I have seen, as an application of the Krein-Milman theorem, that $\text{conv}( \text{ex}B)=B$, where $B$ is the unit ball of $L^p$ (on a compact space of $\mathbb{R}^n$), $p \in (1,\infty)$, $\text{conv}$ means convex hull, $\text{ex}$ are the extremal points.
However I am confused, do we really need to apply Krein-Milman here? Indeed, $\text{ex}B=S$, the unit sphere: this holds as $||\cdot||_p$ strictly convex. Also, $\text{conv}{S}=B$ in any normed space (that is non trivial).
Is there any fallacy in my reasoning?
@JustDroppedIn
Here is a roadmap: using Banach-Alouglu, one has that $B_{X^*}$ is $w^*$ sequentially compact, convex and non empty (provided the normed space $X$ is separable).
Now, $||x^*||:=\sum_n 2^{-n}x^*(x_n)$ is a norm on $X^*$, where $\{x_n\}_n$ form a countable dense subset of $B_X$. It has the properties that $B_{X^*}$ is compact, and that convergence of bounded sequences of functionals is equivalent to to $w^*$ convergence.
Applying Krein Milman with this new norm on $X^*$ you obtain $\overline{\text{conv}( \text{ex}B_{X^*})}=B_{X^*}$, the closure being taken in the new norm. However due to the properties above, that closure is exactly the $w^*$ closure.
And now one can substitute $X$ to $L^q$, where $q$ is the dual index of $p$.
Best Answer
The use of Krein-Milman to prove the result in $L^p$-spaces is a quite over-powered approach. As you show, it can be done much simpler.
I can see two applications of Krein-Milman: (1) as existence result: compact, convex sets do have extreme points, or (2) as prove of non-compactness: if $conv(ex\ B)$ is to small (e.g., empty) such that its closed (in any topology) convex hull can never be equal to $B$ then $B$ cannot be compact in any topology. This is the case for the unit balls of $L^1$, $C(K)$, $c_0$. So existence proofs that rely on compactness of bounded sets are doomed to fail in such spaces.