# Spectral properties and a Parseval’s equality in the singular case for q-Dirac problem

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## Abstract

This paper is devoted to studying a q-analog of the singular Dirac problem. First, we investigate some spectral properties of the problem. Then we prove the existence of a spectral function and establish a Parseval’s equality, for the singular q-Dirac system in a Hilbert space. Although there were given some results for this type of problem, we think that Parseval’s equality has not been studied yet.

## Introduction

In 1910, Jackson introduced the q-derivative operator, $$D_{q}$$, different from the classical derivative, and its right-inverse, the q-integration [19, 20]. Then the q-calculus was based on these notations. This calculus has a lot of applications in different mathematical areas, such as calculus of variations, orthogonal polynomials, theory of relativity, quantum theory and statistical physics (see [1, 23, 29]). Furthermore, there are several physical models involving q-functions, q-derivatives, q-integrals and their related problems [9, 11, 12, 15, 28].

In the present paper, we study an analog of Dirac system when the differential operator is replaced by Jackson’s q-difference operator $$D_{q}$$ (definitions are given in the next section). Let us consider the q-problem which consists of the q-Dirac system

\begin{aligned} \textstyle\begin{cases} -\frac{1}{q} D_{q} Y_{2}(x,\lambda ) +p(x) Y_{1}(x,\lambda ) =\lambda Y_{1}(x,\lambda ), \\ -D_{q^{-1}} Y_{1}(x,\lambda ) +r(x) Y_{2}(x,\lambda ) =\lambda Y_{2}(x, \lambda ), \end{cases}\displaystyle 0\leq x\leq a \leq \infty, \end{aligned}
(1.1)

and the boundary conditions

\begin{aligned} &Y_{1}(0,\lambda ) =Y_{2}(0,\lambda ), \end{aligned}
(1.2)
\begin{aligned} &Y_{1} \bigl(q^{-n-1},\lambda \bigr) =Y_{2} \bigl(q^{-n},\lambda \bigr),\quad n\in \mathbb{N}, \end{aligned}
(1.3)

where λ is the spectral parameter, $$q\in (0,1)$$ is fixed, $$p(x)$$ and $$r(x)$$ are real-valued functions and continuous at zero, and $$p(x),r(x) \in L^{1}_{q}(0,\infty )$$.

The limit-point and limit-circle classifications of the singular point, $$x=\infty$$ of the q-difference equation were defined in , and using Titchmarsh’s technique, sufficient conditions that the singular point is in a limit-point case were given. In the case when $$q=1$$, i.e. $$D_{q}=\frac{d}{dx}$$, the Dirac system (1.1) was studied in many works, and for more details we refer the reader to Refs. [10, 14, 16, 22, 2527, 30]. In [2, 3], a q-analog of one dimensional Dirac system on a finite interval was investigated and the authors studied the existence and uniqueness of its solution, and some spectral properties. Also, the asymptotic formulas for the eigenvalues and the eigenfunctions were obtained in .

The paper is organized as follows. In Sect. 2, we introduce some q-notations and results that will be useful in the next sections. In Sect. 3, we study some spectral properties of the q-problem (1.1)–(1.3) by the theory of q-(basic) Sturm–Liouville problems . Finally, in Sect. 4, we prove the existence of a spectral function for singular q-Dirac system (1.1), and a Parseval’s equality is established for vector functions in a Hilbert space.

## q-notations and results

In this section, we introduce some the required q-notations and q-results which will be used throughout the paper. Hereafter, $$q\in (0,1)$$ is fixed (for some details, see ). We start with the q-shifted factorial for $$\alpha \in \mathbb{R}$$ and $$n=1,2,3, \dots$$:

$$(\alpha;q)_{n}:= \prod^{n-1}_{k=0} \bigl(1-\alpha q^{k} \bigr).$$

A set $$A \subseteq \mathbb{R}$$ is called a q-geometricset if $$qx \in A$$ for all $$x\in A$$.

The q-analogous of sine and cosine functions [4, 6, 13] are defined on $$\mathbb{C}$$ by

\begin{aligned} &\sin (x;q):=\sum^{\infty }_{m=0} \frac{(-1)^{m} q^{m(m+1)} (x(1-q))^{2m+1}}{(q;q)_{2m+1}}, \\ &\cos (x;q):=\sum^{\infty }_{m=0} \frac{(-1)^{m} q^{m^{2}} (x(1-q))^{2m}}{(q;q)_{2m}}. \end{aligned}

Let f be a real- or complex-valued function defined on a q-geometric set A. The q-difference operator $$D_{q}$$, the Jacksonq-derivative, is defined by

$$D_{q}f(x):=\frac{f(x)-f(qx)}{x-qx},\quad x\in A \backslash \{0\}.$$

If $$0\in A$$, the q-derivative at zero is defined by

$$D_{q}f(0):=\lim_{m\rightarrow \infty } \frac{f(xq^{m})-f(0)}{xq^{m}}, \quad x \in A,$$

if the limit exists and does not depend on x. Hence, for $$x \in A$$,

\begin{aligned} D_{q^{-1}} f(x)=(D_{q} f) \bigl(xq^{-1} \bigr). \end{aligned}
(2.1)

In the q-derivative, as $$q\rightarrow 1$$, the q-derivative is reduced to the classical derivative.

The right-inverse to $$D_{q}$$, the Jacksonq-integration , is given by

$$\int _{0}^{x} f(t) \,d_{q}t:= x(1-q) \sum^{\infty }_{m=0} q^{m} f \bigl(xq ^{m} \bigr),\quad x\in A,$$

provided that the series converges, and

$$D_{q} \int _{0}^{x} f(t) \,d_{q}t= f(x),$$

moreover, if $$\lim_{m\rightarrow \infty } f(xq^{m})=f(0)$$ for all $$x\in A$$ (in this case, we say f is q-regular at zero), then

\begin{aligned} \int _{0}^{x} D_{q} f(t) \,d_{q}t= f(x)-f(0). \end{aligned}
(2.2)

Hahn  defined the q-integration for a function f over $$[0,\infty )$$ by

$$\int _{0}^{\infty } f(t) \,d_{q}t=(1-q) \sum^{\infty }_{m=-\infty } q ^{m} f \bigl(q^{m} \bigr).$$

Furthermore, the following non-symmetric Leibniz formula holds:

$$D_{q} (fg) (x)=g(x) D_{q} f(x) + f(qx) D_{q} g(x).$$

If f and g are q-regular at zero, we get

$$\int _{0}^{a} g(x) D_{q} f(x) \,d_{q}x=(fg) (a)-(fg) (0)- \int _{0}^{a} f(qx) D_{q} g(x) \,d_{q}x.$$

The q-Wronskian of $$f(x)$$ and $$g(x)$$ is defined to be

$$W_{q}(f,g) (x):= f(x) D_{q}g(x)-g(x) D_{q}f(x),\quad x\in A.$$

$$\{ Y_{1},Y_{2} \}$$ forms a fundamental set of solutions for (1.1) if their q-Wronskian does not vanish at any point of A.

For more details of q-calculus, we also refer the reader to Refs. [5, 7, 13, 24].

## Spectral properties of the q-Dirac problem

In this section, we investigate some spectral properties of the q-Dirac problem (1.1)–(1.3). Further, the integral equations corresponding to the solution of (1.1) are presented.

### Theorem 3.1

The vector eigenfunctions $$Y(x,\lambda _{1})$$, $$Y(x,\lambda _{2})$$corresponding to the different eigenvalues $$\lambda _{1}$$, $$\lambda _{2}$$are orthogonal.

### Proof

Since the vector eigenfunctions

$$Y(x,\lambda _{1})= \bigl(Y_{1}(x,\lambda _{1}),Y_{2}(x,\lambda _{1}) \bigr), \qquad Y(x, \lambda _{2})= \bigl(Y_{1}(x,\lambda _{2}),Y_{2}(x, \lambda _{2}) \bigr)$$

satisfy the q-system (1.1), then

\begin{aligned} &\frac{1}{q} D_{q}Y_{2}(x,\lambda _{1}) + p(x) Y_{1}(x,\lambda _{1})= \lambda _{1}Y_{1}(x,\lambda _{1}), \end{aligned}
(3.1)
\begin{aligned} &{-}D_{q^{-1}}Y_{1}(x,\lambda _{1}) + r(x) Y_{2}(x,\lambda _{1})=\lambda _{1} Y_{2}(x,\lambda _{1}), \end{aligned}
(3.2)
\begin{aligned} &\frac{1}{q} D_{q}Y_{2}(x,\lambda _{2}) + p(x) Y_{1}(x,\lambda _{2})= \lambda _{2} Y_{1}(x,\lambda _{2}), \end{aligned}
(3.3)
\begin{aligned} &{-}D_{q^{-1}}Y_{1}(x,\lambda _{2}) + r(x) Y_{2}(x,\lambda _{2})=\lambda _{2} Y_{2}(x,\lambda _{2}). \end{aligned}
(3.4)

Multiplying (3.1)–(3.4) by $$Y_{1}(x,\lambda _{2})$$, $$Y_{2}(x,\lambda _{2})$$, $$-Y_{1}(x,\lambda _{1})$$ and $$-Y_{2}(x,\lambda _{1})$$, respectively, and applying (2.1) we have

\begin{aligned} &D_{q} \bigl(Y_{2}(x,\lambda _{1}) Y_{1} \bigl(x q^{-1},\lambda _{2} \bigr) - Y_{1} \bigl(x q ^{-1},\lambda _{1} \bigr) Y_{2}(x,\lambda _{2}) \bigr) \\ &\quad=(\lambda _{1}-\lambda _{2}) \bigl(Y_{1}(x, \lambda _{1})Y_{1}(x,\lambda _{2}) +Y_{2}(x,\lambda _{1})Y_{2}(x,\lambda _{2}) \bigr). \end{aligned}

Thus, for each $$n\in \mathbb{N}$$,

\begin{aligned} & \int _{0}^{q^{-n}} D_{q} \bigl(Y_{2}(x,\lambda _{1}) Y_{1} \bigl(x q^{-1},\lambda _{2} \bigr) - Y_{1} \bigl(x q^{-1},\lambda _{1} \bigr) Y_{2}(x,\lambda _{2}) \bigr) \,d_{q}x \\ &\quad =(\lambda _{1}-\lambda _{2}) \int _{0}^{q^{-n}} Y(x,\lambda _{1}) Y^{T}(x, \lambda _{2}) \,d_{q}x, \end{aligned}

where T is the transposition sign. This together with (2.2) yields

\begin{aligned} & \bigl(Y_{2}(x,\lambda _{1}) Y_{1} \bigl(x q^{-1},\lambda _{2} \bigr) - Y_{1} \bigl(x q^{-1}, \lambda _{1} \bigr) Y_{2}(x,\lambda _{2}) \bigr)\big|_{x=0}^{q^{-n}} \\ &\quad= (\lambda _{1}-\lambda _{2}) \int _{0}^{q^{-n}} Y(x,\lambda _{1}) Y^{T}(x, \lambda _{2}) \,d_{q}x, \end{aligned}

therefore, from (1.2)–(1.3) we obtain

\begin{aligned} (\lambda _{1}-\lambda _{2}) \int _{0}^{q^{-n}} Y(x,\lambda _{1}) Y^{T}(x, \lambda _{2}) \,d_{q}x=0,\quad \forall n \in \mathbb{N}, \end{aligned}
(3.5)

since $$\lambda _{1}\neq \lambda _{2}$$, consequently, $$Y(x,\lambda _{1})$$ and $$Y(x,\lambda _{2})$$ are orthogonal. □

### Theorem 3.2

The eigenvalues of theq-Dirac problem (1.1)(1.3) are real.

### Proof

Suppose, on the contrary that $$\lambda ^{0}$$ is a non-real eigenvalue of (1.1)–(1.3), and $$Y(x,\lambda ^{0})$$ is the corresponding vector eigenfunction of $$\lambda ^{0}$$. Then $$\overline{Y(x,\lambda ^{0})}$$ is the corresponding vector eigenfunction of $$\overline{ \lambda ^{0}}$$. Hence, it follows from $$\lambda ^{0} \neq \overline{ \lambda ^{0}}$$ and (3.5) with $$\lambda _{1}=\lambda ^{0}$$, $$\lambda _{2}=\overline{\lambda ^{0}}$$ that

\begin{aligned} \int _{0}^{q^{-n}} \bigl( \bigl\vert Y_{1} \bigl(x,\lambda ^{0} \bigr) \bigr\vert ^{2} + \bigl\vert Y_{2} \bigl(x,\lambda ^{0} \bigr) \bigr\vert ^{2} \bigr) \,d_{q}x=0, \end{aligned}

i.e., $$Y(x,\lambda ^{0})\equiv 0$$. This contradiction proves the theorem. □

For each $$n\in \mathbb{N}$$, the characteristic function for the problem (1.1)–(1.3) is defined by

\begin{aligned} \Delta _{n}(\lambda ):=Y_{1} \bigl(q^{-n-1},\lambda \bigr)-Y_{2} \bigl(q^{-n}, \lambda \bigr). \end{aligned}
(3.6)

Let $$\zeta (x,\lambda )=(\zeta _{1}(x,\lambda ),\zeta _{2}(x,\lambda ))$$ be the unique solution  of the q-Dirac system (1.1) under the initial conditions

\begin{aligned} \zeta _{1}(0,\lambda )=1=\zeta _{2}(0, \lambda ). \end{aligned}
(3.7)

Obviously, $$\zeta (x,\lambda )$$ satisfies (1.2). In the following lemma, we present the integral equations corresponding to the solution $$\zeta (x,\lambda )$$.

### Lemma 3.3

For the solution $$\zeta (x,\lambda )$$, the following integral equations hold:

\begin{aligned} &\zeta _{1}(x,\lambda )\\ &\quad =\cos (\lambda \sqrt{q}x;q)- \frac{1}{ \sqrt{q}} \sin (\lambda \sqrt{q}x;q) \\ &\qquad{}+ \sqrt{q} \int _{0}^{x} \bigl\{ \cos (\lambda \sqrt{q}t;q) \sin ( \lambda \sqrt{q}x;q) -\sin (\lambda \sqrt{q}t;q)\cos (\lambda \sqrt{q}x;q) \bigr\} r(t) \zeta _{1}(t,\lambda ) \,d_{q}t \\ &\qquad{}- \int _{0}^{x} \bigl\{ \cos (\lambda q t;q) \cos (\lambda \sqrt{q}x;q) + \sqrt{q} \sin (\lambda q t;q)\sin (\lambda \sqrt{q}x;q) \bigr\} p(q t) \zeta _{2}(q t,\lambda ) \,d_{q}t, \\ &\zeta _{2}(x,\lambda )\\ &\quad =q \sin (\lambda x;q)+ \cos (\lambda x;q) \\ &\qquad{}-q \int _{0}^{x} \bigl\{ \cos (\lambda x;q) \cos ( \lambda \sqrt{q}t;q) + \sqrt{q} \sin (\lambda x;q)\sin (\lambda \sqrt{q}t;q) \bigr\} r(t) \zeta _{1}(t,\lambda ) \,d_{q}t \\ &\qquad{}+q \int _{0}^{x} \bigl\{ \cos (\lambda x;q) \sin ( \lambda q t;q) - \sin ( \lambda x;q) \cos (\lambda q t;q) \bigr\} p(q t) \zeta _{2} (q t,\lambda ) \,d _{q}t. \end{aligned}

### Proof

For $$p(x)=r(x)=0$$, the q-system (1.1) has two solutions

\begin{aligned} &\psi _{1}(x,\lambda ) = \bigl(\cos (\lambda \sqrt{q}x;q), q\sin ( \lambda x;q) \bigr), \\ &\psi _{2}(x,\lambda ) = \bigl(-\sqrt{q}\sin (\lambda \sqrt{q}x;q), q \cos (\lambda x;q) \bigr), \end{aligned}

with the q-Wronskian

\begin{aligned} W_{q}(\psi _{1},\psi _{2}) (x,\lambda )=q. \end{aligned}
(3.8)

Therefore, in the case when $$p(x)=r(x)=0$$,

\begin{aligned} \zeta _{g}(x,\lambda )= \begin{bmatrix} c_{1}\cos (\lambda \sqrt{q}x;q)-c_{2} \sqrt{q}\sin (\lambda \sqrt{q}x;q) \\ c_{1}q\sin (\lambda x;q) +c_{2} q\cos (\lambda x;q) \end{bmatrix} ^{T} \end{aligned}
(3.9)

is a fundamental set of (1.1). Moreover, a particular solution $$\zeta _{p}(x,\lambda )=(\zeta _{1}(x,\lambda ),\zeta _{2}(x,\lambda ) )$$ of the q-system (1.1) may be written as

\begin{aligned} \textstyle\begin{cases} \zeta _{1}(x,\lambda )=v_{1}(x)\cos (\lambda \sqrt{q}x;q)-v_{2}(x) \sqrt{q}\sin (\lambda \sqrt{q}x;q), \\ \zeta _{2}(x,\lambda )=v_{1}(x)q\sin (\lambda x;q) +v_{2}(x) q\cos ( \lambda x;q), \end{cases}\displaystyle \end{aligned}
(3.10)

by q-analog of the method of variation of parameters, where $$v_{1}(x), v_{2}(x)$$ are q-regular at zero. Substituting (3.10) into (1.1), we obtain the following system:

\begin{aligned} \textstyle\begin{cases} \cos (\lambda q^{-\frac{1}{2}}x;q) D_{q^{-1}} v_{1}(x) - \sqrt{q} \sin (\lambda q^{-\frac{1}{2}} x;q) D_{q^{-1}} v_{2}(x) =-q p(x) \zeta _{2}(x,\lambda ), \\ q\sin (\lambda x;q) D_{q^{-1}} v_{1}(x) + q\cos (\lambda x;q) D_{q ^{-1}} v_{2}(x) =-q r(x q^{-1}) \zeta _{1}(x,\lambda ), \end{cases}\displaystyle \end{aligned}

and hence by Cramer’s rule and applying (3.8) we have

\begin{aligned} \textstyle\begin{cases} D_{q^{-1}} v_{1}(x) = -q p(x) \cos (\lambda x;q) \zeta _{2}(x,\lambda ) - r(x q^{-1}) \sqrt{q}\sin (\lambda q^{-\frac{1}{2}} x;q) \zeta _{1}(x q^{-1},\lambda ), \\ D_{q^{-1}} v_{2}(x) = q p(x) \sin (\lambda x;q) \zeta _{2}(x,\lambda ) - r(x q^{-1}) \cos (\lambda q^{-\frac{1}{2}} x;q) \zeta _{1}(x q^{-1}, \lambda ). \end{cases}\displaystyle \end{aligned}
(3.11)

Since $$D_{q^{-1}} v_{i}(x)=(D_{q}v_{i})(x q^{-1})$$, it follows from replacing x by xq in (3.11) and integrating from 0 to x that

\begin{aligned} \textstyle\begin{cases} v_{1}(x) = c_{1}-\sqrt{q} \int _{0}^{x}r(t) \sin (\lambda \sqrt{q} t;q) \zeta _{1}(t,\lambda ) \,d_{q}t \\ \phantom{v_{1}(x) = }{} -q \int _{0}^{x} p(qt) \cos (\lambda q t;q) \zeta _{2}(qt,\lambda ) \,d _{q}t, \\ v_{2}(x) = c_{2}- \int _{0}^{x} r(t) \cos (\lambda \sqrt{q} t;q) \zeta _{1}(t,\lambda ) \,d_{q}t \\ \phantom{v_{2}(x) =}{} +q \int _{0}^{x} p(qt) \sin (\lambda q t;q) \zeta _{2}(qt,\lambda ) \,d _{q}t. \end{cases}\displaystyle \end{aligned}
(3.12)

Now, from (3.9) and (3.12), we can write the general solution of (1.1) as

\begin{aligned} \zeta _{1}(x,\lambda )={}&2 c_{1} \cos (\lambda \sqrt{q}x;q)-2 c_{2} \sqrt{q} \sin (\lambda \sqrt{q}x;q) \\ &{}+ \sqrt{q} \int _{0}^{x} \bigl\{ \cos (\lambda \sqrt{q}t;q) \sin ( \lambda \sqrt{q}x;q) \\ &{}-\sin (\lambda \sqrt{q}t;q)\cos (\lambda \sqrt{q}x;q) \bigr\} r(t) \zeta _{1}(t,\lambda ) \,d_{q}t \\ &{}- \int _{0}^{x} \bigl\{ \cos (\lambda q t;q) \cos (\lambda \sqrt{q}x;q) \\ &{}+ \sqrt{q} \sin (\lambda q t;q)\sin (\lambda \sqrt{q}x;q) \bigr\} p(q t) \zeta _{2}(q t,\lambda ) \,d_{q}t, \end{aligned}
(3.13)
\begin{aligned} \zeta _{2}(x,\lambda )={}&2 c_{1} q \sin (\lambda x;q)+ 2 c_{2} q\cos ( \lambda x;q) \\ &{}-q \int _{0}^{x} \bigl\{ \cos (\lambda \sqrt{q}t;q) \cos (\lambda x;q) + \sqrt{q} \sin (\lambda \sqrt{q}t;q) \sin ( \lambda x;q) \bigr\} r(t) \zeta _{1}(t,\lambda ) \,d_{q}t \\ &{}+q \int _{0}^{x} \bigl\{ \sin (\lambda q t;q) \cos (\lambda x;q) \\ &{}- \cos ( \lambda q t;q) \sin (\lambda x;q) \bigr\} p(q t) \zeta _{2}(q t,\lambda ) \,d _{q}t. \end{aligned}
(3.14)

Using (3.7), (3.13) and (3.14) we obtain $$c_{1}= \frac{1}{2}$$, $$c_{2}=\frac{1}{2q}$$, and then the proof is complete. □

In the following theorem, we prove another property of the eigenvalues of (1.1)–(1.3).

### Theorem 3.4

The eigenvalues of the problem (1.1)(1.3) are simple.

### Proof

The eigenvalues of (1.1)–(1.3) are the zeros of $$\Delta _{n}(\lambda )$$. From (3.6) we have

$$\frac{\partial \Delta _{n}(\lambda )}{\partial \lambda }= \frac{\partial Y_{1}(q^{-n-1},\lambda )}{\partial \lambda } - \frac{\partial Y_{2}(q ^{-n},\lambda )}{\partial \lambda },\quad n\in \mathbb{N}.$$

Now, let $$\lambda =\lambda ^{0}$$ be a double eigenvalue of (1.1)–(1.3) with the corresponding vector eigenfunction $$Y(x,\lambda ^{0})$$. Then $$\Delta _{n}(\lambda ^{0})=0$$ and $$\frac{ \partial \Delta _{n}}{\partial \lambda }(\lambda ^{0})=0$$, i.e., for $$n\in \mathbb{N}$$, the system

\begin{aligned} \textstyle\begin{cases} a Y_{1}(q^{-n-1},\lambda ^{0})+b Y_{2}(q^{-n},\lambda ^{0})=0, \\ a \frac{\partial Y_{1}}{\partial \lambda }(q^{-n-1},\lambda ^{0}) + b \frac{\partial Y_{2}}{\partial \lambda }(q^{-n},\lambda ^{0}) =0, \end{cases}\displaystyle \end{aligned}

has the nontrivial solution $$(a,b)=(1,-1)$$. Hence,

\begin{aligned} Y_{1} \bigl(q^{-n-1},\lambda ^{0} \bigr) \frac{\partial Y_{2}}{\partial \lambda } \bigl(q ^{-n},\lambda ^{0} \bigr) - Y_{2} \bigl(q^{-n},\lambda ^{0} \bigr) \frac{\partial Y_{1}}{ \partial \lambda } \bigl(q^{-n-1},\lambda ^{0} \bigr) =0. \end{aligned}
(3.15)

On the other hand, differentiating (1.1) with respect to λ, we get

\begin{aligned} \textstyle\begin{cases} \frac{1}{q} D_{q} (\frac{\partial Y_{2}}{\partial \lambda }) + (p(x) - \lambda ) \frac{\partial Y_{1}}{\partial \lambda } = Y_{1}, \\ -D_{q^{-1}} (\frac{\partial Y_{1}}{\partial \lambda }) +( r(x)- \lambda ) \frac{\partial Y_{2}}{\partial \lambda } = Y_{2}. \end{cases}\displaystyle \end{aligned}
(3.16)

Multiplying (1.1) and (3.16) by $$\frac{\partial Y_{1}}{ \partial \lambda }$$, $$\frac{\partial Y_{2}}{\partial \lambda }$$, $$-Y_{1}$$ and $$-Y_{2}$$, respectively, and applying (2.1), we obtain

\begin{aligned} \frac{1}{q} D_{q} \biggl\{ Y_{1} \bigl(x q^{-1}, \lambda \bigr) \frac{\partial Y_{2}(x, \lambda )}{\partial \lambda } - Y_{2}(x, \lambda )\frac{\partial Y _{1}(x q^{-1}, \lambda )}{\partial \lambda } \biggr\} = Y^{2}_{1}(x, \lambda ) + \frac{1}{q^{2}} Y^{2}_{2}(x, \lambda ). \end{aligned}

Therefore, integrating with respect to x from 0 to $$q^{-n}$$, with applying (2.2), we have

\begin{aligned} &\frac{1}{q} \biggl\{ Y_{1} \bigl(x q^{-1}, \lambda \bigr) \frac{\partial Y_{2}(x, \lambda )}{\partial \lambda } - Y_{2}(x, \lambda )\frac{\partial Y _{1}(x q^{-1}, \lambda )}{\partial \lambda } \biggr\} \bigg|_{x=0}^{q^{-n}} \\ &\quad = \int _{0}^{q^{-n}} \biggl( Y^{2}_{1}(x, \lambda ) + \frac{1}{q^{2}} Y^{2} _{2}(x, \lambda ) \biggr) \,d_{q}x,\quad n\in \mathbb{N}. \end{aligned}
(3.17)

According to Lemma 3.3, $$\frac{\partial Y_{1}}{\partial \lambda } (0, \lambda ^{0})= \frac{\partial Y_{2}}{\partial \lambda } (0, \lambda ^{0}) =0$$. Taking this and (3.15) into the left-side of (3.17), we obtain

$$\int _{0}^{q^{-n}} \biggl( Y^{2}_{1}(x, \lambda ) + \frac{1}{q^{2}} Y^{2} _{2}(x, \lambda ) \biggr) \,d_{q}x =0,\quad n\in \mathbb{N}.$$

Consequently, $$Y_{1}(x, \lambda ^{0})=Y_{2}(x, \lambda ^{0}) \equiv 0$$, i.e. $$Y(x, \lambda ^{0}) \equiv 0$$. Thus, we arrive at the contradiction. The proof is complete. □

## Spectral function and Parseval’s equality

Let $$\lambda _{m,n}$$, $$m\geq 0$$, $$n\in \mathbb{N}$$, be the eigenvalues of the q-Dirac problem (1.1)–(1.3) (i.e. the roots of $$\Delta _{n}(\lambda )$$) with the corresponding eigenfunctions

$$Y_{m,n}(x)=Y(x,\lambda _{m,n}) = \bigl(Y_{1}(x, \lambda _{m,n}), Y_{2}(x, \lambda _{m,n}) \bigr).$$

If $$f(x)=(f_{1} (x), f_{2}(x))$$ is a vector function, $$f_{1},f_{2} \in L^{2}_{q}(0,q^{-n})$$, $$n\in \mathbb{N}$$, $$Y_{m,n,i}(x )=Y_{i}(x, \lambda _{m,n})$$, $$i=1,2$$, and

$$\alpha ^{2}_{m,n,i}= \int _{0}^{q^{-n}}Y^{2}_{m,n,i}(x) \,d_{q}x,\quad i=1,2,$$

then from  we have

\begin{aligned} \int _{0}^{q^{-n}} \bigl(f^{2}_{1} (x)+f^{2}_{2} (x) \bigr) \,d_{q}x= \sum _{m=- \infty }^{\infty } \sum _{i=1}^{2} \frac{1}{\alpha ^{2}_{m,n,i}} \biggl( \int _{0}^{q^{-n}} f_{i} (x) Y_{m,n,i}(x) \,d_{q}x \biggr)^{2}. \end{aligned}
(4.1)

Denote the non-decreasing step function $$\rho _{n}$$ by

\begin{aligned} \rho _{n}(\lambda )= \textstyle\begin{cases} -\sum_{\lambda < \lambda _{m,n} < 0 } \sum_{i=1}^{2} \frac{1}{\alpha ^{2}_{m,n,i}},& \lambda < 0, \\ \sum_{0 \leq \lambda _{m,n} < \lambda } \sum_{i=1}^{2} \frac{1}{\alpha ^{2}_{m,n,i}},& \lambda \geq 0. \end{cases}\displaystyle \end{aligned}

Therefore, (4.1) can be written as

\begin{aligned} \int _{0}^{q^{-n}} \bigl(f^{2}_{1} (x)+f^{2}_{2} (x) \bigr) \,d_{q}x= \int _{-\infty }^{\infty } \bigl(F^{2}_{1,n} (\lambda )+F^{2}_{2,n} (\lambda ) \bigr) \,d\rho _{n}(\lambda ), \end{aligned}
(4.2)

where $$F_{i,n} (\lambda )=\int _{0}^{q^{-n}} f_{i} (x) Y_{i}(x,\lambda ) \,d_{q}x$$, $$i=1,2$$.

### Lemma 4.1

For any positiveτ, the following inequality holds:

\begin{aligned} \sum_{-\tau \leq \lambda _{m,n} < \tau } \sum _{i=1}^{2} \frac{1}{\alpha ^{2}_{m,n,i}}= \rho _{n}(\tau ) -\rho _{n}(-\tau ) < \frac{4}{\tau }. \end{aligned}
(4.3)

### Proof

Since $$\zeta _{1}(x,\lambda )$$ and $$\zeta _{2}(x,\lambda )$$ are continuous at zero, it follows from (3.7) that there is a positive number τ and nearby zero such that

\begin{aligned} \frac{1}{\tau } \sum_{i=1}^{2} \biggl( \int _{0}^{\tau }\zeta _{i}(x,\lambda ) \,d_{q}x \biggr)^{2} > \frac{1}{2}. \end{aligned}
(4.4)

Denote the vector function $${}_{\tau }f(x)=({}_{\tau }f_{1}(x), {}_{ \tau }f_{2}(x))$$ by

\begin{aligned} _{\tau }f_{i}(x)= \textstyle\begin{cases} \frac{1}{\tau },& 0\leq x \leq \tau, \\ 0,& x>\tau. \end{cases}\displaystyle \end{aligned}

Then, from (4.2) and (4.4), we obtain

\begin{aligned} \int _{0}^{\tau } \bigl({}_{\tau }f^{2}_{1}(x)+ {}_{\tau }f^{2}_{2}(x) \bigr) \,d_{q}x &=\frac{2}{\tau } = \int _{-\infty }^{\infty } \sum_{i=1}^{2} \biggl( \int _{0} ^{\tau } \frac{1}{\tau } \zeta _{i}(x,\lambda ) \,d_{q}x \biggr)^{2} \,d \rho _{n}( \lambda ) \\ &\geq \int _{-\tau }^{\tau } \sum _{i=1}^{2} \biggl(\frac{1}{\tau } \int _{0} ^{\tau } \zeta _{i}(x, \lambda ) \,d_{q}x \biggr)^{2} \,d\rho _{n}( \lambda ) \\ &>\frac{1}{2} \bigl( \rho _{n}(\tau )-\rho _{n}(-\tau ) \bigr), \end{aligned}

and we arrive at (4.3). □

The following lemmas were proved in .

### Lemma 4.2

Let $$\{ v_{n} \}_{n=1}^{\infty }$$be a uniformly bounded sequence of real non-decreasing function ofλon a finite interval $$[a,b]$$. Then there exist a subsequence $$\{ v_{n_{k}} \}_{k=1}^{ \infty }$$and a non-decreasing functionvsuch that, for $$\lambda \in [a,b]$$, $$\lim_{k\rightarrow \infty } v_{n_{k}}(\lambda )=v(\lambda )$$.

### Lemma 4.3

Assume $$\{ v_{n} \}_{n=1}^{\infty }$$is a real uniformly bounded sequence of real non-decreasing function ofλon a finite interval $$[a,b]$$, and suppose for $$\lambda \in [a,b]$$, $$\lim_{n\rightarrow \infty } v_{n}(\lambda )=v(\lambda )$$. Ifgis any continuous function ofλon $$[a,b]$$, then $$\lim_{n\rightarrow \infty } \int _{a}^{b} g(\lambda ) \,dv_{n}(\lambda )= \int _{a}^{b} g(\lambda ) \,dv(\lambda )$$.

Now, let ρ be any non-decreasing function of λ on the interval $$(-\infty, \infty )$$. We define by $$L^{2}_{\rho }(-\infty, \infty ) \times L^{2}_{\rho }(-\infty, \infty )$$ the Hilbert space of all vector functions $$g=(g_{1},g_{2}): (-\infty, \infty )\times (- \infty, \infty ) \rightarrow \mathbb{R}$$ which $$g_{1},g_{2}$$ are measurable with respect to the Lebesgue–Stieltjes measure defined by ρ, such that $$\int _{-\infty }^{\infty } g_{i}(\lambda ) \,d\rho ( \lambda ) < \infty$$, $$i=1,2$$, with inner product

$$\langle g,h\rangle _{\rho }:= \int _{-\infty }^{\infty } \bigl( g_{1}(\lambda )h_{1}( \lambda ) +g_{2}(\lambda )h_{2}( \lambda ) \bigr) \,d\rho (\lambda ).$$

In the following theorem, we prove the main result of this section.

### Theorem 4.4

For theq-Dirac problem (1.1)(1.3), there exists a non-decreasing function $$\rho (\lambda )$$on the interval $$(-\infty, \infty )$$such that satisfies the following property:

If $$f=(f_{1},f_{2})$$is a vector function, $$f_{i}\in L^{2}_{q }(0,q ^{-n})$$, $$i=1,2$$, then there exists a function $$F=(F_{1},F_{2}) \in L^{2}_{\rho }(-\infty, \infty ) \times L^{2}_{\rho }(-\infty, \infty )$$such that

\begin{aligned} \lim_{n \rightarrow \infty } \int _{-\infty }^{\infty } \biggl\{ F_{1}( \lambda )+F_{2}(\lambda ) - \int _{0}^{q^{-n}} \bigl(f_{1}(x)Y_{1}(x, \lambda ) +f_{2}(x) Y_{2}(x,\lambda ) \bigr) \,d_{q}x \biggr\} \,d\rho (\lambda )=0, \end{aligned}

and the Parseval’s equality holds:

$$\int _{0 }^{\infty } \bigl( f^{2}_{1}(x) +f^{2}_{2}(x) \bigr) \,d_{q}x= \int _{- \infty }^{\infty } \bigl( F^{2}_{1}( \lambda )+F^{2}_{2}(\lambda ) \bigr) \,d \rho (\lambda ).$$

The function ρ is called the spectral function for the q-problem (1.1)–(1.3).

### Proof

Assume that the vector function $$f_{\eta }(x)=(f_{\eta,1}(x),f_{ \eta,2}(x))$$ satisfies the following conditions:

1. (1)

$$f_{\eta }(0)=(1,1)$$;

2. (2)

$$f_{\eta }(x)$$ vanishes outside $$[0,q^{-\eta }]\times [0,q ^{-\eta }]$$, $$q^{-\eta } < q^{-n }$$;

3. (3)

$$f_{\eta,i}(x)$$ and $$D_{q} f_{\eta,i}(x)$$, $$i=1,2$$, are q-regular at zero.

According to (4.2), we can write

\begin{aligned} \int _{0 }^{q^{-\eta } } \bigl( f^{2}_{\eta,1}(x) +f^{2}_{\eta,1}(x) \bigr)\,d _{q}x= \int _{-\infty }^{\infty } \bigl( F^{2}_{1}( \lambda )+F^{2}_{2}( \lambda ) \bigr) \,d\rho (\lambda ), \end{aligned}
(4.5)

where

$$F_{i} (\lambda )= \int _{0}^{q^{-\eta }} f_{\eta,i} (x) Y_{i}(x,\lambda )\,d_{q}x,\quad i=1,2.$$

Since $$Y(x,\lambda )=(Y_{1}(x,\lambda ),Y_{2}(x,\lambda ))$$ satisfies the q-system (1.1), we have

\begin{aligned} \textstyle\begin{cases} Y_{1}(x,\lambda )=\frac{1}{\lambda } ( \frac{1}{q} D_{q}Y_{2}(x, \lambda ) + p(x) Y_{1}(x,\lambda ) ), \\ Y_{2}(x,\lambda )=\frac{1}{\lambda } ( - D_{q^{-1}}Y_{1}(x,\lambda ) + r(x) Y_{2}(x,\lambda ) ). \end{cases}\displaystyle \end{aligned}

Denote

\begin{aligned} &F_{n,1} (\lambda ):= \frac{1}{\lambda } \int _{0}^{q^{-n } } f_{\eta,1} (x) \biggl( \frac{1}{q} D_{q}Y_{2}(x,\lambda ) + p(x) Y_{1}(x,\lambda ) \biggr) \,d _{q}x, \\ &F_{n,2} (\lambda ):= \frac{1}{\lambda } \int _{0}^{q^{-n }} f_{\eta,2} (x) \bigl( - D_{q^{-1}}Y_{1}(x,\lambda ) + r(x) Y_{2}(x, \lambda ) \bigr) \,d_{q}x . \end{aligned}

Since $$f_{\eta } (x)$$ vanishes in a neighborhood of $$(q^{-n }, q^{-n })$$, and $$f_{\eta } (0)=Y(0,\lambda )=(1,1)$$, using q-integration by parts we get

\begin{aligned} &F_{n,1} (\lambda )= \frac{1}{\lambda } \int _{0}^{q^{-n } } \biggl\{ \frac{1}{q} Y_{2}(x,\lambda ) D_{q} f_{\eta,1} (x)+ p(x) f_{\eta,1} (x) Y_{1}(x,\lambda ) \biggr\} \,d_{q}x, \\ &F_{n,2} (\lambda )= \frac{1}{\lambda } \int _{0}^{q^{-n }} \bigl\{ - Y_{1}(x, \lambda ) D_{q^{-1}} f_{\eta,2} (x)+ r(x) f_{\eta,2} (x) Y_{2}(x, \lambda ) \bigr\} \,d_{q}x. \end{aligned}

Applying (4.2), we have, for any $$\tau >0$$,

\begin{aligned} & \int _{ |\lambda |>\tau } F^{2}_{n,1} (\lambda ) \,d \rho _{n}(\lambda ) \\ &\quad \leq \frac{1}{\tau ^{2}} \int _{|\lambda |>\tau } \biggl\{ \int _{0}^{q^{-n } } \biggl\{ \frac{1}{q} Y_{2}(x,\lambda ) D_{q} f_{\eta,1} (x) \\ &\qquad{}+ p(x) f_{\eta,1} (x) Y_{1}(x,\lambda ) \biggr\} \,d_{q}x \biggr\} ^{2} \,d\rho _{n}( \lambda ) \\ &\quad \leq \frac{1}{\tau ^{2}} \int _{-\infty }^{\infty } \biggl\{ \int _{0}^{q ^{-n } } \biggl\{ \frac{1}{q} Y_{2}(x,\lambda ) D_{q} f_{\eta,1} (x) \\ &\qquad{}+ p(x) f_{\eta,1} (x) Y_{1}(x,\lambda ) \biggr\} \,d_{q}x \biggr\} ^{2} \,d\rho _{n}( \lambda ) \\ & \quad= \frac{1}{\tau ^{2}} \biggl\{ \int _{-\infty }^{\infty } \biggl( \int _{0}^{q^{-n } } \frac{1}{q} Y_{2}(x,\lambda ) D_{q} f_{\eta,1} (x) \,d_{q}x \biggr)^{2} \,d \rho _{n}(\lambda ) \\ &\qquad{} + \int _{-\infty }^{\infty } \biggl( \int _{0}^{q^{-n } } Y_{1}(x,\lambda ) p(x) f_{\eta,1} (x) \,d_{q}x \biggr)^{2} \,d \rho _{n}(\lambda ) + M_{1} \biggr\} , \end{aligned}

where

$$M_{1}=\frac{2}{q} \int _{-\infty }^{\infty } \biggl\{ \int _{0}^{q^{-n } } Y _{1}(x,\lambda ) p(x) f_{\eta,1} (x) \,d_{q}x \biggr\} \biggl\{ \int _{0}^{q^{-n } } Y_{2}(x,\lambda ) D_{q} f_{\eta,1} (x) \,d_{q}x \biggr\} \,d\rho _{n}(\lambda ).$$

Hence, from (4.2) we obtain

\begin{aligned} &\int _{|\lambda |>\tau } F^{2}_{n,1} (\lambda ) \,d \rho _{n}(\lambda ) \\ &\quad \leq \frac{1}{\tau ^{2}} \biggl\{ \int _{0}^{q^{-n } } \bigl( p(x) f_{\eta,1} (x) \bigr)^{2}\,d_{q}x + \int _{0}^{q^{-n } } \biggl( \frac{1}{q} D_{q} f_{\eta,1} (x) \biggr)^{2} \,d_{q}x + M_{1} \biggr\} . \end{aligned}
(4.6)

Similarly, we have

\begin{aligned} &\int _{|\lambda |>\tau } F^{2}_{n,2} (\lambda ) \,d \rho _{n}(\lambda ) \\ &\quad \leq \frac{1}{\tau ^{2}} \biggl\{ \int _{0}^{q^{-n } } \bigl(D_{q^{-1}} f_{\eta,2} (x) \bigr)^{2} \,d_{q}x + \int _{0}^{q^{-n } } \bigl( r(x) f_{\eta,2} (x) \bigr)^{2}\,d _{q}x + M_{2} \biggr\} , \end{aligned}
(4.7)

where

$$M_{2}=-2 \int _{-\infty }^{\infty } \biggl\{ \int _{0}^{q^{-n } } Y_{1}(x, \lambda ) D_{q^{-1}} f_{\eta,2} (x) \,d_{q}x \biggr\} \biggl\{ \int _{0}^{q^{-n } } Y_{2}(x,\lambda ) r(x) f_{\eta,2} (x)\,d_{q}x \biggr\} \,d\rho _{n}( \lambda ).$$

Therefore, it follows from (4.5)–(4.7) that

\begin{aligned} & \biggl\vert \int _{0}^{q^{-\eta } } \bigl( f^{2}_{\eta,1} (x)+f^{2}_{\eta,2} (x) \bigr) \,d _{q}x- \int _{-\tau }^{\tau } \bigl(F^{2}_{n,1} (\lambda ) +F^{2}_{n,2} ( \lambda ) \bigr)\,d\rho _{n}(\lambda ) \biggr\vert \\ & \quad = \int _{ \vert \lambda \vert >\tau } \bigl(F^{2}_{n,1} (\lambda ) +F^{2}_{n,2} ( \lambda ) \bigr)\,d\rho _{n}( \lambda ) \\ &\quad < \frac{1}{\tau ^{2}} \int _{0}^{q^{-\eta } } \biggl\{ \bigl(p(x)f_{\eta,1} (x) \bigr)^{2}+ \biggl(\frac{1}{q}D_{q}f_{\eta,1} (x) \biggr)^{2}+ \bigl(D_{q^{-1}}f_{\eta,2} (x) \bigr)^{2} \\ &\qquad{} + \bigl(r(x)f_{\eta,2} (x) \bigr)^{2} + M_{1}+ M_{2} \biggr\} \,d_{q}x. \end{aligned}
(4.8)

On the other hand, according to Lemma 4.1, the set $$\{ \rho _{n}( \lambda ) \}$$ is bounded. Thus, by Lemmas 4.2 and 4.3, there is a subsequence $$\{ n_{k} \}$$ such that $$\{ \rho _{n_{k}}(\lambda ) \}$$ converges to a monotone function $$\rho (\lambda )$$. Passing to the limit with respect to $$\{ n_{k} \}$$ in (4.8), we obtain

\begin{aligned} & \biggl\vert \int _{0}^{q^{-\eta } } \bigl( f^{2}_{\eta,1} (x)+f^{2}_{\eta,2} (x) \bigr) \,d _{q}x- \int _{-\tau }^{\tau } \bigl(F^{2}_{n,1} (\lambda ) +F^{2}_{n,2} ( \lambda ) \bigr)\,d\rho ( \lambda ) \biggr\vert \\ & \quad < \frac{1}{\tau ^{2}} \int _{0}^{q^{-\eta } } \biggl\{ \bigl(p(x)f_{\eta,1} (x) \bigr)^{2}+ \biggl(\frac{1}{q}D_{q}f_{\eta,1} (x) \biggr)^{2}+ \bigl(D_{q^{-1}}f_{\eta,2} (x) \bigr)^{2} \\ & \qquad{}+ \bigl(r(x)f_{\eta,2} (x) \bigr)^{2}+ M_{1}+ M_{2} \biggr\} \,d_{q}x. \end{aligned}

So,

$$\int _{0}^{q^{-\eta } } \bigl( f^{2}_{\eta,1} (x)+f^{2}_{\eta,2} (x) \bigr) \,d _{q}x = \int _{-\infty }^{\infty } \bigl(F^{2}_{n,1} (\lambda ) +F^{2}_{n,2} (\lambda ) \bigr)\,d\rho (\lambda )$$

as $$\tau \rightarrow \infty$$. Now, let $$f=(f_{1},f_{2})$$ be an arbitrary vector function in $$L^{2}_{q}(0,\infty )\times L^{2}_{q}(0, \infty )$$. We know that there exists a sequence $$\{ f_{\eta }(x)= (f _{\eta,1}(x),f_{\eta,2}(x) ) \}$$ satisfying the conditions (1)–(3) such that

$$\lim_{\eta \rightarrow \infty } \int _{0}^{\infty } \bigl( f_{i}(x) - f_{ \eta,i}(x) \bigr)^{2} \,d_{q}x=0,\quad i=1,2.$$

Then

$$\int _{0}^{\infty } \bigl( f^{2}_{\eta,1}(x) + f^{2}_{\eta,2}(x) \bigr)\,d_{q}x= \int _{-\infty }^{\infty } \bigl(F^{2}_{\eta,1} (\lambda ) +F^{2}_{\eta,2} (\lambda ) \bigr)\,d\rho (\lambda ),$$

where $$F_{\eta,i} (\lambda )=\int _{0}^{\infty } f_{\eta,i} (x) Y _{i}(x,\lambda )\,d_{q}x$$. Since for $$i=1,2$$,

$$\int _{0}^{\infty } \bigl( f_{\eta _{1},i}(x) - f_{\eta _{2},i}(x) \bigr)^{2} \,d _{q}x \rightarrow 0$$

as $$\eta _{1}, \eta _{2}\rightarrow \infty$$, we get

\begin{aligned} \int _{-\infty }^{\infty } \bigl(F_{\eta _{1},i} (\lambda ) -F_{\eta _{2},i} ( \lambda ) \bigr)^{2}\,d\rho (\lambda ) = \int _{0}^{\infty } \bigl( f_{\eta _{1},i}(x) - f_{\eta _{2},i}(x) \bigr)^{2} \,d_{q}x \rightarrow 0,\quad i=1,2, \end{aligned}

as $$\eta _{1},\eta _{2}\rightarrow \infty$$. This is means that there is a limit vector function $$F=(F_{1},F_{2})$$ such that by the completeness of the space $$L^{2}_{\rho }(-\infty,\infty )\times L^{2}_{\rho }(- \infty,\infty )$$,

$$\int _{0 }^{\infty } \bigl( f^{2}_{1}(x) +f^{2}_{2}(x) \bigr) \,d_{q}x= \int _{- \infty }^{\infty } \bigl( F^{2}_{1}( \lambda )+F^{2}_{2}(\lambda ) \bigr) \,d \rho (\lambda ).$$

Now, it remains to show that the function $$\widetilde{F}_{\eta }( \lambda ):=(\widetilde{F}_{\eta,1},\widetilde{F}_{\eta,2})$$ with

$$\widetilde{F}_{\eta,i}(\lambda ):= \int _{0}^{q^{-\eta } } f_{i} (x) Y _{i}(x,\lambda )\,d_{q}x,$$

as $$\eta \rightarrow \infty$$, converges to $$F=(F_{1},F_{2})$$ in $$L^{2}_{\rho }(-\infty,\infty )\times L^{2}_{\rho }(-\infty,\infty )$$. For this purpose, assume that $$s=(s_{1},s_{2})$$ is another function in $$L^{2}_{q}(0,\infty )\times L^{2}_{q}(0,\infty )$$, and by a similar argument, $$S(\lambda )$$ is defined by s. Clearly,

\begin{aligned} \int _{0 }^{\infty } \bigl( f_{i}(x) - s_{i}(x) \bigr)^{2} \,d_{q}x= \int _{-\infty }^{\infty } \bigl( F_{i}(\lambda )-S_{i}(\lambda ) \bigr)^{2} \,d\rho (\lambda ),\quad i=1,2. \end{aligned}

For $$i=1,2$$, set

\begin{aligned} s_{i}(x)= \textstyle\begin{cases} f_{i}(x),& x\in [0, q^{-\eta } ], \\ 0,& x\in (q^{-\eta }, \infty ). \end{cases}\displaystyle \end{aligned}

Then

\begin{aligned} \int _{-\infty }^{\infty } \bigl( F_{i}(\lambda )-\widetilde{F}_{\eta,i}( \lambda ) \bigr)^{2} \,d\rho ( \lambda )= \int _{q^{-\eta } } ^{\infty } f^{2} _{i}(x) \,d_{q}x,\quad i=1,2, \end{aligned}

as $$\eta \rightarrow \infty$$. Consequently, $$\widetilde{F}_{\eta }$$ converges to F in $$L^{2}_{\rho }(-\infty,\infty )\times L^{2}_{ \rho }(-\infty,\infty )$$ as $$\eta \rightarrow \infty$$. This completes the proof. □

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### Acknowledgements

The author gratefully acknowledges that this research is partially supported by the University of Kashan under grant number 682482/10.

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