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Apr 12 at 10:35 history edited Perelman CC BY-SA 4.0
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Apr 9 at 10:12 history edited Hannes CC BY-SA 4.0
typo in Gagliardo norm in last formula
Apr 9 at 10:09 history edited YCor
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Apr 9 at 9:57 answer added Hannes timeline score: 2
Apr 7 at 21:18 vote accept Perelman
Apr 7 at 19:39 answer added Ayman Moussa timeline score: 2
Apr 7 at 19:18 comment added Pietro Majer ok I didn’t realise the norms are different, thankyou
Apr 7 at 18:58 comment added Perelman @PietroMajer This would work if $W_0^{s,p}(\Omega)$ and $W^{s,p}(\mathbb{R}^d)$ have equivalent norms. I think we need to show that $\|Eu\|_{W^{s,p}(\mathbb{R}^d)}\leq C\|u\|_{W^{s,p}(\Omega)}$
Apr 7 at 18:42 comment added Pietro Majer A natural definition of $C^\infty_0(\Omega)$ is: all $C^\infty$ functions $u:\mathbb R^d\to\mathbb R$ with $\text{supp}(u)\subset \Omega$. So $W^{s,p}_0(\Omega)$ is increasing by inclusion wrto $\Omega$, and we don't even have to extend anything.
Apr 7 at 18:19 comment added Perelman @PietroMajer yes
Apr 7 at 18:03 comment added Pietro Majer So by the definition you are adopting $W^{s,p}(\mathbb R^d)=W^{s,p}_0(\mathbb R^d)$, and $W^{s,p}_0$ is defined above, right?
Apr 7 at 17:27 comment added Perelman Thsi defintion holds for arbtrary open $\Omega \subseteq \mathbb{R}^d$
Apr 7 at 17:26 history edited Perelman CC BY-SA 4.0
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Apr 7 at 17:24 comment added Pietro Majer You may add the definition of $W^{s,p}(\mathbb R^d)$ too, by completeness sake
Apr 7 at 17:20 history edited Perelman CC BY-SA 4.0
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Apr 7 at 17:11 history edited Perelman CC BY-SA 4.0
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Apr 7 at 15:53 answer added Liding Yao timeline score: 1
S Apr 7 at 15:07 review First questions
Apr 7 at 19:45
S Apr 7 at 15:07 history asked Perelman CC BY-SA 4.0