设$B$表示$C^{n}$上的单位球, $\partial$B$表示单位球面, $dv$为标准体测度, 满足$v(B)=1$, $d\sigma$为标准面测度, 满足$\sigma(\partial$B$)=1$, $H(B)$代表$B$上的全纯函数类.

定义1.1 设$0 <p<\infty, \alpha>-1, $记

$ \mathcal{A}_\alpha^p=\{f\in H(B):\|f\|_{\mathcal{A}_\alpha^p}=(\int_B|f(z)|^p\text{d}v_\alpha(z))^{\frac{1}{p}}<\infty\} $

表示$B$上的加权Bergman空间$\mathcal{A}_\alpha^p$, 其中$\text{d}v_\alpha(z)=c_\alpha(1-|z|^2)^{\alpha}\text{d}v(z)$, $c_\alpha=\frac{\Gamma(n+\alpha+1)}{n!\Gamma(\alpha+1)}$.

当$1\leq p<\infty$时, $\mathcal{A}_\alpha^p$在范数$\|\cdot\|_{\mathcal{A}^p_\alpha}$下是一个Banach空间.当$0<p<1$时, $\mathcal{A}_\alpha^p$在准范$\|\cdot\|_{\mathcal{A}^p_\alpha}$下构成一个Frechet空间.特别地, $\mathcal{A}^p_0$就是Bergman空间$\mathcal{A}^p$.

定义1.2 设$[0, 1)$上的连续函数$\mu(r)>0$, 如果存在常数$0<a<b$, 使得

(ⅰ) $\frac{\mu(r)}{(1-r)^a}$在$[0, 1)$上单调递减且$\lim\limits_{r\rightarrow 1}\frac{\mu(r)}{(1-r)^a}=0, $

(ⅱ) $\frac{\mu(r)}{(1-r)^b}$在$[0, 1)$上单调递增且$\lim\limits_{r\rightarrow1}\frac{\mu(r)}{(1-r)^b}=+\infty, $则称$\mu$为$[0, 1)$上的正规函数.

设$\mu$为$[0, 1)$上的正规函数, $f\in H(B)$,

$ \|f\|_{\mu, 1}=\sup\limits_{z\in B}\mu(|z|)|\nabla f(z)|, \quad \|f\|_{\mu, 2}=\sup\limits_{z\in B}\mu(|z|)|Rf(z)|, $

这里$\nabla f$表示$f$的复梯度, $\nabla f(z)=(\frac{\partial f(z)}{\partial z_1}, \cdots, \frac{\partial f(z)}{\partial z_n}), Rf(z)=\sum\limits_{j=1}^nz_j\frac{\partial f(z)}{\partial z_j}.$由文献[1]知$\|f\|_{\mu, 1}$, $ \|f\|_{\mu, 2}$等价.

设$\mu$为$[0, 1)$上的正规函数, 称$f$属于$\beta_\mu$空间是指: $\|f\|_{\mu, 1}<\infty, $或者$\|f\|_{\mu, 2}<\infty$. $\beta_\mu$空间在范数

$ \|f\|_{\beta_\mu}=|f(0)|+\|f\|_{\mu, 1}, \quad \|f\|_{\beta_\mu}=|f(0)|+\|f\|_{\mu, 2} $

下是一个Banach空间.

由文献[2]中的引理2.1可知, 设$\mu$为$[0, 1)$上的正规函数, 称$f\in{\beta_\mu}$当且仅当

$ \|f\|_{\mu, 3}=\sup\limits_{u\in C^n-{0}, z\in B}\frac{\mu(|z|)|\langle\nabla f(z), \overline{u}\rangle|} {\{(1-|z|^2)|u|^2+|\langle z, u\rangle|^2\}^\frac{1}{2}}<+\infty. $

进一步有$\|f\|_{\beta_\mu} \approx |f(0)|+ \|f\|_{\mu, 3}.$

定义1.3 设$\mu$为$[0, 1)$上的正规函数, $B$上的全纯函数$f$如果满足

$ \|f\|_{\mathcal{Z}_\mu}=|f(0)|+\sup\limits_{z\in B}\mu(|z|)|R^{(2)}f(z)|<+\infty, $

则称$f$属于Zygmund型空间$\mathcal{Z}_\mu$.若$\lim\limits_{|z|\rightarrow1}\mu(|z|)|R^{(2)}f(z)|=0$, 则称$f$属于小Zygmund型空间$\mathcal{Z}_{\mu, 0}$ (见文献[3-5]).如果$\mu(r)=1-r^2$, Zygmund型空间$\mathcal{Z}_\mu$(小Zygmund型空间$\mathcal{Z}_{\mu, 0}$)就是典型的Zygmund空间$\mathcal{Z}$(小Zygmund空间$\mathcal{Z}_0$), 这里$R^{(2)}f(z)=R(Rf(z))$.

定义1.4 设$\varphi$是$B$上的全纯自映射, $g\in H(B)$且$g(0)=0$, 则由$\varphi$和$g$诱导的$H(B)$上的算子$P_\varphi^g$定义为

$ P_\varphi^g(f)(z)=\int_0^1f(\varphi(tz))g(tz)\frac{\text{d}t}{t}, z\in B. $

文献[3-11]研究了该类算子的性质.若用$Rg$代替$g$, 且$\varphi(z)=z, $算子$P_g$就是加权Cesàro算子, 它在文献[12]被引入并研究, 文献[13-16]对几个全纯函数空间上的加权Cesàro算子的有界性和紧性进行了讨论.关于Bergman空间上的复合算子和Cesàro算子的有界性和紧性问题的讨论得到了一些很好的结论, 如文献[1, 2, 16-18].

本文的主要工作就是在$C^{n}$中的单位球上来给出$P_\varphi^g$为$\mathcal{A}_\alpha^p$空间到$\mathcal{Z}_\mu$型空间上的有界算子和紧算子的充要条件.同时分别得到了单位圆盘$D$上和$\varphi(z)=z$时单位球上的相应结论.本文中用记号$c, c_1, c_2$来表示与变量$z, \omega $无关的正数, $c, c_1, c_2$可以与某些范数或有界量有关, 不同的地方可以表示不同的正常数.

引理2.1 ^[7] 设$f, g\in H(B)$, $g(0)=0$, 则$ R(P_\varphi^gf)(z)=f(\varphi(z))g(z). $

引理2.2 ^[2] 设$0<p<\infty, \alpha>-1, f\in {{\mathcal{A}}^p_\alpha}, \forall z\in B$, 则

(ⅰ) $|f(z)|\leq \frac{\|f\|_{\mathcal{A}_\alpha^p}}{(1-|z|^2)^{\frac{n+1+\alpha}{p}}}, $

(ⅱ) $ f\in\beta_{(1-r^2)^{\frac{n+1+p+\alpha}{p}}}$, 且$\|f\|_{\beta_{(1-r^2)^{\frac{n+1+p+\alpha}{p}}}}\leq c\|f\|_{\mathcal{A}_\alpha^p}.$

引理2.3 设$0<p<+\infty, \alpha>-1$, $\mu$为$[0, 1)$上的正规函数, $\varphi$是$B$上的全纯自映射, $g\in H(B), g(0)=0$, 则$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间上的紧算子的充要条件是对任意在${\mathcal{A}}^p_\alpha$中有界且在$B$的任一紧子集上一致趋于0的序列$\{f_j\}$, 有$\|P_{\varphi}^g(f_j)\|_{\mathcal{Z}_{\mu}}\rightarrow0$ ($j\rightarrow \infty$).

证 由引理2.2和Montel定理按定义可证.

引理2.4 ^[3]  $\mathcal{Z}_{\mu, 0}$中的闭子集$K$是紧子集的充要条件是$K$是有界集, 且满足

$ \lim\limits_{|z|\rightarrow 1}\sup\limits_{f\in K}\mu(|z|)|R^{(2)}f(z)|=0. $

定理3.1 设$0 <p<\infty, \alpha>-1, \mu$是$[0, 1)$上的正规函数, $\varphi$是$B$上的全纯自映射, 且$g\in H(B), g(0)=0$, 则$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的有界算子当且仅当

$ \begin{eqnarray} M=\sup\limits_{z\in B}\frac{\mu(|z|)|g(z)|} {(1-|\varphi(z)|^2)^{\frac{n+1+\alpha}{p}+1}}\cdot \{\frac{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2} {(1-|z|^2)|u|^2+|\langle z, u\rangle|^2}\}^{\frac{1}{2}}<+\infty, \end{eqnarray} $

(3.1)

$ \begin{eqnarray} N=\sup\limits_{z\in B}\frac{\mu(|z|)|Rg(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+1+\alpha}{p}}}<+\infty, \end{eqnarray} $

(3.2)

这里$J\varphi(z)$是$\varphi(z)$的Jacobi矩阵, 且

$ J\varphi(z)=(\frac{\partial\varphi_j(z)}{\partial z_k})_{1\leq j, k\leq n}, J\varphi(z)u=(\sum\limits_{k=1}^n\frac{\partial\varphi_1(z)}{\partial z_k}u_k, \cdots, \sum\limits_{k=1}^n\frac{\partial\varphi_n(z)}{\partial z_k}u_k)^T. $

证  充分性  $\forall f\in \mathcal{A}_\alpha^p$, 设(3.1) - (3.2)成立, 则由$P_\varphi^g(f)(0)=0$和引理2.1-2.2有

$ \begin{eqnarray*} \|P_\varphi^g(f)\|_{\mathcal{Z}_{\mu}}&=&\sup\limits_{z\in B}\mu(|z|)|R^{(2)}[P_{\varphi}^g(f)(z)]|\\ &\leq& c\sup\limits_{z\in B}\mu(|z|)\{|R(g(z))||f(\varphi(z))|+|g(z)||\nabla(f\circ(\varphi(z)))| \}\\ &\leq& c\sup\limits_{z\in B}\frac{\mu(|z|)|Rg(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+1+\alpha}{p}}}\cdot\|f\|_{{\mathcal{A}}^p_\alpha}+c\sup\limits_{z\in B}\frac{\mu(|z|)|g(z)|}{(1-|\varphi(z)|^2)^{\frac{n+1+p+\alpha}{p}}}\\ &&\times\{(\frac{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2} {(1-|z|^2)|u|^2+|\langle z, u\rangle|^2})^{\frac{1}{2}}\\ &&\cdot \frac{(1-|\varphi(z)|^2)^{\frac{n+1+p+\alpha}{p}}|\langle\nabla f(\varphi(z)), J\varphi(z)u\rangle|} {\sqrt{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2}}\}\\ &\leq& c(M+N)\|f\|_{{\mathcal{A}}_\alpha^p}\leq c\|f\|_{{\mathcal{A}}_\alpha^p}. \end{eqnarray*} $

所以$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_\mu$型空间的有界算子.

必要性 设$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_\mu$型空间的有界算子.取$f(z)=1\in {\mathcal{A}}^p_\alpha$和$f_l(z)=z_l\in{\mathcal{A}}^p_\alpha$, 则由${\mathcal{A}}^p_\alpha$和$\beta_\mu$的定义可知$g\in \beta_\mu$, $g\varphi_l\in \beta_\mu (l=1, 2, \cdots, n)$.

(1)  先证(3.1)式成立. $\forall \omega\in B, u\in{C^n-\{0\}}$.

(ⅰ) 当$|\varphi(\omega)|^2\leq \frac{2}{3}$时, 根据$g\in \beta_\mu, g\varphi_l\in\beta_\mu$, 则下列不等式成立

$ \begin{eqnarray*} M&=&\frac{\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \cdot\{\frac{(1-|\varphi(\omega)|^2)|J\varphi(\omega)u|^2+|\langle\varphi(\omega), J\varphi(\omega)u\rangle|^2} {(1-|\omega|^2)|u|^2+|\langle \omega, u\rangle|^2}\}^{\frac{1}{2}}\\ &\leq& c\mu(|\omega|)|g(\omega)|\{\frac{(1-|\varphi(\omega)|^2)|J\varphi(\omega)u|^2+|\langle\varphi(\omega), J\varphi(\omega)u\rangle|^2} {(1-|\omega|^2)|u|^2+|\langle \omega, u\rangle|^2}\}^{\frac{1}{2}}\\ \end{eqnarray*} $

$ \begin{eqnarray} &=&c\mu(|\omega|)|g(\omega)|\frac{|J\varphi(\omega)u|} {\sqrt{(1-|\omega|^2)|u|^2+|\langle\omega, u\rangle|^2}}\nonumber\\ &\leq& c\mu(|\omega|)|g(\omega)| \sum\limits_{l=1}^n\frac{|\nabla\varphi_l(\omega)u|}{\sqrt{(1-|\omega|^2)|u|^2+|\langle\omega, u\rangle|^2}} \leq c\mu(|\omega|) \sum\limits_{l=1}^n|g(\omega)\cdot\nabla\varphi_l(\omega)|\nonumber\\ &\leq& c\mu(|\omega|) \sum\limits_{l=1}^n\{|\nabla(g\varphi_l)(\omega)|+|\nabla g(\omega)\cdot\varphi_l(\omega)|\}\leq c(\sum\limits_{l=1}^n\|g\varphi_l\|_{\beta_\mu}+\|g\|_{\beta_\mu}). \end{eqnarray} $

(3.3)

(ⅱ) 当$|\varphi(\omega)|^2> \frac{2}{3}$时, 假设$\varphi(\omega)=r_\omega e_1, $这里$r_\omega=|\varphi(\omega)|, e_1=(1, 0, \cdots, 0)$.

若$\sqrt{(1-r_\omega^2)(|\xi_2|^2+\cdots+|\xi_n|^2)}\leq |\xi_1|, $这里$(\xi_1, \cdots, \xi_n)^T=J\varphi(\omega)u, $取

$ f_\omega(z)=\frac{z_1-r_\omega}{1-r_\omega z_1}\{\frac{1-r^2_\omega}{(1-r_\omega z_1)^2}\}^{\frac{n+1+\alpha}{p}}, $

则$f_\omega(\varphi(\omega))=0$, 且由文献[19]中的定理1.12有

$ \begin{eqnarray*} \|f_\omega\|_{\mathcal{A}_\alpha^p}&=&\{c_\alpha\int\limits_{B}\frac{|z_1-r_\omega|^p}{|1-r_\omega z_1|^p}(\frac{1-r^2_\omega}{|1-r_\omega z_1|^2})^{n+1+\alpha}(1-|z|^2)^\alpha\text{d}v(z)\}^{\frac{1}{p}}\\ &\leq& \{c_1c_\alpha(1-r^2_\omega)^{n+1+p+\alpha}\int\limits_{B}\frac{(1-|z|^2)^\alpha}{|1-r_\omega z_1|^{p+2(n+1+\alpha)}}\text{d}v(z)\}^{\frac{1}{p}}\\ &\leq& \{c_1c_\alpha(1-r^2_\omega)^{n+1+p+\alpha}\frac{c_2}{(1-r_\omega^2)^{n+1+p+\alpha}}\}^{\frac{1}{p}}\leq c, \end{eqnarray*} $

从而$\|f_\omega\|_{\mathcal{A}_\alpha^p}\leq c, f_\omega\in\mathcal{A}_\alpha^p$.因为

$ \begin{eqnarray*} \nabla f_\omega(z)&=&(\frac{1-r_\omega z_1+(z_1-r_\omega)r_\omega}{(1-r_\omega z_1)^2}\{\frac{1-r^2_\omega}{(1-r_\omega z_1)^2}\}^{\frac{n+1+\alpha}{p}}\\ &&+\frac{z_1-r_\omega}{1-r_\omega z_1}\frac{2(n+1+\alpha)}{p}\frac{(1-r^2_\omega)^{\frac{n+1+\alpha}{p}}r_\omega}{(1-r_\omega z_1)^{\frac{2(n+1+\alpha)}{p}+1}}, 0, \cdots, 0), \end{eqnarray*} $

所以$ \nabla f_\omega(\varphi(\omega))=(\frac{1}{(1-r_\omega^2)^{\frac{n+1+\alpha}{p}+1}}, 0, \cdots, 0), $从而$|\langle\nabla f_\omega(\varphi(\omega)), J\varphi(\omega)u\rangle|=\frac{|\xi_1|}{(1-r_\omega^2)^{\frac{n+1+\alpha}{p}+1}}$.由$P_\varphi^g$是有界算子有

$ \begin{eqnarray} c\cdot\|P_\varphi^g\|&\geq&\|P_\varphi^g(f_\omega)\|_{\mathcal{Z}_\mu} \geq \mu(|\omega|)|f_\omega(\varphi(\omega))Rg(\omega)+g(\omega)\nabla(f_\omega\circ\varphi(\omega))|\nonumber\\ &\approx&\frac{\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \frac{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}|\langle\nabla f_\omega(\varphi(\omega)), J\varphi(\omega)u\rangle|}{\sqrt{(1-|\omega|^2)|u|^2+|\langle\omega, u\rangle|^2}}\nonumber\\ &=&\frac{c\mu(|\omega|)|g(\omega)||\xi_1|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1} \sqrt{(1-|\omega|^2)|u|^2+|\langle\omega, u\rangle|^2}}. \end{eqnarray} $

(3.4)

所以由(3.4)式有

$ \begin{eqnarray} M&=&\frac{\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \cdot\{\frac{(1-|\varphi(\omega)|^2)|J\varphi(\omega)u|^2+|\langle\varphi(\omega), J\varphi(\omega)u\rangle|^2} {(1-|\omega|^2)|u|^2+|\langle \omega, u\rangle|^2}\}^{\frac{1}{2}}\nonumber\\ &\leq&\frac{c\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \cdot\{\frac{(1-|\varphi(\omega)|^2)(|\xi_2|^2+\cdots+|\xi_n|^2)+|\xi_1|^2} {(1-|\omega|^2)|u|^2+|\langle \omega, u\rangle|^2}\}^{\frac{1}{2}}\nonumber\\ &\leq&\frac{\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \cdot\{\frac{\sqrt{2}|\xi_1|} {(1-|\omega|^2)|u|^2+|\langle \omega, u\rangle|^2}\}^{\frac{1}{2}}\leq c\|P_\varphi^g\|. \end{eqnarray} $

(3.5)

若$\sqrt{(1-r_\omega^2)(|\xi_2|^2+\cdots+|\xi_n|^2)}>|\xi_1|, $设$\theta_j=$arg$\xi_j, $且当$\xi_j\neq0$时, $a_j=e^{-i\theta_j}$, 当$\xi_j=0$时, $a_j=0, j=2, \cdots, n$.取

$ f_\omega(z)=(a_2z_2+\cdots+a_nz_n)\{\frac{1-r_\omega^2}{(1-r_\omega z_1)^2}\}^{\frac{n+1+\alpha}{p}}\cdot \frac{1}{1-r_\omega z_1}, $

则$f_\omega(\varphi(\omega))=0$, 且由文献[20]中的命题1.4.10有

$ \begin{eqnarray*} \|f_\omega\|_{\mathcal{A}_\alpha^p}&=&\{c_\alpha\int\limits_{B}|a_2z_2+\cdots+a_nz_n|^p(\frac{1-r_\omega^2} {|1-r_\omega z_1|^2})^{n+1+\alpha}\frac{(1-|z|^2)^\alpha}{|1-r_\omega z_1|^p}\text{d}v(z)\}^{\frac{1}{p}}\\ &\leq& \{c_1c_\alpha(1-r^2_\omega)^{n+1+\alpha}\int\limits_{B}\frac{(|z_2|^2+\cdots+|z_n|^2)^p(1-|z|^2)^\alpha}{|1-r_\omega z_1|^{p+2(n+1+\alpha)}}\text{d}v(z)\}^{\frac{1}{p}}\\ &\leq&\{c_1c_\alpha(1-r^2_\omega)^{n+1+\alpha}\int\limits_{B}\frac{[(n-1)(1-|z_1|^2)]^{\frac{p}{2}}(1-|z|^2)^\alpha}{|1-r_\omega z_1|^{p+2(n+1+\alpha)}}\text{d}v(z)\}^{\frac{1}{p}}\leq\frac{c}{\sqrt{1-r_\omega^2}}, \end{eqnarray*} $

从而$f_\omega\in\mathcal{A}_\alpha^p$.由$P_\omega^g$是有界算子, 类似(3.4)式的证明有

$ \begin{eqnarray} \frac{\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \cdot \frac{|\xi_2|+\cdots+|\xi_n|}{\sqrt{(1-|\omega|^2)|u|^2+|\langle\omega, u\rangle|^2}}\leq\frac{c\cdot\|P_\varphi^g\|} {\sqrt{1-r_\omega^2}}. \end{eqnarray} $

(3.6)

所以类似(3.5)式的证明, 结合(3.6)式有

$ \begin{eqnarray} M&\leq&\frac{\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \cdot\{\frac{2(1-|\varphi(\omega)|^2)(|\xi_2|^2+\cdots+|\xi_n|^2)} {(1-|\omega|^2)|u|^2+|\langle \omega, u\rangle|^2}\}^{\frac{1}{2}}\nonumber\\ &\leq&\frac{\sqrt{2}\mu(|\omega|)|g(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}} \cdot\frac{\sqrt{1-|\varphi(\omega)|^2}(|\xi_2|+\cdots+|\xi_n|)} {\sqrt{(1-|\omega|^2)|u|^2+|\langle \omega, u\rangle|^2}}\leq c\|P_\varphi^g\|. \end{eqnarray} $

(3.7)

一般地, 若$\varphi(\omega)\neq|\varphi(\omega)|e_1, $则可找到酉变换$U_\omega$, 使得$\varphi(\omega)=\rho_\omega e_1U_\omega, $这里$\rho_\omega=|\varphi(\omega)|>\sqrt{\frac{2}{3}}$, 取$g_\omega=f_\omega\circ U_\omega^{-1}$, 经简单计算知$\|g_\omega\|_{\mathcal{A}_\alpha^p}=\|f_\omega\|_{\mathcal{A}_\alpha^p}$, 所以结合(3.5)式及(3.7)式可知(3.1)式成立.

下面证明(3.2)式成立.取

$ h_\omega(z)=(\frac{1-|\varphi(\omega)|^2}{(1-\langle z, \varphi(\omega)\rangle)^2})^{\frac{n+1+\alpha}{p}}, $

则$h_\omega(\varphi(\omega))=\frac{1}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}}}$, 且由文献[20]中的命题1.4.10有

$ \|h_\omega\|_{\mathcal{A}_\alpha^p}=\{c_\alpha\int\limits_{B}(\frac{1-|\varphi(\omega)|^2}{|1-\langle z, \varphi(\omega)\rangle|^2})^{n+1+\alpha}(1-|z|^2)^\alpha\text{d}v(z)\}^{\frac{1}{p}}\\ =\{c_\alpha(1-|\varphi(\omega)|^2)^{n+1+\alpha}\int\limits_{B}\frac{(1-|z|^2)^\alpha} {|1-\langle z, \varphi(\omega)\rangle|^{2(n+1+\alpha)}} \text{d}v(z)\}^{\frac{1}{p}}\leq c, $

从而$h_\omega\in\mathcal{A}_\alpha^p$.由$P_\varphi^g$是有界算子并且$|\langle\nabla h_\omega(\varphi(\omega)), J\varphi(\omega)u\rangle|=\frac{2(n+1+\alpha)|\langle\varphi(\omega), J\varphi(\omega)u\rangle|}{p(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}}$, 结合(3.1)式有

$ \begin{eqnarray*} c\cdot\|P_\varphi^g\|&\geq&\|P_\varphi^g(h_\omega)\|_{\mathcal{Z}_\mu} \geq \mu(|\omega|)|h_\omega(\varphi(\omega))Rg(\omega)+g(\omega)\nabla(h_\omega\circ\varphi(\omega))|\\ &\geq&\frac{\mu(|\omega|)|Rg(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}}}- c_1\mu(|\omega|)|g(\omega)|\frac{|\langle\nabla h_\omega(\varphi(\omega)), J\varphi(\omega)u\rangle|}{\sqrt{(1-|\omega|^2)|u|^2+|\langle\omega, u\rangle|^2}}\\ &=&\frac{\mu(|\omega|)|Rg(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}}}- \frac{c_12(n+1+\alpha)\mu(|\omega|)|g(\omega)||\langle\varphi(\omega), J\varphi(\omega)u\rangle|} {p(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}+1}\sqrt{(1-|\omega|^2)|u|^2+|\langle\omega, u\rangle|^2}}\\ &\geq&\frac{\mu(|\omega|)|Rg(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}}} -Mc_1\frac{2(n+1+\alpha)}{p}, \end{eqnarray*} $

所以

$ \frac{\mu(|\omega|)|Rg(\omega)|}{(1-|\varphi(\omega)|^2)^{\frac{n+1+\alpha}{p}}} \leq c_2M+c\cdot\|P_\varphi^g\|\leq c. $

所以(3.2)式成立.

综上所述, 定理3.1得证.

当$\varphi(z)=z$时, 易知下列结论成立.

推论3.2 设$0<p<\infty, \alpha>-1, \mu$是$[0, 1)$上的正规函数, 且$g\in H(B), g(0)=0$, 则$P_g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的有界算子当且仅当

$ \sup\limits_{z\in B}\frac{\mu(|z|)|g(z)|} {(1-|z|^2)^{\frac{n+1+\alpha}{p}+1}}<+\infty, \quad \sup\limits_{z\in B}\frac{\mu(|z|)|Rg(z)|}{(1-|z|^{2})^{\frac{n+1+\alpha}{p}}}<+\infty. $

而对于单位圆盘则有以下结论成立.

推论3.3 设$0<p<\infty, \alpha>-1, n=1, \mu$是$[0, 1)$上的正规函数, $\varphi$是单位圆盘$D$上的全纯自映射, 且$g\in H(D), g(0)=0$, 则$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的有界算子当且仅当

$ \sup\limits_{z\in D}\frac{\mu(|z|)|g(z)||\varphi'(z)|} {(1-|\varphi(z)|^2)^{\frac{2+\alpha}{p}+1}}<+\infty , \quad \sup\limits_{z\in D}\frac{\mu(|z|)|g'(z)|}{(1-|\varphi(z)|^{2})^{\frac{2+\alpha}{p}}}<+\infty . $

定理3.4 设$0<p<\infty, \alpha>-1, \mu$是$[0, 1)$上的正规函数, $\varphi$是$B$上的全纯自映射, 且$g\in H(B), g(0)=0$, 则$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的紧算子当且仅当$g\in\beta_\mu, g\varphi_l\in\beta_\mu, $且

$ \begin{equation} \lim\limits_{|\varphi(z)|\rightarrow1}\sup\limits_{u\in {C^n-\{0\}}, z\in B}\frac{\mu(|z|)|g(z)|} {(1-|\varphi(z)|^2)^{\frac{n+1+\alpha}{p}+1}}\cdot \{\frac{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2} {(1-|z|^2)|u|^2+|\langle z, u\rangle|^2}\}^{\frac{1}{2}}=0, \end{equation} $

(3.8)

$ \begin{eqnarray} \lim\limits_{|\varphi(z)|\rightarrow1}\sup\limits_{z\in B}\frac{\mu(|z|)|Rg(z)|} {(1-|\varphi(z)|^{2})^{\frac{n+1+\alpha}{p}}}=0, \end{eqnarray} $

(3.9)

证 充分性 假设(3.8)和(3.9)式成立, 则对任意$\varepsilon>0$, 存在$0<\delta<1$, 当$|\varphi(z)|^2>1-\delta$时,

$ \begin{equation} \sup\limits_{u\in {C^n-\{0\}}}\frac{\mu(|z|)|g(z)|} {(1-|\varphi(z)|^2)^{\frac{n+1+\alpha}{p}+1}}\cdot \{\frac{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2} {(1-|z|^2)|u|^2+|\langle z, u\rangle|^2}\}^{\frac{1}{2}}<\varepsilon, \end{equation} $

(3.10)

$ \begin{eqnarray} \frac{\mu(|z|)|Rg(z)|} {(1-|\varphi(z)|^{2})^{\frac{n+1+\alpha}{p}}}<\varepsilon. \end{eqnarray} $

(3.11)

设$\{f_j\}$是在$B$的任一紧子集上一致收敛于0并满足$\|f_j\|_{\mathcal{A}_\alpha^p}\leq1$的全纯函数列, 则$\{f_j\}$和$\{\nabla f_j\}$在$E=\{\omega:|\omega|^2\leq1-\delta\}$上一致收敛于0.

若$|\varphi(z)|^2>1-\delta$, 由(3.10)-(3.11)式及引理2.2, 有

$ \begin{eqnarray} &&\sup\limits_{u\in {C^n-\{0\}}}\mu(|z|)|R^{(2)}[(P_\varphi^g)(f_j)](z)|~\nonumber\\ &\leq&\sup\limits_{u\in {C^n-\{0\}}}\frac{\mu(|z|)|g(z)||\langle\nabla f_j(\varphi(z)), J_{\varphi(z)}u\rangle|}{\sqrt{(1-|z|^2)|u|^2+|\langle z, u\rangle|^2}}+c\mu(|z|)|Rg(z)||f_j(\varphi(z))|\nonumber\\ &\leq&\sup\limits_{u\in {C^n-\{0\}}}\frac{c\mu(|z|)|g(z)|} {(1-|\varphi(z)|^2)^{\frac{n+1+\alpha}{p}+1}}\cdot \{\frac{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2} {(1-|z|^2)|u|^2+|\langle z, u\rangle|^2}\}^{\frac{1}{2}}\nonumber\\ &&\times\frac{(1-|\varphi(z)|^2)^{\frac{n+1+\alpha}{p}+1}\cdot|\langle\nabla f_j(\varphi(z)), J_{\varphi(z)}u\rangle|}{\sqrt{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2}} +\frac{c\mu(|z|)|Rg(z)|\|f_j\|_{\mathcal{A}_\alpha^p}}{(1-|\varphi(z)|^2)^{\frac{n+1+\alpha}{p}}}\nonumber\\ &\leq& c\varepsilon \|f_j\|_{\beta_{(1-r^2)^{(n+1+p+\alpha)/p}}}+c\varepsilon\|f_j\|_{\mathcal{A}_\alpha^p}\leq c\varepsilon \quad (j\rightarrow\infty). \end{eqnarray} $

(3.12)

若$|\varphi(z)|^2\leq1-\delta$, 因为$g\in\beta_\mu, g\varphi_l\in\beta_\mu, $所以由$\{f_j\}$和$\{\nabla f_j\}$在$E=\{\omega:|\omega|^2\leq1-\delta\}$上一致收敛于0, 有

$ \begin{eqnarray} &&\sup\limits_{u\in {C^n-\{0\}}}\mu(|z|)|R^{(2)}[(P_\varphi^g)(f_j)](z)|\nonumber\\ &\leq&\sup\limits_{u\in {C^n-\{0\}}}\frac{\mu(|z|)|g(z)||\nabla f_j(\varphi(z))|(|\nabla\varphi_1(z)u|+\cdots+|\nabla\varphi_n(z)u|)}{\sqrt{(1-|z|^2)|u|^2+|\langle z, u\rangle|^2}}\nonumber\\ &&+\sup\limits_{u\in {C^n-\{0\}}}c\mu(|z|)|Rg(z)||f_j(\varphi(z))|\nonumber\\ &\leq& c\sum\limits_{l=1}^n(\|g\varphi_l\|_{\beta_\mu}+\|g\|_{\beta_\mu})|\nabla f_j(\varphi(z))|+c\|g\|_{\beta_\mu}|f_j(\varphi(z))|\rightarrow0 \quad (j\rightarrow\infty). \end{eqnarray} $

(3.13)

注意到$\{g(0)f_j(\varphi(0))\}$一致收敛于0, 结合(3.12)式和(3.13)式, 有$\lim\limits_{j\rightarrow\infty}\sup\limits_{z\in B}\|P_\varphi^gf_j\|_{\mathcal{Z}_\mu}=0$, 从而根据引理2.3知$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的紧算子.

必要性 设$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的紧算子, 由有界性的证明知$g\in\beta_\mu, g\varphi_l\in\beta_\mu$.

假设(3.8)式不成立, 则存在$\{z^j\}\subset B, \{u^j\}\subset C^n-\{0\}$和常数$\varepsilon_0>0$, 使得$r_j=|\varphi(z^j)|\rightarrow1 ~(j\rightarrow\infty)$, 有

$ \begin{equation} \frac{\mu(|z^j|)|g(z^j)|} {(1-|\varphi(z^j)|^2)^{\frac{n+1+\alpha}{p}+1}}\cdot \{\frac{(1-|\varphi(z^j)|^2)|J\varphi(z^j)u^j|^2+|\langle\varphi(z^j), J\varphi(z^j)u^j\rangle|^2} {(1-|z^j|^2)|u^j|^2+|\langle z^j, u^j\rangle|^2}\}^{\frac{1}{2}}\geq \varepsilon_0. \end{equation} $

(3.14)

(ⅰ) 假设$\varphi(z^j)=r_j e_1, (j=1, 2, \cdots, n), $若$\sqrt{(1-r_j^2)(|\omega_2^j|^2+\cdots+|\omega_n^j|^2)}\leq |\omega_1^j|, $这里$(\omega_1^j, \cdots, \omega_n^j)^T=J\varphi(z^j)u^j, $取

$ f_j(z)=\frac{z_1-r_j}{1-r_j z_1}\{\frac{1-r^2_j}{(1-r_j z_1)^2}\}^{\frac{n+1+\alpha}{p}}, $

则$f_j(\varphi(z^j))=0$.类似定理3.1中的证明易知$\|f_j\|_{\mathcal{A}_\alpha^p}\leq c$.

下面证明$\{f_j\}$在$B$的任一紧子集$E\subseteq\{z:|z|\leq r\}, 0<r<1$上一致趋于0.因为

$ \sup\limits_{z\in E}|f_j(z)|=\sup\limits_{z\in E}|\frac{z_1-r_j}{1-r_j z_1}|\{\frac{1-r^2_j}{|1-r_j z_1|^2}\}^{\frac{n+1+\alpha}{p}}\leq\{\frac{1-r^2_j}{(1-r)^2}\}^{\frac{n+1+\alpha}{p}}\rightarrow0 \quad (j\rightarrow\infty). $

所以$\{f_j\}$在$B$的任一紧子集上一致趋于0.但由(3.14)式, 类似定理3.1中的证明有

$ \begin{eqnarray} \|P_\varphi^g(f_j)\|_{\mathcal{Z}_\mu}&\geq&\frac{c\mu(|z^j|)|g(z^j)||\langle\nabla f_j(\varphi(z^j)), J\varphi(z^j)u^j\rangle|}{\sqrt{(1-|z^j|^2)|u^j|^2+|\langle z^j, u^j\rangle|^2}}-\mu(|z^j|)|Rg(z^j)||f_j(\varphi(z^j))|\nonumber\\ &=&\frac{c\mu(|z^j|)|g(z^j)|} {(1-|\varphi(z^j)|^2)^{\frac{n+1+\alpha}{p}+1}}\cdot \{\frac{(1-|\varphi(z^j)|^2)|J\varphi(z^j)u^j|^2+|\langle\varphi(z^j), J\varphi(z^j)u^j\rangle|^2} {(1-|z^j|^2)|u^j|^2+|\langle z^j, u^j\rangle|^2}\}^{\frac{1}{2}}\nonumber\\ &&\times\frac{|\omega_1^j|}{\sqrt{(1-|\varphi(z^j)|^2)(|\omega_2^j|^2+\cdot+|\omega_n^j|^2)+|\omega_1^j|^2}} \geq \frac{c\varepsilon_0}{\sqrt{2}}. \end{eqnarray} $

(3.15)

根据引理2.3, 这与$P_{\varphi}^g$是紧算子矛盾, 所以(3.8)式成立.

若$\sqrt{(1-r_j^2)(|\omega^j_2|^2+\cdots+|\omega^j_n|^2)}>|\omega^j_1|, $设$\theta_k^j=$arg$\omega_k^j, k=2, \cdots, n$.取

$ f_j(z)=(e^{-i\theta_2^j}z_2+\cdots+e^{-i\theta_n^j}z_n)\{\frac{1-r_j^2}{(1-r_j z_1)^2}\}^{\frac{n+1+\alpha}{p}}\cdot \frac{\sqrt{1-r_j^2}}{1-r_j z_1}, $

则$f_j(\varphi(z^j))=0$.经计算易知$f_j\in\mathcal{A}_\alpha^p$, 且$\{f_j\}$在$B$的任一紧子集$E\subseteq\{z:|z|\leq r\}, 0 <r<1$上一致趋于0.但由(3.14)式有

$ \begin{eqnarray} \|P_\varphi^g(f_j)\|_{\mathcal{Z}_\mu}&\geq&\frac{c\mu(|z^j|)|g(z^j)||\langle\nabla f_j(\varphi(z^j)), J\varphi(z^j)u^j\rangle|}{\sqrt{(1-|z^j|^2)|u^j|^2+|\langle z^j, u^j\rangle|^2}}-\mu(|z^j|)|Rg(z^j)||f_j(\varphi(z^j))|\nonumber\\ &=&\frac{c\mu(|z^j|)|g(z^j)|} {(1-|\varphi(z^j)|^2)^{\frac{n+1+\alpha}{p}+1}}\cdot \{\frac{(1-|\varphi(z^j)|^2)|J\varphi(z^j)u^j|^2+|\langle\varphi(z^j), J\varphi(z^j)u^j\rangle|^2} {(1-|z^j|^2)|u^j|^2+|\langle z^j, u^j\rangle|^2}\}^{\frac{1}{2}}\nonumber\\ &&\times\frac{(|\omega_2^j|+\cdots+|\omega_n^j|)\sqrt{1-r^2_j}}{\sqrt{(1-|\varphi(z^j)|^2)(|\omega_2^j|^2+\cdot+|\omega_n^j|^2)+|\omega_1^j|^2}} \geq \frac{c\varepsilon_0}{\sqrt{2}}. \end{eqnarray} $

(3.16)

根据引理2.3, 这与$P_{\varphi}^g$是紧算子矛盾, 所以此时(3.8)式成立.

(ⅱ) 一般地, 若$\varphi(z^j)\neq|\varphi(z^j)|e_1, $则可找到酉变换$U_j$, 使得$\varphi(z^j)=\rho_j e_1U_j, $这里$\rho_j=|\varphi(z^j)|$, 取$g_j=f_j\circ U_j^{-1}, j\in{\{1, 2, \cdots, n\}}$, 经简单计算知$\|g_\omega\|_{\mathcal{A}_\alpha^p}=\|f_\omega\|_{\mathcal{A}_\alpha^p}$, 所以结合(3.15)式及(3.16)式可知(3.8)式成立.

下面证明(3.9)式成立.假设(3.9)式不成立, 则存在$\{z^j\}\subset B$和常数$\varepsilon_0>0$, 使得$|\varphi(z^j)|\rightarrow1~(j\rightarrow\infty)$, 有

$ \begin{eqnarray} \frac{\mu(|z^j|)|Rg(z^j)|} {(1-|\varphi(z^j)|^2)^{\frac{n+1+\alpha}{p}}}\geq \varepsilon_0. \end{eqnarray} $

(3.17)

取

$ h_j(z)=(\frac{1-|\varphi(z^j)|^2}{(1-\langle z, \varphi(z^j)\rangle)^2})^{\frac{n+1+\alpha}{p}}, $

则$ h_j(\varphi(z^j))=\frac{1}{(1-|\varphi(z^j)|^2)^{\frac{n+1+\alpha}{p}}}$, 且经简单计算易知$h_j\in\mathcal{A}_\alpha^p$, $\{h_j\}$在$B$的任一紧子集$E\subseteq\{z:|z|\leq r\}, 0<r<1$上一致趋于0.所以

$ \begin{eqnarray} &&\|P_\varphi^g(h_j)\|_{\mathcal{Z}_\mu}\nonumber\\ &\geq&\mu(|z^j|)|Rg(z^j)||h_j(\varphi(z^j))| -\sup\limits_{u\in C^n-\{0\}}\frac{c\mu(|z^j|)|g(z^j)||\langle\nabla h_j(\varphi(z^j)), J\varphi(z^j)u\rangle|}{\sqrt{(1-|z^j|^2)|u|^2+|\langle z^j, u\rangle|^2}}\\ &=&\frac{\mu(|z^j|)|Rg(z^j)|}{(1-|\varphi(z^j)|^2)^{\frac{n+1+\alpha}{p}}} -\sup\limits_{u\in C^n-\{0\}}\frac{2(n+1+\alpha)\mu(|z^j|)|g(z^j)||\langle\varphi(z^j), J\varphi(z^j)u\rangle|}{p(1-|\varphi(z^j)|^2)^{\frac{n+1+\alpha}{p}+1}\sqrt{(1-|z^j|^2)|u|^2+|\langle z^j, u\rangle|^2}}\nonumber. \end{eqnarray} $

(3.18)

由(3.17)-(3.18)及(3.8)式有$\lim\limits_{j\rightarrow\infty}\|(P_\varphi^g)(h_j)\|_{\mathcal{Z}_\mu}\neq0$, 根据引理2.3, 这与$P_{\varphi}^g$是紧算子矛盾, 所以(3.9)式成立.综上所述, 定理3.4得证.

当$\varphi(z)=z$时, 易知下列结论成立.

推论3.5 设$0<p<\infty, \alpha>-1, \mu$是$[0, 1)$上的正规函数, 且$g\in H(B), g(0)=0$, 则$P_g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的紧算子当且仅当$g\in\beta_\mu, g\varphi_l\in\beta_\mu, $且

$ \lim\limits_{|z|\rightarrow1}\sup\limits_{z\in B}\frac{\mu(|z|)|g(z)|} {(1-|z|^2)^{\frac{n+1+\alpha}{p}+1}}=0, \quad \lim\limits_{|z|\rightarrow1}\sup\limits_{z\in B}\frac{\mu(|z|)|Rg(z)|} {(1-|z|^{2})^{\frac{n+1+\alpha}{p}}}=0. $

而对于单位圆盘则有以下结论成立.

推论3.6 设$0<p<\infty, \alpha>-1, n=1, \mu$是$[0, 1)$上的正规函数, $\varphi$是单位圆盘$D$上的全纯自映射, 且$g\in H(D), g(0)=0$, 则$P_{\varphi}^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的紧算子当且仅当$g\in\beta_\mu$且

$ \lim\limits_{|\varphi(z)|\rightarrow1}\frac{\mu(|z|)|g(z)||\varphi'(z)|} {(1-|\varphi(z)|^2)^{\frac{2+\alpha}{p}+1}}=0, \quad \lim\limits_{|\varphi(z)|\rightarrow1}\frac{\mu(|z|)|g'(z)|}{(1-|\varphi(z)|^{2})^{\frac{2+\alpha}{p}}}=0. $

定理3.7 设$0<p<\infty, \alpha>-1, \mu$是$[0, 1)$上的正规函数, $\varphi$是$B$上的全纯自映射且$g\in H(B), g(0)=0$, 则下列条件等价

(1) $P_\varphi^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu, 0}$型空间的紧算子;

(2) $P_\varphi^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu, 0}$型空间的有界算子;

(3) $P_\varphi^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu}$型空间的紧算子.

证 因为紧算子是有界算子, 所以$(1)\Rightarrow(2)$成立, 由${\mathcal{Z}}_{\mu, 0}$空间的定义及定理3.4知$(2)\Rightarrow(3)$也成立.再证$(3)\Rightarrow(1)$, 先证$(3)\Rightarrow(2)$.由定理3.4, 对任意的$f\in\mathcal{A}_\alpha^p$, 类似于定理3.1中充分性的证明有

$ \begin{eqnarray} \|P_\varphi^g(f)\|_{\mathcal{Z}_{\mu}}&=&\sup\limits_{z\in B}\mu(|z|)|R^{(2)}[P_{\varphi}^g(f)(z)]|\nonumber\\ &\leq& c\{\sup\limits_{z\in B}\frac{\mu(|z|)|Rg(z)|}{(1-|\varphi(z)|^{2})^{\frac{n+1+\alpha}{p}}}+c\sup\limits_{z\in B}\frac{\mu(|z|)|g(z)|}{(1-|\varphi(z)|^2)^{\frac{n+1+p+\alpha}{p}}}\\ &&\times(\frac{(1-|\varphi(z)|^2)|J\varphi(z)u|^2+|\langle\varphi(z), J\varphi(z)u\rangle|^2} {(1-|z|^2)|u|^2+|\langle z, u\rangle|^2})^{\frac{1}{2}}\}\cdot\|f\|_{{\mathcal{A}}^p_\alpha}\rightarrow0 \quad (|\varphi(z)|\rightarrow1).\nonumber \end{eqnarray} $

(3.19)

于是$P_\varphi^g\in\mathcal{Z}_{\mu, 0}$, 又因为$P_\varphi^g$是${\mathcal{A}}^p_\alpha$空间到 ${\mathcal{Z}}_{\mu}$型空间的紧算子, 所以 $P_\varphi^g$是 ${\mathcal{A}}^p_\alpha$空间到 ${\mathcal{Z}}_{\mu}$型空间的有界算子, 从而$P_\varphi^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu, 0}$空间的有界算子.

再证$(3)\Rightarrow(1)$.若条件(3)成立, 则条件(2)成立.集合$P_\varphi^g\{f\in\mathcal{A}_\alpha^p:\|f\|_{\mathcal{A}_\alpha^p}\leq1\}$在${\mathcal{Z}}_{\mu, 0}$空间中是有界的, 由定理3.4并结合(3.19)式有

$ \lim\limits_{|\varphi(z)|\rightarrow1}\sup\limits_{\|f\|_{\mathcal{A}_\alpha^p}\leq1} \mu(|z|)|R^{(2)}(P_\varphi^g)(f)|=0, $

所以由引理2.4知$P_\varphi^g$是${\mathcal{A}}^p_\alpha$空间到${\mathcal{Z}}_{\mu, 0}$空间的紧算子.定理3.7得证.