1762 CHAPTER 55. COMPLEX MAPPINGS

Lemma 55.5.16 Let Ω be a region and let F be a set of functions analytic on Ω none ofwhich achieve the two distinct values, a and b. If { fn} ⊆F then one of the following hold:Either there exists a function, f analytic on Ω and a subsequence,

{fnk

}such that for any

compact subset, K of Ω,limk→∞

∣∣∣∣ fnk − f∣∣∣∣

K,∞= 0. (55.5.29)

or there exists a subsequence{

fnk

}such that for all compact subsets K,

limk→∞

ρK(

fnk ,∞)= 0. (55.5.30)

Proof: Let B(z0,2R) ⊆ Ω. There are two cases to consider. The first case is thatthere exists a subsequence, nk such that

{fnk (z0)

}is bounded. The second case is that

limn→∞

∣∣ fnk (z0)∣∣= ∞.

Consider the first case. By Theorem 55.5.8{

fnk (z)}

is uniformly bounded on B(z0,R)because by this theorem, and letting θ = 1/2 applied to B(z0,2R) , it follows

∣∣ fnk (z)∣∣ ≤

M(a,b, 1

2 ,β)

where β is an upper bound to the numbers,∣∣ fnk (z0)

∣∣. The Cauchy integralformula implies the existence of a uniform bound on the

{f ′nk

}which implies the func-

tions are equicontinuous and uniformly bounded. Therefore, by the Ascoli Arzela theoremthere exists a further subsequence which converges uniformly on B(z0,R) to a function, fanalytic on B(z0,R). Thus denoting this subsequence by

{fnk

}to save on notation,

limk→∞

∣∣∣∣ fnk − f∣∣∣∣

B(z0,R),∞= 0. (55.5.31)

Consider the second case. In this case, it follows {1/ fn (z0)} is bounded on B(z0,R) andso by the same argument just given {1/ fn (z)} is uniformly bounded on B(z0,R). Therefore,a subsequence converges uniformly on B(z0,R). But {1/ fn (z)} converges to 0 and so thisrequires that {1/ fn (z)} must converge uniformly to 0. Therefore,

limk→∞

ρB(z0,R)

(fnk ,∞

)= 0. (55.5.32)

Now let {Dk} denote a countable set of closed balls, Dk = B(zk,Rk) such that

B(zk,2Rk)⊆Ω

and ∪∞k=1 int(Dk) = Ω. Using a Cantor diagonal process, there exists a subsequence,

{fnk

}of { fn} such that for each D j, one of the above two alternatives holds. That is, either

limk→∞

∣∣∣∣ fnk −g j∣∣∣∣

D j ,∞= 0 (55.5.33)

or,limk→∞

ρD j

(fnk ,∞

). (55.5.34)

Let A ={∪ int(D j) : 55.5.33 holds

}, B =

{∪ int(D j) : 55.5.34 holds

}. Note that the balls

whose union is A cannot intersect any of the balls whose union is B. Therefore, one of A orB must be empty since otherwise, Ω would not be connected.

If K is any compact subset of Ω, it follows K must be a subset of some finite collectionof the D j. Therefore, one of the alternatives in the lemma must hold. That the limit func-tion, f must be analytic follows easily in the same way as the proof in Theorem 55.3.1 onPage 1741. You could also use Morera’s theorem. This proves the lemma.