Example of a weakly separable normed space that is not norm separable

Question

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.

Answer 1

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.

Answer 2

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.