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Mar 4, 2016 at 16:48 comment added Semmel Apologies for the late reply, but I believe that the author was a bit careless about the notation. I think he actually meant to say that "up to the multiplication with an arbitrary smooth function" where he would have spaces ${\rm L}^1(\Omega)$ and ${\rm W}^{-1,p}(\Omega)$. It is the only explanation I have managed to come up with and if I go through the proof with it, it works.
Oct 21, 2015 at 7:15 comment added Jochen Wengenroth In the paper appear expressions like $\|f\|_{L^1_{loc}(\mathbb R^2)}$ which make no sense for the space of locally integrable functions.
Oct 21, 2015 at 7:13 comment added Jochen Wengenroth 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).
Oct 21, 2015 at 6:37 comment added Semmel Could you explain a bit more, please?
Oct 21, 2015 at 0:03 comment added Fan Zheng Then could you just use a sequence of mollifiers exhausting $\Omega$ and the diagonalization argument?
Oct 20, 2015 at 21:13 comment added Semmel 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.
Oct 20, 2015 at 20:48 comment added Fan Zheng How do you define "boundedness" in $L_{loc}^1$, which a priori doesn't have a norm on it?
Oct 20, 2015 at 20:35 history edited Semmel
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Oct 20, 2015 at 16:12 history asked Semmel CC BY-SA 3.0