Representation of support of Gaussian measure by kernels of no-variance functionals Let $\mu$ be a Gaussian measure on a separable Banach space $X$ and $q$ is the covariance operator of $\mu$. I am reading a proof for
$$\operatorname {supp} \mu = \bigcap_{q(f, f) = 0} \ker f =: E$$
but there is a step I don't understand. So far I understand that the intersection is over an uncountable number of sets all having measure $1$ so the real issue is showing that the "measure $1$" property doesn't get lost in the uncountable intersection.
As we have separability, we can find a dense set $\{x_n\}$ of $E^c$, thus: $\forall n\in\mathbb N~ \exists f_n:~ q(f_n,f_n) = 0$ but $f_n(x_n) \neq 0$.
Now as $\{f_n\}_n$ is a subset of all of the linear functionals having the property $q(f,f) = 0$, we obviously have
$$E = \bigcap_{q(f, f) = 0} \ker f  \subset \bigcap_{n} \ker f_n $$
What I don't understand is the other inclusion, i.e.
$$\bigcap_{n} \ker f_n  \subset E$$
I was trying to show the contrapositive, i.e. that for any $y\in E^c$ we find a $f_n$ such that $f_n(y) \neq 0$.
Using separability, for any $\epsilon > 0$ we find a $x_m$ such that $\|y-x_m\| < \epsilon$. Then
$$f_m(y) = f_m(x_m) + f_m(y-x_m)$$
Now I know that the first term is non-zero (and I could have easily set $f_n(x_n) = 1$ above). But how can I be sure that the norms $\|f_m\|$ don't grow like $1/\epsilon$ such that both $f_m(x_m) = 1$ and $\|y-x_m\| < \epsilon$ but $f_m(y) = 0$?
 A: I think I might be to blame for this question.  It looks very similar to something I once wrote, with the same gap.  If so, sorry!
The result is true, but the approach described will not work.  We have to choose the $f_n$ with more care.  
(Indeed, suppose $x$ is outside the linear span of your $x_n$.  Note that $E$ is closed, hence so is $E+x$.  You could thus choose the functionals $f_n$ perversely using Hahn-Banach so that they all vanish on $E +x$; if it does in fact turn out to be the case that $\mu(E)=1$ then you'll still have $q(f_n, f_n)=0$.  Then $x$ is in $\bigcap_n \ker f_n$ but not in $E$.)
I think the following should work instead. Let $F = \{f \in X^* : q(f,f) = 0\}$.  Since $X$ is separable, the unit ball $B^*$ of $X^*$ is weak-* separable metrizable, hence so is $B^* \cap F$.  So pick your sequence $\{f_n\}$ to be weak-* dense in $B^* \cap F$.  Suppose $f_n(x) = 0$ for all $n$ and take any $f \in F$, renormalized so that $\|f\| \le 1$.  Pass to a subsequence so that $f_n \to f$ weak-*.  Then $0 = f_n(x) \to f(x)$.  Since $f$ was arbitrary we have $x \in E$.
