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On the Hermite-Hadamard type inequalities

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.

1 Introduction

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

Theorem 1.1 Let f:[a,b]RR 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)dx 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)dx 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 +(1λ) x 2 ,λ y 1 +(1λ) y 2 D whenever x( x 1 , y 1 ),y( x 2 , y 2 )D and 0λ1.

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

whenever ( x 1 , y 1 ),( x 2 , y 2 )D and 0λ1.

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 ] λf( x 1 , y 1 )+(1λ)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 rR{+,}. 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 p1. 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 ,1p<,

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)dxdy,
(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)dxdy 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 λf ( a , b ) p +(1λ)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 ] λf(a,b)+(1λ)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)dxdyf ( 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)dx,

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

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

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

Theorem 3.2 Let p1 and let a b c d f ( x , y ) p dxdy< and a b c d g ( x , y ) p dxdy<, 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 dxdy

and

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

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,q1, 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].

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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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Correspondence to Chang-Jian Zhao.

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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.

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Zhao, CJ., Cheung, WS. & Li, XY. On the Hermite-Hadamard type inequalities. J Inequal Appl 2013, 228 (2013). https://doi.org/10.1186/1029-242X-2013-228

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Keywords

  • Hermite-Hadamard inequality
  • Barnes-Godunova-Levin inequality
  • Minkowski integral inequality
  • Hölder inequality