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 $\Omega$.
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.