Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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)?

share|cite|improve this question

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.

share|cite|improve this answer
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.

share|cite|improve this answer
Thank you SteveH. – Pacciu Jul 7 '11 at 7:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.