Kenneth Kuttler

1742 CHAPTER 55. COMPLEX MAPPINGS

and let γ (t)≡ z+δ neit , t ∈ [0,2π] . It follows that for z1 ∈ B(z,δ n) , and f ∈F ,

| f (z)− f (z1)| =

∣∣∣∣ 12πi

∫γ

f (w)(

1w− z

− 1w− z1

)dw∣∣∣∣

≤ 12π

∣∣∣∣∫γ

f (w)z− z1

(w− z)(w− z1)dw∣∣∣∣

Letting |z1− z|< δ n2 ,

| f (z)− f (z1)| ≤M2π

2πδ n|z− z1|δ

2n/2

≤ 2M|z− z1|

δ n.

It follows that Fn is equicontinuous and uniformly bounded so by the Arzela Ascoli theo-rem there exists a sequence, { fnk}∞

k=1 ⊆F which converges uniformly on Kn. Let { f1k}∞

k=1converge uniformly on K1. Then use the Arzela Ascoli theorem applied to this sequence toget a subsequence, denoted by { f2k}∞

k=1 which also converges uniformly on K2. Continuein this way to obtain { fnk}∞

k=1 which converges uniformly on K1, · · · ,Kn. Now the sequence{ fnn}∞

n=m is a subsequence of { fmk} ∞k=1 and so it converges uniformly on Km for all m. De-

noting fnn by fn for short, this is the sequence of functions promised by the theorem. It isclear { fn}∞

n=1 converges uniformly on every compact subset of Ω because every such setis contained in Km for all m large enough. Let f (z) be the point to which fn (z) converges.Then f is a continuous function defined on Ω. Is f is analytic? Yes it is by Lemma 51.3.13.Alternatively, you could let T ⊆Ω be a triangle. Then∫

∂Tf (z)dz = lim

n→∞

∫∂T

fn (z)dz = 0.

Therefore, by Morera’s theorem, f is analytic.As for the uniform convergence of the derivatives of f , recall Theorem 51.7.25 about

the existence of a cycle. Let K be a compact subset of int(Kn) and let {γk}mk=1 be closed

oriented curves contained inint(Kn)\K

such that ∑mk=1 n(γk,z) = 1 for every z ∈ K. Also let η denote the distance between ∪ jγ

∗j

and K. Then for z ∈ K,∣∣∣ f (k) (z)− f (k)n (z)∣∣∣ =

∣∣∣∣∣ k!2πi

∑j=1

∫γ j

f (w)− fn (w)

(w− z)k+1 dw

∣∣∣∣∣≤ k!

2π|| fk− f ||Kn

∑j=1

(length of γk)1

ηk+1 .

where here || fk− f ||Kn≡ sup{| fk (z)− f (z)| : z ∈ Kn} . Thus you get uniform convergence

of the derivatives.Since the family, F satisfies the conclusion of Theorem 55.3.1 it is known as a normal

family of functions. More generally,

1742 CHAPTER 55. COMPLEX MAPPINGSand let y(t) =z+6,e",t € [0,27]. It follows that for z, € B(z,6,), and f € F,1 I IIf(2)-fla)l = smi [10 (jest) am1 Z-Z1< aL Gogeaa™Letting |z; —z| < $n,M |z—z1|If(@)—F)l S57 2m6n 52/2|z—z1|< 2M 3It follows that .F,, is equicontinuous and uniformly bounded so by the Arzela Ascoli theo-rem there exists a sequence, { fx };_; C ¥ which converges uniformly on Ky. Let {fix }c_1converge uniformly on K;. Then use the Arzela Ascoli theorem applied to this sequence toget a subsequence, denoted by { f2x };_, which also converges uniformly on K2. Continuein this way to obtain { f,4};_, which converges uniformly on Kj,--- ,K,. Now the sequence{fn };—m iS a subsequence of { fink} £1 and so it converges uniformly on K,,, for all m. De-noting fnn by f, for short, this is the sequence of functions promised by the theorem. It isclear {f,};, converges uniformly on every compact subset of Q because every such setis contained in K,, for all m large enough. Let f (z) be the point to which f,, (z) converges.Then f is a continuous function defined on Q. Is f is analytic? Yes it is by Lemma 51.3.13.Alternatively, you could let T C Q be a triangle. Then[, f@ae= lim | fulg)dz=0.Therefore, by Morera’s theorem, f is analytic.As for the uniform convergence of the derivatives of f, recall Theorem 51.7.25 aboutthe existence of a cycle. Let K be a compact subset of int(K,) and let {y,}7, be closedoriented curves contained inint (K,) \Ksuch that 7.) 1(¥%,z) = 1 for every z € K. Also let 7 denote the distance between Uj;and K. Then for z € K,KS f fr) =f)ye dw2ni Jy; (w—zIATILA Fllay ¥, (length of 7)55 Wk — Sk, ength of ¥;20 ~1J nit .Jwhere here || fx — f||x, = sup {|x (z) — f (z)|: z © Kn}. Thus you get uniform convergenceof the derivatives.Since the family, ¥ satisfies the conclusion of Theorem 55.3.1 it is known as a normalfamily of functions. More generally,