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

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.

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.

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

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 $$y$$.