Inequalities for Riemann–Liouville Fractional Integrals of Strongly -Convex Functions

Abstract

The results of this paper provide two Hadamard-type inequalities for strongly -convex functions via Riemann–Liouville fractional integrals and error estimations of well-known fractional Hadamard inequalities. Their special cases are given and connected with the results of some published papers.

1. Introduction

The most prominent inequality for convex functions is the well-known Hadamard inequality stated in the following.

Theorem 1. (see [1]). Let be a convex function on the interval , where with . Then, the following inequality holds:

Convex functions are extended, generalized, and refined in different ways to define new types of convex functions. For instance, -convex, -convex, -convex, strongly convex, and strongly -convex functions are extensions of convex functions. The aim of this paper is to establish integral inequalities by using the class of strongly -convex functions. We give definitions of -convex and strongly -convex functions as follows.

Definition 1. (see [2]). A function is called -convex in the second sense, if the following inequality holds:for every , and .

Definition 2. (see [3]). A function is called strongly -convex in the second sense with modulus , if the following inequality holds:for every , and .

By setting , , and in (2), we get -convex [4], -convex [5], and convex functions, respectively, while by setting , , and in (3), we get strongly -convex [6], strongly -convex [7], and strongly convex [6] functions, respectively.

Next we give definition of Riemann–Liouville fractional integrals and which are utilized to get the desired results of this paper.

Definition 3. (see [8]). Let . Then, Riemann–Liouville fractional integral operators of order are given bywhere is the gamma function and .

The following special functions are also involved in the findings of this paper.

Definition 4. The beta function, also referred to as first type of Euler integral, is defined bywhere .

Close association of the beta function to the gamma function is an important factor of the beta function

The beta function is symmetric, i.e., . A generalization of the beta function, called the incomplete beta function, is defined bywhere with . The incomplete beta function weakens to the ordinary (beta function) by setting .

In [8], the Hadamard inequality is studied for Riemann–Liouville fractional integrals which is stated in the following theorem.

Theorem 2. Let be a positive function with and . If is convex function on , then the following inequality for fractional integrals holds:with .

The inequality stated in the aforementioned theorem motivates the researchers to work in this direction by establishing other kinds of inequalities for Riemann–Liouville fractional integrals. In the past decade, several classical inequalities have been extended via different kinds of fractional integral operators. The Hadamard inequality is one of the most studied inequalities for fractional integral operators. For some recent work, we refer the readers to [3, 8–17].

This paper is organized as follows. In Section 2, two versions of the Hadamard inequality for strongly -convex functions via Riemann–Liouville fractional integrals are given. Their connection with the well-known results is established in the form of corollaries and remarks. In Section 3, the error estimations of Hadamard inequalities for Riemann–Liouville fractional integrals are obtained by using differentiable strongly -convex functions.

2. Main Results

Theorem 3. Let be a positive function with . If is strongly -convex function on with modulus , , then the following fractional integral inequality holds:with .

Proof. Since is strongly -convex function, for , we haveBy setting and , we haveBy multiplying inequality (11) with on both sides and then integrating over the interval , we getBy change of variables, we will getFurther, the above inequality takes the following form:From the definition of strongly -convex function with modulus , for , we have the following inequality:By multiplying inequality (15) with on both sides and then integrating over the interval , we getBy change of variables, we will getFurther, the above inequality takes the following form:From inequalities (14) and (18), one can get inequality (9).

Remark 1. (i)For in (9), we have the result for strongly -convex function [18].(ii)For and in (9), we have the result for strongly convex function.(iii)For ,, and , we get [[16], Theorem 2](iv)For , , , and , we get the classical Hadamard inequality.(v)For and , we get [[17], Theorem 3].

Corollary 1. For , we have the result for Riemann–Liouville fractional integrals of strongly -convex functions:

Corollary 2. For and , the following inequality holds for strongly -convex function:

In the next theorem, we give another version of the Hadamard inequality.

Theorem 4. Under the assumptions of Theorem 3, the following fractional integral inequality holds:with .

Proof. Let . Using strong -convexity of function for and in inequality (10), we haveBy multiplying (22) with on both sides and making integration over , we getBy using change of variables and computing the last integral, from (23), we getFurther, it takes the following form:The first inequality of (21) can be seen in (25). Now we prove the second inequality of (21). Since is strongly -convex function and , we have the following inequality:By multiplying inequality (26) with on both sides and making integration over , we getBy using change of variables and computing the last integral, from (27), we getFurther, it takes the following form:From inequalities (25) and (29), we have inequality (21).

Remark 2. (i)For in (21), we get the result for strongly -convex function [18].(ii)For , , and , we get [[16], Theorem 2](iii)For , , , and , we get the classical Hadamard inequality.

Corollary 3. For and in (21), we have the result for Riemann–Liouville fractional integrals of strongly convex function:

Corollary 4. For in (21), we get the result for Riemann–Liouville fractional integrals of strongly -convex function:

Corollary 5. For and in (21), we get the result for Riemann–Liouville fractional integrals of -convex function:

Corollary 6. For and in (1), we have the Hadamard inequality for strongly -convex function:

3. Error Estimations of Riemann–Liouville Fractional Integral Inequalities

The following two lemmas are very useful to obtain the results of this section.

Lemma 1. (see [8]). Let be a differentiable mapping on with . If , then the following fractional integral equality holds:

Lemma 2. (see [10]). Let be a differentiable mapping on with . If , then the following equality for fractional integrals holds:

Theorem 5. Let be a differentiable mapping on with . If is a strongly -convex function on with modulus , , and , then the following fractional integral inequality holds:with .

Proof. Since is strongly -convex function on , for , we haveBy using Lemma 1 and (37), we haveAfter simplifying the last inequality of (38), we get (36).

Remark 3. (i)By setting in inequality (36), one can get result for -convex function.(ii)By setting in inequality (36), we get [[18], Theorem 8].

Corollary 7. By taking in (36), we have the result for Riemann–Liouville fractional integrals of strongly -convex function:

Corollary 8. By taking and in inequality (36), we have the result for Riemann–Liouville fractional integrals of strongly convex function:

Corollary 9. By taking and in inequality (36), we get the following inequality:

Inequality (41) provides the refinement of [[19], Theorem 2.2].

Theorem 6. Let be a differentiable mapping on with . If is strongly -convex on with modulus , for , then the following inequality for fractional integrals holds:

Proof. By applying Lemma 2 and strong -convexity of , we haveNow, for strong -convexity of , , using power mean inequality, we getHence, we have inequality (42).

Remark 4. (i)For in inequality (42), we have the result for strongly -convex function [18].(ii)For , , and in inequality (42), we get [[16], Theorem 5].(iii)For , , , and in inequality (42), we get the inequality proved by Kirmaci in [20].

Corollary 10. For and in inequality (42), we have the result for Riemann–Liouville fractional integrals of strongly convex function:

Corollary 11. For in inequality (42), we have the result for Riemann–Liouville fractional integrals of strongly -convex function:

Theorem 7. Let be a differentiable mapping on with . If is strongly -convex function on with modulus , for , then the following fractional integral inequality holds:where .

Proof. By applying Lemma 2 and then using Hölder inequality and strong -convexity of , we getWe have used , for . This completes the proof.

Remark 5. (i)For in inequality (47), we get [[18], Theorem 10].(ii)For and in inequality (47), we get [[10], Theorem 2.7].(iii)For , , and in inequality (47), we get [[16], Theorem 6].

Corollary 12. For and , we have the result for -convex function:

Corollary 13. For and , we have

Corollary 14. For and , we have

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.