On some Hadamard-type inequalities for $(h_1,h_2)$-preinvex functions on the co-ordinates

Abstract

We introduce the class of $(h_1,h_2)$-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

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

(1.1)

holds for all $x,y\in I$ and $t\in [0,1]$. 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:

$$f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le \frac{f(a)+f(b)}{2},$$

(1.2)

where $f:[a,b]\to R$ 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 $f:[0,\mathrm{\infty })\to R$ is said to be s-convex in the second sense if

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

(1.3)

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

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:

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

(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

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

(1.5)

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

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:

$$\frac{1}{2h(\frac{1}{2})}f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}{\int }_a^{b}f(x)dx\le [f(a)+f(b)]\cdot {\int }_0^{1}h(t)dt.$$

(1.6)

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

$$\begin{array}{rcl}\frac{{\int }_a^{b}w(x)dx}{2h(\frac{1}{2})}f\left(\frac{a+b}{2}\right)& \le & {\int }_a^{b}f(x)w(x)dx\\ \le & (b-a)(f(a)+f(b)){\int }_0^{1}h(t)w(ta+(1-t)b)dt,\end{array}$$

(1.7)

where $w:[a,b]\to R$, $w\ge 0$ 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 $\mathrm{\Delta }=[a,b]\times [c,d]$ in $R^2$ with $a<b$ and $c<d$. A mapping $f:\mathrm{\Delta }\to R$ is said to be convex on the co-ordinates on Δ if the partial mappings $f_y:[a,b]\to R$, $f_y(u)=f(u,y)$ and $f_x:[c,d]\to R$, $f_x(v)=f(x,v)$ are convex for all $x\in [a,b]$ and $y\in [c,d]$.

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

$$\begin{array}{rl}f(\frac{a+b}{2},\frac{c+d}{2})& \le \frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(x,y)dxdy\\ \le \frac{f(a,c)+f(b,c)+f(a,d)+f(b,d)}{4}.\end{array}$$

(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 $f_y:[a,b]\to R$ and $f_x:[c,d]\to R$ are s-convex in the second sense.

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

$$\begin{array}{rl}{4}^{s-1}f(\frac{a+b}{2},\frac{c+d}{2})& \le \frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(x,y)dxdy\\ \le \frac{f(a,c)+f(b,c)+f(a,d)+f(b,d)}{{(s+1)}^2}.\end{array}$$

(1.9)

For refinements and counterparts of convex and s-convex functions on the co-ordinates, see [6–10].

The main purpose of this paper is to introduce the class of $(h_1,h_2)$-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 $\eta :X\times X\to R^n$, where X is a nonempty closed set in $R^n$, be continuous functions. First, we recall the following well-known results and concepts; see [11–16] 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

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

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

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

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

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

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 $t\in [0,1]$,

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

Note that for every $x,y\in X$ and every $t_1,t_2\in [0,1]$ from Condition C, we have

$$\eta (y+t_2\eta (x,y),y+t_1\eta (x,y))=(t_2-t_1)\eta (x,y).$$

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

$$f(a+\frac{1}{2}\eta (b,a))\le \frac{1}{\eta (b,a)}{\int }_a^{a+\eta (b,a)}f(x)dx\le \frac{f(a)+f(b)}{2}.$$

(2.1)

Definition 2.3 Let $h:[0,1]\to R$ 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

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

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

Let us note that:

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

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

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

Now let $X_1$ and $X_2$ be nonempty subsets of $R^n$, let ${\eta }_1:X_1\times X_1\to R^n$ and ${\eta }_2:X_2\times X_2\to R^n$.

Definition 2.4 Let $(u,v)\in X_1\times X_2$. We say $X_1\times X_2$ is invex at $(u,v)$ with respect to ${\eta }_1$ and ${\eta }_2$ if for each $(x,y)\in X_1\times X_2$ and $t_1,t_2\in [0,1]$,

$$(u+t_1{\eta }_1(x,u),v+t_2{\eta }_2(y,v))\in X_1\times X_2.$$

$X_1\times X_2$ is said to be an invex set with respect to ${\eta }_1$ and ${\eta }_2$ if $X_1\times X_2$ is invex at each $(u,v)\in X_1\times X_2$.

Definition 2.5 Let $h_1$ and $h_2$ be non-negative functions on $[0,1]$, $h_1\not\equiv 0$, $h_2\not\equiv 0$. The non-negative function f on the invex set $X_1\times X_2$ is said to be co-ordinated $(h_1,h_2)$-preinvex with respect to ${\eta }_1$ and ${\eta }_2$ if the partial mappings $f_y:X_1\to R$, $f_y(x)=f(x,y)$ and $f_x:X_2\to R$, $f_x(y)=f(x,y)$ 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 ${\eta }_1(x,u)=x-u$ and ${\eta }_2(y,v)=y-v$, then the function f is called $(h_1,h_2)$-convex on the co-ordinates.

Remark 1 From the above definition it follows that if f is a co-ordinated $(h_1,h_2)$-preinvex function, then

$$\begin{array}{c}f(x+t_1{\eta }_1(b,x),y+t_2{\eta }_2(d,y))\hfill \\ \le h_1(1-t_1)f(x,y+t_2{\eta }_2(d,y))+h_1(t_1)f(b,y+t_2{\eta }_2(d,y))\hfill \\ \le h_1(1-t_1)h_2(1-t_2)f(x,y)+h_1(1-t_1)h_2(t_2)f(x,d)\hfill \\ +h_1(t_1)h_2(1-t_2)f(b,y)+h_1(t_1)h_2(t_2)f(b,d).\hfill \end{array}$$

Remark 2 Let us note that if ${\eta }_1(x,u)=x-u$, ${\eta }_2(y,v)=y-v$, $t_1=t_2$ and $h_1(t)=h_2(t)=t$, then our definition of a co-ordinated $(h_1,h_2)$-preinvex function reduces to the definition of a convex function on the co-ordinates proposed by Dragomir [6]. Moreover, if $h_1(t)=h_2(t)=t^s$, 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 $f:[a,a+\eta (b,a)]\to R$ is an h-preinvex function, Condition  C for η holds and $a<a+\eta (b,a)$, $h(\frac{1}{2})>0$. Then the following inequalities hold:

$$\frac{1}{2h(\frac{1}{2})}f(a+\frac{1}{2}\eta (b,a))\le \frac{1}{\eta (b,a)}{\int }_a^{a+\eta (b,a)}f(x)dx\le [f(a)+f(b)]\cdot {\int }_0^{1}h(t)dt.$$

(2.2)

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

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

Thus, by integrating, we obtain

$${\int }_0^{1}f(a+t\eta (b,a))dt\le f(a){\int }_0^{1}h(1-t)dt+f(b){\int }_0^{1}h(t)dt=[f(a)+f(b)]{\int }_0^{1}h(t)dt.$$

But

$${\int }_0^{1}f(a+t\eta (b,a))dt=\frac{1}{\eta (b,a)}\cdot {\int }_a^{a+\eta (b,a)}f(x)dx.$$

So,

$$\frac{1}{\eta (b,a)}\cdot {\int }_a^{a+\eta (b,a)}f(x)dx\le [f(a)+f(b)]{\int }_0^{1}h(t)dt.$$

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

That is,

$$\begin{array}{c}\begin{array}{rl}f(a+\frac{1}{2}\eta (b,a))& =f(a+t\eta (b,a)+\frac{1}{2}\eta (a+(1-t)\eta (b,a),a+t\eta (b,a))\\ \le h\left(\frac{1}{2}\right)[f(a+t\eta (b,a))+f(a+(1-t)\eta (b,a))],\end{array}\hfill \\ f(a+\frac{1}{2}\eta (b,a))\le h\left(\frac{1}{2}\right)[{\int }_0^{1}f(a+t\eta (b,a))dt+{\int }_0^{1}f(a+(1-t)\eta (b,a))],\hfill \\ f(a+\frac{1}{2}\eta (b,a))\le 2\cdot h\left(\frac{1}{2}\right)\frac{1}{\eta (b,a)}\cdot {\int }_a^{a+\eta (b,a)}f(x)dx.\hfill \end{array}$$

The proof is complete. □

Theorem 2.2 Suppose that $f:[a,a+{\eta }_1(b,a)]\times [c,c+{\eta }_2(d,c)]\to R$ is an $(h_1,h_2)$-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 $a<a+{\eta }_1(b,a)$, $c<c+{\eta }_2(d,c)$, and $h_1(\frac{1}{2})>0$, $h_2(\frac{1}{2})>0$. Then one has the following inequalities:

(2.3)

Proof Since f is $(h_1,h_2)$-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

$$\begin{array}{rcl}\frac{1}{2h_2(\frac{1}{2})}f(x,c+\frac{1}{2}{\eta }_2(d,c))& \le & \frac{1}{{\eta }_2(d,c)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)dy\\ \le & [f(x,c)+f(x,d)]{\int }_0^1{h}_2(t)dt\end{array}$$

and

$$\begin{array}{rcl}\frac{1}{2h_1(\frac{1}{2})}f(a+\frac{1}{2}{\eta }_1(b,a),y)& \le & \frac{1}{{\eta }_1(b,a)}{\int }_a^{a+{\eta }_1(b,a)}f(x,y)dx\\ \le & [f(a,y)+f(b,y)]{\int }_0^1{h}_1(t)dt.\end{array}$$

Dividing the above inequalities for ${\eta }_1(b,a)$ and ${\eta }_2(d,c)$ and then integrating the resulting inequalities on $[a,a+{\eta }_1(b,a)]$ and $[c,c+{\eta }_2(d,c)]$, respectively, we have

$$\begin{array}{c}\frac{1}{{\eta }_1(b,a)\cdot 2h_2(\frac{1}{2})}{\int }_a^{a+{\eta }_1(b,a)}f(x,c+\frac{1}{2}{\eta }_2(d,c))dx\hfill \\ \le \frac{1}{{\eta }_1(b,a){\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)dxdy\hfill \\ \le \frac{1}{{\eta }_1(b,a)}{\int }_0^1{h}_2(t)dt[{\int }_a^{a+{\eta }_1(b,a)}f(x,c)dx+{\int }_a^{a+{\eta }_1(b,a)}f(x,d)dx]\hfill \end{array}$$

and

$$\begin{array}{c}\frac{1}{{\eta }_2(b,a)\cdot 2h_1(\frac{1}{2})}{\int }_c^{c+{\eta }_2(d,c)}f(a+\frac{1}{2}{\eta }_1(b,a),y)dy\hfill \\ \le \frac{1}{{\eta }_1(b,a){\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)dxdy\hfill \\ \le \frac{1}{{\eta }_2(d,c)}{\int }_0^1{h}_1(t)dt[{\int }_c^{c+{\eta }_2(c,d)}f(a,y)dy+{\int }_c^{c+{\eta }_2(c,d)}f(b,y)dy].\hfill \end{array}$$

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

By the inequality (2.2), we also have

$$\frac{1}{2h_2(\frac{1}{2})}f(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c))\le \frac{1}{{\eta }_2(d,c)}{\int }_c^{c+{\eta }_2(d,c)}f(a+\frac{1}{2}{\eta }_1(b,a),y)dy$$

and

$$\frac{1}{2h_1(\frac{1}{2})}f(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c))\le \frac{1}{{\eta }_1(b,a)}{\int }_a^{a+{\eta }_1(b,a)}f(x,c+\frac{1}{2}{\eta }_2(d,c))dx,$$

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

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

$$\begin{array}{c}\frac{1}{{\eta }_2(d,c)}{\int }_c^{c+{\eta }_2(d,c)}f(a,y)dy\le [f(a,c)+f(a,d)]{\int }_0^1{h}_2(t)dt,\hfill \\ \frac{1}{{\eta }_2(d,c)}{\int }_c^{c+{\eta }_2(d,c)}f(b,y)dy\le [f(b,c)+f(b,d)]{\int }_0^1{h}_2(t)dt,\hfill \\ \frac{1}{{\eta }_1(b,a)}{\int }_a^{a+{\eta }_1(b,a)}f(x,c)dx\le [f(a,c)+f(b,c)]{\int }_0^1{h}_1(t)dt,\hfill \\ \frac{1}{{\eta }_1(b,a)}{\int }_a^{a+{\eta }_1(b,a)}f(x,d)dx\le [f(a,d)+f(b,d)]{\int }_0^1{h}_1(t)dt,\hfill \end{array}$$

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

Remark 3 In particular, for ${\eta }_1(b,a)=b-a$, ${\eta }_2(d,c)=d-c$, $h_1(t_1)=h_2(t_2)=t$, 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 ${\eta }_1(b,a)=b-a$, ${\eta }_2(d,c)=d-c$, and $h_1(t_1)=h_2(t_2)=t^s$, 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 $f,g:[a,a+{\eta }_1(b,a)]\times [c,c+{\eta }_2(d,c)]\to R$ with $a<a+{\eta }_1(b,a)$, $c<c+{\eta }_2(d,c)$. If f is $(h_1,h_2)$-preinvex on the co-ordinates and g is $(k_1,k_2)$-preinvex on the co-ordinates with respect to ${\eta }_1$ and ${\eta }_2$, then

$$\begin{array}{c}\frac{1}{{\eta }_1(b,a)\cdot {\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)g(x,y)dxdy\hfill \\ \le M_1(a,b,c,d){\int }_0^1{\int }_0^1{h}_1(t_1)h_2(t_2)k_1(t_1)k_2(t_2)d{t}_{1}d{t}_2\hfill \\ +M_2(a,b,c,d){\int }_0^1{\int }_0^1{h}_1(t_1)h_2(t_2)k_1(t_1)k_2(1-t_2)d{t}_{1}d{t}_2\hfill \\ +M_3(a,b,c,d){\int }_0^1{\int }_0^1{h}_1(t_1)h_2(t_2)k_1(1-t_1)k_2(t_2)d{t}_{1}d{t}_2\hfill \\ +M_4(a,b,c,d){\int }_0^1{\int }_0^1{h}_1(t_1)h_2(t_2)k_1(1-t_1)k_2(1-t_2)d{t}_{1}d{t}_2,\hfill \end{array}$$

where

$$\begin{array}{c}{M}_1(a,b,c,d)=f(a,c)g(a,c)+f(a,d)g(a,d)+f(b,c)g(b,c)+f(b,d)g(b,d),\hfill \\ M_2(a,b,c,d)=f(a,c)g(a,d)+f(a,d)g(a,c)+f(b,c)g(b,d)+f(b,d)g(b,c),\hfill \\ M_3(a,b,c,d)=f(a,c)g(b,c)+f(a,d)g(b,d)+f(b,c)g(a,c)+f(b,d)g(a,d),\hfill \\ M_4(a,b,c,d)=f(a,c)g(b,d)+f(a,d)g(b,c)+f(b,c)g(a,d)+f(b,d)g(a,c).\hfill \end{array}$$

Proof Since f is $(h_1,h_2)$-preinvex on the co-ordinates and g is $(k_1,k_2)$-preinvex on the co-ordinates with respect to ${\eta }_1$ and ${\eta }_2$, it follows that

$$\begin{array}{c}f(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))\hfill \\ \le h_1(1-t_1)h_2(1-t_2)f(a,c)+h_1(1-t_1)h_2(t_2)f(a,d)\hfill \\ +h_1(t_1)h_2(1-t_2)f(b,c)+h_1(t_1)h_2(t_2)f(b,d)\hfill \end{array}$$

and

$$\begin{array}{c}g(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))\hfill \\ \le k_1(1-t_1)k_2(1-t_2)g(a,c)+k_1(1-t_1)k_2(t_2)g(a,d)\hfill \\ +k_1(t_1)k_2(1-t_2)g(b,c)+k_1(t_1)k_2(t_2)g(b,d).\hfill \end{array}$$

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

$$\begin{array}{c}{\int }_0^1{\int }_0^{1}f(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))\cdot g(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))d{t}_{1}d{t}_2\hfill \\ =\frac{1}{{\eta }_1(b,a)\cdot {\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)g(x,y)dxdy,\hfill \end{array}$$

we obtain our inequality. □

In the next two theorems, we will prove the so-called Hermite-Hadamard-Fejér inequalities for an $(h_1,h_2)$-preinvex function.

Theorem 2.4 Let $f:[a,a+{\eta }_1(b,a)]\times [c,c+{\eta }_2(d,c)]\to R$ be $(h_1,h_2)$-preinvex on the co-ordinates with respect to ${\eta }_1$ and ${\eta }_2$, $a<a+{\eta }_1(b,a)$, $c<c+{\eta }_2(d,c)$, and $w:[a,a+{\eta }_1(b,a)]\times [c,c+{\eta }_2(d,c)]\to R$, $w\ge 0$, symmetric with respect to

$$(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c)).$$

Then

$$\begin{array}{r}\frac{1}{{\eta }_1(b,a)\cdot {\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)w(x,y)dxdy\\ \le [f(a,c)+f(a,d)+f(b,c)+f(b,d)]\\ \cdot {\int }_0^1{\int }_0^1{h}_1(t_1)h_2(t_2)w(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))d{t}_{1}d{t}_2.\end{array}$$

(2.4)

Proof From the definition of $(h_1,h_2)$-preinvex on the co-ordinates with respect to ${\eta }_1$ and ${\eta }_2$, we have

• (a)
  $$\begin{array}{c}f(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))\hfill \\ \le h_1(1-t_1)h_2(1-t_2)f(a,c)+h_1(1-t_1)h_2(t_2)f(a,d)\hfill \\ +h_1(t_1)h_2(1-t_2)f(b,c)+h_1(t_1)h_2(t_2)f(b,d),\hfill \end{array}$$
• (b)
  $$\begin{array}{c}f(a+(1-t_1){\eta }_1(b,a),c+(1-t_2){\eta }_2(d,c))\hfill \\ \le h_1(t_1)h_2(t_2)f(a,c)+h_1(t_1)h_2(1-t_2)f(a,d)\hfill \\ +h_1(1-t_1)h_2(t_2)f(b,c)+h_1(1-t_1)h_2(1-t_2)f(b,d),\hfill \end{array}$$
• (c)
  $$\begin{array}{c}f(a+t_1{\eta }_1(b,a),c+(1-t_2){\eta }_2(d,c))\hfill \\ \le h_1(1-t_1)h_2(t_2)f(a,c)+h_1(1-t_1)h_2(1-t_2)f(a,d)\hfill \\ +h_1(t_1)h_2(t_2)f(b,c)+h_1(t_1)h_2(1-t_2)f(b,d),\hfill \end{array}$$
• (d)
  $$\begin{array}{c}f(a+(1-t_1){\eta }_1(b,a),c+t_2{\eta }_2(d,c))\hfill \\ \le h_1(t_1)h_2(1-t_2)f(a,c)+h_1(t_1)h_2(t_2)f(a,d)\hfill \\ +h_1(1-t_1)h_2(1-t_2)f(b,c)+h_1(1-t_1)h_2(t_2)f(b,d).\hfill \end{array}$$

Multiplying both sides of the above inequalities by $w(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))$, $w(a+(1-t_1){\eta }_1(b,a),c+(1-t_2){\eta }_2(d,c))$, $w(a+t_1{\eta }_1(b,a),c+(1-t_2){\eta }_2(d,c))$, $w(a+(1-t_1){\eta }_1(b,a),c+t_2{\eta }_2(d,c))$, respectively, adding and integrating over ${[0,1]}^2$, we obtain

$$\begin{array}{c}\frac{4}{{\eta }_1(b,a)\cdot {\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)w(x,y)dxdy\hfill \\ \le [f(a,c)+f(a,d)+f(b,c)+f(b,d)]\hfill \\ \cdot 4{\int }_0^1{\int }_0^1{h}_1(t_1)h_2(t_2)w(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))d{t}_{1}d{t}_2,\hfill \end{array}$$

where we use the symmetricity of the w with respect to $(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c))$, which completes the proof. □

Theorem 2.5 Let $f:[a,a+{\eta }_1(b,a)]\times [c,c+{\eta }_2(d,c)]\to R$ be $(h_1,h_2)$-preinvex on the co-ordinates with respect to ${\eta }_1$ and ${\eta }_2$, and $a<a+{\eta }_1(b,a)$, $c<c+{\eta }_2(d,c)$, $w:[a,a+{\eta }_1(b,a)]\times [c,c+{\eta }_2(d,c)]\to R$, $w\ge 0$, symmetric with respect to $(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c))$. Then, if Condition  C for ${\eta }_1$ and ${\eta }_2$ is fulfilled, we have

$$\begin{array}{r}f(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c))\cdot {\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}w(x,y)dxdy\\ \le 4\cdot h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)\cdot {\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)w(x,y)dxdy.\end{array}$$

(2.5)

Proof Using the definition of an $(h_1,h_2)$-preinvex function on the co-ordinates and Condition C for ${\eta }_1$ and ${\eta }_2$, we obtain

$$\begin{array}{c}f(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c))\hfill \\ \le h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)\cdot [f(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))\hfill \\ +f(a+t_1{\eta }_1(b,a),c+(1-t_2){\eta }_2(d,c))+f(a+(1-t_1){\eta }_1(b,a),c+t_2{\eta }_2(d,c))\hfill \\ +f(a+(1-t_1){\eta }_1(b,a),c+(1-t_2){\eta }_2(d,c))].\hfill \end{array}$$

Now, we multiply it by $w(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))$ = $w(a+t_1{\eta }_1(b,c),c+(1-t_2){\eta }_2(d,c))$ = $w(a+(1-t_1){\eta }_1(b,a),c+t_2{\eta }_2(d,c))$ = $w(a+(1-t_1){\eta }_1(b,a),c+(1-t_2){\eta }_2(d,c))$ and integrate over ${[0,1]}^2$ to obtain the inequality

$$\begin{array}{c}f(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c)){\int }_0^1{\int }_0^{1}w(a+t_1{\eta }_1(b,a),c+t_2{\eta }_2(d,c))d{t}_{1}d{t}_2\hfill \\ =f(a+\frac{1}{2}{\eta }_1(b,a),c+\frac{1}{2}{\eta }_2(d,c))\frac{1}{{\eta }_1(b,a)\cdot {\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}w(x,y)dxdy\hfill \\ \le 4\cdot h_1\left(\frac{1}{2}\right)h_2\left(\frac{1}{2}\right)\frac{1}{{\eta }_1(b,a)\cdot {\eta }_2(d,c)}{\int }_a^{a+{\eta }_1(b,a)}{\int }_c^{c+{\eta }_2(d,c)}f(x,y)w(x,y)dxdy,\hfill \end{array}$$

which completes the proof. □

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

$$H(t,r)=\frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(tx+(1-t)\frac{a+b}{2},ry+(1-r)\frac{c+d}{2})dxdy.$$

(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 $(h_1,h_2)$-convex on the co-ordinates functions.

Theorem 2.6 Suppose that $f:[a,b]\times [c,d]$ is $(h_1,h_2)$-convex on the co-ordinates. Then:

• (i) The mapping H is $(h_1,h_2)$-convex on the co-ordinates on ${[0,1]}^2$,

• (ii) $4h_1(\frac{1}{2})h_2(\frac{1}{2})H(t,r)\ge H(0,0)$ for any $(t,r)\in {[0,1]}^2$.

Proof (i) The $(h_1,h_2)$-convexity on the co-ordinates of the mapping H is a consequence of the $(h_1,h_2)$-convexity on the co-ordinates of the function f. Namely, for $r\in [0,1]$ and for all $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$ and $t_1,t_2\in [0,1]$, we have:

$$\begin{array}{c}H(\alpha t_1+\beta t_2,r)\hfill \\ =\frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f((\alpha t_1+\beta t_2,r)x+(1-(\alpha t_1+\beta t_2))\frac{a+b}{2},\hfill \\ ry+(1-r)\frac{c+d}{2})dxdy\hfill \\ =\frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(\alpha (t_{1}x+(1-t_1)\frac{a+b}{2})+\beta (t_{2}x+(1-t_2)\frac{a+b}{2}),\hfill \\ ry+(1-r)\frac{c+d}{2})dxdy\hfill \\ \le h_1(\alpha )\frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(t_{1}x+(1-t_1)\frac{a+b}{2},ry+(1-r)\frac{c+d}{2})dxdy\hfill \\ +h_1(\beta )\frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(t_{2}x+(1-t_2)\frac{a+b}{2},ry+(1-r)\frac{c+d}{2})dxdy\hfill \\ =h_1(\alpha )H(t_1,r)+h_1(\beta )H(t_2,r).\hfill \end{array}$$

Similarly, if $t\in [0,1]$ is fixed, then for all $r_1,r_2\in [0,1]$ and $\alpha ,\beta \ge 0$ with $\alpha +\beta =1$, we also have

$$H(t,\alpha r_1+\beta r_2)\le h_2(\alpha )H(t,r_1)+h_2(\beta )H(t,r_2),$$

which means that H is $(h_1,h_2)$-convex on the co-ordinates.

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

  $$\begin{array}{rcl}H(t,r)& =& \frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(tx+(1-t)\frac{a+b}{2},ry+(1-r)\frac{c+d}{2})dxdy\\ =& \frac{1}{(b-a)(d-c)}{\int }_{u_L}^{u_U}{\int }_{v_L}^{v_U}f(u,v)\frac{b-a}{u_U-u_L}\cdot \frac{d-c}{v_U-v_L}dudv\\ =& \frac{1}{(u_U-u_L)(v_U-v_L)}{\int }_{u_L}^{u_U}{\int }_{v_L}^{v_U}f(u,v)dudv\\ \ge & \frac{1}{4h_1(\frac{1}{2})h_2(\frac{1}{2})}f(\frac{a+b}{2},\frac{c+d}{2}),\end{array}$$

where $u_L=ta+(1-t)\frac{a+b}{2}$, $u_U=tb+(1-t)\frac{a+b}{2}$, $v_L=rc+(1-r)\frac{c+d}{2}$ and $v_U=rd+(1-r)\frac{c+d}{2}$, which completes the proof. □

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