Conditional expectation of product of martingales I did not find the following question addressed nor a counterexample or proof.
For a continious martingale does,
$\mathbb{E}(X_{t}X_{s} \mid \mathcal{F}_{u})=X_{u}\mathbb{E}(X_{s} \mid \mathcal{F}_{u})$
hold for $s,t>u$?
In other words can we apply the martingale property to get a measurable function and pull it out via the property of the conditional?
 A: Since $(X_t)_{t \geq 0}$ is a martingale, we have $\mathbb{E}(X_s \mid \mathcal{F}_u) = X_u$ and therefore your assertion is equivalent to
$$\mathbb{E}(X_t X_s \mid \mathcal{F}_u )=  X_u^2.\tag{1}$$
Note that $(1)$ implies, in particular,
$$\mathbb{E}(X_t X_s) = \mathbb{E}(X_u^2) \qquad \text{for all} \, \, u<s,t.$$
This equation does, in general, fail to hold true and hence $(1)$ does, in general, not hold true. Consider for instance a one-dimensional Brownian motion $(X_t)_{t \geq 0}$, then $(X_t)_{t \geq 0}$ is a martingale with continuous sample paths and
$$\min\{s,t\} = \mathbb{E}(X_s X_t) \neq \mathbb{E}(X_u^2) = u$$
for any $u<s,t$.
Remark: Using the tower property it is not difficult to see that $$\mathbb{E}(X_s X_t \mid \mathcal{F}_u) = \mathbb{E}(X_s^2 \mid \mathcal{F}_u) \qquad \text{for all} \, \, u \leq s \leq t.$$ This means that $(1)$ is actually equivalent to saying that $(X_t^2)_{t \geq 0}$ is a martingale. As $(X_t)_{t \geq 0}$ being a martingale does clearly not imply that $(X_t^2)_{t \geq 0}$ is a martingale, we cannot expect that $(1)$ holds.
