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Here the otherwise undefined printed domain $H$ must mean $D=\mathbb{H}\setminus K$. Let the Brownian motion start at $z\in D$, and let ${\tau }_R$ stop it on reaching height $R$ or leaving $D$. Take $R>\max (\mathrm{Im}z,{\sup }_K\mathrm{Im}w)$. This stopping time is finite almost surely, since it is no later than exit of its imaginary coordinate from $(0,R)$.

By part (d), $0\le v(w)\le R$ in the stopped domain. Its finite-boundary values vanish except on the top boundary, by the argument of part (d). Localization and the optional stopping theorem for this bounded harmonic martingale therefore giveWrite $p_R={\mathbb{P}}_z(\mathrm{Im}{B}_{{\tau }_R}=R)$ and use the uniform displacement bound $M$ from part (c). On the top boundary, $|v(w)-R|\le M$, henceThe stopped imaginary coordinate is itself a bounded martingale. Since its exit height is nonnegative,Consequently $|v(z)-R{p}_R|\le M\mathrm{Im}z/R\to 0$, proving the high-level escape representation of a mapping-out height:If $H$ instead meant the whole upper half-plane, the right side would always be $\mathrm{Im}z$. For example, the mapping-out function of a vertical slit $K=(0,ia]$ is $g_K(w)=\sqrt{w^2+a^2}$ with branch asymptotic to $w$; at $z=iy$, $y>a$, its height is $\sqrt{y^2-a^2}<y$. This demonstrates why killing on the hull is essential to the printed formula.

A nearly closed semicircular slit makes the constant sharp. PutThis is the unit-circle arc attached at $-1$ and ending just short of $1$. Its complement is a simply connected domain. There remains a narrow passage near $1$ connecting the interior bay to infinity. Thus $K_L$ is a compact H-hull inside the closed unit disc.

For an explicit check, let $S=\sqrt{1+L^2}$ and let $q_L(z)=\sqrt{m(z{)}^2+L^2}$ denote the mapping-out function of a vertical slit applied to $m(z)$, with the branch asymptotic to its argument in the slit half-plane. Its normalized map isThe final Möbius transformation takes the image of the original infinity to infinity and gives hydrodynamic normalization at infinity. For a fixed $z$ in the open unit half-disc, $\mathrm{Re}m(z)>0$, so $q_L(z)\sim L$ and $g_{K_L}(z)\to 2$ as $L\to \infty$. Now take $z_{\delta }=-1+\delta +i\delta$, with $0<\delta <1$. First let $L\to \infty$, then let $\delta \downarrow 0$. The nearly closed semicircular slit givesEvery constant smaller than three therefore fails for some member of this family and some interior point. The bound need not be attained at an interior point of a single fixed hull.