= Solution
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 $R>\max(\operatorname{Im}z,\sup_K\operatorname{Im}w)$. This stopping time is finite almost surely, since it is no later than exit of its imaginary coordinate from $(0,R)$.
By part (d), $0\leq v(w)\leq R$ 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 give
$$
v(z)=\mathbb E_z\bigl[v(B_{\tau_R})\,1_{\{\operatorname{Im}B_{\tau_R}=R\}}\bigr].
$$
Write $p_R=\mathbb P_z(\operatorname{Im}B_{\tau_R}=R)$ and use the uniform displacement bound $M$ from part (c). On the top boundary, $|v(w)-R|\leq M$, hence
$$
|v(z)-Rp_R|\leq Mp_R.
$$
The stopped imaginary coordinate is itself a bounded <martingale>. Since its exit height is nonnegative,
$$
\operatorname{Im}z=\mathbb E_z[\operatorname{Im}B_{\tau_R}]\geq Rp_R.
$$
Consequently $|v(z)-Rp_R|\leq M\operatorname{Im}z/R\to0$, proving the <high-level escape representation of a mapping-out height>:
$$
\boxed{\operatorname{Im}g_K(z)=\lim_{R\to\infty}R\,\mathbb P_z(\operatorname{Im}B_{\tau_R}=R).}
$$
If $H$ instead meant the whole upper half-plane, the right side would always be $\operatorname{Im}z$. 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.
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