# Local inverse bound of Cameron Martin and Banach norms

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 $$Y_Z:=\{u\in X: \|u\|_E. Then is there some $$C_Z$$ so that we have the reverse inequality on $$Y_Z$$?

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 $$\|v_n\|_E=Z/2 and $$\|v_n\|_E>n\|v_n\|_X,$$ so that $$\|v_n\|_X<\|v_n\|_E/n. 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 $$\|v\|_E and $$\|v\|_X.