# The Applications of Functional Variants of Jensen's Inequality

Authors: Zlatko Pavić
Journal: Journal of Function Spaces and Applications
 (2013)

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## Abstract

The paper is inspired by McShane's results on the functional form
of Jensen's inequality for convex functions of several variables. The work is
focused on applications and generalizations of this important result. At that,
the generalizations of Jensen's inequality are obtained using the positive linear
functionals.


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## Body

## 1. Introduction

### 1.1. Vector Space of Real Valued Functions and Positive Linear Functionals

Given a nonempty set𝒳 we consider a space of functions g:𝒳→ℝ containing the unit function g=1 defined with 1(x)=1 for every x∈𝒳. Let 𝕏 be a real vector space of real valued functions on 𝒳 containing the unit function. A linear functional M:𝕏→ℝ is said to be positive (nonnegative) or monotone if M(g)≥0 for every nonnegative function g∈𝕏. If M(1)=1, we say that the functional M is unital or normalized.


### 1.2. Functional Forms of Jensen’s Inequality

In 1931, Jessen [1] formulated the functional form of Jensen’s inequality for convex functions of one variable.Theorem A.
Letℐ⊆ℝ be a closed interval. Let g∈𝕏 be a function such that g(x)∈ℐ for every x∈X. Let f:ℐ→ℝ be a continuous convex function such that f(g)∈𝕏.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inequality
(1)f(M(g))≤M(f(g)).The intervalℐ must be closed; otherwise, it could happen that M(g)∉ℐ as mentioned in [2].Example 1.
Let𝒳=ℐ=〈0,1], 𝕏={g:〈0,1]→ℝ∣limx→0+g(x)∈ℝ}, and M:𝕏→ℝ is defined by M(g)=limx→0+g(x). Take the identity function g(x)=x for x∈ℐ. Then g∈𝕏 and g(x)∈ℐ for every x∈𝒳, but M(g)=0∉ℐ.
The functionf must be continuous; otherwise, it could happen that the inequality in (1) does not apply as noted in [3].Example 2.
Consider the previous example with𝒳=ℐ=[0,1]. Take the convex function f:[0,1]→ℝ defined by f(0)=1 and f(x)=0 for x∈〈0,1]. Then f(M(g))=f(0)=1>0=M(f)=M(f(g)).In 1937, McShane [4, Theorems 1-2] 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 (Theorem B) and analytic (Theorem C) formulation of Jensen’s inequality.Theorem B.
Let𝒞⊆ℝn be a closed convex set. Let gi∈𝕏 be functions such that (g1(x),…,gn(x))∈𝒞 for every x∈𝒳.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inclusion
(2)(M(g1),…,M(gn))∈𝒞.Theorem C.
Let𝒞⊆ℝn be a closed convex set. Let gi∈𝕏 be functions such that (g1(x),…,gn(x))∈𝒞 for every x∈𝒳. Let f:𝒞→ℝ be a continuous convex function such that f(g1,…,gn)∈𝕏.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inequality
(3)f(M(g1),…,M(gn))≤M(f(g1,…,gn)).If the functionf is concave, then the reverse inequality is valid in (3). The hyperplanes that contain the set 𝒞 were used in the proof of Theorem B. The epigraph of the function f:
(4)epi(f)={(x1,…,xn,xn+1)∈𝒞×ℝ∣xn+1≥f(x1,…,xn)}
and Theorem B were applied in the proof of Theorem C.


## 1.1. Vector Space of Real Valued Functions and Positive Linear Functionals

Given a nonempty set𝒳 we consider a space of functions g:𝒳→ℝ containing the unit function g=1 defined with 1(x)=1 for every x∈𝒳. Let 𝕏 be a real vector space of real valued functions on 𝒳 containing the unit function. A linear functional M:𝕏→ℝ is said to be positive (nonnegative) or monotone if M(g)≥0 for every nonnegative function g∈𝕏. If M(1)=1, we say that the functional M is unital or normalized.


## 1.2. Functional Forms of Jensen’s Inequality

In 1931, Jessen [1] formulated the functional form of Jensen’s inequality for convex functions of one variable.Theorem A.
Letℐ⊆ℝ be a closed interval. Let g∈𝕏 be a function such that g(x)∈ℐ for every x∈X. Let f:ℐ→ℝ be a continuous convex function such that f(g)∈𝕏.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inequality
(1)f(M(g))≤M(f(g)).The intervalℐ must be closed; otherwise, it could happen that M(g)∉ℐ as mentioned in [2].Example 1.
Let𝒳=ℐ=〈0,1], 𝕏={g:〈0,1]→ℝ∣limx→0+g(x)∈ℝ}, and M:𝕏→ℝ is defined by M(g)=limx→0+g(x). Take the identity function g(x)=x for x∈ℐ. Then g∈𝕏 and g(x)∈ℐ for every x∈𝒳, but M(g)=0∉ℐ.
The functionf must be continuous; otherwise, it could happen that the inequality in (1) does not apply as noted in [3].Example 2.
Consider the previous example with𝒳=ℐ=[0,1]. Take the convex function f:[0,1]→ℝ defined by f(0)=1 and f(x)=0 for x∈〈0,1]. Then f(M(g))=f(0)=1>0=M(f)=M(f(g)).In 1937, McShane [4, Theorems 1-2] 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 (Theorem B) and analytic (Theorem C) formulation of Jensen’s inequality.Theorem B.
Let𝒞⊆ℝn be a closed convex set. Let gi∈𝕏 be functions such that (g1(x),…,gn(x))∈𝒞 for every x∈𝒳.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inclusion
(2)(M(g1),…,M(gn))∈𝒞.Theorem C.
Let𝒞⊆ℝn be a closed convex set. Let gi∈𝕏 be functions such that (g1(x),…,gn(x))∈𝒞 for every x∈𝒳. Let f:𝒞→ℝ be a continuous convex function such that f(g1,…,gn)∈𝕏.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inequality
(3)f(M(g1),…,M(gn))≤M(f(g1,…,gn)).If the functionf is concave, then the reverse inequality is valid in (3). The hyperplanes that contain the set 𝒞 were used in the proof of Theorem B. The epigraph of the function f:
(4)epi(f)={(x1,…,xn,xn+1)∈𝒞×ℝ∣xn+1≥f(x1,…,xn)}
and Theorem B were applied in the proof of Theorem C.


## 2. The Applications and Generalizations of McShane’s Result

This section presents general applications and generalizations of Theorem C. The main generalization is Theorem8.Using McShane’s inequality in (3) it is very easy to get the functional form of Hölder’s and Minkowski’s inequality for n nonnegative functions.Corollary 3 (the functional form of Hölder’s inequality).
Letgi:𝒳→ℝ be nonnegative functions such that g1⋯gn∈𝕏. Let pi>1 be real numbers such that ∑i=1n‍1/pi=1 and all gipi∈𝕏.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inequality
(5)M(g1⋯gn)≤(M(g1p1))1/p1⋯(M(gnpn))1/pn.Proof.
Applying the concave functionf(x1,…,xn)=x11/p1⋯xn1/pn, where all xi≥0, together with the functions gipi to the reverse inequality in (3), it follows that
(6)(M(g1p1))1/p1⋯(M(gnpn))1/pn≥M(g1⋯gn),
which is the required inequality.Corollary 4 (the functional form of Minkowski’s inequality).
Letgi:𝒳→ℝ be nonnegative functions for i=1,…,n. Let p≥1 be a real number such that (g1+⋯+gn)p∈𝕏 and all gip∈𝕏.
Then every unital positive linear functionalM:𝕏→ℝ satisfies the inequality
(7)(M((g1+⋯+gn)p))1/p≤(M(g1p))1/p+⋯+(M(gnp))1/p.Proof.
Applying the concave functionf(x1,…,xn)=(x11/p+⋯+xn1/p)p, where all xi≥0, and the functions gip to the reverse inequality in (3), it follows that
(8)((M(g1p))1/p+⋯+(M(gnp))1/p)p≥M((g1+⋯+gn)p),
and rising to the power of 1/p gives the desired inequality.The McShane inequality in (3) can be easily generalized by introducing the weighted function p∈𝕏 as follows.Theorem 5.
Let𝒞⊆ℝn be a closed convex set. Let gi:𝒳→ℝ be functions such that (g1(x),…,gn(x))∈𝒞 for every x∈𝒳. Let p∈𝕏 be either a nonnegative or nonpositive function such that all pgi∈𝕏. Let f:𝒞→ℝ be a continuous convex function such that pf(g1,…,gn)∈𝕏.
Then every positive linear functionalM:𝕏→ℝ such that M(p)≠0 satisfies the inequality
(9)f(M(pg1)M(p),…,M(pgn)M(p))≤M(pf(g1,…,gn))M(p).Proof.
We use the function vector space𝕐={g:𝒳→ℝ∣pg∈𝕏} and the linear functional L:𝕐→ℝ defined with L(g)=M(pg)/M(p). The functional L is positive and unital. Since all gi∈𝕐 and f(g1,…,gn)∈𝕐, the functional values L(gi) and L(f(g1,…,gn)) may be applied to the inequality in (3).If𝕏 is a vector space of real valued μ-integrable functions on 𝒞, then using the inequality in (9) with the positive linear functional
(10)M(g)=∫𝒞‍gdμ,
we get the following integral form of Jensen’s inequality for convex functions of several variables.Corollary 6.
Letμ be a measure on a closed convex set 𝒞⊆ℝn. Let gi:𝒞→ℝ be μ-measurable functions such that (g1(x),…,gn(x))∈𝒞 for every x∈𝒞. Let p:𝒞→ℝ be either a nonnegative or nonpositive μ-integrable function such that ∫𝒞‍pdμ≠0 and all pgi are μ-integrable.
Then every continuous convex functionf:𝒞→ℝ such that pf(g1,…,gn) is μ-integrable satisfies the integral inequality
(11)f(∫𝒞‍pg1dμ∫𝒞‍pdμ,…,∫𝒞‍pgndμ∫𝒞‍pdμ)≤∫𝒞‍pf(g1,…,gn)dμ∫𝒞‍pdμ.Choosing points(xj1,…,xjn)∈𝒞 for j=1,…,m and applying the inequality in (9) with the positive linear functional
(12)M(g)=∑j=1m‍g(xj1,…,xjn),
projections gi(xj1,…,xjn)=xji, and weighted function p(xj1,…,xjn)=αj, we have the following discrete form of Jensen’s inequality for convex functions of several variables.Corollary 7.
Let𝒞⊆ℝn be a closed convex set. Let (xj1,…,xjn)∈𝒞 be points for j=1,…,m. Let αj∈ℝ be either nonnegative or nonpositive coefficients such that ∑j=1m‍αj≠0.
Then every continuous convex functionf:𝒞→ℝ satisfies the discrete inequality
(13)f(∑j=1mαjxj1∑j=1mαj,…,∑j=1mαjxjn∑j=1mαj)≤∑j=1mαjf(xj1,…,xjn)∑j=1mαj.Now we usem nonempty sets 𝒳j and associated function vector spaces 𝕏j containing the unit function.Theorem 8.
Let𝒞⊆ℝn be a closed convex set. Let gji:𝒳j→ℝ be functions such that (gj1(x),…,gjn(x))∈𝒞 for every x∈𝒳j for j=1,…,m. Let pj∈𝕏j be either nonnegative or nonpositive functions such that pjgji∈𝕏j. Let f:𝒞→ℝ be a continuous convex function such that pjf(gj1,…,gjn)∈𝕏j.
Then everym-tuple of positive linear functionals Mj:𝕏j→ℝ such that all Mj(pj)≠0 satisfies the inequality
(14)f(∑j=1mMj(pjgj1)∑j=1mMj(pj),…,∑j=1mMj(pjgjn)∑j=1mMj(pj))≤∑j=1mMj(pjf(gj1,…,gjn))∑j=1mMj(pj).Proof.
First we apply the inequality in (13) to the numbers xji=Mj(pjgji)/Mj(pj) and coefficients αj=Mj(pj) then the inequality in (9) to every Mj:
(15)f(∑j=1mMj(pjgj1)∑j=1mMj(pj),…,∑j=1mMj(pjgjn)∑j=1mMj(pj))=f(∑j=1mMj(pj)(Mj(pjgj1)/Mj(pj))∑j=1mMj(pj),…,∑j=1mMj(pj)(Mj(pjgjn)/Mj(pj))∑j=1mMj(pj))≤∑j=1mMj(pj)f(Mj(pjgj1)Mj(pj),…,Mj(pjgjn)Mj(pj))×(∑j=1mMj(pj))-1≤∑j=1mMj(pjf(gj1,…,gjn))∑j=1mMj(pj),
and so we get the inequality in (14).The inequality in (14) covers all the inequalities in this section.


## 3. The Generalizations of Jensen’s Inequality Using Functionals

This section presents the actual results of the paper. The main results are the generalizations of Jensen’s inequality written in Theorem12 for convex functions of one variable and Theorem 16 for convex functions of several variables.


### 3.1. Inequalities for Convex Functions with One Variable

In what follows, we use a bounded interval[a,b]⊂ℝ assuming a<b. Every x∈ℝ can be uniquely presented as the affine combination
(16)x=αxa+βxb,
where
(17)αx=|x1b1||a1b1|,βx=-|x1a1||a1b1|.Lemma 9.
Letc∈[a,b] be a point. Let α,β∈[0,1] and γ∈[-1,1] be coefficients of the sum α+β+γ=1.
Then the affine combinationαa+βb+γc belongs to [a,b].Proof.
Sincec∈[a,b], it has to be c=αca+βcb for coefficients αc and βc taken from the formulas in (17). Then we have
(18)αa+βb+γc=αa+βb+(1-α-β)(αca+βcb)=[α(1-αc)+(1-β)αc]a+[β(1-βc)+(1-α)βc]b.
The coefficients in the square brackets are nonnegative with the sum equal to 1, so the observed expression αa+βb+γc belongs to [a,b].Sufficient conditions on the coefficients in Lemma9 are α,β∈[0,1] and α+β+γ=1. Applying these conditions, it follows that γ∈[-1,1].Theorem 10.
Letcj∈[a,b] be points for j=1,…,m. Let α,β,γj∈[0,1] and γ∈[-1,1] be coefficients of the sums α+β+γ=∑j=1m‍γj=1.
Then the affine combinationαa+βb+γ∑j=1m‍γjcj∈[a,b].Proof.
Since∑j=1m‍γjcj∈[a,b], the affine combination αa+βb+γ∑j=1m‍γjcj∈[a,b] by Lemma 9.Corollary 11.
Letg:[a,b]→ℝ be a monotone function, and let xj∈[a,b] be points for j=1,…,m. Let α,β,γj∈[0,1] and γ∈[-1,1] be coefficients of the sums α+β+γ=∑j=1m‍γj=1.
Then the affine combinationαg(a)+βg(b)+γ∑j=1m‍γjg(xj)∈conv{g(a),g(b)}.The convex hull of a set𝒮 will be denoted by conv𝒮.Theorem 12.
Letg:[a,b]→ℝ be a monotone function, and let xj∈[a,b] be points for j=1,…,m. Let α,β,γj∈[0,1] and γ∈[-1,1] be coefficients of the sums α+β+γ=∑j=1m‍γj=1.
Then every continuous convex functionf:conv{g(a),g(b)}→ℝ satisfies the inequality
(19)f(αg(a)+βg(b)+γ∑j=1m‍γjg(xj))≤αf(g(a))+βf(g(b))+γ∑j=1m‍γjf(g(xj)).Proof.
Iff is monotone, then the Jensen inequality in (1) should be applied to the function space 𝕏 generated with g and f(g), that is,
(20)𝕏={h:[a,b]⟶ℝ∣h=c1g+c2f(g)+c3;c1,c2,c3∈ℝ},
and linear functional M on 𝕏 defined by
(21)M(h)=αh(a)+βh(b)+γ∑j=1m‍γjh(xj).
The functional M is unital since α+β+γ∑j=1m‍γj=1 and positive since the function h=(c1+c2f)(g)+c3 is monotone.
Iff is not monotone, then we translate f so that its minimum f(x0)=0 and express f=f1+f2 where
(22)f1(x)={0forx∈[c,x0]f(x)forx∈[x0,d],f2(x)={f(x)forx∈[c,x0]0forx∈[x0,d]
with [c,d]=conv{g(a),g(b)}. The previous procedure should be applied to the monotone convex functions f1 and f2.Putting the identity functiong(x)=x for x∈[a,b], the inequality in (19) is reduced to
(23)f(αa+βb+γ∑j=1m‍γjxj)≤αf(a)+βf(b)+γ∑j=1m‍γjf(xj).
If α=β=0 and γ=1, the above inequality represents the Jensen inequality. If α=β=1 and γ=-1, it represents the Jensen-Mercer inequality.


### 3.2. Inequalities for Convex Functions with Several Variables

We assume thatℝn is the real vector space treating its points as vectors with the standard coordinate addition (x1,…,xn)+(y1,…,yn)=(x1+y1,…,xn+yn) and scalar multiplication α(x1,…,xn)=(αx1,…,αxn). The aim of this subsection is to generalize Theorem 12 to simplexes.To begin let us show how Lemma9 can be generalized to the triangle (n=2). Take the three planar points A1(x1,y1), A2(x2,y2), and A3(x3,y3) that do not belong to one line. Then any point P(x,y)∈ℝ2 is presented by the unique affine combination
(24)P=αP1A1+αP2A2+αP3A3,
where
(25)αP1=|xy1x2y21x3y31||x1y11x2y21x3y31|,αP2=-|xy1x1y11x3y31||x1y11x2y21x3y31|,αP3=|xy1x1y11x2y21||x1y11x2y21x3y31|.So, we suppose thatconv{A1,A2,A3} is the triangle with vertices A1, A2, and A3.Lemma 13.
LetC∈conv{A1,A2,A3} be a point. Let α1,α2,α3∈[0,1] and γ∈[-1,1] be coefficients such that α1,α2,α3≥-γ and α1+α2+α3+γ=1.
Then the affine combinationα1A1+α2A2+α3A3+γC∈conv{A1,A2,A3}.Proof.
Let us show that the affine combinationα1A1+α2A2+α3A3+γC is actually the convex combination. Since the point C∈conv{A1,A2,A3}, the convex combination C=αC1A1+αC2A2+αC3A3 holds with the coefficients αC1, αC2, and αC3 taken from the formulas in (25). Therefore, we have
(26)α1A1+α2A2+α3A3+γC=[α1(1-αC1)+(α1+γ)αC1]A1+[α2(1-αC2)+(α2+γ)αC2]A2+[α3(1-αC3)+(α3+γ)αC3]A3.
The coefficients in the square brackets are nonnegative with the sum equal to 1, so the observed combination is convex, and its center A=α1A1+α2A2+α3A3+γC belongs to conv{A1,A2,A3}.IfA1,…,An+1∈ℝn are points such that the points A1-An+1,…,An-An+1 are linearly independent, then the convex hull conv{A1,…,An+1} is called the n-simplex with vertices A1,…,An+1. Geometrically speaking, all the simplex vertices cannot belong to the same hyperplane. If Ak=Ak(xk1,…,xkn), then any point P(x1,…,xn)∈ℝn is presented by the unique affine combination
(27)P=∑k=1n+1‍αPkAk,
where the numerators αPknum and denominators αPkden of the coefficients αPk are
(28)αPknum=(-1)k+1|x1…xn1x11…x1n1⋮⋱⋮⋮xk-11…xk-1n1xk+11…xk+1n1⋮⋱⋮⋮xn+11…xn+1n1|,αPkden=|x11…x1n1⋮⋱⋮⋮xk-11…xk-1n1xk1…xkn1xk+11…xk+1n1⋮⋱⋮⋮xn+11…xn+1n1|.In what follows we assume thatconv{A1,…,An+1}⊂ℝn is the n-simplex with vertices A1,…,An+1.The generalization of Lemma13 (in n dimensions) to several points Cj stands as follows.Theorem 14.
LetCj∈conv{A1,…,An+1} be points for j=1,…,m. Let αk,γj∈[0,1] and γ∈[-1,1] be coefficients such that all αk≥-γ and ∑k=1n+1‍αk+γ=∑j=1m‍γj=1.
Then the affine combination(29)∑k=1n+1‍αkAk+γ∑j=1m‍γjCj∈conv{A1,…,An+1}.Corollary 15.
Letg:conv{A1,…,An+1}→ℝn be a monotone mapping (either nondecreasing in each variable or nonincreasing in each variable), and let Pj∈conv{A1,…,An+1} be points for j=1,…,m. Let αk,γj∈[0,1] and γ∈[-1,1] be coefficients such that all αk≥-γ and ∑k=1n+1‍αk+γ=∑j=1m‍γj=1.
Then the affine combination(30)∑k=1n+1‍αkg(Ak)+γ∑j=1m‍γjg(Pj)∈conv{g(A1),…,g(An+1)}.Theorem 16.
Letg:conv{A1,…,An+1}→ℝn be a monotone mapping (either nondecreasing in each variable or non-increasing in each variable), and let Pj∈conv{A1,…,An+1} be points for j=1,…,m. Let αk,γj∈[0,1] and γ∈[-1,1] be coefficients such that all αk≥-γ and ∑k=1n+1‍αk+γ=∑j=1m‍γj=1.
Then every continuous convex functionf:conv{g(A1),…,g(An+1)}→ℝ satisfies the inequality
(31)f(∑k=1n+1‍αkg(Ak)+γ∑j=1m‍γjg(Pj))≤∑k=1n+1‍αkf(g(Ak))+γ∑j=1m‍γjf(g(Pj)).Proof.
Let us prove the casen=2.
Iff is monotone (either nondecreasing in both variables or nonincreasing in both variables), then we apply the McShane inequality in (3) to the function space
(32)𝕏={h:conv{A1,A2,A3}⟶ℝ∣h=c1g+c2f(g)+c3;c1,c2,c3∈ℝ}
and linear functional M on 𝕏 defined by
(33)M(h)=∑k=13‍αkh(Ak)+γ∑j=1m‍γjh(Pj).
The functional M is unital since ∑k=13‍αk+γ∑j=1m‍γj=1 and positive since the function h=(c1+c2f)(g)+c3 is monotone.
Iff is not monotone, then we translate f so that its minimum f(x0,y0)=0 and divide the triangle conv{g(A1),g(A2),g(A3)} into 22=4 parts considering the lines x=x0 and y=y0. Then we express f as the sum of the monotone convex functions, f=f1+f2+f3+f4, and apply the previous procedure to these functions.
The proof goes in the same way for any integern.Applying the identity mappingg(P)=P for P∈conv{A1,…,An+1} to the inequality in (31), it reads as follows:
(34)f(∑k=1n+1‍αkAk+γ∑j=1m‍γjPj)≤∑k=1n+1‍αkf(Ak)+γ∑j=1m‍γjf(Pj).


## 3.1. Inequalities for Convex Functions with One Variable

In what follows, we use a bounded interval[a,b]⊂ℝ assuming a<b. Every x∈ℝ can be uniquely presented as the affine combination
(16)x=αxa+βxb,
where
(17)αx=|x1b1||a1b1|,βx=-|x1a1||a1b1|.Lemma 9.
Letc∈[a,b] be a point. Let α,β∈[0,1] and γ∈[-1,1] be coefficients of the sum α+β+γ=1.
Then the affine combinationαa+βb+γc belongs to [a,b].Proof.
Sincec∈[a,b], it has to be c=αca+βcb for coefficients αc and βc taken from the formulas in (17). Then we have
(18)αa+βb+γc=αa+βb+(1-α-β)(αca+βcb)=[α(1-αc)+(1-β)αc]a+[β(1-βc)+(1-α)βc]b.
The coefficients in the square brackets are nonnegative with the sum equal to 1, so the observed expression αa+βb+γc belongs to [a,b].Sufficient conditions on the coefficients in Lemma9 are α,β∈[0,1] and α+β+γ=1. Applying these conditions, it follows that γ∈[-1,1].Theorem 10.
Letcj∈[a,b] be points for j=1,…,m. Let α,β,γj∈[0,1] and γ∈[-1,1] be coefficients of the sums α+β+γ=∑j=1m‍γj=1.
Then the affine combinationαa+βb+γ∑j=1m‍γjcj∈[a,b].Proof.
Since∑j=1m‍γjcj∈[a,b], the affine combination αa+βb+γ∑j=1m‍γjcj∈[a,b] by Lemma 9.Corollary 11.
Letg:[a,b]→ℝ be a monotone function, and let xj∈[a,b] be points for j=1,…,m. Let α,β,γj∈[0,1] and γ∈[-1,1] be coefficients of the sums α+β+γ=∑j=1m‍γj=1.
Then the affine combinationαg(a)+βg(b)+γ∑j=1m‍γjg(xj)∈conv{g(a),g(b)}.The convex hull of a set𝒮 will be denoted by conv𝒮.Theorem 12.
Letg:[a,b]→ℝ be a monotone function, and let xj∈[a,b] be points for j=1,…,m. Let α,β,γj∈[0,1] and γ∈[-1,1] be coefficients of the sums α+β+γ=∑j=1m‍γj=1.
Then every continuous convex functionf:conv{g(a),g(b)}→ℝ satisfies the inequality
(19)f(αg(a)+βg(b)+γ∑j=1m‍γjg(xj))≤αf(g(a))+βf(g(b))+γ∑j=1m‍γjf(g(xj)).Proof.
Iff is monotone, then the Jensen inequality in (1) should be applied to the function space 𝕏 generated with g and f(g), that is,
(20)𝕏={h:[a,b]⟶ℝ∣h=c1g+c2f(g)+c3;c1,c2,c3∈ℝ},
and linear functional M on 𝕏 defined by
(21)M(h)=αh(a)+βh(b)+γ∑j=1m‍γjh(xj).
The functional M is unital since α+β+γ∑j=1m‍γj=1 and positive since the function h=(c1+c2f)(g)+c3 is monotone.
Iff is not monotone, then we translate f so that its minimum f(x0)=0 and express f=f1+f2 where
(22)f1(x)={0forx∈[c,x0]f(x)forx∈[x0,d],f2(x)={f(x)forx∈[c,x0]0forx∈[x0,d]
with [c,d]=conv{g(a),g(b)}. The previous procedure should be applied to the monotone convex functions f1 and f2.Putting the identity functiong(x)=x for x∈[a,b], the inequality in (19) is reduced to
(23)f(αa+βb+γ∑j=1m‍γjxj)≤αf(a)+βf(b)+γ∑j=1m‍γjf(xj).
If α=β=0 and γ=1, the above inequality represents the Jensen inequality. If α=β=1 and γ=-1, it represents the Jensen-Mercer inequality.


## 3.2. Inequalities for Convex Functions with Several Variables

We assume thatℝn is the real vector space treating its points as vectors with the standard coordinate addition (x1,…,xn)+(y1,…,yn)=(x1+y1,…,xn+yn) and scalar multiplication α(x1,…,xn)=(αx1,…,αxn). The aim of this subsection is to generalize Theorem 12 to simplexes.To begin let us show how Lemma9 can be generalized to the triangle (n=2). Take the three planar points A1(x1,y1), A2(x2,y2), and A3(x3,y3) that do not belong to one line. Then any point P(x,y)∈ℝ2 is presented by the unique affine combination
(24)P=αP1A1+αP2A2+αP3A3,
where
(25)αP1=|xy1x2y21x3y31||x1y11x2y21x3y31|,αP2=-|xy1x1y11x3y31||x1y11x2y21x3y31|,αP3=|xy1x1y11x2y21||x1y11x2y21x3y31|.So, we suppose thatconv{A1,A2,A3} is the triangle with vertices A1, A2, and A3.Lemma 13.
LetC∈conv{A1,A2,A3} be a point. Let α1,α2,α3∈[0,1] and γ∈[-1,1] be coefficients such that α1,α2,α3≥-γ and α1+α2+α3+γ=1.
Then the affine combinationα1A1+α2A2+α3A3+γC∈conv{A1,A2,A3}.Proof.
Let us show that the affine combinationα1A1+α2A2+α3A3+γC is actually the convex combination. Since the point C∈conv{A1,A2,A3}, the convex combination C=αC1A1+αC2A2+αC3A3 holds with the coefficients αC1, αC2, and αC3 taken from the formulas in (25). Therefore, we have
(26)α1A1+α2A2+α3A3+γC=[α1(1-αC1)+(α1+γ)αC1]A1+[α2(1-αC2)+(α2+γ)αC2]A2+[α3(1-αC3)+(α3+γ)αC3]A3.
The coefficients in the square brackets are nonnegative with the sum equal to 1, so the observed combination is convex, and its center A=α1A1+α2A2+α3A3+γC belongs to conv{A1,A2,A3}.IfA1,…,An+1∈ℝn are points such that the points A1-An+1,…,An-An+1 are linearly independent, then the convex hull conv{A1,…,An+1} is called the n-simplex with vertices A1,…,An+1. Geometrically speaking, all the simplex vertices cannot belong to the same hyperplane. If Ak=Ak(xk1,…,xkn), then any point P(x1,…,xn)∈ℝn is presented by the unique affine combination
(27)P=∑k=1n+1‍αPkAk,
where the numerators αPknum and denominators αPkden of the coefficients αPk are
(28)αPknum=(-1)k+1|x1…xn1x11…x1n1⋮⋱⋮⋮xk-11…xk-1n1xk+11…xk+1n1⋮⋱⋮⋮xn+11…xn+1n1|,αPkden=|x11…x1n1⋮⋱⋮⋮xk-11…xk-1n1xk1…xkn1xk+11…xk+1n1⋮⋱⋮⋮xn+11…xn+1n1|.In what follows we assume thatconv{A1,…,An+1}⊂ℝn is the n-simplex with vertices A1,…,An+1.The generalization of Lemma13 (in n dimensions) to several points Cj stands as follows.Theorem 14.
LetCj∈conv{A1,…,An+1} be points for j=1,…,m. Let αk,γj∈[0,1] and γ∈[-1,1] be coefficients such that all αk≥-γ and ∑k=1n+1‍αk+γ=∑j=1m‍γj=1.
Then the affine combination(29)∑k=1n+1‍αkAk+γ∑j=1m‍γjCj∈conv{A1,…,An+1}.Corollary 15.
Letg:conv{A1,…,An+1}→ℝn be a monotone mapping (either nondecreasing in each variable or nonincreasing in each variable), and let Pj∈conv{A1,…,An+1} be points for j=1,…,m. Let αk,γj∈[0,1] and γ∈[-1,1] be coefficients such that all αk≥-γ and ∑k=1n+1‍αk+γ=∑j=1m‍γj=1.
Then the affine combination(30)∑k=1n+1‍αkg(Ak)+γ∑j=1m‍γjg(Pj)∈conv{g(A1),…,g(An+1)}.Theorem 16.
Letg:conv{A1,…,An+1}→ℝn be a monotone mapping (either nondecreasing in each variable or non-increasing in each variable), and let Pj∈conv{A1,…,An+1} be points for j=1,…,m. Let αk,γj∈[0,1] and γ∈[-1,1] be coefficients such that all αk≥-γ and ∑k=1n+1‍αk+γ=∑j=1m‍γj=1.
Then every continuous convex functionf:conv{g(A1),…,g(An+1)}→ℝ satisfies the inequality
(31)f(∑k=1n+1‍αkg(Ak)+γ∑j=1m‍γjg(Pj))≤∑k=1n+1‍αkf(g(Ak))+γ∑j=1m‍γjf(g(Pj)).Proof.
Let us prove the casen=2.
Iff is monotone (either nondecreasing in both variables or nonincreasing in both variables), then we apply the McShane inequality in (3) to the function space
(32)𝕏={h:conv{A1,A2,A3}⟶ℝ∣h=c1g+c2f(g)+c3;c1,c2,c3∈ℝ}
and linear functional M on 𝕏 defined by
(33)M(h)=∑k=13‍αkh(Ak)+γ∑j=1m‍γjh(Pj).
The functional M is unital since ∑k=13‍αk+γ∑j=1m‍γj=1 and positive since the function h=(c1+c2f)(g)+c3 is monotone.
Iff is not monotone, then we translate f so that its minimum f(x0,y0)=0 and divide the triangle conv{g(A1),g(A2),g(A3)} into 22=4 parts considering the lines x=x0 and y=y0. Then we express f as the sum of the monotone convex functions, f=f1+f2+f3+f4, and apply the previous procedure to these functions.
The proof goes in the same way for any integern.Applying the identity mappingg(P)=P for P∈conv{A1,…,An+1} to the inequality in (31), it reads as follows:
(34)f(∑k=1n+1‍αkAk+γ∑j=1m‍γjPj)≤∑k=1n+1‍αkf(Ak)+γ∑j=1m‍γjf(Pj).



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*Source: 194830-2013-09-26.xml*