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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.

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 [11], A.E. Farissi improved the Hermite–Hadamard inequality as follows:

Theorem 1.1

([11])

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 [7], 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

([7])

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 [19], 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

([19])

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 \({H(x,y;n)}\).

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 [7].

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).

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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Cao, H. A new Hermite–Hadamard type inequality for coordinate convex function. J Inequal Appl 2020, 162 (2020). https://doi.org/10.1186/s13660-020-02428-3

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MSC

  • 26D15

Keywords

  • Hermite–Hadamard’s inequality
  • Convex function
  • Coordinates