A convex subset of a Banach space is closed if and only if it is weakly closed

Question

I'm looking for a proof that given $(X\textbf{ } \|\cdot\|)$ normed space, $M \subset X$ convex set, $M$ is weakly closed if and only if it's strongly closed as well.

Answer 1

For completeness, I will add a proof of this standard result. It relies on the following form of the Hahn-Banach separation theorem: Let $X$ be a real normed space (or generally, locally convex TVS), and suppose that $A\subset X$ is compact and convex, $B\subset X$ is closed and convex, and $A\cap B$ is empty. Then there exists a linear functional $\phi$ such that $\sup_A \phi <\inf_B \phi$.

Apply the above to a closed convex set $B$ and a one-point set $A=\{x\}$ disjoint from it. The functional $\phi$ provides a weakly open set containing $x$ and disjoint from $B$. Thus, $X\setminus B$ is weakly open, and therefore $B$ is weakly closed.

Answer 2

Hint: one direction is easy; the other uses the Hahn-Banach separation theorem.

Answer 3

Also a quite similar proof is to take a seuqence $y_n$ in $B$ weakly convergent to $y$, then we want to prove $y$ is in $B$. Suppose it is not, then by Hahn-Banach there exists a linear functional $\phi$ such that $\phi(y)<\inf \phi(y_n)$ and this is contradiction, cause if this happens then $y_n$ is not weakly convergent to $y$.