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I wasn't very well introduced to Analytic Continuations, but from what I have seen, showing that the analytic continuation is unique is pretty simple. In Real Analysis, from what I can imagine, there isn't a unique analytic continuation of a function, is that true? If so, why is it that we can so easily restrict the continuation of a complex function to be one and only one thing?

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    $\begingroup$ The identity theorem relies on the complex differentiability. The real question is as to whether the function is analytic, not just differentiable. In the complex case, the two are synonymous. Not so in the real case. $\endgroup$ – Adam Hughes Jul 9 '14 at 0:00
  • $\begingroup$ In real analysis, are analytic continuations of analytic functions unique? $\endgroup$ – user82004 Jul 9 '14 at 0:48
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    $\begingroup$ See unique continuation in this. $\endgroup$ – Tony Piccolo Jul 9 '14 at 7:48
  • $\begingroup$ @AdamHughes Can you point me to a proof that an analytic function (over reals) need not be differentiable. Is analytic same as continuous? (My math teachers were terrible) $\endgroup$ – Jus12 Jul 28 '15 at 2:00
  • $\begingroup$ @jus12 analytic means it has a power series. Necessarily this converges uniformly on compact sets, hence you can differentiate term by term. This implies differentiability in all cases. $\endgroup$ – Adam Hughes Jul 28 '15 at 2:07

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