# Inclusion Properties of New Classes of Analytic Functions

Authors: Mohan Das; Ramesha Chendady; Indira Kasarkod Pal
Journal: Chinese Journal of Mathematics
 (2014)

---

## Abstract

The purpose of the present paper is to introduce certain new subclasses of analytic functions defined by Srivastava-Attiya operator and study their inclusion relationships and to obtain some interesting consequences of the inclusion relations.


---

## Body

## 1. Introduction, Definitions, and Preliminaries

Let𝔸 be the class of analytic functions f
(
z
) defined on the unit disk E
=
{
z
:
|
z
|
<
1
}, normalized by the conditions f
(
0
)
=
f
′
(
0
)
-
1
=
0. Let S be the subclass of 𝔸 consisting of univalent functions. Let H denote the class of analytic convex univalent functions h
(
z
) defined in E satisfying the conditions h
(
0
)
=
1 and Re
{
h
(
z
)
}
>
0 for all z
∈
E.For any two analytic functionsf
(
z
) and g
(
z
) defined in the unit disc E, we say that f
(
z
) is subordinate to g
(
z
) in E, denoted by f
(
z
)
≺
g
(
z
), if g
(
z
) is univalent in E, f
(
0
)
=
g
(
0
) and f
(
E
)
⊆
g
(
E
).The convolution or the Hadamard product of two analytic functions with power series representations(1)
f
(
z
)
=
∑
n
=
0
∞
a
n
z
n
,
g
(
z
)
=
∑
n
=
0
∞
b
n
z
n
convergent in |
z
|
<
1 is the function F
(
z
)
=
f
(
z
)
*
g
(
z
), with power series representation
(2)
F
(
z
)
=
∑
n
=
0
∞
a
n
b
n
z
n
,
|
z
|
<
1
.
The following classes of functions were defined by Shanmugam [1].For a fixed analytic functiong
(
z
)
∈
𝔸 and h
(
z
)
∈
H,
(3)
S
g
(
h
)
=
{
f
(
z
)
∈
𝔸
:
z
(
g
*
f
)
′
(
z
)
(
g
*
f
)
(
z
)
≺
h
(
z
)
,
(
g
*
f
)
(
z
)
z
≠
0
z
(
g
*
f
)
′
(
z
)
(
g
*
f
)
(
z
)
}
,
C
g
(
h
)
=
{
z
(
g
*
f
)
′′
(
z
)
(
g
*
f
)
′
(
z
)
f
∈
𝔸
:
1
+
z
(
g
*
f
)
′′
(
z
)
(
g
*
f
)
′
(
z
)
≺
h
(
z
)
,
(
g
*
f
)
(
z
)
z
≠
0
z
(
g
*
f
)
′′
(
z
)
(
g
*
f
)
′
(
z
)
}
,
K
g
(
h
)
=
{
z
(
g
*
f
)
′
(
z
)
(
g
*
ϕ
)
(
z
)
f
(
z
)
∈
𝔸
:
z
(
g
*
f
)
′
(
z
)
(
g
*
ϕ
)
(
z
)
≺
h
(
z
)
,
ϕ
(
z
)
∈
S
g
(
h
)
,
(
g
*
ϕ
)
(
z
)
z
≠
0
ϕ
(
z
)
z
(
g
*
f
)
′
(
z
)
(
g
*
ϕ
)
(
z
)
}
.
We note that f
(
z
)
∈
C
g
(
h
)
⇔
z
f
′
(
z
)
∈
S
g
(
h
).Wheng
(
z
)
=
z
/
(
1
-
z
) and h
(
z
)
=
(
1
+
z
)
/
(
1
-
z
), S
g
(
h
)
=
S
*, the class of starlike functions; C
g
(
h
)
=
C, the class of convex functions; and K
g
(
h
)
=
K, the class of close to convex functions.The generalized Hurwitz-Lerch Zeta functionϕ
(
z
,
s
,
a
) is defined as
(4)
ϕ
(
z
,
s
,
a
)
=
∑
n
=
0
∞
z
n
(
a
+
n
)
s
,
where
a
∈
ℂ
∖
ℤ
0
-
;
s
∈
ℂ
,
when
|
z
|
<
1
;
Re
s
>
1
,
when
|
z
|
=
1
;
and
ℤ
0
-
=
ℤ
∖
ℕ.This function contains, as its special cases, functions such as the Riemann and Hurwitz zeta function, Lerch zeta function, the polylogarithmic function, and the Lipschitz Lerch zeta function.Several interesting properties and characteristics of the Hurwitz-Lerch zeta functionϕ
(
z
,
s
,
a
) can be found in the recent investigations by Choi and Srivastava [2], Ferreira and López [3], Garg et al. [4], Lin and Srivastava [5], and Lin et al. [6].Using this function, Srivastava and Attiya [7] introduced the following family of linear operator J
μ
,
b
:
𝔸
→
𝔸, defined by
(5)
J
μ
,
b
f
(
z
)
=
G
μ
,
b
*
f
(
z
)
,
(
z
∈
E
,
b
∈
ℂ
∖
ℤ
0
-
;
μ
∈
ℂ
)
,
where the function G
μ
,
b is given by
(6)
G
μ
,
b
(
z
)
=
(
1
+
b
)
μ
[
ϕ
(
z
,
μ
,
b
)
-
b
-
μ
]
=
z
+
∑
n
=
2
∞
(
b
+
1
b
+
n
)
μ
z
n
,
(
z
∈
E
)
.
Using (6) in (5), we get
(7)
J
μ
,
b
f
(
z
)
=
z
+
∑
n
=
2
∞
‍
(
b
+
1
b
+
n
)
μ
a
n
z
n
,
(
z
∈
E
)
.
For f
(
z
)
∈
𝔸 and z
∈
E,
(8)
J
0
,
b
f
(
z
)
=
f
(
z
)
,
J
1,0
f
(
z
)
=
∫
0
z
f
(
t
)
t
d
t
=
A
(
f
)
,
J
1
,
b
f
(
z
)
=
1
+
b
z
b
∫
0
z
t
b
-
1
f
(
t
)
d
t
=
J
b
(
f
)
,
where A
(
f
) denotes Alexander transform [8] and J
b
(
f
) denotes Bernardi integral operator [9].By (7), Srivastava and Attiya obtained the following relation:
(9)
z
(
J
μ
+
1
,
b
f
(
z
)
)
′
=
(
b
+
1
)
J
μ
,
b
f
(
z
)
-
b
J
μ
+
1
,
b
f
(
z
)
.
In order to obtain our main results, we need the following lemmas.Lemma 1 (see [10]).
Letβ and γ be constants. Let ψ be convex univalent in E with ψ
(
0
)
=
1 and Re
(
β
ψ
(
z
)
+
γ
)
>
0 for z in E. Let p
(
z
) be analytic in E with p
(
0
)
=
1. Then
(10)
p
(
z
)
+
z
p
′
(
z
)
β
p
(
z
)
+
γ
≺
ψ
(
z
)
i
m
p
l
i
e
s
p
(
z
)
≺
ψ
(
z
)
.Lemma 2 (see [11]).
Letβ and γ be constants. Let ψ be convex univalent in E with ψ
(
0
)
=
1 and Re
(
β
ψ
(
z
)
+
γ
)
>
0 for z in E. Let q
(
z
) be analytic in E with q
(
z
)
≺
ψ
(
z
). Let p
(
z
) be analytic in E with p
(
0
)
=
1. Then
(11)
p
(
z
)
+
z
p
′
(
z
)
β
q
(
z
)
+
γ
≺
ψ
(
z
)
i
m
p
l
i
e
s
p
(
z
)
≺
ψ
(
z
)
.Lemma 3 (see [12]).
Let a functionp
(
z
)
=
1
+
p
1
z
+
p
2
z
2
+
⋯ be analytic in E and p
(
z
)
≠
0
,
z
∈
E. If there exists a point z
0
∈
E such that |
arg
p
(
z
)
|
<
(
π
/
2
)
α
(
|
z
|
<
|
z
0
|
) and |
arg
p
(
z
0
)
|
=
(
π
/
2
)
α
(
0
<
α
≤
1
), then
(12)
z
0
p
′
(
z
0
)
p
(
z
0
)
=
i
k
α
,
where
(13)
k
≥
1
2
(
a
+
1
a
)
,
w
h
e
n
arg
p
(
z
0
)
=
π
2
α
,
k
≤
-
1
2
(
a
+
1
a
)
,
w
h
e
n
arg
p
(
z
0
)
=
-
π
2
α
,
and (
p
(
z
0
)
)
1
/
α
=
±
i
a
(
a
>
0
).


## 2. Main Results

First we will define a subclassQ
g
(
α
,
h
) as follows.Definition 4.
For a fixed analytic functiong
∈
𝔸 and a real α
≥
0,
(14)
Q
g
(
α
,
h
)
=
{
f
∈
𝔸
:
α
[
z
(
g
*
f
)
′
(
z
)
]
′
(
g
*
ψ
)
′
(
z
)
+
(
1
-
α
)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
≺
h
(
z
)
,
ψ
(
z
)
∈
S
g
(
h
)
[
z
(
g
*
f
)
′
(
z
)
]
′
(
g
*
ψ
)
′
(
z
)
}
.Remark 5.
Wheng
(
z
)
=
z
/
(
1
-
z
), ψ
(
z
)
=
z, and h
(
z
)
=
(
1
+
z
)
/
(
1
-
z
), Q
g
(
α
,
h
) reduces to the class R
(
α
)
=
{
f
∈
𝔸
:
Re
{
α
z
f
′′
(
z
)
+
f
′
(
z
)
}
>
0
}.Theorem 6.
Forα
>
0,
Q
g
(
α
,
h
)
⊆
K
g
(
h
).Proof.
Letf
(
z
)
∈
Q
g
(
α
,
h
) and p
(
z
)
=
z
(
g
*
f
)
′
(
z
)
/
(
g
*
ψ
)
(
z
)
,
ψ
(
z
)
∈
S
g
(
h
).
Note thatp
(
z
) is analytic and p
(
0
)
=
0.
Then,(15)
p
(
z
)
(
g
*
ψ
)
′
(
z
)
+
p
′
(
z
)
(
g
*
ψ
)
(
z
)
=
[
z
(
g
*
f
)
′
(
z
)
]
′
⟹
[
z
(
g
*
f
)
′
(
z
)
]
′
(
g
*
ψ
)
′
(
z
)
=
p
(
z
)
+
p
′
(
z
)
(
g
*
ψ
)
(
z
)
(
g
*
ψ
)
′
(
z
)
.
Hence
(16)
α
[
z
(
g
*
f
)
′
(
z
)
]
′
(
g
*
ψ
)
′
(
z
)
+
(
1
-
α
)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
≺
h
(
z
)
⟹
p
(
z
)
+
(
z
p
′
(
z
)
)
(
1
/
α
)
q
(
z
)
≺
h
(
z
)
,
where
(17)
q
(
z
)
=
(
z
(
g
*
ψ
)
′
(
z
)
)
(
g
*
ψ
)
(
z
)
.
Since ψ
(
z
)
∈
S
g
(
h
)
,
q
(
z
)
≺
h
(
z
).
Thus, by using Lemma2,
(18)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
=
p
(
z
)
≺
h
(
z
)
.
That is, f
(
z
)
∈
K
g
(
h
).In the following, we obtain an inclusion relation of the classQ
g
(
α
,
h
).Theorem 7.
For0
<
β
≤
α
,
Q
g
(
α
,
h
)
⊆
Q
g
(
β
,
h
).Proof.
Letf
(
z
)
∈
Q
g
(
α
,
h
).
Then(19)
α
[
z
(
g
*
f
)
′
(
z
)
]
′
(
g
*
ψ
)
′
(
z
)
+
(
1
-
α
)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
∈
h
(
E
)
,
000000000000000000000000
z
∈
E
,
ψ
∈
S
g
(
h
)
,
and by Theorem 6,
(20)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
≺
h
(
z
)
,
z
∈
E
.
Thus,
(21)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
∈
h
(
E
)
,
z
∈
E
.
Since h
(
E
) is a convex domain,
(22)
t
[
α
[
z
(
g
*
f
)
′
(
z
)
]
′
(
g
*
ψ
)
′
(
z
)
+
(
1
-
α
)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
]
+
(
1
-
t
)
[
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
]
∈
h
(
E
)
,
z
∈
E
,
where 0
<
t
≤
1.
That is,(23)
α
t
[
z
(
g
*
f
)
′
(
z
)
]
′
(
g
*
ψ
)
′
(
z
)
+
(
1
-
α
t
)
z
(
g
*
f
)
′
(
z
)
(
g
*
ψ
)
(
z
)
∈
h
(
E
)
,
(
g
*
ψ
)
′
(
z
)
(
g
*
ψ
)
′
(
z
)
(
g
*
ψ
)
′
(
z
)
(
g
ψ
)
′
(
z
)
z
∈
E
.
Let α
t
=
β. Then 0
<
β
≤
α and (23) implies that f
(
z
)
∈
Q
g
(
β
,
h
).Theorem 8.
Letn
∈
ℕ
0. If f
∈
S
g
(
h
) then J
n
,
b
(
f
)
∈
S
g
(
h
), for Re
b
≥
0.Proof.
Letf
(
z
)
∈
S
g
(
h
).
We prove the result by mathematical induction onn.
Whenn
=
0, J
0
,
b
(
f
)
=
f
∈
S
g
(
h
).
Whenn
=
1, set
(24)
p
(
z
)
=
z
(
J
1
,
b
(
f
)
*
g
)
′
J
1
,
b
(
f
)
*
g
.
Clearly, p
(
z
) is analytic and p
(
0
)
=
1.
From (9), we get
(2)
z
(
J
1
,
b
(
f
)
*
g
)
′
=
(
b
+
1
)
(
f
*
g
)
-
b
(
J
1
,
b
(
f
)
*
g
)
⟹
p
(
z
)
+
b
=
(
b
+
1
)
f
*
g
J
1
,
b
(
f
)
*
g
.
By logarithmic differentiation and multiplication by z, we get
(26)
z
p
′
(
z
)
p
(
z
)
+
b
+
p
(
z
)
=
z
(
f
*
g
)
′
f
*
g
≺
h
(
z
)
.
As Re
b
≥
0, by Lemma 1, we get p
(
z
)
≺
h
(
z
).
That is,J
1
,
b
(
f
)
∈
S
g
(
h
).
Suppose thatJ
m
,
b
(
f
)
∈
S
g
(
h
).
Set(27)
p
(
z
)
=
z
(
J
m
+
1
,
b
(
f
)
*
g
)
′
J
m
+
1
,
b
(
f
)
*
g
.
Clearly, p
(
z
) is analytic and p
(
0
)
=
1.
From (9), we get
(2)
z
(
J
m
+
1
,
b
(
f
)
*
g
)
′
=
(
b
+
1
)
(
J
m
,
b
(
f
)
*
g
)
-
b
(
J
m
+
1
,
b
(
f
)
*
g
)
⟹
p
(
z
)
+
b
=
(
b
+
1
)
J
m
,
b
(
f
)
*
g
J
m
+
1
,
b
(
f
)
*
g
.
By logarithmic differentiation and multiplication by z, we get
(29)
z
p
′
(
z
)
p
(
z
)
+
b
+
p
(
z
)
=
z
(
J
m
,
b
(
f
)
*
g
)
′
J
m
,
b
(
f
)
*
g
≺
h
(
z
)
.
Hence, by Lemma 1, p
(
z
)
≺
h
(
z
), for Re
b
≥
0.
That is,J
m
+
1
,
b
(
f
)
∈
S
g
(
h
), which proves the result.Theorem 9.
Letn
∈
ℕ
0. If f
∈
C
g
(
h
) then J
n
,
b
(
f
)
∈
C
g
(
h
), for Re
b
≥
0.Proof.
Letf
(
z
)
∈
C
g
(
h
).
Then(30)
z
f
′
(
z
)
∈
S
g
(
h
)
⟹
J
n
,
b
(
z
f
′
(
z
)
)
=
z
(
J
n
,
b
(
f
)
)
′
(
z
)
∈
S
g
(
h
)
⟹
J
n
,
b
(
f
)
∈
C
g
(
h
)
.Remark 10.
The Bernardi integral operatorL
c preserves both the classes S
g
(
h
) and C
g
(
h
) whenever Re
c
≥
0.Corollary 11.
Letn
∈
ℕ
0, Re
b
≥
0, and Re
c
≥
0.(i)
Iff
∈
S
g
(
h
) then J
n
,
b
(
L
c
)
(
f
)
∈
S
g
(
h
).
(ii)
Iff
∈
C
g
(
h
) then J
n
,
b
(
L
c
)
(
f
)
∈
C
g
(
h
).Theorem 12.
Letf
∈
S
g
(
h
) and
(31)
α
=
min
⁡
z
∈
E
{
Re
h
(
z
)
}
.
Then
(32)
Re
{
J
n
,
b
(
f
)
*
g
J
n
+
1
,
b
(
f
)
*
g
}
>
b
+
α
b
+
1
f
o
r
b
>
0
.Proof.
Letf
∈
S
g
(
h
).
From (9), we get
(33)
(
b
+
1
)
J
n
,
b
(
f
)
*
g
J
n
+
1
,
b
(
f
)
*
g
=
b
+
z
(
J
n
+
1
,
b
(
f
)
*
g
)
′
J
n
+
1
,
b
(
f
)
*
g
.
From Theorem 8, J
n
,
b
(
f
)
∈
S
g
(
h
)
,for alln.
Hence,(34)
(
b
+
1
)
J
n
,
b
(
f
)
*
g
J
n
+
1
,
b
(
f
)
*
g
-
b
=
z
(
J
n
+
1
,
b
(
f
)
*
g
)
′
J
n
+
1
,
b
(
f
)
*
g
≺
h
(
z
)
.
Let(35)
α
=
min
⁡
z
∈
E
{
Re
h
(
z
)
}
.
Then for z
∈
E,
(36)
Re
{
(
b
+
1
)
J
n
,
b
(
f
)
*
g
J
n
+
1
,
b
(
f
)
*
g
-
b
}
>
α
.
This completes the proof.In the following, we introduce certain classes of analytic functions defined by Srivastava-Attiya operator and investigate their inclusion relationships.Leth
(
z
)
∈
H and g
(
z
)
∈
𝔸.Definition 13.
Forμ
∈
ℂ and b
∈
ℂ
∖
ℤ
0
-, let
(37)
S
g
(
μ
,
b
;
h
)
=
{
f
∈
𝔸
:
J
μ
,
b
f
(
z
)
∈
S
g
(
h
)
}
=
{
f
∈
𝔸
:
z
(
J
μ
,
b
(
f
)
*
g
)
′
J
μ
,
b
(
f
)
*
g
≺
h
(
z
)
}
.Definition 14.
Forμ
∈
ℂ and b
∈
ℂ
∖
ℤ
0
-, let
(38)
C
g
(
μ
,
b
;
h
)
=
{
f
∈
𝔸
:
J
μ
,
b
f
(
z
)
∈
C
g
(
h
)
}
=
{
f
∈
𝔸
:
1
+
z
(
J
μ
,
b
(
f
)
*
g
)
′
′
(
J
μ
,
b
(
f
)
*
g
)
′
≺
h
(
z
)
}
.Remark 15.
Clearly,f
(
z
)
∈
C
g
(
μ
,
b
;
h
) if and only if z
f
′
(
z
)
∈
S
g
(
μ
,
b
;
h
).Definition 16.
Forμ
∈
ℂ and b
∈
ℂ
∖
ℤ
0
-, let
(39)
K
g
(
μ
,
b
;
h
)
=
{
f
∈
𝔸
:
z
(
J
μ
,
b
(
f
)
*
g
)
′
J
μ
,
b
(
ϕ
)
*
g
≺
h
(
z
)
,
ϕ
(
z
)
∈
S
g
(
μ
,
b
;
h
)
}
.Remark 17.
We note that(40)
S
g
(
0
,
b
;
h
)
=
S
g
(
h
)
,
C
g
(
0
,
b
;
h
)
=
C
g
(
h
)
,
K
g
(
0
,
b
;
h
)
=
K
g
(
h
)
.Theorem 18.
IfRe
b
≥
0, then S
g
(
μ
,
b
;
h
)
⊂
S
g
(
μ
+
1
,
b
;
h
).Proof.
Letf
∈
S
g
(
μ
,
b
;
h
). Then,
(41)
z
(
J
μ
,
b
(
f
)
*
g
)
′
J
μ
,
b
(
f
)
*
g
≺
h
(
z
)
.
Let
(42)
p
(
z
)
=
z
(
J
μ
+
1
,
b
(
f
)
*
g
)
′
J
μ
+
1
,
b
(
f
)
*
g
,
where p
(
z
) is analytic and p
(
0
)
=
1.
Using (42) in (9), we get
(43)
b
+
p
(
z
)
=
(
b
+
1
)
(
J
μ
,
b
(
f
)
*
g
)
J
μ
+
1
,
b
(
f
)
*
g
.
On logarithmic differentiation and multiplication by z, we arrive at
(44)
p
(
z
)
+
z
p
′
(
z
)
b
+
p
(
z
)
=
z
(
J
μ
,
b
(
f
)
*
g
)
′
J
μ
,
b
(
f
)
*
g
≺
h
(
z
)
.
Since h
∈
H is convex univalent in E with h
(
0
)
=
1 and Re
h
(
z
)
>
0, Lemma 1 and the subordination relation (44) gives
(45)
p
(
z
)
≺
h
(
z
)
for
Re
b
≥
0
,
proving the theorem.Theorem 19.
IfRe
b
≥
0, then C
g
(
μ
,
b
;
h
)
⊂
C
g
(
μ
+
1
,
b
;
h
).Proof.
Letf
∈
C
g
(
μ
,
b
;
h
).
Then(46)
z
f
′
(
z
)
∈
S
g
(
μ
,
b
;
h
)
⊂
S
g
(
μ
+
1
,
b
;
h
)
⟹
f
(
z
)
∈
C
g
(
μ
+
1
,
b
;
h
)
.Theorem 20.
IfRe
b
≥
0, then K
g
(
μ
,
b
;
h
)
⊂
K
g
(
μ
+
1
,
b
;
h
).Proof.
Letf
(
z
)
∈
K
g
(
μ
,
b
;
h
). Then, by Definition 16, there exists a function ϕ
(
z
)
∈
S
g
(
μ
,
b
;
h
) such that
(47)
z
(
J
μ
,
b
(
f
)
*
g
)
′
J
μ
,
b
(
ϕ
)
*
g
≺
h
(
z
)
.
Using Theorem 18, we get ϕ
(
z
)
∈
S
g
(
μ
+
1
,
b
;
h
).
Thus,(48)
q
(
z
)
=
z
(
J
μ
+
1
,
b
(
ϕ
)
*
g
)
′
J
μ
+
1
,
b
(
ϕ
)
*
g
(
z
)
≺
h
(
z
)
,
where q
(
z
) is analytic.
Let(49)
p
(
z
)
=
z
(
J
μ
+
1
,
b
(
f
)
*
g
)
′
J
μ
+
1
,
b
(
ϕ
)
*
g
,
where p
(
z
) is analytic and p
(
0
)
=
1.
Now, making use of (9),
(50)
z
(
J
μ
,
b
(
f
)
*
g
)
′
J
μ
,
b
(
ϕ
)
*
g
=
J
μ
,
b
(
z
f
′
)
*
g
J
μ
,
b
(
ϕ
)
*
g
=
z
(
J
μ
+
1
,
b
(
z
f
′
)
*
g
)
′
+
b
J
μ
+
1
,
b
(
z
f
′
)
*
g
z
(
J
μ
+
1
,
b
(
ϕ
)
*
g
)
′
+
b
J
μ
+
1
,
b
(
ϕ
)
*
g
=
(
z
(
J
μ
+
1
,
b
(
z
f
′
)
*
g
)
′
J
μ
+
1
,
b
(
ϕ
)
*
g
+
b
J
μ
+
1
,
b
(
z
f
′
)
*
g
J
μ
+
1
,
b
(
ϕ
)
*
g
)
×
(
z
(
J
μ
+
1
,
b
(
ϕ
)
*
g
)
′
J
μ
+
1
,
b
(
ϕ
)
*
g
+
b
)
-
1
.
Simplifying the above, using (48) and (49), and then using (47) we arrive at
(51)
p
(
z
)
+
z
p
′
(
z
)
q
(
z
)
+
b
=
z
(
J
μ
,
b
(
f
)
*
g
)
′
J
μ
,
b
(
ϕ
)
*
g
≺
h
(
z
)
.
Since h
∈
H is convex univalent in E with h
(
0
)
=
1 and Re
h
(
z
)
>
0, Lemma 2 and the subordination relation (51) gives
(52)
p
(
z
)
≺
h
(
z
)
,
for
Re
b
≥
0
,
proving the assertion.Now, we consider certain interesting consequences of the above inclusion relationships.Corollary 21.
Suppose thatf
∈
𝔸.(i)
IfA
(
f
)
∈
S
g
(
h
) then f
∈
S
g
(
n
,
0
;
h
)
,
n
∈
ℕ
∖
{
1
}.
(ii)
IfA
(
f
)
∈
C
g
(
h
) then f
∈
C
g
(
n
,
0
;
h
)
,
n
∈
ℕ
∖
{
1
}.
(iii)
IfA
(
f
)
∈
K
g
(
h
) then f
∈
K
g
(
n
,
0
;
h
)
,
n
∈
ℕ
∖
{
1
}.Corollary 22.
Suppose thatf
∈
𝔸 and Re
c
≥
0. (i)
IfL
c
(
f
)
∈
S
g
(
h
) then f
∈
S
g
(
n
,
c
;
h
)
,
n
∈
ℕ
∖
{
1
}.
(ii)
IfL
c
(
f
)
∈
C
g
(
h
) then f
∈
C
g
(
n
,
c
;
h
)
,
n
∈
ℕ
∖
{
1
}.
(iii)
IfL
c
(
f
)
∈
K
g
(
h
) then f
∈
K
g
(
n
,
c
;
h
)
,
n
∈
ℕ
∖
{
1
}.In a similar manner one can obtain many applications of Theorems18, 19, and 20.Using Srivastava-Attiya integral operator, we introduce some classes of analytic functions, analogous to the class of strongly starlike functions of orderα and the class of strongly convex functions of order α, and study the properties of the newly defined classes.Let0
<
α
≤
1, n
∈
ℕ
0, Re
b
≥
0, g
∈
𝔸, and z
∈
E.Definition 23.
Let(53)
S
~
n
,
b
;
g
(
α
)
=
{
f
∈
𝔸
:
|
arg
(
z
(
J
n
,
b
(
f
)
*
g
)
′
J
n
,
b
(
f
)
*
g
)
|
<
π
2
α
,
z
(
J
n
,
b
(
f
)
*
g
)
′
J
n
,
b
(
f
)
*
g
≠
0
|
arg
(
z
(
J
n
,
b
(
f
)
*
g
)
′
J
n
,
b
(
f
)
*
g
)
|
}
.Definition 24.
Let(54)
C
~
n
,
b
;
g
(
α
)
=
{
f
∈
A
:
|
arg
(
1
+
z
(
J
n
,
b
(
f
)
*
g
)
′′
(
J
n
,
b
(
f
)
*
g
)
′
)
|
<
π
2
α
,
1
+
z
(
J
n
,
b
(
f
)
*
g
)
′′
(
J
n
,
b
(
f
)
*
g
)
′
≠
0
}
.Remark 25.
Clearly,f
∈
C
~
n
,
b
;
g
(
α
) if and only if z
f
′
(
z
)
∈
S
~
n
,
b
;
g
(
α
).Remark 26.
Forn
=
0, g
(
z
)
=
(
1
+
z
)
/
(
1
-
z
), S
~
0
,
b
;
g
(
α
) is the class of strongly starlike functions of order α and C
~
0
,
b
;
g
(
α
) is the class of strongly convex functions of order α.Now we prove inclusion relationships for the above defined classes.Theorem 27.
Suppose thatb
≥
0. Then S
~
n
,
b
;
g
(
α
)
⊂
S
~
n
+
1
,
b
;
g
(
α
) for each n
,
n
∈
ℕ
0.Proof.
Letf
(
z
)
∈
S
~
n
,
b
;
g
(
α
).
We set(55)
(
z
J
n
+
1
,
b
(
f
)
*
g
)
′
J
n
+
1
,
b
(
f
)
*
g
=
p
(
z
)
,
where p
(
z
) is analytic in E, p
(
0
)
=
1 and p
(
z
)
≠
0 for all z
∈
E.
Using (9) and (55), we have
(56)
(
b
+
1
)
J
n
,
b
(
f
)
*
g
J
n
+
1
,
b
(
f
)
*
g
=
b
+
p
(
z
)
.
On logarithmic differentiation and multiplication by z, we get
(57)
z
(
J
n
,
b
(
f
)
*
g
)
′
J
n
,
b
(
f
)
*
g
=
z
(
J
n
+
1
,
b
(
f
)
*
g
)
′
J
n
+
1
,
g
(
f
)
*
g
+
z
p
′
(
z
)
b
+
p
(
z
)
=
p
(
z
)
+
z
p
′
(
z
)
b
+
p
(
z
)
.
Suppose that there exists a point z
0
∈
E such that
(58)
|
arg
p
(
z
)
|
<
π
2
α
(
|
z
|
<
|
z
0
|
)
,
|
arg
p
(
z
0
)
|
=
π
2
α
.
Then, applying Lemma 3, we can write
(59)
z
0
p
′
(
z
0
)
p
(
z
0
)
=
i
k
α
,
(
p
(
z
0
)
)
1
/
α
=
±
i
a
(
a
>
0
)
.
Thus, if arg
p
(
z
0
)
=
-
(
π
/
2
)
α, then from (57)
(60)
z
0
(
J
n
,
b
(
f
(
z
0
)
)
*
g
(
z
0
)
)
′
J
n
,
b
(
f
(
z
0
)
)
*
g
(
z
0
)
=
p
(
z
0
)
[
1
+
z
0
p
′
(
z
0
)
/
p
(
z
0
)
b
+
p
(
z
0
)
]
=
a
α
e
-
i
π
α
/
2
[
1
+
i
k
α
b
+
a
α
e
-
i
π
α
/
2
]
.
This implies that
(61)
arg
{
z
0
(
J
n
,
b
(
f
(
z
0
)
)
*
g
(
z
0
)
)
′
J
n
,
b
(
f
(
z
0
)
)
*
g
(
z
0
)
}
=
-
π
2
α
+
arg
[
1
+
i
k
α
b
+
a
α
e
-
i
π
α
/
2
]
=
-
π
2
α
+
ta
n
-
1
[
k
α
(
b
+
a
α
cos
⁡
(
π
α
/
2
)
)
a
2
α
+
b
2
+
a
α
(
2
b
cos
⁡
(
π
α
/
2
)
-
k
α
sin
(
π
α
/
2
)
)
]
≤
-
π
2
α
(
where
k
≤
-
1
2
(
a
+
1
a
)
≤
-
1
)
,
which contradicts the condition that f
(
z
)
∈
S
~
n
,
b
;
g
(
α
).
Similarly, if argp
(
z
0
)
=
(
π
/
2
)
α, then we obtain that
(62)
arg
{
z
0
(
J
n
,
b
(
f
(
z
0
)
)
*
g
(
z
0
)
)
′
J
n
,
b
(
f
(
z
0
)
)
*
g
(
z
0
)
}
≥
π
2
α
,
which also contradicts the condition that f
(
z
)
∈
S
~
n
,
b
;
g
(
α
).
Thus we have(63)
|
arg
p
(
z
)
|
<
π
2
α
,
z
∈
E
That is,
(64)
|
arg
{
z
(
J
n
+
1
,
b
(
f
)
*
g
)
′
J
n
+
1
,
g
(
f
)
*
g
}
|
<
π
2
α
,
z
∈
E
.
⟹
f
(
z
)
∈
S
~
n
+
1
,
b
;
g
(
α
)
.Theorem 28.
Suppose thatb
≥
0. Then C
~
n
,
b
;
g
(
α
)
⊂
C
~
n
+
1
,
b
;
g
(
α
), for n
∈
ℕ
0.Proof.
Letf
(
z
)
∈
C
~
n
,
b
;
g
(
α
).
Then(65)
z
f
′
(
z
)
∈
S
~
n
,
b
;
g
(
α
)
⟹
z
f
′
(
z
)
∈
S
~
n
+
1
,
b
;
g
(
α
)
⟹
f
(
z
)
∈
C
~
n
+
1
,
b
;
g
(
α
)
.



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*Source: 974175-2014-03-31.xml*