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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.


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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