Journal of Function Spaces

Volume 2016, Article ID 9324804, 6 pages

http://dx.doi.org/10.1155/2016/9324804

## Application of Functionals in Creating Inequalities

^{1}Mechanical Engineering Faculty in Slavonski Brod, University of Osijek, Trg Ivane Brlić Mažuranić 2, 35000 Slavonski Brod, Croatia^{2}Department of Mathematics, Longyan University, Longyan, Fujian 364012, China

Received 14 July 2016; Accepted 27 September 2016

Academic Editor: Calogero Vetro

Copyright © 2016 Zlatko Pavić 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.

#### 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. Let**Then 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 .