记$ {\Bbb B}_n $为$ {\Bbb C}^n $上单位开球, $ \varphi $为$ {\Bbb B}_n $上解析自映射. 记$ H({\Bbb B}_n) $为$ {\Bbb B}_n $上所有解析函数全体, $ H^\infty $为$ {\Bbb B}_n $上所有有界解析函数全体. 设$ u \in H^\infty $, $ f\in H({\Bbb B}_n) $, 定义加权复合算子$ W_{u, \varphi}f = u\cdot(f \circ \varphi) $. 特别地, 当$ u = 1 $时, 即为复合算子$ C_\varphi $. 注意到, $ W_{u, \varphi} = M_uC_\varphi $, 其中, $ M_u $为乘积算子. 记$ {\cal B}(X) $为可分复Hilbert空间上所有有界线性算子全体. 当$ S, T\in {\cal B}(X) $, $ ST $与$ TS $非零且都不是紧时, 若$ [S, T] = ST-TS $紧, 则称交换子$ [S, T] $是非平凡紧. 若$ [T^*, T] = 0 $, 则称$ T\in {\cal B}(X) $正规. 若$ [T^*, T] $紧, 则称$ T\in {\cal B}(X) $本性正规.

许多学者都对一维以及高维线性分式复合算子的交换子感兴趣. Clifford等^[1]研究了当$ \varphi $与$ \psi $是单位圆盘$ {\Bbb D} $的线性分式映射时, Hardy空间$ H^2({\Bbb D}) $上复合算子交换子$ [C^*_\psi, C_\varphi] $非平凡紧性. Jung等^[2]研究了当$ \varphi $与$ \psi $是单位圆盘非自同构线性分式映射时, Hardy空间$ H^2({\Bbb D}) $上加权复合算子交换子$ [W^*_{v, \psi}, W_{u, \varphi}] $紧性. Lacruz等^[3]研究了单位圆盘上线性分式映射复合算子二次交换子性质. 对于多变量情形, 江良英等^[4]研究了线性分式复合算子的本性正规性, 然后第一作者在文献[5] 中研究了单位球Hardy空间$ H({\Bbb B}_n) $上自同构复合算子交换子$ [C^*_\psi, C_\varphi] $紧性.

本文将研究$ \varphi $与$ \psi $是单位球$ {\Bbb B}_n $的非自同构线性分式映射时, Hardy空间$ H({\Bbb B}_n) $上加权复合算子交换子$ [W^*_{v, \psi}, W_{u, \varphi}] $紧性. 第2节给出主要定理的一些准备引理. 第3节给出线性分式加权复合算子交换子$ [W^*_{v, \psi}, W_{u, \varphi}] $的计算公式, 然后讨论当$ \|\varphi\|_\infty = 1 $时$ [W^*_{v, \psi}, W_{u, \varphi}] $的紧性. 本文重点讨论当$ \varphi $在单位球边界$ \partial{\Bbb B}_n $上有一个不动点$ e_1 $情形. 显然, 通过酉变换, 不妨设$ e_1 = (1, 0, \cdots, 0) $.

记$ \sigma $为$ \partial{\Bbb B}_n $上正规Lebesgue测度. 当$ n\geq1 $, 定义Hardy空间为

$ H^2({\Bbb B}_n) = \bigg\{f\in H({\Bbb B}_n): \|f\|^2 = \sup\limits_{0<r<1}\int_{\partial{\Bbb B}_n}|f(r\zeta)|^2{\rm d}\sigma(\zeta)<\infty\bigg\}. $

设$ u $为$ \partial{\Bbb B}_n $上有界可测复值函数, $ H^2({\Bbb B}_n) $上Toeplitz算子定义为$ T_u(f) = P(uf) $, 其中$ P $是$ L^2(\partial{\Bbb B}_n) $到$ H^2({\Bbb B}_n) $的正交投影. 如果$ u $是解析的, 则$ T_u $即$ M_u $. Cowen等^[6]定义单位球$ {\Bbb B}_n $上线性分式映射如下:

定理2.1  设

(2.1) $ \begin{equation} \varphi(z) = \frac{Az+B}{\langle z, C\rangle+D} \end{equation} $

是$ {\Bbb B}_n $的线性分式映射, 其中$ \langle\cdot, \cdot\rangle $表示$ {\Bbb C}^n $中欧式内积, $ A $表示$ n\times n $矩阵, $ B $和$ C $表示$ n\times 1 $列向量, $ D $是一个复数. $ C_\varphi $是$ H^2({\Bbb B}_n) $上由$ \varphi $诱导的复合算子. 则$ C_\varphi $的伴随算子$ C^*_\varphi $为

(2.2) $ \begin{equation} C^*_\varphi = T_gC_{\sigma_\varphi}T^*_h, \end{equation} $

其中

(2.3) $ \begin{equation} \sigma_\varphi(z) = \frac{A^*z-C}{\langle z, -B\rangle+\overline{D}} \end{equation} $

是$ \varphi $的Kreǐn伴随, $ g(z) = (\langle z, -B\rangle+\overline{D})^{-n} $, $ h(z) = (\langle z, C\rangle+D)^n $, $ T_g, T_h $分别是解析Toeplitz算子.

在下文中, 记$ LFM({\Bbb B}_n) $与$ Aut({\Bbb B}_n ) $分别表示$ {\Bbb B}_n $上所有线性分式映射全体与自同构全体. 如果$ \varphi $将$ {\Bbb B}_n $映到自身, 则$ |D| > |C| $, 因此有$ D \neq 0 $, 且$ \varphi $在$ {\Bbb B}_n $的某个邻域内解析. 我们使用下面定义的矩阵验证两个复合算子是否为Kreǐn伴随算子. 设

$ \begin{eqnarray*} m_\varphi = \left(\begin{array}{ccc} A & B\\ C^* & D\end{array}\right), \ \ \ \ \ \ \ \ \ \ m_{\sigma_\varphi} = \left(\begin{array}{ccc} A^* & -C\\ -B^* & \overline{D}\end{array}\right)\\ \end{eqnarray*} $

分别为线性分式映射$ \varphi $及其伴随映射$ \sigma_\varphi $的矩阵表示, 则矩阵$ m_\varphi $与$ m_{\sigma_\varphi} $是Kreǐn伴随矩阵, 即$ m_\varphi = m_{\sigma_\varphi}^\times = Jm^*_{\sigma_\varphi}J $, 其中

$ \begin{eqnarray*} J = \left(\begin{array}{ccc} I& 0\\ 0 & -1\end{array}\right), \end{eqnarray*} $

$ m^*_{\sigma_\varphi} $是Hilbert空间上$ m_\varphi $伴随. 要验证两个复合算子Kreǐn伴随, 只要证明$ m_\varphi $与$ m_{\sigma_\varphi} $是Kreǐn伴随矩阵即可.

下面回顾不动点及其分类.

定理2.2^[7]  设$ \varphi\in LFT({\Bbb B}_n) $在$ {\Bbb B}_n $内部无不动点, 则存在唯一一个不动点$ \zeta\in \partial{\Bbb B}_n $使得$ \varphi(\zeta) = \zeta $以及$ \langle {\rm d}\varphi_\zeta(\zeta), \zeta\rangle = \lambda $其中$ 0<\lambda\leq1 $.

上述定理中的唯一点$ \zeta\in \partial{\Bbb B}_n $称为$ \varphi $的Denjoy-Wolff点, $ \lambda $称为$ \varphi $的边界扩张系数. 根据定理2.2, $ {\Bbb B}_n $的线性分式映射半群分为三类: 如果$ \varphi $在$ {\Bbb B}_n $内部有不动点, 则称$ \varphi $是椭圆型. 如果$ \varphi $在$ {\Bbb B}_n $内部无不动点且$ \lambda $为$ \varphi $的Denjoy-Wolff点的边界扩张系数, 当$ \lambda<1 $时, 称$ \varphi $是双曲型, 当$ \lambda = 1 $时, 称$ \varphi $是抛物型.

江良英等^[4]研究了Hardy空间上复合算子本性正规性时改善文献[8]中其他条件, 获得式(2.4). 本文将在式(2.4)前提下研究加权复合算子交换子.

引理2.1^[4]  设$ \varphi $是$ {\Bbb B}_n $的线性分式映射且满足$ \varphi(e_1) = e_1 $. 对某个$ c > 0 $及$ e_1 $附近一点$ \zeta\in\partial{\Bbb B}_n $, $ \varphi $满足

(2.4) $ \begin{equation} 1-|\varphi(\zeta)|^2\geq c|e_1-\varphi(\zeta)|^2, \end{equation} $

则对每一个复指标$ \alpha $, 算子$ [T^*_{z^\alpha}, C_\varphi] $是$ H^2({\Bbb B}_n) $的紧算子.

为了方便, 下面列出主要定理证明过程中用到的一些引理.

引理2.2^[8]  设$ \varphi $是$ {\Bbb B}_n $的线性分式映射且$ \varphi(e_1) = e_1 $. 对于$ \zeta\in \partial{\Bbb B}_n, \ |\varphi(\zeta)| = 1 $当且仅当$ \zeta = e_1 $. 若$ W(z) $在$ \overline{{\Bbb B}_n} $上连续且$ W(e_1) = 0 $, 则算子$ T_WC_\varphi $是$ H^2({\Bbb B}_n) $的紧算子.

引理2.3^[8]  设$ \varphi $是$ {\Bbb B}_n(n>1) $的线性分式映射, $ \sigma $是$ \varphi $的伴随映射, 由式(2.1)给出. 若$ \zeta, \eta \in\partial{\Bbb B}_n $且$ \varphi(\zeta) = \eta $, 则$ \sigma(\eta) = \zeta. $

引理2.4^[8]  设$ \varphi $是$ {\Bbb B}_n $的线性分式映射且有一个边界不动点. $ [C_\sigma , C_\varphi] $是$ H^2({\Bbb B}_n) $的紧算子当且仅当$ [C_\sigma , C_\varphi] = 0 $, 即$ \varphi\circ\sigma = \sigma\circ\varphi. $

下面用符号$ \psi_\zeta $表示$ \psi $沿$ \zeta $方向坐标, 即$ \psi_\zeta(z) = \langle \psi(z), \zeta\rangle $. 定义$ {\cal D}_\eta\psi_\zeta(z) = \langle \psi'(z)\eta, \zeta\rangle $为$ \psi_\zeta $沿$ \eta $方向导数, 记为$ {\cal D}_\eta\psi_\zeta $. 特别地, 当$ \zeta = \eta = e_1 $, 记为$ {\cal D}_1\psi_1(z) $. 我们给出如下引理:

引理2.5  设$ \varphi\in LFM({\Bbb B}_n)\backslash Aut({\Bbb B}_n) $满足$ \varphi(e_1) = e_1 $, $ e_1\in \partial{\Bbb B}_n $. 设$ \varphi $满足式(2.1), $ \sigma_\varphi $满足式(2.4). 记

$ s = ({\cal D}_1\varphi_1(e_1))^{-n} = \Big(\frac{c_1+\overline{D}}{-\overline{b_1}+\overline{D}}\Big)^{n}, $

其中$ b_1 $, $ c_1 $分别是列向量$ B $和$ C $的首项, $ C_\varphi $是$ H^2({\Bbb B}_n) $上复合算子. 则存在一个紧算子$ K\in H^2({\Bbb B}_n) $使得$ C^*_\varphi = sC_{\sigma_\varphi}+K $, 其中$ \sigma_\varphi $是$ \varphi $的Kreǐn伴随.

证  设$ \sigma_\varphi $如定理2.1中是$ \varphi $伴随, 则$ \sigma_\varphi(e_1) = e_1 $. 由文献[9], 对于$ \overline{\langle z, C\rangle}\in C(\overline{{\Bbb B}_n}) $, 任意非负整数$ j $以及紧算子$ K $, 有$ T_{(\overline{\langle z, C\rangle})^j} = (T_{\overline{\langle z, C\rangle}})^j+K $. 从而有

$ \begin{eqnarray*} T_{\overline{h}}& = &T_{(\overline{\langle z, C\rangle}+{\overline{D}})^n} = T_{C^0_n\overline{D}^n+C^1_n\overline{D}^{n-1}\overline{\langle z, C\rangle}+\cdots+C^n_n(\overline{\langle z, C\rangle})^n}\\ & = &T_{C^0_n\overline{D}^n}+T_{C^1_n\overline{D}^{n-1}\overline{\langle z, C\rangle}}+\cdots+T_{C^n_n(\overline{\langle z, C\rangle})^n}\\ & = &C^0_n(T_{\overline{D}})^n+C^1_n(T_{\overline{D}})^{n-1}T_{\overline{\langle z, C\rangle}}+\cdots+C^n_n(T_{\overline{\langle z, C\rangle}})^n+K\\ & = &(T_{\overline{\langle z, C\rangle}}+T_{\overline{D}})^n+K. \end{eqnarray*} $

于是可得

$ \begin{eqnarray*} & &[C_{\sigma_\varphi}, T^*_h] = C_{\sigma_\varphi}T_{\overline{h}}-T_{\overline{h}}C_{\sigma_\varphi}\\ & = &C_{\sigma_\varphi}((T_{\overline{\langle z, C\rangle}}+T_{\overline{D}})^n+K) -((T_{\overline{\langle z, C\rangle}}+T_{\overline{D}})^n+K)C_{\sigma_\varphi}\\ & = &C_{\sigma_\varphi}\Big(C^0_n\overline{D}^n+C^1_n\overline{D}^{n-1}\sum\limits_{j = 1}^nC_jT^*_{z_j}+\cdots+C^n_n(\sum\limits_{j = 1}^nC_jT^*_{z_j})^n\Big)\\ &&-\Big(C^0_n\overline{D}^n+C^1_n\overline{D}^{n-1}\sum\limits_{j = 1}^nC_jT^*_{z_j}+\cdots+C^n_n(\sum\limits_{j = 1}^nC_jT^*_{z_j})^n\Big)C_{\sigma_\varphi} +[C_{\sigma_\varphi}, K]\\ & = &C^1_n\overline{D}^{n-1}I_1+C^2_n\overline{D}^{n-2}I_2+\cdots+C^n_nI_n+[C_{\sigma_\varphi}, K], \end{eqnarray*} $

其中$ C^j_n\ (j = 1\cdots n) $是二项式展开系数, $ C_j\ (j = 1\cdots n) $是向量$ C $的分向量,

$ I_i = C_{\sigma_\varphi} \bigg(\sum\limits_{j = 1}^nC_jT^*_{z_j}\bigg)^i-\bigg(\sum\limits_{j = 1}^nC_jT^*_{z_j}\bigg)^iC_{\sigma_\varphi}\ (i = 1\cdots n). $

由引理2.1, 可知$ [C_{\sigma_\varphi}, T^*_{z_j}] $是紧算子. 由$ [C_{\sigma_\varphi}, K] $的紧性, 可得$ [C_{\sigma_\varphi}, T^*_h] $是$ H^2({\Bbb B}_n) $上紧算子. 因此由$ T_gT^*_h = T_{g\overline{h}}+K $可得

$ C^*_\varphi = T_gC_{\sigma_\varphi}T^*_h = T_gT^*_hC_{\sigma_\varphi}+K = (T_{g\overline{h}}+K)C_{\sigma_\varphi}+K = T_{g\overline{h}}C_{\sigma_\varphi}+K. $

由引理2.2以及$ W = \overline{h}g-\overline{h(e_1)}g(e_1) $可知

$ T_{\overline{h}g}C_{\sigma_\varphi}-\overline{h(e_1)}g(e_1)C_{\sigma_\varphi} = T_{\overline{h}g-\overline{h(e_1)}g(e_1)}C_{\sigma_\varphi} $

是紧的. 故

$ C^*_\varphi = T_gC_{\sigma_\varphi}T^*_h = \overline{h(e_1)}g(e_1)C_{\sigma_\varphi}+K = sC_{\sigma_\varphi}+K. $

引理2.5证毕.

设$ {\Bbb C}^n $中域$ \Omega $、$ \Omega' $, $ \varphi $、$ \Phi $分别是$ \Omega $、$ \Omega' $上解析映射. 如果存在一个双全纯映射$ \sigma:\Omega\rightarrow \Omega' $使得$ \Phi = \sigma^{-1}\circ\varphi\circ\sigma $, 则称$ \varphi $与$ \Phi $共轭. 我们都知道通过广义Cayley变换$ \sigma_C $, 单位球$ {\Bbb B}_n $双全纯等价于Siegel半平面$ H_n: = \{(w_1, \cdots, w_n) = (w_1, w')\in {\Bbb C}^n, \ \mbox{Re}w_1>|w'|^2\} $, 其中$ \sigma_C $定义为

$ \sigma_C(z) = \Big(\frac{1+z_1}{1-z_1}, \ \frac{z'}{1-z_1}\Big), \ \ \ z = (z_1, z')\in {\Bbb B}_n. $

注意到$ \sigma_C $ ($ \sigma_C(e_1) = \infty $) 是从$ \overline{{\Bbb B}_n} $延拓到$ H_n\bigcup\partial H_n\bigcup\{\infty\} $的双连续映射, $ \overline{H_n} $的一点紧致化.

设$ \varphi\in LFM({\Bbb B}_n) $有唯一一个边界不动点$ e_1 $且边界扩张系数$ \lambda\leq 1 $. 由文献[7, 定理3.2] 知$ \varphi $是具有Denjoy-Wolff点$ e_1 $的非椭圆型. 则由文献[10, 引理4.1]知, $ \varphi $与

(2.5) $ \begin{eqnarray} \Phi(w_1, w') = \frac{1}{\lambda}(w_1+\langle w', b\rangle+c, \ Aw'+d), \ \ \ \ (w_1, w')\in H_n \end{eqnarray} $

共轭, 其中$ c\in{\Bbb C}, b, d\in {\Bbb C}^{n-1}, \lambda \mbox{Re}c\geq |d|^2 $, $ A\in {\Bbb C}^{(n-1)\times(n-1)} $$ \lambda I-A^*A $是Hermitian半正定矩阵. 应用这种形式, 我们得到$ {\Bbb B}_2 $的如下结果.

引理2.6  设$ \varphi, \psi\in LFM({\Bbb B}_2)\backslash Aut({\Bbb B}_2) $是抛物型, 具有相同的边界固定点$ e_1 $. 假设$ \varphi, \psi $分别与$ H_2 $的解析自映射$ \Phi, \Psi $共轭, $ \Phi, \Psi $形式为

$ \begin{eqnarray*} \Phi(w_1, w_2) = (w_1+b_1w_2+c_1, \ a_1w_2+d_1), \\ \Psi(w_1, w_2) = (w_1+b_2w_2+c_2, \ a_2w_2+d_2) , \end{eqnarray*} $

其中$ a_j, b_j, c_j, d_j\in {\Bbb C} $满足

(2.6) $ \begin{eqnarray} |a_2|<1, \ \mbox{Re}c_2>\frac{|b_2|^2}{4}, \ \mbox{Re}(\overline{c_1}-2\overline{d_1}d_2+c_2)>|\overline{a_1}d_2-\frac{\overline{b_1}}{2}|^2. \end{eqnarray} $

则$ \sigma_\psi $与$ \sigma_\varphi\circ\psi $满足式(2.4).

证  直接计算可得

$ \begin{eqnarray*} \varphi(z) = \sigma_C^{-1}\circ\Phi\circ\sigma_C(z) = \frac{A_1z+B_1}{\langle z, C_1\rangle+D_1}, \ \ \ \psi(z) = \sigma_C^{-1}\circ\Psi\circ\sigma_C(z) = \frac{A_2z+B_2}{\langle z, C_2\rangle+D_2}, \end{eqnarray*} $

其中

$ \begin{eqnarray*} A_j = \left(\begin{array}{ccc} 2-c_j & b_j\\ -2d_j & 2a_j\end{array}\right), \ \ \ \ B_j = \left(\begin{array}{ccc} c_j & \\ 2d_j & \end{array}\right), \ \ \ \ C_j = \left(\begin{array}{ccc} -\overline{c_j} & \\ \overline{b_j} & \end{array}\right), \ \ \ \ D_j = c_j+2. \end{eqnarray*} $

于是$ \varphi, \psi $的伴随映射$ \sigma_\varphi, \sigma_\psi $分别为

$ \sigma_\varphi(z) = \frac{A_1^*z-C_1}{\langle z, -B_1\rangle+\overline{D_1}} \ \ \sigma_\psi(z) = \frac{A_2^*z-C_2}{\langle z, -B_2\rangle+\overline{D_2}}, $

我们可以得到分别与$ \sigma_\varphi, \sigma_\psi $共轭的$ H_2 $的解析自映射$ \Delta, \Gamma $, 形式如下

$ \begin{eqnarray*} &&\Delta(w_1, w_2) = \sigma_C\circ\sigma_\varphi\circ\sigma_C^{-1}(w_1, w_2) = (w_1-2\overline{d_1}w_2+\overline{c_1}, \ \overline{a_1}w_2-\frac{\overline{b_1}}{2}), \\ &&\Gamma(w_1, w_2) = \sigma_C\circ\sigma_\psi\circ\sigma_C^{-1}(w_1, w_2) = (w_1-2\overline{d_2}w_2+\overline{c_2}, \ \overline{a_2}w_2-\frac{\overline{b_2}}{2}). \end{eqnarray*} $

由文献[4]中定理4.3, 若$ z = (z_1, z_2) = \sigma_C^{-1}(w)\in\partial{\Bbb B}_2 $, 则$ w = (w_1, w_2) = \sigma_C(z)\in\partial H_2 $且满足$ \mbox{Re} w_1 = |w_2|^2 $, 于是可得

$ \begin{eqnarray*} & &1-|\sigma_\psi(z)|^2-k|e_1-\sigma_\psi(z)|^2 = \frac{4[\mbox{Re}(w_1-2\overline{d_2}w_2+\overline{c_2})-(1+k)|\overline{a_2}w_2- \frac{\overline{b_2}}{2}|^2-k]}{|w_1-2\overline{d_2}w_2+\overline{c_2}+1|^2}\\ & = &\frac{4\{(1-|a_2|^2-k|a_2|^2)|w_2|^2+\mbox{Re}[(\overline{a_2}b_2-2\overline{d_2}+k\overline{a_2}b_2)w_2]+\mbox{Re}\overline{c_2}-\frac{|b_2|^2}{4} -k(1+\frac{|b_2|^2}{4})\}}{|w_1-2\overline{d_2}w_2+\overline{c_2}+1|^2}\\ & = &\frac{4\{(1-|a_2|^2-k|a_2|^2)|z_2|^2+\mbox{Re}[(\overline{a_2}b_2-2\overline{d_2}+k\overline{a_2}b_2)z_2(1-\overline{z_1})]\}} {|1+z_1-2\overline{d_2}z_2+(\overline{c_2}+1)(1-z_1)|^2}\\ &&+\frac{4(\mbox{Re}\overline{c_2}-\frac{|b_2|^2}{4}-k(1+\frac{|b_2|^2}{4}))|1-z_1|^2}{|1+z_1-2\overline{d_2}z_2+(\overline{c_2}+1)(1-z_1)|^2}. \end{eqnarray*} $

与文献[4]中命题3.1和定理4.3的证明类似, 选取

(2.7) $ \begin{eqnarray} k<\min\Big\{\frac{1-|a_2|^2}{|a_2|^2}, \frac{\mbox{Re}\overline{c_2}-\frac{|b_2|^2}{4}}{1+\frac{|b_2|^2}{4}}\Big\}, \end{eqnarray} $

则$ 1-|\sigma_\psi(z)|^2\geq k|e_1-\sigma_\psi(z)|^2 $在$ e_1\in\partial{\Bbb B}_2 $的某个邻域内成立.

通过具体计算, 我们可以得到与$ \sigma_\varphi\circ\psi $共轭的$ H_2 $的解析自映射$ \Lambda $, 形式如下

$ \begin{eqnarray*} \Lambda(w_1, w_2)& = &\sigma_C\circ\sigma_\varphi\circ\psi\circ\sigma_C^{-1}(w_1, w_2) = \sigma_C\circ\sigma_\varphi\circ\sigma_C^{-1}\circ\sigma_C\circ\psi\circ\sigma_C^{-1}(w_1, w_2)\\ & = &\Delta\circ\Psi(w_1, w_2)\\ & = &(w_1+(b_2-2\overline{d_1}a_2)w_2+c_2-2\overline{d_1}d_2+\overline{c_1}, \ \ \overline{a_1}a_2w_2+\overline{a_1}d_2-\frac{\overline{b_1}}{2}). \end{eqnarray*} $

同理可得

$ \begin{eqnarray*} & &1-|\sigma_\varphi\circ\psi(z)|^2-k|e_1-\sigma_\varphi\circ\psi(z)|^2\\ & = &\frac{4[\mbox{Re}(w_1+(b_2-2\overline{d_1}a_2)w_2+c_2-2\overline{d_1}d_2+\overline{c_1})-(1+k)\Big|\overline{a_1}a_2w_2+\overline{a_1}d_2- \frac{\overline{b_1}}{2}\Big|^2-k]}{|w_1+(b_2-2\overline{d_1}a_2)w_2+c_2-2\overline{d_1}d_2+\overline{c_1}+1|^2}\\ & = &\frac{4\{(1-|a_1a_2|^2-k|a_1a_2|^2)|w_2|^2+\mbox{Re}[b_2-2\overline{d_1}a_2-(1+k)(2|a_1|^2a_2\overline{d_2}-\overline{a_1}a_2b_1)]w_2\}} {|w_1+(b_2-2\overline{d_1}a_2)w_2+c_2-2\overline{d_1}d_2+\overline{c_1}+1|^2}\\ &&+\frac{4\{\mbox{Re}(\overline{c_1}-2\overline{d_1}d_2+c_2)-|\overline{a_1}d_2-\frac{\overline {b_1}}{2}|^2-k(1+|\overline{a_1}d_2-\frac{\overline{b_1}}{2}|^2)\}} {|w_1+(b_2-2\overline{d_1}a_2)w_2+c_2-2\overline{d_1}d_2+\overline{c_1}+1|^2}. \end{eqnarray*} $

与文献[4]中命题3.1和定理4.3的证明类似, 选取

(2.8) $ \begin{eqnarray} k<\min\bigg\{\frac{1-|a_1a_2|^2}{|a_1a_2|^2}, \frac{\mbox{Re}(\overline{c_1}-2\overline{d_1}d_2+c_2)-|\overline{a_1}d_2-\frac{\overline{b_1}}{2}|^2} {1+|\overline{a_1}d_2-\frac{\overline{b_1}}{2}|^2}\bigg\}, \end{eqnarray} $

则$ 1-|\sigma_\varphi\circ\psi(z)|^2\geq k|e_1-\sigma_\varphi\circ\psi(z)|^2 $在$ e_1\in\partial{\Bbb B}_2 $的某个邻域内成立.

结合式(2.5), (2.7)与(2.8), 可得

$ |a_2|<1, \ \mbox{Re}c_2>\frac{|b_2|^2}{4}, \ \mbox{Re}(\overline{c_1}-2\overline{d_1}d_2+c_2)>|\overline{a_1}d_2-\frac{\overline{b_1}}{2}|^2. $

引理2.6证毕.

引理2.7  设$ \varphi, \psi\in LFM({\Bbb B}_2)\backslash Aut({\Bbb B}_2) $是双曲型, 具有相同的边界固定点$ e_1 $. 假设$ \varphi, \psi $分别与$ H_2 $的解析自映射$ \Phi, \Psi $共轭, $ \Phi, \Psi $具有如下形式

$ \begin{eqnarray*} &&\Phi(w_1, w_2) = \frac{1}{\lambda_1}(w_1+b_1w_2+c_1, \ a_1w_2+d_1), \\ &&\Psi(w_1, w_2) = \frac{1}{\lambda_2}(w_1+b_2w_2+c_2, \ a_2w_2+d_2), \end{eqnarray*} $

其中$ a_j, b_j, c_j, d_j\in {\Bbb C} $满足

(2.9) $ \begin{eqnarray} |a_2|^2<\lambda_2, \ \lambda_2\mbox{Re}c_2>\frac{|b_2|^2}{4}, \ \frac{\lambda_1}{\lambda_2}\mbox{Re}(\overline{c_1}-2\overline{d_1}d_2+c_2)>|\frac{\overline{a_1}d_2}{\lambda_2}-\frac{\overline{b_1}}{2}|^2, \end{eqnarray} $

则$ \sigma_\psi $与$ \sigma_\varphi\circ\psi $满足式(2.4).

证  与引理2.6同理, 选取

(2.10) $ \begin{equation} k<\min\bigg\{\frac{\lambda_2-|a_2|^2}{|a_2|^2}, \frac{\lambda_2\mbox{Re}\overline{c_2}-\frac{|b_2|^2}{4}}{1+\frac{|b_2|^2}{4}}\bigg\}, \end{equation} $

则$ 1-|\sigma_\psi(z)|^2\geq k|e_1-\sigma_\psi(z)|^2 $在$ e_1\in\partial{\Bbb B}_2 $的某个邻域内成立. 选取

(2.11) $ \begin{equation} k<\min\bigg\{\frac{\lambda_1\lambda_2-\lambda_2|a_1a_2|^2}{|a_1a_2|^2}, \frac{\frac{\lambda_1}{\lambda_2}\mbox{Re}(\overline{c_1}-2\overline{d_1}d_2+c_2)-|\frac{\overline{a_1}d_2}{\lambda_2}-\frac{\overline{b_1}}{2}|^2} {1+|\frac{\overline{a_1}d_2}{\lambda_2}-\frac{\overline{b_1}}{2}|^2}\bigg\}, \end{equation} $

则$ 1-|\sigma_\varphi\circ\psi(z)|^2\geq k|e_1-\sigma_\varphi\circ\psi(z)|^2 $在$ e_1\in\partial{\Bbb B}_2 $的某个邻域内成立. 结合式(2.5), (2.10) 以及(2.11), 可得

$ |a_2|^2<\lambda_2, \ \lambda_2\mbox{Re}c_2>\frac{|b_2|^2}{4}, \ \frac{\lambda_1}{\lambda_2}\mbox{Re}(\overline{c_1}-2\overline{d_1}d_2+c_2)>|\frac{\overline{a_1}d_2}{\lambda_2}-\frac{\overline{b_1}}{2}|^2. $

引理2.7证毕.

我们以定理2.3结束本节, 该定理可在文献[11]中找到. 后面我们将使用定理2.3来证明定理3.2.

定理2.3  假设$ \varphi $与$ \psi $是$ {\Bbb B}_n $的线性分数映射, 且$ C_\varphi-C_\psi $是$ H^2({\Bbb B}_n) $的紧算子, 则$ \varphi = \psi $或者$ \|\varphi\|_\infty<1 $且$ \|\psi\|_\infty<1 $, 因此$ C_\varphi $和$ C_\psi $都是$ H^2({\Bbb B}_n) $的紧算子.

本节给出加权复合算子的交换子方程, 然后研究线性分式映射分别为抛物型和双曲型时交换子的紧性.

定理3.1  设$ \varphi, \psi\in LFM({\Bbb B}_n)\backslash Aut({\Bbb B}_n) $, 且满足$ \varphi(e_1) = \psi(e_1) = e_1 $, $ e_1\in\partial{\Bbb B}_n $. 设$ \sigma_\psi, \sigma_\varphi\circ\psi $满足式(2.4). $ W_{u, \varphi} $与$ W_{v, \psi} $是$ H^2({\Bbb B}_n) $上分别由$ \varphi $与$ \psi $诱导的加权复合算子, 若$ u, v\in H^\infty({\Bbb B}_n) $在$ \partial{\Bbb B}_n $上连续, 则

$ \begin{eqnarray*} [W^*_{v, \psi}, W_{u, \varphi}]\equiv({\cal D}_1\psi_1(e_1))^{-n}u(e_1)\overline{v(e_1)}[C_{\sigma_\psi}, C_\varphi]({\rm mod}\; K), \end{eqnarray*} $

其中$ K $是$ H^2({\Bbb B}_n) $上紧算子.

证  由引理2.5知$ C^*_\psi = ({\cal D}_1\psi_1(e_1))^{-n} C_{\sigma_\psi}+K. $因此

$ \begin{eqnarray*} [W^*_{v, \psi}, W_{u, \varphi}]& = &C^*_\psi T^*_v T_u C_\varphi-T_uC_\varphi C^*_\psi T^*_v\\ & = &({\cal D}_1\psi_1(e_1))^{-n}(C_{\sigma_\psi}T^*_vT_uC_\varphi-T_uC_\varphi C_{\sigma_\psi}T^*_v)+K. \end{eqnarray*} $

由引理2.2知, $ T^*_vC_\varphi-\overline{v(e_1)}C_\varphi $和$ T_uC_\varphi-u(e_1)C_\varphi $是$ H^2({\Bbb B}_n) $上紧算子, 因此

$ \begin{eqnarray*} [W^*_{v, \psi}, W_{u, \varphi}] & = &({\cal D}_1\psi_1(e_1))^{-n}u(e_1)(\overline{v(e_1)}C_{\sigma_\psi}C_\varphi-C_\varphi C_{\sigma_\psi}T^*_v)+K\\ & = &({\cal D}_1\psi_1(e_1))^{-n}u(e_1)(\overline{v(e_1)}C_{\varphi\circ\sigma_\psi}-C_{\sigma_\psi\circ\varphi}T^*_v)+K. \end{eqnarray*} $

通过具体计算得

$ \begin{eqnarray*} m_{\sigma_\psi\circ\varphi}& = &m_{\sigma_\psi}m_\varphi = \left(\begin{array}{ccc} A_2^*{\quad} & -C_2\\ -B^*_2{\quad} & \overline{D_2}\end{array}\right)\left(\begin{array}{ccc} A_1{\quad} & B_1\\ C^*_1{\quad} & D_1\end{array}\right)\\ & = &\left(\begin{array}{ccc} A_2^*A_1-C_2C_1^*{\quad} & A_2^*B_1-C_2D_1\\ -B^*_2A_1+\overline{D_2}C^*_1{\quad} & -B^*_2 B_1+D_1\overline{D_2}\end{array}\right), \\ m_{\sigma_\varphi\circ\psi} & = &\left(\begin{array}{ccc} A_1^*A_2-C_1C_2^*{\quad} & A_1^*B_2-C_1D_2\\ -B^*_1A_2+\overline{D_1}C^*_2{\quad} & -B^*_1 B_2+\overline{D_1}D_2\end{array}\right), \end{eqnarray*} $

从而有$ m_{\sigma_\psi\circ\varphi} = Jm^*_{\sigma_\varphi\circ\psi}J, $即$ m_{\sigma_\psi\circ\varphi} $与$ m_{\sigma_\varphi\circ\psi} $是Kreǐn伴随矩阵. 因此

$ C^*_{\sigma_\psi\circ\varphi} = ({\cal D}_1(\sigma_\psi\circ\varphi)_1(e_1))^{-n}C_{\sigma_\varphi\circ\psi}+K. $

再由引理2.2, 可知$ T_vC_{\sigma_\varphi\circ\psi}-v(e_1)C_{\sigma_\varphi\circ\psi} $是$ H^2({\Bbb B}_n) $上紧算子. 因此

$ \begin{eqnarray*} C_{\sigma_\psi\circ\varphi}T^*_v = (T_vC^*_{\sigma_\psi\circ\varphi})^* & = &({\cal D}_1(\sigma_\psi\circ\varphi)_1(e_1)^{-n}C^*_{\sigma_\varphi\circ\psi}T^*_v+K\\ & = &\overline{v(e_1)}({\cal D}_1(\sigma_\psi\circ\varphi)_1(e_1)^{-n}C^*_{\sigma_\varphi\circ\psi}+K\\ & = &\overline{v(e_1)}C_{\sigma_\psi\circ\varphi}+K. \end{eqnarray*} $

从而得到

$ [W^*_{v, \psi}, W_{u, \varphi}]\equiv({\cal D}_1\psi_1(e_1))^{-n}u(e_1)\overline{v(e_1)}[C_{\sigma_\psi}, C_\varphi]({\rm mod}\; K). $

定理3.1得证.

从上面的定理3.1, 我们得到当$ \varphi\in LFM({\Bbb B}_n)\backslash Aut({\Bbb B}_n) $时加权复合算子自交换子公式.

推论3.1   (ⅰ) 设$ \varphi\in LFM({\Bbb B}_n)\backslash Aut({\Bbb B}_n) $, 且满足$ \varphi(e_1) = e_1 $, $ e_1\in\partial{\Bbb B}_n $. 设$ \sigma_\varphi, \sigma_\varphi\circ\varphi $满足式(2.4). $ W_{u, \varphi} $是$ H^2({\Bbb B}_n) $上由$ \varphi $诱导加权复合算子. 若$ H^\infty({\Bbb B}_n)\ni u $在$ \partial{\Bbb B}_n $上连续, 则

$ \begin{eqnarray*} [W^*_{u, \varphi}, W_{u, \varphi}]\equiv({\cal D}_1\varphi_1(e_1))^{-n}|u(e_1)|^2[C_{\sigma_\varphi}, C_\varphi]({\rm mod}\; K). \end{eqnarray*} $

(ⅱ) 设$ \varphi, \psi\in LFM({\Bbb B}_n)\backslash Aut({\Bbb B}_n) $, 且满足$ \varphi(e_1) = \psi(e_1) = e_1 $, $ e_1\in\partial{\Bbb B}_n $. 设$ \sigma_\psi, \sigma_\varphi\circ\psi $满足式(2.4). $ C_\varphi $与$ C_\psi $是$ H^2({\Bbb B}_n) $上分别由$ \varphi $与$ \psi $诱导的复合算子.则

$ \begin{eqnarray*} [C^*_\psi, C_\varphi]\equiv({\cal D}_1\psi_1(e_1))^{-n}[C_{\sigma_\psi}, C_\varphi]({\rm mod}\; K). \end{eqnarray*} $

证  (ⅰ) 在定理3.1中取$ u = v, \varphi = \psi $即可.

(ⅱ) 在定理3.1中取$ u = v = 1 $即可.

下面我们刻画$ H^2({\Bbb B}_n) $上$ [W^*_{v, \psi}, W_{u, \varphi}] $紧性的充要条件.

定理3.2   设$ \varphi, \psi\in LFM({\Bbb B}_n)\backslash Aut({\Bbb B}_n) $, 且满足$ \varphi(e_1) = \psi(e_1) = e_1 $, $ e_1\in\partial{\Bbb B}_n $. 设$ \sigma_\psi, \sigma_\varphi\circ\psi $满足式(2.4). $ W_{u, \varphi} $与$ W_{v, \psi} $是$ H^2({\Bbb B}_n) $上分别由$ \varphi $与$ \psi $诱导的加权复合算子. 若$ H^\infty({\Bbb B}_n)\ni u, v $在$ \partial{\Bbb B}_n $上连续, 则$ [W^*_{v, \psi}, W_{u, \varphi}] $是$ H^2({\Bbb B}_n) $上紧算子的充要条件是下面两个条件中至少一个成立.

(ⅰ) $ u(e_1)v(e_1) = 0 $,

(ⅱ) $ \varphi\circ\sigma_\psi = \sigma_\psi\circ\varphi $.

证  由定理3.1知, $ [W^*_{v, \psi}, W_{u, \varphi}] $紧当且仅当$ ({\cal D}_1\psi_1(e_1))^{-n}u(e_1)\overline{v(e_1)}[C_{\sigma_\psi}, C_\varphi] $是紧的.

首先, 假设$ ({\cal D}_1\psi_1(e_1))^{-n}u(e_1)\overline{v(e_1)}[C_{\sigma_\psi}, C_\varphi] $是紧的, 则$ u(e_1)\overline{v(e_1)} = 0 $或者$ [C_{\sigma_\psi}, C_\varphi] $是紧的. 于是有$ u(e_1)v(e_1) = 0 $或者$ C_{\varphi\circ\sigma_\psi}-C_{\sigma_\psi\circ\varphi} $是$ H^2({\Bbb B}_n) $上紧算子. 由定理2.3知$ \varphi\circ\sigma_\psi = \sigma_\psi\circ\varphi $或者$ \|\varphi\circ\sigma_\psi\|_\infty<1, \ \|\sigma_\psi\circ\varphi\|_\infty<1. $再由定理条件知$ \varphi\circ\sigma_\psi(e_1) = e_1, \sigma_\psi\circ\varphi(e_1) = e_1 $, 所以$ \|\varphi\circ\sigma_\psi\|_\infty<1 $与$ \|\sigma_\psi\circ\varphi\|_\infty<1 $不成立. 另外, 线性分式映射$ \varphi\circ\sigma_\psi $与$ \sigma_\psi\circ\varphi $的角导数都是有限的. 由文献[12] 的定理3.43知, $ C_{\varphi\circ\sigma_\psi} $与$ C_{\sigma_\psi\circ\varphi} $都不是紧的, 从而必要条件得证.

另一方面, 如果(ⅰ)、(ⅱ)中至少有一个成立, 显然$ ({\cal D}_1\psi_1(e_1))^{-n}u(e_1)\overline{v(e_1)}[C_{\sigma_\psi}, C_\varphi] $是紧的, 从而充分性得证.

推论3.2   设$ \varphi, \psi\in LFM({\Bbb B}_n)\backslash Aut({\Bbb B}_n) $, 且满足$ \varphi(e_1) = \psi(e_1) = e_1 $, $ e_1\in\partial{\Bbb B}_n $. 设$ \sigma_\psi, \sigma_\varphi\circ\psi $满足式(2.4). $ W_{u, \varphi} $与$ W_{v, \psi} $是$ H^2({\Bbb B}_n) $上分别由$ \varphi $与$ \psi $诱导的加权复合算子. 设$ H^\infty({\Bbb B}_n)\ni u, v $在$ \partial{\Bbb B}_n $上连续且满足$ u(e_1)v(e_1)\neq0 $. 若$ \varphi $是抛物型(双曲型), $ \psi $是双曲型(抛物型), 则$ [W^*_{v, \psi}, W_{u, \varphi}] $非紧.

证  假设$ \varphi $与$ \psi $分别是抛物型与双曲型. 由定理3.1, 可得

$ \begin{eqnarray*} [W^*_{v, \psi}, W_{u, \varphi}]\equiv({\cal D}_1\psi_1(e_1))^{-n}u(e_1)\overline{v(e_1)}[C_{\sigma_\psi}, C_\varphi]({\rm mod}\; K). \end{eqnarray*} $

因为$ \psi $是双曲型, 所以$ {\cal D}_1({\sigma_\psi})_1(e_1) = \frac{1}{\lambda}>1 $, 即$ \sigma_\psi $不是双曲型. 由文献[7] 中定理3.2可知, $ \sigma_\psi $有一个孤立不动点$ \alpha\in{\Bbb B}_n $. 若$ \sigma_\psi\circ\varphi = \varphi\circ\sigma_\psi $, 则$ \sigma_\psi\circ\varphi(\alpha) = \varphi\circ\sigma_\psi(\alpha) = \varphi(\alpha), $上式说明$ \varphi(\alpha) $也是$ \sigma_\psi $的一个不动点. 由于$ \varphi $在$ {\Bbb B}_n $内部没有不动点, 即$ \varphi(\alpha)\neq \alpha $. 由引理2.3可知若$ \psi $在$ \partial {\Bbb B}_n $上无不动点, 所以$ \sigma_\psi $在$ \partial {\Bbb B}_n $上也没有不动点. 综上讨论说明$ \sigma_\psi $存在$ \alpha\in\partial{\Bbb B}_n $与$ \varphi(\alpha)\in{\Bbb B}_n $两个不动点, 与$ \sigma_\psi $有一个不动点矛盾. 因此, $ \sigma_\psi\circ\varphi\neq\varphi\circ\sigma_\psi $. 通过定理3.2知$ [W^*_{v, \psi}, W_{u, \varphi}] $非紧. 当$ \varphi $为双曲型, $ \psi $为抛物型时, 同理可证.

注3.1   由推论3.2, 我们判断$ [W^*_{v, \psi}, W_{u, \varphi}] $的紧性, 只要考虑$ \varphi $和$ \psi $都是抛物型或都是双曲型即可.

现在利用引理2.6和引理2.7研究线性分式映射$ \varphi, \psi\in LFM({\Bbb B}_2)\backslash Aut({\Bbb B}_2) $诱导的加权复合算子交换子$ [W^*_{v, \psi}, W_{u, \varphi}] $的紧性.

定理3.3   设$ \varphi, \psi\in LFM({\Bbb B}_2)\backslash Aut({\Bbb B}_2) $都是抛物型且具有相同边界不动点$ e_1 $. 设$ \varphi, \psi $分别与$ H_2 $的解析自映射$ \Phi, \Psi $共轭, $ \Phi, \Psi $具有如下形式

$ \begin{eqnarray*} && \Phi(w_1, w_2) = (w_1+b_1w_2+c_1, \ a_1w_2+d_1), \\ &&\Psi(w_1, w_2) = (w_1+b_2w_2+c_2, \ a_2w_2+d_2), \end{eqnarray*} $

其中$ a_j, b_j, c_j, d_j\in {\Bbb C} $满足式(2.6). $ W_{u, \varphi} $与$ W_{v, \psi} $是$ H^2({\Bbb B}_2) $上分别由$ \varphi $与$ \psi $诱导的加权复合算子. 若$ H^\infty({\Bbb B}_n)\ni u, v $在$ \partial{\Bbb B}_2 $上连续且满足$ u(e_1)v(e_1)\neq 0 $, 则$ [W^*_{v, \psi}, W_{u, \varphi}] $是紧的当且仅当

$ \left\{ \begin{array}{ll} b_1(1-\overline{a_2}) = 2(a_1-1)\overline{d_2}, \\ \overline{b_2}(1-a_1) = 2(\overline{a_2}-1)d_1. \end{array} \right. $

证  由引理2.6可知, $ \sigma_\psi \ \mbox{与}\ \sigma_\varphi\circ\psi $都满足式(2.4). 通过具体计算

$ \begin{eqnarray*} &&\Phi\circ\Gamma(w_1, w_2) = (w_1+(b_1\overline{a_2}-2\overline{d_2})w_2+\overline{c_2}+c_1-\frac{b_1\overline{b_2}}{2}, \ a_1\overline{a_2}w_2-\frac{a_1\overline{b_2}}{2}+d_1), \\ &&\Gamma\circ\Phi(w_1, w_2) = (w_1+(b_1-2\overline{d_2}a_1)w_2+\overline{c_2}+c_1-2d_1\overline{d_2}, \ a_1 \overline{a_2}w_2-\frac{\overline{b_2}}{2}+d_1\overline{a_2}). \end{eqnarray*} $

因为$ u(e_1)v(e_1)\neq0 $, 由定理3.2以及上式, $ [W^*_{v, \psi}, W_{u, \varphi}] $是紧的当且仅当$ \varphi\circ\sigma_\psi = \sigma_\psi\circ\varphi $当且仅当$ \Phi\circ\Gamma = \Gamma\circ\Phi $当且仅当

$ \left\{ \begin{array}{ll} b_1(1-\overline{a_2}) = 2(a_1-1)\overline{d_2}, \\ \overline{b_2}(1-a_1) = 2(\overline{a_2}-1)d_1, \\ b_1\overline{b_2} = 4d_1\overline{d_2}. \end{array} \right. $

上式等价于

$ \left\{ \begin{array}{ll} b_1(1-\overline{a_2}) = 2(a_1-1)\overline{d_2}, \\ \overline{b_2}(1-a_1) = 2(\overline{a_2}-1)d_1. \end{array} \right. $

定理3.3得证.

定理3.4  设$ \varphi, \psi\in LFM({\Bbb B}_2)\backslash Aut({\Bbb B}_2) $都是双曲型, 且具有相同边界不动点$ e_1 $. 设$ \varphi, \psi $分别与$ H_2 $的解析自映射$ \Phi, \Psi $共轭, $ \Phi, \Psi $具有如下形式

$ \begin{eqnarray*} &&\Phi(w_1, w_2) = \frac{1}{\lambda_1}(w_1+b_1w_2+c_1, \ a_1w_2+d_1), \\ &&\Psi(w_1, w_2) = \frac{1}{\lambda_2}(w_1+b_2w_2+c_2, \ a_2w_2+d_2), \end{eqnarray*} $

其中$ \lambda_1 $与$ \lambda_2 $分别是$ \varphi $与$ \psi $的边界扩张系数. $ a_j, b_j, c_j, d_j\in {\Bbb C} $满足式(2.9). $ W_{u, \varphi} $与$ W_{v, \psi} $是$ H^2({\Bbb B}_2) $上分别由$ \varphi $与$ \psi $诱导的加权复合算子. 若$ H^\infty({\Bbb B}_n)\ni u, v $在$ \partial{\Bbb B}_2 $上连续且$ u(e_1)v(e_1)\neq0 $, 则$ [W^*_{v, \psi}, W_{u, \varphi}] $非紧.

证  由引理2.7可知, $ \sigma_\psi $与$ \sigma_\varphi\circ\psi $满足式(2.4). 由定理3.2, 只要证明$ \varphi\circ\sigma_\psi\neq\sigma_\psi\circ\varphi $即可.

假设$ \varphi\circ\sigma_\psi = \sigma_\psi\circ\varphi $. 由定理条件知, $ \psi $只有一个不动点$ e_1\in\partial{\Bbb B}_2 $, 所以$ \psi $的伴随映射$ \sigma_\psi $也只有一个不动点$ e_1 $. 因为$ \psi $是双曲型, 所以它的边界扩张系数$ \lambda_2<1 $. 故$ \sigma_\psi $的边界扩张系数$ \frac{1}{\lambda_2}>1 $. 由文献[7] 定理3.2知, $ \sigma_\psi $有一个孤立不动点$ \alpha\in{\Bbb B}_2 $. 如果$ \sigma_\psi\circ\varphi = \varphi\circ\sigma_\psi $, 则$ \sigma_\psi\circ\varphi(\alpha) = \varphi\circ\sigma_\psi(\alpha) = \varphi(\alpha). $上式说明$ \varphi(\alpha) $是$ \sigma_\psi $一个不动点. 然而, $ \varphi $在$ {\Bbb B}_2 $内无不动点, 因此$ \varphi(\alpha)\neq \alpha $. 由引理2.3, 若$ \psi $没有其它不动点, 则$ \sigma_\psi $也无其它不动点. 通过上述讨论$ \sigma_\psi $有三个不动点$ e_1\in\partial{\Bbb B}_2 $, $ \alpha\in{\Bbb B}_2 $以及$ \varphi(\alpha)\in{\Bbb B}_2 $, 这与$ \sigma_\psi $只有一个不动点$ \alpha\in{\Bbb B}_2 $矛盾. 因此$ \varphi\circ\sigma_\psi\neq\sigma_\psi\circ\varphi $, 由定理3.2可知$ [W^*_{v, \psi}, W_{u, \varphi}] $非紧.