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For a Loewner chain with half-plane capacity $2t$, scale invariance means $({\lambda }^{-1}{K}_{{\lambda }^{2}t}{)}_{t\ge 0}$ has the original law for every $\lambda >0$. Its mapping-out maps are ${\lambda }^{-1}{g}_{{\lambda }^{2}t}(\lambda z)$ and its Loewner driving function is ${\lambda }^{-1}{U}_{{\lambda }^{2}t}$. This follows by differentiating the Chordal Loewner equation. Brownian scaling gives this property for Schramm–Loewner evolution.

After time $t$, map remaining domains by $g_t-U_t$. The new mapping-out function at elapsed time $s$ is ${\hat{g}}_s(z)=g_{t+s}(g_t^{-1}(z+U_t))-U_t$. Direct differentiation of the Chordal Loewner equation proves that its driver is $U_{t+s}-U_t$, and its half-plane capacity is $2s$. Defining transformed hulls as complements of transformed domains avoids ambiguity at the old hull boundary.

For a vertical slit of height $h>0$ attached at $x\in \mathbb{R}$, the mapping-out function of a compact H-hull is $g(z)=x+\sqrt{(z-x{)}^2+h^2}$, with the branch asymptotic to $z-x$ at infinity. Its half-plane capacity is $h^2/2$.

This capacity measures the asymptotic probability that planar Brownian motion hits a compact H-hull before the real axis. It scales linearly with length and is distinct from half-plane capacity. A vertical slit of height $h$ has capacity $2h$; a half-disc of radius $r$ has capacity $4r$. Monotonicity of Brownian hitting probabilities gives $\mathrm{cap}(K)\le 4\mathrm{rad}(K)$, where the radius is that of the least real-centred enclosing half-disc. For finite slit hulls the capacity is the length of the image of the two-sided absorbing boundary under the mapping-out function.

A chordal Loewner chain is an increasing family of compact H-hulls, usually parameterized by half-plane capacity, whose mapping-out functions evolve according to the Chordal Loewner equation.

An increasing family of compact H-hulls of half-plane capacity $2t$ with the Loewner local growth property corresponds uniquely to a continuous real Loewner driving function. Its mapping-out functions solve the Chordal Loewner equation up to the maximal lifetime of each point. Conversely each continuous driver generates that hull family. A continuous Loewner trace is an additional property, not a consequence for every continuous driver. For SLE it is supplied by the trace-existence theorem.

With half-plane capacity $2t$, $f_t=g_t-W_t$ and $G_{\mathbb{H}}(z,w)=\log |(z-\overline{w})/(z-w)|$, differentiating $G_{D_t}(z,w)=G_{\mathbb{H}}(f_t(z),f_t(w))$ gives the displayed identity. The translation by the driver cancels in the Green function. This rank-one covariance loss balances the quadratic covariation of the harmonic mean in the SLE4 coupling with a Gaussian free field.

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.

For the Joukowski map $g(z)=z+z^{-1}$,It maps the exterior upper half-disc $D_0=\{z\in \mathbb{H}:|z|>1\}$ into $\mathbb{H}$. It is injective there: $g(z)=g(w)$ gives $(z-w)(1-1/(zw))=0$, and the second factor cannot vanish when both moduli exceed one. It is onto as well. For $v\in \mathbb{H}$, the two roots of $z^2-vz+1=0$ have product one. Neither lies on the unit circle, since $z+z^{-1}$ is real there. Exactly one root has modulus greater than one, and the displayed imaginary-part formula makes its imaginary part positive. That root belongs to $D_0$ and maps to $v$.

Since $g_{K_0}$ takes values in $\mathbb{H}$, its domain must be contained in $D_0$. The established injectivity and surjectivity of $g:D_0\to \mathbb{H}$ mean that no proper subdomain of $D_0$ can map onto all of $\mathbb{H}$ using this same function. ThereforeThe half-plane capacity is the coefficient of $1/z$ in the normalized expansion. If hulls are regarded as closed subsets of $\overline{\mathbb{H}}$, take the closure of this half-disc, including its diameter; its relative half-plane hull and capacity are unchanged. This is the half-plane capacity of a half-disc.

For chordal SLE with parameter $\kappa$, the Loewner trace $\gamma$ starts at zero and is parametrized by half-plane capacity $2t$. Let $D_t$ be the unbounded component of $\mathbb{H}\setminus \gamma [0,t]$, and let $K_t=\mathbb{H}\setminus D_t$ be its filled compact H-hull. When the trace is not simple, the hull includes the regions it disconnects from infinity. The associated mapping-out functions of compact H-hulls have expansionand satisfy the Chordal Loewner equationwhere $B$ is standard real Brownian motion. The equation is solved until $z$ is swallowed by the hull. The Loewner transform $\xi$ records the real boundary point at which the mapped hull grows; its relation to the trace isEquivalently the tip maps to ${\xi }_t$ in the appropriate boundary prime-end sense. Thus the flow encodes the surviving domains, the transform drives that flow, and the trace generates its hulls. For $\kappa =0$, the driver is identically zero, $g_t(z)=\sqrt{z^2+4t}$ with the branch asymptotic to $z$, and ${\gamma }_t=2i\sqrt{t}$.

A compact H-hull is a bounded relatively closed set $K\subset \mathbb{H}$ whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Equivalently one can retain its compact closure in $\overline{\mathbb{H}}$; real boundary points do not affect the mapping-out function. The Riemann mapping theorem and Schwarz reflection principle near infinity give a unique conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinity. Its Laurent series isThe half-plane capacity is $\mathrm{hcap}(K)=a_K$. The missing constant term is part of the normalization, not a choice to be made again when comparing different hulls.

Here are the conformal map facts we use to justify positivity and comparison. The inverse $f_K:\mathbb{H}\to D\subset \mathbb{H}$ has expansion $f_K(w)=w-a_K/w+O(w^{-2})$. The upper-half-plane Schwarz-Pick theorem, applied to $w$ and $iY$ and then letting $Y\to \infty$, implies $\mathrm{Im}{f}_K(w)\ge \mathrm{Im}w$. Indeed the inequality between pseudohyperbolic distances givesand multiplying by $Y$ gives the height comparison in the limit. At $w=iY$, the inverse expansion reads $\mathrm{Im}{f}_K(iY)=Y+a_K/Y+O(Y^{-2})$, so $a_K\ge 0$.

For $r>0$, uniqueness of the mapping-out function givesConsequentlyTranslation by a real number $b$ similarly gives $g_{K+b}(z)=b+g_K(z-b)$ and leaves half-plane capacity unchanged.

If $K\subset A$ are compact H-hulls, their mapping-out functions compose as $g_A=g_L\circ g_K$, where $L$ is the filled image of $A\setminus K$. Comparing Laurent series yieldsThis proves monotonicity of half-plane capacity, rather than assuming it from inclusion alone. The filled half-disc $A=\{z\in \mathbb{H}:|z-b|\le R\}$ has mapping-out functionwhich takes its semicircular boundary to the real line and is a conformal bijection onto $\mathbb{H}$. Thus its half-plane capacity is $R^2$. Enclose $K$ in such a half-disc, compare capacities, and minimize over real centres and radii. This proves the sharp boundIf one defines the radius by an infimum, take $R$ strictly larger than that infimum and then let $R$ decrease; no attainment assumption is needed.

We use the following standard planar Brownian motion facts: its exit probability distribution is harmonic measure; conformal maps preserve that probability distribution after an increasing time change; and the Poisson kernel for the upper half-plane gives its real-boundary exit density. For rough hulls the boundary is interpreted through prime ends, with boundary correspondence at harmonic-measure-almost every exit point. This avoids imposing a smoothness or local-connectedness assumption on $K$.

Under $g_K$, the exit event through $K$ corresponds to a measurable set $E_K\subset \mathbb{R}$ of boundary images. It is bounded: the mapping-out function extends by Schwarz reflection along the real line outside a sufficiently large interval enclosing $K$, and its boundary near infinity corresponds to the original real boundary near infinity. All prime ends representing $K$ therefore have images in a bounded interval. The boundary correspondence assertion can equivalently be formulated using the almost-everywhere limits of $g_K^{-1}$; changing null sets does not change $E_K$'s measure.

Write $g_K(iy)=u_y+i{v}_y$. The hydrodynamic normalization at infinity gives $u_y=O(y^{-2})$ and $v_y=y-a_K/y+O(y^{-2})$. By conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane,On the bounded set $E_K$, the integrand multiplied by $y$ tends uniformly to one. Therefore the limit exists and is finite:This is the harmonic capacity from infinity in the upper half-plane with the normalization $c(K)=\mathrm{cap}(K)/\pi$; it is different from half-plane capacity. If real boundary points are retained in the definition of $K$, include their boundary images in $E_K$ as well; the argument is unchanged.

For the scaling statement, Brownian scaling says that $r^{-1}{B}_{r^{2}t}$ started from $iy/r$ is again standard planar Brownian motion. The exit event scales with the hull, soSet $v=y/r$, multiply by $y=rv$ and take the limit. It follows thatThe same conclusion follows geometrically from $E_{rK}=r{E}_K$ and the formula above.

First derive the Brownian representation of half-plane capacity with the chosen normalization. SetThe height comparison from part (a), applied to $f_K=g_K^{-1}$, gives $h\ge 0$. On the boundary, $\mathrm{Im}{g}_K$ tends to zero; on $K$ the limiting value of $h$ is therefore $\mathrm{Im}z$, and on the real boundary it is zero. To see the asserted boundary-height limit without assuming a smooth hull, suppose a bounded sequence approaching the boundary had mapped images with imaginary part bounded away from zero. The inverse's expansion near infinity prevents these images from being unbounded, and an interior subsequential limit would map back to an interior point of $D$, a contradiction.

At infinity $h$ tends to zero by the Laurent series. The maximum principle for harmonic functions now bounds $h$ between zero and $R=\mathrm{rad}(K)$, since every point of $K$ has imaginary part at most $R$. Apply the optional sampling theorem to the bounded harmonic function $h(B_{t\wedge T})$. The exit time $T$ is almost surely finite because it is no larger than the first time the imaginary coordinate of planar Brownian motion reaches zero. Taking $t\to \infty$ givesThis uses the bounded harmonic difference, not optional sampling of the unbounded imaginary coordinate at an unbounded stopping time.

At $z=iy$, the expansion gives $yh(iy)\to a_K$. Sincewe obtainThus the quadratic half-plane capacity is controlled by a length times the linearly scaling harmonic capacity from infinity in the upper half-plane. The factor $1/\pi$ is essential if the alternative normalization $\mathrm{cap}(K)=\pi c(K)$ is used.

Write $g_t=g_{K_t}$ and, for $s<t$, let $K_{s,t}$ be the filled image hull $g_s(K_t\setminus K_s)$. The half-plane-capacity composition rule givesThe Loewner local growth property says that these mapped increments shrink uniformly on bounded time intervals: for every $A<\infty$,Using diameter instead of radius is equivalent, since a nonempty compact H-hull has its closure meeting the real axis. This is a condition on the increments after mapping out the past, not merely on the Euclidean size of $K_t\setminus K_s$ in the original domain.

For fixed $t$, the nonempty compact sets $\overline{K_{t,t+h}}$, $h>0$, are nested as $h$ decreases. Their diameters tend to zero by Loewner local growth property. The nested-compact-set property gives a single intersection point, and its imaginary part is zero because $K_{t,t+h}$ lies in a real-centred disc of radius tending to zero. Define the Loewner transform byOne may restrict the intersection to sufficiently small $h$. Every point of $K_{t,t+h}$ is within $2{\omega }_A(h)$ of ${\xi }_t$ when $t+h\le A$.

Here is a direct proof of continuity, using the permitted displacement estimate. Fix $s<t$ and put $\delta =t-s$. Take $0<h\le \delta$ and a point $v\in K_{t,t+h}$. Set $z=g_{K_{s,t}}^{-1}(v)$. The hull composition rule puts $z\in K_{s,t+h}$, while the displacement estimate givesSince $z$ is within $2{\omega }_A(\delta +h)$ of ${\xi }_s$, let $h\downarrow 0$, taking $v\to {\xi }_t$, to obtainUse a slightly larger fixed time interval if $t$ is its endpoint. Both terms tend to zero with $\delta$. This proves continuity from both sides, and indeed uniform continuity on every compact time interval. The real boundary value is obtained as a limit; evaluating the displacement estimate directly on a possibly rough boundary is unnecessary.

The Loewner correspondence theorem states that a capacity-$2t$ increasing hull family with Loewner local growth property has the unique continuous real Loewner driving function just defined, and its mapping-out functions solve the Chordal Loewner equationFor each $z\in \mathbb{H}$, this solution is taken up to its maximal lifetime; $\mathbb{H}\setminus K_t$ is precisely the set of points whose lifetimes exceed $t$. Conversely a continuous real driver gives such a hull family through this ordinary differential equation. A continuous driver alone need not have a continuous Loewner trace; that additional assertion is part of the trace theory of SLE, not of the deterministic correspondence theorem.

A continuous process $\gamma :[0,\infty )\to \overline{\mathbb{H}}$ is chordal Schramm–Loewner evolution with parameter $\kappa$ from $0$ to infinity when ${\gamma }_0=0$, its filled past generates these capacity-$2t$ hulls, and their Loewner driving function iswith $B$ standard real Brownian motion. The filling includes bounded components cut off from infinity. Equivalently its Loewner trace is the continuous extension of ${\gamma }_t={\lim }_{y\downarrow 0}{g}_t^{-1}({\xi }_t+iy)$. This last boundary limit is the trace-existence input for SLE.

Fix $r>0$ and scale space and time together:The half-plane capacity of ${\hat{K}}_t$ is $2t$ by part (a). Differentiation givesBy Brownian scaling, $r^{-1}{B}_{r^{2}t}$ is again standard Brownian motion, so the new driver has the same law as the old one. Uniqueness of the Chordal Loewner equation transfers this equality to the hulls and their continuous Loewner traces. ThusThis is capacity-parametrized scale invariance of a Loewner chain. It includes $\kappa =0$, when the driver is zero and the trace is the deterministic vertical slit ${\gamma }_t=2i\sqrt{t}$.

Let $t(u)$ be the inverse image-capacity clock. Part (a) and the quadratic variation of a stochastic integral giveThe Dambis-Dubins-Schwarz theorem, or the Lévy characterization of Brownian motion, now shows that ${\tilde{\xi }}_u/\sqrt{6}$ is standard Brownian motion up to the terminal image time. If that terminal clock is finite, it can be extended by a fresh Brownian continuation; this does not assert anything about the subsequent image trace. The initial value is zero because $\Phi (0)=0$.

Differentiating the image Loewner chain with respect to its own half-plane capacity time givesThus, until its first hit of $[1,\infty )$, the reparametrized image $\Phi (\gamma )$ has the ordinary chordal SLE law from $0$ to infinity. This follows from the uniqueness of the Chordal Loewner equation for a given continuous driver; the continuous Loewner trace is recovered from the inverse-map limit. Localization can be removed by exhausting times strictly before this boundary contact. The terminal event is defined from the trace itself, so equality of the stopped trace laws follows as well.

On the other hand, the definition of Conformal invariance of SLE says that the full unparameterized image $\Phi (\gamma )$ has the law of chordal ${\mathrm{SLE}}_6$ from $\Phi (0)=0$ to $\Phi (\infty )=1$, namely the law of ${\gamma }^{\prime }$. Its stopping event at $[1,\infty )$ is exactly the image of the original stopping event at $(-\infty ,-1]$. Therefore the same stopped image has both laws: ordinary ${\mathrm{SLE}}_6$ from $0$ to infinity stopped at $T$, and ${\mathrm{SLE}}_6$ from $0$ to $1$ stopped at $T^{\prime }$. We concludeThe statement does not identify $T$ with $T^{\prime }$ in their original clocks. It is target-change locality of SLE6, derived from the zero drift and capacity time change rather than assumed as a theorem.

The conformal automorphism of the upper half-plane with the indicated three boundary values isIts Möbius transformation matrix has determinant one, so it preserves $\mathbb{H}$. The inverse is ${\Phi }^{-1}(w)=w/(1-w)$. In particularon the extended real boundary. Therefore the image trace's first contact with the latter ray corresponds to the original time $S$. The map sends the original target at infinity to the new target $1$; it does not preserve the target of the chordal curve.

We use the image half-plane capacity as its time parameter:Since ${\Phi }_t$ is a conformal automorphism of the upper half-plane analytic at ${\xi }_t$, its real derivative there is strictly positive. Hence $u$ is continuous and strictly increasing before $S$. The original time $t$ and image time $u$ must be kept distinct throughout the comparison.

Work in the intended $\kappa \ge 8$ regime. On one probability-one event, all positive points have finite swallowing times and the strict ordering from part (d) holds for all pairs. Finiteness for arbitrary points follows from finiteness for rational points and monotonicity. Reflection supplies the negative-axis version on the same event.

Fix $r>0$ and suppose the Loewner trace has not visited $r$ by $T(r)$. Its image up to this finite time is compact, so a small interval $I$ about $r$, together with a thin upper collar, is disjoint from the whole trace segment. Every point of that collar belongs to the same component of the complement at every earlier time. Thus all points in a smaller interval are disconnected from infinity at precisely the same time as $r$: before $T(r)$ none is swallowed, and at $T(r)$ all are. This contradicts strict ordering for two positive points in that interval.

Therefore $r$ is visited by time $T(r)$. It cannot be visited before its swallowing time, so the visit occurs at $T(r)$. The same deterministic argument works for every real $r<0$ by reflection, while $0$ is the starting point. HenceThe threshold matters. If part (e) is read as retaining only $\kappa >4$, it is false: for $4<\kappa <8$, part (d) gives positive probability of swallowing a nonempty interval at one time, although a continuous trace can visit at most one of its points at that instant. Interior points of that swallowed boundary interval cannot subsequently be reached in capacity time: a later visit, by continuity, would make the curve stay inside an already filled hull for a time interval, contradicting strictly increasing half-plane capacity. Thus the preceding $\kappa \ge 8$ conclusion is the necessary intended qualification.

The capacity in this question is harmonic capacity from infinity in the upper half-plane, which has units of length; it is distinct from half-plane capacity, which has units of length squared.

Here is a proof of existence that also works with irregular real attachments. The functionis bounded and harmonic on $H$, by the Strong Markov property and the mean-value characterization of harmonic functions. Therefore $u\circ g_K^{-1}$ is a bounded harmonic function on the complex upper half-plane, with a Poisson kernel representationThe boundary function $f$ vanishes outside a bounded interval. Indeed, far enough along either real ray the original domain contains a half-disc neighbourhood, and $g_K$ extends there; the probability of hitting the bounded hull before the real boundary tends to zero as the starting point approaches that ray. Applying the same dominated-limit calculation as in part (i) yieldsWhen the intrinsic boundary pieces landing on the hull are identified, $f$ is their indicator almost everywhere and this is the length of their image under $g_K$. For ordinary finite slit hulls this is exactly the image of $\delta H\setminus H_0$, since real attachment endpoints have zero harmonic measure. The Poisson representation avoids requiring that boundary identification in the general existence argument.

If $K\subset K^{\prime }$, couple the two exit events using the same planar Brownian motion, stopped at its first hit of the real axis. Any path that hits $K$ before the real axis also hits $K^{\prime }$ before the real axis. Consequentlyand taking the limits proves monotonicity of this capacity.

For a half-disc of radius $r$ centred at $b\in \mathbb{R}$, the mapping-out function isIts semicircular boundary maps onto $[b-2r,b+2r]$, of length $4r$. Hence its harmonic capacity from infinity in the upper half-plane is $4r$. Withenclose $K$ in such a half-disc and use monotonicity. Letting the enclosing radius decrease to the infimum provesThe same conclusion holds if radius is instead measured about a specified real centre.

The Phase classification of the SLE trace has its simple rangewith $\kappa =0$ the deterministic vertical slit. For positive $\kappa$, the dividing parameter is precisely $\delta =1+4/\kappa =2$ in the Boundary-point Bessel flow for SLE.

Here is the reason this diffusion threshold controls simplicity. For $\kappa \le 4$, no nonzero real boundary point is swallowed. It suffices to check rational boundary points: a first meeting with either nonzero real half-axis would close a boundary crosscut and swallow a nonempty real interval, including a rational point. Thus the Loewner trace stays in the complex upper half-plane apart from its starting point. By the domain Markov property of a chordal Loewner chain, after any fixed rational time $s$ the future mapped and centred Loewner trace has the same boundary-avoidance property.

Suppose two Loewner trace times $r<r^{\prime }$ had the same image. Positive half-plane capacity growth rules out constancy on a nonempty time interval, so continuity supplies a rational $s\in (r,r^{\prime })$ with ${\gamma }_s\ne {\gamma }_r$. In the mapped future, the point at $r^{\prime }$ is either in the open upper half-plane or at the starting boundary point $0$; it cannot lie at another real point. The first case puts ${\gamma }_{r^{\prime }}$ inside the surviving domain at $s$, whereas ${\gamma }_r$ lies on the past Loewner trace. The second gives ${\gamma }_{r^{\prime }}={\gamma }_s$, also contradicting the choice of $s$. Boundary continuity of the inverse mapping-out function makes these identifications valid. This proves that the Loewner trace has no repeated points.

For $\kappa >4$, part (ii) makes a fixed positive real point have finite swallowing time, so the SLE boundary swallowing criterion ensures that the Loewner trace hits the positive real axis. In fact positive boundary-interval hitting probability for SLE above parameter four holds: any interval $J\subset (0,\infty )$ has positive hitting probability. To prove this, cover the positive axis by countably many dilates of $J$. If $J$ had zero hitting probability, Scaling invariance of SLE would give zero probability for every dilate, contradicting the almost sure hit of the positive axis. Reflection gives the same conclusion for negative intervals.

At a fixed positive time, if the past Loewner trace already repeats a point there is nothing to prove. Otherwise, boundary continuity of the inverse mapping-out function supplies a nonempty real interval, away from the current driving point, mapped back into the earlier Loewner trace in the open upper half-plane. Such an interval exists because positive capacity growth creates a genuine Loewner trace boundary in the interior; choose an accessible boundary point away from the tip and then a small interval around its preimage. Conditional on the past, the domain Markov property of a chordal Loewner chain gives a future centred SLE. With positive conditional probability its Loewner trace hits this interval, by the preceding boundary-interval argument. Mapping back then gives a visit to the earlier Loewner trace. Thus a repeated point occurs by some finite time with positive probability.

Finally let $E_t$ be the event of a repeated point by time $t$. Scaling invariance of SLE makes $\mathbb{P}(E_t)$ the same for every $t>0$. The positive finite-time probability just proved makes this common value positive. Hence ${\bigcap }_n{E}_{1/n}$ also has positive probability. This event belongs to the Brownian germ sigma-field; the Blumenthal zero-one law forces its probability to be one. Consequently the Loewner trace is not simple almost surely for every $\kappa >4$. At $\kappa =4$, the logarithmic scale function gives non-hitting of zero, so equality belongs to the simple range.

A compact H-hull is a bounded relatively closed subset $K$ of the complex upper half-plane such that $\mathbb{H}\setminus K$ is a simply connected domain. This does not require its closure to be connected. Its mapping-out function is the unique conformal map $g_K:\mathbb{H}\setminus K\to \mathbb{H}$ with hydrodynamic normalization at infinity,The half-plane capacity isThe coefficient is zero exactly for the empty hull.

In terms of the least radius of a real-centred enclosing half-disc, the sharp displacement bound for a compact H-hull, also called the continuity estimate, isThe differentiability estimate for a mapping-out function is a uniform small-hull expansion: there is an absolute constant $C$ such that if $K\subset \{z:|z-\xi |\le r\}$ with $\xi \in \mathbb{R}$, thenThese are statements of the two requested estimates. The second is not merely a bound for $g_K^{\prime }$: it controls the first-order change of a mapping-out function when a small hull is removed.