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Fix $\epsilon>0$ and let $(\Omega,\mathcal{F},\mathcal{F}_t,\mathbb{P})$ be a stochastic base, and let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a continous function with $f(0)=0$. Is there a family of Markov diffusion processse $X_t^{x,\epsilon}$ satisfying an SDE of the form: $$ \begin{cases} d X_t^{x,\epsilon} = \mu(t,X_t^{x,\epsilon},\epsilon)dt + \Sigma(t,X_t^{x,\epsilon},\epsilon)dW_t \\ X_0^{x,\epsilon}=f(x) \end{cases} $$ and such that $ \mathbb{P}\left( \sup_{t \in [0,1],x \in \mathbb{R}^n}|X_t^{x,\epsilon} - f(x)| < \epsilon \right) $ holds with high probability?

Note: Of course, here, we require that $\Sigma(t,x,\epsilon)>0$ to avoid trivial solutions.

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  • $\begingroup$ I think Freidlin-Wentzell theory can be used but I'm not sure it will give you a positive probability... $\endgroup$
    – ABIM
    Jun 5, 2020 at 15:47
  • $\begingroup$ What is your initial condition for $X^{x,\epsilon}$? Is it $x$? Then $f$ must go from $\mathbb R^n$ to itself, right? $\endgroup$
    – Pierre PC
    Jun 5, 2020 at 15:56
  • $\begingroup$ You have $x$ under the probability sign. What is the quantifier on $x$ there? $\endgroup$
    – Iosif Pinelis
    Jun 5, 2020 at 15:58
  • $\begingroup$ The supremum should be over both t and x. $\endgroup$
    – ABIM
    Jun 5, 2020 at 16:07
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    $\begingroup$ What about letting $X_t^{x,\epsilon}:=f(x)$ for all $t,x$ (with $\mu=0$), and then modifying $X$ by choosing $\Sigma>0$ to be arbitrarily small? $\endgroup$
    – Iosif Pinelis
    Jun 5, 2020 at 16:55

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