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

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Riemann Mapping Theorem (4/10 - 4/15) 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 comple- ment of the (open) set [a=2GB(a; 1=n). We want to apply the Arzela-Ascoli theorem, hence we need to prove that F is equicontinuous. I.e., for " > 0 and z 2 G we need to show that there is δ > 0 so that if jz − wj < δ, then jf(z) − f(w)j < " for all f 2 F. (We emphasize that δ is allowed to depend on z.) Since every z 2 G is an element of one of the Kn, it suffices to prove this statement for fixed n and z 2 Kn. By assumption F is uniformly bounded on each Kn. Let M be such that jf(z)j ≤ M for all z 2 Kn and all f 2 F. Let δn be such that for all ξ 2 Kn, the relation ◦ B(z; 2δn) ⊆ Kn+1 holds. (Existence follows since Kn ⊆ Kn+1.) Let z; w 2 Kn with jz − wj < 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: jξ −zj = 2δn and jξ −wj > δn. Hence, M jf(z) − f(w)j < jz − wj: 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 2 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 M¨obiustransformations z − a ' (z) = a 1 − az followed by rotations. Moreover, any other one-one function from D into D with f(a) = 0 satisfies 1 jf 0(a)j < = '0 (a) 1 − jaj2 a This generalizes and will lead to a proof of the Riemann mapping theo- rem. Definition 3. Let G 6= C be open, non-empty, and simply connected. De- fine F = ff 2 H(G): f is one-one, f(a) = 0; f 0(a) > 0; f(G) ⊆ Dg: The desired map will be that element f 2 F for which f 0(a) = supfg0(a): g 2 Fg. 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 2 H(G) whose image omits some disk D(z0; r), then we are done: the M¨obiustrans- formation w 7! r maps D(z ; r) to jzj > 1, hence the composition of g w−z0 0 and this transformation, namely r (z) = g(z) − z0 is one-one and has its image in D. Another composition with ' (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 2 H(G) we use the fact that G is simply connected and G 6= C. Let u 2 CnG. The function z 7! z − u has no zero in G, hence an analytic branch of the square root exists, i.e., there exists ' 2 H(G) so that '2(z) = z − u: Note that ' is one-one on G, since even its square is one-one. Moreover, ' is analytic, hence open, so for some r > 0 B('(a); r) ⊂ '(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(−'(a); r) \ '(G) = ;. Assume there is z0 2 G with '(z0) 2 B(−'(a); r). Then −'(z0) 2 B('(a); r), and (1) implies there exists w 2 G with '(w) = −'(z0). It follows that 2 2 w − u = ' (w) = ' (z0) = z0 − u; so z0 = w. Hence, '(z0) = −'(z0), i.e., '(z0) = 0. Plug into the square: 2 z0 − u = ' (z0) = 0, i.e. z0 = u. Contradition, since u2 = G. As an example, if G = Cn(−∞; 0], then can take ! = 0, and ' is the principal branch. In this case the statements in the proof are trivial, since the image of G is fz : <z > 0g, 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. 3 Lemma 2. If f 2 F and ! 2 D is not in the image of f, then there exists g 2 F with g0(a) > f 0(a): Proof. Recall that '!(!) = 0. Hence, '! ◦ f is a function in F that has no 2 zero in G. Hence, there exists h 2 H(G) with h = '! ◦ 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 = '−! ◦ I ◦ h; 2 and '−! ◦ I is not one-one on D. Slight adjustment to replace h by an element from F: We note that 'h(a) ◦ h(a) = 0 0 and we choose jcj = 1 so that c('h(a) ◦ h) (a) > 0. Define Rc(z) = cz, and write f as 2 f = '−! ◦ I ◦ '−h(a) ◦ Rc ◦ Rc ◦ 'h(a) ◦ h: We have now f = F ◦ g 2 with g = Rc ◦ 'h(a) ◦ h 2 F and F = '−! ◦ I ◦ '−h(a) ◦ Rc : D ! D not one-one, hence by Schwarz's lemma jF 0(0)j < 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 2 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 = supff 0(a): f 2 Fg. Since F is normal, there exists (after relabeling) 4 a subsequence ffkg 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 2 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 2 G and set ζ = f(z1) and ζn = fn(z1). Let z2 2 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.
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