I'm trying to solve problem 3.7 (2.) of "Functional analysis, Sobolev Spaces and PDE" by Brezis:
Let $E$ be a Banach space and let $A$, $B$ be two subsets of $E$ s.t. $A$ is weakly-closed and $B$ is weakly-compact.
Prove that $A+B$ is weakly-closed.
This is simple: let $a_k+b_k\in A+B$ a sequence such that $a_n+b_n\rightharpoonup x\in E$ (weakly), then $B$ is weakly compact, so $B$ is limited and so does $(b_n)$, so that it converges to a $b\in B$ up to a subsequence by weak compactness.
So, $a_n\rightharpoonup x-b\in A$ (since $A$ is weakly closed) and this concludes the proof, since $x=(x-b)+b\in A+B$.
Now, let $A$ and $B$ also be convex, nonempty and disjoint. Prove that there exists a closed hyperplane strictly separating $A$ and $B$.
I know that $A$ is strongly closed since it is convex and weakly closed and I know $A+B$ is strongly closed for the very same reason. I think I have to prove that $B$ is somehow strongly compact, in order to apply Hahn-Banach separation theorem, but I can't figure out how.
hint: If $x\not\in C$ and $C$ is a nonempty closed convex set, you can strongly separate $x$ and $C$. This strong separation defines a closed hyperplane. For a particular choice of $x$ and $C$, perhaps you can derive the strong separation of $A$ and $B$ (i.e. Jochen's comment of $0\not\in A-B$ is useful here).
Bauschke & Combettes' book (vol. 2), section 3.4 on separation is also a good reference, although this is for the Hilbert space setting.