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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.

Here the otherwise undefined printed domain $H$ must mean $D=\mathbb{H}\setminus K$. Let the Brownian motion start at $z\in D$, and let ${\tau }_R$ stop it on reaching height $R$ or leaving $D$. Take $R>\max (\mathrm{Im}z,{\sup }_K\mathrm{Im}w)$. This stopping time is finite almost surely, since it is no later than exit of its imaginary coordinate from $(0,R)$.

By part (d), $0\le v(w)\le R$ in the stopped domain. Its finite-boundary values vanish except on the top boundary, by the argument of part (d). Localization and the optional stopping theorem for this bounded harmonic martingale therefore giveWrite $p_R={\mathbb{P}}_z(\mathrm{Im}{B}_{{\tau }_R}=R)$ and use the uniform displacement bound $M$ from part (c). On the top boundary, $|v(w)-R|\le M$, henceThe stopped imaginary coordinate is itself a bounded martingale. Since its exit height is nonnegative,Consequently $|v(z)-R{p}_R|\le M\mathrm{Im}z/R\to 0$, proving the high-level escape representation of a mapping-out height:If $H$ instead meant the whole upper half-plane, the right side would always be $\mathrm{Im}z$. For example, the mapping-out function of a vertical slit $K=(0,ia]$ is $g_K(w)=\sqrt{w^2+a^2}$ with branch asymptotic to $w$; at $z=iy$, $y>a$, its height is $\sqrt{y^2-a^2}<y$. This demonstrates why killing on the hull is essential to the printed formula.

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.