Problems and Solutions in - Mathematics - University of Hawaii

2.2 1991 November 21 2 COMPLEX ANALYSIS
7. Let {fn} be a sequence of analytic functions in the unit disk D, and suppose there exists a positive constant M
such that
|fn(z)| |dz| ≤ M
C
for each fn and for every circle C lying in D. Prove that {fn} has a subsequence converging uniformly on compact
subsets of D.
Solution: We must show that F = {fn} is a normal family. If we can prove that F is a locally bounded family
of holomorphic functions – that is, F ⊂ H(D) and, for any compact set K ⊂ D, there is an MK > 0 such that
|fn(z)| ≤ MK for all z ∈ K and all n = 1, 2, . . . – then the Montel theorem (corollary 2.2) will give the desired
result.
To show F is locally bounded, it is equivalent to show that, for each point zα ∈ D, there is a number Mα and a
neighborhood B(zα, rα) ⊂ D such that |fn(z)| ≤ Mα for all z ∈ B(zα, rα) and all n = 1, 2, . . .. (Why is this
equivalent?) 39 So, fix zα ∈ D. Let Rα > 0 be such that ¯ B(zα, Rα) = {z ∈ C : |z − zα| ≤ Rα} ⊂ D. Then, for
any z ∈ B(zα, Rα/2), Cauchy’s formula gives
|fn(z)| ≤ 1
|fn(ζ)|
2π |ζ−zα|=Rα |ζ − z| |dζ|
≤ 1 1
2π Rα/2
≤ M
.
πRα
|ζ−zα|=Rα
|fn(ζ)| |dζ|
The second inequality holds since |ζ − zα| = Rα and |z − zα| < Rα/2 imply |ζ − z| > Rα/2. The last inequality
follows from the hypothesis
C |fn(z)| |dz| ≤ M for any circle C in D. Letting Mα = M
πRα , and rα = Rα/2, we
have |fn(z)| ≤ Mα for all z ∈ B(zα, rα) and all n = 1, 2, . . . , as desired. ⊓⊔
8. State and prove:
(a) the mean value property for analytic functions
(b) the maximum principle for analytic functions.
39 Answer: If K ⊂ D is compact, we could select a finite covering of K by such neighborhoods B(zαj , rα j ) (j = 1, . . . , J) and then
|fn(z)| ≤ maxj Mα j MK, for all z ∈ K and n = 1, 2, . . . .
46