For \(\Omega\subseteq\CC\), show that \(A(\CC)\definedas \theset{f: \Omega \to \CC \st f\text{ is holomorphic, bounded}}\) is a Banach space.
Hint: Apply Morera’s Theorem and Cauchy’s Theorem
Solution
Interpreting the displayed space as \[ A(\Omega)=\{f:\Omega\to\mathbb C:f\text{ is holomorphic and bounded}\} \] with the supremum norm \(\|f\|_\infty=\sup_{z\in\Omega}|f(z)|\), let \((f_n)\) be Cauchy in \(\|\cdot\|_\infty\). Then for every \(z\in\Omega\) the sequence \((f_n(z))\) is Cauchy in \(\mathbb C\), so define \[ f(z)=\lim_{n\to\infty}f_n(z). \] The Cauchy property in supremum norm shows that \(f_n\to f\) uniformly on all of \(\Omega\): given \(\varepsilon>0\), choose \(N\) such that \(\|f_n-f_m\|_\infty<\varepsilon\) for \(m,n\ge N\) and let \(m\to\infty\).
The limit \(f\) is bounded, since for fixed \(N\), \[ \|f\|_\infty\le \|f-f_N\|_\infty+\|f_N\|_\infty<\infty. \] It is holomorphic because uniform convergence on \(\Omega\) implies uniform convergence on every compact subset, and locally uniform limits of holomorphic functions are holomorphic (equivalently, apply Morera’s theorem on triangles compactly contained in \(\Omega\)). Hence \(f\in A(\Omega)\) and \(\|f_n-f\|_\infty\to0\). Therefore \(A(\Omega)\) is complete.