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

Question

$A$ 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 $u\geq v$?

Answer 1

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 $A$.