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Let

  • $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $k\in\mathbb N$
  • $p\in[1,\infty)$

We say that $\Lambda$ has the $(k,p)$-extension property $:\Leftrightarrow$ There is a bounded linear operator $W^{k,\:p}(\Lambda)\to W^{k,\:p}(\mathbb R^d)$ with $$\left.Eu\right|_\Lambda=u\;\;\;\text{for all }u\in W^{k,\:p}(\Lambda)\;.\tag1$$ My question is: If $\Lambda$ has the $(k,p)$-extension property, are we able to conclude that $\Lambda$ has the $(k',p')$-extension property for some $k'\in\mathbb N$ and $p'\in[1,\infty)$? In particular, if $\Lambda$ has the $(2,2)$-extension property, does it have the $(1,2)$-extension property too?

1 Answers 1

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I think you should consider posting your question in math overflow instead of here. As far as I know the problem of characterizing extension domains is still open except in the case $p=2$ and $d=2$. If $\Lambda$ has finitely many connected component, then in the paper Peter W. Jones, Quasiconformal mappings and extendability of functions in sobolev spaces, Acta Math. Volume 147 (1981), 71-88.

https://projecteuclid.org/euclid.acta/1485890130

it was proved that $\Lambda$ is a $W^{k,2}$ extension domain if and only if it is an $(\varepsilon,\infty)$ domain, that is, there is $\varepsilon>0$ such that for every $x,y\in\Lambda$, there is a rectifiable arc $\gamma$ contained in $\Lambda$ joining $x$ and $y$ with length less than $\frac{1}{\varepsilon}$ and such that $$ dist(z,\partial\Lambda)\geq\frac{\varepsilon|x-z||y-z|}{|x-y|}% $$ for every $z$ in $\gamma$. This answers your second question but only in dimension $d=2$ and for finitely connected domains.

Concerning your first question, the answer in general is no. In $\mathbb{R}% ^{2}$ if you consider a domain with a cusp, $$ \Lambda=\left\{ \left( x,y\right) \in\mathbb{R}^{2}:\,01$, then you can prove that $\mathbb{R}^{2}\setminus \overline{\Lambda}$ is an extension domain for $W^{1,1}$ but not for $W^{1,p}$ for any $p>1$. This example is in Mazya's Sobolev book and in Leoni's Sobolev book.