Let $u \in H^1(\Omega)$, $\Omega \subseteq \mathbb{R}^n$ open, let $\Omega_0 \subseteq \Omega$ be a bounded Lipschitz domain. Then it makes sense to speak about the trace of $u$ on $\partial \Omega_0$.

Is $\text{trace}_{\partial \Omega_0}(u)=u|_{\partial \Omega_0}$?

My idea: yes. I would like to show it by proving that $u \mapsto u|_{\partial \Omega_0}$ is a linear bounded map from $H^1(\Omega) $ to $H^{1/2}(\partial \Omega_0)$, restrict this map and use the characterization of the trace operator. However I'm not sure how to prove the well posedness and moreover I'm not sure the restriction of this operator would be surjective.


As gerw pointed out, the question doesn't really make sense, so I will rephrase it. As I understood, we define traces to be able to speak about the behaviour on the boundary of a Sobolev function.

I'm wondering if we can do away with traces if we look at the same function on the boundary of a sufficiently smooth subdomain, where the function is actually defined ($m_n$ almost everywhere, here I guess hides the difficulty, because this wouldn't imply $\mathcal{H}_{n-1}$ almost everywhere).

  • $\begingroup$ How do you define $u|_{\partial\Omega_0}$? One has to be careful here, since the boundary of $\Omega_0$ is a null set. And your might want to write $\operatorname{trace}_{\partial\Omega_0}(u|_{\Omega_0})$. $\endgroup$ – gerw Jan 13 at 10:10
  • $\begingroup$ Then I guess the whole question is doesn't really make sense, does it? $\endgroup$ – warm_fish Jan 13 at 11:10
  • $\begingroup$ Yes, the question as stated does not make sense. However, you have some question (otherwise you would not try to ask). Hence, we have the meta-question of what are you trying to ask ;) $\endgroup$ – gerw Jan 13 at 12:13
  • $\begingroup$ I think I anwered the meta-question $\endgroup$ – warm_fish Jan 13 at 20:50

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