The classical Hermite-Hadamard inequality, initially introduced by Hermite and Hadamard, offers fundamental bounds for the integral mean of convex functions. For a real-valued convex function $f:\left[a,b\right]\to ℝ$ is a convex then

$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\displaystyle {\int }_a^{b}f\left(x\right)\text{d}x}\le \frac{f\left(a\right)+f\left(b\right)}{2}.$

This inequality has inspired numerous extensions and generalizations across various domains, including operator theory, approximation theory, and fractional calculus.

In 2003, Mercer [1] introduced a refined version of the classical Jensen inequality, now known as the Jensen-Mercer inequality, stated as follows.

For a convex mapping $f:\left[a,b\right]\to ℝ$, the following inequality holds for each $x_j\in \left[a,b\right]$:

$f\left(a+b-\underset{j=1}{\overset{n}{{\displaystyle \sum }}}{u}_j{x}_j\right)\le f\left(a\right)+f\left(b\right)-\underset{j=1}{\overset{n}{{\displaystyle \sum }}}{u}_{j}f\left(x_j\right),$

where $u_j\in \left[0,1\right]$ and ${\displaystyle {\sum }_{j=1}^n{u}_j}=1$.

The Jensen-Mercer inequality prompted significant developments in the analysis of convexity in higher dimensions. More recently, attention has been devoted to the study of coordinate-wise convex functions, i.e., functions convex in each variable separately, which naturally arise in the context of bivariate and multivariate analysis.

Alongside these developments, the framework of fractional calculus has emerged as a powerful tool for extending classical integral inequalities. The incorporation of Riemann-Liouville, Hadamard, and Katugampola-type fractional operators has led to refined inequalities that capture nonlocal and memory-dependent behavior of convex functions. Sarikaya et al., Katugampola, and others have contributed to this area by developing fractional analogues of Hermite-Hadamard-type inequalities. Coordinate convex functions naturally arise in optimization problems, economic modeling, and approximation theory, where convexity along each coordinate direction plays a crucial role in ensuring stability and tractability of solutions.

Motivated by these advances, Toseef et al. [2] recently proposed a Hermite-Hadamard-Mercer-type inequality on the coordinates, which integrates the ideas of Mercer convexity and fractional integrals:

For a convex mapping $f:\left[a,b\right]\times \left[c,d\right]\to ℝ$, the following inequality holds for each $x_j\in \left[a,b\right]$ and $y_j\in \left[c,d\right]$:

$\begin{array}{l}f\left(a+b-\underset{j=1}{\overset{n}{{\displaystyle \sum }}}{u}_j{x}_j,c+d-\underset{j=1}{\overset{n}{{\displaystyle \sum }}}{w}_j{y}_j\right)\\ \le f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)-\underset{j=1}{\overset{n}{{\displaystyle \sum }}}{u}_j{w}_{j}f\left(x_j,y_j\right),\end{array}$

where $u_j,w_j\in \left[0,1\right]$ and ${\displaystyle {\sum }_{j=1}^n{u}_j}=1$ and ${\displaystyle {\sum }_{j=1}^n{w}_j}=1$.

In 2013, Kian and Mosleshian [3] used the new Jensen-Mercer inequality and established the following new version of the Hermite-Hadamard inequality:

For a convex mapping $f:\left[a,b\right]\to ℝ$, the following inequalities hold for all $x,y\in \left[a,b\right]$ and $x<y$:

$\begin{array}{c}f\left(a+b-\frac{x+y}{2}\right)\le f\left(a\right)+f\left(b\right)-\frac{1}{y-x}{\displaystyle {\int }_x^{y}f\left(u\right)\text{d}u}\\ \le f\left(a\right)+f\left(b\right)-f\left(\frac{x+y}{2}\right)\end{array}$

and

$f\left(a+b-\frac{x+y}{2}\right)\le \frac{1}{y-x}{\displaystyle {\int }_{a+b-y}^{a+b-x}f\left(u\right)\text{d}u}$

$\begin{array}{c}\le \frac{f\left(a+b-x\right)+f\left(a+b-y\right)}{2}\\ \le f\left(a\right)+f\left(b\right)-f\left(\frac{x+y}{2}\right).\end{array}$

Building on this foundation, the present paper aims to extend and generalize such inequalities within the framework of generalized fractional integrals, particularly those involving kernels that unify and extend classical operators.

The Riemann-Liouville fractional integral operators are defined as:

For an integrable function $f$ on $\left[a,b\right]$, the left and the right Riemann-Liouville fractional integrals are defined as:

$J_{a^{+}}^{\alpha }f\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_a^x{\left(x-u\right)}^{\alpha -1}f\left(u\right)\text{d}u},$

$J_{b^{-}}^{\alpha }f\left(x\right)=\frac{1}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_x^b{\left(u-x\right)}^{\alpha -1}f\left(u\right)\text{d}u},$

where Γ represents the Gamma function function and is defined as $\Gamma \left(\alpha \right)={\displaystyle {\int }_0^{\infty }{\text{e}}^{-\tau }{\tau }^{\alpha -1}\text{d}\tau }$.

Sarikaya et al. [4] used the Riemann-Liouville fractional integrals and derived the following version of the Hermite-Hadamard-type inequality for convex functions.

For a convex function $f:\left[a,b\right]\to ℝ$, the following inequality holds:

$f\left(\frac{a+b}{2}\right)\le \frac{\Gamma \left(\alpha +1\right)}{2{\left(b-a\right)}^{\alpha }}\left[J_{a^{+}}^{\alpha }f\left(b\right)+J_{b^{-}}^{\alpha }f\left(a\right)\right]\le \frac{f\left(a\right)+f\left(b\right)}{2}.$

For a convex function $f:\left[a,b\right]\to ℝ$, the following inequality holds for all $x,y\in \left[a,b\right]$ and $x<y$:

$\begin{array}{c}f\left(a+b-\frac{x+y}{2}\right)\le \frac{\Gamma \left(\alpha +1\right)}{2{\left(y-x\right)}^{\alpha }}\left[J_{{\left(a+b-y\right)}^{+}}^{\alpha }f\left(a+b-x\right)+J_{{\left(a+b-x\right)}^{-}}^{\alpha }f\left(a+b-y\right)\right]\\ \le \frac{f\left(a+b-x\right)+f\left(a+b-y\right)}{2}\\ \le f\left(a\right)+f\left(b\right)-\frac{f\left(x\right)+f\left(y\right)}{2}.\end{array}$

For an integrable function $f$ on $\left[a,b\right]$, the left and the right $\kappa$-Riemann-Liouville fractional integrals are defined as:

$J_{a^{+}}^{\alpha ,\kappa }f\left(x\right)=\frac{1}{\kappa {\Gamma }_{\kappa }\left(\alpha \right)}{\displaystyle {\int }_a^x{\left(x-u\right)}^{\frac{\alpha }{\kappa }-1}f\left(u\right)\text{d}u},$

$J_{b^{-}}^{\alpha }f\left(x\right)=\frac{1}{\kappa {\Gamma }_{\kappa }\left(\alpha \right)}{\displaystyle {\int }_x^b{\left(u-x\right)}^{\frac{\alpha }{\kappa }-1}f\left(u\right)\text{d}u},$

where ${\Gamma }_{\kappa }\left(\alpha \right)$ is the $\kappa$-Gamma function, and is defined as ${\Gamma }_{\kappa }\left(\alpha \right)={\displaystyle {\int }_0^{\infty }{\text{e}}^{-\frac{{\tau }^{\kappa }}{\kappa }}{\tau }^{\alpha -1}\text{d}\tau }$.

Sarikaya and Ata [5] presented the generalized variant of the Hermite-Hadamard inequality bu using the beta function:

For a convex function $f:\left[a,b\right]\to ℝ$, the following inequality holds:

$f\left(\frac{a+b}{2}\right)\beta \left(m,n\right)\le \frac{1}{2{\left(b-a\right)}^{m+n-1}}{\displaystyle {\int }_a^b\Xi \left(u\right)f\left(u\right)\text{d}u}\le \frac{f\left(a\right)+f\left(b\right)}{2}\beta \left(m,n\right)$

where $\beta \left(m,n\right)={\displaystyle {\int }_0^1{\tau }^{m-1}{\left(1-\tau \right)}^{n-1}\text{d}\tau }$ is the beta function and

$\Xi \left(u\right)={\left(b-u\right)}^{m-1}{\left(u-a\right)}^{n-1}+{\left(b-u\right)}^{n-1}{\left(u-a\right)}^{m-1}.$

Ali, Zhang and Fečkan [6] extended the generalized variant of the Hermite-Hadamard-Mercer inequality by using the beta function:

For a convex function $f:\left[a,b\right]\to ℝ$, the following inequality holds:

$\begin{array}{l}f\left(a+b-\frac{x+y}{2}\right)\beta \left(m,n\right)\\ \le \left[f\left(a\right)+f\left(b\right)\right]\beta \left(m,n\right)-\frac{1}{2{\left(y-x\right)}^{m+n-1}}{\displaystyle {\int }_x^y\Xi \left(u\right)f\left(u\right)\text{d}u}\\ \le \left[f\left(a\right)+f\left(b\right)-\frac{f\left(a\right)+f\left(b\right)}{2}\right]\beta \left(m,n\right),\end{array}$

where $\beta \left(m,n\right)={\displaystyle {\int }_0^1{\tau }^{m-1}{\left(1-\tau \right)}^{n-1}\text{d}\tau }$ is the beta function and

$\Xi \left(u\right)={\left(y-u\right)}^{m-1}{\left(u-x\right)}^{n-1}+{\left(y-u\right)}^{n-1}{\left(u-x\right)}^{m-1}$

and $x,y\in \left[a,b\right]$.

For a convex function $f:\left[a,b\right]\to ℝ$, the following inequality holds:

$\begin{array}{l}f\left(a+b-\frac{x+y}{2}\right)\beta \left(m,n\right)\\ \le \left[f\left(a\right)+f\left(b\right)\right]\beta \left(m,n\right)-\frac{1}{2{\left(y-x\right)}^{m+n-1}}{\displaystyle {\int }_{a+b-y}^{a+b-x}\Xi \left(u\right)f\left(u\right)\text{d}u}\\ \le \frac{f\left(a+b-x\right)+f\left(a+b-y\right)}{2}\beta \left(m,n\right)\\ \le \left[f\left(a\right)+f\left(b\right)-\frac{f\left(a\right)+f\left(b\right)}{2}\right]\beta \left(m,n\right),\end{array}$

where $\beta \left(m,n\right)$ is the beta function and

$\begin{array}{c}\Xi \left(u\right)={\left[\left(a+b-x\right)-u\right]}^{m-1}{\left[u-\left(a+b-y\right)\right]}^{n-1}\\ +{\left[\left(a+b-x\right)-u\right]}^{n-1}{\left[u-\left(a+b-y\right)\right]}^{m-1}\end{array}$

and $x,y\in \left[a,b\right]$.

In [7] , Katugampola introduced a new fractional which generalizes the Riemann-Liouville and the Hadamard fractional integrals into a single form as follow.

Let $\left[a,b\right]\subset ℝ$ be a finite interval. Then, the left- and right-side Katugampola fractional integralsod order $\alpha >0$ of $f\in X_c^p\left(a,b\right)$ are defined by

${}^{\rho }{I}_{a^{+}}^{\alpha }f\left(x\right)=\frac{{\rho }^{1-\alpha }}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_a^x\frac{t^{\rho -1}}{{\left(x^{\rho }-t^{\rho }\right)}^{1-\alpha }}f\left(t\right)\text{d}t}$

and

${}^{\rho }{I}_{b^{-}}^{\alpha }f\left(x\right)=\frac{{\rho }^{1-\alpha }}{\Gamma \left(\alpha \right)}{\displaystyle {\int }_x^b\frac{t^{\rho -1}}{{\left(t^{\rho }-x^{\rho }\right)}^{1-\alpha }}f\left(t\right)\text{d}t}$

where $a<x<b$ and $\rho >0$, if the integral exists.

Let $\alpha >0$ and $\rho >0$. Then for $x>a$,

1) $\underset{\rho \to 1}{\lim }{}^{\rho }I{}_{a^{+}}^{\alpha }f\left(x\right)=J_{a^{+}}^{\alpha }f\left(x\right)$,

2) $\underset{\rho \to 0^{+}}{\lim }{}^{\rho }I{}_{a^{+}}^{\alpha }f\left(x\right)=H_{a^{+}}^{\alpha }f\left(x\right)$.

Similar results also hold for right-sided operators.

In [8] , Sarikaya and Ertuğral gave the definition of generalized fractional integrals (GFIs) as following:

The left-sided and right-sided GFIs are denoted by ${}_{a^{+}}{I}_{\phi }$ and ${}_{b^{-}}{I}_{\phi }$ as followings:

${}_{a^{+}}I{}_{\phi }f\left(x\right)={\displaystyle {\int }_a^x\frac{\phi \left(x-t\right)}{x-t}\text{d}t},x>a,$

and

${}_{b^{-}}I{}_{\phi }f\left(x\right)={\displaystyle {\int }_x^b\frac{\phi \left(t-x\right)}{t-x}\text{d}t},x<b,$

where a function $\phi :\left[0,\infty \right)\to \left[0,\infty \right)$ satisfies the condition ${\displaystyle {\int }_0^1\frac{\phi \left(t\right)}{t}\text{d}t}<\infty$.

The most important feature of generalized fractional integrals is that they generalize some type of fractional integrals such as the Riemann-Liouville fractional integral, k-Riemann-Liouville fractional integral, Katugampola fractional integrals, conformable fractional, and Hadamard feractional integrals. These important special cases of integral operator are mentioned below.

(1) If we choose $\phi \left(x\right)=x$, the operators ${}_{a^{+}}{I}_{\phi }f\left(x\right)$ and ${}_{b^{-}}{I}_{\phi }f\left(x\right)$ are reduce to the Riemann integral.

(2) Considering $\phi \left(x\right)=\frac{x^{\alpha }}{\Gamma \left(\alpha \right)}$ and $\alpha >0$, the operators ${}_{a^{+}}{I}_{\phi }f\left(x\right)$ and ${}_{b^{-}}{I}_{\phi }f\left(x\right)$ are reduce to the Riemann-Liouville fractioal integrals $I_{a^{+}}^{\alpha }f\left(x\right)$ and $I_{b^{-}}^{\alpha }f\left(x\right)$, respectively. Here, Γ is a gamma function.

(3) For $\phi \left(x\right)=\frac{1}{k\Gamma \left(\alpha \right)}{x}^{\frac{\alpha }{k}}$ and $\alpha ,k>0$, the operators ${}_{a^{+}}{I}_{\phi }f\left(x\right)$ and ${}_{b^{-}}{I}_{\phi }f\left(x\right)$ are reduce to the k-Riemann-Liouville fractional integrals $I_{a^{+},k}^{\alpha }f\left(x\right)$ and $I_{b^{-},k}^{\alpha }f\left(x\right)$, respectively. Here, ${\Gamma }_k$ is a k-gamma function.

Further more, for more results in the field if coordinated convex we refer the interested to see [8] .

For a mapping $f:\left[a,b\right]\times \left[c,d\right]\subset {ℝ}^2\to ℝ$ is a convex on the coordinates, the following inequality holds:

$\begin{array}{l}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\\ \le tsf\left(a,c\right)+t\left(1-s\right)f\left(a,d\right)+\left(1-t\right)sf\left(b,c\right)+\left(1-t\right)\left(1-s\right)f\left(b,d\right)\end{array}$

for $t,s\in \left[0,1\right]$.

The following Hermite-Hadamard type inequality for coordinated convex functions on the retangle from the plane ${ℝ}^2$ was proved in [4] , that is:

Supposse that a function $f:\Delta =\left[a,b\right]\times \left[c,d\right]\subset {ℝ}^2\to ℝ$ is a convex on coordinates. Then one has the inequalities:

$\begin{array}{c}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{2}\left[\frac{1}{b-a}{\displaystyle {\int }_a^{b}f}\left(x,\frac{c+d}{2}\right)\text{d}x+\frac{1}{d-c}{\displaystyle {\int }_c^{d}f}\left(\frac{a+b}{2},y\right)\text{d}y\right]\\ \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\displaystyle {\int }_a^b{\displaystyle {\int }_c^{d}f}}\left(x,y\right)\text{d}y\text{d}x\\ \le \frac{1}{4}[\frac{1}{b-a}{\displaystyle {\int }_a^{b}f}\left(x,c\right)\text{d}x+\frac{1}{b-a}{\displaystyle {\int }_a^{b}f}\left(x,d\right)\text{d}x\\ +\frac{1}{d-c}{\displaystyle {\int }_c^{d}f}\left(a,y\right)\text{d}y+\frac{1}{d-c}{\displaystyle {\int }_c^{d}f}\left(b,y\right)\text{d}y]\\ \le \frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}.\end{array}$

In 2024, The Hermite-Hadamard-Mercer type inequalities for coordinated convex functions was recently established by Toseef et al. in [2] , which is stated as:

For a convex mapping $f:\left[a,b\right]\to ℝ$, the following inequality holds for each $x_j\in \left[a,b\right]$, $y_j\in \left[c,d\right]$:

$\begin{array}{l}f\left(a+b-\underset{j=1}{\overset{n}{{\displaystyle \sum }}}{u}_j{x}_j,c+d-\underset{j=1}{\overset{n}{{\displaystyle \sum }}}{w}_j{y}_j\right)\\ \le f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)-\underset{i=1}{\overset{n}{{\displaystyle \sum }}}{u}_j{w}_{j}f\left(x_j,y_y\right),\end{array}$

where $u_j,w_j\in \left[0,1\right]$ and ${\displaystyle {\sum }_{j=1}^n{u}_j}=1$, ${\displaystyle {\sum }_{j=1}^n{w}_j}=1$.

Assume that $f:\Delta =\left[a,b\right]\times \left[c,d\right]\subset {ℝ}^2\to ℝ$ is a convex on coordinates. Then one has the inequalities:

$\begin{array}{l}f\left(a+b-\frac{m+n}{2},c+d-\frac{k+l}{2}\right)\\ \le \frac{1}{2}\left[\frac{1}{m-n}{\displaystyle {\int }_{a+b-n}^{a+b-m}f}\left(x,c+d-\frac{k+l}{2}\right)\text{d}x+\frac{1}{l-k}{\displaystyle {\int }_{c+d-l}^{c+d-k}f}\left(a+b-\frac{m+n}{2},y\right)\text{d}y\right]\\ \le \frac{1}{\left(m-n\right)\left(l-k\right)}{\displaystyle {\int }_{a+b-n}^{a+b-m}{\displaystyle {\int }_{c+d-l}^{c+d-k}f}}\left(x,y\right)\text{d}y\text{d}x\\ \le \frac{1}{2}[\frac{1}{m-n}{\displaystyle {\int }_{a+b-n}^{a+b-m}f}\left(x,c+d-l\right)\text{d}x+\frac{1}{m-n}{\displaystyle {\int }_{a+b-n}^{a+b-m}f}\left(x,c+d-k\right)\text{d}x\\ +\frac{1}{l-k}{\displaystyle {\int }_{c+d-l}^{c+d-k}f}\left(a+b-n,y\right)\text{d}y+\frac{1}{l-k}{\displaystyle {\int }_{c+d-l}^{c+d-k}f}\left(a+b-m,y\right)\text{d}y]\\ -\frac{1}{4}[\frac{1}{m-n}{\displaystyle {\int }_{a+b-n}^{a+b-m}f}\left(x,l\right)\text{d}x+\frac{1}{m-n}{\displaystyle {\int }_{a+b-n}^{a+b-m}f}\left(x,k\right)\text{d}x\\ +\frac{1}{l-k}{\displaystyle {\int }_{c+d-l}^{c+d-k}f}\left(n,y\right)\text{d}y+\frac{1}{l-k}{\displaystyle {\int }_{c+d-l}^{c+d-k}f}\left(m,y\right)\text{d}y]\\ \le \frac{1}{4}[f\left(a+b-m,c+d-k\right)+f\left(a+b-m,c+d-l\right)\end{array}$

$\begin{array}{l}+f\left(a+b-n,c+d-k\right)+f\left(a+b-n,c+d-l\right)]\\ \le f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)-\left[\frac{f\left(m,k\right)+f\left(m,l\right)+f\left(n,k\right)+f\left(n,l\right)}{4}\right],\end{array}$

where $m,n\in \left[a,b\right]$ and $k,l\in \left[c,d\right]$.

We introduce a new class of Hermite-Hadamard-Mercer-type inequalities involving generalized fractional integrals for functions that are convex on the coordinates.

${}_{a^{+},c^{+}}{I}_{\phi }f\left(x,y\right)={\displaystyle {\int }_c^y{\displaystyle {\int }_a^x\frac{\phi \left(x-t\right)\phi \left(y-s\right)}{\left(x-t\right)\left(y-s\right)}f\left(t,s\right)\text{d}t\text{d}s}},$

${}_{a^{+},d^{-}}{I}_{\phi }f\left(x,y\right)={\displaystyle {\int }_y^d{\displaystyle {\int }_a^x\frac{\phi \left(x-t\right)\phi \left(s-y\right)}{\left(x-t\right)\left(s-y\right)}f\left(t,s\right)\text{d}t\text{d}s}},$

${}_{b^{-},c^{+}}{I}_{\phi }f\left(x,y\right)={\displaystyle {\int }_c^y{\displaystyle {\int }_x^b\frac{\phi \left(t-x\right)\phi \left(y-s\right)}{\left(t-x\right)\left(y-s\right)}f\left(t,s\right)\text{d}t\text{d}s}},$

and

${}_{b^{-},d^{-}}{I}_{\phi }f\left(x,y\right)={\displaystyle {\int }_y^d{\displaystyle {\int }_x^b\frac{\phi \left(t-x\right)\phi \left(s-y\right)}{\left(t-x\right)\left(s-y\right)}f\left(t,s\right)\text{d}t\text{d}s}}.$

We derive integral identities that enable us to establish midpoint, trapezoidal inequalities in this generalized setting.

We demonstrate that our results encapsulate several known inequalities as special cases, thus unifying prior literature under a broader and more flexible framework.

Firstly, we establish an integral identity for differential function via general fractional integrals to dirive main results.

Lemma 2.1

Assume that $f:\Delta =\left[a,b\right]\times \left[c,d\right]\subset {ℝ}^2\to ℝ$ is continuous and integrable, the following fractional equality holds:

$\begin{array}{l}[f\left(a+b-n,c+d-l\right)+f\left(a+b-n,c+d-k\right)+f\left(a+b-m,c+d-l\right)+f\left(a+b-m,c+d-k\right)]\\ -\frac{1}{\Lambda \left(1\right)}[{}_{{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ +{}_{{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]\\ -\frac{1}{\Delta \left(1\right)}[{}_{{\left(a+b-m\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(a+b-m\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ +{}_{{\left(a+b-n\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(a+b-n\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]\\ +\frac{1}{\Lambda \left(1\right)\Delta \left(1\right)}[{}_{{\left(a+b-m\right)}^{-},{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(a+b-m\right)}^{-},{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ {+}_{{\left(a+b-n\right)}^{+},{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-m,c+d-l\right){+}_{{\left(a+b-n\right)}^{+},{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]\end{array}$

$\begin{array}{l}=\frac{\left(n-m\right)\left(l-k\right)}{\Lambda \left(1\right)\Delta \left(1\right)}{\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\Omega }}\left(t,s\right)[\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t\\ +\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(sk+\left(1-s\right)l\right)\right)\text{d}s\text{d}t\\ +\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(tm+\left(1-t\right)n\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t\\ +\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(tm+\left(1-t\right)n\right),c+d-\left(sk+\left(1-s\right)l\right)\right)\text{d}s\text{d}t],\end{array}$

where

$\Omega \left(t,s\right)={\displaystyle {\int }_0^t{\displaystyle {\int }_0^s\frac{\phi \left(\left(n-m\right)u\right)\phi \left(\left(l-k\right)\lambda \right)}{u\lambda }\text{d}\lambda \text{d}u}}$

and

$\Lambda \left(t\right)={\displaystyle {\int }_0^t\frac{\phi \left(\left(n-m\right)u\right)}{u}\text{d}u}$ and $\Delta \left(s\right)={\displaystyle {\int }_0^s\frac{\phi \left(\left(l-k\right)\lambda \right)}{\lambda }\text{d}\lambda }$.

Proof:

It suffices to note that

$S_1={\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\Omega }}\left(t,s\right)\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t$

$S_2={\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\Omega }}\left(t,s\right)\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t$

$S_3={\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\Omega }}\left(t,s\right)\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t$

and

$S_4={\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\Omega }}\left(t,s\right)\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t.$

If we add from $S_1-S_2-S_3+S_4$ and multiply by $\frac{\left(n-m\right)\left(l-k\right)}{\Lambda \left(1\right)\Delta \left(1\right)}$, we obtain the proof.

Theorem 2.2

Assume that $f:\Delta =\left[a,b\right]\times \left[c,d\right]\subset {ℝ}^2\to ℝ$ is continuous and integrable and $\left|\frac{{\partial }^2}{\partial t\partial s}f\right|$ satisfies the Mercer’s ineaqality, the following fractional inequality holds:

$\begin{array}{l}|\frac{\left[f\left(a+b-n,c+d-l\right)+f\left(a+b-n,c+d-k\right)+f\left(a+b-m,c+d-l\right)+f\left(a+b-m,c+d-k\right)\right]}{4}\\ -\frac{1}{4\Lambda \left(1\right)}[{}_{{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ +{}_{{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]\\ -\frac{1}{4\Delta \left(1\right)}[{}_{{\left(a+b-m\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(a+b-m\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\end{array}$

$\begin{array}{l}+{}_{{\left(a+b-n\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(a+b-n\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]\\ +\frac{1}{4\Lambda \left(1\right)\Delta \left(1\right)}[{}_{{\left(a+b-m\right)}^{-},{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(a+b-m\right)}^{-},{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ +{}_{{\left(a+b-n\right)}^{+},{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(a+b-n\right)}^{+},{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]|\\ \le \frac{\left(n-m\right)\left(l-k\right)}{\Lambda \left(1\right)\Delta \left(1\right)}{\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\left|\Omega \left(t,s\right)\right|}}[\left|\frac{{\partial }^2}{\partial t\partial s}f\left(a,c\right)\right|+\left|\frac{{\partial }^2}{\partial t\partial s}f\left(a,d\right)\right|+\left|\frac{{\partial }^2}{\partial t\partial s}f\left(b,c\right)\right|+\left|\frac{{\partial }^2}{\partial t\partial s}f\left(b,d\right)\right|\\ -\left|\frac{{\partial }^2}{\partial t\partial s}f\left(m,k\right)\right|-\left|\frac{{\partial }^2}{\partial t\partial s}f\left(m,l\right)\right|-\left|\frac{{\partial }^2}{\partial t\partial s}f\left(n,k\right)\right|-\left|\frac{{\partial }^2}{\partial t\partial s}f\left(n,l\right)\right|]\text{d}s\text{d}t.\end{array}$

Proof:

From the Lemma 2.1, we have

$\begin{array}{l}|\left[f\left(a+b-n,c+d-l\right)+f\left(a+b-n,c+d-k\right)+f\left(a+b-m,c+d-l\right)+f\left(a+b-m,c+d-k\right)\right]\\ -\frac{1}{\Lambda \left(1\right)}[{}_{{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ +{}_{{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]\\ -\frac{1}{\Delta \left(1\right)}[{}_{{\left(a+b-m\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(a+b-m\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ +{}_{{\left(a+b-n\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(a+b-n\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]\\ +\frac{1}{\Lambda \left(1\right)\Delta \left(1\right)}[{}_{{\left(a+b-m\right)}^{-},{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-n,c+d-l\right)+{}_{{\left(a+b-m\right)}^{-},{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-n,c+d-k\right)\\ {+}_{{\left(a+b-n\right)}^{+},{\left(c+d-k\right)}^{-}}{I}_{\phi }f\left(a+b-m,c+d-l\right)+{}_{{\left(a+b-n\right)}^{+},{\left(c+d-l\right)}^{+}}{I}_{\phi }f\left(a+b-m,c+d-k\right)]|\end{array}$

$\begin{array}{l}\le \frac{\left(n-m\right)\left(l-k\right)}{\Lambda \left(1\right)\Delta \left(1\right)}{\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\left|\Omega \left(t,s\right)\right|}}[|\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t\\ +\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(\left(1-t\right)m+tn\right),c+d-\left(sk+\left(1-s\right)l\right)\right)\text{d}s\text{d}t\\ +\frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(tm+\left(1-t\right)n\right),c+d-\left(\left(1-s\right)k+sl\right)\right)\text{d}s\text{d}t\\ \frac{{\partial }^2}{\partial t\partial s}f\left(a+b-\left(tm+\left(1-t\right)n\right),c+d-\left(sk+\left(1-s\right)l\right)\right)\text{d}s\text{d}t|]\\ \le \frac{4\left(n-m\right)\left(l-k\right)}{\Lambda \left(1\right)\Delta \left(1\right)}{\displaystyle {\int }_0^1{\displaystyle {\int }_0^1\left|\Omega \left(t,s\right)\right|}}[\left|\frac{{\partial }^2}{\partial t\partial s}f\left(a,c\right)\right|+\left|\frac{{\partial }^2}{\partial t\partial s}f\left(a,d\right)\right|+\left|\frac{{\partial }^2}{\partial t\partial s}f\left(b,c\right)\right|\\ +\left|\frac{{\partial }^2}{\partial t\partial s}f\left(b,d\right)\right|-\left|\frac{{\partial }^2}{\partial t\partial s}f\left(m,k\right)\right|-\left|\frac{{\partial }^2}{\partial t\partial s}f\left(m,l\right)\right|-\left|\frac{{\partial }^2}{\partial t\partial s}f\left(n,k\right)\right|-\left|\frac{{\partial }^2}{\partial t\partial s}f\left(n,l\right)\right|]\text{d}s\text{d}t.\end{array}$