Math 411: Honours Complex Variables - University of Alberta

106 CHAPTER 15. ANALYTIC CONTINUATION ALONG A CURVE
Proof. Let
I = {t ∈ [0,1] : 〈ft〉γ(t) = 〈gt〉γ(t)},
so that 0 ∈ I.
We first claim that I is closed. Let t ∈ I, and let δ > 0 be such that γ(s) ∈ Dt∩Et
and
〈fs〉γ(s) = 〈ft〉γ(s) and 〈gs〉γ(s) = 〈gt〉γ(s)
for all s ∈ [0,1] with |s−t|< δ. Since t ∈ I, there exists s ∈ I with |s−t|< δ. There
is thus a neighbourhood U ⊂ Dt ∩Ds ∩Et ∩Es of γ(s) such that fs(z) = gs(z) for
all z ∈ U by the definition of I. From the choice of δ, we also have—after possibly
making U smaller—that fs(z) = ft(z) and gs(z) = gt(z) for z ∈ U. It follows that
ft(z) = gt(z) for z ∈ U, so that t ∈ I.
Let t0 := supI. Let δ > 0 be such that γ(s) ∈ Dt0 ∩Et0 and
〈fs〉γ(s) = 〈ft0〉γ(s) and 〈gs〉γ(s) = 〈gt0〉γ(s)
for all s ∈ [0,1] with |s−t|< δ. Since I is closed, we have t0 ∈ I and thus ft0(z) =
gt0(z) for all z in some neighbourhood V of γ(t0) contained in Dt0 ∩Et0. It follows
that 〈ft0〉γ(s) = 〈gt0〉γ(s) for all s ∈ [0,1] such that γ(s) ∈ V. For δ > 0 sufficiently
small, we thus have 〈fs〉γ(s) = 〈gs〉γ(s) for any s ∈ [0,1] with |s − t|< δ. It follows
that [0,1] ∩ (t0 − δ,t0 + δ) ⊂ I. Since t0 = supI, this means that t0 = 1, so that
I = [0,1].