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Suppose $u\in W^{1,1}$ and $\partial u$ is $C^1$. I want to prove the following: $$\int_{\partial\Omega}|u-\bar u|\leq A\int_{\Omega}|\nabla u|$$ where $\bar u=\frac{1}{|\Omega|}\int_{\Omega}u$ and $A>0$. Note that unlike Poincaré inequality, the left integral is over $\partial\Omega$ and not $\Omega$. How can I do that?

I tried to prove in a similar way to Poincare inequality proof (arguing by contradiction), but with no success. I think I am missing something because of the $\partial\Omega$ issue.

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  • $\begingroup$ Why do you think this is true? $\endgroup$ – Tomás Apr 7 '15 at 23:10
  • $\begingroup$ Well, it worked for Poincare inequality. $\endgroup$ – Aron Apr 11 '15 at 18:58
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The answer is yes, and it follows immediately from the usual Poincare inequality: We know that for any function of zero average we have $$ \| g\|_{W^{1,1}} \leq C\| \nabla g \|_{L^1}. $$ Then we know that for such a $g$, if $T$ denotes the trace operator,
$$ \| Tg\|_{L^1(\partial \Omega)} \leq C\| g\|_{W^{1,1}} \leq C\| \nabla g \|_{L^1}. $$ Now just plug $g=f-1/|\Omega|\int_\Omega f$.

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