For an open set \(\Omega\subseteq\CC\), show that \(A(\Omega)\definedas \theset{f: \Omega \to \CC \st f\text{ is holomorphic and bounded}}\), with the supremum norm, is a Banach space.
Hint
Apply Morera’s theorem and Cauchy’s theorem.
Solution
Write \(A(\Omega) \coloneqq \{f: \Omega \to \CC \st f \text{ bounded and holomorphic}\}\) and \(\norm{f}_\infty \coloneqq \sup_{z \in \Omega}\abs{f(z)}\), and let \(\{f_n\}\) be a Cauchy sequence in \(A(\Omega)\).
1. \(A(\Omega)\) is a vector space and \(\norm{\cdot}_\infty\) is a norm on it.
Pointwise addition and scalar multiplication preserve holomorphy (linearity of the derivative) and boundedness (triangle inequality); \(\norm{\cdot}_\infty\) is a norm on bounded functions, so it restricts to a norm on \(A(\Omega)\).
2. There is a function \(f\colon\Omega\to\CC\) with \(f_n\to f\) uniformly on \(\Omega\).
For each \(z \in \Omega\), \(\abs{f_n(z) - f_m(z)} \le \norm{f_n - f_m}_\infty \to 0\), so \(\{f_n(z)\}\) is a Cauchy sequence in \(\CC\); define \(f(z) \coloneqq \lim_n f_n(z)\). Letting \(m\to\infty\) in \(\abs{f_n(z) - f_m(z)} \le \sup_{k\geq n}\norm{f_n - f_k}_\infty\) gives \(\sup_{z\in\Omega}\abs{f_n(z) - f(z)} \le \sup_{k\geq n}\norm{f_n - f_k}_\infty\), which tends to \(0\) because \(\{f_n\}\) is Cauchy.
3. \(f\) is holomorphic.
By step 2, \(f\) is a uniform limit of continuous functions, so \(f\) is continuous on \(\Omega\), and for every closed triangle \(T\subseteq\Omega\), \(\int_{\partial T} f = \lim_n \int_{\partial T} f_n = \lim_n 0 = 0\) (uniform convergence lets the limit pass through the integral; each integral vanishes by Goursat’s theorem). Hence \(f\) is holomorphic by Morera’s theorem T-LHSMY.
4. \(f\in A(\Omega)\) and \(\norm{f_n - f}_\infty \to 0\).
Since \(\{f_n\}\) is Cauchy it is bounded in norm, say \(\norm{f_n}_\infty \le M\); then \(\abs{f(z)} = \lim \abs{f_n(z)} \le M\) for all \(z\), so \(f\) is bounded, and \(f\) is holomorphic by step 3, so \(f \in A(\Omega)\). By step 2, \(\norm{f_n - f}_\infty \to 0\).
5. Q.E.D.
By step 1, \(A(\Omega)\) is a normed space, and by step 4 every Cauchy sequence in it converges in \(A(\Omega)\), so \(A(\Omega)\) is a Banach space.