Revisiting the Hermite-Hadamard fractional integral inequality via a Green function

1.   Introduction

Let $I\subseteq\mathbb{R}$ be an interval. Then a real-valued function $f: I\rightarrow\mathbb{R}$ is said to be convex (concave) if the inequality

holds whenever $x, y\in I$ and $\lambda\in [0, 1]$. It is well-know that the convexity (concavity) has wild applications in pure and applied mathematics ^[1,2,3,4,5], and many inequalities ^{[6,7,8,9,10,11,12,13,14,15,16,17,18,19]} can be found in the literature via the convexity theory. Recently, the generalizations and variants for the convexity have attracted the attention of the researchers, for example, the $GG$- and $GA$-convexity ^[20], $h$-convexity ^[21], qausi-convexity ^[22], $\rho$-convexity ^[23], exponential convexity ^[24], harmonic convexity ^[25], $s$-convexity ^[26,27] and others.

The classical Hermite-Hadamard inequality ^{[28,29,30,31,32,33]} is one of the most famous inequalities in convex function theory, which can be stated as follows:

A real-valued function $\psi:[b_1, b_2]\to\mathbb{R}$ is convex if and only if

If $\psi$ is concave, then the inequalities in (1.1) remain valid in the reversed direction. The Hermite-Hadamard inequality (1.1) provides both upper and lower estimates of the integral mean of any convex function defined on a closed and bounded interval involving the endpoints and midpoints of the function's domain. Because of the excellent significance of the Hermite-Hadamard inequality, the literature is replete with ample amount of research articles dedicated to the generalizations, refinements, and extensions for the Hermite-Hadamard inequality for various families of convexity.

Besides generalization via convexity, great effort has gone into extending (1.1) by means of fractional integral operators. Most popular of them is the Riemann-Liouville fractional integral operators given in the following definition.

Definition 1. Let $\alpha > 0$, $b_1, b_2\in\mathbb{R}$ with $b_1 < b_2$ and $\psi\in L[b_1, b_2]$. Then the left and right Riemann-Liouville fractional integrals $J_{b_1+}^{\alpha}\psi$ and $J_{b_2-}^{\alpha}\psi$ of order $\alpha$ are defined by

and

respectively where $\Gamma(\alpha) = \int_{0}^{\infty}e^{-t}t^{\alpha-1}dt$ is the gamma function.

Sarikaya et al. ^[34] established the following fractional version of the Hermite-Hadamard inequality:

Theorem 2. Let $0\leq b_1 < b_2$ and $\psi:[b_1, b_2]\rightarrow\mathbb{R}$ be a positive convex function such that $\psi\in L[b_1, b_2]$. Then the fractional integrals inequality

holds for $\alpha > 0$.

In view of Theorem 2, it is pertinent to note that the positivity of the function $\psi$ and the numbers $b_1$ and $b_2$ is not necessary. From Definition 1, it is clear that $b_1$ and $b_2$ are any real numbers such that $b_1 < b_2$.

Our main contribution in this article is to use a similar technique used in ^[35] to obtain Theorem 2. This time, our main result contains both sided fractional integral operators in the Riemann-Liouville sense. In this method, we use a green function and in the process, we obtain some identities involving the left and right Riemann-Liouville fractional integral operators. These identities are subsequently employed to establish some new results for the class of convex, concave and monotone functions.

2.   Main results

Our main result will be anchored on the succeeding lemma.

Lemma 3. (See ^[36,37]) Let G be the green function defined on $[b_1, b_2]\times[b_1, b_2]$ by

Then any $\psi\in C^2([b_1, b_2])$ can be expressed as

We are now in a position to frame and prove our results.

Theorem 4. Let $\psi\in C^2([b_1, b_2])$ be a convex function. Then, for any $\alpha > 0$, the following fractional integral inequalities hold:

Proof. Setting $x = \frac{b_1+b_2}{2}$ in (2.1), we get

Equivalently,

Using (2.1), we do the following computations:

Therefore,

Similarly,

So,

Now, adding (2.3) and (2.4) and then multiplying the resultant sum by $\frac{\Gamma(\alpha+1)}{2(b_{2}-b_{1})^{\alpha}}$ to get:

Subtracting (2.5) from (2.2), we obtain

By the definition of the Green function,

we obtain

and

Substituting Eqs (2.6) and (2.7) into (2.6), then we obtain

Let

Here,

Now, if $b_1\leq\mu\leq\frac{b_1+b_2}{2}$, then from (2.9) and (2.10), we have

This shows that $f$ is decreasing and $f(b_1) = 0$ then $f(\mu)\leq 0$ for all $\mu \in [b_{1}, \frac{b_{1}+b_{2}}{2}]$.

If, on the other hand, $\frac{b_{1}+b_{2}}{2} \leq \mu \leq b_{2}$, then

Therefore,

which shows that $f'$ is decreasing and $f'(b_{2}) = 0$ and thus $f'(\mu) \geq 0$. Therefore, $f$ is increasing and $f(b_{2}) = 0$. Hence, $f(\mu) \leq 0$ for all $\mu \in [\frac{b_{1}+b_{2}}{2}, b_{2}]$. Combining the two cases discussed above, we have that

Using (2.8), we deduce the first inequality:

For the right hand side of the inequality, we recall:

Subtracting Eq $(2.5)$ from $(2.11)$, we get:

If we set

then for $b_{1} \leq \mu \leq b_2$,

If $b_{1} \leq \mu \leq \frac{b_{1}+b_{2}}{2}$, then

which proves that $F$ is increasing and $F(b_{1}) = 0$, and hence $F(\mu)\geq 0$.

Suppose also $\frac{b_{1}+b_{2}}{2} \leq \mu \leq b_{2}$. Then,

This implies that $F$ is a decreasing function and $F(b_{2}) = 0$, and thus $F(\mu) \geq 0$. Also, $\psi''(\mu) \geq 0$ since $\psi$ is convex. Hence, $F(\mu) \geq 0$ for all $\mu\in [b_1, b_2]$ and using (2.12) amounts to:

That completes the proof.

Next, we present new Hermite-Hadamard type inequalities for the class of monotone and convex function.

Theorem 5. Let $\psi\in C^2([b_1, b_2])$ and $\alpha > 0$, Then the following statements are true:

(ⅰ) If $|\psi''|$ is an increasing function, then

(ⅱ) If $|\psi''|$ is a decreasing function, then

(ⅲ) If $|\psi''|$ is a convex function, then

Proof. To prove (ⅰ), we use the following identity obtained from (2.12):

Taking absolute values on both sides of (2.14) and using the fact that $\big|\psi''\big|$ is an increasing function, we have

Therefore,

This establishes the inequality in (ⅰ).

Part (ⅱ) can be proved in a similar way. For (ⅲ), we make use of (2.13) and the fact that every convex function $\psi$ defined on the interval $[b_1, b_2]$ is bounded above by $\max\Big\{\big|\psi(b_1)\big|, \big|\psi(b_2)\big|\Big\}$ to get:

Remark 6. By setting $\alpha = 1$ in Theorem 5, we get the following inequalities:

Theorem 7. Let $\psi \in C^2([b_1, b_2])$ and $\alpha > 0$. Then the following statements are true:

$(i)$ If $|\psi''|$ is an increasing function, then

$(ii)$ If $|\psi''|$ is a decreasing function, then

$(iii)$ If $|\psi''|$ is a convex function, then

Proof. The inequality in (ⅰ) is obtained by using (2.8):

Taking absolute values and using triangle inequality, we get

The second part can be deduced by using the same procedure. For Part (ⅲ), we also employ the fact that the convex function $\psi$ is bounded above by $\max\Big\{\big|\psi(b_1)\big|, \big|\psi(b_2)\big|\Big\}$ since it is defined on the interval $[b_1, b_2]$. That is, we obtain from (2.8):

which is the desired inequality.

Remark 8. In Theorem 7, if we take $\alpha$ = 1, then we obtain

Theorem 9. Assume that $\psi \in C^2([b_1, b_2])$ and $|\psi''|$ is a convex function. Then for any $\alpha > 0$ the following inequality holds

Proof. The inequality in (ⅰ) is obtained by using (2.8):

Suppose $\mu = (1-t)b_1+tb_2$ where $d\mu = (b_2-b_1)dt$,

Taking absolute and triangular inequality,

Putting the values of $I_1$ and $I_2$ in above $(2.16)$, we get:

Remark 10. In Theorem 9, if we take $\alpha = 1$, then we obtain

Theorem 11. Assume that $\psi \in C^2([b_1, b_2])$ and $|\psi''|$ is a convex function. Then for any $\alpha > 0$ the following inequality holds

Proof. We start by recalling the following identity from (2.12):

If $\mu = (1-t)b_1+tb_2$ with $t\in[0, 1]$, then we obtain

Taking absolute values and applying the triangle inequality amounts to the intended result.

Remark 12. Let $\alpha = 1$. Then the inequality in Theorem 11 becomes:

Next, we present results associated with the concave functions.

Theorem 13. Let $\psi\in C^2([b_1, b_2])$ and $\big|\psi''\big|$ be a concave function. Then, for any $\alpha > 0$,

Proof. From (2.15), we have:

Now, we use Jensen's integral inequality to get:

For this we calculate:

and

Putting the values of $I_1$, $I_2$, $A_1$ and $A_2$ in (2.17), we obtain:

Remark 14. If we take $\alpha = 1$ in Theorem 13, then we get:

Theorem 15. Let $\psi\in C^2([b_1, b_2])$ and $\big|\psi''\big|$ be a concave function. Then, for any $\alpha > 0$,

Proof. For this, we use (2.14) to obtain the following inequality:

We use Jensen's integral inequality to get:

Remark 16. If we set $\alpha = 1$, then Theorem 15 amounts to:

3.   Conclusion

By means of a green function, we outlined a new method of proving the Hermite-Hadamard inequality involving the Riemann-Liouville fractional integral operators. In the process of doing this, some new identities were obtained. We applied these identities to prove more results in this direction. Results established in this paper can be recast by using the green function $G_2(\lambda, \mu)$ defined on $[b_1, b_2]\times [b_1, b_2]$ by

Acknowledgments

The authors would like to express their sincere thanks to the editor and the anonymous reviewers for their helpful comments and suggestions. The work was supported by the Natural Science Foundation of China (Grant Nos. 61673169, 11301127, 11701176, 11626101, 11601485).

Conflict of interest

There is no conflict of interest to report.