令$ D $为复平面$ C $上的开单位圆盘.它的边界圆周记为$ T $.多圆盘$ D^n $和圆环面$ T^n $分别是$ C^n $中$ n $重$ D $和$ T $的笛卡尔积. $ {\rm d}A(z) $为$ D^n $的标准化的球体积测度. Bergman空间$ L_a^2(D^n) $是$ L^2(D^n, {\rm d}A) $的所有解析函数构成子空间. $ P $为$ L^2(D^n, {\rm d}A) $到$ L_a^2(D^n) $的正交投影.空间$ L_a^2(D^n) $的再生核为

$ K_w{(z)} = \prod\limits_{j = 1}^n\frac{1}{(1-\overline{w}_{j}z_{j})^2} = \prod\limits_{j = 1}^n\bigg[\sum\limits_{k = 1}^{\infty}(k+1)\overline{w_j}^k{z_j}^k\bigg], $

其中$ z = (z_1, z_2, \cdots, z_n), w = (w_1, w_2, \cdots, w_n)\in{D^n} $.对$ \varphi\in{L^\infty(D^n)} $,带有符号函数$ \varphi $的Toeplitz算子$ T_\varphi: L_a^2(D^n)\rightarrow{L_a^2(D^n)} $的定义为

(1.1) $ \begin{equation} T_{\varphi}f(z) = P({\varphi}f)(z) = \int_{D^n}\varphi(w)f(w)\overline{K_z(w)}{\rm d}A(w). \end{equation} $

按此方式定义的算子是最简单最自然的方式.这里我们将考虑更广的一类Toeplitz算子.用$ u $表示$ D^n $上的任意一个有限复测度,与方程(1.1)类似可以定义$ L^2(D^n, {\rm d}A) $上的算子$ T_u $如下

(1.2) $ \begin{equation} T_{u}f(z) = \int_{D^n}f(w)\overline{k_z(w)}{\rm d}u(w). \end{equation} $

对于$ F(z)\in{L^1(D^n, {\rm d}A)} $,如果d$ u(z) = F(z){\rm d}A(z) $,则我们简记为$ T_u = T_F $.这个算子一般是定义在多项式上并且任意多项式在这个算子作用下的像是解析的.我们感兴趣的是在空间$ L_a^2(D^n) $范数下稠定义算子的有界性.这种情形比较常见.例如如果$ u $有紧支集,则$ T_u $不仅是有界的而且是紧的.这样如果$ F\in{L^1(D^n, {\rm d}A)} $并存在一个$ r\in{(0, 1)} $,使得$ F $在环面$ \{z:r < \|z\| = \max\limits_{1\leq{k}\leq{n}}|z_k| < 1\} $是(本质)有界的,则$ F $是等于带有紧支集$ L^1 $函数和$ L^\infty $函数并且$ T_F $是一个有界算子.但在$ {L^1(D^n, {\rm d}A)} $中的函数没有相应于算子有界性的特征.这促使我们给出下面的定义.

定义1.1  取$ F\in{L^1(D^n, {\rm d}A)} $.

(a)在方程(1.1)中取$ u = F $,如果在$ L_a^2(D^n) $上定义一个有界算子,我们称$ F $是$ T $ -函数.

(b)如果$ F $是一个$ T $ -函数,称算子$ T_F $是按方程$ (1.1) $的方式定义的算子的连续扩张.当且仅当$ T_F $用这种形式定义称作Toeplitz算子.

(c)如果存在一个$ r\in{(0, 1)} $使得$ F $是在环面$ \{z:r < \|z\| = \max\limits_{1\leq{k}\leq{n}}|z_k| < 1\} $上(本质)有界,则我们称$ F $几乎有界.

注意到这个$ T $ -函数形成了一个$ {L^1(D^n, {\rm d}A)} $的一个真子集,其中它包含所有界和几乎有界的函数.

本文考虑什么情况下多圆盘Bergman空间上的两个Toeplitz算子$ T_f $和$ T_g $的乘积等于一个Toeplitz算子$ T_h $,以及什么时候两个Toeplitz算子能交换?在Bergman空间中,有关这方面的问题已有了一些成果.在文献[1]中作者得到了Brown-Halmos型结果.文献[2]中Patrick Ahern给出了Berezin变换的值域并应用到了Toeplitz算子乘积$ T_f{T_g} $这个理论.对更广泛的情形,文献[3]中作者讨论了两个带有拟齐次符号的Toeplitz算子的乘积.文献[4]中Louhichi和Zakariasy刻画了带有拟齐次Toeplitz算子的交换性.文献[5-6]中还给出了多元盘Bergman空间中的Toeplitz算子一些相关结果.本文将在多圆盘Bergman空间研究与文献[3-4]相似的问题.

对任何多重指标$ \alpha = (\alpha_1, \alpha_2, \cdots, \alpha_n) $ (其中每个$ \alpha_i $是一个非负整数),我们记$ |\alpha| = \alpha_1+\alpha_2+\cdots+\alpha_n, $和$ \alpha! = \alpha_1!{\alpha_2!}\cdots{\alpha_n!}. $并且对$ z = (z_1, \cdots, z_n)\in{D^n} $,记$ z^\alpha = z_1^{\alpha_1}\cdots{z_n^{\alpha_n}}. $对于取两个多重指标$ \alpha = (\alpha_1, \cdots, \alpha_n) $和$ \beta = (\beta_1, \cdots, \beta_n) $,符号$ \alpha\succeq{\beta} $是指

$ \alpha_i\geq{\beta_i}, i = 1, \cdots, n, $

否则记为$ \alpha\nsucceq{\beta} $.同时$ \alpha\perp{\beta} $表示

$ \alpha_1\beta_1+\cdots+\alpha_n\beta_n = 0. $

另外定义

$ \alpha-\beta = (\alpha_1-\beta_1, \cdots, \alpha_n-\beta_n). $

当$ \alpha\succeq{\beta} $,则有

$ |\alpha-\beta| = |\alpha|-|\beta|. $

在下面的计算过程中Mellin变换是最有用的工具之一.某个函数$ \varphi $的Mellin变换$ \hat{\varphi} $定义形式如下

$ \hat{\varphi}(z) = \int_0^{\infty}\varphi(s)s^{z-1}{\rm d}s. $

对$ L^{1}([0, 1], r{\rm d}r) $ (考虑在区间$ [1, \infty) $等于零的函数)中的函数应用Mellin变换,显而易见对这些函数的Mellin变换在$ \{z:\Re z\geq2\} $上有定义,并且在$ \{z:\Re z\geq2\} $上解析.一个函数是由一部分Mellin系数的值决定的.事实上有下面典型定理.

定理2.1^[7]  若$ f $在$ \{z:\Re(z) > 0\} $上是一个有界解析函数,并且在两两互异的点列$ z_1, z_2, \cdots, z_n, \cdots $上函数值为零,其中

ⅰ) $ \inf\{|z_n|\} > 0 $,

ⅱ) $ \sum\limits_{n\geq1}\Re(\frac{1}{z_n}) = \infty, $

则函数$ f $在$ \{z:\Re(z) > 0\} $上恒等于零.

注2.1  上述定理表明取$ \varphi(r)\in{L^1([0, 1], r{\rm d}r)} $,并且如果存在一列$ \{n_k\}_{k\geq0}\subset{{\Bbb N}} $使得

$ \hat{\varphi}(n_k) = 0 \ \ \mbox{和} \ \ \sum\limits_{k\geq0}\frac{1}{n_k} = \infty, $

则在$ z\in\{z:\Re(z) > 2\} $上$ \hat{\varphi} = 0 $并且$ \varphi = 0 $.

当考虑两个Toeplitz算子的乘积时,经常用到符号函数的Mellin卷积的问题.两个函数$ f $和$ g $的Mellin卷积表示为$ f*{_M}g $,并且它的定义形式如下

$ (f*{_M}g)(r) = \int_r^1f(\frac{r}{t})g(t)\frac{{\rm d}t}{t}. $

由Mellin卷积的定义表明

$ (\widehat{f*{_M}g})(r) = \hat{f}(r)\hat{g}(r). $

并且如果$ f $和$ g $在$ L^1([0, 1], r{\rm d}r) $中,则$ f*{_M}g $也同样属于这个空间.

令$ \varphi_i\in{L^1(D, {\rm d}A)} $为径向函数,并且假定$ \varphi_i(z_i) = \varphi_i(|z_i|) ({z_i}\in{D}). $如果$ \varphi(z)\in{L^1(D^n, {\rm d}A)} $仅依靠$ |z_1|, |z_2|, \cdots, |z_n| $,则称它是可分别径向的.本文研究$ \varphi(z) = \prod\limits_{i = 1}^n\varphi_{i}(z_i) $形式的径向函数$ {\varphi}_i $的乘积.通过直接计算有

(2.1) $ \begin{equation} T_\varphi(z^k) = {2^n}{\prod\limits_{i = 1}^n}(k_i+1)\widehat{\varphi_1}(2k_1+2)\cdots\widehat{\varphi_n}(2k_n+2)z^k. \end{equation} $

现在给出函数$ f\in{L^1(D, {\rm d}A)} $的径向化定义形式

$ rad(f)(z) = \frac{1}{2\pi}\int_0^{2\pi}f({\rm e}^{{\rm i}t}z){\rm d}t. $

明显地一个函数$ f $是径向的当且仅当$ rad(f) = f $.

命题2.1  取$ \varphi\in{L^1(D^n, {\rm d}A)} $,并且$ \varphi(z) = \prod\limits_{i = 1}^n\varphi_{i}(z_i) $.则下面的论断是等价的

(a)存在$ \lambda_k\in C $使得$ T_\varphi(z^k) = \lambda_{k}z^k $,其中$ k = (k_1, k_2, \cdots, k_n), k_i\geq0, i = 1, 2, \cdots, n; $

(b) $ \varphi $是可分离变量的径向函数.

证  取$ \varphi\in{L^1(D^n, {\rm d}A)} $,且$ {{z}^m}, {{z}^l}\in{L_a^2(D^n)} $,则有

$ \begin{eqnarray*} &&\langle T_{rad(\varphi_1)rad(\varphi_2)\cdots{rad(\varphi_n)}}z^l, {z}^m\rangle\\ & = &\int_{D^n}rad(\varphi_1)rad(\varphi_2)\cdots{rad(\varphi_n)}z^l\overline{z}^m{\rm d}A(z)\\ & = &\frac{1}{2\pi}\int_0^{2\pi}\int_D{\varphi_1({\rm e}^{{\rm i}t_1}z_1)}z_1^{l_1}\overline{z_1}^{m_1}{\rm d}A(z_1){\rm d}t_1\cdots \frac{1}{2\pi}\int_0^{2\pi}\int_D{\varphi_n({\rm e}^{{\rm i}t_n}z_n)}z_n^{l_n}\overline{z_n}^{m_n}{\rm d}A(z_n){\rm d}t_n\\ & = &\frac{1}{2\pi} \bigg(\int_0^{2\pi}{\rm e}^{{\rm i}(m_1-l_1)t_1}{\rm d}t_1\bigg)\cdots\frac{1}{2\pi} \bigg(\int_0^{2\pi}{\rm e}^{{\rm i}(m_n-l_n)t_n}{\rm d}t_n\bigg)\langle T_\varphi{z^l}, z^m\rangle\\ & = &\left\{ \begin{array}{ll} 0, & \mbox{当} \ l\neq{m}, \nonumber \\ \langle T_\varphi{z^l}, z^m\rangle , &\mbox{当}\ l = m.\nonumber \end{array} \right. \end{eqnarray*} $

这样$ T_{rad(\varphi_1)rad(\varphi_2)\cdots{rad(\varphi_n)}} = T_\varphi $当且仅当(a)成立.并且$ T_{rad(\varphi_1)rad(\varphi_2)\cdots{rad(\varphi_n)}} = T_\varphi $当且仅当$ rad(\varphi_1)rad(\varphi_2)\cdots{rad(\varphi_n)} = \varphi $.证毕.

命题2.2  设$ \varphi_{1}(z) = \prod\limits_{i = 1}^n\varphi_{1i}(z_i) $和$ \varphi_{2}(z) = \prod\limits_{i = 1}^n\varphi_{2i}(z_i) $是径向的$ T $ -函数乘积.如果$ T_{\varphi_{1}}T_{\varphi_{2}} = T_\psi $,则$ \psi $是可分离变量的径向$ T $ -函数.

证  用方程(2.1)计算$ T_{\varphi_1}T_{\varphi_2}(z^k) $,由命题2.1表明$ \psi $是可分离变量的径向函数.而且$ T_\psi $是有界算子.证毕.

定理2.2  令$ \varphi_{1}(z) = \prod\limits_{i = 1}^n\varphi_{1i}(z_i) $和$ \varphi_{2}(z) = \prod\limits_{i = 1}^n\varphi_{2i}(z_i) $是非零的径向$ T $ -函数乘积.则$ T_{\varphi_1}T_{\varphi_2} $等于Toeplitz算子$ T_\psi $当且仅当$ \psi_i $是方程

(2.2) $ \begin{equation} C_i{{\Bbb I}}*{_M}\psi_i = {\varphi_{1i}}*{_M}\varphi_{2i}, \forall i\in\{1, 2, \cdots, n\} \end{equation} $

的解,其中$ \psi(z) = \prod\limits_{i = 1}^n\psi_{i}(z_i) $, $ {\Bbb I} $表示常值为1的函数, $ C_i $是与$ i $有关的常数.

证  对$ k = (k_1, k_2, \cdots, k_n), k_i\in{{{\Bbb Z}}_{+}} $,由直接计算有

$ T_{\varphi_1}T_{\varphi_2}(z^k) = T_\psi(z^k) $

当且仅当

$ {2^n}{\prod\limits_{i = 1}^n}(k_{i}+1)\widehat{{\varphi_{11}}*{_M}\varphi_{21}}(2k_{1}+2) \cdots\widehat{{\varphi_{1n}}*{_M}\varphi_{2n}}(2k_{n}+2) = \widehat{\psi_{1}}(2k_{1}+2)\cdots\widehat{\psi_{n}}(2k_n+2). $

取定$ k_2^*, k_3^*, \cdots, k_n^* $使得

$ {\prod\limits_{i = 2}^n}({2k_{i}^*}+2) \widehat{{\varphi_{1i}}*{_M}\varphi_{2i}} ({2k_{i}^*}+2)\neq{0}. $

由假设$ {\varphi_{1i}} $和$ {\varphi_{2i}} $不恒等于为零,故满足上述条件的正整数$ {k_i^*}, \ i\geq{2} $存在.因此

$ \widehat{{\varphi_{11}}*{_M}\varphi_{21}} (2k_{1}+2) = \frac{\prod\limits_{i = 2}^n\widehat{\psi_{i}}(2k_{i}^*+2)} {\prod\limits_{i = 2}^n({2k_{i}^*}+2) \widehat{{\varphi_{1i}}*{_M}} \varphi_{2i} (2k_{i}^*+2)} \widehat{{{\Bbb I}}*{_M}}\psi_1(2k_{1}+2), \ k_1\in{{{\Bbb Z}}_{+}}, $

其中$ \frac{1}{2k_1+2} = {\hat{{{\Bbb I}}}(2k_1+2)} $.令$ C_1 = \frac{\prod\limits_{i = 2}^n\widehat{\psi_{i}}(2k_{i}^*+2)} {\prod\limits_{i = 2}^n({2k_{i}^*}+2) \widehat{{\varphi_{1i}}*{_M}\varphi_{2i}}({2k_{i}^*}+2)} $,则有

$ \widehat{\varphi_{11}*{_M}\varphi_{21}}(2k_{1}+2) = C_1{\widehat{{{\Bbb I}}*{_M}}\psi_1(2k_{1}+2)}. $

同理对任意$ {i}\in\{2, 3, \cdots, n\} $,都有

(2.3) $ \begin{equation} C_i\widehat{{{\Bbb I}}*{_M}}\psi_i(2k_{i}+2) = \widehat{\varphi_{1i}*{_M}}\varphi_{2i}(2k_{i}+2), \end{equation} $

由注2.1,方程(2.2)等价于方程(2.3).证毕.

用方程(2.2)可以计算很多Toeplitz算子的乘积.例如

$ T_{|z^n|}T_{|z^m|} = T_{{\psi_1}{\psi_2}\cdots{\psi_n}}, $

其中

$ \begin{eqnarray*} \psi_i = \left\{ \begin{array}{ll} \frac{1}{C_i}\bigg[\frac{n_i}{n_{i}-m_{i}}|z_i|^{n_i}-\frac{m_i}{n_i-m_i}|z_i|^{m_i}\bigg], \; \; \; & n_i\neq{m_i}, \\ \frac{1}{C_i}\big[{|z_i|^{n_{i}}}(1+n_{i}\log|z_i|)\big], \; \; &n_{i} = m_{i}.\nonumber \end{array} \right. \end{eqnarray*} $

接下来将在多圆盘Bergman空间上研究带有可分离径向函数为符号的几个Toeplitz算子零乘积问题.

定理2.3  令$ \varphi_1, \varphi_2, \cdots, \varphi_N $为可分离径向$ T $ -函数, $ \varphi_{ki} $是一个径向函数,其中$ \varphi_k = \varphi_{k1}\varphi_{k2}\cdots\varphi_{kn}, \ k = 1, 2, \cdots {N} $, $ i = 1, 2, \cdots, n $.则$ T_{\varphi_1}T_{\varphi_2}\cdots{T_{\varphi_N}} = 0 $当且仅当存在$ i $和$ k $,有$ \varphi_{ki} = 0 $.

证  充分性显然.为证必要性,假设$ T_{\varphi_1}T_{\varphi_2}\cdots{T_{\varphi_N}} = 0 $,通过用与方程(2.1)相同的计算方法,则有

$ \begin{eqnarray*} T_{\varphi_1}\cdots{T_{\varphi_N}}(z^\alpha) & = &(2\alpha_{1}+2)^{N}\cdots(2\alpha_{n}+2)^{N} \widehat{\varphi_{11}}(2\alpha_{1}+2)\cdots\widehat{\varphi_{1n}}(2\alpha_{n}+2)\\ &&\cdots \widehat{\varphi_{N1}}(2\alpha_{1}+2)\cdots\widehat{\varphi_{Nn}}(2\alpha_{n}+2)z^{\alpha}\\ & = &0. \end{eqnarray*} $

$ \widehat{\varphi_{11}}(2\alpha_{1}+2)\cdots\widehat{\varphi_{1n}}(2\alpha_{n}+2)\cdots \widehat{\varphi_{N1}}(2\alpha_{1}+2)\cdots\widehat{\varphi_{Nn}}(2\alpha_{n}+2) = 0. $

令

$ E_{k, i} = \{\alpha_i\in{{\Bbb N}}:\widehat{\varphi_{k, i}}(2\alpha_i+2) = 0\}, k = 1, \cdots, N; i = 1, 2, \cdots, n. $

注意到

$ E_{11}\cup{E_{12}}\cup\cdots\cup{E_{1n}}\cup\cdots\cup{E_{N1}}\cup\cdots\cup{E_{Nn}} = {\Bbb N}, $

因此存在某个$ k $和$ i $使得

$ \sum\limits_{\alpha_i\in{E_{ki}}}\frac{1}{\alpha_i} = \infty, $

则由注2.1知$ \varphi_{ki} = 0 $.

本段中将要讨论两个带有拟齐次函数为符号Toeplitz算子乘积是否等于一个Toeplitz算子问题.

定义3.1  令$ \varphi $是$ L^1(D^n, {\rm d}A) $中的函数并且具有形式$ {\rm e}^{{\rm i}({k_{1}{\theta}_1}+{k_{2}{\theta}_2}+\cdots{k_{n}{\theta}_n})}f $,其中$ f $是一个径向函数, $ k = ({k_1}, {k_2}, \cdots, k_n) $.则称$ \varphi $是阶为$ k $的拟齐次函数.

定理3.1  令$ p = (p_1, p_2, \cdots, p_n), s = (s_1, s_2, \cdots, s_n) $,其中$ p_i, s_i\in{{{\Bbb Z}}_+}, p\succeq {s} $;并且$ \varphi_1 = \varphi_{11}\varphi_{12}\cdots\varphi_{1n} $, $ \varphi_2 = \varphi_{21}\varphi_{22}\cdots\varphi_{2n} $是$ D^n $上两个可分离变量的径向函数,使得$ {\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}}\varphi_1 $和$ {\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}\varphi_2 $是$ T $ -函数, $ \psi = \psi_1\psi_2\cdots\psi_n $是一个可积的可分离径向函数,则

$ T_{{\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}\varphi_1}}{T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}{\varphi_2}}} = T_{{\rm e}^{{\rm i}(p_1-s_1)\theta_1}{\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}{\psi_2}\cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}{\psi_n}}} $

当且仅当

(a) $ {\rm e}^{{\rm i}(p_1-s_1)\theta_1}{\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}{\psi_2} \cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}{\psi_n}} $是一个$ T $ -函数;

(b)存在某个$ i $,对$ k\nsucceq {s} $有

$ \widehat{\psi_i}(2k_i+p_i-s_i+2) = 0; $

(c)对$ k\succeq{s}, \ i = 1, 2, \cdots, n $, $ \psi_i $是方程

$ {{\Bbb I}}*{_M}r_i^{p_i+s_i}\psi_{i} = {C_i}r_i^{p_i}{\varphi_{1i}}*{_M}r_i^{s_i}\varphi_{2i} $

的解.

证  如果$ {\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}}\varphi_1 $和$ {\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}\varphi_2 $是$ T $ -函数,并且

$ T_{{\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}\varphi_1}}{T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}{\varphi_2}}} = T_{{\rm e}^{{\rm i}(p_1-s_1)\theta_1}{\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}{\psi_2}\cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}{\psi_n}}}, $

则(a)成立.通过直接计算得

$ \begin{eqnarray*} T_{{\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}\varphi_1}}(z^k)& = & 2^n(k_1+p_1+1)\widehat{\varphi_{11}}(2k_1+p_1+2)\cdots\\ &&{(k_n+p_n+1)\widehat{\varphi_{n1}}(2k_n+p_n+2)}z^{k+p}, \end{eqnarray*} $

并且有

$ \begin{eqnarray*} &&T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}\varphi_2}(z^k) \\ & = &\int_{D^{n}}{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}}{\varphi_2}(w)w^{k}{k_{w}(z)}{{\rm d}A(w)}\\ & = &\left\{ \begin{array}{ll} 0, & k\nsucceq {s}, \\ 2^{n}(k_1-s_1+1)\widehat{\varphi_2}(2k_1-s_1+2)\cdots(k_n-s_1+1)\widehat{\varphi_2}(2k_n-s_n+2)z^{k-s}, \ \ &k\succeq{s}.\nonumber \end{array} \right. \end{eqnarray*} $

而且

$ \begin{eqnarray*} &&T_{{\rm e}^{{\rm i}(p_1-s_1)\theta_1}{\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}{\psi_2}\cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}{\psi_n}}}(z^k)\\ & = &{2^n}(k_1+p_1-s_1+1)\widehat{\psi_1}(p_1-s_1+2k_1+2)\cdots\\ &&(k_n+p_n-s_n+1)\widehat{\psi_n}(p_n-s_n+2k_n+2)z^{k-s+p}. \end{eqnarray*} $

因此可得如果$ k\nsucceq {s} $,则存在某个$ i $,使得

$ \widehat{\psi_{i}}(p_i-s_i+2k_i+2) = 0. $

这样有(b)成立.如果$ k\succeq{s} $,则有

(3.1) $ \begin{equation} \prod\limits_{i = 1}^n \widehat{\psi_{i}}(p_i-s_i+2k_i+2) = {2^n} \prod\limits_{i = 1}^n(k_i-s_i+1)\widehat{\varphi_{2i}}(2k_i-s_i+2)\widehat{\varphi_{1i}}(2k_i-2s_i+p_i+2). \end{equation} $

现在取$ k_2^{*}, k_3^{*}, \cdots, k_n^{*} $使得$ \prod\limits_{i = 2}^n \widehat{\psi_{i}}(p_i-s_i+2k_i^{*}+2)\neq{0} $.因为$ \psi_{i} $不恒等于零,因此这样的$ k_i^{*} $存在.故方程(3.1)等价于

$ \widehat{\psi_{1}}(p_1-s_1+2k_i+2) = {C_1}(k_1-s_1+1)\widehat{\varphi_{2i}}(2k_1-s_1+2)\widehat{\varphi_{11}}(2k_1-2s_1+p_1+2), $

其中

$ {C_1} = {2^{n-1}}\frac{{\prod\limits_{i = 2}^n}(k_i^{*}-s_i+1)\widehat{\varphi_{2i}}(2k_i^{*}-s_i+2)\widehat{\varphi_{1i}}(2k_i^{*}-2s_i+p_i+2)} {{\prod\limits_{i = 2}^n}\widehat{\psi_{i}}(p_i-s_i+2k_i^{*}+2)}. $

当$ i = 1 $时有

$ \frac{\widehat{r_1^{p_1+s_1}\psi_i}(2(k_1-s_1)+2)}{{C_1}(2(k_1-s_1)+2)} = \widehat{r_1^{p_1}}\varphi_{11}(2k_1-2s_1+2)\widehat{r_1^{s_1}}\varphi_{21}(2k_1-s_1+2), $

也就是

(3.2) $ \begin{equation} \hat{{\Bbb I}}[2(k_1-s_1+1)]\widehat{r_1^{p_1+s_1}}\psi_1[2(k_1-s_1)+2] = {C_1}\widehat{r_1^{p_1}}\varphi_{11}(2k_1-2s_1+2)\widehat{r_1^{s_1}}\varphi_{21}(2k_1-s_1+2). \end{equation} $

用同样的方法,对$ i = 2, 3, \cdots, n $有

(3.3) $ \begin{equation} \hat{{\Bbb I}}[2(k_i-s_i+1)]\widehat{r_i^{p_i+s_i}}\psi_i[2(k_i-s_i)+2] = {C_i}\widehat{r_i^{p_i}}\varphi_{1i}(2k_i-2s_i+2)\widehat{r_i^{s_i}}\varphi_{2i}(2k_i-s_i+2). \end{equation} $

注2.1表明方程(3.2)和(3.3)等价于条件(c).

相反地如果$ {\rm e}^{{\rm i}(p_1-s_1)\theta_1}{\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}{\psi_2}\cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}{\psi_n}} $是一个T-函数并且$ \psi_i $满足(b)和(c), $ T_{{\rm e}^{{\rm i}(p_1-s_1)\theta_1} {\rm e}^{{\rm i}(p_2-s_2)\theta_2} \cdots {\rm e}^{{\rm i}(p_n-s_n)\theta_n}\psi} $是一个有界Toeplitz算子,并且在多圆盘上与这个算子乘积$ T_{{\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}\varphi_1}} {T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}{\varphi_2}}} $取相同的值.因此这个定理得证.

注3.1  由于$ T_{{\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}\varphi_1}}{T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}{\varphi_2}}} $等于一个Toplitz算子当且仅当它的自伴算子$ T_{{\rm e}^{{\rm i}s_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}}}\overline{\varphi_2}{T_{{\rm e}^{-{\rm i}p_1\theta_1}}}\cdots{{\rm e}^{-{\rm i}p_n\theta_n}\overline{\varphi_1}} $等于一个Toplitz算子,因此对于$ 0\leq{p} < s $的情形,可以由上述定理得到.

对于拟齐次符号的线性组合的情形,我们很容易能获得相应的更复杂的结果.下面应用酉算子

$ U_w:\ \ \ \ L_a^2(D^n)\rightarrow{L_a^2(D^n)}, $

$ U_{w}f = (f\circ{\varphi_w})k_w, \ \ \ \ f\in{L_a^2(D^n)}, $

其中$ \varphi_{w}(z) = (\varphi_{w_1}(z_1), \cdots\varphi_{w_n}(z_n)) $, $ \varphi_{w_i}(z_i) = \frac{w_i-z_i}{1-\overline{w_i}{z_i}} $是单位圆盘的自同构,可以获得定理3.3的推广形式.

推论3.1  令$ p, s $和$ \varphi_1 $, $ \varphi_2 $, $ \psi $与定理3.1中的取值一致,如果$ \widetilde{\varphi_1} = ({\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}}\varphi_1\varphi_1)\circ{\varphi_w} $, $ \widetilde{\varphi_2} = ({\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}\varphi_2)\circ{\varphi_w} $和$ \widetilde{\psi} = ({{\rm e}^{{\rm i}(p_1-s_1)\theta_1}{\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}{\psi_2}\cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}{\psi_n}}})\circ{\varphi_w} $,则Toeplitz算子乘积$ T_{\widetilde{\varphi_1}}T_{\widetilde{\varphi_2}} $是等于一个Toeplitz算子$ T_{\widetilde{\psi}} $.

证  通过定理3.1和经典结果(例如参考文献[6]) $ U_{w}^{-1}T_{f\circ{\varphi_w}}U_w = T_f $,则有

$ \begin{eqnarray*} T_{\widetilde{\varphi_1}}T_{\widetilde{\varphi_2}}& = &U_w^{-1} T_{{\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}\varphi_1}}U_wU_w^{-1}T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}\varphi_2}U_w\\ & = &U_w^{-1}T_{{\rm e}^{{\rm i}p_1{\theta_1}}\cdots{{\rm e}^{{\rm i}p_n{\theta_n}}\varphi_1}}T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}}\varphi_2}U_w\\ & = &U_w^{-1}T_{{{\rm e}^{{\rm i}(p_1-s_1)\theta_1}{\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}{\psi_2}\cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}}\psi}}U_w\\ & = &T_{({\rm e}^{{\rm i}(p_1-s_1)\theta_1\psi_1}{\rm e}^{{\rm i}(p_2-s_2)\theta_2}\psi_2\cdots{{\rm e}^{{\rm i}(p_n-s_n)\theta_n}\psi})\circ{\varphi_w}} = T_{\widetilde{\psi}}. \end{eqnarray*} $

由此结论得证.

在本段中我们刻划多圆盘Bergman空间上带有拟齐次符号Toeplitz算子的交换性.下面定理表明两个带有拟齐次符号的Toeplitz算子交换性仅在平凡情形成立.

定理4.1  令$ \psi = \psi_1\cdots\psi_n $,其中$ \psi_i, i = 1, 2, \cdots, n $是有界非零径向函数并且$ {\rm e}^{{\rm i}p_1\theta_1\cdots{ip_n\theta_n}}\phi $是一个阶数为$ p = (p_1, \cdots, p_n), $$ p_{i} > 0, i = 1, 2, \cdots, n $拟齐次有界函数,如果$ T_{\psi}T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}\phi}} = T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}\phi}}T_{\psi} $,则对某个$ i $有$ \phi_i = 0 $或对所有$ i $有$ \psi_i $为常数.

证  对所有$ k = (k_1, \cdots, k_n), {k_i}\geq0 $,则有

$ T_{\psi}T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}\phi}}(z^k)\\ = \Big(\prod\limits_{i = 1}^n {(2k_i+2p_i+2)^2}\widehat{\phi_i}(2k_i+p_i+2)\widehat{\psi_i}(2k_i+2p_i+2)\Big){z^{k+p}} $

和

$ T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}\phi}}{T_{\psi}}(z^k) = \Big(\prod\limits_{i = 1}^n {(2k_i+2)}{(2k_i+2p_i+2)}\widehat{\psi_i}(2k_i+2)\widehat{\phi_i}(2k_i+p_i+2)\Big){z^{k+p}}. $

如果$ T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}\phi}} $和$ T_\psi $能交换,对所有$ k = (k_1, \cdots, k_n)\in {{\Bbb Z}}^{n}_{+}, $则上面的方程表明

$ \prod\limits_{i = 1}^n{(2k_i+2p_i+2)}\widehat{\psi_i}(2k_i+2p_i+2)\widehat{\phi_i}(2k_i+p_i+2) = \prod\limits_{i = 1}^n{(2k_i+2)}\widehat{\psi_i}(2k_i+2)\widehat{\phi_i}(2k_i+p_i+2). $

令$ E $是由$ k = (k_1, \cdots, k_n)\in {\Bbb Z}^{n}_{+} $中的使得$ \prod\limits_{i = 1}^n\hat{\phi_i}(2k_i+p_i+2)\neq0 $的数组组成的集合,令$ E^c $是$ E $在$ {{\Bbb Z}}^{n}_{+} $中的补集.如果存在$ i^*\in \{1, 2, .\cdots, n\} $使得

$ \sum\limits_{k = (k_1, \cdots , k_{i^*}, \cdots , k_n)\in E^c}\frac{1}{2k_{i^*}+2} = \infty, $

则对这样的$ i^* $有$ \phi_{i^*} = 0 $.如果这样的$ i^* $不存在,则对所有$ i = 1, 2, \cdots, n $,有

$ \sum\limits_{k = (k_1, \cdots , k_i, \cdots , k_n)\in E}\frac{1}{2k_i+2} = \infty. $

因此有

(4.1) $ \begin{equation} \prod\limits_{i = 1}^n(2k_i+2p_i+2)\widehat{\psi_i}(2k_i+2p_i+2) = \prod\limits_{i = 1}^n (2k_i+2)\widehat{\psi_i}(2k_i+2). \end{equation} $

现在取$ k_2^{*}, k_3^{*}, \cdots, k_n^{*} $使得$ \prod\limits_{i = 2}^n\widehat{\psi_i}(2k_i^{*}+2{p_i}+2)\neq{0} $.因为$ \psi_{i} $不恒等于零,因此这样的$ k_i^{*} $存在.因此方程$ (4.1) $等价于

(4.2) $ \begin{equation} C(2k_1+2)\widehat{\psi_1}(2k_1+2) = (2k_1+2p_1+2)\widehat{\psi_1}(2n_1+2p_1+2), \end{equation} $

其中

$ C = \frac{ {\prod\limits_{i = 2}^n}(2k_i^{*}+2)\widehat{\psi_i}{(2k_i^{*}+2)}} {{\prod\limits_{i = 2}^n} (2k_i^{*}+2p_i+2)\widehat{\psi_i}(2k_i^{*}+2p_i+2)} $

和$ \sum\limits_{{k_1}\in{{\Bbb Z}}}\frac{1}{{2k_{1}}+2} = \infty. $表示

$ F(z) = Cz\widehat{\psi_1}(z)-(z+2p_1)\widehat{\psi_1}(z+2p_1), \forall z\in\{\Re z>0\}. $

既然$ \psi_i $是有界的,则$ F $在$ \{z:\in \Re z > 0\} $上是解析的并且有界的.且方程$ (4.2) $表明

$ F(2k_1+2) = 0, \ \forall{k_1\in{{\Bbb N}}}. $

根据定理2.1 $ F $必为零,这样

$ Cz\widehat{\psi_1}(z) = (z+2p_1)\widehat{\psi_1}(z+2p_1), \forall z\in\{{\rm Re}z>0\}. $

上面方程表明,对任何大于零的整数$ n_0 $,有

$ {C^m}n_0\widehat{\psi_1}(n_0) = (n_0+2mp_1)\widehat{\psi_1}(n_0+2mp_1), \forall{m\in{{\Bbb N}}}. $

如果用$ C_1 $表示常数$ {C^m}{n_0\widehat{\psi_1}(n_0)} $,则对所有$ {m\in{{\Bbb N}}} $,有

$ \widehat{\psi_i}(n_{0}+2mp_1) = \frac{C_1}{n_{0}+2mp_1} = C_1{\hat{{\Bbb I}}}(n_0+2mp_1), $

再由注2.1, $ \psi_1 $等于$ C_1{\Bbb I} $.对$ i = 2, 3, \cdots, n $的情形用同样的方法可证得$ \psi_i $为常数.证毕.

注4.1  当$ p < 0 $时,由算子的自伴性可知结果仍成立.

定理4.2  令$ p = (p_1, p_2, \cdots, p_n), s = (s_1, s_2, \cdots, s_n) $, $ p_i, {s_i} $均是整数且$ {p_i}\geq{s_i}\geq1, $$ i = 1, 2, \cdots, n $.令$ \phi = \phi_1\cdots\phi_n $, $ \psi = \psi_1\cdots\psi_n $是径向函数的乘积.如果

$ T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}{\phi}}}T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}{\psi}}} = T_{{\rm e}^{-{\rm i}s_1\theta_1}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}{\psi}}}T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}{\phi}}}, $

则存在某个$ i $使得$ \phi_i = 0 $或者$ \psi_i = 0 $.

证  对任意$ k = (k_1, k_2, \cdots, k_n), \ k_{i}\geq0 $,我们有

$ \begin{eqnarray*} && T_{{\rm e}^{-{\rm i}s_1\theta_1} \cdots {\rm e}^{-{\rm i}s_n\theta_n} \psi} T_{{\rm e}^{{\rm i}p_1\theta_1} \cdots {\rm e}^{{\rm i}p_n\theta_n} \phi} (z^k)\\ & = &\Big(\prod\limits_{i = 1}^n {(2k_i+2p_i+2)(2k_i+2p_i-2s_i+2)}\widehat{\phi_i}(2k_i+p_i+2)\widehat{\psi_i}(2k_i+2p_i-s_i+2)\Big){z^{k+p-s}}, \end{eqnarray*} $

以及

$ \begin{eqnarray*} && T_{{\rm e}^{{\rm i}p_1\theta_1} \cdots {\rm e}^{{\rm i}p_n\theta_n} \phi} T_{{\rm e}^{-{\rm i}s_1\theta_1} \cdots {\rm e}^{-{\rm i}s_n\theta_n} \psi}(z^k) \\ & = &\left\{ \begin{array}{ll} 0, \hskip 8.7cm k\nsucceq {s}, \\ \Big(\prod\limits_{i = 1}^n {(2k_i-2s_i+2)(2k_i+2p_i-2s_i+2)}\widehat{\phi_i}(2k_i-s_i+2)\widehat{\psi_i}(2k_i+2p_i-s_i+2)\Big){z^{k+p-s}}, \\ \hskip 9cm k\succeq{s}.\nonumber \end{array} \right. \end{eqnarray*} $

如果$ T_{{\rm e}^{{\rm i}p_1\theta_1}\cdots{{\rm e}^{{\rm i}p_n\theta_n}{\phi}}} $和$ T_{{\rm e}^{-{\rm i}s_1\theta_1}}\cdots{{\rm e}^{-{\rm i}s_n\theta_n}{\psi}} $交换,则有

(a)存在$ i_0 $, $ \widehat{\phi_{i_0}}(2k_{i_0}+p_{i_0}+2)\widehat{\psi_{i_0}}(2k_{i_0}+2p_{i_0}-s_{i_0}+2) = 0, $如果$ k\nsucceq {s} $;

(b)对所有的$ i $, $ \prod\limits_{i = 1}^n \widehat{\phi_i}(2k_i+p_i+2)\widehat{\psi_i}(2k_i+2p_i-s_i+2) = \prod\limits_{i = 1}^n \frac{k_i-s_i+1}{k_i+p_i+1} \widehat{\phi_i}(2k_i+p_i-2s_i+2)\widehat{\psi_i}(2k_i-s_i+2), $如果$ k\succeq{s} $.

方程(b)中,当$ i\neq{i_0} $时若$ \phi_i = 0 $或$ \psi_i = 0 $,则命题成立;否则取$ k_1^{*}, k_2^{*}, \cdots, k_{i-1}^{*}, \cdots, $$ k_{i+1}^{*}\cdots, $$ k_n^{*} $使

$ \prod\limits_{i\neq{i_0}} \widehat{\phi_i}(2k_i^*+p_i+2)\widehat{\psi_i}(2k_i^*+2p_i-s_i+2)\neq{0}. $

因此对$ i = i_0 $时有

(c) $ \widehat{\phi_{i_0}}(2k_{i_0}+p_{i_0}+2)\widehat{\psi_{i_0}}(2k_{i_0}+2p_{i_0}-s_{i_0}+2) = C\widehat{\phi_{i_0}}(2k_{i_0}+p_{i_0}-2s_{i_0}+2)\widehat{\psi_{i_0}}(2k_{i_0}-s_{i_0}+2), $\\其中

$ C = \frac{\prod\limits_{i\neq{i_0}}{(k_i^*-s_i+1)} \widehat{\phi_i}(2k_i^*+p_i-2s_i+2)\widehat{\psi_i}(2k_i^*-s_i+2)} {\prod\limits_{i\neq{i_0}}{(k_i^*+p_i+1)}\widehat{\phi_i}(2k_i^*+p_i+2)\widehat{\psi_i}(2k_i^*+2p_i-s_i+2)}. $

方程(a)表明$ \widehat{\phi_{i_0}}(2k_{i_0}+p_{i_0}+2) = 0 $或者$ \widehat{\psi_{i_0}}(2k_{i_0}+2p_{i_0}-s_{i_0}+2) = 0, $若$ \widehat{\phi_{i_0}}(2k_{i_0}+p_{i_0}+2) = 0 $,我们令$ k_{i_1} = k_{i_0}+s_{i_0} $,否则$ k_{i_1} = k_{i_0}+p_{i_0} $,代入方程(c)直接计算可得

$ \widehat{\phi_{i_0}}(2k_{i1}+p_{i_0}+2)\widehat{\psi_{i_0}}(2k_{i1}+2p_{i_0}-s_{i_0}+2) = 0. $

依此类推可以找到一列$ k_{{(i_m)}_{m\in{N}}} $定义如下$ k_{i_{m+1}} = k_{i_m}+s_{i_0} $或者$ k_{i_m}+p_{i_0} $,使得

$ \widehat{\phi_{i_0}}(2k_{i_m}+p_{i_0}+2)\widehat{\psi_{i_0}}(2k_{i_m}+2p_{i_0}-s_{i_0}+2) = 0. $

显而易见

$ \sum\limits_{m\in{N}}\frac{1}{2k_{i_m}+1} = \infty. $

令$ E_1 = \{m:\widehat{\phi_{i_0}}(2k_{i_m}+p_{i_0}+2) = 0\} $和$ E_2 = \{m:\widehat{\psi_{i_0}}(2k_{i_m}+2p_{i_0}-s_{i_0}+2) = 0\} $,因此

$ \sum\limits_{m\in{N}}\frac{1}{2k_{i_m}+1}\leq{\sum\limits_{m\in{E_1}}\frac{1}{2k_{i_m}+1}}+{\sum\limits_{m\in{E_2}}\frac{1}{2k_{i_m}+1}}. $

级数$ {\sum\limits_{m\in{E_1}}\frac{1}{2k_{i_m}+1}} $与$ {\sum\limits_{m\in{E_2}}\frac{1}{2k_{i_m}+1}} $中至少有一个发散,则由注2.1得出$ \phi_{i_0} = 0 $或者$ \psi_{i_0} = 0 $.证毕.

注4.2  如果$ 0\leq{p} < s $,由算子$ T_{{\rm e}^{{\rm i}p\theta}{\phi}}T_{{\rm e}^{-{\rm i}s\theta}{\psi}} = (T_{{\rm e}^{{\rm i}s\theta}\bar{\psi}}T_{{\rm e}^{-{\rm i}p\theta}\bar{\phi}})^* = (T_{{\rm e}^{-{\rm i}p\theta}\bar{\phi}}T_{{\rm e}^{{\rm i}s\theta}\bar{\psi}})^* = T_{{\rm e}^{-{\rm i}s\theta}{\psi}}T_{{\rm e}^{{\rm i}p\theta}{\phi}} $的自伴性知此结果同样成立.