Skip to main content

Advertisement

Several integral inequalities and an upper bound for the bidimensional Hermite-Hadamard inequality

Article metrics

  • 2071 Accesses

  • 2 Citations

Abstract

In this paper we prove several integral inequalities and we find an upper bound of the Hermite-Hadamard inequality for a convex function on a bounded area from the plane in special cases.

1 Introduction

Let f be a convex function on [a,b]. Then we have the following inequality, which is called Hermite-Hadamard inequality:

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

There are many extensions, generalizations and similar results of inequality (1.1). In [1], Fejer established the following weighted generalization of inequality (1.1).

Theorem 1.1 If f:[a,b]R is a convex function, then the inequality

f ( a + b 2 ) a b w(x)dx a b f(x)w(x)dx f ( a ) + f ( b ) 2 a b w(x)dx
(1.2)

holds, where w:[a,b]R is non-negative, integrable and symmetric about a + b 2 .

In [2], Yang and Tseng proved the following theorem which refines inequality (1.2).

Theorem 1.2 Let f and w be defined as in Theorem  1.1. If P:[a,b]R is defined by

P(t)= a b f [ t x + ( 1 t ) a + b 2 ] w(x)dx,

then P is convex, increasing on [0,1] and for all t[0,1],

f ( a + b 2 ) a b w(x)dx=P(0)P(t)P(1)= a b f(x)w(x)dx.

In this paper, we find an upper bound for a b f(x)g(x)dx, where f is a convex function on [a,b] and g is non-negative increasing (or decreasing) on [a,b], and a b g(t)dt=1. Finally, in Section 3 we find an upper bound for the following integral:

1 a b ( g ( x ) h ( x ) ) d x a b h ( x ) g ( x ) F(x,y)dydx.

2 Integral inequalities

Theorem 2.1 Let f:[a,b]R be a differentiable convex function and g:[a,b][0,] be a continuous function.

  1. (i)

    If g is decreasing on [a,b], then

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

    If g is increasing on [a,b], then

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

Proof

  1. (i)

    Denote

    H(x)= a x f(t)g(t)dt 1 2 ( f ( a ) + f ( x ) ) a x g(t)dt.

We will show that H (x)0. We have

H ( x ) = f ( x ) g ( x ) 1 2 f ( x ) a x g ( t ) d t 1 2 ( f ( a ) + f ( x ) ) g ( x ) = 1 2 [ g ( x ) ( f ( x ) f ( a ) ) f ( x ) a x g ( t ) d t ] .

By the extended mean value theorem (Cauchy’s theorem), we have

f ( x ) f ( a ) a x g ( t ) d t = f ( ζ ) g ( ζ ) (a<ζ<x).

On the other hand, by the convexity of f and decreasing of g, we obtain

f ( x ) f ( a ) a x g ( t ) d t = f ( ζ ) g ( ζ ) f ( x ) g ( x ) .

Since g is non-negative,

H (x)= 1 2 [ ( f ( x ) f ( a ) ) g ( x ) f ( x ) a x g ( t ) d t ] 0,

which implies that H is decreasing. Hence, H(b)H(a)=0. The proof is complete.

  1. (ii)

    Denote

    H(x)=f ( a + x 2 ) a x g(t)dt a x f(t)g(t)dt.

Then we have

H ( x ) = 1 2 f ( a + x 2 ) a x g ( t ) d t + f ( a + x 2 ) g ( x ) f ( x ) g ( x ) = 1 2 f ( a + x 2 ) a x g ( t ) d t g ( x ) ( f ( x ) f ( a + x 2 ) ) .

By the mean value theorem (Lagrange’s theorem), there exist ζ 1 ( a + x 2 ,x) and ζ 2 (a,x) such that

f ( x ) f ( a + x 2 ) x a + x 2 = f ( ζ 1 )and a x g ( t ) d t 0 x a =g( ζ 2 ).

Hence,

2 ( f ( x ) f ( a + x 2 ) ) a x g ( t ) d t = f ( ζ 1 ) g ( ζ 2 ) .

By the convexity of f and increasing of g, we obtain

2 ( f ( x ) f ( a + x 2 ) ) a x g ( t ) d t = f ( ζ 1 ) g ( ζ 2 ) f ( a + x 2 ) g ( x ) .

So,

H (x)= 1 2 f ( a + x 2 ) a x g(t)dtg(x) ( f ( x ) f ( a + x 2 ) ) 0.

Therefore, H is decreasing and H(b)H(a)=0. The proof is complete. □

Theorem 2.2 Let f:[a,b]R be a convex function and P:[a,b][0,) be an integrable function such that a b P(x)dx=1. Then

a b f(x)P(x)dx b f ( a ) a f ( b ) b a + f ( b ) f ( a ) b a a b xP(x)dx.

Proof

We have

1 b a a b f ( x ) P ( x ) d x = 0 1 f ( t b + ( 1 t ) a ) P ( t b + ( 1 t ) a ) d t f ( b ) 0 1 t P ( b t + ( 1 t ) a ) d t + f ( b ) 0 1 ( 1 t ) P ( b t + ( 1 t ) a ) d t = f ( b ) a b x a b a P ( x ) d x b a + f ( a ) a b b x b a P ( x ) d x b a = f ( b ) ( b a ) 2 [ a b x P ( x ) a a b P ( x ) d x ] + f ( a ) ( b a ) 2 [ b a b P ( x ) d x a b x P ( x ) d x ] = f ( b ) ( b a ) 2 [ a b x P ( x ) d x a ] + f ( a ) ( b a ) 2 [ b a b x P ( x ) d x ] .

So, we get

a b f(x)P(x)dx b f ( a ) a f ( b ) b a + f ( b ) f ( a ) b a a b xP(x)dx.

 □

Corollary 2.1 Let f:[a,b]R be a convex function and g be a non-negative integrable function. Then

a b f(x)dx f ( a ) + f ( b ) 2

and

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

The proof is similar to the proof of theorem.

3 Right bidimensional Hermite-Hadamard inequality

Let us consider the bidimensional interval =[a,b]×[c,d] in R 2 . Recall that the mapping f:R is convex on if

f ( λ x + ( 1 λ ) z , λ y + ( 1 λ ) w ) λf(x,y)+(1λ)f(z,w)

holds for all (x,y),(z,w) and λ[0,1]. A function f:R is called co-ordinated convex on if the partial mappings f y :[a,b]R, f y (u)=f(u,y) and f x :[c,d]R, f x (v)=f(x,v) are convex for all y[c,d] and x[a,b]. Note that every convex function f:R is co-ordinated convex, but the converse is not generally true; see [3].

Dragomir in [4] established the following similar inequality of the Hermite-Hadamard inequality for a co-ordinated convex function on a rectangle from the plane R 2 .

Theorem 3.1 Suppose that f:=[a,b]×[c,d]R is co-ordinated convex on . Then one has the inequalities

f ( a + b 2 , c + d 2 ) 1 ( b a ) ( d c ) a b c d f ( x , y ) d y d x f ( a , c ) + f ( a , d ) + f ( b , c ) + f ( b , d ) 4 .

Now, let be a convex area from the plane R 2 , bounded by a convex function y=h(x) and a concave function y=g(x) and x=a, x=b, such that for any x[a,b], g(x)h(x). Also, let F be a two-variable convex function on . In [5] and [6], the following inequality is proved:

In this paper, we want to find an upper bound for the integral

1 a b ( g ( x ) h ( x ) ) d x a b h ( x ) g ( x ) F(x,y)dydx.
(3.1)

For this purpose, we reach to the following integral:

1 a b ( g ( x ) h ( x ) ) d x a b [ F ( x , g ( x ) ) + F ( x , h ( x ) ) ] ( g ( x ) h ( x ) ) dx.

It is well known that if F(x,y) is increasing relative to y and y=h(x) is convex on [a,b], then F(x,h(x)) is convex on [a,b], but we have no information about the convexity of F(x,h(x)) generally. So, in special cases, we will find an upper bound for the integral (3.1).

Theorem 3.2 Let be a bounded area by a convex function y=h(x) and a concave function y=g(x) on [a,b] such that for any x[a,b], g(x)h(x) and gh is increasing on [a,b]. Also, let F be a two-variable convex function on such that F(x,g(x)) and F(x,h(x)) are convex on [a,b]. Then one has the inequality

Proof Since F is convex on , hence F is co-ordinated convex on . So, F x :[h(x),g(x)]R, F x (y)=F(x,y) is convex on [h(x),g(x)] for all x[a,b]. By the right-hand side of Hermite-Hadamard inequality (1.1), we have

h ( x ) g ( x ) F(x,y)dy ( g ( x ) h ( x ) ) [ F ( x , g ( x ) ) + F ( x , h ( x ) ) 2 ] .

Integrating this inequality on [a,b], we obtain

Since gh is increasing and F(x,g(x)), F(x,h(x)) are convex on [a,b], by Theorem 2.1(i), we have

The proof is complete. □

Theorem 3.3 Let be a bounded area by a convex function h and a concave function g on [a,b] such that for any x[a,b], g(x)h(x). Also, let F be a two-variable convex function on such that F(x,g(x)) and F(x,h(x)) are convex on [a,b]. Then one has the inequality

where α(b)= a b t ( g ( t ) h ( t ) ) d t a b ( g ( t ) h ( t ) ) d t .

Proof

By a similar way to the proof of Theorem 3.2, we have

Since F(x,g(x))+F(x,h(x)) is convex, by Theorem 2.2 (P(x)= g ( x ) h ( x ) a b ( g ( x ) h ( x ) ) d x ), we obtain

The proof is complete. □

In the following theorem, we prove the assertion of Theorem 3.3 with weak conditions.

Theorem 3.4 Let , g and h be defined as in Theorem  3.3. Also, let F be a two-variable convex function on such that

F ( x , g ( x ) ) g ( g ( x ) g ( a ) x a g ( x ) ) + F ( x , h ( x ) ) h ( h ( x ) h ( a ) x a h ( x ) ) 0,

then we have

where α(b)= a b t ( g ( t ) h ( t ) ) d t a b ( g ( t ) h ( t ) ) .

Proof

Denote

H(x)= a x h ( x ) g ( x ) f(t,y)dydt 1 2 K(x) a x ( g ( t ) h ( t ) ) dt,

where

K(x)= ( x α ( x ) x a ) [ F ( a , g ( a ) ) + F ( a , h ( a ) ) ] + ( α ( x ) a x a ) [ F ( x , g ( x ) ) + F ( x , h ( x ) ) ] .

Then we have

H (x)= h ( x ) g ( x ) F(x,y)dy 1 2 K(x) ( g ( x ) h ( x ) ) 1 2 K (x) a x ( g ( t ) h ( t ) ) dt.

Since F is convex, so it is co-ordinated convex. Hence, by the right-hand side of the Hermite-Hadamard inequality, we obtain

H ( x ) 1 2 ( g ( x ) h ( x ) ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) 1 2 K ( x ) ( g ( x ) h ( x ) ) 1 2 K ( x ) a x ( g ( t ) h ( t ) ) d t .

So,

H (x) 1 2 [ ( g ( x ) h ( x ) ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) K ( x ) ) K ( x ) a x ( g ( t ) h ( t ) ) d t ] .

On the other hand, we have

Now, multiplying each term by

a x ( g ( t ) h ( t ) ) dt

and using the fact

a x ( g ( t ) h ( t ) ) dtα(x)= a x t ( g ( t ) h ( t ) ) dt,

we obtain

a x ( g ( t ) h ( t ) ) dt α (x)= ( g ( x ) h ( x ) ) ( x α ( x ) ) .

Therefore,

By a similar way, we obtain

Thus,

So,

H ( x ) 1 2 [ ( g ( x ) h ( x ) ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) K ( x ) ) K ( x ) a x ( g ( t ) h ( t ) ) d t ] = 1 2 ( g ( x ) h ( x ) ) x α ( x ) x a [ F ( x , g ( x ) ) + F ( x , h ( x ) ) F ( a , g ( a ) ) F ( a , h ( a ) ) ] 1 2 [ ( g ( x ) h ( x ) ) ( x α ( x ) ) x a α ( x ) a ( x a ) 2 a x ( g ( t ) h ( t ) ) d t ] × [ F ( x , g ( x ) ) + F ( x , h ( x ) ) F ( a , g ( a ) ) F ( a , h ( a ) ) ] 1 2 a x ( g ( t ) h ( t ) ) d t ( α ( x ) a x a ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) .

Then it follows that

H ( x ) 1 2 [ F ( x , g ( x ) ) + F ( x , h ( x ) ) F ( a , g ( a ) ) F ( a , h ( a ) ) ] × [ ( g ( x ) h ( x ) ) x α ( x ) x a ( g ( x ) h ( x ) ) ( x α ( x ) ) x a + α ( x ) a ( x a ) 2 a x ( g ( t ) h ( t ) ) d t ] 1 2 a x ( g ( t ) h ( t ) ) d t ( α ( x ) a x a ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) .

Thus,

H ( x ) 1 2 ( α ( x ) a x a ) a x ( g ( t ) h ( t ) ) d t × [ F ( x , g ( x ) ) F ( a , g ( a ) ) x a + F ( x , h ( x ) ) F ( x , g ( x ) ) x a F ( x , g ( x ) ) F ( x , h ( x ) ) ] .

Now, notice that if F(x,g(x)), F(x,h(x)) were convex on [a,b], we can deduce the assertion of Theorem 3.3. Since F is convex on , we have

F ( x , g ( x ) ) F ( a , g ( a ) ) F ( x , g ( x ) ) x (xa)+ F ( x , g ( x ) ) g ( g ( x ) g ( a ) )

or

By a similar way, we have

Note that

F ( x , g ( x ) ) = F ( x , g ( x ) ) x + F ( x , g ( x ) ) g g (x)

and

F ( x , h ( x ) ) = F ( x , h ( x ) ) x + F ( x , h ( x ) ) h h (x).

So,

H ( x ) 1 2 α ( x ) a x a a x ( g ( t ) h ( t ) ) d t [ F ( x , g ( x ) ) x + F ( x , g ( x ) ) g g ( x ) g ( a ) x a + F ( x , h ( x ) ) x + F ( x , h ( x ) ) h h ( x ) h ( a ) x a F ( x , g ( x ) ) x F ( x , g ( x ) ) g g ( x ) F ( x , h ( x ) ) x F ( x , h ( x ) ) h h ( x ) ] .

Thus,

H ( x ) 1 2 α ( x ) a x a a x ( g ( t ) h ( t ) ) d t [ F ( x , g ( x ) ) g ( g ( x ) g ( a ) x a g ( x ) ) + F ( x , h ( x ) ) x ( h ( x ) h ( a ) x a h ( x ) ) ] 0 .

Note that α(x)a. Therefore, H is decreasing and

H(b)H(a)=0.

The proof is complete. □

Remark 3.1 Notice that since g is concave and h is convex on [a,b], so g is decreasing and h is increasing on [a,b]. By the mean value theorem, we have

g ( x ) g ( a ) x a g (x)0and h ( x ) h ( a ) x a h (x)0.

In particular, if we have g(x)=mx+n, then g ( x ) g ( a ) x a g (x)=0. So, if F ( x , h ( x ) ) h 0, then

In the following theorem, we find an upper bound of the Hermite-Hadamard inequality for a co-ordinated convex function.

Theorem 3.5 Let , g and h be defined as in Theorem  3.3. Also, let F be a convex function only relative to y, that is, F x :[h(x),g(x)]R, F x (v)=F(x,v) is convex for all x[a,b]. If F (x,g(x))+ F (x,h(x))0, then

1 a b ( g ( t ) h ( t ) ) d t a b h ( x ) g ( x ) F(x,y)dydx 1 2 [ F ( b , g ( b ) ) + F ( b , h ( b ) ) ] .

Proof

Denote

H(x)= a x h ( x ) g ( x ) F(t,y)dydt 1 2 a x ( g ( t ) h ( t ) ) dt [ F ( x , g ( x ) ) + F ( x , h ( x ) ) ] .

We have

H ( x ) = h ( x ) g ( x ) F ( x , y ) d y 1 2 ( g ( x ) h ( x ) ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) 1 2 a x ( g ( t ) h ( t ) ) d t ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) .

Since F is convex relative to y, by the right-hand side of the Hermite-Hadamard inequality, we obtain

H ( x ) 1 2 ( g ( x ) h ( x ) ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) 1 2 ( g ( x ) h ( x ) ) ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) 1 2 a x ( g ( t ) h ( t ) ) d t ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) = 1 2 a x ( g ( t ) h ( t ) ) d t ( F ( x , g ( x ) ) + F ( x , h ( x ) ) ) 0 .

So, H is decreasing on [a,b]. That is, H(b)H(a)=0. □

4 Examples

Example 4.1 Let F(x,y)= x 2 + y 2 and be bounded by g(x)= 1 x 2 , h(x)=x1 on [0,1]. Then g(x)h(x)= 1 x 2 x+1 is decreasing on [0,1] and F(x,g(x))=1, F(x,h(x))= x 2 + ( x 1 ) 2 are convex on [0,1]. By Theorem 3.2, we have

By easy calculation, we see that

0 1 ( 1 x 2 x + 1 ) dx= π 4 + 1 2 = π + 2 4

and

4 π + 2 0 1 x 1 1 x 2 ( x 2 + y 2 ) dydx1.

Example 4.2 Let F, g and h be defined as in Example 4.1. By Theorem 3.3, we have

α ( 1 ) = 0 1 t ( 1 t 2 t + 1 ) d t 0 1 ( 1 t 2 t + 1 ) d t = 5 6 π 4 + 1 2 = 10 3 ( π + 2 ) 1 0 1 ( 1 t 2 t + 1 ) d t 0 1 x 1 1 x 2 ( x 2 + y 2 ) d y d x 1 2 [ 3 π 4 3 ( π + 2 ) ( F ( 0 , g ( 0 ) ) + F ( 0 , h ( 0 ) ) ) + 10 3 ( π + 2 ) ( F ( 1 , g ( 1 ) ) + F ( 1 , h ( 1 ) ) ) ] .

So,

4 π + 2 0 1 x 1 1 x 2 ( x 2 + y 2 ) dydx1.

Example 4.3 Let F(x,y)= x 2 + y 2 and be bounded by g(x)=x+2, h(x)= x 2 on [1,2]. Then gh is not decreasing on [1,2] and also F(x,h(x))= x 2 + x 4 is not convex on [1,2]. So, g, h and F do not hold in the hypothesis of Theorems 3.2 and 3.3. But we have

g ( x ) g ( 1 ) x + 1 g (x)= x + 1 x + 1 1=0, h ( x ) h ( 1 ) x + 1 h (x)=x10

and

F ( x , g ( x ) ) g =2(x+2), F ( x , h ( x ) ) h =2 x 2 .

So,

Thus, we can apply Theorem 3.4

Hence,

2 9 1 2 x 2 x + 2 ( x 2 + y 2 ) dydx11.

Example 4.4 Let F(x,y)=xy and be bounded by g(x)=x+2, and h(x)= x 2 on [1,2]. Then F is not convex on , but it is convex relative to y, we have

F ( x , g ( x ) ) = x 2 +2xandF ( x , h ( x ) ) = x 3 .

So,

F ( x , g ( x ) ) + F ( x , h ( x ) ) =2x+2+3 x 2 >0.

Hence, by Theorem 3.5, we have

1 1 2 ( x + 2 x 2 ) d x 1 2 x 2 x + 2 xydydx 1 2 [ F ( 2 , g ( 2 ) ) + F ( 2 , h ( 2 ) ) ] .

Hence,

2 9 1 2 x 2 x + 2 xydydx8.

References

  1. 1.

    Fejér L: Über die Fourierreihen, II. Math. Naturwiss. Anz Ungar. Akad. Wiss. 1906, 24: 369–390. (In Hungarian)

  2. 2.

    Yang GS, Tseng KH: On certain integral inequalities related to Hermite-Hadamard inequalities. J. Math. Anal. Appl. 1999, 239: 180–187. 10.1006/jmaa.1999.6506

  3. 3.

    Dragomir, SS, M Pearce, CE: Selected Topics on Hermite-Hadamard Inequalities. RGMIA Monographs, Victoria University (2000)

  4. 4.

    Dragomir SS: On the Hadamard’s inequality for convex function on the co-ordinated in a rectangle from the plane. Taiwan. J. Math. 2001, 5: 775–788.

  5. 5.

    Matejika L: Elementary proof of the left multidimensional Hermite-Hadamard inequality on certain convex sets. J. Math. Inequal. 2010, 4(2):259–270.

  6. 6.

    Zabandan G, Kılıçman A: A new version of Jensen’s inequality and related results. J. Inequal. Appl. 2012., 2012: Article ID 238. doi:10.1186/1029–242X-2012–238

Download references

Acknowledgements

The authors express their sincere thanks to the referees for the careful and detailed reading of the manuscript and very helpful suggestions that improved the manuscript substantially. The authors also gratefully acknowledge that this research was partially supported by the University Putra Malaysia under the Research University Grant Scheme 05-01-09-0720RU.

Author information

Correspondence to Adem Kılıçman.

Additional information

Competing interests

Authors declare that they have no competing interest.

Authors’ contributions

Both the authors contributed equally in preparation as well as in typing and further both authors read the proof and approved the modifications.

Rights and permissions

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

Reprints and Permissions

About this article

Cite this article

Zabandan, G., Kılıçman, A. Several integral inequalities and an upper bound for the bidimensional Hermite-Hadamard inequality. J Inequal Appl 2013, 27 (2013) doi:10.1186/1029-242X-2013-27

Download citation

Keywords

  • Continuous Function
  • Convex Function
  • Integrable Function
  • Weak Condition
  • Partial Mapping