Riemann Mapping Theorem (4/10 - 4/15)

Home , Normal family, Region (mathematics)

Math 752 Spring 2015

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/∈GB(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 − w Z 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, M |f(z) − f(w)| < |z − w|. δn

Hence, can choose δ = δn/M, which depends on z, but not on f.

1 We essentially proved that f is Lipschitz continuous with a Lipschitz constant that is independent of f.

1 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 Möbiustransformations z − a ϕ (z) = a 1 − az followed by rotations. Moreover, any other one-one function from D into D with f(a) = 0 satisfies 1 |f 0(a)| < = ϕ0 (a) 1 − |a|2 a 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. De- ﬁne

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{g0(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.

2 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 Möbiustrans- formation w 7→ r maps D(z , r) to |z| > 1, hence the composition of g w−z0 0 and this transformation, namely r ψ(z) = 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 square- root 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

2 2 w − u = ϕ (w) = ϕ (z0) = z0 − u, so z0 = w. Hence, ϕ(z0) = −ϕ(z0), i.e., ϕ(z0) = 0. Plug into the square: 2 z0 − u = ϕ (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 : 0}, and manifestly the negative of a number in the image has

3 Lemma 2. If f ∈ F and ω ∈ D is not in the image of f, then there exists g ∈ F with g0(a) > f 0(a).

Proof. Recall that ϕω(ω) = 0. Hence, ϕω ◦ f is a function in F that has no 2 zero in G. Hence, there exists h ∈ H(G) with h = ϕω ◦ f. As above, h is also one-one and it maps G into D. This is the crucial step; we have now with I2(z) = z2 the identity

2 f = ϕ−ω ◦ I ◦ h,

2 and ϕ−ω ◦ I is not one-one on D. Slight adjustment to replace h by an element from F: We note that

ϕh(a) ◦ h(a) = 0

0 and we choose |c| = 1 so that c(ϕh(a) ◦ h) (a) > 0. Deﬁne Rc(z) = cz, and write f as 2 f = ϕ−ω ◦ I ◦ ϕ−h(a) ◦ Rc ◦ Rc ◦ ϕh(a) ◦ h. We have now f = F ◦ g 2 with g = Rc ◦ ϕh(a) ◦ h ∈ F and F = ϕ−ω ◦ I ◦ 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)g0(a) < g0(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. 0 Take a sequence of elements in F with the property that fn(a) converges to η = sup{f 0(a): f ∈ F}. Since F is normal, there exists (after relabeling)

4 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.