Let $X_t$ be a solution to:
$$\mathrm{d}X_t = F(X_t) \mathrm{d}B_t$$
Where $F(x)$ is a Lipschitz continuous function, $B_t$ a standard Brownian motion, and let $X_0 = x$.
I'm trying to calculate the supremum:
$$\mathbb{E}[\sup_{s \in [0,t]} |X_s - x|^2]$$
I've so far been unable to actually calculate this. I think I should be able to rewrite the equation using $Y_t = X_t - x$, and then use the fact that $F$ is Lipschitz to apply the stochastic exponential defined by:
$$ \mathrm{d}Z_t = C Z_t \mathrm{d}B_t$$ for some $C \in \mathbb{R}$
But I've been unable to find a good way to do this so far.
For fixed $t>0$ set
$$u(t) := \mathbb{E}\left( \sup_{s \leq t} |X_s-x|^2 \right).$$
Since $(X_t)_{t \geq 0}$ is a martingale, it follows from Doob's inequality that
$$u(t)=\mathbb{E}\left( \sup_{s \leq t} |X_s-x|^2 \right) \leq 4 \mathbb{E}(X_t^2).$$
Applying Itô's isometry, we find that
$$u(t) \leq 4 \mathbb{E} \int_0^t (F(X_s))^2 \, ds.$$
By the Lipschitz continuity of $F$, there is a constant $M>0$ such that $|F(x)| \leq M(1+|x|)$ for all $x \in \mathbb{R}$. Thus,
\begin{align*}u(t) &\leq 8M \mathbb{E} \int_0^t (1+|X_s|^2) \, ds \\ &\leq 8Mt + 8M \int_0^t u(s) \, ds \end{align*}
Applying Gronwall's inequality, we find that
$$u(t) \leq (8Mt) \exp(8M t)$$
for all $t \geq 0$, i.e.
$$ \mathbb{E}\left( \sup_{s \leq t} |X_s-x|^2 \right) \leq 8Mt e^{8Mt}.$$