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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)?

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2 Answers 2

hint: Forget the integral. Go directly and prove* the much stronger statement that

If $u\in C^{0,1}(\bar{\Omega})$ has Lipschitz constant at most 1, and $u$ vanishes on the boundary, then $\forall x\in\Omega$, $$u(x) \leq |u(x) - 0| \leq \operatorname{dist}(x,\partial\Omega)$$

Then trivially $I[u] \leq I[\operatorname{dist}(\cdot,\partial\Omega)]$.

*Said proof is one line long.

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Your're right! Thanks a lot. –  Pacciu Jul 7 '11 at 7:53

Not a complete answer but for the case $N=1$ and $\Omega=[-1,1]$, seems that it follows from integration by parts: $$\int_{-1}^1u(x)dx=-\int_{-1}^1xu'(x)dx,$$ since $u(-1)=u(1)=0$, and so $$\left|I(u)\right|\leq \int_{-1}^1|x|dx=\int_{-1}^1(1-|x|)dx.$$ Not sure exactly how to generalize this though.

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Thank you SteveH. –  Pacciu Jul 7 '11 at 7:56

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