0
$\begingroup$

It is well known that given a function $f \in L^p(B_R)$ such that $|\{x \in B_R: f(x) = 0\}|>0$, the following Poincaré inequality holds: $$ \int_{B_R} \left(\frac{|f|}{R}\right)^p \ dx \leq c \int_{B_R} |\nabla f|^p \ dx \, .$$

My question is: does something like this hold on annular regions? More specifically, given $0<r<t<\infty$, and let $f$ be such that $|\{x \in B_t \setminus B_r: f(x) = 0\}| > 0$, then does the following inequality hold? $$ \int_{B_t\setminus B_r} \left(\frac{|f|}{t-r}\right)^p \ dx \leq c \int_{B_t \setminus B_r} |\nabla f|^p \ dx \, .$$

The only reference for inequalities of Poincare type on punctured domains I could find was Lieb–Seiringer–Yngvason (Ann. Math 2003) arXiv link.

I suspect the Poincaré inequality on punctured domains in the way it is asked above might be false. If it is false, then I would like to understand is what sort of functions admit the second inequality?

$\endgroup$
3
  • $\begingroup$ Your question is pretty vague can you explain what you mean $\endgroup$
    – lmao rekt
    Nov 22, 2017 at 10:06
  • $\begingroup$ I have clarified the question a little more now. $\endgroup$
    – Adi
    Nov 22, 2017 at 10:18
  • $\begingroup$ As for poincaré–wirtinger inequality on annular, do you know any references? Thank for you $\endgroup$
    – lic
    May 27, 2021 at 8:13

2 Answers 2

2
$\begingroup$

No. Consider the positive part of a coordinate function on a large but thin annulus.

$\endgroup$
4
  • $\begingroup$ Could you please elaborate a little? I expect there is a problem as $t \rightarrow r$, but I am trying to understand what subclass of functions admits such an inequality. $\endgroup$
    – Adi
    Nov 23, 2017 at 11:05
  • $\begingroup$ Let $r$ be very large and $t=r+1$. Let $f(x_1,\dots,x_n)=x_1$. Then the average of $f$ is $r$ while the average of $\nabla f$ is 1 (no matter which $L^p$ norm you take). This is enough to falsify your proposed inequality. $\endgroup$
    – Fan Zheng
    Nov 27, 2017 at 2:03
  • $\begingroup$ Quick question, the measure condition is not satisfied for this function. It's only zero on the hyperplane $x_1=0$. Am i missing something from your answer? $\endgroup$
    – Adi
    Nov 28, 2017 at 2:56
  • 1
    $\begingroup$ It's me that is missing this condition, but there is an easy fix: let $f=x_1^+:=\max(x_1,0)$. $\endgroup$
    – Fan Zheng
    Nov 28, 2017 at 5:33
0
$\begingroup$

If you assume that $\int_{B_t\setminus B_r} f=0$ (you can pass from this case to the $|\{f=0\}|>0$ case by subtracting an appropriate constant from $f$), you can write $$\left(\int_{B_t\setminus B_r} |f|^p\right)^{1/p} \leq \int_r^t \left(\int_{\partial B_s} \left|f-\frac{1}{\sigma(\partial B_s)} \int_{\partial B_s} f\right|^p\,d\sigma\,ds\right)^{1/p} + \left(\int_r^t \left|\int_{\partial B_s} f\,d\sigma\right|^p\,ds\right)^{1/p}.$$ The Poincare inequality does hold in spheres (you can see this from change of variables arguments), so you can bound the first term using the Poincare inequality in a sphere. You can bound $$\frac{1}{\sigma(\partial B_s)}\int_{\partial B_s} f\,d\sigma-\frac{1}{\sigma(\partial B_q)}\int_{\partial B_q} f\,d\sigma=\int_s^q \frac{d}{dp}\left(\frac{1}{\sigma(\partial B_p)}\int_{\partial B_p} f\,d\sigma\right)\,dp $$ for $r\leq q\leq t$, and from there bound $\frac{1}{\sigma(\partial B_s)}\int_{\partial B_s} f\,d\sigma-\frac{1}{|B_t\setminus B_r|}\int_{B_t\setminus B_r} f\,d\sigma$.

This lets you establish a Poincare inequality in an annulus, but the scaling is not $(t-r)$, it's just $t$: $$\left(\int_{B_t\setminus B_r} |f|^p\right)^{1/p} \leq Ct\left(\int_{B_t\setminus B_r} |\nabla f|^p\right)^{1/p}.$$

You can get the bound you specified for functions $f$ with $f=0$ on $\partial (B_t\setminus B_r)$: just extend $f$ by zero and use a covering lemma and the regular Poincare inequality in balls $B_{2(t-r)}(y)$ for $y\in \partial B_t$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.