Some Parameterized Inequalities for g-Frames
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摘要:
运用算子理论方法得到了Hilbert空间中g-框架的一些带参数的不等式.在引入参数情况下,建立了g-框架及其正则对偶的一个不等式;得到了g-框架及其一般对偶的反向不等式.所得结果推广了现有文献的结果.
Abstract:The notion of g-frame is a generalization of that of frame. It has interested many mathematicians in recent years. In this paper, using the operator theory method, we obtained some parameterized inequalities for g-frames in Hilbert spaces. In the case of introducing parameters, we established an inequality for a g-frame and its canonical dual. And we obtained a converse inequality for a g-frame and its alternate duals. These obtained results generalize the existing ones.
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Keywords:
- g-frame /
- canonical dual g-frame /
- dual g-frame
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框架是标准正交基概念的推广, 它是Duffin等[1]在研究非调和Fourier级数时提出的, 小波分析的产生给框架理论的研究注入了新的活力.目前, 框架理论已被广泛应用于滤波器理论、量子力学、信号处理等领域[2-4].
关于框架的概念有许多推广, Casazza等于2004年在研究重构框架系统时基于对框架整体性与局部性的诠释引入了Fusion框架的概念.之后, Fusion框架的研究引起了不少学者的关注, 在理论和实际应用中取得了许多成果, 详见文献[5-8]. 2006年,Sun[9]在一般Hilbert空间中引入了g-框架的概念, 关于g-框架的研究见文献[10-11]及其参考文献. “没有相位信息, 也能重构信号”是工程领域多年来一个悬而未决的猜想. 2006年, Balan等[12]通过刻画新的Parseval框架类证明了这一猜想的正确性.笔者在处理信号的重构算法时, 发现了一个令人惊奇的结果:
设{fj}j∈J是Hilbert空间U中的Parseval框架.则对任意的K⊂J与f∈U, 有
$$ \begin{array}{*{20}{c}} {\sum\limits_{j \in K} {{{\left| {\left\langle {f,{f_j}} \right\rangle } \right|}^2}} + {{\left\| {\sum\limits_{j \in {K^c}} {\left\langle {f,{f_j}} \right\rangle {f_j}} } \right\|}^2} = }\\ {\sum\limits_{j \in {K^c}} {{{\left| {\left\langle {f,{f_j}} \right\rangle } \right|}^2}} + {{\left\| {\sum\limits_{j \in K} {\left\langle {f,{f_j}} \right\rangle {f_j}} } \right\|}^2}} \end{array} $$ 随后关于框架等式与不等式的研究引起了许多学者的关注, 详见文献[13-17]及其参考文献.令{(Vj, wj):j∈J}是Hilbert空间V中的fusion框架.本文在此基础上运用算子理论得到了Hilbert空间中g-框架及其对偶框架的一些新的不等式.首先运用算子理论将文献[18]定理3.1关于一般Hilbert空间中g-Parseval框架的不等式进行推广, 在引入参数情况下, 建立了g-框架及其正则对偶框架的不等式(见定理3), 特别地, 当wj=1(j∈J)时可得文献[19]定理2.3.其次, 运用算子理论得到了关于文献[20]定理4.1的反向不等式估计, 补充并完善了关于g-框架及其对偶框架的已有不等式(见定理4),推广了文献[19]中定理2.8关于fusion框架及其对偶框架wj=1(j∈J)时不等式.
1. 预备知识
设U、V是2个可分的Hilbert空间, 其内积记作〈·,·〉, 范数记作‖·‖. {Vj:j∈J}是V的一列闭子空间, 记πVj为V到Vj的正交投影, 其中J是整数集$\mathbb{Z}$的一个子集.记L(U, Vj)(L(U))为从U(U)到Vj(U)所有的有界线性算子组成的集合, 记IU为U中的恒等算子, 对任意的T∈L(U, Vj), 记T*为其伴随算子.
定义1[6] 设{wj:j∈J}是一列权重集, 即对所有j∈J, wj>0, 称{(Vj, wj):j∈J}是V关于{wj:j∈J}的fusion框架, 如果存在2个正常数A和B, 对任意的f∈V, 有
$$ A{\left\| f \right\|^2} \le \sum\limits_{j \in J} {w_j^2{{\left\| {{{\rm{ \mathsf{ π} }}_{{V_j}}}f} \right\|}^2}} \le B{\left\| f \right\|^2} $$ 若A=B, 则称{(Vj, wj):j∈J}是V关于{wj:j∈J}的A-紧fusion框架.若A=B=1, 则称{(Vj, wj):j∈J}是V关于{wj:j∈J}的Parseval fusion框架.
定义2[9] 一个序列{Λj∈L(U, Vj):j∈J}称为U关于{Vj:j∈J}的g-框架, 如果存在2个正常数A和B, 对任意的f∈U, 有
$$ A{\left\| f \right\|^2} \le \sum\limits_{j \in J} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} \le B{\left\| f \right\|^2} $$ (1) 式中A、B分别为g-框架的下、上界.若式(1)只有右半不等式成立, 则称序列{Λj:j∈J}为U关于{Vj:j∈J}的g-Bessel序列.若A=B, 则称{Λj:j∈J}为U关于{Vj:j∈J}的A-紧g-框架.若A=B=1, 则称{Λj:j∈J}为U关于{Vj:j∈J}的g-Parseval框架.
定义3[20] 设{Λj:j∈J}和{Γj:j∈J}是U关于{Vj:j∈J}的2个g-框架, 称{Γj:j∈J}为{Λj:j∈J}的对偶g-框架, 如果对任意的f∈U,则
$$ f = \sum\limits_{j \in J} {\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} $$ 成立.若{Λj:j∈J}是U关于{Vj:j∈J}的界为A和B的g-框架,则相应的g-框架算子S:U→U如下:
$$ Sf = \sum\limits_{j \in J} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} ,\forall f \in U $$ 容易证明S是一个正的、自伴的、可逆的有界线性算子, 且满足
$$ \begin{array}{*{20}{c}} {\left\langle {Sf,f} \right\rangle = \left\langle {\sum\limits_{j \in J} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} ,f} \right\rangle = \sum\limits_{j \in J} {\left\langle {{\mathit{\Lambda }_j}f,{\mathit{\Lambda }_j}f} \right\rangle } = }\\ {\sum\limits_{j \in J} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} } \end{array} $$ 因此, A‖f‖2≤〈Sf, f〉≤B‖f‖2, 且重构式子成立.记${\tilde {\mathit{\Lambda}} _j} = {\mathit{\Lambda} _j}{S^{-1}}$, 则${\tilde {\mathit{\Lambda}} _j}$是U关于{Λj:j∈J}的界为B-1和A-1的g-框架, 并称$\left\{ {{{\tilde {\mathit{\Lambda}} }_j}:j \in J} \right\}$是{Λj:j∈J}的正则对偶g-框架.
$$ \begin{array}{l} \forall f \in U\\ \begin{array}{*{20}{c}} {f = S{S^{ - 1}}f = {S^{ - 1}}Sf = \sum\limits_{j \in J} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}{S^{ - 1}}f} = }\\ {\sum\limits_{j \in J} {{S^{ - 1}}\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \end{array} \end{array} $$ 设{Λj∈L(U, Vj):j∈J}是U关于{Vj:j∈J}的g-Bessel序列, K⊂J, Kc=J\K.分别定义2个有界线性算子SK, SKc:U→U如下:
$$ \begin{array}{*{20}{c}} {\forall f \in U}\\ {{S_K}f = \sum\limits_{j \in K} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} ,{S_{{K^c}}}f = \sum\limits_{j \in {K^c}} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \end{array} $$ 下面给出Hilbert空间中正则对偶g-框架的一些结论.
定理1[21] 设{Λj:j∈J}是U关于{Vj:j∈J}的g-框架, 则对任意的K⊂J和f∈U, 有
$$ \begin{array}{*{20}{c}} {\sum\limits_{j \in K} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_{{K^c}}}f} \right)} \right\|}^2}} = }\\ {\sum\limits_{j \in {K^c}} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_K}f} \right)} \right\|}^2}} } \end{array} $$ 定理2[18] 设F={Λj∈L(U, Vj):j∈J}是U关于{Vj:j∈J}的g-Parseval框架, 对任意的K⊂J, 令
$$ {v_ - }\left( {F,K} \right) = \mathop {\sup }\limits_{f \ne 0} \frac{{\sum\limits_{j \in {K^c}} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + {{\left\| {\sum\limits_{j \in K} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2}}}{{{{\left\| f \right\|}^2}}} $$ $$ {v_ + }\left( {F,K} \right) = \mathop {\sup }\limits_{f \ne 0} \frac{{\sum\limits_{j \in {K^c}} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + {{\left\| {\sum\limits_{j \in K} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2}}}{{{{\left\| f \right\|}^2}}} $$ 则
$$ \frac{3}{4} \le {v_ - }\left( {F,K} \right) \le {v_ + }\left( {F,K} \right) \le 1 $$ 成立.
注意到, 若{Λj:j∈J}是一个A-紧g-框架, 则$\left\{ {\frac{1}{{\sqrt A }}{\mathit{\Lambda} _j}:j \in J} \right\}$是g-Parseval框架.由定理2,可得下述推论, 它是文献[21]定理3.1的一个推广.
推论1 设{Λj:j∈J}是U关于{Vj:j∈J}的A-紧g-框架, 则对任意的K⊂J和f∈U, 有
$$ \begin{array}{*{20}{c}} {\frac{{3{A^2}}}{4}{{\left\| f \right\|}^2} \le A\sum\limits_{j \in K} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + }\\ {{{\left\| {\sum\limits_{j \in {K^c}} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} \le {A^2}{{\left\| f \right\|}^2}} \end{array} $$ 定理3[22] 设{Λj:j∈J}是U关于{Vj:j∈J}的g-框架, {Γj:j∈J}是其对偶g-框架, K⊂J.则对任意的f∈U, 有
$$ \begin{array}{*{20}{c}} {{{\left\| {\sum\limits_{j \in K} {\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} + {\mathop{\rm Re}\nolimits} \left( {\sum\limits_{j \in {K^c}} {\left\langle {{\mathit{\Lambda }_j}f,{\mathit{\Gamma }_j}f} \right\rangle } } \right) = }\\ {{{\left\| {\sum\limits_{j \in {K^c}} {\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} + {\mathop{\rm Re}\nolimits} \left( {\sum\limits_{j \in K} {\left\langle {{\mathit{\Lambda }_j}f,{\mathit{\Gamma }_j}f} \right\rangle } } \right) \ge \frac{3}{4}{{\left\| f \right\|}^2}} \end{array} $$ 2. 主要结果及其证明
首先给出一些引理.
引理1[19] 设P、Q∈L(U)是2个自伴算子, 且P+Q=IU, 则对任意的λ∈[0, 1]和f∈U, 有
$$ \begin{array}{*{20}{c}} {{{\left\| {Pf} \right\|}^2} + 2\lambda \left\langle {Qf,f} \right\rangle = }\\ {{{\left\| {Qf} \right\|}^2} + 2\left( {1 - \lambda } \right)\left\langle {Pf,f} \right\rangle + \left( {2\lambda - 1} \right){{\left\| f \right\|}^2} \ge }\\ {\left( {2\lambda - {\lambda ^2}} \right){{\left\| f \right\|}^2}} \end{array} $$ 类似于文献[18]定理2.2的证明, 有下面引理.为方便读者, 本文给出证明.
引理2 设{Λj:j∈J}是U关于{Vj:j∈J}的A-紧g-框架, S为{Λj:j∈J}的g-框架算子且$\left\{ {{{\tilde {\mathit{\Lambda}} }_j}:j \in J} \right\}$为{Λj:j∈J}的正则对偶g-框架, 则对任意的K⊂J和f∈U, 有
$$ \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_K}f} \right)} \right\|}^2}} = \frac{1}{A}{\left\| {\sum\limits_{j \in K} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|^2} $$ $$ \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_{{K^c}}}f} \right)} \right\|}^2}} = \frac{1}{A}{\left\| {\sum\limits_{j \in {K^c}} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|^2} $$ 证明:由于{Λj:j∈J}是U关于{Vj:j∈J}的A-紧g-框架, 又${\tilde {\mathit{\Lambda}} _j} = {\Lambda _j}{S^{-1}}$, 则对任意的f∈U, 有$\sum\limits_{j \in J} {{{\left\| {{{\tilde {\mathit{\Lambda}} }_j}\left( {{S_K}f} \right)} \right\|}^2}} = \frac{1}{A}{\left\| {{S_K}f} \right\|^2}$, 再将SKf的定义代入, 因此有
$$ \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_K}f} \right)} \right\|}^2}} = \frac{1}{A}{\left\| {\sum\limits_{j \in K} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|^2} $$ 同理可得
$$ \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_{{K^c}}}f} \right)} \right\|}^2}} = \frac{1}{A}{\left\| {\sum\limits_{j \in {K^c}} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|^2} $$ 定理4 设{Λj:j∈J}是U关于{Vj:j∈J}的g-框架, S为{Λj:j∈J}的g-框架算子且$\left\{ {{{\tilde {\mathit{\Lambda}} }_j}:j \in J} \right\}$为{Λj:j∈J}的正则对偶g-框架, 则对任意的λ∈[0, 1], K⊂J和f∈U, 有
$$ \begin{array}{*{20}{c}} {\left\langle {Sf,f} \right\rangle \ge \sum\limits_{j \in K} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_{{K^c}}}f} \right)} \right\|}^2}} = }\\ {\sum\limits_{j \in {K^c}} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_K}f} \right)} \right\|}^2}} \ge }\\ {\left( {2\lambda - {\lambda ^2}} \right)\left\langle {{S_K}f,f} \right\rangle + \left( {1 - {\lambda ^2}} \right)\left\langle {{S_{{K^c}}}f,f} \right\rangle } \end{array} $$ (2) 证明:取$P = {S^{-\frac{1}{2}}}{S_K}{S^{-\frac{1}{2}}}, Q = {S^{-\frac{1}{2}}}{S_{{K^c}}}{S^{ - \frac{1}{2}}}$, 则P、Q满足引理1的条件.由于SK+SKc=S, 则
$$ {S^{ - \frac{1}{2}}}{S_K}{S^{ - \frac{1}{2}}} + {S^{ - \frac{1}{2}}}{S_{{K^c}}}{S^{ - \frac{1}{2}}} = {I_U} $$ 以${S^{\frac{1}{2}}}f$代替引理1中的f, 首先证明式(2)的左半不等式.由于
$$ {S_K}{S^{ - 1}}{S_{{K^c}}} = \left( {S - {S_{{K^c}}}} \right){S^{ - 1}}\left( {S - {S_K}} \right) = {S_{{K^c}}}{S^{ - 1}}{S_K} $$ 由此可得PQ=QP.于是,
$$ \begin{array}{*{20}{c}} {0 \le PQ = \left( {{I_U} - Q} \right)Q = Q - {Q^2} = }\\ {{S^{ - \frac{1}{2}}}\left( {{S_{{K^c}}} - {S_{{K^c}}}{S^{ - 1}}{S_{{K^c}}}} \right){S^{ - \frac{1}{2}}}} \end{array} $$ 从而, SKc-SKcS-1SKc≥0.因此,
$$ \begin{array}{*{20}{c}} {\sum\limits_{j \in K} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_{{K^c}}}f} \right)} \right\|}^2}} = }\\ {\left\langle {{S_K}f,f} \right\rangle + \left\langle {{S_{{K^c}}}f,{S^{ - 1}}{S_{{K^c}}}f} \right\rangle = }\\ {\left\langle {{S_K}f,f} \right\rangle + \left\langle {{S_{{K^c}}}{S^{ - 1}}{S_{{K^c}}}f,f} \right\rangle \le }\\ {\left\langle {{S_K}f,f} \right\rangle + \left\langle {{S_{{K^c}}}f,f} \right\rangle = \left\langle {Sf,f} \right\rangle } \end{array} $$ 因此, 式(2)的左半不等式成立.下证式(2)的右半不等式.
$$ \begin{array}{*{20}{c}} {{{\left\| {{S^{ - \frac{1}{2}}}{S_K}f} \right\|}^2} = }\\ {{{\left\| {{S^{ - \frac{1}{2}}}{S_{{K^c}}}f} \right\|}^2} + 2\left( {1 - \lambda } \right)\left\langle {{S^{ - \frac{1}{2}}}{S_K}f,{S^{\frac{1}{2}}}f} \right\rangle + }\\ {\left( {2\lambda - 1} \right){{\left\| {{S^{\frac{1}{2}}}f} \right\|}^2} - 2\lambda \left\langle {{S^{ - \frac{1}{2}}}{S_{{K^c}}}f,{S^{\frac{1}{2}}}f} \right\rangle \ge }\\ {\left( {2\lambda - {\lambda ^2}} \right){{\left\| {{S^{\frac{1}{2}}}f} \right\|}^2} - 2\lambda \left\langle {{S^{ - \frac{1}{2}}}{S_{{K^c}}}f,{S^{\frac{1}{2}}}f} \right\rangle = }\\ {2\lambda \left( {\left\langle {Sf,f} \right\rangle - \left\langle {{S_{{K^c}}}f,f} \right\rangle } \right) - {\lambda ^2}\left\langle {Sf,f} \right\rangle = }\\ {2\lambda \left\langle {{S_K}f,f} \right\rangle - {\lambda ^2}\left\langle {Sf,f} \right\rangle } \end{array} $$ 将上式代入定理1, 便得
$$ \begin{array}{*{20}{c}} {\sum\limits_{j \in K} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_{{K^c}}}f} \right)} \right\|}^2}} = }\\ {\sum\limits_{j \in {K^c}} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + \sum\limits_{j \in J} {{{\left\| {{{\mathit{\tilde \Lambda }}_j}\left( {{S_K}f} \right)} \right\|}^2}} = }\\ {\left\langle {{S_{{K^c}}}f,f} \right\rangle + \left\langle {S{S^{ - 1}}{S_K}f,{S^{ - 1}}{S_K}f} \right\rangle = }\\ {\left\langle {{S_{{K^c}}}f,f} \right\rangle + {{\left\| {{S^{ - \frac{1}{2}}}{S_K}f} \right\|}^2} \ge }\\ {2\lambda \left\langle {{S_K}f,f} \right\rangle - {\lambda ^2}\left\langle {Sf,f} \right\rangle + \left\langle {{S_{{K^c}}}f,f} \right\rangle = }\\ {\left( {2\lambda - {\lambda ^2}} \right)\left\langle {{S_K}f,f} \right\rangle + \left( {1 - {\lambda ^2}} \right)\left\langle {{S_{{K^c}}}f,f} \right\rangle } \end{array} $$ 故式(2)的右半不等式成立.证毕.
注1 取U=V, Λj=πVj, 则由定理4可得文献[19]的定理2.3关于fusion框架不等式wj=1(j∈J)时的结果.
注意到, 若{Λj:j∈J}是一个A-紧g-框架, 作为定理4和引理2的直接推论, 有:
推论2 设{Λj:j∈J}是U关于{Vj:j∈J}的A-紧g-框架, 则对任意的K⊂J和f∈U, 有
$$ \begin{array}{*{20}{c}} {{A^2}{{\left\| f \right\|}^2} \ge A\sum\limits_{j \in K} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + {{\left\| {\sum\limits_{j \in {K^c}} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} = }\\ {A\sum\limits_{j \in {K^c}} {{{\left\| {{\mathit{\Lambda }_j}f} \right\|}^2}} + {{\left\| {\sum\limits_{j \in K} {\mathit{\Lambda }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} \ge }\\ {A\left( {2\lambda - {\lambda ^2}} \right)\left\langle {{S_K}f,f} \right\rangle + A\left( {1 - {\lambda ^2}} \right)\left\langle {{S_{{K^c}}}f,f} \right\rangle } \end{array} $$ 注2 若令推论2中的λ=1/2, 则可得推论1.若又有A=1, 则可得定理2.
定理5 设{Λj:j∈J}是U关于{Vj:j∈J}的g-框架, {Γj:j∈J}是其对偶g-框架, {αj:j∈J}∈l∞(J).定义算子
$$ {T_\alpha }:U \to U,f \to \sum\limits_{j \in J} {{\alpha _j}\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} $$ $$ {T_{1 - \alpha }}:U \to U,f \to \sum\limits_{j \in J} {\left( {1 - {\alpha _j}} \right)\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} $$ 则对任意的f∈U, 有
$$ \begin{array}{*{20}{c}} {\frac{3}{4}{{\left\| f \right\|}^2} \le {{\left\| {\sum\limits_{j \in J} {{\alpha _j}\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} + }\\ {{\mathop{\rm Re}\nolimits} \left( {\sum\limits_{j \in J} {\left( {1 - {\alpha _j}} \right)\left\langle {{\mathit{\Lambda }_j}f,{\mathit{\Gamma }_j}f} \right\rangle } } \right) = }\\ {{{\left\| {\sum\limits_{j \in J} {\left( {1 - {\alpha _j}} \right)\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} + {\mathop{\rm Re}\nolimits} \left( {\sum\limits_{j \in J} {{\alpha _j}\left\langle {{\mathit{\Lambda }_j}f,{\mathit{\Gamma }_j}f} \right\rangle } } \right) \le }\\ {\frac{{3 + {{\left\| {{T_\alpha } - {T_{1 - \alpha }}} \right\|}^2}}}{4}{{\left\| f \right\|}^2}} \end{array} $$ (3) 证明:首先由文献[20]定理4.1, 式(3)的左半不等式已成立.其次证明式(3)的右半不等式.由Tα和T1-α的定义, 易证Tα+T1-α=I, 则对任意的f∈U, 有
$$ \begin{array}{*{20}{c}} {{{\left\| {\sum\limits_{j \in J} {\left( {1 - {\alpha _j}} \right)\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} + {\mathop{\rm Re}\nolimits} \left( {\sum\limits_{j \in J} {{\alpha _j}\left\langle {{\mathit{\Lambda }_j}f,{\mathit{\Gamma }_j}f} \right\rangle } } \right) = }\\ {\left\langle {{T_{1 - \alpha }}f,{T_{1 - \alpha }}f} \right\rangle + {\mathop{\rm Re}\nolimits} \left\langle {{T_\alpha }f,f} \right\rangle = }\\ {\left\langle {{T_{1 - \alpha }}f,{T_{1 - \alpha }}f} \right\rangle + \left\langle {f,f} \right\rangle - {\mathop{\rm Re}\nolimits} \left\langle {{T_{1 - \alpha }}f,f} \right\rangle = }\\ {\left\langle {f,f} \right\rangle - {\mathop{\rm Re}\nolimits} \left\langle {{T_{1 - \alpha }}f,{T_\alpha }f} \right\rangle = }\\ {\frac{3}{4}{{\left\| f \right\|}^2} + \frac{1}{4}\left( {\left\langle {f,f} \right\rangle - 4{\mathop{\rm Re}\nolimits} \left\langle {{T_{1 - \alpha }}f,{T_\alpha }f} \right\rangle } \right) = }\\ {\frac{3}{4}{{\left\| f \right\|}^2} + \frac{1}{4}\left( {\left\langle {f,f} \right\rangle - 2\left\langle {{T_{1 - \alpha }}f,{T_\alpha }f} \right\rangle - 2\left\langle {{T_\alpha }f,{T_{1 - \alpha }}f} \right\rangle } \right) = }\\ {\frac{3}{4}{{\left\| f \right\|}^2} + \frac{1}{4}\left( {\left\langle {\left( {{T_\alpha } + {T_{1 - \alpha }}} \right)f,\left( {{T_\alpha } + {T_{1 - \alpha }}} \right)f} \right\rangle - } \right.}\\ {\left. {2\left\langle {{T_{1 - \alpha }}f,{T_\alpha }f} \right\rangle - 2\left\langle {{T_\alpha }f,{T_{1 - \alpha }}f} \right\rangle } \right) = }\\ {\frac{3}{4}{{\left\| f \right\|}^2} + \frac{1}{4}\left( {\left\langle {{T_\alpha }f,{T_\alpha }f} \right\rangle - \left\langle {{T_{1 - \alpha }}f,{T_\alpha }f} \right\rangle - } \right.}\\ {\left. {\left\langle {{T_\alpha }f,{T_{1 - \alpha }}f} \right\rangle + \left\langle {{T_{1 - \alpha }}f,{T_{1 - \alpha }}f} \right\rangle } \right) = }\\ {\frac{3}{4}{{\left\| f \right\|}^2} + \frac{1}{4}\left\langle {\left( {{T_\alpha } - {T_{1 - \alpha }}} \right)f,\left( {{T_\alpha } - {T_{1 - \alpha }}} \right)f} \right\rangle \le }\\ {\frac{3}{4}{{\left\| f \right\|}^2} + \frac{1}{4}{{\left\| {{T_\alpha } - {T_{1 - \alpha }}} \right\|}^2}{{\left\| f \right\|}^2} = }\\ {\frac{{3 + {{\left\| {{T_\alpha } - {T_{1 - \alpha }}} \right\|}^2}}}{4}{{\left\| f \right\|}^2}} \end{array} $$ 因此, 式(3)的右半不等式得证.证毕.
推论3 设{Λj:j∈J}是U关于{Vj:j∈J}的g-框架, {Γj:j∈J}是其对偶g-框架, K⊂J.定义算子
$$ {L_K}:U \to U,f \to \sum\limits_{j \in K} {\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} $$ $$ {L_{{K^c}}}:U \to U,f \to \sum\limits_{j \in {K^c}} {\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} $$ 则对任意的f∈U, 有
$$ \begin{array}{*{20}{c}} {\frac{3}{4}{{\left\| f \right\|}^2} \le {{\left\| {\sum\limits_{j \in K} {\mathit{\Gamma }_j^ * {\mathit{\Lambda }_j}f} } \right\|}^2} + {\mathop{\rm Re}\nolimits} \left( {\sum\limits_{j \in {K^c}} {\left\langle {{\mathit{\Lambda }_j}f,{\mathit{\Gamma }_j}f} \right\rangle } } \right) \le }\\ {\frac{{3 + {{\left\| {{L_K} - {L_{{K^c}}}} \right\|}^2}}}{4}{{\left\| f \right\|}^2}} \end{array} $$ 证明:在定理5中取${\alpha _j} = \left\{ \begin{gathered} 1, \;\;\;\;j \in K, \hfill \\ 0, \;\;\;\;j \in {K^c}, \hfill \\ \end{gathered} \right.$, 即可得此结论.
注3 推论3建立了定理3的反向不等式.
注4 取U=V, Λj=πVj, 由推论3可得文献[19]中定理2.8关于一般Hilbert空间中fusion框架不等式wj=1(j∈J)时的结果.
3. 结论
1) 运用算子理论在引入参数λ的基础上得到了Hilbert空间中关于正则对偶g-框架的一些新的不等式, 见定理4.
2) 运用算子理论得到了关于文献[20]定理4.1的反向不等式估计, 见定理5.
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