13_747_sharma.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 6, 2010, 1093-1112
ISSN 1307-5543 – www.ejpam.com

SPECIAL ISSUE ON COMPLEX ANALYSIS: THEORY AND APPLICATIONS

DEDICATED TO PROFESSOR HARI M. SRIVASTAVA,

ON THE OCCASION OF HIS 70TH
BIRTHDAY

Some Properties of a Class of p-valent Analytic Functions

Associated with Convolution

Poonam Sharma 1,∗, Prachi Srivastava 2

1 Department of Mathematics and Astronomy, University of Lucknow, Lucknow 226007 India
2 Department of Mathematics, Karamat Hussain Muslim Girls P.G. College, Nishatganj, Lucknow,

India

Abstract. In this paper, we define a class ℜg

h

�

p, m,β
�

associated with convolution of p-valent ana-
lytic functions. Some properties in the form of coefficient inequality, growth and distortion bounds,
sufficient conditions with the help of various lemmas, integral means inequality for convolution of
two functions and a set of class preserving integral operators of functions belonging to this class are
studied.
2000 Mathematics Subject Classifications: Primary 30C45, 30C50, 30C55
Key Words and Phrases: Analytic functions, Convolution, Starlike functions, Convex functions, Close-
to-convex functions

1. Introduction

Let Ap denotes a class of functions of the form:

f (z) = zp +

∞
∑

k=1

ap+kzp+k
�

p ∈ N = 1,2,3 . . .
�

, (1)

which are analytic and p-valent in the open unit disk ∆= {z ∈ C : |z| < 1}. Let g,h ∈ Ap be of
the form:

g(z) = zp +

∞
∑

k=1

bp+kzp+k, bp+k ≥ 0 (2)

∗Corresponding author.

Email addresses: poonambaba�yahoo.
om (P. Sharma), pra
hi2384�gmail.
om (P. Srivastava)

http://www.ejpam.com 1093 c© 2010 EJPAM All rights reserved.



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1094

and

h(z) = zp +

∞
∑

k=1

cp+kzp+k, cp+k ≥ 0. (3)

A function f ∈ Ap is said to be p-valently starlike of order α in ∆, if it satisfies the inequality

Re

(

z f
′
(z)

f (z)

)

> α
�

z ∈∆; 0≤ α < p; p ∈ N
�

.

The class of all p-valent starlike functions of order α is denoted by S∗p (α). On the other hand,
a function f ∈ Ap is said to be p-valently convex of order α in ∆, if it satisfies the inequality

Re

(

1+
z f
′′
(z)

f
′
(z)

)

> α
�

z ∈∆; 0≤ α < p; p ∈ N
�

.

The class of all p-valent convex functions of order α is denoted by Kp (α). Furthermore, a
function f ∈ Ap is said to be p-valently close-to-convex of order α in ∆, if it satisfies the
inequality

Re
¦

z1−p f
′
(z)
©

> α
�

z ∈∆; 0≤ α < p; p ∈ N
�

.

The class of all p-valent close-to-convex functions of order α is denoted by CKp (α). If f ∈ Ap

satisfies
�

�

�

�

�

arg
z f
′
(z)

f (z)

�

�

�

�

�

<
β

p

π

2
(z ∈∆) ,

for some 0 < β ≤ p, then f is said to be p-valently strongly starlike function of order β in ∆
and this class is denoted by S

∗
p

�

β
�

. Further, if f ∈ Ap satisfies

�

�

�

�

�

arg

 

1+
z f
′′
(z)

f
′
(z)

!�

�

�

�

�

<
β

p

π

2
(z ∈∆) ,

for some 0 < β ≤ p, then f is said to be p-valently strongly convex function of order β in ∆
and is denote by K p

�

β
�

, the class of all such functions. Also, if f ∈ Ap satisfies

�

�

�arg
¦

z1−p f
′
(z)
©

�

�

� <
β

p

π

2
(z ∈∆) ,

for some 0 < β ≤ p, then f is said to be p-valently strongly close-to-convex function of order
β in ∆ and denote by CK p

�

β
�

the class of all such functions. A convolution (Hadamard
product) of f ∈ Ap of the form (1) with g ∈ Ap of the form (2) is defined by:

�

f ∗ g
�

(z) = zp +

∞
∑

k=1

ap+k bp+kzp+k =
�

g ∗ f
�

(z) . (4)



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1095

Various convolution operators have been defined so far, which can be obtained by taking suit-
able g in (4). For example the convolution in (4) reduces to the operator W

p
q,s(
�

α1,A1
�

) f (z)

involving a Wright’s generalized hypergeometric function

qΨs [z] ≡ qΨs

�
�

α1,A1
�

,
�

α2,A2
�

, . . . ,
�

αq,Aq

�

�

β1, B1
�

,
�

β2, B2
�

, . . . ,
�

βs, Bs

� ; z

�

if g(z) = zp

s
∏

i=1
Γ(βi)

q
∏

i=1
Γ(αi)

qΨs [z], where for αi ∈ C(
αi

Ai
6= 0,−1,−2, . . .), i = 1,2, . . . ,q,

βi ∈ C(
βi

Bi
6= 0,−1,−2, . . .), i = 1,2, . . . , s and Ai > 0, i = 1,2, . . . ,q, Bi > 0, i = 1,2, . . . , s such

that 1+
s
∑

i=1
Bi −

q
∑

i=1
Ai ≥ 0,

qΨs [z] =

∞
∑

k=0

q
∏

i=1
Γ(αi + Aik)

s
∏

i=1
Γ(βi + Bik) k!

zk, z ∈∆, (5)

(
s
∏

i=1
B

Bi

i
≥

q
∏

i=1
A

Ai

i
in case 1+

s
∑

i=1
Bi−

q
∑

i=1
Ai = 0 [15]). The convolution operator W

p
q,s(
�

α1,A1
�

) f (z),

for which

bp+k =

q
∏

i=1

Γ(αi+Ai k)

Γ(αi)

s
∏

i=1

Γ(βi+Bi k)

Γ(βi)
k!

,

is studied by Aouf and Dziok [3, 4], Dziok and Raina [8], and Dziok et al. [9] and Sharma
[25] in their respective work and taking Ai = 1, i = 1,2, . . . ,q, Bi = 1, i = 1,2, . . . , s, for
q ≤ s+ 1, it reduces to Dziok Srivastava operator [10] which involve a generalized hypergeo-
metric function qFs [z] and is defined by

qHp
s

��

α1
��

f (z) = zp
qFs [z] ∗ f (z) (6)

where

qFs [z] = qFs

�

α1,α2, . . .αq;β1,β2, . . .βs; z
�

=

∞
∑

k=0

q
∏

i=1

�

αi

�

k

s
∏

i=1

�

βi

�

k k!

zk, z ∈∆,

the symbol (α)k is the familiar Pochhammer symbol defined by

(α)k =
Γ(α+ k)

Γ(α)
, k ∈ N0.



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1096

The operator qH
p
s

��

α1
��

f (z) includes Hohlov operator [13] which involve Gaussian hy-
pergeometric function 2F1 as well as Carlson and Shaffer operator [6] defined by Saitoh and
Ruschweyh derivative operator [23] (for detail one may refer to [8, 9]). Also, the convolution
(4) reduces to the Salagean operator [24] if

bp+k =

�

p+ k

p

�n

, n ∈ N0

and to a generalized Salagean operator [2], if

bp+k =

�

p+ δk

p

�n

,δ > 0, n ∈ N0.

Further, the convolution (4) reduces to an integral operator involving generalized fractional

integral operator I
λ,µ,ν
0,z , if

bp+k =

�

p+ 1
�

k

�

p−µ+ ν + 1
�

k
�

p−µ+ 1
�

k

�

p+λ+ ν + 1
�

k

and hence
�

f ∗ g
�

(z) = zµ
Γ
�

p−µ+ 1
�

Γ
�

p+λ+ ν + 1
�

Γ
�

p+ 1
�

Γ
�

p−µ+ ν + 1
� I

λ,µ,ν
0,z f

where

I
λ,µ,ν
0,z zρ =

Γ
�

ρ+ 1
�

Γ
�

ρ−µ+ ν + 1
�

Γ
�

ρ−µ+ 1
�

Γ
�

ρ+λ+ ν + 1
�zρ−µ,

�

0≤ λ < 1,ρ >max
�

0,µ− ν
	

− 1
�

. Again, this convolution (4) reduces to the derivative

operator involving generalized fractional derivative operator J
λ,µ,ν
0,z , if

bp+k =

�

p+ 1
�

k

�

p−µ+ ν + 1
�

k
�

p−µ+ 1
�

k

�

p−λ+ ν + 1
�

k

and hence,
�

f ∗ g
�

(z) = zµ
Γ
�

p−µ+ 1
�

Γ
�

p−λ+ ν + 1
�

Γ
�

p+ 1
�

Γ
�

p−µ+ ν + 1
� J

λ,µ,ν
0,z f ,

where

J
λ,µ,ν
0,z zρ =

Γ
�

ρ+ 1
�

Γ
�

ρ−µ+ ν + 1
�

Γ
�

ρ−µ+ 1
�

Γ
�

ρ−λ+ ν + 1
�zρ−µ.

The generalized fractional calculus operators I
λ,µ,ν
0,z and J

λ,µ,ν
0,z defined above are studied in

[5], [20, 26]. These generalized fractional calculus operators reduce to fractional calculus
operators if we take µ = −λ and µ = λ respectively. Let Tp denotes the subclass of Ap

consisting of functions of the form:

f (z) = zp −
∞
∑

k=1

ap+kzp+k, ap+k ≥ 0. (7)

Motivated with the several work specially the work of Prajapat et al. [21], we consider
ℜg

h

�

p, m,β
�

class defined as follows:



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1097

Definition 1. A function f ∈ Tp is said to be a member of the class ℜg

h

�

p, m,β
�

if and only if

for any g, h ∈ Ap with non-negative coefficients,

�

�

�

�

�

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)
−
�

p−m
�

�

�

�

�

�

< β ,

z ∈∆, p ∈ N, p > m, 0 < β ≤ p, m ∈ N0 = N
⋃

{0}, where
�

f ∗ g
�r
(z) denotes the r th derivative

of
�

f ∗ g
�

and is given by

�

f ∗ g
�r
(z) =

p!
�

p− r
�

!
zp−r +

∞
∑

k=1

�

p+ k
�

!
�

p+ k− r
�

!
ap+k bp+kzp+k−r , r ∈ N0. (8)

Obviously the class ℜg

h

�

p, m,β
�

contains the class S
g

h

�

p, m,β
�

, which is defined as fol-
lows:

Definition 2. A function f (z) ∈ Tp is said to be a member of the class S
g

h

�

p, m,β
�

if and only

if for any g, h ∈ Ap with non-negative coefficients,

Re

(

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

+m

)

> p− β ,

z ∈∆, p ∈ N, p > m, 0 < β ≤ p, m ∈ N0.

Taking m = 0 and 1 respectively and h(z) = g (z) = zp

1−z
, the class ℜg

h

�

p, m,β
�

coincides

with the classes S
∗
p

�

β
�

and K p

�

β
�

respectively and the class S
g

h

�

p, m,β
�

coincides with the

class S∗p
�

p− β
�

and Kp

�

p− β
�

respectively. Also, taking g (z) = zp

1−z
, h(z) = zp and m = 0,

the class ℜg

h

�

p, m,β
�

reduces to the class CK p

�

β
�

and the class S
g

h

�

p, m,β
�

reduces to the
the class CKp

�

p− β
�

.
If h = g, we denote ℜg

h

�

p, m,β
�

≡ ℜg

�

p, m,β
�

. Class ℜg

�

1,0,β
�

for g (z) = z

1−z
, co-

incides with the class studied by Chen et al. [7] as a particular case. In addition, the class
ℜg

�

p, 0, p (1−α)
�

reduces to the class studied by Ali et al. [1]. Taking, for n+ p > 0,

h(z) = g (z) = zp

(1−z)n+p and g (z) = zp

(1−z)n+p ,h(z) = zp respectively, the class ℜg

h

�

p, m,β
�

re-

duces to the classes, which were investigated by Raina and Srivastava [22] and these classes
coincide with the classes, studied by Güney and Breaz [12] if n+ p = 1 and are the gener-
alization of the classes investigated by Murugusundaramoorthi and Srivastava [18]. Further,
taking g ∈ A1 so that bk+1 = (1+ k)n , n ∈ N0, the class ℜg (1,0,1−α) would reduce to the
class studied in [1]. Moreover, a class similar to ℜg

�

p, m,β
�

is studied by Prajapat et al.
[21].

In this paper, we study coefficient inequality, growth and distortion bounds, sufficient
conditions with the help of various lemmas, integral means inequality for convolution of two
functions and a set of class preserving integral operators for functions belonging to the class
ℜg

h

�

p, m,β
�

.



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1098

2. Coefficient Inequality, Growth and Distortion Bounds for the class

ℜg

h

�

p, m,β
�

A necessary and sufficient coefficient condition for a function f ∈ Tp to be in the class
ℜg

h

�

p, m,β
�

is derived in the form of following Theorem:

Theorem 1. Let the function f be of the form (7) and g, h ∈ Ap of the form (2) and (3)
respectively with

�

p+ k−m
�

bp+k >
�

p−m− β
�

cp+k . Then f is in the class ℜg

h

�

p, m,β
�

if

and only if

∞
∑

k=1

�

p+ k
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

�

p+ k−m
�

!
ap+k ≤

βp!
�

p−m
�

!
, (9)

p ∈ N, p > m, 0 < β ≤ p. The result is sharp for the function f given by

fk (z) = zp −
βp!

�

p+ k−m
�

!
�

p+ k
�

!
�

p−m
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�zp+k (k ≥ 1). (10)

Proof. We assume that the inequality (9) holds true, then we have to show that
�

�

�

�

�

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)
−
�

p−m
�

�

�

�

�

�

− β < 0

or,
�

�

�z
�

f ∗ g
�m+1

(z)−
�

p−m
��

f ∗ h
�m
(z)

�

�

�− β
�

�

�

f ∗ h
�m
(z)
�

� < 0.

Using series expansion of
�

f ∗ g
�m+1 and

�

f ∗ g
�m from (8), we have

�

�

�

�

�

−
∞
∑

k=1

�

p+ k
�

!ap+k
�

p+ k−m
�

!

¦
�

p+ k−m
�

bp+k −
�

p−m
�

cp+k

©

zp+k−m

�

�

�

�

�

−β

�

�

�

�

�

p!zp−m

�

p−m
�

!
−
∞
∑

k=1

�

p+ k
�

!ap+kcp+k
�

p+ k−m
�

!
zp+k−m

�

�

�

�

�

≤
∞
∑

k=1

�

p+ k
�

!ap+k
�

p+ k−m
�

!

¦
�

p+ k−m
�

bp+k −
�

p−m
�

cp+k

©

− β

(

p!
�

p−m
�

!
−
∞
∑

k=1

�

p+ k
�

!ap+kcp+k
�

p+ k−m
�

!

)

=

∞
∑

k=1

�

p+ k
�

!ap+k
�

p+ k−m
�

!

¦
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

©

−
βp!

�

p−m
�

!

≤ 0, if (9) holds.

Hence, f ∈ ℜg

h

�

p, m,β
�

. To prove the converse, we suppose that f ∈ ℜg

h

�

p, m,β
�

, that is

�

�

�

�

�

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)
−
�

p−m
�

�

�

�

�

�

< β , (11)



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1099

z ∈ ∆, p ∈ N, p > m, 0 < β ≤ p, m ∈ N0. Since |Re (z)| ≤ |z| for any z. Choosing z to be real
and letting z→ 1− through real values, (11) yields

∞
∑

k=1

�

p+ k
�

!ap+k
�

p+ k−m
�

!

¦
�

p+ k−m
�

bp+k −
�

p−m
�

cp+k

©

−β

(

p!
�

p−m
�

!
−
∞
∑

k=1

�

p+ k
�

!ap+kcp+k
�

p+ k−m
�

!

)

≤ 0

or,
∞
∑

k=1

�

p+ k
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

�

p+ k−m
�

!
ap+k ≤

βp!
�

p−m
�

!

which leads us immediately to the desired inequality (9). Sharpness follows if we take ex-
tremal function given by (10).

Corollary 1. If f ∈ ℜg

h

�

p, m,β
�

, then

ap+k ≤
βp!

�

p+ k−m
�

!
�

p+ k
�

!
�

p−m
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

� , k ≥ 1. (12)

The equality in (12) is attained for the function fk given by ( 10).

Corollary 2. Let f ∈ ℜg

h

�

p, m,β
�

and dp+k :=
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k be such

that dp+k ≥ dp+1,∀ k ≥ 1, then

∞
∑

k=1

ap+k ≤
β
�

p−m+ 1
�

�

p+ 1
�

dp+1
. (13)

Corollary 3. Let f ∈ ℜg

h

�

p, m,β
�

and dp+k :=
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k be such

that dp+k ≥ dp+1,∀ k ≥ 1, then

∞
∑

k=1

�

p+ k
�

ap+k ≤
β
�

p−m+ 1
�

dp+1
.

Corollary 4. Let the function f be of the form (7) and g, h ∈ Ap of the form (2) and (3)
respectively with

�

p+ k−m
�

bp+k >
�

p−m− β
�

cp+k ,if

∞
∑

k=1

�

p+ k
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

�

p+ k−m
�

!
ap+k ≤

βp!
�

p−m
�

!
,

p ∈ N, p > m, 0 < β ≤ p holds, then f ∈ S
g

h

�

p, m,β
�

.

Theorem 2. Let f ∈ Tp of the form (7) be in the class ℜg

h

�

p, m,β
�

and g, h be of the form (2),

(3) respectively with dp+k :=
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k ≥ dp+1,∀ k ≥ 1, then

|zp| −
β
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

� ≤
�

� f (z)
�

� ≤ |zp|+
β
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

� (14)



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1100

and
�

�pzp−1
�

�−
β
�

p−m+ 1
�

dp+1
|zp| ≤

�

�

� f
′
(z)

�

�

� ≤
�

�pzp−1
�

�+
β
�

p−m+ 1
�

dp+1
|zp| . (15)

Also let g (1) be finite and ζ :=max bp+k (k ≥ 1), then

|zp| −
βζ
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

� ≤
�

�

�

f ∗ g
�

(z)
�

� ≤ |zp|+
βζ
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

� . (16)

The bounds are sharp and extremal function may given by

f (z) = zp −
β
�

p−m+ 1
�

�

p+ 1
�

dp+1
zp+1. (17)

Proof. Taking absolute value of f (z) given in (7) and using Corollary 2, we get

�

� f (z)
�

� ≤ |zp|+
∞
∑

k=1

ap+k

�

�zp+k
�

� ≤ |zp|+
β
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

�

and
�

� f (z)
�

�≥ |zp| −
∞
∑

k=1

ap+k

�

�zp+k
�

� ≥ |zp| −
β
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

� ,

which prove assertion (14). Again, taking absolute value of f ′ (z) and using Corollary 3, we
get

�

�

� f
′
(z)

�

�

� ≤
�

�pzp−1
�

�+

∞
∑

k=1

�

p+ k
�

ap+k

�

�zp+k−1
�

�≤
�

�pzp−1
�

�+
β
�

p−m+ 1
�

dp+1
|zp|

and

�

�

� f
′
(z)

�

�

�≥
�

�pzp−1
�

�−
∞
∑

k=1

�

p+ k
�

ap+k

�

�zp+k−1
�

�≥
�

�pzp−1
�

�−
β
�

p−m+ 1
�

dp+1
|zp| ,

which prove assertion (15).
Further, taking absolute value of f ∗ g, where f and g are of the form (7) and (2)

respectively. If ζ :=max bp+k, then using corollary (2), we get

�

�

�

f ∗ g
�

(z)
�

� ≤ |zp|+
∞
∑

k=1

ap+k bp+k

�

�zp+k
�

� ≤ |zp|+
βζ
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

�

and
�

�

�

f ∗ g
�

(z)
�

�≥ |zp| −
∞
∑

k=1

ap+k bp+k

�

�zp+k
�

�≥ |zp| −
βζ
�

p−m+ 1
�

�

p+ 1
�

dp+1

�

�zp+1
�

� ,

which prove (16). The bounds in (14), (15) and (16) are sharp, with extremal function given
by (10).



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1101

3. Sufficient Conditions for Classes ℜg

h

�

p, m,β
�

and S
g

h

�

p, m,β
�

In this section, we obtain sufficient conditions for the classesℜg

h

�

p, m,β
�

and S
g

h

�

p, m,β
�

with the use of following Lemmas:

Lemma 1. [14] Let w (z) be analytic in ∆ and such that w (0) = 0. Then if |w (z)| attains its

maximum value on circle |z| = r < 1 at a point z0 ∈∆, we have

z0w
′ �

z0
�

= kw
�

z0
�

,

where k ≥ 1 is a real number.

Lemma 2. [17] Let φ (u, v) be a complex valued function:

φ : D→ C,
�

D ⊂ C×C; Cis the complex plane
�

,

and let u= u1 + iu2 and v = v1 + iv2. Suppose that the function φ (u, v) satisfies

(i) φ (u, v) is continuous in D;

(ii) (1,0) ∈ D and Re
�

φ (1,0)
�

> 0;

(iii) Re
�

φ
�

iu2, v1
��

≤ 0 for all
�

iu2, v1
�

∈ D and such that v1 ≤ −
�

1+ u2
2

�

/2.

Let p (z) = 1+ p1z + p2z2 + · · · be regular in ∆ such that
�

p (z) , zp
′
(z)
�

∈ D for all z ∈ ∆. If

Re
�

φ
�

p (z) , zp
′
(z)
��

> 0 (z ∈∆), then Re
�

p (z)
�

> 0 (z ∈∆).

Lemma 3. [19] Let a function p (z) be analytic in ∆, p (0) = 1, and p (z) 6= 0 (z ∈∆). If there

exists a point z0 ∈∆ such that

�

�arg p (z)
�

�<
π

2
β for |z| <

�

�z0

�

�

and
�

�arg p
�

z0
�
�

�=
π

2
β

with 0< β ≤ 1, then we have

z0p
′ �

z0
�

p
�

z0
� = ilβ

where

l ≥ 1 when arg p
�

z0
�

=
π

2
β

and

l ≤ −1 when arg p
�

z0

�

= −
π

2
β .



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1102

Theorem 3. Let the function f ∈ Ap, if for g, h ∈ Ap, p ∈ N, p > m, 0< β ≤ p,

�

�

�

�

�

1+
z
�

f ∗ g
�m+2

(z)
�

f ∗ g
�m+1

(z)
−

z
�

f ∗ h
�m+1

(z)
�

f ∗ h
�m
(z)

�

�

�

�

�

<
β

�

p−m
�

+ β
, (18)

holds, then f ∈ ℜg

h

�

p, m,β
�

.

Proof. Let w (z) be defined by

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

=
�

p−m
�

+βw (z) .

Clearly w (z) is analytic in ∆ and w (0) = 0. Differentiating logarithmically, we obtain

1+
z
�

f ∗ g
�m+2

(z)
�

f ∗ g
�m+1

(z)
−

z
�

f ∗ h
�m+1

(z)
�

f ∗ h
�m
(z)

=
zβw

′
(z)

��

p−m
�

+ βw (z)
� .

Suppose that there exists a point z0 ∈∆ such that

max
|z|<|z0|

|w (z)| =
�

�w
�

z0
�
�

� = 1
�

w(z0) 6= 1
�

.

Then using Jack’s Lemma 1, we get z0w
′ �

z0
�

= kw
�

z0
�

(k ≥ 1). Therefore, letting
w
�

z0
�

= eiθ (θ 6= 0),

�

�

�

�

�

1+
z0

�

f ∗ g
�m+2 �

z0

�

�

f ∗ g
�m+1 �

z0

�
−

z0

�

f ∗ h
�m+1 �

z0

�

�

f ∗ h
�m �

z0

�

�

�

�

�

�

=

�

�

�

�

�

z0βw
′ �

z0

�

�

p−m
�

+ βw
�

z0
�

�

�

�

�

�

=
βk

¦
�

p−m
�2
+ β2+ 2β

�

p−m
�

cos θ
©

1
2

≥
β

p−m+ β
,

which contradicts the condition (18), we have |w (z)| < 1 for all z0 ∈ ∆, consequently, we
conclude that f ∈ ℜg

h

�

p, m,β
�

.

Taking h= g, we get following inclusion result with the help of Jack’s Lemma.

Theorem 4. For p > m, ℜg

�

p, m+ 1,β
�

⊂ℜg

�

p, m,α
�

, where

0< α ≤
−
�

p−m− β + 1
�

± 2
Æ

�

p−m− β + 1
�2
+ 4β

�

p−m
�

2
≤ p−m. (19)



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1103

Proof. Let f ∈ ℜg

�

p, m+ 1,β
�

. Then
�

�

�

�

�

z
�

f ∗ g
�m+2

(z)
�

f ∗ g
�m+1

(z)
−
�

p−m− 1
�

�

�

�

�

�

< β (20)

and let w (z) be defined by

z
�

f ∗ g
�m+1

(z)
�

f ∗ g
�m
(z)
−
�

p−m
�

= αw (z) . (21)

Clearly w (z) is analytic in ∆ and w (0) = 0. Differentiating logarithmically, we obtain

z
�

f ∗ g
�m+2

(z)
�

f ∗ g
�m+1

(z)
=

�

p−m− 1
�

+αw (z) +
αzw

′
(z)

�

p−m
�

+αw (z)

z
�

f ∗ g
�m+2

(z)
�

f ∗ g
�m+1

(z)
−
�

p−m− 1
�

= αw (z)



1+
αzw

′
(z)

αw (z)

1
�

p−m
�

+αw (z)



 .

Now, suppose that there exists a point z0 ∈∆ such that

max
|z|<|z0|

|w (z)| =
�

�w
�

z0
�
�

� = 1
�

w(z0) 6= 1
�

.

Using Jack’s Lemma 1, we have z0w
′ �

z0

�

= kw
�

z0

�

(k ≥ 1). Therefore, letting
w
�

z0

�

= eiθ (θ 6= 0),
�

�

�

�

�

z0
�

f ∗ g
�m+2 �

z0
�

�

f ∗ g
�m+1 �

z0

�
−
�

p−m− 1
�

�

�

�

�

�

= α
�

�w(z0)
�

�

�

�

�

�

�

1+
αz0w

′ �
z0
�

αw
�

z0
�

1
�

p−m
�

+αw
�

z0
�

�

�

�

�

�

= α

�

�

�

�

1+
k

�

p−m
�

+αeiθ

�

�

�

�

≥ α











1+
k

�

p−m
�Re







1+ α

(p−m)
cos θ − i α

(p−m)
sin θ

1+
�

α

(p−m)

�2

+ 2α

(p−m)
cos θ

















= α















1+
k

�

p−m
�















1

2+

�

α

(p−m)

�2
−1

1+ α

(p−m)
cos θ





























≥ α



1+
1

�

p−m
�

(

�

p−m+α
��

p−m
�

2
�

p−m+α
��

p−m
�

+α2 −
�

p−m
�2

)



= α

�

p−m+α+ 1

p−m+α

�

,



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1104

on using (19), it gives
�

�

�

�

�

z
�

f ∗ g
�m+2

(z)
�

f ∗ g
�m+1

(z)
−
�

p−m− 1
�

�

�

�

�

�

≥ β

which contradicts (20). Hence |w (z)| < 1 and from (21), it follows that f ∈ ℜg

�

p, m,α
�

.

Theorem 5. Let f ∈ Ap if

Re











δ
z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

+ (1− δ)z

(

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)′












> γ (z ∈∆) ,

for some γ
�

γ < δ
�

p−m
��

, 0≤ δ ≤ 1, then f ∈ S
g

h

�

p, m,β
�

, where β =
2(δ(p−m)−γ)

1+δ
≤ p.

Proof. If δ = 1, the result holds. Let 0≤ δ < 1, define the function p (z) by

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

=
�

p−m− β
�

+ βp (z) . (22)

Then p (z) = 1+ p1z + p2z2 + · · · is regular in ∆. It follows from (22) that

1+ z

�

f ∗ g
�m+2

(z)
�

f ∗ g
�m+1

(z)
− z

�

f ∗ h
�m+1

(z)
�

f ∗ h
�m
(z)

=
βzp

′
(z)

�

p−m− β
�

+ βp (z)
,

or,

z

�

f ∗ h
�m
(z)

�

f ∗ g
�m+1

(z)

(

�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)′

=
βzp

′
(z)−

��

p−m− β
�

+ βp (z)
	

�

p−m− β
�

+ βp (z)
,

or, equivalently

z2

(

�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)′

= βzp
′
(z)−

��

p−m− β
�

+ βp (z)
	

.

Therefore, we have

Re











δ
z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

+ (1− δ) z

(

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)′




− γ







= Re













z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

+ (1− δ) z2

(

�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)′




− γ









P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1105

= Re
¦

δ
�

p−m−β
�

+ βδp (z) + (1− δ)βzp
′
(z)− γ

©

> 0

If we define a function φ (u, v) by

φ (u, v) = δ
�

p−m− β
�

+ βδu+ (1− δ)β v− γ (23)

with u = u1 + iu2 and v = v1 + iv2, then

(i) φ (u, v) is continuous in D ⊂ C×C;

(ii) (1,0) ∈ D and Re φ (1,0) = δ
�

p−m
�

− γ > 0;

(iii) For all
�

iu2, v1
�

∈ D and such that for v1 ≤ −
�

1+ u2
2

�

/2, we get

Re
�

φ
�

iu2, v1
�	

= δ
�

p−m− β
�

+ (1− δ)β v1 − γ

≤ δ
�

p−m− β
�

− (1− δ)β
�

1+ u2
2

�

/2− γ

= − (1− δ)βu2
2/2

≤ 0.

Therefore, φ (u, v) satisfies the conditions of Lemma 2. This show that Re
�

p (z)
�

> 0 (z ∈∆),
i.e.

Re

(

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)

> p−m− β (z ∈∆)

which proves that f (z) ∈ S
g

h

�

p, m,β
�

.

Theorem 6. Let for p ∈ N, p > m, m ∈ N0, 0< β ≤ p, if

�

�

�

�

�

�

�

�

arg







1
�

p−m
�







z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

+ z

 

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

!′













�

�

�

�

�

�

�

�

<
β

p

π

2
+tan−1

�

β

p

�

(z ∈∆) ,

(24)

then

�

�

�

�

arg

�

z( f ∗g)
m+1
(z)

( f ∗h)
m
(z)

�
�

�

�

�

<
β

p

π

2
(z ∈∆) . In particular, if

�

�

�

�

�

arg

(

z f
′
(z)

p f (z)

 

2+
z f
′′
(z)

f
′
(z)
−

z f
′
(z)

f (z)

!)�

�

�

�

�

<
β

p

π

2
+ tan−1

�

β

p

�

(z ∈∆) ,

then f ∈ S
∗
p

�

β
�

.

Proof. Let

p(z) :=
1

�

p−m
�

(

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)

.



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1106

We obtain

zp
′
(z) =

z
�

p−m
�

(

z
�

f ∗ g
�m+1

(z)
�

f ∗ h
�m
(z)

)′

.

Suppose that there exists point z0 ∈∆ such that

�

�arg p (z)
�

� <
β

p

π

2
for |z| <

�

�z0

�

� ,
�

�arg p
�

z0

�
�

� =
β

p

π

2
.

Then applying Lemma 3, we write that

z0p
′ �

z0
�

p
�

z0
� = il

β

p

where

l ≥ 1 when arg p
�

z0
�

=
β

p

π

2

and

l ≤ −1 when arg p
�

z0

�

= −
β

p

π

2
.

Then it follows that

arg







1
�

p−m
�







z0
�

f ∗ g
�m+1 �

z0
�

�

f ∗ h
�m �

z0
� + z0





z0
�

f ∗ g
�m+1 �

z0
�

�

f ∗ h
�m �

z0
�





′











= arg
¦

p
�

z0

�

+ z0p
′ �

z0

�
©

= arg

(

p
�

z0
�

 

1+
z0p

′ �
z0
�

p
�

z0
�

!)

= arg p
�

z0
�

+ arg

�

1+ il
β

p

�

= arg p
�

z0
�

+ tan−1

�

l
β

p

�

.

When arg p
�

z0
�

=
β

p

π

2
, we have

arg







1
�

p−m
�







z0
�

f ∗ g
�m+1 �

z0
�

�

f ∗ h
�m �

z0
� + z0

 

z0
�

f ∗ g
�m+1 �

z0
�

�

f ∗ h
�m �

z0
�

!′













(25)

=
β

p

π

2
+ tan−1

�

l
β

p

�

≥
β

p

π

2
+ tan−1

�

β

p

�

.



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1107

Similarly, if arg p
�

z0

�

= −β
p

π

2
, then we obtain that

arg







1
�

p−m
�







z0
�

f ∗ g
�m+1 �

z0
�

�

f ∗ h
�m �

z0
� + z0





z0
�

f ∗ g
�m+1 �

z0
�

�

f ∗ h
�m �

z0
�





′











(26)

= −
β

p

π

2
+ tan−1

�

l
β

p

�

≤ −
�

β

p

π

2
+ tan−1

�

β

p

��

.

Thus we see that (25) and (26) contradicts the condition (24). Consequently, we conclude
that

�

�arg p(z)
�

� <
β

p

π

2
(z ∈∆) .

This proves Theorem 6.

4. Integral Means Inequality for the Class ℜg

h

�

p, m,β
�

Definition 3. [Subordination Principle]. For two functions f1 and f2, analytic in ∆, we say that

the function f1 (z) is subordinate to f2 (z) in ∆, and write

f1 (z) ≺ f2 (z) (z ∈∆) ,

if there exists a Schwartz function w (z), analytic in ∆ with

w (0) = 0 and |w (z)| < 1,

such that

f1 (z) = f2 (w (z)) (z ∈∆) .

In particular, if the function f2 is univalent in ∆, the subordination is equivalent to

f1 (0) = f2 (0) and f1 (∆)⊂ f2 (∆) .

Littlewood [16] proved the following subordination result (See also Duren [11]).

Lemma 4. [16] If f1 and f2 are analytic in∆ with f1 ≺ f2, then for τ > 0 and z = reiθ (0< r < 1),

2π
∫

0

�

� f1 (z)
�

�

τ
dθ ≤

2π
∫

0

�

� f2 (z)
�

�

τ
dθ .



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1108

Theorem 7. Let g (z), h(z) be of the form (2), (3) respectively and f ∈ ℜg

h

�

p, m,β
�

be of the

form (7) and let for some i ∈ N,
ϕi

bp+i

= min
k≥1

ϕk

bp+k

,

where ϕk :=
(p+k)!dp+k

(p+k−m)!
and dp+k :=

�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

> 0. Also let for

such i ∈ N, functions fi and gi be defined respectively by

fi (z) = zp −
βp!

�

p+ i −m
�

!

dp+i

�

p+ i
�

!
�

p−m
�

!
zp+i , gi = zp + bp+iz

p+i , (27)

If there exists an analytic function w defined by

{w (z)}i =
dp+i

�

p+ i
�

!
�

p−m
�

!

bp+iβp!
�

p+ i −m
�

!

∞
∑

k=1

ap+k bp+kzk

then, for τ > 0 and z = reiθ (0< r < 1),

2π
∫

0

�

�

�

f ∗ g
�

(z)
�

�

τ
dθ ≤

2π
∫

0

�

�

�

fi ∗ gi

�
�

�

τ
dθ (τ > 0) .

Proof. Convolution of f and g is defined as:

�

f ∗ g
�

(z) = zp −
∞
∑

k=1

ap+k bp+kzp+k = zp

 

1−
∞
∑

k=1

ap+k bp+kzk

!

Similarly, from (27), we obtain

�

fi ∗ gi

�

(z) = zp −
bp+iβp!

�

p+ i −m
�

!

dp+i

�

p+ i
�

!
�

p−m
�

!
zp+i

= zp

�

1−
bp+iβp!

�

p+ i −m
�

!

dp+i

�

p+ i
�

!
�

p−m
�

!
z i

�

.

To prove the theorem, we must show that for τ > 0 and z = reiθ (0< r < 1),

2π
∫

0

�

�

�

�

�

1−
∞
∑

k=1

ap+k bp+kzk

�

�

�

�

�

τ

dθ ≤

2π
∫

0

�

�

�

�

�

1−
bp+iβp!

�

p+ i −m
�

!

dp+i

�

p+ i
�

!
�

p−m
�

!
z i

�

�

�

�

�

τ

dθ .

Thus, by applying Lemma 4, it would suffice to show that

1−
∞
∑

k=1

ap+k bp+kzk ≺ 1−
bp+iβp!

�

p+ i −m
�

!

dp+i

�

p+ i
�

!
�

p−m
�

!
z i . (28)



P. Sharma, P. Srivastava / Eur. J. Pure Appl. Math, 3 (2010), 1093-1112 1109

If the subordination (28) holds true, then there exist an analytic function w with w (0) = 0
and |w (z)| < 1 such that

1−
∞
∑

k=1

ap+k bp+kzk = 1−
bp+iβp!

�

p+ i −m
�

!

dp+i

�

p+ i
�

!
�

p−m
�

!
{w (z)}i .

From the hypothesis of the theorem, there exists an analytic function w given by

{w (z)}i =
dp+i

�

p+ i
�

!
�

p−m
�

!

bp+iβp!
�

p+ i −m
�

!

∞
∑

k=1

ap+k bp+kzk

which readily yields w (0) = 0. Thus for such function w, using the hypothesis in the coeffi-
cient inequality for the class ℜg

h

�

p, m,β
�

, we get

|w (z)|r ≤
dp+i

�

p+ i
�

!
�

p−m
�

!

bp+iβp!
�

p+ i −m
�

!

∞
∑

k=1

ap+k bp+k |z|
k

≤ |z|
dp+i

�

p+ i
�

!
�

p−m
�

!

bp+iβp!
�

p+ i −m
�

!

∞
∑

k=1

ap+k bp+k

≤ |z| < 1.

Therefore the subordination (28) holds true, thus the theorem is proved.

5. Class-preserving Integral-Operators for the Class ℜg

h

�

p, m,β
�

In this section, we present several integral operators which preserve class ℜg

h

�

p, m,β
�

.
For f ∈ ℜg

h

�

p, m,β
�

, we define the integral operators by

L1 f (z) =

�

p+ c
�

zc

z
∫

0

t c−1 f (t) d t, c > −p,

L2 f (z) =

�

p+ c
�σ

zcΓ(σ)

z
∫

0

t c−1
�

log
z

t

�σ−1

f (t) d t, c > −p,σ ≥ 0,

L3 f (z) =

�

p+ c +σ− 1

p+ c − 1

�

σ

zc

z
∫

0

�

1−
t

z

�σ−1

t c−1 f (t) d t, c > −p, σ ≥ 0.

Theorem 8. Let f ∈ ℜg

h

�

p, m,β
�

, then for p > m, 0 < β ≤ p, c > −p and

σ ≥ 0, L j f ∈ ℜg

h

�

p, m,β
�

, j = 1,2,3.



REFERENCES 1110

Proof. Let f ∈ Tp of the form (7) be in the class ℜg

h

�

p, m,β
�

, then

L1 f (z) = zp −
∞
∑

k=1

�

c + p

c + p+ k

�

ap+kzp+k,

L2 f (z) = zp −
∞
∑

k=1

�

c + p

c + p+ k

�σ

ap+kzp+k,

L3 f (z) = zp −
∞
∑

k=1

�

p+ c
�

k
�

p+ c +σ
�

k

ap+kzp+k.

Since
�

c+p

c+p+k

�

< 1 , for σ ≥ 0,
�

c+p

c+p+k

�σ

≤ 1 and
(p+c)k
(p+c+σ)k

≤ 1, k ≥ 1, by Theorem 1, we see

that

∞
∑

k=1

�

p+ k
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

�

p+ k−m
�

!

�

c + p

c + p+ k

�

ap+k

≤
∞
∑

k=1

�

p+ k
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

�

p+ k−m
�

!
ap+k ≤

βp!
�

p−m
�

!
.

Hence, by Theorem 1, L1 f (z) ∈ ℜg

h

�

p, m,β
�

. Also

∞
∑

k=1

�

p+ k
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

�

p+ k−m
�

!

�

c + p

c + p+ k

�σ

ap+k

≤
∞
∑

k=1

�

p+ k
�

!
�
�

p+ k−m
�

bp+k −
�

p−m− β
�

cp+k

�

�

p+ k−m
�

!
ap+k ≤

βp!
�

p−m
�

!
.

Hence, L2 f (z) ∈ ℜg

h

�

p, m,β
�

. Similarly, we obtain that L3 f (z) ∈ ℜg

h

�

p, m,β
�

.

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