# Sobolev space and weak maximum principle

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 $$$$ess \sup_{\Omega} u \leq \sup_{\partial\Omega } u.$$$$ From the trace theorem, we know that for any $$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 $$ess \sup_{\partial\Omega} Tr(u)$$.

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 $$H^1(\Omega) \rightarrow L^2(\partial \Omega) : u \mapsto \operatorname{Tr} (u-a)^+$$.
• Thanks, I ignore the continuity of $(u-a)^+$. Commented Aug 27, 2022 at 5:50