9 The Riemann Mapping Theorem References for this section: L. Ahlfors, Complex Analysis, pages 133-136 and 229-233. • J. Shurman, The Riemann Mapping Theorem, available at • http://people.reed.edu/~jerry/311/rmt.pdf R.B. Ash and W.P Novinger, Complex Analysis, (Dover). • The Riemann Mapping Theorem is one of the highlights of complex analysis, and is a really surprising result. It says that all simply connected proper open subsets of the plane are conformally equivalent. Theorem 51. (Riemann Mapping Theorem) Let Ω be a simply connected proper open subset of C. Then there is an analytic bijection f : Ω D. Furthermore given any z Ω, there → 0 ∈ is a unique such map satisfying f(z0)=0, and f ′(z0) is positive. It is worthwhile thinking a little about the hypotheses here. If Ω is a proper subset of C, then there is a point w C Ω. If Ω is also simply connected, this means that C Ω is ∈ \ \ connected, so there is a path connecting w to . ∞ Let us start by looking at the question of uniqueness. Suppose that φ : D D is an → analytic bijection and that that f : Ω D is an analytic bijection. Then f = φ f is an 1 → 2 ◦ 1 analytic bijection from Ω to D too. Question: How many analytic bijections from D to D are there? Example: Obvious ones are φ(z)= eiθz. A more complicated example: fix z D and let 0 ∈ z z φ(z)= − 0 . z z +1 − 0 This M¨obius transformation maps z 0. Furthermore, it is easy to check that φ(1) = 0 7→ | | φ( 1) = φ(i) = 1, and hence φ maps the unit circle onto itself, and hence the open disk | − | | | onto itself. It turns out that these two examples are basically all that you can do. Our first result is Schwarz’s lemma which places limits on the growth of functions on the unit disk, as long as we know that they aren’t too big at the boundary. 70 Theorem 52. Suppose that f : D D is analytic and that f(0) = 0. Then → (a) f(z) z when z < 1, and | |≤| | | | (b) f ′(0) 1. | | ≤ Furthermore: (i) If equality holds at any non-zero point in (a), or in (b), then f(z) = λz for some constant λ. (ii) If such an equality does not hold then, for each R < 1, there exists ρ < 1 such that f(z) ρ z for z R. | | ≤ | | | | ≤ f(z) Proof. has a removable singularity at 0, so define z f(z) , z =0 g(z)= z 6 f ′(0), z =0. Suppose that R< 1. Since g is analytic onz < R and continuous on z R, the maximum | | | | ≤ modulus occurs on the boundary, and only there unless g is a constant. On z = R, | | f(z) 1 g(z) = | | < | | R R and so this bound holds for all z inside this circle. But this is true for all R < 1 and hence we can deduce that g(z) 1 for all z D. This establishes (a) and (b). | | ≤ ∈ If equality holds at some point in (a) then g is a constant and there is a constant λ such that f(z) = λz. If f ′(0) = 1 then it follows that f is constant from a consideration of | | Cauchy’s formula for derivatives. (Alternatively, apply the Maximum Modulus Principle to g(z).) The final assertion follows, for if it were false we would have a sequence z with { n} z R such that f(z )/z 1 . Such a sequence must have a limit point z with z R | n| ≤ | n n|→ | | ≤ and f(z) = z . | | | | Theorem 53. Every conformal map from the unit disk onto itself is of the form z b φ(z)= eiθ − 1 ¯bz − where θ R and b < 1. ∈ | | 71 1 Proof. Let φ be such a conformal map from z 1 onto itself. Then φ− is well-defined | | ≤ 1 1 1 and is itself conformal - remember (φ− )′(φ(z)) = . Let φ− (0) = b. Then b < 1 for φ′(z) | | 1 otherwise φ− is constant by the maximum modulus princple. Consider now z b φ (z)= − . 0 1 ¯bz − This maps the unit circle to itself and, since φ (0) = b, the interior to itself. Now 0 − 1 1 1 consider g = φ φ− ,g− = φ φ− . Both satisfy the conditions of Schwarz’s lemma so ◦ 0 0 ◦ 1 g(z) z , g− (w) w for z 1, w 1. | |≤| | | |≤| | | | ≤ | | ≤ That is g(z) z g(z) for z 1. | |≤| |≤| | | | ≤ iθ iθ Thus g(z)= e z by Schwarz’s lemma. Putting this together, g(φ0(z)) = e φ0(z) or φ(z)= iθ e φ0(z), as claimed. This now allows to to show that there is at most one conformal map from a simply connected proper subset of C onto D with f(z0) = 0 and f ′(z0) positive. Suppose that 1 f , f : Ω D are two such maps. Then φ = f f − is a conformal bijection from D to 1 2 → 2 ◦ 1 itself and hence z b φ(z)= eiθ − . 1 ¯bz − iθ iθ But φ(0) = 0 and so b = 0. That is φ(z)= e z, or f2(z)= e f1(z). Thus iθ f2′ (z0)= e f1′ (z0). iθ But since both derivatives are real and positive at z0, this says that e must be 1, and so f2 = f1. The harder part of the proof of the Riemann Mapping Theorem is to show that any suitable bijection exists at all! Even Riemann fudged when he first ‘proved’ this (in 1851). It took until 1912 before Carath´eodory produced what is now considered a valid proof. Fix then the simply connected proper open subset Ω and the point z Ω. Let 0 ∈ = f : Ω D : f is analytic, 1–1 and f(z )=0 . F { → 0 } Note that we don’t require that elements of are onto. F 72 Lemma 54. Suppose that f : Ω C is an analytic 1–1 function. Then f(Ω) is homeomor- → phic to Ω and hence f(Ω) is simply connected. Proof. The Inverse Function Theorem implies that f is a homeomorphism and hence f preserves simple connectedness. If Ω is bounded, then it is clear that is nonempty: f(z)=(z z )/M will work for F − 0 large enough M. Lemma 55. is nonempty. F Proof. Since Ω is a proper subset of C we can choose a / Ω. Since Ω is simply connected, its ∈ complement is connected as a subset of the Riemann sphere. Thus, there is some continuous path in Ωc connecting a to . This means (think back to Assignment 1!) that we can choose ∞ a branch of the square root function f : Ω C, r(z)= √z a → − which is analytic on Ω. Since each point w C has a unique square, the function r can only have one of w or ∈ w in its range. Pick some w r(Ω). By the Open Mapping Theorem, r(Ω) is open, so − 0 ∈ for some ǫ> 0, B(w , ǫ) r(Ω). But this means that r(Ω) contains none of the elements of 0 ⊂ B( w , ǫ). − 0 Now let’s start composing maps. The range of the map z r(z)+ w now contains none 7→ 0 of the points of B(0, ǫ), and so 2 g(z)= (r(z)+ w ) ǫ 0 completely misses the unit disk. Thus h : Ω D → 1 h(z)= g(z) is well-defined, analytic and 1–1. We can now further compose with an automorphism of the disk to produce a suitable function that sends z0 to 0, and so we have produced an element of . F The next lemma is the clever part. For notational convenience, for w D, lets ∈ z w T (z)= − w 1 wz − 73 denote the natural automorphism of D which maps w to 0. This has inverse 1 z + w T − (z)= . w 1+ wz Lemma 56. Suppose that f satisfies ∈F f ′(z ) g′(z ) , for all g . | 0 |≥| 0 | ∈F Then f is a bijection from Ω to D. Proof. [Remember that we are assuming that elements of are 1–1.] F Suppose that f is not onto. That is, there exists w D which is not in f(Ω). Our ∈F ∈ aim now is to construct a function g with g′(z ) > f ′(z ) . ∈F | 0 | | 0 | The map Ω D, z T (f(z)) has a simply connected image which does not contain 0. → 7→ w As in the proof of Lemma 55, we can define an analytic branch of the square root function r : T (f(Ω)) D. Let w′ = r( w), and let w → − g = T ′ r T f w ◦ ◦ w ◦ which is an analytic 1–1 map from Ω to D. That is, g . ∈F Let s : D D be → 1 1 2 s(z)= Tw− Tw−′ (z) . Then 1 2 1 s(0) = T − (w′) = T − ( w)=0. w w − By Schwarz’s Lemma, s′(0) 1, and if s′(0) = 1 then s(z)= λz for some λ. But s is not | | ≤ | | 1–1, so s is not of this form. Therefore we must have that s′(0) < 1. | | Note now that f = s g. Using the chain rule ◦ f ′(z ) = s′(g(z )) g′(z ) = s′(0) g′(z ) < g′(z ) | 0 | | 0 || 0 | | || 0 | | 0 | as required. To complete the proof of the Riemann Mapping Theorem, we need to show that there is a function f which has a maximal derivative at z .