Let \(\{f_n\}\) be a sequence of analytic functions on a region \(\Omega\) in \(\mathbb{C}\) and assume that for some function \(g\) on \(\Omega\), \(f_n\to g\) uniformly on compact subsets of \(\Omega\). Show that \(f_n'\to g'\) uniformly on compact subsets of \(\Omega\) and that \(g\) is analytic on \(\Omega\).
Explain what part (a) has to do with differentiation of power series.
Solution
- First, \(g\) is holomorphic. If \(T\) is any triangle whose closure lies in \(\Omega\), then uniform convergence on \(\partial T\) gives \[ \int_{\partial T}g(z)\,dz =\lim_{n\to\infty}\int_{\partial T}f_n(z)\,dz=0. \] Since \(g\) is continuous as a locally uniform limit of continuous functions, Morera’s theorem implies that \(g\) is holomorphic.
Now fix \(K\Subset\Omega\). Choose \(r>0\) such that the closed \(r\)-neighborhood \(K_r\) of \(K\) is compactly contained in \(\Omega\). For \(z\in K\), Cauchy’s formula for derivatives on \(|\zeta-z|=r\) gives \[ f_n'(z)-g'(z) =\frac1{2\pi i}\int_{|\zeta-z|=r} \frac{f_n(\zeta)-g(\zeta)}{(\zeta-z)^2}\,d\zeta. \] Hence \[ |f_n'(z)-g'(z)| \le \frac1r\sup_{\zeta\in K_r}|f_n(\zeta)-g(\zeta)|. \] The right side tends to zero independently of \(z\in K\), proving compact uniform convergence of the derivatives.
- A power series converges uniformly on every closed disk strictly inside its disk of convergence. Apply part (a) to its partial sums. Their derivatives therefore converge locally uniformly to the derivative of the sum, which justifies term-by-term differentiation: \[ \left(\sum_{n=0}^\infty a_nz^n\right)' =\sum_{n=1}^\infty n a_n z^{n-1}. \] The differentiated series has the same radius of convergence.