Some New Inequalities for Dual Continuous g-Frames

Abstract

In the present paper, we establish new inequalities for dual continuous generalized frames by adopting operator methods. The inequalities are parameterized by the parameter $\lambda \in \mathbb{R}$. These results generalize those obtained by Balan, Casazza and Găvruţa and cover recently obtained results by Zhang and Li in [Zhang, W.; Li, Y.Z. New inequalities and erasures for continuous g-frames. Math. Rep. 2018, 20, 263–278]. Moreover, we also give an upper bound of inequality for alternate continuous generalized frames in Hilbert spaces. It differs from previous results.

1. Introduction

The frame has a long history, it was first defined by Duffin and Schaeffer in 1952 to study difficult problems in non-harmonic Fourier series ([1]). An at most countable sequence ${\left\{e_i\right\}}_{i\in \mathcal{I}}$ in a separable Hilbert space $\mathcal{H}$ is said to be a frame if the constants $0<C\le D<\infty$ so that

$${C\parallel e\parallel }^2\le \sum _{i\in \mathcal{I}}{\left|\langle e,e_i\rangle \right|}^2\le D{\parallel e\parallel }^2 \mathrm{for} e\in \mathcal{H},$$

(1)

where C, D are called the frame bounds. Particularly, in (1) if $C=D=1$, we call it a Parseval frame. The theory of frames has seen great achievements since the celebrated work [2] by Daubechies, Crossman and Meyer in 1986. It has been widely used in such engineering fields as signal processing, communication and some other fields (see [3,4,5,6]). Due to their nice properties, the concept of a continuous frame (or simply, c-frame) in [7], presented by Ali et al. is an extension of a frame, which was generalized to a family indexed by some locally compact space, which endowed with Radon measure. Some results of c-frames and their adhibition can be seen in ([8,9]). In particular, Sun in 2006 gave the definition of a generalized frame (or simply g-frame) in a Hilbert space in [10], which covers frames of subspaces ([11]), pseudo-frames ([12]), and so on. A continuous generalized frame (or, simply a continuous g-frame or c-g-frame) was firstly introduced by Dehghan and Hasankhani Fard in [13]. The present paper focuses on some inequalities for dual continuous generalized frames.

The research of inequalities and equalities relevant to Parseval frames was done by Balan, Casazza et al. in [14]. It has interested some engineers and mathematicians (see [15,16,17,18,19,20,21,22,23,24,25,26]). In particular, Balan et al. and Găvruţa obtained the following two inequalities:

Proposition 1 ([14]).

Let ${\left\{f_j\right\}}_{j\in J}\subset H$ be a Parseval frame. Then

$$\begin{array}{c}\hfill \sum _{j\in J_1}|\langle f,f_j\rangle {|}^2+\parallel \sum _{j\in J_1^c}\langle f,f_j\rangle f_j{\parallel }^2\ge \frac{3}{4}{\parallel f\parallel }^2\end{array}$$

(2)

for every $f\in H$, $J_1\subset J$.

Proposition 2 ([20]).

Assume that ${\left\{f_j\right\}}_{j\in J}\subset H$ is a frame. Let ${\left\{g_j\right\}}_{j\in J}\subset H$ be a dual frame of ${\left\{f_j\right\}}_{j\in J}$. Then

$$\begin{array}{c}\hfill Re(\sum _{j\in J_1}\langle f,g_j\rangle \overline{\langle f,f_j\rangle })+\parallel \sum _{j\in J_1^c}\langle f,g_j\rangle f_j{\parallel }^2\ge \frac{3}{4}{\parallel f\parallel }^2\end{array}$$

(3)

for any $f\in H$.

In 2010, Xiao and Zeng in [18] generalized Proposition 1 to continuous generalized frames (see [18], Theorem 3.2). In 2018, Zhang and Li in [17] generalized Proposition 2 to continuous generalized frames (see [17], Theorem 3.4), and gave some bi-directional inequalities for continuous generalized frames with more general forms. Motivated by above works, in this paper, using operator techniques we present several new inequalities for dual continuous generalized frames which are parameterized by the parameter $\lambda \in \mathbb{R}$, and we show that our results can lead to the corresponding results in [17,18]. Indeed, by choosing reasonable $\lambda$, we may obtain the previous results as special cases.

We organize this paper as follows. In Section 2, we focus on some basic definitions, properties and essential results. In Section 3, we obtain some important inequalities for canonical dual continuous generalized frames by using the operator techniques. In Section 4, we get some new inequalities for alternate dual continuous generalized frames.

2. Preliminaries

We first reiterate some fundamental concepts, notions and properties of frames in Hilbert space. The scholars may refer to [4,9,10,13] for details.

We denote dual by asterisk, the real part by Re, and the complex conjugation by bar. $\mathcal{H},\mathcal{V}$ are complex Hilbert spaces, $(\mathsf{\Omega },\mu )$ is a measure space and $\mu$ is a positive measure, $I_{\mathcal{H}}$ is the identity operator on $\mathcal{H}$, ${\left\{{\mathcal{V}}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}\subset \mathcal{V}$ is a sequence of closed subspace of $\mathcal{V}$. We denote all bounded linear operators from $\mathcal{H}$ into ${\mathcal{V}}_{\omega }$ by $L(\mathcal{H},{\mathcal{V}}_{\omega })$. Let

$${\left({\oplus }_{\omega \in \mathsf{\Omega }}{\mathcal{V}}_{\omega },\mu \right)}_{L^2}=\left\{f={\left\{f_{\omega }\right\}}_{\omega \in \mathsf{\Omega }},f_{\omega }:\mathsf{\Omega }\to \mathcal{H},{\int }_{\mathsf{\Omega }}{\parallel f_{\omega }\parallel }^{2}d\mu (\omega )<\infty \right\},$$

be the Hilbert space with the inner product:

$$\langle f,g\rangle ={\int }_{\mathsf{\Omega }}\langle f_{\omega },g_{\omega }\rangle d\mu (\omega ),f,g\in {\left({\oplus }_{\omega \in \mathsf{\Omega }}{\mathcal{V}}_{\omega },\mu \right)}_{L^2}.$$

Definition 1 ([10]).

A family ${\{{\mathsf{\Lambda }}_i\in L(\mathcal{H},{\mathcal{V}}_i)\}}_{i\in \mathcal{I}}$ of bounded operator is said to be a g-frame for $\mathcal{H}$ with respect to (or simply w.r.t ) ${\left\{{\mathcal{V}}_i\right\}}_{i\in \mathcal{I}}$ if there exist $0<A_1\le B_1<+\infty$ such that

$$A_1{\parallel g\parallel }^2\le \sum _{i\in \mathcal{I}}\parallel {\mathsf{\Lambda }}_i{g\parallel }^2\le B_1{\parallel g\parallel }^{2}forg\in \mathcal{H},$$

where $A_1,B_1$ are called a lower and upper bounds for the frame, respectively.

Definition 2 ([9]).

Assume that $(X,\mu )$ is a measure space and μ is a positive measure, the mapping $f:X\to \mathcal{H}$ is said to be a continuous frame (or simply c-frame) for $\mathcal{H}$, if

(i)

The mapping $x\to \langle f(x),h\rangle$ is measurable for all $h\in \mathcal{H}$.

(ii)

The constants $0<A_2\le B_2<+\infty$ so that

$$A_2{\parallel h\parallel }^2\le {\int }_X{\left|\langle f(x),h\rangle \right|}^{2}d\mu (x)\le B_2{\parallel h\parallel }^{2}forf\in \mathcal{H},$$

(4)

where $A_2$ and $B_2$ are called c-frame bounds. If only the second inequality in (4) holds, the mapping is call continuous Bessel sequence for $\mathcal{H}$ with Bessel bound $B_2$.

Definition 3 ([13]).

We say that $\mathsf{\Lambda }={\{{\mathsf{\Lambda }}_{\omega }\in L(\mathcal{H},{\mathcal{V}}_{\omega })\}}_{\omega \in \mathsf{\Omega }}$ is a continuous generalized frame (or simply c-g-frame) for $\mathcal{H}$ w.r.t $(\mathsf{\Omega },\mu )$, if

(i)

For fixed $h\in \mathcal{H}$, $\omega \to {\mathsf{\Lambda }}_{\omega }h$ is strongly measurable,

(ii)

The constants $0<A\le B<+\infty$ such that

$${A\parallel h\parallel }^2\le {\int }_{\mathsf{\Omega }}\parallel {\mathsf{\Lambda }}_{\omega }{h\parallel }^{2}d\mu (\omega )\le B{\parallel h\parallel }^{2}forh\in \mathcal{H},$$

(5)

where $A,B$ are called frame bounds for the c-g-frame. If only the second inequality in (5) is satisfied, ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is said to be a Bessel continuous g-mapping for $\mathcal{H}$ w.r.t $(\mathsf{\Omega },\mu )$ with bound B. If $A=B=\lambda$, ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is said to be the λ-tight continuous generalized frame.

Remark 1.

If $\mathsf{\Omega }:=\mathbb{N}$ and μ is a counting measure, the continuous generalized frame will be just the generalized frame.

Definition 4 ([13]).

Assume that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ and ${\left\{{\Theta }_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ are two continuous generalized frames for $\mathcal{H}$ w.r.t $(\mathsf{\Omega },\mu )$ satisfying

$$f={\int }_{\mathsf{\Omega }}{\mathsf{\Lambda }}_{\omega }^{*}{\Theta }_{\omega }fd\mu (\omega )={\int }_{\mathsf{\Omega }}{\Theta }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )forf\in \mathcal{H}.$$

Then ${\left\{{\Theta }_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is called a dual of ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$.

Let ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ be a continuous generalized frame for $\mathcal{H}$. In [13], the continuous generalized frame operator $S:\mathcal{H}\to \mathcal{H}$ is defined by

$$\begin{array}{c}\hfill S(f)={\int }_{\mathsf{\Omega }}{\mathsf{\Lambda }}_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega ),\forall f\in \mathcal{H}.\end{array}$$

(6)

By a standard argument, we can get that the operator S is bounded, positive, invertible and self-adjoint. For $f\in \mathcal{H}$,

$$f=S{S}^{-1}f=S^{-1}Sf={\int }_{\mathsf{\Omega }}{\mathsf{\Lambda }}_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }{S}^{-1}fd\mu (\omega ){\int }_{\mathsf{\Omega }}{S}^{-1}{\mathsf{\Lambda }}_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega ).$$

Denote ${\tilde{\mathsf{\Lambda }}}_{\omega }={\mathsf{\Lambda }}_{\omega }{S}^{-1}$, it is easy to check that ${\left\{{\tilde{\mathsf{\Lambda }}}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a continuous generalized frame for $\mathcal{H}$ related to $(\mathsf{\Omega },\mu )$, $\frac{1}{B},\frac{1}{A}$ are the corresponding lower and upper bounds, $S^{-1}$ is the frame operator. ${\left\{{\tilde{\mathsf{\Lambda }}}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is said to be the canonical dual of ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ (see [13]).

For $\Xi \subset \mathsf{\Omega }$, we define the operators $S_{\Xi }$ and $S_{{\Xi }^c}$ by

$$\begin{array}{c}\hfill S_{\Xi }f={\int }_{\Xi }{\mathsf{\Lambda }}_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega ),\forall f\in \mathcal{H},\end{array}$$

(7)

and

$$\begin{array}{c}\hfill S_{{\Xi }^c}f={\int }_{{\Xi }^c}{\mathsf{\Lambda }}_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega ),\forall f\in \mathcal{H},\end{array}$$

(8)

where ${\Xi }^c=\mathsf{\Omega }\setminus \Xi$. Then $S=S_{\Xi }+S_{{\Xi }^c}$, and $S_{\Xi },S_{{\Xi }^c}$ are positive and self-adjoint operators.

3. Some New Inequalities for Canonical Dual c-g-Frames

We first give the following two essential Theorems in [17,18] respectively.

Theorem 1.

Suppose that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a continuous generalized frame for $\mathcal{H}$. S is the corresponding frame operator. Then

$$\begin{array}{c}0\le {\int }_{{\mathsf{\Omega }}_1}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )-{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\mathsf{\Omega }}_1}{f\parallel }^{2}d\mu (\omega )\le \frac{1}{4}{\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ),\hfill \end{array}$$

$$\begin{array}{c}\hfill \frac{1}{2}{\int }_{\mathsf{\Omega }}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )\le {\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\mathsf{\Omega }}_1}{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\mathsf{\Omega }}_1^c}{f\parallel }^{2}d\mu (\omega )\le \frac{3}{2}{\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ),\end{array}$$

$$\begin{array}{c}\hfill \frac{3}{4}{\int }_{\mathsf{\Omega }}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )\le {\int }_{{\mathsf{\Omega }}_1}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\mathsf{\Omega }}_1^c}{f\parallel }^{2}d\mu (\omega )\le {\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega )\end{array}$$

for ${\mathsf{\Omega }}_1\subset \mathsf{\Omega }$ and $f\in \mathcal{H}$.

Theorem 2.

Assume that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a generalized continuous frame for $\mathcal{H}$ with respect to $(\mathsf{\Omega },\mu )$. Then

$$\begin{array}{c}{\int }_{{\mathsf{\Omega }}_1}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\mathsf{\Omega }}_1^c}f\parallel }^{2}d\mu (\omega )\hfill \\ ={\int }_{{\mathsf{\Omega }}_1^c}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\mathsf{\Omega }}_1}{f\parallel }^{2}d\mu (\omega )\ge \frac{3}{4}{\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega )\hfill \end{array}$$

for ${\mathsf{\Omega }}_1\subset \mathsf{\Omega }$ and $f\in \mathcal{H}$.

In this section, we study some inequalities for canonical dual continuous generalized frames. Applying operator theory method to these two theorems, we obtain some parametric inequalities. For this purpose, we first give some properties of bounded and positive operators.

Lemma 1 ([27]).

Let $T,D\in L(\mathcal{H})$ be positive and self-adjoint operators satisfying $DT=TD$. Then $DT$ is a positive operator.

Proposition 3.

Let $P,Q,T\in L(\mathcal{H})$ and T be positive and invertible operator and satisfy $P+Q=T$. Then

$$\begin{array}{c}\hfill P^{*}{T}^{-1}P+\frac{\lambda }{2}(Q+Q^{*})=Q^{*}{T}^{-1}Q+(1-\frac{\lambda }{2})(P+P^{*})+(1-\lambda )T\ge (\lambda -\frac{{\lambda }^2}{4})T\end{array}$$

(9)

for $\lambda \in \mathbb{R}$.

$$\begin{array}{c}\hfill P^{*}P+\frac{\lambda }{2}(Q^{*}+Q)=Q^{*}Q=(1-\frac{\lambda }{2})(P^{*}+P)+(1-\lambda )I_{\mathcal{H}}\ge (\lambda -\frac{{\lambda }^2}{4})I_{\mathcal{H}}\end{array}$$

(10)

if $T=I_{\mathcal{H}}$.

Proof. 

Note that $T\in L(\mathcal{H})$ is a positive and invertible operator. Then $T^{-1}$ is positive and invertible. Applying ([4], Lemma 2.4.5 ) to T and $T^{-1}$, we see that there exist unique positive and self-adjoint operators $T^{\frac{1}{2}}$ and $T^{-\frac{1}{2}}$ such that ${(T^{\frac{1}{2}})}^2=T$ and ${(T^{-\frac{1}{2}})}^2=T^{-1}$, respectively. Since $P,Q\in L(\mathcal{H})$ and $P+Q=T$, on the one hand,

$$\begin{array}{cc}\hfill P^{*}{T}^{-1}P+\frac{\lambda }{2}(Q^{*}+Q)& =P^{*}{T}^{-1}P+\frac{\lambda }{2}(T-P^{*}+T-P)\hfill \\ & =P^{*}{T}^{-1}P-\frac{\lambda }{2}(P^{*}+P)+\lambda T\hfill \\ & =(T^{-\frac{1}{2}}P)*(T^{-\frac{1}{2}}P)-\frac{\lambda }{2}(P^{*}+P)+\lambda T^{\frac{1}{2}}{T}^{\frac{1}{2}}\hfill \\ & =(T^{-\frac{1}{2}}P-\frac{\lambda }{2}{T}^{\frac{1}{2}})*(T^{-\frac{1}{2}}P-\frac{\lambda }{2}{T}^{\frac{1}{2}})+\lambda T^{\frac{1}{2}}{T}^{\frac{1}{2}}-\frac{{\lambda }^2}{4}{T}^{\frac{1}{2}}{T}^{\frac{1}{2}}\hfill \\ & \ge (\lambda -\frac{{\lambda }^2}{4})T.\hfill \end{array}$$

.

On the other hand,

$$\begin{array}{c}{Q}^{*}{T}^{-1}Q+(1-\frac{\lambda }{2})(P^{*}+P)+(1-\lambda )T\hfill \\ ={(T-P)}^{*}{T}^{-1}(T-P)+(1-\frac{\lambda }{2})(P+P^{*})+(1-\lambda )T\hfill \\ =T-(P^{*}+P)+P^{*}{T}^{-1}P+(1-\frac{\lambda }{2})(P^{*}+P)+(\lambda -1)T\hfill \\ =P^{*}{T}^{-1}P-\frac{\lambda }{2}(P^{*}+P)+\lambda T.\hfill \end{array}$$

Obviously, if $T=I_{\mathcal{H}}$, we obtain

$$P^{*}P+\frac{\lambda }{2}(Q+Q^{*})=Q^{*}Q=(1-\frac{\lambda }{2})(P+P^{*})+(1-\lambda )I_{\mathcal{H}}\ge (\lambda -\frac{{\lambda }^2}{4})I_{\mathcal{H}}.$$

This finishes the proof. ☐

The following corollary is an immediate consequence.

Corollary 1.

Suppose that the operators $P,Q\in L(\mathcal{H})$ are self-adjoint, and satisfy P plus Q is $I_{\mathcal{H}}$. Then

$${\parallel Pf\parallel }^2+\lambda \langle Qf,f\rangle ={\parallel Qf\parallel }^2+(2-\lambda )\langle Pf,f\rangle +{(\lambda -1)\parallel f\parallel }^2\ge (\lambda -\frac{{\lambda }^2}{4}){\parallel f\parallel }^2$$

for every $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$.

Remark 2.

Corollary 1 is Lemma 2.1 in [16]. Taking $2\lambda$ instead of λ in Corollary 1, we can get Proposition 3.4 in [15].

Theorem 3.

Let ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ be a continuous generalized frame for $\mathcal{H}$. S is the corresponding frame operator. Then for $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$, for $\Xi \subset \mathsf{\Omega }$ we have

$$\begin{array}{c}\left(\lambda -\frac{{\lambda }^2}{4}\right){\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+\left(1-\frac{{\lambda }^2}{4}\right){\int }_{{\Xi }^c}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega )\hfill \\ \le {\int }_{{\Xi }^c}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }f\parallel }^{2}d\mu (\omega )\hfill \\ ={\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\Xi }^c}{f\parallel }^{2}d\mu (\omega )\le {\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ).\hfill \end{array}$$

Proof. 

Noting that $S=S_{\Xi }+S_{{\Xi }^c}$, we have

$$I_{\mathcal{H}}=S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}+S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}.$$

Taking $P=S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}$ and $Q=S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}$ and considering Corollary 1 with $S^{\frac{1}{2}}f$ instead of f, we obtain

$${∥P{S}^{\frac{1}{2}}f∥}^2+\lambda 〈Q{S}^{\frac{1}{2}}f,S^{\frac{1}{2}}f〉=〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+\lambda 〈S_{{\Xi }^c}f,f〉,$$

and

$$\begin{array}{c}{∥Q{S}^{\frac{1}{2}}f∥}^2+(2-\lambda )〈P{S}^{\frac{1}{2}}f,S^{\frac{1}{2}}f〉+(\lambda -1){∥S^{\frac{1}{2}}f∥}^2\hfill \\ =〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+(2-\lambda )〈S_{\Xi }f,f〉+(\lambda -1)〈Sf,f〉.\hfill \end{array}$$

Then

$$\begin{array}{c}\hfill 〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+\lambda 〈S_{{\Xi }^c}f,f〉=〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+(2-\lambda )〈S_{\Xi }f,f〉+(\lambda -1)〈Sf,f〉\end{array}$$

(11)

by Corollary 1. Subtracting $\lambda 〈S_{{\Xi }^c}f,f〉$ from both sides of (11),

$$\begin{array}{cc}\hfill 〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉& =〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+(2-\lambda )〈S_{\Xi }f,f〉+(\lambda -1)〈Sf,f〉-\lambda 〈S_{{\Xi }^c}f,f〉\hfill \\ & =〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+2〈S_{\Xi }f,f〉-\lambda 〈(S_{\Xi }+S_{{\Xi }^c})f,f〉+(\lambda -1)〈Sf,f〉\hfill \\ & =〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+2〈S_{\Xi }f,f〉-〈Sf,f〉\hfill \\ & =〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+2〈S_{\Xi }f,f〉-〈(S_{\Xi }+S_{{\Xi }^c})f,f〉\hfill \\ & =〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+〈S_{\Xi }f,f〉-〈S_{{\Xi }^c}f,f〉.\hfill \end{array}$$

And thus

$$\begin{array}{c}\hfill 〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+〈S_{{\Xi }^c}f,f〉=〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+〈S_{\Xi }f,f〉forf\in \mathcal{H}.\end{array}$$

(12)

By the definition of frame operator S, we have

$$\begin{array}{c}\hfill 〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉=〈S^{-1}S(S^{-1}{S}_{\Xi }f),S_{\Xi }f〉=〈S(S^{-1}{S}_{\Xi }f),S^{-1}(S_{\Xi }f)〉={\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }f\parallel }^{2}d\mu (\omega ).\end{array}$$

(13)

Similarly, we get

$$〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉={\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\Xi }^c}f\parallel }^{2}d\mu (\omega ).$$

(14)

$$\begin{array}{c}\hfill 〈S_{\Xi }f,f〉={\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega ),〈S_{{\Xi }^c}f,f〉={\int }_{{\Xi }^c}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ).\end{array}$$

(15)

Combining (12)–(15), we obtain

$${\int }_{{\Xi }^c}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }{f\parallel }^{2}d\mu (\omega )={\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\Xi }^c}f\parallel }^{2}d\mu (\omega ).$$

Next we prove the first “inequality" part. Using $S^{\frac{1}{2}}f$ instead of f in Corollary 1 and Combining (11), we have

$$〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+\lambda 〈S_{{\Xi }^c}f,f〉\ge (\lambda -\frac{{\lambda }^2}{4})\langle Sf,f\rangle .$$

Thus

$$\begin{array}{cc}\hfill 〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉& \ge (\lambda -\frac{{\lambda }^2}{4})\langle Sf,f\rangle -\lambda 〈S_{{\Xi }^c}f,f〉\hfill \\ & =\lambda \langle Sf,f\rangle -\lambda 〈S_{{\Xi }^c}f,f〉-\frac{{\lambda }^2}{4}\langle Sf,f\rangle \hfill \\ & =\lambda 〈S_{\Xi }f,f〉-\frac{{\lambda }^2}{4}\langle S_{\Xi }f,f\rangle -\frac{{\lambda }^2}{4}\langle S_{{\Xi }^c}f,f\rangle \hfill \\ & =(\lambda -\frac{{\lambda }^2}{4})\langle S_{\Xi }f,f\rangle +(1-\frac{{\lambda }^2}{4})\langle S_{{\Xi }^c}f,f\rangle -\langle S_{{\Xi }^c}f,f\rangle .\hfill \end{array}$$

It follows that

$$\begin{array}{c}\hfill 〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+\langle S_{{\Xi }^c}f,f\rangle \ge (\lambda -\frac{{\lambda }^2}{4})\langle S_{\Xi }f,f\rangle +(1-\frac{{\lambda }^2}{4})\langle S_{{\Xi }^c}f,f\rangle .\end{array}$$

(16)

Again combining (12)–(15), we finish the proof of the first inequality.

Now we prove the last inequality. Since $P=S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}$ and $Q=S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}$ are positive operators and $P+Q=I_{\mathcal{H}}$, we have

$$PQ=P(I_{\mathcal{H}}-P)=P-P^2=(I_{\mathcal{H}}-P)P=QP.$$

By Lemma 1, it is easy to get

$$0\le PQ=P(I_{\mathcal{H}}-P)=P-P^2.$$

Hence $Q+PP\le I_{\mathcal{H}}$. And it follows that

$$S_{{\Xi }^c}+S_{\Xi }{S}^{-1}{S}_{\Xi }\le S.$$

By (13) and (14), for $f\in \mathcal{H}$ we have

$$\begin{array}{c}{\int }_{{\Xi }^c}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }f\parallel }^{2}d\mu (\omega )\hfill \\ =\langle S_{{\Xi }^c}f,f\rangle +\langle S^{-1}{S}_{\Xi }f,S_{\Xi }f\rangle \le \langle Sf,f\rangle ={\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ).\hfill \end{array}$$

The proof is completed. ☐

Corollary 2.

Assume that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a Parseval continuous generalized frame for $\mathcal{H}$. Then

$$\begin{array}{c}\left(1-\frac{{\lambda }^2}{4}\right)\langle S_{{\Xi }^c}f,f\rangle +\left(\lambda -\frac{{\lambda }^2}{4}\right)\langle S_{\Xi }f,f\rangle \hfill \\ \le \langle S_{{\Xi }^c}f,f\rangle +\parallel S_{\Xi }{f\parallel }^2=\langle S_{\Xi }f,f\rangle +\parallel S_{{\Xi }^c}{f\parallel }^2\le {\parallel f\parallel }^2\hfill \end{array}$$

for $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$, for all $\Xi \subset \mathsf{\Omega }$.

Corollary 3.

Assume that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a tight continuous generalized frame for $\mathcal{H}$ with bound A. Then for all $\Xi \subset \mathsf{\Omega }$ we have

$$\begin{array}{c}\left(\lambda -\frac{{\lambda }^2}{4}\right)\langle S_{\Xi }f,f\rangle +\left(1-\frac{{\lambda }^2}{4}\right)\langle S_{{\Xi }^c}f,f\rangle \hfill \\ \le \langle S_{{\Xi }^c}f,f\rangle +\frac{1}{A}\parallel S_{\Xi }{f\parallel }^2=\langle S_{\Xi }f,f\rangle +\frac{1}{A}\parallel S_{{\Xi }^c}{f\parallel }^2\le A{\parallel f\parallel }^2\hfill \end{array}$$

for $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$.

Remark 3.

If we take $\lambda =\frac{1}{2}$ in Theorem 3. Then we can get (3.10) in Theorem 3.3 in [17] and Theorem 3.2 in [18]. And we give another proof of the right inequality of (3.10) in Theorem 3.3 in [17].

Theorem 4.

Let ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ be a continuous generalized frame for $\mathcal{H}$. S is the corresponding frame operator. Then for any $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$, for all $\Xi \subset \mathsf{\Omega }$ we have

$$\begin{array}{c}0\le {\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )-{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }f\parallel }^{2}d\mu (\omega )\hfill \\ \le (\lambda -1){\int }_{{\Xi }^c}\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+(1-\frac{\lambda }{2}){\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ).\hfill \end{array}$$

(17)

$$\begin{array}{c}(2\lambda -1-\frac{{\lambda }^2}{2}){\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+(1-\frac{{\lambda }^2}{2}){\int }_{{\Xi }^c}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega )\hfill \\ \le {\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\Xi }^c}{f\parallel }^{2}d\mu (\omega )\le {\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ).\hfill \end{array}$$

(18)

Proof. 

From the proof of Theorem 3, we obtain

$$0\le PQ=P(I_{\mathcal{H}}-P)=P-P^2,$$

where $P=S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}$ and $Q=S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}$ are positive operators and $P+Q=I_{\mathcal{H}}$. It is easy to check that $S_{\Xi }-S_{\Xi }{S}^{-1}{S}_{\Xi }\ge 0$. First we prove (17). We see that

$$\begin{array}{c}{\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )-{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }f\parallel }^{2}d\mu (\omega )\hfill \\ =\langle S_{\Xi }f,f\rangle -\langle S^{-1}{S}_{\Xi }f,S_{\Xi }f\rangle \hfill \\ =\langle S_{\Xi }f,f\rangle -\langle S_{\Xi }{S}^{-1}{S}_{\Xi }f,f\rangle \ge 0\hfill \end{array}$$

for each $f\in \mathcal{H}$.

Next we will prove the right inequality. By (16) we have

$$\begin{array}{c}{\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )-{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }f\parallel }^{2}d\mu (\omega )\hfill \\ =\langle S_{\Xi }f,f\rangle -\langle S^{-1}{S}_{\Xi }f,S_{\Xi }f\rangle \hfill \\ \le \langle S_{\Xi }f,f\rangle -(\lambda -\frac{{\lambda }^2}{4})\langle S_{\Xi }f,f\rangle -(1-\frac{{\lambda }^2}{4})\langle S_{{\Xi }^c}f,f\rangle +\langle S_{{\Xi }^c}f,f\rangle \hfill \\ =\langle S_{\Xi }f,f\rangle -\lambda \langle S_{\Xi }f,f\rangle +\frac{{\lambda }^2}{4}\langle Sf,f\rangle \hfill \\ =(1-\lambda )\langle S_{\Xi }f,f\rangle +\frac{{\lambda }^2}{4}\langle Sf,f\rangle \hfill \\ =(1-\lambda )\langle (S-S_{{\Xi }^c})f,f\rangle +\frac{{\lambda }^2}{4}\langle Sf,f\rangle \hfill \\ =(\lambda -1)\langle S_{{\Xi }^c}f,f\rangle +{(1-\frac{\lambda }{2})}^2\langle Sf,f\rangle .\hfill \end{array}$$

Combining (14) with (15), (17) holds. Finally we will prove (18). From the proof of Theorem 3,

$$\begin{array}{c}\hfill 〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+〈S_{{\Xi }^c}f,f〉=〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+〈S_{\Xi }f,f〉.\end{array}$$

(19)

Using (16), we get

$$〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉+〈S_{\Xi }f,f〉\ge (\lambda -\frac{{\lambda }^2}{4})\langle S_{\Xi }f,f\rangle +(1-\frac{{\lambda }^2}{4})\langle S_{{\Xi }^c}f,f\rangle .$$

Thus

$$\begin{array}{c}\hfill 〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉\ge (\lambda -\frac{{\lambda }^2}{4}-1)\langle S_{\Xi }f,f\rangle +(1-\frac{{\lambda }^2}{4})\langle S_{{\Xi }^c}f,f\rangle .\end{array}$$

(20)

Combining (16) with (20), we obtain

$$\begin{array}{c}{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\Xi }^c}f\parallel }^{2}d\mu (\omega )\hfill \\ =〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉\hfill \\ \ge \lambda 〈S_{\Xi }f,f〉-\frac{{\lambda }^2}{4}\langle Sf,f\rangle +(\lambda -\frac{{\lambda }^2}{4}-1)\langle S_{\Xi }f,f\rangle +(1-\frac{{\lambda }^2}{4})\langle S_{{\Xi }^c}f,f\rangle \hfill \\ =(2\lambda -\frac{{\lambda }^2}{2}-1)\langle S_{\Xi }f,f\rangle +(1-\frac{{\lambda }^2}{2})\langle S_{{\Xi }^c}f,f\rangle \hfill \\ =(2\lambda -1-\frac{{\lambda }^2}{2}){\int }_{\Xi }\parallel {\mathsf{\Lambda }}_{\omega }{f\parallel }^{2}d\mu (\omega )+(1-\frac{{\lambda }^2}{2}){\int }_{{\Xi }^c}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ).\hfill \end{array}$$

Next we prove the right inequality of (18). Note that $P=S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}$ and $Q=S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}$ are positive and self-adjoint operators such that $P+Q=I_{\mathcal{H}}$, we have $0\le P-P^2(0\le Q-Q^2)$,

$${\parallel Pf\parallel }^2\le \langle Pf,f\rangle ,{\parallel Qf\parallel }^2\le \langle Qf,f\rangle forf\in \mathcal{H}.$$

And thus

$$\begin{array}{c}{\int }_{\mathsf{\Omega }}\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{\Xi }{f\parallel }^{2}d\mu (\omega )+{\int }_{\mathsf{\Omega }}{\parallel {\tilde{\mathsf{\Lambda }}}_{\omega }{S}_{{\Xi }^c}f\parallel }^{2}d\mu (\omega )\hfill \\ =〈S^{-1}{S}_{\Xi }f,S_{\Xi }f〉+〈S^{-1}{S}_{{\Xi }^c}f,S_{{\Xi }^c}f〉\hfill \\ =〈S^{-\frac{1}{2}}{S}_{\Xi }f,S^{-\frac{1}{2}}{S}_{\Xi }f〉+〈S^{-\frac{1}{2}}{S}_{{\Xi }^c}f,S^{-\frac{1}{2}}{S}_{{\Xi }^c}f〉\hfill \\ =〈S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}{S}^{\frac{1}{2}}f,S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}{S}^{\frac{1}{2}}f〉\hfill \\ +〈S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}{S}^{\frac{1}{2}}f,S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}{S}^{\frac{1}{2}}f〉\hfill \\ \le 〈S^{-\frac{1}{2}}{S}_{\Xi }{S}^{-\frac{1}{2}}{S}^{\frac{1}{2}}f,S^{\frac{1}{2}}f〉+〈S^{-\frac{1}{2}}{S}_{{\Xi }^c}{S}^{-\frac{1}{2}}{S}^{\frac{1}{2}}f,S^{\frac{1}{2}}f〉\hfill \\ =\langle S_{\Xi }f,f\rangle +\langle S_{{\Xi }^c}f,f\rangle =\langle (S_{\Xi }+S_{{\Xi }^c})f,f\rangle \hfill \\ =\langle Sf,f\rangle ={\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega ).\hfill \end{array}$$

This finishes the proof. ☐

Remark 4.

If we take $\lambda =1$ in Theorem 2, then we can get (3.8) and (3.9) in Theorem 3.3 in [17]. And we give another proof of the right inequality of (3.9) in Theorem 3.3 in [17].

For Parseval continuous generalized frames, we immediately obtain Corollary 4 below.

Corollary 4.

Suppose that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a Parseval continuous generalized frame for $\mathcal{H}$. Then for all $\Xi \subset \mathsf{\Omega }$ we have

$$0\le \langle S_{\Xi }f,f\rangle -\parallel S_{\Xi }{f\parallel }^2\le (\lambda -1)\langle S_{{\Xi }^c}f,f\rangle +{(1-\frac{\lambda }{2})}^2{\parallel f\parallel }^2,$$

and

$$(2\lambda -\frac{{\lambda }^2}{2}-1)\langle S_{\Xi }f,f\rangle +(1-\frac{{\lambda }^2}{2})\langle S_{{\Xi }^c}f,f\rangle \le \parallel S_{\Xi }{f\parallel }^2+\parallel S_{{\Xi }^c}{f\parallel }^2\le {\parallel f\parallel }^2$$

for $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$.

4. Some New Inequalities for Alternate Dual c-g-Frames

In this section we will give some inequalities for alternate dual continuous generalized frames. To this end, some lemmas are established.

Lemma 2.

Let $W,P,Q\in L(\mathcal{V},\mathcal{H})$ satisfy $P+Q=W$, and $T\in L(\mathcal{H})$. Then

$$\begin{array}{c}\hfill W^{*}TP+Q^{*}TQ=Q^{*}TW+P^{*}TP.\end{array}$$

(21)

In addition, if T is positive, for any $f\in \mathcal{H}$, we have

$$\begin{array}{c}Re\langle TPf,Wf\rangle +\langle TQf,Qf\rangle =Re\langle TWf,Qf\rangle +\langle TPf,Pf\rangle \hfill \\ =\frac{3}{4}\parallel T^{\frac{1}{2}}{Wf\parallel }^2+\frac{1}{4}{\parallel T^{\frac{1}{2}}(P-Q)f\parallel }^2.\hfill \end{array}$$

(22)

Proof. 

Noting that $P+Q=W$, by simple calculation we have

$$W^{*}TP-P^{*}TP=(W^{*}-P^{*})TP=Q^{*}T(W-Q)=Q^{*}TW-Q^{*}TQ.$$

Thus (21) holds. For any $f\in \mathcal{H}$,

$$\langle TPf,Wf\rangle +\langle TQf,Qf\rangle =\langle TWf,Qf\rangle +\langle TPf,Pf\rangle .$$

If T is a positive operator, then $\langle TPf,Pf\rangle \ge 0$ and $\langle TQf,Qf\rangle \ge 0$. Thus

$$Re\langle TPf,Wf\rangle +\langle TQf,Qf\rangle =Re\langle TWf,Qf\rangle +\langle TPf,Pf\rangle .$$

Next we prove the “inequality” part. Since T is positive and $P+Q=W$, by ([4], Lemma 2.4.5 ) for any $f\in \mathcal{H}$, we have

$$\begin{array}{c}Re\langle TPf,Wf\rangle =Re\langle T^{\frac{1}{2}}Pf,T^{\frac{1}{2}}Wf\rangle \hfill \\ =\langle T^{\frac{1}{2}}Wf,T^{\frac{1}{2}}Wf\rangle -Re\langle T^{\frac{1}{2}}Qf,T^{\frac{1}{2}}Wf\rangle \hfill \\ =\frac{3}{4}{\parallel T^{\frac{1}{2}}Wf\parallel }^2+\langle \frac{1}{2}{T}^{\frac{1}{2}}Wf,\frac{1}{2}{T}^{\frac{1}{2}}Wf\rangle \hfill \\ -\frac{1}{2}\langle T^{\frac{1}{2}}Qf,T^{\frac{1}{2}}Wf\rangle -\frac{1}{2}\langle T^{\frac{1}{2}}Wf,T^{\frac{1}{2}}Qf\rangle .\hfill \end{array}$$

Thus

$$\begin{array}{cc}\hfill Re\langle TPf,Wf\rangle +\langle TQf,Qf\rangle & =\frac{3}{4}{\parallel T^{\frac{1}{2}}Wf\parallel }^2+{∥\frac{1}{2}{T}^{\frac{1}{2}}Wf-T^{\frac{1}{2}}Qf∥}^2\hfill \\ & =\frac{3}{4}{\parallel T^{\frac{1}{2}}Wf\parallel }^2+\frac{1}{4}{∥T^{\frac{1}{2}}(P-Q)f∥}^2.\hfill \end{array}$$

The proof is completed. ☐

Lemma 3 ([17]).

Assume that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a continuous generalized frame for $\mathcal{H}$. Let ${\left\{{\Gamma }_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ be a dual continuous generalized frame of ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$, and ${\left\{a_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}\in l^{\infty }(\mathsf{\Omega })$. The operator $T_a$ is defined as follows:

$$T_a:\mathcal{H}\to \mathcal{H},T_{a}f={\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )forf\in \mathcal{H},$$

then $T_a$ is bounded and linear,

$$T_a^{*}f={\int }_X{\overline{a}}_{\omega }{\mathsf{\Lambda }}_{\omega }^{*}{\Gamma }_{\omega }fd\mu (\omega ),$$

where $l^{\infty }(\mathsf{\Omega })=\{{\left\{a_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}:\underset{\omega \in \mathsf{\Omega }}{sup}|a_{\omega }|<\infty \}.$

Proposition 4.

Let $P,Q\in \mathcal{H}$ be two linear bounded operator and $P+Q=I_{\mathcal{H}}$. Then

$${\parallel Pf\parallel }^2+Re\langle Qf,f\rangle ={\parallel Qf\parallel }^2+Re\langle Pf,f\rangle +\ge (\lambda -\frac{{\lambda }^2}{4})Re\langle Pf,f\rangle +(1-\frac{{\lambda }^2}{4})Re\langle Qf,f\rangle$$

for $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$.

Proof. 

First, we prove the “equality” part. By (10) in Proposition 3, for $f\in \mathcal{H}$,

$$\begin{array}{c}\langle P^{*}Pf,f\rangle +\frac{\lambda }{2}\langle (Q^{*}+Q)f,f\rangle \hfill \\ =\langle P^{*}Pf,f\rangle +\frac{\lambda }{2}\overline{\langle Qf,f\rangle }+\frac{\lambda }{2}\langle Qf,f\rangle \hfill \\ ={\parallel Pf\parallel }^2+\lambda Re\langle Qf,f\rangle .\hfill \end{array}$$

(23)

$$\begin{array}{c}\langle Q^{*}Qf,f\rangle +(1-\frac{\lambda }{2})\langle (P^{*}+P)f,f\rangle +(\lambda -1){\parallel f\parallel }^2\hfill \\ =\langle Q^{*}Qf,f\rangle +(1-\frac{\lambda }{2})\overline{\langle Pf,f\rangle }+(1-\frac{\lambda }{2})\langle Pf,f\rangle +(\lambda -1){\parallel f\parallel }^2\hfill \\ ={\parallel Qf\parallel }^2+(2-\lambda )Re\langle Pf,f\rangle +(\lambda -1)\langle f,f\rangle .\hfill \end{array}$$

(24)

Combining (23) with (24), we have

$${\parallel Pf\parallel }^2+\lambda Re\langle Qf,f\rangle ={\parallel Qf\parallel }^2+(2-\lambda )Re\langle Pf,f\rangle +(\lambda -1)Re\langle f,f\rangle .$$

Therefore

$$\begin{array}{cc}\hfill {\parallel Pf\parallel }^2& ={\parallel Qf\parallel }^2+(2-\lambda )Re\langle Pf,f\rangle -\lambda Re\langle Qf,f\rangle +(\lambda -1)Re\langle f,f\rangle \hfill \\ & ={\parallel Qf\parallel }^2+2Re\langle Pf,f\rangle -\lambda Re\langle (P+Q)f,f\rangle +(\lambda -1)Re\langle f,f\rangle \hfill \\ & ={\parallel Qf\parallel }^2+2Re\langle Pf,f\rangle -Re\langle f,f\rangle \hfill \\ & ={\parallel Qf\parallel }^2+2Re\langle Pf,f\rangle -Re\langle (P+Q)f,f\rangle \hfill \\ & ={\parallel Qf\parallel }^2+Re\langle Pf,f\rangle -Re\langle Qf,f\rangle ,\hfill \end{array}$$

This implies that

$${\parallel Pf\parallel }^2+Re\langle Qf,f\rangle ={\parallel Qf\parallel }^2+Re\langle Pf,f\rangle .$$

Next we prove the “inequality” part. Again by (10) in Proposition 3, for $f\in \mathcal{H}$,

$$\langle P^{*}Pf,f\rangle +\frac{\lambda }{2}\overline{\langle Qf,f\rangle }+\frac{\lambda }{2}\langle Qf,f\rangle \ge (\lambda -\frac{{\lambda }^2}{4})\langle f,f\rangle .$$

That is

$${\parallel Pf\parallel }^2+\lambda Re\langle Qf,f\rangle \ge (\lambda -\frac{{\lambda }^2}{4})Re\langle f,f\rangle .$$

Thus

$$\begin{array}{cc}\hfill {\parallel Pf\parallel }^2& \ge (\lambda -\frac{{\lambda }^2}{4})Re\langle I_{\mathcal{H}}f,f\rangle -\lambda Re\langle Qf,f\rangle \hfill \\ & =(\lambda -\frac{{\lambda }^2}{4})Re\langle (P+Q)f,f\rangle -\lambda Re\langle Qf,f\rangle \hfill \\ & =(\lambda -\frac{{\lambda }^2}{4})Re\langle Pf,f\rangle -\frac{{\lambda }^2}{4}Re\langle Qf,f\rangle \hfill \\ & =(\lambda -\frac{{\lambda }^2}{4})Re\langle Pf,f\rangle +(1-\frac{{\lambda }^2}{4})Re\langle Qf,f\rangle -Re\langle Qf,f\rangle .\hfill \end{array}$$

The proof is completed. ☐

Remark 5.

Taking $2\lambda$ instead of λ in Proposition 4, we can get Theorm 2.7 in [16].

Theorem 5.

Assume that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a continuous generalized frame for $\mathcal{H}$ with the upper frame bound $B_{\mathsf{\Lambda }}$. Let ${\left\{{\Gamma }_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ be a dual continuous generalized frame of ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ with the upper frame bound $B_{\Gamma }$. Then, for ${\left\{a_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}\in l^{\infty }(\mathsf{\Omega })$, the following hold:

(i)

For $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$,

$$\begin{array}{c}\left(\lambda -\frac{{\lambda }^2}{4}\right)Re\left({\int }_{\mathsf{\Omega }}{a}_{\omega }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)+\left(1-\frac{{\lambda }^2}{4}\right)Re\left({\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\hfill \\ \le {∥{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\hfill \\ ={∥{\int }_{\mathsf{\Omega }}(1-a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\mathsf{\Omega }}{a}_{\omega }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right).\hfill \end{array}$$

(ii)

If $M=\underset{\omega \in \mathsf{\Omega }}{sup}|1-2a_{\omega }|<\infty$, we have

$$\begin{array}{c}\hfill {∥{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\le \frac{3+M^2{B}_{\mathsf{\Lambda }}{B}_{\Gamma }}{4}{\parallel f\parallel }^2\end{array}$$

for $f\in \mathcal{H}$.

Proof. 

By the definition of alternate dual continuous generalized frame, for all $f\in \mathcal{H}$ we have

$$f={\int }_{\mathsf{\Omega }}{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )={\int }_{\mathsf{\Omega }}{\mathsf{\Lambda }}_{\omega }^{*}{\Gamma }_{\omega }fd\mu (\omega ).$$

Define $T_{1-a}$ by

$$T_{1-a}:\mathcal{H}\to \mathcal{H},T_{1-a}f={\int }_{\mathsf{\Omega }}(1-a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega ),\forall f\in \mathcal{H}.$$

By Lemma 3 and a standard argument, $T_{1-a}$ is a linear bounded operator such that $T_a+T_{1-a}=I_{\mathcal{H}}$ and

$$T_{1-a}^{*}f={\int }_{\mathsf{\Omega }}(1-{\overline{a}}_{\omega }){\mathsf{\Lambda }}_{\omega }^{*}{\Gamma }_{\omega }fd\mu (\omega ),\forall f\in \mathcal{H},$$

where ${\overline{a}}_{\omega }$ is the conjugate of $a_{\omega }$. So for any $f\in \mathcal{H}$, we have

$$\begin{array}{c}\hfill {∥{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+{\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )={\parallel T_{a}f\parallel }^2+\langle T_{1-a}f,f\rangle .\end{array}$$

(25)

On the other hand,

$$\begin{array}{c}{∥{\int }_{\mathsf{\Omega }}(1-a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+\overline{{\int }_{\mathsf{\Omega }}{a}_{\omega }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )}\hfill \\ =\parallel T_{1-a}{f\parallel }^2+\overline{\langle T_{a}f,f\rangle }=\langle T_{1-a}f,T_{1-a}f\rangle +\langle f,T_{a}f\rangle \hfill \\ =\langle (I_U-T_a)f,(I_U-T_a)f\rangle +\langle f,T_{a}f\rangle \hfill \\ =\langle f,f\rangle -\langle T_{a}f,f\rangle +\langle T_{a}f,T_{a}f\rangle ={\parallel T_{a}f\parallel }^2+\langle T_{1-a}f,f\rangle .\hfill \end{array}$$

(26)

Combining (25) with (26), we get

$$\begin{array}{c}{∥{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+{\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\hfill \\ ={∥{\int }_{\mathsf{\Omega }}(1-a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+\overline{{\int }_{\mathsf{\Omega }}{a}_{\omega }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )}.\hfill \end{array}$$

Thus

$$\begin{array}{c}{∥{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\hfill \\ ={∥{\int }_{\mathsf{\Omega }}(1-a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\mathsf{\Omega }}{a}_{\omega }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\hfill \\ =\parallel T_a{f\parallel }^2+Re\langle T_{1-a}f,f\rangle .\hfill \end{array}$$

Using Proposition 4, we get

$$\begin{array}{c}\hfill \parallel T_a{f\parallel }^2+Re\langle T_{1-a}f,f\rangle \ge (\lambda -\frac{{\lambda }^2}{4})Re\langle T_{a}f,f\rangle +(1-\frac{{\lambda }^2}{4})Re\langle T_{1-a}f,f\rangle .\end{array}$$

(27)

By a simple calculation, we have

$$\begin{array}{c}\hfill \langle T_{a}f,f\rangle =〈{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega ),f〉={\int }_{\mathsf{\Omega }}{a}_{\omega }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega ).\end{array}$$

(28)

Similarly, we obtain that

$$\begin{array}{c}\hfill \langle T_{1-a}f,f\rangle ={\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega ).\end{array}$$

(29)

Combining (29)–(27), (i) holds.

Now we prove (ii). Since $T_a+T_{1-a}=I_{\mathcal{H}}$. Taking $T=I_{\mathcal{H}},Q=T_{a}andP=T_{1-a}$ in Lemma 2, let f belongs to $\mathcal{H}$, so

$$\begin{array}{c}{∥{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\hfill \\ =\parallel T_a{f\parallel }^2+Re\langle T_{1-a}f,f\rangle ={\parallel T_{1-a}f\parallel }^2+Re\langle f,T_{a}f\rangle \hfill \\ =\frac{3}{4}{\parallel f\parallel }^2+\frac{1}{4}{\parallel (T_a-T_{1-a})f\parallel }^2\hfill \\ =\frac{3}{4}{\parallel f\parallel }^2+\frac{1}{4}{∥{\int }_{\mathsf{\Omega }}(1-2a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2.\hfill \end{array}$$

$$\begin{array}{cc}\hfill {∥{\int }_{\mathsf{\Omega }}(1-2a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2& =\underset{g\in \mathcal{H},\parallel g\parallel =1}{sup}{\left|〈{\int }_{\mathsf{\Omega }}(1-2a_{\omega }){\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega ),g〉\right|}^2\hfill \\ & =\underset{g\in \mathcal{H},\parallel g\parallel =1}{sup}{\left|{\int }_{\mathsf{\Omega }}(1-2a_{\omega })〈{\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }g〉d\mu (\omega )\right|}^2\hfill \\ & =\underset{g\in \mathcal{H},\parallel g\parallel =1}{sup}{\left|{\int }_{\mathsf{\Omega }}|1-2a_{\omega }|\parallel {\mathsf{\Lambda }}_{\omega }f\parallel \parallel {\Gamma }_{\omega }g\parallel d\mu (\omega )\right|}^2\hfill \\ & \le M^2\underset{g\in \mathcal{H},\parallel g\parallel =1}{sup}\left({\int }_{\mathsf{\Omega }}{\parallel {\mathsf{\Lambda }}_{\omega }f\parallel }^{2}d\mu (\omega )\right)\hfill \\ & ·\left({\int }_{\mathsf{\Omega }}{\parallel {\overline{a}}_{\omega }{\Gamma }_{\omega }g\parallel }^{2}d\mu (\omega )\right)\hfill \\ & \le M^2{B}_{\mathsf{\Lambda }}{B}_{\Gamma }{\parallel f\parallel }^2.\hfill \end{array}$$

Thus

$$\begin{array}{c}\hfill {∥{\int }_{\mathsf{\Omega }}{a}_{\omega }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\mathsf{\Omega }}(1-a_{\omega })\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\le \frac{3+M^2{B}_{\mathsf{\Lambda }}{B}_{\Gamma }}{4}{\parallel f\parallel }^2.\end{array}$$

The proof is completed. ☐

Take

$$\begin{array}{c}\hfill a_{\omega }=\left\{1,\omega \in \Xi ;\right.\\ \hfill 0,\omega \in {\Xi }^c\end{array}$$

in Theorem 5, where $\Xi \subset \mathsf{\Omega }$. Then we have

Corollary 5.

Let $\Xi \subset \mathsf{\Omega }$, and assume that ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ is a continuous generalized frame for $\mathcal{H}$ with the upper frame bound $B_{\mathsf{\Lambda }}$. Let ${\left\{{\Gamma }_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ be a dual continuous generalized frame of ${\left\{{\mathsf{\Lambda }}_{\omega }\right\}}_{\omega \in \mathsf{\Omega }}$ with the upper frame bound $B_{\Gamma }$. Then,

$$\begin{array}{c}\left(\lambda -\frac{{\lambda }^2}{4}\right)Re\left({\int }_{\Xi }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)+\left(1-\frac{{\lambda }^2}{4}\right)Re\left({\int }_{{\Xi }^c}\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\hfill \\ \le {∥{\int }_{\Xi }{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{{\Xi }^c}\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\hfill \\ ={∥{\int }_{{\Xi }^c}{\Gamma }_{\omega }^{*}{\mathsf{\Lambda }}_{\omega }fd\mu (\omega )∥}^2+Re\left({\int }_{\Xi }\langle {\mathsf{\Lambda }}_{\omega }f,{\Gamma }_{\omega }f\rangle d\mu (\omega )\right)\le \frac{3+B_{\mathsf{\Lambda }}{B}_{\Gamma }}{4}{\parallel f\parallel }^2\hfill \end{array}$$

for $\lambda \in \mathbb{R}$ and $f\in \mathcal{H}$.