On some new inequalities for differentiable co-ordinated convex functions

Abstract

Several new inequalities for differentiable co-ordinated convex and concave functions in two variables which are related to the left side of Hermite- Hadamard type inequality for co-ordinated convex functions in two variables are obtained.

Mathematics Subject Classification (2000): 26A51; 26D15

1. Introduction

The following definition is well known in literature:

A function f: $I\to ℝ,\mathrm{0̸}\ne I\subseteq ℝ$, is said to be convex on I if the inequality

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

holds for all x, y ∈ I and λ ∈ [0, 1].

Many important inequalities have been established for the class of convex functions, but the most famous is the Hermite-Hadamard's inequality (see for instance [1]). This double inequality is stated as:

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

(1.1)

where f:$I\to ℝ,\mathrm{0̸}\ne I\subseteq ℝ$ a convex function, a, b ∈ I with a < b. The inequalities in (1.1) are in reversed order if f is a concave function.

The inequalities (1.1) have become an important cornerstone in mathematical analysis and optimization and many uses of these inequalities have been discovered in a variety of settings. Moreover, many inequalities of special means can be obtained for a particular choice of the function f. Due to the rich geometrical significance of Hermite-Hadamard's inequality (1.1), there is growing literature providing its new proofs, extensions, refinements and generalizations, see for example [2–5] and the references therein.

Let us consider now a bidimensional interval Δ =: [a, b] × [c, d] in ℝ² with a < b and c < d, a mapping f: Δ → ℝ is said to be convex on Δ if the inequality

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

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

A modification for convex functions on Δ, which are also known as co-ordinated convex functions, was introduced by Dragomir [6, 7] as follows:

A function f: Δ → ℝ is said to be convex on the co-ordinates on Δ 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 x ∈ [a, b], y ∈ [c, d].

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

Definition 1. [8] A function f: Δ → ℝ is said to be convex on the co-ordinates on Δ if the inequality

$$\begin{array}{c}f\left(tx+\left(1-t\right)y,su+\left(1-s\right)w\right)\\ \le tsf\left(x,u\right)+t\left(1-s\right)f\left(x,w\right)+s\left(1-t\right)f\left(y,u\right)+\left(1-t\right)\left(1-s\right)f\left(y,w\right),\end{array}$$

holds for all t, s ∈ [0, 1] and (x, u), (y, w) ∈ Δ.

Clearly, every convex mapping f: Δ → ℝ is convex on the co-ordinates. Furthermore, there exists co-ordinated convex function which is not convex, (see for example [6, 7]). For recent results on co-ordinated convex functions we refer the interested reader to [6, 8–13].

The following Hermite-Hadamrd type inequality for co-ordinated convex functions on the rectangle from the plane ℝ² was also proved in [6]:

Theorem 1. [6] Suppose that f: Δ → ℝ is co-ordinated convex on Δ. Then one has the inequalities:

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

(1.2)

The above inequalities are sharp.

In a recent article [13], Sarikaya et al. proved some new inequalities that give estimate of the difference between the middle and the rightmost terms in (1.2) for differentiable co-ordinated convex functions on rectangle from the plane ℝ². Motivated by notion given in [13], in the present article, we prove some new inequalities which give estimate between the middle and the leftmost terms in (1.2) for differentiable co-ordinated convex functions on rectangle from the plane ℝ².

2. Main results

The following lemma is necessary and plays an important role in establishing our main results:

Lemma 1. Let f: Δ ⊆ ℝ² → ℝ be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If $\frac{{\partial }^{2}f}{\partial s\partial t}\in L\left(\mathrm{\Delta }\right)$, then the following identity holds:

$$\begin{array}{c}\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)\\ -\frac{1}{b-a}\underset{a}{\overset{b}{\int }}f\left(x,\frac{c+d}{2}\right)dx-\frac{1}{d-c}\underset{c}{\overset{d}{\int }}f\left(\frac{a+b}{2},y\right)dy\\ =\left(b-a\right)\left(d-c\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}K\left(t,s\right)\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)dsdt,\end{array}$$

(2.1)

where

$$K\left(t,s\right)=\left\{\begin{array}{cc}ts,\hfill & \left(t,s\right)\in \left[0,\frac{1}{2}\right]\times \left[0,\frac{1}{2}\right]\hfill \\ t\left(s-1\right),\hfill & \left(t,s\right)\in \left[0,\frac{1}{2}\right]\times \left(\frac{1}{2},1\right]\hfill \\ s\left(t-1\right),\hfill & \left(t,s\right)\in \left(\frac{1}{2},1\right]\times \left[0,\frac{1}{2}\right]\hfill \\ \left(t-1\right)\left(s-1\right),\hfill & \left(t,s\right)\in \left(\frac{1}{2},1\right]\times \left(\frac{1}{2},1\right]\hfill \end{array}\right.$$

Proof. Since

$$\begin{array}{c}\left(b-a\right)\left(d-c\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}K\left(t,s\right)\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)dsdt\\ =\left(b-a\right)\left(d-c\right)\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}ts\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)dsdt\\ +\left(b-a\right)\left(d-c\right)\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}t\left(s-1\right)\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)dsdt\\ +\left(b-a\right)\left(d-c\right)\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}s\left(t-1\right)\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)dsdt\\ +\left(b-a\right)\left(d-c\right)\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}\left(t-1\right)\left(s-1\right)\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)dsdt\\ =I_1+I_2+I_3+I_4.\end{array}$$

(2.2)

Now by integration by parts, we have

$$\begin{array}{ll}\hfill I_1& =\left(b-a\right)\left(d-c\right)\underset{0}{\overset{\frac{1}{2}}{\int }}t\left[\underset{0}{\overset{\frac{1}{2}}{\int }}s\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)ds\right]dt\\ =\frac{1}{4}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-\frac{1}{2}\underset{0}{\overset{\frac{1}{2}}{\int }}f\left(ta+\left(1-t\right)b,\frac{c+d}{2}\right)dt\\ -\frac{1}{2}\underset{0}{\overset{\frac{1}{2}}{\int }}f\left(\frac{a+b}{2},sc+\left(1-s\right)d\right)ds+\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)dsdt.\end{array}$$

(2.3)

If we make use of the substitutions x = ta + (1 - t)b and y = sc + (1 - s)d, (t, s) ∈ [0, 1]², in (2.3), we observe that

$$\begin{array}{ll}\hfill I_1& =\frac{1}{4}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-\frac{1}{2\left(b-a\right)}\underset{\frac{a+b}{2}}{\overset{b}{\int }}f\left(x,\frac{c+d}{2}\right)dx\\ -\frac{1}{2\left(d-c\right)}\underset{\frac{c+d}{2}}{\overset{d}{\int }}f\left(\frac{a+b}{2},y\right)dy+\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{\frac{a+b}{2}}{\overset{b}{\int }}\underset{\frac{c+d}{2}}{\overset{d}{\int }}f\left(x,y\right)dydx.\end{array}$$

Similarly, by integration by parts, we also have that

$$\begin{array}{ll}\hfill I_2& =\frac{1}{4}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-\frac{1}{2\left(b-a\right)}\underset{\frac{a+b}{2}}{\overset{b}{\int }}f\left(x,\frac{c+d}{2}\right)dx\\ -\frac{1}{2\left(d-c\right)}\underset{c}{\overset{\frac{c+d}{2}}{\int }}f\left(\frac{a+b}{2},y\right)dy+\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{\frac{a+b}{2}}{\overset{b}{\int }}\underset{c}{\overset{\frac{c+d}{2}}{\int }}f\left(x,y\right)dydx,\\ \hfill I_3& =\frac{1}{4}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-\frac{1}{2\left(b-a\right)}\underset{a}{\overset{\frac{a+b}{2}}{\int }}f\left(x,\frac{c+d}{2}\right)dx\\ -\frac{1}{2\left(d-c\right)}\underset{\frac{c+d}{2}}{\overset{d}{\int }}f\left(\frac{a+b}{2},y\right)dy+\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{\frac{a+b}{2}}{\int }}\underset{\frac{c+d}{2}}{\overset{d}{\int }}f\left(x,y\right)dydx\end{array}$$

and

$$\begin{array}{ll}\hfill I_4& =\frac{1}{4}f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-\frac{1}{2\left(b-a\right)}\underset{a}{\overset{\frac{a+b}{2}}{\int }}f\left(x,\frac{c+d}{2}\right)dx\\ -\frac{1}{2\left(d-c\right)}\underset{c}{\overset{\frac{c+d}{2}}{\int }}f\left(\frac{a+b}{2},y\right)dy+\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{\frac{a+b}{2}}{\int }}\underset{c}{\overset{\frac{c+d}{2}}{\int }}f\left(x,y\right)dydx.\end{array}$$

Substitution of the I₁, I₂, I₃, and I₄ in (2.2) gives the desired identity (2.1).

Theorem 2. Let f: Δ ⊆ ℝ² → ℝ be a partial differentiable mapping on Δ:= [a, b] × [c, d] with a < b, c < d. If $\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|$ is convex on the co-ordinates on Δ, then the following inequality holds:

$$\begin{array}{l}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{16}\left[\frac{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|+\left|\frac{{\partial }^2}{\partial s\partial t}t\left(a,d\right)\right|+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|}{4}\right],\end{array}$$

(2.4)

where

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

Proof. From Lemma 1, we have

$$\begin{array}{l}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \left(b-a\right)\left(d-c\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\end{array}$$

(2.5)

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

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

(2.6)

Substitution of (2.6) in (2.5) gives the following inequality:

$$\begin{array}{c}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \left(b-a\right)\left(d-c\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left[\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|ts+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|t\left(1-s\right)\right.\\ +\left|\frac{{\partial }^2}{\partial s\partial t}\left(b,c\right)\right|s\left(1-t\right)+\left|\frac{{\partial }^2}{\partial s\partial t}\left(b,d\right)\right|\left.\left(1-t\right)\left(1-s\right)\right]dsdt=\left(b-a\right)\left(d-c\right)\\ \times \left\{\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}ts\left[\left|\frac{{\partial }^2}{\partial s\partial t}\left(a,c\right)\right|ts+\left|\frac{{\partial }^2}{\partial s\partial t}\left(a,d\right)\right|t\left(1-s\right)+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|s\left(1-t\right)\right.\right.\\ \left.+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|\left(1-t\right)\left(1-s\right)\right]dsdt+\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}t\left(1-s\right)\left[\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|ts\right.\\ \left.+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|t\left(1-s\right)+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|s\left(1-t\right)+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|\left(1-t\right)\left(1-s\right)\right]dsdt\\ +\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}s\left(1-t\right)\left[\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|ts+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|t\left(1-s\right)+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|s\left(1-t\right)\right.\\ \left.+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|\left(1-t\right)\left(1-s\right)\right]dsdt+\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}\left(1-t\right)\left(1-s\right)\left[\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|ts+\right.\\ \left.\left.\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|t\left(1-s\right)+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|s\left(1-t\right)+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|\left(1-t\right)\left(1-s\right)\right]dsdt\right\}\end{array}$$

(2.7)

Evaluating each integral in (2.7) and simplifying, we get (2.4). Hence the proof of the theorem is complete.

Theorem 3. Let f: Δ ⊆ ℝ² → ℝ be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If ${\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|}^q$ is convex on the co-ordinates on Δ and p, q > 1, $\frac{1}{p}+\frac{1}{q}=1$, then the followin g inequality holds:

$$\begin{array}{c}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{a}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{4{\left(p+1\right)}^{\frac{2}{p}}}{\left[\frac{{\left|\frac{{\partial }^2}{\partial s\partial t}\left(a,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}\left(a,d\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}\left(b,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}\left(b,d\right)\right|}^q}{4}\right]}^{\frac{1}{q}},\end{array}$$

(2.8)

where A is as given in Theorem 2.

Proof. From Lemma 1, we have

$$\begin{array}{c}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \left(b-a\right)\left(d-c\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt.\end{array}$$

(2.9)

Now using the well-known Hölder inequality for double integrals, we obtain

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \le {\left(\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{\left|K\left(t,s\right)\right|}^{p}dsdt\right)}^{\frac{1}{p}}{\left(\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|}^{q}dsdt\right)}^{\frac{1}{q}}.\end{array}$$

(2.10)

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

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|}^{q}dsdt\\ \le \underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left[ts{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|}^q+t\left(1-s\right){\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|}^q\right.\\ \left.+s\left(1-t\right){\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|}^q+\left(1-t\right)\left(1-s\right){\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|}^q\right]dsdt\\ =\frac{{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|}^q}{4}.\end{array}$$

(2.11)

Also, we notice that

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}{\left|K\left(t,s\right)\right|}^{p}dsdt=\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}{t}^p{s}^{p}dsdt+\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}{t}^p{\left(1-s\right)}^{p}dsdt\\ +\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}{s}^p{\left(1-t\right)}^{p}dsdt+\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}{\left(1-t\right)}^p{\left(1-s\right)}^{p}dsdt\\ =\frac{4}{{\left(p+1\right)}^2}{\left(\frac{1}{2}\right)}^{2\left(p+1\right)}.\end{array}$$

(2.12)

Using (2.11) and (2.12) in (2.10), we obtain

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \le \frac{1}{4{\left(p+1\right)}^{\frac{2}{p}}}{\left[\frac{{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|}^q}{4}\right]}^{\frac{1}{q}}.\end{array}$$

Utilizing the last inequality in (2.9) gives us (2.8). This completes the proof of the theorem.

Now we state our next result in:

Theorem 4. Let f: Δ ⊆ ℝ² → ℝ be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If ${\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|}^q$ is convex on the co-ordinates on Δ and q ≥ 1, then the following inequality holds:

$$\begin{array}{c}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{16}{\left[\frac{{\left|\frac{{\partial }^2}{\partial s\partial t}\left(a,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}\left(a,d\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}\left(b,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}\left(b,d\right)\right|}^q}{4}\right]}^{\frac{1}{q}},\end{array}$$

(2.13)

where A is as given in Theorem 2.

Proof. By using Lemma 1, we have that the following inequality:

$$\begin{array}{c}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{a}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \left(b-a\right)\left(d-c\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt.\end{array}$$

(2.14)

By the power mean inequality, we have

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \le {\left(\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|dsdt\right)}^{1-\frac{1}{q}}\\ \times {\left(\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|}^{q}dsdt\right)}^{\frac{1}{q}}\\ ={\left(\frac{1}{16}\right)}^{1-\frac{1}{q}}{\left(\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|}^{q}dsdt\right)}^{\frac{1}{q}}.\end{array}$$

(2.15)

Using the fact that ${\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|}^q$ is convex on the co-ordinates on Δ, we get

$$\begin{array}{c}{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|}^q\\ =ts{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|}^q+t\left(1-s\right){\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|}^q+s\left(1-t\right){\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|}^q\\ +\left(1-t\right)\left(1-s\right){\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|}^q\end{array}$$

and hence, we obtain

$$\begin{array}{l}{\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}\left|K(t,s)\right|}}{\left|\frac{{\partial }^2}{\partial s\partial t}f(ta+(1-t)b,sc+(1-s)d)\right|}^{q}dsdt\\ \le {\displaystyle \underset{0}{\overset{1}{\int }}{\displaystyle \underset{0}{\overset{1}{\int }}\left|K(t,s)\right|}}[ts{\left|\frac{{\partial }^2}{\partial s\partial t}f(a,c)\right|}^q+t(1-s){\left|\frac{{\partial }^2}{\partial s\partial t}f(a,d)\right|}^q\\ +s(1-t){\left|\frac{{\partial }^2}{\partial s\partial t}f(b,c)\right|}^q+(1-t)(1-s){\left|\frac{{\partial }^2}{\partial s\partial t}f(b,d)\right|}^q]dsdt\\ =\frac{1}{64}\left[{\left|\frac{{\partial }^2}{\partial s\partial t}f(a,c)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f(a,d)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f(b,c)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f(b,d)\right|}^q\right].\end{array}$$

Therefore (2.15) becomes

$$\begin{array}{c}\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \le \frac{1}{16}{\left[\frac{{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(a,d\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,c\right)\right|}^q+{\left|\frac{{\partial }^2}{\partial s\partial t}f\left(b,d\right)\right|}^q}{4}\right]}^{\frac{1}{q}}\end{array}$$

(2.16)

Substitution of (2.16) in (2.14), we obtain (2.13). Hence the proof is complete.

Remark 1. Since 2^p> p + 1 if p > 1 and accordingly

$$\frac{1}{4}<\frac{1}{2\left(p+1\right)\frac{1}{p}}$$

and hence we have that the following inequality:

$$\frac{1}{16}<\frac{1}{4}\cdot \frac{1}{4}<\frac{1}{2\left(p+1\right)\frac{1}{p}}\cdot \frac{1}{2\left(p+1\right)\frac{1}{p}}=\frac{1}{4\left(p+1\right)\frac{2}{p}},$$

and as a consequence we get an improvement of the constant in Theorem 3.

Following theorem is about concave functions on the co-ordinates on Δ:

Theorem 5. Let f: Δ ⊆ ℝ² → ℝ be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If ${\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|}^q$ is concave on the co-ordinates on Δ and q ≥ 1, then we have the inequality:

$$\begin{array}{c}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \frac{\left(b-a\right)\left(d-c\right)}{64}\left[\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{a+2b}{3},\frac{c+2d}{3}\right)\right|\right.+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{a+2b}{3},\frac{2c+d}{3}\right)\right|\\ \left.\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{2a+b}{3},\frac{c+2d}{3}\right)\right|+\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{2a+b}{3},\frac{2c+d}{3}\right)\right|\right],\end{array}$$

(2.17)

where A is as defined in Theorem 2.

Proof. By the concavity of ${\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|}^q$ on the co-ordinates on Δ and power mean inequality, we note that the following inequality holds:

$$\begin{array}{c}{\left|\frac{{\partial }^2}{\partial s\partial t}f(\lambda x+(1-\lambda (y,v)\right|}^q\ge \lambda {\left|\frac{{\partial }^2}{\partial s\partial t}f(x,v)\right|}^q+(1-\lambda ){\left|\frac{{\partial }^2}{\partial s\partial t}f(y,v)\right|}^q\\ \ge {\left(\lambda \left|\frac{{\partial }^2}{\partial s\partial t}f(x,v)\right|+(1-\lambda )\left|\frac{{\partial }^2}{\partial s\partial t}f(y,v)\right|\right)}^q,\end{array}$$

for all x, y ∈ [a, b], λ ∈ [0, 1] and for fixed v ∈ [c, d].

Hence,

$$\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\lambda x+\left(1-\lambda \right)y,v\right)\right|\ge \lambda \left|\frac{{\partial }^2}{\partial s\partial t}f\left(x,v\right)\right|+\left(1-\lambda \right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(y,v\right)\right|,$$

for all x, y ∈ [a, b], λ ∈ [0, 1] and for fixed v ∈ [c, d].

Similarly, we can show that

$$\left|\frac{{\partial }^2}{\partial s\partial t}f\left(u,\lambda z+\left(1-\lambda \right)w\right)\right|\ge \lambda \left|\frac{{\partial }^2}{\partial s\partial t}f\left(u,z\right)\right|+\left(1-\lambda \right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(u,w\right)\right|,$$

for all z, w ∈ [c, d], λ ∈ [0, 1] and for fixed u ∈ [a, d], thus $\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|$ is concave on the co-ordinates on Δ.

It is clear from Lemma 1 that

$$\begin{array}{c}\left|\frac{1}{\left(b-a\right)\left(d-c\right)}\underset{a}{\overset{b}{\int }}\underset{c}{\overset{d}{\int }}f\left(x,y\right)dydx+f\left(\frac{a+b}{2},\frac{c+d}{2}\right)-A\right|\\ \le \left(b-a\right)\left(d-c\right)\underset{0}{\overset{1}{\int }}\underset{0}{\overset{1}{\int }}\left|K\left(t,s\right)\right|\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ =\left(b-a\right)\left(d-c\right)\left[\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}st\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\right.\\ +\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}t\left(1-s\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ +\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}s\left(1-t\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \left.+\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}\left(1-t\right)\left(1-s\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\right].\end{array}$$

(2.18)

Since $\left|\frac{{\partial }^{2}f}{\partial s\partial t}\right|$ is concave on the co-ordinates, we have, by Jensen's inequality for integrals, that:

$$\begin{array}{c}\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}st\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ =\underset{0}{\overset{\frac{1}{2}}{\int }}t\left[\underset{0}{\overset{\frac{1}{2}}{\int }}s\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|ds\right]dt\\ \le \underset{0}{\overset{\frac{1}{2}}{\int }}r\left(\underset{0}{\overset{\frac{1}{2}}{\int }}sds\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,\frac{\underset{0}{\overset{\frac{1}{2}}{\int }}s\left(sc+\left(1-s\right)d\right)ds}{\underset{0}{\overset{\frac{1}{2}}{\int }}sds}\right)\right|dt\\ =\frac{1}{8}\underset{0}{\overset{\frac{1}{2}}{\int }}t\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,\frac{c+2d}{3}\right)\right|dt\\ \le \frac{1}{8}\left(\underset{0}{\overset{\frac{1}{2}}{\int }}tdt\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{\underset{0}{\overset{\frac{1}{2}}{\int }}t\left(ta+\left(1-t\right)b\right)dt}{\underset{0}{\overset{\frac{1}{2}}{\int }}tdt},\frac{c+2d}{3}\right)\right|\\ =\frac{1}{64}\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{a+2b}{3},\frac{c+2d}{3}\right)\right|.\end{array}$$

(2.19)

In a similar way, we also have that

$$\begin{array}{c}\underset{0}{\overset{\frac{1}{2}}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}t\left(1-s\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \le \frac{1}{64}\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{a+2b}{3},\frac{2c+d}{3}\right)\right|,\end{array}$$

(2.20)

$$\begin{array}{c}\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{0}{\overset{\frac{1}{2}}{\int }}s\left(1-t\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \le \frac{1}{64}\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{2a+b}{3},\frac{c+2d}{3}\right)\right|\end{array}$$

(2.21)

and

$$\begin{array}{c}\underset{\frac{1}{2}}{\overset{1}{\int }}\underset{\frac{1}{2}}{\overset{1}{\int }}\left(1-t\right)\left(1-s\right)\left|\frac{{\partial }^2}{\partial s\partial t}f\left(ta+\left(1-t\right)b,sc+\left(1-s\right)d\right)\right|dsdt\\ \le \frac{1}{64}\left|\frac{{\partial }^2}{\partial s\partial t}f\left(\frac{2a+b}{3},\frac{c+2d}{3}\right)\right|.\end{array}$$

(2.22)

By making use of (2.19)-(2.22) in (2.18), we get the desired result. This completes the proof.