Chapter 5 Families of Analytic Functions In this chapter we consider the linear space A(Ω) of all analytic functions on an open set Ω and introduce a metric d on A(Ω) with the property that convergence in the d-metric is uniform convergence on compact subsets of Ω. We will characterize the compact subsets of the metric space (A(Ω),d) and prove several useful results on convergence of sequences of analytic functions. After these preliminaries we will present a fairly standard proof of the Riemann mapping theorem and then consider the problem of extending the mapping function to the boundary. Also included in this chapter are Runge’s theorem on rational approximations and the homotopic version of Cauchy’s theorem. 5.1 The Spaces A(Ω) and C(Ω) 5.1.1 Definitions Let Ω be an open subset of C. Then A(Ω) will denote the space of analytic functions on Ω, while C(Ω) will denote the space of all continuous functions on Ω. For n =1, 2, 3 ..., let Kn = D(0,n) ∩{z : |z − w|≥1/n for all w ∈ C \ Ω}. By basic topology of the plane, the sequence {Kn} has the following three properties: (1) Kn is compact, ⊆ o (2) Kn Kn+1 (the interior of Kn+1), (3) If K ⊆ Ω is compact, then K ⊆ Kn for n sufficiently large. Now fix a nonempty open set Ω, let {Kn} be as above, and for f,g ∈ C(Ω), define ∞ 1 f − g d(f,g)= Kn , 2n 1+ f − g n=1 Kn where {| − | ∈ } ∅ − sup f(z) g(z) : z Kn ,Kn = f g Kn = 0,Kn = ∅ 1 2 CHAPTER 5. FAMILIES OF ANALYTIC FUNCTIONS 5.1.2 Theorem The assignment (f,g) → d(f,g) defines a metric on C(Ω). Asequence {fj} in C(Ω) is d- convergent (respectively d-Cauchy) iff {fj} is uniformly convergent (respectively uniformly Cauchy) on compact subsets of Ω. Thus (C(Ω),d) and (A(Ω),d) are complete metric spaces. Proof. That d is a metric on C(Ω) is relatively straightforward. The only troublesome part of the argument is verification of the triangle inequality, whose proof uses the inequality: If a, b and c are nonnegative numbers and a ≤ b + c, then a b c ≤ + . 1+a 1+b 1+c To see this, note that h(x)=x/(1 + x) increases with x ≥ 0, and consequently h(a) ≤ b c ≤ b c { } h(b + c)= 1+b+c + 1+b+c 1+b + 1+c . Now let us show that a sequence fj is d-Cauchy iff {fj} is uniformly Cauchy on compact subsets of Ω. Suppose first that {fj} is d-Cauchy, and let K be any compact subset of Ω. By the above property (3) of the sequence {Kn}, we can choose n so large that K ⊆ Kn. Since d(fj,fk) → 0asj, k →∞, the same is true − − ≤ − { } of fj fk Kn . But fj fk K fj fk Kn , hence fj is uniformly Cauchy on K. { } Conversely, assume that fj is uniformly Cauchy on compact subsets of Ω. Let >0 and ∞ −n { } choose a positive integer m such that n=m+1 2 <. Since fj is uniformly Cauchy ≥ − on Km in particular, there exists N = N(m) such that j, k N implies fj fk Km <, hence m m 1 f − f 1 j k Kn ≤ f − f 2n 1+ f − f 2n j k Kn n=1 j k Kn n=1 m 1 ≤ f − f <. j k Km 2n n=1 It follows that for j, k ≥ N, ∞ 1 f − f d(f ,f )= j k Kn < 2. j k 2n 1+ f − f n=1 j k Kn The remaining statements in (5.1.2) follow from the above, Theorem 2.2.17, and com- pleteness of C. ♣ If {fn} is a sequence in A(Ω) and fn → f uniformly on compact subsets of Ω, then we know that f ∈ A(Ω) also. The next few theorems assert that certain other properties of the limit function f may be inferred from the possession of these properties by the fn. The first results of this type relate the zeros of f to those of the fn. 5.1.3 Hurwitz’s Theorem Suppose that {fn} is a sequence in A(Ω) that converges to f uniformly on compact subsets of Ω. Let D(z0,r) ⊆ Ω and assume that f(z) = 0 for |z − z0| = r. Then there is a positive integer N such that for n ≥ N, fn and f have the same number of zeros in D(z0,r). 5.1. THE SPACES A(Ω) AND C(Ω) 3 Proof. Let = min{|f(z)| : |z−z0| = r} > 0. Then for sufficiently large n, |fn(z)−f(z)| < ≤|f(z)| for |z − z0| = r. By Rouch´e’s theorem (4.2.8), fn and f have the same number of zeros in D(z0,r). ♣ 5.1.4 Theorem Let {fn} be a sequence in A(Ω) such that fn → f uniformly on compact subsets of Ω. If Ω is connected and fn has no zeros in Ω for infinitely many n, then either f has no zeros in Ω or f is identically zero. Proof. Assume f is not identically zero, but f has a zero at z0 ∈ Ω. Then by the identity theorem (2.4.8), there is r>0 such that the hypothesis of (5.1.3) is satisfied. Thus for sufficiently large n, fn has a zero in D(z0,r). ♣ 5.1.5 Theorem Let {fn} be a sequence in A(Ω) such that fn converges to f uniformly on compact subsets of Ω. If Ω is connected and the fn are one-to-one on Ω, then either f is constant on Ω or f is one-to-one. Proof. Assume that f is not constant on Ω, and choose any z0 ∈ Ω. The sequence {fn −fn(z0)} satisfies the hypothesis of (5.1.4) on the open connected set Ω\{z0} (because the fn are one-to-one). Since f − f(z0) is not identically zero on Ω \{z0}, it follows from (5.1.4) that f − f(z0) has no zeros in Ω \{z0}. Since z0 is an arbitrary point of Ω, we conclude that f is one-to-one on Ω. ♣ The next task will be to identify the compact subsets of the space A(Ω) (equipped with the topology of uniform convergence on compact subsets of Ω). After introducing the appropriate notion of boundedness for subsets F⊆A(Ω), we show that each sequence of functions in F has a subsequence that converges uniformly on compact subsets of Ω. This leads to the result that a subset of A(Ω) is compact iff it is closed and bounded. 5.1.6 Definition Aset F⊆C(Ω) is bounded if for each compact set K ⊆ Ω, sup{ f K : f ∈F}< ∞, that is, the functions in F are uniformly bounded on each compact subset of Ω. We will also require the notion of equicontinuity for a family of functions. 5.1.7 Definition Afamily F of functions on Ω is equicontinuous at z0 ∈ Ω if given >0 there exists δ>0 such that if z ∈ Ω and |z − z0| <δ, then |f(z) − f(z0)| <for all f ∈F. We have the following relationship between bounded and equicontinuous subsets of A(Ω). 4 CHAPTER 5. FAMILIES OF ANALYTIC FUNCTIONS 5.1.8 Theorem Let F be a bounded subset of A(Ω). Then F is equicontinuous at each point of Ω. Proof. Let z0 ∈ Ω and choose r>0 such that D(z0,r) ⊆ Ω. Then for z ∈ D(z0,r) and f ∈F, we have − 1 f(w) − 1 f(w) f(z) f(z0)= − dw − dw. 2πi C(z0,r) w z 2πi C(z0,r) w z0 Thus 1 f(w) f(w) |f(z) − f(z0)|≤ sup − : w ∈ C(z0,r) 2πr 2π w − z w − z0 f(w) = r|z − z0| sup : w ∈ C(z0,r) . (w − z)(w − z0) But by hypothesis, there exists Mr such that |f(w)|≤Mr for all w ∈ C(z0,r) and all f ∈F. Consequently, if z ∈ D(z ,r/2) and f ∈F, then 0 | − | f(w) ∈ ≤ | − | Mr r z z0 sup : w C(z0,r) r z z0 2 , (w − z)(w − z0) (r/2) proving equicontinuity of F. ♣ We will also need the following general fact about equicontinuous families. 5.1.9 Theorem Suppose F is an equicontinuous subset of C(Ω) (that is, each f ∈F is continuous on Ω and F is equicontinuous at each point of Ω) and {fn} is a sequence from F such that fn converges pointwise to f on Ω. Then f is continuous on Ω and fn → f uniformly on compact subsets of Ω. More generally, if fn → f pointwise on a dense subset of Ω, then fn → f on all of Ω and the same conclusion holds. Proof. Let >0. For each w ∈ Ω, choose a δw > 0 such that |fn(z) − fn(w)| <for each z ∈ D(w, δw) and all n. It follows that |f(z) − f(w)|≤ for all z ∈ D(w, δw), so f is continuous. Let K be any compact subset of Ω. Since {D(w, δw):w ∈ K} is ∈ ⊆∪m an open cover of K, there are w1,... ,wm K such that K j=1D(wj,δwj ). Now choose N such that n ≥ N implies that |f(wj) − fn(wj)| <for j =1,..