1 Introduction

A frame as a generalization of an orthonormal basis, appeared first in the late 1940s and early 1950s, provides us with a powerful theoretical tool because of its redundancy and flexibility. Now a frame plays an important role in sampling theory [1], compressed sensing [2] and a number of other fields. We refer the readers to [3,4,5] for an introduction to frame theory and its applications. In [6], Sun proposed the notion of g-frame, which generalized the concept of frame extensively. We know that though many basic properties of g-frame can be shared with frame, not all the properties between them are same. For example, an exact frame is equivalent to a Riesz basis, but an exact g-frame is not equivalent to a g-Riesz basis. We refer the reader to the papers [6,7,8,9,10,11] for more information about g-frames.

Being an extension of frame, the concept of K-frame was introduced by Găvruţa [12], which allows an atomic decomposition of elements in the range of K. In fact, a K-frame is a more general version of frame. There are many differences between a K-frame and a frame. For instance, the sequence \(\{f_{j}\}_{j\in J}\) is a frame for \({\mathcal {H}}\) if and only if \(\{f_{j}\}_{j\in J}\) is a Bessel sequence for \({\mathcal {H}}\) and the corresponding synthesis operator is surjective, but the sequence \(\{f_{j}\}_{j\in J}\) is a K-frame for \({\mathcal {H}}\) if and only if \(\{f_{j}\}_{j\in J}\) is a Bessel sequence for \({\mathcal {H}}\) and the range of K is involved in the range of the corresponding synthesis operator. For more details on K-frames, see references in [12,13,14,15,16,17,18].

Recently, Xiao et al. [19] put forward the notion of K-g-frame, which is more general than g-frame and K-frame in Hilbert spaces. Naturally, K-g-frame attracts many scholars’ attention. Now it has been a hot topic to make full use of various conditions to construct a new K-g-frame (see [20,21,22,23]). Hua and others gave several methods to generate tight K-g-frames and tight g-frames (see [24]). For more details on K-g-frame, readers can consult [19,20,21,22,23,24].

In this paper, we first construct a K-g-frame from a given K-g-frame and a g-Bessel sequence. Next, we adopt a novel way to generate a new K-g-frame from two existing K-g-frames. We also give a necessary and sufficient condition to yield a K-g-frame. Finally, we give an equivalent characterization of constructing tight K-g-frames by two given g-Bessel sequences. We correct the results of Theorem 3.4 and Corollary 3.17 in [21] and Theorem 3.10 in [22]. We also generalize and improve some remarkable results.

Throughout this paper, we will adopt such notations. \({\mathcal {H}}\) is a separable Hilbert space, and \(I_{{\mathcal {H}}}\) is the identity operator for \({\mathcal {H}}\). \({\mathbb {C}}\) is the set of all complex numbers. \(B({\mathcal {H}_1},{\mathcal {H}_2})\) is a collection of all bounded linear operators from \({\mathcal {H}_1}\) to \({\mathcal {H}_2}\) , where \({\mathcal {H}_1}\), \({\mathcal {H}_2}\) are Hilbert spaces, and if \({\mathcal {H}_1}\mathrm{{ = }}{\mathcal {H}_2}\mathrm{{ = }}{\mathcal {H}}\), \(B({\mathcal {H}_1},{\mathcal {H}_2})\) is denoted by \(B({\mathcal {H}})\). Let \(K \in B({\mathcal {H}})\) and \(K\ne 0\), the range and the kernel of K are denoted by R(K) and N(K), respectively. \(\{{\mathcal {V}}_{j}\}_{j\in J}\) is a sequence of closed subspaces of \({\mathcal {H}}\), where J is a finite or countable index set. \(\ell ^{2}(\{{\mathcal {V}}_{j}\}_{j\in J})\) is defined by

$$\begin{aligned} \ell ^{2}(\{\mathcal {V}_{j}\}_{j\in J})=\left\{ \{g_{j}\}_{j\in J}:g_{j}\in {\mathcal {V}}_{j},\quad j\in J,\quad \sum _{j\in J}\Vert g_{j}\Vert ^{2}<+\infty \right\} \end{aligned}$$

with the inner product

$$\begin{aligned} \langle \{f_{j}\}_{j\in J},\,\{g_{j}\}_{j\in J}\rangle =\sum _{j\in J}\langle f_{j},\,g_{j}\rangle . \end{aligned}$$

It is trivial that \(\ell ^{2}(\{\mathcal {V}_{j}\}_{j\in J})\) is a Hilbert space.

Definition 1.1

A sequence \(\{\Lambda _{j}:\Lambda _{j}\in B({\mathcal {H}},{\mathcal {V}}_{j})\}_{j\in J}\) is called a g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) if there exist two positive constants A and B such that

$$\begin{aligned} A\Vert f\Vert ^2 \le \sum _{j\in J}\Vert \Lambda _{j}f\Vert ^2 \le B\Vert f\Vert ^2, \quad (\forall f \in {\mathcal {H}}). \end{aligned}$$
(1.1)

The constants A and B are called the lower and upper g-frame bounds, respectively. If only the only right inequality of (1.1) holds, \(\{\Lambda _{j}\}_{j\in J}\) is called a g-Bessel sequence for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with bound B.

If \(\{\Lambda _{j}\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\), we may define the bounded linear operator \(T_{\Lambda }\) by

$$\begin{aligned} T_{\Lambda }:\ell ^{2}(\{\mathcal {V}_{j}\}_{j\in J})\rightarrow {\mathcal {H}}:T_{\Lambda }(\{g_{j}\}_{j\in J})=\sum _{j\in J}\Lambda _{j}^{*}g_{j}, \quad \{g_{j}\}_{j\in J}\in \ell ^{2}(\{\mathcal {V}_{j}\}_{j\in J}. \end{aligned}$$

\(T_{\Lambda }\) is called the synthesis operator. The adjoint operator \(T_{\Lambda }^{*}\) is given as follows:

$$\begin{aligned} T_{\Lambda }^{*}f:{\mathcal {H}}\rightarrow \ell ^{2}(\{\mathcal {V}_{j}\}_{j\in J}):T_{\Lambda }^{*}f=\{\Lambda _{j}f\}_{j\in J}, \quad f\in {\mathcal {H}}. \end{aligned}$$

\(T_{\Lambda }^{*}\) is called the analysis operator. The operator given by

$$\begin{aligned} S_{\Lambda }:{\mathcal {H}}\rightarrow {\mathcal {H}}:S_{\Lambda }f=\sum _{j\in J}\Lambda _{j}^{*}\Lambda _{j}f, \quad f\in {\mathcal {H}} \end{aligned}$$

is called the g-frame operator.

Definition 1.2

Let \(K\in B({\mathcal {H}})\). A sequence \(\{f_j\}_{j\in J}\subset {\mathcal {H}}\) is called a K-frame for \({\mathcal {H}}\) if there exist two positive constants A and B such that

$$\begin{aligned} A\Vert K^*f\Vert ^2 \le \sum _{j\in J}|\langle f, f_j\rangle |^2 \le B\Vert f\Vert ^2, \quad (\forall f \in {\mathcal {H}}). \end{aligned}$$

The constants A and B are called the lower and upper K-frame bounds, respectively.

Definition 1.3

([19]) Let \(K\in B({\mathcal {H}})\). A sequence \(\{\Lambda _{j}:\Lambda _{j}\in B({\mathcal {H}},{\mathcal {V}}_{j})\}_{j\in J}\) is called a K-g-frame for \({\mathcal {H}}\) with respect to \(\{{\mathcal {V}}_{j}\}_{j\in J}\) if there exist two positive constants A and B such that

$$\begin{aligned} A\Vert K^*f\Vert ^2 \le \sum _{j\in J}\Vert \Lambda _{j}f\Vert ^2 \le B\Vert f\Vert ^2, \quad (\forall f \in {\mathcal {H}}). \end{aligned}$$

The constants A and B are called the lower and upper K-g-frame bounds, respectively.

Definition 1.4

([24]) Let \(K\in B({\mathcal {H}})\). A sequence \(\{\Lambda _{j}:\Lambda _{j}\in B({\mathcal {H}},{\mathcal {V}}_{j})\}_{j\in J}\) is called a tight K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\), if there exists a positive constant A such that

$$\begin{aligned} A\Vert {K^*}f\Vert ^2=\sum \limits _{j\in J}\Vert \Lambda _{j}f\Vert ^{2}, \quad (\forall f \in {\mathcal {H}}). \end{aligned}$$

In order to obtain our main results, we need the following lemmas.

Lemma 1.5

Let \({\mathcal {H}}_{1}\) and \({\mathcal {H}}_{2}\) be two Hilbert spaces, and suppose that \(U:{\mathcal {H}_1}\rightarrow {\mathcal {H}_2}\) is a bounded linear operator with closed range R(U). Then, there exists a unique bounded operator \(U^+: {\mathcal {H}_2}\rightarrow {\mathcal {H}_1}\) satisfying

$$\begin{aligned} N_{U^+}=R(U)^\perp , \, R(U^+)=N^\perp _U, \, UU^+f=f, \quad (\forall f\in R(U)). \end{aligned}$$

The operator \(U^+\) is called the pseudo-inverse operator of U.

Lemma 1.6

([13]) Suppose that \(U\in B({\mathcal {H}_1},{\mathcal {H}_2})\) is an operator with closed range, then

$$\begin{aligned} \Vert U^+\Vert ^{-1}\Vert f\Vert \le \Vert U^*f\Vert \le \Vert U\Vert \Vert f\Vert , \quad (\forall f\in R(U)). \end{aligned}$$

Lemma 1.7

([19]) The sequence \(\{\Lambda _j\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) if and only if the synthesis operator \(T_{\Lambda }\) is well defined and bounded, and \(R(K)\subset R(T_{\Lambda })\).

Lemma 1.8

([24]) Let \(\{\Lambda _j\}_{j\in J}\) be a g-Bessel sequence for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Then, \(\{\Lambda _j\}_{j\in J}\) is a tight K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\), if and only if there exists a positive constant A such that \(S_{\Lambda }=AKK^{*}\), where \(S_{\Lambda }\) is the g-frame operator for \(\{\Lambda _j\}_{j\in J}\).

Lemma 1.9

Let \(\{\Lambda _{j}\}_{j\in J}\) and \(\{\Gamma _{j}\}_{j\in J}\) be g-Bessel sequences for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). If \(U_{1},U_{2}\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\), then \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with bound \((\sqrt{B_{1}}\Vert U_{1}\Vert +\sqrt{B_{2}}\Vert U_{2}\Vert )^{2}\).

The proof is easy, we omit it. Later, we will need the following important result from operator theory:

Theorem 1.10

(Douglas’s theorem [25]) Let \(U_{1}\in B({\mathcal {H}}_{1},{\mathcal {H}})\), \(U_{2}\in B({\mathcal {H}}_{2},{\mathcal {H}})\). Then, the following are equivalent:

\(\mathrm {(1)}\):

\(R(U_{1})\subseteq R(U_{2})\);

\(\mathrm {(2)}\):

\(U_{1}U_{1}^{*}\le \alpha ^{2} U_{2}U_{2}^{*}\) for some \(\alpha >0\);

\(\mathrm {(3)}\):

there exists a bounded operator \(X\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\) so that \(U_{1}=U_{2}X\).

Several ways to generate g-frames have been discussed in [9,10,11]. After the notion of K-frame was proposed, there are some references to give a number of construction methods about K-frames (see [14,15,16,17]). Motivated by recent progress in constructions of some new K-g-frames (see [19,20,21,22,23]), we give two different ways to construct new K-g-frames.

Remark 1.11

In [21, Theorem 3.4], the following statement has been formulated: let \(K\in B({\mathcal {H}})\) and \(\{\Lambda _{j}\}_{j\in J}\) be a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with lower and upper bounds A and B, respectively; if \(U\in B({\mathcal {H}})\) has closed range and \(UK=KU\), then \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is a K-g-frame for R(U) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with lower and upper bounds \(A\Vert U^{+}\Vert ^{-2}\) and \(B\Vert U\Vert ^{2}\), respectively. In Example 1.12, we show that this statement is not true in the general case.

Example 1.12

Suppose \({\mathcal {H}}={\mathbb {C}}^{3}\), \(J=\{1,2,3\}\). Let \(\{e_{j}\}_{j\in J}\) be an orthonormal basis of \({\mathcal {H}}\), and \({\mathcal {V}}_{j}=\overline{span}\{e_{j}\}\). Now define \(K\in B({\mathcal {H}})\), \(U\in B({\mathcal {H}})\) and \(\{\Lambda _{j}\}_{j\in J}\) as follows:

$$\begin{aligned} K:{\mathcal {H}}\rightarrow {\mathcal {H}},\, Kf= & {} \langle f,\,e_{3}\rangle e_{1}+\langle f,\,e_{1}+e_{2}\rangle e_{2},\quad f \in {\mathcal {H}},\\ U:{\mathcal {H}}\rightarrow {\mathcal {H}},\, Uf= & {} \langle f,\,e_{3}\rangle (e_{1}-e_{2}),\, \, \, f \in {\mathcal {H}},\\ \Lambda _{1}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{1},\,\Lambda _{1}f= & {} \langle f,\,e_{1}\rangle e_{1},\quad f \in {\mathcal {H}},\\ \Lambda _{2}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{2},\,\Lambda _{2}f= & {} \langle f,\,e_{2}\rangle e_{2},\quad f \in {\mathcal {H}},\\ \Lambda _{3}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{3},\,\Lambda _{3}f= & {} \langle f,\,e_{2}\rangle e_{3},\quad f \in {\mathcal {H}}. \end{aligned}$$

Now we show that \(K^{*}f=\langle f,\,e_{1}\rangle e_{3}+\langle f,\,e_{2}\rangle (e_{1}+e_{2}),\,f\in {\mathcal {H}}\). In fact, for any \(f,m\in {\mathcal {H}}\), we have

$$\begin{aligned} \langle K^{*}f,\,m\rangle= & {} \langle f,\,Km\rangle =\langle f,\,\langle m,\,e_{3}\rangle e_{1}+\langle m,\,e_{1}+e_{2}\rangle e_{2}\rangle \\= & {} \langle f,\,e_{1}\rangle \overline{\langle m,\,e_{3}\rangle }+\langle f,\,e_{2}\rangle \overline{\langle m,\,e_{1}+e_{2}\rangle }\\= & {} \langle f,\,e_{1}\rangle \langle e_{3},\,m\rangle +\langle f,\,e_{2}\rangle \langle e_{1}+e_{2},\,m\rangle \\= & {} \langle \langle f,\,e_{1}\rangle e_{3}+\langle f,\,e_{2}\rangle (e_{1}+e_{2}),\,m\rangle . \end{aligned}$$

Thus, for each \(f\in {\mathcal {H}}\), we obtain

$$\begin{aligned} \Vert K^{*}f\Vert ^{2}= & {} \Vert \langle f,\,e_{1}\rangle e_{3}+\langle f,\,e_{2}\rangle (e_{1}+e_{2})\Vert ^{2}=|\langle f,\,e_{1}\rangle |^{2}+2|\langle f,\,e_{2}\rangle |^{2}\\= & {} \sum _{j=1}^{3}\Vert \Lambda _{j}f\Vert ^{2}\le 3\Vert f\Vert ^{2}. \end{aligned}$$

This implies that \(\{\Lambda _{j}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). It is clear that \(U\in B({\mathcal {H}})\) has closed range. For all \(f\in {\mathcal {H}}\), we get

$$\begin{aligned} UKf= & {} U(\langle f,\,e_{3}\rangle e_{1}+\langle f,\,e_{1}+e_{2}\rangle e_{2})=\langle f,\,e_{3}\rangle Ue_{1}+\langle f,\,e_{1}+e_{2}\rangle Ue_{2}=0\\= & {} \langle f,\,e_{3}\rangle (e_{2}-e_{2})=\langle f,\,e_{3}\rangle (Ke_{1}-Ke_{2})=K(\langle f,\,e_{3}\rangle (e_{1}-e_{2}))=KUf. \end{aligned}$$

Then, \(UK=KU\).

The adjoint operator of U is \(U^{*}\), \(U^{*}f=\langle f,\,e_{1}-e_{2}\rangle e_{3},\,f\in {\mathcal {H}}\). Indeed, for all \(f,m\in {\mathcal {H}}\), we have

$$\begin{aligned} \langle U^{*}f,\,m\rangle= & {} \langle f,\,Um\rangle =\langle f,\langle m,\,e_{3}\rangle (e_{1}-e_{2})\rangle =\langle f,\,e_{1}-e_{2}\rangle \overline{\langle m,\,e_{3}\rangle }\\= & {} \langle f,\,e_{1}-e_{2}\rangle \langle e_{3},\,m\rangle =\langle \langle f,\,e_{1}-e_{2}\rangle e_{3},\,m\rangle . \end{aligned}$$

Choosing \(f=e_{2}-e_{1}\in R(U)=\overline{span}\{e_{1}-e_{2}\}\), we get \(\Vert K^{*}f\Vert ^{2}=3\) and

$$\begin{aligned} \sum _{j=1}^{3}\Vert \Lambda _{j}U^{*}f\Vert ^{2}=\sum _{j=1}^{3}\Vert \Lambda _{j}(\langle f,\,e_{1}-e_{2}\rangle e_{3})\Vert ^{2}=\sum _{j=1}^{3}\Vert \langle f,\,e_{1}-e_{2}\rangle (\Lambda _{j}e_{3})\Vert ^{2}=0. \end{aligned}$$

Hence, \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is not a K-g-frame for R(U) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Furthermore, \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is not a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) and

$$\begin{aligned} \overline{span}\{e_{1}+e_{2}, e_{3}\}=R(K^{*})\nsubseteq R(U)=\overline{span}\{e_{1}-e_{2}\}. \end{aligned}$$

Remark 1.13

In [22, Theorem 3.10], the following statement has been formulated: let \(\{\Lambda _{j}\}_{j\in J}\) be an atomic system for K, and let \(S_{\Lambda }\) be the frame operator of \(\{\Lambda _{j}\}_{j\in J}\); let U be a positive operator, then \(\{\Lambda _{j}+\Lambda _{j}U\}_{j\in J}\) is an atomic system for K. In Example 1.14, we show that this statement is not true in the general case.

Example 1.14

Suppose \({\mathcal {H}}={\mathbb {C}}^{3}\), \(J=\{1,2,3\}\). Assume that \(\{e_{j}\}_{j\in J}\) is an orthonormal basis of \({\mathcal {H}}\), and \({\mathcal {V}}_{1}={\mathcal {V}}_{2}=\overline{span}\{e_{1}\},{\mathcal {V}}_{3}=\overline{span}\{e_{3}\}\). Now define \(K\in B({\mathcal {H}})\), \(U\in B({\mathcal {H}})\) and \(\{\Lambda _{j}\}_{j\in J}\) as follows:

$$\begin{aligned} K:{\mathcal {H}}\rightarrow {\mathcal {H}},\, Kf= & {} \langle f,\,e_{1}\rangle e_{2}, \, \, \, f \in {\mathcal {H}},\\ U:{\mathcal {H}}\rightarrow {\mathcal {H}},\, Uf= & {} \langle f,\,e_{1}\rangle e_{1}+\langle f,\,2e_{2}-e_{3}\rangle e_{2}+\langle f,\,e_{3}-e_{2}\rangle e_{3},\, \, \, f \in {\mathcal {H}},\\ \Lambda _{1}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{1},\,\Lambda _{1}f= & {} \langle f,\,e_{1}\rangle e_{1},\quad f \in {\mathcal {H}},\\ \Lambda _{2}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{2},\,\Lambda _{2}f= & {} \langle f,\,e_{1}\rangle e_{1},\quad f \in {\mathcal {H}},\\ \Lambda _{3}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{3},\,\Lambda _{3}f= & {} \langle f,\,e_{2}\rangle e_{3},\quad f \in {\mathcal {H}}. \end{aligned}$$

Now we show that \(K^{*}f=\langle f,\,e_{2}\rangle e_{1},\,f\in {\mathcal {H}}\). In fact, for all \(f,m\in {\mathcal {H}}\), we get

$$\begin{aligned} \langle K^{*}f,\,m\rangle= & {} \langle f,\,Km\rangle =\langle f,\,\langle m,\,e_{1}\rangle e_{2}\rangle =\langle f,\,e_{2}\rangle \overline{\langle m,\,e_{1}\rangle }\\= & {} \langle f,\,e_{2}\rangle \langle e_{1},\,m\rangle =\langle \langle f,\,e_{2}\rangle e_{1},\,m\rangle . \end{aligned}$$

Hence, for every \(f\in {\mathcal {H}}\), we have

$$\begin{aligned} \Vert K^{*}f\Vert ^{2}= & {} \Vert \langle f,\,e_{2}\rangle e_{1}\Vert ^{2}=|\langle f,\,e_{2}\rangle |^{2}\\\le & {} \sum _{j=1}^{3}\Vert \Lambda _{j}f\Vert ^{2}=2|\langle f,\,e_{1}\rangle |^{2}+|\langle f,\,e_{2}\rangle |^{2}\le 2\Vert f\Vert ^{2}. \end{aligned}$$

Thus, \(\{\Lambda _{j}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\).

By a simple calculation, we can obtain that U is a self-adjoint operator. Then, for each \(f\in {\mathcal {H}}\), we conclude

$$\begin{aligned} \langle Uf,\,f\rangle= & {} \langle \langle f,\,e_{1}\rangle e_{1}+\langle f,\,2e_{2}-e_{3}\rangle e_{2}+\langle f,\,e_{3}-e_{2}\rangle e_{3},\,f\rangle \\= & {} \langle f,\,e_{1}\rangle \overline{\langle f,\,e_{1}\rangle }+2\langle f,\,e_{2}\rangle \overline{\langle f,\,e_{2}\rangle }-\langle f,\,e_{3}\rangle \overline{\langle f,\,e_{2}\rangle }\\&\qquad +\langle f,\,e_{3}\rangle \overline{\langle f,\,e_{3}\rangle }-\langle f,\,e_{2}\rangle \overline{\langle f,\,e_{3}\rangle }\\= & {} |\langle f,\,e_{1}\rangle |^{2}+2|\langle f,\,e_{2}\rangle |^{2}-2\text{ Re }\{\langle f,\,e_{2}\rangle \overline{\langle f,\,e_{3}\rangle }\}+|\langle f,\,e_{3}\rangle |^{2}\\= & {} |\langle f,\,e_{1}\rangle |^{2}+|\langle f,\,e_{2}\rangle |^{2}+|\langle f,\,e_{2}\rangle -\langle f,\,e_{3}\rangle |^{2}\ge 0. \end{aligned}$$

Therefore, U is a positive operator.

It is clear that \(\{\Lambda _{j}+\Lambda _{j}U\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). By a direct computation, we get

$$\begin{aligned}&(\Lambda _{1}+\Lambda _{1}U)f=2\langle f,\,e_{1}\rangle e_{1};\quad (\Lambda _{2}+\Lambda _{2}U)f =2\langle f,\,e_{1}\rangle e_{1}; \\&(\Lambda _{3}+\Lambda _{3}U)f=\langle f,\,3e_{2}-e_{3}\rangle e_{3}. \end{aligned}$$

Choosing \(f=e_{2}+3e_{3}\in {\mathcal {H}}\), we get \(\Vert K^{*}f\Vert ^{2}=|\langle f,\,e_{2}\rangle |^{2}=1\) and

$$\begin{aligned} \sum _{j=1}^{3}\Vert (\Lambda _{j}+\Lambda _{j}U)f\Vert ^{2}=8|\langle f,\,e_{1}\rangle |^{2}+|\langle f,\,3e_{2}-e_{3}\rangle |^{2}=0. \end{aligned}$$

This proves that \(\{\Lambda _{j}+\Lambda _{j}U\}_{j\in J}\) is not a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Let \(S_{\Lambda }\) be the frame operator of \(\{\Lambda _{j}\}_{j\in J}\), then for any \(f\in {\mathcal {H}}\), we have

$$\begin{aligned} S_{\Lambda }f=\sum _{j=1}^{3}\Lambda _{j}^{*}\Lambda _{j}f=2\langle f,\,e_{1}\rangle e_{1}+\langle f,\,e_{2}\rangle e_{2}. \end{aligned}$$

Now we will show that \(US_{\Lambda }\ne S_{\Lambda }U \); indeed, for all \(f\in {\mathcal {H}}\), we obtain

$$\begin{aligned} US_{\Lambda }f=2\langle f,\,e_{1}\rangle e_{1}+\langle f,\,e_{2}\rangle (2e_{2}-e_{3});\quad S_{\Lambda }Uf=2\langle f,\,e_{1}\rangle e_{1}+\langle f,\,2e_{2}-e_{3}\rangle e_{2}. \end{aligned}$$

2 Main Results

Theorem 2.1

Let \(K_{1}\in B({\mathcal {H}}_{1})\) and \(K_{2}\in B({\mathcal {H}}_{2})\). Suppose that \(\{\Lambda _{j}\}_{j\in J}\) is a \(K_{1}\)-g-frame and \(\{\Gamma _{j}\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with the synthesis operators \(T_{\Lambda }\) and \(T_{\Gamma }\), respectively. Assume \(U_{1}, U_{2}\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\) and \(U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}+U_{2}T_{\Gamma }T_{\Gamma }^{*}U_{2}^{*}\ge 0\). If \(U_{1}\) has closed range, \(U_{1}K_{1}=K_{2}U_{1}\) and \(R(K_{2}^{*})\subset R(U_{1})\), then \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\).

Proof

Let \(\{\Lambda _{j}\}_{j\in J}\) be a \(K_{1}\)-g-frame for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with frame bounds \(A_{1}\) and \(B_{1}\). Let \(\{\Gamma _{j}\}_{j\in J}\) be a g-Bessel sequence for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with g-Bessel bound \(B_{2}\). By Lemma 1.9, we conclude that \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with bound \((\sqrt{B_{1}}\Vert U_{1}\Vert +\sqrt{B_{2}}\Vert U_{2}\Vert )^{2}\).

For each \(g\in {\mathcal {H}}_{2}\), we obtain

$$\begin{aligned} \sum _{j\in J}2\mathrm {R}\mathrm {e}\{\langle \Lambda _{j}U_{1}^{*}g,\,\Gamma _{j}U_{2}^{*}g\rangle \}= & {} 2\mathrm {R}\mathrm {e}\left\langle \sum _{j\in J}\Gamma _{j}^{*}\Lambda _{j}U_{1}^{*}g,\,U_{2}^{*}g\right\rangle \\= & {} 2\mathrm {R}\mathrm {e}\langle T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}g,\,U_{2}^{*}g\rangle \\= & {} \langle (U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*})g,\,g\rangle . \end{aligned}$$

Since \(U_{1}\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\) has closed range, and \(U_{1}K_{1}=K_{2}U_{1}\), it is clear that \(K_{1}^{*}U_{1}^{*}=U_{1}^{*}K_{2}^{*}\). According to Lemma 1.6, for any \(g\in {\mathcal {H}}_{2}\), we get

$$\begin{aligned}&\sum _{j\in J}\Vert (\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*})g\Vert ^{2}\\&\quad =\sum _{j\in J}\Vert \Lambda _{j}U_{1}^{*}g\Vert ^{2}+\sum _{j\in J}2\mathrm {R}\mathrm {e}\{\langle \Lambda _{j}U_{1}^{*}g,\,\Gamma _{j}U_{2}^{*}g\rangle \}+\sum _{j\in J}\Vert \Gamma _{j}U_{2}^{*}g\Vert ^{2}\\&\quad =\sum _{j\in J}\Vert \Lambda _{j}U_{1}^{*}g\Vert ^{2}+\langle (U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*})g,\,g\rangle \\&\qquad +\langle T_{\Gamma }T_{\Gamma }^{*}U_{2}^{*}g,\,U_{2}^{*}g\rangle \\&\quad =\sum _{j\in J}\Vert \Lambda _{j}U_{1}^{*}g\Vert ^{2}+\langle (U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}+U_{2}T_{\Gamma }T_{\Gamma }^{*}U_{2}^{*})g,\,g\rangle \\&\quad \ge \sum _{j\in J}\Vert \Lambda _{j}U_{1}^{*}g\Vert ^{2}\ge A_{1}\Vert K_{1}^{*}U_{1}^{*}g\Vert ^{2}=A_{1}\Vert U_{1}^{*}K_{2}^{*}g\Vert ^{2}\\&\quad \ge A_{1}\Vert U_{1}^{+}\Vert ^{-2}\Vert K_{2}^{*}g\Vert ^{2}.\ \end{aligned}$$

Thus, for every \(g\in {\mathcal {H}}_{2}\), we obtain

$$\begin{aligned} A_{1}\Vert U_{1}^{+}\Vert ^{-2}\Vert K_{2}^{*}g\Vert ^{2}\le \sum _{j\in J}\Vert (\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*})g\Vert ^{2}\le (\sqrt{B_{1}}\Vert U_{1}\Vert +\sqrt{B_{2}}\Vert U_{2}\Vert )^{2}\Vert g\Vert ^{2}. \end{aligned}$$

So \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). \(\square \)

Corollary 2.2

Let \(K_{1}\in B({\mathcal {H}}_{1})\) and \(K_{2}\in B({\mathcal {H}}_{2})\). Suppose that \(\{\Lambda _{j}\}_{j\in J}\) is a \(K_{1}\)-g-frame for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). If \(U\in B({\mathcal {H}}_{1}, {\mathcal {H}}_{2})\) has closed range, \(UK_{1}=K_{2}U\) and \(R(K_{2}^{*})\subset R(U)\), then \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\).

Corollary 2.3

Let \(K,U\in B({\mathcal {H}})\). Suppose that \(\{\Lambda _{j}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). If U is a positive operator and \(US_{\Lambda }=S_{\Lambda }U\), where \(S_{\Lambda }\) is the frame operator of \(\{\Lambda _{j}\}_{j\in J}\), then \(\{\Lambda _{j}+\Lambda _{j}U\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\).

Proof

Assume that \(\{\Lambda _{j}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Since

$$\begin{aligned} T_{\Lambda }T_{\Lambda }^{*}U^{*}+UT_{\Lambda }T_{\Lambda }^{*}+UT_{\Lambda }T_{\Lambda }^{*}U^{*}=S_{\Lambda }U+US_{\Lambda }+US_{\Lambda }U, \end{aligned}$$

from Theorem 2.1, we need only to prove that \(S_{\Lambda }U+US_{\Lambda }+US_{\Lambda }U\ge 0\). According to Proposition 4.33 in [26], we obtain that there exists a unique positive operator C such that \(U=C^{2}\). In addition, since \(US_{\Lambda }=S_{\Lambda }U\), we have \(CS_{\Lambda }=S_{\Lambda }C\). It follows that

$$\begin{aligned} \langle (S_{\Lambda }U+US_{\Lambda }+US_{\Lambda }U)f, f\rangle= & {} 2\langle C^2S_\Lambda f, f\rangle +\langle UT_\Lambda T^*_\Lambda Uf, f\rangle \\= & {} 2\langle CS_\Lambda C f, f\rangle +\langle UT_\Lambda T^*_\Lambda Uf, f\rangle \\= & {} 2\Vert T_{\Lambda }^{*}Cf\Vert ^{2}+\Vert T_{\Lambda }^{*}Uf\Vert ^{2}\\\ge & {} 0 \end{aligned}$$

for all \(f\in {\mathcal {H}}\). So from Theorem 2.1, Corollary 2.3 holds. \(\square \)

Remark 2.4

By taking \(U_{1}=U\) and \(U_{2}=0\), we obtain Corollary 2.2. Let \({\mathcal {H}}_{1}={\mathcal {H}}_{2}={\mathcal {H}}\) and \(K_{1}=K_{2}=K\) in Corollary 2.2, we can correct Theorem 3.4 in [21]. In counterexample 1.12, \(R(K^{*})\subset R(U)\) may not be true. Hence, the condition \(R(K^{*})\subset R(U)\) is necessary. From Corollary 2.2, we may obtain Corollary 5.32 in [3], Proposition 2.24 in [9] and Theorem 3.3 in [16], and we also correct Proposition 12 in [17]. In counterexample 1.14, the condition \(US_{\Lambda }=S_{\Lambda }U\) is not true. Hence, this condition is necessary in Corollary 2.3. From Corollary 2.3, we may obtain Theorem 3.11 in [16]. From Theorem 2.1, we improve Theorem 3.2 in [5], Theorem 2.4 in [10], Corollary 4.4 in [11], Theorem 2.12 in [14] and Theorem 3.5 in [22].

Theorem 2.5

Let \(K_{1}\in B({\mathcal {H}}_{1})\) be an operator with closed range, suppose that \(\{\Lambda _{j}\}_{j\in J}\) and \(\{\Gamma _{j}\}_{j\in J}\) are \(K_{1}\)-g-frames for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Assume \(K_{2}\in B({\mathcal {H}}_{2})\), \(U_{1}, U_{2}\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\) and \(U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}\ge 0\). Then, \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\), if one of the following conditions holds:

\(\mathrm {(1)}\):

\(P=U_{1}+U_{2}\), \(R(P^{*})\subset R(K_{1})\), \(R(K_{2})\subset R(P)\).

\(\mathrm {(2)}\):

\(Q=U_{1}-U_{2}\), \(R(Q^{*})\subset R(K_{1})\), \(R(K_{2})\subset R(Q)\).

Proof

Let \(\{\Lambda _{j}\}_{j\in J}\) be a \(K_{1}\)-g-frame for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with frame bounds \(A_{1}\) and \(B_{1}\). Let \(\{\Gamma _{j}\}_{j\in J}\) be a \(K_{1}\)-g-frame for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with frame bounds \(A_{2}\) and \(B_{2}\). From Lemma 1.9, we obtain that \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with bound \((\sqrt{B_{1}}\Vert U_{1}\Vert +\sqrt{B_{2}}\Vert U_{2}\Vert )^{2}\).

According to the proof of Theorem 2.1, for all \(g\in {\mathcal {H}}_{2}\), we get

$$\begin{aligned}&\sum _{j\in J}\Vert (\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*})g\Vert ^{2}\\&\quad =\sum _{j\in J}\Vert \Lambda _{j}U_{1}^{*}g\Vert ^{2}+\sum _{j\in J}2\mathrm {R}\mathrm {e}\{\langle \Lambda _{j}U_{1}^{*}g,\,\Gamma _{j}U_{2}^{*}g\rangle \}+\sum _{j\in J}\Vert \Gamma _{j}U_{2}^{*}g\Vert ^{2}\\&\quad \ge A_{1}\Vert K_{1}^{*}U_{1}^{*}g\Vert ^{2}+\langle (U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*})g,\,g\rangle +A_{2}\Vert K_{1}^{*}U_{2}^{*}g\Vert ^{2}\\&\quad \ge A_{1}\Vert K_{1}^{*}U_{1}^{*}g\Vert ^{2}+A_{2}\Vert K_{1}^{*}U_{2}^{*}g\Vert ^{2}. \end{aligned}$$

Without loss of generality, assume that statement (1) holds; let \(\lambda =\min \{A_{1},A_{2}\}\), by the parallelogram law and Lemma 1.6, for every \(g\in {\mathcal {H}}_{2}\), we obtain

$$\begin{aligned} A_{1}\Vert K_{1}^{*}U_{1}^{*}g\Vert ^{2}+A_{2}\Vert K_{1}^{*}U_{2}^{*}g\Vert ^{2}\ge & {} \lambda (\Vert K_{1}^{*}U_{1}^{*}g\Vert ^{2}+\Vert K_{1}^{*}U_{2}^{*}g\Vert ^{2})\\= & {} \frac{\lambda }{2}(\Vert K_{1}^{*}(U_{1}+U_{2})^{*}g\Vert ^{2}+\Vert K_{1}^{*}(U_{1}-U_{2})^{*}g\Vert ^{2})\\\ge & {} \frac{\lambda }{2}\Vert K_{1}^{*}(U_{1}+U_{2})^{*}g\Vert ^{2}=\frac{\lambda }{2}\Vert K_{1}^{*}P^{*}g\Vert ^{2}\\\ge & {} \frac{\lambda }{2}\Vert K_{1}^{+}\Vert ^{-2}\Vert P^{*}g\Vert ^{2}. \end{aligned}$$

From \(R(K_{2})\subseteq R(P)\), we conclude that there exists \(\alpha >0\) such that \(K_{2}K_{2}^{*}\le \alpha ^{2}PP^{*}\) by Theorem 1.10. It follows that \(\alpha ^{-2}\Vert K_{2}^{*}g\Vert ^{2}\le \Vert P^{*}g\Vert ^{2}\) for all \(g\in {\mathcal {H}}_{2}\). Thus, for each \(g\in {\mathcal {H}}_{2}\), we get

$$\begin{aligned}&\frac{\lambda }{2}\alpha ^{-2}\Vert K_{1}^{+}\Vert ^{-2}\Vert K_{2}^{*}g\Vert ^{2}\\&\quad \le \sum _{j\in J}\Vert (\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*})g\Vert ^{2}\le (\sqrt{B_{1}}\Vert U_{1}\Vert +\sqrt{B_{2}}\Vert U_{2}\Vert )^{2}\Vert g\Vert ^{2}. \end{aligned}$$

This proves that \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). \(\square \)

Remark 2.6

From Theorem 2.5, we can get Proposition 2.24 in [9], Proposition 3.6 in [15], Theorem 3.5 in [20] and Proposition 3.2 in [23]. It is natural to consider whether the conditions \(R(P^{*})\subset R(K_{1})\) and \(R(Q^{*})\subset R(K_{1})\) are not necessary in Theorem 2.5. Now we give an example to illustrate that the conditions are essential.

Example 2.7

Let \({\mathcal {H}}_{1}={\mathbb {C}}^{3}\) and \(J=\{1, 2, 3\}\). Assume that \(\{e_{j}\}_{j\in J}\) is an orthonormal basis for \({\mathcal {H}}_{1}\) and \({\mathcal {V}}_{j}=\overline{span}\{e_{j}\}\). Let \(\{g_{j}\}_{j=1}^{4}\) be an orthonormal basis for \({\mathcal {H}}_{2}={\mathbb {C}}^{4}\). Now define \(K_{1}\in B({\mathcal {H}}_{1})\), \(K_{2}\in B({\mathcal {H}}_{2})\), \(U_{1},U_{2}\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\) and \(\{\Lambda _{j}\}_{j\in J}\) as follows:

$$\begin{aligned} K_{1}:{\mathcal {H}}_{1}\rightarrow {\mathcal {H}}_{1},\, K_{1}f= & {} \langle f,\,e_{1}\rangle e_{1}+\langle f,\,e_{3}\rangle e_{2},\quad f \in {\mathcal {H}}_{1},\\ K_{2}:{\mathcal {H}}_{2}\rightarrow {\mathcal {H}}_{2},\, K_{2}g= & {} \langle g,\,g_{2}\rangle g_{1},\, \, \, g \in {\mathcal {H}}_{2},\\ U_{1}:{\mathcal {H}}_{1}\rightarrow {\mathcal {H}}_{2},\, U_{1}f= & {} \langle f,\,e_{2}\rangle g_{3}+\langle f,\,e_{3}\rangle g_{1},\quad f \in {\mathcal {H}}_{1},\\ U_{2}:{\mathcal {H}}_{1}\rightarrow {\mathcal {H}}_{2},\, U_{2}f= & {} \langle f,\,e_{2}\rangle g_{3},\quad f \in {\mathcal {H}}_{1},\\ \Lambda _{1}:{\mathcal {H}}_{1}\rightarrow {\mathcal {V}}_{1},\,\Lambda _{1}f= & {} \langle f,\,e_{2}\rangle e_{1},\quad f \in {\mathcal {H}}_{1},\\ \Lambda _{2}:{\mathcal {H}}_{1}\rightarrow {\mathcal {V}}_{2},\,\Lambda _{2}f= & {} \langle f,\,e_{1}\rangle e_{2},\quad f \in {\mathcal {H}}_{1},\\ \Lambda _{3}:{\mathcal {H}}_{1}\rightarrow {\mathcal {V}}_{3},\,\Lambda _{3}f= & {} \langle f,\,e_{2}\rangle e_{3},\quad f \in {\mathcal {H}}_{1}.\\ \end{aligned}$$

Let \(\Gamma _{j}=\Lambda _{j},\,j=1,2,3.\) Now we prove that \(K_{1}^{*}f=\langle f,\,e_{1}\rangle e_{1}+\langle f,\,e_{2}\rangle e_{3},\,f\in {\mathcal {H}}_{1}\). Indeed, for all \(f,m\in {\mathcal {H}}_{1}\), we have

$$\begin{aligned} \langle K_{1}^{*}f,\,m\rangle= & {} \langle f,\,K_{1}m\rangle =\langle f,\,\langle m,\,e_{1}\rangle e_{1}+\langle m,\,e_{3}\rangle e_{2}\rangle \\= & {} \langle f,\,e_{1}\rangle \overline{\langle m,\,e_{1}\rangle }+\langle f,\,e_{2}\rangle \overline{\langle m,\,e_{3}\rangle }\\= & {} \langle f,\,e_{1}\rangle \langle e_{1},\,m\rangle +\langle f,\,e_{2}\rangle \langle e_{3},\,m\rangle \\= & {} \langle \langle f,\,e_{1}\rangle e_{1}+\langle f,\,e_{2}\rangle e_{3},\,m\rangle . \end{aligned}$$

Hence, for any \(f\in {\mathcal {H}}_{1}\), we get

$$\begin{aligned} \Vert K_{1}^{*}f\Vert ^{2}= & {} \Vert \langle f,\,e_{1}\rangle e_{1}+\langle f,\,e_{2}\rangle e_{3}\Vert ^{2}=|\langle f,\,e_{1}\rangle |^{2}+|\langle f,\,e_{2}\rangle |^{2}\\\le & {} \sum _{j=1}^{3}\Vert \Lambda _{j}f\Vert ^{2}=\sum _{j=1}^{3}\Vert \Gamma _{j}f\Vert ^{2}=|\langle f,\,e_{1}\rangle |^{2}+2|\langle f,\,e_{2}\rangle |^{2}\le 2\Vert f\Vert ^{2}. \end{aligned}$$

It follows that \(\{\Lambda _{j}\}_{j\in J}\) and \(\{\Gamma _{j}\}_{j\in J}\) are \(K_{1}\)-g-frame for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Let \(T_{\Lambda }\) and \(T_{\Gamma }\) be the corresponding synthesis operators of \(\{\Lambda _{j}\}_{j\in J}\) and \(\{\Gamma _{j}\}_{j\in J}\), respectively. Since \(\Gamma _{j}=\Lambda _{j},\,j=1,2,3\), for every \(f\in {\mathcal {H}}_{1}\), we obtain

$$\begin{aligned} T_{\Lambda }T_{\Gamma }^{*}f=T_{\Gamma }T_{\Lambda }^{*}f=T_{\Lambda }T_{\Lambda }^{*}f=\sum _{j=1}^{3}\Lambda _{j}^{*}\Lambda _{j}f=\langle f,\,e_{1}\rangle e_{1}+2\langle f,\,e_{2}\rangle e_{2}. \end{aligned}$$

Now we show that \(U_{1}^{*}g=\langle g,\,g_{3}\rangle e_{2}+\langle g,\,g_{1}\rangle e_{3},\,\,g\in {\mathcal {H}}_{2}\) and \(U_{2}^{*}g=\langle g,\,g_{3}\rangle e_{2},\,g\in {\mathcal {H}}_{2}\). In fact, for all \(f\in {\mathcal {H}}_{1}\) and \(g\in {\mathcal {H}}_{2}\), we obtain

$$\begin{aligned} \langle U_{1}^{*}g,\,f\rangle= & {} \langle g,\,U_{1}f\rangle =\langle g,\,\langle f,\,e_{2}\rangle g_{3}+\langle f,\,e_{3}\rangle g_{1}\rangle \\= & {} \langle g,\,g_{3}\rangle \overline{\langle f,\,e_{2}\rangle }+\langle g,\,g_{1}\rangle \overline{\langle f,\,e_{3}\rangle }\\= & {} \langle g,\,g_{3}\rangle \langle e_{2},\,f\rangle +\langle g,\,g_{1}\rangle \langle e_{3},\,f\rangle =\langle \langle g,\,g_{3}\rangle e_{2}+\langle g,\,g_{1}\rangle e_{3},\,f\rangle , \end{aligned}$$

and

$$\begin{aligned} \langle U_{2}^{*}g,\,f\rangle= & {} \langle g,\,U_{2}f\rangle =\langle g,\,\langle f,\,e_{2}\rangle g_{3}\rangle =\langle g,\,g_{3}\rangle \overline{\langle f,\,e_{2}\rangle }\\= & {} \langle g,\,g_{3}\rangle \langle e_{2},\,f\rangle =\langle \langle g,\,g_{3}\rangle e_{2},\,f\rangle . \end{aligned}$$

By a direct calculation, we can conclude

$$\begin{aligned} \langle (U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*})g,\,g\rangle =4|\langle g,\,g_{3}\rangle |^{2}\ge 0 \end{aligned}$$

for all \(g\in {\mathcal {H}}_{2}\). This implies that \(U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}\ge 0\).

Now we prove that \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Indeed, for each \(g\in {\mathcal {H}}_{2}\), we get

$$\begin{aligned} \sum _{j=1}^{3}\Vert (\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*})g\Vert ^{2}=\Vert 2\langle g,\,g_{3}\rangle e_{1}\Vert ^{2}+\Vert 2\langle g,\,g_{3}\rangle e_{3}\Vert ^{2}=8|\langle g,\,g_{3}\rangle |^{2}\le 8\Vert g\Vert ^{2}. \end{aligned}$$

The adjoint operator of \(K_{2}\) is \(K_{2}^{*}\), \(K_{2}^{*}g=\langle g,\,g_{1}\rangle g_{2},\,\,g\in {\mathcal {H}}_{2}\). In fact, for any \(g,h\in {\mathcal {H}}_{2}\), we obtain

$$\begin{aligned} \langle K_{2}^{*}g,\,h\rangle= & {} \langle g,\,K_{2}h\rangle =\langle g,\,\langle h,\,g_{2}\rangle g_{1}\rangle =\langle g,\,g_{1}\rangle \overline{\langle h,\,g_{2}\rangle }\\= & {} \langle g,\,g_{1}\rangle \langle g_{2},\,h\rangle =\langle \langle g,\,g_{1}\rangle g_{2},\,h\rangle . \end{aligned}$$

We can choose \(g=g_{1}\in {\mathcal {H}}_{2}\), then we obtain \(\Vert K_{2}^{*}g\Vert ^{2}=|\langle g,\,g_{1}\rangle |^{2}=1\) and

$$\begin{aligned} \sum _{j=1}^{3}\Vert (\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*})g\Vert ^{2}=8|\langle g,\,g_{3}\rangle |^{2}=0. \end{aligned}$$

Therefore, \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is not a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). It is obvious that \(K_{1}\in B({\mathcal {H}}_{1})\) has closed range. Let \(P=U_{1}+U_{2}\) and \(Q=U_{1}-U_{2}\), we have \(Pf=2\langle f,\,e_{2}\rangle g_{3}+\langle f,\,e_{3}\rangle g_{1},\,\,f\in {\mathcal {H}}_{1}\) and \(Qf=\langle f,\,e_{3}\rangle g_{1},\,\,f\in {\mathcal {H}}_{1}\). Hence, we get

$$\begin{aligned} \overline{span}\{g_{1}\}= & {} R(K_{2})\subset R(P)=\overline{span}\{g_{1},2g_{3}\},\\ \overline{span}\{g_{1}\}= & {} R(K_{2})\subset R(Q)=\overline{span}\{g_{1}\}, \end{aligned}$$

but

$$\begin{aligned} \overline{span}\{2e_{2},e_{3}\}= & {} R(P^{*})\nsubseteq R(K_{1})=\overline{span}\{e_{1},e_{2}\},\\ \overline{span}\{e_{3}\}= & {} R(Q^{*})\nsubseteq R(K_{1})=\overline{span}\{e_{1},e_{2}\}. \end{aligned}$$

In the following, we offer an equivalent characterization of generating K-g-frames.

Remark 2.8

In [21, Corollary 3.17], it was stated that \(K\in B({\mathcal {H}})\) is an operator with closed range and \(\{\Lambda _{j}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\); suppose that \(U\in B({\mathcal {H}})\) has closed range and \(UK=KU\), then the following conditions are equivalent: \(\mathrm {(1)}\) U is surjective; \(\mathrm {(2)}\) \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). We announce a counterexample in Example 2.9.

Example 2.9

Assume \({\mathcal {H}}={\mathbb {C}}^{3}\), \(J=\{1,2,3\}\). Let \(\{e_{j}\}_{j\in J}\) be an orthonormal basis of \({\mathcal {H}}\), and \({\mathcal {V}}_{j}=\overline{span}\{e_{j}\}\). Now define \(K\in B({\mathcal {H}})\), \(U\in B({\mathcal {H}})\) and \(\{\Lambda _{j}\}_{j\in J}\) as follows:

$$\begin{aligned} K:{\mathcal {H}}\rightarrow {\mathcal {H}},\, Kf= & {} \langle f,\,e_{1}\rangle e_{2}+\langle f,\,e_{2}\rangle e_{1},\quad f \in {\mathcal {H}},\\ U:{\mathcal {H}}\rightarrow {\mathcal {H}},\, Uf= & {} \sum _{j=1}^{2}\langle f,\,e_{j}\rangle e_{j}\quad f \in {\mathcal {H}},\\ \Lambda _{1}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{1},\,\Lambda _{1}f= & {} \langle f,\,e_{2}\rangle e_{1},\quad f \in {\mathcal {H}},\\ \Lambda _{2}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{2},\,\Lambda _{2}f= & {} \langle f,\,e_{1}\rangle e_{2},\quad f \in {\mathcal {H}},\\ \Lambda _{3}:{\mathcal {H}}\rightarrow {\mathcal {V}}_{3},\,\Lambda _{3}f= & {} \langle f,\,e_{3}\rangle e_{3},\quad f \in {\mathcal {H}}. \end{aligned}$$

It is clear that \(K\in B({\mathcal {H}})\) has closed range. The adjoint operator of K is \(K^{*}\), \(K^{*}f=\langle f,\,e_{1}\rangle e_{2}+\langle f,\,e_{2}\rangle e_{1},\,f\in {\mathcal {H}}\). In fact, for all \(f,m\in {\mathcal {H}}\), we have

$$\begin{aligned} \langle K^{*}f,\,m\rangle= & {} \langle f,\,Km\rangle =\langle f,\langle m,\,e_{1}\rangle e_{2}+\langle m,\,e_{2}\rangle e_{1}\rangle \\= & {} \langle f,\,e_{2}\rangle \overline{\langle m,\,e_{1}\rangle }+\langle f,\,e_{1}\rangle \overline{\langle m,\,e_{2}\rangle }\\= & {} \langle f,\,e_{2}\rangle \langle e_{1},\,m\rangle +\langle f,\,e_{1}\rangle \langle e_{2},\,m\rangle =\langle \langle f,\,e_{1}\rangle e_{2}+\langle f,\,e_{2}\rangle e_{1},\,m\rangle . \end{aligned}$$

For every \(f\in {\mathcal {H}}\), we get

$$\begin{aligned} \Vert K^{*}f\Vert ^{2}= & {} \Vert \langle f,\,e_{1}\rangle e_{2}+\langle f,\,e_{2}\rangle e_{1}\Vert ^{2}=|\langle f,\,e_{1}\rangle |^{2}+|\langle f,\,e_{2}\rangle |^{2}\\\le & {} \sum _{j=1}^{3}\Vert \Lambda _{j}f\Vert ^{2}=\sum _{j=1}^{3}|\langle f,\,e_{j}\rangle |^{2}=\Vert f\Vert ^{2}. \end{aligned}$$

Thus, \(\{\Lambda _{j}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\).

It is obvious that \(U\in B({\mathcal {H}})\) has closed range. By a direct calculation, for each \(f\in {\mathcal {H}}\), we obtain

$$\begin{aligned} UKf= & {} U(\langle f,\,e_{1}\rangle e_{2}+\langle f,\,e_{2}\rangle e_{1})=\langle f,\,e_{1}\rangle Ue_{2}+\langle f,\,e_{2}\rangle Ue_{1}\\= & {} \langle f,\,e_{1}\rangle e_{2}+\langle f,\,e_{2}\rangle e_{1}\\= & {} \langle f,\,e_{1}\rangle Ke_{1}+\langle f,\,e_{2}\rangle Ke_{2}=K\sum _{j=1}^{2}\langle f,\,e_{j}\rangle e_{j}=KUf, \end{aligned}$$

then \(UK=KU\).

By a simple computation, we get \(U^{*}f=\sum _{j=1}^{2}\langle f,\,e_{j}\rangle e_{j},\,\,f\in {\mathcal {H}}\), then for all \(f\in {\mathcal {H}}\), we have

$$\begin{aligned} \Vert K^{*}f\Vert ^{2}= & {} \Vert \langle f,\,e_{1}\rangle e_{2}+\langle f,\,e_{2}\rangle e_{1}\Vert ^{2}=|\langle f,\,e_{1}\rangle |^{2}+|\langle f,\,e_{2}\rangle |^{2}\\= & {} \sum _{j=1}^{3}\Vert \Lambda _{j}U^{*}f\Vert ^{2}=\sum _{j=1}^{2}|\langle f,\,e_{j}\rangle |^{2}\le \Vert f\Vert ^{2}. \end{aligned}$$

Hence, \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is a K-g-frame for \({\mathcal {H}}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) and it is clear that U is not surjective.

Theorem 2.10

Let \(K_{1}\in B({\mathcal {H}}_{1})\) and \(\{\Lambda _{j}\}_{j\in J}\) be a \(K_{1}\)-g-frame for \({\mathcal {H}}_{1}\) with respect to \(\{{\mathcal {V}}_{j}\}_{j\in J}\). Suppose that \(K_{2}\in B({\mathcal {H}}_{2})\) is an operator with dense range, \(U\in B({\mathcal {H}}_{1}, {\mathcal {H}}_{2})\) is an operator with closed range and \(UK_{1}=K_{2}U\), then \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) if and only if U is surjective.

Proof

Suppose that U is surjective, it is obvious that \(R(K_{2}^{*})\subset R(U)\). According to Corollary 2.2, \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\).

On the other hand, let \(T_{\Lambda }\) be the synthesis operator of the \(K_{1}\)-g-frame \(\{\Lambda _{j}\}_{j\in J}\) and L be the synthesis operator of the \(K_{2}\)-g-frame \(\{\Lambda _{j}U^{*}\}_{j\in J}\), then for any \(\{g_{j}\}_{j\in J}\in {\ell ^{2}}(\{\mathcal {V}_{j}\}_{j\in J})\), we obtain

$$\begin{aligned} L\{g_{j}\}_{j\in J}=\sum _{j\in J}(\Lambda _{j}U^{*})^{*}g_{j}=U\sum _{j\in J}\Lambda _{j}^{*}g_{j}=UT_{\Lambda }\{g_{j}\}_{j\in J}. \end{aligned}$$

Hence, we get \(L=UT_{\Lambda }\).

Since \(\{\Lambda _{j}U^{*}\}_{j\in J}\) is a \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\), we have \(R(K_{2})\subset R(L)\) by Lemma 1.7. Hence \(R(K_{2})\subset R(UT_{\Lambda })\subset R(U)\). Then, we obtain \(\overline{R(K_{2})}\subset \overline{R(U)}\). Since \(K_{2}\in B({\mathcal {H}}_{2})\) is an operator with dense range and \(U\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\) is an operator with closed range, then U is surjective. \(\square \)

Remark 2.11

By taking \({\mathcal {H}}_{1}={\mathcal {H}}_{2}={\mathcal {H}}\) and \(K_{1}=K_{2}=K\), we may correct Corollary 3.17 in [21]. In counterexample 2.9, we obtain that statement (2) does not imply statement (1).

In the end, we give a necessary and sufficient condition to yield a series of tight K-g-frames by two existing g-Bessel sequences.

Theorem 2.12

Suppose that \(\{\Lambda _{j}\}_{j\in J}\) and \(\{\Gamma _{j}\}_{j\in J}\) are g-Bessel sequences for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with synthesis operators \(T_{\Lambda }\), \(T_{\Gamma }\) and frame operators \(S_{\Lambda }\), \(S_{\Gamma }\), respectively. Let \(K\in B({\mathcal {H}}_{2})\) and \(U_{1}, U_{2}\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\). Then, \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a tight K-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) if and only if there exists \(A>0\) such that

$$\begin{aligned} AKK^{*}=U_{1}S_{\Lambda }U_{1}^{*}+U_{2}S_{\Gamma }U_{2}^{*}+U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}. \end{aligned}$$

Proof

According to Lemma 1.9, we get that \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a g-Bessel sequence for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\). Let L be the synthesis operator of \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\), then for each \(\{g_{j}\}_{j\in J}\in {\ell ^{2}}(\{\mathcal {V}_{j}\}_{j\in J})\), we obtain

$$\begin{aligned} L\{g_{j}\}_{j\in J}= & {} \sum _{j\in J}(\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*})^{*}g_{j}=U_{1}\sum _{j\in J}\Lambda _{j}^{*}g_{j}+U_{2}\sum _{j\in J}\Gamma _{j}^{*}g_{j}\\= & {} (U_{1}T_{\Lambda }+U_{2}T_{\Gamma })\{g_{j}\}_{j\in J}. \end{aligned}$$

Thus \(L=U_{1}T_{\Lambda }+U_{2}T_{\Gamma }\). Suppose that S is the frame operator of \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\), then

$$\begin{aligned} S= & {} LL^{*}=(U_{1}T_{\Lambda }+U_{2}T_{\Gamma })(U_{1}T_{\Lambda }+U_{2}T_{\Gamma })^{*}\\= & {} U_{1}T_{\Lambda }T_{\Lambda }^{*}U_{1}^{*}+U_{2}T_{\Gamma }T_{\Gamma }^{*}U_{2}^{*}+U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}\\= & {} U_{1}S_{\Lambda }U_{1}^{*}+U_{2}S_{\Gamma }U_{2}^{*}+U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}. \end{aligned}$$

By Lemma 1.8, \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a tight K-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) if and only if there exists \(A>0\) such that

$$\begin{aligned} AKK^{*}=U_{1}S_{\Lambda }U_{1}^{*}+U_{2}S_{\Gamma }U_{2}^{*}+U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}. \end{aligned}$$

This completes the proof. \(\square \)

From Lemma 1.8 and Theorem 2.12, we have the following corollary.

Corollary 2.13

Let \(K_{1}\in B({\mathcal {H}}_{1})\), \(\{\Lambda _{j}\}_{j\in J}\) and \(\{\Gamma _{j}\}_{j\in J}\) be tight \(K_{1}\)-g-frames for \({\mathcal {H}}_{1}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) with frame bounds \(A_{1}\) and \(A_{2}\). Let \(K_{2}\in B({\mathcal {H}}_{2})\) and \(U_{1}, U_{2}\in B({\mathcal {H}}_{1},{\mathcal {H}}_{2})\). Assume that \(T_{\Lambda }\) and \(T_{\Gamma }\) are synthesis operators of \(\{\Lambda _{j}\}_{j\in J}\) and \(\{\Gamma _{j}\}_{j\in J}\), respectively. Then, \(\{\Lambda _{j}U_{1}^{*}+\Gamma _{j}U_{2}^{*}\}_{j\in J}\) is a tight \(K_{2}\)-g-frame for \({\mathcal {H}}_{2}\) with respect to \(\{\mathcal {V}_{j}\}_{j\in J}\) if and only if there exists \(A>0\) such that

$$\begin{aligned} AK_{2}K_{2}^{*}=A_{1}U_{1}K_{1}K_{1}^{*}U_{1}^{*}+A_{2}U_{2}K_{1}K_{1}^{*}U_{2}^{*}+U_{1}T_{\Lambda }T_{\Gamma }^{*}U_{2}^{*}+U_{2}T_{\Gamma }T_{\Lambda }^{*}U_{1}^{*}. \end{aligned}$$

Remark 2.14

We can obtain Theorem 2.1 in [18] and Theorem 10 in [24] from Theorem 2.12. From Corollary 2.13, we may obtain Theorem 2.2, Theorem 2.3, Theorem 2.4 and Theorem 2.5 in [18]; meanwhile, we can also get Theorem 14, Theorem 16, Theorem 18 and Theorem 20 in [24]. We improve Theorem 2.7 in [10] and Theorem 4.7 in [11] by Corollary 2.13.