2 预备知识
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.
3 主要结果及证明
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)~ $\lim\limits_{|\varphi(z)|\rightarrow 1}(1-|z^{2}|)^{\beta}|u(z)||\varphi'(z)|\log\frac{2}{1-|\varphi(z)|^{2}}=0$.
2015, Vol. 35 
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