# Inequalities for convex and s-convex functions on Δ = [a, b] × [c, d]

## Abstract

In this article, two new lemmas are proved and inequalities are established for co-ordinated convex functions and co-ordinated s-convex functions.

Mathematics Subject Classification (2000): 26D10; 26D15.

## 1. Introduction

Let f : I be a convex function defined on the interval I of real numbers and a <b. The following double inequality;

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

is well known in the literature as Hermite-Hadamard inequality. Both inequalities hold in the reversed direction if f is concave.

In [1], Orlicz defined s-convex function in the second sense as following:

Definition 1. A function f : +, where + = [0, ∞), is said to be s-convex in the second sense if

$f\left(\alpha x+\beta y\right)\le {\alpha }^{s}f\left(x\right)+{\beta }^{s}f\left(y\right)$

for all x, y [0, ∞), α, β ≥ 0 with α + β = 1 and for some fixed s (0, 1]. We denote by${K}_{s}^{2}$the class of all s-convex functions.

Obviously one can see that if we choose s = 1, the above definition reduces to ordinary concept of convexity.

For several results related to above definition we refer readers to [210].

In [11], Dragomir defined convex functions on the co-ordinates as following:

Definition 2. Let us consider the bidimensional interval Δ = [a, b] × [c, d] in 2with a <b, c <d. A function f : Δ → will be called convex on the coordinates if the partial mappings f y : [a, b] → , f y (u) = f(u, y) and f x : [c, d] → , f x (v) = f(x, v) are convex where defined for all y [c, d] and x [a, b]. Recall that the mapping f : Δ → is convex on Δ if the following inequality holds,

$f\left(\lambda x+\left(1-\lambda \right)z,\lambda y+\left(1-\lambda \right)w\right)\le \lambda f\left(x,y\right)+\left(1-\lambda \right)f\left(z,w\right)$

for all (x, y), (z, w) Δ and λ [0, 1].

In [11], Dragomir established the following inequalities of Hadamard-type for co-ordinated convex functions on a rectangle from the plane 2.

Theorem 1. Suppose that f : Δ = [a, b] × [c, d] → is convex on the co-ordinates on Δ. Then one has the inequalities;

(1.1)

The above inequalities are sharp.

Similar results can be found in [1214].

In [13], Alomari and Darus defined co-ordinated s-convex functions and proved some inequalities based on this definition. Another definition for co-ordinated s- convex functions of second sense can be found in [15].

Definition 3. Consider the bidimensional interval Δ = [a, b] × [c, d] in [0, ∞)2with a <b and c <d. The mapping f : Δ → is s-convex on Δ if

$f\left(\lambda x+\left(1-\lambda \right)z,\lambda y+\left(1-\lambda \right)w\right)\le {\lambda }^{s}f\left(x,y\right)+{\left(1-\lambda \right)}^{s}f\left(z,w\right)$

holds for all (x, y), (z, w) Δ with λ [0, 1] and for some fixed s (0, 1].

In [16], Sarıkaya et al. proved some Hadamard-type inequalities for co-ordinated convex functions as following:

Theorem 2. Let f : Δ 2 be a partial differentiable mapping on Δ := [a, b] × [c, d] in 2with a <b and c <d. If$\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|$is a convex function on the co-ordinates on Δ, then one has the inequalities:

$\begin{array}{l}\left|J\right|\le \frac{\left(b-a\right)\left(d-c\right)}{16}\phantom{\rule{2em}{0ex}}\\ ×\frac{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(a,c\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(a,d\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(b,c\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(b,d\right)}{4}\phantom{\rule{2em}{0ex}}\end{array}$
(1.2)

where

$J=\frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)dxdy-A$

and

$A=\frac{1}{2}\left[\frac{1}{\left(b-a\right)}{\int }_{a}^{b}\left[f\left(x,c\right)+f\left(x,d\right)\right]dx+\frac{1}{\left(d-c\right)}{\int }_{c}^{d}\left[f\left(a,y\right)+f\left(b,y\right)\right]dy\right].$

Theorem 3. Let f : Δ 2 be a partial differ entiable mapping on Δ := [a, b] × [c, d] in 2with a <b and c <d. If${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}$, q > 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:

$\begin{array}{c}\left|J\right|\le \frac{\left(b-a\right)\left(d-c\right)}{4{\left(p+1\right)}^{\frac{2}{p}}}\\ ×{\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(b,d\right)}{4}\right]}^{\frac{1}{q}}\end{array}$
(1.3)

where A, J are as in Theorem 2 and$\frac{1}{p}+\frac{1}{q}=1$.

Theorem 4. Let f : Δ 2 be a partial differentiable mapping on Δ := [a, b] × [c, d] in 2with a <b and c <d. If${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}$, q ≥ 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:

$\begin{array}{c}\left|J\right|\le \frac{\left(b-a\right)\left(d-c\right)}{16}\\ ×{\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}\left(b,d\right)}{4}\right]}^{\frac{1}{q}}\end{array}$
(1.4)

where A, J are as in Theorem 2.

In [17], Barnett and Dragomir proved an Ostrowski-type inequality for double integrals as following:

Theorem 5. Let f : [a, b] × [c, d] → be continuous on [a, b] × [c, d], ${f}_{x,y}^{″}=\frac{{\partial }^{2}f}{\partial x\partial y}$exists on (a, b) × (c, d) and is bounded, that is

${∥{{f}^{″}}_{x,y}∥}_{\infty }=\underset{\left(x,y\right)\in \left(a,b\right)×\left(c,d\right)}{\text{sup}}\left|\frac{{\partial }^{2}f\left(x,y\right)}{\partial x\partial y}\right|<\infty ,$

then we have the inequality;

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\left|\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(s,t\right)dtds-\left(b-a\right)\underset{c}{\overset{d}{\int }}f\left(x,t\right)dt-\left(d-c\right)\underset{a}{\overset{b}{\int }}f\left(s,y\right)ds-\left(b-a\right)\left(d-c\right)f\left(x,y\right)\right|\phantom{\rule{2em}{0ex}}\\ \le \left[\frac{{\left(b-a\right)}^{2}}{4}+{\left(x-\frac{a+b}{2}\right)}^{2}\right]\left[\frac{{\left(d-c\right)}^{2}}{4}+{\left(y-\frac{c+d}{2}\right)}^{2}\right]{∥{{f}^{″}}_{x,y}∥}_{\infty }\phantom{\rule{2em}{0ex}}\end{array}$
(1.5)

for all (x, y) [a, b] × [c, d].

In [18], Sarıkaya proved an Ostrowski-type inequality for double integrals and gave a corollary as following:

Theorem 6. Let f : [a, b] × [c, d] → be an absolutely continuous function such that the partial derivative of order 2 exists and is bounded, i.e.,

${∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }=\underset{\left(x,y\right)\in \left(a,b\right)×\left(c,d\right)}{\text{sup}}\left|\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}\right|<\infty$

for all (t, s) [a, b] × [c, d]. Then we have,

(1.6)

for all (α1, α2), (β1, β2) [a, b] × [c, d] with α1 <β1, α2 <β2where

$\begin{array}{l}\phantom{\rule{1em}{0ex}}H\left({\alpha }_{1},{\alpha }_{2},{\beta }_{1},{\beta }_{2}\right)\phantom{\rule{2em}{0ex}}\\ =\left({\alpha }_{1}-a\right)\left[\left({\alpha }_{2}-c\right)f\left(a,c\right)+\left(d-{\beta }_{2}\right)f\left(a,d\right)\right]\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(b-{\beta }_{1}\right)\left[\left({\alpha }_{2}-c\right)f\left(b,c\right)+\left(d-{\beta }_{2}\right)f\left(b,d\right)\right]\phantom{\rule{2em}{0ex}}\end{array}$

and

$\begin{array}{l}\phantom{\rule{1em}{0ex}}G\left({\alpha }_{1},{\alpha }_{2},{\beta }_{1},{\beta }_{2}\right)\phantom{\rule{2em}{0ex}}\\ =\left({\beta }_{1}-{\alpha }_{1}\right)\left[\left({\alpha }_{2}-c\right)f\left(\frac{a+b}{2},c\right)+\left(d-{\beta }_{2}\right)f\left(\frac{a+b}{2},d\right)\right]\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left({\beta }_{2}-{\alpha }_{2}\right)\left[\left({\alpha }_{1}-a\right)f\left(a,\frac{c+d}{2}\right)+\left(b-{\beta }_{1}\right)f\left(b,\frac{c+d}{2}\right)\right].\phantom{\rule{2em}{0ex}}\end{array}$

Corollary 1. Under the assumptions of Theorem 6, we have

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\left|\left(b-a\right)\left(d-c\right)f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\right+{\int }_{a}^{b}{\int }_{c}^{d}f\left(t,s\right)dsdt\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\left(d-c\right){\int }_{a}^{b}f\left(t,\frac{c+d}{2}\right)dt-\left(b-a\right)\left|{\int }_{c}^{d}f\left(\frac{a+b}{2},s\right)ds\right|\phantom{\rule{2em}{0ex}}\\ \le \frac{1}{16}{∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }{\left(b-a\right)}^{2}{\left(d-c\right)}^{2}.\phantom{\rule{2em}{0ex}}\end{array}$
(1.7)

In [19], Pachpatte established a new Ostrowski type inequality similar to inequality (1.5) by using elementary analysis.

## 2. Inequalities for co-ordinated convex functions

To prove our main results, we need the following lemma which contains kernels similar to Barnett and Dragomir's kernels in [17], (see the article [17, proof of Theorem 2.1]).

Lemma 1. Let f : Δ = [a, b] × [c, d] → be a partial differentiable mapping on Δ = [a, b] × [c, d]. If$\frac{{\partial }^{2}f}{\partial t\partial s}\in L\left(\Delta \right)$, then the following equality holds:

$\begin{array}{l}\phantom{\rule{1em}{0ex}}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}-\frac{1}{\left(d-c\right)}{\int }_{c}^{d}f\left(\frac{a+b}{2},y\right)dy-\frac{1}{\left(b-a\right)}{\int }_{a}^{b}f\left(x,\frac{c+d}{2}\right)dx\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)dydx\phantom{\rule{2em}{0ex}}\\ =\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}p\left(x,t\right)q\left(y,s\right)\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dsdt\phantom{\rule{2em}{0ex}}\end{array}$

where

and

for each x [a, b] and y [c, d].

Proof. We note that

$B={\int }_{a}^{b}{\int }_{c}^{d}p\left(x,t\right)q\left(y,s\right)\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dsdt.$

Integration by parts, we can write

$\begin{array}{l}B={\int }_{c}^{d}q\left(y,s\right)\left[{\int }_{a}^{\frac{a+b}{2}}\left(t-a\right)\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dt\\ +{\int }_{\frac{a+b}{2}}^{b}\left(t-b\right)\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dt\right]ds\\ ={{\int }_{c}^{d}q\left(y,s\right)\left\{\left[\left(t-a\right)\frac{\partial f}{\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right]}_{a}^{\frac{a+b}{2}}\\ -{\int }_{a}^{\frac{a+b}{2}}\frac{\partial f}{\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dt\\ +{\left[\left(t-b\right)\frac{\partial f}{\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right]}_{\frac{a+b}{2}}^{b}\\ -{\int }_{\frac{a+b}{2}}^{b}\frac{\partial f}{\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dt\right\}ds\\ =\left(b-a\right){\int }_{c}^{d}q\left(y,s\right)\left\{\frac{\partial f}{\partial s}\left(\frac{a+b}{2},\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\\ -{\int }_{a}^{b}\frac{\partial f}{\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dt\right\}ds\\ =\left(b-a\right)\left\{{\int }_{c}^{\frac{c+d}{2}}\left(s-c\right)\frac{\partial f}{\partial s}\left(\frac{a+b}{2},\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)ds\\ +{\int }_{\frac{c+d}{2}}^{d}\left(s-d\right)\frac{\partial f}{\partial s}\left(\frac{a+b}{2},\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)ds\\ -{\int }_{a}^{b}\left[{\int }_{c}^{\frac{c+d}{2}}\left(s-c\right)\frac{\partial f}{\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)ds\\ +{\int }_{\frac{c+d}{2}}^{d}\left(s-d\right)\frac{\partial f}{\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)ds\right]dt\right\}.\end{array}$

By calculating the above integrals, we have

$\begin{array}{c}B=\left(b-a\right)\left(d-c\right)f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ -\left(b-a\right){\int }_{c}^{d}f\left(\frac{a+b}{2},\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)ds\\ -\left(d-c\right){\int }_{a}^{b}f\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{c+d}{2}\right)dt\\ {\int }_{a}^{b}{\int }_{c}^{d}f\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)dsdt.\end{array}$

Using the change of the variable $x=\frac{b-t}{b-a}a+\frac{t-a}{b-a}b$ and $y=\frac{d-s}{d-c}c+\frac{s-c}{d-c}d$, then dividing both sides with (b - a) × (d - c), this completes the proof.

Theorem 7. Let f : Δ = [a, b] × [c, d] → be a partial differentiable mapping on Δ = [a, b] × [c, d]. If$\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|$is a convex function on the co-ordinates on Δ, then the following inequality holds;

Proof. We note that

$\begin{array}{ll}\hfill C& =f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-\frac{1}{\left(d-c\right)}{\int }_{c}^{d}f\left(\frac{a+b}{2},y\right)dy-\frac{1}{\left(b-a\right)}{\int }_{a}^{b}f\left(x,\frac{c+d}{2}\right)dx\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)dydx.\phantom{\rule{2em}{0ex}}\end{array}$

From Lemma 1 and using the property of modulus, we have

$\begin{array}{l}\left|C\right|\le \frac{1}{\left(b-a\right)\left(d-c\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\int }_{a}^{b}{\int }_{c}^{d}\left|p\left(x,t\right)q\left(y,s\right)\right|\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|dsdt\phantom{\rule{2em}{0ex}}\end{array}$

Since $\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|$ is co-ordinated convex, we can write

By computing these integrals, we obtain

Using co-ordinated convexity of $\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|$ again, we get

$\begin{array}{l}|C|\le \frac{\left(b-a}{8\left(d-c\right)}\\ ×\left\{{\int }_{c}^{\frac{c+d}{2}}\left(s-c\right)\left[\frac{d-s}{d-c}|\frac{{\partial }^{2}f}{\partial r\partial s}\left(a,c\right)|\right]ds+{\int }_{c}^{\frac{c+d}{2}}\left(s-c\right)\left[\frac{s-c}{d-c}|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)|\right]ds\\ +{\int }_{\frac{c+d}{2}}^{d}\left(d-s\right)\left[\frac{d-s}{d-c}|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)|\right]ds+{\int }_{\frac{c+d}{2}}^{d}\left(d-s\right)\left[\frac{s-c}{d-c}|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)|\right]ds\\ +{\int }_{c}^{\frac{c+d}{2}}\left(s-c\right)\left[\frac{d-s}{d-c}|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)|\right]ds+{\int }_{d}^{\frac{c+d}{2}}\left(s-c\right)\left[\frac{s-c}{d-c}|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)|\right]ds\\ +{\int }_{\frac{c+d}{2}}^{d}\left(d-s\right)\left[\frac{d-s}{d-c}|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)|\right]ds+{\int }_{\frac{c+d}{2}}^{d}\left(d-s\right)\left[\frac{s-c}{d-c}|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)|\right]ds\right\}.\end{array}$

By a simple computation, we get the required result.

Remark 1. Suppose that all the assumptions of Theorem 7 are satisfied. If we choose$\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|$is bounded, i.e.,

${∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }=\underset{\left(t,s\right)\in \left(a,b\right)×\left(c,d\right)}{\text{sup}}\left|\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}\right|<\infty ,$

we get

$\left|C\right|\le \frac{\left(b-a\right)\left(d-c\right)}{16}{∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }$
(2.1)

which is the inequality in (1.7).

Theorem 8. Let f : Δ = [a, b] × [c, d] → bea partial differentiable mapping on Δ = [a, b] × [c, d]. If${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}$, q > 1, is a convex function on the co-ordinates on Δ, then the following inequality holds;

$\begin{array}{ll}\hfill \left|C\right|& \le \frac{\left(b-a\right)\left(d-c\right)}{4{\left(p+1\right)}^{\frac{2}{p}}}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)\right|}^{q}}{4}\right]}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\end{array}$
(2.2)

where C is in the proof of Theorem 7.

Proof. From Lemma 1, we have

$\begin{array}{ll}\hfill \left|C\right|& \le \frac{1}{\left(b-a\right)\left(d-c\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}x{\int }_{a}^{b}{\int }_{c}^{d}\left|p\left(x,t\right)q\left(y,s\right)\right|\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|dsdt.\phantom{\rule{2em}{0ex}}\end{array}$

By applying the well-known Hölder inequality for double integrals, then one has

$\begin{array}{ll}\hfill \left|C\right|& \le \frac{1}{\left(b-a\right)\left(d-c\right)}\left\{\left({\int }_{a}^{b}{{\int }_{c}^{d}\left|p\left(x,t\right)q\left(y,s\right)\right|}^{p}dtds\right)\right\\frac{1}{p}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\left({\int }_{a}^{b}{\int }_{c}^{d}{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|}^{q}dsdt\right)}^{\frac{1}{q}}.\phantom{\rule{2em}{0ex}}\end{array}$
(2.3)

Since ${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}$ is co-ordinated convex function on Δ, we can write

$\begin{array}{l}\phantom{\rule{1em}{0ex}}{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \le \left(\frac{b-t}{b-a}\right)\left(\frac{d-s}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(\frac{b-t}{b-a}\right)\left(\frac{s-c}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(\frac{t-a}{b-a}\right)\left(\frac{d-s}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(\frac{t-a}{b-a}\right)\left(\frac{s-c}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)\right|}^{q}.\phantom{\rule{2em}{0ex}}\end{array}$
(2.4)

Using the inequality (2.4) in (2.3), we get

$\begin{array}{ll}\hfill \left|C\right|& \le \frac{\left(b-a\right)\left(d-c\right)}{4{\left(p+1\right)}^{\frac{2}{p}}}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}×{\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)\right|}^{q}}{4}\right]}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\end{array}$

where we have used the fact that

${\left({\int }_{a}^{b}{\int }_{c}^{d}{\left|p\left(x,t\right)q\left(y,s\right)\right|}^{p}dtds\right)}^{\frac{1}{p}}=\frac{{\left[\left(b-a\right)\left(d-c\right)\right]}^{1+\frac{1}{p}}}{4{\left(p+1\right)}^{\frac{2}{p}}}.$

This completes the proof.

Remark 2. Suppose that all the assumptions of Theorem 8 are satisfied. If we choose$\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|$is bounded, i.e.,

${∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }=\underset{\left(t,s\right)\in \left(a,b\right)×\left(c,d\right)}{\text{sup}}\left|\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}\right|<\infty ,$

we get

$\left|C\right|\le \frac{\left(b-a\right)\left(d-c\right)}{4{\left(p+1\right)}^{\frac{2}{p}}}{∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }$
(2.5)

which is the inequality in (1.3) with ${∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }.$

Theorem 9. Let f : Δ = [a, b] × [c, d] → bea partial differentiable mapping on Δ = [a, b] × [c, d]. If${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}$, q > 1, is a convex function on the co-ordinates on Δ, then the following inequality holds;

$\begin{array}{l}\left|C\right|\le \frac{\left(b-a\right)\left(d-c\right)}{16}\phantom{\rule{2em}{0ex}}\\ ×{\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)\right|}^{q}+{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)\right|}^{q}}{4}\right]}^{\frac{1}{q}}\phantom{\rule{2em}{0ex}}\end{array}$
(2.6)

where C is in the proof of Theorem 7.

Proof. From Lemma 1 and applying the well-known Power mean inequality for double integrals, then one has

$\begin{array}{l}\phantom{\rule{1em}{0ex}}\left|C\right|\le \frac{1}{\left(b-a\right)\left(d-c\right)}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×{\int }_{a}^{b}{\int }_{c}^{d}\left|p\left(x,t\right)q\left(y,s\right)\right|\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|dsdt\phantom{\rule{2em}{0ex}}\\ \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\left({\int }_{a}^{b}{\int }_{c}^{d}\left|p\left(x,t\right)q\left(y,s\right)\right|dsdt\right)}^{1-\frac{1}{q}}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}×{\left[\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}\left|p\left(x,t\right)q\left(y,s\right)\right|{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|}^{q}dsdt\right]}^{\frac{1}{q}}.\phantom{\rule{2em}{0ex}}\end{array}$
(2.7)

Since ${\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|}^{q}$ is co-ordinated convex function on Δ, we can write

$\begin{array}{l}\phantom{\rule{1em}{0ex}}{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \le \left(\frac{b-t}{b-a}\right)\left(\frac{d-s}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(\frac{b-t}{b-a}\right)\left(\frac{s-c}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(\frac{t-a}{b-a}\right)\left(\frac{d-s}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)\right|}^{q}\phantom{\rule{2em}{0ex}}\\ \phantom{\rule{1em}{0ex}}+\left(\frac{t-a}{b-a}\right)\left(\frac{s-c}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)\right|}^{q}.\phantom{\rule{2em}{0ex}}\end{array}$
(2.8)

If we use (2.8) in (2.7), we get

Computing the above integrals and using the fact that

${\left({\int }_{a}^{b}{\int }_{c}^{d}\left|p\left(x,t\right)q\left(y,s\right)\right|dtds\right)}^{1-\frac{1}{q}}={\left(\frac{{\left(b-a\right)}^{2}{\left(d-c\right)}^{2}}{16}\right)}^{1-\frac{1}{q}},$

we obtained the desired result.

## 3. Inequalities for co-ordinated s-convex functions

To prove our main results we need the following lemma:

Lemma 2. Let f : Δ 2 be an absolutely continuous function on Δ where a <b, c <d and t, λ [0, 1], if$\frac{{\partial }^{2}f}{\partial t\partial \lambda }\in L\left(\Delta \right)$, then the following equality holds:

$D=\frac{\left(b-a\right)\left(d-c\right)}{\left({r}_{1}+1\right)\left({r}_{2}+1\right)}×E$

where

$\begin{array}{l}D=\frac{f\left(a,c\right)+{r}_{2}f\left(a,d\right)+{r}_{1}f\left(b,c\right)+{r}_{1}{r}_{2}f\left(b,d\right)}{\left({r}_{1}+1\left({r}_{2}+1\right)}\phantom{\rule{2em}{0ex}}\\ +\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_{a}^{b}{\int }_{c}^{d}f\left(x,y\right)dxdy\phantom{\rule{2em}{0ex}}\\ -\left(\frac{{r}_{1}}{{r}_{1}+1}\right)\frac{1}{d-c}{\int }_{c}^{d}f\left(b,y\right)dy-\left(\frac{1}{{r}_{1}+1}\right)\frac{1}{d-c}{\int }_{c}^{d}f\left(a,y\right)dy\phantom{\rule{2em}{0ex}}\\ -\left(\frac{{r}_{2}}{{r}_{2}+1}\right)\frac{1}{b-a}{\int }_{a}^{b}f\left(x,d\right)dx-\left(\frac{1}{{r}_{2}+1}\right)\frac{1}{b-a}{\int }_{a}^{b}f\left(x,c\right)dx\phantom{\rule{2em}{0ex}}\end{array}$

and

$E={\int }_{0}^{1}{\int }_{0}^{1}\left(\left({r}_{1}+1\right)t-1\right)\left(\left({r}_{2}+1\right)\lambda -1\right)\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)dtd\lambda$

for some fixed r1, r2 [0, 1].

Proof. Integration by parts, we get

$\begin{array}{l}E={\int }_{0}^{1}\left(\left({r}_{2}+1\right)\lambda -1\right)\\ ×\left[{\int }_{0}^{1}\left(\left({r}_{1}+1\right)t-1\right)\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)dt\right]d\lambda \\ ={\int }_{0}^{1}\left(\left({r}_{2}+1\right)\lambda -1\right)\left[\frac{\left(\left({r}_{1}+1\right)t-1\right)}{\left(b-a\right)}\frac{\partial f}{\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right){|}_{0}^{1}\\ -\frac{{r}_{1}+1}{b-a}{\int }_{0}^{1}\frac{\partial f}{\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)dt\right]d\lambda \\ ={\int }_{0}^{1}\left(\left({r}_{2}+1\right)\lambda -1\left[\frac{{r}_{1}}{b-a}\frac{\partial f}{\partial \lambda }\left(b,\lambda d+\left(1-\lambda \right)c\right)+\frac{1}{b-a}\frac{\partial f}{\partial \lambda }\left(a,\lambda d+\left(1-\lambda \right)c\right)\\ -\frac{{r}_{1}+1}{b-a}{\int }_{0}^{1}\frac{\partial f}{\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)dt\right]d\lambda \\ =\frac{{r}_{1}}{b-a}\frac{\left(\left({r}_{2}+1\right)\lambda -1}{d-c}f\left(b,\lambda d+\left(1-\lambda \right)c\right){|}_{0}^{1}-\frac{{r}_{1}\left({r}_{2}+1\right)}{\left(b-a\right)\left(d-c\right)}{\int }_{0}^{1}f\left(b,\lambda d+\left(1-\lambda \right)c\right)d\lambda \\ +\frac{1}{b-a}\frac{\left(\left({r}_{2}+1\right)\lambda -1}{d-c}f\left(a,\lambda d+\left(1-\lambda \right)c\right){|}_{0}^{1}-\frac{\left({r}_{2}+1\right)}{\left(b-a\right)\left(d-c\right)}{\int }_{0}^{1}f\left(a,\lambda d+\left(1-\lambda \right)c\right)d\lambda \\ -\frac{{r}_{1}+1}{b-a}{\int }_{0}^{1}\left[{\int }_{0}^{1}\left(\left({r}_{2}+1\right)\lambda -1\right)\frac{\partial f}{\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)d\lambda \right]dt.\end{array}$

Computing these integrals, we obtain

Using the change of the variable x = tb + (1 - t) a and y = λd + (1 - λ) c for t, λ [0, 1] and multiplying the both sides by $\frac{\left(b-a\right)\left(d-c\right)}{\left({r}_{1}+1\left({r}_{2}+1\right)}$, we get the required result.

Theorem 10. Let f : Δ = [a, b] × [c, d] [0, ∞)2 → [0, ∞) be an absolutely continuous function on Δ. If$\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|$is s-convex function on the co-ordinates on Δ, then one has the inequality:

(3.1)

where

$\begin{array}{c}M=\left(s+1+2\left({r}_{1}+1\right){\left(\frac{{r}_{1}}{{r}_{1}+1}\right)}^{s+2}-{r}_{1}\right)\\ N=\frac{1}{{\left({r}_{2}+1\right)}^{s+1}}\\ L={r}_{2}\left(s+1\right)+\frac{1}{{\left({r}_{2}+1\right)}^{s+1}}-1\\ R=s+1+{r}_{2}{\left(\frac{{r}_{2}}{{r}_{2}+1}\right)}^{s+1}-{r}_{2}\\ S={r}_{2}{\left(\frac{{r}_{2}}{{r}_{2}+1}\right)}^{s+1}\end{array}$

Proof. From Lemma 2 and by using co-ordinated s-convexity of $\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|$, we have;

By calculating the above integrals, we have

(3.2)

By a similar argument for other integrals, by using co-ordinated s-convexity of $\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|$, we get

By using these in (3.2), we obtain the inequality (3.1).

Corollary 2

1. (1)

If we choose r 1 = r 2 = 1 in (3.1), we have

(3.3)
2. (2)

If we choose r 1 = r 2 = 0 in (3.1), we have

Theorem 11. Let f : Δ = [a, b] × [c, d] [0, ∞)2 → [0, ∞) be an absolutely continuous function on Δ. If${\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^{\frac{p}{p-1}}$is s-convex function on the co-ordinates on Δ, for some fixed s (0, 1] and p > 1, then one has the inequality:

$\begin{array}{c}\left|D\right|\le \frac{\left(b-a\right)\left(d-c\right)}{\left({r}_{1}+1\right)\left({r}_{2}+1\right)}\frac{{\left(1+{r}_{1}^{p+1}\right)}^{\frac{1}{p}}{\left(1+{r}_{2}^{p+1}\right)}^{\frac{1}{p}}}{{\left({r}_{1}+1\right)}^{\frac{1}{p}}{\left({r}_{2}+1\right)}^{\frac{1}{p}}{\left(p+1\right)}^{\frac{2}{p}}}\\ ×{\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^{q}\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^{q}\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^{q}\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^{q}\left(b,d\right)}{{\left(s+1\right)}^{2}}\right]}^{\frac{1}{q}}\end{array}$
(3.4)

for some fixed r1, r2 [0, 1], where$q=\frac{p}{p-1}$.

Proof. From Lemma 2 and using the Hölder inequality for double integrals, we can write

$\begin{array}{l}|D|\le \frac{\left(b-a\right)\left(d-c\right)}{\left({r}_{1}+1\right)\left({r}_{2}+1\right)}{\left({\int }_{0}^{1}{\int }_{0}^{1}{|\left(\left({r}_{1}+1\right)t-1\right)\left(\left({r}_{2}+1\right)\lambda -1\right)|}^{p}dtd\lambda \right)}^{\frac{1}{p}}\\ ×{\left({\int }_{0}^{1}{\int }_{0}^{1}{|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c|}^{q}dtd\lambda \right)}^{\frac{1}{q}}.\end{array}$

In above inequality using the s-convexity on the co-ordinates of ${\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^{q}$ on Δ and calculating the integrals, then we get the desired result.

Corollary 3

1. (1)

Under the assumptions of Theorem 11, if we choose r 1 = r 2 = 1 in (3.4), we have

(3.5)
2. (2)

Under the assumptions of Theorem 11, if we choose r 1 = r 2 = 0 in (3.4), we have