Let $L>0$. Is it possible to find $0<\varepsilon <L$ and a cut off function $\psi: [0,L]\rightarrow \mathbb{R}$, such that
i) $\psi \in C^3([0,L])$,
ii) $\psi(x)=0$ for all $x \in [0,\varepsilon]$,
iii) $\psi(x)=1$ for all $x \in [L-\varepsilon, L]$,
and $$|\psi'(x)||\psi(x)| \leq \frac{1}{2L}, \quad \forall x \in [0,L]?$$
$$1=\int_\varepsilon^{L-\varepsilon}(\psi^2(x))'dx\le (L-2\varepsilon)2\max |\psi(x)|\,|\psi'(x)|\le (L-2\varepsilon)/L<1. $$ Impossible.