I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.6.1 Let $I=(0, 1)$ and $p \in (1, \infty]$.
- Check that $W^{2, p} (I) \subset C^1 (\bar I)$ with compact injection.
- Deduce that for $\varepsilon >0$ there is $C=C(\varepsilon, p)$ such that $$ \| u' \|_{L^\infty} + \| u \|_{L^\infty} \le \varepsilon \| u'' \|_{L^p} + C \| u \|_{L^1}, \quad u \in W^{2, p} (I). $$
- Let $q \in [1, \infty)$. Prove that for $\varepsilon >0$ there is $C=C(\varepsilon, q)$ such that $$ \| u' \|_{L^q} + \| u \|_{L^\infty} \le \varepsilon \| u'' \|_{L^1} + C \| u \|_{L^1}, \quad u \in W^{2, 1} (I). $$
More generally, let $m \ge 2$ be an integer.
- Prove that for $\varepsilon >0$ there is $C=C(\varepsilon, m, p)$ such that $$ \sum_{j=0}^{m-1} \| D^j u \|_{L^\infty} \le \varepsilon \| D^m u \|_{L^p} + C \| u \|_{L^1}, \quad u \in W^{m, p} (I). $$
- Let $q \in [1, \infty)$. Prove that for $\varepsilon >0$ there is $C=C(\varepsilon, q)$ such that $$ \| D^{(m-1)} u \|_{L^q} + \sum_{j=0}^{m-2} \| D^j u \|_{L^\infty} \le \varepsilon \| D^m u \|_{L^1} + C \| u \|_{L^1}, \quad u \in W^{m, 1} (I). $$
There are possibly subtle mistakes that I could not recognize in my below attempt of $(5)$. Could you have a check on it?
- First, we prove that the injection $W^{m, 1}(I) \subset W^{m-1, 1}(I)$ is compact.
We proceed by induction on $m$. The base case $m=2$ has been proved here. Assume that the injection $W^{k, 1}(I) \subset W^{k-1, 1}(I)$ is compact for $2 \le k \le m-1$. We will prove the claim for $k=m$. Let $(u_n)$ be a bounded sequence in $W^{m, 1} (I)$. We have $$ \| u \|_{W^{m, 1}} = \| u \|_{L^{1}} + \| u' \|_{W^{m-1, 1}} = \| u \|_{W^{m-1, 1}} + \| D^{m} u \|_{L^{1}}, $$ where $D^{m} u$ is the $m$-th derivative of $u$. Then $(u_n)$ and $(u'_n)$ are bounded sequences in $W^{m-1, 1} (I)$. By inductive hypothesis, there are $u, v \in W^{m-2, 1}(I)$ and a subsequence $(n_k)$ such that $u_{n_k} \to u$ and $u_{n_k}' \to v$ in $W^{m-2, 1}(I)$. 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 $u, u' \in W^{m-2, 1}(I)$ and thus $u \in W^{m-1, 1}(I)$. Clearly, $u_{n_k} \to u$ and $u_{n_k}' \to u'$ in $W^{m-2, 1}(I)$ imply $u_{n_k} \to u$ in $W^{m-1, 1}(I)$. The claim then follows.
- Second, we prove our desired inequality in $(5)$. We need an auxiliary result, i.e.,
- Exercise 6.12 Let $X,Y,Z$ be real Banach spaces with corresponding norms $|\cdot|_X, |\cdot|_Y, |\cdot|_Z$. Assume that $X \subset Y$ with compact injection and that $Y \subset Z$ with continuous injection. Prove that for every $\varepsilon>0$ there is $C_\varepsilon > 0$ such that $$ |u|_Y \le \varepsilon |u|_X + C_\varepsilon |u|_Z \quad \forall u \in X. $$
- Exercise 8.5.3 Let $q \in [1, \infty)$. For $\varepsilon >0$ there is $C=C(\varepsilon, q)$ such that $$ \| u \|_{L^q} \le \varepsilon \| u' \|_{L^1} + C \| u \|_{L^1}, \quad u \in W^{1, 1} (I). $$
We apply exercise 8.5.3 for $D^{(m-1)} u$ and get $$ \| D^{(m-1)} u \|_{L^q} \le \varepsilon \| D^{m} u \|_{L^1} + C \| D^{(m-1)} u \|_{L^1}. $$
I have proved here that the injection $W^{m-1, 1}(I) \subset C^{m-2} (\bar{I})$ is continuous. Then there is $c>0$ such that $$ \sum_{j=0}^{m-2} \| D^j u \|_{L^\infty} \le c \sum_{j=0}^{m-1} \| D^j u \|_{L^1}, \quad u \in W^{m-1, 1}(I). $$
I have proved above that the injection $W^{m, 1}(I) \subset W^{m-1, 1}(I)$ is compact. We apply exercise 6.12 with $X = W^{m, 1} (I), Y= W^{m-1, 1}(I)$ and $Z= L^1 (I)$ and get $$ \sum_{j=0}^{m-1} \| D^j u \|_{L^1} \le \varepsilon' \sum_{j=0}^{m} \| D^j u \|_{L^1} + C' \| u \|_{L^1}, \quad u \in W^{m, 1} (I). $$
It follows that $$ \sum_{j=0}^{m-1} \| D^j u \|_{L^1} \le \varepsilon_2 \| D^m u \|_{L^1} + C_2 \| u \|_{L^1}, \quad u \in W^{m, 1} (I). $$
Then $$ \begin{align*} \| D^{(m-1)} u \|_{L^q} + \sum_{j=0}^{m-2} \| D^j u \|_{L^\infty} &\le \varepsilon \| D^{m} u \|_{L^1} + C \| D^{(m-1)} u \|_{L^1} + c \sum_{j=0}^{m-1} \| D^j u \|_{L^1} \\ &\le \varepsilon \| D^{m} u \|_{L^1} + (C + c) \sum_{j=0}^{m-1} \| D^j u \|_{L^1} \\ &\le \varepsilon \| D^{m} u \|_{L^1} + (C + c) ( \varepsilon_2 \| D^m u \|_{L^1} + C_2 \| u \|_{L^1}) \\ &= (\varepsilon + \varepsilon_2 C + \varepsilon_2 c) ) \| D^m u \|_{L^1} + (CC_2 + cC_2) \| u \|_{L^1}. \end{align*} $$
Because $\varepsilon_2$ can be chosen independently of $\varepsilon, C$. The quantity $(\varepsilon +C \varepsilon_2+c\varepsilon_2)$ can be arbitrarily small. The claim then follows.
