Hadamard–Mercer, Dragomir–Agarwal–Mercer, and Pachpatte–Mercer Type Fractional Inclusions for Convex Functions with an Exponential Kernel and Their Applications

Abstract

Many scholars have recently become interested in establishing integral inequalities using various known fractional operators. Fractional calculus has grown in popularity as a result of its capacity to quickly solve real-world problems. First, we establish new fractional inequalities of the Hadamard–Mercer, Pachpatte–Mercer, and Dragomir–Agarwal–Mercer types containing an exponential kernel. In this regard, the inequality proved by Jensen and Mercer plays a major role in our main results. Integral inequalities involving convexity have a wide range of applications in several domains of mathematics where symmetry is important. Both convexity and symmetry are closely linked with each other; when working on one of the topics, you can apply what you have learned to the other. We consider a new identity for differentiable mappings and present its companion bound for the Dragomir–Agarwal–Mercer type inequality employing a convex function. Applications involving matrices are presented. Finally, we conclude our article and discuss its future scope.

1. Introduction

The significance of fractional calculus has lately been recognized, and various publications incorporating fractional calculus have recently been published. Fractional operators are particularly important in the development of fractional calculus. Many diverse branches of engineering and science, including physics [1], nanotechnology [2], medicine [3], bioengineering [4], economics [5], fluid mechanics [6], epidemiology [7], and control systems [8] use fractional calculus. Fractional operators have been shown in several studies to accurately explain complicated multiscale phenomena that are difficult to represent using traditional mathematical calculus. The importance of fractional calculus is now widely recognized, and various studies involving fractional calculus have been carried out.

Fractional calculus is well-known for its goal of building mathematical models. Refining well-known integral inequalities using fractional integral operators has recently been a fascinating subject of research among mathematicians. Sarikaya [9] proposed the concept of using a fractional integral operator to present the Hermite–Hadamard inequality. Inspired by this article, Liu et al. [10] generalized the Hermite–Hadamard inequality for the $\psi$-Riemann–Liouville fractional operator. Analogously, to list a few authors: Mumcu et al. [11] used a generalized proportional fractional integral, Gürbüz et al. [12] worked on the Caputo–Fabrizio operator, Fernandez et al. [13] used the concept of the Atangana–Baleanu fractional operator with the Mittag–Leffler kernel, Khan et al. [14] worked on a generalized conformable operator, Mohammed et al. [15] utilized tempered fractional integrals, and Sahoo et al. [16] used $k$-Riemann–Liouville fractional integrals to improve the Hermite–Hadamard inequality. The use of several types of fractional operators and innovative approaches to generalizing some existing integral inequalities, such as Hermite–Hadamard, Simpson, Opial, Fejér, and Ostrowski-type inequalities, led to a revolution in inequality theory.

Based on their basic properties, researchers derive novel fractional operators and use them to tackle a variety of real-world issues. New fractional operators are used in various monographs and papers to improve integral inequalities, such as the Minkowski inequality, the Simpson inequality, the Jensen–Mercer inequality, the Fejér-type inequality, the Hermite–Hadamard-type inequality, and others. These forms of analytic inequalities, as well as the approaches described in this article, have applications in a variety of domains where symmetry is important. For more information on current developments in fractional integral inequalities, see articles [17,18,19,20].

Integral inequalities are widely employed in a variety of fields, such as mathematics, applied sciences, differential equations, and functional analysis. Researchers have been studying these inequalities for the past two decades. Different types of integral inequalities are employed in most mathematical analysis disciplines. Approximation theory and numerical analysis, which estimate error approximation, rely heavily on them.

This article focuses on a new fractional operator having an exponential function in the kernel. In recent years, numerous analysts have looked into many possible approaches to describing new fractional operators using various methodologies that can handle the majority of real-world situations. The kernel structures and other aspects of fractional operators usually differ from one another. The main feature that distinguishes this new fractional operator from others is that it has an exponential function in its kernel, which is useful for finding precise solutions to problems.

Definition 1

([21,22]). A function $\mathcal{S}:\mathbb{X}\to \mathfrak{R}$ said to be a convex function, if

$$\mathcal{S}\left({\sigma }_1\varsigma +\left(1-\varsigma \right){\sigma }_2\right)\leqq \varsigma \mathcal{S}\left({\sigma }_1\right)+\left(1-\varsigma \right)\mathcal{S}\left({\sigma }_2\right),$$

(1)

holds true for all ${\sigma }_1,{\sigma }_2\in \mathbb{X}$ and $\varsigma \in [0,1].$

To encourage further discussion, we first present the classical Hermite–Hadmard inequality that states (see [23]):

If $\mathcal{S}:\mathbb{X}\subseteq \mathfrak{R}\to \mathfrak{R}$ is a convex function in $\mathbb{X}$ for ${\sigma }_1,{\sigma }_2\in \mathbb{X}$ and ${\sigma }_1<{\sigma }_2$, then

$$\mathcal{S}\left(\frac{{\sigma }_1+{\sigma }_2}{2}\right)\le \frac{1}{{\sigma }_2-{\sigma }_1}{\int }_{{\sigma }_1}^{{\sigma }_2}\mathcal{S}\left(x\right)dx\le \frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2}.$$

(2)

We direct interested readers to articles [9,24,25,26,27] for generalizations of the H–H inequality.

The Jensen’s inequality (see [28]) states that:

If the convexity of $\mathcal{S}$ holds true on the interval $\left[{\wp }_1,{\wp }_2\right]$, then

$$\mathcal{S}\left(\underset{k=1}{\sum ^n}{\theta }_k{\eta }_k\right)\leqq \left(\underset{k=1}{\sum ^n}{\theta }_k\mathcal{S}\left({\eta }_k\right)\right),$$

(3)

for all ${\eta }_k\in [\mu ,\nu ],$${\theta }_k\in \left[0,1\right]and\left(k=1,2,\dots ,n\right).$

Mercer in [29] investigated a generalized form of Jensen’s inequality, which is famously known as the J–M inequality, given as:

Let $\mathcal{S}$ be a convex function on $\left[\mu ,\nu \right]$, then

$$\mathcal{S}\left(\mu +\nu -\underset{k=1}{\sum ^n}{\varsigma }_k{\gamma }_k\right)\leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\underset{k=1}{\sum ^n}{\varsigma }_k\mathcal{S}\left({\gamma }_k\right),$$

(4)

holds true for all ${\gamma }_k\in \left[\mu ,\nu \right],$ ${\varsigma }_k\in \left[0,1\right]$ and $\left(k=1,2,\dots ,n\right).$

Then, Matkovic [30] and Vivas et al. [31] published new versions of the Jensen–Mercer inequality in the years 2016 and 2017, respectively. In [32], Kian and Moslehian used classical integrals to prove the following two improved inequalities for convex functions:

Theorem 1

([32]). Let $\mathcal{S}:[{\sigma }_1,{\sigma }_1]⟶\mathbb{R}$ be a convex function. Then the following H–H–M-type fractional integral inequality holds true:

$$\begin{array}{c}\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_1}{2}\right)\leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-{\int }_0^1\mathcal{S}(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2)d\varsigma \hfill \\ \hfill \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\mathcal{S}\left(\frac{{\sigma }_1+{\sigma }_2}{2}\right),\end{array}$$

and

$$\begin{array}{c}\hfill \mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_1}{2}\right)\leqq \frac{1}{{\sigma }_2-{\sigma }_1}{\int }_{{\sigma }_1}^{{\sigma }_2}\mathcal{S}(\mu +\nu -\varsigma )d\varsigma \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2},\end{array}$$

for all ${\sigma }_1,{\sigma }_2\in [\mu ,\nu ]$ and $\zeta >0.$

Much research on the Jensen–Mercer inequality has been conducted recently; see ([33,34,35]). Some researchers worked in this direction to improve the Hermite–Hadamard inequality via the Jensen–Mercer setting (see [36,37,38]). Recently, the Hermite–Hadamard–Mercer type inequality was also improved by You et al. [39] for the harmonically convex function, by Khan et al. [40] for the Csiszár divergence and the Zipf–Mandelbrot entropy, and by Butt et al. [41,42] for fractional operators.

Inspired by the work performed by the above authors, we structure the paper in the following manner. We review some basic definitions and concepts connected to integral inequalities in Section 2. We develop a new variant of the Hermite–Hadamard–Mercer and Pachpatte–Mercer type inequalities employing a novel fractional operator in Section 3. Next, Section 4 is devoted to proving a new fractional integral identity and a Dragomir–Agarwal–Mercer type inequality involving an exponential function in the kernel for differentiable mappings. We derive several concrete examples as applications of our results in Section 5. In the last Section 6, we provide a brief conclusion and discuss some ideas for future research.

2. Preliminaries

To encourage additional conversation about this article, we present the definition of the R–L fractional operator and related Hermite–Hadamard–Mercer type inequalities.

Definition 2

(see [9]). Let $[{\sigma }_1,{\sigma }_1]⟶\mathfrak{R}$. Then, R-L fractional integrals ${\mathrm{I}}_{{\sigma }_1^{+}}^{\zeta }\mathcal{S}\left(x\right)$ and ${\mathrm{I}}_{{\sigma }_2^{-}}^{\zeta }\mathcal{S}\left(x\right)$ of order $\zeta >0$ are defined by

$${\mathrm{I}}_{{\sigma }_1^{+}}^{\zeta }\mathcal{S}\left(z\right)=\frac{1}{\Gamma \left(\zeta \right)}{\int }_{{\sigma }_1}^z{(z-x)}^{\zeta -1}\mathcal{S}\left(\mathsf{x}\right)dx$$

and

$${\mathrm{I}}_{{\sigma }_2^{-}}^{\zeta }\mathcal{S}\left(z\right)=\frac{1}{\Gamma \left(\zeta \right)}{\int }_z^{{\sigma }_2}{(x-z)}^{\zeta -1}\mathcal{S}\left(x\right)dx,$$

where $\Gamma (.)$ is the Gamma function.

In [43], Öğülmüs and Sarikaya employed the R–L fractional integrals to generalize the Hermite–Hadamard–Mercer-type inequalities as follows:

Theorem 2

([43]). Let $\mathcal{S}:[\mu ,\mu ]⟶\mathbb{R}$ be a convex function. Then the following fractional integral inequality holds true:

$$\begin{array}{c}\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_1}{2}\right)\leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\Gamma (\zeta +1)}{2{\left({\sigma }_2-{\sigma }_1\right)}^{\zeta }}\left[{\mathrm{I}}_{{\sigma }_1^{+}}^{\zeta }\mathcal{S}\left({\sigma }_2\right)+{\mathrm{I}}_{{\sigma }_2^{-}}^{\zeta }\mathcal{S}\left({\sigma }_1\right)\right]\hfill \\ \hfill \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\mathcal{S}\left(\frac{{\sigma }_1+{\sigma }_2}{2}\right),\end{array}$$

and

$$\begin{array}{c}\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_1}{2}\right)\hfill \\ \hspace{1em}\hspace{1em}\leqq \frac{\Gamma (\zeta +1)}{2{\left({\sigma }_2-{\sigma }_1\right)}^{\zeta }}\left[{\mathrm{I}}_{{(\mu +\nu -{\sigma }_2)}^{+}}^{\zeta }\mathcal{S}(\mu +\nu -{\sigma }_1)+{\mathrm{I}}_{{(\mu +\nu -{\sigma }_1)}^{-}}^{\zeta }\mathcal{S}(\mu +\nu -{\sigma }_2)\right]\\ \hfill \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2},\end{array}$$

for all ${\sigma }_1,{\sigma }_2\in [\mu ,\nu ]$ and $\zeta >0.$

Many researchers have been interested in applying novel fractional operators to the Jensen–Mercer inequality over the years. Some further interesting fractional refinements on the Hermite–Hadamard–Jensen–Mercer inequalities have been widely studied in the literature; see [44,45,46,47,48,49,50,51]. Researchers have been extremely interested in their generalizations and improvements during the previous few decades, as evidenced by a significant number of publications on the theme.

Before getting into the key conclusions, we go over the definitions of the new fractional operators and some fundamental concepts of inequality theory.

Definition 3

(see [52] for details). Let ${\rm Y}\in \mathcal{L}\left[{\varrho }_1,{\varrho }_2\right]$. Then the fractional integrals ${\mathfrak{J}}_{{\varrho }_1^{+}}^{\alpha }$ and ${\mathfrak{J}}_{{\varrho }_2^{-}}^{\alpha }$ of order $\alpha >0$ are defined as

$${\mathfrak{J}}_{{\varrho }_1}^{\alpha }{\rm Y}\left(x\right):=\frac{1}{\alpha }{\int }_{{\varrho }_1}^x{e}^{-\frac{1-\alpha }{\alpha }\left(x-\mathtt{z}\right)}{\rm Y}\left(\mathtt{z}\right)d\mathtt{z}(0\leqq {\varrho }_1<x<{\varrho }_2)$$

and

$${\mathfrak{J}}_{{\varrho }_2^{-}}^{\alpha }{\rm Y}\left(x\right):=\frac{1}{\alpha }{\int }_x^{{\varrho }_2}{e}^{-\frac{1-\alpha }{\alpha }\left(\mathtt{z}-x\right)}{\rm Y}\left(\mathtt{z}\right)d\mathtt{z}(0\leqq {\varrho }_1<x<{\varrho }_2),$$

respectively.

For brevity, we denote $\rho =\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)$ throughout the manuscript.

3. Hadamard–Jensen–Mercer- and Pachpatte–Mercer-Type Inequalities via Fractional Integrals

Theorem 3.

Let $\mathcal{S}:\mathrm{I}=[{\sigma }_1,{\sigma }_2]\to \mathfrak{R}$ be a convex function on $\mathrm{I}$ such that $({\sigma }_1,{\sigma }_2)\in \mathrm{I}$ and $\mathcal{S}\in \mathcal{L}[{\sigma }_1,{\sigma }_2]$. Then, for $0\leqq {\sigma }_1<{\sigma }_2$ and $\zeta \in [0,1]$ the following fractional inequality holds true:

$$\begin{array}{c}\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}\leqq [\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{{\sigma }_1^{+}}^{\alpha }{\rm Y}\left({\sigma }_2\right)+{\mathfrak{J}}_{{\sigma }_2^{-}}^{\alpha }{\rm Y}\left({\sigma }_1\right)\right]\\ \hfill \leqq [\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-\mathcal{S}\left(\frac{{\sigma }_1+{\sigma }_2}{2}\right).\end{array}$$

(5)

Proof. 

Since $\mathcal{S}$ is a convex function on $[{\sigma }_1,{\sigma }_2]$, by the hypotheses of Jensen–Mercer inequality, we can write

$$\begin{array}{c}\hfill \mathcal{S}\left(\mu +\nu -\frac{{{\rm Y}}_1+{{\rm Y}}_2}{2}\right)\leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\mathcal{S}\left(\frac{{{\rm Y}}_1+{{\rm Y}}_2}{2}\right)\end{array}$$

Now, if we let ${{\rm Y}}_1=\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2$ and ${{\rm Y}}_2=\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1$, we have

$$\begin{array}{c}\hfill \mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)+\mathcal{S}\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)}{2}.\end{array}$$

If we multiply the above inequality by $e^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }$ and then integrate the obtained inequality over [0, 1], we find that

$$\begin{array}{c}{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)d\varsigma \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right){\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }d\varsigma \hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}-\frac{1}{2}{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\left[\mathcal{S}\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)+\mathcal{S}\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)\right]d\varsigma ,\end{array}$$

consequently,

$$\begin{array}{c}\left(\frac{1-e^{-\rho }}{\rho }\right)\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\leqq [\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]\left(\frac{1-e^{-\rho }}{\rho }\right)\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}-\frac{1}{2({\sigma }_2-{\sigma }_1)}\left[{\int }_{{\sigma }_1}^{{\sigma }_2}{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-\mathfrak{u})}\mathcal{S}\left(\mathfrak{u}\right)d\mathfrak{u}+{\int }_{{\sigma }_1}^{{\sigma }_2}{e}^{-\frac{1-\alpha }{\alpha }(\mathfrak{u}-{\sigma }_1)}\mathcal{S}\left(\mathfrak{u}\right)d\mathfrak{u}\right],\end{array}$$

which readily yields

$$\begin{array}{c}\hfill \mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\leqq [\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{{\sigma }_1^{+}}^{\alpha }{\rm Y}\left({\sigma }_2\right)+{\mathfrak{J}}_{{\sigma }_2^{-}}^{\alpha }{\rm Y}\left({\sigma }_1\right)\right].\end{array}$$

(6)

This gives us the first part of the desired result. For the second part, we use the convexity of $\mathcal{S}$.

$$\begin{array}{cc}\hfill \mathcal{S}\left(\frac{{{\rm Y}}_1+{{\rm Y}}_2}{2}\right)& =\mathcal{S}\left(\frac{\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)+\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)}{2}\right)\hfill \\ \hfill & \leqq \frac{\mathcal{S}\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)+\mathcal{S}\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)}{2}.\hfill \end{array}$$

(7)

If we multiply the above Equation (7) by $e^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }$ and then integrate the obtained inequality over [0, 1], we find that

$$\begin{array}{cc}\hfill & \mathcal{S}\left(\frac{{{\rm Y}}_1+{{\rm Y}}_2}{2}\right){\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }d\varsigma \hfill \\ \hfill & \leqq \frac{1}{2}\left[{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\mathcal{S}\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)d\varsigma +{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\mathcal{S}\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)d\varsigma \right].\hfill \end{array}$$

It follows from the above developments that

$$\left(\frac{1-e^{-\rho }}{\rho }\right)\mathcal{S}\left(\frac{{\sigma }_1+{\sigma }_2}{2}\right)\leqq \frac{\alpha }{2({\sigma }_2-{\sigma }_1)}\left[{\mathfrak{J}}_{{\sigma }_1^{+}}^{\alpha }{\rm Y}\left({\sigma }_2\right)+{\mathfrak{J}}_{{\sigma }_2^{-}}^{\alpha }{\rm Y}\left({\sigma }_1\right)\right].$$

This implies

$$-\mathcal{S}\left(\frac{{\sigma }_1+{\sigma }_2}{2}\right)\geqq -\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{{\sigma }_1^{+}}^{\alpha }{\rm Y}\left({\sigma }_2\right)+{\mathfrak{J}}_{{\sigma }_2^{-}}^{\alpha }{\rm Y}\left({\sigma }_1\right)\right].$$

(8)

Now adding $\left[\mathcal{S}\right(\mu )+\mathcal{S}(\nu \left)\right]$ to both sides of the above Equation (8), we have

$$\begin{array}{c}[\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-\mathcal{S}\left(\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\geqq [\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{{\sigma }_1^{+}}^{\alpha }{\rm Y}\left({\sigma }_2\right)+{\mathfrak{J}}_{{\sigma }_2^{-}}^{\alpha }{\rm Y}\left({\sigma }_1\right)\right].\end{array}$$

(9)

From Equations (6) and (9), the proof is completed. □

Theorem 4.

Let $\mathcal{S}:\mathrm{I}=[{\sigma }_1,{\sigma }_2]\to \mathfrak{R}$ be a convex function on $\mathrm{I}$ such that $({\sigma }_1,{\sigma }_2)\in \mathrm{I}$ and $\mathcal{S}\in \mathcal{L}[{\sigma }_1,{\sigma }_2]$. Then, for $0\leqq {\sigma }_1<{\sigma }_2$ and $\zeta \in [0,1]$ the following fractional inequality holds true:

$$\begin{array}{cc}\hfill \mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)& \leqq \frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & \leqq \mathcal{S}(\mu +\nu -{\sigma }_1)+\mathcal{S}(\mu +\nu -{\sigma }_2)\hfill \\ \hfill & \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2}.\hfill \end{array}$$

(10)

Proof. 

Let $\mathcal{S}:[{\sigma }_1,{\sigma }_2]\to \mathfrak{R}$ be a convex function. Then by hypothesis, we have

$$\begin{array}{cc}\hfill \mathcal{S}\left(\mu +\nu -\frac{{{\rm Y}}_1+{{\rm Y}}_2}{2}\right)& =\mathcal{S}\left(\frac{\mu +\nu -{{\rm Y}}_1+\mu +\nu -{{\rm Y}}_2}{2}\right)\hfill \\ \hfill & \leqq \frac{1}{2}\left(\mathcal{S}(\mu +\nu -{{\rm Y}}_1)+\mathcal{S}(\mu +\nu -{{\rm Y}}_2)\right).\hfill \end{array}$$

(11)

Now, if we change the variables as

$$\mathcal{S}(\mu +\nu -{{\rm Y}}_1)=\varsigma (\mu +\nu -{\sigma }_1)+(1-\varsigma )(\mu +\nu -{\sigma }_2),$$

and

$$\mathcal{S}(\mu +\nu -{{\rm Y}}_2)=\varsigma (\mu +\nu -{\sigma }_2)+(1-\varsigma )(\mu +\nu -{\sigma }_1)$$

in Equation (11), we have

$$\begin{array}{cc}\hfill & \mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hfill & \leqq \frac{[\mathcal{S}(\varsigma (\mu +\nu -{\sigma }_1)+(1-\varsigma )(\mu +\nu -{\sigma }_2))+\mathcal{S}(\varsigma (\mu +\nu -{\sigma }_2)+(1-\varsigma )(\mu +\nu -{\sigma }_1))]}{2}.\hfill \end{array}$$

(12)

Multiplying both sides of the above Equation (12) by $e^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }$ and then integrating the obtained result over $[0,1]$, we find that

$$\begin{array}{cc}\hfill & \left(\frac{1-e^{-\rho }}{\rho }\right)\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hfill & \leqq \frac{1}{2({\sigma }_2-{\sigma }_1)}\left[{\int }_{\mu +\nu -{\sigma }_2}^{\mu +\nu -{\sigma }_1}{e}^{-\frac{1-\alpha }{\alpha }(\mathfrak{u}-(\mu +\nu -{\sigma }_2))}\mathcal{S}\left(\mathfrak{u}\right)d\mathfrak{u}+{\int }_{\mu +\nu -{\sigma }_2}^{\mu +\nu -{\sigma }_1}{e}^{-\frac{1-\alpha }{\alpha }(\left(\mu +\nu -{\sigma }_1\right)-\mathfrak{u})}\mathcal{S}\left(\mathfrak{u}\right)d\mathfrak{u}\right]\hfill \\ \hfill & =\frac{\alpha }{2({\sigma }_2-{\sigma }_1)}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right],\hfill \end{array}$$

which readily yields

$$\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\leqq \frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right].$$

This leads us to the first part of the proof. Now, for the second part, we use the convexity of $\mathcal{S}$, given as

$$\mathcal{S}(\varsigma (\mu +\nu -{\sigma }_1)+(1-\varsigma )(\mu +\nu -{\sigma }_2))\leqq \varsigma \mathcal{S}(\mu +\nu -{\sigma }_1)+(1-\varsigma )\mathcal{S}(\mu +\nu -{\sigma }_2).$$

$$\mathcal{S}(\varsigma (\mu +\nu -{\sigma }_2)+(1-\varsigma )(\mu +\nu -{\sigma }_1))\leqq \varsigma \mathcal{S}(\mu +\nu -{\sigma }_2)+(1-\varsigma )\mathcal{S}(\mu +\nu -{\sigma }_1).$$

Adding both the inequalities, we find that

$$\begin{array}{cc}\hfill & \mathcal{S}(\varsigma (\mu +\nu -{\sigma }_1)+(1-\varsigma )(\mu +\nu -{\sigma }_2))+\mathcal{S}(\varsigma (\mu +\nu -{\sigma }_2)+(1-\varsigma )(\mu +\nu -{\sigma }_1))\hfill \\ \hfill & \leqq \mathcal{S}(\mu +\nu -{\sigma }_1)+\mathcal{S}(\mu +\nu -{\sigma }_2)\hfill \\ \hfill & \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\mathcal{S}\left({\sigma }_2\right)\hfill \\ \hfill & =2[\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-[\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)].\hfill \end{array}$$

(13)

Multiplying both sides of the above Equation (13) by $e^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }$ and then integrating over $[0,1]$, we have

$$\begin{array}{c}{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\mathcal{S}(\varsigma (\mu +\nu -{\sigma }_1)+(1-\varsigma )(\mu +\nu -{\sigma }_2))d\varsigma \hfill \\ \hspace{1em}\hspace{1em}\hspace{1em}+{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\mathcal{S}(\varsigma (\mu +\nu -{\sigma }_2)+(1-\varsigma )(\mu +\nu -{\sigma }_1))d\varsigma \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}\leqq 2[\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-[\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)]{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }d\varsigma .\end{array}$$

It follows from the above developments that,

$$\begin{array}{cc}\hfill & \frac{\alpha }{2({\sigma }_2-{\sigma }_1)}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & \leqq \left[\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2}\right]\left[\frac{1-\alpha }{(1-e^{-\rho })}\right],\hfill \end{array}$$

which readily yields

$$\begin{array}{cc}\hfill & \frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2}.\hfill \end{array}$$

This leads us to the proof of the desired Theorem 4. □

Theorem 5.

Let $\mathcal{S}:\mathrm{I}=[{\sigma }_1,{\sigma }_2]\to \mathfrak{R}$ be a convex function on $\mathrm{I}$ such that $({\sigma }_1,{\sigma }_2)\in \mathrm{I}$ and $\mathcal{S}\in \mathcal{L}[{\sigma }_1,{\sigma }_2]$. Then, for $0\leqq {\sigma }_1<{\sigma }_2$ and $\zeta \in [0,1]$ the following fractional inequality holds true:

$$\begin{array}{c}\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hspace{1em}\hspace{1em}\leqq \frac{1-\alpha }{2(1-e^{-\frac{\rho }{2}})}\left[{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\\ \hfill \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2}.\end{array}$$

(14)

Proof. 

Using the convexity of $\mathcal{S}$ on $[{\sigma }_1,{\sigma }_2]$, one has

$$\begin{array}{cc}\hfill \mathcal{S}\left(\mu +\nu -\frac{{{\rm Y}}_1+{{\rm Y}}_2}{2}\right)& =\mathcal{S}\left(\frac{\mu +\nu -{{\rm Y}}_1+\mu +\nu -{{\rm Y}}_2}{2}\right)\hfill \\ \hfill & \leqq \frac{\mathcal{S}(\mu +\nu -{{\rm Y}}_1)+\mathcal{S}(\mu +\nu -{{\rm Y}}_2)}{2}.\hfill \end{array}$$

If we let ${{\rm Y}}_1=\frac{\varsigma }{2}{\sigma }_1+\frac{2-\varsigma }{2}{\sigma }_2$ and ${{\rm Y}}_1=\frac{\varsigma }{2}{\sigma }_2+\frac{2-\varsigma }{2}{\sigma }_1$, we have

$$\begin{array}{c}2\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hfill \hspace{1em}\hspace{1em}\hspace{1em}\leqq \mathcal{S}\left(\mu +\nu -\left(\frac{\varsigma }{2}{\sigma }_1+\frac{2-\varsigma }{2}{\sigma }_2\right)\right)+\mathcal{S}\left(\mu +\nu -\left(\frac{\varsigma }{2}{\sigma }_2+\frac{2-\varsigma }{2}{\sigma }_1\right)\right).\end{array}$$

(15)

If we multiply the above Equation (15) by $e^{-\frac{1-\alpha }{2\alpha }({\sigma }_2-{\sigma }_1)\varsigma }$ and then integrate the obtained inequality over [0, 1]

$$\begin{array}{cc}\hfill & \frac{4(1-e^{-\frac{\rho }{2}})}{\rho }\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hfill & \leqq {\int }_0^1{e}^{-\frac{1-\alpha }{2\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\mathcal{S}\left(\mu +\nu -\left(\frac{\varsigma }{2}{\sigma }_1+\frac{2-\varsigma }{2}{\sigma }_2\right)\right)d\varsigma \hfill \\ & +{\int }_0^1{e}^{-\frac{1-\alpha }{2\alpha }({\sigma }_2-{\sigma }_1)\varsigma }\mathcal{S}\left(\mu +\nu -\left(\frac{\varsigma }{2}{\sigma }_2+\frac{2-\varsigma }{2}{\sigma }_1\right)\right)d\varsigma .\hfill \\ \hfill & \leqq \frac{2}{{\sigma }_2-{\sigma }_1}{\int }_{\mu +\nu -{\sigma }_2}^{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}{e}^{-\frac{1-\alpha }{\alpha }\left(\mathfrak{u}-(\mu +\nu -{\sigma }_2)\right)}\mathcal{S}\left(\mathfrak{u}\right)d\mathfrak{u}\hfill \\ \hfill & +\frac{2}{{\sigma }_2-{\sigma }_1}{\int }_{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}^{\mu +\nu -{\sigma }_1}{e}^{-\frac{1-\alpha }{\alpha }\left((\mu +\nu -{\sigma }_1)-\mathfrak{u}\right)}\mathcal{S}\left(\mathfrak{u}\right)d\mathfrak{u}\hfill \\ \hfill & =\frac{2\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right].\hfill \end{array}$$

It follows from the above developments that

$$\begin{array}{c}\mathcal{S}\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)\hfill \\ \hfill \hspace{1em}\leqq \frac{1-\alpha }{2(1-e^{-\frac{\rho }{2}})}\left[{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right].\end{array}$$

(16)

This leads us to the first part of the proof. Now, for the second part, we use the J–M inequality

$$\begin{array}{c}\mathcal{S}\left(\mu +\nu -\left(\frac{\varsigma }{2}{\sigma }_1+\frac{2-\varsigma }{2}{\sigma }_2\right)\right)+\mathcal{S}\left(\mu +\nu -\left(\frac{\varsigma }{2}{\sigma }_2+\frac{2-\varsigma }{2}{\sigma }_1\right)\right)\hfill \\ \hfill \leqq 2\left[\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)\right]-\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right).\end{array}$$

(17)

If we multiply the above Equation (17) by $e^{-\frac{1-\alpha }{2\alpha }({\sigma }_2-{\sigma }_1)\varsigma }$ and then integrate the obtained result over [0, 1], we find

$$\begin{array}{cc}\hfill & \frac{2\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & \leqq \left[2\left[\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)\right]-\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)\right]{\int }_0^1{e}^{-\frac{1-\alpha }{2\alpha }({\sigma }_2-{\sigma }_1)\varsigma }d\varsigma .\hfill \\ \hfill & =\left[2\left[\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)\right]-\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)\right]\frac{2(1-e^{-\frac{\rho }{2}})}{\rho },\hfill \end{array}$$

from which readily follows

$$\begin{array}{c}\frac{1-\alpha }{2(1-e^{-\frac{\rho }{2}})}\left[{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{{\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill \leqq \mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)-\frac{\mathcal{S}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)}{2}.\end{array}$$

(18)

Reorganizing Equations (16) and (18) completes the proof of Theorem 5. □

Pachpatte–Mercer-Type Fractional Inequality

In this section, we use a novel fractional operator to present an inequality that takes the product of two convex functions in the Jensen–Mercer settings.

Theorem 6.

Let $\mathcal{S},\mathcal{Y}:\mathrm{I}=[{\sigma }_1,{\sigma }_2]\to \mathfrak{R}$ be convex functions on $\mathrm{I}$ such that $({\sigma }_1,{\sigma }_2)\in \mathrm{I}$ with ${\sigma }_1<{\sigma }_2$ and $\mathcal{S}\mathcal{Y}\in \mathcal{L}[{\sigma }_1,{\sigma }_2]$. Then, for $\zeta \in [0,1]$ the following fractional inequality holds true:

$$\begin{array}{cc}\hfill & \frac{\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{{\mu +\nu -{\sigma }_1}^{-}}^{\alpha }\mathcal{S}\mathcal{Y}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{{\mu +\nu -{\sigma }_2}^{+}}^{\alpha }\mathcal{S}\mathcal{Y}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & \leqq \frac{2(1-e^{-\rho })}{\rho }\mathcal{M}(\mu ,\nu )+\left[\frac{{\rho }^2-2\rho +4-({\rho }^2+2\rho +4)e^{-\rho }}{{\rho }^3}\right]{\mathcal{N}}_1({\sigma }_1,{\sigma }_2)\hfill \\ \hfill & +\left[\frac{\rho -2+e^{-\rho }(\rho +2)}{{\rho }^3}\right]{\mathcal{N}}_2({\sigma }_1,{\sigma }_2)-\left(\frac{1-e^{-\rho }}{\rho }\right)\mathcal{K}(\mu ,\nu ,{\sigma }_1,{\sigma }_2).\hfill \end{array}$$

where

$$\begin{array}{cc}\hfill {\mathcal{N}}_1({\sigma }_1,{\sigma }_2)=& \left[\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_2\right)\right]\hfill \\ \hfill {\mathcal{N}}_2({\sigma }_1,{\sigma }_2)=& \left[\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_1\right)\right]\hfill \\ \hfill \mathcal{M}(\mu ,\nu )=& \left[\mathcal{S}\left(\mu \right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left(\nu \right)+\mathcal{S}\left(\mu \right)\mathcal{Y}\left(\nu \right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left(\mu \right)\right],\hfill \\ \hfill \mathcal{K}(\mu ,\nu ,{\sigma }_1,{\sigma }_2)=& \left[\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\nu \right)\right.\hfill \\ \hfill & \left.+\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\nu \right)\right],\hfill \end{array}$$

Proof. 

Since $\mathcal{S}$ and $\mathcal{Y}$ are convex functions on $[{\sigma }_1,{\sigma }_2]$, ∀$\varsigma \in [0,1]$, ${\sigma }_1,{\sigma }_2\in \mathrm{I}$, then using the Jensen–Mercer Inequality, we have

$$\mathcal{S}\left(\mu +\nu -\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)\right)\leqq [\mathcal{S}\left(\mu \right)+\mathcal{S}\left(\nu \right)]-\left(\varsigma \mathcal{S}\left({\sigma }_2\right)+(1-\varsigma )\mathcal{S}\left({\sigma }_1\right)\right),$$

and

$$\mathcal{Y}\left(\mu +\nu -\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)\right)\leqq [\mathcal{Y}\left(\mu \right)+\mathcal{Y}\left(\nu \right)]-\left(\varsigma \mathcal{Y}\left({\sigma }_2\right)+(1-\varsigma )\mathcal{Y}\left({\sigma }_1\right)\right).$$

Now multiplying the above two inequalities, we find

$$\begin{array}{cc}\hfill & \mathcal{S}\left(\mu +\nu -\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)\right)\mathcal{Y}\left(\mu +\nu -\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)\right)\hfill \\ & \leqq \left[\mathcal{S}\right(\mu \left)\mathcal{Y}\right(\mu )+\mathcal{S}(\mu \left)\mathcal{Y}\right(\nu )+\mathcal{S}(\nu \left)\mathcal{Y}\right(\mu )+\mathcal{S}(\nu \left)\mathcal{Y}\right(\nu \left)\right]\hfill \\ \hfill & -\varsigma [\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\nu \right)]\hfill \\ \hfill & -(2-\varsigma )[\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\nu \right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\mu \right)]\hfill \\ \hfill & +\varsigma (1-\varsigma )\left[\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_1\right)\right]+{\varsigma }^2\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_2\right)+{(1-\varsigma )}^2\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_1\right).\hfill \end{array}$$

(19)

Similarly,

$$\begin{array}{cc}\hfill & \mathcal{S}\left(\mu +\nu -\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)\right)\mathcal{Y}\left(\mu +\nu -\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)\right)\hfill \\ & \leqq \left[\mathcal{S}\right(\mu \left)\mathcal{Y}\right(\mu )+\mathcal{S}(\mu \left)\mathcal{Y}\right(\nu )+\mathcal{S}(\nu \left)\mathcal{Y}\right(\mu )+\mathcal{S}(\nu \left)\mathcal{Y}\right(\nu \left)\right]\hfill \\ \hfill & -\varsigma [\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\nu \right)]\hfill \\ \hfill & -(2-\varsigma )[\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\nu \right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\mu \right)]\hfill \\ \hfill & +\varsigma (1-\varsigma )\left[\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_2\right)\right]+{\varsigma }^2\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_1\right)+{(1-\varsigma )}^2\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_2\right).\hfill \end{array}$$

(20)

Adding both the inequalities (19) and (20) and multiplying both sides by $e^{-\frac{1-\alpha }{2\alpha }({\sigma }_2-{\sigma }_1)\varsigma }$, then integrating the obtained result over [0, 1], we find that

$$\begin{array}{cc}\hfill & \left[\mathcal{S}\left(\mu +\nu -\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)\right)\mathcal{Y}\left(\mu +\nu -\left(\varsigma {\sigma }_2+(1-\varsigma ){\sigma }_1\right)\right)\right]{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }d\varsigma \hfill \\ \hfill +& \left[\mathcal{S}\left(\mu +\nu -\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)\right)\mathcal{Y}\left(\mu +\nu -\left(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2\right)\right)\right]{\int }_0^1{e}^{-\frac{1-\alpha }{\alpha }({\sigma }_2-{\sigma }_1)\varsigma }d\varsigma \hfill \\ \hfill & \leqq 2[\mathcal{S}\left(\mu \right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left(\mu \right)\mathcal{Y}\left(\nu \right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left(\nu \right)]\left(\frac{1-e^{-\rho }}{\rho }\right)\hfill \\ \hfill & -\left[\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\mu \right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left(\nu \right)\right.\hfill \\ \hfill & \left.+\mathcal{S}\left(\nu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left(\mu \right)\mathcal{Y}\left({\sigma }_2\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\nu \right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left(\mu \right)\right]\left(\frac{1-e^{-\rho }}{\rho }\right)\hfill \\ \hfill & +\left[\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_2\right)\right]\left[\frac{\rho -2+e^{-\rho }(\rho +2)}{{\rho }^3}\right]\hfill \\ \hfill & +[\mathcal{S}\left({\sigma }_1\right)\mathcal{Y}\left({\sigma }_1\right)+\mathcal{S}\left({\sigma }_2\right)\mathcal{Y}\left({\sigma }_2\right)]\left[\frac{{\rho }^2-2\rho +4-({\rho }^2+2\rho +4)e^{-\rho }}{{\rho }^3}\right],\hfill \end{array}$$

from which readily follows

$$\begin{array}{cc}\hfill & \frac{\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{{\mu +\nu -{\sigma }_1}^{-}}^{\alpha }\mathcal{S}\mathcal{Y}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{{\mu +\nu -{\sigma }_2}^{+}}^{\alpha }\mathcal{S}\mathcal{Y}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & \leqq \frac{2(1-e^{-\rho })}{\rho }\mathcal{M}(\mu ,\nu )+\left[\frac{{\rho }^2-2\rho +4-({\rho }^2+2\rho +4)e^{-\rho }}{{\rho }^3}\right]{\mathcal{N}}_1({\sigma }_1,{\sigma }_2)\hfill \\ \hfill & +\left[\frac{\rho -2+e^{-\rho }(\rho +2)}{{\rho }^3}\right]{\mathcal{N}}_2({\sigma }_1,{\sigma }_2)-\left(\frac{1-e^{-\rho }}{\rho }\right)\mathcal{K}(\mu ,\nu .{\sigma }_1,{\sigma }_2).\hfill \end{array}$$

This leads us to the proof of the desired Theorem 6. □

4. New Fractional Inequalities of Dragomir–Agarwal–Mercer-Type

This section deals with proving a new identity for differentiable mappings that involve fractional integral operators with an exponential kernel. Then, taking this identity into account, we have established a Dragomir–Agarwal–Mercer-type inequality.

Lemma 1.

Let $\mathcal{S}:\mathrm{I}=[{\sigma }_1,{\sigma }_2]\to \mathfrak{R}$ be a differentiable mapping on $\mathrm{I}$ such that $({\sigma }_1,{\sigma }_2)\in \mathrm{I}$ and $0\leqq {\sigma }_1<{\sigma }_2$. If ${\mathcal{S}}^{\prime }\in \mathcal{L}[{\sigma }_1,{\sigma }_2]$. Then,

$$\begin{array}{cc}\hfill & \frac{\mathcal{S}\left(\mu +\nu -{\sigma }_1\right)+\mathcal{S}\left(\mu +\nu -{\sigma }_2\right)}{2}\hfill \\ \hfill & -\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & =\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}{\int }_0^1\left[e^{-\rho (1-\varsigma )}-e^{-\rho \varsigma }\right]{\mathcal{S}}^{\prime }(\mu +\nu -(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2))d\varsigma ,\hfill \end{array}$$

(21)

holds true for $\zeta \in [0,1]$.

Proof. 

$$\begin{array}{cc}\hfill Let{I}_1& ={\int }_0^1{e}^{-\rho \varsigma }{\mathcal{S}}^{\prime }(\mu +\nu -(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2))d\varsigma \hfill \\ \hfill & =\frac{e^{-\rho \varsigma }\mathcal{S}(\mu +\nu -{\sigma }_1)-\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}+\frac{\rho \alpha }{{({\sigma }_2-{\sigma }_1)}^2}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right]\hfill \\ \hfill & =\frac{e^{-\rho \varsigma }\mathcal{S}(\mu +\nu -{\sigma }_1)-\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}+\frac{1-\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right].\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_2& ={\int }_0^1{e}^{-\rho (1-\varsigma )}{\mathcal{S}}^{\prime }(\mu +\nu -(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2))d\varsigma \hfill \\ \hfill & =\frac{\mathcal{S}(\mu +\nu -{\sigma }_1)-e^{-\rho }\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}+\frac{\rho \alpha }{{({\sigma }_2-{\sigma }_1)}^2}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & =\frac{\mathcal{S}(\mu +\nu -{\sigma }_1)-e^{-\rho }\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}-\frac{1-\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right].\hfill \end{array}$$

From the above developments, we have

$$\begin{array}{cc}\hfill I_2-I_1=& \frac{\mathcal{S}(\mu +\nu -{\sigma }_1)-e^{-\rho }\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}-\frac{1-\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]\hfill \\ \hfill & -\left[\frac{e^{-\rho }\mathcal{S}(\mu +\nu -{\sigma }_1)-\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}+\frac{1-\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right]\right]\hfill \\ \hfill & =\frac{\mathcal{S}(\mu +\nu -{\sigma }_1)-e^{-\rho }\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}-\frac{e^{-\rho }\mathcal{S}(\mu +\nu -{\sigma }_1)-\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}\hfill \\ \hfill & -\frac{1-\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)\right]-\frac{1-\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right]\hfill \\ \hfill & =(1-e^{-\rho })\frac{\mathcal{S}(\mu +\nu -{\sigma }_1)+\mathcal{S}(\mu +\nu -{\sigma }_2)}{{\sigma }_2-{\sigma }_1}\hfill \\ \hfill & -\frac{1-\alpha }{{\sigma }_2-{\sigma }_1}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right].\hfill \end{array}$$

(22)

Multiplying both sides of the above equality by $\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}$, we obtain

$$\begin{array}{cc}\hfill & \frac{\mathcal{S}(\mu +\nu -{\sigma }_1)+\mathcal{S}(\mu +\nu -{\sigma }_2)}{2}\hfill \\ \hfill & -\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right]\hfill \\ \hfill & =\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}[I_2-I_1].\hfill \end{array}$$

This leads us to the proof of Lemma 1. □

Theorem 7.

Let $\mathcal{S}:\mathrm{I}=[{\sigma }_1,{\sigma }_2]\to \mathfrak{R}$ be a convex function on $\mathrm{I}$ such that $({\sigma }_1,{\sigma }_2)\in \mathrm{I}$ with ${\sigma }_2>{\sigma }_1$ and $\mathcal{S}\in \mathcal{L}[{\sigma }_1,{\sigma }_2]$. Then for $\zeta \in [0,1]$, we have the following inequality:

$$\begin{array}{cc}\hfill & \frac{\mathcal{S}(\mu +\nu -{\sigma }_1)+\mathcal{S}(\mu +\nu -{\sigma }_2)}{2}\hfill \\ \hfill & -\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right]\hfill \\ \hfill & \leqq \frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}\left(\frac{1+e^{-\rho }-2e^{-\frac{\rho }{2}}}{\rho }\right)\left[\left(\right|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)\left|\right)-\frac{|{\mathcal{S}}^{\prime }\left({\sigma }_1\right)|+|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)|}{2}\right].\hfill \end{array}$$

Proof. 

Employing Lemma 1, using properties of modulus, the convexity of $\left|\mathcal{S}\right|$, and the J–M inequality, we have

$$\begin{array}{cc}\hfill & |\frac{\mathcal{S}(\mu +\nu -{\sigma }_1)+\mathcal{S}(\mu +\nu -{\sigma }_2)}{2}\hfill \\ \hfill & -\frac{1-\alpha }{2(1-e^{-\rho })}\left[{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_1)+{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\mathcal{S}(\mu +\nu -{\sigma }_2)\right]|\hfill \\ \hfill & =\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}{\int }_0^1|e^{-\rho (1-\varsigma )}-e^{-\rho \varsigma }\left|\right|{\mathcal{S}}^{\prime }(\mu +\nu -(\varsigma {\sigma }_1+(1-\varsigma ){\sigma }_2))|d\varsigma \hfill \\ \hfill & \leqq \frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}{\int }_0^1|e^{-\rho (1-\varsigma )}-e^{-\rho \varsigma }\left|\right||{\mathcal{S}}^{\prime }\left(\mu \right)+{\mathcal{S}}^{\prime }\left(\nu \right)-(\varsigma {\mathcal{S}}^{\prime }\left({\sigma }_1\right)+(1-\varsigma ){\mathcal{S}}^{\prime }\left({\sigma }_2\right)|)d\varsigma \hfill \\ \hfill & =\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}{\int }_0^{\frac{1}{2}}(e^{-\rho \varsigma }-e^{-\rho (1-\varsigma )})\left[|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)|-(\varsigma |{\mathcal{S}}^{\prime }\left({\sigma }_1\right)|+(1-\varsigma )|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)\left|\right)\right]d\varsigma \hfill \\ \hfill & +\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}{\int }_{\frac{1}{2}}^1(e^{-\rho (1-\varsigma )}-e^{-\rho \varsigma })\left[|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)|-(\varsigma |{\mathcal{S}}^{\prime }\left({\sigma }_1\right)|+(1-\varsigma )|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)\left|\right)\right]d\varsigma \hfill \\ \hfill & =\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}\left\{\right(|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)\left|\right){\int }_0^{\frac{1}{2}}(e^{-\rho \varsigma }-e^{-\rho (1-\varsigma )})d\varsigma \hfill \\ \hfill & -\left\{|{\mathcal{S}}^{\prime }\left({\sigma }_1\right)|{\int }_0^{\frac{1}{2}}(e^{-\rho \varsigma }-e^{-\rho (1-\varsigma )})\varsigma d\varsigma +\left|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)\right|{\int }_0^{\frac{1}{2}}(e^{-\rho \varsigma }-e^{-\rho (1-\varsigma )})(1-\varsigma )d\varsigma \right\}\}\hfill \\ \hfill & +\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}\left\{\right(|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)\left|\right){\int }_{\frac{1}{2}}^1(e^{-\rho (1-\varsigma )}-e^{-\rho \varsigma })d\varsigma \hfill \\ \hfill & -\left\{|{\mathcal{S}}^{\prime }\left({\sigma }_1\right)|{\int }_{\frac{1}{2}}^1(e^{-\rho (1-\varsigma )}-e^{-\rho \varsigma })\varsigma d\varsigma +\left|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)\right|{\int }_{\frac{1}{2}}^1(e^{-\rho (1-\varsigma )}-e^{-\rho \varsigma })(1-\varsigma )d\varsigma \right\}\}\hfill \\ \hfill & =\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}\left\{\right(|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)\left|\right)\left(\frac{e^{-\rho }(e^{\rho }+1)-2e^{-\frac{\rho }{2}}}{\rho }\right)-\{\left|{\mathcal{S}}^{\prime }\left({\sigma }_1\right)\right|\frac{e^{-\rho }(e^{\rho }-1)-\rho e^{-\frac{\rho }{2}}}{{\rho }^2}\hfill \\ \hfill & +|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)|\left(\frac{e^{-\rho }((\rho -1)e^{\rho }+\rho +1)-\rho e^{-\frac{\rho }{2}}}{{\rho }^2}\right)+\left(\right|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)\left|\right)\left(\frac{e^{-\rho }(e^{\rho }+1)-2e^{-\frac{\rho }{2}}}{\rho }\right)\hfill \\ \hfill & -\left\{|{\mathcal{S}}^{\prime }\left({\sigma }_1\right)|\left(\frac{e^{-\rho }((\rho -1)e^{\rho }+\rho +1)-\rho e^{-\frac{\rho }{2}}}{{\rho }^2}\right)+\left|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)\right|\left(\frac{e^{-\rho }(e^{\rho }-1)-\rho e^{-\frac{\rho }{2}}}{{\rho }^2}\right)\right\}\hfill \\ \hfill & =\frac{{\sigma }_2-{\sigma }_1}{2(1-e^{-\rho })}\left(\frac{1+e^{-\rho }-2e^{-\frac{\rho }{2}}}{\rho }\right)\left[\left(\right|{\mathcal{S}}^{\prime }\left(\mu \right)|+|{\mathcal{S}}^{\prime }\left(\nu \right)\left|\right)-\frac{|{\mathcal{S}}^{\prime }\left({\sigma }_1\right)|+|{\mathcal{S}}^{\prime }\left({\sigma }_2\right)|}{2}\right].\hfill \end{array}$$

This leads us to the desired inequality.  □

5. Applications

Example 1.

Let ${\mathbb{C}}^n$ represent the set of $n\times n$ complex matrices, $\mathbb{M}n$ represent the algebra of $n\times n$ complex matrices, and ${\mathbb{M}}^{+}n$ represent the strictly positive matrices in $\mathbb{M}$. That is, $\mathrm{A}\in {\mathbb{M}}_n^{+}$ for every nonzero $u\in {\mathbb{C}}^n$ if $〈\mathrm{A}u,u〉>0$.

Sababheh [53] proved that $\mathcal{S}\left(\kappa \right)=\Vert {\mathrm{A}}^{\kappa }\mathrm{X}{\mathrm{B}}^{1-\kappa }+{\mathrm{A}}^{1-\kappa }\mathrm{X}{\mathrm{B}}^{\kappa }\Vert$, $\mathrm{A},\mathrm{B}\in {\mathbb{M}}_n^{+},\mathrm{X}\in {\mathbb{M}}_n$ is convex for all $\kappa \in [0,1]$.

Then, by using Theorem 3, we have

$$\begin{array}{cc}\hfill & \Vert {\mathrm{A}}^{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}\mathrm{X}{\mathrm{B}}^{1-\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}+{\mathrm{A}}^{1-\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}\mathrm{X}{\mathrm{B}}^{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}\Vert \hfill \\ \hfill & \leqq \left[\Vert {\mathrm{A}}^{\mu }\mathrm{X}{\mathrm{B}}^{1-\mu }+{\mathrm{A}}^{1-\mu }\mathrm{X}{\mathrm{B}}^{\mu }\Vert +\Vert {\mathrm{A}}^{\nu }\mathrm{X}{\mathrm{B}}^{1-\nu }+{\mathrm{A}}^{1-\nu }\mathrm{X}{\mathrm{B}}^{\nu }\Vert \right]\hfill \\ \hfill & -\frac{1-\alpha }{2(1-e^{-\frac{\rho }{2}})}\left[{\mathfrak{J}}_{{\sigma }_1^{+}}^{\alpha }\Vert {\mathrm{A}}^{{\sigma }_2}\mathrm{X}{\mathrm{B}}^{1-{\sigma }_2}+{\mathrm{A}}^{1-{\sigma }_2}\mathrm{X}{\mathrm{B}}^{{\sigma }_2}\Vert +{\mathfrak{J}}_{{\sigma }_2^{-}}^{\alpha }\Vert {\mathrm{A}}^{{\sigma }_1}\mathrm{X}{\mathrm{B}}^{1-{\sigma }_1}+{\mathrm{A}}^{1-{\sigma }_1}\mathrm{X}{\mathrm{B}}^{{\sigma }_1}\Vert \right]\hfill \\ \hfill & \leqq \Vert {\mathrm{A}}^{\mu }\mathrm{X}{\mathrm{B}}^{1-\mu }+{\mathrm{A}}^{1-\mu }\mathrm{X}{\mathrm{B}}^{\mu }\Vert +\Vert {\mathrm{A}}^{\nu }\mathrm{X}{\mathrm{B}}^{1-\nu }+{\mathrm{A}}^{1-\nu }\mathrm{X}{\mathrm{B}}^{\nu }\Vert \hfill \\ \hfill & -\Vert {\mathrm{A}}^{\frac{{\sigma }_1+{\sigma }_2}{2}}\mathrm{X}{\mathrm{B}}^{1-\frac{{\sigma }_1+{\sigma }_2}{2}}+{\mathrm{A}}^{1-\frac{{\sigma }_1+{\sigma }_2}{2}}\mathrm{X}{\mathrm{B}}^{\frac{{\sigma }_1+{\sigma }_2}{2}}\Vert .\hfill \end{array}$$

Example 2.

Under the assumption of the above example, if we consider Theorem 4, we have

$$\begin{array}{cc}\hfill & \Vert {\mathrm{A}}^{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}\mathrm{X}{\mathrm{B}}^{1-\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}+{\mathrm{A}}^{1-\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}\mathrm{X}{\mathrm{B}}^{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}\Vert \hfill \\ \hfill & \leqq \frac{1-\alpha }{2(1-e^{-\frac{\rho }{2}})}[{\mathfrak{J}}_{\mu +\nu -{\sigma }_1^{-}}^{\alpha }\Vert {\mathrm{A}}^{\mu +\nu -{\sigma }_2}\mathrm{X}{\mathrm{B}}^{1-(\mu +\nu -{\sigma }_2)}+{\mathrm{A}}^{1-(\mu +\nu -{\sigma }_2)}\mathrm{X}{\mathrm{B}}^{\mu +\nu -{\sigma }_2}\Vert \hfill \\ \hfill & +{\mathfrak{J}}_{\mu +\nu -{\sigma }_2^{+}}^{\alpha }\Vert {\mathrm{A}}^{\mu +\nu -{\sigma }_1}\mathrm{X}{\mathrm{B}}^{1-(\mu +\nu -{\sigma }_1)}+{\mathrm{A}}^{1-(\mu +\nu -{\sigma }_1)}\mathrm{X}{\mathrm{B}}^{\mu +\nu -{\sigma }_1}\Vert ]\hfill \\ \hfill & \leqq \Vert {\mathrm{A}}^{\mu }\mathrm{X}{\mathrm{B}}^{1-\mu }+{\mathrm{A}}^{1-\mu }\mathrm{X}{\mathrm{B}}^{\mu }\Vert +\Vert {\mathrm{A}}^{\nu }\mathrm{X}{\mathrm{B}}^{1-\nu }+{\mathrm{A}}^{1-\nu }\mathrm{X}{\mathrm{B}}^{\nu }\Vert \hfill \\ \hfill & -\frac{\Vert {\mathrm{A}}^{{\sigma }_1}\mathrm{X}{\mathrm{B}}^{1-{\sigma }_1}+{\mathrm{A}}^{1-{\sigma }_1}\mathrm{X}{\mathrm{B}}^{{\sigma }_1}\Vert +\Vert {\mathrm{A}}^{{\sigma }_2}\mathrm{X}{\mathrm{B}}^{1-{\sigma }_2}+{\mathrm{A}}^{1-{\sigma }_2}\mathrm{X}{\mathrm{B}}^{{\sigma }_2}\Vert }{2}.\hfill \end{array}$$

Example 3.

Under the assumption of the above example, if we consider Theorem 5, we have

$$\begin{array}{cc}\hfill & \Vert {\mathrm{A}}^{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}\mathrm{X}{\mathrm{B}}^{1-\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}+{\mathrm{A}}^{1-\left(\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}\right)}\mathrm{X}{\mathrm{B}}^{\mu +\nu -\frac{{\sigma }_1+{\sigma }_2}{2}}\Vert \hfill \\ \hfill & \leqq \frac{1-\alpha }{2(1-e^{-\frac{\rho }{2}})}[{\mathfrak{J}}_{\mu +\nu -{\frac{{\sigma }_1+{\sigma }_2}{2}}^{-}}^{\alpha }\Vert {\mathrm{A}}^{\mu +\nu -{\sigma }_2}\mathrm{X}{\mathrm{B}}^{1-(\mu +\nu -{\sigma }_2)}+{\mathrm{A}}^{1-(\mu +\nu -{\sigma }_2)}\mathrm{X}{\mathrm{B}}^{\mu +\nu -{\sigma }_2}\Vert \hfill \\ \hfill & +{\mathfrak{J}}_{\mu +\nu -{\frac{{\sigma }_1+{\sigma }_2}{2}}^{+}}^{\alpha }\Vert {\mathrm{A}}^{\mu +\nu -{\sigma }_1}\mathrm{X}{\mathrm{B}}^{1-(\mu +\nu -{\sigma }_1)}+{\mathrm{A}}^{1-(\mu +\nu -{\sigma }_1)}\mathrm{X}{\mathrm{B}}^{\mu +\nu -{\sigma }_1}\Vert ]\hfill \\ \hfill & \leqq \Vert {\mathrm{A}}^{\mu }\mathrm{X}{\mathrm{B}}^{1-\mu }+{\mathrm{A}}^{1-\mu }\mathrm{X}{\mathrm{B}}^{\mu }\Vert +\Vert {\mathrm{A}}^{\nu }\mathrm{X}{\mathrm{B}}^{1-\nu }+{\mathrm{A}}^{1-\nu }\mathrm{X}{\mathrm{B}}^{\nu }\Vert \hfill \\ \hfill & -\frac{\Vert {\mathrm{A}}^{{\sigma }_1}\mathrm{X}{\mathrm{B}}^{1-{\sigma }_1}+{\mathrm{A}}^{1-{\sigma }_1}\mathrm{X}{\mathrm{B}}^{{\sigma }_1}\Vert +\Vert {\mathrm{A}}^{{\sigma }_2}\mathrm{X}{\mathrm{B}}^{1-{\sigma }_2}+{\mathrm{A}}^{1-{\sigma }_2}\mathrm{X}{\mathrm{B}}^{{\sigma }_2}\Vert }{2}.\hfill \end{array}$$

6. Conclusions

According to recent trends, integral inequalities are one of the most important research areas in the mathematical sciences. Using a novel generalized fractional operator with an exponential kernel, we have presented some new fractional integral inequalities in the context of the Jensen–Mercer inequality. We have also provided a new bound for differentiable convex mapping with the help of a new equality. It will be fascinating to see if we can apply the notions presented in this article to present inequalities in the setting of Jensen–Mercer on coordinates and interval-valued analysis.

We choose to conclude our present investigation by remarking that, in many recent publications, fractional-order analogs of various families of familiar integral inequalities have been routinely derived by using obviously trivial or redundant-parametric variations of widely known and extensively studied operators of fractional integrals and fractional derivatives (for details, see [54]).