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Feb 19, 2017 at 17:16 vote accept Shuhao Cao
Feb 19, 2017 at 16:31 answer added Nate Eldredge timeline score: 4
Feb 19, 2017 at 15:38 comment added Shuhao Cao @almaz Any references? Thanks.
Feb 19, 2017 at 15:36 history edited Shuhao Cao CC BY-SA 3.0
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Feb 19, 2017 at 15:29 comment added Shuhao Cao @NateEldredge Yes.
Feb 19, 2017 at 13:52 comment added anonymous To build a bit upon what @almaz said: In one dimension, we have the embedding $W^{1,p}(\Omega) \subset C(\Omega)$. Point evaluations (i.e., integration w.r.t. a Dirac measure) are continuous functionals on $C(\Omega)$, regardless of the dimension. With the aforementioned embedding, in one dimension, they are also (well-defined) continuous functionals on $W^{1,p}(\Omega)$, so that their respective kernel is closed, and we can factor with respect to those kernels. The functional $f \mapsto \int_U f$ in well-defined and continuous on $W^{1,p}(\Omega)$ regardless of the dimension.
Feb 19, 2017 at 5:54 comment added Nate Eldredge Does $|u|_{W^{1,p}}$ here denote $\left(\int |\nabla u|^p\right)^{1/p}$?
Feb 19, 2017 at 5:29 comment added ABMath The proposition is true only in dimension 1. In higher dimensions, a constant function can be approximated in the Sobolev norm by continuous functions vanishing at finitely many points
Feb 19, 2017 at 4:07 history asked Shuhao Cao CC BY-SA 3.0