Open Access

On the Hermite-Hadamard type inequalities

Journal of Inequalities and Applications20132013:228

DOI: 10.1186/1029-242X-2013-228

Received: 20 October 2012

Accepted: 18 April 2013

Published: 7 May 2013

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 1.1 Let 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 function f is 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., [210]) 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 . 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 r th 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 3.1 Let 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 3.1 Let 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 3.2 Let 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 3.3 If 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 3.4 Let 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].

Declarations

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.

Authors’ Affiliations

(1)
Department of Mathematics, China Jiliang University
(2)
Department of Mathematics, The University of Hong Kong
(3)
Department of Mathematics, Hunan Normal University

References

  1. Mitrinović DS, Pecarič JE, Fink AM Mathematics and Its Applications (East European Series) 61. In Classical and New Inequalities in Analysis. Kluwer Academic, Dordrecht; 1993.View ArticleGoogle Scholar
  2. Kavurmaci H, Avci M, Özdemir ME: New inequalities of Hermite-Hadamard type for convex functions with applications. J. Inequal. Appl. 2011., 2011: Article ID 86Google Scholar
  3. Özdemir ME, Kavurmaci H, Ocak Akdemir A, Avci M: Inequalities for convex and s -convex functions on = [ a , b ] × [ c , d ] . J. Inequal. Appl. 2012., 2012: Article ID 20Google Scholar
  4. Dragomir SS: A sequence of mappings associated with the Hermite-Hadamard inequalities and applications. Appl. Math. 2004, 49: 123–140.MathSciNetView ArticleGoogle Scholar
  5. Kotrys D: Hermite-Hadamard inequality for convex stochastic processes. Aequ. Math. 2012, 83: 143–151. 10.1007/s00010-011-0090-1MathSciNetView ArticleGoogle Scholar
  6. Makó J, Páles Z: Hermite-Hadamard inequalities for generalized convex functions. Aequ. Math. 2005, 69: 32–40. 10.1007/s00010-004-2730-1View ArticleGoogle Scholar
  7. Ger R, Pečarić J: On vector Hermite-Hadamard differences controlled by their scalar counterparts. Int. Ser. Numer. Math. 2010, 161: 165–173.Google Scholar
  8. Xi B-Y, Bai R-F, Qi F: Hermite-Hadamard type inequalities for the m - and ( α , m ) -geometrically convex functions. Filomat 2013, 27: 1–7.MathSciNetView ArticleGoogle Scholar
  9. Dragomir SS, Hunt E, Pearce CEM: Interpolating maps, the modulus map and Hadamard’s inequality. Springer Optimization and Its Applications 32. Optimization, Part 1 2009, 207–223.View ArticleGoogle Scholar
  10. Niculescu CP: The Hermite-Hadamard inequality for convex functions of a vector variable. Math. Inequal. Appl. 2002, 5(4):619–623.MathSciNetGoogle Scholar
  11. Set E, Özdemir ME, Dragomir SS: On the Hermite-Hadamard inequality and other integral inequalities involving two functions. J. Inequal. Appl. 2010., 2010: Article ID 148102Google Scholar
  12. Pachpatte BG: Inequalities for Differentiable and Integral Equations. Academic Press, Boston; 1997.Google Scholar
  13. Bougoffa L: On Minkowski and Hardy integral inequalities. J. Inequal. Pure Appl. Math. 2006., 7(2): Article ID 60MathSciNetGoogle Scholar

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