Let $(X,\tau)$ be a topological vector space such that the associated dual space $X^*$ is separable. Can we say that $X$ is separable ?
I know that this property is valid for Banach spaces but for topological vector spaces, I have no idea.
Let $(X,\tau)$ be a topological vector space such that the associated dual space $X^*$ is separable. Can we say that $X$ is separable ?
I know that this property is valid for Banach spaces but for topological vector spaces, I have no idea.
YCor has given a counterexample for topological vector spaces. The statement is still false for locally convex spaces. Consider the space $X$ defined to be a locally convex coproduct of $\newcommand{\R}{\mathbb{R}}\R$, $\R$-many times. So $X$ is the space of functions $\R \rightarrow \R$ that are zero except for on a finite subset of $\R$, equipped with the locally convex coproduct topology.
The dual space $X^* \cong \R^\R$ by the usual pairing of an element of $X$ with an element of $\R^\R$ by summation, and for the topology it doesn't matter whether we take the strong dual topology or the weak-* topology, either way we get the usual product topology on $\R^\R$. This space is separable, because the polynomial functions with rational coefficients $\R \rightarrow \R$ forms a countable dense subset.
However, $X$ itself is not separable. If $X$ were separable, we would have some countable set $D \subseteq X$ such that each $f \in \R^\R$ is determined by its values when paired with each $\phi \in D$. But since each element of $D$ has finite support, $D$ only determines countably many "coordinates" in $\R^\R$, so we could pick two $f_1,f_2 \in \R^\R$ agreeing on those elements of $\R$ but not being equal. This contradicts $D$ being dense.
Instead of an $\R$-fold coproduct we could have taken any $\kappa$-fold coproduct where $\aleph_0 < \kappa \leq 2^{\aleph_0}$, by the Hewitt-Marczewski-Pondiczery theorem.
Let $\Gamma$ be a set and consider the product measure $\mu^{\otimes \Gamma}$, where $\mu$ is the Lebesgue measure on the unit interval. For any $p\in (0,1)$, the space $L_p(\mu^{\otimes \Gamma})$ has trivial dual because $\mu^{\otimes \Gamma}$ is atomless. However, when $\Gamma$ is uncountable, the space $L_p(\mu^{\otimes \Gamma})$ is non-separable.