Let $u \in H^1(\Omega)$ on a bounded smooth domain $\Omega$. Is it true that if $u \geq 0$ a.e., then $Tu \geq 0$ a.e. on $\partial\Omega$ where $T$ is the trace?
I don't think it is, since $u$ can be negative on the null set $\partial\Omega$. So I suppose the question, under what minimal assumption is it true? Is there something weaker than continuity?
This is true. Recall that the trace operator $T$ is continuous from $H^1(\Omega)$ to $L^2(\partial\Omega)$. Suppose $u\in H^1(\Omega)$ is nonnegative. Consider an approximating sequence $u_n\in C^1(\overline{\Omega})$. These functions can be taken to be nonnegative, for example because mollification of $u$ produces a nonnegative approximating sequence. Since $Tu_n$ is just the boundary values of $u_n$, we have $Tu_n\ge 0$ pointwise. The continuity of $T$ implies $Tu_n\to Tu$; in $L^2(\partial \Omega)$, hence $Tu\ge 0$ a.e. (Indeed, $L^2$ convergence implies there's an a.e. convergent subsequence.)
