Open Access Research

On the Hermite-Hadamard type inequalities

Chang-Jian Zhao1*, Wing-Sum Cheung2 and Xiao-Yan Li3

Author Affiliations

1 Department of Mathematics, China Jiliang University, Hangzhou, 310018, P.R. China

2 Department of Mathematics, The University of Hong Kong, Pokfulam Road, Hong Kong, P.R. China

3 Department of Mathematics, Hunan Normal University, Changsha, 410000, P.R. China

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Journal of Inequalities and Applications 2013, 2013:228  doi:10.1186/1029-242X-2013-228


The electronic version of this article is the complete one and can be found online at: http://www.journalofinequalitiesandapplications.com/content/2013/1/228


Received:20 October 2012
Accepted:18 April 2013
Published:7 May 2013

© 2013 Zhao et al.; 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

In the present paper, we establish some new Hermite-Hadamard type inequalities involving two functions. Our results in a special case yield recent results on Hermite-Hadamard type inequalities.

MSC: 26D15.

Keywords:
Hermite-Hadamard inequality; Barnes-Godunova-Levin inequality; Minkowski integral inequality; Hölder inequality

1 Introduction

The following inequality is well known in the literature as Hermite-Hadamard’s inequality [1].

Theorem 1.1Let f : [ a , b ] R R be a convex function on an interval of real numbers. Then the following Hermite-Hadamard inequality for convex functions holds:

f ( a + b 2 ) 1 b a a b f ( x ) d x f ( a ) + f ( b ) 2 . (1.1)

If the functionfis concave, the inequality (1.1) can be written as follows:

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

Recently, many generalizations, extensions and variants of this inequality have appeared in the literature (see, e.g., [2-10]) and the references given therein. In particular, in 2010, Özdemir and Dragomir [11] established some new Hermite-Hadamard inequalities and other integral inequalities involving two functions in ℝ. Following this work, the main purpose of the present paper is to establish some dual Hermite-Hadamard type inequalities involving two functions in R 2 . Our results provide some new estimates on such type of inequalities.

2 Preliminaries

A region D R 2 is called convex if it contains the close line segment joining any two of its points, or equivalently, if λ x 1 + ( 1 λ ) x 2 , λ y 1 + ( 1 λ ) y 2 D whenever x ( x 1 , y 1 ) , y ( x 2 , y 2 ) D and 0 λ 1 .

Let z = f ( x , y ) be a duality function on the convex region D R 2 . z = f ( x , y ) is called a duality convex function on the convex region D if

f [ λ x 1 + ( 1 λ ) x 2 , λ y 1 + ( 1 λ ) y 2 ] λ f ( x 1 , y 1 ) + ( 1 λ ) f ( x 2 , y 2 ) , (2.1)

whenever ( x 1 , y 1 ) , ( x 2 , y 2 ) D and 0 λ 1 .

If the function f ( x , y ) is concave, the inequality (2.1) can be written as follows:

f [ λ x 1 + ( 1 λ ) x 2 , λ y 1 + ( 1 λ ) y 2 ] λ f ( x 1 , y 1 ) + ( 1 λ ) f ( x 2 , y 2 ) . (2.2)

Let x = ( x 11 , , x 1 n , , x m 1 , , x m n ) and p = ( p 11 , , p 1 n , , p m 1 , , p m n ) be two positive nm-tuples, and let r R { + , } . Then, on putting P m n = k 2 = 1 n k 1 = 1 m p k 1 k 2 , it easy follows that if r < s + , then

M m n [ r ] M m n [ s ] (2.3)

(also see, e.g., [[1], p.15]). Here, the rth power mean of x with weights p is the following: M m n [ r ] = ( 1 P m n k 2 = 1 n k 1 = 1 m p k 1 k 2 x k 1 k 2 r ) 1 / r if r + , 0 , ; M m n [ r ] = ( k 2 = 1 n k 1 = 1 m x k 1 k 2 p k 1 k 2 ) P m n if r = 0 ; M m n [ r ] = min ( x 11 , , x 1 n , , x n 1 , , x m n ) if r = and M m n [ r ] = max ( x 11 , , x 1 n , , x n 1 , , x m n ) if r = + .

Let f ( x , y ) : [ a , b ] × [ c , d ] R , and p 1 . Now, we define the p-norm of the function f ( x , y ) on [ a , b ] × [ c , d ] as follows:

f ( x , y ) p = ( a b c d | f ( x , y ) | p d x d y ) 1 / p , 1 p < ,

and

f ( x , y ) p = sup | f ( x , y ) | , p = ,

and L p ( [ a , b ] × [ c , d ] ) is the set of all functions f ( x , y ) : [ a , b ] × [ c , d ] R such that f ( x , y ) p < .

Lemma 2.1 (see [12]) (Barnes-Godunova-Levin inequality)

Let f ( x , y ) , g ( x , y ) be nonnegative concave functions on [ a , b ] × [ c , d ] , then for p , q > 1 we have

f ( x , y ) p g ( x , y ) q B ( p , q ) a b c d f ( x , y ) g ( x , y ) d x d y , (2.4)

where

B ( p , q ) = 6 [ ( b a ) ( d c ) ] 1 / p + 1 / q 1 ( p + 1 ) 1 / p ( q + 1 ) 1 / q .

Lemma 2.2 (see [1]) (Hermite-Hadamard inequality)

Let f ( x , y ) : [ a , b ] × [ c , d ] R 2 R be a convex function. Then the following dual Hermite-Hadamard inequality for convex functions holds:

f ( a + c 2 , b + d 2 ) 1 ( b a ) ( d c ) a b c d f ( x , y ) d x d y f ( a , b ) + f ( c , d ) 2 . (2.5)

The inequality is reversed if the function f ( x , y ) is concave.

Lemma 2.3 (see [13]) (A reversed Minkowski integral inequality)

Let f ( x , y ) and g ( x , y ) be positive functions satisfying

0 < m f ( x , y ) g ( x , y ) M , ( x , y ) [ a , b ] × [ c , d ] . (2.6)

Then

f ( x , y ) p + g ( x , y ) p c f ( x , y ) + g ( x , y ) p , (2.7)

where c = [ M ( m + 1 ) + ( M + 1 ) ] / [ ( m + 1 ) ( M + 1 ) ] .

3 Main results

Our main results are established in the following theorems.

Theorem 3.1Let p , q > 1 and let f ( x , y ) , g ( x , y ) : [ a , b ] × [ c , d ] R be nonnegative functions such that f ( x , y ) p and g ( x , y ) q are concave on [ a , b ] × [ c , d ] . Then

f ( a , b ) + f ( c , d ) 2 × g ( a , b ) + g ( c , d ) 2 1 [ ( b a ) ( d c ) ] 1 / p + 1 / q B ( p , q ) a b c d f ( x , y ) g ( x , y ) d x d y , (3.1)

where B ( p , q ) is the Barnes-Godunova-Levin constant given by (2.4).

Proof Observe that whenever f p ( x , y ) is concave on [ a , b ] × [ c , d ] , the nonnegative function f ( x , y ) is also concave on [ a , b ] × [ c , d ] . Namely,

f [ λ a + ( 1 λ ) c , λ b + ( 1 λ ) d ] p λ f ( a , b ) p + ( 1 λ ) f ( c , d ) p ,

that is,

f [ λ a + ( 1 λ ) c , λ b + ( 1 λ ) d ] ( ( λ f ( a , b ) p + ( 1 λ ) f ( c , d ) p ) ) 1 / p ,

and p > 1 , using the power-mean inequality (2.3), we obtain

f [ λ a + ( 1 λ ) c , λ b + ( 1 λ ) d ] λ f ( a , b ) + ( 1 λ ) f ( c , d ) .

For q > 1 , similarly, if g q ( x , y ) is concave on [ a , b ] × [ c , d ] , the nonnegative function g ( x , y ) is concave on [ a , b ] × [ c , d ] .

In view that f p ( x , y ) and g q ( x , y ) are concave functions on [ a , b ] × [ c , d ] , from Lemma 2.2, we get

( f ( a , b ) p + f ( c , d ) p 2 ) 1 / p 1 [ ( b a ) ( d c ) ] 1 / p ( a b c d f ( x , y ) p d x d y ) 1 / p f ( a + c 2 , b + d 2 ) , (3.2)

and

( g ( a , b ) p + g ( c , d ) q 2 ) 1 / q 1 [ ( b a ) ( d c ) ] 1 / q ( a b c d g ( x , y ) q d x d y ) 1 / q g ( a + c 2 , b + d 2 ) . (3.3)

By multiplying the above inequalities, we obtain

( f ( a , b ) p + f ( c , d ) p 2 ) 1 / p ( g ( a , b ) p + g ( c , d ) q 2 ) 1 / q 1 [ ( b a ) ( d c ) ] 1 / p + 1 / q ( a b c d f ( x , y ) p d x d y ) 1 / p ( a b c d g ( x , y ) q d x d y ) 1 / q . (3.4)

If p , q > 1 , then it is easy to show that

( f ( a , b ) p + f ( c , d ) p 2 ) 1 / p f ( a , b ) + f ( c , d ) 2 , (3.5)

and

( g ( a , b ) q + g ( c , d ) q 2 ) 1 / q g ( a , b ) + g ( c , d ) 2 . (3.6)

Thus, by applying Barnes-Godunova-Levin inequality to the right-hand side of (3.4) with (3.5), (3.6), we get (3.1).

The proof is complete. □

Remark 3.1 By multiplying inequalities (3.2), (3.3), we obtain

1 [ ( b a ) ( d c ) ] 1 / p + 1 / q ( a b c d f ( x , y ) p d x d y ) 1 / p ( a b c d g ( x , y ) q d x d y ) 1 / q f ( a + c 2 , b + d 2 ) g ( a + c 2 , b + d 2 ) . (3.7)

By applying the Hölder inequality to the left-hand side of (3.7) with ( 1 / p ) + ( 1 / q ) = 1 , we get

1 ( b a ) ( d c ) a b c d f ( x , y ) g ( x , y ) d x d y f ( a + c 2 , b + d 2 ) g ( a + c 2 , b + d 2 ) . (3.8)

Remark 3.2 Let f ( x , y ) and g ( x , y ) change to f ( x ) and g ( x ) , respectively, and with suitable changes in Theorem 3.1 and Remark 3.1, we have the following.

Corollary 3.1Let p , q > 1 and let f ( x ) , g ( x ) : [ a , b ] R , a < b , be nonnegative functions such that f ( x ) p and g ( x ) q are concave on [ a , b ] . Then

f ( a ) + f ( b ) 2 g ( a ) + g ( b ) 2 1 ( b a ) 1 / p + 1 / q B ( p , q ) a b f ( x ) g ( x ) d x ,

and if ( 1 / p ) + ( 1 / q ) = 1 , then one has

1 b a a b f ( x ) g ( x ) d x f ( a + b 2 ) g ( a + b 2 ) .

This is just Theorem 2.1 established by Özdemir and Dragomir [11].

Theorem 3.2Let p 1 and let a b c d f ( x , y ) p d x d y < and a b c d g ( x , y ) p d x d y < , and let f ( x , y ) , g ( x , y ) : [ a , b ] × [ c , d ] R be positive functions with

0 < m f ( x , y ) g ( x , y ) M , ( x , y ) [ a , b ] × [ c , d ] .

Then

f ( x , y ) p 2 + g ( x , y ) p 2 ( ( M + 1 ) ( m + 1 ) M 2 ) f ( x , y ) p g ( x , y ) p . (3.9)

Proof Since f ( x , y ) , g ( x , y ) are positive, as in the proof of Lemma 2.3 (see [[13], p.2]), we have

( a b c d f ( x , y ) p d x d y ) 1 / p M M + 1 ( a b c d ( f ( x , y ) + g ( x , y ) ) p d x d y ) 1 / p

and

( a b c d g ( x , y ) p d x d y ) 1 / p 1 m + 1 ( a b c d ( f ( x , y ) + g ( x , y ) ) p d x d y ) 1 / p .

By multiplying the above inequalities and in view of the Minkowski inequality, we get

( a b c d f ( x , y ) p d x d y ) 1 / p ( a b c d g ( x , y ) p d x d y ) 1 / p M ( M + 1 ) ( m + 1 ) ( a b c d ( f ( x , y ) + g ( x , y ) ) p d x d y ) 2 / p M ( M + 1 ) ( m + 1 ) ( ( a b c d f ( x , y ) p d x d y ) 1 / p + ( a b c d g ( x , y ) p d x d y ) 1 / p ) 2 . (3.10)

Hence

( a b c d f ( x , y ) p d x d y ) 2 / p + ( a b c d g ( x , y ) p d x d y ) 2 / p ( ( M + 1 ) ( m + 1 ) M 2 ) ( a b c d f ( x , y ) p d x d y ) 1 / p ( a b c d g ( x , y ) p d x d y ) 1 / p .

This proof is complete. □

Remark 3.3 Let f ( x , y ) and g ( x , y ) change to f ( x ) and g ( x ) , respectively, and with suitable changes in (3.9), (3.9) reduces to an inequality established by Özdemir and Dragomir [11].

Theorem 3.3If f p ( x , y ) and g q ( x , y ) are as in Theorem 3.1, then the following inequality holds:

1 ( b a ) ( d c ) f ( x , y ) p p g ( x , y ) q q ( f ( a , b ) + f ( c , d ) ) p ( g ( a , b ) + g ( c , d ) ) q 2 p + q . (3.11)

Proof If f p ( x , y ) and g q ( x , y ) are concave on [ a , b ] × [ c , d ] , then from Lemma 2.2, we get

f ( a , b ) p + f ( c , d ) p 2 1 ( b a ) ( d c ) a b c d f ( x , y ) p d x d y

and

g ( a , b ) q + g ( c , d ) q 2 1 ( b a ) ( d c ) a b c d g ( x , y ) q d x d y ,

which imply that

[ f ( a , b ) p + f ( c , d ) p ] [ g ( a , b ) q + g ( c , d ) q ] 4 1 [ ( b a ) ( d c ) ] 2 a b c d f ( x , y ) p d x d y a b c d g ( x , y ) q d x d y . (3.12)

On the other hand, if p , q 1 , from (2.3) we get

f ( a , b ) p + f ( c , d ) p 2 2 p [ f ( a , b ) + f ( c , d ) ] p

and

g ( a , b ) q + g ( c , d ) q 2 2 q [ g ( a , b ) + g ( c , d ) ] q ,

which imply that

[ f ( a , b ) p + f ( c , d ) p ] [ g ( a , b ) p + g ( c , d ) q ] 4 2 p q [ f ( a , b ) + f ( c , d ) ] p [ g ( a , b ) + g ( c , d ) ] q . (3.13)

Combining (3.12) and (3.13), we obtain the desired inequality as

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

This proof is complete. □

Remark 3.4 Let f ( x , y ) and g ( x , y ) change to f ( x ) and g ( x ) , respectively, and with suitable changes in (3.11), (3.11) reduces to an inequality established by Özdemir and Dragomir [11].

Theorem 3.4Let f ( x , y ) , g ( x , y ) : [ a , b ] × [ c , d ] R + be functions such that f ( x , y ) p , g ( x , y ) q and f ( x , y ) g ( x , y ) are in L 1 ( [ a , b ] × [ c , d ] ) , and

0 < m f ( x , y ) g ( x , y ) M , ( x , y ) [ a , b ] × [ c , d ] , a , b , c , d [ 0 , ) .

Then

a b c d f ( x , y ) g ( x , y ) d x d y c 1 ( f ( x , y ) p p + g ( x , y ) p p 2 ) + c 2 ( f ( x , y ) q q + g ( x , y ) q q 2 ) , (3.14)

where

c 1 = 2 p p ( M M + 1 ) p , c 2 = 2 q q ( 1 m + 1 ) q ,

and ( 1 / p ) + ( 1 / q ) = 1 with p > 1 .

Proof Since 0 < m f ( x , y ) g ( x , y ) M , ( x , y ) [ a , b ] × [ c , d ] , we have

f ( x , y ) M M + 1 ( f ( x , y ) + g ( x , y ) )

and

g ( x , y ) 1 m + 1 ( f ( x , y ) + g ( x , y ) ) .

In view of the Young-type inequality and using the elementary inequality

( a + b ) p 2 p 1 ( a p + b p ) , p > 1 , a , b R + ,

we have

a b c d f ( x , y ) g ( x , y ) d x d y 1 p ( M M + 1 ) p a b c d ( f ( x , y ) + g ( x , y ) ) p d x d y + 1 q ( 1 m + 1 ) q a b c d ( f ( x , y ) + g ( x , y ) ) q d x d y 1 p ( M M + 1 ) p 2 p 1 a b c d [ f ( x , y ) p + g ( x , y ) p ] d x d y + 1 q ( 1 m + 1 ) q 2 q 1 a b c d [ f ( x , y ) q + g ( x , y ) q ] d x d y .

This completes the proof. □

Remark 3.5 Let f ( x , y ) and g ( x , y ) change to f ( x ) and g ( x ) , respectively, and with suitable changes in (3.14), (3.14) reduces to an inequality established by Özdemir and Dragomir [11].

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

C-JZ, W-SC and X-YL jointly contributed to the main results Theorems 3.1-3.4. All authors read and approved the final manuscript.

Acknowledgements

The first author’s research is supported by Natural Science Foundation of China (10971205). The second author’s research is partially supported by a HKU Seed Grant for Basic Research.

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