Let $X$ be a Banach space with a centered Gaussian measure $\mu\_0$. Let $E$ be the Cameron-Martin space of $X$. Let the respective norms be $\|\cdot \|\_X$ and $\|\cdot \|\_E$. It is well known (see prop 3.30 here <https://arxiv.org/pdf/0907.4178.pdf>) that


$$\|u\|\_X\leq C\|u\|\_E$$


for all $u\in X$. There is no reverse inequality in general because $E$ is a strictly smaller subset of $X$. But what about locally?


That is, consider the set $Y\_Z:=\{u\in X: \|u\|\_E<Z, \|u\|\_X<Z\}$. Then is there some $C\_Z$ so that we have the reverse inequality on $Y\_Z$?



Local inverse bound of Cameron Martin and Banach norms
If $X$ is infinite dimensional, then the inclusion $E\subset X$ is strict. So, the inequality
$$\|u\|\_X\le C\|u\|\_E \tag1$$
makes sense (and holds) only for $u\in E$, where $C$ is a positive real constant, depending on the measure $\mu\_0$.


Anyhow, the answer to your question (corrected in view of the above remark) is no, if $X$ is infinite dimensional. Indeed, then the inequality reverse to (1) does not hold. That is, for any (say) natural $n$ there is some $u\_n\in E$ such that
$$\|u\_n\|\_E>n\|u\_n\|\_X.$$
Take now any real $Z>0$ and let $v\_n:=(Z/2)u\_n/\|u\_n\|\_E$. Then $\|v\_n\|\_E=Z/2<Z$ and
$$\|v\_n\|\_E>n\|v\_n\|\_X,$$
so that $\|v\_n\|\_X<\|v\_n\|\_E/n<Z/n\le Z$. Thus, there is no real $C>0$ such that the inequality
$$\|v\|\_E\le C\|v\|\_X$$
holds for all $v\in E$ such that $\|v\|\_E<Z$ and $\|v\|\_X<Z$.


