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A nearly closed semicircular slit makes the constant sharp. PutThis is the unit-circle arc attached at $-1$ and ending just short of $1$. Its complement is a simply connected domain. There remains a narrow passage near $1$ connecting the interior bay to infinity. Thus $K_L$ is a compact H-hull inside the closed unit disc.

For an explicit check, let $S=\sqrt{1+L^2}$ and let $q_L(z)=\sqrt{m(z{)}^2+L^2}$ denote the mapping-out function of a vertical slit applied to $m(z)$, with the branch asymptotic to its argument in the slit half-plane. Its normalized map isThe final Möbius transformation takes the image of the original infinity to infinity and gives hydrodynamic normalization at infinity. For a fixed $z$ in the open unit half-disc, $\mathrm{Re}m(z)>0$, so $q_L(z)\sim L$ and $g_{K_L}(z)\to 2$ as $L\to \infty$. Now take $z_{\delta }=-1+\delta +i\delta$, with $0<\delta <1$. First let $L\to \infty$, then let $\delta \downarrow 0$. The nearly closed semicircular slit givesEvery constant smaller than three therefore fails for some member of this family and some interior point. The bound need not be attained at an interior point of a single fixed hull.

In the half-plane-capacity parameterization, let $W$ be standard real Brownian motion and put ${\xi }_t=\sqrt{\kappa }{W}_t$. Solve the Chordal Loewner equationThe points for which this flow ceases to exist form the growing compact H-hulls $K_t$, with half-plane capacity $2t$. The process is an SLE trace when its initial segments generate these hulls andThe limit is the continuous trace of a Loewner chain started at zero. If the curve disconnects a region from infinity, that region is included in $K_t$; the hull is not generally just the visited curve. For $\kappa =0$ the driver is identically zero and the trace is the deterministic vertical slit ${\gamma }_t=2i\sqrt{t}$.

Write $\mathbb{H}=\{z:\mathrm{Im}z>0\}$. A compact H-hull is a bounded, relatively closed set $A\subseteq \mathbb{H}$ for which $D_A=\mathbb{H}\setminus A$ is a simply connected domain. “Compact” refers to its compact closure in $\overline{\mathbb{H}}$; boundary points on $\mathbb{R}$ can be adjoined without changing the remaining domain.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:D_A\to \mathbb{H}$ with hydrodynamic normalization at infinity. By the Schwarz reflection principle outside a large disc, it has the Laurent seriesThe half-plane capacity is the coefficientThe empty compact H-hull has $g_A(z)=z$ and half-plane capacity zero. Nonnegativity follows from the Brownian representation of half-plane capacity proved next.

For $r>0$, the conformal map $h(z)=r{g}_A(z/r)$ maps $\mathbb{H}\setminus rA$ onto $\mathbb{H}$ and has Laurent series $h(z)=z+r^{2}a/z+O(z^{-2})$. The uniqueness in hydrodynamic normalization at infinity givesAt $r=0$ the scaled closure is contained in $\mathbb{R}$, so the interior compact H-hull is empty and both sides are zero. This is the scaling part of scaling and translation of half-plane capacity.

For the vertical slit, choose the branch of $\sqrt{z^2+1}$ asymptotic to $z$ at infinity. It is the mapping-out function of a compact H-hull for $(0,i]$, andThus half-plane capacity of a vertical slit givesThe endpoint $0$ is merely part of the boundary closure convention.

For the filled upper half-disc $K=\{z\in \mathbb{H}:|z|\le 1\}$, the conformal map $g_K(z)=z+1/z$ maps the exterior half-disc onto $\mathbb{H}$. Indeed, it sends the semicircle to $[-2,2]$, the remaining real boundary to the complementary intervals, and its inverse is the branch of $(w+\sqrt{w^2-4})/2$ asymptotic to $w$. Hence half-plane capacity of a half-disc givesIf $\mathbb{D}$ denotes the open unit disc, the printed $\mathbb{H}\cap \mathbb{D}$ is not relatively closed and is literally not a compact H-hull. The intended value is the one for its filled relative closure $K$. With a closed-disc convention the displayed notation already represents that hull.

For a nonempty compact H-hull $A$, its closure meets $\mathbb{R}$: otherwise a nonempty compact set strictly inside $\mathbb{H}$ would be separated from the lower half-plane in the complement, contradicting the simply connected domain condition for $\mathbb{H}\setminus A$. Choose $x\in \overline{A}\cap \mathbb{R}$ and put $d=\mathrm{diam}(A)$. Taking limits in the definition of diameter shows $|z-x|\le d$ for every $z\in A$. Thus $A$ lies in the filled half-disc of radius $d$ centered at $x$. Its half-plane capacity is $d^2$ by scaling and translation of half-plane capacity. Using monotonicity of half-plane capacity,For the empty compact H-hull, use diameter zero. This proves the unit-constant version of half-plane capacity is bounded by squared diameter; no claim of optimality of the constant is needed.

Write $g_s=g_{K_s}$ and let $K_{s,t}$ be the relatively closed hull obtained by mapping the increment $K_t\setminus K_s$ through $g_s$. For a bounded compact H-hull $L$, defineThe Loewner local growth property isOne can equivalently use diameter: for a nonempty compact H-hull, $\mathrm{diam}(L)\le 2\mathrm{rad}(L)\le 2\mathrm{diam}(L)$, using a point of its closure on $\mathbb{R}$ for the second inequality. The condition concerns the mapped increment, not the original difference $K_{t+h}\setminus K_t$, which can be large when a narrow opening closes. It makes growth occur near one continuous Loewner driving function rather than by a sudden addition at separated boundary points.

A family $(K_t{)}_{t\ge 0}$ of compact H-hulls is non-decreasing whenThis permits equal hulls at different times. A Loewner chain with strictly increasing half-plane capacity has strict inclusion for $s<t$. Usually its initial compact H-hull is $K_0=\varnothing$.

The implication is false. Take the thin rectanglesEach is a compact H-hull, and $\mathrm{diam}(A_{\epsilon })=\sqrt{4+{\epsilon }^2}\to 2$. We verify that its half-plane capacity nevertheless tends to zero.

Let $K=\{z\in \mathbb{H}:|z|\le 2\}$, which contains all these rectangles. At the Brownian exit time $\tau$ from $\mathbb{H}\setminus A_{\epsilon }$, the exit height is at most $\epsilon$, and a positive exit height requires hitting $K$ before $\mathbb{R}$. ThereforeFor $y>2$, conformal invariance of planar Brownian motion and $g_K(z)=z+4/z$ identify the probability on the right with the harmonic measure of $[-4,4]$ from $i(y-4/y)$ in $\mathbb{H}$. Integrating the Poisson kernel for the upper half-plane givesThe Brownian representation of half-plane capacity now proves the explicit estimateChoose $\epsilon =1/n$ to obtain the required sequence. This is an instance of half-plane capacity of a low rectangle: shrinking height can make half-plane capacity small while horizontal extent stays fixed.

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.

The chordal restriction property concerns the unparameterized trace law. Let $A$ be a compact H-hull with $0\notin \overline{A}$, and put ${\psi }_A(z)=g_A(z)-g_A(0)$. This conformal map takes $\mathbb{H}\setminus A$ onto $\mathbb{H}$, fixes $0$ and infinity, and has derivative one at infinity. On the event that the curve avoids $A$, its image is a curve from $0$ to infinity in $\mathbb{H}$. Restriction requires positive avoidance probability for every admissible $A$, and the conditional identityup to increasing changes of the curve parameter. Equivalently, conditioning an SLE in a marked simply connected domain to stay in an admissible simply connected domain gives the same SLE law in that subdomain. The property does not assert that the original half-plane-capacity parameterization survives an arbitrary conformal map.

Let ${\psi }_A=g_A-g_A(0)$, and take another admissible compact H-hull $B$ in the image complex upper half-plane. Form the hull $C=A\cup {\psi }_A^{-1}(B)$, including its relative closure and filling bounded complementary components if needed. Its mapping-out function of a compact H-hull isIndeed, this composition maps the correct remaining domain onto $\mathbb{H}$, and its constant term at infinity is zero. Since ${\psi }_A(0)=0$, the chain rule givesBoth derivatives are positive by local Schwarz reflection principle and boundary orientation. Therefore the assumed avoidance formula givesThe conditioning event has positive probability because $g_A^{\prime }(0)>0$.

It remains to justify that these equalities determine the law. For a proper simple curve from $0$ to infinity, its complement has two sides, each connected to an interval of the real boundary. Any point off the curve lies in a thin polygonal tube attached to that boundary which avoids both the curve and zero; its closure, with an end cap, is an admissible compact H-hull. A countable collection of tubes with rational geometry therefore reconstructs the complement as the union of the tubes avoided by the curve.

The joint distribution of their avoidance indicators is also determined. Avoiding finitely many tubes is equivalent to avoiding the filled union when that union is admissible. If the union separates $0$ from infinity, avoidance is impossible and the joint probability is zero. Otherwise a simple curve cannot enter the bounded complementary pockets without crossing the union, so filling does not change the avoidance event. Finite joint avoidance probabilities consequently come from the same single-hull formulas; inclusion-exclusion principle supplies every finite indicator pattern. These patterns determine the random complement and hence the unparameterized simple trace.

Thus avoidance probabilities determine a simple chordal curve law, and the conditional image law above equals the original law. This proves the derivative avoidance criterion for chordal restriction, giving the chordal restriction property. Simplicity is essential to this reconstruction argument; it should not be replaced by a claim that arbitrary random closed sets are determined by these hull-avoidance tests.

A compact H-hull is a bounded subset $A\subset \mathbb{H}$, closed relative to the complex upper half-plane, such that $\mathbb{H}\setminus A$ is a simply connected domain. The word compact refers to its bounded closure in $\overline{\mathbb{H}}$; a nonempty hull need not be a compact subset of the open half-plane.

The Riemann mapping theorem and normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with $g_A(z)-z\to 0$ as $z\to \infty$. This hydrodynamic normalization at infinity yields a Laurent seriesThe coefficient is the half-plane capacity:

There is a small closure issue in the printed set: the unit disc is open, so its half-disc is not relatively closed in $\mathbb{H}$. As written, the set is not a compact H-hull. We compute the intended half-plane capacity after taking its relative closure in $\mathbb{H}$.

First remove the closed unit half-disc $K$. Its mapping-out function of a compact H-hull is the Joukowski mapFor $z=iy$ on the remaining portion of the vertical slit, $1<y\le 2$,Thus the image of the remaining slit is $(0,3i/2]$. By the half-plane capacity of a vertical slit, its capacity is $(3/2{)}^2/2=9/8$. The half-plane-capacity composition rule givesAs a direct check, composing with the slit map gives $g_A(z)=\sqrt{(z+z^{-1}{)}^2+9/4}$, with the branch asymptotic to $z$. Its $z^{-1}$ coefficient is $17/8$.