Analytic Continuations to Two Sets do not necessarily Agree on Intersection

Theorem

Let $U \subseteq \C$ be an open subset of the complex plane.

Let $V_1 \subset \C$ and $V_2 \subset \C$ also be open subsets of the complex plane such that $U \subseteq V_1$ and $U \subseteq V_2$.

Let $f: U \to \C$ be a function defined on $U$.

Let $F_1: V_1 \to \C$ and $F_2: V_2 \to \C$ be the analytic continuations of $f$ to $V_1$ and $V_2$ respectively.

Then it is not necessarily the case that $F_1$ and $F_2$ agree on $V_1 \cap V_2$.

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Proof

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Also see

• Uniqueness of Analytic Continuation

Sources

• 2021: Richard Earl and James Nicholson: The Concise Oxford Dictionary of Mathematics (6th ed.) ... (previous) ... (next): analytic continuation