Positive part of trace of a Sobolev function

Question

Suppose that $\Omega$ has a smooth boundary, $u\in H^1(\Omega)$. I want to know whether the following holds: $$ tr(u^+)=tr(u)^+ $$ where $tr$ denotes the trace of a Soboelv function. In other words, can we prove that $$tr(u)\geq 0\Rightarrow tr(u^+)=tr(u)$$ I do hope it holds, but I don’t know how to prove it. Or any reference is welcoming. Thanks!

Answer 1

If $tr(u)\geq 0$, it follows $tr(u)^-=0$, so by linearity of the trace operator $$tr(u^+)=tr(u+u^-)=tr(u)+tr(u)^-=tr(u).$$