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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!

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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 $\{\psi \in (\ell^\infty)': |\psi(s) - \phi(s)|<\epsilon\}$.

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    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?2012-12-24
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    What is the definition of the weak-* topology?2012-12-24
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    The topology that makes all the linear functionals on the form $x(u) = u(x)$ continuous?2012-12-25
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    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.2012-12-25
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    but $\psi$ is continuous right? so we are done?2012-12-25
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    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.2012-12-25
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    I dont get any of that, how does continuity of $s^*:(l^\infty)'\to \mathbb{R}$ at $\psi$ prove that $\psi\not\in L$?2012-12-31
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    $\{\psi \in (\ell^\infty)': |\psi(s) - \phi(s)| < \epsilon\} = \{\psi \in (\ell^\infty)': s^*(\psi) \in (\phi(s)-\epsilon, \phi(s) + \epsilon)\}$ is an open set in the weak-* topology. If $|m - \phi(s)|>\epsilon$, where $m = \lim_n s_n$, that implies $\psi(s) \ne m$, so $\psi \notin L$.2012-12-31
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    @RobertIsrael, THX2013-01-02