The Hermite-Hadamard inequality for s-Convex functions in the third sense

Abstract: In this paper, the Hermite-Hadamard inequality for s-convex functions in the third sense is provided. In addition, some integral inequalities for them are presented. Also, the new functions based on the integral and double integral of s-convex functions in the third sense are defined and under certain conditions, the third sense s-convexity of these functions are shown and some inequality relations for these are expressed.


Introduction
Among the integral inequalities, Hermite-Hadamard inequality is one of the elegant inequalities involving convex functions, which discloses the relation among the integral mean of a convex function, both the arithmetic mean of the images and the image of the arithmetic mean of the integral limits. Its origin goes back to the studies of Hermite that is published in Mathesis 3 (1883, p.82) and, ten years later, by Hadamard. After it has been called Hadamard's inequality for a long time, the Hermite-Hadamard inequality is commonly used. This inequality asserts that for a convex function defined on the interval [a, b], This inequality attracts special interests of many researchers. They presented various refinements, extensions, generalizations, extensions for differents function types (see [1,2,8,12,13,15,16,22,24,[27][28][29][30][31] and the references therein). Especially, many of the extension studies relates with the new type of convexities. s-convexity, which originates from the studies on modular spaces [6], is one of them and has some applications, especially, in fractal theory [21]. Let us recall the definition of classical convex function for a clear understanding of s-convexity.
Let U be a convex set in a vector space X and f : U → R. f is said to be convex function, if the inequality f (λx + µy) ≤ λ f (x) + µ f (y) (1.2) holds for all x, y ∈ U and the positive numbers λ, µ with λ + µ = 1. s-Convexity is obtained either by replacing λ, µ with λ s , µ s in the condition λ+µ = 1 or by replacing λ, µ with λ s , µ s or λ 1 s , µ 1 s in right handside of (1.2) or by making both properly. According to these replacements, different types of s-convexity are defined. The first sense, the second sense are wellknown ones and given as follows: Let U ⊆ R n and 0 < s ≤ 1. If for each x, y ∈ U, λ, µ ≥ 0 such that λ s + µ s = 1, λx + µy ∈ U, then U is called an s-convex set in R n . This definition is the same as the definition of p-convex set given in article [5]. Therefore, when s-convex set is mentioned in this article, p-convex set will be understood.
Let U ⊆ R n be an s-convex set such that s ∈ (0, 1]. A function f : U → R is said to be s-convex in the first sense if the inequality holds for all x, y ∈ U and λ, µ ≥ 0 with λ s + µ s = 1. Let U ⊆ R n be a convex set and s ∈ (0, 1]. A function f : U → R is said to be s-convex in the second sense if the inequality f (λx + µy) ≤ λ s f (x) + µ s f (y) (1.4) holds for all x, y ∈ U and λ, µ ≥ 0 with λ + µ = 1. Also the third sense s-convexity is introduced [26]. In the literature, there are quite a number of studies on Hermite-Hadamard inequality and extensions for s-convex functions in the first and especially, second sense [3, 7, 9, 10, 14, 18-20, 23, 25].
In this study, we present the Hermite-Hadamard inequality for the s-convex functions in the third sense. Also some integral inequalities are given.

Preliminaries
Throughout this paper, R + and Z ++ denote the set of nonnegative real numbers and the set of positive integers, respectively. Let us express a proposition that will be needed hereafter:  It is clear that, we obtain classical convexity of a function for s = 1 in (2.1). Inequality (2.1) can be expressed in terms of one parameter in two ways: Firstly, since t s + v s = 1, Secondly, replacing the condition t s + v s = 1 with t Gamma, beta functions and some relations used in some of the results are below. These functions are defined as follows, respectively: For x, y > 0, Beta and gamma functions have the following properties: For x, y > 0 and n ∈ Z ++ , Now, let us give an inequality on beta function we will use later: [11] Let p, q > 0. Then

Main results
Throughout this paper, the set U will be taken as an s-convex subset of R.
Theorem 4. Let U ⊆ R + and f : U → R + be an integrable s-convex function in the third sense. For a, b ∈ U with a < b, the following inequality holds Proof. To show the right part of the inequality, we change variable For the first part of the inequality, since f ∈ K 3 s , we have Integrating both side and using the fact that Corollary 5. It is obtained a Hermite-Hadamard type inequality for the convex functions in case s = 1 in Theorem 4.
Theorem 6. Let f : U → R be an integrable s-convex function in the third sense. For a, b ∈ U with a < b, the following inequality holds Proof. Using the s-convexity in the third sense of f , for t ∈ [0, 1] and a, b ∈ U with a < b, we have . By summing these inequalities side by side, it is derived that Using this equality, Using the equality From the inequality (3.2), we obtain The upper bound for the integral at the left side of (3.1) is given in terms of beta function, which is also another integral. We can find the upper bound without beta function.
Theorem 7. Let f : U → R be an integrable s-convex function in the third sense. For a, b ∈ U with a < b, the following inequality holds Proof. From (2.2), we obtain Using this inequality in Theorem 6, we can write Writing x = t s and y = 1 − t s , we have Using the fact that f is nondecreasing on (0, ∞), we obtain If each side of inequality is integrated on [0,1] over t, then Combining Theorem 6 with Theorem 8 and Theorem 7, we conclude the following result.
Corollary 9. Let U ⊆ R + and f : U → R be a nondecreasing s-convex function in the third sense. For a, b ∈ U with a < b, the following inequality holds: Theorem 10. Let f : U → R + be an s-convex function in the third sense. If a, b ∈ U with a < b, then Proof. Using s-convexity in the third sense of f , for all x, y ∈ U and t ∈ [0, 1], Integration on t gives Theorem 11. Let U ⊆ R + and f : U → R be an increasing s-convex function in the third sense. If a, b ∈ U with a < b, then the following inequalities holds: Proof. The fact that h(x) = x So (3.6) yields to the second inequality (from left to right). On the other hand, the s-convexity of f on for all t ∈ [0, 1]. By integrating this inequality over t in [0, 1], one can have that By multiplying each side with 2 1− 1 s 2 , we get the last part of the inequality.
Theorem 12. Let U ⊆ R + and f : U → R + be an integrable s-convex function in the third sense and Proof. Using the s-convexity of f on U for any z, y ∈ U and u ∈ [0, 1], we have By making substitutions z = u 1− 1 s a, u ∈ (0, 1] and y = (1 − u) 1− 1 s b, u ∈ [0, 1), the following is obtained: for all u ∈ (0, 1). To show the validity of the inequality above, first, let us testify that For the integral the change of variable t = 1 − u, u ∈ [0, 1) yields to In a similar way above, by the substitution is deduced. Integrating the inequality (3.7) on (0,1) over u, taking into account that respectively, one can obtain the desired inequality.
Theorem 14. Let f : U → R + be an s-convex function in the third sense. Let a, b ∈ U with a < b. Consider the function (i) If f is decreasing function, then h is also s-convex function in the third sense on [0, 1].
(ii) If f is integrable on [a, b], then the following inequality holds: x + y