Trapezium-Type Inequalities for Raina’s Fractional Integrals Operator Using Generalized Convex Functions

The authors have reviewed a wide production of scientific articles dealing with the evolution of the concept of convexity and its various applications, and based on this they have detected the relationship that can be established between trapezoidal inequalities, generalized convex functions, and special functions, in particular with the so-called Raina function, which generalizes other better known ones such as the hypergeometric function and the Mittag–Leffler function. The authors approach this situation by studying the Hermite–Hadamard inequality, establishing a useful identity using Raina’s fractional integral operator in the setting of φ-convex functions, obtaining some integral inequalities connected with the right-hand side of Hermite–Hadamard-type inequalities for Raina’s fractional integrals. Various special cases have been identified.


Introduction
In recent decades, the concept of convexity has had a more general evolution due to its wide application in various fields of science, as demonstrated in the following works [1][2][3][4]. Among the types of generalized convexity and its applications are some such as h-convexity, MT-convexity, η−convexity, (s, m)-convexity and others, which can be found in the following references [5][6][7][8][9]. Also, the use of special functions, in addition to those involving fractional integral operators, have been related to this topic [6,[10][11][12]. The following inequality, named Hermite-Hadamard inequality, is one of the most famous inequalities in the literature for convex functions.
Theorem 1. Let f : I ⊆ R −→ R be a convex function and p, q ∈ I with p < q. Then the following inequality holds: This inequality has remained an area of great interest due to its wide applications, by example in the field of statistics and probability theory [3] and other fields of mathematical analysis [2]. In recent decades, some researchers have studied (1) under the premises of new definitions that have emerged in the development of the concept of convex function. Interested readers see the references [7,[13][14][15][16][17][18]. Definition 1. Let u ∈ K. The set K is said to be φ-convex, if there exists a function φ such that u + te iφ (u − v) ∈ K for all u, v ∈ K and t ∈ [0, 1].
The function f is said to be φ-concave iff (− f ) is φ-convex. Please note that every convex function is φ-convex but the converse does not hold in general.
The main objective of this paper is to establish Hermite-Hadamard's inequalities for Raina's fractional integral operator using a similar method in [35] via generalized φ-convex functions and we will investigate some integral inequalities connected with the right-hand side of the Hermite-Hadamard type inequalities for Raina's fractional integrals. At the end, a briefly conclusion is given.

Results
We define the following generalized φ-convex set and function, using the class of functions defined by (2).
Please note that the expression F η ν,µ (q − p) indicates the evaluation of q − p under the action of Raina's function. If µ = ν = 1, η(1) = 1 and η(k) = 0 for all k = 1, then the above φ-convex set coincides with the classical convex set.

Definition 4.
A real-valued function f defined on a φ-convex set K is said to be generalized φ-convex, if Again, if µ = ν = 1, η(1) = 1 and η(k) = 0 for all k = 1, then the generalized φ-convex functions coincides with the classical convex function.
Hermite-Hadamard's inequalities using generalized φ-convex function can be represented in Raina's fractional integral form as follows: If f is a generalized φ-convex function on p, p + F η ν,µ (q − p) , then the following inequalities for Raina's fractional integral operator hold for ρ, λ > 0, w ∈ R and a bounded sequence of positive real numbers σ such that the function F σ ρ,λ be a non-negative function.
Proof. Taking into account that the integral of a non-negative function is greater than any value of the function at any point of the integration interval, and since f is a non-negative generalized φ-convex function, then, using the definition of Raina's fractional integral operator and the change of variable Similarly, using the change of variable Adding inequalities (8) and (9) we obtain the left side of the inequality (7): For the proof of the right-side inequality in (7) we first note that if f is generalized φ-convex function, then By adding these inequalities, we have Then multiplying both sides of (10) by ρ t ρ and integrating the resulting inequality with respect to t over [0, 1], we obtain i.e., The proof of this theorem is complete.

Raina's Fractional Integral Inequalities for Generalized φ-Convex Functions
For establishing some new results regarding the right side of Hermite-Hadamard type inequalities for Raina's fractional integrals via generalized φ-convex functions we need to prove the following lemma.
Proof. Integrating by parts, we get and with the change of variable Similarly, Therefore, The proof is complete.
Using Lemma 1, it can be obtained some interesting inequalities.

Remark 2.
If in Theorem 3 we choose λ = α, σ = {1, 0, 0, · · · } and w = 1 it is obtained the following inequality for the Riemann-Liouville fractional integral If α = 1 then it is attained the following inequality for the classical Riemann integral Theorem 4. Let ν, µ > 0, η = {η(k)} ∞ k=0 a bounded sequence of real numbers, and p, q ∈ R with p < q, and f : for r ≥ 1, then the following inequality for Raina's fractional integral operator holds for ρ, λ > 0, w ∈ R and a bounded sequence of positive real numbers σ.
Proof. Let r ≥ 1. Using Lemma 1, the triangular inequality, the power mean inequality and the generalized φ-convexity of | f |, we have Using (12) and (13) ρ the desired result is attained.