# Hermite-Hadamard inequality for functions whose derivatives absolute values are preinvex

Ali Barani1*, Amir G Ghazanfari1 and Sever S Dragomir23

Author Affiliations

1 Department of Mathematics, Lorestan University, P.O. Box 465, Khoramabad, Iran

2 School of Engineering and Science, Victoria University, P.O. Box 14428, Melbourne City, VIC, 8001, Australia

3 School of Computational and Applied Mathematics, University of the Witwatersrand, Private Bag 3, Johannesburg, Wits, 2050, South Africa

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Journal of Inequalities and Applications 2012, 2012:247  doi:10.1186/1029-242X-2012-247

The electronic version of this article is the complete one and can be found online at: http://www.journalofinequalitiesandapplications.com/content/2012/1/247

 Received: 6 August 2011 Accepted: 28 May 2012 Published: 29 October 2012

© 2012 Barani 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 this article, we extend some estimates of the right-hand side of a Hermite-Hadamard-type inequality for preinvex functions. Then, a generalization to functions of several variables on invex subsets of is introduced.

##### Keywords:
Hermite-Hadamard inequality; invex sets; preinvex functions

### 1 Introduction and preliminary

Let I = [ c , d ] be an interval on the real line ℝ, let f : I R be a convex function and let a , b [ c , d ] , a < b . We consider the well-known Hadamard’s inequality

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

Both inequalities hold in the reversed direction if f is concave. We note that Hadamard’s inequality may be regarded as a refinement of the concept of convexity and it follows easily from Jensen’s inequality. Hadamard’s inequality for convex functions has received renewed attention in recent years and a remarkable variety of refinements and generalizations have been found (see, for example, [1-4]).

The classical Hermite-Hadamard inequality provides estimates of the mean value of a continuous convex function f : [ a , b ] R .

Dragomir and Agarwal [5] used the formula,

f ( a ) + f ( b ) 2 1 b a a b f ( x ) d x = b a 2 0 1 ( 1 2 t ) f ( t a + ( 1 t ) b ) d t (2)

to prove the following results.

Theorem 1.1Assume a , b R with a < b and f : [ a , b ] R is a differentiable function on ( a , b ) . If | f | is convex on [ a , b ] then the following inequality holds true

| f ( a ) + f ( b ) 2 1 b a a b f ( x ) d x | ( b a ) ( | f ( a ) | + | f ( b ) | ) 8 . (3)

Theorem 1.2Assume a , b R with a < b and f : [ a , b ] R is a differentiable function on ( a , b ) . Assume p R with p > 1 . If | f | p p 1 is convex on [ a , b ] then the following inequality holds true

| f ( a ) + f ( b ) 2 1 b a a b f ( x ) d x | b a 2 ( p + 1 ) 1 / p [ | f ( a ) | p p 1 + | f ( b ) | p p 1 2 ] p 1 p . (4)

Ion [6] presented some estimates of the right-hand side of a Hermite-Hadamard-type inequality in which some quasi-convex functions are involved.

In recent years, several extensions and generalizations have been considered for classical convexity. A significant generalization of convex functions is that of invex functions introduced by Hanson [7]. Weir and Mond [8] introduced the concept of preinvex functions and applied it to the establishment of the sufficient optimality conditions and duality in nonlinear programming. Aslam Noor [9,10] introduced the Hermite-Hadamard inequality for preinvex and log-preinvex functions.

In this article, we generalize the results in [6] for functions whose first derivatives absolute values are preinvex. Also some results for functions whose second derivatives absolute values are preinvex will be given. Now, we recall some notions in invexity analysis which will be used throughout the article (see [11,12] and references therein).

Definition 1.1 A set is said to be invex with respect to the map , if for every x , y S and t [ 0 , 1 ] ,

y + t η ( x , y ) S . (5)

It is obvious that every convex set is invex with respect to the map η ( x , y ) = x y , but there exist invex sets which are not convex (see [11]). Let be an invex set with respect to . For every x , y S the η-path P x v joining the points x and v : = x + η ( y , x ) is defined as follows

P x v : = { z : z = x + t η ( y , x ) : t [ 0 , 1 ] } .

Definition 1.2 Let be an invex set with respect to . Then, the function is said to be preinvex with respect to η, if for every x , y S and t [ 0 , 1 ] ,

f ( y + t η ( x , y ) ) t f ( x ) + ( 1 t ) f ( y ) . (6)

Every convex function is a preinvex with respect to the map η ( x , y ) = x y but the converse does not holds. For properties and applications of preinvex functions, see [12,13] and references therein.

The organization of the article is as follows: In Section 2, some generalizations of Hermite-Hadamard-type inequality for first-order differentiable functions are given. Section 3 is devoted to a generalization to several variable preinvex functions. Hermite-Hadamard-type inequality for second-order differentiable functions are studied in Section 23.

### 2 First-order differentiable functions

In this section, we introduce some generalizations of Hermite-Hadamard-type inequality for functions whose first derivatives absolute values are preinvex. We begin with the following lemma which is a generalization of Lemma 2.1 in [5] to invex setting.

Lemma 2.1Letbe an open invex subset with respect toand a , b A with θ ( a , b ) 0 . Suppose thatis a differentiable function. If f is integrable on theθ-path P b c , c = b + θ ( a , b ) then, the following equality holds

(7)

Proof Suppose that a , b A . Since A is an invex set with respect to θ, for every t [ 0 , 1 ] we have b + t θ ( a , b ) A . Integrating by parts implies that

(8)

which completes the proof. □

Theorem 2.1Letbe an open invex subset with respect to. Suppose thatis a differentiable function. If | f | is preinvex onAthen, for every a , b A with θ ( a , b ) 0 the following inequality holds

(9)

Proof Suppose that a , b A . Since A is an invex set with respect to θ, for every t [ 0 , 1 ] we have b + t θ ( a , b ) A . By preinvexity of | f | and Lemma 2.1 we get

(10)

where,

0 1 | 1 2 t | ( 1 t ) d t = 0 1 | 1 2 t | t d t = 1 4 .

□

Now, we give an example of an invex set with respect to an θ which is satisfies the conditions of Theorem 2.1.

Example 2.1 Suppose that K : = ( 3 , 1 ) ( 1 , 4 ) and the function is defined by

θ ( x , y ) = { x y , x > 0 , y > 0 , x y , x < 0 , y < 0 , 3 y , x < 0 , y > 0 , 2 y , x > 0 , y < 0 .

Clearly K is an open invex set with respect to θ. Suppose that a ( 3 , 1 ) and b ( 1 , 4 ) , b 3 hence, θ ( a , b ) = 3 b 0 . Now,

P b c = [ b , 3 ] , b < 3 ,

and

P b c = [ 3 , b ] , b > 3 ,

where c = b + θ ( a , b ) .

Another similar result is embodied in the following theorem.

Theorem 2.2Letbe an open invex subset with respect to. Suppose thatis a differentiable function. Assume thatwith p > 1 . If | f | p / p 1 is preinvex onAthen, for every a , b A with θ ( a , b ) 0 the following inequality holds

(11)

Proof Suppose that a , b A . By assumption, Hölder’s inequality and the proof of Theorem 4.1 we have

(12)

where q : = p / ( p 1 ) . □

Note that if A = [ a , b ] and θ ( x , y ) = x y for every x , y A then, we can deduce Theorems 1.1 and 1.2, from Theorems 2.1 and 2.2, respectively.

### 3 An extension to several variables functions

The aim of this section is to extend the Proposition 1 in [6] and Theorem 2.2 to functions of several variables defined on invex subsets of .

The mapping is said to be satisfies the condition C if for every x , y S and t [ 0 , 1 ] ,

Note that, in Example 2.1, θ satisfies the condition C.

For every x , y S and every t 1 , t 2 [ 0 , 1 ] from condition C we have

η ( y + t 2 η ( x , y ) , y + t 1 η ( x , y ) ) = ( t 2 t 1 ) η ( x , y ) , (13)

see [12] for details.

Proposition 3.1Letbe an invex set with respect toandis a function. Suppose thatηsatisfies conditionConS. Then, for every x , y S the functionfis preinvex with respect toηonη-path P x v if and only if the functiondefined by

φ ( t ) : = f ( x + t η ( y , x ) ) ,

is convex on [ 0 , 1 ] .

Proof Suppose that φ is convex on [ 0 , 1 ] and z 1 : = x + t 1 η ( y , x ) P x v , z 2 : = x + t 2 η ( y , x ) P x v . Fix λ [ 0 , 1 ] . Since η satisfies condition C, by (13) we have

f ( z 1 + λ η ( z 2 , z 1 ) ) = f ( x + ( ( 1 λ ) t 1 + λ t 2 ) η ( y , x ) ) = φ ( ( 1 λ ) t 1 + λ t 2 ) ( 1 λ ) φ ( t 1 ) + λ φ ( t 2 ) = ( 1 λ ) f ( z 1 ) + λ f ( z 2 ) . (14)

Hence, f is preinvex with respect to η on η-path P x v .

Conversely, let x , y S and the function f be preinvex with respect to η on η-path P x v . Suppose that t 1 , t 2 [ 0 , 1 ] . Then, for every λ [ 0 , 1 ] we have

φ ( ( 1 λ ) t 1 + λ t 2 ) = f ( x + ( ( 1 λ ) t 1 + λ t 2 ) η ( y , x ) ) = f ( x + t 1 η ( y , x ) ) + λ η ( x + t 2 η ( y , x ) , x + t 1 η ( y , x ) ) λ f ( x + t 2 η ( y , x ) ) + ( 1 λ ) f ( x + t 1 η ( y , x ) ) = λ φ ( t 2 ) + ( 1 λ ) φ ( t 1 ) . (15)

Therefore, φ is quasi-convex on [ 0 , 1 ] . □

The following theorem is a generalization of Proposition 1 in [6].

Theorem 3.1Letbe an open invex set with respect to. Assume thatηsatisfies conditionC. Suppose that for every x , y S the functionis preinvex with respect toηonη-path P x v . Then, for every a , b ( 0 , 1 ) with a < b the following inequality holds,

(16)

Proof Let x , y S and a , b ( 0 , 1 ) with a < b . Since f is preinvex with respect to η on η-path P x v by Proposition 3.1 the function defined by

φ ( t ) : = f ( x + t η ( y , x ) ) ,

is convex on [ 0 , 1 ] . Now, we define the function as follows

ϕ ( t ) : = 0 t φ ( s ) d s = 0 t f ( x + s η ( y , x ) ) d s .

Obviously for every t ( 0 , 1 ) we have

ϕ ( t ) = φ ( t ) = f ( x + t η ( y , x ) ) 0 ,

hence, | ϕ ( t ) | = ϕ ( t ) . Applying Theorem 1.1 to the function ϕ implies that

| ϕ ( a ) + ϕ ( b ) 2 1 b a a b ϕ ( s ) d s | ( b a ) ( ϕ ( a ) + ϕ ( b ) ) 8 ,

and we deduce that (16) holds. □

### 4 Second-order differentiable functions

In this section, we introduce some generalizations of Hermite-Hadamard-type inequality for functions whose second derivatives absolute values are preinvex. We begin with the following lemma (see Lemma 1 in [14] and Lemma 4 in [15]).

Lemma 4.1Letbe an open invex subset with respect toand a , b A with θ ( a , b ) 0 . Suppose thatis a differentiable function. If f is integrable on theθ-path P b c , c = b + θ ( a , b ) then, the following equality holds

(17)

Proof Suppose that a , b A . Since A is an invex set with respect to θ, for every t [ 0 , 1 ] we have b + t θ ( a , b ) A . Integrating by parts implies that

(18)

which completes the proof. □

Theorem 4.1Letbe an open invex subset with respect toand θ ( a , b ) 0 for all a b . Suppose thatis a twice differentiable function onA. If | f | is preinvex onAand f is integrable on theθ-path P b c , c = b + θ ( a , b ) then, the following inequality holds

(19)

Proof Suppose that a , b A . Since A is an invex set with respect to θ, for every t [ 0 , 1 ] we have b + t θ ( a , b ) A . By preinvexity of | f | and Lemma 4.1 we get

(20)

which completes the proof. □

The corresponding version for powers of the absolute value of the second derivative is incorporated in the following theorem.

Theorem 4.2Letbe an open invex subset with respect toand θ ( a , b ) 0 for all a b . Suppose thatis a twice differentiable function onAand | f | p p 1 is preinvex onA, for p > 1 . If f is integrable on theθ-path P b c , c = b + θ ( a , b ) then, the following inequality holds

(21)

Proof By preinvexity of | f | , Lemma 4.1 and using the well-known Hölder integral inequality, we get

which completes the proof. □

A more general inequality is given using Lemma 4.1, as follows:

Theorem 4.3Letbe an open invex subset with respect toand θ ( a , b ) 0 for all a b . Suppose thatis a twice differentiable function onAand | f | q is preinvex onA, for q > 1 . If f is integrable on theθ-path P b c , c = b + θ ( a , b ) then, the following inequality holds

(22)

Proof By preinvexity of | f | , Lemma 4.1 and using the well-known weighted power mean inequality, we get

which completes the proof. □

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

All authors conceived of the study, and participated in the design and coordination. All authors read and approved the final manuscript.

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