Limit of brownian local time

Question

Let $(B_{t})_{t \geq 0}$ be a one-dimensional Brownian motion and $(\phi(t,0))_{t \geq 0}$ be its local time at $0$. Then does it hold that $\lim_{t \uparrow \infty}\phi(t,0)=\infty$ almost surely?

Answer 1

A non-Markovian approach: By Tanaka's formula, assuming $B_0=0$, you have $|B_t| = W_t +\phi(t,0)$, where $W$ is another Brownian motion. Thus $$ \phi(t,0)=|B_t|-W_t\ge -W_t. $$ We know that the Brownian motion $W$ will eventually hit any $b\ll 0$. Because $t\mapsto \phi(t,0)$ is increasing $$ \lim_{t\to\infty}\phi(t,0)\ge -b,\qquad\forall b<0, $$ almost surely.