双全纯映照是一类性质相对较好的映照,因此对其构造及性质的研究是多复变函数论中的主要研究内容之一.对双全纯映照研究的基础是单复变单叶解析函数的理论,在将单复变函数论中的结果向多复变数空间推广的过程中,人们发现单复变中的一些基本结论在多复变中不再成立,于是考虑对双全纯映照加以几何上的限制,例如星形性和凸性^[1],而具有特殊几何性质的双全纯映照,例如星形映照和凸映照,便成为主要的研究对象.

在单复变中比较容易构造具有某种特殊几何性质的双全纯函数,然而在多复变数空间中却相当困难. 1995年Roper-Suffridge算子^[2]的引入解决了这个问题. Roper-Suffridge算子具有很好的性质,人们证明了该算子保持星形性及Bloch性质^{[3, 4]},通过Roper-Suffridge算子我们可以由单复变中的星形函数构造多复变中的星形映照.后来人们将Roper-Suffridge算子进行了推广,发现推广后的Roper-Suffridge算子在一定条件下保持星形性、凸性和螺形性^{[5, 6]}.从而Roper-Suffridge算子可以视为单复变与多复变之间的一座桥梁,利用该算子及其推广我们可以由单复变中具有特殊几何性质的双全纯函数构造出多复变数空间中相应的映照.于是Roper-Suffridge算子成为了多复变函数论中的一个研究热点.而且,随着对具有特殊几何性质的双全纯映照的构造的研究,许多双全纯映照子族涌现了出来,因此讨论这些双全纯映照子族在Roper-Suffridge算子及其推广下的几何不变性便成为了一项重要的工作,于是许多学者在不同空间的不同区域上研究了Roper-Suffridge延拓算子及其推广.到目前为止关于Roper-Suffridge延拓算子已经有了许多很好的结论^[7-10].

2016年,唐言言^[11]在一类Bergman-Hartogs域

$\begin{eqnarray*}\Omega_{p_{1}, \cdots, p_{s}, q}^{B_{n}}&=&\Big\{(w_{(1)}, \cdots, w_{(s)}, z)\in C^{m_{1}}\times\cdots\times C^{m_{s}}\times B^n:\\&&\|w_{(1)}\|^{2p_{1}}+\cdots+\|w_{(s)}\|^{2p_{s}}<K_{B^n}(z, z)^{-q}\Big\}\end{eqnarray*}$

上讨论了推广的Roper-Suffridge延拓算子

(1.1)$F(w, z)=(w_{(1)}(f'(z_{1}))^{\delta_{1}}, \cdots, w_{(s)}(f'(z_{1}))^{\delta_{s}}, f(z_{1}), (f'(z_{1}))^{\gamma}z_{0})'$

保持$\alpha$次殆$\beta$型螺形性, $\alpha$次$\beta$型螺形性和强$\beta$型螺形性.

现在将算子(1.1)进一步推广为

(1.2)$F(w, z)=\left(w_{(1)}\Big(\frac{f(z_{1})}{z_{1}}\Big)^{\delta_{1}}, \cdots, w_{(s)}\Big(\frac{f(z_{1})}{z_{1}}\Big)^{\delta_{s}}, f(z_{1})+G\Big(\Big[\frac{f(z_{1})}{z_{1}}\Big]^\gamma z_0\Big), \Big(\frac{f(z_{1})}{z_{1}}\Big)^{\gamma}z_0\right)', $

其中$(w, z)=(w_{(1)}, \cdots, w_{(n)}, z)\in\Omega_{p_{1}, \cdots, p_{s}, q}^{B_{n}}$, $z=(z_{1}, z_{0})\in B^n$, $f(z_1)$是单位圆盘$D$上正规化的双全纯函数, $G$是$C^{n-1}$中的全纯映照,且$G(0)=0, DG(0)=I, \gamma\geq 0$.幂函数取分支使得$[f'(z_{1})]^{\delta_i}|_{z_{1}=0}=1$$(i=1, \cdots, s), $$[f'(z_{1})]^{\gamma}|_{z_{1}=0}=1$. $G(z)$的齐次展开式是$\sum\limits^{\infty}_{j=0}P_{j}(z)$,其中$P_{j}(z)$是$j$次齐次多项式.

在(1.2)式中令$w_{(1)}=\cdots=w_{(s)}=0$,则得到下述算子

(1.3)$F(z)=\left(f(z_{1})+G\Big(\Big[\frac{f(z_{1})}{z_{1}}\Big]^\gamma z_0\Big), \Big(\frac{f(z_{1})}{z_{1}}\Big)^{\gamma}z_0\right)'.$

本文主要讨论$\alpha$次$\beta$型螺形映照及复数$\lambda$阶殆星映照在Roper-Suffridge延拓算子(1.2)下的几何不变性.在第2节给出所要用到的定义及引理.在第3-5节,应用$\alpha$次$\beta$型螺形映照及复数$\lambda$阶殆星映照的几何性质及增长定理,讨论了算子(1.2)在Bergman-Hartogs域$\Omega_{p_{1}, \cdots, p_{s}, q}^{B_{n}}$上保持$\alpha$次$\beta$型螺形性及复数$\lambda$阶殆星性,从而得到Roper-Suffridge延拓算子(1.3)在一定条件下也保持同样的性质.

定义2.1 (Feng S X, Liu T S^[12])  设$\Omega$是$C^{n}$中的有界星形圆形域,其Minkowski泛函$\rho(z)$除去一个低维流形域外是$C^1$的.设$f:\Omega\rightarrow C^n$是$\Omega$上的正规化局部双全纯映照.若$\alpha\in{(0, 1)}$, $\beta\in{(-\frac{\pi}{2}, \frac{\pi}{2})}$,且

$\Big|(1-i\tan\beta)\frac{2}{\rho(z)}\frac{\partial\rho(z)}{\partial z}[Df(z)]^{-1}f(z)-\displaystyle\frac{1}{2\alpha}+i\tan\beta\Big|<\displaystyle\frac{1}{2\alpha}, \quad z\in\Omega\setminus\{0\}, $

则称$f$是$\Omega$上的$\alpha$次$\beta$型螺形映照.

在定义2.1中令$\beta=0$, $\Omega=B^{n}$,则得到$B^{n}$上$\alpha$次星形映照^[13]的定义, $D$上$\alpha$次星形函数的定义是由M S Robertson引入的^[14].

定义2.2 (赵燕红^[15])  设$\Omega$是$C^{n}$中的有界星形圆形域,其Minkowski泛函$\rho(z)$除去一个低维流形域外是$C^1$的.设$F(z)$是$\Omega$上正规化的局部双全纯映照.若

${\rm Re}\bigg[(1-\lambda)\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)\bigg]\geq-{\rm Re}\lambda, \quad z\in\Omega\setminus\{0\}, $

其中$\lambda\in{\Bbb C}$, ${\rm Re}\lambda\leq$0.则称$F(z)$是$\Omega$上的复数$\lambda$阶殆星映照.

在定义2.2中令$\lambda=\frac{\alpha}{\alpha-1}, \alpha\in[0, 1)$则得到$\Omega$上$\alpha$次殆星形映照的定义.

引理2.3 (Liu T S, Ren G B^[16])  设$\Omega\subset C^n$是有界星形圆形域,其Minkowski泛函$\rho(z)$除去一个低维流行$\Omega_{0}$外一阶可导,则$\forall z=(z_{1}, \cdots, z_{n})\in\Omega\setminus\Omega_{0}$有

$2{\rm Re}\frac{\partial\rho(z)}{\partial z}z=\rho(z), \quad\frac{\partial\rho}{\partial z}(\lambda z)=\frac{\partial\rho(z)}{\partial z}(\lambda\geq0), \quad\frac{\partial\rho}{\partial z}({\rm e}^{{\rm i}\theta}z)={\rm e}^{-{\rm i}\theta}\frac{\partial\rho(z)}{\partial z}(\theta\in R).$

引理2.4 (唐言言^[11])  $\rho(w, z)$是$\Omega_{p_{1}, \cdots, p_{s}, q}^{B_{n}}$上的Minkowski泛函,若$(w, z)\in\partial\Omega_{p_{1}, \cdots, p_{s}, q}^{B_{n}}$,则有$\rho(w, z)=1$及

$\frac{\partial\rho(w, z)}{\partial w_{ij}}=\frac{p_i\|w_{(i)}\|^{2p_i-2}\overline{w_{ij}}}{2\nabla_1+2\nabla_2}, \quad\frac{\partial\rho(w, z)}{\partialz_i}=\frac{\nabla_1\frac{\overline{z_i}}{\|z\|^2}}{2\nabla_1+2\nabla_2}, \quadi=1, \cdots, s, \quad j=1, \cdots, m_i, $

其中$\nabla_1=(n+1)q\pi^{nq}(n!)^{-q}(1-\|z\|^2)^{(n+1)q-1}\|z\|^2, \nabla_2=\Sigma^s_{k=1}p_k\|w_{(k)}\|^{2p_k}.$

引理2.5 (冯淑霞,刘太顺,任广斌^[17])  设$f(z)$是$n$维复Banach空间单位球$B$上的$\alpha$次$\beta$型螺形映照, $\alpha\in(0, 1)$, $\beta\in(-\pi/2, \pi/2)$.则

$\frac{\|x\|}{(1+\|x\|)^{2(1-\alpha)}}\leq\|f(x)\|\leq\frac{\|x\|}{(1-\|x\|)^{2(1-\alpha)}}, \quadx\in B.$

当$n=1$时有

$\frac{|z|}{(1+|z|)^{2(1-\alpha)}}\leq|f(z)|\leq\frac{|z|}{(1-|z|)^{2(1-\alpha)}}, \quadz\in D.$

引理2.6 (Duren P L^[18])  设$f(z)$是单位圆盘$D$上正规化的双全纯函数,则

$\frac{1-|z|}{(1+|z|)^3}\leq|f'(z)|\leq\frac{1+|z|}{(1-|z|)^3}, $

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

定理3.1   设$f(z_{1})$是$D$上正规化的$\alpha$次$\beta$型螺形函数, $\alpha\in[\frac{1}{2}, 1)$, $\beta\in{(-\frac{\pi}{2}, \frac{\pi}{2})}$,且$F(w, z)$是由(1.2)式所定义的映照, $p_i>1, \delta_i\in[0, 1]$$(i=1, \dots, s), (1-\alpha)\gamma<\frac{1}{4}$.幂函数取分支使得$(\frac{f(z_{1})}{z_{1}})^{\delta_{i}}|_{z_1=0}=1, (\frac{f(z_{1})}{z_{1}})^{\gamma}|_{z_1=0}=1$.令$\delta=\max\{p_1\delta_1, \cdots, p_s\delta_s\}$.若$q\geq\frac{\delta}{n+1}$,有

$P_j=0\ (j<\frac{2}{1-4(1-\alpha)\gamma}), \quad\sum\limits^\infty_{j=2}(j-1)\|P_j\|2^{1-2\alpha+\frac{j}{2}} \leq\frac{(1-\alpha)(1-\gamma)\cos\beta}{2\alpha}, $

则$F(w, z)$是$\Omega$上的$\alpha$次$\beta$型螺形映照.

证  由定义2.1下证

(3.1)$\begin{equation}\bigg|2\alpha\ (1-i\tan\beta)\frac{2}{\rho(w, z)}\frac{\partial\rho(w, z)}{\partial(w, z)}(DF(w, z))^{-1}F(w, z)-1+i2\alpha\tan\beta\bigg|<1.\end{equation}$

当$w=z_{0}=0$时(3.1)式显然成立.否则,令$(w, z)=\zeta(\xi, \eta)=|\zeta|{\rm e}^{{\rm i}\theta}(\xi, \eta)$,其中$(\xi, \eta)\in\partial\Omega$, $\zeta\in\bar{D}\setminus\{0\}$,则由引理2.3有

$\begin{eqnarray*} & &\frac{2}{\rho(w, z)}\frac{\partial\rho(w, z)}{\partial(w, z)}(DF(w, z))^{-1}F(w, z)\\ &=&\frac{2}{\rho(|\zeta|{\rm e}^{{\rm i}\theta}(\xi, \eta))}\frac{\partial\rho}{\partial(w, z)}(|\zeta|{\rm e}^{{\rm i}\theta}(\xi, \eta))(DF(\zeta\xi, \zeta\eta))^{-1} F(\zeta\xi, \zeta\eta)\\ &=&\frac{2}{|\zeta|}\frac{{\rm e}^{-{\rm i}\theta}\partial\rho}{\partial(w, z)}(\xi, \eta)(DF(\zeta\xi, \zeta\eta))^{-1}F(\zeta\xi, \zeta\eta)\\ &=&\frac{2\partial\rho}{\partial(w, z)}(\xi, \eta)\frac{(DF(\zeta\xi, \zeta\eta))^{-1}F(\zeta\xi, \zeta\eta)}{\zeta}.\end{eqnarray*}$

若固定$\xi, \eta$,则$\frac{2\partial\rho}{\partial(w, z)}(\xi, \eta)\frac{(DF(\zeta\xi, \zeta\eta))^{-1}F(\zeta\xi, \zeta\eta)}{\zeta}$关于$\zeta$全纯,由全纯函数的最大模原理知(3.1)式左端在$|\zeta|=1$上达到最大值,故只需证明当$(w, z)\in\partial\Omega$时(3.1)式成立即可.此时$\rho(w, z)=1$.令

(3.2)$\begin{equation}h(z_1)=2\alpha(1-i\tan\beta)\frac{f(z_1)}{z_1f'(z_1)}-1+i2\alpha\tan\beta.\end{equation}$

则$h(z_1)\in H(D), \quad|h(z_1)|<1.$令$g(z_1)=\frac{h(z_1)+1-2\alpha}{2\alpha(1-i\tan\beta)}$,则有$g(z_1)\in H(D), g(0)=0$且$|h(z_1)|=|2\alpha(1-i\tan\beta)g(z_1)-(1-2\alpha)|<1$, $\frac{2\alpha}{\cos\beta}|g(z_1)|-|1-2\alpha|<|2\alpha(1-i\tan\beta)g(z_1)-(1-2\alpha)|$.于是当$\alpha\in[\frac{1}{2}, 1)$时有$|g(z_1)|<\frac{\cos\beta}{2\alpha}(1+|1-2\alpha|)\leq\cos\beta\leq1$.由Schwarz引理得$|g(z_1)|\leq|z_1|$,则

(3.3)$\begin{equation}|h(z_1)+1-2\alpha|\leq\frac{2\alpha}{\cos\beta}|z_1|.\end{equation}$

令

$2\alpha(1-i\tan\beta)\frac{2\partial\rho(w, z)}{\partial(w, z)}(DF(w, z))^{-1}F(w, z)-1+i2\alpha\tan\beta=I.$

由(1.2)式经计算可得$(DF(w, z))^{-1}F(w, z)=(h_1, \cdots, h_{s+n})'$,其中

$\left\{\begin{array}{ll} h_i=w_{(i)}\bigg[1-\delta_i\frac{z_1f'-f}{z_1f'}\bigg(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f(z_{1})}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\bigg)\bigg], \ \ \ i=1, \cdots, s, \\[4mm] h_{s+1}=\frac{f}{f'}+\frac{1}{f'}\sum\limits^\infty_{j=2}(\frac{f(z_{1})}{z_{1}})^{\gamma j}(1-j)P_j(z_0), \\[4mm] h_{s+j}=z_j\bigg[1-\gamma\frac{z_1f'-f}{z_1f'}\bigg(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f(z_{1})}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\bigg)\bigg], \ \ \ j=2, \cdots, n.\end{array} \right.$

由引理2.4得

(3.4)$\begin{equation}\frac{2\partial\rho}{\partial(w, z)}(w, z)(DF(w, z))^{-1}F(w, z)=\frac{G(w, z)}{\nabla_{1}+\nabla_{2}}, \end{equation}$

其中

$\begin{eqnarray*} G(w, z)&=&\sum\limits^s_{k=1}p_k\|w_{(k)}\|^{2p_k}\Big[1-\delta_k\frac{z_1f'-f}{z_1f'}\Big(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big)\Big] \\&&+\frac{|z_1|^2\nabla_{1}}{\|z\|^2}\frac{f}{z_1f'}\Big[1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big] \\ &&+\frac{\|z_0\|^2\nabla_{1}}{\|z\|^2}\Big[1-\gamma\frac{z_1f'-f}{z_1f'}\Big(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gammaj}(1-j)P_j(z_0)\Big)\Big].\end{eqnarray*}$

于是由(3.2)及(3.4)式得

$\begin{eqnarray*} (\nabla_1+\nabla_2)I &=&2\alpha(1-i\tan\beta) \bigg\{\sum\limits^s_{k=1}p_k\|w_{(k)}\|^{2p_k}\\ &&\times\Big[1-\delta_k\frac{z_1f'-f}{z_1f'}\Big(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big)\Big]\\&&+\frac{|z_1|^2\nabla_{1}}{\|z\|^2}\frac{f}{z_1f'}\Big[1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big]\\ &&+\frac{\|z_0\|^2\nabla_{1}}{\|z\|^2}\Big[1-\gamma\frac{z_1f'-f}{z_1f'}\Big(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gammaj}(1-j)P_j(z_0)\Big)\Big]\bigg\}\\&&+(i2\alpha\tan\beta-1)(\nabla_1+\nabla_2)\\ &=&2\alpha(1-i\tan\beta)\nabla_2+\widetilde{\nabla_2}[h(z_1)+1-2\alpha] \Big[1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big]\\ &&+\frac{|z_1|^2\nabla_{1}}{\|z\|^2}[h(z_1)+1-i2\alpha\tan\beta]\Big[1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big]\\ &&+(i2\alpha\tan\beta-1)(\nabla_1+\nabla_2)+\frac{\|z_0\|^2\nabla_{1}}{\|z\|^2}\bigg\{2\alpha(1-i\tan\beta)\\&&+\gamma [h(z_1)+1-2\alpha]\Big[1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big]\bigg\}\\ &=&(2\alpha-1)(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}+(2\alpha-1)(\nabla_2-\widetilde{\nabla_2}) \\ && +\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big]h(z_1)\\ &&+\Big[\widetilde{\nabla_2}(h(z_1)+1-2\alpha)+\frac{|z_1|^2\nabla_{1}}{\|z\|^2}(h(z_1)+1-i2\alpha\tan\beta)\\&&+\frac{\gamma\|z_0\|^2\nabla_{1}}{\|z\|^2}(h(z_1)+1-2\alpha)\Big]\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\\ &=&(2\alpha-1)(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}+(2\alpha-1)(\nabla_2-\widetilde{\nabla_2}) \\ && + \bigg[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\bigg]h(z_1)\\&&+\bigg[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\bigg](h(z_1)+1-2\alpha) \frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\\ &&+2\alpha(1-i\tan\beta)\frac{|z_1|^2\nabla_{1}}{\|z\|^2}\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0), \end{eqnarray*}$

其中$\widetilde{\nabla_2}=\sum\limits^s_{k=1}\delta_kp_k\|w_k\|^{2p_k}$.由$(w, z)\in\Omega_{p_{1}, \cdots, p_{s}, q}^{B_{n}}$知$\sum\limits^s_{k=1}\|w_{(k)}\|^{2p_k}< (\frac{\pi^n}{n!})^q(1-\|z\|^2)^{(n+1)q}.$又$\nabla_1=(n+1)q\pi^{nq}(n!)^{-q}(1-\|z\|^2)^{(n+1)q-1}\|z\|^2, \quad\widetilde{\nabla_2}<\delta\sum\limits^s_{k=1}\|w_k\|^{2p_k}, $则$\frac{\widetilde{\nabla_2}}{\nabla_1}<\frac{\delta(1-\|z\|^2)}{(n+1)q\|z\|^2}.$于是当$q\geq\frac{\delta}{n+1}$时有

(3.5)$\begin{eqnarray} \frac{\widetilde{\nabla_2}}{\nabla_1}\|z\|^2 +|z_1|^2+\gamma\|z_0\|^2 &<&\frac{\delta}{(n+1)q}(1-\|z\|^2)+\|z\|^2+(\gamma-1)\|z_0\|^2 \\ &=&\frac{\delta}{(n+1)q}+(1-\frac{\delta}{(n+1)q})\|z\|^2+(\gamma-1)\|z_0\|^2 \\ &<&1.%(3.5)\end{eqnarray}$

由(3.3), (3.5)式及引理2.5知,当$(1-\alpha)\gamma<\frac{1}{4}$且

$P_j=0\ (j<\frac{2}{1-4(1-\alpha)\gamma}), \quad\sum\limits^\infty_{j=2}(j-1)\|P_j\|2^{1-2\alpha+\frac{j}{2}} \leq\frac{(1-\alpha)(1-\gamma)\cos\beta}{2\alpha}$

时有

$\begin{eqnarray*}(\nabla_1+\nabla_2)(|I|-1) &<&(2\alpha-1)(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}+(2\alpha-1)(\nabla_2-\widetilde{\nabla_2}) \\ && +\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big]-(\nabla_1+\nabla_2)\\ &&+\bigg\{\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big]\frac{2\alpha}{\cos\beta}|z_l| +\frac{2\alpha}{\cos\beta}\frac{|z_1|\nabla_{1}}{\|z\|^2}\bigg\}\\ &&\times\frac{1}{|f|}\sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)|P_j(z_0)|\\ &\leq&2(\alpha-1)(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}+2(\alpha-1)(\nabla_2-\widetilde{\nabla_2})\\&&+\frac{2\alpha}{\cos\beta}\frac{\nabla_{1}}{\|z\|^2} \Big[\frac{\widetilde{\nabla_2}}{\nabla_1}\|z\|^2+|z_1|^2 +\gamma\|z_0\|^2+1\Big]\\ &&\times\frac{|z_1|}{|f|}\sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)\|P_j\|\|z_0\|^j\\ &<&2(\alpha-1)(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}\\&& +\frac{4\alpha}{\cos\beta}\frac{\nabla_{1}}{\|z\|^2}\frac{|z_1|}{|f|}\sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)\|P_j\|\|z_0\|^j\\ &\leq&\frac{4\alpha}{\cos\beta}\frac{\|z_0\|^2\nabla_{1}}{\|z\|^2}\Big[\sum\limits^\infty_{j=2}(j-1)\|P_j\|(1+|z_1|)^{2(1-\alpha)+\frac{j}{2}-1}\\&&\times (1-|z_1|)^{2(\alpha-1)\gamma j+\frac{j}{2}-1} -\frac{(1-\alpha)(1-\gamma)\cos\beta}{2\alpha}\Big]\\ &\leq&\frac{4\alpha}{\cos\beta}\frac{\|z_0\|^2\nabla_{1}}{\|z\|^2}\Big[\sum\limits^\infty_{j=2}(j-1)\|P_j\|2^{1-2\alpha+\frac{j}{2}} -\frac{(1-\alpha)(1-\gamma)\cos\beta}{2\alpha}\Big]\\ &\leq&0, \end{eqnarray*}$

故(3.1)式成立,于是定理得证.

在定理3.1中令$w_{(1)}=\cdots=w_{(s)}=0$则有以下结论.

推论3.2  设$f(z_{1})$是$D$上正规化的$\alpha$次$\beta$型螺形函数, $\alpha\in[\frac{1}{2}, 1)$, $\beta\in{(-\frac{\pi}{2}, \frac{\pi}{2})}$. $F(z)$是(1.3)式所定义的映照, $(1-\alpha)\gamma<\frac{1}{4}$.幂函数取分支使得$(\frac{f(z_{1})}{z_{1}})^{\gamma}|_{z_1=0}=1$.若

$P_j=0\ (j<\frac{2}{1-4(1-\alpha)\gamma}), \quad\sum\limits^\infty_{j=2}(j-1)\|P_j\|2^{1-2\alpha+\frac{j}{2}} \leq\frac{(1-\alpha)(1-\gamma)\cos\beta}{2\alpha}, $

则$F(z)$是$B^n$上的$\alpha$次$\beta$型螺形映照.

定理3.3  设$f_1(z_1)\cdots, f_n(z_n)$是$D$上的$\alpha$次$\beta$型螺形函数, $\alpha\in(0, 1)$, $\beta\in{(-\frac{\pi}{2}, \frac{\pi}{2})}$,

$F(w, z)=(w_{(1)}, \cdots, w_{(s)}, f_1(z_{1}), \cdots, f_n(z_{n}))', \quad(w, z)\in\Omega, p_i>1.$

则$F(w, z)$是$\Omega$上的$\alpha$次$\beta$型螺形映照.

证  由$F(w, z)$的表达式经计算可知

$(DF(w, z))^{-1}F(w, z)=\left(w_{(1)}, \cdots, w_{(s)}, \frac{f_1(z_1)}{f_1'(z_1)}, \cdots, \frac{f_1(z_n)}{f_n'(z_n)}\right)'.$

于是由引理2.4得

$\frac{2\partial\rho}{\partial(w, z)}(w, z)(DF(w, z))^{-1}F(w, z)=\frac{1}{\nabla_{1}+\nabla_{2}}\bigg[\sum\limits^s_{k=1}p_k\|w_{(k)}\|^{2p_k}+\sum\limits^n_{m=1}\frac{\nabla_1|z_m|^2}{\|z\|^2}\frac{f_m}{z_mf_m'}\bigg].$

令

$p_m(z_m)=2\alpha(1-i\tan\beta)\frac{f_m(z_m)}{z_mf_m'(z_m)}-1+i2\alpha\tan\beta.$

与定理3.1同理可得结论成立.

推论3.4  设$f_1(z_1)\cdots, f_n(z_n)$是$D$上的$\alpha$次$\beta$型螺形函数, $\alpha\in(0, 1)$, $\beta\in{(-\frac{\pi}{2}, \frac{\pi}{2})}$,

$F(z)=(f_1(z_{1}), \cdots, f_n(z_{n}))', \quad z\in B^n.$

则$F(z)$是$B^n$上的$\alpha$次$\beta$型螺形映照.

注3.5  在定理3.1, 3.3及推论3.2, 3.4中令$\beta=0$,则得到相应的$\alpha$次星形映照的结论.

定理4.1  设$f(z_{1})$是$D$上的复数$\lambda$阶殆星函数, $\lambda\in{\Bbb C}$, ${\rm Re}\lambda\leq$0, $F(w, z)$是(1.2)式所定义的映照, $p_i>1, \delta_i\in[0, 1]$$(i=1, \dots, s), \gamma\in[0, \frac{1}{4})$.幂函数取分支使得$(\frac{f(z_{1})}{z_{1}})^{\delta_{i}}|_{z_1=0}=1, $$(\frac{f(z_{1})}{z_{1}})^{\gamma}|_{z_1=0}=1$.令$\delta=\max\{p_1\delta_1, \cdots, p_s\delta_s\}$.若$q\geq\frac{\delta}{n+1}$, $P_j=0(j<\frac{4}{1-4\gamma})$且

$\sum\limits^\infty_{j=2}2^{\frac{j}{2}+1}(j-1)\|P_j\|\leq\frac{1-\gamma}{2+|1-\lambda|}, $

则$F(w, z)$是$\Omega$上的复数$\lambda$阶殆星映照.

证  由定义2.2下证

(4.1)${\rm Re}\bigg[(1-\lambda)\frac{2}{\rho(w, z)}\frac{\partial\rho}{\partial(w, z)}(w, z)(DF(w, z))^{-1}F(w, z)+\lambda\bigg]\geq0.$

与定理3.1同理可知${\rm Re}\big[(1-\lambda)\frac{2}{\rho(w, z)}\frac{\partial\rho}{\partial(w, z)}(w, z)(DF(w, z))^{-1}F(w, z)+\lambda\big]$是一全纯函数的实部,从而是调和函数.由调和函数的最小值原理知,我们只需证明当$(w, z)\in\partial\Omega$时(4.1)式成立即可.此时$\rho(w, z)=1$.

由于$f(z_{1})$是$D$上的复数$\lambda$阶殆星函数,则${\rm Re}[(1-\lambda)\frac{f(z_{1})}{z_{1}f'(z_{1})}+\lambda]\geq0.$令

(4.2)$h(z_{1})=(1-\lambda)\frac{f(z_{1})}{z_{1}f'(z_{1})}+\lambda.$

则${\rm Re} h(z_{1})>0$, $h(0)=1$.令$g(z_1)=\frac{h(z_{1})-1}{h(z_{1})+1}$.则$g(z_{1})\in H(D), |g(z_{1})|<1, g(0)=0$.由Schwarz引理有$|g(z_{1})|\leq|z_1|$,即$|\frac{h(z_{1})-1}{h(z_{1})+1}|\leq|z_1|$,于是

(4.3)$|h(z_{1})-1|\leq\frac{2|z_1|}{1-|z_1|}.$

由(4.2)及(3.4)式得

$\begin{eqnarray*}& &(\nabla_1+\nabla_2)\bigg[(1-\lambda)\frac{2\partial\rho}{\partial(w, z)}(w, z)(DF(w, z))^{-1}F(w, z)+\lambda\bigg]\\ &=&(1-\lambda)\Big\{\sum\limits^s_{k=1}p_k\|w_{(k)}\|^{2p_k}\Big[1-\delta_k\frac{z_1f'-f}{z_1f'}\Big(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big)\Big]\\ &&+\frac{|z_1|^2\nabla_{1}}{\|z\|^2}\frac{f}{z_1f'}\Big[1+\frac{1}{f} \sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big]\\ &&+\frac{\|z_0\|^2\nabla_{1}}{\|z\|^2}\Big[1-\gamma\frac{z_1f'-f}{z_1f'}\Big(1+\frac{1}{f}\sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gammaj}(1-j)P_j(z_0)\Big)\Big]\Big\}+\lambda(\nabla_1+\nabla_2)\\ &=&(1-\lambda)\nabla_2+\widetilde{\nabla_2}(h(z_1)-1)\Big[1+\frac{1}{f} \sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big]\\ &&+\frac{|z_1|^2\nabla_{1}}{\|z\|^2}(h(z_1)-\lambda)\Big[1+\frac{1}{f} \sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\Big]\\ &&+\frac{\|z_0\|^2\nabla_{1}}{\|z\|^2}\Big[1-\lambda+\gamma(h(z_1)-1) (1+\frac{1}{f} \sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0))\Big]+\lambda(\nabla_1+\nabla_2)\\ &=&(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}+\nabla_2-\widetilde{\nabla_2} +\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big]h(z_1)\\ &&+\Big[\widetilde{\nabla_2}(h(z_1)-1)+\frac{\nabla_1|z_1|^2}{\|z\|^2}(h(z_1)-\lambda) +\frac{\gamma\|z_0\|^2}{\|z\|^2}\nabla_1(h(z_1)-1)\Big]\frac{1}{f} \sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\\ &=&(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}+\nabla_2-\widetilde{\nabla_2} +\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big]h(z_1)\\ &&+\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big](h(z_1)-1)\frac{1}{f} \sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0)\\ &&+(1-\lambda)\frac{\nabla_1|z_1|^2}{\|z\|^2}\frac{1}{f} \sum\limits^\infty_{j=2}(\frac{f}{z_{1}})^{\gamma j}(1-j)P_j(z_0).\end{eqnarray*}$

由(4.3), (3.5)式及引理2.6知,当$P_j=0(j<\frac{4}{1-4\gamma})$及$\sum\limits^\infty_{j=2}2^{\frac{j}{2}+1}(j-1)\|P_j\|\leq\frac{1-\gamma}{2+|1-\lambda|}$时有

$\begin{eqnarray*}& &(\nabla_1+\nabla_2){\rm Re}\bigg[(1-\lambda)\frac{2\partial\rho}{\partial(w, z)}(w, z)(DF(w, z))^{-1}F(w, z)+\lambda\bigg]\\ &\geq&(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}+\nabla_2-\widetilde{\nabla_2} +\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big]{\rm Re} h(z_1)\\ &&-\Big[\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big]\frac{2|z_1|}{1-|z_1|}\frac{1}{|f|} \sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)|P_j(z_0)|\\ &&-|1-\lambda|\frac{\nabla_1|z_1|^2}{\|z\|^2}\frac{1}{|f|} \sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)|P_j(z_0)|\\ &\geq&(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}-\Big[\Big(\widetilde{\nabla_2}+\frac{\nabla_1}{\|z\|^2}(|z_1|^2+\gamma\|z_0\|^2)\Big) \frac{2}{1-|z_1|}+|1-\lambda|\frac{\nabla_1}{\|z\|^2}\Big]\\ &&\cdot|\frac{z_1}{f}|\sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)|P_j(z_0)|\\ &\geq&(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}-\frac{\nabla_1}{\|z\|^2}\Big[2\Big(\frac{\widetilde{\nabla_2}}{\nabla_1}\|z\|^2 +|z_1|^2+\gamma\|z_0\|^2\Big) +|1-\lambda|(1-|z_1|)\Big]\\ &&\times\frac{1}{1-|z_1|}|\frac{z_1}{f}|\sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)\|P_j\|\|z_0\|^j\\ &\geq&(1-\gamma)\frac{\|z_0\|^2\nabla_1}{\|z\|^2}-\frac{\nabla_1}{\|z\|^2}(2+|1-\lambda|)\frac{1}{1-|z_1|} |\frac{z_1}{f}|\sum\limits^\infty_{j=2}|\frac{f}{z_{1}}|^{\gamma j}(j-1)\|P_j\|\|z_0\|^j\\ &\geq&\frac{\|z_0\|^2\nabla_1}{\|z\|^2}\Big[(1-\gamma)-(2+|1-\lambda|)\frac{1}{1-|z_1|} \\ && \times\sum\limits^\infty_{j=2}(1+|z_1|)^2(1-|z_1|)^{-2\gamma j}(j-1)\|P_j\|(1-|z_1|^2)^{\frac{j}{2}-1}\Big]\\ &=&\frac{\|z_0\|^2\nabla_1}{\|z\|^2}\Big[(1-\gamma)-(2+|1-\lambda|) \sum\limits^\infty_{j=2}(1+|z_1|)^{\frac{j}{2}+1}(1-|z_1|)^{\frac{j}{2}-2\gamma j-2}(j-1)\|P_j\|\Big]\\ &\geq&\frac{\|z_0\|^2\nabla_1}{\|z\|^2}\Big[(1-\gamma)-(2+|1-\lambda|) \sum\limits^\infty_{j=2}2^{\frac{j}{2}+1}(j-1)\|P_j\|\Big]\\ &\geq&0.\end{eqnarray*}$

于是(4.1)式成立,则定理得证.

在定理4.1中令$w_{(1)}=\cdots=w_{(s)}=0$,则有以下结论.

推论4.2  设$f(z_{1})$是$D$上的复数$\lambda$阶殆星函数, $\lambda\in{\Bbb C}, {\rm Re}\lambda\leq$0. $F(z)$是(1.3)式所定义的映照, $\gamma\in[0, \frac{1}{4})$.幂函数取分支使得$(\frac{f(z_{1})}{z_{1}})^{\gamma}|_{z_1=0}=1$.若$P_j=0(j<\frac{4}{1-4\gamma})$且

$\sum\limits^\infty_{j=2}2^{\frac{j}{2}+1}(j-1)\|P_j\|\leq\frac{1-\gamma}{2+|1-\lambda|}, $

则$F(z)$是$B^n$上的复数$\lambda$阶殆星映照.

定理4.3  设$f_1(z_1)\cdots, f_n(z_n)$是$D$上的复数$\lambda$阶殆星函数, $\lambda\in{\Bbb C}, $${\rm Re}\lambda\leq$0.

$F(w, z)=(w_{(1)}, \cdots, w_{(s)}, f_1(z_{1}), \cdots, f_n(z_{n}))', \quad(w, z)\in\Omega, p_i>1.$

则$F(w, z)$是$\Omega$上的复数$\lambda$阶殆星映照.

证  令

$p_m(z_m)=(1-\lambda)\frac{f_m(z_m)}{z_mf_m'(z_m)}+\lambda.$

与定理4.1同理可证.

推论4.4  设$f_1(z_1)\cdots, f_n(z_n)$是$D$上的复数$\lambda$阶殆星函数, $\lambda\in{\Bbb C}$, ${\rm Re}\lambda\leq$0.

$F(z)=(f_1(z_{1}), \cdots, f_n(z_{n}))', \quad z\in B^n.$

则$F(z)$是$B^n$上的复数$\lambda$阶殆星映照.

注4.5  在定理4.1, 4.3及推论4.2, 4.4中令$\lambda=\frac{\alpha}{\alpha-1}, \alpha\in[0, 1)$,则得到相应的$\alpha$次殆星形映照的结论.