maximum of trace operator and 0

Question

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?

Answer 1

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 $H^1(\Omega)$.