# Why is there no space whose dual is $C_\mathbb{R}[0,1]$? [duplicate]

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?

-

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

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 at 17:29