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

Question

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.

Answer 1

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 $L^1$.

Answer 2

I am $7+\varepsilon$ years late but I found myself thinking about this, and while I think the answer above is essentially correct, I believe it is very easy to misinterpret (I certainly did misinterpret it). I thought the claim $\lim_{n\to\infty} \|F_n\|_1 = 0$ is coming from Lebesgue's dominated convergence theorem, but I do not think that the condition of being bounded by an integrable function holds here. Indeed, if it did, then Lebesgue's dominated convergence theorem would show that $F_n\to 0$ in $L^1$, which when paired together with $\|F\|_1 = \lim_{n\to\infty} \|F_n\|_1$ produces non-sense, rather than a contradiction which shows that $F_n$ doesn't have a convergent subsequence. (Sorry if something I am saying is obviously wrong here -- but what I am trying to say is that I think Lebesgue's dominated convergence theorem cannot be applied here to conclude that $\lim_{n\to\infty} \|F_n\|_1 = 0$.)

I think the right approach, as well as what the author of the above answer means is the following. Suppose $(F_n)_n$ has a convergent subsequence $(F_{n_k})_k$, which converges to some $G$ in $L^1$. By a well-known result (Theorem 3.12 in Papa Rudin), $\{F_{n_k})_k$ has a subsequence which converges pointwise almost everywhere to $G$. From here, because every subsequence of $(F_n)_n$ converges pointwise to $0$, we have that $G=0$ almost everywhere. Therefore $\|F\|_1 = \lim_{k\to\infty} \|F_{n_k}\|_1 = 0$, which is a contradiction. Therefore, we conclude that $(F_n)_n$ has no convergent subsequence.