Fix a function $ \phi \in C_c^{\infty}(\mathbb{R}), \phi \ne 0$ and set $u_n(x)=\phi(x+n)$. Let $1 \le p \le \infty$. Then
Check that $u_n$ is bounded in $W^{1,p}(\mathbb{R})$
Prove that there exists no subsequence $(u_{n_{k}})$ converging strongly in $L^{q}(\mathbb{R})$, for any $1 \le q \le \infty$
Show that $u_n \rightharpoonup 0$ weakly in $W^{1,p}(\mathbb{R}), \forall p \in (1,\infty)$
The above is a question from Brezis' book on Sobolev Spaces. The first question is easy to solve. I couldn't do 2nd. For the third bit, dual space of $W_0^{1,p}(\mathbb{R})=\text{dual space of} W^{1,p}(\mathbb{R}=W^{-1,p'}(\mathbb{R})$. Hence there exists $f_0,f_1 \in L^{p'}(\mathbb{R})$ such that for $F \in W^{-1,p'}(\mathbb{R})$, $$\langle F,u\rangle=\int_{-\infty}^{\infty}f_0u+\int_{-\infty}^{\infty}f_1u, u \in W^{1,p}(\mathbb{R})$$ . Thus
$$\langle F,u_n\rangle=\int_{-\infty}^{\infty}f_0u_n+\int_{-\infty}^{\infty}f_1u_n=\int_{-\infty}^{\infty}f_0\phi(x+n)+\int_{-\infty}^{\infty}f_1\phi(x+n)$$
Now since $\phi$ is compactly supported, Let's say $\text{supp}{\phi} =[-m,m]$. Then $$\int_{-\infty}^{\infty}|f_0\phi(x+n)|=\int_{-m-n}^{m-n}|f_0(x)\phi(x+n)|dx \le \left(\int_{-m-n}^{m-n}|f_0|^{p'}\right)^{\frac{1}{p'}}||\phi||_{p}$$
Now since $|f_0|^{p'} \in L^{1}$, given $\epsilon \gt 0$, there exists $R \gt 0 $ such that $\int_{|x| \gt R}|f_0|^{p'} \lt \epsilon$. So Choosing $n \gt m-R$, we have $$\int_{-m-n}^{m-n}|f_0|^{p'} \le \int_{-\infty}^{m-n}|f_0|^{p'} \lt \epsilon$$
Is this alright??
Thanks for the help!!
Concerning 3., do not estimate the integrals $\int f_j(x) \phi(x+n)dx$ in your way but approximate $f_j$ by functions with compact support (just multiply with an indicator function). If $f_j$ has compact support the integral is $0$ for $n$ large enough.
Concerning 2. the limit of a convergent subsequence would be necessarily $0$ because of 3. But the norms of $u_n$ are constant because of translation invariance of the Lebesgue measure.