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Let's call a $C^{\infty}$-function $f:\mathbb{R}\rightarrow\mathbb{R}$ Lebesgue supersmooth if whenever $a_{n}\in\mathbb{R}$ for all $n$, then $\lim_{n\rightarrow\infty}a_{n}f^{(n)}(x)\rightarrow 0$ almost everywhere. Similarly, let's call $f:\mathbb{R}\rightarrow\mathbb{R}$ Baire supersmooth if whenever $a_{n}\in\mathbb{R}$ for all $n$, then $\lim_{n\rightarrow\infty}a_{n}f^{(n)}(x)\rightarrow 0$ except on a meager set.

  • Is every Lebesgue supersmooth function a polynomial? $\mathbf{No}$. Answered by Pietro Majer in the comments by referring to this question.

  • Is every Baire supersmooth function a polynomial? $\mathbf{No}$. Answered by Pietro Majer in the comments by referring to this question.

  • Do we get the same solutions if we replace "meager" and "measure zero" with "countable"?

  • What happens in higher dimensions?

I am asking this question mainly out of curiosity and because it seems like such a function would be a good example to have in mind if one exists.

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Joseph Van Name
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  • It might help that already a single $x_0$ that works for $a_n = 1$ suffices to make $f$ analytic, with growth no faster than $\exp|x-x_0|$. – Noam D. Elkies Oct 02 '14 at 00:15
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    @AlexandreEremenko: I don't think it's that obvious, because we get to choose $x$ (or a conull/comeager set of $x$) depending on the sequence $a_n$ – Nate Eldredge Oct 02 '14 at 03:20
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    @Noam D. Elkies: How single $x_0$ and $a_n=1$ implies that $f$ is analytic? – Alexandre Eremenko Oct 02 '14 at 03:30
  • Sorry, it does not, at least not the way I imagined. The Taylor series at $x_0$ converges to a function that's $O(\exp\left|x-x_0\right|)$, but it's not obvious that this function agrees with $f(x)$ on any $x \neq x_0$. – Noam D. Elkies Oct 02 '14 at 04:46
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    As to the first two questions: no: see the examples here, where the exceptional set is the Cantor set http://mathoverflow.net/questions/94038/is-f-a-polynomial-provided-that-it-is-partially-smooth/94157#94157 – Pietro Majer Oct 02 '14 at 06:47
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    @Joseph: a further question would be: is there a Lebesgue (resp., Baire) supersmooth function which is not locally polynomial a.e. (resp., on a residual set)? – Pietro Majer Oct 02 '14 at 07:30
  • If the $a_n$ can be arbitrary, does the limit $\lim_{n\to \infty} a_n f^{(n)}(x) = 0$ not imply that only finitely many $f^{(n)}(x)$ are non-zero, at least at that particular value of $x$? – Igor Khavkine Oct 02 '14 at 08:30
  • @IgorKhavkine: it sure feels that way, if $a_n$ grows quicker than any computable function, then there is no way that $f^{(n)}(x)$ can kill this series unless it is eventually 0 as $n$ grows, (at least not any natural $f$). – Per Alexandersson Oct 02 '14 at 08:55
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    @Pietro Majer. I asked the further question you suggested here http://mathoverflow.net/q/182423/22277. – Joseph Van Name Oct 03 '14 at 00:13

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