On Hermite-Hadamard type inequalities for Sφ−preinvex functions by using Riemann-Liouville fractional integrals

Many authors have been working to fractional integral operators (see [4–7]) due to many applications in differents areas of Mathematics, Engineering and Physics, etc (see [8, 9]). Also, these operators have allow to extended results about integral inequalities of many types (see [4, 10, 11]), for instance, Hermite-Hadamard integral inequalities (see [12–14]), Ostrowski type inequalities (see [7]).


Introduction
Fractional calculus (see [1][2][3]) arise in the mathematical modeling of various problems in sciences and engineering such as mathematics, physics, chemistry and biology.
In particular, in recent years, several extensions and generalizations have been considered for classical convexity (see [13,15,16]). A significant generalizations of convex functions is that of invex functions introduced by Hanson (see [17]).
In this work we derive several new inequalities of Hermite-Hadamard type for s ϕ -preinvex function of first and second sense by using fractional integrals.
In this article, we define and recall some basic concepts and results. Let R n be the finite dimensional Euclidian space, also 0 ≤ ϕ ≤ π 2 be a continuous function.
Definition 2. If K ϕη in R n set, is said to be ϕ−invex at u according to ϕ, if there exists a bifunction η (., .) : K ϕη × K ϕη → R n , so that,

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The ϕ−invex set K ϕη is also called ϕη−connected set. Note that the convex set with ϕ = 0 and η (u, v) = v − u is a ϕ−invex set, but the converse is not true.
Theorem 1. Let f : I ⊆ R → R is a convex function defined on the interval I = [a, b] of real numbers where a < b. Then, the following double inequality the above double inequality is known as Hermite-Hadamard type of inequality in the literature.
Let R be the set of real numbers. During the article I = [a, b] ⊂ R be the interval unless otherwise specified, also let 0 ≤ ϕ ≤ π 2 be a continuous function.
If we take α = 1 in Lemma 1, we obtain to following result.
be a continuous function. We get the following equality for fractional integrals:

Inequalities for S ϕ -preinvex of second sense
In order to obtain main results introduced by [18] the s ϕ −preinvex function of second sense.

Definition 3. [18]
A function f on the set K ϕη is said to be s ϕ −preinvex function of second sense according to ϕ and η, we get Theorem 2. Suppose a function K ϕη ⊆ R n be a open invex set according to bifunction η (., .) : and |f ′′ | is s ϕ -preinvex function of second sense, afterward, we get the following inequality for fractional integrals: Proof. Via Lemma 1 and the fact that |f ′′ | is s ϕ -preinvex function of second sense, we get (1 + τ ) s dτ the above selection will be accepted, namely, which completes the proof of Theorem.
Theorem 3. Suppose a function K ϕη ⊆ R n and ϕ : K ϕη → R, f : [a, b] → R be twice differentiable function on (a, b) with a < b, η (., .) : K ϕη × K ϕη → R n .Also R n be the finite dimensional Euclidian space. The ϕ−invex set K ϕη . Let f ′′ ∈ L [a, b] and |f ′′ | q is s ϕ -preinvex function of second sense, afterward, we get the following inequality for fractional integrals: Proof. From Lemma 1, Holder's inequality and the fact that |f ′′ | q is s ϕ -preinvex function of second sense, we get which completes the proof of Theorem.
Theorem 4. Suppose a functionK ϕη ⊆ R n and ϕ : K ϕη → R , f : [a, b] → R be twice differentiable function on (a, b) with a < b, η (., .) : K ϕη × K ϕη → R n . Also R n be the finite dimensional Euclidian space .The ϕ−invex set K ϕη . Let f ′′ ∈ L [a, b] and |f ′′ | q is s ϕ −preinvex function of second sense, afterward, we get the following inequality for fractional integrals: Proof. From Lemma 1, power-mean inequality and the fact that |f ′′ | q is s ϕ −preinvex function of second sense, we get which completes the proof of Theorem.

Inequalities for S ϕ -convex funstions of first sense
In order to obtain main results introduced by [18] the s ϕ −preinvex function of first sense.

Definition 4. [18]
Suppose a function f on the set K ϕη is said to be s ϕ −preinvex function of first sense according to ϕ and η, let Theorem 5. Suppose a function K ϕη ⊆ R n and ϕ : K ϕη → R, the f : [a, b] → R be twice differentiable function on (a, b) with a < b, η (., .) : and |f ′′ | is s ϕ -preinvex function of second sense, afterward, we get the following inequality for fractional integrals: Proof. From Lemma 1 and the fact that |f ′′ | is s ϕ -preinvex function of first sense, we get which completes the proof of Theorem.. Theorem 6. Suppose a function K ϕη ⊆ R n and ϕ : K ϕη → R, the f : [a, b] → R be twice differentiable function on (a, b) with a < b, η (., .) : K ϕη × K ϕη → R n , Also R n be the finite dimensional Euclidian space. The ϕ−invex set K ϕη . If f ′′ ∈ L [a, b] and |f ′′ | q is s ϕ -preinvex function of second sense, afterward, we get the following inequality for fractional integrals: Proof. From Lemma 1, Hölder inequality and the fact that |f ′′ | q is s ϕ -preinvex function of second sense, we get which completes the proof of Theorem.
Theorem 7. Suppose a function K ϕη ⊆ R n and ϕ : K ϕη → R, the f : [a, b] → R be twice differentiable function on (a, b) with a < b, η (., .) : and |f ′′ | q is s ϕ -preinvex function of second sense, afterward, we get the following inequality for fractional integrals: Proof. From Lemma 1, power-mean inequality and the fact that |f ′′ | q is s ϕ -preinvex function of second sense, we get