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 $$ \| u' \|_{L^q} + \| u \|_{L^\infty} \le \varepsilon \| u'' \|_{L^1} + C \| u \|_{L^1}, \quad u \in W^{2, 1} (I). $$
There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it? My approach is quite long. Is there a shorter alternative?
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 $u'$ and get $$ \| u' \|_{L^q} \le \varepsilon \| u'' \|_{L^1} + C \| u' \|_{L^1}. $$
By Theorem 8.8 (in the same book), the injection $W^{1, 1}(I) \subset C(\bar{I})$ is continuous. Then there is $c>0$ such that $$ \| u \|_{L^\infty} \le c \| u \|_{L^1} + c\| u' \|_{L^1}. $$
I have proved that the injection $W^{2, 1}(I) \subset W^{1, 1}(I)$ is compact. We apply exercise 6.12 with $X = W^{2, 1} (I), Y= W^{1, 1}(I)$ and $Z= L^1 (I)$ and get $$ \begin{align*} \| u \|_{L^1} + \| u' \|_{L^1} &\le \varepsilon' (\| u \|_{L^1} + \| u' \|_{L^1} + \| u'' \|_{L^1}) + C' \| u \|_{L^1}, \quad u \in W^{2, 1} (I). \end{align*} $$
It follows that $$ \| u' \|_{L^1} \le \varepsilon_2 \| u'' \|_{L^1} + C_2 \| u \|_{L^1}, \quad u \in W^{2, 1} (I). $$
Finally, $$ \begin{align*} \| u' \|_{L^q} + \| u \|_{L^\infty} &\le (\varepsilon \| u'' \|_{L^1} + C \| u' \|_{L^1}) + (c \| u \|_{L^1} + c\| u' \|_{L^1}) \\ & = \varepsilon \| u'' \|_{L^1} + c \| u \|_{L^1} +(C+c) \| u' \|_{L^1} \\ &\le \varepsilon \| u'' \|_{L^1} +c \| u \|_{L^1} + (C+c) ( \varepsilon _2 \| u'' \|_{L^1} + C_2 \| u \|_{L^1}) \\ & = (\varepsilon +C \varepsilon_2+c\varepsilon_2) \| u'' \|_{L^1} + (c+cC_2+CC_2) \| u \|_{L^1}, \quad u \in W^{2, 1} (I). \end{align*} $$
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.
