ON THE RESOLVENT OF SINGULAR q-STURM-LIOUVILLE OPERATORS

In this paper, we investigate the resolvent operator of the singular qSturm-Liouville problem dened as 1 q Dq 1 [Dqy (x)] + [r (x) ] y (x) = 0; with the boundary condition y (0; ) cos +Dq 1y (0; ) sin = 0; where 2 C, r is a real-valued function dened on [0;1), continuous at zero and r 2 Lq;loc[0;1). We give a representation for the resolvent operator and investigate some properties of this operator. Furthermore, we obtain a formula for the Titchmarsh-Weyl function of the singular q-Sturm-Liouville problem. 1. Introduction Quantum (or q) calculus is a very interesting eld in mathematics. It has numerous in statistic physics, quantum theory, the calculus of variations and number theory; see, e.g., [12, 1, 11, 14, 15, 18, 21, 24]). The rst results in q-calculus belong to the Euler. In 2005, Annaby and Mansour investigated q-Sturm-Liouville problems [10]: Later in [9], the authors studied the Titchmarsh-Weyl theory for q-Sturm-Liouville equations. In [3,4], the authors proved the existence of a spectral function for q-Sturm-Liouville operator. In this article, we investigate the following q-Sturm-Liouville problem dened as 1 q Dq 1Dqy (x) + u (x) y (x) = y (x) ; (1) 2020 Mathematics Subject Classication. 33D15, 34L40, 39A13, 34L10. Keywords and phrases. q-Sturm-Liouville operator, spectral function, resolvent operator, Titchmarsh-Weyl function. bilenderpasaoglu@sdu.edu.tr; hustuna@gmail.com-Corresponding author 0000-0002-9315-4652; 0000-0001-7240-8687 . c 2021 Ankara University Communications Facu lty of Sciences University of Ankara-Series A1 Mathematics and Statistics 702 ON THE RESOLVENT OF SINGULAR q-STURM-LIOUVILLE OPERATORS 703 where 0 < x < 1: The resolvent operator for this problem is constructed. Using the spectral function, an integral representation is obtained. Furthermore, some properties of this operator are investigated. A formula for the Titchmarsh-Weyl function of Eq. (1) is given. Historically, in 1910, H. Weyl was rst obtained a representation theorem for the resolvent of Sturm-Liouville problem dened by (py0) + qy = y; x 2 (0;1); where p; q are real-valued and p ; q 2 Lloc[0;1). Similar representation theorems were proved in [25,20,2, 5, 6, 7]. 2. Preliminaries In this section, we give a brief introduction to quantum calculus and refer the interested reader to [17,8, 12]. Let 0 < q < 1 and let A R is a q-geometric set, i.e., qx 2 A for all x 2 A: The Jackson q-derivative is dened by Dqy (x) = 1 (x) [y (qx) y (x)] ; where (x) = qx x and x 2 A: We note that there is a connection the Jackson qderivative between and q-deformed Heisenberg uncertainty relation (see [23]). The q-derivative at zero is dened as Dqy (0) = lim n!1 [qx] 1 [y (qx) y (0)] (x 2 A); (2) if the limit in (2) exists and does not depend on x: The Jackson q-integration is given by Z x 0 f (t) dqt = x (1 q) 1 X n=0 qf (qx) (x 2 A); provided that the series converges, and Z b

In this article, we investigate the following q-Sturm-Liouville problem de…ned as ON THE RESOLVENT OF SINGULAR q-STURM -LIOUVILLE OPERATORS 703 where 0 < x < 1: The resolvent operator for this problem is constructed. Using the spectral function, an integral representation is obtained. Furthermore, some properties of this operator are investigated. A formula for the Titchmarsh-Weyl function of Eq. (1) is given. Historically, in 1910, H. Weyl was …rst obtained a representation theorem for the resolvent of Sturm-Liouville problem de…ned by (py 0 ) 0 + qy = y; x 2 (0; 1); where p; q are real-valued and p 1 ; q 2 L 1 loc [0; 1). Similar representation theorems were proved in [25,20,2,5,6,7].

Preliminaries
In this section, we give a brief introduction to quantum calculus and refer the interested reader to [17,8,12].
Let 0 < q < 1 and let A R is a q-geometric set, i.e., qx 2 A for all x 2 A: The Jackson q-derivative is de…ned by where (x) = qx x and x 2 A: We note that there is a connection the Jackson qderivative between and q-deformed Heisenberg uncertainty relation (see [23]). The q-derivative at zero is de…ned as if the limit in (2) exists and does not depend on x: The Jackson q-integration is given by provided that the series converges, and where a; b 2 A: The q-integration for a function over [0; 1) de…ned by the formula ( [13]) Let f be a function on A and let 0 2 A: then f is called q-regular at zero. Throughout the paper, we deal only with functions q-regular at zero. The following relation holds where f and g are q-regular at zero. Let L 2 q [0; 1) be the Hilbert space consisting of all functions f satisfying ( [9]) with the inner product The q-Wronskian of the functions y (:) and z (:) is de…ned by the formula where x 2 [0; 1):
Note that the problem (3)-(5) has a purely discrete spectrum [10]. Let m;q n be the eigenvalues of the problem (3)-(5). Let ' m;q n be the corresponding eigenfunctions and where ' m;q n (x) := ' m;q n x; m;q n and m 2 N: Then we have the following Parseval equality (see [8]) where f (:) 2 L 2 q [0; q n ]: Now, let us de…ne the nondecreasing step function % q n on [0; 1) by where Lemma 2. Let > 0: Then the following relation holds where = ( ) is a positive constant not depending on q n : Proof. Let sin 6 = 0: Since ' (x; ) is continuous at zero, by condition ' (0; ) = sin ; there exists a positive number h and nearby 0 such that Let (10) and (12) that which proves the inequality (11). Let sin = 0 and (10), we can get the desired result.
We now return to the formula (7), whose right-hand side has been called the resolvent. The resolvent is known to exist for all which are not eigenvalues of the problem (3)-(5). Now, we will get the expansion of the resolvent.
Since the function y (x; ) satis…es the equation L(y) y(x) = f (x); x 2 (0; q n ) ( 2 C; 6 = m;q n ; m 2 N) and the boundary conditions (4), (5), via the qintegration by parts, we …nd (the operator A generated by the expression L and the boundary conditions (4), (5) is a self-adjoint (see [10] The set of all eigenfunctions form an orthonormal basis for L 2 q (0; q n ) (see [10]). Then, the function y (:; ) 2 L 2 q (0; q n ) ( 2 C; 6 = m;q n ; m 2 N) can be expanded into Fourier series of eigenfunctions Since y (x; ) ( 2 C; 6 = m;q n ; m 2 N) satis…es the equation we get Then Lemma 3. The following formula holds where x is a …xed number and z is a non-real number.
Proof. Let f (t) = ' m;q n (t) m;q n : By (13), we conclude that Under (15) and (9), we see that It follows from Lemma 1 that the last integral is convergent. The proof is complete Now, we present below for the convenience of the reader. where a b: where a b: Then, we have where f 2 C[a; b]: By Lemma 2 and Theorem 4, one can …nd a sequence fq n k g such that where % ( ) is a monotone function: where x is a …xed number.
Proof. Let > 0: It follows from (14) that Z ' (x; ) z 2 d% q n ( ) < K: Proof. Let sin 6 = 0: From (16), we deduce that Let sin = 0: Hence we see that m;q n m;q n z : It follows from (9) that Proceeding similarly, we can get the desired result.
and z = u + iv: Proof. See [9]. Now we shall state the main result of this paper.
Theorem 8. The following relation holds where f (:) 2 L 2 q [0; 1); and z = 2 R: Proof. De…ne the function f (x) as such that f (x) satis…es (4). By (13), we conclude that where and a > 0. It follows from (13) that Using the q-integration-by-parts formula in the integral below, we have From Lemma 3, we get Application of Bessel inequality yields Likewise, we show that jI 3 j C a : Then I 1 ; I 3 ! 0, as a ! 1; uniformly in q n : By virtue of (19) and Theorem 4, we see that We can …nd a sequence ff (x)g 1 =1 which satis…es the previous conditions and tend to f (x) as ! 1; since f (:) 2 L 2 q [0; 1): It follows from (9) that the sequence of Fourier transform converges to the transform of f (x) : Using Lemmas 5 and 7, one can pass to the limit ! 1 in (22).

Remark 9.
The following formula holds. where and Now, we will study some properties of the resolvent operator. We give the following de…nition and theorems. where then A is a compact operator in the sequence space l 2 : 714 B. P. ALLAHVERD · IEV, H. TUNA Theorem 12. Let the limit circle case holds for Eq. (3) and Then the function G (x; t) de…ned by (26) is a q-Hilbert-Schmidt kernel.
Proof. It follows from (26) that and exist and are a linear combination of the products ' (x) (t) ; and these products belong to L 2 Theorem 13. Let us de…ne the operator R as Under the assumptions of Theorem 12, R is a compact operator.
Proof. Let i = i (t) (i 2 N) be a complete, orthonormal basis of L 2 q [0; 1): By Theorem 12, we can de…ne where i; k 2 N: Then, L 2 q [0; 1) is mapped isometrically l 2 : Therefore, the operator R transforms into A de…ned by (24) in l 2 by this mapping, and (27) is translated into (25). It follows from Theorem 11 that A is compact operator. Consequently, R is a compact operator. Now, we will give some auxiliary lemmas.
Lemma 14. The following equalities hold.
where and 0 are any …xed nonreal numbers.
Using (29) and setting = u + iv and 0 = , we obtain Im fm (u + i )g du = O (1) ; as ! 0: Proof. Let sin 6 = 0: It follows from (9) and (18) that where z = u + iv: Let sin = 0: If the equality (15) is q-di¤erentiated throughout with respect to x; and the limit is taken as n ! 1; then we can get the desired result.
By virtue of (30) and (32), we conclude that Then we have Z u2 Let (a; b) be a …nite interval where a < u 1 and b > u 2 . From (17), we see that Proof. Let f (:) ; g (:) 2 L 2 q [0; 1) vanish outside a …nite interval. By (23), we deduce that It follows from Theorem 16 that Furthermore, we have where (x; u) ; ' (x; u) ; g (x) and f (x) are real-valued functions. It follows from (34) and Lemma 15 that If we choose g (x) and f (x) conveniently, we can make G (u) and F (u) di¤er as little from unity in the …xed interval : From Lemma 15 and (33), we get the desired result. By virtue of (6) and (37), we conclude that G (0; 0; z) = sin fcos + m (z) sin g = Z 1 1 sin 2 z d% ( ) ; i.e., m (z) = cot + Z 1 1 d% ( ) z : Authors Contribution Statement All authors jointly worked on the results and they read and approved the …nal manuscript.

Declaration of Competing Interests
The authors declare that there is no competing interest.