Hermite-Hadamard-Fejer type inequalities via fractional integral of a function concerning another function

1.   Introduction

Fractional calculus is an important research field, not only in pure mathematics but in applied mathematics, physics, biology, engineering, economics, etc., as well. In fact, by considering derivatives and integrals of arbitrary real or complex order, we may model more efficiently certain real phenomena. Applications have been found e.g., in human body modelling ^[1,2], heat conduction ^[3], viscoelasticity ^[4,5], time series analysis ^[6,7], circuits ^[8], material sciences ^[9], shear waves ^[10], etc. Fractional integral operators play a leading and keen role in the development of fractional calculus. The first formulation of a fractional integral operator is due to a continuous study of well-renowned mathematicians and physicists. This fractional integral is well known as RL fractional integral operator. After its existence, many other fractional integral and fractional derivative operators have been introduced. Recently, scientists in their diverse fields are working in the environment of fractional calculus in new directions of respective fields developing rapidly.

Convex functions are very close to the theory of inequalities. Many known and useful inequalities are consequences of convex functions. Some very natural inequalities for example Jensen inequality, Hadamard inequality interpret convex functions beautifully. Fractional integral inequalities are very useful in the study of fractional partial as well as ordinary differential equations. These inequalities establish the uniqueness and bounds of their solutions. In this paper, we study a general form of RL fractional integrals via convex functions. We start with the definition of a convex function.

In $1905$ Jensen ^[11] present the definition of convex function as follows.

Definition 1.1. A function $\psi:[a, b]\to \mathbb{R}$ is said to be convex if the inequality

holds for all $x, y \in [a, b]$ and all $\theta\in [0, 1]$.

Brechner introduce the definition of $s$-convex function in ^[12] which is defined as follows.

Definition 1.2. A function $\psi:(0, \infty)\to\mathbb{R}$ is said to be $s$-convex in the second sense if the inequality

holds for some fixed $s\in(0, 1]$ and for all $x, y \in (0, \infty), , \theta, \zeta > 0,$ where $\theta+\zeta = 1$. The class of $s$-convex functions of the second sense usually denoted by $K_s^2$.

Definition 1.3. The incomplete beta function $B_x(\theta, \zeta)$ ^[13] is defined by

where $0 < x < 1$, $\theta, \zeta > 0$.

Definition 1.4. The left-sided and right-sided RL fractional integrals $I_{\theta^+}^{\varrho}\psi$ and $I_{\zeta^-}^{\varrho}\psi$ of order $\varrho > 0$ on a finite interval $[\theta, \zeta]$ are defined by

and

respectively. Here, $\Gamma(.)$ represents the usual Euler gamma function with integral representation

Now, we recall the definition of fractional integrals of real valued function concerning to another function ^[14].

Definition 1.5. Let $p:[\theta, \zeta]\rightarrow\mathbb{R}$ be an increasing and positive function on $(\theta, \zeta],$ having a continuous derivative $p'$ on $(\theta, \zeta).$ The left and right-sided RL fractional integrals of $\psi$ with respect to the function $p$ on $[\theta, \zeta]$ of order $\varrho > 0$ are defined respectively by

provided that the integrals exists.

Note that

$\textbf{(i)}$ If we take $p(x) = x$, then we get the Definition $1.4$ of RL fractional integrals.

$\textbf{(ii)}$ If we take $p(x) = \frac{x^{v}}{v}$, $v > 0,$ then we get the definition of Katugampola fractional integrals given in ^[15].

$\textbf{(iii)}$ If we take $p(x) = \frac{x^{\nu+s}}{\nu+s}$, then we get the definition of generalized conformable fractional integrals defined by Khan et al. in ^[16].

$\textbf{(iv)}$ If we take $p(x) = \frac{(x-\theta)^{s}}{s}$, $s > 0$ in left and $p(x) = \frac{(\zeta-x)^{s}}{s}$, $s > 0$ in right, then we get the definition of conformable fractional integrals defined by Jarad et al. in ^[17].

2.   Main results

In this section, we develop an integral identity in the form of the left-sided and right-sided RL fractional integrals of a function concerning to another function. With the help of this identity, we further establish some new inequalities for classical convex and $s$-convex function.

Lemma 2.1. Let $p:[\theta, \zeta]\rightarrow\mathbb{R}$ be an increasing and positive function on $(\theta, \zeta],$ having a continuous derivative $p'$ on $(\theta, \zeta)$. Let $\psi:[\theta, \zeta]$ be a differentiable mapping on $(\theta, \zeta)$ with $\theta < \zeta$ and $g:[\theta, \zeta]\rightarrow\mathbb{R}$ be bounded. If $\psi', g\in L[\theta, \zeta],$ then following identity holds:

where

Proof. Consider

Now, integrating $I_{1}$ by parts, we get

Similarly, integrating $I_{2}$ by parts, we get

Therefore, by substituting the values of $I_{1}$ and $I_{2}$ in (2.1), we get the desired result.

Corollary 2.2. If we choose $p(\nu) = \nu$ in Lemma 2.1, then it reduce to [18,Lemma 4], i.e.,

where

Theorem 2.3. Let $\varrho > 0$ and $p:[\theta, \zeta]\rightarrow\mathbb{R}$ be an increasing and positive function on $(\theta, \zeta],$ having a continuous derivative $p'$ on $(\theta, \zeta)$. Let $g:[\theta, \zeta]\rightarrow\mathbb{R}$ be nonnegative, integrable and $\psi$ is a convex function on $[\theta, \zeta]$, then the following Hermite-Hadamard-Fejer type inequality for generalized fractional integrals holds.

where $\Psi$ is defined as $\Psi(\nu) = \psi(\nu)+\tilde{\psi}(\nu)$ and $\tilde{\psi}(\nu) = \psi(\theta+\zeta-\nu)$.

Proof. Since $\psi$ is convex on $[\theta, \zeta]$, for all $x, y\in[\theta, \zeta]$, we have

Now, for $\nu\in[0, 1]$, choose $x = \frac{\nu}{2}\theta+\frac{2-\nu}{2}\zeta$ and $y = \frac{2-\nu}{2}\theta+\frac{\nu}{2}\zeta$, we get

Multiplying both sides of (2.3) by

and integrating the resulting inequality with respect to $\nu$ over $[0, 1]$, we get

Substituting $u = \frac{\nu}{2}\theta+\frac{2-\nu}{2}\zeta$, we obtain

i.e.

Similarly, multiplying both sides of (2.3) by

and integrating the resulting inequality with respect to $\nu$ over $[0, 1]$, we obtain

By adding the inequalities (2.4) and (2.5), we have

which prove the left half part of inequality (2.2).

For the proof of second half, since $\psi$ is convex on $[\theta, \zeta]$, we have

Multiplying both sides of (2.7) by

and integrating the resulting inequality with respect to $\nu$ over $[0, 1]$, we get

Using the change of variable $u = \frac{\nu}{2}\theta+\frac{2-\nu}{2}\zeta$, we get

that is

Similarly, multiplying both sides of (2.7) by

and integrating the resulting inequality with respect to $\nu$ over $[0, 1]$, we get

By adding (2.8) and (2.9), we obtain

From (2.6) and (2.10), we get inequality (2.2).

Corollary 2.4. If we take $p(\nu) = \nu$ and $g$ is symmetric to $\frac{\zeta+\theta}{2}$ in Theorem 2.3, then it reduces to the following inequality

for RL fractional integral.

Theorem 2.5. Let $p:[\theta, \zeta]\rightarrow\mathbb{R}$ be an increasing and positive function on $(\theta, \zeta],$ having a continuous derivative $p'$ on $(\theta, \zeta).$ Let $\psi :I \rightarrow \mathbb{R}$ be a differentiable function on $I^0$ and $\psi'\in L[\theta, \zeta]$ and $g : [\theta, \zeta] \rightarrow \mathbb{R}$ is continuous function. If $|\psi'|\in K^2_s$ on $[\theta, \zeta]$ for some fixed $s\in (0, 1]$, then the following inequality for fractional integrals holds.

where $\Psi$ is defined as $\Psi(\nu) = \psi(\nu)+\tilde{\psi}(\nu)$ and $\tilde{\psi}(\nu) = \psi(\theta+\zeta-\nu)$.

Proof. By using the pre define properties, we can write

By using Lemma 2.1 and inequality (2.11), we get

The proof is done.

Corollary 2.6. If we take $s = 1$ in Theorem 2.5, then we have

Corollary 2.7. If we take $s = 1$, $p(\nu) = \nu$ and $g$ is symmetric to $\frac{\theta+\zeta}{2}$ in Theorem 2.5, then we get [18,Theorem 6], i.e.,

Corollary 2.8. If we choose $p(\nu) = \nu$ and $g$ is symmetric to $\frac{\theta+\zeta}{2}$ in Theorem 2.5, then we get [19,Theorem 4]

Theorem 2.9. Let $p:[\theta, \zeta]\rightarrow\mathbb{R}$ be an increasing and positive function on $(\theta, \zeta],$ having a continuous derivative $p'$ on $(\theta, \zeta).$ Let $\psi :I \rightarrow \mathbb{R}$ be a differentiable function on $I^0$ and $\psi'\in L[\theta, \zeta]$ and $g : [\theta, \zeta] \rightarrow \mathbb{R}$ is continuous function. If $|\psi'|^{q}\in K^2_s$ on $[\theta, \zeta]$ for some fixed $s\in (0, 1]$ and $q\geq 1$, then the following inequality for fractional integrals holds.

Proof. By Lemma 2.1, Power mean inequality and Definition 1.2, we have

The proof is done.

Corollary 2.10. If we choose $s = 1$ in Theorem 2.9, we get

Corollary 2.11. If we choose $s = 1$, and $p(\nu) = \nu$ in Theorem 2.9, then it reduces to [18,Theorem 7], i.e.,

Corollary 2.12. If we take $p(\nu) = \nu$ in Theorem 2.9, then it reduce to [19,Theorem 5], that is

Theorem 2.13. Let $p:[\theta, \zeta]\rightarrow\mathbb{R}$ be an increasing and positive function on $(\theta, \zeta],$ having a continuous derivative $p'$ on $(\theta, \zeta).$ Let $\psi :I \rightarrow \mathbb{R}$ be a differentiable function on $I^0$ and $\psi'\in L[\theta, \zeta]$ and $g : [\theta, \zeta] \rightarrow \mathbb{R}$ is continuous function. If $|\psi'|^{q} \in K^2_s$ on $[\theta, \zeta]$ for some fixed $s\in (0, 1]$ and $q > 1$, then the following inequality for fractional integrals holds.

Proof. By Lemma 2.1, Hölder's inequality and Definition 1.2, we obtain

That completes the proof of the result.

Corollary 2.14. If we choose $s = 1$ in Theorem 2.13, we have

Corollary 2.15. If we take $s = 1$, and $p(\nu) = \nu$ in Theorem 2.13, then it reduces to [18,Theorem 8], that is

where $\frac{1}{p}+\frac{1}{q} = 1$.

Corollary 2.16. If we take $p(\nu) = \nu$ in Theorem 2.13, then we get [19,Theorem 6], that is

where $\frac{1}{p}+\frac{1}{q} = 1$.

3.   Applications to estimations of some special means

This section consists of some particular inequalities which generalizes some classical results such like mid-point inequality.

Proposition 3.1. (Mid-point inequality). By using the assumptions $\varrho = 1,$ $g(\nu) = 1$ and $p(\nu) = \nu$ in Corollary 2.6, we get the following inequality

Proposition 3.2. (Mid-point inequality). By using the assumptions $\varrho = 1$ and $p(\nu) = \nu$ in Corollary 2.10, we get the following inequality

Proposition 3.3. By using the assumptions $\varrho = 1$ and $p(\nu) = \nu$ in Corollary 2.14, we get the following inequality

4.   Conclusions

In the present article, we aim to design the generalized inequalities for RL fractional integral of a function concerning the other function. For this purpose, we use the classical convex and $s$-convex mappings and develop several inequalities. This work includes equality so that, we can make progress in finding more inequalities by using different types of functions in equality. The findings of these investigations complement those of previous studies. Simply, the recent study confirms the earlier results and play an additional role by making generalizations.

Conflict of interest

The authors declares that there is no conflict of interests regarding the publication of this paper.