On the Semi-Group Property of the Perpendicular Bisector in a Normed Space

Abstract

Let (X,d) be a metric linear space and a∈X. The point a divides the space into three sets: H_a = {x ∈ X: d(0,x) < d(x,a)}, M_a = {x ∈ X: d(0,x) = d(x,a)} and L_a = {x ∈ X: d(0,x) > d(x,a)}. If the distance is generated by a norm, H_a is called the Leibnizian halfspace of a, M_a is the perpendicular bisector of the segment 0,a and L_a is the remaining set L_a = X\(H_a ∪ M_a). It is known that the perpendicular bisector of the segment [0,a] is an affine subspace of X for all a ∈ X if, and only if, X is an inner product space, that is, if and only if the norm is generated by an inner product. In this case, it is also true that if x,y ∈ L_a ∪ M_a, then x + y ∈ L_a ∪ M_a. Otherwise written, the set L_a ∪ M_a is a semi-group with respect to addition. We investigate the problem: for what kind of norms in X the pair (L_a ∪ M_a,+) is a semi-group for all a ∈ X? In that case, we say that “$\left(X,\Vert .\Vert \right)$ has the semi-group property” or that “the norm $\Vert .\Vert$ has the semi-group property”. This is a threedimensional property, meaning that if all the three-dimensional subspaces of X have it, then X also has it. We prove that for two-dimensional spaces, (L_a,+) is a semi-group for any norm, that $\left(X,\Vert .\Vert \right)$ has the semi-group property if, and only if, the norm is strictly convex, and, in higher dimensions, the property fails to be true even if the norm is strictly convex. Moreover, studying the L^p norms in higher dimensions, we prove that the semi-group property holds if, and only if, p = 2. This fact leads us to the conjecture that in dimensions greater than three, the semi-group property holds if, and only if, X is an inner-product space.

1. Introduction

Let (X, ‖ ‖) be a real normed space.

In [1], the authors were interested in the following property of X

For every x,y,z ∈ X, the inequalities

(A) For every x,y,z ∈ X, the inequalities

║x║ ≥ ║y + z║, ║y║ ≥ ║y + z║, ║z║ ≥ ║x + y║, imply the equalities

║x║ = ║y + z║, ║y║ = ║y + z║, ║z║ = ║x + y║

The typical examples of spaces with this property are the inner product spaces. Indeed, if there is an inner product <⋅,⋅> such that ‖x‖² = <x,x>, then the inequalities in (A) become ‖x‖² ≥ ‖y‖² + ‖z‖² + 2<x,y>, ‖y‖² ≥ ‖x‖² + ‖z‖² + 2<x,z>, ‖z‖² ≥ ‖y‖² + ‖x‖² + 2<z,y>; if we add them, we obtain ‖x + y + z‖ ≤ 0 ⇔ x + y + z = 0. In this case, we obtain even more than the simple equalities ‖x‖ = ‖y + z‖, ‖y‖ = ‖z + x‖, ‖z‖ = ‖x + y‖, namely, x = −(y + z), y = −(x + z), z = −(y + x).

If we denote a = x + y + z, we can write property (A) as

(A) For every x,y,a ∈ X, the inequalities

║x║ ≥ ║a − x║, ║y║ ≥ ║a − y║, ║a − x − y║ ≥ ║x + y║ imply the equalities

║x║ = ║a − x║, ║y║ = ║a − y║, ║a − x − y║ = ║x + y║

Written in this form, the property (A) receives a geometric flavor.

Notation. 

Let a ∈ X. Denote by L_a, M_a, H_a the sets

L_a = {x ∈ X|║x − a║ < ║x║}

(1)

M_a = {x ∈ X|║x − a║ = ║x║}

(2)

H_a = {x ∈ X|║x − a║ > ║x║}

(3)

Note that M_a is the perpendicular bisector of the segment [0,a], L_a is the set of points closer to a and H_a is the set of points from X closer to 0. Some authors [2] call the set H_a the Leibnizian halfspace of a. Horvath ([2,3]) writes H_0,a instead of H_a and H_a,0 instead of M_a

In the sequel, we always assume that a ≠ 0, since a = 0 is nonsensical.

Using these notations, the property (A) becomes

(A) 

For any 0 ≠ a ∈ X, if x ∈ L_a ∪ M_a, y ∈ L_a ∪ M_a, x + y ∈ H_a ∪ M_a then x,y,x + y ∈ M_a

A weaker form of this property is

(B) 

For any 0 ≠ a ∈ X, if x ∈ L_a ∪ M_a, y ∈ L_a ∪ M_a, x + y ∈ H_a ∪ M_a then x + y ∈ M_a

However, is it possible that x,y,x + y ∈ M_a? In an inner-product space, this is not possible. In this case, x ∈ L_a ∪ M_a, y ∈ L_a ∪ M_a ⇒ x + y ∈ L_a.

In other words, in an inner-product space, (L_a ∪ M_a,+) is a semigroup. Indeed, this is obvious: squaring the inequalities ║x║ ≥ ║a − x║, ║y║ ≥ ║a − y║and adding them, we obtain ║x + y║² ≥ 2║a║² − 2<a,x + y> + ║x + y║² ≥ ║x + y − a║², meaning that x + y is in L_a.

2. The Main Result

A norm space is strictly convex ([1,2,3,4]) if and only if the unity ball B₁ = {x ∈ X: ║x║ ≤ 1} is a strict convex set or, equivalently, the equality

║x + y║ = ║x║ + ║y║ can hold if and only if x = 0 or y = 0 or y = λx for some λ > 0. An equivalent definition is that the norm is a strict convex function, i.e., for x ≠ y, the equality

║(1 − λ)x + λy║ = (1 − λ)║x║ + λ║y║ can hold if, and only if, λ ∈ {0,1}. Here λ ∈ [0,1].

Let us denote by h_a : X → ℜ the mapping h_a(x) = ║x − a║ − ║x║. Then

H_a = {h_a > 0}

M_a = {h_a = 0}

L_a = {h_a < 0}.

As the function h_a is continuous, M_a is a closed set and H_a, L_a are open.

Definition 1.

Say that (X, ║.║) has the semigroup property (or that the norm has the semigroup property) if it satisfies the condition

(M) (L_a ∪ M_a,+) is a semi-group for every a ∈ X

or, equivalently,

(M) (L_a ∪ M_a) + (L_a ∪ M_a) ⊆ (L_a ∪ M_a).

Explicitly, the semi group property means that

║x║ ≥ ║x − a║,║y║ ≥ ║y − a║ ⇒ ║x + y║ ≥ ║x +y − a║ or any x,y,a ∈ X

We shall also consider the following similar property

(M°) (L_a,+) is a semigroup

Proposition 1.

(i)

The property (M) implies the property (B).

(ii)

A space X has the properties (A), (B), (M) or (M°) if, and only if, all three-dimensional subspaces of X have it.

Proof. 

(i)

Obvious. If L_a ∪ M_a is a semigroup, then x ∈ L_a ∪ M_a, y ∈ L_a ∪ M_a ⇒ x + y ∈ L_a ∪ M_a. Thus, if x ∈ L_a ∪ M_a, y ∈ L_a ∪ M_a, x + y ∈ H_a ∪ M_a, then x + y ∈ (H_a ∪ M_a) ∩ (H_a ∪ M_a) = M_a.

(ii)

For instance, if X has the property (A) and Y = span({x,y,z}), then it obviously has the property (A) as well. Conversely, if we know that Y has the property (A), we know that X also has it. □

Remark 1.

Thus, the semigroup property is geometric; it deals with the shape of the unity ball in a three-dimensional subspace of X. In the case of inner product spaces, this is an ellipsoid. In fact, the inner product spaces satisfy a stronger assumption, namely

(C) ‖x‖ ≥ ‖x − a‖, ‖y‖ ≥ ‖y − a‖, ‖x + y‖ = ‖x + y − a‖ ⇒ a = 0

Of course, (C) implies (A).

Any norm space has the following properties:

Proposition 2.

Let X be any norm space and a ∈ X, a ≠ 0.

(i)

If ‖x‖ ≥ ‖x − a‖ and 0 ≤ t ≤ 1 then ‖x‖ ≥ ‖x − ta‖.

(ii)

If x ∈ L_a and t ≥ 1 then tx ∈ L_a.

If x ∈ M_a and 0 ≤ t ≤ 1, then tx ∈ M_a∪ H_a.

If x ∈ M_a, t ≥ 1 then tx ∈ M_a ∪ H_a.

(iii)

If x ∈ L_a, there exists 0 ≤ t ≤ 1 such that tx ∈ M_a

(iv)

If, moreover, the function X is strictly convex, then x ∈ M_a, tx ∈ M_a if, and only if, t = 1. In other words, if x ∈ M_a and t > 1, then tx ∉ M_a.

(v)

If X is strictly convex, the closures are Cl(H_a) = H_a ∪ M_a and Cl(L_a) = L_a ∪ M_a.

Proof. 

(i)

Let f: ℜ → ℜ be defined by f(t) = ║x║ − ║x − ta║. The function f is concave, f(0) = 0 and f(1) ≥ 0. Thus t ∈ [0,1] ⇒ f(t) ≥ 0 or, explicitly, ║x║ ≥ ║x − ta║

(ii)

Obvious if we write the inequality ║x║ ≥ ║x − ta ║ as ║$\frac{1}{t}x$║ ≥ ║$\frac{1}{t}$x − a ║. If t ∈ (0,1], then $\frac{1}{t}$ ≥ 1. If x ∈ M_a, then f(0) = f(1) = 0 ⇒ f(s) ≤ 0 ∀ s ∈ (−∞,0) ∪ (1,∞) ⇔ ║x║ ≤ ║x − ta ║ ∀ t ∈ (0,1). Thus ║tx║ ≤ ║tx − a║ ∀ t ∈ (0,1).

(iii)

The function f from (i) has the property that f(∞) = −∞. As f(1) ≥ 0, there must be t ≥ 1 such that f(t) = 0 ⇔ ║x║ = ║x − ta ║ ⇔ $\frac{1}{t}$x ∈ M_a

(iv)

If the norm is strictly convex, then the function f from 1 is strictly concave. Let s =$\frac{1}{t}$. As x ∈ M_a ⇔ f(0) = f(1) = 0, the strict concavity implies s ∈ (0,1) ⇒ f(s) > 0 ⇔║x║ > ║x − sa║ or ║tx║ > ║tx − a║.

(v)

The inclusion Cl(H_a) ⊆ H_a ∪ M_a holds in any norm space. The problem is to check that M_a ⊆ Cl(H_a), and this is not true in general. However, if the norm ║.║ is strictly convex, and x ∈ M_a, then the points t_nx belong to H_a if t_n > 1. Now, it is obvious that if t_n ↓ 1, then t_nx → x hence x ∈ Cl(H_a). The equality Cl(L_a) = L_a ∪ M_a has the same proof. □

The following result also holds in any norm space and simplifies the issue:

Proposition 3.

If (X, ║⋅║) has the property (M) and T:X → X is a bijective linear operator then (X, ║⋅║_T) has the property (M), too, where the norm is defined by

║x║_T = ║Tx║

Moreover, if (X, ║⋅║) is strictly convex, then (X, ║⋅║_T) is strictly convex, too.

Proof. 

Let H_a^(T) = {x ∈ X: ║x║_T > ║x − a║_T }, M_a^(T) = {x ∈ X: ║x║_T = ║x − a║_T} and L_a^(T) = {x ∈ X: ║x║_T = ║x − a║_T}.

We claim that H_a^(T) = T⁻¹ (H_Ta), M_a^(T) = T⁻¹ (M_Ta) and L_a^(T) = T⁻¹ (L_Ta).

Indeed, H_a^(T) = {x ∈ X: ║Tx║ < ║Tx − Ta║} ⇔ T(H_a^(T)) = {Tx: ║Tx║ < ║Tx − Ta║} = {y ∈ X: ║y║ < ║y − a║} = H_a and the same holds for the other two sets. The property (M) says that H_a ∪ M_a is a semigroup for any a ∈ X. Then H_Ta ∪ M_Ta is also a semigroup. it is obvious that if H ⊆ X is a semigroup and U:X → X is a linear operator, then U(H) is a semigroup, too. Therefore H_a^(T) ∪ M_a^(T) = T⁻¹(H_Ta ∪ M_Ta) is a semigroup.

The last assertion is obvious. □

Example 1.

Ifℜ² endowed with some norm that has the property (M), the same holds for ℜ² endowed with the norm ║(x₁, x₂)║* = ║(ax₁ + bx₂, cx₁ + dx₂)║ provided that $\det \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ ≠ 0. Now we can state our main result.

Theorem 1.

Let X = ℜ² endowed with some norm ║⋅║. Let a ∈ X. Then

(i)

L_a + L_a ⊆ L_a

(ii)

If X is strictly convex, then (L_a ∪ M_a) + (L_a ∪ M_a) ⊆ L_a ∪ M_a

(iii)

Conversely, if the norm has the semi-group property, then it is strictly convex.

Thus, in the two-dimensional case, the semi-group property of a norm is equivalent with its strict convexity.

Proof. 

For the sake of a better understanding, the proof is divided into several steps. □

Step 1. There is no restriction on considering a = (0,1).

Indeed, if a = (a₁, a₂), we can choose a linear operator T such that Ta = (0,1). Next, we replace the norm ║.║ with the norm ║⋅║_T and use Proposition 3.

Step 2. The two-dimensional norms have the following useful property (although it is well known, we prove it here since we did not find appropriate references to it in our research and it can be used to construct many norms on the plane):

Lemma 1.

For any norm on a two dimensional space there exists a convex function f: ℜ → (0,∞), such that

$$\underset{t\to \infty }{\lim }\frac{f\left(t\right)}{t}=\underset{t\to -\infty }{\lim }\frac{f\left(t\right)}{-t}=m>0 \mathrm{and}$$

(4)

$$║(x,y)║=\{\begin{array}{cc}\left|x\right|f\left(\frac{y}{x}\right)& if\hspace{1em}x\ne 0\\ m\left|y\right|& if\hspace{1em}x=0\end{array}. \mathrm{Notice} \mathrm{that} m=║(0,1)║$$

(5)

Moreover, if X is strictly convex, then the function f is also strictly convex.

Conversely, for any convex function f satisfying (4), the equality (5) defines a norm.

All the norms in a two-dimensional space have this form. If f is strictly convex, then the norm given by (5) is also strictly convex.

For the proof of the Lemma, see Appendix A.

Combine Step 1 and Step 2. Thus a = (0,1) and f is a function, as in the above Lemma. Next, we drop the index a and write

L = {(x,y): x ≠ 0, $f\left(\frac{y}{x}\right)>f\left(\frac{y-1}{x}\right)$ or x = 0, m|y| > m|y − 1| ⇔ y ∈(½, ∞)},

M = {(x,y): x ≠ 0, $f\left(\frac{y}{x}\right)=f\left(\frac{y-1}{x}\right)$ or x = 0, m|y| = m|y − 1| ⇔ y = ½}

H = {(x,y): x ≠ 0, $f\left(\frac{y}{x}\right)<f\left(\frac{y-1}{x}\right)$ or x = 0, m|y| < m|y − 1| ⇔ y ∈(−∞, ½)}

Step 3. Describe the sets M, H, L.

Due to the property that $\underset{t\to \infty }{\lim }\frac{f\left(t\right)}{t}=\underset{t\to -\infty }{\lim }\frac{f\left(t\right)}{-t}=m>0$, the convex function f behaves as follows: there exists u₁ ≤ u₂ ∈ ℜ, such that

f is decreasing on (−∞, u₁)

(6)

f is constant on (u₁, u₂)

(7)

f is increasing on (u₂, ∞)

(8)

Note that if f is strictly convex, then u₁ = u₂. Let v = f(u₁) = f(u₂).

Lemma 2.

Let f: ℜ → ℜ be a continuous function which satisfies the assumptions (6), (7), (8). Suppose further that $\underset{t\to \infty }{\lim }\frac{f\left(t\right)}{t}=m_2>0,\underset{t\to -\infty }{\lim }\frac{f\left(t\right)}{-t}=m_1>0$

Let h(x,y) =$f\left(\frac{y}{x}\right)-f\left(\frac{y-1}{x}\right)$, h: ℜ* × ℜ → ℜ. The sets M = {h = 0}, L = {h > 0} and H = {h < 0} may then be characterized as follows:

There are exist two functions φ₁, φ₂: ℜ* → ℜ, φ₁ ≤ φ₂ such that

L = {(x,y): φ₁(x)∨φ₂(x) < y}

(9)

M = {(x,y): φ₁(x)∧φ₂(x) ≤ y ≤ φ₁(x)∨φ₂(x)}

(10)

H= {(x,y): φ₁(x)∧φ₁(x) > y}

(11)

Moreover,

φ_j(−x) = 1 − φ_j(x) ∀ x ∈ ℜ*, j = 1,2

(12)

$${\mathsf{\phi }}_j(0+0)=\frac{m_1}{m_1+m_2},{\mathsf{\phi }}_j(0-0)=\frac{m_2}{m_1+m_2}$$

(13)

$${\mathsf{\phi }}_1\left(x\right)={\mathsf{\phi }}_2\left(x\right) \forall x \in (-\frac{1}{u_2-u_1},\frac{1}{u_2-u_1})\backslash \left\{0\right\}$$

(14)

If f is strictly convex, then φ₁ = φ₂

(15)

$$\mathit{if}(x,y) \in M and(\mathit{tx},\mathit{ty}) \in M \mathit{for} \mathit{some} t > 1, \mathit{then} \mathit{either} x>\frac{1}{u_2-u_1} \mathrm{or} x<-\frac{1}{u_2-u_1}.$$

(16)

For the proof, see Appendix A.

In our case, m₁ = m₂ allows us to extend the mapping of φ_j to 0, defining φ_j(0) = ½.

Recall our claim: we want to prove that

y > φ₁(x)∨φ₂(x), y′ > φ₁(x′)∨φ₂(x′) ⇒ y + y′ > φ₁(x + x′)∨φ₂(x + x′)

(17)

This implication would surely hold if we could prove that φ_j are sub-additive.

Indeed, in this case, y > φ₁(x), y′ > φ₁(x′) ⇒ y + y′ > φ₁(x) + φ₁(x′) ≥ φ₁(x + x′) and, similarly, y + y′ > φ₂(x + x′).

Step 4. The functions φ_j are sub-additive

We shall use the following criterion for sub-additivity:

Lemma 3.

(i)

Let g:ℜ → ℜ be such that g(λs) ≤ λg(s) ∀ λ ≥ 1 ∀ s ∈ ℜ. Next, g(s + t) ≤ g(s) + g(t) ∀ s,t such that st ≥ 0.

If, moreover, λ > 1 ⇒ g(λs) < λg(s) ∀ s ∈ ℜ, then g(s + t) < g(s) + g(t) ∀ s,t such that st ≥ 0.

(ii)

Let g be a function, as before. Suppose that g satisfies the following symmetry property:

(S)

there is a ∈ ℜ, b > 0, such that g(a − s) + g(s) = b.

Consequently, g is sub-additive: g(s + t) ≤ g(s) + g(t).

If, moreover, λ > 1 ⇒ g(λs) < λg(s) ∀ s ∈ ℜ, then g(s + t) < g(s) + g(t) ∀ s,t

The proof in Appendix A.

Proposition 4.

For 1 < p < ∞, the functions φ_j defined by Lemma 2 are sub-additive.

Moreover, if X is strictly convex, they coincide and are strictly sub-additive.

Proof. 

Because of the fact that m₁ = m₂, these two functions are defined on the whole real line (φ_j(0) = ½) and they coincide on the interval [$-\frac{1}{u_2-u_1},\frac{1}{u_2-u_1}$]. The equality (12) demonstrates that the symmetry property (S) is fulfilled with a = 0, b = 1. Thus, we only need to prove that

φ_j(λx) ≤ λφ_j(x) for any x.

(18)

Check that for φ₁. Indeed,

-

if x < $-\frac{1}{u_2-u_1}$, then φ₁(x) = 1 + xu₂, φ₁(tx) = 1 + txu₂ < t + txu₂ = tφ₁(x);

-

if x > $\frac{1}{u_2-u_1}$, φ₁(x) = xu₂ ⇒ φ₁(tx) = tφ₁(x).

-

if x ∈ [$-\frac{1}{u_2-u_1}$,$\frac{1}{u_2-u_1}$], then y = φ₁(x) ⇔ (x,y) ∈ M. Then (tx,ty) ∈ H∪M due to Proposition 2(ii). However, (tx,ty) does not belong to M because of (16); hence, in this case, (tx,ty) ∈ H. According to (9) this means that ty > (φ₁∨φ₂)(tx) ⇒ tφ₁(x) > φ₁(tx).

In the same way, one proves that φ₂(tx) ≤ tφ₂(x) ∀ t ≥ 1.

If X is strictly convex, then if (x,y) ∈ M and t > 1, then (tx, ty) ∈ M ⇔ φ(tx) < ty = tφ(x); hence, Lemma 3 points demonstrates that φ is strictly sub-additive.

This ends the proof. □

The proof of Theorem 1 immediately follows due to Lemma 4.

Step 5. End of proof of Theorem 1

(i)

According to (9), (x,y) ∈ L ⇔ y > φ(x) with φ =φ₁∨φ₂. The functions φ_j are sub-additive; hence, φ is also sub-additive. Therefore y > φ(x), y′ > φ(x′) ⇒ y + y′ > φ(x) + φ(x′) ≥ φ(x + x′) ⇒ (x + x′, y + y′) ∈ L.

(ii)

If the space X is strictly convex, Cl(L) = L ∪ M. However, obviously

L + L ⊆ L ⇒ Cl(L) + Cl(L) ⊆ Cl(L).

(iii)

Suppose that X is not strictly convex. The unity sphere B = {z ∈ X: ║z║ = 1} contains a segment of line I = {(1 − λ)z₁ + λz₂: 0 ≤ λ ≤ 1} for some z₁ ≠ z₂ ∈ B. We suppose these two points to be extreme. Let T:ℜ² → ℜ² a linear operator such that Tz₁ = $\left(\begin{array}{c}1\\ 1\end{array}\right)$, Tz₂ = $\left(\begin{array}{c}1\\ -1\end{array}\right)$. Suppose that (X, ║⋅║) has the semi-group property. Consequently, (X, ║⋅║_T) also has it. The new unity sphere B_T = {z ∈ X: ║z║_T = 1} contains the segment {(1,t)|−1 ≤ t ≤ 1}. Let f(t) = ║$\left(\begin{array}{c}1\\ t\end{array}\right)$ ║ _T. This function is convex and f is decreasing on (−∞,−1), constant on (−1,1), and increasing on (1,∞). Thus, u₁ and u₂ from Lemma 2 are u₁ = −1, u₂ = 1. According to (10), z = (x,y) belongs to M if, and only if, (φ₁∧φ₂)(x) ≤ y ≤ (φ₁∨φ₂)(x). Moreover, we see that x ∈[½, ∞) ⇒ φ₁(x) = x, φ₂(x) = 1 − x and x ∈ (−∞, ½] ⇒ φ₁(x) = 1 + x, φ₂(x) = −x. It follows that there are many pairs of z,z′, such that z,z′ ∈ M but z + z′ = 0 ∈ L (for instance, z = (1,0) and z′ = (−1,0) are both in M). This contradicts the semi-group property: M + M should be included in L ∪ M.

To conclude: if a = T⁻¹e₂, then one can find z ∈ M_a such that −z ∈ M_a, as well.

Furthermore, we can prove the following.

Theorem 2.

The space X = ℜ² endowed with a strict convex norm satisfies the assumption (C)

Proof. 

We have to prove that the relations

║z║≥║z − a║, ║z′║≥║z′ − a║, ║z + z′║=║z +z′ − a║

(19)

for some z, z′ ∈ X can hold if, and only if, a = 0. Suppose, ad absurdum, that this is not true. Let a = (a₁, a₂) ≠ 0. There is a one-to-one and onto linear operator T: X → X, such that Ta = e₂ = (0, 1). Let w = T⁻¹z, w′ = T⁻¹z′. Consequently, (19) becomes

║w║_T ≥ ║w − e₂║_T,║w║_T ≥ ║w′ − e₂║_T,║w + w′║_T = ║w + w′ − e₂║_T

(20)

where ║⋅║ is the norm defined at Proposition 3.

If X is strictly convex, the norm ║⋅║_T is also strictly convex. Let w = (x,y), w′ = (x′,y′) and φ the function defined by Lemma 2. Therefore, (20) becomes

y ≥ φ(x), y′ ≥ φ(x′), y + y′ = φ(x + x′)

(21)

However, this is impossible, since y + y′ ≥ φ(x) + φ(x′) > φ(x + x′), as φ is strictly sub-additive. □

3. Conjectures, Open Problems, and Counterexamples

At a first glance, a possible conjecture would be that a strict convex normed space should have the mediator property. However, this not true: all the spaces L^p({1,2,3}) with 1 < p < ∞ are strictly convex and fail to have the property (M).

Proposition 5.

Let X = ℜ³ endowed with l^p norm, $\Vert x\Vert ={\left({\displaystyle \sum _{j=1}^3|x_j{|}^p}\right)}^{\frac{1}{p}}$.

(i)

If p ∈ (1,2) ∪ (2,∞), then X does not have the mediator property.

(ii)

However, all its proper subspaces have it (due to Theorem 1).

Proof. 

(i)

Case 1: p ∈ (1,2). We choose x = (3,t,−t), y = (3, −t, t), a = (4,4,4). Then

• x + y = (6, 0, 0), x − a = (−1, t − 4, −t − 4), y − a = (−1, −t − 4, t − 4), x + y − a = (2, −4, −4); hence, if t > 4, ║x║^p = ║y║^p = $3^p+2t^p$, ║x − a║^p = ║y − a║^p = 1 + (t − 4)^p + (t + 4)^p, ║x + y║^p = 6^p and ║x + y − a║^p = 2^p + 2⋅4^p. We claim that for any 1 ≤ p < 2, we can choose t > 4, such that ║x║ > ║x − a║and ║y║ > ║y − a║ but ║x + y ║ < ║x + y − a║. Indeed, this is an equivalent to 3^p − 1 > (t − 4)^p + (t + 4)^p − 2t^p but 6^p < 2^p + 2⋅4^p. The function g(t) =(t − 4)^p + (t + 4)^p − 2t^p has the property that g(∞) = 0 for any 1 ≤ p < 2; hence, for any fixed p ∈ [1,2), we can find a t = t(p), such that 3^p − 1 > g(t). As with the second condition, 6^p < 2^p + 2⋅4^p ⇔ 3^p < 1 + 2⋅2^p, it holds for any such p ∈ (1,2), since the function h(p) = 1 + 2⋅2^p − 3^p is decreasing on (1,2) and positive.

(ii)

Case 2: p ∈ (2,∞). Now, we choose x = (1,−1,2), y = (1,2,−1), a = (−t,t,t) for some t ∈ (0,1/2). Therefore, x + y = (2,1,1), x − a = (1 + t,−1 − t,2 − t), y − a = (1 + t,2 − t,−1 − t), x + y − a = (2 + t,1 − t, 1 − t); hence, ║x║^p = ║y║^p = ║x + y║^p = (2^p + 2),║x − a║^p = ║y − a║^p = 2(1 + t)^p+ (2 − t)^p and ║x + y − a║^p = (2 + t)^p + 2(1 − t)^p. We claim that for every p > 2, there is t ∈ (0,1/2), such that ║x║ > ║x − a║, ║y║ > ║y − a║ but ║x + y║ < ║x + y − a║. Indeed, let p > 2 be fixed and let f,g be defined by f(t) = 2^p + 2 − 2(1 + t)^p − (2 − t)^p, g(t) = (2 + t)^p +2(1 − t)^p − 2^p − 2. Note that f(0) = g(0) = 0 and that f′(0) = g′(0) = p(2^p−1 − 2). If p > 2, then 2^p−1 > 2; hence, the derivatives are positive. This means that for small t, we have f(t) > 0, g(t) > 0; this fact agrees with our claim. For p = ∞, it is even simpler, since now ║x║ = ║y║ =║x + y║ = 2,║x − a║ = ║y − a║ = 2 − t and ║x + y − t║ = 2 + t.

Thus, the two-dimensional space ℜ² endowed with the norm l^p, 1 < p < ∞ has the mediator property, while ℜ³ endowed with the same norm does not have it. □

4. Open Problems

• The only three-dimensional spaces that satisfy the mediator property (B) are inner-product spaces. Prove or disprove that if dim(X) ≥ 3, then property (M) implies the fact that X has an inner product. In other problems connected with perpendicular bisectors, this was indeed the case [5,6,7,8,9].
• The only examples of spaces possessing the mediator property also satisfy the property C. Prove or disprove that (M) ⇔ (C).
• Prove or disprove that if M + M ⊆ M ∪ L, then X has the property (M).