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 $1 \leq q < p^*$


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.

  • 1
    $\begingroup$ 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 ... $\endgroup$
    – LL 3.14
    Commented Sep 23, 2022 at 12:18
  • $\begingroup$ I think you can reduce this to $U$ being an open interval. $\endgroup$
    – daw
    Commented Sep 28, 2022 at 8:35

1 Answer 1


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


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