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


*

*Doob's Maximal inequality. This only works for sufficiently large $\epsilon$.

*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 $\mathbb{P}(\sup_{s\leq t} |B_s|\geq x)\leq 2\mathbb{P}(|B_t|\geq x)$ )

*Various fiddling with time change symmetries of Brownian motion.

*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.
 A: 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.
