# Some Applications of Second-Order Differential Subordination on a Class of Analytic Functions Defined by Komatu Integral Operator

Authors: Serap Bulut
Journal: ISRN Mathematical Analysis
 (2014)

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## Abstract

We introduce a new class of analytic functions by using Komatu integral operator and obtain some subordination results.


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## Body

## 1. Introduction, Definitions, and Preliminaries

Letℝ
=
(
-
∞
,
∞
) be the set of real numbers, ℂ the set of complex numbers,
(1)
ℕ
:
=
{
1,2
,
3
,
…
}
=
ℕ
0
∖
{
0
}
be the set of positive integers, and
(2)
ℕ
*
:
=
ℕ
∖
{
1
}
=
{
2,3
,
4
,
…
}
.Letℋ be the class of analytic functions in the open unit disk
(3)
𝕌
=
{
z
∈
ℂ
:
|
z
|
<
1
}
and ℋ
[
a
,
n
] the subclass of ℋ consisting of the functions of the form
(4)
f
(
z
)
=
a
+
a
n
z
n
+
a
n
+
1
z
n
+
1
+
⋯
.Let𝒜
n be the class of all functions of the form
(5)
f
(
z
)
=
z
+
∑
k
=
n
+
1
∞
a
k
z
k
‍
,
which are analytic in the open unit disk 𝕌 with
(6)
𝒜
1
=
𝒜
.
Also let 𝒮 denote the subclass of 𝒜 consisting of functions f which are univalent in 𝕌.A functionf analytic in 𝕌 is said to be convex if it is univalent and f
(
𝕌
) is convex.Let(7)
𝒦
=
{
f
∈
𝒜
:
ℜ
{
1
+
z
f
′′
(
z
)
f
′
(
z
)
}
>
0
,
z
∈
𝕌
}
denote the class of normalized convex functions in 𝕌.Iff and g are analytic in 𝕌, then we say that f is subordinate to g, written symbolically as
(8)
f
≺
g
or
f
(
z
)
≺
g
(
z
)
(
z
∈
𝕌
)
if there exists a Schwarz function w which is analytic in 𝕌 with
(9)
w
(
0
)
=
0
,
|
w
(
z
)
|
<
1
such that
(10)
f
(
z
)
=
g
(
w
(
z
)
)
(
z
∈
𝕌
)
.
Indeed, it is known that
(11)
f
(
z
)
≺
g
(
z
)
(
z
∈
𝕌
)
⟹
f
(
0
)
=
g
(
0
)
,
f
(
𝕌
)
⊂
g
(
𝕌
)
.
Furthermore, if the function g is univalent in 𝕌, then we have the following equivalence [1, page 4]:
(12)
f
(
z
)
≺
g
(
z
)
(
z
∈
𝕌
)
⟺
f
(
0
)
=
g
(
0
)
,
f
(
𝕌
)
⊂
g
(
𝕌
)
.Letψ
:
ℂ
3
×
𝕌
→
ℂ be a function and let h be univalent in 𝕌. If p is analytic in 𝕌 and satisfies the (second-order) differential subordination
(13)
ψ
(
p
(
z
)
,
z
p
′
(
z
)
,
z
2
p
′′
(
z
)
;
z
)
≺
h
(
z
)
,
(
z
∈
𝕌
)
,
then p is called a solution of the differential subordination. The univalent function q is called a dominant of the solutions of the differential subordination, or more simply a dominant, if p
≺
q for all p satisfying (13).A dominantq
~, which satisfies q
~
≺
q for all dominants q of (13), is said to be the best dominant of (13).Recently, Komatu [2] introduced a certain integral operator L
c
δ defined by
(14)
L
c
δ
f
(
z
)
=
c
δ
Γ
(
δ
)
∫
0
1
‍
t
c
-
2
(
log
⁡
1
t
)
δ
-
1
f
(
z
t
)
d
t
,
h
h
(
z
∈
𝕌
;
c
>
0
;
δ
≥
0
;
f
(
z
)
∈
𝒜
)
.Thus, iff
∈
𝒜 is of the form (5), then it is easily seen from (14) that (see [2])
(15)
L
c
δ
f
(
z
)
=
z
+
∑
k
=
2
∞
‍
(
c
c
+
k
-
1
)
δ
a
k
z
k
(
c
>
0
;
δ
≥
0
)
.Using the relation (15), it is easy verify that
(16)
z
(
L
c
δ
+
1
f
(
z
)
)
′
=
c
L
c
δ
f
(
z
)
-
(
c
-
1
)
L
c
δ
+
1
f
(
z
)
,
L
c
δ
(
z
f
′
(
z
)
)
=
z
(
L
c
δ
f
(
z
)
)
′
.We note the following.(i)
Forc
=
1 and δ
=
m (m is any integer), the multiplier transformation L
1
m
f
(
z
)
=
I
m
f
(
z
) was studied by Flett [3] and Sălăgean [4].
(ii)
Forc
=
1 and δ
=
-
m (m
∈
ℕ
0), the differential operator L
1
-
m
f
(
z
)
=
D
m
f
(
z
) was studied by Sălăgean [4].
(iii)
Forc
=
2 and δ
=
m (m is any integer), the operator L
2
m
f
(
z
)
=
L
m
f
(
z
) was studied by Uralegaddi and Somanatha [5].
(iv)
Forc
=
2, the multiplier transformation L
2
δ
f
(
z
)
=
I
δ
f
(
z
) was studied by Jung et al. [6].Using the operatorL
c
δ, we now introduce the following class.Definition 1.
Letℛ
c
,
δ
(
β
) be the class of functions f
∈
𝒜 satisfying
(17)
ℜ
{
(
L
c
δ
f
(
z
)
)
′
}
>
β
,
where z
∈
𝕌, 0
≤
β
<
1, and L
c
δ is the Komatu integral operator.In order to prove our main results, we will make use of the following lemmas.Lemma 2 (see [7]).
Leth be a convex function with h
(
0
)
=
a and let γ
∈
ℂ
*
:
=
ℂ
-
{
0
} be a complex number with ℜ
{
γ
}
≥
0. If p
∈
ℋ
[
a
,
n
] and
(18)
p
(
z
)
+
1
γ
z
p
′
(
z
)
≺
h
(
z
)
(
z
∈
𝕌
)
,
then
(19)
p
(
z
)
≺
q
(
z
)
≺
h
(
z
)
(
z
∈
𝕌
)
,
where
(20)
q
(
z
)
=
γ
n
z
γ
/
n
∫
0
z
‍
t
γ
/
n
-
1
h
(
t
)
d
t
(
z
∈
𝕌
)
.
The function q is convex and is the best dominant.Lemma 3 (see [8]).
Letℜ
{
μ
}
>
0, n
∈
ℕ, and let
(21)
w
=
n
2
+
|
μ
|
2
-
|
n
2
-
μ
2
|
4
n
ℜ
{
μ
}
.
Let h be an analytic function in 𝕌 with h
(
0
)
=
1 and suppose that
(22)
ℜ
{
1
+
z
h
′′
(
z
)
h
′
(
z
)
}
>
-
w
.
If
(23)
p
(
z
)
=
1
+
p
n
z
n
+
p
n
+
1
z
n
+
1
+
⋯
is analytic in 𝕌 and
(24)
p
(
z
)
+
1
μ
z
p
′
(
z
)
≺
h
(
z
)
,
then
(25)
p
(
z
)
≺
q
(
z
)
,
where q is a solution of the differential equation
(26)
q
(
z
)
+
n
μ
z
q
′
(
z
)
=
h
(
z
)
,
q
(
0
)
=
1
,
given by
(27)
q
(
z
)
=
μ
n
z
μ
/
n
∫
0
z
‍
t
μ
/
n
-
1
h
(
t
)
d
t
.
Moreover q is the best dominant.


## 2. Main Results

Theorem 4.
The setℛ
c
,
δ
(
β
) is convex.Proof.
Let(28)
f
j
(
z
)
=
z
+
∑
k
=
2
∞
a
k
,
j
z
k
‍
(
z
∈
𝕌
;
j
=
1,2
,
…
,
l
)
be in the class ℛ
c
,
δ
(
β
). Then, by Definition 1, we have
(29)
ℜ
{
(
L
c
δ
f
j
(
z
)
)
′
}
=
ℜ
{
1
+
∑
k
=
2
∞
‍
k
(
c
c
+
k
-
1
)
δ
a
k
,
j
z
k
-
1
}
>
β
.
For any nonnegative numbers λ
1
,
λ
2
,
…
,
λ
l such that
(30)
λ
1
+
λ
2
+
⋯
+
λ
l
=
1
,
we must show that the function
(31)
h
(
z
)
=
∑
j
=
1
l
λ
j
f
j
(
z
)
‍
is in ℛ
c
,
δ
(
β
); that is,
(32)
ℜ
{
(
L
c
δ
h
(
z
)
)
′
}
>
β
.
By (28) and (31), we have
(33)
h
(
z
)
=
z
+
∑
k
=
2
∞
(
∑
j
=
1
l
λ
j
a
k
,
j
‍
)
z
k
‍
.
Therefore we get
(34)
L
c
δ
h
(
z
)
=
z
+
∑
k
=
2
∞
(
c
c
+
k
-
1
)
δ
(
∑
j
=
1
l
λ
j
a
k
,
j
‍
)
z
k
‍
.
Differentiating (34) with respect to z, we obtain
(35)
(
L
c
δ
h
(
z
)
)
′
=
1
+
∑
k
=
2
∞
k
(
c
c
+
k
-
1
)
δ
(
∑
j
=
1
l
λ
j
a
k
,
j
‍
)
z
k
-
1
‍
.
So we get
(36)
ℜ
{
(
L
c
δ
h
(
z
)
)
′
}
=
1
+
∑
j
=
1
l
‍
λ
j
ℜ
{
∑
k
=
2
∞
k
(
c
c
+
k
-
1
)
δ
a
k
,
j
z
k
-
1
‍
}
>
1
+
∑
j
=
1
l
λ
j
(
β
-
1
)
‍
(
by
(
29
)
)
=
β
since λ
1
+
λ
2
+
⋯
+
λ
l
=
1. Therefore we get the desired result.Theorem 5.
Letq be convex function in 𝕌 with q
(
0
)
=
1 and let
(37)
h
(
z
)
=
q
(
z
)
+
1
γ
+
2
z
q
′
(
z
)
(
z
∈
𝕌
)
,
where γ is a complex number with ℜ
{
γ
}
>
-
2. If f
∈
ℛ
c
,
δ
(
β
) and ℱ
=
ℐ
γ
f, where
(38)
ℱ
(
z
)
=
ℐ
γ
f
(
z
)
=
γ
+
2
z
γ
+
1
∫
0
z
t
γ
f
(
t
)
d
t
‍
,
then
(39)
(
L
c
δ
f
(
z
)
)
′
≺
h
(
z
)
implies
(40)
(
L
c
δ
ℱ
(
z
)
)
′
≺
q
(
z
)
,
and this result is sharp.Proof.
From the equality (38), we get
(41)
z
γ
+
1
ℱ
(
z
)
=
(
γ
+
2
)
∫
0
z
t
γ
f
(
t
)
d
t
‍
.
Differentiating (41) with respect to z, we have
(42)
(
γ
+
1
)
ℱ
(
z
)
+
z
ℱ
′
(
z
)
=
(
γ
+
2
)
f
(
z
)
,
(43)
(
γ
+
1
)
L
c
δ
ℱ
(
z
)
+
z
(
L
c
δ
ℱ
(
z
)
)
′
=
(
γ
+
2
)
L
c
δ
f
(
z
)
.
Differentiating (43) with respect to z, we obtain
(44)
(
L
c
δ
ℱ
(
z
)
)
′
+
1
γ
+
2
z
(
L
c
δ
ℱ
(
z
)
)
′′
=
(
L
c
δ
f
(
z
)
)
′
.
Using the differential subordination (39) in the equality (44), we get
(45)
(
L
c
δ
ℱ
(
z
)
)
′
+
1
γ
+
2
z
(
L
c
δ
ℱ
(
z
)
)
′′
≺
h
(
z
)
.
Let us define
(46)
p
(
z
)
=
(
L
c
δ
ℱ
(
z
)
)
′
.
Then a simple computation yields
(47)
p
(
z
)
=
[
z
+
∑
k
=
2
∞
(
c
c
+
k
-
1
)
δ
γ
+
2
γ
+
k
+
1
a
k
z
k
‍
]
′
=
1
+
p
1
z
+
p
2
z
2
+
⋯
,
(
p
∈
ℋ
[
1,1
]
)
.
Using (46) in the subordination (45), we have
(48)
p
(
z
)
+
1
γ
+
2
z
p
′
(
z
)
≺
h
(
z
)
=
q
(
z
)
+
1
γ
+
2
z
q
′
(
z
)
(
z
∈
𝕌
)
.
Using Lemma 2, we obtain
(49)
p
(
z
)
≺
q
(
z
)
which is desired result. Moreover q is the best dominant.Example 6.
If we take(50)
γ
=
i
,
q
(
z
)
=
1
1
-
z
in Theorem 5, then we have
(51)
h
(
z
)
=
(
2
+
i
)
-
z
(
1
+
i
)
(
2
+
i
)
(
1
-
z
)
2
.
If f
∈
ℛ
c
,
δ
(
β
) and ℱ is given by
(52)
ℱ
(
z
)
=
ℐ
i
f
(
z
)
=
2
+
i
z
1
+
i
∫
0
z
t
i
f
(
t
)
d
t
‍
,
then by Theorem 5, we have
(53)
(
L
c
δ
f
(
z
)
)
′
≺
(
2
+
i
)
-
z
(
1
+
i
)
(
2
+
i
)
(
1
-
z
)
2
⟹
(
L
c
δ
ℱ
(
z
)
)
′
≺
1
1
-
z
(
z
∈
𝕌
)
.Theorem 7.
Letℜ
{
γ
}
>
-
2 and let
(54)
w
=
1
+
|
γ
+
2
|
2
-
|
γ
2
+
4
γ
+
3
|
4
ℜ
{
γ
+
2
}
.
Let h be an analytic function in 𝕌 with h
(
0
)
=
1 and suppose that
(55)
ℜ
{
1
+
z
h
′′
(
z
)
h
′
(
z
)
}
>
-
w
.
If f
∈
ℛ
c
,
δ
(
β
) and ℱ
=
ℐ
γ
δ
f, where ℱ is defined by (38), then
(56)
(
L
c
δ
f
(
z
)
)
′
≺
h
(
z
)
implies
(57)
(
L
c
δ
ℱ
(
z
)
)
′
≺
q
(
z
)
,
where q is the solution of the differential equation
(58)
h
(
z
)
=
q
(
z
)
+
1
γ
+
2
z
q
′
(
z
)
,
q
(
0
)
=
1
,
given by
(59)
q
(
z
)
=
γ
+
2
z
γ
+
2
∫
0
z
t
γ
+
1
h
(
t
)
d
t
‍
.
Moreover q is the best dominant.Proof.
We considern
=
1 and μ
=
γ
+
2 in Lemma 3. Then the proof is easily seen by means of the proof of Theorem 5.Letting(60)
h
(
z
)
=
1
-
(
2
β
-
1
)
z
1
-
z
(
0
≤
β
<
1
)
in Theorem 7, we obtain the following interesting result.Corollary 8.
If0
≤
β
<
1, c
>
0, δ
≥
0, ℜ
{
γ
}
>
-
2, and ℱ
=
ℐ
γ
f is defined by (38), then
(61)
ℐ
γ
(
ℛ
c
,
δ
(
β
)
)
⊂
ℛ
c
,
δ
(
ρ
)
,
where
(62)
ρ
=
min
⁡
|
z
|
=
1
ℜ
{
q
(
z
)
}
=
ρ
(
γ
,
β
)
and this results is sharp. Moreover
(63)
ρ
=
ρ
(
γ
,
β
)
=
2
β
-
1
+
2
(
γ
+
2
)
(
1
-
β
)
τ
(
γ
)
,
where
(64)
τ
(
γ
)
=
∫
0
1
t
γ
+
1
1
-
t
d
t
‍
.Proof.
If we let(65)
h
(
z
)
=
1
-
(
2
β
-
1
)
z
1
-
z
(
0
≤
β
<
1
)
,
then h is convex and by Theorem 7, we deduce
(66)
(
L
c
δ
ℱ
(
z
)
)
′
≺
q
(
z
)
=
γ
+
2
z
γ
+
2
∫
0
z
‍
t
γ
+
1
1
-
(
2
β
-
1
)
t
1
-
t
d
t
=
2
β
-
1
+
2
(
γ
+
2
)
(
1
-
β
)
z
γ
+
2
∫
0
z
t
γ
+
1
1
-
t
d
t
‍
.
On the other hand if ℜ
{
γ
}
>
-
2, then from the convexity of q and the fact that q
(
𝕌
) is symmetric with respect to the real axis, we get
(67)
ℜ
{
(
L
c
δ
ℱ
(
z
)
)
′
}
≥
min
⁡
|
z
|
=
1
ℜ
{
q
(
z
)
}
=
ℜ
{
q
(
1
)
}
=
ρ
(
γ
,
β
)
=
2
β
-
1
+
2
(
γ
+
2
)
(
1
-
β
)
τ
(
γ
)
,
where τ
(
γ
) is given by (64). From (66), we have
(68)
ℐ
γ
(
ℛ
c
,
δ
(
β
)
)
⊂
ℛ
c
,
δ
(
ρ
)
,
where ρ is given by (63).



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*Source: 606235-2014-03-12.xml*