Finding extreme points of closure of convex hull Let H be a infinite dimensional Hilbert space with orthonormal basis $(e_n)_{n\geq 1}$. Let $f_N=N^{-\frac{1}{2}}\sum_{n=1}^Ne_n$ for all $N\geq 1$ and let K be the norm closure of the convex hull of $\{f_N : N\geq 1\}$. I need to show that the extreme points of K are $\{0\} \cup \{f_N:N\geq 1\}$. 
I have already shown that the $f_N$'s are extreme points, but I do not know how to show that 0 is an extreme point. 
I also can't figure out how to show that these are the only extreme points. I have shown that K is weakly compact, so following the converse to Krein-Milman i know that $\text{Ext}(K)\subset \overline{\{f_N:N\geq 1\}}^\tau$, where $\tau$ is the weak topology. But I do not know how to show that $\overline{\{f_N:N\geq 1\}}^\tau\subset \{0\} \cup \{f_N:N\geq 1\}$. One way could be to consider all possible nets in $\{f_N:N\geq 1\}$, but this has not yet proven fruitful. Any help is much appreciated
 A: Since $f_n$ converges weakly to $0$, we have $0$ is in the weak closure of the convex hull of $\{f_n\}$, and hence by a standard result, $0$ is contained in the strong closure of the convex hull, that is, $0\in K$.
Let $E$ denote the extreme points of $K$.
If $0$ were not in $E$ one could write $0$ as a non-trivial convex combination of distinct elements of $K$, that is, $0=pu+qv$ for $u,v\in K$ for some  $u\ne v$, and for some reals $p,q>0$ for which  $p+q=1$.  But each element of $K$ is contained in the intersection of the closed convex half-spaces $H_k=\{u\in H : \langle e_k,u\rangle\ge 0\}$.  So for each $k$ we have $u,v\in H_k$.  But that means $0=p\langle e_k,u\rangle +q\langle e_k,v\rangle$.  Since $0$ is an extreme point of the set of non-negative reals, this means that both $\langle e_k,u\rangle$ and $\langle e_k,v\rangle$ vanish. Since the $e_k$ form a basis, this implies $u=v=0$, contrary to supposition.
Let $F=\{f_n\}\cup\{0\}$; we have just seen that $F\subseteq E$.  Since we know $0\in K$, however, we see that the closed convex hull of $F$ is equal to $K$.  If  there was an extreme point $e\in E$  not in $F$, then there would be a contradiction: one could strictly separate $e$ from the closed convex hull of $F$ with a hyperplane.  But $K$ is at once the closed convex hull of $F$ and (by the Krein-Milman theorem) of $E$, so this cannot be.  
Or, more directly, using Proposition 1.5 in Phelps's Lectures on Choquet Theory, p.6

Suppose that $X$ is a compact convex subset of a locally convex space, that $Z\subset X$, and that $X$ is the closed convex hull of $Z$. Then the extreme points of $X$ are contained in the closure of $Z$.

where the LCS is $H$ with the weak topology, where $\{f_n\}$ plays the role of $Z$, $F$  plays the role of the closure of $Z$, and $K$ plays the role of $X$.
