Stochastic calculus: upper bound given Lipschitz drift and diffusion I am currently stuck on the following problem. Consider the SDE
$$\text{d}X_t=\sigma(X_t)\text{d}W_t+\mu(X_t)\text{d}t,$$ with $|\mu(x)|^2+|\sigma(x)|^2\leq A(1+|x|^2)$, where $A$ is a finite constant, $X_0=0$, and the drift and diffusion terms are Lipschitz functions.
I need to show that

*

*$\mathbb{E}\sup_{t\leq1}|X_t|^2\leq\text{e}^{8A}$,


*the solution $\{X_t\}_{t\leq1}$ exists even if the drift and diffusion terms are only locally Lipschitz.
I'm so confused, I don't know how to begin! I've stared at https://math.stackexchange.com/questions/3251957/calculating-the-expecation-of-the-supremum-of-absolute-value-of-a-brownian-motio in hopes of getting some better insight, but I'm lost -- could someone guide me in the right direction please? Thank you!
 A: The solution satisfies
$$x_t = \int_0^t \mu(X_u)du + \int_0^t \sigma(X_u)dW_u.$$
Using the inequality $|a+b+c|^2 \leq 3|a|^2+3|b|^2+3|c|^2,$ we have
$$|X_t|^2 \leq 3 \left|\int_0^t \mu(X_u)du \right|^2 + 3 \left| \int_0^t \sigma(X_u)dW_u \right|^2.$$
By the Cauchy–Schwarz inequality the first integral above satifies
$$\left|\int_0^t  \mu(X_u)du \right|^2 \leq t \int_0^t \mu^2(X_u)du.$$
So,
\begin{equation} \tag{1}
E \sup_{t \le 1} |x_t|^2 \le 3 \left|\int_0^1 \mu^2(X_u)du \right|^2+3E\sup_{t \leq 1} \left| \int_0^t \sigma(X_u)dW_u \right|^2.
\end{equation}
By Doob's martingale inequality first and the Ito isometry later, the second integral of Equation (1) satisfies
\begin{equation} \tag{2}
E\sup_{t \leq 1} \left| \int_0^t \sigma(X_u)dW_u \right|^2 \leq 4E\left| \int_0^1 \sigma(X_u)dW_u \right|^2=4E \int_0^1 \sigma^2(X_u)du.
\end{equation}
Plugging (2) in (1) and using the linear growth condition (given in the problem), we have
$$E \sup_{t \le 1} |x_t|^2 \leq 3E\int_0^1\mu^2(X_u)du +12E\int_0^1\sigma^2(X_u)du \leq 12AE\int_0^1 (1+ |X_u|^2)du.  $$
Thus
$$1+ E \sup_{t \le 1} |x_t|^2 \leq 1+ 12AE\int_0^1 (1+ |X_u|^2)du$$
and by the Gronwall inequality
$$E \sup_{t \le 1} |x_t|^2 \leq e^{12A}.$$
