On the resolvent of singular q-Sturm-Liouville operators

C om mun.Fac.Sci.U niv.A nk.Ser. A 1 M ath. Stat.

Volum e 70, N umb er 2, Pages 702–718 (2021) D O I: 10.31801/cfsuasm as.866753

ISSN 1303–5991 E-ISSN 2618–6470

Received by the editors: Jan u ary 22, 2021; Accepted: A pril 3, 2021

ON THE RESOLVENT OF SINGULAR q-STURM-LIOUVILLE

OPERATORS

Bilender P. ALLAHVERD·IEV¹ and Hüseyin TUNA²

1Department of Mathematics, Süleyman Demirel University, 32260 Isparta, TURKEY

2Department of Mathematics, Mehmet Akif Ersoy University, 15030 Burdur, TURKEY

Abstract. In this paper, we investigate the resolvent operator of the singular q- Sturm-Liouville problem de…ned as

1

qD_q 1[Dqy (x)] + [r (x) ] y (x) = 0;

with the boundary condition

y (0; ) cos + D_q 1y (0; ) sin = 0;

where 2 C, r is a real-valued function de…ned on [0; 1), continuous at zero and r 2 L¹q;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 nu- merous 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 be- long 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

qD_q 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 2 0 2 1 A n ka ra U n ive rsity C o m m u n ic a t io n s Fa c u lty o f S c ie n c e s U n ive rs ity o f A n ka ra -S e rie s A 1 M a t h e m a tic s a n d S t a t is t ic s

702

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)⁰+ qy = y; x 2 (0; 1);

where p; q are real-valued and p ¹; q 2 L¹loc[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 q- derivative between and q-deformed Heisenberg uncertainty relation (see [23]). The q-derivative at zero is de…ned as

Dqy (0) = lim

n!1[qⁿx] ¹[y (qⁿx) 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) d_qt = x (1 q) X1 n=0

qⁿf (qⁿx) (x 2 A);

provided that the series converges, and Z b

a

f (t) d_qt = Z b

0

f (t) d_qt Z a

0

f (t) d_qt;

where a; b 2 A: The q-integration for a function over [0; 1) de…ned by the formula ( [13])

Z 1 0

f (t) dqt = X1 n= 1

qⁿf (qⁿ) : Let f be a function on A and let 0 2 A: For every x 2 A; if

nlim!1f (xqⁿ) = f (0) ;

then f is called q-regular at zero. Throughout the paper, we deal only with functions q-regular at zero.

The following relation holds Z a

0

g (t) D_qf (t) d_qt + Z a

0

f (qt) D_qg (t) d_qt = f (a) g (a) f (0) g (0) ; where f and g are q-regular at zero.

Let L²_q[0; 1) be the Hilbert space consisting of all functions f satisfying ( [9])

kfk :=

sZ ₁

0 jf (x)j²dqx < +1 with the inner product

(f; g) :=

Z 1 0

f (x) g (x)dqx:

The q-Wronskian of the functions y (:) and z (:) is de…ned by the formula Wq(y; z) (x) := y (x) Dqz (x) z (x) Dqy (x) ;

where x 2 [0; 1):

3. Main Results Consider the q-Sturm-Liouville equation

L(y) := 1

qD_q ¹Dqy (x) + r (x) y (x) = y (x) ; (3) satisfying the conditions

y (0; ) cos + D_q ¹y (0; ) sin = 0; (4) y q ⁿ; cos + D_q ¹y q ⁿ; sin = 0; ; 2 R; n 2 N := f1; 2; :::g; (5) where 2 C; r is a real-valued function de…ned on [0; 1), continuous at zero and r 2 L¹q;loc[0; 1).

Let ' (x; ) and (x; ) be the solutions of the Eq. (3) satisfying the following conditions

' (0; ) = sin ; D_q ¹' (0; ) = cos ;

(0; ) = cos ; D_q ¹ (0; ) = sin : (6)

Lemma 1( [9]). Let 2 R and let=

q ⁿ(x; ) = (x; ) + l ; q ⁿ ' (x; ) 2 L²q(0; 1);

where n 2 N: Then we have

q ⁿ(x; ) ! (x; ) ; Z q ⁿ

0 q ⁿ(qt; )²d_qx ! Z ₁

0 j (x; )j²d_qx; n ! 1:

Putting

G_q ⁿ(x; t; ) = ^{q n}(x; ) ' (t; ) ; t x ' (x; ) _q n(t; ) ; t > x;

y (x; ) := R_q ⁿf (x; ) = Z q ⁿ

0

G_q ⁿ(x; t; ) f (t) dqt; ( 2 C; Im 6= 0); (7) where f 2 L²q[0; q ⁿ]: Now, we shall show that the equality (7) satis…es the equation L(y) y(x) = f (x); x 2 (0; q ⁿ) ( 2 C; Im 6= 0) and the boundary conditions (4)-(5). From (7), we get

y (x; ) = q _q n(x; ) Z x

0

' (qt; ) f (qt) d_qt

+q' (x; ) Z q ⁿ

x q ⁿ(qt; ) f (qt) dqt: (8) From (8), it follows that

D_qy (x; ) = qD_{q q} n(x; ) Z x

0

' (qt; ) f (qt) d_qt

+qDq' (x; ) Z q ⁿ

x q ⁿ(qt; ) f (qt) dqt;

and

D_q ¹D_qy (x; ) = qD_q ¹D_{q q} n(x; ) Z x

0

' (qt; ) f (qt) d_qt

+qD_q ¹Dq' (x; ) Z q ⁿ

x q ⁿ(qt; ) f (qt) dqt qW_{q q} n; ' f (x) :

Hence, by Wq '; _q n = 1 (n 2 N); we deduce that 1

qD_q ¹D_qy (x; )

= ( r (x)) q _q ⁿ(x; ) Z x

0

' (qt; ) f (qt) dqt

+ ( r (x)) q' (x; ) Z q ⁿ

x q ⁿ(qt; ) f (qt) dqt + f (x)

= ( r (x)) y (x; ) + f (x) ;

i.e., the function y (x; ) satis…es the equation L(y) y(x) = f (x); x 2 (0; q ⁿ):

Moreover,

y (0; ) = q' (0; ) Z q ⁿ

0 q ⁿ(qt; ) f (qt) dqt

= q cos Z q ⁿ

0 q ⁿ(qt; ) f (qt) d_qt;

D_q ¹y (0; ) = qD_q ¹' (0; ) Z q ⁿ

0 q ⁿ(qt; ) f (qt) dqt

= q sin Z q ⁿ

0 q ⁿ(qt; ) f (qt) d_qt;

i.e., y (x; ) satis…es (4). Similarly, we may infer that y (x; ) satis…es (5).

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 corre- sponding eigenfunctions and

m;q ⁿ:= '_m;q n =

Z q ⁿ 0

'²_m;q n(x) dqx

!¹₂

;

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]) Z q ⁿ

0 jf (x)j²d_qx = X1 m=1

1

2 m;q ⁿ

(Z q ⁿ 0

f (x) '_m;q n(x) d_qx )2

; (9)

where f (:) 2 L²q[0; q ⁿ]:

Now, let us de…ne the nondecreasing step function %_q n on [0; 1) by

%_q n( ) = 8<

: P

< _m;q n<0 1

2 m;q n

; for 0

P

0 _m;q n< 1

2

m;q n for > 0:

It follows from (9) that Z q ⁿ

0 jf (x)j²dqx = Z ₁

1

F²( ) d%_q ⁿ( ) ; (10) where

F ( ) = Z q ⁿ

0

f (x) ' (x; ) d_qx:

Lemma 2. Let > 0: Then the following relation holds V %_q n( ) = X

m;q n<

1

2 m;q ⁿ

= %_q n( ) %_q n( ) < ; (11)

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

j' (x; )j > 1

p2jsin j ; 0 x h and

1 h

Z h 0

' (x; ) dqx

!2

> 1 p2hsin

Z h 0

dqx

!2

=1

2sin² : (12) Let us de…ne f_h(x) by

fh(x) = 0; x > h

1

h; 0 x h:

It follows from (10) and (12) that Z h

0

f_h²(x) dqx = 1 h=

Z ₁

1

1 h

Z h 0

' (x; ) dqx

!2

d%_q n( )

Z 1

h Z h

0

' (x; ) d_qx

!2

d%_q n( )

> 1

2sin² %_q n( ) %_q n( ) ; which proves the inequality (11).

Let sin = 0 and

f_h(x) = 0; x > h

1

h²; 0 x h:

By (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 q- integration by parts, we …nd (the operator A generated by the expression L and the boundary conditions (4), (5) is a self-adjoint (see [10]))

(Ay; '_m;q n)

= Z q ⁿ

0

1

qD_q ¹Dqy (x; ) + r (x) y (x; ) '_m;q n(x) dqx

= (y; A'_m;q n)

= Z q ⁿ

0

y (x; ) 1

qD_q ¹Dq'_m;q n(x) + r (x) '_m;q n(x) dqx

= _m;q ⁿ Z q ⁿ

0

y (x; ) '_m;q ⁿ(x) dqx:

The set of all eigenfunctions ^{'m;q n(x)}

m;q n (m 2 N) of the self-adjoint operator A form an orthonormal basis for L²_q(0; q ⁿ) (see [10]). Then, the function y (:; ) 2 L²_q(0; q ⁿ) ( 2 C; 6= m;q ⁿ; m 2 N) can be expanded into Fourier series of eigenfunctions ^{'m;q n(x)}

m;q n (m 2 N) of the problem (3)-(5) (or of the operator A).

Then we have

y (x; ) = X1 m=1

tm( )'_m;q n(x)

m;q ⁿ

; where t_m( ) is the Fourier coe¢ cient, i.e.,

t_m( ) = Z q ⁿ

0

y (x; )'_m;q n(x)

m;q ⁿ

d_qx; m 2 N:

Since y (x; ) ( 2 C; 6= m;q ⁿ; m 2 N) satis…es the equation 1

qD_q ¹Dqy (x; ) + (r (x) ) y (x; ) = f (x) ; x 2 (0; q ⁿ);

we get

a_m : = Z q ⁿ

0

f (x)'_m;q n(x)

m;q ⁿ

d_qx

= Z q ⁿ

0

1

qD_q ¹Dqy (x; ) + (r (x) ) y (x; ) '_m;q n(x)

m;q ⁿ

dqx

= Z q ⁿ

0

1

qD_q ¹Dq'_m;q ⁿ(x) + (r (x) )'_m;q ⁿ(x) y (x; )

m;q ⁿ

dqx

= Z q ⁿ

0

m;q ⁿ'_m;q n(x) '_m;q n(x) y (x; )

m;q ⁿ

dqx

= _m;q ⁿtm( ) tm( ) ; m 2 N:

Thus, we have

tm( ) = am m;q ⁿ

; and

y (x; ) =

Z q ⁿ 0

G_q ⁿ(x; t; ) f (t) dqt

= X1 m=1

am m;q ⁿ

'_m;q n(x)

m;q ⁿ ( 2 C; 6= m;q ⁿ; m 2 N):

Then

y (x; z) = R_q ⁿf (x; z)

= X1 m=1

'_m;q n(x)

2

m;q ⁿ m;q ⁿ z Z q ⁿ

0

f (t) '_m;q n(t) dqt

= Z ₁

1

' (x; ) z

(Z q ⁿ 0

f (t) '_m;q n(t; ) dqt )

d%_q n( ) : (13) Lemma 3. The following formula holds

Z 1 1

' (x; ) z

2

d%_q n( ) < K; (14)

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 1

m;q ⁿ

Z q ⁿ 0

G_q ⁿ(x; t; z) '_m;q n(t) dqt = '_m;q n(x)

m;q ⁿ m;q ⁿ z : (15)

Under (15) and (9), we see that Z q ⁿ

0

G_q ⁿ(x; t; z) ²dqt = X1 m=1

'_m;q n(x)²

2

m;q ⁿ m;q ⁿ z²

= Z ₁

1

' (x; ) z

2

d%_q n( ) :

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.

Theorem 4 ( [19]). Let (w_n)_n2N be a uniformly bounded sequence of real non- decreasing function on a …nite interval [a; b]: Then

(i) there exists a subsequence (w_nk)_k2Nand a non-decreasing function w such that

klim!1wnk( ) = w ( ) ;

where a b:

(ii) suppose

nlim!1wn( ) = w ( ) ; where a b: Then, we have

nlim!1

Z b a

f ( ) dw_n( ) = Z b

a

f ( ) dw ( ) ; where f 2 C[a; b]:

By Lemma 2 and Theorem 4, one can …nd a sequence fq ^nkg such that

klim!1%_q nk( ) ! % ( ) ; where % ( ) is a monotone function:

Lemma 5. Let z =2 R: Then we have Z ₁

1

' (x; ) z

2

d% ( ) K; (16)

where x is a …xed number.

Proof. Let > 0: It follows from (14) that Z ' (x; )

z

2

d%_q n( ) < K:

Then Z 1

1

' (x; ) z

2

d% ( ) = lim

!1 n!1

Z ' (x; ) z

2

d%_q n( ) < K:

Lemma 6. Let > 0: Then we have Z

1

d% ( ) j zj² < 1;

Z ₁ d% ( )

j zj² < 1: (17)

Proof. Let sin 6= 0: From (16), we deduce that Z ₁

1

d% ( ) j zj² < 1:

Let sin = 0: Hence we see that 1

m;q ⁿ

Z q ⁿ 0

'_m;q n(t) Dq;x G_q ⁿ(x; t; z) dqt = D_q;x'_m;q n(x)

m;q ⁿ m;q ⁿ z : It follows from (9) that

Z q ⁿ 0

Dq;x G_q ⁿ(x; t; z) ²dqt = Z ₁

1

D_q;x' (x; ) z

2

d%_q n( ) : Proceeding similarly, we can get the desired result.

Lemma 7. Let

G (x; t; z) = (x; z) ' (t; z) ; x t ' (x; z) (t; z) ; x < t;

and let f (:) 2 L²q[0; 1): Then we have Z ₁

0 j(Rf) (x; z)j²dqx 1 v²

Z ₁

0 jf (x)j²dqx;

where

(Rf ) (x; z) = Z ₁

0

G (x; t; z) f (t) dqt;

and z = u + iv:

Proof. See [9].

Now we shall state the main result of this paper.

Theorem 8. The following relation holds (Rf ) (x; z) =

Z ₁

1

' (x; )

z F ( ) d% ( ) ; (18)

where f (:) 2 L²q[0; 1);

F ( ) = lim

!1

Z q 0

f (x) ' (x; ) dqx;

and z =2 R:

Proof. De…ne the function f (x) as

f (x) = f (x) ; x 2 [0; q ];

0; x =2 [0; q ] (q < q ⁿ) such that f (x) satis…es (4). By (13), we conclude that

R_q nf (x; z)

= Z ₁

1

' (x; )

z F ( ) d%_q n( ) = Z a

1

' (x; )

z F ( ) d%_q n( )

+ Z a

a

' (x; )

z F ( ) d%_q n( ) + Z ₁

a

' (x; )

z F ( ) d%_q n( )

= I1+ I2+ I3; (19)

where

F ( ) = Z q

0

f (x) ' (x; ) dqx;

and a > 0.

It follows from (13) that

jI¹j = Z a

1

' (x; )

z F ( ) d%_q n( ) X

k;q n< a

'_k;q n(x) Rq

0 f (t) '_k;q n(t) dqt

2

k;q ⁿ k;q n z 0

@ X

k;q n< a

'²_k;q n(x)

2

k;q ⁿ k;q ⁿ z ² 1 A

1=2

0

@ X

k;q n< a

1

2 k;q ⁿ

"Z q 0

f (x) '_k;q n(x) dqx

#21 A

1=2

: (20)

Using the q-integration-by-parts formula in the integral below, we have Z q

0

f (x) '_k;q n(x) d_qx

= 1

k;q ⁿ

Z q 0

f (x) 1

qD_q ¹D_q'_k;q n(x) + r (x) '_k;q n(x) d_qx

= 1

k;q ⁿ

Z q 0

1

qD_q ¹Dqf (x) + r (x) f (x) '_k;q n(x) dqx: (21) From Lemma 3, we get

jI¹j ^K1=2a

0

@

P

k;q n< a 1

2 k;q n

hRq 0

n 1

qD_q ¹D_qf (x) + r (x) f (x)o

'_k;q n(x) d_qxi² 1 A

1=2

:

Application of Bessel inequality yields jI¹j K¹⁼²

a

"Z q 0

1

qD_q ¹Dqf (x) + r (x) f (x)

2

dqx

#1=2

= C a: Likewise, we show that jI3j ^C_a: Then I₁; I₃! 0, as a ! 1; uniformly in q ⁿ: By virtue of (19) and Theorem 4, we see that

(Rf ) (x; z) = Z ₁

1

' (x; )

z F ( ) d% ( ) : (22)

We can …nd a sequence ff (x)g¹=1which satis…es the previous conditions and tend to f (x) as ! 1; since f (:) 2 L²q[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.

Z ₁

0

(Rf ) (x; z) g (x) dqx = Z ₁

1

F ( ) G ( )

z d% ( ) ; (23)

where

G ( ) = lim

!1

Z q 0

g (x) ' (x; ) dqx;

and

F ( ) = lim

!1

Z q 0

f (x) ' (x; ) dqx:

Now, we will study some properties of the resolvent operator. We give the fol- lowing de…nition and theorems.

De…nition 10. Let M (x; t) be a complex-valued function, where x; t 2 (a; b): If Z b

a

Z b

a jM (x; t)j²dqxdqt < +1;

then M (x; t) is called the q-Hilbert-Schmidt kernel.

Theorem 11 ( [22]). Let us de…ne the operator A as A fxⁱg = fyⁱg ; where

yi= X1 k=1

aikxk; i 2 N: (24)

If

X1 i;k=1

ja^ikj²< +1 (25)

then A is a compact operator in the sequence space l²:

Theorem 12. Let the limit circle case holds for Eq. (3) and G (x; t) = G (x; t; 0) = ' (x) (t) ; x < t

(x) ' (t) ; x t: (26)

Then the function G (x; t) de…ned by (26) is a q-Hilbert-Schmidt kernel.

Proof. It follows from (26) that Z ₁

0

d_qx Z x

0 jG (x; t)j²d_qt < +1;

and Z 1

0

dqx Z 1

x jG (x; t)j²dqt < +1;

since the integrals Z ₁

0 jG (x; t)j²dqx

and Z ₁

0 jG (x; t)j²d_qt

exist and are a linear combination of the products ' (x) (t) ; and these products belong to L²_q[0; 1) L²_q[0; 1): Then

Z ₁

0

Z ₁

0 jG (x; t)j²d_qxd_qt < +1: (27)

Theorem 13. Let us de…ne the operator R as (Rf ) (x) =

Z ₁

0

G (x; t) f (t) dqt

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 L²q[0; 1): By Theorem 12, we can de…ne

xi = (f; _i) = Z ₁

0 i(t)f (t) dqt;

yi = (g; _i) = Z ₁

0 i(t)g (t) dqt;

a_ik = Z ₁

0

Z ₁

0 k(t) _i(x)G (x; t) d_qxd_qt;

where i; k 2 N: Then, L²q[0; 1) is mapped isometrically l²: Therefore, the operator R transforms into A de…ned by (24) in l² 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.

xlim!1Wq (x; ) ; x; ⁰ = 0; (28)

Z 1 0

(x; ) ; x; ⁰ dqx = m ( ) m ⁰

0 ; (29)

where and ⁰ are any …xed nonreal numbers.

Proof. See [9].

Using (29) and setting = u + iv and ⁰= , we obtain Z ₁

0 j (x; )j²dqx = Im fm ( )g

v : (30)

Lemma 15. For …xed u₁ and u₂; we have Z u2

u1

Im fm (u + i )g du = O (1) ; as ! 0: (31) Proof. Let sin 6= 0: It follows from (9) and (18) that

Z 1

0 j (t; z)j²dqt = Z 1

1

d% ( )

(u )²+ v²; (32)

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 Im fm (u + i )g =

Z ₁

1

d% ( ) (u )²+ ²: Then we have

Z u2

u1

Im fm (u + i )g du = Z u2

u1

du Z ₁

1

d% ( ) (u )²+ ²:

Let (a; b) be a …nite interval where a < u1and b > u2. From (17), we see that Z u2

u1

du Z a

1

d% ( )

(u )²+ ² = O (1) ; Z u2

u1

du Z ₁

b

d% ( )

(u )²+ ² = O (1) : Hence, we get

Z u2

u1

du Z b

a

d% ( ) (u )²+ ² =

Z b a

d% ( ) Z ^u2

u1

dv

1 + v² = O (1) :

Assume that ( ) = ₁( ) + i ₂( ) is a complex bounded variation on the entire line. Set

' (z) = Z ₁

1

d ( )

z; ( ; ) = sgn ' (z) ' (z) 2i

= 1Z ₁

1

j j d ( )

( )²+ ²; z = + i :

Theorem 16( [20]). Let the points a; b are points of continuity of ( ) : Then we obtain

(b) (a) = lim

!0

Z b a

( ; ) d :

Theorem 17. Let the endpoints of = ( ; + ) be the points of continuity of

% ( ) : Then, we deduce that

% ( + ) % ( ) = 1 lim!0

Z

Im fm (u + i )g du: (33) Proof. Let f (:) ; g (:) 2 L²q[0; 1) vanish outside a …nite interval. By (23), we deduce that

y ( ) = Z 1

0

(Rf ) (x; z) g (x) dqx

= Z ₁

1

F ( ) G ( )

z d% ( ) = Z ₁

1

d ( ) z; where

( ) = Z

F ( ) G ( ) d% ( ) : It follows from Theorem 16 that

( ) = 1 lim!0

Z

Im f (u + i )g du: (34)

Furthermore, we have Im f (u + i )g =

Z ₁

0

g (x) dqx

f Z x

0

[ (x; u + i ) + m (u + i ) ' (x; u + i )] ' (t; u + i ) f (t) d_qt

+ Z 1

x

[ (t; u + i ) + m (u + i ) ' (t; u + i )] ' (x; u + i ) f (t) dqtg;

where (x; u) ; ' (x; u) ; g (x) and f (x) are real-valued functions. It follows from (34) and Lemma 15 that

( ) = 1 lim!0

Z

Im fm (u + i )g G (u) F (u) du: (35)

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.

Theorem 18. Let z =2 R. Then we have m (z) = cot +

Z ₁

1

d% ( )

z: (36)

Proof. It follows from (18) that G (x; t; z) =

Z ₁

1

' (x; ) ' (t; ) d% ( )

z ; (37)

since f (x) is an arbitrary function. By de…nition, we get

G (x; t; z) = [ (t; z) + m (z) ' (t; z)] ' (x; z) ; t > x [ (x; z) + m (z) ' (x; z)] ' (t; z) ; t x:

By virtue of (6) and (37), we conclude that

G (0; 0; z) = sin fcos + m (z) sin g

= Z ₁

1

sin²

zd% ( ) ; i.e.,

m (z) = cot + Z ₁

1

d% ( ) z:

Authors Contribution StatementAll authors jointly worked on the results and they read and approved the …nal manuscript.

Declaration of Competing InterestsThe authors declare that there is no com- peting interest.

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