# 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 $u$.

• can u write down the inequality , just for reference ! – Theorem May 23 '12 at 9:56
• @Ananda I have added inequality, thanks! – Lilly May 23 '12 at 10:20
• Start with $\Omega=\mathbb{R}$ and consider a sequence of functions which are constant on $[-n, n]$ then decay smoothly and vanish outside of $[-n-1, n+1]$. Suppose by contradiction that the inequality holds and see what happens for $n \to \infty$. – Giuseppe Negro May 23 '12 at 10:26

• 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).