 Solution

Set $v(z)=\mathrm{Im}{g}_K(z)$, 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 $\mathrm{Im}w>0$, continuity of the inverse inside $\mathbb{H}$ would give $p=g_K^{-1}(w)\in D$, a contradiction. ThusThis boundary degeneration under a mapping-out function is valid without a smooth or locally connected hull boundary.

The harmonic function $u(z)=v(z)-\mathrm{Im}z$ 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 $\max (0,{\epsilon }_R)$, where ${\epsilon }_R={\sup }_{D\cap \{|z|=R\}}|g_K(z)-z|\to 0$. Let $R\to \infty$ with $z$ fixed. We obtainThis 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.