Sharp Estimation Type Inequalities for Lipschitzian Mappings in Euclidean Sense on a Disk

Rostamian Delavar, M.; Dragomir, S. S.; De La Sen, M.

doi:https://doi.org/10.1155/2021/6615626

Journal of Function Spaces

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Abstract Introduction and Preliminaries Data Availability Conflicts of Interest References Copyright Related Articles

Research Article | Open Access

Volume 2021 | Article ID 6615626 | https://doi.org/10.1155/2021/6615626

Sharp Estimation Type Inequalities for Lipschitzian Mappings in Euclidean Sense on a Disk

M. Rostamian Delavar,¹S. S. Dragomir,²and M. De La Sen³

Academic Editor: Ismat Beg

Received14 Dec 2020

Accepted01 Feb 2021

Published08 Feb 2021

Abstract

Some sharp trapezoid and midpoint type inequalities for Lipschitzian bifunctions defined on a closed disk in Euclidean sense are obtained by the use of polar coordinates. Also, bifunctions whose partial derivative is Lipschitzian are considered. A new presentation of Hermite-Hadamard inequality for convex function defined on a closed disk and its reverse are given. Furthermore, two mappings and are considered to give some generalized Hermite-Hadamard type inequalities in the case that considered functions are Lipschitzian in Euclidean sense on a disk.

1. Introduction and Preliminaries

Consider that is a closed disk in the plane centered at the point having the radius . In [1] (see also [2]), the Hermite-Hadamard inequality for a convex function defined on has been obtained as follows:

Theorem 1. If the mapping is convex on , then one has the inequalitywhere is the circle centered at the point with radius . The above inequalities are sharp.

First of all, we give the following result which is including a new presentation of (1) and its reverse as well:

Theorem 2. For a continuous function defined on a convex subset ,(1)if is convex on , then for any , we havewhere is the boundary of (2)if (2) holds for all , then is convex

Proof. (1)Consider the change of coordinates defined asIt follows thatNow consider in above integrals with to obtain the desired result(2)Suppose that there exist and such thatSince is continuous on , then there exists and a point in convex combination of and such that (5) holds on whole of . Now, if we follow the proof of part (1) for by the use of (5) on and , then we haveThis contradiction proves the convexity of on .

We remind that the classic form of Hermite-Hadamard inequality (see [3–5]) for a real valued convex function defined on is the following:

Generally, in the literature associated to any Hermite-Hadamard type inequality, there exist two inequalities which we call them trapezoid and mid-point type inequalities. The names “trapezoid” and “midpoint” comes from two classic inequalities (due to their geometric interpretation) related to the Hermite-Hadamard inequality obtained in [6, 7], respectively:where is a differentiable mapping on with and is convex on . For more results about convex functions, related inequalities, and generalizations of (7)–(9), see [8–23] and references therein.

Recently, in [17], the authors obtained the trapezoid and midpoint type inequality related to (1) as follows, respectively:

Theorem 3. Consider a set with . Suppose that the mapping has continuous partial derivatives in the disk with respect to the variables and in polar coordinates. If for any constant , the function is convex with respect to the variable on then

Note that inequality (10) is sharp.

As we can see in (8) and (9), the classic trapezoid and midpoint type inequalities have been obtained for the functions whose the first derivative absolute values are convex. In [22, 24], the authors considered Lipschitzian mappings instead of those whose the first derivative absolute values are convex to obtain some midpoint and trapezoid type inequalities:

Theorem 4 [24]. Let be an -Lipschitzian mapping on and with . Then, we have the inequalities

Motivated by above works and results, we obtain some trapezoid and midpoint type inequalities related to (1) for Lipschitzian mappings (in Euclidean sense) defined on the disk in a plane. Also we investigate trapezoid and mid-point type inequalities in the case that in polar coordinates , the derivative of considered function with respect to the variable is Lipschitzian. Furthermore, two mappings and are considered to give some generalized Hermite-Hadamard type inequalities in the case that the functions are Lipschitzian on a disk .

Here, we should mention that in [25], we can find some inequalities for the integral mean of Hölder continuous functions defined on disks in a plane which in a special case leads to trapezoid and midpoint type inequalities for a kind of Lipschitzian mappings as the following:

Theorem 5. If satisfies the conditionwhere , then we have the inequalities

The main point is that the Euclidean Lipschitz condition used in this paper is a stronger condition than (13) in the case that and so our results obtained in (20) and (29) will provide more accurate estimation compared to (14) and (15). Furthermore, we obtain new trapezoid and midpoint type inequalities of our function is Lipschitzian.

2. Main Results

In this section, first, we obtain some trapezoid and midpoint type inequalities related to (1) for the case that our considered function is Lipschitzian (in Euclidean sense). Second, we obtain some trapezoid and mid-point type inequalities related to (1) for the case that the partial derivative of our function with respect to the variable in polar coordinates is Lipschitzian (in Euclidean sense).

Definition 6 [26]. A function is said to satisfy a Lipschitz condition (briefly -Lipschitzian) on with respect to a norm , if there exists a constant such thatfor any .

If is Lipschitzian with respect to a constant and the Euclidean norm , then for any and , we havefor any and . Also, it is obvious that if is Lipschitzian with respect to a constant on , then, it is continuous and so integrable on .

2.1. Is Lipschitzian

The first result of this section is the trapezoid type inequality related to (1) for the case that our considered function is Lipschitzian. We start with a lemma.

Lemma 7. Define a function asfor fixed and all , . Then, the function is -Lipschitzian.

Proof. Consider and , for and . Sofor all .

Theorem 8. Suppose that the mapping is Lipschitzian with respect to a constant and the Euclidean norm . Then,where is the boundary of and is its corresponding curve. Also, inequality (20) is sharp.

Proof. Since is Lipschitzian with respect to and the Euclidean norm on , then, we haveNow, consider the constant and the curve defined byIt is clear that , and then by integrating, we obtain thatwhere , and are derivatives of “” and “” with respect to , respectively. So the fact thatimplies thatAlso, by the use of polar coordinates, we get toFinally, by replacing (25) and (26) in (21) and then dividing the result with “,” we deduce the desired result. To prove sharpness of (20), consider the function defined byfor fixed and all , . The function is -Lipschitzian by Lemma 7. It is not hard to see that for all and also for the case that , we have . Now applying these results in (21) implies that

The following result is the midpoint type inequality related to (1) for Lipschitzian functions defined on a closed disk.

Theorem 9. Suppose that the mapping is Lipschitzian with respect to a constant and the Euclidean norm . Then,

Furthermore, inequality (29) is sharp.

Proof. Since the mapping satisfies a Lipschitz condition with respect to a constant and the Euclidean norm on , we havefor all and . It follows thatBy the use of identity (26) in inequality (31), we obtain thatFinally, it is enough to divide (32) with “ to get the result. For the sharpness of (29), consider the function defined byfor , and . By a similar method used in the proof of Lemma 7, the function is -Lipschitzian. Also, it is obvious that and . So, we haveshowing that inequality (29) is sharp.

Corollary 10. Suppose that is an open set with . If is a convex function defined on , then Theorem D of Section 41 in [26] implies that satisfies a Lipschitz condition on with respect to a constant and so from inequalities (20) and (29) along with inequality (1), we have the following results:

In the following example, for a given function, it is illustrated how we can obtain a Lipschitz constant for a real valued bifunction defined on a disk.

Example 11. Consider , , . We find a Lipschitz constant for as follows:
For , consider the path from to in asfor . The fundamental theorem of calculus implies thatAlso, the chain rule for differentiation implies thatwhere is the gradient vector of . So,which implies that

Now, we conclude that (if exists) is a Lipschitz constant for . Therefore, for any , we haveand then by the use of polar transformation, we get

So, we can choose as a Lipschitz constant for on .

Remark 12. According to the above example, if we have a function such that with respect to the Euclidean norm , then we can consider as a Lipschitz constant and then obtain inequalities (20) and (29).

2.2. Is Lipschitzian

In this part, we investigate the trapezoid and midpoint type inequalities in the case that in polar coordinates , the partial derivative of considered function with respect to the variable is Lipschitzian in the Euclidean norm .

Theorem 13. Consider a set with and a mapping such that (partial derivative of with respect to the variable in polar coordinates) is Lipschitzian with respect to a constant and the Euclidean norm . Then,

Proof. For any fixed , if we setthen we obtain that , where , are the derivatives of , respectively, with respect to the variable in . By the above facts, using integration by parts and identities (25) and (26) obtained in Theorem 8, we getOn the other hand, we haveSo from (45) and (46), we obtain thatFinally, it is enough to divide (47) with “” to get the desired result.

The following is a trapezoid type inequality for the case that the partial derivative of considered function with respect to the variable “” is Lipschitzian with respect to the Euclidean norm .

Theorem 14. Consider a set with and a mapping such that (partial derivative of with respect to the variable in polar coordinates) is Lipschitzian with respect to a constant and the Euclidean norm . Then,

Proof. Using the description provided in the beginning of Theorem 13, it is not hard to see thatAlso by the use of (23) in Theorem 8, we haveOn the other hand, we haveThen,Since is -Lipschitzian, we have thatNow triangle inequality and inequality (43) imply that

Here, we provide two examples in connection with results obtained in this subsection.

Example 15. Define a function byfor , , and . Since , then according to Remark 12, we can consider as a Lipschitz constant with respect to the Euclidean norm . On the other hand,So,which shows that (43) is sharp. Also,which implies that (53) is sharp.

Example 16. Consider , and . For and polar function which is defined on , by some calculations we can conclude thatand thenis a Lipschitz constant for . Then, from inequality (53), we have thatwhere is arithmetic mean of and . Also for the function , by the use of (43) and (48), we can obtain other arithmetic mean type inequalities.

3. Mappings and

In this section, by the use of two mappings and defined in [1], we give some generalized Hermite-Hadamard type inequalities in the case that considered functions are Lipschitzian with respect to Euclidean norm on a disk :

By the use of some properties for the mappings and , we give some refinements for trapezoid and midpoint type inequalities obtained in previous sections for -Lipschitzian mappings .

Theorem 17. Suppose that the mapping is Lipschitzian with respect to a constant and the Euclidean norm . Then, the mapping is Lipschitzian with respect to “” and the mapping is Lipschitzian with respect to “.” The following inequalities also for all hold.

Proof. Consider the following relations for , which prove the first part of this theorem:Also,For inequality (63), we use the definition of and the fact that is -Lipschitzian:To prove (64), if , we consider the following identity presented in [1],Now, consider transformationThis implies thatAlso, we haveSo, we conclude thatInequality (65) is a consequence of the fact thatThe details are omitted.

The following results also are of interest:

Theorem 18. Suppose that the mapping is Lipschitzian with respect to a constant and the Euclidean norm . The following inequalities hold:for all .for all .for all , andfor all .

Proof. It is enough to consider special cases for and in two inequalities (66) and (67) obtained in the proof of previous theorem.

Remark 19. (1)For a convex function with, if we consider (1), some results obtained in [1], and the following inequalitythen, we deduce that (64) and (75)–(78) hold without using absolute value symbol.(2)If we consider in inequality (75) or consider in inequality (76), then, we recapture inequality (29) in Theorem 9. Also from inequality (77), we obtain this new inequalitywhere is Lipschitzian with respect to a constant and the Euclidean norm .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest regarding this article.

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Copyright © 2021 M. Rostamian Delavar et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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