# On some Hadamard-type inequalities for $\left({h}_{1},{h}_{2}\right)$-preinvex functions on the co-ordinates

## Abstract

We introduce the class of $\left({h}_{1},{h}_{2}\right)$-preinvex functions on the co-ordinates, and we prove some new inequalities of Hermite-Hadamard and Fejér type for such mappings.

MSC:26A15, 26A51, 52A30.

## 1 Introduction

A function $f:I\to R$, $I\subseteq R$ is an interval, is said to be a convex function on I if

$f\left(tx+\left(1-t\right)y\right)\le tf\left(x\right)+\left(1-t\right)f\left(y\right)$
(1.1)

holds for all $x,y\in I$ and $t\in \left[0,1\right]$. If the reversed inequality in (1.1) holds, then f is concave.

Many important inequalities have been established for the class of convex functions, but the most famous is the Hermite-Hadamard inequality. This double inequality is stated as follows:

$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{f\left(a\right)+f\left(b\right)}{2},$
(1.2)

where $f:\left[a,b\right]\to R$ is a convex function. The above inequalities are in reversed order if f is a concave function.

In 1978, Breckner introduced an s-convex function as a generalization of a convex function [1].

Such a function is defined in the following way: a function $f:\left[0,\mathrm{\infty }\right)\to R$ is said to be s-convex in the second sense if

$f\left(tx+\left(1-t\right)y\right)\le {t}^{s}f\left(x\right)+{\left(1-t\right)}^{s}f\left(y\right)$
(1.3)

holds for all $x,y\in \mathrm{\infty }$, $t\in \left[0,1\right]$ and for fixed $s\in \left(0,1\right]$.

Of course, s-convexity means just convexity when $s=1$.

In [2], Dragomir and Fitzpatrick proved the following variant of the Hermite-Hadamard inequality, which holds for s-convex functions in the second sense:

${2}^{s-1}f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{f\left(a\right)+f\left(b\right)}{s+1}.$
(1.4)

In the paper [3] a large class of non-negative functions, the so-called h-convex functions, is considered. This class contains several well-known classes of functions such as non-negative convex functions and s-convex in the second sense functions. This class is defined in the following way: a non-negative function $f:I\to R$, $I\subseteq R$ is an interval, is called h-convex if

$f\left(tx+\left(1-t\right)y\right)\le h\left(t\right)f\left(x\right)+h\left(1-t\right)f\left(y\right)$
(1.5)

holds for all $x,y\in I$, $t\in \left(0,1\right)$, where $h:J\to R$ is a non-negative function, $h\not\equiv 0$ and J is an interval, $\left(0,1\right)\subseteq J$.

In the further text, functions h and f are considered without assumption of non-negativity.

In [4] Sarikaya, Saglam and Yildirim proved that for an h-convex function the following variant of the Hadamard inequality is fulfilled:

$\frac{1}{2h\left(\frac{1}{2}\right)}f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_{a}^{b}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \left[f\left(a\right)+f\left(b\right)\right]\cdot {\int }_{0}^{1}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$
(1.6)

In [5] Bombardelli and Varošanec proved that for an h-convex function the following variant of the Hermite-Hadamard-Fejér inequality holds:

$\begin{array}{rcl}\frac{{\int }_{a}^{b}w\left(x\right)\phantom{\rule{0.2em}{0ex}}dx}{2h\left(\frac{1}{2}\right)}f\left(\frac{a+b}{2}\right)& \le & {\int }_{a}^{b}f\left(x\right)w\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\\ \le & \left(b-a\right)\left(f\left(a\right)+f\left(b\right)\right){\int }_{0}^{1}h\left(t\right)w\left(ta+\left(1-t\right)b\right)\phantom{\rule{0.2em}{0ex}}dt,\end{array}$
(1.7)

where $w:\left[a,b\right]\to R$, $w\ge 0$ and symmetric with respect to $\frac{a+b}{2}$.

A modification for convex functions, which is also known as co-ordinated convex functions, was introduced by Dragomir [6] as follows.

Let us consider a bidimensional $\mathrm{\Delta }=\left[a,b\right]×\left[c,d\right]$ in ${R}^{2}$ with $a and $c. A mapping $f:\mathrm{\Delta }\to R$ is said to be convex on the co-ordinates on Δ if the partial mappings ${f}_{y}:\left[a,b\right]\to R$, ${f}_{y}\left(u\right)=f\left(u,y\right)$ and ${f}_{x}:\left[c,d\right]\to R$, ${f}_{x}\left(v\right)=f\left(x,v\right)$ are convex for all $x\in \left[a,b\right]$ and $y\in \left[c,d\right]$.

In the same article, Dragomir established the following Hadamard-type inequalities for convex functions on the co-ordinates:

$\begin{array}{rl}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\\ \le \frac{f\left(a,c\right)+f\left(b,c\right)+f\left(a,d\right)+f\left(b,d\right)}{4}.\end{array}$
(1.8)

The concept of s-convex functions on the co-ordinates was introduced by Alomari and Darus [7]. Such a function is defined in following way: the mapping $f:\mathrm{\Delta }\to R$ is s-convex in the second sense if the partial mappings ${f}_{y}:\left[a,b\right]\to R$ and ${f}_{x}:\left[c,d\right]\to R$ are s-convex in the second sense.

In the same paper, they proved the following inequality for an s-convex function:

$\begin{array}{rl}{4}^{s-1}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\\ \le \frac{f\left(a,c\right)+f\left(b,c\right)+f\left(a,d\right)+f\left(b,d\right)}{{\left(s+1\right)}^{2}}.\end{array}$
(1.9)

For refinements and counterparts of convex and s-convex functions on the co-ordinates, see [610].

The main purpose of this paper is to introduce the class of $\left({h}_{1},{h}_{2}\right)$-preinvex functions on the co-ordinates and establish new inequalities like those given by Dragomir in [6] and Bombardelli and Varošanec in [5].

Throughout this paper, we assume that considered integrals exist.

## 2 Main results

Let $f:X\to R$ and $\eta :X×X\to {R}^{n}$, where X is a nonempty closed set in ${R}^{n}$, be continuous functions. First, we recall the following well-known results and concepts; see [1116] and the references therein.

Definition 2.1 Let $u\in X$. Then the set X is said to be invex at u with respect to η if

$u+t\eta \left(v,u\right)\in X$

for all $v\in X$ and $t\in \left[0,1\right]$.

X is said to be an invex set with respect to η if X is invex at each $u\in X$.

Definition 2.2 The function f on the invex set X is said to be preinvex with respect to η if

$f\left(u+t\eta \left(v,u\right)\right)\le \left(1-t\right)f\left(u\right)+tf\left(v\right)$

for all $u,v\in X$ and $t\in \left[0,1\right]$.

We also need the following assumption regarding the function η which is due to Mohan and Neogy [11].

Condition C Let $X\subseteq R$ be an open invex subset with respect to η. For any $x,y\in X$ and any $t\in \left[0,1\right]$,

$\begin{array}{c}\eta \left(y,y+t\eta \left(x,y\right)\right)=-t\eta \left(x,y\right),\hfill \\ \eta \left(x,y+t\eta \left(x,y\right)\right)=\left(1-t\right)\eta \left(x,y\right).\hfill \end{array}$

Note that for every $x,y\in X$ and every ${t}_{1},{t}_{2}\in \left[0,1\right]$ from Condition C, we have

$\eta \left(y+{t}_{2}\eta \left(x,y\right),y+{t}_{1}\eta \left(x,y\right)\right)=\left({t}_{2}-{t}_{1}\right)\eta \left(x,y\right).$

In [12], Noor proved the Hermite-Hadamard inequality for preinvex functions

$f\left(a+\frac{1}{2}\eta \left(b,a\right)\right)\le \frac{1}{\eta \left(b,a\right)}{\int }_{a}^{a+\eta \left(b,a\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \frac{f\left(a\right)+f\left(b\right)}{2}.$
(2.1)

Definition 2.3 Let $h:\left[0,1\right]\to R$ be a non-negative function, $h\not\equiv 0$. The non-negative function f on the invex set X is said to be h-preinvex with respect to η if

$f\left(u+t\eta \left(v,u\right)\right)\le h\left(1-t\right)f\left(u\right)+h\left(t\right)f\left(v\right)$

for each $u,v\in X$ and $t\in \left[0,1\right]$.

Let us note that:

− if $\eta \left(v,u\right)=v-u$, then we get the definition of an h-convex function introduced by Varošanec in [3];

− if $h\left(t\right)=t$, then our definition reduces to the definition of a preinvex function;

− if $\eta \left(v,u\right)=v-u$ and $h\left(t\right)=t$, then we obtain the definition of a convex function.

Now let ${X}_{1}$ and ${X}_{2}$ be nonempty subsets of ${R}^{n}$, let ${\eta }_{1}:{X}_{1}×{X}_{1}\to {R}^{n}$ and ${\eta }_{2}:{X}_{2}×{X}_{2}\to {R}^{n}$.

Definition 2.4 Let $\left(u,v\right)\in {X}_{1}×{X}_{2}$. We say ${X}_{1}×{X}_{2}$ is invex at $\left(u,v\right)$ with respect to ${\eta }_{1}$ and ${\eta }_{2}$ if for each $\left(x,y\right)\in {X}_{1}×{X}_{2}$ and ${t}_{1},{t}_{2}\in \left[0,1\right]$,

$\left(u+{t}_{1}{\eta }_{1}\left(x,u\right),v+{t}_{2}{\eta }_{2}\left(y,v\right)\right)\in {X}_{1}×{X}_{2}.$

${X}_{1}×{X}_{2}$ is said to be an invex set with respect to ${\eta }_{1}$ and ${\eta }_{2}$ if ${X}_{1}×{X}_{2}$ is invex at each $\left(u,v\right)\in {X}_{1}×{X}_{2}$.

Definition 2.5 Let ${h}_{1}$ and ${h}_{2}$ be non-negative functions on $\left[0,1\right]$, ${h}_{1}\not\equiv 0$, ${h}_{2}\not\equiv 0$. The non-negative function f on the invex set ${X}_{1}×{X}_{2}$ is said to be co-ordinated $\left({h}_{1},{h}_{2}\right)$-preinvex with respect to ${\eta }_{1}$ and ${\eta }_{2}$ if the partial mappings ${f}_{y}:{X}_{1}\to R$, ${f}_{y}\left(x\right)=f\left(x,y\right)$ and ${f}_{x}:{X}_{2}\to R$, ${f}_{x}\left(y\right)=f\left(x,y\right)$ are ${h}_{1}$-preinvex with respect to ${\eta }_{1}$ and ${h}_{2}$-preinvex with respect to ${\eta }_{2}$, respectively, for all $y\in {X}_{2}$ and $x\in {X}_{1}$.

If ${\eta }_{1}\left(x,u\right)=x-u$ and ${\eta }_{2}\left(y,v\right)=y-v$, then the function f is called $\left({h}_{1},{h}_{2}\right)$-convex on the co-ordinates.

Remark 1 From the above definition it follows that if f is a co-ordinated $\left({h}_{1},{h}_{2}\right)$-preinvex function, then

$\begin{array}{c}f\left(x+{t}_{1}{\eta }_{1}\left(b,x\right),y+{t}_{2}{\eta }_{2}\left(d,y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left(1-{t}_{1}\right)f\left(x,y+{t}_{2}{\eta }_{2}\left(d,y\right)\right)+{h}_{1}\left({t}_{1}\right)f\left(b,y+{t}_{2}{\eta }_{2}\left(d,y\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left(1-{t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(x,y\right)+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(x,d\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{h}_{1}\left({t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(b,y\right)+{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(b,d\right).\hfill \end{array}$

Remark 2 Let us note that if ${\eta }_{1}\left(x,u\right)=x-u$, ${\eta }_{2}\left(y,v\right)=y-v$, ${t}_{1}={t}_{2}$ and ${h}_{1}\left(t\right)={h}_{2}\left(t\right)=t$, then our definition of a co-ordinated $\left({h}_{1},{h}_{2}\right)$-preinvex function reduces to the definition of a convex function on the co-ordinates proposed by Dragomir [6]. Moreover, if ${h}_{1}\left(t\right)={h}_{2}\left(t\right)={t}^{s}$, then our definition reduces to the definition of an s-convex function on the co-ordinates proposed by Alomari and Darus [7].

Now, we will prove the Hadamard inequality for the new class functions.

Theorem 2.1 Suppose that $f:\left[a,a+\eta \left(b,a\right)\right]\to R$ is an h-preinvex function, Condition  C for η holds and $a, $h\left(\frac{1}{2}\right)>0$. Then the following inequalities hold:

$\frac{1}{2h\left(\frac{1}{2}\right)}f\left(a+\frac{1}{2}\eta \left(b,a\right)\right)\le \frac{1}{\eta \left(b,a\right)}{\int }_{a}^{a+\eta \left(b,a\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \left[f\left(a\right)+f\left(b\right)\right]\cdot {\int }_{0}^{1}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$
(2.2)

Proof From the definition of an h-preinvex function, we have that

$f\left(a+t\eta \left(b,a\right)\right)\le h\left(1-t\right)f\left(a\right)+h\left(t\right)f\left(b\right).$

Thus, by integrating, we obtain

${\int }_{0}^{1}f\left(a+t\eta \left(b,a\right)\right)\phantom{\rule{0.2em}{0ex}}dt\le f\left(a\right){\int }_{0}^{1}h\left(1-t\right)\phantom{\rule{0.2em}{0ex}}dt+f\left(b\right){\int }_{0}^{1}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt=\left[f\left(a\right)+f\left(b\right)\right]{\int }_{0}^{1}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$

But

${\int }_{0}^{1}f\left(a+t\eta \left(b,a\right)\right)\phantom{\rule{0.2em}{0ex}}dt=\frac{1}{\eta \left(b,a\right)}\cdot {\int }_{a}^{a+\eta \left(b,a\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.$

So,

$\frac{1}{\eta \left(b,a\right)}\cdot {\int }_{a}^{a+\eta \left(b,a\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx\le \left[f\left(a\right)+f\left(b\right)\right]{\int }_{0}^{1}h\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.$

The proof of the second inequality follows by using the definition of an h-preinvex function, Condition C for η and integrating over $\left[0,1\right]$.

That is,

$\begin{array}{c}\begin{array}{rl}f\left(a+\frac{1}{2}\eta \left(b,a\right)\right)& =f\left(a+t\eta \left(b,a\right)+\frac{1}{2}\eta \left(a+\left(1-t\right)\eta \left(b,a\right),a+t\eta \left(b,a\right)\right)\\ \le h\left(\frac{1}{2}\right)\left[f\left(a+t\eta \left(b,a\right)\right)+f\left(a+\left(1-t\right)\eta \left(b,a\right)\right)\right],\end{array}\hfill \\ f\left(a+\frac{1}{2}\eta \left(b,a\right)\right)\le h\left(\frac{1}{2}\right)\left[{\int }_{0}^{1}f\left(a+t\eta \left(b,a\right)\right)\phantom{\rule{0.2em}{0ex}}dt+{\int }_{0}^{1}f\left(a+\left(1-t\right)\eta \left(b,a\right)\right)\right],\hfill \\ f\left(a+\frac{1}{2}\eta \left(b,a\right)\right)\le 2\cdot h\left(\frac{1}{2}\right)\frac{1}{\eta \left(b,a\right)}\cdot {\int }_{a}^{a+\eta \left(b,a\right)}f\left(x\right)\phantom{\rule{0.2em}{0ex}}dx.\hfill \end{array}$

The proof is complete. □

Theorem 2.2 Suppose that $f:\left[a,a+{\eta }_{1}\left(b,a\right)\right]×\left[c,c+{\eta }_{2}\left(d,c\right)\right]\to R$ is an $\left({h}_{1},{h}_{2}\right)$-preinvex function on the co-ordinates with respect to ${\eta }_{1}$ and ${\eta }_{2}$, Condition  C for ${\eta }_{1}$ and ${\eta }_{2}$ is fulfilled, and $a, $c, and ${h}_{1}\left(\frac{1}{2}\right)>0$, ${h}_{2}\left(\frac{1}{2}\right)>0$. Then one has the following inequalities:

(2.3)

Proof Since f is $\left({h}_{1},{h}_{2}\right)$-preinvex on the co-ordinates, it follows that the mapping ${f}_{x}$ is ${h}_{2}$-preinvex and the mapping ${f}_{y}$ is ${h}_{1}$-preinvex. Then, by the inequality (2.2), one has

$\begin{array}{rcl}\frac{1}{2{h}_{2}\left(\frac{1}{2}\right)}f\left(x,c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)& \le & \frac{1}{{\eta }_{2}\left(d,c\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dy\\ \le & \left[f\left(x,c\right)+f\left(x,d\right)\right]{\int }_{0}^{1}{h}_{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\end{array}$

and

$\begin{array}{rcl}\frac{1}{2{h}_{1}\left(\frac{1}{2}\right)}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),y\right)& \le & \frac{1}{{\eta }_{1}\left(b,a\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\\ \le & \left[f\left(a,y\right)+f\left(b,y\right)\right]{\int }_{0}^{1}{h}_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt.\end{array}$

Dividing the above inequalities for ${\eta }_{1}\left(b,a\right)$ and ${\eta }_{2}\left(d,c\right)$ and then integrating the resulting inequalities on $\left[a,a+{\eta }_{1}\left(b,a\right)\right]$ and $\left[c,c+{\eta }_{2}\left(d,c\right)\right]$, respectively, we have

$\begin{array}{c}\frac{1}{{\eta }_{1}\left(b,a\right)\cdot 2{h}_{2}\left(\frac{1}{2}\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}f\left(x,c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)\phantom{\rule{0.2em}{0ex}}dx\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\eta }_{1}\left(b,a\right){\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\eta }_{1}\left(b,a\right)}{\int }_{0}^{1}{h}_{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\left[{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}f\left(x,c\right)\phantom{\rule{0.2em}{0ex}}dx+{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}f\left(x,d\right)\phantom{\rule{0.2em}{0ex}}dx\right]\hfill \end{array}$

and

$\begin{array}{c}\frac{1}{{\eta }_{2}\left(b,a\right)\cdot 2{h}_{1}\left(\frac{1}{2}\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),y\right)\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\eta }_{1}\left(b,a\right){\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le \frac{1}{{\eta }_{2}\left(d,c\right)}{\int }_{0}^{1}{h}_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt\left[{\int }_{c}^{c+{\eta }_{2}\left(c,d\right)}f\left(a,y\right)\phantom{\rule{0.2em}{0ex}}dy+{\int }_{c}^{c+{\eta }_{2}\left(c,d\right)}f\left(b,y\right)\phantom{\rule{0.2em}{0ex}}dy\right].\hfill \end{array}$

Summing the above inequalities, we get the second and the third inequalities in (2.3).

By the inequality (2.2), we also have

$\frac{1}{2{h}_{2}\left(\frac{1}{2}\right)}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)\le \frac{1}{{\eta }_{2}\left(d,c\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),y\right)\phantom{\rule{0.2em}{0ex}}dy$

and

$\frac{1}{2{h}_{1}\left(\frac{1}{2}\right)}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)\le \frac{1}{{\eta }_{1}\left(b,a\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}f\left(x,c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)\phantom{\rule{0.2em}{0ex}}dx,$

which give, by addition, the first inequality in (2.3).

Finally, by the same inequality (2.2), we ca also state

$\begin{array}{c}\frac{1}{{\eta }_{2}\left(d,c\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(a,y\right)\phantom{\rule{0.2em}{0ex}}dy\le \left[f\left(a,c\right)+f\left(a,d\right)\right]{\int }_{0}^{1}{h}_{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ \frac{1}{{\eta }_{2}\left(d,c\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(b,y\right)\phantom{\rule{0.2em}{0ex}}dy\le \left[f\left(b,c\right)+f\left(b,d\right)\right]{\int }_{0}^{1}{h}_{2}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ \frac{1}{{\eta }_{1}\left(b,a\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}f\left(x,c\right)\phantom{\rule{0.2em}{0ex}}dx\le \left[f\left(a,c\right)+f\left(b,c\right)\right]{\int }_{0}^{1}{h}_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\hfill \\ \frac{1}{{\eta }_{1}\left(b,a\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}f\left(x,d\right)\phantom{\rule{0.2em}{0ex}}dx\le \left[f\left(a,d\right)+f\left(b,d\right)\right]{\int }_{0}^{1}{h}_{1}\left(t\right)\phantom{\rule{0.2em}{0ex}}dt,\hfill \end{array}$

which give, by addition, the last inequality in (2.3). □

Remark 3 In particular, for ${\eta }_{1}\left(b,a\right)=b-a$, ${\eta }_{2}\left(d,c\right)=d-c$, ${h}_{1}\left({t}_{1}\right)={h}_{2}\left({t}_{2}\right)=t$, we get the inequalities obtained by Dragomir [6] for functions convex on the co-ordinates on the rectangle from the plane ${R}^{2}$.

Remark 4 If ${\eta }_{1}\left(b,a\right)=b-a$, ${\eta }_{2}\left(d,c\right)=d-c$, and ${h}_{1}\left({t}_{1}\right)={h}_{2}\left({t}_{2}\right)={t}^{s}$, then we get the inequalities obtained by Alomari and Darus in [7] for s-convex functions on the co-ordinates on the rectangle from the plane ${R}^{2}$.

Theorem 2.3 Let $f,g:\left[a,a+{\eta }_{1}\left(b,a\right)\right]×\left[c,c+{\eta }_{2}\left(d,c\right)\right]\to R$ with $a, $c. If f is $\left({h}_{1},{h}_{2}\right)$-preinvex on the co-ordinates and g is $\left({k}_{1},{k}_{2}\right)$-preinvex on the co-ordinates with respect to ${\eta }_{1}$ and ${\eta }_{2}$, then

$\begin{array}{c}\frac{1}{{\eta }_{1}\left(b,a\right)\cdot {\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)g\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le {M}_{1}\left(a,b,c,d\right){\int }_{0}^{1}{\int }_{0}^{1}{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right){k}_{1}\left({t}_{1}\right){k}_{2}\left({t}_{2}\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2}\hfill \\ \phantom{\rule{2em}{0ex}}+{M}_{2}\left(a,b,c,d\right){\int }_{0}^{1}{\int }_{0}^{1}{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right){k}_{1}\left({t}_{1}\right){k}_{2}\left(1-{t}_{2}\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2}\hfill \\ \phantom{\rule{2em}{0ex}}+{M}_{3}\left(a,b,c,d\right){\int }_{0}^{1}{\int }_{0}^{1}{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right){k}_{1}\left(1-{t}_{1}\right){k}_{2}\left({t}_{2}\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2}\hfill \\ \phantom{\rule{2em}{0ex}}+{M}_{4}\left(a,b,c,d\right){\int }_{0}^{1}{\int }_{0}^{1}{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right){k}_{1}\left(1-{t}_{1}\right){k}_{2}\left(1-{t}_{2}\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2},\hfill \end{array}$

where

$\begin{array}{c}{M}_{1}\left(a,b,c,d\right)=f\left(a,c\right)g\left(a,c\right)+f\left(a,d\right)g\left(a,d\right)+f\left(b,c\right)g\left(b,c\right)+f\left(b,d\right)g\left(b,d\right),\hfill \\ {M}_{2}\left(a,b,c,d\right)=f\left(a,c\right)g\left(a,d\right)+f\left(a,d\right)g\left(a,c\right)+f\left(b,c\right)g\left(b,d\right)+f\left(b,d\right)g\left(b,c\right),\hfill \\ {M}_{3}\left(a,b,c,d\right)=f\left(a,c\right)g\left(b,c\right)+f\left(a,d\right)g\left(b,d\right)+f\left(b,c\right)g\left(a,c\right)+f\left(b,d\right)g\left(a,d\right),\hfill \\ {M}_{4}\left(a,b,c,d\right)=f\left(a,c\right)g\left(b,d\right)+f\left(a,d\right)g\left(b,c\right)+f\left(b,c\right)g\left(a,d\right)+f\left(b,d\right)g\left(a,c\right).\hfill \end{array}$

Proof Since f is $\left({h}_{1},{h}_{2}\right)$-preinvex on the co-ordinates and g is $\left({k}_{1},{k}_{2}\right)$-preinvex on the co-ordinates with respect to ${\eta }_{1}$ and ${\eta }_{2}$, it follows that

$\begin{array}{c}f\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left(1-{t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(a,c\right)+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(a,d\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{h}_{1}\left({t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(b,c\right)+{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(b,d\right)\hfill \end{array}$

and

$\begin{array}{c}g\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {k}_{1}\left(1-{t}_{1}\right){k}_{2}\left(1-{t}_{2}\right)g\left(a,c\right)+{k}_{1}\left(1-{t}_{1}\right){k}_{2}\left({t}_{2}\right)g\left(a,d\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{k}_{1}\left({t}_{1}\right){k}_{2}\left(1-{t}_{2}\right)g\left(b,c\right)+{k}_{1}\left({t}_{1}\right){k}_{2}\left({t}_{2}\right)g\left(b,d\right).\hfill \end{array}$

Multiplying the above inequalities and integrating over ${\left[0,1\right]}^{2}$ and using the fact that

$\begin{array}{c}{\int }_{0}^{1}{\int }_{0}^{1}f\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\cdot g\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2}\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{{\eta }_{1}\left(b,a\right)\cdot {\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)g\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy,\hfill \end{array}$

we obtain our inequality. □

In the next two theorems, we will prove the so-called Hermite-Hadamard-Fejér inequalities for an $\left({h}_{1},{h}_{2}\right)$-preinvex function.

Theorem 2.4 Let $f:\left[a,a+{\eta }_{1}\left(b,a\right)\right]×\left[c,c+{\eta }_{2}\left(d,c\right)\right]\to R$ be $\left({h}_{1},{h}_{2}\right)$-preinvex on the co-ordinates with respect to ${\eta }_{1}$ and ${\eta }_{2}$, $a, $c, and $w:\left[a,a+{\eta }_{1}\left(b,a\right)\right]×\left[c,c+{\eta }_{2}\left(d,c\right)\right]\to R$, $w\ge 0$, symmetric with respect to

$\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right).$

Then

$\begin{array}{r}\frac{1}{{\eta }_{1}\left(b,a\right)\cdot {\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)w\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\\ \phantom{\rule{1em}{0ex}}\le \left[f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)\right]\\ \phantom{\rule{2em}{0ex}}\cdot {\int }_{0}^{1}{\int }_{0}^{1}{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)w\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2}.\end{array}$
(2.4)

Proof From the definition of $\left({h}_{1},{h}_{2}\right)$-preinvex on the co-ordinates with respect to ${\eta }_{1}$ and ${\eta }_{2}$, we have

1. (a)
$\begin{array}{c}f\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left(1-{t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(a,c\right)+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(a,d\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{h}_{1}\left({t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(b,c\right)+{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(b,d\right),\hfill \end{array}$
2. (b)
$\begin{array}{c}f\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(a,c\right)+{h}_{1}\left({t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(a,d\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(b,c\right)+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(b,d\right),\hfill \end{array}$
3. (c)
$\begin{array}{c}f\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left(1-{t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(a,c\right)+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(a,d\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(b,c\right)+{h}_{1}\left({t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(b,d\right),\hfill \end{array}$
4. (d)
$\begin{array}{c}f\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left({t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(a,c\right)+{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(a,d\right)\hfill \\ \phantom{\rule{2em}{0ex}}+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left(1-{t}_{2}\right)f\left(b,c\right)+{h}_{1}\left(1-{t}_{1}\right){h}_{2}\left({t}_{2}\right)f\left(b,d\right).\hfill \end{array}$

Multiplying both sides of the above inequalities by $w\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)$, $w\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)$, $w\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)$, $w\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)$, respectively, adding and integrating over ${\left[0,1\right]}^{2}$, we obtain

$\begin{array}{c}\frac{4}{{\eta }_{1}\left(b,a\right)\cdot {\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)w\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le \left[f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)\right]\hfill \\ \phantom{\rule{2em}{0ex}}\cdot 4{\int }_{0}^{1}{\int }_{0}^{1}{h}_{1}\left({t}_{1}\right){h}_{2}\left({t}_{2}\right)w\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2},\hfill \end{array}$

where we use the symmetricity of the w with respect to $\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)$, which completes the proof. □

Theorem 2.5 Let $f:\left[a,a+{\eta }_{1}\left(b,a\right)\right]×\left[c,c+{\eta }_{2}\left(d,c\right)\right]\to R$ be $\left({h}_{1},{h}_{2}\right)$-preinvex on the co-ordinates with respect to ${\eta }_{1}$ and ${\eta }_{2}$, and $a, $c, $w:\left[a,a+{\eta }_{1}\left(b,a\right)\right]×\left[c,c+{\eta }_{2}\left(d,c\right)\right]\to R$, $w\ge 0$, symmetric with respect to $\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)$. Then, if Condition  C for ${\eta }_{1}$ and ${\eta }_{2}$ is fulfilled, we have

$\begin{array}{r}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)\cdot {\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}w\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\\ \phantom{\rule{1em}{0ex}}\le 4\cdot {h}_{1}\left(\frac{1}{2}\right){h}_{2}\left(\frac{1}{2}\right)\cdot {\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)w\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy.\end{array}$
(2.5)

Proof Using the definition of an $\left({h}_{1},{h}_{2}\right)$-preinvex function on the co-ordinates and Condition C for ${\eta }_{1}$ and ${\eta }_{2}$, we obtain

$\begin{array}{c}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left(\frac{1}{2}\right){h}_{2}\left(\frac{1}{2}\right)\cdot \left[f\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}+f\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)+f\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\hfill \\ \phantom{\rule{2em}{0ex}}+f\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)\right].\hfill \end{array}$

Now, we multiply it by $w\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)$ = $w\left(a+{t}_{1}{\eta }_{1}\left(b,c\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)$ = $w\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)$ = $w\left(a+\left(1-{t}_{1}\right){\eta }_{1}\left(b,a\right),c+\left(1-{t}_{2}\right){\eta }_{2}\left(d,c\right)\right)$ and integrate over ${\left[0,1\right]}^{2}$ to obtain the inequality

$\begin{array}{c}f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right){\int }_{0}^{1}{\int }_{0}^{1}w\left(a+{t}_{1}{\eta }_{1}\left(b,a\right),c+{t}_{2}{\eta }_{2}\left(d,c\right)\right)\phantom{\rule{0.2em}{0ex}}d{t}_{1}\phantom{\rule{0.2em}{0ex}}d{t}_{2}\hfill \\ \phantom{\rule{1em}{0ex}}=f\left(a+\frac{1}{2}{\eta }_{1}\left(b,a\right),c+\frac{1}{2}{\eta }_{2}\left(d,c\right)\right)\frac{1}{{\eta }_{1}\left(b,a\right)\cdot {\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}w\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le 4\cdot {h}_{1}\left(\frac{1}{2}\right){h}_{2}\left(\frac{1}{2}\right)\frac{1}{{\eta }_{1}\left(b,a\right)\cdot {\eta }_{2}\left(d,c\right)}{\int }_{a}^{a+{\eta }_{1}\left(b,a\right)}{\int }_{c}^{c+{\eta }_{2}\left(d,c\right)}f\left(x,y\right)w\left(x,y\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy,\hfill \end{array}$

which completes the proof. □

Now, for a mapping $f:\left[a,b\right]×\left[c,d\right]\to R$, let us define a mapping $H:{\left[0,1\right]}^{2}\to R$ in the following way:

$H\left(t,r\right)=\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(tx+\left(1-t\right)\frac{a+b}{2},ry+\left(1-r\right)\frac{c+d}{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy.$
(2.6)

Some properties of this mapping for a convex on the co-ordinates function and an s-convex on the co-ordinates function are given in [6, 7], respectively. Here we investigate which of these properties can be generalized for $\left({h}_{1},{h}_{2}\right)$-convex on the co-ordinates functions.

Theorem 2.6 Suppose that $f:\left[a,b\right]×\left[c,d\right]$ is $\left({h}_{1},{h}_{2}\right)$-convex on the co-ordinates. Then:

1. (i)

The mapping H is $\left({h}_{1},{h}_{2}\right)$-convex on the co-ordinates on ${\left[0,1\right]}^{2}$,

2. (ii)

$4{h}_{1}\left(\frac{1}{2}\right){h}_{2}\left(\frac{1}{2}\right)H\left(t,r\right)\ge H\left(0,0\right)$ for any $\left(t,r\right)\in {\left[0,1\right]}^{2}$.

Proof (i) The $\left({h}_{1},{h}_{2}\right)$-convexity on the co-ordinates of the mapping H is a consequence of the $\left({h}_{1},{h}_{2}\right)$-convexity on the co-ordinates of the function f. Namely, for $r\in \left[0,1\right]$ and for all $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$ and ${t}_{1},{t}_{2}\in \left[0,1\right]$, we have:

$\begin{array}{c}H\left(\alpha {t}_{1}+\beta {t}_{2},r\right)\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(\left(\alpha {t}_{1}+\beta {t}_{2},r\right)x+\left(1-\left(\alpha {t}_{1}+\beta {t}_{2}\right)\right)\frac{a+b}{2},\hfill \\ \phantom{\rule{2em}{0ex}}ry+\left(1-r\right)\frac{c+d}{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}=\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(\alpha \left({t}_{1}x+\left(1-{t}_{1}\right)\frac{a+b}{2}\right)+\beta \left({t}_{2}x+\left(1-{t}_{2}\right)\frac{a+b}{2}\right),\hfill \\ \phantom{\rule{2em}{0ex}}ry+\left(1-r\right)\frac{c+d}{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}\le {h}_{1}\left(\alpha \right)\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left({t}_{1}x+\left(1-{t}_{1}\right)\frac{a+b}{2},ry+\left(1-r\right)\frac{c+d}{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{2em}{0ex}}+{h}_{1}\left(\beta \right)\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left({t}_{2}x+\left(1-{t}_{2}\right)\frac{a+b}{2},ry+\left(1-r\right)\frac{c+d}{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\hfill \\ \phantom{\rule{1em}{0ex}}={h}_{1}\left(\alpha \right)H\left({t}_{1},r\right)+{h}_{1}\left(\beta \right)H\left({t}_{2},r\right).\hfill \end{array}$

Similarly, if $t\in \left[0,1\right]$ is fixed, then for all ${r}_{1},{r}_{2}\in \left[0,1\right]$ and $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$, we also have

$H\left(t,\alpha {r}_{1}+\beta {r}_{2}\right)\le {h}_{2}\left(\alpha \right)H\left(t,{r}_{1}\right)+{h}_{2}\left(\beta \right)H\left(t,{r}_{2}\right),$

which means that H is $\left({h}_{1},{h}_{2}\right)$-convex on the co-ordinates.

1. (ii)

After changing the variables $u=tx+\left(1-t\right)\frac{a+b}{2}$ and $v=ry+\left(1-r\right)\frac{c+d}{2}$, we have

$\begin{array}{rcl}H\left(t,r\right)& =& \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(tx+\left(1-t\right)\frac{a+b}{2},ry+\left(1-r\right)\frac{c+d}{2}\right)\phantom{\rule{0.2em}{0ex}}dx\phantom{\rule{0.2em}{0ex}}dy\\ =& \frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{{u}_{L}}^{{u}_{U}}{\int }_{{v}_{L}}^{{v}_{U}}f\left(u,v\right)\frac{b-a}{{u}_{U}-{u}_{L}}\cdot \frac{d-c}{{v}_{U}-{v}_{L}}\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ =& \frac{1}{\left({u}_{U}-{u}_{L}\right)\left({v}_{U}-{v}_{L}\right)}{\int }_{{u}_{L}}^{{u}_{U}}{\int }_{{v}_{L}}^{{v}_{U}}f\left(u,v\right)\phantom{\rule{0.2em}{0ex}}du\phantom{\rule{0.2em}{0ex}}dv\\ \ge & \frac{1}{4{h}_{1}\left(\frac{1}{2}\right){h}_{2}\left(\frac{1}{2}\right)}f\left(\frac{a+b}{2},\frac{c+d}{2}\right),\end{array}$

where ${u}_{L}=ta+\left(1-t\right)\frac{a+b}{2}$, ${u}_{U}=tb+\left(1-t\right)\frac{a+b}{2}$, ${v}_{L}=rc+\left(1-r\right)\frac{c+d}{2}$ and ${v}_{U}=rd+\left(1-r\right)\frac{c+d}{2}$, which completes the proof. □

Remark 5 If f is convex on the co-ordinates, then we get $H\left(t,r\right)\ge H\left(0,0\right)$. If f is s-convex on the co-ordinates in the second sense, then we have the inequality $H\left(t,r\right)\ge {4}^{s-1}H\left(0,0\right)$.

## References

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Matłoka, M. On some Hadamard-type inequalities for $\left({h}_{1},{h}_{2}\right)$-preinvex functions on the co-ordinates. J Inequal Appl 2013, 227 (2013). https://doi.org/10.1186/1029-242X-2013-227

• $\left({h}_{1},{h}_{2}\right)$-preinvex function on the co-ordinates