ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

32 ANALYSIS QUALS
form e iθ z for some θ.
Thus, we have that
iθ z − i
g(z) = h ◦ f(z) = e
z + i
for some θ.
(b) Since f is holomorphic, its imaginary part is a harmonic function on
the disk. Then by Harnack’s inequality, we have that
1 − |z|
1 + |z|
Im(f(0)) ≤ Im(f(z)) ≤
1 + |z| 1 − |z| Im(f(0)).
Hence, as Im(f(0)) = 1, we have that Im(f(x)) ≥ 1−x
1+x .
Problem 10: Suppose U is a bounded connected open set in C and z0 ∈ U.
Let F = {f : U → D, f holomorphic, f(z0) = 0}.
(a) Show that if K is a compact subset of U, then there is a constant
MK > 0 such that |f ′ (z)| ≤ MK for all z ∈ K, f ∈ F .
(b) Use part (a) to show that if {fn} is a sequence in F , then there is a
subsequence {fnj } which converges uniformly on every compact subset
of U to a function f0 ∈ F .
(Note: Part of this is to show f0(U) ⊂ D.)
Solution. (a) Let r = dist(K, U c ). Let γ be a closed curve in U around K
with dist(K, γ) ≥ r
2 . Then by Cauchy Integral formula, we have that
f ′ (w) = 1
f(z)
dz.
2πi (z − w) 2
Hence for w ∈ K, we have that
|f ′ (w)| ≤ 1
1 1
dz ≤
2π |z − w| 2 2π
γ
γ
4
4M
length(γ) ≤
r2 r2 where M is the bound on |z| in U. Thus, let MK = 4M
r2 .
(b) As f : U → D, |f(z)| < 1 for all z ∈ U and f ∈ F . Hence by Montel’s
theorem, there exists a subsequence {fnj } converging uniformly on
compact subsets of U to a function f0.
But then fnj (z0) → f0(z0), so f0(z0) = 0. Since |fnj (z)| < 1 for all
z ∈ U, we must have that |f0(z)| ≤ 1 for all z ∈ U. But as f0 is
holomorphic, it is an open mapping by the Open Mapping theorem,
so as U is open, f0(U) is open. Thus, f0(U) ⊂ D, so f0 ∈ F .
Problem 11: Let D denote the open unit disk in the complex plane let D
denote its closure. Suppose f : D → C is continuous on D and analytic in
its interior. Show that if f takes only real values on ∂D, then f must be
constant.
Solution. Let g(z) be defined to be f(z) for z ∈ D, and f
1 1
z for z ∈ D.
Note that this is well-defined: if both conditions are satisfied, then z must
be of the form eiθ , in which case f
1
z = f(eiθ iθ ) = f(e ) since f takes real
values on the circle.
Thus, by Schwarz Reflection Principle, g is entire.