Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 203 / 1 / b

Let $d=\mathrm{diam}(A)$. Unless $A$ is empty, its closure meets the real axis; otherwise a loop in $\mathbb{H}\setminus A$ surrounding $A$ could not contract, contrary to $\mathbb{H}\setminus A$ being a simply connected domain. Choose $x\in \overline{A}\cap \mathbb{R}$. Then $A\subseteq \{z\in \mathbb{H}:|z-x|\le d\}$.

The unit half-disc is a compact H-hull with mapping-out function $z+z^{-1}$, so its half-plane capacity is one. The monotonicity of half-plane capacity and its scaling rule now giveThus the assertion holds with the universal constant $c=1$ under this normalization.

For $r\ge 1$, let $R_r=[-r,r]\times (0,1]$. The Brownian representation of half-plane capacity givesOn hitting $R_r$ the imaginary part is at most one, while the harmonic measure estimate supplied in the question shows that the probability of reaching a disc of radius $O(r)$ containing $R_r$ is $O(r/y)$. Hence $\mathrm{hcap}(R_r)\le Cr$, the half-plane capacity of a low rectangle bound.

SetThe scaling and translation of half-plane capacity giveswhereas $\mathrm{diam}(A_n)=\sqrt{4+n^{-2}}\to 2$. This supplies the required sequence.