# Functions Like Convex Functions

Authors: Zlatko Pavić
Journal: Journal of Function Spaces
 (2015)

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## 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.


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## Body

## 1. Introduction

Introduction is intended to be a brief overview of the concept of convexity and affinity. LetX be a real linear space. Let a,b∈X be points and let α,β∈R be coefficients. Their binomial combination(1)αa+βbis convex if α,β≥0 and if(2)α+β=1.If c=αa+βb, then the point c itself is called the combination center.A setS⊆X is convex if it contains all binomial convex combinations of its points. The convex hull convS of the set S is the smallest convex set containing S, and it consists of all binomial convex combinations of points of S.LetC⊆X be a convex set. A function f:C→R is convex if the inequality(3)f(αa+βb)≤αf(a)+βf(b)holds for all binomial convex combinations αa+βb of pairs of points a,b∈C.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 ton-membered combinations for any positive integer n. 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

LetX be a nonempty set, and let X be a subspace of the linear space of all real functions on the domain X. We assume that X contains the unit function 1 defined by 1(x)=1 for every x∈X.LetI⊆R be an interval, and let XI be the set containing all functions g∈X with the image in I. Then, XI is convex set in the space X. The same is true for convex sets of Euclidean spaces. Let C⊆Rk be a convex set, and let (Xk)C be the set containing all function k-tuples g=(g1,…,gk)∈Xk with the image in C. Then, (Xk)C is convex set in the space Xk.A linear functionalL:X→R is positive (nonnegative) if L(g)≥0 for every nonnegative function g∈X, and L is unital (normalized) if L(1)=1. If g∈X, then for every unital positive functional L the number L(g) is in the closed interval of real numbers containing the image of g. Through the paper, the space of all linear functionals on the space X will be denoted with L(X).Letf:R→R be an affine function, that is, the function of the form f(x):=κx+λ where κ and λ are real constants. If g1,…,gn∈X are functions and if L1,…,Ln∈L(X) are positive functionals providing the unit equality(4)∑i=1n‍Li(1)=1,then(5)f∑i=1n‍Ligi=κ∑i=1n‍Ligi+λ∑i=1n‍Li1=∑i=1n‍Liκgi+λ1=∑i=1n‍Lifgi.Respecting the requirement of unit equality in (4), the sum ∑i=1n‍Li(gi) could be called the functional convex combination. In the case n=1, the functional L=L1 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.
LetI⊆R be a closed interval, and let g∈XI be a function.
Then, a unital positive functionalL∈L(X) ensures the inclusion(6)L(g)∈Iand satisfies the inequality(7)f(L(g))≤L(f(g))for every continuous convex function f:I→R providing that f(g)∈X.
Iff is concave, then the reverse inequality is valid in (7). If f is affine, then the equality is valid in (7).The intervalI must be closed, otherwise it could happen that L(g)∉I. The function f 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.
LetC⊆Rk be a closed convex set, and let g=(g1,…,gk)∈(Xk)C be a function.
Then, a unital positive functionalL∈L(X) ensures the inclusion(8)(L(g1),…,L(gk))∈Cand satisfies the inequality(9)f(L(g1),…,L(gk))≤L(f(g1,…,gk))for every continuous convex function f:C→R providing that f(g1,…,gk)∈X.
Iff is concave, then the reverse inequality is valid in (9). If f 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 Theorem1 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 intervalI⊆R and a bounded closed subinterval [a,b]⊆I with endpoints a<b.Every numberx∈R can be uniquely presented as the binomial affine combination(10)x=b-xb-aa+x-ab-ab,which is convex if and only if the number x belongs to the interval [a,b]. Let f:I→R be a function, and let f{a,b}line:R→R be the function of the line passing through the points A(a,f(a)) and B(b,f(b)) of the graph of f. Applying the affinity of the function f{a,b}line to the combination in (10), we obtain its equation(11)f{a,b}line(x)=b-xb-af(a)+x-ab-af(b).The consequence of the representations in (10) and (11) is the fact that every convex function f:I→R satisfies the inequality(12)f(x)≤f{a,b}line(x)forx∈[a,b]and the reverse inequality(13)f(x)≥f{a,b}line(x)forx∈I∖(a,b).In the following consideration, we use continuous functions satisfying the inequalities in (12)-(13).Theorem 1.
LetI⊆R be a closed interval, let [a,b]⊆I be a bounded closed subinterval, and let g∈X[a,b] and h∈XI∖(a,b) be functions.
Then, a pair of unital positive functionalsL,H∈L(X) such that(14)L(g)=H(h),satisfies the inequality(15)L(f(g))≤H(f(h))for every continuous function f:I→R satisfying (12)-(13) and providing that f(g),f(h)∈X.Proof.
The numberL(g) belongs to the interval [a,b] by the inclusion in (6). Using the features of the function f and applying the affinity of the function f{a,b}line, we get(16)Lfg≤L(f{a,b}line(g))=f{a,b}line(L(g))=f{a,b}line(H(h))=H(f{a,b}line(h))≤H(f(h))because f{a,b}line(h(x))≤f(h(x)) for every x∈X.It is obvious that a continuous convex functionf:I→R satisfies Theorem 1 for every subinterval [a,b]⊆I with endpoints a<b. The function used in Theorem 1 is shown in Figure 1. Such a function satisfies only the global property of convexity on the sets [a,b] and I∖(a,b).Figure 1
A continuous function satisfying (12)-(13).Involving the binomial convex combinationαa+βb with the equality in (14) by assuming that(17)L(g)=αa+βb=H(h)and inserting the term αf(a)+βf(b) in (16) via the double equality(18)f{a,b}line(L(g))=αf(a)+βf(b)=f{a,b}line(H(h))which is true because f{a,b}line(αa+βb)=αf(a)+βf(b), we achieve the double inequality(19)L(f(g))≤αf(a)+βf(b)≤H(f(h)).The functions used in Theorem1 satisfy the functional form of Jensen’s inequality in the following case.Corollary 2.
LetI⊆R be a closed interval, let [a,b]⊆I be a bounded closed subinterval, and let h∈XI∖(a,b) be a function.
Then, a unital positive functionalH∈L(X) such that(20)H(h)∈[a,b]satisfies the inequality(21)f(H(h))≤H(f(h))for every continuous function satisfying (12)-(13) and providing that f(h)∈X.Proof.
Puttingαa+βb=H(h), it follows that(22)fHh=fαa+βb≤fa,blineαa+βb=αfa+βfb≤Hfhby the right inequality in (19).Now, we give a characterization of continuous convex functions by using unital positive functionals.Proposition 3.
LetI⊆R be a closed interval. A continuous function f:I→R is convex if and only if it satisfies the inequality(23)L(f(g))≤f{a,b}line(L(g))for every pair of interval endpoints a,b∈I, every function g∈X[a,b] such that f(g)∈X, and every unital positive functional L∈L(X).Proof.
Let us prove the sufficiency. Letc:=αa+βb be a convex combination of points a,b∈I where a<b. We take the constant function g=c1 in X[a,b] (actually g(x)=c for every x∈X) and a unital positive functional L. Then, connecting(24)Lfg=Lfc1=fc=fαa+βb,fa,blineLg=fa,blineαa+βb=αfa+βfbvia (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 Theorem6 generalizing Theorem 1 to functions of several variables.LetC⊆R2 be a convex set, let △⊆C be a triangle with vertices A, B, and C, and let △o be its interior. In the following observation, we assume that f:C→R is a continuous function satisfying the inequality(25)fP≤fA,B,CplanePforP∈△and the reverse inequality(26)f(P)≥f{A,B,C}plane(P)forP∈C∖△o,where f{A,B,C}plane is the function of the plane passing through the corresponding points of the graph of f.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 functionf(x,y)=x2+y2, the triangle with vertices A(0,0), B(1,0), and C(0,2), and the outside point P(1,1).
The valuation of functionsf and f{A,B,C}plane(x,y)=x+2y at the point P is(27)2=f(P)<f{A,B,C}plane(P)=3as opposed to (26).The generalization of Theorem1 to two dimensions is as follows.Lemma 5.
LetC⊆R2 be a closed convex set, let △⊆C be a triangle, and let g=(g1,g2)∈(X2)△ and h=(h1,h2)∈(X2)C∖△o be functions.
Then, a pair of unital positive functionalsL,H∈L(X) such that(28)Lg1,Lg2=Hh1,Hh2satisfies the inequality(29)Lfg1,g2≤Hfh1,h2for every continuous function satisfying (25)-(26) and providing that f(g1,g2),f(h1,h2)∈X.Proof.
The proof is similar to that of Theorem1. Using the triangle vertices A, B, and C, we apply the plane function f{A,B,C}plane instead of the line function f{a,b}line.The previous lemma suggests how the results of the previous subsection can be transferred to higher dimensions.LetS1,…,Sk+1∈Rk be points. Their convex hull(30)S=conv{S1,…,Sk+1}is the k-simplex in Rk if the points S1-Sk+1,…,Sk-Sk+1 are linearly independent.LetC⊆Rk be a convex set, and let S⊆C be a k-simplex with vertices S1,…,Sk+1. In the consideration that follows, we use a function f:C→R satisfying the inequality(31)f(P)≤f{S1,…,Sk+1}hyperplane(P)forP∈Sand the reverse inequality(32)fP≥fS1,…,Sk+1hyperplanePforP∈C∖So,where f{S1,…,Sk+1}hyperplane is the function of the hyperplane passing through the corresponding points of the graph of f.Theorem 6.
LetC⊆Rk be a closed convex set, let S⊆C be a k-simplex, and let g=(g1,…,gk)∈(Xk)S and h=(h1,…,hk)∈(Xk)C∖So be functions.
Then, a pair of unital positive functionalsL,H∈L(X) such that(33)(L(g1),…,L(gk))=(H(h1),…,H(hk))satisfies the inequality(34)Lfg1,…,gk≤Hfh1,…,hkfor every continuous function satisfying (31)-(32) and providing that f(g1,…,gk),f(h1,…,hk)∈X.Proof.
Relying on the hyperplane functionf{S1,…,Sk+1}hyperplane where S1,…,Sk+1 are the simplex vertices, we can apply the proof similar to that of Theorem 1.Including the(k+1)-membered convex combination ∑p=1k+1‍γpSp with the equality in (33) in a way that(35)(L(g1),…,L(gk))=∑p=1k+1‍γpSp=(H(h1),…,H(hk))and using the double equality(36)fS1,…,Sk+1hyperplaneLg1,…,Lgk=∑p=1k+1‍γpfSp=f{S1,…,Sk+1}hyperplane(H(h1),…,H(hk)),we can derive the double inequality(37)Lfg1,…,gk≤∑p=1k+1‍γpfSp≤Hfh1,…,hk.The following functional form of Jensen’s inequality is true for functions of several variables.Corollary 7.
LetC⊆Rk be a closed convex set, let S⊆C be a k-simplex, and let h=(h1,…,hk)∈(Xk)C∖So be a function.
Then, a unital positive functionalH∈L(X) such that(38)(H(h1),…,H(hk))∈Ssatisfies the inequality(39)f(H(h1),…,H(hk))≤H(f(h1,…,hk))for every continuous function satisfying (25)-(26) and providing that f(h1,…,hk)∈X.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.
LetC⊆Rk be a closed convex set of dimension k. A continuous function f:C→R is convex if and only if it satisfies the inequality(40)L(f(g1,…,gk))≤f{S1,…,Sk+1}hyperplane(L(g1),…,L(gk))for every (k+1)-tuple of k-simplex vertices S1,…,Sk+1∈C, every function g=(g1,…,gk)∈(Xk)S such that f(g1,…,gk)∈X, and every unital positive functional L∈L(X).Proof.
To prove the sufficiency, we take a convex combinationC=∑p=1k+1‍γpSp of k-simplex vertices S1,…,Sk+1∈C. If C=(c1,…,ck), we take the constant mapping g=(g1,…,gk)∈(Xk)S consisting of constant functions gi=ci1 and continue the proof in the same way as in Proposition 3. Finally, we get Jensen’s inequality(41)f∑p=1k+1‍γpSp≤∑p=1k+1‍γpf(Sp)confirming the convexity of the function f.


## 3.1. Functions of One Variable

The main result of this subsection is Theorem1 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 intervalI⊆R and a bounded closed subinterval [a,b]⊆I with endpoints a<b.Every numberx∈R can be uniquely presented as the binomial affine combination(10)x=b-xb-aa+x-ab-ab,which is convex if and only if the number x belongs to the interval [a,b]. Let f:I→R be a function, and let f{a,b}line:R→R be the function of the line passing through the points A(a,f(a)) and B(b,f(b)) of the graph of f. Applying the affinity of the function f{a,b}line to the combination in (10), we obtain its equation(11)f{a,b}line(x)=b-xb-af(a)+x-ab-af(b).The consequence of the representations in (10) and (11) is the fact that every convex function f:I→R satisfies the inequality(12)f(x)≤f{a,b}line(x)forx∈[a,b]and the reverse inequality(13)f(x)≥f{a,b}line(x)forx∈I∖(a,b).In the following consideration, we use continuous functions satisfying the inequalities in (12)-(13).Theorem 1.
LetI⊆R be a closed interval, let [a,b]⊆I be a bounded closed subinterval, and let g∈X[a,b] and h∈XI∖(a,b) be functions.
Then, a pair of unital positive functionalsL,H∈L(X) such that(14)L(g)=H(h),satisfies the inequality(15)L(f(g))≤H(f(h))for every continuous function f:I→R satisfying (12)-(13) and providing that f(g),f(h)∈X.Proof.
The numberL(g) belongs to the interval [a,b] by the inclusion in (6). Using the features of the function f and applying the affinity of the function f{a,b}line, we get(16)Lfg≤L(f{a,b}line(g))=f{a,b}line(L(g))=f{a,b}line(H(h))=H(f{a,b}line(h))≤H(f(h))because f{a,b}line(h(x))≤f(h(x)) for every x∈X.It is obvious that a continuous convex functionf:I→R satisfies Theorem 1 for every subinterval [a,b]⊆I with endpoints a<b. The function used in Theorem 1 is shown in Figure 1. Such a function satisfies only the global property of convexity on the sets [a,b] and I∖(a,b).Figure 1
A continuous function satisfying (12)-(13).Involving the binomial convex combinationαa+βb with the equality in (14) by assuming that(17)L(g)=αa+βb=H(h)and inserting the term αf(a)+βf(b) in (16) via the double equality(18)f{a,b}line(L(g))=αf(a)+βf(b)=f{a,b}line(H(h))which is true because f{a,b}line(αa+βb)=αf(a)+βf(b), we achieve the double inequality(19)L(f(g))≤αf(a)+βf(b)≤H(f(h)).The functions used in Theorem1 satisfy the functional form of Jensen’s inequality in the following case.Corollary 2.
LetI⊆R be a closed interval, let [a,b]⊆I be a bounded closed subinterval, and let h∈XI∖(a,b) be a function.
Then, a unital positive functionalH∈L(X) such that(20)H(h)∈[a,b]satisfies the inequality(21)f(H(h))≤H(f(h))for every continuous function satisfying (12)-(13) and providing that f(h)∈X.Proof.
Puttingαa+βb=H(h), it follows that(22)fHh=fαa+βb≤fa,blineαa+βb=αfa+βfb≤Hfhby the right inequality in (19).Now, we give a characterization of continuous convex functions by using unital positive functionals.Proposition 3.
LetI⊆R be a closed interval. A continuous function f:I→R is convex if and only if it satisfies the inequality(23)L(f(g))≤f{a,b}line(L(g))for every pair of interval endpoints a,b∈I, every function g∈X[a,b] such that f(g)∈X, and every unital positive functional L∈L(X).Proof.
Let us prove the sufficiency. Letc:=αa+βb be a convex combination of points a,b∈I where a<b. We take the constant function g=c1 in X[a,b] (actually g(x)=c for every x∈X) and a unital positive functional L. Then, connecting(24)Lfg=Lfc1=fc=fαa+βb,fa,blineLg=fa,blineαa+βb=αfa+βfbvia (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 Theorem6 generalizing Theorem 1 to functions of several variables.LetC⊆R2 be a convex set, let △⊆C be a triangle with vertices A, B, and C, and let △o be its interior. In the following observation, we assume that f:C→R is a continuous function satisfying the inequality(25)fP≤fA,B,CplanePforP∈△and the reverse inequality(26)f(P)≥f{A,B,C}plane(P)forP∈C∖△o,where f{A,B,C}plane is the function of the plane passing through the corresponding points of the graph of f.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 functionf(x,y)=x2+y2, the triangle with vertices A(0,0), B(1,0), and C(0,2), and the outside point P(1,1).
The valuation of functionsf and f{A,B,C}plane(x,y)=x+2y at the point P is(27)2=f(P)<f{A,B,C}plane(P)=3as opposed to (26).The generalization of Theorem1 to two dimensions is as follows.Lemma 5.
LetC⊆R2 be a closed convex set, let △⊆C be a triangle, and let g=(g1,g2)∈(X2)△ and h=(h1,h2)∈(X2)C∖△o be functions.
Then, a pair of unital positive functionalsL,H∈L(X) such that(28)Lg1,Lg2=Hh1,Hh2satisfies the inequality(29)Lfg1,g2≤Hfh1,h2for every continuous function satisfying (25)-(26) and providing that f(g1,g2),f(h1,h2)∈X.Proof.
The proof is similar to that of Theorem1. Using the triangle vertices A, B, and C, we apply the plane function f{A,B,C}plane instead of the line function f{a,b}line.The previous lemma suggests how the results of the previous subsection can be transferred to higher dimensions.LetS1,…,Sk+1∈Rk be points. Their convex hull(30)S=conv{S1,…,Sk+1}is the k-simplex in Rk if the points S1-Sk+1,…,Sk-Sk+1 are linearly independent.LetC⊆Rk be a convex set, and let S⊆C be a k-simplex with vertices S1,…,Sk+1. In the consideration that follows, we use a function f:C→R satisfying the inequality(31)f(P)≤f{S1,…,Sk+1}hyperplane(P)forP∈Sand the reverse inequality(32)fP≥fS1,…,Sk+1hyperplanePforP∈C∖So,where f{S1,…,Sk+1}hyperplane is the function of the hyperplane passing through the corresponding points of the graph of f.Theorem 6.
LetC⊆Rk be a closed convex set, let S⊆C be a k-simplex, and let g=(g1,…,gk)∈(Xk)S and h=(h1,…,hk)∈(Xk)C∖So be functions.
Then, a pair of unital positive functionalsL,H∈L(X) such that(33)(L(g1),…,L(gk))=(H(h1),…,H(hk))satisfies the inequality(34)Lfg1,…,gk≤Hfh1,…,hkfor every continuous function satisfying (31)-(32) and providing that f(g1,…,gk),f(h1,…,hk)∈X.Proof.
Relying on the hyperplane functionf{S1,…,Sk+1}hyperplane where S1,…,Sk+1 are the simplex vertices, we can apply the proof similar to that of Theorem 1.Including the(k+1)-membered convex combination ∑p=1k+1‍γpSp with the equality in (33) in a way that(35)(L(g1),…,L(gk))=∑p=1k+1‍γpSp=(H(h1),…,H(hk))and using the double equality(36)fS1,…,Sk+1hyperplaneLg1,…,Lgk=∑p=1k+1‍γpfSp=f{S1,…,Sk+1}hyperplane(H(h1),…,H(hk)),we can derive the double inequality(37)Lfg1,…,gk≤∑p=1k+1‍γpfSp≤Hfh1,…,hk.The following functional form of Jensen’s inequality is true for functions of several variables.Corollary 7.
LetC⊆Rk be a closed convex set, let S⊆C be a k-simplex, and let h=(h1,…,hk)∈(Xk)C∖So be a function.
Then, a unital positive functionalH∈L(X) such that(38)(H(h1),…,H(hk))∈Ssatisfies the inequality(39)f(H(h1),…,H(hk))≤H(f(h1,…,hk))for every continuous function satisfying (25)-(26) and providing that f(h1,…,hk)∈X.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.
LetC⊆Rk be a closed convex set of dimension k. A continuous function f:C→R is convex if and only if it satisfies the inequality(40)L(f(g1,…,gk))≤f{S1,…,Sk+1}hyperplane(L(g1),…,L(gk))for every (k+1)-tuple of k-simplex vertices S1,…,Sk+1∈C, every function g=(g1,…,gk)∈(Xk)S such that f(g1,…,gk)∈X, and every unital positive functional L∈L(X).Proof.
To prove the sufficiency, we take a convex combinationC=∑p=1k+1‍γpSp of k-simplex vertices S1,…,Sk+1∈C. If C=(c1,…,ck), we take the constant mapping g=(g1,…,gk)∈(Xk)S consisting of constant functions gi=ci1 and continue the proof in the same way as in Proposition 3. Finally, we get Jensen’s inequality(41)f∑p=1k+1‍γpSp≤∑p=1k+1‍γpf(Sp)confirming the convexity of the function f.


## 4. Applications to Functional Quasiarithmetic Means

Functions investigated in Subsection3.1 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 Theorem1 will be applied to the consideration of functional quasiarithmetic means.Corollary 9.
LetI⊆R be a closed interval, let [a,b]⊆I be a bounded closed subinterval, and let g1,…,gn∈X[a,b] and h1,…,hm∈XI∖(a,b) be functions.
Then, a pair of collections of positive functionalsLi,Hj∈L(X) providing the unit equalities ∑i=1n‍Li(1)=∑j=1m‍Hj(1)=1 and the equality(42)∑i=1n‍Li(gi)=∑j=1m‍Hj(hj)satisfies the inequality(43)∑i=1n‍Li(f(gi))≤∑j=1m‍Hj(f(hj))for every continuous function satisfying (12)-(13) and providing that all functions f(gi),f(hj)∈X.Now, we present a way of introducing the functional quasiarithmetic means. Letg1,…,gn∈XI be functions, and let φ:I→R be a strictly monotone continuous function such that all φ(gi)∈X. Let L1,…,Ln:X→R be positive linear functionals providing the unit equality ∑i=1n‍Li(1)=1. The quasiarithmetic mean of functions gi respecting the function φ and functionals Li can be defined by(44)Mφ(L1…,Ln;g1,…,gn)=φ-1∑i=1n‍Li(φ(gi)).In what follows, we will use the abbreviation Mφ(Li,gi) for the above mean. The term in parentheses belongs to the interval φ(I), and therefore the quasiarithmetic mean Mφ(Li,gi) belongs to the interval I.In applications of the function convexity, we use a pair of strictly monotone continuous functionsφ,ψ:I→R such that ψ is convex with respect to φ (it also says that ψ is φ-convex), which means that the function f=ψ(φ-1) is convex on the interval φ(I). A similar notation is used for the concavity.Instead of the convexity off, we will apply the conditions in (12)-(13) via Corollary 9 as follows.Theorem 10.
LetI⊆R be a closed interval, let [a,b]⊆I be a bounded closed subinterval, and let g1,…,gn∈X[a,b] and h1,…,hm∈XI∖(a,b) be functions. Let Li,Hj∈L(X) be a pair of collections of positive functionals providing the unit equalities ∑i=1n‍Li(1)=∑j=1m‍Hj(1)=1. Let φ,ψ:I→R be strictly monotone continuous functions such that all functions φ(gi),φ(hj),ψ(gi),ψ(hj)∈X, and let f=ψ(φ-1) be the composite function.
Iff satisfies (12)-(13) and ψ is increasing and if the equality(45)Mφ(Li,gi)=Mφ(Hj,hj)is valid, then we have the inequality(46)Mψ(Li,gi)≤Mψ(Hj,hj).Proof.
We takeJ=φ(I) and [c,d]=φ([a,b]). We will apply Corollary 9 to the functions ui=φ(gi)∈X[c,d] and vj=φ(hj)∈XJ∖(c,d) and the function f:J→R.
Using the equalityφ(Mφ(Li,gi))=φ(Mφ(Hj,hj)) and including the functions ui and vj, we have(47)∑i=1n‍Li(ui)=∑j=1m‍Hj(vj).Then, the inequality(48)∑i=1n‍Li(f(ui))≤∑j=1m‍Hj(f(vj))follows from Corollary 9, and applying the increasing function ψ-1, we get(49)ψ-1∑i=1n‍Li(f(ui))≤ψ-1∑j=1m‍Hj(f(vj)).The above inequality is actually the inequality in (46) because f(ui)=ψ(gi) and f(vj)=ψ(hj).All the cases of the above theorem are as follows.Corollary 11.
Letf=ψ(φ-1) be the composite function satisfying the conditions of Theorem 10.
If eitherf satisfies (12)-(13) and ψ is increasing or -f satisfies (12)-(13) and ψ is decreasing and if the equality in (45) is valid, then the inequality holds in (46).
If eitherf satisfies (12)-(13) and ψ is decreasing or -f satisfies (12)-(13) and ψ 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 r. Thus, using the functions(50)φr(x)=xr,r≠0ln⁡x,r=0,where x∈(0,∞), we get the power means of order r in the form(51)Mr(Li,gi)=∑i=1n‍Ligir1/r,r≠0exp⁡∑i=1n‍Li(ln⁡gi),r=0.To apply Theorem1 to the power means, we use a closed interval I=[ε,∞) where ε is a positive number and the equality(52)M1(L,gi)=∑i=1n‍Li(gi).Corollary 12.
LetI=[ε,∞) be an unbounded closed interval where ε>0, let [a,b]⊂I be a bounded closed subinterval, and let g1,…,gn∈X[a,b] and h1,…,hm∈XI∖(a,b) be functions. Let Li,Hj∈L(X) be a pair of collections of positive functionals providing the unit equalities ∑i=1n‍Li(1)=∑j=1m‍Hj(1)=1.
If(53)M1(Li,gi)=M1(Hj,hj),then(54)Mr(Li,gi)≤Mr(Hj,hj)forr≥1,Mr(Li,gi)≥Mr(Hj,hj)forr≤1.Proof.
The proof follows from Theorem10 and Corollary 11 by using convex and concave functions such as φ(x)=x and ψ(x)=xr for r≠0, and ψ(x)=ln⁡x for r=0.


## 5. Applications to Discrete and Integral Inequalities

Our aim is to use Theorem6 to obtain certain discrete and integral inequalities concerning functions of several variables. The following is the application to discrete inequalities.Proposition 13.
LetC⊆Rk be a closed convex set, let S⊆C be a k-simplex, let ∑i=1n‍αiAi be a convex combination of points Ai∈S, and let ∑j=1m‍βjBj be a convex combination of points Bj∈C∖So.
If the above convex combinations have the common center(55)∑i=1n‍αiAi=∑j=1m‍βjBj,then the inequality(56)∑i=1n‍αif(Ai)≤∑j=1m‍βjf(Bj)holds for every continuous function f:C→R satisfying (31)-(32).Proof.
We take the setX=C and the space X containing all real functions on C. We also take any simplex vertex S and its coordinates (s1,…,sk).
Letgp,hp∈X (p=1,…,k) be functions defined by(57)gp(x1,…,xk)=xp,(x1,…,xk)∈Ssp,x1,…,xk∈C∖S,(58)hp(x1,…,xk)=sp,(x1,…,xk)∈Soxp,x1,…,xk∈C∖So.Then, g=(g1,…,gk)∈(Xk)S and h=(h1,…,hk)∈(Xk)C∖So.
LetL,H∈L(X) be summarizing unital positive functionals defined by(59)Lg=∑i=1n‍αigAi,H(h)=∑j=1m‍βjh(Bj).
Applying the functionalL to the functions gp and the functional H to the functions hp, we obtain(60)∑i=1n‍αiAi=Lg1,…,Lgk=(H(h1),…,H(hk))=∑j=1m‍βjBj.Now, we can apply Theorem 6 and get the inequality(61)∑i=1n‍αifAi=Lfg1,…,gk≤Hfh1,…,hk=∑j=1m‍βjf(Bj)which concludes the proof.Proposition13 does not generally hold for convex functions. The next example demonstrates a concrete planar case of k=2.Example 14.
We take the convex functionf(x,y)=x2+y2, the triangle with vertices A1(-3,0), A2(3,0), and A3(0,3), and the outside points B1(-2,2), B2(0,-2), and B3(2,2).
Then, we have(62)13A1+13A2+13A3=38B1+28B2+38B3,9=13fA1+13fA2+13fA3>38f(B1)+28f(B2)+38f(B3)=7.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. Letμ be a positive measure on Rk, and let A⊆Rk be a μ-measurable set with μ(A)>0. Given the positive integer n, let A=∪i=1nAni be the partition of pairwise disjoint μ-measurable sets Ani. Taking points Ani∈Ani, we determine the convex combination(63)An=∑i=1n‍μ(Ani)μ(A)Aniwhose center An belongs to convA. The μ-barycenter of the set A can be defined as the limit of the sequence (An)n; that is,(64)MA,μ=limn→∞∑i=1n‍μAniμAAni=1μ(A)∫Ax1dμ,…,∫Axkdμ.As defined above, the point M(A,μ) is in convA. So, the convex sets contain its barycenters.The application of Theorem6 to integral inequalities is as follows.Proposition 15.
Letμ be a positive measure on Rk. Let C⊆Rk be a closed convex set, let S⊆C be a k-simplex, and let A⊆S and B⊆C∖So be sets of positive μ-measures.
If the above sets have the commonμ-barycenter(65)M(A,μ)=M(B,μ),then the inequality(66)1μ(A)∫Af(x1,…,xk)dμ≤1μ(B)∫Bf(x1,…,xk)dμholds for every continuous function f:C→R satisfying (31)-(32).Proof.
The proof is similar to that of Proposition13 by using X as the space of all μ-integrable functions on C. We apply the integrating unital positive functional L defined by(67)L(g)=1μ(A)∫Ag(x1,…,xk)dμto the functions gp of (57), as well as the integrating unital positive functional H defined by(68)H(h)=1μ(B)∫Bh(x1,…,xk)dμto the functions hp of (58).IfS1,…,Sk+1 are the simplex vertices, then using the unique convex combination ∑p=1k+1‍γpSp satisfying(69)M(A,μ)=∑p=1k+1‍γpSp=M(B,μ)and applying (37), we obtain the extension of (66) as the double inequality(70)1μA∫A‍fx1,…,xkdμ≤∑p=1k+1‍γpfSp≤1μ(B)∫Bf(x1,…,xk)dμ.The 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.
Letμ be a positive measure on Rk. Let S⊂Rk be a k-simplex of positive μ-measure, let S1,…,Sk+1 be simplex vertices, and let ∑p=1k+1‍γpSp be their convex combination.
If the convex combination center and theμ-barycenter of S both fall at the same point(71)∑p=1k+1‍γpSp=M(S,μ),then the double inequality(72)f∑p=1k+1‍γpSp≤1μ(S)∫Sf(x1,…,xk)dμ≤∑p=1k+1‍γpf(Sp)holds for every μ-integrable convex function f:S→R.More on the important and interesting Hermite-Hadamard’s inequality, including historical facts about its name, can be found in [10, 11].



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