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

Transcription

1 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 K n Kn+1 be a sequence of compact sets whose union is G. Construction for these: K n 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 K n, it suffices to prove this statement for fixed n and z K n. By assumption F is uniformly bounded on each K n. Let M be such that f(z) M for all z K n and all f F. Let δ n be such that for all ξ K n, the relation B(z, 2δ n ) K n+1 holds. (Existence follows since K n Kn+1.) Let z, w K n with z w < δ n. Let γ be the circle with center z and radius 2δ n. Then Cauchy s formula gives f(z) f(w) = z w 2πi γ f(ξ) (ξ z)(ξ w) dξ. Estimate the denominator terms from below: ξ z = 2δ n and ξ w > δ n. Hence, f(z) f(w) < M δ n z w. Hence, can choose δ = δ n /M, which depends on z, but not on f. 1

2 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 G 1 and G 2 conformally equivalent if there exists f H(G 1 ) which is one-one on G 1 and satisfies f(g 1 ) = G 2. 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 ϕ a (z) = z a 1 az followed by rotations. Moreover, any other one-one function from D into D with f(a) = 0 satisfies f (a) < 1 1 a 2 = ϕ a(a) This generalizes and will lead to a proof of the Riemann mapping theorem. Definition 3. Let G C be open, non-empty, and simply connected. Define F = {f H(G) : f is one-one, f(a) = 0, f (a) > 0, f(g) D}. The desired map will be that element f F for which f (a) = sup{g (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

3 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(z 0, r), then we are done: the Möbius transformation w r w z 0 maps D(z 0, r) to z > 1, hence the composition of g and this transformation, namely r ψ(z) = g(z) z 0 is one-one and has its image in D. Another composition with ϕ ψ(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 C. Let u C\G. The function z 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 ϕ 2 (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 z 0 G with ϕ(z 0 ) B( ϕ(a), r). Then ϕ(z 0 ) B(ϕ(a), r), and (1) implies there exists w G with ϕ(w) = ϕ(z 0 ). It follows that w u = ϕ 2 (w) = ϕ 2 (z 0 ) = z 0 u, so z 0 = w. Hence, ϕ(z 0 ) = ϕ(z 0 ), i.e., ϕ(z 0 ) = 0. Plug into the square: z 0 u = ϕ 2 (z 0 ) = 0, i.e. z 0 = 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 : Rz > 0}, and manifestly the negative of a number in the image has Rz < 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. 3

4 Lemma 2. If f F and ω D is not in the image of f, then there exists g F with g (a) > f (a). Proof. Recall that ϕ ω (ω) = 0. Hence, ϕ ω f is a function in F that has no zero in G. Hence, there exists h H(G) with h 2 = ϕ ω 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 ϕ h(a) h(a) = 0 and we choose c = 1 so that c(ϕ h(a) h) (a) > 0. Define R c (z) = cz, and write f as f = ϕ ω I 2 ϕ h(a) R c R c ϕ h(a) h. We have now f = F g with g = R c ϕ h(a) h F and F = ϕ ω I 2 ϕ h(a) R c : D D not one-one, hence by Schwarz s lemma F (0) < 1. Thus, chain rule gives as was to be shown. f (a) = F (0)g (a) < g (a) Theorem 3 (Riemann mapping theorem). Let G 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 (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 f n(a) converges to η = sup{f (a) : f F}. Since F is normal, there exists (after relabeling) 4

5 a subsequence {f k } that converges uniformly on compact subsets of G, hence its limit function f is analytic and satisfies f (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 f n (G) D, hence f(g) D, and the open mapping theorem implies that f(g) D. Since f n (a) = 0 we obtain f(a) = 0. It remains to show that f is one-one. Fix z 1 G and set ζ = f(z 1 ) and ζ n = f n (z 1 ). Let z 2 G, distinct from z 1. Let K G be a closed disk about z 2 that does not contain z 1. Since f n is one-one, g n (z) := f n (z) ζ n 0 on K for all n. From the assumptions, g n (z) converges uniformly on K to f(z) ζ. To summarize these properties: 1. g n g uniformly on K, 2. g δ > 0 for some δ on K, 3. g n 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(z 2 ) f(z 1 ). 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 cz with c = 1, and since the derivatives are positive at z = a, c = 1. 5