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Timeline for Question about measure lemma?

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Jul 2, 2019 at 9:34 vote accept Vrouvrou
Apr 27, 2015 at 21:40 comment added Nate Eldredge @Vrouvrou: All I am using here is that $L^1(\Omega, m)$ is continuously (indeed, isometrically) embedded into $C_0(\Omega)^*$. The embedding $T$ is the natural one: for $f \in L^1(\Omega, m)$ and $g \in C_0(\Omega)$, let $(Tf)(g) = \int_0^1 f g\,dm$. As I am not really interested in writing a functional analysis textbook for you in this comment thread, I think I will leave it at that.
Apr 27, 2015 at 21:37 comment added Vrouvrou and how we prove that $W^{1,p}$ or $L^{p^*}$ is cintinuouly embeded into $C_{0}(\Omega)^*$ ?
Apr 27, 2015 at 21:35 comment added Nate Eldredge @Vrouvrou: $C_0(\Omega)$ is the space of continuous functions on $\Omega$ which vanish at "infinity", i.e. the uniform closure of the continuous compactly supported functions $C_c(\Omega)$. For a bounded open set $\Omega$, this means continuous functions on $\bar{\Omega}$ that vanish on $\partial \Omega$. $C_0(\Omega)^*$ is the dual of this space, which as I said, by the Riesz representation theorem, is isometrically isomorphic to the space of signed Radon measures on $\Omega$, with the total variation norm.
Apr 27, 2015 at 21:32 comment added Vrouvrou what is the space $C_0(\Omega)^*$ ? thank you
Apr 27, 2015 at 21:30 comment added Nate Eldredge @Vrouvrou: Yes, the compact embedding is not used here. But your other space will also need to be continuously embedded in $C_0(\Omega)^*$.
Apr 27, 2015 at 21:29 comment added Vrouvrou can i replace $L^{p^*}$ by an other space such that $W^{1,p}$ is continuously embeded to it but not compactly ?
Apr 27, 2015 at 21:19 comment added Nate Eldredge @Vrouvrou: This is such a standard argument that I do not think you will find a reference that discusses it specifically. The results I mentioned will be in any elementary textbook on functional analysis and Sobolev spaces, such as Folland's Real Analysis for example.
Apr 27, 2015 at 21:09 history answered Nate Eldredge CC BY-SA 3.0