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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.

This is Schwarz lemma. Define $g(z)=f(z)/z$ for $z\ne 0$ and $g(0)=f^{\prime }(0)$. The removable singularity makes $g$ holomorphic. Applying the maximum modulus principle to $g$ on disks of radius $r<1$, then letting $r\uparrow 1$, gives $|g(z)|\le 1$. Hence

Because $f(D)$ avoids the closed negative real axis, its principal square root $g=\sqrt{f}$ is holomorphic, satisfies $g(0)=1$, and has $\mathrm{Re}g>0$. The Cayley transform between the half-plane and diskmaps the right half-plane into the unit disk and has $h(0)=0$. By Schwarz lemma, $|h(z)|\le |z|$. When $|z|\le 1/2$,Thusso one may take the universal constant $C=9$.

In local coordinates around $p\in R$ and $f(p)\in S$, a nonconstant analytic map has the formAfter shrinking the chart, $h$ has an analytic $m$th root, so a change of coordinate makes the map $z\mapsto z^m$. It maps small discs onto neighborhoods of the origin; hence every nonconstant analytic map of Riemann surfaces is open. If $R$ is compact and $S$ connected, $f(R)$ is compact and therefore closed, and it is also nonempty and open. Thus $f(R)=S$.

For $f:\mathbb{D}\to \mathbb{D}$ with $f(0)=0$, the removable-singularity theorem makes $g=f/z$ analytic. On $|z|=r$, $|g(z)|=|f(z)|/r\le 1/r$, and the maximum principle gives the same bound throughout ${\mathbb{D}}_r$. Applying the Schwarz lemma to an automorphism $h$ and to $h^{-1}$ shows that every automorphism fixing zero is

If $F:\mathbb{C}\to \mathbb{C}$ is an analytic isomorphism fixing zero, continuity of $F^{-1}$ implies that the preimage of every closed disc is compact. Hence $|F(z)|\to \infty$ as $|z|\to \infty$, and defining $F(\infty )=\infty$ gives an analytic isomorphism of the Riemann sphere.

Consequently an automorphism of $\mathbb{D}$ fixing zero does extend to $\mathbb{C}$. A general disc automorphism need not: for $a\ne 0$,has a finite pole and cannot extend to an entire isomorphism.