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

Journal of Inequalities and Applications20132013:443

https://doi.org/10.1186/1029-242X-2013-443

Received: 5 August 2013

Accepted: 23 August 2013

Published: 23 September 2013

Abstract

By using the definition of geometric-arithmetically s-convex functions in (Analysis 33:197-208, 2013) and first-order fractional integral identities in (Math. Comput. Model. 57:2403-2407, 2013; J. Appl. Math. Stat. Inform. 8:21-28, 2012; Comput. Math. Appl. 63:1147-1154, 2012), we present some interesting Riemann-Liouville fractional Hermite-Hadamard inequalities for differentiable geometric-arithmetically s-convex functions. Both beta function and incomplete beta function are used in the desired estimations.

MSC:26A33, 26A51, 26D15.

Keywords

fractional Hermite-Hadamard inequalitiesRiemann-Liouville fractional integralsgeometric-arithmetically s-convex functions

1 Introduction

Fractional integrals and derivatives arise in many engineering and scientific disciplines as the mathematical modeling of systems and processes in the fields of physics, chemistry, aerodynamics, electrodynamics of a complex medium. In particular, the subject of fractional calculus has become a rapidly growing area and has found applications in diverse fields ranging from physical sciences and engineering to biological sciences and economics [18].

Fractional Hermite-Hadamard inequalities involving all kinds of fractional integrals have attracted by many researchers. Many authors paid attention to the study of fractional Hermite-Hadamard inequalities according to the first-order integral equalities and convex functions of different classes. For instance, the readers can refer to [911] for convex functions and [12] for nondecreasing functions, [1315] for m-convex functions and [16] for ( s , m ) -convex functions, [17, 18] for functions satisfying s-e-condition, [19] for ( α , m ) -logarithmically convex functions and the references therein.

Very recently, the authors [20] introduced the new concept of geometric-arithmetically s-convex functions and established some interesting Hermite-Hadamard type inequalities for integer integrals of such functions. However, to our knowledge, fractional Hermite-Hadamard inequalities for geometric-arithmetically s-convex functions have not been reported. Motivated by [9, 10, 13, 20], 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 [20])

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 [19])

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

Lemma 2.5 For t [ 0 , 1 ] , x , y > 0 , we have
t x + ( 1 t ) y y 1 t x t .

Proof If x y , then we consider the function f ( z ) = t z + 1 t z t , z 1 . If y x , then we consider the function g ( z ) = t + ( 1 t ) z z 1 t , z 1 . Clearly f ( ) and g ( ) are increasing functions for all z 1 , then f ( z ) f ( 1 ) = 0 and g ( z ) g ( 1 ) = 0 . So, f ( x y ) 0 and g ( y x ) 0 , i.e., the statement holds. □

We collect the following first-order fractional integrals identities.

Lemma 2.6 ([[9], Lemma 2])

Let f : [ a , b ] R be a 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 0 1 [ ( 1 t ) α t α ] f ( t a + ( 1 t ) b ) d t .

Lemma 2.7 ([[10], Lemma 2.1])

Let f : [ a , b ] R be a differentiable mapping on ( a , b ) . If f L [ a , b ] , then the following equality 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 0 1 [ h ( t ) ( 1 t ) α + t α ] f ( t a + ( 1 t ) b ) d t ,
where
h ( t ) = { 1 , 0 < t < 1 2 , 1 , 1 2 < t < 1 .

Lemma 2.8 ([[13], Lemma 2])

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

3 Main results

In this section, we use Lemmas 2.6, 2.7 and 2.8 via geometric-arithmetically s-convex functions to derive the main results in this paper.

3.1 The first 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 α + s | f ( b ) | | f ( a ) | | f ( b ) | ) ( α + s + 1 ) 2 α + s + 1 + ( b a ) | f ( a ) | [ 0.5 B ( s + 1 , α + 1 ) B 0.5 ( α + 1 , s + 1 ) ] + ( b a ) | f ( b ) | [ B 0.5 ( s + 1 , α + 1 ) 0.5 B ( s + 1 , α + 1 ) ] ,
where
B ( s + 1 , α + 1 ) = 0 1 t s ( 1 t ) α d t ,
and
B 0.5 ( s + 1 , α + 1 ) = 0 0.5 t s ( 1 t ) α d t .
Proof By using Definition 2.2, Definition 2.3, 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 0 1 | ( 1 t ) α t α | | f ( t a + ( 1 t ) b ) | d t b a 2 1 2 1 ( t α ( 1 t ) α ) | f ( t a + ( 1 t ) b ) | d t + b a 2 0 1 2 ( ( 1 t ) α t α ) | f ( t a + ( 1 t ) b ) | d t b a 2 1 2 1 ( t α ( 1 t ) α ) | f ( a t b 1 t ) | d t + b a 2 0 1 2 ( ( 1 t ) α t α ) | f ( a t b 1 t ) | d t b a 2 1 2 1 ( t α ( 1 t ) α ) [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + b a 2 0 1 2 ( ( 1 t ) α t α ) [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) | f ( a ) | 2 1 2 1 t α + s d t ( b a ) | f ( a ) | 2 1 2 1 t s ( 1 t ) α d t ( b a ) | f ( b ) | 2 1 2 1 ( 1 t ) α + s d t + ( b a ) | f ( b ) | 2 1 2 1 t α ( 1 t ) s d t ( b a ) | f ( a ) | 2 0 1 2 t α + s d t + ( b a ) | f ( a ) | 2 0 1 2 t s ( 1 t ) α d t + ( b a ) | f ( b ) | 2 0 1 2 ( 1 t ) α + s d t ( b a ) | f ( b ) | 2 0 1 2 t α ( 1 t ) s d t ( b a ) | f ( a ) | 1 2 1 t α + s d t ( b a ) | f ( a ) | 1 2 1 t s ( 1 t ) α d t ( b a ) | f ( b ) | 1 2 1 ( 1 t ) α + s d t + ( b a ) | f ( b ) | 1 2 1 t α ( 1 t ) s d t ( b a ) | f ( a ) | 0 1 t α + s d t + ( b a ) | f ( a ) | 2 0 1 t s ( 1 t ) α d t + ( b a ) | f ( b ) | 2 0 1 ( 1 t ) α + s d t ( b a ) | f ( b ) | 2 0 1 t α ( 1 t ) s d t ( b a ) | f ( a ) | 1 ( α + s + 1 ) 2 α + s + 1 + ( b a ) | f ( a ) | 1 α + s + 1 ( b a ) | f ( a ) | B 0.5 ( α + 1 , s + 1 ) ( b a ) | f ( b ) | 1 ( α + s + 1 ) 2 α + s + 1 + ( b a ) | f ( b ) | B 0.5 ( s + 1 , α + 1 ) ( b a ) | f ( a ) | 1 α + s + 1 + ( b a ) | f ( a ) | 2 B ( s + 1 , α + 1 ) + ( b a ) | f ( b ) | 2 1 α + s + 1 ( b a ) | f ( b ) | 2 B ( s + 1 , α + 1 ) ( b a ) ( 2 α + s | f ( b ) | | f ( a ) | | f ( b ) | ) ( α + s + 1 ) 2 α + s + 1 + ( b a ) | f ( a ) | [ 0.5 B ( s + 1 , α + 1 ) B 0.5 ( α + 1 , s + 1 ) ] + ( b a ) | f ( b ) | [ B 0.5 ( s + 1 , α + 1 ) 0.5 B ( s + 1 , α + 1 ) ] .

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 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 2 2 1 p α p α + 1 ) 1 p ,

where 1 p + 1 q = 1 .

Proof By using Definition 2.2, 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 0 1 | ( 1 t ) α t α | | f ( t a + ( 1 t ) b ) | d t b a 2 ( 0 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 ( 0 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 ( 0 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 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 | ( 1 t ) α t α | p d t ) 1 p b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 ( t α ( 1 t ) α ) p d t + 0 1 2 ( ( 1 t ) α t α ) p d t ) 1 p b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 ( t p α ( 1 t ) p α ) d t + 0 1 2 ( ( 1 t ) p α t p α ) d t ) 1 p b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 2 2 1 p α p α + 1 ) 1 p .

The proof is done. □

3.2 The second results

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

Theorem 3.3 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:
Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] f ( a + b 2 ) 0.5 ( b a ) ( | f ( a ) | + | f ( b ) | ) × ( B 0.5 ( α + 1 , s + 1 ) B 0.5 ( s + 1 , α + 1 ) + 2 α 1 α + s + 1 + 1 s + 1 ) .
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 0 1 | h ( t ) ( 1 t ) α + t α | | f ( t a + ( 1 t ) b ) | d t b a 2 1 2 1 | 1 ( 1 t ) α + t α | | f ( t a + ( 1 t ) b ) | d t + b a 2 0 1 2 | 1 ( 1 t ) α + t α | | f ( t a + ( 1 t ) b ) | d t b a 2 1 2 1 ( 1 + ( 1 t ) α t α ) | f ( t a + ( 1 t ) b ) | d t + b a 2 0 1 2 ( 1 ( 1 t ) α + t α ) | f ( t a + ( 1 t ) b ) | d t b a 2 1 2 1 ( 1 + ( 1 t ) α t α ) | f ( a t b 1 t ) | d t + b a 2 0 1 2 ( 1 ( 1 t ) α + t α ) | f ( a t b 1 t ) | d t b a 2 1 2 1 ( 1 + ( 1 t ) α t α ) [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t + b a 2 0 1 2 ( 1 ( 1 t ) α + t α ) [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t ( b a ) | f ( a ) | 2 1 2 1 t α + s d t + ( b a ) | f ( a ) | 2 1 2 1 t s d t + ( b a ) | f ( a ) | 2 1 2 1 t s ( 1 t ) α d t + ( b a ) | f ( b ) | 2 1 2 1 ( 1 t ) α + s d t ( b a ) | f ( b ) | 2 1 2 1 t α ( 1 t ) s d t + ( b a ) | f ( b ) | 2 1 2 1 ( 1 t ) s d t + ( b a ) | f ( a ) | 2 0 1 2 t α + s d t ( b a ) | f ( a ) | 2 0 1 2 t s ( 1 t ) α d t + ( b a ) | f ( a ) | 2 0 1 2 t s d t ( b a ) | f ( b ) | 2 0 1 2 ( 1 t ) α + s d t + ( b a ) | f ( b ) | 2 0 1 2 t α ( 1 t ) s d t + ( b a ) | f ( b ) | 2 0 1 2 ( 1 t ) s d t ( b a ) | f ( a ) | 2 1 2 α s 1 α + s + 1 + ( b a ) | f ( a ) | 2 1 2 s 1 s + 1 + ( b a ) | f ( a ) | 2 B 0.5 ( α + 1 , s + 1 ) + ( b a ) | f ( b ) | 2 2 α s 1 α + s + 1 ( b a ) | f ( b ) | 2 B 0.5 ( s + 1 , α + 1 ) + ( b a ) | f ( b ) | 2 2 s 1 s + 1 + ( b a ) | f ( a ) | 2 2 α s 1 α + s + 1 ( b a ) | f ( a ) | 2 B 0.5 ( s + 1 , α + 1 ) + ( b a ) | f ( a ) | 2 2 s 1 s + 1 ( b a ) | f ( b ) | 2 1 2 α s 1 α + s + 1 + ( b a ) | f ( b ) | 2 1 2 s 1 s + 1 + ( b a ) | f ( b ) | 2 B 0.5 ( α + 1 , s + 1 ) ( b a ) | f ( a ) | 2 2 α s 1 α + s + 1 + ( b a ) | f ( a ) | 2 [ B 0.5 ( α + 1 , s + 1 ) B 0.5 ( s + 1 , α + 1 ) ] + ( b a ) | f ( a ) | 2 ( s + 1 ) + ( b a ) | f ( b ) | 2 2 α s 1 α + s + 1 + ( b a ) | f ( b ) | 2 [ B 0.5 ( α + 1 , s + 1 ) B 0.5 ( s + 1 , α + 1 ) ] + ( b a ) | f ( b ) | 2 ( s + 1 ) 0.5 ( b a ) ( | f ( a ) | + | f ( b ) | ) × ( B 0.5 ( α + 1 , s + 1 ) B 0.5 ( s + 1 , α + 1 ) + 2 α 1 α + s + 1 + 1 s + 1 ) .

The proof is done. □

Theorem 3.4 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:
| Γ ( α + 1 ) 2 ( b a ) α [ J a + α f ( b ) + J b α f ( a ) ] f ( a + b 2 ) | max { b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( ( 1 + 2 1 α ) p 2 2 p ( 1 2 p α ) ( p α + 1 ) ) 1 p , ( b a ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 2 p 1 2 p ( 1 2 p α 1 ) p α + 1 ) 1 p } ,

where 1 p + 1 q = 1 .

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 2.7, 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 0 1 | h ( t ) ( 1 t ) α + t α | | f ( t a + ( 1 t ) b ) | d t b a 2 ( 0 1 | h ( t ) ( 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 ( 0 1 | h ( t ) ( 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 ( 0 1 | h ( t ) ( 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 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 0 1 | h ( t ) ( 1 t ) α + t α | p d t ) 1 p b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 ( 1 t α + ( 1 t ) α ) p d t + 0 1 2 ( 1 ( 1 t ) α + t α ) p d t ) 1 p b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 [ 1 t α + 2 1 α t α ] p d t + 0 1 2 [ ( 1 + t α ) ( 1 t α ) ] p d t ) 1 p b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 [ ( 1 + 2 1 α ) p 2 p t p α ] d t + 0 1 2 2 p t p α d t ) 1 p b a 2 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( ( 1 + 2 1 α ) p 2 2 p ( 1 2 p α ) ( p α + 1 ) ) 1 p .
Case 2: α [ 1 , ) . By using Definition 2.2, Lemma 2.4 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 ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 ( 1 t α + ( 1 t ) α ) p d t + 0 1 2 ( 1 ( 1 t ) α + t α ) p d t ) 1 p ( b a ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 ( 1 t α + 1 t α ) p d t ) 1 p ( b a ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 1 2 1 2 p ( 1 t p α ) d t ) 1 p ( b a ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ( 2 p 1 2 p ( 1 2 p α 1 ) p α + 1 ) 1 p .

The proof is done. □

3.3 The third results

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

Theorem 3.5 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 < x < b , then the following inequality for fractional integrals holds:
| ( x a ) α + ( b x ) α b a f ( x ) Γ ( α + 1 ) b a [ J x α f ( a ) + J x + α f ( b ) ] | | ( x a ) α + 1 ( b x ) α + 1 | b a [ | f ( a ) | α + s + 1 + | f ( b ) | B ( α + 1 , s + 1 ) ] ,
where
B ( s + 1 , α + 1 ) = 0 1 t s ( 1 t ) α d t .
Proof By using Definition 2.2, Lemma 2.5 and Lemma 2.8, we have
| ( x a ) α + ( b x ) α b a f ( x ) Γ ( α + 1 ) b a [ J x α f ( a ) + J x + α f ( b ) ] | | ( x a ) α + 1 ( b x ) α + 1 | b a 0 1 t α | f ( t x + ( 1 t ) a ) | d t | ( x a ) α + 1 ( b x ) α + 1 | b a 0 1 t α | f ( x t a 1 t ) | d t | ( x a ) α + 1 ( b x ) α + 1 | b a 0 1 t α [ t s | f ( a ) | + ( 1 t ) s | f ( b ) | ] d t | ( x a ) α + 1 ( b x ) α + 1 | b a 0 1 [ t α + s | f ( a ) | + t α ( 1 t ) s | f ( b ) | ] d t | ( x a ) α + 1 ( b x ) α + 1 | b a [ | f ( a ) | α + s + 1 + | f ( b ) | B ( α + 1 , s + 1 ) ] .

The proof is done. □

Theorem 3.6 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:
| ( x a ) α + ( b x ) α b a f ( x ) Γ ( α + 1 ) b a [ J x α f ( a ) + J x + α f ( b ) ] | b a 2 ( p α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q .
Proof By using Definition 2.2, Lemma 2.5, Hölder’s inequality and Lemma 2.8, we have
| ( x a ) α + ( b x ) α b a f ( x ) Γ ( α + 1 ) b a [ J x α f ( a ) + J x + α f ( b ) ] | | ( x a ) α + 1 ( b x ) α + 1 | b a 0 1 t α | f ( t x + ( 1 t ) a ) | d t b a 2 ( 0 1 t p α d t ) 1 p ( 0 1 | f ( t a + ( 1 t ) b ) | q d t ) 1 q b a 2 ( p α + 1 ) ( 0 1 | f ( t a + ( 1 t ) b ) | q d t ) 1 q b a 2 ( p α + 1 ) ( 0 1 | f ( a t b 1 t ) | q d t ) 1 q b a 2 ( p α + 1 ) ( 0 1 [ t s | f ( a ) | q + ( 1 t ) s | f ( b ) | q ] d t ) 1 q b a 2 ( p α + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q ,

where 1 p + 1 q = 1 . The proof is done. □

4 Applications to special means

Consider the following special means (see Pearce and Pec̆arić [22]) for arbitrary real numbers α, β, α β as follows:
  1. (i)

    H ( α , β ) = 2 1 α + 1 β , α , β R { 0 } ;

     
  2. (ii)

    A ( α , β ) = α + β 2 , α , β R ;

     
  3. (iii)

    L ( α , β ) = β α ln | β | ln | α | , | α | | β | , α β 0 ;

     
  4. (iv)

    L n ( α , β ) = [ β n + 1 α n + 1 ( n + 1 ) ( β α ) ] 1 n , n Z { 1 , 0 } , α , β R , α β .

     

Next, we apply the obtained results to give some applications to special means of real numbers.

Theorem 4.1 For some s ( 0 , 1 ] , n Z { 1 , 0 } , 0 a < b , the following inequality for fractional integrals holds:
| A ( a n , b n ) L n n ( a n , b n ) | n ( b a ) 2 ( a ( n 1 ) q + b ( n 1 ) q s + 1 ) 1 q ( 2 2 1 p p + 1 ) 1 p ,
where
1 p + 1 q = 1 , 1 < q < .

Proof Applying Theorem 3.2 for f ( x ) = x n , α = 1 , one can obtain the result immediately. □

Theorem 4.2 For some s ( 0 , 1 ] , n Z { 1 , 0 } , 0 a < b , the following inequality for fractional integrals holds:
| L n n ( a n , b n ) A n ( a , b ) | ( b a ) ( a ( n 1 ) q + b ( n 1 ) q s + 1 ) 1 q max { ( 2 p 2 2 p 1 0.5 p + 1 ) 1 p , ( 2 p 1 2 p 0.5 p + 1 ) 1 p } ,
where
1 p + 1 q = 1 , 1 < q < .

Proof Applying Theorem 3.4 for f ( x ) = x n , α = 1 , one can obtain the result immediately. □

Theorem 4.3 For some s ( 0 , 1 ] , n Z { 1 , 0 } , 1 < q < , 0 a < x < b , the following inequality for fractional integrals holds:
| L n n ( a n , b n ) x n | b a 2 ( p + 1 ) ( | f ( a ) | q + | f ( b ) | q s + 1 ) 1 q .

Proof Applying Theorem 3.6 for f ( x ) = x n , α = 1 , one can obtain the result immediately. □

Theorem 4.4 For some s ( 0 , 1 ] , 0 a < b , the following inequality for fractional integrals holds:
| 1 H ( a , b ) 1 L ( a , b ) | b a 2 ( a 2 q + b 2 q ( s + 1 ) a 2 q b 2 q ) 1 q ( 2 2 1 p p + 1 ) 1 p ,
where
1 p + 1 q = 1 , 1 < q < .

Proof Applying Theorem 3.2 for f ( x ) = x 1 , α = 1 , one can obtain the result immediately. □

Theorem 4.5 For some s ( 0 , 1 ] , 0 a < b , the following inequality for fractional integrals holds:
| 1 H ( a , b ) 1 A ( a , b ) | ( b a ) ( a 2 q + b 2 q ( s + 1 ) a 2 q b 2 q ) 1 q max { ( ( 1 + 2 1 α ) p 4 2 p 1 ( 1 2 p α ) ( p α + 1 ) ) 1 p , ( 2 p 1 2 p ( 1 2 p α 1 ) p α + 1 ) 1 p } ,
where
1 p + 1 q = 1 , 1 < q < .

Proof Applying Theorem 3.4 for f ( x ) = x 1 , α = 1 , one can obtain the result immediately. □

Theorem 4.6 For some s ( 0 , 1 ] , 0 a < x < b , the following inequality for fractional integrals holds:
| 1 H ( a , b ) 1 x | b a 2 ( p + 1 ) ( a 2 q + b 2 q ( s + 1 ) a 2 q b 2 q ) 1 q ,
where
1 p + 1 q = 1 , 1 < q < .

Proof Applying Theorem 3.6 for f ( x ) = x 1 , α = 1 , one can obtain the result immediately. □

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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