The question is as in the title:


Is there a nonpolynomial $C^\infty$ function $f$ on $\mathbb{R}$ such that $\sup\_{x \in \mathbb{R}} \lvert f^{(q)}(x) \rvert \leq (\ln q)^{-q}$ for every natural number $q >1$?


Here, "nonpolynomial" excludes constant functions as well, of course.


I think nonpolynomial functions "very uniformly approximating" a constant function" might satisfy such a bound, but I cannot find such a nice example.


Could anyone please help me?



Is there a nonpolynomial $C^\infty$ function $f$ such that $\sup_{x \in \mathbb{R}} \lvert f^{(q)}(x) \rvert \leq (\ln q)^{-q}$ for every $q >1$?
The answer is no. From [Taylor's theorem with remainder](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder), we see that for any integer $q>2$, we have
$$ f^{(2)}(x) = \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} x^j + O( \frac{(\ln q)^{-q} |x|^{q-2}}{(q-2)!} )$$
and thus for $|x| \leq \frac{q \ln q}{10}$ (say) we have from Stirling's formula that
$$ \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} x^j = O( 1 ).$$
The key point here is that we have bounded control on a polynomial on an interval of length much wider than its degree (a situation in which one expects favorable estimates thanks to the uncertainty principle).
Making the trigonometric substitution $x = \frac{q \ln q}{10} \sin \theta$ we conclude
$$ \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} (\frac{q \ln q}{10})^j \sin^j \theta = O( 1 )$$
for all $\theta$. The LHS is a trigonometric polynomial of degree $O(q)$, so by the [Bernstein inequality](https://en.wikipedia.org/wiki/Bernstein%27s_theorem_(polynomials)#Bernstein%27s_inequality) we have
$$ \frac{d}{d\theta} \sum\_{j=0}^{q-3} \frac{f^{(2+j)}(0)}{j!} (\frac{q \ln q}{10})^j \sin^j \theta = O( q )$$
for all $\theta$. Evaluating this at $\theta=0$ we conclude that
$$ f^{(3)}(0) \frac{q \ln q}{10} = O(q).$$
Dividing by $q \ln q$ and sending $q \to \infty$ we conclude that $f^{(3)}(0)$ vanishes. Applying a translation to this argument, we see that $f^{(3)}$ vanishes everywhere, thus $f$ is a polynomial (in fact it is quadratic).


