= Solution
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 $(0,i]$, and
$$
\sqrt{z^2+1}=z+\frac1{2z}+O(z^{-3}).
$$
Thus <half-plane capacity of a vertical slit> gives
$$
\boxed{\operatorname{hcap}([0,i])=\frac12.}
$$
The endpoint $0$ is merely part of the <boundary> <closure> convention.
For the filled upper half-disc $K=\{z\in\mathbb H:|z|\leq1\}$, 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> gives
$$
\boxed{\operatorname{hcap}(K)=1.}
$$
If $\mathbb D$ 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 $d=\operatorname{diam}(A)$. Taking limits in the definition of <diameter> shows $|z-x|\leq d$ 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>,
$$
\boxed{\operatorname{hcap}(A)\leq\operatorname{diam}(A)^2.}
$$
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.
Back to article page
