设$ {\bf C}^{n} $表示$ n $维复欧几里得空间, $ {\bf C}^{n} $中两点$ z = (z_{1}, \cdots, z_{n}) $和$ w = (w_{1}, \cdots, w_{n}) $的内积定义为$ \langle z, w\rangle = z_{1}\overline{w_{1}}+\cdots+z_{n}\overline{w_{n}} $,因而$ |z| = \sqrt{|z_{1}|^{2}+\cdots+|z_{n}|^{2}} = \sqrt{\langle z, z\rangle} $.设$ B $表示$ {\bf C}^{n} $中的单位球, $ D $表示$ {\bf C} $中的单位圆盘, $ H(B) $表示$ B $上的全纯函数函数的全体.对$ f\in H(B) $,其复梯度$ \nabla f $和径向导数$ Rf $分别定义为

$ \nabla f(z) = \left(\frac{\partial f}{\partial z_{1}}(z), \cdots, \frac{\partial f}{\partial z_{n}}(z)\right) \ \mbox{和} \ R f(z) = \langle \nabla f(z), \overline{z}\rangle = \sum\limits_{j = 1}^{n}z_{j}\frac{\partial f}{\partial z_{j}}(z). $

定义1.1  $ [0, 1) $上一个正的连续函数$ \mu $被称为正规函数是指：存在常数$ 0 < a\leq b $以及常数$ 0\leq r_{0} < 1 $使得

(1) $ {\frac{\mu(r)}{(1-r)^{a}}} $在$ [r_{0}, 1) $上单调递减; (2) $ {\frac{\mu(r)}{(1-r)^{b}}} $在$ [r_{0}, 1) $上单调递增.

在本文中, $ a $和$ b $总表示正规函数$ \mu $定义中的那两个参数.为了简化论证,不失一般性,本文中总设$ r_{0} = 0 $.

例如$ \mu(r) = (1-r)^{\alpha}\log^{\beta}\frac{e}{1-r} $$ (\alpha > 0, \beta \ \mbox{为实数}), $$ \mu(r) = (1-r)^{\alpha}\log^{\beta}\left\{\frac{e}{1-r}\log\log\frac{e^{2}}{1-r}\right\}^{p} $$ (\alpha > 0, \beta\和p为实数), $

$ \mu(r) = \left\{ \begin{array}{ll} {\frac{(2n-2)!!}{(2n-1)!!}(1-r)^{1/2}}, & {1-\frac{1}{n}\leq r<1-\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)}, \\ {\frac{(2n)!!(n+1)}{(2n+1)!!}(1-r)^{3/2}}, & {1-\frac{1}{2}\left(\frac{1}{n}+\frac{1}{n+1}\right)\leq r<1-\frac{1}{n+1}}, \end{array}\right. $

$ (n = 1, 2, \cdots) $等都是此种形式的正规函数.尤其上述第一个权在各种加权型函数空间中频繁出现(如文献[1-3]等).

单位圆盘上的解析函数$ f $若满足条件

$ b_{Z}(f): = \sup\limits_{z\in D}(1-|z|^{2})|f''(z)|<\infty, $

我们称$ f $属于Zygmund类(可参见文献[4]),而以$ \| f\| _{Z}: = |f(0)|+|f'(0)|+b_{Z}(f) $为范数的Zygmund空间,则首先是在文献[3]中被引进.很快从Zygmund空间到本身或者从Zygmund空间到其他函数空间的一些算子被研究,如文献[5-8].

具有正规权的Zygmund型空间在文献[9]中被提到,这是一种包含Zygmund型空间的更一般的函数空间,在一些文献中被介绍和研究,例如文献[10-11].实际上, $ 1-|z|^2 $可以看成是一种权函数,在本文中,我们把权函数推广到了正规函数$ \mu(|z|) $.同时,我们把变量从单复变量推广到了多复变量.

定义1.2  设$ \mu $是$ [0, 1) $上的一个正规函数.函数$ f $说成是属于正规权Zygmund空间$ {\cal Z}_{\mu}(B) $是指: $ f\in H(B) $且

$ {\| f\| _{\mu}: = \sup\limits_{z\in B}\mu(|z|)\ \sum\limits_{k = 1}^{n}\sum\limits_{j = 1}^{n} \left|\frac{\partial^{2}f}{\partial z_{j}\partial z_{k}}(z)\right|}<\infty . $

我们记

$ {\| f\| _{{\cal Z}_{\mu}}: = |f(0)|+\sum\limits_{k = 1}^{n}\left|\frac{\partial f}{\partial z_{k}}(0)\right|+\| f\| _{\mu}}. $

在范数$ \| \cdot\| _{{\cal Z}_{\mu}} $下, $ {\cal Z}_{\mu}(B) $构成一个Banach空间.特别地,当$ \mu(r) = 1-r^{2} $时, $ {\cal Z}_{\mu}(B) $恰好是Zygmund空间$ {\cal Z}(B) $.当$ n > 1 $时,我们在文献[12]和文献[13]中分别给出过空间$ {\cal Z}_{\mu}(B) $的几个等价范数.

定义1.3  设$ \mu $是$ [0, 1) $上的一个正规函数.函数$ f $说成是属于正规权Bloch空间$ {\cal B}_{\mu}(B) $是指: $ f\in H(B) $且

$ {\| f\| _{{\cal B}_{\mu}}: = |f(0)|+ \sup\limits_{z\in B}\mu(|z|)|\nabla f(z)|}<\infty. $

特别地,当$ \mu(r) = 1-r^{2} $时, $ {\cal B}_{\mu}(B) $就是Bloch空间$ {\cal B}(B) $.当$ n > 1 $时,空间$ {\cal B}_{\mu}(B) $也有几个等价范数(可参见文献[14-16]).

定义1.4  设$ X $和$ Y $是$ B $上两个全纯函数空间, $ \varphi = (\varphi_{1}, \cdots, \varphi_{n}) $是$ B $上的一个全纯自映射且$ \psi\in H(B) $.从空间$ X $到$ Y $的加权复合算子$ \psi C_{\varphi} $定义为

$ (\psi C_{\varphi})f: = \psi\cdot f\circ\varphi\in Y \ \ \ (f\in X). $

过去几十年内,在单位圆盘或单位球上各种解析或者全纯函数空间之间的许多具体算子已经被广泛研究,例如文献[2-6, 8-11, 14-15, 17-30, 37-41]等等.在单位圆盘上Zygmund型空间之间$ C_{\varphi} $或者$ \psi C_{\varphi} $有界或紧的充要条件被给出,如文献[4-5, 22].然而,在多复变情形处理全纯函数空间上复合算子的问题要比单复变困难很多.在2009年,借助于精细的技巧,陈和Gauthier在文献[14]中获得了$ {\cal B}_{\mu}(B) $上复合算子有界或紧的充要条件,几乎同时,张和李在文献[15]中获得了正规权Bloch空间上加权复合算子有界或紧的充要条件.人们很自然会在以单位球为支撑集的Zygmund型空间上考虑复合算子和其他算子的刻画问题.在2012年,戴在文献[23]中就$ {\cal Z}(B) $上的复合算子$ C_{\varphi} $给了如下刻画：

命题A  设$ \varphi $是$ B $上的一个全纯自映射且$ n > 1 $.

(1) $ C_{\varphi} $是$ {\cal Z}(B) $上有界算子当且仅当对所有$ l\in \{1, \cdots, n\} $都有$ \varphi_{l}\in {\cal Z}(B) $且下列两条同时成立：

$ \sup\limits_{z\in B}\frac{(1-|z|^{2})\ |\langle R\varphi(z), \varphi(z)\rangle|^{2}}{1-|\varphi(z)|^{2}}<\infty, $

$ \sup\limits_{z\in B}(1-|z|^{2})\ |\langle R^{(2)}\varphi(z), \varphi(z)\rangle|\log\frac{1}{1-|\varphi(z)|^{2}}<\infty. $

(2) $ C_{\varphi} $在$ {\cal Z}(B) $上紧当且仅当$ \| \varphi\| _{\infty} < 1 $且对所有$ l\in \{1, \cdots, n\} $有$ \varphi_{l}\in {\cal Z}(B) $.

在单位圆盘上,叶和胡在文献[22]中考虑了$ {\cal Z}(D) $上加权复合算子$ \psi C_{\varphi} $的有界性和紧性问题,给出如下结果:

命题B  设$ u\in H(D) $且$ \varphi $是单位圆盘$ D $上的一个解析自映射.

(1) $ uC_{\varphi} $在$ {\cal Z}(D) $上有界当且仅当$ u\in {\cal Z}(D) $且下列两条同时成立:

$ \sup\limits_{z\in D}\frac{(1-|z|^{2})\ |u(z)[\varphi'(z)]|^{2}}{1-|\varphi(z)|^{2}}<\infty, $

$ \sup\limits_{z\in D}(1-|z|^{2})\ |2u'(z)\varphi'(z)+u(z)\varphi''(z)|\log\frac{1}{1-|\varphi(z)|^{2}}<\infty. $

(2)若$ uC_{\varphi} $在$ {\cal Z}(D) $上有界,则$ uC_{\varphi} $在$ {\cal Z}(D) $上紧当且仅当下列两条同时成立:

$ \lim\limits_{|\varphi(z)|\rightarrow 1^{-}}\frac{(1-|z|^{2})\ |u(z)[\varphi'(z)]|^{2}}{1-|\varphi(z)|^{2}} = 0, $

$ \lim\limits_{|\varphi(z)|\rightarrow 1^{-}}(1-|z|^{2})\ |2u'(z)\varphi'(z)+u(z)\varphi''(z)|\log\frac{1}{1-|\varphi(z)|^{2}} = 0. $

很自然的问题是:当具体的权函数$ 1-|z|^{2} $被推广为抽象的正规函数$ \mu(|z|) $以及$ D $被推广到$ B $后是否能获得类似结果?在高维,刻画这类问题最主要的难度在目标空间.在2015年,梁玉霞等在文献[25]中给出了从$ {\cal Z}(B) $到$ {\cal B}(B) $之加权复合算子$ \psi C_{\varphi} $有界性和紧性条件的刻画.然而,如果目标空间是$ {\cal Z}_{\mu}(B) $,那么对同样问题的刻画会更困难.在文献[13]中,张和黎讨论了$ {\cal Z}_{\mu}(B) $上$ C_{\varphi} $有界或紧的条件,给出了部分充要条件.本文的主要目的就是考虑当$ n > 1 $时$ {\cal Z}_{\mu}(B) $上加权复合算子有界或紧的一些充要条件.为了解决这个问题,我们必须寻找不同的处理方法.

在本文中,我们将用记号$ c, c_{1}, c_{2}, c_{3}, \cdots $来表示与变量$ z $、$ w $无关的正数,当然$ c, c_{1}, $$ c_{2} $, $ c_{3}, $$ \cdots $可以与某些参数有关,不同的地方可以代表不同的数; $ "E\approx F" $表示比较,称为$ E $等价于$ F $,即存在正的常数$ A_{1} $和$ A_{2} $使得$ A_{1}E \leq F \leq A_{2}E $.

设$ \mu $是$ [0, 1) $上的一个正规函数且

$ {\sigma_{\mu}(t) = \left\{\frac{1}{\mu(0)}+\int_{0}^{t}\frac{{\rm d}\tau}{(1-\tau)^{\frac{1}{2}}\ \mu(\tau)}\right\}^{-1}\ \ (0\leq t<1)} $

(参见文献[14-15]).

对任意$ u\in{{\bf C}^{n}} $,我们记$ { G^{\mu}_{0}(u) = \frac{|u|^{2}}{\mu^{2}(0)}} $.若$ 0\neq z\in B $,我们记

$ {G^{\mu}_{z}(u) = \frac{1}{\mu^{2}(|z|)}\left\{\frac{\mu^{2}(|z|)}{\sigma_{\mu}^{2}(|z|)}\ |u|^{2}+\left(1-\frac{\mu^{2}(|z|)}{\sigma_{\mu}^{2}(|z|)}\right)\ \frac{|\langle z, u\rangle|^{2}}{|z|^{2}}\right\}}. $

若$ z\neq 0 $,可以将$ u $分解成$ {u = u_1\frac{z}{|z|}+u_2\xi} $,这里$ \langle z, \xi\rangle = 0 $且$ \xi\in\partial B $.经计算得

$ u_1 = \frac{\langle u, z\rangle}{|z|}, \ |u|^2 = |u_{1}|^2+ |u_2|^2, \ G_z^\mu(u) = \frac{|u_1|^2}{\mu^{2}(|z|)}+ \frac{|u_2|^2}{\sigma_{\mu}^{2}(|z|)} $

且

(2.1) $ \begin{equation} (1-t)^{b}\left(1+\int_{0}^{t}\frac{{\rm d}\tau}{(1-\tau)^{b+\frac{1}{2}}}\right)\leq\frac{\mu(t)}{\sigma_{\mu}(t)}\leq (1-t)^{a}\left(1+\int_{0}^{t}\frac{{\rm d}\tau}{(1-\tau)^{a+\frac{1}{2}}}\right). \end{equation} $

实际上,上述(2.1)式可以来自于不等式

$ \frac{(1-t)^{b}}{(1-\tau)^{b}}\leq \frac{\mu(t)}{\mu(\tau)}\leq \frac{(1-t)^{a}}{(1-\tau)^{a}} \ \ \ \ (0\leq \tau\leq t<1). $

众所周知$ (p^{\frac{1}{2}}+q^{\frac{1}{2}})/2\leq (p+q)^{^{\frac{1}{2}}}\leq p^{\frac{1}{2}}+q^{\frac{1}{2}} $对所有$ p\geq 0 $和$ q\geq 0 $成立.因此,由(2.1)式可知存在常数$ 1/2 < t_{0} < 1 $使得

(2.2) $ \begin{equation} \frac{1}{4}\left(\frac{|\langle z, u\rangle|}{\mu(|z|)}+\frac{|u|}{\sigma_{\mu}(|z|)}\right)\leq\sqrt{G_{z}^{\mu}(u)} \leq \frac{3}{2}\left(\frac{|\langle z, u\rangle|}{\mu(|z|)}+\frac{|u|}{\sigma_{\mu}(|z|)}\right) \end{equation} $

对所有$ t_{0} < |z| < 1 $成立.实际上,由(2.1)式明显有

$ G_{z}^{\mu}(u) = \frac{|u|^{2}}{\sigma^{2}_{\mu}(|z|)}+\frac{|\langle z, u\rangle|^{2}}{\mu^{2}(|z|)}\left\{1-\left(\frac{\mu(|z|)}{\sigma_{\mu}(|z|)}\right)^{2}\right\}\frac{1}{|z|^{2}} $

和

$ \lim\limits_{|z|\rightarrow 1^{-}}\left\{1-\left(\frac{\mu(|z|)}{\sigma_{\mu}(|z|)}\right)^{2}\right\}\frac{1}{|z|^{2}} = 1. $

为了给出主要结果,我们首先给出一些引理.

引理2.1  设$ \mu $是$ [0, 1) $上一个正规函数且$ f\in H(B) $,则下列条件是等价的:

(1) $ f\in {\cal Z}_{\mu}(B) $;

(2) $ {I_{1} = |f(0)|+\sup\limits_{z\in B}\mu(|z|)|R^{(2)}f(z)|} < \infty $,这里$ R^{(2)}f = R(Rf) $;

(3) $ {I_{2} = |f(0)|+\sum\limits_{j = 1}^{n}\left|\frac{\partial f}{\partial z_{j}}(0)\right|+\sup\limits_{z\in B}W_{f}^{\mu}(z) < \infty} $,这里

$ W_{f}^{\mu}(z) = \sup\limits_{u\in {\bf C}^{n}\mbox{-}\{0\}}\sum\limits_{j = 1}^{n}\frac{|\langle \nabla (D_{j}f)(z), \overline{u}\rangle|}{\sqrt{G_{z}^{\mu}(u)}}, \ \ D_{j} = \frac{\partial}{\partial z_{j}} \ \ \ \mbox{$(j = 1, 2, \cdots, n)$}; $

(4) $ {I_{3} = |f(0)|+\sum\limits_{j = 1}^{n}\left|\frac{\partial f}{\partial z_{j}}(0)\right|+\sup\limits_{z\in B}P_{f}^{\mu}(z)} < \infty $,这里

$ {P_{f}^{\mu}(z) = \sup\limits_{u\in {\bf C}^{n}\mbox{-}\{0\}}\frac{|\langle uH_{f}(z), \overline{u}\rangle|}{E_{z}^{\mu}(u)}}, \ H_{f}(z) \ \mbox{表示矩阵} \ \left(\frac{\partial^{2}f(z)}{\partial z_{i}\partial z_{j}}\right)_{n\times n}, $

$ E_{z}^{\mu}(u) = \frac{|u|^{2}}{\sigma_{\mu}(|z|)}+\frac{|\langle z, u\rangle|^{2}}{\mu(|z|)}. $

进一步, $ I_{1}\asymp I_{2}\asymp I_{3}\asymp \| f\| _{{\cal Z}_{\mu}} $,其中控制常数与$ f $无关.

证  这些结果来自文献[13]中的引理2.1.

引理2.2  设$ \mu $是$ [0, 1) $上的一个正规函数.

(1)如果$ f\in {\cal Z}_{\mu}(B) $,则有

$ |Rf(z)|\leq |\nabla f(z)|\leq c\bigg(1+\int_{0}^{|z|}\frac{1}{\mu(t)}\ {\rm d}t\bigg)\| f\| _{{\cal Z}_{\mu}} $

且

$ |f(z)|\leq c\bigg\{1+\int_{0}^{1}\bigg(\int_{0}^{\rho|z|}\frac{1}{\mu(t)}\ {\rm d}t\bigg){\rm d}\rho\bigg\}\| f\| _{{\cal Z}_{\mu}} \ \ (z\in B ). $

(2)如果$ f\in {\cal B}_{\mu}(B) $,则有

$ |f(z)|\leq c\bigg(1+\int_{0}^{|z|}\frac{{\rm d}t}{\mu(t)}\bigg)\| f\| _{{\cal B}_{\mu}} \ \ (z\in B ). $

证  结果(1)来自文献[17],结果(2)来自文献[31]中的引理2.2.

引理2.3  设$ \mu $是$ [0, 1) $上的一个正规函数且

$ g(\xi) = 1+\sum\limits_{s = 1}^{\infty}2^{s}\ \xi^{n_{s}}\ \ (\xi\in D). $

那么,下列结果成立: (1) $ g(r) $在$ [0, 1) $上严格单调递增且

$ \inf\limits_{r\in [0, 1)}\mu(r)g(r) = N_{0}>0, \ \ \ \ \sup\limits_{\xi\in D}\mu(|\xi|)|g(\xi)| = M_{0}<\infty. $

(2)设$ l $是一个正整数,则存在常数$ M_{1} > 0 $和$ N_{1} > 0 $使得$ {\mu(\rho)g^{(l)}(\rho)\leq \frac{N_{1}}{(1-\rho)^{l}}} $对所有$ {0\leq \rho < 1} $成立且$ {\mu(\rho)g^{(l)}(\rho)\geq \frac{M_{1}} {(1-\rho)^{l}}} $对所有$ {\max\{r_{1}, 1-1/2l\} < \rho < 1} $成立.

其中$ n_{s} $为$ (1-r_{s})^{-1} $的整数部分, $ r_{0} = 0, \ \ \mu(r_{s}) = 2^{-s}\ \ (s = 1, 2, \cdots) $.

证  这些结果来自文献[32]中的引理2.4.其他与引理2.3有关的结果也能在文献[16, 27, 33]中找到.

引理2.4  设$ \mu $是$ [0, 1) $上的一个正规函数, $ k $为一个正整数.给定$ 0 < r_{0} < 1 $,则有

$ \int_{0}^{\rho|w|}\frac{{\rm d}t}{\mu(t)} \asymp \int_{0}^{\rho|w|^{k}}\frac{{\rm d}t}{\mu(t)} \ \mbox{和} \ \int_{0}^{|w|}\frac{{\rm d}t}{\sqrt{1-t}\mu(t)}+\frac{1}{\mu(0)} \asymp \int_{0}^{|w|^{k}}\frac{{\rm d}t}{\sqrt{1-t}\mu(t)} $

对一切$ r_{0} < |w| < 1 $和$ 0 < \rho\leq 1 $成立.

证  这些结果来自文献[13]中的引理2.4.

引理2.5  设$ \mu $是$ [0, 1) $上的一个正规函数,函数列$ \{f_{j}(z)\} $在$ {\cal Z}_{\mu}(B) $上有界且在$ B $中任何紧子集上一致收敛于0.

(1)若$ {\int_{0}^{1}\frac{{\rm d}t}{\mu(t)} < \infty} $,则$ {\lim\limits_{j\rightarrow\infty}\sup\limits_{z\in B}|\nabla f_{j}(z)| = 0 = \lim\limits_{j\rightarrow\infty}\sup\limits_{z\in B}| f_{j}(z)|.} $

(2)若$ {\int_{0}^{1}\left(\int_{0}^{\rho}\frac{{\rm d}t}{\mu(t)}\right){\rm d}\rho < \infty} $,则$ {\lim\limits_{j\rightarrow\infty}\sup\limits_{z\in B}| f_{j}(z)| = 0.} $

证  结果来源于文献[34].

引理2.6  设$ \mu $是$ [0, 1) $上的一个正规函数满足

$ {\int_{0}^{1}\left(\int_{0}^{\rho}\frac{{\rm d}t}{\mu(t) \sqrt{1-t}}\right){\rm d}\rho<\infty}. $

对$ 0 < r_{0} < 1 $和$ f\in {\cal Z}_{\mu}(B) $,若$ |z|\leq r_{0} $时有$ |\nabla f(z)|\leq m $,则有常数$ c > 0 $使得

$ |\langle \nabla f(z), \overline{\xi}\rangle|\leq m+c\| f\| _{{\cal Z}_{\mu}}\int_{r_{0}}^{1}\left(\int_{0}^{\rho}\frac{{\rm d}t}{\mu(t)\sqrt{1-t}}\right){\rm d}\rho $

对所有$ r_{0} < |z| < 1 $成立,其中$ \xi\in \partial B $满足$ \langle z, \xi\rangle = 0 $.

证  利用酉变换,我们可设$ z = (|z|, 0, \cdots, 0) $满足$ |z| < 1 $和$ \xi = (0, 1, 0, \cdots, 0). $就固定的$ 0\leq\rho < 1 $,我们设$ h(\eta) = D_{1}(Rf)(\rho, \eta, 0, \cdots, 0) $.如果$ f\in {\cal Z}_{\mu}(B) $,则有

$ |h(z_{2})|\leq \frac{c_{1}\| f\| _{{\cal Z}_{\mu}}}{\mu(\sqrt{\rho^{2}+|z_{2}|^{2}})}\leq \frac{c_{1}\| f\| _{{\cal Z}_{\mu}}}{\mu(\sqrt{\frac{\rho^{2}+1}{2}})}\leq \frac{c_{1}4^{b}\| f\| _{{\cal Z}_{\mu}}}{\mu(\rho)} $

对所有$ |z_{2}|^{2}\leq (1-\rho^{2})/2 $成立.

因此,对任意$ r_{0} < |z| < 1 $和$ 0\leq t\leq |z| $,我们有

$ \begin{eqnarray*} |D_{2}(Rf)(t, 0, \cdots, 0)-D_{2}(Rf)(0, 0, \cdots, 0)| & = &\left|\int_{0}^{t}h'(0)\ {\rm d}\rho\right| \\ & = &\frac{1}{2\pi}\left|\int_{0}^{t}\left(\int_{|w| = \sqrt{\frac{1-\rho^{2}}{2}}}\frac{h(w) {\rm d}w}{w^{2}}\right){\rm d}\rho\right|\\ &\leq &c_{2}\| f\| _{{\cal Z}_{\mu}} \int_{0}^{t}\frac{{\rm d}\rho}{\sqrt{1-\rho} \mu(\rho)}. \end{eqnarray*} $

当$ r_{0} < |z| < 1 $时,我们可得

$ \begin{eqnarray*} |\langle \nabla f(z), \overline{\xi}\rangle|& = &|D_{2}f(|z|, 0, \cdots, 0)|\\ & = & \frac{1}{|z|}\left|r_{0}D_{2}f(r_{0}, 0, \cdots, 0)+\int_{r_{0}}^{|z|}D_{2}(Rf)(t, 0, \cdots, 0){\rm d}t\right|\\ &\leq& m+c\| f\| _{{\cal Z}_{\mu}}\int_{r_{0}}^{1}\left(\int_{0}^{\rho}\frac{{\rm d}t}{\mu(t)\sqrt{1-t}}\right){\rm d}\rho. \end{eqnarray*} $

本引理证完.

定理3.1  设$ n > 1 $, $ \mu $是$ [0, 1) $上的一个正规函数, $ \psi\in H(B) $且$ \varphi $为$ B $上满足对所有$ l\in \{1, \cdots, n\} $都有$ \varphi_{l}\in {\cal Z}_{\mu}(B) $的一个全纯自映射.

(1)如果$ 0 < a\leq b < 1 $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上有界算子的充要条件为$ \psi\in {\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $都有$ \psi\varphi_{l}\in{\cal Z}_{\mu}(B) $.

(2)如果$ {1/2 < a\leq 1\leq b < a/2+3/4 } $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上有界算子的充要条件为$ \psi\in {\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $都有$ \psi\varphi_{l}\in{\cal Z}_{\mu}(B) $以及下列两个条件同时成立：

(3.1) $ \begin{equation} \sup\limits_{z\in B}\frac{\mu(|z|)|\psi(z)\| \langle R\varphi(z), \varphi(z)\rangle|^{2}}{\mu(|\varphi(z)|)}<\infty, \end{equation} $

(3.2) $ \begin{equation} \sup\limits_{z\in B}\mu(|z|)\{|\langle 2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z), \varphi(z)\rangle|\int_{0}^{|\varphi(z)|}\frac{{\rm d}\rho}{\mu(\rho)}<\infty. \end{equation} $

(3)如果$ b\geq a > 1 $,则$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界当且仅当下列三条同时成立：

(3.3) $ \begin{equation} \sup\limits_{z\in B}\frac{\mu(|z|)|\psi(z)|(1-|\varphi(z)|^{2})|\langle R\varphi(z), \varphi(z)\rangle|}{(1-|z|^{2})\mu(|\varphi(z)|)}<\infty, \end{equation} $

(3.4) $ \begin{equation} \sup\limits_{z\in B}\frac{\mu(|z|)|\psi(z)\| R\varphi(z)|}{1-|z|^{2}}\bigg(1+\int_{0}^{|\varphi(z)|}\frac{\sqrt{1-\rho^{2}} }{\mu(\rho)}\ {\rm d}\rho\bigg)<\infty, \end{equation} $

(3.5) $ \begin{equation} \sup\limits_{z\in B}\frac{\mu(|z|)|R\psi(z)|}{1-|z|^{2}}\bigg(1+\int_{0}^{|\varphi(z)|}\frac{1-\rho^{2} }{\mu(\rho)}\ {\rm d}\rho\bigg)<\infty. \end{equation} $

证  首先,如果$ f\in H(B) $,则有

(3.6) $ \begin{eqnarray} R^{(2)}[(\psi C_{\varphi})f](z)& = &\langle (\nabla f)[\varphi(z)], \overline{2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)}\rangle \\ &&+ R^{(2)}\psi(z)f[\varphi(z)]+\psi(z)\langle R\varphi(z)H_{f}[\varphi(z)], \overline{R\varphi(z)}\rangle. \end{eqnarray} $

(1)如果$ \psi\in {\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $都有$ \psi\varphi_{l}\in{\cal Z}_{\mu}(B) $,通过$ R^{(2)}(\psi \varphi_{l})(z) $$ = R^{(2)}\psi(z)\varphi_{l}(z) $$ + $$ 2R\psi(z)R\varphi_{l}(z)+\psi(z)R^{(2)}\varphi_{l}(z) $和引理2.1可得

(3.7) $ \begin{eqnarray} \mu(|z|)|2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)| &\leq& \sum\limits_{l = 1}^{n}\mu(|z|)|2R\psi(z)R\varphi_{l}(z)+\psi(z)R^{(2)}\varphi_{l}(z)| \\ & \leq& \sum\limits_{l = 1}^{n}\mu(|z|)\{|R^{(2)}(\psi \varphi_{l})(z)|+|R^{(2)}\psi(z)\varphi_{l}(z)|\} \\ &\leq& \sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}}). \end{eqnarray} $

当$ b < 1 $时,通过$ \psi, \varphi_{l}\in {\cal Z}_{\mu}(B) $以及引理2.2和不等式$ (1-|z|^{2})/(1-|\varphi(z)|^{2})\leq (1+|\varphi(0)|)/(1-|\varphi(0)|) $ (这个不等式在很多文献中可见,例如文献[35,引理2.4]或文献[36])结合$ {\int_{0}^{1}\frac{{\rm d}t}{\mu(t)} < \infty} $和$ \sigma_{\mu} $的定义,我们可以得到

(3.8) $ \begin{eqnarray} \frac{\mu(|z|)|\psi(z)\| R\varphi(z)|^{2}}{\sigma_{\mu}(|\varphi(z)|)} &\leq& \frac{c\mu(|z|)|R\varphi(z)|^{2}\| \psi\| _{{\cal Z}_{\mu}}}{\sigma_{\mu}(|\varphi(z)|)} \\ &\leq& \frac{c_{1}\mu(|z|)}{\sigma_{\mu}(|\varphi(z)|)} \bigg(\int_{0}^{|z|}\frac{{\rm d}t}{\mu(t)}\bigg)^{2}\| \psi\| _{{\cal Z}_{\mu}} \bigg(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| ^{2}_{{\cal Z}_{\mu}}\bigg) \\ & \leq& c_{2}\mu(|z|)\bigg(\frac{1}{\mu(0)}+\int_{0}^{|\varphi(z)|}\frac{{\rm d}t}{\sqrt{1-t}\ \mu(t)}\bigg)\| \psi\| _{{\cal Z}_{\mu}}\bigg(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| ^{2}_{{\cal Z}_{\mu}}\bigg) \\ & \leq& c_{3}\bigg\{1+\int_{0}^{|\varphi(z)|}\left(\frac{(1-|z|)^{a}}{(1-t)^{a+\frac{1}{2}}}+\frac{(1-|z|)^{b}} {(1-t)^{b+\frac{1}{2}}}\right){\rm d}t\bigg\}\| \psi\| _{{\cal Z}_{\mu}}\bigg(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| ^{2}_{{\cal Z}_{\mu}}\bigg) \\ & \leq& c_{4}\bigg\{1+\left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{b}\bigg\}\| \psi\| _{{\cal Z}_{\mu}}\bigg(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| ^{2}_{{\cal Z}_{\mu}}\bigg). \end{eqnarray} $

类似地,我们也可得到

(3.9) $ \begin{equation} \frac{\mu(|z|)|\psi(z)\| \langle R\varphi(z), \varphi(z)\rangle|^{2}}{\mu(|\varphi(z)|)}\leq c\left(\frac{1+|\varphi(0)|}{1-|\varphi(0)|}\right)^{b}\| \psi\| _{{\cal Z}_{\mu}}\sum\limits_{l = 1}^{n}\| \varphi_{l}\| _{{\cal Z}_{\mu}}^{2}. \end{equation} $

对任意$ f\in {\cal Z}_{\mu}(B) $,如果$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, 2, \cdots, n\} $有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $,通过(3.6)–(3.9)式和引理2.1,引理2.2,我们有

$ \begin{eqnarray*} \mu(|z|)|R^{(2)}[(\psi C_{\varphi})f](z)| &\leq &c_{1}\| \psi\| _{{\cal Z}_{\mu}}\| f\| _{{\cal Z}_{\mu}}+c_{2}\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}})\| f\| _{{\cal Z}_{\mu}}\\ && + c_{3}\mu(|z|)|\psi(z)|E_{\varphi(z)}^{\mu}[R\varphi(z)] \frac{|\langle R\varphi(z)H_{f}[\varphi(z)], \overline{R\varphi(z)}\rangle|}{E_{\varphi(z)}^{\mu}[R\varphi(z)]}\\ & \leq &c_{1}\| \psi\| _{{\cal Z}_{\mu}}\| f\| _{{\cal Z}_{\mu}}+c_{2}\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}}) \| f\| _{{\cal Z}_{\mu}}\\ && + c_{4}\mu(|z|)|\psi(z)|\left\{\frac{|R\varphi(z)|^{2}}{\sigma_{\mu}(|\varphi(z)|)}+\frac{|\langle R\varphi(z), \varphi(z)\rangle|^{2}}{\mu(|\varphi(z)|)}\right\}\| f\| _{{\cal Z}_{\mu}}\\ &\leq &c_{5}\| f\| _{{\cal Z}_{\mu}}. \end{eqnarray*} $

再利用引理2.1就有$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界.

反过来,如果$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界,则通过取$ f(z) = 1 $可得$ \psi\in{\cal Z}_{\mu}(B) $,通过取$ f(z) = z_{l} $就可得到$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $ ($ l\in \{1, \cdots, n\} $).

(2)如果$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, 2, \cdots, n\} $都有$ \varphi_{l}\in {\cal Z}_{\mu}(B) $,则由引理2.2和不等式$ (1-|z|^{2})/(1-|\varphi(z)|^{2})\leq (1+|\varphi(0)|)/(1-|\varphi(0)|) $可知:当$ 1/2 < a\leq 1\leq b < a/2+3/4 $时,我们可得到

(3.10) $ \begin{eqnarray} \frac{\mu(|z|)|\psi(z)\| R\varphi(z)|^{2}}{\sigma_{\mu}(|\varphi(z)|)} & \leq &\frac{c\mu(|z|)}{\sigma_{\mu}(|\varphi(z)|)}\left(\int_{0}^{|z|}\frac{{\rm d}t}{\mu(t)} \right)^{2}\left(\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}})\right) \left(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| _{{\cal Z}_{\mu}}\right) \\ &\leq&\frac{c_{1}(1-|z|)^{a}}{\mu(0)\sigma_{\mu}(|\varphi(z)|)}\left(\int_{0}^{|z|}\frac{{\rm d}t}{(1-t)^{a}} \right)\int_{0}^{|z|}\frac{{\rm d}t}{(1-t)^{b}} \\ &&\times\left(\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}})\right)\left(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| _{{\cal Z}_{\mu}}\right) \\ &\leq&\frac{c_{2}(1-|z|)^{a}}{(1-|\varphi(z)|)^{b-a}}\left\{1+\int_{0}^{|\varphi(z)|}\frac{{\rm d}t}{(1-t)^{a+\frac{1}{2}}} \right\}\left(\int_{0}^{|z|}\frac{{\rm d}t}{(1-t)^{a}}\right) \\ &&\times\int_{0}^{|z|}\frac{{\rm d}t}{(1-t)^{b}} \left(\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}})\right) \left(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| _{{\cal Z}_{\mu}}\right) \\ &\leq &c_{3}. \end{eqnarray} $

同时还有当$ b < 2 $时

(3.11) $ \begin{eqnarray} \mu(|z|)\{| 2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)|\} \int_{0}^{|\varphi(z)|}\frac{{\rm d}\rho}{\sigma_{\mu}(\rho)} \leq c\sum\limits_{l = 1}^{n}(\| \varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}) . \end{eqnarray} $

另一方面,对任意$ f\in {\cal Z}_{\mu}(B) $,我们有

(3.12) $ \begin{eqnarray} &&\langle(\nabla f)[\varphi(z)], \overline{2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)}\rangle - \langle(\nabla f)(0), \overline{2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)}\rangle \\ & = &\sum\limits_{j = 1}^{n}\{2R\psi(z)R\varphi_{j}(z)+\psi(z)R^{(2)}\varphi_{j}(z)\}\int_{0}^{1}\frac{\rm d}{{\rm d}t}\{D_{j}f[t\varphi(z)]\} {\rm d}t \\ & = &\sum\limits_{l = 1}^{n}\varphi_{l}(z)\int_{0}^{1}\langle \nabla (D_{l}f)[t\varphi(z)], \overline{2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)}\rangle {\rm d}t. \end{eqnarray} $

通过(3.7)式和(3.12)式以及引理2.1和(2.2)式有

(3.13) $ \begin{eqnarray} && \mu(|z|)|\langle(\nabla f)[\varphi(z)], \overline{2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)}\rangle| \\ &\leq &c_{1}\left(\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+ \| \psi\| _{{\cal Z}_{\mu}})\right)\| f\| _{{\cal Z}_{\mu}} \\ && + c_{2}|\varphi(z)|\mu(|z|)\int_{0}^{1}\left(\sum\limits_{l = 1}^{n}|\langle \nabla (D_{l}f)[t\varphi(z)], \overline{2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)}\rangle|\right){\rm d}t \\ & \leq& c_{1}\left(\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}})\right)\| f\| _{{\cal Z}_{\mu}} \\ &&+ c_{3}|\varphi(z)|\mu(|z|)\left(\int_{0}^{1}\sqrt{G_{t\varphi(z)}^{\mu}[2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)]}\ {\rm d}t\right)\| f\| _{{\cal Z}_{\mu}} \\ &\leq &c_{4}\| f\| _{{\cal Z}_{\mu}}+ c_{5}\mu(|z|)|2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)|\left(\int_{0}^{|\varphi(z)|} \frac{{\rm d}t}{\sigma_{\mu}(t)}\right)\| f\| _{{\cal Z}_{\mu}} \\ &&+ c_{6}\mu(|z|)|\langle 2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z), \varphi(z)\rangle|\left(\int_{0}^{|\varphi(z)|}\frac{{\rm d}t}{\mu(t)}\right)\| f\| _{{\cal Z}_{\mu}}. \end{eqnarray} $

如果(3.1)–(3.2)式成立,由(3.6)式, (3.10)–(3.11)式和(3.13)式以及引理2.1,我们可得$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界.

反过来,如果$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界,立即可得$ \psi, \ \psi\varphi_{l}, \ \psi\varphi_{l}^{2}\in {\cal Z}_{\mu}(B) $成立.

对任意$ w\in B $,若$ |\varphi(w)|\leq 1/2 $,则有

(3.14) $ \begin{equation} \frac{\mu(|w|)|\psi(w)\| \langle R\varphi(w), \varphi(w)\rangle|^{2}}{\mu(|\varphi(w)|)}\leq \frac{c}{\mu(1/2)}\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}^{2}\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}). \end{equation} $

若$ |\varphi(w)| > 1/2 $,设$ g $为引理2.3中的那个函数,我们选取示性函数

$ \begin{eqnarray*} f_{w}(z)& = &\int_{0}^{\langle z, \varphi(w)\rangle^{3}}\left(\int_{0}^{\rho}g(t){\rm d}t\right){\rm d} \rho-2\int_{0}^{|\varphi(w)|^{2}\langle z, \varphi(w)\rangle^{2}}\left(\int_{0}^{\rho}g(t){\rm d}t\right){\rm d}\rho \\ &&+ \int_{0}^{|\varphi(w)|^{4}\langle z, \varphi(w)\rangle}\left(\int_{0}^{\rho}g(t){\rm d}t\right){\rm d}\rho \ \ (z\in B). \end{eqnarray*} $

通过引理2.1和引理2.3不难证明$ \| f_{w}\| _{{\cal Z}_{\mu}}\leq c $.同时有$ C_{\varphi}f_{w}(w) = 0 $, $ R[C_{\varphi}f_{w}](w) = 0 $以及

(3.15) $ \begin{equation} R^{(2)}[C_{\varphi}f_{w}](w) = 2\left(|\varphi(w)|^{8}g(|\varphi(w)|^{6})+|\varphi(w)|^{2}\int_{0}^{|\varphi(w)|^{6}}g(t) {\rm d}t\right)\langle R\varphi(w), \varphi(w)\rangle^{2}. \end{equation} $

由引理2.1和(3.15)式以及引理2.3可得

(3.16) $ \begin{eqnarray} c\| \psi C_{\varphi}\| &\geq &\mu(|w|)|R^{(2)}[(\psi C_{\varphi})f_{w}](w)| = \mu(|w|)|\psi(w)R^{(2)}[C_{\varphi}f_{w}](w)| \\ & \geq& 2\mu(|w|)|\psi(w)\| \varphi(w)|^{8}g(|\varphi(w)|^{6})|\langle R\varphi(w), \varphi(w)\rangle|^{2} \\ & \geq& \frac{c_{1}\mu(|w|)|\psi(w)\| \langle R\varphi(w), \varphi(w)\rangle|^{2}}{\mu(|\varphi(w)|)}. \end{eqnarray} $

(3.14)式和(3.16)式表明(3.1)式成立.我们再取示性函数

$ h_{w}(z) = \int_{0}^{\langle z, \varphi(w)\rangle}\left(\int_{0}^{\rho}g(t){\rm d}t\right){\rm d}\rho \ \ (z\in B). $

由引理2.1和引理2.3可得$ \| h_{w}\| _{{\cal Z}_{\mu}}\leq c $.

若$ |\varphi(w)| > 1/2 $,则由引理2.1 –引理2.4和(3.1)式有

(3.17) $ \begin{eqnarray} c\| \psi C_{\varphi}\| &\geq& \mu(|w|)|R^{(2)}[(\psi C_{\varphi})h_{w}](w)| \\ &\geq &-\mu(|w|)|R^{(2)}\psi(w)\| h_{w}[\varphi(w)]| \\ &&+|2R\psi(w)\langle R\varphi(w), \varphi(w)\rangle+\psi(w)\langle R^{(2)}\varphi(w), \varphi(w)\rangle|\int_{0}^{|\varphi(w)|^{2}}g(t){\rm d}t \\ && -\mu(|w|)g(|\varphi(w)|^{2})|\langle R\varphi(w), \varphi(w)\rangle|^{2} \\ & \geq& -c_{1}\| \psi\| _{{\cal Z}_{\mu}}-\frac{c_{2}\mu(|w|)|\psi(w)\| \langle R\varphi(w), \varphi(w)\rangle|^{2}}{\mu(|\varphi(w)|)} \\ &\;& + c_{3}\mu(|w|)|\langle 2R\psi(w)R\varphi(w)+\psi(w)R^{(2)}\varphi(w), \varphi(w)\rangle|\int_{0}^{|\varphi(w)|}\frac{{\rm d}t}{\mu(t)} \\ &\Rightarrow & \mu(|w|)|\langle 2R\psi(w)R\varphi(w)+\psi(w)R^{(2)}\varphi(w), \varphi(w)\rangle|\int_{0}^{|\varphi(w)|}\frac{{\rm d}t}{\mu(t)} \\ &\leq& c_{4}. \end{eqnarray} $

(3.11)式和(3.17)式表明(3.2)式成立.

(3)如果$ b\geq a > 1 $,则空间$ {\cal Z}_{\mu}(B) $上的复合算子问题恰好是权为$ \nu(r) = \mu(r)/(1-r) $的Bloch型空间$ {\cal B}_{\nu}(B) $上的复合算子问题.实际上,对任意$ f\in {\cal B}_{\nu}(B) $,如果在文献[35]的定理3.1(1)中取$ h = Rf $和$ t = 0 $,则有$ f\in {\cal Z}_{\mu}(B) $且$ \| f\| _{{\cal Z}_{\mu}}\leq c\| f\| _{{\cal B}_{\nu}} $.反过来,对任意$ f\in {\cal Z}_{\mu}(B) $,由引理2.2可得

$ \nu(|z|)|Rf(z)|\leq \frac{c\| f\| _{{\cal Z}_{\mu}}\mu(|z|)}{1-|z|}\int_{0}^{|z|}\frac{{\rm d}t}{\mu(t)}\leq c\| f\| _{{\cal Z}_{\mu}}\int_{0}^{|z|}\frac{(1-|z|)^{a-1}\ {\rm d}t}{(1-t)^{a}}\leq \frac{c\| f\| _{{\cal Z}_{\mu}}}{a-1}. $

通过文献[14]中的定理2可得$ f\in {\cal B}_{\nu}(B) $且$ \| f\| _{{\cal B}_{\nu}}\leq c\| f\| _{{\cal Z}_{\mu}} $.

对任意$ f\in {\cal Z}_{\mu}(B) $,我们有$ f\in {\cal B}_{\nu}(B) $.如果(3.3)–(3.5)式成立,由引理2.2以及文献[14]中的定理2和(2.2)式,我们可得

$ \begin{eqnarray*} \sup\limits_{z\in B}\mu(|z|)|R^{(2)}[(\psi C_{\varphi})f](z)| &\leq& c_{1}\sup\limits_{z\in B}\nu(|z|)|R[(\psi C_{\varphi})f](z)|\\ & \leq& c_{1}\| f\| _{{\cal B}_{\nu}}\sup\limits_{z\in B}\bigg\{\nu(|z|)|R\psi(z)|\bigg(1+\int_{0}^{|\varphi(z)|}\frac{{\rm d}t}{\nu(t)}\bigg)\bigg\}\\ &\;& + c_{2}\sup\limits_{z\in B}\nu(|z|)|\psi(z)|\sqrt{G^{\nu}_{\varphi(z)}[R\varphi(z)]} \| f\| _{{\cal B}_{\nu}} \\ &\leq &c_{3}\| f\| _{{\cal B}_{\nu}} \leq c_{4}\| f\| _{{\cal Z}_{\mu}}. \end{eqnarray*} $

这表明$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界.

反过来,设$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界.对任意$ w\in B $,仅需考虑情形$ |\varphi(w)| > 1/2 $.取

$ f_{w}(z) = \int_{0}^{\langle z, \varphi(w)\rangle^{2}}g_{1}(t){\rm d}t- \int_{0}^{|\varphi(w)|^{2}\langle z, \varphi(w)\rangle}g_{1}(t){\rm d}t, $

其中$ g_{1} $是引理2.3中与$ \nu $对应的函数.我们可以证明(3.3)式成立.如果我们取示性函数

$ f_{w}(z) = \int_{0}^{\langle z, \varphi(w)\rangle^{2}}g_{1}(t){\rm d}t- 2\int_{0}^{|\varphi(w)|^{2}\langle z, \varphi(w)\rangle}g_{1}(t){\rm d}t, $

则可以证明(3.5)式成立.设$ \langle \varphi(w), \xi\rangle = 0 $满足$ \xi\in\partial B $.我们再取

$ f_{w}(z) = \langle z, \xi\rangle\int_{0}^{\langle z, \varphi(w)\rangle}\frac{g_{1}(t){\rm d}t}{\sqrt{1-t}}. $

通过这个示性函数, (2.1)式和(3.3)式,我们可以证明(3.4)式成立.本定理证完.

定理3.2  设$ n > 1 $, $ \mu $是$ [0, 1) $上的一个正规函数, $ \varphi $是$ B $上的一个全纯自映射且满足对一切$ l\in \{1, \cdots, n\} $都有$ \varphi_{l}\in {\cal Z}_{\mu}(B) $.

(1)如果$ {\| \varphi\| _{\infty} < 1} $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上紧算子当且仅当$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $都有$ \psi\varphi_{l}, \psi\varphi_{l}^{2}\in {\cal Z}_{\mu}(B) $.

(2)如果$ {\| \varphi\| _{\infty} = 1} $, $ 0 < a\leq b < 1 $或$ 1/2 < a\leq 1\leq b < a/2+3/4 $或$ \int_{0}^{1}\mu^{-1}(t){\rm d}t < \infty $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上紧算子当且仅当对所有$ l\in \{1, \cdots, n\} $都有$ \psi, \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $且

(3.18) $ \begin{equation} \lim\limits_{|\varphi(z)|\rightarrow 1^{-}}\frac{\mu(|z|)|\psi(z)\| \langle R\varphi(z), \varphi(z)\rangle|^{2}}{\mu(|\varphi(z)|)} = 0. \end{equation} $

(3)如果$ {\| \varphi\| _{\infty} = 1} $, $ 1/2 < a\leq 1\leq b < a/2+3/4 $和$ \int_{0}^{1}\mu^{-1}(t){\rm d} t = \infty $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上紧算子当且仅当对所有$ l\in \{1, \cdots, n\} $都有$ \psi, \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $, (3.18)式成立且

(3.19) $ \begin{equation} \lim\limits_{|\varphi(z)|\rightarrow 1^{-}}\mu(|z|)|\langle 2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z), \varphi(z)\rangle|\int_{0}^{|\varphi(z)|}\frac{{\rm d}\rho}{\mu(\rho)} = 0. \end{equation} $

(4)如果$ {\| \varphi\| _{\infty} = 1} $, $ b\geq a > 1 $且$ \int_{0}^{1}\mu^{-1}(t)\sqrt{1-t}{\rm d}t < \infty $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上紧算子当且仅当$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $都有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $以及

(3.20) $ \begin{equation} \lim\limits_{|\varphi(z)|\rightarrow 1^{-}}\frac{\mu(|z|)|\psi(z)|(1-|\varphi(z)|^{2})|\langle R\varphi(z), \varphi(z)\rangle|}{(1-|z|^{2})\mu(|\varphi(z)|)} = 0. \end{equation} $

(5)如果$ {\| \varphi\| _{\infty} = 1} $, $ b\geq a > 1 $且$ \int_{0}^{1}\mu^{-1}(t)\sqrt{1-t}{\rm d}t = \infty > \int_{0}^{1}(1-t)\mu^{-1}(t){\rm d}t $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上紧算子当且仅当$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $都有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $, (3.20)式成立且还满足

(3.21) $ \begin{equation} \lim\limits_{|\varphi(z)|\rightarrow 1^{-}}\frac{\mu(|z|)|\psi(z)\| R\varphi(z)|}{1-|z|^{2}}\int_{0}^{|\varphi(z)|}\frac{\sqrt{1-\rho^{2}} }{\mu(\rho)}\ {\rm d}\rho = 0. \end{equation} $

(6)如果$ {\| \varphi\| _{\infty} = 1} $, $ b\geq a > 1 $且$ \int_{0}^{1}(1-t)\mu^{-1}(t){\rm d}t = \infty $,则$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上紧算子当且仅当$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $, (3.20)–(3.21)式成立且

(3.22) $ \begin{equation} \lim\limits_{|\varphi(z)|\rightarrow 1^{-}}\frac{\mu(|z|)|R\psi(z)|}{1-|z|^{2}}\int_{0}^{|\varphi(z)|}\frac{1-\rho^{2} }{\mu(\rho)}\ {\rm d}\rho = 0. \end{equation} $

证   (1)设$ \{f_{j}(z)\} $是任一满足在$ B $上内闭一致收敛于0且$ \| f_{j}\| _{{\cal Z}_{\mu}}\leq 1 $的函数列,则$ \{|\nabla f_{j}(z)|\} $和$ \{|\nabla (D_{l}f_{j})(z)|\} \ (l = 1, \cdots, n) $也在$ B $上内闭一致收敛于0.

如果$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $有$ \psi\varphi_{l}, \psi\varphi_{l}^{2}\in {\cal Z}_{\mu}(B) $,那么就有

(3.23) $ \begin{eqnarray} \mu(|z|)|\psi(z)\| R\varphi(z)|^{2}&\leq&\sum\limits_{l = 1}^{n}\mu(|z|)|\psi(z)\| R\varphi_{l}(z)|^{2} \\ &\leq&\sum\limits_{l = 1}^{n}\frac{\mu(|z|)\{2|R^{(2)}[\psi(z)\varphi_{l}(z)]|+|R^{(2)}[\psi(z)\varphi_{l}^{2}(z)]|+|R^{(2)}\psi(z)|\}}{2} \ \ \\ &\leq& c\sum\limits_{l = 1}^{n}(\| \psi\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}^{2}\| _{{\cal Z}_{\mu}}) . \end{eqnarray} $

由$ \| \varphi\| _{\infty} < 1 $以及(3.23)式和引理2.1可得

$ \begin{eqnarray*} c\| (\psi C_{\varphi})f_{j}\| _{{\cal Z}_{\mu}} &\leq &|\psi(0)f_{j}[\varphi(0)]|+\sup\limits_{z\in B}\mu(|z|)|R^{(2)}[(\psi C_{\varphi}) f_{j}](z)|\\ & = &|\psi(0)f_{j}[\varphi(0)]| + \sup\limits_{z\in B} \mu(|z|)| \langle (\nabla f_{j})[\varphi(z)], \overline{2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z)}\rangle\\ &\;&+ R^{(2)}\psi(z)f_{j}[\varphi(z)] + \psi(z)\sum\limits_{l = 1}^{n}\langle \nabla (D_{l}f_{j})[\varphi(z)], \overline{R\varphi_{l}(z)}\ \overline{R\varphi(z)}\rangle |\\ & \leq& c_{1}\| \psi\| _{{\cal Z}_{\mu}}\sup\limits_{|w|\leq \| \varphi\| _{\infty}}|f_{j}(w)|+ c_{2}\sum\limits_{l = 1}^{n}(\| \psi\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}})\sup\limits_{|w|\leq\| \varphi\| _{\infty}}|\nabla f_{j}(w)|\\ &\;& + c_{3}\sum\limits_{l = 1}^{n}(\| \psi\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}^{2}\| _{{\cal Z}_{\mu}})\sup\limits_{|w|\leq \| \varphi\| _{\infty}}\sum\limits_{l = 1}^{n}|\nabla (D_{l}f_{j})(w)|\\ &\;& + |\psi(0)f_{j}[\varphi(0)]| \rightarrow 0 \ \ (j\rightarrow\infty). \end{eqnarray*} $

这意味着$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上是紧算子.

反之是显然的.

(2)如果(3.18)式成立,则对任意$ \varepsilon > 0 $,存在$ 0 < \delta_{1} < 1 $,当$ |\varphi(z)| > \delta_{1} $时有

(3.24) $ \begin{equation} \frac{\mu(|z|)|\psi(z)\| \langle R\varphi(z), \varphi(z)\rangle|^{2}}{\mu(|\varphi(z)|)}<\varepsilon. \end{equation} $

另一方面,当$ 1/2 < a\leq 1\leq b < a/2+ 3/4 $或$ 0 < a\leq b < 1 $时有

(3.25) $ \begin{eqnarray} \frac{\mu(|z|)|\psi(z)\| R\varphi(z)|^{2}}{\sigma_{\mu}(|\varphi(z)|)} &\leq &\frac{\mu(|z|)(|R(\psi\varphi)(z)|+|R\psi(z)|)|R\varphi(z)|}{\sigma_{\mu}(|\varphi(z)|)} \\ &\leq& \frac{c\mu(|z|)}{\sigma_{\mu}(|\varphi(z)|)} \bigg(\int_{0}^{|z|}\frac{{\rm d}t}{\mu(t)}\bigg)^{2}\bigg(\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}})\bigg) \bigg(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| _{{\cal Z}_{\mu}}\bigg) \\ &\leq&\frac{c_{2}(1-|z|)^{a}}{(1-|\varphi(z)|)^{b-a}}\bigg\{1+\int_{0}^{|\varphi(z)|}\frac{{\rm d}t}{(1-t)^{a+\frac{1}{2}}} \bigg\}\bigg(\int_{0}^{|z|}\frac{{\rm d}t}{(1-t)^{a}}\bigg) \\ &&\times \int_{0}^{|z|}\frac{{\rm d}t}{(1-t)^{b}} \bigg(\sum\limits_{l = 1}^{n}(\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}}+\| \psi\| _{{\cal Z}_{\mu}})\bigg)\bigg(\sum\limits_{l = 1}^{n}\| \varphi_{l}\| _{{\cal Z}_{\mu}}\bigg) \\ & \rightarrow & 0 \ \ (|\varphi(z)|\rightarrow 1^{-}). \end{eqnarray} $

因此,存在$ 0 < \delta_{1}\leq \delta < 1 $,当$ |\varphi(z)| > \delta $时有

(3.26) $ \begin{equation} \frac{\mu(|z|)|\psi(z)\| R\varphi(z)|^{2}}{\sigma_{\mu}(|\varphi(z)|)}<\varepsilon. \end{equation} $

设$ \{f_{j}(z)\} $是任一满足在$ B $上内闭一致收敛于0且$ \| f_{j}\| _{{\cal Z}_{\mu}}\leq 1 $的函数列.

如果$ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, \cdots, n\} $都有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $以及(3.18)式成立,由引理2.1和引理2.5, (3.24)式和(3.26)式,我们可得

$ \begin{eqnarray*} c\| (\psi C_{\varphi})f_{j}\| _{{\cal Z}_{\mu}} &\leq& |\psi(0)f_{j}[\varphi(0)]|+c_{1}\sum\limits_{l = 1}^{n}(\| \psi\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}})\sup\limits_{w\in B}|\nabla f_{j}(w)|\\ &\;& + c_{2}\| \psi\| _{{\cal Z}_{\mu}}\sup\limits_{w\in B}|f_{j}(w)|+c_{3}\| f_{j}\| _{{\cal Z}_{\mu}}\sup\limits_{|\varphi(z)|>\delta}\frac{\mu(|z|)|\psi(z)\| R\varphi(z)|^{2}}{\sigma_{\mu}(|\varphi(z)|)}\\ &\;& + c_{4}\| f_{j}\| _{{\cal Z}_{\mu}}\sup\limits_{|\varphi(z)|>\delta}\frac{\mu(|z|)|\psi(z)\| \langle R\varphi(z), \varphi(z)\rangle|^{2}}{\mu(|\varphi(z)|)} \\ &\;& + c_{5}\sum\limits_{l = 1}^{n}(\| \psi\| _{{\cal Z}_{\mu}}+\| \psi\varphi_{l}\| _{{\cal Z}_{\mu}})^{2}\sup\limits_{|w|\leq \delta}\sum\limits_{l = 1}^{n}|\nabla (D_{l}f_{j})(w)|\\ & \Rightarrow & \overline{\lim\limits_{j\rightarrow\infty}}\ \| (\psi C_{\varphi})f_{j}\| _{{\cal Z}_{\mu}}\leq c_{6}\varepsilon. \end{eqnarray*} $

根据$ \varepsilon $的任意性,这意味着$ \psi C_{\varphi} $是$ {\cal Z}_{\mu}(B) $上的紧算子.

反过来,设$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上是紧算子,则$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上有界,这样就有$ \psi\in {\cal Z}_{\mu}(B) $且对任意$ l\in \{1, 2, \cdots, n\} $都有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $.对任意满足$ {\lim\limits_{j\rightarrow\infty}|\varphi(z^{j})| = 1} $且$ |\varphi(z^{j})| > t_{0} \ \ (j = 1, 2, \cdots) $的点列$ \{z^{j}\}\subset B $,我们选取函数列

$ \begin{eqnarray*} f_{j}(z)& = &(1-|\varphi(z^{j})|^{2})\bigg(\int_{0}^{\langle z, \varphi(z^{j})\rangle^{3}}g(t){\rm d}t+\int_{0}^{|\varphi(z^{j})|^{4}\langle z, \varphi(z^{j})\rangle}g(t){\rm d}t\bigg) \\ && - \ 2(1-|\varphi(z^{j})|^{2})\int_{0}^{|\varphi(z^{j})|^{2}\langle z, \varphi(z^{j})\rangle^{2}}g(t){\rm d}t. \end{eqnarray*} $

容易验证$ \{f_{j}(z)\} $在$ B $的任一紧子集上一致收敛于0.此外,利用引理2.1和引理2.3可得$ \| f_{j}\| _{{\cal Z}_{\mu}}\leq c $.这意味着

(3.27) $ \begin{equation} \lim\limits_{j\rightarrow\infty}\| C_{\varphi}f_{j}\| _{{\cal Z}_{\mu}} = 0. \end{equation} $

对所有$ j\in \{1, 2, \cdots\} $,我们有$ C_{\varphi}f_{j}(z^{j}) = 0 = R[C_{\varphi}f_{j}](z^{j}) $,

(3.28) $ \begin{equation} R^{(2)}[C_{\varphi}f_{j}](z^{j}) = 2(1-|\varphi(z^{j})|^{2})\{|\varphi(z^{j})|^{8}g'(|\varphi(z^{j})|^{6})+|\varphi(z^{j})|^{2}g(|\varphi(z^{j})|^{6})\}\langle R\varphi(z^{j}), \varphi(z^{j})\rangle^{2}. \end{equation} $

通过引理2.1和引理2.3以及(3.27)–(3.28)式有

$ \begin{eqnarray*} c\| (\psi C_{\varphi})f_{j}\| _{{\cal Z}_{\mu}} &\geq &\mu(|z^{j}|)|R^{(2)}[(\psi C_{\varphi})f_{j}](z^{j})| = \mu(|z^{j}|)|\psi(z^{j})\| R^{(2)}[ C_{\varphi}f_{j}](z^{j})|\\ & \geq& 2(1-|\varphi(z^{j})|^{2})|\varphi(z^{j})|^{8}\mu(|z^{j}|)|\psi(z^{j})|g'(|\varphi(z^{j})|^{6})|\langle R\varphi(z^{j}), \varphi(z^{j})\rangle|^{2}\\ & \geq& \frac{c_{1}\mu(|z^{j}|)|\psi(z^{j})\| \langle R\varphi(z^{j}), \varphi(z^{j})\rangle|^{2}}{\mu(|\varphi(z^{j})|)} \ \Rightarrow \ \mbox{(3.18)式成立.} \end{eqnarray*} $

(3)由$ b < 5/4 $可得$ {\int_{0}^{1}\frac{{\rm d}t}{\sigma_{\mu}(t)} < \infty} $.如果(3.18)–(3.19)式成立, $ \psi\in{\cal Z}_{\mu}(B) $且对所有$ l\in \{1, 2, \cdots, n\} $有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $,则同样可以有(3.24)式和(3.26)式成立.同时,对上述$ \varepsilon > 0 $,存在$ \delta_{1} < \delta_{2} < 1 $,当$ |\varphi(z)| > \delta_{2} $时有

(3.29) $ \begin{equation} \mu(|z|)|\langle 2R\psi(z)R\varphi(z)+\psi(z)R^{(2)}\varphi(z), \varphi(z)\rangle|\int_{0}^{|\varphi(z)|}\frac{{\rm d}\rho}{\mu(\rho)}<\varepsilon. \end{equation} $

通过$ {\int_{0}^{1}\frac{{\rm d}t}{\sigma_{\mu}(t)} < \infty} $和引理2.6可知,存在$ \delta_{2} < r_{0} < 1 $使得

(3.30) $ \begin{equation} |\langle\nabla f_{j}[\varphi(z)], \overline{\xi}\rangle|\leq \sup\limits_{|z|\leq r_{0}}|\nabla f_{j}[\varphi(z)]|+c\| f_{j}\| _{{\cal Z}_{\mu}}\varepsilon, \end{equation} $

其中$ \xi\in \partial B $且$ \langle \varphi(z), \xi\rangle = 0 $.

由(3.24)式和(3.26)式, (3.29)–(3.30)式,引理2.1和引理2.5,我们可以证明$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上是紧算子.

反过来,设$ \psi C_{\varphi} $在$ {\cal Z}_{\mu}(B) $上是紧算子,可以得到$ \psi\in{\cal Z}_{\mu}(B) $且对任意$ l\in \{1, 2, \cdots, n\} $有$ \psi\varphi_{l}\in {\cal Z}_{\mu}(B) $.在情形(2)中我们已证(3.18)式成立,接下来只需证明(3.19)式成立.

对任意满足$ {\lim\limits_{j\rightarrow\infty}|\varphi(z^{j})| = 1} $且$ |\varphi(z^{j})| > t_{0} \ (j = 1, 2, \cdots) $的点列$ \{z^{j}\}\subset B $,取

$ h_{j}(z) = \int_{0}^{1}\frac{H_{j}(\rho z)}{\rho}\ {\rm d}\rho/\int_{0}^{|\varphi(z^{j})|^{2}}g(t){\rm d}t, $

其中$ H_{j}(z) = \left(\int_{0}^{\langle z, \varphi(z^{j})\rangle}g(t){\rm d}t\right)^{2}. $

通过简单计算可以验证$ \| h_{j}\| _{{\cal Z}_{\mu}}\leq c $且$ \{h_{j}(z)\} $在$ B $上内闭一致收敛于0.

同时,我们又有

$ R[C_{\varphi}h_{j}](z^{j}) = \frac{\langle R\varphi(z^{j}), \varphi(z^{j})\rangle}{|\varphi(z^{j})|^{2}}\int_{0}^{|\varphi(z^{j})|^{2}}g(t){\rm d}t, \ \ \ \ \ $

(3.31) $ \begin{eqnarray} R^{(2)}[C_{\varphi}h_{j}](z^{j})& = &\frac{\langle R^{(2)}\varphi(z^{j}), \varphi(z^{j})\rangle}{|\varphi(z^{j})|^{2}}\int_{0}^{|\varphi(z^{j})|^{2}}g(t){\rm d}t \\ && + \frac{\langle R\varphi(z^{j}), \varphi(z^{j})\rangle^{2}}{|\varphi(z^{j})|^{4}}\bigg(2g(|\varphi(z^{j})|^{2})-\int_{0}^{|\varphi(z^{j})|^{2}}g(t) {\rm d}t\bigg). \end{eqnarray} $

通过引理2.1以及(3.31)式可得

(3.32) $ \begin{eqnarray} c\| (\psi C_{\varphi})h_{j}\| _{{\cal Z}_{\mu}} &\geq& -\| \psi\| _{{\cal Z}_{\mu}}\sup\limits_{w\in B}|h_{j}(w)|-\frac{3\mu(|z^{j}|) g(|\varphi(z^{j})|^{2})|\psi(z^{j})|\langle R\varphi(z^{j}), \varphi(z^{j})\rangle^{2}}{|\varphi(z^{j})|^{4}} \\ &&+ \frac{\mu(|z^{j}|)|\langle 2R\psi(z^{j})R\varphi(z^{j})+\psi(z^{j})R^{(2)}\varphi(z^{j}), \varphi(z^{j})\rangle|}{|\varphi(z^{j})|^{2}}\int_{0}^{|\varphi(z^{j})|^{2}}g(t){\rm d}t. \end{eqnarray} $

由(3.18)式和(3.32)式以及引理2.3–2.5可得(3.19)式成立.

(4)–(6)如果$ b\geq a > 1 $,那么在$ {\cal Z}_{\mu}(B) $上的加权复合算子问题恰好与权为$ \nu(r) = \mu(r)/(1-r) $的Bloch型空间$ {\cal B}_{\nu}(B) $上的加权复合算子问题相同.通过(2.2)式和文献[15]中的定理B,我们可以获得这些结果.

本定理证完.

注  定理3.1和定理3.2含有下列情形:

$ \mu(r) = (1-r)^{s}\log^{t}\frac{e}{1-r} \ \ \mbox{或} \ \ \mu(r) = (1-r)^{s}\log^{t}\frac{e}{1-r}\left(\log\log\frac{e^{2}}{1-r}\right)^{p} $

($ s > 0 $, $ t $和$ p $是实数)等.

对于上述注记中的$ \mu $,存在许多$ \psi $和$ \varphi $满足定理3.1的条件,例如, $ \psi(z) = z_{1} $和$ \varphi $是$ B $上的任何自同构就满足.至于紧性,当$ \| \varphi\| _{\infty} = 1 $时会复杂些.