Let $B\_t$ denote a standard Brownian motion, and $0 < l < u$. I am wondering if the law of $\sup\_{l \leq t \leq u} \frac{|B\_t|}{\sqrt{t}}$ is atomless, that is, $\mathbb{P}\left(\sup\_{l \leq t \leq u} \frac{|B\_t|}{\sqrt{t}} = c \right) = 0$ for all constant $c$.


Any help will be greatly appreciated! 



Is the law of $\sup_{l \leq t \leq u} \frac{|B_t|}{\sqrt{t}}$ atomless?
This seems to follow easily from the strong Markov property. Informally: if this probability were positive for some $c$, then we could stop the process when $t^{-1/2} |B\_t| = c$ for the first time, and get a contradiction with the oscillatory character of $B\_t$ for small times.




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Write $$M = \sup\_{l\le t\le u} t^{-1/2} |B\_t|$$ and $$\tau = \inf \{t \ge l : t^{-1/2} |B\_t| = c\}.$$ Observe that if $M = c$, then $\tau \le u$.


We have $$\mathbb{P}(M = c, \tau = u) \le \mathbb{P}(M = c, u^{-1/2} |B\_u| = c) \le \mathbb{P}(u^{-1/2} |B\_u| = c) = 0.$$ It is thus enough to show that $\mathbb{P}(M = c, \tau < u) = 0$. To this end, we show that for any $\varepsilon > 0$, $\mathbb{P}(M = c, \tau < u - \varepsilon) = 0$.


We have
$$
 \mathbb{P}(M = c, \tau < u - \varepsilon) \le \mathbb{P}(\sup\_{\tau \le t\le \tau + \varepsilon} t^{-1/2} |B\_t| = c, \tau < u) .
$$
By the strong Markov property, the right-hand side is equal to $\mathbb{E}(\phi(\tau, X\_\tau))$, where
$$
 \phi(t, x) = \mathbb{P}(\sup\_{0 \le s\le \varepsilon} (t + s)^{-1/2} |x + B\_s| = c) .
$$
However, $\phi(t, x) = 0$ if $t^{-1/2} |x| = c$ (by the law of the iterated logarithm, for example), and therefore $\phi(\tau, X\_\tau) = 0$ almost surely.


