Suppose \(f\) is analytic, defined on all of \(\CC\), and for each \(z_0 \in \CC\) there is at least one coefficient in the expansion \(f(z) = \sum_{n=0}^\infty c_n(z-z_0)^n\) is zero. Prove that \(f\) is a polynomial.
Hint: use the fact that \(c_n n! = f^{(n)}(z_0)\) and use a countability argument.
Solution
Write \(Z_n \da \ts{z\in \CC \st f^{(n)}(z) = 0 }\), then by hypothesis \(\Union_{n\geq 0} Z_n = \CC\). A version of the Baire category theorem is that if \(X\) is a complete metric space and \(X\) is a countable union of closed sets, then at least one such set has a nonempty interior. Thus some \(Z_n\) has an interior point \(z_0\), and as a result there is some disc \(\DD_\eps(z_0)\) on which \(f^{(n)}(z_0) \equiv 0\). This implies that \(f^{(k)}(z_0) \equiv 0\) on \(\DD_\eps(z_0)\) for every \(k\geq n\), so \(f\) is a polynomial of degree at most \(n\).