Does the strong law of Large Number hold for an infinite dimensional Brownian motion?

Question

For finite-dimensional Brownian motion $W_t$, it is well known that \begin{equation} \lim_{t\to \infty}\frac{W_t}{t}=0,\text{ a.s. }\ \ \ \ \hspace{1cm} \langle 1\rangle \end{equation}

Now suppose we are given an $L^2(\mathcal{D})$-valued Brownian motion $W_t$ defined by $$W_t:=\sum_{k=1}^{\infty}\sqrt{\sigma_k}W_t^k\phi_k(x),$$ where $\mathcal{D}$ is bounded domain in $\mathbb{R}^d$, $\{\phi_k(x)\}$ forms the complete orthogonal basis of $L^2(\mathcal{D})$, $\{W_t^k\}_{k\in \mathbb{N}^{+}}$ are mutually independent one-dimensional standard BM, and $\sigma_k$ satisfies $$\sum_{k=1}^{\infty}\sigma_k<\infty.$$

I wonder does $\langle1\rangle$ still holds for the infinite-dimensional BM introduced above?

Answer 1

Yes.

One way to prove it is using Fernique's theorem:

Let $\mu$ be a centered Gaussian measure on a separable Banach space $(X, \|\cdot\|)$. Then there exists $\alpha > 0$ such that $$\int_X e^{\alpha \|x\|^2} \,\mu(dx) < \infty.$$

Using Chebyshev's inequality, it follows that for all $r$,

$$\mu(\{x : \|x\| > r\}) \le C e^{-\alpha r^2}$$

where $C := \int_X e^{\alpha \|x\|^2} \,\mu(dx) < \infty$.

Let's replace $L^2(\mathcal{D})$ with an arbitrary separable Banach space $Y$. Now, the law of your Brownian motion $\{W_t : 0 \le t \le 1\}$ is a centered Gaussian measure on the separable Banach space $X = C([0,1]; Y)$ equipped with the sup norm. Fernique's theorem then tells us that

$$\mathbb{P}\left(\sup_{0 \le s \le 1} \|W_s\|_{Y} > r\right) \le C e^{-\alpha r^2}$$ for appropriate constants $C,\alpha$. Using the Brownian scaling, we thus have for any integer $n$, $$\mathbb{P} \left( \frac{1}{n} \sup_{0 \le s \le n} \|W_s\|_Y > \epsilon \right) = \mathbb{P} \left( \sup_{0 \le s \le 1} \|W_s\|_Y > \epsilon \sqrt{n} \right) \le C e^{-\alpha \epsilon^2 n}.$$

By the Borel-Cantelli lemma, it follows that $$\mathbb{P} \left( \frac{1}{n} \sup_{0 \le s \le n} \|W_s\|_Y > \epsilon \quad \text{i.o. $n$}\right) = 0$$ and letting $\epsilon \downarrow 0$ along a sequence, $$\frac{1}{n} \sup_{0 \le s \le n} \|W_s\|_Y \to 0, \quad \text{a.s.}$$

Noting that when $n-1 \le t \le n$, we have $\frac{\|W_t\|_Y}{t} \le \frac{1}{n-1} \sup_{0 \le t \le n} \|W_t\|_Y$, it follows that $\frac{\|W_t\|_Y}{t} \to 0$ almost surely. That is $\frac{W_t}{t} \to 0$ in $Y$-norm almost surely.

Answer 2

One can see it much easier by using the subadditive ergodic theorem, which is a standard tool for dealing with the law of large numbers.

Since the second moment of $W_t$ is obviously finite (this is precisely summability of $\sigma_k$), its first moment is also finite, so that the subadditive ergodic theorem implies that there exists a constant $C$ such that $\|W_t\|/t \to C$ a.e. and in $L^1$. In particular, $\|W_t\|^2/t^2\to C^2$ in probability. On the other hand, $\mathbb E\|W_t\|^2=t\,\mathbb E\|W_1\|^2$ grows linearly, so that $C=0$.