Local inverse bound of Cameron Martin and Banach norms

Question

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 $Y_Z$?

Answer 1

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 $\|v\|_X<Z$.