TITLE: Show that the closures in the topology weak and the norm are coincide:
QUESTION [5 upvotes]: Let $X$ be a Banach space and $Y \subset  X$a vector subspace. Let $Y_f$ and $Y_F$ are the closures of $Y$ in the topology weak and the norm, respectively. Prove that $Y_F = Y_f$.

REPLY [6 votes]: If $x$ is not in $Y_F$, there is a member $f^*\in X^*$ such that $f^*(x)=1$ and $f(Y_F)=\{0\}$. The weak nhood of $x$ determined by $f^*$ and $\epsilon=1/2$ contains $x$ and is disjoint from $Y_F$. It follows that $x$ is not in $Y_f$. So the weak closure of $Y$ is contained in the norm closure of $Y$.
The reverse inclusion is obvious.