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 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.