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Given that a function $f$ is analytic on $[-1,1]$, that is, for any $s ∈ [−1, 1]$, $f$ has a Taylor series about $s$ that converges to $f$ in a neighborhood of $s$. Can we conclude that $f$ is analytic on some larger region, for example a rectangle with corners $\pm (1+\epsilon)\pm i\epsilon$, or a Bernstein Ellipse (parameterized by some $\rho >1$)?

I think the answer is yes. For each $s \in [-1,1]$ we can find a ball $B(s,r_s)$, these balls form an open subcover of the compact interval $[-1,1]$, giving a finite subcover. I am then not sure how to justify being able to find a $\epsilon$ (or $\rho$) that works throughout the whole of the interval.

Does the same reasoning extend to any closed, bounded subset of the complex plane on which $f$ is analytic?

Thanks

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    $\begingroup$ Your argument about the finite subcover should give you an $\epsilon$: just take the minimum of the finitely many $r_s$ you have remaining. In general, the set on which a function $f$ is analytic is an open set (since it's the union of all the local open balls on which its power series converge), and thus if $f$ is analytic on a closed set then it is automatically analytic on an open neighborhood of that set. $\endgroup$ – Greg Martin May 25 at 20:24
  • $\begingroup$ @GregMartin Thanks for your response. I don't think taking the minimum $r_s$ can always work though? Consider two open balls centred at $\pm a$ that only just overlap at 0. $\endgroup$ – H Park May 25 at 20:34
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    $\begingroup$ @HPark two open balls centered at $\pm a$ can not "just overlap" at $0$, you are thinking of a closed ball. $\endgroup$ – pre-kidney May 26 at 0:30
  • $\begingroup$ @pre-kidney That's not what I meant. Suppose $a, \epsilon' > 0$. They can overlap to give a small interval $(-\epsilon',\epsilon')$ on the real line covered. Taking the point at zero, we can't say that $B(0,a+\epsilon')$ lies within these two balls, as suggested above. $\endgroup$ – H Park May 26 at 10:34
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    $\begingroup$ @HPark fair point, Greg Martin's comment appears incorrect as literally stated, just use the Lebesgue number lemma in this case en.wikipedia.org/wiki/Lebesgue%27s_number_lemma $\endgroup$ – pre-kidney May 26 at 10:53
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Your intuition is correct and the same reasoning extends to any closed and bounded subset of the complex plane on which $f$ is analytic. Indeed, this follows from a routine application of the Lebesgue number lemma, which is exactly the tool you need to extract a single small radius that works uniformly across your finite subcover of the compact set in question.

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