Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 36 / 1

Write $D=\mathbb{H}\setminus K$. By definition of a compact H-hull, $D$ is a proper simply connected domain, and it agrees with the complex upper half-plane outside a sufficiently large disc. The Riemann mapping theorem supplies a conformal isomorphism $f:D\to \mathbb{H}$. Choose its boundary normalization so that the prime end at infinity maps to infinity. This prime end is unambiguous because the domain has an ordinary straight real boundary near infinity.

The Schwarz reflection principle, applied there in the coordinate $1/z$, extends $f$ across the real boundary near infinity. It has a simple pole there, with Laurent seriesThe pole is simple because the reflected map is locally conformal at infinity; its leading coefficient is positive because it maps the upper side to the upper side. Postcomposing with the half-plane automorphism $w\mapsto (w-b)/a$ givesThis is hydrodynamic normalization at infinity.

If $g_1,g_2$ both satisfy it, $g_2\circ g_1^{-1}$ is a conformal automorphism of the upper half-plane. The asymptotics force it to fix infinity, so it has the form $w\mapsto \alpha w+\beta$ with $\alpha >0$ and $\beta \in \mathbb{R}$. The same asymptotics force $\alpha =1$, $\beta =0$. Hence the normalized mapping-out function exists and is unique. The argument requires reflection only outside a large disc, not regularity of the hull's entire boundary.

For the Joukowski map $g(z)=z+z^{-1}$,It maps the exterior upper half-disc $D_0=\{z\in \mathbb{H}:|z|>1\}$ into $\mathbb{H}$. It is injective there: $g(z)=g(w)$ gives $(z-w)(1-1/(zw))=0$, and the second factor cannot vanish when both moduli exceed one. It is onto as well. For $v\in \mathbb{H}$, the two roots of $z^2-vz+1=0$ have product one. Neither lies on the unit circle, since $z+z^{-1}$ is real there. Exactly one root has modulus greater than one, and the displayed imaginary-part formula makes its imaginary part positive. That root belongs to $D_0$ and maps to $v$.

Since $g_{K_0}$ takes values in $\mathbb{H}$, its domain must be contained in $D_0$. The established injectivity and surjectivity of $g:D_0\to \mathbb{H}$ mean that no proper subdomain of $D_0$ can map onto all of $\mathbb{H}$ using this same function. ThereforeThe half-plane capacity is the coefficient of $1/z$ in the normalized expansion. If hulls are regarded as closed subsets of $\overline{\mathbb{H}}$, take the closure of this half-disc, including its diameter; its relative half-plane hull and capacity are unchanged. This is the half-plane capacity of a half-disc.

At a real point $x$ outside the hull's closure, the Schwarz reflection principle makes the mapping-out function analytic across the boundary and $g_K^{\prime }(x)>0$. Use the course's Brownian excursion restriction identity: for an excursion from $x$ to infinity in the complex upper half-plane,One way to obtain this identity is to start its Doob h-transform at $x+i\epsilon$: its avoidance probability is $\mathrm{Im}{g}_K(x+i\epsilon )/\epsilon$, which tends to $g_K^{\prime }(x)$ by reflection. Thus the boundary derivative is an excursion avoidance probability, and in particular it is at most one.

Let $D$ be the filled unit half-disc. Since $K\subseteq D$, avoiding $D$ implies avoiding $K$. The monotonicity of boundary derivatives of mapping-out functions and part (b) give, for $|x|>1$,The upper bound is attained by the empty hull and the lower by the filled half-disc, so the comparison is sharp on either real tail.