最近几年, 单圆盘Dirichlet空间上的Toeplitz算子得到深入研究, 取得了丰富的成果^[1-8].同时调和Dirichlet空间上Toeplitz算子的研究也受到关注^[9-12].文[10, 11]分别给出了调和Dirichlet空间上调和符号Toeplitz算子的(半)交换性与乘积为零的刻画.文[12]推广了[10, 11]中的结论, 给出了一般符号Toeplitz算子一些代数性质及紧性的刻画.本文, 我们考虑调和Dirichlet空间上调和符号的Toeplitz算子与小Hankel算子的交换性.

首先我们回顾调和Dirichlet空间及其上Toeplitz算子和小Hankel算子的基本知识.

设 $\mathbb{D}$是复平面 $\mathbb{C}$上的单位开圆盘, $dA$表示 $\mathbb{D}$上正规化的面积测度.称 $\mathbb{D}$上光滑函数 $f$在如下范数

$ \begin{eqnarray*}\|f\|=\bigg{\{}\bigg{|}\int_\mathbb{D}f dA\bigg{|}^2+\int_\mathbb{D}\bigg{(}\bigg{|}\frac{\partial f}{\partial z}\bigg{|}^2+\bigg{|}\frac{\partial f}{\partial\bar{z}}\bigg{|}^2\bigg{)}dA\bigg{\}}^\frac{1}{2}<\infty, \end{eqnarray*} $

取闭包所得到的空间为Sobolev空间, 记为 $\mathcal{S}$.则 $\mathcal{S}$是一个Hilbert空间, 其上内积为

$ \langle f, g\rangle=\int_\mathbb{D} f dA\int_\mathbb{D} \overline{g}dA + \int_\mathbb{D} \bigg{(}\frac{\partial f}{\partial z}\overline{\frac{\partial g}{\partial z}}+\frac{\partial f}{\partial \bar{z}}\overline{\frac{\partial g}{\partial \bar{z}}}\bigg{)}dA, \ \ f, g\in S. $

Dirichlet空间是由 $\mathcal{S}$中所有满足条件 $f(0)=0$的解析函数 $f$构成的闭子空间, 记为 $\mathcal{D}$. $\mathcal{D}$的一组标准正交基为 $\{\frac{z^n}{\sqrt{n}}\}_{n=1}^\infty$.

$\mathcal{S}$中所有调和函数构成的闭子空间称为调和Dirichlet空间, 记为 $\mathcal{D}_h$.显然有

$ \mathcal{D}_h=\mathcal{D}\oplus \mathbb{C}\oplus \overline{\mathcal{D}}, $

其中 $\overline{\mathcal{D}}=\{\bar{f}\ \mid \ f\in\mathcal{D}\}$. $\mathcal{D}_h$的一组标准正交基为 $\{\frac{z^n}{\sqrt{n}}\}_{n=1}^\infty\cup\{1\}\cup\{\frac{\bar{z}^n}{\sqrt{n}}\}_{n=1}^\infty$.

$\mathbb{D}$上的一个非负测度 $\mu$称为 $\mathcal{D}$-Carleson测度如果存在非负常数 $c$, 使得

$ \int_\mathbb{D}|f|^2d\mu\leq c\|f\|^2, \ f\in\mathcal{D}. $

$\mathbb{D}$上所有有界解析函数全体记为 $H^\infty(\mathbb{D})$.令

$ \mathcal{M}=\bigg{\{}u\ \text{是}\ \mathbb{D} \text{上的调和函数}\ \bigg{|} \begin{array}{c}u=f+\bar{g}, \quad f, g\in H^\infty(\mathbb{D}), \;\;\;\;\\ |f^\prime|^2dA, \ |g^\prime|^2dA\ \text{是}\ \mathcal{D}-\text{Carleson 测度}\;\;\;\;\end{array}\bigg{\}}. $

若 $\phi\in \mathcal{M}$, $\mathcal{D}_h$上的Toeplitz算子 $\widetilde{T}_\phi$定义为

$ \widetilde{T}_\phi(f)=Q(\phi f), \quad f\in\mathcal{D}_h, $

其中 $Q$是从 $S$到 $\mathcal{D}_h$上的正交投影. $\widetilde{T}_\phi$是 $\mathcal{D}_h$上的有界算子^[12].

定义 $\mathcal{S}$上的算子 $U:$ $(Uf)(z)=f(\bar{z})$, $f\in\mathcal{S}$.则 $U$是 $\mathcal{S}$上的酉算子, $U^*=U=U^{-1}$^[9].

对于 $\phi\in \mathcal{M}$, 定义 $\mathcal{D}_h$上的小Hankel算子 $\widetilde{\mathit{\Gamma}}_\phi$为

$ \widetilde{\mathit{\Gamma}}_\phi(f)=Q(U(\phi f)), \quad f\in\mathcal{D}_h. $

容易验证 $\widetilde{\mathit{\Gamma}}_\phi$是有界算子.实际上, 如果记 $\hat{f}(z)=f(\bar{z})$, $f\in\mathcal{S}$, 则有 $\widetilde{\mathit{\Gamma}}_\phi=\widetilde{T}_{\hat{\phi}}U$.

在Hardy空间上, 文[13, 14]分别给出了Toeplitz算子与Hankel算子交换性和本质交换性的刻画.文[8]将Dirichlet空间上调和符号Toeplitz算子与小Hankel算子的交换性转化为Hardy空间上相应问题的研究.但目前关于调和函数空间上Toeplitz算子与(小) Hankel算子的交换性研究较少.文[15, 16]研究了调和Bergman空间上Toeplitz算子的交换性.

利用Sobolev空间的正交分解, 文[7]和[12]分别刻画了Dirichlet空间和调和Dirichlet空间上一般符号Toeplitz算子的代数性质等, 其证明过程表明, 在Dirichlet空间和调和Dirichlet空间上一般符号定义的Toeplitz算子与调和符号定义的Toeplitz算子有着密切的联系, 一般符号Toeplitz算子的研究可转化为相应调和符号Toeplitz算子的研究上.因此本文集中研究调和符号定义的Toeplitz算子和Hankel算子.主要结论如下.

定理1.1  设 $\varphi$, $\psi\in\mathcal{M}$.在 $\mathcal{D}_h$上 $\widetilde{T}_\varphi\widetilde{\mathit{\Gamma}}_\psi=\widetilde{\mathit{\Gamma}}_\psi\widetilde{T}_\varphi$当且仅当下列条件之一成立:

$1^\circ$ $\varphi$是常数;

$2^\circ$ $\varphi$不是常数, 存在常数 $\alpha$, $\beta$使得 $\psi=\alpha\varphi+\beta$, 且当 $\alpha\neq 0$或 $\beta\neq 0$时, $\varphi=U\varphi$.

本节, 我们给出定理1.1的证明.

设 $P$是从 $S$到 $\mathcal{D}$上的正交投影, $P_1$是从 $S$到 $\bar{\mathcal{D}}$上的正交投影, 则 $P_1=UPU$^[9].对于从 $S$到 $\mathcal{D}_h$上的正交投影 $Q$, 我们有下面的结论.

引理2.1   $UQ=QU$.即 $\forall\ f\in\mathcal{S}, \ (UQ)f=(QU)f.$

证  设 $f\in S$, 则

$ Q(f)=P(f)+\langle f, 1\rangle + P_1(f). $

因此

$ \begin{eqnarray*} && (UQ)f=UPf+U\langle f, 1\rangle+UP_{1}f =UPf+\langle f, 1\rangle+PUf, \\ && (QU)f=PUf+\langle Uf, 1\rangle+P_{1}Uf=PUf+\langle\hat{f}, 1\rangle+UPf. \end{eqnarray*} $

因为

$ \langle f, 1\rangle=\int_{\mathbb{D}}f(z)dA(z)=\int_{\mathbb{D}}f(\bar{z})dA(z)=\langle \hat{f}, 1\rangle, $

所以由上面两个等式可得 $(UQ)f=(QU)f, $ $f\in S$.证毕.

容易验证 $U$将 $\overline{\mathcal{D}}$映到 $\mathcal{D}$上.定义算子 $\widetilde{U}:$ $\mathcal{D}_h=\mathcal{D}\oplus \mathbb{C}\oplus \overline{\mathcal{D}}\rightarrow \mathcal{D}\oplus \mathbb{C}\oplus \mathcal{D}$, 其中

$ \tilde{U}=\left( \begin{array}{ccc} I&0&0\\ 0&1&0\\ 0&0&U \end{array} \right). $

显然, $\tilde{U}^*$将 $\mathcal{D}\oplus \mathbb{C}\oplus \mathcal{D}$映到 $\mathcal{D}_h$上, 且

$ \tilde{U}^*=\left( \begin{array}{ccc} I&0&0\\ 0&1&0\\ 0&0&U \end{array} \right). $

设 $\phi\in\mathcal{M}$, 定义

$ (1\otimes \phi)(f)=\int_{\mathbb{D}}f(z)\phi(z)dA(z), \ f\in\mathcal{D}_h, P_\phi(a)=aP(\phi), \ a\in\mathbb{C}. $

$T_\phi$表示 $\mathcal{D}$上的Toeplitz算子, $T_\phi(f)=P(\phi f), \ f\in\mathcal{D}.$ $\mathit{\Gamma}_\phi$表示 $\mathcal{D}$上的小Hankel算子

$ \mathit{\Gamma}_\phi(f)=P(U(\phi f)), \ \ f\in\mathcal{D}. $

下面的引理分别给出了调和Dirichlet空间上Topelitz算子与小Hankel算子的矩阵表示.它们表明调和Dirichlet空间上Topelitz算子与小Hankel算子与Dirichlet空间上Topelitz算子与小Hankel算子有着密切的联系.这一思想来自于文[17].

引理2.2^[9]  设 $\phi\in\mathcal{M}$, 则在 $\mathcal{D}\oplus\mathbb{C}\oplus\mathcal{D}$上,

$ \begin{eqnarray*} \tilde{U}\tilde{T}_\phi\tilde{U}^*=\left( \begin{array}{ccc} T_\phi&P_\phi&\mathit{\Gamma}_{\hat{\phi}}\\ 1\otimes \phi&1\otimes \phi&1\otimes \hat{\phi}\\ \mathit{\Gamma}_{\phi}&P_{\hat{\phi}}&T_{\hat{\phi}} \\ \end{array} \right). \end{eqnarray*} $

引理2.3  设 $\phi\in\mathcal{M}$.则在 $\mathcal{D}\oplus\mathbb{C}\oplus\mathcal{D}$上,

$ \begin{eqnarray*} \tilde{U}\tilde{\mathit{\Gamma}}_\phi\tilde{U}^*=\left( \begin{array}{ccc} \mathit{\Gamma}_\phi&P_{\hat{\phi}}&T_{\hat{\phi}}\\ 1\otimes \phi&1\otimes {\hat{\phi}}&1\otimes {\hat{\phi}}\\ T_\phi&P_\phi&\mathit{\Gamma}_{\hat{\phi}} \end{array} \right). \end{eqnarray*} $

证  设 $(f_1, \ a, \ f_2)\in \mathcal{D}\oplus\mathbb{C}\oplus\mathcal{D}$, 直接计算可得

$ \begin{eqnarray*} \widetilde{\mathit{\Gamma}}_{\phi}f_{1}&=&Q(U(\phi f_{1})) =P(U(\phi f_{1}))+\langle\hat{\phi}\hat{f_{1}}, 1\rangle+P_{1}(U(\phi f_{1}))\\ &=&\mathit{\Gamma}_{\phi}f_{1}+(1\otimes\hat{\phi})\hat{f}_{1}+UPU(U(\phi f_{1})) =\mathit{\Gamma}_{\phi}f_{1}+(1\otimes\hat{\phi})\hat{f}_{1}+UP(\phi f_{1})\\ &=&\mathit{\Gamma}_{\phi}f_{1}+(1\otimes\phi)f_{1}+UT_{\phi}f_{1}, \\ \widetilde{\mathit{\Gamma}}_{\phi}a &=&Q(U(\phi a)) =P(\hat{\phi}a)+\langle\hat{\phi}a, 1\rangle+P_{1}(\hat{\phi}a) =P_{\hat{\phi}}(a)+(1\otimes\hat{\phi})(a)+UP_{\phi}(a), \\ \widetilde{\mathit{\Gamma}}_{\phi}\hat{f_{2}}&=&Q(U(\phi\hat{f_{2}}) ) =P(\hat{\phi}f_{2})+\langle\hat{\phi}f_{2}, 1\rangle+P_{1}(\hat{\phi}f_{2})\\ &=&P(\hat{\phi}f_{2})+\langle\hat{\phi}f_{2}, 1\rangle+UPU(\hat{\phi}f_{2}) =T_{\hat{\phi}}f_{2}+(1\otimes\hat{\phi})f_{2}+U\mathit{\Gamma}_{\hat{\phi}}f_{2}. \end{eqnarray*} $

因此

$ \begin{eqnarray*} \widetilde{U}\widetilde{\mathit{\Gamma}}_{\phi}\widetilde{U}^{*}\left( \begin{array}{c} f_{1} \\ a \\ f_{2} \end{array} \right) &=&\widetilde{U}\widetilde{\mathit{\Gamma}}_{\phi}\left( \begin{array}{c} f_{1} \\ a \\ \hat{f_{2}} \end{array} \right) =\widetilde{U}\left( \begin{array}{c} \mathit{\Gamma}_{\phi}f_{1} +P_{\hat{\phi}}(a)+T_{\hat{\phi}}f_{2}\\ (1\otimes\phi)f_{1}+(1\otimes\hat{\phi})(a)+(1\otimes\hat{\phi})f_{2}\\ UT_{\phi}f_{1}+UP_{\phi}(a)+U\mathit{\Gamma}_{\hat{\phi}}f_{2} \end{array} \right)\\ &=&\left( \begin{array}{c} \Gamma_{\phi}f_{1} +P_{\hat{\phi}}(a)+T_{\hat{\phi}}f_{2}\\ (1\otimes\phi)f_{1}+(1\otimes\hat{\phi})(a)+(1\otimes\hat{\phi})f_{2}\\ T_{\phi}f_{1}+P_{\phi}(a)+\mathit{\Gamma}_{\hat{\phi}}f_{2} \end{array} \right)\\ &=&\left( \begin{array}{ccc} \mathit{\Gamma}_\phi&P_{\hat{\phi}}&T_{\hat{\phi}}\\ 1\otimes \phi&1\otimes {\hat{\phi}}&1\otimes {\hat{\phi}}\\ T_\phi&P_\phi&\mathit{\Gamma}_{\hat{\phi}} \end{array} \right)\left( \begin{array}{c} f_{1} \\ a \\ f_{2} \\ \end{array} \right). \end{eqnarray*} $

证毕.

为了方便后面的应用, 下面给出 $\mathcal{D}$上Toeplitz算子与Hankel算子及其它算子在标准正交基 $\{e_n=\frac{z^n}{\sqrt{n}}\}_{n=1}^\infty$下的系数.

引理2.4  设 $\varphi(z)=\sum\limits_{k<0}a_k\bar{z}^{-k}+\sum\limits_{k\geq 0}a_kz^k, $ $\psi(z)=\sum\limits_{k<0}b_k\bar{z}^{-k}+\sum\limits_{k\geq 0}b_kz^k\in\mathcal{M}$, 则

$1^\circ$ $\langle T_\varphi e_n, e_m\rangle=\frac{\sqrt{m}}{\sqrt{n}}a_{m-n}, \ \ \langle \mathit{\Gamma}_\varphi e_n, e_m\rangle=\frac{\sqrt{m}}{\sqrt{n}}a_{-(m+n)}$;

$2^\circ$ $\langle T_{\varphi}\mathit{\Gamma}_{\psi}e_{n}, e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{m-k}b_{-n-k}$, $\langle\mathit{\Gamma}_{\varphi}T_{\psi}e_{n}, e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{-m-k}b_{-n+k}$;

$3^\circ$ $\langle\mathit{\Gamma}_{\varphi}\mathit{\Gamma}_{\psi}e_{n}, e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{-m-k}b_{-n-k}$, $\langle T_{\varphi}T_{\psi}e_{n}, e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{m-k}b_{-n+k}$;

$4^\circ$ $(1\otimes \varphi) e_n =\frac{a_{-n}}{\sqrt{n}(n+1)}, $ $\langle P_{\varphi}(1\otimes\psi)e_{n}, e_{m}\rangle =\frac{\sqrt{m}~a_{m}b_{-n}}{\sqrt{n}(1+n)}$;

$5^\circ$ $\langle T_{\varphi}P_{\psi}1, e_{m}\rangle =\sqrt{m}\sum\limits_{k=1}^\infty a_{m-k}b_{k}$, $\langle P_{\varphi}(1\otimes\psi)1, e_{m}\rangle =\sqrt{m}~a_{m}b_{0}$,

$ \langle\mathit{\Gamma}_{\varphi}P_{\psi}1, e_{m}\rangle =\sqrt{m}\sum\limits_{k=1}^\infty a_{-m-k}b_{k}. $

证  直接计算可知 $\langle T_\varphi e_n, e_m\rangle=\frac{\sqrt{m}}{\sqrt{n}}a_{m-n}, \ \ \langle \mathit{\Gamma}_\varphi e_n, e_m\rangle=\frac{\sqrt{m}}{\sqrt{n}}a_{-(m+n)}. $因此

$\begin{eqnarray*} T_\varphi e_n&=&\sum\limits_{k=1}^\infty\frac{\sqrt{k}}{\sqrt{n}}a_{k-n}e_k, \mathit{\Gamma}_\varphi e_n=\sum\limits_{k=1}^\infty\frac{\sqrt{k}}{\sqrt{n}}a_{-(k+n)}e_k.\\ \langle T_{\varphi}\mathit{\Gamma}_{\psi}e_{n}, e_{m}\rangle &=&\langle T_{\varphi}(\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n-k}e_{k}), e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n-k}\langle T_{\varphi}e_{k}, e_{m}\rangle\\ &=&\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{m-k}b_{-n-k}, \\ \langle\mathit{\Gamma}_{\varphi}T_{\psi}e_{n}, e_{m}\rangle &=&\langle\mathit{\Gamma}_{\varphi}(\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n+k}e_{k}), e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n+k}\langle\mathit{\Gamma}_{\varphi}e_{k}, e_{m}\rangle\\ &=&\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{-m-k}b_{-n+k}, \\ % \nonumber to remove numbering (before each equation) \langle\mathit{\Gamma}_{\varphi}\mathit{\Gamma}_{\psi}e_{n}, e_{m}\rangle &=&\langle\mathit{\Gamma}_{\varphi}(\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n-k}e_{k}), e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n-k}\langle\mathit{\Gamma}_{\varphi}e_{k}, e_{m}\rangle\\ &=&\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{-m-k}b_{-n-k}, \\ % \nonumber to remove numbering (before each equation) \langle T_{\varphi}T_{\psi}e_{n}, e_{m}\rangle &=&\langle T_{\varphi}(\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n+k}e_{k}), e_{m}\rangle =\sum\limits_{k=1}^{\infty}\frac{\sqrt{k}}{\sqrt{n}}b_{-n+k}\langle T_{\varphi}e_{k}, e_{m}\rangle\\ &=&\sum\limits_{k=1}^{\infty}\frac{\sqrt{m}}{\sqrt{n}}a_{m-k}b_{-n+k}. \end{eqnarray*} $

同样计算可得

$ \begin{eqnarray*} (1\otimes \varphi) e_n&=& \int_{\mathbb{D}}\varphi e_ndA =\int_{\mathbb{D}}\big{(}\sum\limits_{k<0}a_k\bar{z}^{-k}+\sum\limits_{k\geq 0}a_kz^k\big{)}\frac{z^n}{\sqrt{n}}dA(z) =\frac{a_{-n}}{\sqrt{n}(n+1)}, \\ \langle P_{\varphi}(1\otimes\psi)e_{n}, e_{m}\rangle &=&(1\otimes\psi)e_{n}\langle P(\varphi), e_{m}\rangle =\frac{b_{-n}}{\sqrt{n}(1+n)}\langle \sum\limits_{k>0}a_kz^k, e_m\rangle =\frac{\sqrt{m}~a_{m}b_{-n}}{\sqrt{n}(1+n)}, \\ \langle T_{\varphi}P_{\psi}1, e_{m}\rangle &=&\langle T_{\varphi}P(\psi), e_{m}\rangle =\sum\limits_{k>0}b_{k}\sqrt{k}~\langle T_{\varphi}e_{k}, e_{m}\rangle =\sqrt{m}\sum\limits_{k=1}^\infty a_{m-k}b_{k}, \\ \langle P_{\varphi}(1\otimes\psi)1, e_{m}\rangle &=&((1\otimes\psi)1)\langle P(\varphi), e_{m}\rangle =b_{0}~\langle\sum\limits_{k>0}a_{k}z^{k}~, ~e_{m}\rangle =\sqrt{m}~a_{m}b_{0}, \\ \langle\mathit{\Gamma}_{\varphi}P_{\psi}1, e_{m}\rangle &=&\langle\mathit{\Gamma}_{\varphi}P(\psi), e_{m}\rangle =\sum\limits_{k>0}b_{k}\sqrt{k}~\langle\mathit{\Gamma}_{\varphi}e_{k}~, ~e_{m}\rangle =\sqrt{m}\sum\limits_{k=1}^\infty a_{-m-k}b_{k}. \end{eqnarray*} $

下面我们给出本文主要定理1.1的证明.

证  充分性的证明.

当 $\varphi$是常数时, 则 $\widetilde T_{\varphi}\widetilde{\mathit{\Gamma}}_{\psi}=\widetilde{\mathit{\Gamma}}_{\psi}\widetilde T_{\varphi}$显然成立.

当 $\varphi$不为常数, $\psi=\alpha\varphi+\beta, $且当 $\alpha\neq 0$或 $\beta\neq0$时, $\varphi=\hat{\varphi}$, 则 $\forall~f\in\mathcal{D}_{h}$,

$ \begin{eqnarray*} \widetilde{T}_{\varphi}\widetilde{\mathit{\Gamma}}_{\psi}f &=&\widetilde{T}_{\varphi}\widetilde{\mathit{\Gamma}}_{\alpha\varphi}f +\widetilde{T}_{\varphi}\widetilde{\mathit{\Gamma}}_{\beta}f =\widetilde{T}_{\varphi}Q(U(\alpha\varphi f)) +\widetilde{T}_{\varphi}Q(U(\beta f))\\ &=&\alpha\widetilde{T}_{\varphi}UQ(\varphi f)+\beta\widetilde{T}_{\varphi}Uf =\alpha Q(\varphi UQ(\varphi f))+\beta Q(\varphi Uf) =\alpha\widetilde{\mathit{\Gamma}}_{\hat{\varphi}}\widetilde{T}_{\varphi}f+\beta\widetilde{\mathit{\Gamma}}_{\hat{\varphi}}f, \\ \widetilde{\mathit{\Gamma}}_{\psi}\widetilde{T}_{\varphi}f &=&\widetilde{\mathit{\Gamma}}_{\alpha\varphi}\widetilde{T}_{\varphi}f+\widetilde{\mathit{\Gamma}}_{\beta}\widetilde{T}_{\varphi}f=Q(U(\alpha\varphi \widetilde{T}_{\varphi}f))+Q(\beta UQ(\varphi f))\\ &=&\alpha Q(U(\varphi \widetilde{T}_{\varphi}f))+Q(\beta QU(\varphi f)) =\alpha\widetilde{\mathit{\Gamma}}_{\varphi}\widetilde{T}_{\varphi}f+\beta\widetilde{\mathit{\Gamma}}_{\varphi}f. \end{eqnarray*} $

在上面两式中用到引理2.1.

比较上面两式可知, 当 $\alpha\neq 0$或 $\beta\neq0, $且 $\varphi=\hat{\varphi}$时, $\widetilde T_{\varphi}\widetilde{\mathit{\Gamma}}_{\psi}=\widetilde{\mathit{\Gamma}}_{\psi}\widetilde T_{\varphi}$.

必要性的证明.

若 $\widetilde T_{\varphi}\widetilde{\mathit{\Gamma}}_{\psi}=\widetilde{\mathit{\Gamma}}_{\psi}\widetilde T_{\varphi}, $则 $\widetilde{U}\widetilde T_{\varphi}\widetilde{U}^{*}\widetilde{U}\widetilde{\mathit{\Gamma}}_{\psi}\widetilde{U}^{*}=\widetilde{U}\widetilde{\mathit{\Gamma}}_{\psi}\widetilde{U}^{*}\widetilde{U}\widetilde T_{\varphi}\widetilde{U}^{*}$, 由引理2.2与引理2.3得

$ \begin{eqnarray*}&& \left( \begin{array}{ccc} T_{\varphi}&P_{\varphi}&\mathit{\Gamma}_{\hat{\varphi}} \\ 1\otimes\varphi&1\otimes\varphi&1\otimes\hat{\varphi} \\ \mathit{\Gamma}_{\varphi}&P_{\hat{\varphi}}&T_{\hat\varphi} \end{array} \right)\left( \begin{array}{ccc} \mathit{\Gamma}_{\psi}&P_{\hat{\psi}}&T_{\hat\psi} \\ 1\otimes\psi&1\otimes\hat{\psi}&1\otimes\hat{\psi} \\ T_{\psi}& P_{\psi}&\mathit{\Gamma}_{\hat{\psi}} \end{array} \right)\\ &=&\left( \begin{array}{ccc} \mathit{\Gamma}_{\psi}&P_{\hat{\psi}}&T_{\hat\psi} \\ 1\otimes\psi&1\otimes\hat{\psi}& 1\otimes\hat{\psi} \\ T_{\psi}& P_{\psi}&\mathit{\Gamma}_{\hat{\psi}} \end{array} \right)\left( \begin{array}{ccc} T_{\varphi}&P_{\varphi}&\mathit{\Gamma}_{\hat{\varphi}} \\ 1\otimes\varphi&1\otimes\varphi&1\otimes\hat{\varphi} \\ \mathit{\Gamma}_{\varphi}&P_{\hat{\varphi}}&T_{\hat\varphi} \end{array} \right). \end{eqnarray*} $

因此

$ T_{\varphi}\mathit{\Gamma}_{\psi}+P_{\varphi}(1\otimes\psi)+\mathit{\Gamma}_{\hat{\varphi}}T_{\psi} =\mathit{\Gamma}_{\psi}T_{\varphi}+P_{\hat{\psi}}(1\otimes\varphi)+T_{\hat{\psi}}\mathit{\Gamma}_{\varphi}, $

(2.1)

$ T_{\varphi}T_{\hat{\psi}}+P_{\varphi}(1\otimes\hat{\psi})+\mathit{\Gamma}_{\hat{\varphi}}\mathit{\Gamma}_{\hat{\psi}} =\mathit{\Gamma}_{\psi}\mathit{\Gamma}_{\hat{\varphi}}+P_{\hat{\psi}}(1\otimes\hat{\varphi})+T_{\hat{\psi}}T_{\hat{\varphi}}, $

(2.2)

$ \mathit{\Gamma}_{\varphi}\mathit{\Gamma}_{\psi}+P_{\hat{\varphi}}(1\otimes\psi)+T_{\hat{\varphi}}T_{\psi} =T_{\psi}T_{\varphi}+P_{\psi}(1\otimes\varphi)+\mathit{\Gamma}_{\hat{\psi}}\mathit{\Gamma}_{\varphi}, $

(2.3)

$ \mathit{\Gamma}_{\varphi}T_{\hat{\psi}}+P_{\hat{\varphi}}(1\otimes\hat{\psi})+T_{\hat{\varphi}}\mathit{\Gamma}_{\hat{\psi}} =T_{\psi}\mathit{\Gamma}_{\hat{\varphi}}+P_{\psi}(1\otimes\hat{\varphi})+\mathit{\Gamma}_{\hat{\psi}}T_{\hat{\varphi}}, $

(2.4)

$ T_{\varphi}P_{\hat{\psi}}+P_{\varphi}(1\otimes\hat{\psi})+\mathit{\Gamma}_{\hat{\varphi}}P_{\psi} =\mathit{\Gamma}_{\psi}P_{\varphi}+P_{\hat{\psi}}(1\otimes\varphi)+T_{\hat{\psi}}P_{\hat{\varphi}}, $

(2.5)

$ \mathit{\Gamma}_{\varphi}P_{\hat{\psi}}+P_{\hat{\varphi}}(1\otimes\hat{\psi})+T_{\hat{\varphi}}P_{\psi} =T_{\psi}P_{\varphi}+P_{\psi}(1\otimes\varphi)+\mathit{\Gamma}_{\hat{\psi}}P_{\hat{\varphi}}, $

(2.6)

$ (1\otimes\varphi)P_{\hat\psi}+(1\otimes\varphi)(1\otimes\hat{\psi})+(1\otimes\hat{\varphi})P_{\psi} =(1\otimes\psi)P_{\varphi}+(1\otimes\hat{\psi})(1\otimes\varphi)+(1\otimes\hat{\psi})P_{\hat{\varphi}}. $

(2.7)

设 $\varphi(z)=\sum\limits_{k<0}a_{k}\bar{z}^{-k}+\sum\limits_{k\geq0}a_{k}z^{k}, ~~\psi(z)=\sum\limits_{k<0}b_{k}\bar{z}^{-k}+\sum\limits_{k\geq0}b_{k}z^{k}$, 则

$ \hat{\varphi}(z)=\sum\limits_{k<0}a_{-k}\bar{z}^{-k}+\sum\limits_{k\geq0}a_{-k}z^{k}, ~~\hat{\psi}(z)=\sum\limits_{k<0}b_{-k}\bar{z}^{-k}+\sum\limits_{k\geq0}b_{-k}z^{k}. $

由引理2.4计算得

$ \begin{eqnarray*}&&\langle (T_{\varphi}\mathit{\Gamma}_{\psi}+P_{\varphi}(1\otimes\psi)+\mathit{\Gamma}_{\hat{\varphi}}T_{\psi})e_{n}, e_{m}\rangle \\ &=&\sum\limits_{k=1}^\infty \frac{\sqrt{m}}{\sqrt{n}}a_{m-k}b_{-n-k}+\frac{\sqrt{m}}{\sqrt{n}}\frac{a_mb_{-n}}{1+n}+\sum\limits_{k=1}^\infty \frac{\sqrt{m}}{\sqrt{n}}a_{m+k}b_{k-n}, \\ && \langle(\mathit{\Gamma}_{\psi}T_{\varphi}+P_{\hat{\psi}}(1\otimes\varphi)+T_{\hat{\psi}}\mathit{\Gamma}_{\varphi})e_{n}, e_{m}\rangle\\ &=&\sum\limits_{k=1}^\infty \frac{\sqrt{m}}{\sqrt{n}}a_{-n+k}b_{-m-k}+\frac{\sqrt{m}}{\sqrt{n}}\frac{a_{-n}b_{-m}}{1+n}+\sum\limits_{k=1}^\infty \frac{\sqrt{m}}{\sqrt{n}}a_{-n-k}b_{-m+k}.\end{eqnarray*} $

由(2.1) 式得

$ \sum\limits_{k=-\infty}^{\infty}a_{m+k}b_{-n+k}+(\frac{1}{1+n}-1)a_{m}b_{-n} =\sum\limits_{k=-\infty}^{\infty}a_{-n+k}b_{-m-k}+(\frac{1}{1+n}-1)a_{-n}b_{-m}. $

(2.1')

同样计算(2.2), (2.3), (2.4) 式得

$ \sum\limits_{k=-\infty}^{\infty}a_{m+k}b_{n+k}+(\frac{1}{1+n}-1)a_{m}b_{n} =\sum\limits_{k=-\infty}^{\infty}a_{n+k}b_{-m-k}+(\frac{1}{1+n}-1)a_{n}b_{-m}, $

(2.2')

$ \sum\limits_{k=-\infty}^{\infty}a_{-m+k}b_{-n+k}+(\frac{1}{1+n}-1)a_{-m}b_{-n} =\sum\limits_{k=-\infty}^{\infty}a_{-n+k}b_{m-k}+(\frac{1}{1+n}-1)a_{-n}b_{m}, $

(2.3')

$ \sum\limits_{k=-\infty}^{\infty}a_{-m+k}b_{n+k}+(\frac{1}{1+n}-1)a_{-m}b_{n} =\sum\limits_{k=-\infty}^{\infty}a_{n+k}b_{m-k}+(\frac{1}{1+n}-1)a_{n}b_{m}. $

(2.4')

由引理2.4得

$ \langle (T_{\varphi}P_{\hat{\psi}}+P_{\varphi}(1\otimes\hat{\psi})+\mathit{\Gamma}_{\hat{\varphi}}P_{\psi})1~, e_{m}\rangle =\\ \sqrt{m}\sum\limits_{k=1}^\infty a_{m-k}b_{-k}+\sqrt{m}~a_{m}b_{0}+\sqrt{m}\sum\limits_{k=1}^\infty a_{m+k}b_{k}, \\ \langle(\mathit{\Gamma}_{\psi}P_{\varphi} +P_{\hat{\psi}}(1\otimes\varphi)+T_{\hat{\psi}}P_{\hat{\varphi}})1~, e_{m}\rangle=\sqrt{m}\sum\limits_{k=1}^\infty a_{k}b_{-m-k}+\\ \sqrt{m}~a_{0}~b_{-m}+\sqrt{m}~\sum\limits_{k=1}^\infty a_{-k}b_{-m+k}. $

由(2.5) 式得

$ \sum\limits_{k=-\infty}^{\infty}a_{m+k}b_{k}=\sum\limits_{k=-\infty}^{\infty}a_{k}b_{-m-k}. $

(2.5')

同样计算(2.6) 式可得

$ \sum\limits_{k=-\infty}^{\infty}a_{-m+k}b_{k}=\sum\limits_{k=-\infty}^{\infty}a_{k}b_{m-k}. $

(2.6')

因为

$ \begin{eqnarray*} \langle(1\otimes\varphi)P_{\hat\psi}1, 1\rangle&=&\langle(1\otimes\varphi)P(\hat{\psi}), 1\rangle =\langle(1\otimes\varphi)\sum\limits_{k>0}b_{-k}z^{k}, 1\rangle \\ &=&\sum\limits_{k>0}b_{-k}\sqrt{k}\langle(1\otimes\varphi)e_{k}, 1\rangle =\sum\limits_{k=1}^\infty {\frac{a_{-k}b_{-k}}{1+k}}, \\ \langle(1\otimes\varphi)(1\otimes\hat{\psi})1, 1\rangle&=&a_{0}b_{0}, \\ \langle(1\otimes\hat{\varphi})P_{\psi}1, 1\rangle &=&\langle(1\otimes\hat{\varphi})P(\psi), 1\rangle =\langle(1\otimes\hat{\varphi})\sum\limits_{k>0}b_{k}z^{k}, 1\rangle\\ &=&\sum\limits_{k>0}b_{k}\sqrt{k}~\langle(1\otimes\hat{\varphi})e_{k}, 1\rangle =\sum\limits_{k=1}^\infty{\frac{a_{k}b_{k}}{1+k}}, \end{eqnarray*} $

所以

$ \langle((1\otimes\varphi)P_{\hat\psi}+(1\otimes\varphi)(1\otimes\hat{\psi}) +\\ (1\otimes\hat{\varphi})P_{\psi})1, 1\rangle =\sum\limits_{k=1}^\infty{\frac{a_{-k}b_{-k}}{1+k}}+a_{0}b_{0}+\sum\limits_{k=1}^\infty{\frac{a_{k}b_{k}}{1+k}}, \\ \langle((1\otimes\psi)P_{\varphi} +(1\otimes\hat{\psi})(1\otimes\varphi)+(1\otimes\hat{\psi})P_{\hat{\varphi}})1, 1\rangle=\sum\limits_{k=1}^\infty{\frac{a_{k}b_{-k}}{1+k}}+\\ a_{0}b_{0}+\sum\limits_{k=1}^\infty{\frac{a_{-k}b_{k}}{1+k}}. $

由(2.7) 式得

$ \sum\limits_{k=1}^\infty{\frac{a_{-k}b_{-k}}{1+k}}+a_{0}b_{0}+\sum\limits_{k=1}^\infty{\frac{a_{k}b_{k}}{1+k}} =\sum\limits_{k=1}^\infty{\frac{a_{k}b_{-k}}{1+k}}+a_{0}b_{0}+\sum\limits_{k=1}^\infty{\frac{a_{-k}b_{k}}{1+k}}. $

(2.7')

在(2.1')与(2.3')式中令 $-n+k=j, $则 $k=n+j, $

$ \sum\limits_{j=-\infty}^{\infty}a_{m+n+j}b_{j}+(\frac{1}{1+n}-1)a_{m}b_{-n} =\sum\limits_{j=-\infty}^{\infty}a_{j}b_{-m-n-j}+(\frac{1}{1+n}-1)a_{-n}b_{-m}, $

(2.1")

$ \sum\limits_{j=-\infty}^{\infty}a_{n-m+j}b_{j}+(\frac{1}{1+n}-1)a_{-m}b_{-n} =\sum\limits_{j=-\infty}^{\infty}a_{j}b_{m-n-j}+(\frac{1}{1+n}-1)a_{-n}b_{m}. $

(2.3")

在(2.2')与(2.4')中令 $n+k=j, $则 $k=-n+j, $

$ \sum\limits_{j=-\infty}^{\infty}a_{m-n+j}b_{j}+(\frac{1}{1+n}-1)a_{m}b_{n} =\sum\limits_{j=-\infty}^{\infty}a_{j}b_{n-m-j}+(\frac{1}{1+n}-1)a_{n}b_{-m}, $

(2.2")

$ \sum\limits_{j=-\infty}^{\infty}a_{-m-n+j}b_{j}+(\frac{1}{1+n}-1)a_{-m}b_{n} =\sum\limits_{j=-\infty}^{\infty}a_{j}b_{m+n-j}+(\frac{1}{1+n}-1)a_{n}b_{m}. $

(2.4")

对任意正整数 $m, n$, 由(2.1")和(2.5')式结合可得

$ a_{m}b_{-n}=a_{-n}b_{-m}. $

(2.8)

由(2.3")和(2.5'), (2.6')式结合可得

$ a_{-m}b_{-n}=a_{-n}b_{m}~~(m\neq n). $

(2.9)

由(2.2")和(2.5'), (2.6')式结合可得

$ a_{m}b_{n}=a_{n}b_{-m}~~(m\neq n). $

(2.10)

由(2.4")和(2.6')式结合可得

$ a_{-m}b_{n}=a_{n}b_{m}. $

(2.11)

在(2.2"), (2.3")式中令 $m=n$, 则

$ \begin{eqnarray*}&& \sum\limits_{j=-\infty}^{\infty}a_{j}b_{j}+(\frac{1}{n+1}-1)a_{-n}b_{-n} =\sum\limits_{j=-\infty}^{\infty}a_{j}b_{-j}+(\frac{1}{n+1}-1)a_{-n}b_{n}, \\ && \sum\limits_{j=-\infty}^{\infty}a_{j}b_{j}+(\frac{1}{n+1}-1)a_{n}b_{n} =\sum\limits_{j=-\infty}^{\infty}a_{j}b_{-j}+(\frac{1}{n+1}-1)a_{n}b_{-n}.\end{eqnarray*} $

即有

$ \begin{eqnarray*}&& \sum\limits_{j=-\infty}^{\infty}a_{j}b_{j}-\sum\limits_{j=-\infty}^{\infty}a_{j}b_{-j} =(\frac{1}{1+n}-1)(a_{-n}b_{n}-a_{-n}b_{-n}), \\ && \sum\limits_{j=-\infty}^{\infty}a_{j}b_{j}-\sum\limits_{j=-\infty}^{\infty}a_{j}b_{-j}= (\frac{1}{1+n}-1)(a_{n}b_{-n}-a_{n}b_{n}).\end{eqnarray*} $

对比以上两式可得 $a_{-n}b_{n}-a_{-n}b_{-n}=a_{n}b_{-n}-a_{n}b_{n}, $即

$ (a_{-n}+a_{n})(b_{n}-b_{-n})=0. $

(2.12)

从而 $a_{-n}=-a_{n}$或 $b_{-n}=b_{n}.$

设 $\varphi$不是常数, 则存在正整数 $m_0$使得 $a_{m_0}\neq 0$或 $a_{-m_0}\neq 0$.由对称性, 不妨设 $a_{m_0}\neq 0$.

对任意正整数 $n$, 由 $(2.8)$式得

$ b_{-n}=\frac{b_{-m_{0}}}{a_{m_{0}}}a_{-n}. $

(2.13)

由(2.10) 式得

$ b_{n}=\frac{b_{-m_{0}}}{a_{m_{0}}}a_{n}(n\neq m_{0}). $

(2.14)

若 $b_{-m_0}\neq 0$, 则由(2.13) 式得 $a_{-m_{0}}=a_{m_{0}}\neq 0$, 由(2.12) 式得 $b_{m_0}=b_{-m_0}=\frac{b_{-m_0}}{a_{m_0}}a_{m_0}$.令 $\alpha_1=\frac{b_{-m_{0}}}{a_{m_{0}}}$, 则存在常数 $\beta_1$, 使得 $\psi=\alpha_1\varphi+\beta_1$.

若 $b_{-m_0}=0$, 则由(2.13), (2.14) 式得 $b_{-n}=0, $ $b_n=0(n\neq m_0)$.由(2.7')式得

$ a_0b_0+\frac{a_{m_0}b_{m_0}}{1+m_0}=a_0b_0+\frac{a_{-m_0}b_{m_0}}{1+m_0}. $

因此有 $(a_{m_0}-a_{-m_0})b_{m_0}=0.$因为 $b_{-m_0}= 0$, 由(2.12) 式得 $(a_{-{m_0}}+a_{m_0})b_{m_0}=0.$若 $b_{m_0}\neq 0$, 则 $a_{-{m_0}}+a_{m_0}=0$, 且 $a_{m_0}-a_{-{m_0}}=0$, 因此有 $a_{m_0}=0$, 矛盾.所以 $b_{m_0}=0$.因此 $\psi$是一个常数.设 $\psi=\beta_2$, 令 $\alpha_2=0$, 则有 $\psi=\alpha_2\varphi+\beta_2$.

以下设 $\psi=\alpha\varphi+\beta$, 则对于任意正整数 $n$, $b_n=\alpha a_n$, $b_{-n}=\alpha a_{-n}.$

当 $\alpha\neq0$时, 代入(2.11) 式得 $a_{-m}a_{n}=a_{n}a_{m}$.因为 $a_{m_0}\neq 0$, 所以对任意正整数 $n$, 有 $a_n=a_{-n}$.因此 $\varphi=U\varphi$.

当 $\alpha=0$, $\beta\neq 0$时, 由 $\widetilde{T}_{\varphi}\widetilde{\mathit{\Gamma}}_{\psi}=\widetilde{\mathit{\Gamma}}_{\psi}\widetilde{T}_{\varphi}$得 $\widetilde{T}_{\varphi}\widetilde{\mathit{\Gamma}}_{\psi}(1)=\widetilde{\mathit{\Gamma}}_{\psi}\widetilde{T}_{\varphi}(1)$.直接计算得 $\varphi=U\varphi.$证毕.