# Fractional Calculus of Analytic Functions Concerned with Möbius Transformations

Authors: Nicoleta Breaz; Daniel Breaz; Shigeyoshi Owa
Journal: Journal of Function Spaces
 (2016)

---

## Abstract

LetA be the class of functions f
(
z
) in the open unit disk U with f
(
0
)
=
0 and f
′
(
0
)
=
1. Also, let w
(
ζ
) be a Möbius transformation in U for some z
∈
U. Applying the Möbius transformations, we consider some properties of fractional calculus (fractional derivatives and fractional integrals) of f
(
z
)
∈
A. Also, some interesting examples for fractional calculus are given.


---

## Body

## 1. Introduction

LetA denote the class of functions f
(
z
) of the form (1)
f
z
=
z
+
∑
k
=
2
∞
a
k
z
kwhich are analytic in the open unit disk U
=
{
z
∈
C
:
|
z
|
<
1
}.Iff
(
z
)
∈
A satisfies(2)
R
e
z
f
′
z
f
z
>
α
z
∈
Ufor some real α
(
0
≦
α
<
1
), then f
(
z
) is said to be starlike of order α in U. We denote by S
∗
(
α
) the class of all starlike functions f
(
z
) of order α in U and S
∗
(
0
)
≡
S
∗. Furthermore, if f
(
z
)
∈
A satisfies (3)
R
e
1
+
z
f
′
′
z
f
′
z
>
α
z
∈
Ufor some real α
(
0
≦
α
<
0
), then we say that f
(
z
) is convex of order α in U. We also denote by K
(
α
) the class of all such functions f
(
z
) and K
(
0
)
≡
K. In view of definitions for the classes S
∗
(
α
) and K
(
α
), we know that(i)
f
(
z
)
∈
K
(
α
) if and only if z
f
′
(
z
)
∈
S
∗
(
α
);
(ii)
f
(
z
)
∈
S
∗
(
α
) if and only if ∫
0
z
(
f
t
/
t
)
d
t
∈
K
(
α
).Further, MacGregor [1] and Wilken and Feng [2] have the sharp inclusion relation that K
(
α
)
⊂
S
∗
(
β
) for each α
(
0
≦
α
<
1
) with (4)
β
=
1
-
2
α
2
2
1
-
α
1
-
2
2
α
-
1
α
≠
1
2
1
2
log
⁡
2
α
=
1
2
.For α
=
0, Marx [3] and Strohhäcker [4] showed that K
⊂
S
∗
(
1
/
2
). Also, by Robertson [5], we know that the extremal function f
(
z
) for the class S
∗
(
α
) is (5)
f
z
=
z
1
-
z
2
1
-
α
=
z
+
∑
k
=
2
∞
∏
j
=
2
k
j
-
2
α
k
-
1
!
z
kand the extremal function f
(
z
) for the class K
(
α
) is(6)
f
z
=
1
-
1
-
z
2
α
-
1
2
α
-
1
=
z
+
∑
k
=
2
∞
∏
j
=
2
k
j
-
2
α
k
!
z
k
α
≠
1
2
-
log
⁡
1
-
z
=
z
+
∑
k
=
2
∞
1
k
z
k
α
=
1
2
.Forf
(
z
)
∈
A, we apply the following Möbius transformation:(7)
w
ζ
=
z
+
ζ
1
+
z
¯
ζ
ζ
∈
Ufor a fixed z
∈
U. This Möbius transformation w
(
ζ
) maps U onto itself and ζ
=
0 to w
(
0
)
=
z.


## 2. Fractional Calculus

From among the various definitions for fractional calculus (i.e., fractional derivatives and fractional integrals) given in the literature, we have to recall here the following definitions for fractional calculus which are used by Owa [6, 7] and by Owa and Srivastava [8].Definition 1.
The fractional integral of orderλ is defined, for f
(
z
)
∈
A, by (8)
D
z
-
λ
f
z
=
1
Γ
λ
∫
0
z
f
ζ
z
-
ζ
1
-
λ
d
ζ
,where λ
>
0 and the multiplicity of (
z
-
ζ
)
λ
-
1 is removed by requiring log
⁡
(
z
-
ζ
) to be real when z
-
ζ
>
0.Definition 2.
The fractional derivative of orderλ is defined, for f
(
z
)
∈
A, by(9)
D
z
λ
f
z
=
d
d
z
D
z
λ
-
1
f
z
=
1
Γ
1
-
λ
d
d
z
∫
0
z
f
ζ
z
-
ζ
λ
d
ζ
,where 0
≦
λ
<
1 and the multiplicity of (
z
-
ζ
)
-
λ is removed as in Definition 1 above.Definition 3.
Under the hypotheses of Definition2, the fractional derivative of order n
+
λ is defined by (10)
D
z
n
+
λ
f
z
=
d
n
d
z
n
D
z
λ
f
z
,where 0
≦
λ
<
1 and n
∈
N
0
=
0,1
,
2
,
….Remark 4.
In view of definitions for the fractional calculus off
(
z
)
∈
A, we see that (11)
D
z
-
λ
f
z
=
1
Γ
2
+
λ
z
1
+
λ
+
2
!
Γ
3
+
λ
a
2
z
2
+
λ
+
⋯
+
k
!
Γ
k
+
1
+
λ
a
k
z
k
+
λ
+
⋯
=
1
Γ
2
+
λ
z
1
+
λ
+
∑
k
=
2
∞
k
!
Γ
k
+
1
+
λ
a
k
z
k
+
λ
λ
>
0
,
D
z
λ
f
z
=
1
Γ
2
-
λ
z
1
-
λ
+
2
!
Γ
3
-
λ
a
2
z
2
-
λ
+
⋯
+
k
!
Γ
k
+
1
-
λ
a
k
z
k
-
λ
+
⋯
=
1
Γ
2
-
λ
z
1
-
λ
+
∑
k
=
2
∞
k
!
Γ
k
+
1
-
λ
a
k
z
k
-
λ
0
≦
λ
<
1
,
D
z
n
+
λ
f
z
=
d
n
d
z
n
1
Γ
2
-
λ
z
1
-
λ
+
∑
k
=
2
∞
k
!
Γ
k
+
1
-
λ
a
k
z
k
-
λ
=
1
Γ
2
-
n
-
λ
z
1
-
n
-
λ
+
∑
k
=
2
∞
k
!
Γ
k
+
1
-
n
-
λ
a
k
z
k
-
n
-
λfor 0
≦
λ
<
1 and n
∈
N
0.
Therefore, we can write that(12)
D
z
n
+
λ
f
z
=
d
n
d
z
n
D
z
λ
f
z
=
D
z
λ
d
n
d
z
n
f
z
,
(13)
D
z
λ
f
z
=
1
Γ
2
-
λ
z
1
-
λ
+
∑
k
=
2
∞
k
!
Γ
k
+
1
-
λ
a
k
z
k
-
λfor any real number λ.
Using the fractional calculus (13), we define(14)
F
z
=
Γ
2
-
λ
z
λ
D
z
λ
f
z
=
z
+
∑
k
=
2
∞
k
!
Γ
2
-
λ
Γ
k
+
1
-
λ
a
k
z
k
λ
∈
R
,
λ
≠
2
.
If we takeλ
=
-
1 in (14), then(15)
F
z
=
Γ
3
z
-
1
D
z
-
1
f
z
=
2
z
∫
0
z
f
t
d
t
=
z
+
∑
k
=
2
∞
2
k
+
1
a
k
·
z
kimplies the Libera integral operator defined by Libera [9]. Therefore, F
(
z
) given by (14) is the generalization operator of Libera integral operator.Let us give two examples for the fractional operatorF
(
z
) defined in (14).Example 5.
Let us definef
(
z
) by(16)
f
z
=
z
+
2
-
λ
6
z
2
∈
A
-
1
≦
λ
<
2
.Then, we have that(17)
Re
⁡
z
f
′
z
f
z
=
Re
⁡
2
-
1
1
+
M
z
=
2
-
1
+
M
cos
⁡
θ
1
+
M
2
+
2
M
cos
⁡
θ
z
=
e
i
θ
,where M
=
(
2
-
λ
)
/
6
>
0. If we define(18)
h
t
=
1
+
M
t
1
+
M
2
+
2
M
t
t
=
cos
⁡
θ
,then(19)
h
′
t
=
M
M
+
1
M
-
1
1
+
M
2
+
2
M
t
2
<
0
0
<
M
≦
1
2
.This shows us that(20)
h
t
≦
h
-
1
=
1
1
-
M
.That is, (21)
Re
⁡
z
f
′
z
f
z
>
2
-
1
1
-
M
=
2
+
2
λ
4
+
λ
>
0
z
∈
U
.Therefore, f
(
z
)
∈
S
∗
(
2
+
2
λ
)
/
(
4
+
λ
).
Forf
(
z
), given by (16), F
(
z
) becomes(22)
F
z
=
Γ
2
-
λ
z
λ
D
z
λ
f
z
=
z
+
1
3
z
2
-
1
≦
λ
<
2
.Then, we see that F
(
z
)
∈
S
∗
1
/
2.
Next, let us consider the functiong
(
ζ
) given by(23)
g
ζ
=
F
∘
w
ζ
-
F
z
1
-
z
2
F
′
z
ζ
∈
Ufor a fixed z
∈
U, where w
(
ζ
) is given by (7). Then, it is easy to see that g
(
ζ
)
∈
A. Taking z
=
1
/
2 in (23), we have that(24)
g
ζ
=
ζ
11
ζ
+
16
4
ζ
+
2
2
ζ
∈
U
,
Re
⁡
ζ
g
′
ζ
g
ζ
=
Re
⁡
1
-
ζ
11
ζ
+
10
ζ
+
2
11
ζ
+
16
=
1
-
704
cos
2
⁡
θ
+
848
cos
⁡
θ
+
149
1408
cos
2
⁡
θ
+
3268
cos
⁡
θ
+
1885
θ
=
e
i
θ
.Letting(25)
H
t
=
704
t
2
+
848
t
+
149
1408
t
2
+
3268
t
+
1885
t
=
cos
⁡
θ
,we obtain that(26)
H
′
t
=
12
9224
t
2
+
186208
t
+
92629
1408
t
2
+
3268
t
+
1885
2
.This shows that H
′
(
-
1
)
<
0, H
′
(
0
)
>
0, and H
′
(
1
)
>
0. Therefore, there exists some t
0 such that H
′
(
t
0
)
=
0 for -
1
<
t
0
<
0. It follows that(27)
max
-
1
≦
t
≦
1
⁡
H
t
=
max
⁡
H
-
1
,
H
1
=
H
1
=
7
27
.Thus, we say that (28)
R
e
ζ
g
′
ζ
g
ζ
>
1
-
7
27
=
20
27
ζ
∈
U
.Consequently, we say that F
(
z
)
∈
S
∗
1
/
2, g
(
ζ
)
∈
S
∗
20
/
27 for f
(
z
)
∈
S
∗
(
2
+
2
λ
)
/
(
4
+
λ
) given by (16).
Ifλ
=
-
1
/
2, then(29)
f
z
=
z
+
5
12
z
2
∈
S
∗
2
7
.The open unit disk U is mapped on the starlike domain of order 2
/
7 in Figure 1.
Ifλ
=
1
/
3, then(30)
f
z
=
z
+
5
18
z
2
∈
S
∗
8
13
.Thus, f
(
z
) maps U onto the starlike domain of order 8
/
13 in Figure 2.Figure 1Figure 2Example5 means that there is some function f
(
z
)
∈
S
∗
(
α
) such that F
(
z
)
∈
S
∗
(
β
) and g
(
ζ
)
∈
S
∗
(
γ
).Next, we consider the following.Example 6.
Let a functionf
(
z
) be given by (31)
f
z
=
z
+
2
-
λ
12
z
2
∈
A
-
1
≦
λ
<
2
.Then, we have that(32)
Re
⁡
1
+
z
f
′
′
z
f
′
z
=
Re
⁡
2
-
1
1
+
2
M
z
=
2
-
1
+
2
M
cos
⁡
θ
1
+
4
M
2
+
4
M
cos
⁡
θ
z
=
e
i
θ
,where M
=
(
2
-
λ
)
/
12
>
0. Defining h
(
t
) by (33)
h
t
=
1
+
2
M
t
1
+
4
M
2
+
4
M
t
t
=
cos
⁡
θ
,we have that (34)
h
′
t
=
2
M
2
M
+
1
2
M
-
1
1
+
4
M
2
+
4
M
t
2
<
0
0
<
M
≦
1
4which shows us that (35)
h
t
≦
h
-
1
=
1
1
-
2
M
.Thus, we obtain that(36)
R
e
1
+
z
f
′
′
z
f
′
z
>
2
-
1
1
-
2
M
=
2
+
2
λ
4
+
λ
>
0
z
∈
U
.This gives us that f
(
z
)
∈
K
(
2
+
2
λ
)
/
(
4
+
λ
).
Forf
(
z
), given by (31), F
(
z
) becomes(37)
F
z
=
Γ
2
-
λ
z
λ
D
z
λ
f
z
=
z
+
1
6
z
2
-
1
≦
λ
<
2
.Then, it is easy to see that F
(
z
)
∈
K
1
/
2.
For thisF
(
z
), we consider g
(
ζ
) defined by (23). If we take z
=
1
/
2 for g
(
ζ
), we have that (38)
g
ζ
=
ζ
17
ζ
+
28
7
ζ
+
2
2
ζ
∈
U
,
Re
⁡
1
+
ζ
g
′
′
ζ
g
′
ζ
=
Re
⁡
1
-
ζ
10
ζ
+
11
5
ζ
+
7
ζ
+
2
=
1
-
280
cos
2
⁡
θ
+
379
cos
⁡
θ
+
97
280
cos
2
⁡
θ
+
646
cos
⁡
θ
+
370
ζ
=
e
i
θ
.If we write that (39)
H
t
=
280
t
2
+
379
t
+
97
280
t
2
+
646
t
+
370
t
=
cos
⁡
θ
,then(40)
H
′
t
=
24
3115
t
2
+
6370
t
+
3232
280
t
2
+
646
t
+
370
2
.Since H
′
(
-
1
)
<
0, H
′
(
0
)
>
0, and H
′
(
1
)
>
0, there exists some t
0 such that H
′
(
t
0
)
=
0 for -
1
<
t
0
<
0. This gives us that (41)
max
-
1
≦
t
≦
1
⁡
H
t
=
max
⁡
H
-
1
,
H
1
=
H
1
=
7
12
.It follows that(42)
Re
⁡
1
+
ζ
g
′
′
ζ
g
′
ζ
>
1
-
7
12
=
5
12
ζ
∈
U
.Therefore, we say that F
(
z
)
∈
K
1
/
2, g
(
ζ
)
∈
K
5
/
12 for f
(
z
)
∈
K
(
2
+
2
λ
)
/
(
4
+
λ
).
Ifλ
=
-
2
/
3, then(43)
f
z
=
z
+
2
9
z
2
∈
K
1
5maps U onto the convex domain of order 1
/
5 in Figure 3.
Ifλ
=
3
/
2, then (44)
f
z
=
z
+
1
24
z
2
∈
K
10
11
.This function f
(
z
) maps U onto the convex domain of order 10
/
11 in Figure 4.Figure 3Figure 4Example6 says that there exists some function f
(
z
)
∈
K
(
α
) such that F
(
z
)
∈
K
(
β
) and g
(
ζ
)
∈
K
(
γ
).In view of Examples5 and 6, we introduce the following.Definition 7.
Letf
(
z
)
∈
A, F
(
z
)
=
Γ
(
2
-
λ
)
z
λ
D
z
λ
f
(
z
) with -
1
≦
λ
<
2 and let g
(
ζ
) be defined by (23) for a fixed z
∈
U. Then, we say that(i)
f
(
z
)
∈
S
0 if g
(
ζ
) is univalent in U,
(ii)
f
(
z
)
∈
S
0
∗
(
α
) if g
(
ζ
)
∈
S
∗
(
α
),
(iii)
f
(
z
)
∈
K
0
(
α
) if g
(
ζ
)
∈
K
(
α
).
Also, we write thatS
0
∗
(
0
)
≡
S
0
∗ and K
0
(
0
)
≡
K
0 when α
=
0.In order to discuss our classesS
0, S
0
∗
(
α
), and K
0
(
α
), we need the following lemma due to Robertson [5] (also see Duren [10]).Lemma 8.
Iff
(
z
)
∈
S
∗
(
α
), then(45)
a
k
≦
∏
j
=
2
k
j
-
2
α
k
-
1
!
k
=
2,3
,
4
,
…with the equality in (45) with f
(
z
) given by (5). If f
(
z
)
∈
K
(
α
), then(46)
a
k
≦
∏
j
=
2
k
j
-
2
α
k
!
k
=
2,3
,
4
,
…with the equality in (46) with f
(
z
) given by (6).We also need the following lemma.Lemma 9.
Ifg
(
ζ
) is defined by(47)
g
ζ
=
f
∘
w
ζ
-
f
z
1
-
z
2
f
′
z
ζ
∈
Ufor a fixed z
∈
U for f
(
z
)
∈
A, then(48)
d
n
/
d
ζ
n
f
∘
w
ζ
f
′
z
=
n
!
n
-
1
!
1
+
z
¯
ζ
2
n
1
-
z
2
n
-
1
∑
j
=
0
n
-
1
g
n
-
j
ζ
z
¯
j
n
-
j
!
n
-
j
-
1
!
j
!
1
+
z
¯
ζ
jfor n
=
1,2
,
3
,
…, where w
(
ζ
) is given by (7).Proof.
We use the mathematical induction to prove (48). For n
=
1, the right-hand side of (48) becomes (49)
1
+
z
¯
ζ
2
g
′
ζ
=
d
/
d
ζ
f
∘
w
ζ
f
′
z
.Therefore, (48) holds true for n
=
1.
Assume that relation (48) is true for a fixed positive integer n. Then, some calculations lead us to (50)
d
n
+
1
/
d
ζ
n
+
1
f
∘
w
ζ
1
-
z
2
1
+
z
¯
ζ
2
f
′
z
=
n
!
n
-
1
!
1
+
z
¯
ζ
2
n
1
-
z
2
n
-
1
∑
j
=
0
n
-
1
1
n
-
j
!
n
-
j
-
1
!
j
!
g
n
+
1
-
j
ζ
+
2
n
-
j
g
n
-
j
ζ
z
¯
1
+
z
¯
ζ
z
¯
j
1
+
z
¯
ζ
j
=
n
+
1
!
n
!
1
+
z
¯
ζ
2
n
1
-
z
2
n
-
1
∑
j
=
0
n
g
n
+
1
-
j
ζ
z
¯
j
n
+
1
-
j
!
n
-
j
!
j
!
1
+
z
¯
ζ
j
.This means that relation (48) holds true for n
+
1. Thus, by applying the mathematical induction, we complete the proof of the lemma.Takingζ
=
0 in Lemma 9, we have the following.Corollary 10.
Ifg
(
ζ
) is defined by (47) for f
(
z
)
∈
A, then one has(51)
f
n
z
f
′
z
≦
n
!
n
-
1
!
1
-
z
2
n
-
1
∑
j
=
0
n
-
1
g
n
-
j
0
z
j
n
-
j
!
n
-
j
-
1
!
j
!for z
∈
U. Furthermore, we have(52)
f
′
′
z
f
′
z
≦
g
′
′
0
+
2
g
′
0
z
1
-
z
2
z
∈
U
.Applying Corollary10, we have the following.Theorem 11.
LetF
(
z
) be defined by (14) for f
(
z
)
∈
A with -
1
≦
λ
<
2.(i)
Iff
(
z
)
∈
S
0, then (53)
F
n
z
F
′
z
≦
n
!
n
+
z
1
-
z
n
-
1
1
+
z
n
=
1,2
,
3
,
…with the equality for g
(
ζ
) given by (54)
g
ζ
=
ζ
1
+
e
i
θ
ζ
2
θ
∈
R
.
(ii)
Iff
(
z
)
∈
S
0
∗
(
α
), then (55)
F
n
z
F
′
z
≦
n
!
n
-
1
!
1
-
z
2
n
-
1
∑
j
=
0
n
-
1
∏
k
=
2
n
-
j
k
-
2
α
j
!
n
-
j
-
1
!
2
z
jwith the equality for g
(
ζ
) given by (56)
g
ζ
=
ζ
1
+
e
i
θ
ζ
2
1
-
α
θ
∈
R
.
(iii)
Iff
(
z
)
∈
K
0
(
α
), then (57)
F
n
z
F
′
z
≦
n
!
n
-
1
!
1
-
z
2
n
-
1
∑
j
=
0
n
-
1
∏
k
=
2
n
-
j
k
-
2
α
j
!
n
-
j
!
n
-
j
-
1
!
z
j
n
=
1,2
,
3
,
…with the equality for g
(
ζ
) given by(58)
g
ζ
=
1
-
1
-
ζ
2
α
-
1
2
α
-
1
α
≠
1
2
,
-
log
⁡
1
-
ζ
α
=
1
2
.Proof.
Note that(59)
g
ζ
=
F
∘
w
ζ
-
F
z
1
-
z
2
F
′
z
ζ
∈
Ufor F
(
z
)
=
Γ
(
2
-
λ
)
z
λ
D
z
λ
f
(
z
). Therefore, Corollary 10 gives us that (60)
F
n
z
F
′
z
≦
n
!
n
-
1
!
1
-
z
2
n
-
1
∑
j
=
0
n
-
1
g
n
-
j
0
z
j
n
-
j
!
n
-
j
-
1
!
j
!
.According to Lemma 8, we have (61)
g
n
-
j
0
≦
n
-
j
!
n
-
j
,if f
(
z
)
∈
S
0, then we calculate that (62)
F
n
z
F
′
z
≦
n
!
n
-
1
!
1
-
z
2
n
-
1
∑
j
=
0
n
-
1
n
-
j
n
-
j
-
1
!
j
!
z
j
=
n
!
n
+
z
1
-
z
n
-
1
1
+
z
,because (63)
∑
j
=
0
n
-
1
n
-
j
n
-
j
-
1
!
j
!
z
j
=
n
+
z
1
+
z
n
-
2
n
-
1
!
.If f
(
z
)
∈
S
0
∗
(
α
), then(64)
g
n
-
j
0
≦
n
-
j
∏
k
=
2
n
-
j
k
-
2
α
g
′
0
=
1by means of (45). This implies inequality (55) for f
(
z
)
∈
S
0
∗
(
α
). Furthermore, if f
(
z
)
∈
K
0
(
α
), then g
(
ζ
) satisfies(65)
g
n
-
j
0
≦
∏
k
=
2
n
-
j
k
-
2
α
g
′
0
=
1
,which implies inequality (57). Consequently, we complete the proof of the theorem.Sinceg
′
(
0
)
=
1, letting n
=
2 in Theorem 11, we have the following.Corollary 12.
Letf
(
z
)
∈
A with -
1
≦
λ
<
2.(i)
Iff
(
z
)
∈
S
0, then (66)
λ
λ
-
1
D
z
λ
f
z
+
2
λ
z
D
z
λ
+
1
f
z
+
z
2
D
z
λ
f
z
z
λ
D
z
λ
f
z
+
z
D
z
λ
+
1
f
z
≦
2
2
+
z
1
-
z
2for z
∈
U.
(ii)
Iff
(
z
)
∈
S
0
∗
(
α
), then (67)
λ
λ
-
1
D
z
λ
f
z
+
2
λ
z
D
z
λ
+
1
f
z
+
z
2
D
z
λ
f
z
z
λ
D
z
λ
f
z
+
z
D
z
λ
+
1
f
z
≦
2
2
1
-
α
+
z
1
-
z
2for z
∈
U.
(iii)
Iff
(
z
)
∈
K
0
(
α
), then (68)
λ
λ
-
1
D
z
λ
f
z
+
2
λ
z
D
z
λ
+
1
f
z
+
z
2
D
z
λ
f
z
z
λ
D
z
λ
f
z
+
z
D
z
λ
+
1
f
z
≦
2
1
-
α
+
z
1
-
z
2for z
∈
U.Takingλ
=
0 in Corollary 12, we have the following.Corollary 13.
Iff
(
z
)
∈
S
0, then(69)
f
z
f
′
z
≦
2
2
+
z
1
-
z
2
z
∈
U
,if f
(
z
)
∈
S
0
∗
(
α
), then (70)
f
z
f
′
z
≦
2
2
1
-
α
+
z
1
-
z
2
z
∈
U
,and if f
(
z
)
∈
K
0
(
α
), then (71)
f
z
f
′
z
≦
2
1
-
α
+
z
1
-
z
2
z
∈
U
.


## 3. Univalence of Fractional Calculus

Letf
(
z
) and g
(
z
) be analytic in U. Then, f
(
z
) is said to be subordinate to g
(
z
), written f
(
z
)
≺
g
(
z
), if there exists a function w
(
z
) analytic in U with w
(
0
)
=
0 and |
w
(
z
)
|
<
1 (z
∈
U), such that f
(
z
)
=
g
(
w
(
z
)
). Furthermore, if g
(
z
) is univalent in U, then the subordination f
(
z
)
≺
g
(
z
) is equivalent to f
(
0
)
=
g
(
0
) and f
(
U
)
⊂
g
(
U
) (cf. Miller and Mocanu [11]).To discuss the univalence of fractional calculusF
(
z
) given by (14), we need the following lemma due to Miller and Mocanu [12] (or due to Jack [13]).Lemma 14.
Let the functionw
(
z
) be analytic in U with w
(
0
)
=
0. If there exists a point z
0
∈
U such that(72)
max
z
≦
z
0
⁡
w
z
=
w
z
0
,then(73)
z
0
w
′
z
0
w
z
0
=
k
,
R
e
1
+
z
0
w
′
′
z
0
w
′
z
0
≧
k
,where k
≧
1.Now, we derive the following.Theorem 15.
IfF
(
z
) defined by (14) for f
(
z
)
∈
A satisfies (74)
1
+
1
2
z
F
′
′
z
F
′
z
-
z
F
′
z
F
z
<
2
-
α
4
α
z
∈
Ufor some real α which satisfies 2
(
2
-
1
)
≦
α
<
1, then(75)
z
2
F
′
z
F
z
2
≺
1
+
1
-
α
z
1
-
z
z
∈
U
.Proof.
We define the functionw
(
z
) by(76)
z
2
F
′
z
F
z
2
=
1
+
1
-
α
w
z
1
-
w
z
z
∈
Uwith 2
(
2
-
1
)
≦
α
<
1. Then, we see that w
(
z
) is analytic in U with w
(
0
)
=
0 and that(77)
1
+
1
2
z
F
′
′
z
F
′
z
-
z
F
′
z
F
z
=
z
w
′
z
2
1
-
α
1
+
1
-
α
w
z
+
1
1
-
w
z
.If there exists a point z
0
∈
U such that (78)
max
z
≦
z
0
⁡
w
z
=
w
z
0
=
1
,then Lemma 14 gives us that (79)
z
0
w
′
z
0
=
k
w
z
0
k
≧
1and w
(
z
0
)
=
e
i
θ. This implies that(80)
1
+
1
2
z
0
F
′
′
z
0
F
′
z
0
-
z
0
F
′
z
0
F
z
0
=
k
2
1
-
α
1
+
1
-
α
e
i
θ
+
1
1
-
e
i
θ
≧
2
-
α
2
1
1
-
e
i
θ
1
+
1
-
α
e
i
θ
=
2
-
α
2
1
2
1
-
cos
⁡
θ
1
+
1
-
α
2
+
2
1
-
α
cos
⁡
θ
.
Letting(81)
g
t
=
1
-
t
1
+
1
-
α
2
+
2
1
-
α
t
t
=
cos
⁡
θ
,we have that(82)
g
′
t
=
-
α
2
+
4
1
-
α
t
≦
-
α
2
+
4
α
-
4
≦
0for 2
(
2
-
1
)
≦
α
<
1. Thus, (83)
g
t
≦
g
-
1
=
2
α
2
.Therefore, F
(
z
) satisfies(84)
1
+
1
2
z
0
F
′
′
z
0
F
′
z
0
-
z
0
F
′
z
0
F
z
0
≧
2
-
α
4
αfor 2
(
2
-
1
)
≦
α
<
1. This contradicts our condition (74) for F
(
z
). Thus, w
(
z
) satisfies |
w
(
z
)
|
<
1 for all z
∈
U. With this reason above, we conclude that (85)
z
2
F
′
z
F
z
2
≺
1
+
1
-
α
z
1
-
z
z
∈
U
.Next, we show the following.Theorem 16.
IfF
(
z
) defined by (14) for f
(
z
)
∈
A satisfies (86)
1
+
1
2
z
F
′
′
z
F
′
z
-
z
F
′
z
F
z
<
α
2
1
+
α
z
∈
Ufor some real α
>
0, then(87)
z
2
F
′
z
F
z
2
-
1
<
α
z
∈
U
.Proof.
Let us define the functionw
(
z
) by (88)
z
2
F
′
z
F
z
2
-
1
=
α
w
z
z
∈
U
.Then,(89)
1
+
1
2
z
F
′
′
z
F
′
z
-
z
F
′
z
F
z
=
α
2
z
w
′
z
1
+
α
w
z
<
α
2
1
+
α
z
∈
U
.Suppose that there exists a point z
0
∈
U such that (90)
max
z
≦
z
0
⁡
w
z
=
w
z
0
=
1
,then we can write that(91)
z
0
w
′
z
0
=
k
w
z
0
k
≧
1and w
(
z
0
)
=
e
i
θ. Therefore, we have that (92)
1
+
1
2
z
0
F
′
′
z
0
F
′
z
0
-
z
0
F
′
z
0
F
z
0
=
k
α
2
1
1
+
α
e
i
θ
≧
α
2
1
+
α
,which contradicts our condition (86). Thus, we say that there is no z
0
∈
U such that |
w
(
z
0
)
|
=
1. Consequently, letting |
w
(
z
)
|
<
1 for all z
∈
U, we prove the theorem.Takingα
=
1 in Theorem 16, we haveCorollary 17.
IfF
(
z
) defined by (14) for f
(
z
)
∈
A satisfies(93)
1
+
1
2
z
F
′
′
z
F
′
z
-
z
F
′
z
F
z
<
1
4
z
∈
U
,then(94)
z
2
F
′
z
F
z
2
-
1
<
1
z
∈
U
.Remark 18.
In view of the result for the univalence of analytic functions due to Ozaki and Nunokawa [14], we see that F
(
z
) satisfying inequality (94) is univalent in U.Example 19.
Let us consider the functionf
(
z
) given by (95)
f
z
=
D
z
-
λ
z
1
-
λ
Γ
2
-
λ
e
z
/
2
-
1
≦
λ
<
2
.Then, we have that(96)
F
z
=
Γ
2
-
λ
z
λ
D
z
λ
f
z
=
z
e
z
/
2
,
(97)
z
F
′
z
F
z
=
1
+
1
2
z
,
(98)
z
F
′
′
z
F
′
z
=
1
2
z
+
z
2
+
z
.Therefore, F
(
z
) satisfies(99)
1
+
1
2
z
F
′
′
z
F
′
z
-
z
F
′
z
F
z
=
1
4
z
2
2
+
z
<
1
4
z
∈
U
.For such a function F
(
z
), we see that(100)
z
2
F
′
z
F
z
2
-
1
=
e
-
z
/
2
1
+
1
2
z
-
1
≦
c
z
∈
U
.By using the computer, we know that c
<
0.18
<
1. Indeed, the function F
(
z
) satisfying (97) implies that(101)
Re
⁡
z
F
′
z
F
z
>
1
2
z
∈
U
.This shows us that F
(
z
)
∈
S
∗
1
/
2.



---

*Source: 6086409-2016-03-22.xml*