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数学物理学报, 2021, 41(3): 595-602 doi:

论文

Dirichlet空间上一类积分算子的约化子空间

张核心,

On Reducing Subspaces of a Class of Integral Operator on Dirichlet Spaces

Zhang Hexin,

收稿日期: 2020-03-20  

基金资助: 河北师范大学研究生创新资助项目.  CXZZSS2020052

Received: 2020-03-20  

Fund supported: the Postgraduate Innovative Funding Project of Hebei Normal University.  CXZZSS2020052

作者简介 About authors

张核心,E-mail:corehebtu@126.com , E-mail:corehebtu@126.com

Abstract

We characterize the similarity of n-shift plus certain weighted Volterra operator denoted by T1 on the Dirichlet space D, and prove that T1 acting on D is similar to the multiplication operator Mp acting on S(D) by using some techniques in operator theory. Furthermore, we consider the case of p(z)=zn corresponding to the operator T2, and prove that the operator T2 has exactly 2n reducing subspaces.

Keywords: Dirichlet space ; Volterra operator ; n-Shift operator ; Reducing subspaces

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本文引用格式

张核心. Dirichlet空间上一类积分算子的约化子空间. 数学物理学报[J], 2021, 41(3): 595-602 doi:

Zhang Hexin. On Reducing Subspaces of a Class of Integral Operator on Dirichlet Spaces. Acta Mathematica Scientia[J], 2021, 41(3): 595-602 doi:

1 引言

不变子空间和约化子空间问题一直是算子理论中的一个基本问题. 自上世纪中叶Beurling研究了Hardy空间上移位算子的格以后, 许多学者研究了在Bergman空间和Dirichlet空间上算子的不变子空间问题, 见文献[1-10]. 复平面C上的单位圆盘记作D, H(D)为其上的全纯函数构成的空间, 令D表示Dirichlet空间, D:={fH(D):D|f(z)|2dA(z)<,f(0)=0}, dA表示复平面上的面积测度, dA(z):=1πdxdy=rπdrdθ.其范数. f, g\in{\cal D} , f(z) = \sum\limits_{k = 1}^{\infty}a_{k}z^{k} , g(z) = \sum\limits_{k = 1}^{\infty}b_{k}z^{k} , 则 f(z) g(z) 的内积为 \langle f, g\rangle = \sum\limits_{k = 1}^{\infty}ka_{k}{\bar{b}_k} .

2 算子 T_{1} 的相似性

算子的相似性有助于分析算子的结构, 故很多学者研究了算子的相似性, 见文献[11-13]. 设 {\cal H} 是Hilbert空间, A B 分别是 {\cal H} 上的有界线性算子, 若存在 {\cal H} 上的有界可逆线性算子 T 使得 T^{-1}AT = B , 则称 A B 相似. 设 p(z) (n\geq 2) 是一个 n 次多项式, 即 p(z) = \sum\limits_{k = 0}^{n}d_{k}z^{k} , 并且满足 \left| d_{1} \right|>\sum\limits_{k = 2}^{n}k\left| d_{k}\right| . 定义 {\cal D } 上带特定权的Volterra算子 (V_{p}f)(z): = \int_{0}^{z}p'(w)f(w){\rm d}w , {\cal D} 上乘法算子 M_{p(z)}f(z): = p(z)f(z) , 微分算子 Dh: = \frac{{\rm d}h}{{\rm d}z} . 定义算子 T_{1} 如下

\begin{eqnarray*} (T_{1}f)(z):& = &M_{p(z)}f(z)+(V_{p}f)(z)\nonumber\\ & = &p(z)f(z)+\int_{0}^{z}p'(w)f(w){\rm d}w, f\in {\cal D }.\nonumber \end{eqnarray*}

定义空间

S({\Bbb D}): = \{h\in H({\Bbb D}): h^{(l)} (0) = 0 (l = 0, 1), Dh\in {\cal D}\}.

引理2.1  设 h\in S(\mathbb D) , p(z) 为满足上述条件的多项式, 则 \frac{Dh}{p'(z)}\in{\cal D } .

   设 h(z) = \sum\limits_{k = 2}^{\infty}a_{k}z^{k} , 则 h'(z) = \sum\limits_{k = 2}^{\infty}ka_{k}z^{k-1} ,

h''(z) = \sum\limits_{k = 2}^{\infty}k(k-1)a_{k}z^{k-2} = \sum\limits_{k = 0}^{\infty}(k+2)(k+1)a_{k+2}z^{k}\nonumber,

Dh\in{\cal D} , 即

\begin{eqnarray*} \int_{\Bbb D} |(Dh)'|^2{\rm d}A(z) & = &\int_{\Bbb D} |2a_{2}+6a_{3}z+\cdots|^2{\rm d}A(z)\nonumber\\ & = &\sum\limits_{k = 1}^{\infty}(k+1)^2{k}|a_{k+1}|^2 <\infty.\nonumber \end{eqnarray*}

注意到

\|h\|_{\cal D }^2 = \int_{\Bbb D} |h'|^2{\rm d}A(z) = \int_{\Bbb D} |\sum\limits_{k = 1}^{\infty}(k+1)a_{k+1}z^{k}|^2{\rm d}A(z) = \sum\limits_{k = 1}^{\infty}(k+1)|a_{k+1}|^2\leq\|Dh\|_{\cal D }^2,

\|h\|^{2}_{{\cal D}}\leq \|Dh\|^{2}_{{\cal D}}.

由儒歇定理可知满足上述条件的多项式 p'(z) \overline{{\Bbb D}} 内无零点, 从而 \frac{1}{|p'(z)|} {\Bbb D} 内上界是有限的. 于是

\begin{eqnarray*} \int_{\Bbb D}|(\frac{Dh}{p'(z)})'|^2{\rm d}A(z) & = &\int_{\mathbb D}|\frac{(Dh)'p'(z)-(Dh)p''(z)}{(p'(z))^2}|^2{\rm d}A(z)\nonumber\\ &\leq& \sup\limits_{z\in{\Bbb D}} \frac{1}{|p'(z)|^4} \int_{\Bbb D} |(Dh)'p'(z)-(Dh)p''(z)|^2{\rm d}A(z)\nonumber\\ &\leq & 2\sup\limits_{z\in{\Bbb D}} \frac{1}{|p'(z)|^4} \int_{\Bbb D} (|h''p'(z)|^2+|h'p''(z)|^2){\rm d}A(z)\nonumber\\ &\leq &C_{1}\|Dh\|^{2}_{{\cal D}}+C_{2}\|h\|^{2}_{{\cal D}}\nonumber\\ &\leq& (C_{1}+C_{2})\|Dh\|^{2}_{{\cal D}} <\infty \nonumber, \end{eqnarray*}

其中 C_{1} , C_{2} 为正常数. 易知, \frac{Dh}{p'(z)}|_{z = 0} = 0 , 所以 \frac{Dh}{p'(z)}\in{\cal D} . 证毕.

定义 S({\Bbb D}) 的范数为

\begin{eqnarray} \|h\|^{2}_{S({\Bbb D})}: = \|h\|^{2}_{{\cal D }}+\|Dh\|^{2}_{{\cal D}}. \end{eqnarray}
(2.1)

S({\Bbb D}) 中任意向量 h_{1} h_{2} , 相应的内积定义为

\langle h_{1}, h_{2}\rangle_{S({\Bbb D})}: = \langle h_{1}, h_{2}\rangle_{{\cal D}}+\langle Dh_{1}, Dh_{2}\rangle_{{\cal D }}.

定理2.1  设 p(z) (n\geq 2) 是一个 n 次多项式, 即 p(z) = \sum\limits_{k = 0}^{n}d_{k}z^{k} , 并且满足 \left| d_{1} \right|>\sum\limits_{k = 2}^{n}k\left| d_{k}\right| , V_{p} 是在 {\cal D} 上的带特定权的Volterra算子, 则以下结论成立:

(1) V_{p} 的值域恰为 S({\Bbb D}) ;

(2) V_{p} 是从 {\cal D} S({\Bbb D}) 上的有界同构, 其逆算子为 \frac{1}{p'}D ;

(3) T_{1} 相似于 M_{p} .

   (1) 一方面, 任取 f\in{\cal D} , 设 {h(z) = (V_{p}f)(z) = \int_{0}^{z}p'(w)f(w){\rm d}w} , 则

\int_{\Bbb D}|(Dh(z))'|^{2}{\rm d}A(z) = \int_{\Bbb D}|(p'(z)f(z))'|^{2}{\rm d}A(z) <C_3\|f(z)\|_{\cal D}^{2},

其中 C_3 为正常数, 又 h(0) = 0 , h'(0) = p'(z)f(z)|_{z = 0} = 0 , 因此 h(z)\in S({\Bbb D}) . 另一方面, 设 h\in S({\Bbb D}) , 下证存在 f\in{\cal D} , 使得 V_{p}(f) = h . 由引理2.1知, 只要令 f(z) = \frac{Dh}{p'(z)} 即可. 事实上

{V_{p}(\frac{Dh}{p'(z)})(z) = \int_{0}^{z}p'(w)\frac{Dh}{p'(w)}(w){\rm d}w = h(z)-h(0) = h(z)}.

(2) 首先, 证明 V_{p} {\cal D} S({\Bbb D}) 的有界算子. 设 f\in {\cal D } , 则由(2.1) 式知

\begin{eqnarray*} \|V_{p}f\|_{S({\Bbb D})}^{2}& = &\|V_{p}f\|^{2}_{{\cal D }}+\|D(V_{p}f)\|^{2}_{{\cal D}}\nonumber\\ &\leq&2\|D(V_{p}f)\|^{2}_{{\cal D}}\nonumber\\ &\leq&2 \sup\limits_{z\in{\Bbb D}}|p'(z)|^{2}\|f(z)\|_{{\cal D}}^{2}.\nonumber\\ & = &C_4\|f(z)\|^2_{{\cal D}}, \nonumber \end{eqnarray*}

其中 C_4 为正常数. 所以, V_{p} 是有界线性算子. 下证 V_{p} 是一一的, 设 f_{1}, f_{2}\in {\cal D} , 满足

\int_{0}^{z}p'(w)f_{1}(w){\rm d}w = \int_{0}^{z}p'(w)f_{2}(w){\rm d}w, z\in {\Bbb D}.\nonumber

对等式两边求导数得, f_{1}(z) = f_{2}(z) . V_{p} 的定义可得, 对任意 h\in S({\Bbb D})

V_{p}(\frac{1}{p'(z)}Dh)(z) = h(z),

\frac{1}{p'(z)}Dh(z) = f(z)\in {\cal D} . 因此, V_{p} 是从 {\cal D} S({\Bbb D}) 上的双射, 而且 V_{p}^{-1} = \frac{1}{p'}D .

(3) f\in{\cal D} , (V_{p}f)(z) = h(z) , 注意到

(T_{1}f)(z) = (M_{p}f)(z)+(V_{p}f)(z) = p(z)\frac{h'(z)}{p'(z)}+h(z).

把算子 V_{p} 作用到等式的两边, 得

\begin{eqnarray*} (V_{p}T_{1}f)(z)& = &\int_{0}^{z}p'(w)[p(w)\frac{h'(w)}{p'(w)}+h(w)]{\rm d}w\nonumber\\ & = &\int_{0}^{z}D(p(w)h(w)){\rm d}w\nonumber\\ & = &p(z)h(z)\nonumber\\ & = &M_{p}(V_{p}f)(z)\nonumber \end{eqnarray*}

所以, V_{p}T_{1} = M_{p}V_{p} , 即 V_{p}T_{1}V_{p}^{-1} = M_{p} , 故 T_{1} 相似于 M_{p} .

3 算子 T_{2} 的约化子空间

{\cal X} 是Hilbert空间 {\cal H} 的一个非平凡闭线性子空间, T 是一个有界线性算子, 若 T({\cal X })\subseteq {\cal X} , 则称 {\cal X} T 的不变子空间. 若 {\cal X}^{\perp} 也是 T 的不变子空间, 则称 {\cal X} T 的约化子空间. 算子 T 的约化子空间 {\cal X } 称为极小的, 如果它不真包含任何非零的约化子空间. 所有和算子 T 可交换的有界线性算子的全体称作 T 的换位子, 记作 {\cal A}'(T) , 即 {\cal A }'(T): = \{P\in L({\cal H})| PT = TP\} . 算子 P 满足即自伴又幂等, 则称 P 是投影算子. 算子的约化子空间是由其换位代数里的投影算子决定的(见文献[14]), 若 P_{{\cal X}} 是到 {\cal X} 上的投影算子, 则 {\cal X} T 的约化子空间当且仅当 P_{{\cal X}}T = TP_{{\cal X}} . 本部分将讨论当 p(z) = z^{n} 的情形, 其中 n\geq 2 , 此时定义算子 T_{2}

\begin{eqnarray*} (T_{2}f)(z):& = &M_{z^n}f(z)+(V_{z^{n}}f)(z) = z^nf(z)+\int_{0}^{z}nw^{n-1}f(w){\rm d}w, \ f\in {\cal D }\nonumber. \end{eqnarray*}

引理3.1  令 {\cal D} 记Dirichlet空间, P 是其上的有界算子, 则 P\in{\cal A}'(M_{z}) 当且仅当 P {\cal D} 的规范正交基 \{ e_{k}(z) = \frac{z^{k}}{\gamma_{k}}\}_{k = 1}^{\infty} 下可表示为

\begin{equation} P = \left(\begin{array}{cccccc} p_{11}&0&0&\cdots &0&\cdots \\ p_{21}&p_{11}&0&\cdots &0&\cdots \\ p_{31}&{ } \frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{3}}{\gamma_{2}}p_{21}&p_{11}&\cdots &0&\cdots \\ \vdots&\vdots&\vdots &\ddots &\vdots &\ddots \\ p_{k1}&{ }\frac{\gamma_{1}}{\gamma_{2}}\frac{\gamma_{k}}{\gamma_{k-1}}p_{k-1, 1}& { }\frac{\gamma_{1}}{\gamma_{3}}\frac{\gamma_{k}}{\gamma_{k-2}}p_{k-2, 1}&\cdots &p_{11}&\cdots \\ \vdots&\vdots&\vdots &\ddots&\vdots&\ddots\\ \end{array}\right), \end{equation}
(3.1)

其中 p_{jk}\in {\Bbb C} (j, k\geq 1) , \gamma_{k} = \sqrt{k} . P 是投影算子, 则 P\in{\cal A}'(M_{z}) 当且仅当 P = I 0 .

   由 M_{z}e_{k} = z\frac{z^{k}}{\gamma_{k}} = \frac{\gamma_{k+1}}{\gamma_{k}}e_{k+1}\nonumber , 得 M_{z} {\cal D} 的这组规范正交基下的矩阵表示为

M_{z} = \left(\begin{array}{cccccc} 0&0&\cdots &0&0&\cdots \\ { }\frac{\gamma_{2}}{\gamma_{1}}&0&\cdots &0&0&\cdots \\ 0&{ }\frac{\gamma_{3}}{\gamma_{2}}&\cdots &0&0&\cdots \\ \vdots&\vdots& \ddots&\vdots&\vdots &\ddots \nonumber\\ 0&0&\cdots &{ }\frac{\gamma_{k+1}}{\gamma_{k}}&0&\cdots \\ \vdots&\vdots&\ddots&\vdots&\vdots&\ddots\\ \end{array}\right).

假设 P 在该组规范正交基下的矩阵表示为

P = \left(\begin{array}{cccccc} p_{11}&p_{12}&p_{13}&\cdots &p_{1k}&\cdots \\ p_{21}&p_{22}&p_{23}&\cdots &p_{2k}&\cdots \\ p_{31}&p_{32}&p_{33}&\cdots &p_{3k}&\cdots \\ \vdots&\vdots&\vdots &\ddots&\vdots &\ddots \nonumber\\ p_{k1}&p_{k2}&p_{k3}&\cdots &p_{kk}&\cdots \\ \vdots&\vdots&\vdots&\ddots&\vdots&\ddots\\ \end{array}\right),

其中 p_{ij} = <Pe_{j}, e_{i}> . M_{z}P = PM_{z} , 得

\left\{\begin{array}{lll} p_{j, j+k+1} = 0, \\ { } \frac{\gamma_{j}}{\gamma_{j+l}} p_{j+l, j} = \frac{\gamma_{j+k}}{\gamma_{j+l+k}} p_{j+l+k, j+k}, &(k\geq 0, l\geq 0, j\geq 1).\nonumber \end{array}\right.

于是, (3.1) 式成立. 反之, 如果 P 在上述规范正交基下有如(3.1) 式的矩阵表示, 易证 M_{z}P = PM_{z} , 即 P\in{\cal A}'(M_{z}) . 进而, 如果 P 是一个投影算子当且仅当 P = {\rm diag}(p_{11}, p_{11}, \cdots , p_{11}, \cdots ) , 其中 p_{11} = 1 0 .

引理3.2   若 {\cal H }_{j} = \overline{\rm span}\{e_{(k-1)n+j}:k\geq 1\} (j = 1, 2, \cdots , n) , 则

(1) \{e_{(k-1)n+j}\}_{k = 1}^{\infty} {\cal H}_{j} 的规范正交基;

(2) \rm{ } {\cal D} = {\cal H}_{1}\oplus {\cal H}_{2} \oplus \cdots \oplus {\cal H}_{n} ;

(3) \rm{ } {\cal H}_{j} T_{2} 的约化子空间.

   (1) , (2) 显然. 下证 (3) .

\begin{eqnarray} T_{2}e_{(k-1)n+j}(z)& = &M_{z^n}e_{(k-1)n+j}(z)+V_{z^{n}}e_{(k-1)n+j}(z){}\\ & = &{z^n}\frac{z^{(k-1)n+j}}{\gamma_{(k-1)n+j}}+\int_{0}^{z}nw^{n-1}\frac{w^{(k-1)n+j}}{\gamma_{(k-1)n+j}}{\rm d}w{}\\ & = &\Big(1+\frac{n}{kn+j}\Big)\frac{\gamma_{kn+j}}{\gamma_{(k-1)n+j}}e_{kn+j}. \end{eqnarray}
(3.2)

所以 T_{2}{\cal H}_{j}\subset {\cal H}_{j}

\begin{eqnarray*} &&T_{2}({\cal H}_{1}\bigoplus {\cal H}_{2} \bigoplus \cdots \bigoplus {\cal H}_{j-1}\bigoplus {\cal H}_{j+1}\bigoplus \cdots \bigoplus {\cal H}_{n})\nonumber\\ && \subset{\cal H}_{1}\bigoplus {\cal H}_{2} \bigoplus \cdots \bigoplus {\cal H}_{j-1}\bigoplus {\cal H}_{j+1}\bigoplus \cdots \bigoplus {\cal H}_{n}. \end{eqnarray*}

{\cal H}_{j} T_{2} 的约化子空间.

定理3.1  设 Q:{\cal D}\rightarrow {\cal D} 是一个投影算子, T_{2j} = T_{2}|_{{\cal H}_j} (j = 1, 2, \cdots , n) , 则

Q\left(\begin{array}{cccccc} T_{21}\\ &T_{22} \\ & &\ddots& \\ & & &T_{2n}& \\ \end{array}\right) = \left(\begin{array}{cccccc} T_{21}\\ &T_{22} \\ & &\ddots& \\ & & &T_{2n}& \\ \end{array}\right) Q

当且仅当 Q = {\rm diag}(Q_1, Q_2, \cdots , Q_{n}) , 其中 Q_{i} (i = 1, 2, \cdots , n) I_{{\cal H}_i} 0 .

   若 Q\in{\cal A}'({\rm diag }(T_{21}, T_{22}, \cdots , T_{2n})) , 注意到 {\cal D} = {\cal H}_{1}\bigoplus {\cal H}_{2}\bigoplus\cdots\bigoplus {\cal H}_{n} , 则算子 Q 能分块为如下形式

Q = \left(\begin{array}{cccccc} Q_{11}&Q_{12}&\cdots &Q_{1n}\\ Q_{21}&Q_{22}&\cdots &Q_{2n}\\ \vdots&\vdots&\ddots&\vdots & \\ Q_{n1}&Q_{n2}&\cdots &Q_{nn}\nonumber\\ \end{array}\right),

其中 Q_{ij}: {\cal H}_{j}\rightarrow {\cal H}_{i} (i, j = 1, 2, \cdots , n) . Q\in{\cal A}'({\rm diag }(T_{21}, T_{22}, \cdots , T_{2n}))

\begin{equation} \begin{array}{lll} &\left(\begin{array}{cccccc} Q_{11}T_{21}&Q_{12}T_{22}&\cdots &Q_{1n}T_{2n}\\ Q_{21}T_{21}&Q_{22}T_{22}&\cdots &Q_{2n}T_{2n}\\ \vdots&\vdots&\ddots&\vdots & \\ Q_{n1}T_{21}&Q_{n2}T_{22}&\cdots &Q_{nn}T_{2n}\\ \end{array}\right) = \left(\begin{array}{cccccc} T_{21}Q_{11}&T_{21}Q_{12}&\cdots &T_{21}Q_{1n}\\ T_{22}Q_{21}&T_{22}Q_{22}&\cdots &T_{22}Q_{2n}\\ \vdots&\vdots&\ddots&\vdots & \\ T_{2n}Q_{n1}&T_{2n}Q_{n2}&\cdots &T_{2n}Q_{nn}\\ \end{array}\right) \end{array}. \end{equation}
(3.3)

假设 Q_{ij} {\cal H}_{j} 的规范正交基下的矩阵表示如下

Q_{ij} = \left(\begin{array}{cccccc} q_{11}^{ij}&q_{12}^{ij}&q_{13}^{ij}&\cdots &q_{1k}^{ij}&\cdots \\ q_{21}^{ij}&q_{22}^{ij}&q_{23}^{ij}&\cdots &q_{2k}^{ij}&\cdots \\ q_{31}^{ij}&q_{32}^{ij}&q_{33}^{ij}&\cdots &q_{3k}^{ij}&\cdots \\ \vdots&\vdots&\vdots &\ddots&\vdots &\ddots \\ q_{k1}^{ij}&q_{k2}^{ij}&q_{k3}^{ij}&\cdots &q_{kk}^{ij}&\cdots \nonumber\\ \vdots&\vdots&\vdots&\ddots&\vdots&\ddots\\ \end{array}\right),

其中 q_{t_{1}t_{2}}^{ij} = \langle Q_{ij}e_{(t_{2}-1)n+j}, e_{(t_{1}-1)n+i}\rangle (t_{1}, t_{2} = 1, 2, \cdots ) . 从(3.2) 式可知算子 T_{2j} {\cal H}_{j} 的这组规范正交基下的矩阵表示为

T_{2j} = \left(\begin{array}{cccccc} 0&0&\cdots &0&0&\cdots \\ { }(1+\frac{n}{n+j})\frac{\gamma_{n+j}}{\gamma_{j}}&0&\cdots &0&0&\cdots \\ 0&{ }(1+\frac{n}{2n+j})\frac{\gamma_{2n+j}}{\gamma_{n+j}}&\cdots &0&0&\cdots \\ \vdots&\vdots& \ddots&\vdots&\vdots &\ddots \\ 0&0&\cdots &{ }(1+\frac{n}{kn+j})\frac{\gamma_{kn+j}}{\gamma_{(k-1)n+j}}&0&\cdots \\ \vdots&\vdots&\ddots&\vdots&\vdots&\ddots\nonumber\\ \end{array}\right).

1) i = j . 从(3.3) 式, 得 Q_{jj}T_{2j} = T_{2j}Q_{jj} (j = 1, 2, \cdots, n) . 由引理3.1可知 Q_{jj} 有形如(3.1)式的表示. 又 Q 是一个投影算子, 所以 Q_{jj} = Q_{jj}^{\ast} . 于是得 Q_{jj} = q_{11}^{jj}I_{{\cal H }_j} (q_{11}^{jj}\in {\mathbb R}) .

2) i\neq j . 由(3.3) 式, 得 T_{2i}Q_{ij} = Q_{ij}T_{2j} , 即

\begin{eqnarray*} & &\left(\begin{array}{cccccc} 0&0&\cdots&0&\cdots\\ (1+\frac{n}{n+i})\frac{\gamma_{n+i}}{\gamma_{i}}q_{11}^{ij}&(1+\frac{n}{n+i})\frac{\gamma_{n+i}}{\gamma_{i}}q_{12}^{ij}&\cdots&(1+\frac{n}{n+i})\frac{\gamma_{n+i}}{\gamma_{i}}q_{1k}^{ij}&\cdots\\ (1+\frac{n}{2n+i})\frac{\gamma_{2n+i}}{\gamma_{n+i}}q_{21}^{ij}&(1+\frac{n}{2n+i})\frac{\gamma_{2n+i}}{\gamma_{n+i}}q_{22}^{ij}&\cdots&(1+\frac{n}{2n+i})\frac{\gamma_{2n+i}}{\gamma_{n+i}}q_{2k}^{ij}&\cdots\\ \vdots&\vdots& \ddots&\vdots& \ddots\\ \frac{kn+i}{(k-1)n+i}\frac{\gamma_{(k-1)n+i}}{\gamma_{(k-2)n+i}}q_{k-1, 1}^{ij}&\frac{kn+i}{(k-1)n+i}\frac{\gamma_{(k-1)n+i}}{\gamma_{(k-2)n+i}}q_{k-1, 2}^{ij} &\cdots&\frac{kn+i}{(k-1)n+i}\frac{\gamma_{(k-1)n+i}}{\gamma_{(k-2)n+i}}q_{k-1, k}^{ij}&\cdots\\ \vdots&\vdots&\ddots&\vdots&\ddots\nonumber\\ \end{array}\right){\nonumber}\\ & = & \left(\begin{array}{cccccc}(1+\frac{n}{n+j})\frac{\gamma_{n+j}}{\gamma_{j}}q_{12}^{ij}&(1+\frac{n}{2n+j})\frac{\gamma_{2n+j}}{\gamma_{n+j}}q_{13}^{ij}&\cdots&(1+\frac{n}{(k-1)n+j})\frac{\gamma_{(k-1)n+j}}{\gamma_{(k-2)n+j}}q_{1k}^{ij}&\cdots\\ (1+\frac{n}{n+j})\frac{\gamma_{n+j}}{\gamma_{j}}q_{22}^{ij}&(1+\frac{n}{2n+j})\frac{\gamma_{2n+j}}{\gamma_{n+j}}q_{23}^{ij}&\cdots&(1+\frac{n}{(k-1)n+j})\frac{\gamma_{(k-1)n+j}}{\gamma_{(k-2)n+j}}q_{2k}^{ij}&\cdots\\ (1+\frac{n}{n+j})\frac{\gamma_{n+j}}{\gamma_{j}}q_{32}^{ij}&(1+\frac{n}{2n+j})\frac{\gamma_{2n+j}}{\gamma_{n+j}}q_{33}^{ij}&\cdots&(1+\frac{n}{(k-1)n+j})\frac{\gamma_{(k-1)n+j}}{\gamma_{(k-2)n+j}}q_{3k}^{ij}&\cdots\\ \vdots&\vdots&\ddots &\vdots&\ddots \\ (1+\frac{n}{n+j})\frac{\gamma_{n+j}}{\gamma_{j}}q_{k2}^{ij}&(1+\frac{n}{2n+j})\frac{\gamma_{2n+j}}{\gamma_{n+j}}q_{k3}^{ij} &\cdots&(1+\frac{n}{(k-1)n+j})\frac{\gamma_{(k-1)n+j}}{\gamma_{(k-2)n+j}}q_{kk}^{ij}&\cdots\\ \vdots&\vdots&\ddots&\vdots&\ddots\nonumber\\ \end{array}\right). \end{eqnarray*}

由此得

Q_{ij} = \left(\begin{array}{cccccc} q_{11}^{ij}&0&0&0&\cdots \\ q_{21}^{ij}&{ }\frac{1+\frac{n}{n+i}}{1+\frac{n}{n+j}}\frac{\gamma_{j}}{\gamma_{i}}\frac{\gamma_{n+i}}{\gamma_{n+j}}q_{11}^{ij}&0&0&\cdots \\ q_{31}^{ij}&{ }\frac{1+\frac{n}{2n+i}}{1+\frac{n}{n+j}}\frac{\gamma_{j}}{\gamma_{n+i}}\frac{\gamma_{2n+i}}{\gamma_{n+j}}q_{21}^{ij}& { }\frac{1+\frac{n}{n+i}}{1+\frac{n}{n+j}}\frac{1+\frac{n}{2n+i}}{1+\frac{n}{2n+j}}\frac{\gamma_{j}}{\gamma_{i}}\frac{\gamma_{2n+i}}{\gamma_{2n+j}}q_{11}^{ij}&0&\cdots \\ \vdots&\vdots& \vdots&\vdots&\ddots\nonumber \end{array}\right).

另一方面, 由(3.3) 式也有 T_{2j}Q_{ji} = Q_{ji}T_{2i} . 类似对式 T_{2i}Q_{ij} = Q_{ij}T_{2j} 的处理, 得到

Q_{ji} = \left(\begin{array}{cccccc} q_{11}^{ji}&0&0&0&\cdots \\ q_{21}^{ji}&{ }\frac{1+\frac{n}{n+j}}{1+\frac{n}{n+i}}\frac{\gamma_{i}}{\gamma_{j}}\frac{\gamma_{n+j}}{\gamma_{n+i}}q_{11}^{ji}&0&0&\cdots \\ q_{31}^{ji}&{ }\frac{1+\frac{n}{2n+j}}{1+\frac{n}{n+i}}\frac{\gamma_{i}}{\gamma_{n+j}}\frac{\gamma_{2n+j}}{\gamma_{n+i}}q_{21}^{ji}& { }\frac{1+\frac{n}{n+j}}{1+\frac{n}{n+i}}\frac{1+\frac{n}{2n+j}}{1+\frac{n}{2n+i}}\frac{\gamma_{i}}{\gamma_{j}}\frac{\gamma_{2n+j}}{\gamma_{2n+i}}q_{11}^{ji}&0&\cdots \\ \vdots&\vdots& \vdots&\vdots&\ddots\nonumber \end{array}\right).

Q {\cal D} 上的一个投影算子, 则 Q 是自伴的, 即 Q_{ij}^{\ast} = Q_{ji} . 因此, q_{l1}^{ij} = 0 (l = 2, 3, \cdots ) . 解方程组

\left\{\begin{array}{ll} {\bar q}_{11}^{ij} = q_{11}^{ji}, \\ { } \frac{1+\frac{n}{n+i}}{1+\frac{n}{n+j}}\frac{\gamma_{j}}{\gamma_{i}}\frac{\gamma_{n+i}}{\gamma_{n+j}}{\bar q}_{11}^{ij} = \frac{1+\frac{n}{n+j}}{1+\frac{n}{n+i}}\frac{\gamma_{i}}{\gamma_{j}}\frac{\gamma_{n+j}}{\gamma_{n+i}}q_{11}^{ji}, \nonumber \end{array} \right.

q_{11}^{ij} = 0 . 因此, Q_{ij} = 0 . Q^{2} = Q 蕴含 Q = {\rm diag}(Q_1, Q_2, \cdots , Q_{n}) , 其中 Q_{i}(i = 1, 2, \cdots , n) I_{{\cal H}_i} 0 . 反之, 若 Q = {\rm diag}(Q_1, Q_2, \cdots , Q_n) , 其中 Q_{i} (i = 1, 2, \cdots , n) I_{{\cal H}_i} 0 , 显然 Q\in{\cal A}' (diag (T_{21}, T_{22}, \cdots , T_{2n})) .

定理3.2  在Dirichlet空间 {\cal D } 中, 算子 T_{2} 2^{n} 个约化子空间, 极小约化子空间为 {\cal H}_{1}, {\cal H}_{2}, \cdots , {\cal H}_{n} .

   设 Q:{\cal D}\rightarrow {\cal D } 是一个投影算子, 满足 QT_{2} = T_{2}Q . 注意到 T_{2}|_{\cal D } = T_{21}\bigoplus T_{22} \bigoplus\cdots \bigoplus T_{2n} . 因此

Q\left(\begin{array}{cccccc} T_{21}\\ &T_{22} \\ & &\ddots& \\ & & &T_{2n}& \\ \end{array}\right) = \left(\begin{array}{cccccc} T_{21}\\ &T_{22} \\ & &\ddots& \\ & & &T_{2n}& \\ \end{array}\right) Q.\nonumber

由定理3.1, 投影算子 Q = {\rm diag}(Q_1, Q_2, \cdots , Q_n) , Q_{i} (i = 1, 2, \cdots , n) I_{{\cal H}_i} 0 . 由引理3.2得 T_{2} 的约化子空间为 c_{1}{\cal H}_{1}\bigoplus c_{2}{\cal H}_{2} \bigoplus \cdots \bigoplus c_{n}{\cal H}_{n} , c_{i} = 0 1 , (i = 1, 2, \cdots , n) , 即 T_{2} 2^{n} 个约化子空间, 极小约化子空间为 {\cal H}_{1}, {\cal H}_{2}, \cdots , {\cal H}_{n} .

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