Let $I_t=\int_0^t f_tdB_t,$ where $(f_t,t\ge 0)$ is a bounded process, $|f_t|\leq M$ almost surely for all $t \ge 0.$ Show that $\mathcal{P}\left[\sup_{0\leq t\leq T}|I_t|>\lambda\right]\leq \exp\left(-\frac{\lambda^2}{2M^2T}\right).$
Hint: Define $Y_t^{\alpha}=\exp\left(\alpha I_t-\frac{1}{2}\int_{0}^t f^{2}(s)ds\right)$, where $\alpha\in \mathbb{R}.$ First use $Y^{\alpha}$ to get an upper bound, then optimize over choice of $\alpha$ to get the smallest bound.