a "formal version" (PDF)

Generalizations of Popoviciu’s inequality

Darij Grinberg

Formal version; 4 March 2009

This is the ”formal” version of my noteGeneralizations of Popoviciu’s inequality. It contains the proofs with more details, but is much more burdensome to read because of this. I advise you to use this formal version only if you have troubles with understanding the standard version.

UPDATE: A glance into the survey [10], Chapter XVIII has revealed that most theorems in this paper are far from new. For instance, Theorem 5b was proven under weaker conditions (!) by Vasi´c and Stankovi´c in [11]. Unfortunately, I have no access to [11] and the other references related to these inequalities.

Notation

First, we introduce a notation that we will use in the following paper: For any set S of numbers, we denote by maxS the greatest element of the set S,and by minS the smallest element ofS.

1. Introduction

In the last few years there was some activity on the MathLinks forum related to the Popoviciu inequality on convex functions. A number of generalizations were conjectured and subsequently proven using majorization theory and (mostly) a lot of computations. In this note I am presenting a probably new approach that proves these generalizations as well as some additional facts with a lesser amount of computation and avoiding the combinatorial difficulties of majorization theory (we will prove a version of the Karamata inequality on the way, but no prior knowledge of majorization theory is required - what we actually avoid is the asymmetric definition of majorization).

The very starting point of the whole theory is the following famous fact:

Theorem 1a, the Jensen inequality. Let f be a convex function from an interval I ⊆ R to R. Let x1, x2, ..., xn be finitely many points from I.

Then,

f(x₁) +f(x₂) +...+f(x_n)

n ≥f

x₁+x₂+...+x_n n

.

In words, the arithmetic mean of the values of f at the points x1, x2, ..., xn

is greater or equal to the value off at the arithmetic mean of these points.

We can obtain a ”weighted version” of this inequality by replacing arithmetic means by weighted means with some nonnegative weightsw1, w2, ..., wn:

Theorem 1b, the weighted Jensen inequality. Let f be a convex function from an interval I ⊆ R to R. Let x₁, x₂, ..., x_n be finitely many points from I. Let w₁, w₂, ..., w_n be n nonnegative reals which are not all equal to 0. Then,

w₁f(x₁) +w₂f(x₂) +...+w_nf(x_n) w₁+w₂+...+w_n ≥f

w₁x₁+w₂x₂+...+w_nx_n w₁+w₂+...+w_n

.

Obviously, Theorem 1a follows from Theorem 1b applied tow1 =w2 =...=wn = 1, so that Theorem 1b is more general than Theorem 1a.

We won’t stop at discussing equality cases here, since they can depend in various ways on the input (i. e., on the function f, the reals w1, w2, ..., wn and the points x1, x₂, ..., x_n) - but each time we use a result like Theorem 1b, with enough patience we can extract the equality case from the proof of this result and the properties of the input.

The Jensen inequality, in both of its versions above, is applied often enough to be called one of the main methods of proving inequalities. Now, in 1965, a similarly styled inequality was found by the Romanian Tiberiu Popoviciu:

Theorem 2a, the Popoviciu inequality. Let f be a convex function from an interval I ⊆ R to R, and let x1, x2, x3 be three points from I.

Then,

f(x₁)+f(x₂)+f(x₃)+3f

x₁+x₂+x₃ 3

≥2f

x₂+x₃ 2

+2f

x₃+x₁ 2

+2f

x₁+x₂ 2

. Again, a weighted version can be constructed:

Theorem 2b, the weighted Popoviciu inequality. Let f be a convex function from an interval I ⊆ R to R, let x₁, x₂, x₃ be three points from I, and let w1, w2, w3 be three nonnegative reals such that w2 +w3 6= 0, w₃ +w₁ 6= 0 andw₁+w₂ 6= 0.Then,

w₁f(x₁) +w₂f(x₂) +w₃f(x₃) + (w₁+w₂+w₃)f

w₁x₁+w₂x₂+w₃x₃ w₁+w₂ +w₃

≥(w₂+w₃)f

w₂x₂+w₃x₃ w₂+w₃

+ (w₃+w₁)f

w₃x₃+w₁x₁ w₃+w₁

+ (w₁+w₂)f

w₁x₁+w₂x₂ w₁+w₂

. Now, the really interesting part of the story began when Vasile Cˆırtoaje - alias

”Vasc” on the MathLinks forum - proposed the following two generalizations of Theo- rem 2a ([1] and [2] for Theorem 3a, and [1] and [3] for Theorem 4a):

Theorem 3a (Vasile Cˆırtoaje). Let f be a convex function from an interval I ⊆ R to R. Let x₁, x₂, ..., x_n be finitely many points from I.

Then,

n

X

i=1

f(x_i)+n(n−2)f

x₁+x₂+...+x_n n

≥

n

X

j=1

(n−1)f





 P

1≤i≤n; i6=j

x_i n−1





.

Theorem 4a (Vasile Cˆırtoaje). Let f be a convex function from an interval I ⊆ R to R. Let x1, x2, ..., xn be finitely many points from I.

Then, (n−2)

n

X

i=1

f(x_i) +nf

x₁+x₂+...+x_n n

≥ X

1≤i<j≤n

2f

x_i+x_j 2

.

In [1], both of these facts were nicely proven by Cˆırtoaje. I gave a different and rather long proof of Theorem 3a in [2]. All of these proofs use the Karamata inequality.

Of course, Theorem 2a follows from each of the Theorems 3a and 4a upon settingn = 3.

It is pretty straightforward to obtain generalizations of Theorems 3a and 4a by putting in weights as in Theorems 1b and 2b. A more substantial generalization was given by Yufei Zhao - alias ”Billzhao” on MathLinks - in [3]:

Theorem 5a (Yufei Zhao). Let f be a convex function from an interval I ⊆RtoR. Letx₁, x₂, ..., x_n be finitely many points fromI,and let m be an integer. Then,

n−2 m−1

n

X

i=1

f(x_i) +

n−2 m−2

nf

x₁+x₂+...+x_n n

≥ X

1≤i1<i2<...<im≤n

mf

x_i1 +x_i2 +...+x_im m

.

Note that if m≤0 orm > n, the sum P

1≤i1<i2<...<im≤n

mf

x_i1 +x_i2 +...+x_im m

is empty, so that its value is 0.

It is left to the reader to verify that Theorems 3a and 4a both are particular cases of Theorem 5a (in fact, set m =n−1 to get Theorem 3a and m = 2 to get Theorem 4a).

An elaborate proof of Theorem 5a was given by myself in [3]. After some time, the MathLinks user ”Zhaobin” proposed a weighted version of this result:

Theorem 5b (Zhaobin). Let f be a convex function from an interval I ⊆ R to R. Let x₁, x₂, ..., x_n be finitely many points from I, let w₁, w₂, ..., w_n be nonnegative reals, and let m be an integer. Assume that w1+w2+...+wn 6= 0,and thatwi1+wi2+...+wim 6= 0 for anym integers i₁, i₂, ..., i_m satisfying 1≤i₁ < i₂ < ... < i_m ≤n.

Then, n−2

m−1 ⁿ

X

i=1

w_if(x_i) +

n−2 m−2

(w₁+w₂+...+w_n)f

w1x1+w2x2+...+wnxn

w₁+w₂+...+w_n

≥ X

1≤i1<i2<...<im≤n

(w_i1+w_i2 +...+w_im)f

w_i1x_i1 +w_i2x_i2 +...+w_imx_im wi1 +wi2 +...+wim

.

If we set w₁ = w₂ =... =w_n = 1 in Theorem 5b, we obtain Theorem 5a. On the other hand, putting n = 3 andm= 2 in Theorem 5b, we get Theorem 2b.

In this note, I am going to prove Theorem 5b (and therefore also its particular cases - Theorems 2a, 2b, 3a, 4a and 5a). The proof is going to use no preknowledge - in particular, classical majorization theory will be avoided. Then, we are going to discuss an assertion analogous to Theorem 5b with its applications.

2. Absolute values interpolate convex functions

We start preparing for our proof by showing a classical property of convex func- tions¹:

Theorem 6 (Hardy, Littlewood, P´olya). Let f be a convex function from an intervalI ⊆RtoR.Letx₁, x₂, ..., x_n be finitely many points from I.Then, there exist two real constantsuandv andnnonnegative constants a₁, a₂, ..., a_n such that

f(t) = vt+u+

n

X

i=1

a_i|t−x_i| holds for every t∈ {x₁, x₂, ..., x_n}.

In brief, this result states that every convex function f(x) on n reals x₁, x₂, ..., x_n can be interpolated by a linear combination with nonnegative coefficients of a linear function and the n functions |x−x_i|.

The proof of Theorem 6, albeit technical, will be given here for the sake of com- pleteness: First, we need a (very easy to prove) fact which I use to call the max{0, x}

formula: For any real number x,we have max{0, x}= 1

2(x+|x|). Furthermore, we denote f[y, z] = f(y)−f(z)

y−z for any two points y and z from I satisfying y 6= z. Then, we have (y−z)·f[y, z] = f(y)−f(z) for any two points y and z fromI satisfying y6=z.

We can assume that all points x₁, x₂, ..., x_n are pairwisely distinct (in fact, if we can find two different integers i₁ and i₂ from the set {1,2, ..., n} such that x_i1 = x_i2, then we can just remove x_i2 from the list (x₁, x₂, ..., x_n) and set a_i2 = 0, and since we have {x₁, x₂, ..., x_n} = {x₁, x₂, ..., x_i2−1, x_i2+1, ...x_n}, it remains to prove Theorem 6 for the n −1 points x₁, x₂, ..., x_i2−1, x_i2+1, ..., x_n instead of all the n points x₁, x₂, ..., x_n; we can repeat this procedure as long as there are two equal points in the list (x₁, x₂, ..., x_n), until we have reduced Theorem 6 to the case of a list of pairwisely distinct points). Therefore, we can WLOG assume that x₁ < x₂ < ... < x_n. Then, for

1This property appeared as Proposition B.4 in [8], which refers to [9] for its origins. It was also mentioned by a MathLinks user called ”Fleeting Guest” in [4], post #18 as a known fact, albeit in a slightly different (but equivalent) form.

every j ∈ {1,2, ..., n},we have f(xj) =f(x1) +

j−1

X

k=1

(f(xk+1)−f(xk)) =f(x1) +

j−1

X

k=1

(xk+1−xk)·f[xk+1, xk]

=f(x₁) +

j−1

X

k=1

(x_k+1−x_k)· f[x₂, x₁] +

k

X

i=2

(f[x_i+1, x_i]−f[x_i, x_i−1])

!

=f(x₁) +

j−1

X

k=1

(x_k+1−x_k)·f[x₂, x₁] +

j−1

X

k=1

(x_k+1−x_k)·

k

X

i=2

(f[x_i+1, x_i]−f[x_i, xi−1])

=f(x₁) +f[x₂, x₁]·

j−1

X

k=1

(x_k+1−x_k) +

j−1

X

k=1 k

X

i=2

(f[x_i+1, x_i]−f[x_i, xi−1])·(x_k+1−x_k)

=f(x₁) +f[x₂, x₁]·

j−1

X

k=1

(x_k+1−x_k) +

j−1

X

i=2 j−1

X

k=i

(f[x_i+1, x_i]−f[x_i, xi−1])·(x_k+1−x_k)

=f(x₁) +f[x₂, x₁]·

j−1

X

k=1

(x_k+1−x_k) +

j−1

X

i=2

(f[x_i+1, x_i]−f[x_i, x_i−1])·

j−1

X

k=i

(x_k+1−x_k)

=f(x₁) +f[x₂, x₁]·(x_j −x₁) +

j−1

X

i=2

(f[x_i+1, x_i]−f[x_i, xi−1])·(x_j−x_i). Now we set

α₁ =α_n = 0;

α_i =f[x_i+1, x_i]−f[x_i, xi−1] for all i∈ {2,3, ..., n−1}.

Using these notations, the above computation becomes f(x_j) =f(x₁) +f[x₂, x₁]·(x_j −x₁) +

j−1

X

i=2

α_i ·(x_j −x_i)

=f(x₁) +f[x₂, x₁]·(x_j −x₁) + 0

|{z}

=α1

·max{0, x_j−x₁}

+

j−1

X

i=2

α_i· (x_j −x_i)

| {z }

=max{0,x_j−x_i},sincexj−x_i≥0,asxi≤x_j

+

n

X

i=j

α_i· 0

|{z}

=max{0,x_j−x_i},sincexj−x_i≤0,asxj≤x_i

=f(x₁) +f[x₂, x₁]·(x_j −x₁) +α₁·max{0, x_j−x₁} +

j−1

X

i=2

α_i·max{0, x_j−x_i}+

n

X

i=j

α_i·max{0, x_j −x_i}

=f(x₁) +f[x₂, x₁]·(x_j −x₁) +

n

X

i=1

α_i ·max{0, x_j−x_i}

=f(x₁) +f[x₂, x₁]·(x_j −x₁) +

n

X

i=1

α_i ·1

2((x_j −x_i) +|x_j −x_i|)

since max{0, x_j−x_i}= 1

2((x_j−x_i) +|x_j−x_i|) by the max{0, x} formula

=f(x₁) +f[x₂, x₁]·(x_j −x₁) +

n

X

i=1

α_i ·1

2(x_j−x_i) +

n

X

i=1

α_i· 1

2|x_j −x_i|

=f(x₁) + (f[x₂, x₁]x_j −f[x₂, x₁]x₁) + 1 2

n

X

i=1

α_ix_j − 1 2

n

X

i=1

α_ix_i

! +

n

X

i=1

1

2α_i|x_j−x_i|

= f[x2, x1] +1 2

n

X

i=1

αi

!

xj + f(x1)−f[x2, x1]x1− 1 2

n

X

i=1

αixi

! +

n

X

i=1

1

2αi|xj−xi|. Thus, if we denote

v =f[x₂, x₁] + 1 2

n

X

i=1

α_i; u=f(x₁)−f[x₂, x₁]x₁− 1 2

n

X

i=1

α_ix_i; a_i = 1

2α_i for all i∈ {1,2, ..., n}, then we have

f(xj) =vxj +u+

n

X

i=1

ai|xj −xi|.

Since we have shown this for every j ∈ {1,2, ..., n}, we can restate this as follows: We have

f(t) = vt+u+

n

X

i=1

a_i|t−x_i| for every t∈ {x₁, x₂, ..., x_n}.

Hence, in order for the proof of Theorem 6 to be complete, it is enough to show that the n reals a₁, a₂, ..., a_n are nonnegative. Since a_i = 1

2α_i for every i ∈ {1,2, ..., n},

this will follow once it is proven that the n reals α₁, α₂, ..., α_n are nonnegative. Thus, we have to show that αi is nonnegative for every i ∈ {1,2, ..., n}. This is trivial for i = 1 and for i = n (since α₁ = 0 and α_n = 0), so it remains to prove that α_i is nonnegative for every i ∈ {2,3, ..., n−1}. Now, since α_i = f[x_i+1, x_i]− f[x_i, xi−1] for every i ∈ {2,3, ..., n−1}, we thus have to show that f[xi+1, xi]−f[xi, xi−1] is nonnegative for every i ∈ {2,3, ..., n−1}. In other words, we have to prove that f[x_i+1, x_i]≥f[x_i, xi−1] for every i∈ {2,3, ..., n−1}. But since xi−1 < x_i < x_i+1, this follows from the next lemma:

Lemma 7. Letf be a convex function from an intervalI ⊆R toR.Letx, y, z be three points from I satisfying x < y < z. Then, f[z, y]≥f[y, x].

Proof of Lemma 7. Since the functionf is convex onI,and sincez andxare points fromI, the definition of convexity yields

1

z−yf(z) + 1

y−xf(x) 1

z−y+ 1 y−x

≥f







 1

z−yz+ 1 y−xx 1

z−y + 1 y−x









(here we have used that 1

z−y > 0 and 1

y−x > 0, what is clear from x < y < z).

Since 1

z−yz+ 1 y−xx 1

z−y + 1 y−x

=y, this simplifies to

1

z−yf(z) + 1

y−xf(x) 1

z−y + 1 y−x

≥f(y), so that 1

z−yf(z) + 1

y−xf(x)≥ 1

z−y + 1 y−x

f(y), so that 1

z−yf(z) + 1

y−xf(x)≥ 1

z−yf(y) + 1

y−xf(y), so that 1

z−yf(z)− 1

z−yf(y)≥ 1

y−xf(y)− 1

y−xf(x), so that f(z)−f(y)

z−y ≥ f(y)−f(x) y−x .

This becomes f[z, y] ≥ f[y, x], and thus Lemma 7 is proven. Thus, the proof of Theorem 6 is completed.

3. The Karamata inequality in symmetric form

Now as Theorem 6 is proven, it becomes easy to prove the Karamata inequality in the following form:

Theorem 8a, the Karamata inequality in symmetric form. Let f be a convex function from an interval I ⊆ R to R, and let n be a positive integer. Let x₁, x₂, ..., x_n, y₁, y₂, ..., y_n be 2n points from I.Assume that

|x₁−t|+|x₂−t|+...+|x_n−t| ≥ |y₁−t|+|y₂−t|+...+|y_n−t|

holds for every t∈ {x₁, x₂, ..., x_n, y₁, y₂, ..., y_n}. Then,

f(x1) +f(x2) +...+f(xn)≥f(y1) +f(y2) +...+f(yn). This is a particular case of the following result:

Theorem 8b, the weighted Karamata inequality in symmetric form. Let f be a convex function from an interval I ⊆ R to R, and let N be a positive integer. Let z₁, z₂, ..., z_N be N points from I, and let w₁, w₂, ..., w_N beN reals. Assume that

N

X

k=1

w_k= 0, (1)

and that

N

X

k=1

w_k|z_k−t| ≥0 holds for every t∈ {z₁, z₂, ..., z_N}. (2) Then,

N

X

k=1

w_kf(z_k)≥0. (3)

It is very easy to conclude Theorem 8a from Theorem 8b; we postpone this argument until Theorem 8b is proven.

Time for a remark to readers familiar with majorization theory. One may wonder why I call the two results above ”Karamata inequalities”. In fact, the Karamata inequality in its most known form claims:

Theorem 9, the Karamata inequality. Letf be a convex function from an interval I ⊆R to R, and let n be a positive integer. Let x₁, x₂, ..., x_n, y₁, y₂, ..., y_n be 2n points from I such that (x₁, x₂, ..., x_n)(y₁, y₂, ..., y_n). Then,

f(x₁) +f(x₂) +...+f(x_n)≥f(y₁) +f(y₂) +...+f(y_n).

According to [2], post #11, Lemma 1, the condition (x₁, x₂, ..., x_n)(y₁, y₂, ..., y_n) yields that |x₁−t|+|x₂−t|+...+|x_n−t| ≥ |y₁−t|+|y₂ −t|+...+|y_n−t| holds for every real t - and thus, in particular, for everyt∈ {z₁, z₂, ..., z_n}. Hence, whenever the condition of Theorem 9 holds, the condition of Theorem 8a holds as well. Thus, Theorem 9 follows from Theorem 8a. With just a little more work, we could also derive Theorem 8a from Theorem 9, so that Theorems 8a and 9 are equivalent.

Note that Theorem 8b is more general than the Fuchs inequality (a more well- known weighted version of the Karamata inequality). See [5] for a generalization of majorization theory to weighted families of points (apparently already known long time ago), with a different approach to this fact.

As promised, here is aproof of Theorem 8b: First, substitutingt= max{z1, z2, ..., zN} into (2) (it is clear that this t satisfies t ∈ {z1, z2, ..., zN}), we get

N

P

k=1

wk|zk−t| ≥ 0, what is equivalent to −

N

P

k=1

wkzk ≥ 0 (since t = max{z1, z2, ..., zN} yields zk ≤ t for every k ∈ {1,2, ..., N}, so that z_k−t ≤ 0 and thus |z_k−t| = −(z_k−t) = t−z_k for every k ∈ {1,2, ..., N}, so that

N

X

k=1

w_k|z_k−t|=

N

X

k=1

w_k(t−z_k) =t

N

X

k=1

w_k

| {z }

=0

−

N

X

k=1

w_kz_k =t·0−

N

X

k=1

w_kz_k=−

N

X

k=1

w_kz_k

). Hence,

N

P

k=1

w_kz_k ≤0.

On the other hand, substituting t = min{z1, z2, ..., zN} into (2) (again, it is clear that thist satisfiest∈ {z1, z2, ..., zN}), we get

N

P

k=1

wk|zk−t| ≥0, what is equivalent to

N

P

k=1

wkzk ≥0 (since t = min{z1, z2, ..., zN} yields zk ≥t for every k ∈ {1,2, ..., N}, so that z_k−t≥0 and thus|z_k−t|=z_k−t for every k ∈ {1,2, ..., N}, so that

N

X

k=1

w_k|z_k−t|=

N

X

k=1

w_k(z_k−t) =

N

X

k=1

w_kz_k−t

N

X

k=1

w_k

| {z }

=0

=

N

X

k=1

w_kz_k−t·0 =

N

X

k=1

w_kz_k

).

Combining

N

P

k=1

w_kz_k ≤0 with

N

P

k=1

w_kz_k ≥0, we get

N

P

k=1

w_kz_k= 0.

The functionf :I →Ris convex, andz₁, z₂, ..., z_N are finitely many points fromI.

Hence, Theorem 6 yields the existence of two real constantsuandv andN nonnegative constants a₁, a₂, ..., a_N such that

f(t) =vt+u+

N

X

i=1

a_i|t−z_i| holds for every t∈ {z₁, z₂, ..., z_N}. Thus,

f(z_k) =vz_k+u+

N

X

i=1

a_i|z_k−z_i| for every k ∈ {1,2, ..., N}

(since z_k ∈ {z₁, z₂, ..., z_N}). Hence,

N

X

k=1

w_kf(z_k) =

N

X

k=1

w_k vz_k+u+

N

X

i=1

a_i|z_k−z_i|

!

=v

N

X

k=1

w_kz_k

| {z }

=0

+u

N

X

k=1

w_k

| {z }

=0

+

N

X

k=1

w_k

N

X

i=1

a_i|z_k−z_i|

=

N

X

k=1

w_k

N

X

i=1

a_i|z_k−z_i|=

N

X

i=1

a_i

N

X

k=1

w_k|z_k−z_i|

| {z }

≥0 according to (2) fort=zi

≥0.

Thus, Theorem 8b is proven.

Now, as Theorem 8b is verified, let us conclude Theorem 8a:

Proof of Theorem 8a: Set N = 2n and z_k =

x_k for all k∈ {1,2, ..., n};

y_k−n for all k ∈ {n+ 1, n+ 2, ...,2n} ; w_k=

1 for allk ∈ {1,2, ..., n};

−1 for all k ∈ {n+ 1, n+ 2, ...,2n} . That is,

z₁ =x₁, z₂ =x₂, ..., z_n=x_n; z_n+1 =y₁, z_n+2 =y₂, ..., z_2n=y_n;

w₁ =w₂ =...=w_n = 1; w_n+1 =w_n+2 =...=w_2n =−1.

Then, the conditions of Theorem 8b are fulfilled: In fact, (1) is fulfilled because

N

X

k=1

w_k =

2n

X

k=1

w_k=

n

X

k=1

w_k+

2n

X

k=n+1

w_k =

n

X

k=1

1 +

2n

X

k=n+1

(−1) =n·1 +n·(−1) = 0.

Also, (2) is fulfilled, because for everyt ∈ {z₁, z₂, ..., z_N},we havet∈ {x₁, x₂, ..., x_n, y₁, y₂, ..., y_n} (because {z1, z2, ..., zN}={x1, x2, ..., xn, y1, y2, ..., yn}) and thus, after the condition of

Theorem 8a, we have

|x1−t|+|x2−t|+...+|xn−t| ≥ |y1−t|+|y2−t|+...+|yn−t|, so that

n

X

k=1

|x_k−t| ≥

n

X

k=1

|y_k−t|, so that

n

X

k=1

|x_k−t| −

n

X

k=1

|y_k−t| ≥0, and thus

N

X

k=1

w_k|z_k−t|=

2n

X

k=1

w_k|z_k−t|=

n

X

k=1

w_k|z_k−t|+

2n

X

k=n+1

w_k|z_k−t|

=

n

X

k=1

1· |x_k−t|+

2n

X

k=n+1

(−1)· |yk−n−t|

=

n

X

k=1

|x_k−t| −

2n

X

k=n+1

|yk−n−t|=

n

X

k=1

|x_k−t| −

n

X

k=1

|y_k−t| ≥0,

what proves (2).

Hence, we can apply Theorem 8b and obtain

N

X

k=1

w_kf(z_k)≥0.

That is, 0≤

N

X

k=1

w_kf(z_k) =

2n

X

k=1

w_kf(z_k) =

n

X

k=1

w_kf(z_k) +

2n

X

k=n+1

w_kf(z_k) =

n

X

k=1

1f(x_k) +

2n

X

k=n+1

(−1)f(yk−n)

=

n

X

k=1

f(x_k)−

2n

X

k=n+1

f(yk−n) =

n

X

k=1

f(x_k)−

n

X

k=1

f(y_k),

so that

n

P

k=1

f(x_k)≥

n

P

k=1

f(y_k), and Theorem 8a is proven.

4. A property of zero-sum vectors Next, we are going to show some properties of real vectors.

If k is an integer and v ∈ R^k is a vector, then, for any i∈ {1,2, ..., k}, we denote byv_i the i-th coordinate of the vector v.Then, v =







 v₁ v₂ ...

v_k







 .

Let n be a positive integer. We consider the vector space Rⁿ. Let (e1, e2, ..., en) be the standard basis of this vector spaceRⁿ; in other words, for everyi∈ {1,2, ..., n},let e_i be the vector fromRⁿsuch that (e_i)_i = 1 and (e_i)_j = 0 for everyj ∈ {1,2, ..., n}\{i}. LetVn be the subspace ofRⁿ defined by

V_n ={x∈Rⁿ | x₁ +x₂+...+x_n = 0}.

For any u ∈ {1,2, ..., n} and any two distinct numbers i and j from the set {1,2, ..., n}, we have

(ei−ej)_u =







1, if u=i;

−1, if u=j;

0, if u6=i and u6=j

. (4)

We haveei−ej ∈Vn for any two numbersiandj from the set{1,2, ..., n}(in fact, if the numbersiandj are distinct, then (4) yields (e_i−e_j)₁+(e_i−e_j)₂+...+(e_i−e_j)_n = 0 and thus e_i−e_j ∈V_n, and if not, then i=j and thus e_i−e_j =e_j −e_j = 0∈V_n).

For any vector t ∈ Rⁿ, we denote I(t) = {k ∈ {1,2, ..., n} |tk>0} and J(t) = {k ∈ {1,2, ..., n} |t_k <0}. Obviously, for every t ∈ Rⁿ, the sets I(t) and J(t) are disjoint (since there does not exist any k satisfying both t_k >0 andt_k<0).

Now we are going to show:

Theorem 10. Let n be a positive integer. Let x∈V_n be a vector. Then, there exist nonnegative realsa_i,j for all pairs (i, j)∈I(x)×J(x) such that

x= X

(i,j)∈I(x)×J(x)

a_i,j(e_i−e_j).

Proof of Theorem 10. We will prove Theorem 10 by induction over|I(x)|+|J(x)|. The basis of the induction - the case when |I(x)| +|J(x)| = 0 - is trivial: If

|I(x)|+|J(x)|= 0,then I(x) =J(x) = ∅,so thatx= P

(i,j)∈I(x)×J(x)

a_i,j(e_i−e_j) holds because x = 0 (because if x were different from 0, then there would exist at least one k ∈ {1,2, ..., n} such that x_k 6= 0, so that either x_k > 0 or x_k < 0, but x_k > 0 is impossible because {k ∈ {1,2, ..., n} |x_k >0} = I(x) = ∅, and x_k < 0 is impossible because {k ∈ {1,2, ..., n} |x_k <0} = J(x) = ∅) and P

(i,j)∈I(x)×J(x)

a_i,j(e_i−e_j) = 0 (since I(x) = J(x) = ∅ yields I(x)×J(x) = ∅, so that P

(i,j)∈I(x)×J(x)

ai,j(ei−ej) is an empty sum and thus equals 0).

Now we come to the induction step: Let r be a positive integer. Assume that Theorem 10 holds for all x ∈ V_n with |I(x)|+ |J(x)| < r. We have to show that Theorem 10 holds for all x∈V_n with |I(x)|+|J(x)|=r.

In order to prove this, we letz ∈V_nbe an arbitrary vector with|I(z)|+|J(z)|=r.

We then have to prove that Theorem 10 holds for x =z. In other words, we have to show that there exist nonnegative realsa_i,j for all pairs (i, j)∈I(z)×J(z) such that

z = X

(i,j)∈I(z)×J(z)

a_i,j(e_i−e_j). (5)

First, |I(z)|+|J(z)| =r and r > 0 yield|I(z)|+|J(z)| >0. Hence, at least one of the sets I(z) and J(z) is non-empty.

Now, since z ∈ V_n, we have z₁ +z₂ +...+z_n = 0. Hence, either z_k = 0 for every k ∈ {1,2, ..., n}, or there is at least one positive number and at least one negative number in the set{z₁, z₂, ..., z_n}.The first case is impossible (in fact, ifz_k = 0 for every k ∈ {1,2, ..., n}, then I(z) ={k ∈ {1,2, ..., n} |z_k >0} =∅ and similarly J(z) =∅, contradicting the fact that at least one of the setsI(z) andJ(z) is non-empty). Thus, the second case must hold - i. e., there is at least one positive number and at least one negative number in the set {z₁, z₂, ..., z_n}. In other words, there exists a number u ∈ {1,2, ..., n} such that zu > 0, and a number v ∈ {1,2, ..., n} such that zv <0. Of course,z_u >0 yieldsu∈I(z),andz_v <0 yieldsv ∈J(z). Needless to say thatu6=v (since z_u >0 andz_v <0).

Now, we distinguish between two cases: thefirst case will be the case whenzu+zv ≥ 0,and the second case will be the case when z_u+z_v ≤0.

Let us consider thefirst case: In this case,z_u+z_v ≥0.Then, letz⁰ =z+z_v(e_u−e_v). Sincez ∈Vnandeu−ev ∈Vn,we havez+zv(eu−ev)∈Vn (sinceVn is a vector space), so that z⁰ ∈V_n. Fromz⁰ =z+z_v(e_u−e_v),the coordinate representation of the vector z⁰ is easily obtained:

z⁰ =







 z⁰₁ z⁰₂ ...

z_n⁰









, where







z_k⁰ =z_k for all k ∈ {1,2, ..., n} \ {u, v}; z_u⁰ =z_u+z_v;

z_v⁰ = 0

.

Thus,

I(z⁰) = {k∈ {1,2, ..., n} |z⁰_k>0}

={k∈ {1,2, ..., n} \ {u, v} |z⁰_k>0} ∪ {k =u|z_k⁰ >0} ∪ {k =v |z_k⁰ >0}

| {z }

=∅,sincez_v⁰ is not>0,but =0

={k∈ {1,2, ..., n} \ {u, v} |z⁰_k>0} ∪ {k =u|z_k⁰ >0}

={k∈ {1,2, ..., n} \ {u, v} |z_k>0}

| {z }

subset of{k∈{1,2,...,n}|z_k>0}=I(z)

∪ {k =u|z_k⁰ >0}

| {z }

this is either{u}or∅,anyway a subset ofI(z) sinceu∈I(z)

(we have replacedz_k⁰ byz_k here, since z_k⁰ =z_k for all k ∈ {1,2, ..., n} \ {u, v})

⊆I(z)

(since the union of two subsets of I(z) must be a subset of I(z)). Thus, |I(z⁰)| ≤

|I(z)|.Besides, z_u⁰ ≥0 (sincez_u⁰ =zu+zv ≥0), so that J(z⁰) = {k∈ {1,2, ..., n} |z⁰_k<0}

={k∈ {1,2, ..., n} \ {u, v} |z_k⁰ <0} ∪ {k=u|z_k⁰ <0}

| {z }

=∅,sincez⁰_uis not<0,but≥0

∪ {k=v |z_k⁰ <0}

| {z }

=∅,sincez⁰_vis not<0,but =0

={k∈ {1,2, ..., n} \ {u, v} |z_k⁰ <0}

={k∈ {1,2, ..., n} \ {u, v} |z_k <0}

(we have replacedz_k⁰ byzk here, since z_k⁰ =zk for all k ∈ {1,2, ..., n} \ {u, v})

⊆ {k ∈ {1,2, ..., n} |z_k<0}=J(z).

Moreover, J(z⁰) is a proper subset of J(z), because v /∈ J(z⁰) (since z_v⁰ is not < 0, but = 0) but v ∈J(z).Hence, |J(z⁰)|<|J(z)|. Combined with |I(z⁰)| ≤ |I(z)|, this yields|I(z⁰)|+|J(z⁰)|<|I(z)|+|J(z)|.In view of |I(z)|+|J(z)|=r, this becomes

|I(z⁰)|+|J(z⁰)|< r.Thus, since we have assumed that Theorem 10 holds for allx∈V_n with |I(x)|+|J(x)| < r, we can apply Theorem 10 to x =z⁰, and we see that there exist nonnegative reals a⁰_i,j for all pairs (i, j)∈I(z⁰)×J(z⁰) such that

z⁰ = X

(i,j)∈I(z⁰)×J(z⁰)

a⁰_i,j(e_i−e_j).

Now, z⁰ =z+z_v(e_u−e_v) yields z =z⁰−z_v(e_u−e_v). Since z_v <0, we have−z_v >0, so that, particularly, −z_v is nonnegative.

Since I(z⁰)⊆I(z) andJ(z⁰)⊆J(z),we have I(z⁰)×J(z⁰)⊆I(z)×J(z).Also, (u, v) ∈ I(z)× J(z) (because u ∈ I(z) and v ∈ J(z)) and (u, v) ∈/ I(z⁰)×J(z⁰) (because v /∈J(z⁰)).

Hence, the setsI(z⁰)×J(z⁰) and {(u, v)}are two disjoint subsets of the set I(z)× J(z). We can thus define nonnegative reals a_i,j for all pairs (i, j) ∈ I(z)×J(z) by setting

a_i,j =







a⁰_i,j, if (i, j)∈I(z⁰)×J(z⁰) ;

−z_v, if (i, j) = (u, v) ;

0, if neither of the two cases above holds

(these a_i,j are all nonnegative because a⁰_i,j, −z_v and 0 are nonnegative). Then, X

(i,j)∈I(z)×J(z)

ai,j(ei−ej)

= X

(i,j)∈I(z⁰)×J(z⁰)

a_i,j(e_i−e_j) + X

(i,j)=(u,v)

a_i,j(e_i−e_j) + X

(i,j)∈(I(z)×J(z))\((I(z⁰)×J(z⁰))∪{(u,v)})

a_i,j(e_i−e_j)

= X

(i,j)∈I(z⁰)×J(z⁰)

a⁰_i,j(e_i−e_j) + X

(i,j)=(u,v)

(−z_v) (e_i−e_j) + X

(i,j)∈(I(z)×J(z))\((I(z⁰)×J(z⁰))∪{(u,v)})

0 (e_i−e_j)

= X

(i,j)∈I(z⁰)×J(z⁰)

a⁰_i,j(e_i−e_j) + (−z_v) (e_u −e_v) + 0 =z⁰+ (−z_v) (e_u−e_v) + 0

= (z+zv(eu −ev)) + (−zv) (eu−ev) + 0 =z.

Thus, (5) is fulfilled.

Similarly, we can fulfill (5) in the second case, repeating the arguments we have done for the first case while occasionally interchanginguwithv,as well asI withJ,as well as<with >. Here is a brief outline of how we have to proceed in the second case:

Denote z⁰ =z−zu(eu−ev). Show that z⁰ ∈Vn (as in the first case). Notice that

z⁰ =







 z⁰₁ z⁰₂ ...

z_n⁰









, where







z_k⁰ =z_k for all k ∈ {1,2, ..., n} \ {u, v}; z_u⁰ = 0;

z_v⁰ =zu+zv

.

Prove thatu /∈I(z⁰) (as we provedv /∈J(z⁰) in the first case). Prove thatJ(z⁰)⊆J(z) (similarly to the proof ofI(z⁰)⊆I(z) in the first case) and thatI(z⁰) is a proper subset of I(z) (similarly to the proof that J(z⁰) is a proper subset of J(z) in the first case).

Show that there exist nonnegative realsa⁰_i,j for all pairs (i, j)∈I(z⁰)×J(z⁰) such that

z⁰ = X

(i,j)∈I(z⁰)×J(z⁰)

a⁰_i,j(e_i −e_j)

(as in the first case). Note that zu is nonnegative (since zu > 0). Prove that the sets I(z⁰)×J(z⁰) and {(u, v)} are two disjoint subsets of the set I(z)×J(z) (as in the first case). Define nonnegative reals a_i,j for all pairs (i, j)∈I(z)×J(z) by setting

a_i,j =







a⁰_i,j, if (i, j)∈I(z⁰)×J(z⁰) ; z_u, if (i, j) = (u, v) ;

0, if neither of the two cases above holds .

Prove that these nonnegative reals a_i,j fulfill (5).

Thus, in each of the two cases, we have proven that there exist nonnegative reals a_i,j for all pairs (i, j) ∈ I(z)×J(z) such that (5) holds. Hence, Theorem 10 holds for x = z. Thus, Theorem 10 is proven for all x ∈ V_n with |I(x)|+|J(x)| = r. This completes the induction step, and therefore, Theorem 10 is proven.

As an application of Theorem 10, we can now show:

Theorem 11. Let n be a positive integer. Let a₁, a₂, ..., a_n be n nonneg- ative reals. Let S be a finite set. For every s ∈ S, let r_s be an element of

(Rⁿ)^∗ (in other words, a linear transformation from Rⁿ toR), and let b_s be a nonnegative real. Define a function f :Rⁿ →Rby

f(x) =

n

X

u=1

a_u|x_u| −X

s∈S

b_s|r_sx|, wherex=







 x1

x₂ ...

xn









∈Rⁿ.

Then, the following two assertions are equivalent:

Assertion A1: We have f(x)≥0 for every x∈Vn.

Assertion A₂: We have f(e_i−e_j) ≥ 0 for any two distinct integers i and j from{1,2, ..., n}.

Proof of Theorem 11. We have to prove that the assertionsA1andA2are equivalent.

In other words, we have to prove that A₁ =⇒ A₂ and A₂ =⇒ A₁.Actually, A₁ =⇒ A₂ is trivial (we just have to use that e_i −e_j ∈ V_n for any two numbers i and j from {1,2, ..., n}). It remains to show thatA2 =⇒ A1.So assume that Assertion A2 is valid, i. e. we havef(e_i−e_j)≥0 for any two distinct integers iand j from{1,2, ..., n}. We have to prove that Assertion A₁ holds, i. e. that f(x)≥0 for every x∈V_n.

So let x ∈ Vn be some vector. According to Theorem 10, there exist nonnegative reals a_i,j for all pairs (i, j)∈I(x)×J(x) such that

x= X

(i,j)∈I(x)×J(x)

a_i,j(e_i−e_j). We will now show that

|xu|= X

(i,j)∈I(x)×J(x)

ai,j

(ei−ej)_u

for every u∈ {1,2, ..., n}. (6) Here, of course, (e_i−e_j)_u means the u-th coordinate of the vector e_i−e_j.

In fact, two cases are possible: the case when x_u ≥ 0, and the case when x_u <0.

We will consider these cases separately.

Case 1: We havex_u ≥0.Then,|x_u|=x_u.Hence, in this case, we have (e_i−e_j)_u ≥0 for any two numbers i ∈ I(x) and j ∈ J(x) (in fact, j ∈ J(x) yields x_j < 0, so that u 6= j (because x_j < 0 and x_u ≥ 0) and thus (e_j)_u = 0, so that (e_i−e_j)_u = (e_i)_u−(e_j)_u = (e_i)_u−0 = (e_i)_u =

1, if u=i;

0, if u6=i ≥0). Thus, (e_i−e_j)_u =

(e_i−e_j)_u for any two numbers i∈I(x) and j ∈J(x). Thus,

|xu|=xu = X

(i,j)∈I(x)×J(x)

ai,j(ei−ej)_u



since x= X

(i,j)∈I(x)×J(x)

ai,j(ei−ej)





= X

(i,j)∈I(x)×J(x)

a_i,j

(e_i−e_j)_u , and (6) is proven.

Case 2: We have x_u < 0. Then, u ∈ J(x) and |x_u| = −x_u. Hence, in this case, we have (ei−ej)_u ≤ 0 for any two numbers i∈ I(x) and j ∈ J(x) (in fact, i ∈I(x) yields x_i > 0, so that u 6= i (because x_i > 0 and x_u < 0) and thus (e_i)_u = 0, so that (e_i−e_j)_u = (e_i)_u −(e_j)_u = 0−(e_j)_u = −(e_j)_u =−

1, if u=j;

0, if u6=j ≤ 0). Thus,

−(e_i−e_j)_u =

(e_i−e_j)_u

for any two numbers i∈I(x) and j ∈J(x). Thus,

|x_u|=−x_u =− X

(i,j)∈I(x)×J(x)

a_i,j(e_i−e_j)_u



since x= X

(i,j)∈I(x)×J(x)

a_i,j(e_i−e_j)





= X

(i,j)∈I(x)×J(x)

a_i,j −(e_i−e_j)_u

= X

(i,j)∈I(x)×J(x)

a_i,j

(e_i−e_j)_u , and (6) is proven.

Hence, in both cases, (6) is proven. Thus, (6) always holds. Now let us continue our proof of A₂ =⇒ A₁:

We have X

s∈S

bs|rsx|=X

s∈S

bs

rs

X

(i,j)∈I(x)×J(x)

ai,j(ei−ej)



since x= X

(i,j)∈I(x)×J(x)

ai,j(ei−ej)





=X

s∈S

b_s

X

(i,j)∈I(x)×J(x)

a_i,jr_s(e_i−e_j)

≤X

s∈S

b_s X

(i,j)∈I(x)×J(x)

a_i,j|r_s(e_i−e_j)|

(by the triangle inequality, since all a_i,j and allb_s are nonnegative). Thus,

f(x) =

n

X

u=1

a_u|x_u| −X

s∈S

b_s|r_sx| ≥

n

X

u=1

a_u|x_u| −X

s∈S

b_s X

(i,j)∈I(x)×J(x)

a_i,j|r_s(e_i−e_j)|

=

n

X

u=1

a_u· X

(i,j)∈I(x)×J(x)

a_i,j

(e_i−e_j)_u −X

s∈S

b_s X

(i,j)∈I(x)×J(x)

a_i,j|r_s(e_i−e_j)| (by (6))

= X

(i,j)∈I(x)×J(x)

a_i,j

n

X

u=1

a_u

(e_i−e_j)_u

− X

(i,j)∈I(x)×J(x)

a_i,jX

s∈S

b_s|r_s(e_i−e_j)|

= X

(i,j)∈I(x)×J(x)

a_i,j ·

n

X

u=1

a_u

(e_i−e_j)_u −X

s∈S

b_s|r_s(e_i−e_j)|

!

= X

(i,j)∈I(x)×J(x)

a_i,j

|{z}≥0

·f(e_i−e_j)

| {z }

≥0

≥0.

(Here, f(e_i−e_j) ≥ 0 because i and j are two distinct integers from {1,2, ..., n}; in fact, i and j are distinct because i∈I(x) and j ∈J(x), and the sets I(x) and J(x) are disjoint.)