Some Properties of K-Frames in Quaternionic Hilbert Spaces

Abstract

In the present paper, we discuss some properties of K-frames in quaternionic Hilbert spaces such as the invertibility of the frame operator as well as the interchangeability of two Bessel sequences. Further, we propose several approaches to construct K-frames and we show that a T-frame can be constructed from a K-frame by the perturbation of a bounded linear operator T. Finally, we study the stability of K-frames under some perturbations.

1 Introduction

The theory of frames, dates backs to 1952, was introduced in the pioneeristic paper of Duffin and Schaeffer [11] in the context of nonharmonic Fourier series and popularized later by Daubechies et al. [10] in 1986. A frame is a countable family of elements in a separable Hilbert space which allows stable and not necessarily unique decompositions of arbitrary elements in an expansion of frame elements. It can be viewed as a generalization of Riesz basis intensely studied in literature [4, 5, 12, 13, 18]. The redundancy and flexibility offered by frames have spurred their applications in a variety of areas such as wavelet and frequency analysis theories, filter bank theory, signal and image processing.

In the last two decades, the theory of frames has been a focus of study for several authors such as Christensen [8], Jeribi [18] and Young [22]. Further, many generalizations of this notion have been introduced and developed in numerous paper as [2, 3, 6, 14, 15, 21].

Mainly in [15], Găvruţa introduced the concept of K-frames to study the atomic systems with respect to a bounded linear operator K in separable Hilbert spaces. This generalization allows to write every element of the range of K, which need not be closed, as a superposition of elements which do not necessarily belong to the range of K. However, since a K-frame may not be in general a frame, unless if $K=I$, many properties of ordinary frames such as the surjectivity of the corresponding synthesis operator, the isomorphism of the frame operator, the interchangeability of the alternate dual reconstruction pair; may not hold for K-frames as shown in [21]. Several methods to construct K-frames and the stability of perturbations for the K-frames have been discussed in [17, 19].

Recently, Charfi and Ellouz [7] extend the concepts of atomic systems for operators and K-frames in separable complex Hilbert spaces to separable quaternionic Hilbert spaces. They have studied the existence of atomic systems for operators and they gave a formal definition of K-frames. Moreover, they have characterized them. These concepts lead to a generalization of frames which were recently studied in [20] and allow us to reconstruct elements from the range of a linear and bounded operator in a quaternionic Hilbert space.

In the present paper, we investigate some properties of K-frames in quaternionic Hilbert spaces. More precisely, we will focus on the study of the isomorphism of the frame operator as well as the interchangeability of the two Bessel sequences in the alternate dual reconstruction for K-frames. Further, we give several approaches for constructing new K-frames from given ones. Finally, we discuss the stability of perturbations for the K-frames.

Throughout this paper, we will adopt such notations. $V_{R}(\mathfrak {Q})$ is a separable right quaternionic Hilbert space; I is the identity operator for $V_{R}(\mathfrak {Q})$; $\mathcal{L}(V_{R}(\mathfrak {Q}),V_{1,R}(\mathfrak {Q}))$ is a collection of all bounded linear operators from $V_{R}(\mathfrak {Q})$ to $V_{1,R}(\mathfrak {Q})$ , where $V_{R}(\mathfrak {Q})$, $V_{1,R}(\mathfrak {Q})$ are two quaternionic Hilbert spaces, and if $V_{R}(\mathfrak {Q})=V_{1,R}(\mathfrak {Q})$, $\mathcal{L}(V_{R}(\mathfrak {Q}),V_{1,R}(\mathfrak {Q}))$ is denoted by $\mathcal{L}(V_{R}(\mathfrak {Q}))$. Let $K \in \mathcal{L}(V_{R}(\mathfrak {Q}))$. The range of K is denoted by R(K) and the pseudo-inverse of K is denoted by $K^{\dagger }$.

2 Quaternionic Hilbert Space

As the quaternions are non-commutative in nature therefore there are two different types of quaternionic Hilbert spaces, the left quaternionic Hilbert space and the right quaternionic Hilbert space depending on positions of quaternions. This fact can entail several problems. For example, when a Hilbert space H is one-sided (either left or right) the set of linear operators acting on it does not have a linear structure. In this section, we will study some basic notations about the algebra of quaternions, right quaternionic Hilbert spaces and operators acting on these spaces.

In the next, let $\mathfrak {Q}$ denotes the skew field of quaternions, whose elements are in the form $q = x_0 + x_1i + x_2j + x_3k$, where $x_0,x_1,x_2$ and $x_3$ are real numbers and i, j, k are called imaginary units and obey the following multiplication rules:

$$\begin{aligned} i^2 = j^2 = k^2 =-1, ~~ij =-ji= k, ~~jk =-kj = i, ~~\hbox {and}~~ ki =-ik=j. \end{aligned}$$

For more information about the properties of quaternions, we refer the readers to [1, 9, 16].

Let $V_{R}(\mathfrak {Q})$ be a linear vector space over the skew field of quaternions under right scalar multiplication. It is called a quaternionic pre-Hilbert space if there exists a Hermitian quaternionic scalar product; that is, a map

$$\begin{aligned} \langle , \,\rangle :V_{R}(\mathfrak {Q})\times V_{R}(\mathfrak {Q})\rightarrow \mathfrak {Q} \end{aligned}$$

satisfying, for every $u,v,w \in V_{R}(\mathfrak {Q})$ and $p,q \in \mathfrak {Q}$, the following properties:

• (i) $\langle u,v\rangle =\overline{\langle v,u\rangle }$,

• (ii) $\langle u,u\rangle > 0$ unless $u = 0$,

• (iii) $\langle u,vp+ wq\rangle =\langle u,v\rangle p +\langle u,w\rangle q$.

Suppose that $V_{R}(\mathfrak {Q})$ is equipped with such a Hermitian quaternionic scalar product. Then, we can define the quaternionic norm $\Vert \, . \, \Vert :V_{R}(\mathfrak {Q})\rightarrow \mathbb {R}_{+}$ by setting

$$\begin{aligned} \Vert u\Vert = \sqrt{\langle u,u\rangle },~~u \in V_{R}(\mathfrak {Q}). \end{aligned}$$

(2.1)

It has been mentioned, in [9, 16], that the quaternionic norm satisfies all properties of a norm, including Cauchy-Schwarz inequality and parallelogram identity.

The right quaternionic pre-Hilbert space $V_{R}(\mathfrak {Q})$ is said to be a right quaternionic Hilbert space, if it is complete with respect to the norm (2.1).

It should be noted here that quaternionic Hilbert spaces share many of the standard properties of complex Hilbert spaces such as Hilbert basis. So, let us recall the following results from [9, 16].

Proposition 2.1

Let $V_{R}(\mathfrak {Q})$ be a right quaternionic Hilbert space and N be a subset of $V_{R}(\mathfrak {Q})$ such that, for $z,z'\in N$, $\langle z,z'\rangle =0$ if $z \ne z'$ and $\langle z,z'\rangle =1$. Then, the following assertions are equivalent:

• (i) For every $u,v\in V_{R}(\mathfrak {Q})$, the series $\sum _{z\in N}\langle u,z\rangle \langle z,v\rangle$ converges absolutely and it holds:

  $$\begin{aligned} \langle u,v\rangle = \displaystyle \sum _{z\in N}\langle u,z\rangle \langle z,v\rangle . \end{aligned}$$

• (ii) $\Vert u\Vert ^2 = \sum _{z\in N}\vert \langle z,u\rangle \vert ^2$ for every $u\in V_{R}(\mathfrak {Q})$.

• (iii) $N^{\bot } :=\left\{ v \in V_{R}(\mathfrak {Q}) :\langle v,z\rangle =0 , ~~\forall z \in N\right\} =\{0\}$.

• (iv) Span(N) is dense in $V_{R}(\mathfrak {Q})$.

$\diamondsuit$

Remark 2.1

The subset N in Proposition 2.1 is called a Hilbert basis.$\diamondsuit$

Proposition 2.2

Every right quaternionic Hilbert space admits a Hilbert basis, and two Hilbert bases have the same cardinality. Furthermore, if N is a Hilbert basis of $V_{R}(\mathfrak {Q})$, then every $u\in V_{R}(\mathfrak {Q})$ can be uniquely decomposed as follows

$$\begin{aligned} u = \displaystyle \sum _{z\in N} z\langle z,u\rangle , \end{aligned}$$

where the series $\sum _{z\in N} z\langle z,u\rangle$ converges absolutely in $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Now, we shall define right $\mathfrak {Q}$-linear operators and recall some basic properties.

A mapping $T : V_{R}(\mathfrak {Q}) \longrightarrow V_{R}(\mathfrak {Q})$ is said to be a right linear operator if for all $u,v\in V_{R}(\mathfrak {Q})$ and $p\in \mathfrak {Q}$,

$$\begin{aligned} T(up + v)=(Tu)p + Tv. \end{aligned}$$

Such an operator is called bounded if there exists $K \ge 0$ such that for all $u\in V_{R}(\mathfrak {Q})$,

$$\begin{aligned} \Vert Tu\Vert \le K\Vert u\Vert . \end{aligned}$$

As in the complex case, the norm of a bounded right linear operator T is defined by

$$\begin{aligned} \Vert T\Vert =\left\{ \sup \displaystyle \frac{\Vert Tu\Vert }{\Vert u\Vert },~~0\ne u\in V_{R}(\mathfrak {Q})\right\} . \end{aligned}$$

(2.2)

The set of all bounded right linear operators on $V_{R}(\mathfrak {Q})$ is denoted by $\mathcal{L}(V_{R}(\mathfrak {Q}))$, which is a complete normed space with the norm defined by (2.2) (see [9, 16]). For every $T \in \mathcal{L}(V_{R}(\mathfrak {Q}))$, there exists a unique operator $T^* \in \mathcal{L}(V_{R}(\mathfrak {Q}))$, which is called the adjoint of T, such that, for all $u,v \in V_{R}(\mathfrak {Q})$,

$$\begin{aligned} \langle Tu,v\rangle = \langle u,T^*v\rangle ~~\hbox {and} ~~\Vert T\Vert = \Vert T^*\Vert . \end{aligned}$$

3 Properties of K-Frames

In this part, we explore some properties of K-frame. To this interest, let us introduce some basic facts about the concept of atomic systems and K-frames.

Throughout this paper, $I\subseteq \mathbb {N}$ denotes a finite or countable index set.

Let’s begin with the definition of frame and Bessel sequence generalized by Sharma and Goel in [20] to separable right quaternionic Hilbert spaces $V_{R}(\mathfrak {Q})$.

Definition 3.1

[20] A family $\{f_n\}_{n\in I}$ is said to be a frame for $V_{R}(\mathfrak {Q})$, if there exist two positive constants $0< A \le B<\infty$ such that

$$\begin{aligned} A\Vert x\Vert ^2\le \displaystyle \sum _{n\in I} \vert \langle f_n,x\rangle \vert ^2\le B\Vert x\Vert ^2, ~~\hbox {for all} ~~x \in V_{R}(\mathfrak {Q}). \end{aligned}$$

(3.1)

The numbers A and B are called lower and upper frame bounds. If only the right inequality of Eq. (3.1) holds, $\{f_n\}_{n\in I}$ is called a Bessel sequence.$\diamondsuit$

For a Bessel sequence $\{f_n\}_{n\in I}$, we define its synthesis operator $T: l^2(\mathfrak {Q}) \rightarrow V_{R}(\mathfrak {Q})$ by

$$\begin{aligned}Tq = \displaystyle \sum _{n\in I}f_nq_n,~~q=\{q_n\}\in l^2(\mathfrak {Q}), \end{aligned}$$

where

$$\begin{aligned}l^2(\mathfrak {Q}):=\left\{ \{q_i\}_{i\in I}\subset \mathfrak {Q}~~\hbox {such that}~~\displaystyle \sum _{i\in I}\vert q_i\vert ^2<+\infty \right\} \end{aligned}$$

equipped with the following quaternionic inner product

$$\begin{aligned} \langle p,q\rangle =\displaystyle \sum _{i\in I}\overline{p_i}q_i,~~p=\{p_i\}_{i\in I}~~\hbox {and}~~q=\{q_i\}_{i\in I} \in l^2(\mathfrak {Q}) \end{aligned}$$

defines a right quaternionic Hilbert space.

The adjoint operator of T, $T^* : V_{R}(\mathfrak {Q})\rightarrow l^2(\mathfrak {Q})$ defined by $T^*f=\{\langle f_n,f\rangle \}_{n\in I}$ for $f\in V_{R}(\mathfrak {Q})$, is called the analysis operator. By composing T with its adjoint $T^*$, we obtain the frame operator

$$\begin{aligned} S:V_{R}(\mathfrak {Q})\rightarrow V_{R}(\mathfrak {Q}),~~Sf=TT^*f=\displaystyle \sum _{n\in I}f_n\langle f_n,f\rangle . \end{aligned}$$

Now, we recall the concept of atomic systems and K-frames introduced in [7].

Definition 3.2

[7] A family $\{f_n\}_{n\in I}$ of $V_{R}(\mathfrak {Q})$ is called an atomic system for $K\in \mathcal{L} (V_{R}(\mathfrak {Q}))$ if the following statements hold:

• (i) the series $\sum _{n\in I}f_nc_n$ converges for all $c=\{c_n\}_{n\in I}\in l^2(\mathfrak {Q})$;

• (ii) there exists $C>0$ such that for every $x\in V_{R}(\mathfrak {Q})$ there exists $a_x=\{a_{n}\}_{n\in I}\in l^2(\mathfrak {Q})$ such that $\Vert a_x\Vert _{l^2(\mathfrak {Q})}\le C\Vert x\Vert$ and $Kx=\sum _{n\in I}f_na_n$.$\diamondsuit$

Proposition 3.1

[7] Let $\{f_n\}_{n\in I}\subset V_{R}(\mathfrak {Q})$. Then, the following statements are equivalent:

• (i) $\{f_n\}_{n\in I}$ is an atomic system for K;

• (ii) there exist $A, B>0$ such that

  $$\begin{aligned} A\Vert K^*x\Vert ^2\le \displaystyle \sum _{n\in I}\vert \langle f_n,x\rangle \vert ^2\le B\Vert x\Vert ^2,~~\forall x\in V_{R}(\mathfrak {Q}); \end{aligned}$$

  (3.2)

• (iii) $\{f_n\}_{n\in I}$ is a Bessel sequence and there exists a Bessel sequence $\{g_n\}_{n\in I}$ such that

  $$\begin{aligned} Kx=\displaystyle \sum _{n\in I}f_n\langle g_n,x\rangle . \end{aligned}$$

  (3.3)

$\diamondsuit$

Definition 3.3

[7] A family $\{f_n\}_{n\in I}$ is said to be a K-frame for $V_{R}(\mathfrak {Q})$ with lower and upper K-frame bounds A and B, if $\{f_n\}_{n\in I}$ satisfies Eq. (3.2).$\diamondsuit$

Remark 3.1

Due to Definition (3.3), a sequence satisfying any of the conditions in Proposition 3.1 is also called a K-frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

In the next, we give an example of a K-frame.

Example 3.1

Let $\{e_n\}_{n=1}^3$ be a Hilbert basis for a three dimensional right quaternionic Hilbert space $V_{R}(\mathfrak {Q})$ and $K\in \mathcal{L}(V_{R}(\mathfrak {Q}))$ be defined as

$$\begin{aligned} Ke_1=e_1,~~Ke_2=e_1,~~Ke_3=e_2. \end{aligned}$$

Let $\{f_n\}_{n=1}^3=\{e_1,e_1,e_2\}$. Clearly, we have

$$\begin{aligned} \displaystyle \sum _{n=1}^3\vert \langle f_n,f\rangle \vert ^2\le 2\Vert f\Vert ^2. \end{aligned}$$

(3.4)

Further, by simple calculations, we can see that the adjoint of K is given as

$$\begin{aligned} K^{*}e_1=e_1+e_2,~~K^*e_2=e_3,~~K^*e_3=0. \end{aligned}$$

Then, we get

$$\begin{aligned} \Vert K^{*} f\Vert ^2{=}\Vert (e_1+e_2)\langle e_1,f\rangle +e_3\langle e_2,f\rangle \Vert ^2=2\vert \langle e_1,f\rangle \vert ^2+\vert \langle e_2,f\rangle \vert ^2=\displaystyle \sum _{n=1}^3\vert \langle f_n,f\rangle \vert ^2.\nonumber \\ \end{aligned}$$

(3.5)

So, Eqs. (3.4) and (3.5) imply that $\{f_n\}_{n=1}^3$ is a K-frame for $V_{R}(\mathfrak {Q})$. However, $\{f_n\}_{n=1}^3$ is not a frame for $V_{R}(\mathfrak {Q})$ since it does not possess a lower frame bound.

Remark 3.2

Since K-frames are not frames in general, many properties of ordinary frames may not hold for K-frames, such as the corresponding synthesis operator for K-frames is not surjective, the frame operator for K-frames is not isomorphic and the alternate dual reconstruction pair for K-frames is not interchangeable in the quaternionic Hilbert space $V_{R}(\mathfrak {Q})$. However, they are satisfied on a subspace R(K) of $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Proposition 3.2

Let K be with a closed range and $\{f_n\}_{n\in I}$ be a K-frame for $V_{R}(\mathfrak {Q})$. Then, the frame operator for the K-frame $\{f_n\}_{n\in I}$ is invertible on a subspace R(K) of $V_{R}(\mathfrak {Q})$. $\diamondsuit$

To prove this result, we need the following lemma which is a slight modification of [8,  Lemma  2.5.1]. The proof of this Lemma is similar to the one in complex case.

Lemma 3.1

Let $V_{R}(\mathfrak {Q})$ and $V_{1,R}(\mathfrak {Q})$ be two right quaternionic Hilbert spaces and suppose that $U:V_{R}(\mathfrak {Q})\rightarrow V_{1,R}(\mathfrak {Q})$ is a bounded operator with closed range R(U). Then, there exists a bounded operator $U^{\dagger }:V_{1,R}(\mathfrak {Q})\rightarrow V_{R}(\mathfrak {Q})$ for which

$$\begin{aligned} UU^{\dagger }x=x,~~\forall x\in R(U). \end{aligned}$$

$\diamondsuit$

Proof of Proposition 3.2

As $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$, then there exist $A, B>0$ such that

$$\begin{aligned} A\Vert K^*f\Vert ^2\le \displaystyle \sum _{n\in I}\vert \langle f_n,f\rangle \vert ^2\le B\Vert f\Vert ^2,~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

(3.6)

Further, since $K\in \mathcal{L}(V_{R}(\mathfrak {Q}))$ and R(K) is closed, then in view of Lemma 3.1, there exists a pseudo-inverse $K^{\dagger }$ of K such that

$$\begin{aligned} KK^{\dagger }f=f,~~\forall f\in R(K), ~~i.e.,~~KK^{\dagger }_{|R(K)}=I_{R(K)}. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} I_{R(K)}^*=(K^{\dagger }_{|R(K)})^*K^*. \end{aligned}$$

Hence, for $f\in R(K)$ we have

$$\begin{aligned} \Vert f\Vert =\Vert (K^{\dagger })^*K^*f\Vert \le \Vert K^{\dagger }\Vert \Vert K^*f\Vert , \end{aligned}$$

which implies that

$$\begin{aligned} \Vert K^*f\Vert ^2\ge \Vert K^{\dagger }\Vert ^{-2}\Vert f\Vert ^2. \end{aligned}$$

(3.7)

So, using Eq. (3.7) we get

$$\begin{aligned} \langle Sf,f\rangle= & {} \sum _{n\in I}\vert \langle f_n,f\rangle \vert ^2\nonumber \\\ge & {} A\Vert K^*f\Vert ^2\nonumber \\\ge & {} A\Vert K^{\dagger }\Vert ^{-2}\Vert f\Vert ^2,~~\forall f\in R(K). \end{aligned}$$

(3.8)

Combining Eqs. (3.6) and (3.8), we obtain

$$\begin{aligned} A\Vert K^{\dagger }\Vert ^{-2}\Vert f\Vert \le \Vert Sf\Vert \le B\Vert f\Vert ,~~\forall f\in R(K). \end{aligned}$$

(3.9)

Hence, $S:R(K)\rightarrow S(R(K))$ is a homeomorphism; that is if $f\in S(R(K))$ then $S^{-1}f\in R(K)$ and consequently, Eq. (3.9) yields

$$\begin{aligned} A\Vert K^{\dagger }\Vert ^{-2}\Vert S^{-1}f\Vert \le \Vert f\Vert \le B\Vert S^{-1}f\Vert ,~~\forall f\in S(R(K)). \end{aligned}$$

Therefore, we get

$$\begin{aligned} B^{-1}\Vert f\Vert \le \Vert S^{-1}f\Vert \le A^{-1}\Vert K^{\dagger }\Vert ^{2}\Vert f\Vert ,~~\forall f\in S(R(K)). \end{aligned}$$

$\square$

Proposition 3.3

Suppose that K is with closed range and $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$ are as in Eq. (3.3). Then there exists a sequence $\{h_n\}_{n\in I}=\{(K_{|R(K)}^{\dagger })^*g_n\}_{n\in I}$ derived by $\{g_n\}_{n\in I}$ such that

$$\begin{aligned} f=\displaystyle \sum _{n\in I}f_n\langle h_n,f\rangle ,~~\forall f \in R(K). \end{aligned}$$

Moreover, $\{h_n\}_{n\in I}$ and $\{f_n\}_{n\in I}$ are interchangeable for any $f\in R(K)$.$\diamondsuit$

Proof

As $K\in \mathcal{L}(V_{R}(\mathfrak {Q}))$ and R(K) is closed, then it follows from Lemma 3.1 that there exists a pseudo-inverse $K^{\dagger }$ of K such that $f = KK^{\dagger }f$, $\forall f\in R(K)$. So, Eq. (3.3) yields

$$\begin{aligned} f= & {} KK^{\dagger }f \\= & {} \sum _{n\in I}f_n\langle g_n,K^{\dagger }f\rangle \\= & {} \sum _{n\in I}f_n\langle g_n,K_{|R(K)}^{\dagger }f\rangle \\= & {} \sum _{n\in I}f_n\langle (K_{|R(K)}^{\dagger })^* g_n,f\rangle ,~~\forall f\in R(K), \end{aligned}$$

where $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$ are Bessel sequences in $V_{R}(\mathfrak {Q})$ satisfying Eq. (3.3).

Now let $h_n =( K_{|R(K)}^{\dagger })^*g_n$. Clearly, $\{h_n\}_{n \in I}\subset R(K)$. In fact, it suffices to see that

$$\begin{aligned} K_{|R(K)}^{\dagger } : R(K)\rightarrow V_{R}(\mathfrak {Q}) \end{aligned}$$

and so we may obtain

$$\begin{aligned} ( K_{|R(K)}^{\dagger })^* : V_{R}(\mathfrak {Q})\rightarrow R(K). \end{aligned}$$

Further, $\{h_n\}_{n\in I}$ is a Bessel sequence in R(K). Indeed, let $f\in R(K)\subset V_{R}(\mathfrak {Q})$ we have

$$\begin{aligned} \sum _{n\in I}\vert \langle h_n,f\rangle \vert ^2= & {} \sum _{n\in I}\vert \langle (K_{|R(K)}^{\dagger })^* g_n,f\rangle \vert ^2\\= & {} \sum _{n\in I}\vert \langle g_n,K_{|R(K)}^{\dagger }f\rangle \vert ^2. \end{aligned}$$

As $\{g_n\}_{n\in I}$ is a Bessel sequence, then there exists $B>0$ such that

$$\begin{aligned} \sum _{n\in I}\vert \langle h_n,f\rangle \vert ^2\le & {} B\Vert K^{\dagger }\Vert ^2\Vert f\Vert ^2,~~\forall f\in R(K). \end{aligned}$$

To complete our proof, it suffices to show that $\{h_n\}_{n\in I}$ and $\{f_n\}_{n\in I}$ are interchangeable on R(K). To this interest, let $f,g\in R(K)\subset V_{R}(\mathfrak {Q})$. We have

$$\begin{aligned} \langle f,g\rangle= & {} \left\langle \sum _{n\in I}f_n\langle h_n,f\rangle ,g\right\rangle \\= & {} \sum _{n\in I}\overline{\langle h_n,f\rangle }\langle f_n,g\rangle \\= & {} \sum _{n\in I}\langle f,h_n\rangle \langle f_n,g\rangle \\= & {} \left\langle f,\sum _{n\in I}h_n\langle f_n,g\rangle \right\rangle , \end{aligned}$$

that is, $f = \sum \nolimits _{n\in I}h_n\langle f_n,f\rangle ,$$\forall f \in R(K)$.

$\square$

4 Construction of K-Frames

In this part, we are concerning with the construction of new K-frames. We begin first by obtaining a K-frame from ordinary frame.

Proposition 4.1

Suppose that $\{f_n\}_{n\in I}$ is an ordinary frame for $V_{R}(\mathfrak {Q})$, then $\{Kf_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Proof

By Theorem 3.1, it suffices to show that $\{Kf_n\}_{n\in I}$ is an atomic system for K.

Since $\{f_n\}_{n\in I}$ is an ordinary frame for $V_{R}(\mathfrak {Q})$, then for any $f \in V_{R}(\mathfrak {Q})$ we have

$$\begin{aligned} f = \sum _{n\in I}f_n\langle S^{-1}f_n,f\rangle , \end{aligned}$$

(4.1)

where S is the frame operator for $\{f_n\}_{n\in I}$. Thus, Eq. (4.1) implies the following equality

$$\begin{aligned} Kf = \displaystyle \sum _{n\in I}Kf_n\langle S^{-1}f_n,f\rangle ,~~ \forall f \in V_{R}(\mathfrak {Q}). \end{aligned}$$

Now, it remains to show that $\{Kf_n\}_{n\in I}$ is a Bessel sequence and $\Vert \{\langle S^{-1}f_n,f\rangle \}\Vert _{l^2(\mathfrak {Q})} \le C \Vert f\Vert$, where C is a positive constant. In fact, since $\{f_n\}_{n\in I}$ is an ordinary frame for $V_{R}(\mathfrak {Q})$, we let A, B be its lower and upper frame bounds, then it follows from [20] that $\{S^{-1}f_n\}_{n\in I}$ is a frame for $V_{R}(\mathfrak {Q})$ with bounds $B^{-1},A^{-1}>0$. Therefore, for any $f \in V_{R}(\mathfrak {Q})$ we obtain

$$\begin{aligned} \displaystyle \sum _{n\in I}\vert \langle S^{-1}f_n,f\rangle \vert ^2\le A^{-1}\Vert f\Vert ^2. \end{aligned}$$

On the other hand, we have

$$\begin{aligned} \displaystyle \sum _{n\in I}\vert \langle Kf_n,f\rangle \vert ^2= & {} \displaystyle \sum _{n\in I}\vert \langle f_n,K^*f\rangle \vert ^2\\\le & {} B\Vert K^*f\Vert ^2\\\le & {} B\Vert K\Vert ^2\Vert f\Vert ^2,~~\forall f \in V_{R}(\mathfrak {Q}). \end{aligned}$$

Consequently, $\{Kf_n\}_{n\in I}$ is an atomic system for K.

$\square$

Corollary 4.1

Assume that $\{e_n\}_{n\in I}$ is a Hilbert basis for $V_{R}(\mathfrak {Q})$, then $\{Ke_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Proposition 4.2

If $T \in \mathcal{L}(V_{R}(\mathfrak {Q}))$ and $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$, then $\{Tf_n\}_{n\in I}$ is a TK-frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Remark 4.1

It should be noted that the result of Proposition 4.2 does not hold for ordinary frames in general. More precisely, if we assume that $T \in \mathcal{L}(V_{R}(\mathfrak {Q}))$ and $\{f_n\}_{n\in I}$ is an ordinary frame for $V_{R}(\mathfrak {Q})$, then $\{Tf_n\}_{n\in I}$ is not an ordinary frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Proof of Proposition 4.2

Let $T \in \mathcal{L}(V_{R}(\mathfrak {Q}))$ and $\{f_n\}_{n\in I}$ be a K-frame for $V_{R}(\mathfrak {Q})$ with frame bounds A and B, i.e.,

$$\begin{aligned} A\Vert K^*f\Vert ^2\le \displaystyle \sum _{n\in I}\vert \langle f_n,f\rangle \vert ^2\le B\Vert f\Vert ^2,~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

(4.2)

By observing that

$$\begin{aligned} \displaystyle \sum _{n\in I}\vert \langle Tf_n,f\rangle \vert ^2=\displaystyle \sum _{n\in I}\vert \langle f_n,T^*f\rangle \vert ^2, \end{aligned}$$

Eq. (4.2) yields

$$\begin{aligned} A\Vert K^*T^*f\Vert ^2\le \displaystyle \sum _{n\in I}\vert \langle Tf_n,f\rangle \vert ^2\le B\Vert T^* f\Vert ^2\le B\Vert T\Vert ^2\Vert f\Vert ^2,~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

Hence, $\{Tf_n\}_{n\in I}$ is a TK-frame for $V_{R}(\mathfrak {Q})$.

$\square$

Corollary 4.2

If $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$, then $\{K^Nf_n\}_{n\in I}$ is a $K^N$-frame for $V_{R}(\mathfrak {Q})$, where $N \ge 1$ is a fixed integer. $\diamondsuit$

Now, we construct new K-frames from old ones by taking sums.

Theorem 4.1

Let $\{f_n\}_{n\in I}$ be a K-frame for $V_{R}(\mathfrak {Q})$, $T \in \mathcal{L}(V_{R}(\mathfrak {Q}))$ and $TKK^*$ be a positive operator. Then $\{f_n + Tf_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Proof

Suppose that $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$ with frame bounds A and B. Then, for $f \in V_{R}(\mathfrak {Q})$, we obtain

$$\begin{aligned} A\Vert K^*f\Vert ^2 \le \displaystyle \sum _{n\in I} \vert \langle f_n,f\rangle \vert ^2\le B\Vert f\Vert ^2. \end{aligned}$$

(4.3)

Now, for $f\in V_{R}(\mathfrak {Q})$ we have

$$\begin{aligned} \displaystyle \sum _{n\in I} \vert \langle (I+T)f_n,f\rangle \vert ^2=\displaystyle \sum _{n\in I} \vert \langle f_n,(I+T)^*f\rangle \vert ^2. \end{aligned}$$

(4.4)

Using Eqs. (4.3) and (4.4), we get

$$\begin{aligned} A\Vert K^*(I+T)^*f\Vert ^2\le \displaystyle \sum _{n\in I} \vert \langle (I+T)f_n,f\rangle \vert ^2\le B\Vert I+T\Vert ^2\Vert f\Vert ^2. \end{aligned}$$

(4.5)

On the other hand, we have

$$\begin{aligned} A\Vert K^*(I+T)^*f\Vert ^2= & {} A(\Vert K^*f\Vert ^2 +2Re \langle K^*f,K^*T^*f\rangle +\Vert K^*T^*f\Vert ^2)\nonumber \\= & {} A(\Vert K^*f\Vert ^2 +2Re \langle TKK^*f,f\rangle +\Vert K^*T^*f\Vert ^2). \end{aligned}$$

Since $TKK^*$ is a positive operator, then

$$\begin{aligned} \langle TKK^*f,f\rangle \ge 0, ~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

So, we obtain

$$\begin{aligned} A\Vert K^*(I+T)^*f\Vert ^2\ge & {} A(\Vert K^*f\Vert ^2 +\Vert K^*T^*f\Vert ^2)\nonumber \\\ge & {} A\Vert K^*f\Vert ^2,~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

(4.6)

Combining Eqs. (4.5) and (4.6), we get that $\{f_n + Tf_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$.

$\square$

Theorem 4.2

Suppose that $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$ are K-frames for $V_{R}(\mathfrak {Q})$. Let $U_1$ and $U_2$ be the synthesis operators of $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$, respectively. If $U_1U_2^*=0$ and T or L is surjective operator in $\{K\}'$, then $\{Tf_n+Lg_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$. $\diamondsuit$

Proof

Let $f\in V_{R}(\mathfrak {Q})$. We have

$$\begin{aligned} \displaystyle \sum _{n\in I}\vert \langle Tf_n+Lg_n,f\rangle \vert ^2= & {} \displaystyle \sum _{n\in I}\vert \langle Tf_n,f\rangle +\langle Lg_n,f\rangle \vert ^2\nonumber \\= & {} \displaystyle \sum _{n\in I}\vert \langle f_n,T^*f\rangle +\langle g_n,L^*f\rangle \vert ^2\nonumber \\= & {} \displaystyle \sum _{n\in I}\vert \langle T^*f,f_n\rangle +\langle L^*f,g_n\rangle \vert ^2. \end{aligned}$$

As $U_1U_2^*=0$, then for $f,g\in V_{R}(\mathfrak {Q})$ we get

$$\begin{aligned} \displaystyle \sum _{n\in I}\langle f,f_n\rangle \langle g_n,g\rangle =\displaystyle \sum _{n\in I}\langle g,g_n\rangle \langle f_n,f\rangle =0. \end{aligned}$$

Therefore, we obtain

$$\begin{aligned} \displaystyle \sum _{n\in I}\vert \langle Tf_n+Lg_n,f\rangle \vert ^2= & {} \displaystyle \sum _{n\in I}\vert \langle f_n,T^*f\rangle \vert ^2+\displaystyle \sum _{n\in I}\vert \langle g_n,L^*f\rangle \vert ^2. \end{aligned}$$

Since $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$ are K-frames for $V_{R}(\mathfrak {Q})$, then there exist $A_i,B_i>0,i=1,2,$ such that

$$\begin{aligned} A_1\Vert K^*f\Vert ^2\le \displaystyle \sum _{n\in I}\vert \langle f_n,f\rangle \vert ^2\le B_1\Vert f\Vert ^2 \end{aligned}$$

and

$$\begin{aligned} A_2\Vert K^*f\Vert ^2\le \displaystyle \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2\le B_2\Vert f\Vert ^2. \end{aligned}$$

So, we get

$$\begin{aligned} \displaystyle \sum _{n\in I}\vert \langle Tf_n+Lg_n,f\rangle \vert ^2\le & {} B_1\Vert T^*f\Vert ^2+B_2\Vert L^*f\Vert ^2\nonumber \\\le & {} (B_1\Vert T\Vert ^2+B_2\Vert L\Vert ^2)\Vert f\Vert ^2,~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

(4.7)

Without loss of generality, assume that T is surjective. Then, there exists $C>0$ such that

$$\begin{aligned} \Vert T^*f\Vert ^2\ge C\Vert f\Vert ^2,~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

Since $TK=KT$, thus $T^*K^*=K^*T^*$. So,

$$\begin{aligned} \displaystyle \sum _{n\in I}\vert \langle Tf_n+Lg_n,f\rangle \vert ^2\ge & {} \displaystyle \sum _{n\in I}\vert \langle f_n,T^*f\rangle \vert ^2\nonumber \\\ge & {} A_1\Vert K^*T^*f\Vert ^2\nonumber \\= & {} A_1\Vert T^*K^*f\Vert ^2\nonumber \\\ge & {} A_1C\Vert K^*f\Vert ^2. \end{aligned}$$

(4.8)

Hence, Eqs. (4.7) and (4.8) entail that $\{Tf_n+Lg_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$.

$\square$

Let $L=0$. We get the following corollary.

Corollary 4.3

Suppose that $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$. If T is surjective and $TK=KT$, then $\{Tf_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

Let $T=L=I$. The following result holds.

Corollary 4.4

Suppose that $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$ are K-frames for $V_{R}(\mathfrak {Q})$. Let $U_1$ and $U_2$ be the synthesis operators of $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$, respectively. If $U_1U_2^*=0$, then $\{f_n+g_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$.$\diamondsuit$

We close this part by showing that a T-frame can be constructed from a K-frame by the perturbation of a bounded right linear operator T.

Theorem 4.3

Let K be with a closed range and let $\{f_n\}_{n\in I}$ be a K-frame for $V_{R}(\mathfrak {Q})$. Let $T\in \mathcal{L}(V_{R}(\mathfrak {Q}),V_{1,R}(\mathfrak {Q}))$. If $R(T^*)\subset R(K)$, then $\{Tf_n\}_{n\in I}$ is a T-frame for $V_{1,R}(\mathfrak {Q})$.$\diamondsuit$

The following result extends [19,  Lemma 2.2] to quaternionic Hilbert spaces. We omit the proof since it is similar to the complex case.

Lemma 4.1

Let $V_{R}(\mathfrak {Q})$, $V_{1,R}(\mathfrak {Q})$ be two right quaternionic Hilbert spaces, and suppose that $T \in \mathcal{L} (V_{R}(\mathfrak {Q}),V_{1,R}(\mathfrak {Q}))$ has a closed range, then

$$\begin{aligned} \Vert T^{\dagger }\Vert ^{-1}\Vert f\Vert \le \Vert T^*f\Vert \le \Vert T\Vert \Vert f\Vert ,~~\forall f \in R(T). \end{aligned}$$

$\diamondsuit$

Proof of Theorem 4.3

Assume that $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$ with frame bounds A and B. Hence, we have

$$\begin{aligned} A\Vert K^*f\Vert ^2 \le \displaystyle \sum _{n\in I} \vert \langle f_n,f\rangle \vert ^2\le B\Vert f\Vert ^2,~~\forall f \in V_{R}(\mathfrak {Q}). \end{aligned}$$

Now, let $g\in V_{1,R}(\mathfrak {Q})$. As $T\in \mathcal{L}(V_{R}(\mathfrak {Q}),V_{1,R}(\mathfrak {Q}))$, then $T^*g\in V_{R}(\mathfrak {Q})$. So, we obtain

$$\begin{aligned} A\Vert K^*T^*g\Vert ^2\le & {} \displaystyle \sum _{n\in I}\vert \langle f_n,T^*g\rangle \vert ^2\\= & {} \displaystyle \sum _{n\in I} \vert \langle Tf_n,g\rangle \vert ^2\\\le & {} B\Vert T^*g\Vert ^2\\\le & {} B\Vert T\Vert ^2\Vert g\Vert ^2. \end{aligned}$$

Since $K \in \mathcal{L}(V_{R}(\mathfrak {Q}))$ has a closed range and $R(T^*) \subset R(K)$, by Lemma 4.1, we have

$$\begin{aligned} A\Vert K^*T^*g\Vert ^2\ge A\Vert K^{\dagger }\Vert ^{-2}\Vert T^*g\Vert ^2,~~\forall g \in V_{1,R}(\mathfrak {Q}). \end{aligned}$$

This implies that

$$\begin{aligned} A\Vert K^{\dagger }\Vert ^{-2}\Vert T^*g\Vert ^2\le \displaystyle \sum _{n\in I} \vert \langle Tf_n,g\rangle \vert ^2\le B\Vert T\Vert ^2\Vert g\Vert ^2,~~\forall g \in V_{1,R}(\mathfrak {Q}). \end{aligned}$$

Consequently, $\{Tf_n\}_{n\in I}$ is a T-frame for $V_{1,R}(\mathfrak {Q})$.

$\square$

5 Stability of Perturbation of K-Frames

In this section, we prove a stability result for K-frames. To this interest, we recall the following lemma due to Sharma and Goel [20].

Lemma 5.1

[20] A sequence of vectors $\{f_n\}_{n\in I}\subset V_{R}(\mathfrak {Q})$ is a Bessel sequence for $V_{R}(\mathfrak {Q})$ with bound B if and only if the right linear operator $T : l^2(\mathfrak {Q})\rightarrow V_{R}(\mathfrak {Q})$ given by

$$\begin{aligned} T (\{q_n\}_{n\in I}) =\displaystyle \sum _{n\in I} f_nq_n, ~~\{q_n\}_{n\in I} \in l^2(\mathfrak {Q}) \end{aligned}$$

is a well defined and bounded operator with $\Vert T\Vert \le \sqrt{B}$.$\diamondsuit$

Now, we are ready to state our result.

Theorem 5.1

Suppose that K is with closed range, $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$ with frame bounds A and B and $\alpha ,\beta ,\gamma \in [0,\infty )$, such that $\max \{\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert ,\beta \}<1$. If $\{g_n\}_{n\in I}\subset V_{R}(\mathfrak {Q})$ and satisfy

$$\begin{aligned} \left\| \displaystyle \sum _{n\in J}(f_n-g_n)q_n\right\| \le \alpha \left\| \displaystyle \sum _{n\in J}f_nq_n\right\| +\beta \left\| \displaystyle \sum _{n\in J}g_nq_n\right\| +\gamma \left( \displaystyle \sum _{n\in J}\left| q_n\right| ^2\right) ^{\frac{1}{2}} \end{aligned}$$

(5.1)

for all finite quaternions $q_n\in l^2(\mathfrak {Q})$, $n\in J\subseteq I$ with $\vert J\vert <+\infty$, then $\{g_n\}_{n\in I}$ is a $P_{Q(R(K))}K$-frame for $V_{R}(\mathfrak {Q})$, with frame bounds

$$\begin{aligned} \displaystyle \frac{\left( \sqrt{A}\Vert K^{\dagger }\Vert ^{-1}(1-\alpha )-\gamma \right) ^2}{(1+\beta )^2\Vert K\Vert ^2},~~\displaystyle \frac{\left( \sqrt{B}(1+\alpha )+\gamma \right) ^2}{(1-\beta )^2}, \end{aligned}$$

where $P_{Q(R(K))}$ is the orthogonal projection operator from $V_R(\mathfrak {Q})$ to Q(R(K), $Q=UT^*,T,U$ are the synthesis operators for $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$, respectively.$\diamondsuit$

Proof

Let $J\subseteq I$ with $|J|<+\infty$. Then, we have

$$\begin{aligned} \left\| \displaystyle \sum _{n\in J} g_nq_n\right\|\le & {} \left\| \displaystyle \sum _{n\in J} (f_n-g_n)q_n\right\| +\left\| \displaystyle \sum _{n\in J} f_nq_n\right\| \nonumber \\\le & {} (\alpha +1)\left\| \displaystyle \sum _{n\in J}f_nq_n\right\| +\beta \left\| \displaystyle \sum _{n\in J}g_nq_n\right\| +\gamma \left( \displaystyle \sum _{n\in J}\left| q_n\right| ^2\right) ^{\frac{1}{2}}\nonumber \\\le & {} \frac{\alpha +1}{1-\beta }\left\| \displaystyle \sum _{n\in J}f_nq_n\right\| +\frac{\gamma }{1-\beta }\left( \displaystyle \sum _{n\in J}\left| q_n\right| ^2\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.2)

Furthermore, we have

$$\begin{aligned} \left\| \displaystyle \sum _{n\in J}f_nq_n\right\|= & {} \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\left| \left\langle \displaystyle \sum _{n\in J}f_nq_n,g\right\rangle \right| \nonumber \\\le & {} \left( \displaystyle \sum _{n\in J}\left| q_n\right| ^2\right) ^{\frac{1}{2}}\sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\left( \displaystyle \sum _{n\in J}\vert \left\langle f_n,g\right\rangle \vert ^2\right) ^{\frac{1}{2}}\nonumber \\\le & {} \sqrt{B}\left( \displaystyle \sum _{n\in J}\left| q_n\right| ^2\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.3)

Hence, using Eqs. (5.2) and (5.3), we obtain

$$\begin{aligned} \left\| \displaystyle \sum _{n\in J} g_nq_n\right\| \le \left( \frac{\alpha +1}{1-\beta }\sqrt{B}+\frac{\gamma }{1-\beta }\right) \left( \displaystyle \sum _{n\in J}\left| q_n\right| ^2\right) ^{\frac{1}{2}}, \end{aligned}$$

for all finite quaternions $q_n\in l^2(\mathfrak {Q})$, $n\in J$. So, we can define a bounded operator

$$\begin{aligned} U:l^2(\mathfrak {Q})\rightarrow V_{R}(\mathfrak {Q}),~~Uq=\displaystyle \sum _{n\in I}g_nq_n,~~q=\{q_n\}\in l^2(\mathfrak {Q}). \end{aligned}$$

Clearly, U is well defined and bounded with $\Vert U\Vert \le \frac{\alpha +1}{1-\beta }\sqrt{B}+\frac{\gamma }{1-\beta }$. Therefore, it follows from Lemma 5.1 that $\{g_n\}_{n\in I}$ is a Bessel sequence for $V_{R}(\mathfrak {Q})$.

Further, as $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$, then we consider its synthesis operator

$$\begin{aligned} T:l^2(\mathfrak {Q})\rightarrow V_{R}(\mathfrak {Q}),~~Tq=\displaystyle \sum _{n\in I}f_nq_n,~~q=\{q_n\}\in l^2(\mathfrak {Q}). \end{aligned}$$

Setting $Q=UT^*:R(K)\rightarrow R(Q)$. We prove that Q is invertible. To this interest, let begin with the closure of R(Q). Let $\{y_n\}_{n\in I}\subset R(Q)$ such that $y_n\rightarrow y,$$y\in V_{R}(\mathfrak {Q})$, then there exists $x_n\in R(K)$ such that

$$\begin{aligned} y_n=Q(x_n). \end{aligned}$$

(5.4)

Although, it follows from Eq. (5.1) that

$$\begin{aligned} \Vert Tq-Uq\Vert \le \alpha \Vert Tq\Vert +\beta \Vert Uq\Vert +\gamma \Vert q\Vert _{l^2(\mathfrak {Q})},~~q\in l^2(\mathfrak {Q}). \end{aligned}$$

(5.5)

So, if we consider $q_f=T^*f\in l^2(\mathfrak {Q})$, $f\in R(K)$, Eq. (5.5) implies that

$$\begin{aligned} \Vert Sf-UT^*f\Vert= & {} \Vert TT^*f-UT^*f\Vert \nonumber \\\le & {} \alpha \Vert Sf\Vert +\beta \Vert UT^*f\Vert +\gamma \Vert T^*f\Vert _{l^2(\mathfrak {Q})}. \end{aligned}$$

(5.6)

On the other hand, by Eq. (3.9) we have

$$\begin{aligned} \Vert T^*f\Vert ^2= & {} \langle T^*f,T^*f\rangle \nonumber \\= & {} \langle Sf,f\rangle \nonumber \\\le & {} \Vert Sf\Vert \Vert f\Vert \nonumber \\\le & {} A^{-1}\Vert K^{\dagger }\Vert ^2\Vert Sf\Vert ^2,~~\forall f\in R(K). \end{aligned}$$

(5.7)

Combining Eqs. (5.6) and (5.7), we obtain

$$\begin{aligned} \Vert Sf-UT^*f\Vert \le \left( \alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert \right) \Vert Sf\Vert +\beta \Vert UT^*f\Vert ,~~\forall f\in R(K). \end{aligned}$$

(5.8)

Using the triangle inequality, Eq. (5.8) yields

$$\begin{aligned} \displaystyle \frac{1-(\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert )}{1+\beta }\Vert Sf\Vert \le \Vert UT^*f\Vert \le \displaystyle \frac{1+\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert }{1-\beta }\Vert Sf\Vert . \end{aligned}$$

(5.9)

Hence, it follows from Eqs. (3.9) and (5.9) that for any $f\in R(K)$ we have

$$\begin{aligned} \displaystyle \frac{(1-(\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert ))A\Vert K^{\dagger }\Vert ^{-2}}{1+\beta }\Vert f\Vert \le \Vert UT^*f\Vert \le \displaystyle \frac{(1+\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert )B}{1-\beta }\Vert f\Vert . \end{aligned}$$

(5.10)

Consequently, Eqs. (5.4) and (5.10) imply that

$$\begin{aligned} \Vert x_{n+m}-x_n\Vert\le & {} C^{-1}\Vert Q(x_{n+m}-x_n)\Vert \nonumber \\\le & {} C^{-1}\Vert y_{n+m}-y_n\Vert , \end{aligned}$$

(5.11)

where $C=\frac{(1-(\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert ))A\Vert K^{\dagger }\Vert ^{-2}}{1+\beta }$. Since $\{y_n\}_{n\in I}$ is a cauchy sequence, Eq. (5.11) entails that $\{x_n\}_{n\in I}$ is a cauchy sequence. Hence, there exists $x\in R(K)$ such that $x_n\rightarrow x$. As Q is bounded, we have $y_n=Q(x_n)\rightarrow Q(x)$. By the uniqueness of the limit, we obtain $y=Q(x)\in R(Q)$ which implies that R(Q) is closed. On the other hand, it follows from Eq. (5.1) that Q is injective on R(K). As a consequence, we conclude that $Q:R(K)\rightarrow R(Q)$ is invertible. Taking into accounts Eqs. (5.9) and (5.10), we get

$$\begin{aligned} \Vert SQ^{-1}y\Vert \le \displaystyle \frac{1+\beta }{1-(\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert )}\Vert y\Vert ,~~\forall y\in Q(R(K)), \end{aligned}$$

(5.12)

which implies that

$$\begin{aligned} \Vert Q^{-1}y\Vert \le \displaystyle \frac{1+\beta }{(1-(\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert ))A\Vert K^{\dagger }\Vert ^{-2}}\Vert y\Vert ,~~\forall y\in Q(R(K)). \end{aligned}$$

(5.13)

Now, let $f\in V_{R}(\mathfrak {Q})$. Since

$$\begin{aligned} P_{Q(R(K))}f= & {} QQ^{-1}P_{Q(R(K))}f\\= & {} U(T^*Q^{-1}P_{Q(R(K))}f)\\= & {} \sum _{n\in I}g_n(T^*Q^{-1}P_{Q(R(K))}f)_n, \end{aligned}$$

then

$$\begin{aligned} \Vert K^*(P_{Q(R(K))})^*f\Vert= & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\vert \langle g,K^*(P_{Q(R(K))})^*f\rangle \vert \nonumber \\= & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\vert \langle P_{Q(R(K))}Kg,f\rangle \vert \nonumber \\= & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\left| \left\langle \sum _{n\in I}g_n(T^*Q^{-1}P_{Q(R(K))}Kg)_n,f\right\rangle \right| \nonumber \\= & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\left| \sum _{n\in I}\overline{(T^*Q^{-1}P_{Q(R(K))}Kg)_n}\langle g_n,f\rangle \right| \nonumber \\\le & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\left( \sum _{n\in I}\vert (T^*Q^{-1}P_{Q(R(K))}Kg)_n\vert ^2\right) ^{\frac{1}{2}}\nonumber \\&\times \left( \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2\right) ^{\frac{1}{2}}\nonumber \\= & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\Vert T^*Q^{-1}P_{Q(R(K))}Kg\Vert _{l^2(\mathfrak {Q})}\left( \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2\right) ^{\frac{1}{2}}\nonumber \\= & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\left( \langle SQ^{-1}P_{Q(R(K))}Kg,Q^{-1}P_{Q(R(K))}Kg\rangle \right) ^{\frac{1}{2}}\nonumber \\&\times \left( \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2\right) ^{\frac{1}{2}}\nonumber \\\le & {} \displaystyle \sup _{g\in V_{R}(\mathfrak {Q}),\Vert g\Vert =1}\Vert SQ^{-1}P_{Q(R(K))}Kg\Vert ^{\frac{1}{2}}\Vert Q^{-1}P_{Q(R(K))}Kg\Vert ^{\frac{1}{2}}\nonumber \\&\times \left( \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2\right) ^{\frac{1}{2}}. \end{aligned}$$

(5.14)

Using Eqs. (5.12), (5.13) and (5.14), we get

$$\begin{aligned} \Vert K^*(P_{Q(R(K))})^*f\Vert\le & {} \frac{1+\beta }{(1-(\alpha +\gamma \sqrt{A^{-1}}\Vert K^{\dagger }\Vert ))\sqrt{A}\Vert K^{\dagger }\Vert ^{-1}}\Vert K\Vert \left( \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2\right) ^{\frac{1}{2}}\\= & {} \frac{(1+\beta )\Vert K\Vert }{\sqrt{A}\Vert K^{\dagger }\Vert ^{-1}(1-\alpha )-\gamma }\left( \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2\right) ^{\frac{1}{2}},~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

Consequently, we obtain

$$\begin{aligned} \displaystyle \frac{(\sqrt{A}\Vert K^{\dagger }\Vert ^{-1}(1-\alpha )-\gamma )^2}{(1+\beta )^2\Vert K\Vert ^2}\Vert K^*(P_{Q(R(K))})^*f\Vert ^2\le \sum _{n\in I}\vert \langle g_n,f\rangle \vert ^2,~~\forall f\in V_{R}(\mathfrak {Q}). \end{aligned}$$

$\square$

Corollary 5.1

Suppose that K is with closed range and $\{f_n\}_{n\in I}$ is a K-frame for $V_{R}(\mathfrak {Q})$ with frame bounds A and B and assume further that there exists $0<R<A$. If $\{g_n\}_{n\in I}\subset V_{R}(\mathfrak {Q})$ and satisfies

$$\begin{aligned} \left\| \displaystyle \sum _{n\in I}(f_n-g_n)q_n\right\| \le \sqrt{R}\left( \displaystyle \sum _{n\in I}\vert q_n\vert ^2\right) ^{\frac{1}{2}}, \end{aligned}$$

for all quaternions $q_n\in l^2(\mathfrak {Q})$,then $\{g_n\}_{n\in I}$ is a $P_{Q(R(K))}K$-frame for $V_{R}(\mathfrak {Q})$ with frame bounds

$$\begin{aligned} \displaystyle \frac{\left( \sqrt{A}\Vert K^{\dagger }\Vert ^{-1}-\sqrt{R}\right) ^2}{\Vert K\Vert ^2},~~\left( \sqrt{B}+\sqrt{R}\right) ^2, \end{aligned}$$

where $P_{Q(R(K))}$ is the orthogonal projection operator from $V_{R}(\mathfrak {Q})$ to Q(R(K), $Q=UT^*,T,U$ are the synthesis operators for $\{f_n\}_{n\in I}$ and $\{g_n\}_{n\in I}$, respectively.$\diamondsuit$

Proof

It suffices to take $\alpha =\beta =0$ and $\gamma =\sqrt{R}$ in Theorem 5.1.$\square$