A proof of this theorem can be found in Evans (p.272, 2nd Ed.). This question addresses persons who are familiar with the proof, since I have not been able to formulate a question that generalizes my problem.

THEOREM 1 (Trace Theorem). Assume $U$ is bounded and $\partial U$ is $C^1$. Then there exists a bounded linear operator $$T : W^{1,p}(U) \rightarrow L^p(\partial U)$$ such that

$\quad$(i) $Tu=u|_{\partial U}$ if $u \in W^{1,p}(U) \cap C(\bar{U})$


$\quad$(ii) $$\|Tu\|_{L^p(\partial U)} \le C \| u\|_{W^{1,p} (U)},$$ for each $u \in W^{1,p}(U)$, with the constant $C$ depending only on $p$ and $U$.

The problem lies in the second part (2.) of the proof.

Applying estimate (1) and changing variables, we obtain the bound $$ \int_\Gamma |u|^p dS \leq C \int_U |u|^p + |Du|^p dx. $$

Let $\Gamma$ denote the portion of the boundary of $U$ and let $\Gamma' = \Phi(\Gamma)$ the straightened portion of the boundary.

Using the change of variables formula we calculate $$ \int_\Gamma |u|^p dS = \int_{\Gamma'} |u \circ \Phi|^p dS' \leq \int_{\Phi(U)} |u \circ \Phi|^p + |D(u \circ \Phi)|^p dx $$

The first part can be transformed back as usual but we don't know how to transform the second part involving $D(u \circ \Phi)$.

  • 1
    $\begingroup$ You have $D(u \circ \Phi) = (Du(\Phi) )D\Phi$. Since the boundary is nice there is some uniform control on $D\Phi$. $\endgroup$
    – Umberto P.
    Sep 28, 2016 at 10:35
  • 2
    $\begingroup$ I agree that what you say is valid but I don't see how this is useful. $\endgroup$ Sep 28, 2016 at 11:02
  • 1
    $\begingroup$ Small typo in (ii): $U$ on the right inside the norm should be $u$. $\endgroup$
    – Glitch
    Sep 28, 2016 at 22:54


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