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Several integral inequalities and an upper bound for the bidimensional Hermite-Hadamard inequality
Journal of Inequalities and Applications volume 2013, Article number: 27 (2013)
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 . Then we have the following inequality, which is called Hermite-Hadamard inequality:
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 is a convex function, then the inequality
holds, where is non-negative, integrable and symmetric about .
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 is defined by
then P is convex, increasing on and for all ,
In this paper, we find an upper bound for , where f is a convex function on and g is non-negative increasing (or decreasing) on , and . Finally, in Section 3 we find an upper bound for the following integral:
2 Integral inequalities
Theorem 2.1 Let be a differentiable convex function and be a continuous function.
-
(i)
If g is decreasing on , then
-
(ii)
If g is increasing on , then
Proof
-
(i)
Denote
We will show that . We have
By the extended mean value theorem (Cauchy’s theorem), we have
On the other hand, by the convexity of f and decreasing of g, we obtain
Since g is non-negative,
which implies that H is decreasing. Hence, . The proof is complete.
-
(ii)
Denote
Then we have
By the mean value theorem (Lagrange’s theorem), there exist and such that
Hence,
By the convexity of f and increasing of g, we obtain
So,
Therefore, H is decreasing and . The proof is complete. □
Theorem 2.2 Let be a convex function and be an integrable function such that . Then
Proof
We have
So, we get
□
Corollary 2.1 Let be a convex function and g be a non-negative integrable function. Then
and
The proof is similar to the proof of theorem.
3 Right bidimensional Hermite-Hadamard inequality
Let us consider the bidimensional interval in . Recall that the mapping is convex on △ if
holds for all and . A function is called co-ordinated convex on △ if the partial mappings , and , are convex for all and . Note that every convex function 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 .
Theorem 3.1 Suppose that is co-ordinated convex on △. Then one has the inequalities
Now, let △ be a convex area from the plane , bounded by a convex function and a concave function and , , such that for any , . 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
For this purpose, we reach to the following integral:
It is well known that if is increasing relative to y and is convex on , then is convex on , but we have no information about the convexity of 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 and a concave function on such that for any , and is increasing on . Also, let F be a two-variable convex function on △ such that and are convex on . Then one has the inequality

Proof Since F is convex on △, hence F is co-ordinated convex on △. So, , is convex on for all . By the right-hand side of Hermite-Hadamard inequality (1.1), we have
Integrating this inequality on , we obtain

Since is increasing and , are convex on , 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 such that for any , . Also, let F be a two-variable convex function on △ such that and are convex on . Then one has the inequality

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

Since is convex, by Theorem 2.2 , 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
then we have

where .
Proof
Denote
where
Then we have
Since F is convex, so it is co-ordinated convex. Hence, by the right-hand side of the Hermite-Hadamard inequality, we obtain
So,
On the other hand, we have

Now, multiplying each term by
and using the fact
we obtain
Therefore,

By a similar way, we obtain

Thus,

So,
Then it follows that
Thus,
Now, notice that if , were convex on , we can deduce the assertion of Theorem 3.3. Since F is convex on △, we have
or

By a similar way, we have

Note that
and
So,
Thus,
Note that . Therefore, H is decreasing and
The proof is complete. □
Remark 3.1 Notice that since g is concave and h is convex on , so is decreasing and is increasing on . By the mean value theorem, we have
In particular, if we have , then . So, if , 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, , is convex for all . If , then
Proof
Denote
We have
Since F is convex relative to y, by the right-hand side of the Hermite-Hadamard inequality, we obtain
So, H is decreasing on . That is, . □
4 Examples
Example 4.1 Let and △ be bounded by , on . Then is decreasing on and , are convex on . By Theorem 3.2, we have

By easy calculation, we see that
and
Example 4.2 Let F, g and h be defined as in Example 4.1. By Theorem 3.3, we have
So,
Example 4.3 Let and △ be bounded by , on . Then is not decreasing on and also is not convex on . So, g, h and F do not hold in the hypothesis of Theorems 3.2 and 3.3. But we have
and
So,

Thus, we can apply Theorem 3.4

Hence,
Example 4.4 Let and △ be bounded by , and on . Then F is not convex on △, but it is convex relative to y, we have
So,
Hence, by Theorem 3.5, we have
Hence,
References
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Dragomir, SS, M Pearce, CE: Selected Topics on Hermite-Hadamard Inequalities. RGMIA Monographs, Victoria University (2000)
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.
Matejika L: Elementary proof of the left multidimensional Hermite-Hadamard inequality on certain convex sets. J. Math. Inequal. 2010, 4(2):259–270.
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
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.
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Both the authors contributed equally in preparation as well as in typing and further both authors read the proof and approved the modifications.
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Zabandan, G., Kılıçman, A. Several integral inequalities and an upper bound for the bidimensional Hermite-Hadamard inequality. J Inequal Appl 2013, 27 (2013). https://doi.org/10.1186/1029-242X-2013-27
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DOI: https://doi.org/10.1186/1029-242X-2013-27
Keywords
- Continuous Function
- Convex Function
- Integrable Function
- Weak Condition
- Partial Mapping