Let $I=(0, 1)$ and $m \ge 2$. Recently, I have proved here that the injection $W^{m, 1}(I) \subset W^{m-1, 1}(I)$ is compact. I would like to generalize this result, i.e.,
Let $p \in [1, \infty]$. The injection $W^{m, p}(I) \subset W^{m-1, p}(I)$ is compact.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it?
We proceed by induction on $m$.
- First, we prove base case $m=2$.
Let $(u_n)$ be a bounded sequence in $W^{2, p} (I)$. Then $(u_n)$ and $(u'_n)$ are bounded sequences in $W^{1, p} (I)$. I have proved here that the injection $W^{1, p}(I) \subset L^{p}(I)$ is compact. Then there are $u, v \in L^p(I)$ and a subsequence $(n_k)$ such that $u_{n_k} \to u$ and $u'_{n_k} \to v$ in $L^p (I)$. First, we prove that $u' =v$. Indeed, $$ \int_I u_n \varphi' = - \int_I u_n' \varphi, \quad \varphi \in C^1_c (I), $$ and at the limit $$ \int_I u \varphi' = - \int_I v \varphi, \quad \varphi \in C^1_c (I). $$
It follows that $u, u' \in L^p (I)$ and thus $u \in W^{1, p}(I)$. Clearly, $u_{n_k} \to u$ and $u_{n_k}' \to u'$ in $L^p (I)$ imply $u_{n_k} \to u$ in $W^{1, p}(I)$. The claim then follows.
- Assume that the injection $W^{k, p}(I) \subset W^{k-1, p}(I)$ is compact for all $1 \le k \le m$. We will prove that the claim holds for $k=m+1$.
Let $(u_n)$ be a bounded sequence in $W^{m+1, p} (I)$. Then $(u_n)$ and $(u'_n)$ are bounded sequences in $W^{m, p} (I)$. By inductive hypothesis, there are $u, v \in W^{m-1, p}(I)$ and a subsequence $(n_k)$ such that $u_{n_k} \to u$ and $u_{n_k}' \to v$ in $W^{m-1, p}(I)$. First, we prove that $u' =v$. Indeed, $$ \int_I u_n \varphi' = - \int_I u_n' \varphi, \quad \varphi \in C^1_c (I), $$ and at the limit $$ \int_I u \varphi' = - \int_I v \varphi, \quad \varphi \in C^1_c (I). $$
It follows that $u, u' \in W^{m-1, p}(I)$ and thus $u \in W^{m, p}(I)$. Clearly, $u_{n_k} \to u$ and $u_{n_k}' \to u'$ in $W^{m-1, p}(I)$ imply $u_{n_k} \to u$ in $W^{m, p}(I)$. The claim then follows.
