I am looking for an example of a normed space which is separable with respect to the topology induced by all continuous linear functionals, but not separable with respect to the norm topology.
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My first guess for a place to look would be $L^\infty([0,1])$; have you tried seeing if that works? – Ian Jun 29 '15 at 19:04
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Yes, I am pretty sure that a simple variation of the standard argument that $\ell^1$ is not the dual of $\ell^\infty$ shows that $\ell^\infty$ will not work. I didn't consider other $\infty$-spaces after that. – Jonas Dahlbæk Jun 29 '15 at 19:17
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I think that does not exist. Indeed, if
$X$ is a normed space, then $X$ is weak separable if and only if $X$ is norm separable.
Proof.
"$\Leftarrow$": Let $X$ be norm separable. Let $A$ be countable and norm dense in $X$. Then, we use the fact that the weak topology is indeed weaker than the norm topology to see that $\overline{A}^w\supset \overline{A}^{\|\cdot\|} = X$.
"$\Rightarrow$": Now, let $A$ be countable and weakly dense in $X$. Then, clearly we have that $C:={\text{span}} ( A)$ is weakly dense in $X$ and convex. Moreover, due to Mazur's theorem we have that $ \overline{C}^{\|\cdot\|}= \overline{C}^{w} =X$. Now, the rational span of $A$ will be a countable and dense set in the norm topology as $\overline {\text{span}_\mathbb{Q}} A = \overline{\text{span}} A$ in any LCS as addition and multiplication are continuous.
Peter
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Thank you for the quick answer! I think David's answer below fixes the problem by pointing out that the span with rational coefficients is norm dense in the full span, which is then convex and weakly dense. – Jonas Dahlbæk Jun 29 '15 at 19:38
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If $X$ is a Banach space then a subspace is norm-closed if and only if it is weakly closed. (This is a major difference between weak and weak*). Say $S$ is a weakly dense countable set. Let $V$ be the norm-closed span of $S$. Then $V$ is weakly closed, hence $V=X$. So the linear combinations of elements of $S$ with rational coefficients are dense in $X$, hence $X$ is norm-separable.
David C. Ullrich
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Ah, perfect, thank you. @Ian It is a classical theorem that norm closure and weak closure coincides for convex sets in locally convex spaces. Check out Mazur's lemma. – Jonas Dahlbæk Jun 29 '15 at 19:31