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This question is an exact duplicate of:

i have this question : in an example of the compact embedding, the autor gives a demonstration of : the sobolev space $W^{1,1}(\mathbb{R}^n)$ is not compactly embedded in $L^1(\mathbb{R}^n)$ and it is this one :

So let $F\in D(\mathbb{R}^n)$(=the space of smooth functions with a compact support in $\mathbb{R}^n)$ ., not identically equal to zero and $\{x_n\}$ a sequence such that lim $x_n=+\infty$ when $n\rightarrow \infty$. so $F_n(x)=F(x-x_n)$ is bounded in $W^{1,1}(\mathbb{R}^n)$ and it converge a.e. to 0.

so if it converge strongly in $L^1$ we will have :$||F_n||_{L^1}=||F||_{L^1}=0$, an this is a contradiction .

my question is : where is the contradiction and how to prove that the embedding is compact in "this case or in normed (Banach) spaces (general case)"?

thank you very much.

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marked as duplicate by user98602, Joel Reyes Noche, user223391, Harish Chandra Rajpoot, Claude Leibovici Dec 6 '15 at 5:55

This question was marked as an exact duplicate of an existing question.

  • $\begingroup$ Because $\|F_n\| \neq 0$. $\endgroup$ – user98602 Dec 5 '15 at 18:07

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