Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 27 / 4

A compact H-hull is a bounded relatively closed subset $K$ of the complex upper half-plane such that $\mathbb{H}\setminus K$ is a simply connected domain. This does not require its closure to be connected. Its mapping-out function is the unique conformal map $g_K:\mathbb{H}\setminus K\to \mathbb{H}$ with hydrodynamic normalization at infinity,The half-plane capacity isThe coefficient is zero exactly for the empty hull.

In terms of the least radius of a real-centred enclosing half-disc, the sharp displacement bound for a compact H-hull, also called the continuity estimate, isThe differentiability estimate for a mapping-out function is a uniform small-hull expansion: there is an absolute constant $C$ such that if $K\subset \{z:|z-\xi |\le r\}$ with $\xi \in \mathbb{R}$, thenThese are statements of the two requested estimates. The second is not merely a bound for $g_K^{\prime }$: it controls the first-order change of a mapping-out function when a small hull is removed.

Take increasing to mean strict growth on every nonempty time interval; otherwise a constant family has no uniquely determined driving point. The half-plane-capacity parameterization imposed below guarantees strict growth. For $s<t$, define the increment hull in the mapped domain byThis definition includes filling and boundary conventions automatically. Its mapping-out function is $g_{s,t}=g_t\circ g_s^{-1}$. The Loewner local growth property meansfor every finite $T$ within the parameter range. Equivalently, the mapped increments have uniformly vanishing diameter on compact time intervals.

For fixed $t$, the nonempty compact Euclidean closures ${\overline{K}}_{t,t+h}$ are nested as $h$ decreases. Their diameters tend to zero, so their intersection is a singleton. Its point lies on the real axis: the imaginary part of any point in a hull is at most its enclosing radius. Define the Loewner transform bySince ${\xi }_t$ lies in each closure, every point of $K_{t,t+h}$ is at distance at most $2\mathrm{rad}(K_{t,t+h})$ from it.

To prove continuity, choose $z\in K_{t+2h}\setminus K_{t+h}$ and write $w=g_t(z)$, $w^{\prime }=g_{t+h}(z)$. Then $w\in K_{t,t+2h}$, $w^{\prime }\in K_{t+h,t+2h}$ and $w^{\prime }=g_{t,t+h}(w)$. The continuity estimate from part (i) givesThe right-hand side tends to zero uniformly on compact time intervals. Applying the same inequality with the earlier time as the base proves left continuity as well. Thus the Loewner transform is continuous.

Now impose $\mathrm{hcap}(K_t)=2t$. The half-plane-capacity composition rule follows by composing Laurent expansions at infinity and givesFor a point not yet swallowed, put $z_s=g_s(z)$ and $r_{s,t}=2\mathrm{rad}(K_{s,t})$. The increment is contained in the half-disc of radius $r_{s,t}$ centred at ${\xi }_s$. The continuity estimate first proves continuity of $s\mapsto g_s(z)$: its increment is bounded by $3\mathrm{rad}(K_{s,t})$. The differentiability estimate for a mapping-out function then gives, when $t-s$ is small enough,On a compact interval before swallowing the denominator stays away from zero. Divide by $t-s$ and let $t\downarrow s$. The local-growth property makes the error tend to zero. The analogous backward quotient has the same limit, using continuity of $g_s(z)$ and ${\xi }_s$. Thus the Chordal Loewner equation follows:The initial value follows from $\mathrm{hcap}(K_0)=0$, hence $K_0=\varnothing$. Without the capacity parameterization, the same argument gives $d{g}_t(z)=d\mathrm{hcap}(K_t)/(g_t(z)-{\xi }_t)$, interpreted with the capacity clock.