Normal Families and Montel's Theorem

Each $K_n$ is bounded by $\mathrm{\mid }z\mathrm{\mid }\le n$ and closed because limits preserve both the radius bound and the distance-to-boundary inequality, so [L1] makes each $K_n$ compact; early members are allowed to be empty.

If $z\in K_n$, then $\mathrm{\mid }z\mathrm{\mid }<n+1$ and $\mathrm{dist}(z,\mathrm{\partial }\mathrm{\Omega })>1\mathrm{/}(n+1)$, so a small disc about $z$ stays inside $K_{n+1}$. Hence $K_n\subseteq \mathrm{int}{K}_{n+1}$.

If $z\in \mathrm{\Omega }$, openness gives a closed disc $\bar{D}(z,r)\subseteq \mathrm{\Omega }$; choosing $n>\mathrm{\mid }z\mathrm{\mid }$ and $1\mathrm{/}n<r$ puts $z$ in $K_n$. Therefore ${\bigcup }_{n\ge 1}{K}_n=\mathrm{\Omega }$.

If $d_K(f_m,f)\to 0$, then each summand $2^{-n}\min (1,s_n(f_m,f))$ tends to $0$, so $f_m\to f$ uniformly on every $K_n$.

Any compact set $L\subseteq \mathrm{\Omega }$ is contained in some $K_N$ because the interiors of the $K_n$ cover $\mathrm{\Omega }$ by [L1]. Step 1.1 then gives uniform convergence on every compact $L$, and [L2] makes the convergence locally uniform.

Conversely, if $f_m\to f$ locally uniformly, then [L2] gives uniform convergence on every $K_n$, including the empty stages with zero supremum. A finite-head plus geometric-tail estimate then gives $d_K(f_m,f)\to 0$.

By [L1], each exhaustion metric induces the same convergence notion, namely local uniform convergence on $\mathrm{\Omega }$.

Fact [L2] identifies that convergence notion with compact convergence and hence with the compact-open topology on $C(\mathrm{\Omega },\mathbb{C})$. Therefore the induced topology is independent of the chosen exhaustion.

On each $K_n$, the $d_K$-Cauchy property makes $(f_m{\mathrm{\mid }}_{K_n})$ uniformly Cauchy; if $K_n=\mathrm{\varnothing }$ take the zero function there, and otherwise [L1] gives a uniform limit $g_n:K_n\to \mathbb{C}$, which is continuous by [L2].

The nesting of [L3] makes these limits compatible on overlaps, so they glue to a single function $f:\mathrm{\Omega }\to \mathbb{C}$. Because every point of $\mathrm{\Omega }$ lies in the interior of some $K_n$ by [L3], the glued function is continuous there.

Uniform convergence $f_m{\mathrm{\mid }}_{K_n}\to f{\mathrm{\mid }}_{K_n}$ on each $K_n$, together with the usual finite-head and geometric-tail estimate in the defining series, gives $d_K(f_m,f)\to 0$. Thus $C(\mathrm{\Omega },\mathbb{C})$ is complete for the exhaustion metric.

Fact [L1] turns the metric convergence into local uniform convergence on the domain.

Applying [L2] shows that the limit function $f$ is holomorphic. Hence holomorphic functions form a closed subspace for local uniform convergence.

Fix $a\in \mathrm{\Omega }$. Local boundedness gives a closed disc $\bar{D}(a,R)\subseteq \mathrm{\Omega }$ and a common bound $\mathrm{\mid }f\mathrm{\mid }\le M$ there for every $f\in \mathcal{F}$. Applying [L1] to the inner disc $\bar{D}(a,R\mathrm{/}2)$ gives the uniform derivative bound $\mathrm{\mid }{f}^{\mathrm{\prime }}(z)\mathrm{\mid }\le 4M\mathrm{/}R$ on that smaller disc.

The smaller closed disc is convex, so integrating $f^{\mathrm{\prime }}$ along the line segment from $w$ to $z$ yields $\mathrm{\mid }f(z)-f(w)\mathrm{\mid }\le (4M\mathrm{/}R)\mathrm{\mid }z-w\mathrm{\mid }$ for every $f\in \mathcal{F}$. This is the local equicontinuity estimate.

On each compact stage $K_n$ of [L3], local boundedness plus [L4] makes the restricted family equicontinuous and pointwise relatively compact; hence [L1] makes its uniform closure compact.

Using [L5], choose successively a subsequence converging uniformly on $K_1$, then a further subsequence converging uniformly on $K_2$, and so on, and take the diagonal subsequence. This is the explicit Choice step named in the Statement.

For each fixed $m$, the diagonal subsequence eventually lies in the $m$th chosen subsequence, so it converges uniformly on $K_m$. Because the interiors of the $K_m$ cover $\mathrm{\Omega }$ by [L3], this gives local uniform convergence on all of $\mathrm{\Omega }$.

Fact [L2] makes the local uniform limit holomorphic, so every sequence in $\mathcal{F}$ has a locally uniformly convergent subsequence in $H(\mathrm{\Omega })$. That is exactly normality.

If $\mathcal{F}$ were not locally bounded at some point, then on some closed disc there would be a sequence $(f_n)$ in $\mathcal{F}$ with $\sup \mathrm{\mid }{f}_n\mathrm{\mid }>n$.

By [L1], a subsequence would converge uniformly on that disc to a holomorphic limit, and uniform convergence on a compact disc forces that subsequence to be uniformly bounded there. This contradicts step 1.1, so the family is locally bounded.

By [L1], every subsequence of $(f_n)$ has a further subsequence converging locally uniformly to a holomorphic limit. Any two such subsequential limits agree on $E$, because the scalar sequence $f_n(z)$ has a fixed pointwise limit there, so [L2] makes them equal on all of $\mathrm{\Omega }$.

If the whole sequence did not converge locally uniformly to that common limit, then some subsequence would stay a definite distance away on a compact set. Applying [L1] again to that subsequence would produce a further subsequence converging locally uniformly to the same limit from step 1.1, which is impossible.

Therefore the full sequence converges locally uniformly on $\mathrm{\Omega }$ to a holomorphic function.

Fact [L1] gives $f_n^{(k)}\to f^{(k)}$ locally uniformly on $\mathrm{\Omega }$ for every natural $k$.

This is exactly continuity of the operator $D^k$, and the case $k=0$ is included because $D^0$ is the identity.

A local uniform limit of continuous maps into the metric space $(\widehat{\mathbb{C}},\chi )$ is continuous, so $f$ is chordally continuous. If $f(a)\in \mathbb{C}$, choose a chordal neighbourhood $U$ of $f(a)$ whose closure misses $\mathrm{\infty }$. On $U$ the chordal and Euclidean metrics are comparable, and continuity of $f$ together with chordal local uniform convergence gives a neighbourhood $V$ of $a$ with $f(V)\cup f_n(V)\subseteq U$ for all large $n$. On $V$ the maps $f_n$ have no poles, hence are holomorphic there, and [L1] makes the Euclidean local uniform limit $f$ holomorphic near $a$.

If $f(a)=\mathrm{\infty }$, choose a chordal neighbourhood $U$ of $\mathrm{\infty }$ whose complement is a closed Euclidean disc. Continuity of $f$ and chordal local uniform convergence give a neighbourhood $V$ of $a$ with $f(V)\cup f_n(V)\subseteq U$ for all large $n$. The infinity-chart expressions $g=1\mathrm{/}f$ and $g_n=1\mathrm{/}{f}_n$ are then well-defined holomorphic maps on $V$, and the same metric comparison turns $g_n\to g$ into Euclidean local uniform convergence. Fact [L1] makes $g$ holomorphic, so $f$ is meromorphic at $a$.

When every $f_n$ is holomorphic, the functions $g_n$ of step 1.2 are holomorphic and nowhere zero on $V$. By [L2], their limit $g$ is either identically $0$ on $V$ or nowhere zero. Because $g(a)=0$, one gets $g\equiv 0$ on $V$, so $f\equiv \mathrm{\infty }$ on $V$. Thus the $\mathrm{\infty }$-value set of $f$ is open in the holomorphic-input case.

If $f$ takes some finite value, then steps 1.1 and 1.2 show that it is meromorphic at every point of $\mathrm{\Omega }$; otherwise $f\equiv \mathrm{\infty }$. In the holomorphic-input case, the finite-value set is open by step 1.1 and the $\mathrm{\infty }$-value set is open by step 2.1, so connectedness leaves only the two possibilities: $f$ is holomorphic on all of $\mathrm{\Omega }$, or $f\equiv \mathrm{\infty }$.

If the compact-set equicontinuity condition holds, then on each compact stage $K_n$ of [L2] the restricted family is equicontinuous. Pointwise relative compactness is automatic because the target sphere is compact, so [L1] makes the uniform closure on $K_n$ compact.

Using [L3], choose successively a subsequence converging uniformly on $K_1$, then on $K_2$, and so on, and take the diagonal subsequence. By [L2] that diagonal converges chordally locally uniformly on $\mathrm{\Omega }$, and [L4] makes its limit meromorphic or identically $\mathrm{\infty }$. Thus compact-set chordal equicontinuity implies meromorphic normality.

Conversely, if $\mathcal{F}$ is meromorphically normal, then every sequence of restrictions to a fixed $K_n$ has a uniformly convergent subsequence. Fact [L3] turns that sequential compactness into compactness of the restriction closure, and [L1] then gives equicontinuity on $K_n$. Because every compact subset of $\mathrm{\Omega }$ lies in some stage of the exhaustion by [L2], the family is chordally equicontinuous on every compact subset.

The first two steps prove the forward implication and step 1.3 proves the reverse implication, so the two conditions are equivalent.