在单复变几何函数论中有许多优美的结果. 很自然的,人们讨论是否可以将 这些结论推广到多复变数中. 1933年,Cartan^[1]建议考虑具有特殊几何性质的双全纯映照, 例如星形映照及凸映照. 之后许多人开始研究这两类映照. 到目前为止, 关于星形映照和凸映照已经有了许多很好的结论. 许多学者开始讨论它们的子族.

Rønning在1991年引入了抛物星形函数的概念.

定义 1.1 (Rønning F^[2])设$f(z)$是单位圆盘$D$上正规化的解析函数,若 $$\left|\frac{zf'(z)}{f(z)}-1\right| <\Re\frac{zf'(z)}{f(z)}-\alpha,~~z\in D,\alpha\in{[0,1]}, $$ 则称$f(z)$是$D$上的抛物星形函数.

1993年Rønning在研究单位圆盘$D$上的一致凸函数的性质时将抛物星形函数的定义做了进一步的修改.

定义 1.2 (Rønning F^[3])设$f(z)$是单位圆盘$D$上正规化的解析函数,若 $$\Big|\frac{zf'(z)}{f(z)}-1\Big| <\Re\frac{zf'(z)}{f(z)},~~ z\in D, $$ 则称$f(z)$是$D$上的抛物星形函数.

后来Ali 对文献^[1]中定义的参数做了适当的修改,定义了$\rho$次抛物星形函数$(\rho\in[0,1))$.

定义 1.3 (Ali R M^[4])设$f(z)$是单位圆盘$D$上正规化的解析函数,若 $$\left|\frac{zf'(z)}{f(z)}-1\right| <(1-2\rho)+\Re\frac{zf'(z)}{f(z)},~~z\in D,\rho\in{(0,1)}, $$ 则称$f(z)$是$D$上的$\rho$次抛物星形函数.

Hamada,Honda,Kohr将Ali的定义推广到$C^{n}$中单位球$B^{n}$上^[5],冯淑霞,张晓飞和陈慧勇^[6]对Ali所给出的抛物星形函数和$\rho$次的抛物星形函数做了合理的修改, 引入了抛物形$\beta$型螺形函数和 $\rho$次的抛物形$\beta$型螺形函数的定义,并讨论了在不同空间上Roper-Suffridge算子保持相应的抛物形$\beta$型螺形映照和 $\rho$次的抛物形$\beta$型螺形映照的性质.

定义 1.4 (冯淑霞,张晓飞,陈慧勇^[6]) 设$f(z)$是单位圆盘$D$上正规化的解析函数,若 $\beta\in(-\pi/2,\pi/2),\cos\beta>1/(1+\rho)$,且 $$\left|\textrm{e}^{-\textrm{i}\beta}\frac{f(z)}{zf'(z)}-(1-\textrm{i}\sin\beta)\right| <(1-2\rho)+\Re\left(\textrm{e}^{-\textrm{i}\beta}\frac{f(z)}{zf'(z)}\right). $$ 则称$f(z)$是$D$上的$\rho\in{(0,1)}$次抛物形$\beta$型螺形函数.

1995年Roper-Suffridge算子^[7]的引入使得我们可以由单复变中具有特殊几 何性质的双全纯映照构造多复变中相应的映照,因此许多学者开始在不同空间的不同区域 上研究Roper-Suffridge 延拓算子. 到目前为止关于Roper-Suffridge延拓算子已经有了许多很好的结论^{[8, 9, 10, 11, 12]}.

本文是对文献^[6]中定义的抛物形$\beta$型螺形函数和 $\rho$次的抛物形$\beta$型螺形函数做适当的修改,从而定义螺形映照的新子族. 在第2章,从映照的几何意义出发,给出单复变中$\rho$次椭圆星形函数和$\rho$次椭圆形$\beta$型 螺形函数的定义,并将它们推广到多复变数不同空间的不同区域中. 在第3--5章,本文 分别讨论了在$C^{n}$中单位球$B^{n}$上、在复Banach空间及复Hilbert 空间单位球上、 在Reinhardt域上推广的Roper-Suffridge延拓算子保持$\rho$次椭圆形$\beta$型螺形性.

定义 2.1 若$f(z)$是单位圆盘$D$上正规化的解析函数,$\rho\in{(0,1)}$,

由定义2.1知$\Re\frac{f(z)}{zf' (z)}>0$,此时$f(z)$是星形函数,从而是双全纯的. (2.1)式表明$\frac{f(z)}{zf' (z)}$将单位圆盘映为右半平面内由椭圆 $\{w=u+{\rm i}v|(u-\frac{1}{1-\rho^{2}})+\frac{v^{2}}{1-\rho^{2}}< (\frac{\rho}{1-\rho^{2}})^{2}\}$所围成的椭圆区域,并且经简单计算可知该区域相对于 $\zeta=\frac{1}{1-\rho^{2}}$是星形域,因此这里我们称这类域为椭圆星形域. 而螺形函数是星形函数的一种扩充函数,那么是否可以将$\rho$次椭圆星形函数与螺形函 数类结合起来呢？ 若将定义2.1中的不等式改变为 $$ \bigg|{\rm e}^{-{\rm i}\beta}\frac{f(z)}{zf' (z)}-(1-{\rm i}\sin\beta)\bigg| <\rho\Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z)}{zf' (z)}\bigg],\quad \beta\in\Big(-\frac{\pi}{2},\frac{\pi}{2}\Big),\quad z\in D.$$ 则显然有$\Re[{\rm e}^{-{\rm i}\beta}\frac{f(z)}{zf' (z)}]>0$,此时$f(z)$是螺形函数, 从而也是双全纯的.经计算可知${\rm e}^{-{\rm i}\beta}\frac{f(z)}{zf' (z)}$将单位 圆盘映为右半平面内由椭圆 $\{w=u+{\rm i}v|(u-\frac{1}{1-\rho^{2}})+\frac{(v+\sin\beta)^{2}}{1-\rho^{2}}< (\frac{\rho}{1-\rho^{2}})^{2}\}$所围成的椭圆区域中,并且该区域相对于 $\zeta=\frac{1}{1-\rho^{2}}-{\rm i}\sin\beta$是星形域. 又由于当$z\rightarrow 0$时${\rm e}^{-{\rm i}\beta}\frac{f(z)}{zf' (z)}\rightarrow {\rm e}^{-{\rm i}\beta}$,于是${\rm e}^{-{\rm i}\beta}$在上述椭圆区域, 这就要求$\cos\beta>\frac{1}{1+\rho}$. 由此我们给出如下螺形函数的子类.

定义 2.2 若$f(z)$是单位圆盘$D$上正规化的解析函数, $\rho\in{(0,1)}$,$\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$,且 $$\bigg|{\rm e}^{-{\rm i}\beta}\frac{f(z)}{zf' (z)}-(1-{\rm i}\sin\beta)\bigg| <\rho\Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z)}{zf' (z)}\bigg],\quad z\in D.$$ 则称$f(z)$是$D$上$\rho$次椭圆形$\beta$型螺形函数.

在定义2.2中若$\beta=0$,即为定义2.1.

定理 2.3 令$\rho\in{(0,1)}$, $\beta\in(-\pi/2,\pi/2),\cos\beta>1/(1+\rho)$,且

证 令 $$w_{1}(z)=\frac{b-\overline{b}z}{1-z},\quad w_{2}(z)=\frac{\sqrt{w_{1}(z)}+1}{\sqrt{w_{1}(z)}-1},\quad w(z)=\frac{2}{\textrm{i}}w_{2}(z),\quad z\in D.$$ 则 $w_{1}(z)$将单位圆盘映到上半平面,且$0$被映成$b$. $w_{2}(z)$将上半平面映成单位圆盘的上半部分,且$b$被映成$\textrm{i}/2$. $w(z)$ 将单位圆盘的上半部分映成半径为2的圆盘的 右半部分,且$\textrm{i}/2$被映成$1$. 所以$|w|<2$且$\arg w\in(-\pi/2,\pi/2)$. 由(2.2)式有 $$f'(z)=\exp\int^{z}_{0}\left\{\textrm{e}^{-\textrm{i}\beta}\left[1-\textrm{i}\sin\beta+\frac{a}{\pi^{2}}\left(\ln w(x)\right)^{2}\right]^{-1}-1\right\}\frac{\textrm{d}x}{x} \cdot\frac{\textrm{e}^{-\textrm{i}\beta}}{1-\textrm{i}\sin\beta+\frac{a}{\pi^{2}}\left(\ln w(z)\right)^{2}}.$$ 故 $$\frac{f(z)}{zf'(z)}=\textrm{e}^{\textrm{i}\beta}(1-\textrm{i}\sin\beta)+\textrm{e}^{\textrm{i}\beta}\frac{a}{\pi^{2}}\left(\ln w(z)\right)^{2}.$$ 于是

又显然$f(z)$是正规化的, 由定义2.2可知$f(z)$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数.

注 2.4 若在定理2.3中令$\beta=0$,则有 $$f(z)=z\exp\int^{z}_{0}\left\{\left[1+\frac{a}{\pi^{2}}\left(\ln w(x)\right)^{2}\right]^{-1}-1\right\}\frac{\textrm{d}x}{x}$$ 是$D$上的$\rho$次椭圆星形函数.

如果将定义2.1推广到不同空间不同区域上,则有以下定义.

定义 2.5 若$F(z)$是$B^{n}$上正规化的双全纯映照, $\rho\in{(0,1)}$,$\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$, $$|{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle-(1-{\rm i}\sin\beta)\|z\|^{2}| <\rho {\rm Re}\{{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle\},\quad z\in B^{n},$$ 则称$F(z)$是$B^{n}$上$\rho$次椭圆形$\beta$型螺形映照.

定义 2.6 若$F(x)$是Banach空间单位球$B$上正规化的双全纯映照, $\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$,且 $$\bigg|{\rm e}^{-{\rm i}\beta}\frac{1}{\|x\|}T_{x}[(DF(x))^{-1}F(x)]-(1-{\rm i}\sin\beta)\bigg| <\rho \Re\bigg\{{\rm e}^{-{\rm i}\beta}\frac{1}{\|x\|}T_{x}[(DF(x))^{-1}F(x)]\bigg\},\ x\in B,$$ 则称$F(x)$是$B$上$\rho$次椭圆形$\beta$型螺形映照.

定义 2.7 若$F(z)$是有界完全Reinhardt域$\Omega$上正规化的局部双全纯映照, $\rho(z)$表示$\Omega$上的Minkowski泛函,$\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$,且 $$ \bigg|{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)-(1-{\rm i}\sin\beta)\bigg|<\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)\bigg],$$ 则称 $F(z)$是$\Omega$上的$\rho$次椭圆形$\beta$型螺形映照.

定理3.1 设$f_{j}(1\leq j\leq n)$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,$\rho\in{(0,1)}$,$\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$, $$F(z)=(f_{1}(z_{1}),f_{2}(z_{2}),\cdots,f_{n}(z_{n}))' ,$$ 则$F(z)$是$C^{n}$中单位球$B^{n}$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

证 由于$f_{j}(1\leq j\leq n)$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,则 $$\bigg|{\rm e}^{-{\rm i}\beta}\frac{f_{j}(z_{j})}{z_{j}f_{j}' (z_{j})}-(1-{\rm i}\sin\beta)\bigg| <\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f_{j}(z_{j})}{z_{j}f_{j}' (z_{j})}\bigg],$$ 由$F(z)$的表达式可知$F(z)$是正规化的,且 \begin{eqnarray*} &&|{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle-(1-{\rm i}\sin\beta)\|z\|^{2}|\\ &=&\bigg|{\rm e}^{-{\rm i}\beta}\sum\limits^{n}_{j=1}|z_{j}|^{2}\frac{f_{j}(z_{j})}{z_{j}f_{j}' (z_{j})} -(1-{\rm i}\sin\beta)\sum\limits^{n}_{j=1}|z_{j}|^{2}\bigg| \\ &=&\sum\limits^{n}_{j=1}|z_{j}|^{2} \bigg|{\rm e}^{-{\rm i}\beta}\frac{f_{j}(z_{j})}{z_{j}f_{j}' (z_{j})}-(1-{\rm i}\sin\beta)\bigg| <\rho\sum\limits^{n}_{j=1}|z_{j}|^{2} \Re \bigg[{\rm e}^{-{\rm i}\beta}\frac{f_{j}(z_{j})}{z_{j}f_{j}' (z_{j})}\bigg] \\ &=&\rho\Re\{{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle\}, \end{eqnarray*} 由定义2.5知$F(z)$是$C^{n}$中单位球$B^{n}$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

定理3.2 设$f(z_{1})$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,$\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$,且 $$F(z)= \bigg(f(z_{1}),\Big(\frac{f(z_{1})}{z_{1}}\Big)^{\beta_{2}}z_{2},\cdots, \Big(\frac{f(z_{1})}{z_{1}}\Big)^{\beta_{n}}z_{n}\bigg)' ,\quad z\in B^{n},$$ 其中$\beta_{j}\in[0, 1]$,幂级数取分支使得$(\frac{f(z_{1})}{z_{1}})^{\beta_{j}}\mid_{z_{1}=0}=1(j=2,\cdots,n)$, 则$F(z)$是$C^{n}$中单位球$B^{n}$上的$\rho$次椭圆形$\beta$型螺形映照.

证 由于$f(z_{1})$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,则 $$\bigg|{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg| <\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg],$$ 又由$F(z)$的表达式经简单计算可知$F(z)$是正规化的,且 $$J^{-1}_{F}(z)F(z)=\bigg(\Big(\frac{f(z_{1})}{f' (z_{1})}\Big), \cdots,z_{j}\Big(1-\beta_{j}+\beta_{j}\frac{f(z_{1})}{z_{1}f' (z_{1})}\Big), \cdots\bigg)' ,\quad 2\leq j\leq n,$$ 则 $$\langle J^{-1}_{F}(z)F(z),z\rangle=|z_{1}|^{2} \frac{f(z_{1})}{z_{1}f' (z_{1})}+\sum\limits^{n}_{j=2}|z_{j}|^{2}\bigg(1-\beta_{j}+ \beta_{j}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg),$$ 则 \begin{eqnarray*} &&|{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle-(1-{\rm i}\sin\beta)\|z\|^{2}| \\ &=&\bigg||z_{1}|^{2} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg] +\displaystyle\sum\limits^{n}_{j=2}|z_{j}|^{2}\beta_{j} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg] \\ &&+\displaystyle\sum\limits^{n}_{j=2}|z_{j}|^{2}\beta_{j}(1-{\rm i}\sin\beta)+\sum\limits^{n}_{j=2}|z_{j}|^{2}{\rm e}^{-{\rm i}\beta}(1-\beta_{j})-\sum\limits^{n}_{j=2}(1-{\rm i}\sin\beta)|z_{j}|^{2} \bigg| \\ &=&\bigg||z_{1}|^{2}\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta) \bigg]+\displaystyle\sum\limits^{n}_{j=2}|z_{j}|^{2}\beta_{j} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg] \\ &&+\displaystyle\sum\limits^{n}_{j=2}(\cos\beta-1)(1-\beta_{j})|z_{j}|^{2}\bigg| \\ &\leq& |z_{1}|^{2}\bigg|{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})} -(1-{\rm i}\sin\beta)\bigg|+\displaystyle\sum\limits^{n}_{j=2}|z_{j}|^{2}\beta_{j} \bigg|{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg| \\ &&+\displaystyle\sum\limits^{n}_{j=2}(1-\cos\beta)(1-\beta_{j})|z_{j}|^{2} \\ &<&\bigg(|z_{1}|^{2}+\displaystyle\sum\limits^{n}_{j=2}|z_{j}|^{2}\beta_{j}\bigg)\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] +\displaystyle\sum\limits^{n}_{j=2}(1-\cos\beta)(1-\beta_{j})|z_{j}|^{2}, \end{eqnarray*} 而 \begin{eqnarray*} &&\rho\Re\{{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle\}\\ &=&\rho \Re\bigg\{{\rm e}^{-{\rm i}\beta}|z_{1}|^{2}\frac{f(z_{1})}{z_{1}f' (z_{1})} +{\rm e}^{-{\rm i}\beta}\sum\limits^{n}_{j=2}|z_{j}|^{2}\Big(1-\beta_{j}+ \beta_{j}\frac{f(z_{1})}{z_{1}f' (z_{1})}\Big)\bigg\} \\ &=&\rho\bigg\{|z_{1}|^{2}\Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})} {z_{1}f' (z_{1})}\bigg] +\sum\limits^{n}_{j=2}|z_{j}|^{2} \Re\bigg[{\rm e}^{-{\rm i}\beta}\Big((1-\beta_{j})+\beta_{j} \frac{f(z_{1})}{z_{1}f' (z_{1})}\Big)\bigg] \bigg\} \\ &=&\rho\bigg(|z_{1}|^{2}+\displaystyle\sum\limits^{n}_{j=2}\beta_{j}|z_{j}|^{2}\bigg) \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] +\rho\displaystyle\sum\limits^{n}_{j=2}\cos\beta(1-\beta_{j})|z_{j}|^{2}, \end{eqnarray*} 于是当$\cos\beta>\frac{1}{1+\rho}$时有 $$|{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle-(1-{\rm i}\sin\beta)\|z\|^{2}|<\rho \Re\{{\rm e}^{-{\rm i}\beta}\langle J^{-1}_{F}(z)F(z),z\rangle\},$$ 由定义2.5知$F(z)$是$C^{n}$中单位球$B^{n}$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

以下$X$表示复Banach空间,$B=\{x\in X:\|x\|<1\}$表示$X$中单位球,$X^{\ast}$是$X$的对偶空间, 对任意的$x\in X\backslash\{0\}$,$T_{x}=\{T_{x}\in X^{\ast}:\|T_{x}\|=1,T_{x}(x)=\|x\|\}$为连续线性泛函, 由Hahn-Banach定理知此集合非空.

引理4.1 (崔艳艳,王朝君^[13]) 设$f$是$D$上正规化双全纯映照, $$F(x)=\Phi_{\beta_{2},\cdots,\beta_{n-1},0}(f)(x)=\sum_{j=1}^{n-1} \bigg(\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)}\bigg)^{\beta_{j}} T_{x_{j}}(x)x_{j}+x-\sum_{j=1}^{n-1}T_{x_{j}}(x)x_{j},$$ 其中$n\in N(n\geq2),\beta_{1}=1,0\leq\beta_{j}\leq1,\quad j=2,\cdots,n-1$, 且$(\frac{f(z)}{z})^{\beta_{j}}|_{z=0}=1$, $j=1,\cdots,n-1$,$x_{1}\in\bar B$,$\|x_{1}\|=1$, 并且$x_{1},\cdots,x_{n}\in X$线性无关,对任意$x_{i}$, 选取$T_{x_{i}}\in X^{\ast}$,使$\|T_{x_{i}}\|=1$,且 $T_{x_{i}}(x_{i})=1$,$T_{x_{i}}(x_{j})=0,(i\neq j)$ (由Hahn-Banach定理及其推论知此条件可取到), 则 $$\|x\|T_{x}[(DF(x))^{-1}F(x)]=\|x\|^{2}+\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2} \bigg[\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}-1\bigg].$$

定理4.2 设$f$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数, $\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$, $F(x)$为引理4.1中所定义的映照,则$F(x)$是$B$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

证 由定义2.6需证 $$ \bigg|{\rm e}^{-{\rm i}\beta}\frac{1}{\|x\|}T_{x}[(DF(x))^{-1}F(x)]-(1-{\rm i}\sin\beta)\bigg| <\rho \Re\bigg\{{\rm e}^{-{\rm i}\beta}\frac{1}{\|x\|}T_{x}[(DF(x))^{-1}F(x)]\bigg\}, $$ 即 $$ |{\rm e}^{-{\rm i}\beta}\|x\|T_{x}[(DF(x))^{-1}F(x)]-(1-{\rm i}\sin\beta)\|x\|^{2}| <\rho\Re\{{\rm e}^{-{\rm i}\beta}\|x\|T_{x}[(DF(x))^{-1}F(x)]\}.$$ 由于$f$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,则 $$ \bigg|{\rm e}^{-{\rm i}\beta}\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}-(1-{\rm i}\sin\beta)\bigg| <\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}\bigg]. $$ 由引理4.1知 \begin{eqnarray*} &&|{\rm e}^{-{\rm i}\beta}\|x\|T_{x}[(DF(x))^{-1}F(x)]-(1-{\rm i}\sin\beta)\|x\|^{2}| \\ &=&\bigg|{\rm e}^{-{\rm i}\beta}\bigg\{\|x\|^{2}+\displaystyle\sum_{j=1}^{n-1} \beta_{j}|T_{x_{j}}(x)|^{2} \bigg[\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}-1\bigg]\bigg\} -(1-{\rm i}\sin\beta)\|x\|^{2}\bigg| \\ &=&\bigg|\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}-(1-{\rm i}\sin\beta) \bigg] \\ &&+ \sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2}(1-\cos\beta) +(\cos\beta-1)\|x\|^{2}\bigg| \\ &=&\bigg|\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}-(1-{\rm i}\sin\beta) \bigg] \\ &&+(1-\cos\beta)\bigg[\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2}-\|x\|^{2}\bigg] \bigg| \\ &<&\rho\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2}\Re \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}\bigg]+ (1-\cos\beta)\bigg[\|x\|^{2}-\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2}\bigg], \end{eqnarray*} 而 \begin{eqnarray*} &&\rho\Re\{{\rm e}^{-{\rm i}\beta}\|x\|T_{x}[(DF(x))^{-1}F(x)]\} \\ &=&\rho \Re\bigg\{{\rm e}^{-{\rm i}\beta}\bigg[\|x\|^{2}+\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2} \Big(\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}-1\Big)\bigg]\bigg\} \\ &=&\rho\cos\beta\|x\|^{2} +\rho\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2}\Re \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}\bigg] -\rho \cos\beta\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2} \\ &=&\rho\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2} \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)f' (T_{x_{1}}(x))}\bigg] +\rho \cos\beta\bigg(\|x\|^{2}-\displaystyle\sum_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2}\bigg), \end{eqnarray*} 这里由$\beta_{1}=1,0\leq\beta_{j}\leq1~(j=2,\cdots,n-1)$ 知$\|x\|^{2}-\sum\limits_{j=1}^{n-1}\beta_{j}|T_{x_{j}}(x)|^{2}\geq 0$,于是当$\cos\beta>\frac{1}{1+\rho}$时有 $|{\rm e}^{-{\rm i}\beta}\|x\|T_{x}[(DF(x))^{-1}F(x)]-(1-{\rm i}\sin\beta)\|x\|^{2}| <\rho\Re\{{\rm e}^{-{\rm i}\beta}\|x\|T_{x}[(DF(x))^{-1}F(x)]\},$ 则$F(x)$是$B$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

若$X$是$n$维复Hilbert空间, 则由Riesz表示定理知$T_{x_{1}}(x)=\langle x,x_{1}\rangle$, 取$x_{1}=(1,\cdots,0)$, 则$\|x\|=1$,$x=(z_{1},\cdots,z_{n})=(z_{1},z_{0})$, 则有$T_{x_{1}}(x)=z_{1}$,于是引理4.1所定义的函数为

推论4.3 令$F(z)$为式(4.1)定义的函数,其中$n\in N(n\geq2),\beta_{1}=1,0\leq\beta_{j}\leq1,$ $j=2,\cdots,n-1$, 且$(\frac{f(z)}{z})^{\beta_{j}}|_{z=0}=1$ $(j=1,\cdots,n-1)$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$, 则$F(z)$在复Hilbert空间单位球上保持$\rho$次椭圆形$\beta$型螺形性.

与定理4.2同理可得以下结论.

定理4.4 设$f$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,$\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$,且 $$F(x)=f(T_{x_{1}}(x))x_{1}+\Big(\frac{f(T_{x_{1}}(x))}{T_{x_{1}}(x)}\Big)^{\gamma}(x-T_{x_{1}}(x)x_{1}),$$ 其中$0\leq\gamma\leq1$, 且$(\frac{f(z)}{z})^{\gamma}|_{z=0}=1$,$x_{1}\in\bar B$,$\|x_{1}\|=1$, 则$F(x)$是复Banach空间单位球$B$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

推论4.5 设$f$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,$\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$, $0\leq\gamma\leq1$,$z=(z_{1},z_{0})\in B$, $F(z)=(f(z_{1}),(\frac{f(z_{1})}{z_{1}})^{\gamma}z_{0}),$ 其中$(\frac{f(z_{1})}{z_{1}})^{\gamma}|_{z=0}=1$, 则$F(z)$是复Hilbert空间单位球上正规化的$\rho$次椭圆形$\beta$型螺形映照.

引理5.1 (Liu T S,Zhang W J^[14])设$\rho(z)$是 $D_{p}=\{z=(z_{1},\cdots,z_{n})\in C^{n}:\sum\limits^{n}_{j=1}|z_{j}|^{p_{j}}<1\}$ 上的Minkowski泛函,$z\in D_{p}\setminus\{0\}$,则 $$\frac{\partial\rho}{\partial z_{j}}(z)= \frac{p_{j}\bar{z}_{j}|\frac{z_{j}}{\rho(z)}|^{p_{j}-2}}{2\rho(z) \bigg[\sum\limits^{n}_{k=1}p_{k} |\frac{z_{k}}{\rho(z)}|^{p_{k}}\bigg]}.$$ 引理5.2 (刘名生,朱玉灿^[15]) 设有界完全Reinhardt域$\Omega$的Minkowski泛函$\rho(z)$是$\bar{\Omega}\setminus\{0\}$上的一个$C^{1}$函数,则 对于$z=(z_{1},\cdots,z_{n})' \in\Omega\setminus\{0\}$有$\frac{\partial\rho (z)}{\partial z_{j}}z_{j}\geq 0(j=1,\cdots,n)$,且$\rho (z)=2\sum\limits^{n}_{j=1}\frac{\partial\rho (z)}{\partial z_{j}}z_{j}$.

定理5.3 设 $f$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,$\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$,$r=\sup\{|z_{1}|:(z_{1},\cdots,z_{n})' \in\Omega\}$, 且 $$F(z)=\bigg(rf\Big(\frac{z_{1}}{r}\Big), \Big(\frac{rf(\frac{z_{1}}{r})}{z_{1}}\Big)^{\beta_{2}}z_{2},\cdots, \Big(\frac{rf(\frac{z_{1}}{r})}{z_{1}}\Big)^{\beta_{n}}z_{n}\bigg)' ,$$ 其中$0\leq\beta_{j}\leq1$, 且$(\frac{rf(\frac{z_{1}}{r})}{z_{1}})^{\beta_{j}}|_{z_{1}=0}=1,j=2,\cdots,n$, 则$F(z)$是Reinhardt域$D_{p}=\{z=(z_{1},\cdots,z_{n})\in C^{n}:\sum\limits^{n}_{j=1}|z_{j}|^{p_{j}}<1\}$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

证 令$\omega=\frac{z_{1}}{r}$, 由于$f$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,则 $$ \bigg|{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}-(1-{\rm i}\sin\beta)\bigg| <\rho\Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}\bigg]. $$ 由$F(z)$的表达式可知$F(z)$是正规化的,且 $$J^{-1}_{F}(z)F(z)= \bigg(\frac{rf(\omega)}{f' (\omega)}, \Big(1-\beta_{2}+\beta_{2}\frac{f(\omega)}{\omega f' (\omega)}\Big)z_{2},\cdots, \Big(1-\beta_{n}+\beta_{n}\frac{f(\omega)}{\omega f' (\omega)}\Big)z_{n}\bigg)' .$$ 若记$\sigma=\sum\limits^{n}_{k=1}p_{k}| \frac{z_{k}}{\rho(z)}|^{p_{k}}$,则由引理5.1及引理5.2有 \begin{eqnarray*} &&\bigg|{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)-(1-{\rm i}\sin\beta)\bigg| \\ &=&\bigg|\frac{1}{\sigma}\bigg\{p_{1} \bigg|\frac{z_{1}}{\rho(z)}\bigg|^{p_{1}}{\rm e}^{-{\rm i}\beta}\displaystyle \frac{f(\omega)}{\omega f' (\omega)}+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}} \bigg[(1-\beta_{j}){\rm e}^{-{\rm i}\beta}+\beta_{j}{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}\bigg]\bigg\}\\ &&-(1-{\rm i}\sin\beta)\bigg| \\ &=&\bigg|\frac{1}{\sigma} \bigg\{p_{1}\bigg|\frac{z_{1}}{\rho(z)} \bigg|^{p_{1}}{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}} \bigg[(1-\beta_{j}){\rm e}^{-{\rm i}\beta}+\beta_{j}{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}\bigg]\bigg\} \\ &&-\frac{1}{\sigma}\displaystyle\sum\limits^{n}_{j=1}p_{j}\Big|\frac{z_{j}}{\rho(z)}\Big|^{p_{j}}(1-\textrm{i}\sin\beta)\Big| \\ &=&\bigg|\frac{1}{\sigma}\bigg\{p_{1} \bigg|\frac{z_{1}}{\rho(z)}\bigg|^{p_{1}} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}-(1-{\rm i}\sin\beta)\bigg]+\sum\limits^{n}_{j=2} p_{j}\bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}}\beta_{j} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}-(1-{\rm i}\sin\beta)\bigg] \\ &&+\displaystyle\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}} \big[(1-\beta_{j}){\rm e}^{-{\rm i}\beta}+\beta_{j}(1-{\rm i}\sin\beta)\big] \bigg\} -\frac{1}{\sigma}\displaystyle\sum\limits^{n}_{j=1}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}}(1-{\rm i}\sin\beta)\bigg| \\ &\leq& \frac{1}{\sigma}\bigg\{ \bigg[p_{1}\bigg|\frac{z_{1}}{\rho(z)} \bigg|^{p_{1}}+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}}\beta_{j}\bigg] \bigg|{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}-(1-{\rm i}\sin\beta)\bigg|\\ &&+\sum\limits^{n}_{j=2}p_{j}\bigg|\frac{z_{j}}{\rho(z)} \bigg|^{p_{j}}(1-\beta_{j}) (1-\cos\beta)\bigg\} \\ &<&\frac{1}{\sigma}\bigg\{ \bigg[p_{1}\bigg|\frac{z_{1}}{\rho(z)} \bigg|^{p_{1}}+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}}\beta_{j}\bigg] \rho\Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}\bigg]+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}} (1-\beta_{j})(1-\cos\beta)\bigg\}, \end{eqnarray*} 而 \begin{eqnarray*} &&\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)\bigg] \\ &=&\rho\frac{1}{\sigma}\Re \bigg\{p_{1}\bigg|\frac{z_{1}}{\rho(z)}\bigg|^{p_{1}}{\rm e}^{-{\rm i}\beta} \frac{f(\omega)}{\omega f' (\omega)}+\sum\limits^{n}_{j=2}p_{j}\bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}} \bigg[(1-\beta_{j}){\rm e}^{-{\rm i}\beta}+\beta_{j}{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}\bigg]\bigg\} \\ &=&\rho\frac{1}{\sigma}\bigg\{p_{1} \bigg|\frac{z_{1}}{\rho(z)}\bigg|^{p_{1}}\Re\bigg[{\rm e}^{-{\rm i}\beta}\displaystyle \frac{f(\omega)}{\omega f' (\omega)}\bigg]+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}}(1-\beta_{j})\cos\beta+\beta_{j}\Re \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}\bigg]\bigg\} \\ &=&\rho\frac{1}{\sigma} \bigg\{\bigg[p_{1}\bigg|\frac{z_{1}}{\rho(z)}\bigg|^{p_{1}}+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}}\beta_{j}\bigg] \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(\omega)}{\omega f' (\omega)}\bigg]+\sum\limits^{n}_{j=2}p_{j} \bigg|\frac{z_{j}}{\rho(z)}\bigg|^{p_{j}} (1-\beta_{j})\cos\beta\bigg\}, \end{eqnarray*} 于是当$\cos\beta>\frac{1}{1+\rho}$时有 $$\bigg|{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)-(1-{\rm i}\sin\beta)\bigg|<\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)\bigg],$$ 由定义2.7知 $F(z)$是$D_{p}$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

2009年阮林要^[16]将Roper-Suffridge算子推广为 $$F(z)=\bigg(f(z_{1}),\Big(\frac{f(z_{1})}{z_{1}}\Big)^{\beta_{2}}\widetilde{z_{2}},\cdots, \Big(\frac{f(z_{1})}{z_{1}}\Big)^{\beta_{k}}\widetilde{z_{k}}\bigg)' ,$$ 并证明了该算子在$\Omega' _{N}=\{z=(z_{1},\widetilde{z_{2}},\cdots,\widetilde{z_{k}})\in C\times C^{n_{2}}\times \cdots\times C^{n_{k}}:|z_{1}|^{p_{1}}+\sum\limits^{k}_{j=2}\|\widetilde{z_{j}}\|^{p_{j}}_{j}<1\}$上保持$\alpha$次殆 $\beta$型螺形性及$\alpha$次 $\beta$型螺形性, 其中$\|\cdot\|_{j}$是$C^{n_{j}}~(n_{j}\in N,j=2,\cdots,k)$上的Banach范数, 且$p_{j}\geq1~(j=1,\cdots,k)$, $N=1+\sum\limits^{k}_{j=2}n_{j}$. 下面我们讨论该算子在$\Omega' _{N}$上也保持$\rho$次椭圆形$\beta$型螺形性.

定理5.4 设 $f(z_{1})$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,$\rho\in{(0,1)}$, $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$, $\cos\beta>\frac{1}{1+\rho}$,且 $$F(z)=\bigg(f(z_{1}),\Big( \frac{f(z_{1})}{z_{1}}\Big)^{\beta_{2}}\widetilde{z_{2}},\cdots, \Big(\frac{f(z_{1})}{z_{1}}\Big)^{\beta_{k}}\widetilde{z_{k}}\bigg)' ,$$ 其中$0\leq\beta_{j}\leq1$, 且$(\frac{f(z_{1})}{z_{1}})^{\beta_{j}}|_{z_{1}=0}=1,j=2,\cdots,k$, 则$F(z)$在Reinhardt域$\Omega' _{N}=\{z=(z_{1},\widetilde{z_{2}},\cdots, \widetilde{z_{k}})\in C\times C^{n_{2}}\times \cdots\times C^{n_{k}}:|z_{1}|^{p_{1}}+\sum\limits^{k}_{j=2}\|\widetilde{z_{j}} \|^{p_{j}}_{j}<1\}$上是正规化的$\rho$次椭圆形$\beta$型螺形映照.

证 由于$f(z_{1})$是$D$上正规化的$\rho$次椭圆形$\beta$型螺形函数,则 $$\bigg|{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg| <\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg],$$ 由$F(z)$的表达式可知$F(z)$是正规化的,且 \begin{eqnarray*} J^{-1}_{F}(z)= \left( \begin{array}{cccc} \frac{1}{f' (z_{1})} &0&\cdots&0\\[3mm] s_{2}&\Big(\frac{f(z_{1})}{z_{1}}\Big)^{-\beta_{2}}I^{n_{2}}_{2}&\cdots&0 \\[3mm] \vdots&\vdots&\ & \vdots \\[3mm] s_{n}&0& \cdots&\Big (\frac{f(z_{1})}{z_{1}}\Big)^{-\beta_{k}}I^{n_{k}}_{k} \end{array} \right). \end{eqnarray*} 其中$s_{j}=\beta_{j}(\frac{1}{z_{1}f' (z_{1})}-\frac{1}{f(z_{1})})\widetilde{z_{j}}$, $I^{n_{j}}_{j}$表示第$j$行第$j$列的$n_{j}$阶单位方阵, 由引理5.1经简单计算可知 $$\frac{\partial\rho(z)}{\partial z}J^{-1}_{F}(z)F(z) =\frac{f(z_{1})}{z_{1}f' (z_{1})}\frac{\partial\rho(z)}{\partial z_{1}}z_{1} +\sum\limits^{k}_{j=2}\bigg[1-\beta_{j}+\beta_{j}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] \frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}},$$ 则由引理5.2有 \begin{eqnarray*} &&\bigg|{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)-(1-{\rm i}\sin\beta)\bigg| \\ &=& \bigg|{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)} \bigg\{\frac{f(z_{1})}{z_{1}f' (z_{1})}\frac{\partial\rho(z)}{\partial z_{1}}z_{1} +\sum\limits^{k}_{j=2} \bigg[1-\beta_{j}+\beta_{j}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg]\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg\}-(1-{\rm i}\sin\beta)\bigg| \\ &=&\bigg|\frac{2}{\rho(z)}\bigg\{\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] \frac{\partial\rho(z)}{\partial z_{1}}z_{1}+\sum\limits^{k}_{j=2}\beta_{j}\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] \frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\\ &&+{\rm e}^{-{\rm i}\beta}\sum\limits^{k}_{j=2} (1-\beta_{j})\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg\} -\frac{2}{\rho(z)}(1-{\rm i}\sin\beta)\bigg[\frac{\partial\rho(z)}{\partial z_{1}}z_{1}+\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg]\bigg| \\ &=&\bigg|\frac{2}{\rho(z)}\bigg\{\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})} -(1-{\rm i}\sin\beta)\bigg]\frac{\partial\rho(z)}{\partial z_{1}}z_{1}\\ &&+\sum\limits^{k}_{j=2}\beta_{j} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg] \frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}} +{\rm e}^{-{\rm i}\beta}\displaystyle\sum\limits^{k}_{j=2}(1-\beta_{j}) \frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\\ && +\displaystyle\sum\limits^{k}_{j=2} (\beta_{j}-1)(1-{\rm i}\sin\beta)\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg\}\bigg| \\ &=&\bigg|\frac{2}{\rho(z)}\bigg\{\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}- (1-{\rm i}\sin\beta)\bigg]\frac{\partial\rho(z)}{\partial z_{1}}z_{1}\\ &&+\sum\limits^{k}_{j=2}\beta_{j} \bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}-(1-{\rm i}\sin\beta)\bigg] \frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}} +\displaystyle\sum\limits^{k}_{j=2}(1-\beta_{j})(\cos\beta-1)\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg\}\bigg| \\ &<&\frac{2}{\rho(z)}\bigg\{\bigg[\frac{\partial\rho(z)}{\partial z_{1}}z_{1}+\sum\limits^{k}_{j=2}\beta_{j}\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg]\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] \\ && +\displaystyle\sum\limits^{k}_{j=2}(1-\beta_{j}) (1-\cos\beta)\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg\} \\ &=&\frac{2}{\rho(z)}\bigg[\frac{\partial\rho(z)}{\partial z_{1}}z_{1}+\sum\limits^{k}_{j=2}\beta_{j}\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg]\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] \\ && +\displaystyle\sum\limits^{k}_{j=2}\frac{2}{\rho(z)}(1-\beta_{j})(1-\cos\beta) \frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}, \end{eqnarray*} 而 \begin{eqnarray*} &&\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)\bigg] \\ &=&\rho \Re\bigg\{{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)} \bigg[\frac{f(z_{1})}{z_{1}f' (z_{1})}\frac{\partial\rho(z)}{\partial z_{1}}z_{1} +\sum\limits^{k}_{j=2}(1-\beta_{j}+\beta_{j}\frac{f(z_{1})}{z_{1}f' (z_{1})}) \frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg]\bigg\} \\ &=&\frac{2}{\rho(z)}\bigg[\frac{\partial\rho(z)}{\partial z_{1}}z_{1}+\sum\limits^{k}_{j=2}\beta_{j}\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}\bigg]\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{f(z_{1})}{z_{1}f' (z_{1})}\bigg] +\displaystyle\sum\limits^{k}_{j=2}2 \cos\beta(1-\beta_{j})\frac{\partial\rho(z)}{\partial \widetilde{z_{j}}}\widetilde{z_{j}}, \end{eqnarray*} 于是当$\cos\beta>\frac{1}{1+\rho}$时有 $$ \bigg|{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)-(1-{\rm i}\sin\beta)\bigg|<\rho \Re\bigg[{\rm e}^{-{\rm i}\beta}\frac{2}{\rho(z)}\frac{\partial\rho}{\partial z}(z)J^{-1}_{F}(z)F(z)\bigg], $$ 由定义2.7知 $F(z)$是$\Omega' _{N}$上正规化的$\rho$次椭圆形$\beta$型螺形映照.

注 5.5 与定理5.3同理可证定理3.1中所讨论的算子 $F(z)$在$D_{p}$上也保持$\rho$次椭圆形$\beta$型螺形性,且在上述所有结论中,若令$\beta=0$, 即得到相应的关于$\rho$次椭圆星形映照的结论.