In the Krein-Milman theorem, can the weak closure of the convex hull be replaced by norm-closure? I have a question on the following formulation of the Krein-Milman theorem:

Consider a vector space $X$ equipped with the  weak topology induced by a separating space $X^*$ of functionals on $X$. Then for each convex, compact subset $D$ of $X$, the convex hull of the extremal boundary $\partial D$ of $D$ is dense in $D$.

Now I want to apply this result to $X^*$, the dual of a Banach space $X$, equipped with the wk* topology. 
What I'm interested in is whether or not we can say that the convex hull of $\partial D$ is also norm-dense in $D$? Since the wk* topology is weaker then the norm topology on $X^*$ this is not a given, and I suspect it's generally not true, but I'm curious about this, and don't see a way of proving/disproving it.
The book I'm reading (by Pedersen) proves density by taking $x\in D\setminus\overline{\operatorname{convhull}(\partial D)}$. Then since $x$ is part of an open set in a LCS there is an open convex nbd $U$ of $x$ disjoint from $\overline{\operatorname{convhull}(\partial D)}$. Then there is a functional separating $U$ from $\overline{\operatorname{convhull}(\partial D)}$, from which a contradiction is derived.
However, there are more open nbds of $x$ in the norm topology then in the wk* topology, so although there cannot be a wk*-open nbd of $x$ in $D\setminus\overline{\operatorname{convhull}(\partial D)}$, there might be such norm-open nbds of $x$.
Again, I don't see why denisty in the norm topology should hold (it's not an exercise to prove this; it might very well not be true), but I'm wondering if there is something to be said here.
 A: The statement of the Krein-Milman theorem refers to the topology of $X$ twice: in "compact", and in "dense". The topology can be weak*, norm, or any topology making $X$ a locally convex TVS, but it needs to be the same on both occasions. If you assume compactness with respect to a weaker topology and take closure with respect to stronger one, the statement may fail. 
Example: let $X=C[0,1]$; then $X^*$ is the space of (complex) Radon measures on $[0,1]$. Let $D$ be the closed unit ball of $X^*$, which is weak* compact by Banach-Alaoglu. The extreme points of $D$ are of the form $c\delta_a$ where $|c|=1$ and $\delta_a$ is the Dirac mass at $a\in[0,1]$. The convex hull of extreme points consists of all finite linear combinations $\sum_k c_k \delta_{a_k}$ with $\sum |c_k|\le 1$. This is weak* dense in $D$, but is not norm-dense. 
Indeed, let $\mu$ be the Lebesgue measure on $[0,1]$. For any finite set $A=\{a_k\}$ there is a continuous function $f$ on $[0,1]$ that vanishes on $A$ and has $\int_0^1 f(x)\,dx\ge 1/2$. Indeed, $f(x)=\operatorname{dist}(x,A)^\epsilon$ will do for $\epsilon$ small enough. Hence, 
$$\|\mu - \sum_k c_k \delta_{a_k}\|\ge 
\int f\,d\mu - 0 \ge \frac12$$ which shows that $\mu$ cannot be approximated by $\sum_k c_k \delta_{a_k}$ in the norm.
