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Riemann Mapping Theorem RIEMANN MAPPING THEOREM VED V. DATAR∗ Recall that two domains are called conformally equivalent if there exists a holomorphic bijection from one to the other. This automatically implies that there is an inverse holomorphic function. The aim of this lecture is to prove the following deep theorem due to Riemann. Denote by D the unit disc centered at the origin. Theorem 0.1. Let Ω ⊂ C be a simply connected set that is not all of C. Then for any z0 2 Ω, there exists a unique biholomorphism F :Ω ! D such that 0 F (z0) = 0; and F (z0) > 0: 0 0 Here F (z0) > 0 stands for F (z0) being real and positive. Note that by Liouville's theorem, such a statement is patently false if Ω = C, and so the hypothesis that Ω is a proper subset is a necessary condition. As a consequence of the Theorem, we have the following corollary. Corollary 0.1. Any two proper, simply connected subsets for C are confor- mally equivalent. Uniqueness The uniqueness follows easily from the following lemma. Lemma 0.1 (Schwarz's lemma). Let f : D ! D be a holomorphic map such that f(0) = 0. Then jf(z)j < jzj 0 for all z 2 D. Moreover, if jf(z)j = jzj for some non-zero z, or jf (0)j = 1, then f(z) = az for some a 2 C with jaj = 1. Proof. Consider the holomorphic function on D n f0g defined by f(z) g(z) := : z Then since f(0) = 0, g(z) has a removable singularity at z = 0, and hence extends to a holomorphic function on all of D. Note that the extension, which we continue to call g(z), satisfies g(0) = f 0(0): Date: 24 August 2016. 1 One way to see this to consider the power series of f(z). Since f(0) = 0, this has the form 2 f(z) = a1z + a2z + ··· = z(a1 + a2z + ··· ): Dividing by z we see that the Laurent series for g(z) is actually a power series, and hence g has a removable singularity at z = 0. Moreover the first term in the power series is f 0(0), and hence the extension satisfies 0 g(0) = f (0). Now, for any z 2 D, and any jzj < r < 1, by the maximum modulus principle, sup jf(w)j 1 jg(z)j < sup jg(w)j = jwj=r ≤ : jwj=r r r But this is true for any r 2 (jzj; 1), and so letting r ! 1− we see that jg(z)j ≤ 1; or equivalently jf(z)j ≤ jzj: Now suppose that jf(z)j = jzj for some non-zero 0 z, or jf (0)j = 1, then jg(z)j = 1 at some interior point of D. Then by the maximum modulus principle, g(z) = a for some a with jaj = 1. This completes the proof. Proof of uniqueness in Theorem 0.1. Let F1 :Ω ! D and F2 :Ω ! D −1 be two such mappings. Then f = F2 ◦ F1 satisfies the following properties • f : D ! D is injective and onto. • f(0) = 0. • f 0(0) > 0. • f −1 also satisfies both these properties. −1 By Schwarz lemma, jf(z)j ≤ jzj for all z 2 D and jf (w)j ≤ jwj for all w 2 D. Let w = f(z), then second inequality gives jzj ≤ jf(z)j, and hence jzj = jf(z)j. But then by the equality part of Schwarz lemma, we see that 0 f(z) = az for some a 2 C with jaj = 1. But then f (0) = a, which forces 0 a = 1 (since f (0) > 0). Hence f(z) = z for all z 2 D or equivalently F2(w) = F1(w) for all w 2 Ω. Montel's and Hurwitz's theorems The proof relies on two theorems on sequences of holomorphic functions. A family of continuous functions F on an open set Ω is called normal if every sequence of functions in F has a subsequence that converges uniformly on compact subsets of Ω to a function in F. The family is said to be locally uniformly bounded if for any K ⊂ Ω compact, there exists a constant MK such that sup jf(z)j < MK z2K for all f 2 F. 2 Theorem 0.2 (Montel's theorem). Suppose F is the family of functions defined by F := ff :Ω ! C hol. j 8K ⊂ Ω compact 9MK > 0 s.t. sup jf(z)j < MK ; 8f 2 Fg: K Then F is normal. To prove this, we first recall the Arzela-Ascoli theorem. Recall that family F of continuous functions on Ω is said to locally equicontinuous of for all a 2 Ω and all " > 0 there exists a δ = δ(a; ") such that z; w 2 Dδ(a) =) jf(z) − f(w)j < "; for all f 2 F: Then we have the following basic theorem, which we state without proof. Theorem 0.3 (Arzela-Ascoli). If a family of functions is locally equicon- tinuous and locally uniformly bounded, then for every sequence of functions ffng 2 F, there exists a continuous function f and a subsequence ffnk g which converges to f uniformly on compact subsets. Proof of Montel's theorem. By the Arzela-Ascoli theorem, if we show that a locally uniformly bounded family of holomorphic functions is auto- matically locally equicontinuous, then for every sequence, there will exist a subsequence which converges uniformly on compact subsets to a continuous function f. The limit function will then be holomorphic, and will automati- cally satisfy the required estimate on compact subsets, and hence will belong to the family F. Hence it is enough to show that the family F is locally equicontinuous. To do this, we use the Cauchy integral formula. Fix an a 2 Ω and " > 0. We need to choose a delta that works. Let r > 0 be such that D2r(a) ⊂ Ω, and let Mr such that jf(ζ)j ≤ Mr; for all ζ 2 D2r(a) and all f 2 F. By the Cauchy integral formula, for any z; w 2 Dr(a), 1 Z 1 1 f(z) − f(w) = f(ζ) − dζ 2πi jζ−aj=2r ζ − z ζ − w z − w Z f(ζ) = dζ: 2πi jζ−aj=2r (ζ − z)(ζ − w) Now, since z; w 2 Dr(a) and jζ − aj = 2r, by the triangle inequality, jζ − zj; jζ − wj > r: Then by the triangle inequality for line integrals jz − wj jf(ζ)j 2Mr jf(z) − f(w)j ≤ · sup 2 · 4πr ≤ jz − wj 2π jζ−aj=r r r 3 So letting r" δ = min r; ; 4Mr we see that if z; w 2 Dδ(a), then jz − wj < 2δ < r"=2Mr, and so jf(z) − f(w)j < " for all f 2 F, and F is a locally equicontinuous family. We also need the following theorem due to Hurwitz on the limit of injective holomorphic functions. Theorem 0.4 (Hurwitz). Let fn :Ω ! C be a sequence of holomorphic, injective functions on an open connected subset, which converge uniformly on compact subsets to F :Ω ! C. Then either F is injective, or is a constant. Proof. We argue by contradiction. So suppose F is non constant and not injective. Then for some w 2 C, there exists a; b 2 Ω such that F (a) = F (b) = w. Let fn(a) = wn, then wn ! w. Choose an r > 0 small enough so that there does not exist any z 2 Dr(b) such that F (z) = w. This is possible by the principle of analytic continuation since we are assuming that F is non-constant. In particular a2 = Dr(b). Since fn is injective for any n, there exists no solution to fn(z) = wn in the closure of the disc Dr(b), and so by the argument principle applied to fn(z) − wn, we see that 1 Z f 0 (ζ) n dζ = 0: 2πi jζ−bj=r fn(ζ) − wn But since fn ! F uniformly on compact sets, in particular, on the compact 0 0 set Dr(a) we have fn(ζ) ! F (ζ) and fn(ζ) − wn ! F (ζ) − w uniformly. Hence the integral also converges uniformly, and from this we conclude that 1 Z F 0(ζ) dζ = 0: 2πi jζ−bj=r F (ζ) − w This integral calculates the number of zeroes of F (ζ)−w = 0 in Dr(b) which we know is at least one (counting multiplicity) since F (b) = w. This is a contradiction, and hence if F is non-constant, it has to be injective. Proof of Riemann mapping theorem We define a family F of holomorphic function functions by F = ff :Ω ! D j f holomorphic and injective; f(z0) = 0g: The required biholomorphic map will be obtained by maximizing the mod- ulus of the derivative at z0, amongst all functions in this family. We first show that this family is non-empty. Lemma 0.2. There is an injective function f :Ω ! D such f(z0) = 0. Or equivalently, that F 6= φ. 4 Proof. Since Ω 6= C, there is an a 2 C n Ω. Then z − a is never zero on Ω. Since Ω is simply connected, we can choose a holomorphic branch of log (z − a), or in other words, there is a holomorphic function l :Ω ! C such that el(z) = z − a for all z 2 Ω. Clearly l(z) is injective. Moreover, if z1; z2 2 Ω and z1 6= z2 then l(z2) − l(z1) 2= 2πiZ, i.e their difference cannot be an integral multiple of 2πi.
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