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 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 infnite-dimensional BM introduced above?

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?

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

Now suppose 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 $<1>$ still holds for the infnite-dimensional BM introduced above?

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 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 infnite-dimensional BM introduced above?