Hilbert空间中的框架概念是在1952年由Duffin和Schaeffer^[1]在研究非调和Fourier级数时首先提出的. 1986年Daubechies^[2]等突破性的研究引起学者对框架的极大关注和兴趣.到目前为止,框架已经广泛应用于图像处理^[3]、无线通讯^[4]、神经网络^[5]、量化测度^[6]等领域.因此,有关框架的研究成果十分丰富^[7-10].

随着对框架研究的不断深入,出现了许多推广形式,比如子空间框架或Fusion框架^[11]、有界拟投影^[12]和伪框架^[13]等. 2006年,孙文昌教授^[14]对现有几种推广的框架进行统一处理,引入了广义框架的概念(详见定义2.1),广义框架是将框架概念从有界线性泛函推广到了算子形式. Hilbert空间上不相交性概念^[15]是由Han和Larson在2000年首先提出的. 2014年, Guo^[8]将通常框架的不相交性推广到广义框架,引入了广义框架的不相交性、强不相交性和弱不相交性的概念.本文受Han及Guo在文献[15]和[8]中工作的启发,继续研究广义框架的不相交性.首先,我们给出了广义框架的不相交性、强不相交性和弱不相交性充分必要条件；利用这些充分必要条件,对于文献[5]中定理4.2,我们给出了新的证明方法；接着建立了对偶广义框架之间强不相交相和弱不相交性之间的关系；最后利用给定的强不相交的广义框架构造新的(超)对偶广义框架.

本部分主要回顾本文所用到的一些记号、概念和基本性质(详见文献[8, 14, 16-17]).采用如下记号：$ H, \, K, \, V $表示复可分Hilbert空间, $ I $表示整数集$ \mathbb{Z} $的子集, $ \{V_i\}_{i\in I} $表示$ V $的一列闭子空间, $ L(H, \, V_{i}) $表示由$ H $到$ V_{i} $的所有有界线性算子的集合,特别地,如果对于任意$ i\in I $,有$ V_{i} = H $,则记$ L(H, \, V_{i}) $为$ L(H) $, $ I_{H} $表示$ H $的恒等算子.

定义2.1^[14]  给定算子列$ \Lambda = \{\Lambda_{i}\in L(H, \, V_i):i\in I\} $,如果存在常数$ 0<A\leq B<\infty $,使得

(2.1) $ \begin{equation} A\|f\|^{2}\leq\sum\limits_{i\in I}\|\Lambda_{i}f\|^{2}\leq B\|f\|^{2}, \forall f\in H, \end{equation} $

则称$ \Lambda $为$ H $关于$ \{V_i:i\in I\} $的广义框架, $ A $和$ B $分别称为广义框架的下界和上界.如果(2.1)式仅对右半不等式成立,则称$ \Lambda $为$ H $关于$ \{V_i:i\in I\} $的广义Bessel序列;如果(2.1)式中$ A = B $,则称$ \Lambda $为$ H $关于$ \{V_i:i\in I\} $的紧广义框架;如果(2.1)式中$ A = B = 1 $,则称$ \Lambda $为$ H $关于$ \{V_i:i\in I\} $的Parseval广义框架.

对上述$ \{V_i:i\in I\} $,定义线性空间$ l^2(\{V_i\}_{i\in I}) $如下:

$ l^2(\{V_i\}_{i\in I}) = \left\{ \{g_{i}\}_{i\in I}: g_{i}\in V_{i}, \sum\limits_{i\in I}\|g_{i}\|^{2}<\infty\right\}, $

在其上定义内积

$ \langle \{f_{i}\}_{i\in I}, \, \{g_{i}\}_{i\in I}\rangle = \sum\limits_{i\in I}\langle f_{i}, \, g_{i}\rangle, $

则$ l^2(\{V_i\}_{i\in I}) $是复可分Hilbert空间.

对于广义Bessel序列$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $,称有界算子

$ T_{\Lambda}: l^2(\{V_i\}_{i\in I})\rightarrow H, T_{\Lambda}(\{f_{i}\}_{i\in I}) = \sum\limits_{i\in I} \Lambda_{i}^{\ast}f_{i}, $

为$ \{\Lambda_{i}\in L( H, \, V_{i}):i\in I\} $的合成算子.易证$ T_{\Lambda} $是线性有界算子,它的共轭算子

$ T_{\Lambda}^{\ast}: H\rightarrow l^2(\{V_i\}_{i\in I}), T_{\Lambda}^{\ast}f = \{\Lambda_{i}f\}_{i\in I}, $

称为$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $的分析算子.对于广义框架$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $称有界算子

$ S_{\Lambda}: H\rightarrow H, S_{\Lambda}f = \sum\limits_{i\in I} \Lambda_{i}^{\ast}\Lambda_{i}f $

为框架算子, $ S_{\Lambda} $是一个线性有界、自伴、正的可逆算子^[15].

定义2.2^[15]  设$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $是$ H $关于$ \{V_i:i\in I\} $的广义框架,如果存在$ H $关于$ \{V_i:i\in I\} $的广义框架$ \{\Gamma_{i}\in L(H, \, V_{i}):i\in I\} $,使得对任意的$ f\in H $,有

$ f = \sum\limits_{i\in I} \Gamma_{i}^{\ast}\Lambda_{i}f, $

则称$ \{\Gamma_{i}\in L(H, \, V_{i}):i\in I\} $为$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $的对偶广义框架.

易知$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $也为$ \{\Gamma_{i}\in L(H, \, V_{i}):i\in I\} $的对偶广义框架.它们相互对偶.如果$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $的框架下上界分别为$ A, \, B $,框架算子为$ S_{\Lambda} $,令$ \widetilde{\Lambda}_{i} = \Lambda_{i}S_{\Lambda}^{-1} $,则$ \{\widetilde{\Lambda}_{i}\}_{i\in I} $也是$ H $关于$ \{V_i:i\in I\} $的广义框架,框架下上界分别为$ \frac{1}{B}, \, \frac{1}{A} $,框架算子为$ S_{\Lambda}^{-1} $, $ \{\widetilde{\Lambda}_{i}\}_{i\in I} $称为$ \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $的典则对偶广义框架.

定义2.3^[8]  设$ \Lambda = \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $和$ \Gamma = \{\Gamma_{i}\in L(K, \, V_{i}):i\in I\} $分别是$ H $和$ K $关于$ \{V_i:i\in I\} $的广义框架.则$ \Lambda $和$ \Gamma $被称为

(1)不相交的,如果$ Range T_{\Lambda}^{\ast}\cap Range T_{\Gamma}^{\ast} = \{0\} $并且$ Range T_{\Lambda}^{\ast}+Range T_{\Gamma}^{\ast} $为$ l^2(\{V_i\}_{i\in I}) $的闭子空间;

(2)强不相交的,如果$ Range T_{\Lambda}^{\ast}\perp Range T_{\Gamma}^{\ast} $;

(3)弱不相交的,如果$ Range T_{\Lambda}^{\ast}\cap Range T_{\Gamma}^{\ast} = \{0\} $.

下面的定理利用超广义框架刻画了广义框架的不相交性、强不相交性及弱不相交性.

定理3.1  设$ \Lambda = \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $和$ \Gamma = \{\Gamma_{i}\in L(K, \, V_{i}):i\in I\} $分别是$ H $和$ K $关于$ \{V_i:i\in I\} $的广义框架.考虑$ \Delta = \{\Delta_{i} = \Lambda_{i}\oplus\Gamma_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $,其中

$ \Delta_{i}(h\oplus k) = \Lambda_{i}h+\Gamma_{i}k, i\in I, h\in H, k\in K, $

则$ \Lambda $与$ \Gamma $是

(i)强不相交的当且仅当存在可逆算子$ T_1\in L(H), \, T_2\in L(K) $,使得$ \{\Lambda_{i}T_1\in L(H, \, V_{i}):i\in I\} $, $ \{\Gamma_{i}T_2\in L(K, \, V_{i}):i\in I\} $和$ \Theta = \{\Theta_{i} = \Lambda_{i}T_1\oplus\Gamma_{i}T_2\in L(H\oplus K, \, V_{i}):i\in I\} $均是Parseval广义框架,其中

$ \Theta_{i}(h\oplus k) = \Lambda_{i}T_1h+\Gamma_{i}T_2k, i\in I, h\in H, k\in K. $

(ii)不相交的当且仅当$ \Delta $是一个广义框架.

(iii)弱不相交的当且仅当

$ \{h\oplus k: \Delta_{i}(h\oplus k) = 0: i\in I\} = \{0\}. $

证   (i)  设$ \Lambda $与$ \Gamma $是强不相交的.由定义可知$ Range T_{\Lambda}^{\ast}\perp Range T_{\Gamma}^{\ast} $.注意到$ \widetilde{\Lambda} = \{\Lambda_{i}S_{\Lambda}^{-\frac{1}{2}}\in L(H, \, V_{i}):i\in I\} $和$ \widetilde{\Gamma} = \{\Gamma_{i}S_{\Gamma}^{-\frac{1}{2}}\in L(K, \, V_{i}):i\in I\} $都是Parseval广义框架,并且

$ Range T_{\Lambda}^{\ast} = Range T_{\widetilde{\Lambda}}^{\ast}, Range T_{\Gamma}^{\ast} = Range T_{\widetilde{\Gamma}}^{\ast}. $

对任意的$ h\in H, k\in K $,有

$ \begin{eqnarray*} &&\sum\limits_{i\in I}\|\Lambda_{i}S_{\Lambda}^{-\frac{1}{2}}h+\Gamma_{i} S_{\Gamma}^{-\frac{1}{2}}k\|^{2}\\ & = & \sum\limits_{i\in I}\|\Lambda_{i}S_{\Lambda}^{-\frac{1}{2}}h\|^{2}+\sum\limits_{i\in I}\|\Gamma_{i}S_{\Gamma}^{-\frac{1}{2}}k\|^{2}+ 2{\rm Re}\sum\limits_{i\in I}\langle\Lambda_{i}S_{\Lambda}^{-\frac{1}{2}}h, \, \Gamma_{i}S_{\Gamma}^{-\frac{1}{2}}k\rangle\\ & = &\sum\limits_{i\in I}\|\Lambda_{i}S_{\Lambda}^{-\frac{1}{2}}h\|^{2}+\sum\limits_{i\in I}\|\Gamma_{i}S_{\Gamma}^{-\frac{1}{2}}k\|^{2}\\ & = & \|h\|^{2}+\|k\|^{2} = \|h\oplus k\|^{2}. \end{eqnarray*} $

取$ T_1 = S_{\Lambda}^{-\frac{1}{2}}, \, T_2 = S_{\Gamma}^{-\frac{1}{2}} $即可.

反之,对任意的$ h\in H, k\in K $,有

$ \begin{eqnarray*} \|h\oplus k\|^{2}& = &\sum\limits_{i\in I}\|\Theta_{i}(h\oplus k)\|^{2}\\& = & \sum\limits_{i\in I}\|\Lambda_{i}T_1h\|^{2}+\sum\limits_{i\in I}\|\Gamma_{i}T_2k\|^{2}+ 2{\rm Re}\sum\limits_{i\in I}\langle\Lambda_{i}T_1h, \, \Gamma_{i}T_2k\rangle\\& = & \|h\|^{2}+\|k\|^{2}+2{\rm Re}\sum\limits_{i\in I}\langle\Lambda_{i}T_1h, \, \Gamma_{i}T_2k\rangle. \end{eqnarray*} $

因此

$ {\rm Re}\sum\limits_{i\in I}\langle\Lambda_{i}T_1h, \, \Gamma_{i}T_2k\rangle = 0, h\in H, k\in K. $

再由极化恒等式,有

$ {\rm Im}\sum\limits_{i\in I}\langle\Lambda_{i}T_1h, \, \Gamma_{i}T_2k\rangle = {\rm Re}\sum\limits_{i\in I}\langle\Lambda_{i}T_1h, \, i\Gamma_{i}T_2k\rangle = 0, $

所以

$ \sum\limits_{i\in I}\langle\Lambda_{i}T_1h, \, \Gamma_{i}T_2k\rangle = 0, h\in H, k\in K. $

注意到$ T_1, \, T_2 $均是可逆算子,因此$ Range T_{\Lambda}^{\ast}\perp Range T_{\Gamma}^{\ast} $.

(ii)   设$ \Delta $是一个$ H\oplus K $关于$ \{V_i:i\in I\} $的广义框架,框架界为$ A_\Delta, \, B_\Delta $.对任意的$ h\in H, k\in K $,有

$ A_\Delta(\|h\|^{2}+\|k\|^{2})\leq\|T_{\Lambda}^{\ast}h +T_{\Gamma}^{\ast}k\|^2\leq B_\Delta(\|h\|^{2}+\|k\|^{2}). $

若存在$ h_1\in H, k_1\in K $满足$ T_{\Lambda}^{\ast}h_1 = -T_{\Gamma}^{\ast}k_1 $.利用上述的左半不等式可知, $ h_1 = k_1 = 0 $,因此$ T_{\Lambda}^{\ast}h_1 = T_{\Gamma}^{\ast}k_1 = 0 $.所以$ Range T_{\Lambda}^{\ast}\cap Range T_{\Gamma}^{\ast} = \{0\} $,又有$ Range T_{\Lambda}^{\ast}+ Range T_{\Gamma}^{\ast} = Range T_{\Delta}^{\ast} $是$ l^2(\{V_i\}_{i\in I}) $的闭子空间.

反之,考虑算子$ L $

$ L:Range T_{\Lambda}^{\ast}\oplus Range T_{\Gamma}^{\ast}\rightarrow Range T_{\Lambda}^{\ast}+ Range T_{\Gamma}^{\ast}, L(F\oplus G) = F+G. $

注意到

$ \begin{eqnarray*} \|L(F\oplus G)\|^{2} = \|F+G\|^{2}\leq(\|F\|+\|G\|)^{2} \leq 2(\|F\|^{2}+\|G\|^{2}) = 2\|F\oplus G)\|^{2}, \end{eqnarray*} $

则$ L $是一个有界双射.对任意的$ h\in H, k\in K $,有

$ \begin{eqnarray*} \|L^{-1}\|^{-2}\min\{A_{\Lambda}, \, A_{\Gamma}\}\|h\oplus k\|^{2}&\leq&\sum\limits_{i\in I}\|\Delta_{i}(h\oplus k)\|^{2} = \|L(T_{\Lambda}^{\ast}h\oplus T_{\Gamma}^{\ast}k)\|^{2}\\ &\leq&\|L\|^{2}\max\{B_{\Lambda}, \, B_{\Gamma}\}\|h\oplus k\|^{2}. \end{eqnarray*} $

(iii)   设$ \Lambda $与$ \Gamma $是弱不相交的, $ h\in H, k\in K $满足对所有的$ i\in I $有$ \Delta_{i}(h\oplus k) = 0 $.则

$ T_{\Lambda}^{\ast}h = T_{\Gamma}^{\ast}(-k)\in Range T_{\Lambda}^{\ast}\cap Range T_{\Gamma}^{\ast} = \{0\}, $

因此$ h = k = 0 $.反之,设$ \psi\in Range T_{\Lambda}^{\ast}\cap Range T_{\Gamma}^{\ast} $,则存在$ h\in H, k\in K $满足$ T_{\Lambda}^{\ast}h = T_{\Gamma}^{\ast}k = \psi $,因而

$ T_{\Delta}^{\ast}(h\oplus(-k)) = T_{\Lambda}^{\ast}h-T_{\Gamma}^{\ast}k = 0. $

这样可以得到$ h = k = 0 $,所以$ \psi = 0 $.证毕.

下面利用广义框架不相交性构造新的广义框架的定理是已知的,借助定理3.1给出了另外一种证明方法.首先给出一个引理.

引理3.2^[7]  设$ T: K\rightarrow H $是一个线性有界满射算子,存在一个线性有界算子$ T^{\dagger}: H\rightarrow K $,使得

$ TT^{\dagger}f = f, \forall f\in H, $

则称$ T^{\dagger} $为$ T $的伪逆.

定理3.3^{[8, Theorem 4.2]}  设$ \Lambda = \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $和$ \Gamma = \{\Gamma_{i}\in L(H, \, V_{i}):i\in I\} $是$ H $关于$ \{V_i:i\in I\} $不相交的广义框架, $ T_1, \, T_2\in L(H) $.如果$ T_1 $或者$ T_2 $是满射,则$ \{\Lambda_{i}T_{1}^{\ast}+\Gamma_{i}T_{2}^{\ast}\in L(H, \, V_{i}):i\in I\} $是$ H $关于$ \{V_i:i\in I\} $的广义框架.

证  不失一般性,假设$ T_1 $是满射.由引理3.2,存在$ T_{1}^{\dagger}\in L(H) $满足$ T_{1}T_{1}^{\dagger} = I_H $,则$ (T_{1}^{\dagger}) ^{\ast}T_{1}^{\ast} = I_H $.对任意的$ h\in H $,有

$ \|h\| = \|(T_{1}^{\dagger}) ^{\ast}T_{1}^{\ast}h\|\leq\|T_{1}^{\dagger} \|\|T_{1}^{\ast}h\|, $

可知

$ \|T_{1}^{\ast}h\|\geq\frac{\|h\|}{\|T_{1}^{\dagger} \|}. $

假设

$ (\Lambda_{i}\oplus\Gamma_{i})(h\oplus k) = \Lambda_{i}h+\Gamma_{i}k, i\in I, h\in H, k\in K. $

依据定理3.1(ii), $ \Delta = \{\Delta_{i} = \Lambda_{i}\oplus\Gamma_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $是$ H\oplus K $关于$ \{V_i:i\in I\} $的广义框架,因此,对任意的$ h\in H $,有

$ \begin{eqnarray*} \frac{A_{\Delta}}{\|T_{1}^{\dagger}\|^2}\|h\|^2&\leq& A_{\Delta}\|T_{1}^{\ast}h\|^2\leq A_{\Delta}(\|T_{1}^{\ast}h\|^2+\|T_{2}^{\ast}h\|^2)\\& = & A_{\Delta}\|T_{1}^{\ast}h\oplus T_{2}^{\ast}h\|^2\leq \sum\limits_{i\in I}\|\Delta_{i}(T_{1}^{\ast}h\oplus T_{2}^{\ast}h)\|^2\\ &\leq& B_{\Delta}\|T_{1}^{\ast}h\oplus T_{2}^{\ast}h\|^2 = B_{\Delta}(\|T_{1}^{\ast}h\|^2+\|T_{2}^{\ast}h\|^2)\\&\leq& 2B_{\Delta}\max\{\|T_{1}\|^2, \, \|T_{2}\|^2\}\|h\|^2, \end{eqnarray*} $

其中$ A_{\Delta}, \, B_{\Delta} $为$ \{\Delta_{i}:i\in I\} $的下界和上界.因此,对任意的$ h\in H $,有

$ \frac{A_{\Delta}}{\|T_{1}^{\dagger}\|^2}\|h\|^2 \leq\sum\limits_{i\in I}\|\Lambda_{i}T_{1}^{\ast}h+\Gamma_{i} T_{2}^{\ast}h)\|^2\leq2B_{\Delta}\max\{\|T_{1}\|^2, \, \|T_{2}\|^2\}\|h\|^2. $

类似可证$ T_2 $是满射的情形.证毕.

注3.4  由定理3.1易知,广义框架对是强不相交的一定是不相交的,因此在定理3.3中将条件改为两个广义框架是强不相交的,也能推出相同的结果(详见文献[8, Corollary 4.4]).

下面定理给出对偶广义框架之间强不相交相和弱不相交性之间的关系,为此首先给出一个引理.

引理3.5^[8]  设$ \Lambda = \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $和$ \Gamma = \{\Gamma_{i}\in L(K, \, V_{i}):i\in I\} $分别是$ H $和$ K $关于$ \{V_i:i\in I\} $的广义框架.则$ \Lambda $与$ \Gamma $是强不相交的当且仅当对任意的$ h\in H, k\in K $,有

$ \sum\limits_{i\in I}\Gamma_{i}^{\ast}\Lambda_{i}h = 0 \mbox {或} \sum\limits_{i\in I}\Lambda_{i}^{\ast}\Gamma_{i}k = 0. $

定理3.6  设$ \Lambda = \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $和$ \Gamma = \{\Gamma_{i}\in L(H, \, V_{i}):i\in I\} $是$ H $关于$ \{V_i:i\in I\} $的对偶广义框架, $ \Xi = \{\Xi_{i}\in L(K, \, V_{i}):i\in I\} $是$ K $关于$ \{V_i:i\in I\} $的广义框架.如果$ \Xi $和$ \Lambda $是强不相交的,则$ \Xi $和$ \Gamma $是弱不相交的.

证  设$ \Lambda $, $ \Gamma $和$ \Xi $的分析算子分别为$ T_{\Lambda}^{\ast} $, $ T_{\Gamma}^{\ast} $和$ T_{\Xi}^{\ast} $.由于$ \Lambda = \{\Lambda_{i}:i\in I\} $和$ \Gamma = \{\Gamma_{i}:i\in I\} $是$ H $关于$ \{V_i:i\in I\} $的对偶广义框架,对任意的$ h\in H $,有

$ h = \sum\limits_{i\in I} \Gamma_{i}^{\ast}\Lambda_{i}h \mbox{或} h = \sum\limits_{i\in I} \Lambda_{i}^{\ast}\Gamma_{i}h. $

即有$ T_{\Lambda}T_{\Gamma}^{\ast} = I_{U} $.又由$ \Xi $和$ \Lambda $是强不相交的,依据引理3.5,有$ T_{\Lambda}T_{\Xi}^{\ast} = 0 $.设$ \psi\in Range T_{\Gamma}^{\ast}\cap Range T_{\Xi}^{\ast} $,则存在$ h\in H, k\in K $满足$ \psi = T_{\Gamma}^{\ast}h = T_{\Xi}^{\ast}k $,这样可以得到

$ T_{\Lambda}\psi = T_{\Lambda}T_{\Gamma}^{\ast}h = T_{\Lambda}T_{\Xi}^{\ast}k. $

因此$ T_{\Lambda}\psi = h = 0 $,所以$ \psi = 0 $.即$ Range T_{\Gamma}^{\ast}\cap Range T_{\Xi}^{\ast} = \{0\} $, i.e. $ \Xi $和$ \Gamma $是弱不相交的.证毕.

下面定理提供了从给定的强不相交的广义框架构造新的对偶广义框架的方法.

定理3.7  设$ \{\Lambda_{i_{1}}\in L(H, \, V_{i}):i_{1}\in I\}, \, \cdot\cdot\cdot, \, \{\Lambda_{i_{k}}\in L(H, \, V_{i}):i_{k}\in I\} $都是$ H $关于$ \{V_i:i\in I\} $的广义框架,满足任意一对都是强不相交的广义框架.如果$ T_1, \, \cdot\cdot\cdot, \, T_k\in L(H) $,并且对某个$ j, \, 1\leq j\leq k $, $ T_j $是满射,则$ \{\widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}:i_{j}\in I\} $是$ \{\Lambda_{i_{1}}T_{1}^{\ast}+ \cdot\cdot\cdot+\Lambda_{i_{k}}T_{k}^{\ast}\} $和$ \{\Lambda_{i_{j}}T_{j}^{\ast} +\sum\limits_{m\neq j}\varepsilon_{m}\Lambda_{i_{m}}T_{m}^{\ast}\} $的一个对偶广义框架,其中$ \varepsilon_{m} = 0 $或者$ \varepsilon_{m} = \pm 1 $.

证  依据注解3.4,可以得到$ \{\Lambda_{i_{1}}T_{1}^{\ast}+ \cdot\cdot\cdot+\Lambda_{i_{k}}T_{k}^{\ast}\} $$ H $关于$ \{V_i:i\in I\} $的广义框架.又因为$ \{\Lambda_{i_{1}}\in L(H, \, V_{i}):i_{1}\in I\}, \, \cdot\cdot\cdot, \, \{\Lambda_{i_{k}}\in L(H, \, V_{i}):i_{k}\in I\} $中任意一对都是强不相交的广义框架.特别地,当$ k\neq j $时,广义框架$ \{\Lambda_{i_{k}}\in L(H, \, V_{i}):i_{k}\in I\} $和$ \{\Lambda_{i_{j}}\in L(H, \, V_{i}):i_{j}\in I\} $是强不相交的.由引理3.5可知,对任意的$ h\in H $,有

$ \sum\limits_{i_{j}, i_{k}\in I}\Lambda_{i_{j}}^{\ast}\Lambda_{i_{k}}h = 0 \mbox{或} \sum\limits_{i_{j}, i_{k}\in I}\Lambda_{i_{k}}^{\ast}\Lambda_{i_{j}}h = 0. $

注意到$ T_j $是满射,由引理3.2,存在有界算子$ T_{j}^{\dagger}\in L(H) $ (也即是$ T_j $的伪逆算子)满足$ T_{j}T_{j}^{\dagger} = I_H $.则$ (T_{j}^{\dagger}) ^{\ast}T_{j}^{\ast} = I_H $, $ (T_{j}^{\dagger}) ^{\ast} $也是满射.所以$ \{\widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}:i_{j}\in I\} $是$ H $关于$ \{V_i:i\in I\} $的广义框架.设$ \{\Lambda_{i_{j}}:i_{j}\in I\} $的框架算子为$ S $,对任意的$ h\in H $,有

$ \begin{eqnarray*} &&\sum\limits_{i_{1}, \cdot\cdot\cdot, i_{k}\in I}(\Lambda_{i_{1}}T_{1}^{\ast}+ \cdot\cdot\cdot+\Lambda_{i_{k}}T_{k}^{\ast})^{\ast} \widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}h\\& = &\sum\limits_{i_{1}, \cdot\cdot\cdot, i_{k}\in I}(T_{1}\Lambda_{i_{1}}^{\ast}+ \cdot\cdot\cdot+T_{k}\Lambda_{i_{k}}^{\ast}) \widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}h\\& = & \sum\limits_{i_{1}, i_{j}\in I}T_{1}\Lambda_{i_{1}}^{\ast}\widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}h +\cdot\cdot\cdot+\sum\limits_{i_{j}\in I}T_{j}\Lambda_{i_{j}}^{\ast}\widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}h +\cdot\cdot\cdot+\sum\limits_{i_{j}, i_{k}\in I}T_{k}\Lambda_{i_{k}}^{\ast}\widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}h\\& = & T_{1}\sum\limits_{i_{1}, i_{j}\in I}\Lambda_{i_{1}}^{\ast}\Lambda_{i_{j}}S^{-1}T_{j}^{\dagger}h +\cdot\cdot\cdot+T_{j}\sum\limits_{i_{j}\in I}\Lambda_{i_{j}}^{\ast}\Lambda_{i_{j}}S^{-1}T_{j}^{\dagger}h \\&& +\cdot\cdot\cdot+T_{k}\sum\limits_{i_{j}, i_{k}\in I}\Lambda_{i_{k}}^{\ast}\Lambda_{i_{j}}S^{-1}T_{j}^{\dagger}h\\& = & 0+\cdot\cdot\cdot+T_{j}\sum\limits_{i_{j}\in I}\Lambda_{i_{j}}^{\ast}\Lambda_{i_{j}}S^{-1}T_{j}^{\dagger}h +\cdot\cdot\cdot+0\\ & = &0+\cdot\cdot\cdot+T_{j}T_{j}^{\dagger}h+\cdot\cdot\cdot+0\\& = & 0+\cdot\cdot\cdot+h+\cdot\cdot\cdot+0 = h. \end{eqnarray*} $

因此, $ \{\widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}:i_{j}\in I\} $是$ \{\Lambda_{i_{1}}T_{1}^{\ast}+ \cdot\cdot\cdot+\Lambda_{i_{k}}T_{k}^{\ast}\} $的对偶广义框架.类似可以证明, $ \{\Lambda_{i_{j}}T_{j}^{\ast} +\sum\limits_{m\neq j}\varepsilon_{m}\Lambda_{i_{m}}T_{m}^{\ast}\} $是$ H $关于$ \{V_i:i\in I\} $的广义框架,并且对任意的$ h\in H $,有

$ \sum\limits_{i_{1}, \cdot\cdot\cdot, i_{k}\in I}(\Lambda_{i_{j}}T_{j}^{\ast} +\sum\limits_{m\neq j}\varepsilon_{m}\Lambda_{i_{m}}T_{m}^{\ast})^{\ast} \widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}h = h, $

其中$ \varepsilon_{m} = 0 $或者$ \varepsilon_{m} = \pm 1 $.所以, $ \{\widetilde{\Lambda}_{i_{j}}T_{j}^{\dagger}:i_{j}\in I\} $是$ \{\Lambda_{i_{j}}T_{j}^{\ast} +\sum\limits_{m\neq j}\varepsilon_{m}\Lambda_{i_{m}}T_{m}^{\ast}\} $的对偶广义框架,其中$ \varepsilon_{m} = 0 $或者$ \varepsilon_{m} = \pm 1 $.证毕.

注3.8  在定理3.7中,如果$ k = 2, \, \varepsilon_{m} = 0 $或者$ \varepsilon_{m} = 1 $可以恢复到文献[8]中的定理4.6.

定理3.9  设$ \Lambda = \{\Lambda_{i}\in L(H, \, V_{i}):i\in I\} $和$ \Gamma = \{\Gamma_{i}\in L(H, \, V_{i}):i\in I\} $都是$ H $关于$ \{V_i:i\in I\} $的广义Bessel序列, $ \Phi = \{\Phi_{i}\in L(H, \, V_{i}):i\in I\} $和$ \Psi = \{\Psi_{i}\in L(H, \, V_{i}):i\in I\} $都是$ K $关于$ \{V_i:i\in I\} $的广义Bessel序列,则$ \Theta = \{\Theta_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $和$ \Delta = \{\Delta_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $是$ H\oplus K $关于$ \{V_i:i\in I\} $的对偶广义框架,其中

$ \Theta_{i}(h\oplus k) = \Lambda_{i}h+\Phi_{i}k, \Delta_{i}(h\oplus k) = \Gamma_{i}h+\Psi_{i}k, i\in I, h\in H, k\in K, $

当且仅当$ \Lambda, \, \Gamma $和$ \Phi, \, \Psi $分别是$ H $和$ K $关于$ \{V_i:i\in I\} $的偶广义框架,并且$ \Lambda, \, \Psi $和$ \Gamma, \, \Phi $都是强不相交的广义框架对.

证   $ (\Rightarrow) $设$ \Theta = \{\Theta_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $和$ \Delta = \{\Delta_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $是$ H\oplus K $关于$ \{V_i:i\in I\} $的对偶广义框架,对任意的$ h\in H $,有

$ \begin{eqnarray*} h\oplus 0 = \sum\limits_{i\in I}\Delta_{i}^{\ast}\Theta_{i}(h\oplus 0) = \sum\limits_{i\in I}\Delta_{i}^{\ast}\Lambda_{i}h = \sum\limits_{i\in I}\Gamma_{i}^{\ast}\Lambda_{i}h\oplus\sum\limits_{i\in I}\Psi_{i}^{\ast}\Lambda_{i}h. \end{eqnarray*} $

因此有

$ \sum\limits_{i\in I}\Gamma_{i}^{\ast}\Lambda_{i}h = h, \, \sum\limits_{i\in I}\Psi_{i}^{\ast}\Lambda_{i}h = 0, $

也即是$ \Lambda $和$ \Gamma $是$ H $关于$ \{V_i:i\in I\} $的对偶广义框架,且$ \Lambda $和$ \Psi $是强不相交的广义框架对.类似地,对任意的$ k\in K $,有

$ \begin{eqnarray*} 0\oplus k = \sum\limits_{i\in I}\Delta_{i}^{\ast}\Theta_{i}(0\oplus k) = \sum\limits_{i\in I}\Gamma_{i}^{\ast}\Phi_{i}h\oplus\sum\limits_{i\in I}\Psi_{i}^{\ast}\Phi_{i}h. \end{eqnarray*} $

由此可知, $ \Phi $和$ \Psi $是$ K $关于$ \{V_i:i\in I\} $的对偶广义框架,且$ \Gamma $和$ \Phi $是强不相交的广义框架对.

$ (\Leftarrow) $设$ \Lambda, \, \Gamma $和$ \Phi, \, \Psi $分别是$ H $和$ K $关于$ \{V_i:i\in I\} $的对偶广义框架,并且$ \Lambda, \, \Psi $和$ \Gamma, \, \Phi $都是强不相交的广义框架对.对任意的$ h\in H, k\in K $,有

$ \begin{eqnarray*} \sum\limits_{i\in I}\Delta_{i}^{\ast}\Theta_{i}(h\oplus k)& = & \sum\limits_{i\in I}\Delta_{i}^{\ast}(\Lambda_{i}h+\Phi_{i}k)\\& = & \sum\limits_{i\in I}\Gamma_{i}^{\ast}(\Lambda_{i}h+\Phi_{i}k)\oplus\sum\limits_{i\in I}\Psi_{i}^{\ast}(\Lambda_{i}h+\Phi_{i}k)\\ & = &\sum\limits_{i\in I}\Gamma_{i}^{\ast}\Lambda_{i}h\oplus\sum\limits_{i\in I}\Psi_{i}^{\ast}\Phi_{i}k = h\oplus k. \end{eqnarray*} $

所以$ \Theta = \{\Theta_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $和$ \Delta = \{\Delta_{i}\in L(H\oplus K, \, V_{i}):i\in I\} $是$ H\oplus K $关于$ \{V_i:i\in I\} $的对偶广义框架.证毕.