Cutoff function with controlled derivative Let $U\subset\Bbb R^n$ be open, bounded, and $V\subset\subset U$ be open with $\mathrm{dist}(V,\partial U)\ge \delta\ne 0$. Does there exist a function $\eta\in C_c^\infty(U)$ with $0\le \eta\le 1$, $\eta=1$ on $V$, and with $$\sup_{x\in U}||D\eta(x)||\le \frac{2}{\delta},$$
where $D\eta=(\partial_1\eta,\dotsc,\partial_n\eta)?$ The usual existence proof of cutoff functions (using partition of unity) gives no information about the derivative, so how does one get an estimate on it/control it?
This result is invoked in section 8.3 of Gilbarg-Trudinger with no explanation.
 A: One way to construct cutoff functions is mollification. Let $\zeta:\mathbb{R}^n \to \mathbb{R}$ be smooth and non-negative with compact support in $B(0,1)$ and
$$\int_{B(0,1)}\zeta(x) \, dx = 1. \ \ \ \ \ \ (*)$$
Then define 
$$\zeta_\delta(x) = \frac{1}{\delta^n}\zeta\left(\frac{x}{\delta}\right).$$
Then you can define the cutoff function $\eta$ as
$$\eta =\chi_V*\zeta_\delta,$$
where $\chi_V$ is the indicator (or characteristic) function of $V$ (i.e., $\chi_V(x)=1$ for $x \in V$ and $\chi_V(x)=0$ for $x \not\in V$).
Then we have
$$D\eta(x) = (\chi_V * D\zeta_\delta)(x) = \int_{B(x,\delta)}\chi_V(y) \frac{1}{\delta^{n+1}} D\zeta\left(\frac{x-y}{\delta}\right) \, dy.$$
Making the change of variables $z=(x-y)/\delta$ we have
$$|D\eta(x)| \leq \frac{1}{\delta}\int_{B(0,1)} |D\zeta(z)| \, dz.$$
Now the question is how small can we make the norm on the right hand side. Presumably we should just look for radially symmetric mollifiers $\zeta(z) = h(|z|)$ for some smooth $h$ with $h(r)=0$ for $r\geq 1$. Then
$$\int_{B(0,1)} |D\zeta(z)| \, dz = \int_0^1 \int_{\partial B(0,r)} |h'(r)| \, dS(z) dr = \int_0^1 n\alpha(n)r^{n-1}|h'(r)| \, dr,  \ \ \ \ \ \ (**)$$
where $\alpha(n)$ is the volume of the $n$-dimensional unit ball. Now by (*) $h$ satisfies
$$\int_0^1 n\alpha(n)r^{n-1} h(r) \, dr = 1.$$
For the moment, make the (nonsmooth) choice of $h(r)=C(1-r)$, and plug this into the constraint above to find that
$$C = \frac{n+1}{\alpha(n)}.$$
Notice that $|h'(r)|=C$. If we smooth this out a bit, we can find a smooth $h$ with let's say $|h'(r)| \leq 2C$. Plugging this into (**) we find that
$$\int_{B(0,1)}|D\zeta(z)| \, dz \leq 2\frac{n+1}{\alpha(n)}\int_0^1 n\alpha(n)r^{n-1}\, dr = 2(n+1).$$
With this construction we are getting an extra factor of $n+1$, but this should be sufficient for any of the proofs in Gilbarg-Trudinger. Perhaps there is another way to prove the sharper condition.
