由于Roper-Suffridge算子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函数构造多复变中相应的映照,因此许多人都致力于研究该算子[1, 2, 3]. 后来人们发现推广的Roper-Suffridge算子与Loewner链有着非常紧密的联系.Hamada和Kohr[4]利用Loewner链得到了复Banach空间单位球上$\beta$型螺形映照的解析特征. Graham,Hamada和Kohr[5, 6]证明了一些推广的Roper-Suffridge算子在${{\Bbb C}}^{n}$中单位球$B^n$上能够嵌入Loewner链,并得到了一些推广的Roper-Suffridge算子在$B^n$上保持星形性和$\beta$型螺形性. 刘太顺和徐庆华[7]得到了$\alpha$次殆星形映照的解析特征. 冯淑霞,刘小松和徐庆华[8]得到了$\alpha$次殆$\beta$型螺形映照的解析特征,并证明了一些推广的Roper-Suffridge算子能够嵌入Loewner链.
本文是在前人工作的基础上讨论一类推广的Roper-Suffridge算子与Loewner链的关系,从而讨论推广的Roper-Suffridge算子的性质.
在本文中$D$表示${\Bbb C}$中的单位圆盘,$X$表示具有范数$\|\cdot\|$的复Banach空间,$X^*$是$X$的对偶空间. 对于$x\in X\setminus\{0\}$,定义
$$T(x)=\{T_{x}\in X^*:\|T_{x}\|=1,T_{x}(x)=\|x\|\}.
$$
$H(\Omega)$ 表示$\Omega$上的全纯映照. $Df(z)=(\frac{\partial f_{j}}{\partial z_{k}})_{1\leq j,k\leq n}$表示$f(z)\in H(\Omega)$的Fréchet导数. $B$表示复Banach空间中的单位球. 令
$$P=\{p\in H(D):p(0)=1,\Re ep(z)>0,z\in D\},
$$
$$M=\{h\in H(B): h(0)=0,Dh(0)=I,\Re e[T_{z}h(z)]\geq 0,z\in B,T_{z}\in T(z)\}.
$$
定义 2.1 (Poreda[9]) 令$\{f(z,t)\}_{t\geq 0}={\rm e}^t z+\cdots:B\times[0,+\infty)\rightarrow X$,$h(z,t):B\times[0,+\infty)\rightarrow X$. 如果
(1) 对于$\forall t\geq 0$,有$f(\cdot,t)\in H(B)$且$h(\cdot,t)\in M$;
(2) 对于$\forall z\in B$,$f(z,\cdot)$在$[0,+\infty)$上绝对连续,$h(z,\cdot)$在$[0,+\infty)$上可测且对于$z\in B$有
$$\frac{\partial}{\partial t}f(z,t)=Df(z,t)h(z,t),\ \ {\rm a.e.}\ t\geq 0,
$$
并且存在$G\in H(B)$和一个非负递增趋于$+\infty$的序列$\{t_{m}\}$使得$\lim\limits_{m\rightarrow+\infty}{\rm e}^{-t_{m}}f(z,t_{m})=G(z)$在$B$上局部一致地成立.
则$f(z,t)$是一个Loewner链.
引理 2.2 (Pommerenke[10]) 函数族$\{f(z,t)\}_{t\geq 0}$满足$f(0,t)=0,f'(0,t)={\rm e}^t$. 则$f(z,t)$是一个Loewner链当且仅当
(1) 对于$\forall t\geq 0$,$f(z,t)={\rm e}^t z+\cdots$在$\{z:|z|< r(r\in(0,1))\}$上全纯; 对于$\forall z(|z|< r)$,$f(z,\cdot)$在$[0,+\infty)$上绝对连续; 对于$\forall r\in (0,1)$,存在$M(r)>0$使得$|f(z,t)|\leq M(r){\rm e}^t$ $(t\geq0,|z|\leq r)$;
(2) 存在$p(z,t)\in H(D)$使得$p(\cdot,t)\in P$,$p(z,\cdot)$在$[0,+\infty)$上是可测的,且
$$\frac{\partial}{\partial t}f(z,t)=zf'(z,t)p(z,t),{\rm a.e.}\ t\geq 0.
$$
引理 2.3 (Graham,Kohr[11]) 令$p(z)\in P$,则
$$\frac{1-|z|}{1+|z|}\leq|p(z)|\leq\frac{1+|z|}{1-|z|},\frac{1-|z|}{1+|z|}\leq \Re ep(z)\leq\frac{1+|z|}{1-|z|},|p'(z)|\leq\frac{2\Re ep(z)}{1-|z|^2}.
$$
引理 2.4 (Liu,Zhu[12]) 令
$$\Omega=\Omega_{p_{1},p_{2},\cdots,p_{n}}=\bigg\{z\in {{\Bbb C}}^{n}:\sum^{n}_{j=1}|z_{j}|^{p_{j}}<1,p_{j}\geq1\bigg\},
$$
其Minkowski泛函是$\rho(z)$. 如果$\rho(z)$在$\Omega\setminus\{0\}$上可微,则$T_{z}(\cdot)=2\langle\cdot,\overline{\frac{\partial\rho(z)}{\partial z}}\rangle$,其中$\frac{\partial\rho(z)}{\partial z}=(\frac{\partial\rho(z)}{\partial z_{1}},\frac{\partial\rho(z)}{\partial z_{2}},\cdots,\frac{\partial\rho(z)}{\partial z_{n}})$.
引理 2.5 (Liu,Zhu[12]) 令$\Omega\in {{\Bbb C}}^{n}$是一个有界完全Reinhardt域,其Minkowski泛函$\rho(z)$在除去一个低微流行外是$C^1$的. 则
$$\frac{\partial\rho(z)}{\partial z_{j}}z_{j}\geq 0,\rho(z)=2\sum^{n}_{j=1}\frac{\partial\rho(z)}{\partial z_{j}}z_{j}.
$$
引理 2.6 (Muir[13]) 令$f(z)$是$D$上的一个正规化双全纯函数. 则
$$\left|\frac{1-|z|^2}{\alpha}\frac{f"(z)}{f'(z)}-\bar{z}\right|\leq\frac{4+(\alpha-2)|z|}{\alpha},z\in D,
$$
其中$\alpha\geq 2$.
{引理 2.7 (Feng,Liu,Xu[8]) 令$f(z)$是$B$上的一个正规化局部双全纯函数,$\alpha\in[0,1),$ $\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$,$a=\tan\beta$. 则$f(z)$是$B$上的$\alpha$次殆$\beta$型螺形映照的充要条件是$g(z,t)={\rm e}^{\frac{1-{\rm i}a}{1-\alpha}t}f({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}z)\ (t\geq0)$是一个Loewner链.
定理3.1 令
$$\Omega=\Omega_{p_{1},p_{2},\cdots,p_{n}}=\bigg\{z\in {{\Bbb C}}^{n}:\sum^n_{j=1}|z_{j}|^{p_{j}}<1,p_{j}\geq 1\bigg\},
$$
其Minkowski泛函为$\rho(z)$. 令
$$F(z)=\left(rf\Big(\frac{z_{1}}{r}\Big),\Big(\frac{rf(\frac{z_{1}}{r})}{z_{1}}\Big)^{\beta_{2}}\Big(f'\Big(\frac{z_{1}}{r}\Big)\Big)^{\gamma_{2}}z_{2},\cdots,\Big(\frac{rf(\frac{z_{1}}{r})}{z_{1}}\Big)^{\beta_{n}}\Big(f'\Big(\frac{z_{1}}{r}\Big)\Big)^{\gamma_{n}}z_{n}\right)',
$$
其中$f$是$D$上正规化的双全纯函数,$r=\sup\{|z_{1}|:z=(z_{1},\cdots,z_{n})\in\Omega\}\ (n\geq2,n\in{\Bbb Z}^+)$,$\beta_{j},\gamma_{j}\in[0, 1]$,$\beta_{j}+\gamma_{j}\leq1$. 幂函数取分支使得
$$\Big(\frac{rf(\frac{z_{1}}{r})}{z_{1}}\Big)^{\beta_{j}}\Big|_{z_{1}=0}=1,\Big(f'\Big(\frac{z_{1}}{r}\Big)\Big)^{\gamma_{j}}\Big|_{z_{1}=0}=1.
$$
如果$r\leq\frac{1}{2}$,$\beta_{j}+M\gamma_{j}\leq1(M=\max\{M_{1},M_{2}\})$,其中
$$M_{1}=\sup_{0\leq|z_{1}|\leq(-1+\sqrt{2})r}\bigg\{\frac{r^2-|z_{1}|^2-2r|z_{1}|}{r(r+|z_{1}|)^2}-\frac{(1-r)(4r^2+2(1-r)|z_{1}|)|z_{1}|}{r^2(r-|z_{1}|)^2}\bigg\},
$$
$$M_{2}=\sup_{(-1+\sqrt{2})r<|z_{1}|< r}\frac{(3r-2r^2-2)|z_{1}|^2+2r^2(2r-3)|z_{1}|+r^3}{r^2(r-|z_{1}|)^2},
$$
则$F(z)$在$\Omega$上能嵌入loewner链.
证 由于$f$是一个正规化的双全纯函数,则存在一个Loewner链$\{g(\frac{z_{1}}{r},t)\}_{t\geq 0}$使得$f(\frac{z_{1}}{r})=g(\frac{z_{1}}{r},0)$. 令
$$\begin{array}[b]{rl}F_{\psi}(z,t)&=\Big(rg\Big(\frac{z_{1}}{r},t\Big),{\rm e}^{(1-\beta_{2}-\gamma_{2})t}\Big(\frac{rg(\frac{z_{1}}{r},t)}{z_{1}}\Big)^{\beta_{2}}\Big(g'\Big(\frac{z_{1}}{r},t\Big)\Big)^{\gamma_{2}}z_{2},\cdots\\ [4mm]& {\rm e}^{(1-\beta_{n}-\gamma_{n})t}\Big(\frac{rg(\frac{z_{1}}{r},t)}{z_{1}}\Big)^{\beta_{n}}\Big(g'\Big(\frac{z_{1}}{r},t\Big)\Big)^{\gamma_{n}}z_{n}\Big)',\end{array}
$$其中$z=(z_{1},\cdots,z_{n})\in\Omega,t\geq 0$. 由引理2.2,我们有以下结论.
(ⅰ) 对于$\forall\frac{z_{1}}{r}\in D_{r'}(D_{r'}=\{z_{1}\in C:|\frac{z_{1}}{r}|< r'\})$,$g(\frac{z_{1}}{r},\cdot)$在$[0,+\infty)$上是局部绝对连续的. 而且,对于$\forall r\in(0,1)$,存在$M(r')$使得
$$\Big|g\Big(\frac{z_{1}}{r},t\Big)\Big|\leq M(r'){\rm e}^t,\ \ t\geq 0.
$$
(ⅱ) 存在$D$上的全纯函数$p(\frac{z_{1}}{r},t)$使得$p(\cdot,t)\in P$,$p(\frac{z_{1}}{r},\cdot)$在$[0,+\infty)$上是可测的,且
$$\frac{\partial}{\partial t}g\Big(\frac{z_{1}}{r},t\Big)=\frac{z_{1}}{r}g'\Big(\frac{z_{1}}{r},t\Big)p\Big(\frac{z_{1}}{r},t\Big).
$$
显然$F_{\psi}(z,t)$在$\Omega$上全纯且在$[0,+\infty)$上局部绝对连续. 经简单计算可知$F_{\psi}(0,t)=0,$ $DF_{\psi}(0,t)={\rm e}^tI$,且
\begin{eqnarray*}\frac{\partial F_{\psi}}{\partial t}(z,t)&=&\Big(z_{1}g'\Big(\frac{z_{1}}{r},t\Big)p\Big(\frac{z_{1}}{r},t\Big),\\&& {\rm e}^{(1-\beta_{2}-\gamma_{2})t}\Big(\frac{rg}{z_{1}}\Big)^{\beta_{2}}(g')^{\gamma_{2}}z_{2} \Big[(1-\beta_{2}-\gamma_{2})+\beta_{2}\frac{z_{1}}{rg}g'p+\gamma_{2}\frac{1}{g'}\frac{\partial g'}{\partial t}\Big],\cdots,\\ && {\rm e}^{(1-\beta_{n}-\gamma_{n})t}\Big(\frac{rg}{z_{1}}\Big)^{\beta_{n}}(g')^{\gamma_{n}}z_{n}\Big[(1-\beta_{n}-\gamma_{n})+\beta_{n}\frac{z_{1}}{rg}g'p+\gamma_{n}\frac{1}{g'}\frac{\partial g'}{\partial t}\Big]\Big)',\end{eqnarray*}
$$D F_{\psi}(z,t)= \left( \begin{array}{cccc} g'\Big(\frac{z_{1}}{r},t\Big) &0 &\cdots &~0\\ v_{2} & ~ {\rm e}^{(1-\beta_{2}-\gamma_{2})t}\Big(\frac{r g}{z_{1}}\Big)^{\beta_{2}}(g')^{\gamma_{2}} ~ &\cdots &~0\\ \cdots &\cdots &\cdots &~\cdots\\ v_{n} &0 &\cdots &~ {\rm e}^{(1-\beta_{n}-\gamma_{n})t}\Big(\frac{r g}{z_{1}}\Big)^{\beta_{n}}(g')^{\gamma_{n}} \end{array} \right),
$$
其中
$$v_{k}={\rm e}^{(1-\beta_{k}-\gamma_{k})t}\Big(\frac{rg}{z_{1}}\Big)^{\beta_{k}}(g')^{\gamma_{k}}z_{k}\Big[\beta_{k}\frac{z_{1}g'-rg}{rgz_{1}}+\frac{\gamma_{k}g"}{rg'}\Big],k=2,\cdots,n.
$$
于是
$$(D F_{\psi}(z,t))^{-1}= \left( \begin{array}{cccc} \frac{1}{g'(\frac{z_{1}}{r},t)} &0 &\cdots &~0\\ w_{2} & ~ {\rm e}^{(\beta_{2}+\gamma_{2}-1)t} \Big(\frac{r g}{z_{1}}\Big)^{-\beta_{2}}(g')^{-\gamma_{2}} ~ &\cdots &~0\\ \cdots &\cdots &\cdots &~\cdots\\ w_{n} &0 &\cdots & ~ {\rm e}^{(\beta_{n}+\gamma_{n}-1)t} \Big(\frac{r g}{z_{1}}\Big)^{-\beta_{n}}(g')^{-\gamma_{n}} \end{array} \right),
$$
其中
$$w_{k}=-\frac{z_{k}}{rg'}\Big[\beta_{k}\frac{z_{1}g'-rg}{gz_{1}}+\frac{\gamma_{k}g"}{g'}\Big],k=2,\cdots,n.
$$
由于
$$\frac{\partial g'}{\partial t}=\frac{1}{r^2}(rg'p+z_{1}g"p+z_{1}g'p'),
$$
则
$$(D F_{\psi}(z,t))^{-1}\frac{\partial F_{\psi}}{\partial t}(z,t)=\left(z_{1}p\Big(\frac{z_{1}}{r},t\Big),A_{2}z_{2},\cdots,A_{n}z_{n}\right)',
$$
其中
$$A_{k}=\beta_{k}p+(1-\beta_{k}-\gamma_{k})+\frac{\gamma_{k}}{r}p+\frac{1-r}{r^2}\gamma_{k}\frac{g"}{g'}z_{1}p+\frac{\gamma_{2}}{r^2}z_{1}p',k=2,\cdots,n.
$$
令
|
$h(z,t)=\left(z_{1}p\Big(\frac{z_{1}}{r},t\Big),A_{2}z_{2},\cdots,A_{n}z_{n}\right)'.$
| (3.1) |
显然$h(z,t)$在$\Omega$上全纯且$h(0,t)=0,Dh(0,t)=I$.又由引理2.3可得
|
$\Re e(z_{1}p')\geq-|z_{1}p'|\geq-|z_{1}|\frac{2r^2\Re ep}{r^2-|z_{1}|^2}.$
| (3.2) |
由(3.1),(3.2)式和引理2.4,2.5,有
|
\begin{eqnarray} & &\Re e\bigg\{\Big\langle h(z,t),\overline{\frac{\partial \rho(z)}{\partial z}}\Big\rangle\bigg\} \\ &=&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\Re eplaystyle\sum^{n}_{j=2}\Big(\beta_{j}+\frac{\gamma_{j}}{r}\Big)\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\sum^{n}_{j=2}(1-\beta_{j}-\gamma_{j})\frac{\partial \rho(z)}{\partial z_{j}}z_{j} \\&&+laystyle\frac{1-r}{r^2}\Re e\Big[\frac{g"}{g'}z_{1}p\Big]laystyle\sum^{n}_{j=2}\gamma_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\frac{\Re e(z_{1}p')}{r^2}\sum^{n}_{j=2}\gamma_{j}\frac{\partial\rho(z)}{\partial z_{j}}z_{j} \\ &\geq&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\Re ep laystyle\sum^{n}_{j=2}\Big(\beta_{j}+\frac{\gamma_{j}}{r}\Big)\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\sum^{n}_{j=2}(1-\beta_{j}-\gamma_{j})\frac{\partial \rho(z)}{\partial z_{j}}z_{j} \\&&+laystyle\frac{1-r}{r^2}\Re e\Big[\frac{g"}{g'}z_{1}p\Big] laystyle\sum^{n}_{j=2}\gamma_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}-\frac{2|z_{1}|\Re ep}{r^2-|z_{1}|^2}laystyle\sum^{n}_{j=2}\gamma_{j}\frac{\partial\rho(z)}{\partial z_{j}}z_{j} \\ &=&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\Re ep laystyle\sum^{n}_{j=2}\beta_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\Big[\frac{1}{r}-\frac{2|z_{1}|}{r^2-|z_{1}|^2}\Big]\Re ep\sum^{n}_{j=2}\gamma_{j}\frac{\partial\rho(z)}{\partial z_{j}}z_{j} \\&&+laystyle\sum^{n}_{j=2}(1-\beta_{j}-\gamma_{j})\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\frac{1-r}{r^2}\Re e\Big[\frac{g"}{g'}z_{1}p\Big]\sum^{n}_{j=2}\gamma_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}.\end{eqnarray}
| (3.3) |
由引理2.3和2.6,我们有
|
\begin{eqnarray} \Re e\Big[laystyle\frac{g"}{g'}z_{1}p\Big]&\geq&-\Big|laystyle\frac{g"}{g'}z_{1}p\Big| \\ &=&-\Big|\Big[\Big(laystyle\frac{r^2-|z_{1}|^2}{r}\frac{g"}{g'}-\frac{\overline{z_{1}}}{r}\Big)+\frac{\overline{z_{1}}}{r}\Big]\frac{r}{r^2-|z_{1}|^2}z_{1}p\Big| \\ &\geq&-\Big[\Big|laystyle\frac{r^2-|z_{1}|^2}{r}\frac{g"}{g'}-\frac{\overline{z_{1}}}{r}\Big|+\frac{|z_{1}|}{r}\Big]\frac{r}{r^2-|z_{1}|^2}|z_{1}p| \\ &\geq&-\Big[laystyle\frac{4r^2+(1-2r)|z_{1}|}{r}+\frac{|z_{1}|}{r}\Big]\frac{r}{r^2-|z_{1}|^2}|z_{1}p| \\ &=&-laystyle\frac{4r^2+2(1-r)|z_{1}|}{r^2-|z_{1}|^2}|z_{1}||p| \\ &\geq&-laystyle\frac{(4r^2+2(1-r)|z_{1}|)|z_{1}|}{(r-|z_{1}|)^2}.\end{eqnarray}
| (3.4) |
(1) 当$0\leq|z_{1}|\leq(-1+\sqrt{2})r$时,由(3.3),(3.4)式和引理2.3有
\begin{eqnarray*}& &\Re e\bigg\{\Big\langle h(z,t),\overline{laystyle\frac{\partial \rho(z)}{\partial z}}\Big\rangle\bigg\}\\ &\geq&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\Re ep\sum^{n}_{j=2}\beta_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\frac{r^2-|z_{1}|^2-2r|z_{1}|}{r(r+|z_{1}|)^2}\sum^{n}_{j=2}\gamma_{j}\frac{\partial\rho(z)}{\partial z_{j}}z_{j}\\ &&+laystyle\sum^{n}_{j=2}(1-\beta_{j}-\gamma_{j})\frac{\partial \rho(z)}{\partial z_{j}}z_{j}-\frac{1-r}{r^2}\frac{(4r^2+2(1-r)|z_{1}|)|z_{1}|}{(r-|z_{1}|)^2}\sum^{n}_{j=2}\gamma_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &\geq&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\sum^{n}_{j=2}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\Big[\frac{r-|z_{1}|}{r+|z_{1}|}-1\Big] \sum^{n}_{j=2}\beta_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &&+\Big[laystyle\frac{r^2-|z_{1}|^2-2r|z_{1}|}{r(r+|z_{1}|)^2}-\frac{(1-r)(4r^2+2(1-r)|z_{1}|)|z_{1}|}{r^2(r-|z_{1}|)^2}\Big]\sum^{n}_{j=2}\gamma_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &=&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\sum^{n}_{j=2} \bigg\{1-\frac{2|z_{1}|}{r+|z_{1}|}\beta_{j}\\ [4mm]&&-\Big[laystyle\frac{r^2-|z_{1}|^2-2r|z_{1}|}{r(r+|z_{1}|)^2}-\frac{(1-r)(4r^2+2(1-r)|z_{1}|)|z_{1}|}{r^2(r-|z_{1}|)^2}\Big]\gamma_{j}\bigg\}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &\geq&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\sum^{n}_{j=2}(1-\beta_{j}-M_{1}\gamma_{j})\frac{\partial\rho(z)}{\partial z_{j}}z_{j} \geq 0,\end{eqnarray*}
其中$\beta_{j}+M_{1}\gamma_{j}\leq 1~ (j=2,\cdots,n)$和
$$M_{1}=\sup_{0\leq|z_{1}|\leq(-1+\sqrt{2})r}\bigg\{\frac{r^2-|z_{1}|^2-2r|z_{1}|}{r(r+|z_{1}|)^2}-\frac{(1-r)(4r^2+2(1-r)|z_{1}|)|z_{1}|}{r^2(r-|z_{1}|)^2}\bigg\}.
$$
(2) 当$(-1+\sqrt{2})r<|z_{1}|< r$时,由(3.3),(3.4)式和引理2.3得
\begin{eqnarray*} & &\Re e\bigg\{\Big\langle h(z,t),\overline{laystyle\frac{\partial \rho(z)}{\partial z}}\Big\rangle\bigg\}\\ &\geq&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\Re ep\sum^{n}_{j=2}\beta_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\frac{r^2-|z_{1}|^2-2r|z_{1}|}{r(r-|z_{1}|)^2}\sum^{n}_{j=2}\gamma_{j}\frac{\partial\rho(z)}{\partial z_{j}}z_{j}\\ &&+laystyle\sum^{n}_{j=2}(1-\beta_{j}-\gamma_{j})\frac{\partial \rho(z)}{\partial z_{j}}z_{j}-\frac{1-r}{r^2}\frac{(4r^2+2(1-r)|z_{1}|)|z_{1}|}{(r-|z_{1}|)^2}\sum^{n}_{j=2}\gamma_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &\geq&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\sum^{n}_{j=2}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}+\Big[\frac{r-|z_{1}|}{r+|z_{1}|}-1\Big]\sum^{n}_{j=2}\beta_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &&+laystyle\frac{(3r-2r^2-2)|z_{1}|^2+2r^2(2r-3)|z_{1}|+r^3}{r^2(r-|z_{1}|)^2}\sum^{n}_{j=2}\gamma_{j}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &=&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\sum^{n}_{j=2} \bigg\{1-\frac{2|z_{1}|}{r+|z_{1}|}\beta_{j} \\&&-laystyle\frac{(3r-2r^2-2)|z_{1}|^2+2r^2(2r-3)|z_{1}|+r^3}{r^2(r-|z_{1}|)^2}\gamma_{j}\bigg\}\frac{\partial \rho(z)}{\partial z_{j}}z_{j}\\ &\geq&laystyle\frac{\partial \rho(z)}{\partial z_{1}}z_{1}\Re ep+\sum^{n}_{j=2}(1-\beta_{j}-M_{2}\gamma_{j})\frac{\partial\rho(z)}{\partial z_{j}}z_{j} \geq 0,\end{eqnarray*}
其中$\beta_{j}+M_{2}\gamma_{j}\leq 1~(j=2,\cdots,n)$,且
$$M_{2}=\sup_{(-1+\sqrt{2})r<|z_{1}|< r}\frac{(3r-2r^2-2)|z_{1}|^2+2r^2(2r-3)|z_{1}|+r^3}{r^2(r-|z_{1}|)^2}.
$$
对于$z_{1}< r$,当$\beta_{j}+M\gamma_{j}\leq 1~(M=\max\{M_{1},M_{2}\})$时,(1)和(2)表明
$$\Re e\bigg\{\Big\langle h(z,t),\overline{\frac{\partial \rho(z)}{\partial z}}\Big\rangle\bigg\}\geq 0.
$$
因此$h(z,t)\in M(z\in\Omega)$.
由于${\rm e}^{-t}g(\frac{z_{1}}{r},t)(t\geq 0)$在$D$上局部一致有界,则存在一个非负递增趋于$+\infty$的序列$\{t_{m}\}$ 使得
$$\lim_{m\rightarrow+\infty}{\rm e}^{-t_{m}}g\Big(\frac{z_{1}}{r},t_{m}\Big)=f\Big(\frac{z_{1}}{r}\Big)
$$
在$D$上局部一致地成立. 而且,显然$h(z,t)$是可测的. 因此
\begin{eqnarray*}&&laystyle\lim_{m\rightarrow+\infty}{\rm e}^{-t_{m}}F_{\psi}(z,t_{m})\\ &=&\Big(r{\rm e}^{-t_{m}}g\Big(laystyle\frac{z_{1}}{r},t_{m}\Big),\Big({\rm e}^{-t_{m}}\frac{rg(\frac{z_{1}}{r},t_{m})}{z_{1}}\Big)^{\beta_{2}}\Big({\rm e}^{-t_{m}}g'\Big(\frac{z_{1}}{r},t_{m}\Big)\Big)^{\gamma_{2}}z_{2},\cdots\\ &&\Big({\rm e}^{-t_{m}}\frac{rg(\frac{z_{1}}{r},t_{m})}{z_{1}}\Big)^{\beta_{n}}\Big({\rm e}^{-t_{m}}g'\Big(laystyle\frac{z_{1}}{r},t_{m}\Big)\Big)^{\gamma_{n}}z_{n}\Big)'\\ &=&\Big(rf\Big(laystyle\frac{z_{1}}{r}\Big),\Big(\frac{rf(\frac{z_{1}}{r})}{z_{1}}\Big)^{\beta_{2}}\Big(f'\Big(\frac{z_{1}}{r}\Big)\Big)^{\gamma_{2}}z_{2},\cdots,\Big(\frac{rf(laystyle\frac{z_{1}}{r})}{z_{1}}\Big)^{\beta_{n}}\Big(f'\Big(laystyle\frac{z_{1}}{r}\Big)\Big)^{\gamma_{n}}z_{n}\Big)\\ &=&F(z).\end{eqnarray*}
由定义2.1知$F_{\psi}(z,t)$是一个Loewner链. 令$F_{\psi}(z)=F_{\psi}(z,0)$. 则$F_{\psi}(z)=F(z)$. 故$F(z)$在$\Omega$上能够嵌入Loewner链.
注3.2 在定理3.1中分别令$\beta_{j}=0$和$\gamma_{j}=0$,则可以得到相应的结论. 若$\gamma_{j}=0$,则得到文献[14]中的部分结论.
定理3.3 若$f$是$D$上的$\alpha$次殆$\beta$型螺形函数,$\alpha\in[0,1),\beta\in(-\frac{\pi}{2},\frac{\pi}{2})$且$a=\tan\beta$. $F(z)$和$\Omega$为定理3.1中所定义,$\beta_{j}$,$\gamma_{j}$及$r$满足定理3.1中的条件,则$F(z)$是$\Omega$上的$\alpha$次殆$\beta$型螺形映照.
证 由于$f$是$D$上的$\alpha$次殆$\beta$型螺形函数,由引理2.7可得
|
\begin{equation}g\Big(\frac{z_{1}}{r},t\Big)={\rm e}^{\frac{1-{\rm i}a}{1-\alpha}t}f\Big({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}\frac{z_{1}}{r}\Big)\end{equation}
| (3.5) |
是一个Loewner链.由定理3.1得
|
\begin{eqnarray}F_{\psi}(z,t)&=&\Big(rg\Big(laystyle\frac{z_{1}}{r},t\Big),{\rm e}^{(1-\beta_{2}-\gamma_{2})t}\Big(\frac{rg(\frac{z_{1}}{r},t)}{z_{1}}\Big)^{\beta_{2}}\Big(g'\Big(\frac{z_{1}}{r},t\Big)\Big)^{\gamma_{2}}z_{2},\cdots\\&&{\rm e}^{(1-\beta_{n}-\gamma_{n})t}\Big(\frac{rg(\frac{z_{1}}{r},t)}{z_{1}}\Big)^{\beta_{n}}\Big(g'\Big(laystyle\frac{z_{1}}{r},t\Big)\Big)^{\gamma_{n}}z_{n}\Big)'\end{eqnarray}
| (3.6) |
是一个Loewner链. 由(3.5),(3.6)式及$F_{\psi}(z)=F(z)$得
\begin{eqnarray*}F_{\psi}(z,t)&=&{\rm e}^{\frac{1-{\rm i}a}{1-\alpha}t}\Big(rf\Big({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}laystyle\frac{z_{1}}{r}\Big),\Big(\frac{rf({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}\frac{z_{1}}{r})}{{\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}z_{1}}\Big)^{\beta_{2}}\Big(f'\Big({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}\frac{z_{1}}{r}\Big)\Big)^{\gamma_{2}}{\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}z_{2},\cdots\\ &&\Big(\frac{rf({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}\frac{z_{1}}{r})}{{\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}z_{1}}\Big)^{\beta_{n}}\Big(f'\Big({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}laystyle\frac{z_{1}}{r}\Big)\Big)^{\gamma_{n}}{\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}z_{n}\Big)'\\ &=&{\rm e}^{\frac{1-{\rm i}a}{1-\alpha}t}F\Big({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}z\Big)\\ &=&{\rm e}^{\frac{1-{\rm i}a}{1-\alpha}t}F_{\psi}\Big({\rm e}^{\frac{{\rm i}a-\alpha}{1-\alpha}t}z\Big).\end{eqnarray*}
由引理2.7知$F(z)$是$\Omega$上的$\alpha$次殆$\beta$型螺形映照.
注3.4 在定理3.3中分别令$\beta_{j}=0$及$\gamma_{j}=0$可得相应的结论. 当$\gamma_{j}=0$时,则得到文献[14]中的部分结论.在定理3.3中分别令$\alpha=0$和$\beta=0$,则得到关于$\beta$型螺形映照及$\alpha$次殆星形映照的相应结论.
2016, Vol. 36 
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