Is it true that
\begin{equation}
M_t = \sup_{s \leq t} B_s > 0 \ \ \text{a.s.}
\end{equation}
for all $t>0$? I remember reading this somewhere, but intuitively, can't the Brownian motion B stay below 0 for some time with probability $>0$?
Supremum of brownian motion almost surely > 0
brownian motionstochastic-processes
Related Solutions
Fix $\delta>0$. By the very definition of the supremum, there exists for any $n \in \mathbb{N}$ some $s_n \in (0,s_0]$ such that
$$\sup_{0<s \leq s_0} \left| \frac{W(t_n(s))-W(s)}{t_n(s)^{1/2-\epsilon}} \right| \leq \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| + \delta.$$
If we can show that
$$\limsup_{n \to \infty} \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| =0 \tag{4}$$
then this proves the assertion.
Proof of $(4)$: There exists a subsequence $(s_n')_{n \in \mathbb{N}}$ of $(s_n)_{n \in \mathbb{N}}$ such that the $\limsup$ in $(4)$ is attained, i.e.
$$\lim_{n \to \infty} \left| \frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}} \right| = \limsup_{n \to \infty} \left| \frac{W(t_n(s_n))-W(s_n)}{t_n(s_n)^{1/2-\epsilon}} \right| \tag{5}$$
Since $(s_n')_{n \in \mathbb{N}}$ is contained in the compact interval $[0,s_0]$, we may assume without loss of generality that $u := \lim_{n \to \infty} s_n'$ exists (otherwise we take another subsequence). In order to prove $(4)$, we will show that the left-hand side of $(5)$ is zero, and to this end we consider two cases separately.
Case 1: $u>0$. Since $s_n' \to 0$ implies, by $(1)$, $t_n(s_n') \to u$ it follows from the continuity of the sample paths of Brownian motion that
$$\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}} \xrightarrow[]{n \to \infty} \frac{W(u)-W(u)}{u^{1/2-\epsilon}} = 0.$$
Case 2: $u=0$. Fix some $\gamma>0$. Because of $(2)$, there exists $r>0$ such that $$\sup_{0<s \leq r} \left| \frac{W(s)}{s^{1/2-\epsilon}} \right| \leq \gamma.$$
As in Case 1 we have $t_n(s_n') \to u$ and so $t_n(s_n') \to 0$. In particular, we can choose $N \in \mathbb{N}$ such that $$|t_n(s_n')| \leq r \quad \text{and} \quad |s_n'| \leq r $$ for all $n \geq N$. Hence,
$$\begin{align*} \left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| &\leq \left| \frac{W(t_n(s_n'))}{t_n(s_n')^{1/2-\epsilon}} \right| + \frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \left| \frac{W(s_n')}{|s_n'|^{1/2-\epsilon}} \right| \\ &\leq \gamma + \frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \gamma \end{align*}$$
for all $n \geq N$. It is immediate from $(1)$ that
$$\frac{|s_n'|^{1/2-\epsilon}}{t_n(s_n')^{1/2-\epsilon}} \leq 1+ \gamma$$
for $n$ sufficiently large, and therefore we conclude that
$$\left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| \leq \gamma + (1+\gamma) \gamma$$
for all $n$ sufficiently large. As $\gamma>0$ was arbitrary, this proves
$$\lim_{n \to \infty} \left|\frac{W(t_n(s_n'))-W(s_n')}{t_n(s_n')^{1/2-\epsilon}}\right| =0.$$
Remark: Note that we have actually shown that $(4)$ holds for any sequence $(s_n)_{n \in \mathbb{N}} \subseteq (0,s_0]$ (... and not only for the sequence which we picked at the very beginning of this answer).
A good first guess is that $B_t$ is unlikely to become negative on $[a,a+1]$ if $B_a$ is already a large positive number. Since your $a$ is allowed to depend on the realisation of the Brownian path, this means it's reasonable to consider the stopping times $T_n = \inf\{t: B_t = n\}$ for $n \in \mathbb{N}$.
Now you should try to compute $\mathbb{P}(B_t = 0 \text{ for some } t \in [T_n,T_{n} + 1])$ (use the strong markov property and a standard result about Brownian hitting times). Once you've done this, it should be easy to show $$\mathbb{P}(\forall n, B_t < 0 \text{ for some } t \in [T_n, T_n + 1]) = 0$$ which implies the desired result.
Best Answer
One way to argue is using Blumenthal's zero one law. Define $A_n=\{B_{1/n}>\frac{1}{\sqrt{n}}\}$, and set $B=\{B_{1/n}>\frac{1}{\sqrt{n}} ~\text{i.o.}\} $
Then, \begin{align*} \mathbb{P}(B) &=\mathbb{P}(\limsup_n A_n)\\ &\geq \limsup_n\mathbb{P}( A_n)\\ &= \limsup_n\mathbb{P}(B_{1/n}>\frac{1}{\sqrt{n}})\\ &=\limsup_n\mathbb{P}(N(0,1)>1)\\ &=\mathbb{P}(N(0,1)>1)=M>0 \end{align*}
By Blumenthal's zero one law $\mathbb{P}(B)= 1 ~\text{or}~0$. So $\mathbb{P}(B)=1$.