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On some Hadamard-type inequalities for product of two s-convex functions on the co-ordinates

Abstract

In this article, Hadamard-type inequalities for product of s-convex in the second sense on the co-ordinates in a rectangle from the plane are established.

1. Introduction

A function f : I, I is an interval, is said to be a convex function on I if

f ( t x + ( 1 - t ) y ) tf ( x ) + ( 1 - t ) f ( y )
(1.1)

holds for all x, y I and t [0, 1]. If the reversed inequality in (1.1) holds, then f is concave. Let f : I be a convex function and a, b I with a < b. Then the following double inequality:

f a + b 2 1 b - a a b f ( x ) d x f ( a ) + f ( b ) 2
(1.2)

is known as Hermite-Hadamard inequality for convex mappings. For particular choice of the function f in (1.2) yields some classical inequalities of means. Both inequalities in (1.2) hold in reversed direction if f is concave.

Some basic definitions can be given as followings:

Definition 1. (See [1], [2, p. 410]) We say that f : I is a Godunova-Levin function or that f belongs to the class Q(I) if f is non-negative and for all x, y I and t (0, 1) we have

f ( t x + ( 1 - t ) y ) f ( x ) t + f ( y ) 1 - t .

Definition 2. (See [3]) We say that f : I is a P-function or that f belongs to the class P (I) if f is non-negative and for all x, y I and t [0, 1], we have

f ( t x + ( 1 - t ) y ) f ( x ) +f ( y ) .

Definition 3. (See [4]) Let s (0, 1]. A function f : [0, ∞) → [0, ∞) is said to be s-convex in the second sense if

f ( t x + ( 1 - t ) y ) t s f ( x ) + ( 1 - t ) s f ( y )

for all x, y (0, b) and t [0, 1].

In [5], Hudzik and Maligranda considered among others the class of functions which are s-convex in the first sense. This class is defined in the following way:

Definition 4. A function f : +, where + = [0, ∞), is said to be s-convex in the first sense if

f ( α x + β y ) α s f ( x ) + β s f ( y )

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

In 1978, Breckner introduced s-convex functions as a generalization of convex functions in [4]. Also, he proved the important fact that the set valued map is s-convex only if the associated support function is s-convex function [6]. Of course, s-convexity means just convexity when s = 1. The definition of s-convexity of real valued functions are very important for Orlicz spaces and Banach normed spaces (see [79]). A number of properties of s-convex functions are discussed in articles [5, 1013].

In article [14] the following generalization of the previously described functions was given.

Definition 5. (See [14, 15]) Let I, J be intervals , (0, 1) J and let h : J. A function f : I is called an h-convex function, or that f belongs to the class SX(h, I), if for all x, y I and t (0, 1) we have

f ( t x + ( 1 - t ) y ) h ( t ) f ( x ) + h ( 1 - t ) f ( y ) .
(1.3)

If inequality in (1.3) is reversed, then f is said to be h-concave.

Obviously, if h(t) = t, for all t [0, 1] J, then all convex functions belong to SX (h, I) and all concave functions are h-concave; if h ( t ) = 1 t , for all t (0, 1), then SX(h, I) = Q(I); if h(t) = 1, SX(h, I) P(I); and if h(t) = ts, where s (0, 1), then SX ( h , I ) K s 2 . For some recent results about h-convex functions we refer the reader to articles [1518].

In [10], Dragomir and Fitzpatrick proved the following variant of Hermite-Hadamard inequality which hold for s-convex functions in the second sense:

Theorem 1. Suppose that f : [0, ∞) → [0, ∞) is an s-convex function in the second sense, where s (0, 1) and let a, b [0, ∞), a < b. If f L1([a, b]), then the following inequalities hold:

2 s - 1 f a + b 2 1 b - a a b f ( x ) d x f ( a ) + f ( b ) s + 1
(1.4)

The constant k= 1 s + 1 is the best possible in the second inequality in (1.4).

Again in [10], Dragomir and Fitzpatrick also proved the following Hadamard-type inequality for s-convex functions in the first sense:

Theorem 2. Suppose that f : [0, ∞) → [0, ∞) is an s-convex function in the first sense, where s (0, 1) and let a, b [0, ∞). If f L1([a, b]) then the following inequalities hold:

f a + b 2 1 b - a a b f ( x ) d x f ( a ) + s f ( b ) s + 1
(1.5)

The above inequalities are sharp.

A modification for convex functions which is also known as co-ordinated convex functions was introduced as following by Dragomir [19]:

Let us consider a bidimensional interval Δ =: [a, b] × [c, d] in 2 with a < b and c < d. A mapping f : Δ → is said to be convex on Δ if the following inequality:

f ( α x + ( 1 - α ) z , α y + ( 1 - α ) w ) α f ( x , y ) + ( 1 - α ) f ( z , w )

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

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

In the same article, Dragomir established the following Hadamard-type inequalities for convex functions on the co-ordinates in a rectangle from the plane 2:

Theorem 3. Suppose f : Δ = [a, b] × [c, d] [0, ∞) → is convex function on the co-ordinates on Δ. Then one has the inequalities:

f a + b 2 , c + d 2 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x f ( a , c ) + f ( b , c ) + f ( a , d ) + f ( b , d ) 4
(1.6)

The concept of s-convex functions on the co-ordinates in both sense was introduced by Alomari and Darus [20, 21]:

Definition 6. Consider the bidimensional interval Δ =: [a, b] × [c, d] in [0, ∞)2 with a < b and c < d. The mapping f : Δ → is s-convex in the first sense (in the second sense) on Δ if

f ( α x + β z , α y + β w ) α s f ( x , y ) + β s f ( z , w )

holds for all (x, y), (z, w) Δ, α, β ≥ 0 with αs + βs = 1 (α + β = 1) and for some fixed s (0, 1]. We write f K s i ( i = 1 , 2 ) which means that f is s-convex in the first sense when i = 1, (in the second sense when i = 2).

A function f : Δ =: [a, b] × [c, d] [0, ∞)2 is called s-convex in first sense (in the second sense) on the co-ordinates on Δ 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 s-convex in the first sense (in the second sense) for all y [c, d], x [a, b], and s (0, 1], i.e, the partial mappings f y and f x are s-convex in the first sense (second sense) with some fixed s (0, 1].

In [20], Alomari and Darus proved the following inequalities for s-convex functions (in the second sense) on the co-ordinates in a rectangle from the plane 2:

Theorem 4. Suppose f : Δ = [a, b] × [c, d] [0, ∞) → is s-convex function (in the second sense) on the co-ordinates on Δ. Then one has the inequalities:

4 s - 1 f a + b 2 , c + d 2 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x f ( a , c ) + f ( b , c ) + f ( a , d ) + f ( b , d ) ( s + 1 ) 2
(1.7)

Also in [21] (see also [22]), Alomari and Darus established the following inequalities for s-convex functions (in the first sense) on the co-ordinates in a rectangle from the plane 2:

Theorem 5. Suppose f : Δ = [a, b] × [c, d] [0, ∞) → is s-convex function (in the first sense) on the co-ordinates on Δ. Then one has the inequalities:

f a + b 2 , c + d 2 1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x f ( a , c ) + s f ( b , c ) + s f ( a , d ) + s 2 f ( b , d ) ( s + 1 ) 2
(1.8)

The above inequalities are sharp.

In [23], Sarikaya et al. proved some Hadamard-type inequalities for co-ordinated convex functions as followings:

Theorem 6. Let f : Δ 2 be a partial differentiable mapping on Δ := [a, b] × [c, d] in 2 with 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:

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 ( 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

where

A = 1 2 1 ( b - a ) a b [ f ( x , c ) + f ( x , d ) ] d x + 1 ( d - c ) c d [ f ( a , y ) d y + f ( b , y ) ] d y .

Theorem 7. Let f : Δ 2 be a partial differentiable mapping on Δ := [a, b] × [c, d] in 2 with 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:

f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 1 ( b - a ) ( d - c ) a b c d ( x , y ) d x d y - A ( 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

where

A = 1 2 1 ( b - a ) a b [ f ( x , c ) + f ( x , d ) ] d x + 1 ( d - c ) c d [ f ( a , y ) d y + f ( b , y ) ] d y

and 1 p + 1 q = 1 .

Theorem 8. Let f : Δ 2 be a partial differentiable mapping on Δ := [a, b] × [c, d] in 2 with 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:

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 ( 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

where

A = 1 2 1 ( b - a ) a b [ f ( x , c ) + f ( x , d ) ] d x + 1 ( d - c ) c d [ f ( a , y ) d y + f ( b , y ) ] d y .

For refinements, counterparts, generalizations and new Hadamard-type inequalities see the articles [3, 5, 1013, 1928]. In [25] (see also [22]), Alomari and Darus introduced new classes of s-convex functions on the co-ordinates as following:

Definition 7. Consider the bidimensional interval Δ =: [a, b] × [c, d] in [0, ∞)2 with a < b and c < d. The mapping f : Δ → is s-convex in the first sense on Δ if there exist s1, s2 (0, 1] such that s = s 1 + s 2 2 ,

f ( α x + β z , α y + β w ) α s 1 f ( x , y ) + β s 2 f ( z , w )

holds for all (x, y), (z, w) Δ with α, β ≥ 0 with α s 1 + β s 2 =1 and for some fixed s1, s2 (0, 1]. We denote this class of functions by MW O s 1 , s 2 1 .

Definition 8. Consider the bidimensional interval Δ =: [a, b] × [c, d] in [0, ∞)2 with a < b and c < d. The mapping f : Δ → is s-convex in the second sense on Δ if there exist s1, s2 (0, 1] such that s = s 1 + s 2 2 ,

f ( α x + β z , α y + β w ) α s 1 f ( x , y ) + β s 2 f ( z , w )

holds for all (x, y), (z, w) Δ with α, β ≥ 0 with α + β = 1 and for all fixed s1, s2 (0, 1]. We denote this class of functions by MW O s 1 , s 2 2 .

In [26], Pachpatte established some inequalities for product of convex functions as follows:

Theorem 9. Let f, g : [a, b] → [0, ∞) be convex functions on [a, b], a < b.

Then

1 b - a a b f ( x ) g ( x ) d x 1 3 M ( a , b ) + 1 6 N ( a , b )
(1.9)

and

2 f a + b 2 g a + b 2 1 b - a a b f ( x ) g ( x ) d x + 1 6 M ( a , b ) + 1 3 N ( a , b )
(1.10)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

Similar results for s-convex functions is due to Kirmaci et al. [13] as follows:

Theorem 10. Let f, g : [a, b] a, b [0, ∞), a < b, be functions such that g and fg are in L1([a, b]). If f is convex and non-negative on [a, b] and if g is s-convex on [a, b], for some s (0, 1), then

2 s f a + b 2 g a + b 2 - 1 b - a a b f ( x ) g ( x ) d x 1 ( s + 1 ) ( s + 2 ) M ( a , b ) + 1 s + 2 N ( a , b )
(1.11)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

Theorem 11. Let f, g : [a, b] a, b [0, ∞), a < b, be functions such that g and fg are in L1([a, b]). If f is convex and non-negative on [a, b] and if g is s-convex on [a, b] for some s (0, 1), then

1 b - a a b f ( x ) g ( x ) d x 1 s + 2 M ( a , b ) + 1 ( s + 1 ) ( s + 2 ) N ( a , b )
(1.12)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

Theorem 12. Let f, g : [a, b] a, b [0, ∞), a < b, be functions such that f, g and fg are in L1([a, b]). If f is s1-convex and g is s2-convex on [a, b] for some fixed s1, s2 (0, 1), then

1 b - a a b f ( x ) g ( x ) d x 1 s 1 + s 2 + 1 M ( a , b ) + B ( s 1 + 1 , s 2 + 1 ) N ( a , b ) = 1 s 1 + s 2 + 1 M ( a , b ) + s 1 s 2 Γ ( s 1 ) Γ ( s 2 ) Γ ( s 1 + s 2 + 1 ) N ( a , b )
(1.13)

where M(a, b) = f(a)g(a) + f(b)g(b) and N(a, b) = f(a)g(b) + f(b)g(a).

In the last theorem Beta function of Euler type, defined by

B ( x , y ) = 0 t t x - 1 ( 1 - t ) y - 1 dt= Γ ( x ) Γ ( y ) Γ ( x + y )

has been used.

The main purpose of the present article is to establish new Hadamard-type inequalities similar to the above inequalities, but now for product of s-convex functions (in the second sense) on the co-ordinates in a rectangle from the plane 2.

2. Main results

We will start with the following theorem.

Theorem 13. Let f, g : Δ = [a, b] × [c, d] [0, ∞)2, a < b, c < d, be functions such that f, g and fg are in L2(Δ). If f is non-negative and convex on the coordinates on Δ and if g is s-convex in the second sense on the co-ordinates on Δ, for all s1, s2 (0, 1), such that s = s 1 + s 2 2 , then one has the inequality;

1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x 1 2 ( p 2 + r 2 ) L ( a , b , c , d ) + 1 2 ( p q + r t ) M ( a , b , c , d ) + 1 2 ( q 2 + t 2 ) N ( a , b , c , d )
(2.1)

where

L ( a , b , c , d ) = f ( a , c ) g ( a , c ) + f ( b , c ) g ( b , c ) + f ( a , d ) g ( a , d ) + f ( b , d ) g ( b , d ) , M ( a , b , c , d ) = f ( a , c ) g ( a , d ) + f ( a , d ) g ( a , c ) + f ( b , c ) g ( b , d ) + f ( b , d ) g ( b , c ) + f ( b , c ) g ( a , c ) + f ( b , d ) g ( a , d ) + f ( a , c ) g ( b , c ) + f ( a , d ) g ( b , d ) , N ( a , b , c , d ) = f ( b , c ) g ( a , d ) + f ( b , d ) g ( a , c ) + f ( a , c ) g ( b , d ) + f ( a , d ) g ( b , c )

and

p = 1 s 2 + 2 , q = 1 ( s 2 + 1 ) ( s 2 + 2 ) , r = 1 s 1 + 2 , t = 1 ( s 1 + 1 ) ( s 1 + 2 ) .

Proof. Since f is convex and g is s-convex in the second sense on the co-ordinates on Δ. Therefore the partial mappings

f y : [ a , b ] [ 0 , ) , f y ( x ) =f ( x , y )

and

f x : [ c , d ] [ 0 , ) , f x ( y ) =f ( x , y )

are convex and non-negative on [a, b] and [c, d], respectively. The partial mappings

g y : [ a , b ] [ 0 , ) , g y ( x ) =g ( x , y )

and

g x : [ c , d ] [ 0 , ) , g x ( y ) =g ( x , y )

are s1-, s2-convex on [a, b] and [c, d], respectively, for all x [a, b], y [c, d], for all s1, s2 (0, 1], such that s = s 1 + s 2 2 . Now by applying f x (y)g x (y) to (1.12) on [c, d], we get

1 d - c c d f x ( y ) g x ( y ) d y p [ f x ( c ) g x ( c ) + f x ( d ) g x ( d ) ] + q [ f x ( c ) g x ( d ) + f x ( d ) g x ( c ) ] .

That is

1 d - c c d f ( x , y ) g ( x , y ) d y p [ f ( x , c ) g ( x , c ) + f ( x , d ) g ( x , d ) ] + q [ f ( x , c ) g ( x , d ) + f ( x , d ) g ( x , c ) ] .

Integrating over [a, b] with respect to x and dividing both sides by b - a, we have

1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x p 1 b - a a b f ( x , c ) g ( x , c ) d x + 1 b - a a b f ( x , d ) g ( x , d ) d x + q 1 b - a a b f ( x , c ) g ( x , d ) d x + 1 b - a a b f ( x , d ) g ( x , c ) d x .
(2.2)

Now by applying (1.12) to each integral on right-hand side of (2.2) again, we get

1 b - a a b f ( x , c ) g ( x , c ) d x p [ f ( a , c ) g ( a , c ) + f ( b , c ) g ( b , c ) ] + q [ f ( a , c ) g ( b , c ) + f ( b , c ) g ( a , c ) ] .
1 b - a a b f ( x , d ) g ( x , d ) d x p [ f ( a , d ) g ( a , d ) + f ( b , d ) g ( b , d ) ] + q [ f ( a , d ) g ( b , d ) + f ( b , d ) g ( a , d ) ] .
1 b - a a b f ( x , c ) g ( x , d ) d x p [ f ( a , c ) g ( a , d ) + f ( b , c ) g ( b , d ) ] + q [ f ( a , c ) g ( b , d ) + f ( b , c ) g ( a , d ) ] .
1 b - a a b f ( x , d ) g ( x , c ) d x p [ f ( a , d ) g ( a , c ) + f ( b , d ) g ( b , c ) ] + q [ f ( a , d ) g ( b , c ) + f ( b , d ) g ( a , c ) ] .

On substitution of these inequalities in (2.2), we obtain

1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x p 2 L ( a , b , c , d ) + p q M ( a , b , c , d ) + q 2 N ( a , b , c , d ) .
(2.3)

Similarly, if we apply f y (x)g y (x) to (1.12) on [a, b], we get the following result:

1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x r 2 L ( a , b , c , d ) + r t M ( a , b , c , d ) + t 2 N ( a , b , c , d )
(2.4)

Adding the inequalities (2.3), (2.4) and dividing by 2, we get (2.1).   □

Theorem 14. Let f, g : Δ = [a, b] × [c, d] 2, a < b, c < d, be functions such that f, g, and fg are in L2(Δ). If f is non-negative and convex on the co-ordinates on Δ and if g is s-convex on the co-ordinates on Δ, for all s1, s2 (0, 1), such that s = s 1 + s 2 2 , then one has the inequality;

2 s 1 + s 2 f a + b 2 , c + d 2 g a + b 2 , c + d 2 1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x + 1 2 ( q 2 + t 2 + 2 p t + 2 q r ) L ( a , b , c , d ) + 1 2 ( p q + r t + 2 q t + 2 r p ) M ( a , b , c , d ) + 1 2 ( p 2 + r 2 + 2 p t + 2 r q ) N ( a , b , c , d )
(2.5)

where L(a, b, c, d), M(a, b, c, d), N(a, b, c, d), p, q, r and t as in Theorem 13.

Proof. Applying 2 s 1 f a + b 2 , c + d 2 g a + b 2 , c + d 2 to (1.11) and multiplying both sides by 2 s 2 , we get

2 s 1 + s 2 f a + b 2 , c + d 2 g a + b 2 , c + d 2 2 s 2 b - a a b f x , c + d 2 g x , c + d 2 d x + t 2 s 2 f a , c + d 2 g a , c + d 2 + 2 s 2 f b , c + d 2 g b , c + d 2 + r 2 s 2 f a , c + d 2 g b , c + d 2 + 2 s 2 f b , c + d 2 g a , c + d 2 .
(2.6)

Similarly, by applying 2 s 2 f a + b 2 , c + d 2 g a + b 2 , c + d 2 to (1.11) and multiplying both sides by 2 s 1 , we get

2 s 1 + s 2 f a + b 2 , c + d 2 g a + b 2 , c + d 2 2 s 1 d - c c d f a + b 2 , y g a + b 2 , y d y + q 2 s 1 f a + b 2 , c g a + b 2 , c + 2 s 1 f a + b 2 , d g a + b 2 , d + p 2 s 1 f a + b 2 , c g a + b 2 , d + 2 s 1 f a + b 2 , d g a + b 2 , c .
(2.7)

Adding (2.6) and (2.7), we have

2 s 1 + s 2 + 1 f a + b 2 , c + d 2 g a + b 2 , c + d 2 2 s 2 b - a a b f x , c + d 2 g x , c + d 2 d x + 2 s 1 d - c c d f a + b 2 , y g a + b 2 , y d y + t 2 s 2 f a , c + d 2 g a , c + d 2 + 2 s 2 f b , c + d 2 g b , c + d 2 + r 2 s 2 f a , c + d 2 g b , c + d 2 + 2 s 2 f b , c + d 2 g a , c + d 2 + q 2 s 1 f a + b 2 , c g a + b 2 , c + 2 s 1 f a + b 2 , d g a + b 2 , d + p 2 s 1 f a + b 2 , c g a + b 2 , d + 2 s 1 f a + b 2 , d g a + b 2 , c .
(2.8)

Applying (1.11) to each term within the brackets, we get

2 s 2 f a , c + d 2 g a , c + d 2 1 d - c c d f ( a , y ) g ( a , y ) d y + t [ f ( a , c ) g ( a , c ) + f ( a , d ) g ( a , d ) ] + r [ f ( a , c ) g ( a , d ) + f ( a , d ) g ( a , c ) ] .
2 s 2 f b , c + d 2 g b , c + d 2 1 d - c c d f ( b , y ) g ( b , y ) d y + t [ f ( b , c ) g ( b , c ) + f ( b , d ) g ( b , d ) ] + r [ f ( b , c ) g ( b , d ) + f ( b , d ) g ( b , c ) ] .
2 s 2 f a , c + d 2 g b , c + d 2 1 d - c c d f ( a , y ) g ( b , y ) d y + t [ f ( a , c ) g ( b , c ) + f ( a , d ) g ( b , d ) ] + r [ f ( a , c ) g ( b , d ) + f ( a , d ) g ( b , c ) ] .
2 s 2 f b , c + d 2 g a , c + d 2 1 d - c c d f ( b , y ) g ( a , y ) d y + t [ f ( b , c ) g ( a , c ) + f ( b , d ) g ( a , d ) ] + r [ f ( b , c ) g ( a , d ) + f ( b , d ) g ( a , c ) ] .
2 s 1 f a + b 2 , c g a + b 2 , c 1 b - a a b f ( x , c ) g ( x , c ) d x + q [ f ( a , c ) g ( a , c ) + f ( b , c ) g ( b , c ) ] + p [ f ( a , c ) g ( b , c ) + f ( b , c ) g ( a , c ) ] .
2 s 1 f a + b 2 , d g a + b 2 , d 1 b - a a b f ( x , d ) g ( x , d ) d x + q [ f ( a , d ) g ( a , d ) + f ( b , d ) g ( b , d ) ] + p [ f ( a , d ) g ( b , d ) + f ( b , d ) g ( a , d ) ] .
2 s 1 f a + b 2 , c g a + b 2 , d 1 b - a a b f ( x , c ) g ( x , d ) d x + q [ f ( a , c ) g ( a , d ) + f ( b , c ) g ( b , d ) ] + p [ f ( a , c ) g ( b , d ) + f ( b , c ) g ( a , d ) ] .
2 s 1 f a + b 2 , d g a + b 2 , c 1 b - a a b f ( x , d ) g ( x , c ) d x + q [ f ( a , d ) g ( a , c ) + f ( b , d ) g ( b , c ) ] + p [ f ( a , d ) g ( b , c ) + f ( b , d ) g ( a , c ) ] .

Substituting these inequalities in (2.8) and simplifying, we obtain

2 s 1 + s 2 + 1 f a + b 2 , c + d 2 g a + b 2 , c + d 2 2 s 2 b - a a b f x , c + d 2 g x , c + d 2 d x + 2 s 1 d - c c d f a + b 2 , y g a + b 2 , y d y + t 1 d - c c d f ( a , y ) g ( a , y ) d y + t 1 d - c c d f ( b , y ) g ( b , y ) d y + r 1 d - c c d f ( a , y ) g ( b , y ) d y + r 1 d - c c d f ( b , y ) g ( a , y ) d y + q 1 b - a a b f ( x , c ) g ( x , c ) d x + q 1 b - a a b f ( x , d ) g ( x , d ) d x + p 1 b - a a b f ( x , c ) g ( x , d ) d x + p 1 b - a a b f ( x , d ) g ( x , c ) d x + ( q 2 + t 2 ) L ( a , b , c , d ) + ( p q + r t ) M ( a , b , c , d ) + ( p 2 + r 2 ) N ( a , b , c , d ) .
(2.9)

Now by applying 2 s 1 f a + b 2 , y g a + b 2 , y to (1.11), integrating over [c, d] and dividing both sides by d - c, we get

2 s 1 d - c c d f a + b 2 , y g a + b 2 , y d y 1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d x d y + t 1 d - c c d f ( a , y ) g ( a , y ) d y + t 1 d - c c d f ( b , y ) g ( b , y ) d y + r 1 d - c c d f ( a , y ) g ( b , y ) d y + r 1 d - c c d f ( b , y ) g ( a , y ) d y .
(2.10)

Again by applying 2 s 2 f x , c + d 2 g x , c + d 2 to (1.11), integrating over [a, b] and dividing both sides by b - a, we get

2 s 2 b - a a b f x , c + d 2 g x , c + d 2 d x 1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x + q 1 b - a a b f ( x , c ) g ( x , c ) d x + q 1 b - a a b f ( x , d ) g ( x , d ) d x + p 1 b - a a b f ( x , c ) g ( x , d ) d x + p 1 b - a a b f ( x , d ) g ( x , c ) d x .
(2.11)

Adding (2.10) and (2.11), we have

2 s 2 b - a c d f x , c + d 2 g x , c + d 2 d x + 2 s 1 d - c c d f a + b 2 , y g a + b 2 , y d y 2 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x + t 1 d - c c d f ( a , y ) g ( a , y ) d y + t 1 d - c c d f ( b , y ) g ( b , y ) d y + r 1 d - c c d f ( a , y ) g ( b , y ) d y + r 1 d - c c d f ( b , y ) g ( a , y ) d y + q 1 b - a a b f ( x , c ) g ( x , c ) d x + q 1 b - a a b f ( x , d ) g ( x , d ) d x + p 1 b - a a b f ( x , c ) g ( x , d ) d x + p 1 b - a a b f ( x , d ) g ( x , c ) d x .
(2.12)

Therefore from (2.9) and (2.12), we obtain

2 s 1 + s 2 + 1 f a + b 2 , c + d 2 g a + b 2 , c + d 2 2 ( b - a ) ( d - c ) a b f ( x , y ) g ( x , y ) d x d y + 2 t 1 d - c c d f ( a , y ) g ( a , y ) d y + 2 t 1 d - c c d f ( b , y ) g ( b , y ) d y + 2 r 1 d - c c d f ( a , y ) g ( b , y ) d y + 2 r 1 d - c c d f ( b , y ) g ( a , y ) d y + 2 q 1 b - a a b f ( x , c ) g ( x , c ) d x + 2 q 1 b - a a b f ( x , d ) g ( x , d ) d x + 2 p 1 b - a a b f ( x , c ) g ( x , d ) d x + 2 p 1 b - a a b f ( x , d ) g ( x , c ) d x + ( q 2 + t 2 ) L ( a , b , c , d ) + ( p q + r t ) M ( a , b , c , d ) + ( p 2 + r 2 ) N ( a , b , c , d ) .
(2.13)

By applying (1.12) to each of the integral in (2.13) and simplifying, we get

2 s 1 + s 2 + 1 f a + b 2 , c + d 2 g a + b 2 , c + d 2 2 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x + ( q 2 + t 2 + 2 p t + 2 q r ) L ( a , b , c , d ) + ( p q + r t + 2 q t + 2 r p ) M ( a , b , c , d ) + ( p 2 + r 2 + 2 p t + 2 r q ) N ( a , b , c , d )

Dividing both sides by 2, we get required result.   □

Theorem 15. Let f, g : Δ = [a, b] × [c, d] 2, a < b, c < d, be functions such that f, g, and fg are in L2(Δ). If f is s1-convex on the co-ordinates on Δ and g is s2-convex on the co-ordinates on Δ, for some fixed s11, s12, s21, s22 (0, 1), such that s 1 = s 11 + s 12 2 , s 2 = s 21 + s 22 2 , then one has the following inequality;

1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x 1 2 ( p 1 2 + r 1 2 ) L ( a , b , c , d ) + 1 2 ( p 1 q 1 + r 1 t 1 ) M ( a , b , c , d ) + 1 2 ( q 1 2 + t 1 2 ) N ( a , b , c , d )
(2.14)

where L(a, b, c, d), M(a, b, c, d), and N(a, b, c, d) as defined in Theorem 13 and p 1 = 1 s 12 + s 22 + 1 , q1 = B(s12 + 1, s22 + 1), r 1 = 1 s 11 + s 21 + 1 , t1 = B(s11 + 1, s21 + 1).

Proof. By a similar way to Theorem 13 with p 1 = 1 s 12 + s 22 + 1 , q1 = B(s12+1, s22+1), r 1 = 1 s 11 + s 21 + 1 , t1 = B(s11 + 1, s21 + 1) and thus (2.14) is established.   □

Theorem 16. Let f, g : Δ = [a, b] × [c, d] 2, a < b, c < d, be functions such that f, g, and fg are in L2(Δ). If f is s1-convex on the co-ordinates on Δ and g is s2-convex on the co-ordinates on Δ, for some fixed s11, s12, s21, s22 (0, 1), such that s 1 = s 11 + s 12 2 , s 2 = s 21 + s 22 2 , then one has the following inequality;

2 s 11 + s 21 + s 12 + s 22 - 2 f a + b 2 , c + d 2 g a + b 2 , c + d 2 1 ( b - a ) ( d - c ) a b c d f ( x , y ) g ( x , y ) d y d x + 1 2 ( q 1 2 + t 1 2 + 2 p 1 t 1 + 2 q 1 r 1 ) L ( a , b , c , d ) + 1 2 ( p 1 q 1 + r 1 t 1 + 2 q 1 t 1 + 2 r 1 p 1 ) M ( a , b , c , d ) + 1 2 ( p 1 2 + r 1 2 + 2 p 1 t 1 + 2 r 1 q 1 ) N ( a , b , c , d )
(2.15)

where L(a, b, c, d), M(a, b, c, d), and N(a, b, c, d) as defined in Theorem 13 and p 1 = 1 s 12 + s 22 + 1 , q1 = B(s12 + 1, s22 + 1), r 1 = 1 s 11 + s 21 + 1 , t1 = B(s11 + 1, s21 + 1).

Proof. By a similar way to Theorem 13 with p 1 = 1 s 12 + s 22 + 1 , q1 = B(s12+1, s22+1), r 1 = 1 s 11 + s 21 + 1 , t1 = B(s11 + 1, s21 + 1) the proof is completed   □

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Correspondence to Ahmet Ocak Akdemir.

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MA and AOA carried out the design of the study and performed the analysis. MEO (adviser) participated in its design and coordination. All authors read and approved the final manuscript.

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Özdemir, M.E., Latif, M.A. & Akdemir, A.O. On some Hadamard-type inequalities for product of two s-convex functions on the co-ordinates. J Inequal Appl 2012, 21 (2012). https://doi.org/10.1186/1029-242X-2012-21

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Keywords

  • Reversed Direction
  • Present Article
  • Recent Result
  • Convex Function
  • Support Function