Let $H$ be a Hausdorff topological vector space. A completely metrizable topological vector space $X \subseteq H$ is an (FH)-space if the topology of $X$ is finer than that induced by $H$ (that is the inclusion $X \subseteq H$ is continuous). By the closed graph theorem, any linear subspace of $H$ has at most one (FH)-space topology (see Wilansky, "Modern Methods in Topological Vector Spaces", Section 5.5). If $Y \subseteq X$ are two (FH)-spaces then the topology of $Y$ is finer than that induced by $X$ and it is strictly finer if and only if $Y$ is not closed in $X$.
As an example, if $K$ is compact Hausdoff and $H := \mathbb{R}^K$ the product space then the usual sup-norm topology on $C(K) \subseteq H$ is quite natural since it is the unique linear topology which is completely metrizable and such that all the evaluations $f \mapsto f(x)$, $x \in K$ are continuous.
Questions:
- Are there methods available to detect whether a given linear subspace $X \subseteq H$ can be turned into an (FH)-space? More generally, is there a known characterization of all (FH)-spaces $X \subseteq H$? I am particularly interested in $H = C[0,1]$.
As an example, no linear topology for the polynomials $\varphi \subseteq C[0,1]$ is completely metrizable due to the Baire category theorem ($\varphi$ is the countable union of its closed subspaces consisting of polynomials of degree $\leq n$).
- Let $H$ be a Banach space and $X \subseteq H$ a dense subspace that can not be turned into an (FH)-space. Then the restriction for $X$ to be completely metrizable is too strong. Can in this case $X$ still be equipped with a "nice" and in some sense (unique) canonical linear topology that is possibly "not too far away" from being completely metrizable? This is a rather vaguely formulated question.