I was given the definition of anlytic continuation as in Stein's book- given two regions $\Omega\subset \Omega'$ for analytic functions $F:\Omega'\to \mathbb{C}$ is an analytic continuation of $f:\Omega \to \mathbb{C}$ if $F$ agrees with $f$ on $\Omega$. I know that $\sin:\mathbb{C}\to \mathbb{C}$ is an analytic continuation of $\sin:\mathbb{R}\to \mathbb{R}$ because they agree on $\mathbb{R}$, but this is not a region in $\mathbb{C}$, so how does that add up with the defintion?


There is some ambiguity here (not in what you wrote, which is perfectly clear, but in Stein's textbook). Sometimes, while writing about analytic continuation, Stein indeed assumes that $\Omega$ and $\Omega'$ are regions. But when finally Stein states a theorem about analytic continuation, the statement is:

Theorem: Suppose $f$ is a holomorphic function in a region $\Omega$ that vanishes on a sequence of distinct points with a limit point in $\Omega$. Then $f$ is identically $0$.

As you can see, when it is stated like this, the theorem on analytic continuation applies indeed to the situation that you described.

  • $\begingroup$ I think it answered my question; We know that $\sin(z)$ agrees with $\sin(x)$ on $\mathbb{R}$, and I want to show that this extantion is unique- this is where I use the theorem you mentioned, correct? Thanks for your answer! $\endgroup$ – Simon Green Jan 18 at 11:10
  • 1
    $\begingroup$ Yes, that is correct. $\endgroup$ – José Carlos Santos Jan 18 at 11:20

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