On Upper Estimations of Hermite–Hadamard Inequalities

Abstract

Convex functions play a key role in many branches of pure and applied mathematics. In this paper, we prove that if a convex function is not continuous, then the classical Hermite–Hadamard inequality, the Hermite–Hadamard inequality for the Riemann–Liouville fractional integral and the Hermite–Hadamard inequality for the left variable order Riemann–Liouville fractional integral can be improved with slightly sharper bounds. In particular, we give new upper bounds of Hermite–Hadamard inequalities in terms of right-hand limit $g\left(a+\right)$  and left-hand limit $g\left(b-\right)$  values. Furthermore, classical Hermite–Hadamard inequalities only applied to closed bounded intervals, but our new improved inequalities can be applied to open bounded and half-open bounded intervals. As a consequence of our method, we also show some nonconvex functions that satisfy our new improvement of Hermite–Hadamard inequalities.

1. Introduction

It is not always possible to evaluate definite integrals. Therefore, we use inequalities which estimate the value of integrals. Good examples of these are Hermite–Hadamard inequalities. The Hermite–Hadamard inequality gives a connection between the mean values of a convex function over a closed bounded interval and the function values at the endpoints. Convex functions have a great geometric insight. Convex functions are a fundamental concept in convex optimization. They form an important collection of functions. The classical inequality of Hermite–Hadamard states that the following inequality is valid for convex functions $g:\left[a,b\right]\to \mathbb{R}$ 

$$g\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx\le {\displaystyle \frac{g\left(a\right)+g\left(b\right)}{2}}.$$

(1)

Actually, this inequality makes an estimate of the value of the integral

$$\underset{a}{\overset{b}{\int }}g\left(x\right)dx.$$

It can be easily seen that function $f:\left[0,2\right]\to \mathbb{R}$  defined by $f\left(x\right)=e^{x^2}$  for $x\in \left[0,2\right]$  is a convex function. Also, consider function $g:\left[0,2\right]\to \mathbb{R}$, defined by

$$g\left(x\right)=\left\{\begin{array}{cc}\mathrm{10,000},\hfill & \hfill x=0\\ e^{x^2},\hfill & \hfill 0<x<2\\ \mathrm{30,000},\hfill & \hfill x=2\end{array}\right.$$

Since the set of points above the graph of function $g$  is a convex subset of the real plane, this function is convex, too. The difference between these two functions is only on the values of the two endpoints. From Lebesgue’s integration theory, for functions $f$  and $g$, we have equality of

$${\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx.$$

Estimates of Hermite–Hadamard inequality show that for $f$, we have

$$e\le {\displaystyle \frac{1}{2}}\underset{a}{\overset{b}{\int }}f\left(x\right)dx\le {\displaystyle \frac{1+e^4}{2}}.$$

This inequality is sharp. Again, the estimate of Hermite–Hadamard inequality shows that for $g$, we have

$$e\le {\displaystyle \frac{1}{2}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx\le \mathrm{20,000}.$$

This is not a sharp inequality. It is obvious that there is a significant difference between the upper values, $\mathrm{20,000}$  and $\left(1+e^4\right)/2$. The worse news is that we can choose the endpoint values that are bigger than $g\left(0\right)=\mathrm{10,000}$  and $g\left(2\right)=\mathrm{30,000}$. As these examples show, in some cases, usage of Hermite–Hadamard inequality is not useful for the right-side estimate of mean value function. Many extensions, generalizations, refinements and variants of the Hermite–Hadamard inequality (1) have been studied by researchers: see [1,2,3] and references therein. One important extension of Hermite–Hadamard is given for the fractional integral. Fractional calculus is a field of mathematical analysis and applied mathematics that studies various properties and applications of integrals and derivatives of arbitrary order, which can be real or complex numbers.

Therefore, fractional calculus is an extension of the integer-order calculus. Recently, due to its broader scope and applications, fractional calculus has become one of the most actively developing fields of mathematical analysis.

Throughout this paper, we assume that $a,b\in \mathbb{R}$  and $a\le b$. Unless stated explicitly otherwise, we will assume that $J$  is a bounded interval in this paper.

2. Preliminaries and Lemmas

There are many ways in which convex functions can be defined and conceptualized. The following describes the classical case.

Definition 1.

A real-valued function $g$, on an arbitrary interval (bounded or unbounded, open or closed or neither)  $J$  is called convex if it satisfies the following condition: for every  $x,y\in J$  and for each $\rho \in \left[0,1\right]$ 

$$\left(1-\rho \right)g\left(x\right)+\rho g\left(y\right)\ge g\left(\left(1-\rho \right)x+\rho y\right).$$

If $-g:J\to \mathbb{R}$  is a convex function, then the function $g:J\to \mathbb{R}$  is called concave. Geometrically, a function is convex on $J$  $⟺$  the set of points above graph of $g:J\to \mathbb{R}$  is a convex subset of the real plane.

It is important to note that a convex function may not be continuous at endpoints of interval definition $\left[a,b\right]$. A convex function may have upward jumps at points $a$  and $b$.

Lemma 1 

(see [4]). Let $g:J\to \mathbb{R}$ be convex function. Then, $g$ is continuous on the interior of $J$.

Lemma 2 

(see Proposition 1.3.5 of [5]). Suppose that $g:J\to \mathbb{R}$ is a convex function. Then, either $g$ is a monotone function on the interior of interval $J$, or there exists a point $C$ in the interior of $J$, such that $g$ is nondecreasing on the interval $J\cap \left[\left.C,\infty \right)\right.$ and nonincreasing on the interval $J\cap \left(\left.-\infty ,C\right]\right.$.

There are four types of finite intervals. These are $\left[a,b\right]$, $\left(a,b\right)$, $\left[\left.a,b\right)\right.$  and $\left(\left.a,b\right]\right.$. The following theorem actually satisfies all these intervals. We only prove this for interval $\left(a,b\right)$. The proofs for the other cases are similar. Without full proof, a case of the following lemma is stated in (see Proposition 1.3.4 of [5]).

Lemma 3.

If  $g:\left(a,b\right)\to \mathbb{R}$ is a bounded convex function, then right-hand limit  $g\left(a+\right)$ and left-hand limit $g\left(b-\right)$ exists in $\left(-\infty ,\infty \right)$, and the function is defined by

$$G\left(x\right)=\left\{\begin{array}{c}g\left(a+\right),x=a\\ g\left(x\right),\\ g\left(b-\right),x=b\end{array}\right.a<x<b$$

also is convex. Moreover, $G$ is continuous on $\left[a,b\right]$.

Proof. 

We use the sequential definition of the limit. From Lemma 2, a convex function is monotone or can be divided into a monotone nondecreasing and nonincreasing function. Due to on interior of the $J$  function $g$  is continuous, by monotone convergence theorem $g\left(a+\right)$  and $g\left(b-\right)$  exists. It is now obvious that $G$  is continuous on $\left[a,b\right]$.

By the definition of $G$  for all $x,y\in \left(a,b\right)$, we have $G\left(x\right)=g\left(x\right)$, then for these values, it follows that

$$\left(1-\rho \right)G\left(x\right)+\rho G\left(y\right)\ge G\left(\left(1-\rho \right)x+\rho y\right).$$

Now, let’s show for $y\in \left(a,b\right)$, we obtain

$$\left(1-\rho \right)G\left(a\right)+\rho G\left(y\right)\ge G\left(\left(1-\rho \right)a+\rho y\right).$$

Let $a_n$  be any sequence that satisfies $a_n>a$  and $a_n\to a$. Then, we obtain

$$G\left(\left(1-\rho \right)a+\rho y\right)=\underset{n\to \infty }{\lim }G\left(\left(1-\rho \right)a_n+\rho y\right)=\underset{n\to \infty }{\lim }g\left(\left(1-\rho \right)a_n+\rho y\right)$$

$$\le \underset{n\to \infty }{\lim }\left(1-\rho \right)g\left(a_n\right)+\rho g\left(y\right)=\left(1-\rho \right)G\left(a\right)+\rho G\left(y\right)$$

In a similar manner, we can prove that for $x\in \left(a,b\right)$, we obtain

$$\left(1-\rho \right)G\left(x\right)+\rho G\left(b\right)\ge G\left(\left(1-\rho \right)x+\rho b\right)$$

Finally, we show that

$$\left(1-\rho \right)G\left(a\right)+\rho G\left(b\right)\ge G\left(\left(1-\rho \right)a+\rho b\right)$$

Let $a_n$, $b_n$  be any sequences that satisfy $a_n>a$, $a_n\to a$  and $b_n<b$, $b_n\to b$. Then, we have

$$G\left(\left(1-\rho \right)a+\rho b\right)=\underset{n\to \infty }{\lim }G\left(\left(1-\rho \right)a_n+\rho b_n\right)=\underset{\mathrm{n}\to \mathrm{\infty }}{\lim }g\left(\left(1-\rho \right)a_n+\rho b_n\right)$$

$$\le \underset{n\to \infty }{\lim }\left(1-\rho \right)g\left(a_n\right)+\rho g\left(b_n\right)=\left(1-\rho \right)G\left(a\right)+\rho G\left(b\right)$$

This ends the proof of Lemma 3. □

Corollary 1.

Under the conditions of Lemma 3, we have the following equality

$${\displaystyle \underset{\left(a,b\right)}{\int }}g\left(x\right)dx={\displaystyle \underset{\left[a,b\right]}{\int }}G\left(x\right)dx.$$

Proof. 

Due to continuity and the monotonicity of function $g$, integral ${\int }_{\left(a,b\right)}g\left(x\right)dx$  is convergent. Therefore, by Lebesgue’s integration theory, we obtain the equality of integrals

$${\displaystyle \underset{\left(a,b\right)}{\int }}g\left(x\right)dx={\displaystyle \underset{\left(a,b\right)}{\int }}g\left(x\right)dx+{\displaystyle \underset{\left\{a,b\right\}}{\int }}G\left(x\right)dx={\displaystyle \underset{\left[a,b\right]}{\int }}G\left(x\right)dx.$$

This ends the proof of Corollary 1. □

3. Main Results

It is important to be aware that our results about the upper estimation of Hermite–Hadamard inequalities never make an improvement in the lower estimation, but for the sake of completeness, we write the lower estimation in proofs. As is known, inequalities are more useful when they are sharper. The following right-side improvement of Hermite–Hadamard inequality for a convex function is our first result.

Theorem 1.

Let  $g:\left[a,b\right]\to \mathbb{R}$ be a convex function. Then, the following inequality holds

$$g\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx\le {\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}.$$

Proof. 

The left side of the inequality is exactly the Hermite–Hadamard inequality.

Let $\mathrm{G}$  be defined as it is in Lemma 3. Therefore, the function $G\left(x\right)$  is continuous and convex. Then, we can write

$$G\left({\displaystyle \frac{a+b}{2}}\right)=g\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx={\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}G\left(x\right)dx$$

$$\le {\displaystyle \frac{G\left(a\right)+G\left(b\right)}{2}}={\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}$$

This ends the proof of Theorem 1. □

Remark 1.

If the convex functions  $g:\left[a,b\right]\to \mathbb{R}$ are continuous, then our improved inequality coincides with Hermite–Hadamard inequality. However, if the function is not continuous, then we see the benefits of our Theorem 1. In particular, we can see the difference in example ($g:\left[0,2\right]\to \mathbb{R}$, defined by  $g\left(0\right)=\mathrm{10,000}$,  $g\left(2\right)=\mathrm{30,000}$  and  $g\left(x\right)=e^{x^2}$  for  $x\in \left(0,2\right)$), given in the introduction. If we use the classical Hermite–Hadamard inequality, we obtain the upper bound of the inequality as being

$${\displaystyle \frac{1}{2}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx\le \mathrm{20,000}.$$

However, if we use our new upper bound formula we obtain

$${\displaystyle \frac{1}{2}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx\le {\displaystyle \frac{1+e^4}{2}}.$$

Hence, we achieve a better upper bound than the inequality of Hermite–Hadamard.

It is worth noting that classical Hermite–Hadamard inequalities are only applied to intervals $\left[a,b\right]$. In contrast, our following inequality is applied to intervals $\left(a,b\right)$  (and similarly, can be applied to half-open bounded intervals).

Theorem 2.

Let  $g:\left(a,b\right)\to \mathbb{R}$ be a convex and bounded function. Then, the following inequality holds

$$g\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx\le {\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}.$$

Proof. 

Let $\mathrm{G}$  be defined as it is in Lemma 3. Therefore, function $\mathrm{G}\left(\mathrm{x}\right)$  is continuous and convex. Then, we have

$$g\left({\displaystyle \frac{a+b}{2}}\right)=G\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}G\left(x\right)dx={\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx$$

$$\le {\displaystyle \frac{G\left(a\right)+g\left(b\right)}{2}}={\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}$$

This ends the proof of Theorem 2. □

Corollary 2.

Suppose  $h:\left[a,b\right]\to \mathbb{R}$ is a function whose restriction to open interval  $\left(\mathrm{a},\mathrm{b}\right)$  is convex and integrable. Then, the following inequality holds.

$$h\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}h\left(x\right)dx\le {\displaystyle \frac{h\left(a+\right)+h\left(b-\right)}{2}}.$$

Proof. 

If function $h$  is unbounded, then the corollary is obvious. If $h$  is bounded, then the corollary is obtained from Theorem 2. □

It is important to be aware of the following remark, which goes beyond convexity. In other words, in the following remark, we show that some nonconvex functions satisfy our improved Hermite–Hadamard inequality.

Remark 2.

Although the function  $h:\left[0,2\right]\to \mathbb{R}$ is defined by  $h\left(0\right)=-17$  ,  $h\left(2\right)=-7$  and for  $x\in \left(0,2\right)$,  $h\left(x\right)=e^{x^2}$  is not convex, because of

$$h\left({\displaystyle \frac{1}{2}}0+{\displaystyle \frac{1}{2}}2\right)=e\nleqq {\displaystyle \frac{1}{2}}h\left(0\right)+{\displaystyle \frac{1}{2}}h\left(2\right)=-12.$$

However, it satisfies our following inequality

$$h\left({\displaystyle \frac{0+2}{2}}\right)=e\le {\displaystyle \frac{1}{2-0}}\underset{0}{\overset{2}{\int }}h\left(x\right)dx\le {\displaystyle \frac{h\left(0+\right)+h\left(2-\right)}{2}}={\displaystyle \frac{1+e^4}{2}}$$

Now, let us justify this inequality. By Lemma 3, the function $H:\left[0,2\right]\to \mathbb{R}$  is defined as

$$H\left(x\right)=\left\{\begin{array}{c}h\left(0+\right)=1,x=0\\ h\left(x\right)=e^{x^2},0<x<2\\ h\left(2-\right)=e^4,x=2\end{array}\right.$$

which is equivalent to $H\left(x\right)=e^{x^2}$  for $x\in \left[0,2\right]$. It is now obvious that $H$  is a continuous convex function. So, by the properties of the integral, we obtain

$$h\left({\displaystyle \frac{0+2}{2}}\right)=H\left({\displaystyle \frac{0+2}{2}}\right)\le {\displaystyle \frac{1}{2-0}}\underset{0}{\overset{2}{\int }}H\left(x\right)dx={\displaystyle \frac{1}{2-0}}\underset{0}{\overset{2}{\int }}h\left(x\right)dx$$

$$\le {\displaystyle \frac{H\left(0\right)+H\left(2\right)}{2}}={\displaystyle \frac{h\left(0+\right)+h\left(2-\right)}{2}}.$$

As another example, consider function $h:\left[0,2\right]\to \mathbb{R}$, defined by $h\left(x\right)=7$  for $x\in \left(0,2\right)$  and $f\left(0\right)=f\left(2\right)=-4$. It is obvious that $h$  is not convex. So, we can not apply Hermite–Hadamard inequality, but we can apply our inequality

$$h\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}h\left(x\right)dx\le {\displaystyle \frac{h\left(a+\right)+h\left(b-\right)}{2}}$$

which gives $7\le 7\le 7.$ 

Hereafter, $\mathsf{\Gamma }\left(\theta \right)$  denotes the gamma function evaluated at $\theta$.

Definition 2.

Left-sided Riemann–Liouville fractional integral of order  $\theta >0$ is defined by

$$I_{a^{+}}^{\theta }g\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\theta \right)}}\underset{a}{\overset{x}{\int }}{\left(x-w\right)}^{\theta -1}g\left(w\right)dw,\hspace{1em}x>a.$$

Right-sided Riemann–Liouville fractional integral of order $\theta >0$  is defined by

$$I_{b^{-}}^{\theta }g\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\theta \right)}}\underset{x}{\overset{b}{\int }}{\left(w-x\right)}^{\theta -1}g\left(w\right)dw,x<b.$$

Hermite–Hadamard inequality for Riemann–Liouville fractional integrals is showed by Sarikaya et al. in [6], as follows.

Theorem 3.

Let function  $g:\left[a,b\right]\to \mathbb{R}$ be convex and integrable. Then, the following inequalities for the Riemann–Liouville fractional integral satisfy

$$g\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta +1\right)}{2{\left(b-a\right)}^{\theta }}}\left[I_{a^{+}}^{\theta }g\left(b\right)+I_{b^{-}}^{\theta }g\left(a\right)\right]\le {\displaystyle \frac{g\left(a\right)+g\left(b\right)}{2}}$$

(2)

with $\theta >0$.

Theorem 4.

Suppose  $g:\left[a,b\right]\to \mathbb{R}$ is a function whose restriction to open interval  $\left(\mathrm{a},\mathrm{b}\right)$  is convex and integrable. Then, the following inequalities for the Riemann–Liouville fractional integral satisfy

$$g\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta +1\right)}{2{\left(b-a\right)}^{\theta }}}\left[I_{a^{+}}^{\theta }g\left(b\right)+I_{b^{-}}^{\theta }g\left(a\right)\right]\le {\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}$$

with $\theta >0$.

Proof. 

Let $\mathrm{G}$  be defined as it is in Lemma 3. Therefore, function $G\left(x\right)$  is continuous and convex. Therefore, by Theorem 3, we can write

$$g\left({\displaystyle \frac{a+b}{2}}\right)=G\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta +1\right)}{2{\left(b-a\right)}^{\theta }}}\left[I_{a^{+}}^{\theta }G\left(b\right)+I_{b^{-}}^{\theta }G\left(a\right)\right]$$

$$={\displaystyle \frac{\mathsf{\Gamma }\left(\theta +1\right)}{2{\left(b-a\right)}^{\theta }}}\left[I_{a^{+}}^{\theta }g\left(b\right)+I_{b^{-}}^{\theta }g\left(a\right)\right]\le {\displaystyle \frac{G\left(a\right)+G\left(b\right)}{2}}={\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}$$

This ends the proof of Theorem 4. □

Remark 3.

If the convex functions  $g:[a,b]\to \mathbb{R}$ are continuous, then our improved inequality coincides with Theorem 3.

Theorem 5.

Let  $g:\left(a,b\right)\to \mathbb{R}$ be a convex and integrable function. Let  $\mathrm{G}$  be defined as it is in Lemma 3. Then, the following inequalities for fractional integral satisfy

$$g\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta +1\right)}{2{\left(b-a\right)}^{\theta }}}\left[I_{a^{+}}^{\theta }G\left(b\right)+I_{b^{-}}^{\theta }G\left(a\right)\right]\le {\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}$$

with $\theta >0$.

Proof. 

Let $\mathrm{G}$  be defined as it is in Lemma 3. Therefore, function $G\left(x\right)$  is continuous and convex. Therefore, by Theorem 3, we can write

$$g\left({\displaystyle \frac{a+b}{2}}\right)=G\left({\displaystyle \frac{a+b}{2}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta +1\right)}{2{\left(b-a\right)}^{\theta }}}\left[I_{a^{+}}^{\theta }G\left(b\right)+I_{b^{-}}^{\theta }G\left(a\right)\right]$$

$$\le {\displaystyle \frac{G\left(a\right)+G\left(b\right)}{2}}={\displaystyle \frac{g\left(a+\right)+g\left(b-\right)}{2}}$$

This ends the proof of Theorem 5. □

Definition 3.

For all  $x\in \left[a,b\right]$, we have  $\theta \left(x\right)\in \left(0,1\right)$  . The left-sided Riemann–Liouville fractional integral variable order,  $\theta \left(x\right)$, is defined by

$$I_{a^{+}}^{\theta \left(x\right)}g\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\theta \left(x\right)\right)}}\underset{a}{\overset{x}{\int }}{\left(x-w\right)}^{\theta \left(x\right)-1}g\left(w\right)dw$$

and the right-sided Riemann–Liouville fractional integral variable order $\alpha \left(x\right)$  is defined by

$$I_{b^{-}}^{\theta \left(x\right)}g\left(x\right)={\displaystyle \frac{1}{\mathsf{\Gamma }\left(\theta \left(x\right)\right)}}\underset{x}{\overset{b}{\int }}{\left(w-x\right)}^{\theta \left(x\right)-1}g\left(w\right)dw.$$

We refer the reader to [7] for the details of variable order fractional integrals.

Theorem 6 

(see [8]). Let $a\ge 0$ and a convex function $g:\left[a,b\right]\to \mathbb{R}$ be integrable and positive. Then, the following inequalities for the fractional integral of variable order satisfy

$$g\left({\displaystyle \frac{\theta \left(b\right)a+b}{\theta \left(b\right)+1}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta \left(b\right)+1\right)}{{\left(b-a\right)}^{\theta \left(b\right)}}}{I}_{a^{+}}^{\theta \left(b\right)}g\left(b\right)\le {\displaystyle \frac{\theta \left(b\right)g\left(a\right)+g\left(b\right)}{\theta \left(b\right)+1}}$$

(3)

with $\theta \left(b\right)>0$.

Again, we see that if the convex function $g$  is not continuous at the end points of an interval $\left[a,b\right]$, the inequality $\left(3\right)$  will not be sharp, (or is not valid, as in the case of Remark 2). As a result, we obtain the following improvement of the preceding Theorem 6.

Theorem 7.

Let  $a\ge 0$ and a convex function  $g:\left[a,b\right]\to \mathbb{R}$  be integrable, positive. Then, the following inequalities for the fractional integral of variable order satisfy

$$g\left({\displaystyle \frac{\theta \left(b\right)a+b}{\theta \left(b\right)+1}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta \left(b\right)+1\right)}{{\left(b-a\right)}^{\theta \left(b\right)}}}{I}_{a^{+}}^{\theta \left(b\right)}g\left(b\right)\le {\displaystyle \frac{\theta \left(b\right)g\left(a+\right)+g\left(b-\right)}{\theta \left(b\right)+1}}$$

with $\theta \left(b\right)>0$.

Proof. 

Let $\mathrm{G}$  be defined as it is in Lemma 3. Since $\alpha \left(b\right)>0$, we have

$${\displaystyle \frac{\theta \left(b\right)a+b}{\theta \left(b\right)+1}}={\displaystyle \frac{\theta \left(b\right)a+a+b-a}{\theta \left(b\right)+1}}=a+{\displaystyle \frac{b-a}{\theta \left(b\right)+1}}\in \left(a,b\right).$$

From the result of Theorem 6 and the convexity of function $\mathrm{G}$, we observe that

$$g\left({\displaystyle \frac{\theta \left(b\right)a+b}{\theta \left(b\right)+1}}\right)=G\left({\displaystyle \frac{\theta \left(b\right)a+b}{\theta \left(b\right)+1}}\right)\le {\displaystyle \frac{\mathsf{\Gamma }\left(\theta \left(b\right)+1\right)}{{\left(b-a\right)}^{\theta \left(b\right)}}}{I}_{a^{+}}^{\theta \left(b\right)}G\left(b\right)$$

$$={\displaystyle \frac{\mathsf{\Gamma }\left(\theta \left(b\right)+1\right)}{{\left(b-a\right)}^{\theta \left(b\right)}}}{I}_{a^{+}}^{\theta \left(b\right)}g\left(b\right)\le {\displaystyle \frac{\theta \left(b\right)G\left(a\right)+G\left(b\right)}{\theta \left(b\right)+1}}={\displaystyle \frac{\theta \left(b\right)g\left(a+\right)+g\left(b-\right)}{\theta \left(b\right)+1}}$$

This ends the proof of Theorem 7. □

Remark 4.

If the convex functions  $g:\left[a,b\right]\to \mathbb{R}$ are continuous, then our improved inequality coincides with Theorem 6.

4. Conclusions

We have proved new sharp upper bounds for the classical Hermite–Hadamard inequality and its Riemann–Liouville fractional versions. By using our sharp inequality methods, we see even some nonconvex functions satisfy the improved Hermite–Hadamard inequality. Lemma 3 is essential for all the work in this paper. Lemma 2 plays a key role in the proof of Lemma 3. It is well known amongst researchers that there are many different notions of convexity. Therefore, this paper may give rise to restudy Hermite–Hadamard-type inequalities for these notions of convexities. It is important to note that our results in some cases also include open interval $\left(a,b\right)$, which is a new development. Our last result is an open problem.

Open Problem 1.

Let  $g:\left(a,b\right)\to \mathbb{R}$ be a convex unbounded continuous and integrable function. Then, can Hermite–Hadamard-like inequality bounds for the value of the following integral be found?

$${\displaystyle \frac{1}{b-a}}\underset{a}{\overset{b}{\int }}g\left(x\right)dx?$$

If it is possible, I think the solution to the problem will contain a series.