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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.

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  • $\begingroup$ Is $W_0^{1,p}(\Omega)$ here the closure of $C_c^{\infty}(\Omega)$ under the Sobolev norm? $\endgroup$ – detnvvp Oct 16 '13 at 4:08
  • $\begingroup$ yes that's right. $\endgroup$ – digiboy1 Oct 16 '13 at 4:16
  • $\begingroup$ I think the answer would be false if $\Omega$ is not bounded. Let $\Omega =\mathbb R$ and $u$ be piecewise linear such that $u(0)=1$, $u(-r) = u(r) = 0$. Then $||u|| = 1$ but $||Du|| = 1/r$. $\endgroup$ – user99914 Oct 16 '13 at 4:30
  • $\begingroup$ What if I assume that $\Omega$ is bounded? $\endgroup$ – digiboy1 Oct 16 '13 at 4:58
  • $\begingroup$ The assumption that $\Omega$ has finite width implies that $\Omega$ is bounded in dimension $1$. $\endgroup$ – Tomás Oct 16 '13 at 12:15
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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

$$ \| u\|_{\infty} \leq D \| \nabla u\|_{\infty}. $$

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  • $\begingroup$ How do you get around the fact that $C^{\infty}_{c}(\Omega)$ isn't dense in $W^{1,\infty}_{c}(\Omega)$? $\endgroup$ – fourierwho Apr 13 '18 at 2:36
  • $\begingroup$ @fourierwho: We can treat functions in $W^{1,\infty}_c(\mathbb{R}^n)$ directly since these are Lipschitz and the argument above still holds. $\endgroup$ – Jose27 Apr 13 '18 at 2:52
  • $\begingroup$ We need to assume the boundary of $U$ is sufficiently regular (e.g. $C^{1}$). Otherwise, $W^{1,\infty}(U)$ functions aren't necessarily Lipschitz. $\endgroup$ – fourierwho Apr 13 '18 at 3:00
  • $\begingroup$ @fourierwho: We're working with $W^{1,\infty}_0$ which has no such problems (although maybe I'm using a nonstandard definition of this space?). $\endgroup$ – Jose27 Apr 13 '18 at 3:03
  • $\begingroup$ That's a fair point. Can one prove that $W^{1,\infty}_{0}$ functions are Lipschitz regardless of the boundary regularity? I don't know the answer. I was only thinking of the fact that $W^{1,\infty}$ doesn't imply Lipschitz. $\endgroup$ – fourierwho Apr 13 '18 at 3:08
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$W^{1,\infty}$ 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.

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