Chapter 7: Riemann Mapping Theorem
Course 414, 2003–04 March 30, 2004
C 7.1 Theorem (Hurwitz’ Theorem). Let G be a connected open set in and (fn)n a sequence in H(G) which converges to f H(G) (uniformly on compact subsets of G). Suppose f 0, ¯ ∈ 6≡ D(a, R) G and f(z) is never zero on z a = R. Then there exist n0 such that for n n0, ⊂ | − | ≥ fn and f have the same number of zeros (counting multiplicities) in D(a, R). Proof. Since f(z) is never zero on the circle, we have inf f(z) = δ > 0. z a =R | | | − |
For n large enough (say for n n0) ≥ sup fn(z) f(z) < δ/2 z a =R | − | | − | and thus on the circle z a = R we have | − | f(z) fn(z) < δ/2 < δ f(z) . | − | ≤ | | By Rouches´ theorem, fn and f have the same total number of zeros inside the circle z a = R | − | (counting multiplicity) for n n0. ≥ 7.2 Corollary. Let G be a connected open set in C and (fn) a sequence in H(G) such that each fn is never zero in G. Suppose fn f in H(G). If f(z) is ever zero in G, then f 0. → ≡ Proof. This follows immediately from 7.1. If f is not identically zero but f(a) = 0, then by the identity theorem we can choose R > 0 sufficiently small that z = a is the only zero of f in the closed disk D¯(z, R) and also D¯(z, R) G. Counting multiplicity, f will have a positive number of zeros in D(a, R) and by Hurwitz so⊂ must fn for n large. 7.3 Definition. If G C is open and f : G C is an injective analytic function, then f is called a conformal mapping⊂ from G to f(G). →
1 Recall that f(G) is necessarily open and f − : f(G) G is automatically analytic by the → open mapping theorem. Also, f 0(z) is never zero in G and this leads to the angle-preserving property of conformal mapping that gives them their name:
1 2 414 2003–04 R. Timoney
1 If γ1 and γ2 are two C curves in G which meet at an angle θ to one another at a G (say γ1(0) = γ2(0) = a and arg(γ0 (0)) arg(γ0 (0)) = θ), then f γ1 and ∈ 1 − 2 ◦ f γ2 also meet at an angle θ at f(a). ◦ To avoid worrying about ambiguity of the argument, we could restate
arg(γ0 (0)) arg(γ0 (0)) = θ 1 − 2 as iθ γ10 (0) e− > 0 γ20 (0) (with θ restricted to a range such as [0, 2π) — which means θ is the angle you would need to turn
γ20 (0) anticlockwise to align it with γ10 (0)). The reasoning for the angle preserving property is that
iθ (f γ1)0(0) iθ f 0(a)(γ1)0(0) iθ γ10 (0) e− ◦ = e− = e− > 0. (f γ2)0(0) f (a)(γ2)0(0) γ0 (0) ◦ 0 2 The term conformal really means angle-preserving at each point, but it is usual in complex analysis to use it for injective analytic functions.
7.4 Corollary. If G C is a connected open set and (fn) is a sequence of injective functions ⊂ n in H(G) such that fn f H(G), then the limit f is either constant or injective. → ∈ Proof. Suppose f is not constant and not injective. Then there exist a, b G, a = b, with ∈ 6 f(a) = f(b) = w. Taking fn(z) w and f(z) w instead of the original fn and f, we can assume that w = 0. − − We can find a positive δ < a b /2 so that D¯(a, δ) G, D¯(b, δ) G and f(z) never 0 on the circles z a = δ, z b |=−δ. Applying| Hurwitz’⊂ theorem (7.1),⊂ we find that for all large | − | | − | n there exists an D(a, δ) with fn(an) = 0 = w. Similarly, for all large n there is bn D(b, δ) ∈ ∈ with fn(bn) = 0 = w. By the choice of δ, D(a, δ) D(b, δ) = and therefore an = bn. Thus ∩ ∅ 6 fn(an) = fn(bn) = 0 and this contradicts injectivity of fn.
7.5 Examples. (i) The map
f : z C: π/2 < (z) < π/2 z C: (z) > 0 { ∈ − = } → { ∈ < } f(z) = ez
is a conformal mapping (of its domain onto its target set, the right half plane).
(ii) There is no conformal map φ of the unit disc D(0, 1) onto the whole complex plane C 1 (because the inverse function φ− : C D(0, 1) would be a bounded entire function which was not constant, contradicting Liouvilles→ theorem). Chapter 7 — Riemann Mapping Theorem 3
(iii) If ad bc = 0, then the function − 6 az + b φ(z) = cz + d is called a Mobius¨ transformation or a linear fractional transformation. Since these maps are rational functions, we can (and often do) regard them as analytic functions
φ: Cˆ Cˆ. → The condition ad bc = 0 is there to rule out constant functions and the ‘missing’ values are φ( ) = a/c,−φ( d/c6 ) = (unless c = 0, in which case a = 0 and d = 0, and we have φ∞(z) = (a/d)z +− (b/d), φ∞( ) = .) 6 6 ∞ ∞ 7.6 Proposition. Let φ, ψ be Mobius¨ transformations. Then
(i) φ ψ is a Mobius¨ transformation. ◦ ˆ ˆ 1 ˆ ˆ (ii) φ: C C is a bijection and its inverse φ− : C C is also a Mobius¨ transformation. → → (iii) The Mobius¨ transformations form a group (under composition).
Proof. Exercise.
7.7 Proposition. Every Mobius¨ transformation can be expressed as a composition of Mobius¨ transformations of the following kinds:
(i) z z + a (translation) 7→ (ii) z λz (λ > 0) (dilation) 7→ (iii) z eiαz (α R) (rotation) 7→ ∈ (iv) z 1 . 7→ z
Proof. If φ(z) = az+b and c = 0, we can write cz+d 6 a (cz + d) + b ad φ(z) = c − c cz + d a b ad 1 = + c c c2 d − z + c = φ5 φ4 φ3 φ2 φ1(z), ◦ ◦ ◦ ◦ 4 414 2003–04 R. Timoney where d φ (z) = z + 1 c 1 φ (z) = 2 z b ad φ3(z) = z c − c2
iα b ad φ4(z) = e z α = arg c − c2 a φ (z) = + z 5 c In the case c = 0, a iβ b φ(z) = (a/d)z + (b/d) = e z + , d d with β = arg(a/d), and thus φ is a composition of a rotation, a dilation and a translation.
7.8 Proposition. Let P : C S2 denote the stereographic projection map from C to the Rie- mann sphere. If C is a circle→ in C, then its image P (C) is a circle on S2. If L is a straight line in C, then P (L) (0, 0, 1) is also a circle on S2. ∪ { } 2 1 Conversely, if C1 is a circle on S , then P − (C1) is either a circle or a line in C. In other words, circles on the Riemann sphere correspond to circles and lines in the plane under stereographic projection (but the point at or the north pole has to be added to lines to close the circle on the sphere). ∞
Proof. A circle (or line) in C has an equation
A(x2 + y2) + 2βx + 2γy + C = 0 with A, β, γ, C R. In complex terms, this can be rewritten in the form ∈ Azz¯ + Bz¯ + Bz¯ + C = 0.
(B = β + iγ.) If A = 0 we can multiply across by A and rearrange this as 6 Az + B 2 = (Az + B)(Az¯ + B¯) = A2 z 2 + ABz¯ + ABz¯ + B 2 = B 2 AC | | | | | | | | − and the condition B 2 > AC corresponds to a positive radius (not an empty ‘circle’ or a single point). If A = 0, then| | B 2 > AC = 0 is the condition for the equation to be that of a genuine line. | | Now ξ + iη P 1(ξ, η, ζ) = , − 1 ζ − Chapter 7 — Riemann Mapping Theorem 5 and we can use this in the equation of the circle/line to say that the points (ξ, η, ζ) on the stereo- graphic projection of the circle/line onto the sphere must be those that satisfy (ξ + iη)(ξ iη) ξ + iη ξ iη A − + B¯ + B − + C = 0 (1 ζ)2 1 ζ 1 ζ − − − (and for lines the north pole P ( ) as well). Observe that ∞ (ξ + iη)(ξ iη) ξ2 + η2 1 ζ2 1 + ζ − = = − = (1 ζ)2 (1 ζ)2 (1 ζ)2 1 ζ − − − − because ξ2 + η2 + ζ2 = 1. Using this, we can simplify the equation to get 2βξ + 2γη + (A C)ζ + (A + C) = 0. − This is the equation of a plane in R3, which intersects the sphere in a circle. Hence P maps circles and lines to circles. Conversely, starting with the equation of any plane in R3
a1ξ + a2η + a3ζ + c = 0.
We can choose β = a1/2, γ = a2/2, A = (a3 + c)/2 and C = (c a3)/2. The point on the plane closest to the origin is − c (ξ, η, ζ) = 2 −2 2 (a1, a2, a3) a1 + a2 + a3 and this is inside the sphere exactly when
a2 + a2 + a2 > c . q 1 2 3 | | So the condition for the plane to intersect the sphere in more than one point is c2 = (A + C)2 < a2 + a2 + a2 = 4β2 + 4γ2 + (A C)2 1 2 3 − or 4AC < 4(β2 + γ2). This is exactly what is needed to get a genuine circle or line. 7.9 Theorem. Let denote the class of all circles in Cˆ (by which we mean all circles in C together with the setsC L with L a line in C). Then Mobius¨ transformations map to . ∪ {∞} C C Proof. The equation of a circle or line in C can be written in the form Azz¯ + Bz¯ + Bz¯ + C = 0, with A, C R, B C, and B 2 > AC. It is quite∈ easy∈ to verify| that| an equation of this kind transforms to another equation of the same type under the basic kinds of Mobius¨ transformations (translations, rotations, dilations and inversion in the unit circle). Since every Mobius¨ transformation is a composition of these, the result follows. 6 414 2003–04 R. Timoney
ˆ 7.10 Proposition. Given two lists of three distinct point in C, say z1, z2, z3 and w1, w2, w3, there exists a unique Mobius¨ transformation φ satisfying φ(zj) = wj (j = 1, 2, 3). Proof. It is sufficient (for the existence part) to show that we can always find φ with
φ(z1) = 0, φ(z2) = 1, φ(z3) = , ∞ because if we can do this, we can find ψ which does the same for w1, w2, w3 and then we will 1 have (ψ− φ)(zj) = wj. ◦ If z1, z2, z3 C, the required φ is provided by ∈ z2 z3 z z1 φ(z) = − − . z2 z1 z z3 − − If one of the zj is , we have the following formulae which work ∞ 1 (z2 z3) if z1 = − z z3 ∞ z z1 − φ(z) = − if z2 = z z3 ∞ −1 (z z1) if z3 = z2 z1 − ∞ − To establish the uniqueness part of the result, suppose φ and ψ are two Mobius¨ transforma- 1 tions that map the zj’s to the corresponding wj’s. Then ψ− φ is a Mobius¨ transformation that ◦ fixes the zj, j = 1, 2, 3. But if 1 az + b (ψ− φ)(z) = , ◦ cz + d the equation az + b = z cz + d multiplies out to az + b = cz2 + dz. This is a quadratic equation which can only have two solutions unless it simplifies to 0 = 0 — which means b = c = 0, a = d and az + b az = = z. cz + d d
This reasoning works too if we consider the case where one of the zj is — if (az+b)/(cz+d) = z has as a solution, then c = 0 and then the equation we get on multiplying∞ out is linear (only one finite∞ solution unless it reduces to 0 = 0). 7.11 Theorem. If a < 1 and λ = 1, then | | | | z a φ(z) = λ − 1 az¯ − is a Mobius¨ transformation that maps the unit disc D(0, 1) bijectively to itself and also maps the unit circle z = 1 to itself. | | Chapter 7 — Riemann Mapping Theorem 7
Proof. If z = 1, then | | z a φ(z) = λ − | | 1 az¯ − z a = − 1 az¯ − 1 z a = − since z = z¯ = 1 z¯ 1 az¯ | | | | − z a = − z¯ a¯ z 2 − | | z a = − since z = 1 z¯ a¯ | |
= 1 −
Thus φ maps the unit circle into itself. The image of the unit circle under φ must be the whole circle by 7.9, and thus φ maps the unit circle to itself bijectively. Since φ: Cˆ Cˆ is a bijection, we must also have a bijection φ: Cˆ z = 1 Cˆ z = 1 . → Note that φ is continuous and C\ˆ {| | z =} → 1 has\ {| two| connected} components — the disc D(0, 1) and the complement of the closed\ {| | unit disc} in Cˆ. Now φ(0) = λa D(0, 1), and hence φ(D(0, 1)) D(0, 1). Since φ(1/a¯) = , the same reasoning− tells us∈ that φ maps the exterior of the⊂ closed disc into itself. Since φ∞is a bijection of Cˆ to Cˆ, we must have that φ(D(0, 1)) = D(0, 1) (and also that φ maps the exterior of the unit disc onto itself).
7.12 Corollary. If a, b D(0, 1), then there exist a conformal map φ: D(0, 1) D(0, 1) with φ(a) = b. ∈ → In fact the set of conformal maps of the unit disc onto itself forms a group under composition (easy to check) and this corollary can be stated as saying that this group of self-transformations of the unit disc acts transitively on the disc.
Proof. Take z a z b φ1(z) = − , φ2(z) = − . 1 az¯ 1 ¯bz − − Then φ1(a) = 0, φ2(b) = 0 and φ1, φ2 are conformal maps of the disc to itself. The one we want 1 is φ2− φ1, which satisfies ◦ 1 (φ− φ1)(a) = b. 2 ◦
7.13 Theorem (Schwarz Lemma). Let f : D(0, 1) D(0, 1) be an analytic function which maps the unit disc D(0, 1) to itself. If f(0) = 0, then →
(i) f(z) z for 0 < z < 1; | | ≤ | | | |
(ii) f 0(0) 1; | | ≤ 8 414 2003–04 R. Timoney
(iii) If equality holds in (ii) or if equality holds in (i) for any single z = 0, then 6 f(z) λz ≡ where λ is a constant of modulus λ = 1. | |
f(z) Proof. Let g(z) = z . Then g(z) is analytic for 0 < z < 1 and it is has a removable singularity at z = 0 since f(0) = 0. It becomes analytic at 0 |if| we define
f(z) g(0) = lim = f 0(0). z 0 z → Fix a number 0 < r < 1. For z = r | | f(z) 1 g(z) = < . | | z r
By the maximum modulus theorem for the analytic function g, it follows that g(z) < 1/r for | | z r. Fix z D(0, 1) and let r 1− to get | | ≤ ∈ → g(z) 1. | | ≤ This is true for all z D(0, 1) and we deduce ∈ f(z) 1, (0 < z < 1) z ≤ | |
and from this f(z) z (0 = z D(0, 1)). | | ≤ | | 6 ∈ This proves (i) and the fact that g(0) 1 proves (ii). | | ≤ To prove (iii), observe that if we get equality, then we must have a pointz D(0, 1) where g(z) = 1 — in other words, a point where the maximum modulus of g is∈ attained. By the maximum| | modulus theorem again, this means that g must be a constant function — g(z) λ and then we must have λ = 1. Hence f(z) = λz. ≡ | | 7.14 Corollary. Let φ: D(0, 1) D(0, 1) be any conformal map of the unit disc onto itself. Then there exist λ = 1 and a D→(0, 1) such that | | ∈ z a φ(z) = λ − 1 az¯ − (that is, φ is a Mobius¨ transformation of the kind studied above). Chapter 7 — Riemann Mapping Theorem 9
Proof. Suppose φ(0) = a. Then let
φ(z) a ψ(z) = − . 1 aφ¯ (z) − Being the composition of two conformal mappings, ψ : D(0, 1) D(0, 1) is a conformal map and ψ(0) = 0. → By the Schwarz lemma (7.13), ψ0(0) 1. 1 | | ≤ 1 But also, ψ− : D(0, 1) D(0, 1) is conformal and has ψ− (0) = 0. Again by the Schwarz lemma, →
1 (ψ− )0(0) 1 | | ≤ 1 1 ψ (ψ 1(0)) ≤ 0 − 1 1 ψ (0) ≤ 0 ψ0(0) 1 | | ≥
Combining with ψ0(0) 1 gives ψ0(0) = 1. Therefore, by the Schwarz lemma, | | ≤ | | ψ(z) λz ( λ = 1) ≡ | | φ(z) a − λz 1 aφ¯ (z) ≡ − λz + a φ(z) = 1 + λaz¯ z + λa¯ = λ 1 + λaz¯
7.15 Theorem (Riemann Mapping Theorem). Let G C be a simply connected connected open set with G = C. Let a G be arbitrary. Then⊂ there exists a (unique) conformal map f : G D(0, 1) of6 G onto the∈ unit disc which satisfies →
f(a) = 0 and f 0(a) > 0.
Proof of uniqueness. If there was a second conformal map g : G D(0, 1) with the given properties, then → 1 f g− : D(0, 1) D(0, 1) ◦ → would be a conformal map and it would satisfy
1 (f g− )(0) = f(a) = 0. ◦ 10 414 2003–04 R. Timoney
Hence by 7.14, 1 (f g− )(z) λz for some λ with λ = 1. ◦ ≡ | | Looking at the derivative at the origin, we find
1 1 1 1 f 0(a) 0 0 (f g− ) (0) = f 0(g− (0))(g− ) (0) = f 0(a) 1 = > 0. ◦ g0(g− (0)) g0(a) But, calculating it another way, we get λ. Hence λ is a positive real number. But also λ = 1 1 | | and that means λ = 1. Thus f g− is the identity function z and f = g. The proof of existence will◦ be divided into several steps.
Lemma 7.16. If G C is a simply connected connected open set and f : G C is analytic and never vanishes, then⊂f has an analytic square root — that is, there exists an→ analytic g : G C with g(z)2 f(z). → ≡ Proof. There exists a branch of log f in G (see chapter 2).
g(z) = e(1/2) log f(z) will do.
7.17 Lemma. If G C (G = C) is a connected open set with the property that every nowhere- vanishing analytic function⊂ on6 G has an analytic square root, then there exists an injective ana- lytic function f on G with f(G) D(0, 1). ⊂ Proof. Pick b C G. Then there exists g : G C analytic with g(z)2 = z b. This g is injective since ∈ \ → −
2 2 g(z1) = g(z2) z1 b = g(z1) = g(z2) = z2 b z1 = z2. ⇒ − − ⇒
Now, by the open mapping theorem, g(G) is open. Pick w0 g(G) and choose r > 0 ∈ so that D(w0, r) g(G). Then D( w0, r) C g(G) because if there exists a point w ⊂ − ⊂ \ ∈ D( w0, r) g(G), then w = g(z1) for some z1 G and also w D(w0, r) g(G) which − ∩ ∈ − ∈ ⊂ tells us that w = g(z2) for some z2 G. Now − ∈ 2 2 g(z1) = g(z2) g(z1) = g(z2) − ⇒ z1 b = z2 b ⇒ − − z1 = z2 ⇒ g(z2) = g(z1) = g(z2) ⇒ − g(z2) = 0 ⇒ 2 0 = g(z2) = z2 b ⇒ − z2 = b C G ⇒ ∈ \ and this contradicts z2 G. ∈ Chapter 7 — Riemann Mapping Theorem 11
Put r f(z) = . 2(g(z) + w0)
Then f is analytic on G ( g(z) + w0 r for all z G is proved above, and shows that the denominator is never 0), injective| on|G ≥since g is and∈ also f satisfies f(z) 1/2 < 1 for z G. | | ≤ ∈ 7.18 Lemma. With the same hypothesis on G as in Lemma 7.17, and for a G a given point, there exists an injective analytic function on G satisfying ∈
f(G) D(0, 1) ⊂ f(a) = 0
f 0(a) > 0
Proof. Replace the f(z) obtained in Lemma 7.17 by
f(z) f(a) λ − 1 f(a)f(z) − with λ = 1, λ suitably chosen to make the derivative at a positive. | | 7.19 Lemma. Let G C be a connected open set , G = C, with the property that every nowhere- vanishing analytic function⊂ has an analytic square root.6 Let a G be fixed. Then there exists a conformal map f : G D(0, 1) of G onto the unit disc with∈ the properties f(a) = 0 and → f 0(a) > 0. Observe that this lemma will complete the proof of the Riemann mapping theorem (7.15).
Proof. Let denote the family of all analytic functions f : G C such that either f 0 or else F → ≡
f is injective, f(G) D(0, 1), f(a) = 0 and f 0(a) > 0. ⊂ By Lemma 7.18, contains nonzero functions. By Montels theoremF is relatively compact in H(G). By Corollary 7.4 to Hurwitz’ theorem, we can see that all functionsF in the closure of in H(G) are either constant or injective. Since all the function in have the value 0 at a, thisF must be true of all functions in the closure of . Hence the onlyF constant function in the closure is 0 (which is in ).The other functions fF in the closure must have f(G) D¯(0, 1). Since f(G) is open (openF mapping theorem), ⊂ f(G) D(0, 1). Also, because the map f f 0(a): H(G) C is continuous, all functions in ⊂ 7→ → the closure of must satisfy f 0(a) 0. Since they are injective if not identically zero, f 0(a) = 0 F ≥ 6 and so f 0(a) > 0 (unless f 0). The conclusion from these observations is that is actually closed in H(G). Hence is≡ compact in H(G). F F 12 414 2003–04 R. Timoney
Since the map f f 0(a): R is a continuous function on a compact set, it must attain 7→ F → its maximum value. Let f be a function with f 0(a) as large as possible. As remarked earlier, ∈ F f is not the zero function and f 0(a) > 0. We now show that this f maps G onto the unit disc (and so it is the required Riemann mapping). Suppose for the sake of obtaining a contradiction that f(G) = D(0, 1) and choose w D(0, 1) f(G). By the assumption on G about analytic square roots,6 we can find h H(G) with∈ \ ∈ f(z) w h(z)2 = − . (1) 1 wf¯ (z) − Let h(z) h(a) g(z) = λ − 1 h(a)h(z) − where λ = 1 and λ is chosen to ensure that g0(a) > 0. Now| |g and ∈ F g0(a) = g0(a) = (φ h)0(a) , | | | ◦ | where φ(ζ) = (ζ h(a))/(1 h(a)ζ). This leads to − − 1 g0(a) = φ0(h(a))h0(a) = h0(a) . | | 1 h(a) 2 | | − | | Differentiating (1) gives
1 wf¯ (z) +w ¯(f(z) w) 2h(z)h0(z) = − − f 0(z). (1 wf¯ (z))2 − Evaluating at z = a we get 2 2h(a)h0(a) = (1 w )f 0(a). − | |
2 2 f 0(a)(1 w ) f 0(a)(1 w ) h0(a) = − | | = − | | | | 2 h(a) 2 w | | p| | (recall h(a)2 = (f(a) w)/(1 wf¯ (a)) = w). Thus − − − 1 g0(a) = h0(a) 1 h(a) 2 | | − | | 1 1 w 2 = − | | f 0(a) 1 w 2 w − | | p| | 1 + w = | |f 0(a) 2 w p| | > f 0(a). Chapter 7 — Riemann Mapping Theorem 13
The last step is justified because
1 + w > 2 w | | p| | 1 2 w + w > 0 − p| | | | (1 w )2 > 0. − p| |
The fact that g0(a) > f 0(a) contradicts the choice of f as maximising f 0(a). This contra- diction shows that f(G) = D(0, 1). ∈ F
7.20 Theorem. Let G C be a connected open set. Then the following are equivalent properties for G ⊂
(i) G is simply connected
1 (ii) Indγ(a) = 0 for each piecewise C closed curve in G and each a C G. ∈ \ (iii) For each analytic function f : G C and each piecewise C1 closed curve γ in G, f(z) dz = 0. → Rγ (iv) Every analytic function f : G C has an antiderivative. → (v) For every analytic function f : G C which never vanishes in G, there is a branch of log f in G. →
(vi) Every never-vanishing analytic f : G C has an analytic square root. → (vii) G is homeomorphic to the unit disc D(0, 1) (that is, there exists a continuous bijection 1 f : G D(0, 1) with a continuous inverse f − ). → (viii) Cˆ G is connected (this says exactly that G has no “holes” — no bounded connected components\ of C G). \ Proof. We know (i) (ii) (iii) (iv) (v) ⇒ ⇐⇒ ⇐⇒ ⇐⇒ (see Chapter 2). (v) (vi) is easy — e(1/2) log f is an analytic square root of f. (vi)⇒ (vii): In the case G = C, Lemma 7.19 gives this. For G = C, f(z) = z/(1 + z ) is a homeomorphism⇒ of C to the unit6 disc. | | (vii) (i) is straightforward — since the disc is simply connected, so is anything homeo- morphic⇒ to the disc. It remains therefore to show that (viii) is equivalent to the rest. (viii) (ii): First, the assumption implies that every connected component of C G is unbounded.⇒ A bounded connected component E of C G would be open and closed in C\ G. Hence E would be closed in C since C G is closed. As\ E is also bounded, it would be closed\ in Cˆ. Hence E is closed in Cˆ G. E open\ in C G implies E = (C G) U with U C open. \ \ \ ∩ ⊂ 14 414 2003–04 R. Timoney
Hence E = (Cˆ G) U is open in Cˆ G. Being an open and closed subset it must be a union of connected components\ ∩ of Cˆ G, but\ that is impossible if Cˆ G is connected and contains / E. \ \ ∞ ∈ 1 To show (ii) take any piecewise C closed curve γ in G. We know Indγ(w) is constant on connected components of C γ. Hence constant on connected components of C G C γ. (Each connected component\ of the smaller set is contained in a connected component\ ⊂ of\ the larger one.) But we also know that there is just one unbounded connected component of C γ \ and Indγ = 0 there. As all components of C G are unbounded, they must all be in the unbounded component of C γ. Hence C G is contained\ in the unbounded component of C γ and so we \ \ \ must have Indγ(w) = 0 for all w C G. ∈ \ (iv) (viii): Suppose that (iv) holds but Cˆ G is not connected. Then we can write ⇒ \ Cˆ G = K L \ ∪ where K and L are disjoint nonempty relatively closed subsets of Cˆ G. Since G is open in C, it is also open as a subset of Cˆ and hence Cˆ G is closed in Cˆ. Hence\ K and L are closed in Cˆ (even compact). Now K or L \— say L. Then K C and K is compact. G˜ = G K is open in C.∞ ∈ ∞ ∈ ∞ ∈ ⊂ ∪ ˜ Essentially, the idea is that we can find a finite number of closed curves γj in G K = G with \ Ind (a) = 1 for all a K X γj j ∈ This is contradicts (iv) because f(z) = 1/(z a) is analytic on G, hence has an antiderivative by (iv) and so − 1 1 Ind (a) = dz = 0 γj 2πi Z z a γj − for each j. The proof that we can find such γj is a bit tricky and so we will check out something a little simpler which is good enough for our purpose. We use here the notation that for a, b C, [a, b] denotes the line segment from a to b, or the parameterised path Γ(t) = (1 t)a∈+ tb (0 t 1). − ≤ ≤ Lemma 7.21. If G˜ C is open and K G˜ is compact then we can find a finite number of ⊂ ⊂ ˜ closed line segments Γj = [aj, bj] (j = 1, 2, . . . , n) in G K such that \ 1. For each f : G˜ C analytic and a K, → ∈ n f(z) dz = 2πif(a) (2) X Z z a j=1 Γj −
2. The number of times any given point w occurs in the list a1, a2, . . . , an of initial points of the line segments is the same as the number of times w occurs in the list b1, b2, . . . , bn of the end points. Chapter 7 — Riemann Mapping Theorem 15
Armed with this Lemma, we can use it for f(z) = 1 to get
n 1 Z dz = 2πi (a K). X z a ∈ j=1 Γj − Using our assumption (iv) tells us that 1/(z a) has an antiderivative (call it g) on G = G˜ K and so we have − \ n 1 n n n Z dz = g(bj) g(aj) = g(bj) g(ak) = 0 X z a X − X − X j=1 Γj − j=1 j=1 k=1 by the second property in the Lemma. This contradiction shows that (iv) (viii), once we verify the Lemma. ⇒ Proof. (of Lemma 7.21) ˜ If G = C we can take Γ1, Γ2, Γ3, Γ4 to be the four sides of a large square, where the square contains K in its interior and the sides are traversed anticlockwise. Statement 1 of the Lemma is true by the Cauchy integral formula. The case where K is empty is also trivial. Assuming that G˜ = C and K = . there is a positive shortest distance 6 6 ∅ δ = inf a w : a K, w C G˜ {| − | ∈ ∈ \ } from K to the complement of G˜. We now consider an infinite grid on the plane of squares with sides parallel to the real and imaginary axes and a fixed side length r < δ/√2. To be exact, take all the squares of the grid with vertices at the points nr + imr (n, m Z). By boundedness of K only a finite number of ∈ these square can have a point of K in them. We list all the squares S1,S2,...,SN of the grid which intersect K at any point (inside or on the edge of Sj). To make things more precise let us say that Sj means the set of points inside or on the edge of the square. Because the squares have diagonal r√2 < δ, no two points of Sj can be separated by as much as distance δ. Since there is ˜ one of the points of Sj that belongs to K, we can conclude from the choice of δ that Sj G. ⊂ Now write down for each Sj the 4 sides of it so that the boundary square is traversed anti- clockwise. We end up with 4N line segments Γ1∗, Γ2∗,..., Γ4∗N and let us write
Γj∗ = [aj∗, bj∗] From Cauchys integral formula we have that f(z) dz = 2πif(a) Z z a ∂Sj − ˜ for a in the interior of Sj (and any f H(G)). The integral is zero for a outside Sj and so ∈ N f(z) 4N f(z) Z dz = Z dz = 2πif(a) X ∂S z a X Γ z a j=1 j − k=1 k∗ − 16 414 2003–04 R. Timoney
holds for a in the union of the interiors of the Sj. Also observe that each corner of Sj occurs once as a starting point and once as an ending point of a side of Sj. Hence the number of times any given point occurs among a∗ (1 k 4N) k ≤ ≤ is the same as the number of times it occurs among b∗ (1 k 4N). k ≤ ≤ Some segments may occur twice in the list Γ1∗, Γ2∗,..., Γ4∗N . This happens if two adjacent squares of the original grid meet K, but then the two segments will be traversed in opposite
directions. We can omit both from the list Γ1∗, Γ2∗,..., Γ4∗N without disturbing either the property f(z) Z dz = 2πif(a) X Γ z a k k∗ − or the property that every endpoint occurs as often as every initial point. Removing all such pairs of segments, we end up with a list Γ1, Γ2,... Γn (where Γj = [aj, bj]). We then have (2) for a in the union of the interiors of the Sj. Both sides of (2) give continuous functions of a. The left side is continuous for a not on Γj while the right hand side is continu- Sj ous on G˜. It follows that the equality persists at all points a G˜ which are limits of points in the ∈ union of the interiors of the Sj, but not in Γj. This will include all a K because if a K Sj ∈ ∈ then a belongs in at least one Sj. If a is an interior point of Sj, then we already know (2) holds. If a is on the edge of Sj but not a corner of Sj, the adjacent square on the original grid must also be one of the Sk and the side containing a would be canceled out. If a is a corner of Sj then all 4 squares of the grid that meet at that corner are among the Sk and all 4 segments that contain a (East, North, West and South of a) must have been canceled out. ˜ Thus all remaining segments Γj are in G K and (2) holds for a K. \ ∈ 7.22 Remark. As a consequence of the Riemann mapping theorem any questions about analytic functions on a simply connected connected open set G C can be reduced to questions on two simple domains — G = C and G = D(0, 1). However,⊂ translating some problems from G to the unit disc may require some specific knowledge about the “Riemann mapping function” 1 f : G D(0, 1), or equivalently information about its inverse f − : D(0, 1) G. This→ leads to the study of problems concerning injective analytic functions→ on the unit disc. One such problem attracted a lot of attention since the early part of the 20th century, although it actually does not seem to have much practical value. It was known as the Bieberbach conjecture and relates to the power series coefficients of injective functions on the unit disc. It said that if n ∞ anz is the power series of an injective analytic function on the disc, then Pn=0
an | | n a1 ≤ | | It is possible to have equality. The function
∞ z k(z) = nzn = X (1 z)2 n=1 − is injective on the unit disc. It is known as the Koebe function. Chapter 7 — Riemann Mapping Theorem 17
Bieberbach proved that a2 2 a1 for any injective analytic function on the disc and that if | | ≤ | | a2 = 2 a1 , then | | | |
∞ a zn = a + a αk¯ (αz) (3) X n 0 1 n=0 for some α, α = 1. In other words he proved his conjecture for the case n = 2 and showed that equality holds| | only if the function is closely related to the Koebe function. He conjectured that if an = n a1 for any n 2, then the function has to be as in (3). | | | | ≥ A lot of work was done, much of it related to the Bieberbach conjecture in some way, where the Koebe function (or really the functions (3)) was the extremal case. The Bieberbach conjecture itself was making slow progress (it was proved for most values of n up to n = 7) until Louis de Branges proved the whole conjecture in 1985 (Acta Math. volume 154, pages 137-152). It does not seem that knowing the Bieberbach conjecture has any important consequence, but the result attracted a lot of attention because the conjecture remained unsolved for so long. The original result of Bieberbach for n = 2 can be used to prove the following elegant result (which is quite useful).
7.23 Theorem (Koebe 1/4-theorem). If f is an injective analytic function on the unit disc, then the range of f contains the disc 1 D f(0), f 0(0) . 4| |
Unless f has the form (3), the range of f will contain a disc about f(0) of radius r > (1/4) f 0(0) . | | I omit the proof of this as I have also omitted the (reasonably elementary) material necessary to show a2 2 a1 . | | ≤ | | One other theorem that was mentioned earlier, but we have not covered, is Runges theorem.
Theorem 7.24 (Runges theorem). Let G C be open and E Cˆ G a subset with the property that the closure of E intersects every connected⊂ component of⊂Cˆ \ G. Then the set of rational functions with poles only at points of E (here we include z = as\ a possible pole) is dense in H(G). ∞
Again we omit the full proof, although Lemma 7.21 helps quite a lot in the proof. Here is a sketch. If K G is compact, and f H(G), then Lemma 7.21 expresses f(a) for a K as an integral (hence⊂ approximable by a Riemann∈ sum) of functions ∈
1 f(z) g (a) = z 2πi z a − with poles at points z G K. The essential part of the proof (not a very short part) consists in ∈ \ showing the functions gz : K C can be approximated uniformly on K by rational functions with poles in E. → 18 414 2003–04 R. Timoney
The conclusion of this argument is that for f H(G) and K G compact, it is possible to find a rational function r(z) with poles only in E ∈so that ⊂
sup f(z) r(z) z K | − | ∈
is small. Finally, we do this for an exhaustive sequence K1 K2 K2 of compact ⊂ ⊂ ⊂ · · · subsets of G to produce a sequence of rational functions rn(z), each with poles only on E so that 1 sup f(z) rn(z) < . z Kn | − | n ∈
It follows then that rn f in H(G). → Corollary 7.25. Let G C be open and assume Cˆ G is connected. The the polynomial functions are dense in H(⊂G). \
Proof. Apply Runges theorem with E = and note that rational functions with poles only at are polynomials. {∞} ∞ Remark 7.26. One reason to advertise the existence of Runges theorem in advance was to state that it can be used to exhibit pointwise convergent sequences of analytic functions with discon- tinuous limits. Let Gn = z C : z = 1/(2n) (the complex plane minus a vertical line) and let Kn Gn be the compact{ set∈ < 6 } ⊂
Kn = Ln Rn,Ln = z C : z n, z 0 ,Rn = z C : z n, z 1/n ∪ { ∈ | | ≤ < ≤ } { ∈ | | ≤ < ≥ }
Let fn : Gn C be the analytic function → 0 z < 1/(2n) f (z) = n z
Richard M. Timoney March 30, 2004