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Apr 23, 2019 at 15:07 comment added Denis Serre @KwekuA When $s=-\sigma<0$, $H^s(\Omega)$ is the dual of the space $H^\sigma_0(\Omega)$, the closure of ${\cal D}(\Omega)$ in $H^\sigma(\Omega)$. Thus $H^s(\Omega)$ ``ignores '' the boundary.
Apr 23, 2019 at 14:03 comment added Kweku A How are you defining $H^s(\Omega)$ when $s<0$?
Apr 19, 2019 at 13:59 history answered Denis Serre CC BY-SA 4.0