Let $(M_t)$ a continuous martingale. Prove that $$\mathbb E\left[\sup_{s\in [0,t]}M_s^2\right]\leq 4\mathbb E[ M_t^2].$$
I already proved that $$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E|M_t|}{y},$$ and I know that $$\mathbb E Y^2=\int_0^\infty 2y\mathbb P\{Y>y\}d y,$$ when $Y\geq 0$ a.s. I tried to use this last formula for $Y=|M_t|$ but it's not conclusif. How can I conclude ?
You should use a finer version of $$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E|M_t|}{y}$$
which is $$\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}\leq \frac{\mathbb E(|M_t|1_{\{\sup_{s\in [0,t]}|M_s|\geq y\}})}{y}$$
By integrating both sides with respect to $y$ , we have
$$\int_{0}^{\infty}{y\mathbb P\left\{\sup_{s\in [0,t]}|M_s|\geq y\right\}dy}\leq\int_{0}^{\infty}{\mathbb E(|M_t|1_{\{\sup_{s\in [0,t]}|M_s|\geq y\}})}dy$$
Using Fubini theorem on both sides, we have $$\mathbb E\left(\frac{\left(\sup_{s\in [0,t]}|M_s|\right)^2}{2}\right) \leq E\left(|M_t|\sup_{s\in [0,t]}|M_s|\right) $$
or
$$\mathbb E\left({\left(\sup_{s\in [0,t]}|M_s|\right)^2}\right)^2 \leq 4 E\left(|M_t|\sup_{s\in [0,t]}|M_s|\right)^2 $$
Using Cauchy-Schwarz inequality, and the fact that $\left(\sup_{s\in [0,t]}|M_s|\right)^2=\sup_{s\in [0,t]}|M_s|^2$ a.s., we conclude that $$E\left(|M_t|\sup_{s\in [0,t]}|M_s|\right) \leq \sqrt{E\left(M_t^2\right)E\left(\left(\sup_{s\in [0,t]}|M_s|\right)^2\right)} $$
and finally,
$$\mathbb E\left[\sup_{s\in [0,t]}M_s^2\right]\leq 4\mathbb E[ M_t^2]$$