Poincaré inequality in unbounded domain Help me please, how can I show that Poincaré inequality doesn't hold in an unbounded domain?
Thanks a lot!
If $\Omega$ is a bounded domain and $u \in H_{0}^{1}(\Omega)$ the following inequality holds :
$\left \| u \right \|_{L^{2}(\Omega) }\leq c \left \| \nabla u \right \|_{L^{2}(\Omega) } $
where $c$ depends only on $\Omega$ and not on $u$.

*

*Poincaré inequality is true if $\Omega$ is bounded in a direction or of finite measure in a direction. 

*But not in general: if $\Omega=\mathbb R$, $\varphi$ smooth with compact support and such that $\varphi=1$ on $[0,1]$, $\varphi(x)=0$ if $x\geq 2$ (bump function), $\varphi_n(t)=\varphi\left(\frac tn\right)$, we have
$$\lVert \varphi_n\rVert_{L^2}^2=\int_0^{+\infty}\varphi\left(\frac tn\right)^2dt=
n\int_0^{+\infty}\varphi(s)^2ds\geq n$$
and $$\lVert\varphi'_n\rVert^2_{L^2}=\frac 1{n^2}\int_0^{+\infty}\varphi'\left(\frac tn\right)^2dt=\frac 1n\int_0^{+\infty}\varphi'(s)^2ds$$
so Poincaré inequality cannot be true (if it was, we would be able to find $c>0$ such that $\lVert \varphi_n\rVert^2_{L^2}\leq c^2\lVert \varphi'_n\rVert^2_{L^2}$ hence $n\leq c^2\frac 1n$ for each $n$, which obviously can't hold). 

*For necessary and sufficient conditions, I don't know.

