Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 29 / 3

Write $D=\mathbb{H}\setminus K$. The mapping-out function is a conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinityEquivalently, $g_K(z)-z\to 0$. The Riemann mapping theorem gives a map onto the half-plane. Since $K$ is bounded, infinity has an analytic real-boundary neighborhood. After sending its image to infinity, the Schwarz reflection principle gives a Laurent series $az+b+O(z^{-1})$ there, with $a>0$ and $b$ real. A real affine automorphism removes $a$ and $b$, giving the stated normalization. Reflection also makes $a_K$ real.

If $g_1,g_2$ both have this normalization, $g_2{g}_1^{-1}$ is a conformal automorphism of the upper half-plane. Such maps are real Möbius transformations. The expansion forces it to fix infinity, have leading coefficient $1$, and have zero translation. It is therefore the identity. Thus the normalized mapping-out function exists and is unique.

The half-plane capacity is the Laurent coefficientIt is real and nonnegative. For planar Brownian motion starting at $z\in D$, let ${\tau }_D$ be its first exit from $D$. The Brownian representation of half-plane capacity isOne may also state the underlying harmonic identity, valid throughout $D$,The process is killed at the hull or real boundary; real-boundary exits contribute zero. These are the Brownian characterizations, with the normalization that a Chordal Loewner equation at speed $2$ produces capacity $2t$.

Let $f=g_K^{-1}$. The reflection at infinity used in part (a) gives analytic expansions there for both inverse maps. In particular,uniformly for large $w\in \mathbb{H}$, including approach close to the real axis.

Fix $R<\infty$. If $g_K$ were unbounded on $D\cap \{|z|\le R\}$, there would be points $z_n$ there with $w_n=g_K(z_n)$ tending to infinity in modulus. The inverse expansion would imply $z_n=f(w_n)=w_n+O(1/w_n)$, contradicting boundedness of $z_n$. Thus $g_K$ is bounded on every bounded portion of its domain, including points arbitrarily near a rough hull boundary. This is the inverse-at-infinity criterion for local boundedness of a mapping-out function.

On the region $|z|>R$ for sufficiently large $R$, the expansion of $g_K$ gives $g_K(z)-z=O(1/z)$. On the remaining bounded region, the preceding bound for $g_K$ and the bound for $z$ give a finite bound for their difference. ThereforeThe argument uses reflection near infinity only. It does not assume that the real part of the map extends continuously at every point of an arbitrary hull; the sharp displacement bound for a compact H-hull is a further quantitative version of this boundedness.

Set $v(z)=\mathrm{Im}{g}_K(z)$, a positive harmonic function on $D$. First justify its boundary behavior. For $z_n\to p\in \partial D$ with $p$ finite, part (c) makes $g_K(z_n)$ bounded. Any subsequential limit $w$ lies in $\overline{\mathbb{H}}$. If $\mathrm{Im}w>0$, continuity of the inverse inside $\mathbb{H}$ would give $p=g_K^{-1}(w)\in D$, a contradiction. ThusThis boundary degeneration under a mapping-out function is valid without a smooth or locally connected hull boundary.

The harmonic function $u(z)=v(z)-\mathrm{Im}z$ consequently has nonpositive finite-boundary values. Its value tends uniformly to zero at infinity, by the Laurent series. On the bounded domain $D\cap \{|z|<R\}$, the maximum principle for harmonic functions bounds $u$ by $\max (0,{\epsilon }_R)$, where ${\epsilon }_R={\sup }_{D\cap \{|z|=R\}}|g_K(z)-z|\to 0$. Let $R\to \infty$ with $z$ fixed. We obtainThis is the height contraction of a hydrodynamically normalized mapping-out function. The exhaustion controls infinity explicitly, which is necessary when using a maximum principle on an unbounded domain.

Here the otherwise undefined printed domain $H$ must mean $D=\mathbb{H}\setminus K$. Let the Brownian motion start at $z\in D$, and let ${\tau }_R$ stop it on reaching height $R$ or leaving $D$. Take $R>\max (\mathrm{Im}z,{\sup }_K\mathrm{Im}w)$. This stopping time is finite almost surely, since it is no later than exit of its imaginary coordinate from $(0,R)$.

By part (d), $0\le v(w)\le R$ in the stopped domain. Its finite-boundary values vanish except on the top boundary, by the argument of part (d). Localization and the optional stopping theorem for this bounded harmonic martingale therefore giveWrite $p_R={\mathbb{P}}_z(\mathrm{Im}{B}_{{\tau }_R}=R)$ and use the uniform displacement bound $M$ from part (c). On the top boundary, $|v(w)-R|\le M$, henceThe stopped imaginary coordinate is itself a bounded martingale. Since its exit height is nonnegative,Consequently $|v(z)-R{p}_R|\le M\mathrm{Im}z/R\to 0$, proving the high-level escape representation of a mapping-out height:If $H$ instead meant the whole upper half-plane, the right side would always be $\mathrm{Im}z$. For example, the mapping-out function of a vertical slit $K=(0,ia]$ is $g_K(w)=\sqrt{w^2+a^2}$ with branch asymptotic to $w$; at $z=iy$, $y>a$, its height is $\sqrt{y^2-a^2}<y$. This demonstrates why killing on the hull is essential to the printed formula.