Growth rate of Lipschitz constants for derivatives of $C^\infty$ functions 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 $\lim_{x\to\pm\infty}f(x)=0$.
 A: 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.

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.
