Brownian motion property on interval

Question

Say $B$ is a Brownian motion. Knowing that for some constant $a\gt0$ $$P(\sup_{[0,n]} |B|\leq 1) \leq e^{-{an}}$$ how can I deduce that

$$P\left(\sup_{[0,1]} |B| \leq \frac{1}{n}\right)\leq e^{-{an^2}}$$

Answer 1

First, rewriting the inequality for $n' \equiv n^2$, we have that $$ \mathbb P(\sup_{[0,n^2]} |B|\leq 1) \leq e^{-{an^2}} \tag1$$ Now, if $(B_t)_{t\ge0}$ is a Brownian Motion, then $W \equiv \left(nB_{t/n^2}\right)_{t\ge0}$ is also a Brownian Motion for all $n >0$. Therefore, applying inequality $(1)$ to the process $W$, we get : $$\begin{align}\mathbb P(\sup_{[0,n^2]} |W|\leq 1) &= \mathbb P(\sup_{t\in[0,n^2]} n|B_{t/n^2}|\leq 1)\\ &=\mathbb P\left(\sup_{t\in[0,n^2]} |B_{t/n^2}|\leq \frac1 n\right)\\ &=\mathbb P\left(\sup_{s\in[0,1]} |B_{s}|\leq \frac1 n\right) \\ &\le e^{-{an^2}} \end{align} $$