12
$\begingroup$

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.

Edit: (t.b.) A version of this question now is on MathOverflow and already has a few answers.

$\endgroup$
5
  • 1
    $\begingroup$ Also, de Branges's proof of the Stone-Weierstrass theorem comes to mind which is of course crucial for the theory of Fourier series. $\endgroup$
    – t.b.
    May 2, 2012 at 5:30
  • $\begingroup$ @t.b. That's a nice application, which deserves to be better known, but I would suggest that in the applications to Fourier series one may as well use other approximation theorems (Bernstein; Fejer) $\endgroup$
    – user16299
    May 2, 2012 at 6:21
  • $\begingroup$ That is indeed a good example, but it doesn't meet my "hard to achieve any other way" standard: after all, I believe you need to prove the Riesz representation theorem to get de Brange's argument off the ground. It's possible that I've simply set the bar too high. $\endgroup$
    – Paul Siegel
    May 4, 2012 at 18:40
  • $\begingroup$ What about Liapounov's convexity theorem on the range of a vector-valued measure? Or the so-called Birkhoff-von Neumann theorem on doubly-stochastic matrices? $\endgroup$
    – Etienne
    Jun 12, 2013 at 11:44
  • $\begingroup$ There's also this problem about dividing pizzas math.stackexchange.com/questions/2290402/… $\endgroup$
    – D.R.
    Apr 15 at 19:33

1 Answer 1

Reset to default
1
$\begingroup$

The Krein-Milman theorem is one way to prove De Finnetti's theorem: that every exchangeable sequence of random variables can be seen as a random draw among i.i.d. random variables.

The proof still involves the nontrivial step of showing that the i.i.d. distributions are the extreme points of that set, so it may not be as elementary as you want. It can also be proven in other ways; what's harder is ultimately subjective.

Nevertheless, I think it's a great way to illustrate the power of the theorem, because the statement itself is very easy to understand, and the result is surprising.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .