加权Bergman空间上Toeplitz算子的乘积的有限和问题

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

摘要: 函数空间算子理论一直是泛函分析研究中的一个重要分支之一。本文研究了加权Bergman空间上Toeplitz算子T_φ²，其中φ(re)^iθ=e^isθφ₀(r)，且δ∈Z，δ<0，φ_α(r)∈ℜ_α为亚正规算子的一个必要条件。

Abstract: Function space operator theory has always been one of the important branches in functional analysis research. This paper studies Toeplitz operator in weighted Bergman space T_φ², in which φ(re)^iθ=e^isθφ₀(r), and δ∈Z, δ<0; φ_α(r)∈ℜ_α is a necessary condition for subnormal operator.

1. 引言

关于函数空间上以有界调和函数为符号的Toeplitz算子拟正规性的研究，主要集中在Hardy空间和Bergman空间上。文献 [1] 中给出了Hardy空间上以三角多项式为符号的Toeplitz算子的亚正规性的结果，在2013年，Phukon和Hazarika [2] 刻画了Bergman空间上Toeplitz算子 $T_{\phi }$ 是亚正规的必要条件。本文思考将某符号的Toeplitz算子亚正规性问题推广到加权Bergman空间上进行研究。

2. 预备知识

在这一部分中，我们介绍了Mellin变换我们将在下一节计算中运用这个公式。

定义1.1 若 $\phi$ 是 $\left[0,1\right]$ 上的可积函数(即 ${\displaystyle {\int }_{\left[0,1\right]}\left|\phi \left(t\right)\right|\text{d}t}<\infty$ )，可定义Mellin变换 $\hat{\phi }$ ：

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

Mellin变换是半平面 $\left\{z:\mathrm{Re}z>\text{1}\right\}$ 上的有界解析函数。

定义1.2 对于 $-1<p<+\infty$， $\alpha >-1$，D上加权Bergman空间 $A_{\alpha }^2$ 是解析函数空间在 $L^2\left(D,\text{d}{A}_{\alpha }\right)$ 上。此时，

$\text{d}{A}_{\alpha }\left(z\right)=\left(\alpha +1\right){\left(1-{\left|z\right|}^2\right)}^{\alpha }\text{d}A\left(z\right)$.

显而易见， $A_{\alpha }^2$ 是Hilbert空间 $L^2\left(D,\text{d}{A}_{\alpha }\right)$ 中的闭子空间。

定义1.3 设 $T\in B\left(H\right)$，若 $T^{\ast }T=T{T}^{\ast }$，称T为正规算子。若满足

$T^{\ast }T-T{T}^{\ast }\ge 0$,

引理1.1 [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1.1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 $\phi \equiv 0$。

重要的是，一个函数是由Mellin变换的零点决定的。

引理1.2 [4] 设 $\phi \in L^1\left(\left[0,1\right],r\text{d}r\right)$。如果存在 $N_0$， $p\in N$ 使得

$\hat{\phi }\left(n_0+pk\right)=0\forall k\in N$,

则 $\phi \equiv 0$。

3. 主要结果

引理2.1 [5] $\phi \left(r{\text{e}}^{i\theta }\right)={\text{e}}^{i\delta \theta }{\phi }_0\left(r\right)\in L^{\infty }\left(D\right)$， $\delta \in {\rm Z}$， ${\phi }_0\left(r\right)\in {\Re }_a$，

其中

${\Re }_a=\left\{a:D\to C:a\left(z\right)=a\left(\left|z\right|\right),且{\displaystyle {\int }_0^1{\left|a\left(r\right)\right|}^{2}r{\left(1-r^2\right)}^a\text{d}r<\infty }\right\}$,

那么有对于非负整数n，有

1) $T_{\phi }=\{\begin{array}{l}\frac{\left(2n+2\delta +2\right)}{\left(n+\delta +1\right)!}\frac{\Gamma \left(n+\delta +2+\alpha \right)}{\Gamma \left(1+\alpha \right)}{\hat{\phi }}_{\alpha ,0}\left(2n+\delta +2\right)z^{n+\delta },n+\delta \ge 0\\ 0,n+\delta <0\end{array}$

2) $T_{\bar{\phi }}{z}^{\alpha }=\{\begin{array}{l}\frac{\left(2n-2\delta +2\right)}{\left(n-\delta +1\right)!}\frac{\Gamma \left(n-\delta +2+\alpha \right)}{\Gamma \left(1+\alpha \right)}{\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right)z^{n-\delta },n-\delta \ge 0\\ 0,n-\delta <0\end{array}$

其中

${\hat{\phi }}_{\alpha ,0}\left(z\right)={\displaystyle {\int }_0^1{\phi }_0\left(r\right)r^{z-1}}{\left(1-r^2\right)}^{\alpha }\text{d}r$ 为 ${\phi }_0\left(r\right){\left(1-r^2\right)}^{\alpha }$

的Mellin变换。

定理2.1 $\phi \left(r{\text{e}}^{i\theta }\right)={\text{e}}^{i\delta \theta }{\phi }_0\left(r\right)\in L^{\infty }\left(D\right)$， $\delta \in {\rm Z}$ 且 $\delta <0$， ${\phi }_{\alpha }\left(r\right)\in {\Re }_a$。若 $T_{\phi }^2$ 为亚正规算子，则

1) $-\delta \le n<-2\delta$ 时， ${\hat{\phi }}_{\alpha ，0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ，0}\left(2n-3\delta +2\right)=0$。

2)时， $\begin{array}{l}\left|{\hat{\phi }}_{\alpha ，0}\left(2n+2\delta +2\right){\hat{\phi }}_{\alpha ，0}\left(2n-3\delta +2\right)\right|\\ \ge \frac{\Gamma \left(n-\delta +2+\alpha \right)}{\Gamma \left(n+\delta +2+\alpha \right)}\frac{\left(n+\delta \right)!}{\left(n-\delta \right)!}\cdot \sqrt{\frac{\Gamma \left(n-2\delta +\alpha +2\right)}{\Gamma \left(n+2\delta +\alpha +2\right)}\frac{\left(n+2\delta \right)!}{\left(n-2\delta \right)!}}\\ \times \left|{\hat{\phi }}_{\alpha ，0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ，0}\left(2n-3\delta +2\right)\right|\end{array}$。

证明：由引理2.1，当 $n+\delta \ge 0$ 时， $n\in N_0$。

$T_{\phi }{z}^n=\{\begin{array}{l}\begin{array}{l}\frac{\left(2n+2\delta +2\right)}{\left(n+\delta +1\right)!}\frac{\Gamma \left(n+\delta +2+\alpha \right)}{\Gamma \left(1+\alpha \right)}\frac{\left(2n+4\delta +2\right)}{\left(n+2\delta +1\right)!}\frac{\Gamma \left(n+2\delta +2+\alpha \right)}{\Gamma \left(1+\alpha \right)}\hfill \\ {\hat{\phi }}_{\alpha ，0}\left(2n+\delta +2\right){\hat{\phi }}_{\alpha ，0}\left(2n+3\delta +2\right)z^{n+2\delta }\hfill \end{array}n+\delta \ge 0且n+2\delta \ge 0\\ 0n-2\delta <0\end{array}$

又时， $T_{\phi }{z}^n=\frac{\left(2n-2\delta +2\right)}{\left(n-\delta +1\right)!}\frac{\Gamma \left(n-\delta +2+\alpha \right)}{\Gamma \left(1+\alpha \right)}{\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right)z^{n-\delta }$

$T_{\bar{\phi }}{z}^n=\{\begin{array}{l}\begin{array}{l}\frac{\left(2n-2\delta +2\right)}{\left(n-\delta +1\right)!}\frac{\Gamma \left(n-\delta +2+\alpha \right)}{\Gamma \left(1+\alpha \right)}\frac{\left(2n+4\delta +2\right)}{\left(n-2\delta +1\right)!}\frac{\Gamma \left(n-2\delta +2+\alpha \right)}{\Gamma \left(1+\alpha \right)}\hfill \\ {\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)z^{n-2\delta }\hfill \end{array}n-\delta \ge 0且n-2\delta \ge 0\\ 0n-2\delta <0\end{array}$

若 $\delta =0$， ${\Vert T_{\phi }^2{z}^n\Vert }^2={\Vert T_{\bar{\phi }}^2{z}^n\Vert }^2$。

若 $\delta <0$，当 $0\le n\le -2\delta -1$ 时， ${\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)=0$。进一步有 ${\hat{\phi }}_{\alpha ,0}\left(p\right){\hat{\phi }}_{\alpha ,0}\left(q\right)=0$，其中，

$p\in M_0=\left\{2\left(n+1\right)-\delta :0\le n\le -2\delta -1,n\in N_0\right\}$,

$q\in M_1=\left\{2\left(n+1-\delta \right)-\delta :0\le n\le -2\delta -1,n\in N_0\right\}$.

选取 $f\left(z\right)={\displaystyle \underset{n=0}{\overset{\infty }{\sum }}{a}_n}{z}^n$，使得f的幂级数展开式中前 $2\delta$ 项为0，则

$\begin{array}{l}\langle T_{\phi }^2,T_{\phi }^2\rangle -\langle T_{\bar{\phi }}^2,T_{\bar{\phi }}^2\rangle \\ ={\displaystyle \underset{n=-2\delta }{\overset{\infty }{\sum }}16{\left|a_n\right|}^2}\left(\alpha +1\right)\{{\left[\frac{\Gamma \left(n+\delta +2+\alpha \right)}{\left(n+\delta \right)!\Gamma \left(1+\alpha \right)}\right]}^2\frac{\Gamma \left(n+2\delta +2+\alpha \right)}{\left(n+2\delta \right)!\Gamma \left(1+\alpha \right)}\left|{\hat{\phi }}_{\alpha ,0}\left(2n+\delta +2\right){\hat{\phi }}_{\alpha ,0}{\left(2n+3\delta +2\right)}^2\right|\\ -{\left[\frac{\Gamma \left(n-\delta +2+\alpha \right)}{\left(n-\delta \right)!\Gamma \left(1+\alpha \right)}\right]}^2\frac{\Gamma \left(n-2\delta +2+\alpha \right)}{\left(n-2\delta \right)!\Gamma \left(1+\alpha \right)}{\left|{\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)\right|}^2\}\end{array}$

由于 $2n+\delta +2=2\left(n+1+\delta \right)-\delta$，。那么有

${\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)=0$.

在 $-2\delta \le n\le -4\delta -1$ 时，故有

${\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)=0$.

如此下去， ${\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)=0$。 $\forall n\in N_0$。而 $2n-3\delta +2=2\left(n+1-\delta \right)-\delta$， $2n-\delta +2=2\left(n+1\right)-\delta$， $n=p\cdot 2\delta +q$，故 ${\phi }_0=0$。

若 $\delta <0$，

$\begin{array}{l}\langle T_{\phi }^{2}f,T_{\phi }^2\rangle -\langle T_{\bar{\phi }}^{2}f,T_{\bar{\phi }}^2\rangle \\ ={\displaystyle \underset{n=0}{\overset{2\delta -1}{\sum }}16{\left|a_n\right|}^2}\left(\alpha +1\right){\left[\frac{\Gamma \left(n+\delta +2+\alpha \right)}{\left(n+\delta \right)!\Gamma \left(1+\alpha \right)}\right]}^2\frac{\Gamma \left(n+2\delta +2+\alpha \right)}{\left(n+2\delta \right)!\Gamma \left(1+\alpha \right)}{\left|{\hat{\phi }}_{\alpha ,0}\left(2n+\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n+3\delta +2\right)\right|}^2\\ +{{\displaystyle \underset{n=2\delta }{\overset{+\infty }{\sum }}16\left|a_n\right|}}^2\left(\alpha +1\right)\{{\left[\frac{\Gamma \left(n+\delta +2+\alpha \right)}{\left(n+\delta \right)!\Gamma \left(1+\alpha \right)}\right]}^2\frac{\Gamma \left(n+2\delta +2+\alpha \right)}{\left(n+2\delta \right)!\Gamma \left(1+\alpha \right)}{\left|{\hat{\phi }}_{\alpha ,0}\left(2n+\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n+3\delta +2\right)\right|}^2\\ -{\left[\frac{\Gamma \left(n-\delta +2+\alpha \right)}{\left(n-\delta \right)!\Gamma \left(1+\alpha \right)}\right]}^2\frac{\Gamma \left(n-2\delta +2+\alpha \right)}{\left(n-2\delta \right)!\Gamma \left(1+\alpha \right)}{\left|{\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)\right|}^2\}\end{array}$

则 $T_{\phi }^2$ 亚正规当且仅当

$\begin{array}{l}\left|{\hat{\phi }}_{\alpha ,0}\left(2n+\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n+3\delta +2\right)\right|\\ \ge \frac{\Gamma \left(n-\delta +2+\alpha \right)}{\Gamma \left(n-\delta +2+\alpha \right)}\frac{\left(n+\delta \right)!}{\left(n-\delta \right)!}\frac{\Gamma \left(n-2\delta +\alpha +2\right)}{\Gamma \left(n+2\delta +\alpha +2\right)}\frac{\left(n+2\delta \right)!}{\left(n-2\delta \right)!}\\ \times \left|{\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)\right|\end{array}$

4. 结论

本文研究了加权Bergman空间上Toeplitz算子 $T_{\phi }^2$，其中 $\phi \left(r{\text{e}}^{i\theta }\right)={\text{e}}^{i\delta \theta }{\phi }_0\left(r\right)$，且 $\delta \in {\rm Z}$，， ${\phi }_{\alpha }\left(r\right)\in {\Re }_a$ 为亚正规算子的一个必要条件：

若 $T_{\phi }^2$ 为亚正规算子，则有

1) $-\delta \le n<-2\delta$ 时， ${\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)=0$。

2) $n\ge -2\delta$ 时， $\begin{array}{l}\left|{\hat{\phi }}_{\alpha ,0}\left(2n+2\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)\right|\\ \ge \frac{\Gamma \left(n-\delta +2+\alpha \right)}{\Gamma \left(n+\delta +2+\alpha \right)}\frac{\left(n+\delta \right)!}{\left(n-\delta \right)!}\cdot \sqrt{\frac{\Gamma \left(n-2\delta +\alpha +2\right)}{\Gamma \left(n+2\delta +\alpha +2\right)}\frac{\left(n+2\delta \right)!}{\left(n-2\delta \right)!}}\\ \times \left|{\hat{\phi }}_{\alpha ,0}\left(2n-\delta +2\right){\hat{\phi }}_{\alpha ,0}\left(2n-3\delta +2\right)\right|\end{array}$。

显然上述讨论结果对于这样函数的线性组合仍然成立。我们希望得出加权Bergman空间上Toeplitz算子的乘积有限和为亚正规算子的更普遍的结论，但是由于其广泛性和复杂性，目前只能得到上述结论。