Small-ball ratios for Cameron–Martin translates of a Banach Gaussian

Put $A_r={\bar{B}}_B(0,r)$. By [F4], $\mu (A_r)>0$ because it contains the open ball of radius $r$ about the support point $0$. Let $({\mathrm{\ell }}_j)$ be the norming family from [F6], and define $A_{r,n}={\bigcap }_{j\le n}\{\mathrm{\mid }{\mathrm{\ell }}_j(W)\mathrm{\mid }\le r\}$. The sets decrease to $A_r$. For fixed $n$, the vector $(z_s,{\mathrm{\ell }}_1(W),\dots ,{\mathrm{\ell }}_n(W))$ is centered Gaussian: [F3] makes $z_s$ an $L^2$ limit of continuous-linear Gaussian coordinates, so adjoining the finitely many displayed coordinates gives a centered Gaussian limit. Its law is invariant under simultaneous sign reversal, and $A_{r,n}$ is invariant under that reversal. Hence $\mathbb{E}[e^{z_s}{\mathbf{1}}_{A_{r,n}}]=\mathbb{E}[e^{-z_s}{\mathbf{1}}_{A_{r,n}}],\mathbb{E}[z_s{\mathbf{1}}_{A_{r,n}}]=0.$ The variables $e^{\pm z_s}$ are integrable by the Gaussian moment formula in [F3]. Dominated convergence as $n\to \mathrm{\infty }$ therefore gives conditional symmetry and conditional mean zero on $A_r$. Thus $1\le {\mathbb{E}}_{\mu }[e^{-z_s}\mid A_r]\le e^{\mathrm{\parallel }s{\mathrm{\parallel }}_E^2\mathrm{/}2}\mathrm{.}$ For the lower bound, conditional Jensen applies. For the upper bound, [F3] applied to $-s$ gives ${\mathbb{E}}_{\mu }[e^{-z_s}{\mathbf{1}}_{A_r}]=e^{\mathrm{\parallel }s{\mathrm{\parallel }}_E^2\mathrm{/}2}\mu ({\bar{B}}_B(is,r)),$ and [F5] bounds the last probability by $\mu (A_r)$.

For any $h\in E$, choose ${\mathrm{\ell }}_n\in B^{\ast }$ with $h_n:=C{\mathrm{\ell }}_n\to h$ in $E$, possible by the completion in [F2], and put $s_n=h-h_n$. The Cameron–Martin formula [F3], applied to the translate by $-h$, gives $\frac{\mu ({\bar{B}}_B(ih,r))}{\mu (A_r)}=e^{-\mathrm{\parallel }h{\mathrm{\parallel }}_E^2\mathrm{/}2}{\mathbb{E}}_{\mu }[e^{-z_h}\mid A_r]\mathrm{.}$ Since $z_{h_n}={\mathrm{\ell }}_n(W)$ almost surely and $z_h=z_{h_n}+z_{s_n}$, the bound $\mathrm{\mid }{\mathrm{\ell }}_n(W)\mathrm{\mid }\le r\mathrm{\parallel }{\mathrm{\ell }}_n{\mathrm{\parallel }}_{B^{\ast }}$ on $A_r$ and step 1.1 imply $e^{-r\mathrm{\parallel }{\mathrm{\ell }}_n{\mathrm{\parallel }}_{B^{\ast }}}\le {\mathbb{E}}_{\mu }[e^{-z_h}\mid A_r]\le e^{r\mathrm{\parallel }{\mathrm{\ell }}_n{\mathrm{\parallel }}_{B^{\ast }}}{e}^{\mathrm{\parallel }{s}_n{\mathrm{\parallel }}_E^2\mathrm{/}2}\mathrm{.}$ For fixed $n$, let $r\downarrow 0$; then let $n\to \mathrm{\infty }$, so $\mathrm{\parallel }{s}_n{\mathrm{\parallel }}_E\to 0$. The two bounds show ${\mathbb{E}}_{\mu }[e^{-z_h}\mid A_r]\to 1$.

Applying step 2.1 separately to $h_1$ and $h_2$ yields $\frac{\mu ({\bar{B}}_B(i{h}_1,r))}{\mu ({\bar{B}}_B(i{h}_2,r))}=e^{(\mathrm{\parallel }{h}_2{\mathrm{\parallel }}_E^2-\mathrm{\parallel }{h}_1{\mathrm{\parallel }}_E^2)\mathrm{/}2}\frac{\mathbb{E}[e^{-z_{h_1}}\mid A_r]}{\mathbb{E}[e^{-z_{h_2}}\mid A_r]}\mathrm{.}$ Both conditional expectations tend to $1$, so the ratio has the stated limit. Positivity of the denominator follows from [F4], since each closed ball contains the corresponding open neighborhood of a support point.