I am trying to find a way of extending functions in the Sobolev Space $W^{1, \infty}(U)$ to $W^{1, \infty}(\mathbb{R}^n)$ where $U\subset\mathbb{R}^{n}$ is open such that $U\subset\subset V$ for $V\subset\mathbb{R}^{n}$ also open and bounded. Furthermore, assume $\partial U$ is $C^1$.
Extension Thoerem for the Sobolev Space $W^{1, \infty}(U)$
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sobolev-spaces
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1Your space $W^{1,\infty}(U)$ is in fact the space of Lipschitz continuous functions defined on $U$. It looks like you need something like this: http://en.wikipedia.org/wiki/Kirszbraun_theorem – 2012-06-15
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1Kirszbraun's theorem works, but for scalar-valued functions. McShane-Whitney formula $U(y)=\inf_{x\in U}u(x)+L|x-y|$ is all one needs. Either way, one first needs to know that $W^{1,\infty}(U)=\mathrm{Lip}(U)$ for the indicated class of domains $U$. If this nontrivial fact cannot be taken for granted, it will be the main part of the proof. – 2012-06-15
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0@LeonidKovalev : What should one take care of in this case compared to extension of $W^{1,p}$ ? why does it not work for $p=\infty $ – 2012-06-15
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0@Theorem The Whitney-type extension, as described, e.g., in Stein's 1970 book, works fine for all $1\le p\le \infty$. But when $p=\infty$, one is tempted to give a shorter proof using the identification of $W^{1,\infty}$ with the Lipschitz space, see the comment by Beni Bogosel. I made a remark to the effect that the identification is not entirely trivial. ... Side remark: it's hard to answer question like this without knowing the tools that the questioner has in disposal. I can't tell if Nirav knows about Whitney decomposition etc. – 2012-06-15