23|

≤ |τ |2B21

32(1 + γ)(1 + 3γ)

4p|B3B1 − pB22 | − 4(1− p)B1[|B2|(3− 2p) + B1]

− 4B22(1− p)2 −B2

1(1− 2p)2

|B3B1 − pB22 | − (1− p)B1(2|B2|+ B1)

.

Proof. For f ∈ Rτγ(ϕ), there exists an analytic function w with w(0) = 0 and


1 +1

τ(f ′(z) + γzf ′′(z)− 1) = ϕ(w(z)). (3.21)

Since f has the Maclaurin series given by (3.1), a computation shows that

1+1

τ(f ′(z)+γzf ′′(z)−1) = 1+

2a2(1 + γ)

τz+

3a3(1 + 2γ)

τz2+

4a4(1 + 3γ)

τz3+· · · .

(3.22)

52

It follows from (3.21), (3.8) and (3.22) that

a2 =τB1c1

4(1 + γ),

a3 =τB1

12(1 + 2γ)

[2c2 + c21

(B2B1

− 1

)],

a4 =τ

32(1 + 3γ)[B1(4c3 − 4c1c2 + c31) + 2B2c1(2c2 − c21) + B3c

31].

Therefore

a2a4 − a23 =

τ2B1c1128(1 + γ)(1 + 3γ)

[B1(4c3 − 4c1c2 + c31) + 2B2c1(2c2 − c21) + B3c

31

]

− τ2B21

144(1 + 2γ)2

[4c22 + c41

(B2B1

− 1

)2+ 4c2c

21

(B2B1

− 1

)]

=τ2B2

1128(1 + γ)(1 + 3γ)

[(4c1c3 − 4c21c2 + c41) +

2B2c21

B1(2c2 − c21) +

B3B1

c41

]

−8

9

(1 + γ)(1 + 3γ)

(1 + 2γ)2

[4c22 + c41

(B2B1

− 1

)2+ 4c2c

21

(B2B1

− 1

)],

which yields

|a2a4 − a23| = T

∣∣∣∣∣4c1c3 + c41

[1− 2

B2B1

− p

(B2B1

− 1

)2+

B3B1

]− 4pc22

−4c21c2

[1− B2

B1+ p

(B2B1

− 1

)]∣∣∣∣ , (3.23)

where

T =τ2B2

1128(1 + γ)(1 + 3γ)

and p =8

9

(1 + γ)(1 + 3γ)

(1 + 2γ)2.

It can be easily verified that p ∈[

6481 , 8

9

]for 0 ≤ γ ≤ 1.

Let

d1 = 4, d2 = −4[1− B2

B1+ p

(B2B1− 1

)],

d3 = −4p, d4 = 1− 2B2B1− p

(B2B1− 1

)2+ B3

B1.

(3.24)

53

Then (3.23) becomes

|a2a4 − a23| = T |d1c1c3 + d2c

21c2 + d3c

22 + d4c

41|.

It follows that

|a2a4 − a23| =

T

4

∣∣∣c4(d1 + 2d2 + d3 + 4d4) + 2xc2(4− c2)(d1 + d2 + d3)

+(4− c2)x2(−d1c2 + d3(4− c2)) + 2d1c(4− c2)(1− |x|2)z

∣∣∣ .

An application of triangle inequality, replacement of |x| by µ and substitution of

the values of d1, d2, d3 and d4 from (3.24) yield

|a2a4 − a23| ≤

T

4

[4c4

∣∣∣∣∣B3B1

− pB2

2B2

1

∣∣∣∣∣ + 8

∣∣∣∣B2B1

∣∣∣∣µc2(4− c2)(1− p)

+ (4− c2)µ2(4c2 + 4p(4− c2)) + 8c(4− c2)(1− µ2)]

= T[c4

∣∣∣∣∣B3B1

− pB2

2B2

1

∣∣∣∣∣ + 2c(4− c2) + 2µ

∣∣∣∣B2B1

∣∣∣∣ c2(4− c2)(1− p)

+ µ2(4− c2)(1− p)(c− α)(c− β)]

≡ F (c, µ)

where α = 2 and β = 2p/(1− p) > 2.

Similarly as in the previous proofs, it can be shown that F (c, µ) is an increasing

function of µ for 0 < µ < 1. So for fixed c ∈ [0, 2], let

max F (c, µ) = F (c, 1) ≡ G(c),

which is

G(c) = T

c4

[∣∣∣∣∣B3B1

− pB2

2B2

1

∣∣∣∣∣− (1− p)

(2

∣∣∣∣B2B1

∣∣∣∣ + 1

)]

54

+4c2[2

∣∣∣∣B2B1

∣∣∣∣ (1− p) + 1− 2p

]+ 16p

.

Let

P =

∣∣∣∣∣B3B1

− pB2

2B2

1

∣∣∣∣∣− (1− p)

(2

∣∣∣∣B2B1

∣∣∣∣ + 1

),

Q = 4

[2

∣∣∣∣B2B1

∣∣∣∣ (1− p) + 1− 2p

], (3.25)

R = 16p.

Using (3.4), we have

|a2a4 − a23| ≤ T

R, Q ≤ 0, P ≤ −Q4 ;

16P + 4Q + R, Q ≥ 0, P ≥ −Q8 or Q ≤ 0, P ≥ −Q

4 ;

4PR−Q2

4P , Q > 0, P ≤ −Q8


For the choice ϕ(z) := (1 + Az)/(1 + Bz) with −1 ≤ B < A ≤ 1, Theorem 3.3

reduces to the following corollary.

Corollary 3.4 [18, Theorem 2.1] Let the function f be in the class Rτγ(A,B)

where 0 ≤ γ ≤ 1, τ ∈ C \ 0 and −1 ≤ B < A ≤ 1, then

|a2a4 − a23| ≤

|τ |2(A−B)2

9(1 + 2γ)2.

In [2], Al-Amiri and Reade introduced the class Gα := Gα((1 + z)/(1 − z))

and showed that Gα ⊂ S for α < 0. Univalence of the functions in the class Gα

was also investigated in [111, 112]. Verma et al. [130] obtained the bound for the

second Hankel determinant of functions in Gα. Now we define the following class.

55

Definition 3.4 Let ϕ : D → C be analytic and as given in (3.5). For a fixed

real number α, a function f ∈ A is in the class Gα(ϕ) if it satisfies the following

subordination:

(1− α)f ′(z) + α

(1 +

zf ′′(z)

f ′(z)

)≺ ϕ(z).

The following theorem provides a bound for the second Hankel determinant of

the functions in the class Gα(ϕ).

Theorem 3.4 Let the function f given by (3.1) be in the class Gα(ϕ), 0 ≤ α ≤ 1.

Also, let

p =8

9

(1 + 2α)

(1 + α).


B21α(3− 2p) + 2|B2|(1 + α− p) + B1(1 + α− 2p) ≤ 0,

|B41α(2α− 1− pα) + αB2

1B2(3− 2p) + (α + 1)B1B3 − pB22 | − pB2

1 ≤ 0,


|a2a4 − a23| ≤

B21

9(1 + α)2.


B21α(3− 2p) + 2|B2|(1 + α− p) + B1(1 + α− 2p) ≥ 0,

2|B41α(2α− 1− pα) + αB2

1B2(3− 2p) + (α + 1)B1B3 − pB22 | −B3

1α(3− 2p)

−2(1 + α− p)B1|B2| − (α + 1)B21 ≥ 0,

or

B21α(3− 2p) + 2|B2|(1 + α− p) + B1(1 + α− 2p) ≤ 0,

56

|B41α(2α− 1− pα) + αB2

1B2(3− 2p) + (α + 1)B1B3 − pB22 | − pB2

1 ≥ 0,


|a2a4 − a23| ≤

|B41α(2α− 1− pα) + αB2

1B2(3− 2p) + (α + 1)B1B3 − pB22 |

8(1 + α)(1 + 2α).


B21α(3− 2p) + 2|B2|(1 + α− p) + B1(1 + α− 2p) > 0,

2|B41α(2α− 1− pα) + αB2

1B2(3− 2p) + (α + 1)B1B3 − pB22 | −B3

1α(3− 2p)

−2(1 + α− p)B1|B2| − (α + 1)B21 ≤ 0,


|a2a4 − a23| ≤

B21

32(1 + α)(1 + 2α)

×

4p−

[B2

1α(3− 2p) + 2|B2|(1 + α− p) + B1(1 + α− 2p)]2

|B41α(2α− 1− pα) + αB2

1B2(3− 2p) + (α + 1)B1B3 − pB22 |

−B31α(3− 2p)− (1 + α− p)B1(2|B2|+ B1)

.

Proof. For f ∈ Gα(ϕ), some calculations shows that

a2 =B1c1

4,

a3 =1

12(1 + α)

[(αB2

1 −B1 + B2)c21 + 2B1c2

],

a4 =1

32(α + 1)(2α + 1)

[−3αB2

1 + α(2α− 1)B31 + B1(1 + α + 3αB2)

+ (α + 1)(B3 − 2B2)]c31 + 2[−2(1 + α)B1 + 3αB2

1 + 2(1 + α)B2]c1c2

+ 4(1 + α)B1c3

.

57

Therefore

a2a4 − a23

=B1

128(1 + α)(1 + 2α)

[[− 3αB2

1 + α(2α− 1)B31 + B1(1 + α) + 3αB1B2

+ (1 + α)(B3 − 2B2)]c41 + 2[−2(1 + α)B1 + 3αB2

1 + 2(1 + α)B2]c21c2

+ 4(1 + α)B1c1c3

− 8

9

(1 + 2α)

(1 + α)

[(αB2

1 −B1 + B2)2

B1c41 + 4(αB2

1 −B1 + B2)c2c21 + 4B1c

22

]]

which yields

|a2a4 − a23|

= T∣∣∣4(1 + α)B1c1c3 + c41

[− 3αB2

1 + α(2α− 1)B31 + B1(1 + α) + 3αB1B2

+ (1 + α)(B3 − 2B2)− p(αB2

1 −B1 + B2)2

B1

]− 4pB1c

22

+ 2c21c2

[−2(1 + α)B1 + 3αB2

1 + 2(1 + α)B2 − 2p(αB21 −B1 + B2)

] ∣∣∣ (3.26)

where

T =B1

128(1 + α)(1 + 2α)and p =

8

9

(1 + 2α)

(1 + α).

It can be easily verified that for 0 ≤ α ≤ 1, p ∈[

89 , 4

3

]. Let

d1 = 4(1 + α)B1,

d2 = 2[−2(1 + α)B1 + 3αB2

1 + 2(1 + α)B2 − 2p(αB21 −B1 + B2)

],

d3 = −4pB1,

d4 = −3αB21 + α(2α− 1)B3

1 + B1(1 + α) + 3αB1B2 + (1 + α)(B3 − 2B2)

− p(αB2

1 −B1 + B2)2

B1,

58

Then (3.26) becomes

|a2a4 − a23| = T |d1c1c3 + d2c

21c2 + d3c

22 + d4c

41|.

Similar as in earlier theorems, it follows that

|a2a4 − a23| =

T

4

∣∣∣c4(d1 + 2d2 + d3 + 4d4) + 2xc2(4− c2)(d1 + d2 + d3)

+(4− c2)x2(−d1c2 + d3(4− c2)) + 2d1c(4− c2)(1− |x|2)z

∣∣∣

≤ T[c4

∣∣∣∣∣B31α(2α− 1− pα) + αB1B2(3− 2p) + (α + 1)B3 − p

B22

B1

∣∣∣∣∣

+ µc2(4− c2)[B21α(3− 2p) + 2|B2|(1 + α− p)] + 2c(4− c2)B1(1 + α)

+ µ2(4− c2)B1(1 + α− p)(c− 2)

(c− 2p

1 + α− p

) ]

≡ F (c, µ),

and for fixed c ∈ [0, 2], max F (c, µ) = F (c, 1) ≡ G(c) with

G(c) = T

[c4

[∣∣∣B31α(2α− 1− pα) + αB1B2(3− 2p) + (α + 1)B3 − p

B22

B1

∣∣∣

−B21α(3− 2p)− (1 + α− p)(2|B2|+ B1)

]+ 4c2[B2

1α(3− 2p)

+ 2|B2|(1 + α− p) + B1(1 + α− 2p)] + 16pB1

].

Let

P =∣∣∣B3

1α(2α− 1− pα) + αB1B2(3− 2p) + (α + 1)B3 − pB2

2B1

∣∣∣

−B21α(3− 2p)− (1 + α− p)(2|B2|+ B1)

59

Q = 4[B2

1α(3− 2p) + 2|B2|(1 + α− p) + B1(1 + α− 2p)], (3.27)

R = 16pB1,

By using (3.4), we have

|a2a4 − a23| ≤ T

R, Q ≤ 0, P ≤ −Q4 ;

16P + 4Q + R, Q ≥ 0, P ≥ −Q8 or Q ≤ 0, P ≥ −Q

4 ;

4PR−Q2

4P , Q > 0, P ≤ −Q8


Remark 3.1 For α = 1, Theorem 3.4 reduces to Theorem 3.2.

For 0 ≤ α < 1, let ϕ(z) := (1 + (1 − 2α)z)/(1 − z). For this function ϕ,

B1 = B2 = B3 = 2(1 − α). In this case, Theorem 3.4 reduces to the following

corollary.

Corollary 3.5 [130, Theorem 3.1] If f ∈ Gα(ϕ), then

|a2a4 − a23| ≤

4(1− α)2

9(1 + α)2.

Remark 3.2 By letting γ = 0 and τ = 1 in Theorem 3.3 and α = 0 in Theo-

rem 3.4, the two results coincide.

60

CHAPTER 4

APPLICATIONS OF DIFFERENTIAL SUBORDINATION

FOR FUNCTIONS WITH FIXED SECOND COEFFICIENT


For univalent functions f(z) = z +∑∞

n=2 anzn defined on D := z ∈ C : |z| < 1,the famous Bieberbach theorem (Theorem 1.2) shows that |a2| ≤ 2 and this bound

yields the growth and distortion bounds as well as the covering theorem. In view

of the influence of the second coefficient in the properties of univalent functions,

several authors have investigated functions with fixed second coefficient. For a brief

survey of the various developments, mainly on radius problems, from 1920 to this

date, see the recent work by Ali et al. [13]. The theory of first-order differential

subordination was developed by Miller and Mocanu, and a very comprehensive

account of the theory and numerous applications can be found in their monograph

[61]. Ali et al. [10] extended this well-known theory of differential subordination to

the functions with preassigned second coefficient. Nagpal and Ravichandran [68]

then applied the results in [10] to obtain several extensions of well-known results

to the functions with fixed second coefficient. In this chapter, we continue their

investigation by deriving several sufficient conditions for starlikeness of functions

with fixed second coefficient.

For convenience, let An,b denote the class of all functions f(z) = z + bzn+1 +

an+2zn+2 + · · · where n ∈ N = 1, 2, . . . and b is a fixed non-negative real

number. For fixed µ ≥ 0, and n ∈ N, let Hµ,n be the class consists of analytic

functions p on D of the form

p(z) = 1 + µzn + pn+1zn+1 + · · · (z ∈ D).

Let Ω be a subset of C and the class Ψµ,n[Ω] consists of those functions ψ : C2 → C

61

that are continuous in a domain D ⊂ C2 with (1, 0) ∈ D, ψ(1, 0) ∈ Ω, and satisfy

the admissibility condition: ψ(iρ, σ) 6∈ Ω whenever (iρ, σ) ∈ D, ρ ∈ R, and

σ ≤ −1

2

(n +

2− µ

2 + µ

)(1 + ρ2). (4.1)

When Ω = w : Re w > 0, let Ψµ,n := Ψµ,n[Ω]. The following theorem is needed

to prove our main results.

Theorem 4.1 [10, Theorem 3.4] Let p ∈ Hµ,n with 0 < µ ≤ 2. Let ψ ∈ Ψµ,n

with associated domain D. If (p(z), zp′(z)) ∈ D and Re ψ(p(z), zp′(z)) > 0, then

Re p(z) > 0 for z ∈ D.

For α 6= 1, let

ST (α) :=

f ∈ A :

zf ′(z)

f(z)≺ 1 + (1− 2α)z

1− z

.

The function pα(z) := (1 + (1 − 2α)z)/(1 − z) maps D onto w ∈ C : Re w > αfor α < 1 and onto w ∈ C : Re w < α for α > 1. Therefore, for α < 1, ST (α) is

the class of starlike functions of order α that consist of functions f ∈ A for which

Re(zf ′(z)/f(z)) > α.

Motivated by the works of Lewandowski et al. [50], several authors [52,55,56,

70,87,90,96,101,103,110,137] investigated the functions f for which zf ′(z)/f(z)·(αzf ′′(z)/f ′(z) + 1) lies in certain region in the right half-plane. For α ≥ 0 and

β < 1, Ravichandran et al. [102] showed that a function f of the form f(z) =

z + an+1zn+1 + · · · satisfying

Re

(zf ′(z)

f(z)

(α

zf ′′(z)

f ′(z)+ 1

))> αβ

(β +

n

2− 1

)+ β − αn

2

is starlike of order β. In Theorem 4.4, we will obtain the corresponding result for

f ∈ An,b.

62

For function p of the form p(z) = 1 + p1z + p2z2 + · · · , Nunokawa et al. [88]

showed that for certain analytic function w with w(0) = α, αp2(z)+βzp′(z) ≺ w(z)

implies Re p(z) > 0, where β > 0, α ≥ −β/2. See also [104]. Lemma 4.5 investi-

gates the conditions for the subordination p2(z)+γzp′(z) ≺ (1 + (1− 2δ)z)/(1− z).

For complex numbers β and γ, the differential subordination

q(z) +zq′(z)

βq(z) + γ≺ h(z),

where q is analytic and h is univalent with q(0) = h(0), is popularly known as

Briot-Bouquet differential subordination. This particular differential subordina-

tion has a significant number of important applications in the theory of analytic

functions (for details see [61]). The importance of Briot-Bouquet differential sub-

ordination inspired many researchers to work in this area and many generalizations

and extensions of the Briot-Bouquet differential subordination have recently been

obtained. Ali et al. [6] obtained several results related to the Briot-Bouquet differ-

ential subordination. In Lemma 4.1, the Briot-Bouquet differential subordination

is investigated for functions with fixed second coefficient.

4.2 Subordinations for starlikeness

In this section, several sufficient conditions are given for the starlikeness of order

β of functions f ∈ An,b. We need following lemmas to prove the theorems.

Lemma 4.1 Let n ∈ N be fixed. For α > 0, 0 ≤ β < 1, γ + αβ > 0 and

0 < µ ≤ 2(1− β), let

δ =

−12

(1−β)(αβ+γ)

(n +

2(1−β)−µ2(1−β)+µ

)+ β, if γ + αβ ≥ α(1− β),

−12

(αβ+γ)α2(1−β)

(n +

2(1−β)−µ2(1−β)+µ

)+ β, if γ + αβ ≤ α(1− β).

63

If the function p ∈ Hµ,n satisfies the subordination

p(z) +zp′(z)

αp(z) + γ≺ 1 + (1− 2δ)z

1− z, (4.2)

then

p(z) ≺ 1 + (1− 2β)z

1− z.

Proof. Let 0 ≤ β < 1. Note that

δ :=−1

2

(1− β)

(αβ + γ)

(n +

2(1− β)− µ

2(1− β) + µ

)+ β

= −(1− β)

[1

2

1

(αβ + γ)

(n +

2(1− β)− µ

2(1− β) + µ

)+ 1

]+ 1 < 1,

and

δ :=−1

2

αβ + γ

α2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ β

= −(1− β)

[1

2

αβ + γ

α2(1− β)2

(n +

2(1− β)− µ

2(1− β) + µ

)+ 1

]+ 1 < 1.

Then for p ∈ Hµ,n, (4.2) can be rewritten as

Re

(p(z) +

zp′(z)

αp(z) + γ

)> δ, (4.3)

and we want to show that Re p(z) > β. Define the function q : D→ C by

q(z) =p(z)− β

1− β= 1 +

µ

1− βzn +

an+11− β

zn+1 + · · · , n ∈ N.

Then q is analytic on D and q ∈ Hµ′,n, where µ′ = µ/(1−β). Since (1−β)q(z)+β =

p(z), if we can show that Re q(z) > 0, then it will imply Re p(z) > β. Note that

p(z) +zp′(z)

αp(z) + γ= (1− β)q(z) + β +

(1− β)

α[(1− β)q(z) + β] + γzq′(z).

64

Then by (4.3),

Re

((1− β)q(z) +

(1− β)

α(1− β)q(z) + αβ + γzq′(z) + β

)> δ. (4.4)

Define ψ : C2 → C by

ψ(r, s) = (1− β)r +(1− β)

α(1− β)r + αβ + γs + β − δ.

Then ψ is a continuous function of r and s on D := C\−(αβ + γ)/α(1− β)×Cand Re(ψ(1, 0)) = 1− δ > 0. Note that (1, 0) ∈ D because (αβ + γ)/α(1−β) > 0.

Also for ρ ∈ R, and by (4.1), σ in our case satisfying

σ ≤ −1

2

(n +

2− µ′2 + µ′

)(1 + ρ2), (4.5)

where µ′ = µ/(1− β), it follows that

Re ψ(iρ, σ) = Re

[(1− β)iρ +

(1− β)

α(1− β)iρ + αβ + γσ + β − δ

]

=(1− β)(αβ + γ)

(αβ + γ)2 + α2(1− β)2ρ2σ + β − δ

≤ (1− β)(αβ + γ)

(αβ + γ)2 + α2(1− β)2ρ2

[−1

2

(n +

2− µ′2 + µ′

)(1 + ρ2)

]+ β − δ

=−1

2(1− β)(αβ + γ)

(n +

2(1− β)− µ

2(1− β) + µ

)(1 + ρ2

(αβ + γ)2 + α2(1− β)2ρ2

)

+ β − δ.

Thus, for αβ + γ ≥ α(1− β),

Re ψ(iρ, σ) ≤ −1

2(1− β)(αβ + γ)

(n +

2(1− β)− µ

2(1− β) + µ

)(1 + ρ2

(αβ + γ)2 + α2(1− β)2ρ2

)

+ β − δ

≤ −1

2(1− β)(αβ + γ)

(n +

2(1− β)− µ

2(1− β) + µ

)(1 + ρ2

(αβ + γ)2 + (αβ + γ)2ρ2

)

65

+ β − δ

=−1

2

(1− β)

(αβ + γ)

(n +

2(1− β)− µ

2(1− β) + µ

)+ β − δ

= 0.

Similarly, for αβ + γ ≤ α(1− β),

Re ψ(iρ, σ) ≤ −1

2(1− β)(αβ + γ)

(n +

2(1− β)− µ

2(1− β) + µ

)(1 + ρ2

α2(1− β)2 + α2(1− β)2ρ2

)

+ β − δ

=−1

2

(αβ + γ)

α2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ β − δ

= 0.

Hence ψ ∈ Ψµ′,n, where µ′ = µ/(1 − β). Since q ∈ Hµ′,n, by Theorem 4.1,

Re q(z) > 0 if (q(z), zq′(z)) ∈ D and Re(ψ(q(z), zq′(z)) > 0. The former is true

by (4.2) and the latter is true by (4.4). So Re q(z) > 0 as desired.

Lemma 4.2 For 0 ≤ β < 1, λ > 0, and 0 < µ ≤ 2(1− β), let

δ := −λ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ β.

If p ∈ Hµ,n satisfies the subordination

p(z) + λzp′(z) ≺ 1 + (1− 2δ)z

1− z,

then

p(z) ≺ 1 + (1− 2β)z

1− z.

Proof. Let α = 0 and γ = 1/λ in Lemma 4.1.

66

Theorem 4.2 Let n ∈ N be fixed, 0 ≤ β < 1 and 0 < µ = nb ≤ 2(1− β). Let δ1

be given by

δ1 = −1

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ β.

If f ∈ An,b satisfies the subordination

zf ′(z)

f(z)

(2 +

zf ′′(z)

f ′(z)− zf ′(z)

f(z)

)≺ 1 + (1− 2δ1)z

1− z,

then

zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

Proof. For a given function f ∈ An,b, define the function p : D → C by p(z) =

zf ′(z)/f(z). A computation shows that p(z) = 1+nbzn + · · · , so p ∈ Hµ,n, where

µ = nb. Further calculations yield

zf ′(z)

f(z)

(2 +

zf ′′(z)

f ′(z)− zf ′(z)

f(z)

)= p(z) + zp′(z).

Hence by hypothesis and Lemma 4.2, it follows that

p(z) =zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

Lemma 4.3 Let n ∈ N. For ε > 0, 0 ≤ β < 1 and 0 < µ ≤ 2(1− β), let

δ =

−εβ2(1−β)

(n +

2(1−β)−µ2(1−β)+µ

)+ β, if 0 ≤ β ≤ 1

2 ,

−ε2β (1− β)

(n +

2(1−β)−µ2(1−β)+µ

)+ β, if 1

2 ≤ β < 1.


p(z) + εzp′(z)

p(z)≺ 1 + (1− 2δ)z

1− z(4.6)

67

then

p(z) ≺ 1 + (1− 2β)z

1− z.

Proof. Let α = 1/ε and γ = 0 in Lemma 4.1.

Theorem 4.3 Let n ∈ N, ε > 0, 0 ≤ β < 1 and 0 < µ = nb ≤ 2(1− β). Let δ2 be

given by

δ2 =

−εβ2(1−β)

(n +

2(1−β)−µ2(1−β)+µ

)+ β, if 0 ≤ β ≤ 1

2 ,

−ε2β (1− β)

(n +

2(1−β)−µ2(1−β)+µ

)+ β, if 1

2 ≤ β < 1.

If f ∈ An,b satisfies the following subordination

(1− ε)zf ′(z)

f(z)+ ε

(1 +

zf ′′(z)

f ′(z)

)≺ 1 + (1− 2δ2)z

1− z,

then

zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

Proof. Let the function p : D → C be defined by p(z) = zf ′(z)/f(z). Then

p ∈ Hµ,n and some calculations yield

(1− ε)zf ′(z)

f(z)+ ε

(1 +

zf ′′(z)

f ′(z)

)= p(z) + ε

zp′(z)

p(z).


p(z) =zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

Lemma 4.4 Let n ∈ N. For α ≥ 0, 0 ≤ β < 1, γ > 0, and 0 < µ ≤ 2(1− β), let

δ := −γ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ (1− α)β + αβ2.

68

If p ∈ Hµ,n satisfies the subordination

(1− α)p(z) + αp2(z) + γzp′(z) ≺ 1 + (1− 2δ)z

1− z, (4.7)

then

p(z) ≺ 1 + (1− 2β)z

1− z.


δ := −γ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ (1− α)β + αβ2

= −γ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)− αβ(1− β)− (1− β) + 1

= −(1− β)

[γ

2

(n +

2(1− β)− µ

2(1− β) + µ

)+ αβ + 1

]+ 1 < 1.

Then for p ∈ Hµ,n, Re((1 − α)p(z) + αp2(z) + γzp′(z)) > δ. To show that

Re p(z) > β, consider the function q : D→ C given by

q(z) =p(z)− β

1− β.

Then q is analytic on D, q ∈ Hµ′,n where µ′ = µ/(1− β), and

Re[(1− β)(1− α + 2αβ)q(z) + α(1− β)2q2(z)

+γ(1− β)zq′(z) + (1− α)β + αβ2]

> δ.

Define the function ψ : C2 → C by

ψ(r, s) = (1− β)(1− α + 2αβ)r + α(1− β)2r2 + γ(1− β)s + (1− α)β + αβ2 − δ.

69

For ρ ∈ R and σ satisfying (4.5), it follows that

Re ψ(iρ, σ)

= Re[(1− β)(1− α + 2αβ)iρ− α(1− β)2ρ2 + γ(1− β)σ + (1− α)β + αβ2 − δ

]

= γ(1− β)σ − α(1− β)2ρ2 + (1− α)β + αβ2 − δ

≤ γ(1− β)

[−1

2

(n +

2− µ′2 + µ′

)(1 + ρ2)

]− α(1− β)2ρ2 + (1− α)β + αβ2 − δ

= −γ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)(1 + ρ2)− α(1− β)2(ρ2 + 1) + α(1− β)2

+ (1− α)β + αβ2 − δ

= −(1 + ρ2)

[γ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ α(1− β)2

]

+ α(1− β)2 + (1− α)β + αβ2 − δ

≤ −γ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ (1− α)β + αβ2 − δ = 0

Hence ψ ∈ Ψµ′,n. By Theorem 4.1, Re q(z) > 0 or equivalently Re p(z) > β.

Remark 4.1 Lemma 4.2 can also be obtained by letting α = 0 in Lemma 4.4.

Theorem 4.4 Let n ∈ N. For α > 0, 0 ≤ β < 1 and 0 < µ = nb ≤ 2(1− β), let

δ3 be given by

δ3 = −α

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ (1− α)β + αβ2.


zf ′(z)

f(z)

(α

zf ′′(z)

f ′(z)+ 1

)≺ 1 + (1− 2δ3)z

1− z,

then

zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

70

Proof. For a given function f ∈ An,b, let the function p : D → C be defined by

p(z) = zf ′(z)/f(z). Then p(z) = 1 + nbzn + · · · . So p ∈ Hµ,n where µ = nb.

Some simple calculations yield

zf ′(z)

f(z)

(α

zf ′′(z)

f ′(z)+ 1

)= (1− α)p(z) + αp2(z) + αzp′(z).


p(z) =zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

If µ = 2(1− β), Theorem 4.4 reduces to the following result.

Corollary 4.1 [102, Theorem 2.1] If f of the form f(z) = z + an+1zn+1 +

an+2zn+2 + . . . satisfies

Re

zf ′(z)

f(z)

(α

zf ′′(z)

f ′(z)+ 1

)> αβ

(β +

n

2− 1

)+

(β − αn

2

)α ≥ 0, β ≤ 1,

then f ∈ ST (β).

Lemma 4.5 Let n ∈ N. For 0 ≤ β < 1, γ > 0, and 0 < µ ≤ 2(1− β), let

δ := −γ

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)+ β2.


p2(z) + γzp′(z) ≺ 1 + (1− 2δ)z

1− z, (4.8)

then

p(z) ≺ 1 + (1− 2β)z

1− z.

Proof. Let α = 1 in Lemma 4.4.

71

Lemma 4.6 Let n ∈ N. For 0 ≤ β < 1 and 0 < µ ≤ 2(1− β), let

δ =

−β2(1−β)

(n +

2(1−β)−µ2(1−β)+µ

), if 0 ≤ β < 1

2 ,

−12β (1− β)

(n +

2(1−β)−µ2(1−β)+µ

), if 1

2 ≤ β < 1.


zp′(z)

p(z)≺ 1 + (1− 2δ)z

1− z,

then

p(z) ≺ 1 + (1− 2β)z

1− z.


δ :=−β

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)< 1,

and

δ :=−1

2β(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)< 1.

Thus for p ∈ Hµ,n,

Rezp′(z)

p(z)> δ

and we want to show Re p(z) > β. Considering the function q(z) = (p(z)−β)/(1−β) or equivalently (1 − β)q(z) + β = p(z). Then q is analytic on D, q ∈ Hµ′,n

where µ′ = µ/(1− β), and

Rezp′(z)

p(z)= Re

((1− β)zq′(z)

(1− β)q(z) + β

)> δ.

72

Define ψ : C2 → C by

ψ(r, s) =(1− β)s

(1− β)r + β− δ.

Then ψ(r, s) is continuous on (C−−β/(1−β))×C. For ρ ∈ R and σ satisfying

(4.5), it follows that

Re ψ(iρ, σ) = Re

((1− β)

(1− β)iρ + βσ − δ

)

= Re

(β(1− β)

β2 + (1− β)2ρ2σ − (1− β)2iρ

β2 + (1− β)2ρ2σ − δ

)

=β(1− β)

β2 + (1− β)2ρ2σ − δ

≤ β(1− β)

β2 + (1− β)2ρ2

[−1

2

(n +

2− µ′2 + µ′

)(1 + ρ2)

]− δ

= −β

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)(1 + ρ2

β2 + (1− β)2ρ2

)− δ.

For 1/2 ≤ β < 1,

Re ψ(iρ, σ) ≤ −β

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)1

β2 − δ

=−1

2β(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)− δ

= 0.

For 0 ≤ β < 1/2,

Re ψ(iρ, σ) ≤ −β

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)1

(1− β)2− δ

=−β

2(1− β)

(n +

2(1− β)− µ

2(1− β) + µ

)− δ

= 0.

Hence ψ ∈ Ψµ′,n and then Theorem 4.1 implies Re q(z) > 0 or equivalently

Re p(z) > β.

73

Theorem 4.5 Let n ∈ N. For 0 ≤ β < 1 and 0 < µ = nb ≤ 2(1 − β). Let δ4 be

given by

δ4 =

−β2(1−β)

(n +

2(1−β)−µ2(1−β)+µ

), if 0 ≤ β < 1

2 ,

−12β (1− β)

(n +

2(1−β)−µ2(1−β)+µ

), if 1

2 ≤ β < 1.


1 +zf ′′(z)

f ′(z)− zf ′(z)

f(z)≺ 1 + (1− 2δ4)z

1− z

then

zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

Proof. Let the function p : D → C be defined by p(z) = zf ′(z)/f(z). Then

p ∈ Hµ,n. Also we have

1 +zf ′′(z)

f ′(z)− zf ′(z)

f(z)=

zp′(z)

p(z).


p(z) =zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

The above technique for analytic functions also works for meromorphic func-

tions. Let Σn,b be the class of normalized meromorphic functions, of the form

f(z) =1

z+ bzn + an+1z

n+1 + · · · , (b ≤ 0)

74

that are analytic in the punctured unit disk D∗ := z ∈ C : 0 < |z| < 1. For

0 ≤ β < 1, the class of meromorphic starlike functions of order β is defined by

−zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

The following theorems gives sufficient condition for meromorphic functions to be

starlike of order β in D∗.

Theorem 4.6 Let n ∈ N. For 0 ≤ β < 1 and 0 < µ = −(n + 1)b ≤ 2(1− β). Let

δ1 be given as in Theorem 4.2. If f ∈ Σn,b satisfies the following subordination

− zf ′(z)

f(z)

(2 +

zf ′′(z)

f ′(z)− zf ′(z)

f(z)

)≺ 1 + (1− 2δ1)z

1− z,

then

−zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

Proof. Let f ∈ Σn,b, and define the function p : D → C be defined by p(z) =

−zf ′(z)/f(z) for z ∈ D∗ and p(0) = 1. Then p(z) = 1− (n+1)bzn+1 + · · · ∈ Hµ,n

with µ = −(n + 1)b. Simple computations shows that

− zf ′(z)

f(z)

(2 +

zf ′′(z)

f ′(z)− zf ′(z)

f(z)

)= p(z) + zp′(z).

Hence the result follows from Lemma 4.2.

Theorem 4.7 Let n ∈ N. For ε > 0, 0 ≤ β < 1 and 0 < µ = −(n+1)b ≤ 2(1−β).

Let δ2 be given as in Theorem 4.3. If f ∈ Σn,b satisfies the following subordination

ε

(1 +

zf ′′(z)

f ′(z)

)− (1 + ε)

zf ′(z)

f(z)≺ 1 + (1− 2δ2)z

1− z,

then

−zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

75

Proof. Let p : D→ C be defined by p(z) = −zf ′(z)/f(z) for z ∈ D∗ and p(0) = 1.

Then p ∈ Hµ,n. Simple computations shows that

ε

(1 +

zf ′′(z)

f ′(z)

)− (1 + ε)

zf ′(z)

f(z)= p(z) + ε

zp′(z)

p(z).


Theorem 4.8 Let n ∈ N, α ≥ 0, 0 ≤ β < 1 and 0 < µ = −(n + 1)b ≤ 2(1 − β).

Let δ3 be given as in Theorem 4.4. If f ∈ Σn,b satisfies

zf ′(z)

f(z)

(2α

zf ′(z)

f(z)− α

zf ′′(z)

f ′(z)− 1

)≺ 1 + (1− 2δ3)z

1− z,

then

−zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

Proof. Let f ∈ Σn,b, and define the function p : D → C given by p(z) =

−zf ′(z)/f(z) for z ∈ D∗ and p(0) = 1. Then p ∈ Hµ,n with µ = −(n + 1)b.

A simple computation shows that

zf ′(z)

f(z)

(2α

zf ′(z)

f(z)− α

zf ′′(z)

f ′(z)− 1

)= (1− α)p(z) + αp2(z) + αzp′(z).


Theorem 4.9 Let n ∈ N, 0 ≤ β < 1 and 0 < µ = −(n + 1)b ≤ 2(1 − β). Let δ4

be given as in Theorem 4.5. If f ∈ Σn,b satisfies

1 +zf ′′(z)

f ′(z)− zf ′(z)

f(z)≺ 1 + (1− 2δ4)z

1− z

then

−zf ′(z)

f(z)≺ 1 + (1− 2β)z

1− z.

76

Proof. Let p(z) = −zf ′(z)/f(z) for z ∈ D∗. It is clear that

1 +zf ′′(z)

f ′(z)− zf ′(z)

f(z)=

zp′(z)

p(z).


4.3 Subordinations for univalence

Theorem 4.10–4.13 give sufficient conditions for the subordination f ′(z) ≺ (1 +

(1 − 2β)z)/(1 − z) to hold. For β = 0, this latter condition is sufficient for the

close-to-convexity and hence univalence of the function f .

Theorem 4.10 Let n ∈ N, 0 ≤ β < 1 and 0 < µ = (n + 1)b ≤ 2(1 − β). Let δ1

be given as in Theorem 4.2. If f ∈ An,b satisfies following subordination

f ′(z) + zf ′′(z) ≺ 1 + (1− 2δ1)z

1− z,

then

f ′(z) ≺ 1 + (1− 2β)z

1− z.

Proof. For f ∈ An,b, let the function p : D→ C be defined by p(z) = f ′(z). Then

p(z) = 1 + (n + 1)bzn + (n + 2)an+2zn+1 + · · · ∈ Hµ,n, with µ = (n + 1)b. Also,

we have

f ′(z) + zf ′′(z) = p(z) + zp′(z).


Theorem 4.11 Let n ∈ N, ε ≥ 0, 0 ≤ β < 1 and 0 < µ = (n + 1)b ≤ 2(1 − β).

Let δ2 be given as in Theorem 4.3. If f ∈ An,b satisfies following subordination

εzf ′′(z)

f ′(z)+ f ′(z) ≺ 1 + (1− 2δ2)z

1− z,

77

then

f ′(z) ≺ 1 + (1− 2β)z

1− z.

Proof. For p(z) = f ′(z), we have

εzf ′′(z)

f ′(z)+ f ′(z) = p(z) + ε

zp′(z)

p(z),

and the result follows from Lemma 4.3.

Theorem 4.11 reduces to the following result in the case when µ = 2(1 − β),

n = 1 and β = 1/2.

Corollary 4.2 [112, Theorem 2, p. 182] For α ≥ 0, if

Re

((1− α)f ′(z) + α

(1 +

zf ′′(z)

f ′(z)

))>

1

2,

then Re f ′(z) > 1/2.

Theorem 4.12 Let n ∈ N, α ≥ 0, 0 ≤ β < 1 and 0 < µ = (n + 1)b ≤ 2(1 − β).

Let δ3 be given as in Theorem 4.4. If f ∈ An,b satisfies

f ′(z)

[α

(zf ′′(z)

f ′(z)+ f ′(z)− 1

)+ 1

]≺ 1 + (1− 2δ3)z

1− z,

then

f ′(z) ≺ 1 + (1− 2β)z

1− z.

Proof. For f ∈ An,b, let the function p : D→ C be defined by p(z) = f ′(z). Since

f ′(z)

(α

(zf ′′(z)

f ′(z)+ f ′(z)− 1

)+ 1

)= (1− α)p(z) + αp2(z) + αzp′(z),

the result follows from Lemma 4.4.

78

Theorem 4.13 Let n ∈ N, 0 ≤ β < 1 and 0 < µ = (n + 1)b ≤ 2(1 − β). Let δ4

be given as in Theorem 4.5. If f ∈ An,b satisfies the subordination

zf ′′(z)

f ′(z)≺ 1 + (1− 2δ4)z

1− z

then

f ′(z) ≺ 1 + (1− 2β)z

1− z.

Proof. For p(z) = f ′(z),

zf ′′(z)

f ′(z)=

zp′(z)

p(z),

hence the result follows from Lemma 4.6.

If µ = 2(1 − β), n = 1 and β = (α + 1)/2, Theorem 4.13 reduces to the

following result.

Corollary 4.3 [94, Theorem 1] Let the function f ∈ A satisfy the inequality

Re

(1 +

zf ′′(z)

f ′(z)

)>

1 + 3α

2(1 + α), 0 ≤ α < 1,

then Re f ′(z) > (α + 1)/2.

79

CHAPTER 5

CONVEXITY OF FUNCTIONS SATISFYING CERTAIN

DIFFERENTIAL INEQUALITIES AND INTEGRAL OPERATORS


Let H(D) denote the class of analytic functions f defined in the open unit disk D.

In this chapter, instead of considering functions with fixed second coefficient as in

Chapter 4, we consider the more general function. For a ∈ C, and n a positive

integer, let

H[a, n] =

f ∈ H(D) : f(z) = a +

∞∑

k=n

akzk

,

and

An =

f ∈ H(D) : f(z) = z +

∞∑

k=n+1

akzk

.

For 0 ≤ β < 1, recall ST (β) the subclass of A consisting of functions starlike of

order β satisfying

Re

(zf ′(z)

f(z)

)> β, z ∈ D.

Also for 0 ≤ β < 1, CV(β) is the subclass of A consisting of functions convex of

order β satisfying

Re

(1 +

zf ′′(z)

f ′(z)

)> β, z ∈ D.

Several authors investigated the sufficient conditions to ensure starlikeness of

functions. These include conditions in terms of differential inequalities, see for

example [4,5,12,15,32,89,109,135]. Miller and Mocanu [62], Kuroki and Owa [49],

and Ali et al. [9], determined conditions for starlikeness of functions defined by an

integral operator of the form

f(z) =

∫ 1

0W (r, z)dr,

80

or by the double integral operator

f(z) =

∫ 1

0

∫ 1

0W (r, s, z)drds.

More recently, Chandra et al. [28], obtained sufficient conditions for starlikeness

of positive order for analytic functions satisfying certain third-order differential

inequalities.

In this chapter, conditions that would imply convexity of positive order for

functions satisfying certain second-order and third-order differential inequalities

are determined. As a consequence, conditions on the kernel of certain integral

operators are also obtained to ensure functions defined by these operators are

convex.

We are going to use the concepts of subordination to help us to determine the

convexity conditions. The following properties of differential subordination are

needed.

Lemma 5.1 [38, Theorem 1, p. 192](see also [61, Theorem 3.1b, p. 71]) Let h be

convex in D with h(0) = a, γ 6= 0 and Re γ ≥ 0. If p ∈ Hn(a) and

p(z) +zp′(z)

γ≺ h(z),

then

p(z) ≺ q(z) ≺ h(z),

where

q(z) =γ

nzγ/n

∫ z

0h(t)t(γ/n)−1dt.

The function q is convex and is the best (a, n)−dominant.

Lemma 5.2 [123](see also [61, Theorem 3.1d, p. 76]) Let h be a starlike function

81

with h(0) = 0. If p ∈ Hn(a) satisfies

zp′(z) ≺ h(z),

then

p(z) ≺ q(z) = a +1

n

∫ z

0

h(t)

tdt.

The function q is convex and is the best (a, n)−dominant.

Lemma 5.3 [3, Theorem 1, p. 13] Let n be a positive integer and α real, with

0 ≤ α < n. Let q be analytic with q(0) = 0, q′(0) 6= 0 and

Rezq′′(z)

q′(z)+ 1 >

α

n.

If p ∈ Hn(0) satisfies

zp′(z)− αp(z) ≺ nzq′(z)− αq(z),

then p(z) ≺ q(z) and this result is sharp.

5.2 Convexity of functions satisfying second-order differential inequal-

ities

The following theorem discusses on the function F (f ′, f ′′) = δzf ′′(z)−α(f ′(z)−1).

Theorem 5.1 Let f ∈ An, δ > 0, 0 ≤ α < nδ and 0 ≤ β < 1. If

|δzf ′′(z)− α(f ′(z)− 1)| < (1− β)(nδ − α)

n + 1− β, (5.1)

then f ∈ CV(β).

82

Proof. We want to show that Re(1 +

zf ′′(z)f ′(z)

)> β. For notation simplicity, let

Q(z) = 1 +zf ′′(z)

f ′(z).

Then

f ′(z)[δ(Q(z)− 1)− α] = δzf ′′(z)− α(f ′(z)− 1)− α.

So by the triangle inequality and (5.1),

|f ′(z)||δ(Q(z)− 1)− α| ≤ |δzf ′′(z)− α(f ′(z)− 1|+ α

<(1− β)(nδ − α)

n + 1− β+ α. (5.2)

To determine the bound of |f ′(z)|, note that the inequality (5.1) can be expressed

in the subordination form

δzf ′′(z)− α(f ′(z)− 1

) ≺ (1− β)(nδ − α)

n + 1− βz, z ∈ D.

By writing

P (z) = δf ′(z)− (δ + α)f(z)

z= −α + (nδ − α)an+1z

n + · · · ∈ Hn(−α),

it follows that

P (z) + zP ′(z) = z

[δf ′′(z)− (δ + α)

(f ′(z)

z− f(z)

z2

)]+ δf ′(z)− (δ + α)

f(z)

z

= δzf ′′(z)− αf ′(z)

≺ (1− β)(nδ − α)

n + 1− βz − α.

83

Let

h(z) =(1− β)(nδ − α)

n + 1− βz − α.

Then h is convex and h(0) = −α, and hence Lemma 5.1 with γ = 1 yields

P (z) = δf ′(z)− (δ + α)f(z)

z

≺ 1

nz1/n

∫ z

0

[(1− β)(nδ − α)

n + 1− βt− α

]t(1/n)−1dt

=1

nz1/n

∫ z

0

[(1− β)(nδ − α)

n + 1− βt(1/n) − αt(1/n)−1

]dt

=(1− β)(nδ − α)

(n + 1)(n + 1− β)z − α. (5.3)

Now consider

p(z) = δ

(f(z)

z− 1

)and q(z) =

δ(1− β)

(n + 1)(n + 1− β)z.

Then

p ∈ Hn(0), q(0) = 0, q′(0) =δ(1− β)

(n + 1)(n + 1− β)6= 0, Re

(zq′′(z)

q′(z)+ 1

)= 1 >

α

δn.

Also

zp′(z)− α

δp(z) = zδ

(f ′(z)

z− f(z)

z2

)− α

δδ

(f(z)

z− 1

)

= δf ′(z)− (δ + α)f(z)

z+ α,

and

nzq′(z)− α

δq(z) = nz

δ(1− β)

(n + 1)(n + 1− β)− α

δ

δ(1− β)

(n + 1)(n + 1− β)z

=(1− β)(nδ − α)

(n + 1)(n + 1− β)z.

84

Then the subordination (5.3) can be written as

zp′(z)− α

δp(z) ≺ nzq′(z)− α

δq(z).

Applying Lemma 5.3 gives p ≺ q, which implies

f(z)

z≺ 1 +

1− β

(n + 1)(n + 1− β)z.

By the definition of subordination, we have

f(z)

z= 1 +

1− β

(n + 1)(n + 1− β)w(z) (|w(z)| < 1),

∣∣∣∣f(z)

z

∣∣∣∣ =

∣∣∣∣1 +1− β

(n + 1)(n + 1− β)w(z)

∣∣∣∣

> 1− 1− β

(n + 1)(n + 1− β)|w(z)|

> 1− 1− β

(n + 1)(n + 1− β).

From subordination (5.3) and the inequality −∣∣a− b

∣∣ ≤ |a| − |b|,

∣∣∣∣δf ′(z)− (δ + α)f(z)

z

∣∣∣∣ <

∣∣∣∣(1− β)(nδ − α)

(n + 1)(n + 1− β)z − α

∣∣∣∣ ,

|δf ′(z)| −∣∣(δ + α)

f(z)

z

∣∣ > −∣∣∣∣

(1− β)(nδ − α)

(n + 1)(n + 1− β)z − α

∣∣∣∣

δ|f ′(z)| > (δ + α)

∣∣∣∣f(z)

z

∣∣∣∣−∣∣∣∣

(1− β)(nδ − α)

(n + 1)(n + 1− β)z − α

∣∣∣∣

> (δ + α)

(1− (1− β)

(n + 1)(n + 1− β)

)−

((1− β)(nδ − α)

(n + 1)(n + 1− β)+ α

),

|f ′(z)| > n

n + 1− β.

Substituting this bound of |f ′(z)| in (5.2), we get

n

n + 1− β|δ(Q(z)− 1)− α| < (1− β)(nδ − α)

(n + 1− β)+ α,

85

|Q(z)− 1− α

δ| < 1− β +

α

δ.

Since −|w| ≤ Re w, it follows that

Re(Q(z)− 1− α

δ

)> −

(1− β +

α

δ

).

Hence Re Q(z) > β, i.e.,

Re

(1 +

zf ′′(z)

f ′(z)

)> β,

as desired.

Theorem 5.2 Let δ > 0, 0 ≤ α < nδ, 0 ≤ β < 1 and g ∈ H. If

|g(z)| < (1− β)(nδ − α)

n + 1− β,

then f ∈ An given by

f(z) = z +zn+1

δ

∫ 1

0

∫ 1

0g(rsz)r[(n−1)δ−α]/δsndrds

is a convex function of order β.

Proof. Let f ∈ An satisfying the differential equation

δzf ′′(z)− α(f ′(z)− 1) = zng(z). (5.4)

From Theorem 5.1, it is clear that the solution f of (5.4) is a convex function of

order β. We find the closed form of f . Let φ(z) = f ′(z)− 1, then (5.4) becomes

δzφ′(z)− αφ(z) = zng(z). (5.5)

86

Using the integrating factor e∫ −α

δz dz = z−αδ , (5.5) becomes

z−αδ φ′(z)− α

δz−αδ −1φ(z) =

zn−1−αδ

δg(z)

d

dz

[z−αδ φ(z)

]=

zn−1−αδ

δg(z)

z−αδ φ(z) =

1

δ

∫ z

0ξn−1−α

δ g(ξ)dξ

φ(z) =z

αδ

δ

∫ z

0g(ξ)ξn−1−α

δ dξ

=z

αδ

δ

∫ 1

0g(rz)(rz)n−1−α

δ zdr

=zn

δ

∫ 1

0g(rz)rn−1−α

δ dr.

Since φ(z) = f ′(z)− 1, we have

f ′(z)− 1 =zn

δ

∫ 1

0g(rz)rn−1−α

δ dr

f(z) = z +

∫ z

0

ξn

δ

∫ 1

0g(rξ)rn−1−α

δ drdξ

= z +

∫ 1

0

(sz)n

δ

∫ 1

0g(rsz)rn−1−α

δ zdrds

= z +zn+1

δ

∫ 1

0

∫ 1

0g(rsz)r[(n−1)δ−α]/δsndrds.


The next theorem discusses on the function F (f, f ′, f ′′) = δzf ′′(z) − α(f ′(z)

−f(z)/z).

Theorem 5.3 Let f ∈ An, δ > 0, and 0 ≤ α < δ, with 0 ≤ β < 1. If

∣∣∣∣δzf ′′(z)− α

(f ′(z)− f(z)

z

)∣∣∣∣ <n(1− β)

(δ(n + 1)− α

)

(n + 1)(n + 1− β), (5.6)

then f ∈ CV(β).

87

Proof. We again want to show that Re(1 +

zf ′′(z)f ′(z)

)> β. Let

Q(z) = 1 +zf ′′(z)

f ′(z).

Then rewriting (5.6), we have

∣∣∣∣δf ′(z)(Q(z)− 1)− α

(f ′(z)− f(z)

z

)∣∣∣∣ <n(1− β)[δ(n + 1)− α]

(n + 1)(n + 1− β)

and after some computation we have

δ|f ′(z)||Q(z)− 1| < n(1− β)[δ(n + 1)− α]

(n + 1)(n + 1− β)+ α

∣∣∣∣f ′(z)− f(z)

z

∣∣∣∣ . (5.7)

So we need to determine the bounds for |f ′(z)| and∣∣f ′(z)− f(z)/z

∣∣.First, to determine the bound of

∣∣f ′(z)− f(z)/z∣∣, note that the inequality

(5.6) can be expressed in the subordination form

δzf ′′(z)− α

(f ′(z)− f(z)

z

)≺ n(1− β)(δ(n + 1)− α)

(n + 1)(n + 1− β)z, z ∈ D. (5.8)

Writing

P (z) = δ

(f ′(z)− f(z)

z

)= δnan+1z

n + · · · ∈ Hn(0),

it follows that

(δ − α

δ

)P (z) + zP ′(z)

= z

[δf ′′(z)− δ

(f ′(z)

z− f(z)

z2

)]+

δ − α

δδ

(f ′(z)− f(z)

z

)

= δzf ′′(z)− α

(f ′(z)− f(z)

z

)

≺ n(1− β)(δ(n + 1)− α)

(n + 1)(n + 1− β)z.

88

By letting

h(z) =δn(1− β)(δ(n + 1)− α)

(δ − α)(n + 1)(n + 1− β)z,

then h is convex and h(0) = 0, and hence by applying Lemma 5.1 with γ =

(δ − α)/δ yields

P (z) = δ

(f ′(z)− f(z)

z

)

≺ (δ − α)

δ

1

nz(δ−α)/δn

∫ z

0

(δn(1− β)(δ(n + 1)− α)

(δ − α)(n + 1)(n + 1− β)t

)t(δ−α)/δn−1dt

=(1− β)(δ(n + 1)− α)

(n + 1)(n + 1− β)

1

z(δ−α)/δn

∫ z

0t(δ−α)/δndt

=δn(1− β)

(n + 1)(n + 1− β)z,

which implies

f ′(z)− f(z)

z≺ n(1− β)

(n + 1)(n + 1− β)z, (5.9)

or equivalently ∣∣∣∣f ′(z)− f(z)

z

∣∣∣∣ <n(1− β)

(n + 1)(n + 1− β). (5.10)

Next consider

p(z) =f(z)

z, zp′(z) = f ′(z)− f(z)

z.

Then (5.9) can be rewritten as

zp′(z) ≺ n(1− β)

(n + 1)(n + 1− β)z.

Applying Lemma 5.2,

p(z) =f(z)

z≺ 1 +

1

n

∫ z

0

n(1− β)

(n + 1)(n + 1− β)dt = 1 +

(1− β)

(n + 1)(n + 1− β)z,

89

and hence (same computation as in the proof of Theorem 5.1)

∣∣∣∣f(z)

z

∣∣∣∣ > 1− (1− β)

(n + 1)(n + 1− β). (5.11)

Using −|a− b| ≤ |a| − |b|, and (5.11), (5.10) becomes

∣∣f ′(z)∣∣−

∣∣∣∣f(z)

z

∣∣∣∣ >−n(1− β)

(n + 1)(n + 1− β)

|f ′(z)| >(

1− (1− β)

(n + 1)(n + 1− β)

)−

(n(1− β)

(n + 1)(n + 1− β)

)

=n

n + 1− β. (5.12)

Substituting (5.12) and (5.10) in (5.7) yields

δn

n + 1− β|Q(z)− 1| < n(1− β)[δ(n + 1)− α]

(n + 1)(n + 1− β)+ α

(n(1− β)

(n + 1)(n + 1− β)

),

=δn(1− β)

n + 1− β,

|Q(z)− 1| < 1− β.

Since −|w| ≤ Re w, it follows that

Re (Q(z)− 1) > −(1− β),

and hence Re Q(z) > β. This completes the proof.

Theorem 5.4 Let δ > 0, 0 ≤ α < δ, 0 ≤ β < 1 and g ∈ H. If

|g(z)| < n(1− β)(δ(n + 1)− α)

(n + 1)(n + 1− β),

then f ∈ An defined by

f(z) = z +zn+1

δ

∫ 1

0

∫ 1

0g(rsz)r(nδ−α)/δsn−1drds

90


Proof. Let f ∈ An satisfying


(f ′(z)− f(z)

z

)= zng(z). (5.13)

From Theorem 5.3, it is clear that the solution f of (5.13) is a convex function of

order β. Let

φ(z) = δ

(f ′(z)− f(z)

z

),

then

zφ′(z) +(δ − α)

δφ(z) = zδ

[f ′′(z)−

(f ′(z)

z− f(z)

z2

)]+

δ − α

δδ

(f ′(z)− f(z)

z

)

= δzf ′′(z)− α

(f ′(z)− f(z)

z

)

= zng(z). (5.14)

Using the integrating factor e∫

δ−αδz dz = z

δ−αδ , (5.14) becomes

zδ−α

δ φ′(z) +(δ − α)

δz

δ−αδ −1φ(z) = zn−1+ δ−α

δ g(z)

d

dz

[z

δ−αδ φ(z)

]= zn−1+ δ−α

δ g(z)

zδ−α

δ φ(z) =

∫ z

0ξn−1+ δ−α

δ g(ξ)dξ

φ(z) = z−δ−α

δ

∫ z

0g(ξ)ξn−1+ δ−α

δ dξ

= z−δ−α

δ

∫ 1

0g(rz)(rz)n−1+ δ−α

δ zdr

= zn∫ 1

0g(rz)rn−1+ δ−α

δ dr.

91

Substituting φ(z) = δ(f ′(z)− f(z)/z

), we have

δ

(f ′(z)− f(z)

z

)= zn

∫ 1


δ dr

δz

(f(z)

z

)′= zn

∫ 1


δ dr

(f(z)

z

)′=

zn−1

δ

∫ 1


δ dr.

Another integration gives

f(z)

z− 1 =

∫ z

0

ξn−1

δ

∫ 1

0g(rξ)rn−1+ δ−α

δ drdξ

=1

δ

∫ 1

0(zs)n−1

∫ 1

0g(rsz)rn−1+ δ−α

δ zdrds

=1

δ

∫ 1

0(zs)n−1

∫ 1

0g(rsz)rn−1+ δ−α

δ zdrds

f(z) = z +zn+1

δ

∫ 1

0

∫ 1

0g(rsz)r(nδ−α)/δsn−1drds.


5.3 Convexity of functions satisfying third-order differential inequali-

ties

Theorem 5.5 Let f ∈ An, 0 < α < nν, µ > 0 and 0 ≤ β < 1. If

|γz2f ′′′(z) + δzf ′′(z)− α(f ′(z)− 1)| < (1 + nµ)(1− β)(nν − α)

n + 1− β, (5.15)

where ν − αµ = δ − γ and νµ = γ, then f ∈ CV(β).

Proof. Let

p(z) = νzf ′′(z)− α(f ′(z)− 1).

92

Then

p(z) + µzp′(z) = νzf ′′(z)− α(f ′(z)− 1) + µz[νf ′′(z) + νzf ′′′(z)− αf ′′(z)]

= µνz2f ′′′(z) + zf ′′[ν − µα + µν]− α(f ′(z)− 1)

= γz2f ′′′(z) + δzf ′′(z)− α(f ′(z)− 1), (5.16)

where µν = γ, ν − µα + µν = δ. Hence, by (5.16) and (5.15), we have

p(z) + µzp′(z) ≺ (1 + nµ)(1− β)(nν − α)

n + 1− βz,

p(z) +zp′(z)

1µ

≺ (1 + nµ)(1− β)(nν − α)

n + 1− βz.

Apply Lemma 5.1 to obtain

p(z) = νzf ′′(z)− α(f ′(z)− 1)

≺ 1

µ.

1

nz1/nµ

∫ z

0

(1 + nµ)(1− β)(nν − α)

n + 1− βt.t(1/nµ)−1dt

=1

µnz1/nµ

∫ z

0

(1 + nµ)(1− β)(nν − α)

n + 1− βt(1/nµ)dt

=1

µnz1/nµ

[nµ(1 + nµ)(1− β)(nν − α)

(1 + nµ)(n + 1− β)z(1/nµ)+1

]

=(1− β)(nν − α)

n + 1− βz

which yields

|νzf ′′(z)− α(f ′(z)− 1)| < (1− β)(nν − α)

n + 1− β.

Hence, applying Theorem 5.1, f ∈ CV(β).

Example 5.1 For the function

f(z) = z +(1− β)

(n + 1)(n + 1− β)zn+1, 0 ≤ β < 1,

93

we have

|γz2f ′′′(z) + δzf ′′(z)− α(f ′(z)− 1)|

=

∣∣∣∣γn(n− 1)(1− β)zn

n + 1− β+

δn(1− β)zn

n + 1− β− α(1− β)zn

n + 1− β

∣∣∣∣

=

∣∣∣∣(1− β)(1 + nµ)(nν − α)

n + 1− β

∣∣∣∣ |z|n

<(1− β)(1 + nµ)(nν − α)

n + 1− β.

So by Theorem 5.5, f ∈ CV(β). Indeed we have

Re

(1 +

zf ′′(z)

f ′(z)

)= Re

(1 + ((n + 1)(1− β)/(n + 1− β))zn

1 + ((1− β)/(n + 1− β))zn

)

>1− ((n + 1)(1− β)/(n + 1− β))

1− ((1− β)/(n + 1− β))= β.

Theorem 5.6 Let 0 < α < nν, µ > 0, 0 ≤ β < 1 and g ∈ H. If

|g(z)| < (1 + µn)(1− β)(nν − α)

n + 1− β,

then f ∈ An defined by

f(z) = z +zn+1

µν

∫ 1

0

∫ 1

0

∫ 1

0g(rstz)rn−1−α

ν sntn−1+ 1

µ drdsdt


Proof. Let f ∈ An satisfying

γz2f ′′′(z) + δzf ′′(z)− α(f ′(z)− 1)

)= zng(z) (5.17)

where ν−αµ = δ−γ and νµ = γ. From Theorem 5.5, we note that the solution of

differential equation (5.17) is convex of order β. Let p(z) = νzf ′′(z)−α(f ′(z)−1).

94

Then equation (5.17) becomes

p(z) + µzp′(z) = zng(z). (5.18)

Using the integrating factor e∫

1µz dz

= z1µ , equation (5.18) simplifies to

zp′(z) +1

µp(z) =

zn

µg(z)

z1µ p′(z) +

(1

µ

)z

1µ−1

p(z) =zn−1+ 1

µ

µg(z)

d

dz

[z

1µ p(z)

]=

zn−1+ 1

µ

µg(z)

z1µ p(z) =

1

µ

∫ z

0ξn−1+ 1

µ g(ξ)dξ

p(z) =z− 1

µ

µ

∫ z

0g(ξ)ξ

n−1+ 1µ dξ

=z− 1

µ

µ

∫ 1

0g(tz)(tz)

n+ 1µ−1

zdt

=zn

µ

∫ 1

0g(tz)t

n+ 1µ−1

dt

= znφ(z),

where

φ(z) =1

µ

∫ 1

0g(tz)t

n+ 1µ−1

dt. (5.19)

Note that the function f in Theorem 5.2 satisfies δzf ′′(z) − α(f ′(z) − 1) =

zng(z). Then by replacing the appropriate parameters, the equation

νzf ′′(z)− α(f ′(z)− 1) = znφ(z)

95

has a solution

f(z) = z +zn+1

ν

∫ 1

0

∫ 1

0φ(rsz)rn−α

ν−1sndrds

= z +zn+1

ν

∫ 1

0

∫ 1

0

[1

µ

∫ 1

0g(trsz)t

n+ 1µ−1

dt

]rn−α

ν−1sndrds

= z +zn+1

µν

∫ 1

0

∫ 1

0

∫ 1

0g(rstz)rn−α

ν−1sntn+ 1

µ−1drdsdt.


Theorem 5.7 Let f ∈ An, 0 < µ < 1, 0 ≤ α < (1− µ)ν and 0 ≤ β < 1. If

∣∣∣∣γz2f ′′′(z) + δzf ′′(z)− α

(f ′(z)− f(z)

z

)∣∣∣∣ <n[(n + 1)(ν − γ)− α](1− β)(1 + nµ)

(1− µ)(n + 1)(n + 1− β),

(5.20)

where ν − αµ1−µ = δ − γ and νµ = γ, then f ∈ CV(β).

Proof. Let p(z) = νzf ′′(z)− α1−µ

(f ′(z)− f(z)

z

). Then

p(z) + µzp′(z) = νzf ′′(z)− α

1− µ

(f ′(z)− f(z)

z

)+ µνz2f ′′′(z) + µνzf ′′(z)

− µα

1− µzf ′′(z) +

µα

1− µ

(f ′(z)− f(z)

z

)

= µνz2f ′′′(z) +

(µν − αµ

1− µ+ ν

)zf ′′(z)− α

(f ′(z)− f(z)

z

)

= γz2f ′′′(z) + δzf ′′(z)− α

(f ′(z)− f(z)

z

), (5.21)

where µν = γ and µν − αµ1−µ + ν = δ.

By (5.21) and (5.20),

p(z) + µzp′(z) ≺ n(1− β)(1 + nµ)[(n + 1)(ν − γ)− α]

(1− µ)(n + 1)(n + 1− β)z.

96

By applying Lemma 5.1, we obtain

p(z) ≺ 1

µ

1

nz1

nµ

∫ z

0

(1 + nµ)(1− β)

(n + 1− β)(n + 1)(1− µ)n[(n + 1)(ν − γ)− α]tt

1nµ−1

dt

=1

µnz1

nµ

∫ z

0

(1 + nµ)(1− β)

(n + 1− β)(n + 1)(1− µ)n[(n + 1)(ν − γ)− α]t

1nµ dt

=n(1− β)[(n + 1)(ν − γ)− α]

(1− µ)(n + 1)(n + 1− β)z,

=n(1− β)

(n + 1)(n + 1− β)

[(n + 1)(ν − νµ)

(1− µ)− α

(1− µ)

]z

=n(1− β)

(n + 1)(n + 1− β)

[ν(n + 1)− α

(1− µ)

]z.

Since p(z) = νzf ′′(z)− α1−µ

(f ′(z)− f(z)

z

), the above subordination implies

∣∣∣∣νzf ′′(z)− α

1− µ

(f ′(z)− f(z)

z

)∣∣∣∣ <n(1− β)

(ν(n + 1)− α

1−µ

)

(n + 1)(n + 1− β).

Hence, by Theorem 5.3, f ∈ CV(β).

Theorem 5.8 Let 0 < µ < 1, 0 ≤ α < (1− µ)ν, 0 ≤ β < 1 and g ∈ H. If

|g(z)| < n(1− β)(1 + nµ)[(n + 1)(ν − γ)− α]

(1− µ)(n + 1)(n + 1− β),

where νµ = γ, then

f(z) = z +zn+1

γ

∫ 1

0

∫ 1

0

∫ 1

0g(rstz)r

1ν [nν− α

1−µ ]snt

n−1+ 1µ drdsdt


Proof. Suppose f ∈ An satisfies

γz2f ′′′(z) + δzf ′′(z)− α

(f ′(z)− f(z)

z

)= zng(z), (5.22)

97

where γ + ν − αµ1−µ = δ. Then by Theorem 5.7, f is a convex function of order β.

Let

p(z) = νzf ′′(z)− α

1− µ

(f ′(z)− f(z)

z

).

Now equation (5.22) can be written as

p(z) + µzp′(z) = zng(z). (5.23)

Similarly as in the proof of Theorem 5.6, equation (5.23) has the solution

p(z) = znφ(z),

that is,

νzf ′′(z)− α

1− µ

(f ′(z)− f(z)

z

)= znφ(z),

where

φ(z) =1

µ

∫ 1

0g(tz)t

n+ 1µ−1

dt.

On the other hand, note that the function f in Theorem 5.4 satisfies


(f ′(z)− f(z)

z

)= zng(z).

Hence by Theorem 5.4, the equation

νzf ′′(z)− α

1− µ

(f ′(z)− f(z)

z

)= znφ(z)

has the solution

f(z) = z +zn+1

ν

∫ 1

0

∫ 1

0φ(rsz)r

1ν [nν− α

1−µ ]sn−1drds

= z +zn+1

ν

∫ 1

0

∫ 1

0

[1

µ

∫ 1

0g(trsz)t

n+ 1µ−1

dt

]r

1ν [nν− α

1−µ ]sn−1drds

98

= z +zn+1

γ

∫ 1

0

∫ 1

0

∫ 1

0g(rstz)r

1ν [nν− α

1−µ ]sn−1t

n−1+ 1µ drdsdt.


99

CHAPTER 6

CLOSE-TO-CONVEXITY AND STARLIKENESS OF

ANALYTIC FUNCTIONS


Let D := z ∈ C : |z| < 1 be the open unit disk. For a fixed p ∈ N :=

1, 2, . . . , let Ap be the class of all analytic functions of the form f(z) = zp +

ap+1zp+1 + ap+2z

p+2 + . . . , that are p-valent (multivalent) in the open unit disk,

with A := A1. Let Ap,n be the class of all analytic functions f : D → C of the

form f(z) = zp+an+pzn+p+an+p+1z

n+p+1+. . . with Ap := Ap,1 and A := A1,1.

For studies related to multivalent functions, see [57,83–86]. Also recall CCV(α),

the subclass of A consisting of functions which are close-to-convex of order α in D

(0 ≤ α < 1) defined by

CCV(α) :=

f : f ∈ A and Re

(f ′(z)

g′(z)

)> α, g ∈ CV

.

Singh and Singh [113] obtained several interesting conditions for functions f ∈ Asatisfying inequalities involving f ′(z) and zf ′′(z) to be univalent or starlike in D.

Owa et al. [94] generalized the results of Singh and Singh [113] and also obtained

several sufficient conditions for close-to-convexity, starlikeness and convexity of

functions f ∈ A. In fact, they proved the following theorems.

Theorem 6.1 [94, Theorems 1-3] Let 0 ≤ α < 1 and β, γ ≥ 0. If f ∈ A, then

Re

(1 +

zf ′′(z)

f ′(z)

)>

1 + 3α

2(1 + α)=⇒ Re

(f ′(z)

)>

1 + α

2,

Re

(1 +

zf ′′(z)

f ′(z)

)<

3 + 2α

(2 + α)=⇒

∣∣f ′(z)− 1∣∣ < 1 + α,

∣∣f ′(z)− 1∣∣β |zf ′′(z)|γ <

(1− α)β+γ

2β+2γ =⇒ Re(f ′(z)

)>

1 + α

2.

100

Theorem 6.2 [94, Theorem 4] Let 1 < λ < 3. If f ∈ A, then

Re

(1 +

zf ′′(z)

f ′(z)

)<

5λ−12(λ+1) , 1 < λ ≤ 2;

λ+12(λ−1) , 2 < λ < 3,

=⇒ zf ′(z)

f(z)≺ λ(1− z)

λ− z.

In this chapter, the above results are extended to functions f ∈ Ap,n.

6.2 Close-to-convexity and Starlikeness

For the proof of our main results, we need the following lemma.

Lemma 6.1 [61, Lemma 2.2a] Let z0 ∈ D and r0 = |z0|. Let f(z) = anzn +

an+1zn+1 + · · · be continuous on Dr0 and analytic on Dr0 ∪ z0 with f(z) 6≡ 0

and n ≥ 1. If

|f(z0)| = max|f(z)| : z ∈ Dr0,

then there exists an m ≥ n such that

1.z0f

′(z0)

f(z0)= m, and

2. Rez0f

′′(z0)

f ′(z0)+ 1 ≥ m.

Recall that for f ∈ A, the condition Re f ′(z) > 0 implies the close-to-convexity

and univalence of f . Similarly, for f ∈ Ap, the inequality Re(f ′(z)/zp−1) > 0

implies p-valency of f , see [127,128]. A function f ∈ Ap is close-to-convex if there

is a p-valent convex function φ such that Re(f ′(z)/φ(z)) > 0. In particular, they

are all close-to-convex with respect to φ(z) = zp.

Theorem 6.3 Let 0 ≤ α < 1. If the function f ∈ Ap,n satisfies the inequality

Re

(1 +

zf ′′(z)

f ′(z)

)>

(2p− n) + α(2p + n)

2(α + 1), (6.1)

101

then

Re

(f ′(z)

pzp−1

)>

1 + α

2.

Proof. Let the function w be defined by

f ′(z)

pzp−1 =1 + αw(z)

1 + w(z). (6.2)

Then w can be written as

w(z) =1

α− 1

[(n + p)

pan+pz

n − (n + p)2

p2(1− α)a2n+pz

2n + · · ·]

,

hence it is analytic in D with w(0) = 0. From (6.2), logarithmic differentiation

yield

1 +zf ′′(z)

f ′(z)= p +

αzw′(z)

1 + αw(z)− zw′(z)

1 + w(z). (6.3)

Suppose there exists a point z0 ∈ D such that

|w(z0)| = 1 and |w(z)| < 1 when |z| < |z0|.

Applying Lemma 6.1, there exists m ≥ n such that

z0w′(z0) = mw(z0), (w(z0) = eiθ; θ ∈ R). (6.4)

Thus, by using (6.3) and (6.4), it follows that

Re

(1 +

z0f′′(z0)

f ′(z0)

)= p + Re

(αmw(z0)

1 + αw(z0)

)− Re

(mw(z0)

1 + w(z0)

)

= p + Re

(αmeiθ

1 + αeiθ

)− Re

(meiθ

1 + eiθ

).

102

Note that

αmeiθ

1 + αeiθ=

αmeiθ(1 + αe−iθ)

(1 + αeiθ)(1 + αe−iθ)=

α2m + αm cos θ + iαm sin θ

1 + α2 + 2α cos θ,

so

Re

(αmeiθ

1 + αeiθ

)=

α2m + αm cos θ

1 + α2 + 2α cos θ.

Similarly,

Re

(meiθ

1 + eiθ

)=

m

2.

Hence

Re

(1 +

z0f′′(z0)

f ′(z0)

)= p +

αm(α + cos θ)

1 + α2 + 2α cos θ− m

2

= p− m(α + 1)(1− α)

2(1 + α2 + 2α cos θ)

≤ p− m(1− α)

2(1 + α)

≤ p− n(1− α)

2(1 + α)

=(2p− n) + α(2p + n)

2(α + 1),

which contradicts the hypothesis (6.1). It follows that |w(z)| < 1, for all z ∈ D,

that is, ∣∣∣∣∣∣1− f ′(z)

pzp−1

f ′(z)pzp−1 − α

∣∣∣∣∣∣< 1,

or equivalently

Re

(f ′(z)

pzp−1

)>

1 + α

2.

A function f ∈ Ap is starlike if

Re1

p

zf ′(z)

f(z)> 0.

103

Owa [91] shows that a function f ∈ Ap satisfying Re(1 + zf ′′(z)/f ′(z)) < p + 1/2

implies f is p-valently starlike (that is, f ∈ Ap is starlike). Our next theorem

investigates the close-to-convexity of this type of functions. For related results,

see [54, 93,136].

Theorem 6.4 For 0 ≤ α < 1, if the function f ∈ Ap,n satisfies the inequality

Re

(1 +

zf ′′(z)

f ′(z)

)<

(p + n)α + (2p + n)

(α + 2), (6.5)

then ∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣ < 1 + α.

Proof. Consider the function w defined by

f ′(z)

pzp−1 = (1 + α)w(z) + 1. (6.6)

It can be checked similarly as in Theorem 6.3 that w is analytic in D with w(0) = 0.

From (6.6), some computation yields

1 +zf ′′(z)

f ′(z)= p +

(1 + α)zw′(z)

(1 + α)w(z) + 1. (6.7)


|w(z0)| = 1 and |w(z)| < 1 when |z| < |z0|.

By applying Lemma 6.1, there exists m ≥ n such that

z0w′(z0) = mw(z0), (w(z0) = eiθ; θ ∈ R). (6.8)

104

Thus, by using (6.7) and (6.8), it follows that

Re

(1 +

z0f′′(z0)

f ′(z0)

)= p + Re

((1 + α)mw(z0)

(1 + α)w(z0) + 1

)

= p + Re

((1 + α)meiθ

(1 + α)eiθ + 1

).

Note that

(1 + α)meiθ

(1 + α)eiθ + 1=

(1 + α)2m + (1 + α)m cos θ + i(1 + α)m sin θ

1 + (1 + α)2 + 2 cos θ(1 + α),

so

Re

((1 + α)meiθ

(1 + α)eiθ + 1

)=

m(1 + α)(1 + α + cos θ)

1 + (1 + α)2 + 2 cos θ(1 + α).

Hence

Re

(1 +

z0f′′(z0)

f ′(z0)

)= p +

m(1 + α)(1 + α + cos θ)

1 + (1 + α)2 + 2 cos θ(1 + α)

≥ p +m(1 + α)

(α + 2)≥ p +

n(1 + α)

(α + 2)

=(p + n)α + (2p + n)

(α + 2),

which contradicts the hypothesis (6.5). It follows that |w(z)| < 1, for all z ∈ D,

that is, ∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣ < 1 + α.


Owa [92] has also showed that a function f ∈ A satisfying |f ′(z)/g′(z) −1|β |zf ′′(z)/g′(z) − zf ′(z)g′′(z)/(g′(z))2|γ < (1 + α)β+α, for 0 ≤ α < 1, β ≥ 0,

γ ≥ 0 and g a convex function, is close-to-convex. Also, see [53]. Our next theorem

investigates the close-to-convexity of similar class of functions.

105

Theorem 6.5 Let 0 ≤ α < 1 and β, γ ≥ 0. If f ∈ Ap,n, then

∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣β ∣∣∣∣

f ′′(z)

zp−2 − (p− 1)f ′(z)

zp−1

∣∣∣∣γ

<(pn)γ(1− α)β+γ

2β+2γ (6.9)

implies

Re

(f ′(z)

pzp−1

)>

1 + α

2,

and ∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣β ∣∣∣∣

f ′′(z)

zp−2 − (p− 1)f ′(z)

zp−1

∣∣∣∣γ

< (pn)γ |1− α|β+γ (6.10)

implies ∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣ < 1− α.

Proof. For the analytic function w defined by

f ′(z)

pzp−1 =1 + αw(z)

1 + w(z), (6.11)

we can rewrite (6.11) as

f ′(z)

pzp−1 − 1 =(α− 1)w(z)

1 + w(z),

which yields ∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣β

=|w(z)|β |1− α|β|1 + w(z)|β . (6.12)

By using logarithmic differentiation on (6.11) and some computation,

f ′′(z)

zp−2 − (p− 1)f ′(z)

zp−1 =p(α− 1)zw′(z)

(1 + w(z))2

or equivalently

∣∣∣∣f ′′(z)

zp−2 − (p− 1)f ′(z)

zp−1

∣∣∣∣γ

=pγ |zw′(z)|γ |1− α|γ

|1 + w(z)|2γ. (6.13)

106

From (6.12) and (6.13), it follows that

∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣β ∣∣∣∣

f ′′(z)

zp−2 − (p− 1)f ′(z)

zp−1

∣∣∣∣γ

=pγ |w(z)|β(1− α)β+γ |zw′(z)|γ

|1 + w(z)|β+2γ .


|w(z0)| = 1 and |w(z)| < 1 when |z| < |z0|.

Then (6.4) and Lemma 6.1 yield

∣∣∣∣∣f ′(z0)

pzp−10

− 1

∣∣∣∣∣β ∣∣∣∣∣

f ′′(z0)

zp−20

− (p− 1)f ′(z0)

zp−10

∣∣∣∣∣γ

=pγ(1− α)β+γ |w(z0)|β |mw(z0)|γ

|1 + eiθ|β+2γ

=pγmγ(1− α)β+γ

(2 + 2 cos θ)(β+2γ)/2

≥ pγnγ(1− α)β+γ

2β+2γ ,

which contradicts the hypothesis (6.9). Hence |w(z)| < 1, for all z ∈ D, that is

∣∣∣∣∣∣1− f ′(z)

pzp−1

f ′(z)pzp−1 − α

∣∣∣∣∣∣< 1,

or equivalently

Re

(f ′(z)

pzp−1

)>

1 + α

2.

For the second implication, now consider the function w defined by

f ′(z)

pzp−1 = 1 + (1− α)w(z). (6.14)

Then ∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣β

= |1− α|β |w(z)|β . (6.15)

107

Also with some computation on (6.14), we get

∣∣∣∣f ′′(z)

zp−2 − (p− 1)f ′(z)

zp−1

∣∣∣∣γ

= pγ |zw′(z)|γ |1− α|γ . (6.16)

From (6.15) and (6.16), it is clear that

∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣β ∣∣∣∣

f ′′(z)

zp−2 − (p− 1)f ′(z)

zp−1

∣∣∣∣γ

= pγ |w(z)|β |1− α|β+γ |zw′(z)|γ .


|w(z0)| = 1 and |w(z)| < 1 when |z| < |z0|.

Then by applying Lemma 6.1 and using (6.4), it follows that

∣∣∣∣∣f ′(z0)

pzp−10

− 1

∣∣∣∣∣β ∣∣∣∣∣

f ′′(z0)

zp−20

− (p− 1)f ′(z0)

zp−10

∣∣∣∣∣γ

= pγ |w(z0)|β |1− α|β+γ |z0w′(z0)|γ

= pγmγ |1− α|β+γ

≥ (pn)γ |1− α|β+γ ,

but now this contradicts the hypothesis (6.10). Hence |w(z)| < 1 and this implies

∣∣∣∣f ′(z)

pzp−1 − 1

∣∣∣∣ < 1− α, or Re

(f ′(z)

pzp−1

)> α.

Therefore the proof is complete.

In the next theorem, we will use the concept of subordination. Again, recall

that for f and g analytic on D, we say f is subordinate to g, written f ≺ g, if there

is an analytic function w : D→ D with w(0) = 0 such that f = g w.

108

Recall that a function f ∈ Ap is starlike if

Re1

p

zf ′(z)

f(z)> 0.

Note that for f ∈ Ap the subordination

1

p

zf ′(z)

f(z)≺ λ(1− z)

λ− z

implies f ∈ Ap is starlike as

Re1

p

zf ′(z)

f(z)= Re

λ(1− w(z))

λ− w(z)> 0.

Theorem 6.6 Let λ1 and λ2 be given by

λ1 =n + 2

4p + n− 2p,

λ2 =n + 2

2− n,

and 1 ≤ λ1 < λ < λ2 ≤ 3. If the function f ∈ Ap,n satisfies the inequality

Re

(1 +

zf ′′(z)

f ′(z)

)<

(4p+n)λ−n2(λ+1) , λ1 < λ ≤ p+n

p ;

n(λ+1)2(λ−1) ,

p+np < λ < λ2,

(6.17)

then

1

p

zf ′(z)

f(z)≺ λ(1− z)

λ− z. (6.18)

The result is sharp for the function f given by

f(z) = zp (λ− z)p(λ−1) . (6.19)

109

Proof. Let w be

1

p

zf ′(z)

f(z)=

λ(1− w(z))

λ− w(z). (6.20)

By doing the logarithmic differentiation on (6.20), we get

1 +zf ′′(z)

f ′(z)=

pλ(1− w(z))

λ− z− zw′(z)

1− w(z)+

zw′(z)

λ− w(z).

Assume that there exists a point z0 ∈ D such that |w(z0)| = 1 and |w(z)| < 1

when |z| < |z0|. By applying Lemma 6.1, it follows that

Re

(1 +

z0f′′(z0)

f ′(z0)

)= Re

(pλ(1− eiθ)

λ− eiθ

)− Re

(meiθ

1− eiθ

)+ Re

(meiθ

λ− eiθ

)

Since

1− eiθ

λ− eiθ=

λ + 1− (1 + λ) cos θ + i(1− λ) sin θ

λ2 + 1− 2λ cos θ,

eiθ

1− eiθ=

cos θ − 1 + i sin θ

2− 2 cos θ,

and

eiθ

λ− eiθ=

λ cos θ − 1 + iλ sin θ

λ2 + 1− 2λ cos θ,

it follows that

Re

(1 +

z0f′′(z0)

f ′(z0)

)=

pλ(λ + 1)(1− cos θ)

λ2 + 1− 2λ cos θ+

m

2+

m(λ cos θ − 1)

λ2 + 1− 2λ cos θ

=2pλ(λ + 1)(1− cos θ) + m(λ2 − 1)

2(λ2 + 1− 2λ cos θ)

=λ + 1

2p +

(λ2 − 1)[(p + m)− pλ]

2(λ2 + 1− 2λ cos θ)

≥ λ + 1

2p +

(λ2 − 1)[(p + n)− pλ]

2(λ2 + 1− 2λ cos θ).

110

For λ1 < λ ≤ (p + n)/p, some computations yield

Re

(1 +

z0f′′(z0)

f ′(z0)

)≥ (4p + n)λ− n

2(λ + 1), (6.21)

and for (p + n)/p < λ < λ2, we get

Re

(1 +

z0f′′(z0)

f ′(z0)

)≥ n(λ + 1)

2(λ− 1). (6.22)

Since (6.21) and (6.22) obviously contradict hypothesis (6.17), it follows that

|w(z)| < 1. Hence by the definition of subordination, (6.20) becomes (6.18).

Finally, for (6.18) to be sharp, consider

1

p

zf ′(z)

f(z)=

λ(1− z)

λ− z. (6.23)

By integrating both sides of the equality and after some arrangement, we get

f(z) = zp (λ− z)p(λ−1) . This completes the proof.

Remark 6.1 The subordination (6.18) can be written in equivalent form as

∣∣∣∣λ(zf ′(z)/pf(z)− 1)

zf ′(z)/pf(z)− λ

∣∣∣∣ < 1,

or by further computation, as

∣∣∣∣1

p

zf ′(z)

f(z)− λ

λ + 1

∣∣∣∣ <λ

λ + 1.

The last inequality shows that f is p-valent starlike in D.

Remark 6.2 When p = 1 and n = 1, Theorems 6.3–6.6 reduce to Theorems 6.1

and 6.2.

111

CONCLUSION

This thesis investigates complex-valued analytic (and meromorphic) functions

on simply connected proper domains. By the Riemann Mapping Theorem, these

functions f can be assumed to take the form f(z) = z +∑∞

n=2 anzn defined on

the unit disk D. The collection of these functions is denoted by A.

The subclass S of A consists of univalent functions. When f ∈ S, then f−1

need not belong to S. Bi-univalent functions are those functions f ∈ S satisfying

both f and f−1 are in S. This thesis finds initial coefficient estimates for the

Ma Minda class of bi-univalent convex and starlike functions. Under this general

framework, it gets more computationally involved, but on the other hand, several

earlier known results are obtained as simple consequences. The results obtained

in this thesis rest on using known coefficient bounds of functions with positive real

part. However the expressions to be optimized involved non-linear combinations

of these coefficients, of which no known estimates are available. Further work

on getting good estimates of these non-linear expressions would lead to better

estimates for the initial coefficients of bi-univalent functions.

This thesis also determines the bound for the second Hankel determinant

H2(2) = a2a4 − a23 of analytic Ma Minda starlike and convex functions. Similar

problems were also treated for related classes defined by subordination. Since the

classes introduced by subordination naturally include several well known classes of

univalent functions, earlier known results for these classes are simple consequences

of the theorems obtained in this thesis. A good continuation to the work done

here would be to investigate the Hankel determinant for transforms of analytic

functions.

For f(z) = z +∑∞

n=2 anzn ∈ S, the sharp estimate |a2| ≤ 2 yields the growth,

distortion as well as the Koebe disk for the class S. In view of the influence

of the second coefficient on the geometric properties of univalent functions, this

112

thesis examines the class of functions with fixed second coefficient. By extending

the theory of differential subordination, certain known properties for the class

S are shown to extend to the class of functions with fixed second coefficient.

In particular, several sufficient conditions related to starlikeness, meromorphic

starlikeness and univalence of normalized analytic functions are derived.

Integral operators involving double and triple integrals from the class A to Aare discussed in this thesis. For the classes of functions treated here, the impor-

tant technique of duality is not applicable. Thus the technique used in this thesis

involves differential subordination and exploited the properties of best dominant

solutions to these differential equations. Convexity conditions for analytic func-

tions defined in the open unit disk satisfying certain second-order and third-order

differential inequalities are obtained. As a consequence, sufficient conditions are

determined that ensure that the double and triple integral operators map functions

in A into convex functions of positive order.

The thesis concludes by finding sufficient conditions for close-to-convexity and

starlikeness of a subclass of multivalent functions. Relevant connections with pre-

viously known results are also made.

113

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PUBLICATIONS

1. R. M. Ali, S. K. Lee, V. Ravichandran and S. Supramaniam, Coefficient

estimates for bi-univalent Ma-Minda starlike and convex functions, Appl.

Math. Lett., 25 (2012), Pages 344-351. (ISI)

2. S. K. Lee, V. Ravichandran and S. Supramaniam, Bounds for the second

Hankel determinant of certain univalent functions, J. Inequal. Appl. 2013,

2013:281, 17 pp. (ISI)

3. S. K. Lee, V. Ravichandran and S. Supramaniam, Applications of differen-

tial subordination for functions with fixed second coefficient to geometric

function theory, Tamsui Oxf. J. Inf. Math. Sci., 29 (2013), no. 2, 267-284.

(SCOPUS)

4. S. K. Lee, V. Ravichandran and S. Supramaniam, Initial coefficients of bi-

univalent functions, Abstr. Appl. Anal. Volume 2014 (2014), Article ID

640856, 6 pages. (ISI)

5. S. K. Lee, V. Ravichandran and S. Supramaniam, Close-to-convexity and

starlikeness of analytic functions, Tamkang Journal of Mathematics, ac-

cepted. (SCOPUS)

6. R. M. Ali, R. Chandrashekar, S. K. Lee and S. Supramaniam, Convexity of

functions defined by differential inequalities and integral operators, submit-

ted.

128

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