The theorem states that if $f$ and $g$ are analytic functions and their values agree on an open set that is contained in a larger, connected domain, then $f$ must equal $g$ on the entire domain. (The domain need not be simply connected, which is the part that confuses me most.)

Instead of the set-theoretic proof (the clopen set proof), can we understand it from the point of view of power series expansions?

Let $f = g$ on an some open set $S$, which is contained in larger connected domain $\Omega$. $f$ and $g$ are analytic on this domain.

Then look at the function $f-g = 0$ on this open set. Its Taylor expansion is identically zero.

Can I now use this argument for the whole domain? And this is what is challenging - there can be a hole in the domain. So, I can't argue "$f-g = 0$ converges in the largest Taylor disk possible, so its Taylor series is identically zero everywhere in the domain; therefore, $f=g$ everywhere on the domain".


EDIT: I'm sketching some small, overlapping disks of convergence of $f-g = 0$, on a connected domain with a hole in it. If I start at the first known disk of convergence, the disk that overlaps with the first disk must also have all zero terms in its Taylor expansion, i.e., $f-g$ is again identically zero, but we've extended the region in the domain where we know the values of $f-g = 0$. And iterate this process, until we have enough overlapping disks to cover the connected domain - all disks will have Taylor expansion identically zero.

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    $\begingroup$ Are you aware of analytic continuation? What you're suggesting is very closely related to that. $\endgroup$ – Cameron Williams Aug 2 '15 at 1:57
  • $\begingroup$ Hi @CameronWilliams - I just edited my post, which is probably exactly what you're referring to. Does my sketch of the (visual) proof...work? $\endgroup$ – user258344 Aug 2 '15 at 1:59
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    $\begingroup$ More or less, that is the right idea. Formalizing it is a much more arduous task however. $\endgroup$ – Cameron Williams Aug 2 '15 at 1:59
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    $\begingroup$ You're welcome! I presume you're using Greene and Krantz. Chapter 10 deals exactly with what you're interested in. $\endgroup$ – Cameron Williams Aug 2 '15 at 2:01
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    $\begingroup$ In that clopen set proof, the proof that said set is open uses power series. At least the proof I know does... $\endgroup$ – David C. Ullrich Aug 2 '15 at 3:39

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