Let $I := (0,1)$. I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,
Exercise 8.2.1 Assume that $(u_n)$ is a bounded sequence in $W^{1, p}(I)$ with $1<p \le \infty$. Show that there exist a subsequence $(u_{n_k})$ and some $u \in W^{1, p}(I)$ such that $\|u_{n_k}-u\|_{L^{\infty}} \xrightarrow{k \to \infty} 0$. Moreover, $u_{n_k}' \xrightarrow{k \to \infty} u'$ in the weak topology $\sigma (L^p, L^{p^*})$ if $1<p<\infty$, and $u_{n_k}' \xrightarrow{k \to \infty} u^{\prime}$ in the weak$^*$ topology $\sigma (L^{\infty}, L^1)$ if $p=\infty$. Here $p^*$ is the Hölder conjugate of $p$.
There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it?
Let $C := \sup_n \|u_n\|_{W^{1, p}} < \infty$. By Theorem 8.8 (in the same book), the injection $W^{1, p}(I) \subset C (\bar I)$ is compact because $I$ is bounded and $p \in (1, \infty]$. Then there is a subsequence $(u_{n_k})$ and $u \in C (\bar I)$ such that $\|u_{n_k}-u\|_{L^{\infty}} \xrightarrow{k \to \infty} 0$.
First, we prove that $u \in W^{1, p}(I)$. Because $p \in (1, \infty]$, we can apply Proposition 8.3 and get $$ \left|\int_I u_n \varphi' \right| \leq C \|\varphi\|_{L^{p^*}}, \quad n \in \mathbb N,\varphi \in C^1_c (I). $$
This together with $\|u_{n_k}-u\|_{L^{\infty}} \xrightarrow{k \to \infty} 0$ implies $$ \left|\int_I u \varphi' \right| \leq C \|\varphi\|_{L^{p^*}}, \quad \varphi \in C^1_c (I). $$
By Proposition 8.3 again, we get $u \in W^{1, p}(I)$.
Let $T:W^{1, p}(I) \to L^p (I) \times L^p (I), v \mapsto (v, v')$ be the canonical isometric homomorphism. We have $$ (L^{p^*} (I) \times L^{p^*} (I))^* = L^p (I) \times L^p (I). $$
Because $p^* \in [1, \infty)$, the space $L^{p^*} (I) \times L^{p^*} (I)$ is separable. Then the closed unit ball of $L^p (I) \times L^p (I)$ is sequentially compact in its weak$^*$ topology. Then there is a subsequence (WLOG denoted by $(n_k)$) and $(v, w) \in L^p (I) \times L^p (I)$ such that $(u_{n_k}, u_{n_k}') \xrightarrow{k \to \infty} (v, w)$ in the weak$^*$ topology of $L^p (I) \times L^p (I)$. The weak$^*$ topology of product space coincides with the product of weak$^*$ topologies. Then $u_{n_k} \xrightarrow{k \to \infty} v$ and $u_{n_k}' \xrightarrow{k \to \infty} w$ both in $\sigma (L^p, L^{p^*})$.
Because $\|u_{n_k}-u\|_{L^{\infty}} \xrightarrow{k \to \infty} 0$ and $I$ is bounded, $\|u_{n_k}-u\|_{L^p} \xrightarrow{k \to \infty} 0$. Then $u_{n_k} \xrightarrow{k \to \infty} u$ in $\sigma (L^p, L^{p^*})$ and thus $u=v$ a.e.
Let $R(T)$ be the range of $T$. Then $R(T)$ is a norm-closed subspace of $L^p (I) \times L^p (I)$. However, $R(T)$ is not necessarily closed in the weak$^*$ topology of $L^p (I) \times L^p (I)$. So we can not immediately conclude that $(v, w) \in R(T)$. On the other hand, $$ \int_I w \varphi = \lim_k \int_I u_{n_k}' \varphi = \lim_k \int_I u_{n_k} \varphi' = \int_I v \varphi' = \int_I u \varphi' = \int_I u' \varphi, \quad \varphi \in C^1_c (I). $$
Then $w = u'$ a.e. This completes the proof.
