Equivalence of Gaussian measures Let $H$ be a separable Hilbert space and $N(0, C)$ and $N(0, D)$ be Gaussian measures on it. Further, for each $v \in H$, define $R_v = \frac{\left\langle v,Cv \right\rangle}{\left\langle v,Dv \right\rangle}$. Basically, $R_v$ is the ratio of Covariance of one dimensional Gaussian measures induced by $N(0, C)$ and $N(0, D)$ along $v$ respectively. Are the following statements equivalent-
1) $N(0, C)$ and $N(0, D
)$ are equivalent to each other.
2) $ 0 < \inf_{v \in H} R_v \leq \sup_{v \in H} R_v < \infty.$
 A: They are not equivalent.
For an explicit counterexample, let $\{e_1, e_2, \dots\}$ be an orthonormal basis for $H$, and let $C$ be the diagonal operator $C e_n = \frac{1}{n^2} e_n$.  Let $D =2C$.  Then $R_v = 1/2$ for every $v$ so (2) is satisfied.
Define random variables $X_n$ on $H$ by $X_n(v) = {n} \langle v, e_n \rangle$.  Then under $N(0,C)$, the $X_n$ are iid $N(0,1)$, and under $N(0,D)$ they are iid $N(0,2)$.  So by the strong law of large numbers, we have $\frac{1}{n} (X_1(v)^2 + \dots + X_n(v)^2) \to 1$ for $N(0,C)$-a.e. $v$, but likewise we have $\frac{1}{n} (X_1(v)^2 + \dots + X_n(v)^2) \to 2$ for $N(0,D)$-a.e. $v$.  So $N(0,C)$ and $N(0,D)$ are mutually singular and (1) fails.
In H. H. Kuo's book Gaussian Measures in Banach Spaces, section II.3, you can find the following theorem:

Theorem. $N(0,C)$ is equivalent to $N(0,D)$ iff there exists a bounded operator $T$ which is positive definite and invertible and such that $I-T$ is Hilbert–Schmidt, with $D = \sqrt{C} T \sqrt{C}$.  Otherwise, $N(0,C)$ and $N(0,D)$ are mutually singular.

(Kuo credits this result to Feldman–Hajek, but the reference given is to Varadhan's Stochastic Processes.)
This shows that we do have (1) implies (2), since if $D = \sqrt{C} T \sqrt{C}$ we can write $R_v = \frac{\|\sqrt{C} v\|^2}{\|\sqrt{T} \sqrt{C} v\|^2}$, and from that you can conclude $\|T\|^{-1} \le R_v \le \|T^{-1}\|$.
