TITLE: If $x_n \rightarrow x_0$ weakly then show that lim inf $\|x_n\| \ge \|x_0\|$
QUESTION [8 upvotes]: How does one show that if $x_n \to x_0$ weakly, then $\liminf ||x_n|| \geq ||x_0||$?
I'm just doing additional problems from a book to try and prepare for a final. Unfortunately, I have really gotten nowhere with this one, so I'm afraid I just have to ask for help from the get go. Any suggestions?

REPLY [9 votes]: Suppose $x_n \to x$ weakly. Fix $x^* \in X^*$. Then
$$
 |x^*(x)| = \lim_{n \to \infty}  | x^*(x_n)| \leq \liminf_{n \to \infty}  ||x^*||\ ||x_n|| = ||x^*||\liminf_{n \to \infty} ||x_n|| .
$$
Now take the supremum of both sides over all $x^* \in X^*$ with $||x^*||\leq 1$, giving
$$
||x|| \leq \liminf_{n \to \infty}||x_n||.
$$
(Recall that $||x|| = \sup \{|x^*(x)| : x^* \in X^*, ||x|| \leq 1\}$; this follows quickly from the Hahn-Banach theorem. This is not definition, it is a (very useful) theorem which one proves after one has Hahn-Banach at one's disposal).
Suggestion: To see if you really understand this proof, try this similar problem: If $x_n^* \to x^*$ weakly$^*$, then $||x^*|| \leq \liminf_{n \to \infty} ||x_n^*||$.