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I was reading one preprint and stumbled upon a part in the proof where one particular embedding was used. Namely:

Let $\Omega\subset{\bf R}^2$ be open and bounded and let $p\in\langle 1,2\rangle$. Then ${\rm L}^1_{loc}(\Omega)$ is compactly embedded into ${\rm W}^{-1,p}_{loc}(\Omega)$.

Since I never saw ${\rm L}^1_{loc}$ being embedded into anything "useful", I have decided to check it. My only, and so far unsuccessful, attempt was to use a version of Schauder's theorem for locally convex topological vector spaces (it can be found, for example, in Grothendieck's book "Topological Vector Spaces"), but since ${\rm L}^1_{loc}$ and ${\rm L}^\infty_c$ are not reflexive, it got me nowhere.

Does anyone have any idea why this embedding would be valid?

Preprint with the claim: click here (one sentence before formula (2.23))

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  • $\begingroup$ How do you define "boundedness" in $L_{loc}^1$, which a priori doesn't have a norm on it? $\endgroup$
    – Fan Zheng
    Commented Oct 20, 2015 at 20:48
  • $\begingroup$ If it were normed, I could just use the standard Schauder's theorem. ${\rm L}^1_{loc}$ is complete metrizable locally convex space (the topology is generated by a countable family of seminorms). So, to say that a set is bounded it is to say that it is bounded with respect to all continuous seminorms. $\endgroup$
    – Semmel
    Commented Oct 20, 2015 at 21:13
  • $\begingroup$ Then could you just use a sequence of mollifiers exhausting $\Omega$ and the diagonalization argument? $\endgroup$
    – Fan Zheng
    Commented Oct 21, 2015 at 0:03
  • $\begingroup$ Could you explain a bit more, please? $\endgroup$
    – Semmel
    Commented Oct 21, 2015 at 6:37
  • $\begingroup$ If, as usual, $L^1_{loc}$ is the spaces of (equivalence classes of) measurable functions $f$ with $\int_k |f| <\infty$ for all compact sets $K$, it is not clear to me what compactness means. As Semmel comments, it is a (complete metrizable) locally convex space and in that category there are two candidates for a definition: Either a $0$-neighborhood is mapped to a relatively compact set (this is a very strong requirement) or all bounded sets are mapped to relatively compact ones (I know the name Montel-operator for the latter case). $\endgroup$
    – Jochen Wengenroth
    Commented Oct 21, 2015 at 7:13

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