56.2. THE MITTAG-LEFFLER THEOREM 1781

5. Every function analytic on Ω can be uniformly approximated by polynomials on com-pact subsets.

6. For every f analytic on Ω and every closed continuous bounded variation curve, γ,∫γ

f (z)dz = 0.

7. Every function analytic on Ω has a primitive on Ω.

8. If f ,1/ f are both analytic on Ω, then there exists an analytic, g on Ω such thatf = exp(g) .

9. If f ,1/ f are both analytic on Ω, then there exists φ analytic on Ω such that f = φ2.

Proof: 1⇒2. Assume 1 and let γ be a closed curve in Ω. Let h be the homeomorphism,h : B(0,1)→Ω. Let H (α, t) = h

(α(h−1γ (t)

)). This works.

2⇒3 This is Lemma 56.2.6.3⇒4. Suppose 3 but 4 fails to hold. Then if Ĉ\Ω is not connected, there exist disjoint

nonempty sets, A and B such that A∩B = A∩B = /0. It follows each of these sets must beclosed because neither can have a limit point in Ω nor in the other. Also, one and only oneof them contains ∞. Let this set be B. Thus A is a closed set which must also be bounded.Otherwise, there would exist a sequence of points in A, {an} such that limn→∞ an = ∞

which would contradict the requirement that no limit points of A can be in B. Therefore, Ais a compact set contained in the open set, BC ≡ {z ∈ C : z /∈ B} . Pick p ∈ A. By Lemma56.2.4 there exist continuous bounded variation closed curves {Γk}m

k=1 which are containedin BC, do not intersect A and such that

1 =m

∑k=1

n(p,Γk)

However, if these curves do not intersect A and they also do not intersect B then theymust be all contained in Ω. Since p /∈ Ω, it follows by 3 that for each k, n(p,Γk) = 0, acontradiction.

4⇒5 This is Corollary 56.1.12 on Page 1775.5⇒6 Every polynomial has a primitive and so the integral over any closed bounded

variation curve of a polynomial equals 0. Let f be analytic on Ω. Then let { fn} be asequence of polynomials converging uniformly to f on γ∗. Then

0 = limn→∞

∫γ

fn (z)dz =∫

γ

f (z)dz.

6⇒7 Pick z0 ∈ Ω. Letting γ (z0,z) be a bounded variation continuous curve joining z0to z in Ω, you define a primitive for f as follows.

F (z) =∫

γ(z0,z)f (w)dw.