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 $f\_m(y) = 0$?



Representation of support of Gaussian measure by kernels of no-variance functionals
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 $x \in E$.


