Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 203 / 1

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ such that $\mathbb{H}\setminus A$ is a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe nonnegative coefficient $a$ is the half-plane capacity $\mathrm{hcap}(A)$.

Put $F(z)=z+z^{-1}$. On the unit semicircle, $F(e^{i\theta })=2\cos \theta$, andThe conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane therefore give the exit density with respect to $d\theta$:As $z\to \infty$ in $\mathbb{H}$,uniformly in $\theta \in [0,\pi ]$. Hence

Let $\sigma$ be the first time Brownian motion started outside the unit half-disc reaches its semicircular boundary. A path that reaches $A$ must first cross that semicircle. The Strong Markov property at $\sigma$ givesTake $z=iy$, multiply by $y$, and let $y\to \infty$. The Brownian representation of half-plane capacity identifies the left side with $\mathrm{hcap}(A)$, while part b givesSince the exit height is between zero and one, the dominated convergence theorem applies and yields

Write $A_r=[-r,r]\times (0,1]$ and take $r\ge 1$. By the Brownian representation of half-plane capacity,On hitting $A_r$, the exit height is at most one. Moreover, $A_r$ lies in the half-disc of radius $\sqrt{r^2+1}\le 2r$. The harmonic measure of that semicircle as viewed from $iy$ is $O(r/y)$: mapping its exterior to $\mathbb{H}$ by $z\mapsto z+(2r{)}^2/z$ reduces the estimate to the Poisson kernel for the upper half-plane on an interval of length $O(r)$. Consequentlyfor large $y$, and $\mathrm{hcap}(A_r)\le Cr$. This is the half-plane capacity of a low rectangle.

Let $R_r=[-r,r]\times (0,1]$ and setThe scaling and translation of half-plane capacity and part i giveOn the other hand,Thus very long, very shallow hulls can have vanishing half-plane capacity.