$\limsup_{t \to 0} {L_t}/\sqrt{t} = \infty$ with probability one?

Question

Let $B_t$ be a standard Brownian motion, $L(x, t)$ be the local time $x$ at time $t$, and $L_t = L(0, t)$. Do we have$$\limsup_{t \to 0} {{L_t} \over{\sqrt{t}}} = \infty$$with probability one?

Answer 1

It is hard to understand why don't you use Alex R.'s hint, so let me write it down.

By Levy's theorem, $L_t$ has the same distribution as $M_t = \sup_{s\in[0,t]} B_t$. So by the law of the iterated logarithm, $$ \limsup_{t\to 0+} \frac{L_t}{\sqrt{2t\log \log \frac1t}} = \limsup_{t\to 0+} \frac{M_t}{\sqrt{2t\log \log \frac1t}}\ge \limsup_{t\to 0+} \frac{B_t}{\sqrt{2t\log \log \frac1t}} = 1. $$ It follows that the answer to your question is positive.