Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 0 down vote favorite

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

share|improve this question
5  
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
2  
@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
1  
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
show 2 more comments

1 Answer

up vote 1 down vote

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.

share|improve this answer
2  
+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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.