Weak convergence in a subspace Let $V$ be a normed linear space and $W$ a closed subspace of $V$. Suppose a sequence $\{w_{n}\} \subset W$ and $w \in W$ with $w_{n}$ converges to $w$ weakly in $V$. Why must $w_{n}$ converge weakly to $w$ in $W$?
 A: One way to see this is by arguing that the weak topology on $W$ is just the subspace topology $W$ inherits from $V$ (with its weak topology).  To see this, note that a basic open set in the weak topology for $W$ is $\phi^{-1}(U)$ for $\phi \in W^*$, $U \subseteq \mathbb{C}$ open (where $W^*$ is the continuous dual space of $W$, of course).
So, consider any such $\phi:W \to \mathbb{C}$ and open subset $U \subseteq \mathbb{C}$.  Note that for all $w \in W$, $|\phi(w)| \le \|\phi\| \|w\|$.  Thus, by the Hahn-Banach theorem, $\phi$ extends to a linear functional $\Phi: V \to \mathbb{C}$ with $|\Phi(v)| \le \|\phi\| \|v\|$ for all $v \in V$, and so $\phi^{-1}(U) = \Phi^{-1}(U) \cap W$ is open in the subspace topology that $W$ inherits.
Conversely, a basic open set in the subspace topology is $\Psi^{-1}(U') \cap W$ for some $\Psi \in V^*$, $U' \subseteq \mathbb{C}$ open; now, $\Psi^{-1}(U') \cap W = (\Psi|W)^{-1}(U')$ is an open set in the weak topology on $W$, and the restriction $(\Psi|W)$ is a continuous linear functional on $W$, i.e. $(\Psi|W) \in W^*$.  Thus, each basic set $\Psi^{-1}(U') \cap W$ for the subspace topology is in fact open in the weak topology on $W$.
So, the weak topology on $W$ is in fact the subspace topology that $W$ inherits from $V$ (with its weak topology).  So, if $W \owns w_n \to w \in W$ in the weak topology on $V$, then $w_n \to w$ in the subspace topology that $W$ inherits, and hence $w_n \to w$ in the weak topology on $W$.
As an aside, I was hoping to find a proof that didn't rely on the Hahn-Banach theorem, but none came to mind.  Perhaps someone else can find a proof without using such powerful theorems?
