 Solution

Let $f=g_K^{-1}$. The reflection at infinity used in part (a) gives analytic expansions there for both inverse maps. In particular,uniformly for large $w\in \mathbb{H}$, including approach close to the real axis.

Fix $R<\infty$. If $g_K$ were unbounded on $D\cap \{|z|\le R\}$, there would be points $z_n$ there with $w_n=g_K(z_n)$ tending to infinity in modulus. The inverse expansion would imply $z_n=f(w_n)=w_n+O(1/w_n)$, contradicting boundedness of $z_n$. Thus $g_K$ is bounded on every bounded portion of its domain, including points arbitrarily near a rough hull boundary. This is the inverse-at-infinity criterion for local boundedness of a mapping-out function.

On the region $|z|>R$ for sufficiently large $R$, the expansion of $g_K$ gives $g_K(z)-z=O(1/z)$. On the remaining bounded region, the preceding bound for $g_K$ and the bound for $z$ give a finite bound for their difference. ThereforeThe argument uses reflection near infinity only. It does not assume that the real part of the map extends continuously at every point of an arbitrary hull; the sharp displacement bound for a compact H-hull is a further quantitative version of this boundedness.