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Let $(X,g)$ be a non compact Riemannian manifold, such that its closure $\bar X=X\cup Y$ is a compact manifold with boundary $Y$.

Q: For the Poincare inequality $$\|u\|_{L^2}\leq C \|\nabla u\|_{L^2},$$ for any $u\in C^1_c(X)$, how to determine the Constant $C$ ?(what will it be related to ? e.g. dimension, diameter or volume?)

This is a revised version.

Thanks for the Arun Debray's comments.

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On $\mathbb R$, one can construct a sequence of $C_c^1$ functions $u_n$ such that $\|u_n\| > n-1$ but $\|u_n'\| \le 2$, providing a counterexample for any $C$, so this inequality cannot hold.

The idea is to let $u_n$ be $1$ on $[1,n]$, be $0$ on $\mathbb R\setminus (0, n+1)$, and not change too quickly in between. An explicit example is given by

$$ u_n(x) = \begin{cases} 1, &x\in[0,n]\\ x^2, &x\in[0,1/2)\\ 1 - (x-1)^2, &x\in [1/2, 1)\\1-(x-n)^2, &x\in(n, n+1/2)\\(x-(n+1))^2, &x\in[n+1/2, n+1)\\0, &\text{otherwise}.\end{cases}$$

$u_n$ is supported in $[0,n+1]$, and since $u_n\ge\chi_{[1,n]}$, then $\|u_n\|_{L^2}\ge \|\chi_{[1,n]}\|_{L^2} = n-1$.

$u_n'$ is supported in $[0,1]\cup[n,n+1]$, is continuous, and is bounded above by $u'(1/2) = 1$, so $\|u_n'\|_{L^2}\le 2$.

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  • $\begingroup$ Thank you for your answer. But if $X\cup Y\cup \bar Y$ is a compact manifold with boundary $Y\amalg\bar Y$, does the inequality hold? $\endgroup$
    – DLIN
    Commented Oct 10, 2017 at 3:49
  • $\begingroup$ I think the answer is yes: take $X\cup Y\cup \overline Y$ and embed it in its double $D$. Since $D$ is closed, the Poincaré inequality is true on $D$, so you can extend $u$ to a function on $D$, obtain the inequality there, and use it to deduce the inequality on the manifold with boundary. There are some details to fill in (e.g. what's the metric on $D$?), but it should be possible to turn this into a rigorous proof. $\endgroup$
    – Arun Debray
    Commented Oct 10, 2017 at 4:17

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