TITLE: If a subspace of $X^*$ separates points, is it weak-* dense?
QUESTION [13 upvotes]: Let $X$ be a Banach space, $X^*$ its dual.  Suppose $E$ is a linear subspace of $X^*$ which separates points (i.e. if $f(x)=0$ for all $f \in E$, then $x=0$).
Must $E$ be weak-* dense in $X^*$?
In all the examples I can think of, it is, but this seems too good to be true.
If not, does it help if $X$ is separable?  Reflexive?

REPLY [13 votes]: Yes, it is weak-* dense.  The weak-* continuous linear functionals on $X^*$ are, by definition, evaluation at the members of $X$.  If $E$ was not weak-* dense in $X^*$, then by the separation theorem in topological vector spaces there would be such a functional that was $0$ on $E$ and not identically $0$.  Since $E$ separates points, that is not the case.