Definition of regular point of a boundary with planar brownian motion

Question

This is an exercise in G.Lawler's book Conformally invariant processes in the plane.

First he defined regular point of a boundary using brownian motion:

Suppose $D$ is a domain in $\mathbb{C}$ with boundary $\partial D$. A point $z\in\partial D$ is called regular if $\mathbb{P}^z[\tilde{\tau}_D=0]=1$, where $\tilde{\tau}_D=\inf\{t>0:B_t\not\in D\}$.

The exercise is to show that $z\in\partial D$ is regular if and only if for every $\delta>0$, $$ \lim_{\epsilon\to0+}\sup_{|w-z|<\epsilon}\mathbb{P}^w[|B_{\tau_D}-z|\ge\delta]=0, $$ where $\tau_D=\inf\{t\ge0:B_t\not\in D\}$.

This seems intuitively to be right because if a brownian motion is started very near to $z$, it also exits the domain at a point near $z$. I know how to prove this when the boundary is a straight line. However here we don't know the exact form of the boundary. How can we use the stopping time information to show the equivalence?

Answer 1

"$\Rightarrow$" Let $h>0$ arbritrary, then clearly

$$\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta) = \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D \leq h)}_{=:p_1} + \underbrace{\mathbb{P}^w(|B_{\tau_D}-z| \geq \delta, \tau_D > h)}_{=:p_2}$$

We have $$\begin{align*} p_1 &\leq \mathbb{P}^w \left( \sup_{t \leq h} |B_t-z| \geq \delta \right) \stackrel{\varepsilon \leq \frac{\delta}{2}}{\leq} \mathbb{P}^w \left( \sup_{t \leq h} |B_t-w| \geq \frac{\delta}{2} \right) \\ &= \mathbb{P}^0 \left( \sup_{t \leq h} |B_t| \geq \frac{\delta}{2} \right) \leq \left( \frac{2}{\delta} \right)^2 \cdot \mathbb{E} \left( \sup_{t \leq h} |B_t|^2 \right) \\ &\leq 4 \cdot \left( \frac{2}{\delta} \right)^2 \cdot \mathbb{E}(B_h^2) = 4 \cdot \left( \frac{2}{\delta} \right)^2 \cdot d \cdot h \to 0 \qquad (h \to 0)\end{align*}$$ by applying Markov's inequality and Doob's maximal inequality.

For the second addend, we use that $$\forall h>0: \lim_{w \to z} \mathbb{P}^w(\tau_D > h) = 0$$ since $z$ is a regular point (see the theorem below).

"$\Leftarrow$" This is the easier direction, I leave it to you.

Theorem Let $z_0 \in \partial D$ a regular point. Then $$\lim_{w \to z_0} \mathbb{P}^w(\tau_D > h) = 0$$ for all $h>0$.

Proof: Let $w \in \mathbb{R}^d$, then $$\begin{align*} \mathbb{P}^w(\tau_D > h) &= \mathbb{P}^w(\forall s \in (0,h]: B_s \in D) \\ &= \inf_{n \in \mathbb{N}} \mathbb{P}^w(\forall s \in [1/n,h]: B_s \in D) \\ &= \inf_n \mathbb{E}^w \bigg( \mathbb{P}^{B_{1/n}}(\tau_D > h-1/n) \bigg) \\ &= \inf_n \mathbb{E}^w \big( \phi(B(1/n)) \big)\end{align*}$$ where $\phi(y) := \mathbb{P}^y(\tau_D>h-1/n)$ is bounded and borel-measurable. Hence $$w \mapsto \mathbb{E}^w(\phi(B(1/n)))$$ is continuous. This implies the assumption.

(cf. René L. Schilling/Lothar Partzsch, Brownian motion - An Introduction to Stochastic Processes, Theorem 8.16)