If there is a bounded linear extension $W^{k,\:p}(Λ)\to W^{k,\:p}(ℝ^d)$, are we able to conclude the existence of such an extension for other $k,p$? 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?
 A: 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,
$$
\Lambda=\left\{  \left(  x,y\right)  \in\mathbb{R}^{2}:\,0<x<1,\,0<y<x^{\alpha
}\right\}  ,
$$
where $\alpha>1$, 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.
