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Article

# New Conformable Fractional Integral Inequalities of Hermite–Hadamard Type for Convex Functions

Department of Mathematics, College of Education, University of Sulaimani, Sulaimani 46001, Kurdistan Region, Iraq
*
Author to whom correspondence should be addressed.
Symmetry 2019, 11(2), 263; https://doi.org/10.3390/sym11020263
Received: 28 December 2018 / Revised: 1 February 2019 / Accepted: 8 February 2019 / Published: 20 February 2019

## Abstract

:
In this work, we established new inequalities of Hermite–Hadamard type for convex functions via conformable fractional integrals. Through the conformable fractional integral inequalities, we found some new inequalities of Hermite–Hadamard type for convex functions in the form of classical integrals.

## 1. Introduction

A function $h : I ⊆ R → R$ is said to be convex on the interval $I$, if the inequality
$h ( ζ x + ( 1 − ζ ) y ) ≤ ζ h ( x ) + ( 1 − ζ ) h ( y )$
holds for all $x , y ∈ I$ and $ζ ∈ [ 0 , 1 ]$. We say that h is concave if $− h$ is convex.
For convex functions, many equalities or inequalities have been established by many authors; for example the Ostrowski type inequality [1], Hardy type inequality [2], Olsen type inequality [3], and Gagliardo–Nirenberg type inequality [4] but the most common and significant inequality is the Hermite–Hadamard type inequality [5,6], which is defined as:
where the function $h : I ⊆ R → R$ is assumed to be a convex function with $u < v$ and $u , v ∈ I$.
A number of mathematicians in the field of applied and pure mathematics have dedicated their efforts to extend, generalize, counterpart, and refine the Hermite–Hadamard inequality (2) for different classes of convex functions and mappings. For more recent results obtained on inequality (2), we refer the reader to References [5,7,8,9,10].
Definition 1
([11]). Suppose that $h ∈ L ( [ u , v ] )$. The left and right Riemann–Liouville fractional integrals $J u + α h$ and $J v − α h$ of order $α > 0$ are defined by
$J u + α h ( x ) = 1 Γ ( α ) ∫ u x ( x − ζ ) α − 1 h ( ζ ) d ζ , x > u ,$
and
$J v − α h ( x ) = 1 Γ ( α ) ∫ x v ( ζ − x ) α − 1 h ( ζ ) d ζ , x < v ,$
respectively, where $Γ ( α )$ is the standard gamma function defined by $Γ ( α ) = ∫ 0 ∞ e − ζ ζ α − 1 d ζ$ and $J u + 0 h ( x ) = J v − 0 h ( x ) = h ( x )$.
As for classical integrals, many Hermite–Hadamard type inequalities have been established for the Riemann–Liouville fractional integrals; for more details and interesting applications see References [6,12,13].
Now, we give the definition of the conformable fractional derivative with its important properties which are useful in order to obtain our main results (see References [14,15,16,17,18,19,20,21]). In our study, we use the Katugampola derivative formulation of conformable derivative which is explained in the following definition:
Definition 2
([20]). Given a function $h : [ 0 , ∞ ) → R$. Then, the conformable fractional derivative of h of order α of h at ζ is defined by
If h is $α$-differentiable in some $( 0 , α ) , α > 0$, $lim ζ → 0 + h ( α ) ( ζ )$ exist, then define
$h ( α ) ( 0 ) = lim ζ → 0 + h ( α ) ( ζ ) .$
Additionally, note that if h is differentiable, then
We can write $h ( α ) ( ζ )$ for $D α ( h ) ( ζ )$ or $d α d α ζ ( h ( ζ ) )$ to denote the conformable fractional derivatives of h of order $α$ at $ζ$. In addition, if the conformable fractional derivative of h of order $α$ exists, then we simply say h is $α$-differentiable.
Theorem 1
([20]). Let $α ∈ ( 0 , 1 ]$ and $h , g$ be α-differentiable at a point $ζ > 0$. Then,
1.
$D α ( u f + v g ) = u D α ( h ) + v D α ( g )$ for all $u , v ∈ R$,
2.
$D α ( h g ) = h D α ( g ) + g D α ( h )$,
3.
,
4.
$D α ( c ) = 0$ for all constant function $h ( ζ ) = c$,
5.
$D α 1 = 0$,
6.
.
Now, we give the definition of conformable fractional integral:
Definition 3
([16]). Let $α ∈ ( 0 , 1 ]$ and $0 ≤ u ≤ v$. We say that a function $h : [ u , v ] → R$ is α-fractional integrable on $[ u , v ]$, if the integral
$∫ u v h ( ζ ) d α ζ = ∫ u v h ( ζ ) ζ α − 1 d ζ$
exists and is finite.
Remark 1.
(u) All α-fractional integrable functions on $[ u , v ]$ are indicated by $L α 1 ( [ u , v ] )$.
(v) For the usual Riemann improper integral and $α ∈ ( 0 , 1 ]$, we have
The aim of our article is to establish some new inequalities connected with the Hermite–Hadamard inequalities (2) via conformable fractional integral.

## 2. Main Results

Our main results depend on the following equality:
Lemma 1.
Let $h : [ u , v ] ⊆ R → R$ be an α-fractional differentiable mapping on $( u , v )$ with $0 ≤ u < v$. If $D α ( h ) ∈ L α 1 ( [ u , v ] )$, then the following identity for conformable fractional integral holds:
$Φ α ( u , v ) = ∑ i = 1 4 δ i ,$
where
and
Proof.
By using the definition of the conformable fractional derivative (4), we have
On integrating by parts, one can have
Using the change of the variable $x : = u ζ + ( 1 − ζ ) 3 u + v 4 , ζ ∈ [ 0 , 1 ]$ and definition of conformable fractional integral (5), we obtain
Similarly, we get
and
Adding $δ 1 , δ 2 , δ 3$ and $δ 4$ together, we obtain the desired identity (7). This completes the proof of Lemma 1. □
Remark 2.
With the similar assumptions of Lemma 1, if $α = 1$, then identity (7) reduces to the following identity:
where
which is obtained by Shi et al. [12].
Theorem 2.
Let $h : [ u , v ] ⊆ R → R$ be an α-fractional differentiable mapping on $( u , v )$ with $0 ≤ u < v$. If $D α ( h ) ∈ L α 1 ( [ u , v ] )$ and is convex on $[ u , v ]$, then the following inequality for conformable fractional integral holds:
where
Proof.
Using Lemma 1 and the property (4), we have
By using the convexity of $x α − 1$ for $x > 0 , α ∈ ( 0 , 1 ]$, we have
and
Using (11), (12), (13), and (14) in (10) and using the properties of modulus, we get
Since $| h ′ |$ is convex on $[ u , v ]$ for any $ζ ∈ [ 0 , 1 ]$, so (17) becomes
Simple calculation gives
This completes the proof of Theorem 2. □
Corollary 1.
With the similar assumptions of Theorem 2, if $α = 1$, then
Theorem 3.
Let $h : [ u , v ] ⊆ R → R$ be an α-fractional differentiable mapping on $( u , v )$ with $0 ≤ u < v$. If $D α ( h ) ∈ L α 1 ( [ u , v ] )$ and is convex on $[ u , v ]$, then the following inequality for conformable fractional integral holds:
where
$A 1 ( α ) = 4 u ( 3 u + v ) α − ( 4 u ) α + 1 4 α ( v − u ) ( α + 1 ) − α ( 3 u + v ) α 4 α , A 2 ( α ) = ( 3 u + v ) α + 2 − ( 4 u ) α + 1 [ α ( v − u ) + ( u + v ) ] 4 α ( v − u ) 2 ( α + 1 ) ( α + 2 ) − ( 3 u + v ) α 2 2 α + 1 , A 3 ( α ) = ( 4 u ) α + 2 − ( 3 u + v ) α + 2 ( 5 u − v ) − α ( 3 u + v ) α ( 3 u − v ) ( u + v ) 4 α ( v − u ) 2 ( α + 1 ) ( α + 2 ) − ( 3 u + v ) α 2 2 α + 1 ,$
$B 1 ( α ) = ( 2 ( u + v ) ) α + 2 − ( 3 u + v ) α + 1 4 α ( v − u ) ( α + 1 ) − ( 3 u + v ) α 4 α , B 2 ( α ) = ( 2 ( u + v ) ) α + 2 − ( 3 u + v ) α + 1 ( u + 3 v ) − α ( 3 u + v ) α + 1 ( v − u ) 4 α ( v − u ) 2 ( α + 1 ) ( α + 2 ) − ( 3 u + v ) α 2 2 α + 1 , B 3 ( α ) = ( 3 u + v ) α + 2 + 4 ( 2 ( u + v ) ) α + 1 + α ( 2 ( u + v ) ) α + 1 ( v − u ) 4 α ( v − u ) 2 ( α + 1 ) ( α + 2 ) − ( 3 u + v ) α 2 2 α + 1 ,$
$D 1 ( α ) = ( 4 v ) α + 1 − ( u + 3 v ) α + 1 4 α ( α + 1 ) ( v − u ) − ( u + 3 v ) α 2 2 α + 1 , D 2 ( α ) = ( 4 v ) α + 2 + ( u + 3 v ) α + 1 ( a − 5 v ) + α ( u + 3 v ) α + 1 ( v − u ) 4 α ( v − u ) 2 ( α + 1 ) ( α + 2 ) , D 3 ( α ) = α ( 4 v ) α + 2 ( v − u ) − 2 ( 4 v ) α + 1 ( u + v ) + ( u + 3 v ) α + 2 4 α ( v − u ) 2 ( α + 1 ) ( α + 2 ) .$
Proof.
Using Lemma 1 and the property (4), we have
It follows from the power–mean inequality that
Since $| h ′ | q$ is convex on $[ u , v ]$ for any $ζ ∈ [ 0 , 1 ]$, we obtain
where we have used the facts that
and
Analogously
and
Using $J 1 , J 2 , J 3$, and $J 4$ in (18), we obtain the desired inequality (17). This completes the proof of Theorem 3. □
Corollary 2.
With the similar assumptions of Theorem 3, if $α = 1$, then
Theorem 4.
Let $h : [ u , v ] ⊆ R → R$ be an α-fractional differentiable mapping on $( u , v )$ with $0 ≤ u < v$. If $D α ( h ) ∈ L α 1 ( [ u , v ] )$ and is concave on $[ u , v ]$, then the following inequality for conformable fractional integral holds:
where $A 1 ( α ) , B 1 ( α ) , C 1 ( α )$, and $D 1 ( α )$ are given in Theorem 3 and
$P 1 ( α ) = 2 1 − α ( 3 u + v ) α + 2 − 2 2 α + 5 − 2 2 α + 1 ( 7 u + 6 v ) ( v − u ) ( 3 u + v ) α − α ( 7 u + v ) ( v − u ) ( 3 u + v ) α 2 2 α + 3 ( v − u ) ( α + 2 ) , P 2 ( α ) = 2 ( 2 ( u + v ) ) α + 2 − 8 ( u + v ) 2 ( 3 u + v ) α − α ( 5 u + 3 v ) ( v − u ) ( 3 u + v ) α 2 2 α + 3 ( v − u ) ( α + 2 ) , P 3 ( α ) = 8 ( u + v ) 2 ( u + 3 v ) α − 4 α ( u + v ) α ( u + 3 v ) α − α ( u + 3 v ) α + 2 − 2 α + 3 ( u + v ) α + 2 2 2 α + 1 ( v − u ) ( α + 2 ) , P 4 ( α ) = 2 2 α − 1 α ( u + 3 v ) + 4 α ( u + 3 v ) 3 + 4 ( 4 v ) α + 2 − 4 α + 2 v 2 ( u + 3 v ) − 4 ( u + 3 v ) α + 2 − 2 2 α + 3 α v 2 ( u + 3 v ) 4 α + 2 ( v − u ) ( α + 2 ) .$
Proof.
By using the power–mean inequality and the concavity of for any $ζ ∈ [ 0 , 1 ]$, we have
This implies that
This means that $| h ′ |$ is also concave. Using inequality (20) in (18) and then applying the Jensen’s integral inequality, we get
where we have used the facts that
Similarly, we get , and .
Using $J 1 , J 2 , J 3$, and $J 4$ in (18), we obtain the required inequality (19). This completes the proof of Theorem 4. □
Corollary 3.
With the similar assumptions of Theorem 4, if $α = 1$, then
where
$A 1 ( 1 ) = u − v 8 , B 1 ( 1 ) = v − u 8 , D 1 ( 1 ) = u − v 8 , D 1 ( 1 ) = v − u 8 ,$
and
$P 1 ( 1 ) = ( u − v ) ( 11 u + v ) 96 , P 2 ( 1 ) = ( v − u ) ( 7 u + 5 v ) 96 , P 3 ( 1 ) = ( u − v ) ( 5 u + 7 v ) 96 , P 4 ( 1 ) = ( v − u ) ( u + 11 v ) 96 .$

## 3. Conclusions

In this work, we have established new conformable fractional integral inequalities of Hermite–Hadamard type for convex functions. As a special case, if we substitute $α = 1$ into the general definition of conformable fractional integrals (Definition 2), we obtain the classical integrals. In view of this, we obtained some new inequalities of Hermite–Hadamard type for convex functions involving classical integrals.

## Author Contributions

The authors contributed equally to this work. All authors read and approved the final manuscript.

## Funding

This research received no external funding.

## Acknowledgments

The authors are thankful to the anonymous referees for their careful corrections to and valuable comments on the original version of this paper.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Mohammed, P.O.; Hamasalh, F.K. New Conformable Fractional Integral Inequalities of Hermite–Hadamard Type for Convex Functions. Symmetry 2019, 11, 263. https://doi.org/10.3390/sym11020263

AMA Style

Mohammed PO, Hamasalh FK. New Conformable Fractional Integral Inequalities of Hermite–Hadamard Type for Convex Functions. Symmetry. 2019; 11(2):263. https://doi.org/10.3390/sym11020263

Chicago/Turabian Style

Mohammed, Pshtiwan Othman, and Faraidun Kadir Hamasalh. 2019. "New Conformable Fractional Integral Inequalities of Hermite–Hadamard Type for Convex Functions" Symmetry 11, no. 2: 263. https://doi.org/10.3390/sym11020263

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