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 ⊆ ℝ → ℝ 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 : ℝ⁺ → ℝ, where ℝ⁺ = [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 [2–10].

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 ℝ²with a <b, c <d. A function f : Δ → ℝ will be called convex on the coordinates if the partial mappings f _y : [a, b] → ℝ, f _y (u) = f(u, y) and f _x : [c, d] → ℝ, 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 → ℝ 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 ℝ².

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

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

(1.1)

The above inequalities are sharp.

Similar results can be found in [12–14].

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 → ℝ 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 ⊂ ℝ² → ℝ be a partial differentiable mapping on Δ := [a, b] × [c, d] in ℝ²with 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:

$$\begin{array}{l}\left|J\right|\le \frac{\left(b-a\right)\left(d-c\right)}{16}\\ \times \frac{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(a,c\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(a,d\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(b,c\right)+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\right|\left(b,d\right)}{4}\end{array}$$

(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 ⊂ ℝ² → ℝ be a partial differ entiable mapping on Δ := [a, b] × [c, d] in ℝ²with 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:

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

(1.3)

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

Theorem 4. Let f : Δ ⊂ ℝ² → ℝ be a partial differentiable mapping on Δ := [a, b] × [c, d] in ℝ²with 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:

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

(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] → ℝ 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

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

then we have the inequality;

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

(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] → ℝ be an absolutely continuous function such that the partial derivative of order 2 exists and is bounded, i.e.,

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

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

$$\begin{array}{l}\left|\left({\beta }_1-{\alpha }_1\right)\left({\beta }_2-{\alpha }_2\right)f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\right.+H\left({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\right)+G\left({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\right)\\ -\left({\beta }_2-{\alpha }_2\right){\int }_a^{b}f\left(t,\frac{c+d}{2}\right)dt-\left({\beta }_1-{\alpha }_1\right){\int }_c^{d}f\left(\frac{a+b}{2},s\right)ds\\ -{\int }_a^b\left[\left({\alpha }_2-c\right)f\left(t,c\right)+\left(d-{\beta }_2\right)f\left(t,d\right)\right]dt\\ \left.-{\int }_c^d\left[\left({\alpha }_1-a\right)f\left(a,s\right)+\left(b-{\beta }_1\right)f\left(b,s\right)\right]ds+{\int }_a^b{\int }_c^{d}f\left(t,s\right)dsdt\right|\\ \le \left[\frac{{\left({\alpha }_1-a\right)}^2+{\left(b-{\beta }_1\right)}^2}{2}+\frac{{\left(a+b-2{\alpha }_1\right)}^2+{\left(a+b-2{\beta }_1\right)}^2}{8}\right]\\ \times \left[\frac{{\left({\alpha }_2-c\right)}^2+{\left(d-{\beta }_2\right)}^2}{2}+\frac{{\left(c+d-2{\alpha }_2\right)}^2+{\left(c+d-2{\beta }_2\right)}^2}{8}\right]{∥\frac{{\partial }^{2}f\left(t,s\right)}{\partial t\partial s}∥}_{\infty }\end{array}$$

(1.6)

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

$$\begin{array}{l}H\left({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\right)\\ =\left({\alpha }_1-a\right)\left[\left({\alpha }_2-c\right)f\left(a,c\right)+\left(d-{\beta }_2\right)f\left(a,d\right)\right]\\ +\left(b-{\beta }_1\right)\left[\left({\alpha }_2-c\right)f\left(b,c\right)+\left(d-{\beta }_2\right)f\left(b,d\right)\right]\end{array}$$

and

$$\begin{array}{l}G\left({\alpha }_1,{\alpha }_2,{\beta }_1,{\beta }_2\right)\\ =\left({\beta }_1-{\alpha }_1\right)\left[\left({\alpha }_2-c\right)f\left(\frac{a+b}{2},c\right)+\left(d-{\beta }_2\right)f\left(\frac{a+b}{2},d\right)\right]\\ +\left({\beta }_2-{\alpha }_2\right)\left[\left({\alpha }_1-a\right)f\left(a,\frac{c+d}{2}\right)+\left(b-{\beta }_1\right)f\left(b,\frac{c+d}{2}\right)\right].\end{array}$$

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

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

(1.7)

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

The main purpose of this article is to establish inequalities of Hadamard-type for co-ordinated convex functions by using Lemma 1 and to establish some new Hadamard-type inequalities for co-ordinated s-convex functions by using Lemma 2.

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] → ℝ 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:

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

where

$$p\left(x,t\right)=\left\{\begin{array}{cc}\hfill \left(t-a\right),\hfill & \hfill t\in \left[a,\frac{a+b}{2}\right]\hfill \\ \hfill \left(t-b\right),\hfill & \hfill t\in \left(\left.\frac{a+b}{2},b\right]\right.\hfill \end{array}\right.$$

and

$$q\left(y,s\right)=\left\{\begin{array}{cc}\hfill \left(s-c\right),\hfill & \hfill s\in \left[c,\frac{c+d}{2}\right]\hfill \\ \hfill \left(s-d\right),\hfill & \hfill s\in \left(\left.\frac{c+d}{2},d\right]\right.\hfill \end{array}\right.$$

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={\displaystyle {\int }_c^{d}q(y,s)[{\displaystyle {\int }_a^{\frac{a+b}{2}}(t-a)\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}\\ +{\displaystyle {\int }_{\frac{a+b}{2}}^b(t-b)\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]ds\\ ={{\displaystyle {\int }_c^{d}q(y,s)\{\left[(t-a)\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}}\\ -{\displaystyle {\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[(t-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)\right]}_{\frac{a+b}{2}}^b\\ -{\displaystyle {\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}\}ds\\ =(b-a){\displaystyle {\int }_c^{d}q(y,s)\{\frac{\partial f}{\partial s}\left(\frac{a+b}{2},\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)}\\ -{\displaystyle {\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\}ds\\ =(b-a)\{{\displaystyle {\int }_c^{\frac{c+d}{2}}(s-c)\frac{\partial f}{\partial s}\left(\frac{a+b}{2},\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)ds}\\ +{\displaystyle {\int }_{\frac{c+d}{2}}^d(s-d)\frac{\partial f}{\partial s}\left(\frac{a+b}{2},\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)}ds\\ -{\displaystyle {\int }_a^b[{\displaystyle {\int }_c^{\frac{c+d}{2}}(s-c)\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}}\\ +{\displaystyle {\int }_{\frac{c+d}{2}}^d(s-d)\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]dt\}.\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] → ℝ 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;

$$\begin{array}{l}\left|f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\right.-\frac{1}{\left(d-c\right)}{\int }_c^{d}f\left(\frac{a+b}{2},y\right)dy-\frac{1}{\left(b-a\right)}{\int }_a^{b}f\left(x,\frac{c+d}{2}\right)dx\\ \left.+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dydx\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{64}\left[\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)\right|+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)\right|+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)\right|+\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)\right|\right].\end{array}$$

Proof. We note that

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

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

$$\begin{array}{l}\left|C\right|\le \frac{1}{\left(b-a\right)\left(d-c\right)}\\ \times {\int }_a^b{\int }_c^d\left|p\left(x,t\right)q\left(y,s\right)\right|\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|dsdt\end{array}$$

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

$$\begin{array}{ll}\hfill \left|C\right|& \le \frac{1}{\left(b-a\right)\left(d-c\right)}\\ \times {\int }_c^d\left|q\left(y,s\right)\right|\left\{{\int }_a^{\frac{a+b}{2}}\left(t-a\right)\left[\frac{b-t}{b-a}\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|\right]\right.dt\\ +{\int }_a^{\frac{a+b}{2}}\left(t-a\right)\left[\frac{t-a}{b-a}\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|\right]dt\\ +{\int }_{\frac{a+b}{2}}^b\left(b-t\right)\left[\frac{b-t}{b-a}\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|\right]dt\\ \left.+{\int }_{\frac{a+b}{2}}^b\left(b-t\right)\left[\frac{t-a}{b-a}\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|\right]dt\right\}ds.\end{array}$$

By computing these integrals, we obtain

$$\begin{array}{ll}\hfill \left|C\right|& \le \frac{\left(b-a\right)}{8\left(d-c\right)}\left[{\int }_c^d\left|q\left(y,s\right)\right|\right.\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|\\ \left.+{\int }_c^d\left|q\left(y,s\right)\right|\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|\right]ds.\end{array}$$

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

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

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.,

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

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] → ℝ 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;

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

(2.2)

where C is in the proof of Theorem 7.

Proof. From Lemma 1, we have

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

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

$$\begin{array}{ll}\hfill \left|C\right|& \le \frac{1}{\left(b-a\right)\left(d-c\right)}\left\{\left({\int }_a^b{{\int }_c^d\left|p\left(x,t\right)q\left(y,s\right)\right|}^{p}dtds\right)\right.\frac{1}{p}\\ \times {\left({\int }_a^b{\int }_c^d{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|}^{q}dsdt\right)}^{\frac{1}{q}}.\end{array}$$

(2.3)

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

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

(2.4)

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

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

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.,

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

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] → ℝ 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;

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

(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

$$\begin{array}{l}\left|C\right|\le \frac{1}{\left(b-a\right)\left(d-c\right)}\\ \times {\int }_a^b{\int }_c^d\left|p\left(x,t\right)q\left(y,s\right)\right|\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|dsdt\\ \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\left({\int }_a^b{\int }_c^d\left|p\left(x,t\right)q\left(y,s\right)\right|dsdt\right)}^{1-\frac{1}{q}}\\ \times {\left[\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}\left|p\left(x,t\right)q\left(y,s\right)\right|{\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(\frac{b-t}{b-a}a+\frac{t-a}{b-a}b,\frac{d-s}{d-c}c+\frac{s-c}{d-c}d\right)\right|}^{q}dsdt\right]}^{\frac{1}{q}}.\end{array}$$

(2.7)

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

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

(2.8)

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

$$\begin{array}{ll}\hfill \left|C\right|& \le \frac{1}{\left(b-a\right)\left(d-c\right)}{\left\{\left({\int }_a^b{\int }_c^d\left|p\left(x,t\right)q\left(y,s\right)\right|dsdt\right)\right.}^{1-\frac{1}{q}}\\ \times {\int }_a^b{\int }_c^d\left|p\left(x,t\right)q\left(y,s\right)\right|\left[\left.\left(\frac{b-t}{b-a}\right)\left(\frac{d-s}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,c\right)\right|}^q+\left(\frac{b-t}{b-a}\right)\left(\frac{s-c}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(a,d\right)\right|}^q\right)\right.\\ \left.{\left.\left.+\left(\frac{t-a}{b-a}\right)\left(\frac{d-s}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,c\right)\right|}^q+\left(\frac{t-a}{b-a}\right)\left(\frac{s-c}{d-c}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial s}\left(b,d\right)\right|}^q\right]\right)}^{\frac{1}{q}}\right\}.\end{array}$$

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 ⊂ ℝ² → ℝ 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:

$$D=\frac{\left(b-a\right)\left(d-c\right)}{\left(r_1+1\right)\left(r_2+1\right)}\times E$$

where

$$\begin{array}{l}D=\frac{f\left(a,c\right)+r_{2}f\left(a,d\right)+r_{1}f\left(b,c\right)+r_1{r}_{2}f\left(b,d\right)}{(r_1+1\left(r_2+1\right)}\\ +\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy\\ -\left(\frac{r_1}{r_1+1}\right)\frac{1}{d-c}{\int }_c^{d}f\left(b,y\right)dy-\left(\frac{1}{r_1+1}\right)\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)dy\\ -\left(\frac{r_2}{r_2+1}\right)\frac{1}{b-a}{\int }_a^{b}f\left(x,d\right)dx-\left(\frac{1}{r_2+1}\right)\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)dx\end{array}$$

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

$$\begin{array}{l}E={\displaystyle {\int }_0^1((r_2+1)\lambda -1)}\\ \times \left[{\displaystyle {\int }_0^1((r_1+1)t-1})\frac{{\partial }^{2}f}{\partial t\partial \lambda }(tb+(1-t)a,\lambda d+(1-\lambda )c)dt\right]d\lambda \\ ={\displaystyle {\int }_0^1((r_2+1)\lambda -1})[\frac{((r_1+1)t-1)}{(b-a)}\frac{\partial f}{\partial \lambda }(tb+(1-t)a,\lambda d+(1-\lambda )c){|}_0^1\\ -\frac{r_1+1}{b-a}{\displaystyle {\int }_0^1\frac{\partial f}{\partial \lambda }(tb+(1-t)a,\lambda d+(1-\lambda )c)dt}]d\lambda \\ ={\displaystyle {\int }_0^1((r_2+1)\lambda -1}[\frac{r_1}{b-a}\frac{\partial f}{\partial \lambda }(b,\lambda d+(1-\lambda )c)+\frac{1}{b-a}\frac{\partial f}{\partial \lambda }(a,\lambda d+(1-\lambda )c)\\ -\frac{r_1+1}{b-a}{\displaystyle {\int }_0^1\frac{\partial f}{\partial \lambda }}(tb+(1-t)a,\lambda d+(1-\lambda )c)dt]d\lambda \\ =\frac{r_1}{b-a}\frac{((r_2+1)\lambda -1}{d-c}f(b,\lambda d+(1-\lambda )c){|}_0^1-\frac{r_1(r_2+1)}{(b-a)(d-c)}{\displaystyle {\int }_0^{1}f}(b,\lambda d+(1-\lambda )c)d\lambda \\ +\frac{1}{b-a}\frac{((r_2+1)\lambda -1}{d-c}f(a,\lambda d+(1-\lambda )c){|}_0^1-\frac{(r_2+1)}{(b-a)(d-c)}{\displaystyle {\int }_0^{1}f(a,\lambda d+(1-\lambda )c)d\lambda }\\ -\frac{r_1+1}{b-a}{\displaystyle {\int }_0^1\left[{\displaystyle {\int }_0^1((r_2+1})\lambda -1)\frac{\partial f}{\partial \lambda }(tb+(1-t)a,\lambda d+(1-\lambda )c)d\lambda \right]dt.}\end{array}$$

Computing these integrals, we obtain

$$\begin{array}{ll}\hfill E& =\frac{1}{\left(b-a\right)\left(d-c\right)}[f\left(a,c\right)+r_{2}f\left(a,d\right)+r_{1}f\left(b,c\right)+r_1{r}_{2}f\left(b,d\right)\\ -r_1\left(r_2+1\right){\int }_0^{1}f\left(b,\lambda d+\left(1-\lambda \right)c\right)d\lambda -\left(r_2+1\right){\int }_0^{1}f\left(a,\lambda d+\left(1-\lambda \right)c\right)d\lambda \\ -r_2\left(r_1+1\right){\int }_0^{1}f\left(tb+\left(1-t\right)a,d\right)dt-\left(r_1+1\right){\int }_0^{1}f\left(tb+\left(1-t\right)a,c\right)dt\\ \left.+\left(r_1+1\right)\left(r_2+1\right){\int }_0^1{\int }_0^{1}f\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)dtd\lambda \right].\end{array}$$

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 $\frac{(b-a)(d-c)}{(r_1+1(r_2+1)}$, 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:

$$\begin{array}{c}D|\le \frac{\left(b-a\right)\left(d-c\right)}{\left(r_1+1\right)\left(r_2+1\right){\left(s+1\right)}^2{\left(s+2\right)}^2}\\ \times \left[MS\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,c\right)\right|+ML\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,d\right)\right|\right.\\ \left.+KR\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|+KN\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|\right]\end{array}$$

(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;

$$\begin{array}{ll}\hfill \left|D\right|& \le \frac{\left(b-a\right)\left(d-c\right)}{\left(r_1+1\right)\left(r_2+1\right)}\\ \times {\int }_0^1{\int }_0^1\left|\left(\left(r_1+1\right)t-1\right)\left(\left(r_2+1\right)\lambda -1\right)\right|\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)\right|dtd\lambda \\ \le \frac{\left(b-a\right)\left(d-c\right)}{\left(r_1+1\right)\left(r_2+1\right)}\\ \times {\int }_0^1\left[{\int }_0^1\left|\left(\left(r_1+1\right)t-1\right)\left(\left(r_2+1\right)\lambda -1\right)\right|\right.\\ \left.\left\{t^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,\lambda d+\left(1-\lambda \right)c\right)\right|+{\left(1-t\right)}^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,\lambda d+\left(1-\lambda \right)c\right)\right|\right\}dt\right]d\lambda .\end{array}$$

By calculating the above integrals, we have

$$\begin{array}{l}{\int }_0^1\left|\left(\left(r_1+1\right)t-1\right)\right|\left\{t^s\right.\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,\lambda d+\left(1-\lambda \right)c\right)\right|\\ \left.+{\left(1-t\right)}^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,\lambda d+\left(1-\lambda \right)c\right)\right|\right\}dt\\ ={\int }_0^{\frac{1}{r_1+1}}\left(1-\left(r_1+1\right)t\right)\left\{t^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,\lambda d+\left(1-\lambda \right)c\right)\right|\right.\\ \left.+{\left(1-t\right)}^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,\lambda d+\left(1-\lambda \right)c\right)\right|\right\}dt\\ +{\int }_{\frac{1}{r_1+1}}^1\left(\left(r_1+1\right)t-1\right)\left\{t^2\right.\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,\lambda d+\left(1-\lambda \right)c\right)\right|\\ \left.+{\left(1-t\right)}^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,\lambda d+\left(1-\lambda \right)c\right)\right|\right\}dt\\ =\frac{1}{\left(s+1\right)\left(s+2\right)}\left[\left(r_1\left(s+1\right)+2{\left(\frac{1}{r_1+1}\right)}^{s+1}-1\right)\right.\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,\lambda d+\left(1-\lambda \right)c\right)\right|\\ \left.+\left(s+1+2\left(r_1+1\right){\left(\frac{r_1}{r_1+1}\right)}^{s+2}-r_1\right)\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,\lambda d+\left(1-\lambda \right)c\right)\right|\right].\end{array}$$

(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

$$\begin{array}{c}{\int }_0^1\left|\left(\left(r_2+1\right)\lambda -1\right)\right|\left\{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,\lambda d+\left(1-\lambda \right)c\right)\right|+\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,\lambda d+\left(1-\lambda \right)c\right)\right|\right\}d\lambda \\ \le {\int }_0^{\frac{1}{r_2+1}}\left(1-\left(r_2+1\right)\lambda \right)\left\{{\lambda }^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|+{\left(1-\lambda \right)}^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|\right\}d\lambda \\ +{\int }_{\frac{1}{r_2+1}}^1\left(\left(r_2+1\right)\lambda -1\right)\left\{{\lambda }^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,d\right)\right|+{\left(1-\lambda \right)}^s\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,c\right)\right|\right\}d\lambda \\ =\frac{1}{\left(s+1\right)\left(s+2\right)}\left\{\frac{1}{{\left(r_2+1\right)}^{s+1}}\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|\right.\\ +\left[r_2\left(s+1\right)+\frac{1}{{\left(r_2+1\right)}^{s+1}}-1\right]\left|\frac{{\partial }^{2}f}{\partial t\partial f}\left(a,d\right)\right|\\ +\left[s-r_2+1+r_2{\left(\frac{r_2}{r_2+1}\right)}^{s+1}\right]\left|\frac{{\partial }^{2}f}{\partial t\partial f}\left(b,c\right)\right|\\ \left.+r_2{\left(\frac{r_2}{r_2+1}\right)}^{s+1}\left|\frac{{\partial }^{2}f}{\partial t\partial f}\left(a,c\right)\right|\right\}.\end{array}$$

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

Corollary 2

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

  $$\begin{array}{l}\left|\frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}\right.-\frac{1}{2}\left[\frac{1}{d-c}{\int }_c^d\left[f\left(b,y\right)+f\left(a,y\right)\right]dy\right]\\ \left.-\frac{1}{2}\left[\frac{1}{b-a}{\int }_a^b\left[f\left(x,d\right)+f\left(x,c\right)\right]dx\right]+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{{4\left(s+1\right)}^2{\left(s+2\right)}^2}{\left(s+\frac{1}{2^s}\right)}^2\\ \times \left[\frac{1}{2^{s+1}}\left(\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,c\right)\right|+\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|\right)+\left(s+\frac{1}{2^{s+1}}\right)\left(\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,d\right)\right|+\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|\right)\right].\end{array}$$

  (3.3)

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

  $$\begin{array}{l}\left|f\left(a,c\right)-\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)dy-\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)dx\right.\\ \left.+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{{\left(s+1\right)}^2{\left(s+2\right)}^2}\\ \times \left[\left(s+1\right)\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|+\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|\right].\end{array}$$

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:

$$\begin{array}{c}\left|D\right|\le \frac{\left(b-a\right)\left(d-c\right)}{\left(r_1+1\right)\left(r_2+1\right)}\frac{{\left(1+r_1^{p+1}\right)}^{\frac{1}{p}}{\left(1+r_2^{p+1}\right)}^{\frac{1}{p}}}{{\left(r_1+1\right)}^{\frac{1}{p}}{\left(r_2+1\right)}^{\frac{1}{p}}{\left(p+1\right)}^{\frac{2}{p}}}\\ \times {\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,d\right)}{{\left(s+1\right)}^2}\right]}^{\frac{1}{q}}\end{array}$$

(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

$$\begin{array}{l}\left|D\right|\le \frac{(b-a)(d-c)}{(r_1+1)(r_2+1)}{\left({\displaystyle {\int }_0^1{\displaystyle {\int }_0^1{\left|((r_1+1)t-1)((r_2+1)\lambda -1)\right|}^p}dtd\lambda }\right)}^{\frac{1}{p}}\\ \times {\left({\displaystyle {\int }_0^1{\displaystyle {\int }_0^1{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }(tb+(1-t)a,\lambda d+(1-\lambda )c\right|}^q}dtd\lambda }\right)}^{\frac{1}{q}}.\end{array}$$

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

  $$\begin{array}{l}\left|\frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}\right.\\ -\frac{1}{2}\left[\frac{1}{d-c}{\int }_c^d\left[f\left(b,y\right)+f\left(a,y\right)\right]dy+\frac{1}{b-a}{\int }_a^b\left[f\left(x,d\right)+f\left(x,c\right)\right]dx\right]\\ \left.+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{4{\left(p+1\right)}^{\frac{2}{p}}}\\ \times {\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,d\right)}{{\left(s+1\right)}^2}\right]}^{\frac{1}{q}}.\end{array}$$

  (3.5)

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

  $$\begin{array}{l}\left|f\left(a,c\right)-\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)dy-\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)dx\right.\\ \left.+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy\right|\\ =\frac{\left(b-a\right)\left(d-c\right)}{{\left(p+1\right)}^{\frac{2}{p}}}\\ \times {\left[\frac{{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,d\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,c\right)+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,d\right)}{{\left(s+1\right)}^2}\right]}^{\frac{1}{q}}.\end{array}$$

Remark 4. If we choose s = 1 in (3.5), we obtain the inequality in (1.3)

Theorem 12. Let f : Δ = [a, b] × [c, d] ⊂ [0, ∞)² → [0, ∞) be an absolutely continuous function on Δ. If${\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q$is s-convex function on the co-ordinates on Δ, for some fixed s ∈ (0, 1] and q ≥ 1, then one has the inequality:

$$\begin{array}{c}\left|D\right|\le \frac{\left(b-a\right)\left(d-c\right)}{\left(r_1+1\right)\left(r_2+1\right)}{\left(\frac{\left(1+r_1^2\right)\left(1+r_2^2\right)}{4\left(r_1+1\right)\left(r_2+1\right)}\right)}^{1-\frac{1}{q}}\\ \times {\left(\frac{MS{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,c\right)\right|}^q+ML{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,d\right)\right|}^q+KR{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|}^q+KN{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|}^q}{{\left(s+1\right)}^2{\left(s+2\right)}^2}\right)}^{\frac{1}{q}}\end{array}$$

for some fixed r₁, r₂ ∈ [0, 1].

Proof. From Lemma 2 and using the well-known Power-mean inequality, we can write

$$\begin{array}{l}\left|D\right|\le \frac{\left(b-a\right)\left(b-c\right)}{\left(r_1+1\right)\left(r_2+1\right)}{\left({\int }_0^1{\int }_0^1\left|\left(\left(r_1+1\right)t-1\right)\left(\left(r_2+1\right)\lambda -1\right)\right|dtd\lambda \right)}^{1-\frac{1}{q}}\\ \times {\left[{\int }_0^1{\int }_0^1\left|\left(\left(r_1+1\right)t-1\right)\left(\left(r_2+1\right)\lambda -1\right)\right|{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(tb+\left(1-t\right)a,\lambda d+\left(1-\lambda \right)c\right)\right|}^{q}dtd\lambda \right]}^{\frac{1}{q}}.\end{array}$$

Since ${\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q$ is s-convex function on the co-ordinates on Δ, we have

$$\begin{array}{l}{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }(tb+(1-t)a,\lambda d+(1-\lambda )c\right|}^q\\ \le t^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }(b,\lambda d+(1-\lambda )c)\right|}^q+{(1-t)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }(a,\lambda d+(1-\lambda )c)\right|}^q\end{array}$$

and

$$\begin{array}{l}{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }(tb+(1-t)a,\lambda d+(1-\lambda )c\right|}^q\\ \le t^s{\lambda }^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q(b,d)+t^s{(1-\lambda )}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q(b,c)\\ +{\lambda }^s{(1-t)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q(a,d)+{(1-\lambda )}^s{(1-t)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q(a,c)\end{array}$$

hence, it follows that

$$\begin{array}{l}\left|D\right|\le \frac{\left(b-a\right)\left(d-c\right)}{\left(r_1+1\right)\left(r_2+1\right)}{\left(\frac{\left(1+r_1^2\right)\left(1+r_2^2\right)}{4\left(r_1+1\right)\left(r_2+1\right)}\right)}^{1-\frac{1}{q}}\\ \times \left({\int }_0^1{\int }_0^1\left|\left(\left(r_1+1\right)t-1\right)\left(\left(r_2+1\right)\lambda -1\right)\right|\right.\left\{t^s{\lambda }^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\right.\left(b,d\right)\\ +t^s{\left(1-\lambda \right)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,c\right)+{\lambda }^s{\left(1-t\right)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,d\right)\\ {\left.+\left.{\left(1-\lambda \right)}^s{\left(1-t\right)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,c\right)\right\}dtd\lambda \right)}^{\frac{1}{q}}\end{array}$$

(3.6)

By a simple computation, one can see that

$$\begin{array}{c}\times \left({\int }_0^1{\int }_0^1\left|\left(\left(r_1+1\right)t-1\right)\left(\left(r_2+1\right)\lambda -1\right)\right|\right.\left\{t^s{\lambda }^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\right.\left(b,d\right)\\ +t^s{\left(1-\lambda \right)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(b,c\right)+{\lambda }^s{\left(1-t\right)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,d\right)\\ {\left.+\left.{\left(1-\lambda \right)}^s{\left(1-t\right)}^s{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\right|}^q\left(a,c\right)\right\}dtd\lambda \right)}^{\frac{1}{q}}\\ ={\left(\frac{MS{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,c\right)\right|}^q+ML{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,d\right)\right|}^q+KR{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|}^q+KN{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|}^q}{{\left(s+1\right)}^2{\left(s+2\right)}^2}\right)}^{\frac{1}{q}}\end{array}$$

where K, L, M, N, R, and S as in Theorem 10. By substituting these in (3.6) and simplifying we obtain the required result.

Corollary 4

• (1) Under the assumptions of Theorem 12, if we choose r ₁ = r ₂ = 1, we have

  $$\begin{array}{l}\left|\frac{f\left(a,c\right)+f\left(a,d\right)+f\left(b,c\right)+f\left(b,d\right)}{4}\right.\\ -\frac{1}{2}\left[\frac{1}{d-c}{\int }_c^d\left[f\left(b,y\right)+f\left(a,y\right)\right]dy+\frac{1}{b-a}{\int }_a^b\left[f\left(x,d\right)+f\left(x,c\right)\right]dx\right]\\ \left.+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{4}{\left(\frac{1}{4}\right)}^{1-\frac{1}{q}}{\left(s+\frac{1}{2^s}\right)}^{\frac{1}{q}}\\ \times {\left[\frac{\frac{1}{2^{s+1}}{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,c\right)\right|}^q+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|}^q+\left(s+\frac{1}{2^{s+1}}\right){\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(a,d\right)\right|}^q+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|}^q}{{\left(s+1\right)}^2{\left(s+2\right)}^2}\right]}^{\frac{1}{q}}\end{array}$$

• (2) Under the assumptions of Theorem 12, if we choose r ₁ = r ₂ = 0, we have

  $$\begin{array}{l}\left|f\left(a,c\right)+\frac{1}{\left(b-a\right)\left(d-c\right)}{\int }_a^b{\int }_c^{d}f\left(x,y\right)dxdy\right.\\ \left.-\frac{1}{d-c}{\int }_c^{d}f\left(a,y\right)dy-\frac{1}{b-a}{\int }_a^{b}f\left(x,c\right)dx\right|\\ \le \left(b-a\right)\left(d-c\right){\left(\frac{1}{4}\right)}^{1-\frac{1}{q}}\\ \times {\left[\frac{\left(s+1\right){\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,c\right)\right|}^q+{\left|\frac{{\partial }^{2}f}{\partial t\partial \lambda }\left(b,d\right)\right|}^q}{{\left(s+1\right)}^2{\left(s+2\right)}^2}\right]}^{\frac{1}{q}}\end{array}$$

Remark 5. Under the assumptions of Theorem 1.2., if we choose r₁ = r₂ = 1 and s = 1, we get the inequality in (1.4).