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I know that if $u:\Omega\longrightarrow\mathbb{R}$ is Lipschitz continuous, by the Rademacher Theorem it is differentiable almost everywhere in $\Omega$. Moreover $$ \|Du\|_{L^{\infty}(\Omega)}\leq\operatorname{Lip}(u, \Omega), $$ where $\operatorname{Lip}(u, \Omega)$ is the Lipschitz constant of $u$ in $\Omega$. I do not understand why if $\Omega$ is convex, then $\|Du\|_{L^{\infty}(\Omega)}=\operatorname{Lip}(u, \Omega)$. I think that $u$ should be convex in order to get the equality. Am I right?

Thank You

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If $\Omega$ is convex, then you can use the mean value theorem and you get the other inequality, that is $\operatorname{Lip}(u, \Omega) \leq \|Du\|_{L^{\infty}(\Omega)}$.

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  • $\begingroup$ Ok, but how can I obtain this inequality by the Mean Value Theorem? Thank You $\endgroup$ – Redeldio Jun 5 '18 at 11:31
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(1) Recall definition $\|Du\|_\infty = \sup_x\ \sup_{|v|=1}\ \lim_t\ \bigg|\frac{u(x+tv)-u(x) }{t} \bigg| $

Here $|u(x+tv)-u(x) |\leq C|tv|$ where $C$ is a Lipschitz constant. That is $ \|Du\|_\infty \leq C$.

(2) Assume that $\Omega$ is convex. If $|u(x)-u(y)| =C|x-y|$, then there is a path $c(t)=tx+(1-t)y$. Here $$ \|Du\|_\infty |c'|\geq |\int_0^1\ (u\circ c) '(t)\ dt |= |u(x)-u(y)| $$

Hence $\|Du\|_\infty \geq C$.

(3) $\Omega =\mathbb{R}^2-B(o,1)$ is not convex. Then define $u(x)$ to be an infimum of lengths of curves from $p=(1,0)$ to $x$ in $\Omega$.

By considering a point $q=(-1,0)$, i.e., $ |u(p)-u(q) | \leq C|p-q|$, we have a Lipschitz constant $ C=\frac{\pi}{2}$.

If $u(p)<u(q)$, then $u(p)-u(q)\leq |p-q|$ so that $\| Du\|_\infty =1<C$.

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