# Does there exist a supersmooth non-polynomial function?

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.

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