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?

  • $\begingroup$ 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

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

  • $\begingroup$ 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

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