I want to show that: $$ \sup_{u \geq 0} \frac{1}{u} \left( | B_u | - 1 \right) = \sup_{u \geq 0} \left( |B_u| - u \right) = \sup_{0 \leq u \leq 1 } b_u^2 $$ in distribution; with $ B_u = (1+u)b_{\frac{u+1}{u}}$ where $b_u$ is a Brownian bridge.
I already proved that $ B_u $ is a Brownian motion.
From the book: "Continuous Martingales and Brownian Motion" by Revuz and Yor, I concluded that I must first show that:
$$ \sup_{t \geq 0}( |B_t| - t^{\frac{p}{2}}) = \sup_{t \geq 0} \left( |B_t| \cdot \frac{1}{1 + t^\frac{p}{2}} \right)^q \text{ in distribution with} \frac{1}{p} + \frac{1}{q} = 1 \text{ and } B_t \text{ a Brownian motion.} $$
Then I can take $p = q = 2$ .Which then easily gives me:
$$ \sup_{t \geq 0} ( |B_t| - t ) = \sup_{t \geq 0} \left( b_\frac{t}{1 + t} \right)^2 \text{ in distribution. Which is equal to }\sup_{0 \leq u \leq 1} ( b_u )^2 \text{in distribution} $$
which proves the second equality.
But I'm stuck proving that: $$ \sup_{t \geq 0}( |B_t| - t^{\frac{p}{2}}) = \sup_{t \geq 0} \left( |B_t| \cdot \frac{1}{1 + t^\frac{p}{2}} \right)^q \text{ in distribution} $$
Can anybody help me?