在单复变函数中, 众所周知双全纯函数的偏差定理是成立的, 如下所示.

定理1.1^[1]  若 $f$为复平面单位圆盘 $U$上正规化双全纯函数, 则 $\frac{1-|z|}{(1+|z|)^{3}}\leq|f^{'}(z)|\leq\frac{1+|z|}{(1-|z|)^{3}}.$

然而, 在多复变量的情况下, Cartan^[2]指出正规化的双全纯映照相应的偏差定理不再成立.而且他建议人们去研究双全纯星形映照, 双全纯凸映照及其它重要的双全纯映照子族精确的偏差定理.遗憾的是, 目前, 我们仅能获得星形映照, 螺形映照及其扩充子族的增长与掩盖定理(相关结果可见文献[3-4]), 但对应的偏差结果仍处于猜测阶段.近年来, 双全纯凸(准凸)映照的偏差定理的研究已经获得一些可喜成果(如文献[5-9]).但目前星形映照与螺形映照的偏差估计研究成果还较少.在这篇文章中, 针对一类 $\alpha$次殆 $\beta$型螺形映照, 我们建立了相应的偏差上界估计.

全文中用 $C$表示复平面, $U$表示复平面上的单位圆盘; $B_{n}$为复向量空间 $C_{n}$中的开单位球, $X$为具有任意范数 $\|\cdot\|$的复Banach空间, $B$为 $X$中的开单位球, $X^{\ast}$为 $X$的对偶空间.对每一 $x\in X\backslash\{0\}$, 定义 $T(x)=\{T_{x}\in X^{\ast}:\|T_{x}\|=1, T_{x}(x)=\|x\|\}$.由Hahn-Banach定理 $T(x)$非空.对任意 $\alpha(\neq0)\in C$, 对应于 $T_{x}\in T(x)$, $\frac{|\alpha|}{\alpha}T_{x}\in T(\alpha x)$, 我们总是记 $\frac{|\alpha|}{\alpha}T_{x}=T_{\alpha x}$.用 $H(U, C)$表示 $U$到 $C$的全纯函数的全体, $H(B)$表示 $B$到 $X$的全纯映射的全体.若Fréchet导数 $Df(x)$在每一点 $x\in B$有有界逆, 称 $f$为 $B$上的局部双全纯映射; 若逆映射 $f^{-1}$存在, 且在 $f(B)$上全纯, 称 $f$为 $B$上的双全纯映射; 若 $f\in H(B)$满足 $f(0)=0, Df(0)=I$, 其中 $I$为单位矩阵, 称 $f$为 $B$上正规化全纯映射.

对于Reinhard域 $\Omega_{p_{1}, \cdot\cdot\cdot, p_{n}} =\{z\in C^{n}:\sum\limits_{j=1}^{n}|z_{j}|^{p_{j}}<1\}, p_{j}\geq1, j=1, \cdots, n. $容易知道, 它是 $C_{n}$中的有界凸圆型域, 即 $\Omega_{p_{1}, \cdots, p_{n}}$满足:对 $\forall z\in \Omega_{p_{1}, \cdots, p_{n}}, \xi_{j}\in \overline{D}, $都有 $(\xi_{1}z_{1}, \ldots, \xi_{n}z_{n})\in \Omega_{p_{1}, \cdots, p_{n}}$. $\Omega_{p_{1}, \cdots, p_{n}}$上的Minkowski泛函 $\rho(z)$定义为 $\rho(z)=inf\{t>0:\frac{z}{t}\in \Omega_{p_{1}, \cdots, p_{n}}\}.$易知, Minkowski泛函 $\rho(z)$是 $C_{n}$的一种Banach范数, 且 $\Omega_{p_{1}, \cdots, p_{n}}$是 $C_{n}$在此Banach范数下空间的单位球^[10].在 $\overline{\Omega}_{p_{1}, \cdots, p_{n}}$上, 除去一个低维流形外, Minkowski泛函 $\rho(z)$是 $C^{1}$的.

定义2.1^[11]  设 $f$是 $B_{n}$上的正规化局部双全纯映射, 若 $\alpha\in (0, 1], \beta\in(-\frac{\pi}{2}, \frac{\pi}{2})$, 且

$ {\hbox{Re}}\{e^{-i\beta}\frac{\overline{z^{'}}}{\|z\|^{2}}J_{f}^{-1}(z)f(z)\}\geq \alpha\cos\beta, z\in B_{n}, $

则称 $f$为 $B_{n}$上的 $\alpha$次殆 $\beta$型螺形映射.

当 $n=1$时, 定义转化为Re $\{e^{-i\beta}\frac{f(z)}{zf^{'}(z)}\}\geq \alpha\cos\beta.$

定义2.2  设 $f$是 $B$上的正规化局部双全纯映射, 若 $\alpha\in (0, 1], \beta\in(-\frac{\pi}{2}, \frac{\pi}{2})$, 且

$ {\hbox{Re}}\{e^{-i\beta}\frac{1}{\|x\|}T_{x}[(DF(x))^{-1}F(x)]\}\geq\alpha\cos\beta, $

则称 $f$为 $B$上的 $\alpha$次殆 $\beta$型螺形映射.

定义2.3^[11]  设 $G$为 $C_{n}$中的有界星形圆型域, 其Minkowski泛函 $\rho(z)$除去一个低维流形外是一个 $C^{1}$函数, $f(z)$为 $G$上正规化局部双全纯映射, 若 $\alpha\in (0, 1], \beta\in(-\frac{\pi}{2}, \frac{\pi}{2})$, 且

$ {\hbox{Re}}\{e^{-i\beta}\frac{2}{\rho(z)}\frac{\partial\rho(z)}{\partial z}J_{f}^{-1}(z)f(z)\}\geq \alpha\cos\beta, $

则称 $f(z)$为 $G$上 $\alpha$次殆 $\beta$型螺形映射.此处 $\frac{\partial\rho(z)}{\partial z} =(\frac{\partial\rho(z)}{\partial z_{1}}, \cdots, \frac{\partial\rho(z)}{\partial z_{n}}).$

本文的研究结果要用到下述引理, 陈述如下

引理2.1^[12]  设 $p\in H(U)$, 且 $p(0)=1$, 若Re $p(z)\geq0, $则 $\frac{1-|z|}{1+|z|}\leq {\hbox{Re}}p(z)\leq|p(z)|\leq\frac{1+|z|}{1-|z|}.$

引理2.2  若 $\tau\neq0$为一复数, 则Re $\tau\geq\lambda\Longleftrightarrow |\frac{1}{\tau}-\frac{1}{2\lambda}|\leq\frac{1}{2\lambda}, \lambda>0$.

证  简单计算可得.

引理2.3  若 $f$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, 则

$ |\frac{zf^{'}(z)}{f(z)}|\leq \frac{1+|z|}{\cos\beta[1-(1-2\alpha)|z|]}. $

证  记 $p(z)=\frac{1-i\tan\beta}{1-\alpha}\frac{f(z)}{zf^{'}(z)}-\frac{\alpha-i\tan\beta}{1-\alpha}, z\in U$.因 $f$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, 则 $p(z)\in H(U), p(0)=1, {\hbox{Re}}p(z)\geq0.$由引理2.1 $\frac{1-|z|}{1+|z|}\leq {\hbox{Re}}p(z)\leq\frac{1+|z|}{1-|z|}.$整理得

$ \frac{1-(1-2\alpha)|z|}{1+|z|}\cos\beta\leq {\hbox{Re}}\{e^{-i\beta}\frac{f(z)}{zf^{'}(z)}\} \leq\frac{1+(1-2\alpha)|z|}{1-|z|}\cos\beta. $

对上式左端利用引理2.2, 即有

$ \begin{eqnarray*}&& |\frac{zf^{'}(z)}{f(z)}|-\frac{1+|z|}{2\cos\beta[1-(1-2\alpha)|z|]} =|e^{i\beta}\frac{zf^{'}(z)}{f(z)}|-\frac{1+|z|}{2\cos\beta[1-(1-2\alpha)|z|]}\\ &\leq& |e^{i\beta}\frac{zf^{'}(z)}{f(z)}-\frac{1+|z|}{2\cos\beta[1-(1-2\alpha)|z|]}| \leq\frac{1+|z|}{2\cos\beta[1-(1-2\alpha)|z|]}\end{eqnarray*} $

或

$ |\frac{zf^{'}(z)}{f(z)}|\leq\frac{1+|z|}{\cos\beta[1-(1-2\alpha)|z|]}. $

引理2.4  设 $f_{j}$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, $\lambda_{j}\geq0$且 $\sum\limits_{j=1}^{n}\lambda_{j}=1$, 则 $F(z)=z\prod\limits_{j=1}^{n}(\frac{f_{j}(z_{j})}{z_{j}})^{\lambda_{j}}$为 $B_{n}$上 $\alpha$次殆 $\beta$型螺形映照.

证  计算易得 $\frac{\overline{z^{'}}}{\|z\|^{2}}J_{F}^{-1}(z)F(z) =\frac{1}{\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}}, $由已知条件Re $\{e^{-i\beta}\frac{f(z)}{zf^{'}(z)}\}\geq \alpha\cos\beta, $利用引理2.2, 可知其等价于 $|e^{i\beta}\frac{zf^{'}(z)}{f(z)}-\frac{1}{2\alpha\cos\beta}|\leq\frac{1}{2\alpha\cos\beta}, $而

$ \begin{eqnarray*}&& |e^{i\beta}\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})} -\frac{1}{2\alpha\cos\beta}| =|\sum\limits_{j=1}^{n}\lambda_{j}[e^{i\beta}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})} -\frac{1}{2\alpha\cos\beta}]|\\ &\leq& \sum\limits_{j=1}^{n}\lambda_{j}|e^{i\beta}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})} -\frac{1}{2\alpha\cos\beta}]| <\sum\limits_{j=1}^{n}\lambda_{j}\frac{1}{2\alpha\cos\beta}=\frac{1}{2\alpha\cos\beta}.\end{eqnarray*} $

再次利用引理2.2可知

$ {\hbox{Re}}\{e^{-i\beta}\frac{\overline{z^{'}}}{\|z\|^{2}}J_{F}^{-1}(z)F(z)\} ={\hbox{Re}}\{e^{-i\beta}\frac{1}{\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}}\} \geq\alpha\cos\beta. $

故本结论成立.

引理2.5^[11]  若 $f$为 $B_{n}$上 $\alpha$次殆 $\beta$型螺形映照, 则

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

可见若 $f$为单位圆盘 $U$上 $\alpha$次殆 $\beta$型螺形映照, 则

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

引理2.6  设 $f_{j}$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, $\lambda_{j}\geq0$且 $\sum\limits_{j=1}^{n}\lambda_{j}=1$, 则

$ F(x)=x\prod\limits_{j=1}^{n}(\frac{f_{j}({T_{u_{j}}(x)})}{T_{u_{j}}(x)})^{\lambda_{j}} $

为 $B$上 $\alpha$次殆 $\beta$型螺形映照.

证  记 $g(x)=\prod\limits_{j=1}^{n}(\frac{f_{j}({T_{u_{j}}(x)})}{T_{u_{j}}(x)})^{\lambda_{j}}$, 则 $DF(x)\eta=(Dg(x)\eta)x+g(x)\eta, $且

$ \frac{Dg(x)x}{g(x)}= \sum\limits_{j=1}^{n}\lambda_{j}(\frac{x_{j}f_{j}^{'}(x_{j})}{f_{j}(x_{j})}-1) =\sum\limits_{j=1}^{n}\lambda_{j}\frac{x_{j}f_{j}^{'}(x_{j})}{f_{j}(x_{j})}-1. $

此处记 $x_{j}=T_{u_{j}}(x)$.计算易得 $(DF(x))^{-1}\eta=\frac{1}{g(x)}[\eta-\frac{(Dg(x)\eta)x}{g(x)+(Dg(x)x)}].$进而

$ \frac{1}{\|x\|}T_{x}[(DF(x))^{-1}F(x)]=\frac{1}{\sum\limits_{j=1}^{n}\lambda_{j}\frac{x_{j}f_{j}^{'}(x_{j})}{f_{j}(x_{j})}}. $

其它类似引理2.4可得.

对于Reinhard域 $\Omega_{p_{1}, \cdot\cdot\cdot, p_{n}}$, 其Minkowski泛函 $\rho(z)$有下述引理

引理2.7^[13]  设 $\rho(z)$是域 $\Omega_{p_{1}, \cdots, p_{n}}$上的Minkowski泛函, $z\in \Omega_{p_{1}, \cdots, p_{n}}\backslash \{0\}$, 则

$ \frac{\partial\rho(z)}{\partial z_{j}}=\frac{p_{j}\overline{z_{j}}|\frac{z_{j}}{\rho(z)}|^{p_{j}-2}}{2\rho(z)\sum\limits_{i=1}^{n}p_{i}|\frac{z_{i}}{\rho(z)}|^{p_{i}}}, j=1, \cdots, n. $

引理2.8  设 $f_{j}$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, $\lambda_{j}\geq0$且 $\sum\limits_{j=1}^{n}\lambda_{j}=1$, 则 $F(z)=z\prod\limits_{j=1}^{n}(\frac{f_{j}(z_{j})}{z_{j}})^{\lambda_{j}}$为域 $\Omega_{p_{1}, \cdots, p_{n}}$上 $\alpha$次殆 $\beta$型螺形映照.

证  利用引理2.7简单计算可得 $\frac{2}{\rho(z)}\frac{\partial\rho(z)}{\partial z}J_{F}^{-1}(z)F(z) =\frac{1}{\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}}$, 其它类似引理2.4.

定理3.1  设 $f_{j}$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, $\lambda_{j}\geq0$且 $\sum\limits_{j=1}^{n}\lambda_{j}=1$, 则

$ |{\hbox{det}}J_{F}(z)| \leq\frac{1+\|z\|}{\cos\beta[1-(1-2\alpha)\|z\|]^{\frac{2(\alpha-1)}{2\alpha-1}n+1}}, $

其中 $F(z)=z\prod\limits_{j=1}^{n}(\frac{f_{j}(z_{j})}{z_{j}})^{\lambda_{j}}, z\in B_{n}.$

证  记 $g(z)=\prod\limits_{j=1}^{n}(\frac{f_{j}(z_{j})}{z_{j}})^{\lambda_{j}}$, 则

$ J_{F}(z)=g(z)\\ \left( \begin{array}{cccc} \displaystyle 1+z_{1}\lambda_{1}(\frac{f_{1}^{'}(z_{1})}{f_{1}(z_{1})}-\frac{1}{z_{1}})&z_{1}\lambda_{2}(\frac{f_{2}^{'}(z_{2})}{f_{2}(z_{2})}-\frac{1}{z_{2}}) &\ldots&z_{1}\lambda_{n}(\frac{f_{n}^{'}(z_{n})}{f_{n}(z_{n})}-\frac{1}{z_{n}}) \\ z_{2}\lambda_{1}(\frac{f_{1}^{'}(z_{1})}{f_{1}(z_{1})}-\frac{1}{z_{1}})& 1+z_{2}\lambda_{2}(\frac{f_{2}^{'}(z_{2})}{f_{2}(z_{2})}-\frac{1}{z_{2}})&\ldots& z_{2}\lambda_{n}(\frac{f_{n}^{'}(z_{n})}{f_{n}(z_{n})}-\frac{1}{z_{n}}) \\ \cdot\cdot\cdot&\cdot\cdot\cdot&\cdot\cdot\cdot&\cdot\cdot\cdot \\ z_{n}\lambda_{1}(\frac{f_{1}^{'}(z_{1})}{f_{1}(z_{1})}-\frac{1}{z_{1}}) &z_{n}\lambda_{2}(\frac{f_{2}^{'}(z_{2})}{f_{2}(z_{2})}-\frac{1}{z_{2}})&\ldots&1+z_{n}\lambda_{n}(\frac{f_{n}^{'}(z_{n})}{f_{n}(z_{n})}-\frac{1}{z_{n}}) \end{array}\right). $

进而

$ |{\hbox{det}}J_{F}(z)|=|g(z)|^{n}|1+\sum\limits_{j=1}^{n}z_{j}\lambda_{j}(\frac{f_{j}^{'}(z_{j})}{f_{j}(z_{j})}-\frac{1}{z_{j}})| =\prod\limits_{j=1}^{n}|\frac{f_{j}(z_{j})}{z_{j}}|^{n\lambda_{j}}|\sum\limits_{j=1}^{n}\lambda_{j} \frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}|, $

由引理2.3,

$ |\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}| \leq\sum\limits_{j=1}^{n}\lambda_{j}|\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}| \leq\sum\limits_{j=1}^{n}\lambda_{j}\frac{1+|z_{j}|}{\cos\beta[1-(1-2\alpha)|z_{j}|]}. $

注意到 $|z_{j}|\leq\|z\|, \frac{1+t}{1-(1-2\alpha)t}$分别为递增函数, 上式变为

$ |\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}| \leq\frac{1+\|z\|}{\cos\beta[1-(1-2\alpha)\|z\|]}. $

另一方面由引理2.5

$ \prod\limits_{j=1}^{n}|\frac{f_{j}(z_{j})}{z_{j}}|^{n\lambda_{j}} \leq\prod\limits_{j=1}^{n}\frac{1}{[1-(1-2\alpha)|z_{j}|]^{\frac{2(\alpha-1)}{2\alpha-1}n\lambda_{j}}}, $

而 $\frac{1}{[1-(1-2\alpha)t]^{\frac{2(\alpha-1)}{2\alpha-1}}}$为递增函数, 又 $|z_{j}|\leq\|z\|$, 上式化为

$ \prod\limits_{j=1}^{n}|\frac{f_{j}(z_{j})}{z_{j}}|^{n\lambda_{j}} \leq\frac{1}{[1-(1-2\alpha)\|z\|]^{\frac{2(\alpha-1)n}{2\alpha-1}}}. $

综上可知

$ |{\hbox{det}}J_{F}(z)| \leq\frac{1+\|z\|}{\cos\beta[1-(1-2\alpha)\|z\|]^{\frac{2(\alpha-1)}{2\alpha-1}n+1}}. $

定理3.2  设 $f_{j}$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, $\lambda_{j}\geq0$且 $\sum\limits_{j=1}^{n}\lambda_{j}=1$, 则

$ \|J_{F}(z)z\| \leq\frac{\|z\|(1+\|z\|)}{\cos\beta[1-(1-2\alpha)\|z\|]^{\frac{4\alpha-3}{2\alpha-1}}}, $

其中 $F(z)=z\prod\limits_{j=1}^{n}(\frac{f_{j}(z_{j})}{z_{j}})^{\lambda_{j}}, z\in B_{n}.$

证  记 $g(z)=\prod\limits_{j=1}^{n}(\frac{f_{j}(z_{j})}{z_{j}})^{\lambda_{j}}$, 计算易得

$ J_{F}(z)z =zg(z)\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})} =z\prod\limits_{j=1}^{n}(\frac{f_{j}(z_{j})}{z_{j}})^{\lambda_{j}}\sum\limits_{j=1}^{n}\lambda_{j}\frac{z_{j}f_{j}^{'}(z_{j})}{f_{j}(z_{j})}. $

类似定理3.1的证明可得.

定理3.3  设 $f_{j}$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, $\lambda_{j}\geq0$且 $\sum\limits_{j=1}^{n}\lambda_{j}=1$, 则

$ |{\hbox{det}}DF(x)|\leq\frac{1+\|x\|}{\cos\beta[1-(1-2\alpha)\|x\|]^{\frac{2(\alpha-1)}{2\alpha-1}n+1}}, $

其中 $F(x)=x\prod\limits_{j=1}^{n}(\frac{f_{j}({T_{u_{j}}(x)})}{T_{u_{j}}(x)})^{\lambda_{j}}, x\in B.$

证  记 $g(x)=\prod\limits_{j=1}^{n}(\frac{f_{j}({T_{u_{j}}(x)})}{T_{u_{j}}(x)})^{\lambda_{j}}$, 则 $DF(x)=g(x)\{I+x\sum\limits_{j=1}^{n}\lambda_{j}[\frac{f_{j}^{'}(T_{u_{j}(x)})}{f_{j}(T_{u_{j}}(x))}-\frac{1}{T_{u_{j}}(x)}]T_{u_{j}}(\cdot)\}, $此处 $T_{u_{j}}(\cdot)$是行向量.因此

$ \begin{eqnarray*}&& {\hbox{det}}DF(x)=|g(x)|^{n}|1+\sum\limits_{j=1}^{n}\lambda_{j}[\frac{f_{j}^{'}(T_{u_{j}(x)})}{f_{j}(T_{u_{j}}(x))}-\frac{1}{T_{u_{j}}(x)}]T_{u_{j}}(x)|\\ &=& \prod\limits_{j=1}^{n}|\frac{f_{j}(T_{u_{j}}(x))}{T_{u_{j}}(x)}|^{n\lambda_{j}} |\sum\limits_{j=1}^{n}\lambda_{j}\frac{T_{u_{j}}(x)f_{j}^{'}(T_{u_{j}(x)})}{f_{j}(T_{u_{j}}(x))}|. \end{eqnarray*} $

记 $x_{j}=T_{u_{j}}(x)$, 上式化为

$ {\hbox{det}}DF(x)=|g(x)|^{n}|1+\sum\limits_{j=1}^{n}\lambda_{j}[\frac{f_{j}^{'}(x_{j})}{f_{j}(x_{j})}-\frac{1}{x_{j}}]x_{j}| =\prod\limits_{j=1}^{n}|\frac{f_{j}(x_{j})}{x_{j}}|^{n\lambda_{j}} |\sum\limits_{j=1}^{n}\lambda_{j}\frac{x_{j}f_{j}^{'}(x_{j})}{f_{j}(x_{j})}|. $

其它类似定理3.1, 可证本命题成立.

定理3.4  设 $f_{j}$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, $\lambda_{j}\geq0$且 $\sum\limits_{j=1}^{n}\lambda_{j}=1$, 则

$ \|DF(x)x\|\leq\frac{\|x\|(1+\|x\|)}{\cos\beta[1-(1-2\alpha)\|x\|]^{\frac{4\alpha-3}{2\alpha-1}}}, $

其中 $F(x)=x\prod\limits_{j=1}^{n}(\frac{f_{j}({T_{u_{j}}(x)})}{T_{u_{j}}(x)})^{\lambda_{j}}, x\in B.$

证  类似定理3.1和3.2可证.

注1  当 $\beta=0$时, 定理3.1-定理3.4即为 $\alpha$次殆星形映照的偏差上界, 因而推广了文^[11]中的结果.

注2  复Bananch空间单位球 $B$上的偏差上界和复 $C^{n}$空间单位球 $B_{n}$上的结果一致, 正如预期的一样.同样对Reinhard域 $\Omega_{p_{1}, \cdots, p_{n}}$执行类似讨论, 有与定理3.1和定理3.2同样的结论, 文中不再赘述.

对于更一般的有界星形(或凸)圆型域, $\alpha$次殆 $\beta$型螺形映照相应的偏差上界本文没有给出.但引理2.3中的下界本文猜测如下

猜想4.1  若 $f$为单位圆盘 $U$上的 $\alpha$次殆 $\beta$型螺形函数, 则

$ {\hbox{Re}}\frac{zf^{'}(z)}{f(z)}\geq \frac{1-|z|}{\cos\beta[1+(1-2\alpha)|z|]}. $

若下界可得, 则复欧式空间 $C_{n}$中的开单位球 $B_{n}$和Reinhard域 $\Omega_{p_{1}, \cdots, p_{n}}$的偏差下界估计即可顺利解决.盼望对此文感兴趣的读者能做出进一步的研究, 不断丰富多复变函数论的内容.