# Convex Combination Inequalities of the Line and Plane

Authors: Zlatko Pavić; Shanhe Wu; Štefanija Klarić
Journal: Abstract and Applied Analysis
 (2014)

---

## Abstract

The paper deals with convex combinations, convex functions, and
Jensen’s functionals. The main idea of this work is to present the given convex
combination by using two other convex combinations with minimal number of
points. For example, as regards the presentation of the planar combination, we
use two trinomial combinations. Generalizations to higher dimensions are also
considered.


---

## Body

## 1. Introduction

LetX be a real vector space. A set A⊆X is affine if it contains the lines passing through all pairs of its points (all binomial affine combinations in A, i.e., the combinations p1P1+p2P2 of points P1,P2∈A and coefficients p1,p2∈R of the sum p1+p2=1). A function f:A→R is affine if it satisfies the equality f(p1P1+p2P2)=p1f(P1)+p2f(P2) for all binomial affine combinations in A.A setC⊆X is convex if it contains the line segments connecting all pairs of its points (all binomial convex combinations in C, i.e., the combinations p1P1+p2P2 of points P1,P2∈C and nonnegative coefficients p1,p2∈R of the sum p1+p2=1). A function f:C→R is convex if it satisfies the inequality f(p1P1+p2P2)≤p1f(P1)+p2f(P2) for all binomial convex combinations in C.Using the mathematical induction, it can be proved that every affine functionf:A→R satisfies the equality
(1)f∑i=1n‍piPi=∑i=1n‍pif(Pi)
for all affine combinations in A and that every convex function f:C→R satisfies the Jensen inequality
(2)f∑i=1n‍piPi≤∑i=1n‍pif(Pi)
for all convex combinations in C.For an affine or a convex combinationP=∑i=1n‍piPi the point P itself is called the combination center, and it is important to mathematical inequalities. Recognizing the importance of the combination center, the authors (see [1]) have recently considered inequalities on simplexes and their cones.A general overview of convex sets, convex functions, and its applications can be found in [2]. In working with means and their inequalities, we can rely on the book in [3]. Many details of the branch of mathematical inequalities are written in [4].


## 2. Convex Combinations of the Line

The section shows the importance of convex combination centers in deriving inequalities. The main result is Theorem2.Ifa,b∈R are different numbers, say a<b, then every number x∈R can be uniquely presented as the affine combination
(3)x=b-xb-aa+x-ab-ab.
The above binomial combination is convex if and only if the number x belongs to the interval [a,b]. Given the function f:R→R, let f{a,b}line:R→R be the function of the line passing through the points (a,f(a)) and (b,f(b)) of the graph of f. Using the affinity of f{a,b}line, we get the equation
(4)f{a,b}line(x)=b-xb-af(a)+x-ab-af(b).
If the function f is convex, then, using the definition of convexity, we obtain the inequality
(5)fx≤fa,blinex,ifx∈[a,b],
and the reverse inequality
(6)fx≥fa,blinex,ifx∉(a,b).By the end of this section we will use an intervalI⊆R with the nonempty interior I0.The following lemma represents a systematised version of [5, Proposition 2] and deals with two convex combinations having the same center. Assigning the convex function to such convex combinations, we obtain the following Jensen type inequality.Lemma 1.
Leta,b∈I⊆R be points such that a≤b. Let ∑i=1n‍pixi be a convex combination with points xi∈[a,b], and let ∑j=1m‍qjyj be a convex combination with points yj∈I∖(a,b).
If the above convex combinations have the same center(7)∑i=1n‍pixi=∑j=1m‍qjyj,
then every convex function f:I→R satisfies the inequality
(8)∑i=1n‍pif(xi)≤∑j=1m‍qjf(yj).
Iff is concave, then the reverse inequality is valid in (8).Proof.
Assumef is convex. If a<b, the right-hand side follows from the series of inequalities
(9)∑i=1n‍pifxi≤∑i=1n‍pif{a,b}line(xi)=f{a,b}line∑i=1n‍pixi=f{a,b}line∑j=1m‍qjyj=∑j=1m‍qjf{a,b}line(yj)≤∑j=1m‍qjf(yj)
derived applying the inequality in (5) to xi and the inequality in (6) to yj. If a=b, we use any support line f{a}line instead of the chord line f{a,b}line.Lemma1 is the generalization of Jensen’s inequality: applying this lemma to the convex combination center equality
(10)1c=∑i=1n‍pixi,
with the assumption a=b=c, we come to the Jensen inequality
(11)f∑i=1n‍pixi=1f(c)≤∑i=1n‍pif(xi).So, the discrete form of the famous Jensen inequality (discrete form in [6] and integral form in [7]) can be derived applying the convexity definition and the affinity of the chord or support line. The different forms of Jensen’s inequality can be seen in [8].Theorem 2.
Leta,b,a1,b1∈I⊆R be points such that a1<a<b<b1. Let c=∑i=1n‍pixi be a convex combination of the points xi∈[a1,b1]∖(a,b) with the center c∈(a,b).
Then there exist two binomial convex combinationsαa+βb and α1a1+β1b1 so that
(12)αa+βb=∑i=1n‍pixi=α1a1+β1b1,
and consequently, every convex function f:I→R satisfies the inequality
(13)αf(a)+βf(b)≤∑i=1n‍pif(xi)≤α1f(a1)+β1f(b1).
Iff is concave, then the reverse inequality is valid in (13).Proof.
We use the formula in (3) to calculate the coefficients α=(b-c)/(b-a) and β=(c-a)(b-a) that satisfy c=αa+βb and also for α1 and β1. Now we need to apply Lemma 1 to both sides of the obtained equality in (12).The graphical representation of the equality in (12) and the inequality in (13) is shown in Figure 1.Figure 1
Line convex combinations with the same centerc.Binomial convex combinations are included into the definition of convexity. The following corollary demonstrates how the binomial combinations may be assigned to each convex combination.Corollary 3.
Let∑i=1n‍pixi be a convex combination in I⊆R. Then there exist two binomial convex combinations p0x0+q0y0 and px+qy with points x0,y0,x,y from the set {x1,…,xn} so that
(14)p0x0+q0y0=∑i=1n‍pixi=px+qy,
and consequently, every convex function f:I→R satisfies the inequality
(15)p0f(x0)+q0f(y0)≤∑i=1n‍pif(xi)≤pf(x)+qf(y).Proof.
Putc=∑i=1n‍pixi and suppose x1≤⋯≤xn. Take x=x1 and y=xn. If x=y, we take p=q=1/2. If x<y, we calculate the coefficients p and q by the formula in (3) to get c=px+qy.
Ifc is equal to some xi0, then we take x0=y0=xi0 and p0=q0=1/2. Otherwise, it must be c∈(xi,xi+1) for some pair, in which case we take x0=xi and y0=xi+1. Calculating p0 and q0 by (3), we also get c=p0x0+q0y0.
It remains to apply Lemma1 to both sides of the obtained equality in (14).Respecting the Jensen inequality, the formula in (15) can be expressed in the extended form:
(16)f∑i=1n‍pixi≤p0f(x0)+q0f(y0)≤∑i=1n‍pif(xi)≤pf(x)+qf(y).Let us show how the intermediate member in (13) can be transformed into integral. Let L⊆I be a bounded set with the length |L|>0, and let f be an integrable (in the sense of Riemann or Lebesgue) function on L. Given the positive integer n, we employ a partition
(17)L=⋃i=1n‍Lni,
where each of pairwise disjoint subsets Lni contracts to the point as n approaches infinity. Take one point xni∈Lni for every i=1,…,n and then compose the convex combination cn of the points f(xni) with the coefficients pni=|Lni|/|L|; that is,
(18)cn=1L∑i=1n‍Lnif(xni).
Letting n to infinity, the sequence (cn)n approaches the point
(19)c=1|L|∫L‍f(x)dx.Using the integral method with convex combinations, we obtain the mixed discrete-integral form of Theorem2 as follows.Corollary 4.
Leta,b,a1,b1∈I⊆R be points such that a1<a<b<b1. Let the barycenter c of the set [a1,b1]∖(a,b) belongs to (a,b); that is,
(20)c=∫[a1,b1]∖(a,b)‍xdxb1-a1-b+a∈(a,b).
Then there exist two binomial convex combinationsαa+βb and α1a1+β1b1 so that
(21)αa+βb=c=α1a1+β1b1,
and consequently, every convex function f:I→R satisfies the inequality
(22)αfa+βfb≤∫[a1,b1]∖(a,b)‍f(x)dxb1-a1-b+a≤α1f(a1)+β1f(b1).
Iff is concave, then the reverse inequality is valid in (22).The summarizing Jensen’s functional of a functionf:I→R for the given convex combination ∑i=1n‍pixi in I is defined with
(23)Jp1x1+⋯+pnxn(f)=∑i=1n‍pif(xi)-f∑i=1n‍pixi.
If the conditions of Theorem 2 hold, we get the functional inequality
(24)Jαa+βb(f)≤Jp1x1+⋯+pnxn(f)≤Jα1a1+β1b1(f).The integrating Jensen’s functional of an integrable functionf:I→R for the given subset L⊆I with the length |L|>0 is defined with
(25)JLf=1L∫L‍fxdx-f1L∫L‍xdx.
If the conditions of Corollary 4 are satisfied, we get
(26)Jαa+βb(f)≤J[a1,b1]∖(a,b)(f)≤Jα1a1+β1b1(f).The inequalities in (24) and (26) offer the local bounds of Jensen’s functionals. The global bounds of Jensen’s summarizing functional were investigated in [9]. Some new Jensen type inequalities were obtained in [10].


## 3. Convex Combinations of the Plane

This section contains the main results, Theorem6, and its consequences.We assume thatR2 is the real vector space treating its points as the vectors with the standard coordinate addition (x1,y1)+(x2,y2)=(x1+x2,y1+y2) and the scalar multiplication α(x,y)=(αx,αy).IfA(xA,yA), B(xB,yB), and C(xC,yC) are the planar points that do not belong to one line, respectively, the convex hull conv{A,B,C} is a real triangle, and then every point P(x,y)∈R2 can be presented by the unique affine combination
(27)P=αA+βB+γC,
where
(28)α=xy1xByB1xCyC1xAyA1xByB1xCyC1,β=-xy1xAyA1xCyC1xAyA1xByB1xCyC1,γ=xy1xAyA1xByB1xAyA1xByB1xCyC1.
The above trinomial combination is convex if and only if the point P belongs to the triangle conv{A,B,C}.Given the triangle with verticesA, B, and C, the convex cone CA with the vertex at A is the set spanned by the vectors A-B and A-C (similarly CB and CC; all three cones can be viewed in Figure 2); that is,
(29)CA=A+pA-B+qA-C:p,q∈R,p,q≥0.Figure 2
Plane convex combinations with the same centerP.Given the functionf:R2→R, let f{A,B,C}plane:R2→R be the function of the plane passing through the points (A,f(A)), (B,f(B)), and (C,f(C)) of the graph of f. Due to the affinity of f{A,B,C}plane, it follows
(30)f{A,B,C}plane(P)=αf(A)+βf(B)+γf(C).
For a convex function f, using the convexity definition, we get the inequality
(31)fP≤fA,B,CplaneP,ifP∈conv{A,B,C},
and using the affinity of f{A,B,C}plane, the reverse inequality
(32)fP≥fA,B,CplaneP,ifP∈CA∪CB∪CC.By the end of the section we will use a planar convex setC⊆R2 with the nonempty interior C0. The area of a planar set A will be denoted with ar(A).Lemma 5.
LetA,B,C∈C⊆R2 be points, Δ=conv{A,B,C}, and CΔ=CA∪CB∪CC be the cone union. Let ∑i=1n‍αiAi be a convex combination of the points Ai∈Δ, and let ∑j=1m‍βjBj be a convex combination of the points Bj∈CΔ∩C.
If the above convex combinations have the same center(33)∑i=1n‍αiAi=∑j=1m‍βjBj,
then every convex function f:C→R satisfies the inequality
(34)∑i=1n‍αif(Ai)≤∑j=1m‍βjf(Bj).Proof.
If the setconv{A,B,C} is a real triangle, we can apply the proof of Lemma 1 using f{A,B,C}plane instead of f{a,b}line respecting the plane inequalities in (31)-(32). If conv{A,B,C}=conv{A,B}, then we rely on the proof of Lemma 1 with the chord line f{A,B}line. If conv{A,B,C}={A}, we use any support line f{A}line at the point A.Theorem 6.
LetA,B,C;A1,B1,C1∈C⊆R2 be points such that
(35)Δ=conv{A,B,C}⊂conv{A1,B1,C1}=Δ1
with Δ0≠∅, and let CΔ=CA∪CB∪CC be the cone union. Let P=∑i=1n‍piPi be a convex combination of the points Pi∈CΔ∩Δ1 with the center P∈Δ0.
Then there exist two trinomial convex combinationsαA+βB+γC and α1A1+β1B1+γ1C1 so that
(36)αA+βB+γC=∑i=1n‍piPi=α1A1+β1B1+γ1C1,
and consequently, every convex function f:C→R satisfies the inequality
(37)αfA+βfB+γfC≤∑i=1n‍pif(Pi)≤α1f(A1)+β1f(B1)+γ1f(C1).Proof.
First we calculate the coefficients by the formula in (28) to get the equality in (36) and then apply Lemma 5 to its both sides.The graphical representation of the equality in (36) can be seen in Figure 2.Applying the integral method with convex combinations, we obtain the form as follows.Corollary 7.
LetA,B,C;A1,B1,C1∈C⊆R2 be points such that
(38)Δ=conv{A,B,C}⊂conv{A1,B1,C1}=Δ1
with Δ0≠∅, and let CΔ=CA∪CB∪CC be the cone union. Let the barycenter P of the set CΔ∩Δ1 belong to Δ0; that is,
(39)P∫CΔ∩Δ1‍xdxdyar(CΔ∩Δ1),∫CΔ∩Δ1‍ydxdyar(CΔ∩Δ1)∈Δ0.
Then there exist two trinomial convex combinationsαA+βB+γC and α1A1+β1B1+γ1C1 so that
(40)αA+βB+γC=P=α1A1+β1B1+γ1C1,
and consequently, every convex function f:C→R satisfies the inequality
(41)αf(A)+βf(B)+γf(C)≤∫CΔ∩Δ1‍f(x,y)dxdyar(CΔ∩Δ1)≤α1f(A1)+β1f(B1)+γ1f(C1).
Iff is concave, then the reverse inequality is valid in (41).If the conditions of Theorem6 are valid, then using the summarizing Jensen functional of f for the given convex combination ∑i=1n‍piPi,
(42)Jp1P1+⋯+pnPn(f)=∑i=1n‍pif(Pi)-f∑i=1n‍piPi,
we get the functional inequality
(43)JαA+βB+γCf≤Jp1P1+⋯+pnPn(f)≤Jα1A1+β1B1+γ1C1(f).
If the conditions of Corollary 7 are satisfied, then applying another integrating Jensen functional of f for the given subset A⊆C with the area ar(A)>0,
(44)JAf=∫A‍f(x,y)dxdyar(A)-f∫A‍xdxdyar(A),∫A‍ydxdyar(A),
we have the inequality
(45)JαA+βB+γC(f)≤JCΔ∩Δ1(f)≤Jα1A1+β1B1+γ1C1(f).


## 4. Generalization

The aim of the section is to generalize Theorem6 and Corollary 7 to more dimensions.IfA1,…,Am+1∈Rm are points such that the vectors A1-A2,…,A1-Am+1 are linearly independent, then the convex hull
(46)Δ=conv{A1,…,Am+1}
is called the m-simplex with the vertices A1,…,Am+1. All the simplex vertices can not belong to the same hyperplane in Rm. Any point P∈Rm can be presented by the unique affine combination
(47)P=∑j=1m+1‍αjAj.
If we use the point coordinates, then the coefficients αj can be determined by the generalized coefficient formula in (28). The combination in (47) is convex if and only if the point P belongs to the m-simplex Δ.Given them-simplex with vertices A1,…,Am+1, let CA1 be the convex cone with the vertex at A1 spanned by the vectors A1-A2,…,A1-Am+1 (similarly CA2,…,CAm+1); that is,
(48)CA1=A1+∑k=2m+1‍pk(A1-Ak):pk∈R,pk≥0.Given the functionf:Rm→R, let fΔhp:Rm→R be the function of the hyperplane (in Rm+1) passing through the points (Aj,f(Aj)) of the graph of f. Applying the affinity of fΔhp to the combination in (47), it follows
(49)fΔhp(P)=∑j=1m+1‍αjf(Aj).
If we use the convex function f, then we get the inequality
(50)fP≤fΔhpP,ifP∈Δ,
and the reverse inequality
(51)fP≥fΔhpP,ifP∈⋃j=1m+1CAj
which can be proved in the same way as the inequality in (32).By the end of the section we will use a convex setC⊆Rm with the nonempty interior C0. The volume of a set V⊂Rm will be denoted with vol(V).The generalization of Theorem6 applied to m-simplexes is as follows.Theorem 8.
LetA1,…,Am+1; B1,…,Bm+1∈C⊆Rm be points such that
(52)Δ=conv{A1,…,Am+1}⊂conv{B1,…,Bm+1}=Δ1
with Δ0≠∅, and let CΔ=∪j=1m+1CAj be the cone union. Let P=∑i=1n‍piPi be a convex combination of the points Pi∈CΔ∩Δ1 with the center P∈Δ0.
Then there exist two(m+1)-membered convex combinations ∑j=1m+1‍αjAj and ∑j=1m+1‍βjBj so that
(53)∑j=1m+1‍αjAj=∑i=1n‍piPi=∑j=1m+1‍βjBj,
and consequently every convex function f:C→R satisfies the inequality
(54)∑j=1m+1‍αjf(Aj)≤∑i=1n‍pif(Pi)≤∑j=1m+1‍βjf(Bj).Using the integrals, we get the form as follows.Corollary 9.
LetA1,…,Am+1;B1,…,Bm+1∈C⊆Rm be points such that
(55)Δ=conv{A1,…,Am+1}⊂conv{B1,…,Bm+1}=Δ1
with Δ0≠∅, and let CΔ=∪j=1m+1CAj be the cone union. Let the barycenter P of the set CΔ∩Δ1 belong to Δ0; that is,
(56)P∫CΔ∩Δ1‍x1dx1,…,dxmvol(CΔ∩Δ1),…,∫CΔ∩Δ1‍xmdx1,…,dxmvol(CΔ∩Δ1)∈Δ0.
Then there exist two(m+1)-membered convex combinations ∑j=1m+1‍αjf(Aj) and ∑j=1m+1‍βjf(Bj) so that
(57)∑j=1m+1‍αjAj=P=∑j=1m+1‍βjBj,
and consequently every convex function f:C→R satisfies the inequality
(58)∑j=1m+1‍αjfAj≤∫CΔ∩Δ1‍f(x1,…,xm)dx1,…,dxmvol(CΔ∩Δ1)≤∑j=1m+1‍βjf(Bj).
Iff is concave, then the reverse inequality is valid in (58).If the conditions of Theorem8 are valid, then using the summarizing Jensen functional of f for the given convex combination ∑i=1n‍piPi, we get the functional inequality
(59)Jα1A1+⋯+αm+1Am+1f≤Jp1P1+⋯+pnPn(f)≤Jβ1B1+⋯+βm+1Bm+1(f).
If the conditions of Corollary 9 are satisfied, then applying another integrating Jensen functional of f for the given subset V⊆C with the volume vol(V)>0,
(60)JVf=∫V‍fdx1,…,dxmvol(V)-f∫V‍x1dx1,…,dxmvol(V),…,∫V‍xmdx1,…,dxmvol(V),
we have the inequality
(61)Jα1A1+⋯+αm+1Am+1f≤JCΔ∩Δ1(f)≤Jβ1B1+⋯+βm+1Bm+1(f).



---

*Source: 916243-2014-11-16.xml*