Suppose \(f\) is analytic on a region \(\Omega\) such that \(\DD \subseteq \Omega \subseteq \CC\) and \(f(z) = \sum_{n=0}^\infty a_n z^n\) is a power series with radius of convergence exactly 1.
Give an example of such an \(f\) that converges at every point of \(S^1\).
Give an example of such an \(f\) which is analytic at \(1\) but \(\sum_{n=0}^\infty a_n\) diverges.
Prove that \(f\) can not be analytic at every point of \(S^1\).
Solution
For (a), take \[ f(z)=\sum_{n=1}^\infty {z^n\over n^2}. \] Its radius of convergence is \(1\), since \[ \limsup_{n\to\infty}\left({1\over n^2}\right)^{1/n}=1. \] For every \(|z|=1\) the series converges absolutely by comparison with \(\sum n^{-2}\).
For (b), take \[ f(z)={1\over1+z}=\sum_{n=0}^\infty(-1)^nz^n. \] The Taylor series at \(0\) has radius exactly \(1\), because the nearest singularity is at \(z=-1\). But \(f\) is holomorphic in a neighborhood of \(z=1\), while \[ \sum_{n=0}^\infty a_n =\sum_{n=0}^\infty(-1)^n \] diverges.
For (c), suppose instead that \(f\) were analytic at every point of \(S^1\) as well as on \(\DD\). Then the closed unit disk \(\overline\DD\) would be a compact subset of the open domain of analyticity. Hence there is \(\varepsilon>0\) such that \[ \{z:|z|<1+\varepsilon\} \] is still contained in that domain. The Taylor series of \(f\) at \(0\) would therefore have radius of convergence at least \(1+\varepsilon\), contradicting the hypothesis that its radius is exactly \(1\).