Some Inequalities for g-Frames in Hilbert C*-Modules

Abstract

In this paper, we obtain new inequalities for g-frames in Hilbert $C^{*}$-modules by using operator theory methods, which are related to a scalar $\lambda \in \mathbb{R}$ and an adjointable operator with respect to two g-Bessel sequences. It is demonstrated that our results can lead to several known results on this topic when suitable scalars and g-Bessel sequences are chosen.

1. Introduction

Since their appearance in the literature [1] on nonharmonic Fourier series, frames for Hilbert spaces have been a useful tool and applied to different branches of mathematics and other fields. For details on frames, the reader can refer to the papers [2,3,4,5,6,7,8,9,10,11]. The author in [12] extended the concept of frames to bounded linear operators and thus gave us the notion of g-frames, which possess some properties that are quite different from those of frames (see [13,14]).

In the past decade, much attention has been paid to the extension of frame and g-frame theory from Hilbert spaces to Hilbert $C^{*}$-modules, and some significant results have been presented (see [15,16,17,18,19,20,21,22,23]). It should be pointed out that, due to the essential differences between Hilbert spaces and Hilbert $C^{*}$-modules and the complex structure of the $C^{*}$-algebra involved in a Hilbert $C^{*}$-module, the problems on frames and g-frames for Hilbert $C^{*}$-modules are expected to be more complicated than those for Hilbert spaces. Also, increasingly more evidence is indicating that there is a close relationship between the theory of wavelets and frames and Hilbert $C^{*}$-modules in many aspects. This suggests that the discussion of frame and g-frame theory in Hilbert $C^{*}$-modules is interesting and important.

The authors in [24] provided a surprising inequality while further discussing the remarkable identity for Parseval frames derived from their research on effective algorithms to compute the reconstruction of a signal, which was later generalized to the situation of general frames and dual frames [25]. Those inequalities have already been extended to several generalized versions of frames in Hilbert spaces [26,27,28]. Moreover, the authors in [29,30,31] showed that g-frames in Hilbert $C^{*}$-modules have their inequalities based on the work in [24,25]; it is worth noting that the inequalities given in [30] are associated with a scalar in $[0,1]$ or $[\frac{1}{2},1]$. In this paper, we establish several new inequalities for g-frames in Hilbert $C^{*}$-modules, where a scalar $\lambda$ in $\mathbb{R}$, the real number set, and an adjointable operator with respect to two g-Bessel sequences are involved. Also, we show that some corresponding results in [29,31] can be considered a special case of our results.

We continue with this section for a review of some notations and definitions.

This paper adopts the following notations: $\mathbb{J}$ and $\mathcal{A}$ are, respectively, a finite or countable index set and a unital $C^{*}$-algebra; $\mathcal{H}$, $\mathcal{K}$, and ${\mathcal{K}}_j$’s ($j\in \mathbb{J}$) are Hilbert $C^{*}$-modules over $\mathcal{A}$ (or simply Hilbert $\mathcal{A}$-modules), setting $\langle f,f\rangle ={\left|f\right|}^2$ for any $f\in \mathcal{H}$. The family of all adjointable operators from $\mathcal{H}$ to $\mathcal{K}$ is designated ${\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H},\mathcal{K})$, which is abbreviated to ${\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H})$ if $\mathcal{K}=\mathcal{H}$.

A sequence $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\in {\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H},{\mathcal{K}}_j)\}}_{j\in \mathbb{J}}$ denotes a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ if there are real numbers $0<C\le D<\infty$ satisfying

$$C\langle f,f\rangle \le \sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle \le D\langle f,f\rangle ,\forall f\in \mathcal{H}.$$

(1)

If only the second inequality in Equation (1) is required, then $\mathsf{\Lambda }$ is said to be a g-Bessel sequence.

For a given g-frame $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\in {\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H},{\mathcal{K}}_j)\}}_{j\in \mathbb{J}}$, there is always a positive, invertible, and self-adjoint operator in ${\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H})$, which we call the g-frame operator of $\mathsf{\Lambda }$, defined by

$$S_{\mathsf{\Lambda }}:\mathcal{H}\to \mathcal{H},S_{\mathsf{\Lambda }}f=\sum _{j\in \mathbb{J}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f.$$

(2)

For any $\mathbb{I}\subset \mathbb{J}$, let ${\mathbb{I}}^c$ be the complement of $\mathbb{I}$. We define a positive and self-adjoint operator in ${\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H})$ related to $\mathbb{I}$ and a g-frame $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\in {\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H},{\mathcal{K}}_j)\}}_{j\in \mathbb{J}}$ in the following form

$$S_{\mathbb{I}}^{\mathsf{\Lambda }}:\mathcal{H}\to \mathcal{H},S_{\mathbb{I}}^{\mathsf{\Lambda }}f=\sum _{j\in \mathbb{I}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f.$$

(3)

Recall that a g-Bessel $\mathsf{\Gamma }={\{{\mathsf{\Gamma }}_j\in {\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H},{\mathcal{K}}_j)\}}_{j\in \mathbb{J}}$ is an alternate dual g-frame of $\mathsf{\Lambda }$ if, for every $f\in \mathcal{H}$, we have $f={\sum }_{j\in \mathbb{J}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Gamma }}_{j}f$.

Let $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ and $\mathsf{\Gamma }={\{{\mathsf{\Gamma }}_j\}}_{j\in \mathbb{J}}$ be g-Bessel sequences for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$. We observe from the Cauchy–Schwarz inequality that the operator

$$S_{\mathsf{\Gamma }\mathsf{\Lambda }}:\mathcal{H}\to \mathcal{H},S_{\mathsf{\Gamma }\mathsf{\Lambda }}f=\sum _{j\in \mathbb{J}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f$$

(4)

is well defined, and a direct calculation shows that $S_{\mathsf{\Gamma }\mathsf{\Lambda }}\in {\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H})$.

2. The Main Results

The following result for operators is used to prove our main results.

Lemma 1.

Suppose that $U,V,L\in {\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H})$ and that $U+V=L$. Then, for any $\lambda \in \mathbb{R}$, we have

$$U^{*}U+\frac{\lambda }{2}(V^{*}L+L^{*}V)=V^{*}V+(1-\frac{\lambda }{2})(U^{*}L+L^{*}U)+(\lambda -1)L^{*}L\ge (\lambda -\frac{{\lambda }^2}{4})L^{*}L.$$

Proof. 

On the one hand, we obtain

$$U^{*}U+\frac{\lambda }{2}(V^{*}L+L^{*}V)=U^{*}U+\frac{\lambda }{2}((L^{*}-U^{*})L+L^{*}(L-U))=U^{*}U-\frac{\lambda }{2}(U^{*}L+L^{*}U)+\lambda L^{*}L.$$

On the other hand, we have

$$\begin{array}{c}{V}^{*}V+(1-\frac{\lambda }{2})(U^{*}L+L^{*}U)+(\lambda -1)L^{*}L\hfill \\ =(L^{*}-U^{*})(L-U)+(U^{*}L+L^{*}U)-\frac{\lambda }{2}(U^{*}L+L^{*}U)+(\lambda -1)L^{*}L\hfill \\ =L^{*}L-(U^{*}L+L^{*}U)+U^{*}U+(U^{*}L+L^{*}U)-\frac{\lambda }{2}(U^{*}L+L^{*}U)+(\lambda -1)L^{*}L\hfill \\ =U^{*}U-\frac{\lambda }{2}(U^{*}L+L^{*}U)+\lambda L^{*}L=(U-\frac{\lambda }{2}L{)}^{*}(U-\frac{\lambda }{2}L)+(\lambda -\frac{{\lambda }^2}{4})L^{*}L\ge (\lambda -\frac{{\lambda }^2}{4})L^{*}L.\hfill \end{array}$$

This completes the proof. □

Theorem 1.

Let $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ be a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$. Suppose that $\mathsf{\Gamma }={\{{\mathsf{\Gamma }}_j\}}_{j\in \mathbb{J}}$ and $\mathsf{\Theta }={\{{\mathsf{\Theta }}_j\}}_{j\in \mathbb{J}}$ are two g-Bessel sequences for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$, and that the operator $S_{\mathsf{\Gamma }\mathsf{\Lambda }}$ is defined in Equation (4). Then, for any $\lambda \in \mathbb{R}$ and any $f\in \mathcal{H}$, we have

$$\begin{array}{c}|\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2+\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle =|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2+\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle \hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +(1+\frac{\lambda }{2}-\frac{{\lambda }^2}{4})\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ -\frac{\lambda }{2}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle .\hfill \end{array}$$

(5)

Proof. 

We let

$$Uf=\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f\mathrm{and}Vf=\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f$$

(6)

for each $f\in \mathcal{H}$. Then, $U,V\in {\mathrm{End}}_{\mathcal{A}}^{*}(\mathcal{H})$ and, further,

$$Uf+Vf=\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f+\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f=\sum _{j\in \mathbb{J}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f=S_{\mathsf{\Gamma }\mathsf{\Lambda }}f.$$

By Lemma 1, we get

$$\begin{array}{c}{\left|Uf\right|}^2+\frac{\lambda }{2}(\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle )\hfill \\ ={\left|Vf\right|}^2+(1-\frac{\lambda }{2})(\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle )+(\lambda -1){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2.\hfill \end{array}$$

Hence,

$$\begin{array}{cc}\hfill {\left|Uf\right|}^2& ={\left|Vf\right|}^2+(1-\frac{\lambda }{2})(\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle )+(\lambda -1){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2\hfill \\ & -\frac{\lambda }{2}(\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle )\hfill \\ & ={\left|Vf\right|}^2+\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\frac{\lambda }{2}(\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle )\hfill \\ & -\frac{\lambda }{2}(\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle )+(\lambda -1){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2\hfill \\ & ={\left|Vf\right|}^2+\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\frac{\lambda }{2}(\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle )\hfill \\ & -\frac{\lambda }{2}(\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle )+(\lambda -1){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2\hfill \\ & ={\left|Vf\right|}^2+\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\lambda |S_{\mathsf{\Gamma }\mathsf{\Lambda }}{f|}^2+(\lambda -1){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2\hfill \\ & ={\left|Vf\right|}^2+\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle -\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle .\hfill \end{array}$$

It follows that

$${\left|Uf\right|}^2+\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle ={\left|Vf\right|}^2+\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle ,$$

(7)

from which we arrive at

$$|\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2+\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle =|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2+\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle .$$

We are now in a position to prove the inequality in Equation (5).

Again by Lemma 1,

$$\begin{array}{cc}\hfill {\left|Uf\right|}^2& \ge (\lambda -\frac{{\lambda }^2}{4}){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2-\frac{\lambda }{2}(\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle )\hfill \\ & =(\lambda -\frac{{\lambda }^2}{4})\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +(\lambda -\frac{{\lambda }^2}{4})\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle -\frac{\lambda }{2}\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle -\frac{\lambda }{2}\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle \hfill \\ & =(\lambda -\frac{{\lambda }^2}{4})\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +(\frac{\lambda }{2}-\frac{{\lambda }^2}{4})\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle -\frac{\lambda }{2}\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle .\hfill \end{array}$$

(8)

Therefore,

$$\begin{array}{c}|\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2+\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle ={\left|Uf\right|}^2+\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +(1+\frac{\lambda }{2}-\frac{{\lambda }^2}{4})\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle -\frac{\lambda }{2}\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle \hfill \\ =(\lambda -\frac{{\lambda }^2}{4})\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +(1+\frac{\lambda }{2}-\frac{{\lambda }^2}{4})\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle -\frac{\lambda }{2}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle \hfill \end{array}$$

for any $f\in \mathcal{H}$. □

Corollary 1.

Suppose that $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ is a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ with g-frame operator $S_{\mathsf{\Lambda }}$ and that ${\tilde{\mathsf{\Lambda }}}_j={\mathsf{\Lambda }}_j{S}_{\mathsf{\Lambda }}^{-1}$ for each $j\in \mathbb{J}$. Then, for any $\lambda \in \mathbb{R}$, for all $\mathbb{I}\subset \mathbb{J}$ and all $f\in \mathcal{H}$, we have

$$\begin{array}{c}\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle =\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f\rangle \hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +(1-\frac{{\lambda }^2}{4})\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .\hfill \end{array}$$

Proof. 

Taking ${\mathsf{\Gamma }}_j={\mathsf{\Lambda }}_j{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}$ for any $j\in \mathbb{J}$, then it is easy to see that $S_{\mathsf{\Gamma }\mathsf{\Lambda }}=S_{\mathsf{\Lambda }}^{\frac{1}{2}}$. For each $j\in \mathbb{J}$, let

$${\mathsf{\Theta }}_j=\left\{\begin{array}{cc}{\mathsf{\Gamma }}_j,\hfill & j\in \mathbb{I},\hfill \\ 0,\hfill & j\in {\mathbb{I}}^c.\hfill \end{array}\right.$$

Now, for each $f\in \mathcal{H}$,

$$\begin{array}{cc}\hfill |\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2& =|\sum _{j\in {\mathbb{I}}^c}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2={|S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f|}^2=\langle S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle \hfill \\ & =\langle S_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,S_{\mathsf{\Lambda }}^{-1}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle =\langle S_{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-1}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,S_{\mathsf{\Lambda }}^{-1}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle \hfill \\ & =\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_j{S}_{\mathsf{\Lambda }}^{-1}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_j{S}_{\mathsf{\Lambda }}^{-1}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle =\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle .\hfill \end{array}$$

(9)

Since $|{\sum }_{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_j{f|}^2=|{\sum }_{j\in \mathbb{I}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_j{f|}^2={|{\sum }_{j\in \mathbb{I}}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f|}^2$, a replacement of ${\mathbb{I}}^c$ by $\mathbb{I}$ in the last item of Equation (9) leads to

$$|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2=\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f\rangle .$$

(10)

We also have

$$\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle =\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle ,\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle =\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .$$

(11)

Hence, the conclusion follows from Theorem 1. □

Let $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ be a Parseval g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$; then, $S_{\mathsf{\Lambda }}={\mathrm{Id}}_{\mathcal{H}}$. Thus, for any $\mathbb{I}\subset \mathbb{J}$,

$$\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle =\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle ={\left|S_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\right|}^2=|\sum _{j\in {\mathbb{I}}^c}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2.$$

Similarly,

$$\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f\rangle =|\sum _{j\in \mathbb{I}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2.$$

This fact, together with Corollary 1, yields

Corollary 2.

Suppose that $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ is a Parseval g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$. Then, for any $\lambda \in \mathbb{R}$, for all $\mathbb{I}\subset \mathbb{J}$ and all $f\in \mathcal{H}$, we have

$$\begin{array}{c}\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +|\sum _{j\in {\mathbb{I}}^c}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2=\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +|\sum _{j\in \mathbb{I}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2\hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +(1-\frac{{\lambda }^2}{4})\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .\hfill \end{array}$$

Corollary 3.

Suppose that $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ is a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ with an alternate dual g-frame $\mathsf{\Gamma }={\{{\mathsf{\Gamma }}_j\}}_{j\in \mathbb{J}}$. Then, for any $\lambda \in \mathbb{R}$, for all $\mathbb{I}\subset \mathbb{J}$ and all $f\in \mathcal{H}$, we have

$$\begin{array}{c}|\sum _{j\in \mathbb{I}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2+\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Gamma }}_{j}f\rangle =|\sum _{j\in {\mathbb{I}}^c}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2+\sum _{j\in \mathbb{I}}\langle {\mathsf{\Gamma }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle \hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Gamma }}_{j}f\rangle +(1+\frac{\lambda }{2}-\frac{{\lambda }^2}{4})\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Gamma }}_{j}f\rangle -\frac{\lambda }{2}\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Gamma }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .\hfill \end{array}$$

Proof. 

We conclude first that $S_{\mathsf{\Gamma }\mathsf{\Lambda }}={\mathrm{Id}}_{\mathcal{H}}$. Now, the result follows immediately from Theorem 1 if, for any $\mathbb{I}\subset \mathbb{J}$, we take ${\mathsf{\Theta }}_j=\left\{\begin{array}{cc}{\mathsf{\Gamma }}_j,\hfill & j\in {\mathbb{I}}^c,\hfill \\ 0,\hfill & j\in \mathbb{I}.\hfill \end{array}\right.$ □

Remark 1.

Theorems 4.1 and 4.2 in [31] can be obtained if we take $\lambda =1$, respectively, in Corollaries 1 and 2.

Theorem 2.

Let $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ be a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$. Suppose that $\mathsf{\Gamma }={\{{\mathsf{\Gamma }}_j\}}_{j\in \mathbb{J}}$ and $\mathsf{\Theta }={\{{\mathsf{\Theta }}_j\}}_{j\in \mathbb{J}}$ are two g-Bessel sequences for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ and that the operator $S_{\mathsf{\Gamma }\mathsf{\Lambda }}$ is defined in Equation (4). Then, for any $\lambda \in \mathbb{R}$ and any $f\in \mathcal{H}$, we have

$$\begin{array}{cc}\hfill |\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2+|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2& \ge (\lambda -\frac{{\lambda }^2}{2})|\sum _{j\in \mathbb{J}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2-(1-\lambda )\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle \hfill \\ & +(1-\lambda )\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle .\hfill \end{array}$$

Moreover, if $U^{*}V$ is positive, where U and V are given in Equation (6), then

$$\begin{array}{c}|\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2+|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2\hfill \\ \le \sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle +\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle .\hfill \end{array}$$

(12)

Proof. 

Combining Equation (7) with Lemma 1, we obtain

$$\begin{array}{c}|\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2+|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2\hfill \\ ={\left|Uf\right|}^2+{\left|Vf\right|}^2=2{\left|Vf\right|}^2+\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ \ge (2-\frac{{\lambda }^2}{2}){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2-(2-\lambda )(\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle +\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle )+\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ =(2-\frac{{\lambda }^2}{2}){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2-(2-\lambda )\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -(2-\lambda )\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ -(2-\lambda )\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +(1-\lambda )\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle \hfill \\ =(2-\frac{{\lambda }^2}{2})|S_{\mathsf{\Gamma }\mathsf{\Lambda }}{f|}^2-(1-\lambda )\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -(2-\lambda ){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2+(1-\lambda )\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ =(\lambda -\frac{{\lambda }^2}{2}){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2-(1-\lambda )\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle +(1-\lambda )\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ =(\lambda -\frac{{\lambda }^2}{2})|\sum _{j\in \mathbb{J}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2-(1-\lambda )\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle +(1-\lambda )\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \end{array}$$

for any $f\in \mathcal{H}$. We next prove Equation (12). Since $U^{*}V$ is positive, we see from Equation (7) that

$${\left|Uf\right|}^2={\left|Vf\right|}^2+\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle =\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle -\langle Vf,Uf\rangle \le \langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle$$

for each $f\in \mathcal{H}$. A similar discussion gives ${\left|Vf\right|}^2\le \langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle$. Thus,

$$\begin{array}{cc}\hfill |\sum _{j\in \mathbb{J}}{({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)}^{*}{\mathsf{\Lambda }}_{j}f{|}^2+|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2& ={\left|Uf\right|}^2+{\left|Vf\right|}^2\le \langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle +\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ & =\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle +\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle .\hfill \end{array}$$

□

Corollary 4.

Let $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ be a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ with g-frame operator $S_{\mathsf{\Lambda }}$, and ${\tilde{\mathsf{\Lambda }}}_j={\mathsf{\Lambda }}_j{S}_{\mathsf{\Lambda }}^{-1}$ for each $j\in \mathbb{J}$. Then, for any $\lambda \in \mathbb{R}$, for all $\mathbb{I}\subset \mathbb{J}$ and all $f\in \mathcal{H}$, we have

$$\begin{array}{c}(\lambda -\frac{{\lambda }^2}{2})\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle -(1-\lambda )\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +(1-\lambda )\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle \hfill \\ \le \sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f\rangle +\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}f\rangle \le \sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .\hfill \end{array}$$

Proof. 

For every $j\in \mathbb{J}$, taking ${\mathsf{\Gamma }}_j={\mathsf{\Lambda }}_j{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}$ and ${\mathsf{\Theta }}_j=\left\{\begin{array}{cc}{\mathsf{\Gamma }}_j,\hfill & j\in \mathbb{I},\hfill \\ 0,\hfill & j\in {\mathbb{I}}^c,\hfill \end{array}\right.$ then the operators U and V defined in Equation (6) can be expressed as $U=S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}$ and $V=S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{\mathbb{I}}^{\mathsf{\Lambda }}$, respectively. Hence, $U^{*}V=S_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-1}{S}_{\mathbb{I}}^{\mathsf{\Lambda }}$. Since $S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{\mathbb{I}}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}$ and $S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}$ are positive and commutative, it follows that

$$0\le S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{\mathbb{I}}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}}=S_{\mathsf{\Lambda }}^{-\frac{1}{2}}{S}_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-1}{S}_{\mathbb{I}}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-\frac{1}{2}},$$

and, consequently, $S_{{\mathbb{I}}^c}^{\mathsf{\Lambda }}{S}_{\mathsf{\Lambda }}^{-1}{S}_{\mathbb{I}}^{\mathsf{\Lambda }}\ge 0$. Note also that

$$|\sum _{j\in \mathbb{J}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2=|S_{\mathsf{\Lambda }}^{-\frac{1}{2}}\sum _{j\in \mathbb{J}}{\mathsf{\Lambda }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2={|S_{\mathsf{\Lambda }}^{\frac{1}{2}}f|}^2=\langle S_{\mathsf{\Lambda }}f,f\rangle =\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .$$

Now, the result follows by combining Theorem 2 and Equations (9)–(11). □

Theorem 3.

Let $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ be a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ with g-frame operator $S_{\mathsf{\Lambda }}$. Suppose that $\mathsf{\Gamma }={\{{\mathsf{\Gamma }}_j\}}_{j\in \mathbb{J}}$ and $\mathsf{\Theta }={\{{\mathsf{\Theta }}_j\}}_{j\in \mathbb{J}}$ are two g-Bessel sequences for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ and that the operator $S_{\mathsf{\Gamma }\mathsf{\Lambda }}$ is defined in Equation (4). Then, for any $\lambda \in \mathbb{R}$ and any $f\in \mathcal{H}$, we have

$$\begin{array}{cc}\hfill \sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Lambda }}^{\frac{1}{2}}f\rangle -|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2& \le \sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle -\frac{\lambda }{2}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ & +(1-\frac{\lambda }{2})\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle +\frac{{\lambda }^2}{4}|\sum _{j\in \mathbb{J}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2.\hfill \end{array}$$

Moreover, if $U^{*}V$ is positive, where U and V are given in Equation (6), then

$$\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Lambda }}^{\frac{1}{2}}f\rangle -|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2\ge \sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle .$$

Proof. 

Combining Equations (7) and (8) leads to

$$\begin{array}{c}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Lambda }}^{\frac{1}{2}}f\rangle -|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2=\langle S_{\mathsf{\Lambda }}^{\frac{1}{2}}Vf,f\rangle -{\left|Vf\right|}^2\hfill \\ \le \langle S_{\mathsf{\Lambda }}^{\frac{1}{2}}Vf,f\rangle -(\lambda -\frac{{\lambda }^2}{4})\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle -(\frac{\lambda }{2}-\frac{{\lambda }^2}{4})\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ +\frac{\lambda }{2}\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle -\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle \hfill \\ =\langle Vf,(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle -(\frac{\lambda }{2}-\frac{{\lambda }^2}{4})(\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle )-\frac{\lambda }{2}\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ +\frac{\lambda }{2}(\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Vf\rangle +\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle )+(1-\frac{\lambda }{2})\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle \hfill \\ =\langle Vf,(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle -(\frac{\lambda }{2}-\frac{{\lambda }^2}{4}){\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2\hfill \\ -\frac{\lambda }{2}\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +\frac{\lambda }{2}{\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2+(1-\frac{\lambda }{2})\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle \hfill \\ =\langle Vf,(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle +\frac{{\lambda }^2}{4}{\left|S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\right|}^2-\frac{\lambda }{2}\langle Uf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle +(1-\frac{\lambda }{2})\langle S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,Uf\rangle \hfill \\ =\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle -\frac{\lambda }{2}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ +(1-\frac{\lambda }{2})\sum _{j\in \mathbb{J}}\langle ({\mathsf{\Gamma }}_j-{\mathsf{\Theta }}_j)S_{\mathsf{\Gamma }\mathsf{\Lambda }}f,{\mathsf{\Lambda }}_{j}f\rangle +\frac{{\lambda }^2}{4}|\sum _{j\in \mathbb{J}}{\mathsf{\Gamma }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2,\forall f\in \mathcal{H}.\hfill \end{array}$$

Suppose that $U^{*}V$ is positive; then, ${\left|Vf\right|}^2\le \langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle$. Now, the “Moreover” part follows from the following inequality:

$$\begin{array}{cc}\hfill \sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Lambda }}^{\frac{1}{2}}f\rangle -|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2& =\langle S_{\mathsf{\Lambda }}^{\frac{1}{2}}Vf,f\rangle -{\left|Vf\right|}^2\ge \langle S_{\mathsf{\Lambda }}^{\frac{1}{2}}Vf,f\rangle -\langle Vf,S_{\mathsf{\Gamma }\mathsf{\Lambda }}f\rangle \hfill \\ & =\langle Vf,(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle =\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle .\hfill \end{array}$$

□

Corollary 5.

Let $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_j\}}_{j\in \mathbb{J}}$ be a g-frame for $\mathcal{H}$ with respect to ${\{{\mathcal{K}}_j\}}_{j\in \mathbb{J}}$ with g-frame operator $S_{\mathsf{\Lambda }}$. Then, for any $\lambda \in \mathbb{R}$, for all $\mathbb{I}\subset \mathbb{J}$ and all $f\in \mathcal{H}$, we have

$$\begin{array}{cc}\hfill 0& \le \sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle -\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f\rangle \hfill \\ & \le (1-\lambda )\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +\frac{{\lambda }^2}{4}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .\hfill \end{array}$$

Proof. 

For each $j\in \mathbb{J}$, let ${\mathsf{\Gamma }}_j$ and ${\mathsf{\Theta }}_j$ be the same as in the proof of Corollary 4. By Theorem 3, we have

$$\begin{array}{c}\sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle -\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f\rangle =\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Lambda }}^{\frac{1}{2}}f\rangle -|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2\hfill \\ \le -\frac{\lambda }{2}\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +(1-\frac{\lambda }{2})\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +\frac{{\lambda }^2}{4}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle \hfill \\ =(1-\lambda )\sum _{j\in {\mathbb{I}}^c}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle +\frac{{\lambda }^2}{4}\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle .\hfill \end{array}$$

By Theorem 3 again,

$$\begin{array}{cc}\hfill \sum _{j\in \mathbb{I}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Lambda }}_{j}f\rangle -\sum _{j\in \mathbb{J}}\langle {\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f,{\tilde{\mathsf{\Lambda }}}_j{S}_{\mathbb{I}}^{\mathsf{\Lambda }}f\rangle & =\sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j{S}_{\mathsf{\Lambda }}^{\frac{1}{2}}f\rangle -|\sum _{j\in \mathbb{J}}{\mathsf{\Theta }}_j^{*}{\mathsf{\Lambda }}_{j}f{|}^2\hfill \\ & \ge \sum _{j\in \mathbb{J}}\langle {\mathsf{\Lambda }}_{j}f,{\mathsf{\Theta }}_j(S_{\mathsf{\Lambda }}^{\frac{1}{2}}-S_{\mathsf{\Gamma }\mathsf{\Lambda }})f\rangle =0,\hfill \end{array}$$

and the proof is finished. □

Remark 2.

Taking $\lambda =1$ in Corollaries 4 and 5, we can obtain Theorem 2.4 in [29].