TITLE: Lim inf with norm and weak convergence
QUESTION [9 upvotes]: The following is an real analysis qualifying exam problem that I cannot solve:
Suppose $X$ is a Banach space and that $(x_n)$ converges weakly to $x$. Show that $\liminf ||x_n|| \geq ||x||$.
Using the Uniform Boundedness Principle I can show that $\sup_{n \in \mathbb{N}} ||x_n -x||$ is finite. Using Alaoglu's Theorem I can show that some subset of $(x_n)$ converges in norm to $x$. I feel like I am close with this but cannot seem to finish the problem.

REPLY [13 votes]: Take $\phi \in X^{*}$ such that $\|\phi \|_{\text{op}}=1$ and $|\phi (x)|=\|x\| $ (this is a basic application of the Hahn-Banach theorem)
then $\phi(x_n) \rightarrow \phi (x)$ so $|\phi (x_n)|  \rightarrow |\phi (x)| =\|x\|$
\begin{align}
\|x\|= |\phi (x)| &= \lim _{n \rightarrow \infty} |\phi (x_n) |\\
&\leq\liminf _{n \rightarrow \infty} |\phi (x_n) |\\
&\leq \liminf _{n \rightarrow \infty} (\|\phi\|_{\text{op}} \|x_n\| )\leq \liminf _{n \rightarrow \infty}  \|x_n\|
\end{align}