Does the Doob inequality hold for local martingales? Let $M$ be a real-valued continuous local martingale with $M_0=0$, and  fix $t_0\geq 0$. Do we then have the inequality $$\mathbb{E}[~\sup_{t\leq t_0} M_t^2~]\leq C\cdot \mathbb{E}[M_{t_0}^2],$$
where $C$ is a universal constant (i.e. not depending on $t_0$ or $M$)?
The inequality is trivial in case $M_{t_0}\notin L^2$, so we may assume $\mathbb{E}[M_{t_0}^2]<\infty$.  This inequality holds when $M$ is a true martingale and $C=4$, in which case it is known as the Doob inequality. If we localize the inequality and let the stopping times tend to infinity, the left hand side is a monotone limit, but it's not clear what to do with the limit of the right hand side. If somehow the assumption $\mathbb{E}[M_{t_0}^2]<\infty$ implied that $\mathbb{E}[\langle M, M \rangle _{t_0}]<\infty$ (as it would if $M^2-\langle M, M \rangle$ were a true martingale rather than just a local martingale), then it is a theorem that $M_{t\wedge t_0}$ is a true martingale and the normal Doob inequality would apply.
 A: No, this is not true for local martingales.  The closest we have is the BDG inequalities: for any $p \in [1,\infty)$ there exist universal constants $c_p$ and $C_p$ such that for any continuous local $M$ we have $$c_p \mathbb{E}[\langle M,M \rangle_t^{p/2}] \le \mathbb{E}\left[\sup_{s \le t} |M_s|^p\right] \le C_p \mathbb{E}[\langle M,M \rangle_t^{p/2}].$$
For a counterexample where the classical Doob inequality you wrote does not hold, let $B_t$ be a Brownian motion in $\mathbb{R}^3$ with $B_0 = (1,1,1)$.  Then $M_t := \frac{1}{|B_t|}$ is a continuous local martingale and bounded in $L^2$ (I'm sure you can find a proof of that either on this site or at the blog TheBridge linked in the comments).  Since $(M_t) \rightarrow 0$ almost surely and is bounded in $L^2$, we have $\mathbb{E}[M_t^2] \rightarrow 0$.  If we had something like the Doob inequality for local martingales, this would imply
\begin{align*}
1 \le \mathbb{E}\left[\sup_{s \le t}|M_s|^2 \right] \le C \mathbb{E}[|M_t|^2].
\end{align*}
Sending $t \rightarrow \infty$ gives $0$ on the right, which is a contradiction.
