New Inequalities of Weaving K-Frames in Subspaces

Abstract

In the present paper, we obtain some new inequalities for weaving K-frames in subspaces based on the operator methods. The inequalities are associated with a sequence of bounded complex numbers and a parameter $\lambda \in \mathbb{R}$. We also give a double inequality for weaving K-frames with the help of two bounded linear operators induced by K-dual. Facts prove that our results cover those recently obtained on weaving frames due to Li and Leng, and Xiang.

1. Introduction

This paper adopts the following notations: $\mathbb{J}$ is a countable index set, $\mathbb{H}$ and $\mathbb{K}$ are complex Hilbert spaces, and ${\mathrm{Id}}_{\mathbb{H}}$ and $\mathbb{R}$ are used to denote respectively the identical operator on $\mathbb{H}$ and the set of real numbers. As usual, we denote by $B(\mathbb{H},\mathbb{K})$ the set of all bounded linear operators on $\mathbb{H}$ and, if $\mathbb{H}=\mathbb{K}$, then $B(\mathbb{H},\mathbb{K})$ is abbreviated to $B\left(\mathbb{H}\right)$.

Frames were introduced by Duffin and Schaeffer [1] in their study of nonharmonic Fourier series, which have now been used widely not only in theoretical work [2,3], but also in many application areas such as quantum mechanics [4], sampling theory [5,6,7], acoustics [8], and signal processing [9]. As a generalization of frames, the notion of K-frames (also known as frames for operators) was proposed by L. Găvruţa [10] when dealing with atomic decompositions for a bounded linear operator K. Please check the papers [11,12,13,14,15,16,17] for further information of K-frames.

Recall that a family ${\left\{{\psi }_j\right\}}_{j\in \mathbb{J}}\subset \mathbb{H}$ is called a K-frame for $\mathbb{H}$, if there exist two positive numbers A and B satisfying

$$A\parallel K^{\ast }{f\parallel }^2\le \sum _{j\in \mathbb{J}}|\langle f,{\psi }_j\rangle {|}^2\le B{\parallel f\parallel }^2,\forall f\in \mathbb{H}.$$

The constants A and B are called K-frame bounds. If $K={\mathrm{Id}}_{\mathbb{H}}$, then a K-frame turns to be a frame. In addition, if only the right-hand inequality holds, then we call ${\left\{{\psi }_j\right\}}_{j\in \mathbb{J}}$ a Bessel sequence.

Inspired by a question arising in distributed signal processing, Bemrose et al. [18] introduced the concept of weaving frames, which have interested many scholars because of their potential applications such as in wireless sensor networks and pre-processing of signals; see [19,20,21,22,23,24]. Later on, Deepshikha and Vashisht [25] applied the idea of L. Găvruţa to the case of weaving frames and thus providing us the notion of weaving K-frames.

Balan et al. [26] obtained an interesting inequality when they further examined the remarkable identity for Parseval frames deriving from their work on signal reconstruction [27]. The inequality was then extended to alternate dual frames and general frames by P. Găvruţa [28], the results in which have already been applied in quantum information theory [29]. Recently, those inequalities have been extended to some generalized versions of frames such as continuous g-frames [30], fusion frames and continuous fusion frames [31,32], Hilbert–Schmidt frames [33], and weaving frames [34,35].

Motivated by the above-mentioned works, in this paper, we establish several new inequalities for weaving K-frames in subspaces from the operator-theoretic point of view, and we show that our results can naturally lead to some corresponding results in [34,35].

One says that two frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are woven, if there are universal constants $C_{\mathsf{\Psi }}$ and $D_{\mathsf{\Psi }}$ such that, for any $\sigma \subset \mathbb{J}$, ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$ is a frame for $\mathbb{H}$ with bounds $C_{\mathsf{\Psi }}$ and $D_{\mathsf{\Psi }}$. If $C_{\mathsf{\Psi }}=D_{\mathsf{\Psi }}=1$, then we call ${\mathsf{\Psi }}_1$ and ${\mathsf{\Psi }}_2$ 1-woven. Each family ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$ is said to be a waving frame, related to which there is an invertible operator $S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}:\mathbb{H}\to \mathbb{H}$, called the frame operator, given by

$$S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}f=\sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle {\psi }_{1j}+\sum _{j\in {\sigma }^c}\langle f,{\psi }_{2j}\rangle {\psi }_{2j}.$$

Recall also that a frame ${\mathsf{\Psi }}_3={\left\{{\psi }_{3j}\right\}}_{j\in \mathbb{J}}$ is called an alternate dual frame of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$, if for each $f\in \mathbb{H}$ we have

$$f=\sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle {\psi }_{3j}+\sum _{j\in {\sigma }^c}\langle f,{\psi }_{2j}\rangle {\psi }_{3j},\forall f\in \mathbb{H}.$$

Lemma 1.

Suppose that P, Q, and K are bounded linear operators on $\mathbb{H}$ and $P+Q=K$. Then, for each $f\in \mathbb{H}$,

$${\parallel Pf\parallel }^2+\mathrm{Re}\langle Qf,Kf\rangle \ge \frac{3}{4}{\parallel Kf\parallel }^2.$$

Proof. 

We have

$$\begin{array}{cc}\hfill {\parallel Pf\parallel }^2+\mathrm{Re}\langle Qf,Kf\rangle & =\langle (K-Q)f,(K-Q)f\rangle +\frac{1}{2}(\langle Qf,Kf\rangle +\langle Kf,Qf\rangle )\hfill \\ \hfill & =\langle (Q^{\ast }Q-(K^{\ast }Q+Q^{\ast }K)+\frac{1}{2}(K^{\ast }Q+Q^{\ast }K))f,f\rangle +\langle K^{\ast }Kf,f\rangle \hfill \\ \hfill & =\langle {(Q-\frac{1}{2}K)}^{\ast }(Q-\frac{1}{2}K)f,f\rangle +\frac{3}{4}\langle K^{\ast }Kf,f\rangle \ge \frac{3}{4}{\parallel Kf\parallel }^2\hfill \end{array}$$

for any $f\in \mathbb{H}$. □

The next two lemmas are collected from the papers [36] and [32], respectively.

Lemma 2.

If $\mathsf{\Phi }\in B(\mathbb{H},\mathbb{K})$ has a closed range, then there is the pseudo-inverse ${\mathsf{\Phi }}^{†}\in B(\mathbb{K},\mathbb{H})$ of Φ such that

$$\mathsf{\Phi }{\mathsf{\Phi }}^{†}\mathsf{\Phi }=\mathsf{\Phi },{\mathsf{\Phi }}^{†}\mathsf{\Phi }{\mathsf{\Phi }}^{†}={\mathsf{\Phi }}^{†},{\left(\mathsf{\Phi }{\mathsf{\Phi }}^{†}\right)}^{\ast }=\mathsf{\Phi }{\mathsf{\Phi }}^{†},{\left({\mathsf{\Phi }}^{†}\mathsf{\Phi }\right)}^{\ast }={\mathsf{\Phi }}^{†}\mathsf{\Phi }.$$

Lemma 3.

If P and Q in $B\left(\mathbb{H}\right)$ satisfy $P+Q={\mathrm{Id}}_{\mathbb{H}}$, then, for any $\lambda \in \mathbb{R}$, we have

$$P^{\ast }P+\lambda (Q^{\ast }+Q)=Q^{\ast }Q+(1-\lambda )(P^{\ast }+P)+(2\lambda -1){\mathrm{Id}}_{\mathbb{H}}\ge (2\lambda -{\lambda }^2){\mathrm{Id}}_{\mathbb{H}}.$$

2. Main Results

We start with the definition on weaving K-frames due to Deepshikha and Vashisht [25].

Definition 1.

Two K-frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are said to be K-woven, if there are universal constants $C_{\mathsf{\Psi }}$ and $D_{\mathsf{\Psi }}$ such that, for any $\sigma \subset \mathbb{J}$, the family ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$ is a K-frame for $\mathbb{H}$ with K-frame bounds $C_{\mathsf{\Psi }}$ and $D_{\mathsf{\Psi }}$. In this case, the family ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$ is called a weaving K-frame.

Given a weaving K-frame ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$ for $\mathbb{H}$, recall that a Bessel sequence $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ for $\mathbb{H}$ is said to be a K-dual of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$, if

$$Kf=\sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}\langle f,{\psi }_{2j}\rangle {\varphi }_j,\forall f\in \mathbb{H}.$$

Let ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ be a given K-frame for $\mathbb{H}$. For any $\sigma \subset \mathbb{J}$, we can define a positive operator $S_{{\mathsf{\Psi }}_1}^{\sigma }$ in the following way:

$$S_{{\mathsf{\Psi }}_1}^{\sigma }:\mathbb{H}\to \mathbb{H},S_{{\mathsf{\Psi }}_1}^{\sigma }f=\sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle {\psi }_{1j}.$$

In the following, we show that, for given two K-woven frames, we can get some inequalities under the condition that K has a closed range, which are related to a sequence of bounded complex numbers, the corresponding K-dual and a parameter $\lambda \in \mathbb{R}$.

Theorem 1.

Suppose that $K\in B\left(\mathbb{H}\right)$ has a closed range and K-frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are K-woven. Then,

(i) for any $f\in Range\left(K\right)$, for all $\sigma \subset \mathbb{J}$, ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$, and $\lambda \in \mathbb{R}$,

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ =\parallel \sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ +(1-\frac{{\lambda }^2}{4})\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle ),\hfill \end{array}$$

(1)

where $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ is a K-dual of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$.

(ii) for any $f\in Range\left(K^{\ast }\right)$, for all $\sigma \subset \mathbb{J}$, ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$, and $\lambda \in \mathbb{R}$,

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle {\psi }_{1j}+\sum _{j\in {\sigma }^c}{a}_j\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle {\psi }_{2j}{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{2j},f\rangle )\hfill \\ =\parallel \sum _{j\in \sigma }(1-a_j)\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle {\psi }_{1j}+\sum _{j\in {\sigma }^c}(1-a_j)\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle {\psi }_{2j}{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{2j},f\rangle )\hfill \\ \ge (2\lambda -{\lambda }^2)\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{2j},f\rangle )\hfill \\ +(1-{\lambda }^2)\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle {\left(K^{\ast }\right)}^{†}f,{\varphi }_j\rangle \langle {\psi }_{2j},f\rangle ),\hfill \end{array}$$

where $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ is a K-dual of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$.

Proof. 

We define two bounded linear operators $P_1$ and $P_2$ on $\mathbb{H}$ as follows:

$$\begin{array}{c}\hfill P_{1}f=\sum _{j\in \sigma }{a}_j\langle f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle f,{\psi }_{2j}\rangle {\varphi }_j,\\ \hfill P_{2}f=\sum _{j\in \sigma }(1-a_j)\langle f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}(1-a_j)\langle f,{\psi }_{2j}\rangle {\varphi }_j.\end{array}$$

(2)

Then, clearly, $P_{1}f+P_{2}f=Kf$ for each $f\in \mathbb{H}$ and thus $P_1+P_2=K$. Since K has a closed range, by Lemma 2, we have

$$P_1{K}^{†}+P_2{K}^{†}=K{K}^{†}=P_{Range\left(K\right)},$$

where $P_{Range\left(K\right)}$ is the orthogonal projection onto $Range\left(K\right)$. Thus,

$$P_1{K}^{†}{\mid }_{Range\left(K\right)}+P_2{K}^{†}{\mid }_{Range\left(K\right)}={\mathrm{Id}}_{Range\left(K\right)}.$$

By Lemma 3 (taking $\frac{\lambda }{2}$ instead of $\lambda$), we get

$$\parallel P_1{K}^{†}{f\parallel }^2+\lambda \mathrm{Re}\langle P_2{K}^{†}f,f\rangle =\parallel P_2{K}^{†}{f\parallel }^2+(2-\lambda )\mathrm{Re}\langle P_1{K}^{†}f,f\rangle +(\lambda -1){\parallel f\parallel }^2,$$

for any $f\in Range\left(K\right)$. Hence,

$$\begin{array}{cc}\hfill \parallel P_1{K}^{†}{f\parallel }^2& =\parallel P_2{K}^{†}{f\parallel }^2+2\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -\lambda (\mathrm{Re}\langle P_1{K}^{†}f,f\rangle +\mathrm{Re}\langle P_2{K}^{†}f,f\rangle )+(\lambda -1){\parallel f\parallel }^2\hfill \\ \hfill & =\parallel P_2{K}^{†}{f\parallel }^2+2\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -{\lambda \parallel f\parallel }^2+(\lambda -1){\parallel f\parallel }^2\hfill \\ \hfill & =\parallel P_2{K}^{†}{f\parallel }^2+2\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -\mathrm{Re}\langle P_2{K}^{†}f,f\rangle .\hfill \end{array}$$

It follows that

$$\parallel P_1{K}^{†}{f\parallel }^2+\mathrm{Re}\langle P_2{K}^{†}f,f\rangle ={\parallel P_2{K}^{†}f\parallel }^2+\mathrm{Re}\langle P_1{K}^{†}f,f\rangle ,$$

(3)

from which we arrive at

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ =\parallel \sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle ).\hfill \end{array}$$

For the inequality in Equation (1), we apply Lemma 3 again,

$$\begin{array}{cc}\hfill \parallel P_1{K}^{†}{f\parallel }^2& \ge (\lambda -\frac{{\lambda }^2}{4}){\parallel f\parallel }^2-\lambda \mathrm{Re}\langle P_2{K}^{†}f,f\rangle \hfill \\ \hfill & =(\lambda -\frac{{\lambda }^2}{4})\mathrm{Re}\langle P_1{K}^{†}f+P_2{K}^{†}f,f\rangle -\lambda \mathrm{Re}\langle P_2{K}^{†}f,f\rangle \hfill \\ \hfill & =(\lambda -\frac{{\lambda }^2}{4})\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -\frac{{\lambda }^2}{4}\mathrm{Re}\langle P_2{K}^{†}f,f\rangle .\hfill \end{array}$$

(4)

Thus, for any $f\in Range\left(K\right)$,

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\mathrm{Re}\langle P_1{K}^{†}f,f\rangle +(1-\frac{{\lambda }^2}{4})\mathrm{Re}\langle P_2{K}^{†}f,f\rangle \hfill \\ =(\lambda -\frac{{\lambda }^2}{4})\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ +(1-\frac{{\lambda }^2}{4})\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle ).\hfill \end{array}$$

(ii) The proof is similar to (i), so we omit the details. □

Corollary 1.

Suppose that two frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are woven. Then, for any $f\in \mathbb{H}$, for all $\sigma \subset \mathbb{J}$ and all $\lambda \in \mathbb{R}$, we have

$$\begin{array}{c}\sum _{j\in {\sigma }^c}|\langle f,{\psi }_{2j}\rangle {|}^2+\sum _{j\in \sigma }|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{1j}\rangle {|}^2+\sum _{j\in {\sigma }^c}{|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{2j}\rangle |}^2\hfill \\ =\sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2+\sum _{j\in \sigma }|\langle S_{{\mathsf{\Psi }}_2}^{{\sigma }^c}f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{1j}\rangle {|}^2+\sum _{j\in {\sigma }^c}{|\langle S_{{\mathsf{\Psi }}_2}^{{\sigma }^c}f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{2j}\rangle |}^2\hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2+(1-\frac{{\lambda }^2}{4})\sum _{j\in {\sigma }^c}{|\langle f,{\psi }_{2j}\rangle |}^2.\hfill \end{array}$$

Proof. 

Letting $K^{†}={\mathrm{Id}}_{\mathbb{H}}$ and

$${\varphi }_j=\left\{\begin{array}{cc}{S}_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j},\hfill & j\in \sigma ,\hfill \\ S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j},\hfill & j\in {\sigma }^c.\hfill \end{array}\right.$$

In addition, taking $S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j}$, $S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j}$ and $S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{1/2}f$ instead of ${\psi }_{1j}$, ${\psi }_{2j}$ and f respectively in (i) of Theorem 1 leads to

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle f,{\psi }_{1j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j}+\sum _{j\in {\sigma }^c}{a}_j\langle f,{\psi }_{2j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j}{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle f,{\psi }_{1j}\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle f,{\psi }_{2j}\rangle \langle {\psi }_{2j},f\rangle )\hfill \\ =\parallel \sum _{j\in \sigma }(1-a_j)\langle f,{\psi }_{1j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j}+\sum _{j\in {\sigma }^c}(1-a_j)\langle f,{\psi }_{2j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j}{\parallel }^2\hfill \\ +\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle f,{\psi }_{1j}\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle f,{\psi }_{2j}\rangle \langle {\psi }_{2j},f\rangle )\hfill \\ \ge (\lambda -\frac{{\lambda }^2}{4})\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle f,{\psi }_{1j}\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle f,{\psi }_{2j}\rangle \langle {\psi }_{2j},f\rangle )\hfill \\ +(1-\frac{{\lambda }^2}{4})\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle f,{\psi }_{1j}\rangle \langle {\psi }_{1j},f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle f,{\psi }_{2j}\rangle \langle {\psi }_{2j},f\rangle ).\hfill \end{array}$$

(5)

A direction calculation shows that

$$\begin{array}{cc}\hfill \parallel \sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j}{\parallel }^2& =\parallel S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}\sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle {\psi }_{1j}{\parallel }^2={\parallel S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f\parallel }^2\hfill \\ \hfill & =\langle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f\rangle =\langle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}{S}_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f\rangle \hfill \\ \hfill & =\sum _{j\in \sigma }\langle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f,{\psi }_{1j}\rangle \langle {\psi }_{1j},S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f\rangle +\sum _{j\in {\sigma }^c}\langle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f,{\psi }_{2j}\rangle \langle {\psi }_{2j},S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{S}_{{\mathsf{\Psi }}_1}^{\sigma }f\rangle \hfill \\ \hfill & =\sum _{j\in \sigma }|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{1j}\rangle {|}^2+\sum _{j\in {\sigma }^c}{|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{2j}\rangle |}^2,\hfill \end{array}$$

(6)

and, similarly,

$$\parallel \sum _{j\in {\sigma }^c}\langle f,{\psi }_{2j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j}{\parallel }^2=\sum _{j\in \sigma }|\langle S_{{\mathsf{\Psi }}_2}^{{\sigma }^c}f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{1j}\rangle {|}^2+\sum _{j\in {\sigma }^c}{|\langle S_{{\mathsf{\Psi }}_2}^{{\sigma }^c}f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{2j}\rangle |}^2.$$

(7)

Thus, the result follows if, in Equation (5), we take $a_j=\left\{\begin{array}{cc}1,\hfill & j\in \sigma ,\hfill \\ 0,\hfill & j\in {\sigma }^c.\hfill \end{array}\right.$ □

Corollary 2.

Suppose that two frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are woven. Then, for any $\sigma \subset \mathbb{J}$, for all $\lambda \in \mathbb{R}$ and all $f\in \mathbb{H}$, we have

$$\begin{array}{c}\parallel \sum _{j\in \sigma }\langle f,{\varphi }_j\rangle {\psi }_{1j}{\parallel }^2+\mathrm{Re}\sum _{j\in {\sigma }^c}\langle f,{\varphi }_j\rangle \langle {\psi }_{2j},f\rangle \hfill \\ =\parallel \sum _{j\in {\sigma }^c}\langle f,{\varphi }_j\rangle {\psi }_{2j}{\parallel }^2+\mathrm{Re}\sum _{j\in \sigma }\langle f,{\varphi }_j\rangle \langle {\psi }_{1j},f\rangle \hfill \\ \ge (2\lambda -{\lambda }^2)\mathrm{Re}\sum _{j\in \sigma }\langle f,{\varphi }_j\rangle \langle {\psi }_{1j},f\rangle +(1-{\lambda }^2)\mathrm{Re}\sum _{j\in {\sigma }^c}\langle f,{\varphi }_j\rangle \langle {\psi }_{2j},f\rangle ,\hfill \end{array}$$

where $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ is an alternate dual of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$.

Proof. 

The result follows immediately from (ii) in Theorem 1 when taking $K^{†}={\mathrm{Id}}_{\mathbb{H}}$ and

$$a_j=\left\{\begin{array}{cc}1,\hfill & j\in \sigma ,\hfill \\ 0,\hfill & j\in {\sigma }^c.\hfill \end{array}\right.$$

 □

Suppose that two frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are 1-woven. For any $\sigma \subset \mathbb{J}$ and any $j\in \mathbb{J}$, taking ${\varphi }_j=\left\{\begin{array}{cc}{\psi }_{1j},\hfill & j\in \sigma ,\hfill \\ {\psi }_{2j},\hfill & j\in {\sigma }^c.\hfill \end{array}\right.$ Then, obviously, $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ is an alternate dual of the frame ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$. Thus, Corollary 2 provides us a direct consequence as follows.

Corollary 3.

Let the two frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ be 1-woven. Then, for any $\sigma \subset \mathbb{J}$, for all $\lambda \in \mathbb{R}$ and all $f\in \mathbb{H}$, we have

$$\begin{array}{cc}\hfill \parallel \sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle {\psi }_{1j}{\parallel }^2+\sum _{j\in {\sigma }^c}{|\langle f,{\psi }_{2j}\rangle |}^2& =\parallel \sum _{j\in {\sigma }^c}\langle f,{\psi }_{2j}\rangle {\psi }_{2j}{\parallel }^2+\sum _{j\in \sigma }{|\langle f,{\psi }_{1j}\rangle |}^2\hfill \\ \hfill & \ge (2\lambda -{\lambda }^2)\sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2+(1-{\lambda }^2)\sum _{j\in {\sigma }^c}{|\langle f,{\psi }_{2j}\rangle |}^2.\hfill \end{array}$$

Remark 1.

Corollaries 1 and 2 are respectively Theorems 7 and 9 in [34], and Theorem 5 in [34] can be obtained if we put $\lambda =\frac{1}{2}$ in Corollary 3.

Theorem 2.

Suppose that $K\in B\left(\mathbb{H}\right)$ has a closed range and that K-frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are K-woven. Then, for any $f\in Range\left(K\right)$, for all $\sigma \subset \mathbb{J}$, ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$, and $\lambda \in \mathbb{R}$,

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2+\parallel \sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ \ge (2\lambda -\frac{{\lambda }^2}{2}-1)\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ +(1-\frac{{\lambda }^2}{2})\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle ),\hfill \end{array}$$

where $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ is a K-dual of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$.

Moreover, if ${\left(P_1{K}^{†}\right)}^{\ast }{P}_2{K}^{†}$ is a positive operator, then

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ +\parallel \sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\le {\parallel f\parallel }^2\hfill \end{array}$$

for any $f\in Range\left(K\right)$, where $P_1$ and $P_2$ are given in Equation (2).

Proof. 

For any $f\in Range\left(K\right)$, for all $\sigma \subset \mathbb{J}$, ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$, and $\lambda \in \mathbb{R}$, we know, by combining Equation (3) and Lemma 3, that

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2+\parallel \sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ =\parallel P_1{K}^{†}{f\parallel }^2+\parallel P_2{K}^{†}{f\parallel }^2=2{\parallel P_2{K}^{†}f\parallel }^2+\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -\mathrm{Re}\langle P_2{K}^{†}f,f\rangle \hfill \\ \ge (2-\frac{{\lambda }^2}{2}){\parallel f\parallel }^2-(4-2\lambda )\mathrm{Re}\langle P_1{K}^{†}f,f\rangle +\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -\mathrm{Re}\langle P_2{K}^{†}f,f\rangle \hfill \\ =(2\lambda -\frac{{\lambda }^2}{2}-1)\mathrm{Re}\langle P_1{K}^{†}f,f\rangle +(1-\frac{{\lambda }^2}{2})\mathrm{Re}\langle P_2{K}^{†}f,f\rangle \hfill \\ =(2\lambda -\frac{{\lambda }^2}{2}-1)\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ +(1-\frac{{\lambda }^2}{2})\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle ).\hfill \end{array}$$

For the “Moreover” part, we have for any $f\in Range\left(K\right)$ that

$$\begin{array}{cc}\hfill \parallel P_1{K}^{†}{f\parallel }^2& =\parallel P_2{K}^{†}{f\parallel }^2-\mathrm{Re}\langle P_2{K}^{†}f,f\rangle +\mathrm{Re}\langle P_1{K}^{†}f,f\rangle \hfill \\ \hfill & =\mathrm{Re}\langle P_2{K}^{†}f,P_2{K}^{†}f\rangle -\mathrm{Re}\langle P_2{K}^{†}f,f\rangle +\mathrm{Re}\langle P_1{K}^{†}f,f\rangle \hfill \\ \hfill & =-(\mathrm{Re}\langle P_2{K}^{†}f,P_1{K}^{†}f+P_2{K}^{†}f\rangle -\mathrm{Re}\langle P_2{K}^{†}f,P_2{K}^{†}f\rangle )+\mathrm{Re}\langle P_1{K}^{†}f,f\rangle \hfill \\ \hfill & =-\mathrm{Re}\langle P_2{K}^{†}f,P_1{K}^{†}f\rangle +\mathrm{Re}\langle P_1{K}^{†}f,f\rangle \le \mathrm{Re}\langle P_1{K}^{†}f,f\rangle .\hfill \end{array}$$

With a similar discussion, we can show that $\parallel P_2{K}^{†}{f\parallel }^2\le \mathrm{Re}\langle P_2{K}^{†}f,f\rangle$. Thus,

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2+\parallel \sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ \le \mathrm{Re}\langle P_1{K}^{†}f,f\rangle +\mathrm{Re}\langle P_2{K}^{†}f,f\rangle =\mathrm{Re}\langle P_1{K}^{†}f+P_2{K}^{†}f,f\rangle ={\parallel f\parallel }^2.\hfill \end{array}$$

 □

Corollary 4.

Suppose that two frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are woven. Then, for any $\sigma \subset \mathbb{J}$, for all $\lambda \in \mathbb{R}$ and all $f\in \mathbb{H}$, we have

$$\begin{array}{c}(2\lambda -\frac{{\lambda }^2}{2}-1)\sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2+(1-\frac{{\lambda }^2}{2})\sum _{j\in {\sigma }^c}{|\langle f,{\psi }_{2j}\rangle |}^2\hfill \\ \le \sum _{j\in \sigma }|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{1j}\rangle {|}^2+\sum _{j\in {\sigma }^c}{|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{2j}\rangle |}^2\hfill \\ +\sum _{j\in \sigma }|\langle S_{{\mathsf{\Psi }}_2}^{{\sigma }^c}f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{1j}\rangle {|}^2+\sum _{j\in {\sigma }^c}{|\langle S_{{\mathsf{\Psi }}_2}^{{\sigma }^c}f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{2j}\rangle |}^2\hfill \\ \le \sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2+\sum _{j\in {\sigma }^c}{|\langle f,{\psi }_{2j}\rangle |}^2.\hfill \end{array}$$

(8)

Proof. 

Letting $K^{†}={\mathrm{Id}}_{\mathbb{H}}$ and for any $\sigma \subset \mathbb{J}$, taking

$$a_j=\left\{\begin{array}{cc}1,\hfill & j\in \sigma ,\hfill \\ 0,\hfill & j\in {\sigma }^c,\hfill \end{array}\right.{\varphi }_j=\left\{\begin{array}{cc}{S}_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j},\hfill & j\in \sigma ,\hfill \\ S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j},\hfill & j\in {\sigma }^c.\hfill \end{array}\right.$$

If, now, we replace ${\psi }_{1j}$, ${\psi }_{2j}$ and f in the left-hand inequality of Theorem 2 respectively by $S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j}$, $S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j}$ and $S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{1/2}f$, then

$$\begin{array}{c}\parallel \sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{1j}{\parallel }^2+\parallel \sum _{j\in {\sigma }^c}\langle f,{\psi }_{2j}\rangle S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1/2}{\psi }_{2j}{\parallel }^2\hfill \\ \ge (2\lambda -\frac{{\lambda }^2}{2}-1)\mathrm{Re}\sum _{j\in \sigma }\langle f,{\psi }_{1j}\rangle \langle {\psi }_{1j},f\rangle +(1-\frac{{\lambda }^2}{2})\mathrm{Re}\sum _{j\in {\sigma }^c}\langle f,{\psi }_{2j}\rangle \langle {\psi }_{2j},f\rangle \hfill \\ =(2\lambda -\frac{{\lambda }^2}{2}-1)\sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2+(1-\frac{{\lambda }^2}{2})\sum _{j\in {\sigma }^c}{|\langle f,{\psi }_{2j}\rangle |}^2.\hfill \end{array}$$

This along with Equations (6) and (7) gives the left-hand inequality in Equation (8), and the proof of the right-hand inequality is similar and we omit the details. □

Theorem 3.

Suppose that $K\in B\left(\mathbb{H}\right)$ has a closed range and that K-frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are K-woven. Then, for all $\sigma \subset \mathbb{J}$, for any ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$, $\lambda \in \mathbb{R}$ and $f\in Range\left(K\right)$,

$$\begin{array}{c}\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )-\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ \le {(1-\frac{\lambda }{2})}^2\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ +\frac{{\lambda }^2}{4}\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle ),\hfill \end{array}$$

where $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ is a K-dual of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$.

Moreover, if ${\left(P_1{K}^{†}\right)}^{\ast }{P}_2{K}^{†}\ge 0$, then

$$\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )-\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\ge 0$$

for any $f\in Range\left(K\right)$, where $P_1$ and $P_2$ are given in Equation (2).

Proof. 

For all $\sigma \subset \mathbb{J}$, for any ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$, $\lambda \in \mathbb{R}$ and $f\in Range\left(K\right)$, we see from Equation (4) that

$$\begin{array}{c}\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )-\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ =\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -{\parallel P_1{K}^{†}f\parallel }^2\hfill \\ \le \mathrm{Re}\langle P_1{K}^{†}f,f\rangle -(\lambda -\frac{{\lambda }^2}{4})\mathrm{Re}\langle P_1{K}^{†}f,f\rangle +\frac{{\lambda }^2}{4}\mathrm{Re}\langle P_2{K}^{†}f,f\rangle \hfill \\ ={(1-\frac{\lambda }{2})}^2\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )\hfill \\ +\frac{{\lambda }^2}{4}\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle ).\hfill \end{array}$$

Suppose now that ${\left(P_1{K}^{†}\right)}^{\ast }{P}_2{K}^{†}$ is a positive operator. Then

$$\begin{array}{c}\mathrm{Re}(\sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle \langle {\varphi }_j,f\rangle +\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle \langle {\varphi }_j,f\rangle )-\parallel \sum _{j\in \sigma }{a}_j\langle K^{†}f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle K^{†}f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ =\mathrm{Re}\langle P_1{K}^{†}f,f\rangle -{\parallel P_1{K}^{†}f\parallel }^2=\mathrm{Re}\langle P_1{K}^{†}f,P_1{K}^{†}f+P_2{K}^{†}f\rangle -\mathrm{Re}\langle P_1{K}^{†}f,P_1{K}^{†}f\rangle \hfill \\ =\mathrm{Re}\langle P_1{K}^{†}f,P_2{K}^{†}f\rangle =\mathrm{Re}\langle f,{\left(P_1{K}^{†}\right)}^{\ast }{P}_2{K}^{†}f\rangle \ge 0.\hfill \end{array}$$

 □

Corollary 5.

Let the two frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ be woven. Then, for any $\sigma \subset \mathbb{J}$, for all $\lambda \in \mathbb{R}$ and all $f\in \mathbb{H}$, we have

$$\begin{array}{cc}\hfill 0& \le \sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2-\sum _{j\in \sigma }|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{1j}\rangle {|}^2-\sum _{j\in {\sigma }^c}{|\langle S_{{\mathsf{\Psi }}_1}^{\sigma }f,S_{{\mathsf{\Psi }}_1{\mathsf{\Psi }}_2}^{-1}{\psi }_{2j}\rangle |}^2\hfill \\ \hfill & \le {(1-\frac{\lambda }{2})}^2\sum _{j\in \sigma }|\langle f,{\psi }_{1j}\rangle {|}^2+\frac{{\lambda }^2}{4}\sum _{j\in {\sigma }^c}{|\langle f,{\psi }_{2j}\rangle |}^2.\hfill \end{array}$$

Proof. 

The proof is similar to Corollary 4 by using Theorem 3, so we omit it. □

Remark 2.

Corollaries 4 and 5 are respectively Theorems 15 and 14 in [34].

We conclude the paper with a double inequality for K-weaving frames stated as follows.

Theorem 4.

Suppose that K-frames ${\mathsf{\Psi }}_1={\left\{{\psi }_{1j}\right\}}_{j\in \mathbb{J}}$ and ${\mathsf{\Psi }}_2={\left\{{\psi }_{2j}\right\}}_{j\in \mathbb{J}}$ in $\mathbb{H}$ are K-woven. Then, for any $\sigma \subset \mathbb{J}$, for all ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$ and all $f\in \mathbb{H}$, we have

$$\begin{array}{cc}\hfill \frac{3}{4}{\parallel Kf\parallel }^2& \le \parallel \sum _{j\in \sigma }{a}_j\langle f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2\hfill \\ & +\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle f,{\psi }_{1j}\rangle \langle {\varphi }_j,Kf\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle f,{\psi }_{2j}\rangle \langle {\varphi }_j,Kf\rangle )\hfill \\ & \le \frac{{3\parallel K\parallel }^2+{\parallel P_1-P_2\parallel }^2}{4}{\parallel f\parallel }^2,\hfill \end{array}$$

where $P_1$ and $P_2$ are given in Equation (2), and $\mathsf{\Phi }={\left\{{\varphi }_j\right\}}_{j\in \mathbb{J}}$ is a K-dual of ${\left\{{\psi }_{1j}\right\}}_{j\in \sigma }\cup {\left\{{\psi }_{2j}\right\}}_{j\in {\sigma }^c}$.

Proof. 

For any $\sigma \subset \mathbb{J}$, for all ${\left\{a_j\right\}}_{j\in \mathbb{J}}\in {\ell }^{\infty }\left(\mathbb{J}\right)$ and all $f\in \mathbb{H}$, it is easy to check that $P_1+P_2=K$. By Lemma 1, we get

$$\begin{array}{c}\parallel \sum _{j\in \sigma }{a}_j\langle f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2+\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle f,{\psi }_{1j}\rangle \langle {\varphi }_j,Kf\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle f,{\psi }_{2j}\rangle \langle {\varphi }_j,Kf\rangle )\hfill \\ =\parallel P_1{f\parallel }^2+\mathrm{Re}\langle P_{2}f,Kf\rangle \ge \frac{3}{4}{\parallel Kf\parallel }^2.\hfill \end{array}$$

We also have

$$\begin{array}{l}\parallel \sum _{j\in \sigma }{a}_j\langle f,{\psi }_{1j}\rangle {\varphi }_j+\sum _{j\in {\sigma }^c}{a}_j\langle f,{\psi }_{2j}\rangle {\varphi }_j{\parallel }^2+\mathrm{Re}(\sum _{j\in \sigma }(1-a_j)\langle f,{\psi }_{1j}\rangle \langle {\varphi }_j,Kf\rangle +\sum _{j\in {\sigma }^c}(1-a_j)\langle f,{\psi }_{2j}\rangle \langle {\varphi }_j,Kf\rangle )\hfill \\ =\langle P_{1}f,P_{1}f\rangle +\frac{1}{2}\langle P_{2}f,Kf\rangle +\frac{1}{2}\langle Kf,P_{2}f\rangle \hfill \\ =\langle P_{1}f,P_{1}f\rangle +\frac{1}{2}\langle (K-P_1)f,Kf\rangle +\frac{1}{2}\langle Kf,(K-P_1)f\rangle \hfill \\ =\langle Kf,Kf\rangle -\frac{1}{2}[\langle P_{1}f,Kf\rangle -\langle P_{1}f,P_{1}f\rangle ]-\frac{1}{2}[\langle Kf,P_{1}f\rangle -\langle P_{1}f,P_{1}f\rangle ]\hfill \\ =\langle Kf,Kf\rangle -\frac{1}{2}\langle P_{1}f,P_{2}f\rangle -\frac{1}{2}\langle P_{2}f,P_{1}f\rangle \hfill \\ =\frac{3}{4}\langle Kf,Kf\rangle +\frac{1}{4}\langle P_{1}f+P_{2}f,P_{1}f+P_{2}f\rangle -\frac{1}{2}\langle P_{1}f,P_{2}f\rangle -\frac{1}{2}\langle P_{2}f,P_{1}f\rangle \hfill \\ =\frac{3}{4}\langle Kf,Kf\rangle +\frac{1}{4}\langle (P_1-P_2)f,(P_1-P_2)f\rangle \hfill \\ \le \frac{3}{4}{\parallel K\parallel }^2{\parallel f\parallel }^2+\frac{1}{4}\parallel P_1-P_2{\parallel }^2{\parallel f\parallel }^2=\frac{{3\parallel K\parallel }^2+{\parallel P_1-P_2\parallel }^2}{4}{\parallel f\parallel }^2,\hfill \end{array}$$

and the proof is over. □

Remark 3.

Theorem 3 in [35] can be obtained when taking $K={\mathrm{Id}}_{\mathbb{H}}$ in Theorem 4.