Let $(B(t))_{t\geq0}$ be a standard Brownian Motion. Show $$\lim_{t\to\infty}\frac{B(t)}{t}=0\ a.s.$$
A hint is given by $\sup_{0\leq s\leq1}B(s)\overset{d}{=}|B(1)|$.
I only know it holds $\lim_{n\to\infty}\frac{B_n}{n}=0$ a.s. because of the law of large numbers. Now let $t\in[n,n+1)$, then I tried to estimate
$$\left | \frac{B_t}{t} - \frac{B_n}{n} \right | \leq\frac{\sup_{s\in[n,n+1)}|B_s-B_n|}{n}.$$ It is $\sup_{s\in[n-1,n)}|B_s-B_n| \overset{d}{=} \sup_{s\in[0,1]}|B_s|$ but I don't know how to use the hint.
Maybe if I could show $\mathbb{E}(\sup_{s\in[n-1,n)}|B_s-B_n|)<\infty$ for every $n\in\mathbb{N}$, I could use the law of large numbers again to obtain the claim (with $\sum_{i=1}^n\sup_{s\in[i-1,i)}|B_s-B_i|$).
But if I show $\frac{|B_t|}{t}\to 0\ (t\to\infty)$, $\frac{B_t}{t}\to 0\ (t\to\infty)$ doesn't follow, right?
I am grateful for any help!
Since Brownian motion has independent and stationary increments, the random variables $D_n := \sup_{s\in[n-1,n)}|B_s-B_n|$ are i.i.d., and by time reversal and symmetry they have the same distribution as $D := \sup_{s\in[0,1]}|B_s|$
But $$D := \sup_{s\in[0,1]}|B_s| \le \sup_{s\in[0,1]}B_s + \sup_{s\in[0,1]}-B_s.$$ $$D^2 \le 2\Big(\sup_{s\in[0,1]}B_s\Big)^2 + 2\Big(\sup_{s\in[0,1]}-B_s\Big)^2.$$ Taking expectations, using symmetry again, and the hint, we get $$\mathbb{E}(D^2) \le 4\mathbb{E}\Big((\sup_{s\in[0,1]}B_s)^2\Big) = 4\mathbb{E}(|B_1|^2) = 4.$$ Thus $$\sum_{n \ge 1}\mathbb{P}(D_n^2 \ge n) = \sum_{n \ge 1}\mathbb{P}(D^2 \ge n) = \mathbb{E}\Big(\sum_{n \ge 1}1_{(D^2 \ge n)}\Big) = \mathbb{E}(\lfloor D^2 \rfloor) \le 4.$$ Borel - Cantelli Lemma applies, so almost surely, $|D_n| < \sqrt{n}$ for every large enough $n$. This gives you the missing argument.