# Locally Lipschitz sufficiently implies a Gronwall inequality?

In the paper [1], it seems to me the authors implicitly use a local Lipschitz property to deduce a Gronwall's inequality. I am not able to justify/show that this is indeed the case and perhaps someone here can help enlighten how.

Copying from (page 9) 1, we consider a stochastic process $x(t)$ and a controlled process $X(t)$ satisfying the SDE and controlled ODE $$x(t) = y + \int_0^t Y(x(s)) ds + \Sigma W(t), \\ X(t) = y + \int_0^t Y(X(s)) ds + \Sigma U(t).$$ where the drift $Y$ is assumed to be locally Lipschitz, $W$ is a Brownian motion and $U$ is (smooth) chosen control such that $X(0)=y$ and $X(t)=z$ for some interested point $z$ and $\Sigma$ is a matrix with linear independent columns hence invertible. By a Stroock and Varadhan theorem or a Levy Forgery type theorem, one has with positive probability that the event

$$\sup_{0\leq s \leq t} ||U(t) - W(t)|| \leq \epsilon$$

occurs for any $\epsilon > 0$.

If this event occurs and taking the difference, we have $$||x(t) - X(t)|| \leq \int_0^t ||Y(x(s)) - Y(X(t))|| ds + ||\Sigma||\epsilon.$$ From here, it was concluded because $Y$ is locally Lipschitz then for some terminal time $T$
$$\sup_{0 \leq t \leq T} ||x(t) - X(t)|| \rightarrow 0 \quad as \quad \epsilon \rightarrow 0.$$ I do not see where this conclusion comes from. If indeed $Y$ is globally Lipschitz with constant $L$, I would be happy with the claim by Gronwall's inequality i.e. $$||x(t) - X(t)|| \leq \int_0^t ||Y(x(s)) - Y(X(t))|| ds + ||\Sigma||\epsilon \\ \leq L \int_0^t ||x(s) - X(t)|| ds + ||\Sigma||\epsilon \\ \leq \epsilon ||\Sigma|| e^{Lt} \quad \text{Gronwall}$$ Hence $$\sup_{0 \leq t \leq T} ||x(t) - X(t)|| \leq \epsilon ||\Sigma|| e^{LT} \rightarrow 0 \quad \epsilon \rightarrow 0$$.

If the global Lipschitz condition is relaxed to a local Lipschitz property, it does not seem clear that one can apply Gronwall's inequality and indeed did not seem to find much during Google searches.

Perhaps, the authors repeatedly apply Gronwall inequality every small time-step to deduce a more global result an somehow make an argument continuously in time by taking the time steps to zero. In part, I think there may be a snag with the local Lipschitz property akin only local existence as seen in Picard-Lindelof theorem. Hence, it seems to me that further constraints may be required perhaps norm of $Y$ is small compared to the terminal time $T$ to resolve the local existence issue.

Aside, it does not help that 1 cited Theorem 5.2 from Stroock and Varadhan 2 where it was assumed the drift term is uniformly Lipschitz for their support theorem.

[1] "Ergodicity for SDEs and Approximations - Locally Lipschitz Vector Fields and Degenerate Noise" by J.C. Mattingly, A.M. Stuart and D.J. Higman (2002)

[2] "On The Support of Diffusion Processes with Applications to the Strong Maximum Principle" by D. Stroock and S.R. Varadhan (1972)

• At worst, you can condition on $x(t)$ and $X(t)$ never leaving some ball $B_R(0)$ and use the local Lipschitz constant on that ball. This should be enough because in context they are only trying to show some probability is positive, and don't need a quantitative lower bound. – user101142 Aug 12 '18 at 15:08