Let $(M_t)_t$ and $(M_t)_t$ be two càdlàg martingales on the same filtered probability space. We know that $M_{\infty}$ and $N_{\infty}$ are orthogonal in $L^2$. Is it true that $(M_t N_t)_t$ is a martingale? If so, is it uniformly integrable?
I suspect the answers are yes and no, but not sure on how to prove.
On the first statement: $MN$ is clearly adapted and càdlàg and the martingale equality should be equivalent to $$ \mathbb{E}_t [(M_{t+h}-M_t)(N_{t+h}-N_t)] = 0 \quad \forall t,h >0 $$
If possible, provide an answer that does not involve the process $[M,N]$.
No, in general $(M_t \cdot N_t)_{t \geq 0}$ is not a martingale. Consider the (continous) $L^2$-martingales
$$M_t := \int_0^t f(s) \, dB_s \qquad N_t := \int_0^t g(s) \, dB_s$$
where $f,g$ are deterministic functions such that $$\int_0^{\infty} f(s)^2 \, ds < \infty \quad \int_0^{\infty} g(s)^2 \, ds < \infty$$
Then, by Itô's formula
$$\mathbb{E}(M_t \cdot N_t) = \mathbb{E} \left( \int_0^t f(s) \cdot g(s) \, ds \right) = \int_0^t f(s) \cdot g(s) \,ds$$
for any $t \in [0,\infty]$. This means that $M_{\infty}$ and $N_{\infty}$ are orthogonal if, and only if, $$\int_0^{\infty} f(s) \cdot g(s) \, ds = 0 \tag{1}$$
Obviously, we can choose $f,g$ such that $(1)$ is satisfied and
$$t \mapsto \int_0^t f(s) \cdot g(s) \, ds$$
is not constant. But this means that $\mathbb{E}(M_t \cdot N_t)$ is not constant; hence $(M_t \cdot N_t)_{t \geq 0}$ is not a martingale.