# On Newp-Valent Meromorphic Function Involving Certain Differential and Integral Operators

Authors: Aabed Mohammed; Maslina Darus
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We define new subclasses of meromorphicp-valent functions by using certain differential operator. Combining the differential operator and certain integral operator, we introduce a general p-valent meromorphic function. Then we prove the sufficient conditions for the function in order to be in the new subclasses.


---

## Body

## 1. Introduction

LetΣ
p denote the class of meromorphic functions of the form
(1)
f
(
z
)
=
1
z
p
+
∑
n
=
p
+
1
∞
a
n
z
n
(
p
∈
ℕ
=
{
1,2
,
…
}
)
,
which are analytic and p-valent in the punctured unit disc:
(2)
𝕌
*
=
{
z
∈
ℂ
:
0
<
|
z
|
<
1
}
=
𝕌
-
{
0
}
.
A function f
∈
Σ
p is said to be in the class Σ
p
⋆
(
δ
) of meromorphic p-valent starlike of order δ
(
0
≤
δ
<
p
) if it satisfies the following inequality:
(3)
-
ℜ
(
z
f
′
(
z
)
f
(
z
)
)
>
δ
.Forf
∈
Σ
p, Saif and Kılıçman [1] introduced the linear operator 𝒟
λ
k, as follows:
(4)
𝒟
λ
f
(
z
)
=
(
1
+
p
λ
)
f
(
z
)
+
λ
z
f
′
(
z
)
,
λ
≥
0
,
𝒟
λ
0
f
(
z
)
=
f
(
z
)
,
𝒟
λ
1
f
(
z
)
=
𝒟
λ
f
(
z
)
,
𝒟
λ
2
f
(
z
)
=
𝒟
λ
(
𝒟
λ
1
f
(
z
)
)
,
and in general, for k
=
0,1
,
2
,
…, we can write
(5)
𝒟
λ
k
f
(
z
)
=
1
z
p
+
∑
n
=
p
+
1
∞
(
1
+
p
λ
+
n
λ
)
k
a
n
z
n
,
(
k
∈
ℕ
0
=
ℕ
∪
{
0
}
,
p
∈
ℕ
)
.
It is easy to see that, for f
∈
Σ
p, we have
(6)
λ
z
(
𝒟
λ
k
f
(
z
)
)
′
=
𝒟
λ
k
+
1
f
(
z
)
-
(
1
+
p
λ
)
𝒟
λ
k
f
(
z
)
,
(
k
∈
ℕ
0
,
p
∈
ℕ
)
.Meromorphically multivalent functions have been extensively studied by several authors; see, for example, Uralegaddi and Somanatha [2, 3], Liu and Srivastava [4, 5], Mogra [6, 7], Srivastava et al. [8], Aouf et al. [9, 10], Joshi and Srivastava [11], Owa et al. [12], and Kulkarni et al. [13].Now, forf
∈
Σ
p, we define the following new subclasses.Definition 1.
Let a functionf
∈
Σ
p be analytic in 𝕌
*. Then f is in the class Σ
p
S
k
(
δ
,
b
,
λ
) if, and only if, f satisfies
(7)
ℜ
{
p
-
1
b
(
𝒟
λ
k
+
1
f
(
z
)
𝒟
λ
k
f
(
z
)
-
1
)
}
>
δ
,
whereδ
∈
[
0
,
p
),   b
∈
ℂ
∖
{
0
},  λ
≥
0,   k
∈
ℕ
0.From (6), one can see that (7) is equivalent to
(8)
ℜ
{
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
}
>
δ
.Remark 2.
In Definition1, if we set(i)
k
=
0 and p
=
λ
=
1, then we have [14, Definition 1.1];
(ii)
k
=
0 and p
=
λ
=
b
=
1, then we have Σ
p
⋆
(
δ
), the class of meromorphic p-valent starlike of order δ;
(iii)
k
=
1 and p
=
λ
=
1, then we have [14, Definition 1.7].Definition 3.
Let a functionf
∈
Σ
p be analytic in 𝕌
*. Then f is in the class Σ
p
U
S
k
(
α
,
δ
,
b
,
λ
) if, and only if, f satisfies
(9)
ℜ
{
p
-
1
b
(
𝒟
λ
k
+
1
f
(
z
)
𝒟
λ
k
f
(
z
)
-
1
)
}
>
α
|
1
b
(
𝒟
λ
k
+
1
f
(
z
)
𝒟
λ
k
f
(
z
)
-
1
)
|
+
δ
,
where α
≥
0,  δ
∈
[
-
1
,
p
),  b
∈
ℂ
∖
{
0
},  λ
≥
0,  k
∈
ℕ
0.Inequality (9) is equivalent to
(10)
ℜ
{
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
}
>
α
|
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
|
+
δ
.Remark 4.
In Definition3, if we set(i)
k
=
0 and p
=
λ
=
1, then we have [14, Definition 1.3];
(ii)
fork
=
1 and p
=
λ
=
1, then we have [14, Definition 1.8].Definition 5.
Let a functionf
∈
Σ
p be analytic in 𝕌
*. Then f is in the class Σ
p
S
H
k
(
α
,
b
,
λ
), if, and only if, f satisfies
(11)
|
p
-
1
b
(
𝒟
λ
k
+
1
f
(
z
)
𝒟
λ
k
f
(
z
)
-
1
)
-
2
α
(
2
-
1
)
|
<
2
ℜ
{
p
-
1
b
(
𝒟
λ
k
+
1
f
(
z
)
𝒟
λ
k
f
(
z
)
-
1
)
}
+
2
α
(
2
-
1
)
,
where α
>
0,  b
∈
ℂ
∖
{
0
},  λ
≥
0,  k
∈
ℕ
0.Inequality (11) is equivalent to
(12)
|
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
-
2
α
(
2
-
1
)
|
<
2
ℜ
{
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
}
+
2
α
(
2
-
1
)
.Remark 6.
In Definition5, if we set(i)
k
=
0 and p
=
λ
=
1, then we have [14, Definition 1.5];
(ii)
fork
=
1 and p
=
λ
=
1, then we have [14, Definition 1.9].Recently, Mohammed and Darus [15] introduced the following p-valent meromorphic function:
(13)
G
(
z
)
=
z
ℱ
p
,
γ
1
,
…
,
γ
n
′
(
z
)
+
(
p
+
1
)
ℱ
p
,
γ
1
,
…
,
γ
n
(
z
)
,
where ℱ
p
,
γ
1
,
…
,
γ
n is the integral operator introduced and studied by the authors [15, 16] and defined by
(14)
ℱ
p
,
γ
1
,
…
,
γ
n
(
z
)
=
1
z
p
+
1
∫
0
z
(
u
p
f
1
(
u
)
)
γ
1
⋯
(
u
p
f
n
(
u
)
)
γ
n
d
u
,
where
(15)
n
,
p
∈
ℕ
,
j
∈
{
1,2
,
3
,
…
,
n
}
,
γ
j
>
0
.
For p
=
1 we obtain [17]. It is clear that
(16)
G
(
z
)
=
1
z
p
(
z
p
f
1
(
z
)
)
γ
1
⋯
(
z
p
f
n
(
z
)
)
γ
n
.By using the differential operator given by (4), we introduce the following p-valent meromorphic function.Definition 7.
Letk
∈
ℕ
0, l
=
(
l
1
,
…
,
l
n
)
∈
ℕ
0
n   and γ
j
>
0,  1
≤
j
≤
n. One defines the p-valent meromorphic function I
k
,
n
,
l
,
γ:Σ
p
n
→
Σ
p,
(17)
I
k
,
n
,
l
,
γ
(
f
1
,
…
,
f
n
)
=
Φ
,
(18)
𝒟
λ
k
Φ
(
z
)
=
1
z
p
[
(
z
p
𝒟
λ
l
1
f
1
(
z
)
)
γ
1
⋯
(
z
p
𝒟
λ
l
n
f
n
(
z
)
)
γ
n
]
,
where f
1
,
…
,
f
n
∈
Σ
p
, and 𝒟
λ is the differential operator given by (4).Remark 8.
If we setλ
=
1,  k
=
0
, and l
1
=
⋯
=
l
n
=
0, then we have the p-valent meromorphic function given by (13).


## 2.  Main Results

To prove our main results, we need the following lemma.Lemma 9.
For thep-valent meromorphic function I
k
,
n
,
l
,
γ
(
f
1
,
…
,
f
n
)
=
Φ given by (18), one has
(19)
-
λ
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
=
-
∑
j
=
1
n
γ
j
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
+
p
λ
+
∑
j
=
1
n
γ
j
.Proof.
From (18), we have
(20)
z
p
𝒟
λ
k
Φ
(
z
)
=
[
(
z
p
𝒟
λ
l
1
f
1
(
z
)
)
γ
1
⋯
(
z
p
𝒟
λ
l
n
f
n
(
z
)
)
γ
n
]
.
Differentiating (20) logarithmically and then by simple computation, we get
(21)
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
=
∑
j
=
1
n
‍
γ
j
(
z
(
𝒟
λ
l
j
f
j
(
z
)
)
′
+
p
𝒟
λ
l
j
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
)
-
p
.
From (6), we obtain
(22)
(
𝒟
λ
l
j
f
j
(
z
)
)
′
=
𝒟
λ
l
j
+
1
f
j
(
z
)
-
(
1
+
p
λ
)
𝒟
λ
l
j
f
j
(
z
)
λ
z
.
Then using (22) on the right-hand side of (21), one gets
(23)
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
=
∑
j
=
1
n
γ
j
(
𝒟
λ
l
j
+
1
f
j
(
z
)
λ
𝒟
λ
l
j
f
j
(
z
)
-
1
λ
)
-
p
.
Multiplying (23) by λ yields that
(24)
λ
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
=
∑
j
=
1
n
γ
j
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
-
p
λ
,
or, equivalently, we can write that
(25)
-
λ
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
=
-
∑
j
=
1
n
γ
j
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
i
(
z
)
+
p
λ
+
∑
j
=
1
n
γ
j
,
which is the desired result.Our first theorem is as follows.Theorem 10.
Letα
j
≥
0,   δ
j
∈
[
-
1
,
p
),  α
j
+
δ
j
≥
0
,
(
1
≤
j
≤
n
) and b
∈
ℂ
∖
{
0
},  λ
≥
0. Suppose that
(26)
∑
j
=
1
n
γ
j
(
p
-
δ
j
α
j
+
1
)
≤
p
.
If f
j
∈
Σ
p
U
S
l
j
(
α
j
,
δ
j
,
b
,
λ
)  (
1
≤
j
≤
n
), then the function 𝒟
λ
k
Φ
(
z
) defined by (18) is in the class Σ
p
S
k
(
μ
,
b
,
λ
), where
(27)
μ
=
p
-
∑
j
=
1
n
γ
j
(
p
-
δ
j
α
j
+
1
)
.Proof.
Sincef
j
∈
Σ
p
U
S
l
j
(
α
j
,
δ
j
,
b
,
λ
)   (
1
≤
j
≤
n
), by (9), we have
(28)
ℜ
{
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
}
>
p
α
j
+
δ
j
1
+
α
j
.
By (19), we get
(29)
-
λ
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
-
p
λ
=
-
∑
j
=
1
n
γ
j
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
.
This is equivalent to
(30)
p
-
λ
b
(
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
+
p
)
=
p
-
1
b
∑
j
=
1
n
γ
j
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
=
∑
j
=
1
n
γ
j
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
p
-
p
∑
j
=
1
n
γ
j
.
From (28) together with (30), we can get
(31)
ℜ
{
p
-
λ
b
(
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
+
p
)
}
=
∑
j
=
1
n
γ
j
ℜ
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
p
-
p
∑
j
=
1
n
γ
j
>
∑
j
=
1
n
γ
j
(
p
α
j
+
δ
j
1
+
α
j
)
-
p
∑
j
=
1
n
γ
j
+
p
=
p
-
∑
j
=
1
n
γ
j
(
p
-
δ
j
1
+
α
j
)
.
Hence, we obtain 𝒟
λ
k
Φ
(
z
)
∈
Σ
p
S
k
(
μ
,
b
,
λ
), where μ
=
p
-
∑
j
=
1
n
γ
j
(
(
p
-
δ
j
)
/
(
α
j
+
1
)
).Corollary 11.
Letα
j
≥
0,   δ
j
∈
[
-
1
,
p
),  α
j
+
δ
j
≥
0,  (
1
≤
j
≤
n
), and b
∈
ℂ
∖
{
0
},  λ
≥
0. Suppose that
(32)
∑
j
=
1
n
γ
j
(
p
-
δ
j
α
j
+
1
)
≤
p
.
If f
j
∈
Σ
p
U
S
l
j
(
α
j
,
δ
j
,
b
,
1
)  (
1
≤
j
≤
n
), then the function 𝒟
λ
k
Φ
(
z
), defined by (18), is in the class Σ
p
S
k
+
1
(
μ
,
b
,
1
), where μ is defined as in (27).Proof.
In Theorem10, we consider λ
=
1.By Corollary11, we easily get the following.Corollary 12.
Letα
j
≥
0,   δ
j
∈
[
-
1
,
p
),  α
j
+
δ
j
≥
0,  (
1
≤
j
≤
n
), and b
∈
ℂ
∖
{
0
},  λ
≥
0. Suppose that
(33)
∑
j
=
1
n
γ
j
(
p
-
δ
j
α
j
+
1
)
≤
p
.
If f
j
∈
Σ
p
U
S
l
j
(
α
j
,
δ
j
,
b
,
1
)  (
1
≤
j
≤
n
), then the function 𝒟
λ
k
Φ
(
z
), defined by (18), is in the class Σ
p
S
k
+
1
(
0
,
b
,
1
).Now, we prove a sufficient condition for the function𝒟
λ
k
Φ
(
z
) defined by (18) to belong to the class Σ
p
U
S
k
(
α
,
δ
,
b
,
λ
).Theorem 13.
Letα
≥
0,   δ
∈
[
-
1
,
p
), α
+
δ
≥
0  (
1
≤
j
≤
n
), and b
∈
ℂ
∖
{
0
},  λ
≥
0. Suppose that
(34)
∑
j
=
1
n
γ
j
≤
1
.
If f
j
∈
Σ
p
U
S
l
j
(
α
,
δ
,
b
,
λ
)  (
1
≤
j
≤
n
), then the function 𝒟
λ
k
Φ
(
z
) defined by (18) is in the class Σ
p
U
S
k
(
α
,
δ
,
b
,
λ
).Proof.
Sincef
j
∈
Σ
p
U
S
l
j
(
α
,
δ
,
b
,
λ
)  (
1
≤
j
≤
n
), by (9), we have
(35)
ℜ
{
p
-
1
b
(
𝒟
λ
l
j
+
1
f
(
z
)
𝒟
λ
l
j
f
(
z
)
-
1
)
}
>
α
|
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
|
+
δ
.
On the other hand, from (19), we obtain the following:
(36)
p
-
λ
b
(
z
(
𝒟
λ
k
Φ
(
z
)
)
′
𝒟
λ
k
Φ
(
z
)
+
p
)
=
∑
j
=
1
n
γ
j
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
p
-
p
∑
j
=
1
n
γ
j
.
Considering (10) with the above equality, we find
(37)
ℜ
{
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
}
-
α
|
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
|
-
δ
=
p
-
p
∑
j
=
1
n
γ
j
+
∑
j
=
1
n
γ
j
ℜ
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
-
α
|
∑
j
=
1
n
γ
j
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
|
-
δ
≥
p
-
p
∑
j
=
1
n
γ
j
+
∑
j
=
1
n
γ
j
ℜ
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
-
α
∑
j
=
1
n
γ
j
|
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
|
-
δ
>

p
-
p
∑
j
=
1
n
γ
j
+
∑
j
=
1
n
γ
j
[
α
|
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
|
+
δ
]
-
α
∑
j
=
1
n
γ
j
|
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
|
-
δ
=
(
p
-
δ
)
(
1
-
∑
j
=
1
n
γ
j
)
≥
0
.
The proof is complete.Corollary 14.
Letα
≥
0,   δ
∈
[
-
1
,
p
), α
+
δ
≥
0  (
1
≤
j
≤
n
), and b
∈
ℂ
∖
{
0
}. Suppose that
(38)
∑
j
=
1
n
γ
j
≤
1
.
If f
j
∈
Σ
p
U
S
l
j
(
α
,
δ
,
b
,
1
)  (
1
≤
j
≤
n
), then the function 𝒟
λ
k
Φ
(
z
) defined by (18) is in the class Σ
p
U
S
k
+
1
(
α
,
δ
,
b
,
1
).Proof.
In Theorem13, we consider that λ
=
1Next, for the function𝒟
λ
k
Φ defined by (18) to belong to the class Σ
p
S
H
k
(
α
,
b
,
λ
), we have the following result.Theorem 15.
Letα
≥
0,   λ
≥
0, and b
∈
ℂ
∖
{
0
}. Suppose that
(39)
∑
j
=
1
n
γ
j
≤
1
.
If f
j
∈
Σ
p
S
H
l
j
(
α
,
b
,
λ
), then the function 𝒟
λ
k
Φ
(
z
)
∈
Σ
p
S
H
k
(
α
,
b
,
λ
).Proof.
Sincef
j
∈
Σ
p
S
H
l
j
(
α
,
b
,
λ
), by (11), we have
(40)
2
ℜ
{
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
}
+
2
α
(
2
-
1
)
-
|
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
-
2
α
(
2
-
1
)
|
>
0
.
Combining (12), (30), and the above inequality, we obtain
(41)
2
ℜ
{
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
}
+
2
α
(
2
-
1
)
-
|
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
-
2
α
(
2
-
1
)
|
=
2
ℜ
{
∑
j
=
1
n
γ
j
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
p
-
p
∑
j
=
1
n
γ
j
}
+
2
α
(
2
-
1
)
-
|
∑
j
=
1
n
γ
j
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
p
-
p
∑
j
=
1
n
γ
j
-
2
α
(
2
-
1
)
|
=
∑
j
=
1
n
γ
j
{
2
ℜ
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
+
2
α
(
2
-
1
)
2
ℜ
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
}
-
2
α
(
2
-
1
)
∑
j
=
1
n
γ
j
+
2
(
p
-
p
∑
j
=
1
n
γ
j
)
+
2
α
(
2
-
1
)
-
|
∑
j
=
1
n
γ
j
{
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
-
2
α
(
2
-
1
)
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
}
+
2
α
(
2
-
1
)
∑
j
=
1
n
γ
j
-
2
α
(
2
-
1
)
+
p
-
p
∑
j
=
1
n
γ
j
|
=
∑
j
=
1
n
γ
j
{
2
ℜ
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
+
2
α
(
2
-
1
)
2
ℜ
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
}
+
[
2
p
+
2
α
(
2
-
1
)
]
(
1
-
∑
j
=
1
n
γ
j
)
-
|
∑
j
=
1
n
γ
j
{
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
-
2
α
(
2
-
1
)
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
}
+
[
p
-
2
α
(
2
-
1
)
]
(
1
-
∑
j
=
1
n
γ
j
)
|
,
which is
(42)
≥
∑
j
=
1
n
γ
j
{
2
ℜ
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
+
2
α
(
2
-
1
)
2
ℜ
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
}
+
[
2
p
+
2
α
(
2
-
1
)
]
(
1
-
∑
j
=
1
n
γ
j
)
-
∑
j
=
1
n
γ
j
|
{
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
-
2
α
(
2
-
1
)
{
(
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
)
}
|
-
|
p
-
2
α
(
2
-
1
)
|
(
1
-
∑
j
=
1
n
γ
j
)
=
∑
j
=
1
n
γ
j
{
2
ℜ
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
2
α
(
2
-
1
)
-
|
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
-
2
α
(
2
-
1
)
|
}
+
[
2
p
+
2
α
(
2
-
1
)
-
|
p
-
2
α
(
2
-
1
)
|
]
×
(
1
-
∑
j
=
1
n
γ
j
)
>
[
2
p
+
2
α
(
2
-
1
)
-
|
p
-
2
α
(
2
-
1
)
|
]
×
(
1
-
∑
j
=
1
n
γ
j
)
,
and finally
(43)
>

(
1
-
∑
j
=
1
n
γ
j
)
min
⁡
{
(
2
-
1
)
(
p
+
4
α
)
,
p
(
2
+
1
)
}
≥
0
.
Hence, by (12), we have 𝒟
λ
k
Φ
(
z
)
∈
Σ
p
S
H
k
(
α
,
b
,
λ
).Corollary 16.
Letα
≥
0 and b
∈
ℂ
∖
{
0
}. Suppose that
(44)
∑
j
=
1
n
γ
j
≤
1
.
If f
j
∈
Σ
p
S
H
l
j
(
α
,
b
,
1
), then the function 𝒟
λ
k
Φ
(
z
) defined by (18) is in the class Σ
p
S
H
k
+
1
(
α
,
b
,
1
).Proof.
In Theorem15, we consider λ
=
1.Finally, we end this paper by the following theorem and its consequence.Theorem 17.
Letα
≥
0,   λ
≥
0, and b
∈
ℂ
∖
{
0
}. Suppose that
(45)
(
p
+
2
α
(
2
-
1
)
)
∑
j
=
1
n
γ
j
<
p
.
If f
j
∈
Σ
p
S
H
l
j
(
α
,
b
,
λ
), then the function 𝒟
λ
k
Φ
(
z
) defined by (18) is in the class Σ
p
S
H
k
(
0
,
b
,
λ
).Proof.
Sincef
j
∈
Σ
p
S
H
l
j
(
α
,
b
,
λ
), by (11), we have
(46)
2
ℜ
{
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
}
+
2
α
(
2
-
1
)
>
|
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
-
2
α
(
2
-
1
)
|
.
Considering this inequality and (30), we obtain
(47)
2
ℜ
{
p
-
λ
b
(
z
(
𝒟
λ
k
f
(
z
)
)
′
𝒟
λ
k
f
(
z
)
+
p
)
}
=
2
ℜ
{
∑
j
=
1
n
γ
j
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
p
-
p
∑
j
=
1
n
γ
j
∑
j
=
1
n
γ
j
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
}
=
∑
j
=
1
n
γ
j
{
2
ℜ
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
+
2
α
(
2
-
1
)
2
ℜ
[
p
-
1
b
(
𝒟
λ
l
j
+
1
f
j
(
z
)
𝒟
λ
l
j
f
j
(
z
)
-
1
)
]
}
+
2
p
(
1
-
∑
j
=
1
n
γ
j
)
-
2
α
(
2
-
1
)
∑
j
=
1
n
γ
j
>
2
p
(
1
-
∑
j
=
1
n
γ
j
)
-
2
α
(
2
-
1
)
∑
j
=
1
n
γ
j
=
2
(
p
-
(
p
+
2
α
(
2
-
1
)
)
∑
j
=
1
n
γ
j
)
>
0
.
Hence, we have 𝒟
λ
k
Φ
(
z
)
∈
Σ
p
S
H
k
(
0
,
b
,
λ
).Corollary 18.
Letα
≥
0 and b
∈
ℂ
∖
{
0
}. Suppose that
(48)
(
p
+
2
α
(
2
-
1
)
)
∑
j
=
1
n
γ
j
<
p
.
If f
j
∈
Σ
p
S
H
l
j
(
α
,
b
,
1
), then the function 𝒟
λ
k
Φ
(
z
) defined by (18) is in the class Σ
p
S
H
k
+
1
(
0
,
b
,
1
).Proof.
In Theorem17, we consider that λ
=
1.For other work that we can look at regarding differential and integral operators, see [14, 18–24].



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*Source: 208530-2014-02-17.xml*