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.

  • $\begingroup$ 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

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

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

  • $\begingroup$ 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

$W^{1,\infty}$ 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.


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.