# A New Roper-Suffridge Extension Operator on a Reinhardt Domain

Authors: Jianfei Wang; Cailing Gao
Journal: Abstract and Applied Analysis
 (2011)

---

## Abstract

We introduce a new Roper-Suffridge extension operator on the following Reinhardt domainΩ
n
,
p
2
,
…
,
p
n
=
{
z
∈
ℂ
n
:
|
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
j
<
1
}
given by F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
, where f is a normalized locally biholomorphic function on the unit disc D, p
j are positive integer, a
j are complex constants, and j
=
2
,
…
,
n. Some conditions for a
j are found under which the operator preserves almost starlike mappings of order α and starlike mappings of order α, respectively. In particular, our results reduce to many well-known results when all α
j
=
0.


---

## Body

## 1. Introduction

In 1995, Roper and Suffridge [1] introduced an extension operator. This operator is defined as follows: (1.1)
Φ
n
(
f
)
(
z
)
=
(
f
(
z
1
)
,
f
′
(
z
1
)
z
0
′
)
′
,
where f is a normalized locally biholomorphic function on the unit disk D in ℂ, z
=
(
z
1
,
z
0
′
)
′ belonging to the unit ball ℬ
n in ℂ
n, z
0
=
(
z
2
,
…
,
z
n
)
′
∈
ℂ
n
-
1 and the branch of the square root is chosen such that f
′
(
0
)
=
1.It is well known that the Roper-Suffridge extension operator has the following remarkable properties:(i)
iff is a normalized convex function on D, then Φ
n
(
f
) is a normalized convex mapping on ℬ
n;
(ii)
iff is a normalized starlike function on D, then Φ
n
(
f
) is a normalized starlike mapping on ℬ
n;
(iii)
iff is a normalized Bloch function on D, then Φ
n
(
f
) is a normalized Bloch mapping on ℬ
n.The above result (i) was proved by Roper and Suffridge [1] and the result (ii) and (iii) was proved by Graham and Kohr [2, 3]. Until now, it is difficult to construct the concrete convex mappings, starlike mappings, and Bloch mappings on ℬ
n. By making use of the Roper-Suffridge extension operator, we may easily give many concrete examples about these mappings. This is one important reason why people are interested in this extension operator.In 2005, Muir [4] modified the Roper-Suffridge extension operator as follows: (1.2)
F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
P
(
z
0
)
,
f
′
(
z
1
)
z
0
′
)
′
,
where P
(
z
0
) is a homogeneous polynomial of degree 2 with respect to z
0, and f
,
z
1, and z
0 are defined as above. They proved that this operator preserves starlikeness and convexity if and only if ∥
P
∥
⩽
1
/
4 and ∥
P
∥
⩽
1
/
2, respectively. The modified operator plays a key role to study the extreme points of convex mappings on ℬ
n (see [5, 6]). Later, Kohr [7] and Muir [8] used the Loewner chain to study the modified Roper-Suffridge extension operator. Recently, the modified Roper-Suffridge extension operator on the unit ball is also studied by Wang and Liu [9] and Feng and Yu [10].On the other hand, people also considered the generalized Roper-Suffridge extension operator on the general Reinhardt domains. For example, Gong and Liu [11, 12] induced the definition of ɛ starlike mappings and obtained that the operator (1.3)
Φ
n
,
(
1
/
p
)
(
f
)
(
z
)
=
(
f
(
z
1
)
,
(
f
′
(
z
1
)
)
1
/
p
z
0
′
)
′
maps the ɛ starlike functions on D to the ɛ starlike mappings on the Reinhardt domain Ω
n
,
p
=
{
z
∈
ℂ
n
:
|
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
<
1
}, where p
⩾
1
,
f
,
z
1, and z
0 are defined as above. When ɛ
=
0 and ɛ
=
1, Φ
n
,
(
1
/
p
) maps the starlike function and convex function on D to the starlike mapping and the convex mapping on Ω
n
,
p, respectively.Furthermore, Gong and  Liu [13] proved that the operator (1.4)
Φ
n
,
(
1
/
p
2
)
,
…
,
1
/
p
n
(
f
)
(
z
)
=
(
f
(
z
1
)
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
maps the ɛ starlike functions on D to the ɛ starlike mappings on the domain Ω
n
,
p
2
,
…
,
p
n
=
{
z
∈
ℂ
n
:
|
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
j
<
1
}, where p
j
⩾
1
,
j
=
2
,
…
,
n
,
f
,
z
1, and z
0 are defined as above. Liu and Liu [14] proved that this operator preserves starlikeness of order α on the domain Ω
n
,
p
2
,
…
,
p
n. On the other hand, Feng and Liu [15] proved that this operator preserves almost starlikeness of order α on the domain Ω
n
,
p
2
,
…
,
p
n.In contrast to the modified Roper-Suffridge extension operator in the unit ball, it is natural to ask if we can modify the Roper-Suffridge extension operator on the Reinhardt domains. In this paper, we will introduce the following modified operator:(1.5)
F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
on the Reinhardt domain Ω
n
,
p
2
,
…
,
p
n. We will give some sufficient conditions for a
j under which the above Roper-Suffridge operator preserves an almost starlike mappings of order α and starlike mappings of order α, respectively.In the following, we give some notation and definitions. Letℂ
n be the space of n complex variables z
=
(
z
1
,
…
,
z
n
)
′ with the Euclidean inner product 〈
z
,
w
〉
=
∑
i
=
1
n
z
i
w
¯
i and the Euclidean norm ∥
z
∥
=
〈
z
,
z
〉
1
/
2, where z
,
w
∈
ℂ
n and the symbol “′” means transpose. The unit ball of ℂ
n is the set ℬ
n
=
{
z
∈
ℂ
n
:
∥
z
∥
<
1
}, and the unit sphere is denoted by ∂
ℬ
n
=
{
z
∈
ℂ
n
:
∥
z
∥
=
1
}. In the case of one complex variable, ℬ
1 is the unit disk, usually denoted by D. Let Ω be a domain in ℂ
n. Denote H
(
Ω
) by the space of all holomorphic mappings from Ω into ℂ
n. A mapping f
∈
H
(
ℬ
n
) is called normalized if f
(
0
)
=
0 and J
f
(
0
)
=
I
n, where J
f
(
0
) is the complex Jacobian matrix of f at the origin and I
n is the identity operator on ℂ
n. A mapping f
∈
H
(
ℬ
n
) is said to be locally biholomorphic if det
J
f
(
z
)
≠
0 for every z
∈
ℬ
n. A normalized mapping f
∈
H
(
ℬ
n
) is said to be convex if λ
ω
1
+
(
1
-
λ
)
ω
2
∈
f
(
ℬ
n
) for arbitrary ω
1
,
ω
2
∈
f
(
ℬ
n
) and 0
⩽
λ
⩽
1. A normalized mapping f
∈
H
(
ℬ
n
) is said to be starlike with respect to the origin if λ
f
(
ℬ
n
)
⊂
f
(
ℬ
n
)
,
0
⩽
λ
⩽
1. A normalized mapping f
∈
H
(
ℬ
n
) is said to be ɛ starlike if there exists a positive number ɛ, 0
⩽
ɛ
⩽
1, such that f
(
ℬ
n
) is starlike with respect to every point in ɛ
f
(
ℬ
n
).A domainΩ is called a Reinhardt domain if (
e
i
θ
1
z
1
,
e
i
θ
2
z
2
,
…
,
e
i
θ
n
z
n
)
′
∈
Ω holds for any z
=
(
z
1
,
z
2
,
…
,
z
n
)
′
∈
Ω and θ
1
,
θ
2
,
…
,
θ
n
∈
ℝ. A domain Ω is called a circular domain if e
i
θ
z
∈
Ω holds for any z
∈
Ω and θ
∈
ℝ. The Minkowski functional ρ
(
z
) of the Reinhardt domain (1.6)
Ω
n
,
p
2
,
…
,
p
n
=
{
z
∈
C
n
:
|
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
j
<
1
}
,
p
j
⩾
1
,
j
=
2
,
…
,
n
is defined as (1.7)
ρ
(
z
)
=
inf
⁡
{
t
>
0
,
z
t
∈
Ω
n
,
p
2
,
…
,
p
n
}
,
z
∈
C
n
.
Then, the Minkowski functional ρ
(
z
) is a norm of ℂ
n and Ω
n
,
p
2
,
⋯
,
p
n is the unit ball in the Banach space ℂ
n with respect to this norm. The Minkowski functional ρ
(
z
) is C
1 on Ω
¯
n
,
p
2
,
…
,
p
n except for a lower-dimensional manifold Ω
0. Moreover, we give the following properties of the Minkowski functional ρ
(
z
) (see [16]): (1.8)
2
∂
ρ
∂
z
(
z
)
z
=
ρ
(
z
)
,
∀
z
∈
C
n
∖
Ω
0
,
2
∂
ρ
∂
z
(
z
)
z
=
1
,
∀
z
∈
∂
Ω
n
,
p
2
,
…
,
p
n
∖
Ω
0
,
∂
ρ
∂
z
(
λ
z
)
=
∂
ρ
∂
z
(
z
)
,
∀
λ
∈
[
0
,
∞
)
,
z
∈
C
n
∖
Ω
0
,
∂
ρ
∂
z
(
e
i
θ
z
)
=
e
-
i
θ
∂
ρ
∂
z
(
z
)
,
∀
z
∈
C
n
∖
Ω
0
,
θ
∈
R
.Definition 1.1 (see [17]).
Suppose thatΩ is a bounded starlike circular domain in ℂ
n. Its Minkowski functional ρ
(
z
) is C
1 except for a lower-dimensional manifold. Let 0
⩽
α
<
1. We say that a normalized locally biholomorphic mapping f
∈
H
(
Ω
) is an almost starlike mapping of order α if the following condition holds:
(1.9)
R
2
ρ
(
z
)
∂
ρ
∂
z
(
z
)
J
f
-
1
(
z
)
f
(
z
)
⩾
α
,
z
∈
Ω
∖
{
0
}
.
When Ω
=
ℬ
n, its Minkowski functional ρ
(
z
)
=
∥
z
∥, the above inequality becomes
(1.10)
R
z
′
¯
J
f
-
1
(
z
)
f
(
z
)
⩾
α
‖
z
‖
2
,
z
∈
B
n
.
In particular, whenα
=
0, f reduces to a starlike mapping on Ω.Definition 1.2 (see [18]).
Suppose thatΩ
∈
ℂ
n is a bounded starlike circular domain. Its Minkowski functional ρ
(
z
) is C
1 except for a lower-dimensional manifold. Let 0
<
α
<
1. We say that a normalized locally biholomorphic mapping f
∈
H
(
Ω
) is a starlike mapping of order α if the following condition holds:
(1.11)
|
2
ρ
(
z
)
∂
ρ
∂
z
(
z
)
J
f
-
1
(
z
)
f
(
z
)
-
1
2
α
|
<
1
2
α
,
z
∈
Ω
∖
{
0
}
.
When Ω
=
ℬ
n, the above inequality reduces to
(1.12)
|
1
‖
z
‖
2
z
′
¯
J
f
-
1
(
z
)
f
(
z
)
-
1
2
α
|
<
1
2
α
,
z
∈
B
n
∖
{
0
}
.


## 2. Some Lemmas

In order to prove the main results, we need the following three lemmas.Lemma 2.1 (see [19]).
Letp be a holomorphic function on D. If  ℜ
p
(
z
)
>
0 and p
(
0
)
>
0, then
(2.1)
|
p
′
(
z
)
|
⩽
2
R
p
(
z
)
1
-
|
z
|
2
.Lemma 2.2 (see [19]).
Letf be a normalized biholomorphic function on D. Then,
(2.2)
|
(
1
-
|
z
|
2
)
f
′
′
(
z
)
f
′
(
z
)
-
2
z
¯
|
⩽
4
holds for all z
∈
D.Lemma 2.3 (see [20]).
Ifρ
(
z
) is a Minkowski function of the domain Ω
n
,
p
2
,
…
,
p
n, z
≠
0, then
(2.3)
∂
ρ
∂
z
1
(
z
)
=
z
¯
1
ρ
(
z
)
[
2
|
z
1
/
ρ
(
z
)
|
2
+
∑
j
=
2
n
p
j
|
z
j
/
ρ
(
z
)
|
p
j
]
,
∂
ρ
∂
z
j
(
z
)
=
p
j
z
¯
j
|
z
j
/
ρ
(
z
)
|
p
j
-
2
2
ρ
(
z
)
[
2
|
z
1
/
ρ
(
z
)
|
2
+
∑
j
=
2
n
p
j
|
z
j
/
ρ
(
z
)
|
p
j
]
,
j
=
2
,
…
,
n
.


## 3. Main Results

Theorem 3.1.
Let0
⩽
α
<
1 and let f be an almost starlike function of order α on the unit disc D. If complex numbers a
j satisfy the condition |
a
j
|
⩽
(
1
-
α
)
/
4
,
j
=
2
,
…
,
n, then
(3.1)
F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
is an almost starlike mapping of order α on the domain Ω
n
,
p
2
,
…
,
p
n, where p
j are positive integer and p
j
⩾
2; the branches are chosen such that (
f
′
(
z
1
)
)
1
/
p
j
∣
z
1
=
0
=
1.Proof.
By the definition of almost starlike mapping of orderα, we need only to prove that the following inequality:
(3.2)
R
2
ρ
(
z
)
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
⩾
α
holds for all z
∈
Ω
n
,
p
2
,
…
,
p
n and z
≠
0.
The case ofz
0
=
0 is trivial. So, we need only consider that z
=
(
z
1
,
z
0
′
)
′
∈
Ω
¯
n
,
p
2
,
…
,
p
n
,
z
0
≠
0.  Let z
=
ζ
u
=
|
ζ
|
e
i
θ
u
,
u
∈
∂
Ω
n
,
p
2
,
…
,
p
n, and ζ
∈
D
¯
∖
{
0
}, then we have
(3.3)
R
2
ρ
(
z
)
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
⩾
α
⟺
R
2
ρ
(
|
ζ
|
e
i
θ
u
)
∂
ρ
∂
z
(
|
ζ
|
e
i
θ
u
)
J
F
-
1
(
|
ζ
|
e
i
θ
u
)
F
(
|
ζ
|
e
i
θ
u
)
⩾
α
⟺
R
2
|
ζ
|
e
-
i
θ
∂
ρ
∂
z
(
u
)
J
F
-
1
(
|
ζ
|
e
i
θ
u
)
F
(
|
ζ
|
e
i
θ
u
)
⩾
α
⟺
R
2
∂
ρ
∂
z
(
u
)
J
F
-
1
(
ζ
u
)
F
(
ζ
u
)
ζ
⩾
α
.
For a fixed u, the expression ℜ
(
2
∂
ρ
/
∂
z
)
(
u
)
(
J
F
-
1
(
ζ
u
)
F
(
ζ
u
)
/
ζ
)
-
α is the real part of a holomorphic function with respect to ζ, so it is a harmonic function. By the minimum of harmonic function principle, we know that it attains its minimum on |
ζ
|
=
1, so we need only to prove for all z
∈
∂
Ω
n
,
p
2
,
…
,
p
n and z
0
≠
0. Hence, ρ
(
z
)
=
1 and inequality (3.2) becomes
(3.4)
R
2
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
⩾
α
,
z
∈
∂
Ω
n
,
p
2
,
…
,
p
n
,
z
0
≠
0
.
In the following, we will prove inequality (3.4).
Since(3.5)
F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
,
we have(3.6)
J
F
(
z
)
=
(
f
′
(
z
1
)
+
f
′′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
a
2
p
2
f
′
(
z
1
)
z
2
p
2
-
1
⋯
a
n
p
n
f
′
(
z
1
)
z
n
p
n
-
1
1
p
2
(
f
′
(
z
1
)
)
(
1
/
p
2
)
-
1
f
′′
(
z
1
)
z
2
(
f
′
(
z
1
)
)
1
/
p
2
⋯
0
⋯
⋯
⋯
⋯
1
p
n
(
f
′
(
z
1
)
)
(
1
/
p
n
)
-
1
f
′′
(
z
1
)
z
n
0
⋯
(
f
′
(
z
1
)
)
1
/
p
n
)
.
Suppose that J
F
-
1
(
z
)
F
(
z
)
=
A
=
(
x
1
,
x
2
,
…
,
x
n
)
′, then F
(
z
)
=
J
F
(
z
)
A; that is,
(3.7)
x
1
[
f
′
(
z
1
)
+
f
′′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
]
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
p
j
x
j
z
j
p
j
-
1
=
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
x
1
f
′′
(
z
1
)
p
2
f
′
(
z
1
)
z
2
+
x
2
=
z
2
,
⋮
x
1
f
′′
(
z
1
)
p
n
f
′
(
z
n
)
z
n
+
x
n
=
z
n
.
Some computation shows that
(3.8)
x
1
=
f
(
z
1
)
f
′
(
z
1
)
-
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
,
x
2
=
[
1
-
f
(
z
1
)
f
′′
(
z
1
)
p
2
(
f
′
(
z
1
)
)
2
+
f
′′
(
z
1
)
p
2
f
′
(
z
1
)
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
]
z
2
,
⋮
x
n
=
[
1
-
f
(
z
1
)
f
′′
(
z
1
)
p
n
(
f
′
(
z
1
)
)
2
+
f
′′
(
z
1
)
p
n
f
′
(
z
1
)
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
]
z
n
.
From Lemma 2.3, we obtain
(3.9)
∂
ρ
∂
z
1
(
z
)
=
z
¯
1
ρ
(
z
)
[
2
|
z
1
/
ρ
(
z
)
|
2
+
∑
j
=
2
n
p
j
|
z
j
/
ρ
(
z
)
|
p
j
]
=
z
¯
1
2
|
z
1
|
2
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
,
∂
ρ
∂
z
j
(
z
)
=
p
j
z
¯
j
|
z
j
|
p
j
-
2
2
ρ
(
z
)
[
2
|
z
1
/
ρ
(
z
)
|
2
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
]
=
p
j
z
¯
j
|
z
j
|
p
j
-
2
2
[
2
|
z
1
|
2
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
]
.
In terms of (3.8) and (3.9), we obtain
(3.10)
2
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
=
G
(
z
)
2
|
z
1
|
2
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
,
where
(3.11)
G
(
z
)
=
2
z
¯
1
[
f
(
z
1
)
f
′
(
z
1
)
-
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
]
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
[
1
-
f
(
z
1
)
f
′′
(
z
1
)
p
j
(
f
′
(
z
1
)
)
2
+
f
′′
(
z
1
)
p
j
f
′
(
z
1
)
∑
k
=
2
n
a
k
(
p
k
-
1
)
z
k
p
k
]
=
2
|
z
1
|
2
f
(
z
1
)
z
1
f
′
(
z
1
)
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
[
1
-
f
(
z
1
)
f
′′
(
z
1
)
p
j
(
f
′
(
z
1
)
)
2
]
+
∑
k
=
2
n
a
k
(
p
k
-
1
)
z
k
p
k
[
f
′′
(
z
1
)
f
′
(
z
1
)
∑
j
=
2
n
|
z
j
|
p
j
-
2
z
¯
1
]
.
By making use of the equality |
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
j
=
1, we then get
(3.12)
G
(
z
)
=
2
|
z
1
|
2
f
(
z
1
)
z
1
f
′
(
z
1
)
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
[
1
-
f
(
z
1
)
f
′′
(
z
1
)
p
j
(
f
′
(
z
1
)
)
2
]
+
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
[
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
1
¯
]
.
Let h
(
z
1
)
=
(
f
(
z
1
)
/
z
1
f
′
(
z
1
)
)
-
α. Notice that f is an almost starlike function of order α on the unit disc; hence, ℜ
h
(
z
1
)
>
0 and h
(
0
)
=
1
-
α
>
0. By Lemma 2.1, we can obtain that
(3.13)
|
h
′
(
z
)
|
⩽
2
R
h
(
z
)
1
-
|
z
|
2
.
Furthermore, we get
(3.14)
f
(
z
1
)
f
′′
(
z
1
)
[
f
′
(
z
1
)
]
2
=
1
-
α
-
h
(
z
1
)
-
z
1
h
′
(
z
1
)
.
Substituting (3.14) into (3.12), we have
(3.15)
G
(
z
)
=
2
|
z
1
|
2
(
h
(
z
1
)
+
α
)
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
(
1
-
1
p
j
+
α
p
j
+
h
(
z
1
)
p
j
+
z
1
p
j
h
′
(
z
1
)
)
+
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
[
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
]
=
h
(
z
1
)
(
2
|
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
j
)
+
2
α
|
z
1
|
2
+
∑
j
=
2
n
(
p
j
-
1
+
α
)
|
z
j
|
p
j
+
∑
j
=
2
n
z
1
|
z
j
|
p
j
h
′
(
z
1
)
+
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
[
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
]
.
Hence,
(3.16)
R
G
(
z
)
⩾
(
2
|
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
j
)
R
h
(
z
1
)
+
2
α
|
z
1
|
2
+
∑
j
=
2
n
(
p
j
-
1
+
α
)
|
z
j
|
p
j
-
∑
j
=
2
n
|
z
j
|
p
j
|
z
1
h
′
(
z
1
)
|
-
∑
j
=
2
n
|
a
j
|
(
p
j
-
1
)
|
z
j
|
p
j
|
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
|
.
By Lemma 2.2 and (3.13), we can get that
(3.17)
R
G
(
z
)
⩾
(
1
+
|
z
1
|
2
)
R
h
(
z
1
)
+
2
α
|
z
1
|
2
+
∑
j
=
2
n
(
p
j
-
1
+
α
)
|
z
j
|
p
j
-
(
1
-
|
z
1
|
2
)
2
|
z
1
|
R
h
(
z
1
)
1
-
|
z
1
|
2
-
4
∑
j
=
2
n
|
a
j
|
(
p
j
-
1
)
|
z
j
|
p
j
=
(
1
+
|
z
1
|
2
)
R
h
(
z
1
)
+
2
α
|
z
1
|
2
+
∑
j
=
2
n
(
p
j
-
1
+
α
)
|
z
j
|
p
j
-
2
|
z
1
|
R
h
(
z
1
)
-
4
∑
j
=
2
n
|
a
j
|
(
p
j
-
1
)
|
z
j
|
p
j
=
(
1
-
|
z
1
|
)
2
R
h
(
z
1
)
+
2
α
|
z
1
|
2
+
∑
j
=
2
n
|
z
j
|
p
j
[
α
+
(
1
-
|
4
a
j
|
)
(
p
j
-
1
)
]
.
Hence, when |
a
j
|
⩽
(
1
-
α
)
/
4, we have
(3.18)
R
G
(
z
)
⩾
(
1
-
|
z
1
|
)
2
R
h
(
z
1
)
+
2
α
|
z
1
|
2
+
α
∑
j
=
2
n
p
j
|
z
j
|
p
j
⩾
2
α
|
z
1
|
2
+
α
∑
j
=
2
n
p
j
|
z
j
|
p
j
.
In terms of (3.10) and (3.18), we obtain
(3.19)
R
2
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
⩾
α
,
which completes the proof of Theorem 3.1.Remark 3.2.
Whena
2
=
a
3
=
⋯
=
a
n
=
0, the result of Theorem 3.1 has been obtained by Liu and Liu [14].Corollary 3.3.
Letf be a normalized biholomorphic starlike function on the unit disc D. If |
a
j
|
⩽
1
/
4
,
j
=
2
,
…
,
n, then
(3.20)
F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
is a normalized biholomorphic starlike mapping on the domain Ω
n
,
p
2
,
…
,
p
n, where p
j are positive integer and p
j
⩾
2; the branches are chosen such that (
f
′
(
z
1
)
)
1
/
p
j
∣
z
1
=
0
=
1.Theorem 3.4.
Let0
<
α
<
1 and let f be a starlike function of order α on the unit disc D. If complex numbers a
j satisfy the condition |
a
j
|
⩽
(
1
-
|
2
α
-
1
|
)
/
8
α
,
j
=
2
,
…
,
n, then
(3.21)
F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
is a starlike mapping of order α on the domain Ω
n
,
p
2
,
…
,
p
n, where p
j are positive integer and p
j
⩾
1; the branches are chosen such that (
f
′
(
z
1
)
)
1
/
p
j
∣
z
1
=
0
=
1.Proof.
By the definition of starlike mapping of orderα, we need only to prove that the following inequality:
(3.22)
|
2
ρ
(
z
)
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
-
1
2
α
|
<
1
2
α
holds for all z
∈
Ω
n
,
p
2
,
…
,
p
n and z
0
≠
0.
Similar to the proof of Theorem3.1, we need only to prove that (3.22) holds for ρ
(
z
)
=
1 and z
0
≠
0 according to the maximum modulus theorem for analytic functions. So, it is suffice to show that
(3.23)
|
2
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
-
1
2
α
|
<
1
2
α
.
From the proof of Theorem 3.1, we can get
(3.24)
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
=
1
2
[
2
|
z
1
|
2
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
]
{
2
|
z
1
|
2
f
(
z
1
)
z
1
f
′
(
z
1
)
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
[
1
-
f
(
z
1
)
f
′′
(
z
1
)
p
j
(
f
′
(
z
1
)
)
2
]
+
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
[
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
]
}
.
Hence,
(3.25)
2
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
-
1
2
α
=
H
(
z
)
2
α
[
2
|
z
1
|
2
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
]
,
where
(3.26)
H
(
z
)
=
2
|
z
1
|
2
[
2
α
f
(
z
1
)
z
1
f
′
(
z
1
)
-
1
]
+
2
α
∑
j
=
2
n
p
j
|
z
j
|
p
j
[
1
-
1
2
α
-
f
(
z
1
)
f
′′
(
z
1
)
p
j
(
f
′
(
z
1
)
)
2
]
+
2
α
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
[
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
]
.
Let h
(
z
1
)
=
2
α
(
f
(
z
1
)
/
z
1
f
′
(
z
1
)
)
-
1. Then, |
h
(
z
1
)
|
<
1 because f is a starlike function of order α on the unit disc D. By the Schwarz-Pick lemma, we obtain that
(3.27)
|
h
′
(
z
1
)
|
⩽
1
-
|
h
(
z
1
)
|
2
1
-
|
z
1
|
2
.
On the other hand, we can get
(3.28)
f
(
z
1
)
f
′′
(
z
1
)
[
f
′
(
z
1
)
]
2
=
1
-
1
2
α
-
h
(
z
1
)
2
α
-
z
1
h
′
(
z
1
)
2
α
.
Substituting (3.28) into (3.26), we have
(3.29)
H
(
z
)
=
2
|
z
1
|
2
[
2
α
f
(
z
1
)
z
1
f
′
(
z
1
)
-
1
]
+
2
α
∑
j
=
2
n
p
j
|
z
j
|
p
j
[
1
-
1
2
α
-
(
1
p
j
-
1
2
α
p
j
-
h
(
z
1
)
2
α
p
j
-
z
1
h
′
(
z
1
)
2
α
p
j
)
]
+
2
α
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
[
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
]
=
2
|
z
1
|
2
h
(
z
1
)
+
h
(
z
1
)
∑
j
=
2
n
|
z
j
|
p
j
+
∑
j
=
2
n
z
1
h
′
(
z
1
)
|
z
j
|
p
j
+
(
2
α
-
1
)
∑
j
=
2
n
(
p
j
-
1
)
|
z
j
|
p
j
+
2
α
∑
j
=
2
n
a
j
(
p
j
-
1
)
z
j
p
j
[
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
]
.
Hence,
(3.30)
|
H
(
z
)
|
⩽
(
1
+
|
z
1
|
2
)
|
h
(
z
1
)
|
+
∑
j
=
2
n
|
z
1
h
′
(
z
1
)
|
|
z
j
|
p
j
+
|
2
α
-
1
|
∑
j
=
2
n
(
p
j
-
1
)
|
z
j
|
p
j
+
2
α
∑
j
=
2
n
|
a
j
|
(
p
j
-
1
)
|
z
j
|
p
j
|
f
′′
(
z
1
)
f
′
(
z
1
)
(
1
-
|
z
1
|
2
)
-
2
z
¯
1
|
.
By Lemma 2.2 and (3.27), we have
(3.31)
|
H
(
z
)
|
⩽
(
1
+
|
z
1
|
2
)
|
h
(
z
1
)
|
+
|
z
1
|
1
-
|
h
(
z
1
)
|
2
1
-
|
z
1
|
2
(
1
-
|
z
1
|
2
)
+
|
2
α
-
1
|
∑
j
=
2
n
(
p
j
-
1
)
|
z
j
|
p
j
+
8
α
∑
j
=
2
n
|
a
j
|
(
p
j
-
1
)
|
z
j
|
p
j
⩽
(
1
+
|
z
1
|
2
)
(
|
h
(
z
1
)
|
-
1
)
+
(
1
+
|
z
1
|
2
)
+
2
|
z
1
|
(
1
-
|
h
(
z
1
)
|
)
+
|
2
α
-
1
|
∑
j
=
2
n
(
p
j
-
1
)
|
z
j
|
p
j
+
8
α
∑
j
=
2
n
|
a
j
|
(
p
j
-
1
)
|
z
j
|
p
j
⩽
(
1
+
|
z
1
|
2
)
+
(
|
h
(
z
1
)
|
-
1
)
(
1
-
|
z
1
|
)
2
+
∑
j
=
2
n
(
|
2
α
-
1
|
+
8
α
|
a
j
|
)
(
p
j
-
1
)
|
z
j
|
p
j
.
If |
a
j
|
⩽
(
1
-
|
2
α
-
1
|
)
/
8
α, then we obtain
(3.32)
|
H
(
z
)
|
<
1
+
|
z
1
|
2
+
(
|
2
α
-
1
|
+
8
α
1
-
|
2
α
-
1
|
8
α
)
∑
j
=
2
n
(
p
j
-
1
)
|
z
j
|
p
j
≤
1
+
|
z
1
|
2
+
∑
j
=
2
n
(
p
j
-
1
)
|
z
j
|
p
j
=
2
|
z
1
|
2
+
∑
j
=
2
n
p
j
|
z
j
|
p
j
.
The equality (3.25) and (3.32) show that
(3.33)
|
2
∂
ρ
∂
z
(
z
)
J
F
-
1
(
z
)
F
(
z
)
-
1
2
α
|
<
1
2
α
,
which completes the proof of Theorem 3.4.


## 4. Problem

In 2003, Gong and Liu [13] proved that the Roper-Suffridge extension operator (4.1)
Φ
n
,
(
1
/
p
2
)
,
…
,
(
1
/
p
n
)
(
f
)
(
z
)
=
(
f
(
z
1
)
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
does preserve convexity on Ω
n
,
p
2
,
…
,
p
n, which solved the open problem posed by Graham and Kohr [2]. Naturally, we will propose the following problem on the new Roper-Suffridge extension operator.Problem 1.
Letp
j be positive integer. Under what conditions for a
j such that if f is a convex function in the disc D, then the mapping defined by the new Roper-Suffridge extension operator
(4.2)
F
(
z
)
=
(
f
(
z
1
)
+
f
′
(
z
1
)
∑
j
=
2
n
a
j
z
j
p
j
,
(
f
′
(
z
1
)
)
1
/
p
2
z
2
,
…
,
(
f
′
(
z
1
)
)
1
/
p
n
z
n
)
′
is a convex mapping in the Reinhardt domain Ω
n
,
p
2
,
…
,
p
n?



---

*Source: 865496-2011-11-22.xml*