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 each 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?

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?

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 $q >1$?

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?

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 each 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?