Characterization for strongly orthogonal martingales

Question

In Limit Theorems for Stochastic Processes by Jacod and Shiryaev we have the following statement:

For two square-integrable martingales $X,Y$ starting in zero we have $$ \langle X,Y\rangle =0 \iff \text{ XY is a local martingale} $$

My question is if $XY$ is a martingale is equivalent to $\langle X,Y\rangle=0$ holds. Obviously, we have that if $XY$ is a martingale, then $\langle X,Y\rangle=0$ holds.

But if $\langle X,Y\rangle=0$, then $XY$ is clearly a local martingale. In order to prove that $XY$ is a martingale, we can show that $E[\sup_{t\geq 0}(X_tY_t)]<\infty]$ holds. Does this hold?

Answer 1

In order to prove that $X Y$ is a martingale, it actually suffices to show that

$$\mathbb{E} \left( \sup_{t \leq T} |X_t Y_t| \right) < \infty$$

for any $T>0$. To this end, note that

$$|X_t Y_t| \leq 2 |X_t|^2 + 2 |Y_t|^2$$

and therefore by the square-integrability of $(X_t)_{t \geq 0}$ and $(Y_t)_{t \geq 0}$ and Doob's maximal inequality

$$\mathbb{E} \left( \sup_{t \leq T} |X_t Y_t| \right) \leq 2 \mathbb{E} \left( \sup_{t \leq T} |X_t|^2 \right) +2 \mathbb{E} \left( \sup_{t \leq T} |Y_t|^2 \right) \leq 8 \mathbb{E}(|X_T|^2) + 8 \mathbb{E}(|Y_T|^2)<\infty.$$