In solving exercise 8.6.3 in Brezis' Functional Analysis, I have to prove a result, i.e.,
Let $I=(0, 1)$. The injection $W^{2, 1}(I) \subset W^{1, 1}(I)$ is compact.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it?
Let $(u_n)$ be a bounded sequence in $W^{2, 1} (I)$. Then $(u_n)$ and $(u'_n)$ are bounded sequences in $W^{1, 1} (I)$. By Theorem 8.8 (in the same book), the injection $W^{1, 1}(I) \subset L^1(I)$ is compact. Then there are $u, v \in L^1(I)$ and a subsequence $(n_k)$ such that $u_{n_k} \to u$ and $u'_{n_k} \to v$ in $L^1(I)$.
Remark 4. It is convenient to keep in mind the following fact, which we have used in the proof of Proposition 8.1: let $\left(u_n\right)$ be a sequence in $W^{1, p}$ such that $u_n \rightarrow u$ in $L^p$ and $\left(u_n^{\prime}\right)$ converges to some limit in $L^p$; then $u \in W^{1, p}$ and $\left\|u_n-u\right\|_{W^{1, p}} \rightarrow 0$. In fact, when $1<p \leq \infty$ it suffices to know that $u_n \rightarrow u$ in $L^p$ and $\left\|u_n^{\prime}\right\|_{L^p}$ stays bounded to conclude that $u \in W^{1, p}$ (see Exercise 8.2).
By above remark (at page 204 in the same book), we get $u \in W^{1, 1} (I)$ and $\|u_{n_k}-u\|_{W^{1, 1}} \to 0$.
