How to show that $P( \sup_{0 \leq s \leq 1} |B_s| \leq \epsilon) > 0$ for any $\epsilon > 0$?

Question

Let $(B_s)_{s \geq 0}$ be a standard Brownian motion.

In order to solve an interesting exercise, I need to show that $P( \sup_{0 \leq s \leq 1} |B_s| \leq \epsilon) > 0$ for any $\epsilon$.

I've tried the following general methods, and failed at them:

1. Doob's Maximal inequality. This only works for sufficiently large $\epsilon$.
2. Reflection principle / using what I know about $B_t^*$. Again, this only works for sufficiently large $\epsilon$. (Essentially what I did boiled down to this result, which is not strong enough: Show that $\mathbb{P}(\sup_{s\leq t} |B_s|\geq x)\leq 2\mathbb{P}(|B_t|\geq x)$ )
3. Various fiddling with time change symmetries of Brownian motion.
4. Using the distributions of the stopping time for one sided barriers. (Similar to 2.)

Perhaps one of these approaches works, but I'm totally stuck. I would appreciate a hint.

Answer 1

I learned how to do this from David Freedman's book "Brownian motion and Diffusion" (Lemma 37)

Let $T = \inf \{ t : B(t) = \pm \epsilon /2 \}$.

There is a $\delta > 0$ so that $P (T \geq \delta) > 0$. (For example, one can argue using Doob's maximal inequality: $PP ( T \geq \delta) = P( \sup_{0 \leq t \leq \delta} |B_t| < \epsilon/2 ) \geq 1 - \frac{ 2 E |B_{\delta}|}{\epsilon} = 1 - \frac{ 2 \delta E |B_1| }{\epsilon}$.)

$P ( T \geq \delta , B(T) = \epsilon /2) = 1/2 P( T \geq \delta$ by symmetry. (That is, because $\{ T \geq \delta , B(T) = \epsilon /2 \} \cup \{ T \geq \delta , B(T) = -\epsilon /2 \} = \{ T \geq \delta \}$.)

Choose (even) $n$ large enough so that $n \delta > 1$.

Let $T_0 = 0$ and $T_{j+1} = \inf \{ t > T_j : |B(t) - B(T_j)| = \epsilon /2 \}$.

(So $T = T_1$.)

(Now this is the crucial idea) consider this event:

$E = \{ T_1 \geq \delta, T_2 - T_1 \geq \delta,\ldots, T_n - T_{n-1} \geq \delta, B(T_1) = \epsilon /2, B(T_2) = 0, B(T_3) = \epsilon/2, \ldots, B(T_n) = 0\}$.

In this event, $B_t$ goes up to $\epsilon/2$ before $-\epsilon/2$ in in time $\geq \delta$, then back $0$ before getting to $\epsilon$ (taking time $\geq \delta$). Then it repeats this $n/2$ times. Hence, $T_n \geq n \delta > 1$ on $E$. Because of our description, on the event $E$, $B_s$ always stays within $(- \epsilon/2, \epsilon)$ up to time $T_n > 1$. So in particular:

$E \subset \{ \sup_{0 \leq s \leq t} |B(s)| < \epsilon \}$.

Moreover, by the strong Markov property, we have $P(E) = (1/2 P( T \geq \delta))^n > 0$.

This proves the claim.