Incremental ratios (alias difference quotients) of Lipschitz functions

Question

Let $u$ be Lipschitz on (the closure of)¹ a bounded domain $\Omega\subseteq\mathbb{R}^n$. Set:

$$I(u,x,y):=\frac{|u(x)-u(y)|}{|x-y|},$$

which I call incremental ratio of $u$ w.r.t. $x,y$. My teacher states that:

$$\sup_{x,y\in\overline\Omega}I(u,x,y)=\hspace{-.3cm}\sup_{\substack{x,y\in\overline\Omega \\ |x-y|<\epsilon}}\hspace{-.25cm}I(u,x,y),$$

for any $\epsilon>0$. How do I prove that? Googling for the title doesn't seem to retrieve much…

¹ $u$ is Lipschitz, so as soon as it is defined on any subset of $\mathbb{R}^n$, it can be extended to the whole of $\mathbb{R}^n$, preserving the Lipschitz constant: thus sayeth Kirszbraun. For the sups below we will need it defined on the closure, but by Kirszbraun it is enough to only define it on $\Omega$.

Is $u$ defined only in $\Omega?$ If so, you need to change those sups. — zhw., Nov 12 '16 at 18:37
@zhw. Right, but $u$ is Lipschitz, so Kirszbraun tells us we can extend it to the whole of $\mathbb{R}^n$ with the same Lipschitz constant. — MickG, Nov 12 '16 at 18:41
@zhw. Well, that is a start. And the context in which I need this is minimal area, and the final existence theorem we proved is for domains with positive curvature, which means convex, right? So that could be a start for an answer, and enough as a reference in my notes. We could agree that you answer with a proof for convex domains, and then we see if someone else comes in and gives a proof or hint about how to generalize, and if not, I just accept your answer after some time. What say you? — MickG, Nov 13 '16 at 20:53
I added to my answer. The result can fail for general $\Omega.$ — zhw., Nov 15 '16 at 18:52

zhw. · Accepted Answer · 2016-11-16T00:25:43.173

The result is false for general $\Omega.$ Consider in the plane the U-shaped closed region $\overline {\Omega} = [0,1]\times [0,3] \cup [1,2]\times [0,1] \cup [2,3]\times [0,3].$

Define:

$$ u(x,y) = \begin{cases} 0 & 0\le y \le 2 \\ 2-y & 0\le x\le 1, 2\le y \le 3\\y-2 & 2\le x\le 3, 2\le y \le 3 \end{cases}.$$

Then $u$ is Lipshitz in $\overline {\Omega}.$ Now take $\epsilon=1/2.$ Then $p,q\in \overline {\Omega}, |p-q|<1/2$ implies $|u(p)-u(q)| \le |p-q|.$ But note $|u(1,3)-u(2,3)| = |(-1)-1| = 2 > 1 = |(1,3)-(2,3)|.$

Proof for convex $\Omega:$ Let $\epsilon > 0.$ Suppose $|u(y)-u(x)|\le C|y-x|$ for $|y-x| <\epsilon.$ Let $a,b\in \overline {\Omega}.$ Then the line segment $[a,b]\subset \overline {\Omega}.$ Choose $m$ such that $|b-a|/m<\epsilon.$ Let $x_k = a +(k/m)(b-a), k = 0, \dots , m.$ Then

$$\begin{split} |u(b)- u(a)| &= \left|\sum_{k=1}^{m}\big(u(x_k)-u(x_{k-1})\big)\right|\\ &\le\sum_{k=1}^{m}|u(x_k)-u(x_{k-1})| \\ &\le \sum_{k=1}^{m}C|x_k-x_{k-1}|\\ & = C|b-a|. \end{split}$$

That gives the result.

Thanks for the drawings, but I didn't get too much out of the second one, so I deleted it. I'm going to leave the first drawing in - it's a great help. Thank you. — zhw., Nov 16 '16 at 00:26
The idea of the second one was that the black lines were the region, and the red ones were the function's value. The plane axes were y and z, whereas the x axis coincided with the $z=-y$ line. Such a way of drawing is something I learnt way back in high school. It had been a while since I lasy drew such a drawing :). Anyways you're welcome. In fact, thank YOU for the answer. — MickG, Nov 16 '16 at 06:01

Incremental ratios (alias difference quotients) of Lipschitz functions

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