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The Poincare inequality states that if domain $\Omega$ is bounded in one direction by length $d>0$ then for any $u\in W_0^{1,p}(\Omega)$ we have $$ \int_\Omega|u|^p\,dx\leq \frac{d^p}{p}\int_\Omega |\nabla u|^p\,dx $$

Now I assume the domain $U\subset \mathbb R^N$ contains a sequence $B(x_n,r_n)$, where $x_n\in U$ and $r_n\to \infty$, then I want to prove that the previous Poincare's inequality fails on $W_0^{1,p}(U)$.

Yes, of course, if I have such balls in $U$ then the domain $U$ can not be bound in one direction, and hence I done. But honest, is that it? I feel uncomfortable with my argument... Is there something more going on? Could you help me to write a more serious argument?

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Actually you can build an example out of almost any function: Just notice that if $u: B(0,1)\to \mathbb{R}$, then $u_r(x)=u(x/r)$ is defined in the ball of radius $r$ and you get $$ \| u_r\|_{p,B(0,r)} = r^{n/p}\| u\|_{p,B(0,1)}, \quad \| \nabla u_r\|_{p,B(0,r)} =r^{-1} r^{n/p} \| \nabla u\|_{p, B(0,1)}. $$ Translating appropriately you get an example in the balls $B(x_k, r_k)$.

Also notice that your argument doesn't work: What you have is "$U$ bounded in one direction, then we have Poincare in the domain". However you're actually using the other direction (which you don't know), i.e. "If Poincare holds in $U$, then $U$ is bounded in one direction" (actually what you want to prove is a weaker version of this statement).

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  • $\begingroup$ +1 - it's a lot simpler than what I did (even though it is essentially equivalent). $\endgroup$ – Stephen Montgomery-Smith Jan 1 '15 at 15:07
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Find functions $u$ compactly supported inside $B(x,r)$ so that $\int_{B(x,r)} |u|^p \, dx / \int_{B(x,r)} |\nabla u|^p \, dx$ goes to infinity as $r \to \infty$. Probably something like $$ u(y) = \prod_{k=1}^N \cos\bigl((y_k-x_k)\sqrt N \pi/(2r)\bigr) I_{|x_k-y_k| < r/\sqrt N} $$

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