Proving martingale inequality $\mathbb{E}[\sup_t M_t^2] \leq 4\mathbb{E}[Z^2]$ Let $Z$ be a square int'ble rv. on a filtered probability space with filtration $(\mathfrak{F}_t)_{t \in \mathbb{R}_+}$, we define the martingale $M_t = \mathbb{E} [Z|\mathfrak{F}_t] $. I now want to show the inequality:
$$\mathbb{E}[\sup_{t \in \mathbb{R} _+} M_t^2] \leq 4\mathbb{E}[Z^2]$$
This looks a lot like Doob's martingale inequality, however there we need a non-negative sub-martingale, which is not the case here. What could I try here?
 A: I would use an adaptation of this theorem here below.
Theorem.
If $X$ is a nonnegative submartingale and $p>1$, then one has Doob's $L^p$ inequality
$$
\left\|\sup _{s \leq u \leq t} X_u\right\|_p \leq \frac{p}{p-1}\left\|X_t\right\|_p
$$
where for random variables $\xi$ we put $\|\xi\|_p=\left(\mathbb{E}|\xi|^p\right)^{1 / p}$. In particular, if $M$ is a right-continuous martingale with $\mathbb{E} M_t^2<\infty$ for all $t>0$, then
$$
\mathbb{E}\left(\sup _{s \leq u \leq t} M_u^2\right) \leq 4 \mathbb{E} M_t^2 .
$$
I will just add the following: the proofs of these results are essentially the same as in the discrete time case. The basic argument to justify this claim is to consider $X$ restricted to a finite set $F$ in the interval $[s, t]$. With $X$ restricted to such a set, the above inequalities are valid as we know them from discrete time theory. Take then a sequence of such $F$, whose union is a dense subset of $[s, t]$, then the inequalities keep on being valid. By right-continuity of $X$ we can extend the validity of the inequalities to the whole interval $[s, t]$.
