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For a compact H-hull and a Borel subset of the intrinsic boundary of a simply connected domain, hydrodynamic normalization at infinity and the Poisson kernel for the upper half-plane give this limit as $y\to \infty$ with $x/y\to 0$. After multiplying the kernel by $\pi y$, it tends pointwise to one and is eventually uniformly bounded. Dominated convergence proves the finite-measure case and Fatou's lemma proves the infinite-measure case.

The mapping-out function is the unique conformal map $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ with hydrodynamic normalization at infinity

For the compact H-hull $(x,x+ih]$, the square-root branch asymptotic to $z-x$ gives the hydrodynamic normalization at infinity of its mapping-out function of a compact H-hull. Its expansion has coefficient $h^2/2$, the half-plane capacity of a vertical slit.

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.

Write $D=\mathbb{H}\setminus K$. By definition of a compact H-hull, $D$ is a proper simply connected domain, and it agrees with the complex upper half-plane outside a sufficiently large disc. The Riemann mapping theorem supplies a conformal isomorphism $f:D\to \mathbb{H}$. Choose its boundary normalization so that the prime end at infinity maps to infinity. This prime end is unambiguous because the domain has an ordinary straight real boundary near infinity.

The Schwarz reflection principle, applied there in the coordinate $1/z$, extends $f$ across the real boundary near infinity. It has a simple pole there, with Laurent seriesThe pole is simple because the reflected map is locally conformal at infinity; its leading coefficient is positive because it maps the upper side to the upper side. Postcomposing with the half-plane automorphism $w\mapsto (w-b)/a$ givesThis is hydrodynamic normalization at infinity.

If $g_1,g_2$ both satisfy it, $g_2\circ g_1^{-1}$ is a conformal automorphism of the upper half-plane. The asymptotics force it to fix infinity, so it has the form $w\mapsto \alpha w+\beta$ with $\alpha >0$ and $\beta \in \mathbb{R}$. The same asymptotics force $\alpha =1$, $\beta =0$. Hence the normalized mapping-out function exists and is unique. The argument requires reflection only outside a large disc, not regularity of the hull's entire boundary.

Let $g_t$ map out the original filled hull and let $g_t^{\ast }$ map out the filled hull generated by $\Phi (\gamma [0,t])$. Both have hydrodynamic normalization at infinity. WriteAlthough the original $\Phi$ is specified only on $N$, this conjugacy is defined in the surviving neighborhood of the tip. Boundary preservation gives a real analytic extension there by the Schwarz reflection principle, with $a_t>0$. The identities below are first applied on compactly localized intervals before exit and then continued up to $T$.

Set $u(t)=\mathrm{hcap}(K_t^{\ast })/2$, the capacity clock of the image hull. The Chordal Loewner equation for each flow and the chain rule giveAnalyticity at the driving point cancels the coefficient of $(z-{\xi }_t{)}^{-1}$, forcing $\dot{u}(t)=a_t^2$. The constant term, computed as in Question 3, is then ${\partial }_t{h}_t({\xi }_t)=-3b_t$. For a general SLE driver $d{\xi }_t=\sqrt{\kappa }d{B}_t$, the Itô formula consequently gives the transformed SLE driving functionAt $\kappa =6$, this becomesThus $W$ is a continuous local martingale up to the exit time. If the source notation ${\xi }_t^{\ast }$ indexes the image driver by original time, it denotes $W_t$. If it uses the image's own half-plane-capacity time, it denotes $W_{t(u)}$, with $t(u)$ the inverse clock; that time-changed process is also a local martingale. This distinguishes the two clocks without changing the requested conclusion.

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.

As printed, the lower bound needs a connectedness hypothesis. This illustrates that radius gives no positive lower bound for disconnected harmonic hull capacity. A compact H-hull need not have connected closure. To see the obstruction, take $0<\epsilon <1/12$, set $c=\sqrt{1-{\epsilon }^2}$, and use three disjoint vertical slits,Their complement in the complex upper half-plane is a simply connected domain: the three slits attach to the real boundary, and create no interior holes. Symmetry shows that the smallest enclosing half-disc is centred at zero and has radius one. More explicitly, any real centre $b$ has maximum distance to the two outer tips at least $\sqrt{c^2+{\epsilon }^2}=1$, while the unit half-disc contains every slit. Also $0\in {\overline{K}}_{\epsilon }$ and $c+i\epsilon$ has modulus one.

For a vertical slit of height $\epsilon$, the mapping-out function of a vertical slit maps its two faces onto an interval of length $2\epsilon$, so its harmonic capacity from infinity in the upper half-plane is $2\epsilon$. Subadditivity of harmonic hull capacity, from the union bound for the Brownian hitting events, givesThus the requested conclusion does not follow from the printed assumptions.

The intended reflection lower bound for harmonic hull capacity works when the closure is a connected continuum joining the two specified points, as for a slit hull. Reflect in the vertical line through $x$, using $\rho (z)=2x-\bar{z}$. The reflected continuum joins $2x$ to $x+iy$. Together the original and reflected continua form a barrier between infinity and the segment $I$ together with the real interval between $0$ and $2x$. One can first verify this separation for polygonal simple arcs, then use decreasing connected neighbourhoods of the continuum. Hence, for Brownian motion started at $x+iY$, $Y$ large, reaching $I$ or the real interval $J$ between $0$ and $x$ requires a hit of the original or reflected barrier. Reflection symmetry and the union bound giveFor the connected slit setting, the real attachment endpoints have zero harmonic measure, so the last event may be written with $K$.

For $y>0$, use the branch of the square root fixed by hydrodynamic normalization at infinity:The two slit faces together have image length $2y$. The interval $J$ has image length $\sqrt{x^2+y^2}-y=1-y$, by the square-root formula on the appropriate real side. Thus the image length of $I\cup J$ is $1+y$. Multiply the probability inequality by $\pi Y$ and use part (i); the allowed starting approach includes $x+iY$. We obtainunder the stated connected-barrier interpretation. If $y=0$, the model slit is empty and $J$ has length $|x|=1$, giving the same lower bound by the real-interval argument. The connectedness repair is essential, as the explicit three-slit counterexample demonstrates.

Use the intrinsic boundary of a simply connected domain, so different approaches to the two sides of a slit remain distinct. The mapping-out function $g_K$ extends as a homeomorphism from this intrinsic compactification to that of the complex upper half-plane. Set $T=T(H)$. The imaginary coordinate of planar Brownian motion hits zero in finite time almost surely, and $T$ is no larger than that time. Thus $T<\infty$ and continuity gives a finite Euclidean exit point.

By conformal invariance of planar Brownian motion, $g_K(B_t)$ is a planar Brownian motion in the complex upper half-plane, run with clockThe terminal clock cannot be infinite: that would make the transformed Brownian motion stay in the upper half-plane forever. It cannot stop while the transformed path is in the interior either, since continuity of $g_K^{-1}$ would then put the original exit point inside $H$. Hence the terminal clock is precisely the transformed Brownian exit time. The transformed path converges to a real boundary point, and applying the extended inverse proves almost sure convergence to a point of the intrinsic boundary.

Write $g_K(x+iy)=u+iv$, and let $E=g_K(S)\cap \mathbb{R}$. The point at infinity has zero harmonic measure. Conformal invariance of planar Brownian motion and the Poisson kernel for the upper half-plane giveThe hydrodynamic normalization at infinity gives $g_K(z)=z+O(1/z)$, so along the specified approach $v/y\to 1$ and $u/y\to 0$. For each fixed real $t$,If $\mathrm{Leb}(E)<\infty$, dominated convergence applies because $y/v$ is eventually bounded. If $\mathrm{Leb}(E)=\infty$, Fatou's lemma makes the limit infinite. This proves the harmonic-measure asymptotic at infinitywith the equality understood in the extended nonnegative reals.

The complex form of the Chordal Loewner equation, driven by a continuous real Loewner driving function, isFor $z\in \mathbb{C}\setminus \{{\xi }_0\}$ this is an ordinary differential equation up to its maximal lifetime, when the solution reaches the driving singularity. The coefficients respect complex conjugation, so the lower-half-plane flow is the conjugate of the upper-half-plane flow. On surviving real points it is the real boundary flow. The hydrodynamic normalization at infinity and factor two correspond to half-plane-capacity parameterization $\mathrm{hcap}(K_t)=2t$.

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.

A compact H-hull is a bounded set $K\subset \mathbb{H}$ that is closed relative to the complex upper half-plane and whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Its Euclidean closure is compact in $\overline{\mathbb{H}}$; “compact” here does not require separation from the real axis. One may equivalently describe the closed hull in $\overline{\mathbb{H}}$, with its real-boundary convention understood. Its mapping-out function is the unique conformal map to $\mathbb{H}$ with hydrodynamic normalization at infinity.

Use the capacity convention $\mathrm{hcap}(K_t)=2t$. A chordal Schramm–Loewner evolution from $0$ to infinity is the hull family of the Chordal Loewner equationwhere $W$ is a real standard Brownian motion and $\kappa \ge 0$. The points with interior-point swallowing time for a Loewner chain at most $t$ constitute $K_t$. For each $t$, $g_t$ maps $\mathbb{H}\setminus K_t$ conformally onto $\mathbb{H}$ and has expansion $g_t(z)=z+2t/z+O(z^{-2})$. The case $\kappa =0$ is the deterministic vertical-slit chain.

First identify the relevant transformed Loewner driving functions directly. For $\lambda >0$, defineDifferentiation gives ${\partial }_t{\tilde{g}}_t=2/({\tilde{g}}_t-{\lambda }^{-1}{U}_{{\lambda }^{2}t})$. Its initial value is $z$, its domain is $\mathbb{H}\setminus {\lambda }^{-1}{K}_{{\lambda }^{2}t}$, and its expansion is $z+2t/z+O(z^{-2})$. Thus its driving function is ${\lambda }^{-1}{U}_{{\lambda }^{2}t}$, by uniqueness of the differential equation and of the hydrodynamically normalized map. Brownian scaling provesThis is capacity-parametrized scale invariance of a Loewner chain.

For a fixed time $t$, map the future remaining domains by $z\mapsto g_t(z)-U_t$. The corresponding hulls ${\hat{K}}_s$ are specified without any boundary-image ambiguity byTheir mapping-out functions areAt $s=0$ this is the identity. Differentiating at a point in its domain yieldsThe Laurent series is $z+2s/z+O(z^{-2})$. Hence the transformed Loewner driver is exactly $U_{t+s}-U_t$, with the original capacity clock unchanged. This proves the composition rule for chordal Loewner driving functions, rather than assuming an identification of transforms.

The Brownian motion increments after $t$ are independent of its past and have the original law. Since past hulls are measurable functions of the past Loewner driver, conditionally on the past, the centered mapped future is an independent copy of the original chain. This is the domain Markov property of a chordal Loewner chain. At an almost surely finite stopping time it follows likewise from the Strong Markov property.

For the converse, a precise class is essential: take capacity-parametrized Loewner chains generated by a continuous real Loewner driving function $U$, with $U_0=0$, and require the two displayed hull properties, including independence from the entire hull past in the domain Markov property. Within this class they characterize ${\mathrm{SLE}}_{\kappa }$, for some $\kappa \ge 0$.

To see why the hull assertions determine the Loewner driver assertions, the hull domains determine their normalized maps uniquely. For any surviving point,The left time derivative suffices for $t>0$, since $U$ is continuous. A sufficiently high point survives up to any given finite time, so points $in$, $n\in \mathbb{N}$, suffice to reconstruct the Loewner driver locally and measurably from the hull past. Thus the driving-function reconstruction for a chordal Loewner chain identifies the two past filtrations and makes the preceding scale and composition computations reversible.

The domain Markov property therefore gives stationary independent increments for $U$. Its continuity makes it a continuous Lévy process. Its classification gives $U_t=bt+\sqrt{\kappa }{W}_t$: in the Lévy–Khintchine formula continuity removes the jump measure, leaving characteristic function $\exp (t(ibu-\kappa u^2/2))$. Driver scaling would turn the drift $b$ into $\lambda b$, so invariance for every $\lambda$ forces $b=0$. The diffusion coefficient is unchanged. This proves the scale-and-domain-Markov characterization of SLE with its regularity and parametrization hypotheses stated explicitly.

Write $D=\mathbb{H}\setminus K$. The mapping-out function is a conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinityEquivalently, $g_K(z)-z\to 0$. The Riemann mapping theorem gives a map onto the half-plane. Since $K$ is bounded, infinity has an analytic real-boundary neighborhood. After sending its image to infinity, the Schwarz reflection principle gives a Laurent series $az+b+O(z^{-1})$ there, with $a>0$ and $b$ real. A real affine automorphism removes $a$ and $b$, giving the stated normalization. Reflection also makes $a_K$ real.

If $g_1,g_2$ both have this normalization, $g_2{g}_1^{-1}$ is a conformal automorphism of the upper half-plane. Such maps are real Möbius transformations. The expansion forces it to fix infinity, have leading coefficient $1$, and have zero translation. It is therefore the identity. Thus the normalized mapping-out function exists and is unique.

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 $(1,\infty )$.

Set $g_K(iy)=u_y+i{v}_y$. The hydrodynamic normalization at infinity givesThe real interval under consideration lies outside the hull, so the reflected boundary map sends it monotonically to $(g_K(x),g_K(b))$. Conformal invariance of planar Brownian motion transports its harmonic measure to the half-plane. By the Poisson kernel for the upper half-plane,The endpoints are fixed and finite. On this interval,uniformly, using $v_y/y\to 1$ and $u_y\to 0$. Integrating gives the interval-length limit:

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.

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 $x\in \mathbb{R}$, the conformal mapmaps $\mathbb{H}\setminus (A+x)$ onto $\mathbb{H}$ and has hydrodynamic normalization at infinity. Its Laurent series is $h(z)=z+a/z+O(z^{-2})$. Uniqueness of the mapping-out function of a compact H-hull therefore givesThis is the translation part of scaling and translation of half-plane capacity.