Closed ball is weakly closed The problem is in a Banach space, if $||x_n||\leq 1$ and $x_n\to x$ weakly, then $||x||\leq 1$
This question has an answer here: math.stackexchange.com/questions/714049/closed-unit-ball-in-a-banach-space-is-closed-in-the-weak-topology
But for convenience, I will repost the answer.
"If $x_n \to x$ weakly then we have that $\lambda x_n \to \lambda x$ for all $\lambda \in V^*$and $|\lambda x_n| \leq \|\lambda\| \|x_n\|$. Dividing both sides by $\|\lambda\|$ gives $$ \frac{|\lambda x_n|}{\|\lambda \|} \leq \|x_n\|.$$ Taking $n \to \infty$ and substituting in $\|x\| = \sup_{\lambda \in V^*} \frac{|\lambda x|}{\|\lambda \|}$ gives $$\|x\| \leq \liminf_{n \to \infty} \|x_n\|.$$Then any limit $x$ of $x_n$ with $\|x_n \| \leq 1$ for all $n$ will necessarily have $\|x\| \leq 1$. "
My question is, how does "liminf" appears? My understanding is that when we take $n\to \infty$, the left side is $||x||$, but would the right side be $\lim ||x_n||$?

Weak convergence does not guarantee existence of $\lim \|x\|$. So we cannot take limit on both sides of the inequality $|\lambda x_n | \leq \|\lambda\| \|x_n\|$. But we can always take $\lim \inf $ or  $\lim \sup$. When you  take $\lim \inf$ on both sides LHS becomes $|\lambda x|$ because if  a sequence is convergent then its $\lim \inf$ is same as its limit.
