Regular and irregular points on the boundary for Brownian motion

Question

If we take unit ball which is punctured at the origin and consider the origin as the boundary point, then why is it so that the origin is a irregular point if d>1 but regular for d=1?

Answer 1

You know that the zero set of Brownian Motion intersects $[0,t]$ at infinitely many points (this follows, for instance, by the law of the iterated logarithm). This proves that $(-a,b)\setminus \{0\}$ is a regular set for one-dimensional Brownian Motion.

For two dimensions or higher, you can simply pick a sequence $t_n\to 0$ and note that $\mathbb{P}^0(X_t\neq 0\;\forall t>0)=\lim_{n\to\infty}\mathbb{P}^0(X_t\neq 0\;\forall t>t_n)$. By the Markov Property, we have $$ \mathbb{P}^0(X_t\neq 0\;\forall t>t_n)=\mathbb{E}^0 [\mathbb{P}^{X_{t_n}}[X_t\neq 0\; \forall t>0]] $$ Now, by Brownian scaling and rotation invariance, we have $$ \mathbb{P}^{x}[X_t\neq 0\; \forall t>0]=\mathbb{P}^{e_1}[X_t\neq 0\; \forall t>0] $$ for all $x\neq 0$. Hence, since $X_{t_n}\neq 0$ almost surely, we get that $ \mathbb{P}^0(X_t\neq 0\;\forall t>t_n)=\mathbb{P}^{e_1}[X_t\neq 0\; \forall t>0]=1$, where the last equality follows from the fact that we're in dimension two or greater. If you don't know this result, you can see the following for a solution in three dimensions or higher (Modulus 3-dim. Brownian motion convergence). The two dimensional case is similar but the harmonic function involved is a logarithm.

Hence, if $\tau_{B'}$ and $\tau_B$ denote the first exit times from the punctured and the unpunctured balls respectively, we get $$ \mathbb{P}^0[\tau_{B'}=\tau_B]=1 $$ and $\mathbb{P}^0[\tau_B=0]=0$. Since $0$ is on the boundary of the punctured ball, this implies that $B'$ is not regular for Brownian Motion.