Sharp Inequalities of Homogeneous Expansions of Almost Starlike Mappings of Order Alpha

Abstract

In this paper, we first obtain several sharp inequalities of homogeneous expansion for the subclass of all normalized almost starlike mappings of order $\alpha$ defined on the unit ball B of a complex Banach space X. Then, with these sharp inequalities, we derive the sharp estimates of the third and fourth homogeneous expansions for the above mappings defined on the unit polydisk $D^n$ in $\mathbb {C}^n$.

1 Introduction

In classical complex analysis, the following result is well known.

Branges Theorem [3]   If $f(z)=z+\sum _{n=2}^{\infty }a_{n}z^n$ is a biholomorphic function on the unit disk $D=\{z\in \mathbb {C}{:} |z|<1\}$, then

$$\begin{aligned} |a_{n}|\leqslant n,\quad n=2,3,\ldots . \end{aligned}$$

The above estimations are sharp with the Koebe function $K(z)=\frac{z}{(1-z)^2}=z+\sum _{n=2}^{\infty }n\, z^n$ as their extremal function.

However, Cartan [1] had pointed out that the above theorem does not hold in the case of several complex variables. Therefore, it is necessary to add some additional properties for a class of mappings in order to obtain the analogous estimations, for instance, the starlikeness, the convexity, and so on. In 1998, Sheng Gong posed the following conjecture, which we will refer as Bieberbach–Gong Sheng Conjecture.

Bieberbach–Gong Sheng Conjecture   If $f{:}D^n\rightarrow \mathbb {C}^n$ is a normalized biholomorphic starlike mapping, where $D^n=\{z=(z_1,\ldots , z_n)\in \mathbb {C}^n{:}\, |z_k|<1,\, k=1, 2, \ldots , n\}$ is the unit polydisk in $\mathbb {C}^n$, then

$$\begin{aligned} \frac{\Vert D^{m}f(0)(z^m)\Vert }{m!}\le m\Vert z\Vert ^m, z\in D^n, m=2,3,\ldots . \end{aligned}$$

At the present time, only the case of $m = 2$ (see [2]) has been shown. For the related works, we may consult [2,3,4,5,6,7,8,9,10,11,12,13,14,15]. On the other hand, with respect to the estimation of homogeneous expansion for normalized biholomorphic starlike mappings on the Euclidean unit ball $B^n$ in $\mathbb {C}^n$, Roper and Suffridge in [17, 18] have provided a counter example to verify that the above conjecture does not hold for $m = 2$. Then, in this paper, we extend some results in one complex variable to the case in several complex variables on the unit polydisk $D^n$ in $\mathbb {C}^n$. From these results, we also obtain the sharp third and fourth estimates of homogeneous expansion for biholomorphic almost starlike mappings of order $\alpha$ on the unit polydisk $D^n$ in $\mathbb {C}^n$.

We first recall the following results in the case of one complex variable.

Theorem A

[3] If $f(z)=z+\sum _{n=2}^{\infty }a_{n}z^n$ is a univalent function on the unit disk D in the complex plane $\mathbb {C}$, then

$$\begin{aligned} |a_3-a_{2}^2|\le 1. \end{aligned}$$

Theorem B

[3]   If $f(z)=z+\sum _{n=2}^{\infty }a_{n}z^n$ is a univalent convex function on the unit disk D, then

$$\begin{aligned} |a_3-a_{2}^2|\le \frac{1-|a_2|^2}{3}. \end{aligned}$$

Corollary C

[13]   If $f(z)=z+\sum _{n=2}^{\infty }a_{n}z^n$ is a univalent starlike function on the unit disk D in $\mathbb {C}$, then

$$\begin{aligned} |2a_3-a_{2}^2|\le 2 . \end{aligned}$$

Now we recall some notations and concepts. Let X be a complex Banach space with norm $\Vert \cdot \Vert$, $B=\{x\in X{:}\Vert x\Vert <1\}$ be the open unit ball in X; let $\partial _0B$ be the boundary of B, ${\bar{B}}$ be the closure of B, and $\partial _0D^n$ be the characteristic boundary (i.e., the boundary on which the maximum modulus of the holomorphic function can be attained) of the unit polydisk $D^n$. $\mathbb {N}_+$ represents the set of positive integers. Let H(B) be the set of all holomorphic mappings from B into X. It is well known that if $f\in H(B)$, then

$$\begin{aligned} f(y)=\sum _{n=0}^{\infty }\frac{1}{n!}D^nf(x)((y-x)^n)=\sum _{n=0}^{\infty }\frac{1}{n!}D^nf(x)(\underbrace{y-x,\ldots ,y-x}_{n}) \end{aligned}$$

for all y in the neighborhood of $x\in B$, where $D^nf(x)$ is the nth-$Fr\acute{e}chet$ derivative of f at x. Moreover, $D^nf(x)$ is a bounded symmetric n-linear mapping from $\prod _{j=1}^{n}X$ into X.

A holomorphic mapping $f{:}B\rightarrow X$ is said to be biholomorphic if the inverse $f^{-1}$ exists and is holomorphic on the open set f(B). A mapping $f\in H(B)$ is said to be locally biholomorphic if the $Fr\acute{e}chet$ derivative Df(x) has a bounded inverse for each $x\in B$. If $f{:}B\rightarrow X$ is a holomorphic mapping, we say that f is normalized if $f(0) = 0$ and $Df(0)=I$, where I represents the identity operator from X into X.

If $X^*$ is the dual space of X, for each $x\in X\backslash \{0\}$, we define

$$\begin{aligned} T(x)=\{T_x\in X^*{:}\Vert T_x\Vert =1, T_x(x)=\Vert x\Vert \}. \end{aligned}$$

According to Hahn–Banach theorem, T(x) is nonempty. For any $\alpha (\ne 0)\in \mathbb {C}$, since $\frac{|\alpha |}{\alpha }T_x\in T(\alpha x)$ corresponding to each $T_x\in T(x)$, we always denote $\frac{|\alpha |}{\alpha }T_x$ by $T_{\alpha x}$.

Let $\Omega \subset \mathbb {C}^n$ be a bounded circular domain. The first order Fr$\acute{e}$chet derivative and the $m(m\ge 2)$ order Fr$\acute{e}$chet derivatives of the mapping $f\in H(\Omega )$ were denoted by Df(z) and $D^mf(z)(b^{m-1},\cdot )$, respectively. The corresponding matrixes become

$$\begin{aligned} Df(z)= & {} \Big (\frac{\partial f_p(z)}{\partial z_k}\Big )_{1\le p,k\le n},\\ D^mf(z)(a^{m-1},\cdot )= & {} \Big (\sum _{l_1.l_2,\ldots ,l_{m-1}}^n\frac{\partial ^mf_p(z)}{\partial z_k\partial z_{l_1}\ldots \partial z_{l_{m-1}}}a_{l_1}\ldots a_{l_{m-1}}\Big )_{1\le p,k\le n}, \end{aligned}$$

where $f(z)=(f_1(z),f_2(z),\ldots ,f_n(z))', a=(a_1,a_2,\ldots ,a_n)'\in \mathbb {C}^n$.

Definition 1.1

[7]   Suppose that $\alpha \in [0,1)$, and $f{:}B\rightarrow X$ is a locally biholomorphic mapping. If

$$\begin{aligned} \mathrm{Re}\{T_x[(Df(x))^{-1}f(x)]\}\ge {\alpha \Arrowvert x\Arrowvert }, \quad x\in B, \end{aligned}$$

then f is said to be an almost starlike of order $\alpha$ on B.

Definition 1.2

[5]   Suppose $f\in H(B)$. We say that $x = 0$ is the zero of order k of f(x) if $f(0) = 0,\ldots ,D^{k-1}f(0) = 0$, but $D^kf(0)\ne 0$, where $k\in \mathbb {N}_+$. Note that the definition is the same as that in the case $X =\mathbb {C}$.

Definition 1.3

[16, 19]   Suppose that $L{:}X^m\rightarrow \mathbb {C}$ is a continuous $m-$linear form, if

$$\begin{aligned} L(x_1,x_2,\ldots ,x_m)=L(x_{\sigma (1)},x_{\sigma (2)},\ldots ,x_{\sigma (m)}),\quad \forall x_1,\ldots ,x_m\in X, \end{aligned}$$

for any $x_1,\ldots ,x_n$ in X and any permutation $\sigma$ of the first m natural numbers, and

$$\begin{aligned} \Vert L\Vert =\sup {\{|L(x_1,x_2,\ldots ,x_m)|{:}\Vert x_1\Vert \le 1,\ldots ,\Vert x_m\Vert \le 1\}}, \end{aligned}$$

then L is said to be continuous symmetric m-linear form. Denote $\mathcal {L}^s(^mX)$ to be the space of all continuous symmetric m-linear forms.

Definition 1.4

[16, 19]   Suppose that $L{:}\,X^m\rightarrow \mathbb {C}$ is a continuous symmetric $m-$linear form, if

$$\begin{aligned} P(x)=L(x,\ldots ,x),\quad \forall x\in X, \end{aligned}$$

then $P{:}X\rightarrow \mathbb {C}$ is said to be continuous homogeneous polynomial of degree m. Let $\Vert P\Vert =\sup {\{|P(x)|{:}\Vert x\Vert \le 1\}}$. Take the family as $\mathcal {P}^{s}(^mX)$.

For the sake of convenience, we let ${\hat{L}}=P$.

2 Some Lemmas

In order to prove the main results of this paper, we need the following lemmas.

Lemma 2.1

[3]   If $f(z)=a_0+\sum _{n=1}^{\infty }a_{n}z^n\in H(D)$, and $f(D)\subset D$, then

$$\begin{aligned} |a_n|\le 1-|a_0|^2, n=1,2,\ldots , \end{aligned}$$

when $n=1$, the above estimate is sharp.

Lemma 2.2

Let $p(z)=1+\sum _{n=1}^{\infty }b_{n}z^n\in H(D)$, and $\alpha \in [0,1)$. If ${Re}\, p(z)\ge \alpha , z\in D$, then

$$\begin{aligned}&\Big |b_2-\frac{b_{1}^2}{2(1-\alpha )}\Big |\le 2(1-\alpha )-\frac{|b_{1}|^2}{2(1-\alpha )}, \end{aligned}$$

(2.1)

$$\begin{aligned}&\Big |b_3-\frac{b_{1}b_2}{1-\alpha }+\frac{b_{1}^3}{4(1-\alpha )^2}\Big |\le 2(1-\alpha )-\frac{|b_{1}|^2}{2(1-\alpha )}. \end{aligned}$$

(2.2)

Furthermore, we have

$$\begin{aligned} |b_2|\le 2(1-\alpha ), \end{aligned}$$

(2.3)

and

$$\begin{aligned} \Big |b_2-\frac{b_{1}^2}{1-\alpha }\Big |\le 2(1-\alpha ). \end{aligned}$$

(2.4)

Proof

Let $h(z)=\frac{1-\frac{p(z)-\alpha }{1-\alpha }}{1+\frac{p(z)-\alpha }{1-\alpha }}=\frac{1-p(z)}{1-2\alpha +p(z)}, z\in D$, then $h(0)=0,h(D)\subset D,h(z)\in H(D)$, that is, h(z) is just a Schwarz function, moreover,

$$\begin{aligned} h(z)= & {} \frac{-\sum _{n=1}^{\infty }b_{n}z^n}{2(1-\alpha )+\sum _{n=1}^{\infty }b_{n}z^n}\\= & {} \frac{-b_1}{2(1-\alpha )}z- \Big [\frac{b_2}{2(1-\alpha )}-\frac{b_{1}^2}{4(1-\alpha )^2}\Big ]z^2\\&-\,\Big [\frac{b_3}{2(1-\alpha )}-\frac{b_{1}b_2}{2(1-\alpha )^2}+\frac{b_{1}^3}{8(1-\alpha )^3}\Big ]z^3+\cdots . \end{aligned}$$

According to Schwarz Lemma, we have

$$\begin{aligned} |h(z)/z|=1 \ or \ |h(z)/z|<1. \end{aligned}$$

When $|h(z)/z|=1$, we have

$$\begin{aligned}&\Big |\frac{b_{1}}{2(1-\alpha )}\Big |=1 , \frac{b_2}{2(1-\alpha )}-\frac{b_{1}^2}{4(1-\alpha )^2}=0 ,\nonumber \\&\quad \frac{b_3}{2(1-\alpha )}-\frac{b_{1}b_2}{2(1-\alpha )^2}+\frac{b_{1}^3}{8(1-\alpha )^3}=0. \end{aligned}$$

(2.5)

When $|h(z)/z|<1$, by Lemma 2.1, we have

$$\begin{aligned}&\Big |\frac{b_2}{2(1-\alpha )}-\frac{b_{1}^2}{4(1-\alpha )^2}\Big |\le 1-\Big |\frac{b_1}{2(1-\alpha )}\Big |^2 , \end{aligned}$$

(2.6)

$$\begin{aligned}&\Big |\frac{b_3}{2(1-\alpha )}-\frac{b_{1}b_2}{2(1-\alpha )^2}+\frac{b_{1}^3}{8(1-\alpha )^3}\Big |\le 1-\Big |\frac{b_1}{2(1-\alpha )}\Big |^2. \end{aligned}$$

(2.7)

From (2.5), (2.6) and (2.7), we obtain that

$$\begin{aligned}&\Big |b_2-\frac{b_{1}^2}{2(1-\alpha )}\Big |\le 2(1-\alpha )-\frac{|b_{1}|^2}{2(1-\alpha )},\\&\Big |b_3-\frac{b_{1}b_2}{1-\alpha }+\frac{b_{1}^3}{4(1-\alpha )^2}\Big |\le 2(1-\alpha )-\frac{|b_{1}|^2}{2(1-\alpha )}. \end{aligned}$$

Finally, we have

$$\begin{aligned} |b_2|\leqslant \Big |b_2-\frac{b_{1}^2}{2(1-\alpha )}\Big |+\frac{|b_{1}|^2}{2(1-\alpha )}\leqslant 2(1-\alpha ), \end{aligned}$$

and

$$\begin{aligned} \Big |b_2-\frac{b_{1}^2}{1-\alpha }\Big |\leqslant \Big |b_2-\frac{b_{1}^2}{2(1-\alpha )}\Big |+\frac{|b_{1}|^2}{2(1-\alpha )}\le 2(1-\alpha ). \end{aligned}$$

This completes the proof. $\square$

Lemma 2.3

[7]   If $f(x){:}B\rightarrow X$ is a normalized locally biholomorphic mapping, and $g(x)=(Df(x))^{-1}f(x)$, then

$$\begin{aligned} \frac{D^2g(0)(x^2)}{2!}= & {} -\frac{D^2f(0)(x^2)}{2!},\\ \frac{D^3g(0)(x^3)}{3!}= & {} -2\frac{D^3f(0)(x^3)}{3!}+2\frac{D^2f(0)}{2!}\left( x,\frac{D^2f(0)(x^2)}{2!}\right) ,\\ \frac{D^4g(0)(x^4)}{4!}= & {} -3\frac{D^4f(0)(x^4)}{4!}+3\frac{D^3f(0)}{3!}\left( x^2,\frac{D^2f(0)(x^2)}{2!}\right) \\&+\,4\frac{D^2f(0)}{2!}\left( x,\frac{D^3f(0)(x^3)}{3!}\right) \\&-\,4\frac{D^2f(0)}{2!}\left( x,\frac{D^2f(0)}{2!}\left( x,\frac{D^2f(0)(x^2)}{2!}\right) \right) . \end{aligned}$$

Lemma 2.4

[10]   Suppose $0\le \alpha <1, f{:}D^n\rightarrow X$ is a normalized locally biholomorphic mapping, then f is an almost starlike mapping of order $\alpha$ if and only if

$$\begin{aligned} \text{ Re }\Big \{\frac{g_j(z)}{z_j}\Big \}\ge \alpha , \forall z\in D^n, \end{aligned}$$

where $g(z)=(g_1(z),g_2(z),\ldots ,g_n(z))'=(Df(z))^{-1}f(z))$ is a column vector in $\mathbb {C}^n$, $|z_j|=\Vert z\Vert =\max _{1\le k\le n}\{|z_k|\}$.

Lemma 2.5

[7]   Suppose $g\in H(D^n), g(0)=0, Dg(0)=I,\alpha \in [0,1)$. If $\text{ Re }\frac{g_j(z)}{z_j}\ge \alpha , z\in D^n$, where $|z_j|=\Vert z\Vert =\max _{1\le k\le n}\{|z_k|\}$, then

$$\begin{aligned} \frac{\Vert D^mg(0)(z^m)\Vert }{m!}\le 2(1-\alpha )\Vert z\Vert ^m, z\in D^n, m=2,3,\ldots . \end{aligned}$$

These estimates are sharp.

Lemma 2.6

If

$$\begin{aligned}&\left\| \left( z_{1}^q\left( \sum _{l=1}^{n}a_{p1l}z_l^p\right) ,z_{2}^q\left( \sum _{l=1}^{n}a_{p2l}z_l^p\right) ,\ldots ,z_{n}^q\left( \sum _{l=1}^{n}a_{pnl}z_l^p\right) \right) '\right\| \le C_0\Vert z\Vert ^m,\\&z\in D^n,p+q=m,p,q\in \mathbb {N}^+,m=2,3,\ldots , \end{aligned}$$

where each $a_{pkl}(p=1,2,\ldots ,m-1,k,l=1,2,\ldots ,n)$ is a complex number independent of $z_k(k=1,2,\ldots ,n)$, $\Vert z\Vert =\max _{1\le k\le n}\{|z_k|\}$, $C_0$ is a nonnegative real constant. Then

$$\begin{aligned} M=\max _{1\le k\le n}\left\{ \sum _{l=1}^n|a_{pkl}|\right\} \le C_0,p=1,2,\ldots ,m-1. \end{aligned}$$

Proof

For every $z\in D^n\backslash \{0\}$, taking into the hypothesis of Lemma 2.6, we have

$$\begin{aligned} \left| \frac{z_{k}^q}{\Vert z\Vert ^q}\left( \sum _{l=1}^{n}{a_{pkl}\left( \frac{z_l}{\Vert z\Vert }\right) ^p}\right) \right| \le C_0. \end{aligned}$$

In particular, for each k, if $a_{pkl}\ne 0$, taking $z_l=e^{-i\frac{\arg {a_{pkl}}}{p}}\Vert z\Vert , l=1,2,\ldots ,n$, where $i^2=-1$. Then, we conclude that

$$\begin{aligned} \sum _{l=1}^{n}{|a_{pkl}|}\le C_0, k=1,2,\ldots ,n, \end{aligned}$$

that is

$$\begin{aligned} M=\max _{1\le k\le n}{\sum _{l=1}^n{\{|a_{pkl}|}\}}\le C_0,p=1,2,\ldots ,m-1. \end{aligned}$$

This completes the proof. $\square$

Lemma 2.7

[19]   Suppose $L\in \mathcal {L}^s(^m \ l_{n}^{\infty })$, ${\hat{L}}\in \mathcal {P}^s(^m\ l_{n}^{\infty })$, where $\mathcal {L}^s(^m \ l_{n}^{\infty }),\mathcal {P}^s(^m\ l_{n}^{\infty })$ are defined by Definitions 1.3 and 1.4, then

$$\begin{aligned}&\Vert L\Vert =\Vert {\hat{L}}\Vert ,n=2,\\&\Vert L\Vert \le m^{m/2}(m+1)^{(m+1)/2}/2^{m}m!\Vert {\hat{L}}\Vert ,n\ge 3,n\in \mathbb {N}_+. \end{aligned}$$

In particular, when $m=2$, according to Lemma 2.7, we have

Lemma 2.8

Suppose $L\in \mathcal {L}^s(^2 \ l_{n}^{\infty })$, ${\hat{L}}\in \mathcal {P}^s(^2\ l_{n}^{\infty })$, where $\mathcal {L}^s(^2 \ l_{n}^{\infty }),\mathcal {P}^s(^2\ l_{n}^{\infty })$ are defined by Definitions 1.3 and 1.4, then

$$\begin{aligned}&\Vert L\Vert =\Vert {\hat{L}}\Vert ,n=2,\\&\Vert L\Vert \le \frac{3}{4}\sqrt{3}\Vert {\hat{L}}\Vert ,n\ge 3,n\in \mathbb {N}_+. \end{aligned}$$

By Lemma 2.8, we can obtain the following lemma.

Lemma 2.9

Suppose $f{:}D^n\rightarrow X$ is a holomorphic mapping. Define

$$\begin{aligned} L(x,y)= & {} D^2f(0)(x,y),x,y\in X,\ \ \Vert L\Vert =\sup {\{\Vert L(x,y)\Vert {:}\Vert x\Vert \le 1,\Vert y\Vert \le 1\}},\\ {\hat{L}}(x,x)= & {} D^2f(0)(x^2),x\in X,\ \ \ \ \Vert {\hat{L}}\Vert =\sup {\{\Vert {\hat{L}}(x,x)\Vert {:}\Vert x\Vert \le 1\}}, \end{aligned}$$

where $\Vert x\Vert =\max _{1\le k\le n}{\{|x_k|\}},\ \Vert y\Vert =\max _{1\le k\le n}{\{|y_k|\}}$. If L and ${\hat{L}}$ are bounded linear operators, then

$$\begin{aligned} \Vert L\Vert= & {} \Vert {\hat{L}}\Vert ,\, \text{ for } \, n=2,\\ \Vert L\Vert\le & {} \frac{3}{4}\sqrt{3}\Vert {\hat{L}}\Vert ,\, \text{ for } \, n\ge 3,\, n\in \mathbb {N}_+. \end{aligned}$$

Furthermore

$$\begin{aligned} \Vert D^2f(0)(x,y)\Vert \le \Vert L\Vert \Vert x\Vert \Vert y\Vert ,\, \text{ for } \, x,y\in X. \end{aligned}$$

Proof

Since L and ${\hat{L}}$ are bounded linear operators, according to Lemma 2.8, we have

$$\begin{aligned} \Vert L\Vert= & {} \sup {\{\Vert L(x,y)\Vert {:}\Vert x\Vert \le 1,\Vert y\Vert \le 1\}}=\sup _{x,y\ne 0,x,y\in X}{\frac{\Vert D^2f(0)(x,y)\Vert }{\Vert x\Vert \Vert y\Vert }},\\ \Vert {\hat{L}}\Vert= & {} \sup {\{\Vert {\hat{L}}(x,x)\Vert {:}\Vert x\Vert \le 1\}}=\sup _{x\ne 0,x\in X}{\frac{\Vert D^2f(0)(x,x)\Vert }{\Vert x\Vert ^2}}, \end{aligned}$$

then

$$\begin{aligned} \Vert D^2f(0)(x,y)\Vert\le & {} \Vert L\Vert \Vert x\Vert \Vert y\Vert ,x,y\in X,\\ \Vert D^2f(0)(x,x)\Vert\le & {} \Vert {\hat{L}}\Vert \Vert x\Vert ^2,x,\in X. \end{aligned}$$

Hence, when $\Vert x\Vert \le 1,\Vert y\Vert \le 1$, $\Vert D^2f(0)(x,y)\Vert ,\Vert D^2f(0)(x,x)\Vert$ are all bounded, where

$$\begin{aligned} \Vert D^2f(0)(x,y)\Vert= & {} \max _{1\le k\le n}{\{|D^2f_{k}(0)(x,y)|,\Vert x\Vert \le 1,\Vert y\Vert \le 1\}},\\ \Vert D^2f(0)(x,x)\Vert= & {} \max _{1\le k\le n}{\{|D^2f_{k}(0)(x^2)|,\Vert x\Vert \leqslant 1\}}. \end{aligned}$$

From Lemma 2.8, we have

$$\begin{aligned} L(x,y)= & {} \Big (\sum _{i,j=1}^{n}{a_{ij}^{1}x_{i}y_{j}},\ldots ,\sum _{i,j=1}^{n}{a_{ij}^{n}x_{i}y_{j}}\Big )',\\ {\hat{L}}(x,x)= & {} \Big (\sum _{i,j=1}^{n}{a_{ij}^{1}x_{i}x_{j}},\ldots ,\sum _{i,j=1}^{n}{a_{ij}^{n}x_{i}x_{j}}\Big )', \end{aligned}$$

where $a_{ij}^{k}=\frac{\partial ^2f_{k}(0)}{\partial x_i\partial x_j},\ i,j,k=1,2,\ldots ,n$. Let

$$\begin{aligned} L(x,y)= & {} (L_1(x,y),\ldots ,L_{n}(x,y))',\\ {\hat{L}}(x,x)= & {} ({\hat{L}}_{1}(x,x),\ldots ,\hat{L_n}(x,x))', \end{aligned}$$

where $L_k=\sum _{i,j=1}^{n}{a_{ij}^{k}x_{i}y_{j}},\ {\hat{L}}_{k}=\sum _{i,j=1}^{n}{a_{ij}^{k}x_{i}x_{j}},\ k=1,2,\ldots ,n$. By Definitions 1.3 and 1.4, it is easy to obtain that $L_k\in \mathcal {L}^s(^2\ l_{n}^{\infty })$, ${\hat{L}}_{k}\in \mathcal {P}^s(^2\ l_{n}^{\infty }),\ k=1,2,\ldots ,n$, thus $L_k$ and ${\hat{L}}_{k}$ satisfy the conditions of Lemma 2.8. Therefore,

$$\begin{aligned} \Vert L_k\Vert= & {} \Vert {\hat{L}}_{k}\Vert ,\, \text{ for } \, n=2, \end{aligned}$$

(2.8)

$$\begin{aligned} \Vert L_k\Vert\le & {} \frac{3}{4}\sqrt{3}\Vert {\hat{L}}_{k}\Vert ,\, \text{ for } \, n\ge 3,\, n\in \mathbb {N}_+. \end{aligned}$$

(2.9)

Because

$$\begin{aligned} \Vert L\Vert= & {} \sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\{\Vert L(x,y)\Vert \}}\\= & {} \sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\max _{1\le k\le n}{\{|L_{k}(x,y)|\}}}\\\ge & {} \sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\{|L_{k}(x,y)|\}}\\= & {} \Vert L_{k}\Vert ,k=1,2,\ldots , n, \end{aligned}$$

we have

$$\begin{aligned} \Vert L\Vert \ge \max _{1\le k\le n}{\{\Vert L_{k}\Vert \}}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \Vert L\Vert= & {} \sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\{\Vert L(x,y)\Vert \}}\\= & {} \sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\max _{1\le k\le n}{\{|L_{k}(x,y)|\}}}\\\le & {} \sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\max _{1\le k\le n}{\sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\{|L_{k}(x,y)|\}}}}\\= & {} \sup _{\Vert x\Vert \le 1,\Vert y\Vert \le 1}{\max _{1\le k\le n}{\{\Vert L_{k}\Vert \}}}\\= & {} \max _{1\le k\le n}{\{\Vert L_{k}\Vert \}}. \end{aligned}$$

Hence,

$$\begin{aligned} \Vert L\Vert =\max _{1\le k\le n}{\{\Vert L_{k}\Vert \}}. \end{aligned}$$

Similarly, we have

$$\begin{aligned} \Vert {\hat{L}}\Vert =\max _{1\le k\le n}{\{\Vert {\hat{L}}_{k}\Vert \}}. \end{aligned}$$

According to (2.8) and (2.9), we obtain

$$\begin{aligned} \Vert L\Vert= & {} \Vert {\hat{L}}\Vert , \text{ if } n=2,\\ \Vert L\Vert\le & {} \frac{3}{4}\sqrt{3}\Vert {\hat{L}}\Vert , \text{ if } n\ge 3,\, n\in \mathbb {N}_+. \end{aligned}$$

This completes the proof. $\square$

Lemma 2.10

Assume that $\alpha \in [0,\frac{37-\sqrt{505}}{72}]$, and

$$\begin{aligned} h(x)=\frac{12\alpha ^2-10\alpha +1}{4(1-\alpha )^2}x^3-\frac{1}{2(1-\alpha )}x^2+(5-7\alpha )x+2(1-\alpha ), \end{aligned}$$

then h(x) is strictly increasing on $x\in [0,2(1-\alpha )]$, and

$$\begin{aligned} \max _{x\in [0,2(1-\alpha )]}{\{h(x)\}}=h(2(1-\alpha ))=(1-\alpha )(3-4\alpha )(4-6\alpha ). \end{aligned}$$

Proof

Since $h(x)=\frac{12\alpha ^2-10\alpha +1}{4(1-\alpha )^2}x^3-\frac{1}{2(1-\alpha )}x^2+(5-7\alpha )x+2(1-\alpha )$, we have

$$\begin{aligned} h'(x)= & {} \frac{3(12\alpha ^2-10\alpha +1)}{4(1-\alpha )^2}x^2-\frac{1}{1-\alpha }x+5-7\alpha \\= & {} \frac{3(12\alpha ^2-10\alpha +1)}{4(1-\alpha )^2}\Big (x-\frac{2(1-\alpha )}{3(12\alpha ^2-10\alpha +1)}\Big )^2\\&+\,5-7\alpha -\frac{1}{3(12\alpha ^2-10\alpha +1)}. \end{aligned}$$

Now we split into four cases to prove h(x) is strictly increasing on $x\in [0,2(1-\alpha )]$.

• (1) When $\alpha \in [0,\frac{15-\sqrt{153}}{36})$, since

  $$\begin{aligned} 12\alpha ^2-10\alpha +1>0,\ \ 0<\frac{2(1-\alpha )}{3(12\alpha ^2-10\alpha +1)}<2(1-\alpha ),\\ h'(\frac{2(1-\alpha )}{3(12\alpha ^2-10\alpha +1)})=5-7\alpha -\frac{1}{3(12\alpha ^2-10\alpha +1)}>0, \end{aligned}$$

  we get that $h'(x)\geqslant h'(\frac{2(1-\alpha )}{3(12\alpha ^2-10\alpha +1)})>0$ for each $x\in [0,2(1-\alpha )]$. Hence h(x) is strictly increasing on $x\in [0,2(1-\alpha )]$.

• (2) When $\alpha \in [\frac{15-\sqrt{153}}{36},\frac{10-\sqrt{52}}{24})$, since

  $$\begin{aligned}&12\alpha ^2-10\alpha +1>0,\ \ \frac{2(1-\alpha )}{3(12\alpha ^2-10\alpha +1)}\ge 2(1-\alpha ),\ \ h'(0)\\&\quad =5-7\alpha>0,\ \ h'(2(1-\alpha ))>0, \end{aligned}$$

  we get that $h'(x)>0$ for every $x\in [0,2(1-\alpha )]$. So h(x) is strictly increasing on $x\in [0,2(1-\alpha )]$.

• (3) When $\alpha =\frac{10-\sqrt{52}}{24}$, since

  $$\begin{aligned} h'(x)=-\frac{1}{1-\alpha }x+5-7\alpha ,\ \ h'(0)=5-7\alpha>0,\ \ h'(2(1-\alpha ))=3-7\alpha >0, \end{aligned}$$

  we get that $h'(x)>0$ for all $x\in [0,2(1-\alpha )]$, that is, h(x) is strictly increasing on $x\in [0,2(1-\alpha )]$.

• (4) When $\alpha \in (\frac{10-\sqrt{52}}{24},\frac{37-\sqrt{505}}{72}]$, since

  $$\begin{aligned} 12\alpha ^2-10\alpha +1<0,\ \ \frac{2(1-\alpha )}{3(12\alpha ^2-10\alpha +1)}<0,\ \ h'(2(1-\alpha ))\ge 0, \end{aligned}$$

  we get that $h'(x)>0$ for all $x\in [0,2(1-\alpha )]$, that is, h(x) is strictly increasing on $x\in [0,2(1-\alpha )]$.

Hence, for all $\alpha \in [0,\frac{37-\sqrt{505}}{36}]$, where $\frac{37-\sqrt{505}}{36}\approx 0.2018$, h(x) is strictly increasing on $x\in [0,2(1-\alpha )]$, thus

$$\begin{aligned} \max _{x\in [0,2(1-\alpha )]}{\{h(x)\}}=h(2(1-\alpha ))=(1-\alpha )(3-4\alpha )(4-6\alpha ). \end{aligned}$$

This completes the proof. $\square$

3 Main results

We first establish the sharp inequalities of homogeneous expansion for the almost starlike mappings of order $\alpha$.

Theorem 3.1

If f(x) is an almost starlike of order $\alpha$ on B, $0\le \alpha <1$, then

$$\begin{aligned}&\left| 2T_x\left( \frac{D^3f(0)(x^3)}{3!}\right) \Vert x\Vert -2T_x\left( \frac{D^2f(0)}{2!}\left( x,\frac{D^2f(0)(x^2)}{2!}\right) \right) \Vert x\Vert +\frac{\left[ T_x\left( \frac{D^2f(0)(x^2)}{2!}\right) \right] ^2}{1-\alpha }\right| \\&\quad \leqslant 2(1-\alpha )\Vert x\Vert ^4, \forall x\in B. \end{aligned}$$

The above inequality is sharp.

Proof

For fixed $x\in B\backslash \{0\}$, let $x_0=\frac{x}{\Vert x\Vert }$. Define $p(\xi )=\frac{T_{x_0}(g(\xi x_0))}{\xi }, \xi \in D$, where

$$\begin{aligned} g(x)=(Df(x))^{-1}f(x). \end{aligned}$$

By the hypothesis of Theorem 3.1, we have $p(\xi )\in H(D)$, $\mathrm{Re}p(\xi )=Re\frac{1}{|\xi |}T_{\xi x_0}(g(\xi x_0))\ge \alpha$,

$$\begin{aligned} p(0)=\lim _{\xi \rightarrow 0}\frac{1}{\xi }T_{x_0}(g(\xi x_0))=T_{x_0}(x_0)=1, \end{aligned}$$

and

$$\begin{aligned} p(\xi )=1+\frac{T_{x_0}(D^2g(0)(x_{0}^2))}{2!}\xi +\frac{T_{x_0}(D^3g(0)(x_{0}^3))}{3!}\xi ^2+\cdots . \end{aligned}$$

Consequently, by (2.4) in Lemma 2.2, we obtain

$$\begin{aligned} \Big |T_{x_0}\Big (\frac{D^3g(0)(x_{0}^3)}{3!}\Big )-\frac{1}{1-\alpha }\Big [T_{x_0}\Big (\frac{D^2g(0)(x_{0}^2)}{2!}\Big )\Big ]^2\Big |\le 2(1-\alpha ). \end{aligned}$$

From Lemma 2.3, we conclude that

$$\begin{aligned} \frac{D^2g(0)(x_{0}^2)}{2!}= & {} -\frac{D^2f(0)(x_{0}^2)}{2!},\\ \frac{D^3g(0)(x_{0}^3)}{3!}= & {} -2\frac{D^3f(0)(x_{0}^3)}{3!}+2\frac{D^2f(0)}{2!}\Big (x_0,\frac{D^2f(0)(x_{0}^2)}{2!}\Big ), \end{aligned}$$

that is

$$\begin{aligned}&\left| 2T_{x_0}\left( \frac{D^3f(0)(x_{0}^3)}{3!}\right) -2T_{x_0}\left( \frac{D^2f(0)}{2!}\left( x_0,\frac{D^2f(0)(x_{0}^2)}{2!}\right) \right) +\frac{\left[ T_{x_0}\left( \frac{D^2f(0)(x_{0}^2)}{2!}\right) \right] ^2}{1-\alpha }\right| \\&\quad \leqslant 2(1-\alpha ), \forall x_0\in \partial B, \end{aligned}$$

For $x_0=\frac{x}{\Vert x\Vert }$, so we have $T_x(\cdot )=T_{x_0}(\cdot )$. Thus we obtain

$$\begin{aligned}&\left| 2T_x\left( \frac{D^3f(0)}{3!}\left( \frac{x^3}{\Vert x\Vert ^3}\right) \right) -2T_x\left( \frac{D^2f(0)}{2!}\left( \frac{x}{\Vert x\Vert }, \frac{D^2f(0)}{2!}\left( \frac{x^2}{\Vert x\Vert ^2}\right) \right) \right) \right. \\&\left. \quad +\,\frac{\left[ T_x\left( \frac{D^2f(0)}{2!}\left( \frac{x^2}{\Vert x\Vert ^2}\right) \right) \right] ^2}{1-\alpha }\right| \\&\quad \leqslant 2(1-\alpha ), \forall x\in B, \end{aligned}$$

that is,

$$\begin{aligned}&\left| 2T_x\left( \frac{D^3f(0)(x^3)}{3!}\right) \Vert x\Vert -2T_x\left( \frac{D^2f(0)}{2!}\left( x,\frac{D^2f(0)(x^2)}{2!}\right) \right) \Vert x\Vert \right. \\&\quad \qquad +\left. \,\frac{\left[ T_x\left( \frac{D^2f(0)(x^2)}{2!}\right) \right] ^2}{1-\alpha }\right| \\&\qquad \leqslant 2(1-\alpha )\Vert x\Vert ^4, \forall x\in B. \end{aligned}$$

Finally, it is not difficult to check that the function

$$\begin{aligned} f_1(x)=\left\{ \begin{array}{lll} \frac{x}{(1-(1-2\alpha )T_u(x))^{\frac{2(1-\alpha )}{1-2\alpha }}},&{}\quad \alpha \in [0,1)\backslash \left\{ \frac{1}{2}\right\} ,\\ xe^{T_u(x)}, &{}\quad \alpha =\frac{1}{2} \end{array} \right. \end{aligned}$$

satisfies the hypothesis of Theorem 3.1, where $u\in \partial B$. We set $x=ru, \Vert u\Vert =1, 0\le r<1$. By direct computation, we obtain that

$$\begin{aligned}&\left| 2T_x\left( \frac{D^3f(0)(x^3)}{3!}\right) \Vert x\Vert -2T_x\left( \frac{D^2f(0)}{2!}\left( x,\frac{D^2f(0)(x^2)}{2!}\right) \right) \Vert x\Vert \right. \\&\quad \left. +\,\frac{\left[ T_x\left( \frac{D^2f(0)(x^2)}{2!}\right) \right] ^2}{1-\alpha }\right| = 2(1-\alpha )\Vert r\Vert ^4. \end{aligned}$$

Hence, the inequality in Theorem 3.1 is sharp.$\square$

Setting $\alpha =0$ in Theorem 3.1, we can obtain the following corollary.

Corollary 3.2

[13] If $f(x)\in S^*(B)$. Then

$$\begin{aligned}&\left| 2T_x\left( \frac{D^3f(0)(x^3)}{3!}\right) \Vert x\Vert -2T_x\left( \frac{D^2f(0)}{2!}\left( x,\frac{D^2f(0)(x^2)}{2!}\right) \right) \Vert x\Vert \right. \\&\quad \left. +\,\left[ T_x\left( \frac{D^2f(0)(x^2)}{2!}\right) \right] ^2\right| \\&\quad \leqslant 2\Vert x\Vert ^4, \forall x\in B. \end{aligned}$$

The above inequality is sharp.

When $\alpha =0, \ n=1$, we obtain Corollary C by Corollary 3.2.

Now we will establish the sharp estimates of the third and fourth homogeneous expansions for the almost starlike mappings of order $\alpha$ on $D^n$ in $\mathbb {C}^n$.

Theorem 3.3

Suppose $0\le \alpha \le \frac{1}{2}$, f is an almost starlike of order $\alpha$ in $D^n$, and

$$\begin{aligned}&2z_k\frac{D^2f_k(0)}{2!}\left( z_0,\frac{D^2f(0)(z_0^2)}{2!}\right) \Vert z\Vert -\frac{1}{1-\alpha }\left[ \frac{D^2f_k(0)(z_{0}^2)}{2!}\right] ^2\Vert z\Vert ^2\\&=\frac{1-2\alpha }{1-\alpha }z_{k}^2\left( \sum _{k=1}^na_{km}\frac{z_k}{\Vert z\Vert }\left( \sum _{l=1}^na_{ml}\frac{z_l}{\Vert z\Vert }\right) \right) ,\, \text{ for } \, z\in D^n\backslash \{0\}. \end{aligned}$$

where $a_{ml}=\frac{1}{2!}\frac{\partial ^2f_m(0)}{\partial z_m\partial z_l}, m,l=1,2,\ldots ,n$, then

$$\begin{aligned} \frac{\Vert D^3f(0)(z^3)\Vert }{3!}\le (1-\alpha )(3-4\alpha )\Vert z\Vert ^3,\, \text{ for } \, z\in D^n. \end{aligned}$$

The above estimate is sharp.

Proof

For fixed $z\in D^n\setminus \{0\}$, define $z_0=\frac{z}{\Arrowvert z\Arrowvert }$. Taking $T_z=(0,\ldots ,0,\frac{|z_j|}{z_j},\ldots ,0)$, where $|z_j|=\Vert z\Vert =\max _{1\le k\le n}\{|z_k|\}$. Applying Theorem 3.1, we have

$$\begin{aligned}&\left| 2\frac{D^3f_j(0)(z_{0}^3)}{3!}\frac{\Vert z\Vert }{z_j}-2\frac{D^2f_j(0)}{2!}\left( z_0,\frac{D^2f(0)(z_{0}^2)}{2!}\right) \frac{\Vert z\Vert }{z_j}\right. \\&\quad \left. +\,\frac{1}{1-\alpha }\left[ \frac{D^2f_j(0)(z_{0}^2)}{2!}\right] ^2\left( \frac{\Vert z\Vert }{z_j}\right) ^2\right| \le 2(1-\alpha ). \end{aligned}$$

According to the hypothesis of Theorem 3.3, we obtain

$$\begin{aligned} \frac{|D^3f_j(0)(z_{0}^3)|}{3!}\le 1-\alpha +\frac{1-2\alpha }{1-\alpha }\frac{M^2}{2}, \end{aligned}$$

where $|z_j|=\Vert z\Vert =\max _{1\le k\le n}\{|z_k|\},\ M=\max _{1\le k\le n}\{\sum _{l=1}^n|a_{kl}|\}$. Applying the maximum modulus principle of holomorphic functions, and Lemmas 2.4, 2.5 and 2.6 with $C_0=2(1-\alpha )$, we have

$$\begin{aligned} \frac{\Vert D^3f(0)(z^3)\Vert }{3!}\le (1-\alpha +\frac{1-2\alpha }{1-\alpha }\frac{M^2}{2})\Vert z\Vert ^3\le (1-\alpha )(3-4\alpha )\Vert z\Vert ^3, z\in D^n. \end{aligned}$$

Finally, it is easy to prove that the function

$$\begin{aligned} f_2(z)=\left\{ \begin{array}{lll} \Big (\frac{z_1}{(1-(1-2\alpha )z_1)^{\frac{2(1-\alpha )}{1-2\alpha }}},\ldots ,\frac{z_n}{(1-(1-2\alpha )z_n)^{\frac{2(1-\alpha )}{1-2\alpha }}}\Big )',&{}0\leqslant \alpha <\frac{1}{2},\\ (z_1e^{z_1},\ldots ,z_ne^{z_n})',&{} \alpha =\frac{1}{2} \end{array} \right. \end{aligned}$$

(3.1)

meets the hypothesis of Theorem 3.3, where $z\in D^n$. Taking $z=(r,0,\ldots ,0)'(0\le r<1)$,

$$\begin{aligned} D^3f(0)(z^3)=(6(1-\alpha )(3-4\alpha )r^3,0,\ldots ,0)'. \end{aligned}$$

Then

$$\begin{aligned} \frac{\Vert D^3f(0)(z^3)\Vert }{3!}=(1-\alpha )(3-4\alpha )r^3, \end{aligned}$$

This completes the proof of Theorem 3.3. $\square$

Corollary 3.4

Suppose $0\le \alpha \le \frac{1}{2}$, f is an almost starlike mapping of order $\alpha$ in $D^n$, and $\frac{D^2f_k(0)(z^2)}{2!}=z_k(\sum _{l=1}^n{a_{kl}z_l}), k=1,2,\ldots$, where $a_{ml}=\frac{1}{2!}\frac{\partial ^2f_m(0)}{\partial z_m\partial z_l}, m,l=1,2,\ldots ,n$, then

$$\begin{aligned} \frac{\Vert D^3f(0)(z^3)\Vert }{3!}\le (1-\alpha +\frac{1-2\alpha }{1-\alpha }\frac{M^2}{2})\Vert z\Vert ^3\le (1-\alpha )(3-4\alpha )\Vert z\Vert ^3, z\in D^n. \end{aligned}$$

where $M=\max _{1\le k\le n}\{\sum _{l=1}^n|a_{kl}|\}$. The above estimate is sharp.

Remark 3.5

Setting $\alpha =0$ and $f\in S^*(D^n)$ in Corollary 3.4, we can obtain Theorem 1.3 in [12].

Theorem 3.6

Suppose that $0\le \alpha <1$. If f is an almost starlike of order $\alpha$ in $D^n$, then

$$\begin{aligned}&\frac{\Vert D^3f(0)(z^3)\Vert }{3!}\le (1-\alpha )(5-4\alpha )\Vert z\Vert ^3,z\in D^n, \text{ if } n=2,\\&\frac{\Vert D^3f(0)(z^3)\Vert }{3!}\le (1-\alpha )(3\sqrt{3}+1-3\sqrt{3}\alpha )\Vert z\Vert ^3,z\in D^n, \text{ if } n\ge 3,\, n\in \mathbb {N}_+. \end{aligned}$$

Proof

For fixed $z\in D^n\setminus \{0\}$, define $z_0=\frac{z}{\Arrowvert z\Arrowvert }$. Since f is an almost starlike of order $\alpha$ in $D^n$, by Lemmas 2.3, 2.4 and 2.5, we have

$$\begin{aligned} \Arrowvert D^2f(0)(z^{2}_0)\Arrowvert \le 4(1-\alpha ),\Vert z_0\Vert =1. \end{aligned}$$

(3.2)

Apparently, ${\hat{L}}(z^2)=D^2f(0)(z^{2})$ is a bounded linear operator, we have

$$\begin{aligned} \Vert {\hat{L}}\Vert =\sup {\{\Vert {\hat{L}}(z^2)\Vert {:}\Vert z\Vert \le 1\}}=\sup {\{\Vert {\hat{L}}(z^2)\Vert {:}\Vert z\Vert =1\}}. \end{aligned}$$

(3.3)

According to (3.2) and (3.3), we obtain

$$\begin{aligned} \Vert {\hat{L}}\Vert \leqslant 4(1-\alpha ). \end{aligned}$$

Let $L(x,y)=D^2f(0)(x,y),x,y\in X$, by Lemma 2.9, we have

$$\begin{aligned} \Vert L\Vert\leqslant & {} 4(1-\alpha ),\, \text{ for } \, n=2,\\ \Vert L\Vert\le & {} 3\sqrt{3}(1-\alpha ),\, \text{ for } \, n\ge 3,n\in \mathbb {N}_+. \end{aligned}$$

Let $w=\frac{D^2f(0)(z^2)}{2!}$ and $\Vert w_0\Vert =\frac{\Vert D^2f(0)(z_{0}^2)\Vert }{2!}\le 2(1-\alpha )$, again by Lemma 2.9, we get that

$$\begin{aligned} \Vert D^2f(0)(z_0,w_0)\Vert\le & {} \Vert L\Vert \Vert z_0\Vert \Vert w_0\Vert \leqslant 8(1-\alpha )^2, \text{ if } n=2,\\ \Vert D^2f(0)(z_0,w_0)\Vert\le & {} \Vert L\Vert \Vert z_0\Vert \Vert w_0\Vert \leqslant 6\sqrt{3}(1-\alpha )^2, \text{ if } n\geqslant 3,\, n\in \mathbb {N}_+. \end{aligned}$$

According to Lemmas 2.2–2.4, we conclude that

$$\begin{aligned} \Big |-2\frac{D^3f(0)(z_0^3)}{3!}+2\frac{D^2f(0)}{2!}(z_0,w_0)\Big |\le 2(1-\alpha ). \end{aligned}$$

(3.4)

By (3.2) and (3.4), we have

$$\begin{aligned}&\frac{\Arrowvert D^3f(0)(z_0^3)\Arrowvert }{3!}\le {1-\alpha +\frac{1}{2}\Vert D^2f(0)(z_0,w_0)\Vert }\le (1-\alpha )(5-4\alpha ), \text{ if } n=2,\\&\frac{\Arrowvert D^3f(0)(z_0^3)\Arrowvert }{3!}\le (1-\alpha )(3\sqrt{3}+1-3\sqrt{3}\alpha ), \text{ if } n\ge 3,\, n\in \mathbb {N}_+, \end{aligned}$$

that is

$$\begin{aligned}&\frac{\Vert D^3f(0)(z^3)\Vert }{3!}\le (1-\alpha )(5-4\alpha )\Vert z\Vert ^3,z\in D^n, \text{ if } n=2,\\&\frac{\Vert D^3f(0)(z^3)\Vert }{3!}\le (1-\alpha )(3\sqrt{3}+1-3\sqrt{3}\alpha )\Vert z\Vert ^3,z\in D^n, \text{ if } n\ge 3,\, n\in \mathbb {N}_+. \end{aligned}$$

This completes the proof. $\square$

Theorem 3.7

Suppose $0\le \alpha \le \frac{37-\sqrt{505}}{36}$. If f is an almost starlike mapping of order $\alpha$ in $D^n$, and

$$\begin{aligned} \frac{D^2f_k(0)(z^2)}{2!}=a_kz_{k}^2,\quad \frac{D^3f_k(0)(z^3)}{3!}=b_kz_{k}^3, \qquad k=1,2,\ldots ,n, z\in D^n, \end{aligned}$$

where

$$\begin{aligned} a_{k}=\frac{1}{2!}\frac{\partial ^2f_k(0)}{\partial ^2 z_k},\quad b_{kl}=\frac{1}{3!}\frac{\partial ^3f_k(0)}{\partial ^3 z_k}, \qquad k,l=1,2,\ldots ,n, \end{aligned}$$

then

$$\begin{aligned} \frac{\Vert D^4f(0)(z^4)\Vert }{4!}\le \frac{(1-\alpha )(3-4\alpha )(4-6\alpha )}{3}\Vert z\Vert ^4, z\in D^n. \end{aligned}$$

The above estimate is sharp.

Proof

For fixed $z\in D^n\backslash \{0\}$, let $z_0=\frac{z}{\Vert z\Vert }$. Define $p_j(\xi )=\frac{g_j(\xi z_0)\Vert z\Vert }{\xi z_j}, \xi \in D$, where

$$\begin{aligned} g(z)=(g_1(z),g_2(z),\ldots ,g_n(z))'=(Df(z))^{-1}f(z) \end{aligned}$$

is a column vector in $\mathbb {C}^n$, $|z_j|=\Vert z\Vert =\max _{1\le k\le n}\{|z_k|\}$. Since f(z) is an almost starlike mapping of order $\alpha$ in $D^n$, we have $p_j(\xi )\in H(D), \mathrm{Re}p_j(\xi )\ge \alpha , p_j(0)=1$, and

$$\begin{aligned} p_j(\xi )=1+\frac{D^2g_j(0)(z_{0}^2)}{2!}\frac{\Vert z\Vert }{z_j}\xi +\cdots +\frac{D^{m}g_j(0)(z_{0}^{m})}{m!}\frac{\Vert z\Vert }{z_j}\xi ^{m-1}+\cdots ,\xi \in D. \end{aligned}$$

According to Lemma 2.2, we have

$$\begin{aligned}&\left| \frac{D^{4}g_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}-\frac{\frac{D^{2}g_j(0)(z_{0}^{2})}{2!}\frac{\Vert z\Vert }{z_j}\frac{D^{3}g_j(0)(z_{0}^{3})}{3!} \frac{\Vert z\Vert }{z_j}}{1-\alpha } +\frac{\left( \frac{D^{2}g_j(0)(z_{0}^{2})}{2!}\frac{\Vert z\Vert }{z_j}\right) ^3}{4(1-\alpha )^2}\right| \nonumber \\\le & {} 2(1-\alpha )-\frac{\left| \frac{D^{2}g_j(0)(z_{0}^{2})}{2!}\frac{\Vert z\Vert }{z_j}\right| ^2}{2(1-\alpha )}. \end{aligned}$$

(3.5)

From Lemma 2.4 and the conditions of Theorem 3.7, we have

$$\begin{aligned}&\frac{D^{2}g_j(0)(z_{0}^{2})}{2!}\frac{\Vert z\Vert }{z_j}=-a_j\frac{z_j}{\Vert z\Vert },\\&\frac{D^{3}g_j(0)(z_{0}^{3})}{3!}\frac{\Vert z\Vert }{z_j}=2a_{j}^2\left( \frac{z_j}{\Vert z\Vert }\right) ^2-2b_j\left( \frac{z_j}{\Vert z\Vert }\right) ^2,\\&\frac{D^{4}g_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}=-3\frac{D^{4}f_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}-4a_{j}^3\left( \frac{z_j}{\Vert z\Vert }\right) ^3 +7a_{j}b_{j}\left( \frac{z_j}{\Vert z\Vert }\right) ^3. \end{aligned}$$

Consequently, we have

$$\begin{aligned} -3\frac{D^{4}f_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}=\frac{D^{4}g_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}+4a_{j}^3\left( \frac{z_j}{\Vert z\Vert }\right) ^3 -7a_{j}b_{j}\left( \frac{z_j}{\Vert z\Vert }\right) ^3, \end{aligned}$$

(3.6)

and

$$\begin{aligned}&\Big |\frac{D^{4}g_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}+\frac{7-8\alpha }{4(1-\alpha )^2}a_{j}^3\left( \frac{z_j}{\Vert z\Vert }\right) ^3- \frac{2}{1-\alpha }a_{j}b_{j}\left( \frac{z_j}{\Vert z\Vert }\right) ^3\Big |\nonumber \\&\quad \le 2(1-\alpha )-\frac{|a_j|^2}{2(1-\alpha )}. \end{aligned}$$

(3.7)

According to Theorem 3.1, we have

$$\begin{aligned} \left| \frac{3-4\alpha }{2(1-\alpha )}a_{j}^2\left( \frac{z_j}{\Vert z\Vert }\right) ^2-2b_j\left( \frac{z_j}{\Vert z\Vert }\right) ^2\right| \le 2(1-\alpha )-\frac{|a_j|^2}{2(1-\alpha )}. \end{aligned}$$

(3.8)

Connecting (3.6), (3.7) and (3.8), we have

$$\begin{aligned} \left| -3\frac{D^{4}f_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}\right|= & {} \left| \frac{D^{4}g_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}+ 4a_{j}^3\left( \frac{z_j}{\Vert z\Vert }\right) ^3-7a_{j}b_{j}\left( \frac{z_j}{\Vert z\Vert }\right) ^3\right| \nonumber \\= & {} \left| \left( \frac{D^{4}g_j(0)(z_{0}^{4})}{4!}\frac{\Vert z\Vert }{z_j}\right. \right. \nonumber \\&\left. +\, \frac{7-8\alpha }{4(1-\alpha )^2}a_{j}^3\left( \frac{z_j}{\Vert z\Vert }\right) ^3-\frac{2}{1-\alpha }a_{j}b_{j}\left( \frac{z_j}{\Vert z\Vert }\right) ^3\right) \nonumber \\&+\,\frac{5-7\alpha }{2(1-\alpha )}a_j\frac{z_j}{\Vert z\Vert }\left[ \frac{3-4\alpha }{2(1-\alpha )}a_{j}^2\left( \frac{z_j}{\Vert z\Vert }\right) ^2 -2b_j\left( \frac{z_j}{\Vert z\Vert }\right) ^2\right] \\&\left. -\,\frac{12\alpha ^2-17\alpha +6}{4(1-\alpha )^2}a_{j}^3\left( \frac{z_j}{\Vert z\Vert }\right) ^3\right| \nonumber \\= & {} 2(1-\alpha )-\frac{|a_j|^2}{2(1-\alpha )}+\frac{5-7\alpha }{2(1-\alpha )}|a_j|\Big (2(1-\alpha ) -\frac{|a_j|^2}{2(1-\alpha )}\Big )\nonumber \\&+\,\frac{12\alpha ^2-17\alpha +6}{4(1-\alpha )^2}|a_{j}|^3\nonumber \\= & {} \frac{12\alpha ^2-10\alpha +1}{4(1-\alpha )^2}|a_{j}|^3-\frac{|a_j|^2}{2(1-\alpha )}+(5-7\alpha )|a_j|+2(1-\alpha ). \end{aligned}$$

By Lemma 2.10, we obtain

$$\begin{aligned} \Big |\frac{D^{4}f_j(0)(z_{0}^{4})}{4!}\Big |\leqslant & {} \frac{1}{3}\Big [\frac{12\alpha ^2-10\alpha +1}{4(1-\alpha )^2}M^3 -\frac{1}{2(1-\alpha )}M^2\\&+\,(5-7\alpha )M|+2(1-\alpha )\Big ]\\\leqslant & {} \frac{(1-\alpha )(3-4\alpha )(4-6\alpha )}{3}. \end{aligned}$$

Notice that $M=\max _{\{1\le k\le n\}}\{|a_k|\}\le 2(1-\alpha )$, by Lemma 2.6,

$$\begin{aligned} \frac{\Vert D^{4}f(0)(z^{4})\Vert }{4!}\le \frac{(1-\alpha )(3-4\alpha )(4-6\alpha )}{3}\Vert z\Vert ^4, z\in D^n. \end{aligned}$$

By applying the mapping $f_2(z)$ given by (3.1), we may verifies the accuracy of Theorem 3.7. This completes the proof. $\square$