= Solution
{c}
The <Cameron-Martin space of a Gaussian measure> $\mu=\mathcal N(0,\Sigma)$ is
$$
\boxed{E=\operatorname{Ran}\Sigma^{1/2},\qquad
\langle h,g\rangle_E=\langle\Sigma^{-1/2}h,\Sigma^{-1/2}g\rangle_{\mathcal H}.}
$$
The inverse is defined on the range of $\Sigma^{1/2}$. In this infinite-dimensional setting “positive definite” must mean $\langle\Sigma h,h\rangle>0$ 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 $\Sigma e_j=\lambda_je_j$, $\lambda_j>0$, and $\sum_j\lambda_j<\infty$. (A strictly positive <trace-class operator> also forces the ambient <Hilbert space> to be separable.) In coordinates $h_j=\langle h,e_j\rangle$,
$$
E=\left\{h\in\mathcal H:\sum_j\frac{h_j^2}{\lambda_j}<\infty\right\},\qquad
\|h\|_E^2=\sum_j\frac{h_j^2}{\lambda_j}.
$$
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> is
$$
\boxed{\frac{d\mu_h}{d\mu}(x)=\exp\left(\ell_h(x)-\tfrac12\|h\|_E^2\right),\qquad
\ell_h(x)=\lim_{N\to\infty}\sum_{j=1}^N\frac{h_jx_j}{\lambda_j}.}
$$
The series has <mean-square convergence> and converges <almost surely> under $\mu$, since its independent summands have total <variance> $\sum_jh_j^2/\lambda_j$. Its law is $\mathcal N(0,\|h\|_E^2)$, so the density has <expected value> one. It is sometimes formally written $\ell_h(x)=\langle h,x\rangle_E$, 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>.
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