Does someone know any inequality between $||u||_{p}$ and $||\gamma (u)||_{p, \partial \Omega}$, where $\gamma$ is the Trace Operator? I need to find something like

$||u||_{p}\leq C||\gamma (u)||_{p, \partial \Omega}$ or $||u||_{p}\leq C(||\gamma (u)||_{p, \partial \Omega} + ||\nabla u||_{p}) $ ,

but I couldn't find any in the bibliography.

Thanks in advice.

Here $ ||u||_{p}=||u||_{L^{p}} $, $\Omega $ Lipschitz's Domain.


Not a proof, but an idea too long for comments that could maybe become a proof...I think you can let $\phi_n\ge 0$ be a smooth function with $\phi_n = 1 $ on $\partial \Omega$ and support in $\{x\in\Omega : d(x,\Omega) < 1/n\}$ with $|\nabla \phi_n| \sim Cn^\alpha d(x,\Omega)$ there, for some constants $C,\alpha$ I don't know yet. Then by Poincare for functions in $W^{1,p}_0(\Omega)$, $$ \|u\|_p \le \| u \phi_n \|_p + \|u(1-\phi_n)\|_p \lesssim \| u \phi_n \|_p + \|\nabla u(1-\phi_n)\|_p + \|u\nabla\phi_n\|_p$$ $ \| u \phi_n \|_p \to 0$, $\|\nabla u (1-\phi_n)\|_p \to \|\nabla u\|_p$, and I'm guessing that for the correct choice of $C,\alpha$ that might depend on $p$, $$ \|u\nabla \phi_n \|_p \to \|\gamma (u)\|_{L^p(\partial \Omega)}$$ with the hope that $\nabla \phi_n $ "converges in some sense" to the surface measure on $\partial \Omega$?


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