Brezis' exercise 8.4: weak convergence of a sequence in $W^{1, p} (\mathbb R)$

Question

I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Exercise 8.4 Fix a function $\varphi \in C^\infty_c (\mathbb R)$ such that $\varphi \neq 0$. Let $u_n (x) := \varphi (x+n)$ for $n \in \mathbb N$. Let $p \in [1, \infty]$.

1. Check that $(u_n)$ is bounded in $W^{1, p} (\mathbb R)$.
2. Prove that there is no subsequence $(u_{n_k})$ that converges strongly in $L^q (\mathbb R)$ for any $q \in [1, \infty]$.
3. Show that $u_n \to 0$ in the weak topology of $W^{1, p} (\mathbb R)$ for any $p \in (1, \infty)$.

There are possibly subtle mistakes that I could not recognize in below attempt. Could you have a check on it?

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1. We have $u'_n (x) = \varphi'(x+n)$. Then $\| u_n \|_{L^p} = \| \varphi \|_{L^p}$ and $\| u'_n \|_{L^p} = \| \varphi' \|_{L^p}$ for all $n$. Then $\sup_n \|u_n\|_{W^{1, p}} = \|\varphi\|_{W^{1, p}} < \infty$.

2. It's clear that $u_n \to 0$ everywhere. Assume the contrary that $(u_{n_k})$ converges strongly to some $u$ in $L^q (\mathbb R)$. If $q=\infty$, then $u_{n_k} \xrightarrow{k \to \infty} u$ a.e. If $q \in [1, \infty)$, then there is a further subsequence (WLOG denoted by $(u_{n_k})$) such that $u_{n_k} \xrightarrow{k \to \infty} u$ a.e. In any case, $u=0$ a.e. and thus $\| u_{n_k} \|_{L^q} \xrightarrow{k \to \infty} 0$ which is a contradiction.

3. Let $i:W^{1, p} (\mathbb R) \to L^p (\mathbb R) \times L^p (\mathbb R), v \mapsto (v, v')$ be the canonical isometric homomorphism. We have $$ (L^p (\mathbb R) \times L^p (\mathbb R))^* = L^{p^*} (\mathbb R) \times L^{p^*} (\mathbb R), $$ where $p^*$ is the Hölder conjugate of $p$. Let $T \in (W^{1, p} (\mathbb R))^*$. Then there is $(f,g) \in L^{p^*} (\mathbb R) \times L^{p^*} (\mathbb R)$ such that $Tv = \int_{\mathbb R} (vf+v'g)$. It follows from $u_n \to 0$ and $u'_n \to 0$ everywhere that $Tu_n \to 0$. This completes the proof.

Answer 1

1. and 2) look fine.

2. An easier proof would be to show that $\|u_n - u_m\|_{W^{1,p}}= 2\|\phi\|_{W^{1,p}}$ if $|m-n|$ is large enough. Then $(u_n)$ cannot be a Cauchy sequence.

3. I do not see how you argue that $Tu_n \to0$. Up to now, you only have pointwise convergence. There is no dominating function to apply dominated convergence theorem. One possibility to prove the claim would be to use that $\int_{|x|>R} |f(x)|^pdx$ converges to zero for $R\to \infty$ and $f\in L^p(\mathbb R)$.