4

Possible Duplicate:
$C_0(X)$ is not the dual of a complete normed space
Is any Banach space a dual space?

While studying for a course of functional analysis I read somewhere that there is no normed vector space $X$ with $X^*=C_\mathbb{R}[0,1]$. I also found what at first glance seems like a complete proof of this fact:

Assume there is such a space $X$, then by Alaoglu's theorem the closed unit ball $B^*$ in $X^*$ is weak* compact. The unique extremal points of $B^*$ are the constant functions $f(x) = \pm 1$, and their closed convex hull is not all of $B^*$. This is a contradiction to Krein-Milman's theorem.

Now, I have a problem with this proof: to apply Krein-Milman to a set you need it to be convex and compact with respect to the topology induced by the norm, at least according to how it is usually stated. My hypothesis is that actually you can apply it to sets which are only compact in the weak* topology. Is this true? How do you prove it?

Community
  • 1
Daniel Robert-Nicoud
  • 30,663
  • 1
    It seems http://en.wikipedia.org/wiki/Krein-Milman_theorem contains the usual statement of the Krein-Milman theorem, valid for compact convex sets in locally convex spaces. The proof is the same as the one you probably know: 1. Zorn's lemma shows that every extremal set contains a minimal extremal set. 2. Hahn-Banach shows that a minimal extremal set is reduced to a point, hence there are extremal points. 3. If the closed convex hull of the extremal points were not everything, there would have to be another extremal point. – Martin Feb 02 '13 at 21:28
  • You can look at this question: http://math.stackexchange.com/questions/242034/is-any-banach-space-a-dual-space – Seirios Feb 02 '13 at 22:05

1 Answers1

2

Krein-Milman is a theorem about locally convex topological vector spaces. The weak and weak* topologies are topological vector spaces of this kind. To wit, you do not need norm compactness.

ncmathsadist
  • 50,127
  • I know this is an old post but I have a question about this: from what I understand the Krein-Milman theorem asserts that the convex hull is wk-dense, not norm-dense. So while the constant functions are not norm-dense in $C[0,1]$, they might be wk-dense right? Hence I do not see a contradiction, since we do not know anything about the wk* topology. – user2520938 Nov 23 '15 at 17:29
  • A finite-dimensional subspace of any Hausdorff topological vector space is closed. The constants are a 1-dimensional subspace, hence weak-*closed, and clearly they are not all of $C([0,1])$, hence not weak-* dense. – Nate Eldredge Jun 18 '19 at 21:33