# If a subspace of $X^*$ separates points, is it weak-* dense?

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?

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If $X$ is reflexive, we can show that a continuous linear functional on $X^*$ which vanishes on $E$ is the identically vanishing functional. A corollary of Hahn-Banach theorem show that $E$ is dense in $X^*$. I guess the case non-reflexive should be more difficult. – Davide Giraudo Sep 2 '11 at 17:58
If $X$ and $Y$ are in duality via a non-degenerate bilinear form then $E \subset Y$ is $\sigma(Y,X)$-dense if and only if $E$ separates points of $X$. The argument of the non-trivial direction is the same as Robert's below. – t.b. Sep 3 '11 at 4:45

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.
For the sake of having a reference: the fact that the dual of the weak-* topology on $X^*$ is again $X$ appears as Theorem V.1.3 in Conway's A Course in Functional Analysis, 2e. I'm not sure it's quite "by definition" but it is pretty simple. The "separation theorem" alluded to is a consequence of Hahn-Banach in locally convex spaces and appears in Conway as Corollary IV.3.14. Thanks again, Robert! – Nate Eldredge Jul 19 '12 at 16:36