Let \(\Omega\subset\mathbb C\) be a bounded region and let \((f_n)\) be continuous on \(\overline\Omega\) and holomorphic in \(\Omega\). If \((f_n)\) converges uniformly on \(\partial\Omega\), prove that it converges uniformly on \(\Omega\).
Solution
1. Boundary differences control differences on the entire closure.
For each \(n,m\), the function \(h=f_n-f_m\) is continuous on the compact set \(\overline\Omega\) and holomorphic on \(\Omega\). Its modulus attains a maximum. If that maximum is attained in \(\Omega\), the maximum modulus principle makes \(h\) constant on the region, and continuity gives the same value on its boundary [@SS03]. Otherwise the maximum is already a boundary value. Thus in every case \[ \sup_{\overline\Omega}|f_n-f_m| \leq\sup_{\partial\Omega}|f_n-f_m|. \]
2. The sequence has a uniform limit on \(\overline\Omega\).
Uniform convergence on \(\partial\Omega\) makes the sequence uniformly Cauchy there. Given \(\varepsilon>0\), choose \(N\) so that the right-hand side of step 1 is less than \(\varepsilon\) for \(n,m\geq N\). The same estimate holds on \(\overline\Omega\). Completeness of \(\mathbb C\) gives a pointwise limit \(F(z)\) for every \(z\in\overline\Omega\). For fixed \(n\geq N\), let \(m\to\infty\) in the estimate to obtain \(|f_n(z)-F(z)|\leq\varepsilon\) simultaneously for all \(z\in\overline\Omega\). This proves uniform convergence on the closure, hence also on \(\Omega\).
The limit is continuous on \(\overline\Omega\) as a uniform limit of continuous functions. It is holomorphic on \(\Omega\) by the local uniform limit theorem [@SS03].