Lecture 15 { The Riemann mapping theorem MATH-GA 2451.001 Complex Variables The point of this lecture is to prove that the unit disk can be mapped conformally onto any simply connected open set in the plane, other than the plane itself. We will then know that any two such simply connected open set can be mapped conformally onto each other, since we can map either of these sets to the unit disk. The result above is known as the Riemann mapping theorem. We will prove it using basic theory of normal families. We start this lecture with that. 1 Normal families 1.1 Definition Let Ω be an open connected set in C and F a family of analytic functions on Ω. F is said to be (i) normal or relatively compact if every sequence from F has a subsequence that converges uniformly on every compact subset of Ω (ii) locally uniformly bounded if for any compact subset K of Ω, there exists M > 0 such that jf(z)j ≤ M 8f 2 F , 8z 2 K. (iii) equicontinuous on a compact set K of Ω if 8 > 0 ; 9δ > 0 such that 8(z; w) 2 K2 ; jz − wj < δ ) jf(z) − f(w)j < 8f 2 F Observe that if there exists M > 0 such that 8 z 2 Ω ; 8 f 2 F, jf 0(z)j ≤ M, then F is equicontinuous. 1.2 Equivalence of the definitions Theorem (Montel's theorem): Suppose F is a locally uniformly bounded family of analytic functions in an open connected set Ω. Then F is normal. Proof : The proof of Montel's theorem can be decomposed into the following two steps: (1) Lemma: Suppose F is a family of analytic functions uniformly bounded on compact subsets of Ω. Then F is equicontinuous on every compact subset of Ω. To show this, let us consider a compact subset K of Ω. There exists r > 0 such that the set K~ defined by K~ := fz 2 C : d(z; K) ≤ 2rg is contained in Ω. Let M = sup jf(z)j f2F;z2K~ and (z; w) 2 K2 such that jz − wj < r. The closed disk D2r(z) is contained in Ω. If we call γ the boundary of this disk, we may write 1 Z f(ζ) f(ζ) 1 Z f(ζ)(z − w) f(z) − f(w) = − dζ = dζ 2πi γ ζ − z ζ − w 2πi γ (ζ − w)(ζ − z) 8 > 0, let δ = min (r=M; r). 1 4πr jz − wj < δ ) jf(z) − f(w)j ≤ Mjz − wj ≤ 2π 2r2 Hence F is equicontinuous on K. (2) Now that we know that F in Montel's theorem is equicontinuous on every compact subset of Ω, we can prove the theorem. 1 Let (fn)n2N be a sequence from F. Consider a sequence (wj)j2N which is everywhere dense in Ω. The sequence (fn(w1))n2N is bounded, so we can extract a convergent subsequence, and write the correspond- ing functions (fn;1)n2N.(fn;1(w2))n2N is bounded, so we can extract another convergent subsequence ∗ (fn;2)n2N. Repeating the process, we obtain the nested sequence (fn;k)k2N ;n2N such that (fn;k(wk))n2N converges for every wj with j 2 〚1; k〛. Let gn = fn;n be the diagonal sequence of functions. gn converges at all the points (wn)n2N. We now show that (gn)n 2 N converges uniformly on every compact subset K of Ω. Let K be such a compact subset. Since F is equicontinuous on K, 8 > 0 9δ > 0 such that 8(z; w) 2 K2 ; jz − wj < δ ) jf(z) − f(w)j < 3 N Since K is compact, we can select N points from (wj)j2N such that K ⊂ [p=1Dδ(wp). By construction of the functions gn, for that choice of , there exists N 2 N such that m; n ≥ N ) 8p 2 〚1;N〛; jgn(wp) − gm(wp)j < 3 Now, 8w 2 K, 9p0 such that w 2 Dδ(wp0 ). Then, if m; n ≥ N, jg (w) − g (w)j ≤ jg (w) − g (w )j + jg (w ) − g (w )j + jg (w ) − g (w)j ≤ + + = n m n n p0 n p0 m p0 m p0 m 3 3 3 from which we conclude that gn converges uniformly on K, as it is uniformly Cauchy on K. 2 The Riemann mapping theorem 2.1 Statement of the theorem Theorem (Riemann mapping theorem): Given any simply connected open set Ω which is not the whole plane, and a point z0 in Ω, there exists a unique analytic function f in Ω, normalized by the conditions 0 f(z0) = 0, f (z0) > 0, such that f defines a one-to-one mapping of Ω onto the disk jwj < 1. 2.2 Proving the theorem −1 Uniqueness: Suppose there are two such functions f1 and f2. Then f1 ◦ f2 is a one-to-one mapping of the −1 unit disk to itself. You will prove in Homework 9, using the lemma of Schwarz, that this implies that f1 ◦ f2 is a M¨obiustransformation of the form z − a S(z) = λ 1 − az with jλj = 1 and a 2 C such that jaj < 1. Then the normalization conditions S(0) = 0 ;S0(0) > 0 imply that S(z) = z, so that f1 ≡ f2. Existence: • Lemma: If Ω is a simply connected open set which is not C, there exists a one-to-one conformal map h :Ω 7! h(Ω) such that h(Ω) does not intersect a disk Dδ(w0) for some w0 2 C and δ > 0. To prove the lemma, observe that by hypothesis, there exists a 2 C n Ω. Hence '(z) = z − a is a nonvanishing analytic function on Ω. From Lecture 9, we then know that we can define an analytic function h on Ω such that h2(z) := z − a. 2 Observe that h does not take the same value twice, nor opposite values. Indeed if (z1; z2) 2 Ω are such that h(z1) = ±h(z2), then z1 − a = z2 − a ) z1 = z2. Now, by the open mapping theorem, 8z0 2 Ω, h(Ω) contains a disk Dδ(h(z0)). This means that h(Ω) does not contain the disk Dδ(−h(z0)), which proves the lemma, with w0 = −h(z0). • We now show that there are one-to-one, analytic functions g from Ω to D1(0) such that g(z0) = 0 and 0 g (z0) > 0. 2 To see that there is at least one such function, consider the function G0 on Ω defined by 0 δ jh (z0)j h(z0) h(z) − h(z0) G0(z) = 2 0 4 jh(z0)j h (z0) h(z) + h(z0) with δ and h as defined in the previous lemma. G0 is one-to-one since h is and G0 is obtained from h by a linear fractional transformation. 0 0 δ jh (z0)j G0(z0) = 0 ;G0(z0) = 2 > 0 8 jh(z0)j Finally, observe that δ 1 h(z) − h(z0) δ 1 2 8z 2 Ω ; jG0(z)j = = − < 1 4 jh(z0)j h(z) + h(z0) 4 h(z0) h(z) + h(z0) This proves our point. • The last step of the proof is to show that within the family F of functions g discussed previously, there exists an f with maximal derivative, and that this f has all the properties required for the mapping theorem. The key is to show that f is surjective. Observe that by Cauchy's estimate on Dr(z0) ⊂ Ω, 8 g 2 F, 1 jg0(z )j ≤ 0 r 0 0 The set fjg (z0)j ; g 2 Fg is therefore bounded, and has a supremum B = supg2F jg (z0)j. There exists 0 a sequence (gn)n2N 2 F such that gn(z0) ! B as n ! +1. Now, since jgnj < 1 on Ω, by Montel's theorem we know that there exists a subsequence (gnk ) of gn converging to a function f uniformly on every compact subset of Ω. By Weierstrass' theorem, f is analytic. Furthermore, f(z ) = lim g (z ) = 0 ; jf 0(z )j = lim jg0 (z )j = B > 0 0 nk 0 0 nk 0 nk!1 nk!1 From the latter result, we can say that f is not a constant function. Now, we take some z1 in Ω, and defineg ~nk := gnk (z) − gnk (z1).g ~nk is nonzero on Ω n fz1g. Hence, by ~ Hurwitz' theorem, f := f(z) − f(z1) is nonzero on Ω n fz1g. We have just proved that f is one-to-one on Ω. The remaining question is: is f onto? Suppose it is not: 9w0 2 D1(0) such that w0 2= f(Ω). Then, as before, we can consider the single-valued function F on Ω such that f(z) − w F 2(z) := 0 (1) 1 − w0f(z) F is one-to-one and satisfies jF (z)j < 1 for z 2 Ω. F can be normalized as follows: 0 jF (z0)j F (z) − F (z0) G(z) := 0 F (z0) 1 − F (z0)F (z) G is one-to-one, satisfies jG(z)j < 1 8z 2 Ω, and G(z0) = 0. Furthermore, jF 0(z )j jf 0(z )j 1 1 + jw j G0(z ) = 0 = 0 (1 − jw j2) = jf 0(z )j 0 > jf 0(z )j 0 2 p 0 0 p 0 1 − jF (z0)j 2 jw0j 1 − jw0j 2 jw0j This is a contradiction, so we may say that f is onto, which concludes the proof of the theorem 3 2.3 Returning to our definition of simple connectedness You may recall that in Lecture 9, we introduced an unusual definition of simple connectedness, only valid in R2, but very convenient for our purposes at the time.