1768 CHAPTER 56. APPROXIMATION BY RATIONAL FUNCTIONS

This theorem implies the following.

Theorem 56.1.3 Let K ⊆Ω where K is compact and Ω is open. Then there exist orientedclosed curves, γk such that γ∗k ∩K = /0 but γ∗k ⊆Ω, such that for all z ∈ K,

f (z) =1

2πi

p

∑k=1

∫γk

f (w)w− z

dw. (56.1.1)

Proof: This follows from Theorem 51.7.25 and the Cauchy integral formula. As shownin the proof, you can assume the γk are linear mappings but this is not important.

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

Lemma 56.1.4 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 ||K,∞ < ε .

Proof: By Theorem 56.1.3 there are oriented curves, γk described there such that forall z ∈ K,

f (z) =1

2πi

p

∑k=1

∫γk

f (w)w− z

dw. (56.1.2)

Defining g(w,z) ≡ f (w)w−z for (w,z) ∈ ∪p

k=1γ∗k ×K, it follows since the distance between Kand ∪kγ∗k is positive that g is uniformly continuous and so there exists a δ > 0 such that if||P||< δ , then for all z ∈ K,∣∣∣∣∣ f (z)− 1

2πi

p

∑k=1

n

∑j=1

f (γk (τ j))(γk (ti)− γk (ti−1))

γk (τ j)− z

∣∣∣∣∣< ε

2.

The complicated expression is obtained by replacing each integral in 56.1.2 with a Riemannsum. Simplifying the appearance of this, it follows there exists a rational function of theform

R(z) =M

∑k=1

Ak

wk− z

where the wk are elements of components of C \K and Ak are complex numbers or in thecase where f has values in X , these would be elements of X such that

||R− f ||K,∞ <ε

2.

This proves the lemma.

56.1.2 Moving The Poles And Keeping The ApproximationLemma 56.1.4 is a nice lemma but needs refining. In this lemma, the Riemann sum handedyou the poles. It is much better if you can pick the poles. The following theorem fromadvanced calculus, called Merten’s theorem, will be used

1768 CHAPTER 56. APPROXIMATION BY RATIONAL FUNCTIONSThis theorem implies the following.Theorem 56.1.3 Let K C Q where K is compact and Q is open. Then there exist orientedclosed curves, Y;, such that y,K = 90 but yj, © Q, such that for allz € K,r@=s5y [|20i (Jy, WZdw. (56.1.1)Proof: This follows from Theorem 51.7.25 and the Cauchy integral formula. As shownin the proof, you can assume the y, are linear mappings but this is not important.Next I will show how the Cauchy integral formula leads to approximation by rationalfunctions, quotients of polynomials.Lemma 56.1.4 Let K be a compact subset of an open set, Q and let f be analytic on Q.Then there exists a rational function, Q whose poles are not in K such thatIlQ— fl lx. <€.Proof: By Theorem 56.1.3 there are oriented curves, y, described there such that forallz EK,= — dw. (56.1.2)201 (Jy, WZDefining g(w,z) = om) for (w,z) € Up_1¥% x K, it follows since the distance between Kand U;¥% is positive that g is uniformly continuous and so there exists a 6 > 0 such that if||P || < 6, then for all zE K,<y Sf (Me (Ti)) (Ye (ti) = Ye i-1)1 €f 2)-aS 5°®) 2nt % (Tj) —Z 2The complicated expression is obtained by replacing each integral in 56.1.2 with a Riemannsum. Simplifying the appearance of this, it follows there exists a rational function of theform MAkR(z)=@= yewhere the w; are elements of components of C \ K and Ax are complex numbers or in thecase where f has values in X, these would be elements of X such thatER- <x.IR— Flaw <5This proves the lemma.56.1.2 Moving The Poles And Keeping The ApproximationLemma 56.1.4 is a nice lemma but needs refining. In this lemma, the Riemann sum handedyou the poles. It is much better if you can pick the poles. The following theorem fromadvanced calculus, called Merten’s theorem, will be used