# Some Subordination Results onq-Analogue of Ruscheweyh Differential Operator

Authors: Huda Aldweby; Maslina Darus
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

We derive theq-analogue of the well-known Ruscheweyh differential operator using the concept of q-derivative. Here, we investigate several interesting properties of this q-operator by making use of the method of differential subordination.


---

## Body

## 1. Introduction

Recently, the area ofq-analysis has attracted the serious attention of researchers. This great interest is due to its application in various branches of mathematics and physics. The application of q-calculus was initiated by Jackson [1, 2]. He was the first to develop q-integral and q-derivative in a systematic way. Later, from the 80s, geometrical interpretation of q-analysis has been recognized through studies on quantum groups. It also suggests a relation between integrable systems and q-analysis. In ([3–5]) the q-analogue of Baskakov Durrmeyer operator has been proposed, which is based on q-analogue of beta function. Another important q-generalization of complex operators is q-Picard and q-Gauss-Weierstrass singular integral operators discussed in [6–8]. The authors studied approximation and geometric properties of these q-operators in some subclasses of analytic functions in compact disk. Very recently, other q-analogues of differential operators have been introduced in [9]; see also ([10, 11]). These q-operators are defined by using convolution of normalized analytic functions and q-hypergeometric functions, where several interesting results are obtained. From this point, it is expected that deriving q-analogues of operators defined on the space of analytic functions would be important in future. A comprehensive study on applications of q-analysis in operator theory may be found in [12].We provide some notations and concepts ofq-calculus used in this paper. All the results can be found in [12–14]. For n
∈
ℕ
,
0
<
q
<
1, we define
(1)
[
n
]
q
=
1
-
q
n
1
-
q
,
[
n
]
q
!
=
{
[
n
]
q
[
n
-
1
]
q
⋯
[
1
]
q
,
n
=
1,2
,
…
;
;
1
,
n
=
0
.Asq
→
1
,
[
n
]
q
→
n, and this is the bookmark of a q-analogue: the limit as q
→
1 recovers the classical object.For complex parametersa
,
b
,
c
,
q
(
c
∈
ℂ
∖
{
0
,
-
1
,
-
2
,
…
}
,
|
q
|
<
1
), the q-analogue of Gauss’s hypergeometric function 2
Φ
1
(
a
,
b
;
c
,
q
,
z
) is defined by
(2)
2
Φ
1
(
a
,
b
;
c
,
q
,
z
)
=
∑
k
=
0
∞
‍
(
a
,
q
)
k
(
b
,
q
)
k
(
q
,
q
)
k
(
c
,
q
)
k
z
k
,
(
z
∈
𝕌
)
,
where (
n
,
q
)
k is the q-analogue of Pochhammer symbol defined by
(3)
(
n
,
q
)
k
=
{
1
,
k
=
0
;
(
1
-
n
)
(
1
-
n
q
)
(
1
-
n
q
2
)
⋯
(
1
-
n
q
k
-
1
)
,
k
∈
ℕ
.Theq-derivative of a function h
(
x
) is defined by
(4)
D
q
(
h
(
x
)
)
=
h
(
q
x
)
-
h
(
x
)
(
q
-
1
)
x
,
q
≠
1
,
x
≠
0
,
and D
q
(
h
(
0
)
)
=
f
′
(
0
). For a function h
(
z
)
=
z
k observe that
(5)
D
q
(
h
(
z
)
)
=
D
q
(
z
k
)
=
1
-
q
k
1
-
q
z
k
-
1
=
[
k
]
q
z
k
-
1
;
;
then lim
⁡
q
→
1
D
q
(
h
(
z
)
)
=
lim
⁡
q
→
1
[
k
]
q
z
k
-
1
=
k
z
k
-
1
=
h
′
(
z
), where h
′
(
z
) is the ordinary derivative.Next, we state the class𝒜 of all functions of the following form:
(6)
f
(
z
)
=
z
+
∑
k
=
2
∞
‍
a
k
z
k
,
which are analytic in the open unit disk 𝕌
=
{
z
∈
ℂ
:
|
z
|
<
1
}. If f and g are analytic functions in 𝕌, we say that f is subordinate to g; written f
≺
g, if there is a function w analytic in 𝕌, with w
(
0
)
=
0
,
|
w
(
z
)
|
<
1, for all z
∈
𝕌, such that f
(
z
)
=
g
(
w
(
z
)
) for all z
∈
𝕌. If g is univalent, then f
≺
g if and only if f
(
0
)
=
g
(
0
) and f
(
𝕌
)
⊆
g
(
𝕌
).For eachA and B such that -
1
≤
B
<
A
≤
1, we define the function
(7)
h
(
A
,
B
;
z
)
=
1
+
A
z
1
+
B
z
,
(
z
∈
𝕌
)
.
It is well known that h
(
A
,
B
;
z
) for -
1
≤
B
≤
1 is the conformal map of the unit disk onto the disk symmetrical with respect to the real axis having the center (
1
-
A
B
)
/
(
1
-
B
2
) for B
≠
±
1 and radius (
A
-
B
)
/
(
1
-
B
2
). The boundary circle cuts the real axis at the points (
1
-
A
)
/
(
1
-
B
) and (
1
+
A
)
/
(
1
+
B
).Definition 1.
Letf
∈
𝒜. Denote by ℛ
q
λ the q-analogue of Ruscheweyh operator defined by
(8)
ℛ
q
λ
f
(
z
)
=
z
+
∑
k
=
2
∞
‍
[
k
+
λ
-
1
]
q
!
[
λ
]
q
!
[
k
-
1
]
q
!
a
k
z
k
,
where [
a
]
q and [
a
]
q
! are defined in (1).From the definition we observe that, ifq
→
1, we have
(9)
lim
⁡
q
→
1
ℛ
q
λ
f
(
z
)
=
z
+
lim
⁡
q
→
1
[
∑
k
=
2
∞
‍
[
k
+
λ
-
1
]
q
!
[
λ
]
q
!
[
k
-
1
]
q
!
a
k
z
k
]
=
z
+
∑
k
=
2
∞
‍
(
k
+
λ
-
1
)
!
(
λ
)
!
(
k
-
1
)
!
a
k
z
k
=
ℛ
λ
f
(
z
)
,
where ℛ
λ is Ruscheweyh differential operator which was defined in [15] and has been studied by many authors, for example [16–18].It can also be shown that thisq-operator is hypergeometric in nature as
(10)
ℛ
q
λ
f
(
z
)
=
z
2
Φ
1
(
q
λ
+
1
,
q
,
q
,
q
;
z
)
*
f
(
z
)
,
where 2
Φ
1 is the q-analogue of Gauss hypergeometric function defined in (2), and the symbol (
*
) stands for the Hadamard product (or convolution).The following identity is easily verified for the operatorℛ
q
λ:
(11)
q
λ
z
(
D
q
(
ℛ
q
λ
f
(
z
)
)
)
=
[
λ
+
1
]
q
ℛ
q
λ
+
1
f
(
z
)
-
[
λ
]
q
ℛ
q
λ
f
(
z
)
.


## 2. Main Results

Before we obtain our results, we state some known lemmas.LetP
(
β
) be the class of functions of the form
(12)
ϕ
(
z
)
=
1
+
c
1
z
+
c
2
z
2
+
⋯
,
which are analytic in 𝕌 and satisfy the following inequality:
(13)
Re
(
ϕ
(
z
)
)
>
β
,
(
0
≤
β
<
1
;
z
∈
𝕌
)
.Lemma 2 (see [19]).
Letϕ
j
∈
P
(
β
j
) be given by (12), where (
0
≤
β
j
<
1
;
j
=
1,2
); then
(14)
(
ϕ
1
*
ϕ
2
)
∈
P
(
β
3
)
,
(
β
3
=
1
-
2
(
1
-
β
1
)
(
1
-
β
2
)
)
,
and the bound β
3 is the best possible.Lemma 3 (see [20]).
Let the functionϕ, given by (12), be in the class P
(
β
). Then
(15)
Re
ϕ
(
z
)
>
2
β
-
1
+
2
(
1
-
β
)
1
+
|
z
|
,
(
0
≤
β
<
1
)
.Lemma 4 (see [21]).
The function(
1
-
z
)
γ
≡
e
γ
log
⁡
(
1
-
z
)
,
γ
≠
0, is univalent in 𝕌 if and only if γ is either in the closed disk |
γ
-
1
|
≤
1 or in the closed disk |
γ
+
1
|
≤
1.We now generalize the lemmas introduced in [22] and [23], respectively, using q-derivative.Lemma 5.
Leth
(
z
) be analytic and convex univalent in 𝕌 and h
(
0
)
=
1 and let g
(
z
)
=
1
+
b
1
z
+
b
2
z
2
+
⋯ be analytic in 𝕌. If
(16)
g
(
z
)
+
z
D
q
(
g
(
z
)
)
c
≺
h
(
z
)
,
(
z
∈
𝕌
;
c
≠
0
)
,
Then, for Re
(
c
)
≥
0,
(17)
g
(
z
)
≺
c
z
c
∫
0
z
‍
t
c
-
1
h
(
t
)
d
t
.Proof.
Suppose thath is analytic and convex univalent in 𝕌 and g is analytic in 𝕌. Letting q
→
1 in (16), we have
(18)
g
(
z
)
+
z
g
′
(
z
)
c
≺
h
(
z
)
,
(
z
∈
𝕌
;
c
≠
0
)
.
Then, from Lemma in [22], we obtain
(19)
g
(
z
)
≺
c
z
c
∫
0
z
‍
t
c
-
1
h
(
t
)
d
t
.Lemma 6.
Letq
(
z
) be univalent in 𝕌 and let θ
(
w
) and ϕ
(
w
) be analytic in a domain D containing q
(
𝕌
) with ϕ
(
w
)
≠
0 when w
∈
q
(
𝕌
). Set Q
(
z
)
=
z
D
q
(
q
(
z
)
)
ϕ
(
q
(
z
)
)
,
h
(
z
)
=
θ
(
q
(
z
)
+
Q
(
z
)
) and suppose that (1)
Q
(
z
) is starlike univalent in 𝕌;
(2)
Re
(
z
D
q
(
h
(
z
)
)
/
Q
(
z
)
)
=
Re
(
(
D
q
(
θ
q
(
z
)
)
)
/
ϕ
(
q
(
z
)
)
)
+
(
z
D
q
(
Q
(
z
)
)
/
Q
(
z
)
)
)
>
0
(
z
∈
𝕌
).
Ifp
(
z
) is analytic in 𝕌, with p
(
0
)
=
q
(
0
)
,
p
(
𝕌
)
⊂
D, and
(20)
θ
(
p
(
z
)
)
+
z
D
q
(
p
(
z
)
)
ϕ
(
p
(
z
)
)
≺
θ
(
q
(
z
)
)
+
z
D
q
(
q
(
z
)
)
ϕ
(
q
(
z
)
)
=
h
(
z
)
,
then p
(
z
)
≺
q
(
z
) and q
(
z
) is the best dominant.The proof is similar to the proof of Lemma5.Theorem 7.
Let>
0, α
>
0, and -
1
≤
B
<
A
≤
1. If f
∈
𝒜 satisfies
(21)
(
1
-
α
)
ℛ
q
λ
f
(
z
)
z
+
α
ℛ
q
λ
+
1
f
(
z
)
z
≺
h
(
A
,
B
;
z
)
,
then
(22)
Re
(
(
ℛ
q
λ
f
(
z
)
z
)
1
/
n
)
>
(
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
1
-
A
u
1
-
B
u
)
d
u
)
1
/
n
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
G
R
K
R
R
R
R
Q
Q
Q
I
Q
Z
M
(
n
≥
1
)
.
The result is sharp.Proof.
Let(23)
g
(
z
)
=
ℛ
q
λ
f
(
z
)
z
,
for f
∈
𝒜. Then the function g
(
z
)
=
1
+
b
1
z
+
⋯ is analytic in 𝕌. By using logarithmic q-differentiation on both sides of (23) and multiplying by z, we have
(24)
z
D
q
(
g
(
z
)
)
g
(
z
)
=
z
R
q
λ
f
(
z
)
R
q
λ
f
(
z
)
-
1
;
;
by making use of identity (11), we obtain
(25)
z
D
q
(
g
(
z
)
)
g
(
z
)
=
[
λ
+
1
]
q
q
λ
R
q
λ
+
1
f
(
z
)
R
q
λ
f
(
z
)
-
[
λ
]
q
q
λ
-
1
.
Taking into account that[
λ
+
1
]
q
=
[
λ
]
q
+
q
λ, we obtain
(26)
q
λ
[
λ
+
1
]
q
z
D
q
(
g
(
z
)
)
+
g
(
z
)
=
ℛ
λ
+
1
f
(
z
)
z
.
From (11), (23), and (26), we get
(27)
g
(
z
)
+
q
λ
α
[
λ
+
1
]
q
z
D
q
(
g
(
z
)
)
≺
h
(
A
,
B
;
z
)
.
Now, applying Lemma5, we have
(28)
g
(
z
)
≺
[
λ
+
1
]
q
q
λ
α
z
-
[
λ
+
1
]
q
/
q
λ
α
∫
0
1
‍
t
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
1
+
A
t
1
+
B
t
)
d
t
,
or by the concept of subordination
(29)
ℛ
q
λ
f
(
z
)
z
=
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
1
+
A
u
w
(
z
)
1
+
B
u
w
(
z
)
)
d
u
.
In view of-
1
≤
B
<
A
≤
1 and λ
>
0, it follows from (29) that
(30)
Re
(
ℛ
q
λ
f
(
z
)
z
)
>
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
1
-
A
u
1
-
B
u
)
d
u
,
with the aid of the elementary inequality Re
(
w
1
/
n
)
≥
(
Re
w
)
1
/
n for Re
w
>
0 and n
≥
1. Hence, inequality (22) follows directly from (30). To show the sharpness of (22), we define f
∈
𝒜 by
(31)
ℛ
q
λ
f
(
z
)
z
=
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
1
+
A
u
z
1
+
B
u
z
)
d
u
.
For this function, we find that(32)
(
1
-
α
)
ℛ
q
λ
f
(
z
)
z
+
α
ℛ
q
λ
+
1
f
(
z
)
z
=
1
+
A
z
1
-
B
z
,
ℛ
q
λ
f
(
z
)
z
⟶
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
1
-
A
u
1
-
B
u
)
d
u
ℛ
q
λ
f
(
z
)
z
⟶
[
λ
+
1
]
q
q
λ
α
RSQRRRRRRK
as
z
⟶
-
1
.
This completes the proof.Corollary 8.
LetA
=
2
β
-
1
and
B
=
-
1, where 0
≤
β
<
1
and
α
,
λ
>
1. If f satisfies
(33)
(
1
-
α
)
ℛ
q
λ
f
(
z
)
z
+
α
ℛ
q
λ
+
1
f
(
z
)
z
≺
h
(
2
β
-
1
,
-
1
;
z
)
,
then
(34)
Re
(
(
ℛ
q
λ
f
(
z
)
z
)
1
/
n
)
>
(
2
(
1
-
β
)
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
1
+
u
d
u
(
2
β
-
1
)
u
[
λ
+
1
]
q
/
q
λ
α
+
2
(
1
-
β
)
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
1
+
u
d
u
)
1
/
n
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
G
R
K
R
R
R
R
R
R
Q
I
Q
Z
M
(
n
≥
1
)
.Proof.
Following the same steps as in the proof of Theorem7 and considering g
(
z
)
=
ℛ
q
λ
f
(
z
)
/
z, the differential subordination (27) becomes
(35)
g
(
z
)
+
q
λ
α
[
λ
+
1
]
q
z
D
q
(
g
(
z
)
)
≺
1
+
(
2
β
-
1
)
z
1
+
z
.
Therefore,(36)
Re
(
(
ℛ
q
λ
f
(
z
)
z
)
1
/
n
)
>
(
[
λ
+
1
]
q
q
λ
α
∫
0
1
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
‍
(
1
+
(
1
-
2
β
)
u
1
+
u
)
d
u
)
1
/
n
=
(
[
λ
+
1
]
q
q
λ
α
∫
0
1
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
‍
×
(
(
2
β
-
1
)
+
2
(
1
-
β
)
1
+
u
)
d
u
)
1
/
n
=
(
∫
0
1
u
[
λ
+
1
]
q
/
(
q
λ
α
-
1
)
1
+
u
d
u
(
2
β
-
1
)
u
[
λ
+
1
]
q
/
q
λ
α
+
2
(
1
-
β
)
[
λ
+
1
]
q
q
λ
∫
0
1
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
1
+
u
d
u
)
1
/
n
.Theorem 9.
Letλ
>
0 and 0
≤
ρ
<
1. Let γ be a complex number with γ
≠
0 and satisfy either |
2
γ
(
1
-
ρ
)
(
[
λ
+
1
]
q
/
q
λ
)
-
1
|
≤
1 or |
2
γ
(
1
-
ρ
)
(
[
λ
+
1
]
q
/
q
λ
)
+
1
|
≤
1. If f
∈
𝒜 satisfies the condition
(37)
Re
(
ℛ
q
λ
+
1
f
(
z
)
ℛ
q
λ
f
(
z
)
)
>
ρ
,
(
z
∈
𝕌
)
,
then
(38)
(
ℛ
q
λ
f
(
z
)
z
)
γ
≺
1
(
1
-
z
)
2
γ
(
1
-
ρ
)
(
[
λ
+
1
]
q
/
q
λ
)
,
(
z
∈
𝕌
)
,
where q
(
z
) is the best dominant.Proof.
Let(39)
p
(
z
)
=
(
ℛ
q
λ
f
(
z
)
z
)
γ
,
(
z
∈
𝕌
)
.Then, by making use of (11), (37), and (39), we obtain
(40)
1
+
q
λ
z
D
q
(
p
(
z
)
)
γ
[
λ
+
1
]
q
p
(
z
)
≺
1
+
(
1
-
2
ρ
)
z
1
-
z
,
(
z
∈
𝕌
)
.
We now assume that(41)
q
(
z
)
=
1
(
1
-
z
)
2
γ
(
1
-
ρ
)
[
λ
+
1
]
q
/
q
λ
,
θ
(
w
)
=
1
,
ϕ
(
w
)
=
q
λ
γ
[
λ
+
1
]
q
w
;
then q
(
z
) is univalent by condition of the theorem and Lemma 4. Further, it is easy to show that q
(
z
)
,
θ
(
w
), and ϕ
(
w
) satisfy the conditions of Lemma 6. Note that the function
(42)
Q
(
z
)
=
z
D
q
(
q
(
z
)
)
ϕ
(
q
(
z
)
)
=
2
(
1
-
ρ
)
z
1
-
z
is univalent starlike in 𝕌 and
(43)
h
(
z
)
=
θ
(
q
(
z
)
)
+
Q
(
z
)
=
1
+
(
1
-
2
ρ
)
z
1
-
z
.
Combining (40) and Lemma 6 we get the assertion of Theorem 9.Theorem 10.
Letα
<
1
,
λ
>
0 and -
1
≤
B
i
<
A
i
≤
1. If each of the functions f
i
∈
𝒜 satisfies the following subordination condition,
(44)
(
1
-
α
)
ℛ
q
λ
f
i
(
z
)
z
+
α
ℛ
q
λ
+
1
f
i
(
z
)
z
≺
h
(
A
i
,
B
i
;
z
)
,
then,
(45)
(
1
-
α
)
ℛ
q
λ
Θ
(
z
)
z
+
α
ℛ
q
λ
+
1
Θ
(
z
)
z
≺
h
(
1
-
2
γ
,
-
1
;
z
)
,
where
(46)
Θ
(
z
)
=
ℛ
q
λ
(
f
1
*
f
2
)
(
z
)
,
(47)
γ
=
1
-
4
(
A
1
-
B
1
)
(
A
2
-
B
2
)
(
1
-
B
1
)
(
1
-
B
2
)
×
(
1
-
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
1
+
u
d
u
)
.Proof.
we define the functionh
i by
(48)
h
i
(
z
)
=
(
1
-
α
)
ℛ
q
λ
f
i
(
z
)
z
+
α
ℛ
q
λ
+
1
f
i
(
z
)
z
(
1
-
α
)
ℛ
q
λ
f
i
z
(
f
i
∈
𝒜
,
i
=
1,2
)
;
we have h
i
(
z
)
∈
P
(
β
i
), where β
i
=
(
1
-
A
i
)
/
(
1
-
B
i
)
(
i
=
1,2
). By making use of (11) and (48), we obtain
(49)
ℛ
q
λ
f
i
(
z
)
=
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
t
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
h
i
(
t
)
d
t
,
(
i
=
1,2
)
,
which, in the light of (46), can show that
(50)
ℛ
q
λ
Θ
(
z
)
=
[
λ
+
1
]
q
q
λ
α
z
1
-
(
[
λ
+
1
]
q
/
q
λ
α
)
∫
0
1
‍
t
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
h
0
(
t
)
d
t
,
where, for convenience,
(51)
h
0
(
z
)
=
(
1
-
α
)
ℛ
q
λ
Θ
(
z
)
z
+
α
ℛ
q
λ
+
1
Θ
(
z
)
z
=
[
λ
+
1
]
q
q
λ
α
z
1
-
(
[
λ
+
1
]
q
/
q
λ
α
)
×
∫
0
1
‍
t
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
h
1
*
h
2
)
(
t
)
d
t
.
Note that, by using Lemma2, we have (
h
1
*
h
2
)
∈
P
(
β
3
), where β
3
=
1
-
2
(
1
-
β
1
)
(
1
-
β
2
).
Now, with an application of Lemma3, we have
(52)
Re
(
h
0
(
z
)
)
=
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
Re
(
(
h
1
*
h
2
)
(
u
z
)
)
d
u
≥
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
2
β
3
-
1
+
2
(
1
-
β
3
)
1
+
u
|
z
|
)
d
u
>
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
(
2
β
3
-
1
+
2
(
1
-
β
3
)
1
+
u
)
d
u
=
1
-
4
(
A
1
-
B
1
)
(
A
2
-
B
2
)
(
1
-
B
1
)
(
1
-
B
2
)
×
(
1
-
[
λ
+
1
]
q
q
λ
α
∫
0
1
‍
u
(
[
λ
+
1
]
q
/
q
λ
α
)
-
1
1
+
u
d
u
)
=
γ
,
which shows that the desired assertion of Theorem 10 holds.



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*Source: 958563-2014-04-14.xml*