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From the Evans'PDE book I learned $W^{1,\infty}(U)$ coincide with Lipschitz continuous $C^{0,1}(U)$ with $U\in\Bbb{R}^n(n\geqslant1)$ is bounded and $\partial U\in C^1$. I wonder the counterpart result for $U$ is unbounded, e.g.$U=\Bbb{R}_+^n$(the half space). More precisely,

1)How to prove $W^{1,\infty}(\Bbb{R}_+^1)=C^{0,1}(\Bbb{R}_+^1)$(I'm not sure that the statement must be right);

2)How to prove $n\geqslant 2,W^{1,\infty}(\Bbb{R}_+^n)\hookrightarrow C^{0,\alpha}(\Bbb{R}_+^n),\forall 0<\alpha<1$;

3)Please give a counterexample to demonstrate that $n\geqslant 2,W^{1,\infty}(\Bbb{R}_+^n)\nsubseteq C^{0,1}(\Bbb{R}_+^n)$.

Every comment，hint and answer will be appreciated!

Darry
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1 Answers1

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Unboundedness is not a problem. The equality $W^{1,\infty}=C^{0,1}$ holds in any quasiconvex domain $U$. A domain is quasiconvex if there exists a constant $M$ such that any two points $a,b\in U$ can be joined by a curve $\gamma$ of length at most $M|a-b|$. In particular, convex domains have this property with $M=1$.

Proof: let $\gamma:[0,1]\to U$ be a curve as above. Given $u\in W^{1,\infty}(U)$, consider the composition $u\circ \gamma$. By the chain rule $|D(u\circ \gamma)|\le L|D\gamma|$ where $L=\operatorname{ess\,sup}_U |Du|$. Hence, $$|u(b)-u(a)| = |u(\gamma(1))-u(\gamma(0))|\le L\int_0^1 |D\gamma|\le LM|a-b|$$

But, one doesn't simply use the chain rule for functions that are only differentiable a.e. But since $U$ is open, we can consider a family of perturbations of $\gamma$ (translates by a small vector), and use Fubini to justify saying that for a.e. translate the function $u$ will be differentiable a.e. on the image of $\gamma$. This is easier to do when $\gamma$ is a line segment, which is all you need to handle the convex domains in your question.

  • @ Pavel M: Thank you! I learn the new knowledge from your answer. By the way，is there a counterexample for a non-quasiconvex domain? In addition, could you recommend me some good books cover the main topics on sobolev spaces which are useful for PDE? Hum,I maybe some greedy,thanks again! – Darry Jan 03 '13 at 03:29
  • @Darry Counterexample: $U$ is the plane $\mathbb R^2$ with negative real axis removed; $u(r,\theta)=r\theta$ where $r,\theta$ are polar coordinates and $-\pi<\theta<\pi$. As for a reference, I think nothing beats Brezis. –  Jan 03 '13 at 04:01
  • $r\theta$ is not Lipschitz, but neither bounded, so it isn't Sobolev. Unboundedness is a problem an example: take $\pi(x)=x_1$ for a Lipschitz function not Sobolev on the whole space. – Lolman Sep 24 '18 at 12:07
  • What is the sense of $u \in W^{1,\infty(U)}$ and $Du$? $u$ admits a weak derivative ? $u$ is differentiable almost everywhere ? – Liam Mar 16 '22 at 06:52