Norm is weakly lower semicontinuous

Solution 1:

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 $$f(x) \leq \liminf_{n \to \infty} f(x_n).$$

Solution 2:

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.

Solution 3:

Let X be a normed space. We want to show that, for any λ≥0, the set {x∈X:∥x∥≤λ} is weakly closed.Firstly,the set {x∈X:∥x∥≤λ} is convex. Secondly, {x∈X:∥x∥≤λ} is closed in the norm topology.Hence,the result follows from the fact that a convex set's closure is the closure in the weak topology.