I'm reading Kuo's lectures notes, Gaussian measures on Banach spaces, and I'm having trouble justifying a few lines in a proof. $\mu$ and $\nu$ are Gaussian measures on a Hilbert space $H$, $\mu = [0, S]$ and $\nu = [a, S]$ - where $0$ and $a$ are the respective means and $S$ the covariance operator (assumed to be trace class and self adjoint) of the Gaussian measures.
$\mu_h$, $\nu_h$ are measures on $\mathbb{R}$ representing the pushforward measure of $\langle h, \cdot \rangle$ with respect to $(H, \mu)$ and $(H, \nu)$ respectively.
This is what I want to justify:
If $\mu$ and $\nu$ are not orthogonal measures (mutually singular), then $\mu_h$ and $\nu_h$ are not orthogonal to each other for any $h \in H$.
As $\mu$ and $\nu$ are Gaussian measures, by definition, $\mu_h$ is $N(0, \langle Sh, h\rangle)$ and $\nu_h$ in $N(\langle a, h \rangle, \langle Sh, h\rangle)$. Now, $\mu_h$ and $\nu_h$ are just Gaussian measures on $\mathbb{R}$, so they are mutually orthogonal only if $\langle a, h \rangle \neq 0$ and $\langle Sh, h \rangle = 0$. Trying to prove the constrapositive, if there is such an $h \in H$, this should mean that $\mu \perp \nu$. But I'm not sure how we get there from here.
EDIT: This part above is fine, as answered by Kavi Rama Murthy in the comments.
I still need to justify the following statement:
We have $ \sup_{|h| = 1} \frac{|\langle a, h \rangle|}{\sqrt{\langle Sh, h \rangle}} < \infty$
Given that $\mu_h$ is not orthogonal to $\mu_h$ for all $h$, this should be true. I don't know how this follows, it isn't obvious to me.
Note that this is a part of the proof showing that $\mu$ and $\nu$ are NOT orthogonal to each other implies $a \in \sqrt{S}(H)$. So, we can't use this fact here.
This is immediate from the definition of pushfoward measure. Suppose $\mu_h \perp \nu_h$. Then there is a Borel set $A$ such that $\mu_h(A)=0$ and $\nu_h(A^{c})=0$. Let $E=\{x: \langle h, x \rangle \in A\}$. Then $\mu(E)=\mu_h(E)=0$ and $\nu(E^{c})=\nu_h(E^{c})=0$, so $\mu \perp \nu$.