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Let $ n \in \mathbb{N}, \ \Omega \subset \mathbb{R}^n $ be a bounded domain with Lipschitz-boundary and $ S:H^1(\Omega) \to L^2(\partial \Omega) $ the trace Operator.

For $ u \in H^1(\Omega): \quad$ $S$(max{u,0}) = max{$Su$,0} in $L^2(\partial\Omega$).

Does someone have an idea why this is true?

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By definition of the trace operator, this equality is true for $u \in H^1(\Omega) \cap C(\bar\Omega)$. By density, you can extend this to the entire space $H^1(\Omega)$.

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  • $\begingroup$ It seems that you're implicitly relying on the continuity of $u \mapsto \max(u,0)$ in $H^1(\Omega)$. However, this is not a continuous operator. $\endgroup$
    – Michał Miśkiewicz
    Commented Feb 4, 2019 at 14:05
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    $\begingroup$ @MichałMiśkiewicz: This operator is continuous. This follows from Stampacchia's lemma on $\nabla\max(u,0)$ and the dominated convergence theorem. $\endgroup$
    – gerw
    Commented Feb 4, 2019 at 14:07
  • $\begingroup$ Can you add a proof of this fact then? I was pretty sure I had a counterexample in mind, so it'd be very instructive for me to see it in detail. $\endgroup$
    – Michał Miśkiewicz
    Commented Feb 4, 2019 at 14:10
  • $\begingroup$ I remembered now that my example only shows that $u \mapsto \max(u,0)$ is not uniformly continuous. $\endgroup$
    – Michał Miśkiewicz
    Commented Feb 4, 2019 at 15:22

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