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

Muhamet Emin Özdemir; Havva Kavurmaci; Ahmet Ocak Akdemir; Merve Avci

doi:10.1186/1029-242X-2012-20

Research

Open Access

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

Muhamet Emin Özdemir¹,
Havva Kavurmaci¹Email author,
Ahmet Ocak Akdemir² and
Merve Avci³

Journal of Inequalities and Applications20122012:20

https://doi.org/10.1186/1029-242X-2012-20

Received: 1 May 2011

Accepted: 1 February 2012

Published: 1 February 2012

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.

Keywords

Hadamard-type inequalityco-ordinatess-convex functions.

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 (\frac{a + b}{2}) \leq \frac{1}{b - a} \int_{a}^{b} f (x) d x \leq \frac{f (a) + f (b)}{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 (α x + β y) \leq α^{s} f (x) + β^{s} f (y)

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 (λ x + (1 - λ) z, λ y + (1 - λ) w) \leq λ f (x, y) + (1 - λ) f (z, w)

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{align} f (\frac{a + b}{2}, \frac{c + d}{2}) \\ \leq \frac{1}{2} [\frac{1}{b - a} \int_{a}^{b} f (x, \frac{c + d}{2}) d x + \frac{1}{d - c} \int_{c}^{d} f (\frac{a + b}{2}, y) d y] \\ \leq \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y \\ \leq \frac{1}{4} [\frac{1}{(b - a)} \int_{a}^{b} f (x, c) d x + \frac{1}{(b - a)} \int_{a}^{b} f (x, d) d x \\ + \frac{1}{(d - c)} \int_{c}^{d} f (a, y) d y + \frac{1}{(d - c)} \int_{c}^{d} f (b, y) d y] \\ \leq \frac{f (a, c) + f (a, d) + f (b, c) + f (b, d)}{4} . \end{align}

(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 (λ x + (1 - λ) z, λ y + (1 - λ) w) \leq λ^{s} f (x, y) + {(1 - λ)}^{s} f (z, w)

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

|\frac{\partial^{2} f}{\partial t \partial s}|

is a convex function on the co-ordinates on Δ, then one has the inequalities:

\begin{align} |J| \leq \frac{(b - a) (d - c)}{16} \\ \times \frac{|\frac{\partial^{2} f}{\partial t \partial s}| (a, c) + |\frac{\partial^{2} f}{\partial t \partial s}| (a, d) + |\frac{\partial^{2} f}{\partial t \partial s}| (b, c) + |\frac{\partial^{2} f}{\partial t \partial s}| (b, d)}{4} \end{align}

(1.2)

where

J = \frac{f (a, c) + f (a, d) + f (b, c) + f (b, d)}{4} + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y - A

and

A = \frac{1}{2} [\frac{1}{(b - a)} \int_{a}^{b} [f (x, c) + f (x, d)] d x + \frac{1}{(d - c)} \int_{c}^{d} [f (a, y) + f (b, y)] d y] .

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

{|\frac{\partial^{2} f}{\partial t \partial s}|}^{q}

, q > 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:

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

(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

{|\frac{\partial^{2} f}{\partial t \partial s}|}^{q}

, q ≥ 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:

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

(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} = sup_{(x, y) \in (a, b) \times (c, d)} |\frac{\partial^{2} f (x, y)}{\partial x \partial y}| < \infty,

then we have the inequality;

\begin{align} |\int_{a}^{b} \int_{c}^{d} f (s, t) d t d s - (b - a) \int_{c}^{d} f (x, t) d t - (d - c) \int_{a}^{b} f (s, y) d s - (b - a) (d - c) f (x, y)| \\ \leq [\frac{{(b - a)}^{2}}{4} + {(x - \frac{a + b}{2})}^{2}] [\frac{{(d - c)}^{2}}{4} + {(y - \frac{c + d}{2})}^{2}] {∥{f^{″}}_{x, y}∥}_{\infty} \end{align}

(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 (t, s)}{\partial t \partial s}∥}_{\infty} = sup_{(x, y) \in (a, b) \times (c, d)} |\frac{\partial^{2} f (t, s)}{\partial t \partial s}| < \infty

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

\begin{align} |(β_{1} - α_{1}) (β_{2} - α_{2}) f (\frac{a + b}{2}, \frac{c + d}{2}) + H (α_{1}, α_{2}, β_{1}, β_{2}) + G (α_{1}, α_{2}, β_{1}, β_{2}) \\ - (β_{2} - α_{2}) \int_{a}^{b} f (t, \frac{c + d}{2}) d t - (β_{1} - α_{1}) \int_{c}^{d} f (\frac{a + b}{2}, s) d s \\ - \int_{a}^{b} [(α_{2} - c) f (t, c) + (d - β_{2}) f (t, d)] d t \\ - \int_{c}^{d} [(α_{1} - a) f (a, s) + (b - β_{1}) f (b, s)] d s + \int_{a}^{b} \int_{c}^{d} f (t, s) d s d t| \\ \leq [\frac{{(α_{1} - a)}^{2} + {(b - β_{1})}^{2}}{2} + \frac{{(a + b - 2 α_{1})}^{2} + {(a + b - 2 β_{1})}^{2}}{8}] \\ \times [\frac{{(α_{2} - c)}^{2} + {(d - β_{2})}^{2}}{2} + \frac{{(c + d - 2 α_{2})}^{2} + {(c + d - 2 β_{2})}^{2}}{8}] {∥\frac{\partial^{2} f (t, s)}{\partial t \partial s}∥}_{\infty} \end{align}

(1.6)

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

\begin{align} H (α_{1}, α_{2}, β_{1}, β_{2}) \\ = (α_{1} - a) [(α_{2} - c) f (a, c) + (d - β_{2}) f (a, d)] \\ + (b - β_{1}) [(α_{2} - c) f (b, c) + (d - β_{2}) f (b, d)] \end{align}

and

\begin{align} G (α_{1}, α_{2}, β_{1}, β_{2}) \\ = (β_{1} - α_{1}) [(α_{2} - c) f (\frac{a + b}{2}, c) + (d - β_{2}) f (\frac{a + b}{2}, d)] \\ + (β_{2} - α_{2}) [(α_{1} - a) f (a, \frac{c + d}{2}) + (b - β_{1}) f (b, \frac{c + d}{2})] . \end{align}

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

\begin{align} |(b - a) (d - c) f (\frac{a + b}{2}, \frac{c + d}{2}) + \int_{a}^{b} \int_{c}^{d} f (t, s) d s d t \\ - (d - c) \int_{a}^{b} f (t, \frac{c + d}{2}) d t - (b - a) |\int_{c}^{d} f (\frac{a + b}{2}, s) d s| \\ \leq \frac{1}{16} {∥\frac{\partial^{2} f (t, s)}{\partial t \partial s}∥}_{\infty} {(b - a)}^{2} {(d - c)}^{2} . \end{align}

(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 (Δ)

, then the following equality holds:

\begin{align} f (\frac{a + b}{2}, \frac{c + d}{2}) \\ - \frac{1}{(d - c)} \int_{c}^{d} f (\frac{a + b}{2}, y) d y - \frac{1}{(b - a)} \int_{a}^{b} f (x, \frac{c + d}{2}) d x \\ + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d y d x \\ = \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} p (x, t) q (y, s) \frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d) d s d t \end{align}

where

p (x, t) = \{\begin{matrix} (t - a), & t \in [a, \frac{a + b}{2}] \\ (t - b), & t \in (\frac{a + b}{2}, b] \end{matrix}

and

q (y, s) = \{\begin{matrix} (s - c), & s \in [c, \frac{c + d}{2}] \\ (s - d), & s \in (\frac{c + d}{2}, d] \end{matrix}

for each x ∈ [a, b] and y ∈ [c, d].

Proof. We note that

B = \int_{a}^{b} \int_{c}^{d} p (x, t) q (y, s) \frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d) d s d t .

Integration by parts, we can write

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

By calculating the above integrals, we have

\begin{gathered} B = (b - a) (d - c) f (\frac{a + b}{2}, \frac{c + d}{2}) \\ - (b - a) \int_{c}^{d} f (\frac{a + b}{2}, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d) d s \\ - (d - c) \int_{a}^{b} f (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{c + d}{2}) d t \\ \int_{a}^{b} \int_{c}^{d} f (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d) d s d t . \end{gathered}

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

|\frac{\partial^{2} f}{\partial t \partial s}|

is a convex function on the co-ordinates on Δ, then the following inequality holds;

\begin{align} |f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{1}{(d - c)} \int_{c}^{d} f (\frac{a + b}{2}, y) d y - \frac{1}{(b - a)} \int_{a}^{b} f (x, \frac{c + d}{2}) d x \\ + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d y d x| \\ \leq \frac{(b - a) (d - c)}{64} [|\frac{\partial^{2} f}{\partial t \partial s} (a, c)| + |\frac{\partial^{2} f}{\partial t \partial s} (b, c)| + |\frac{\partial^{2} f}{\partial t \partial s} (a, d)| + |\frac{\partial^{2} f}{\partial t \partial s} (b, d)|] . \end{align}

Proof. We note that

\begin{align} C & = f (\frac{a + b}{2}, \frac{c + d}{2}) - \frac{1}{(d - c)} \int_{c}^{d} f (\frac{a + b}{2}, y) d y - \frac{1}{(b - a)} \int_{a}^{b} f (x, \frac{c + d}{2}) d x \\ + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d y d x . \end{align}

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

\begin{align} |C| \leq \frac{1}{(b - a) (d - c)} \\ \times \int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| |\frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)| d s d t \end{align}

Since

|\frac{\partial^{2} f}{\partial t \partial s}|

is co-ordinated convex, we can write

\begin{align} |C| & \leq \frac{1}{(b - a) (d - c)} \\ \times \int_{c}^{d} |q (y, s)| \{\int_{a}^{\frac{a + b}{2}} (t - a) [\frac{b - t}{b - a} |\frac{\partial^{2} f}{\partial t \partial s} (a, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|] d t \\ + \int_{a}^{\frac{a + b}{2}} (t - a) [\frac{t - a}{b - a} |\frac{\partial^{2} f}{\partial t \partial s} (b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|] d t \\ + \int_{\frac{a + b}{2}}^{b} (b - t) [\frac{b - t}{b - a} |\frac{\partial^{2} f}{\partial t \partial s} (a, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|] d t \\ + \int_{\frac{a + b}{2}}^{b} (b - t) [\frac{t - a}{b - a} |\frac{\partial^{2} f}{\partial t \partial s} (b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|] d t\} d s . \end{align}

By computing these integrals, we obtain

\begin{align} |C| & \leq \frac{(b - a)}{8 (d - c)} [\int_{c}^{d} |q (y, s)| |\frac{\partial^{2} f}{\partial t \partial s} (a, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)| \\ + \int_{c}^{d} |q (y, s)| |\frac{\partial^{2} f}{\partial t \partial s} (b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|] d s . \end{align}

Using co-ordinated convexity of

|\frac{\partial^{2} f}{\partial t \partial s}|

again, we get

\begin{array}{l} | C | \leq \frac{(b - a}{8 (d - c)} \\ \times {\int_{c}^{\frac{c + d}{2}} (s - c) [\frac{d - s}{d - c} | \frac{\partial^{2} f}{\partial r \partial s} (a, c) |] d s + \int_{c}^{\frac{c + d}{2}} (s - c) [\frac{s - c}{d - c} | \frac{\partial^{2} f}{\partial t \partial s} (a, d) |] d s \\ + \int_{\frac{c + d}{2}}^{d} (d - s) [\frac{d - s}{d - c} | \frac{\partial^{2} f}{\partial t \partial s} (a, c) |] d s + \int_{\frac{c + d}{2}}^{d} (d - s) [\frac{s - c}{d - c} | \frac{\partial^{2} f}{\partial t \partial s} (a, d) |] d s \\ + \int_{c}^{\frac{c + d}{2}} (s - c) [\frac{d - s}{d - c} | \frac{\partial^{2} f}{\partial t \partial s} (b, c) |] d s + \int_{d}^{\frac{c + d}{2}} (s - c) [\frac{s - c}{d - c} | \frac{\partial^{2} f}{\partial t \partial s} (b, d) |] d s \\ + \int_{\frac{c + d}{2}}^{d} (d - s) [\frac{d - s}{d - c} | \frac{\partial^{2} f}{\partial t \partial s} (b, c) |] d s + \int_{\frac{c + d}{2}}^{d} (d - s) [\frac{s - c}{d - c} | \frac{\partial^{2} f}{\partial t \partial s} (b, d) |] d s} . \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

|\frac{\partial^{2} f}{\partial t \partial s}|

is bounded, i.e.,

{∥\frac{\partial^{2} f (t, s)}{\partial t \partial s}∥}_{\infty} = sup_{(t, s) \in (a, b) \times (c, d)} |\frac{\partial^{2} f (t, s)}{\partial t \partial s}| < \infty,

we get

|C| \leq \frac{(b - a) (d - c)}{16} {∥\frac{\partial^{2} f (t, s)}{\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

{|\frac{\partial^{2} f}{\partial t \partial s}|}^{q}

, q > 1, is a convex function on the co-ordinates on Δ, then the following inequality holds;

\begin{align} |C| & \leq \frac{(b - a) (d - c)}{4 {(p + 1)}^{\frac{2}{p}}} \\ \times {[\frac{{|\frac{\partial^{2} f}{\partial t \partial s} (a, c)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (b, c)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (a, d)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (b, d)|}^{q}}{4}]}^{\frac{1}{q}} \end{align}

(2.2)

where C is in the proof of Theorem 7.

Proof. From Lemma 1, we have

\begin{array}{l} |C| & \leq \frac{1}{(b - a) (d - c)} \\ x \int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| |\frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)| d s d t . \end{array}

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

\begin{align} |C| & \leq \frac{1}{(b - a) (d - c)} \{(\int_{a}^{b} {\int_{c}^{d} |p (x, t) q (y, s)|}^{p} d t d s) \frac{1}{p} \\ \times {(\int_{a}^{b} \int_{c}^{d} {|\frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|}^{q} d s d t)}^{\frac{1}{q}} . \end{align}

(2.3)

Since

{|\frac{\partial^{2} f}{\partial t \partial s}|}^{q}

is co-ordinated convex function on Δ, we can write

\begin{align} {|\frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|}^{q} \\ \leq (\frac{b - t}{b - a}) (\frac{d - s}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (a, c)|}^{q} \\ + (\frac{b - t}{b - a}) (\frac{s - c}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (a, d)|}^{q} \\ + (\frac{t - a}{b - a}) (\frac{d - s}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (b, c)|}^{q} \\ + (\frac{t - a}{b - a}) (\frac{s - c}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (b, d)|}^{q} . \end{align}

(2.4)

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

\begin{align} |C| & \leq \frac{(b - a) (d - c)}{4 {(p + 1)}^{\frac{2}{p}}} \\ \times {[\frac{{|\frac{\partial^{2} f}{\partial t \partial s} (a, c)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (b, c)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (a, d)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (b, d)|}^{q}}{4}]}^{\frac{1}{q}} \end{align}

where we have used the fact that

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

This completes the proof.

Remark 2. Suppose that all the assumptions of Theorem 8 are satisfied. If we choose

|\frac{\partial^{2} f}{\partial t \partial s}|

is bounded, i.e.,

{∥\frac{\partial^{2} f (t, s)}{\partial t \partial s}∥}_{\infty} = sup_{(t, s) \in (a, b) \times (c, d)} |\frac{\partial^{2} f (t, s)}{\partial t \partial s}| < \infty,

we get

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

(2.5)

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

Theorem 9. Let f : Δ = [a, b] × [c, d] → ℝ bea partial differentiable mapping on Δ = [a, b] × [c, d]. If

{|\frac{\partial^{2} f}{\partial t \partial s}|}^{q}

, q > 1, is a convex function on the co-ordinates on Δ, then the following inequality holds;

\begin{align} |C| \leq \frac{(b - a) (d - c)}{16} \\ \times {[\frac{{|\frac{\partial^{2} f}{\partial t \partial s} (a, c)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (b, c)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (a, d)|}^{q} + {|\frac{\partial^{2} f}{\partial t \partial s} (b, d)|}^{q}}{4}]}^{\frac{1}{q}} \end{align}

(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{align} |C| \leq \frac{1}{(b - a) (d - c)} \\ \times \int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| |\frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)| d s d t \\ \leq \frac{1}{(b - a) (d - c)} {(\int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| d s d t)}^{1 - \frac{1}{q}} \\ \times {[\int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| {|\frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|}^{q} d s d t]}^{\frac{1}{q}} . \end{align}

(2.7)

Since

{|\frac{\partial^{2} f}{\partial t \partial s}|}^{q}

is co-ordinated convex function on Δ, we can write

\begin{align} {|\frac{\partial^{2} f}{\partial t \partial s} (\frac{b - t}{b - a} a + \frac{t - a}{b - a} b, \frac{d - s}{d - c} c + \frac{s - c}{d - c} d)|}^{q} \\ \leq (\frac{b - t}{b - a}) (\frac{d - s}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (a, c)|}^{q} \\ + (\frac{b - t}{b - a}) (\frac{s - c}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (a, d)|}^{q} \\ + (\frac{t - a}{b - a}) (\frac{d - s}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (b, c)|}^{q} \\ + (\frac{t - a}{b - a}) (\frac{s - c}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (b, d)|}^{q} . \end{align}

(2.8)

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

\begin{align} |C| & \leq \frac{1}{(b - a) (d - c)} {\{(\int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| d s d t)}^{1 - \frac{1}{q}} \\ \times \int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| [(\frac{b - t}{b - a}) (\frac{d - s}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (a, c)|}^{q} + (\frac{b - t}{b - a}) (\frac{s - c}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (a, d)|}^{q}) \\ {+ (\frac{t - a}{b - a}) (\frac{d - s}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (b, c)|}^{q} + (\frac{t - a}{b - a}) (\frac{s - c}{d - c}) {|\frac{\partial^{2} f}{\partial t \partial s} (b, d)|}^{q}])}^{\frac{1}{q}}\} . \end{align}

Computing the above integrals and using the fact that

{(\int_{a}^{b} \int_{c}^{d} |p (x, t) q (y, s)| d t d s)}^{1 - \frac{1}{q}} = {(\frac{{(b - a)}^{2} {(d - c)}^{2}}{16})}^{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 λ} \in L (Δ)

, then the following equality holds:

D = \frac{(b - a) (d - c)}{(r_{1} + 1) (r_{2} + 1)} \times E

where

\begin{align} D = \frac{f (a, c) + r_{2} f (a, d) + r_{1} f (b, c) + r_{1} r_{2} f (b, d)}{(r_{1} + 1 (r_{2} + 1)} \\ + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y \\ - (\frac{r_{1}}{r_{1} + 1}) \frac{1}{d - c} \int_{c}^{d} f (b, y) d y - (\frac{1}{r_{1} + 1}) \frac{1}{d - c} \int_{c}^{d} f (a, y) d y \\ - (\frac{r_{2}}{r_{2} + 1}) \frac{1}{b - a} \int_{a}^{b} f (x, d) d x - (\frac{1}{r_{2} + 1}) \frac{1}{b - a} \int_{a}^{b} f (x, c) d x \end{align}

and

E = \int_{0}^{1} \int_{0}^{1} ((r_{1} + 1) t - 1) ((r_{2} + 1) λ - 1) \frac{\partial^{2} f}{\partial t \partial λ} (t b + (1 - t) a, λ d + (1 - λ) c) d t d λ

for some fixed r₁, r₂ ∈ [0, 1].

Proof. Integration by parts, we get

\begin{array}{l} E = \int_{0}^{1} ((r_{2} + 1) λ - 1) \\ \times [\int_{0}^{1} ((r_{1} + 1) t - 1) \frac{\partial^{2} f}{\partial t \partial λ} (t b + (1 - t) a, λ d + (1 - λ) c) d t] d λ \\ = \int_{0}^{1} ((r_{2} + 1) λ - 1) [\frac{((r_{1} + 1) t - 1)}{(b - a)} \frac{\partial f}{\partial λ} (t b + (1 - t) a, λ d + (1 - λ) c) |_{0}^{1} \\ - \frac{r_{1} + 1}{b - a} \int_{0}^{1} \frac{\partial f}{\partial λ} (t b + (1 - t) a, λ d + (1 - λ) c) d t] d λ \\ = \int_{0}^{1} ((r_{2} + 1) λ - 1 [\frac{r_{1}}{b - a} \frac{\partial f}{\partial λ} (b, λ d + (1 - λ) c) + \frac{1}{b - a} \frac{\partial f}{\partial λ} (a, λ d + (1 - λ) c) \\ - \frac{r_{1} + 1}{b - a} \int_{0}^{1} \frac{\partial f}{\partial λ} (t b + (1 - t) a, λ d + (1 - λ) c) d t] d λ \\ = \frac{r_{1}}{b - a} \frac{((r_{2} + 1) λ - 1}{d - c} f (b, λ d + (1 - λ) c) |_{0}^{1} - \frac{r_{1} (r_{2} + 1)}{(b - a) (d - c)} \int_{0}^{1} f (b, λ d + (1 - λ) c) d λ \\ + \frac{1}{b - a} \frac{((r_{2} + 1) λ - 1}{d - c} f (a, λ d + (1 - λ) c) |_{0}^{1} - \frac{(r_{2} + 1)}{(b - a) (d - c)} \int_{0}^{1} f (a, λ d + (1 - λ) c) d λ \\ - \frac{r_{1} + 1}{b - a} \int_{0}^{1} [\int_{0}^{1} ((r_{2} + 1) λ - 1) \frac{\partial f}{\partial λ} (t b + (1 - t) a, λ d + (1 - λ) c) d λ] d t . \end{array}

Computing these integrals, we obtain

\begin{align} E & = \frac{1}{(b - a) (d - c)} [f (a, c) + r_{2} f (a, d) + r_{1} f (b, c) + r_{1} r_{2} f (b, d) \\ - r_{1} (r_{2} + 1) \int_{0}^{1} f (b, λ d + (1 - λ) c) d λ - (r_{2} + 1) \int_{0}^{1} f (a, λ d + (1 - λ) c) d λ \\ - r_{2} (r_{1} + 1) \int_{0}^{1} f (t b + (1 - t) a, d) d t - (r_{1} + 1) \int_{0}^{1} f (t b + (1 - t) a, c) d t \\ + (r_{1} + 1) (r_{2} + 1) \int_{0}^{1} \int_{0}^{1} f (t b + (1 - t) a, λ d + (1 - λ) c) d t d λ] . \end{align}

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

|\frac{\partial^{2} f}{\partial t \partial λ}|

is s-convex function on the co-ordinates on Δ, then one has the inequality:

\begin{gathered} D | \leq \frac{(b - a) (d - c)}{(r_{1} + 1) (r_{2} + 1) {(s + 1)}^{2} {(s + 2)}^{2}} \\ \times [M S |\frac{\partial^{2} f}{\partial t \partial λ} (a, c)| + M L |\frac{\partial^{2} f}{\partial t \partial λ} (a, d)| \\ + K R |\frac{\partial^{2} f}{\partial t \partial λ} (b, c)| + K N |\frac{\partial^{2} f}{\partial t \partial λ} (b, d)|] \end{gathered}

(3.1)

where

\begin{gathered} M = (s + 1 + 2 (r_{1} + 1) {(\frac{r_{1}}{r_{1} + 1})}^{s + 2} - r_{1}) \\ N = \frac{1}{{(r_{2} + 1)}^{s + 1}} \\ L = r_{2} (s + 1) + \frac{1}{{(r_{2} + 1)}^{s + 1}} - 1 \\ R = s + 1 + r_{2} {(\frac{r_{2}}{r_{2} + 1})}^{s + 1} - r_{2} \\ S = r_{2} {(\frac{r_{2}}{r_{2} + 1})}^{s + 1} \end{gathered}

Proof. From Lemma 2 and by using co-ordinated s-convexity of

|\frac{\partial^{2} f}{\partial t \partial λ}|

, we have;

\begin{align} |D| & \leq \frac{(b - a) (d - c)}{(r_{1} + 1) (r_{2} + 1)} \\ \times \int_{0}^{1} \int_{0}^{1} |((r_{1} + 1) t - 1) ((r_{2} + 1) λ - 1)| |\frac{\partial^{2} f}{\partial t \partial λ} (t b + (1 - t) a, λ d + (1 - λ) c)| d t d λ \\ \leq \frac{(b - a) (d - c)}{(r_{1} + 1) (r_{2} + 1)} \\ \times \int_{0}^{1} [\int_{0}^{1} |((r_{1} + 1) t - 1) ((r_{2} + 1) λ - 1)| \\ \{t^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (b, λ d + (1 - λ) c)| + {(1 - t)}^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (a, λ d + (1 - λ) c)|\} d t] d λ . \end{align}

By calculating the above integrals, we have

\begin{align} \int_{0}^{1} |((r_{1} + 1) t - 1)| \{t^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (b, λ d + (1 - λ) c)| \\ + {(1 - t)}^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (a, λ d + (1 - λ) c)|\} d t \\ = \int_{0}^{\frac{1}{r_{1} + 1}} (1 - (r_{1} + 1) t) \{t^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (b, λ d + (1 - λ) c)| \\ + {(1 - t)}^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (a, λ d + (1 - λ) c)|\} d t \\ + \int_{\frac{1}{r_{1} + 1}}^{1} ((r_{1} + 1) t - 1) \{t^{2} |\frac{\partial^{2} f}{\partial t \partial λ} (b, λ d + (1 - λ) c)| \\ + {(1 - t)}^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (a, λ d + (1 - λ) c)|\} d t \\ = \frac{1}{(s + 1) (s + 2)} [(r_{1} (s + 1) + 2 {(\frac{1}{r_{1} + 1})}^{s + 1} - 1) |\frac{\partial^{2} f}{\partial t \partial λ} (b, λ d + (1 - λ) c)| \\ + (s + 1 + 2 (r_{1} + 1) {(\frac{r_{1}}{r_{1} + 1})}^{s + 2} - r_{1}) |\frac{\partial^{2} f}{\partial t \partial λ} (a, λ d + (1 - λ) c)|] . \end{align}

(3.2)

By a similar argument for other integrals, by using co-ordinated s-convexity of

|\frac{\partial^{2} f}{\partial t \partial λ}|

, we get

\begin{gathered} \int_{0}^{1} |((r_{2} + 1) λ - 1)| \{|\frac{\partial^{2} f}{\partial t \partial λ} (b, λ d + (1 - λ) c)| + |\frac{\partial^{2} f}{\partial t \partial λ} (a, λ d + (1 - λ) c)|\} d λ \\ \leq \int_{0}^{\frac{1}{r_{2} + 1}} (1 - (r_{2} + 1) λ) \{λ^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (b, d)| + {(1 - λ)}^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (b, c)|\} d λ \\ + \int_{\frac{1}{r_{2} + 1}}^{1} ((r_{2} + 1) λ - 1) \{λ^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (a, d)| + {(1 - λ)}^{s} |\frac{\partial^{2} f}{\partial t \partial λ} (a, c)|\} d λ \\ = \frac{1}{(s + 1) (s + 2)} \{\frac{1}{{(r_{2} + 1)}^{s + 1}} |\frac{\partial^{2} f}{\partial t \partial λ} (b, d)| \\ + [r_{2} (s + 1) + \frac{1}{{(r_{2} + 1)}^{s + 1}} - 1] |\frac{\partial^{2} f}{\partial t \partial f} (a, d)| \\ + [s - r_{2} + 1 + r_{2} {(\frac{r_{2}}{r_{2} + 1})}^{s + 1}] |\frac{\partial^{2} f}{\partial t \partial f} (b, c)| \\ + r_{2} {(\frac{r_{2}}{r_{2} + 1})}^{s + 1} |\frac{\partial^{2} f}{\partial t \partial f} (a, c)|\} . \end{gathered}

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
$\begin{array}{l} |\frac{f (a, c) + f (a, d) + f (b, c) + f (b, d)}{4} - \frac{1}{2} [\frac{1}{d - c} \int_{c}^{d} [f (b, y) + f (a, y)] d y] \\ - \frac{1}{2} [\frac{1}{b - a} \int_{a}^{b} [f (x, d) + f (x, c)] d x] + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y| \\ \leq \frac{(b - a) (d - c)}{{4 (s + 1)}^{2} {(s + 2)}^{2}} {(s + \frac{1}{2^{s}})}^{2} \\ \times [\frac{1}{2^{s + 1}} (|\frac{\partial^{2} f}{\partial t \partial λ} (a, c)| + |\frac{\partial^{2} f}{\partial t \partial λ} (b, d)|) + (s + \frac{1}{2^{s + 1}}) (|\frac{\partial^{2} f}{\partial t \partial λ} (a, d)| + |\frac{\partial^{2} f}{\partial t \partial λ} (b, c)|)] . \end{array}$

(3.3)
(2)

If we choose r ₁ = r ₂ = 0 in (3.1), we have
$\begin{array}{l} |f (a, c) - \frac{1}{d - c} \int_{c}^{d} f (a, y) d y - \frac{1}{b - a} \int_{a}^{b} f (x, c) d x \\ + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y| \\ \leq \frac{(b - a) (d - c)}{{(s + 1)}^{2} {(s + 2)}^{2}} \\ \times [(s + 1) |\frac{\partial^{2} f}{\partial t \partial λ} (b, c)| + |\frac{\partial^{2} f}{\partial t \partial λ} (b, d)|] . \end{array}$

Theorem 11. Let f : Δ = [a, b] × [c, d] ⊂ [0, ∞)² → [0, ∞) be an absolutely continuous function on Δ. If

{|\frac{\partial^{2} f}{\partial t \partial λ}|}^{\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{gathered} |D| \leq \frac{(b - a) (d - c)}{(r_{1} + 1) (r_{2} + 1)} \frac{{(1 + r_{1}^{p + 1})}^{\frac{1}{p}} {(1 + r_{2}^{p + 1})}^{\frac{1}{p}}}{{(r_{1} + 1)}^{\frac{1}{p}} {(r_{2} + 1)}^{\frac{1}{p}} {(p + 1)}^{\frac{2}{p}}} \\ \times {[\frac{{|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (a, c) + {|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (a, d) + {|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (b, c) + {|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (b, d)}{{(s + 1)}^{2}}]}^{\frac{1}{q}} \end{gathered}

(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 | \leq \frac{(b - a) (d - c)}{(r_{1} + 1) (r_{2} + 1)} {(\int_{0}^{1} \int_{0}^{1} {| ((r_{1} + 1) t - 1) ((r_{2} + 1) λ - 1) |}^{p} d t d λ)}^{\frac{1}{p}} \\ \times {(\int_{0}^{1} \int_{0}^{1} {| \frac{\partial^{2} f}{\partial t \partial λ} (t b + (1 - t) a, λ d + (1 - λ) c |}^{q} d t d λ)}^{\frac{1}{q}} . \end{array}

In above inequality using the s-convexity on the co-ordinates of ${|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q}$ 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
$\begin{align} |\frac{f (a, c) + f (a, d) + f (b, c) + f (b, d)}{4} \\ - \frac{1}{2} [\frac{1}{d - c} \int_{c}^{d} [f (b, y) + f (a, y)] d y + \frac{1}{b - a} \int_{a}^{b} [f (x, d) + f (x, c)] d x] \\ + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y| \\ \leq \frac{(b - a) (d - c)}{4 {(p + 1)}^{\frac{2}{p}}} \\ \times {[\frac{{|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (a, c) + {|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (a, d) + {|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (b, c) + {|\frac{\partial^{2} f}{\partial t \partial λ}|}^{q} (b, d)}{{(s + 1)}^{2}}]}^{\frac{1}{q}} . \end{align}$

(3.5)
(2)

Under the assumptions of Theorem 11, if we choose r ₁ = r ₂ = 0 in (3.4), we have
$\begin{align} |f (a, c) - \frac{1}{d - c} \int_{c}^{d} f (a, y) d y - \frac{1}{b - a} \int_{a}^{b} f (x, c) d x \\ + \frac{1}{(b - a) (d - c)} \int_{a}^{b} \int_{c}^{d} f (x, y) d x d y| \\ = (b - a) (d - c \end{align}$