Some New Quantum Hermite–Hadamard Inequalities for Co-Ordinated Convex Functions

Abstract

In this paper, we establish some new versions of Hermite–Hadamard type inequalities for co-ordinated convex functions via $q_1,q_2$-integrals. Since the inequalities are newly proved, we therefore consider some examples of co-ordinated convex functions and show their validity for particular choices of $q_1,q_2\in (0,1)$. We hope that the readers show their interest in these results.

1. Introduction

Quantum calculus (sometimes is called q-calculus) is known as the study of calculus with no limits. Note that q-calculus can be reduced to ordinary calculus if we take ${lim}_{q\to 1}$. It was firstly studied by the famous mathematician Euler (1707–1783). In 1910, F. H. Jackson [1] determined the definite q-integral known as the q-Jackson integral. Quantum calculus has many applications in several mathematical areas such as combinatorics, number theory, orthogonal polynomials, basic hypergeometric functions, mechanics, quantum theory, and theory of relativity, see, for instance, [2,3,4,5,6,7] and the references therein. The book by V. Kac and P. Cheung [8] covers the fundamental knowledge and also the basic theoretical concepts of quantum calculus.

In 2013, J. Tariboon and S. K. Ntouyas [9] defined the q-derivative and q-integral of a continuous function on finite intervals and proved some of its properties. Many well-known integral inequalities such as Hölder, Hermite–Hadamard, trapezoid, Ostrowski, Cauchy–Bunyakovsky–Schwarz, Grüss, and Grüss–Čebyšev inequalities have been studied in the concept of q-calculus, see [10] for more details. Based on these results, there are many outcomes concerning q-calculus. For example, in [11], some new Hermite–Hadamard type inequalities were established for co-ordinated convex functions and Simpson’s type inequalities for co-ordinated convex functions were established in [12]. In [13], Kalsoom et al. used co-ordinated n-polynomial preinvexity and proved some Ostrowski type inequalities for quantum integrals. In [14,15], the authors used quantum integrals for the functions of two variables and proved some new Hermite–Hadamard type inequalities for co-ordinated convex functions.

In 2020, S. Bermudo et al. [16] newly defined the q-derivative and q-integral of a continuous function on finite intervals, called $q^b$-calculus, while the definition of J. Tariboon and S. K. Ntouyas is called $q_a$-calculus. Moreover, in their paper, they proved Hermite–Hadamard inequalities for convex functions and h-convex functions by using such the new definition.

The Hermite–Hadamard inequality is a classical inequality stated as: If $f:[a,b]\to \mathbb{R}$ is a convex function, then

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

(1)

Inequality (1) was introduced by C. Hermite [17] in 1883 and was investigated by J. Hadamard [18] in 1893.

In [19], Alp et al. proved the following quantum Hermite–Hadamard inequality for convex functions using the following quantum integrals:

Theorem 1.

If $f:[a,b]\to \mathbb{R}$ is a convex function, then we have

$$\begin{array}{c}\hfill f\left(\frac{qa+b}{1+q}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}_a{d}_{q}x\le \frac{qf\left(a\right)+f\left(b\right)}{1+q}.\end{array}$$

(2)

Bermudo et al. also proved the corresponding Hermite–Hadamard inequality for $q^b$-integrals, as follows:

Theorem 2.

If $f:[a,b]\to \mathbb{R}$ is a convex function, then we have

$$\begin{array}{c}\hfill f\left(\frac{a+qb}{1+q}\right)\le \frac{1}{b-a}{\int }_a^{b}f\left(x\right){}^b{d}_{q}x\le \frac{f\left(a\right)+qf\left(b\right)}{1+q}.\end{array}$$

(3)

Recently, Sitthiwirattham et al. proved some new quantum Hermite–Hadamard inequalities for convex functions by using their new techniques.

Theorem 3.

If $f:[a,b]\to \mathbb{R}$ is a convex function, then we have

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}f\left(x\right){}^{\frac{a+b}{2}}{d}_{q}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x\right){}_{\frac{a+b}{2}}{d}_{q}x\right]\le \frac{f\left(a\right)+f\left(b\right)}{2}.\end{array}$$

(4)

Moreover, Ali et al. proved the following new version of quantum Hermite–Hadamard inequality involving a $q_a$-integral and $q^b$-integral. They also proved some inequalities for estimations of the left and right hand sides of this inequality.

Theorem 4.

If $f:[a,b]\to \mathbb{R}$ is a convex function, then we have

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}f\left(x\right){}_a{d}_{q}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x\right){}^b{d}_{q}x\right]\le \frac{f\left(a\right)+f\left(b\right)}{2}.\end{array}$$

(5)

When $f:△\to \mathbb{R}$ is a co-ordinated convex function, S. Dragomir [20] presented the Hermite–Hadamard type inequalities in 2001 as follows:

Theorem 5.

If $f:△\to \mathbb{R}$ is a co-ordinated convex function, then we have

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

(6)

Inspired by the ongoing studies, we prove some new versions of quantum Hermite–Hadamard inequalities for co-ordinated convex functions. We also show the validity of newly established inequalities with some examples for particular choices of $q_1,q_2\in (0,1)$.

The structure of this paper is as follows: The fundamentals of q-calculus for one and two variable functions, as well as other relevant topics in this field, are briefly discussed in Section 2. In Section 3, we establish new variants of the q-Hermite–Hadamard inequality for co-ordinated convex functions. We present some examples in Section 4 to illustrate the newly established inequalities. Section 5 concludes with some research suggestions for the future.

2. Preliminaries

Throughout this paper, we let $△:=[a,b]\times [c,d]\subseteq \mathbb{R}\times \mathbb{R}$, $0<q<1$ and $0<q_i<1$ for $i=1,2$. The definitions of q-calculus, co-ordinated functions, and q-calculus for co-ordinates are given in [9,14,15,16,20].

Definition 1

([9]). Let $f:[a,b]\to \mathbb{R}$ be a continuous function. Then, the $q_a$-derivative of f at $x\in (a,b]$ is defined by

$${}_a{D}_{q}f\left(x\right)=\frac{f\left(x\right)-f\left(qx+(1-q)a\right)}{(1-q)(x-a)}.$$

The $q_a$-integral is defined by

$${\int }_a^{x}f\left(t\right){}_a{d}_{q}t=(1-q)(x-a)\sum _{n=0}^{\infty }{q}^{n}f\left(q^{n}x+\left(1-q^n\right)a\right).$$

Example 1.

Let $f\left(x\right)=x^2$ for $x\in [0,1]$. Then, we have

$$\begin{array}{cc}\hfill {}_0{D}_{q}f\left(x\right)& ={\displaystyle \frac{f\left(x\right)-f(qx+(1-q\left)0\right))}{(1-q)(x-0)}}\hfill \\ \hfill & ={\displaystyle \frac{f\left(x\right)-f\left(qx\right)}{(1-q)x}}\hfill \\ \hfill & ={\displaystyle \frac{x^2-q^2{x}^2}{(1-q)x}}\hfill \\ \hfill & =(1+q)x,forx\in (0,1].\hfill \end{array}$$

Example 2.

Let $f\left(x\right)=x$ for $x\in [0,1]$. Then, we have

$$\begin{array}{cc}\hfill {\int }_0^{1/2}f\left(x\right){}_0{d}_{q}x& =(1-q)\left(\frac{1}{2}-0\right)\sum _{n=0}^{\infty }{q}^{n}f\left(q^n\left(\frac{1}{2}\right)+\left(1-q^n\right)0\right)\hfill \\ \hfill & ={\displaystyle \frac{1-q}{2}}\sum _{n=0}^{\infty }{q}^n\left({\displaystyle \frac{q^n}{2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1-q}{4}}\sum _{n=0}^{\infty }{q}^{2n}\hfill \\ \hfill & ={\displaystyle \frac{1-q}{4}}\left({\displaystyle \frac{1}{1-q^2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{4(1+q)}}.\hfill \end{array}$$

Definition 2

([16]). Let $f:[a,b]\to \mathbb{R}$ be a continuous function. Then, the $q^b$-derivative of f at $x\in [a,b)$ is defined by

$${}^b{D}_{q}f\left(x\right)=\frac{f\left(x\right)-f\left(qx+(1-q)b\right)}{(1-q)(x-b)}.$$

The$q^b$-integral is defined by

$${\int }_x^{b}f\left(t\right){}^b{d}_{q}t=(1-q)(b-x)\sum _{n=0}^{\infty }{q}^{n}f\left(q^{n}x+\left(1-q^n\right)b\right).$$

Example 3.

Let $f\left(x\right)=x^2$ for $x\in [0,1]$. Then, we have

$$\begin{array}{cc}\hfill {}^1{D}_{q}f\left(x\right)& ={\displaystyle \frac{f\left(x\right)-f(qx+(1-q\left)1\right))}{(1-q)(x-1)}}\hfill \\ \hfill & ={\displaystyle \frac{x^2-{(qx-q+1)}^2}{(1-q)(x-1)}}\hfill \\ \hfill & =(1+q)x+1-q,forx\in [0,1).\hfill \end{array}$$

Example 4.

Let $f\left(x\right)=x$ for $x\in [0,1]$. Then, we have

$$\begin{array}{cc}\hfill {\int }_{1/2}^{1}f\left(x\right){}^1{d}_{q}x& =(1-q)\left(1-\frac{1}{2}\right)\sum _{n=0}^{\infty }{q}^{n}f\left(q^n\left(\frac{1}{2}\right)+\left(1-q^n\right)1\right)\hfill \\ \hfill & ={\displaystyle \frac{1-q}{2}}\sum _{n=0}^{\infty }{q}^n\left(1-{\displaystyle \frac{q^n}{2}}\right)\hfill \\ \hfill & ={\displaystyle \frac{1-q}{2}}\left[\sum _{n=0}^{\infty }{q}^n-\frac{1}{2}\sum _{n=0}^{\infty }{q}^{2n}\right]\hfill \\ \hfill & ={\displaystyle \frac{1-q}{2}}\left[{\displaystyle \frac{1}{1-q}}-\frac{1}{2}\left({\displaystyle \frac{1}{1-q^2}}\right)\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4(1+q)}}.\hfill \end{array}$$

Definition 3

([20]). A function $f:△\to \mathbb{R}$ is said to be co-ordinated convex (or convex on co-ordinates) if the partial mappings

$$f_x:[c,d]\ni v\mapsto f(x,v)\in \mathbb{R}\mathrm{and}{f}_y:[a,b]\ni u\mapsto f(u,y)\in \mathbb{R}$$

are convex for all $x\in (a,b)$ and $y\in (c,d)$.

A formal definition for co-ordinated convex functions may be stated as follows:

Definition 4

([21]). A function $f:△\to \mathbb{R}$ is said to be convex on co-ordinates if

$$\begin{array}{cc}\hfill f(tx+(1-t)z,\lambda y+(1-\lambda \left)w\right)& \le t\lambda f(x,y)+t(1-\lambda )f(x,w)+(1-t)\lambda f(z,y)\hfill \\ \hfill & +(1-t)(1-\lambda )f(z,w)\hfill \end{array}$$

(7)

holds for all $t,\lambda \in [0,1]\mathrm{and}(x,y),(z,w)\in △$.

Definition 5

([14]). Suppose that $f:△\to \mathbb{R}$ is a continuous function of two variables. Then, the definite integral is given by

$$\begin{array}{cc}\hfill {\int }_a^x{\int }_c^{y}f(t,s){}_c{d}_{q_2}s{}_a{d}_{q_1}t& =(1-q_1)(1-q_2)(x-a)(y-c)\hfill \\ \hfill & \times \sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f(q_1^{n}x+(1-q_1^n)a,q_2^{m}y+(1-q_2^m)c).\hfill \end{array}$$

Definition 6

([15]). Suppose that $f:△\to \mathbb{R}$ is a continuous function of two variables. Then, the definite integrals are given by

$$\begin{array}{cc}\hfill {\int }_a^x{\int }_y^{d}f(t,s){}^d{d}_{q_2}s{}_a{d}_{q_1}t& =(1-q_1)(1-q_2)(x-a)(d-y)\hfill \\ \hfill & \times \sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f(q_1^{n}x+(1-q_1^n)a,q_2^{m}y+(1-q_2^m)d),\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\int }_x^b{\int }_c^{y}f(t,s){}_c{d}_{q_2}s{}^b{d}_{q_1}t& =(1-q_1)(1-q_2)(b-x)(y-c)\hfill \\ \hfill & \times \sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f(q_1^{n}x+(1-q_1^n)b,q_2^{m}y+(1-q_2^m)c)\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {\int }_x^b{\int }_y^{d}f(t,s){}^d{d}_{q_2}s{}^b{d}_{q_1}t& =(1-q_1)(1-q_2)(b-x)(d-y)\hfill \\ \hfill & \times \sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f(q_1^{n}x+(1-q_1^n)b,q_2^{m}y+(1-q_2^m)d).\hfill \end{array}$$

3. Main Results

In this section, we prove some new Hermite–Hadamard inequalities for co-ordinated convex functions.

Theorem 6.

Let $f:△\to \mathbb{R}$ be a co-ordinated convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x\right.\hfill \\ \hfill & +{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.\hfill \end{array}$$

(8)

Proof. 

Since f is co-ordinated convex, we have

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& =f\left(\frac{1}{2}\left(\frac{1+t}{2}a+\frac{1-t}{2}b\right)+\frac{1}{2}\left(\frac{1-t}{2}a+\frac{1+t}{2}b\right),\frac{1}{2}\left(\frac{1+s}{2}c+\frac{1-s}{2}d\right)+\frac{1}{2}\left(\frac{1-s}{2}c+\frac{1+s}{2}d\right)\right)\hfill \\ \hfill & \le \frac{1}{4}\left[f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right)+f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right)\right.\hfill \\ \hfill & +\left.f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right)+f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right)\right],\hfill \end{array}$$

(9)

where $t,s\in [0,1]$.

$q_{a,c}$-Integrating both sides of (9) over $[0,1]\times [0,1]$, we obtain

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{4}\left[{\int }_0^1{\int }_0^{1}f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\right.\hfill \\ \hfill & +{\int }_0^1{\int }_0^{1}f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & +{\int }_0^1{\int }_0^{1}f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & +\left.{\int }_0^1{\int }_0^{1}f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\right]\hfill \\ \hfill & =:\frac{1}{4}\left(I_1+I_2+I_3+I_4\right).\hfill \end{array}$$

By Definitions 5 and 6, we have

$$\begin{array}{cc}\hfill I_1& ={\int }_0^1{\int }_0^{1}f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =(1-q_1)(1-q_2)\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f\left(\frac{1+q_1^n}{2}a+\frac{1-q_1^n}{2}b,\frac{1+q_2^m}{2}c+\frac{1-q_2^m}{2}d\right)\hfill \\ \hfill & =(1-q_1)(1-q_2)\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f\left(q_1^{n}a+(1-q_1^n)\frac{a+b}{2},q_2^{m}c+(1-q_2^m)\frac{c+d}{2}\right)\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x.\hfill \end{array}$$

Similarly, we obtain

$$\begin{array}{cc}\hfill I_2& ={\int }_0^1{\int }_0^{1}f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x,\hfill \\ \hfill I_3& ={\int }_0^1{\int }_0^{1}f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_4& ={\int }_0^1{\int }_0^{1}f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x.\hfill \end{array}$$

Replacing $I_1,I_2,I_3$, and $I_4$ in (10), we obtain

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x\right.\hfill \\ \hfill & +{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\right].\hfill \end{array}$$

Thus, the first inequality of (8) holds.

Next, by co-ordinated convexity of f again, we have

$$\begin{array}{cc}\hfill f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right)& \le \frac{(1+t)(1+s)}{4}f(a,c)+\frac{(1+t)(1-s)}{4}f(a,d)\hfill \\ \hfill & +\frac{(1-t)(1+s)}{4}f(b,c)+\frac{(1-t)(1-s)}{4}f(b,d),\hfill \end{array}$$

(10)

$$\begin{array}{cc}\hfill f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right)& \le \frac{(1+t)(1-s)}{4}f(a,c)+\frac{(1+t)(1+s)}{4}f(a,d)\hfill \\ \hfill & +\frac{(1-t)(1-s)}{4}f(b,c)+\frac{(1-t)(1+s)}{4}f(b,d),\hfill \end{array}$$

(11)

$$\begin{array}{cc}\hfill f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right)& \le \frac{(1-t)(1+s)}{4}f(a,c)+\frac{(1-t)(1-s)}{4}f(a,d)\hfill \\ \hfill & +\frac{(1+t)(1+s)}{4}f(b,c)+\frac{(1+t)(1-s)}{4}f(b,d)\hfill \end{array}$$

(12)

and

$$\begin{array}{cc}\hfill f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right)& \le \frac{(1-t)(1-s)}{4}f(a,c)+\frac{(1-t)(1+s)}{4}f(a,d)\hfill \\ \hfill & +\frac{(1+t)(1-s)}{4}f(b,c)+\frac{(1+t)(1+s)}{4}f(b,d).\hfill \end{array}$$

(13)

Summing (10)–(13), we obtain

$$\begin{array}{cc}\hfill & f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right)+f\left(\frac{1+t}{2}a+\frac{1-t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right)\hfill \\ \hfill & +f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1+s}{2}c+\frac{1-s}{2}d\right)+f\left(\frac{1-t}{2}a+\frac{1+t}{2}b,\frac{1-s}{2}c+\frac{1+s}{2}d\right)\hfill \\ \hfill & \le f(a,c)+f(a,d)+f(b,c)+f(b,d).\hfill \end{array}$$

(14)

$q_{a,c}$-Integrating both sides of (14) over $[0,1]$, the second inequality of (8) holds. The proof is completed. □

Theorem 7.

Let $f:△\to \mathbb{R}$ be a co-ordinated convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}_a{d}_{q_1}x\right.\hfill \\ \hfill & +{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}^b{d}_{q_1}x\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}^b{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.\hfill \end{array}$$

(15)

Proof. 

Since f is co-ordinated convex, we have

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& =f\left(\frac{1}{2}\left(\frac{t}{2}a+\frac{2-t}{2}b\right)+\frac{1}{2}\left(\frac{2-t}{2}a+\frac{t}{2}b\right),\frac{1}{2}\left(\frac{s}{2}c+\frac{2-s}{2}d\right)+\frac{1}{2}\left(\frac{2-s}{2}c+\frac{s}{2}d\right)\right)\hfill \\ \hfill & \le \frac{1}{4}\left[f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right)+f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right)\right.\hfill \\ \hfill & +\left.f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right)+f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right)\right],\hfill \end{array}$$

(16)

where $t,s\in [0,1]$.

$q_{a,c}$-Integrating both sides of (16) over $[0,1]\times [0,1]$, we obtain

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{4}\left[{\int }_0^1{\int }_0^{1}f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\right.\hfill \\ \hfill & +{\int }_0^1{\int }_0^{1}f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & +{\int }_0^1{\int }_0^{1}f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & +\left.{\int }_0^1{\int }_0^{1}f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\right]\hfill \\ \hfill & =:\frac{1}{4}\left(I_5+I_6+I_7+I_8\right).\hfill \end{array}$$

(17)

By Definitions 5 and 6, we have

$$\begin{array}{cc}\hfill I_5& ={\int }_0^1{\int }_0^{1}f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =(1-q_1)(1-q_2)\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f\left(\frac{q_1^n}{2}a+\frac{2-q_1^n}{2}b,\frac{q_2^m}{2}c+\frac{2-q_2^m}{2}d\right)\hfill \\ \hfill & =(1-q_1)(1-q_2)\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{q}_1^n{q}_2^{m}f\left(q_1^n\left(\frac{a+b}{2}\right)+(1-q_1^n)b,q_2^m\left(\frac{c+d}{2}\right)+(1-q_2^m)d\right)\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}^b{d}_{q_1}x.\hfill \end{array}$$

Similarly, we obtain

$$\begin{array}{cc}\hfill I_6& ={\int }_0^1{\int }_0^{1}f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}^b{d}_{q_1}x,\hfill \\ \hfill I_7& ={\int }_0^1{\int }_0^{1}f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}_a{d}_{q_1}x\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill I_8& ={\int }_0^1{\int }_0^{1}f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right){}_0{d}_{q_2}s{}_0{d}_{q_1}t\hfill \\ \hfill & =\frac{4}{(b-a)(d-c)}{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}_a{d}_{q_1}x.\hfill \end{array}$$

Substituting $I_5,I_6,I_7$, and $I_8$ in (17), we obtain

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}_a{d}_{q_1}x\right.\hfill \\ \hfill & +{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}^b{d}_{q_1}x\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}^b{d}_{q_1}x\right].\hfill \end{array}$$

Thus, the first inequality of (15) holds.

Next, by co-ordinated convexity of f again, we have

$$\begin{array}{cc}\hfill f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right)& \le \frac{ts}{4}f(a,c)+\frac{t(2-s)}{4}f(a,d)\hfill \\ \hfill & +\frac{(2-t)s}{4}f(b,c)+\frac{(2-t)(2-s)}{4}f(b,d),\hfill \end{array}$$

(18)

$$\begin{array}{cc}\hfill f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right)& \le \frac{t(2-s)}{4}f(a,c)+\frac{ts}{4}f(a,d)\hfill \\ \hfill & +\frac{(2-t)(2-s)}{4}f(b,c)+\frac{(2-t)s}{4}f(b,d),\hfill \end{array}$$

(19)

$$\begin{array}{cc}\hfill f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right)& \le \frac{(2-t)s}{4}f(a,c)+\frac{(2-t)(2-s)}{4}f(a,d)\hfill \\ \hfill & +\frac{ts}{4}f(b,c)+\frac{t(2-s)}{4}f(b,d)\hfill \end{array}$$

(20)

and

$$\begin{array}{cc}\hfill f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right)& \le \frac{(2-t)(2-s)}{4}f(a,c)+\frac{(2-t)s}{4}f(a,d)\hfill \\ \hfill & +\frac{t(2-s)}{4}f(b,c)+\frac{ts}{4}f(b,d).\hfill \end{array}$$

(21)

Summing (18)–(21), we obtain

$$\begin{array}{cc}\hfill & f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right)+f\left(\frac{t}{2}a+\frac{2-t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right)\hfill \\ \hfill & +f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{s}{2}c+\frac{2-s}{2}d\right)+f\left(\frac{2-t}{2}a+\frac{t}{2}b,\frac{2-s}{2}c+\frac{s}{2}d\right)\hfill \\ \hfill & \le f(a,c)+f(a,d)+f(b,c)+f(b,d).\hfill \end{array}$$

(22)

$q_{a,c}$-Integrating both sides of (22) over $[0,1]$ and then multiplying by $\frac{1}{4}$, the second inequality of (15) holds. The proof is completed. □

Theorem 8.

Let $f:△\to \mathbb{R}$ be a co-ordinated convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{2(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{2(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}_{\frac{c+d}{2}}{d}_{q_2}y\right]\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x\right.\hfill \\ \hfill & +{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f(x,c)+f(x,d){}^{\frac{a+b}{2}}{d}_{q_1}x\right]+\frac{1}{4(b-a)}\left[{\int }_{\frac{a+b}{2}}^{b}f(x,c)+f(x,d){}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f(a,y)+f(b,y){}^{\frac{c+d}{2}}{d}_{q_2}y\right]+\frac{1}{4(d-c)}\left[{\int }_{\frac{c+d}{2}}^{d}f(a,y)+f(b,y){}_{\frac{c+d}{2}}{d}_{q_2}y\right]\hfill \\ \hfill & \le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.\hfill \end{array}$$

(23)

Proof. 

Let $g_y:[a,b]\to \mathbb{R}$ be a function defined by $g_y\left(x\right)=f(x,y)$. Then, $g_y$ is convex on $[a,b]$ because f is co-ordinated convex on △. By Theorem 3, we can write

$$\begin{array}{c}\hfill g_y\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}{g}_y\left(x\right){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{g}_y\left(x\right){}_{\frac{a+b}{2}}{d}_{q_1}x\right]\le \frac{g_y\left(a\right)+g_y\left(b\right)}{2}.\end{array}$$

That is,

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2},y\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}f(x,y){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f(x,y){}_{\frac{a+b}{2}}{d}_{q_1}x\right]\le \frac{f(a,y)+f(b,y)}{2}.\end{array}$$

(24)

$q^d$-Integrating both sides of (24) over $[c,\frac{c+d}{2}]$ and then dividing by $(d-c)$, we have

$$\begin{array}{cc}\hfill \frac{1}{d-c}& {\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}^{\frac{c+d}{2}}{d}_{q_2}y\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{2(d-c)}{\int }_c^{\frac{c+d}{2}}f(a,y)+f(b,y){}^{\frac{c+d}{2}}{d}_{q_2}y.\hfill \end{array}$$

(25)

Similarly, $q_c$-integrating both sides of (24) over $[\frac{c+d}{2},d]$ and then dividing by $(d-c)$, we have

$$\begin{array}{cc}\hfill \frac{1}{d-c}& {\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}_{\frac{c+d}{2}}{d}_{q_2}y\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{2(d-c)}{\int }_{\frac{c+d}{2}}^{d}f(a,y)+f(b,y){}_{\frac{c+d}{2}}{d}_{q_2}y.\hfill \end{array}$$

(26)

Let $h_x:[c,d]\to \mathbb{R}$ be defined by $h_x\left(y\right)=f(x,y)$. Then, $h_x$ is convex on $[c,d]$ because f is co-ordinated convex on △. By Theorem 3, we can write

$$\begin{array}{c}\hfill h_x\left(\frac{c+d}{2}\right)\le \frac{1}{d-c}\left[{\int }_c^{\frac{c+d}{2}}{h}_x\left(y\right){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^d{h}_x\left(y\right){}_{\frac{c+d}{2}}{d}_{q_2}y\right]\le \frac{h_x\left(c\right)+h_x\left(d\right)}{2}.\end{array}$$

That is,

$$\begin{array}{c}\hfill f\left(x,\frac{c+d}{2}\right)\le \frac{1}{d-c}\left[{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y\right]\le \frac{f(x,c)+f(x,d)}{2}.\end{array}$$

(27)

$q^b$-Integrating both sides of (27) over $[a,\frac{a+b}{2}]$ and then dividing by $(b-a)$, we have

$$\begin{array}{cc}\hfill \frac{1}{b-a}& {\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}^{\frac{a+b}{2}}{d}_{q_1}x\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{2(b-a)}{\int }_a^{\frac{a+b}{2}}f(x,c)+f(x,d){}^{\frac{a+b}{2}}{d}_{q_1}x.\hfill \end{array}$$

(28)

Similarly, $q_a$-integrating both sides of (27) over $[\frac{a+b}{2},b]$ and then dividing by $(b-a)$, we have

$$\begin{array}{cc}\hfill \frac{1}{b-a}& {\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}_{\frac{a+b}{2}}{d}_{q_1}x=\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\right]=\hfill \\ \hfill & \le \frac{1}{2(b-a)}{\int }_{\frac{a+b}{2}}^{b}f(x,c)+f(x,d){}_{\frac{a+b}{2}}{d}_{q_1}x.\hfill \end{array}$$

(29)

Summing (25), (26), (28), and (29), we derive

$$\begin{array}{cc}\hfill & \frac{1}{2(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{2(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}_{\frac{c+d}{2}}{d}_{q_2}y\right]\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}^{\frac{a+b}{2}}{d}_{q_1}x\right.\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}^{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}_{\frac{c+d}{2}}{d}_{q_2}y{}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f(x,c)+f(x,d){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f(x,c)+f(x,d){}_{\frac{a+b}{2}}{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f(a,y)+f(b,y){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f(a,y)+f(b,y){}_{\frac{c+d}{2}}{d}_{q_2}y\right].\hfill \end{array}$$

(30)

Now, the second and the third inequalities of (23) hold.

For the first inequality, by the left hand side of the inequality (24) with $y=\frac{c+d}{2}$, we have

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}_{\frac{a+b}{2}}{d}_{q_1}x\right]\end{array}$$

(31)

and by the left hand side of the inequality (27) with $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{d-c}\left[{\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}_{\frac{c+d}{2}}{d}_{q_2}y\right].\end{array}$$

(32)

Combining (31) and (32), we obtain the first inequality of (23).

Using the right hand side of (24) with $y=c$ and $y=d$, we obtain

$$\begin{array}{cc}\hfill \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f(x,c){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f(x,c){}_{\frac{a+b}{2}}{d}_{q_1}x\right]& \le \frac{f(a,c)+f(b,c)}{8}\hfill \end{array}$$

(33)

and

$$\begin{array}{cc}\hfill \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f(x,d){}^{\frac{a+b}{2}}{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f(x,d){}_{\frac{a+b}{2}}{d}_{q_1}x\right]& \le \frac{f(a,d)+f(b,d)}{8},\hfill \end{array}$$

(34)

respectively. Using the right hand side of (27) with $x=a$ and $x=b$, we have

$$\begin{array}{cc}\hfill \frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f(a,y){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f(a,y){}_{\frac{c+d}{2}}{d}_{q_2}y\right]& \le \frac{f(a,c)+f(a,d)}{8}\hfill \end{array}$$

(35)

and

$$\begin{array}{cc}\hfill \frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f(b,y){}^{\frac{c+d}{2}}{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f(b,y){}_{\frac{c+d}{2}}{d}_{q_2}y\right]& \le \frac{f(b,c)+f(b,d)}{8},\hfill \end{array}$$

(36)

respectively. Replacing (33)–(36) in (30), we obtain the last inequality of (23). The proof is completed. □

Theorem 9.

Let $f:△\to \mathbb{R}$ be a co-ordinated convex function. Then, we have

$$\begin{array}{cc}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)& \le \frac{1}{2(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}^b{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{2(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}^d{d}_{q_2}y\right]\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}_a{d}_{q_1}x\right.\hfill \\ \hfill & +{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}^b{d}_{q_1}x\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}^b{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}\left[f(x,c)+f(x,d)\right]{}_a{d}_{q_1}x\right]+\frac{1}{4(b-a)}\left[{\int }_{\frac{a+b}{2}}^b\left[f(x,c)+f(x,d)\right]{}^b{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}\left[f(a,y)+f(b,y)\right]{}_c{d}_{q_2}y\right]+\frac{1}{4(d-c)}\left[{\int }_{\frac{c+d}{2}}^d\left[f(a,y)+f(b,y)\right]{}^d{d}_{q_2}y\right]\hfill \\ \hfill & \le \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}.\hfill \end{array}$$

(37)

Proof. 

Let $g_y:[a,b]\to \mathbb{R}$ be a function defined by $g_y\left(x\right)=f(x,y)$. Then, $g_y$ is convex on $[a,b]$ because f is co-ordinated convex on △. By Theorem 4, we can write

$$\begin{array}{c}\hfill g_y\left(\frac{a+b}{2}\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}{g}_y\left(x\right){}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{g}_y\left(x\right){}^b{d}_{q_1}x\right]\le \frac{g_y\left(a\right)+g_y\left(b\right)}{2}.\end{array}$$

That is,

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2},y\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}f(x,y){}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f(x,y){}^b{d}_{q_1}x\right]\le \frac{f(a,y)+f(b,y)}{2}.\end{array}$$

(38)

$q_c$-Integrating both sides of (38) over $[c,\frac{c+d}{2}]$ and then dividing by $(d-c)$, we have

$$\begin{array}{cc}\hfill \frac{1}{d-c}& {\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}_c{d}_{q_2}y\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}^b{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{2(d-c)}{\int }_c^{\frac{c+d}{2}}\left[f(a,y)+f(b,y)\right]{}_c{d}_{q_2}y.\hfill \end{array}$$

(39)

Similarly, $q^d$-integrating both sides of (38) over $[\frac{c+d}{2},d]$ and then dividing by $(d-c)$, we have

$$\begin{array}{cc}\hfill \frac{1}{d-c}& {\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}^d{d}_{q_2}y\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}^b{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{2(d-c)}{\int }_{\frac{c+d}{2}}^d\left[f(a,y)+f(b,y)\right]{}^d{d}_{q_2}y.\hfill \end{array}$$

(40)

On the other hand, let $h_x:[c,d]\to \mathbb{R}$ be defined by $h_x\left(y\right)=f(x,y)$. Then, $h_x$ is convex on $[c,d]$ since f is co-ordinated convex on △. By Theorem 4, we can write

$$\begin{array}{c}\hfill h_x\left(\frac{c+d}{2}\right)\le \frac{1}{d-c}\left[{\int }_c^{\frac{c+d}{2}}{h}_x\left(y\right){}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^d{h}_x\left(y\right){}^d{d}_{q_2}y\right]\le \frac{h_x\left(c\right)+h_x\left(d\right)}{2}.\end{array}$$

That is,

$$\begin{array}{c}\hfill f\left(x,\frac{c+d}{2}\right)\le \frac{1}{d-c}\left[{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y\right]\le \frac{f(x,c)+f(x,d)}{2}.\end{array}$$

(41)

$q_a$-Integrating both sides of (41) over $[a,\frac{a+b}{2}]$ and then dividing by $(b-a)$, we have

$$\begin{array}{cc}\hfill \frac{1}{b-a}& {\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}_a{d}_{q_1}x\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}_a{d}_{q_1}x+{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}_a{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{2(b-a)}{\int }_a^{\frac{a+b}{2}}\left[f(x,c)+f(x,d)\right]{}_a{d}_{q_1}x.\hfill \end{array}$$

(42)

Similarly, $q^b$-integrating both sides of (41) over $[\frac{a+b}{2},b]$ and then dividing by $(b-a)$, we have

$$\begin{array}{cc}\hfill \frac{1}{b-a}& {\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}^b{d}_{q_1}x\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}^b{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}^b{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{2(b-a)}{\int }_{\frac{a+b}{2}}^b\left[f(x,c)+f(x,d)\right]{}^b{d}_{q_1}x.\hfill \end{array}$$

(43)

Summing (39), (40), (42), and (43), we derive

$$\begin{array}{cc}\hfill & \frac{1}{2(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}^b{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{2(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}^d{d}_{q_2}y\right]\hfill \\ \hfill & \le \frac{1}{(b-a)(d-c)}\left[{\int }_a^{\frac{a+b}{2}}{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}_a{d}_{q_1}x+{\int }_a^{\frac{a+b}{2}}{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}_a{d}_{q_1}x\right.\hfill \\ \hfill & \left.+{\int }_{\frac{a+b}{2}}^b{\int }_c^{\frac{c+d}{2}}f(x,y){}_c{d}_{q_2}y{}^b{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b{\int }_{\frac{c+d}{2}}^{d}f(x,y){}^d{d}_{q_2}y{}^b{d}_{q_1}x\right]\hfill \\ \hfill & \le \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}\left[f(x,c)+f(x,d)\right]{}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^b\left[f(x,c)+f(x,d)\right]{}^b{d}_{q_1}x\right]\hfill \\ \hfill & +\frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}\left[f(a,y)+f(b,y)\right]{}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^d\left[f(a,y)+f(b,y)\right]{}^d{d}_{q_2}y\right].\hfill \end{array}$$

(44)

Now, the second and the third inequalities of (37) hold.

For the first inequality, by the left hand side of the inequality (38) with $y=\frac{c+d}{2}$, we have

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{b-a}\left[{\int }_a^{\frac{a+b}{2}}f\left(x,\frac{c+d}{2}\right){}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f\left(x,\frac{c+d}{2}\right){}^b{d}_{q_1}x\right]\end{array}$$

(45)

and by the left hand side of the inequality (41) with $x=\frac{a+b}{2}$, we obtain

$$\begin{array}{c}\hfill f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\le \frac{1}{d-c}\left[{\int }_c^{\frac{c+d}{2}}f\left(\frac{a+b}{2},y\right){}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f\left(\frac{a+b}{2},y\right){}^d{d}_{q_2}y\right].\end{array}$$

(46)

By combining (45)–(46), we obtain the first inequality of (37).

Using the right hand side of (38) with $y=c$ and $y=d$, we obtain

$$\begin{array}{c}\hfill \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f(x,c){}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f(x,c){}^b{d}_{q_1}x\right]\le \frac{f(a,c)+f(b,c)}{8}\end{array}$$

(47)

and

$$\begin{array}{c}\hfill \frac{1}{4(b-a)}\left[{\int }_a^{\frac{a+b}{2}}f(x,d){}_a{d}_{q_1}x+{\int }_{\frac{a+b}{2}}^{b}f(x,d){}^b{d}_{q_1}x\right]\le \frac{f(a,d)+f(b,d)}{8},\end{array}$$

(48)

respectively. Using the right hand side of (41) with $x=a$ and $x=b$, we have

$$\begin{array}{c}\hfill \frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f(a,y){}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f(a,y){}^d{d}_{q_2}y\right]\le \frac{f(a,c)+f(a,d)}{8}\end{array}$$

(49)

and

$$\begin{array}{c}\hfill \frac{1}{4(d-c)}\left[{\int }_c^{\frac{c+d}{2}}f(b,y){}_c{d}_{q_2}y+{\int }_{\frac{c+d}{2}}^{d}f(b,y){}^d{d}_{q_2}y\right]\le \frac{f(b,c)+f(b,d)}{8},\end{array}$$

(50)

respectively. Substituting (47)–(50) in (44), we obtain the last inequality of (37). The proof is completed. □

Remark 1.

If we take the limit $q_1\to 1^{-}$ and $q_2\to 1^{-}$ in Theorem 8 and 9, then Theorem 8 and 9 reduce to Theorem 5.

4. Examples

Now, we give some examples of our main results to demonstrate our theorems.

Example 5.

Let $f:[0,1]\times [0,1]\to \mathbb{R}$ be a function defined by $f(x,y)=x^2{y}^2$. Then, f is co-ordinated convex on $[0,1]\times [0,1]$. By applying Theorem 6 with $q_1=\frac{1}{4}$ and $q_2=\frac{3}{4}$, the first inequality of (8) becomes

$$\begin{array}{cc}\hfill \frac{1}{16}& ={\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^2=f\left(\frac{1}{2},\frac{1}{2}\right)=f\left(\frac{0+1}{2},\frac{0+1}{2}\right)\hfill \\ \hfill & \le \frac{1}{(1-0)(1-0)}\left[{\int }_0^{\frac{0+1}{2}}{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}^{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\right.\hfill \\ \hfill & +{\int }_0^{\frac{0+1}{2}}{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}_{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{0+1}{2}}{d}_{\frac{1}{4}}x+{\int }_{\frac{0+1}{2}}^1{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}^{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\hfill \\ \hfill & \left.+{\int }_{\frac{0+1}{2}}^1{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}_{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & ={\int }_0^{\frac{1}{2}}{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}^{\frac{1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}_{\frac{1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{1}{2}}{d}_{\frac{1}{4}}x\hfill \\ \hfill & +{\int }_{\frac{1}{2}}^1{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}^{\frac{1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}_{\frac{1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{1}{2}}{d}_{\frac{1}{4}}x\hfill \\ \hfill & =\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1-{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1-{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1-{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1+{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1+{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1-{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1+{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1+{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & =\frac{85}{\mathrm{116,032}}+\frac{\mathrm{11,339}}{\mathrm{1,740,480}}+\frac{1765}{\mathrm{116,032}}+\frac{\mathrm{235,451}}{\mathrm{1,740,480}}=\frac{53}{336}.\hfill \end{array}$$

We also have

$$\begin{array}{c}\hfill \frac{f(0,0)+f(0,1)+f(1,0)+f(1,1)}{4}=\frac{1}{4}.\end{array}$$

It is clear that

$$\begin{array}{c}\hfill \frac{1}{16}\le \frac{53}{336}\le \frac{1}{4},\end{array}$$

which demonstrates the result described in Theorem 6.

Example 6.

Let $f:[0,1]\times [0,1]\to \mathbb{R}$ be a function defined by $f(x,y)=x^2{y}^2$. Then, f is co-ordinated convex on $[0,1]\times [0,1]$. By applying Theorem 7 with $q_1=\frac{1}{4}$ and $q_2=\frac{3}{4}$, the first inequality of (15) becomes

$$\begin{array}{cc}\hfill \frac{1}{16}& ={\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^2=f\left(\frac{1}{2},\frac{1}{2}\right)=f\left(\frac{0+1}{2},\frac{0+1}{2}\right)\hfill \\ \hfill & \le \frac{1}{(1-0)(1-0)}\left[{\int }_0^{\frac{0+1}{2}}{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x\right.\hfill \\ \hfill & +{\int }_0^{\frac{0+1}{2}}{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x+{\int }_{\frac{0+1}{2}}^1{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x\hfill \\ \hfill & \left.+{\int }_{\frac{0+1}{2}}^1{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & ={\int }_0^{\frac{1}{2}}{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x\hfill \\ \hfill & +{\int }_{\frac{1}{2}}^1{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x\hfill \\ \hfill & =\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(\frac{1}{4}\right)}^{2n}{\left(\frac{3}{4}\right)}^{2m}\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}{\left(\frac{1}{4}\right)}^{2n}{\left(1-\frac{1}{2}{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}{\left(1-\frac{1}{2}{\left(\frac{1}{4}\right)}^n\right)}^2{\left(\frac{3}{4}\right)}^{2m}\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m{\left(1-\frac{1}{2}{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1-\frac{1}{2}{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & =\frac{4}{777}+\frac{139}{5439}+\frac{41}{3885}+\frac{5699}{\mathrm{108,780}}=\frac{\mathrm{10,187}}{\mathrm{108,780}}.\hfill \end{array}$$

We also have

$$\begin{array}{c}\hfill \frac{f(0,0)+f(0,1)+f(1,0)+f(1,1)}{4}=\frac{1}{4}.\end{array}$$

It is clear that

$$\begin{array}{c}\hfill \frac{1}{16}\le \frac{\mathrm{10,187}}{\mathrm{108,780}}\le \frac{1}{4},\end{array}$$

which demonstrates the result described in Theorem 7.

Example 7.

Let $f:[0,1]\times [0,1]\to \mathbb{R}$ be a function defined by $f(x,y)=x^2{y}^2$. Then, f is co-ordinated convex on $[0,1]\times [0,1]$. By applying Theorem 8 with $q_1=\frac{1}{4}$ and $q_2=\frac{3}{4}$, the first inequality of (23) becomes

$$\begin{array}{cc}\hfill \frac{1}{16}& ={\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^2=f\left(\frac{1}{2},\frac{1}{2}\right)=f\left(\frac{0+1}{2},\frac{0+1}{2}\right)\hfill \\ \hfill & \le \frac{1}{2(1-0)}\left[{\int }_0^{\frac{0+1}{2}}f\left(x,\frac{0+1}{2}\right){}^{\frac{0+1}{2}}{d}_{\frac{1}{4}}x+{\int }_{\frac{0+1}{2}}^{1}f\left(x,\frac{0+1}{2}\right){}_{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & +\frac{1}{2(1-0)}\left[{\int }_0^{\frac{0+1}{2}}f\left(\frac{0+1}{2},y\right){}^{\frac{0+1}{2}}{d}_{\frac{3}{4}}y+{\int }_{\frac{0+1}{2}}^{1}f\left(\frac{0+1}{2},y\right){}_{\frac{0+1}{2}}{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{2}\left[{\int }_0^{\frac{1}{2}}\frac{x^2}{4}{}^{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1\frac{x^2}{4}{}_{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}\frac{y^2}{4}{}^{\frac{1}{2}}{d}_{\frac{3}{4}}y+{\int }_{\frac{1}{2}}^1\frac{y^2}{4}{}_{\frac{1}{2}}{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{2}\left[\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n\frac{1}{4}·\frac{1}{4}{\left(1-{\left(\frac{1}{4}\right)}^n\right)}^2+\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n\frac{1}{4}·\frac{1}{4}{\left(1+{\left(\frac{1}{4}\right)}^n\right)}^2\right.\hfill \\ \hfill & +\left.\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n\frac{1}{4}·\frac{1}{4}{\left(1-{\left(\frac{3}{4}\right)}^n\right)}^2+\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n\frac{1}{4}·\frac{1}{4}{\left(1+{\left(\frac{3}{4}\right)}^n\right)}^2\right]\hfill \\ \hfill & =\frac{1}{2}\left[\frac{17}{3360}+\frac{353}{3360}+\frac{75}{8288}+\frac{667}{8288}\right]=\frac{1241}{\mathrm{12,432}}.\hfill \end{array}$$

The third inequality of (23) becomes

$$\begin{array}{cc}\hfill \frac{53}{336}& =\frac{85}{\mathrm{116,032}}+\frac{\mathrm{11,339}}{\mathrm{1,740,480}}+\frac{1765}{\mathrm{116,032}}+\frac{\mathrm{235,451}}{\mathrm{1,740,480}}\hfill \\ \hfill & =\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1-{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1-{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1-{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1+{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1+{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1-{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(1+{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1+{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & ={\int }_0^{\frac{1}{2}}{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}^{\frac{1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}_{\frac{1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{1}{2}}{d}_{\frac{1}{4}}x\hfill \\ \hfill & +{\int }_{\frac{1}{2}}^1{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}^{\frac{1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}_{\frac{1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{1}{2}}{d}_{\frac{1}{4}}x\hfill \\ \hfill & =\frac{1}{(1-0)(1-0)}\left[{\int }_0^{\frac{0+1}{2}}{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}^{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\right.\hfill \\ \hfill & +{\int }_0^{\frac{0+1}{2}}{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}_{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}^{\frac{0+1}{2}}{d}_{\frac{1}{4}}x+{\int }_{\frac{0+1}{2}}^1{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}^{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\hfill \\ \hfill & \left.+{\int }_{\frac{0+1}{2}}^1{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}_{\frac{0+1}{2}}{d}_{\frac{3}{4}}y{}_{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & \le \frac{1}{4(1-0)}\left[{\int }_0^{\frac{0+1}{2}}f(x,0)+f(x,1){}^{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\right]+\frac{1}{4(1-0)}\left[{\int }_{\frac{0+1}{2}}^{1}f(x,0)+f(x,1){}_{\frac{0+1}{2}}{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & +\frac{1}{4(1-0)}\left[{\int }_0^{\frac{0+1}{2}}f(0,y)+f(1,y){}^{\frac{0+1}{2}}{d}_{\frac{3}{4}}y\right]+\frac{1}{4(1-0)}\left[{\int }_{\frac{0+1}{2}}^{1}f(0,y)+f(1,y){}_{\frac{0+1}{2}}{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{4}\left[{\int }_0^{\frac{1}{2}}{x}^2{}^{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1{x}^2{}_{\frac{1}{2}}{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}{y}^2{}^{\frac{1}{2}}{d}_{\frac{3}{4}}y+{\int }_{\frac{1}{2}}^1{y}^2{}_{\frac{1}{2}}{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{4}\left[\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n\frac{1}{4}{\left(1-{\left(\frac{1}{4}\right)}^n\right)}^2+\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n\frac{1}{4}{\left(1+{\left(\frac{1}{4}\right)}^n\right)}^2\right.\hfill \\ \hfill & +\left.\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n\frac{1}{4}{\left(1-{\left(\frac{3}{4}\right)}^n\right)}^2+\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n\frac{1}{4}{\left(1+{\left(\frac{3}{4}\right)}^n\right)}^2\right]\hfill \\ \hfill & =\frac{1}{4}\left[\frac{68}{3360}+\frac{1412}{3360}+\frac{300}{8288}+\frac{2668}{8288}\right]=\frac{1241}{6216}.\hfill \end{array}$$

We also have

$$\begin{array}{c}\hfill \frac{f(0,0)+f(0,1)+f(1,0)+f(1,1)}{4}=\frac{1}{4}.\end{array}$$

It is clear that

$$\begin{array}{c}\hfill \frac{1}{16}\le \frac{1241}{\mathrm{12,432}}\le \frac{53}{336}\le \frac{1241}{6216}\le \frac{1}{4},\end{array}$$

which demonstrates the result described in Theorem 8.

Example 8.

Let $f:[0,1]\times [0,1]\to \mathbb{R}$ be a function defined by $f(x,y)=x^2{y}^2$. Then, f is co-ordinated convex on $[0,1]\times [0,1]$. By applying Theorem 9 with $q_1=\frac{1}{4}$ and $q_2=\frac{3}{4}$, the first inequality of (37) becomes

$$\begin{array}{cc}\hfill \frac{1}{16}& ={\left(\frac{1}{2}\right)}^2{\left(\frac{1}{2}\right)}^2=f\left(\frac{1}{2},\frac{1}{2}\right)=f\left(\frac{0+1}{2},\frac{0+1}{2}\right)\hfill \\ \hfill & \le \frac{1}{2(1-0)}\left[{\int }_0^{\frac{0+1}{2}}f\left(x,\frac{0+1}{2}\right){}_0{d}_{\frac{1}{4}}x+{\int }_{\frac{0+1}{2}}^{1}f\left(x,\frac{0+1}{2}\right){}^1{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & +\frac{1}{2(1-0)}\left[{\int }_0^{\frac{0+1}{2}}f\left(\frac{0+1}{2},y\right){}_0{d}_{\frac{3}{4}}y+{\int }_{\frac{0+1}{2}}^{1}f\left(\frac{0+1}{2},y\right){}^1{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{2}\left[{\int }_0^{\frac{1}{2}}\frac{x^2}{4}{}_0{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1\frac{x^2}{4}{}^1{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}\frac{y^2}{4}{}_0{d}_{\frac{3}{4}}y+{\int }_{\frac{1}{2}}^1\frac{y^2}{4}{}^1{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{2}\left[\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n\frac{1}{4}·\frac{1}{4}{\left(\frac{1}{4}\right)}^{2n}+\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n\frac{1}{4}{\left(1-\frac{1}{2}{\left(\frac{1}{4}\right)}^n\right)}^2\right.\hfill \\ \hfill & +\left.\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n\frac{1}{4}·\frac{1}{4}{\left(\frac{3}{4}\right)}^{2n}+\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n\frac{1}{4}{\left(1-\frac{1}{2}{\left(\frac{3}{4}\right)}^n\right)}^2\right]\hfill \\ \hfill & =\frac{1}{2}\left[\frac{3}{126}+\frac{41}{840}+\frac{1}{74}+\frac{139}{2072}\right]=\frac{2381}{\mathrm{31,080}}.\hfill \end{array}$$

The third inequality of (37) becomes

$$\begin{array}{cc}\hfill \frac{10,187}{108,780}& =\frac{4}{777}+\frac{139}{5439}+\frac{41}{3885}+\frac{5699}{\mathrm{108,780}}\hfill \\ \hfill & =\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}·\frac{1}{4}{\left(\frac{1}{4}\right)}^{2n}{\left(\frac{3}{4}\right)}^{2m}\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}{\left(\frac{1}{4}\right)}^{2n}{\left(1-\frac{1}{2}{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m\frac{1}{4}{\left(1-\frac{1}{2}{\left(\frac{1}{4}\right)}^n\right)}^2{\left(\frac{3}{4}\right)}^{2m}\hfill \\ \hfill & +\frac{1}{4}·\frac{3}{4}·\frac{1}{2}·\frac{1}{2}\sum _{n=0}^{\infty }\sum _{m=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(\frac{3}{4}\right)}^m{\left(1-\frac{1}{2}{\left(\frac{1}{4}\right)}^n\right)}^2{\left(1-\frac{1}{2}{\left(\frac{3}{4}\right)}^m\right)}^2\hfill \\ \hfill & ={\int }_0^{\frac{1}{2}}{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x\hfill \\ \hfill & +{\int }_{\frac{1}{2}}^1{\int }_0^{\frac{1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1{\int }_{\frac{1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x\hfill \\ \hfill & =\frac{1}{(1-0)(1-0)}\left[{\int }_0^{\frac{0+1}{2}}{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x\right.\hfill \\ \hfill & +{\int }_0^{\frac{0+1}{2}}{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}_0{d}_{\frac{1}{4}}x+{\int }_{\frac{0+1}{2}}^1{\int }_0^{\frac{0+1}{2}}{x}^2{y}^2{}_0{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x\hfill \\ \hfill & \left.+{\int }_{\frac{0+1}{2}}^1{\int }_{\frac{0+1}{2}}^1{x}^2{y}^2{}^1{d}_{\frac{3}{4}}y{}^1{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & \le \frac{1}{4(1-0)}\left[{\int }_0^{\frac{0+1}{2}}\left[f(x,0)+f(x,1)\right]{}_0{d}_{\frac{1}{4}}x\right]+\frac{1}{4(1-0)}\left[{\int }_{\frac{0+1}{2}}^1\left[f(x,0)+f(x,1)\right]{}^1{d}_{\frac{1}{4}}x\right]\hfill \\ \hfill & +\frac{1}{4(1-0)}\left[{\int }_0^{\frac{0+1}{2}}\left[f(0,y)+f(1,y)\right]{}_0{d}_{\frac{3}{4}}y\right]+\frac{1}{4(1-0)}\left[{\int }_{\frac{0+1}{2}}^1\left[f(0,y)+f(1,y)\right]{}^1{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{4}\left[{\int }_0^{\frac{1}{2}}{x}^2{}_0{d}_{\frac{1}{4}}x+{\int }_{\frac{1}{2}}^1{x}^2{}^1{d}_{\frac{1}{4}}x+{\int }_0^{\frac{1}{2}}{y}^2{}_0{d}_{\frac{3}{4}}y+{\int }_{\frac{1}{2}}^1{y}^2{}^1{d}_{\frac{3}{4}}y\right]\hfill \\ \hfill & =\frac{1}{4}\left[\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n\frac{1}{4}{\left(\frac{1}{4}\right)}^{2n}+\frac{3}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{1}{4}\right)}^n{\left(1-\frac{1}{2}{\left(\frac{1}{4}\right)}^n\right)}^2\right.\hfill \\ \hfill & +\left.\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n\frac{1}{4}{\left(\frac{3}{4}\right)}^{2n}+\frac{1}{4}·\frac{1}{2}\sum _{n=0}^{\infty }{\left(\frac{3}{4}\right)}^n{\left(1-\frac{1}{2}{\left(\frac{3}{4}\right)}^n\right)}^2\right]\hfill \\ \hfill & =\frac{1}{4}\left[\frac{12}{126}+\frac{164}{840}+\frac{4}{74}+\frac{556}{2072}\right]=\frac{2381}{\mathrm{15,540}}.\hfill \end{array}$$

We also have

$$\begin{array}{c}\hfill \frac{f(0,0)+f(0,1)+f(1,0)+f(1,1)}{4}=\frac{1}{4}.\end{array}$$

It is clear that

$$\begin{array}{c}\hfill \frac{1}{16}\le \frac{2381}{\mathrm{31,080}}\le \frac{\mathrm{10,187}}{\mathrm{108,780}}\le \frac{2381}{\mathrm{15,540}}\le \frac{1}{4},\end{array}$$

which demonstrates the result described in Theorem 9.

5. Conclusions

In this paper, we proved some new Hermite–Hadamard inequalities for co-ordinated convex functions in q-calculus. We also gave some examples in order to demonstrate our main results. We can extend these results further to another convexities, post-quantum calculus, and fractional calculus. We can also use other techniques to improve the outcomes.