5.2 Normal Families
A collection of functions is better known as a family of functions. One important distinguishing property of families of functions, as opposed to sequences, is that families may be uncountable and may not be indexed by the natural numbers. We will now introduce the following classification of families of functions:
Definition 5.2.1 (Normal Family).
A family of holomorphic functions defined on a region is said to be normal if every sequence of functions in has a locally uniformly (compactly) convergent subsequence on .
The following notion was introduced and formalized by the Italian mathematicians Cesare Arzelà and Giulio Ascoli to formulate a clear distinction in how uniformity is applied.
Definition 5.2.2 (Equicontinuity).
A family of functions defined on a region is said to be equicontinuous at a point if for every , there exists a (that may depend on ) such that for all and all with , we have .
In contrast, the uniform continuity of a function guarantees that may be chosen independently of . In the case of (pointwise) equicontinuity, it is chosen independently of . A family of functions is said to be uniformly equicontinuous on if can be chosen independently of both and (in other words, it attains a positive infimum in ). Similar to Theorem 1.2.15
Theorem 5.2.1.
A family of functions that is pointwise equicontinuous on every point for a compact set is uniformly equicontinuous on .
Proof.
Fix . By pointwise equicontinuity, , such that , ,
The collection forms an open cover of , and by the Heine–Borel Theorem, it admits a finite subcover for some finite . Let .
For any such that , such that . Evidently,
Therefore, from (5.2.1), we have ,
Hence, , we have
which proves the uniform equicontinuity of .
The following theorem is important in many areas of mathematical analysis and has a plethora of generalizations. It was first introduced by Ascoli (who proved the sufficiency of compactness) and later formalized by Arzelà, who proved the necessity of uniform equicontinuity and uniform boundedness.
Theorem 5.2.2 (Arzelà–Ascoli).
Let be a family of complex continuous functions defined on a compact subset . Then, is uniformly bounded and uniformly equicontinuous on iff is normal on .
Proof.
We will first prove the sufficiency of uniform boundedness and uniform equicontinuity. Let be any sequence in . By the uniform boundedness of , there exists a constant such that for all and all .
Let be a countably dense subset of . By the Bolzano–Weierstrass Theorem (Theorem 1.1.2), there exists a subsequence of , namely , such that is convergent. The set is also bounded by , and hence, by the Bolzano–Weierstrass Theorem, it too has a convergent subsequence . Similarly, there exists a subsequence of , namely , such that is convergent.
By the method of construction, we have:
and furthermore, the sequence in each row is a subsequence of the previous row. As a result, we have
We will now invoke a diagonalization argument. Since the sequences above in (5.2.2) are strictly increasing and from the results of (5.2.3), it follows that is strictly increasing. Let be denoted by . Since is uniformly equicontinuous on , , such that satisfying , , we have
Since each is convergent at (for a fixed ) by construction, and since is a subsequence of , it is evident that is convergent at each . We then have that , such that ,
For the fixed value of , the collection forms an open cover of , and by the Heine–Borel Theorem (Theorem 1.1.3), it admits finite subcovering for some finite .
Hence, such that any point lies in . By (5.2.4), we have that
Letting , we have that , ,
Hence, the sequence is uniformly convergent on by the Cauchy Criterion (Theorem 2.3.1).
For the proof of the necessity, we will first assume the normality of in .
For the sake of contradiction, assume that is not uniformly bounded. Then , and such that . By assumption, this sequence has a subsequence that uniformly converges. Hence, such that , , . By the reverse triangle inequality, it follows that . Since is continuous on by Theorem 2.3.5, it is bounded by some (Theorem 1.2.13). Let . It follows that this subsequence is uniformly bounded by . However, since for any , this subsequence cannot be uniformly bounded, and hence we have a contradiction.
We will now assume that is not pointwise equicontinuous at some arbitrary point . In other words, such that , , such that satisfying
Let us define sequences and such that and
Since is assumed to be normal, the sequence has a uniformly convergent subsequence converging to a continuous function . In particular, since uniform convergence preserves continuity (Theorem 2.3.5), the limit is continuous at , and hence,
where the rightmost inequality is derived from the fact that on . Thus,
which contradicts the result that for all .
Hence, by contradiction, is pointwise equicontinuous on all of . By Theorem 5.2.1, must be uniformly equicontinuous on .
The notions and results introduced have profound implications and uses in the theory of differential equations and harmonic analysis.
In the definition of equicontinuity used in the Arzelà–Ascoli theorem, the distance is taken with respect to the Euclidean metric. However, the theorem continues to hold for other metrics as well, with the proof requiring little modification. We will rely on this formulation in Section 8.4.
Lastly, we will prove Montel’s Theorem in preparation of the Riemann Mapping Theorem (Theorem 5.3.1).
Definition 5.2.3.
Let be a family of functions defined on an open set . The family is said to be locally uniformly bounded if, for every point , there exists a neighborhood of such that is uniformly bounded on . This condition is equivalent to the condition that is uniformly bounded on all compact subsets of .
Obviously, the equivalence is established similarly to local finiteness and locally uniform convergence.
Theorem 5.2.3 (Montel's Theorem).
Let be open, and suppose that is a family of holomorphic functions on . Then, is locally uniformly bounded on iff is a normal family.
Proof.
Obviously, if is normal on , for any compact , it follows that is normal on , and the uniform boundedness on follows from the Arzelà–Ascoli Theorem (Theorem 5.2.2).
Conversely, we will first assume that is locally uniformly bounded. Let be arbitrary, and choose such that . Therefore, it follows that is closed and disjoint from and the distance between them is positive. Let this distance be
It follows that the disk is relatively compact in . By Corollary 3.2.1, there exists a finite constant independent of such that
where the maximum on the right-hand side is finite by assumption of the locally uniform boundedness of . For simplicity, let
Let be arbitrary and distinct, and let be the straight curve from to . For an arbitrary function , we have that
Therefore, is uniformly equicontinuous in (and also in ). Indeed, , we can choose and the assertion follows.
Let be compact and arbitrary. The collection forms an open cover of and by the Heine–Borel Theorem (Theorem 1.1.3) admits a finite subcover for some finite . If we let , it follows that is uniformly equicontinuous on . By the Arzelà–Ascoli Theorem (Theorem 5.2.2), any sequence has a uniformly convergent subsequence on .
Let be arbitrary. Let be exhausted by the compact sets . By the argument above, we may extract a subsequence that uniformly converges on . By the same argument, there exists a subsequence that uniformly converges on . Let .
We will now invoke the same diagonalization argument as in the proof of the Arzelà–Ascoli Theorem (Theorem 5.2.2). Let be an arbitrary compact set. It follows that for some , . Since is the subsequence of a sequence that converges on , the assertion follows.