Let $f\in C^\infty$ have bounded derivatives, i.e.
$$ \sup\_{x\in\mathbb{R}}|f^{(p)}(x)| = B\_p < \infty$$
for every $p\ge 1$.


I would like to find a proof or a counterexample for the following conjecture:



> 
> For every such $f$ there exists $C\_f>0$ such that $B\_p \le (p+1)^{C\_f p}$ for all $p\ge1$
> 
> 
> 


That is, if all derivatives of a function are bounded, the bound cannot grow too quickly.


I am also curious about the converse statement



> 
> If $f$ is not constant and $\lim\_{x\to\pm\infty}f(x)=0$, there exist $c\_f, c\_f'>0$ such that $B\_p \ge c\_f'p^{c\_f p}$ 
> 
> 
> 


That is, if all derivatives of a non-constant function are bounded, the bound cannot grow too slowly. Note that polynomials and sine functions are not counterexamples due to the requirements $f$ is not constant and $\lim\_{x\to\pm\infty}f(x)=0$.



Growth rate of Lipschitz constants for derivatives of $C^\infty$ functions
The answer to the second question is also *no.* If $f$ is a Schwartz function, then $B\_p\le C^p \| t^p\widehat{f}\|\_{L^1}$, and this increases at most exponentially in $p$ if $\widehat{f}$ has compact support.




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Let me also add some detail to what I said about the first part of your question in my comments above: We can build an $f=\sum f\_n$, with $f\_n$ supported near $x=n$, say, such that $|f^{(n)}(n)|\ge M\_n\equiv (n+1)^{(n+1)n}$ (showing that $C\_f=n$ doesn't work in your desired bound), but $f^{(j)}$ still stays bounded on $\mathbb R$ for each $j$.


To do this, take $f\_n^{(n)}(x)= \pm M\_n$, $\int f\_n^{(n)}=0$, on a tiny symmetric interval $(n-\delta,n+\delta)$, or rather, take $f^{(n)}$ as a smooth version of this. The first property of $f\_n$ is already built into this, and $f\_n^{(j)}(x)$ for $j=0,1, \ldots , n-1$ can be kept as small as desired by taking $\delta>0$ sufficiently small, so if we keep those derivatives smaller than what we already had, then indeed $\|f^{(j)}\|\_{\infty}\le \max\_{0\le k\le j} \|f\_k^{(j)}\|\_{\infty}$ will be finite for each fixed $j$.


Finally, we now see that we can also produce a compactly supported counterexample in just the same way.


