Let $B_t$ be $d$-dim Brownian motion. Let $U \subset \mathbb{R}^d$ be a bounded domain and denote by $\tau_U = \inf\{t \mid B_t\notin U\}$ the exit time. For $x \in U$ and $A \subset \partial U$ let $\mu_{\partial U}^x (A)= \mathbb{P}^x(B_{\tau_{\partial U}}\in A)$ be the harmonic measure on $\partial U$. I have two questions:
- If $\sigma$ denotes the surface measure on $\partial U$, then there exists a Radon Nikodym derivative $f_{\partial U}^x= \frac{d\mu_{\partial U}^x}{d \sigma}$ such that $$\mu_{\partial U}^x(A) = \int_A f_{\partial U}^x(y) d\sigma$$ for any measurable $A \subset \partial U$. This follows by the Radon Nikodym theorem. Is that correct?
- What do we know about $f$? Are there any explicit formulae or is it possible to show that $f$ is bounded? In my case I would need an $L^2$ estimate, i.e. an upper bound for $\lVert f_{\partial U}^x\rVert_{\partial U}$. Does anybody have an idea for this? Or a good reference?
Considering the question whether $\frac{\mu^x_{\partial U}}{\sigma}$ is bounded I think I already found a case where this is not true, namely when $U$ is the unit ball. Observe the following statement, copied from Durrett - Brownian motion and martingales in Analysis, p. 36,
where $D$ is the unit sphere.
If I am not misunderstanding things, this shows that in this case $\frac{\mu^x_{\partial U}}{\sigma} = \frac{1-|x|^2}{|x-y|^d}$ (up to some multiplicative constant, since $\nu$ is normalized), which is not bounded. Hence, in general, $\frac{\mu^x_{\partial U}}{\sigma}$ is not bounded. Right?