Does there exist a smooth compactly supported function $$f \in C^{\infty}\_c((0,1))$$
such that 
$$ \|D^k f\|\_{L^{2}(0,1)} \leq \left\lfloor{\alpha\,k}\right \rfloor! \quad \forall\, k\in \mathbb N$$
for some $\alpha \in (0,1)$?



Existence of a function with slow growth on derivatives
For all natural $k$, we have $\|D^kf\|\_1\le\|D^kf\|\_2\le k!$, where $\|\cdot\|\_p:=\|\cdot\|\_{L^p(0,1)}$. So, for all $x\in(0,1)$ we have 
$$|(D^kf)(x)|\le\int\_0^x |(D^{k+1}f)(u)|\,du\le\|D^{k+1}f\|\_1\le(k+1)!. 
$$
So, for the Lagrange remainder 
$$R\_n(a,x)=\int\_a^x (D^{n+1}f)(u)\frac{(x-u)^n}{n!}\,du
$$
for the $n$th-order Taylor approximation of $f$ at $a\in(0,1)$ we have $|R\_n(a,x)|\le(n+2)|x-a|^{n+1}\to0$ as $n\to\infty$, for any $a$ and $x$ in $(0,1)$. So, $f$ is real analytic on $(0,1)$. Since $f$ is compactly supported, it follows that $f=0$. 


