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In a problem I ended up with the vector space $V=\{f\in H^1(\Omega):\int_\Omega f=0\}$. I think it can be proven (and it would be really helpful) that $||f||_{L^2(\Omega)}\le c ||\nabla f||_{L^2(\Omega)}$ because this would show norm equivalence. But how do I show that?

When I try to bound $f$ by $\nabla f$ I end up with the integral of $f$ over $\partial\Omega$ and I cannot see how this bounds the integral of $f$ in $\Omega$. Is there something I don't see or is there an other way to do it?

EDIT: $\Omega$ is a bounded subset of $\mathbb{R}^n$.

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up vote 2 down vote accepted

If you know about Sobolev embedding theorems (and specifically the Rellich compactness theorem) then a simple proof can be found on page 1 of this note.

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Thanks, it was actually quite simple. – tst Nov 6 '12 at 15:18
Hmm, that's actually a quite nice proof. +1 even though your answer competes with mine. ☺ – Harald Hanche-Olsen Nov 6 '12 at 15:28

That is the Poincaré inequality if $\Omega$ is bounded. The result is false in general when $\Omega$ is unbounded.

To see the latter, consider $\Omega=\mathbb{R}$ and let $$f(x)=\begin{cases}x&|x|\le n\\2n-x&n<x<2n\\x-2n&-2n<x<n\\0&|x|\ge 2n\end{cases}$$

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Oh, yes I forgot to say that $\Omega$ is bounded and $\partial \Omega$ is smooth. – tst Nov 6 '12 at 15:07
Poincaré inequality cannot be applied directly because the function is not necessarily $0$ on the boundary. – tst Nov 6 '12 at 15:17
The Poincaré inequality that I know does not require the function to vanish on the boundary. (But it does require the boundary to be at least Lipshitz, and I had forgotten that.) – Harald Hanche-Olsen Nov 6 '12 at 15:25

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