I am studying Freidlin and Wentzell large deviation theory from their 2012 book "Random perturbations of dynamical systems" and I have a problem in justifying two steps of the proof of theorem 1.3. The theorem states
Theorem
We assume that there exists a continuous function $\bar{b}(t, x), t > 0,x \in \mathbb{R}^{r}$ such that for any $δ > 0, T > 0, x ∈ R^r$ we have \begin{equation} \label{eqn: hp1} \lim_{\epsilon \to 0} \mathbb{P} \Big( \Big| \int_{t_0}^{t_0+T}b(\epsilon,t,x,\omega)dt - \int_{t_0}^{t_0+T} \bar{b}(t,x)dt \Big| > \delta \Big) = 0 \end{equation} uniformly in $t_0 \ge 0$. Then the equation \begin{equation} \label{eqn: hp2} \dot{\bar{x}}_t=\bar{b}(t,\bar{x}_t), \qquad \bar{x}_0=x \end{equation} has a unique solution and \begin{equation} \lim_{\epsilon \to 0} \mathbb{P} \big( \max_{0 \le t \le T}|X_{t}^{\epsilon}-\bar{x}_t| > \delta \big) =0 \end{equation} for every $T>0$ and $\delta>0$.
Proof Since the function $\bar{b}(t,x)$ is continuous, by the mean value theorem we have $ \int_{t}^{t+\Delta}\bar{b}(s,x) ds =\bar{b}(t,x)\Delta+o(\Delta), \qquad \Delta \to 0 $ Taking into account the first hypothesis, we get \begin{equation} \begin{split} |\bar{b}(t,x)-\bar{b}(t,y)| &= \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} \bar{b}(s,x)ds - \int_{t}^{t+\Delta} \bar{b}(s,y)ds \Big| + \frac{o(\Delta)}{\Delta}\\ & \le \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} b(\epsilon,s,x,\omega)ds - \int_{t}^{t+\Delta} b(\epsilon,s,y,\omega)ds \Big| \\ & + \frac{o(\Delta)}{\Delta} +\delta_{\epsilon} \\ & \le K|x-y|+\frac{o(\Delta)}{\Delta} +\delta_{\epsilon} \end{split} \end{equation} where $\delta_{\epsilon}=\delta_{\epsilon}(t,\omega) \to 0$ in probability as $\epsilon \to 0$. And from this follows the Lipschitz condition on $\bar{b}$ and so the existence and uniqueness.
Here I don't understand how to get $\delta_{\epsilon}$, which I don't even know how it is defined.
I tried in this way: from the hypotesis I remark that $-\delta \le \int_{t_0}^{t_0+T}b(\epsilon,t,x,\omega)dt - \int_{t_0}^{t_0+T} \bar{b}(t,x)dt \le \delta$ and so, after adding and subtracting both $\int_{t}^{t+\Delta}b(\epsilon,t,x,\omega)dt$ and $\int_{t}^{t+\Delta}b(\epsilon,t,y,\omega)dt$ and use the remark above and I obtain \begin{equation} \begin{split} |\bar{b}(t,x)-\bar{b}(t,y)| &= \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} \bar{b}(s,x)ds - \int_{t}^{t+\Delta} \bar{b}(s,y)ds \Big| + \frac{o(\Delta)}{\Delta}\\ & \le \frac{1}{\Delta} \Big| \int_{t}^{t+\Delta} b(\epsilon,s,x,\omega)ds - \int_{t}^{t+\Delta} b(\epsilon,s,y,\omega)ds \Big| \\ & + \frac{o(\Delta)}{\Delta} +\frac{2}{\Delta}\delta \\ & \le K|x-y|+\frac{o(\Delta)}{\Delta} +\delta_{\epsilon} \end{split} \end{equation} where $\delta_{\epsilon}=\frac{2}{\Delta}\delta$. But it doesn't seem right since it is not random and so how can I say that it converges in probability as the quantity of the authors?
The other doubt is some steps after this. I have the estimate that follows: \begin{equation} \begin{split} & \int_{0}^{t} [b(\epsilon,s,\bar{x}_s,\omega)-\bar{b}(s,\bar{x}_s)]ds \\ &= \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_s,\omega)-\bar{b}(s,\bar{x}_s)]ds \\ &= \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_{kt/n},\omega)-\bar{b}(s,\bar{x}_{kt/n})]ds \\ &+ \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_s,\omega)-b(\epsilon,s,\bar{x}_{kt/n},\omega)]ds \\ &+ \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [\bar{b}(s,\bar{x}_{kt/n},)-\bar{b}(s,\bar{x}_s)]ds \\ &= \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} [b(\epsilon,s,\bar{x}_{kt/n},\omega)-\bar{b}(s,\bar{x}_{kt/n})]ds + \rho_{n,t}^{\epsilon} \end{split} \end{equation} where they say that $|\rho_{n,t}^{\epsilon}| \le C/n$ with $C=C(T,K)$. I tried in this way but I'm stuck and don't know how to go on \begin{equation} \begin{split} |\rho_{n,t}^{\epsilon}| &\le \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} |b(\epsilon,s,\bar{x}_s,\omega)-b(\epsilon,s,\bar{x}_{kt/n},\omega)|ds \\ &+ \sum_{k=0}^{n-1} \int_{kt/n}^{(k+1)t/n} |\bar{b}(s,\bar{x}_{kt/n},)-\bar{b}(s,\bar{x}_s|ds \\ & \le 2K \sum_{k=0}^{n-1}\int_{kt/n}^{(k+1)t/n}|\bar{x}_s - \bar{x}_{kt/n}|ds \end{split} \end{equation} Now I don't know how to procede: how can I get the same result as in the book?
Questions
- is what I did at the beginning correct? If not, how do I change it?
- how do I go on in the second part? Thanks to anyone who will have the patience to read and answer.
For the first question, I think you are almost there. When you wrote something like $$-\delta \le \int_{t_0}^{t_0+T}b(\epsilon,t,x,\omega)dt - \int_{t_0}^{t_0+T} \bar{b}(t,x)dt \le \delta,$$ you should be aware of the randomness contained in $\omega$, so this statement can only hold with certain probability (here with high probability), so the bound on $|\bar{b}(t,x) - \bar{b}(t,y)|$ you wrote only hold with high probability. For the second question I am not completely sure as well. I guess the authors used something like $|\bar{x}_s - \bar{x}_{kt/n}| \leq C/n$ for $s \in [kt/n,(k+1)t/n]$, this can probably be justified by the Lipschitz continuity enjoyed by $\bar{b}$ (maybe we also need to use that $\bar{b}$ is uniformly bounded on a compact time-space region). Hope this helps!