# Biholomorphic Convex Mappings of Order $\alpha$ on the Unit Ball in Hilbert
Spaces

Ming-Sheng Liu
School of Mathematical Sciences, South China Normal University,
Guangzhou, 510631 Guangdong, P.R.China
<EMAIL_ADDRESS>
Yu-Can Zhu
Department of Mathematics, Fuzhou University, Fuzhou,350002 Fujian, P.R.China
<EMAIL_ADDRESS>Corresponding author.

Abstract. In this paper, we first introduce the concept of biholomorphic
convex mapping of order $\alpha$ on the unit ball $B$ in a complex Hilbert
space $X$. Next we provide some sufficient conditions that a locally
biholomorphic mapping $f$ is a biholomorphic convex mapping of order $\alpha$
and give an Alexander’s theorem between the subclass of convex mappings and
the subclass of starlike mappings on $B$ in Hilbert space. We also obtain the
order of starlikeness of biholomorphic convex mappings of order $\alpha$ on
$B$ in Hilbert spaces. Finally, we construct some concrete examples of
biholomorphic convex mappings of order $\alpha$ on $B$ in Hilbert spaces by
means of a linear operator.

Keywords. Biholomorphic convex mapping; Biholomorphic starlike mapping;
Locally biholomorphic mapping; biholomorphic convex mapping of order $\alpha$.

2000 MR Subject Classification 32H02, 30C45

##  1  Introduction

The holomorphic functions of one complex variable which map the unit disk onto
starlike or convex domains have been extensively studied. These functions are
easily characterized by simple analytic or geometric conditions. In moving to
higher dimensions, several difficulties arise. In the case of one complex
variable, the following well known theorems had been established(cf. [3]).

Theorem A Suppose that $\alpha\in[0,1)$ and
$f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n}$ is a holomorphic function on the
unit disk $U=\\{z\in\mathbb{C}:|z|<1\\}$ in the complex plane $\mathbb{C}$.

(1) If $\sum\limits_{n=2}^{\infty}n^{2}|a_{n}|\leq 1$, then $f$ is a convex
function in the unit disk $U$.

(2) If $\sum\limits_{n=2}^{\infty}n(n-\alpha)|a_{n}|\leq 1-\alpha$, then $f$
is a convex function of order $\alpha$ in $U$.

Theorem B Suppose that $\alpha\in[0,1)$ and
$f(z)=z+\sum\limits_{n=2}^{\infty}a_{n}z^{n}$ is a holomorphic function on the
unit disk $U$ in the complex plane $\mathbb{C}$.

(1) If $\sum\limits_{n=2}^{\infty}n|a_{n}|\leq 1$, then $f$ is a starlike
function in the unit disk $U$.

(2) If $\sum\limits_{n=2}^{\infty}(n-\alpha)|a_{n}|\leq 1-\alpha$, then $f$ is
a starlike function of order $\alpha$ in $U$.

Roper and Suffridge established the n-dimensional version of Theorem A(1), and
we[9] established the n-dimensional version of Theorem B(1)(2) as follows.

Theorem C(Roper and Suffridge[12]) Let
$f(z)=z+\sum\limits_{k=2}^{\infty}A_{k}(z^{k})$ be a holomorphic mapping on
the unit ball $B_{2}^{n}$. If $\sum\limits_{k=2}^{\infty}k^{2}\|A_{k}\|\leq
1$, then $f(z)$ is a convex mapping on $B_{2}^{n}$.

Theorem D(Liu and Zhu[9]) Suppose that $\alpha\in[0,1)$. Let
$f(z)=z+\sum\limits_{k=2}^{\infty}A_{k}(z^{k})$ be a holomorphic mapping on
the unit ball $B$ in Hilbert space. If
$\sum\limits_{k=2}^{\infty}(k-\alpha)\|A_{k}\|\leq A(\alpha)$, where
$A(\alpha)$ is defined by

$A(\alpha)=\left\\{\begin{array}[]{lll}\frac{(2-\alpha)\sqrt{1-2\alpha}}{\sqrt{5-2\alpha}},&0\leq\alpha\leq\frac{1}{4},\\\
\frac{(2-\alpha)(1-\alpha)}{2+\alpha},&\frac{1}{4}<\alpha\leq\frac{2}{5},\\\
\alpha,&\frac{2}{5}<\alpha<\frac{1}{2},\\\
1-\alpha,&\frac{1}{2}\leq\alpha<1.\end{array}\right.$

Then $f(z)$ is a starlike mapping of order $\alpha$ on $B$ in Hilbert space.

A problem is naturally proposed: can we establish the n-dimensional version
for Theorem A(2)?

##  2 Preliminaries

In order to state and prove our main results, we recall some definitions and
notations. Suppose that $X$ is a complex Hilbert space with inner product
$\langle\cdot,\cdot\rangle$ and norm
$\|\cdot\|=\sqrt{\langle\cdot,\cdot\rangle}$, and $G$ is a domain in $X$. A
mapping $f:G\to X$ is said to be holomorphic on $G$, if for any $z\in G$,
there exists a linear operator $Df(z):X\to X$ such that

$\lim_{h\to 0}\frac{\|f(z+h)-f(z)-Df(z)h\|}{\|h\|}=0.$

The linear operator $Df(z)$ is called the Fr$\acute{e}$chet derivative of $f$
at $z\in G$.

If $f$ is holomorphic on $G$, then for every $k=1,2,\cdots$, and every
$z_{0}\in G$, there is a bounded symmetric $k-$ linear operator
$D^{k}f(z_{0}):X\times X\times\cdots\times X\to X$ such that

$f(z)=\sum\limits_{k=0}^{+\infty}\frac{1}{k!}D^{k}f(z_{0})((z-z_{0})^{k})$

for all $z$ in some neighborhood of $z_{0}$, where
$D^{0}f(z_{0})((z-z_{0})^{0})=f(z_{0})$ and

$D^{k}f(z_{0})((z-z_{0})^{k})=D^{k}f(z_{0})(z-z_{0},z-z_{0},\cdots,z-z_{0})$

for $k\geq 1$.

A mapping $f:G\to X$ is said to be biholomorphic on $G$ if $f$ is holomorphic
on $G$, $f(G)$ is a domain, and the inverse $f^{-1}$ exists and is holomorphic
on $f(G)$. A mapping $f:G\to X$ is said to be locally biholomorphic on $G$, if
for any $z\in G$, there exists a neighborhood $U$ of $z$ such that $f|_{U}$ is
biholomorphic on $U$. Then $f$ is locally biholomorphic on $G$ if and only if
its Fréchet derivative $Df(z)$ has a bounded inverse at each $z\in G$.

The unit ball in $X$ is $B=\\{z\in X:\|z\|<1\\}$. Let $N(B)$ denote the class
of all local biholomorphic mappings $f:B\rightarrow X$ such that
$f(0)=0,Df(0)=I$, where $I$ is the identity operator in $X$. A biholomorphic
mapping $f:B\to X$ is called a biholomorphic starlike mapping if $tf(B)\subset
f(B)$ for $0\leq t\leq 1$ with $f(0)=0$. Let $S^{\ast}(B)$ be the subclass of
$N(B)$ consisting of starlike mappings on $B$. Then $f\in S^{\ast}(B)$ if and
only if $f$ is locally biholomorphic such that

$\mbox{Re}\langle Df(z)^{-1}f(z),z\rangle>0$

for all $z\in B\backslash\\{0\\}$(cf.[1, 3, 6]).

A mapping $f\in N(B)$ is called starlike of order $\alpha\in(0,1)$ on $B$ if

$\bigg{|}\langle
Df(z)^{-1}f(z),z\rangle-\frac{1}{2\alpha}\|z\|^{2}\bigg{|}<\frac{1}{2\alpha}\|z\|^{2}\quad\mbox{for
all }\quad z\in B\backslash\\{0\\},$

Let $S^{\ast}(B,\alpha)$ denote the class of starlike mappings of order
$\alpha$ on $B$ for $0<\alpha<1$ and let $S^{\ast}(B,0)\equiv S^{\ast}(B)$. It
is obvious that $S^{\ast}(B,\alpha)\subset S^{\ast}(B)$ for $0\leq\alpha<1$.

A biholomorphic mapping $f:B\to X$ is said biholomorphic convex mapping if

$(1-t)f(z_{1})+tf(z_{2})\in f(B)$

for all $z_{1},z_{2}\in B$ and $0\leq t\leq 1$. The class of all biholomorphic
convex mappings on $B$ with $f(0)=0,Df(0)=I$ is denoted by $K(B)$.

We[16] obtained a necessary and sufficient condition that a locally
biholomorphic mapping was a biholomorphic convex mapping on B in the Hilbert
space $X$ as follows.

Theorem A(Zhu and Liu[16]). Let $f:B\to X$ be a locally biholomorphic mapping.
Then $f$ is a biholomorphic convex mapping on $B$ if and only if

$\displaystyle\|x\|^{2}-\mbox{Re}\Big{\langle}Df(z)^{-1}D^{2}f(z)(x,x),z\Big{\rangle}>0$
(2.1)

for $z\in B\setminus\\{0\\}$ and $x\in X\setminus\\{0\\}$ with
$\mbox{Re}\langle x,z\rangle=0$.

Remark 1. Theorem A had improved a result of Hamada and Kohr[4]. Setting
$X=C^{n}$ in Theorem A, we also obtain Theorem 2 in [2].

Corollary 1. Let $0\leq\alpha<1$ and $f:B\to X$ be a locally biholomorphic
mapping. If $f$ satisfies the following inequality

$\displaystyle\|x\|^{2}-\mbox{Re}\Big{\langle}Df(z)^{-1}D^{2}f(z)(x,x),z\Big{\rangle}>\alpha\cdot\|x\|^{2}$
(2.2)

for $z\in B\setminus\\{0\\}$ and $x\in X\setminus\\{0\\}$ with
$\mbox{Re}\langle x,z\rangle=0$. Then $f$ is a biholomorphic convex mapping on
$B$.

We call such mapping $f$, which satisfies the hypothesis of Corollary 1, a
biholomorphic convex mapping of order $\alpha$ on $B$. We let $K(B,\alpha)$
denote the subclass of all biholomorphic convex mappings of order $\alpha$ on
$B$ with $f(0)=0,Df(0)=I$.

In this paper, we provide some sufficient conditions for biholomorphic convex
mapping of order $\alpha$ and an Alexander’s theorem between the subclass of
convex mappings and the subclass of starlike mappings on $B$ in Hilbert space.
We also obtain the order of starlikeness of biholomorphic convex mappings of
order $\alpha$ on $B$ in Hilbert spaces. Finally, we introduce a linear
operator in purpose to construct some concrete examples of biholomorphic
convex mappings on $B$ in Hilbert spaces. From these, we give some examples of
biholomorphic convex mappings on $B$ in Hilbert spaces.

##  3 Main Results

We first establish some sufficient conditions for biholomorphic convex mapping
of order $\alpha$ on $B$.

Theorem 1. Suppose that $0\leq\alpha<1$ and $f:B\to X$ is a locally
biholomorphic mapping. If $f$ satisfies

$\displaystyle\Big{\|}Df(z)^{-1}D^{2}f(z)(x,x)\Big{\|}\leq 1-\alpha$ (3.1)

for $z\in B\setminus\\{0\\}$ and $x\in X$ with $\|x\|=1$ and $\mbox{Re}\langle
x,z\rangle=0$, then $f$ is a biholomorphic convex mapping of order $\alpha$ on
$B$.

Proof. Since $f:B\to X$ is a locally biholomorphic mapping, for any $z\in
B\setminus\\{0\\}$ and $x\in X\setminus\\{0\\}$ with $\|x\|=1$ and
$\mbox{Re}\langle x,z\rangle=0$, we have

$\begin{array}[]{lll}\|x\|^{2}-\mbox{Re}\Big{\langle}Df(z)^{-1}D^{2}f(z)(x,x),z\Big{\rangle}&\geq&\|x\|^{2}-|\langle
Df(z)^{-1}D^{2}f(z)(x,x),z\rangle|\\\
&\geq&\|x\|^{2}-\|Df(z)^{-1}D^{2}f(z)(x,x)\|\|z\|\\\
&>&\|x\|^{2}-\|Df(z)^{-1}D^{2}f(z)(x,x)\|\\\
&=&\|x\|^{2}-\|Df(z)^{-1}D^{2}f(z)(\frac{x}{\|x\|},\frac{x}{\|x\|})\|\|x\|^{2}.\end{array}$

Notice that $\|\frac{x}{\|x\|}\|=1$, we conclude from (3.1) that

$\begin{array}[]{lll}\|x\|^{2}-\mbox{Re}\Big{\langle}Df(z)^{-1}D^{2}f(z)(x,x),z\Big{\rangle}&>&\|x\|^{2}-\|Df(z)^{-1}D^{2}f(z)(\frac{x}{\|x\|},\frac{x}{\|x\|})\|\|x\|^{2}\\\
&\geq&\alpha\cdot\|x\|^{2}.\end{array}$

Hence by Corollary 1, we obtain that $f\in K(B,\alpha)$, and the proof is
complete.

Corollary 2. Suppose that $0\leq\alpha<1$ and $f:B\to X$ is a locally
biholomorphic mapping with $\|Df(z)-I\|\leq c<1$ for each $z\in B$, where $I$
is the identity operator in $X$. If $f$ satisfies

$\|D^{2}f(z)(x,x)\|\leq(1-c)(1-\alpha)$

for all $x\in X$ with $\|x\|=1$ and $z\in B\setminus\\{0\\}$ such that
$\mbox{Re}\langle x,z\rangle=0$, then $f$ is a biholomorphic convex mapping of
order $\alpha$ on $B$.

Proof. Since $\|Df(z)-I\|\leq c<1$ for any $z\in B$, we obtain that
$Df(z)=I-(I-Df(z))$ is an invertible linear operator(see [13], P192), and

$\|Df(z)^{-1}\|\leq\frac{1}{1-\|I-Df(z)\|}\leq\frac{1}{1-c}$

for all $z\in B$.

Thus for any $x\in X$ with $\|x\|=1$ and $z\in B\setminus\\{0\\}$ such that
$\mbox{Re}\langle x,z\rangle=0$, by the hypothesis of Corollary 2, we have

$\displaystyle\|Df(z)^{-1}D^{2}f(z)(x,x)\|$ $\displaystyle\leq$
$\displaystyle\|Df(z)^{-1}\|\|D^{2}f(z)(x,x)\|$ $\displaystyle\leq$
$\displaystyle\frac{1}{1-c}\cdot(1-c)(1-\alpha)=1-\alpha.$

Hence by Theorem 1, we obtain that $f$ is a biholomorphic convex mapping of
order $\alpha$ on $B$, and the proof is complete.

Remark 2. Setting $\alpha=0$ in Theorem 1, we get Corollary 1 in [16]; Setting
$\alpha=0$ in Corollary 2, we get Corollary 2 in [16].

Theorem 2. Let $0\leq\alpha<1$ and
$f(z)=z+\sum\limits_{k=2}^{+\infty}A_{k}(z^{k}):B\to X$ be a holomorphic
mapping. If $f$ satisfies
$\displaystyle\sum\limits_{k=2}^{+\infty}k(k-\alpha)\|A_{k}\|\leq 1-\alpha$,
then $f\in K(B,\alpha)$.

Proof. By direct calculating the Fr$\acute{e}$chet derivatives of $f(z)$, we
obtain

$\displaystyle Df(z)=I+\sum\limits_{k=2}^{+\infty}kA_{k}(z^{k-1},\cdot),$
$\displaystyle
D^{2}f(z)(x,x)=\sum\limits_{k=2}^{+\infty}k(k-1)A_{k}(z^{k-2},x^{2})$

and

$\|Df(z)-I\|\leq\sum\limits_{k=2}^{+\infty}k\|A_{k}\|\leq\frac{1}{2-\alpha}\sum\limits_{k=2}^{+\infty}k(k-\alpha)\|A_{k}\|\leq\frac{1-\alpha}{2-\alpha}<1$

for $z\in B,\,x\in X$. Hence we obtain that $Df(z)=I-(I-Df(z))$ is an
invertible linear operator(see [13],P192), and

$\displaystyle\|D^{2}f(z)(x,x)\|$ $\displaystyle\leq$
$\displaystyle\sum\limits_{k=2}^{+\infty}(k^{2}-k)\|A_{k}\|\|z\|^{k-2}\|\|x\|^{2}$
$\displaystyle\leq$ $\displaystyle
1-\alpha+\alpha\sum\limits_{k=2}^{+\infty}k\|A_{k}\|-\sum\limits_{k=2}^{+\infty}k\|A_{k}\|$
$\displaystyle=$
$\displaystyle(1-\sum\limits_{k=2}^{+\infty}k\|A_{k}\|)(1-\alpha)$

for $z\in B\setminus\\{0\\}$ and $\|x\|=1$ with $\mbox{Re}\langle
x,z\rangle=0$. By Corollary 2 for $c=\sum\limits_{k=2}^{+\infty}k\|A_{k}\|$,
we obtain that $f\in K(B,\alpha)$, and the proof is complete.

Remark 3. Setting $X=\mathbb{C}^{n},\alpha=0$ in Theorem 2, we may obtain
Theorem 2.1 in [12]. Our proof is more simple than theirs. Setting
$X=\mathbb{C}$ in Theorem 2, we get Theorem A(2).

Example 1. Let $0\leq\alpha<1$ and $A$ be a symmetric bilinear operator from
$X\times X$ to $X$ with $\|A\|\leq\frac{1-\alpha}{4-2\alpha}$. If we let
$f(z)=z+A(z,z)$, then $f\in K(B,\alpha)$.

Proof. Some straightforward computations yield the relations

$Df(z)=I+2A(z,\cdot),\quad D^{2}f(z)(x,y)=2A(x,y)$

for $z\in B,x,y\in X$. It implies

$Df(0)=I,\quad D^{2}f(0)(\cdot,\cdot)=2A(\cdot,\cdot)\quad\mbox{and}\quad
D^{k}f(0)=0$

for $k=3,4,\cdots$. Hence we obtain

$\sum\limits_{k=2}^{+\infty}\frac{k(k-\alpha)\|D^{k}f(0)\|}{k!}=(2-\alpha)\|D^{2}f(0)\|=2(2-\alpha)\|A\|\leq
1-\alpha.$

By Theorem 2, we conclude that $f\in K(B,\alpha)$, and the proof is complete.

Example 2. Let $0\leq\alpha<1$, $0<|a|\leq 1/2$ and $u\in X$ with $\|u\|=1$.
Then

$f(z)=z+a\langle z,u\rangle^{2}u\in
K(B,\alpha)\Longleftrightarrow|a|\leq\frac{1-\alpha}{4-2\alpha}.$

Proof. Let $c=1-\frac{2|a|}{1-\alpha}$. If
$0<|a|\leq\frac{1-\alpha}{4-2\alpha}$, then we have
$\frac{1-\alpha}{2-\alpha}\leq c<1$ and $2|a|=(1-c)(1-\alpha)\leq c$. Short
computations yield the relations

$\displaystyle Df(z)=I+2a\langle z,u\rangle\langle\cdot,u\rangle u,\quad
D^{2}f(z)(x,x)=2a\langle x,u\rangle^{2}u.$ (3.2)

It implies

$\|Df(z)-I\|\leq 2|a|\|z\|<2|a|\leq c,\quad\|D^{2}f(z)(x,x)\|\leq
2|a|=(1-c)(1-\alpha)$

for all $x\in X$ with $\|x\|=1$ and $z\in B$ such that $\mbox{Re}\langle
x,z\rangle=0$.

By Corollary 2, we obtain that $f\in K(B,\alpha)$.

Conversely, we shall prove that $0<|a|\leq\frac{1-\alpha}{4-2\alpha}$ when
$f\in K(B,\alpha)$.

If not, then $|a|>\frac{1-\alpha}{4-2\alpha}$. Let $\theta=\arg
a,\,x=ie^{-i\theta}u$ and $z_{0}=-re^{-i\theta}u$ for
$\frac{1-\alpha}{(4-2\alpha)|a|}<r<1$, where $u\in X$ with $\|u\|=1$. Then
$\|x\|=1,z_{0}\in B\setminus\\{0\\}$ and $\mbox{Re}\langle
x,z_{0}\rangle=\mbox{Re}\\{-ir\\}=0$.

Some straightforward computations from (3.2) yield the relations

$\displaystyle Df(z_{0})^{-1}=I+\frac{2|a|r}{1-2|a|r}\langle\cdot,u\rangle
u,\quad D^{2}f(z_{0})(x,x)=-2|a|e^{-i\theta}u$ $\displaystyle
Df(z_{0})^{-1}D^{2}f(z_{0})(x,x)=-\frac{2|a|e^{-i\theta}}{1-2|a|r}u.$

Hence we obtain

$\|x\|^{2}-\mbox{Re}\Big{\langle}Df(z_{0})^{-1}D^{2}f(z_{0})(x,x),z_{0}\Big{\rangle}=\frac{1-4|a|r}{1-2|a|r}<\alpha.$

This contradicts (2.2), and the proof is complete.

Next, we provide an Alexander’s theorem between the subclass of convex
mappings and the subclass of starlike mappings on $B$ in Hilbert space. In the
case of one complex variable, Alexander’s theorem told us that $f(z)$ is a
convex function on the unit disc $U$ if and only if $zf^{\prime}(z)$ is a
starlike function on the unit disc $U$. This theorem is no longer true in
several complex variables(see [3]). However, we have the following Alexander’s
theorem.

Theorem 3(Alexander’s Theorem). Suppose that $0\leq\alpha<1$ and $A(\alpha)$
is defined by Theorem D. Let

$SK(B,\alpha)=\\{f(z):f(z)=z+\sum\limits_{k=2}^{\infty}A_{k}(z^{k})\in
H(B)\mbox{ such that }\sum\limits_{k=2}^{\infty}k(k-\alpha)\parallel
A_{k}\parallel\leq A(\alpha)\\},$

and

$SS^{*}(B,\alpha)=\\{f(z):f(z)=z+\sum\limits_{k=2}^{\infty}A_{k}(z^{k})\in
H(B)\mbox{ such that }\sum\limits_{k=2}^{\infty}(k-\alpha)\parallel
A_{k}\parallel\leq A(\alpha)\\}.$

Then $SK(B,\alpha)\subset K(B)$, $SS^{*}(B,\alpha)\subset S^{*}(B)$, and
$f(z)\in SK(B,\alpha)$ if and only if $Df(z)(z)\in SS^{*}(B,\alpha)$.

Notice that $A(\alpha)\leq 1-\alpha$ for $0\leq\alpha<1$, by applying Theorem
D and Theorem 2, we can prove this theorem easily.

Now we establish a result on the order of starlikeness of function class
$K(B,\alpha)$.

Theorem 4. Suppose that $0\leq\alpha<1$. Then $K(B,\alpha)\subset
S^{*}(B,\beta)$, where

$\beta=\beta(\alpha)=\frac{2\alpha-1+\sqrt{(2\alpha-1)^{2}+8}}{4}.$

In order to prove the above theorem, we need the following lemmas.

Lemma 1.([10]) Let $g(z)=a+a_{1}z+\cdots$ is analytic in $U$ and
$g(z)\not\equiv a$. If there exists $z_{0}\in U\backslash\\{0\\}$ such that
$|g(z_{0})|=\max\limits_{|z|\leq|z_{0}|}|g(z)|$, then there exists real number
$t\geq\frac{|g(z_{0})|-|a|}{|g(z_{0})|+|a|}$ such that
$z_{0}g^{{}^{\prime}}(z_{0})=t\ g(z_{0})$.

Lemma 2. Let $f\in N(B),\,0<\rho<1$. If there exists $z_{0}\in
B\backslash\\{0\\}$ such that

${\rm Re}\ \frac{\|z\|^{2}}{\langle
Df(z_{0})^{-1}f(z_{0}),z_{0}\rangle}=\rho,$

and ${\rm Re}\ \frac{\|z\|^{2}}{\langle Df(z)^{-1}f(z),z\rangle}\geq\rho$ for
all $\|z\|<\|z_{0}\|$. Then there exist real numbers $\theta,\ t,\,m$ such
that:

(i) $\langle h(z_{0}),z_{0}\rangle=\frac{\|z_{0}\|^{2}}{2\rho}\
(1+e^{i\theta})$, where $h(z)=Df(z)^{-1}f(z)$ ;

(ii) $\langle Dh(z_{0})(z_{0}),z_{0}\rangle=\frac{\|z_{0}\|^{2}}{2\rho}\
[1+(1+t)e^{i\theta}]$, where $t\geq\frac{1-|2\rho-1|}{1+|2\rho-1|}$ ;

(iii) $e^{i\theta}\overline{Dh(z_{0})}(z_{0})+e^{-i\theta}h(z_{0})=mz_{0}$,
where $m=\frac{2\cos\theta+2+t}{2\rho}$.

Proof. Since $h:B\to X$ is a holomorphic mapping and $h(0)=0,Dh(0)=I$, and

$\bigg{|}\langle
h(z_{0}),z_{0}\rangle-\frac{\|z_{0}\|^{2}}{2\rho}\bigg{|}=\frac{\|z_{0}\|^{2}}{2\rho},$
(3.3) $\bigg{|}\langle
h(z),z\rangle-\frac{\|z\|^{2}}{2\rho}\bigg{|}\leq\frac{\|z\|^{2}}{2\rho},\quad\|z\|<\|z_{0}\|.$
(3.4)

Let

$\psi(\xi)=\frac{2\rho}{\|\xi z_{0}\|^{2}}\langle h(\xi z_{0}),\xi
z_{0}\rangle-1=\frac{2\rho}{\|z_{0}\|^{2}}\langle\frac{h(\xi
z_{0})}{\xi},z_{0}\rangle-1,$

then $\psi(\xi)$ is analytic in $\overline{U}$ and $|\psi(\xi)|\leq
1=|\psi(1)|$ for $\xi\in\overline{U}$, $\psi(0)=2\rho-1$. By Lemma 1, there is
a real number $t\geq\frac{1-|2\rho-1|}{1+|2\rho-1|}$ such that
$\psi^{{}^{\prime}}(1)=t\psi(1)$.

Let $\psi(1)=e^{i\theta}$ for some real number $\theta$, then we obtain (i)
holds, and

$\psi^{{}^{\prime}}(1)=\frac{2\rho}{\|z_{0}\|^{2}}\langle
Dh(z_{0})(z_{0})-h(z_{0}),z_{0}\rangle=te^{i\theta},$

which implies (ii) holds.

From (3.3) and (3.4), we obtain that

${\rm Re}[e^{-i\theta}\langle
h(z),z\rangle]\leq\frac{\|z_{0}\|^{2}}{2\rho}(1+\cos\theta)={\rm
Re}[e^{-i\theta}\langle h(z_{0}),z_{0}\rangle],\quad\|z\|<\|z_{0}\|.$ (3.5)

Let $r=\|z_{0}\|,B(r)=\\{z\in X:\|z\|<r\\}$, then the tangent hyperplane of
$\partial B(r)$ at $z_{0}$ is

$T_{z_{0}}(\partial B(r))=\\{b\in X:\langle b,z_{0}\rangle=0\\}.$

For any tangent vector $a\in T_{z_{0}}(\partial B(r))$ with $\|a\|=1$, set
$\gamma(t)=\sqrt{1-t^{2}}\ z_{0}+t\ ra$, then $\|\gamma(t)\|=r$ for
$t\in(-1,1)$ and $\gamma(0)=z_{0},\gamma^{\prime}(0)=ra$. Let

$\varphi(t)={\rm Re}[e^{-i\theta}\langle h(\gamma(t)),\gamma(t)\rangle].$

From (3.5), we obtain $\varphi(t)\leq\varphi(0)$ for $t\in(-1,1)$, so that
$\varphi(0)=\max\limits_{t\in(-1,1)}\varphi(t)$. Hence

$\displaystyle 0=\varphi^{\prime}(0)$ $\displaystyle=$ $\displaystyle{\rm
Re}[e^{-i\theta}\langle Dh(z_{0})ra,z_{0}\rangle+e^{-i\theta}\langle
h(z_{0}),ra\rangle]=r{\rm Re}\langle v,a\rangle,$

where $v=e^{i\theta}\overline{Dh(z_{0})}(z_{0})+e^{-i\theta}h(z_{0})$. This
implies that $v$ is a normal vector of $\partial B(r)$ at $z_{0}$, thus there
exists a real number $m$ such that $v=mz_{0}$. Since

$\displaystyle m\|z_{0}\|^{2}$ $\displaystyle=$ $\displaystyle{\rm Re}\langle
z_{0},mz_{0}\rangle={\rm Re}\langle z_{0},v\rangle$ $\displaystyle=$
$\displaystyle{\rm Re}\ \langle
z_{0},e^{i\theta}\overline{Dh(z_{0})}(z_{0})\rangle+{\rm Re}\ \langle
z_{0},e^{-i\theta}h(z_{0})\rangle$ $\displaystyle=$ $\displaystyle{\rm Re}\
[e^{-i\theta}\langle Dh(z_{0})(z_{0}),z_{0}\rangle]+{\rm Re}\
\langle[e^{-i\theta}\langle h(z_{0}),z_{0}\rangle]$ $\displaystyle=$
$\displaystyle\frac{\|z_{0}\|^{2}}{2\rho}\
(\cos\theta+1+t)+\frac{\|z_{0}\|^{2}}{2\rho}\ (\cos\theta+1),$

we obtain that $m=\frac{2\cos\theta+2+t}{2\rho}$, and this completes the proof
of Lemma 2.

Proof of Theorem 4. Let $h(z)=[Df(z)]^{-1}f(z)$ with $f(z)\in K(B,\alpha)$,
and $g(z)=\frac{\|z\|^{2}}{\langle h(z),z\rangle}$, then
$g(0)=1>\beta=\beta(\alpha)$.

Suppose $f\not\in S^{\ast}(B,\beta)$, then by the continuity of $g(z)$, there
exists $z_{0}\in B\setminus\\{0\\}$ such that ${\rm Re}\ g(z_{0})=\beta$ and
${\rm Re}\ g(z)\geq\beta$ for all $\|z\|<\|z_{0}\|$. Thus it follows from
Lemma 2 that there exist real numbers $\theta$, $t\geq\frac{1-\beta}{\beta}$
and $m=\frac{2\cos\theta+2+t}{2\beta}$ such that (i)-(iii) of Lemma 2 hold.

Let $b=e^{-i\theta}h(z_{0})-\frac{z_{0}}{2\beta}(1+\cos\theta)$, then it
follows from Lemma 2(i) that

${\rm Re}\ \langle b,z_{0}\rangle={\rm Re}\ [e^{-i\theta}\langle
h(z_{0}),z_{0}\rangle]-\frac{\|z_{0}\|^{2}}{2\beta}(1+\cos\theta)=0,$

so that

${\rm Re}\langle[Df(z_{0})]^{-1}D^{2}f(z_{0})(b,b),z_{0}\rangle<(1-\alpha)\
\|b\|^{2}.$ (3.6)

Let $b_{1}=iz_{0}$, then ${\rm Re}\ \langle b_{1},z_{0}\rangle={\rm
Re}[i\|z_{0}\|^{2}]=0$, so we conclude from the fact $f(z)\in K(B,\alpha)$
that

${\rm
Re}\langle[Df(z_{0})]^{-1}D^{2}f(z_{0})(iz_{0},iz_{0}),z_{0}\rangle<(1-\alpha)\
\|z_{0}\|^{2}.$ (3.7)

On the other hand, by Lemma 2, we have

$\|b\|^{2}=\|h(z_{0})\|^{2}-\frac{\|z_{0}\|^{2}}{4\beta^{2}}(1+\cos\theta)^{2},$
(3.8)

and
$\|h(z_{0})\|\geq\frac{\|z_{0}\|}{2\beta}|1+e^{i\theta}|=\frac{\|z_{0}\|}{2\beta}\sqrt{2+2\cos\theta}$,
and

$\displaystyle m\ e^{-i\theta}\langle h(z_{0}),z_{0}\rangle$ $\displaystyle=$
$\displaystyle\langle e^{-i\theta}h(z_{0}),mz_{0}\rangle$ $\displaystyle=$
$\displaystyle\langle
e^{-i\theta}h(z_{0}),e^{i\theta}\overline{Dh(z_{0})}(z_{0})\rangle+\langle
e^{-i\theta}h(z_{0}),e^{-i\theta}h(z_{0})\rangle$ $\displaystyle=$
$\displaystyle e^{-i2\theta}\ \langle
Dh(z_{0})h(z_{0}),z_{0}\rangle+\|h(z_{0})\|^{2},$

so that

$\displaystyle{\rm Re}\ [e^{-i2\theta}\ \langle
Dh(z_{0})h(z_{0}),z_{0}\rangle]$ $\displaystyle=$ $\displaystyle m{\rm Re}\
[e^{-i\theta}\langle h(z_{0}),z_{0}\rangle]-\|h(z_{0})\|^{2}$ $\displaystyle=$
$\displaystyle\frac{m\|z_{0}\|^{2}}{2\beta}(1+\cos\theta)-\|h(z_{0})\|^{2}.$

By computing the Frechet derivatives for both sides of equation
$Df(z)h(z)=f(z)$, we obtain

$[Df(z)]^{-1}D^{2}f(z)(h(z),h(z))+Dh(z)h(z)=h(z),$

and

$[Df(z)]^{-1}D^{2}f(z)(h(z),z)+Dh(z)(z)=z,$

thus we can obtain from the above equalities that

$\displaystyle{\rm
Re}\langle[Df(z_{0})]^{-1}D^{2}f(z_{0})(b,b),z_{0}\rangle={\rm
Re}[e^{-i2\theta}\langle[Df(z_{0})]^{-1}D^{2}f(z_{0})(h(z_{0}),h(z_{0})),z_{0}\rangle$
$\displaystyle\hskip 85.35826pt-\frac{1+\cos\theta}{\beta}{\rm
Re}[e^{-i\theta}\langle[Df(z_{0})]^{-1}D^{2}f(z_{0})(h(z_{0}),z_{0}),z_{0}\rangle]$
$\displaystyle\hskip 85.35826pt+\frac{(1+\cos\theta)^{2}}{4\beta^{2}}{\rm
Re}[e^{-i\theta}\langle[Df(z_{0})]^{-1}D^{2}f(z_{0})(z_{0},z_{0}),z_{0}\rangle]$
$\displaystyle\hskip 85.35826pt=-{\rm Re}[e^{-i2\theta}\ \langle
Dh(z_{0})h(z_{0}),z_{0}\rangle]+{\rm Re}[e^{-i2\theta}\ \langle
h(z_{0}),z_{0}\rangle]$ $\displaystyle\hskip
85.35826pt+\frac{1+\cos\theta}{\beta}\\{{\rm Re}[e^{-i\theta}\langle
Dh(z_{0})(z_{0}),z_{0}\rangle]-{\rm Re}[e^{-i\theta}\langle
z_{0},z_{0}\rangle]\\}$ $\displaystyle\hskip
85.35826pt-\frac{(1+\cos\theta)^{2}}{4\beta^{2}}{\rm
Re}[e^{-i\theta}\langle[Df(z_{0})]^{-1}D^{2}f(z_{0})(iz_{0},iz_{0}),z_{0}\rangle]$
$\displaystyle\hskip
85.35826pt\geq-\frac{m\|z_{0}\|^{2}}{2\beta}(1+\cos\theta)+\|h(z_{0})\|^{2}-\frac{\|z_{0}\|^{2}}{2\beta}(1+\cos\theta)$
$\displaystyle\hskip
85.35826pt+\frac{\|z_{0}\|^{2}}{2\beta^{2}}(1+\cos\theta)(\cos\theta+1+t)+\frac{(1+\cos\theta)^{2}}{4\beta^{2}}(\alpha-1)\|z_{0}\|^{2}$
$\displaystyle\hskip 85.35826pt\geq(1-\alpha)\
\|b\|^{2}+\alpha\|h(z_{0})\|^{2}-\alpha\frac{\|z_{0}\|^{2}}{4\beta^{2}}(2+2\cos\theta)$
$\displaystyle\hskip
85.35826pt+\frac{1+\cos\theta}{4\beta^{2}}\|z_{0}\|^{2}(2\alpha-2\beta+\frac{1-\beta}{\beta})$
$\displaystyle\hskip 85.35826pt\geq(1-\alpha)\ \|b\|^{2},$

which contradicts (3.6). Hence $f\in S^{\ast}(B,\beta)$, and the proof is
complete.

Remark 4. Setting $X=\mathbb{C}^{n}$ in Theorem 4, we obtain the related
result in [7, 8, 15]

By applying the growth theorem[5] of starlike mappings of order $\rho$ and
Theorem 4, we have the following corollary.

Corollary 3. Let
$0\leq\alpha<1,\beta=\beta(\alpha)=\frac{2\alpha-1+\sqrt{(2\alpha-1)^{2}+8}}{4}$.
If $f(z)\in K(B,\alpha)$, then for $z\in B$, we have

$\frac{\|z\|}{(1+\|z\|)^{2(1-\beta)}}\leq\|f(z)\|\leq\frac{\|z\|}{(1-\|z\|)^{2(1-\beta)}},$

and $f(B)\supset\frac{1}{2^{2(1-\beta)}}B$.

Remark 5. Setting $\alpha=0$ in Corollary 3, we obtain the growth theorem of
convex mappings[1, 3].

Finally, we introduce a linear operator[16] in purpose to construct some
concrete examples of biholomorphic convex mappings of order $\alpha$ on $B$ in
a Hilbert space $X$.

Let

$H(U)=\\{f:U\rightarrow\mathbb{C}\mbox{ are analytic in }U\mbox{ with
}f(0)=0,f^{\prime}(0)=1\\},$

then

$f\in K(\alpha)\Longleftrightarrow f\in H(U)\mbox{ and
}\mbox{Re}\bigg{\\{}1+\frac{zf^{\prime\prime}(z)}{f^{\prime}(z)}\bigg{\\}}>\alpha\quad\mbox{for
all}\quad z\in U.$

Let

$SK(B,\alpha)=\\{f\in N(B):\|Df(z)^{-1}D^{2}f(z)(\cdot,\cdot)\|\leq
1-\alpha\mbox{ for all }z\in B\\}.$

From Theorem 1, we have $SK(B,\alpha)\subset K(B,\alpha),SK(U,\alpha)\subset
K(\alpha)$ and

$SK(U,\alpha)=\bigg{\\{}f\in
H(U):\bigg{|}\frac{f^{\prime\prime}(z)}{f^{\prime}(z)}\bigg{|}\leq
1-\alpha\mbox{ for all }z\in U\bigg{\\}}.$

Let $m$ be a positive integer and $\dim X\geq m\geq 2$. Then there exist
$u_{1},u_{2},\cdots,u_{m}\in X$ with $\|u_{j}\|=1(j=1,2,\cdots,m)$ such that
$\langle u_{j},u_{k}\rangle=0(j\neq k)$. For $g_{1},g_{2},\cdots,g_{m}\in
H(U)$, we define the operator $\Phi$ as

$\displaystyle\Phi_{u_{1},u_{2},\cdots,u_{m}}(g_{1},g_{2},\cdots,g_{m})(z)=z-\sum\limits_{j=1}^{m}\langle
z,u_{j}\rangle u_{j}+\sum\limits_{j=1}^{m}g_{j}(\langle z,u_{j}\rangle)u_{j}$
(3.9)

for $z\in B$.

Theorem 5. Suppose that
$0\leq\alpha<1,\Phi_{u_{1},u_{2},\cdots,u_{m}}(g_{1},g_{2},\cdots,g_{m})$ is
defined by (3.9), where $g_{1},g_{2},\cdots,g_{m}\in H(U)$ are locally
univalent functions on $\Delta$.

(1) If $\Phi_{u_{1},\cdots,u_{m}}(g_{1},\cdots,g_{m})\in K(B,\alpha)$, then
$g_{1},g_{2},\cdots,g_{m}\in K(\alpha)$.

(2) If
$h(\xi)=\left\\{\begin{array}[]{lll}\frac{1-(1-\xi)^{2\alpha-1}}{2\alpha-1},&\alpha\neq\frac{1}{2}\\\
-\ln(1-\xi),&\alpha=\frac{1}{2}\end{array}\right.$, then $h\in K(\alpha)$, but
$\Phi_{u_{1},\cdots,u_{m}}(h,\cdots,h)\notin K(B,\alpha)$.

(3) $\Phi_{u_{1},u_{2},\cdots,u_{m}}(g_{1},g_{2},\cdots,g_{m})\in
SK(B,\alpha)$ if and only if $g_{1},g_{2},\cdots,g_{m}\in SK(U,\alpha)$.

Proof. Let
$f(z)=\Phi_{u_{1},u_{2},\cdots,u_{m}}(g_{1},g_{2},\cdots,g_{m})(z)$, where
$g_{1},g_{2},\cdots,g_{m}\in H(U)$ are locally univalent functions on $U$. By
some straightforward computations, we obtain

$\displaystyle Df(z)=I-\sum\limits_{j=1}^{m}\langle\cdot,u_{j}\rangle
u_{j}+\sum\limits_{j=1}^{m}g^{\prime}_{j}(\langle
z,u_{j}\rangle)\langle\cdot,u_{j}\rangle u_{j},$ $\displaystyle
Df(z)^{-1}=I-\sum\limits_{j=1}^{m}\bigg{(}1-\frac{1}{g^{\prime}_{j}(\langle
z,u_{j}\rangle)}\bigg{)}\langle\cdot,u_{j}\rangle u_{j},$ $\displaystyle
D^{2}f(z)(x,x)=\sum\limits_{j=1}^{m}g^{\prime\prime}_{j}(\langle
z,u_{j}\rangle)[\langle x,u_{j}\rangle]^{2}u_{j}$

for $z\in B$ and $x\in X$. Hence we have

$\displaystyle
Df(z)^{-1}D^{2}f(z)(x,x)=\sum\limits_{j=1}^{m}\frac{g^{\prime\prime}_{j}(\langle
z,u_{j}\rangle)}{g^{\prime}_{j}(\langle z,u_{j}\rangle)}[\langle
x,u_{j}\rangle]^{2}u_{j}.$ (3.10)

(1) Assume that
$f=\Phi_{u_{1},u_{2},\cdots,u_{m}}(g_{1},g_{2},\cdots,g_{m})\in K(B)$, for
every $\xi\in U\setminus\\{0\\}$ and $k$ fixed, we let $z=\xi u_{k}$ and
$x=i\xi u_{k}$, then $\mbox{Re}\langle x,z\rangle=\mbox{Re}\\{i|\xi|^{2}\\}=0$
and $z\in B\setminus\\{0\\}$. Note that $\langle u_{j},u_{k}\rangle=0(j\neq
k)$, from (3.10), we obtain

$\displaystyle\|x\|^{2}-\mbox{Re}\langle Df(z)^{-1}D^{2}f(z)(x,x),z\rangle$
$\displaystyle=$ $\displaystyle|\xi|^{2}+|\xi|^{2}\mbox{Re}\bigg{\\{}\frac{\xi
g^{\prime\prime}_{k}(\xi)}{g^{\prime}_{k}(\xi)}\bigg{\\}}=|\xi|^{2}\mbox{Re}\bigg{\\{}1+\frac{\xi
g^{\prime\prime}_{k}(\xi)}{g^{\prime}_{k}(\xi)}\bigg{\\}}>\alpha\|x\|^{2}=\alpha|\xi|^{2}$

for $\xi\in U\setminus\\{0\\}$. This implies $g_{k}\in K(\alpha)$ for
$k=1,2,\cdots,m$.

(2) A simple computation yields

$\mbox{Re}\bigg{\\{}1+\frac{\xi
h^{\prime\prime}(\xi)}{h^{\prime}(\xi)}\bigg{\\}}=\alpha+(1-\alpha)\mbox{Re}\bigg{\\{}\frac{1+\xi}{1-\xi}\bigg{\\}}>\alpha$

for all $\xi\in U$. It follows $h\in K(\alpha)$.

Let
$\displaystyle\sqrt{\frac{1-\alpha}{2}}<a<1,\displaystyle\frac{\sqrt{1-\alpha}}{\sqrt{2}a}<r<1$,

$z=ru_{1}+a\sqrt{1-r^{2}}u_{2}\quad\mbox{and}\quad
x=a\sqrt{1-r^{2}}u_{1}-ru_{2}.$

Then we have $\|x\|^{2}=a^{2}(1-r^{2})+r^{2}>0,\mbox{Re}\langle x,z\rangle=0$
and

$0<\|z\|^{2}=r^{2}+a^{2}(1-r^{2})<1.$

Notice that $\langle x,u_{j}\rangle=0(j\geq 3)$, from (3.10) , we obtain

$\displaystyle\|x\|^{2}$ $\displaystyle-$ $\displaystyle\mbox{Re}\langle
DF(z)^{-1}D^{2}F(z)(x,x),z\rangle$ $\displaystyle=$
$\displaystyle\|x\|^{2}-\mbox{Re}\bigg{\\{}\frac{2}{1-\langle
z,u_{1}\rangle}[\langle x,u_{1}\rangle]^{2}\langle
u_{1},z\rangle+\frac{2}{1-\langle z,u_{2}\rangle}[\langle
x,u_{2}\rangle]^{2}\langle u_{2},z\rangle\bigg{\\}}$ $\displaystyle=$
$\displaystyle
a^{2}(1-r^{2})+r^{2}-\bigg{\\{}\frac{2r}{1-r}a^{2}(1-r^{2})+\frac{2a\sqrt{1-r^{2}}}{1-a\sqrt{1-r^{2}}}r^{2}\bigg{\\}}$
$\displaystyle<$ $\displaystyle 1-2r^{2}a^{2}<\alpha,$

where $F=\Phi_{u_{1},u_{2},\cdots,m}(h,h,\cdots,h)$. By Corollary 1, we have
$F\not\in K(B,\alpha)$.

(3) Assume that $g_{1},g_{2},\cdots,g_{m}\in
SK(U,\alpha),f=\Phi_{u_{1},u_{2},\cdots,u_{m}}(g_{1},g_{2},\cdots,g_{m})$,
from (3.10), we obtain

$\displaystyle\Big{\|}Df(z)^{-1}D^{2}f(z)(x,x)\Big{\|}$ $\displaystyle=$
$\displaystyle\bigg{\|}\sum\limits_{j=1}^{m}\frac{g^{\prime\prime}_{j}(\langle
z,u_{j}\rangle)}{g^{\prime}_{j}(\langle z,u_{j}\rangle)}[\langle
x,u_{j}\rangle]^{2}u_{j}\bigg{\|}$ (3.11) $\displaystyle\leq$
$\displaystyle\sum\limits_{j=1}^{m}\bigg{|}\frac{g^{\prime\prime}_{j}(\langle
z,u_{j}\rangle)}{g^{\prime}_{j}(\langle z,u_{j}\rangle)}\bigg{|}|\langle
x,u_{j}\rangle|^{2}$ $\displaystyle\leq$
$\displaystyle(1-\alpha)\sum\limits_{j=1}^{m}|\langle x,u_{j}\rangle|^{2}$

for $z\in B$ and $x\in X$.

Fix $x\in X$, let $x_{0}=\sum\limits_{j=1}^{m}\langle x,u_{j}\rangle u_{j}$, a
simple computation yields

$\langle x-x_{0},u_{j}\rangle=\langle
x,u_{j}\rangle-\sum\limits_{k=1}^{m}\langle x,u_{k}\rangle\langle
u_{k},u_{j}\rangle=\langle x,u_{j}\rangle-\langle x,u_{j}\rangle=0,$

for $j=1,2,\cdots,m$. This leads to $\langle x-x_{0},x_{0}\rangle=0$. Hence we
conclude that

$\displaystyle\|x\|^{2}$ $\displaystyle=$
$\displaystyle\|(x-x_{0})+x_{0}\|^{2}=\|x-x_{0}\|^{2}+\|x_{0}\|^{2}$ (3.12)
$\displaystyle=$ $\displaystyle\|x-x_{0}\|^{2}+\sum\limits_{j=1}^{m}|\langle
x,u_{j}\rangle|^{2}$ $\displaystyle\geq$
$\displaystyle\sum\limits_{j=1}^{m}|\langle x,u_{j}\rangle|^{2}.$

From (3.11) and (3.12), we obtain

$\Big{\|}Df(z)^{-1}D^{2}f(z)(x,x)\Big{\|}\leq(1-\alpha)\sum\limits_{j=1}^{m}|\langle
x,u_{j}\rangle|^{2}\leq(1-\alpha)\|x\|^{2}\leq 1-\alpha$

for all $z\in B$ and $x\in X$ with $\|x\|=1$. Since $X$ is a Hilbert space, by
the result in [13](see P.342), we have

$\displaystyle\|Df(z)^{-1}D^{2}f(z)(\cdot,\cdot)\|$ $\displaystyle=$
$\displaystyle\sup_{\|x\|=1,\ \|y\|=1}\|Df(z)^{-1}D^{2}f(z)(x,y)\|$
$\displaystyle=$ $\displaystyle\sup_{\|x\|=1}\|Df(z)^{-1}D^{2}f(z)(x,x)\|\leq
1-\alpha.$

It follows that $f\in SK(B,\alpha)$.

Conversely, suppose that
$f=\Phi_{u_{1},u_{2},\cdots,u_{m}}(g_{1},g_{2},\cdots,g_{m})\in SK(B,\alpha)$.
For every $\xi\in U$ and $k$ fixed ($1\leq k\leq m$), we let $z=\xi u_{k}$ and
$x=u_{k}$, then we have $z\in B,\langle z,u_{k}\rangle=\xi$ and $\|x\|=1$.
Note that $\langle u_{j},u_{k}\rangle=0(j\neq k)$ and $\|u_{k}\|=1$, from
(3.10), we obtain

$\bigg{|}\frac{g^{\prime\prime}_{k}(\xi)}{g^{\prime}_{k}(\xi)}\bigg{|}=\|Df(z)^{-1}D^{2}f(z)(x,x)\|\leq\|Df(z)^{-1}D^{2}f(z)(\cdot,\cdot)\|\|x\|^{2}\leq
1-\alpha$

for $\xi\in U$. That is, $g_{k}\in SK(U,\alpha)$ for $k=1,2,\cdots,m$, and the
proof is complete.

Remark 6. Let $X=\mathbb{C}^{n},\alpha=0$. If we choose
$u_{1}=(1,0,\cdots,0),u_{2}=(0,1,\cdots,0),\cdots$,
$u_{n}=(0,0,\cdots,1)\in\mathbb{C}^{n}$, then we have
$z=\sum\limits_{j=1}^{n}\langle z,u_{j}\rangle u_{j}$ for
$z\in\mathbb{C}^{n}$. From Theorem 5, we obtain a result, which is Theorem 3
and Theorem 4 in [14] for case $p=2$. Part (2) was obtained by Roper and
Suffridge [11], [12] using a different method.

Example 3. Let $0\leq\alpha<1,\dim X\geq m\geq 2$ and
$\lambda_{j}\in\mathbb{C}$ with $\lambda_{j}\neq 0(j=1,2,\cdots,m)$, then

$f(z)=z-\sum\limits_{j=1}^{m}\langle z,u_{j}\rangle
u_{j}+\sum\limits_{j=1}^{m}\frac{e^{\lambda_{j}\langle
z,u_{j}\rangle}-1}{\lambda_{j}}u_{j}\in
K(B,\alpha)\Longleftrightarrow|\lambda_{j}|\leq 1-\alpha(j=1,2,\cdots,m),$

where $u_{j}\in X$ with $\|u_{j}\|=1$ such that $\langle
u_{j},u_{k}\rangle=0(j,k=1,2,\cdots,m,j\neq k)$.

Proof. If $|\lambda_{j}|\leq 1(j=1,2,\cdots,m)$, setting
$g_{j}(\xi)=\displaystyle\frac{e^{\lambda_{j}\xi}-1}{\lambda_{j}}$, then we
have that $g_{j}$ is analytic on $U$ with $g_{j}(0)=0,g^{\prime}_{j}(0)=1$
such that
$\displaystyle\frac{g^{\prime\prime}_{j}(\xi)}{g^{\prime}_{j}(\xi)}=\lambda_{j}$
for $\xi\in U(j=1,2,\cdots,m)$. Hence $g_{j}\in SK(U)$. From Theorem 5, we
obtain $f\in SK(B,\alpha)$.

Conversely, we shall prove that $|\lambda_{j}|\leq 1-\alpha$ for all
$j=1,2,\cdots,m$ when $f$ is a biholomorphic convex mapping of order $\alpha$
on $B$.

If not, then there exists $k$ such that $|\lambda_{k}|>1-\alpha$. Let
$\displaystyle\frac{1-\alpha}{|\lambda_{k}|}<r<1,\theta=\arg\lambda_{k},z_{0}=-re^{-i\theta}u_{k}$
and $x=ie^{-i\theta}u_{k}$, then $\|x\|=1,\mbox{Re}\langle
x,z_{0}\rangle=\mbox{Re}\\{-ir\\}=0$. Using the fact that $\langle
u_{j},u_{k}\rangle=0(j\neq k)$, from (3.10), we obtain

$\displaystyle Df(z_{0})^{-1}D^{2}f(z_{0})(x,x)$ $\displaystyle=$
$\displaystyle\sum\limits_{j=1}^{m}\frac{g^{\prime\prime}_{j}(\langle
z_{0},u_{j}\rangle)}{g^{\prime}_{j}(\langle z_{0},u_{j}\rangle)}[\langle
x,u_{j}\rangle]^{2}u_{j}$ $\displaystyle=$ $\displaystyle\lambda_{k}[\langle
x,u_{k}\rangle]^{2}u_{k}=-|\lambda_{k}|e^{-i\theta}u_{k}.$

Hence we have

$\|x\|^{2}-\mbox{Re}\langle
Df(z_{0})^{-1}D^{2}f(z_{0})(x,x),z_{0}\rangle=1-r|\lambda_{k}|<\alpha,$

which contradicts (2.2). This completes the proof.

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