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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 $iy$.

The Brownian representation of half-plane capacity is $\mathrm{hcap}(K)={\lim }_{y\to \infty }y{\mathbb{E}}_{iy}\mathrm{Im}{B}_T$. The harmonic capacity from infinity in the upper half-plane uses $\mathrm{cap}(K)={\lim }_{y\to \infty }\pi y{\mathbb{P}}_{iy}(B_T\in K)$. Since $0\le \mathrm{Im}{B}_T\le \mathrm{rad}(K)1_{\{B_T\in K\}}$, the displayed bound follows. The normalization factor $\pi$ cannot be omitted. Independently, enclosure in a half-disc and monotonicity of half-plane capacity give the sharp bound $\mathrm{hcap}(K)\le \mathrm{rad}(K{)}^2$.

If compact H-hulls satisfy $A\subseteq C$, thenThis follows from the Brownian representation of half-plane capacity or from composition of their mapping-out functions.

First derive the Brownian representation of half-plane capacity with the chosen normalization. SetThe height comparison from part (a), applied to $f_K=g_K^{-1}$, gives $h\ge 0$. On the boundary, $\mathrm{Im}{g}_K$ tends to zero; on $K$ the limiting value of $h$ is therefore $\mathrm{Im}z$, and on the real boundary it is zero. To see the asserted boundary-height limit without assuming a smooth hull, suppose a bounded sequence approaching the boundary had mapped images with imaginary part bounded away from zero. The inverse's expansion near infinity prevents these images from being unbounded, and an interior subsequential limit would map back to an interior point of $D$, a contradiction.

At infinity $h$ tends to zero by the Laurent series. The maximum principle for harmonic functions now bounds $h$ between zero and $R=\mathrm{rad}(K)$, since every point of $K$ has imaginary part at most $R$. Apply the optional sampling theorem to the bounded harmonic function $h(B_{t\wedge T})$. The exit time $T$ is almost surely finite because it is no larger than the first time the imaginary coordinate of planar Brownian motion reaches zero. Taking $t\to \infty$ givesThis uses the bounded harmonic difference, not optional sampling of the unbounded imaginary coordinate at an unbounded stopping time.

At $z=iy$, the expansion gives $yh(iy)\to a_K$. Sincewe obtainThus the quadratic half-plane capacity is controlled by a length times the linearly scaling harmonic capacity from infinity in the upper half-plane. The factor $1/\pi$ is essential if the alternative normalization $\mathrm{cap}(K)=\pi c(K)$ is used.

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 $2t$.

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.

Put $u(z)=\mathrm{Im}(z-g_A(z))$ on $D_A$. This is a harmonic function, tends to zero at infinity, and is bounded. To justify the boundary behavior without requiring a locally connected space as its boundary, let $f_A=g_A^{-1}$. Its expansion at infinity implies that bounded $z$ cannot have $|g_A(z)|\to \infty$. If $z_n$ approaches a finite point of $\partial D_A$ and $g_A(z_n)$ had a subsequential limit inside $\mathbb{H}$, continuity of $f_A$ would force that point to lie inside $D_A$, a contradiction. Thus boundary degeneration under a mapping-out function gives $\mathrm{Im}{g}_A(z_n)\to 0$. Consequently $u$ extends continuously to the finite boundary with value $\mathrm{Im}z$.

Let $B$ be planar Brownian motion started at $z\in D_A$, and $\tau$ its Brownian exit time from $D_A$. This time is finite almost surely: it is at most the first time the imaginary coordinate, a one-dimensional Brownian motion, hits zero. By the Itô formula, $u(B_{t\wedge \tau })$ is a bounded martingale. The optional stopping theorem and dominated convergence theorem therefore giveBoundedness is important here: directly stopping the unbounded imaginary-coordinate martingale would require an unjustified uniform integrability assertion.

At $z=iy$, the Laurent series gives $u(iy)=a/y+O(y^{-2})$. HenceThe sign follows because the exit point lies on $\partial D_A\subseteq \overline{\mathbb{H}}$. This proves the Brownian representation of half-plane capacity.

Suppose $A\subseteq B$. Map out $A$ and let $C$ be the relatively closed image hull corresponding to $g_A(B\setminus A)$, with bounded complementary components filled if needed. Then $g_A$ maps $\mathbb{H}\setminus B$ onto $\mathbb{H}\setminus C$. Uniqueness of the mapping-out function of a compact H-hull gives $g_B=g_C\circ g_A$. Comparing their Laurent series yields the half-plane-capacity composition ruleThe Brownian representation of half-plane capacity makes the last term nonnegative, proving monotonicity of half-plane capacity:

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 statement is true. In the notation of part (ii), equality of the capacities forces $\mathrm{hcap}(D)=0$. Every nonempty compact H-hull has strictly positive half-plane capacity: by the Brownian representation of half-plane capacity, Brownian motion started sufficiently high has positive harmonic measure of a boundary portion of positive height. Hence $D=\varnothing$, so $C\setminus A=\varnothing$ and

Let $g=g_A$ and define $u(z)=\mathrm{Im}(z-g(z))$. This is harmonic on $\mathbb{H}\setminus A$, has boundary values $\mathrm{Im}z$ on the hull boundary and zero on the real boundary, and tends to zero at infinity. The representation by harmonic measure and optional sampling theorem therefore giveAt $z=iy$, the hydrodynamic expansion givesConsequently the Brownian representation of half-plane capacity is

If $A\subseteq C$, Brownian motion exits $\mathbb{H}\setminus C$ no later than it exits $\mathbb{H}\setminus A$. The Strong Markov property and the nonnegative harmonic function $h_A$ show that the expected exit height for $C$ is at least that for $A$. Taking the limits in the Brownian representation of half-plane capacity proves the monotonicity of half-plane capacity

Set $h_A(z)=\mathrm{Im}(z-g_A(z))$. This is a nonnegative harmonic function on $\mathbb{H}\setminus A$, has boundary value $\mathrm{Im}z$ on the hull boundary, and tends to zero on the real boundary. If $\tau$ is the first exit time of planar Brownian motion from $\mathbb{H}\setminus A$, the optional stopping theorem givesThe hydrodynamic expansion yields $h_A(iy)=\mathrm{hcap}(A)/y+O(y^{-2})$, and therefore the Brownian representation of half-plane capacity

Let $\sigma$ be the first time Brownian motion started outside the unit half-disc reaches its semicircular boundary. A path that reaches $A$ must first cross that semicircle. The Strong Markov property at $\sigma$ givesTake $z=iy$, multiply by $y$, and let $y\to \infty$. The Brownian representation of half-plane capacity identifies the left side with $\mathrm{hcap}(A)$, while part b givesSince the exit height is between zero and one, the dominated convergence theorem applies and yields

Write $A_r=[-r,r]\times (0,1]$ and take $r\ge 1$. By the Brownian representation of half-plane capacity,On hitting $A_r$, the exit height is at most one. Moreover, $A_r$ lies in the half-disc of radius $\sqrt{r^2+1}\le 2r$. The harmonic measure of that semicircle as viewed from $iy$ is $O(r/y)$: mapping its exterior to $\mathbb{H}$ by $z\mapsto z+(2r{)}^2/z$ reduces the estimate to the Poisson kernel for the upper half-plane on an interval of length $O(r)$. Consequentlyfor large $y$, and $\mathrm{hcap}(A_r)\le Cr$. This is the half-plane capacity of a low rectangle.

For $r\ge 1$, let $R_r=[-r,r]\times (0,1]$. The Brownian representation of half-plane capacity givesOn hitting $R_r$ the imaginary part is at most one, while the harmonic measure estimate supplied in the question shows that the probability of reaching a disc of radius $O(r)$ containing $R_r$ is $O(r/y)$. Hence $\mathrm{hcap}(R_r)\le Cr$, the half-plane capacity of a low rectangle bound.

SetThe scaling and translation of half-plane capacity giveswhereas $\mathrm{diam}(A_n)=\sqrt{4+n^{-2}}\to 2$. This supplies the required sequence.