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Let $\Omega \subseteq \mathbb{R}^n$ be a nice convex domain (bounded,open,connected, with smooth boundary...)

Let $u:\Omega \to \mathbb{R} \in W^{1,\infty}(\Omega)$.

Define $c_u=\text{ess} \sup_{\Omega}{\|Du\|_{op}}. \, \,$ More precisely, for every $x \in \Omega$ we consider $Du(x)=\big(\partial_1 u(x),\partial_2 u(x),...,\partial_n u(x)\big)$ (all the derivatives are weak of course).

Now we take the operator norm (this is merely the usual Euclidean norm) of $Du(x)$, and so we get a function $x \to \|Du(x)\|_{op} $ from $\Omega$ to $\mathbb{R}$. $\, \,c_u$ is defined to be its essential supremum.

Question: Is it true that every $u \in W^{1,\infty}(\Omega)$ is $c_u$-Lipschitz?

Remarks:

$(1)\,$ The statement clearly holds in the case where $u$ is continuously differentiable everywhere,by the mean value theorem. (It even holds if $u$ is only differentiable everywhere, when we take $c_u$ to be the actual supremum, instead of the essential supremum).

$(2) \,$ It is a well-known theorem that every $u \in W^{1,\infty}(\Omega)$ is $\|Du\|_{\infty}$-Lipschitz where $\|Du\|_{\infty}$ is defined as $\sum_{i=1}^n \text{ess} \sup_{\Omega}|\partial_i u|$.

The classic proofs (see Evans PDE book for instance, theorem 4, pg 279) goes as follows:

Let $\epsilon >0$, and let $\eta_{\epsilon}$ be the usual mollifier, and define $u^{\epsilon}(x)=u * \eta_{\epsilon}(x)= \int_{\Omega} \eta_{\epsilon}(x-y)u(y)$.

It can be proven that $ \partial_i u^{\epsilon}=\partial_i u * \eta_{\epsilon}$, hence (since $\int_{\mathbb{R}^n} \eta_{\epsilon}=1$, and $\partial_i u(y) \le \text{ess} \sup_{\Omega}|\partial_i u|$ a.e)

$|\partial_i u^{\epsilon}(x)| \le \text{ess} \sup_{\Omega}|\partial_i u|$ for all $x \in \Omega$. Thus,

$$ (1) \, \, \|Du^{\epsilon}(x)\|_{op}=\sqrt{\sum_{i=1}^n |\partial_i u^{\epsilon}(x)|^2} \le \sum_{i=1}^n |\partial_i u^{\epsilon}(x)| \le \sum_{i=1}^n \text{ess} \sup_{\Omega}|\partial_i u|=\|Du\|_{\infty}$$

holds for every $x \in \Omega$.

From here, the proof proceeds as follows:

$$ u^{\epsilon}(x) - u^{\epsilon}(y)=\int_0^1 \frac{d}{dt}u^{\epsilon}(tx+(1-t)y)dt=\int_0^1 Du^{\epsilon}(tx+(1-t)y)\big((x-y)\big)dt \Rightarrow$$

$$ |u^{\epsilon}(x) - u^{\epsilon}(y)| \le \int_0^1 \|Du^{\epsilon}(tx+(1-t)y)\|_{op}\cdot\|x-y\|\big)dt \stackrel{(1)}{\le} \|Du\|_{\infty} \cdot\|x-y\|$$

Now letting $\epsilon \to 0$ (recalling $u^{\epsilon} \to u$ a.e), we conclude that

$$ |u(x) - u(y)| \le \|Du\|_{\infty} \cdot\|x-y\|$$

as required.

The key point is that in estimate $(1)$ we had to pass to the essential supremum of each component separately.

If we could prove $\text{ess} \sup_{\Omega}{\|Du^{\epsilon}\|_{op}} \le \text{ess} \sup_{\Omega}{\|Du\|_{op}}$, then the above proof would produce the stronger result.

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Since $\partial_i u^\epsilon = \partial_i u * \eta^\epsilon$, you have that $D_v u^\epsilon = D_v u * \eta^\epsilon$ for every vector $v$. Therefore, you have $$|D_vu^\epsilon(x)| \le \text{ess} \sup |D_v u|\le \|Du\|_{op}|v|.$$ By dividing by $|v|$ and taking supremum over $x$, you obtain the wanted bound.

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  • $\begingroup$ I am not sure I am following, when I try to open it, I get the following: Fix $v =(v^1,...,v^n) \in \mathbb{R}^n$. $D_vu^\epsilon(x) =\big(Du^\epsilon(x)\big) (v)= \sum_{i=1}^n v^i \partial_i u^{\epsilon} (x)=\sum_{i=1}^n v^i (\partial_i u * \eta_{\epsilon})(x)$ $=\sum_{i=1}^n v^i \int_{\Omega} \eta_{\epsilon}(x-y)\partial_i u(y)dy$ So, $|D_vu^\epsilon(x)| \le \sum_{i=1}^n |v^i| \int_{\Omega} \eta_{\epsilon}(x-y)|\partial_i u(y)|dy \le ?$ I do not see how do you finish to get your claim. $\endgroup$ – Asaf Shachar Dec 11 '16 at 16:48
  • $\begingroup$ $ֿ\sum v^i \int_\Omega \eta_\epsilon(x-y) \partial_i u(y) dy = \int_\Omega \eta_\epsilon(x-y) \left(\sum v^i \partial_i u(y)\right) dy$.. $\endgroup$ – C M Dec 11 '16 at 16:54
  • $\begingroup$ Sorry, published the previous comment accidentally. I follows that $|D_vu^\epsilon(x)| \le \int_\Omega \eta_\epsilon(x-y) \|Du(y)\|_{op} |v| dy \le \text{ess}\sup \|Du\|_{op} |v|$. $\endgroup$ – C M Dec 11 '16 at 16:58
  • $\begingroup$ Another viewpoint -- there is nothing unique in the coordinate directions, for which you obviously get the wanted bound. That is, given $x$, choose a unit vector $v$ for which $D_vu^\epsilon(x)=\|Du^\epsilon(x)\|_{op}$. Now choose coordinates at a ball around $x$ such that $v=x^1$. $\endgroup$ – C M Dec 11 '16 at 17:05

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