Let \(\Omega \subseteq \mathbb{C}\) be a connected open region, and let \(f: \Omega \to \mathbb{C}\) be a holomorphic function. Suppose that for every point \(z_0 \in \Omega\), at least one coefficient in the Taylor series expansion: \[f(z) = \sum_{n=0}^\infty c_n(z_0) (z - z_0)^n\] is zero (i.e. for every \(z_0 \in \Omega\), there exists some \(n \in \mathbb{N}\) such that \(f^{(n)}(z_0) = 0\)). Prove that \(f\) is a polynomial.
Solution
For \(n\ge0\), set \[ E_n=\{z\in\Omega:f^{(n)}(z)=0\}. \] Each \(E_n\) is closed in \(\Omega\) because \(f^{(n)}\) is continuous, and the hypothesis says \[ \Omega=\bigcup_{n=0}^{\infty}E_n. \]
1. The region \(\Omega\) is a locally compact Hausdorff space, hence a Baire space. Therefore a countable union of closed subsets with empty interior cannot equal \(\Omega\). Thus some \(E_N\) has nonempty interior.
2. Hence \(f^{(N)}\) vanishes on a nonempty open subset of \(\Omega\). Since \(f^{(N)}\) is holomorphic and \(\Omega\) is connected, the identity theorem gives\[ f^{(N)}\equiv0\quad\text{on }\Omega. \]
3. If \(N=0\), then \(f\equiv0\), which is a polynomial. If \(N\ge1\), then \(f^{(N)}\equiv0\), so the Taylor expansion of \(f\) at any point terminates after degree at most \(N-1\). Equivalently, repeated integration gives a polynomial \(P\) of degree at most \(N-1\) such that \(f=P\) on \(\Omega\).
Therefore \(f\) is a polynomial.