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 be a convex function defined on the interval I of real numbers and a <b. The following double inequality;
f a + b 2 1 b - a a b f ( x ) d x 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 : +, where + = [0, ∞), is said to be s-convex in the second sense if
f ( α x + β y ) 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 [210].

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

Definition 2. Let us consider the bidimensional interval Δ = [a, b] × [c, d] in 2with a <b, c <d. A function f : Δ → will be called convex on the coordinates if the partial mappings f y : [a, b] → , f y (u) = f(u, y) and f x : [c, d] → , f x (v) = f(x, v) are convex where defined for all y [c, d] and x [a, b]. Recall that the mapping f : Δ → is convex on Δ if the following inequality holds,
f ( λ x + ( 1 - λ ) z , λ y + ( 1 - λ ) w ) 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 2.

Theorem 1. Suppose that f : Δ = [a, b] × [c, d] → is convex on the co-ordinates on Δ. Then one has the inequalities;
f a + b 2 , c + d 2 1 2 1 b - a a b f x , c + d 2 d x + 1 d - c c d f a + b 2 , y d y 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d x d y 1 4 1 ( b - a ) a b f ( x , c ) d x + 1 ( b - a ) a b f ( x , d ) d x + 1 ( d - c ) c d f ( a , y ) d y + 1 ( d - c ) c d f ( b , y ) d y f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 .
(1.1)

The above inequalities are sharp.

Similar results can be found in [1214].

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

Definition 3. Consider the bidimensional interval Δ = [a, b] × [c, d] in [0, ∞)2with a <b and c <d. The mapping f : Δ → is s-convex on Δ if
f ( λ x + ( 1 - λ ) z , λ y + ( 1 - λ ) w ) 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 2 be a partial differentiable mapping on Δ := [a, b] × [c, d] in 2with a <b and c <d. If 2 f t s is a convex function on the co-ordinates on Δ, then one has the inequalities:
J ( b - a ) ( d - c ) 16 × 2 f t s ( a , c ) + 2 f t s ( a , d ) + 2 f t s ( b , c ) + 2 f t s ( b , d ) 4
(1.2)
where
J = f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d x d y - A
and
A = 1 2 1 ( b - a ) a b [ f ( x , c ) + f ( x , d ) ] d x + 1 ( d - c ) c d [ f ( a , y ) + f ( b , y ) ] d y .
Theorem 3. Let f : Δ 2 be a partial differ entiable mapping on Δ := [a, b] × [c, d] in 2with a <b and c <d. If 2 f t s q , q > 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:
J ( b - a ) ( d - c ) 4 ( p + 1 ) 2 p × 2 f t s q ( a , c ) + 2 f t s q ( a , d ) + 2 f t s q ( b , c ) + 2 f t s q ( b , d ) 4 1 q
(1.3)

where A, J are as in Theorem 2 and 1 p + 1 q = 1 .

Theorem 4. Let f : Δ 2 be a partial differentiable mapping on Δ := [a, b] × [c, d] in 2with a <b and c <d. If 2 f t s q , q ≥ 1, is a convex function on the co-ordinates on Δ, then one has the inequalities:
J ( b - a ) ( d - c ) 16 × 2 f t s q ( a , c ) + 2 f t s q ( a , d ) + 2 f t s q ( b , c ) + 2 f t s q ( b , d ) 4 1 q
(1.4)

where A, J are as in Theorem 2.

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

Theorem 5. Let f : [a, b] × [c, d] → be continuous on [a, b] × [c, d], f x , y = 2 f x y exists on (a, b) × (c, d) and is bounded, that is
f x , y = sup ( x , y ) ( a , b ) × ( c , d ) 2 f ( x , y ) x y < ,
then we have the inequality;
a b c d f ( s , t ) d t d s - ( b - a ) c d f ( x , t ) d t - ( d - c ) a b f ( s , y ) d s - ( b - a ) ( d - c ) f ( x , y ) ( b - a ) 2 4 + x - a + b 2 2 ( d - c ) 2 4 + y - c + d 2 2 f x , y
(1.5)

for all (x, y) [a, b] × [c, d].

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

Theorem 6. Let f : [a, b] × [c, d] → be an absolutely continuous function such that the partial derivative of order 2 exists and is bounded, i.e.,
2 f ( t , s ) t s = sup ( x , y ) ( a , b ) × ( c , d ) 2 f ( t , s ) t s <
for all (t, s) [a, b] × [c, d]. Then we have,
( β 1 - α 1 ) ( β 2 - α 2 ) f a + b 2 , 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 ) a b f t , c + d 2 d t - ( β 1 - α 1 ) c d f a + b 2 , s d s - a b [ ( α 2 - c ) f ( t , c ) + ( d - β 2 ) f ( t , d ) ] d t - c d [ ( α 1 - a ) f ( a , s ) + ( b - β 1 ) f ( b , s ) ] d s + a b c d f ( t , s ) d s d t ( α 1 - a ) 2 + ( b - β 1 ) 2 2 + ( a + b - 2 α 1 ) 2 + ( a + b - 2 β 1 ) 2 8 × ( α 2 - c ) 2 + ( d - β 2 ) 2 2 + ( c + d - 2 α 2 ) 2 + ( c + d - 2 β 2 ) 2 8 2 f ( t , s ) t s
(1.6)
for all (α1, α2), (β1, β2) [a, b] × [c, d] with α1 <β1, α2 <β2where
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 ) ]
and
G ( α 1 , α 2 , β 1 , β 2 ) = ( β 1 - α 1 ) ( α 2 - c ) f a + b 2 , c + ( d - β 2 ) f a + b 2 , d + ( β 2 - α 2 ) ( α 1 - a ) f a , c + d 2 + ( b - β 1 ) f b , c + d 2 .
Corollary 1. Under the assumptions of Theorem 6, we have
( b - a ) ( d - c ) f a + b 2 , c + d 2 + a b c d f ( t , s ) d s d t - ( d - c ) a b f t , c + d 2 d t - ( b - a ) c d f a + b 2 , s d s 1 16 2 f ( t , s ) t s ( b - a ) 2 ( d - c ) 2 .
(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] → be a partial differentiable mapping on Δ = [a, b] × [c, d]. If 2 f t s L ( Δ ) , then the following equality holds:
f a + b 2 , c + d 2 - 1 ( d - c ) c d f a + b 2 , y d y - 1 ( b - a ) a b f x , c + d 2 d x + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x = 1 ( b - a ) ( d - c ) a b c d p ( x , t ) q ( y , s ) 2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d d s d t
where
p ( x , t ) = ( t - a ) , t a , a + b 2 ( t - b ) , t a + b 2 , b
and
q ( y , s ) = ( s - c ) , s c , c + d 2 ( s - d ) , s c + d 2 , d

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

Proof. We note that
B = a b c d p ( x , t ) q ( y , s ) 2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d d s d t .
Integration by parts, we can write
B = c d q ( y , s ) [ a a + b 2 ( t a ) 2 f t s ( b t b a a + t a b a b , d s d c c + s c d c d ) d t + a + b 2 b ( t b ) 2 f t s ( b t b a a + t a b a b , d s d c c + s c d c d ) d t ] d s = c d q ( y , s ) { [ ( t a ) f s ( b t b a a + t a b a b , d s d c c + s c d c d ) ] a a + b 2 a a + b 2 f s ( b t b a a + t a b a b , d s d c c + s c d c d ) d t + [ ( t b ) f s ( b t b a a + t a b a b , d s d c c + s c d c d ) ] a + b 2 b a + b 2 b f s ( b t b a a + t a b a b , d s d c c + s c d c d ) d t } d s = ( b a ) c d q ( y , s ) { f s ( a + b 2 , d s d c c + s c d c d ) a b f s ( b t b a a + t a b a b , d s d c c + s c d c d ) d t } d s = ( b a ) { c c + d 2 ( s c ) f s ( a + b 2 , d s d c c + s c d c d ) d s + c + d 2 d ( s d ) f s ( a + b 2 , d s d c c + s c d c d ) d s a b [ c c + d 2 ( s c ) f s ( b t b a a + t a b a b , d s d c c + s c d c d ) d s + c + d 2 d ( s d ) f s ( b t b a a + t a b a b , d s d c c + s c d c d ) d s ] d t } .
By calculating the above integrals, we have
B = ( b - a ) ( d - c ) f a + b 2 , c + d 2 - ( b - a ) c d f a + b 2 , d - s d - c c + s - c d - c d d s - ( d - c ) a b f b - t b - a a + t - a b - a b , c + d 2 d t a b c d f b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d d s d t .

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

Theorem 7. Let f : Δ = [a, b] × [c, d] → be a partial differentiable mapping on Δ = [a, b] × [c, d]. If 2 f t s is a convex function on the co-ordinates on Δ, then the following inequality holds;
f a + b 2 , c + d 2 - 1 ( d - c ) c d f a + b 2 , y d y - 1 ( b - a ) a b f x , c + d 2 d x + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x ( b - a ) ( d - c ) 64 2 f t s ( a , c ) + 2 f t s ( b , c ) + 2 f t s ( a , d ) + 2 f t s ( b , d ) .
Proof. We note that
C = f a + b 2 , c + d 2 - 1 ( d - c ) c d f a + b 2 , y d y - 1 ( b - a ) a b f x , c + d 2 d x + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x .
From Lemma 1 and using the property of modulus, we have
C 1 ( b - a ) ( d - c ) × a b c d p ( x , t ) q ( y , s ) 2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d d s d t
Since 2 f t s is co-ordinated convex, we can write
C 1 ( b - a ) ( d - c ) × c d q ( y , s ) a a + b 2 ( t - a ) b - t b - a 2 f t s a , d - s d - c c + s - c d - c d d t + a a + b 2 ( t - a ) t - a b - a 2 f t s b , d - s d - c c + s - c d - c d d t + a + b 2 b ( b - t ) b - t b - a 2 f t s a , d - s d - c c + s - c d - c d d t + a + b 2 b ( b - t ) t - a b - a 2 f t s b , d - s d - c c + s - c d - c d d t d s .
By computing these integrals, we obtain
C ( b - a ) 8 ( d - c ) c d q ( y , s ) 2 f t s a , d - s d - c c + s - c d - c d + c d q ( y , s ) 2 f t s b , d - s d - c c + s - c d - c d d s .
Using co-ordinated convexity of 2 f t s again, we get
| C | ( b a 8 ( d c ) × { c c + d 2 ( s c ) [ d s d c | 2 f r s ( a , c ) | ] d s + c c + d 2 ( s c ) [ s c d c | 2 f t s ( a , d ) | ] d s + c + d 2 d ( d s ) [ d s d c | 2 f t s ( a , c ) | ] d s + c + d 2 d ( d s ) [ s c d c | 2 f t s ( a , d ) | ] d s + c c + d 2 ( s c ) [ d s d c | 2 f t s ( b , c ) | ] d s + d c + d 2 ( s c ) [ s c d c | 2 f t s ( b , d ) | ] d s + c + d 2 d ( d s ) [ d s d c | 2 f t s ( b , c ) | ] d s + c + d 2 d ( d s ) [ s c d c | 2 f t s ( b , d ) | ] d s } .

By a simple computation, we get the required result.

Remark 1. Suppose that all the assumptions of Theorem 7 are satisfied. If we choose 2 f t s is bounded, i.e.,
2 f ( t , s ) t s = sup ( t , s ) ( a , b ) × ( c , d ) 2 f ( t , s ) t s < ,
we get
C ( b - a ) ( d - c ) 16 2 f ( t , s ) t s
(2.1)

which is the inequality in (1.7).

Theorem 8. Let f : Δ = [a, b] × [c, d] → bea partial differentiable mapping on Δ = [a, b] × [c, d]. If 2 f t s q , q > 1, is a convex function on the co-ordinates on Δ, then the following inequality holds;
C ( b - a ) ( d - c ) 4 ( p + 1 ) 2 p × 2 f t s ( a , c ) q + 2 f t s ( b , c ) q + 2 f t s ( a , d ) q + 2 f t s ( b , d ) q 4 1 q
(2.2)

where C is in the proof of Theorem 7.

Proof. From Lemma 1, we have
C 1 ( b - a ) ( d - c ) x a b c d p ( x , t ) q ( y , s ) 2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d d s d t .
By applying the well-known Hölder inequality for double integrals, then one has
C 1 ( b - a ) ( d - c ) a b c d p ( x , t ) q ( y , s ) p d t d s 1 p × a b c d 2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d q d s d t 1 q .
(2.3)
Since 2 f t s q is co-ordinated convex function on Δ, we can write
2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d q b - t b - a d - s d - c 2 f t s ( a , c ) q + b - t b - a s - c d - c 2 f t s ( a , d ) q + t - a b - a d - s d - c 2 f t s ( b , c ) q + t - a b - a s - c d - c 2 f t s ( b , d ) q .
(2.4)
Using the inequality (2.4) in (2.3), we get
C ( b - a ) ( d - c ) 4 ( p + 1 ) 2 p × 2 f t s ( a , c ) q + 2 f t s ( b , c ) q + 2 f t s ( a , d ) q + 2 f t s ( b , d ) q 4 1 q
where we have used the fact that
a b c d p ( x , t ) q ( y , s ) p d t d s 1 p = [ ( b - a ) ( d - c ) ] 1 + 1 p 4 ( p + 1 ) 2 p .

This completes the proof.

Remark 2. Suppose that all the assumptions of Theorem 8 are satisfied. If we choose 2 f t s is bounded, i.e.,
2 f ( t , s ) t s = sup ( t , s ) ( a , b ) × ( c , d ) 2 f ( t , s ) t s < ,
we get
C ( b - a ) ( d - c ) 4 ( p + 1 ) 2 p 2 f ( t , s ) t s
(2.5)

which is the inequality in (1.3) with 2 f ( t , s ) t s .

Theorem 9. Let f : Δ = [a, b] × [c, d] → bea partial differentiable mapping on Δ = [a, b] × [c, d]. If 2 f t s q , q > 1, is a convex function on the co-ordinates on Δ, then the following inequality holds;
C ( b - a ) ( d - c ) 16 × 2 f t s ( a , c ) q + 2 f t s ( b , c ) q + 2 f t s ( a , d ) q + 2 f t s ( b , d ) q 4 1 q
(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
C 1 ( b - a ) ( d - c ) × a b c d p ( x , t ) q ( y , s ) 2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d d s d t 1 ( b - a ) ( d - c ) a b c d p ( x , t ) q ( y , s ) d s d t 1 - 1 q × a b c d p ( x , t ) q ( y , s ) 2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d q d s d t 1 q .
(2.7)
Since 2 f t s q is co-ordinated convex function on Δ, we can write
2 f t s b - t b - a a + t - a b - a b , d - s d - c c + s - c d - c d q b - t b - a d - s d - c 2 f t s ( a , c ) q + b - t b - a s - c d - c 2 f t s ( a , d ) q + t - a b - a d - s d - c 2 f t s ( b , c ) q + t - a b - a s - c d - c 2 f t s ( b , d ) q .
(2.8)
If we use (2.8) in (2.7), we get
C 1 ( b - a ) ( d - c ) a b c d p ( x , t ) q ( y , s ) d s d t 1 - 1 q × a b c d p ( x , t ) q ( y , s ) b - t b - a d - s d - c 2 f t s ( a , c ) q + b - t b - a s - c d - c 2 f t s ( a , d ) q + t - a b - a d - s d - c 2 f t s ( b , c ) q + t - a b - a s - c d - c 2 f t s ( b , d ) q 1 q .
Computing the above integrals and using the fact that
a b c d p ( x , t ) q ( y , s ) d t d s 1 - 1 q = ( b - a ) 2 ( d - c ) 2 16 1 - 1 q ,

we obtained the desired result.

3. Inequalities for co-ordinated s-convex functions

To prove our main results we need the following lemma:

Lemma 2. Let f : Δ 2 be an absolutely continuous function on Δ where a <b, c <d and t, λ [0, 1], if 2 f t λ L ( Δ ) , then the following equality holds:
D = ( b - a ) ( d - c ) ( r 1 + 1 ) ( r 2 + 1 ) × E
where
D = 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 ) + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d x d y - r 1 r 1 + 1 1 d - c c d f ( b , y ) d y - 1 r 1 + 1 1 d - c c d f ( a , y ) d y - r 2 r 2 + 1 1 b - a a b f ( x , d ) d x - 1 r 2 + 1 1 b - a a b f ( x , c ) d x
and
E = 0 1 0 1 ( ( r 1 + 1 ) t - 1 ) ( ( r 2 + 1 ) λ - 1 ) 2 f t 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
E = 0 1 ( ( r 2 + 1 ) λ 1 ) × [ 0 1 ( ( r 1 + 1 ) t 1 ) 2 f t λ ( t b + ( 1 t ) a , λ d + ( 1 λ ) c ) d t ] d λ = 0 1 ( ( r 2 + 1 ) λ 1 ) [ ( ( r 1 + 1 ) t 1 ) ( b a ) f λ ( t b + ( 1 t ) a , λ d + ( 1 λ ) c ) | 0 1 r 1 + 1 b a 0 1 f λ ( t b + ( 1 t ) a , λ d + ( 1 λ ) c ) d t ] d λ = 0 1 ( ( r 2 + 1 ) λ 1 [ r 1 b a f λ ( b , λ d + ( 1 λ ) c ) + 1 b a f λ ( a , λ d + ( 1 λ ) c ) r 1 + 1 b a 0 1 f λ ( t b + ( 1 t ) a , λ d + ( 1 λ ) c ) d t ] d λ = r 1 b a ( ( r 2 + 1 ) λ 1 d c f ( b , λ d + ( 1 λ ) c ) | 0 1 r 1 ( r 2 + 1 ) ( b a ) ( d c ) 0 1 f ( b , λ d + ( 1 λ ) c ) d λ + 1 b a ( ( r 2 + 1 ) λ 1 d c f ( a , λ d + ( 1 λ ) c ) | 0 1 ( r 2 + 1 ) ( b a ) ( d c ) 0 1 f ( a , λ d + ( 1 λ ) c ) d λ r 1 + 1 b a 0 1 [ 0 1 ( ( r 2 + 1 ) λ 1 ) f λ ( t b + ( 1 t ) a , λ d + ( 1 λ ) c ) d λ ] d t .
Computing these integrals, we obtain
E = 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 ) 0 1 f ( b , λ d + ( 1 - λ ) c ) d λ - ( r 2 + 1 ) 0 1 f ( a , λ d + ( 1 - λ ) c ) d λ - r 2 ( r 1 + 1 ) 0 1 f ( t b + ( 1 - t ) a , d ) d t - ( r 1 + 1 ) 0 1 f ( t b + ( 1 - t ) a , c ) d t + ( r 1 + 1 ) ( r 2 + 1 ) 0 1 0 1 f ( t b + ( 1 - t ) a , λ d + ( 1 - λ ) c ) d t d λ .

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 ( 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 2 f t λ is s-convex function on the co-ordinates on Δ, then one has the inequality:
D | ( b - a ) ( d - c ) ( r 1 + 1 ) ( r 2 + 1 ) ( s + 1 ) 2 ( s + 2 ) 2 × M S 2 f t λ ( a , c ) + M L 2 f t λ ( a , d ) + K R 2 f t λ ( b , c ) + K N 2 f t λ ( b , d )
(3.1)
where
M = s + 1 + 2 ( r 1 + 1 ) r 1 r 1 + 1 s + 2 - r 1 N = 1 ( r 2 + 1 ) s + 1 L = r 2 ( s + 1 ) + 1 ( r 2 + 1 ) s + 1 - 1 R = s + 1 + r 2 r 2 r 2 + 1 s + 1 - r 2 S = r 2 r 2 r 2 + 1 s + 1
Proof. From Lemma 2 and by using co-ordinated s-convexity of 2 f t λ , we have;
D ( b - a ) ( d - c ) ( r 1 + 1 ) ( r 2 + 1 ) × 0 1 0 1 ( ( r 1 + 1 ) t - 1 ) ( ( r 2 + 1 ) λ - 1 ) 2 f t λ ( t b + ( 1 - t ) a , λ d + ( 1 - λ ) c ) d t d λ ( b - a ) ( d - c ) ( r 1 + 1 ) ( r 2 + 1 ) × 0 1 0 1 ( ( r 1 + 1 ) t - 1 ) ( ( r 2 + 1 ) λ - 1 ) t s 2 f t λ ( b , λ d + ( 1 - λ ) c ) + ( 1 - t ) s 2 f t λ ( a , λ d + ( 1 - λ ) c ) d t d λ .
By calculating the above integrals, we have
0 1 ( ( r 1 + 1 ) t - 1 ) t s 2 f t λ ( b , λ d + ( 1 - λ ) c ) + ( 1 - t ) s 2 f t λ ( a , λ d + ( 1 - λ ) c ) d t = 0 1 r 1 + 1 ( 1 - ( r 1 + 1 ) t ) t s 2 f t λ ( b , λ d + ( 1 - λ ) c ) + ( 1 - t ) s 2 f t λ ( a , λ d + ( 1 - λ ) c ) d t + 1 r 1 + 1 1 ( ( r 1 + 1 ) t - 1 ) t 2 2 f t λ ( b , λ d + ( 1 - λ ) c ) + ( 1 - t ) s 2 f t λ ( a , λ d + ( 1 - λ ) c ) d t = 1 ( s + 1 ) ( s + 2 ) r 1 ( s + 1 ) + 2 1 r 1 + 1 s + 1 - 1 2 f t λ ( b , λ d + ( 1 - λ ) c ) + s + 1 + 2 ( r 1 + 1 ) r 1 r 1 + 1 s + 2 - r 1 2 f t λ ( a , λ d + ( 1 - λ ) c ) .
(3.2)
By a similar argument for other integrals, by using co-ordinated s-convexity of 2 f t λ , we get
0 1 ( ( r 2 + 1 ) λ - 1 ) 2 f t λ ( b , λ d + ( 1 - λ ) c ) + 2 f t λ ( a , λ d + ( 1 - λ ) c ) d λ 0 1 r 2 + 1 ( 1 - ( r 2 + 1 ) λ ) λ s 2 f t λ ( b , d ) + ( 1 - λ ) s 2 f t λ ( b , c ) d λ + 1 r 2 + 1 1 ( ( r 2 + 1 ) λ - 1 ) λ s 2 f t λ ( a , d ) + ( 1 - λ ) s 2 f t λ ( a , c ) d λ = 1 ( s + 1 ) ( s + 2 ) 1 ( r 2 + 1 ) s + 1 2 f t λ ( b , d ) + r 2 ( s + 1 ) + 1 ( r 2 + 1 ) s + 1 - 1 2 f t f ( a , d ) + s - r 2 + 1 + r 2 r 2 r 2 + 1 s + 1 2 f t f ( b , c ) + r 2 r 2 r 2 + 1 s + 1 2 f t f ( a , c ) .

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

Corollary 2
  1. (1)
    If we choose r 1 = r 2 = 1 in (3.1), we have
    f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 - 1 2 1 d - c c d [ f ( b , y ) + f ( a , y ) ] d y - 1 2 1 b - a a b [ f ( x , d ) + f ( x , c ) ] d x + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d x d y ( b - a ) ( d - c ) 4 ( s + 1 ) 2 ( s + 2 ) 2 s + 1 2 s 2 × 1 2 s + 1 2 f t λ ( a , c ) + 2 f t λ ( b , d ) + s + 1 2 s + 1 2 f t λ ( a , d ) + 2 f t λ ( b , c ) .
    (3.3)
     
  2. (2)
    If we choose r 1 = r 2 = 0 in (3.1), we have
    f ( a , c ) - 1 d - c c d f ( a , y ) d y - 1 b - a a b f ( x , c ) d x + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d x d y ( b - a ) ( d - c ) ( s + 1 ) 2 ( s + 2 ) 2 × ( s + 1 ) 2 f t λ ( b , c ) + 2 f t λ ( b , d ) .
     
Theorem 11. Let f : Δ = [a, b] × [c, d] [0, ∞)2 → [0, ∞) be an absolutely continuous function on Δ. If 2 f t λ 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:
D ( b - a ) ( d - c ) ( r 1 + 1 ) ( r 2 + 1 ) 1 + r 1 p + 1 1 p 1 + r 2 p + 1 1 p ( r 1 + 1 ) 1 p ( r 2 + 1 ) 1 p ( p + 1 ) 2 p × 2 f t λ q ( a , c ) + 2 f t λ q ( a , d ) + 2 f t λ q ( b , c ) + 2 f t λ q ( b , d ) ( s + 1 ) 2 1 q
(3.4)

for some fixed r1, r2 [0, 1], where q = p p - 1 .

Proof. From Lemma 2 and using the Hölder inequality for double integrals, we can write
| D | ( b a ) ( d c ) ( r 1 + 1 ) ( r 2 + 1 ) ( 0 1 0 1 | ( ( r 1 + 1 ) t 1 ) ( ( r 2 + 1 ) λ 1 ) | p d t d λ ) 1 p × ( 0 1 0 1 | 2 f t λ ( t b + ( 1 t ) a , λ d + ( 1 λ ) c | q d t d λ ) 1 q .

In above inequality using the s-convexity on the co-ordinates of 2 f t λ q on Δ and calculating the integrals, then we get the desired result.

Corollary 3
  1. (1)
    Under the assumptions of Theorem 11, if we choose r 1 = r 2 = 1 in (3.4), we have
    f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 - 1 2 1 d - c c d [ f ( b , y ) + f ( a , y ) ] d y + 1 b - a a b [ f ( x , d ) + f ( x , c ) ] d x + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d x d y ( b - a ) ( d - c ) 4 ( p + 1 ) 2 p × 2 f t λ q ( a , c ) + 2 f t λ q ( a , d ) + 2 f t λ q ( b , c ) + 2 f t λ q ( b , d ) ( s + 1 ) 2 1 q .
    (3.5)
     
  2. (2)
    Under the assumptions of Theorem 11, if we choose r 1 = r 2 = 0 in (3.4), we have
    f ( a , c ) - 1 d - c c d f ( a , y ) d y - 1 b - a a b f ( x , c ) d x + 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d x d y = ( b - a ) ( d - c ) ( p + 1 ) 2 p × 2 f t λ q ( a , c ) + 2 f t λ q ( a , d ) + 2 f t λ q ( b , c ) + 2 f t λ q ( b , d ) ( s + 1 ) 2 1 q .
     

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 2 f t λ 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:
D ( b - a ) ( d - c ) ( r 1 + 1 ) ( r 2 + 1 ) ( 1 +