Next Article in Journal
Anomalous Diffusion with an Apparently Negative Diffusion Coefficient in a One-Dimensional Quantum Molecular Chain Model
Previous Article in Journal
Application of Wind Tunnel Device for Evaluation of Biokinetic Parameters of Running
 
 
Font Type:
Arial Georgia Verdana
Font Size:
Aa Aa Aa
Line Spacing:
Column Width:
Background:
Article

Fractional (p,q)-Calculus on Finite Intervals and Some Integral Inequalities

1
Department of Mathematics, Khon Kaen University, Khon Kaen 40002, Thailand
2
Department of Mathematics, Faculty of Applied Science, King Mongkut’s University of Technology North Bangkok, Bangkok 10800, Thailand
3
Department of Mathematics, University of Ioannina, 451 10 Ioannina, Greece
4
Nonlinear Analysis and Applied Mathematics (NAAM)-Research Group, Department of Mathematics, Faculty of Science, King Abdulaziz University, P.O. Box 80203, Jeddah 21589, Saudi Arabia
*
Author to whom correspondence should be addressed.
Symmetry 2021, 13(3), 504; https://doi.org/10.3390/sym13030504
Submission received: 19 February 2021 / Revised: 16 March 2021 / Accepted: 17 March 2021 / Published: 19 March 2021
(This article belongs to the Section Mathematics)

Abstract

:
Fractional q-calculus has been investigated and applied in a variety of fields in mathematical areas including fractional q-integral inequalities. In this paper, we study fractional ( p , q ) -calculus on finite intervals and give some basic properties. In particular, some fractional ( p , q ) -integral inequalities on finite intervals are proven.

1. Introduction

In mathematics, quantum calculus or q-calculus is the study of calculus without limits. In the early Eighteenth Century, the well-known mathematician Leonhard Euler (1707–1783) established q-calculus in the way of Newton’s work for infinite series. Yet, q-calculus was known to be initiated by F. H. Jackson in 1910, who introduced the q-derivative and q-integral in [1] (see also [2]).
As a connection between the fields of mathematics and physics, q-calculus has played a significant role in physics phenomena; for instance, Fock [3] studied the symmetry of hydrogen atoms using the q-difference equation. Furthermore, in modern mathematical analysis, q-calculus has many applications such as combinatorics, orthogonal polynomials, basic hypergeometric functions, number theory, quantum theory, mechanics, and the theory of relativity; see also [4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24] and the references cited therein. The book by V. Kac and P. Cheung [25] covers the basic theoretical concept of q-calculus.
In 2013, being one of the most attractive areas, some new researchers are interested in q-calculus; in particular, J. Tariboon and S.K. Ntouyas [26] defined the q k -calculus and proved some of its significant properties. Next, J. Tariboon and S.K. Ntouyas [27] extended some of the important integral inequalities to q-calculus. Moreover, in 2016, J. Necmettin, Z.S. Mehmet, and İ. İmdat [28] proved the correctness of the left part of the q-Hermite–Hadamard inequality and the generalized q-Hermite–Hadamard inequality. With these results, many researchers have extended some important topics of q-calculus together with applications in many fields, such as q-integral inequalities; see [29,30,31,32,33,34,35,36,37] for more details.
Fractional calculus is the field of mathematical analysis that deals with the investigation and applications of integrals and derivatives of arbitrary order. In 2015, J. Tariboon, S.K. Ntouyas, and P. Agarwal [38] proposed a new q-shifting operator a Φ q ( m ) = q m + ( 1 q ) a for studying new concepts of fractional q-calculus. Furthermore, in 2016, since inequalities play a vital role in modern analysis, as well as mathematical analysis depends on many inequalities, W. Sudsutad, S.K Ntouyas, and J. Tariboon [39] studied some fractional q-integral inequalities such as the fractional q-Hermite–Hadamard integral inequality, the q-Hölder integral inequality, the q-Korkine equality, the q-Grüss integral inequality, the q-Grüss–Chebyshev integral inequality, and the q-Polya–Szeqö integral inequality on finite intervals.
Inspired and motivated by some of the above applications, in 2016, M. Tunç and E. Göv [40,41] defined the ( p , q ) -derivative and the ( p , q ) -integral on finite intervals and proved some of its properties. Later on, they also extended some of the new important integral inequalities on finite intervals to ( p , q ) -calculus. In 2017, M. Kun, İ. İmdat, N. Alp, and M.Z. Sarikaya [42] proved the correctness of the left part of the ( p , q ) -Hermite–Hadamard inequality and the generalized ( p , q ) -Hermite–Hadamard inequality. In addition, in 2020, J. Soontharanon and T. Sitthiwirattham [43] introduced the new concept of ( p , q ) -difference operators on [ 0 , T ] , where T > 0 , and studied some fractional ( p , q ) -calculus properties in the sense of ( p , q ) -difference operators; especially, they proposed the fractional ( p , q ) -difference operator of the Riemann–Liouville and Caputo types.
Shortly afterward, many authors generalized and developed the q-calculus theory into a two-parameter ( p , q ) -integer, which is used efficiently in many fields, and some results on the study of ( p , q ) -calculus can be found in [44,45,46,47,48,49,50,51,52,53,54,55,56,57,58,59,60,61,62,63,64,65,66,67,68].
However, fractional ( p , q ) -calculus on finite intervals via some integral inequalities has not been studied yet. This gap is the motivation and inspiration for this research. The main purpose of this paper is to study the fractional ( p , q ) -calculus on finite intervals and to give some of its important properties. Then, we prove many fractional ( p , q ) -integral inequalities on finite intervals, for instance the fractional ( p , q ) -Hölder integral inequality, the ( p , q ) -Hermite–Hadamard integral inequality, the ( p , q ) -Korkine equality, the ( p , q ) -Grüss integral inequality, the ( p , q ) -Grüss–Chebyshev integral inequality, and the ( p , q ) -Polya–Szeqö integral inequality.

2. Preliminaries

In this section, we would like to recall some well-known facts on fractional ( p , q ) -calculus, which can be found in [10,11,38,53,55]. Throughout this paper, let [ a , b ] R be an interval with a < b and 0 < q < p 1 be constants,
[ k ] p , q = p k q k p q , k N , [ k ] p , q ! = [ k ] p , q [ k 1 ] p , q [ 1 ] p , q = i = 1 k p i q i p q , k N , 1 , k = 0 .
A q-shifting operator is defined as:
a Φ q ( m ) = q m + ( 1 q ) a ,
where m R . For any positive integer k, we have:
a Φ q k ( m ) = a Φ q k 1 ( a Φ q ( m ) ) and a Φ q 0 ( m ) = m .
By computing directly, we get the following results.
Property 1.
For any m , n R and for all positive integers j , k , we have:
(i) 
a Φ q k ( m ) = a Φ q k ( m ) ;
(ii) 
a Φ q j ( a Φ q k ( m ) ) = a Φ q k ( a Φ q j ( m ) ) = a Φ q j + k ( m ) ;
(iii) 
a Φ q ( a ) = a ;
(iv) 
a Φ q k ( m ) a = q k ( m a ) ;
(v) 
m a Φ q k ( m ) = ( 1 q k ) ( m a ) ;
(vi) 
a Φ q k ( m ) = m a / m Φ q k ( 1 ) , for m 0 ;
(vii) 
a Φ q ( m ) a Φ q k ( n ) = q m a Φ q k 1 ( n ) .
For m , n R and k N { 0 } , the q-Pochhammer symbol or the q-shifted factorial is a q-analogue of the Pochhammer symbol, which is defined by:
( m ; q ) 0 = 1 , ( m ; q ) k = i = 0 k 1 ( 1 q i m ) ,
and the new power of q-shifting operator is defined by:
( n m ) a ( 0 ) = 1 , ( n m ) a ( k ) = i = 0 k 1 ( n a Φ q i ( m ) ) .
More generally, if γ R , then:
( n m ) ( γ ) = n γ i = 0 1 m n q i 1 m n q γ + i ,
and:
( n m ) a ( γ ) = i = 0 n a Φ q i ( m ) n a Φ q γ + i ( m ) , n 0 .
The ( p , q ) -derivative of function f is defined on [ a , b ] at t [ a , b ] as follows.
Definition 1 
([40]). Let f : [ a , b ] R be a continuous function. Then:
a D p , q f ( t ) = f p t + ( 1 p ) a f q t + ( 1 q ) a ( p q ) ( t a ) , t a ;
a D p , q f ( a ) = lim t a a D p , q f ( t )
is called the ( p , q ) -derivative of a function f at a.
Obviously, f is ( p , q ) -differentiable on [ a , b ] if a D p , q f ( t ) exists for all t [ a , b ] . In Definition 1, if p = 1 , then a D p , q f = a D q f , which is the q-derivative of function f on [ a , b ] , and if a = 0 , then (8) reduces to the q-derivative of the function f on [ 0 , b ] ; see [25,41] for more details.
Definition 2 
([40]). Let f : [ a , b ] R be a continuous function. Then:
a t f ( s ) a d p , q s = ( p q ) ( t a ) n = 0 q n p n + 1 f q n p n + 1 t + 1 q n p n + 1 a
is called the ( p , q ) -integral of f for t [ a , b ] .
Furthermore, if c ( a , t ) , then the ( p , q ) -integral is defined by:
c t f ( s ) a d p , q s = a t f ( s ) a d p , q s a c f ( s ) a d p , q s .
Note that if a = 0 and p = 1 , then (10) reduces to the q-integral of the function, which can be found in [26].
Theorem 1 
([40]). The following formulas hold for t [ a , b ] :
(i) 
a D p , q a t f ( s ) a d p , q s = f ( t ) f ( a ) ;
(ii) 
a b a D p , q f ( s ) a d p , q s = f ( t ) ;
(iii) 
c t a D p , q f ( s ) a d p , q s = f ( t ) f ( c ) , for c ( a , t ) .
Theorem 2 
([40]). If f , g : [ a , b ] R are continuous functions, t [ a , b ] and λ R , then the following formulas hold:
(i) 
a t f ( s ) + g ( s ) a d p , q s = a t f ( s ) a d p , q s + a t g ( s ) a d p , q s ;
(ii) 
a t λ f ( s ) a d p , q s = λ a t f ( s ) a d p , q s ;
(iii) 
a t f ( p s + ( 1 p ) s ) a D p , q g ( s ) a d p , q s = f g ( s ) | a t a t g ( q s + ( 1 q ) a ) a D p , q ( f ( s ) ) a d p , q s .
Let us define the new ( p , q ) -analogue of the power function a ( m n ) p , q k with k N { 0 } and m , n R as the following:
a ( m n ) p , q ( 0 ) : = 1 , a ( m n ) p , q ( k ) : = i = 0 k 1 a Φ p i ( m ) a Φ q i ( n ) .
It is easy to see that:
a ( m n ) p , q ( k ) = ( m a ) k i = 0 k 1 p i 1 n a m a q p i .
More generally, if α R , then:
a ( m n ) p , q ( α ) = ( m a ) α i = 0 p i p α + i 1 n a m a ( q p ) i 1 n a m a ( q p ) ( α + i ) .
Property 2.
For α > 0 , the following formulas hold:
(i) 
a Φ q / p k ( m ) a = q p k ( m a ) ;
(ii) 
a m a Φ q / p k ( m ) p , q ( α ) = ( m a ) α i = 0 p i p α + i 1 q p k ( q p ) i 1 q p k ( q p ) ( α + i ) = ( m a ) α 1 q p k p , q ( α ) .
Proof. 
(i) For α > 0 , we have:
a Φ q / p k ( m ) a = q p k m + 1 q p k a a = q p k ( m a ) .
To prove (ii), we use (i) and let n = a Φ q / p k ( m ) in (14); we have:
a ( m a Φ q / p k ( m ) ) p , q ( α ) = ( m a ) α i = 0 p i p α + i 1 q p k ( q p ) i 1 q p k ( q p ) ( α + i ) = ( m a ) α 1 q p k p , q ( α ) .
Lemma 1.
If f : [ a , b ] R is continuous at a, then:
a t a s f ( r ) a d p , q r a d p , q s = a t / p + 1 1 / p a p q r + ( 1 p q ) a t f ( r ) a d p , q s a d p , q r .
Proof. 
By Definition 2, we have:
a t a s f ( r ) a d p , q r a d p , q s = a t ( p q ) ( s a ) n = 0 q n p n + 1 f q n p n + 1 s + 1 q n p n + 1 a a d p , q s = ( p q ) n = 0 a t q n p n + 1 s q n p n + 1 a f q n p n + 1 s + 1 q n p n + 1 a a d p , q s = ( p q ) n = 0 a t q n p n + 1 s + 1 q n p n + 1 a f q n p n + 1 s + 1 q n p n + 1 a a f q n p n + 1 s + 1 q n p n + 1 a a d p , q s .
By letting u = q n p n + 1 s + 1 q n p n + 1 a and using Definition 2, we get:
a t a s f ( r ) a d p , q r a d p , q s = ( p q ) p n + 1 q n n = 0 a q n p n + 1 t + 1 q n p n + 1 a ( u a ) f ( u ) a d p , q u = ( p q ) 2 ( t a ) n = 0 m = 0 q m p m + 1 q m + n p m + n + 2 t + 1 q m + n p m + n + 2 a a × f q m + n p m + n + 2 t + 1 q m + n p m + n + 2 a = ( p q ) 2 ( t a ) 2 m = 0 n = 0 q 2 m + n p 2 m + n + 3 f q m + n p m + n + 2 t + 1 q m + n p m + n + 2 a = ( p q ) 2 ( t a ) 2 m = 0 n = m q m + n p m + n + 3 f q n p n + 2 t + 1 q n p n + 2 a = ( p q ) 2 ( t a ) 2 n = 0 m = 0 n q m + n p m + n + 3 f q n p n + 2 t + 1 q n p n + 2 a = ( p q ) ( t a ) 2 n = 0 p n + 1 q n + 1 p n + 1 q n p n + 2 f q n p n + 2 t + 1 q n p n + 2 a = ( p q ) ( t a ) n = 0 1 q p n + 1 q n p n + 2 ( t a ) f q n p n + 2 t + 1 q n p n + 2 a = ( p q ) t p + 1 1 p a a n = 0 1 q p n + 1 q n p n + 1 ( t a ) × f q n p n + 1 t p + 1 1 p a + 1 q n p n + 1 a = a t / p + 1 1 / p a t p q r ( 1 p q ) a f ( r ) a d p , q r = a t / p + 1 1 / p a p q r + ( 1 p q ) a t f ( r ) a d p , q s a d p , q r .
The proof is complete. □
For t R { 0 , 1 , 2 , } , the ( p , q ) -gamma function is defined by:
Γ p , q ( t ) = p q p , q ( t 1 ) ( p q ) t 1 ,
and an equivalent definition of (16) was given in [56] as:
Γ p , q ( t ) = p t ( t 1 ) 2 0 x t 1 E p , q q x d p , q x ,
where:
E p , q q x = n = 0 q n 2 [ n ] p , q q x n .
Obviously, Γ p , q ( t + 1 ) = [ t ] p , q Γ p , q ( t ) . For s , t > 0 , the definition of the ( p , q ) -beta function is defined by:
B p , q ( s , t ) = 0 1 u s 1 0 1 0 Φ q ( u ) p , q ( t 1 ) 0 d p , q u ,
and (18) can also be written as:
B p , q ( s , t ) = p t 1 2 s + t 2 / 2 Γ p , q ( s ) Γ p , q ( t ) Γ p , q ( s + t ) ;
see [43,69] for more details.

3. Main Results

From Lemma 1, we shall give that which leads to a definition of the fractional ( p , q ) -integral of the Riemann–Liouville type with the consideration of the n-time as follows:
a I p , q n f ( t ) = a t a r n 1 a r 2 f ( r 1 ) a d p , q r 1 a d p , q r 2 a d p , q r n 2 a d p , q r n 1 .
The function f in (20) will be assumed to be continuous on [ a , b ] . From (15), we have:
a I p , q 2 f ( t ) = a t a s f ( r ) a d p , q r a d p , q s = a t / p + 1 1 / p a p q r + ( 1 p q ) a t f ( r ) a d p , q s a d p , q r = a t / p + 1 1 / p a t p q r ( 1 p q ) a f ( r ) a d p , q r = a t / p + 1 1 / p a t q ( p r + ( 1 p ) a ) 1 1 q a f ( r ) a d p , q r = 1 p a t t q r ( 1 q ) a f r p + 1 1 p a a d p , q r = 1 p 2 2 [ 1 ] p , q a t a t a Φ q ( r ) p , q ( 2 1 ) f r p + 1 1 p a a d p , q r .
Integrating n-times reduces (20) to a single ( p , q ) -integral on [ a , b ] as follows.
a I p , q n f ( t ) = 1 p n 2 [ n 1 ] p , q a t a t a Φ q ( r ) p , q ( n 1 ) f r p n 1 + 1 1 p n 1 a a d p , q r .
Definition 3.
Let f be defined on [ a , b ] , and let α > 0 . The Riemann–Liouville fractional ( p , q ) -integral is defined by:
a I p , q α f ( t ) = 1 p α 2 Γ p , q ( α ) a t a t a Φ q ( s ) p , q ( α 1 ) f s p α 1 + 1 1 p α 1 a a d p , q s = ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q ( α 1 ) f q n p α + n t + 1 q n p α + n a .
Definition 4.
The fractional ( p , q ) -derivative of the Riemann–Liouville type of order α > 0 of a continuous function f on [ a , b ] is defined by ( a D p , q 0 f ) ( t ) = f ( t ) and:
( a D p , q α f ) ( t ) = a D p , q v a I p , q v α ( t ) ,
where α > 0 and v is the smallest integer greater than or equal to α.
The basic q-hypergeometric function is defined in [70] as:
r F s a 1 , , a r ; b 1 , , b s ; x = n = 0 ( a 1 ; q ) n ( a r ; q ) n ( q ; q ) n ( b 1 ; q ) n ( b s ; q ) n x n ,
and q-Vandermonde reversing the order of summation is defined as:
2 F 1 q n , b ; c ; c q n b = c / b ; q n c ; q n .
Theorem 3.
If f is a continuous function on [ a , b ] and α , β > 0 , then the Riemann–Liouville fractional ( p , q ) -integral has the following semi-group property:
a I p , q β a I p , q α f ( t ) = a I p , q α a I p , q β f ( t ) = a I p , q α + β f ( t ) .
Proof. 
For t [ a , b ] , we have:
a I p , q β a I p , q α f ( t ) = 1 p α 2 + β 2 Γ p , q ( α ) Γ p , q ( β ) a t a x p α 1 + 1 1 p α 1 a a t a Φ q ( x ) p , q ( α 1 ) a x p α 1 + 1 1 p α 1 a a Φ q ( s ) p , q ( β 1 ) × f s p β 1 + 1 1 p β 1 a a d p , q s a d p , q x = 1 p α 1 p α 2 + β 2 Γ p , q ( α ) Γ p , q ( β ) a t ( p q ) ( x a ) j = 0 q j p j + 1 1 p α 1 β 1 x a β 1 × 1 q p j + 1 p , q β 1 a t a Φ q ( x ) p , q ( α 1 ) f q j p α + β + j 1 ( x a ) + 1 1 p α + β + j 1 a a d p , q x = ( p q ) 2 ( t a ) α + β p α + β 2 + 2 Γ p , q ( α ) Γ p , q ( β ) i = 0 j = 0 q p i + i β q p j 1 q p j + 1 p , q β 1 1 q p i + 1 p , q α 1 × f q i + j p α + β + i + j ( t a ) + 1 1 p α + β + i + j a .
Applying the ( p , q ) -gamma function in (16), we obtain:
a I p , q β a I p , q α f ( t ) = ( p q ) α + β ( t a ) α + β p α + β 2 + 2 i = 0 j = 0 q p i + i β q p j 1 q p j + 1 p , q β 1 ( p q ) p , q ( β 1 ) 1 q p i + 1 p , q ( α 1 ) ( p q ) p , q ( α 1 ) × f q i + j p α + β + i + j ( t a ) + 1 1 p α + β + i + j a = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β i = 0 q p i + i β q p α ; q p i q p ; q p i j = 0 q p j q p β ; q p j q p ; q p j × f q i + j p α + β + i + j ( t a ) + 1 1 p α + β + i + j a .
Taking m = i + j and interchanging the order of summation, we obtain:
a I p , q β a I p , q α f ( t ) = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β i = 0 q p i + i β q p α ; q p i q p ; q p i m = i q p m i q p β ; q p m i q p ; q p m i × f q m p α + β + m ( t a ) + 1 1 p α + β + m a = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β m = 0 q p i + i β q p α ; q p i q p ; q p i i = 0 m q p m i q p β ; q p m i q p ; q p m i × f q m p α + β + m ( t a ) + 1 1 p α + β + m a = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β m = 0 q p m i = 0 m q p i β q p α ; q p i q p ; q p i q p β ; q p m i q p ; q p m i × f q m p α + β + m ( t a ) + 1 1 p α + β + m a = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β m = 0 q p m f q m p α + β + m ( t a ) + 1 1 p α + β + m a × i = 0 m q p i β q p α ; q p i q p ; q p i q p β ; q p m i q p ; q p m i .
On the other hand,
q p β ; q p m i q p ; q p m i = 1 q p β q p m i 1 q p q p m i = 1 q p β q p m 1 q p β + m q p i 1 q p q p m 1 q p m + 1 q p i ,
and:
1 q p β + m q p i 1 q p m + 1 q p i = 1 q p m i + 1 q p i 1 q p β + m i q p i = k = 0 i 1 1 q p m i + 1 q p k k = 0 i 1 1 q p β + m i q p k = q p i i β k = 0 i 1 1 q p i 1 + m q p k 1 q p β m + i q p k = q p m ; q p i q p i i β q p 1 β m ; q p i .
Substituting (28) into (27), we get:
q p β ; q p m i q p ; q p m i = q p β ; q p m q p ; q p m q p m ; q p i q p 1 β m ; q p i q p i i β .
Therefore, we obtain:
a I p , q β a I p , q α f ( t ) = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β m = 0 q p m f q m p α + β + m ( t a ) + 1 1 p α + β + m a × q p β ; q p m q p ; q p m i = 0 m q p α ; q p i q p ; q p i q p m ; q p i q p 1 β m ; q p i q p i .
By (24), we have:
a I p , q β a I p , q α f ( t ) = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β m = 0 q p m f q m p α + β + m ( t a ) + 1 1 p α + β + m a × q p β ; q p m q p ; q p m 2 F 1 q p m , q p α ; q p 1 β m ; q p .
From (25), we obtain:
2 F 1 q p m , q p α ; q p 1 β m ; q p = q p 1 β m α ; q p m q p 1 β m ; q p m q p m α = q p α + β ; q p m q p β ; q p m .
Substituting (32) into (31), we get:
a I p , q β a I p , q α f ( t ) = ( p q ) α + β ( t a ) α + β p α + β 2 p α + β 1 m = 0 q m p m + 1 q p α + β ; q p m q p ; q p m × f q m p α + β + m ( t a ) + 1 1 p α + β + m a ,
which is the series representation of a I p , q α + β f ( t ) . Therefore, (26) holds. □
Throughout this paper, the variable s is shown inside the fractional integral, which is denoted as:
( a I p , q α f ( s ) ) ( t ) = 1 p α 2 Γ p , q ( α ) a t a t a Φ q ( s ) p , q ( α 1 ) f s p α 1 + 1 1 p α 1 a a d p , q s .
Lemma 2.
If s ¯ = p α 1 s + ( 1 p α 1 ) a , α , β > 0 , and t [ a , p α b + ( 1 p α a ) ] , then the following formula holds:
a I p , q α ( s ¯ a ) β ( t ) = p β α 1 Γ p , q ( β + 1 ) Γ p , q ( β + α + 1 ) ( t a ) β + α .
Proof. 
Applying Definition 2, Property 2, and (18) and (19), we have:
a I p , q α ( s ¯ a ) β ( t ) = 1 p α 2 Γ p , q ( α ) a t a t a Φ q ( s ) p , q ( α 1 ) s a β a d p , q s = ( p a ) ( t a ) p α 2 Γ p , q ( α ) i = 0 q i p i + 1 a ( t a ) q i + 1 p i + 1 ( t a ) p , q ( α 1 ) q i p i + 1 ( t a ) β = ( p a ) ( t a ) β + 1 p α 2 Γ p , q ( α ) i = 0 q i p i + 1 i = 0 1 q p i + 1 q p i 1 q p i + 1 q p α 1 + i q i p i + 1 β = ( p a ) ( t a ) α + β p α 2 Γ p , q ( α ) i = 0 q i p i + 1 1 q p i + 1 p , q ( α 1 ) q i p i + 1 β = ( t a ) α + β p α 2 Γ p , q ( α ) 0 1 s β 1 0 Φ q ( s ) p , q ( α 1 ) 0 d p , q s = ( t a ) α + β p α 2 Γ p , q ( α ) B p , q ( β + 1 , α ) = p β α 1 Γ p , q ( β + 1 ) Γ p , q ( β + α + 1 ) ( t a ) β + α .
The proof of Lemma 2 is complete. □
Remark 1.
Define functions f , g : [ a , b ] R by f ( t ) = t and g ( t ) = t 2 . For α > 0 , we get:
(i) 
( a I p , q α f ( s ¯ ) ) ( t ) = ( t a ) α Γ p , q ( α + 2 ) [ p α 1 t a + ( [ α + 1 ] p , q ) a ] ;
(ii) 
( a I p , q α g ( s ¯ ) ) ( t ) = ( t a ) α Γ p , q ( α + 3 ) p 2 ( α 1 ) p + q ( t a ) 2 + 2 a p α 1 ( t a ) [ α + 2 ] p , q + a 2 [ α + 2 ] p , q [ α + 1 ] p , q .
Next, we study some fractional ( p , q ) integral inequalities on finite intervals.
  • The fractional ( p , q ) -Hölder inequality on [ a , p α b + ( 1 p α ) ] :
Theorem 4.
Let f : [ a , b ] R be a continuous function, s ¯ = p α 1 s + ( 1 p α 1 ) a , and α > 0 . If p 1 , p 2 0 , such that 1 / p 1 + 1 / p 2 = 1 , then we have:
( a I p , q α | f ( s ¯ ) | | g ( s ¯ ) | ) ( t ) ( a I p , q α | f ( s ¯ ) | p 1 ( t ) ) 1 / p 1 ( a I p , q α | g ( s ¯ ) | p 2 ( t ) ) 1 / p 2 .
Proof. 
Using Definition 3 and the discrete Hölder inequality, we have:
( a I p , q α | f ( s ¯ ) | | g ( s ¯ ) | ) ( t ) = 1 p α 2 Γ p , q ( α ) a t ( t a Φ q ( s ) ) p , q ( α 1 ) f ( s ) g ( s ) a d p , q s = ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 t a Φ q / p n + 1 ( t ) p , q ( α 1 ) f q n p n + 1 t + 1 q n p n + 1 a g q n p n + 1 t + 1 q n p n + 1 a = ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 t a Φ q / p n + 1 ( t ) p , q ( α 1 ) q n p n + 1 1 p 1 f q n p n + 1 t + 1 q n p n + 1 a × q n p n + 1 1 p 2 g q n p n + 1 t + 1 q n p n + 1 a ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 t a Φ q / p n + 1 ( t ) p , q ( α 1 ) q n p n + 1 f q n p n + 1 t + 1 q n p n + 1 a p 1 1 / p 1 × ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 t a Φ q / p n + 1 ( t ) p , q ( α 1 ) q n p n + 1 g q n p n + 1 t + 1 q n p n + 1 a p 2 1 / p 2 = a I p , q α | f ( s ¯ ) | p 1 ( t ) 1 / p 1 a I p , q α | g ( s ¯ ) | p 2 ( t ) 1 / p 2 .
Therefore, the proof of Theorem 4 is complete. □
Remark 2.
If p = 1 , then (35) reduces to the fractional q-Hölder inequality on [ a , b ] as:
( a I q α | f ( s ) | | g ( s ) | ) ( t ) ( a I q α | f ( s ) | p 1 ( t ) ) 1 / p 1 ( a I q α | g ( s ) | p 2 ( t ) ) 1 / p 2 ,
which appeared in [39]. However, if p = 1 and a = 0 , then (36) reduces to the q-Hölder inequality as:
0 t f ( s ) g ( s ) d q s 0 t f ( s ) p 1 d q s 1 / p 1 0 t g ( s ) p 2 d q s 1 / p 2 ,
which can be found in [71].
  • The fractional ( p , q ) -Hermite–Hadamard integral inequalities on [ a , p α b + ( 1 p α ) a ] :
Theorem 5.
If f : [ a , b ] R is a convex differentiable function and α > 0 , then we have:
f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q Γ p , q ( α + 1 ) p α 2 ( b a ) α a I p , q α f ( s ) ( p α b + ( 1 p α ) a ) [ α + 1 ] p , q p α f ( a ) + p α f ( b ) [ α + 1 ] p , q .
Proof. 
From the left-hand side of the proof in Theorem 3 and Theorem 5.1 in [15,42], respectively, there is a one line support:
f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q + m s p α 1 + 1 1 p α 1 a [ α + 1 ] p , q p α f ( a ) + p α f ( b ) [ α + 1 ] p , q f s p α 1 + 1 1 p α 1 a ,
for all s [ a , b ] and m f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q , f + [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q .
Multiplying by a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) / p α 2 Γ p , q ( α ) on both sides of (38), we obtain:
1 p α 2 Γ p , q ( α ) a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q + m 1 p α 2 Γ p , q ( α ) a Φ p α ( b ) a Φ q ( s ) p , q α 1 s p α 1 + 1 1 p α 1 a [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q 1 p α 2 Γ p , q ( α ) a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) f s p α 1 + 1 1 p α 1 a .
Taking the fractional ( p , q ) -integration of order α > 0 with respect to s on (39), where s ( a , p α b + ( 1 p α ) a ) , we have:
1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q a d p , q s + m 1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) × s p α 1 + 1 1 p α 1 a [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q a d p , q s 1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) f s p α 1 + 1 1 p α 1 a a d p , q s = a I p , q α f ( s ) p α b + ( 1 p α ) a .
Moreover, from the left-hand side of (40), we have:
1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q a d p , q s = ( p q ) p α ( b a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a Φ p α ( b ) q n + 1 p n ( b a ) + a p , q ( α 1 ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q = ( p q ) p α 2 ( b a ) α p α 2 Γ p , q ( α ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q n = 0 q n p n + 1 1 q p n + 1 p , q ( α 1 ) = p α 2 ( b a ) α p α 2 Γ p , q ( α ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q 0 1 1 0 Φ q ( s ) p , q ( α 1 ) 0 d p , q s = p α 2 ( b a ) α p α 2 Γ p , q ( α ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q B p , q 1 , α = p α 2 ( b a ) α Γ p , q ( α + 1 ) f [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q
and similar to the computation of (41) above, we also get:
1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) × s p α 1 + 1 1 p α 1 a [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q a d p , q s = ( p q ) p α 2 + 1 ( b a ) α + 1 p α 2 Γ p , q ( α ) n = 0 q n p n + 1 1 q p n + 1 p , q ( α 1 ) q i p i + 1 + ( p q ) p α 2 ( b a ) α p α 2 Γ p , q ( α ) n = 0 q n p n + 1 1 q p n + 1 p , q ( α 1 ) a [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q = p α 2 ( b a ) α p α 2 Γ p , q ( α ) p ( b a ) B p , q ( 2 , α ) + a B p , q ( 1 , α ) [ α + 1 ] p , q p α a + p α b [ α + 1 ] p , q B p , q ( 1 , α ) = p α 2 ( b a ) α [ α + 1 ] p , q p α a + p α b Γ p , q ( α + 2 ) [ α + 1 ] p , q p α a + p α b Γ p , q ( α + 2 ) = 0 .
By substituting (42) and (41) into (40), we obtain the first part of (37). On the other hand, from the proof of the right-hand side of Theorem 3 in [42], we have:
f s p α 1 + 1 1 p α 1 a f ( a ) + f ( b ) f ( a ) b a s p α 1 + 1 1 p α 1 a a .
Multiplying by a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) / p α 2 Γ p , q ( α ) on both sides of (43) and taking the ( p , q ) -integral of order α > 0 with respect to s [ a , p α b + ( 1 p α ) a ] , we obtain:
1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) f s p α 1 + 1 1 p α 1 a a d p , q s 1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) × f ( a ) + f ( b ) f ( a ) b a s p α 1 + 1 1 p α 1 a a a d p , q s .
By using the same computation as in (41) and (42) for the left-hand side of (44), we obtain:
1 p α 2 Γ p , q ( α ) a p α b + ( 1 p α ) a a a Φ p α ( b ) a Φ q ( s ) p , q ( α 1 ) × f ( a ) + f ( b ) f ( a ) b a s p α 1 + 1 1 p α 1 a a a d p , q s p α 2 ( b a ) α Γ p , q ( α + 1 ) f a + p α 2 + α ( b a ) α Γ p , q ( α + 2 ) f ( b ) f ( a )
= p α 2 ( b a ) α Γ p , q ( α + 1 ) [ α + 1 ] p , q p α f ( a ) + p α f ( b ) [ α + 1 ] p , q .
Substituting (45) into (44), we derive the second part of (37). Therefore, the proof of Theorem 5 is complete. □
Remark 3.
If p = 1 , then (37) reduces to the fractional q-Hermite–Hadamard integral inequality as:
f [ α + 1 ] q 1 a + b [ α + 1 ] q Γ q ( α + 1 ) ( b a ) α a I q α f ( s ) ( b ) [ α + 1 ] q 1 f ( a ) + f ( b ) [ α + 1 ] q ,
which appeared in [15].
If α = 1 , then (37) reduces to the ( p , q ) -Hermite–Hadamard integral inequality on [ a , b ] as:
f q a + p b p + q 1 p ( b a ) a p b + ( 1 p ) a f ( s ) a d p , q s q f ( a ) + p f ( b ) p + q ,
which appeared in [42]. Moreover, if p = 1 and q 1 , then (47) reduces to the classical Hermite–Hadamard integral inequality as:
f a + b 2 1 b a a b f ( s ) d s f ( a ) + f ( b ) 2 ,
which can be found in [72,73].
  • The fractional ( p , q ) -Korkine equality on [ a , p α b + ( 1 p α ) a ] :
Theorem 6.
If f , g : [ a , b ] R are continuous functions, s ¯ = p α 1 s + ( 1 p α 1 ) a , r ¯ = p α 1 r + ( 1 p α 1 ) a , and α > 0 , then we have:
1 2 a I p , q 2 α f ( s ¯ ) f ( r ¯ ) g ( s ¯ ) g ( r ¯ ) p α b + ( 1 p α ) a = p α ( b a ) α Γ p , q ( α + 1 ) a I p , q α f ( s ¯ ) g ( s ¯ ) p α b + ( 1 p α ) a a I p , q α f ( s ¯ ) p α b + ( 1 p α ) a a I p , q α g ( s ¯ ) p α b + ( 1 p α ) a .
Proof. 
Using Definition 3 and applying (18), we have:
a I p , q 2 α f ( s ¯ ) f ( r ¯ ) g ( s ¯ ) g ( r ¯ ) ( t ) = 1 p α 2 p α 2 Γ p , q 2 ( α ) a t a t a t a Φ q ( s ) p , q α 1 a t a Φ q ( r ) p , q α 1 × f ( s ) g ( s ) f ( s ) g ( r ) f ( r ) g ( s ) + f ( r ) g ( r ) a d p , q s a d p , q r = ( t a ) α Γ p , q ( α + 1 ) ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 × f q n p n + 1 t + 1 q n p n + 1 a g q n p n + 1 t + 1 q n p n + 1 a ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 f q n p n + 1 t + 1 q n p n + 1 a × ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 g q n p n + 1 t + 1 q n p n + 1 a ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 f q n p n + 1 t + 1 q n p n + 1 a × ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 g q n p n + 1 t + 1 q n p n + 1 a + ( t a ) α Γ p , q ( α + 1 ) ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 × f q n p n + 1 t + 1 q n p n + 1 a g q n p n + 1 t + 1 q n p n + 1 a = 2 ( t a ) α Γ p , q ( α + 1 ) ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 × f q n p n + 1 t + 1 q n p n + 1 a g q n p n + 1 t + 1 q n p n + 1 a 2 ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 f q n p n + 1 t + 1 q n p n + 1 a × ( p q ) ( t a ) p α 2 Γ p , q ( α ) n = 0 q n p n + 1 a t a Φ q / p n + 1 ( t ) p , q α 1 g q n p n + 1 t + 1 q n p n + 1 a = 2 ( t a ) α Γ p , q ( α + 1 ) 1 p α 2 Γ p , q ( α ) a t a t a Φ q ( s ) p , q α 1 f ( s ) g ( s ) a d p , q s 2 1 p α 2 Γ p , q ( α ) a t a t a Φ q ( s ) p , q α 1 f ( s ) a d p , q s × 1 p α 2 Γ p , q ( α ) a t a t a Φ q ( s ) p , q α 1 g ( s ) a d p , q s = 2 ( t a ) α Γ p , q ( α + 1 ) a I p , q α f ( s ¯ ) g ( s ¯ ) ( t ) 2 a I p , q α f ( s ¯ ) ( t ) a I p , q α g ( r ¯ ) ( t ) = p α ( b a ) α Γ p , q ( α + 1 ) a I p , q α f ( s ¯ ) g ( s ¯ ) p α b + ( 1 p α ) a a I p , q α f ( s ¯ ) p α b + ( 1 p α ) a a I p , q α g ( s ¯ ) p α b + ( 1 p α ) a .
Therefore, the proof is complete. □
Remark 4.
If p = 1 , then (48) reduces to the fractional q-Korkine equality on [ a , b ] as:
1 2 ( a I q 2 α f ( s ) f ( r ) g ( s ) g ( r ) ) ( b ) = ( b a ) α Γ q ( α + 1 ) ( a I q α f ( s ) g ( s ) ) ( b ) ( a I q α f ( s ) ) ( b ) ( a I q α g ( s ) ) ( b ) ,
which appeared in [39]. Moreover, if α = 1 , then (48) reduces to the q-Korkine equality as:
1 2 a b a b f ( s ) g ( r ) g ( s ) g ( r ) a d q s a d q r = b a a b f ( s ) g ( s ) a d q s a b f ( s ) a d q s a b g ( s ) a d q s ,
which appeared in [27].
  • The fractional ( p , q ) -Cauchy–Bunyakovsky–Schwarz integral inequality on [ a , p α b + ( 1 p α ) a ] :
Theorem 7.
If f , g : [ a , b ] R are continuous functions, s ¯ = p α 1 s + ( 1 p α 1 ) a , r ¯ = p β 1 r + ( 1 p β 1 ) a , and α , β > 0 , then we have:
a I p , q α + β f ( s ¯ , r ¯ ) g ( s ¯ , r ¯ ) p α b + ( 1 p α ) a a I p , q α + β f 2 ( s ¯ , r ¯ ) p α b + ( 1 p α ) a a I p , q α + β g 2 ( s ¯ , r ¯ ) p α b + ( 1 p α ) a .
Proof. 
Using Theorem 3 and Definition 3, we have:
a I p , q α + β f ( s ¯ , r ¯ ) g ( s ¯ , r ¯ ) ( t ) = 1 p α 2 p β 2 Γ p , q 2 ( α ) a t a t a t a Φ q ( s ) p , q α 1 a t a Φ q ( r ) p , q β 1 f ( s , r ) a d p , q s a d p , q r = ( p q ) 2 ( t a ) 2 p α 2 p β 2 Γ p , q ( α ) Γ p , q ( β ) i = 0 j = 0 q i + j p i + j + 2 a t a Φ q / p i + 1 ( t ) p , q α 1 a t a Φ q / p j + 1 ( t ) p , q β 1 × f q i p i + 1 t + 1 q i p i + 1 a , q j p j + 1 t + 1 q j p j + 1 a .
Applying the classical discrete Cauchy–Schwarz inequality, we have:
a I p , q α + β f ( s ¯ , r ¯ ) g ( s ¯ , r ¯ ) ( t ) 2 = [ ( p q ) 2 ( t a ) 2 p α 2 p β 2 Γ p , q ( α ) Γ p , q ( β ) i = 0 j = 0 q i + j p i + j + 2 a t a Φ q / p i + 1 ( t ) p , q α 1 a t a Φ q / p j + 1 ( t ) p , q β 1 × f q i p i + 1 t + 1 q i p i + 1 a , q j p j + 1 t + 1 q j p j + 1 a × g q i p i + 1 t + 1 q i p i + 1 a , q j p j + 1 t + 1 q j p j + 1 a ] 2 ( ( p q ) 2 ( t a ) 2 p α 2 p β 2 Γ p , q ( α ) Γ p , q ( β ) i = 0 j = 0 q i + j p i + j + 2 a t a Φ q / p i + 1 ( t ) p , q α 1 a t a Φ q / p j + 1 ( t ) p , q β 1 × f 2 q i p i + 1 t + 1 q i p i + 1 a , q j p j + 1 t + 1 q j p j + 1 a ) × ( ( p q ) 2 ( t a ) 2 p α 2 p β 2 Γ p , q ( α ) Γ p , q ( β ) i = 0 j = 0 q i + j p i + j + 2 a t a Φ q / p i + 1 ( t ) p , q α 1 a t a Φ q / p j + 1 ( t ) p , q β 1 × g 2 q i p i + 1 t + 1 q i p i + 1 a , q j p j + 1 t + 1 q j p j + 1 a ) = a I p , q α + β f 2 ( s ¯ , r ¯ ) ( t ) a I p , q α + β f 2 ( s ¯ , r ¯ ) ( t ) = a I p , q α + β f 2 ( s ¯ , r ¯ ) p α b + ( 1 p α ) a a I p , q α + β f 2 ( s ¯ , r ¯ ) p α b + ( 1 p α ) a .
Consequently, the proof of (49) is finished. □
Remark 5.
If p = 1 , then (49) reduces to the fractional q-Cauchy–Bunyakovsky–Schwarz integral inequality on [ a , b ] as:
a I q α + β f ( s , r ) g ( s , r ) ( b ) a I q α + β f 2 ( s , r ) ( b ) a I q α + β g 2 ( s , r ) ( b ) ,
which appeared in [39]. Moreover, if α = 1 , then (49) reduces to the q-Cauchy–Bunyakovsky–Schwarz integral inequality as:
a b a b f ( s , r ) g ( s , r ) a d q s a d q r a b a b f 2 ( s , r ) a d q s a d q r 1 / 2 a b a b g 2 ( s , r ) a d q s a d q r 1 / 2 ,
which can be found in [27].
  • The fractional ( p , q ) -Grüss integral inequality on [ a , p α b + ( 1 p α ) a ] :
Theorem 8.
If f , g : [ a , b ] R are continuous functions satisfying:
ϕ f ( s ) Φ and ψ g ( s ) Ψ ,
for all s [ a , b ] , ϕ , Φ , ψ , Ψ R , s ¯ = p α 1 s + ( 1 p α 1 ) a , r ¯ = p α 1 r + ( 1 p α 1 ) a , and α > 0 , then we have:
Γ p , q ( α + 1 ) p α ( b a ) α a I p , q α f ( s ¯ ) g ( s ¯ ) p α b + ( 1 p α ) a Γ p , q ( α + 1 ) p α ( b a ) α a I p , q α f ( s ¯ ) p α b + ( 1 p α ) a × Γ p , q ( α + 1 ) p α ( b a ) α a I p , q α f ( s ¯ ) p α b + ( 1 p α ) a 1 4 ( Φ ϕ ) ( Ψ ψ ) .
Proof. 
Applying Theorem 7, we have:
| ( a I p , q 2 α f ( s ¯ ) f ( r ¯ ) g ( s ¯ ) g ( r ¯ ) ) ( t ) | a I p , q 2 α f ( s ¯ ) f ( r ¯ ) 2 ( t ) × a I p , q 2 α g ( s ¯ ) g ( r ¯ ) 2 ( t ) .
By Theorem 6, we obtain:
1 2 Γ p , q ( α + 1 ) ( t a ) α 2 a I p , q 2 α f ( s ¯ ) f ( r ¯ ) 2 ( t ) = Γ p , q ( α + 1 ) ( t a ) α a I p , q α f 2 ( s ¯ ) ( t ) Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( r ¯ ) ( t ) 2 .
On the other hand, we have:
Γ p , q ( α + 1 ) ( t a ) α a I p , q α f 2 ( s ¯ ) ( t ) Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( r ¯ ) ( t ) 2 = Φ Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( s ¯ ) ( t ) Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( s ¯ ) ( t ) ϕ Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( s ¯ ) ϕ Φ f ( s ¯ ) ( t ) .
From (50), we have f ( s ) ϕ Φ f ( s ) 0 . It follows that:
a I p , q α f ( s ¯ ) ϕ Φ f ( s ¯ ) ( t ) 0 .
Applying (52) and using the truth that K + K 2 2 K K , where K , K R , we have:
Γ p , q ( α + 1 ) ( b a ) α a I p , q α f 2 ( s ¯ ) ( t ) Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( r ¯ ) ( b ) 2 Φ Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( s ¯ ) ( t ) Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( s ¯ ) ( t ) ϕ 1 4 Φ Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( s ¯ ) ( t ) + Γ p , q ( α + 1 ) ( t a ) α a I p , q α f ( s ¯ ) ( t ) ϕ 2 = 1 4 Φ ϕ 2 .
Similarly,
1 2 Γ p , q ( α + 1 ) ( t a ) α 2 a I p , q 2 α g ( s ¯ ) g ( r ¯ ) 2 ( t ) 1 4 Ψ ψ 2 .
From (50), it follows that:
a I p , q 2 α f ( s ¯ ) f ( r ¯ ) g ( s ¯ ) g ( r ¯ ) ( t ) 1 2 ( t a ) α Γ p , q ( α + 1 ) 2 Φ ϕ Ψ ψ = a I p , q 2 α f ( s ¯ ) f ( r ¯ ) g ( s ¯ ) g ( r ¯ ) p α b + ( 1 p α ) a 1 2 p α ( b a ) α Γ p , q ( α + 1 ) 2 Φ ϕ Ψ ψ
Applying Theorem 6 again for (57), we have that Theorem 8 holds. □
Remark 6.
If p = 1 , then (51) reduces to the fractional q-Grüss integral inequality as:
Γ q ( α + 1 ) ( b a ) α a I q α f ( s ) g ( s ) ( b ) Γ q ( α + 1 ) ( b a ) α a I q α f ( s ) ( b ) × Γ q ( α + 1 ) ( b a ) α a I q α f ( s ) ( b ) 1 4 ( Φ ϕ ) ( Ψ ψ ) ,
which appeared in [39]. Moreover, if α = 1 , p = 1 , and q 1 , then (51) reduces to the classical Grüss integral as:
1 b a a b f ( s ) g ( s ) d s 1 b a a b f ( s ) d s 1 b a a b g ( s ) d s 1 4 Φ ϕ Ψ ψ ,
which can be found in [72,73].
  • The fractional ( p , q ) -Grüss–Chebyshev integral inequality on [ a , p α b + ( 1 p α ) a ] :
Theorem 9.
If f , g : [ a , b ] R are L 1 , L 2 -Lipschitzian continuous functions such that:
f ( s ) f ( r ) L 1 s r , g ( s ) g ( r ) L 2 s r , for all s , r [ a , b ] ,
s ¯ = p α 1 s + ( 1 p α 1 ) a , r ¯ = p α 1 r + ( 1 p α 1 ) a , L 1 , L 2 0 , and α 0 , then we have:
p α ( b a ) α Γ p , q ( α + 1 ) ( a I p , q α f ( s ¯ ) g ( s ¯ ) ) p α b + ( 1 p α ) a ( a I p , q α f ( s ¯ ) ) p α b + ( 1 p α ) a ( a I p , q α g ( s ¯ ) ) p α b + ( 1 p α ) a | L 1 L 2 p 2 α + 2 ( b a ) 2 α + 2 p 2 ( α 1 ) Γ p , q ( α + 2 ) Γ p , q ( α + 3 ) ( p + q ) [ α + 1 ] p , q [ α + 2 ] p , q .
Proof. 
By Theorem 6, we have:
( t a ) α Γ p , q ( α + 1 ) ( a I p , q α f ( s ¯ ) g ( s ¯ ) ) ( t ) ( a I p , q α f ( s ¯ ) ) ( t ) ( a I p , q α g ( s ¯ ) ) ( t ) = 1 2 ( a I p , q 2 α f ( s ¯ ) f ( r ¯ ) g ( s ¯ ) g ( r ¯ ) ) ( t ) .
From the condition in (58), we obtain:
f ( s ) f ( r ) g ( s ) g ( r ) L 1 L 2 s r 2 , for all s , r [ a , b ] .
Taking double fractional ( p , q ) -integration of order α > 0 with respect to s , r [ a , t ] with t = p α b + ( 1 p α ) a , we have:
a I p , q 2 α f ( s ¯ ) f ( r ¯ ) g ( s ¯ ) g ( r ¯ ) ( t ) = 1 p α 2 p α 2 Γ p , q 2 ( α ) a t a t a t a Φ q ( s ) p , q α 1 a t a Φ q ( r ) p , q α 1 × f ( s ) g ( s ) f ( s ) g ( r ) f ( r ) g ( s ) + f ( s ) g ( s ) a d p , q s a d p , q r L 1 L 2 p α 2 p α 2 Γ p , q 2 ( α ) a t a t a t a Φ q ( s ) p , q α 1 a t a Φ q ( r ) p , q α 1 s r 2 a d p , q s a d p , q r . = L 1 L 2 p α 2 p α 2 Γ p , q 2 ( α ) a t a t a t a Φ q ( s ) p , q α 1 a t a Φ q ( r ) p , q α 1 s 2 a d p , q s a d p , q r 2 L 1 L 2 p α 2 p α 2 Γ p , q 2 ( α ) a t a t a t a Φ q ( s ) p , q α 1 a t a Φ q ( r ) p , q α 1 s r a d p , q s a d p , q r + L 1 L 2 p α 2 p α 2 Γ p , q 2 ( α ) a t a t a t a Φ q ( s ) p , q α 1 a t a Φ q ( r ) p , q α 1 r 2 a d p , q s a d p , q r = 2 L 1 L 2 ( t a ) α Γ p , q ( α + 1 ) a I p , q α s ¯ 2 ( t ) a I p , q α s ¯ ( t ) 2 .
From Remark 1, we get:
a I p , q α s ¯ ( t ) = ( t a ) α Γ p , q ( α + 2 ) p α 1 t a + a [ α + 1 ] p , q
and:
a I p , q α s ¯ 2 ( t ) = ( t a ) α Γ p , q ( α + 3 ) p 2 ( α 1 ) p + q ( t a ) 2 + 2 a p α 1 ( t a ) [ α + 2 ] p , q + a 2 [ α + 2 ] p , q [ α + 1 ] p , q .
Moreover, we have:
( t a ) α Γ p , q ( α + 1 ) a I p , q α s ¯ 2 ( t ) a I p , q α s ¯ ( t ) 2 = ( t a ) 2 α Γ p , q ( α + 1 ) Γ p , q ( α + 3 ) p 2 ( α 1 ) ( p + q ) ( t a ) 2 + 2 a p α 1 t a [ α + 2 ] p , q + a 2 [ α + 2 ] p , q [ α + 1 ] p , q ( t a ) 2 α Γ p , q 2 ( α + 2 ) p α 1 t a + a [ α + 1 ] p , q 2 = ( t a ) 2 α + 2 p 2 ( α 1 ) Γ p , q ( α + 2 ) Γ p , q ( α + 3 ) ( p + q ) [ α + 1 ] p , q [ α + 2 ] p , q . = p 2 α + 2 ( b a ) 2 α + 2 p 2 ( α 1 ) Γ p , q ( α + 2 ) Γ p , q ( α + 3 ) ( p + q ) [ α + 1 ] p , q [ α + 2 ] p , q .
Substituting (63) into (62) and applying to (60), we complete the proof. □
Remark 7.
If p = 1 , then (59) reduces to the fractional q-Grüss–Chebyshev integral inequality on [ a , b ] as:
( b a ) α Γ q ( α + 1 ) a I q α f ( s ) g ( s ) ( b ) a I q α f ( s ) ( b ) a I q α g ( s ) ( b ) L 1 L 2 ( b a ) 2 α + 2 Γ q ( α + 2 ) Γ q ( α + 3 ) ( 1 + q ) [ α + 1 ] q [ α + 2 ] q ,
which appeared in [39]. Moreover, if α = 1 , p = 1 and q 1 , then (59) reduces to the classical Grüss–Chebyshev integral as:
1 b a a b f ( s ) g ( s ) d s 1 b a a b f ( s ) d s 1 b a a b L 1 L 2 12 b a 2 ,
which can be found in [72,73].
  • The fractional ( p , q ) -Polya–Szeqö integral inequality on [ a , p α b + ( 1 p α ) a ] :
Theorem 10.
If f , g : [ a , b ] R are two positive integrable functions satisfying:
0 < ϕ f ( s ) Φ and 0 < ψ g ( s ) Ψ ,
for all s [ a , b ] , ϕ , Φ , ψ , Ψ R + , and α > 0 , then we have:
a I p , q α f 2 p α 1 s + ( 1 p α 1 ) a p α b + ( 1 p α ) a a I p , q α g 2 p α 1 s + ( 1 p α 1 ) a p α b + ( 1 p α ) a a I p , q α f p α 1 s + ( 1 p α 1 ) a g p α 1 s + ( 1 p α 1 ) a p α b + ( 1 p α ) a 2 1 4 ϕ ψ Φ Ψ + Φ Ψ ϕ ψ .
Proof. 
For s [ a , b ] , from (64), we obtain:
ϕ Ψ f ( s ) g ( s ) Φ ψ ,
which yields:
Φ ψ f ( s ) g ( s ) 0
and:
f ( s ) g ( s ) ϕ Ψ 0 .
Multiplying (66) and (67), we get:
Φ ψ f ( s ) g ( s ) f ( s ) g ( s ) ϕ Ψ 0
or:
Φ ψ + ϕ Ψ f ( s ) g ( s ) f 2 ( s ) g 2 ( s ) + ϕ Φ ψ Ψ .
Therefore, (69) can be written as:
ϕ ψ + Φ Ψ f ( s ) g ( s ) ψ Ψ f 2 ( s ) + ϕ Φ g 2 ( s ) .
Multiplying a b a Φ p , q ( s ) q α 1 / p α 2 Γ p , q ( α ) and integrating of order α 0 on both sides of (70), where s [ a , t ] with t = p α b + ( 1 p α ) a , we obtain:
ϕ ψ + Φ Ψ a I p , q α f p α 1 s + ( 1 p α 1 ) a g p α 1 s + ( 1 p α 1 ) a ( t ) ψ Ψ a I p , q α f 2 p α 1 s + ( 1 p α 1 ) a ( t ) + ϕ Φ a I p , q α g 2 p α 1 s + ( 1 p α 1 ) a ( t ) .
Applying the AM-GMinequality, we get:
ϕ ψ + Φ Ψ a I p , q α f p α 1 s + ( 1 p α 1 ) a g p α 1 s + ( 1 p α 1 ) a ( t ) 2 ϕ ψ Φ Ψ a I p , q α f 2 p α 1 s + ( 1 p α 1 ) a ( t ) a I p , q α g 2 p α 1 s + ( 1 p α 1 ) a ( t ) ,
which turns into:
ϕ ψ Φ Ψ a I p , q α f 2 p α 1 s + ( 1 p α 1 ) a ( t ) a I p , q α g 2 p α 1 s + ( 1 p α 1 ) a ( t ) 1 4 ϕ ψ + Φ Ψ a I p , q α f p α 1 s + ( 1 p α 1 ) a g p α 1 s + ( 1 p α 1 ) a ( t ) 2 .
Therefore, Theorem 10 is proven. □
Remark 8.
If p = 1 , then (65) reduces to the fractional q-Polya–Szeqö integral inequality on [a, b] as:
a I q α f 2 ( s ) ( b ) a I q α g 2 ( s ) ( b ) a I q α f ( s ) g ( s ) ( b ) 2 1 4 ϕ ψ Φ Ψ + Φ Ψ ϕ ψ ,
which appeared in [39]. Moreover, if α = 1 , p = 1 , and q 1 , then (65) reduces to the classical Polya–Szeqö integral inequality as:
a b f 2 ( s ) d s a b g 2 ( s ) d s a b f ( s ) g ( s ) d s 2 1 4 ϕ ψ + Φ Ψ 2 ,
which can be found in [74].

4. Conclusions

In this work, we studied the fractional ( p , q ) -calculus on finite intervals. We also gave some of its significant properties. Furthermore, we proved some fractional ( p , q ) -integral inequalities on finite intervals. For the ideas, as well as the techniques of this paper, we hope that it will inspire interested readers working in this field.

Author Contributions

Conceptualization, J.T. and S.K.N.; investigation, P.N. and K.N.; methodology, K.N.; validation, P.N., K.N., J.T. and S.K.N.; visualization, P.N. and S.K.N.; writing—original draft, P.N.; writing—review and editing, K.N. All authors have read and agreed to the published version of the manuscript.

Funding

This research received no external funding.

Institutional Review Board Statement

Not applicable.

Informed Consent Statement

Not applicable.

Data Availability Statement

Not applicable.

Conflicts of Interest

The authors declare no conflict of interest.

References

  1. Jackson, F.H. On a q-definite integrals. Quart. J. Pure Appl. Math. 1910, 41, 193–203. [Google Scholar]
  2. Jackson, F.H. q-Difference equations. Am. J. Math. 1910, 32, 305–314. [Google Scholar] [CrossRef]
  3. Fock, V. Zur theorie des wasserstoffatoms. Z. Physik. 1935, 98, 145–154. [Google Scholar] [CrossRef]
  4. Bangerezaka, G. Variational q-calculus. J. Math. Anal. Appl. 2004, 289, 650–665. [Google Scholar] [CrossRef] [Green Version]
  5. Asawasamrit, S.; Sudprasert, C.; Ntouyas, S.; Tariboon, J. Some results on quantum Hanh integral inequalities. J. Inequal. Appl. 2019, 2019, 154. [Google Scholar] [CrossRef] [Green Version]
  6. Bangerezako, G. Variational calculus on q-nonuniform. J. Math. Anal. Appl. 2005, 306, 161–179. [Google Scholar] [CrossRef] [Green Version]
  7. Exton, H. q-Hypergeometric Functions and Applications; Hastead Press: New York, NY, USA, 1983. [Google Scholar]
  8. Annyby, H.M.; Mansour, S.K. q-Fractional Calculus and Equations; Springer: Helidelberg, Germany, 2012. [Google Scholar]
  9. Ernst, T. A Comprehensive Treatment of q-Calculus; Springer: Basel, Switzerland, 2012. [Google Scholar]
  10. Ernst, T. A History of q-Calculus and a New Method; UUDM Report 2000:16; Department of Mathematics, Uppsala University: Uppsala, Sweden, 2000. [Google Scholar]
  11. Ferreira, R. Nontrivial solutions for fractional q-difference boundary value problems. Electron. J. Qual. Theory Differ. Equ. 2010, 2010, 1–10. [Google Scholar] [CrossRef]
  12. Noor, M.A.; Awan, M.U.; Noor, K.I. Quantum Ostrowski inequalities for q-differentiable convex function. J. Math. Inequal. 2016, 10, 1013–1018. [Google Scholar] [CrossRef]
  13. Aral, A.; Gupta, V.; Agarwal, R.P. Applications of q-Calculus in Operator Theory; Springer Science + Business Media: New York, NY, USA, 2013. [Google Scholar]
  14. Gauchman, H. Integral inequalities in q-calculus. J. Comput. Appl. Math. 2002, 47, 281–300. [Google Scholar] [CrossRef] [Green Version]
  15. Kunt, M.; Aljasem, M. Riemann–Liouville fractional quantum Hermite–Hadamard type inequalities for convex functions. Konuralp J. Math. 2020, 8, 122–136. [Google Scholar]
  16. Dobrogowska, A.; Odzijewicz, A. A second order q-difference equation solvable by factorization method. J. Comput. Appl. Math. 2006, 193, 319–346. [Google Scholar] [CrossRef] [Green Version]
  17. Gasper, G.; Rahman, M. Some systems of multivariable orthogonal q-Racah polynomials. Ramanujan J. 2007, 13, 389–405. [Google Scholar] [CrossRef] [Green Version]
  18. Ismail, M.E.H.; Simeonov, P. q-Difference operators for orthogonal polynomials. J. Comput. Appl. Math. 2009, 233, 749–761. [Google Scholar] [CrossRef] [Green Version]
  19. Bohner, M.; Guseinov, G.S. The h-Laplace and q-Laplace transforms. J. Comput. Appl. Math. 2010, 365, 75–92. [Google Scholar] [CrossRef] [Green Version]
  20. El-Shahed, M.; Hassan, H.A. Positive solutions of q-difference equation. Proc. Am. Math. Soc. 2010, 138, 1733–1738. [Google Scholar] [CrossRef] [Green Version]
  21. Ahmad, B. Boundary-value problems for nonlinear third-order q-difference equations. Electron. J. Differ. Equ. 2011, 2011, 94. [Google Scholar]
  22. Ahmad, B.; Alsaedi, A.; Ntouyas, S.K. A study of second-order q-difference equations with boundary conditions. Adv. Differ. Equ. 2012, 2012, 35. [Google Scholar] [CrossRef] [Green Version]
  23. Ahmad, B.; Ntouyas, S.K.; Purnaras, I.K. Existence results for nonlinear q-difference equations with nonlocal boundary conditions. Commun. Appl. Nonlinear Anal. 2012, 19, 59–72. [Google Scholar]
  24. Ahmad, B.; Nieto, J.J. On nonlocal boundary value problems of nonlinear q-difference equation. Adv. Differ. Equ. 2012, 2012, 81. [Google Scholar] [CrossRef] [Green Version]
  25. Kac, V.; Cheung, P. Quantum Calculus; Springer: New York, NY, USA, 2002. [Google Scholar]
  26. Tariboon, J.; Ntouyas, S.K. Quantum calculus on finite intervals and applications to impulsive difference equations. Adv. Differ. Equ. 2013, 2013, 282. [Google Scholar] [CrossRef] [Green Version]
  27. Tariboon, J.; Ntouyas, S.K. Quantum integral inequalities on finite intervals. Adv. Differ. Equ. 2014, 2014, 121. [Google Scholar] [CrossRef] [Green Version]
  28. Necmettin, A.; Mehmet, Z.S.; Mehmet, K.; İmdat, İ. q-Hermite Hadamard inequalities and quantum estimates for midpoint type inequalities via convex functions and quasi-convex functions. J. King Saud Univ. Sci. 2018, 30, 193–203. [Google Scholar]
  29. Sudsudat, W.; Ntouyas, S.K.; Tariboon, J. Quantum integral inequalities for convex functions. J. King Saud Univ. Sci. 2015, 9, 781–793. [Google Scholar]
  30. Muhammad, A.N.; Khalida, I.N.; Muhammad, U.A. Some quantum integral inequalities via preinvex functions. Appl. Math. Comput. 2015, 269, 242–251. [Google Scholar]
  31. Muhammad, A.K.; Noor, M.; Eze, N.R.; Yu-Ming, C. Quantum Hermite–Hadamard inequality by means of a Green function. Adv. Differ. Equ. 2020, 2020, 99. [Google Scholar]
  32. Muhammad, A.N.; Khalida, I.N.; Muhammad, U.A. Quantum analogues of Hermite–Hadamard type inequalities for generalized convexity. In Computation, Cryptography, and Network Security; Springer: Cham, Switzerland, 2015; pp. 413–439. [Google Scholar]
  33. Humaira, K.; Muhammad, I.; Baleanu, D.; Yu-Ming, C. New estimates of-Ostrowski-type inequalities within a class of-polynomial prevexity of functions. J. Funct. Spaces 2020, 2020, 13. [Google Scholar]
  34. Wenjun, L.; Hefeng, Z. Some quantum estimates of Hermite–Hadamard inequalities for convex functions. Appl. Math. Comput. 2015, 251, 675–679. [Google Scholar]
  35. Tun, M.; Gov, E.; Balgeçti, S. Simpson type quantum integral inequalities for convex functions. Miskolc Math. Notes 2018, 9, 649–664. [Google Scholar]
  36. Latif, M.A.; Dragomir, S.S.; Momoniat, E. Some φ-analogues of Hermite–Hadamard inequality for s-convex functions in the second sense and related estimates. Punjab Univ. J. Math. 2016, 48, 147–166. [Google Scholar]
  37. Hefeng, Z.; Wenjun, L.; Jaekeun, P. Some quantum estimate of Hermite–Hadamard inequalities for quasi-convex functions.