Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
    
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

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.