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Before the Loewner swallowing time of $z$, set $Z_t=g_t(z)-U_t=X_t+i{Y}_t$ and ${{\rm Y}}_t=Y_t/|g_t^{\prime }(z)|$. Since the conformal radius of the complex upper half-plane at $Z_t$ is $2Y_t$, the transformation rule gives $2{{\rm Y}}_t={\mathrm{crad}}_{D_t}(z)$. The Koebe quarter theorem and the interior-disc comparison give $\mathrm{dist}(z,\partial D_t)/2\le {{\rm Y}}_t\le 2\mathrm{dist}(z,\partial D_t)$. Differentiating the Chordal Loewner equation gives $\boxed{{\partial }_t\log {{\rm Y}}_t=-4Y_t^2/|Z_t{|}^4}$. Thus this quantity decreases as the remaining domain shrinks.

If $f$ is a univalent function on the unit disc, then $B(f(0),|f^{\prime }(0)|/4)\subseteq f(\mathbb{D})$. The constant $1/4$ is sharp, as shown by the Koebe function. This gives useful comparisons between the derivative of a conformal map, its conformal radius, and its distance to the domain boundary.

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.

Set $F(w)=(\varphi (w)-\varphi (0))/w$ for $w\ne 0$. The singularity is removable, with $F(0)={\varphi }^{\prime }(0)$. Injectivity of the conformal map makes $F(w)\ne 0$ away from zero, and conformality makes $F(0)\ne 0$ as well. Since the disc is simply connected, $F$ has a holomorphic logarithm. Therefore $\psi =\log |F|$ is a harmonic function, including at zero, and $\psi (0)=\log |{\varphi }^{\prime }(0)|$.

The mean value property for harmonic functions gives, for every $0<r<1$,For general $D$, boundary values mean radial limits, rather than a continuous extension of $\varphi$ to every point of the circle. The boundary logarithmic mean of a univalent function justifies taking $r\uparrow 1$: the Koebe distortion theorem bounds $|F|$ below by $|{\varphi }^{\prime }(0)|/4$, while the standard integral-mean bound for a univalent function, ${\sup }_{r<1}\int |\varphi (r{e}^{i\theta }){|}^{p}d\theta <\infty$ for $0<p<1/2$, gives uniform integrability of its positive logarithm. Radial limits exist almost everywhere, and passage to the integral follows. Thus, writing $dw=d\theta$ for arc length on the unit circle,The expression on the left is the logarithm of the conformal radius of $D$ at $z=\varphi (0)$.

In the unit disc, the relevant pointed Brownian loop measure is rooted at $0$ and normalized by its conformal deficit: for simply connected $U$ containing $0$, its mass of loops not contained in $U$ is minus the logarithm of the conformal radius. Its outer-boundary pushforward gives the corresponding pointed simple-loop restriction measure. It is not the full unrooted loop measure or a probability law of a fixed-duration Brownian bridge.

Radial SLE grows from a boundary point to an interior target. In the unit disc from $1$ to $0$, its conformal-radius parameterization satisfiesThe target remains in the remaining simply connected component; the time parameter records its decreasing conformal radius.