Let \((f_n)\) be a sequence of holomorphic functions in a domain \(D\). Suppose that \(f_n \to f\) uniformly on each compact subset of \(D\). Show that
\(f\) is holomorphic on \(D\).
\(f_n' \to f'\) uniformly on each compact subset of \(D\).
Solution
Fix a closed triangle \(T\subset D\). Since \(f_n\to f\) uniformly on the compact set \(\partial T\), \[ \int_{\partial T}f(z)\,dz =\lim_{n\to\infty}\int_{\partial T}f_n(z)\,dz=0. \] The locally uniform limit \(f\) is continuous, so Morera’s theorem implies that \(f\) is holomorphic on \(D\).
Now let \(K\subset D\) be compact. Choose \(r>0\) so that the closed \(r\)-neighborhood \[ K_r=\{z\in D:\operatorname{dist}(z,K)\le r\} \] is compact and contained in \(D\). For every \(z\in K\), Cauchy’s formula for the derivative on the circle \(|\zeta-z|=r\) gives \[ f_n'(z)-f'(z) =\frac{1}{2\pi i}\int_{|\zeta-z|=r} \frac{f_n(\zeta)-f(\zeta)}{(\zeta-z)^2}\,d\zeta. \] Hence \[ |f_n'(z)-f'(z)| \le \frac1r\sup_{\zeta\in K_r}|f_n(\zeta)-f(\zeta)|. \] The right-hand side tends to \(0\) independently of \(z\in K\). Therefore \(f_n'\to f'\) uniformly on \(K\). Since \(K\) was arbitrary, the convergence is uniform on every compact subset of \(D\).