Some New Results on Hermite–Hadamard–Mercer-Type Inequalities Using a General Family of Fractional Integral Operators

: The aim of this article is to obtain new Hermite–Hadamard–Mercer-type inequalities using Raina’s fractional integral operators. We present some distinct and novel fractional Hermite– Hadamard–Mercer-type inequalities for the functions whose absolute value of derivatives are convex. Our main ﬁndings are generalizations and extensions of some results that existed in the literature.


Introduction
Convex functions have a very useful structure in terms of both definition and properties. This concept has an important role in the theory of inequality. This class of functions has many applications in the different branches of mathematics, and many important inequalities are obtained with the help of this class of functions. Hermite-Hadamard inequality, Jensen inequality, and Mercer inequality, which are well known in the literature, are some of them. Jensen inequality has been caught attention of many researchers, and many articles related to different versions of this inequality have been found in the literature. Jensen's famous inequality can be given as follows: Let 0 < x 1 ≤ x 2 ≤ . . . ≤ x n and µ = (µ 1 , µ 2 , . . . , µ n ) be non-negative weights such that ∑ n k=1 µ k = 1. The famous Jensen inequality (see [1]) in the literature states that f is convex function on the interval [a, b]; then where ∀x k ∈ [a, b] and all µ k ∈ [0, 1], (k = 1, n).
Let us now recall another important inequality obtained by using convex functions. The Hermite-Hadamard inequality has been the focus of many researchers in the fields of inequality theory, numerical analysis, and applied mathematics for nearly a hundred years. A great number of generalizations, expansions, new variants, and improvements have been made regarding this inequality (see, e.g., [12]). The following inequality holds for convex functions and known as Hermite-Hadamard inequality. If f is concave, both inequalities hold as a reverse direction.
Recently, a modern direction of research has been to investigate various likely ways to define fractional integrals and derivatives in fractional calculus. Fractional operators differ from each other with their kernel structures and further properties. Most of them have a general form of the previous operators. Motivated by this, several fractional operators are introduced that generalize ordinary integral operators.
Let us recall the fractional integral of Riemann-Liouville and its general form, which is called Raina's fractional integral operator. and In addition, Raina [13] defined the following results related to the general class of fractional integral operators.
where the coefficients σ(k) (k ∈ N = N ∪ {0}) are a bounded sequence of positive real numbers and R is the set of real numbers. With the help of (4), Raina [13] and Agarwal et al. [14] defined the following left-sided and right-sided fractional integral operators, respectively, as follows: where λ, ρ > 0, ω ∈ R and f (t) is such that the integral on the right-hand side exits.
It is easy to verify that J σ ρ,λ,a+;ω f (x) and J σ ρ,λ,b−;ω f (x) are bounded integral operators on L(a, b) if Here, many useful fractional integral operators can be obtained by customizing the coefficient σ(k).
In this article, motivated by the Jensen-Mercer inequality and Raina's fractional integral operator, we establish new Hermite-Hadamard-Mercer-type integral inequalities for convex functions.
Multiplying both sides of (10) by and then integrating the resulting inequality with respect to t over [0, 1], we have and so the first inequality of (7) is proven. For the proof of the second inequality (7), we first note that if f is a convex function, then, for t ∈ [0, 1], it yields Multiplying both sides of (12) by and then integrating the resulting inequality with respect to t over [0, 1], we obtain and so Adding f (a) + f (b) to both sides of (13), we find the second inequality of (7). Now, we prove inequality (8). From the convexity of f , we have for all 1] in (14), we find that Multiplying both sides of (15) by and then integrating the resulting inequality with respect to t over [0, 1], we have and so The proof of the first inequality of (8) is completed. On the other hand, using the convexity of f , we can write By adding these inequalities and using the Jensen-Mercer inequality, we have Multiplying both sides of (16) by t λ−1 F σ ρ,λ [ω(y − x) ρ t ρ ] and then integrating the resulting inequality with respect to t over [0, 1], we obtain the second and third inequalities of (8).
Proof. To prove the first inequality of (17), by writing x 1 = t 2 x + 2−t 2 y and y 1 = 2−t 2 x + t 2 y for x, y ∈ [a, b] and t ∈ [0, 1] in the inequality (14), we get Then, multiplying both sides of (18) by t λ−1 F σ ρ,λ ω(y − x) ρ t 2 ρ and then integrating the resulting inequality with respect to t over [0, 1], we have The first inequality of (17) is proven. For the proof of the second inequality of (17), by using Jensen-Mercer inequality, we obtain By adding these inequalities, we have Multiplying both sides of (19) by t λ−1 F σ ρ,λ ω(y − x) ρ t 2 ρ and then integrating the resulting inequality with respect to t over [0, 1], we find the second inequality of (17).
, then the following equality for fractional integral holds: Proof. It suffices to note that where and By combining (22) and (23) with (21), we get (20).

Proof. Let
Note that By substituting u = a + b − t 2 x + 2−t 2 y , we get, after some computations, By proceeding with a similar process, we obtain By using (25) and (26), it follows that Thus, by multiplying y−x 4 on both sides of the above equality, we get (24).
By adding L 1 and L 2 , we obtain the inequality (27).
Theorem 5. Let f : [a, b] → R be a differentiable mapping on (a, b) with a < b. If | f | is convex on [a, b], then the following inequality holds for fractional integral operators: for all x, y ∈ [a, b], x < y, λ, ρ, ω > 0, and 1 p + 1 q = 1.
Proof. From Lemma 2, using Hölder's inequality, we have Using the Jensen-Mercer inequality and taking into account the convexity of | f | q , we have