 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 $|f^{\prime }|$ 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 $t>0$,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 $z_j=x_j+i{y}_j\to 0$ with $|x_j|\le A{y}_j$ and $y_j>0$. Put $n_j=\lfloor 1/y_j\rfloor$. Then $n_j\to \infty$, $n_j{y}_j\to 1$, and the points ${\zeta }_j=n_j{z}_j$ lie, for large $j$, in a fixed compact rectangle inside the half-plane. ThereforeThis 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, useIf $z=x+iy$ with $y>0$, then $\mathrm{Re}(-i/z)=-y/(x^2+y^2)<0$, so $g$ is a holomorphic function and $|g|<1$. Also $g(iy)=e^{-1/y}\to 0$. ButTheir ratio $|\mathrm{Re}{z}_n|/\mathrm{Im}{z}_n=2\pi n^2$ tends to infinity. Thus the nontangential limit is zero while the unrestricted boundary limit does not exist.