$W^{1,1}(\mathbb{R}^n)$ is not compactly embedded in $L^1(\mathbb{R}^n)$?

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.

Solutions Collecting From Web of "$W^{1,1}(\mathbb{R}^n)$ is not compactly embedded in $L^1(\mathbb{R}^n)$?"

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 $L^1$.