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 $\phi' \le c_\epsilon \phi$?

## 2 Answers

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. $\quad\Box$

Under the described hypotheses, the function $f: x \in \mathbf{R} \mapsto e^{-cx} \phi(x)$ would be decreasing, because $f'(x) = e^{-cx} (-c \phi(x) + \phi'(x)) \leq 0$ for all $x \in \mathbf{R}$. As $f(0) = 0$ and $f \geq 0$, one would have $f(x) = 0$ for all $x \geq 0$. For this reason a cutoff function cannot have $\phi' \leq c \phi$.

Here for simplicity, we let $\overline{x} = 0$, $\epsilon = 1$, and wrote $c = c_{\epsilon}$.