Using $\mathcal{B}$ for the standard Brownian Motion. I`m supposed to prove the following:
$$\limsup_{\delta\to0^+}\frac{1}{\delta}\mathbb{P}\Bigg(\sup_{v\in[0,1]}|\mathcal{B}(v)|>\frac{\eta}{\sqrt{\delta}}\Bigg)=0,$$ for every $\eta>0$. I`ve tried to use the distribution of the $\sup$ of a Brownian Motion together with Chebyshev's Inequality, but I couldn't just get to this result. Can anyone give a hint?
Can use Doob's submartingale inequality and a trick to get $$ \frac{1}{\delta}P\left(\sup_{0<t<1}|B_t|>\frac{\eta}{\sqrt{\delta}}\right) = \frac{1}{\delta}P\left(\sup_{0<t<1}|B_t|^3>\frac{\eta^3}{\delta^{3/2}}\right) \le \frac{1}{\delta}\frac{E(|B_1|^3)\delta^{3/2}}{\eta^3}$$