Cut-off function My question is: For $R>0$, Can we choose a family of functions $\eta_R\in C_c^1(\mathbf{R}^N)$ satisfying
$0\leq\eta_R\leq 1$ in $\mathbf{R}^N$, $\eta_R=1$ in $B_R(0)$ and
$\eta_R=0$ in $\mathbf{R}^N \setminus B_{2R}(0)$ with $|\nabla\eta|\leq\frac{C}{R}$ for $C>0$ a positive constant independent of $R$.
I know these type of functions can be chosen for any $R$, but don't know whether there is a choice of such functions for which the constant is independent of $R$, as asked in the question.
Can anyone give a proper explanation to this question?
Thanks...
 A: Let $\eta$ be a smooth function on $\mathbb{R}^N$ satisfying $\eta \equiv 1$ in $B_1 (0)$ and $\eta (x) = 0$ for all $x \in \mathbb{R}^N \setminus B_2 (0)$. For $R > 0$, define $\eta_R$ by $\eta_R (x) := \eta (\frac{x}{R})$.  Then $\eta_R$ is smooth, supported on the ball of radius $2R$, and we have by the chain rule
\begin{equation*}
\nabla \eta_R (x) = \nabla \eta_R (\frac{x}{R}) = \frac{1}{R} \nabla \eta (x).
\end{equation*}
Thus, we have your desired inequality with $C := \max_{B_2 (0)} |\nabla \eta|$, which does not depend on $R$.
A: The usual construction for such fonctions is to define the cut-off as a radial function, that is, $$\eta_R(x) = \eta\left(\frac{|x|}R\right)$$ where $\eta\in c^\infty(\mathbb R)$ is given by
$$
x\to \begin{cases}1 &\text{ for }x\leq1 \\ 0&\text{ for } x\geq2
\end{cases}
$$
and monotone. The dependence on $R$ of the derivative comes from the chain rule. The standard construction for $\eta$ is to mollify an indicator function. You can also do it this way, to minimise the constant $C$.
