Topics In Analysis

55.1.1 Approximation With Rational Functions

It turns out you can approximate analytic functions by rational functions, quotients of polynomials. The resulting theorem is one of the most profound theorems in complex analysis. The basic idea is simple. The Riemann sums for the Cauchy integral formula are rational functions. The idea used to implement this observation is that if you have a compact subset, K of an open set, Ω there exists a cycle composed of closed oriented curves

_j=1ⁿ which are contained in Ω ∖ K such that for every z ∈ K,∑ _k=1ⁿn

= 1. One more ingredient is needed and this is a theorem which lets you keep the approximation but move the poles.

To begin with, consider the part about the cycle of closed oriented curves. Recall Theorem 50.7.25 which is stated for convenience.

Theorem 55.1.2 Let K be a compact subset of an open set, Ω. Then there exist continuous, closed, bounded variation oriented curves

_j=1^m for which γ_j^∗∩ K = ∅ for each j, γ_j^∗⊆ Ω, and for all p ∈ K,

\summ n (p,γk) = 1. k=1

and

\summ n (z,γk) = 0 k=1

for all z

Ω.

This theorem implies the following.

Theorem 55.1.3 Let K ⊆ Ω where K is compact and Ω is open. Then there exist oriented closed curves, γ_k such that γ_k^∗∩ K = ∅ but γ_k^∗⊆ Ω, such that for all z ∈ K,

\sump \int f (z) = -1- f (w)dw. (55.1.1) 2πik=1 γk w- z

(55.1.1)

Proof: This follows from Theorem 50.7.25 and the Cauchy integral formula. As shown in the proof, you can assume the γ_k are linear mappings but this is not important.

Next I will show how the Cauchy integral formula leads to approximation by rational functions, quotients of polynomials.

Lemma 55.1.4 Let K be a compact subset of an open set, Ω and let f be analytic on Ω. Then there exists a rational function, Q whose poles are not in K such that

||Q - f|| < ε. K,\infty

Proof: By Theorem 55.1.3 there are oriented curves, γ_k described there such that for all z ∈ K,

\int -1-\sump -f (w) f (z) = 2πi γkw - zdw. (55.1.2) k=1

(55.1.2)

Defining g

≡

for

∈∪_k=1^pγ_k^∗×K, it follows since the distance between K and ∪_kγ_k^∗ is positive that g is uniformly continuous and so there exists a δ > 0 such that if

< δ, then for all z ∈ K,

| | || 1 \sump \sumn f (γ (τ))(γ (t)- γ (t ))|| ε ||f (z)- --- ---k--j---k--i---k--i--1-||< -. | 2πik=1j=1 γk (τj)- z | 2

The complicated expression is obtained by replacing each integral in 55.1.2 with a Riemann sum. Simplifying the appearance of this, it follows there exists a rational function of the form

M R (z) = \sum --Ak-- k=1wk - z

where the w_k are elements of components of ℂ ∖ K and A_k are complex numbers or in the case where f has values in X, these would be elements of X such that

ε ||R - f||K,\infty < 2.

This proves the lemma.