# How to prove Poincaré-like inequality for the integral over the boundary?

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.

• Why do you think this is true? – Tomás Apr 7 '15 at 23:10
• Well, it worked for Poincare inequality. – Aron Apr 11 '15 at 18:58

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 $g=f-1/|\Omega|\int_\Omega f$.