ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

Problem 10: Let the power series f(z) =
ANALYSIS QUALS 37
∞
anz n have radius of conver-
gence r > 0. For each ρ with 0 < ρ < r let Mf (ρ) := sup{|f(z)|; |z| = ρ}.
Show that the following holds for each such ρ:
∞
n=0
n=0
|an| 2 ρ 2n ≤ Mf (ρ) 2 .
Solution. Let φ be a function on the unit circle T defined by φ(z) = f(ρz).
Then the Fourier expansion of φ is anρnz n , so φ 2
2 = 2π |an| 2ρ2n , by
Parseval. But we have that
φ 2
2 ≤ 2π φ2 ∞
and since φ ∞ = Mf (ρ), we have that
∞
n=0
|an| 2 ρ 2n ≤ Mf (ρ) 2 .
Problem 11: Let f : D → D be a holomorphic map having two unequal fixed
points a, b ∈ D. Show that f(z) = z for all z ∈ D. (Hint: use Schwarz’s
lemma.)
Solution. Let φ(z) = z−a
1−az . Then φ is an automorphism of the disc. Let
g = φ ◦ f ◦ φ−1 . Then g : D → D, g(0) = φ ◦ f(a) = φ(a) = 0, and
g(φ(b)) = φ ◦ f ◦ φ−1 (φ(b)) = φ(b). Furthermore, g is holomorphic. Thus,
we may apply Schwarz Lemma, and as g(φ(b)) = φ(b), we may conclude
that g(z) = z for all z ∈ D.
Then f = φ−1 ◦ g ◦ φ is also the identity, f(z) = z.
Problem 12: Consider the annulus A := {z ∈ C : r < |z| < R}, where
0 < r < R. Show that the function f(z) = 1/z cannot be uniformly
approximated in A by complex polynomials.
Solution. Let P (z) =
arbitrary. Then
2π
0
n
ckz k be a complex polynomial. Let r < η < R be
k=0
f(ηe it )P (ηe it )dt = 1
η
= c0
η
= 0
2π
e
0
it (c0 + ηc1e it + c2η 2 e 2it + . . .)dt
2π 2π
0
e it dt + c1
0
e 2it dt + . . .