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If f is infinitely differentiable then f coincides with a polynomial

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). 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, 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!

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    $\begingroup$ The question has been asked and answered on MO more than once. See mathoverflow.net/questions/34059/… $\endgroup$ May 8 '11 at 0:34
  • $\begingroup$ Yes, thank you. It seems my search wasn't broad enough. $\endgroup$ May 8 '11 at 0:58
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    $\begingroup$ Voting to close as a duplicate question (although I much prefer the phrasing of the question here, with context and commentary, to the version which Andrey links to) $\endgroup$
    – Yemon Choi
    May 8 '11 at 1:45
  • $\begingroup$ (Oops, finger slipped, so the reason given for closing may not be the one I intended.) $\endgroup$
    – Yemon Choi
    May 8 '11 at 1:46
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    $\begingroup$ +1, I agree with Yemon. $\endgroup$
    – S. Carnahan
    May 8 '11 at 1:57