TITLE: Does the strong law of Large Number hold for an infinite dimensional Brownian motion?
QUESTION [10 upvotes]: 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?

REPLY [8 votes]: 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 $C=0$.