Norm is weakly lower semicontinuous Is it the case that for every normed space, the norm is always weakly lower semicontinuous? Does it also hold for topologies other than the weak one?
 A: Let $X$ be a normed space. We want to show that, for any $\lambda \ge 0$, the set $\{ x \in X: \| x \| \le \lambda \}$ is weakly closed.
Firstly, the set $\{x \in X: \| x \| \le \lambda \}$ is convex. Secondly, $\{x \in X: \| x \| \le \lambda \}$ is closed in the norm topology.
Hence the result follows from the fact that a convex sets closure is the closure in the weak topology.
A: One can show that for a given real-valued function $f$ the equivalence $$f \, \text{convex, lower semicontinuous} \, \Leftrightarrow \, f \, \text{convex, weakly lower semicontinuous}$$
holds. Since the norm $f(x) := \|x\|$ is convex and continuous (by the triangle inequality), the claim follows.
Moreover, for any topology $\mathcal{S}$ finer as the weak topology $\mathcal{T}$, $\mathcal{T}$-lower semicontinuity implies $\mathcal{S}$-lower semicontinuity right from the definition: Let $(x_n)_n$ a sequence such that $x_n \to x$ in $(X,\mathcal{S})$, then $x_n \to x$ in $(X,\mathcal{T})$ since $\mathcal{T} \subseteq \mathcal{S}$. Consequently, by the $\mathcal{T}$-lower semicontinuity $$f(x) \leq \liminf_{n \to \infty} f(x_n).$$
A: Here's a direct argument. Let $X$ be a normed space. We want to show that, for any $\lambda \geq 0$, the set $\{ x \in X : \|x\| \leq \lambda\}$ is weakly closed. Let $x_i$  be a net in this set converging weakly to $x \in X$. By Hahn-Banach, there is a $\varphi \in X^*$ with $\|\varphi\| = 1$ such that $\varphi(x) = \|x\|$. By weak convergence,
$$\|x\| = \varphi(x) =\lim_i \varphi(x_i) = \lim_i |\varphi(x_i)| \leq \sup_i \|x_i\| \leq \lambda$$
so the claim holds.
