设$D$为复平面$C$上的开单位圆盘,$H(D)$表示$D$ 上所有全纯函数的集合.

对于$0<\alpha<\infty$, $${B^\alpha}=\Big\{f\in H(D):\| f\|_{B^{\alpha}}=|f(0)|+\sup_{z\in D}(1-|z|^{2})^{\alpha}|f'(z)|<\infty\Big\}$$ 称为$\alpha$-Bloch 空间; $${B_{0}^\alpha}=\Big\{f\in H(D):\lim_{|z|\rightarrow1}\sup_{z\in D}(1-|z|^{2})^{\alpha}|f'(z)|=0\Big\}$$ 称为小$\alpha$-Bloch 空间.

当$\alpha$={1}时，$B^{\alpha}$、$B_{0}^{\alpha}$ 空间即为经典的Bloch 空间和小Bloch 空间. $${Z^\alpha}=\Big\{f\in H(D):\| f\|_{Z^{\alpha}}=|f(0)|+\sup_{z\in D}(1-|z|^{2})^{\alpha}|f''(z)|<\infty\Big\}$$ 称为$\alpha$-Zygmund空间.

在范数$\|f\|_{Z^{\alpha}}=|f(0)|+\sup\limits_{z\in D}(1-|z|^{2})^{\alpha}|f''(z)|$下, $\alpha$-Zygmund 空间成为Banach空间. 若函数$f\in H(D)$ 满足关系式 $\lim\limits_{|Z|\rightarrow1}(1-|z|^{2})^{\alpha}|f''(z)|=0$,则称属于小$\alpha$-Zygmund空间, 记为$f\in Z_{0}^{\alpha}$. 显然$Z_{0}^{\alpha}$ 是 $Z^{\alpha}$ 的一个闭子空间. 当$\alpha=1$ 时，$Z_{0}^{\alpha}$和$Z^{\alpha}$空间即为经典的Zygmund空间$Z$和 小Zygmund空间$Z_{0}$.

设$u\in H(D)$,$\varphi$ 是 $D\rightarrow D$ 的全纯自映射, 定义$H(D)$上的加权复合算子$uC_{\varphi}$如下: $(uC_{\varphi})(f)(z)=u(z)f(\varphi(z)), z\in D,f\in H(D)$. 它同时是乘积算子和复合算子的推广, 当 $u(z)=1$ 时上述算子就是由 $\varphi$ 诱导的复合算子 $C_{\varphi}$. 对于这方面的基础知识,可参见文献^{[1, 2, 3]}. 文献^{[4, 5, 6, 7]} 研究了 $\alpha$-Bloch空间之间的复合算子和加权复合算子的有界性和紧性. 文献^{[8, 9]}研究了 $\alpha$-Zygmund空间之间的加权复合算子的有界性和紧性. 最近,李颂孝和 Stevic^[10]研究了从Zygmund空间到Bloch空间的加权复合算子, 分别得到加权复合算子成为有界算子的充分且必要条件和成为紧算子的充分且必要条件. 本文推广了文献^[10] 中的结果,对于 $0<\alpha,\beta<\infty$,得到了单位圆 $D$ 上从 $\alpha$-Zygmund空间到 $\beta$-Bloch空间的加权复合算子的有界性和紧性的充 分且必要条件,详见本文定理3.1,定理3.2(第3节).

引理2.1^[8] 设$\alpha>0$,若$f\in Z^{\alpha}$,则

1)~ $|f'(z)|\leq\frac{2}{1-\alpha}\|f\|_{Z^{\alpha}}$,$|f(z)|\leq\frac{2}{1-\alpha}\|f\|_{Z^{\alpha}}$,$0<\alpha<1$;

2)~ $|f'(z)|\leq2\|f\|_{Z}\log\frac{2}{1-|z|}$,$|f(z)|\leq\|f\|_{Z^{\alpha}}$,$\alpha=1$;

3)~ $|f'(z)|\leq\frac{2}{\alpha-1}\|f\|_{Z^{\alpha}}(1-|z|)^{1-\alpha}$,$\alpha>1$;

4)~ $|f(z)|\leq\frac{2}{(\alpha-1)(2-\alpha)}\|f\|_{Z^{\alpha}}$,$1<\alpha<2$;

5)~ $|f(z)|\leq2\|f\|_{Z^{\alpha}}\log\frac{2}{1-|z|}$,$\alpha=2$;

6)~ $|f(z)|\leq\frac{2}{(\alpha-1)(\alpha-2)}\|f\|_{Z^{\alpha}}(1-|z|)^{2-\alpha}$,$\alpha>2$.

引理2.2 ^[6] 设$\alpha>0$,若$f\in B^{\alpha}$,则

1)~ $|f(z)|\leq\frac{2-\alpha}{1-\alpha}\|f\|_{B^{\alpha}}$,$0<\alpha<1$;

2)~ $|f(z)|\leq\frac{\|f\|_{B^{\alpha}}}{\log2}\log\frac{2}{1-|z|^{2}}$,$\alpha=1$;

3)~ $|f(z)|\leq(1+\frac{1}{(\alpha-1)2^{\alpha-1}})\frac{\|f\|_{B^{\alpha}}}{(1-|z|^{2})^{\alpha-1}}$,$\alpha>1$.

引理2.3 算子$uC_{\varphi}: Z^{\alpha}\rightarrow B^{\beta}$或 $Z_{0}^{\alpha}\rightarrow B^{\beta}$ 是紧的当且仅当 $uC_{\varphi}: Z^{\alpha}\rightarrow B^{\beta}$ 或 $Z_{0}^{\alpha}\rightarrow B^{\beta}$ 是有界的,且对 $Z^{\alpha}$ 或 $Z_{0}^{\alpha}$ 中的任意范数有界的函数序列 $\{f_{n}\}_{n\in N}$,$f_{n}$ 在 $D$ 上内闭一致收敛到0$(n\rightarrow \infty)$, 蕴含$\|uC_{\varphi}f_{n}\|\rightarrow 0$,$n\rightarrow \infty$.

证 利用引理2.1和引理2.2,仿造文献^[1]的弱收敛定理(第2.4 节,29-30 页) 或文献^[5]中引理3.7的证明过程可以证得. 具体略.

定理3.1 设$0<\alpha,\beta<\infty$,$\varphi$是$D$的全纯自映射,$u\in H(D)$,则下列表述等价

i)~ $uC_{\varphi}: Z^{\alpha}\rightarrow B^{\beta}$是有界的;

ii)~ $uC_{\varphi}: Z_{0}^{\alpha}\rightarrow B^{\beta}$是有界的;

iii) a)~ $0<\alpha<1$时,有$u\in B^{\beta}$,且

$$\begin{equation}%\tag{3.1} \sup_{z\in D}(1-|z^{2}|)^{\beta}|u(z)\varphi'(z)|<\infty, \end{equation}$$

(3.1)

b)~ $\alpha=1$时,有$u\in B^{\beta}$,且

$$\begin{equation}%\tag{3.2} \sup_{z\in D}(1-|z^{2}|)^{\beta}|u(z)\varphi'(z)|\log\frac{2}{1-|\varphi(z)|^{2}} <\infty, \end{equation}$$

(3.2)

c)~ $1<\alpha<2$时,有$u\in B^{\beta}$,且

$$\begin{equation}%\tag{3.3} \sup_{z\in D}\frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-1}}|u(z)\varphi'(z)|<\infty, \end{equation}$$

(3.3)

d)~ $\alpha=2$时,有(3.3)式成立,且

$$\begin{equation}%\tag{3.4} \sup_{z\in D}(1-|z^{2}|)^{\beta}|u'(z)|\log\frac{2}{1-|\varphi(z)|^{2}} <\infty, \end{equation}$$

(3.4)

e)~ $\alpha>2$时,有(3.3)式成立,且

$$\begin{equation}%\tag{3.5} \sup_{z\in D}\frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-2}}|u'(z)|<\infty. \end{equation}$$

(3.5)

证 iii)$\Rightarrow$ i)~ 假设iii)中的条件成立,对于任意$z\in D$, 任意$f\in Z^{\alpha}$, $$\begin{eqnarray*} I&\triangleq & (1-|z|^{2})^{\beta} |(uC_{\varphi}f)'(z)|=(1-|z|^{2})^{\beta}|u'(z)f(\varphi(z))+u(z)f'(\varphi(z))\varphi'(z)|\nonumber\\ &\leq & (1-|z|^{2})^{\beta}|u'(z)||f(\varphi(z))|+(1-|z|^{2})^{\beta}|u(z)\varphi'(z)||f'(\varphi(z))|. \end{eqnarray*}$$ 由引理2.1,当$\alpha>2$时,

$$\begin{equation}%\tag{3.6} I\leq \frac{2}{(\alpha-1)(\alpha-2)}\frac{(1-|z|^{2})^{\beta}}{(1-|\varphi(z)|)^{\alpha-2}}|u'(z)|\|f\|_{z^{\alpha}}+ \frac{2}{\alpha-1}\frac{(1-|z|^{2})^{\beta}}{(1-|\varphi(z)|)^{\alpha-1}}|u(z)\varphi'(z)|\|f\|_{Z^{\alpha}}. \end{equation}$$

(3.6)

对(3.6)式两边取上确界并结合(3.3)和(3.5)式,可证得$uC_{\varphi}: Z^{\alpha}\Rightarrow B^{\beta}$是有界的.

对于$0<\alpha\leq2$的情况,类似可证.

i)$\Rightarrow$ ii)~ 显然成立.

ii)$\Rightarrow$ iii)~ 假设$uC_{\varphi}:Z^{\alpha}\Rightarrow B^{\beta}$是有界的.

情况a)实际上,当$0<\alpha<\infty$时,取$f=1\in Z_{0}^{\alpha}$,则有$u\in B^{\beta}$; 取$f=z\in Z_{0}^{\alpha}$,有 $$\begin{eqnarray*} \sup_{z\in D}(1-|z^{2}|)^{\beta}|u(z)\varphi'(z)+u'(z)\varphi(z)|<\infty, \end{eqnarray*}$$ 由上式以及$\varphi$的有界性,有(3.1)式成立;

对于$1\leq\alpha<\infty$的情况,取$a\in D$,使得$\frac{1}{2}<|a|<1$,

情况b)当$\alpha=1$ 时,取 $$f_{a}(z)=\frac{\overline{a}z-1}{\overline{a}} \bigg[\Big(1+\log\frac{2}{1-\overline{a}z}\Big)^{2}+1\bigg] \Big(\log\frac{2}{1-|a|^{2}}\Big)^{-1}\in Z_{0},$$ $$f_{a}(0)=-\frac{1}{\overline{a}}[(1+\log2)^{2}+1]\Big(\log\frac{2}{1-|a|^{2}}\Big)^{-1},$$ 则 $$f_{a}(z)=\Big(\log\frac{2}{1-\overline{a}z}\Big)^{2}\Big(\log\frac{2}{1-|a|^{2}}\Big)^{-1},$$ $$f'_{a}(0)=(\log2)^{2}\Big(\log\frac{2}{1-|a|^{2}}\Big)^{-1},$$ $$f''_{a}(z)=\frac{2\overline{a}}{1-\overline{a}z}\Big(\log\frac{2}{1-\overline{a}z}\Big) \Big(\log\frac{2}{1-|a|^{2}}\Big)^{-1},$$ $$|f''_{a}(z)|\leq\frac{2}{1-|z|}\Big(\log\frac{2}{1-|a|}+2\pi\Big) \Big(\log\frac{2}{1-|a|^{2}}\Big)^{-1}\leq\frac{2}{1-|z|}\Big(1+\frac{2\pi}{\log4}\Big),$$ $$\sup_{\frac{1}{\sqrt2}<|a|<1}\|f_{a}\|_{Z}<M,$$ 其中$M=\frac{\sqrt{2}}{\log4}[(1+\log2)^{2}+1]+\frac{\log2}{2}+4(1+\frac{2\pi}{\log4})$, 所以$f_{a}\in Z$. 而且,因$f_{a}$在$|z|<\frac{1}{|a|}$ 内解析,从而$f'_{a}$在闭圆$\overline{D}={|z|\leq1}$内解析,故$f_{a}\in Z_{0}$.\\ 所以$\forall\lambda\in D$,使得$\frac{1}{\sqrt2}<\varphi(\lambda)<1$,有 $$\begin{eqnarray*} M\|uC_{\varphi}\|&\geq&\|f_{\varphi(\lambda)}\|_{Z}\|uC_{\varphi}\|\geq\|uC_{\varphi}f_{ \varphi(z)}\|_{B^{\beta}}\\ &\geq&-M(1-|\lambda|^{2})^{\beta}|u'(\lambda)|+(1-|\lambda|^{2})^{\beta}|u(\lambda)\varphi'(\lambda)|\log\frac{2}{1-|\varphi(\lambda)|^{2}}, \end{eqnarray*}$$ 所以有

$$\begin{equation}%\tag{3.7} \sup_{\frac{1}{\sqrt2}<\varphi(\lambda)<1}(1-|\lambda|^{2})^{\beta}|u(\lambda)\varphi'(\lambda)|\log\frac{2}{1-|\varphi(\lambda)|^{2}}. \end{equation}$$

(3.7)

对$\forall\lambda\in D$,使得$\varphi(\lambda)<\frac{1}{\sqrt2}$,有

$$\begin{equation}%\tag{3.8} \sup_{z\in D}(1-|z^{2}|)^{\beta}|u(z)\varphi'(z)|\log\frac{2}{1-|\varphi(z)|^{2}} <\sup_{z\in D}(1-|z^{2}|)^{\beta}|u(z)\varphi'(z)|\log4<\infty. \end{equation}$$

(3.8)

由式(3.1),(3.7),(3.8),得到式(3.2);

情况c)对于$1<\alpha<\infty$,取 $$f_{a}(z)=\frac{1}{\overline{a}} \bigg[\frac{(1-|a|^{2})^{2}}{(1-\overline{a}z)^{\alpha}}-\frac{1-|a|^{2}}{(1-\overline{a}z)^{\alpha-1}}\bigg]\in Z_{0}^{\alpha},$$ $$f_{a}(0)=\frac{1}{\overline{a}}[(1-|a|^{2})^{2}-(1-|a|^{2})],$$ $$f'_{a}(z)=\frac{\alpha(1-|a|^{2})^{2}} {(1-\overline{a}z)^{\alpha+1}}-\frac{(\alpha-1)(1-|a|^{2})}{1-\overline{a}z},$$ $$f'_{a}(0)=\alpha(1-|a|^{2})^{2}-(\alpha-1)(1-|a|^{2}),$$ $$f''_{a}(z)=\frac{\alpha(\alpha+1)\overline{a}(1-|a|^{2})^{2}}{(1-\overline{a}z)^{\alpha+2}} -\frac{(\alpha-1)\alpha\overline{a}(1-|a|^{2})}{(1-\overline{a}z)^{\alpha+1}},$$ $$|f''_{a}(z)|\leq\alpha(\alpha+1) \bigg|\frac{(1-|a|^{2})^{2}}{(1-|\overline{a}|^{2})(1-|z|)^{\alpha}}\bigg| +\alpha(\alpha-1)\bigg| \frac{1-|a|^{2}}{(1-|\overline{a}|)(1-|z|)^{\alpha}}\bigg| \leq\frac{6\alpha^{2}+2\alpha}{(1-|z|)^{\alpha}},$$ $$\|f_{a}\|_{Z^{\alpha}}\leq M,$$ 其中$M=\frac{\sqrt{2}}{2}+\frac{3\alpha-2}{4}+2^{\alpha}(6\alpha^{2}+2\alpha)$, 所以$f_{a}\in Z^{\alpha}$. 而且类似情况b)的证明可得式(3.3); 情况d)$\alpha=2$,令$f_{a}(z)=\log\frac{2}{1-\overline{a}z}\in Z_{0}^{\alpha}$, $f_{a}(0)=\log2$,则 $$f'_{a}(z)=\frac{\overline{a}}{1-\overline{a}z},\quad f'_{a}(0)=\overline{a}, \quad f''_{a}(z)=\frac{(\overline{a})^{2}}{(1-\overline{a}z)^{2}},$$ $$|f''_{a}(z)|<\frac{1}{(1-|z|^{2})},\quad \sup_{\frac{1}{\sqrt{2}}<|a|<1}{\|f_{a}\|_{Z^{\alpha}}}\leq M,$$ 其中$M=\log2+5$,$f_{a}\in Z^{\alpha}$. 结合(3.3)式,类似于情况b)可证得式(3.4)成立; 情况e) $\alpha>2$,令 $$f_{a}{z}=\frac{1-|a|^{2}}{(1-\overline{a}z)^{\alpha-1}}-\frac{(1-|a|^{2})^{\alpha}}{2(1-\overline{a}z)^{2\alpha-2}}\in Z_{0}^{\alpha},$$ $$f_{a}(0)=1-|a|^{2}-(1-|a|^{2})^{\alpha},$$ $$f'_{a}(z)=\frac{(\alpha-1)(1-|a|^{2})\overline{a}}{(1-\overline{a}z)^{\alpha}}-\frac{(\alpha-1)(1-|a|^{2}) ^{\alpha}\overline{a}}{(1-\overline{a}z)^{2\alpha-1}},$$ $$f'_{a}(0)=(\alpha-1)(1-|a|^{2})\overline{a}-(\alpha-1)(1-|a|^{2})^{\alpha}\overline{a},$$ $$f''_{a}(z)=\frac{(\alpha-1)\alpha(1-|a|^{2})(\overline{a})^{2}}{(1-\overline{a}z)^{\alpha+1}} -\frac{(\alpha-1)(2\alpha-1)(1-|a|^{2})^{\alpha}(\overline{a})^{2}}{(1-\overline{a}z)^{2\alpha}},$$ $$|f''_{a}(z)|<\frac{\alpha( \alpha+1)+(\alpha-1)(2\alpha-1)2^{\alpha-1}}{(1-|z|)^{\alpha}},$$ $$\sup_{\frac{1}{\sqrt{2}}<|a|<1}{\|f_{a}\|_{Z^{\alpha}}}\leq M,$$ 其中$M=\alpha+2^{\alpha}(\alpha+1)+(\alpha-1)(2\alpha-1)$,同样类似情况b) 的证明, 可得(3.5)式,定理3.1证毕.

定理3.2 设$0<\alpha,\beta<\infty$,$\varphi$是$D$的全纯自映射,$u\in H(D)$,则下列表述等价

i)~ $uC_{\varphi}:Z^{\alpha}\rightarrow B^{\beta}$是紧的;

ii)~ $uC_{\varphi}:Z_{0}^{\alpha}\rightarrow B^{\beta}$是紧的;

iii) a)~ $0<\alpha<1$时,有

$$\begin{equation}%\tag{3.9} \lim_{|\varphi(z)|\rightarrow 1}(1-|z^{2}|)^{\beta}|u'(z)|=0 \end{equation}$$

(3.9)

且

$$\begin{equation}%\tag{3.10} \lim_{|\varphi(z)|\rightarrow 1}(1-|z^{2}|)^{\beta}|u(z)||\varphi'(z)|=0; \end{equation}$$

(3.10)

b)~ $\alpha=1$时,有(3.9)式成立,且

$$\begin{equation}%\tag{3.11} \lim_{|\varphi(z)|\rightarrow 1}(1-|z^{2}|)^{\beta}|u(z)||\varphi'(z)|\log\frac{2}{1-|\varphi(z)|^{2}}=0; \end{equation}$$

(3.11)

c)~ $1<\alpha<2$时,有(3.9)式成立,且

$$\begin{equation}%\tag{3.12} \lim_{|\varphi(z)|\rightarrow 1}\frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-1}}|u(z)||\varphi'(z)|=0; \end{equation}$$

(3.12)

d)~ $\alpha=2$时,有(3.12)式成立,且

$$\begin{equation}%\tag{3.13} \lim_{|\varphi(z)|\rightarrow 1}(1-|z^{2}|)^{\beta}|u'(z)|\log\frac{2}{1-|\varphi(z)|^{2}} =0; \end{equation}$$

(3.13)

e)~ $\alpha>2$时,有(3.12)式成立,且

$$\begin{equation}%\tag{3.14} \lim_{|\varphi(z)|\rightarrow 1}\frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-2}}|u'(z)|=0. \end{equation}$$

(3.14)

证 iii)$\Rightarrow$ i)~ 先考虑情况e),由式(3.12),(3.14) 易证(3.3),(3.5)式成立,故由定理3.1知$uC_{\varphi}:Z^{\alpha}\rightarrow B^{\beta}$ 是有界的. 为了证明$uC_{\varphi}$ 是紧的,对于$\forall \|f_{n}\|_{Z^{\alpha}}\leq1$,且$f_{n}(z)$ 在$D$内闭一致收敛到$0(n\rightarrow\infty)$,由引理2.3,只需证明$\|uC_{\varphi}f_{n}\|\rightarrow 0$,$n\rightarrow \infty$.\\ 由(3.12)和(3.14)式,对$\forall \varepsilon>0,\exists \delta\in (0,1)$,使得$\delta<\varphi(z)<1$时,有 $$\frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-1}}|u(z)||\varphi'(z)|<\varepsilon; \qquad \frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-2}}|u'(z)|<\varepsilon.$$ 由上面两式以及引理2.1有 $$\begin{eqnarray*} &&\|uC\varphi(f_{n})\|_{B^{\beta}}\\ &=&\sup_{z\in D}(1-|z|^{2})^{\beta}|uC\varphi(f_{n})(z)|+|u(0)f_{n}(\varphi(0))|\\ &\leq&\sup_{z\in D;|\varphi(z)|\leq\delta}(1-|z|^{2})^{\beta}|u'(z)f_{n}(\varphi(z))|+\sup_{z\in D;\delta<|\varphi(z)|\leq1}(1-|z|^{2})^{\beta}|u'(z)f_{n}(\varphi(z))|\\ &&+\sup_{z\in D;|\varphi(z)|\leq\delta}(1-|z|^{2})^{\beta}|u(z)\varphi'(z)f'_{n} (\varphi(z))|\\ &&+\sup_{z\in D;\delta<|\varphi(z)|\leq1}(1-|z|^{2})^{\beta}|u(z)\varphi'(z)f'_{n}(\varphi(z))| +|u(0)f_{n}(\varphi(0))|\\ &=&L\sup_{|\omega|\leq\delta}|f'_{n}(\varphi(z))|+\frac{\varepsilon_{1}2^{\alpha}}{(\alpha-1)M}\|f\|_{Z^{\alpha}}+|u(0)f_{n}(\varphi(0))|, \end{eqnarray*}$$ 其中$L=\sup\limits_{z\in D}(1-|z^{2}|)^{\beta}|u(z)||\varphi'(z)|$. 由(3.12)式知(3.10)式成立,从而$L<\infty$. 又因为在$D$上$f_{n}$内闭一致收敛到0, 特别地,在圆盘$|\omega|\leq\delta$ 上有$f_{n}$内闭一致收敛到0; 由柯西高阶导数公 式及估值定理可知当$n\rightarrow\infty$时,$f'_{n}$在$D$上内闭一致收敛到到0,特别地, 在圆盘$|\omega|\leq\delta$上有$f'_{n}$一致收敛到0; 显然, 有$|u(0)f_{n}(\varphi(0))|\rightarrow 0$. 由这些事实,当$n\rightarrow\infty$时, 有$\lim\limits_{n\rightarrow\infty}\|uC_{\varphi}f_{n}\|_{B^{\beta}}=0$. 所以对$\forall\varepsilon>0$,$\limsup\limits_{n\rightarrow\infty} \|uC_{\varphi}f_{n}\|_{B^{\beta}}=0$. 所以$uC_{\varphi}: Z^{\alpha}\rightarrow B^{\beta}$ 是紧的. 对于$0<\alpha\leq2$的情况,类似可证.

i)$\Rightarrow$ii)~ 显然.

ii)$\Rightarrow$iii)~ 情况a)当$0<\alpha<\infty$时,假设$uC_{\varphi}: Z_{0}^{\alpha}\rightarrow B^{\beta}$是紧的,则类似定理3.1,有$\sup\limits_{n\in N}\|f_{n}\|_{Z^{\alpha}}\leq M<\infty$, 当$n\rightarrow\infty$时,$f_{n}$ 内闭一致收敛到0.

因为$uC_{\varphi}: Z_{0}^{\alpha}\rightarrow B^{\beta}$是紧的, 也就是$\limsup\limits_{n\rightarrow\infty}\|uC_{\varphi}f_{n}\|_{B^{\beta}}=0$, 令$\{Z_{n}\}_{n\in N}$ 是$D$中一序列,$|\varphi(z_{n})\rightarrow1|$,$n\rightarrow\infty$,取 $$f_{n}(z)=3\Big(\log\frac{2}{1-|\overline{\varphi(z_{n})}|^{2}}\Big)^{-1} \Big(\log\frac{2}{1-\overline{\varphi(z_{n})}z}\Big)^{2} -2\Big(\log\frac{2}{1-|\overline{\varphi(z_{n})}|^{2}}\Big)^{-2} \Big(\log\frac{2}{1-\overline{\varphi(z_{n})}z}\Big)^{3},$$ 则$f_{n}(z)\in Z_{0}^{\alpha}$,且 $$\|uC_{\varphi}f_{n}\|_{B^{\beta}}\geq (1-|z_{n}^{2}|)^{\beta}|u'(z_{n})|\log\frac{2}{1-|\varphi(z_{n})|^{2}},$$ 因为$\varphi(z_{n})<1$,所以(3.9) 式成立. 另一方面,取 $$f_{n}(z)=\frac{\overline{\varphi(z_{n})}z-1}{\overline{\varphi(z_{n})}} \bigg[\Big(1+\log\frac{2}{1-\overline{\varphi(z_{n})}z}\Big)^{2}+1\bigg] \Big(\log\frac{2}{1-|\varphi(z_{n})|^{2}} \Big)^{-1}\in Z_{0}^{\alpha},$$ 有 $$\begin{eqnarray}%\tag{3.15} &&\|uC_{\varphi}f_{n}\|_{B^{\beta}} \geq (1-|z_{n}^{2}|)^{\beta}|u(z_{n})\varphi'(z_{n})|\log\frac{2}{1-|\varphi(z_{n})|^{2}} \nonumber\\ &&-\frac{1-|\varphi(z_{n})|^{2}}{|\varphi(z_{n})|} \bigg[\Big(1+\log\frac{2}{1-\overline{\varphi(z_{n})}z}\Big)^{2}+1\bigg] \Big(\log\frac{2}{1-|\varphi(z_{n})|^{2}}\Big)^{-1} (1-|z_{n}^{2}|)^{\beta}|u'(z_{n})|. \qquad \end{eqnarray}$$ 因为$\lim\limits_{x\rightarrow 1}\frac{1-x^{2}}{x}[(1+\log\frac{2}{1-x^{2}})^{2}+1](\log\frac{2}{1-x^{2}})^{-1}=0$,由(3.9)和(3.15)式可知 $$\lim_{|\varphi(z)|\rightarrow 1}(1-|z^{2}|)^{\beta}|u(z)||\varphi'(z) |\log\frac{2}{1-|\varphi(z)|^{2}}=0$$ 成立.

同样因为$|\varphi(z_{n})|<1$,有(3.10)式成立.

情况b)当$\alpha=1$时,(3.9)式成立,取 $$f_{n}(z)=\frac{\overline{\varphi(z_{n})}z-1}{\overline{\varphi(z_{n})}} \bigg[\Big(1+\log\frac{2}{1-\overline{\varphi(z_{n})}z}\Big)^{2}+1\bigg] \Big(\log\frac{2}{1-|\varphi(z_{n})|^{2}} \Big)^{-1}\in Z_{0}^{\alpha},$$ 类似情况a),由(3.9),(3.15)式可知(3.11)式成立.

情况c)当$1<\alpha<\infty$时,(3.9)式成立,取 $$f_{n}(z)=\frac{1}{\overline{\varphi(z_{n})}}\bigg[\frac{(1-|\varphi(z_{n})|^{2})^{2}}{(1-\overline{\varphi(z_{n})}z)^{\alpha}}-\frac{1-|\varphi(z_{n})|^{2}}{(1- \overline{\varphi(z_{n})}z)^{\alpha-1}}\bigg]\in Z_{0}^{\alpha},$$ 则类似情况a)的证明,有 $$\limsup_{n\rightarrow\infty}\|uC_{\varphi}f_{n}\|_{B^{\beta}}=0,\quad \|uC_{\varphi}f_{n}\|_{B^{\beta}}\geq \frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-1}}|u(z)\varphi'(z)|,$$ 由上式可得(3.12)式成立;

情况d) $\alpha=2$时,易知(3.12)式成立,取 $$\begin{eqnarray*} f_{n}(z)&=&3\Big(\log\frac{2}{1-|\overline{\varphi(z_{n})}|^{2}}\Big)^{-1} \Big(\log\frac{2}{1-\overline{\varphi(z_{n})}z}\Big)^{2} -2\Big(\log\frac{2}{1-|\overline{\varphi(z_{n})} |^{2}}\Big)^{-2}\Big(\log\frac{2}{1-\overline{\varphi(z_{n})}z}\Big)^{3} \\ &\in & Z_{0}^{\alpha}, \end{eqnarray*}$$ 类似情况a)的证明,$\limsup\limits_{n\rightarrow\infty}\|uC_{\varphi}f_{n}\|_{B^{\beta}}=0$,所以 $$\|uC_{\varphi}f_{n}\|_{B^{\beta}}\geq (1-|z^{2}|)^{\beta}|u'(z)|\log\frac{2}{1-|\varphi(z)|^{2}},$$ 所以(3.13)式成立;

情况e)$\alpha>2$时,易知(3.12)式成立,下面取 $$f_{n}(z)=\frac{(1-|\varphi(z_{n})|^{2})^{2}}{(1-\overline{\varphi(z_{n})}z)^{\alpha-1}}-\frac{(1-|\varphi(z_{n})|^{2})^{\alpha}} {(1-\overline{\varphi(z_{n})}z)^{2\alpha-2}}\in Z_{0}^{\alpha}.$$ 类似情况a)的证明,有$\limsup\limits_{n\rightarrow\infty}\|uC_{\varphi}f_{n}\|_{B^{\beta}}=0$,而 $$\|uC_{\varphi}f_{n}\|_{B^{\beta}}\geq (1-|z^{2}|)^{\beta}|u'(z)|\frac{1}{2(1-|\varphi(z)|^{2})^{\alpha-2}}+(2\alpha-1)\frac{(1-|z^{2}|)^{\beta}}{(1-|\varphi(z)|^{2})^{\alpha-1}}|u(z)\varphi'(z)|,$$ 所以由(3.12)式证得(3.14)式成立. 定理3.2证毕.

推论3.1 设$0<\alpha,\beta<\infty$,$\varphi$是$D$的全纯自映射,$u\in H(D)$, 当$0<\alpha\leq1$时,$uC_{\varphi}: Z^{\alpha}\rightarrow B^{\beta}$ 或者 $uC_{\varphi}:Z_{0}^{\alpha}\rightarrow B^{\beta}$ 是紧的充分且必要条件为

i)~ $\lim\limits_{|\varphi(z)|\rightarrow 1}(1-|z^{2}|)^{\beta}|u'(z)|\log\frac{2}{1-|\varphi(z)|^{2}} =0$,

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