Poincaré inequality for $W_0^{1,\infty}$ In the book A first course in Sobolev spaces by Leoni, the following Poincaré inequality for $W_0^{1,p}(\Omega)$ is stated:
Suppose $\Omega\subset \mathbb{R}^n$ has finite width (lies between two parallel hyperplanes) and $p\in [1,\infty)$. Then for all $u\in W^{1,p}_0(\Omega)$,
$$\|u\|_{L^p} \leq C \|\nabla u\|_{L^p}$$
My question is: Does the Poincaré inequality above still hold for $p=\infty$?
If yes, how to prove it? And if no, what is the counterexample?
Thank you.
 A: By the fundamental theorem of calculus we have that, if $u\in C_c^\infty(\mathbb{R}^n)$ and $x_0$ is such that $u(x_0)=0$ then 
$$
u(x)=\int_0^1 \nabla u(tx+(1-t)x_0)\cdot (x-x_0)dt
$$
from which we get
$$
|u(x)|\leq \| \nabla u \|_\infty |x-x_0|
$$
Now if $\Omega$ has finite width and $u\in C_c^\infty(\Omega)$, then for every $x\in \Omega$ there is an $x_0 \notin \Omega$ such that $|x-x_0| \leq D$ (where $D$ is the distance between the two parallel hyperplanes bounding $\Omega$). We conclude that
$$
\| u\|_{\infty} \leq D \| \nabla u\|_{\infty}.
$$
A: $W^{1,\infty}$ is indeed larger than the bounded Lipschitz class. For example, if the domain has an internal cusp, say $U$ is the planar domain given by 
$U=\{(x,y)\in\mathbb{R}^2\, :\, |y|>e^{-1/x^2}\text{ when }x>0\}$, consider the function $f$ given by $f(x,y)$ is the inner distance (with respect to $U$) between $(1/2, e^{-4})$ and $(x,y)$, we have that $f$ is (at least locally) in $W^{1,\infty}(U)$ but fails to be Lipschitz. Functions in $W^{1,\infty}$ are locally Lipschitz though. 
