Example of cut off function For any $R>0,$ I need an example of a family of functions $\phi_R\in C_c^{1}(\mathbb{R}^N)$ such that $0\leq \phi_R\leq 1$ in $\mathbb{R}^N$ satifying
$\phi_R=1$ in $B(0,R)$ and $\phi_R=0$ in $\mathbb{R}^N\setminus B(0,2R)$ such that 
$$
|\nabla\phi_R|\leq \frac{c}{R}
$$
for some positive constant $c$ independent of $R$.
I tried in the following way:
Let $\phi\in C_c^{1}{\mathbb{R}^N}$ such that $0\leq \phi\leq 1$ in $\mathbb{R}^N$ and 
$\phi(t)=1$ if $|t|\leq 1$ and $\phi(t)=0$ if $|t|\geq 2$.
Then define for any $R>0$ 
$$
\phi_R=\phi(\frac{|x|}{R})
$$
Then it follows from the property of $\phi$ that
$0\leq \phi_R\leq 1$ in $\mathbb{R}^N$ together with $\phi_R=1$ in $B(0,R)$ and $\phi_R=0$ in $\mathbb{R}^N\setminus B(0,2R)$.
But I am unable to prove the last condition
$$
|\nabla\phi_R|\leq \frac{c}{R}
$$
for $c$ independent of $R$.
Please help me.
Thank you.
 A: In your example, notice first of all that $\phi_R$ is differentiable, since it is equal to 1 on a neighbourhood of the origin, so the absolute value in the argument poses no problem. 
Furthermore, $\phi_R$ vanishes outside of the ball $B(0,2R)$, so our bound really only depends on the values of $\phi_R$ inside the ball.
The chain rule gives
$$|\partial_j \phi_R(x)| = |\partial_j \big(\phi\left(|x|/R\right)| \big) = |\partial_j \phi(|x|/R)| \cdot \frac{1}{R}. $$
where the equalities follow since we are taking the absolute value anyway (otherwise there could be problems with the signs).
Now since $\phi \in C^1_c$, all of its partial derivatives are in particular continuous, and hence bounded on the compact set $B(0,2R)$, so that $|\partial_j \phi(|x|/R)| \leq C_j$ for each $j = 1, \ldots, N$.
Hence,
$$ |\nabla \phi_R| \leq \sum_{j=1}^N |\partial_j \phi_R| \leq \sum_{j=1}^N \frac{C_j}{R} = \frac{C}{R}, $$
with $C = \sum_{j=1}^N C_j.$
A: Choose $\epsilon>0$, and use the construction given here and use $$\phi(x)= f\left(\frac{|x|}{R}\right).$$ The gradient is bounded by $\frac{1+\epsilon}R$.
