The support of a Banach Gaussian is the closure of its RKHS

First, $0\in \mathrm{supp}\mu$. Fix $\epsilon >0$. By [F2] choose compact $K$ with $\mu (K)>1\mathrm{/}2$ and cover it by finitely many balls of radius $\rho =\epsilon \mathrm{/}(2\sqrt{2})$. One ball $B(x,\rho )$ has positive $\mu$-measure. On $(B\times B,\mathcal{B}(B)\otimes \mathcal{B}(B),\mu \otimes \mu )$ let $W,W^{\mathrm{\prime }}$ be the independent coordinate copies. By [F5], this product sigma-algebra is the Borel sigma-algebra of $B\times B$; the maps $(x,y)\mapsto x-y$ and $x\mapsto \sqrt{2}x$ are continuous, so $W-W^{\mathrm{\prime }}$ and $\sqrt{2}W$ are Borel random elements. The event that both copies lie in $B(x,\rho )$ has positive probability and implies $\mathrm{\parallel }W-W^{\mathrm{\prime }}\mathrm{\parallel }<2\rho$. Choose a countable norming family using countable choice and HB from [A2]. For any finite subfamily, the coordinate vectors of $W,W^{\mathrm{\prime }}$ are independent centered multivariate normals with common covariance $\mathrm{\Sigma }$; [F6] makes their stack multivariate Gaussian, so its linear image giving the difference is multivariate Gaussian. The vector for $\sqrt{2}W$ is also multivariate Gaussian, and both vectors have covariance $2\mathrm{\Sigma }$. Thus [F3] makes their finite laws equal. Countable norming-law determination in [F3] gives . Hence $\mu (B(0,\epsilon \mathrm{/}2))>0$, and therefore $\mu (B(0,\epsilon ))>0$.

Suppose $x\mathrm{\notin }{\bar{H}}^B$. By [F4] choose $\mathrm{\ell }\in B^{\ast }$ with $\mathrm{\ell }(H)=0$ and $\mathrm{\ell }(x)\mathrm{\ne }0$. For every $m\in B^{\ast }$, $\mathrm{\ell }(Cm)=E[\mathrm{\ell }(W)m(W)]=0$, since $Cm\in H$. Taking $m=\mathrm{\ell }$ gives $E[\mathrm{\ell }(W{)}^2]=0$, so $\mathrm{\ell }(W)=0$ almost surely. Hence $\mu (\ker \mathrm{\ell })=1$. Because $\ker \mathrm{\ell }$ is closed and does not contain $x$, some open neighborhood of $x$ is disjoint from $\ker \mathrm{\ell }$ and has measure zero. Thus $x\mathrm{\notin }\mathrm{supp}\mu$.

For $h\in H$, [F1] gives equivalence of the law of $W-h$ with $\mu$. Thus its mass on $B(0,\epsilon )$ is positive by step 1.1. That mass is $\mu (B(h,\epsilon ))$, so every $h\in H$ lies in the support. Since the support is closed, ${\bar{H}}^B\subseteq \mathrm{supp}\mu$.

Step 2.1 gives ${\bar{H}}^B\subseteq \mathrm{supp}\mu$, and step 1.2 gives the reverse inclusion. Therefore the two sets are equal.