This is an exercise from Brezis's functional analysis where we're trying to show that the set $$ B = \{f \in W^{1,p}((0,1)): ||f||_{W^{1,p}((0,1))} \le 1\} $$ is compact in $L^{p}((0,1))$, for $p > 1$.
To do this, we use the fact that any sequence $\{f_{n}\}_{n \in \mathbb{N}}$ in $B$ is bounded, so it has a subsequence that converges to something ($f$) in $W^{1,p}((0,1))$ in $||\cdot||_{L^{\infty}}$. I want to use this fact to show that $f_{n}$ (not re-labelling the subsequence) converges to $f$ in $W^{1,p}$ norm.
It is easy to show that $f_{n} \to f$ in $L^{p}$, but I also need to show that $f_{n}' \to f_{n}'$ in $L^{p}$ (where the derivatives are taken in the weak sense), which I am unable to do. It is fine even if $f_{n}' \to f'$ in $L^{p}$ up to a subsequence, since we can extract a further subsequence without any worry here. How would one do this?
Edit: Can something like the following work? We know that $W^{1,p}$ is isomorphic to a subspace of $L^{\infty}$, call it S, so we have a bijective continuous map $T: S \to W^{1,p}((0,1))$, which should give us the assertion without having to go through the process of showing $f_n, f_n'$ converging separately in $L^{p}$?
Let $ 1<p < \infty $. Recall that the identity operator $$ id \colon(W^{1,p}(0,1),||\cdot ||_{1,p} ) \to (L^p(0,1),||\cdot ||_p) $$ is compact and thus completely continuous (that is, it maps weakly convergent sequences to norm convergent ones). Let $$ B_1(p) =\{ u \in W^{1,p} (0,1) \colon ||{u}||_{p} + ||u'||_p \leq 1\}.$$ Let $ (u_n)_n \subset B_1(p) $. By the reflexivity of $ W^{1,p} (0,1)$, there exists $ u \in B_1(p) $ such that $ u_n \xrightarrow{weakly} u $ in $ W^{1,p} (0,1)$ (atleast for a subsequence). Thus, (by the complete continuity of the identity operator) $ u_n \to u $ in $ L^p(0,1)$. In other words, $ B_1(p) $ is compact in $ L^p (0,1)$.
Now, let $p= \infty$. Let $(u_n) \subset B_1 (\infty)$. By the compactness of the identity operator, there exists $u \in L^\infty(0,1)$ such that $||u_n-u||_\infty \to 0 $ (for a subsequence). We wish to show that $u \in B_1(\infty)$. Since $(u_n') \subset L^\infty(0,1) $ is bounded, invoking Alaoglou's theorem, we infer that $u_n' \xrightarrow{weak^*} g $ for some $g \in L^\infty(0,1)$. It is easy to see that $u \in W^{1,\infty}(0,1)$ and $u'=g$. Recall that the norm is weak-star lower semi continuous and so $$ ||u'||_\infty \leq \liminf ||u_n'||_\infty .$$ In conclusion, $$||u||_\infty + ||u'||_\infty \leq \liminf (||u_n'||_\infty + ||u_n||_\infty) \leq 1 $$ and thus $u \in B_1(\infty).$
For $p=1$ things are not as good. You can show that $B_1(1) \subset L^1(0,1) $ is not closed (and thus not compact) in $L^1(0,1)$. To see this, consider the following sequence in $ W^{1,1}(0,1)$: $$ u_n(t) = \left\{\begin{array}{lr} 0 , ~~ 0 \leq t \leq \frac 12\\ n(t-1/2), ~~ \frac 12 <t \leq \frac 12 + \frac 1n \\ 1, ~~ \frac 12 + \frac 1n < t \leq 1. \end{array}\right. $$
with $$ u_n'(t) = \left\{\begin{array}{lr} 0 , ~~ 0 \leq t \leq \frac 12\\ n,~~ \frac 12 <t \leq \frac 12 + \frac 1n \\ 0, ~~ \frac 12 + \frac 1n < t \leq 1 \end{array}\right. $$ We have that $ ||{u_n}||_1=\frac 12 - \frac{1}{2n} $ and $ ||{u_n'}||_1 =1$ and thus $ ||{u_n}||_{1,1} \leq 3/2 $. Let $ u = \mathcal X_{(\frac 12,1)} $, then $$ ||u_n-u ||_1= \int |{u_n-u}| = \frac{1}{2n} \to 0 $$ and thus $ \frac 23 u_n \to \frac 23 u $ in $ L^1(0,1) $ but $ \frac23 u \notin W^{1,1}(0,1) $ and in particular $ \frac23 u \notin B_1(1) $.