Let \(\Omega \subset \CC\) be a connected open subset. Let \(\left\{f_{n}: \Omega \rightarrow \CC\right\}_{n=1}^{\infty}\) be a sequence of holomorphic functions uniformly bounded on compact subsets of \(\Omega\). Let \(f: \Omega \rightarrow \CC\) be a holomorphic function such that the set \[ \left\{z \in \Omega \mid \lim _{n \rightarrow \infty} f_{n}(z)=f(z)\right\} \] has a limit point in \(\Omega\). Show that \(f_{n}\) converges to \(f\) uniformly on compact subsets of \(\Omega\).
Solution
Write \(g(z) \da \lim_{n\to\infty}f_n(z)\) for the pointwise limit, then \(g(z) = f(z)\) on a set with a limit point. By the identity principle, \(g\equiv f\) on \(\Omega\), making \(f\) the pointwise limit of the \(f_n\).
By Montel, locally uniformly bounded implies normal and locally equicontinuous. So \(\ts{f_n}\) is normal, and thus has a locally uniformly convergent subsequence \(\ts{f_{n_k}}\). Since singletons \(\ts{z}\) are compact, \(f_{n_k}(z) \to g(z)\) pointwise, and by uniqueness of limits, \(\lim_{k\to\infty } f_{n_k} = g = f\) on any compact \(K \subseteq \Omega\).
It remains to show that the original sequence \(\ts{f_n}\) converges locally uniformly to \(f\), not just the subsequence. Suppose not, then there exists a compact \(K \subseteq \Omega\) and \(\eps>0\) so that \(\norm{f_n - f}_{K, \infty} > \eps\) for infinitely many \(n\). This produces a subsequence \(\ts{f_{n_j}}\) with \(\norm{f_{n_j} - f} > \eps\) for all \(j\). However, since \(\mcf\) was normal, every subsequence has a locally uniformly convergent subsequence, so this has a further subsequence \(f_{n_{j'}}\) uniformly converging to \(f\), a contradiction.