Let \(f\) be an entire function. Suppose that for each \(z_0 \in \mathbb{C}\), the power series expansion
\[f(z) = \sum_{n=0}^\infty c_n(z - z_0)^n\]
has at least one coefficient \(c_n = 0\). Show that \(f\) is a polynomial.
Solution
1. One fixed derivative of \(f\) vanishes identically.
For each integer \(n\geq0\), set \[ Z_n=\{z\in\mathbb C:f^{(n)}(z)=0\}, \qquad f^{(0)}=f. \] The Taylor coefficient of order \(n\) at \(z_0\) is \(f^{(n)}(z_0)/n!\) [@SS03]. Thus the hypothesis says \(\mathbb C=\bigcup_{n\geq0}Z_n\). Each \(Z_n\) is closed, since the derivative is continuous.
By the Baire category theorem, the nonempty complete metric space \(\mathbb C\) cannot be a countable union of closed sets with empty interior [@Fol13]. Hence some \(Z_m\) has nonempty interior. The entire function \(f^{(m)}\) vanishes on this open set. The identity theorem on the connected plane gives \(f^{(m)}\equiv0\) [@SS03].
2. The entire Taylor series terminates.
If \(m=0\), step 1 already says \(f=0\), a polynomial. If \(m\geq1\), every derivative of order \(n\geq m\) vanishes identically by differentiating \(f^{(m)}=0\). The Taylor series at zero, which represents the entire function on all of \(\mathbb C\), is therefore \[ f(z)=\sum_{n=0}^{m-1}\frac{f^{(n)}(0)}{n!}z^n. \] This is a polynomial of degree at most \(m-1\).