Poincaré inequality on annular regions

Question

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?

Answer 1

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

Answer 2

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 $y\in \partial B_t$.