If the dual of a topological vector space separates points, does it separate a point and a closed subspace?

Question

The Hahn-Banach Theorem implies that if $X$ is a normed vector space, then the dual space $X^*$, consisting of continuous linear functionals on $X$, has the following two properties:

1. $X^*$ separates points, i.e. if $x_1,x_2\in X$ with $x_1\neq x_2$, then there exists an $f\in X^*$ such that $f(x_1)=0$ and $f(x_2)=1$.

2. $X^*$ separates points from closed subspaces, i.e. if $Y$ is a closed subspace of $X$ and $x_0\in X$ with $x_0\not\in Y$, then there exists an $f\in X^*$ such that $f(Y)=\{0\}$ and $f(x_0)=1$.

But this answer shows that there are topological vector spaces which do not satisfy property 1. And property 2 clearly implies property 1, so such spaces satisfy neither one of the two properties.

But my question is, if a topological vector space satisfies property 1, does it necessarily satisfy property 2? To put it another way, are separating points and separating points from closed subspaces equivalent?

If not, does anyone know of a counterexample? It would have to be a topological vector space that isn't normable.

Answer 1

The Hardy space $H^p$ for $0<p<1$ provides a counterexample.

$H^p$ is a certain quasi-Banach space of analytic functions on the open unit disc $\{z\in\mathbb C\mid |z|<1\}.$ For $|z|<1$ the evaluation functional $\delta_z(f)=f(z)$ is continuous (e.g. see p. 39 of the reference below) so the continuous dual $(H^p)^*$ separates points. However, there is a closed subspace $SH^p\subsetneq H^p$ such that no non-zero functional in $(H^p)^*$ annihilates $SH^p.$ See Theorem 13 on page 53 here:

Shields, A.L., Duren, P.L., and Romberg, B.W. "Linear functionals on $H^p$ spaces with $0 <p<1$." Journal für die reine und angewandte Mathematik 238 (1969): 32-60.

They then show the stronger result that the quotient space $H^p/SH^p$ has a trivial continuous dual, and reference an earlier relevant result of Peck.