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For $0<\kappa <4$ and $x>0$, stop the simple chordal SLE trace on first entering the radius-$r$ half-disc about $x$, with $r<x/4$. A crosscut from its tip to $x$ separates the positive-side boundary arc containing $x/2$ from infinity. Harmonic measure comparison with the half-disc bounds the length of that arc's mapped interval, $g_t(x/2)-{\xi }_t$, by a universal constant times $r$. This centered boundary image is a positive Bessel process of dimension $1+4/\kappa >2$ and has a strictly positive all-time minimum. Hence the full trace stays a positive distance from each fixed nonzero real point. Reflection gives the negative-side version.

If $B$ is planar Brownian motion in a domain $D$ and $f:D\to D^{\prime }$ is a nonconstant holomorphic function, then $f(B)$, run with the clock ${\int }_0^t|f^{\prime }(B_s){|}^{2}ds$, is planar Brownian motion in $D^{\prime }$ until the relevant exit time. In particular, conformal maps transport harmonic measure between planar domains.

There is a universal constant $C$ such that every compact H-hull satisfiesTranslate and scale so the hull lies in a unit half-disc, then use the Brownian representation of half-plane capacity and the $O(y^{-1})$ harmonic measure of that half-disc as viewed from $iy$.

Write $f=u+iv$. The Cauchy-Riemann equations give $v_x=-u_y$, $v_y=u_x$ and $\Delta u=\Delta v=0$. Since the two Brownian coordinates are independent, $[X{]}_t=[Y{]}_t=t$ and $[X,Y{]}_t=0$. The Itô formula therefore becomesStop in balls on which the derivatives are bounded. The stopped integrals are martingales, proving that $U$ and $V$ are local martingales. Their coefficient rows have equal squared length and are orthogonal. ThusThese are quadratic covariations of an analytic Brownian image; orthogonality does not in general imply independence of the two transformed coordinates.

For the exit law, put $c=2a^2$. Squaring sends the positive quadrant to the upper half-plane, sends $a+ia$ to $ic$, and the Möbius transformation $w\mapsto (w-ic)/(w+ic)$ sends the upper half-plane to the unit disk with the starting point sent to zero. Its composition is the displayed conformal map. The exit time is finite: it is the minimum of the two almost surely finite one-dimensional hitting times of zero. There is no hit at the quadrant's corner, since a planar Brownian motion started away from zero avoids that point.

We can justify the uniform image exit law without assuming a conformal-time-change theorem. For each integer $n\ge 1$, the real and imaginary parts of $f(Z_{t\wedge T}{)}^n$ are bounded martingales obtained from harmonic functions. At the start their values are zero. The bounded convergence theorem at the finite exit time gives $\mathbb{E}f(Z_T{)}^n=0$. The boundary image has modulus one, and conjugation gives the vanishing negative Fourier moments as well. Trigonometric polynomials are dense among continuous functions on the unit circle, so these moments characterize the uniform distribution there.

Consequently the real variable $W=Z_T^2$ is the inverse image of a uniform point $e^{i\Theta }$ under that Möbius transformation:The Jacobian of this change of variable gives the centered Cauchy distribution of scale $c$:Positive $w$ corresponds to exit at $\sqrt{w}$ on the real axis; negative $w$ corresponds to exit at $i\sqrt{-w}$ on the imaginary axis. Pulling back by $w=\pm r^2$ therefore gives the complete boundary lawHere $dr$ on each axis denotes length along that axis, not two-dimensional area measure. Each density integrates to $1/2$, with no remaining atoms. Equivalently the chosen axis is a fair Bernoulli choice independent of the radius, andThis is Brownian exit law from a quadrant, obtained as a harmonic measure and not as an absolutely continuous planar density.

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.

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.

Here the map direction is opposite to Question 2: write $\varphi :U\to D$, with $\varphi (0)=0$ and $\varphi (1)=1$. The separation condition makes $U$ agree with $D$ near $1$. Schwarz reflection across that analytic boundary arc therefore defines the boundary derivative, whose modulus is positive.

For ordinary Brownian motion $W$ started at $0$, let ${\sigma }_U$ be its first exit from $U$. Take sufficiently short arcs $I_{ϵ}$ around $1$. They are also arcs of $\partial U$. The event that $W$ stays in $U$ until its disc exit and exits in $I_{ϵ}$ is exactly the event $W_{{\sigma }_U}\in I_{ϵ}$. Therefore the conditional avoidance probability isConformal invariance of planar Brownian motion gives ${\omega }_U(0,I_{ϵ})={\omega }_D(0,\varphi (I_{ϵ}))$. At the disc center harmonic measure is normalized arc length, so the ratio tends to $|{\varphi }^{\prime }(1)|$.

To justify that this limit is the avoidance probability for the point-conditioned diffusion, one may use the stopped transform directly. Its probability of reaching a smooth boundary arc at $1$ before any other boundary of $U$ is the mass at that pole in the transformed harmonic measure. All other exit points have weights $h(w)$ and represent failure. Equivalently integrate the Poisson-kernel density at $1$ for $U$ and divide by that for $D$. This gives the boundary-density ratio above, without assuming that a full-path avoidance event is a continuity set for arbitrary weak convergence. The kernels transform by the boundary Jacobian:ConsequentlyThere is no extra factor of $|{\Phi }_U^{\prime }(0)|$. The starting point is fixed and the conditioning concerns boundary harmonic-measure density. In this normalization the boundary derivative is a positive real, so it may also be written ${\Phi }_U^{\prime }(1)$. It is at most one by the probability interpretation. Given avoidance, the mapped path is the same conditioned Brownian motion up to conformal time change; this is radial restriction for a conditioned Brownian path.

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.

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.

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.

For a simple Loewner trace, the domain mapped by $g_t$ is $D_t=\mathbb{H}\setminus \gamma [0,t]$. For a non-simple Loewner trace, use instead the unbounded component $D_t=\mathbb{H}\setminus K_t$; the map $g_t$ is not defined on swallowed bounded components. This is the necessary domain interpretation when $\kappa >4$.

The function $\mathrm{Im}\log (g_t(z)-{\xi }_t)$ is harmonic in $D_t$ because the logarithm is holomorphic on the complex upper half-plane. Let $S_t^{-}$ and $S_t^{+}$ be the intrinsic boundary sets mapped to $(-\infty ,{\xi }_t)$ and $({\xi }_t,\infty )$ respectively. The Dirichlet problem has boundary dataFor a simple Loewner trace these are the left bank of the Loewner trace together with the negative real boundary, and the right bank together with the positive real boundary. Left and right refer to the orientation from the starting point towards the tip. The tip and infinity correspond to discontinuities of the data; no unique limit is imposed there. They have zero harmonic measure.

More precisely, the bounded solution isby the Poisson kernel for the upper half-plane and conformal invariance of planar Brownian motion. This establishes the boundary values in the intrinsic sense and uniqueness among bounded solutions of the Dirichlet problem, without treating the two banks as one Euclidean boundary point.

Let $E_{+}$ and $E_{-}$ denote positive and negative real part at the circular exit. On $\{S<T(r)\}$, reflect the portion of the planar Brownian motion after $S$ by $z\mapsto -\overline{z}$. This reflection fixes the imaginary axis and preserves distances from the origin. The Strong Markov property and reflection symmetry show that the resulting path has the same law, its circular exit time is unchanged, and $E_{+}$ is exchanged with $E_{-}$. ThereforeAn exit with negative real part must first cross the imaginary axis. An exit before $S$ has positive real part. The two points $ir,-ir$ have zero circular exit probability, since circular harmonic measure has no atoms. HenceSubtracting provesThis is the reflection identity for Brownian exit from a half-disc.

Use $U_t=2W_t$ and write $Z_t=X_t+i{Y}_t$ up to its interior-point swallowing time for a Loewner chain $T_z$. The Chordal Loewner equation givesThe branch of the complex logarithm with argument in $(0,\pi )$ is well-defined while $t<T_z$. The Itô formula yields the crucial cancellationThe $2/Z_t^2$ drift is cancelled by the quadratic variation term, because $\kappa =4$. Thus both $L_t=\log |Z_t|$ and ${\theta }_t=\arg Z_t$ are continuous local martingales, withThis is the logarithmic martingale for SLE4.

We justify absence of a finite swallowing time rather than presuming that the logarithm survives forever. On a finite horizon $H$, $0<Y_t\le \mathrm{Im}z$. Also $|X_t|$ is bounded pathwise before $T_z\wedge H$. Indeed, while $|X_t|\ge 1$ its drift $2X_t/|Z_t{|}^2$ has absolute value at most $2$; on each excursion outside $[-1,1]$, integrate from its starting point and bound the Brownian oscillation on $[0,H]$. For example,Therefore $L_t$ is bounded above pathwise on this interval.

Suppose $T_z\le H$. By the Dambis-Dubins-Schwarz theorem, $L$ is a Brownian motion run at its own quadratic variation. If that clock diverged as $t\uparrow T_z$, Brownian oscillation would make $L$ unbounded above, contradicting the preceding bound. The one-sided bound criterion for a martingale clock therefore gives a finite clock limit and a finite real limit for $L_t$. Hence $|Z_t|$ is bounded away from zero near $T_z$.

Now $Y_t=Y_0\exp (-2{\int }_0^t|Z_s{|}^{-2}ds)$ has a strictly positive limit at $T_z$. The drift in $X$ is integrable there, so continuity of $W$ gives a finite limit for $X$ as well. The limiting point is in $\mathbb{H}$ and away from the Loewner driver singularity, and the differential equation extends past $T_z$, a contradiction. Thus $T_z=\infty$ almost surely. A point of the Loewner trace at a finite time belongs to that time's hull, soThis proves the fixed-interior-point avoidance of SLE4 without using simplicity as an input.

The angle remains in $(0,\pi )$ at all finite times. The bounded local martingale criterion upgrades its local martingale equation to a genuine martingale:In particular this is the SLE4 angle martingale, and it converges almost surely and in $L^1$ by bounded martingale convergence theorem.

It remains to identify the limiting angle using the assumed simple path tending to infinity. Orient that path from $0$ to infinity. Its left component is the one adjacent to the negative real half-axis. Under $g_t$, the left boundary of the slit domain maps to $(-\infty ,U_t)$ and the right boundary to $(U_t,\infty )$. The harmonic measure of the former as seen from $z$ isThis follows by conformal invariance of planar Brownian motion: in the upper half-plane, $\arg (w-U_t)/\pi$ is the bounded harmonic function with values $1$ on the left half-axis and $0$ on the right.

To justify the limiting boundary classification, condition on a simple proper realization of the path and use an independent planar Brownian motion from $z$. It exits the upper half-plane in finite time almost surely, so its path up to that time is compact. The curve tends to infinity, so its intersection with this compact set is contained in a finite initial curve segment. Once that segment has been drawn, the Brownian path exits the slit domain through its left boundary exactly when $z$ is in the final left component: a path from that component cannot reach the right boundary without crossing the curve, and the reverse assertion holds on the right. Endpoints have zero harmonic measure. Bounded convergence of these exit indicators provesTaking expectations in the bounded angle martingale gives the SLE4 left-passage probabilityFor the imaginary axis it is $1/2$; near the negative real axis it tends to $1$, fixing the orientation of “left”.

A standard form of the Kakutani solution of the Dirichlet problem uses a bounded domain $D\subset {\mathbb{R}}^d$, continuous Dirichlet boundary data $f\in C(\partial D)$, and regularity of every boundary point for the Dirichlet problem. The boundedness of $D$ makes $\partial D$ a compact set, so $f$ is a bounded function and a uniformly continuous function. A bounded domain with $C^2$ boundary is a sufficient geometric case; one must not omit boundary regularity for an arbitrary bounded domain.

One standard analytic characterization of a regular boundary point on a bounded domain is the existence of a positive harmonic barrier: for each $\xi \in \partial D$ there is a harmonic function $h_{\xi }\in C(\overline{D})$ with $h_{\xi }(\xi )=0$ and $h_{\xi }(x)>0$ for $x\in \overline{D}\setminus \{\xi \}$. This is the harmonic form of a barrier for the Dirichlet problem; the barrier characterization of regularity is a standard fact of potential theory.

Let $B$ be $d$-dimensional Brownian motion started at $x\in D$ and let ${\tau }_D=\inf \{t\ge 0:B_t\notin D\}$ be its Brownian exit time. Then the unique solution in $C(\overline{D})\cap C^2(D)$ isEquivalently, $u(x)={\int }_{\partial D}f(\xi ){\omega }_D(x,d\xi )$, where ${\omega }_D$ is harmonic measure, the exit probability distribution of Brownian motion. The harmonic function equation is $\Delta u=0$, with the Laplace operator convention $\Delta ={\sum }_j{\partial }_j^2$.

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:

Compare the hull with the empty hull and the filled unit half-disc $J=\{z\in \mathbb{H}:|z|\le 1\}$. Their mapping-out functions of compact H-hulls are $g_{\varnothing }(z)=z$ and $g_J(z)=z+1/z$, the latter also giving the half-plane capacity of a half-disc.

Use one planar Brownian motion started at $iy$, with $y>1$. For $x>1$, exiting $\mathbb{H}\setminus J$ through $(x,\infty )$ entails avoiding $K$, while exiting $\mathbb{H}\setminus K$ through this interval entails reaching it before exit from the whole half-plane. Thus the tail harmonic measures satisfyFor any of these fixed hulls $L$, the Poisson kernel for the upper half-plane gives the exact tail formulaSubtract from $1/2$, multiply by $\pi y$, and take the limit. The inequalities reverse, yielding the real boundary bounds:Applying this result to the reflected hull $-\overline{K}$ also givesThe comparison is of Brownian exit events, so it does not assume that arbitrary conformal maps are pointwise ordered by domain inclusion.

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.