Half-plane capacity ($\mathrm{hcap}(A)$)

The half-plane capacity of a compact H-hull is the nonnegative coefficient $a$ in

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

For $x\in \mathbb{R}$ and $R>0$, the filled half-disc $K=\{z\in \mathbb{H}:|z-x|\le R\}$ is a compact H-hull. Its mapping-out function of a compact H-hull is $g_K(z)=z+R^2/(z-x)$. On the semicircle the image is real, and the inverse branch asymptotic to the identity maps the complex upper half-plane onto the exterior half-disc. Its Laurent series has coefficient $R^2/z$, so $\boxed{\mathrm{hcap}(K)=R^2}$. Using an open half-disc instead requires taking its relative closure first.

Suppose a capacity-parameterized Loewner chain lies in a domain on which $\psi$ is a conformal map. For its image chain set ${\psi }_t={\tilde{g}}_t\circ \psi \circ g_t^{-1}$. Then the image Loewner driving function is ${\tilde{U}}_t={\psi }_t(U_t)$ andThe derivative is real by the Schwarz reflection principle. Cancellation of the pole in the differentiated conjugacy identity proves the squared-derivative rule.

For conjugated Loewner chains, ${\varphi }_t={\tilde{g}}_t\circ \varphi \circ g_t^{-1}$ and ${\tilde{\xi }}_t={\varphi }_t({\xi }_t)$. The image capacity speed is $2{\varphi }_t^{\prime }({\xi }_t{)}^2$. Differentiating this identity gives ${\dot{\varphi }}_t(z)=2{\varphi }_t^{\prime }({\xi }_t{)}^2/({\varphi }_t(z)-{\varphi }_t({\xi }_t))-2{\varphi }_t^{\prime }(z)/(z-{\xi }_t)$. The removable singularity at the driver has value $-3{\varphi }_t^{\prime \prime }({\xi }_t)$.

Let $h_t$ intertwine two Loewner chains, with driver ${\xi }_t=\sqrt{\kappa }{B}_t$, and set $a_t=h_t^{\prime }({\xi }_t)$, $b_t=h_t^{\prime \prime }({\xi }_t)$, $c_t=h_t^{\prime \prime \prime }({\xi }_t)$. Expanding the conjugacy equation at the driver gives ${\partial }_t{h}_t^{\prime }({\xi }_t)=b_t^2/(2a_t)-4c_t/3$; the Itô formula gives the displayed diffusion. For a real exponent $p$, the drift of $a_t^p$ is $p[1+\kappa (p-1)]a_t^{p-2}{b}_t^2/2+p(\kappa /2-4/3)a_t^{p-1}{c}_t$. At $\kappa =8/3$, $p=5/8$ cancels both terms and yields the SLE eight-thirds restriction martingale.

Apply the Itô formula to the transformed Loewner driving function ${\tilde{\xi }}_t={\varphi }_t(\sqrt{\kappa }{W}_t)$. The conformal conjugacy derivative for the chordal Loewner equation cancels the Itô drift exactly when $\kappa =6$. Its quadratic variation is $\kappa {\int }_0^t{\varphi }_s^{\prime }({\xi }_s{)}^{2}ds$.

For a vertical slit of height $h>0$ attached at $x\in \mathbb{R}$, the mapping-out function of a compact H-hull is $g(z)=x+\sqrt{(z-x{)}^2+h^2}$, with the branch asymptotic to $z-x$ at infinity. Its half-plane capacity is $h^2/2$.

For $\lambda >0$ and $x\in \mathbb{R}$,

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.

If compact H-hulls satisfy $A\subseteq C$ and $D$ is the bounded filling of $g_A(C\setminus A)$, thenIn particular, inclusion is strict exactly when the capacity increase is positive.

If $\tau$ is the first exit time of planar Brownian motion from $\mathbb{H}\setminus A$, thenIt follows by applying the optional sampling theorem for a supermartingale to the harmonic function $\mathrm{Im}(z-g_A(z))$.

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 $iy$.

For $R_r=[-r,r]\times (0,1]$ with $r\ge 1$,Indeed, the imaginary part at the Brownian exit point is at most one and the probability of reaching the radius-$O(r)$ neighbourhood containing the rectangle from $iy$ is $O(r/y)$. Consequently $r^{-1}{R}_r=[-1,1]\times (0,r^{-1}]$ has capacity $O(r^{-1})$ although its diameter tends to two.

A growing hull is parameterized by half-plane capacity when $\mathrm{hcap}(K_t)=2t$. The factor two makes its Chordal Loewner equation take the conventional form ${\partial }_t{g}_t=2/(g_t-U_t)$.