17

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?

user90823
  • 173

3 Answers3

24

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

saz
  • 120,083
13

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.

Mike F
  • 22,196
  • 2
    The infimum in the inequality $\lim_i |\varphi(x_i)| \leq \inf_i |x_i|$ should be replaced by $\liminf_i$ because nothing prevents a few $x_i$ from having small norm. Another direct argument would be to write ${x \in X : |x| \leq \lambda} = \bigcap_{|\varphi| = 1} \varphi^{-1}([-\lambda,\lambda])$ and the right hand side is obviously weakly closed. – commenter Aug 18 '13 at 07:54
  • Sure, that's also fine, so +1 :-) As a matter of principle I'd slightly prefer stating it in the form $| x | \leq \liminf_{i} |x_i|$ for weakly convergent nets $x_i \to x$, because it's sharper at no cost and actually is an equivalent and useful form of stating weak lower semicontinuity. – commenter Aug 18 '13 at 08:37
  • Could you explain the application of Hahn Banach here? The set on the left hand side does not seem to necessarily be linear, and hence I don't see how Hahn Banach can be invoked. Thanks!!!

    Mark

     – Mark Oct 26 '17 at 03:23
  • @Mark: I guess technically what I'm using is a standard corollary of the Hahn-Banach theorem. See the 3rd bullet point in this wikipedia article: https://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem#Important_consequences – Mike F Oct 26 '17 at 13:16
  • Please, where is $|\varphi| = 1$ used? – Danilo Gregorin Afonso Aug 27 '20 at 14:10
  • 1
    @DaniloGregorin: It's used to get $\lim |\varphi(x_i)| \leq \sup |x_i|$. The assumption on $\varphi$ gives $|\varphi(x_i)| \leq |x_i|$. – Mike F Aug 27 '20 at 17:21
7

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.

ViktorStein
  • 4,838
David Lee
  • 1,072