Let $f:\mathbb{R} \to \mathbb{R}$ be infinitely differentiable. Does $\forall x \in \mathbb{R}, \exists n \in \mathbb{Z}\ \text{s.t.}\ \forall m \ge n, f^{(m)}(x) = 0 $ imply that $f$ is a polynomial? Of course if the first two quantifiers are reversed, integration gives the result trivially. The Baire Category Theorem applies immediately to say (roughly) that $f(x)$ is a polynomial at densely many points. I've tried and haven't gotten any further. This was originally discussed on the Wikipedia math reference desk several months back (archive [here][1]). There was virtually no progress towards a solution after several days and several pages, though there were plenty of false starts.

A counterexample would have to be a non-analytic function of a strange sort--one where at each point the derivatives eventually stabilize to 0. All the non-analytic smooth functions I've seen are essentially of [another sort][2], which do not have this property. They "sacrifice" having non-zero derivatives of all orders in a neighborhood around the non-analytic point in order to make the derivatives agree at all orders at that point. Being a polynomial at densely many points seems awfully inconvenient for a non-analytic smooth function as well. At this point, I think my knowledge of such functions is too incomplete to know how to construct a counterexample or to know what properties prevent such a construction.

I'm very curious about an answer. If the result is true, it would be a cute characterization of polynomials. If the result is false and a counterexample is given, it would probably have interest in its own right.

Thanks for any thoughts!


  [1]: http://en.wikipedia.org/wiki/Wikipedia%3AReference_desk/Archives/Mathematics/2010_November_13#Infinitely_differentiable_implies_polynomial
  [2]: http://en.wikipedia.org/wiki/Non-analytic_smooth_function