17.6.1 Montel’s Theorem
Theorem 17.6.1 Let Ω be an open set in ℂ and let ℱ denote a set of analytic functions mapping
Ω to B
⊆ ℂ. Then there exists a sequence of functions from ℱ,
n=1∞ and an analytic
function f such that for each k ∈ ℕ, fn
converges uniformly to f
on every compact subset of
Ω. Here f
denotes the kth derivative.
Proof: First note there exists a sequence of compact sets, Kn such that Kn ⊆ intKn+1 ⊆ Ω for all n
where here intK denotes the interior of the set K, the union of all open sets contained in K and
∪n=1∞Kn = Ω. In fact, you can verify that B
exist positive numbers, δn
such that if z ∈ Kn,
Now denote by ℱn
the set of
restrictions of functions of ℱ
Then let z ∈ Kn
and let γ
+ δneit,t ∈
It follows that
for z1 ∈ B
and f ∈ℱ,
It follows that ℱn
is equicontinuous and uniformly bounded so by the Arzela Ascoli theorem there exists a
which converges uniformly on Kn.
converge uniformly on K1.
Then use the Arzela Ascoli theorem applied to this sequence to get a subsequence, denoted by
which also converges uniformly on K2.
Continue in this way to obtain
uniformly on K1,
Now the diagonal sequence
is a subsequence of
it converges uniformly on Km
for all m.
for short, this is the sequence of
functions promised by the theorem. It is clear
converges uniformly on every compact
subset of Ω because every such set is contained in Km
for all m
large enough. (Why?) Let
be the point to which
is a continuous function defined on Ω.
analytic? Yes it is by Lemma 15.5.5
. Alternatively, you could let T ⊆
Ω be a triangle.
Therefore, by Morera’s theorem, f is analytic.
As for the uniform convergence of the derivatives of f, recall Theorem 16.5.2 about the existence of a
cycle. Let K be a compact subset of int
be closed oriented curves contained
such that ∑
= 1 for every
z ∈ K
. Also let η
denote the distance between ∪jγj∗
It follows that η > 0. (Why? In general, two disjoint compact sets are at a positive distance from each
other. Show this is so.) Then for z ∈ K,
Thus you get uniform convergence of the
Another surprising consequence of this theorem is that the property of being one to one is
Lemma 17.6.2 Suppose hn is one to one, analytic on Ω, and converges uniformly to h on compact
subsets of Ω along with all derivatives. Then h is also one to one.
Proof: Pick z1 ∈ Ω and suppose z2 is another point of Ω. Since the zeros of h − h
have no limit
point, there exists a circular contour bounding a circle which has
on the inside of this circle but not z1
such that γ∗
contains no zeros of h − h
Using the theorem on counting zeros, Theorem 17.3.1, and the fact that hn is one to one,
which shows that h−h
has no zeros in
In particular z2
is not a zero of h−h
is one to one since z2≠z1
was arbitrary. ■
Since the family, ℱ satisfies the conclusion of Theorem 17.6.1 it is known as a normal family of
functions. More generally,
Definition 17.6.3 Let ℱ denote a collection of functions which are analytic on Ω, a region. Then ℱ
is normal if every sequence contained in ℱ has a subsequence which converges uniformly on compact
subsets of Ω.