Consequence of trace theorem

Asked Oct 17 '20 at 16:33

Active Oct 17 '20 at 17:01

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Trace theorem says that there exists unique continous linear mapping $\gamma : W^{k,p}(\varOmega) \rightarrow L^q(\varOmega)$ that $\gamma(u) = u|_{\partial \varOmega}$.

It is correct to write $||u||_{L_q(\partial \varOmega)} = ||\gamma(u)||_{W^{k,p}(\varOmega)}$? I think that that is what the theorem says, if I understand right. By the way, I am trying to prove $||u||_{L^q(\partial \varOmega)} \leq c ||u||_{W^{k,p}(\varOmega)}$ and I am using the equality above to prove.

edited Oct 17 '20 at 17:01

asked Oct 17 '20 at 16:33

Cygne

$||\gamma(u)||{W^{k,p}(\varOmega)}$ is wierd since $\gamma(u)$ is defined on $\partial \Omega$ and the $||.||{W^{k,p}(\varOmega)}$ norm is not defined there. Maybe some typo somewhere? – Korf Oct 27 '20 at 12:42

Consequence of trace theorem

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