Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 203 / 3

The Koebe quarter theorem says that if $F:\mathbb{D}\to \mathbb{C}$ is a univalent function with $F(0)=0$ and $F^{\prime }(0)=1$, thenEquivalently, an arbitrary univalent function $f$ on the unit disc has $B(f(0),|f^{\prime }(0)|/4)\subseteq f(\mathbb{D})$. The constant cannot be increased: the Koebe function $F(w)=w/(1-w{)}^2$ omits the ray $(-\infty ,-1/4]$.

Assume the domains are proper, so $d$ and $\tilde{d}$ are finite and positive, and interpret the conformal transformation as a bijection. Apply the Koebe quarter theorem toThe open ball $B(z,d)$ lies in $D$, so $F$ is a normalized univalent function. Its image inclusion shows that $B(\tilde{z},d|f^{\prime }(z)|/4)\subseteq \tilde{D}$. Therefore $d|f^{\prime }(z)|/4\le \tilde{d}$, or $|f^{\prime }(z)|\le 4\tilde{d}/d$.

Apply the same argument to the inverse conformal map, whose derivative at $\tilde{z}$ is $1/f^{\prime }(z)$. It gives $\tilde{d}/(4|f^{\prime }(z)|)\le d$. Thus the boundary-distance derivative bound for a conformal bijection isThe full domains need not be simply connected domains: only the two interior discs are used. If the whole plane is allowed as a domain, a conformal bijection onto another plane domain is affine and both domains are the whole plane. For example, $D=\tilde{D}=\mathbb{C}$ and $f(z)=z$ make both boundary distances infinite and the printed ratios $\infty /\infty$ undefined. This is a necessary proper-domain convention. Bijectivity also matters: the injective conformal map $f(w)=w/8$ from the unit disc into itself has $d=\tilde{d}=1$ at zero but $|f^{\prime }(0)|=1/8<1/4$. Thus an interpretation allowing a map into, rather than onto, the second domain makes the lower bound false.

Use the Chordal Loewner equation ${\partial }_t{g}_t=2/(g_t-U_t)$ with $U_t=\sqrt{\kappa }{B}_t$. For a fixed $z\in \mathbb{H}$, let $T_z$ be its Loewner swallowing time, and write $Z_t=X_t+i{Y}_t=g_t(z)-U_t$ and $J_t=|g_t^{\prime }(z)|$ before that time. The Loewner conformal radius is ${{\rm Y}}_t=Y_t/J_t$, half of the conformal radius of $D_t=\mathbb{H}\setminus K_t$ at $z$. The Koebe quarter theorem bounds the conformal radius above by four times the distance to the boundary. For the reverse comparison, if $\varphi :\mathbb{D}\to D_t$ maps zero to $z$, apply the Schwarz lemma to ${\varphi }^{-1}(z+dw)$ on the unit disc, where $d=\mathrm{dist}(z,\partial D_t)$; it gives $d\le |{\varphi }^{\prime }(0)|$. ThusWe use the basic trace theorems that, for $\kappa >8$, the Loewner trace is continuous and Transience of chordal SLE gives $|\gamma (t)|\to \infty$. These facts make its image relatively closed in $\mathbb{H}$; they do not assume the space-filling conclusion. The continuity and transience statements are available in Rohde and Schramm's basic trace theorems.

Take $\rho =\kappa -8>0$ in the supplied SLE interior-point martingale. The derivative exponent vanishes, leavingA nonnegative local martingale is a supermartingale. Applying the optional stopping theorem after localization at its first hit of a level $R$ gives the maximal inequality for a nonnegative supermartingaleThus $C={\sup }_{t<T_z}{M}_t<\infty$ almost surely.

Suppose the Loewner trace avoids some open ball $B(z,r)$ whose closure is inside $\mathbb{H}$. Before $T_z$, the whole open ball is in $D_t$: a connected open ball disjoint from the trace cannot be partly in the unbounded connected component. Hence ${{\rm Y}}_t\ge r/2$. The bound on $M$ forces $S_t\ge c>0$ throughout this interval, for a positive random constant $c$.

The Chordal Loewner equation and its derivative yieldThe first identity forces $T_z<\infty$, since otherwise $Y_t^2\le Y_0^2-4c^{2}t$ becomes negative. At a finite maximal lifetime, $Y_t\to 0$; otherwise the continuous Loewner driving function and the ordinary differential equation continue past that time. Integrating the last identity, using $S_t\ge c$, givescontradicting ${{\rm Y}}_t\ge r/2$. Thus every fixed rational ball inside $\mathbb{H}$ is hit almost surely. A countable intersection makes the trace a dense subset almost surely, and its relatively closed image, established by continuity and transience, then contains all of $\mathbb{H}$:This proves space-filling SLE above parameter eight. It rules out unvisited open regions, rather than inferring visits merely from membership in the filled compact H-hulls. The choice $\rho =\kappa -8$ does not address the critical value eight.