Math 752
Spring 2015
Riemann Mapping Theorem (4/10 - 4/15)
Definition 1. A class F of continuous functions defined on an open set
G is called a normal family if every sequence of elements in F contains a
subsequence that converges uniformly on compact subsets of G.
The limit function is not required to be in F. By the above theorems
we know that if F ⊆ H(G), then the limit function is in H(G). We require
one direction of Montel’s theorem, which relates normal families to families
that are uniformly bounded on compact sets.
Theorem 1. Let G be an open, connected set. Suppose F ⊂ H(G) and F
is uniformly bounded on each compact subset of G. Then F is a normal
family.
◦
Proof. Let Kn ⊆ Kn+1
be a sequence of compact sets whose union is G.
Construction for these: Kn is the intersection of B(0, n) with the complement of the (open) set ∪a∈G
/ B(a, 1/n).
We want to apply the Arzela-Ascoli theorem, hence we need to prove
that F is equicontinuous. I.e., for ε > 0 and z ∈ G we need to show that
there is δ > 0 so that if |z − w| < δ, then |f (z) − f (w)| < ε for all f ∈ F.
(We emphasize that δ is allowed to depend on z.) Since every z ∈ G is an
element of one of the Kn , it suffices to prove this statement for fixed n and
z ∈ Kn .
By assumption F is uniformly bounded on each Kn . Let M be such that
|f (z)| ≤ M
for all z ∈ Kn and all f ∈ F. Let δn be such that for all ξ ∈ Kn , the relation
◦ .)
B(z, 2δn ) ⊆ Kn+1 holds. (Existence follows since Kn ⊆ Kn+1
Let z, w ∈ Kn with |z − w| < δn . Let γ be the circle with center z and
radius 2δn . Then Cauchy’s formula gives
Z
z−w
f (ξ)
f (z) − f (w) =
dξ.
2πi γ (ξ − z)(ξ − w)
Estimate the denominator terms from below: |ξ − z| = 2δn and |ξ − w| >
δn . Hence,
|f (z) − f (w)| <
M
|z − w|.
δn
Hence, can choose δ = δn /M , which depends on z, but not on f .
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We essentially proved that f is Lipschitz continuous with a Lipschitz
constant that is independent of f .
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Riemann mapping theorem
Definition 2. We call two regions G1 and G2 conformally equivalent if there
exists f ∈ H(G1 ) which is one-one on G1 and satisfies f (G1 ) = G2 .
Theorem 2 (Riemann mapping theorem). Every simply connected region
G ( C is conformally equivalent to the open unit disk.
Before the proof, form some intuition. Consider one-one maps from D
to D. On first sight this is a boring situation, since f (z) = z is such a map,
but does not tell us anything worthwhile about the situation G → D.
It turns out that something more specific needs to be considered. Fix a
point a in G. Is it possible to find an analytic one-one map f : G → D so
that f (a) = 0 ?
Here the unit disk situation is very helpful; we know all maps that send
D to D and given a to the origin, namely the Möbius transformations
ϕa (z) =
z−a
1 − az
followed by rotations. Moreover, any other one-one function from D into D
with f (a) = 0 satisfies
|f 0 (a)| <
1
= ϕ0a (a)
1 − |a|2
This generalizes and will lead to a proof of the Riemann mapping theorem.
Definition 3. Let G 6= C be open, non-empty, and simply connected. Define
F = {f ∈ H(G) : f is one-one, f (a) = 0, f 0 (a) > 0, f (G) ⊆ D}.
The desired map will be that element f ∈ F for which f 0 (a) = sup{g 0 (a) :
g ∈ F}.
Problems: Is there even one function in F? Is the supremum above even
finite? (We will see that we don’t need to show that it is finite.) For a
sequence of functions whose derivatives converge to the supremum, is there
a limit function and is it analytic? Is the image of the limit function already
all of D? We start with the first question.
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Lemma 1. F is not empty.
Proof. We observe first that if we can find a one-one function g ∈ H(G)
whose image omits some disk D(z0 , r), then we are done: the Möbius transr
formation w 7→ w−z
maps D(z0 , r) to |z| > 1, hence the composition of g
0
and this transformation, namely
ψ(z) =
r
g(z) − z0
is one-one and has its image in D. Another composition with ϕψ(a) (to
obtain that a is mapped to the origin) and a rotation (to obtain a positive
derivative at z = a) gives an element in F.
To find such g ∈ H(G) we use the fact that G is simply connected and
G 6= C. Let u ∈ C\G. The function z 7→ z − u has no zero in G, hence an
analytic branch of the square root exists, i.e., there exists ϕ ∈ H(G) so that
ϕ2 (z) = z − u.
Note that ϕ is one-one on G, since even its square is one-one. Moreover, ϕ
is analytic, hence open, so for some r > 0
B(ϕ(a), r) ⊂ ϕ(G).
(1)
Finally, we use the fact that the square-root restricted to G is one-one.
Intuitively speaking, (1) gives a disk in the domain of an injective squareroot function, hence the negative of that disk, should be a subset of the
complement, otherwise the function would not be injective. It remains to
formalize this argument.
We claim that B(−ϕ(a), r) ∩ ϕ(G) = ∅. Assume there is z0 ∈ G with
ϕ(z0 ) ∈ B(−ϕ(a), r). Then −ϕ(z0 ) ∈ B(ϕ(a), r), and (1) implies there
exists w ∈ G with ϕ(w) = −ϕ(z0 ). It follows that
w − u = ϕ2 (w) = ϕ2 (z0 ) = z0 − u,
so z0 = w. Hence, ϕ(z0 ) = −ϕ(z0 ), i.e., ϕ(z0 ) = 0. Plug into the square:
z0 − u = ϕ2 (z0 ) = 0, i.e. z0 = u. Contradition, since u ∈
/ G.
As an example, if G = C\(−∞, 0], then can take ω = 0, and ϕ is the
principal branch. In this case the statements in the proof are trivial, since
the image of G is {z : <z > 0}, and manifestly the negative of a number in
the image has <z < 0 and is therefore itself not in the image.
We prove next the crucial step that shows that considering the derivatives
at z = a is the correct object to consider.
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Lemma 2. If f ∈ F and ω ∈ D is not in the image of f , then there exists
g ∈ F with
g 0 (a) > f 0 (a).
Proof. Recall that ϕω (ω) = 0. Hence, ϕω ◦ f is a function in F that has no
zero in G. Hence, there exists h ∈ H(G) with h2 = ϕω ◦ f . As above, h is
also one-one and it maps G into D. This is the crucial step; we have now
with I 2 (z) = z 2 the identity
f = ϕ−ω ◦ I 2 ◦ h,
and ϕ−ω ◦ I 2 is not one-one on D. Slight adjustment to replace h by an
element from F: We note that
ϕh(a) ◦ h(a) = 0
and we choose |c| = 1 so that c(ϕh(a) ◦ h)0 (a) > 0. Define Rc (z) = cz, and
write f as
f = ϕ−ω ◦ I 2 ◦ ϕ−h(a) ◦ Rc ◦ Rc ◦ ϕh(a) ◦ h.
We have now
f =F ◦g
with g = Rc ◦ ϕh(a) ◦ h ∈ F and F = ϕ−ω ◦ I 2 ◦ ϕ−h(a) ◦ Rc : D → D not
one-one, hence by Schwarz’s lemma |F 0 (0)| < 1. Thus, chain rule gives
f 0 (a) = F 0 (0)g 0 (a) < g 0 (a)
as was to be shown.
Theorem 3 (Riemann mapping theorem). Let G 6= C be a simply connected,
open, non-empty set, and let a ∈ G. There exists a unique analytic function
f with the properties
1. f (a) = 0 and f 0 (a) > 0,
2. f is one-one,
3. f (G) = D.
Proof. Consider F defined above. We have shown that F is not empty. Note
that each element of F is in absolute values bounded by 1 on G, hence F is
a normal family.
Take a sequence of elements in F with the property that fn0 (a) converges
to η = sup{f 0 (a) : f ∈ F}. Since F is normal, there exists (after relabeling)
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a subsequence {fk } that converges uniformly on compact subsets of G, hence
its limit function f is analytic and satisfies f 0 (a) = η. (We get here for free
that η is finite.) Note that η > 0, hence f is not constant.
We need to prove that f ∈ F. We have that fn (G) ⊆ D, hence f (G) ⊆
D, and the open mapping theorem implies that f (G) ⊆ D. Since fn (a) = 0
we obtain f (a) = 0. It remains to show that f is one-one.
Fix z1 ∈ G and set ζ = f (z1 ) and ζn = fn (z1 ). Let z2 ∈ G, distinct from
z1 . Let K ⊆ G be a closed disk about z2 that does not contain z1 .
Since fn is one-one, gn (z) := fn (z) − ζn 6= 0 on K for all n. From the
assumptions, gn (z) converges uniformly on K to f (z) − ζ.
To summarize these properties:
1. gn → g uniformly on K,
2. |g| ≥ δ > 0 for some δ on ∂K,
3. gn is never zero on K.
We will prove next time that under these assumptions the limit function
g is either constant or never zero on K, and since g has positive derivative
at a, it will follow that g is never zero, i.e., f (z2 ) 6= f (z1 ).
Since f ∈ F and by construction there exist no element in F with a
larger derivative at z = a, Lemma 2 implies that f (G) = D.
Uniqueness follows from the fact that the composition of such a ’maximal’ map with an inverse from another maximal map is by Schwarz’s lemma
necessarily of the form z 7→ cz with |c| = 1, and since the derivatives are
positive at z = a, c = 1.
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