4_sokol.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 5, No. 4, 2012, 469-479

ISSN 1307-5543 – www.ejpam.com

A Family of Convolution Operators for Multivalent Analytic

Functions

Janusz Sokół1,∗, Khalida Inayat Noor 2, Hari Mohan Srivastava3

1 Department of Mathematics, Rzeszów University of Technology, Al. Powstańców Warszawy 12,

35-959 Rzeszów, Poland
2 Department of Mathematics, COMSATS Institute of Information Technology, Park Road,

Islamabad, Pakistan
3Department of Mathematics and Statistics, University of Victoria, Victoria, British Columbia V8W

3R4, Canada

Abstract. In this paper, we consider a family of multiplier transformations and several subclasses

of multivalent functions which are defined by means of convolution. Several interesting results are

derived. Some (known or new) special cases of the multivalent function classes, which are investigated

here, are also discussed.

2010 Mathematics Subject Classifications: 30C45; 30C80, 33C05, 33C20

Key Words and Phrases: Univalent functions, Convex functions, Starlike functions, Subordination be-

tween analytic functions, Hadamard product (or convolution), Gauss and generalized hypergeometric

functions

1. Introduction and Definitions

LetA (p) denote the class of functions of the following form:

f (z) = zp +

∞
∑

n=1

ap+nzp+n (p ∈ N := {1,2,3, . . .}), (1)

which are analytic in the open unit disk

U = {z : z ∈ C and |z| < 1}.
∗Corresponding author.

Email addresses: jsokol�prz.edu.pl (J. Sokół), khalidanoor�hotmail.
om (K. Noor),

harimsri�math.uvi
.
a (H. M. Srivastava)

http://www.ejpam.com 469 c© 2012 EJPAM All rights reserved.



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 470

Let f , g ∈A (p), f be given by (1) and

g(z) = zp +

∞
∑

n=1

bp+nzp+n. (2)

Then the Hadamard product (or convolution) of f and g is defined by

( f ∗ g)(z) := zp +

∞
∑

n=1

ap+n bp+nzp+n =: (g ∗ f )(z) (p ∈ N). (3)

Also, if f and g are analytic in U, we say that f is subordinate to g in U, and we write

f ≺ g (z ∈ U), (4)

if there exists a Schwarz function w such that

f (z) = g
�

w(z)
�

and |w(z)| ≤ |z| (z ∈ U).
We now define a linear operator Lc

k
:A (p)→A (p) as follows: let the linear operator L0

L0 :A (p)→A (p) (k ∈ N; c ∈ C \ {0}) (5)

be given and

cLc
k f (z) = z

�

Lc
k−1 f (z)

�′
+ (c − p)Lc

k−1 f (z) (6)

with

Lc
0 := L0. (7)

It can easily be seen from (6) that the operator Lc
k

is linear and it satisfies the following

property:

Lc
0 f (z) = zp +

∞
∑

n=1

Ap+nzp+n, (8)

which implies that

Lc
k

f (z) = zp +

∞
∑

n=1

(1+ n/c)kAp+nzp+n. (9)

We also have

cLc
1 f = z(Lc

0 f )′ + (c − p)Lc
0 f , (10)

cLc
k

f = zp+1(z−p Lc
k−1

f )′+ cLc
k−1

f (11)

and
Lc

k
f

zp
=

z

c

�

Lc
k−1

f

zp

�′
+

Lc
k−1

f

zp
. (12)

By appropriately choosing Lc
k

given by (8), we obtain several applications studied by var-

ious earlier authors (see, for example, [2, 3, 4, 5, 6, 8, 14, 9, 10, 11, 15, 20]; see also

[13, 17, 18, 21]). We now define the following analytic function class.



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 471

Definition 1. Let q and h be analytic in U. Also let the function h be convex univalent in U with

h(0) = q(0) = 1. Then q ∈ P (h) if and only if

q(z) ≺ h(z) (z ∈ U). (13)

Some well-known examples of the convex function h are listed below.

(i) If

h(z) =
1+ (1− 2α)z

1− z
and 0≦ α < 1,

then

ℜ�h(z)�> α (z ∈ U; 0≦ α < 1).

(ii) If

h(0) = 1 and h(z) =

�

1+ z

1− z

�β

(0< β < 1),

then
�

�arg
�

h(z)
�
�

� <
βπ

2
(z ∈ U).

(iii) Let

h(z) =
M(1+ z)

M + (1−M)z

�

M >
1

2

�

.

Also

h(U) = {w : |w −M | < M}.

(iv) If

h(z) =
p

z + 1 and ℜ
�p

z + 1
�

≧ 0 (z ∈ U),
then h(U) is the interior of the right part of the Bernoulli lemniscate [see 1].

(v) If

h(z) = 1+
2

π2

�

log

�

1+
p

z

1−pz

��2

and ℑ
�p

z
�

> 0 (z ∈ U),

then h(U) is the interior of the parabola given by

¦

w : [ℑ(w)]2 = 2ℜ(w)− 1
©

.

Definition 2. Let L0 be a linear operator on A (p) and let Lc
k

be given by (6). Then, for λ ≧ 0,

a function f ∈A (p) is said to be in the class S c
k
(p,λ; h) if and only if

�

(1−λ) Lc
k
f (z)

zp
+λ

Lc
k+1

f (z)

zp

�

∈ P (h). (14)



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 472

2. Preliminary Results

We need each of the following lemmas in our present investigation.

Lemma 1 (see [7] and [12]). Let h be an analytic and convex univalent function in U. Let the

function f be analytic in U with h(0) = f (0) = 1. If

f (z) +
z f ′(z)
γ
≺ h(z)

�

z ∈ U; ℜ(γ)≧ 0; γ 6= 0
�

, (15)

then

f (z) ≺ g(z) =
γ

zγ

∫ z

0

tγ−1h(t) dt ≺ h(z) (z ∈ U).

Moreover, the function g is convex univalent in U and it is the best dominant of the subordination

(15) in the sense that in the sense that f ≺ g for all f satisfying (15), and if there exists q such

that f ≺ q for all f satisfying (15), then g ≺ q.

Lemma 2 (see [19]). Let the functions q and h be analytic in U with q(0) = 1. Suppose also

that

ℜ�q(z)�> 1

2
(z ∈ U).

Then

(q ∗ h)(U)⊂ co {h(U)} ,
where co {h(U)} is the convex hull of h(U).

Lemma 3 (see [16]). Let

f (z) ≺ F(z) (z ∈ U) and g(z)≺ G(z) (z ∈ U).

If the functions F and G are convex in U, then

( f ∗ g)(z)≺ (F ∗ G)(z) (z ∈ U).

Unless otherwise stated, we shall assume throughout this paper that

λ≧ 0, c ∈ C \ {0}, ℜ(c) > 0, k, p ∈ N, and z ∈ U.

3. Main Results

Our first main result in this paper is contained in Theorem 1 below.

Theorem 1. If the function f belongs to the class S c
k
(p,λ; h), then

Lc
k

f (z)

zp
∈ P (h).



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 473

Moreover, if λ > 0, then
Lc

k
f (z)

zp
∈ P (g), (16)

where

g(z) =
c

λ
z−

c

λ

∫ z

0

t
c

λ
−1h(t) dt ≺ h(z) (z ∈ U), (17)

the function g is convex univalent in U and g is the best dominant of the subordination

Lc
k

f (z)

zp
≺ g (z ∈ U).

Proof. The proof for the case when λ= 0 is trivial. We, therefore, suppose that λ > 0. Let

f ∈ S c
k
(p,λ; h). (18)

Then, by (12), we have

(1−λ) Lc
k

f (z)

zp
+λ

Lc
k+1

f (z)

zp
=

Lc
k

f (z)

zp
+
λz

c

�

Lc
k

f (z)

zp

�′
∈ P (h). (19)

Let the function H(z) be given by

H(z) :=
Lc

k
f (z)

zp
(z ∈ U). (20)

Then, by (19), it follows that

�

H(z) +
λ

c
zH ′(z)

�

∈ P (h)

and
�

H(z) +
λ

c
zH ′(z)

�

≺ h(z) (z ∈ U). (21)

Now, using Lemma 1 in (21) with

γ=
c

λ
and λ > 0, (22)

we obtain (17). This shows that H ∈ P (g), where the function g is given by (17). Conse-

quently, the proof of Theorem 1 is complete.

We take

L0 f (z) = f (z) ∗φ(a, c, z), (23)

where

φ(a, c, z) =

∞
∑

n=0

(a)n

(c)n
zp+n (c 6= 0,−1,−2,−3, . . . ; z ∈ U)



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 474

and (λ)n is the Pochhammer symbol defined, in terms of the familiar Gamma function, by

(λ)n =
Γ(λ+ n)

Γ(λ)
=

(

1 (n= 0; λ 6= 0),

λ(λ+ 1) . . . (λ+ n− 1) (n ∈ N),

it being understood conventionally that (0)0 := 1. We also let

h(z) =
1+ Az

1+ Bz
(−1≦ B < A≦ 1). (24)

Then, by applying Theorem 1, we obtain the subordination (17) with

g(z) =







A

B
+

�

1− A

B

�

(1+ Bz)−1
2 F1

�

1,1;
c − 1

λ
+ 1;

Bz

Bz + 1

�

(B 6= 0)′

1−
�

c − 1

c − 1+λ

�

Az (B = 0),

where 2F1 is the Gauss hypergeometric function defined by

2F1(α,β ;γ; z) :=

∞
∑

n=0

(α)n(β)n

(γ)n

zn

n!
(z ∈ U; γ 6= 0,−1,−2,−3, . . .). (25)

Theorem 2. Let 0≦ λ1 ≦ λ2. Then

S c
k
(p,λ2; h)⊂ S c

k
(p,λ1; h). (26)

Proof. Suppose that f ∈ S c
k
(p,λ2; h). A simple computation will then yield

(1−λ1)
Lc

k
f (z)

zp
+λ1

Lc
k+1

f (z)

zp

=

�

1− λ1

λ2

�

Lc
k

f (z)

zp
+
λ1

λ2

�

(1−λ2)
Lc

k
f (z)

zp
+λ2

Lc
k+1

f (z)

zp

�

. (27)

It can now be easily shown that the class P (h) is a convex set. We can write (27) as follows:

(1−λ1)
Lc

k
f (z)

zp
+λ1

Lc
k+1

f (z)

zp
=

�

1− λ1

λ2

�

h1(z) +
λ1

λ2

h2(z) =ψ(z), (28)

where h1 ∈ P (h), by Theorem 1, and h2 ∈ P (h), since f ∈ S c
k
(p,λ2; h). We thus find that

ψ ∈ P (h). Consequently, f ∈ S c
k
(p,λ1; h). This proves Theorem 2.

Theorem 3. The following inclusion relationship holds true:

S c
k
(p,λ; h)⊂ S c

k−1
(p,λ; h). (29)



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 475

Proof. Let f ∈ S c
k
(p,λ; h) and suppose that
�

(1−λ) Lc
k−1

f (z)

zp
+λ f racLc

k f (z)zp

�

= H(z).

Then, from (12), we have

�

(1−λ) Lc
k−1

f (z)

zp
+λ

Lc
k

f (z)

zp

�

+
z

c

�

(1−λ) Lc
k−1

f (z)

zp
+λ

Lc
k

f (z)

zp

�′

=H(z) +
1

c
zH ′(z)

=(1−λ)




Lc
k−1

f (z)

zp
+

z

c

�

Lc
k−1

f (z)

zp

�′


+λ





Lc
k

f (z)

zp
+

z

c

�

Lc
k

f (z)

zp

�′


=

�

(1−λ) Lc
k

f (z)

zp
+λ

Lc
k+1

f (z)

zp

�

∈ P (h).

We thus find that
�

H(z) +
1

c
zH ′(z)

�

≺ h(z) (z ∈ U). (30)

By applying Lemma 1, it follows that

H(z) ≺ c

zc

∫ z

0

t c−1h(t) dt ≺ h(z) (z ∈ U),

which shows that H ∈ P (h). Consequently, we have
�

(1−λ) Lc
k−1

f (z)

zp
+λ

Lc
k

f (z)

zp

�

∈ P (h). (31)

This evidently proves that f ∈ S c
k−1
(p,λ; h).

Corollary 1. For ℜ(c) > 0, let f ∈ S c
k
(p,λ; h). Then

Lc
s f (z)

zp
∈ P (h) (s ∈ {0,1,2, . . . , k}). (32)

Proof. We can readily deduce the assertion (32) of the above Corollary from the assertion

(17) of Theorem 1. The details involved are being omitted here.

In order to get the convolution results of the multivalent analytic function classS c
k
(p,λ; h),

it is necessary to put the following restrictions on the operator Lc
k
:

Lc
k
( f ∗ g) = (Lc

k
f ) ∗ g = f ∗ (Lc

k
g), (33)

where f , g ∈ S c
k
(p,λ; h) (k ∈ N). We now prove our next result contained in Theorem 4

below.



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 476

Theorem 4. Let the operator Lc
k

satisfy the condition (33). If f j ∈ S c
k
(p,λ; h j) ( j = 1,2), then

each of the following inclusion relationships holds true:

G(z) = (1−λ)Lc
k
( f1 ∗ f2)(z) +λLc

k+1
( f1 ∗ f2)(z) ∈ S c

k
(p,λ,h1 ∗ h2), (34)

Lc
k
( f1 ∗ f2)(z) ∈ S c

k
(p,λ; h1 ∗ h2) (35)

and
Lc

k

�

Lc
k
( f1 ∗ f2)(z)

�

zp
∈ P (h1 ∗ h2). (36)

Proof. Since

f1 ∈ S c
k
(p,λ; h1) and f2 ∈ S c

k
(p,λ; h2), (37)

it follows that
�

(1−λ) Lc
k

f1(z)

zp
+λ

Lc
k+1

f1(z)

zp

�

∈ P (h1) (38)

and
�

(1−λ) Lc
k
f2(z)

zp
+λ

Lc
k+1

f2(z)

zp

�

∈ P (h2). (39)

Also, from (38), (39) and Theorem 1, we have

Lc
k

f1(z)

zp
∈ P (h1) (40)

and
Lc

k
f2(z)

zp
∈ P (h2). (41)

Thus, by making use of (33), (38), (39) and Lemma 3, in conjunction with the technique used

before, we have

(1−λ) Lc
k

�

(1−λ)Lc
k
( f1 ∗ f2)(z) +λLc

k+1
( f1 ∗ f2)(z)

�

zp

+λ
Lc

k+1

�

(1−λ)Lc
k
( f1 ∗ f2)(z) +λLc

k+1
( f1 ∗ f2)(z)

�

zp

=

�

(1−λ) Lc
k
g(z)

zp
+λ

Lc
k+1

g(z)

zp

�

∈ P (h1 ∗ h2),

that is, G ∈ S c
k
(p,λ; h1 ∗ h2). This proves the first assertion (34) of Theorem 4. In order to

demonstrate the second assertion (35) of Theorem 4, we again proceed in a similar manner

and apply Lemma 3 to (38) and (41). We thus obtain

 

(1−λ)
Lc

k

�

Lc
k
( f1 ∗ f2)(z)

�

zp
+λ

Lc
k+1

�

Lc
k
( f1 ∗ f2)(z)

�

zp

!

∈ P (h1 ∗ h2), (42)



J. Sokół, K. Noor, H. M. Srivastava / Eur. J. Pure Appl. Math, 5 (2012), 469-479 477

which clearly implies (35). Finally, from (42) and Theorem 1, we obtain the third assertion

(36) of Theorem 4.

As a special case of Theorem 4, we obtain a result proved in [11] (where c = c1 and k = 0)

for

h j(z) =
1+ A jz

1+ B jz
(z ∈ U, j = 1,2)

and

Lc
k

f (z) = f (z) ∗ qFr(z),

where qFr is the generalized hypergeometric function defined by (see also [4] and [5])

qFr (z) = qFr(α1, . . . ,αq;β1, . . . ,βr ; z) :=

∞
∑

n=0

(α1)n . . . (αq)n

(β1)n . . . (βr)n

zn

n!

(q, r ∈ N0 = N∪ {0}; q ≦ r + 1) (43)

for complex parameters

α1, . . . ,αq and β1, . . . ,βr (β j 6= 0,−1,−2, . . . ; j = 1, . . . , r). (44)

Theorem 5. Let the operator Lc
k

satisfy the condition (33). If f ∈ S c
k
(p,λ; h) and q ∈ A (p)

with

ℜ
�

q(z)

zp

�

≧
1

2
(z ∈ U), (45)

then f ∗ q ∈ S c
k
(p,λ; h).

Proof. By using the properties of convolution and (33), we have

(1−λ) Lc
k
( f ∗ q)(z)

zp
+λ

Lc
k+1
( f ∗ q)(z)
zp

=

�

(1−λ) Lc
k

f (z)

zp
+λ

Lc
k+1

f (z)

zp

�

∗ q(z)
zp

=H(z) ∗ q(z)
zp

�

H ∈ P (h)).
Now, by using Lemma 2, we get

�

H(z) ∗ q(z)
zp

�

∈ P (h),

which implies that
�

(1−λ) Lc
k
( f ∗ q)(z)

zp
+λ

Lc
k+1
( f ∗ q)(z)
zp

�

∈ P (h). (46)

By means of (46), we have thus proved the assertion of Theorem 5 that

f ∗ q ∈ S c
k (p,λ; h). (47)



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