If $\Omega$ is a sufficiently nice bounded open set in $\mathbb{R}^d$, it's known that there exists a continuous linear operator $$\mathcal{E}:W^{1,p}(\Omega)\rightarrow W^{1,p}(\mathbb{R}^d)$$ such that $\mathcal{E}(f)=f$ almost surely on $\Omega$. What do we know regarding explicit estimates of the operator norm of $\mathcal{E}$? As in, how does it depend on $p,d,$ the regularity of $\partial \Omega$, etc.
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1$\begingroup$ For a fixed open subset with Lipschitz boundary, the norm of the extension operator from $W^{1,p}(\Omega)$ depends only on $d$. As $\Omega$ varies, you should get dependence on its diameter and the Lipschitz constant on the boundary $\endgroup$– Piero D'AnconaCommented Oct 30, 2019 at 7:12
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$\begingroup$ Look at Thm. 9.7 in Brezis' book Functional Analysis, Sobolev Speces ... and keep trak of the constants there. This should not be too hard since the proof is very well written. $\endgroup$– Liviu NicolaescuCommented Oct 30, 2019 at 15:14
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