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On some new inequalities for differentiable co-ordinated convex functions

MA Latif1* and SS Dragomir2

Author Affiliations

1 College of Science, Department of Mathematics, University of Hail, Hail-2440, Saudi Arabia

2 School of Engineering & Science, Victoria University, PO Box 14428, Melbourne City, MC 8001, Australia

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Journal of Inequalities and Applications 2012, 2012:28  doi:10.1186/1029-242X-2012-28


The electronic version of this article is the complete one and can be found online at: http://www.journalofinequalitiesandapplications.com/content/2012/1/28


Received:13 October 2011
Accepted:15 February 2012
Published:15 February 2012

© 2012 Latif and Dragomir; licensee Springer.

This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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

Keywords:
convex function; co-ordinated convex function; Hermite-Hadamard's inequality; Jensen's integral inequality

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 [2-5] and the references therein.

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

f ( λ x + ( 1 - λ ) z , λ y + ( 1 - λ ) w ) λ f ( 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 → ℝ is said to be convex on the co-ordinates on Δ if the partial mappings fy: [a, b] → ℝ, fy(u) = f(u, y) and fx: [c, d] → ℝ, fx(v) = f(x, v) are convex where defined for all x ∈ [a, b], y ∈ [c, d].

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

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

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

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

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

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 ℝ2. Motivated by notion given in [13], in the present article, we prove some new inequalities which give estimate between the middle and the leftmost terms in (1.2) for differentiable co-ordinated convex functions on rectangle from the plane ℝ2.

2. Main results

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

Lemma 1. Let f: Δ ⊆ ℝ2 → ℝ be a partial differentiable mapping on Δ: = [a, b] × [c, d] with a < b, c < d. If 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 ⊆ ℝ2 → ℝ 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 ⊆ ℝ2 → ℝ 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 following 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 ⊆ ℝ2 → ℝ 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

01 01 | K ( t , s ) | | 2 s t f ( t a + ( 1 t ) b , s c + ( 1 s ) d ) | q d s d t 01 01 | 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 ⊆ ℝ2 → ℝ 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.

Competing interests

The authors declare that they have no competing interests.

Authors' contributions

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.

Acknowledgements

This article is in final form and no version of it will be submitted for publication elsewhere.

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