If Gaussian measures on a Hilbert space converge weakly to 0, how do their covariance operators converge?

Question

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?

Answer 1

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").

Answer 2

Let $X$ be a separable Banach space, and let $\theta \mapsto \mathbb P_\theta$ be a weakly-continuous family of Radon probability measures. Suppose that each $\mathbb P_\theta$ admits a mean vector $\mu_\theta \in X$ and a covariance operator $k_\theta : X^* \to X$, in the sense that

$$\varphi(\mu_\theta) = \mathbb E_\theta[\varphi] \quad \mbox{and} \quad k_\theta = \int_X \varphi(x)x \, \mathbb P_\theta(\mathrm d x) - \varphi(\mu_\theta) \mu_\theta$$

for all continuous linear functionals $\varphi \in X^*$. The integrals are weak integrals (i.e., Pettis integrals). In your case, $X$ is a Hilbert space $H$, so $X^* \cong H$.

Under the weak condition of finite variance (i.e., $\operatorname{var}_\theta(\varphi) < \infty$ for all $\varphi \in X^*$), you can always represent a covariance structure as a covariance operator in this way (ref. Theorem 2.1, p. 3). Everything works fine for Fréchet spaces or more general locally convex spaces (ref. Theorem 3.ii of Vakhania-Tarieladze).

So now you can prove weak continuity of $\theta \mapsto \mu_\theta$ and $\theta \mapsto k_\theta$ without too much trouble. I'll do the case for the mean; you can write up the result for the covariance operator.

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Let $\varphi \in X^*$ be a continuous linear functional. Suppose that $\theta^n \to \theta$ is a convergence sequence. By tightness of the sequence of measures $\mathbb P_{\theta^n}$, for all $\epsilon > 0$, there exists a compact $K \subseteq X$ so that $\mathbb P_{\theta^n}(K) \ge 1 - \epsilon$.

Let $\varphi_K$ be a mollification of $\varphi$ which is continuous, agrees with $\varphi$ on $K$, and is compactly supported. Since $\varphi_K$ is continuous and bounded, $\mathbb E_{\theta^n}[\varphi_K] \to \mathbb E_\theta[\varphi_K]$.

Let $M < \infty$ denote the supremum of $\varphi$ on the (pre-)compact set $\operatorname{supp} \varphi_K - K$. Consequently, $\big| \mathbb E_\theta[\varphi] - \mathbb E_\theta[\varphi_K] \big| \le M \epsilon$. There is an identical estimate replacing $\theta^n$ with $\theta$.

Using the mean representation and the triangle inequality, we have $$\big| \varphi(\mu_\theta) - \varphi(\mu_{\theta^n}) \big| \le \big| \mathbb E_\theta[\varphi] - \mathbb E_\theta[\varphi_K] \big| + \big| \mathbb E_\theta[\varphi_K] - \mathbb E_{\theta^n}[\varphi_K] \big| + \big| \mathbb E_{\theta^n}[\varphi_K] - \mathbb E_{\theta^n}[\varphi] \big|.$$ The first and last terms are bounded by $M\epsilon$, and the middle term converges to $0$. Therefore the mean vector varies weakly continuously.

Try the same exercise for the covariance operator and let me know if you're having any trouble.