Determine the extreme points of the unit balls of $l^\infty$, and $C([0,1])$ for real-valued functions, with the uniform norm. Is $C([0,1])$ the dual of a Banach space?
I've found the extreme points of $C([0,1])$, but I'm not sure if it is the dual or not.
And can anyone help me with finding the extreme points of the unit balls of $l^\infty$?
Thank you.
Best Answer
I assume that you are interested in the space of all real-valued continuous functions defined on the real interval $[0,1]$. There are only two extreme points of its closed unit ball $B:=B_{C[0,1]}$, namely the constant function $1$ and the constant function $-1$. According to the Krein--Milman theorem, $C[0,1]$ cannot be isometric to a dual Banach space. To be precise, assume that $X$ is a Banach space whose dual is isometric to $C[0,1]$. Then $B\ (=B_{X^*})$ is the $w^*$-closed convex hull of its extreme points. This is false, since certainly $C[0,1]$ has dimension greater than $2$. Regarding the possibility that $C[0,1]$ should be isomorphic to a dual Banach space, the answer is a little bit more sophisticated (and again negative), as $C[0,1]$ is separable, and, if isomorphic to a dual space, then every nonempty closed, bounded and convex subset would be the closed convex hull of the set of all its strongly exposed points (all of them are extreme). This is a result of Namioka and Phelps.
Regarding $\ell^{\infty}$, observe that the only way a vector in $B_{\ell^{\infty}}$ cannot be located in between two other vectors in $B_{\ell^{\infty}}$ is that all its coordinates are $\pm 1$.