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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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    $\begingroup$ @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 $a_n$ $\endgroup$ – Nate Eldredge Oct 2 '14 at 3:20
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    $\begingroup$ @Noam D. Elkies: How single $x_0$ and $a_n=1$ implies that $f$ is analytic? $\endgroup$ – Alexandre Eremenko Oct 2 '14 at 3:30
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    $\begingroup$ As to the first two questions: no: see the examples here, where the exceptional set is the Cantor set mathoverflow.net/questions/94038/… $\endgroup$ – Pietro Majer Oct 2 '14 at 6:47
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    $\begingroup$ @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)? $\endgroup$ – Pietro Majer Oct 2 '14 at 7:30
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    $\begingroup$ @Pietro Majer. I asked the further question you suggested here mathoverflow.net/q/182423/22277. $\endgroup$ – Joseph Van Name Oct 3 '14 at 0:13

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