I have (presumably non-degenerate) Gaussian $\mu$ over separable Banach space $X$. I would like to prove that for any ball of radius $r$ centered at $x$, $\mu(B_r(x))$. I know how to prove this in the Hilbert space case using the series representation of samples from $\mu$, but I am stumped on the Banach space case. I would be satisfied to understand the centered case ($\mu$ centered and $x=0$).
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$\begingroup$ I'm sure this is in Bogachev's Gaussian Measures. $\endgroup$– user479223Commented Apr 21 at 18:55
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$\begingroup$ Do you have a response to the answer below? $\endgroup$– Iosif PinelisCommented Apr 26 at 1:45
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1 Answer
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By (say) Lemma 5.1, the support of a mean zero Gaussian measure $\mu$ is the closure in $X$ of the reproducing kernel Hilbert space of $\mu$.
See the beginning of the proof of that lemma for a very simple proof of the fact that $\mu(B_r(0))>0$ for all $r>0$.
answered Apr 21 at 21:57