Some results on frames by pre-frame operators in Q-Hilbert spaces

1.   Introduction

The concept of frames, which generalizes that of bases, was first introduced in the 1950s when Duffin and Schaeffer ^[1] studied some ongoing problems in the nonharmonic Fourier series. Looking back upon the a sequence $\{e_{j}:\; j\in J\}\subseteq\mathcal{H}$ (Hilbert space), we call $\{e_{j}:\; j\in J\}$ is a frame for $\mathcal{H}$ if the following inequality holds,

where positive constants $A, \, B$ are called the frame bounds. Frames have turned into a hot issue even since 1986 when Daubechies, Crossman and Meyer published their pioneering work ^[2]. Nowadays, great achievements have been made in the research of frame theory ^[3,4], and frames have been heavily used in numerous fields, such as coding and wireless communication ^[5], image and signal processing ^[6], sampling theory ^[7], quantum measurements ^[8], and so on (^[9,10]).

Hilbert space can be defined not only in real field and complex field, but also in quaternion field ^[11,12]. In 1936, Birkhoff and von Neumann ^[13] in their famous pioneering work on quantum logic commented that quantum mechanics can also be formulated in Hilbert space where the ground field of complex numbers is replaced by divisible algebras of quaternions ^[14]. By now, this opinion has been confirmed in the reference ^[15]. However, it is worth noting that most existing works on the frame theory only focus on real or complex Hilbert spaces instead of quaternionic Hilbert (Q-Hilbert) spaces. Note that both the real field and the complex field are associative and commutative, while the quaternion field only constitutes noncommutative associative algebra. This key characteristic greatly limited mathematicians to establish a complete theory of functional analysis in Q-Hilbert spaces ^[16], which affected the development of quantum physics in Q-Hilbert space. Luckily, the study on quaternion field has been developed from the mathematical point of view, and achievements in the frames in Q-Hilbert space especially have been obtained in recent. For example, Khokulan, Thirulogasanthar, Srisatkunarajah ^[17] and Sharma, Virender ^[18] introduced and studied frames for finite dimensional Q-Hilbert spaces, Sharma, Goel ^[19] and Sharma, Singh, Sahu ^[20] studied frames and dual frames for separable Q-Hilbert spaces, and Ellouz ^[21] introduced K-Frames and Zhang, Li ^[22] characterized Riesz bases in Q-Hilbert spaces.

When characterizing dual frames of a frame, constructing new frames is a big issues in frame theory. Finding suitable frames are of great significance in applications, and plenty of achievements have been acquired with regard to such issues. For instance, in ^[7] Li proved that for a given frame, one could obtain all its dual frames by seeking the left inverse of the invertible operator and then giving the accurate expression of all dual frames of the given frame. In ^[23], Guo looked to ways of constructing ($\Omega$, $\mu$)-frames, consisting of the structures of new ($\Omega$, $\mu$)-frames and the dual ($\Omega$, $\mu$)-frames in some conditions. In ^[24], Obeidat, Samarah, Casazza and Tremain went into the sums of frames in Hilbert spaces, and gave simple necessary and sufficient conditions on Bessel sequences $\{x_{i}\}_{i\in I}$ and $\{y_{i}\}_{i\in I}$ as well as the operators $Q_{1}, Q_{2}$ on $\mathcal{H}$ so that $\{Q_{1}x_{i}+Q_{2}y_{i}\}_{i\in I}$ formed a new frame for $\mathcal{H}$. In ^[25,26]], the authors discussed the sums of g-frames in Hilbert spaces, it was a simple and effective method to construct new frames by using the sums of known frames. Inspired by these works on frames, and aided by the pre-frame operators, we discuss analogous problems on frames in Q-Hilbert spaces. Especially, we obtain some more general construction methods by means of pre-frame operators (see Theorems 3.2 and 4.2), and other current methods. Usually, a new frame can be constructed by using the frame operator and the synthesis (analysis) operator to satisfy certain conditions, such as in ^[24] where $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are frames for $\mathcal{H}$ with analysis operators $T_{1}, T_{2}$ and frame operators $S_{1}, S_{2}$, respectively. $\{x_{j}+y_{j}\}_{j\in J}$ is a new frame for $\mathcal{H}$ if and only if $S_{1}+S_{2}+T_{1}^{\ast}T_{2}+T_{2}^{\ast}T_{1} > 0$. Comparatively, the methods we use are more convenient and direct.

In section two, we give some essential notions and existing results for later use. In section there, we first introduce the notion of the pre-frame operator, which is an important class of operators in frame theory, and characterize orthonormal bases, frames, dual frames and Riesz bases in terms of pre-frame operators. We then we obtain the accurate expressions of all dual frames of a given frame by taking advantage of pre-frame operators. With the help of an operator equation, we also give the characterization of the dual frames in Q-Hilbert spaces. In section four, we discuss the sum of frames and Bessel sequences. By means of pre-frame operators, we gain some more general methods to construct new frames. Moreover, we obtain a necessary and sufficient condition for the finite sum of frames to be a (tight) frame.

2.   Preliminaries

In this section we arrange some notions and results of frames in Q-Hilbert spaces (see ^[19,20] for details), which are necessary for below. $\mathfrak{Q}$ denotes a noncommutative quaternion field, and $J$ is an index set. Let $\mathcal{H}_{R}(\mathfrak{Q}), \; \mathcal{K}_{R}(\mathfrak{Q})$ be right Q-Hilbert spaces (or simply R-Q-Hilbert spaces) and $\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}), \mathcal{K}_{R}(\mathfrak{Q}))$ denote the collection of all bounded right $\mathfrak{Q}$-linear operators from $\mathcal{H}_{R}(\mathfrak{Q})$ to $\mathcal{K}_{R}(\mathfrak{Q})$, as a special case, $\mathcal{H}_{R}(\mathfrak{Q}) = \mathcal{K}_{R}(\mathfrak{Q})$, $\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}), \mathcal{K}_{R}(\mathfrak{Q})) = \mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$, and $I_{\mathcal{H}_{R}}$ be the identity operator in $\mathcal{H}_{R}(\mathfrak{Q})$. For $K\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$, the range of $K$ is represented by $R(K)$, and the pseudo-inverse of $K$ is represented by $K^{\dagger}$ if $R(K)$ is closed.

The noncommutative field of quaternions $\mathfrak{Q}$ is a four-dimensional real algebra with unity. In $\mathfrak{Q}$, zero denotes the null element and one denotes the identity with respect to multiplication. It also includes three so-called imaginary units, denoted by $\vec{i}, \, \vec{j}, \, \vec{k}$, i.e.,

where $\vec{i}^{2} = \vec{j}^{2} = \vec{k}^{2} = -1, \vec{i}\cdot\vec{j} = -\vec{j}\cdot\vec{i} = \vec{k}, \vec{j}\cdot\vec{k} = -\vec{k}\cdot\vec{j} = \vec{i}$ and $\vec{i}\cdot\vec{k} = -\vec{k}\cdot\vec{i} = \vec{j}$. For more information about the properties of quaternions, we refer the readers to [11–16].

Let $\mathbb{H}_{R}(\mathfrak{Q})$ be a linear vector space under right scalar multiplication over the field of quaternions $\mathfrak{Q}$. $\mathbb{H}_{R}(\mathfrak{Q})$ is called a right quaternionic pre-Hilbert space or right quaternionic inner product space if it is equipped with a Hermitian quaternionic inner product (or simply the inner product)

satisfying the following conditions:

(a)$\; \; \overline{\langle\phi\; |\; \psi\rangle} = \langle\psi\; |\; \phi\rangle$ for all $\phi, \, \psi\in\mathbb{H}_{R}(\mathfrak{Q})$;

(b)$\; \; \langle\phi\; |\; \phi\rangle > 0$ unless $\phi = 0$;

(c)$\; \; \langle\phi\; |\; \psi+\omega\rangle = \langle\phi\; |\; \psi\rangle+ \langle\phi\; |\; \omega\rangle$ for all $\phi, \, \psi, \, \omega\in\mathbb{H}_{R}(\mathfrak{Q})$;

(d)$\; \; \langle\phi\; |\; \psi q\rangle = \langle\phi\; |\; \psi\rangle q$, $\langle \phi q\; |\; \psi\rangle = \bar{q}\langle\phi\; |\; \psi\rangle$ for all $\phi, \, \psi\in\mathbb{H}_{R}(\mathfrak{Q})$ and $q\in\mathfrak{Q}$.

Let $\mathbb{H}_{R}(\mathfrak{Q})$ be a right quaternionic pre-Hilbert space with the inner product $\langle \cdot\; |\; \cdot\rangle$. We define the quaternionic norm $\|\cdot\|:\; \mathbb{H}_{R}(\mathfrak{Q})\rightarrow \mathbb{R}^{+}$ on $\mathbb{H}_{R}(\mathfrak{Q})$ by $\|\phi\| = \sqrt{\langle\phi\; |\; \phi\rangle}, \; \; \phi\in\mathbb{H}_{R}(\mathfrak{Q})$. If $\mathbb{H}_{R}(\mathfrak{Q})$ is complete with respect to the norm $\|\cdot\|$, it is called an R-Q-Hilbert space and is denoted by $\mathcal{H}_{R}(\mathfrak{Q})$.

Proposition 2.1. (^[12]) Let $\mathcal{H}_{R}(\mathfrak{Q})$ be an R-Q-Hilbert space and $\mathcal{N}\subseteq\mathcal{H}_{R}(\mathfrak{Q})$ meet that for $z, \, z^{\prime}\in\mathcal{N}$ $\left\langle z \mid z^{\prime}\right\rangle= \begin{cases}1 & \text { if } z=z^{\prime}; \\ 0 & \text { if } z \neq z^{\prime}.\end{cases}$ Then the following assertions are equivalent:

(ⅰ) $\forall x, \, y\in\mathcal{H}_{R}(\mathfrak{Q})$, the progression $\sum\limits_{z\in\mathcal{N}}\langle x|z\rangle\langle z|y\rangle$ is absolute convergence in $\mathcal{H}_{R}(\mathfrak{Q})$ and it possess:

(ⅱ) $\|x\| = \sum\limits_{z\in\mathcal{N}}|\langle z|x\rangle|^{2}$, $\forall x\in\mathcal{H}_{R}(\mathfrak{Q})$.

(ⅲ) $\mathcal{N}^{\bot} = \{u\in\mathcal{H}_{R}(\mathfrak{Q}):\; \langle u|z\rangle = 0, \forall z\in\mathcal{N}\} = \{0\}$.

(ⅳ) $span{\mathcal{N}}$ is dense in $\mathcal{H}_{R}(\mathfrak{Q})$.

Definition 2.1. (^[12]) Let $\mathcal{H}_{R}(\mathfrak{Q})$ be an R-Q-Hilbert space, and $\mathcal{N}\subseteq\mathcal{H}_{R}(\mathfrak{Q})$ is called a Hilbert basis or orthonormal basis of $\mathcal{H}_{R}(\mathfrak{Q})$ if it satisfies $\left\langle z \mid z^{\prime}\right\rangle= \begin{cases}1 & \text { if } z=z^{\prime}; \\ 0 & \text { if } z \neq z^{\prime}.\end{cases}$ for $z, \, z^{\prime}\in\mathcal{N}$ and all the conditions in Proposition 2.1. What is more, if $\mathcal{N}$ is a Hilbert basis of $\mathcal{H}_{R}(\mathfrak{Q})$, then arbitrary $x\in\mathcal{H}_{R}(\mathfrak{Q})$, the decomposition $x = \sum\limits_{z\in\mathcal{N}}z\langle z|x\rangle$ is unique and the progression $\sum_{z\in\mathcal{N}}z\langle z|x\rangle$ is an absolute convergence in $\mathcal{H}_{R}(\mathfrak{Q})$.

Compared with complex Hilbert spaces, Q-Hilbert spaces inherit a great deal of standard properties (see ^[12,16]).

Definition 2.2. (^[11]) An operator $T:\; \mathcal{H}_{R}(\mathfrak{Q})\rightarrow \mathcal{H}_{R}(\mathfrak{Q})$, for arbitrary $\phi, \, \psi\in\mathcal{H}_{R}(\mathfrak{Q})$ and $\alpha, \, \beta\in\mathfrak{Q}$, if $T(\phi\alpha+\psi\beta) = T(\phi)\alpha+T(\psi)\beta$, then $T$ is called right $\mathfrak{Q}$-linear; if there is a constant $M > 0$ such that $\|T\phi\|\leq M\|\phi\|$, then $T$ is bounded.

Proposition 2.2. (^[12]) Let $T\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$, and satisfy $T = T^{\ast}$, then the norm of $T$ is defined as follows

Proposition 2.3. (^[12]) Let $\mathcal{H}_{R}(\mathfrak{Q})$ be an R-Q-Hilbert space, $U, \, V\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$. Then

(ⅰ) $U+V$ and $UV\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$. In addition,

(ⅱ) $(U+V)^{\ast} = U^{\ast}+V^{\ast};$

(ⅲ) $(UV)^{\ast} = V^{\ast}U^{\ast}, (U^{\ast})^{\ast} = U;$

(ⅳ) if the operator $U$ is invertible, then $(U^{-1})^{\ast} = (U^{\ast})^{-1};$

(ⅴ) $I^{\ast}_{\mathcal{H}_{R}} = I_{\mathcal{H}_{R}}$, where $I_{\mathcal{H}_{R}}$ is the identity operator in $\mathcal{H}_{R}(\mathfrak{Q})$.

For more background information on Q-Hilbert spaces, see ^[11,12].

In ^[19], Sharma and Goel extended the concept of frame in Hilbert space to the Q-Hilbert space, as described next.

Definition 2.3. (^[19]) Let $\mathcal{H}_{R}(\mathfrak{Q})$ be an R-Q-Hilbert space. A sequence $\{x_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ is called a frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if there are two finite constants with $0 < A\leq B < \infty$ such that

The numbers $A, \, B$ are called frame bounds of $\{x_{j}\}_{j\in J}$. We call $\{x_{j}\}_{j\in J}$ a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$ if only the righthand inequality of (2.1) is established in these circumstances, $B$ is called Bessel bound. We call $\{x_{j}\}_{j\in J}$ a $\lambda$-tight frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if $A = B = \lambda$. What is more, we call $\{x_{j}\}_{j\in J}$ a Parseval frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if $\lambda = 1$.

Now define the space $l^{2}(\mathfrak{Q})$ by

and endow $l^{2}(\mathfrak{Q})$ with the inner product

Then $l^{2}(\mathfrak{Q})$ is an R-Q-Hilbert space.

If $\{x_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$, then the operator $S:\; \mathcal{H}_{R}(\mathfrak{Q})\rightarrow \mathcal{H}_{R}(\mathfrak{Q})$ defined by

and $S$ is called the (right) frame operator related to $\{f_{j}\}_{j\in J}$. It is understood that $S$ is a right linear bounded invertible operator (see ^[19]).

Definition 2.4. (^[20]) Let $\{x_{j}\}_{j\in J}$ be a frame for $\mathcal{H}_{R}(\mathfrak{Q})$. A sequence $\{y_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ fulfills

Then $\{y_{j}\}_{j\in J}$ is commonly known as the alternate dual for $\{x_{j}\}_{j\in J}$ in $\mathcal{H}_{R}(\mathfrak{Q})$.

Definition 2.5. (^[22]) Let $\mathcal{H}_{R}(\mathfrak{Q})$ be an R-Q-Hilbert space and $\{x_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$. $\{x_{j}\}_{j\in J}$ has been described as a Riesz basis for $\mathcal{H}_{R}(\mathfrak{Q})$ if the following conditions are met

(ⅰ) $\{x_{j}\}_{j\in J}$ is complete, that is, for $f\in\mathcal{H}_{R}(\mathfrak{Q})$, if $\langle x_{j}\, |\, f\rangle = 0, \forall j\in J$, then $f = 0$.

(ⅱ) There are two positive finite constants $A$ and $B$ such that

where $q_j\in\mathfrak{Q}, j\in J_{1}$, $J_{1}$ is any finite subset of $J$. $A$ and $B$ are called Riesz bounds of $\{x_{j}\}_{j\in J}$.

3.   Characterizations of (dual) frames

In this section, we introduce the definition of the pre-frame operators, and utilize the pre-frame operators for characterizing frames and dual frames in the R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. For a given frame in $\mathcal{H}_{R}(\mathfrak{Q})$, we also obtain the accurate expression formula about the dual frames. For these purposes, we first introduce a lemma, which was given by Sharma and Goel in ^[19].

Lemma 3.1. (^[19]) Let $\mathcal{H}_{R}(\mathfrak{Q})$ be an R-Q-Hilbert space and $\{x_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$. Then $\{f_{j}\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$ with bound $B$ if and only if the right linear operator $T:\; l^{2}(\mathfrak{Q})\rightarrow \mathcal{H}_{R}(\mathfrak{Q})$ defined by

is well defined, and $\|T\|_{op}\leq\sqrt{B}$.

Proposition 3.1. Let $\mathcal{H}_{R}(\mathfrak{Q})$ be an R-Q-Hilbert space and $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for $\mathcal{H}_{R}(\mathfrak{Q})$. Then $\{x_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if there exists a unique bounded right linear operator $V:\; \mathcal{H}_{R}(\mathfrak{Q})\rightarrow \mathcal{H}_{R}(\mathfrak{Q})$ such that $x_{j} = Vz_{j}$ for all $j\in J$.

Proof. ($\Rightarrow$). Note that $\mathcal{N} = \{z_{j}\}_{j\in J}$ is a Hilbert basis for $\mathcal{H}_{R}(\mathfrak{Q})$, and therefore $\{\langle z_{j}|f\rangle\}_{j\in J}\in l^{2}(\mathfrak{Q})$ for each $f\in \mathcal{H}_{R}(\mathfrak{Q})$. If $\{x_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ is a Bessel sequence, then the operator

is well defined by Lemma 3.1. We obtain for each $f\in \mathcal{H}_{R}(\mathfrak{Q})$ that

where $B$ is Bessel bound of $\{x_{j}\}_{j\in J}$. It follows that $V$ is a bounded right linear operator on $\mathcal{H}_{R}(\mathfrak{Q})$. By the definition of Hilbert basis, for an arbitrary $f\in\mathcal{H}_{R}(\mathfrak{Q})$, we obtain that $f = \sum\limits_{j\in J}z_{j}q_{j}$, where $\{q_{j}\}_{j\in J}\in l^{2}(\mathfrak{Q})$ is unique, and

Hence $\sum\limits_{j\in J}Vz_{j}q_{j} = \sum\limits_{j\in J}f_{j}q_{j}$, which implies that $x_{j} = Vz_{j}$. Suppose that $V_{1}, \, V_{2}\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ and $V_{1}z_{j} = V_{2}z_{j} = x_{j}$ for all $j\in J$. For $f\in\mathcal{H}_{R}(\mathfrak{Q})$, we have $\langle(V_{1}-V_{2})z_{j}\big|f\rangle = 0$ for all $j\in J$. It follows that

and thus $V_{1} = V_{2}$. Hence the operator $V$ is unique.

($\Leftarrow$). If $V\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ satisfies $x_{j} = Vz_{j}$ for arbitrary $j\in J$, then

This shows that $\{x_{j}\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$. □

Definition 3.1. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$ and $\{x_{j}\}_{j\in J}$ be a Bessel sequence in $\mathcal{H}_{R}(\mathfrak{Q})$. The operator $V$ in Proposition 3.1 is called the (right) pre-frame operator associated with $\{x_{j}\}_{j\in J}$.

Lemma 3.2. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$ and $\{x_{j}\}_{j\in J}$ be a Bessel sequence in $\mathcal{H}_{R}(\mathfrak{Q})$. If $V$ is the pre-frame operator associated with $\{x_{j}\}_{j\in J}$ and $S$ is the frame operator associated with $\{x_{j}\}_{j\in J}$, then $S = VV^{\ast}$.

Proof. By the definition of $V$, $x_{j} = Vz_{j}$ for $j\in J$, then

for $f\in\mathcal{H}_{R}(\mathfrak{Q})$. Hence $S = VV^{\ast}$. □

Now, we characterize frames and dual frames in terms of pre-frame operators. We begin with a lemma.

Lemma 3.3. (^[21]) Let $\mathcal{H}_{R}(\mathfrak{Q}), \mathcal{K}_{R}(\mathfrak{Q})$ be two R-Q-Hilbert spaces, and $K: \mathcal{H}_{R}(\mathfrak{Q})\rightarrow \mathcal{K }_{R}(\mathfrak{Q})$ be a bounded operator. If $R(K)$ is closed, then, there is a bounded operator $K^{\dagger}: \mathcal{K}_{R}(\mathfrak{Q})\rightarrow \mathcal{H}_{R}(\mathfrak{Q})$ for which

Theorem 3.1. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$ and $\{x_{j}\}_{j\in J}$ be a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$. If $V$ and $S$ denote the pre-frame operator and frame operator of $\{x_{j}\}_{j\in J}$, respectively, then

(i) $\{x_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if $V$ is onto.

(ii) $\{x_{j}\}_{j\in J}$ is a Parseval frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if $V$ is coisometry (i.e., $V^{\ast}$ is isometry).

(iii) $\{x_{j}\}_{j\in J}$ is a Riesz basis for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if $V$ is invertible.

(iv) $\{x_{j}\}_{j\in J}$ is a Hilbert basis for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if $V$ is unitary.

Proof. (ⅰ) If $\{x_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$, then $S$ is invertible by (Theorem 3.5 in ^[19]). By Lemma 3.2, we have $S = VV^{\ast}$, so $V$ is onto. On the other hand, if $V$ is onto, then $\{x_{j}\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$ by Lemma 3.1. Next we only need to show the existence of lower frame bound. Note that $V$ is onto, we have $VV^{\dagger} = I_{\mathcal{H}_{R}}$ by Lemma 3.3. It follows that $(V^{\dagger})^{\ast}V^{\ast} = I_{\mathcal{H}_{R}}$. Accordingly,

Thus,

(ⅱ) It is easy to check that $\{x_{j}\}_{j\in J}$ is a Parseval frame for $\mathcal{H}_{R}(\mathfrak{Q})$ iff $S$ is an identity operator on $\mathcal{H}_{R}(\mathfrak{Q})$. By Lemma 3.2, we have $S = VV^{\ast}$, so $S = I_{\mathcal{H}_{R}}$ if and only if $V$ is a coisometry.

(ⅲ) See Theorem 3.7 in ^[22].

(ⅳ) If $\{x_{j}\}_{j\in J}$ is a Hilbert basis for $\mathcal{H}_{R}(\mathfrak{Q})$, then we have for any $f\in\mathcal{H}_{R}(\mathfrak{Q})$ that

Therefore, $VV^{\ast} = I_{\mathcal{H}_{R}}$. It follows that $V$ is a unitary operator. On the contrary, if $V$ is a unitary operator, then for any $f\in\mathcal{H}_{R}(\mathfrak{Q})$, by simple calculation we have

So, $\{x_{j}\}_{j\in J}$ is a Hilbert basis for $\mathcal{H}_{R}(\mathfrak{Q})$. □

Theorem 3.2. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Let $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ be Bessel sequences for $\mathcal{H}_{R}(\mathfrak{Q})$, and let the pre-frame operators related with $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ be $V$ and $W$, respectively. Then, $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are dual frames if and only if $VW^{\ast} = I_{\mathcal{H}_{R}}$ or $WV^{\ast} = I_{\mathcal{H}_{R}}$.

Proof. Note that $V$ and $W$ are pre-frame operators related with $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$, respectively, so we have

Hence, for $f\in \mathcal{H}_{R}(\mathfrak{Q})$, we have

Similarly,

It can be seen from this that $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are dual frames if and only if $VW^{\ast} = I_{\mathcal{H}_{R}}$ or $WV^{\ast} = I_{\mathcal{H}_{R}}$. □

If $U, V\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ and $UV = I_{\mathcal{H}_{R}}$, then $U$ is called a left inverse operator of $V$. Our next goal is to characterize dual frames for the existing frame in an R-Q-Hilbert space. The following theorem gave the characterization of the right linear bounded left inverses of the existing pre-frame operator.

Theorem 3.3. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$ and $\{x_{j}\}_{j\in J}$ be a frame for $\mathcal{H}_{R}(\mathfrak{Q})$. Suppose that $\{y_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ and the pre-frame operator of $\{x_{j}\}_{j\in J}$ is $V$, then $\{y_{j}\}_{j\in J}$ is a dual frame of $\{x_{j}\}_{j\in J}$ if and only if $y_{j} = Wz_{j}$ for an arbitrary $j\in J$, where $W$ is a right linear bounded left inverse of $V^{\ast}$.

Proof. ($\Rightarrow$). Let $\{y_{j}\}_{j\in J}$ be an arbitrary dual frame of $\{x_{j}\}_{j\in J}$, and $W$ be the pre-frame operator of $\{y_{j}\}_{j\in J}$. There is a bounded right linear operator $W$ such that $y_{j} = Wz_{j}$ for arbitrary $j\in J$ by Proposition 3.1, and so for $f\in\mathcal{H}_{R}(\mathfrak{Q})$, we have

Note as $V$ as the pre-frame operator of $\{x_{j}\}_{j\in J}$, we have $x_{j} = Vz_{j}$ for arbitrary $j\in J$, so

This implies that $VW^{\ast} = I_{\mathcal{H}_{R}}$. Therefore, $WV^{\ast} = I_{\mathcal{H}_{R}}$, as required.

($\Leftarrow$). Let $y_{j} = Wz_{j}$ for any $j\in J$, where the bounded right linear operator $W$ is a left inverse of $V^{\ast}$. Since $WV^{\ast} = I_{\mathcal{H}_{R}}$, $W\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ is surjective. Hence $\{y_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$ by Theorem 3.1 (i). For all $f\in\mathcal{H}_{R}(\mathfrak{Q})$, we have

Thus, $\{y_{j}\}_{j\in J}$ is an arbitrary dual frame of $\{x_{j}\}_{j\in J}$ by Definition 2.4. □

Theorem 3.3 suggests the operator $W$ has great independent interest. To have a better understanding of $W$, we prove the following lemma.

Lemma 3.4. Let $\{x_{j}\}_{j\in J}$ be a frame for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$, and its pre-frame operator and frame operator are $V$ and $S$, respectively. Then $W\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ is a left invertible operator of $V^{\ast}$ if and only if

where $U\in \mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$.

Proof. Suppose that $W\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ is an arbitrary left invertible of $V^{\ast}$. Let $U = W$, then

Conversely, we suppose that $W = S^{-1}V+U(I_{\mathcal{H}_{R}}-V^{\ast}S^{-1}V)$, by Lemma 3.2, and we have

Hence $W$ is a bounded right linear left inverse of $V^{\ast}$. □

Based on Theorem 3.3 and Lemma 3.4, we characterize all dual frames for an arbitrarily given frame in R-Q-Hilbert spaces, and give the accurate expressions of all dual frames by taking advantage of pre-frame operators.

Theorem 3.4. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. If $\{x_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$, $V$ and $S$ are the pre-frame operator and frame operator of $\{x_{j}\}_{j\in J}$, respectively, then the sequence $\{y_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ is a dual frame for $\{x_{j}\}_{j\in J}$ if and only if

where $U\in \mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ is a right linear operator.

Proof. ($\Rightarrow$). Suppose that the sequence $\{y_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ is an arbitrary dual frame for $\{x_{j}\}_{j\in J}$. The results in Theorem 3.3 show that $y_{j} = Wz_{j}$ for arbitrary $j\in J$, where $W$ is a left inverse of $V^{\ast}$. By Lemma 3.4, we have

for some right linear operator $U\in \mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$. Hence, for any $j\in J$, we have

($\Leftarrow$). Assume that

where $U\in \mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$ is a right linear operator. Next to prove that $\{y_{j}\}_{j\in J}$ is an arbitrary dual frame for $\{x_{j}\}_{j\in J}$, note that $V$ is the pre-frame operator of $\{x_{j}\}_{j\in J}$, then

It is easy to prove that $\{y_{j}\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$. Let $W$ denote the pre-frame operator of $\{y_{j}\}_{j\in J}$, then

Thus, $W$ is a bounded right linear left inverse of $V^{\ast}$ by Lemma 3.4. We conclude that $\{y_{j}\}_{j\in J}$ is what we are looking for by Theorem 3.3. □

At the end of this section, we give some characterizations of dual frames by taking advantage of operator equations.

Theorem 3.5. Let $\{x_{j}\}_{j\in J}$ be a Parseval frame for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$, and $\{y_{j}\}_{j\in J}$ be a frame for $\mathcal{H}_{R}(\mathfrak{Q})$. Use $T_{x}, \, T_{y}$ to denote the pre-frame operators of $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$, respectively. Suppose that $P_{x}$ is the orthogonal projection: $l^{2}(\mathfrak{Q})\rightarrow R(T_{x}^{\ast})$, then $\{y_{j}\}_{j\in J}$ is a dual frame of $\{x_{j}\}_{j\in J}$ if and only if $P_{x}T_{y}^{\ast} = T_{x}^{\ast}$.

Proof. ($\Rightarrow$). Note that $\{x_{j}\}_{j\in J}$ is a Parseval frame for $\mathcal{H}_{R}(\mathfrak{Q})$, $T_{x}T_{x}^{\ast} = I_{\mathcal{H}_{R}}$ by Lemma 3.2. Hence $T_{x}^{\ast}T_{x} = P_{x}$. If $\{y_{j}\}_{j\in J}$ is a dual frame of $\{x_{j}\}_{j\in J}$, then $T_{x}T_{y}^{\ast} = I_{\mathcal{H}_{R}}$. It follows that $P_{x}T_{y}^{\ast} = T_{x}^{\ast}T_{x}T_{y}^{\ast} = T_{x}^{\ast}$.

($\Leftarrow$). Since $T_{x}^{\ast} = P_{x}T_{y}^{\ast} = T_{x}^{\ast}T_{x}T_{y}^{\ast}$, $T_{x}^{\ast}-T_{x}^{\ast}T_{x}T_{y}^{\ast} = 0$, i.e., $T_{x}^{\ast}(I_{\mathcal{H}_{R}}-T_{x}T_{y}^{\ast}) = 0$. By Theorem 3.12 in ^[19], we know that $T_{x}^{\ast}$ is onto, so we have $T_{x}T_{y}^{\ast} = I_{\mathcal{H}_{R}}$. Therefore, $\{y_{j}\}_{j\in J}$ is a dual frame of $\{x_{j}\}_{j\in J}$. □

Theorem 3.6. Let $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ be frames for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$, and their pre-frame operators be $T_{x}$ and $T_{y}$, respectively. If $P_{x}$ is an orthogonal projection from $l^{2}(\mathfrak{Q})$ to $R(T_{x}^{\ast})$, then $\{y_{j}\}_{j\in J}$ is a dual frame of $\{x_{j}\}_{j\in J}$ if and only if $P_{x}T_{y}^{\ast} = T_{x}^{\ast}S_{x}^{-1}$, where $S_{x}$ denotes the frame operator of $\{x_{j}\}_{j\in J}$.

Proof. In accordance with Theorem 3.9 in ^[19], we have $\{S_{x}^{-\frac{1}{2}}f_{x}\}_{j\in J}$ is a Parseval frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if $\{x_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$. The rest is similar to the proof of Theorem 3.5. □

4.   Sums of frames in R-Q-Hilbert spaces

In application, constructing new frames is one of the active research directions. In ^[24], the authors debated the constructions of frames by means of the sum of frames in Hilbert spaces. Inspired by their work, in this section, we will apply the pre-frame operators to discuss the finite sum of frames in R-Q-Hilbert spaces, which generalize the corresponding results on general frames in Hilbert spaces. At first, we give an example.

Example 4.1. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Define two sequences $\{x_{j}\}_{j\in J}, \; \{y_{j}\}_{j\in J}\subset\mathcal{H}_{R}(\mathfrak{Q})$ by

and $y_{j} = -x_{j}$ for all $j\in J$. Through simple calculation, we know that $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are frames for $\mathcal{H}_{R}(\mathfrak{Q})$, but $\{x_{j}+y_{j}\}_{j\in J}$ is not frame for $\mathcal{H}_{R}(\mathfrak{Q})$. Define $x_{j} = z_{j}$ for every $j\in J$ and $y_{j} = \frac{1}{j}z_{j}$ for every $j\in J$, then $\{x_{j}+y_{j}\}_{j\in J}$ is frame for $\mathcal{H}_{R}(\mathfrak{Q})$. However, $\{y_{j}\}_{j\in J}$ is not a frame but a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$.

By Example 4.1, it shows that the sum of frames for $\mathcal{H}_{R}(\mathfrak{Q})$ is not a new frame. It is natural to ask for some proper conditions, and when the conditions have been established, the sum of frames is a frame in R-Q-Hilbert spaces. The following theorems give some sufficient conditions on the frame $\{x_{j}\}_{j\in J}$ and Bessel sequence $\{y_{j}\}_{j\in J}$, which lead to new frames of the form $\{\alpha x_{j}+\beta y_{j}\}_{j\in J}$ or $\{\alpha_{j}x_{j}+\beta_{j}y_{j}\}_{j\in J}$.

Theorem 4.1. Suppose that $\{x_{j}\}_{j\in J}$ is a frame for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$, and its frame bounds are $A$ and $B$; $\{y_{j}\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$ and its Bessel bound is $B_{1}$. If $A, \; B$ and $B_{1}$ satisfy

for non-zero constants $\alpha, \, \beta\in\mathfrak{Q}$, then a new frame of the form $\{\alpha x_{j}+\beta y_{j}\}_{j\in J}$ can be constructed for $\mathcal{H}_{R}(\mathfrak{Q})$.

Proof. To prove that $\{\alpha x_{j}+\beta y_{j}\}_{j\in J}$ is a newly constructed frame for $\mathcal{H}_{R}(\mathfrak{Q})$, we must find the upper and lower bounds of $\{\alpha x_{j}+\beta y_{j}\}_{j\in J}$. For $\forall f\in\mathcal{H}_{R}(\mathfrak{Q})$, we have

Similarly,

and it follows that

Thus, we have

Observe that $A|\alpha|^{2}-2B_{1}|\beta|^{2} > 0,$ then $\{\alpha x_{j}+\beta y_{j}\}_{j\in J}$ is a newly constructed frame for $\mathcal{H}_{R}(\mathfrak{Q})$. □

Theorem 4.2. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Suppose that $\{x_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$, and its frame bounds are $A$ and $B$; $\{y_{j}\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$ and its pre-frame operator is $V$. For any two families $\{\alpha_{j}\}_{j\in J}$ and $\{\beta_{j}\}_{j\in J}$ ($\alpha_{j}, \, \beta_{j}\in\mathfrak{Q}, \; \; j\in J$), if

then $\{\alpha_{j}x_{j}+\beta_{j}y_{j}\}_{j\in J}$ is a newly constructed frame for $\mathcal{H}_{R}(\mathfrak{Q})$.

Proof. For all $f\in\mathcal{H}_{R}(\mathfrak{Q})$, we have

On the other hand, since

then,

If $\|V\|^{2} < \frac{A\inf\limits_{j\in J}|\alpha_{j}|^{2}}{2\sup\limits_{j\in J}|\beta_{j}|^{2}}$, then $A\left(\inf\limits_{j\in J}|\alpha_{j}|^{2}\right)-2\left(\sup\limits_{j\in J}|\beta_{j}|^{2}\right)\|V\|^{2} > 0$, so $\{\alpha_{j}x_{j}+\beta_{j}y_{j}\}_{j\in J}$ is a newly constructed frame for $\mathcal{H}_{R}(\mathfrak{Q})$. □

From Theorem 4.2, we have the following corollary.

Corollary 4.1. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Suppose that $\{x_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$, and its frame bounds are $A$ and $B$; $\{y_{j}\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$ with the pre-frame operator $V$. If $\|V\|^{2} < \frac{A} {2}$, then $\{x_{j}+y_{j}\}_{j\in J}$ is a new frame for $\mathcal{H}_{R}(\mathfrak{Q})$.

Theorem 4.3. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Assume that $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are both frames for $\mathcal{H}_{R}(\mathfrak{Q})$, and $V_{x}$, $V_{y}$ are their pre-frame operators, respectively. If the condition $V_{y}V_{x}^{\ast} = 0$ is met, then $\{x_{j}+y_{j}\}_{j\in J}$ is a newly constructed frame for $\mathcal{H}_{R}(\mathfrak{Q})$. Moreover, if $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are both one-tight frames for $\mathcal{H}_{R}(\mathfrak{Q})$ and $V_{y}V_{x}^{\ast} = 0$, then $\{x_{j}+y_{j}\}_{j\in J}$ is a two-tight frame for $\mathcal{H}_{R}(\mathfrak{Q})$.

Proof. Note that the pre-frame operators of $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are $V_{x}$ and $V_{y}$, respectively. It can be seen from their definitions

Hence $x_{j}+y_{j} = V_{x}z_{j}+V_{y}z_{j} = (V_{x}+V_{y})z_{j}$ for any $j\in J$. To show $\{x_{j}+y_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$, it is sufficient to show $V_{x}+V_{y}$ is onto by Theorem 3.1. Using $V_{y}V_{x}^{\ast} = 0$, we have

Once again, to utilize the invertibility of $V_{x}V_{x}^{\ast}$, for an arbitrary element $g$ in $\mathcal{H}_{R}(\mathfrak{Q})$, taking $f = V_{x}^{\ast}(V_{x}V_{x}^{\ast})^{-1}g$, undoubtedly, $f\in\mathcal{H}_{R}(\mathfrak{Q})$ satisfies

Thus $V_{x}+V_{y}$ is onto.

Especially, if $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are both one-tight frames for $\mathcal{H}_{R}(\mathfrak{Q})$, and their pre-frame satisfies operators $V_{y}V_{x}^{\ast} = 0$, then $\{x_{j}+y_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$ by the proof of the previous part. Letting $S_{\langle x+y\rangle}$ denote the frame operator of $\{x_{j}+y_{j}\}_{j\in J}$, for any $f\in\mathcal{H}_{R}(\mathfrak{Q})$, we know that

Thus,

It follows that $\{x_{j}+y_{j}\}_{j\in J}$ is a two-tight frame for $\mathcal{H}_{R}(\mathfrak{Q})$. □

Extend the number of frames to a finite number and we have the following corollary.

Corollary 4.2. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Suppose that $\{x_{1, j}\}_{j\in J}$, $\{x_{2, j}\}_{j\in J}, \, \cdot\cdot\cdot, \, \{x_{l, j}\}_{j\in J}$ are frames for $\mathcal{H}_{R}(\mathfrak{Q})$, and $V_{1}, \, V_{2}, \, \cdot\cdot\cdot, \, V_{l}$ are pre-frame operators associated with $\{x_{1, j}\}_{j\in J}, \, \{x_{2, j}\}_{j\in J}, \, \cdot\cdot\cdot, \, \{x_{l, j}\}_{j\in J}$, respectively. If $V_{m}V_{n}^{\ast} = 0, \, m, \, n = 1, \, 2, \, \cdot\cdot\cdot, \, l$, then $\{x_{1, j}+x_{2, j}+\cdot\cdot\cdot+x_{l, j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$. Moreover, if $\{x_{1, j}\}_{j\in J}, \, \{x_{2, j}\}_{j\in J}, \, \cdot\cdot\cdot, \, \{x_{l, j}\}_{j\in J}$ are one-tight frames for $\mathcal{H}_{R}(\mathfrak{Q})$ and $V_{m}V_{n}^{\ast} = 0, \, m, \, n = 1, \, 2, \, \cdot\cdot\cdot, \, l$, then $\{x_{1, j}+x_{2, j}+\cdot\cdot\cdot+x_{l, j}\}_{j\in J}$ is an $l$-tight frame for $\mathcal{H}_{R}(\mathfrak{Q})$.

More generally, we have the following theorem.

Theorem 4.4. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Assume that $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are both frames for $\mathcal{H}_{R}(\mathfrak{Q})$, and $V_{x}$, $V_{y}$ are their pre-frame operators, respectively, and satisfy $V_{y}V_{x}^{\ast} = 0$. If $P, \, Q\in\mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$, and $P$ or $Q$ is onto, then $\{Px_{j}+Qy_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$.

Proof. Note that the pre-frame operators of $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are $V_{x}$ and $V_{y}$, respectively. It can be seen from their definitions

After a simple calculation,

To show $\{Px_{j}+Qy_{j}\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$, it is sufficient to show that the operator $PV_{x}+QV_{y}$ is onto by Theorem 3.1. By Lemma 3.2, we know that $V_{x}V_{x}^{\ast}$ is invertible. Without loss of generality, let us suppose $P$ is onto. For an arbitrary element $g\in\mathcal{H}_{R}(\mathfrak{Q})$, there is always $f\in\mathcal{H}_{R}(\mathfrak{Q})$ meets $Pf = g$. Thus, for any $g\in\mathcal{H}_{R}(\mathfrak{Q})$ and taking $h = V_{x}^{\ast}(V_{x}V_{x}^{\ast})^{-1}f$, undoubtedly, $h\in\mathcal{H}_{R}(\mathfrak{Q})$ satisfies

So $PV_{x}+QV_{y}$ is onto. □

In particular, to two one-tight frames in an R-Q-Hilbert space, a necessary and sufficient condition is given, for which the new frame is tight.

Theorem 4.5. Let $\mathcal{N} = \{z_{j}\}_{j\in J}$ be a Hilbert basis for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$. Assume that $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$ are two one-tight frames for $\mathcal{H}_{R}(\mathfrak{Q})$, $V_{x}$, $V_{y}$ are their pre-frame operators, respectively, and satisfy $V_{y}V_{x}^{\ast} = 0$. Let $U_{1}, \, U_{2}\in \mathfrak{B}(\mathcal{H}_{R}(\mathfrak{Q}))$, then $\{U_{1}x_{j}+U_{2}y_{j}\}_{j\in J}$ is a $\lambda$-tight frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if $U_{1}U_{1}^{\ast}+U_{2}U_{2}^{\ast} = \lambda I_{\mathcal{H}_{R}}$.

Proof. Note that $V_{x}$ and $V_{y}$ are pre-frame operators associated with $\{x_{j}\}_{j\in J}$ and $\{y_{j}\}_{j\in J}$, respectively. For every $j\in J$, we have

For any $f\in\mathcal{H}_{R}(\mathfrak{Q})$, we have

It follows that $\{U_{1}x_{j}+U_{2}y_{j}\}_{j\in J}$ is a $\lambda$-tight frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if $U_{1}U_{1}^{\ast}+U_{2}U_{2}^{\ast} = \lambda I_{\mathcal{H}_{R}}$. □

In the end, a necessary and sufficient condition is given, for which the finite sum of frames to be a frame in an R-Q-Hilbert space.

Theorem 4.6. Let $\{x_{1, j}\}_{j\in J}$, $\{x_{2, j}\}_{j\in J}, \, \cdot\cdot\cdot, \, \{x_{l, j}\}_{j\in J}$ be frames for an R-Q-Hilbert space $\mathcal{H}_{R}(\mathfrak{Q})$, and $A_{i}$ and $B_{i}$ be the lower and upper bounds of the frame $\{x_{i, J}\}_{j\in J}$ for each $i\in\{1, \, 2, \, \cdot\cdot\cdot, \, l\}$, respectively. Let $\{\alpha_{1}, \, \alpha_{2}, \, \cdot\cdot\cdot, \alpha_{l}\}$ ($\alpha_{i}\in\mathfrak{Q}, \; \; i = 1, \, 2, \, \cdot\cdot\cdot, \, l$) be any given scalars, then $\left\{\sum\limits_{i = 1}^{l}\alpha_{i}x_{i, j}\right\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$ if and only if there exists an $M > 0$ and some $p\in\{1, \, 2, \, \cdot\cdot\cdot, \, l\}$ such that

Proof. ($\Rightarrow$). Note that $\{x_{i, j}\}_{j\in J}\; (i = 1, \, 2, \, \cdot\cdot\cdot, \, l)$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$ with frame bounds $A_{i}$ and $B_{i}$. We have for some $p\in\{1, \, 2, \, \cdot\cdot\cdot, \, l\}$ that

It follows that

Assume that $\left\{\sum\limits_{i = 1}^{l}\alpha_{i}x_{i, j}\right\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$ with the lower and upper bounds $A$ and $B$, respectively. We have for each $f\in\mathcal{H}_{R}(\mathfrak{Q})$ that

so

for $f\in\mathcal{H}_{R}(\mathfrak{Q})$. Therefore, we can conclude that

for $f\in\mathcal{H}_{R}(\mathfrak{Q})$. Taking $M = \frac{A}{B_{p}} > 0$, we have for any $f\in\mathcal{H}_{R}(\mathfrak{Q})$ that

($\Leftarrow$). For each $i\in\{1, \, 2, \, \cdot\cdot\cdot, \, l\}$, let $M > 0$ be a constant such that for some $p\in\{1, \, 2, \, \cdot\cdot\cdot, \, l\}$,

Since $\{x_{i, j}\}_{j\in J}\; (i = 1, \, 2, \, \cdot\cdot\cdot, \, l)$ is a frame, we have

so

and the lower bound of $\left\{\sum\limits_{i = 1}^{l}\alpha_{i}x_{i, j}\right\}_{j\in J}$ exists. Next, we look for the upper bound of $\left\{\sum\limits_{i = 1}^{l}\alpha_{i}x_{i, j}\right\}_{j\in J}$, we will show that $\left\{\sum\limits_{i = 1}^{l}a_{i}x_{i, j}\right\}_{j\in J}$ is a Bessel sequence for $\mathcal{H}_{R}(\mathfrak{Q})$, we have for all $f\in\mathcal{H}_{R}(\mathfrak{Q})$ that

Therefore, $\left\{\sum\limits_{i = 1}^{l}a_{i}x_{i, j}\right\}_{j\in J}$ is a frame for $\mathcal{H}_{R}(\mathfrak{Q})$. □

5.   Conclusions

Frames in Q-Hilbert spaces both retain the frame properties, and also have some advantages, such as simple structure for approximation. In this paper, the definition of pre-frame operator was introduced. We characterized the Hilbert (orthonormal) bases, frames, dual frames and Riesz bases, and obtained the accurate expressions of all dual frames of a given frame by taking advantage of pre-frame operators. We also discussed the constructions of frames with the help of the pre-frame operators, and gained some more general methods to construct new frames. The obtained results further enriched the frame theory in Q-Hilbert spaces.

Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

The authors would like to express their sincere thanks to the anonymous reviewers for their helpful comments.

The work was supported the Key Scientific Research Projects of Henan Colleges and Universities (Grant No. 21A110004).

Conflict of interest

The authors declare that there are no conflicts of interest.