Let $\Omega$ be a smooth bounded domain, $H^1(\Omega) :=\{u: u, Du\in L^2(\Omega)\},$
and $H^1\_0(\Omega)$ is the closure of $C^{\infty}\_{c}(\Omega)$ in $H^1(\Omega)$. Define:


* $\sup\_{\partial\Omega } u:=\inf\{a :(u-a)^+\in H^1\_0(\Omega)\}$
* $ess \sup\_{\Omega} u:= \inf\{a :(u-a)^+=0, a.e.~ in~ \Omega\}$


The weak maximum principle tells me that if $-\Delta u \leq 0$ and $u\in H^1(\Omega)$ then \begin{equation}ess \sup\_{\Omega} u \leq \sup\_{\partial\Omega } u.\end{equation}
From the trace theorem, we know that for any $u\in H^1(\Omega),$ the trace of $u$ called $Tr(u)$ is a measurable function on $\partial \Omega$, hence we can define $ess \sup\_{\partial\Omega} Tr(u).$


* I'm confused about the difference of $ess \sup\_{\partial\Omega} Tr(u)$ and $\sup\_{\partial\Omega } u$, could the same be true in the weak maximum principle by replacing $\sup\_{\partial\Omega } u$ with $ess \sup\_{\partial\Omega} Tr(u)$.



Sobolev space and weak maximum principle
I may have overlooked something, but I think $\operatorname{esssup}\_{\partial \Omega} \operatorname{Tr} u = \sup\_{\partial \Omega} u$ for all $u \in H^1(\Omega)$.
This follows at once from the well-known characterization $H^1\_0(\Omega) = \{u \in H^1(\Omega) \mid \operatorname{Tr} u = 0\}$ and the fact that $(\operatorname{Tr}u - a)^+ = \operatorname{Tr} ((u-a)^+)$ for all $u \in H^1(\Omega)$. The last equation is easily verified for $u \in C^{\infty}(\Omega)$, then extended to $u \in H^1(\Omega)$ by continuity of $H^1(\Omega) \rightarrow L^2(\partial \Omega) : u \mapsto (\operatorname{Tr}u - a)^+$ and $H^1(\Omega) \rightarrow L^2(\partial \Omega) : u \mapsto \operatorname{Tr} (u-a)^+$.


