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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.

For the vertical slit, choose the branch of $\sqrt{z^2+1}$ asymptotic to $z$ at infinity. It is the mapping-out function of a compact H-hull for $(0,i]$, andThus half-plane capacity of a vertical slit givesThe endpoint $0$ is merely part of the boundary closure convention.

For the filled upper half-disc $K=\{z\in \mathbb{H}:|z|\le 1\}$, the conformal map $g_K(z)=z+1/z$ maps the exterior half-disc onto $\mathbb{H}$. Indeed, it sends the semicircle to $[-2,2]$, the remaining real boundary to the complementary intervals, and its inverse is the branch of $(w+\sqrt{w^2-4})/2$ asymptotic to $w$. Hence half-plane capacity of a half-disc givesIf $\mathbb{D}$ denotes the open unit disc, the printed $\mathbb{H}\cap \mathbb{D}$ is not relatively closed and is literally not a compact H-hull. The intended value is the one for its filled relative closure $K$. With a closed-disc convention the displayed notation already represents that hull.

For a nonempty compact H-hull $A$, its closure meets $\mathbb{R}$: otherwise a nonempty compact set strictly inside $\mathbb{H}$ would be separated from the lower half-plane in the complement, contradicting the simply connected domain condition for $\mathbb{H}\setminus A$. Choose $x\in \overline{A}\cap \mathbb{R}$ and put $d=\mathrm{diam}(A)$. Taking limits in the definition of diameter shows $|z-x|\le d$ for every $z\in A$. Thus $A$ lies in the filled half-disc of radius $d$ centered at $x$. Its half-plane capacity is $d^2$ by scaling and translation of half-plane capacity. Using monotonicity of half-plane capacity,For the empty compact H-hull, use diameter zero. This proves the unit-constant version of half-plane capacity is bounded by squared diameter; no claim of optimality of the constant is needed.

False. LetAfter filling bounded complementary components if necessary, this is a compact H-hull with diameter asymptotic to $2n$. The half-plane capacity of a low rectangle estimate, together with scaling and translation of half-plane capacity, givesThus unbounded diameter need not force unbounded capacity.

Uniqueness under hydrodynamic normalization at infinity givesandIndeed, each right-hand side maps the required complement conformally onto $\mathbb{H}$ and has expansion $z+O(1/z)$. This is the scaling and translation of half-plane capacity.

Let $R_r=[-r,r]\times (0,1]$ and setThe scaling and translation of half-plane capacity and part i giveOn the other hand,Thus very long, very shallow hulls can have vanishing half-plane capacity.

A compact H-hull is a bounded relatively closed set $A\subset \mathbb{H}$ for which $\mathbb{H}\setminus A$ is a simply connected domain. The Riemann mapping theorem and hydrodynamic normalization at infinity give a unique mapping-out function of a compact H-hull $g_A:\mathbb{H}\setminus A\to \mathbb{H}$ withThe coefficient $a\ge 0$ is the half-plane capacity $\mathrm{hcap}(A)$.

For $\lambda >0$ and $x\in \mathbb{R}$, uniqueness of the normalized map givesSubstituting the expansion of $g_A$ yieldsTherefore the scaling and translation of half-plane capacity is

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.

Let $S_r=\{s{e}^{i\theta }:0<s\le r\}$. Since $S_r=r{S}_1$, the scaling and translation of half-plane capacity givesThe half-plane-capacity parameterization requires $ar(t{)}^2=2t$, so