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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 $\mathrm{hcap}(K)\le \mathrm{rad}(K{)}^2$.

A compact H-hull is a bounded relatively closed set $K\subset \mathbb{H}$ whose complement $D=\mathbb{H}\setminus K$ is a simply connected domain. Equivalently one can retain its compact closure in $\overline{\mathbb{H}}$; real boundary points do not affect the mapping-out function. The Riemann mapping theorem and Schwarz reflection principle near infinity give a unique conformal map $g_K:D\to \mathbb{H}$ with hydrodynamic normalization at infinity. Its Laurent series isThe half-plane capacity is $\mathrm{hcap}(K)=a_K$. The missing constant term is part of the normalization, not a choice to be made again when comparing different hulls.

Here are the conformal map facts we use to justify positivity and comparison. The inverse $f_K:\mathbb{H}\to D\subset \mathbb{H}$ has expansion $f_K(w)=w-a_K/w+O(w^{-2})$. The upper-half-plane Schwarz-Pick theorem, applied to $w$ and $iY$ and then letting $Y\to \infty$, implies $\mathrm{Im}{f}_K(w)\ge \mathrm{Im}w$. Indeed the inequality between pseudohyperbolic distances givesand multiplying by $Y$ gives the height comparison in the limit. At $w=iY$, the inverse expansion reads $\mathrm{Im}{f}_K(iY)=Y+a_K/Y+O(Y^{-2})$, so $a_K\ge 0$.

For $r>0$, uniqueness of the mapping-out function givesConsequentlyTranslation by a real number $b$ similarly gives $g_{K+b}(z)=b+g_K(z-b)$ and leaves half-plane capacity unchanged.

If $K\subset A$ are compact H-hulls, their mapping-out functions compose as $g_A=g_L\circ g_K$, where $L$ is the filled image of $A\setminus K$. Comparing Laurent series yieldsThis proves monotonicity of half-plane capacity, rather than assuming it from inclusion alone. The filled half-disc $A=\{z\in \mathbb{H}:|z-b|\le R\}$ has mapping-out functionwhich takes its semicircular boundary to the real line and is a conformal bijection onto $\mathbb{H}$. Thus its half-plane capacity is $R^2$. Enclose $K$ in such a half-disc, compare capacities, and minimize over real centres and radii. This proves the sharp boundIf one defines the radius by an infimum, take $R$ strictly larger than that infimum and then let $R$ decrease; no attainment assumption is needed.

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:

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.

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

Let $d=\mathrm{diam}(A)$. Unless $A$ is empty, its closure meets the real axis; otherwise a loop in $\mathbb{H}\setminus A$ surrounding $A$ could not contract, contrary to $\mathbb{H}\setminus A$ being a simply connected domain. Choose $x\in \overline{A}\cap \mathbb{R}$. Then $A\subseteq \{z\in \mathbb{H}:|z-x|\le d\}$.

The unit half-disc is a compact H-hull with mapping-out function $z+z^{-1}$, so its half-plane capacity is one. The monotonicity of half-plane capacity and its scaling rule now giveThus the assertion holds with the universal constant $c=1$ under this normalization.