Let $\Omega \subseteq \mathbb{R}^N$ be open and bounded, let $\mathcal{I}:C(\overline{\Omega}) \ni u\mapsto \int_\Omega u(x)\ \text{d} x \in \mathbb{R}$ and set:
$\phi(x):=\text{dist} (x,\partial \Omega) =\inf_{y\in \partial \Omega} |x-y|$
for $x\in \overline{\Omega}$. Function $\phi$ is Lipschitz continuous in $\overline{\Omega}$ with Lipschitz constant equal to $1$, hence it is a.e. differentiable and $\lVert \nabla \phi \rVert_{\infty ,\Omega}\leq 1$.
Is it true that $\phi$ maximize $\mathcal{I}[u]$ in the Lipschitz class $C^{0,1}(\overline{\Omega})$ under constraints $\lVert \nabla u\rVert_{\infty ,\Omega} \leq 1$ and $u(x)=0$ for each $x\in \partial \Omega$?
In other words, is it true that $\phi$ solves:
$\tag{1} \max \left\{ \mathcal{I}[u],\ \text{with } u\in C^{0,1}(\overline{\Omega}),\ u\big|_{\partial \Omega}=0\ \text{and } \lVert \nabla u\rVert_{\infty ,\Omega} \leq 1 \right\} \; \text{?}$
I'm almost sure $\phi$ does solve (1), but I don't know how to prove it. Actually, I'm in trouble here, because the constraint is not in integral form, hence I don't know if Lagrange's multipliers apply...
Even in the simpler case when $N=1$ solution of problem (1) eludes me. In this case, up to translation and scaling, one can assume $\Omega =]-1,1[$ so that problem (1) becomes:
$\tag{2} \max \left\{ \int_{-1}^1 u(x)\ \text{d} x,\ \text{with } u\in C^{0,1}([-1,1]),\ u(-1)=0=u(1) \text{ and } \sup_{-1\leq x\leq 1}|u^\prime(x)|\leq 1\right\} \; .$
How can I prove that $\phi (x)=1-|x|$ (which is the distance from the boundary of $[-1,1]$) solves (2)?