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May 29, 2019 at 17:13 vote accept asv
May 29, 2019 at 9:55 answer added Hannes timeline score: 4
May 29, 2019 at 8:27 answer added Mateusz Kwaśnicki timeline score: 2
May 29, 2019 at 5:28 comment added Jochen Glueck @MikeMiller: I'm not sure I follow your argument. The closure of the compactly supported smooth functions within the space $W^{1,2}(\Omega) \cap C^0(\overline{\Omega})$ might be smaller than $W^{1,2}_0(\Omega)$.
May 29, 2019 at 0:06 comment added mme Sure, the trace (restriction to boundary) map $r: C^\infty(\overline \Omega) \to C^\infty(\partial \Omega)$ is continuous when you equip the domain with the $W^{1,2}$ norm and the codomain with the $L^2$ norm, or when you equip both with the $C^0$ norm, and hence extends to a continuous map $(W^{1,2} \cap C^0)(\overline \Omega) \to (L^2 \cap C^0)(\partial \Omega)$. Because the map $r$ vanishes on the set of compactly supported smooth functions in $\Omega$, it also vanishes on its closure. The continuity in $C^0$ is obvious, while the continuity in $W^{1,2} \to L^2$ is in many PDE books.
May 29, 2019 at 0:04 history edited asv CC BY-SA 4.0
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May 28, 2019 at 22:00 history asked asv CC BY-SA 4.0