Math 411: Honours Complex Variables - University of Alberta

108 CHAPTER 16. MONTEL’S THEOREM
• (fn,k+1) ∞ n=1 is a subsequence of (fn,k) ∞ n=1 , and
• (fn,k(xk)) ∞ n=1 converges.
For n ∈ N, set gn := fn,n. Then (gn) ∞ n=1 is a subsequence of (fn) ∞ n=1, and
(gn(xk)) ∞ n=1 converges for each k ∈ N.
We show that (gn) ∞ n=1 is a uniform Cauchy sequence on K (and thus convergent).
Let ǫ > 0. Choose δ > 0 such that |f(x) − f(y)|< ǫ
for all f ∈ F and for all
3
x,y ∈ K with |x−y|< δ. Since K is compact, there exist y1,...,yν ∈ K such that
K ⊂ �ν j=1Bδ (yj). Since {x1,x2,x3,...} is dense in K, there exist k1,...,kν ∈ N such
2
that xkj ∈ Bδ(yj).
It follows that K ⊂
2
�ν Bδ(xkj j=1 ).
By construction, (gn(xk)) ∞ n=1 is a Cauchy sequence for each k ∈ N. Choose N ∈ N
such that
ǫ
|gn(xkj )−gm(xkj )|<
3
for n,m ≥ N and j = 1,...,ν. Let x ∈ K be arbitrary, and let n,m ≥ N. Choose
j ∈ {1,...,ν} such that x ∈ Bδ(xkj ), and note that
|gn(x)−gm(x)|≤ |gn(x)−gn(xkj )| +|gn(xkj )−gm(xkj )|
Hence, (gn) ∞ n=1
� �� �
< ǫ
3
� �� �
< ǫ
3
is a uniform Cauchy sequence on K.
+|gm(xkj )−gm(x)|
� �� �
< ǫ
< ǫ.
3
Proposition 16.1. Let D ⊂ R N be open, and let F be a family of functions from D
to R M that is equicontinuous and uniformly bounded on compact subsets of D. Then
every sequence in F has a compactly convergent subsequence.
Proof. For each k ∈ N, define Kk := Bk[0] if D = R N and Kk := Bk[0] ∩{x ∈ D :
dist(x,∂D) ≥ 1
k } if D �= RN . Notice that
• � ∞
k=1 Kk = D and
• Kk ⊂ intKk+1 for n ∈ N.
Let (fn) ∞ n=1 be a sequence in F. By the Arzelà–Ascoli Theorem, there exists a
subsequence (fn,1) ∞ n=1 of (fn) ∞ n=1 and a function g1: K1 → RM such that fn,1|K1→ g1
uniformlyonK1. Invoking theArzelà–AscoliTheoremagain,weobtainasubsequence
(fn,2) ∞ n=1 of (fn,1) ∞ n=1 and a function g2: K2 → RM such that fn,2|K2→ g2 uniformly
on K2. Inductively, we thus obtain, for each k ∈ N, a subsequence (fn,k) ∞ n=1 of (fn) ∞ n=1
and a function gk: Kk → RM such that, for each k ∈ N,
• (fn,k+1) ∞ n=1 is a subsequence of (fn,k) ∞ n=1 , and
• fn,k|Kk → gk uniformly on Kk.