Past exam of the mathematics course of the University of Cambridge / 2010 / iii / Paper 27 / 1 / a / Solution

A compact H-hull is a bounded relatively closed set $K\subset \mathbb{H}$ whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Equivalently one can retain its compact closure in $\overline{\mathbb{H}}$; real boundary points do not affect the mapping-out function. The Riemann mapping theorem and Schwarz reflection principle near infinity give a unique conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinity. Its Laurent series isThe half-plane capacity is $\mathrm{hcap}(K)=a_K$. The missing constant term is part of the normalization, not a choice to be made again when comparing different hulls.

Here are the conformal map facts we use to justify positivity and comparison. The inverse $f_K:\mathbb{H}\to D\subset \mathbb{H}$ has expansion $f_K(w)=w-a_K/w+O(w^{-2})$. The upper-half-plane Schwarz-Pick theorem, applied to $w$ and $iY$ and then letting $Y\to \infty$, implies $\mathrm{Im}{f}_K(w)\ge \mathrm{Im}w$. Indeed the inequality between pseudohyperbolic distances givesand multiplying by $Y$ gives the height comparison in the limit. At $w=iY$, the inverse expansion reads $\mathrm{Im}{f}_K(iY)=Y+a_K/Y+O(Y^{-2})$, so $a_K\ge 0$.

For $r>0$, uniqueness of the mapping-out function givesConsequentlyTranslation by a real number $b$ similarly gives $g_{K+b}(z)=b+g_K(z-b)$ and leaves half-plane capacity unchanged.

If $K\subset A$ are compact H-hulls, their mapping-out functions compose as $g_A=g_L\circ g_K$, where $L$ is the filled image of $A\setminus K$. Comparing Laurent series yieldsThis proves monotonicity of half-plane capacity, rather than assuming it from inclusion alone. The filled half-disc $A=\{z\in \mathbb{H}:|z-b|\le R\}$ has mapping-out functionwhich takes its semicircular boundary to the real line and is a conformal bijection onto $\mathbb{H}$. Thus its half-plane capacity is $R^2$. Enclose $K$ in such a half-disc, compare capacities, and minimize over real centres and radii. This proves the sharp boundIf one defines the radius by an infimum, take $R$ strictly larger than that infimum and then let $R$ decrease; no attainment assumption is needed.