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The time-changed diffusion has generator $\mathcal{A}=(\kappa /2){\partial }_{zz}+2(1/z+1/(z-1)){\partial }_z$. For $1<z<R$, its boundary hitting probability from a diffusion scale function isTo see this directly, stop the scale local martingale on $[1+ϵ,R]$ and pass to $ϵ\downarrow 0$. Exit from $(1,R)$ occurs in finite clock time: the scale density behaves as $(z-1{)}^{-4/\kappa }$ and the speed density as $(z-1{)}^{4/\kappa }$, so the usual scale-times-speed exit integral is finite at $1$. More explicitly, with $m(v)=2/(\kappa s_{\kappa }^{\prime }(v))$ the finite-interval diffusion exit Green kernel givesThe equation $\mathcal{A}u=-1$ follows by differentiation and the derivative jump at $v=z$, so stopping the corresponding Itô identity gives this exit bound. The integrand near $1$ is $O(v-1)$; elsewhere it is bounded. At large $z$ the drift is bounded, excluding explosion at infinity in finite clock time.

Here is why these exits describe the original swallowing event. In clock time,If $\hat{Z}$ hits $1$ at finite $u$, its integrated stochastic differential equation shows ${\int }_0^{{\tau }_1}({\hat{Z}}_v-1{)}^{-1}dv<\infty$: the Brownian term has a finite limit and both positive drift integrals must then have finite limits. Consequently $D({\tau }_1)>0$, $X$ hits zero and $Y$ stays positive. This gives $T(x)<T(y)$.

Conversely $T(x)$ is finite by part (b). If its clock ends at a finite value, the diffusion must hit $1$; an interior endpoint with positive $D$ would leave $X$ positive. If the clock runs forever without hitting $1$, the standard one-dimensional scale classification gives escape to infinity whenever $S_{\kappa }<\infty$. This cannot represent a strict swallowing event, which has $D\to Y_{T(x)}>0$ and $Z\to 1$. Hence the surviving event represents $T(x)=T(y)$. This reasoning also explains why a simultaneous collision must be treated separately rather than assigned the boundary value at $1$.

For $\kappa \ge 8$, the scale is unbounded. The maximal inequality for a nonnegative supermartingale gives $\mathbb{P}(\sup s_{\kappa }(\hat{Z})\ge s_{\kappa }(R))\le s_{\kappa }(z)/s_{\kappa }(R)$; letting $R\to \infty$ rules out escape. Finite-interval exit then forces a hit at $1$. For $4<\kappa <8$, taking $R\to \infty$ givesThe strict-event probability is also positive, since $0<s_{\kappa }(z)<S_{\kappa }$ for finite $z$.

For $\kappa \ge 8$, first intersect the probability-one strict-order events over rational $0<a<b$. The real boundary flow gives nondecreasing swallowing times on $(0,\infty )$. For any $0<x<y$, choose rational $x<a<b<y$; then $T(x)\le T(a)<T(b)\le T(y)$. This proves the simultaneous assertionThe range is positive points, as in the definition of $F$; reflection reverses the ordering on the negative axis. At the printed endpoint $\kappa =4$, each fixed positive point has $T=\infty$. Thus $\mathbb{P}(T(x)=T(y))=1$ if equality of extended times is allowed, but there is no finite simultaneous-swallowing assertion there. The preceding formulas assume $\kappa >4$.

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.

For $\kappa >8$, the Loewner trace visits every point of the complex upper half-plane almost surely. Choose $\rho =\kappa -8$ in the SLE interior-point martingale to obtain $M_t={{\rm Y}}_t^{(\kappa -8)/8}{S}_t^{-(\kappa -8)/\kappa }$. The maximal inequality for a nonnegative supermartingale bounds its running supremum. If a fixed interior ball were avoided, its center would have ${{\rm Y}}_t\ge c_1>0$ before swallowing, forcing $S_t\ge c_2>0$. But the Chordal Loewner equation gives $d{Y}_t^2/dt=-4S_t^2$, forcing a finite lifetime with $Y_t\to 0$, and $d\log {{\rm Y}}_t/d\log Y_t=2S_t^2$ then forces ${{\rm Y}}_t\to 0$, a contradiction. Countably many rational balls give a dense trace; continuity and Transience of chordal SLE make its image relatively closed, so it is all of the complex upper half-plane. This argument does not cover the critical parameter eight.