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 $$ \| u' \|_{L^\infty} + \| u \|_{L^\infty} \le \varepsilon \| u'' \|_{L^p} + C \| u \|_{L^1}, \quad u \in W^{2, p} (I). $$
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 $W^{2, p} (I)$. Then $(u_n)$ and $(u'_n)$ are bounded sequences in $W^{1, p} (I)$. By Theorem 8.8 (in the same book), the injection $W^{1, p}(I) \subset C(\bar{I})$ is compact. Then there are $u, v \in C(\bar{I})$ and a subsequence $(n_k)$ such that $u_{n_k} \to u$ and $u'_{n_k} \to v$ uniformly. By this well-known result, $v=u'$. This implies $\|u_{n_k} - u\|_{C^1 (\bar I)} \to 0$.
2.
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 $$ |u|_Y \le \varepsilon |u|_X + C_\varepsilon |u|_Z \quad \forall u \in X. $$
We apply above exercise with $X = W^{2, p} (I), Y=C^1 (\bar I)$ and $Z= L^1 (I)$ and get $$ \| u' \|_{L^\infty} + \| u \|_{L^\infty} \le \varepsilon (\| u'' \|_{L^p} + \| u' \|_{L^p} + \| u \|_{L^p}) + C \| u \|_{L^1}, \quad u \in W^{2, p} (I). $$
We have $\| u' \|_{L^p} \le \| u' \|_{L^\infty}$ and $\| u \|_{L^p} \le \| u \|_{L^\infty}$. Then $$ \| u' \|_{L^\infty} + \| u \|_{L^\infty} \le \frac{\varepsilon}{1-\varepsilon} \| u'' \|_{L^p} + \frac{C}{1-\varepsilon} \| u \|_{L^1}, \quad u \in W^{2, p} (I). $$
The quantity $\frac{\varepsilon}{1-\varepsilon}$ can be arbitrarily small. The claim then follows.
