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For a continuous Loewner trace generating its hulls, the inverse mapping-out function of a compact H-hull $g_t^{-1}$ extends continuously to the closed complex upper half-plane. The finite trace is a continuous map image of an interval, and the boundary of its unbounded complementary component has local connectedness; the Caratheodory boundary extension theorem gives the extension. In particular $g_t^{-1}(U_t)=\gamma (t)$, while $g_t^{-1}$ maps the open complex upper half-plane into $D_t$. This is the deterministic boundary-extension input of Rohde and Schramm, Theorem 4.1. Under a Conformal Markov property of SLE restart, this relates boundary contacts of the mapped future to contacts with the old hull, without assuming the old trace is simple.

A Jordan domain is the bounded complementary component of a closed simple curve in the plane. The Jordan curve theorem separates the plane into this inside component and one unbounded outside component. A conformal bijection from the disc or complex upper half-plane to this simply connected domain extends to a homeomorphism of their closed conformal boundaries by the Caratheodory boundary extension theorem.

A compact H-hull is a bounded relatively closed subset $K$ of $\mathbb{H}$ with simply connected complement, considered together with its compact closure in $\overline{\mathbb{H}}$. Its unique mapping-out function has hydrodynamic normalization at infinityThe nonnegative coefficient $a_K$ is its half-plane capacity. Thus the half-plane-capacity parameterization here gives $g_t(z)=z+2t/z+O(z^{-2})$. Strict increase means $K_s⊊K_t$ whenever $s<t$; the strictly increasing capacity already excludes equality.

The Loewner local growth property means that the new growth becomes small after removing the old compact H-hull. More precisely set $K_{t,t+h}=\overline{g_t(K_{t+h}\setminus K_t)}$, taking the closure from the complex upper half-plane. On each bounded time interval, the diameters of these image compact H-hulls tend uniformly to zero as $h\downarrow 0$. Equivalently the new growth can be separated from infinity by crosscuts of vanishing diameter in the mapped domain. It is this single-point local growth, together with capacity continuity, that gives a continuous real Loewner driver.

The Loewner transform ${\xi }_t$ is the limiting point at which this mapped growth occurs:The limit is independent of the chosen new-growth points. The Loewner correspondence theorem says that the resulting mapping-out functions satisfyConversely, given a continuous real $\xi$, solve this ordinary differential equation until the maximal lifetime $T_z$ of each $z\in \mathbb{H}$. The surviving set $H_t=\{z:T_z>t\}$ is mapped conformally onto $\mathbb{H}$ by $g_t$; its complement, with the appropriate boundary closure, is $K_t$. The Laurent coefficient is $2t$, and the composition law reconstructs the local growth. A continuous Loewner driver always determines the compact H-hulls; generation by a continuous Loewner trace is an additional property, not a conclusion for every continuous Loewner driver.

A chordal SLE Loewner trace in the complex upper half-plane with parameter $\kappa \ge 0$ is the continuous curve associated with the Loewner driver ${\xi }_t=\sqrt{\kappa }{B}_t$, where $B$ is standard real Brownian motion started at zero. It starts at zero, its compact H-hull at time $t$ is the curve's past with all pockets cut off from infinity filled, andThe capacity clock has $\mathrm{hcap}(K_t)=2t$. For $\kappa =0$ this gives the vertical slit $\gamma (t)=2i\sqrt{t}$.

To prove Scaling invariance of SLE, fix $r>0$ and putThe Laurent series of ${\hat{g}}_t$ has coefficient $2t$, and direct differentiation givesBrownian scaling says $(r^{-1}{B}_{r^{2}t}{)}_{t\ge 0}$ has the same law as $B$. The Loewner drivers therefore agree in law, and uniqueness of the Chordal Loewner equation gives equality in law of the maps, compact H-hulls, and continuous Loewner traces:This identifies the time change $t\mapsto r^{2}t$ as well as the spatial scaling.

For a simply connected Jordan domain $D$ with distinct marked boundary points, choose a conformal bijection $f:\mathbb{H}\to D$ taking zero to the initial point and infinity to the target, and use $f(\gamma )$ as an unparameterized curve. The Caratheodory boundary extension theorem supplies the boundary values of $f$. Any second such map is $f\circ (z\mapsto rz)$ for $r>0$, because the automorphisms of $\mathbb{H}$ fixing zero and infinity are positive dilations. The Scaling invariance of SLE just proved makes the two image laws identical, up to the corresponding deterministic change of capacity clock. Thus this defines SLE consistently in marked Jordan domains. The assumed $|\gamma (t)|\to \infty$ and the continuous boundary extension give $f(\gamma (t))\to f(\infty )$, the specified target. The law is conformally natural and does not impose a distinguished half-plane-capacity parameterization on an arbitrary domain.

Let ${\xi }_t=\sqrt{\kappa }{B}_t$ and $0<\kappa <4$. We use these precise Bessel process facts, which the question permits: for dimension $\delta >2$ a process started at $r>0$ exists for all time, never reaches zero, and has a strictly positive all-time minimum. More quantitatively,No assertion about SLE simplicity or transience is being assumed.

For a real point $x>0$ before its Loewner lifetime, set $Z_t^x=g_t(x)-{\xi }_t$. Its equation isThus $Z^x/\sqrt{\kappa }$ is a Bessel process of dimension $1+4/\kappa >2$. For each rational $x>0$ it stays positive on every finite interval, almost surely. These countably many events hold together. They cover every real $y>0$ as well: choose rational $0<x<y$ and observe that the noise cancels in the difference,Consequently $Z^y$ cannot reach zero while $Z^x$ remains positive. The negative side has the identical argument with the reflected Brownian motion.

This excludes Loewner trace contact with every real point other than zero, rather than just with each fixed rational point. On any finite interval a positive gap at a real point stays bounded away from zero. Continuous dependence of the Chordal Loewner equation then solves it on a complex neighborhood of that point, and the analytic flow extends the map conformally across a real interval there. Its real derivative is $\exp (-2{\int }_0^t(Z_s^x{)}^{-2}ds)>0$, and its image stays away from the driving point. The Loewner trace tip therefore cannot be there. We have proved $\gamma [0,\infty )\cap \mathbb{R}\subseteq \{0\}$.

Now use Brownian motion independent increments to restart at every deterministic rational time $s>0$. The centered future maps have Loewner driver ${\xi }_{s+t}-{\xi }_s$ and therefore the original SLE law, independently of the past. Its Loewner trace stays in $\mathbb{H}\cup \{0\}$ by the preceding argument. The inverse map $g_s^{-1}$ extends continuously to the real boundary for a continuous trace-generated compact H-hull: the finite Loewner trace gives a locally connected boundary, and the Caratheodory boundary extension theorem applies. Its value at ${\xi }_s$ is $\gamma (s)$. Thus the future in the original plane can meet $K_s\cup \mathbb{R}$ only at $\gamma (s)$. All these restart conclusions hold simultaneously for rational $s$.

Suppose $\gamma (u)=\gamma (v)$ with $u<v$. The Loewner trace cannot be constant throughout $(u,v)$, since its compact H-hull capacity increases strictly. By continuity there is a rational $s\in (u,v)$ with $\gamma (s)\ne \gamma (u)$. But $\gamma (v)=\gamma (u)$ lies in $K_s\cup \mathbb{R}$, contradicting the restart conclusion at $s$. This also excludes any return to the starting point. Hence the Loewner trace is simple and lies in the open complex upper half-plane at every positive time.

For transience, first strengthen the boundary conclusion to avoidance of the closure at each fixed real point. Fix $x>0$ and $0<r<x/4$. If the now-simple curve first enters the closed half-disc of radius $r$ about $x$ at time $\sigma$, join its tip to $x$ by the segment inside that disc. Before $\sigma$ the segment misses the Loewner trace. Together with the Loewner trace and the real interval from zero to $x$, it bounds a pocket whose boundary includes the positive side of the Loewner trace and $[0,x/2]$. A Brownian motion path from high on the imaginary axis reaching that boundary arc must enter the pocket through the segment. The maximum principle for harmonic functions bounds its harmonic measure by the chance of hitting the entire half-disc.

The boundary arc maps under $g_{\sigma }$ to $[{\xi }_{\sigma },g_{\sigma }(x/2)]$. Its harmonic measure from $iY$, as $Y\to \infty$, isThe half-disc maps out by $z\mapsto z+r^2/(z-x)$; its semicircle becomes an interval of length $4r$. Its harmonic measure has asymptotic $4r/(\pi Y)$. Comparing and letting $Y\to \infty$ yieldsBut $m_x={\inf }_{t\ge 0}(g_t(x/2)-{\xi }_t)>0$ by the all-time minimum of a transient Bessel process fact. The curve cannot enter any such disc with $4r<m_x$. Therefore it stays a strictly positive distance from $x$ for all time. Reflection proves the same for fixed $x<0$. This is the boundary approach forces a small SLE Bessel gap argument; the crosscut comparison is also described in the boundary-closure lemma in the primary SLE paper.

Map out the simple initial slit at time one. The starting point zero has two distinct real prime end images $b_{-},b_{+}$ under $g_1$, lying on opposite sides of ${\xi }_1$. Conditionally on the initial slit, the image future is a fresh independent SLE translated by ${\xi }_1$. The just-proved fixed-point closure avoidance applies to each of $b_{-}-{\xi }_1$ and $b_{+}-{\xi }_1$. Therefore neither $b_{-}$ nor $b_{+}$ lies in the closure of the mapped future. If a sequence of future Loewner trace points approached zero in the original domain, their images would approach one of these two prime ends. This is impossible. ThusFinally Scaling invariance of SLE gives $R_T:={\inf }_{t\ge T}|\gamma (t)|\stackrel{d}{=}\sqrt{T}{R}_1$. For every fixed $R>0$,The events of returning to the closed radius-$R$ disc after time $T$ decrease as $T$ increases. Continuity of probability on decreasing events implies that the probability of returns at arbitrarily late times is zero. Taking the countable intersection over positive integer $R$ provesThis proves actual eventual escape, not merely unboundedness of the growing compact H-hulls.

The boundary assertion is interpreted at positive finite times: a chordal SLE already starts on the domain boundary, and its marked target is another boundary point. The precise conclusion in the complex upper half-plane is $\gamma (0,\infty )\subseteq \mathbb{H}$ almost surely.

For $0<\kappa \le 4$ and a fixed real $x>0$, the centered Boundary-point Bessel flow for SLE $V_t=g_t(x)-U_t$ solvesThus $R_t=V_t/\sqrt{\kappa }$ is a Bessel process driven by $-B$ of dimensionThe Hitting-zero classification for a Bessel process says that, started positively, it never hits zero when $\delta \ge 2$. For $x<0$, reflect the equation and apply the same result. Therefore every fixed nonzero boundary point has infinite swallowing time. A countable intersection gives this simultaneously for all nonzero rational boundary points.

To extend this countable conclusion to every real $x>0$, choose rational $q$ with $0<q<x$. The difference of two centered boundary solutions satisfiesso it remains positive while both flows exist. On any finite time interval, $V_t(q)$ is bounded away from zero, hence so is $V_t(x)$. Moreover $V_t(x)=x-U_t+{\int }_0^{t}2/V_s(x)ds$ remains bounded there, so the differential equation continues for the whole interval. Its uniform separation from the Loewner driving function also allows each flow map to continue as a holomorphic function on a complex neighborhood of $x$. Thus $x$ is outside $K_t$ and cannot be visited by the Loewner trace. Reflection gives the negative real axis. This uses countably many Bessel processes followed by deterministic flow comparison, rather than an uncountable union of probability-zero events.

To exclude a return to the starting point, use the boundary extension of the inverse map for a continuous Loewner trace: $g_s^{-1}$ is continuous on the closed complex upper half-plane and maps $U_s$ to $\gamma (s)$. This follows from the continuous trace theorem and the Caratheodory boundary extension theorem; it does not assume simplicity. At each deterministic rational time $s$, the Conformal Markov property of SLE makes the mapped future a fresh SLE, so its trace lies in $\mathbb{H}\cup \{0\}$ by the preceding argument. Mapping back with $g_s^{-1}$ therefore showsIf a point $q=\gamma (r)$ were revisited at $t>r$, choose rational $s\in (r,t)$ with $\gamma (s)\ne q$. Such an $s$ exists because continuity and strictly increasing half-plane capacity forbid a constant trace on an interval. But $q\in K_s\cup \mathbb{R}$, contradicting the displayed inclusion. This excludes all self-contacts, including returns to zero, and proves the simple curve property needed when discussing later swallowing times.

For $\kappa =0$, the deterministic Loewner trace is $\gamma (t)=2i\sqrt{t}$, so the conclusion follows directly. Hence the boundary-intersection threshold for SLE givesMapping to another marked simply connected domain carries positive-time points into its interior. The literal statement including the starting point is false, since $\gamma (0)$ is prescribed to be on the domain boundary.