# Extension of function analytic on [-1,1]

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

• 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. – Greg Martin May 25 at 20:24
• @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. – H Park May 25 at 20:34
• @HPark two open balls centered at $\pm a$ can not "just overlap" at $0$, you are thinking of a closed ball. – pre-kidney May 26 at 0:30
• @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. – H Park May 26 at 10:34
• @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 – pre-kidney May 26 at 10:53

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.