ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

6 ANALYSIS QUALS
Problem 10: Let D = {z : |z| < 1} and let Ω = {z ∈ D : Imz > 0}.
Evaluate
sup{Ref ′ ( i
)|f : Ω → D is holomorphic}.
2
Proof. Note: When I did this on the qual, I misread the problem and
thought that Ω was the upper half-plane instead of merely the upper halfdisc,
which caused the already awful calculations to get worse. Here is my
attempt at the correct calculations, but it seems too ugly to be correct.
Follow at your own peril.
First, note that by rotation, it suffices to find the sup of f ′ ( i
2 ) instead
of Re f ′ ( i
2 ).
The idea is to use Schwarz Lemma. That is, we want g : D → Ω in order
to have f ◦ g map D to D, and then tweak by automorphisms of the disc to
get the map to send 0 to 0.
We have that z ↦→ i 1+z
1−z takes D to the upper half-plane, z ↦→ √ z maps
the upper half-plane to the first quadrant, and z ↦→ z−1
z+1 maps the first
quadrant to Ω. Thus, let
1+z i 1−z − 1
g(z) = .
i + 1
1+z
1−z
Let f1 be the automorphism of the disc taking g(0) to i
2 , and f2 be the
automorphism of the disc taking f( i
2
explicitly; for example we recall that f2(z) =
) to 0. These guys can be written out
i z−f( 2 )
1−f( i
2
Let h = f2 ◦ f ◦ f1 ◦ g. Then h : D → D with h(0) = 0. By Schwarz
Lemma, |h ′ (0) ≤ 1. But we note that
By calculation, we have that
and
and
where we note that
h ′ (0) = g ′ (0)f ′ 1(g(0))f ′ ( i
2 )f ′ 2(f( i
2 )).
f ′ 2(f( i 1
)) =
2 1 − f( i
2 ) ,
g ′ (0) =
f ′ 1(g(0)) =
|g(0)| =
1
√ i( √ i + 1) 2 ,
i 1 + 2
2 ,
1 − |g(0)|
√ i − 1
√ i + 1 .
)z .