Let \(\Omega \subseteq \mathbb{C}\) be an open domain, and let \(\{f_n\}\) be a sequence of holomorphic functions on \(\Omega\) that converges uniformly to a function \(f\) on every compact subset \(K \subset \Omega\). Prove that: (1) The limit function \(f\) is holomorphic on \(\Omega\). (2) The sequence of derivatives \(\{f_n'\}\) converges uniformly to \(f'\) on every compact subset \(K \subset \Omega\) (Weierstrass Convergence Theorem).
Solution
1. The limit \(f\) is holomorphic. Fix a closed triangle \(T\subset\Omega\). Since \(T\) is compact, \(f_n\to f\) uniformly on \(T\), and therefore \(f\) is continuous there. Moreover,\[ \int_{\partial T}f(z)\,dz =\lim_{n\to\infty}\int_{\partial T}f_n(z)\,dz =0, \]because every \(f_n\) is holomorphic on a neighborhood of \(T\). Since this holds for every such triangle, Morera’s theorem gives that \(f\) is holomorphic on \(\Omega\).
2. Let \(K\subset\Omega\) be compact. Choose \(r>0\) such that the closed \(r\)-neighborhood\[ K_r=\{z:\operatorname{dist}(z,K)\le r\} \]is contained in \(\Omega\). Then \(K_r\) is compact.
3. For every \(z\in K\), Cauchy’s derivative formula on \(|\zeta-z|=r\) gives\[ f_n'(z)-f'(z) =\frac1{2\pi i}\int_{|\zeta-z|=r} \frac{f_n(\zeta)-f(\zeta)}{(\zeta-z)^2}\,d\zeta. \]Hence\[ |f_n'(z)-f'(z)| \le\frac1r\sup_{\zeta\in K_r}|f_n(\zeta)-f(\zeta)|. \]Taking the supremum over \(z\in K\) yields\[ \sup_K|f_n'-f'| \le\frac1r\sup_{K_r}|f_n-f|\longrightarrow0. \]Thus \(f_n'\to f'\) uniformly on every compact subset of \(\Omega\).