International Journal of Analysis and Applications

Volume 17, Number 4 (2019), 674-685

URL: https://doi.org/10.28924/2291-8639

DOI: 10.28924/2291-8639-17-2019-674

APPLICATION OF SRIVASTAVA-ATTIYA OPERATOR TO THE GENERALIZATION

OF MOCANU FUNCTIONS

KHALIDA INAYAT NOOR AND SHUJAAT ALI SHAH∗

COMSATS University Islamabad, Pakistan

∗Corresponding author: shahglike@yahoo.com

Abstract. In this paper we introduce certain subclasses of analytic functions by applying Srivastava-Attiya

operator. Our main purpose is to derive inclusion results by using concept of conic domain and subordination

techniques. We also deduce some new as well as well-known results from our investigations.

1. Introduction

Let χ denotes the class of analytic functions f(z) in the open unit disk

f = {z : |z| < 1} such that

f(z) = z +

∞∑
n=2

anz
n. (1.1)

Subordination of two functions f and g is denoted by f ≺ g and defined as f(z) = g(w(z)), where w(z) is

schwarz function in f. Let S, S∗ and C denotes the subclasses of χ of univalent functions, starlike functions

and convex functions respectively. For 0 ≤ δ < 1, S∗(δ) and C(δ) are the subclasses of S of functions f

satisfies;

zf′(z)

f(z)
≺

1 + (1 − 2δ) z
1 −z

, (z ∈ f) , (1.2)

Received 2019-04-24; accepted 2019-05-27; published 2019-07-01.

2010 Mathematics Subject Classification. 30C45, 30C55.

Key words and phrases. Srivastava-Attiya operator; Mocanu functions; conic domains.

c©2019 Authors retain the copyrights

of their papers, and all open access articles are distributed under the terms of the Creative Commons Attribution License.

674

https://doi.org/10.28924/2291-8639
https://doi.org/10.28924/2291-8639-17-2019-674


Int. J. Anal. Appl. 17 (4) (2019) 675

(zf′(z))
′

f′(z)
≺

1 + (1 − 2δ) z
1 −z

, (z ∈ f) , (1.3)

respectively. Mocanu [13] introduced the class Mα of α−convex functions f ∈ S satisfies;

(
(1 −α)

zf′(z)

f(z)
+ α

(zf′(z))
′

f′(z)

)
≺

1 + z

1 −z
, (1.4)

where α ∈ [0, 1], f(z)
z
f′(z) 6= 0. and z ∈ f. We see that M0 = S∗ and M1 = C. This class is vastly studied

by several authors. See [4, 15, 17–19]. For k ∈ [0,∞) , Kanas and Wisniowska [8, 9] introduced the classes

k − UCV of k-uniformly convex functions and k − ST of k-starlike functions. The analytic conditions for

these classes are given [6–9] as;

k −UCV =
{
f ∈ S : Re

(
1 +

zf′′(z)

f′(z)

)
> k

∣∣∣∣zf′′(z)f′(z)
∣∣∣∣
}
, (z ∈ f) . (1.5)

k −ST =
{
f ∈ S : Re

(
zf′(z)

f(z)

)
> k

∣∣∣∣zf′(z)f(z) − 1
∣∣∣∣
}
, (z ∈ f) . (1.6)

We can rewrite the above relations easily as;

Re (p(z)) > k |p(z) − 1| , (1.7)

where p(z) = 1 +
zf′′(z)
f′(z)

or p(z) =
zf′(z)
f(z)

. It is clear that p(f) is conic domain defined as;

Ωk = {w ∈ C : Re (w) > k |w − 1|} , (1.8)

or

Ωk =

{
u + iv : u > k

√
(u− 1)2 + v2

}
, (0 ≤ k < ∞) . (1.9)

These conic domains are being studied by several authors. See [2, 6, 14, 16]. Sokol and Nonukawa [23]

introduced the class defined as;

MN =

{
f ∈ S:Re

(
1 +

zf′′(z)

f′(z)

)
>

∣∣∣∣zf′(z)f(z) − 1
∣∣∣∣
}
, (z ∈ f) . (1.10)

It is obvius that MN ⊂ C. Recently S. Sivasubramanian et al. [22] extend the Sokol and Nonukawa’s work

in terms of conic domains. They introduced a new class k −MN of functions f ∈ S such that

Re

(
1 +

zf′′(z)

f′(z)

)
> k

∣∣∣∣zf′(z)f(z) − 1
∣∣∣∣ , (z ∈ f) . (1.11)

In motivation of the work [23], A. Rasheed et al. [21], introduced an interesting class k −UMα (0 ≤ α ≤ 1)

of functions f ∈ S such that



Int. J. Anal. Appl. 17 (4) (2019) 676

Re

[
(1 −α)

zf′(z)

f(z)
+ α

(zf′(z))
′

f′(z)

]
> k

∣∣∣∣zf′(z)f(z) − 1
∣∣∣∣ , (z ∈ f) . (1.12)

Obviously, we can see k −UM1 = k −MN and 1 −UM1 = MN.

We recall a Hurwitz-Lerch Zeta function Φ(s,b; z) [25] defined by

Φ(s,a; z) =

∞∑
n=2

zn

(n + b)
s , (1.13)

( b � C\Z−0 ; s � C when |z| < 1; Re(s) > 1 when |z| = 1),

where C and Z−0 denotes the set of complex numbers and the set of negative integers respectively.

Srivastava and Attiya [24] introduced the linear operator Js,b : χ → χ defined in terms of the convolution

( or Hadamard product ), by

Js,bf(z) = Gs,b(z) ∗f(z), (1.14)

where

Gs,b(z) = (1 + b)
s [Φ(s,b; z) − bs] , (1.15)

with z � f, b � C\Z−0 and s � C. Therefore, using (1.13) to (1.15), we have

Js,bf(z) = z +

∞∑
n=2

(
1 + b

n + b

)s
anz

n, (1.16)

where z � f, b � C\Z−0 and s � C.

The srivastava-Attiya operator generalizes the integral operators introduced by Alexandar [1], Libera [10],

Bernardi [3] and Jung et al. [5].

In 2007, Raducanu and Srivastava [20] introduced and studied the class S∗s,b(δ) of functions f � χ satisfies

Js,bf(z) � S
∗(δ).

Now by using concepts of conic domains and Srivastava-Attiya integral operator, we introduce new classes

as following.

Definition 1.1. Let k ∈ [0,∞) and α,β ∈ [0, 1]. Then f ∈ k −UM(α,β) if and only if

Re

[
(1 −α)

zf′(z)

f(z)
+ α

(zf′(z))
′

f′(z)

]
> k

∣∣∣∣(1 −β)zf′(z)f(z) + β (zf
′(z))

′

f′(z)
− 1
∣∣∣∣ , (z ∈ f) .

Some of the special cases are given below and we refer to [8, 9, 21–23].

Special cases:

(i) For β = 0, the class k −UM(α,β) reduces to the class k −UMα. See [21].

(ii) For α = 1 and β = 0, the class k −UM(α,β) reduces to the class k −MN. See [22].

(iii) For α = 1, β = 0 and k = 1, the class k −UM(α,β) reduces to the class MN. See [23].

(iv) For α = 1 and β = 1 , the class k −UM(α,β) reduces to the class k −UCV . See [9].



Int. J. Anal. Appl. 17 (4) (2019) 677

(v) For α = 0 and β = 0 , the class k −UM(α,β) reduces to the class k −ST. See [8].

Definition 1.2. Let α,β � [0, 1], k � [0,∞), b � C\Z−0 and s � C. Then

f � k −UMsb (α,β) if and only if Js,bf(z) � k −UM(α,β).

Clearly, for s = 0 the classes k −UMsb (α,β) and k −UM(α,β) coincides.

2. Preliminaries

Lemma 2.1. [12] Let ~ be an analytic function on f except for at most one pole on ∂f and univalent on

f, ℘ be an analytic function in f with ℘(0) = ~(0) and ℘(z) 6= ℘(0), z ∈ f. If ℘ is not subordinate to ~,

then there exist points z0 ∈ f, ξ0 ∈ ∂f and ε ≥ 1 for which

℘ (|z| < |z0|) ⊂ ~ (f) , ℘(z0) = ~(ξ0), z0℘′(z0) = εξ0℘′(ξ0).

Lemma 2.2. [6] If f ∈ S∗(α) for some α ∈
[
1
2
, 0
)
, then

Re

(
f(z)

z

)
>

1

3 − 2α
.

Lemma 2.3. [6] If Re
(√

f′(z)
)
> α for some α ∈

[
1
2
, 0
]
, then

Re

(
f(z)

z

)
>

2α2 + 1

3
.

3. Main Results

Theorem 3.1. Let k ∈ [0,∞) and α,β ∈ [0, 1]. Also, let p be a function analytic in the unit disk such that

p(0) = 1. If

Re

[
p(z) + α

zp′(z)

p(z)

]
−k

∣∣∣∣p(z) − 1 + βzp′(z)p(z)
∣∣∣∣ > 0,

then

p(z) ≺
1 + (1 − 2γ)z

1 −z
:= h(z),

where γ = γ(k,α,β) is given by

γ(k,α,β) =
1

4

[√
(α− 2k + βk)2

(1 + k)
2

+
8 (α + βk)

(1 + k)
−

(α− 2k + βk)
(1 + k)

]
. (3.1)

Proof. We may assume that γ ≥ 1
2

since the condition Re
(
p(z) +

zp′(z)
p(z)

)
> 0

implies at least Re (p(z)) > 1
2
. (See [11]). Suppose now, on the contrary that p ⊀ h. Then, by Lemma 2.1,

there exist z0 ∈ f, ξ0 ∈ ∂f and m ≥ 1 such that



Int. J. Anal. Appl. 17 (4) (2019) 678

p(z0) = γ + ix, z0p
′(z0) = my, where y ≤−

(1 −γ)2 + x2

2 (1 −γ)
, (x,y ∈ R) .

Using these relations, we have

Re

[
p(z0) + α

z0p
′(z0)

p(z0)

]
−k

∣∣∣∣p(z0) − 1 + βz0p′(z0)p(z0)
∣∣∣∣ > 0,

or

0 < Re

[
p(z0) + α

z0p
′(z0)

p(z0)

]
−k

∣∣∣∣p(z0) − 1 + βz0p′(z0)p(z0)
∣∣∣∣

= Re

[
γ + ix + α

my

γ + ix

]
−k

∣∣∣∣γ + ix− 1 + β myγ + ix
∣∣∣∣

= γ +
αmyγ

γ2 + x2
−k

∣∣∣∣∣(γ + ix)
2 − (γ + ix) + βmy
γ + ix

∣∣∣∣∣
≤ γ −

αγ

2(1 −γ)

(
(1 −γ)2 + x2

γ2 + x2

)
−k

√
(X + Y x2)

2
+ Tx2

γ2 + x2
= R(x),

where X =
(2γ+β)(1−γ)

2
, Y =

2(1−γ)+β
2(1−γ) and T = (2γ − 1)

2
. The function R(x) is even in regard of x. Now

we have to show that R(x) has maximum value at x = 0 when α,β ∈ [0, 1] and γ ∈
[
1
2
, 1
)
. We can easily

check

R′(x) = −x


 αγ (2γ − 1)

(1 −γ) (γ2 + x2)
−k


2Y (X + Y x2) + T − 2

√
(X + Y x2)

2
+ Tx2

γ2 + x2




 .

Then R′(x) = 0, if and only if, x = 0. Since α,β ∈ [0, 1] and γ ∈
[
1
2
, 1
)
. So one can see

R′′(x) = −
[
α (2γ − 1)
γ(1 −γ)

−
k

2

{
(2(1 −γ) + β) (2γ + β) + 2 (2γ − 1)2

}]
< 0.

Thus R(x) has maximum value at x = 0, that is

R(x) ≤ R(0) = γ −
αγ(1 −γ)

2γ
−
k(1 −γ) (2γ + β)

2γ
= 0,

for γ = γ(k,α,β) is given by (3.1), which contradicts the assumption. Hence

p(z) ≺
1 + (1 − 2γ)z

1 −z
:= h(z),

where γ = γ(k,α,β) is given by (3.1). �

Theorem 3.2. Let α,β � [0, 1], k � [0,∞), b � C\Z−0 and s � C. Then

k −UMsb (α,β) ⊂ S
∗
s,b(γ),

where γ = γ(k,α,β) is given by (3.1).



Int. J. Anal. Appl. 17 (4) (2019) 679

Proof. Let f ∈ k −UMsb (α,β). Then, by Definition 1.2, Js,bf(z) � k −UM(α,β), that is

Re

[
(1 −α)

z (Js,bf(z))
′

Js,bf(z)
+ α

(
z (Js,bf(z))

′)′
(Js,bf(z))

′

]
> k

∣∣∣∣∣(1 −β)z (Js,bf(z))
′

Js,bf(z)
+ β

(
z (Js,bf(z))

′)′
(Js,bf(z))

′ − 1

∣∣∣∣∣ , (z ∈ f) .
Putting p(z) =

z(Js,bf(z))
′

Js,bf(z)
, we have

Re

[
p(z) + α

zp′(z)

p(z)

]
−k

∣∣∣∣p(z) − 1 + βzp′(z)p(z)
∣∣∣∣ > 0.

Our required result follows easily by using Theorem 3.1. �

When s = 0, then we have the following new result for class k −UM(α,β)

Theorem 3.3. Let k ∈ [0,∞) and α,β ∈ [0, 1]. Then

k −UM(α,β) ⊂ S∗(γ),

where γ = γ(k,α,β) is given by (3.1).

The proof is straight forward by putting p(z) =
zf′(z)
f(z)

and using Theorem 3.1.

When k = 0, then we have the following result for a class 0−UM(α,β) = Mα, introduced by Mocanu [13].

Corollary 3.1. Let f ∈ Mα. Then f ∈ S∗(γ), where

γ (α) =
−α +

√
α2 + 8α

4
. (3.2)

When β = 0, then we have the following result, proved in [21].

Corollary 3.2. Let f ∈ k −UM(α, 0) = k −UMα. Then f ∈ S∗(γ), where

γ (α,k) =
(2ϑ−η) +

√
(2ϑ−η)2 + 8η
4

, (3.3)

where ϑ = k
k+1

, η = α+k
k+1

.

When α = 1, β = 0, then we have the following result, proved in [22].

Corollary 3.3. Let f ∈ k −UM(1, 0) = k −MN. Then f ∈ S∗(γ), where

γ(k) =
1

4



√(

1 − 2k
1 + k

)2
+

8

(1 + k)
−
(

1 − 2k
1 + k

) . (3.4)
When α = 1, β = 0, and k = 1, then we have the following result, proved in [22].

Corollary 3.4. Let f ∈ 1 −UM(1, 0) = MN. Then f ∈ S∗(γ), where γ ' 0.6403.

When α = β = 1, then we have the following result, proved in [6].



Int. J. Anal. Appl. 17 (4) (2019) 680

Corollary 3.5. Let f ∈ k −UM(1, 1) = k −UCV . Then f ∈ S∗(γ), where

γ(k) =
1

4



√(

1 −k
1 + k

)2
+ 8 −

(
1 −k
1 + k

) . (3.5)
Theorem 3.4. Let f ∈ k −UMsb (α,β). Then

Js,bf(z)

z
≺

1 + (1 − 2η)z
1 −z

,

where η = 1
3−2γ and γ = γ(k,α,β) is given by (3.1).

Proof. Let f ∈ k −UMsb (α,β). Then by Theorem 3.2 we have

z (Js,bf(z))
′

Js,bf(z)
≺

1 + (1 − 2γ)z
1 −z

,

where γ = γ(k,α,β) is given by (3.1). Using Lemma 2.2, we get

Js,bf(z)

z
≺

1 + (1 − 2η)z
1 −z

,

where η = 1
3−2γ . �

When s = 0, then one can prove the following result by using Theorem 3.3 together with Lemma 2.2.

Theorem 3.5. Let f ∈ k −UM(α,β). Then

f(z)

z
≺

1 + (1 − 2η)z
1 −z

,

where η = 1
3−2γ and γ = γ(k,α,β) is given by (3.1).

When α = 1, β = 0, then we have the following result, proved in [22].

Corollary 3.6. Let f ∈ k −UM(1, 0) = k −MN. Then

f(z)

z
≺

1 + (1 − 2η)z
1 −z

,

where η = 1
3−2γ and γ = γ(k) is given by (3.4).

When α = 1, β = 0 and k = 1, then we have the following result, proved in [22].

Corollary 3.7. Let f ∈ 1 −UM(1, 0) = MN. Then

f(z)

z
≺

1 + (1 − 2η)z
1 −z

, where η ' 0.58159.

When α = β = 1, then we have the following result, proved in [6].

Corollary 3.8. Let f ∈ k −UM(1, 1) = k −UCV . Then

f(z)

z
≺

1 + (1 − 2η)z
1 −z

,

where η = 1
3−2γ and γ = γ(k) is given by (3.5).



Int. J. Anal. Appl. 17 (4) (2019) 681

When α = β = k = 1, then we have the following result, proved in [6].

Corollary 3.9. Let f ∈ 1 −UM(1, 1) = 1 −UCV . Then

Re

(
f(z)

z

)
> 0.6289.

Theorem 3.6. Let α,β � [0, 1], k � [0,∞), b � C\Z−0 and s � C. If

Re

[√
(Js,bf(z))

′
+ α

z (Js,bf(z))
′′

2 (Js,bf(z))
′

]
> k

∣∣∣∣(Js,bf(z))′ + βz (Js,bf(z))′′2 (Js,bf(z))′ − 1
∣∣∣∣ ,

then √
(Js,bf(z))

′ ≺
1 + (1 − 2γ)z

1 −z
⇒

Js,bf(z)

z
≺

1 + (1 − 2η)z
1 −z

where η = 2γ
2+1
3

and γ = γ(k,α,β) is given by (3.1).

Proof. If we put p(z) =
√

(Js,bf(z))
′
, then

zp′(z)

p(z)
=
z (Js,bf(z))

′′

2 (Js,bf(z))
′ .

The proof follows easily by using Theorem 3.1 along with Lemma 2.3. �

We can deduce the following result from Theorem 3.6 by choosing s = 0.

Theorem 3.7. Let k ∈ [0,∞) and α,β ∈ [0, 1]. If

Re

[√
f′(z) + α

zf′′(z)

2f′(z)

]
> k

∣∣∣∣√f′(z) + βzf′′(z)2f′(z) − 1
∣∣∣∣ ,

then √
f′(z) ≺

1 + (1 − 2γ)z
1 −z

⇒
f(z)

z
≺

1 + (1 − 2η)z
1 −z

where η = 2γ
2+1
3

and γ = γ(k,α,β) is given by (3.1).

When α = 1, β = 0, then we have the following result, proved in [22].

Corollary 3.10. If

Re

[√
f′(z) +

zf′′(z)

2f′(z)

]
> k

∣∣∣√f′(z) − 1∣∣∣ ,
then

Re
(√

f′(z)
)
> γ ⇒ Re

(
f(z)

z

)
> η,

where η = 2γ
2+1
3

and γ = γ(k) is given by (3.4).

When k = 1, then we have the following result, proved in [22].



Int. J. Anal. Appl. 17 (4) (2019) 682

Corollary 3.11. If

Re

[√
f′(z) +

zf′′(z)

2f′(z)

]
>
∣∣∣√f′(z) − 1∣∣∣ ,

then

Re
(√

f′(z)
)
> γ ' 0.64 ⇒ Re

(
f(z)

z

)
> η ' 0.60.

For k = 0, we have the following result, refer to [22].

Corollary 3.12. If

Re

[√
f′(z) +

zf′′(z)

2f′(z)

]
> 0,

then

Re
(√

f′(z)
)
> γ ' 0.64 ⇒ Re

(
f(z)

z

)
> η ' 0.60.

Theorem 3.8. Let α,β � [0, 1], k � [0,∞), b � C\Z−0 and s � C. If

Re

[
Js,bf(z)

z
+ α

(
z (Js,bf(z))

′

Js,bf(z)
− 1
)]

> k

∣∣∣∣Js,bf(z)z + β
(
z (Js,bf(z))

′

Js,bf(z)
− 1
)
− 1
∣∣∣∣ ,

then

Js,bf(z)

z
≺

1 + (1 − 2γ)z
1 −z

,

where γ = γ(k,α,β) is given by (3.1).

The proof follows easily by substituting p(z) =
Js,bf(z)

z
in Theorem 3.1.

For s = 0, we can easily deduce the following result.

Theorem 3.9. Let k ∈ [0,∞) and α,β ∈ [0, 1]. If

Re

[
f(z)

z
+ α

(
zf′(z)

f(z)
− 1
)]

> k

∣∣∣∣f(z)z + β
(
zf′(z)

f(z)
− 1
)
− 1
∣∣∣∣ ,

then

f(z)

z
≺

1 + (1 − 2γ)z
1 −z

,

where γ = γ(k,α,β) is given by (3.1).

When α = 1, β = 0, then we have the following result, proved in [22].

Corollary 3.13. If

Re

[
zf′(z)

f(z)
+
f(z)

z
− 1
]
> k

∣∣∣∣f(z)z − 1
∣∣∣∣ ⇒ f(z)z ≺ 1 + (1 − 2γ)z1 −z

where γ = γ(k) is given by (3.4).

When α = 1, β = 0 and k = 1, then we have the following result, proved in [22].



Int. J. Anal. Appl. 17 (4) (2019) 683

Corollary 3.14. If

Re

[
zf′(z)

f(z)
+
f(z)

z
− 1
]
>

∣∣∣∣f(z)z − 1
∣∣∣∣ ⇒ Re

(
f(z)

z

)
> γ ' 0.64.

When α = 1, β = 0 and k = 0, then we have following result.

Corollary 3.15. If

Re

[
zf′(z)

f(z)
+
f(z)

z
− 1
]
> 0 ⇒ Re

(
f(z)

z

)
>

1

2
.

If we substitute p(z) = (Js,bf(z))
′

in Theorem 3.1, then we have the following result.

Theorem 3.10. Let α,β � [0, 1], k � [0,∞), b � C\Z−0 and s � C. If

Re

[
(Js,bf(z))

′
+ α

z (Js,bf(z))
′′

(Js,bf(z))
′

]
> k

∣∣∣∣(Js,bf(z))′ + βz (Js,bf(z))′′(Js,bf(z))′ − 1
∣∣∣∣ ,

then

Re
(
(Js,bf(z))

′)
> γ

where γ = γ(k,α,β) is given by (3.1).

For s = 0, we have the following result.

Theorem 3.11. Let k ∈ [0,∞) and α,β ∈ [0, 1]. If

Re

[
f′(z) + α

zf′′(z)

f′(z)

]
> k

∣∣∣∣f′(z) + βzf′′(z)f′(z) − 1
∣∣∣∣ ,

then

Re (f′(z)) > γ

where γ = γ(k,α,β) is given by (3.1).

When α = 1, β = 0, then we have the following result, proved in [22].

Corollary 3.16. If

Re

[
f′(z) +

zf′′(z)

f′(z)

]
> k |f′(z) − 1|⇒ Re (f′(z)) > γ ' 0.64.

When α = 1, β = k = 0, then we have the following result.

Corollary 3.17. If Re
[
f′(z) +

zf′′(z)
f′(z)

]
> 0, then Re (f′(z)) > 1

2
.



Int. J. Anal. Appl. 17 (4) (2019) 684

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Academic Publishers, 2001.


	1. Introduction
	2. Preliminaries
	3. Main Results
	References