# counter-example for the Poincaré's inequality

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 $\Omega= \mathbb R^n .$

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 $\epsilon^{-1}\operatorname{supp}u$.)