Journal of Function Spaces

Volume 2018, Article ID 5357463, 12 pages

https://doi.org/10.1155/2018/5357463

## Generalization of Hermite-Hadamard Type Inequalities via Conformable Fractional Integrals

^{1}College of Science, Hunan City University, Yiyang 413000, China^{2}Department of Mathematics, University of Peshawar, Peshawar 25000, Pakistan^{3}Department of Mathematics, College of Science, China Three Gorges University, Yichang 443002, China^{4}Department of Mathematics, Huzhou University, Huzhou 313000, China

Correspondence should be addressed to Yu-Ming Chu; nc.ude.uhjz@gnimuyuhc

Received 20 May 2018; Revised 6 July 2018; Accepted 24 July 2018; Published 5 August 2018

Academic Editor: Lishan Liu

Copyright © 2018 Muhammad Adil Khan et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

#### Abstract

We establish a Hermite-Hadamard type identity and several new Hermite-Hadamard type inequalities for conformable fractional integrals and present their applications to special bivariate means.

#### 1. Introduction

In the field of nonlinear programming and optimization theory, no one can ignore the role of convex sets and convex functions. For the class of convex functions, many inequalities have been introduced such as Jensen’s, Hermite-Hadmard, and Slater’s inequalities. Among those inequalities, the most famous and important inequality is the Hermite-Hadamard’s inequality [1] which can be stated as follows.

Let be an interval and be a convex function defined on . Then the double inequalityholds for all with . Both inequalities in (1) hold in the reverse direction if the function is concave on .

In the last 60 years, many efforts have gone on generalizations, extensions, variants, and applications for the Hermite-Hadamard’s inequality (see [2–13]). Anderson [14] and Sarikaya et al. [15] provide the important variants for the Hermite-Hadamard’s inequality.

Recently, the author in [16] gave a new definition for the (conformable) fractional derivative as follows.

Let and be a real-valued function. Then the -order (conformable) fractional derivative of at is defined by is said to be -differentiable if the -order (conformable) fractional derivative of exists, and the -order (conformable) fractional derivative of at is defined as .

Now we discuss some theorems for the (conformable) fractional derivative.

Theorem 1. *Let and be -differentiable at . Then one has the following:*(i)* for all .*(ii)* for all constants .*(iii)* for all constants .*(iv)*.*(v)*(vi)** if is differentiable at .** In addition,if is differentiable.*

*Definition 2 (conformable fractional integral). *Let and . Then the function is said to be -fractional integrable on if the integralexists and is finite. All -fractional integrable functions on are indicated by

*Remark 3. * where the integral is the usual Riemann improper integral and

*Recently, the conformable integrals and derivatives have attracted the attention of many researchers, and many remarkable properties and inequalities for the conformable integrals and derivatives can be found in the literature [17–24]. Anderson [14] found the conformable integral version of the Hermite-Hadamard inequality as follows.*

*Theorem 4 (see [14]). If and is an -fractional differentiable function such that is increasing, then we have the following inequality:Moreover if the function is decreasing on , then we haveIf , then inequalities (6) and (7) reduce to the classical Hermite-Hadamard’s inequalities.*

*The main purpose of the article is to present the conformable fractional integrals version of the Hermite-Hadamard’s inequality. We first establish an identity for the conformable fractional integrals (Lemma 5) and discuss their special cases. Then applying Jensen’s inequality, power mean inequality, Hölder inequality, the convexity of the functions and , and the identity given by Lemma 5, we obtain inequalities for conformable fractional integrals version of the Hermite-Hadamard’s inequality. At last, using particular classes of convex functions we find several new inequalities for some special bivariate means. For some related results, see [25, 26].*

*2. Main Results*

*The main results of our work can be calculated with the help of the following lemma associated with inequality (8).*

*Lemma 5. Let with , , and be an -fractional differentiable function on . Then the identityholds for any if .*

*Proof. *It follows from Theorem 1, Definition 2, and integrating by parts that where we have used the changes of variable and to get the desired result.

*Remark 6. *Let . Then identity (8) becomes which was proved by Kavurmaci et al. in [2].

*Theorem 7. Let with , , and be an -differentiable function on . Then the inequality holds for any if and is convex on .*

*Proof. *Let , , and . Then we clearly see that both the functions and are convex. From Lemma 5 and the convexity of , , and , we haveFrom the final upper bound above, we have the following:

*Remark 8. *Let . Then inequality (12) leads to which was proved by Kavurmaci et al. in [2].

*Theorem 9. Let with , , such that and be an -differentiable function on . Then the inequalityholds for any if and is convex on , where *

*Proof. *It follows from inequality (13) that Making use of Hölder’s inequality, one hasSimilarly, we have Hence, we have the result in (16).

*Remark 10. *By putting in (16), we obtain the inequality which was proved by Kavurmaci et al. in [2].

*Theorem 11. Let with , , and be an -differentiable function on . Then the inequalityholds for any if and is convex on , where *

*Proof. *It follows from inequality (13) that Making use of the power mean inequality, we get Similarly, we have From the convexity of , we have and where we have also used the facts that Hence, we have the result in (22).

*Remark 12. *If , then inequality (22) becomes which can be found in [2].

*Theorem 13. Let with , , and be an -differentiable function on . Then the inequalityis valid for any if and is convex on , where *

*Proof. *From Theorem 1, Definition 2, and Lemma 5, we get Making use of power mean inequality, we get Similarly, we have It follows from the convexity of that and