= Solution
Set $v(z)=\operatorname{Im}g_K(z)$, a positive <harmonic function> on $D$. First justify its boundary behavior. For $z_n\to p\in\partial D$ with $p$ finite, part (c) makes $g_K(z_n)$ bounded. Any subsequential limit $w$ lies in $\overline{\mathbb H}$. If $\operatorname{Im}w>0$, continuity of the inverse inside $\mathbb H$ would give $p=g_K^{-1}(w)\in D$, a contradiction. Thus
$$
v(z)\longrightarrow0\quad\text{as }z\to p\in\partial D.
$$
This <boundary degeneration under a mapping-out function> is valid without a smooth or locally connected hull boundary.
The <harmonic function> $u(z)=v(z)-\operatorname{Im}z$ consequently has nonpositive finite-boundary values. Its value tends uniformly to zero at infinity, by the <Laurent series>. On the bounded domain $D\cap\{|z|<R\}$, the <maximum principle for harmonic functions> bounds $u$ by $\max(0,\varepsilon_R)$, where $\varepsilon_R=\sup_{D\cap\{|z|=R\}}|g_K(z)-z|\to0$. Let $R\to\infty$ with $z$ fixed. We obtain
$$
\boxed{\operatorname{Im}g_K(z)\leq\operatorname{Im}z.}
$$
This is the <height contraction of a hydrodynamically normalized mapping-out function>. The exhaustion controls infinity explicitly, which is necessary when using a maximum principle on an unbounded domain.
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