Suppose \(f\) is analytic on a region \(\Omega\) in \(\CC\) containing the open unit disk \(\DD\) and we have \(f(z) = \sum a_n z^n\) with this power series having radius of convergence \(1\).
Give an example of such an \(f\) so that the series converges at every point on the unit circle \(\mathbb{T}\).
Give an example of such an \(f\) which is analytic at \(1\), but \(\sum a_n\) diverges.
Prove that \(f\) cannot be analytic at each point of \(\mathbb{T}\).
Solution
1. For part (a), one example is\[ \boxed{ f_a(z)=\sum_{n=0}^{\infty}\frac{z^n}{(n+1)^2}. } \]
If \(\abs{z}<1\), the series converges absolutely by comparison with the geometric series \(\sum \abs{z}^n\). If \(\abs{z}>1\), then \[ \frac{\abs{z}^{n+1}/(n+2)^2} {\abs{z}^n/(n+1)^2} = \abs{z}\left(\frac{n+1}{n+2}\right)^2 \longrightarrow \abs{z}>1, \] so the terms \(\abs{z}^n/(n+1)^2\) do not tend to zero. Thus its radius of convergence is exactly \(1\), and the sum defines an analytic function on the open unit disk.
If \(\abs{z}=1\), then \[ \sum_{n=0}^{\infty} \abs{\frac{z^n}{(n+1)^2}} = \sum_{n=0}^{\infty}\frac1{(n+1)^2} <\infty. \] Hence the power series converges at every point of the unit circle.
2. For part (b), one example is\[ \boxed{ f_b(z)=\frac1{1+z}. } \]
The function \(f_b\) is analytic on \[ \CC\sm\{-1\}, \] a region containing the open unit disk, and in particular it is analytic at \(z=1\). At the origin, \[ f_b(z) = \sum_{n=0}^{\infty}(-1)^n z^n \] for \(\abs{z}<1\). The nearest singularity to the origin is \(-1\), so this Taylor series has radius of convergence \(1\). Its coefficient sum is \[ \sum_{n=0}^{\infty}a_n = \sum_{n=0}^{\infty}(-1)^n, \] whose partial sums alternate between \(1\) and \(0\). Thus the coefficient sum diverges even though \(f_b\) is analytic at \(1\).
3. For part (c), \(f\) cannot be analytic at every point of the unit circle.
Suppose instead that \(f\) were analytic at every point of the unit circle. Together with the given analyticity on a region containing the open unit disk, this gives an open set \(U\) on which \(f\) is analytic and which contains the closed unit disk \[ K=\{z\in\CC:\abs{z}\leq1\}. \] Since \(K\) is compact and \(U\) is open, there is an \(\varepsilon>0\) such that every point at distance less than \(\varepsilon\) from \(K\) lies in \(U\). Consequently, \[ \{z\in\CC:\abs{z}<1+\varepsilon\} \subseteq U. \]
The Taylor series of a function analytic on the disk \(\abs{z}<1+\varepsilon\) converges there to the function. Hence the Taylor series of \(f\) at the origin has radius of convergence at least \(1+\varepsilon\), contradicting the hypothesis that its radius is exactly \(1\).
4. Q.E.D.
Steps 1 and 2 give the requested examples, and step 3 proves the impossibility in part (c).