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Apr 16 at 7:56 history edited Hannes CC BY-SA 4.0
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Apr 12 at 11:43 comment added Hannes Bof, I need to think about this for a second and possibly revert my answer to its previous state :-D It IS delicate..
Apr 12 at 11:42 comment added Ayman Moussa @Perelman, I think you're right, I just took for granted that $E$ arrives in $W^{s,p}(\mathbf{R}^d)$ because of the last sentence of your post. My answer only tells you that if $E$ is well-defined then it has to be bounded.
Apr 12 at 10:46 comment added Perelman May I add to "...Zero extension is still a bounded linear operator for the given $W^{s,p}_0(\Omega)$ by Ayman's abstract answer...": Thats unfortunately not necessarily the case, since we do not know if $E$ is even a well defined map into $W^{s,p}(\mathbb{R}^d)$ (see (1) in my post). I did not realize that too in the first view.
Apr 12 at 9:13 history edited Hannes CC BY-SA 4.0
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Apr 12 at 7:56 comment added Jochen Glueck This is great, thanks a lot for the reference!
Apr 12 at 6:54 comment added Hannes @JochenGlueck Sure! (Else I would not have dared to say anything about "classical" :-)) How about Theorem 5.3.4 in Edmunds and Evans: Spectral Theory and Differential Operators?
Apr 11 at 16:14 comment added Jochen Glueck Do you happen to know a reference for an embedding result that relates that weight $\operatorname{dist}^{-k}_{\partial \Omega}$ to $W_0^{k,p}(\Omega)$ (for $k=1$ or maybe even for larger $k$)? I tried to search for it, but did not find much.
Apr 9 at 9:57 history answered Hannes CC BY-SA 4.0