Hermite–Hadamard and Fejér Inequalities for Co-Ordinated (F,G)-Convex Functions on a Rectangle

Abstract

We introduce the notion of a co-ordinated $(F,G)$-convex function defined on an interval in ${\mathbb{R}}^2$ and we prove the Hermite–Hadamard and Fejér type inequalities for such functions.

1. Introduction

The celebrated inequality states that, if $f:[a,b]\to \mathbb{R}$ is a convex function, then

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

Furthermore, if $p:[a,b]\to [0,\infty )$ is an integrable function symmetric with respect to $\frac{a+b}{2}$, that is

$$p(a+b-x)=p\left(x\right)\mathrm{for}x\in [a,b],$$

then the following weighted generalization of the Hermite–Hadamard inequality is known as the Fejér inequality

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

Dragomir [1] established a counterpart of the Hermite–Hadamard inequality for co-ordinated convex functions, that is functions $f:[a,b]\times [c,d]\to \mathbb{R}$ which are convex with respect to each variable separately. It has been proven in [1] that for such functions, the following inequalities hold

$$f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{2}\left[\frac{1}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)dx+\frac{1}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)dy\right]$$

$$\le \frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(x,y)dydx$$

$$\le \frac{1}{4}\left[\frac{1}{b-a}{\int }_a^{b}f(x,c)dx+\frac{1}{b-a}{\int }_a^{b}f(x,d)dx+\frac{1}{d-c}{\int }_c^{d}f(a,y)dy+\frac{1}{d-c}{\int }_c^{d}f(b,y)dy\right]$$

$$\le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.$$

Refinement versions of these inequalities have been presented in [1,2,3].

A counterpart of the Fejér inequality for co-ordinated convex functions has been formulated by Alomari and Darus [4]. They proved that if $p:[a,b]\times [c,d]\to [0,\infty )$ is an integrable function symmetric with respect to the lines $x=\frac{a+b}{2}$ and $y=\frac{c+d}{2}$, i.e.,

$$p(a+b-x,y)=p(x,y)\mathrm{for}x\in [a,b],y\in [c,d]$$

(1)

and

$$p(x,c+d-y)=p(x,y)\mathrm{for}x\in [a,b],y\in [c,d],$$

(2)

then for every co-ordinated convex function the following inequalities hold

$$f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{{\int }_a^b{\int }_c^{d}f(x,y)p(x,y)dydx}{{\int }_a^b{\int }_c^{d}p(x,y)dydx}\le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.$$

In recent years, several modifications of the notion of convexity were studied by many authors (see e.g., [5,6,7,8,9]). The following general definition was introduced in [10].

Definition 1.

Let$F:[0,1]\times [a,b]\times [a,b]\to \mathbb{R}$be a continuous function. A function$f:[a,b]\to \mathbb{R}$is said to be convex with respect to F, or briefly F-convex, provided

$$f(tx+(1-t\left)y\right)\le tf\left(x\right)+(1-t)f\left(y\right)+F(t,x,y)\mathrm{for}x,y\in [a,b],t\in [0,1].$$

(3)

In particular, if F is of the form

$$F(t,x,y)=Ct(1-t)|x-y|\mathrm{for}x,y\in [a,b],t\in [0,1],$$

(4)

where $C\in (0,\infty )$, then any function $f:[a,b]\to \mathbb{R}$ satisfying (3) is called approximately convex. Furthermore, if $f:[a,b]\to \mathbb{R}$ satisfies (3) with F given by

$$F(t,x,y)=-Ct(1-t){(x-y)}^2\mathrm{for}x,y\in [a,b],t\in [0,1],$$

(5)

where $C\in (0,\infty )$, then it is called strongly convex with modulus C. For some applications of F-convex functions in the optimization theory and in the theory of partial differential equations we refer to [11] and [12], respectively.

It should be noted here that, although a definition of the F-convex function does not require any additional properties of F, it is reasonable to assume that F is symmetric, that is

$$F(1-t,y,x)=F(t,x,y)\mathrm{for}x,y\in [a,b],t\in [0,1].$$

(6)

In fact, if f is F-convex then there exists a symmetric function $F_s$ such that f is $F_s$-convex and

$$F_s(t,x,y)\le F(t,x,y)\mathrm{for}x,y\in [a,b],t\in [0,1].$$

To find this, one could take

$$F_s(t,x,y):=min\{F(t,x,y),F(1-t,y,x)\}\mathrm{for}x,y\in [a,b],t\in [0,1].$$

Note that F given by (4) or (5) is symmetric. Moreover, a symmetry of F is a necessary condition for the existence of an F-affine function, i.e., a function satisfying equation

$$f(tx+(1-t\left)y\right)=tf\left(x\right)+(1-t)f\left(y\right)+F(t,x,y)\mathrm{for}x,y\in [a,b],t\in [0,1].$$

In what follows we deal with the functions of two variables, which are F-convex with respect to each variable.

Definition 2.

Let$F:[c,d]\times [0,1]\times [a,b]\times [a,b]\to \mathbb{R}$,$G:[a,b]\times [0,1]\times [c,d]\times [c,d]\to \mathbb{R}$be continuous functions. We call a function$f:[a,b]\times [c,d]\to \mathbb{R}$co-ordinated$(F,G)$-convex, provided

$$f(t{x}_1+(1-t)x_2,y)\le tf(x_1,y)+(1-t)f(x_2,y)+F(y,t,x_1,x_2),$$

$$f(x,t{y}_1+(1-t)y_2)\le tf(x,y_1)+(1-t)f(x,y_2)+G(x,t,y_1,y_2)$$

for $t\in [0,1]$, $x_1,x_2\in [a,b]$, $y_1,y_2\in [c,d]$, $x\in [a,b]$, $y\in [c,d]$.

Following the remark formulated above, we restrict our attention to the case where $F(y,·,·,·)$ for $y\in [c,d]$ and $G(x,·,·,·)$ for $x\in [a,b]$ are symmetric functions, i.e.,

$$F(y,1-t,x_2,x_1)=F(y,t,x_1,x_2)\mathrm{for}{x}_1,x_2\in [a,b],y\in [c,d],t\in [0,1]$$

and

$$G(x,1-t,y_2,y_1)=G(x,t,y_1,y_2)\mathrm{for}x\in [a,b],y_1,y_2\in [c,d],t\in [0,1],$$

respectively. This assumption will not be repeated. Our main aim is to present the Hermite–Hadamard and the Fejér type inequalities for co-ordinated $(F,G)$-convex functions.

2. Results

2.1. Hermite–Hadamard Type Inequalities

In this section, we prove the Hermite–Hadamard type inequalities for $(F,G)$-convex functions. Our proof is based on some methods used in [1,3]. We begin with the result establishing the Hermite–Hadamard type inequalities for F-convex functions. It will be useful in further considerations.

Theorem 1.

Let$F:[0,1]\times [a,b]\times [a,b]\to \mathbb{R}$be a continuous symmetric function (cf. (6)). If$f:[a,b]\to \mathbb{R}$is an integrable F-convex function then

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

(7)

and

$$\frac{1}{b-a}{\int }_a^{b}f\left(t\right)dt\le \frac{f\left(a\right)+f\left(b\right)}{2}+{\int }_0^{1}F(t,a,b)dt.$$

(8)

Proof. 

Assume that $f:[a,b]\to \mathbb{R}$ is an integrable F-convex function. In view of (3), we obtain

$$\frac{1}{b-a}{\int }_a^{b}f\left(s\right)ds={\int }_0^{1}f(ta+(1-t)b)dt\le \frac{1}{2}f\left(a\right)+\frac{1}{2}f\left(b\right)+{\int }_0^{1}F(t,a,b)dt,$$

which gives (8). Note also that, as f is F-convex, we have

$$f\left(\frac{x+y}{2}\right)\le \frac{f\left(x\right)+f\left(y\right)}{2}+F\left(\frac{1}{2},x,y\right)\mathrm{for}x,y\in [a,b].$$

(9)

Setting in (9) $x=ta+(1-t)b$, $y=tb+(1-t)a$, where $t\in [0,1]$, and integrating obtained in this way inequality with respect to t, we obtain (7). □

Now, we are going to formulate and prove the Hermite–Hadamard type inequalities for co-ordinated $(F,G)$-convex functions.

Theorem 2.

Assume that$f:[a,b]\times [c,d]\to \mathbb{R}$is an integrable co-ordinated$(F,G)$-convex function. Then:

$$f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{2}\left[\frac{1}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)dx+\frac{1}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)dy\right]+R_1,$$

(10)

where

$$R_1=\frac{1}{2}\left[\frac{1}{b-a}{\int }_a^{b}F\left(\frac{c+d}{2},\frac{1}{2},x,a+b-x\right)dx+\frac{1}{d-c}{\int }_c^{d}G\left(\frac{a+b}{2},\frac{1}{2},y,c+d-y\right)dy\right];$$

$$\frac{1}{2}\left[\frac{1}{b-a}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)dx+\frac{1}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)dy\right]\le \frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(x,y)dydx+R_2,$$

(11)

where

$$R_2=\frac{1}{2(b-a)(d-c)}\left[{\int }_a^b{\int }_c^{d}G\left(x,\frac{1}{2},y,c+d-y\right)dydx+{\int }_a^b{\int }_c^{d}F\left(y,\frac{1}{2},x,a+b-x\right)dydx\right];$$

$$\begin{array}{c}\frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}f(x,y)dydx\\ \le \frac{1}{4}\left[\frac{1}{b-a}{\int }_a^{b}f(x,c)dx+\frac{1}{b-a}{\int }_a^{b}f(x,d)dx+\frac{1}{d-c}{\int }_c^{d}f(a,y)dy+\frac{1}{d-c}{\int }_c^{d}f(b,y)dy\right]+R_3,\end{array}$$

(12)

where

$$R_3=\frac{1}{2}\left[\frac{1}{b-a}{\int }_a^b{\int }_0^{1}G(x,t,c,d)dtdx+\frac{1}{d-c}{\int }_c^d{\int }_0^{1}F(y,t,a,b)dtdy\right];$$

and

$$\begin{array}{c}\frac{1}{4}\left[\frac{1}{b-a}{\int }_a^{b}f(x,c)dx+\frac{1}{b-a}{\int }_a^{b}f(x,d)dx+\frac{1}{d-c}{\int }_c^{d}f(a,y)dy+\frac{1}{d-c}{\int }_c^{d}f(b,y)dy\right]\\ \le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}+R_4,\end{array}$$

(13)

where

$$R_4=\frac{1}{4}\left[{\int }_0^{1}F(c,t,a,b)dt+{\int }_0^{1}F(d,t,a,b)dt+{\int }_0^{1}G(a,t,c,d)dt+{\int }_0^{1}G(b,t,c,d)dt\right].$$

Proof. 

Note that, for every $x\in [a,b]$, the function $f(x,·)$ is $G(x,·,·,·)$-convex. Thus, applying Theorem 1, we obtain

$$f\left(x,\frac{c+d}{2}\right)\le \frac{1}{d-c}{\int }_c^{d}f(x,y)dy+\frac{1}{d-c}{\int }_c^{d}G\left(x,\frac{1}{2},y,c+d-y\right)dy$$

$$\le \frac{f(x,c)+f(x,d)}{2}+{\int }_0^{1}G(x,t,c,d)dt+\frac{1}{d-c}{\int }_c^{d}G\left(x,\frac{1}{2},y,c+d-y\right)dy.$$

Integrating this inequality with respect to x, we find

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

$$\le \frac{1}{(b-a)(d-c)}\left[{\int }_a^b{\int }_c^{d}f(x,y)dydx+{\int }_a^b{\int }_c^{d}G\left(x,\frac{1}{2},y,c+d-y\right)dydx\right]$$

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

$$+\frac{1}{b-a}{\int }_a^b{\int }_0^{1}G(x,t,c,d)dtdx+\frac{1}{(b-a)(d-c)}{\int }_a^b{\int }_c^{d}G\left(x,\frac{1}{2},y,c+d-y\right)dydx.$$

Moreover, since for every $y\in [c,d]$, $f(·,y)$ is $F(y,·,·,·)$-convex, using the similar arguments, we conclude that

$$\frac{1}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)dy$$

$$\le \frac{1}{(b-a)(d-c)}\left[{\int }_c^d{\int }_a^{b}f(x,y)dxdy+{\int }_c^d{\int }_a^{b}F\left(y,\frac{1}{2},x,a+b-x\right)dxdy\right]$$

$$\le \frac{1}{2(d-c)}\left[{\int }_c^{d}f(a,y)dy+{\int }_c^{d}f(b,y)dy\right]$$

$$+\frac{1}{d-c}{\int }_c^d{\int }_0^{1}F(y,t,a,b)dtdy+\frac{1}{(b-a)(d-c)}{\int }_c^d{\int }_a^{b}F\left(y,\frac{1}{2},x,a+b-x\right)dxdy.$$

Adding up these inequalities, we obtain (11) and (12).

Since $f(·,\frac{c+d}{2})$ is $F(\frac{c+d}{2},·,·,·)$-convex and $f(\frac{a+b}{2},·)$ is $G(\frac{a+b}{2},·,·,·)$-convex, taking into account the first inequality in Theorem 1, we have

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

and

$$f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{d-c}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)dy+\frac{1}{d-c}{\int }_c^{d}G\left(\frac{a+b}{2},\frac{1}{2},y,c+d-y\right)dy.$$

Adding them up we obtain (10).

Finally, as $f(·,c)$, $f(·,d)$, $f(a,·)$ and $f(b,·)$ are $F(c,·,·,·)$-, $F(d,·,·,·)$-, $G(a,·,·,·)$- and $G(b,·,·,·)$-convex, respectively, applying the second inequality in Theorem 1, we find

$$\frac{1}{b-a}{\int }_a^{b}f(x,c)dx\le \frac{f(a,c)+f(b,c)}{2}+{\int }_0^{1}F(c,t,a,b)dt,$$

$$\frac{1}{b-a}{\int }_a^{b}f(x,d)dx\le \frac{f(a,d)+f(b,d)}{2}+{\int }_0^{1}F(d,t,a,b)dt,$$

$$\frac{1}{d-c}{\int }_c^{d}f(a,y)dy\le \frac{f(a,c)+f(a,d)}{2}+{\int }_0^{1}G(a,t,c,d)dt$$

and

$$\frac{1}{d-c}{\int }_c^{d}f(b,y)dy\le \frac{f(b,c)+f(b,d)}{2}+{\int }_0^{1}G(b,t,c,d)dt.$$

Adding up these inequalities, we obtain (13). □

2.2. Fejér Type Inequalities

In order to prove the Fejér type inequalities for co-ordinated $(F,G)$-convex functions we need the following auxiliary result.

Lemma 1.

Assume that$f:[a,b]\times [c,d]\to \mathbb{R}$is a co-ordinated$(F,G)$-convex function.

(i) 

If$[x_1,x_2]\subset [x_1^{\prime },x_2^{\prime }]\subset [a,b]$and$x_1+x_2=x_1^{\prime }+x_2^{\prime }$then

$$f(x_1,y)+f(x_2,y)\le f(x_1^{\prime },y)+f(x_2^{\prime },y)+F\left(y,\frac{x_2^{\prime }-x_1}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right)+F\left(y,\frac{x_2^{\prime }-x_2}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right)$$

for$y\in [c,d]$.

(ii) 

If$[y_1,y_2]\subset [y_1^{\prime },y_2^{\prime }]\subset [c,d]$and$y_1+y_2=y_1^{\prime }+y_2^{\prime }$then

$$f(x,y_1)+f(x,y_2)\le f(x,y_1^{\prime })+f(x,y_2^{\prime })+G\left(x,\frac{y_2^{\prime }-y_1}{y_2^{\prime }-y_1^{\prime }},y_1^{\prime },y_2^{\prime }\right)+G\left(x,\frac{y_2^{\prime }-y_2}{y_2^{\prime }-y_1^{\prime }},y_1^{\prime },y_2^{\prime }\right)$$

for$x\in [a,b]$.

Proof. 

We prove only the first part of the lemma since the proof of the second part is similar. Assume that $[x_1,x_2]\subset [x_1^{\prime },x_2^{\prime }]\subset [a,b]$ and $x_1+x_2=x_1^{\prime }+x_2^{\prime }$. Since

$$x_1=\frac{x_2^{\prime }-x_1}{x_2^{\prime }-x_1^{\prime }}{x}_1^{\prime }+\frac{x_1-x_1^{\prime }}{x_2^{\prime }-x_1^{\prime }}{x}_2^{\prime }$$

and

$$x_2=\frac{x_2^{\prime }-x_2}{x_2^{\prime }-x_1^{\prime }}{x}_1^{\prime }+\frac{x_2-x_1^{\prime }}{x_2^{\prime }-x_1^{\prime }}{x}_2^{\prime },$$

for every $y\in [c,d]$, we obtain

$$f(x_1,y)+f(x_2,y)\le \frac{x_2^{\prime }-x_1}{x_2^{\prime }-x_1^{\prime }}f(x_1^{\prime },y)+\frac{x_1-x_1^{\prime }}{x_2^{\prime }-x_1^{\prime }}f(x_2^{\prime },y)+F\left(y,\frac{x_2^{\prime }-x_1}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right)$$

$$+\frac{x_2^{\prime }-x_2}{x_2^{\prime }-x_1^{\prime }}f(x_1^{\prime },y)+\frac{x_2-x_1^{\prime }}{x_2^{\prime }-x_1^{\prime }}f(x_2^{\prime },y)+F\left(y,\frac{x_2^{\prime }-x_2}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right)$$

$$=\frac{2x_2^{\prime }-(x_1+x_2)}{x_2^{\prime }-x_1^{\prime }}f(x_1^{\prime },y)+\frac{x_1+x_2-2x_1^{\prime }}{x_2^{\prime }-x_1^{\prime }}f(x_2^{\prime },y)+F\left(y,\frac{x_2^{\prime }-x_1}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right)+F\left(y,\frac{x_2^{\prime }-x_2}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right)$$

$$=f(x_1^{\prime },y)+f(x_2^{\prime },y)+F\left(y,\frac{x_2^{\prime }-x_1}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right)+F\left(y,\frac{x_2^{\prime }-x_2}{x_2^{\prime }-x_1^{\prime }},x_1^{\prime },x_2^{\prime }\right).$$

□

In the next theorem we establish the Fejér type inequalities for $(F,G)$-convex functions.

Theorem 3.

Assume that$p:[a,b]\times [c,d]\to \mathbb{R}$is a positive integrable function symmetric with respect to the lines$x=\frac{a+b}{2}$and$y=\frac{c+d}{2}$(cf. (1) and (2)). If$f:[a,b]\times [c,d]\to \mathbb{R}$is a continuous co-ordinated$(F,G)$-convex function such that$fp$is integrable on$[a,b]\times [c,d]$then

$$f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{{\int }_a^b{\int }_c^{d}f(x,y)p(x,y)dydx+K}{{\int }_a^b{\int }_c^{d}p(x,y)dydx},$$

(14)

where

$$K=2{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}G\left(x,\frac{1}{2},y,c+d-y\right)p(x,y)dydx$$

$$+2{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}G\left(a+b-x,\frac{1}{2},y,c+d-y\right)p(x,y)dydx$$

$$+4{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}F\left(\frac{c+d}{2},\frac{1}{2},x,a+b-x\right)p(x,y)dydx$$

and

$$\frac{{\int }_a^b{\int }_c^{d}f(x,y)p(x,y)dydx-L}{{\int }_a^b{\int }_c^{d}p(x,y)dydx}\le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4},$$

(15)

where

$$L=L_1+L_2+L_3$$

$$\genfrac{}{}{0pt}{}{:={\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}\left[F\left(y,\frac{b-x}{b-a},a,b\right)+F\left(y,\frac{x-a}{b-a},a,b\right)+F\left(c+d-y,\frac{b-x}{b-a},a,b\right)+F\left(c+d-y,\frac{x-a}{b-a},a,b\right)\right]}{p(x,y)dydx}$$

$$+{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}\left[G\left(a,\frac{d-y}{d-c},c,d\right)+G\left(a,\frac{y-c}{d-c},c,d\right)\right]p(x,y)dydx$$

$$+{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}\left[G\left(b,\frac{d-y}{d-c},c,d\right)+G\left(b,\frac{y-c}{d-c},c,d\right)\right]p(x,y)dydx.$$

Proof. 

Assume that $f:[a,b]\times [c,d]\to \mathbb{R}$ is an integrable co-ordinated $(F,G)$-convex function such that $fp$ is integrable. Then, for every $x\in [a,b]$ and $y\in [c,d]$, we have

$$f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{2}f\left(x,\frac{c+d}{2}\right)+\frac{1}{2}f\left(a+b-x,\frac{c+d}{2}\right)+F\left(\frac{c+d}{2},\frac{1}{2},x,a+b-x\right)$$

$$\le \frac{1}{4}f(x,y)+\frac{1}{4}f(x,c+d-y)+\frac{1}{4}f(a+b-x,y)+\frac{1}{4}f(a+b-x,c+d-y)$$

$$+\frac{1}{2}G\left(x,\frac{1}{2},y,c+d-y\right)+\frac{1}{2}G\left(a+b-x,\frac{1}{2},y,c+d-y\right)+F\left(\frac{c+d}{2},\frac{1}{2},x,a+b-x\right).$$

Therefore, as p is symmetric with respect to the lines $x=\frac{a+b}{2}$ and $y=\frac{c+d}{2}$, we obtain

$$f\left(\frac{a+b}{2},\frac{c+d}{2}\right){\int }_a^b{\int }_c^{d}p(x,y)dydx=4{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)p(x,y)dydx$$

$$\le {\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}[f(x,y)+f(a+b-x,c+d-y)]p(x,y)dydx$$

$$+{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}[f(x,c+d-y)+f(a+b-x,y)]p(x,y)dydx+K$$

$$={\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}[f(x,y)+f(a+b-x,c+d-y)]p(x,y)dydx$$

$$+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}[f(a+b-x,c+d-y)+f(x,y)]p(a+b-x,y)dydx+K$$

$$={\int }_a^b{\int }_c^{\frac{c+d}{2}}[f(x,y)+f(a+b-x,c+d-y)]p(x,y)dydx+K$$

$$={\int }_a^b{\int }_c^{\frac{c+d}{2}}f(x,y)p(x,y)dydx+{\int }_a^b{\int }_c^{\frac{c+d}{2}}f(a+b-x,c+d-y)p(x,y)dydx+K$$

$$={\int }_a^b{\int }_c^{\frac{c+d}{2}}f(x,y)p(x,y)dydx+{\int }_a^b{\int }_{\frac{c+d}{2}}^{d}f(x,y)p(a+b-x,c+d-y)dydx+K$$

$$={\int }_a^b{\int }_c^{d}f(x,y)p(x,y)dydx+K.$$

Thus, (14) holds.

Furthermore, using again the symmetry of p and applying Lemma 1 to $[y,c+d-y]\subset [c,d]$ and $[x,a+b-x]\subset [a,b]$, where $x\in [a,\frac{a+b}{2}]$, $y\in [c,\frac{c+d}{2}]$, we have

$$\frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}{\int }_a^b{\int }_c^{d}p(x,y)dydx$$

$$={\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}[f(a,c)+f(a,d)+f(b,c)+f(b,d)]p(x,y)dydx$$

$$\ge {\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}\left[f(a,y)+f(a,c+d-y)-G\left(a,\frac{d-y}{d-c},c,d\right)-G\left(a,\frac{y-c}{d-c},c,d\right)\right.$$

$$\left.+f(b,y)+f(b,c+d-y)-G\left(b,\frac{d-y}{d-c},c,d\right)-G\left(b,\frac{y-c}{d-c},c,d\right)\right]p(x,y)dydx$$

$$\ge {\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}\left[f(x,y)+f(a+b-x,y)-F\left(y,\frac{b-x}{b-a},a,b\right)-F\left(y,\frac{x-a}{b-a},a,b\right)\right.$$

$$+f(x,c+d-y)+f(a+b-x,c+d-y)$$

$$\left.-F\left(c+d-y,\frac{b-x}{b-a},a,b\right)-F\left(c+d-y,\frac{x-a}{b-a},a,b\right)\right]p(x,y)dydx-(L_2+L_3)$$

$$={\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}[f(x,y)+f(a+b-x,c+d-y)]p(x,y)dydx$$

$$+{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}[f(a+b-x,y)+f(x,c+d-y)]p(x,y)dydx-(L_1+L_2+L_3)$$

$$={\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}[f(x,y)+f(a+b-x,c+d-y)]p(x,y)dydx$$

$$+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}[f(a+b-x,c+d-y)+f(x,y)]p(a+b-x,y)dydx-L$$

$$={\int }_a^b{\int }_c^{\frac{c+d}{2}}[f(x,y)+f(a+b-x,c+d-y)]p(x,y)dydx-L$$

$$={\int }_a^b{\int }_c^{\frac{c+d}{2}}f(x,y)p(x,y)dydx+{\int }_a^b{\int }_c^{\frac{c+d}{2}}f(a+b-x,c+d-y)p(x,y)dydx-L$$

$$={\int }_a^b{\int }_c^{\frac{c+d}{2}}f(x,y)p(x,y)dydx+{\int }_a^b{\int }_{\frac{c+d}{2}}^{c}f(x,y)p(a+b-x,c+d-y)dydx-L$$

$$={\int }_a^b{\int }_c^{d}f(x,y)p(x,y)dydx-L,$$

which gives (15). □

3. Discussion

In this paper the Hermite–Hadamard and Fejér type inequalities for co-ordinated $(F,G)$-convex functions are proved. Since every co-ordinated convex function is co-ordinated $(F,G)$-convex (with F and G being identically 0), from our results, one can easily deduce the results by Dragomir [1] and Alomari and Darus [4]. Furthermore, applying Theorems 2 and 3, one can obtain the Hermite–Hadamard and Fejér type inequalities for co-ordinated $(C,D)$-approximately convex functions and co-ordinated $(C,D)$-strongly convex functions defined by

$$f(t{x}_1+(1-t)x_2,y)\le tf(x_1,y)+(1-t)f(x_2,y)+D\left(y\right)t(1-t)|x_1-x_2|,$$

$$f(x,t{y}_1+(1-t)y_2)\le tf(x,y_1)+(1-t)f(x,y_2)+C\left(x\right)t(1-t)|y_1-y_2|$$

for $t\in [0,1]$, $x_1,x_2\in [a,b]$, $y_1,y_2\in [c,d]$, $x\in [a,b]$, $y\in [c,d]$; and

$$f(t{x}_1+(1-t)x_2,y)\le tf(x_1,y)+(1-t)f(x_2,y)-D\left(y\right)t(1-t){(x_1-x_2)}^2,$$

$$f(x,t{y}_1+(1-t)y_2)\le tf(x,y_1)+(1-t)f(x,y_2)-C\left(x\right)t(1-t){(y_1-y_2)}^2$$

for $t\in [0,1]$, $x_1,x_2\in [a,b]$, $y_1,y_2\in [c,d]$, $x\in [a,b]$, $y\in [c,d]$, respectively, where $C:[a,b]\to (0,\infty )$ and $D:[c,d]\to (0,\infty )$ are given functions.

Note also that from Theorem 1 the Hermite–Hadamard inequalities for approximately convex functions and strongly convex functions can be derived. Finally, applying Theorem 1, with $F\equiv 0$, we obtain the classical Hermite–Hadamard inequality.