Cutoff and Poincare inequality

Question

I want to prove the following result. Let $\psi$ be in $C^\infty_0(\mathbb{R}^n)$, $ n\geq 2$ satisfying $\psi=1$ if $|x|\leq \frac {1}{2}$ and $\psi=0$ if $|x|\geq 1$.

For $N>0$, set $\psi_N (x)=\psi (x/N)$. Then the author claim we have the following inequality $$\Vert (\nabla \psi_N)v \Vert \leq C (n) \Vert \nabla v \Vert$$ for all $v \in C^\infty_0(\mathbb{R}^n)$.

The norm is $L^2 (\mathbb{R}^n) $.

For the case $n\geq 3$, I can prove it by using Holder and Gagliardo-Nirenberg-Sobolev inequality. But I have a problem when $n=2$.

The author mentioned that reader use Poincare inequality but the problem is the support of $v$.

How can I prove it?

The inequality comes from the paper Form boundedness of general second-order differential operator (Maz'ya, Verbitsky) on page 17.

Answer 1

Looks like the claim is false for $n=2$. For small $\epsilon>0$ let $v(x) = (1-\epsilon \log^+|x|)^+$ where $a^+=\max(a,0)$. Then $v =1$ on the unit ball, so the norm $\|(\nabla \psi_N )v\|$ is independent ot $\epsilon$. On the other hand, $$ \|\nabla v\|^2 = 2\pi \int_1^{\exp(1/\epsilon)} (\epsilon/r)^2\,r\,dr = 2\pi \epsilon $$ which can be arbitrarily small. This $v$ is not $C^\infty$, but it is Lipschitz with compact support, which is just as good in this context (it can be smoothed without changing either norm much).

I think the inequality still works for $n\ge 3$, but since the claim is explicitly made for $n\ge 2$, I would

1. Check the journal form of the article in case the argument is changed in the final version.
2. Ask one of the authors.

Answer 2

Here I give a proof in the case $n\geq 3$.

Note that $\mathrm{supp}\left(\nabla\psi_{N}\right)\subset B_{N}\left(0\right)$. For any $v\in C_{0}^{\infty}\left(\mathbb{R}^{n}\right)$, Holder's inequality gives $$ \left\Vert \left(\nabla\psi_{N}\right)v\right\Vert _{L^{2}\left(\mathbb{R}^{n}\right)}\le\left\Vert \nabla\psi_{N}\right\Vert _{L^{n}\left(\mathbb{R}^{n}\right)}\left\Vert v\right\Vert _{L^{\frac{2n}{n-2}}\left(\mathbb{R}^{n}\right)}. $$ Now note that $$ \left\Vert \nabla\psi_{N}\right\Vert _{L^{n}\left(\mathbb{R}^{n}\right)}\le\frac{C}{N}\left\Vert \chi_{B_{N}\left(0\right)}\right\Vert _{L^{n}\left(\mathbb{R}^{n}\right)}=\frac{C}{N}N=C(\psi) $$ and the Gagliardo-Nirenberg-Sobolev inequality gives $$ \left\Vert v\right\Vert _{L^{\frac{2n}{n-2}}\left(\mathbb{R}^{n}\right)}\le C\left(n,\psi\right)\left\Vert \nabla v\right\Vert _{L^{2}\left(\mathbb{R}^{n}\right)}. $$ Hence $$ \left\Vert \left(\nabla\psi_{N}\right)v\right\Vert _{L^{2}\left(\mathbb{R}^{n}\right)}\le C\left(n,\psi\right)\left\Vert \nabla v\right\Vert _{L^{2}\left(\mathbb{R}^{n}\right)}, $$ which completes the proof when $n\ge3$.