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 $u$.

  • $\begingroup$ can u write down the inequality , just for reference ! $\endgroup$ – Theorem May 23 '12 at 9:56
  • $\begingroup$ @Ananda I have added inequality, thanks! $\endgroup$ – Lilly May 23 '12 at 10:20
  • 1
    $\begingroup$ 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 $n \to \infty$. $\endgroup$ – 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).
  • For necessary and sufficient conditions, I don't know.
  • $\begingroup$ Thanks a lot, very good explanation and example. $\endgroup$ – Lilly May 25 '12 at 9:56
  • $\begingroup$ @Davide Giraudo what do you mean by bounded in a direction ? $\endgroup$ – Vrouvrou May 4 at 9:27

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