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$C_0(X)$ 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?

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marked as duplicate by Seirios, Davide Giraudo, 5PM, Michael Greinecker, Ittay Weiss Feb 2 '13 at 22:55

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It seems 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 2 '13 at 21:28
    
You can look at this question: math.stackexchange.com/questions/242034/… –  Seirios Feb 2 '13 at 22:05
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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.

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