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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.

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.

When a finite union is a compact H-hull, hitting that union before the real axis implies hitting at least one member before the real axis. The union bound followed by the defining limit proves subadditivity of harmonic capacity from infinity in the upper half-plane.