Bergman空间上的Toeplitz算子的乘积有限和问题

The Finite Sum Problem of the Product of Toeplitz Operators on Bergman Spaces

摘要: 本文讨论了Bergman空间上两个形如T_fk，T_gk的Toeplitz算子，其中假设g_k=g₁^(k)+g₂^(k)，g₁^(k)=a_j^(k) z^j，g₂^(k)=b_j^(k)z^j∈H^∞(D)；f_k∈L^∞(D,dA)。f_k(re^iθ)=f_p^(k)(r)e^ipθ(1≤k≤N)。探究Toeplitz算子Tfk，Tgk的有限乘积有限和的相关问题，分析计算得到了其为零算子的一个必要条件。

Abstract: This paper discusses two Toeplitz operators as T_fk, T_gk in Bergman space. In case g_k=g₁^(k)+g₂^(k), g₁^(k)=a_j^(k) z^j, g₂^(k)=b_j^(k)z^j∈H^∞(D); f_k∈L^∞(D,dA). f_k(re^iθ)=f_p^(k)(r)e^ipθ(1≤k≤N). This paper explores the related problems of the sum of the finite products of the two Toeplitz operators as Tfk, Tgk under a large amount of data and calculations. A necessary condition of zero operator is obtained (N).

1. 引言

Toeplitz算子的有限乘积有限和问题在算子理论研究中具备着重要的代数意义，算子的乘积问题，是算子理论探究中的热点问题，由于其复杂多变性，通过查阅资料发现，国内许多学者对算子的有限乘积有限和问题的研究有很大兴趣。在文献 [1] 中证明了下面刻画了两个以有界调和函数为符号的一类算子 $T_{u_j}$，

$T_{v_j}$，并探究了这类算子的有限乘积相加之和 ${\displaystyle \underset{j=1}{\overset{N}{\sum }}{T}_{u_j}{T}_{v_j}}$ 为零算子和紧算子的等价条件。将上述问题推广到

调和Bergman空间上的情况相对复杂，Yufeng Lu，Qian Ding等人 [2] 在2017年经过计算与推广发现，得到关于两类特殊符号的Toeplitz算子乘积的有限秩的结论。而对于Bergman空间上的有限乘积有限和问题，将其推广到某类特殊符号的Toeplitz算子上还有待探究，因此本文对Bergman空间上某类符号Toeplitz算子的乘积有限和问题进行了推广与计算，进而分析证明得到其为零算子的一个必要条件。

2. 预备知识

在这一部分中，我们介绍了Berezin变换和Mellin变换，并用第二个变换来表示第一个变换，我们将在下一节中使用这个公式。

定义2.1 若 $\phi$ 是 $\left[0,1\right]$ 上的可积函数，可定义Mellin变换：

$\hat{\phi }\left(z\right)={\displaystyle {\int }_0^1\phi \left(r\right)r^{z-1}}\text{d}r$.

Mellin变换是半平面 $\left\{z:\mathrm{Re}z>\text{1}\right\}$ 上的有界解析函数。由于三角多项式在 $L^2\left(D,\text{d}A\right)$ 中是稠密的，所以有如下极分解表示

$L^2\left(D,\text{d}A\right)=\underset{k=-\infty }{\overset{+\infty }{\oplus }}{\text{e}}^{ik\theta }\Re$,

其中 $\Re =\left\{u:D\to C:u\left(z\right)=u\left(\left|z\right|\right),且{\displaystyle {\int }_0^{1}r{\left|u\left(r\right)\right|}^2\text{d}r<\infty }\right\}$，因此对任 $f\in L^2\left(D,\text{d}A\right)$，有 $z=r{\text{e}}^{i\theta }\in D$，则f的极分解 $f\left(r{\text{e}}^{i\theta }\right)$ 为

$f\left(r{\text{e}}^{i\theta }\right)={\displaystyle \underset{k=-\infty }{\overset{\infty }{\sum }}{f}_k\left(r\right){\text{e}}^{ik\theta }},f_k\in \Re$

Mellin变换是由其零点所唯一确定，并有以下结论。

引理2.1 [3] 若 $\phi$ 为 $\left\{z:\mathrm{Re}z>1\right\}$ 上的有界解析函数，在两两不相同的点列 $z_1,z_2,\cdots$ 上取值为零，若

1) $\inf \left\{\left|z_n\right|\right\}>0$，

2) ${\displaystyle \underset{n=1}{\overset{\infty }{\sum }}\mathrm{Re}\left(\frac{1}{z_n}\right)}=\infty$，

则 $\phi$ 在半平面 $\left\{z:\mathrm{Re}z>1\right\}$ 上恒等于零。

注解：我们经常用到引理2.1的一种特殊情况：若 $\phi$ 为 $\left\{z:\mathrm{Re}z>1\right\}$ 上的有界解析函数，若存在自然数序列 ${\left\{n_k\right\}}_{k=0}^{\infty }$ 使得

$\hat{\phi }\left(n_k\right)=0$ 且 ${\displaystyle \underset{k=0}{\overset{\infty }{\sum }}\frac{1}{n_k}}=\infty$,

则 $\phi \equiv 0$。

定义2.2 Mellin卷积的定义，可用 $f{\ast }_{M}g$ 表示f和g的Mellin卷积：

$\left(f{\ast }_{M}g\right)\left(r\right)={\displaystyle {\int }_r^{1}f\left(\frac{r}{t}\right)g\left(t\right)\frac{\text{d}t}{t}}$.

其中乘法 ${\ast }_M$ 是通过上述变量的变化与正规卷积有关的。很容易看出Mellin变换将卷积转化为点积，即：

$\left(f{\ast }_{M}g\right)\left(r\right)=\hat{f}\left(r\right)\hat{g}\left(r\right)$,

如果f和g在中，则 $f{\ast }_{M}g$ 也是。

定义2.3 若 $f\in L^1\left(D,\text{d}A\right)$，f的Berezin变换记为Bf：

$\left(Bf\right)\left(z\right)=\langle f{k}_z,k_z\rangle ={\displaystyle {\int }_{D}f\left(\omega \right)\frac{{\left(1-{\left|z\right|}^2\right)}^2}{{\left|1-\bar{z}\omega \right|}^4}\text{d}A\left(\omega \right)}$, $\left(z\in D\right)$.

本文要用到Berezin变换的极坐标表示，Čučković Ž给出下述定理。

引理2.2 [4] 若 $f\in L^2\left(D,\text{d}A\right)$，对 $z=R{e}^{i\theta }$，则f的Berezin变换Bf有如下的极坐标表达式：

$\left(Bf\right)\left(z\right)=\langle f{k}_z,k_z\rangle =2{\left(1-R^2\right)}^2{\displaystyle \underset{k=-\infty }{\overset{\infty }{\sum }}{R}^{\left|k\right|}\left[{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}n\left(n+\left|k\right|\right){\hat{f}}_k\left(2n+\left|k\right|\right)R^{2\left(n-1\right)}}\right]{\text{e}}_{.}^{ik\theta }}$

下面我们来描述 $L^1\left(D,\text{d}A\right)$ 上关于Berezin变换的不动点问题，这比在 $C\left(\bar{D}\right)$ 上的不动点问题困难的多。回顾复平面上的Laplacian (拉普拉斯)变换为：

$\Delta =\frac{{\partial }^2}{\partial x^2}+\frac{{\partial }^2}{\partial y^2}=4\frac{{\partial }^2}{\partial z\partial \bar{z}}$.

在处理Berezin变换时，使用如下的不变拉普拉斯变换将更为方便：

$\tilde{\Delta }f\left(z\right)={\left(1-{\left|z\right|}^2\right)}^2\Delta f\left(z\right)$,

其中f是D上的任意一个二次可微函数。

引理2.3 若f在D上是二次可微的，则

$\tilde{\Delta }\left(f\circ \phi \right)=\left(\tilde{\Delta }f\right)\circ \phi$, 对任意 $\phi \in Aut\left(D\right)$.

其中 $Aut\left(D\right)$ 表示D的全纯自同构群，任给 $z\in D$，存在唯一 $\phi \in Aut\left(D\right)$ 满足： $\phi$ 是对合的，即 $\phi \circ \phi =I$ 为恒等映射，并且 $\phi \left(0\right)=z$， $\phi \left(z\right)=0$。

定义2.4 一个解析函数 $f:D\to C$，如果满足

$\Vert f\Vert =\sup \left\{\left|f^{\prime }\left(z\right)\right|\left(1-{\left|z\right|}^2\right)|z\in D\right\}<\infty$.

则称为函数f为Bloch函数。

为了避免下文中的重复，我们认为如果函数的极分解只有负幂项，则称它具有性质(N)。

3. 主要结果

定理3.1 假设 $g_k=g_1^{\left(k\right)}+g_2^{\left(k\right)}$， $g_1^{\left(k\right)}={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}{a}_j^{\left(k\right)}{z}^j}$， $g_2^{\left(k\right)}={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}{b}_j^{\left(k\right)}{z}^j}\in H^{\infty }\left(D\right)$ ； $f\in L^{\infty }\left(D,\text{d}A\right)$ 且 $f_k\left(r{\text{e}}^{i\theta }\right)={\displaystyle \underset{p<0}{\sum }{f}_p^{\left(k\right)}\left(r\right){\text{e}}^{ip\theta }}\left(1\le k\le N\right)$。若 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{T}_{f_k}{T}_{g_k}}=0$，则。

证明：对 $1\le k\le N$，

$\begin{array}{c}{T}_{f_k}{T}_{g_k}{k}_{\omega }\left(z\right)=T_{f_k}{T}_{g_k^{\left(1\right)}+\overline{g_k^{\left(2\right)}}}{k}_{\omega }\left(z\right)\\ =T_{f_k}{T}_{g_k^{\left(1\right)}}{k}_{\omega }\left(z\right)+T_{f_k}{T}_{\overline{g_k^{\left(2\right)}}}{k}_{\omega }\left(z\right)\\ =T_{f{g}_k^{\left(1\right)}}{k}_{\omega }\left(z\right)+\overline{g_2^{\left(k\right)}\left(\omega \right)}{T}_{f_k}{k}_{\omega }\left(z\right)\end{array}$.

进而，

$\begin{array}{c}\langle {\displaystyle \underset{k=1}{\overset{N}{\sum }}{T}_{f_k}{T}_{g_k}{k}_{\omega },k_{\omega }}\rangle ={\displaystyle \underset{k=1}{\overset{N}{\sum }}\langle T_{f_k}{T}_{g_k}{k}_{\omega },k_{\omega }\rangle }\\ ={\displaystyle \underset{k=1}{\overset{N}{\sum }}\left[\langle T_{f_k{g}_k^{\left(1\right)}}{k}_{\omega },k_{\omega }\rangle +\overline{g_2^{\left(k\right)}\left(\omega \right)}\langle T_{f_k}{k}_{\omega },k_{\omega }\rangle \right]}\\ ={\displaystyle \underset{k=1}{\overset{N}{\sum }}\left[B\left(f_k{g}_1^{\left(k\right)}\right)+\overline{g_2^{(k)}}B{f}_k\right]}\left(\omega \right)\end{array}$.

由 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{T}_{f_k}{T}_{g_k}}=0$，得

.

由引理2.2，令 $\omega =R{e}^{i\theta }$，得

$B\left(f_k\right)\left(\omega \right)=2{\left(1-R^2\right)}^2{\displaystyle \underset{P=-\infty }{\overset{+\infty }{\sum }}{R}^{\left(p\right)}\left[{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}n\left(n+\left|p\right|\right){\hat{f}}_p^k\left(2n+\left|p\right|\right)R^{2n-1}}\right]{\text{e}}^{ip\theta }}$,

而 $f_k\left(1\le k\le N\right)$ 的极分解式中关于 ${\text{e}}^{i\theta }$ 的负幂项系数为0，故系数函数对应的Mellin变换为0，进而关于 ${\text{e}}^{ip\theta }\left(p\ge 0\right)$ 的系数为0，即

$B\left(f_k\right)\left(\omega \right)=2{\left(1-R^2\right)}^2{\displaystyle \underset{p<0}{\sum }{R}^{\left|p\right|}\left[{\displaystyle \underset{n=1}{\overset{\infty }{\sum }}n\left(n+\left|p\right|\right){\hat{f}}_p^{\left(k\right)}\left(2n+\left|p\right|\right)R^{2n-2}}\right]{\text{e}}^{ip\theta }}$,

进而 $B\left(f_k\right)$ 具有性质(N) ( $1\le k\le N$ )。

另一方面，记

$\overline{g_2^{(k)}}\left(R{e}^{i\theta }\right)={\displaystyle \underset{n\ge 0}{\sum }\overline{b_n^{(k)}}{R}^n{\text{e}}^{-in\theta }}={\displaystyle \underset{n\le 0}{\sum }\overline{b_{-n}^{(k)}}{R}^{-n}{\text{e}}^{in\theta }},R{e}^{i\theta }\in D,1\le k\le N$,

综上， $\overline{g_2^{\left(k\right)}}B\left(f_k\right)$ 具有性质(N)，因而 ${\displaystyle {\sum }_{k=1}^N\overline{g_2^{\left(k\right)}}B\left(f_k\right)}$ 具有性质(N)。由 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}B\left(f_k{g}_1^{\left(k\right)}\right)}=-{\displaystyle \underset{k=1}{\overset{N}{\sum }}\overline{g_2^{\left(k\right)}}B\left(f_k\right)}$，则 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}B\left(f_k{g}_1^{\left(k\right)}\right)}$ 具有性质(N)，往证 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{f}_k{g}_1^{\left(k\right)}}$ 具有性质(N)。

令 $G_m\left(z\right)={\displaystyle \underset{j=1}{\overset{m}{\prod }}\left(1-\frac{z}{4j\left(j+1\right)}\right)}$，由引理2.3， $G_m\left(\tilde{\Delta }\right)B\left({\displaystyle \underset{k=1}{\overset{N}{\sum }}{f}_k{g}_1^{\left(k\right)}}\right)$ 在 $L^1\left(D,\text{d}A\right)$ 范数下收敛到 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{f}_k{g}_1^{\left(k\right)}}$，对 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}B\left(f_k{g}_1^{\left(k\right)}\right)}=-{\displaystyle \underset{k=1}{\overset{N}{\sum }}\overline{g_2^{\left(k\right)}}B\left(f_k\right)}$ 两边求不变Laplacian变换 $\tilde{\Delta }$ 得，

$\tilde{\Delta }B\left({\displaystyle \underset{k=1}{\overset{N}{\sum }}{f}_k{g}_1^{\left(k\right)}}\right)=-\tilde{\Delta }\left[{\displaystyle \underset{k=1}{\overset{N}{\sum }}\overline{g_2^{\left(k\right)}}B\left(f_k\right)}\right]=-{\displaystyle \underset{k=1}{\overset{N}{\sum }}\tilde{\Delta }\left[\overline{g_2^{\left(k\right)}}B\left(f_k\right)\right]}$.

计算可得，

$\tilde{\Delta }\left[\overline{g_2^{\left(k\right)}}B\left(f_k\right)\right]=\left(1-{\left|z\right|}^2\right)\overline{g_2^{\left(k\right)}}\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial z}B\left(f_k\right)+\overline{g_2^{\left(k\right)}}\tilde{\Delta }B\left(f_k\right)$.

由 $g_2^{\left(k\right)}$ 解析，故具有性质(N)；

又

$\begin{array}{l}\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial z}B\left(f_k\right)\\ =\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial z}\left[{\left(1-{\left|z\right|}^2\right)}^2{\displaystyle {\int }_D\frac{f_k\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3{\left(1-\xi \bar{z}\right)}^2}\text{d}A\left(\xi \right)}\right]\\ =\left(-2\bar{z}\right){\left(1-{\left|z\right|}^2\right)}^2{\displaystyle {\int }_D\frac{f_k\left(\xi \right)}{{\left|1-\bar{\xi }z\right|}^4}\text{d}A\left(\xi \right)}+{\left(1-{\left|z\right|}^2\right)}^3{\displaystyle {\int }_D\frac{2\bar{\xi }{f}_k\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3\left(1-\xi \bar{z}\right){}^2}\text{d}A(\; \xi \; )}\end{array}$

因为 $B\left(f_k\right)$ 具有性质(N)。故 $\left(-2\bar{z}\right)B\left(f_k\right)$ 极分解表达式中有 ${\text{e}}^{-i2\theta },{\text{e}}^{-i3\theta },\cdots ,{\text{e}}^{-ip\theta },\cdots$ ( $p\ge 2$ )。

$\begin{array}{l}{\displaystyle {\int }_D\frac{2\bar{\xi }{f}_k\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3{\left(1-\xi \bar{z}\right)}^2}\text{d}A\left(\xi \right)}\\ ={\displaystyle {\int }_0^1{\displaystyle {\int }_0^{2\pi }\left(2r{\text{e}}^{-i\theta }\right)\left({\displaystyle \underset{p<0}{\sum }{f}_p^{\left(k\right)}\left(r\right){\text{e}}^{ip\theta }}\right)}}\left[{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{2}\left(n+1\right)\left(n+2\right)z^n{r}^n{\text{e}}^{-in\theta }}\right]\left[{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}\left(m+1\right){\bar{z}}^m{r}^m{\text{e}}^{im\theta }}\right]\frac{r}{\pi }\text{d}r\text{d}\theta \\ ={\displaystyle \underset{p<0}{\sum }{\displaystyle \underset{m,n=0}{\overset{\infty }{\sum }}\left(n+1\right)\left(n+2\right)\left(m+1\right)z^n{\bar{z}}^m}}{\displaystyle {\int }_0^1{r}^{2+m+n}{f}_p^{\left(k\right)}\left(r\right)\text{d}r\frac{1}{\pi }{\displaystyle {\int }_0^{2\pi }{\text{e}}^{i\left[-\left(n+1\right)+p+m\right]}}\text{d}\theta }\\ ={\displaystyle \underset{p<0}{\sum }{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\left(n+1\right)}}\left(n+2\right)\left(n+2-p\right)z^n{\bar{z}}^{\left(n+1\right)-p}{\hat{f}}_p^{\left(k\right)}\left(2n+4-p\right)\end{array}$

所以 $\tilde{\Delta }\left[\overline{g_2^{\left(k\right)}}B\left(f_k\right)\right]$ 具有性质(N)。 $=\left(-2\bar{z}\right){\left(1-{\left|z\right|}^2\right)}^{2}B\left(f_k\right)\left(z\right)+{\left(1-{\left|z\right|}^2\right)}^3{\displaystyle {\int }_D\frac{2\bar{\xi }{f}_k\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3\left(1-\xi \bar{z}\right){}^2}\text{d}A(\; \xi \; )}$

$\begin{array}{c}\tilde{\Delta }B\left(f_k\right)\left(z\right)=\tilde{\Delta }\langle f_k{k}_z,k_z\rangle =\tilde{\Delta }{\displaystyle \underset{D}{\int }\frac{f_k\left(\xi \right)}{{\left(1-z\bar{\xi }\right)}^2{\left(1-\bar{z}\xi \right)}^2}\text{d}A\left(\xi \right)}\\ ={\left(1-{\left|z\right|}^2\right)}^2{\displaystyle \underset{D}{\int }{f}_k\left(\xi \right)\cdot \frac{\partial }{\partial z}}\left[\frac{1}{{\left(1-z\bar{\xi }\right)}^2}\right]\cdot \frac{\partial }{\partial z}\left[\frac{1}{{\left(1-\bar{z}\xi \right)}^2}\right]\text{d}A\left(\xi \right)\\ ={\left(1-{\left|z\right|}^2\right)}^2{\displaystyle \underset{D}{\int }{f}_k\left(\xi \right)\frac{2\bar{\xi }}{{\left(1-z\bar{\xi }\right)}^3}\cdot \frac{2\xi }{{\left(1-\bar{z}\xi \right)}^3}\text{d}A(\; \xi \; )}\end{array}$

$\begin{array}{l}={\left(1-{\left|z\right|}^2\right)}^2{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{2\pi }{\int }}4r^2\left({\displaystyle \underset{p<0}{\sum }{f}_p^{\left(k\right)}\left(r\right){\text{e}}^{ip\theta }}\right)}}\left[{\displaystyle \underset{n=0}{\overset{\infty }{\sum }}\frac{1}{2}\left(n+1\right)\left(n+2\right){\bar{z}}^n{r}^n{\text{e}}^{in\theta }}\right]\\ \times \left[{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}\frac{1}{2}\left(m+1\right)\left(m+2\right){\bar{z}}^m{r}^m{\text{e}}^{-im\theta }}\right]\frac{r}{\pi }\text{d}r\text{d}\theta \\ ={\left(1-{\left|z\right|}^2\right)}^2{\displaystyle \underset{p<0}{\sum }{\displaystyle \underset{m,n=0}{\overset{\infty }{\sum }}\left(n+1\right)\left(n+2\right)\left(m+1\right)\left(m+2\right){\bar{z}}^n{z}^m}{\displaystyle \underset{0}{\overset{1}{\int }}{r}^{n+m+3}}{f}_p^{\left(k\right)}\left(r\right)\text{d}r\frac{1}{\pi }{\displaystyle \underset{0}{\overset{2\pi }{\int }}{\text{e}}^{i\left(n-m\right)+p}}\text{d}\theta }\\ ={\left(1-{\left|z\right|}^2\right)}^2{\displaystyle \underset{p<0}{\sum }{\displaystyle \underset{m=0}{\overset{\infty }{\sum }}\left(-p+m+2\right)\left(m+1\right)\left(m+2\right){\bar{z}}^{\left(-p\right)+m}{z}^m}{\hat{f}}_p^{\left(k\right)}}\left(2m-p+4\right)\end{array}$

进一步， $\overline{g_2^{\left(k\right)}}\tilde{\Delta }B\left(f_k\right)$ 具有性质(N)。其极分解为关于 ${\text{e}}^{-i2\theta },{\text{e}}^{-i3\theta },\cdots ,{\text{e}}^{-ip\theta }\left(p\ge 2\right),\cdots$ 的表达式。综合以上计算结果， $\tilde{\Delta }\left[\overline{g_2^{\left(k\right)}}B\left(f_k\right)\right]$ 具有性质(N)。且极分解表达式的首项为关于 $\tilde{\Delta }\left[\overline{g_2^{\left(k\right)}}B\left(f_k\right)\right]$ 的单项式。进而 $-{\displaystyle \underset{k=1}{\overset{N}{\sum }}\tilde{\Delta }\left(\overline{g_2^{\left(k\right)}}B\left(f_k\right)\right)}$ 具有性质(N)且首项为关于 ${\text{e}}^{-i2\theta }$ 的单项式。

下证 $\tilde{\Delta }\overline{g_2^{\left(k\right)}}B\left(f_k\right)\in L^2\left(D,dA\right)\left(1\le k\le N\right)$。

因为 $g_2^{\left(k\right)}$ 为Bloch函数，故 $\left(1-{\left|z\right|}^2\right)\overline{g_2^{\left(k\right)}}$ 有界。因为Berezin变换把 $L^2\left(D,\text{d}A\right)$ 映成 $L^2\left(D,\text{d}A\right)$。故

$-2\bar{z}B\left(f_k\right)\in L^2\left(D,\text{d}A\right)\left(1\le k\le N\right)$。

另一方面，

$\begin{array}{l}{\left(1-{\left|z\right|}^2\right)}^3\left|{\displaystyle \underset{D}{\int }\frac{\bar{\xi }f\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3{\left(1-\xi \bar{z}\right)}^2}\text{d}A\left(\xi \right)}\right|\\ \le {\left(1-{\left|z\right|}^2\right)}^3{\displaystyle \underset{D}{\int }\frac{\left|f\left(\xi \right)\right|}{{\left(1-\bar{\xi }z\right)}^3{\left(1-\xi \bar{z}\right)}^2}\text{d}A\left(\xi \right)}\\ \le {\left(1-{\left|z\right|}^2\right)}^3{\displaystyle \underset{D}{\int }\frac{\left|f\left(\xi \right)\right|}{\left(1-\left|z\right|\right){\left(1-\bar{\xi }z\right)}^4}\text{d}A\left(\xi \right)}\\ \le 2B\left(\left|f\right|\right)(\; z\; )\end{array}$

故 $\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial z}B\left(f\right)\in L^2\left(D,\text{d}A\right)$。

$\begin{array}{c}\tilde{\Delta }B\left(f_k\right)\left(z\right)={\left(1-{\left|z\right|}^2\right)}^2\frac{\partial }{\partial z}\frac{\partial }{\partial z}B\left(f_k\right)\left(z\right)\\ =\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial \bar{z}}\left[-2\bar{z}B\left(f_k\right)\left(z\right)+2{\left(1-{\left|z\right|}^2\right)}^3{\displaystyle \underset{D}{\int }\frac{\bar{\xi }f\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3{\left(1-\xi \bar{z}\right)}^2}\text{d}A\left(\xi \right)}\right]\\ =\left(1-{\left|z\right|}^2\right)[\left(-2\right)B\left(f_k\right)\left(z\right)+\left(-2\bar{z}\right)\frac{\partial }{\partial \bar{z}}B{f}_k\left(z\right)\\ +\left(-6z\right)\left(1-{\left|z\right|}^2\right){\displaystyle \underset{D}{\int }\frac{\bar{\xi }f\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3{\left(1-\xi \bar{z}\right)}^2}\text{d}A\left(\xi \right)}\\ +2{\left(1-{\left|z\right|}^2\right)}^3{\displaystyle \underset{D}{\int }\frac{2{\left|\xi \right|}^{2}f\left(\xi \right)}{{\left|1-\bar{\xi }z\right|}^6}\text{d}A\left(\xi \right)}]\end{array}$

$\begin{array}{l}=\left(-2\right)\left(1-{\left|z\right|}^2\right)B\left(f_k\right)\left(z\right)+\left(-2\bar{z}\right)\left(1-{\left|z\right|}^2\right)\frac{\partial }{\partial z}B\left(f_k\right)\left(z\right)\\ +\left(-6z\right){\left(1-{\left|z\right|}^2\right)}^3{\displaystyle \underset{D}{\int }\frac{\bar{\xi }f\left(\xi \right)}{{\left(1-\bar{\xi }z\right)}^3{\left(1-\xi \bar{z}\right)}^2}\text{d}A\left(\xi \right)}\\ +2{\left(1-{\left|z\right|}^2\right)}^4{\displaystyle \underset{D}{\int }\frac{2{\left|\xi \right|}^{2}f\left(\xi \right)}{{\left|1-\bar{\xi }z\right|}^6}\text{d}A(\; \xi \; )}\end{array}$

由前面已证结果中的前三项都在 $L^2\left(D,\text{d}A\right)$ 中，而

$\begin{array}{l}\left|{\left(1-{\left|z\right|}^2\right)}^4{\displaystyle \underset{D}{\int }\frac{2{\left|\xi \right|}^2{f}_k\left(\xi \right)}{{\left|1-\bar{\xi }z\right|}^6}\text{d}A\left(\xi \right)}\right|\\ \le 2{\left(1-{\left|z\right|}^2\right)}^4{\displaystyle \underset{D}{\int }\frac{\left|f_k\left(\xi \right)\right|}{{\left(1-\left|z\right|\right)}^2{\left|1-\bar{\xi }z\right|}^4}\text{d}A\left(\xi \right)}\\ \le 8{\left(1-{\left|z\right|}^2\right)}^2{\displaystyle \underset{D}{\int }\frac{\left|f_k\left(\xi \right)\right|}{{\left|1-\bar{\xi }z\right|}^4}\text{d}A\left(s\right)}=8B\left|f_k\right|\left(z\right)\end{array}$.

所以 $\tilde{\Delta }\left[\overline{g_2^{\left(k\right)}}B\left(f_k\right)\right]\in L^2\left(D,\text{d}A\right)\left(1\le k\le N\right)$，故。

进一步可知， $F_m\left(z\right)=G_m\left(\tilde{\Delta }\right)B\left[{\displaystyle \underset{k=1}{\overset{N}{\sum }}{f}_k{g}_1^{\left(h\right)}}\right]\in L^2\left(D,\text{d}A\right)$ 且具有性质(N)。进而，在 $L^1\left(D,\text{d}A\right)$ 中， $F_m\left(z\right)$ 依范数收敛 $\left(f{g}_1\right)\left(z\right)$，而 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{f}_k{g}_1^{\left(k\right)}}$ 的极分解式中，只有 ${\text{e}}^{ik\theta }$ 负幂项，进而 $f{g}_1$ 具有性质(N)。而 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{f}_k{g}_1^{\left(k\right)}\left(z\right)}={\displaystyle \underset{p<0}{\sum }{\displaystyle \underset{j\ge 0}{\sum }{\displaystyle \underset{k=1}{\overset{N}{\sum }}{a}_j^{\left(k\right)}{f}_p^{\left(k\right)}{r}^j{\text{e}}^{i\left(j+p\right)\theta }}}}\left(z=r{\text{e}}^{i\theta }\right)$ 只有负幂项，即当 $j+p\le 0$ 时，有

${\displaystyle \underset{k=1}{\overset{N}{\sum }}\left[{\displaystyle \underset{p<0}{\sum }{f}_p^{\left(k\right)}\left(r\right)}\right]}\left[{\displaystyle \underset{0\le j\le -p}{\sum }{a}_j^{\left(k\right)}{r}^j}\right]\equiv 0,\left(0\le r<1\right)$

上面证明了某一类符号的Toeplitz算子有限乘积有限和为零算子的必要条件。下面我们来思考其逆否命题是否成立。

推论3.2 假设 $g_k=g_1^{\left(k\right)}+\overline{g_2^{\left(k\right)}}$，其中 $g_1^{\left(k\right)}\left(z\right)={\displaystyle \sum a_j^{\left(k\right)}{z}^j}$， $g_2^{\left(k\right)}\left(z\right)={\displaystyle \sum b_j^{\left(k\right)}{z}^j}\in H\left(D\right)$，

$f_k\left(r{\text{e}}^{i\theta }\right)={\displaystyle \underset{p<0}{\sum }{f}_p^{\left(k\right)}\left(r\right){\text{e}}^{ip\theta }}\left(1\le k\le N\right)$。若 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}\left[{\displaystyle \underset{P<0}{\sum }{f}_p^{\left(k\right)}\left(r\right)}\right]}\left[{\displaystyle \underset{0\le j\le p}{\sum }{a}_j^{\left(k\right)}{r}^j}\right]\ne 0$，则 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{T}_{f_k}{T}_{g_k}}\ne 0$。

例3.1 若 $f\left(z\right)=z^2{\bar{z}}^4+z{\bar{z}}^4,g=g_1+\overline{g_2}$，其中 $g_1\left(z\right)=1+z,g_{\text{2}}\left(z\right)\in H^{\infty }\left(D\right)$，则 $T_f{T}_g\ne 0$。

4. 结论

本文讨论Bergman空间上两个形如 $T_{f_k}$， $T_{g_k}$ 的Toeplitz算子，其中 $g_1^{\left(k\right)}\left(z\right)={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}{a}_j^{\left(k\right)}{z}^j}$，

$g_2^{\left(k\right)}\left(z\right)={\displaystyle \underset{j=0}{\overset{\infty }{\sum }}{b}_j^{\left(k\right)}{z}^j}$， $f_k\in L^{\infty }\left(D,\text{d}A\right)$， $f_k\left(r{\text{e}}^{i\theta }\right)={\displaystyle \underset{p<0}{\sum }{f}_p^{\left(k\right)}\left(r\right){\text{e}}^{ip\theta }}\left(1\le k\le N\right)$。在进行大量的查阅资料和推算

下，探究Toeplitz算子 $T_{f_k}$， $T_{g_k}$ 的乘积有限和的相关问题，得到以下结论：

1) 假设 $g_k=g_1^{\left(k\right)}+g_2^{\left(k\right)}$，若 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{T}_{f_k}{T}_{g_k}}=0$，则 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}\left[{\displaystyle \underset{p<0}{\sum }{f}_p^{\left(k\right)}\left(r\right)}\right]}\left[{\displaystyle \underset{0\le j\le -p}{\sum }{a}_j^{\left(k\right)}{r}^j}\right]\equiv 0$。

2) 假设 $g_k=g_1^{\left(k\right)}+\overline{g_2^{\left(k\right)}}$，若 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}\left[{\displaystyle \underset{P<0}{\sum }{f}_p^{\left(k\right)}\left(r\right)}\right]}\left[{\displaystyle \underset{0\le j\le p}{\sum }{a}_j^{\left(k\right)}{r}^j}\right]\ne 0$，则 ${\displaystyle \underset{k=1}{\overset{N}{\sum }}{T}_{f_k}{T}_{g_k}}\ne 0$。

参考文献