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 $ess \sup_{\partial\Omega} Tr(u)$.