Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 202 / 6

For Brownian motion started at $x\in D$, the Itô formula and $\Delta u=0$ show thatis a local martingale. Since $D$ is bounded and $u$ is continuous on the compact set $\overline{D}$, the stopped process is bounded and hence a true martingale. Brownian motion exits every bounded domain almost surely, so $t\wedge \tau \to \tau$. The dominated convergence theorem, continuity at the boundary, and $u=f$ on $\partial D$ give the Brownian representation of the Dirichlet problem

The assertion is false without a boundedness or uniform integrability condition. Take the upper half-planeand $u(x,y)=y$. This function is harmonic and continuous on $\overline{D}$, with boundary value $f=0$. The second coordinate of planar Brownian motion is one-dimensional Brownian motion, so its first hitting time $\tau$ of zero is finite almost surely. NeverthelessThe stopped local martingale $u(B_{t\wedge \tau })$ is not uniformly integrable, which is exactly why optional stopping fails in the limit.

For each $x$, the bounded local martingale $u(x+B_t)$ is a true martingale. Taking expectations givesfor every $t>0$. Thus $u$ equals its convolution with every heat kernel. The convolution is smooth, so the originally Borel function $u$ is smooth. Differentiating the heat semigroup identity at $t=0$ gives $\Delta u=0$, so $u$ is a bounded harmonic function on the plane. The Harmonic Liouville theorem now implies that $u$ is constant.