Let $\bar x \in \mathbb R$. Is there a cut-off function such that $\phi\_\epsilon \in C^\infty(\mathbb R)$, $0 \le \phi \le 1$, and
$$\phi\_\epsilon(x) = \begin{cases}
1 &\text{ if } |x - \bar x| \ge \epsilon\\\
0 &\text{ if }|x-\bar x|\le \epsilon/2
\end{cases}
$$
and
$\phi' \le c\_\epsilon \phi$?



Construct suitable cutoff function
The answer is no.


Indeed, let $f:=\phi=\phi\_\epsilon$. Let $a:=\sup\{x\colon f(x)=0\}$. Then $a$ is real, $f(a)=0$ and $f>0$ on the interval $(a,\infty)$. Without loss of generality, $a=0$, so that $f(0)=0$ and $f>0$ on the interval $(0,\infty)$.


Suppose now that for some real $c>0$ we have $f'\le cf$. Then $(\ln f)'\le c$ on $(0,\infty)$, whence $\ln f(1)-\ln f(0+)\le c$, that is, $0=f(0+)\ge f(1)e^{-c}>0$, a contradiction. $\quad\Box$


