Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\ell^\infty$ be the Banach space of bounded sequences with the usual norm. and let $\ell_0(x) = \lim_{n \rightarrow \infty} x_n$, for convergent sequences. Show that the sett L consisting of all continuous extensions of $\ell_0$ to $\ell^\infty$ is closed in the weak* topology on $(\ell^\infty)'$.

The extension is done by Hahn-Banach, but how do I show that something is closed in the weak*? Can I use sequentially closed here? are the topology metriceble? I can show that this is not a Hilbert space. Is it reflexive? Im a little bit unsure about all this weak* stuff, Please help me out and merry christmas!

share|cite|improve this question
up vote 2 down vote accepted

To show $L$ is weak-* closed, you want to show that its complement is weak-* open, i.e. any $\phi \in (\ell^\infty)' \backslash L$ has a weak-* neighborhood disjoint from $L$. In fact, if $\phi \in (\ell^\infty)' \backslash L$ there is some sequence $s$ that converges to a limit $m$ that is not $\phi(s)$. Consider $\{\psi \in (\ell^\infty)': |\psi(s) - \phi(s)|<\epsilon\}$.

share|cite|improve this answer
Great, I did not know this was the way to solve it, but how do I know that the set is open? Since the $\psi$ do not converge to m either? – Johan Dec 24 '12 at 22:49
What is the definition of the weak-* topology? – Robert Israel Dec 24 '12 at 23:07
The topology that makes all the linear functionals on the form $x(u) = u(x)$ continuous? – Johan Dec 25 '12 at 9:27
Not quite: it's the weakest topology that does that. But $\psi \to \psi(s)$ being continuous is all you need to show that $\{\psi: |\psi(s) - \phi(s)|<\epsilon\}$ is open. – Robert Israel Dec 25 '12 at 20:35
The continuity of $\psi$ is not an issue here. The map $\psi \to \psi(s)$ is continuous (because that's what the weak-* topology says is continuous), so we are done. – Robert Israel Dec 25 '12 at 21:22

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.