Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 220 / 3

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityIts half-plane capacity is $\mathrm{hcap}(A)=a$.

Set $h_A(z)=\mathrm{Im}(z-g_A(z))$. This is a nonnegative harmonic function on $\mathbb{H}\setminus A$, has boundary value $\mathrm{Im}z$ on the hull boundary, and tends to zero on the real boundary. If $\tau$ is the first exit time of planar Brownian motion from $\mathbb{H}\setminus A$, the optional stopping theorem givesThe hydrodynamic expansion yields $h_A(iy)=\mathrm{hcap}(A)/y+O(y^{-2})$, and therefore the Brownian representation of half-plane capacity

If $A\subseteq C$, Brownian motion exits $\mathbb{H}\setminus C$ no later than it exits $\mathbb{H}\setminus A$. The Strong Markov property and the nonnegative harmonic function $h_A$ show that the expected exit height for $C$ is at least that for $A$. Taking the limits in the Brownian representation of half-plane capacity proves the monotonicity of half-plane capacity

On $\mathbb{H}\setminus (A\cup C)$, the nonnegative harmonic function $h_A+h_C$ dominates the boundary data $\mathrm{Im}z$ on $A\cup C$: at a point of $A$, for example, $h_A(z)=\mathrm{Im}z$ and $h_C(z)\ge 0$. The maximum principle for harmonic functions, or equivalently Brownian motion stopped on the union, givesComparing the coefficients at infinity proves

The assertion is false. LetAfter filling bounded complementary components if necessary, this is a compact H-hull with diameter asymptotic to $2n$. The half-plane capacity of a low rectangle estimate, together with scaling and translation of half-plane capacity, givesThus unbounded diameter need not force unbounded capacity.

The assertion is true. The standard estimate half-plane capacity is bounded by squared diameter givesConsequently $\mathrm{hcap}(A_n)\to \infty$ forces $\mathrm{diam}(A_n)\to \infty$.

Fix $x\in \mathbb{R}\setminus \{0\}$ and, before $x$ is swallowed, put $X_t=(g_t(x)-U_t)/\sqrt{\kappa }$. The Chordal Loewner equation and $U_t=\sqrt{\kappa }{B}_t$ show, after changing the sign of the Brownian motion, thatThus $X$ is a Bessel process of dimensionA Bessel process hits zero exactly when $\delta <2$, which here is equivalent to $\kappa >4$. Hitting zero is precisely the swallowing of a nonzero boundary point by the SLE hull. For $\kappa \le 4$ the trace is simple and swallows no such point; for $\kappa >4$ swallowing occurs through a boundary contact. Hence the trace intersects $\partial \mathbb{H}\setminus \{0\}$ exactly when $\kappa >4$.

For $\kappa \le 4$, part i gives no nonzero boundary intersection. For $\kappa >4$, part i gives a boundary hit almost surely. After any hit, map out the past hull and recenter at the current tip. The Conformal Markov property of SLE says that the future is again an SLE in the remaining domain. Applying part i repeatedly and using Scaling invariance of SLE produces another boundary hit after every finite number of hits. Therefore there are almost surely infinitely many.

For $0<\kappa \le 4$, the trace meets the real boundary only at its starting point, so the intersection has zero Lebesgue measure. Suppose $4<\kappa <8$. A boundary point swallowed by an interval need not itself lie on the trace. The stated fact that $\mathbb{P}({\tau }_x={\tau }_y)>0$ for $0<x<y$, combined with the Strong Markov property and Scaling invariance of SLE at successively nested swallowed intervals, implies that a fixed deterministic $x\ne 0$ is swallowed in an interval rather than hit by the trace with probability one. Equivalently,Applying the Tonelli theorem to the random indicator of the boundary trace givesThe nonnegative random measure is therefore zero almost surely.