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Let $\Omega\subset\mathbb{R}^N$ be a bounded regular domain and consider the Sobolec space $W^{1,p}(\Omega)$ for $p\in (1,\infty)$. I have some doubts with th trace theorem.

Roughly speaking, the Trace theorem states that there is a linear functional $$T:W^{1,p}(\Omega)\to L^p(\partial \Omega)$$

such that $Tu=u_{|\partial\Omega}$ for every $u\in C^\infty(\overline{\Omega})$.

My questions is:

The only functions for which $Tu=u_{|\partial\Omega}$ are the $C^\infty(\overline{\Omega})$ functions, or can we find, for instance, functions not continuous in $W^{1,p}(\Omega)$ such that the same equality holds?

Thank you

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  • $\begingroup$ You can relax the requirement of $u \in C^{\infty}(\bar{\Omega})$ to be $C(\bar{\Omega})$, but your question is essentially the same. $\endgroup$ – al0 May 28 '13 at 12:38
  • $\begingroup$ Out of personal curiosity: What measure is used on $\partial\Omega$? $N-1$-dimensional Hausdorff measure? Surely it can't be $N$-dimensional Lebesgue measure, because $\partial\Omega$ will be a Lebesgue(-$N$) null set for most "nice" $\Omega$ making $L^p(\partial\Omega)$ rather boring. $\endgroup$ – kahen May 28 '13 at 13:37
  • $\begingroup$ @kahen, yes it is surface measure. $\endgroup$ – Tomás May 28 '13 at 14:39
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The reason that the equality $Tu=u_{|\partial \Omega}$ is stated for functions in $C(\overline{\Omega})$ is that for other functions it is not clear what $u_{|\partial \Omega}$ means. Of course, we can take any function $u\in W^{1,p}(\Omega)$ and extend its domain to $\overline{\Omega}$ by letting $u$ be equal to $Tu$ on the boundary. This will be an instance of $Tu=u_{|\partial \Omega}$ being true despite $u$ not necessarily being in $C(\overline{\Omega})$. But this is merely a tautology that does not teach us anything new.

But there are useful and nontrivial results along the lines of your question: starting with $\phi\in L^p(\partial \Omega)$, one can extend $\phi$ to a function in $\Omega$ (by solving the Dirichlet problem for some elliptic operator) and then recover $\phi$ from $u$ via nontangential limits. This does not require $\phi$ to be continuous. For a simple example, take $\Omega=\{z\in\mathbb C: |z|<1\}$ and $u=\log|z-1|$. Then $u\in W^{1,p}(\Omega)$ for $p<2$. The trace operator agrees with the boundary values of $u$ understood in the sense of nontangential limits.

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  • $\begingroup$ Interesting. So if $\Omega=B(0,1)=B$, can I conclude from here math.stackexchange.com/questions/404142/… that for almost $t\in (0,1]$ the trace of $u$ in $W^{1,p}(B_t)$ is the restriction of $u$ to $\partial B_t$? $\endgroup$ – Tomás May 28 '13 at 14:48
  • $\begingroup$ @Tomás For almost every $t$, almost every point of $\partial B_t$ is a Lebesgue point of $u$. Does the trace agree with the values at Lebesgue points on the boundary? I'm not sure. $\endgroup$ – ˈjuː.zɚ79365 May 28 '13 at 23:44
  • $\begingroup$ This is really a annoying thing and I am having a hard time in trying to understand it. If you have access to this paper, please take a look on it Proposition 1: Zhikov, V. V.; Pastukhova, S. E. The compensated compactness principle. (Russian) Dokl. Akad. Nauk 433 (2010), no. 5, 590--595; translation in Dokl. Math. 82 (2010), no. 1, 590–595 $\endgroup$ – Tomás May 29 '13 at 11:16

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