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A normal family of holomorphic functions on a domain is one from which every sequence has a subsequence converging locally uniformly to a holomorphic function, or locally uniformly to infinity in the extended convention. For a bounded family, only finite limits occur.

For functions mapping a plane domain $D$ into the unit disc, the Cauchy integral formula supplies uniform derivative bounds on each compact subset. More explicitly, if a closed disc of radius $r$ around $z$ lies in $D$, the Cauchy estimate gives $|f^{\prime }(z)|\le 1/r$. A slightly larger compact neighborhood of any given compact set therefore gives uniform boundedness and equicontinuity there. Apply the Arzelà-Ascoli theorem on a compact exhaustion and take a diagonal subsequence. Its limit is locally uniform; the Cauchy integral formula shows that the limit is holomorphic. This proves normality rather than merely invoking it.

Now let $|g|\le M$ on the complex upper half-plane. Every dilation $g_t(w)=g(tw)$ has the same bound, so the argument just given, after scaling the bound if necessary, proves that $\{g_t:t\ge 1\}$ is a normal family. For any sequence $t_j\to \infty$, every locally uniform subsequential limit $G$ satisfiesThe identity theorem for holomorphic functions forces $G\equiv \ell$. It follows that $g_t\to \ell$ locally uniformly as $t\to \infty$: otherwise a sequence violating convergence on one compact set would have a convergent subsequence with this same constant limit, a contradiction.

For $0<\epsilon <\pi /2$, the arc $K_{\epsilon }=\{e^{i\theta }:\epsilon \le \theta \le \pi -\epsilon \}$ is a compact subset of the complex upper half-plane. Write $z=|z|w$ with $w\in K_{\epsilon }$. Local uniform convergence of the dilations givesFor $\epsilon \ge \pi /2$ the specified sector is empty. This is the same compactness mechanism as the normal-family proof of angular boundary convergence.

To transfer the result to the unit disc, use the Cayley transform between the half-plane and diskIt sends $r\omega$ to $iy$ with $y=(1+r)/(1-r)$, so $g(iy)\to \ell$ as $y\to \infty$. Choose the curvature-minus-one normalization of the hyperbolic metric, for which the disc line element is $2|dz|/(1-|z{|}^2)$ and the half-plane line element is $|dw|/\mathrm{Im}w$. The Cayley transform between the half-plane and disk preserves these metrics. For $w\in \mathbb{H}$ and $y>0$,Thus ${\rho }_{\mathbb{D}}(z,r\omega )<k$ implies $\mathrm{Im}w/|w|>\mathrm{sech}k$. This is the hyperbolic tube around a radial segment estimate. It places $w$ in a fixed sector bounded away from the real axis. Moreover $z\to \omega$ implies $|C_{\omega }(z)|\to \infty$. Applying the sector result givesA different constant normalization of the hyperbolic metric only rescales $k$ and gives the same conclusion.