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.. Let Ω C be a simply connected set that is not all of C. Then for any z 0 Ω, there exists a unique biholomorphism F : Ω D such that F (z 0 ) = 0, and F (z 0 ) > 0. Here F (z 0 ) > 0 stands for F (z 0 ) 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.. Any two proper, simply connected subsets for C are conformally equivalent. Uniqueness The uniqueness follows easily from the following lemma. Lemma 0. (Schwarz s lemma). Let f : D D be a holomorphic map such that f(0) = 0. Then f(z) < z for all z D. Moreover, if f(z) = z for some non-zero z, or f (0) =, then f(z) = az for some a C with a =. Proof. Consider the holomorphic function on D \ {0} defined by g(z) := f(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). Date: 24 August 206.
One way to see this to consider the power series of f(z). Since f(0) = 0, this has the form f(z) = a z + a 2 z 2 + = z(a + a 2 z + ). 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), and hence the extension satisfies g(0) = f (0). Now, for any z D, and any z < r <, by the maximum modulus principle, g(z) < sup g(w) = sup w =r f(w) w =r r r. But this is true for any r ( z, ), and so letting r we see that g(z), or equivalently f(z) z. Now suppose that f(z) = z for some non-zero z, or f (0) =, then g(z) = at some interior point of D. Then by the maximum modulus principle, g(z) = a for some a with a =. This completes the proof. Proof of uniqueness in Theorem 0.. Let F : Ω D and F 2 : Ω D be two such mappings. Then f = F 2 F satisfies the following properties f : D D is injective and onto. f(0) = 0. f (0) > 0. f also satisfies both these properties. By Schwarz lemma, f(z) z for all z D and f (w) w for all w D. Let w = f(z), then second inequality gives z f(z), and hence z = f(z). But then by the equality part of Schwarz lemma, we see that f(z) = az for some a C with a =. But then f (0) = a, which forces a = (since f (0) > 0). Hence f(z) = z for all z D or equivalently F 2 (w) = F (w) for all w 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 M K such that sup f(z) < M K z K for all f F. 2
Theorem 0.2 (Montel s theorem). Suppose F is the family of functions defined by F := {f : Ω C hol. K Ω compact M K > 0 s.t. sup f(z) < M K, f F}. 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 Ω and all ε > 0 there exists a δ = δ(a, ε) such that z, w D δ (a) = f(z) f(w) < ε, for all f 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 equicontinuous and locally uniformly bounded, then for every sequence of functions {f n } F, there exists a continuous function f and a subsequence {f nk } 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 automatically 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 automatically 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 Ω and ε > 0. We need to choose a delta that works. Let r > 0 be such that D 2r (a) Ω, and let M r such that f(ζ) M r, for all ζ D 2r (a) and all f F. By the Cauchy integral formula, for any z, w D r (a), f(z) f(w) = ( f(ζ) ζ z ) dζ ζ w = z w ζ a =2r ζ a =2r f(ζ) (ζ z)(ζ w) dζ. Now, since z, w D r (a) and ζ a = 2r, by the triangle inequality, ζ z, ζ w > r. Then by the triangle inequality for line integrals f(z) f(w) z w 2π f(ζ) sup ζ a =r r 2 3 4πr 2M r z w r
So letting ( rε ) δ = min r,, 4M r we see that if z, w D δ (a), then z w < 2δ < rε/2m r, and so f(z) f(w) < ε for all f 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 f n : 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 C, there exists a, b Ω such that F (a) = F (b) = w. Let f n (a) = w n, then w n w. Choose an r > 0 small enough so that there does not exist any z D r (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 a / D r (b). Since f n is injective for any n, there exists no solution to f n (z) = w n in the closure of the disc D r (b), and so by the argument principle applied to f n (z) w n, we see that ζ b =r f n(ζ) f n (ζ) w n dζ = 0. But since f n F uniformly on compact sets, in particular, on the compact set D r (a) we have f n(ζ) F (ζ) and f n (ζ) w n F (ζ) w uniformly. Hence the integral also converges uniformly, and from this we conclude that ζ b =r F (ζ) dζ = 0. F (ζ) w This integral calculates the number of zeroes of F (ζ) w = 0 in D r (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 = {f : Ω D f holomorphic and injective, f(z 0 ) = 0}. The required biholomorphic map will be obtained by maximizing the modulus of the derivative at z 0, 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(z 0 ) = 0. Or equivalently, that F φ. 4
Proof. Since Ω C, there is an a C \ 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 e l(z) = z a for all z Ω. Clearly l(z) is injective. Moreover, if z, z 2 Ω and z z 2 then l(z 2 ) l(z ) / Z, i.e their difference cannot be an integral multiple of. In particular, l(z) l(z 0 ) +. We in fact claim that f(z) (f(z 0 ) + ) is bounded strictly away from zero. That is, Claim.There exists an ε > 0 such that l(z) (l(z 0 )+) > ε for all z Ω. To see this, assume the claim is false. Then there is a sequence {z n } Ω such that l(z n ) l(z 0 ) +. But then exponentiating, since the exponential function is continuous, we see that z n z 0. But then, since l(z) is continuous, this implies that l(z n ) l(z 0 ) contradicting the assumption that l(z n ) l(z) +. This proves the claim. Now consider the function f(z) = l(z) l(z 0 ). by the above observations, this is a bounded, injective and holomorphic function on Ω, and hence f : Ω D R (0), where R can be taken to be R = /ε where ε is from the claim above. Suppose f(z 0 ) = a, then is the required function. f(z) = f(z) a R + a Next, let λ = sup f (z 0 ). f F We claim that λ > 0. To see this, consider the f F constructed above. If f (z 0 ) = 0, then since f(z 0 ) = 0 and z 0 is a zero of multiplicity at least two, by the argument principle, there exist small neighborhoods U and V of z 0 and 0 respectively such that for all w V, w 0, there are at least two distinct solutions to f(z) = w, which contradicts the injectivity of f. Hence f (z 0 ) > 0 and hence λ > 0. Lemma 0.3. There is a function F F such that F (z 0 ) = λ. Proof. Let f n F be a sequence of functions that maximize f (z 0 ) ; that is lim f n n(z 0 ) = λ. Since f n (z) <, by Montel s theorem there is a subsequence that converges uniformly on compact sets to a holomorphic function F satisfying F (z), and F (z 0 ) = 0. Moreover since the derivatives also converge, we must have f (z 0 ) = λ. To show that F maps into the unit disc, i.e. F (z) < for all 5
z Ω, we use the open mapping theorem. Firstly, since f (z 0 ) = λ > 0, F cannot be a constant. Now suppose there is a point a Ω such that F (a) =. Then by the open mapping, for small r > 0, F (D r (a)) is an open set containing the point F (a) in the interior. But this open set will contain many points with modulus strictly greater than one, contradicting F (z). Finally to show that F F, we need to show that F is injective. But this follows from Hurwitz s theorem since F is non-constant and all f n F are injective. Proof of Riemann mapping. Since F (z 0 ) 0, by composing with a suitable rotation, we can assume that F (z 0 ) is real and positive. We claim that this F is the required bi-holomorphism. We already know that F : Ω D, F (z 0 ) = 0 and F is injective. To complete the proof, we need to show that F is surjective. If not, then there exists a α D such that F (z) = α has no solution in Ω. We then exhibit a G F with G (z 0 ) > F (z 0 ) contradicting the choice of F. To do this, consider ψ α : D D defined by ψ α (w) = α z ᾱz, and let g(z) = ψ α F (z). Since ψ α (z) = 0 if and only if z = α, we see that ψ α F is always zero free, and so a holomorphic branch of log ψ α F can be defined since Ω is simply connected. We can then choose a holomorphic branch for g(z) be letting g(z) = e 2 log ψα F (z). Note that g(z 0 ) = α. To construct a member of the family, we need to bring this back to the origin, and hence we define G(z) = ψ g(z0 ) g(z). Then G(z 0 ) = 0. Moreover, G(z) is also injective since ψ g(z0 ) and g(z) are injective, and so G F. Claim. G (z 0 ) > F (z 0 ). To see this, observe that where s(w) = w 2 D D. ψ α F (z) = ψ α s ψ g(z 0 ) G(z) = Φ G(z), is the squaring function and Φ = ψα s ψ g(z 0 ) : We compute that Φ(0) = ψα (s(ψ g(z 0 )(0)) = ψ α (s(g(z 0 ))) = (ψ α (F (z 0 ))) = F (z 0 ) = 0. By Schwarz lemma, Φ(z) z, and so Φ (0). We claim that Φ (0) <. Suppose, Φ (0) =, then by the second part of Schwarz lemma, Φ(z) = az for some unit complex number a. In particular, Φ(z) is injective. But Φ cannot be injective since s(z) is a 2 function 6
ie. sends two points to a single point, and ψ α and ψ g(z0 ) are injective. This shows that Φ (0) <. But then F (z 0 ) = Φ (G(z 0 )) G (z 0 ) = Φ (0) G (z 0 ), and hence F (z 0 ) < G (z 0 ) which proves the claim, and completes the proof of the theorem. Department of Mathematics, UC Berkeley E-mail address: vvdatar@berkeley.edu 7