## Some Classes of Function Spaces, Their Properties, and Their Applications 2014

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Zlatko Pavić, "Functions Like Convex Functions", *Journal of Function Spaces*, vol. 2015, Article ID 919470, 8 pages, 2015. https://doi.org/10.1155/2015/919470

# Functions Like Convex Functions

**Academic Editor:**Janusz Matkowski

#### Abstract

The paper deals with convex sets, functions satisfying the global convexity property, and positive linear functionals. Jensen's type inequalities can be obtained by using convex combinations with the common center. Following the idea of the common center, the functional forms of Jensen's inequality are considered in this paper.

#### 1. Introduction

Introduction is intended to be a brief overview of the concept of convexity and affinity. Let be a real linear space. Let be points and let be coefficients. Their binomial combinationis convex if and ifIf , then the point itself is called the combination center.

A set is convex if it contains all binomial convex combinations of its points. The convex hull of the set is the smallest convex set containing , and it consists of all binomial convex combinations of points of .

Let be a convex set. A function is convex if the inequalityholds for all binomial convex combinations of pairs of points .

Requiring only the condition in (2) for coefficients and requiring the equality in (3), we get a characterization of the affinity.

Implementing mathematical induction, we can prove that all of the above applies to -membered combinations for any positive integer . In that case, the inequality in (3) is the famous Jensen’s inequality obtained in [1]. Numerous papers have been written on Jensen’s inequality; different types and variants can be found in [2, 3].

#### 2. Positive Linear Functionals and Convex Sets of Functions

Let be a nonempty set, and let be a subspace of the linear space of all real functions on the domain . We assume that contains the unit function defined by for every .

Let be an interval, and let be the set containing all functions with the image in . Then, is convex set in the space . The same is true for convex sets of Euclidean spaces. Let be a convex set, and let be the set containing all function -tuples with the image in . Then, is convex set in the space .

A linear functional is positive (nonnegative) if for every nonnegative function , and is unital (normalized) if . If , then for every unital positive functional the number is in the closed interval of real numbers containing the image of . Through the paper, the space of all linear functionals on the space will be denoted with .

Let be an affine function, that is, the function of the form where and are real constants. If are functions and if are positive functionals providing the unit equalitythenRespecting the requirement of unit equality in (4), the sum could be called the functional convex combination. In the case , the functional must be unital by the unit equality in (4).

In 1931, Jessen stated the functional form of Jensen’s inequality for convex functions of one variable; see [4]. Adapted to our purposes, that statement is as follows.

Theorem A. *Let be a closed interval, and let be a function.**Then, a unital positive functional ensures the inclusion**and satisfies the inequality**for every continuous convex function providing that .**If is concave, then the reverse inequality is valid in (7). If is affine, then the equality is valid in (7).*

The interval must be closed, otherwise it could happen that . The function must be continuous, otherwise it could happen that the inequality in (7) does not apply. Such boundary cases are presented in [5].

In 1937, McShane extended the functional form of Jensen’s inequality to convex functions of several variables. He has covered the generalization in two steps, calling them the geometric (the inclusion in (8)) and analytic (the inequality in (9)) formulation of Jensen’s inequality; see [6, Theorems 1 and 2]. Summarized in a theorem, that generalization is as follows.

Theorem B. *Let be a closed convex set, and let be a function.**Then, a unital positive functional ensures the inclusion**and satisfies the inequality**for every continuous convex function providing that .**If is concave, then the reverse inequality is valid in (9). If is affine, then the equality is valid in (9).*

#### 3. Main Results

##### 3.1. Functions of One Variable

The main result of this subsection is Theorem 1 relying on the idea of a convex function graph and its secant line. Using functions that are more general than convex functions and positive linear functionals, we obtain the functional Jensen’s type inequalities.

Through the paper, we will use an interval and a bounded closed subinterval with endpoints .

Every number can be uniquely presented as the binomial affine combinationwhich is convex if and only if the number belongs to the interval . Let be a function, and let be the function of the line passing through the points and of the graph of . Applying the affinity of the function to the combination in (10), we obtain its equationThe consequence of the representations in (10) and (11) is the fact that every convex function satisfies the inequalityand the reverse inequality

In the following consideration, we use continuous functions satisfying the inequalities in (12)-(13).

Theorem 1. *Let be a closed interval, let be a bounded closed subinterval, and let and be functions.**Then, a pair of unital positive functionals such that**satisfies the inequality**for every continuous function satisfying (12)-(13) and providing that .*

*Proof. *The number belongs to the interval by the inclusion in (6). Using the features of the function and applying the affinity of the function , we getbecause for every .

It is obvious that a continuous convex function satisfies Theorem 1 for every subinterval with endpoints . The function used in Theorem 1 is shown in Figure 1. Such a function satisfies only the global property of convexity on the sets and .

Involving the binomial convex combination with the equality in (14) by assuming thatand inserting the term in (16) via the double equalitywhich is true because , we achieve the double inequality

The functions used in Theorem 1 satisfy the functional form of Jensen’s inequality in the following case.

Corollary 2. *Let be a closed interval, let be a bounded closed subinterval, and let be a function.**Then, a unital positive functional such that**satisfies the inequality**for every continuous function satisfying (12)-(13) and providing that .*

*Proof. *Putting , it follows thatby the right inequality in (19).

Now, we give a characterization of continuous convex functions by using unital positive functionals.

Proposition 3. *Let be a closed interval. A continuous function is convex if and only if it satisfies the inequality**for every pair of interval endpoints , every function such that , and every unital positive functional .*

*Proof. *Let us prove the sufficiency. Let be a convex combination of points where . We take the constant function in (actually for every ) and a unital positive functional . Then, connectingvia (23), we get the convexity inequality in (3).

##### 3.2. Functions of Several Variables

We want to transfer the results of the previous subsection to higher dimensions. The main result in this subsection is Theorem 6 generalizing Theorem 1 to functions of several variables.

Let be a convex set, let be a triangle with vertices , , and , and let be its interior. In the following observation, we assume that is a continuous function satisfying the inequalityand the reverse inequalitywhere is the function of the plane passing through the corresponding points of the graph of .

It should be noted that convex functions of two variables do not generally satisfy (26). The next example confirms this claim.

*Example 4. *We take the convex function , the triangle with vertices , , and , and the outside point .

The valuation of functions and at the point isas opposed to (26).

The generalization of Theorem 1 to two dimensions is as follows.

Lemma 5. *Let be a closed convex set, let be a triangle, and let and be functions.**Then, a pair of unital positive functionals such that**satisfies the inequality**for every continuous function satisfying (25)-(26) and providing that .*

*Proof. *The proof is similar to that of Theorem 1. Using the triangle vertices , , and , we apply the plane function instead of the line function .

The previous lemma suggests how the results of the previous subsection can be transferred to higher dimensions.

Let be points. Their convex hullis the -simplex in if the points are linearly independent.

Let be a convex set, and let be a -simplex with vertices . In the consideration that follows, we use a function satisfying the inequalityand the reverse inequalitywhere is the function of the hyperplane passing through the corresponding points of the graph of .

Theorem 6. *Let be a closed convex set, let be a -simplex, and let and be functions.**Then, a pair of unital positive functionals such that**satisfies the inequality**for every continuous function satisfying (31)-(32) and providing that .*

*Proof. *Relying on the hyperplane function where are the simplex vertices, we can apply the proof similar to that of Theorem 1.

Including the -membered convex combination with the equality in (33) in a way thatand using the double equalitywe can derive the double inequality

The following functional form of Jensen’s inequality is true for functions of several variables.

Corollary 7. *Let be a closed convex set, let be a -simplex, and let be a function.**Then, a unital positive functional such that**satisfies the inequality**for every continuous function satisfying (25)-(26) and providing that .*

Continuous convex functions of several variables can be characterized by unital positive functionals in the following way. The dimension of a convex set is defined as the dimension of its affine hull.

Proposition 8. *Let be a closed convex set of dimension . A continuous function is convex if and only if it satisfies the inequality**for every -tuple of -simplex vertices , every function such that , and every unital positive functional .*

*Proof. *To prove the sufficiency, we take a convex combination of -simplex vertices . If , we take the constant mapping consisting of constant functions and continue the proof in the same way as in Proposition 3. Finally, we get Jensen’s inequalityconfirming the convexity of the function .

#### 4. Applications to Functional Quasiarithmetic Means

Functions investigated in Subsection 3.1 can be included to quasiarithmetic means by applying methods such as those for convex functions. The basic facts relating to quasiarithmetic and power means can be found in [7]. For more details on different forms of quasiarithmetic and power means, as well as their refinements, see [8].

The next generalization of Theorem 1 will be applied to the consideration of functional quasiarithmetic means.

Corollary 9. *Let be a closed interval, let be a bounded closed subinterval, and let and be functions.**Then, a pair of collections of positive functionals providing the unit equalities and the equality**satisfies the inequality**for every continuous function satisfying (12)-(13) and providing that all functions .*

Now, we present a way of introducing the functional quasiarithmetic means. Let be functions, and let be a strictly monotone continuous function such that all . Let be positive linear functionals providing the unit equality . The quasiarithmetic mean of functions respecting the function and functionals can be defined byIn what follows, we will use the abbreviation for the above mean. The term in parentheses belongs to the interval , and therefore the quasiarithmetic mean belongs to the interval .

In applications of the function convexity, we use a pair of strictly monotone continuous functions such that is convex with respect to (it also says that is -convex), which means that the function is convex on the interval . A similar notation is used for the concavity.

Instead of the convexity of , we will apply the conditions in (12)-(13) via Corollary 9 as follows.

Theorem 10. *Let be a closed interval, let be a bounded closed subinterval, and let and be functions. Let be a pair of collections of positive functionals providing the unit equalities . Let be strictly monotone continuous functions such that all functions , and let be the composite function.**If satisfies (12)-(13) and is increasing and if the equality**is valid, then we have the inequality*

*Proof. *We take and . We will apply Corollary 9 to the functions and and the function .

Using the equality and including the functions and , we haveThen, the inequalityfollows from Corollary 9, and applying the increasing function , we getThe above inequality is actually the inequality in (46) because and .

All the cases of the above theorem are as follows.

Corollary 11. *Let be the composite function satisfying the conditions of Theorem 10.**If either satisfies (12)-(13) and is increasing or satisfies (12)-(13) and is decreasing and if the equality in (45) is valid, then the inequality holds in (46).**If either satisfies (12)-(13) and is decreasing or satisfies (12)-(13) and is increasing and if the equality in (45) is valid, then the reverse inequality holds in (46).*

A special case of the quasiarithmetic means in (44) is power means depending on real exponents . Thus, using the functionswhere , we get the power means of order in the form

To apply Theorem 1 to the power means, we use a closed interval where is a positive number and the equality

Corollary 12. *Let be an unbounded closed interval where , let be a bounded closed subinterval, and let and be functions. Let be a pair of collections of positive functionals providing the unit equalities .**If**then*

*Proof. *The proof follows from Theorem 10 and Corollary 11 by using convex and concave functions such as and for , and for .

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

Our aim is to use Theorem 6 to obtain certain discrete and integral inequalities concerning functions of several variables. The following is the application to discrete inequalities.

Proposition 13. *Let be a closed convex set, let be a -simplex, let be a convex combination of points , and let be a convex combination of points .**If the above convex combinations have the common center**then the inequality**holds for every continuous function satisfying (31)-(32).*

*Proof. *We take the set and the space containing all real functions on . We also take any simplex vertex and its coordinates .

Let () be functions defined byThen, and .

Let be summarizing unital positive functionals defined byApplying the functional to the functions and the functional to the functions , we obtainNow, we can apply Theorem 6 and get the inequalitywhich concludes the proof.

Proposition 13 does not generally hold for convex functions. The next example demonstrates a concrete planar case of .

*Example 14. *We take the convex function , the triangle with vertices , , and , and the outside points , , and .

Then, we have

More details on the behavior of a convex function of two variables on the triangle and outside the triangle can be found in [9, Theorem 3.2]. Triangle cones have a prominent part in these considerations.

The integral analogy of the concept of convex combination is the concept of barycenter. Let be a positive measure on , and let be a -measurable set with . Given the positive integer , let be the partition of pairwise disjoint -measurable sets . Taking points , we determine the convex combinationwhose center belongs to . The -barycenter of the set can be defined as the limit of the sequence ; that is,As defined above, the point is in . So, the convex sets contain its barycenters.

The application of Theorem 6 to integral inequalities is as follows.

Proposition 15. *Let be a positive measure on . Let be a closed convex set, let be a -simplex, and let and be sets of positive -measures.**If the above sets have the common -barycenter**then the inequality**holds for every continuous function satisfying (31)-(32).*

*Proof. *The proof is similar to that of Proposition 13 by using as the space of all -integrable functions on . We apply the integrating unital positive functional defined byto the functions of (57), as well as the integrating unital positive functional defined byto the functions of (58).

If are the simplex vertices, then using the unique convex combination satisfyingand applying (37), we obtain the extension of (66) as the double inequalityThe above inequality is reminiscent of Hermite-Hadamard’s inequality where discrete and integral terms are replaced, see the below inequality in (72).

Implementing convex combinations to the integral method, one may derive the following version of the Hermite-Hadamard inequality for convex functions on simplexes.

Proposition 16. *Let be a positive measure on . Let be a -simplex of positive -measure, let be simplex vertices, and let be their convex combination.**If the convex combination center and the -barycenter of both fall at the same point**then the double inequality**holds for every -integrable convex function .*

More on the important and interesting Hermite-Hadamard’s inequality, including historical facts about its name, can be found in [10, 11].

#### Conflict of Interests

The author declares that there is no conflict of interests regarding the publication of this paper.

#### Acknowledgments

This work has been fully supported by Mechanical Engineering Faculty in Slavonski Brod and Croatian Science Foundation under Project 5435. The author thanks Velimir Pavić (graphic designer at Školska knjiga Zagreb) who has graphically prepared Figure 1.

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#### Copyright

Copyright © 2015 Zlatko Pavić. 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.