TITLE: If Gaussian measures on a Hilbert space converge weakly to 0, how do their covariance operators converge?
QUESTION [6 upvotes]: Suppose we have a sequence of Gaussian measures $N(0, S(n))$ supported on a Hilbert space $H$ and we know that the sequence converges weakly to the delta measure at $0$, what are the necessary and sufficient conditions on convergence of the sequence of operators(trace class) $S(n)$? They should converge to $0$. Should the convergence be uniform, strong, weak?

REPLY [5 votes]: Denote the Gaussian random vectors by $X(n)$.
Clearly, $\mathrm{tr} \, S(n) = \mathsf{E} \, \Vert X(n) \Vert^2$, so $\mathrm{tr} \, S(n) \to 0$ is certainly sufficient for weak convergence to $0$. And in fact it's also necessary. Indeed, $\Vert X(n) \Vert^2$ is a quadratic functional of a Gaussian, and for those convergence in probability is known to be equivalent to convergence in $L^2$ (this is also true for any polynomials of bounded degree, cf. Corollary 6.11 of Janson "Gaussian Hilbert spaces").