= Solution
A <normal family> is a family such that every sequence has a subsequence converging locally uniformly in the spherical metric to a meromorphic limit, possibly the constant infinity. For a family uniformly bounded in modulus, the possible limits are finite <holomorphic functions>, and ordinary locally uniform convergence suffices. The limit need not take values in the open <unit disc>; a sequence of constants can approach its boundary.
Here is a direct compactness proof for disc-valued maps. For each <compact subset> $K\subset\Omega$, choose a slightly larger compact neighborhood contained in $\Omega$. The <Cauchy estimate> on a fixed-radius disk around each point bounds $|f'|$ independently of $f$ because $|f|<1$. A finite covering of $K$ yields uniform boundedness and equicontinuity there. The <Arzelà-Ascoli theorem>, applied successively to a <compact exhaustion> and followed by a diagonal subsequence, gives a limit uniformly on every compact. The <Cauchy integral formula> passes to that limit on small circles, proving it is a <holomorphic function>. This proves the required <normal family> assertion rather than just invoking <Montel theorem>.
Every $g_n(z)=g(z/n)$ is a <holomorphic function> on the <complex upper half-plane> with values in the <unit disc>, so the same argument makes $(g_n)$ a <normal family>. If a subsequence converges locally uniformly to $h$, then for every $t>0$,
$$
h(it)=\lim_j g(it/n_j)=\ell.
$$
The <identity theorem for holomorphic functions> gives $h\equiv\ell$. Since every convergent subsequence has this same limit, the whole sequence converges locally uniformly to $\ell$: otherwise a compact set and a subsequence with a fixed positive discrepancy would have a further locally uniformly convergent subsequence, a contradiction.
To obtain the <nontangential limit>, take any $z_j=x_j+iy_j\to0$ with $|x_j|\leq A y_j$ and $y_j>0$. Put $n_j=\lfloor1/y_j\rfloor$. Then $n_j\to\infty$, $n_jy_j\to1$, and the points $\zeta_j=n_jz_j$ lie, for large $j$, in a fixed compact rectangle inside the half-plane. Therefore
$$
\boxed{g(z_j)=g_{n_j}(\zeta_j)\longrightarrow\ell.}
$$
This is the <normal-family proof of angular boundary convergence>; using integer rescalings requires this step because the points need not lie exactly on one fixed scaled ray.
For a tangential counterexample, use
$$
\boxed{g(z)=e^{-i/z}.}
$$
If $z=x+iy$ with $y>0$, then $\operatorname{Re}(-i/z)=-y/(x^2+y^2)<0$, so $g$ is a <holomorphic function> and $|g|<1$. Also $g(iy)=e^{-1/y}\to0$. But
$$
z_n=\frac1{2\pi n-i/n}\in\mathbb H,\qquad z_n\to0,\qquad g(z_n)=e^{-1/n}\to1.
$$
Their ratio $|\operatorname{Re}z_n|/\operatorname{Im}z_n=2\pi n^2$ tends to infinity. Thus the <nontangential limit> is zero while the unrestricted boundary limit does not exist.
Back to article page
