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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,I, is said to be convex on I if the inequality

f ( λ x + ( 1 - λ ) y ) λ f ( x ) + ( 1 - λ ) f ( y ) ,

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 a + b 2 1 b - a a b f ( x ) d x f ( a ) + f ( b ) 2 ,
(1.1)

where f:I,I 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 [25] and the references therein.

Let us consider now 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 inequality

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

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 → 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].

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

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

f ( t x + ( 1 - t ) y , s u + ( 1 - s ) w ) t s f ( x , u ) + t ( 1 - s ) f ( x , w ) + s ( 1 - t ) f ( y , u ) + ( 1 - t ) ( 1 - s ) f ( y , w ) ,

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

Clearly, every convex mapping f: Δ → 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, 813].

The following Hermite-Hadamrd type inequality for co-ordinated convex functions on the rectangle from the plane 2 was also proved in [6]:

Theorem 1. [6] Suppose that f: Δ → is co-ordinated convex 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 y d x 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 .
(1.2)

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 2. 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 2.

2. Main results

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

Lemma 1. Let f: Δ 2 be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If 2 f s t L ( Δ ) , then the following identity holds:

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 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 = ( b - a ) ( d - c ) 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t ,
(2.1)

where

K ( t , s ) = t s , ( t , s ) 0 , 1 2 × 0 , 1 2 t ( s - 1 ) , ( t , s ) 0 , 1 2 × 1 2 , 1 s ( t - 1 ) , ( t , s ) 1 2 , 1 × 0 , 1 2 ( t - 1 ) ( s - 1 ) , ( t , s ) 1 2 , 1 × 1 2 , 1

Proof. Since

( b - a ) ( d - c ) 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t = ( b - a ) ( d - c ) 0 1 2 0 1 2 t s 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t + ( b - a ) ( d - c ) 0 1 2 1 2 1 t ( s - 1 ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t + ( b - a ) ( d - c ) 1 2 1 0 1 2 s ( t - 1 ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t + ( b - a ) ( d - c ) 1 2 1 1 2 1 ( t - 1 ) ( s - 1 ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t = I 1 + I 2 + I 3 + I 4 .
(2.2)

Now by integration by parts, we have

I 1 = ( b - a ) ( d - c ) 0 1 2 t 0 1 2 s 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t = 1 4 f a + b 2 , c + d 2 - 1 2 0 1 2 f t a + ( 1 - t ) b , c + d 2 d t - 1 2 0 1 2 f a + b 2 , s c + ( 1 - s ) d d s + 0 1 2 0 1 2 f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t .
(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

I 1 = 1 4 f a + b 2 , c + d 2 - 1 2 ( b - a ) a + b 2 b f x , c + d 2 d x - 1 2 ( d - c ) c + d 2 d f a + b 2 , y d y + 1 ( b - a ) ( d - c ) a + b 2 b c + d 2 d f ( x , y ) d y d x .

Similarly, by integration by parts, we also have that

I 2 = 1 4 f a + b 2 , c + d 2 - 1 2 ( b - a ) a + b 2 b f x , c + d 2 d x - 1 2 ( d - c ) c c + d 2 f a + b 2 , y d y + 1 ( b - a ) ( d - c ) a + b 2 b c c + d 2 f ( x , y ) d y d x , I 3 = 1 4 f a + b 2 , c + d 2 - 1 2 ( b - a ) a a + b 2 f x , c + d 2 d x - 1 2 ( d - c ) c + d 2 d f a + b 2 , y d y + 1 ( b - a ) ( d - c ) a a + b 2 c + d 2 d f ( x , y ) d y d x

and

I 4 = 1 4 f a + b 2 , c + d 2 - 1 2 ( b - a ) a a + b 2 f x , c + d 2 d x - 1 2 ( d - c ) c c + d 2 f a + b 2 , y d y + 1 ( b - a ) ( d - c ) a a + b 2 c c + d 2 f ( x , y ) d y d x .

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

Theorem 2. Let f: Δ 2 be a partial differentiable mapping on Δ:= [a, b] × [c, d] with a < b, c < d. If 2 f s t is convex on the co-ordinates on Δ, then the following inequality holds:

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 16 2 s t f ( a , c ) + 2 s t t ( a , d ) + 2 s t f ( b , c ) + 2 s t f ( b , d ) 4 ,
(2.4)

where

A = 1 b - a a b f x , c + d 2 d x + 1 d - c c d f a + b 2 , y d y .

Proof. From Lemma 1, we have

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t
(2.5)

Since 2 f s t is convex on the co-ordinates on Δ, we have

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

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

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 0 1 0 1 K ( t , s ) 2 s t f ( a , c ) t s + 2 s t f ( a , d ) t ( 1 - s ) + 2 s t ( b , c ) s ( 1 - t ) + 2 s t ( b , d ) ( 1 - t ) ( 1 - s ) d s d t = ( b - a ) ( d - c ) × 0 1 2 0 1 2 t s 2 s t ( a , c ) t s + 2 s t ( a , d ) t ( 1 - s ) + 2 s t f ( b , c ) s ( 1 - t ) + 2 s t f ( b , d ) ( 1 - t ) ( 1 - s ) d s d t + 0 1 2 1 2 1 t ( 1 - s ) 2 s t f ( a , c ) t s + 2 s t f ( a , d ) t ( 1 - s ) + 2 s t f ( b , c ) s ( 1 - t ) + 2 s t f ( b , d ) ( 1 - t ) ( 1 - s ) d s d t + 1 2 1 0 1 2 s ( 1 - t ) 2 s t f ( a , c ) t s + 2 s t f ( a , d ) t ( 1 - s ) + 2 s t f ( b , c ) s ( 1 - t ) + 2 s t f ( b , d ) ( 1 - t ) ( 1 - s ) d s d t + 1 2 1 1 2 1 ( 1 - t ) ( 1 - s ) 2 s t f ( a , c ) t s + 2 s t f ( a , d ) t ( 1 - s ) + 2 s t f ( b , c ) s ( 1 - t ) + 2 s t f ( b , d ) ( 1 - t ) ( 1 - s ) d s d t
(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 2 be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If 2 f s t q is convex on the co-ordinates on Δ and p, q > 1, 1 p + 1 q =1, then the followin g inequality holds:

1 ( b - a ) ( d - c ) a b a d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 4 ( p + 1 ) 2 p 2 s t ( a , c ) q + 2 s t ( a , d ) q + 2 s t ( b , c ) q + 2 s t ( b , d ) q 4 1 q ,
(2.8)

where A is as given in Theorem 2.

Proof. From Lemma 1, we have

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t .
(2.9)

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

0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 0 1 0 1 K ( t , s ) p d s d t 1 p 0 1 0 1 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) q d s d t 1 q .
(2.10)

Since 2 f s t q is convex on the co-ordinates on Δ, we have

0 1 0 1 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) q d s d t 0 1 0 1 t s 2 s t f ( a , c ) q + t ( 1 - s ) 2 s t f ( a , d ) q + s ( 1 - t ) 2 s t f ( b , c ) q + ( 1 - t ) ( 1 - s ) 2 s t f ( b , d ) q d s d t = 2 s t f ( a , c ) q + 2 s t f ( a , d ) q + 2 s t f ( b , c ) q + 2 s t f ( b , d ) q 4 .
(2.11)

Also, we notice that

0 1 0 1 K ( t , s ) p d s d t = 0 1 2 0 1 2 t p s p d s d t + 0 1 2 1 2 1 t p ( 1 - s ) p d s d t + 1 2 1 0 1 2 s p ( 1 - t ) p d s d t + 1 2 1 1 2 1 ( 1 - t ) p ( 1 - s ) p d s d t = 4 ( p + 1 ) 2 1 2 2 ( p + 1 ) .
(2.12)

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

0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 1 4 ( p + 1 ) 2 p 2 s t f ( a , c ) q + 2 s t f ( a , d ) q + 2 s t f ( b , c ) q + 2 s t f ( b , d ) q 4 1 q .

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 2 be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If 2 f s t q is convex on the co-ordinates on Δ and q ≥ 1, then the following inequality holds:

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 16 2 s t ( a , c ) q + 2 s t ( a , d ) q + 2 s t ( b , c ) q + 2 s t ( b , d ) q 4 1 q ,
(2.13)

where A is as given in Theorem 2.

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

1 ( b - a ) ( d - c ) a b a d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t .
(2.14)

By the power mean inequality, we have

0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 0 1 0 1 K ( t , s ) d s d t 1 - 1 q × 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) q d s d t 1 q = 1 16 1 - 1 q 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) q d s d t 1 q .
(2.15)

Using the fact that 2 f s t q is convex on the co-ordinates on Δ, we get

2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) q = t s 2 s t f ( a , c ) q + t ( 1 - s ) 2 s t f ( a , d ) q + s ( 1 - t ) 2 s t f ( b , c ) q + ( 1 - t ) ( 1 - s ) 2 s t f ( b , d ) q

and hence, we obtain

0 1 0 1 | K ( t , s ) | | 2 s t f ( t a + ( 1 t ) b , s c + ( 1 s ) d ) | q d s d t 0 1 0 1 | K ( t , s ) | [ t s | 2 s t f ( a , c ) | q + t ( 1 s ) | 2 s t f ( a , d ) | q + s ( 1 t ) | 2 s t f ( b , c ) | q + ( 1 t ) ( 1 s ) | 2 s t f ( b , d ) | q ] d s d t = 1 64 [ | 2 s t f ( a , c ) | q + | 2 s t f ( a , d ) | q + | 2 s t f ( b , c ) | q + | 2 s t f ( b , d ) | q ] .

Therefore (2.15) becomes

0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 1 16 2 s t f ( a , c ) q + 2 s t f ( a , d ) q + 2 s t f ( b , c ) q + 2 s t f ( b , d ) q 4 1 q
(2.16)

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

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

1 4 < 1 2 ( p + 1 ) 1 p

and hence we have that the following inequality:

1 16 < 1 4 1 4 < 1 2 ( p + 1 ) 1 p 1 2 ( p + 1 ) 1 p = 1 4 ( p + 1 ) 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 2 be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If 2 f s t q is concave on the co-ordinates on Δ and q ≥ 1, then we have the inequality:

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 64 2 s t f a + 2 b 3 , c + 2 d 3 + 2 s t f a + 2 b 3 , 2 c + d 3 2 s t f 2 a + b 3 , c + 2 d 3 + 2 s t f 2 a + b 3 , 2 c + d 3 ,
(2.17)

where A is as defined in Theorem 2.

Proof. By the concavity of 2 f s t q on the co-ordinates on Δ and power mean inequality, we note that the following inequality holds:

| 2 s t f ( λ x + ( 1 λ ( y , v ) | q λ | 2 s t f ( x , v ) | q + ( 1 λ ) | 2 s t f ( y , v ) | q ( λ | 2 s t f ( x , v ) | + ( 1 λ ) | 2 s t f ( y , v ) | ) q ,

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

Hence,

2 s t f ( λ x + ( 1 - λ ) y , v ) λ 2 s t f ( x , v ) + ( 1 - λ ) 2 s t f ( y , v ) ,

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

Similarly, we can show that

2 s t f ( u , λ z + ( 1 - λ ) w ) λ 2 s t f ( u , z ) + ( 1 - λ ) 2 s t f ( u , w ) ,

for all z, w [c, d], λ [0, 1] and for fixed u [a, d], thus 2 f s t is concave on the co-ordinates on Δ.

It is clear from Lemma 1 that

1 ( b - a ) ( d - c ) a b c d f ( x , y ) d y d x + f a + b 2 , c + d 2 - A ( b - a ) ( d - c ) 0 1 0 1 K ( t , s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t = ( b - a ) ( d - c ) 0 1 2 0 1 2 s t 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t + 0 1 2 1 2 1 t ( 1 - s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t + 1 2 1 0 1 2 s ( 1 - t ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t + 1 2 1 1 2 1 ( 1 - t ) ( 1 - s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t .
(2.18)

Since 2 f s t is concave on the co-ordinates, we have, by Jensen's inequality for integrals, that:

0 1 2 0 1 2 s t 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t = 0 1 2 t 0 1 2 s 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 0 1 2 r 0 1 2 s d s 2 s t f t a + ( 1 - t ) b , 0 1 2 s ( s c + ( 1 - s ) d ) d s 0 1 2 s d s d t = 1 8 0 1 2 t 2 s t f t a + ( 1 - t ) b , c + 2 d 3 d t 1 8 0 1 2 t d t 2 s t f 0 1 2 t ( t a + ( 1 - t ) b ) d t 0 1 2 t d t , c + 2 d 3 = 1 64 2 s t f a + 2 b 3 , c + 2 d 3 .
(2.19)

In a similar way, we also have that

0 1 2 1 2 1 t ( 1 - s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 1 64 2 s t f a + 2 b 3 , 2 c + d 3 ,
(2.20)
1 2 1 0 1 2 s ( 1 - t ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 1 64 2 s t f 2 a + b 3 , c + 2 d 3
(2.21)

and

1 2 1 1 2 1 ( 1 - t ) ( 1 - s ) 2 s t f ( t a + ( 1 - t ) b , s c + ( 1 - s ) d ) d s d t 1 64 2 s t f 2 a + b 3 , c + 2 d 3 .
(2.22)

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

References

  1. Pečarić JE, Proschan F, Tong YL: Convex Functions, Partial Ordering and Statistical Applications. Academic Press, New York; 1991.

    Google Scholar 

  2. Alomari M, Darus M, Dragomir SS: Inequalities of Hermite-Hadamard's type for functions whose derivatives absolute values are quasi-convex. RGMIA Res Rep Collect 2009., 12(suppl. 14):

  3. Dragomir SS, Agarwal RP: Two inequalities for differentiable mappings and applications to special means of real numbers and to Trapezoidal formula. Appl Math Lett 1998, 11(5):91–95. 10.1016/S0893-9659(98)00086-X

    Article  MathSciNet  Google Scholar 

  4. Kirmaci US: Inequalities for differentiable mappings and applications to special means of real numbers to midpoint formula. Appl Math Comput 2004, 147: 137–146. 10.1016/S0096-3003(02)00657-4

    Article  MathSciNet  Google Scholar 

  5. Pearce CME, Pečarić JE: Inequalities for differentiable mappings with applications to special means and quadrature formula. Appl Math Lett 2000, 13: 51–55.

    Article  Google Scholar 

  6. Dragomir SS: On Hadamard's inequality for convex functions on the co-ordinates in a rectangle from the plane. Taiwanese J Math 2001, 4: 775–788.

    Google Scholar 

  7. Dragomir SS, Pearce CEM: Selected Topics on Hermite-Hadamard Inequalities and Applications. RGMIA Monographs, Victoria University; 2000.http://www.staff.vu.edu.au/RGMIA/monographs/hermite_hadamard.html Online:

    Google Scholar 

  8. Latif MA, Alomari M: Hadamard-type inequalities for product two convex functions on the co-ordinetes. Int Math Forum 2009, 4(47):2327–2338.

    MathSciNet  Google Scholar 

  9. Alomari M, Darus M: Co-ordinated s-convex function in the first sense with some Hadamard-type inequalities. Int J Contemp Math Sci 2008, 3(32):1557–1567.

    MathSciNet  Google Scholar 

  10. Alomari M, Darus M: On the Hadamard's inequality for log-convex functions on the coordinates. J Inequal Appl 2009., 13: (Article ID 283147)

    Google Scholar 

  11. Latif MA, Alomari M: On the Hadamard-type inequalities for h-convex functions on the co-ordinetes. Int J Math Anal 2009, 3(33):1645–1656.

    MathSciNet  Google Scholar 

  12. Özdemir ME, Set E, Sarikaya MZ: New some Hadamard's type inequalities for coordinated m -convex and ( α, m )-convex functions. RGMIA Res Rep Collect 2010., 13: Supplement, (Article 4)

    Google Scholar 

  13. Sarikaya MZ, Set E, Özdemir ME, Dragomir SS: Some new Hadamard's type inequalities for co-ordinated convex functions. arXiv:1005.0700v1 [math.CA]

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MAL and SSD carried out the design of the study and performed the analysis. Both of the authors read and approved the final version of the manuscript.

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Latif, M., Dragomir, S. On some new inequalities for differentiable co-ordinated convex functions. J Inequal Appl 2012, 28 (2012). https://doi.org/10.1186/1029-242X-2012-28

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