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$ \mathbb C $上的单位圆盘记作$ {\Bbb D} $, $ H({{\Bbb D}}) $为其上的全纯函数构成的空间, 令$ {\cal D} $表示Dirichlet空间, $ {\cal D}: = \{f\in H({\Bbb D}):\int_{\Bbb D}|f'(z)|^{2}{\rm d}A(z) <\infty, $$ f(0) = 0\} $, $ {\rm d}A $表示复平面上的面积测度, $ {\rm d}A(z): = \frac{1}{\pi}{\rm d}x {\rm d}y = \frac{r}{\pi}{\rm d}r{\rm d}\theta. $其范数$ \|f\|^2_{{\cal D}}: = \int_{\Bbb D}|f'(z)|^{2}{\rm d}A(z) $. 设$ 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) $与$ g(z) $的内积为$ \langle f, g\rangle = \sum\limits_{k = 1}^{\infty}ka_{k}{\bar{b}_k} $.

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 $与$ 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 } $.

证   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, $

由$ 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, $

即

$ \|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) $在$ \overline{{\Bbb D}} $内无零点, 从而$ \frac{1}{|p'(z)|} $在$ {\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

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

对$ 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} $到$ 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} $.

证   $ (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} $是$ {\cal D} $到$ 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) $. 从$ V_{p} $的定义可得, 对任意$ h\in S({\Bbb D}) $有

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

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

$ (3) $设$ 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} $.

设$ {\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} $为$ T $的不变子空间. 若$ {\cal X}^{\perp} $也是$ T $的不变子空间, 则称$ {\cal X} $为$ 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} $是$ 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} $为

$ \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 $在$ {\cal D} $的规范正交基$ \{ e_{k}(z) = \frac{z^{k}}{\gamma_{k}}\}_{k = 1}^{\infty} $下可表示为

(3.1) $ \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} $

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

证   由$ M_{z}e_{k} = z\frac{z^{k}}{\gamma_{k}} = \frac{\gamma_{k+1}}{\gamma_{k}}e_{k+1}\nonumber $, 得$ M_{z} $在$ {\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}> $. 由$ 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 $或$ 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} $是$ {\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} $是$ T_{2} $的约化子空间.

证   $ (1) $, $ (2) $显然. 下证$ (3) $.

(3.2) $ \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} $

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

$ \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*} $

即$ {\cal H}_{j} $是$ 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) $为$ I_{{\cal H}_i} $或$ 0 $.

证   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) $. 由$ Q\in{\cal A}'({\rm diag }(T_{21}, T_{22}, \cdots , T_{2n})) $得

(3.3) $ \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} $

假设$ Q_{ij} $在$ {\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} $在$ {\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 $是$ {\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. $

得$ 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) $是$ I_{{\cal H}_i} $或$ 0 $. 反之, 若$ Q = {\rm diag}(Q_1, Q_2, \cdots , Q_n) $, 其中$ Q_{i} (i = 1, 2, \cdots , n) $是$ I_{{\cal H}_i} $或$ 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} $有$ 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} $.

证   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) $为$ I_{{\cal H}_i} $或$ 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 $或$ 1 $, $ (i = 1, 2, \cdots , n) $, 即$ T_{2} $有$ 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} $.