Weak Convergence of sequence in a Sobolov Space. Consider the question asked in here. 
I understood most of the answer in the question but the part about the weak convergence I did not get. To show that $u'_{n_k}$ converges weakly to $u$ in $L^p$, we have to show that for every linear continuous functional $J:L^p\to \mathbb{R}$, we have $J(u'_{n_k})\to J(u')$ in $\mathbb{R}$. Any hints on how to solve this one?
My Try 
By the Riez representation theorem we can find $f\in L^{p'}$ such that for all $v\in L^p$,
 $$J(v)=\int_0^1 vf.$$
We can apply this to $u'_{n_k}$ to obtain
$$J(u'_{n_k})=\int_0^1u'_{n_k}f,\;\;\;\text{ for all }k.$$
But then how to proceed from here?
 A: What you want to show is that the convergence holds for all functionals $J$ induced by an element $f\in L^{p'}$.
What Umberto P. has shown in the answer that you linked is that the convergence holds for all functionals $J$
induced by an element $\psi\in C_0^\infty(0,1)$.
There is a result in functional analysis that says that in order to show weak convergence of a bounded sequence in a space $X$
it suffices to test the sequence with functionals from a dense set of the dual space
proof: Let $u_n$ be a sequence in $X$ that is bounded by $C>0$
and let $M\subset X^*$ be dense.
Suppose we already know that
$f(u_n)\to f(u)$ is true for all $f\in M$.
Then for arbitrary $g\in X^*$ and $\varepsilon>0$
there is an $f\in M$ such that $\|f-g\|<\varepsilon$.
We have
$$
|g(u_n)-g(u)|\leq |g(u_n)-f(u_n)|+|f(u_n)-f(u)|+|f(u)-g(u)|
\leq
\varepsilon \|u_n\|+\varepsilon \|u\| + |f(u_n)-f(u)|
\leq
2C\varepsilon+ |f(u_n)-f(u)|
\to 2C\varepsilon.
$$
Since $\varepsilon>0$ can be arbitrarilly small, the convergence $g(u_n)\to g(u)$ follows.
q.e.d.
Since $C_0^\infty(0,1)$ is dense in $L^{p'}$
(and you already use the representation $L^{p'}$ for the dual space of $L^p$)
and the sequence $u_n$ is bounded, we can apply this result.
