- Suppose \(f\) is an analytic function defined everywhere in \(\mathbb { C }\) and such that for each \(z _ { 0 } \in \mathbb { C }\) at least one coeficient in the expansion
\[ f (z) = \sum_ {n = 0} ^ {\infty} c _ {n} (z - z _ {0}) ^ {n} \]
is equal to 0. Prove that f is a polynomial.
[Hint: Use the fact that \(c _ { n } n ! = f ^ { ( n ) } ( z _ { 0 } )\) and use a countability argument.]
Solution
For each \(n\ge0\), let \[ Z_n=\{z\in\mathbb C:f^{(n)}(z)=0\}. \] By hypothesis, for every \(z_0\in\mathbb C\) at least one Taylor coefficient of \(f\) at \(z_0\) vanishes. Since that coefficient is \(f^{(n)}(z_0)/n!\) for some \(n\), we have \[ \mathbb C=\bigcup_{n=0}^\infty Z_n. \tag{1} \]
Suppose \(f\) were not a polynomial. Then no derivative \(f^{(n)}\) could vanish identically, since \(f^{(n)}\equiv0\) would imply that \(f\) is a polynomial of degree at most \(n-1\). Hence, by the isolated-zero theorem, each \(Z_n\) is a discrete subset of \(\mathbb C\).
Every discrete subset of \(\mathbb C\) is countable: for each of its points choose a disc with rational center and rational radius containing no other point of the set, and assign one such disc to that point; there are only countably many rational discs. Thus every \(Z_n\) is countable, and so the right side of (1) is countable. This contradicts the uncountability of \(\mathbb C\).
Therefore some \(f^{(n)}\) is identically zero, and consequently \(f\) is a polynomial.