Entire functions representable in power series

How to prove that an entire function f, which is representable in power series with at least one coefficient is 0, is a polynomial?

• Welcome to Math.SE. Your proposition is false as stated: $\exp x -1$ is entire, can be represented with the power series $x+ x^2/2! + x^3/3! + \cdots$ which has a 0 coefficient, yet is not a polynomial. Are you sure this is the correct version of the problem you intended to ask? – Ragib Zaman Sep 16 '12 at 17:47
• @Ragib: Presumably the question is: if $f$ is entire and at each point $z$ there is some derivative $n = n(z)$ such that $f^{(n)}(z) = 0$ then $f$ must be a polynomial (this is an easy consequence of the fact that a non-constant entire function can have at most countably many zeros). This statement is in fact true (but much harder to prove) even for $C^\infty$-functions on intervals of $\mathbb{R}$, see this MO thread. – t.b. Sep 16 '12 at 18:15
• This question might be (the contrapositive of) what is intended -- ignore the part on Casorati-Weierstrass. – t.b. Sep 16 '12 at 18:24
• @t.b. : How can then I prove that for an entire polynomial function the set Dn is uncountable? – abby Sep 16 '12 at 18:40
• @abby: If $p$ is a polynomial of degree $n$ then $p^{(n+1)}(z) \equiv 0$, so $D_{n+1} = \mathbb{C}$. – t.b. Sep 16 '12 at 18:44

Define $F_n:=\{z\in \Bbb C, f^{(n)}(z)=0\}$. Since for each $n$, $f^{(n)}$ is holomorphic it's in particular continuous, hence $F_n$ is closed. Since we can write at each $z_0$, $f(z)=\sum_{k=0}^{+\infty}\frac{f^{(k)}(z_0)}{k!}(z-z_0)^k$, the hypothesis implies that $\bigcup_{n\geq 0}F_n=\Bbb C$. As $\Bbb C$ is complete, by Baire's categories theorem, one of the $F_n$ has a non empty interior, say $F_N$. Then $f^{(N)}(z)=0$ for all $z\in B(z_0,r)$, for some $z_0\in \Bbb C$ and some $r>0$. As $B(z_0,r)$ is not discrete and $\Bbb C$ is connected, we have $f^{(N)}\equiv 0$, hence $f$ is a polynomial.
• +1, although I think appealing to Baire is a bit of an overkill: if $f$ is not a polynomial, none of the $f^{(n)}$ is constant, hence each $F_n$ is countable, so $\bigcup_{n = 0}^\infty F_n \subsetneqq \mathbb{C}$. – t.b. Sep 16 '12 at 19:01