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?


$(1)\,$ 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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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