Kakutani's solution of the Dirichlet problem

Remarks

Proof idea: $\tau$ is a stopping time, and $\tau < \infty$ because $D$ is bounded. Take a ball $\overline{B}(x, r) \subset D$. The strong Markov property at the exit time of the ball and rotation invariance show that $u(x)$ is the average of $u$ over the sphere $\partial B(x, r)$. A bounded measurable function with this mean value property is harmonic. At a regular point $\zeta$, a path started near $\zeta$ leaves $D$ quickly and close to $\zeta$, with high probability, which gives clause 2. Example: for the unit disc and $x = 0$, the exit point $B_\tau$ is uniform on the circle, so $u(0)$ is the average of $\varphi$, as the mean value property requires. Regularity is needed: on the punctured disc $D = \set{0 < \abs{x} < 1}$ in $\R^2$, Brownian motion never hits $0$, so $u$ ignores $\varphi(0)$. With $\varphi(0) = 1$ and $\varphi = 0$ on the unit circle one gets $u \equiv 0$, which does not tend to $1$ at $0$.

Going deeper

For a student

The theorem identifies classical potential theory with the theory of Brownian motion. The probabilistic regularity criterion, $P^\zeta(\tau = 0) = 1$, agrees with the analytic notion of a regular point given by barriers or Wiener's criterion. The formula also gives Monte Carlo methods for harmonic functions, such as the walk-on-spheres method. The distribution of $B_\tau$ under $P^x$ is the harmonic measure of $D$ at $x$.