Open Access

q-Fractional calculus for Rubin’s q-difference operator

Advances in Difference Equations20132013:276

https://doi.org/10.1186/1687-1847-2013-276

Received: 10 June 2013

Accepted: 15 August 2013

Published: 23 September 2013

Abstract

In this paper we introduce a fractional q-integral operator and derivative as a generalization of Rubin’s q-difference operator. We also reformulate the definition of the q 2 -Fourier transform and the q-analogue of the Fourier multiplier introduced by Rubin in (J. Math. Anal. Appl. 212(2):571-582, 1997; Proc. Am. Math. Soc. 135(3):777-785, 2007). As applications, we give summation formulas for ϕ 1 2 finite series, we also use the q 2 -Fourier transform and Hahn q-Laplace transform to solve a fractional q-diffusion equation.

MSC:39A12, 33D15, 42A38, 35R11.

Keywords

q 2 -Fourier transformRubin’s q-difference operatorHahn’s q-Laplace transformfractional q-integral operatorfractional q-diffusion equation

1 Introduction and preliminaries

Let q be a positive number, 0 < q < 1 . In the following, we follow the notations and notions of q-hypergeometric functions, the q-gamma function Γ q ( x ) , Jackson q-exponential functions E q ( x ) , and the q-shifted factorial as in [1, 2]. The q-difference operator is defined by
D q f ( z ) : = f ( z ) f ( q z ) z q z ( z 0 ) .
(1.1)
Jackson [3] introduced an integral denoted by
a b f ( x ) d q x
as a right inverse of the q-derivative. It is defined by
a b f ( t ) d q t : = 0 b f ( t ) d q t 0 a f ( t ) d q t ( a , b C ) ,
(1.2)
where
0 x f ( t ) d q t : = ( 1 q ) n = 0 x q n f ( x q n ) ( x C ) ,
(1.3)

provided that the series on the right-hand side of (1.3) converges at x = a  and  b .

There is no unique canonical choice for the q-integration over [ 0 , ) . In [4], Hahn defined the q-integration for a function f over [ 0 , ) by
0 f ( t ) d q t = ( 1 q ) n = q n f ( q n ) ,
while in [5] Matsuo defined q-integrations on the interval [ 0 , ) and ( , ) by
0 / b f ( t ) d q t : = 1 q b n = q n f ( q n / b ) ( b > 0 ) ,
(1.4)
/ b / b f ( t ) d q t = 1 q b q n ( f ( q n / b ) + f ( q n / b ) ) ,
(1.5)
respectively, provided that the series converges absolutely. For any q ( 0 , 1 ) and 0 < b < , we define the spaces
L p ( R b , q ) : = { f : / b / b | f ( x ) | p d q x < , p 1 } , R b , q : = { ± q n / b : n Z } , L ( R b , q ) : = { f : f : = sup { f ( ± q n / b ) n Z } < } .
We shall use the particular notation R q , R ˜ q and R ˜ q , + to denote R 1 , q , R 1 q , q and { q k 1 q , k Z } , respectively. One can verify that L 2 ( R b , q ) associated with the inner product
f , g : = / b / b f ( t ) g ( t ) ¯ d q t , f , g L 2 ( R b , q ) ,
is a Hilbert space. The Riemann-Liouville fractional q-integral operator is introduced by Al-Salam in [6] and later by Agarwal in [7] and defined by
I q α f ( x ) : = x α 1 Γ q ( α ) 0 x ( q t / x ; q ) α 1 f ( t ) d q t , α { 1 , 2 , } .
(1.6)
Using (1.3), (1.6) reduces to
I q α f ( x ) = x α ( 1 q ) α n = 0 q n ( q α ; q ) n ( q ; q ) n f ( x q n ) ,
(1.7)
which is valid for all α. The Riemann-Liouville fractional q-derivative of order α, α > 0 , is defined by
D q α = D q k I q k α ( k = α ) .
Rubin in [8, 9] introduced the q-difference operator
q f ( z ) = f ( q 1 z ) + f ( q 1 z ) f ( q z ) + f ( q z ) 2 f ( z ) 2 ( 1 q ) z ( z 0 ) .
(1.8)
It is straightforward to prove that if a function f is differentiable at a point z, then
lim q 1 q f ( z ) = f ( z ) .
Also,
δ q f ( z ) = { D q f ( z ) if  f  is odd , 1 q D q 1 f ( z ) if  f  is even .
Let f and g be functions defined on a set A, where A satisfies
z A ± q ± 1 z A ,
and let f e and f o be the even and odd parts of f, respectively. The following properties of the q operator are from [9, 10] and hold for all z A { 0 } .
  1. (i)

    q f ( z ) = 1 q D q 1 f e ( z ) + D q f o ( z ) .

     
  2. (ii)

    For two functions f and g,

     
  • if f is even and g is odd, then
    q ( f g ) ( z ) = q g ( z ) ( q f ) ( q z ) + f ( q z ) q g ( z ) ;
  • if f and g are even, then
    q ( f g ) ( z ) = q f ( z ) g ( z ) + f ( z / q ) q g ( z ) ;
  • if f and g are odd, then
    q ( f g ) ( z ) = 1 q ( f ( z ) ( q g ) ( z / q ) + ( q f ) ( z / q ) g ( z / q ) ) .
The q-translation ε y is introduced by Ismail in [2] and is defined on monomials by
ε y x n : = x n ( y / x ; q ) n ,
(1.9)
and it is extended to polynomials as a linear operator. Thus
ε y ( n = 0 m f n x n ) : = n = 0 m f n x n ( y / x ; q ) n .
(1.10)
The q-translation operator is defined for x a , a > 0 , to be
ε y x a : = x a ( y / x ; q ) a .
(1.11)
In [4], Hahn defined the following q-analogue of the Laplace transform:
L s q f ( x ) = ϕ ( s ) = 1 1 q 0 s 1 E q ( q s x ) f ( x ) d q x ( Re ( s ) > 0 ) .
(1.12)
Abdi [11] studied certain properties of these q-transforms. In [12], he used these analogues to solve linear q-difference equations with constant coefficients and certain allied equations. In [[4], equation (9.5)], Hahn defined the convolution of two functions F, G to be
( F G ) ( x ) = x 1 q 0 1 F ( t x ) G [ x t q x ] d q t ,
(1.13)
where G [ x y ] , for
G ( x ) : = n = 0 a n x n ,
is defined to be
G [ x y ] : = n = 0 a n [ x y ] n , with  [ x y ] n : = x n ( y / x ; q ) n .
Using the definition of q-integration, ( F G ) is nothing but
( F G ) ( x ) = 1 1 q 0 x F ( t ) ε q t G ( x ) d q t ,
(1.14)
where ε is the translation operator (1.10). It is remarked by Hahn [[4], p.373] that the convolution theorem
L s q ( F G ) = q L s F q L s G
(1.15)
holds. One can verify that if Φ ( s ) : = q L s F ( x ) and 0 < α < 1 , then
L s q D q α F ( x ) = s α ( 1 q ) α Φ ( s ) I q 1 α F ( 0 + ) 1 ( 1 q ) ;
(1.16)

see [13].

2 Orthogonality relations and completeness criteria

Koornwinder and Swarttouw introduced a q-analogue of the cosine and sine Fourier transform in [14] with the functions Cos ( z ; q 2 ) and Sin ( z ; q 2 ) defined by
Cos ( z ; q 2 ) = k = 0 ( 1 ) k q k ( k + 1 ) ( z ( 1 q ) ) 2 k ( q ; q ) 2 k Cos ( z ; q 2 ) = 1 ϕ 1 ( 0 ; q ; q 2 , q 2 z 2 ( 1 q ) 2 ) , Sin ( z ; q 2 ) = k = 0 ( 1 ) k q k ( k + 1 ) ( z ( 1 q ) ) 2 k + 1 ( q ; q ) 2 k + 1 Sin ( z ; q 2 ) = z 1 ϕ 1 ( 0 ; q 3 ; q 2 , q 2 z 2 ( 1 q ) 2 ) .
(2.1)
A q-analogue of the exponential function is introduced in [8, 9] and defined by
e ( z ; q 2 ) : = Cos ( i z ; q 2 ) i Sin ( i z ; q 2 ) .
Straightforward calculations give
δ q Cos ( λ x ; q 2 ) = λ Sin ( λ x ; q 2 ) , δ q Sin ( λ x ; q 2 ) = λ Cos ( λ x ; q 2 )
and
δ q e ( λ x ; q 2 ) = λ e ( λ x ; q 2 ) ,
where x C and λ is a fixed complex number. Fitouhi et al. in [15] proved that
| ( z ; q ) ( q ; q ) 1 ϕ 1 ( 0 ; z , q , q 1 + n ) | ( | z | , q ; q ) ( q ; q ) { 1 if  n 0 , | z | n q n ( n + 1 ) 2 if  n < 0 .
Hence,
| Sin ( q n 1 q ; q 2 ) | ( q 2 ; q 2 ) ( q ; q 2 ) { 1 if  n 0 , q n 2 if  n < 0
(2.2)
and
| Cos ( q n 1 q ; q 2 ) | ( q 2 ; q 2 ) ( q ; q 2 ) { 1 if  n 0 , q n 2 2 n if  n < 0 .
(2.3)
Consequently,
| e ( q n 1 q ; q 2 ) | 2 ( q 2 ; q 2 ) ( q ; q 2 ) { 1 , n 0 , q n 2 , n < 0 .
(2.4)

The following orthogonality relation is proved in [14].

Theorem 2.1 Let | z | < 1 and n, m be integers. Then
δ m n = k = z k + n ( z 2 ; q ) ( q ; q ) 1 ϕ 1 ( 0 ; z 2 ; q , q n + k + 1 ) z k + m ( z 2 ; q ) ( q ; q ) 1 ϕ 1 ( 0 ; z 2 ; q , q m + k + 1 ) ,
(2.5)

where the sum converges absolutely and uniformly on compact subsets of the open unit disc.

The following identity, which follows from (2.5) when we replace q by q 2 and z by q α , α > 0 , is essential in our investigations.
q α ( n + m ) ( q 2 ; q 2 ) 2 ( q 2 α ; q 2 ) 2 δ m n = k = q 2 k α 1 ϕ 1 ( 0 ; q 2 α ; q 2 , q 2 n + 2 k + 2 ) 1 ϕ 1 ( 0 ; q 2 α ; q 2 , q 2 m + 2 k + 2 ) .
(2.6)
Theorem 2.2 For 0 < q < 1 ,
0 / 1 q Sin ( q n z 1 q ; q 2 ) Sin ( q m z 1 q ; q 2 ) d q z = 1 q ( q 2 ; q 2 ) 2 ( q ; q 2 ) 2 q n δ n , m ,
(2.7)
0 / 1 q Cos ( q n z 1 q ; q 2 ) Cos ( q m z 1 q ; q 2 ) d q z = 1 q ( q 2 ; q 2 ) 2 ( q ; q 2 ) 2 q n δ n , m
(2.8)
and
/ 1 q / 1 q e ( i q n z 1 q ; q 2 ) e ( i q m z 1 q ; q 2 ) d q z = 4 1 q ( q 2 ; q 2 ) 2 ( q ; q 2 ) 2 q n δ n , m .
(2.9)
Proof We start with proving (2.7). Since
0 / 1 q Sin ( q n 1 q z ; q 2 ) Sin ( q m 1 q z ; q 2 ) d q z = k = q k 1 q Sin ( q n + k 1 q ; q 2 ) Sin ( q m + k 1 q ; q 2 ) = q n + m ( 1 q ) 3 / 2 q 3 k 1 ϕ 1 ( 0 ; q 3 , q 2 ; q 2 + 2 n + 2 k ) 1 ϕ 1 ( 0 ; q 3 , q 2 ; q 2 + 2 m + 2 k ) .
(2.10)
In (2.6), set α = 3 / 2 to obtain
q 3 k 1 ϕ 1 ( 0 ; q 3 ; q 2 , q 2 + 2 n + 2 k ) 1 ϕ 1 ( 0 ; q 3 , q 2 ; q 2 + 2 m + 2 k ) = q 3 ( n + m ) / 2 ( q 2 ; q 2 ) 2 ( q 3 ; q 2 ) 2 δ n , m .
(2.11)

Combining (2.10) and (2.11) yields (2.7). The proof of (2.8) follows similarly and the proof of (2.9) follows by combining (2.7) and (2.8). □

Theorem 2.3 For any q ( 0 , 1 ) ,
  1. (a)

    the set { e ( ± q n 1 q x ; q 2 ) , n Z } is a complete orthogonal set in L q 2 ( R ˜ q ) ,

     
  2. (b)

    both of the sets { Cos ( x q n 1 q ; q 2 ) , n Z } and { Sin ( x q n 1 q ; q 2 ) , n Z } are complete orthogonal sets in L q 2 ( R ˜ q , + ) .

     
Proof We only proove (a). The proof of (b) is similar and is omitted. From Theorem 2.2, it remains only to prove that the set { e ( ± q n 1 q x ; q 2 ) , n Z } is complete in L q 2 ( R ˜ q ) . This is equivalent to proving that if there exists a function f L 2 ( R ˜ q ) such that
f , e ( ± q n 1 q x ; q 2 ) = 0 ( n Z ) ,
(2.12)
then
f ( ± q n 1 q ) = 0 ( n Z ) .
From (2.12) we deduce
0 / 1 q f e ( t ) Cos ( q n 1 q t ; q 2 ) d q t = 0 ( n Z ) , 0 / 1 q f o ( t ) Sin ( q n 1 q t ; q 2 ) d q t = 0 ( n Z ) ,

where f e and f o are the even and odd parts of the function f. Then from (3.8)-(3.9) we obtain f e ( t ) = f o ( t ) = f ( t ) = 0 for all t R ˜ q . Hence { e ( ± q n 1 q x ; q 2 ) , n Z } is a complete orthogonal set in L 2 ( R ˜ q ) . □

Theorem 2.4
  1. (1)
    If f L 2 ( R ˜ q ) , then
    f ( x ) = c n e ( i x q n 1 q ; q 2 ) + d n e ( i x q n 1 q ; q 2 ) ( x R ˜ q ) ,
    (2.13)
     
where
c n = q n C / 1 q / 1 q f ( t ) e ( i t q n 1 q ; q 2 ) d q t ( n Z ) , d n = q n C / 1 q / 1 q f ( t ) e ( i t q n 1 q ; q 2 ) d q t ( n Z ) ,
and C : = 4 1 q ( q 2 ; q 2 ) 2 ( q ; q 2 ) 2 .
  1. (2)
    If f L 2 ( R ˜ q ) , then
    f ( x ) = n = a n Cos ( x q n 1 q ; q 2 ) + n = b n Sin ( x q n 1 q ; q 2 ) ( x R ˜ q ) ,
    (2.14)
     
where
a n = 4 C 0 / 1 q f e ( t ) Cos ( t q n 1 q ; q 2 ) d q t ( n Z )
and
b n = 4 C 0 / 1 q f o ( t ) Sin ( t q n 1 q ; q 2 ) d q t ( n Z ) .
Proof The proof of (1) follows directly from Theorem 2.3 and the orthogonality relations (2.9). In the following we give in detail the proof of (2). Let f = f e + f o be any function in L 2 ( R ˜ q ) . Clearly both f e and f o belong to L 2 ( R ˜ q ) . The restriction of f e to R ˜ q , + can be represented in the complete orthogonal set { Cos ( x q n 1 q ) , n Z } as
f e ( x ) = n = a n Cos ( x q n 1 q ; q 2 ) ( x R ˜ q , + ) ,
(2.15)
where
a n = 4 C 0 / 1 q f e ( t ) Cos ( t q n 1 q ; q 2 ) d q t ( n Z ) .
The orthogonal set { Sin ( x q n 1 q ) , n Z } also spans L 2 ( R ˜ q , + ) , hence
f o ( x ) = n = b n Sin ( x q n 1 q ; q 2 ) ( x R ˜ q , + ) ,
(2.16)
where
b n = 4 C 0 / 1 q f o ( t ) Sin ( t q n 1 q ; q 2 ) d q t ( n Z ) .

Because both sides of (2.15) are even functions on R ˜ q , the equality extends on R ˜ q ; and similarly the two sides of (2.16). Hence we have the representation (2.14) of any f L 2 ( R ˜ q ) . □

3 Rubin’s q 2 -Fourier transform

Koornwinder and Swarttouw [14] introduced the pair of q-transforms
g ( q n ) = ( q ; q 2 ) ( q 2 ; q 2 ) k = q k { Cos ( q k + n 1 q ; q 2 ) or Sin ( q k + n 1 q ; q 2 ) f ( q k ) , f ( q k ) = ( q ; q 2 ) ( q 2 ; q 2 ) n = q n { Cos ( q k + n 1 q ; q 2 ) or Sin ( q k + n 1 q ; q 2 ) g ( q n ) ,
(3.1)
where 0 < q < 1 and f, g are in the space L 2 ( R q ) . Now assume that log ( 1 q ) log q 2 Z or, equivalently,
q { q ( 0 , 1 ) : 1 q = q 2 m  for some integer  m } .
(3.2)
Then, by replacing q k and q n in (3.1) by q k 1 q and q n 1 q , and then f ( q k 1 q ) and g ( q n 1 q ) by f ( q k ) and g ( q k ) , Koornwinder and Swarttouw obtained the following q-analogue of the cosine and sine Fourier transforms:
g ( λ ) = 1 + q Γ q 2 ( 1 / 2 ) 0 f ( t ) { Cos ( t λ ; q 2 ) or Sin ( t λ ; q 2 ) d q t , f ( x ) = 1 + q Γ q 2 ( 1 / 2 ) 0 f ( t ) { Cos ( x λ ; q 2 ) or Sin ( x λ ; q 2 ) d q λ .
(3.3)
Therefore, if we let q 1 for such q’s that satisfy (3.2), we obtain the cosine and sine Fourier transforms
g ( λ ) = 2 π 0 f ( t ) cos ( λ t ) d t , f ( t ) = 2 π 0 g ( λ ) cos ( t λ ) d λ ,
(3.4)
g ( λ ) = 2 π 0 f ( t ) sin ( λ t ) d t , f ( t ) = 2 π 0 g ( λ ) sin ( t λ ) d λ .
(3.5)
The pair of functions Cos ( λ x ; q 2 ) and Sin ( λ x ; q 2 ) satisfy
1 q D q 1 D q y ( x ) = { λ 2 y ( x ) if  y ( x ) = Sin ( λ x ; q 2 ) , q λ 2 y ( x ) if  y ( x ) = Cos ( λ x ; q 2 ) .
Therefore, the eigenfunctions { Cos ( λ x ; q 2 ) , Sin ( λ x ; q 2 ) } have two different eigenvalues. Consequently, as remarked by Koornwinder and Swarttouw in [14], no q-exponential functions built from { Cos ( x ; q 2 ) , Sin ( x ; q 2 ) } will satisfy an eigenfunction problem. This motivated Rubin [8] to define the q-difference operator (1.8) since for this operator, the functions { Cos ( λ x ; q 2 ) , Sin ( λ x ; q 2 ) } are solutions of the eigenvalue problem
δ q 2 y ( x ) = λ 2 y ( x ) .
Rubin [8] introduced a q 2 -analogue of the Fourier transform in the form
f ˆ ( x ; q 2 ) : = F q ( f ) ( x ) = 1 + q 2 Γ q 2 ( 1 / 2 ) f ( t ) e ( i t x ; q 2 ) d q t ,
(3.6)

where f L 1 ( R q ) and q satisfies condition (3.2).

Remark 3.1 Rubin [9] proved that
  1. (1)

    the q 2 -Fourier transform defines a bounded linear operator from L 1 ( R q ) to L ( R q ) ,

     
  2. (2)

    the q 2 -Fourier transform is defined and bounded on L 1 ( R q ) L 2 ( R q ) ,

     
  3. (3)

    L 1 ( R q ) L 2 ( R q ) is dense in L 2 ( R q ) (consider the functions with finite support).

     

Consequently, the q 2 -Fourier transform defines a bounded extension to L 2 ( R q ) .

Koornwinder and Swarttouw introduced the q-Hankel transforms (3.1) which can be written in the form (3.3) only if q satisfies condition (3.2). In fact, we can write the q-transforms in (3.1) as q-integral on ( , ) by using Matsuo definition (1.4) as in the following. Rewrite the transform pair in (3.1) as
g ( q n 1 q ) = ( q ; q 2 ) ( q 2 ; q 2 ) k = q k { Cos ( q k + n 1 q ; q 2 ) or Sin ( q k + n 1 q ; q 2 ) f ( q k 1 q ) , f ( q k 1 q ) = ( q ; q 2 ) ( q 2 ; q 2 ) n = q n { Cos ( q k + n 1 q ; q 2 ) or Sin ( q k + n 1 q ; q 2 ) g ( q n 1 q ) ,
(3.7)
where we assume that the functions f and g are in the space L 1 ( R ˜ q ) L 2 ( R ˜ q ) . Using Matsuo definition of the q-integration on ( 0 , ) , (1.4) with b = 1 q , the transformations in (3.7) can be written as
g ( x ) = 1 + q Γ q 2 ( 1 / 2 ) 0 / 1 q f ( t ) Cos ( x t ; q 2 ) d q t , f ( t ) = 1 + q Γ q 2 ( 1 / 2 ) 0 / 1 q g ( x ) Cos ( x t ; q 2 ) d q x
(3.8)
and
g ( x ) = 1 + q Γ q 2 ( 1 / 2 ) 0 / 1 q f ( t ) Sin ( x t ; q 2 ) d q t , f ( t ) = 1 + q Γ q 2 ( 1 / 2 ) 0 / 1 q g ( x ) Sin ( x t ; q 2 ) d q x ,
(3.9)

where x , t R ˜ q and f, g are in L 1 ( R ˜ q ) L 2 ( R ˜ q ) . This is similar to Rubin’s work in [9]. Consequently, we set the following reformulation of Rubin’s definition of the q 2 -Fourier transform (3.6).

Definition 3.2 Let 0 < q < 1 . We define the q 2 -Fourier transform for any function f L 1 ( R ˜ q ) to be
f ˆ ( x ; q 2 ) : = F q ( f ) ( x ) = 1 + q 2 Γ q 2 ( 1 / 2 ) / 1 q / 1 q f ( t ) e ( i t x ; q 2 ) d q t .
(3.10)
It is clear that Rubin’s definition of the q 2 -Fourier transform is a special case of (3.2) because if 1 q = q 2 m for some m Z , then
0 / 1 q f ( t ) d q t = 0 / q m f ( t ) d q t = 0 f ( t ) d q t .

However, we get the classical Fourier transform only when q 1 and q satisfies (3.2). Similar to Rubin’s results mentioned in Remark 3.1, we can prove that the q 2 -Fourier transform defines a bounded linear operator from L 1 ( R ˜ q ) to L ( R ˜ q ) , and L 1 ( R ˜ q ) L 2 ( R ˜ q ) is dense in L 2 ( R ˜ q ) . Therefore, the q 2 -Fourier transform in (3.2) defines a bounded extension to L 2 ( R ˜ q ) .

The proofs of the following results, which are valid for any q ( 0 , 1 ) , are similar to the proofs in [8]. Therefore, we state them without proofs.
  1. (1)
    If f L 2 ( R ˜ q ) , then
    f ( t ) = 1 + q 2 Γ q 2 ( 1 / 2 ) / 1 q / 1 q F ( f ) ( x ) e ( i t x ( 1 q ) ; q 2 ) d q x , t R ˜ q .
    (3.11)
     
  2. (2)
    If f ( u ) and u f ( u ) L q 1 ( R ˜ q ) , then
    q ( F q f ) ( x ) = F q ( i u f ( u ) ) ( x ) .
     
  3. (3)
    If f and q f L q 1 ( R ˜ q ) , then
    F q ( q f ) ( x ) = i x F q ( f ) ( x ) .
    (3.12)
     

We reformulate the definitions of q 2 -Fourier multiplier and the q 2 -Fourier convolution formula introduced by Rubin in [9] with the restriction (3.2) to any q ( 0 , 1 ) .

Definition 3.3 Let q ( 0 , 1 ) . We define the q 2 -Fourier multiplier operator corresponding to translation by y to be
( T y f ) ( x ) = 1 + q 2 Γ q 2 ( 1 / 2 ) / 1 q / 1 q e ( i t y ; q 2 ) ( F q f ) ( t ) e ( i t x ; q 2 ) d q t ,
(3.13)
whenever the q-integral makes sense. If f L 2 ( R ˜ q ) and g L 1 ( R ˜ q ) , we define the multiplier corresponding to Fourier convolution of f with g to be
( f g ) ( z ) = / 1 q / 1 q [ T λ f ] ( z ) g ( λ ) d q λ .
(3.14)
Theorem 3.4 Let f and g be two functions in ( L 1 L 2 ) ( R ˜ q ) . Then
F q ( f g ) ( x ) = F q ( f ) ( x ) F q ( g ) ( x ) ( x R ˜ q ) .
(3.15)

Proof The proof of (3.15) is completely similar to the proof of [[9], Theorem 8] and is omitted. □

4 Fractional q-operator as a generalization of a q-difference operator

Let f be an integrable function of period 2π. Weyl, see Zygmund’s book [16], introduced a fractional operator which is more convenient for trigonometric series than the Riemann-Liouville fractional operator. This operator is defined by
( I α f ) ( x ) n = c n e i n x ( i n ) α if  f ( x ) c n e i n x , c 0 = 0 ,
(4.1)
where i α = e i α π / 2 . Zygmund [[16], p.133] pointed out that
I α f ( x ) = 1 2 π 0 2 π f ( t ) Ψ α ( x t ) d t , Ψ α ( x ) = n 0 e i n x ( i n ) α .
He also proved the semigroup identity
I α I β = I α + β , α , β > 0 .
In [17], Ismail and Rahman defined a q-analogue of the fractional operator I α , so that α = 1 represents a right inverse of the Askey-Wilson operator D q which is defined by
( D q f ) ( x ) : = f ˘ ( q 1 / 2 e i θ ) f ˘ ( q 1 / 2 e i θ ) ( q 1 / 2 q 1 / 2 ) sin θ , x = cos θ ,

where f ( x ) = f ˘ ( z ) with x = ( z + 1 / z ) / 2 .

In this section, we introduce a q-analogue of the fractional operator (4.1) as a generalization of the q-difference operator defined by Rubin in [8]. From Theorem 2.3, consequently,
f ( x ) = / 1 q / 1 q f ( t ) Ψ 0 ( x , t ) d q t ,
where
Ψ 0 ( x , t ) = q n e ( i x q n 1 q ) e ( i t q n 1 q ) + q n e ( i x q n 1 q ) e ( i t q n 1 q ) = q n Cos ( x q n 1 q ; q 2 ) Cos ( t q n 1 q ) + q n Sin ( x q n 1 q ; q 2 ) Sin ( t q n 1 q ; q 2 ) .
Lemma 4.1 The series
n = q n ( 1 α ) e ( i x q n 1 q ; q 2 ) e ( i t q n 1 q ; q 2 ) ( x , t R ˜ q )
(4.2)

is absolutely convergent only when Re α < 1 .

Proof The series in (4.2) can be written as
( n = 1 + n = 0 ) q n ( 1 α ) e ( i x q n 1 q ; q 2 ) e ( i t q n 1 q ; q 2 ) .

From (2.4), the series n = 0 q n ( 1 α ) e ( i x q n 1 q ; q 2 ) e ( i t q n 1 q ; q 2 ) is absolutely convergent for Re α < 1 and diverges for Re α 1 , while the series n = 1 q n ( 1 α ) e ( i x q n 1 q ; q 2 ) e ( i t q n 1 q ; q 2 ) is absolutely convergent for all α C . □

Set
Ψ α ( x , t ) : = ( 1 q ) α / 2 n = q n e ( i x q n 1 q ; q 2 ) e ( i t q n 1 q ; q 2 ) ( i q n ) α + ( 1 q ) α / 2 n = q n e ( i x q n 1 q ; q 2 ) e ( i t q n 1 q ; q 2 ) ( i q n ) α ,

where x , t R ˜ q and i α is defined with respect to the principal branch, i.e., i α = e i π 2 α .

Lemma 4.2 For k N and Re α < 1 ,
δ q , x k Ψ α ( x , t ) = Ψ α k ( x , t ) .
Proof The proof follows directly by using that
δ q , x k e ( i x q n 1 q ; q 2 ) = ( i q n 1 q ) k e ( i x q n 1 q ; q 2 ) .

 □

A direct calculation yields the following identity, which holds for Re α < 1 ,
( 1 q ) α / 2 Ψ α ( x , t ) = 2 cos ( π 2 α ) A α ( x , t ) + 2 sin ( π 2 α ) B α ( x , t ) ,
(4.3)
where
A α ( x , t ) : = k = q k ( 1 α ) Cos ( q n + k 1 q ; q 2 ) Cos ( q m + k 1 q ; q 2 ) + q k ( 1 α ) Sin ( q n + k 1 q ; q 2 ) Sin ( q m + k 1 q ; q 2 )
(4.4)
and
B α ( x , t ) : = k = q k ( 1 α ) Cos ( q m + k 1 q ; q 2 ) Sin ( q n + k 1 q ; q 2 ) q k ( 1 α ) Cos ( q n + k 1 q ; q 2 ) Sin ( q m + k 1 q ; q 2 ) .
(4.5)
Theorem 4.3 For Re ( α ) < 1 ,
( 1 q ) α / 2 Ψ α ( q n 1 q , q m 1 q ) = { 2 cos ( π 2 α ) q m ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r 2 sin ( π 2 α ) q m ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r if n > m + [ 1 Re α 2 ] , 2 cos ( π 2 α ) q n ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r 2 sin ( π 2 α ) q n ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r if m > n + [ 1 Re α 2 ] ,
(4.6)

where τ r = { q r 2 q 1 α 2 , r is odd , q r 2 , r is even .

Moreover,
( 1 q ) α / 2 Ψ α ( q n 1 q , q m 1 q ) = { 2 cos ( π 2 α ) q m ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 ( 1 ) r q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r + 2 sin ( π 2 α ) q m ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 ( 1 ) r q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r , n > m + [ 1 Re α 2 ] , 2 cos ( π 2 α ) q n ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 ( 1 ) r q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r + 2 sin ( π 2 α ) q n ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 ( 1 ) r q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r , m > n + [ 1 Re α 2 ] ,
(4.7)
( 1 q ) α / 2 Ψ α ( q n 1 q , q m 1 q ) = { 2 cos ( π 2 α ) q m ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 ( 1 ) r q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r 2 sin ( π 2 α ) q m ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r if n > m + [ 1 Re α 2 ] , 2 cos ( π 2 α ) q n ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 ( 1 ) r q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r 2 sin ( π 2 α ) q n ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r if m > n + [ 1 Re α 2 ] ,
(4.8)
( 1 q ) α / 2 Ψ α ( q n 1 q , q m 1 q ) = { 2 cos ( π 2 α ) q m ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r + 2 sin ( π 2 α ) q m ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r if n > m + [ 1 Re α 2 ] , 2 cos ( π 2 α ) q n ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r + 2 sin ( π 2 α ) q n ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r if m > n + [ 1 Re α 2 ] .
(4.9)
Proof Using the following formula from [[14], p.455]
k = s k y n + k ( y 2 ; q 2 ) ( q 2 ; q 2 ) 1 ϕ 1 ( 0 ; y 2 ; q 2 , q 2 n + 2 k + 2 ) x m + k ( x 2 ; q 2 ) ( q 2 ; q 2 ) 1 ϕ 1 ( 0 ; x 2 ; q 2 , q 2 m + 2 k + 2 ) = s m y n m ( s 1 x y 1 , y 2 ; q 2 ) ( s x y , q 2 ; q 2 ) 2 ϕ 1 ( q 2 s x 1 y , s x y ; y 2 ; q 2 , q 2 n 2 m s 1 x y 1 ) = s n x m n ( s 1 y x 1 , x 2 ; q 2 ) ( s x y , q 2 ; q 2 ) 2 ϕ 1 ( q 2 s x y 1 , s x y ; x 2 ; q 2 , q 2 m 2 n s 1 y x 1 ) ,
where | s x y | < 1 , we can prove that
k = q k ( 1 α ) Cos ( q m + k 1 q ; q 2 ) Cos ( q n + k 1 q ; q 2 ) = { q m ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) 2 ϕ 1 ( q 2 α , q 1 α ; q ; q 2 , q 2 n 2 m + α ) , n m + [ Re α / 2 ] , q n ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) 2 ϕ 1 ( q 2 α , q 1 α ; q ; q 2 , q 2 m 2 n + α ) , m n + [ Re α / 2 ] ,
where Re ( 1 α ) > 0 and
k = q k ( 1 α ) Sin ( q m + k 1 q ; q 2 ) Sin ( q n + k 1 q ; q 2 ) = { q n m q m ( 1 α ) 1 q ( q α , q 2 ; q 2 ) ( q 3 α , q ; q 2 ) 2 ϕ 1 ( q 2 α , q 3 α ; q 3 ; q 2 , q 2 n 2 m + α ) , n > m + [ Re α / 2 ] , q m n q n ( 1 α ) 1 q ( q α , q 2 ; q 2 ) ( q 3 α , q ; q 2 ) 2 ϕ 1 ( q 2 α , q 3 α ; q 3 ; q 2 , q 2 m 2 n + α ) , m > n + [ Re α / 2 ] .
Also,
k = q k ( 1 α ) Sin ( q m + k 1 q ; q 2 ) Cos ( q n + k 1 q ; q 2 ) = { q m ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) 2 ϕ 1 ( q 1 α , q 2 α ; q ; q 2 , q 2 n 2 m + α + 1 ) , q m n q n ( 1 α ) 1 q ( q α 1 , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) 2 ϕ 1 ( q 2 α , q 3 α ; q 3 ; q 2 , q 2 m 2 n + α 1 ) .
Hence, if x : = q n 1 q and t : = q m 1 q , then
A α ( x , t ) = { t ( 1 α ) ( 1 q ) ( 1 α ) / 2 ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r ( x t ) r , n > m + [ Re α / 2 ] , x ( 1 α ) ( 1 q ) ( 1 α ) / 2 ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r ( t x ) r , m > n + [ Re α / 2 ] .
(4.10)
Also,
A α ( x , t ) = { t ( 1 α ) ( 1 q ) ( 1 α ) / 2 ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 ( 1 ) r q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r ( x t ) r , n > m + [ Re α / 2 ] , x ( 1 α ) ( 1 q ) ( 1 α ) / 2 ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) r = 0 ( 1 ) r q α [ r 2 ] ( q 1 α ; q ) r ( q ; q ) r ( t x ) r , m > n + [ Re α / 2 ] ,
(4.11)
B α ( x , t ) = { q m ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 ( 1 ) r q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r , n m + [ 1 Re α 2 ] , q n ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 ( 1 ) r q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r , m > n + [ 1 Re α 2 ] ,
(4.12)
B α ( x , t ) = { q m ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 ( 1 ) r q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( n m ) r , n m + [ 1 Re α 2 ] , q n ( 1 α ) ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) r = 0 ( 1 ) r q α r 2 τ r ( q 1 α ; q ) r ( q ; q ) r q ( m n ) r , m > n + [ 1 Re α 2 ] .
(4.13)

Substituting from (4.10)-(4.13) into (4.3) yields the values Ψ α ( ± x , ± t ) and the theorem follows. □

Remark 4.4 In the previous theorem, we calculated the value of Ψ α ( x , t ) , x , t R ˜ q and for specific values of x, t. We can calculate the values of Ψ α ( x , t ) for all x, t by using the identity
s k + m y k + n ( y 2 ; q 2 ) ( q 2 ; q 2 ) 1 ϕ 1 ( 0 ; y 2 ; q 2 , q 2 k + 2 n + 2 ) x k + m ( x 2 ; q 2 ) ( q 2 ; q 2 ) 1 ϕ 1 ( 0 ; x 2 ; q 2 , q 2 k + 2 m + 2 ) = y n m ( s 1 x y 1 , q 2 n 2 m + 2 ; q 2 ) ( q 2 n 2 m s 1 x y 1 , q 2 ; q 2 ) 2 ϕ 1 ( q 2 n 2 m s 1 x y 1 , s 1 y x 1 ; q 2 n 2 m + 2 ; q 2 , s x y )

for | s x y | < 1 . See Proposition 4.1 of [14].

In this case we have
CC α ( q n 1 q , q m 1 q ) : = r = q r ( 1 α ) Cos ( q n + r 1 q ; q 2 ) Cos ( q m + r 1 q ; q 2 ) = q m ( 1 α ) ( q α , q 2 n 2 m + 2 , q 2 ; q 2 ) ( q 2 n 2 m + α , q , q ; q 2 ) 2 ϕ 1 ( q 2 n 2 m + α , q α ; q 2 n 2 m + 2 ; q 2 , q 1 α ) ,
(4.14)
SS α ( q n 1 q , q m 1 q ) : = r = q r ( 1 α ) Sin ( q n + r 1 q ; q 2 ) Sin ( q m + r 1 q ; q 2 ) = q n m q m ( 1 α ) ( q α , q 2 n 2 m + 2 , q 2 ; q 2 ) ( q 2 n 2 m + α , q , q ; q 2 ) 2 ϕ 1 ( q 2 n 2 m + α , q α ; q 2 n 2 m + 2 ; q 2 , q 3 α ) , SC α ( q n 1 q , q m 1 q ) : = r = q r ( 1 α ) Sin ( q n + r 1 q ; q 2 ) Cos ( q m + r 1 q ; q 2 ) = q n m q m ( 1 α ) ( q α 1 , q 2 n 2 m + 2 , q 2 ; q 2 ) ( q 2 n 2 m + α 1 , q , q ; q 2 ) 2 ϕ 1 ( q 2 n 2 m + α 1 , q α + 1 ; q 2 n 2 m + 2 ; q 2 , q 2 α ) , CS α ( q n 1 q , q m 1 q ) : = r = q r ( 1 α ) Sin ( q m + r 1 q ; q 2 ) Cos ( q n + r 1 q ; q 2 ) = q m ( 1 α ) ( q α + 1 , q 2 n 2 m + 2 , q 2 ; q 2 ) ( q 2 n 2 m + α + 1 , q , q ; q 2 ) 2 ϕ 1 ( q 2 n 2 m + α + 1 , q α 1 ; q 2 n 2 m + 2 ; q 2 , q 2 α ) .
(4.15)

Corollary 4.5 For each fixed x , t R ˜ q , the function Ψ α ( x , t ) as a function of α can be extended to an entire function on .

Proof If α is a positive integer, then the series on the right-hand sides of (4.6)-(4.9) are finite sums and hence are convergent. Since the zeros of the function cos ( π 2 α ) are the poles of the function ( q 1 α ; q 2 ) with the same orders. In fact
lim α ( 2 j + 1 ) cos π 2 α ( q 1 α ; q 2 ) = π 2 ln q q j 2 + j ( q 2 ; q 2 ) j ( q 2 ; q 2 ) ( j N 0 ) .
Similarly, the zeros of the function sin ( π 2 α ) are the poles of the function ( q 2 α ; q 2 ) with the same orders and
lim α ( 2 j ) sin π 2 α ( q 2 α ; q 2 ) = π 2 ln q q j 2 j ( q 2 ; q 2 ) j 1 ( q 2 ; q 2 ) ( j N ) .

Then the left-hand sides of equations (4.6)-(4.9) are entire functions. Hence, as a function of α, the functions Ψ α ( x , t ) ( x , t R ˜ q ) can be analytically extended by defining its values when α 1 by the left-hand sides of (4.6)-(4.9). □

It also should be noted that for Re ( α ) 1 , the left-hand sides of (4.6) and (4.7) determine Ψ α ( x , t ) for all x , t R ˜ q , which is different from the case of Re ( α ) < 1 ; see Remark 4.4.

Definition 4.6 For Re α > 0 , we define a fractional q-integral operator J q α on L 2 ( R ˜ q ) L 1 ( R ˜ q ) by
J q α f ( x ) : = 1 C / 1 q / 1 q f ( t ) Ψ α ( x , t ) d q t .

The following properties follow at once from (4.3) and their analytic continuation on .

  • If f L 2 ( R ˜ q ) L ( R ˜ q ) and f is even, then
    J q α f ( x ) = 2 0 / 1 q f ( t ) φ α ( x , t ) d q t ,
where
( 1 q ) α / 2 φ α ( q n 1 q , q m 1 q ) = { 2 q m ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q 1 α ; q ) 2 j ( q ; q ) 2 j q j ( 2 n 2 m + α ) 2 q n m ( m + 1 ) ( 1 α ) sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 n 2 m + α 1 ) , 2 q n ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q 1 α ; q ) 2 j ( q ; q ) 2 j q j ( 2 m 2 n + α ) + 2 q n ( 1 α ) sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j ( q ; q ) 2 j q j ( 2 m 2 n + α + 1 )
and
( 1 q ) α / 2 φ α ( q n 1 q , q m 1 q ) = { 2 q m ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q α + 1 ; q ) 2 j ( q ; q ) 2 j q i ( 2 n 2 m + α ) + 2 q n m m ( 1 α ) sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 n 2 m + α 1 ) , 2 q n ( 1 α ) ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q 1 α ; q ) 2 j ( q ; q ) 2 j q j ( 2 m 2 n + α ) 2 q n ( 1 α ) sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j ( q ; q ) 2 j q j ( 2 m 2 n + α + 1 ) .

Hence, if f is an even function, then δ q 2 k f ( x ) is an even function and δ q 2 k + 1 f ( x ) is an odd function for all k N 0 .

  • If f L 2 ( R ˜ q ) L ( R ˜ q ) and f is odd, then
    J q α f ( x ) = 2 0 / 1 q f ( t ) ψ α ( x , t ) d q t ,
where
( 1 q ) α / 2 ψ α ( q n 1 q , q m 1 q ) = { 2 q m ( 1 α ) q n m ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 n 2 m + α ) 2 q m ( 1 α ) sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j ( q ; q ) 2 j q j ( 2 n 2 m + α + 1 ) , 2 q n ( 1 α ) q m n ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 m 2 n + α ) + 2 q ( n + 1 ) ( 1 α ) q m n sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 m 2 n + α 1 )
and
( 1 q ) α / 2 ψ α ( q n 1 q , q m 1 q ) = { 2 q m ( 1 α ) q n m ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 n 2 m + α ) 2 q m ( 1 α ) sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j ( q ; q ) 2 j q j ( 2 n 2 m + α + 1 ) , 2 q n ( 1 α ) q m n ( q α , q 2 ; q 2 ) ( q 1 α , q ; q 2 ) cos π 2 α j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 m 2 n + α ) 2 q ( n + 1 ) ( 1 + α ) q m n sin π 2 α ( q 1 + α , q 2 ; q 2 ) ( q 2 α , q ; q 2 ) j = 0 ( q 1 α ; q ) 2 j + 1 ( q ; q ) 2 j + 1 q j ( 2 m 2 n + α 1 ) .

Hence, if f is an odd function, then δ q 2 k f ( x ) is an odd function and δ q 2 k + 1 f ( x ) is an even function for all k N 0 .

Theorem 4.7 Let f L 2 ( R ˜ q ) . Then
δ q ( J q f ) ( x ) = f ( x ) for all x R ˜ q .
(4.16)
Moreover, if f is q-regular at zero, then
J q ( δ q f ) ( x ) = f ( x ) for all x R ˜ q .
(4.17)
Proof If f L 2 ( R ˜ q ) , then
J q f ( x ) = 2 C 0 / 1 q f e ( t ) φ 1 ( x , t ) d q t + 2 C 0 / 1 q f e ( t ) ψ 1 ( x , t ) d q t ,
where
φ 1 ( q n 1 q , q m 1 q ) = { π 1 q ln q , n > m , π 1 q ln q + C 2 , m n
and
ψ 1 ( q n 1 q , q m 1 q ) = { C 2 , n > m 1 , 0 , m > n .
Hence,
J q f ( x ) = ( 1 + 2 π 1 q C ln q ) 0 x f e ( t ) d q t + 2 π 1 q C ln q x f e ( t ) d q t q x f 0 ( t ) d q t .
(4.18)
One can verify that the first and second q-integrals of (4.18) are odd functions, while the last q-integral is an even function. Consequently,
δ q J q f ( x ) = ( 1 + 2 π 1 q C ln q ) D q , x 0 x f e ( t ) d q t + 2 π 1 q C ln q D q , x x f e ( t ) d q t D q , x x f 0 ( t ) d q t .
(4.19)
Using the fundamental theorem of q-calculus, see [13], we obtain (4.16). To prove (4.17), we assume that f is q-regular at zero,
J q δ q f ( x ) = J q ( δ q f e ) ( x ) + J q ( δ q f 0 ) ( x ) = 2 C 0 / 1 q δ q f e ( t ) ψ 1 ( x , t ) d q t + 2 C 0 / 1 q δ q f 0 ( t ) φ 1 ( x , t ) d q t .
But
0 / 1 q δ q f e ( t ) ψ 1 ( x , t ) d q t = 0 / 1 q D q , t f e ( t / q ) ψ 1 ( x , t ) d q t = C 2 q x / 1 q D q , t f e ( t / q