Some examples of compact convex sets

Due to reasons (connected to incomplete von-neumann morgenstern orders if you must know) I’ve been thinking about the Krein-Milman theorem recently, and I realised that I had some misconceptions in my head that were obviously false when I thought about them for five minutes. Here are some examples to illustrate its boundaries.

The Krein-Milman theorem says that every compact convex subset of a locally convex vector space is the closed convex hull of its extreme points.

Consider \(\mathbb{R}^n\) with the normal euclidean norm. Then the unit ball \(\overline{B}(0, 1)\) is a compact convex set with infinitely many extreme points.

The same is true of any \(l^p\) norm with \(1 < p < \infty\), because such spaces are strictly convex.

It is not true for \(p = 1\) (whose extreme points are the points \(\pm e_n\)), or for \(l^\infty\) (whose extreme points are the set of points with \(x_i = \pm 1\)), both of whose unit balls have finitely many extreme points.

Moreover you cannot drop the “closed” part. Not every point is a convex combination of finitely many extreme points:

Consider \(A \subseteq l^\infty\) defined by \(A = \{x : |x_n| \leq \frac{1}{n}\}\). This is convex (because it’s a product of convex sets) and compact (it’s closed, so complete, and you can construct \(\epsilon\)-nets explicitly). Then the extreme points are the set of all points \(\{x : |x_n| = \frac{1}{n}\}\).

Let \(y = \sum\limits_n \lambda_i x_i\) be a finite combination of extreme points. Then \(n y_i\) can only take the \(2^n\) (and in particular finitely many) values \(\sum\limits_n \pm \lambda i\).

Consider \(y_n = 2^{-n}\). Then \(y \in A\) and \(n y_n\) takes infinitely many values and thus \(y\) cannot be a combination of finitely many extreme points of \(A\).

Note that it is true that every point in a compact convex set is a sum \(\sum\limits_{n=0}^\infty p_n x_n\) where \(\sum p_n = 1\) and the \(x_n\) are extreme points, you just need the sum to be infinite.

Edit: Actually I think it’s not, though I don’t yet have a counterexample. In general you may need a full integral over a probability measure on the extreme points. I’m going to have a thing about how to construct such.

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  1. Pingback: Another example of a compact convex set | David R. MacIver

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