Let $ f: [0,1] \rightarrow \mathbb{R} $ be analytic. Let $ r_f(x) $ be the radius of convergence of $ f $ at $ x $. Is $ r_x(f) $ continuous?
Alternatively, is there an $ r_{min} $ I can choose so that the power series of $ f $ about $ x $ converges in $ (x-r_{min},\; x+ r_{min}) $ for all $ x $. Obviously if $ r_f(x) $ is continuous, then this will be true.
Also does this hold in higher dimensions, ie $ f: [0,1] \times [0,1] \rightarrow \mathbb{R} $?
Thank you for reading!
If the power series at $x_0$ has radius of convergence $>r$, then the same is true in a neighborhood of $x_0$ (you can look at the standard proof of the fact that a power series at $x_0$ with positive radius of convergence is analytic at any point of the open interval of convergence).
Thus $r_f(x)$ is lower semicontinuous, hence it has a (positive) minimum on $[0,1]$ since it is compact. The same reasoning works for higher dimensions.
If you want to prove the full continuity of $r_f(x)$, the only proof I know relies on (basic) complex analysis.