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?