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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 $L^1(\mathbb{R}^n)$

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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  • $\begingroup$ What is meant by $D(\mathbb{R}^n)$? $\endgroup$ – charlestoncrabb Dec 2 '15 at 23:05
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The idea is that if the sequence $F_n$ converges strongly, then it has to converge to zero since $F_n(x)\rightarrow0$ a.e., but as we know since we are just shifting the original function $F$, so $\|F_n\|_1=\|F\|_1$ for all $n$, and the contradiction is $0\neq \|F\|_1=\lim_{n\rightarrow\infty}\|F_n\|_1=0$. Thus we conclude that the sequence $F_n$ does not converge and has no convergent subsequence.

A simple, illustrative example: consider the $1-$D traveling hat: $F(x)=\begin{cases}2x & x\in[0,1/2]\\ 3-2x & x\in[1/2,1]\end{cases}$, zero everywhere else, and set $x_n=n$. Then $F_n=\chi_{[n,n+1]}$. Then we have $\|F_n\|_{W^{1,1}(\mathbb{R})}=\|F\|_{W^{1,1}(\mathbb{R})}$, and $\|F_n\|_1=1$ for all $n$, so $F_n$ is a bounded sequence in $W^{1,1}(\mathbb{R})$, but again $F_n(x)\rightarrow0$ for all $x$ and so $F_n$ does not have any convergent subsequence in $L^1$.

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  • $\begingroup$ cf en.wikipedia.org/wiki/Sobolev_inequality $\endgroup$ – charlestoncrabb Dec 2 '15 at 23:30
  • $\begingroup$ thank you very much for unswering @charlestotcrabb , the ambiguity that i have is : a sequence $U_n$ that is not convergent means that $U_n\rightarrow \infty$ or that has no limit for example $(-1)^n$ , so to prove that the injection is not compact we have to find a sequence that it is bounded in the first space here $W^{1,1}(\mathbb{R}^n)$ and prove that it has no convergent subsequence in (here) $L^1$ , is it sufficient to prove that the squence is not convergent to say that it has no convergent subsequence , (for example $(-1)^n$ is not convergent and it has a subsequence that converge $\endgroup$ – kawazaki Dec 12 '15 at 4:55
  • $\begingroup$ sorry for my english , but this idea disturbs me too much $\endgroup$ – kawazaki Dec 12 '15 at 4:58
  • $\begingroup$ @kawazaki, you are thinking correctly; we need a bounded sequence in $W^{1,1}$ that has no convergent subsequence (so $(-1)^n$ does not fall into that category). $\endgroup$ – charlestoncrabb Dec 16 '15 at 0:20

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