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Suppose $X_t$ is the solution to $$ d X_t=b(X_t)dt+dL_t,\quad X_0=x. $$ where $L$ is a rotational symmetric $\alpha-$stable process with $\alpha\in (0,1]$, $b$ is Lipchitz.

Assume $\Gamma\subseteq \mathbb{R}^d$ and $\ dim(\Gamma)<d-1$. Is it possible to prove $\Gamma$ is a polar set, which means $$ {\bf P}_x(T_\Gamma<\infty)=0, \quad \forall x\in \mathbb{R}^d? $$ where $T_{\Gamma}:=\inf\{t>0: X_t\in\Gamma \}$.

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  • $\begingroup$ What if $X_t = X_0 + b t$ for some $b \ne 0$? Then even single points are non-polar. (They are semipolar, though). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jan 23, 2018 at 13:16
  • $\begingroup$ Thank you so much for your comments. I didn't explain my question clear before, so I reedit it in an easier way. Thanks again if you can give me any comment or reference. $\endgroup$
    – Guohuan Zhao
    Commented Jan 24, 2018 at 22:17
  • $\begingroup$ I am aware of numerous papers dealing with $\alpha \in (1, 2]$ (by Bogdan–Jakubowski, Kim–Song–Vondraček, Chen–Wang). The case $\alpha \in (0, 1]$ that you are interested in seems to be more difficult, at least with this approach, and I do not know any results in this direction. $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jan 24, 2018 at 22:27
  • $\begingroup$ @MateuszKwaśnicki: Maybe some literatures about general Markov process will talk about this topic, but I have not find any related results. Any way, thank you again for your help. $\endgroup$
    – Guohuan Zhao
    Commented Jan 25, 2018 at 1:03

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