# Differential Subordination with Generalized Derivative Operator of Analytic Functions

Authors: Entisar El-Yagubi; Maslina Darus
Journal: Chinese Journal of Mathematics
 (2014)

---

## Abstract

Motivated by generalized derivative operator defined by the authors (El-Yagubi and Darus, 2013) and the technique of differential subordination, several interesting properties of the operatorD
λ
1
,
λ
2
,
δ
m
,
b are given.


---

## Body

## 1. Introduction

LetA denote the class of functions of the form
(1)
f
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
z
n
,
which are analytic in the open unit disk U
=
{
z
:
z
∈
C
,
|
z
|
<
1
}.Also letS be the the subclass of A consisting of all functions which are univalent in U. We denote by S
*
(
α
) and C
(
α
)
(
0
≤
α
<
1
) the familiar subclasses of A consisting of functions which are, respectively, starlike of order α and convex of order α in U:. 						(2)
S
*
(
α
)
=
{
f
∈
A
:
R
{
z
f
′
(
z
)
f
(
z
)
}
>
α
,
z
∈
U
}
,
C
(
α
)
=
{
f
∈
A
:
R
{
1
+
z
f
′′
(
z
)
f
′
(
z
)
}
>
α
,
z
∈
U
}
.LetH
(
U
) be the class of holomorphic function in unit disk U
=
{
z
:
z
∈
C
,
|
z
|
<
1
}. For a
∈
C and n
∈
N we let
(3)
H
[
a
,
n
]
=
{
a
n
+
1
z
n
+
1
f
∈
H
(
U
)
,
f
(
z
)
=
a
+
a
n
z
n
+
a
n
+
1
z
n
+
1
+
⋯
}
,
(
z
∈
U
)
.Let two functions given byf
(
z
)
=
∑
n
=
2
∞
‍
a
n
z
n and g
(
z
)
=
∑
n
=
2
∞
‍
b
n
z
n be analytic in U. Then the Hadamard product (or convolution) f
*
g of the two functions f, g is defined by
(4)
f
(
z
)
*
g
(
z
)
=
z
+
∑
n
=
2
∞
‍
a
n
b
n
z
n
.Recall that the functionf is subordinate to g if there exists the Schwarz function ω, analytic in U, with ω
(
0
)
=
0 and |
ω
(
z
)
|
<
1 such that f
(
z
)
=
g
(
ω
(
z
)
), z
∈
U. We denote this subordination by f
(
z
)
≺
g
(
z
). If g
(
z
) is univalent in U, then the subordination is equivalent to f
(
0
)
=
g
(
0
) and f
(
U
)
⊂
g
(
U
).Letψ
:
C
3
×
U
→
C and h be univalent in U. If p is analytic in U and satisfies the (second order) differential subordination
(5)
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 functionq is called a dominant of the solutions of the differential subordination, or more simply a dominant, if p
≺
q for all p satisfying (5).A dominantq
~ that satisfies q
~
≺
q for all dominants q of (5) is said to be the best dominant of (5) (note that the best dominant is unique up to a rotation of U).In order to prove the original results we need the following lemmas.Lemma 1 (see [1]).
Leth be a convex function with h
(
0
)
=
a and let γ
∈
C
-
{
0
} be a complex number with R
{
γ
}
≥
0. If p
∈
H
[
a
;
n
] and
(6)
p
(
z
)
+
z
p
′
(
z
)
γ
≺
h
(
z
)
,
(
z
∈
U
)
,
then
(7)
p
(
z
)
≺
q
(
z
)
≺
h
(
z
)
,
(
z
∈
U
)
,
where
(8)
q
(
z
)
=
γ
n
z
γ
/
n
∫
0
z
‍
h
(
t
)
t
(
γ
/
n
)
-
1
d
t
,
(
z
∈
U
)
.
The function q is convex and is the best dominant.Lemma 2 (see [2]).
Letg be a convex function in U and let
(9)
h
(
z
)
=
g
(
z
)
+
n
α
z
g
′
(
z
)
,
where α
>
0 and n is a positive integer. If
(10)
p
(
z
)
=
g
(
0
)
+
p
n
z
n
+
p
n
+
1
z
n
+
1
+
⋯
,
(
z
∈
U
)
is analytic in U and
(11)
p
(
z
)
+
α
z
p
′
(
z
)
≺
h
(
z
)
,
(
z
∈
U
)
,
then
(12)
p
(
z
)
≺
g
(
z
)
,
and this result is sharp.Lemma 3 (see [3]).
Letf
∈
A; if
(13)
R
{
1
+
z
f
′′
(
z
)
f
′
(
z
)
}
>
-
1
2
,
then
(14)
2
z
∫
0
z
‍
f
(
t
)
d
t
,
(
z
∈
U
,
z
≠
0
)
belongs to the class of convex functions.We now state the following generalized derivative operator [4]:
(15)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
=
z
+
∑
n
=
2
∞
‍
[
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
1
+
λ
2
(
n
-
1
)
+
b
]
m
C
(
δ
,
n
)
a
n
z
n
,
where λ
2
≥
λ
1
≥
0
,
C
(
δ
,
n
)
=
(
δ
+
1
)
n
-
1
/
(
n
-
1
)
!, for δ
,
m
,
b
∈
N
0
=
{
0,1
,
2
,
…
}, and (
x
)
n is the Pochhammer symbol defined by
(16)
(
x
)
n
=
Γ
(
x
+
n
)
Γ
(
x
)
=
{
1
,
n
=
0
,
x
(
x
+
1
)
⋯
(
x
+
n
-
1
)
,
n
=
{
1,2
,
3
,
…
}
.HereD
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
) can also be written in terms of convolution as
(17)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
=
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
D
λ
2
m
,
b
(
z
)
.To prove our results, we need the following inclusion relation:(18)
(
1
+
b
)
D
λ
1
,
λ
2
,
δ
m
+
1
,
b
f
(
z
)
=
(
1
-
(
λ
1
+
λ
2
)
+
b
)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
+
(
λ
1
+
λ
2
)
z
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
,
where φ
λ
2
b
(
z
) is analytic function given by φ
λ
2
b
(
z
)
=
z
+
∑
n
=
2
∞
‍
(
z
n
/
(
1
+
λ
2
(
n
-
1
)
+
b
)
).


## 2. Main Results

In the present paper, we will use the method of differential subordination to derive certain properties of generalised derivative operatorD
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
). Note that differential subordination has been studied by various authors, and here we follow similar works done by Oros [5] and G. Oros and G. I. Oros [6].Definition 4.
Forλ
2
≥
λ
1
≥
0
,
δ
,
m
,
b
∈
N
0, and 0
≤
α
<
1, let R
λ
1
,
λ
2
,
δ
m
,
b
(
α
) denote the class of functions f
∈
A which satisfy the condition
(19)
R
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
)
′
>
α
,
(
z
∈
U
)
.
Also, letK
λ
1
,
λ
2
,
δ
m
,
b
(
β
) denote the class of functions f
∈
A which satisfy the condition
(20)
R
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
>
β
,
(
z
∈
U
)
.Remark 5.
It is clear thatR
λ
1
,
0,0
1,0
(
α
)
≡
R
(
λ
1
,
α
), and the class of functions f
∈
A satisfy
(21)
R
(
λ
1
z
f
′′
(
z
)
+
f
′
(
z
)
)
>
α
,
(
z
∈
U
)
,
studied by Ponnusamy [7] and others.Theorem 6.
Let(22)
h
(
z
)
=
1
+
(
2
α
-
1
)
z
1
+
z
,
(
z
∈
U
)
,
be convex in U, with h
(
0
)
=
1 and 0
≤
α
<
1.
Ifλ
2
≥
λ
1
≥
0
,
δ
,
m
,
b
∈
N
0, and f
∈
A and satisfies the differential subordination
(23)
(
D
λ
1
,
λ
2
,
δ
m
+
1
,
b
f
(
z
)
)
′
≺
h
(
z
)
,
(
z
∈
U
)
,
then
(24)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
≺
q
(
z
)
=
2
α
-
1
+
2
(
1
-
α
)
(
1
+
b
)
(
λ
1
+
λ
2
)
z
(
1
+
b
)
/
(
λ
1
+
λ
2
)
σ
(
1
+
b
λ
1
+
λ
2
)
,
where σ is given by
(25)
σ
(
x
)
=
∫
0
z
‍
t
x
-
1
1
+
t
d
t
,
(
z
∈
U
)
.
The function q is convex and is the best dominant.Proof.
By differentiating (18), with respect to z, we obtain
(26)
(
1
+
b
)
(
D
λ
1
,
λ
2
,
δ
m
+
1
,
b
f
(
z
)
)
′
=
(
1
+
b
)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
+
(
λ
1
+
λ
2
)
z
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′′
.
Using (26) in (23), the differential subordination (23) becomes
(27)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
+
(
λ
1
+
λ
2
)
1
+
b
z
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′′
≺
h
(
z
)
=
1
+
(
2
α
-
1
)
z
1
+
z
.
Let
(28)
p
(
z
)
=
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
=
(
z
+
∑
n
=
2
∞
‍
(
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
m
(
1
+
λ
2
(
n
-
1
)
+
b
)
m
+
1
)
C
(
δ
,
n
)
a
n
z
n
)
′
=
1
+
p
1
z
+
p
2
z
2
+
⋯
,
(
p
∈
H
[
1,1
]
,
z
∈
U
)
.
Using (28) in (27), the differential subordination becomes
(29)
p
(
z
)
+
(
λ
1
+
λ
2
)
1
+
b
z
p
′
(
z
)
≺
h
(
z
)
=
1
+
(
2
α
-
1
)
z
1
+
z
.
By using Lemma1, we have
(30)
p
(
z
)
≺
q
(
z
)
=
1
+
b
(
λ
1
+
λ
2
)
z
(
1
+
b
)
/
(
λ
1
+
λ
2
)
∫
0
z
‍
h
(
t
)
t
(
(
1
+
b
)
/
(
λ
1
+
λ
2
)
)
-
1
d
t
=
1
+
b
(
λ
1
+
λ
2
)
z
(
1
+
b
)
/
(
λ
1
+
λ
2
)
×
∫
0
z
‍
(
1
+
(
2
α
-
1
)
t
1
+
t
)
t
(
(
1
+
b
)
/
(
λ
1
+
λ
2
)
)
-
1
d
t
=
2
α
-
1
+
2
(
1
-
α
)
(
1
+
b
)
(
λ
1
+
λ
2
)
z
(
1
+
b
)
/
(
λ
1
+
λ
2
)
σ
(
1
+
b
λ
1
+
λ
2
)
,
where σ is given by (25); that is,
(31)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
≺
q
(
z
)
=
2
α
-
1
+
2
(
1
-
α
)
(
1
+
b
)
(
λ
1
+
λ
2
)
z
(
1
+
b
)
/
(
λ
1
+
λ
2
)
σ
(
1
+
b
λ
1
+
λ
2
)
.
The function q is convex and is the best dominant. The proof is complete.Theorem 7.
Ifλ
2
≥
λ
1
≥
0
,
δ
,
m
,
b
∈
N
0, and 0
≤
α
<
1, then one has
(32)
R
λ
1
,
λ
2
,
δ
m
+
1
,
b
(
α
)
⊂
K
λ
1
,
λ
2
,
δ
m
,
b
(
β
)
,
where
(33)
β
=
2
α
-
1
+
2
(
1
-
α
)
(
1
+
b
)
(
λ
1
+
λ
2
)
σ
(
1
+
b
λ
1
+
λ
2
)
,
and σ is given by (25).Proof.
Letf
∈
R
λ
1
,
λ
2
,
δ
m
+
1
,
b
(
α
), and then from (19) we have
(34)
R
(
D
λ
1
,
λ
2
,
δ
m
+
1
,
b
f
(
z
)
)
′
>
α
,
(
z
∈
U
)
,
which is equivalent to
(35)
(
D
λ
1
,
λ
2
,
δ
m
+
1
,
b
f
(
z
)
)
′
≺
h
(
z
)
=
1
+
(
2
α
-
1
)
z
1
+
z
.
Using Theorem6, we have
(36)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
≺
q
(
z
)
=
2
α
-
1
+
2
(
1
-
α
)
(
1
+
b
)
(
λ
1
+
λ
2
)
z
(
1
+
b
)
/
(
λ
1
+
λ
2
)
σ
(
1
+
b
λ
1
+
λ
2
)
.
Sinceq is convex and q
(
U
) is symmetric with respect to the real axis, we deduce that
(37)
R
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
>
R
q
(
1
)
=
β
(
α
,
λ
1
,
λ
2
,
b
)
=
2
α
-
1
+
2
(
1
-
α
)
(
1
+
b
)
(
λ
1
+
λ
2
)
σ
(
1
+
b
λ
1
+
λ
2
)
,
for which we deduce R
λ
1
,
λ
2
,
δ
m
+
1
,
b
(
α
)
⊂
K
λ
1
,
λ
2
,
δ
m
,
b
(
β
). The proof is complete.Theorem 8.
Letq be a convex function in U, with q
(
0
)
=
1, and let
(38)
h
(
z
)
=
q
(
z
)
+
(
λ
1
+
λ
2
)
1
+
b
z
q
′
(
z
)
,
(
z
∈
U
)
.
If λ
2
≥
λ
1
≥
0
,
δ
,
m
,
b
∈
N
0, and f
∈
A satisfies the differential subordination
(39)
(
D
λ
1
,
λ
2
,
δ
m
+
1
,
b
f
(
z
)
)
′
≺
h
(
z
)
,
then
(40)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
≺
q
(
z
)
,
(
z
∈
U
)
,
and the result is sharp.Proof.
Using (28) in (26), the differential subordination (39) becomes
(41)
p
(
z
)
+
(
λ
1
+
λ
2
)
1
+
b
z
p
′
(
z
)
≺
h
(
z
)
=
q
(
z
)
+
(
λ
1
+
λ
2
)
1
+
b
z
q
′
(
z
)
(
z
∈
U
)
.
Using Lemma2, we have
(42)
p
(
z
)
≺
q
(
z
)
,
(
z
∈
U
)
;
that is,
(43)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
≺
q
(
z
)
,
(
z
∈
U
)
,
and the result is sharp. The proof of Theorem 8 is complete.Example 9.
Form
=
1
,
δ
=
0
,
λ
2
≥
λ
1
≥
0
,
b
∈
N
0
,
q
(
z
)
=
(
1
+
z
)
/
(
1
-
z
)
,
f
∈
A, and z
∈
U, by applying Theorem 8, we have
(44)
h
(
z
)
=
1
+
z
1
-
z
+
(
λ
1
+
λ
2
)
1
+
b
z
(
1
+
z
1
-
z
)
′
=
(
1
+
b
)
+
2
(
λ
1
+
λ
2
)
z
-
(
1
+
b
)
z
2
(
1
+
b
)
(
1
-
z
)
2
.
By using equality (18) we find that
(45)
(
1
+
b
)
D
λ
1
,
λ
2
,
0
1
,
b
=
(
1
-
(
λ
1
+
λ
2
)
+
b
)
×
(
f
(
z
)
*
φ
λ
2
b
(
z
)
)
+
(
λ
1
+
λ
2
)
z
(
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
.
Now,(46)
(
1
+
b
)
(
D
λ
1
,
λ
2
,
0
1
,
b
*
φ
λ
2
b
(
z
)
)
=
(
1
-
(
λ
1
+
λ
2
)
+
b
)
(
z
+
∑
n
=
2
∞
‍
a
n
z
n
(
1
+
λ
2
(
n
-
1
)
+
b
)
2
)
+
(
λ
1
+
λ
2
)
(
z
+
∑
n
=
2
∞
‍
n
a
n
z
n
(
1
+
λ
2
(
n
-
1
)
+
b
)
2
)
.
A straightforward calculation gives the following:(47)
(
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
=
1
+
∑
n
=
2
∞
‍
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
(
1
+
λ
2
(
n
-
1
)
+
b
)
2
)
n
a
n
z
n
-
1
=
(
z
+


∑
n
=
2
∞
‍
(
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
)
(
1
+
λ
2
(
n
-
1
)
+
b
)
2
)
a
n
z
n
)
×
(
z
)
-
1
=
(
(
z
+
∑
n
=
2
∞
‍
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
(
1
+
λ
2
(
n
-
1
)
+
b
)
z
n
)
)
(
f
(
z
)
*
φ
λ
2
b
(
z
)
)
*
(
z
+


∑
n
=
2
∞
‍
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
(
1
+
λ
2
(
n
-
1
)
+
b
)
)
a
n
z
n
)
)
×
(
z
)
-
1
.
Similarly, using (18), we see that
(48)
(
1
+
b
)
D
λ
1
,
λ
2
,
0
2
,
b
f
(
z
)
=
(
1
-
(
λ
1
+
λ
2
)
+
b
)
×
(
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
+
(
λ
1
+
λ
2
)
z
(
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
,
and then
(49)
(
1
+
b
)
(
D
λ
1
,
λ
2
,
0
2
,
b
f
(
z
)
)
′
=
(
1
+
b
)
(
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′
+
(
λ
1
+
λ
2
)
z
(
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′′
.
By using (47) we have
(50)
(
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
*
φ
λ
2
b
(
z
)
)
′′
=
∑
n
=
2
∞
‍
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
(
1
+
λ
2
(
n
-
1
)
+
b
)
2
)
n
(
n
-
1
)
a
n
z
n
-
2
;
we deduce(51)
(
1
+
b
)
(
D
λ
1
,
λ
2
,
0
2
,
b
f
(
z
)
)
′
=
(
1
+
b
)
(
1
+
∑
n
=
2
∞
‍
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
(
1
+
λ
2
(
n
-
1
)
+
b
)
2
)
n
a
n
z
n
-
1
)
+
(
λ
1
+
λ
2
)
×
∑
n
=
2
∞
‍
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
(
1
+
λ
2
(
n
-
1
)
+
b
)
2
)
n
(
n
-
1
)
a
n
z
n
-
1
;
that is,
(52)
(
D
λ
1
,
λ
2
,
0
2
,
b
f
(
z
)
)
′
=
1
+
∑
n
=
2
∞
‍
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
1
+
λ
2
(
n
-
1
)
+
b
)
2
n
a
n
z
n
-
1
=
(
(
z
+
∑
n
=
2
∞
‍
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
2
(
1
+
λ
2
(
n
-
1
)
+
b
)
z
n
)
)
(
f
(
z
)
*
φ
λ
2
b
(
z
)
)
*
(
z
+
∑
n
=
2
∞
‍
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
2
(
1
+
λ
2
(
n
-
1
)
+
b
)
)
a
n
z
n
)
)
×
(
z
)
-
1
.
From Theorem8 we get
(53)
(
(
z
+
∑
n
=
2
∞
‍
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
2
(
1
+
λ
2
(
n
-
1
)
+
b
)
z
n
)
)
(
f
(
z
)
*
φ
λ
2
b
(
z
)
)
*
(
z
+
∑
n
=
2
∞
‍
(
n
(
1

+

(
λ
1

+

λ
2
)
(
n
-
1
)

+

b
)
2
(
1
+
λ
2
(
n
-
1
)
+
b
)
)
a
n
z
n
)
)
×
(
z
)
-
1
≺
(
1
+
b
)
+
2
(
λ
1
+
λ
2
)
z
-
(
1
+
b
)
z
2
(
1
+
b
)
(
1
-
z
)
2
,
which implies that
(54)
(
(
z
+
∑
n
=
2
∞
‍
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
(
1
+
λ
2
(
n
-
1
)
+
b
)
z
n
)
)
(
f
(
z
)
*
φ
λ
2
b
(
z
)
)
*
(
z
+




∑
n
=
2
∞
‍
(
n
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
(
1
+
λ
2
(
n
-
1
)
+
b
)
)
a
n
z
n
)
)
×
(
z
)
-
1
≺
1
+
z
1
-
z
,
(
z
∈
U
)
.Theorem 10.
Letq be a convex function in U, with q
(
0
)
=
1, and let
(55)
h
(
z
)
=
q
(
z
)
+
z
q
′
(
z
)
,
(
z
∈
U
)
.
If λ
2
≥
λ
1
≥
0
,
δ
,
m
,
b
∈
N
0, and f
∈
A satisfies the differential subordination
(56)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
)
′
≺
h
(
z
)
,
then
(57)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
≺
q
(
z
)
,
(
z
∈
U
)
,
and the result is sharp.Proof.
Let(58)
p
(
z
)
=
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
,
(
z
∈
U
)
.
Differentiating (58), with respect to z, we obtain
(59)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
)
′
=
p
(
z
)
+
z
p
′
(
z
)
,
(
z
∈
U
)
.
Using (58), the differential subordination (56) becomes
(60)
p
(
z
)
+
z
p
′
(
z
)
≺
h
(
z
)
=
q
(
z
)
+
z
q
′
(
z
)
,
(
z
∈
U
)
.
Using Lemma2, we deduce that
(61)
p
(
z
)
≺
q
(
z
)
,
(
z
∈
U
)
.
By using (58), we have
(62)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
≺
q
(
z
)
,
(
z
∈
U
)
.
The proof of Theorem 10 is complete.Example 11.
Forδ
=
0
,
m
=
1
,
λ
2
≥
λ
1
≥
0
,
b
∈
N
0
,
q
(
z
)
=
1
/
(
1
-
z
)
,
f
∈
A, and z
∈
U, from Theorem 10 we obtain
(63)
h
(
z
)
=
1
1
-
z
+
z
(
1
1
-
z
)
′
=
1
(
1
-
z
)
2
.
From Example 9, we have
(64)
(
1
+
b
)
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
=
(
1
-
(
λ
1
+
λ
2
)
+
b
)
(
f
(
z
)
*
φ
λ
2
b
)
+
(
λ
1
+
λ
2
)
z
(
f
(
z
)
*
φ
λ
2
b
)
′
,
and then
(65)
(
1
+
b
)
(
D
λ
1
,
λ
2
,
0
1
,
b
f
(
z
)
)
′
=
(
1
+
b
)
(
f
(
z
)
*
φ
λ
2
b
)
′
+
(
λ
1
+
λ
2
)
z
(
f
(
z
)
*
φ
λ
2
b
)
′′
.
From Theorem 10 we deduce that
(66)
(
f
(
z
)
*
φ
λ
2
b
)
′
+
(
λ
1
+
λ
2
)
(
1
+
b
)
z
(
f
(
z
)
*
φ
λ
2
b
)
′′
≺
1
(
1
-
z
)
2
implies that
(67)
(
1
-
(
λ
1
+
λ
2
)
+
b
)
(
f
(
z
)
*
φ
λ
2
b
)
+
(
λ
1
+
λ
2
)
z
(
f
(
z
)
*
φ
λ
2
b
)
′
z
(
1
+
b
)
≺
1
1
-
z
.Theorem 12.
Leth be a convex function in U, with h
(
0
)
=
1
,
0
≤
α
<
1, and let
(68)
h
(
z
)
=
1
+
(
2
α
-
1
)
z
1
+
z
,
(
z
∈
U
)
.
If λ
2
≥
λ
1
≥
0
,
δ
,
m
,
b
∈
N
0, and f
∈
A satisfies the differential subordination
(69)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
)
′
≺
h
(
z
)
,
then
(70)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
≺
q
(
z
)
=
2
α
-
1
+
2
(
1
-
α
)
ln
⁡
(
1
+
z
)
z
,
(
z
∈
U
)
.
The function q is convex and is the best dominant.Proof.
Let(71)
p
(
z
)
=
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
=
(
z
+
∑
n
=
2
∞
‍
(
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
m
(
1
+
λ
2
(
n
-
1
)
+
b
)
m
)
×
C
(
δ
,
n
)
a
n
z
n
∑
n
=
2
∞
‍
(
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
m
(
1
+
λ
2
(
n
-
1
)
+
b
)
m
)
)
×
(
z
)
-
1
=
1
+
p
1
z
+
p
2
z
2
+
⋯
,
(
p
∈
H
[
1,1
]
,
z
∈
U
)
.
Differentiating (71), with respect to z, we obtain
(72)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
)
′
=
p
(
z
)
+
z
p
′
(
z
)
,
(
z
∈
U
)
.
Using (72), the differential subordination (69) becomes
(73)
p
(
z
)
+
z
p
′
(
z
)
≺
h
(
z
)
=
1
+
(
2
α
-
1
)
z
1
+
z
,
(
z
∈
U
)
.
Using Lemma1, we deduce that
(74)
p
(
z
)
≺
q
(
z
)
=
1
z
∫
0
z
‍
h
(
t
)
d
t
=
1
z
∫
0
z
‍
(
1
+
(
2
α
-
1
)
t
1
+
t
)
d
t
=
1
z
(
∫
0
z
‍
1
1
+
t
d
t
+
(
2
α
-
1
)
∫
0
z
‍
t
1
+
t
d
t
)
=
2
α
-
1
+
2
(
1
-
α
)
ln
⁡
(
1
+
z
)
z
.
By using (71), we have
(75)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
≺
q
(
z
)
=
2
α
-
1
+
2
(
1
-
α
)
ln
⁡
(
1
+
z
)
z
.
The proof of Theorem 12 is complete.Corollary 13.
Iff
∈
R
λ
1
,
λ
2
,
δ
m
,
b
(
α
), then
(76)
R
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
)
>
(
2
α
-
1
)
+
2
(
1
-
α
)
ln
⁡
2
,
(
z
∈
U
)
.Proof.
Sincef
∈
R
λ
1
,
λ
2
,
δ
m
,
b
(
α
), from Definition 4 we have
(77)
R
(
D
λ
1
,
λ
2
,
δ
m
,
b
(
α
)
)
′
>
α
,
(
z
∈
U
)
,
which is equivalent to
(78)
(
D
λ
1
,
λ
2
,
δ
m
,
b
(
α
)
)
′
≺
h
(
z
)
=
1
+
(
2
α
-
1
)
z
1
+
z
.
Using Theorem12, we obtain
(79)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
≺
q
(
z
)
=
2
α
-
1
+
2
(
1
-
α
)
ln
⁡
(
1
+
z
)
z
.
Sinceq is convex and q
(
U
) is symmetric with respect to the real axis, we have that
(80)
R
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
)
>
R
q
(
1
)
=
(
2
α
-
1
)
+
2
(
1
-
α
)
ln
⁡
2
,
(
z
∈
U
)
.Theorem 14.
Leth
∈
H
(
U
), with h
(
0
)
=
1, h
′
(
0
)
≠
0, which satisfies the inequality
(81)
R
(
1
+
z
h
′′
(
z
)
h
′
(
z
)
)
>
-
1
2
,
(
z
∈
U
)
.
If λ
2
≥
λ
1
≥
0
,
δ
,
m
,
b
∈
N
0, and f
∈
A satisfies the differential subordination
(82)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
)
′
≺
h
(
z
)
,
then
(83)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
≺
q
(
z
)
=
1
z
∫
0
z
‍
h
(
t
)
d
t
.Proof.
Let(84)
p
(
z
)
=
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
=
(
z
+
∑
n
=
2
∞
‍
(
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
m
(
1
+
λ
2
(
n
-
1
)
+
b
)
m
)
×
C
(
δ
,
n
)
a
n
z
n
∑
n
=
2
∞
‍
(
(
1
+
(
λ
1
+
λ
2
)
(
n
-
1
)
+
b
)
m
(
1
+
λ
2
(
n
-
1
)
+
b
)
m
)
)
×
(
z
)
-
1
=
1
+
p
1
z
+
p
2
z
2
+
⋯
,
(
p
∈
H
[
1,1
]
,
z
∈
U
)
.
Differentiating (84), with respect to z, we obtain
(85)
(
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
)
′
=
p
(
z
)
+
z
p
′
(
z
)
,
(
z
∈
U
)
.
Using (85), the differential subordination (82) becomes
(86)
p
(
z
)
+
z
p
′
(
z
)
≺
h
(
z
)
,
(
z
∈
U
)
.
Using Lemma1, we deduce that
(87)
p
(
z
)
≺
q
(
z
)
=
1
z
∫
0
z
‍
h
(
t
)
d
t
;
by using (84), we have
(88)
D
λ
1
,
λ
2
,
δ
m
,
b
f
(
z
)
z
≺
q
(
z
)
=
1
z
∫
0
z
‍
h
(
t
)
d
t
.
From Lemma 3, we see that the function q is convex, and from Lemma 1, q is the best dominant for subordination (82). The proof of Theorem 14 is complete.Note that other work related to differential operators and differential subordination can be seen in [8–13].



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*Source: 656258-2014-08-20.xml*