Hilbert空间中的框架是标准正交基的一种推广, 框架有些性质和标准正交基相似, 但是也有明显的区别的性质, 比如: Hilbert空间中的元素用框架展开的表达式是不唯一的, 而用标准正交基展开的表达式却是唯一的.近年来框架理论不仅成为数学理论研究的重点内容之一, 而且在一些应用领域中诸如采样理论^[1]、压缩感知^[2]等也起着重要的作用.有关框架理论及其应用的介绍可以参见文献[3-4].

在2012年, Găvruƫa^[5]引入了一种称之为$ K $-框架的推广框架, 它可以重构有界线性算子$ K $值域中的元素, 并且证明了序列$ \{f_{j}\}_{j\in J} $是$ {\cal H} $的$ K $-框架当且仅当$ \{f_{j}\}_{j\in J} $是$ {\cal H} $的Bessel序列且其合成算子$ T_{f} $满足$ R(K)\subset R(T_{f}) $, 但是文献[3]已经指出序列$ \{f_{j}\}_{j\in J} $是$ {\cal H} $的框架当且仅当$ \{f_{j}\}_{j\in J} $是$ {\cal H} $的Bessel序列且其合成算子$ T_{f} $为满射的.由此可见框架的有些性质对$ K $-框架而言并不成立, 从而对$ K $-框架的一些性质进行研究具有深远的意义.而框架的具体构造是实现从数学理论到实际应用的最关键的一个环节, 因此许多学者对框架、$ K $-框架和紧$ K $-框架的具体构造进行一些深入研究, 已经得到了$ K $-框架和紧$ K $-框架算子扰动的稳定性的一些结论, 文献[6-11]中也已经给出了构造$ K $-框架的几种方法.

本文将继续对$ K $-框架和紧$ K $-框架算子扰动的稳定性进一步研究.文献[12]给出框架经有界线性算子$ T $扰动后为框架当且仅当其扰动算子$ T $为满算子, 但是这一结论对于$ K $-框架而言并不成立.我们在此基础上给出$ K $-框架经有界线性算子$ T $扰动后为$ K $-框架的一个充要条件, 其结果可以得到文献[9]和文献[12]的部分结果.文献[13]讨论了用一个框架和一个Bessel序列构造框架的方法, 郭训香^[14]使用强分离性框架的概念给出用两个已知的框架构造新的框架的方法, 随后文献[7-8]讨论了用两个已有的$ K $-框架构造新的$ K $-框架的几种方法.同时, 郭训香^[15]提供了一种从给定的强分离性紧框架构造出新的紧框架的方法, 丁明玲等人^[11]提出了紧$ K $-框架的概念后, 给出了几种构造紧$ K $-框架和紧框架的方法.在此基础上, 本文给出用两个Bessel序列或者$ K $-框架构造新的$ K $-框架的方法, 给出了用两个Bessel序列构造紧$ K $-框架的充要条件.我们的结果推广并改进了由Casazza和Christensen等得到的著名结果.

本文采用如下记号:设$ {\cal H} $为可分的复Hilbert空间; $ I_{{\cal H}} $是$ {\cal H} $的恒等算子; $ J $为一个可数指标集; $ \ell^{2}(J) $表示满足$ \sum\limits_{j\in J}|a_{j}|^{2}<+\infty $的复数列$ \{a_{j}\}_{j\in J} $全体所成的线性空间.设$ {{\cal H}_1} $和$ {{\cal H}_2} $为两个复Hilbert空间, 记$ B({{\cal H}_1}, {{\cal H}_2}) $表示从$ {{\cal H}_1} $到$ {{\cal H}_2} $的所有有界线性算子的集合, 若$ {{\cal H}_1}{\rm{ = }}{{\cal H}_2}{\rm{ = }}{\cal H} $, 则记$ B({{\cal H}_1}, {{\cal H}_2}) = B({\cal H}) $.设$ K \in B({\cal H}) $且$ K\neq 0 $, $ K $的值域记为$ R(K) $.

首先介绍Hilbert空间中的框架和$ K $-框架等概念.

定义1.1^[3]  设序列$ \{f_j\}_{j\in J}\subset {\cal H} $, 若存在正数$ A $和$ B $, 使得对于$ \forall{f}\in{{\cal H}} $, 有

(1.1) $ \begin{eqnarray} A\|f\|^2 \leq\sum\limits_{j\in J} |\langle f, f_j\rangle|^2 \leq B\|f\|^2, \end{eqnarray} $

则称$ \{f_j\}_{j\in J} $为$ {\cal H} $的框架, 其中$ A $和$ B $分别称为框架的下界和上界.

如果只有不等式$ (1.1) $的右边成立, 则称$ \{f_j\}_{j\in J} $为$ {\cal H} $的界为$ B $的Bessel序列, 从而可定义有界线性算子$ T_{f} $, $ T_{f}^{*} $如下

$ T_{f}:\ell^{2}(J)\rightarrow {\cal H}, \, T_{f}a = \sum\limits_{j\in J}a_{j}f_{j}, \, \, \, a = \{a_{j}\}_{j\in J}\in \ell^{2}(J), $

$ T_{f}^{*}:{\cal H}\rightarrow \ell^{2}(J), T_{f}^{*}f = \{\langle{f}, {f_{j}}\rangle\}_{j\in J}, \, \, \, f \in {\cal {\cal H}}. $

称$ T_{f} $为$ \{f_j\}_{j\in J} $的合成算子, 称有界线性算子$ S = T_{f}T_{f}^{*} $为$ \{f_j\}_{j\in J} $的框架算子

$ \begin{eqnarray*} S:{\cal H}\rightarrow {\cal H}, \, \;Sf = \sum\limits_{j\in J} \langle {f}, \;{f_j}\rangle f_{j}, \, \, \, f \in {\cal {\cal H}}. \end{eqnarray*} $

定义1.2  设$ K\in{B({\cal H})} $, 序列$ \{f_j\}_{j\in J}\subset{{\cal H}} $, 若存在正数$ A $和$ B $, 使得对于$ \forall{f}\in{{\cal H}} $, 有

$ \begin{eqnarray*} A\|K^{*}f\|^2\leq\sum\limits_{j\in J}|\langle{f}, \, {f_{j}}\rangle|^{2}\leq{B\|f\|^{2}}, \end{eqnarray*} $

称$ \{f_j\}_{j\in J} $为$ {\cal H} $的$ K $-框架, 其中$ A $和$ B $分别称为$ K $-框架的下界和上界.特别地, 当$ K = I_{{\cal H}} $时，$ K $-框架就是框架.

定义1.3^[11]  设$ K\in{B({\cal H})} $, 如果存在正数$ A $, 使得对于$ \forall{f}\in{{\cal H}} $, 有

$ \begin{eqnarray*} A\|{K^*}f\|^2 = \sum\limits_{j\in J}{|\langle f, \;{f_j}\rangle {|^2}}, \end{eqnarray*} $

称$ \{f_j\}_{j\in J} $为$ {\cal H} $的紧$ K $-框架, 其中$ A $称为紧$ K $-框架的界.

为了证明本文的定理, 我们先给出一些引理.

引理2.1^[3]  设$ {{\cal H}_1} $和$ {{\cal H}_2} $是两个复Hilbert空间, $ T:{{\cal H}_1}\rightarrow {{\cal H}_2} $是具有闭值域的有界线性算子, 则存在唯一的有界线性算子$ T^+: {{\cal H}_2}\rightarrow {{\cal H}_1} $满足

$ \begin{eqnarray*} N_{T^+} = R(T)^\perp, \, R(T^+) = N^\perp_T, \, TT^+f = f, \, f\in R(T), \end{eqnarray*} $

称算子$ T^+ $为$ T $的伪逆算子.

引理2.2^[6]  设$ T:{{\cal H}_1}\rightarrow {{\cal H}_2} $是具有闭值域的有界线性算子, 则

$ \begin{eqnarray*} \|T^+\|^{-1}\|f\|\leq\|T^*f\|\leq\|T\|\|f\|, \, \, \, f\in R(T). \end{eqnarray*} $

引理2.3^[5]  设$ K\in B({\cal H}) $, 则序列$ \{f_j\}_{j\in J} $为$ {\cal H} $的$ K $-框架当且仅当$ \{f_j\}_{j\in J} $为$ {\cal H} $的Bessel序列且其合成算子$ T_{f} $满足$ {R(K)}\subset{R(T_{f})} $.

引理2.4^[16]  设$ T_{1}\in B({\cal H}_{1}, {\cal H}) $, $ T_{2}\in B({\cal H}_{2}, {\cal H}) $, 则下列叙述等价

$ \mathrm{(1)} $$ R(T_{1})\subset R(T_{2}) $;

$ \mathrm{(2)} $存在$ \alpha>0 $使得$ T_{1}T_{1}^{*}\leq \alpha^{2} T_{2}T_{2}^{*} $;

$ \mathrm{(3)} $存在有界线性算子$ X\in B({\cal H}_{1}, {\cal H}_{2}) $使得$ T_{1} = T_{2}X $.

引理2.5  设$ T_{1}, T_{2}\in B({\cal H}_{1}, {\cal H}_{2}) $, $ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的Bessel序列, 则$ \{ {T_{1}f_j+T_{2}g_j}\} _{j\in J} $是$ {\cal H}_{2} $的Bessel序列, 其界为$ (\sqrt{B_{1}}\|T_{1}\|+\sqrt{B_{2}}\|T_{2}\|)^{2} $.

证  设$ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的Bessel序列, 其界分别为$ B_{1} $和$ B_{2} $, 从而对于$ \forall f\in {\cal H}_{1} $, 有

$ \sum\limits_{j\in J}{|\langle f, \;{f_j}\rangle {|^2}}\leq B_{1}\|f\|^{2}, \quad \sum\limits_{j\in J}{|\langle f, \;{g_j}\rangle {|^2}}\leq B_{2}\|f\|^{2}. $

由闵可夫斯基不等式得, 对于$ \forall g\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} \bigg(\sum\limits_{j\in J}{|\langle g, \;{T_{1}f_j+T_{2}g_j}\rangle {|^2}}\bigg)^{\frac 12}& = &\bigg(\sum\limits_{j\in J}{|\langle T_{1}^{*}g, \;{f_j}\rangle + \langle T_{2}^{*}g, \;{g_j}\rangle {|^2}}\bigg)^{\frac 12}\\ &\leq& \bigg(\sum\limits_{j\in J}{|\langle T_{1}^{*}g, \;{f_j}\rangle{|^2}}\bigg)^{\frac 12}+\bigg(\sum\limits_{j\in J}{|\langle T_{2}^{*}g, \;{g_j}\rangle{|^2}}\bigg)^{\frac 12}\\ &\leq&\sqrt{B_{1}}\|T_{1}^{*}g\|+\sqrt{B_{2}}\|T_{2}^{*}g\|\\ &\leq&(\sqrt{B_{1}}\|T_{1}\|+\sqrt{B_{2}}\|T_{2}\|)\|g\|, \end{eqnarray*} $

从而$ \{ {T_{1}f_j+T_{2}g_j}\} _{j\in J} $是$ {\cal H}_{2} $的Bessel序列, 其界为$ (\sqrt{B_{1}}\|T_{1}\|+\sqrt{B_{2}}\|T_{2}\|)^{2} $.证毕.

引理2.6^[11]  设$ \{ {f_j}\} _{j\in J} $是$ {\cal H} $的Bessel序列, 其合成算子为$ T_{f} $.则$ \{ {f_j}\} _{j\in J} $是$ {\cal H} $的紧$ K $-框架当且仅当存在$ A>0 $使得$ AKK^{*} = T_{f}T_{f}^{*} $.

文献[7, 9-10]已经讨论了框架或者$ K $-框架经过有界线性算子扰动后的序列为$ K $-框架的几种方法.现在我们给出$ K $-框架经过有界线性算子$ T $扰动后的序列为$ K $-框架的一个充要条件.

定理2.1  设$ K_{1}\in B({\cal H}_{1}) $具有闭值域, $ K_{2}\in B({\cal H}_{2}) $, 且$ T\in B({\cal H}_{1}, {\cal H}_{2}) $和$ R(T^{*})\subset R(K_{1}) $.若序列$ \{f_j\}_{j\in J} $为$ {\cal H}_{1} $的$ K_{1} $-框架, 则$ \{Tf_{j}\}_{j\in J} $是$ {\cal H}_{2} $的$ K_{2} $-框架当且仅当$ R(K_{2})\subset R(T) $.

证  设$ \{f_j\}_{j\in J} $为$ {\cal H}_{1} $的$ K_{1} $-框架, 不妨设其框架界为$ A $和$ B $, 则对于$ \forall f\in {\cal H}_{1} $, 有

$ A\|K_{1}^{*}f\|^{2}\leq\sum\limits_{j\in J}{|\langle f, \;{f_j}\rangle {|^2}}\leq B\|f\|^{2}. $

因此对于$ \forall g\in {\cal H}_{2} $, 有

$ A\|K_{1}^{*}T^{*}g\|^{2}\leq\sum\limits_{j\in J}|\langle T^{*}g, \, f_{j}\rangle|^{2} = \sum\limits_{j\in J}|\langle g, \, Tf_{j}\rangle|^{2}\leq B\|T^{*}g\|^{2}\leq B\|T\|^{2}\|g\|^{2}. $

因为$ K_{1}\in B({\cal H}_{1}) $具有闭值域且$ R(T^{*})\subset R(K_{1}) $, 由引理2.2, 得对于$ \forall g\in {\cal H}_{2} $, 有

$ A\|K_{1}^{*}T^{*}g\|^{2}\geq A\|K_{1}^{+}\|^{-2}\|T^{*}g\|^{2}. $

设$ R(K_{2})\subset R(T) $, 根据引理2.4, 存在$ \alpha>0 $使得$ K_{2}K_{2}^{*}\leq \alpha^{2} TT^{*} $, 对于$ \forall g\in {\cal H}_{2} $, 有

$ A\|K_{1}^{+}\|^{-2}\|T^{*}g\|^{2}\geq A\alpha^{-2}\|K_{1}^{+}\|^{-2}\|K_{2}^{*}g\|^{2}. $

综上所述, 对于$ \forall g\in {\cal H}_{2} $, 有

$ A\alpha^{-2}\|K_{1}^{+}\|^{-2}\|K_{2}^{*}g\|^{2}\leq\sum\limits_{j\in J}|\langle g, \, Tf_{j}\rangle|^{2}\leq B\|T\|^{2}\|g\|^{2}. $

因此$ \{Tf_{j}\}_{j\in J} $是$ {\cal H}_{2} $的$ K_{2} $-框架.

另一方面, 若$ \{f_j\}_{j\in J} $是$ {\cal H}_{1} $的$ K_{1} $-框架, 记其合成算子为$ T_{f} $; $ \{Tf_{j}\}_{j\in J} $是$ {\cal H}_{2} $的$ K_{2} $-框架, 记其合成算子为$ L $, 则对于$ \forall a = \{a_{j}\}_{j\in J}\in{\ell^{2}}(J) $, 有

$ La = \sum\limits_{j\in J}a_{j}(T{f_j}) = T\sum\limits_{j\in J}a_{j}f_{j} = TT_{f}a. $

因此$ L = TT_{f} $, 从而根据引理2.3, 有$ R(K_{2})\subset R(TT_{f})\subset R(T) $.证毕.

注2.1  由定理2.1, 我们可得文献[9]的命题3.6, 同时可得文献[12]的命题2.3、命题2.4和推论2.5.在定理2.1中, 若$ R(T^{*})\nsubseteq R(K_{1}) $, 则$ \{Tf_{j}\}_{j\in J} $不一定是$ {\cal H}_{2} $的$ K_{2} $-框架, 现在给出反例如下.

例2.1  设$ \{e_{j}\}_{j = 1}^{3} $是$ {\cal H}_{1} = {\mathbb{C}}^{3} $的标准正交基, $ \{h_{j}\}_{j = 1}^{4} $是$ {\cal H}_{2} = {\mathbb{C}}^{4} $的标准正交基.设$ f_{1} = e_{1}, \, f_{2} = e_{1}, \, f_{3} = e_{2} $.定义有界线性算子$ K_{1}\in B({\cal H}_{1}) $, $ K_{2}\in B({\cal H}_{2}) $和$ T\in B({\cal H}_{1}, {\cal H}_{2}) $如下

$ K_{1}:{\cal H}_{1}\rightarrow {\cal H}_{1}, \, K_{1}f = \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{3}\rangle e_{2}, \, \, \, f \in {\cal H}_{1}, $

$ K_{2}:{\cal H}_{2}\rightarrow {\cal H}_{2}, \, K_{2}g = \langle g, \, h_{2}+h_{3}\rangle h_{1}, \, \, \, g \in {\cal H}_{2}, $

$ T:{\cal H}_{1}\rightarrow {\cal H}_{2}, \, Tf = \langle f, \, e_{2}\rangle h_{3}+\langle f, \, e_{3}\rangle h_{1}, \, \, \, f \in {\cal H}_{1}. $

显然$ K_{1}\in B({\cal H}_{1}) $具有闭值域.下面我们计算$ K_{1}^{*}f $, 对于$ \forall f, m\in {\cal H}_{1} $, 有

$ \begin{eqnarray*} \langle K_{1}^{*}f, \, m\rangle& = &\langle f, \, K_{1}m\rangle = \langle f, \langle m, \, e_{1}\rangle e_{1}+\langle m, \, e_{3}\rangle e_{2}\rangle\\ & = &\langle f, \, e_{1}\rangle\overline{\langle m, \, e_{1}\rangle}+\langle f, \, e_{2}\rangle\overline{\langle m, \, e_{3}\rangle}\\ & = &\langle f, \, e_{1}\rangle\langle e_{1}, \, m\rangle+\langle f, \, e_{2}\rangle\langle e_{3}, \, m\rangle\\ & = &\langle \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{2}\rangle e_{3}, \, m\rangle. \end{eqnarray*} $

即$ K_{1}^{*}f = \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{2}\rangle e_{3}, \, f\in {\cal H}_{1} $.从而对于$ \forall f\in {\cal H}_{1} $, 有

$ \begin{eqnarray*} \|K_{1}^{*}f\|^{2}& = &\|\langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{2}\rangle e_{3}\|^{2} = |\langle f, \, e_{1}\rangle|^{2}+|\langle f, e_{2}\rangle|^{2}\\&\leq& \sum\limits_{j = 1}^{3}|\langle f, \, f_{j}\rangle|^{2} = 2|\langle f, \, e_{1}\rangle|^{2}+|\langle f, e_{2}\rangle|^{2}\leq2\|f\|^{2}, \end{eqnarray*} $

因此$ \{f_{j}\}_{j = 1}^{3} $是$ {\cal H}_{1} $的$ K_{1} $-框架, 且$ \overline{span}\{h_{1}\} = R(K_{2})\subset R(T) = \overline{span}\{h_{1}, h_{3}\} $.

下面我们计算$ K_{2}^{*}g $, 对于$ \forall g, h\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} \langle K_{2}^{*}g, \, h\rangle& = &\langle g, \, K_{2}h\rangle = \langle g, \, \langle h, \, h_{2}+h_{3}\rangle h_{1}\rangle = \langle g, \, h_{1}\rangle\overline{\langle h, \, h_{2}+h_{3}\rangle}\\ & = &\langle g, \, h_{1}\rangle\langle h_{2}+h_{3}, \, h\rangle = \langle\langle g, \, h_{1}\rangle h_{2}+ \langle g, \, h_{1}\rangle h_{3}, \, h\rangle. \end{eqnarray*} $

即$ K_{2}^{*}g = \langle g, \, h_{1}\rangle h_{2}+\langle g, \, h_{1}\rangle h_{3}, \, g\in {\cal H}_{2} $.因此对于$ \forall g\in {\cal H}_{2} $, 有

$ \|K_{2}^{*}g\|^{2} = \|\langle g, \, h_{1}\rangle h_{2}+\langle g, \, h_{1}\rangle h_{3}\|^{2} = 2|\langle g, \, h_{1}\rangle|^{2}. $

通过简单计算可得$ Tf_{1} = 0, \, Tf_{2} = 0, \, Tf_{3} = h_{3} $, 因此对于$ \forall g\in {\cal H}_{2} $, 有

$ \sum\limits_{j = 1}^{3}|\langle g, \, Tf_{j}\rangle|^{2} = |\langle g, \, h_{3}\rangle|^{2}. $

取$ h_{1} = g\in {\cal H}_{2} $, 我们得到$ \|K_{2}^{*}g\|^{2} = 2 $和$ \sum\limits_{j = 1}^{3}|\langle g, \, Tf_{j}\rangle|^{2} = 0 $, 因此$ \{Tf_{j}\}_{j = 1}^{3} $不是$ {\cal H}_{2} $的$ K_{2} $-框架.

现在我们计算$ T^{*}g $, 对于$ \forall f\in {\cal H}_{1} $和$ \forall g\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} \langle T^{*}g, \, f\rangle& = &\langle g, \, Tf\rangle = \langle g, \, \langle f, \, e_{2}\rangle h_{3}+\langle f, \, e_{3}\rangle h_{1}\rangle = \langle g, \, h_{3}\rangle\overline{\langle f, \, e_{2}\rangle}+\langle g, \, h_{1}\rangle\overline{\langle f, \, e_{3}\rangle}\\ & = &\langle g, \, h_{3}\rangle\langle e_{2}, \, f\rangle+\langle g, \, h_{1}\rangle\langle e_{3}, \, f\rangle = \langle\langle g, \, h_{3}\rangle e_{2}+\langle g, \, h_{1}\rangle e_{3}, \, f\rangle. \end{eqnarray*} $

因此$ T^{*}g = \langle g, \, h_{3}\rangle e_{2}+\langle g, \, h_{1}\rangle e_{3}, \, g\in {\cal H}_{2} $, 这时

$ \overline{span}\{e_{2}, e_{3}\} = R(T^{*})\nsubseteq R(K_{1}) = \overline{span}\{e_{1}, e_{2}\}. $

文献[7-8]已经给出了用两个$ K $-框架构造新的$ K $-框架的方法, 现在我们进一步给出用两个Bessel序列或者$ K $-框架构造新的$ K $-框架的方法.

定理2.2  设$ K\in B({\cal H}_{2}) $和$ T_{1}, T_{2}\in B({\cal H}_{1}, {\cal H}_{2}) $.又设$ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的Bessel序列, 其合成算子分别为$ T_{f} $和$ T_{g} $, 其框架算子分别为$ S_{1} $和$ S_{2} $, 且$ T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}\geq 0 $.设$ R(K)\subset R(T_{1}T_{f}) $, 则$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的$ K $-框架, 其框架算子为

$ S = T_{1}S_{1}T_{1}^{*}+T_{2}S_{2}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}. $

证  设$ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的Bessel序列, 其界分别为$ B_{1} $和$ B_{2} $.由引理2.5, 我们得到$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的Bessel序列, 其界为$ (\sqrt{B_{1}}\|T_{1}\|+\sqrt{B_{2}}\|T_{2}\|)^{2} $, 设其合成算子为$ Q $, 则对于$ \forall a = \{a_{j}\}_{j\in J}\in{\ell^{2}}(J) $, 有

$ Qa = \sum\limits_{j\in J}a_{j}(T_{1}{f_j}+T_{2}{g_j}) = T_{1}\sum\limits_{j\in J}a_{j}f_{j}+T_{2}\sum\limits_{j\in J}a_{j}g_{j} = (T_{1}T_{f}+T_{2}T_{g})a. $

因此$ Q = T_{1}T_{f}+T_{2}T_{g} $, 从而得其框架算子为

$ \begin{eqnarray*} S& = &QQ^{*} = (T_{1}T_{f}+T_{2}T_{g})(T_{1}T_{f}+T_{2}T_{g})^{*}\\ & = &T_{1}T_{f}T_{f}^{*}T_{1}^{*}+T_{2}T_{g}T_{g}^{*}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}\\ & = &T_{1}S_{1}T_{1}^{*}+T_{2}S_{2}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}. \end{eqnarray*} $

因为$ T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}\geq 0 $, 所以对于$ \forall g\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} \langle Sg, \;g\rangle& = &\langle (T_{1}S_{1}T_{1}^{*}+T_{2}S_{2}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*})g, \;g\rangle\\ & = &\langle T_{1}S_{1}T_{1}^{*}g, \;g\rangle+\langle T_{2}S_{2}T_{2}^{*}g, \;g\rangle +\langle (T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*})g, \;g\rangle\\ &\geq& \langle T_{1}S_{1}T_{1}^{*}g, \;g\rangle+\langle T_{2}S_{2}T_{2}^{*}g, \;g\rangle. \end{eqnarray*} $

若$ R(K)\subset R(T_{1}T_{f}) $, 则由引理2.4, 存在$ \beta>0 $使得

$ KK^{*}\leq \beta^{2}(T_{1}T_{f})(T_{1}T_{f})^{*} = \beta^{2}T_{1}T_{f}T_{f}^{*}T_{1}^{*} = \beta^{2}T_{1}S_{1}T_{1}^{*}, $

因此对于$ \forall g\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} \sum\limits_{j\in J}{|\langle g, \;{T_{1}f_j+T_{2}g_j}\rangle {|^2}}& = &\langle Sg, \;g\rangle\geq\langle T_{1}S_{1}T_{1}^{*}g, \;g\rangle+\langle T_{2}S_{2}T_{2}^{*}g, \;g\rangle\\ & = &\langle T_{1}S_{1}T_{1}^{*}g, \;g\rangle+\langle T_{2}T_{g}T_{g}^{*}T_{2}^{*}g, \;g\rangle\\ & = &\langle T_{1}S_{1}T_{1}^{*}g, \;g\rangle+\|T_{g}^{*}T_{2}^{*}g\|^{2}\\ &\geq&\langle T_{1}S_{1}T_{1}^{*}g, \;g\rangle\geq\beta^{-2}\|K^{*}g\|^{2}. \end{eqnarray*} $

综上所述对于$ \forall g\in {\cal H}_{2} $, 有

$ \beta^{-2}\|K^{*}g\|^{2}\leq\sum\limits_{j\in J} |\langle g, \;T_{1}{f_j}+T_{2}{g_j}\rangle{|^2}\leq(\sqrt{B_{1}}\|T_{1}\|+\sqrt{B_{2}}\|T_{2}\|)^{2}\|g\|^{2}. $

因此$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的$ K $-框架, 其框架算子为

$ S = T_{1}S_{1}T_{1}^{*}+T_{2}S_{2}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}. $

证毕.

现在我们给出例子说明若$ {\cal H}_{1} = {\cal H}_{2} = {\cal H} $, $ K, T_{1}, T_{2}\in B({\cal H}) $, 确实存在$ {\cal H} $的Bessel序列$ \{f_{j}\}_{j\in J} $和$ \{g_{j}\}_{j\in J} $, $ T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}>0 $, $ R(K)\subset R(T_{1}T_{f}) $, 使得结论成立.

例2.2  设$ \{e_{j}\}_{j = 1}^{3} $是$ {\cal H} = {\mathbb{C}}^{3} $的标准正交基.设$ T_{1}, T_{2}\in B({\cal H}) $和$ T_{1} = T_{2} = I_{{\cal H}} $.令$ f_{j} = g_{j} = e_{j}, \, j = 1, 2, 3 $.定义有界线性算子$ K\in B({\cal H}) $如下

$ K:{\cal H}\rightarrow {\cal H}, \, Kf = \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{3}\rangle e_{2}, \, \, \, f \in {\cal H}. $

显然$ \{f_{j}\}_{j = 1}^{3} $和$ \{g_{j}\}_{j = 1}^{3} $是$ {\cal H} $的Bessel序列, 设其合成算子分别为$ T_{f} $和$ T_{g} $, 则对于$ \forall f\in {\cal H} $, 有

$ T_{f}T_{g}^{*}f = T_{g}T_{f}^{*}f = T_{f}T_{f}^{*}f = \sum\limits_{j = 1}^{3}\langle f, \, e_{j}\rangle e_{j} = f. $

因此$ T_{f}T_{g}^{*} = T_{g}T_{f}^{*} = T_{f}T_{f}^{*} = I_{{\cal H}} $, 从而$ T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*} = 2I_{{\cal H}}>0 $.

由例2.1, 有$ K^{*}f = \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{2}\rangle e_{3}, \, f\in {\cal H} $, 则对于$ \forall f\in {\cal H} $, 有

$ \begin{eqnarray*} \langle KK^{*}f, \, f\rangle& = &\|K^{*}f\|^{2} = |\langle f, \, e_{1}\rangle|^{2}+|\langle f, \, e_{2}\rangle|^{2}\\ &\leq&\|f\|^{2} = \langle T_{1}T_{f}T_{f}^{*}T_{1}^{*}f, \, f\rangle = \langle (T_{1}T_{f})(T_{1}T_{f})^{*}f, \, f\rangle. \end{eqnarray*} $

因此$ KK^{*}\leq (T_{1}T_{f})(T_{1}T_{f})^{*} $, 则由引理2.4, 可得$ R(K)\subset R(T_{1}T_{f}) $.

注2.2  由定理2.2证明过程得, 当$ R(K)\subset R(T_{2}T_{g}) $时, 结论也成立.由定理2.2, 我们改进了文献[7]的定理4.1、定理4.6, 文献[8]的定理2.12, 文献[10]的定理3.3, 文献[13]的定理3.2和文献[14]的定理2.1.

定理2.3  设$ K_{1}\in B({\cal H}_{1}) $, $ K_{2}\in B({\cal H}_{2}) $和$ T_{1}, T_{2}\in B({\cal H}_{1}, {\cal H}_{2}) $.又设$ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的$ K_{1} $-框架, 其合成算子分别为$ T_{f} $和$ T_{g} $, 且$ T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}\geq 0 $.若$ T_{1}+T_{2} $具有闭值域, $ (T_{1}+T_{2})K_{1} = K_{2}(T_{1}+T_{2}) $且$ R(K_{2}^{*})\subset R(T_{1}+T_{2}) $, 则$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal {\cal H}}_{2} $的$ K_{2} $-框架.

证  设$ \{ {f_j}\} _{j\in J} $是$ {\cal H}_{1} $的$ K_{1} $-框架, 其界为$ A_{1} $和$ B_{1} $, 且$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的$ K_{1} $-框架, 其界为$ A_{2} $和$ B_{2} $.由引理2.5, 我们得到$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的Bessel序列, 其界为$ (\sqrt{B_{1}}\|T_{1}\|+\sqrt{B_{2}}\|T_{2}\|)^{2} $.

设$ S_{1} $和$ S_{2} $分别为$ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $的框架算子, 由$ T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}\geq 0 $, 根据定理2.2的证明过程可得, 对于$ \forall g\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} \sum\limits_{j\in J}{|\langle g, \;{T_{1}f_j+T_{2}g_j}\rangle {|^2}}&\geq&\langle T_{1}S_{1}T_{1}^{*}g, \;g\rangle+\langle T_{2}S_{2}T_{2}^{*}g, \;g\rangle\\ & = &\langle S_{1}T_{1}^{*}g, \;T_{1}^{*}g\rangle+\langle S_{2}T_{2}^{*}g, \;T_{2}^{*}g\rangle\\ & = &\sum\limits_{j\in J} |\langle T_{1}^{*}g, \;{f_j}\rangle{|^2}+\sum\limits_{j\in J}|\langle T_{2}^{*}g, \;{g_j}\rangle{|^2}\\ &\geq& A_{1}\|K_{1}^{*}T_{1}^{*}g\|^{2}+A_{2}\|K_{1}^{*}T_{2}^{*}g\|^{2}. \end{eqnarray*} $

若$ (T_{1}+T_{2})K_{1} = K_{2}(T_{1}+T_{2}) $, 则$ K_{1}^{*}(T_{1}+T_{2})^{*} = (T_{1}+T_{2})^{*}K_{2}^{*} $.取$ \lambda = \min\{A_{1}, A_{2}\} $, 由内积空间中的平行四边形公式和引理2.2, 得对于$ \forall g\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} A_{1}\|K_{1}^{*}T_{1}^{*}g\|^{2}+A_{2}\|K_{1}^{*}T_{2}^{*}g\|^{2}&\geq& \lambda(\|K_{1}^{*}T_{1}^{*}g\|^{2}+\|K_{1}^{*}T_{2}^{*}g\|^{2})\\ & = &\frac\lambda2(\|K_{1}^{*}(T_{1}+T_{2})^{*}g\|^{2}+\|K_{1}^{*}(T_{1}-T_{2})^{*}g\|^{2})\\ &\geq& \frac\lambda2 \|K_{1}^{*}(T_{1}+T_{2})^{*}g\|^{2}\\ & = &\frac\lambda2\|(T_{1}+T_{2})^{*}K_{2}^{*}g\|^{2}\\ &\geq& \frac\lambda2 \|(T_{1}+T_{2})^{+}\|^{-2}\|K_{2}^{*}g\|^{2}. \end{eqnarray*} $

综上所述对于$ \forall g\in {\cal H}_{2} $, 有

$ \frac\lambda2\|(T_{1}+T_{2})^{+}\|^{-2}\|K_{2}^{*}g\|^{2}\leq \sum\limits_{j\in J}{|\langle g, \;{T_{1}f_j+T_{2}g_j}\rangle {|^2}}\leq(\sqrt{B_{1}}\|T_{1}\|+\sqrt{B_{2}}\|T_{2}\|)^{2}\|g\|^{2}. $

因此$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal {\cal H}}_{2} $的$ K_{2} $-框架.证毕.

注2.3  由定理2.3证明过程, 当$ T_{1}-T_{2} $具有闭值域, $ (T_{1}-T_{2})K_{1} = K_{2}(T_{1}-T_{2}) $, $ R(K_{2}^{*})\subset R(T_{1}-T_{2}) $时, 结论也成立.由定理2.3, 我们可推出文献[3]的推论5.3.2和文献[10]的定理3.3, 改进文献[7]的定理4.1.比较这两个定理条件, 自然想到定理2.3是否是定理2.2的特殊情况?现在我们给出例子说明存在满足定理2.3条件但不满足定理2.2条件的情况.

例2.3  设$ \{e_{j}\}_{j = 1}^{3} $是$ {\cal H}_{1} = {\mathbb{C}}^{3} $的标准正交基, $ \{h_{j}\}_{j = 1}^{4} $是$ {\cal H}_{2} = {\mathbb{C}}^{4} $的标准正交基.设$ f_{j} = g_{j} = e_{j}, \, j = 1, 2, 3 $.定义有界线性算子$ K_{1}\in B({\cal H}_{1}) $, $ K_{2}\in B({\cal H}_{2}) $和$ T_{1}, T_{2}\in B({\cal H}_{1}, {\cal H}_{2}) $如下

$ K_{1}:{\cal H}_{1}\rightarrow {\cal H}_{1}, \, K_{1}f = \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{3}\rangle e_{2}, \, \, \, f \in {\cal H}_{1}, $

$ K_{2}:{\cal H}_{2}\rightarrow {\cal H}_{2}, \, K_{2}g = \langle g, \, h_{1}\rangle h_{1}+\langle g, \, h_{3}\rangle h_{2}, \, \, \, g \in {\cal H}_{2}, $

$ T_{1}:{\cal H}_{1}\rightarrow {\cal H}_{2}, \, T_{1}f = \langle f, \, e_{1}\rangle h_{1}, \, \, \, f \in {\cal H}_{1}, $

$ T_{2}:{\cal H}_{1}\rightarrow {\cal H}_{2}, \, T_{2}f = \langle f, \, e_{2}\rangle h_{2}+\langle f, \, e_{3}\rangle h_{3}, \, \, \, f \in {\cal H}_{1}. $

现在计算$ K_{1}^{*}f $, 对于$ \forall f, m\in {\cal H}_{1} $, 有

$ \begin{eqnarray*} \langle K_{1}^{*}f, \, m\rangle& = &\langle f, \, K_{1}m\rangle = \langle f, \langle m, \, e_{1}\rangle e_{1}+\langle m, \, e_{3}\rangle e_{2}\rangle = \langle f, \, e_{1}\rangle\overline{\langle m, \, e_{1}\rangle}+\langle f, \, e_{2}\rangle\overline{\langle m, \, e_{3}\rangle}\\ & = &\langle f, \, e_{1}\rangle\langle e_{1}, \, m\rangle+\langle f, \, e_{2}\rangle\langle e_{3}, \, m\rangle = \langle \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{2}\rangle e_{3}, \, m\rangle. \end{eqnarray*} $

即$ K_{1}^{*}f = \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{2}\rangle e_{3}, \, \, f\in {\cal H}_{1} $.从而对于$ \forall f\in {\cal H}_{1} $, 有

$ \begin{eqnarray*} \|K_{1}^{*}f\|^{2}& = &\|\langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{2}\rangle e_{3}\|^{2} = |\langle f, \, e_{1}\rangle|^{2}+|\langle f, e_{2}\rangle|^{2}\\ &\leq&\sum\limits_{j = 1}^{3}|\langle f, \, f_{j}\rangle|^{2} = \sum\limits_{j = 1}^{3}|\langle f, \, g_{j}\rangle|^{2} = \sum\limits_{j = 1}^{3}|\langle f, \, e_{j}\rangle|^{2} = \|f\|^{2}. \end{eqnarray*} $

因此$ \{f_{j}\}_{j = 1}^{3} $和$ \{g_{j}\}_{j = 1}^{3} $是$ {\cal H}_{1} $的$ K_{1} $-框架, 设其合成算子分别为$ T_{f} $和$ T_{g} $, 则对于$ \forall f\in {\cal H}_{1} $, 有

$ T_{f}T_{g}^{*}f = T_{g}T_{f}^{*}f = T_{f}T_{f}^{*}f = \sum\limits_{j = 1}^{3}\langle f, \, f_{j}\rangle f_{j} = \sum\limits_{j = 1}^{3}\langle f, \, e_{j}\rangle e_{j} = f. $

因此$ T_{f}T_{g}^{*} = T_{g}T_{f}^{*} = I_{{\cal H}_{1}} $.下面计算$ T_{1}^{*}g $, 对于$ \forall f\in {\cal H}_{1} $和$ \forall g\in {\cal H}_{2} $, 有

$ \langle T_{1}^{*}g, \, f\rangle = \langle g, \, T_{1}f\rangle = \langle g, \, \langle f, \, e_{1}\rangle h_{1}\rangle = \langle g, \, h_{1}\rangle\overline{\langle f, \, e_{1}\rangle} = \langle g, \, h_{1}\rangle\langle e_{1}, \, f\rangle = \langle\langle g, \, h_{1}\rangle e_{1}, \, f\rangle. $

即$ T_{1}^{*}g = \langle g, \, h_{1}\rangle e_{1}, \, \, g\in {\cal H}_{2} $.

同理可得$ T_{2}^{*}g = \langle g, \, h_{2}\rangle e_{2}+\langle g, \, h_{3}\rangle e_{3}, \, \, g\in {\cal H}_{2} $.又对于$ \forall g\in {\cal H}_{2} $, 有

$ \begin{eqnarray*} \langle(T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*})g, \, g\rangle& = &\langle T_{1}T_{f}T_{g}^{*}T_{2}^{*}g, \, g\rangle+\langle T_{2}T_{g}T_{f}^{*}T_{1}^{*}g, \, g\rangle\\ & = &\langle T_{1}T_{2}^{*}g, \, g\rangle+\langle T_{2}T_{1}^{*}g, \, g\rangle\\ & = &\langle T_{1}(\langle g, \, h_{2}\rangle e_{2}+\langle g, \, h_{3}\rangle e_{3}), \, g\rangle+\langle T_{2}(\langle g, \, h_{1}\rangle e_{1}), \, g\rangle\\ & = &\langle\langle g, \, h_{2}\rangle T_{1} e_{2}+\langle g, \, h_{3}\rangle T_{1} e_{3}, \, g\rangle+\langle\langle g, \, h_{1}\rangle T_{2}e_{1}, \, g\rangle \\& = &0. \end{eqnarray*} $

则$ T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*} = 0 $.通过简单计算得$ (T_{1}+T_{2})f = \langle f, e_{1}\rangle h_{1}+\langle f, e_{2}\rangle h_{2}+\langle f, e_{3}\rangle h_{3}, $$ f\in {\cal H}_{1} $, 因此$ T_{1}+T_{2} $具有闭值域.又对于$ \forall f\in {\cal H}_{1} $, 有

$ \begin{eqnarray*} (T_{1}+T_{2})K_{1}f& = &\sum\limits_{j = 1}^{3}\langle K_{1}f, \, e_{j}\rangle h_{j} = \sum\limits_{j = 1}^{3}\langle \langle f, \, e_{1}\rangle e_{1}+\langle f, \, e_{3}\rangle e_{2}, e_{j}\rangle h_{j}\\ & = &\langle f, \, e_{1}\rangle h_{1}+\langle f, \, e_{3}\rangle h_{2} = \sum\limits_{j = 1}^{3}\langle f, \, e_{j}\rangle K_{2}h_{j} = K_{2}(T_{1}+T_{2})f. \end{eqnarray*} $

则$ (T_{1}+T_{2})K_{1} = K_{2}(T_{1}+T_{2}) $, 且可得$ \overline{span}\{h_{1}, h_{3}\} = R(K_{2}^{*})\subset R(T_{1}+T_{2}) = \overline{span}\{h_{1}, h_{2}, h_{3}\} $.

由于$ T_{1}f_{1} = h_{1}, \, T_{1}f_{j} = 0, \, j = 2, 3 $, 显然可得$ \{T_{1}f_{j}\}_{j = 1}^{3} $是$ {\cal H}_{2} $的Bessel序列.取$ h_{2} = g\in {\cal H}_{2} $, 可得$ \|K_{2}^{*}g\|^{2} = 1 $和

$ \sum\limits_{j = 1}^{3}|\langle g, \, T_{1}f_{j}\rangle|^{2} = |\langle g, \, h_{1}\rangle|^{2} = 0. $

因此$ \{T_{1}f_{j}\}_{j = 1}^{3} $不是$ {\cal H}_{2} $的$ K_{2} $-框架, 从而由定理2.1的证明过程和引理2.3得$ R(K_{2})\nsubseteq R(T_{1}T_{f}) $.同理可得$ R(K_{2})\nsubseteq R(T_{2}T_{g}) $.

最后我们给出用两个Bessel构造紧$ K $-框架的充要条件.

定理2.4  设$ K\in B({\cal H}_{2}) $和$ T_{1}, T_{2}\in B({\cal H}_{1}, {\cal H}_{2}) $.又设$ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的Bessel序列, 其合成算子分别是$ T_{f} $和$ T_{g} $, 则$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的紧$ K $-框架当且仅当存在$ A>0 $使得

$ AKK^{*} = T_{1}T_{f}T_{f}^{*}T_{1}^{*}+T_{2}T_{g}T_{g}^{*}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}. $

证  设$ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的Bessel序列, 则根据定理2.2的证明过程可得$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的Bessel序列, 且其框架算子$ S $为

$ S = T_{1}T_{f}T_{f}^{*}T_{1}^{*}+T_{2}T_{g}T_{g}^{*}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}. $

根据引理2.6, 得$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的紧$ K $-框架当且仅当存在$ A>0 $使得

$ AKK^{*} = T_{1}T_{f}T_{f}^{*}T_{1}^{*}+T_{2}T_{g}T_{g}^{*}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}. $

证毕.

推论2.1  设$ K_{1}\in B({\cal H}_{1}) $, $ \{ {f_j}\} _{j\in J} $和$ \{ {g_j}\} _{j\in J} $是$ {\cal H}_{1} $的紧$ K_{1} $-框架, 其合成算子分别为$ T_{f} $和$ T_{g} $, 其界分别为$ A_{1} $和$ A_{2} $.又设$ K_{2}\in B({\cal H}_{2}) $和$ T_{1}, T_{2}\in B({\cal H}_{1}, {\cal H}_{2}) $, 则$ \{ T_{1}{f_j}+T_{2}{g_j}\} _{j\in J} $是$ {\cal H}_{2} $的紧$ K_{2} $-框架当且仅当存在$ A>0 $使得

$ AK_{2}K_{2}^{*} = A_{1}T_{1}K_{1}K_{1}^{*}T_{1}^{*}+A_{2}T_{2}K_{1}K_{1}^{*}T_{2}^{*}+T_{1}T_{f}T_{g}^{*}T_{2}^{*}+T_{2}T_{g}T_{f}^{*}T_{1}^{*}. $

注2.4  由引理2.6和定理2.4, 可得推论2.1.由定理2.4可推出文献[11]的定理2.1.由推论2.1, 可推出文献[11]的定理2.2, 定理2.3, 定理2.4和定理2.5, 改进文献[7]的定理4.5和文献[15]的定理3.1.