Cut off function on complete Riemannian manifolds Let $(M,g)$ be a complete Riemannian manifold. For every $R>0$, does there always exists a nonnegative smooth function on $M$ with the following properties
$$
 \varphi= \begin{cases}1 & \text { on } B(R) \\ 0 & \text { on } M \backslash B(2 R)\end{cases}
 $$
and $|\nabla \varphi| \leq \frac{2}{R}$ ?
If not, which condition do we need to assume in order to have this function?
 A: Yes, it is always possible.
Note that a function $f:M\to\mathbb{R}$, the condition $\|\nabla f\|\le K$ is satisfied iff $\operatorname{Lip}(f)\le K$, where $\operatorname{Lip}(f)$ denotes the Lipschitz constant of $f$. It is easy to find a function satisfying the correponding Lipschitz condition, and this function can then be "nicely approximated" by a smooth one. The key theorem to use is the following:

Theorem: Let $M$ be a complete Riemannian manifold, $f:M\to\mathbb{R}$ be a $K$-Lipschitz function (not necessarily differentiable), and $\epsilon>0$. There exists a $C^\infty$ function $\hat{f}:M\to\mathbb{R}$ satisfying $|f(p)-\hat{f}(p)|<\epsilon$ and $\operatorname{Lip}(\hat{f})\le K+\epsilon$.

For deteails, see theorem 1 in this paper.
Using this, we can construct the desired function. Fix $p\in M$ and $R,\epsilon>0$. Define $f:\mathbb{R}\to\mathbb{R}$ by
$$
f(x)=\begin{cases}
1 & x\le R \\
2-\frac{x}{R} & R<x<2R \\
0 & x\ge 2R
\end{cases}
$$
and define $\theta(q)=f(d(p,q))$. Note that $\theta$ need not be differentiable, but it satisfies the following conditions:
$$
\theta|_{B(p,R)}=1,\ \ \ \theta|_{M\setminus B(p,2R)}=0,\ \ \ \operatorname{Lip}(\theta)\le \frac{1}{R}
$$
Applying the theorem above, there exists a $C^\infty$ function $\psi:M\to\mathbb{R}$ satisfying the following:
$$
|\psi(q)-\theta(q)|<\epsilon,\ \ \ \operatorname{Lip}(\psi)<\frac{1}{R}+\epsilon
$$
Additionally, for $\epsilon<1/4$ one can construct a smooth function $h:\mathbb{R}\to\mathbb{R}$ with the following properties:
$$
h|_{(1-\epsilon,\infty)}=1,\ \ \ h|_{(-\infty,\epsilon)}=0,\ \ \ \operatorname{Lip}(h)\le\frac{1}{1-4\epsilon}
$$
It can then be shown that for sufficiently small $\epsilon$, $\varphi:=h\circ\psi$ satisfies all of the conditions.
