Let $X_{t}$ a right continuous $\textbf{F}_{t}$ martingale and $\textbf{F}_{t}$ satisfying the usual condition
Show that $ \sup_{t\geq 0}\textbf{E}(X^{2}_{t})<\infty$.
I know that $X^{2}_{t}$ is a submartingale and by Doob's inequality we have $\textbf{E}(\sup_{s\leq t}X^{2}_{s}) \leq 4\textbf{E}(X^{2}_{t})$
but i don't know how continue. Some help would be appreciated
Please note that this does not hold. @Stephen Montgomery-Smith already mentioned the "trivial" counterexample $$X_t := X, \qquad t \geq 0$$ for some $X \notin L^2$. Another counterexample is a one-dimensional Brownian motion $(X_t)_{t \geq 0}$; it is a martingale with continuous sample paths and satisfies $\mathbb{E}(X_t^2)=t$, hence $$\sup_{t \geq 0} \mathbb{E}(X_t^2) = \infty.$$ (In both cases, the usual hypotheses on the filtration are satisfied.)