55.4. ANALYTIC CONTINUATION 1749

55.4.2 Continuation Along A CurveNext I will describe what is meant by continuation along a curve. The following definitionis standard and is found in Rudin [113].

Definition 55.4.3 A function element is an ordered pair, ( f ,D) where D is an open balland f is analytic on D. ( f0,D0) and ( f1,D1) are direct continuations of each other ifD1 ∩D0 ̸= /0 and f0 = f1 on D1 ∩D0. In this case I will write ( f0,D0) ∼ ( f1,D1) . Achain is a finite sequence, of disks, {D0, · · · ,Dn} such that Di−1 ∩Di ̸= /0. If ( f0,D0) is agiven function element and there exist function elements, ( fi,Di) such that {D0, · · · ,Dn}is a chain and

(f j−1,D j−1

)∼ ( f j,D j) then ( fn,Dn) is called the analytic continuation of

( f0,D0) along the chain {D0, · · · ,Dn}. Now suppose γ is an oriented curve with parameterinterval [a,b] and there exists a chain, {D0, · · · ,Dn} such that γ∗ ⊆ ∪n

k=1Dk,γ (a) is thecenter of D0,γ (b) is the center of Dn, and there is an increasing list of numbers in [a,b] ,a=s0 < s1 · · · < sn = b such that γ ([si,si+1]) ⊆ Di and ( fn,Dn) is an analytic continuation of( f0,D0) along the chain. Then ( fn,Dn) is called an analytic continuation of ( f0,D0) alongthe curve γ . (γ will always be a continuous curve. Nothing more is needed. )

In the above situation it does not follow that if Dn∩D0 ̸= /0, that fn = f0! However, thereare some cases where this will happen. This is the monodromy theorem which follows.This is as far as I will go on the subject of analytic continuation. For more on this subjectincluding a development of the concept of Riemann surfaces, see Alfors [3].

Lemma 55.4.4 Suppose ( f ,B(0,r)) for r < 1 is a function element and ( f ,B(0,r)) can beanalytically continued along every curve in B(0,1) that starts at 0. Then there exists ananalytic function, g defined on B(0,1) such that g = f on B(0,r) .

Proof: Let

R = sup{r1 ≥ r such that there exists gr1

analytic on B(0,r1) which agrees with f on B(0,r) .}

Define gR (z)≡ gr1 (z) where |z|< r1. This is well defined because if you use r1 and r2, bothgr1 and gr2 agree with f on B(0,r), a set with a limit point and so the two functions agree atevery point in both B(0,r1) and B(0,r2). Thus gR is analytic on B(0,R) . If R < 1, then bythe assumption there are no singular points on B(0,R) and so Theorem 55.4.2 implies theradius of convergence of the power series for gR is larger than R contradicting the choiceof R. Therefore, R = 1 and this proves the lemma. Let g = gR.

The following theorem is the main result in this subject, the monodromy theorem.

Theorem 55.4.5 Let Ω be a simply connected proper subset of C and suppose ( f ,B(a,r))is a function element with B(a,r) ⊆ Ω. Suppose also that this function element can beanalytically continued along every curve through a. Then there exists G analytic on Ω suchthat G agrees with f on B(a,r).

Proof: By the Riemann mapping theorem, there exists h : Ω→ B(0,1) which is ana-lytic, one to one and onto such that f (a) = 0. Since h is an open map, there exists δ > 0such that

B(0,δ )⊆ h(B(a,r)) .