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While going through some old notes from a PDE's course I came across Poincaré's inequality. In the simplest case for functions in $H^1_0=W^{1,2}_0$ states that

Let $\Omega \subset \mathbb R^n$ be open and bounded at least in one direction. Then there is a constant $C>0$ such that

$$ \| u \| \leq C \| Du \|, \hspace{0.1in} \text{for all } \hspace{0.1in} u \in H^1_0 ,$$

where $ \| \cdot \| = \| \cdot \|_2 .$

I want to find a counter-example for the upper half-plane, namely for the set

$$ \Omega=\{ (x,y) : y > 0 \} \subset \mathbb R^2.$$

I thought to pick a distribution $ u $ and consider the dilation $ v = u(\epsilon x)$ for small $ \epsilon > 0.$ If, the inequality was true in this case, then after substituting we would end up with something like $ 1/ \epsilon \leq C \epsilon$ which obviously cannot be true for all $ \epsilon >0.$

My questions are the following:

  1. Does the above sound fine? In particular, I am not sure whether I actually use the fact that we are in the half-plane or the above arguement works also in the case where $\Omega= \mathbb R^n .$

  2. What about higher dimensions?

Any help would be really appreciated.

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Your approach is correct; this is what I would do myself. It uses the structure of the half-plane in that we need to know that $u(\epsilon x)$ vanishes on the boundary. We know that $u$ does, but to conclude the same about $u(\epsilon x)$ we need to know that
$$ x\in\Omega \implies \epsilon^{-1}x\in\Omega $$

(In case this is unclear: consider the case of continuous compactly supported $u$, and note that the support of $u(\epsilon x)$ is exactly $\epsilon^{-1}\operatorname{supp}u$.)

So, any domain that is invariant under scaling will work. It can be quarter-plane or some smaller sector. In higher dimensions, it could be an infinite cone.

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  • $\begingroup$ Dear Crazy Ivan, thank you very much for your reply! $\endgroup$ – passenger Dec 21 '17 at 12:32

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