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This is a cross-post from MSE: http://math.stackexchange.com/questions/139754/elementary-applications-of-krein-milman. I'm starting to suspect that the question just doesn't really have a great answer, it's worth a try.

Recall that the Krein-Milman theorem asserts that a compact convex set in a LCTVS is the closed convex hull of its extreme points. This has lots of applications to areas of mathematics that use analysis: the existence of pure states in C*-algebra theory, the existence of irreducible representations of groups, the existence of ergodic measures...

I'm interested in applications of the theorem which are very easy to state but hard to achieve any other way. When I say "very easy to state" I mean the result should be expressible in the language of elementary Banach space or Hilbert space theory - no C*-algebras, representation theory, or measures. For an example of what I have in mind, the Krein-Milman theorem implies that C[0,1] is not the dual of any Banach space. If anyone knows an application of Krein-Milman to the theory of Fourier series, that would be ideal.

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Oops, this should have been a community wiki. –  Paul Siegel May 19 '12 at 22:27
    
Perhaps, it is also worth mentioning that a large part of the hard work towards applying KM can lie in identifying the extreme points of the convex set being considered. –  Suvrit May 20 '12 at 23:34
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The convexity of the range of a vector measure would be off-topic then I guess. Lindenstrauss proof (iumj.indiana.edu/IUMJ/FULLTEXT/1966/15/15064) was described by Halmos as "the slickest proof to end all proofs". –  Michael Greinecker Jan 11 '13 at 8:14
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4 Answers 4

Krein-Milman theorem can be used to prove the following: suppose we associate to each point of the integer lattice in the plane a real number in $[0, 1]$, such that the number corresponding to each point is the average of the numbers corresponding to the nearest four points of the lattice. Then all numbers must necessarily be equal.

Although the related problem for a finite numbers of points is trivial to prove by looking at the smallest number, this trick fails in the infinite version since there might not be a smallest number at all. But one can consider the unit ball of the space $L^{\infty}(\mathbf{Z} \times \mathbf{Z})$ and the shift operators acting on the space by shifting to the right or to the top. The condition in the problem can then be stated as an equality involving these operators and their inverses, and is not difficult to see that the set of functions in the space satisfying that condition is weak* compact (by using Alaoglu's theorem) and convex. It is also easy to see that extreme points are necessarily constant functions, and deduce from Krein-Milman the general case.

This same idea can be used to prove Liouville's theorem for harmonic functions in the plane, of which the previous problem was a sort of discrete version.

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That is very nice, thanks! –  Paul Siegel May 19 '12 at 22:26
    
Welcome! You can find this and other few applications of KM here: charlydif.wordpress.com/category/analisis-funcional The blog is not mine, though, and it is in Spanish, but I suppose the mathematics behind can be read without much effort. –  godelian May 19 '12 at 22:58
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Very cool argument! Is there a reference? –  Igor Rivin May 19 '12 at 23:12
    
Oops, you already gave a reference... –  Igor Rivin May 19 '12 at 23:13
    
A reference for that proof is the book "A Problem Seminar" by Donald J. Newman. –  Daniel m3 May 20 '12 at 13:07
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You mentioned you were interested in results like "the Krein-Milman theorem implies that $C[0,1]$ is not the dual of any Banach space". The follow results are closely related, so I hope they're not boring:

  1. If $\Omega\subseteq \mathbb{R}^d$, then $L^1(\Omega)$ is not the dual of any normed space since the closed unit ball of $L^1(\Omega)$ is convex with no extreme points. I'm sure you're familiar with the usual argument, but for interested readers who aren't: If $L^1(\Omega)$ were the dual of a normed space, then Alagolu's Theorem shows that the closed unit ball of $L^1(\Omega)$ is wk*-compact. But then by Krein-Milman, since the closed unit ball of $L^1(\Omega)$ is convex and (wk*-)compact, it admits an extreme point. But, since the closed unit ball of $L^1(\Omega)$ doesn't admit any extreme points, it must be the case that $L^1(\Omega)$ is not the dual of some normed space.

  2. It follows similarly that $C_0(\mathbb{R}^d)$, the family of continuous real-valued functions that vanish at infinity, is not the dual of any normed space. The closed unit ball of $C_0(\mathbb{R}^d)$ is convex with no extreme points. The proof of this is pretty slick (just add on small bumps when $f$ is small enough).

(Edit: If $\Omega\subseteq \mathbb{R}^d$ is open endowed with Lebesgue measure, then the closed unit ball of $L^1(\Omega)$ is convex with no extreme points)

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$L^1$ does have extreme points. Think about $|\Omega|=2$. –  Valerio Capraro May 20 '12 at 0:56
    
I should have mentioned the additional hypothesis that $\Omega$ is open and endowed with Lebesgue measure. Thank you. I've included the note in the main post as well. –  Adam Azzam May 20 '12 at 1:06
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One can prove by using repeatedly the Krein-Milman theorem that

  • If $T : C[0,1] \rightarrow X$ is an operator of norm at most one, with $T$ isometric on the $2$-dimensional subspace spanned by $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is an isometry.
  • If $T$ is an endomorphism of $C[0,1]$ of norm at most one, which fixes the functions $x \mapsto \cos(\pi x) $ and $x \mapsto \sin(\pi x)$, then $T$ is the identity operator.
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The Krein-Milman Theorem is used in the proof of Birkhoff's Theorem that the set of bistochastic matrices is the convex envelop of permutations matrices.

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That's a finite dimensional statement, and the finite-dimensional version of K-M is easy and precedes the K-M theorem by (at least) a few decades, I would think... –  Igor Rivin May 20 '12 at 18:26
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