以$U^{n}=\{z=(z_{1}, z_{2}, \cdots, z_{n}):|z_{i}|<1, i=1, 2, \cdots, n\}$表示$n$维复空间$C^{n}$中的单位多圆柱; $H(U^{n})$和$H(U^{n}, U^{n})$分别表示$U^{n}$上全纯函数和全纯自映射的全体; $\partial U^{n}=\{z=(z_{1}, z_{2}, \cdots, z_{n})\in C^{n}:|z_{i}|=1, i=1, 2, \cdots, n\}$表示$U^{n}$的边界; $U^{n}$上加对数权的Bloch空间是满足

$ B_{{\log}}(U^{n})=\left\{f\in H(U^{n}):\|f\|_{\log}=\sup\limits_{z\in U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial f}{\partial z_{k} }(z)\right|<\infty\right\} $

的函数$f$的全体, 在范数$\|f\|_{B_{\log}}=|f(0)|+\|f\|_{\log}$下, $B_{\log}(U^{n})$空间是Banach空间.加对数权的小Bloch空间是满足

$ B_{0, \log}(U^{n})=\left\{f\in H(U^{n}):\lim\limits_{z\rightarrow \partial U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial f}{\partial z_{k} }(z)\right|=0\right\} $

的函数$f$的全体, 由于$B_{0, \log}(U^{n})\subseteq B_{\log}(U^{n})$, 所以$B_{0, \log}(U^{n})$是$B_{\log}(U^{n})$的子空间.设$\psi\in H(U^{n}), \varphi\in H(U^{n}, U^{n}), z\in U^{n}, f\in H(U^{n})$, 加权复合算子定义为

$ (W_{\psi, \varphi}f)(z)=\psi(z)f(\varphi(z)). $

显然$W_{\psi, \varphi}$是线性算子.特殊地, 当$\psi(z)=1$时, $W_{1, \varphi}$即为通常的复合算子; 当$\psi(z)=z$时, $W_{\psi, z}$即为通常的点乘算子, 因此$W_{\psi, \varphi}$可以看成是复合算子和点乘算子的推广.在$n$维空间的单位多圆柱上, 文献[1, 2]研究了Bloch空间上的复合算子和加权复合算子, 文献[3]探讨了不同的$p$-Bloch和小$p$-Bloch空间上的加权复合算子, 文献[4]中给出了一个定义在$[0, 1)$上的非负函数$\mu$, 并研究了加正规权的Bloch型空间$B_{\omega}$到$B_{\mu}$以及加正规权的小Bloch型空间$B_{\omega, 0}$到$B_{\mu, 0}$上的复合算子, 他们分别给出了算子的有界性和紧性的充要条件.关于Bloch空间、加权Bloch空间以及其上的相关算子的其他研究成果见文献[6-10].

本文把文献[5]中的单位圆盘$D$推广到了$n$维空间的单位多圆柱上, 讨论了单位多圆柱上加对数权的Bloch空间和加对数权的小Bloch空间上的加权复合算子的有界性和紧性问题.

在本文中, $C$表示正的常数, 不同的地方可以表示不一样的正常数.

为了得到本文的主要结果, 需要用到下面的几个引理.

引理2.1 如果$f\in B_{\log}(U^{n})$, 则$|f(z)|\leq\left(2+\sum\limits_{k=1}^{n}\ln(\ln\frac{2}{1-|z_{k}|})\right)\|f\|_{B_{\log}}$.特别地, 当$\displaystyle|z_{k}|>1-\frac{2}{e^{e^{2}}}~(k=1, 2, \cdots, n)$时, $|f(z)|\leq2\sum\limits_{k=1}^{n}\ln(\ln\frac{2}{1-|z_{k}|})\|f\|_{B_{\log}}$.

证 对任意的$f\in B_{\log}(U^{n})$, 有

$ \begin{align*} |f(z)|&\leq|f(0)|+\left|\int_{0}^{1}\frac{d}{dt}f(tz)dt\right|\leq|f(0)| +\sum\limits_{k=1}^{n}\int_{0}^{1}|z_{k}|\left|\frac{\partial f}{\partial z_{k}}(tz)\right|dt\\ &\leq\|f\|_{B_{\log}}+\|f\|_{B_{\log}}\sum\limits_{k=1}^{n}\int_{0}^{|z_{k}|}\frac{1}{(1-x)\ln\frac{2}{1-x}}dx\\ &\leq\|f\|_{B_{\log}}+\|f\|_{B_{\log}}\left[\sum\limits_{k=1}^{n}\ln(\ln\frac{2}{1-|z_{k}|})-\ln(\ln2)\right]\\ &\leq\left (2+\sum\limits_{k=1}^{n}\ln(\ln\frac{2}{1-|z_{k}|})\right )\|f\|_{B_{\log}}. \end{align*} $

引理2.2 设$\psi\in H(U^{n}), \varphi\in H(U^{n}, U^{n})$.则$W_{\psi, \varphi}$为$B_{\log}(U^{n})$ ($B_{0, \log}(U^{n})$)上的紧算子当且仅当$W_{\psi, \varphi}$为有界算子并且对于$B_{\log}(U^{n})$ ($B_{0, \log}(U^{n})$)中在$U^{n}$的任意紧子集上一致趋于零的有界序列$\{f_{n}\}$, 当$n\rightarrow\infty$时, $\|W_{\psi, \varphi}f_{n}\|_{B_{\log}}\rightarrow0$.

证 只证$B_{\log}(U^{n})$空间的情形, $B_{0, \log}(U^{n})$空间的情形可类似证明.

设$W_{\psi, \varphi}$为$B_{\log}(U^{n})$上的紧算子, 显然$W_{\psi, \varphi}$为$B_{\log}(U^{n})$上的有界算子.设$\{f_{n}\}$是$B_{\log}(U^{n})$中的一有界列且当$n\rightarrow\infty$时, $\{f_{n}\}$在$U^{n}$的任意紧子集上一致趋于零, 由紧算子的定义, $\{W_{\psi, \varphi}f_{n}\}$存在一个在$B_{\log}(U^{n})$中收敛的子列$\{W_{\psi, \varphi}f_{n_{k}}\}$.不妨设当$k\rightarrow\infty$时, $\{W_{\psi, \varphi}f_{n_{k}}\}$收敛于$f$, 于是$\|W_{\psi, \varphi}f_{n_{k}}-f\|_{B_{\log}}\rightarrow0$.由引理2.1, 对$U^{n}$的任意紧子集$K$都存在正常数$C$, 使得$\forall z\in K$都有

$|(W_{\psi, \varphi}f_{n_{k}})(z)-f(z)|\leq C\|W_{\psi, \varphi}f_{n_{k}}-f\|_{B_{\log}}. $

因此当$k\rightarrow\infty$时, $(W_{\psi, \varphi}f_{n_{k}})(z)-f(z)\rightarrow0$, 又由于$f_{n_{k}}\rightarrow0$, 所以$W_{\psi, \varphi}f_{n_{k}}\rightarrow0$, $f\rightarrow0$, 由$\{f_{n}\}$的任意性, 当$n\rightarrow\infty$时, $\|W_{\psi, \varphi}f_{n}\|_{B_{\log}}\rightarrow0$.

反之, 假设$W_{\psi, \varphi}$在$B_{\log}(U^{n})$上不是紧的, 则存在一个$B_{\log}(U^{n})$中在$U^{n}$上的有界序列$\{g_{n}\}$, 使得$\{W_{\psi, \varphi}g_{n}\}$在$B_{\log}(U^{n})$中没有收敛的子列.进一步, 再设$\{g_{n}\}$在$U^{n}$的任意紧子集上一致有界, 那么由Montel定理知$\{g_{n}\}$是正规的, $\{g_{n}\}$存在一个在$U^{n}$的任意紧子集上一致收敛的子列$\{f_{n}\}$.不妨设$\{f_{n}\}$一致收敛于$f$, 那么$f'_{n}\rightarrow f'$, 并且$f\in B_{\log}(U^{n})$.从而序列$\{f_{n}-f\}$在$B_{\log}(U^{n})$中有界且在$U^{n}$的任意紧子集上一致趋于零, 由题设当$n\rightarrow\infty$时, $\|W_{\psi, \varphi}(f_{n}-f)\|_{B_{\log}}\rightarrow0$, 这意味着$\{W_{\psi, \varphi}g_{n}\}$的子列$\{W_{\psi, \varphi}f_{n}\}$收敛于$W_{\psi, \varphi}f$, 矛盾.证毕.

引理2.3 如果$f\in B_{0, \log}(U^{n})$, 令$\displaystyle Q(z)=\sum\limits_{k=1}^{n}\ln(\ln\frac{2}{1-|z_{k}|})$, 那么$\displaystyle\lim\limits_{z\rightarrow\partial U^{n}}\frac{|f(z)|}{Q(z)}=0$.

证 由于$f\in B_{0, \log}(U^{n})$, 则$\forall \varepsilon>0$, $\exists~\delta\in (0, 1)$, 使得当$|z_{k}|>\delta~(k=1, 2, \cdots, n)$时,

$ \sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial f}{\partial z_{k} }(z)\right|<\varepsilon. $

以$U^{n}(0, \delta)$表示以原点为中心的$\delta$邻域, 令

$ \displaystyle I=\sum\limits_{k=1}^{n}\int_{\delta}^{1}\frac{|z_{k}|}{(1-t|z_{k}|)\ln\frac{2}{1-t|z_{k}|}} \left(1-t^{2}|z_{k}|^{2}\right)\ln\frac{2}{1-t|z_{k}|}\left|\frac{\partial f}{\partial z_{k}}(tz)\right|dt. $

那么对任意的$f\in B_{0, \log}(U^{n})$, 有

$ \begin{equation} \begin{split} |f(z)|&\leq|f(0)|+\left|\int_{0}^{1}\frac{d}{dt}f(tz)dt\right|\leq|f(0)| +\sum\limits_{k=1}^{n}\int_{0}^{1}|z_{k}|\left|\frac{\partial f}{\partial z_{k}}(tz)\right|dt\\ &\leq|f(0)|+\sum\limits_{k=1}^{n}\max\limits_{\omega\in U^{n}(0, \delta)}\left|\frac{\partial f}{\partial z_{k}}(\omega)\right|+I\leq C+I. \end{split} \end{equation} $

(2.1)

因$\displaystyle\lim\limits_{z\rightarrow\partial U^{n}}\frac{1}{Q(z)}=0$, 所以对上述$\varepsilon>0$, $\exists ~\eta\in (0, 1)$, 当${\rm dist}(z, \partial U^{n})<\eta$时,

$ \begin{equation} \begin{split} \frac{C}{Q(z)}<\varepsilon, \frac{C\|f\|_{B_{\log}}}{Q(z)}<\varepsilon. \end{split} \end{equation} $

(2.2)

固定$k~(1\leq k\leq n)$, 当$|z_{k}|<\delta$时,

$ \begin{align*} I&\leq\|f\|_{B_{\log}}\sum\limits_{k=1}^{n}\int_{\delta}^{1}\frac{|z_{k}|} {(1-t|z_{k}|)\ln\frac{2}{1-t|z_{k}|}}dt\\ &=\|f\|_{B_{\log}}\sum\limits_{k=1}^{n} \left(\ln(\ln\frac{2}{1-|z_{k}|})-\ln(\ln\frac{2}{1-\delta|z_{k}|})\right) <C\|f\|_{B_{\log}}, \end{align*} $

所以

$ \begin{equation} \begin{split} \frac{I}{Q(z)}<\frac{C\|f\|_{B_{\log}}}{Q(z)}<\varepsilon, \end{split} \end{equation} $

(2.3)

所以当$|z_{k}|>\delta$时,

$\begin{align*} I=\sum\limits_{k=1}^{n}\int_{\delta}^{1}\frac{|z_{k}|}{(1-t|z_{k}|)\ln\frac{2}{1-t|z_{k}|}} \left(1-t^{2}|z_{k}|^{2}\right) \ln\frac{2}{1-t|z_{k}|}\left|\frac{\partial f}{\partial z_{k}}(tz)\right|dt<C\varepsilon, \end{align*} $

从而

$ \begin{equation} \begin{split} \frac{I}{Q(z)}<C\varepsilon, \end{split} \end{equation} $

(2.4)

于是由(2.1)-(2.4)式, 结论成立.证毕.

本部分给出了加权复合算子$W_{\psi, \varphi}:B_{\log}(U^{n})\rightarrow B_{\log}(U^{n})$为有界算子和紧算子的充要条件.

定理3.1 设$\psi\in H(U^{n})$, $\varphi\in H(U^{n}, U^{n})$.则$W_{\psi, \varphi}$为$B_{\log}(U^{n})$上的有界算子的充要条件是

(ⅰ) $\displaystyle\sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|} $$\ln(\ln\frac{2}{1-|\varphi_{l}(z)|})\left|\frac{\partial\psi}{\partial z_{k}}(z)\right|<\infty$;

(ⅱ) $\displaystyle\sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2}) \ln\frac{2}{1-|z_{k}|}}{(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}}$$\left|\psi(z)\frac{\partial\varphi_{l}}{\partial z_{k}}(z)\right|<\infty$.

证 对任意的$f\in B_{\log}(U^{n})$, 由结论(ⅰ)-(ⅱ)及引理2.1得

$ \begin{align*} \|W_{\psi, \varphi}f\|_{\log}&\leq\sup\limits_{z\in U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)f(\varphi(z))\right|\\ &~~~~~+\sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n} \left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial f}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\\ &\leq\sup\limits_{z\in U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|\left (2+\sum\limits_{l=1}^{n}\ln(\ln\frac{2}{1-|\varphi_{l}(z)|})\right )\|f\|_{B_{\log}}\\ &~~~~~+\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}} {(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\|f\|_{B_{\log}}\\ &\leq C\|f\|_{B_{\log}}. \end{align*} $

所以$\|W_{\psi, \varphi}f\|_{B_{\log}}=|\psi(0)f(\varphi(0))|+\|W_{\psi, \varphi}f\|_{\log}\leq C\|f\|_{B_{\log}}$, 于是$W_{\psi, \varphi}$是$B_{\log}(U^{n})$上的有界算子.

反之, 设$W_{\psi, \varphi}$为$B_{\log}(U^{n})$上的有界算子, 则对任意的$f\in B_{\log}(U^{n})$, 都存在常数$C$, 使得$\|W_{\psi, \varphi}\|_{B_{\log}}\leq C\|f\|_{B_{\log}}$.分别取$f(z)=1$和$f(z)=z_{l}~(l=1, 2, \cdots, n)$, 则$\psi\in B_{\log}(U^{n})$, $\psi\varphi_{l}\in B_{\log}(U^{n})$, 结合$|\varphi_{l}(z)|<1$, 有

$ \begin{equation} \begin{split} \sup\limits_{z\in U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|<\infty. \end{split} \end{equation} $

(3.1)

对任意固定的$\omega\in U^{n}$, $|\omega|<1$和固定的$l~(1\leq l\leq n)$, 取测试函数

$ \begin{equation} \begin{split} f_{\omega}(z)=2\ln(\ln\frac{4}{1-\overline{\varphi_{l}(\omega)}z_{l}})- \left(\ln(\ln\frac{4}{1-|\varphi_{l}(\omega)|^{2}})\right)^{-1} \left(\ln(\ln\frac{4}{1-\overline{\varphi_{l}(\omega)}z_{l}})\right)^{2}, \end{split} \end{equation} $

(3.2)

通过计算$f_{\omega}\in B_{\log}(U^{n})$, $\|f_{\omega}\|_{B_{\log}}\leq C$, $\displaystyle f_{\omega}(\varphi(\omega))=\ln(\ln\frac{4}{1-|\varphi_{l}(\omega)|^{2}})$, $\displaystyle\frac{\partial f_{\omega}}{\partial z_{l}}(\varphi(\omega))=0$, 则有

$ \begin{align*} \|W_{\psi, \varphi}f_{\omega}\|_{B_{\log}}&\geq\sum\limits_{k=1}^{n} \left(1-|\omega_{k}|^{2}\right)\ln\frac{2}{1-|\omega_{k}|}\left|\frac{\partial \psi}{\partial \omega_{k}}(\omega)\right|\ln(\ln\frac{4}{1-|\varphi_{l}(\omega)|^{2}})\\ &\geq\sum\limits_{k=1}^{n} \left(1-|\omega_{k}|^{2}\right)\ln\frac{2}{1-|\omega_{k}|}\left|\frac{\partial \psi}{\partial \omega_{k}}(\omega)\right|\ln(\ln\frac{2}{1-|\varphi_{l}(\omega)|}). \end{align*} $

由$\omega$的任意性, 对上式关于$\omega$取上确界, 可得结论(ⅰ)成立.同样的再取测试函数

$ \begin{equation} \begin{split} f_{\omega}(z)=\ln(\ln\frac{2}{1-\overline{\omega} z_{l}}), \end{split} \end{equation} $

(3.3)

通过计算$f_{\omega}\in B_{\log}(U^{n})$, $\|f_{\omega}\|_{B_{\log}}\leq C$, 如果$\forall z\in U^{n}$, $\varphi_{l}(z)\neq0$.令$\omega=\varphi_{l}(z)$, 那么由结论(ⅰ)得到

$ \begin{align*} \sum\limits_{k=1}^{n}\frac{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}} {(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}} \left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|<\infty. \end{align*} $

如果$\forall z\in U^{n}$, $\varphi_{l}(z)=0$, 那么由(3.1)式

$ \begin{align*} &~~~~~\sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n} \frac{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}}{(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}} \left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\\ &=\frac{1}{\ln2}\sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n} \left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|<\infty. \end{align*} $

因此结论(ⅱ)成立.证毕.

定理3.2 设$\psi\in H(U^{n})$, $\varphi\in H(U^{n}, U^{n})$.则$W_{\psi, \varphi}$为$B_{\log}(U^{n})$上的紧算子的充要条件是

(ⅰ) $\displaystyle\lim\limits_{\varphi(z)\rightarrow\partial U^{n}}\sum\limits_{k, l=1}^{n} \left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\ln(\ln\frac{2}{1-|\varphi_{l}(z)|}) \left|\frac{\partial \psi}{\partial z_{k}}(z)\right|=0$;

(ⅱ) $\displaystyle\lim\limits_{\varphi(z)\rightarrow\partial U^{n}}\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2}) \ln\frac{2}{1-|z_{k}|}}{(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}} \left|\psi(z)\frac{\partial\varphi_{l}}{\partial z_{k}}(z)\right|=0$;

(ⅲ) $\psi\in B_{\log}(U^{n})$;

(ⅳ)对任意的$l=1, 2, \cdots, n$, 都有$\displaystyle\sup\limits_{z\in U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right) \ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial\varphi_{l}}{\partial z_{k}}(z)\right|<\infty$.

证 设结论(ⅰ)-(ⅳ)成立.假设$\{f_{n}\}$是$B_{\log}(U^{n})$中在$U^{n}$的任意紧子集上一致趋于零的有界序列, 令$\sup\limits_{n}\|f_{n}\|_{B_{\log}}\leq L$, 由引理2, 仅需证明$\lim\limits_{n\rightarrow\infty}\|W_{\psi, \varphi}f_{n}\|=0$.而当$n\rightarrow\infty$时, $\psi(0)f_{n}(\varphi(0))\rightarrow0$, 因此这等价于证明下面的(3.4), (3.5)式同时成立.

$ \sup\limits_{z\in U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)f_{n}(\varphi(z))\right|\rightarrow 0, n\rightarrow\infty, $

(3.4)

$ \sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial f_{n}}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\rightarrow 0, n\rightarrow\infty. $

(3.5)

首先由结论(ⅰ)-(ⅱ), $\forall\varepsilon>0$, 都$\exists~\delta>0~(0<\delta<1)$, 当${\rm dist}(\varphi(z), \partial U^{n})<\delta$时,

$ \sum\limits_{k, l=1}^{n} \left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\ln(\ln\frac{2}{1-|\varphi_{l}(z)|})\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|<\varepsilon, $

(3.6)

$ \sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2}) \ln\frac{2}{1-|z_{k}|}}{(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}} \left|\psi(z)\frac{\partial\varphi_{l}}{\partial z_{k}}(z)\right|<\varepsilon. $

(3.7)

再设$K=\{\omega\in U^{n}:{\rm dist}(\omega, \partial U^{n})\geq\delta\}$, 则$K$是$U^{n}$的一个紧集, 由假设及文献[4]中的引理3知$\{f_{n}\}$及$\{\frac{\partial f_{n}}{\partial \omega_{l}}\}$在$K$上收敛于零.当$z\in K$时, 结合结论(ⅲ), (ⅳ), 有

$ \sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|} \left|\frac{\partial \psi}{\partial z_{k}}(z)f_{n}(\varphi(z))\right|\leq\|\psi\|_{B_{\log}}\sup\limits_{z\in K}|f_{n}(\varphi(z))|<\varepsilon, $

(3.8)

$ \ \ \ \ \sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial f_{n}}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\nonumber\\ \leq\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\sup\limits_{z\in K}\left|\frac{\partial f_{n}}{\partial \omega_{l}}(\varphi(z))\right|<\varepsilon. $

(3.9)

当$z\in U^{n}\backslash K$时, 由(3.6), (3.7)式并结合引理2.1, 有

$ \ \ \ \ \sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)f_{n}(\varphi(z))\right|\nonumber\\ \leq2\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|\ln(\ln\frac{2}{1-|\varphi_{l}(z)|})\|f_{n}\|_{B_{\log}}<\varepsilon, $

(3.10)

$ \ \ \ \ \sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|} \left|\psi(z)\frac{\partial f_{n}}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\nonumber\\ \leq L\sum\limits_{k, l=1}^{n} \frac{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}}{(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}} \left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|<\varepsilon, $

(3.11)

由(3.8)-(3.11)式可得(3.4), (3.5)式成立.

反之, 如果$W_{\psi, \varphi}$是紧算子, 则$W_{\psi, \varphi}$是有界算子, 分别取函数$f(z)=1$及$f(z)=z_{l}$, 可得结论(ⅲ), (ⅳ)成立, 下证结论(ⅰ)成立.采用反证法, 假设(ⅰ)不成立, 则存在$\varepsilon_{0}>0$, $\{z^{m}\}\subset U^{n}$, $\varphi(z^{m})=(\varphi_{1}(z^{m}), \varphi_{2}(z^{m}), \cdots, \varphi_{n}(z^{m}))$, 当$m\rightarrow\infty$时, $\varphi(z^{m})\rightarrow \partial U^{n}$, 使得

$ \begin{align*} \sum\limits_{k, l=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \ln(\ln\frac{2}{1-|\varphi_{l}(z_k^m)|})\left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|\geq n\varepsilon_{0}. \end{align*} $

不妨设

$ \begin{align*} \sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \ln(\ln\frac{2}{1-|\varphi_{1}(z_k^m)|})\left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|\geq \varepsilon_{0}. \end{align*} $

情形1 如果$\lim\limits_{m\rightarrow\infty}|\varphi_{1}(z^{m})|=1$, 那么取函数

$ \begin{equation} \begin{split} f_{m}(z)=3\frac{(\ln(\ln\frac{4}{1-\overline{\varphi_{1}(z^{m})}z_{1}}))^{2}} {\ln(\ln\frac{4}{1-|\varphi_{1}(z^{m})|^{2}})}-2\frac{(\ln(\ln\frac{4}{1-\overline{\varphi_{1}(z^{m}})z_{1}}))^{3}} {(\ln(\ln\frac{4}{1-|\varphi_{1}(z^{m})|^{2}}))^{2}}, \end{split} \end{equation} $

(3.12)

通过计算$\|f_{m}\|_{B_{\log}}\leq C$且$\{f_{m}\}$在$U^{n}$的任意紧子集上一致收敛于零, 由引理2.2, 当$m\rightarrow\infty$时, $\|W_{\psi, \varphi}f_{m}\|_{B_{\log}}\rightarrow0$.但是

$ \begin{align*} \|W_{\psi, \varphi}f_{m}\|_{B_{\log}}&\geq\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \ln(\ln\frac{4}{1-|\varphi_{1}(z^{m})|^{2}})\left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|\\ &\geq\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \ln(\ln\frac{2}{1-|\varphi_{1}(z^{m})|^{2}})\left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|. \end{align*} $

令$m\rightarrow\infty$得到$0\geq\varepsilon_{0}$, 这与$\varepsilon_{0}>0$矛盾.

情形2 如果$\lim\limits_{m\rightarrow\infty}|\varphi_{1}(z^{m})|<1$, 不妨设$|\varphi_{1}(z^{m})|\leq \lambda<1$, 由于当$m\rightarrow\infty$时, $\varphi(z^{m})\rightarrow\partial U^{n}$, 那么必存在$s~(1<s\leq n)$, 使得$\lim\limits_{m\rightarrow\infty}|\varphi_{s}(z^{m})|=1$.那么取函数

$ \begin{equation} \begin{split} \displaystyle f_{m}(z)=3\frac{(\ln(\ln\frac{4}{1-\overline{\varphi_{s}(z^{m})}z_{s}}))^{2}} {\ln(\ln\frac{4}{1-|\varphi_{s}(z^{m})|^{2}})}-2\frac{(\ln(\ln\frac{4}{1-\overline{\varphi_{s}(z^{m}})z_{s}}))^{3}} {(\ln(\ln\frac{4}{1-|\varphi_{s}(z^{m})|^{2}}))^{2}}, \end{split} \end{equation} $

(3.13)

由情形1的讨论可得

$ \begin{align*} \lim\limits_{m\rightarrow\infty}\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \ln(\ln\frac{2}{1-|\varphi_{s}(z_k^m)|})\left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|=0. \end{align*} $

再取函数$g_{m}(z)$为(3.12), (3.13)式中两个函数之和, 所以$\|g_{m}\|_{B_{\log}}\leq C$且$\{g_{m}\}$在$U^{n}$的任意紧子集上一致收敛于零, 而

$ \begin{align*} \|W_{\psi, \varphi}g_{m}\|_{B_{\log}}&\geq\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \ln(\ln\frac{2}{1-|\varphi_{1}(z^{m})|})\left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|\\ &-\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \ln(\ln\frac{2}{1-|\varphi_{s}(z^{m})|})\left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|. \end{align*} $

令$m\rightarrow\infty$仍然得到矛盾, 因此结论(ⅰ)成立.同样地, 假设结论(ⅱ)不成立.不妨设

$ \begin{align*} \sum\limits_{k=1}^{n}\frac{(1-|z_k^m|^{2}) \ln\frac{2}{1-|z_k^m|}}{(1-|\varphi_{l}(z^{m})|^{2})\ln\frac{2}{1-|\varphi_{l}(z^{m})|}} \left|\psi(z^{m})\frac{\partial\varphi_{l}}{\partial z_{k}}(z^{m})\right|\geq\varepsilon_{0}. \end{align*} $

对于情形1, 设$\varphi_{1}(z^{m})=r_{1}e^{i\theta_{1}}$, 令函数

$ \begin{align*} h_{m}(z)=\int_{0}^{z_{1}}\left(\frac{r_{1}}{1-r_{1}e^{-i\theta_{1}}z}-\frac{r_1^2}{1-r_1^2e^{-i\theta_{1}}z}\right) \left(\ln\frac{4}{1-r_1^2e^{-i\theta_{1}}z}\right)^{-1}dz, \end{align*} $

通过计算$\|h_{m}\|_{B_{\log}}\leq C$且$\{h_{m}\}$在$U^{n}$的任意紧子集上一致收敛于零, 而由引理2.1, 有

$ \begin{align*} \|W_{\psi, \varphi}h_{m}\|_{B_{\log}} &\geq\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|}\left|\psi(z^{m})\frac{\partial \varphi_{1}}{\partial z_{k}}(z^{m})\right|\left(\frac{r_{1}}{1-r_1^2}-\frac{r_1^2}{1-r_1^3}\right)\\& \left(\ln\frac{4}{1-r_1^3}\right)^{-1}\\ &~~~~~-2\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|}\ln(\ln\frac{2}{1-|\varphi_{1}(z^{m})|}) \left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|\|h_{m}\|_{B_{\log}}\\ &\geq\sum\limits_{k=1}^{n}\frac{(1-|z_k^m|^{2})\ln\frac{2}{1-|z_k^m|}|\psi(z^{m})\frac{\partial \varphi_{1}}{\partial z_{k}}(z^{m})||\varphi_{1}(z^{m})|}{6(1-|z_k^m|^{2})\ln\frac{2}{1-|z_k^m|}}\\ &~~~~~-C\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|}\ln(\ln\frac{2}{1-|\varphi_{1}(z^{m})|}) \left|\frac{\partial \psi}{\partial z_{k}}(z^{m})\right|. \end{align*} $

令$m\rightarrow\infty$, 由引理2.2并结合结论(ⅰ)得到矛盾.对于情形2, 由于$|\varphi_{1}(z^{m})|\leq\lambda<1$, 所以

$ \begin{align*} &~~~~~\sum\limits_{k=1}^{n}\frac{(1-|z_k^m|^{2})\ln\frac{2}{1-|z_k^m|}}{(1-|\varphi_{1}(z^{m})|^{2}) \ln\frac{2}{1-|\varphi_{1}(z^{m})|}}\left|\psi(z^{m})\frac{\partial \varphi_{1}}{\partial z_{k}}(z^{m})\right|\\ &\leq\frac{1}{\ln2(1-\lambda^{2})}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|}\left|\psi(z^{m})\frac{\partial \varphi_{1}}{\partial z_{k}}(z^{m})\right|. \end{align*} $

由于紧算子必是有界算子, 则由定理3.1的结论(ⅱ)得到

$ \begin{align*} \lim\limits_{m\rightarrow\infty}\sum\limits_{k=1}^{n}\left(1-|z_k^m|^{2}\right)\ln\frac{2}{1-|z_k^m|} \left|\psi(z^{m})\frac{\partial \varphi_{1}}{\partial z_{k}}(z^{m})\right|=0. \end{align*} $

这也得到了矛盾, 结论(ⅱ)成立.证毕.

本部分给出了加权复合算子$W_{\psi, \varphi}:B_{0, \log}(U^{n})\rightarrow B_{0, \log}(U^{n})$为有界算子和紧算子的充要条件.

定理4.1 设$\psi\in H(U^{n})$, $\varphi\in H(U^{n}, U^{n})$.则$W_{\psi, \varphi}$为$B_{0, \log}(U^{n})$上的有界算子的充要条件是

(ⅰ) $\displaystyle\sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|} \ln(\ln\frac{2}{1-|\varphi_{l}(z)|})\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|<\infty;$

(ⅱ) $\displaystyle\sup\limits_{z\in U^{n}}\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2}) \ln\frac{2}{1-|z_{k}|}}{(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}} \left|\psi(z)\frac{\partial\varphi_{l}}{\partial z_{k}}(z)\right|<\infty;$

(ⅲ) $\psi\in B_{0, \log}(U^{n});$

(ⅳ)对任意的$l=1, 2, \cdots, n$都有$\displaystyle\lim\limits_{z\rightarrow\partial U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|} \left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|=0.$

证 设结论(ⅰ)-(ⅳ)成立.若$f\in B_{0, \log}(U^{n})$, 根据引理2.3, $\forall\varepsilon>0$, $\exists~\delta_{1}\in (0, 1)$, 当$|z_{k}|>\delta_{1}$, $z_{k}\in U^{n}$, $k=1, 2, \cdots, n$时, 有$\displaystyle|f(z)|<\varepsilon\sum\limits_{k=1}^{n}\left|\ln(\ln\frac{2}{1-|z_{k}|})\right|$.因此当$|\varphi_{l}(z)|>\delta_{1}$时, 根据结论(ⅰ)得到

$ \begin{equation} \begin{split} &\sum\limits_{k=1}^{n}(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)f(\varphi(z))\right|\\ <&\varepsilon\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|\left|\ln(\ln\frac{2}{1-|\varphi_{l}(z)|})\right|<C\varepsilon. \end{split} \end{equation} $

(4.1)

再由结论(ⅲ)知, 对上述$\varepsilon>0$, $\exists~\delta_{2}\in (0, 1)$, 当$|z_{k}|>\delta_{2}$, $z_{k}\in U^{n}, k=1, 2, \cdots, n$时,

$ \begin{align*} \sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|<\varepsilon. \end{align*} $

因此当$|\varphi_{l}(z)|\leq\delta_{1}$且$z_{k}>\delta_{2}$时, 结合引理2.1得到

$ \begin{equation} \begin{split} &\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)f(\varphi(z))\right|\\ \leq&\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|\left(2+\sum\limits_{k=1}^{n}\ln(\ln\frac{2}{1-\delta_{1}})\right)\|f\|_{B_{\log}} <C\varepsilon. \end{split} \end{equation} $

(4.2)

因此由(4.1), (4.2)式, 当$|z_{k}|>\delta^{*}=\max\{\delta_{1}, \delta_{2}\}$时,

$ \begin{equation} \begin{split} \sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial \psi}{\partial z_{k}}(z)f(\varphi(z))\right|<C\varepsilon. \end{split} \end{equation} $

(4.3)

又由于$f\in B_{0, \log}$, 因此对上述$\varepsilon>0$, $\exists~\delta_{3}\in (0, 1)$, 当$|z_{k}|>\delta_{3}$, $z_{k}\in U^{n}$, $k=1, 2, \cdots, n$时, 有

$\begin{align*} \displaystyle\left|\frac{\partial f}{\partial z_{k}}(z)\right|<\frac{\varepsilon}{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}}. \end{align*} $

因此当$|\varphi_{l}(z)|>\delta_{3}$时, 根据结论(ⅱ)得到

$ \begin{equation} \begin{split} &\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial f}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\\ <&\varepsilon\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}} {(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|<C\varepsilon. \end{split} \end{equation} $

(4.4)

再由结论(ⅳ)知, 对上述$\varepsilon>0$, $\exists~\delta_{4}\in (0, 1)$, 当$|z_{k}|>\delta_{4}$, $z_{k}\in U^{n}$, $k=1, 2, \cdots, n$时, 有

$ \begin{align*} \sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|<\varepsilon, \end{align*} $

因此当$|\varphi_{l}(z)|\leq\delta_{3}$且$|z_{k}|>\delta_{4}$, 得到

$ \begin{equation} \begin{split} &\sum\limits_{k, l=1}^{n}(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial f}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\\ \leq&\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}} {(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\|f\|_{B_{\log}}\\ \leq&\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2})\ln\frac{2}{1-|z_{k}|}} {(1-\delta_{3}^{2})\ln2}\left|\psi(z)\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\|f\|_{B_{\log}}\leq C\varepsilon. \end{split} \end{equation} $

(4.5)

因此由(4.4), (4.5)式, 当$|z_{k}|>\delta^{**}=\max\{\delta_{3}, \delta_{4}\}$时, 有

$ \begin{equation} \begin{split} \sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial f}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\leq C\varepsilon. \end{split} \end{equation} $

(4.6)

因此由(4.3)和(4.6)式, 令$\delta=\max\{\delta^{*}, \delta^{**}\}$, 当$|z_{k}|>\delta$时, 有

$ \begin{align*} \sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left(\left|\frac{\partial \psi}{\partial z_{k}}(z)f(\varphi(z))\right| +\sum\limits_{l=1}^{n}\left|\psi(z)\frac{\partial f}{\partial \omega_{l}}(\varphi(z))\frac{\partial \varphi_{l}}{\partial z_{k}}(z)\right|\right)<C\varepsilon. \end{align*} $

即

$ \begin{align*} \lim\limits_{z\rightarrow\partial U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\frac{\partial (W_{\psi, \varphi }f)}{\partial z_{k}}(z)\right|=0. \end{align*} $

因此$W_{\psi, \varphi}f\in B_{0, \log}$.

反之, 假设$W_{\psi, \varphi}$为$B_{0, \log}(U^{n})$上的有界算子.分别取$f(z)=1$和$f(z)=z_{l}$, 得到$\psi, \psi\varphi_{l}\in B_{0, \log}(U^{n})$, 从而结论(ⅲ), (ⅳ)成立, 其余同定理3.1必要性的证明, 此处略去.证毕.

定理4.2 设$\psi\in H(U^{n})$, $\varphi\in H(U^{n}, U^{n})$.则$W_{\psi, \varphi}$为$B_{0, \log}(U^{n})$上的紧算子的充要条件是

(ⅰ) $\displaystyle\lim\limits_{\varphi(z)\rightarrow\partial U^{n}}\sum\limits_{k, l=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|} \ln(\ln\frac{2}{1-|\varphi_{l}(z)|})\left|\frac{\partial \psi}{\partial z_{k}}(z)\right|=0;$

(ⅱ) $\displaystyle\lim\limits_{\varphi(z)\rightarrow\partial U^{n}}\sum\limits_{k, l=1}^{n}\frac{(1-|z_{k}|^{2}) \ln\frac{2}{1-|z_{k}|}}{(1-|\varphi_{l}(z)|^{2})\ln\frac{2}{1-|\varphi_{l}(z)|}} \left|\psi(z)\frac{\partial\varphi_{l}}{\partial z_{k}}(z)\right|=0;$

(ⅲ) $\psi\in B_{0, \log}(U^{n})$;

(ⅳ)对任意的$l=1, 2, \cdots, n$都有$\displaystyle\lim\limits_{z\rightarrow\partial U^{n}}\sum\limits_{k=1}^{n}\left(1-|z_{k}|^{2}\right)\ln\frac{2}{1-|z_{k}|}\left|\psi(z)\frac{\partial\varphi_{l}}{\partial z_{k}}(z)\right|=0.$

证 假设结论(ⅰ)-(ⅳ)成立, 由定理3.2知$W_{\psi, \varphi}$在$B_{\log}(U^{n})$上是紧的, 又$B_{0, \log}(U^{n})$是$B_{\log}(U^{n})$的闭子空间, 由定理4.1的证明知$\forall f\in B_{0, \log}(U^{n})$都有$W_{\psi, \varphi}f\in B_{0, \log}(U^{n})$, 因此$W_{\psi, \varphi}f$在$B_{0, \log}(U^{n})$上是紧的.

反之, 设$W_{\psi, \varphi}f$在$B_{0, \log}(U^{n})$上是紧的, 则结论(ⅲ), (ⅳ)显然成立, 并且由于定理3.2必要性的证明中所取的测试函数(3.12), (3.13)均是$B_{0, \log}(U^{n})$空间上一致有界的函数列, 与定理3.2必要性的证明类似仍能够得到矛盾, 因此结论(ⅰ), (ⅱ)成立.证毕.