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Innovative Applications of Fractional Calculus

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Volume 2021 |Article ID 5516987 | https://doi.org/10.1155/2021/5516987

Hua Wang, Jamroz Khan, Muhammad Adil Khan, Sadia Khalid, Rewayat Khan, "The Hermite–Hadamard–Jensen–Mercer Type Inequalities for Riemann–Liouville Fractional Integral", Journal of Mathematics, vol. 2021, Article ID 5516987, 18 pages, 2021. https://doi.org/10.1155/2021/5516987

The Hermite–Hadamard–Jensen–Mercer Type Inequalities for Riemann–Liouville Fractional Integral

Academic Editor: Ahmet Ocak Akdemir
Received01 Feb 2021
Revised22 Mar 2021
Accepted10 Apr 2021
Published15 May 2021

Abstract

In this paper, we give Hermite–Hadamard type inequalities of the Jensen–Mercer type for Riemann–Liouville fractional integrals. We prove integral identities, and with the help of these identities and some other eminent inequalities, such as Jensen, Hölder, and power mean inequalities, we obtain bounds for the difference of the newly obtained inequalities.

1. Introduction

The concept of convex functions plays a vital role in both pure and applied mathematics. Convex functions also have many applications in other branches of science such as finance, economics, and engineering.

Definition 1. (see [1]). A function is convex iffor all and .
If the inequality in (1) is strict for , then is said to be a strictly convex function, and if is convex, then is said to be a concave function [2, 3].
Many important inequalities such as Jensen, Jensen–Mercer, Hermite–Hadamard, and support line inequalities hold for convex functions. The classical Jensen’s inequality is among the most prominent inequalities stated as follows [4, 5].
If is convex, thenfor all and with .
In [6], Mercer presented a type of Jensen’s inequality called Jensen–Mercer inequality.

Theorem 1. If is convex, thenfor each and with .
For a convex function, there exist at least one line lies on or below the graph of the function.

Definition 2. (see [7]). A function has a support at iffor all and for each . Inequality (4) is said to be the support line inequality.
The following theorem connects the support line inequality with convex functions.

Theorem 2. (see [7]). The function is convex if and only if has at least one line of support at each .

The Hermite–Hadamard inequality is one of the most investigated inequality in the theory of convex functions due to its geometrical significance and applications. Because of the importance of Hermite–Hadamard inequality, there is an ample amount of research work dedicated to the extensions, generalizations, refinements, and applications of the Hermite–Hadamard inequality. The Hermite–Hadamard inequality is given below [8].

Let be a convex function, where is an interval and such that . Then,

If is concave, then (5) holds in the reversed direction. For more results associated with Hermite–Hadamard inequality, see [918].

The Hermite–Hadamard inequality has been extended by means of fractional integral operators. Most popular of them is the Riemann–Liouville fractional operator given in the following definition [1922].

Definition 3. (see [23, 24]). Let be an integrable function defined on . Then, the integrals and defined byare called the left and right Riemann–Liouville fractional integrals of order respectively. Here, represents gamma function defined by .
In [25, 26], authors used the following lemmas to obtain trapezoidal and midpoint type inequalities.

Lemma 1. (see [25]). Let (where is the interior of ) be a differentiable function and such that . If , then

Lemma 2. (see [26]). Let all the assumptions of Lemma 1 hold. Then,In this article, we establish fractional Hermite–Hadamard–Jensen–Mercer type inequalities. We give identities involving fractional integrals, and from these identities, we derive trapezoidal and midpoint type inequalities.

Throughout this article, represents a positive real number.

2. Main Results

We begin this section with our first main result.

Theorem 3. Suppose is a convex function and such that . Then,

Proof. Since is convex, it has support line at each point , that is,for each . Substituting and , where , in inequality (12), we obtainMultiplying (13) with and integrate with respect to , we obtainUsing Mercer’s inequality, we obtainSince is convex, we have and (15) becomesSubstituting in (16), we obtainNow, we prove the second inequality of (10). Put and in (12), we obtainMultiplying the above inequality with and integrating and using and , we obtainPut , and we obtainAdding on both sides of (20), we obtainand on combining (17) and (21), we obtain (10).
Now, we prove the inequalities in (11). Let and (14) becomeNow, we prove the other two inequalities of (11). As is a convex function, we haveMultiplying with and integrating, we obtainBy changing of variable, (24) becomesand on combining (22) and (25), we obtain (11).

Remark 1. If we put in Theorem 3 and in the obtained expressions substitute  =  and , respectively, we obtainrespectively. Inequalities (26) and (27) have been proved in [27].

Remark 2. Substituting , , and in (11), one can obtain Hermite–Hadamard inequality.

3. Bounds for the Difference of Hermite–Hadamard–Jensen–Mercer Type Inequalities

Throughout this section, we consider is a differentiable function. To give the bounds for the difference of Hermite–Hadamard–Jensen–Mercer type inequalities, first, we present the following lemmas.

Lemma 3. Let such that and let . Then,

Proof. Using the techniques of integration, we have

Remark 3. Substituting , , and in (28), we obtain (8).

Lemma 4. Let all the assumptions of Lemma 3 hold. Then,

Proof. Using techniques of integration, we have

Remark 4. If we put , , and in (31), we obtain (9).
We use Lemmas 3 and 4 and obtain bounds for the difference of the inequalities in (11).

Theorem 4. Let be a convex function defined on and let such that . Then,where

Proof. From Lemma 3, we haveSince is convex, using Mercer’s inequality, we obtainequivalent towhereSubstituting these values in (37), we get (33).

Remark 5. If we put , , and in (33), we get the inequality given in Theorem 2.2 of [25].

Theorem 5. Let be a convex function for and let such that . Then,where , , and are the same as defined in Theorem 4.

Proof. From Lemma 3, we have (35). Applying power mean inequality, we obtainSince is convex, using Mercer’s inequality, we haveThis implies thatSubstituting the values of , , , , , and as given in the proof of Theorem 4 in (42), we get (39).

Remark 6. If we put , , and in (39), we obtain the inequality proved in Theorem 1 of [28].
In the following theorem, we derive trapezoidal type inequality.

Theorem 6. Let , and be a convex function, and let such that . Then,where such that .

Proof. Using Lemma 3, we have (35). Applying Hölder’s inequality, we obtain

Remark 7. Substituting , , and in Theorem 6, we obtain Theorem 2.3 of [25].

Theorem 7. Let be a convex function and let such that . Then,where

Proof. Using Lemma 4, we haveSince is convex, using Mercer’s inequality, we obtainwhich implies thatwhere