In the case that I'm working with a separable Hilbert space, $H$, on which I have a trace class operator, $K$, that's coming from a Gaussian (i.e., $K$ is self-adjoint, and for simplicity, has trivial kernel), how can I see the following two properties:


1. The Cameron-Martin space, defined as $K^{1/2}(H)$ in this case, is dense in $H$
2. The Cameron-Martin space is compactly embedded in $H$.



Properties of Cameron Martin Space
1) To see that $K^{1/2}(H)$ is dense in $H$: if not, there is some nonzero $v$ orthogonal to it. But since $K^{1/2}$ is self-adjoint, that says $0 = (K^{1/2})^\* v = K^{1/2} v$, and then $K v = K^{1/2} K^{1/2} v = 0$, violating your assumption that the kernel is trivial.


2) Since the embedding $K^{1/2}$ is compact, it is compactly embedded.


