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The half-plane capacity is the Laurent coefficientIt is real and nonnegative. For planar Brownian motion starting at $z\in D$, let ${\tau }_D$ be its first exit from $D$. The Brownian representation of half-plane capacity isOne may also state the underlying harmonic identity, valid throughout $D$,The process is killed at the hull or real boundary; real-boundary exits contribute zero. These are the Brownian characterizations, with the normalization that a Chordal Loewner equation at speed $2$ produces capacity $2t$.

The mapping-out function of a compact H-hull is the unique conformal bijection from $H=\mathbb{H}\setminus K$ onto the complex upper half-plane with hydrodynamic normalization at infinity:The nonnegative coefficient $a_K$ is the half-plane capacity of $K$. The leading coefficient one and constant term zero remove the affine ambiguity of a half-plane map. In the compact H-hull convention, $K$ is bounded and relatively closed in $\mathbb{H}$, and its complement is a simply connected domain; boundary attachment points can be included in its closure.

We will use these standard properties explicitly: $g_K$ and its inverse extend holomorphically by Schwarz reflection principle across real intervals outside the hull's real attachment region; their real boundary restrictions are strictly increasing; and their normalized Laurent series hold in a full reflected neighbourhood of infinity. In particular, when $K$ lies in the unit disc, these extensions exist across $(-\infty ,-1)$ and $(1,\infty )$.

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 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.

With the normalization used in the Chordal Loewner equation, half-plane-capacity parameterization meansThus $K_0$ is empty and the half-plane-capacity composition rule gives $\mathrm{hcap}(K_{s,t})=2(t-s)$. The factor two produces ${\partial }_t{g}_t(z)=2/(g_t(z)-U_t)$. A convention using half-plane capacity $t$ instead changes that coefficient to one and rescales time; the convention must be kept consistent.

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.

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.

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 $17/8$.

Interpret the printed $A_0=0$ as the empty initial hull. Write $a(t)=\mathrm{hcap}({\tilde{A}}_t)$ and $F_t={\psi }_t$. Conformal maps preserve inclusion and simple connectedness of the complementary domains, so the image sets form an increasing hull family. Boundedness of $\psi$ on bounded sets ensures that these image hulls are bounded. The Loewner local growth property is preserved under conformal transport: after mapping out the hull at time $t$, the small new hull is transported by $F_t$ near its single boundary growth point. The Schwarz reflection principle extends $F_t$ analytically across that point, with real positive derivative. The image diameters therefore tend to zero as the original ones do. These statements use the usual hull closures and prime ends; the initial image hull is empty.

For completeness, the infinitesimal capacity rule underlying this argument is that a shrinking hull $J$ attached near $u$, transported by a map $F$ analytic there, hasOne obtains this by rescaling at $u$: the transported map tends to its linear part, and half-plane capacity scales by the square of the dilation. Uniform analytic distortion near the growth point controls the error. Applying it to the mapped-out increments gives local absolute continuity of $a$ and $a^{\prime }(t)=2F_t^{\prime }(U_t{)}^2$.

The coefficient can also be read directly from the Chordal Loewner equation. The image driver is ${\tilde{U}}_t=F_t(U_t)$. Differentiate $F_t={\tilde{g}}_t\circ \psi \circ g_t^{-1}$, at a fixed point $w$, to obtainThe left side is regular at $w=U_t$. On the right, the coefficient of $(w-U_t{)}^{-1}$ is $a^{\prime }(t)/F_t^{\prime }(U_t)-2F_t^{\prime }(U_t)$, so it must vanish. This proves the conformal change of half-plane capacity rule and its integrated version:If the derivative at the initial boundary point is singular, the formula is interpreted by integrating from a positive time and taking the lower limit to zero; finite image capacity and kernel continuity give this limit. The duplicated incomplete normalization sentence in the TeX is absent from the PDF.

A compact H-hull is a bounded, relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe half-plane capacity is

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityIts half-plane capacity is $\mathrm{hcap}(A)=a$.

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ such that $\mathbb{H}\setminus A$ is a simply connected domain. Its mapping-out function of a compact H-hull is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinityThe nonnegative coefficient $a$ is the half-plane capacity $\mathrm{hcap}(A)$.

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. The Riemann mapping theorem and hydrodynamic normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ withThe coefficient $a\ge 0$ is the half-plane capacity $\mathrm{hcap}(A)$.

For $\lambda >0$ and $x\in \mathbb{R}$, uniqueness of the normalized map givesSubstituting the expansion of $g_A$ yieldsTherefore the scaling and translation of half-plane capacity is

Let $d=\mathrm{diam}(A)$. Unless $A$ is empty, its closure meets the real axis; otherwise a loop in $\mathbb{H}\setminus A$ surrounding $A$ could not contract, contrary to $\mathbb{H}\setminus A$ being a simply connected domain. Choose $x\in \overline{A}\cap \mathbb{R}$. Then $A\subseteq \{z\in \mathbb{H}:|z-x|\le d\}$.

The unit half-disc is a compact H-hull with mapping-out function $z+z^{-1}$, so its half-plane capacity is one. The monotonicity of half-plane capacity and its scaling rule now giveThus the assertion holds with the universal constant $c=1$ under this normalization.

A holomorphic self-map of the unit disc contracts its pseudohyperbolic distance. Conjugating by Möbius transformations that send one chosen source point and its image to zero reduces the statement to the Schwarz lemma. In the complex upper half-plane this givesIf $f(iY)=iY+o(Y)$ at infinity, multiply the equivalent inequality between one minus the squared distances by $Y$ and let $Y\to \infty$. It yields $\mathrm{Im}f(z)\ge \mathrm{Im}z$. Applied to the inverse mapping-out function of a compact H-hull, this proves the nonnegativity of its half-plane capacity.

For a chordal SLE trace $\gamma$, the Scaling invariance of SLE makes $\mathbb{P}({\sup }_t\mathrm{Im}\gamma (t)>r)$ independent of $r>0$. The trace must enter the upper half-plane because its positive-time hull has positive half-plane capacity. Letting $r\downarrow 0$ therefore makes this probability one, and continuity implies that every positive height is reached. This argument does not assume Transience of chordal SLE.