Kenneth Kuttler

56.1. RUNGE’S THEOREM 1775

Since D contains z /∈Ω, it must contain the component, Hz, determined by this point. Thereason for this is that

Hz ⊆ Ĉ\Ω⊆ Ĉ\Kn

and Hz is connected. Therefore, Hz can only have points in one component of Ĉ\Kn. Sinceit has a point in D, it must therefore, be totally contained in D. This verifies the desiredcondition in the case where ∞ /∈ D.

Now suppose that ∞∈D. ∞ /∈Ω because Ω is given to be a set inC. Letting H∞ denotethe component of Ĉ\Ω determined by ∞, it follows both D and H∞ contain ∞. Therefore,the connected set, H∞ cannot have any points in another component of Ĉ\Kn and it is a setwhich is contained in Ĉ\Kn so it must be contained in D. This proves the lemma.

The following picture is a very simple example of the sort of thing considered byRunge’s theorem. The picture is of a region which has a couple of holes.

a1 a2Ω

However, there could be many more holes than two. In fact, there could be infinitelymany. Nor does it follow that the components of the complement of Ω need to have anyinterior points. Therefore, the picture is certainly not representative.

Theorem 56.1.11 (Runge) Let Ω be an open set, and let A be a set which has one pointin each component of Ĉ \Ω and let f be analytic on Ω. Then there exists a sequence ofrational functions, {Rn} having poles only in A such that Rn converges uniformly to f oncompact subsets of Ω.

Proof: Let Kn be the compact sets of Lemma 56.1.10 where each component of Ĉ\Kn

contains a component of Ĉ\Ω. It follows each component of Ĉ\Kn contains a point of A.Therefore, by Theorem 56.1.9 there exists Rn a rational function with poles only in A suchthat

||Rn− f ||Kn,∞<

1n.

It follows, since a given compact set, K is a subset of Kn for all n large enough, that Rn→ funiformly on K. This proves the theorem.

Corollary 56.1.12 Let Ω be simply connected and f analytic on Ω. Then there exists asequence of polynomials, {pn} such that pn→ f uniformly on compact sets of Ω.

Proof: By definition of what is meant by simply connected, Ĉ \Ω is connected andso there are no bounded components of Ĉ\Ω. Therefore, in the proof of Theorem 56.1.11when you use Theorem 56.1.9, you can always have Rn be a polynomial by Lemma 56.1.8.

56.1. RUNGE’S THEOREM 1775Since D contains z ¢ Q, it must contain the component, H,, determined by this point. Thereason for this is thatH.C C\QCC\K,and H, is connected. Therefore, H, can only have points in one component of c \ Kn. Sinceit has a point in D, it must therefore, be totally contained in D. This verifies the desiredcondition in the case where oo ¢ D.Now suppose that oo € D. «© ¢ Q because Q is given to be a set in C. Letting H.. denotethe component of Cc \ Q determined by , it follows both D and H.. contain ce. Therefore,the connected set, H.. cannot have any points in another component of c \ K, and it is a setwhich is contained in C \ Ky so it must be contained in D. This proves the lemma.The following picture is a very simple example of the sort of thing considered byRunge’s theorem. The picture is of a region which has a couple of holes.ea: 5 ean:However, there could be many more holes than two. In fact, there could be infinitelymany. Nor does it follow that the components of the complement of Q need to have anyinterior points. Therefore, the picture is certainly not representative.Theorem 56.1.11 (Runge) Let Q be an open set, and let A be a set which has one pointin each component of C \ Q and let f be analytic on Q. Then there exists a sequence ofrational functions, {R,} having poles only in A such that Ry, converges uniformly to f oncompact subsets of Q..Proof: Let K,, be the compact sets of Lemma 56.1.10 where each component of Cc \ Kncontains a component of C \ Q. It follows each component of C \ K;, contains a point of A.Therefore, by Theorem 56.1.9 there exists R, a rational function with poles only in A suchthat1R,- <r.Rn — flag <—It follows, since a given compact set, K is a subset of K,, for all n large enough, that R, - funiformly on K. This proves the theorem.Corollary 56.1.12 Let Q be simply connected and f analytic on Q. Then there exists asequence of polynomials, {p,} such that py — f uniformly on compact sets of Q.Proof: By definition of what is meant by simply connected, Cc \ Q is connected andso there are no bounded components of C \ Q. Therefore, in the proof of Theorem 56.1.11when you use Theorem 56.1.9, you can always have R,, be a polynomial by Lemma 56.1.8.