I am wondering if the Brownian motion is unifomly continuous? I suspect it isn't, but I am not sure how to show this. Specifically I am wondering if
$$P(\{\omega: B_t(\omega)\text{ is uniformly continuous on }[0,\infty)\})=0,$$
is this statement true? Since the brownian motion is continuous and all continuous functions are uniforlmy continuous on compact intervals, something has to happen when $t$ goes to infinity in order for this statement to be true.
attempt
My idea to show this was to prove that for any given $\delta$ we have
$$P(\exists n\text{ such that }|B_{n+\delta}-B_n|>1)=1.$$
In order to prove this I was thinking about looking at the random variable $\sum_{n=1}^K|B_{n+\delta}-B_n|/K$ and noting that all the terms are independent and distributed as $|B_\delta|$. However, this idea didn't lead anywhere.
Do you have any idea on how to prove this?
Consider $X_k = B_{\delta(k+1)} - B_{\delta k}$. It is a collection of iid $\mathscr N(0, \sqrt \delta)$ random variables. You can then easily compute probability that $\{|X_k| \leq 1, k = 1..n\}$ for each $n$, and note that it goes to $0$ as $n\to\infty$. As a result, you get that $\mathsf P(|X_k| \leq 1 \quad \forall k) = 0$ which means that the complementary probability (something you're after) is $1$.