I am trying to solve this exercise from Brezis.
I proceeded as follow.
For 1) I have observed that $$\int_{\mathbb{R}}|u_n|^p(x) dx=\int_{a-n}^{b-n}|\varphi(x)|^p dx\le |(\max\varphi)|^p|a-b|\ \ \ \ \forall n$$ and the same holds for $u'_n$.
For 2) I have that for $n$ large enough $\varphi(x+n)=0$ for all $x$, so if it exists $u_{n_k}$ converging to some limit $u$, then $u=0$. But having for all $n$, $||u_n||_p=||\varphi||_p\ne0$ (since $\varphi\not\equiv0$) it holds for every subsequence which cannot converge.
For 3) I thought this way: from Banach-Alaoglu (since $W^{1,p}$ is reflexive for $p\in(1,\infty)$ it exists a subsquence $u_{n_k}$ converging weakly to 0. Since this argument could be performed for all subsequences of $u_n$ I deduce that the whole sequence $u_n\rightharpoonup 0$.
It is my solution correct? Thank you.
You did 1) and 2) correctly and for the last question you only to show that for all $v\in C_0^{\infty}(\Bbb R)$
$$\int_{\Bbb R} u_n vdx\to 0~~~~ and~~~~~\int_{\Bbb R} u'_n vdx\to 0$$
which is straightforward since for $n$ large enough we have
$$ u_n(x)=u'_n(x)=0$$ since $supp \phi\subset (a,b)$