# A new Hermite–Hadamard type inequality for coordinate convex function

## Abstract

In the article, we establish a new Hermite–Hadamard type inequality for the coordinate convex function by constructing two monotonic sequences. The given result is the generalization and improvement of some previously obtained results.

## 1 Introduction

Let $$I \subseteq \mathbb{R}$$ be an interval. Then a real-valued function $$f: I \to \mathbb{R}$$ is said to be convex (concave) if the inequality

$$f\bigl(ta + (1-t)b\bigr) \leq (\geq )tf(a) + (1-t)f(b)$$

holds for all $$a, b \in I$$ and $$t \in [0, 1]$$. Recently, the generalizations, extensions, variants and applications of convexity have attracted the attention of many researchers (e.g., [4, 2022]). In particular, many inequalities can be found in the literature (e.g., [13, 15, 17]) via the convexity theory.

The well known Hermite–Hadamard inequality for convex function is formulated as follows:

Let $$f: I\subseteq \mathbb{R}\to \mathbb{R}$$ be a convex function defined on the interval $$I=[a,b]$$ with $$a< b$$. Then the following inequality holds:

\begin{aligned} f \biggl(\frac{a+b}{2} \biggr) \leq \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \leq \frac{f(a)+f(b)}{2}. \end{aligned}
(1)

In recent years, more and more refinements of the Hermite–Hadamard inequality for convex functions have been extensively investigated by a number of authors (e.g., [13, 5, 6, 810, 12, 14, 16, 18, 23]).

In , A.E. Farissi improved the Hermite–Hadamard inequality as follows:

### Theorem 1.1

()

Let$$f:I\to \mathbb{R}$$be a convex function on$$I=[a,b]$$with$$a< b$$. Then for all$$\lambda \in [0,1]$$,

\begin{aligned} f \biggl(\frac{a+b}{2} \biggr) \leq l(\lambda ) \leq \frac{1}{b-a} \int _{a}^{b}f(x)\,dx \leq L(\lambda ) \leq \frac{f(a)+f(b)}{2}, \end{aligned}
(2)

where

$$l(\lambda )=\lambda f \biggl(\frac{\lambda b+(2-\lambda )a}{2} \biggr)+(1- \lambda )f \biggl( \frac{(1+\lambda )b+(1-\lambda )a}{2} \biggr)$$

and

$$L(\lambda )=\frac{1}{2} \bigl(f\bigl(\lambda b+(1-\lambda )a\bigr)+\lambda f(a)+(1- \lambda ) f(b) \bigr).$$

Consider the two-dimensional interval $$\Delta :=[a,b]\times [c,d]$$ with $$a < b$$ and $$c< d$$. A function $$f:\Delta \to \mathbb{R}$$ is said to be coordinate convex on Δ if the partial mappings $$f_{y}:[a,b] \to \mathbb{R}$$, $$f_{y}(u)=f(u,y)$$ and $$f_{x}:[c,d] \to \mathbb{R}$$, $$f_{x}(v)=f(x,v)$$, are convex for all $$y \in [c,d]$$ and $$x \in [a,b]$$.

In , S.S. Dragomir established the following Hadamard-type inequalities for coordinate convex functions in a rectangle from the plane $$\mathbb{R}^{2}$$.

### Theorem 1.2

()

Let$$f: \Delta =[a,b]\times [c,d] \to \mathbb{R}$$be a coordinate convex function on Δ. Then

\begin{aligned} \begin{aligned}[b] f \biggl(\frac{a+b}{2}, \frac{c+d}{2} \biggr)&\leq \frac{1}{2} \biggl[ \frac{1}{b-a} \int _{a}^{b}f \biggl(x, \frac{c+d}{2} \biggr)\,dx+ \frac{1}{d-c} \int _{c}^{d}f \biggl(\frac{a+b}{2}, y \biggr)\,dy \biggr] \\ &\leq \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d}f(x,y)\,dy \,dx \\ &\leq \frac{1}{4} \biggl[\frac{1}{b-a} \int _{a}^{b}\bigl[f(x,c)+f(x,d)\bigr]\,dx+ \frac{1}{d-c} \int _{c}^{d}\bigl[f(a,y)+f(b,y)\bigr]\,dy \biggr] \\ &\leq \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}. \end{aligned} \end{aligned}
(3)

In , M.E. Özdemir defined a new mapping associated with coordinate convexity and proved the following inequalities based on the properties of this mapping.

### Theorem 1.3

()

Let$$f: \Delta \subset \mathbb{R}^{2}\to \mathbb{R}$$be a coordinate convex function on$$\Delta =[a,b]\times [c,d]$$. Then

\begin{aligned} \begin{aligned}[b] &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\quad \leq \frac{1}{4} \biggl[\frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4} \\ &\qquad{} + \frac{f (\frac{a+b}{2},c )+f (\frac{a+b}{2},d )+f (a,\frac{c+d}{2} ) +f (b,\frac{c+d}{2} )}{2}+f \biggl(\frac{a+b}{2}, \frac{c+d}{2} \biggr) \biggr]. \end{aligned} \end{aligned}
(4)

In this paper, we present some new Hermite–Hadamard inequalities for coordinate convex function by defining two sequences $${F(x,y;n)}$$ and $${H(x,y;n)}$$, which also are generalizations of some existing results. Moreover, we also discuss the monotonicity of the sequences $${F(x,y;n)}$$ and $${H(x,y;n)}$$.

## 2 Main results

In this section, a refinement of the Hermite–Hadamard inequality by defining two sequences $${F(x,y;n)}$$ and $${H(x,y;n)}$$ is presented.

### Theorem 2.1

Let$$f: \Delta \subset \mathbb{R}^{2}\to \mathbb{R}$$be a coordinate convex function on$$\Delta =[a,b]\times [c,d]$$. Then

\begin{aligned} \begin{aligned}[b] f\biggl(\frac{a+b}{2},\frac{c+d}{2} \biggr)&\leq H(x,y;n)\leq \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\leq F(x,y;n) \leq \frac{f(a,c)+f(b,c)+f(a,d)+f(b,d)}{4} \end{aligned} \end{aligned}
(5)

for all$$x\in [a, b]$$, $$y \in [c, d]$$and$$n \in \mathbb{N}$$, where

\begin{aligned} H(x,y;n)&=\frac{1}{2^{n+1}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{b-a} \int _{a}^{b} f \biggl(x,c+i\frac{d-c}{2^{n}}- \frac{d-c}{2^{n+1}} \biggr)\,dx\\ &\quad {}+ \frac{1}{d-c} \int _{c}^{d} f \biggl(a+i\frac{b-a}{2^{n}}- \frac{b-a}{2^{n+1}},y \biggr)\,dy \biggr] \end{aligned}

and

\begin{aligned} &F(x,y;n)\\ &\quad =\frac{1}{2^{n+2}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{b-a} \int _{a}^{b} \biggl[ f \biggl(x,\biggl(1- \frac{i}{2^{n}}\biggr)c+\frac{i}{2^{n}}d \biggr)+f \biggl(x,\biggl(1- \frac{i-1}{2^{n}}\biggr)c+\frac{i-1}{2^{n}}d \biggr) \biggr]\,dx \\ &\qquad{}+ \frac{1}{d-c} \int _{c}^{d} \biggl[ f \biggl(\biggl(1- \frac{i}{2^{n}}\biggr)a+\frac{i}{2^{n}}b,y \biggr)+f \biggl(\biggl(1- \frac{i-1}{2^{n}}\biggr)a+\frac{i-1}{2^{n}}b,y \biggr) \biggr]\,dy \biggr]. \end{aligned}

### Proof

Since f is coordinate convex on $$\Delta =[a,b]\times [c,d]$$, its partial mapping $$g_{x}(y)=f(x,y)$$ is convex on $$[c,d]$$ for all $$x\in [a,b]$$, and so, applying (1) to $$g_{x}(y)$$,

\begin{aligned} g_{x}\biggl(\frac{c+d}{2}\biggr)\leq \frac{1}{d-c} \int _{c}^{d} g_{x}(y)\,dy\leq \frac{g_{x}(c)+g_{x}(d)}{2}. \end{aligned}
(6)

On the one hand, by (6), we have

\begin{aligned} \begin{aligned}[b] \frac{1}{d-c} \int _{c}^{d} g_{x}(y)\,dy&= \frac{1}{d-c}\sum_{i=1}^{2^{n}} \int _{c+(i-1)\frac{d-c}{2^{n}}}^{c+i\frac{d-c}{2^{n}}}g_{x}(y)\,dy \\ &\leq \frac{1}{2^{n+1}}\sum_{i=1}^{2^{n}} \biggl[g_{x} \biggl(\biggl(1- \frac{i}{2^{n}}\biggr)c+ \frac{i}{2^{n}}d \biggr)+g_{x} \biggl(\biggl(1- \frac{i-1}{2^{n}} \biggr)c+\frac{i-1}{2^{n}}d \biggr) \biggr] \\ &=y(x;n). \end{aligned} \end{aligned}
(7)

On the other hand, by the convexity of $$g_{x}(y)$$, we obtain

\begin{aligned} \begin{aligned}[b] y(x;n)&\leq \frac{1}{2^{n+1}}\sum _{i=1}^{2^{n}} \biggl[\biggl(1- \frac{i}{2^{n}} \biggr)g_{x}(c)+\frac{i}{2^{n}}g_{x}(d)+\biggl(1- \frac{i-1}{2^{n}}\biggr)g_{x}(c)+ \frac{i-1}{2^{n}}g_{x}(d) \biggr] \\ &=\frac{1}{2^{n+1}} \Biggl[g_{x}(c)\sum _{i=1}^{2^{n}}\biggl(2- \frac{i}{2^{n-1}}+ \frac{1}{2^{n}}\biggr)+g_{x}(d)\sum_{i=1}^{2^{n}} \biggl( \frac{i}{2^{n-1}}-\frac{1}{2^{n}}\biggr) \Biggr] \\ &=\frac{g_{x}(c)+g_{x}(d)}{2}. \end{aligned} \end{aligned}
(8)

By (7) and (8), we have

\begin{aligned} \frac{1}{d-c} \int _{c}^{d} g_{x}(y)\,dy\leq y(x;n)\leq \frac{g_{x}(c)+g_{x}(d)}{2}. \end{aligned}
(9)

Integrating both sides of (9) with respect to x on $$[a,b]$$, we have

\begin{aligned} \begin{aligned}[b] &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\quad \leq \frac{1}{2^{n+1}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{b-a} \int _{a}^{b} f \biggl(x,\biggl(1-\frac{i}{2^{n}} \biggr)c+\frac{i}{2^{n}}d \biggr)\,dx\\ &\qquad {}+\frac{1}{b-a} \int _{a}^{b} f \biggl(x,\biggl(1-\frac{i-1}{2^{n}} \biggr)c+\frac{i-1}{2^{n}}d \biggr)\,dx \biggr] \\ &\quad \leq \frac{1}{2} \biggl[\frac{1}{b-a} \int _{a}^{b} f(x,c)\,dx+ \frac{1}{b-a} \int _{a}^{b} f(x,d)\,dx \biggr]. \end{aligned} \end{aligned}
(10)

By a similar process, we can obtain

\begin{aligned} \begin{aligned}[b] &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\quad \leq \frac{1}{2^{n+1}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{d-c} \int _{c}^{d} f \biggl(\biggl(1-\frac{i}{2^{n}} \biggr)a+\frac{i}{2^{n}}b,y \biggr)\,dy\\ &\qquad {}+\frac{1}{d-c} \int _{c}^{d} f \biggl(\biggl(1-\frac{i-1}{2^{n}} \biggr)a+\frac{i-1}{2^{n}}b,y \biggr)\,dy \biggr] \\ &\quad \leq \frac{1}{2} \biggl[\frac{1}{d-c} \int _{c}^{d} f(a,y)\,dy+ \frac{1}{d-c} \int _{c}^{d} f(b,y)\,dy \biggr]. \end{aligned} \end{aligned}
(11)

By (10) and (11), we have

\begin{aligned} &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\quad \leq \frac{1}{2^{n+2}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{b-a} \int _{a}^{b} f \biggl(x,\biggl(1-\frac{i}{2^{n}} \biggr)c+\frac{i}{2^{n}}d \biggr)\,dx\\ &\qquad {}+\frac{1}{b-a} \int _{a}^{b} f \biggl(x,\biggl(1-\frac{i-1}{2^{n}} \biggr)c+\frac{i-1}{2^{n}}d \biggr)\,dx \\ &\qquad{}+ \frac{1}{d-c} \int _{c}^{d} f \biggl(\biggl(1-\frac{i}{2^{n}} \biggr)a+ \frac{i}{2^{n}}b,y \biggr)\,dy\\ &\qquad {}+\frac{1}{d-c} \int _{c}^{d} f \biggl(\biggl(1- \frac{i-1}{2^{n}} \biggr)a+\frac{i-1}{2^{n}}b,y \biggr)\,dy \biggr] \\ &\quad =F(x,y;n) \\ &\quad \leq \frac{1}{4} \biggl[\frac{1}{b-a} \int _{a}^{b} f(x,c)\,dx+ \frac{1}{b-a} \int _{a}^{b} f(x,d)\,dx\\ &\qquad {}+\frac{1}{d-c} \int _{c}^{d} f(a,y)\,dy+ \frac{1}{d-c} \int _{c}^{d} f(b,y)\,dy \biggr]. \end{aligned}

Furthermore, by the convexity of $$f(x,y)$$, we have

\begin{aligned}& \frac{1}{b-a} \int _{a}^{b} f(x,c)\,dx\leq \frac{f(a,c)+f(b,c)}{2}, \\& \frac{1}{b-a} \int _{a}^{b} f(x,d)\,dx\leq \frac{f(a,d)+f(b,d)}{2}, \\& \frac{1}{d-c} \int _{c}^{d} f(a,y)\,dy\leq \frac{f(a,c)+f(a,d)}{2}, \\& \frac{1}{d-c} \int _{c}^{d} f(b,y)\,dy\leq \frac{f(b,c)+f(b,d)}{2}. \end{aligned}

Therefore,

\begin{aligned}[b] &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx\\ &\quad \leq F(x,y;n) \leq \frac{f(a,c)+f(b,c)+f(a,d)+f(b,d)}{4}. \end{aligned}
(12)

Moreover, by (1), we have

\begin{aligned} \begin{aligned}[b] \frac{1}{d-c} \int _{c}^{d} g_{x}(y)\,dy&= \frac{1}{d-c}\sum_{i=1}^{2^{n}} \int _{c+(i-1)\frac{d-c}{2^{n}}}^{c+i\frac{d-c}{2^{n}}}g_{x}(y)\,dy \\ &\geq \frac{1}{2^{n}}\sum_{i=1}^{2^{n}}g_{x} \biggl(c+i \frac{d-c}{2^{n}}-\frac{d-c}{2^{n+1}} \biggr) \\ &=x(x;n). \end{aligned} \end{aligned}
(13)

By the convexity of $$g_{x}(y)$$ and Jensen’s inequality, we obtain

\begin{aligned} x(x;n)\geq g_{x} \Biggl[\frac{1}{2^{n}}\sum _{i=1}^{2^{n}} \biggl(c+i \frac{d-c}{2^{n}}- \frac{d-c}{2^{n+1}} \biggr) \Biggr]=g_{x} \biggl( \frac{c+d}{2} \biggr). \end{aligned}
(14)

It follows from (13) and (14) that

\begin{aligned} \frac{1}{d-c} \int _{c}^{d} g_{x}(y)\,dy\geq x(x;n)\geq g_{x} \biggl( \frac{c+d}{2} \biggr). \end{aligned}
(15)

Integrating both sides of (15) with respect to x on $$[a,b]$$, we have

\begin{aligned} \begin{aligned}[b] &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\quad \geq \frac{1}{2^{n}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{b-a} \int _{a}^{b} f \biggl(x,c+i\frac{d-c}{2^{n}}- \frac{d-c}{2^{n+1}} \biggr)\,dx \biggr] \\ &\quad \geq \frac{1}{b-a} \int _{a}^{b} f\biggl(x,\frac{c+d}{2}\biggr)\,dx. \end{aligned} \end{aligned}
(16)

By a similar process, we can obtain

\begin{aligned} &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \end{aligned}
(17)
\begin{aligned} &\quad \geq \frac{1}{2^{n}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{d-c} \int _{c}^{d} f \biggl(a+i\frac{b-a}{2^{n}}- \frac{b-a}{2^{n+1}},y \biggr)\,dy \biggr] \end{aligned}
(18)
\begin{aligned} &\quad \geq \frac{1}{d-c} \int _{c}^{d} f\biggl(\frac{a+b}{2},y\biggr)\,dy. \end{aligned}
(19)

By (16) and (17), we have

\begin{aligned} &\frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\quad \geq \frac{1}{2^{n+1}}\sum_{i=1}^{2^{n}} \biggl[\frac{1}{b-a} \int _{a}^{b} f \biggl(x,c+i\frac{d-c}{2^{n}}- \frac{d-c}{2^{n+1}} \biggr)\,dx\\ &\qquad {}+ \frac{1}{d-c} \int _{c}^{d} f \biggl(a+i\frac{b-a}{2^{n}}- \frac{b-a}{2^{n+1}},y \biggr)\,dy \biggr] \\ &\quad =H(x,y;n) \\ &\quad \geq \frac{1}{2} \biggl[\frac{1}{b-a} \int _{a}^{b} f\biggl(x,\frac{c+d}{2}\biggr)\,dx+ \frac{1}{d-c} \int _{c}^{d} f\biggl(\frac{a+b}{2},y\biggr)\,dy \biggr]. \end{aligned}

Moreover, by the convexity of $$f(x,y)$$, we have

\begin{aligned}& \frac{1}{b-a} \int _{a}^{b} f\biggl(x,\frac{c+d}{2}\biggr)\,dx \geq f\biggl(\frac{a+b}{2}, \frac{c+d}{2}\biggr), \\& \frac{1}{d-c} \int _{c}^{d} f\biggl(\frac{a+b}{2},y\biggr)\,dy \geq f\biggl(\frac{a+b}{2}, \frac{c+d}{2}\biggr). \end{aligned}

Therefore,

\begin{aligned} \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx\geq H(x,y;n) \geq f\biggl( \frac{a+b}{2},\frac{c+d}{2}\biggr). \end{aligned}
(20)

By (12) and (20), we have

\begin{aligned} \begin{aligned} f\biggl(\frac{a+b}{2},\frac{c+d}{2}\biggr)&\leq H(x,y;n)\leq \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\leq F(x,y;n) \leq \frac{f(a,c)+f(b,c)+f(a,d)+f(b,d)}{4}. \end{aligned} \end{aligned}

□

### Remark 2.1

Let $$n = 0$$. Then inequality (5) reduces to (3). Therefore, our Theorem 1.2 is a generalization of Theorem 1.2 of .

In the following, we discuss the monotonicity of $$F(x; y; n)$$ and $$H(x; y; n)$$ which are defined as in Theorem 2.1.

### Theorem 2.2

Let$$f: \Delta \subset \mathbb{R}^{2}\to \mathbb{R}$$be a coordinate convex function on$$\Delta =[a,b]\times [c,d]$$. Then$${F(x,y;n)}$$decreasing, $${H(x,y;n)}$$is increasing and

$$\lim_{n\rightarrow \infty }F(x,y;n)=\lim_{n\rightarrow \infty }H(x,y;n)= \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx.$$

### Proof

On the one hand, we have

\begin{aligned} x(x;n)&=\frac{1}{2^{n}}\sum_{i=1}^{2^{n}}g_{x} \biggl(c+i \frac{d-c}{2^{n}}-\frac{d-c}{2^{n+1}} \biggr) \\ &=\frac{1}{2^{n}}\sum_{i=1}^{2^{n}}g_{x} \biggl(\frac{1}{2} \frac{(2^{n+2}-4i+3)c+(4i-3)d+(2^{n+2}-4i+1)c+(4i-1)d}{2^{n+2}} \biggr) \\ &\leq \frac{1}{2^{n+1}}\sum_{i=1}^{2^{n}}g_{x} \biggl( \frac{(2^{n+2}-4i+3)c+(4i-3)d}{2^{n+2}} \biggr) \\ &\quad +\frac{1}{2^{n+1}}\sum_{i=1}^{2^{n}}g_{x} \biggl( \frac{(2^{n+2}-4i+1)c+(4i-1)d}{2^{n+2}} \biggr). \end{aligned}

Setting $$A=\{1,3,\ldots , 2^{n+1}-1\}$$ and $$B=\{2,4,\ldots , 2^{n+1}\}$$, thus we obtain

\begin{aligned}& \sum_{i=1}^{2^{n}}g_{x} \biggl( \frac{(2^{n+2}-4i+3)c+(4i-3)d}{2^{n+2}} \biggr)=\sum_{A}g_{x} \biggl(\frac{(2^{n+2}-2i+1)c+(2i-1)d}{2^{n+2}} \biggr), \\& \sum_{i=1}^{2^{n}}g_{x} \biggl( \frac{(2^{n+2}-4i+1)c+(4i-1)d}{2^{n+2}} \biggr)=\sum_{B}g_{x} \biggl(\frac{(2^{n+2}-2i+1)c+(2i-1)d}{2^{n+2}} \biggr), \end{aligned}

which implies that

$$x(x;n)\leq \frac{1}{2^{n+1}}\sum_{A\cup B}g_{x} \biggl( \frac{(2^{n+2}-2i+1)c+(2i-1)d}{2^{n+2}} \biggr)=x(x;n+1).$$

Since integration is sign-preserving, we know

$$H(x,y;n)\leq H(x,y;n+1).$$

So $${H(x,y;n)}$$ is increasing.

On the other hand, we have

\begin{aligned} y(x;n+1)&= \frac{1}{2^{n+2}} \Biggl[f(a)+f(b)+2\sum _{i=1}^{2^{n+1}-1}f \biggl[ \biggl(1-\frac{i}{2^{n+1}} \biggr)a+\frac{i}{2^{n+1}}b \biggr] \Biggr] \\ &=\frac{1}{2^{n+2}} \Biggl[f(a)+f(b)+2\sum_{i=1}^{2^{n+1}-1}f \biggl( \frac{(2^{n+1}-i)a+ib}{2^{n+1}} \biggr) \Biggr]. \end{aligned}

Setting $$C =\{2, 4, 6, \dots , 2^{n+1}-2\}$$, we obtain

\begin{aligned} y(x;n+1)&= \frac{1}{2^{n+2}} \biggl[f(a)+f(b)+2\sum _{i\in C}f \biggl( \frac{(2^{n+1}-i)a+ib}{2^{n+1}} \biggr)+2\sum _{i\in A}f \biggl( \frac{(2^{n+1}-i)a+ib}{2^{n+1}} \biggr) \biggr] \\ &=\frac{1}{2^{n+2}} \Biggl[f(a)+f(b)+2\sum_{i=1}^{2^{n}-1}f \biggl( \frac{(2^{n}-i)a+ib}{2^{n}} \biggr) \\ &\quad +2\sum_{i=1}^{2^{n}}f \biggl( \frac{1}{2} \frac{(2^{n}-i)a+ib+(2^{n}-i+1)a+(i-1)b}{2^{n}} \biggr) \Biggr] \\ &\leq \frac{1}{2^{n+2}} \Biggl[f(a)+f(b)+2\sum_{i=1}^{2^{n}-1}f \biggl( \frac{(2^{n}-i)a+ib}{2^{n}} \biggr)+\sum_{i=1}^{2^{n}}f \biggl( \frac{(2^{n}-i)a+ib}{2^{n}} \biggr) \\ &\quad +\sum_{i=1}^{2^{n}}f \biggl( \frac{(2^{n}-i+1)a+(i-1)b}{2^{n}} \biggr) \Biggr] \\ &=\frac{1}{2^{n+1}} \Biggl[f(a)+f(b)+2\sum_{i=1}^{2^{n}-1}f \biggl( \frac{(2^{n}-i)a+ib}{2^{n}} \biggr) \Biggr] \\ &=y(x;n). \end{aligned}

So $${y(x;n)}$$ is decreasing.

Since integration is sign-preserving,we know

$$F(x,y;n)\geq F(x,y;n+1).$$

For the proof of the last assertions, since $$f(x,y)$$ is continuous on $$[a,b]\times [c,d]$$, we use the following well known equalities:

\begin{aligned}& \lim_{n\rightarrow \infty }\frac{b-a}{n}\sum_{i=1}^{n} f \biggl(a+i \frac{b-a}{n},y \biggr)= \int _{a}^{b} f(x,y)\,dx, \\& \lim_{n\rightarrow \infty }\frac{d-c}{n}\sum_{i=1}^{n} f \biggl(x,c+i \frac{d-c}{n} \biggr)= \int _{c}^{d} f(x,y)\,dy. \end{aligned}

So we obtain

$$\lim_{n\rightarrow \infty }F(x,y;n)=\lim_{n\rightarrow \infty }H(x,y;n)= \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx.$$

□

By the above theorems, the following corollary can be easily obtained:

### Corollary 2.1

Let$$f: \Delta =[a,b]\times [c,d] \to \mathbb{R}$$be a coordinate convex on Δ. Then

\begin{aligned} \begin{aligned}[b] &f \biggl(\frac{a+b}{2}, \frac{c+d}{2} \biggr)\\ &\quad \leq H(x,y;0)= \frac{1}{2} \biggl[\frac{1}{b-a} \int _{a}^{b}f \biggl(x, \frac{c+d}{2} \biggr)\,dx+\frac{1}{d-c} \int _{c}^{d}f \biggl(\frac{a+b}{2}, y \biggr)\,dy \biggr] \\ &\quad \leq H(x,y;1) \leq \cdots \leq H(x,y;n) \leq \cdots \\ &\quad \leq \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d}f(x,y)\,dy \,dx \\ &\quad \leq \cdots \leq F(x,y;n) \leq \cdots \leq F(x,y;1) \\ &\quad \leq F(x,y;0)= \frac{1}{4} \biggl[\frac{1}{b-a} \int _{a}^{b}\bigl[f(x,c)+f(x,d)\bigr]\,dx+ \frac{1}{d-c} \int _{c}^{d}\bigl[f(a,y)+f(b,y)\bigr]\,dy \biggr] \\ &\quad \leq \frac{f(a,c)+f(a,d)+f(b,c)+f(b,d)}{4}. \end{aligned} \end{aligned}
(21)

### Remark 2.2

Corollary 2.1 shows that inequalities (21) are better than (3) and (4).

## 3 Conclusions

In this paper, we present some new Hermite–Hadamard inequalities for coordinate convex functions by defining two sequences $${F(x,y;n)}$$ and $${H(x,y;n)}$$,

\begin{aligned} f\biggl(\frac{a+b}{2},\frac{c+d}{2}\biggr)&\leq H(x,y;n)\leq \frac{1}{(b-a)(d-c)} \int _{a}^{b} \int _{c}^{d} f(x,y)\,dy \,dx \\ &\leq F(x,y;n) \leq \frac{f(a,c)+f(b,c)+f(a,d)+f(b,d)}{4}, \end{aligned}

which also are generalizations of some existing results. Moreover, we show the monotonicity of the sequences $${F(x,y;n)}$$ and $${H(x,y;n)}$$ in Theorem 2.2.

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The author provided the questions and gave the proof for all results. He read and approved this manuscript.

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Correspondence to Haisong Cao.

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