# Does $W^{1,1}([a,b])$ compactly embed in $L^1([a,b])$?

I'm trying to prove the Remark at page 274 of the second edition of the Book "Partial Differential Equations" by Evans Lawrence, the Remark uses the Theorem 1 at page 272 (Rellich-Kondrachov Compcactness Theorem), here I quote the theorem, the remark and the definition of compactly embedding

Definition of compactly embedding

Let $$X$$ be a Banach space with norm $$||\,\cdot\,||_X$$, let $$Y$$ be a Banach space with norm $$||\,\cdot\,||_Y$$. One says that $$X$$ compactly embeds into $$Y$$, which is denoted by the simbol $$X \subset\subset Y$$, if and only if

1. $$X \subset Y$$

2. there exists $$C > 0$$ such that $$||u||_Y \leq C||u||_X$$ $$\forall u \in X$$

3. each bounded sequence in $$X$$ has a convergent subsequence in $$Y$$

(Rellich-Kondrachov Compactness Theorem) Assume $$U$$ is a bounded $$C^1$$ open set of $$\mathbb{R}^n$$, suppose $$1 \leq p < n$$, then

$$W^{1,p} \subset\subset L^q(U)$$

for each $$1 \leq q < p^*$$

Remark.

Observe that since $$p^* > p$$ and $$p^* \to \infty$$ as $$p \to n$$, we have in particular

$$W^{1,p}(U) \subset\subset L^p(U)$$

for all $$1 \leq p \leq \infty$$

Observe that if $$n < p \leq \infty$$, this follows from Morrey's inequality and the Ascoli-Arzela compactness criterion)

The problem is that the proof of the Remark doesn't work when $$p = n = 1$$, this is because you can't use Rellich Kondrachov Theorem because there is no $$p$$ such that $$1 \leq p < n$$ because $$n = 1$$, I asked to my professor how to prove the Remark in this case and he told me that the proof of the Rellich-Kondrachov Compactness Theorem (the one on the Evans' Book) can be modified to prove the Remark in this case, but I don't get how to do it.

Therefore my question is :

Is it true that $$W^{1,1}(U)$$ compactly embeds in $$L^1(U)$$ when $$U$$ is an open $$C^1$$ bounded subset of $$\mathbb{R}^1$$?? How can I prove it or disprove it??

Clearly it would be enough to prove the statement in the case where $$U = (a,b)$$, so you can assume it if you wish.

• What did you try? Try following the proof of Rellich-Kondrachov Theorem with $n=p=1$ and $p^* = \infty$. Where do you have a difficulty? Then we can help you on this specific difficulty ... Commented Sep 23, 2022 at 12:18
• I think you can reduce this to $U$ being an open interval.
– daw
Commented Sep 28, 2022 at 8:35

Just use Fréchet-Kolmogorov compactness criterion: let us assume $$[a,b]=[0,1]$$ and $$||f_n||_{W^{1,1}}\leq M$$. We just need to verify the equicontinuity assumption. We set $$\tau_h f(x)=f(x+h)$$ (I won't discuss $$x+h \notin [0,1]$$ but you can imagine what to do in that case) and estimate $$\int_0^1|\tau_h f_n(x)-f_n(x)|dx\leq\int_0^1\int_x^{x+h}|f_n^\prime(t)|dtdx=\int_0^1\int_{t-h}^t|f_n^\prime(t)|dxdt = h\cdot||f_n^\prime||_{L^1}\leq M h.$$ In the previous estimates I used the Fundamental Theorem of Calculus and Fubini Theorem
Therefore $$\lim_{h\to 0}\sup_{n\in\mathbb{N}}||\tau_hf_n-f_n||_{L^1}=0$$ and Fréchet-Kolmogorov criterion applies, so that you have compactness of the embedding