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.


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.

  • $\begingroup$ Yes, I think this works for G-T proofs because they want $|D\zeta|$ to be bounded using only the constants of the problem. $\endgroup$ – Ryan Unger Nov 30 '16 at 14:25
  • $\begingroup$ Hmm, what are the conditions on $h(r)$? After smoothing it out a bit, we don't necessarily have $\int \zeta=1$, so we'd have to modify it by an overall constant, which was not accounted for, and would have to be taken into account when estimating $|h'(r)|\le 2C$. $\endgroup$ – Ryan Unger Nov 30 '16 at 18:13
  • $\begingroup$ Yes, you have to smooth it out, and then renormalize slightly to make the integral one. There is plenty of room to do this by allowing $h'$ to be magnified by a factor of $2$. $\endgroup$ – Jeff Nov 30 '16 at 20:23
  • $\begingroup$ I noticed my argument is not exactly correct. I constructed a smooth cutoff function that is identically 1 on the set $\{x \in V\, : \, \text{dist}(x,\partial V)\geq \delta\}$. To get the cutoff function you want, you have to insert a second set $V \subset\subset W \subset\subset U$, and mollify the characteristic function of $W$ with a smaller mollifier ($\delta/2$ might work). The basic idea is in my post though. $\endgroup$ – Jeff Nov 30 '16 at 20:25
  • $\begingroup$ I've done some thinking, and came up with this. Does it make sense? We take the continuous function $$ \eta_0(x) = \begin{cases} 1 & \text {if $\text{dist}(x,V)\le \frac{\delta}{8}$} \\ 0 & \text{if $\text{dist}(x,V)\ge \frac{7\delta}{8}$} \\ \frac{7}{6}-\frac{4}{3\delta}\text{dist}(x,V) & \text{if $\frac{\delta}{8}\le\text{dist}(x,V)\le\frac{7\delta}{8}$}. \\ \end{cases}$$ $\endgroup$ – Ryan Unger Nov 30 '16 at 20:34

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