# Why if $\nabla(u-v)=0$ in $\{u<v\}$ imply $u\geq v$?

$A$ is a bounded domain in $\mathbb{R}^N$ with $\partial A$ of class $C^2$;

$u,v:\overline{A}\to\mathbb{R}$ - we write $u\leq v$ if $u(x)\leq v(x)$ for a.e. $x\in A$ and $u(x)<v(x)$ if $u(x)\leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure;

$\|\nabla u\|_\infty<1$, $\|\nabla v\|_\infty<1$;

$u,v\in C^{0,1}(\overline{A})$ and $u=0$ on $\partial A$ and $v\leq 0$ on $\partial A$.

Why $\nabla(u-v)=0$ in the set $\{u<v\}$ imply that $u\geq v$?

• Please clarify what is meant with ${u<v}$. Also, what are $u$ and $v$? Are those functions or variables? – Mefitico Jul 16 '18 at 19:16
• Assuming that $u$ and $v$ are functions of $x,y,z$ in the $3$-dimensional real space and that $\{u<v\}$ is the set of $(x,y,z)$ such that $u(x,y,z)<v(x,y,z)$, what about $u=0$ and $v=1$ (i.e, they are constant functions)? – Batominovski Jul 16 '18 at 19:19
• $A$ is a bounded domain in $\mathbb{R}^N$ with $\partial A$ of class $C^2$; $u,v:\overline{A}\to\mathbb{R}$ - we write $u\leq v$ if $u(x)\leq v(x)$ for a.e. $x\in A$ and $u(x)<v(x)$ if $u(x)\leq v(x)$ and $u(x)<v(x)$ in a subset of $A$ having positive measure. – BlackHawk Jul 16 '18 at 19:26
• I think then u and v must vanish at the boundary then. @BlackHawk please give the precise space for u and v – Calvin Khor Jul 16 '18 at 21:05
• There must be more information missing. I can imagine either $u=v$ on the boundary, or $u$ is greater than $v$ somewhere in the domain... – Jeff Jul 16 '18 at 21:05

Write $w=(v-u)_+$, where $t_+=\max\{t,0\}$. Then $w\in C^{0,1}(\bar{A})$, $\nabla w=0$ on $A$ and $w\leq 0$ on $\partial A$. It follows that $w\leq 0$ on $A$.