Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Question: Let $S$ be a convex weakly compact set in Banach space $H$. Propose a natural way to define the unique center $O \in S$.

Motivation: A lot! For example, in game theory $S$ can be a set of possible (fair) allocations, and we need to suggest a natural method to choose one.

Discussion. If S consists of 2 points, or line in $\mathbb R^2$, we have no natural way to select a center, thus $S$ should be convex and weakly compact. To define these properties, we need vector space and topology, thus the natural setting is Bahach space. If $H$ is $\mathbb R^n$, the natural choice is centroid (center mass), but to define it for general case, we need a natural notion of "uniform density" in a Banach space. Is this someting standard which I do not know? My main example is $H = L^1$, space of all intergable functions $[0,1]\to \mathbb R$. In this case if $S$ consists of all functions with range in $[a,b]$, the center should naturally be a constant function $f(x)\equiv (a+b)/2$. Also, $O$ should be tractable to compute, at least for such a simple examples of $H$ and $S$. A good axiomatic foundation ($O$ is the unique point saisfying axioms A1, A2, and A3) would be a plus.

share|improve this question
    
You might also ask the same question for norm-compact subsets of Banach spaces. The definition of centroid is somewhat more likely to exist in that more restricted case (but I still don't know how it would go). –  André Henriques Jun 14 '12 at 18:06
2  
In the L1 setting you can still define a Chebyshev center, but it may sometimes lie outside of your closed convex set, and it may not be unique. If you want the center to be preserved by isometries then likely the best that can be done is Theorem A in: Bader, Gelander, Monod: A fixed point theorem for L1 spaces. (springerlink.com/content/fv17437n8v104867) –  Jesse Peterson Jun 14 '12 at 23:40
1  
For my applications, it is absolutely crutial for the center to belong to the set. This is not the case in Theorem A you mention, as well as for Chebyshev center... –  Bogdan Jun 15 '12 at 8:10
add comment

2 Answers

I propose the axiom that any isometry between two such sets must take the center of one onto the center of the other. This axiom by itself is consistent with the existence of a center for every weakly compact convex set by the Ryll-Nardzewski fixed point theorem (we need the group of isometries of $S$ onto itself to always have a fixed point), and it alone already uniquely determines the center in some cases. For example, let $S$ be the positive part of the unit ball of $l^p({\bf Z})$ for $1 < p < \infty$. Translation (taking the sequence $(a_n)$ to the sequence $(a_{n-1})$) is an isometry of this set onto itself, and the only fixed point is the origin.

This example shows that the center will sometimes be an extreme point, which may be counterintuitive. But it's actually reasonable if you look at the center of mass of the positive part of the unit ball of $l^p_n$; as $n \to \infty$ it does converge to the origin (weakly, regarding $l^p_n$ as sitting inside $l^p$).

share|improve this answer
1  
Yes, this axiom makes sence. However, the non-uniqueness is very serious. Actually, for generic convex compact set S in R^2 there is no nontrivial isometry preserving S, thus any point can be a center! Theorem 1 in Teck-Cheong Lim: The center of a convex set (1981) ams.org/journals/proc/1981-081-02/S0002-9939-1981-0593489-7/… defines the unique center, and it satisfies your axiom. What I do not like is that 1) this construction uses transivite induction and thus can be hardly applied on practice; 2) in case of R^n it does not coincide with centroid. –  Bogdan Jun 15 '12 at 8:26
add comment

For the sake of readability, I am going to make this a separate answer. In response to my other answer, Bogdan points out that preservation under isometries need not determine the center. I suppose in most cases it wouldn't.

What I want to say here is that it doesn't seem possible to have a good infinite dimensional generalization of the centroid in ${\bf R}^n$, even for norm compact convex sets, for the following reason: already in finite dimensions we can have convex sets $A \subset B$ such that $B$ is contained in the $\epsilon$-neighborhood of $A$ but the centroid of $B$ is far away from the centroid of $A$. For instance, let $A$ be the line segment $[0,1]$ in ${\bf R}^1$ cross a ball about the origin of radius $\epsilon^2$ in ${\bf R}^n$, and let $B$ be the convex hull of $A$ together with the ball about the origin of radius $\epsilon$ in ${\bf R}^{n+1}$. If $n >> 1/\epsilon$ then every point of $B$ is close to $A$, but because of the way volume scales in large dimensions (the volume of a ball goes like its radius to the $n$) almost all of the mass of $B$ is near the origin. But the centroid of $A$ is halfway along the line segment.

We can use this phenomenon to construct a sequence of finite dimensional sets $A_1 \subset A_2 \subset \cdots$ in $l^2$ such that each $A_{n+1}$ is contained in the $2^{-n}$-neighborhood of $A_n$, but the centroid of $A_{n+1}$ is far from the centroid of $A_n$. We can make the centroid bounce back and forth on a line segment. So where should the centroid of ${\overline{\bigcup A_n}}$ be? It looks like we have to take the limit of a sequence that doesn't converge. There won't be any canonical way to do this.

I could add that there is a simple procedure that will sometimes work. For each finite dimensional subspace $V$ let $x_V$ be the centroid of $S \cap V$. Then consider the net $(x_V)$ with the finite dimensional subspaces ordered by inclusion. If this net converges, that seems like a good definition for the centroid of $S$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.