1776 CHAPTER 56. APPROXIMATION BY RATIONAL FUNCTIONS

56.2 The Mittag-Leffler Theorem56.2.1 A Proof From Runge’s TheoremThis theorem is fairly easy to prove once you have Theorem 56.1.9. Given a set of complexnumbers, does there exist a meromorphic function having its poles equal to this set ofnumbers? The Mittag-Leffler theorem provides a very satisfactory answer to this question.Actually, it says somewhat more. You can specify, not just the location of the pole but alsothe kind of singularity the meromorphic function is to have at that pole.

Theorem 56.2.1 Let P ≡ {zk}∞

k=1 be a set of points in an open subset of C, Ω. Supposealso that P⊆ Ω ⊆ C. For each zk, denote by Sk (z) a function of the form

Sk (z) =mk

∑j=1

akj

(z− zk)j .

Then there exists a meromorphic function, Q defined on Ω such that the poles of Q are thepoints, {zk}∞

k=1 and the singular part of the Laurent expansion of Q at zk equals Sk (z) . Inother words, for z near zk,Q(z) = gk (z)+Sk (z) for some function, gk analytic near zk.

Proof: Let {Kn} denote the sequence of compact sets described in Lemma 56.1.10.Thus ∪∞

n=1Kn = Ω, Kn ⊆ int(Kn+1) ⊆ Kn+1 · · · , and the components of Ĉ\Kn contain thecomponents of Ĉ \Ω. Renumbering if necessary, you can assume each Kn ̸= /0. Also letK0 = /0. Let Pm ≡ P∩ (Km \Km−1) and consider the rational function, Rm defined by

Rm (z)≡ ∑zk∈Km\Km−1

Sk (z) .

Since each Km is compact, it follows Pm is finite and so the above really is a rational func-tion. Now for m > 1,this rational function is analytic on some open set containing Km−1.There exists a set of points, A one point in each component of Ĉ \Ω. Consider Ĉ \Km−1.Each of its components contains a component of Ĉ\Ω and so for each of these componentsof Ĉ \Km−1, there exists a point of A which is contained in it. Denote the resulting set ofpoints by A′. By Theorem 56.1.9 there exists a rational function, Qm whose poles are allcontained in the set, A′ ⊆ΩC such that

||Rm−Qm||Km−1,∞<

12m .

The meromorphic function is

Q(z)≡ R1 (z)+∞

∑k=2

(Rk (z)−Qk (z)) .

It remains to verify this function works. First consider K1. Then on K1, the above sumconverges uniformly. Furthermore, the terms of the sum are analytic in some open setcontaining K1. Therefore, the infinite sum is analytic on this open set and so for z ∈ K1