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

1.   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 implies that the mean value of the integral of the convex function $f$ on $[a, b]$ interpolates between the image of the arithmetic mean of the endpoints of $[a, b]$ and the arithmetic mean of the images of the endpoints of $[a, b]$.

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 $f:U\rightarrow \mathbb{R}$. $f$ is said to be convex function, if the inequality

holds for all $x, y\in U$ and the positive numbers $\lambda, \mu$ with $\lambda +\mu = 1.$

$s$-Convexity is obtained either by replacing $\lambda, \mu$ with $\lambda ^{s}, \mu^{s}$ in the condition $\lambda+\mu = 1$ or by replacing $\lambda, \mu$ with $\lambda^{s}, \mu^{s}$ or $\lambda^{\frac{1}{s}}, \mu^{\frac{1}{s}}$\ 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 well-known ones and given as follows:

Let $U\subseteq{\mathbb{R}}^{n}$ and $0 < s\leq1$. If for each $x, y\in U$, $\lambda, \mu\geq0$ such that $\lambda^{s}+\mu^{s} = 1,$ $\lambda x+\mu y\in U$, then $U$ is called an $s$-convex set in $\mathbb{R}^{n}$. 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 $U\subseteq \mathbb{R}^{n}$ be an $s$-convex set such that $s\in(0, 1]$. A function $f:U\rightarrow \mathbb{R}$ is said to be $s$-convex in the first sense if the inequality

holds for all $x, y\in U$ and $\lambda, \mu\geq0$ with $\lambda^{s}+\mu^{s} = 1$.

Let $U\subseteq\mathbb{R}^{n}$ be a convex set and $s\in(0, 1]$. A function $f:U\rightarrow \mathbb{R}$ is said to be $s$-convex in the second sense if the inequality

holds for all $x, y\in U$ and $\lambda, \mu\geq0$ with $\lambda+\mu = 1$.

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 ^{[3,7,9,10,14,18,19,20,23,25]}.

In this study, we present the Hermite-Hadamard inequality for the $s$-convex functions in the third sense. Also some integral inequalities are given.

2.   Preliminaries

Throughout this paper, $\mathbb{R}_{+}$ and $\mathbb{Z}_{++}$ denote the set of nonnegative real numbers and the set of positive integers, respectively. Let us express a proposition that will be needed hereafter:

Proposition 1. ^[4] For $a > 0,$ $[0, a)$ is an $s$-convex set.

Definition 2. ^[17] Let $s \in (0, 1]$, $U\subseteq \mathbb{R}^n$ be an $s$-convex set and $f:U\rightarrow \mathbb{R}$. If for all $x, y\in U$ and $t, v\geq 0$ such that $t^{s}+v^{s} = 1$,

then $f$ is said to be $s$-convex function in the third sense. The class of these functions are denoted by $K_{s}^{3}$.

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 $t^{s}+v^{s} = 1$, hence, $v = (1-t^{s})^{\frac{1}{s}}$, we can write

Secondly, replacing the condition $t^{s}+v^{s} = 1$ with $\left(t^{\frac{1}{s} }\right) ^{s}+\left(v^{\frac{1}{s}}\right) ^{s} = 1$, we have

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 $n\in\mathbb{Z}_{++},$

Now, let us give an inequality on beta function we will use later:

Theorem 3. ^[11] Let $p, q > 0$. Then

3.   Main results

Throughout this paper, the set $U$ will be taken as an $s$-convex subset of $\mathbb{R}$.

Theorem 4. Let $U\subseteq \mathbb{R}_+$ and $f: U\rightarrow \mathbb{R}_+$ be an integrable $s$-convex function in the third sense. For $a, b\in U$ with $a < b,$ the following inequality holds

Proof. To show the right part of the inequality, we change variable $x = t^{\frac{1}{s}}b+(1-t)^{\frac{1}{s}}a$, then we have

Using the $s$-convexity of $f$, we can write

For the first part of the inequality, since $f\in K_s^3$, we have

for all $x, y > 0$. Let $x = ta+(1-t)b$ and $y = tb+(1-t)a$ for $t\in(0, 1]$. Then

Integrating both side and using the fact that

one can have

Corollary 5. It is obtained a Hermite-Hadamard type inequality for the convex functions in case $s = 1$ in Theorem 4.

Theorem 6. Let $f: U\rightarrow \mathbb{R}$ be an integrable $s$-convex function in the third sense. For $a, b\in U$ with $a < b$, the following inequality holds

Proof. Using the $s$-convexity in the third sense of $f$, for $t\in\lbrack0, 1]$ and $a, b\in U$ with $a < b$, we have

and

By summing these inequalities side by side, it is derived that

Using the definite integration on $[0, 1]$ yields

By changing variable $v = \left(1-t^{s}\right) ^{\frac{1}{s}}$ at first integral of left side of the inequality above, $t^{s} = 1-v^{s}$, $v\in\lbrack0, 1]$ and $t^{s-1}dt = -v^{s-1}dv$, hence

Using this equality,

Using the equality $\int\limits_{0}^{1}\left(1-t^{s}\right) ^{\frac{1}{s^{2}}}dt = \frac{1}{s(s+1)}B(\frac{1}{s^{2}}, \frac{1}{s})$ related on beta function, we have

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 $f: U \rightarrow \mathbb{R}$ be an integrable $s$-convex function in the third sense. For $a, b\in U$ with $a < b$, the following inequality holds

Proof. From (2.2), we obtain

Using this inequality in Theorem 6, we can write

Theorem 8. Let $U\subseteq\mathbb{R}_{+}$ and $f:U\rightarrow \mathbb{R}$ be a nondecreasing $s$-convex function in the third sense. For $a, b\in U$ with $a < b,$ the following inequality holds:

Proof. Since $h(x) = x^{\frac{1}{s}}$ is a convex function on $[0, \infty)$ for $0 < s < 1,$ we know that

Writing $x = t^{s}$ and $y = 1-t^{s}$, we have

hence,

Using the fact that $f$ is nondecreasing on $(0, \infty),$ we obtain

Integration on $[0, 1]$ yields to the upper inequality in (3.3). From $f$ being $s$-convex, it is known that for all $x, y\in\lbrack0, \infty),$

Putting $x = ta+\left(1-t^{s}\right) ^{\frac{1}{s}}b$ and $y = \left(1-t^{s}\right) ^{\frac{1}{s}}a+tb$ with $t\in\lbrack0, 1]$ above, we deduce

If each side of inequality is integrated on [0, 1] over $t$, then

By changing variable $v = \left(1-t^{s}\right) ^{\frac{1}{s}}$ as indicated in the proof of Theorem 6 we have,

Thus

is obtained.

Combining Theorem 6 with Theorem 8 and Theorem 7, we conclude the following result.

Corollary 9. Let $U\subseteq\mathbb{R}_{+}$ and $f:U\rightarrow \mathbb{R}$ be a nondecreasing $s$-convex function in the third sense. For $a, b\in U$ with $a < b,$ the following inequality holds:

Theorem 10. Let $f: U\rightarrow \mathbb{R}_{+}$ be an $s$-convex function in the third sense. If $a, b\in U$ with $a < b$, then

Proof. Using $s$-convexity in the third sense of $f$, for all $x, y\in U$ and $t\in [0, 1]$,

Since $t^{\frac{1}{s^{2}}}\leq t$ and $(1-t)^{\frac{1}{s^{2}}}\leq 1-t$, we have

Integration on $t$ gives

Theorem 11. Let $U\subseteq \mathbb{R}_+$ and $f: U\rightarrow \mathbb{R}$ be an increasing $s$-convex function in the third sense. If $a, b\in U$ with $a < b$, then the following inequalities holds:

Proof. The fact that $h(x) = x^{\frac{1}{s}}$ is a convex function for $s\in(0, 1]$ on $[0, \infty)$ implies

for all $t\in [0, 1]$. From the monotonicity of $f$, one can get

for all $t\in\lbrack0, 1]$. Thus the first inequality (from left to right) is obtained.

By using the $s$-convexity of $f$, we have

From the changing of variables, it is clear that

So (3.6) yields to the second inequality (from left to right). On the other hand, the $s$-convexity of $f$ on $[0, \infty)$ implies

for all $t\in\lbrack0, 1]$. By integrating this inequality over $t$ in $[0, 1],$ one can have that

By multiplying each side with $2^{1-\frac{1}{s^{2}}}$, we get the last part of the inequality.

Theorem 12. Let $U\subseteq \mathbb{R}_+$ and $f:U\rightarrow \mathbb{R}_{+}$ be an integrable $s$-convex function in the third sense and let $0 < a < b$. If

is finite, then

Proof. Using the $s$-convexity of $f$ on $U$ for any $z, y\in U$ and $u\in\lbrack0, 1]$, we have

By making substitutions $z = u^{1-\frac{1}{s}}a, u\in(0, 1]$ and $y = (1-u)^{1-\frac {1}{s}}b, u\in\lbrack0, 1)$, the following is obtained:

for all $u\in(0, 1)$. To show the validity of the inequality above, first, let us testify that $\int_{0}^{1}u^{\frac{1}{s^{2}}}f\left(u^{1-\frac{1}{s} }a\right) du$ is finite. The change of the variable $x = u^{1-\frac{1}{s} }a, u\in(0, 1]$ yields

and

Thus,

For the integral

the change of variable $t = 1-u,$ $u\in\lbrack0, 1)$ yields to

In a similar way above, by the substitution $x = t^{1-\frac{1}{s}}b,$ $t\in(0, 1]$,

is deduced. Integrating the inequality (3.7) on (0, 1) over $u,$ taking into account that

and

respectively, one can obtain the desired inequality.

Theorem 13. Let $U\subseteq \mathbb{R}_+$ and $f: U \rightarrow \mathbb{R}_{+}$ be an $s$-convex function in the third sense and let $0 < a < b$. Then

(i) Let $g:[0, 1]\rightarrow \mathbb{R}$ be a function defined as follows

If $f$ is decreasing (or noninreasing) function then $g \in K^3_s$.

(ii) If $f$ is integrable on $[a, b]$, then the following inequality holds for $t\in(0, 1]$, $g(t)\geq2^{\frac{1}{s^{2}}-1}f\left(\frac{a+b}{2^{\frac{1}{s}}}\right).$

Proof. $(i)$ Let $x, y\in\lbrack0, 1]$ and $\lambda, \mu\geq 0$ with $\lambda^{s}+\mu ^{s} = 1$. Using $\lambda+\mu\leq 1$, monotonicity and $s$-convexity of $f$, we can write

$(ii)$ Assume that $t\in(0, 1]$. By changing variable $u = tx+(1-t)\frac{a+b}{2}$, we have

where $m = tb+(1-t)\frac{a+b}{2}$ and $n = ta+(1-t)\frac{a+b}{2}$. The left part of the Hermite-Hadamard inequality gives

hence, we get the required inequality.

Theorem 14. Let $f: U \rightarrow \mathbb{R}_{+}$ be an $s$-convex function in the third sense. Let $a, b\in U$ 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:

Proof. (i) Let $x, y\in\lbrack0, 1]$ and $\lambda, \mu\geq0$ with $\lambda^{s}+\mu^{s} = 1$. Taking into account that $\lambda^{s-1}-t_{1}\geq1-t_{1}$ for $t_{1}, \lambda\in\lbrack0, 1]$ and $\mu^{s-1}-t_{2}\geq1-t_{2}$ for $t_{2}, \mu\in\lbrack0, 1]$, then using the monotonicity and $s$-convexity of $f$, respectively, we have

(ii) From (3.4), we can write

for all $t\in\lbrack0, 1]$ and $x, y\in\lbrack a, b].$ Integrating this inequality on $[a, b]^{2}$ we get

since

the above inequality gives us the desired result.

Acknowledgement

The author is very grateful to the Referee and Gültekin Tınaztepe for their valuable comments and contributions.

Conflict of interest

The authors declare no conflict of interest.