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Let $\Omega \subseteq \mathbb{R}^n$ be a bounded domain with smooth boundary and suppose $u\in W_0^{1,p}(\Omega)$. If $v\in W^{1,p}(\Omega)$ satisfies $\lvert v \rvert \leq \lvert u \rvert$ in all of $\Omega$, can we conclude that $v\in W_0^{1,p}(\Omega)$.

Clearly, the statement is true if we assume that $u$ and $v$ are continuous up to the boundary of $\Omega$. More generally, the statement "feels" true and almost intuitive, but I have no idea how to solve or approach the problem.

Thanks in advance for any idea or references about this problem!

Quoka
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1 Answers1

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Firstly, we will use that there for $u\geq0\Rightarrow tr(u)\geq0$. See here for ideas on a proof.

Secondly, we will use that for $u\in W^{1,p}$ also $u^+=\max(0,u)\in W^{1,p}$ and similarly $u^-=\min(0,u)\in W^{1,p}$.

Thirdly, we will use the trace Operator which is a continuous linear Operator from $W^{1,p}\to L^p$ which allows the following identification $W^{1,p}_0=\{u\in W^{1,p}|tr(u)=0\}$.

Now, we start by splitting the functions into $$ u=u^++u^-\\ v=v^++v^-\\ |v|\leq|u|\Leftrightarrow v^+\leq u^+\wedge u^-\leq v^- $$

Now since $$0\leq tr(u^+-v^+)=tr(u^+)-tr(v^+)=-tr(v^+)\leq0$$ and $$0\leq tr(v^--u^-)=tr(v^-)-tr(u^-)=tr(v^-)\leq0$$ we can deduct that $$tr(u)=tr(u^++u^-)=tr(u^+)+tr(u^-)=0$$

maxmilgram
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