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Research Article | Open Access

Volume 2016 |Article ID 9324804 | https://doi.org/10.1155/2016/9324804

Zlatko Pavić, Shanhe Wu, Vedran Novoselac, "Application of Functionals in Creating Inequalities", Journal of Function Spaces, vol. 2016, Article ID 9324804, 6 pages, 2016. https://doi.org/10.1155/2016/9324804

# Application of Functionals in Creating Inequalities

Accepted27 Sep 2016
Published19 Oct 2016

#### Abstract

The paper deals with the fundamental inequalities for convex functions in the bounded closed interval. The main inequality includes convex functions and positive linear functionals extending and refining the functional form of Jensen’s inequality. This inequality implies the Jensen, Fejér, and, thus, Hermite-Hadamard inequality, as well as their refinements.

#### 1. Introduction

In our research, we apply the theory of positive linear functionals to convex analysis. Let us remember the initial notions related to positive linear functionals on the space of real functions.

Let be a nonempty set, and let be a subspace of the linear space of all real functions on domain . We assume that space contains unit function defined by for every . Such space contains every real constant within the meaning of and every composite function of a function and an affine function . Actually, if , then the compositionbelongs to .

Let be the space of all linear functionals on space . Functional is said to be unital (normalized) if . Such functional has property for every real constant . If is a function and if is a unital functional, then affine function satisfies equalityFunctional is said to be positive (nonnegative) if inequality holds for every nonnegative function . If a pair of functions satisfies inequality for every , then it follows thatIf is a function with the image in interval (i.e., ), then every positive unital functional meets inclusion (i.e., ). The same is true for each closed interval .

Introducing a continuous convex function, we can expose the functional form of Jensen’s inequality.

Theorem A. Let be a function with the image in closed interval , and let be a positive unital functional.
Then each continuous convex function such that satisfies inequality

We will consider convex functions in bounded closed interval with endpoints . Each point can be represented by the unique binomial convex combinationwhere

Convex function is bounded by two lines. The secant line of function passes through graph points and , and its equation isLet be an interior point. The support lines of function pass through graph point . Each support line is specified by slope coefficient , and its equation isThe support-secant line inequalityholds for every .

In 1931, Jessen (see [1, 2]) stated the functional form of Jensen’s inequality for convex functions in interval . In 1988, I. Rasa and I. Raşa (see [3]) pointed out that must be closed otherwise it could happen that and that must be continuous otherwise it could happen that the inequality in formula (4) does not apply. In Theorem A, we have taken into account I. Rasa and I. Raşa’s remarks. Some generalizations of the functional form of Jensen’s inequality can be found in [4].

A concise book on functional analysis, which contains an essential overview of operator theory and indicates the importance of positive linear functionals, is certainly the book in [5].

#### 2. Main Results

We firstly present the extension of the inequality in formula (4) concerning interval .

Lemma 1. Let be a function with the image in , and let be a positive unital functional.
Then each continuous convex function such that satisfies double inequality

Proof. The point is in interval . We sketch the proof in two steps depending on the position of .
If , we take support line of at . By applying positive functional to the support-secant inequality in formula (9) with , where , we get By utilizing the affinity of functions and via formula (2), the above inequality takes the formwhere the first term If , we rely on the continuity of using a support line at a point of open interval that is close enough to . Given , we can find so that By combining the above inequality and the inequality in formula (12) with the support line at , we obtain Letting approach zero, we reach the conclusion . In this case, trivial inequality represents formula (10).

Formula (10) can be expressed in the form which includes the convex combination of interval endpoints and . The respective form of Lemma 1 is as follows.

Corollary 2. Let be a function with the image in , and let be a positive unital functional. LetThen each continuous convex function such that satisfies double inequality

Proof. As regards to the last terms of formulae (10) and (17), we have because of the affinity of and its coincidence with at endpoints.

In order to refine the inequality in formula (10), we will combine the secant lines of convex function with positive unital functionals.

Lemma 3. Let be a point.
Then each convex function satisfies the secant lines inequalityfor every .

Proof. Cases and should be considered.

Suppose that function has the image in and is not identically equal to or . Such function satisfies inequality for some number and some pair of points . In that case, we can find a pair of functionals meeting related inequalityFor example, we can take the point evaluations at and , that is, the functionals defined by and for every function .

In the main theorem, we use functionals and satisfying the inequality in formula (20).

Theorem 4. Let be a point. Let be a function with the image in , and let be positive unital functionals such that and . Let be a convex combination of and .
Then each continuous convex function such that satisfies the series of inequalities

Proof. By applying the convexity of to convex combination , we getBy applying the left-hand side of formula (10) to and , we obtainAs the right-hand side of formula (19) with , the inequality holds for every . By acting with in the above inequality and using assumption , we find and similarly, by acting with and using the assumption , we find Multiplication by and and then summation yieldUsing main secant , we reach conclusionPutting together the inequalities in formulae (22), (23), (27), and (28) into a series, we achieve the inequality in formula (21).

The geometric presentation of the series of inequalities in formula (21) is created in Figure 1. The inequality terms are represented by five black dots above point .

To emphasize interval endpoints and , we present the following version of Theorem 4.

Corollary 5. Let be a point. Let be a function with the image in , and let be positive unital functionals such that and . Let be a convex combination of and , and let .
Then each continuous convex function such that satisfies the series of inequalitieswhere

Proof. To calculate coefficients , , and , we include convex combinations and . Then the fourth term of formula (21) takes the form Taking the coefficient of and using formula (6), we calculate Similarly we determine and .

Let us finish the section by presenting the generalization of Theorem 4 that uses several secant lines.

Corollary 6. Let be points. Let be a function with the image in , and let be positive unital functionals such that for . Let be a convex combination of functionals .
Then each continuous convex function such that satisfies the series of inequalities

#### 3. Applications to Integral and Discrete Inequalities

We firstly utilize Lemma 1 to obtain a very general integral inequality.

Corollary 7. Let be an integrable function with the image in , and let be a positive integrable function.
Then each convex function satisfies double inequality

Proof. Let be the space of all integrable functions over domain . Composition is bounded in and continuous almost everywhere in . Therefore is integrable over , that is, .
The integrating linear functional defined byfor every is positive and unital. The first term of formula (34) is equal to , the second term is equal to , and the third term is equal to . Thus, formula (34) fits into the frame of formula (10), and it is valid for a continuous convex function .
Let us verify that the inequality in formula (34) applies to a convex function which is not continuous at endpoints. We observe the position of pointIf , then we may utilize continuous extension of to in formula (34). The first two terms are the same as we use , and the last terms satisfy inequalitySo, formula (34) applies to in this case.
If , then either or for every . We rearrange formula (36) to be integral equationfrom which it follows that for almost every . Thus we have that for almost every , and inequality represents formula (34).
Respecting all considerations, we may conclude that the inequality in formula (34) applies to any convex function .

The inequality in formula (34) is the extended version of Jensen’s inequality for the ratio of integrals in interval , as well as the generalized form of the Fejér and Hermite-Hadamard inequality.

Let us demonstrate the simplifications of the inequality in formula (34) relating to the identity, unit, and symmetric function.

Using , we get the extension of the classical integral form of Jensen’s inequality (see [6])

Using identity function and function , satisfying equation , which represents the symmetry with a center at midpoint , we have the classical form of Fejér inequality (see [7])Namely, as a consequence of the symmetry we havebecause Following formula (41), we can conclude that the Fejér inequality is valid for function satisfying weaker condition .

Using in Fejér’s inequality in formula (40), we obtain the classical form of Hermite-Hadamard inequality (see [8, 9])

To obtain refinements of the inequality in formula (34), we use point and apply Theorem 4.

Corollary 8. Let be a point. Let be an integrable function such that for and for , and let be a positive integrable function.
Then each convex function satisfies the series of inequalities

Proof. Just as in Corollary 7, we use as the space of all integrable functions in interval . In order to apply Theorem 4, we define integrating linear functionalsfor every . Functionals and are positive and unital.
Since for , point falls into , and similarly point falls into . Using coefficientsand functional convex combination , we getBy further calculating the functional terms according to formula (21), we obtain the integral terms of formula (44).

The series of inequalities in formula (44) with and gives the refinement of the Hermite-Hadamard inequality in formula (43) as follows:The above refinement holds for each . The version of the above refinement was obtained in [10] by using the Jensen type inequality for convex combinations with the common center. That inequality was used to refine some important means.

The inequality in formula (44) with identity function and a symmetric function satisfying equation provides the refinements of the Fejér inequality in formula (40).

At the end, let us present the discrete version of Corollary 8. Point evaluations and will be shortened by and , respectively.

Corollary 9. Let be a point. Let be a function such that for and for , and let be a positive function. Let and be points.
Then each convex function satisfies the series of inequalities

Proof. Let be the space of all real functions on domain . We can apply the proof of Corollary 8 to summarizing linear functionalsacting to every , and coefficientsFunctionals and are certainly positive and unital.

The inequality in formula (49) is the extension and refinement of the famous discrete form of Jensen’s inequality (see [11]).

#### Competing Interests

The authors declare that they have no competing interests.

#### Acknowledgments

The work of the first and third author has been fully supported by Mechanical Engineering Faculty in Slavonski Brod and Croatian Science Foundation under Project HRZZ-5435. The work of the second author has been supported by the Natural Science Foundation of Fujian province of China under Grant no. 2016J01023. The authors would like to thank Velimir Pavić who has graphically prepared Figure 1.

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