Chapter 6: The metric space M(G) and normal families

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Chapter 6: The metric space MG) and normal families Course 414, 003 04 March 9, 004 Remark 6.1 For G C open, we recall the notation MG) for the set algebra) of all meromorphic functions on G. We now consider convergence in MG) is a way analogous to what we did for HG) in Chapter 5. Before we do that, we explain how meromorphic functions on G can be regarded as functions or maps from G to an extended complex plane C { } with one extra point at infinity added. While the process of adding a point at infinity can be carried out very abstractly Alexandroff one point compactification of a locally compact topological space), for the complex plane we can visualise it rather geometrically via stereographic projection. Stereographic Projection 6. There is a transformation which maps the complex plane C bijectively to a sphere in R 3 with one point removed. To explain it we consider C as embedded in space R 3 in the most obvious way: and we consider the unit sphere S in R 3 z x + iy C x, y, 0) R 3 x, y R) S {ξ, η, ζ) R 3 : ξ + η + ζ 1}. The North Pole of the sphere 0, 0, 1) will be a special point in our discussion and we sometimes write N P for 0, 0, 1). Define a mapping P : C S by the geometrical rules that P z) is the point other than the North Pole N P) where the line joining N P to z x, y, 0) intersects S. This map P is called stereographic projection. From a picture you can see that if z > 1, then P z) is on the upper hemisphere. If z 1, then P z) will be the same as z or x, y, 0) while for z < 1, P z) will be in the lower hemisphere. 1

414 003 04 R. Timoney P z) z z P z) Proposition 6.3 The stereographic projection map P : C S \ {N P} S \ {0, 0, 1)} is a bijection and is given by ) x P z) P x + iy) x + y + 1, y x + y + 1, x + y 1. 1) x + y + 1 The inverse map is P 1 ξ, η, ζ) ) ξ 1 ζ, η 1 ζ, 0 Proof. The line in R 3 we used to define P has parametric equations ξ, η, ζ) 0, 0, 1) + tx, y, 0) 0, 0, 1)) tx, ty, 1 t) and this meets S at the values of t where ξ + η + ζ 1 t x + t y + 1 t) 1 t x + t y + 1 t + t 1 t x + y + 1) t 0 ttx + y + 1) ) 0 and so where t 0 the North Pole N P) and where t /x + y + 1). This must be the value of t for the point P z), and so ) x P z) x + y + 1, y x + y + 1, 1 x + y + 1 ) x x + y + 1, y x + y + 1, x + y 1. x + y + 1 One can check in a straightforward way that the map Q: S \ {N P} C given by ) ξ Qξ, η, ζ) 1 ζ, η 1 ζ, 0 is the inverse by verifying Q P id C and P Q id are both identity maps. )

Chapter 6 the metric space MG) and normal families 3 Definition 6.4 The extended complex plane Ĉ defined as the complex plane C with one extra point, denoted, adjoined. If P ) is defined to be the North Pole N P 0, 0, 1) in S, then stereographic projection P : Ĉ S is a bijection. The sphere S or Ĉ which we can identify with S via the stereographic projection map P ) is called the Riemann sphere. For z, w Ĉ we introduce a notation σz, w) for the Euclidean distance between P z) and P w). This means the straight line distance in R 3, as opposed to the length of the shortest path on the sphere.) σ is called the chordal distance or chordal metric on the Riemann sphere. Lemma 6.5 For z, w C, σz, w) σz, ) z w 1 + z )1 + w ) 1 + z ) Also if z, w are not zero Proof. We have and we want to compute P x + iy) Observe that for unit vectors v, w R 3 and so 1 σ z, 1 ) w ) 1 σ z, x x + y + 1, σz, w) σz, 0) P x + iy) P r + is) ) y x + y + 1, x + y 1 x + y + 1 v w v w).v w) v + w v.w v.w P x + iy) P r + is) x + y + 1)r + s + 1) [ x + y + 1)r + s + 1) 4xr + 4ys + x + y 1)r + s 1)) ] [x + y )r + s ) + r + s ) + x + y ) + 1 4xr + ys) x + y )r + s ) + r + s ) + x + y ) 1] [ x + y ) + r + s ) 4xr + ys) ] 4[x xr + r ) + y ys + s )] 4[x r) + y s) ] 4 x + iy) r + is)

4 414 003 04 R. Timoney Thus σz, w) z + 1) w + 1) 4 z w and the formula for σz, w) is as claimed. All the other statements in the proposition are quite easy to check. σz, ) P x + iy) 0, 0, 1) x x + y + 1, x x + y + 1, y x + y + 1, 1 x + y + 1) 4x + 4y + 4) 4 x + y + 1 4 z + 1 1 σ z, 1 ) w 1 1 z w ) ) 1 + 1 1 + 1 z w ) x + y + 1 1 ) y x + y + 1, x + y 1 x + y + 1 1 + z ) w z z + 1) w + 1) where we multiplied above and below by zw z w. Finally, ) 1 σ z, 1 + 1 ) z σz, 0). z + 1) z σz, w) Remark 6.6 Now Ĉ, σ) is a metric space in fact the same as the sphere S in R 3 with the subset metric, or the distance from R 3 restricted to S ) and so we can look at open balls in Ĉ, open sets in C, limits of sequences in C, continuous functions and so on. Proposition 6.7 A subset U Ĉ is open in the metric space Ĉ, σ) is satisfies both i) U C is open in C in the usual sense) ii) if U then there is some r > 0 so that {z C : z > r} U Proof. Note that the map P C : C S \ {N P} is continuous because the formula 1) for P in Proposition 6.3 is clearly continuous from R to R 3. Hence if U is open in Ĉ then P U) is open in S because the metrics on Ĉ and S are copies of one another) and the inverse image P C ) 1 P U)) U C is therefore open in C. If U U Ĉ open), then U contains a ball of some positive radius δ > 0 about. That is B σ, δ) {z Ĉ : σ, z) < δ} U.

Chapter 6 the metric space MG) and normal families 5 Assuming δ < we can make δ smaller if necessary) we have for z C σ, z) 1 + z < δ 1 + z > 4 δ z > 4 δ 1 and so with r 1/δ) 4 δ we have {z C : z > r} U This was under the assumption U Ĉ open. Conversely assume U Ĉ satisfies the two conditions. The second condition tells us that is an interior point of U if it is in U at all. U and {z C : z > r} U B σ, δ) U with δ / 1 + r. The first condition U C open in C plus continuity of P 1 : S \ {N P} C which is clear from the formula ) in Proposition 6.3) implies that P 1 ) 1 U C) P U C) is open in S \ {N P}. It follows that every point of P U C) is an interior point relative to S and so since P transforms the distance σ to the distances on S ) all points of U C are interior points with respect to Ĉ, σ)). Definition 6.8 If G C is open and f : G Ĉ is a function, then we say that f is analytic if it satisfies a) f it is continuous from G with its usual metric to Ĉ, σ)); b) f f 1 C) : f 1 C) C is analytic in the usual sense); [Note that f 1 C) is open in G by continuity of f, hence f 1 C) is open in C and so there is no problem looking at analyticity of this function.] and 1 1 c) the restriction of to f Ĉ) \ {0} is analytic when we define 1/fz) 0 at points fz) where fz). Note that 1/fz) is continuous because of Lemma 6.5 and so the third condition means that f and 1/f are treated equally. A reasonable way to look at it is that the two maps z Ĉ \ { } C z C and w Ĉ \ {0} 1/w C with 1/ interpreted as 0) are two coordinate charts on Ĉ. In this

6 414 003 04 R. Timoney way we can regard Ĉ as a -dimensional manifold1. Because the transition from coordinate z to coordinate w is given by an analytic transition function w 1/z on the intersection C Ĉ\{0}) where both coordinates can be used, and the opposite w z 1/w is also analytic, we say that Ĉ is a complex analytic manifold of complex dimension 1, and these are actually called Riemann surfaces). A map f with values in a complex manifold is analytic if f is continuous and each coordinate map composed with f is analytic on the open set where it makes sense). Our definition of f : G Ĉ analytic fits the pattern that is used for maps with values in a general Riemann surface. Proposition 6.9 If G C is open and connected and f : G Ĉ is analytic but not identically, then the points of f 1 ) are isolated in G. That is z 0 f 1 ) implies there is a punctured disc Dz 0, r) \ {z 0 } G \ f 1 ). Proof. Note that f 1 Ĉ \ {0}) is open and 1/f is analytic there. If z 0 f 1 ), then z 0 f 1 Ĉ\{0}) and so there is r > 0 with Dz 0, r) f 1 Ĉ\{0}) and 1/f is analytic on Dz 0, r). Notice 1/fz 0 ) 1/ 0. If z 0 is not an isolated point of f 1 ) then z 0 is not an isolated zero of 1/f in Dz 0, r). Then 1/f is identically 0 on Dz 0, r) by the identity theorem for analytic function, and in fact 1/f is identically 0 on the connected component of z 0 in f 1 Ĉ \ {0}). Call this connected component G 0. Then f on G 0 and because connected components are open and closed) G 0 is open in f 1 Ĉ \ {0}), hence in G. But G 0 is also closed in G because if z n ) n1 is a sequence in G 0 converging to a limit z G, then continuity of f implies fz) lim n fz n ). So z f 1 Ĉ \ {0}). Since G 0 is closed in f 1 Ĉ \ {0}) being a connected component), we have z G 0. Thus z 0 G 0 G is open in G, closed in G and nonempty. As G is connected, G 0 G and so f. But this contradicts the assumptions. Theorem 6.10 If G C is open and f is a meromorphic function on G, then we can define an analytic Ĉ-valued function f : G Ĉ be setting fz) at poles of f. Conversely, if G C is open and connected and f : G Ĉ is analytic but not identically, then f is a meromorphic function on G with poles at the points in f 1 ). Proof. Starting with f meromorphic on G, we have an open set G 0 G on which f is analytic in the usual sense with finite values) and so that each point of G \ G 0 is a pole of f. If we define fz) for each z G \ G 0, then f : G Ĉ will be continuous. To verify continuity at points z 0 G where fz 0 ) C is finite so that z 0 G 0 ) use lim z z0 fz) fz 0 ) or lim z z0 fz) fz 0 ) 0) to deduce fz) fz 0 ) lim σfz), fz 0 )) lim z z 0 z z0 1 + fz) )1 + fz 0 ) 0. 1 a metric space or topological space where in an open set around each point there is a coordinate system that identifies it with an open piece of R ; we assume the coordinate functions are continuous from their domain to their range in R and have continuous inverses; usually we also make the slightly technical assumption that there is a countable dense subset in a manifold or that the topology is second countable if we don t use a metric)

Chapter 6 the metric space MG) and normal families 7 At poles z 0 we know lim z z0 fz) and so lim σfz), fz 0 )) lim σfz), ) lim 0 z z 0 z z0 z z0 1 + fz) Hence f : G Ĉ is continuous. As f 1 C) is the complement of the poles of f in G, we know that the restriction of f to f 1 C) is analytic. Since 1/f is meromorphic on each connected component of G unless f 0 1 on the component by Corollary 4.5) we can also see that 1/f is analytic on the set f Ĉ\{0}. Hence f : G Ĉ is analytic. For the converse, if G is connected, f : G Ĉ is analytic but f, then the points of f 1 ) are isolated in G by Proposition 6.9. Thus f is analytic on H f 1 C), the points of G \ H f 1 ) are isolated in G and at points z 0 G \ H 0 lim z z0 σfz), fz 0 )) lim z z0 σfz), ) lim z z0 1 + fz) implies lim z z0 fz). Thus z 0 is a pole of f by Proposition 4.16. Remark 6.11 We now extend the notion of analyticity one step further to functions defined on open subsets of) Ĉ and still allowing values in the extended complex plane). Definition 6.1 If G Ĉ is open we say that a function f : G Ĉ is analytic if i) f is continuous from G with the metric σ to Ĉ with the metric σ) ii) f is analytic on G C iii) gz) f1/z) is analytic on {z C : 1/z G} we include z 0 if G and g0) means f ) in that case). Lemma 6.13 If G C, the same functions f : G Ĉ are analytic according to Definition 6.8 and Definition 6.1. Proof. The type of continuity required in each case is different, as in one case we consider G with the usual metric on C while in the other we use the metric σ. However, the open subsets V G are the same in either case by Proposition 6.7. Thus looking at continuity in the form U Ĉ open f 1 U) G open, we can see that the continuity requirements are equivalent. The rest of the proof is straightforward. Example 6.14 All rational functions rz) pz) qz) with p, q polynomials and q 0) are analytic r : Ĉ Ĉ.

8 414 003 04 R. Timoney Proof. By the Fundamental theorem of algebra, we can factor p and q and cancel common factors to write j i1 rz) λ z α i) k i1 z β i) where j 0, k 0 and {α 1, α,..., α j } {β 1, β,..., β k }. It follows that rz) is meromorphic on C with poles at z β i 1 i k). When suitably interpreted at z β i, rz) therefore becomes analytic r : C Ĉ. r1/z) is also rational and so there is a suitable value for r1/0) r ) to make gz) r1/z) analytic g : C Ĉ. One can verify that r : Ĉ Ĉ is continuous. Remark 6.15 It is in fact the case that rational functions and the constant are the only analytic functions f : Ĉ Ĉ. The first step of the proof is to observe that f 1 ) cannot be an infinite set if f is not identically. Proof. As Ĉ is compact, an infinite number of distinct points z n Ĉ would have to have a convergent subsequence. If the limit was in C, we would have a contradiction to Proposition 6.9. If the limit was, we can switch that to 0 by looking at f1/z) instead. For each of points a C with fa), we have a pole of the meromorphic function on C. Hence a Laurent series about a with a finite number of negative terms fz) n p a n z a) n 0 < z a < δ) We call the sum of the negative terms the principal part of f at a and write it as P f,a z) 1 n p a n z a) n The sum of the principal parts of f at the finitely many) points a C where fa) produces a rational function rz). Subtracting from f, gives fz) rz) analytic at all points of C. Hence it has a power series about the origin. As f1/z) r1/z) has a pole or a removable singularity at z 0, it follows as in Remark 4.18) that fz) rz) pz) a polynomial. Hence fz) rz) + pz) is rational. Remark 6.16 We now describe the version of convergence appropriate for functions. Ĉ-valued analytic Definition 6.17 For G C is open we let CG, Ĉ) the set of all continuous functions : G Ĉ, and HG, Ĉ) the analytic or holomorphic) functions. A sequence f n) n1 in CG, Ĉ) is said to converge uniformly with respect to σ on compact subsets of G if for each K G compact we have lim σf n z), fz)) 0 sup n z K

Chapter 6 the metric space MG) and normal families 9 Lemma 6.18 If f n ) n1 is a sequence in CG, Ĉ) that converges uniformly with respect to σ on compact subsets of G to a limit function f : G Ĉ, then f CG, Ĉ). If f n ) n1 is a sequence in HG, Ĉ) that converges uniformly with respect to σ on compact subsets of G to a limit function f : G Ĉ, then f HG, Ĉ). Proof. Omitted. The first part is a standard fact from metric space theory. For the second, analyticity of f at points z 0 G where fz) C is finite can be shown by establishing that for n large f n has finite values near z 0 and f n f uniformly with respect to absolute value distance) on some disk about z 0. For points where fz 0 ), we can consider 1/f n 1/f. Proposition 6.19 Let G C be open. There is a metric ρ on CG, Ĉ) so that convergence of sequences in ρ corresponds to uniform convergence with respect to σ on compact subsets of G. Proof. Choose an exhaustive sequence K n ) n1 of compact subsets of G and define ρf, g) n1 1 n sup z K n σfz), gz)) The rest is not much different from the earlier case where we constructed ρ on CG). One simplification here results from sup z Kn σfz), gz)). Definition 6.0 Let G C be open. A family F of meromorphic functions on G is called a normal family if each sequence f n ) n1 in F has a subsequence that converges uniformly with respect to σ on compact subsets of G to a limit function that is either a meromorphic f on G, or the constant function Proposition 6.1 If G C is connected then a family F of meromorphic functions on G is a normal family F is relatively compact when considered as a subset of HG, Ĉ), ρ). Proof. In the case of G connected HG, Ĉ) the meromorphic functions on G together with the constant by Theorem 6.10). Proposition 6.19 implies the result. Definition 6. Let G C be open. A family F CG, Ĉ) is called equicontinuous with respect to σ) at a point z 0 G if for each ε > 0 it is possible to find δ > 0 so that z z 0 < δ, f F σfz), fz 0 )) < ε. Theorem 6.3 version of Ascoli) For G C open, a family F CG, Ĉ) is relatively compact in CG, Ĉ), ρ) if and only if it is equicontinuous at each point of G. Proof. Omitted. It is not so different from the Ascoli theorem we had before 5.0), where we left the proof to an appendix. The pointwise boundedness condition is not needed now because Ĉ is compact.

10 414 003 04 R. Timoney Remark 6.4 Montel s theorem does not work for meromorphic functions. There is a theorem for meromorphic functions that takes the place of Montels theorem. It relies on the notion of a spherical derivative µf) of a meromorphic function f. At points where fz) C is finite, the spherical derivative is µf)z) f z) 1 + fz) and the definition at poles where fz) ) relies on taking reciprocals. When fz) is neither 0 nor one can see that ) 1 µ z) µf)z) f Theorem 6.5 Marti criterion) If G C is open and connected and F is a family of meromorphic functions on G, then F is a normal family if and only if it satisfies sup f F sup µf)z) < z K for each compact K G. This means uniform boundedness of the spherical derivative on each compact subset of G. Proof. Omitted. Remark 6.6 One application of these ideas or variants of them) is in the topic of iteration theory for analytic functions. The most commonly studied case is the situation of rational maps f : Ĉ Ĉ and iteration means considering the composites f, f f f, f 3 f f,... Slightly more precisely, one considers a point z Ĉ and the long term behaviour of the iterated images z, fz), f z) ffz)), f 3 z),... and examines whether there is a trend. The well-behaved set of z for a given rational f) is called the Fatou set F f) of f. It is the largest open subset of Ĉ on which the family of iterates {f, f, f 3,...} is an equicontinuous family. This means that a small change in z has a small long term effect on the iterates f n z). The Julia set Jf) is the complement in Ĉ of the Fatou set. The Julia set is the bad set where a small change in z can produce a major change in the behaviour of the sequence f n z) of iterates. One basic theorem is that for polynomials f of degree or more, is in the Fatou set and the Julia set is not empty. The Mandelbrot set consists of those values of a parameter c C for which the the Julia set Jp c ) is connected, where p c z) z + c. For lack of time, we won t deal with this topic. Richard M. Timoney March 9, 004