Sobolev space and weak maximum principle

Question

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

Answer 1

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 $H^1(\Omega) \rightarrow L^2(\partial \Omega) : u \mapsto \operatorname{Tr} (u-a)^+$.