 Solution

The Cameron-Martin space of a Gaussian measure $\mu =\mathcal{N}(0,\Sigma )$ isThe inverse is defined on the range of ${\Sigma }^{1/2}$. In this infinite-dimensional setting “positive definite” must mean $⟨\Sigma h,h⟩>0$ for every nonzero $h$, rather than a uniform lower bound: a trace-class operator cannot be uniformly positive on an infinite-dimensional Hilbert space.

By the spectral theorem for compact Hermitian operators, choose an orthonormal basis $(e_j)$ with $\Sigma e_j={\lambda }_j{e}_j$, ${\lambda }_j>0$, and ${\sum }_j{\lambda }_j<\infty$. (A strictly positive trace-class operator also forces the ambient Hilbert space to be separable.) In coordinates $h_j=⟨h,e_j⟩$,This is a Hilbert space with its indicated norm, even though its range need not be closed in the ambient norm.

The Cameron-Martin theorem for a Gaussian measure states that the translated law ${\mu }_h(B)=\mu (B-h)=\mathcal{N}(h,\Sigma )(B)$ is an equivalent probability measure to $\mu$ exactly when $h\in E$. For such $h$, its Radon-Nikodym derivative isThe series has mean-square convergence and converges almost surely under $\mu$, since its independent summands have total variance ${\sum }_j{h}_j^2/{\lambda }_j$. Its law is $\mathcal{N}(0,\Vert h{\Vert }_E^2)$, so the density has expected value one. It is sometimes formally written ${\ell }_h(x)=⟨h,x{⟩}_E$, but a typical infinite-dimensional Gaussian sample is not in $E$; the series interpretation is essential. If $h\notin E$, the two laws are mutually singular measures.