本注记中, $X, Y$表示复无限维Banach空间, $M^{\perp}$表示子空间$M\subseteq X$的零化子.用$\mathcal{B}(X, Y)$表示从$X$到$Y$的有界线性算子全体, $\mathcal{B}(X):=\mathcal{B}(X, X)$.对$T\in \mathcal{B}(X), T^*$表示$T$的共轭算子, $\mathcal{R}(T)$或$T(X)$表示$T$的值域, $\mathcal{N}(T)$表示$T$的零空间, $\alpha(T)$和$\beta(T)$分别表示$T$的零维和亏维, 即$\alpha(T)={\rm{dim}}\mathcal{N}(T), \ \beta(T)={\rm{dim}}X/\mathcal{R}(T)$. Fredholm算子集$\Phi(X) = \{T \in \mathcal{B}(X): \alpha(T) < \infty {且\ } \beta(T) < \infty\}, $上半Fredholm算子集$\Phi_{+}(X) = \{T \in {B}(X): \alpha(T) < \infty {且\ } \mathcal{R}(T) {闭} \}, $下半Fredholm算子集$\Phi_{-}(X) = \{T \in \mathcal{B}(X): \beta(T) < \infty \}, $半Fredholm算子集$\Phi_{\pm}(X)=\Phi_+(X)\cup \Phi_-(X)$.易知, $T^*\in \Phi_{-}(X^*)\Leftrightarrow T\in \Phi_{+}(X)$, $T^*\in \Phi_{+}(X^*)\Leftrightarrow T\in \Phi_{-}(X).$对$T\in\Phi_{\pm}(X)$, 定义指标${\rm{ind}}(T)=\alpha(T)-\beta(T)$.记$\Phi_{m, n}(X)=\{T \in \mathcal{B}(X): \mathcal{R}(T)$闭, $\alpha (T)=m, \beta (T)=n$}, 此处$m, n \in \mathbb{N}\cup \{\infty \}$.

用$Lat(T)$表示$T\in \mathcal{B}(X)$的不变子空间格.对闭子空间$M\in Lat(T)$, $T$可相应地诱导出如下的两个映射: $T$在$M$上的限制$T|_M\in \mathcal{B}(M)$和$T_M\in \mathcal{B}(X/M):x+M\longrightarrow Tx+M$.本文主要由闭子空间$M \in Lat(T)$和相应的诱导映射$T|_M\in \mathcal{B}(M)$和$T_M\in \mathcal{B}(X/M)$, 给出(上,下)半Fredholm算子的等价刻画, 深化对半Fredholm算子结构的理解.

首先, 利用一个诱导映射$T|_M\in \mathcal{B}(M)$ (或$T_M\in \mathcal{B}(X/M)$), 给出(上, 下)半Fredholm算子的等价刻画.

命题1  设$T\in \mathcal{B}(X)$, 则以下陈述等价:

$(1)$ $T\in \Phi_{+}(X).$

$(2)$对任意闭子空间$M\in Lat(T), $有$T|_M\in \Phi_{+}(M)$.

$(3)$存在$M\in Lat(T), {\rm{dim}}X/M < \infty, $使得$T|_M\in \Phi_{+}(M)$.

$(4)$对任意的$M\in Lat(T), {\rm{dim}}M < \infty, $有$T_M\in \Phi_{+}(X/M)$.

$(5)$存在$M\in Lat(T), {\rm{dim}}M < \infty, $使得$T_M\in \Phi_{+}(X/M)$.

证  $(1)\Rightarrow (2)$对任意闭子空间$M\in Lat(T), $由文献[4]的引理4.3.1可知, $R(T|_M)$闭.又$\alpha (T|_M)={\rm{dim}}\mathcal{N}(T)\cap M \leq \alpha (T) < \infty, $从而$T|_M\in \Phi_{+}(M)$.

$(2)\Rightarrow (3)$显然.

$(3)\Rightarrow (1)$存在闭子空间$N\subseteq X, $使得$X=N\oplus M$, 则$T = \left( {\begin{array}{*{20}{c}} {T{|_M}}&{{T_1}}\\ 0&{{T_2}} \end{array}} \right) \in \mathcal{B}(M\oplus N).$由

$ {\rm{dim}}N={\rm{dim}}X/M <\infty, $

则$T_2\in \Phi(N)$.由文献[2]的引理2.7可知, $T|_M\in \Phi_{+}(M)\Rightarrow T\in \Phi_{+}(X)$.

$(1)\Rightarrow (4)$对任意的$M\in Lat(T), {\rm{dim}}M < \infty, $则存在闭子空间$N\subseteq X, $使得$X=N\oplus M$, 从而$X^*=M^{\bot}\oplus N^{\bot}, {\rm{dim}}N^{\bot}={\rm{dim}}(X/N)^*={\rm{dim}}X/N={\rm{dim}}M < \infty, $所以

$ \begin{eqnarray*} \beta (T^*|_{M^{\bot}})&=&{\rm{dim}}M^{\bot}/T^*({M^{\bot}}) ={\rm{dim}}X^*/T^*({M^{\bot}})-{\rm{dim}}X^*/M^{\bot}\\ & =&{\rm{dim}}X^*/\mathcal{R}(T^{*})+{\rm{dim}} \mathcal{R}(T^{*})/T^*({M^{\bot}})-{\rm{dim}}X^*/M^{\bot}\\ &=&\beta(T^{*}) +{\rm{dim}}(T^{*}({M^{\bot}})+T^{*}({N^{\bot}}))/T^*({M^{\bot}})-{\rm{dim}}N^{\bot}\\ &\leq& \beta(T^{*})+{\rm{dim}}T^{*}({N^{\bot}})-{\rm{dim}}N^{\bot}\leq \beta(T^{*}).\end{eqnarray*} $

由$T\in \Phi_{+}(X), $可得$T^*\in \Phi_{-}(X^*), $即$\beta(T^{*}) < \infty, $因此$\beta (T^*|_{M^{\bot}}) < \infty $, 即$T^*|_{M^{\bot}}\in \Phi_{-}(M^{\bot}).$而$T_M^*=JT^*|_{M^{\bot}}J^{-1}$, 其中$J\in \mathcal{B}(M^{\bot}, (X/M)^*)$是同构映射, 所以$T_M^*\in \Phi_{-}((X/M)^*), $从而$T_M\in \Phi_{+}(X/M)$.

$(4)\Rightarrow (5)$显然.

$(5)\Rightarrow (1)$存在闭子空间$N\subseteq X, $使得$X=N\oplus M$, 则$X^*=M^{\bot}\oplus N^{\bot}.$由$M\in Lat(T)$可知$M^{\bot} \in Lat(T^*)$, 从而$T^*=\left(\begin{array}{cc} T^*|_{M^{\bot}} & T_3 \\ 0 & T_4 \\ \end{array} \right)\in \mathcal{B}(M^{\bot}\oplus N^{\bot}).$由

$ {\rm{dim}}N^{\bot}={\rm{dim}}(X/N)^*={\rm{dim}}X/N={\rm{dim}}M <\infty, $

可得$T_4\in \Phi(N^{\bot}).$由文献[2]的引理2.7, 可知$T^*|_{M^{\bot}}\in \Phi_{-}(M^{\bot})\Leftrightarrow T^*\in \Phi_{-}(X^*).$由$T_M\in \Phi_{+}(X/M), $ 可得$T_M^*\in \Phi_{-}((X/M)^*).$又$ T_M^*=JT^*|_{M^{\bot}}J^{-1}$, 其中$J\in \mathcal{B}(M^{\bot}, (X/M)^*)$是同构映射, 从而$T^*|_{M^{\bot}}\in \Phi_{-}(M^{\bot}), $所以$T^*\in \Phi_{-}(X^*)$, 故$T\in \Phi_{+}(X).$

对偶地, 由命题1我们有如下的结果.

命题2  设$T\in \mathcal{B}(X)$, 则以下陈述等价:

$(1)$ $T\in \Phi_{-}(X).$

$(2)$对任意闭子空间$M\in Lat(T), $有$T_M\in \Phi_{-}(X/M)$.

$(3)$存在$M\in Lat(T), {\rm{dim}}M < \infty, $使得$T_M\in \Phi_{-}(X/M)$.

$(4)$对任意的$M\in Lat(T), {\rm{dim}}X/M < \infty, $有$T|_M\in \Phi_{-}(M)$.

$(5)$存在$M\in Lat(T), {\rm{dim}}X/M < \infty, $使得$T|_M\in \Phi_{-}(M)$.

下面的反例说明, 命题1的(3) 和(5), 命题2的(3) 和(5) 中有限维的条件是必不可少的.

例3  设$X=M\oplus N, {\rm{dim}}M={\rm{dim}}N=\infty, $定义$T=I\oplus 0\in \mathcal{B}(M\oplus N), $易知

$ T|_{M\oplus \{0\}}=I|_{M\oplus \{0\}}, T_{\{0\}\oplus N}=I_{\{0\}\oplus N}, $

从而$T|_{M\oplus \{0\}}\in \Phi({M\oplus \{0\}}), T_{\{0\}\oplus N}\in \Phi(X/({\{0\}\oplus N}))$, 但$T \not\in \Phi_{+}(X)\cup \Phi_{-}(X). $

在给出进一步的结果之前, 需要如下的一些概念和事实.

对$T\in \mathcal{B}(X)$, 定义升指数${\rm{asc}}(T)=\inf\{n\in \mathbb{N}:\ \mathcal{N}(T^n)=\mathcal{N}(T^{n+1})\}$, 降指数

$ {\rm{des}}(T)=\inf\{n\in \mathbb{N}:\ \mathcal{R}(T^n)=\mathcal{R}(T^{n+1})\}, $

规定$\inf\emptyset=\infty$.上半Browder算子集$B_{+}(X) = \{T \in \mathcal{B}(X): T \in \Phi_{+}(X), {且\ }{\rm{asc}}(T) < \infty \}, $下半Browder算子集$B_{-}(X) = \{T \in \mathcal{B}(X): T \in \Phi_{-}(X), {且\ } {\rm{des}}(T) < \infty \}, $半正则算子集$\mathcal{S}(X)=\{T \in \mathcal{B}(X): \mathcal{R}(T)$闭且$\mathcal{N}(T^{n})\subseteq \mathcal{R}(T)$, 对任意$n\in \mathbb{N}$}.

闭子空间对$(M, N)$称为$T\in \mathcal{B}(X)$的约化子空间对(记为$(M, N)\in Red(T)$), 如果$X=M\oplus N$且$M, N\in Lat(T).$称$T$有广义Kato分解(记为{GKD}), 如果存在$(M, N)\in Red(T)$使得$T|_M\in \mathcal{S}(M)$且$T|_N$是拟幂零; 称$T$是Kato型算子(记为$T\in \mathcal{K}_a (X)$), 若$T$有${GKD}(M, N)$使得$T|_N$是幂零; 称$T$是本性半正则算子(记为$T\in \mathcal{S}_e(X)$), 若$T$有${GKD}(M, N)$且${\rm{dim}}N < \infty.$由文献[3]的定理1.62可知, $\Phi_{-}(X)\cup \Phi_{+}(X)= \Phi_{\pm}(X)\subseteq \mathcal{S}_e(X)$.

对$T\in \mathcal{B}(X)$, 定义超值域$\mathcal{R}(T^{\infty})=\bigcap\limits_{n=1}^{\infty}\mathcal{R}(T^n)$, 超核$\mathcal{N}(T^{\infty})=\bigcup\limits_{n=1}^{\infty}\mathcal{N}(T^n), $易见$\mathcal{R}(T^{\infty}), $ $ \mathcal{N}(T^{\infty})\in Lat(T)$.对子空间$M_1, M_2\subseteq X$, 称$M_1$本性包含于$M_2$ (记为$M_1\subseteq _e M_2$), 若存在有限维子空间$F \subseteq X$, 使得$M_1\subseteq M_2+F$.易见, $M\subseteq _e N \Leftrightarrow {\rm{dim}}M/(M\cap N) < \infty $.由文献[3]的定理1.48可知, $T\in \mathcal{S}_e(X)\Leftrightarrow \mathcal{R}(T)$闭且$\mathcal{N}(T^{\infty})\subseteq _e \mathcal{R}(T) \Leftrightarrow \mathcal{R}(T)$闭且$\mathcal{N}(T^{\infty})\subseteq _e \mathcal{R}(T^{\infty}).$

Grabiner^[6]定义了拓扑一致降指数算子:称$T\in \mathcal{B}(X)$对$n\geq d$有拓扑一致降指数(记为$T\in {TUD}_d(X)$), 若存在$d\in\mathbb{N}$使得$\mathcal{R}(T)+\mathcal{N}(T^d)=\mathcal{R}(T)+\mathcal{N}(T^{\infty})$且$\mathcal{R}(T)+\mathcal{N}(T^d)$闭.拓扑一致降指数算子集${TUD}(X)=\bigcup\limits_{d=0}^{\infty}{TUD}_d(X)$涵盖了Fredholm理论中的诸多算子类, 如${TUD}_0(X)=\mathcal{S}(X), B_{+}(X)\cup B_{-}(X)\subseteq \Phi_{\pm}(X)\subseteq \mathcal{S}_e(X)\subseteq \mathcal{K}_a (X)\subseteq {TUD}(X)$.更多关于此类算子的摄动理论可参阅文献[6].

Fang^[5]定义了$T\in \Phi_{\pm}(X)$的左(右)移位Samuel重数:对$T\in \Phi_{\pm}(X), $它的右移位Samuel重数(shift Samuel multiplicity)和左移位Samuel重数(backward shift Samuel multiplicity), 分别定义为

$ s_{-}{mul}(T)=\lim\limits_{k \rightarrow \infty} \frac{\beta(T^{k})}{k}; \ \ \ b.s._{-}{mul}(T)=\lim\limits_{k \rightarrow \infty} \frac{\alpha(T^{k})}{k}. $

由文献[5]的定理2可知:对$T\in \Phi_{\pm}(X)$, $T$的左、右移位Samuel重数$b.s._{-}{mul}(T), $ $s_{-}{mul}(T)\in \mathbb{N} \cup \{\infty \}$, 且指标${\rm{ind}}(T)=b.s._{-}{mul}(T)-s_{-}{mul}(T)$.同时, 当$n\in \mathbb{N}$足够大时, 有

$ \begin{eqnarray*}&&b.s._{-}{mul}(T)={\rm{dim}}\mathcal{N}(T)\cap \mathcal{R}(T^{n}) ={\rm{dim}}\mathcal{N}(T)\cap \mathcal{R}(T^{\infty}), \\ &&s_{-}{mul}(T)={\rm{dim}} X/(\mathcal{R}(T)+\mathcal{N}(T^{n})) ={\rm{dim}}X/(\mathcal{R}(T)+\mathcal{N}(T^{\infty})).\end{eqnarray*} $

接下来, 利用两个诱导映射$T|_M\in \mathcal{B}(M)$和$T_M\in \mathcal{B}(X/M)$, 给出(上, 下)半Fredholm算子的等价刻画.

定理4  设$T\in \mathcal{B}(X)$, 则以下陈述等价:

(1) $ T\in \Phi_{+}(X)$.

(2) 存在闭子空间$M\in Lat(T), m \in \mathbb{N}$, 使得$T|_M\in \Phi_{m, 0}(M)$, $T_M\in B_{+}(X/M)$.

(3) 存在闭子空间$M\in Lat(T)$, 使得$T|_M\in \Phi_{+}(M)$, $T_M\in \Phi_{+}(X/M)$.

证   $(1)\Rightarrow (2)$取$M=\mathcal{R}(T^{\infty}), $易知$M$闭.由文献[5]的引理7可知, $T|_M$满射, 即$\beta (T|_M)=0 $.又$\alpha (T|_M)\leq \alpha (T) < \infty $, 且

$ \alpha (T|_M)={\rm{dim}}\mathcal{N}(T)\cap M={\rm{dim}}\mathcal{N}(T)\cap \mathcal{R}(T^{\infty}) =b.s._{-}mul(T), $

则$T|_M\in \Phi_{m, 0}(X)$, 其中$m=b.s._{-}mul(T) \in \mathbb{N}.$由$\mathcal{R}(T)$闭, $M=T(M)\subseteq \mathcal{R}(T)$, 且$\mathcal{R}(T_M)=(\mathcal{R}(T)+M)/M=\mathcal{R}(T)/M$, 可见$\mathcal{R}(T_M)$闭.由文献[6]的引理2.1可得

$ T^{-1}(M)/M=T^{-1}(T(M))/M= (\mathcal{N}(T)+M)/M=\mathcal{N}(T)/(\mathcal{N}(T)\cap M), $

从而$\alpha (T_M)\leq \alpha (T) < \infty, $则$T_M\in \Phi_{+}(X/M)$.由$T\in \Phi_{+}(X)\subseteq \mathcal{S}_e(X)$, 则$\mathcal{N}(T^{\infty})\subseteq _e \mathcal{R}(T^{\infty})=M$, 即

$ {\rm{dim}}\mathcal{N}(T^{\infty})/(\mathcal{N}(T^{\infty})\cap \mathcal{R}(T^{\infty})) ={\rm{dim}}(\mathcal{N}(T^{\infty})+\mathcal{R}(T^{\infty}))/\mathcal{R}(T^{\infty})< \infty . $

又由$T(M)=M, $易见对$k\in \mathbb{N}, $有$T^{k}(M)=M, $即$T^{k}(\mathcal{R}(T^{\infty}))=\mathcal{R}(T^{\infty})$.结合文献[6]的引理2.1, 可得

$ \begin{eqnarray*}N(T_M^{\infty})&=&\bigcup\limits_{k=1}^{\infty} T^{-k}(M)/M =\bigcup\limits_{k=1}^{\infty} T^{-k}(\mathcal{R}(T^{\infty}))/\mathcal{R}(T^{\infty}) =\bigcup\limits_{k=1}^{\infty} T^{-k}(T^{k}(\mathcal{R}(T^{\infty})))/\mathcal{R}(T^{\infty})\\ &=&\bigcup\limits_{k=1}^{\infty} (\mathcal{N}(T^k)+\mathcal{R}(T^{\infty}))/\mathcal{R}(T^{\infty}) =(\mathcal{N}(T^{\infty})+\mathcal{R}(T^{\infty}))/\mathcal{R}(T^{\infty}), \end{eqnarray*} $

从而${\rm{dim}}\mathcal{N}(T_M^{\infty}) < \infty, $因此${\rm{asc}}(T_M) < \infty, $那么$T_M \in B_{+}(X/M) $.

$(2)\Rightarrow (3)$显然.

$(3)\Rightarrow (1)$由假设, 存在闭子空间$M\in Lat(T)$, 使得$T(M)$闭, $\alpha (T|_M)={\rm{dim}}\mathcal{N}(T)\cap M < \infty, \mathcal{R}(T_M)$闭, 且$\alpha (T_M) < \infty$.由$T(M)\subseteq M$可得, $M\subseteq \mathcal{N}(T)+ M\subseteq T^{-1}(M)$, 所以存在子空间$F, {\rm{dim}}F=\alpha (T_M)={\rm{dim}}T^{-1}(M)/M$, 使得$T^{-1}(M)=M\oplus F, $显然${\rm{dim}}T(F)\leq {\rm{dim}}F < \infty.$由文献[6]的引理2.1(d)可得

$ \mathcal{R}(T)\cap M=T(X)\cap M=T(X\cap T^{-1}(M))=T(T^{-1}(M))=T(M\oplus F)=T(M)+T(F), $

故$\mathcal{R}(T)\cap M$闭.又$\mathcal{R}(T_M)=(\mathcal{R}(T)+M)/M$闭, 则$\mathcal{R}(T)+M$闭, 从而由文献[1]的定理2, 可得$\mathcal{R}(T)$闭.由文献[6]的引理2.1(b)可得

$ \begin{eqnarray*}&&\alpha (T)={\rm{dim}}\mathcal{N}(T)={\rm{dim}}\mathcal{N}(T)\cap M+{\rm{dim}}\mathcal{N}(T)/(\mathcal{N}(T)\cap M)\\ & =&\alpha (T|_M)+{\rm{dim}}(\mathcal{N}(T)+ M)/M\leq \alpha (T|_M)+{\rm{dim}}T^{-1}(M)/M= \alpha (T|_M)+\alpha (T_M)<\infty, \end{eqnarray*} $

故$T\in \Phi_{+}(X).$

定理5   设$T\in \mathcal{B}(X)$,

则以下陈述等价:

(1) $T\in \Phi_{-}(X)$.

(2) 存在闭子空间$M\in Lat(T), m\in \mathbb{N}$, 使得$T_M\in \Phi_{0, m}(X/M)$, $T|_M\in B_{-}(M)$.

(3) 存在闭子空间$M\in Lat(T)$, 使得$T_M\in \Phi_{-}(X/M)$, $T|_M\in \Phi_{-}(M)$.

证  $(1)\Rightarrow (2)$取$M =\overline{\mathcal{N}(T^{\infty})}, $易知$M\in Lat(T)$.由文献[6]的定理3.4, 可得$T_M$是下有界算子, 即$\alpha (T_M)=0, T^{-1}(M)=M.$又

$ \beta (T_M)={\rm{dim}}(X/M)/((\mathcal{R}(T)+M)/M) ={\rm{dim}} X/(\mathcal{R}(T)+M)\leq {\rm{dim}}X/\mathcal{R}(T)=\beta (T), $

即$\beta (T_M)\leq \beta (T) < \infty$, 又由$T\in \Phi_{-}(X)\subseteq \Phi_{\pm}(X)\subseteq {TUD}(X), $不妨设$T\in {TUD}_d(X)$ $(d\in \mathbb{N})$, 可知$\mathcal{R}(T)+\mathcal{N}(T^{\infty})$闭.又

$ \mathcal{R}(T)+\mathcal{N}(T^{\infty})\subseteq \mathcal{R}(T)+\overline{\mathcal{N}(T^{\infty})}\subseteq \overline{\mathcal{R}(T)+\mathcal{N}(T^{\infty})}=\mathcal{R}(T)+\mathcal{N}(T^{\infty}), $

则$\mathcal{R}(T)+\mathcal{N}(T^{\infty})= \mathcal{R}(T)+\overline{\mathcal{N}(T^{\infty})}$, 从而

$ \begin{eqnarray*}\beta (T_M)&=&{\rm{dim}}X/(\mathcal{R}(T)+M)={\rm{dim}}X/(\mathcal{R}(T)+\overline{\mathcal{N}(T^{\infty})})\\ &=&{\rm{dim}}X/(\mathcal{R}(T)+\mathcal{N}(T^{\infty}))=s_{-}mul(T), \end{eqnarray*} $

故$T_M\in \Phi_{0, m}(X/M)$, 其中$m=s_{-}mul(T) \in \mathbb{N}.$由$T\in \Phi_{-}(X)\subseteq \mathcal{S}_e(X), $可得$\mathcal{N}(T^{\infty})\subseteq _e \mathcal{R}(T)$, 易见$\overline{\mathcal{N}(T^{\infty})}\subseteq _e \mathcal{R}(T)$, 即${\rm{dim}}\overline{\mathcal{N}(T^{\infty})}/(\overline{\mathcal{N}(T^{\infty})}\cap \mathcal{R}(T)) < \infty$.由$T^{-1}(M)=M, $可得$T(M)=\mathcal{R}(T)\cap M.$又

$ \beta (T|_M)={\rm{dim}}M/T(M)={\rm{dim}}M/(\mathcal{R}(T)\cap M) ={\rm{dim}}\overline{\mathcal{N}(T^{\infty})}/(\overline{\mathcal{N}(T^{\infty})}\cap \mathcal{R}(T))< \infty, $

因此$T|_M\in \Phi_{-}(M)$.由$T^{-1}(M)=M, $可得$T^{-k}(M)=M, $从而$T^{k}(M)=\mathcal{R}(T^{k})\cap M, $对任意的$k\in \mathbb{N}$.又由$T\in {TUD}_d(X)$和文献[6]的引理3.6(c)可知

$ \mathcal{R}(T^{d})\cap \overline{\mathcal{N}(T^{\infty})} =\mathcal{R}(T^{\infty})\cap \overline{\mathcal{N}(T^{\infty})}, $

则$ \mathcal{R}(T^{d})\cap M=\mathcal{R}(T^{d})\cap \overline{\mathcal{N}(T^{\infty})} =\mathcal{R}(T^{d+1})\cap \overline{\mathcal{N}(T^{\infty})}=\mathcal{R}(T^{d+1})\cap M$, 因此$T^{d}(M)=T^{d+1}(M), $从而${\rm{des}}(T|_M)\leq d < \infty, $故$T|_M\in B_{-}(M) $.

$(2)\Rightarrow (3)$显然.

$(3)\Rightarrow (1)$由假设, 存在闭子空间$M\in Lat(T)$, 使得$\beta(T|_M)={\rm{dim}}M/T(M) < \infty, $ $ \beta(T_M)={\rm{dim}}(X/M)/((\mathcal{R}(T)+M)/M)={\rm{dim}}X/(\mathcal{R}(T)+M) < \infty.$存在子空间$F\subseteq X$, 使得${\rm{dim}}F=\beta(T|_M), M=T(M)\oplus F$.由

$ \beta(T_M)={\rm{dim}}X/(\mathcal{R}(T)+M) ={\rm{dim}}X/(\mathcal{R}(T)+T(M)+F)={\rm{dim}}X/(\mathcal{R}(T)+F), $

可得

$ \begin{eqnarray*}\beta(T)&=&{\rm{dim}}X/\mathcal{R}(T)\\ &=&{\rm{dim}}X/(\mathcal{R}(T)+F)+ {\rm{dim}}(\mathcal{R}(T)+F)/\mathcal{R}(T)\leq \beta(T_M)+\beta(T|_M)<\infty, \end{eqnarray*} $

因此$T\in \Phi_{-}(X).$

注这里对定理4和定理5的“$(3)\Rightarrow (1)$”进行一些说明.

当定理4的条件$(3)$中$M\in Lat(T)$是可补的, 则“$(3)\Rightarrow (1)$”是易知的.事实上, 不妨设$X=M\oplus N$, 此时$T=\left(\begin{array}{cc} A & C \\ 0 & B \\ \end{array} \right)\in \mathcal{B}(M\oplus N), $其中$A\in \mathcal{B}(M), B\in \mathcal{B}(N), C\in \mathcal{B}(N, M).$由$A=T|_{M}, B=J^{-1}T_{M}J, $这里$J\in \mathcal{B}(N, X/M)$是自然同构映射, 可知$A\in \Phi_{+}(M), B\in \Phi_{+}(N), $从而由上三角算子矩阵的知识可知, $T\in \Phi_{+}(X).$

同样地, 当定理5的条件$(3)$中$M\in Lat(T)$是可补的, 则“$(3)\Rightarrow(1)$”也是易知的.

当定理4和定理5的条件$(3)$中$M\in Lat(T)$不可补时, 上述矩阵分块的证明方法将失效.