Let \(a_n(z)\) be an analytic sequence in a domain \(D\) such that \(\displaystyle \sum_{n=0}^\infty |a_n(z)|\) converges uniformly on bounded and closed sub-regions of \(D\). Show that \(\displaystyle \sum_{n=0}^\infty |a'_n(z)|\) converges uniformly on bounded and closed sub-regions of \(D\).
Solution
Let \(K\Subset D\) be compact. Choose \(\rho>0\) so that the closed \(\rho\)-neighborhood \[ K_\rho=\{w:\operatorname{dist}(w,K)\le\rho\} \] is compactly contained in \(D\). For \(z\in K\), Cauchy’s formula on \(|w-z|=\rho\) gives \[ |a_n'(z)| \le \frac1{2\pi\rho} \int_{|w-z|=\rho}|a_n(w)|\,|dw|. \] Therefore for integers \(N\le M\), \[ \sum_{n=N}^M|a_n'(z)| \le \frac1{2\pi\rho} \int_{|w-z|=\rho} \sum_{n=N}^M|a_n(w)|\,|dw|. \] Since \(\sum |a_n|\) converges uniformly on \(K_\rho\), its tails are uniformly small there. Hence \[ \sup_{z\in K}\sum_{n=N}^M|a_n'(z)|\to0 \qquad(N,M\to\infty). \] Thus \(\sum |a_n'|\) is uniformly Cauchy, hence uniformly convergent, on \(K\). Since \(K\Subset D\) was arbitrary, the conclusion holds on every bounded closed subregion compactly contained in \(D\).