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 Poincare 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 https://arxiv.org/abs/math/0205088.

I suspect the Poincare 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?

$\endgroup$
  • $\begingroup$ Your question is pretty vague can you explain what you mean $\endgroup$ – lmao rekt Nov 22 '17 at 10:06
  • $\begingroup$ I have clarified the question a little more now. $\endgroup$ – Adi Nov 22 '17 at 10:18
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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 '17 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 '17 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 '17 at 2:56
  • $\begingroup$ 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 '17 at 5:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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