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Article

Generalized Integral Inequalities for Hermite–Hadamard-Type Inequalities via s-Convexity on Fractal Sets

by 1,† and 2,*,†
1
Department of Mathematics, University of Hafr Al-Batin, Hafr Al-Batin 31991, Saudi Arabia
2
Department of Mathematics and Institute for Mathematical Research, University Putra Malaysia, Serdang 43400, Malaysia
*
Author to whom correspondence should be addressed.
These authors contributed equally to this work.
Mathematics 2019, 7(11), 1065; https://doi.org/10.3390/math7111065
Received: 23 September 2019 / Revised: 31 October 2019 / Accepted: 1 November 2019 / Published: 6 November 2019

Abstract

In this article, we establish new Hermite–Hadamard-type inequalities via Riemann–Liouville integrals of a function ψ taking its value in a fractal subset of R and possessing an appropriate generalized s-convexity property. It is shown that these fractal inequalities give rise to a generalized s-convexity property of ψ . We also prove certain inequalities involving Riemann–Liouville integrals of a function ψ provided that the absolute value of the first or second order derivative of ψ possesses an appropriate fractal s-convexity property.
Keywords: s-convex function; Hermite–Hadamard inequalities; Riemann–Liouville fractional integrals; fractal space s-convex function; Hermite–Hadamard inequalities; Riemann–Liouville fractional integrals; fractal space

1. Introduction

Convexity is considered to be an important property in mathematical analysis. The applications of convex functions can be found in many fields of studies including economics, engineering and optimization (see for example [1,2]). A well-known result which was identified as Hermite–Hadamard inequalities is the reformulation through convexity. These inequalities, widely reported in the literature, can be defined as follows:
Theorem 1.
Suppose that ψ : [ u , v ] R R is a convex function on [ u , v ] with u < v , then
ψ u + v 2 1 v u u v ψ ( x ) d x ψ ( u ) + ψ ( v ) 2 .
These two inequalities, which are refinement of convexity, can be held in reverse order as concave. Following this, many refinements of convex functions using Hermite–Hadamard inequalities have been continuously studied [3,4,5,6]. Given the variation of Hermite–Hadamard inequalities, Dragomir and Fitzpatrick [7] established a new generalization of s-convex functions in the second sense.
Theorem 2.
Suppose that ψ : R + R + is a s-convex function in the second sense, where 0 < s 1 , u , v R + and u < v . If ψ L 1 ( [ u , v ] ) , then
2 s 1 ψ u + v 2 1 v u u v ψ ( x ) d x ψ u + ψ v s + 1 .
Though the Hermite–Hadamard inequalities were established for classical integrals [8], the inequalities can also hold for fractional calculus, such as Riemann–Liouville [9,10,11], Katugampola [12] and local fractional integrals [13]. Some of these were studied through Mittag–Leffler function [14,15]. Other important generalizations include the work of Sarikaya et al. [16], who proved the Hermite–Hadamard inequalities through fractional integrals as follows:
Theorem 3.
Suppose that ψ : [ u , v ] R is a non-negative function with 0 u < v and ψ L 1 [ u , v ] . If ψ is convex function on [ u , v ] , we have:
ψ u + v 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) ψ ( u ) + ψ ( v ) 2 ,
where 0 < α 1 .
The s-convexity mentioned in Hudzik and Maligranda [2] was also given as the generalization on fractal sets.
Definition 1
([17]). A function ψ : R + R α is called generalized s-convex in the second sense if
ψ ( γ 1 u + γ 2 v ) γ 1 α s ψ ( u ) + γ 2 α s ψ ( v ) ,
holds for all u , v R + , γ 1 , γ 2 0 , with γ 1 + γ 2 = 1 and for some fixed s ( 0 , 1 ] . The symbol G K s 2 denotes the class of this functions.
The Riemann–Liouville fractional integral is introduced here due to its importance.
Definition 2
([18]). Suppose that ψ L 1 [ u , v ] . The Riemann–Liouville integrals J u + α ψ and J v α ψ of order α R + are defined by
J u + α ψ ( x ) = 1 Γ ( α ) u x ( x γ ) α 1 ψ ( γ ) d γ , x > u ,
and
J v α ψ ( x ) = 1 Γ ( α ) x v ( γ x ) α 1 ψ ( γ ) d γ , x < v ,
respectively.
The following lemma for differentiable function is given by Sarikaya et al. [16].
Lemma 1.
Let ψ : [ u , v ] R be a differentiable function on ( u , v ) with u < v . If ψ L 1 [ u , v ] , then we have:
ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) = v u 2 0 1 ( 1 γ ) α γ α ψ ( γ u + ( 1 γ ) v ) d γ .
Wang et al. [9] extended Lemma 1 to include two cases, one of which involves the second derivative of Riemann–Liouville fractional integrals.
Lemma 2.
Let ψ : [ u , v ] R be a twice-differentiable function on ( u , v ) with u < v . If ψ L 1 [ u , v ] , then
ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) = ( v u ) 2 2 0 1 1 ( 1 γ ) α + 1 γ α + 1 α + 1 ψ ( γ u + ( 1 γ ) v ) d γ ,
holds.
Even though studies were conducted on generalized Hermite–Hadamard inequality via Riemann–Liouville fractional integrals for s-convexity [16,19,20,21], inequalities of this type for generalized s-convexity are lacking. Therefore, this paper is aimed at establishing some new integral inequalities via generalized s-convexity on fractal sets. We show that the newly established inequalities are generalizations of Theorem 2. The new Hermite–Hadamard-type inequalities in the class of functions with derivatives in absolute values are shown to be s-convex function on fractal sets. This was achieved using Riemann–Liouville fractional integrals inequalities.

2. Main Results

Our first main result is obtained in the following theorem.
Theorem 4.
Suppose that ψ : [ u , v ] R + R + α is a generalized s-convex on [ u , v ] , where 0 < s < 1 , u , v R + and u v . If ψ L 1 [ u , v ] , then we obtain
2 α ( s 1 ) ψ u + v 2 Γ ( α + 1 ) 2 ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) [ ψ ( u ) + ψ ( v ) ] 2 .
Proof. 
Since ψ G K s 2 , we get
ψ x + y 2 ψ ( x ) + ψ ( y ) 2 α s .
Substituting x = γ u + ( 1 γ ) v and y = ( 1 γ ) u + γ v with γ [ 0 , 1 ] in inequality (5), we obtain
2 α s ψ x + y 2 ψ ( γ u + ( 1 γ ) v ) + ψ ( γ v + ( 1 γ ) u ) .
Multiplying both sides of (6) by γ α 1 and integrating the resulting inequality with respect to γ over [ 0 , 1 ] yields
2 α s α ψ x + y 2 0 1 γ α 1 ψ ( γ u + ( 1 γ ) v ) + 0 1 γ α 1 ψ ( γ v + ( 1 γ ) u ) = Γ ( α ) ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] .
Then the first inequality in (4) is proved.
To prove the second inequality in (4), since ψ G K s 2 , we get
ψ ( γ u + ( 1 γ ) v ) γ α s ψ ( u ) + ( 1 γ ) α s ψ ( v ) ,
and
ψ ( γ v + ( 1 γ ) u ) γ α s ψ ( v ) + ( 1 γ ) α s ψ ( u ) .
Combining the inequalities (8) and (9), we obtain
ψ ( γ u + ( 1 γ ) v ) + ψ ( ( 1 γ ) u + γ v ) γ α s ψ ( u ) + ( 1 γ ) α s ψ ( v ) + γ α s ψ ( u ) + ( 1 γ ) α s ψ ( u ) = [ γ α s + ( 1 γ ) α s ] [ ψ ( u ) + ψ ( v ) ] .
A similar technique used in (6) is applied to inequality (10) to get the following:
Γ ( α ) ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] 0 1 γ α 1 [ γ α s + ( 1 γ ) α s ] [ ψ ( u ) + ψ ( v ) ] d γ 1 α ( s + 1 ) + Γ ( α s + 1 ) Γ ( α ) Γ ( α ( s + 1 ) + 1 ) [ ψ ( u ) + ψ ( v ) ,
where
0 1 γ α s + s 1 d γ = 1 α s + α ,
and
0 1 γ α 1 ( 1 γ ) α s d γ = Γ ( α s + 1 ) Γ ( α ) Γ ( α s + α + 1 ) .
Using inequalities (7) and (11), we prove Theorem 4. □
Remark 1.
In the second inequality of Theorem 4, the expression 1 2 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) for 0 < s 1 is the best possible. The map ψ : [ 0 , 1 ] 0 α , 1 α given by ψ ( z ) = z s α is generalized s-convex in the second sense, and it satisfies the following equalities:
Γ ( α + 1 ) 2 [ J u + α ψ ( 1 ) + J v α ψ ( 0 ) ] = Γ ( α + 1 ) 2 1 Γ ( α ) 1 α ( s + 1 ) + Γ ( α s + 1 ) Γ ( α ) Γ ( α s + α + 1 ) = Γ ( α + 1 ) 2 1 α Γ ( α ) ( s + 1 ) + Γ ( α s + 1 ) Γ ( α ) Γ ( α ) Γ ( α s + α + 1 ) = 1 2 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) ,
and
1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) [ ψ ( 0 ) + ψ ( 1 ) ] 2 = 1 2 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) .
Corollary 1.
By taking α = 1 in Theorem 4, the inequalities in (2) of Theorem 2 are recovered.
This result is the same as Theorem 2.1 in Dragomir and Fitzpatrick [7].
Remark 2.
The equality
β ( u , v ) = Γ ( u ) Γ ( v ) Γ ( u + v )
implies
2 α s 1 ψ ( u + v 2 ) Γ ( α + 1 ) 2 ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) [ ψ ( u ) + ψ ( v ) ] 2 1 s + 1 + β ( α s + 1 , α + 1 ) ( α ( s + 1 ) + 1 ) × [ ψ ( u ) + ψ ( v ) ] 2 .
Theorem 5.
Suppose that M : [ 0 , 1 ] R α is the mapping given by
M ( γ ) = Γ ( α + 1 ) ( v u ) α J u + α ψ γ v + ( 1 γ ) u + v 2 + J v α ψ γ u + ( 1 γ ) u + v 2 , γ ( 0 , 1 ) ,
where ψ : [ u , v ] R α belongs to G K s 2 , s ( 0 , 1 ] , u , v R + , u < v and ψ L 1 ( [ u , v ] ) . Then
(i) 
M G K s 2 on [ 0 , 1 ] .
(ii) 
We have the following inequality:
M ( γ ) 2 α s ψ u + v 2 .
(iii) 
We have the following inequality:
M ( γ ) min { M 1 ( γ ) , M 2 ( γ ) } , γ [ 0 , 1 ] ,
where
M 1 ( γ ) = γ α s Γ ( α + 1 ) ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] + ( 1 γ ) α s ψ u + v 2 ,
and
M 2 ( γ ) = 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 2 ) ψ γ u + ( 1 γ ) u + v 2 + ψ γ v + ( 1 γ ) u + v 2 .
(iv) 
If M ˜ = max { M 1 ( γ ) , M 2 ( γ ) } , γ [ 0 , 1 ] , then we have
M ˜ 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) γ α s [ ψ ( u ) + ψ ( v ) ] + 2 α s ( 1 γ ) α s ψ u + v 2 .
Proof. 
(i)
Let γ 1 , γ 2 [ 0 , 1 ] and μ 1 , μ 2 0 with μ 1 + μ 2 = 1 , then
M ( μ 1 γ 1 + μ 2 γ 2 ) = Γ ( α + 1 ) ( v u ) α [ J u + α ψ ( ( μ 1 γ 1 + μ 2 γ 2 ) u + v 2 + ( 1 ( μ 1 γ 1 + μ 2 γ 2 ) ) v ) + J v α ψ ( μ 1 γ 1 + μ 2 γ 2 ) u + ( 1 ( μ 1 γ 1 + μ 2 γ 2 ) ) u + v 2 ] Γ ( α + 1 ) ( v u ) α [ μ 1 α s J u + α ψ γ 1 u + v 2 + ( 1 γ 1 ) v + J v α ψ γ 1 u + ( 1 γ 1 ) u + v 2 + μ 2 α s J u + α ψ γ 2 u + v 2 + ( 1 γ 2 ) v + J v α ψ γ 2 u + ( 1 γ 2 ) u + v 2 ] = μ 1 α s M ( γ 1 ) + μ 2 α s M ( γ 2 ) .
(ii)
Assume that γ ( 0 , 1 ] . Then by the change of variables q = γ v + ( 1 γ ) u + v 2 and p = γ u + ( 1 γ ) u + v 2 , we have
M ( γ ) = Γ ( α + 1 ) γ α ( v u ) α [ J ( γ v + ( 1 γ ) u + v 2 ) + α ψ γ u + ( 1 γ ) u + v 2 + J ( γ u + ( 1 γ ) u + v 2 ) α ψ γ v + ( 1 γ ) u + v 2 ] = Γ ( α + 1 ) ( p q ) α [ J q + α ψ ( p ) + J p α ψ ( q ) ] .
Applying the first generalized Hermite–Hadamard inequality, we obtain
Γ ( α + 1 ) ( p q ) α [ J q + α ψ ( p ) + J p α ψ ( q ) ] 2 α s ψ q + p 2 = 2 α s ψ u + v 2 ,
and inequality (12) is obtained.
If γ = 0 , the inequality
ψ u + v 2 2 α s 1 ψ u + v 2 ,
also holds.
(iii)
Applying the second generalized Hermite–Hadamard inequality, we obtain
Γ ( α + 1 ) ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) [ ψ ( u ) + ψ ( v ) ] = 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) [ ψ γ u + ( 1 γ ) u + v 2 + ψ γ v + ( 1 γ ) u + v 2 ] = A 2 ( γ ) , γ [ 0 , 1 ] .
Please note that if γ = 0 , then the inequality
ψ u + v 2 = M ( 0 ) M 2 ( 0 ) = 2 α 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) ψ u + v 2 ,
holds as it is equivalent to
Γ ( α s + α + 1 ) ( s + 1 ) Γ ( α s + 1 ) Γ ( α + 1 ) 2 α ψ u + v 2 0 α ,
which is known to hold for s ( 0 , 1 ] .
Since for all γ [ 0 , 1 ] and x [ u , v ] the inequalities
ψ γ u + ( 1 γ ) u + v 2 γ α s ψ ( u ) + ( 1 γ ) α s ψ u + v 2 ,
and
ψ γ v + ( 1 γ ) u + v 2 γ α s ψ ( v ) + ( 1 γ ) α s ψ u + v 2 .
are true, we obtain
M ( γ ) = Γ ( α + 1 ) ( v u ) α J u + α ψ γ v + ( 1 γ ) u + v 2 + J v α ψ γ u + ( 1 γ ) u + v 2 γ α s Γ ( α + 1 ) ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) + ( 1 γ ) α s ψ u + v 2 = M 1 ( γ )
and the inequality (13) is proved.
(iv)
We have
M 2 ( γ ) = 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 2 ) [ ψ γ u + ( 1 γ ) u + v 2 + ψ γ v + ( 1 γ ) u + v 2 ] 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) [ γ α s ψ ( u ) + ( 1 γ ) α s ψ u + v 2 + γ α s ψ ( v ) + ( 1 γ ) α s ψ u + v 2 ] = 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) γ α s [ ψ ( u ) + ψ ( v ) ] + 2 α s ( 1 γ ) α s ψ u + v 2 .
Since
Γ ( α + 1 ) 2 ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) [ ψ ( u ) + ψ ( v ) ] 2 ,
and
( 1 γ ) α s ψ u + v 2 2 α ( 1 γ ) α s 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) ψ u + v 2 ,
then
M 1 ( γ ) γ α s 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) ( ψ ( u ) + ψ ( v ) ) + 2 α ( 1 γ ) α s 1 s + 1 + Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 1 ) ψ u + v 2 ,
and the proof of Theorem 5 is complete.
 □
Corollary 2.
Choosing s = 1 in Theorem 5, we have
(i) 
Γ ( α + 1 ) ( v u ) α [ J u + α ψ γ v + ( 1 γ ) u + v 2 + J v α ψ γ u + ( 1 γ ) u + v 2 ] min { γ α Γ ( α + 1 ) ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] + ( 1 γ ) α ψ u + v 2 , 1 2 + ( Γ ( α + 1 ) ) 2 Γ ( 2 ( α + 1 ) ) × ψ γ u + ( 1 γ ) u + v 2 + ψ γ v + ( 1 γ ) u + v 2 } .
(ii) 
Since
M ˜ = max { γ α Γ ( α + 1 ) ( v u ) α [ J u + α ψ ( v ) + J v α ψ ( u ) ] + ( 1 γ ) α ψ ( u + v 2 ) , 1 2 + ( Γ ( α + 1 ) ) 2 Γ ( 2 ( α + 1 ) ) ψ ( γ u + ( 1 γ ) u + v 2 ) + ψ γ v + ( 1 γ ) u + v 2 } ,
we get
M ˜ = 1 2 + Γ ( α + 1 ) Γ ( α + 1 ) Γ ( 2 ( α + 1 ) ) γ α ( ψ ( u ) + ψ ( v ) ) + 2 α ( 1 γ ) α ψ u + v 2 .
Theorem 6.
Let ψ : [ u , v ] R + R α be a differentiable function on ( u , v ) where 0 u < v . For some fixed q 1 , if | ψ | q is generalized s-convex on ( u , v ) , we obtain
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | v u 2 2 α + 1 1 1 q × ( β ( α + 1 , α s + 1 ) + 1 α ( s + 1 ) + 1 × [ | ψ ( u ) | q + | ψ ( v ) | q ] ) 1 q .
Proof. 
Applying Lemma 1, we obtain
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | = v u 2 | 0 1 ( 1 γ ) α γ α × ψ ( γ u + ( 1 γ ) v ) d γ | .
First, suppose q = 1 . Since the function | ψ | is generalized s-convex on ( u , v ) , we obtain
| ψ ( γ u + ( 1 γ ) v γ | γ α s | ψ ( u ) | + ( 1 γ ) α s | ψ ( v ) | .
Therefore,
| 0 1 ( 1 γ ) α γ α ψ ( γ u + ( 1 γ ) v ) d γ | | ψ ( u ) | 0 1 [ ( 1 γ ) α γ α s + γ α s + α ] d γ + | ψ ( v ) | 0 1 [ ( 1 γ ) α s + α + ( 1 γ α s ) γ α ] d γ = β ( α + 1 , α s + 1 ) + 1 α s + α + 1 [ ψ ( u ) + ψ ( v ) ] .
Next suppose that q > 1 . From the power mean inequality and the generalized s-convexity of the function | ψ | q we obtain
| 0 1 ( 1 γ ) α γ α ψ ( γ u + ( 1 γ ) v ) d γ | | 0 1 ( 1 γ ) α γ α 1 1 q ( 1 γ ) α γ α 1 q × ψ ( γ u + ( 1 γ ) v ) d γ | 0 1 | ( 1 γ ) α γ α | d γ 1 1 q × 0 1 | ( 1 γ ) α γ α | | ψ ( γ u + ( 1 γ ) v ) | q d γ 1 q 0 1 ( 1 γ ) α + γ α d γ 1 1 q × 0 1 [ ( 1 γ ) α + γ α ] [ γ α s | ψ ( u ) | q + ( 1 γ ) α s | ψ ( v ) | q ] d γ 1 q = 2 α + 1 1 1 q ( β ( α + 1 , α s + 1 ) + 1 α s + α + 1 × [ | ψ ( u ) | q + | ψ ( v ) | q ] ) 1 q .
In view of inequalities (14), (16) and (17) the proof of Theorem 6 is complete now. □
Corollary 3.
Under the conditions of Theorem 6, we get
(i) 
If q = s = 1 , then
ψ u + ψ v 2 Γ ( α + 1 ) 2 v u α J u + α ψ v + J v α ψ u v u 2 1 2 α + 1 + β ( α + 1 , α + 1 ) ψ u + ψ v .
(ii) 
If q = α = s = 1 , then
| ψ ( u ) + ψ ( v ) 2 1 v u u v ψ ( x ) d x | v u 4 ( | ψ ( u ) | + | ψ ( v ) | ) .
(iii) 
If q > 1 and s = 1
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | v u 2 2 α + 1 1 1 q ( β ( α + 1 , α s + 1 ) + 1 α s + α + 1 × [ | ψ ( u ) | q + | ψ ( v ) | q ] ) 1 q .
(iv) 
If q > 1 and α = s = 1 then
| ψ ( u ) + ψ ( v ) 2 1 v u u v ψ ( x ) d x | v u 2 | ψ ( u ) | q + | ψ ( v ) | q 2 1 q .
Theorem 7.
Let ψ : [ u , v ] R + R α be a differentiable function on ( u , v ) where 0 u < v .
If | ψ | q is generalized s-convex on ( u , v ) for q > 1 , we get
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | v u 2 2 α + 1 1 1 2 α 1 1 q × Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α s + α + 2 ) + 1 α s + α + 1 1 q × [ | ψ ( u ) | q + | ψ ( v ) | q ] 1 q .
Proof. 
Since | ψ | is generalized s-convex on ( u , v ) , we obtain
| ψ ( γ u + ( 1 γ ) v γ | γ α s | ψ ( u ) | + ( 1 γ ) α s | ψ ( v ) | .
From this fact and applying the Hölder’s inequality, we have
| 0 1 ( 1 γ ) α γ α ψ ( γ u + ( 1 γ ) v ) d γ | | 0 1 ( 1 γ ) α γ α 1 1 q ( 1 γ ) α γ α 1 q ψ ( γ u + ( 1 γ ) v ) d γ | 0 1 2 [ ( 1 γ ) α γ α ] d γ + 1 2 1 [ γ α ( 1 γ ) α ] d γ 1 1 q × 0 1 | ( 1 γ ) α γ α | | ψ ( γ u + ( 1 γ ) v ) | q d γ 1 q 0 1 2 [ ( 1 γ ) α γ α ] d γ + 1 2 1 [ γ α ( 1 γ ) α ] d γ 1 1 q × 0 1 [ ( 1 γ ) α + γ α ] [ γ α s | ψ ( u ) | q + ( 1 γ ) α s | ψ ( v ) | q ] d γ 1 q = 2 α + 1 1 1 2 α 1 1 q Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α s + α + 2 ) + 1 α s + α + 1 1 q × [ | ψ ( u ) | q + | ψ ( v ) | q ] 1 q .
Thus, the inequalities (14) and (18) complete the proof of Theorem 7. □
Theorem 8.
Let ψ : [ u , v ] R + R α be a differentiable mapping on ( u , v ) with 0 u < v . For some fixed q > 1 , if | ψ | q is generalized s-convex on [ u , v ] , then we have:
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v + α ψ ( u ) | v u 2 2 α p + 1 1 1 2 α p 1 p 1 α s + 1 1 q × | ψ ( u ) | q + | ψ ( v ) | q 1 q .
Proof. 
By applying Hölder’s inequality and (15), we obtain
| 0 1 ( 1 γ ) α γ α ψ ( γ u + ( 1 γ ) v ) d γ | 0 1 | ( 1 γ ) α γ α | p d γ 1 p × 0 1 | ψ ( γ u + ( 1 γ ) v ) | q d γ 1 q 0 1 2 [ ( 1 γ ) α γ α ] p d γ + 1 2 1 [ γ α ( 1 γ ) α ] p d γ 1 p × 0 1 [ γ α s | ψ ( u ) | q + ( 1 γ ) α s | ψ ( v ) | q ] d γ 1 q 2 α p + 1 1 1 2 α p 1 p | ψ ( u ) | q + | ψ ( v ) | q α s + 1 1 q .
Finally, from (14) and (19) we get the desired result. □
Remark 3.
From Theorems 6–8, we obtain the following inequality for q > 1
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | min { S 1 , S 2 , S 3 } ( v u ) 2 [ | ψ ( u ) | q + | ψ ( v ) | q ] 1 q ,
where
  • S 1 = 2 α + 1 1 1 q β ( α + 1 , α s + 1 ) + 1 α ( s + 1 ) + 1 1 q ,
  • S 2 = 2 α + 1 1 1 2 α 1 1 q Γ ( α s + 1 ) Γ ( α + 1 ) Γ ( α ( s + 1 ) + 2 ) + 1 α ( s + 1 ) + 1 1 q ,
  • S 3 = 2 α p + 1 1 1 2 α p 1 p 1 α s + 1 1 q .
Theorem 9.
Let ψ : [ u , v ] R + R α be a twice-differentiable function on ( u , v ) with 0 u < v . If, for some fixed q 1 , the function | ψ | q is generalized s-convex on the interval [ u , v ] , then we have
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | ( v u ) 2 2 ( α + 1 ) α α + 2 1 1 q × 1 α s + 1 β ( α s + 1 , α + 2 ) 1 α s + α + 2 1 q × | ψ ( u ) | q + | ψ ( v ) | q 1 q .
Proof. 
Applying Lemma 2, we have
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | = ( v u ) 2 2 0 1 1 ( 1 γ ) α + 1 γ α + 1 α + 1 | ψ ( γ u + ( 1 γ ) v ) | d γ .
First, suppose that q = 1 . Since the mapping | ψ | is generalized s-convex on ( u , v ) , we obtain
ψ ( γ u + ( 1 γ ) v ) γ α s ψ ( u ) + ( 1 γ ) α s ψ ( v ) .
Therefore,
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | ( v u ) 2 2 0 1 1 ( 1 γ ) α + 1 γ α + 1 α + 1 γ α s | ψ ( u ) | + ( 1 γ ) α s | ψ ( v ) | d γ = ( v u ) 2 2 ( α + 1 ) 1 α s + 1 β ( α s + 1 , α + 2 ) 1 α s + α + 2 ( | ψ ( u ) | + | ψ ( v ) | ,
where
β ( α s + 1 , α + 2 ) = β ( α + 2 , α s + 1 ) .
Secondly, for q > 1 . From Lemma 2 and the power mean inequality, we have
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | ( v u ) 2 2 ( α + 1 ) 0 1 ( 1 ( 1 γ ) α + 1 γ α + 1 ) d γ 1 1 q × 0 1 1 ( 1 γ ) α + 1 γ α + 1 | ψ ( γ u + ( 1 γ ) v ) | q d γ 1 q .
Hence, from inequalities (21) and (22), we obtain
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | ( v u ) 2 2 ( α + 1 ) 0 1 1 ( 1 γ ) α + 1 γ α + 1 1 1 q 0 1 1 ( 1 γ ) α + 1 γ α + 1 | ψ ( γ u + ( 1 γ ) v ) | q d γ 1 q ( v u ) 2 2 ( α + 1 ) 0 1 1 ( 1 γ ) α + 1 γ α + 1 1 1 q × 0 1 1 ( 1 γ ) α + 1 γ α + 1 [ γ α s | ψ ( u ) | q + ( 1 γ ) α s | ψ ( v ) | q ] d γ 1 q ( v u ) 2 2 ( α + 1 ) α α + 2 1 1 q 1 α s + 1 β ( α s + 1 , α + 2 ) 1 α s + α + 2 1 q | ψ ( u ) | q + | ψ ( v ) | q 1 q .
This completes the proof of Theorem 9. □
Theorem 10.
Let 0 u < v < and let the function ψ : [ u , v ] R α be twice-differentiable on the open interval ( u , v ) , and fix s ( 0 , 1 ] and fix q > 1 . If, in addition, the function | ψ | q is generalized s-convex on [ u , v ] , then
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | ( v u ) 2 2 ( α + 1 ) 1 2 p ( α + 1 ) + 1 p × | ψ ( u ) | q + | ψ ( v ) | q α s + 1 1 q ,
where 1 p + 1 q = 1 .
Proof. 
From (20), (21) and the Hölder’s inequality, we have
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J b α ψ ( u ) | ( v u ) 2 2 ( α + 1 ) 0 1 ( 1 ( 1 γ ) α + 1 γ α + 1 ) p d γ 1 p | ψ ( γ u + ( 1 γ ) v ) | q d γ 1 q ( v u ) 2 2 ( α + 1 ) 0 1 ( 1 ( 1 γ ) p ( α + 1 ) γ p ( α + 1 ) ) d γ 1 p | ψ ( u ) | q 0 1 γ α s d γ + | ψ ( v ) | q 0 1 ( 1 γ ) α s d γ 1 q ( v u ) 2 2 ( α + 1 ) 1 2 p ( α + 1 ) + 1 p | ψ ( u ) | q + | ψ ( v ) | q α s + 1 1 q .
We use
1 ( 1 γ ) α + 1 γ α + 1 q 1 ( 1 γ ) q ( α + 1 ) t q ( α + 1 ) ,
for any γ [ 0 , 1 ] , which follows from
( V N ) q V q N q ,
where
V > N 0 and q 1 .
The proof of Theorem 10 is complete now. □
The following result exhibits another Hermite–Hadamard type inequality in terms of the second derivative of a function.
Theorem 11.
Under the same assumptions of Theorem 10, we have
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | ( v u ) 2 2 ( α + 1 ) × ( 1 α s + 1 β ( α s + 1 , q ( α + s ) + 1 ) 1 ( α + 1 ) q + α s + 1 ) 1 q × | ψ ( u ) | q + | ψ ( v ) | q 1 q .
Proof. 
By applying Lemma 2 and the Hölder’s inequality, we obtain
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v b ) α J u + α ψ ( v ) + J v α ψ ( u ) | ( v u ) 2 2 ( α + 1 ) 0 1 1 d γ 1 p 0 1 ( 1 ( 1 γ ) α + 1 γ α + 1 ) q | ψ ( γ u + ( 1 γ ) v ) | q d γ 1 q ( v u ) 2 2 ( α + 1 ) ( | ψ ( u ) | q 0 1 ( γ α s ( 1 γ ) q ( α + s ) γ α s γ q ( α + 1 ) + α s ) d γ + | ψ ( v ) | q 0 1 ( ( 1 γ ) α s ( 1 γ ) q ( α + 1 ) + α s γ q ( α + 1 ) ( 1 γ ) α s ) d γ ) 1 q ( v u ) 2 2 ( α + 1 ) × 1 α s + 1 β ( α s + 1 , 1 + q ( α + s ) ) 1 ( α + 1 ) q + α s + 1 1 q × | ψ ( u ) | q + | ψ ( v ) | q 1 q .
This completes the proof of Theorem 11. □
Remark 4.
From Theorems 9, 10 and 11, we have
| ψ ( u ) + ψ ( v ) 2 Γ ( α + 1 ) 2 ( v u ) α J u + α ψ ( v ) + J v α ψ ( u ) | min { K 1 , K 2 , K 3 } ,
where
  • K 1 = ( v u ) 2 2 ( α + 1 ) α α + 2 1 1 q 1 α s + 1 β ( α s + 1 , α + 2 ) 1 α s + α + 2 1 q | ψ ( u ) | q + | ψ ( v ) | q 1 q ,
  • K 2 = ( v u ) 2 2 ( α + 1 ) 1 2 p ( α + 1 ) + 1 p | ψ ( u ) | q + | ψ ( v ) | q α s + 1 1 q ,
  • K 3 = ( v u ) 2 2 ( α + 1 ) 1 α s + 1 β ( α s + 1 , q ( α + s ) + 1 ) 1 q ( α + 1 ) + α s + 1 1 q | ψ ( u ) | q + | ψ ( v ) | q 1 q .

3. Applications to Special Means

Using the obtained results, we examine some applications to special means of non-negative numbers u and v.
  • The arithmetic mean:
    A = A ( u , v ) = u + v 2 ; u , v R , with u , v > 0 .
  • The logarithmic mean:
    L ( u , v ) = v u log v log u ; u , v R , with u , v > 0 .
  • The generalized logarithmic mean:
    L r ( u , v ) = v r + 1 u r + 1 ( v u ) ( r + 1 ) 1 r ; r Z { 1 , 0 } u , v R , with u , v > 0 .
Using the results obtained in Section 2, and the above applications of means, we get the following proposition.
Proposition 1.
Suppose that r Z , | r | 2 and u , v R such that 0 < u < v . Then we get the following inequality:
| A ( u r , v r ) L r r ( u , v ) | ( v u ) | r | 2 A ( | u | r 1 , | v | r 1 ) .
Proof. 
This result follows Corollary 3 (ii) applied to the function ψ ( x ) = x r . □
Proposition 2.
Suppose that n Z , | r | 2 and u , v R such that 0 < u < v . Then for q 1 , we get the following:
| A ( u r , v r ) L r r ( u , v ) | ( v u ) | r | 2 A 1 q ( | u | q ( r 1 ) , | v | q ( r 1 ) ) .
Proof. 
This result follows from Corollary 3 (iv) applied to the function ψ ( x ) = x r . □
Proposition 3.
Suppose that u , v R such that 0 < u < v , then
| A ( u 1 , v 1 ) L ( u , v ) | ( v u ) | 2 A ( | u | 2 , | v | 2 ) .
Proof. 
This result follows from Corollary 3 (ii) applied to the function ψ ( x ) = x 1 . □
Proposition 4.
Suppose that u , v R such that 0 < u < v , then
| A ( u 1 , v 1 ) L ( u , v ) | ( v u ) | 2 A 1 q ( | u | 2 q , | v | 2 q ) .
Proof. 
This result follows from Corollary 3 (iv) applied to the function ψ ( x ) = x 1 . □

Author Contributions

O.A.: writing—original draft preparation, visualization, A.K.: writing–review and editing, supervision.

Funding

This research received no external funding.

Acknowledgments

The authors would like to thank to referees and editors for their very useful and constructive comments and remarks that improved the present manuscript substantially.

Conflicts of Interest

The authors declare no conflict of interest.

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