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Riemann-Liouville fractional Hermite-Hadamard inequalities. Part II: for twice differentiable geometric-arithmetically s-convex functions

Journal of Inequalities and Applications20132013:517

https://doi.org/10.1186/1029-242X-2013-517

Received: 5 August 2013

Accepted: 30 August 2013

Published: 8 November 2013

Abstract

Motivated by the definition of geometric-arithmetically s-convex functions in (Shuang et al. in Analysis 33:197-208, 2013) and second-order fractional integral identities in (Zhang and Wang in J. Inequal. Appl. 2013:220, 2013; Wang et al. in Appl. Anal. 2012, doi:10.1080/00036811.2012.727986), we establish some interesting Riemann-Liouville fractional Hermite-Hadamard inequalities for twice differentiable geometric-arithmetically s-convex functions via beta function and incomplete beta function.

MSC:26A33, 26A51, 26D15.

Keywords

fractional Hermite-Hadamard inequalities Riemann-Liouville fractional integrals geometric-arithmetically s-convex functions

1 Introduction

Fractional calculus, which is a generalization of classical differentiation and integration to arbitrary order, was born in 1695. In the past three hundred years, fractional calculus developed not only in pure theoretical field but also in diverse fields ranging from physical sciences and engineering to biological sciences and economics [18].

The classical Hermite-Hadamard inequalities have attracted many researchers since 1893. Researchers investigated Hermite-Hadamard inequalities involving fractional integrals according to the associated fractional integral equalities and different types of convex functions. For instance, one can refer to [911] for convex functions and to [12] for nondecreasing functions, to [1315] for m-convex functions and to [16] for ( s , m ) -convex functions, to [17] for functions satisfying s-e-condition, to [18] for ( α , m ) -logarithmically convex functions and see the references therein.

In [19], Shuang et al. introduced a new concept of geometric-arithmetically s-convex functions and presented interesting Hermite-Hadamard type inequalities for integer integrals of such functions. In [20], the authors used the definition of geometric-arithmetically s-convex functions in [19] and applied first-order fractional integral identities in [9, 10, 14] to establish some interesting Riemann-Liouville fractional Hermite-Hadamard inequalities for once differentiable geometric-arithmetically s-convex functions.

However, fractional Hermite-Hadamard inequalities for twice geometric-arithmetically s-convex functions have not been reported. In this work, we continue the development in [20]. Note that Wang et al. [13, 16] presented some elementary fractional integral equalities for twice differential functions. Motivated by [13, 16, 19], we study Riemann-Liouville fractional Hermite-Hadamard type inequalities for geometric-arithmetically s-convex functions by means of first-order fractional integral equalities.

2 Preliminaries

In this section, we introduce notations, definitions, and preliminary facts.

Definition 2.1 (see [3])

Let f L [ a , b ] . The symbols J a + α f and J b α f denote the left-sided and right-sided Riemann-Liouville fractional integrals of the order α R + and are defined by
( J a + α f ) ( x ) = 1 Γ ( α ) a x ( x t ) α 1 f ( t ) d t ( 0 a < x b )
and
( J b α f ) ( x ) = 1 Γ ( α ) x b ( t x ) α 1 f ( t ) d t ( 0 a x < b ) ,

respectively, here Γ ( ) is the gamma function.

Definition 2.2 (see [19])

Let f : I R + R + and s ( 0 , 1 ] . A function f ( x ) is said to be geometric-arithmetically s-convex on I if for every x , y I and t [ 0 , 1 ] , we have
f ( x t y 1 t ) t s ( f ( x ) ) + ( 1 t ) s f ( y ) .

Definition 2.3 (see [21])

The incomplete beta function is defined as follows:
B x ( a , b ) = 0 x t a 1 ( 1 t ) b 1 d t ,

where x [ 0 , 1 ] , a , b > 0 .

The following inequality will be used in the sequel.

Lemma 2.4 (see [18])

For t [ 0 , 1 ] , we have
( 1 t ) n 2 1 n t n for n [ 0 , 1 ] , ( 1 t ) n 2 1 n t n for n [ 1 , ) .

The following elementally inequality was used in the proof directly in [19]. Here, we revisit this inequality from the point of our view and give a proof in [20].

Lemma 2.5 (see [20])

For t [ 0 , 1 ] , x , y > 0 , we have
t x + ( 1 t ) y y 1 t x t .

We introduce the following integral identities.

Lemma 2.6 (see [16])

Let f : [ a , b ] R be a twice differentiable mapping on ( a , b ) with a < b . If f L [ a , b ] , then the following equality for fractional integrals holds:
f ( a ) + f ( b ) 2 Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] = ( b a ) 2 2 0 1 1 ( 1 t ) α + 1 t α + 1 α + 1 f ( t a + ( 1 t ) b ) d t .

Lemma 2.7 (see [13])

Let f : [ a , b ] R be a twice differentiable mapping on ( a , b ) with a < b . If f L [ a , b ] , then
Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] f ( a + b 2 ) = ( b a ) 2 2 0 1 m ( t ) f ( t a + ( 1 t ) b ) d t ,
where
m ( t ) = { t 1 ( 1 t ) α + 1 t α + 1 α + 1 , t [ 0 , 1 2 ) , 1 t 1 ( 1 t ) α + 1 t α + 1 α + 1 , t [ 1 2 , 1 ) .

Lemma 2.8 (see [13])

Let f : [ a , b ] R be a twice differentiable mapping on ( a , b ) with a < b . If f L [ a , b ] , r > 0 , then
f ( a ) + f ( b ) r ( r + 1 ) + 2 r + 1 f ( a + b 2 ) Γ ( α + 1 ) r ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] = ( b a ) 2 0 1 k ( t ) f ( t a + ( 1 t ) b ) d t ,
where
k ( t ) = { 1 ( 1 t ) α + 1 t α + 1 r ( α + 1 ) t r + 1 , t [ 0 , 1 2 ) , 1 ( 1 t ) α + 1 t α + 1 r ( α + 1 ) 1 t r + 1 , t [ 1 2 , 1 ) .

3 The first main results

By using Lemma 2.6, we can obtain the main results in this section.

Theorem 3.1 Let f : [ 0 , b ] R be a differentiable mapping. If | f | is measurable and | f | is decreasing and geometric-arithmetically s-convex on [ 0 , b ] for some fixed α ( 0 , ) , s ( 0 , 1 ] , 0 a < b , then the following inequality for fractional integrals holds:
| f ( a ) + f ( b ) 2 Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 ( | f ( a ) | + | f ( b ) | ) 2 ( α + 1 ) ( 1 s + 1 1 α + s + 2 ) .
Proof By using Definition 2.2, Lemma 2.5 and Lemma 2.6, we have
| f ( a ) + f ( b ) 2 Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 2 0 1 | 1 ( 1 t ) α + 1 t α + 1 α + 1 | | f ( t a + ( 1 t ) b ) | d t ( b a ) 2 2 ( α + 1 ) 0 1 ( 1 ( 1 t ) α + 1 t α + 1 ) | f ( t a + ( 1 t ) b ) | d t ( b a ) 2 2 ( α + 1 ) 0 1 ( 1 ( 1 t ) α + 1 t α + 1 ) | f ( a t b 1 t ) | d t ( b a ) 2 2 ( α + 1 ) 0 1 ( 1 ( 1 t ) α + 1 t α + 1 ) [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) 2 | f ( a ) | 2 ( α + 1 ) 0 1 t α + s + 1 d t + ( b a ) 2 | f ( a ) | 2 ( α + 1 ) 0 1 t s d t ( b a ) 2 | f ( a ) | 2 ( α + 1 ) 0 1 t s ( 1 t ) α + 1 d t ( b a ) 2 | f ( b ) | 2 ( α + 1 ) 0 1 ( 1 t ) α + s + 1 d t ( b a ) 2 | f ( b ) | 2 ( α + 1 ) 0 1 t α + 1 ( 1 t ) s d t + ( b a ) 2 | f ( b ) | 2 ( α + 1 ) 0 1 ( 1 t ) s d t ( b a ) 2 | f ( a ) | 2 ( α + 1 ) 1 α + s + 2 + ( b a ) 2 | f ( a ) | 2 ( α + 1 ) 1 s + 1 ( b a ) 2 | f ( a ) | 2 ( α + 1 ) B ( s + 1 , α + 2 ) ( b a ) 2 | f ( b ) | 2 ( α + 1 ) 1 α + s + 2 + ( b a ) 2 | f ( b ) | 2 ( α + 1 ) 1 s + 1 ( b a ) 2 | f ( b ) | 2 ( α + 1 ) B ( s + 1 , α + 2 ) ( b a ) 2 ( | f ( a ) | + | f ( b ) | ) 2 ( α + 1 ) ( 1 s + 1 1 α + s + 2 ) .

The proof is done. □

Theorem 3.2 Let f : [ 0 , b ] R be a differentiable mapping and 1 < q < . If | f | q is measurable and | f | q is decreasing and geometric-arithmetically s-convex on [ 0 , b ] for some fixed α ( 0 , ) , s ( 0 , 1 ] , 0 a < b , then the following inequality for fractional integrals holds:
| f ( a ) + f ( b ) 2 Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 max { 1 2 1 α , 2 1 α 1 } 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q .

Proof To achieve our aim, we divide our proof into two cases.

Case 1: α ( 0 , 1 ) . By using Definition 2.2, Lemma 2.4, Lemma 2.5, Hölder’s inequality and Lemma 2.6, we have
| f ( a ) + f ( b ) 2 Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 2 0 1 | 1 ( 1 t ) α + 1 t α + 1 α + 1 | | f ( t a + ( 1 t ) b ) | d t ( b a ) 2 2 ( α + 1 ) ( 0 1 | 1 ( 1 t ) α t α | p d t ) 1 p ( 0 1 | f ( t a + ( 1 t ) b ) | q d t ) 1 q ( b a ) 2 2 ( α + 1 ) ( 0 1 | 1 ( 1 t ) α t α | p d t ) 1 p ( 0 1 | f ( a t b 1 t ) | q d t ) 1 q ( b a ) 2 2 ( α + 1 ) ( 0 1 | 1 ( 1 t ) α t α | p d t ) 1 p ( 0 1 [ t s | f ( a ) | q + ( 1 t ) s | f ( b ) | q ] d t ) 1 q ( b a ) 2 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 [ ( 1 t ) α + t α 1 ] p d t ) 1 p ( b a ) 2 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 [ 2 1 α 1 ] p d t ) 1 p ( b a ) 2 ( 2 1 α 1 ) 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ,

where 1 p + 1 q = 1 .

Case 2: α [ 1 , ) . By using Definition 2.2, Lemma 2.4, Lemma 2.5, Hölder’s inequality and Lemma 2.6, we have
| f ( a ) + f ( b ) 2 Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 [ 1 ( 1 t ) α t α ] p d t ) 1 p ( b a ) 2 ( 1 2 1 α ) 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q .

The proof is done. □

4 The second main results

By using Lemma 2.7, we can obtain the main results in this section.

Theorem 4.1 Let f : [ 0 , b ] R be a differentiable mapping. If | f | is measurable and | f | is decreasing geometric-arithmetically s-convex functions on [ 0 , b ] for some fixed α ( 0 , ) , s ( 0 , 1 ] , 0 a < b , then the following inequality for fractional integrals holds:
| Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] f ( a + b 2 ) | ( b a ) 2 | f ( a ) | 2 ( α + 1 ) [ α α 2 s 1 2 s 1 1 + s α + 1 2 + s + 2 B ( s + 1 , α + 2 ) + 1 α + s + 2 ] + ( b a ) 2 | f ( b ) | 2 ( α + 1 ) [ α 2 s 1 + 2 s 1 1 1 + s + 1 α + s + 2 + 2 B ( α + 2 , s + 1 ) ] ,
where
0 1 t s ( 1 t ) α + 1 d t = B ( s + 1 , α + 2 ) .
Proof By using Definition 2.2, Lemma 2.5 and Lemma 2.7, we have
| Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] f ( a + b 2 ) | ( b a ) 2 2 0 1 | m ( t ) | | f ( t a + ( 1 t ) b ) | d t ( b a ) 2 2 0 1 | m ( t ) | | f ( a t b 1 t ) | d t ( b a ) 2 2 0 1 | m ( t ) | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) 2 2 1 2 1 | 1 t 1 ( 1 t ) α + 1 t α + 1 α + 1 | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + ( b a ) 2 2 0 1 2 | t 1 ( 1 t ) α + 1 t α + 1 α + 1 | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) 2 2 ( α + 1 ) 1 2 1 | α t α t + ( 1 t ) α + 1 + t α + 1 | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + ( b a ) 2 2 ( α + 1 ) 0 1 2 | t α + t 1 + ( 1 t ) α + 1 + t α + 1 | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) 2 2 ( α + 1 ) | f ( a ) | 1 2 1 [ α t s ( α + 1 ) t s + 1 + t s ( 1 t ) α + 1 + t α + s + 1 ] d t + ( b a ) 2 2 ( α + 1 ) | f ( b ) | 1 2 1 [ α ( 1 t ) s ( α + 1 ) t ( 1 t ) s + ( 1 t ) α + s + 1 + t α + 1 ( 1 t ) s ] d t + ( b a ) 2 2 ( α + 1 ) | f ( a ) | 0 1 2 [ t s + ( α + 1 ) t s + 1 + t s ( 1 t ) α + 1 + t α + s + 1 ] d t + ( b a ) 2 2 ( α + 1 ) | f ( b ) | 0 1 2 [ ( 1 t ) s + ( α + 1 ) t ( 1 t ) s + ( 1 t ) α + s + 1 + t α + 1 ( 1 t ) s ] d t ( b a ) 2 2 ( α + 1 ) | f ( a ) | [ α 1 2 s 1 1 + s ( α + 1 ) 1 2 s 2 2 + s + B ( s + 1 , α + 2 ) + 1 2 α s 2 α + s + 2 ] + ( b a ) 2 2 ( α + 1 ) | f ( b ) | [ α 2 s 1 s + 1 ( α + 1 ) B ( 2 , s + 1 ) + 2 α s 2 α + s + 2 + B ( α + 2 , s + 1 ) ] + ( b a ) 2 2 ( α + 1 ) | f ( a ) | [ 2 s 1 s + 1 + ( α + 1 ) 2 s 2 s + 2 + B ( α + 2 , s + 1 ) + 2 α s 2 α + s + 2 ] + ( b a ) 2 2 ( α + 1 ) | f ( b ) | [ 1 2 s 1 1 + s + ( α + 1 ) B ( 2 , s + 1 ) + 1 2 α s 2 α + s + 2 + B ( α + 2 , s + 1 ) ] ( b a ) 2 | f ( a ) | 2 ( α + 1 ) [ α α 2 s 1 2 s 1 1 + s α + 1 2 + s + 2 B ( s + 1 , α + 2 ) + 1 α + s + 2 ] + ( b a ) 2 | f ( b ) | 2 ( α + 1 ) [ α 2 s 1 + 2 s 1 1 1 + s + 1 α + s + 2 + 2 B ( α + 2 , s + 1 ) ] .

The proof is done. □

Theorem 4.2 Let f : [ 0 , b ] R be a differentiable mapping. | f | is measurable and 1 < q < . If | f | q is decreasing and geometric-arithmetically s-convex on [ 0 , b ] for some fixed α ( 0 , ) , s ( 0 , 1 ] , 0 a < b , then the following inequality for fractional integrals holds:
Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] f ( a + b 2 ) ( b a ) 2 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( ( α + 1 ) 2 p 1 + ( α + 0.5 ) p + 1 α p + 1 p + 1 ) 1 p ,

where 1 p + 1 q = 1 .

Proof By using Definition 2.2, Lemma 2.5, Hölder’s inequality and Lemma 2.7, we have
| Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] f ( a + b 2 ) | ( b a ) 2 2 0 1 | m ( t ) | | f ( t a + ( 1 t ) b ) | d t ( b a ) 2 2 ( 0 1 | m ( t ) | p d t ) 1 p ( 0 1 | f ( t a + ( 1 t ) b ) | q d t ) 1 q ( b a ) 2 2 ( 0 1 | m ( t ) | p d t ) 1 p ( 0 1 | f ( a t b 1 t ) | q d t ) 1 q ( b a ) 2 2 ( 0 1 | m ( t ) | p d t ) 1 p ( 0 1 [ t s | f ( a ) | q + ( 1 t ) s | f ( b ) | q ] d t ) 1 q ( b a ) 2 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 | m ( t ) | p d t ) 1 p ( b a ) 2 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 2 | t 1 ( 1 t ) α + 1 t α + 1 α + 1 | p d t + 1 2 1 | 1 t 1 ( 1 t ) α + 1 t α + 1 α + 1 | p d t ) 1 p ( b a ) 2 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 2 | α t + t 1 + ( 1 t ) α + 1 + t α + 1 | p d t + 1 2 1 | α t + ( 1 t ) α + 1 + t α + 1 | p d t ) 1 p ( b a ) 2 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( ( α + 1 ) 0 1 2 t p d t + 1 2 1 ( α t + 1 ) p d t ) 1 p ( b a ) 2 2 ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( ( α + 1 ) 2 p 1 + ( α + 0.5 ) p + 1 α p + 1 p + 1 ) 1 p .

The proof is done. □

5 The third main results

By using Lemma 2.8, we can obtain the main results in this section.

Theorem 5.1 Let f : [ 0 , b ] R be a differentiable mapping. If | f | is measurable and | f | is decreasing and geometric-arithmetically s-convex on [ 0 , b ] for some fixed α ( 0 , ) , s ( 0 , 1 ] , 0 a < b , then the following inequality for fractional integrals holds:
| f ( a ) + f ( b ) r ( r + 1 ) + 2 r + 1 f ( a + b 2 ) Γ ( α + 1 ) r ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 r ( r + 1 ) ( α + 1 ) max { [ r + 1 ( r + 1 ) 2 α ] [ 2 s 1 | f ( a ) | s + 1 + ( 1 2 s 1 ) | f ( b ) | s + 1 ] r ( α + 1 ) [ 2 s 2 | f ( a ) | s + 2 + B 0.5 ( 2 , s + 1 ) | f ( b ) | ] , r ( α + 1 ) [ 2 s 2 | f ( a ) | s + 2 + B 0.5 ( 2 , s + 1 ) | f ( b ) | ] } + ( b a ) 2 r ( r + 1 ) ( α + 1 ) max { [ r + 1 ( r + 1 ) 2 α r ( α + 1 ) ] × [ ( 1 2 s 1 ) | f ( a ) | s + 1 2 s 1 | f ( b ) | s + 1 ] + r ( α + 1 ) [ ( 1 2 s 2 ) | f ( a ) | s + 2 + B 0.5 ( s + 1 , 2 ) | f ( b ) | ] , r ( α + 1 ) [ ( 1 2 s 1 ) | f ( a ) | 2 s 1 | f ( b ) | s + 1 ( 1 2 s 2 ) | f ( a ) | s + 2 B 0.5 ( s + 1 , 2 ) | f ( b ) | ] } .
Proof By using Definition 2.2, Definition 2.3, Lemma 2.5 and Lemma 2.8, we have
| f ( a ) + f ( b ) r ( r + 1 ) + 2 r + 1 f ( a + b 2 ) Γ ( α + 1 ) r ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 0 1 | k ( t ) | | f ( t a + ( 1 t ) b ) | d t ( b a ) 2 0 1 | k ( t ) | | f ( a t b 1 t ) | d t ( b a ) 2 0 1 | k ( t ) | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) 2 0 1 2 | 1 ( 1 t ) α + 1 t α + 1 r ( α + 1 ) t r + 1 | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + ( b a ) 2 1 2 1 | 1 ( 1 t ) α + 1 t α + 1 r ( α + 1 ) 1 t r + 1 | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) 2 r ( r + 1 ) ( α + 1 ) 0 1 2 | r + 1 ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) tr ( α + 1 ) | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + ( b a ) 2 r ( r + 1 ) ( α + 1 ) 1 2 1 | r + 1 + tr ( α + 1 ) ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) r ( α + 1 ) | [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) 2 r ( r + 1 ) ( α + 1 ) max { [ r + 1 ( r + 1 ) 2 α ] [ 2 s 1 | f ( a ) | s + 1 + ( 1 2 s 1 ) | f ( b ) | s + 1 ] r ( α + 1 ) [ 2 s 2 | f ( a ) | s + 2 + B 0.5 ( 2 , s + 1 ) | f ( b ) | ] , r ( α + 1 ) [ 2 s 2 | f ( a ) | s + 2 + B 0.5 ( 2 , s + 1 ) | f ( b ) | ] } + ( b a ) 2 r ( r + 1 ) ( α + 1 ) max { [ r + 1 ( r + 1 ) 2 α r ( α + 1 ) ] × [ ( 1 2 s 1 ) | f ( a ) | s + 1 2 s 1 | f ( b ) | s + 1 ] + r ( α + 1 ) [ ( 1 2 s 2 ) | f ( a ) | s + 2 + B 0.5 ( s + 1 , 2 ) | f ( b ) | ] , r ( α + 1 ) [ ( 1 2 s 1 ) | f ( a ) | 2 s 1 | f ( b ) | s + 1 ( 1 2 s 2 ) | f ( a ) | s + 2 B 0.5 ( s + 1 , 2 ) | f ( b ) | ] } ,
where we have used the following inequalities:
0 1 2 [ r + 1 ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) tr ( α + 1 ) ] [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t [ r + 1 ( r + 1 ) 2 α ] 0 1 2 [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t r ( α + 1 ) 0 1 2 [ t s + 1 | f ( a ) | + t ( 1 t ) s | f ( b ) | ] d t [ r + 1 ( r + 1 ) 2 α ] [ 2 s 1 | f ( a ) | s + 1 + ( 1 2 s 1 ) | f ( b ) | s + 1 ] r ( α + 1 ) [ 2 s 2 | f ( a ) | s + 2 + B 0.5 ( 2 , s + 1 ) | f ( b ) | ] ,
and
0 1 2 [ r 1 + ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) + tr ( α + 1 ) ] [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t [ r 1 + ( r + 1 ) ] 0 1 2 [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + r ( α + 1 ) 0 1 2 [ t s + 1 | f ( a ) | + t ( 1 t ) s | f ( b ) | ] d t r ( α + 1 ) [ 2 s 2 | f ( a ) | s + 2 + B 0.5 ( 2 , s + 1 ) | f ( b ) | ] ,
and
1 2 1 [ r + 1 + tr ( α + 1 ) ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) r ( α + 1 ) ] [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t [ r + 1 ( r + 1 ) 2 α r ( α + 1 ) ] 1 2 1 [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + r ( α + 1 ) 1 2 1 [ t s + 1 | f ( a ) | + t ( 1 t ) s | f ( b ) | ] d t [ r + 1 ( r + 1 ) 2 α r ( α + 1 ) ] [ ( 1 2 s 1 ) | f ( a ) | s + 1 2 s 1 | f ( b ) | s + 1 ] + r ( α + 1 ) [ ( 1 2 s 2 ) | f ( a ) | s + 2 + B 0.5 ( s + 1 , 2 ) | f ( b ) | ] ,
and
1 2 1 [ r 1 tr ( α + 1 ) + ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) + r ( α + 1 ) ] [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t [ r 1 + ( r + 1 ) + r ( α + 1 ) ] 1 2 1 [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t r ( α + 1 ) 1 2 1 [ t s + 1 | f ( a ) | + t ( 1 t ) s | f ( b ) | ] d t r ( α + 1 ) [ ( 1 2 s 1 ) | f ( a ) | s + 1 2 s 1 | f ( b ) | s + 1 ] r ( α + 1 ) [ ( 1 2 s 2 ) | f ( a ) | s + 2 + B 0.5 ( s + 1 , 2 ) | f ( b ) | ] r ( α + 1 ) [ ( 1 2 s 1 ) | f ( a ) | 2 s 1 | f ( b ) | s + 1 ( 1 2 s 2 ) | f ( a ) | s + 2 B 0.5 ( s + 1 , 2 ) | f ( b ) | ] .

The proof is done. □

Theorem 5.2 Let f : [ 0 , b ] R be a differentiable mapping. | f | is measurable and 1 < q < . If | f | q is decreasing and geometric-arithmetically s-convex on [ 0 , b ] for some fixed α ( 0 , ) , s ( 0 , 1 ] , 0 a < b , then the following inequality for fractional integrals holds:
| f ( a ) + f ( b ) r ( r + 1 ) + 2 r + 1 f ( a + b 2 ) Γ ( α + 1 ) r ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 [ r ( r + 1 ) ( α + 1 ) ] 1 + p 1 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q × ( max { [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) ( 1 + r ) 2 α ] p + 1 , [ r ( α + 1 ) ] p + 1 2 p 1 } + max { [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) + ( 1 + r ) 2 α ] p + 1 , [ 0.5 r ( α + 1 ) ] p + 1 } ) 1 p ,

where 1 p + 1 q = 1 .

Proof By using Definition 2.2, Lemma 2.5, Hölder’s inequality and Lemma 2.8, we have
| f ( a ) + f ( b ) r ( r + 1 ) + 2 r + 1 f ( a + b 2 ) Γ ( α + 1 ) r ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] | ( b a ) 2 0 1 | k ( t ) | | f ( t a + ( 1 t ) b ) | d t ( b a ) 2 ( 0 1 | k ( t ) | p d t ) 1 p ( 0 1 | f ( t a + ( 1 t ) b ) | q d t ) 1 q ( b a ) 2 ( 0 1 | k ( t ) | p d t ) 1 p ( 0 1 | f ( a t b 1 t ) | q d t ) 1 q ( b a ) 2 ( 0 1 | k ( t ) | p d t ) 1 p ( 0 1 [ t s | f ( a ) | q + ( 1 t ) s | f ( b ) | q ] d t ) 1 q ( b a ) 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 | k ( t ) | p d t ) 1 p ( b a ) 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 2 | 1 ( 1 t ) α + 1 t α + 1 r ( α + 1 ) t r + 1 | p d t + 1 2 1 | 1 ( 1 t ) α + 1 t α + 1 r ( α + 1 ) 1 t r + 1 | p d t ) 1 p ( b a ) 2 r ( r + 1 ) ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 2 | r + 1 ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) tr ( α + 1 ) | p d t + 1 2 1 | r + 1 + tr ( α + 1 ) ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) r ( α + 1 ) | p d t ) 1 p ( b a ) 2 r ( r + 1 ) ( α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q × ( max { [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) ( 1 + r ) 2 α ] p + 1 r ( α + 1 ) ( p + 1 ) , [ r ( α + 1 ) ] p + 1 2 p 1 r ( α + 1 ) ( p + 1 ) } + max { [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) + ( 1 + r ) 2 α ] p + 1 r ( α + 1 ) ( p + 1 ) , [ 0.5 r ( α + 1 ) ] p + 1 r ( α + 1 ) ( p + 1 ) } ) 1 p ( b a ) 2 [ r ( r + 1 ) ( α + 1 ) ] 1 + p 1 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q × ( max { [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) ( 1 + r ) 2 α ] p + 1 , [ r ( α + 1 ) ] p + 1 2 p 1 } + max { [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) + ( 1 + r ) 2 α ] p + 1 , [ 0.5 r ( α + 1 ) ] p + 1 } ) 1 p ,
where we have used the following inequalities:
0 1 2 [ r + 1 ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) tr ( α + 1 ) ] p d t [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) ( 1 + r ) 2 α ] p + 1 r ( α + 1 ) ( p + 1 ) ,
and
0 1 2 [ r 1 + ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) + tr ( α + 1 ) ] p d t [ r ( α + 1 ) ] p + 1 2 p 1 r ( α + 1 ) ( p + 1 ) ,
and
1 2 1 [ r + 1 + tr ( α + 1 ) ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) r ( α + 1 ) ] p d t [ r + 1 ( r + 1 ) 2 α ] p + 1 [ 1 + 0.5 r ( 1 α ) + ( 1 + r ) 2 α ] p + 1 r ( α + 1 ) ( p + 1 ) ,
and
1 2 1 [ r 1 tr ( α + 1 ) + ( r + 1 ) ( t α + 1 + ( 1 t ) α + 1 ) + r ( α + 1 ) ] p d t [ 0.5 r ( α + 1 ) ] p + 1 r ( α + 1 ) ( p + 1 ) .

The proof is done. □

Declarations

Acknowledgements

This work is supported by the National Natural Science Foundation of China (11201091), Key Projects of Science and Technology Research in the Chinese Ministry of Education (211169), Key Support Subject (Applied Mathematics) and Key project on the reforms of teaching contents and course system of Guizhou Normal College.

Authors’ Affiliations

(1)
School of Mathematics and Computer Science, Guizhou Normal College
(2)
Department of Mathematics, Guizhou University

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