Past exam of the mathematics course of the University of Cambridge / 2022 / 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 simply connected. Its mapping-out function of a compact H-hull is the unique conformal bijectionwith hydrodynamic normalization

Uniqueness under hydrodynamic normalization at infinity givesandIndeed, each right-hand side maps the required complement conformally onto $\mathbb{H}$ and has expansion $z+O(1/z)$. This is the scaling and translation of half-plane capacity.

The assertion is false. Here $\mathbb{H}\setminus S=\{z:\mathrm{Im}z>1\}$. If a normalized map $g_S$ existed, thenwould be a conformal automorphism of $\mathbb{H}$. Hencewith real coefficients and $ad-bc=1$. Hydrodynamic normalization forces $h(w)\to \infty$ linearly, so $c=0$ and $h(w)=\alpha w+\beta$ with $\alpha >0$ and $\beta \in \mathbb{R}$. Butand $g_S(z)-z\to 0$ would require $\alpha =1$ and $\beta =i$, contradicting $\beta \in \mathbb{R}$.

Suppose a subsequence had $\mathrm{Im}{g}_A(z_n)\ge ϵ>0$. The images cannot tend to infinity, because the inverse mapping-out function satisfies $g_A^{-1}(w)=w+o(1)$ at infinity while $z_n$ remains bounded. A further subsequence therefore converges to some $w\in \mathbb{H}$. Continuity of $g_A^{-1}$ inside $\mathbb{H}$ would givecontradicting $z\in A$. Thus $\mathrm{Im}{g}_A(z_n)\to 0$.

Convergence of the real parts can fail. For the vertical slitthe two sides of the slit at $z=iy$, $0<y<1$, map to the two boundary values $\pm \sqrt{1-y^2}$. Alternating sequences approaching $iy$ from the two sides make $g_A(z_n)$ alternate between neighborhoods of these distinct limits.