Does it follow that $u_n \rightharpoonup 0$ weakly in $W^{1, p}(\mathbb{R})$ for all $p \in (1, \infty)$?

Question

Fix a function $\varphi \in C_c^\infty$, $\varphi \not\equiv 0$, and set $u_n(x) = \varphi(x + n)$. Does it follow that $u_n \rightharpoonup 0$ weakly in $W^{1, p}(\mathbb{R})$ for all $p \in (1, \infty)$?

Answer 1

Yes. The sequence is bounded and, hence, has a weakly convergent subsequence. This subsequence converges strongly in $L^q([-L,L])$ for any $L > 0$. Moreover, it converges to zero pointwise. Hence, the limit must be zero.

Finally, it is straightforward to show that the entire sequence converges towards zero weakly.

Answer 2

Let $T\in W^{-1,p'}(\mathbb{R})$. Then, there are functions $f_1,f_2\in L^{p'}(\mathbb{R})$ such that $$T(u)=\int f_1u+\int f_2u',\ \forall\ u\in W^{1,p}(\mathbb{R}).$$

We have to prove that $T(u_n)\to 0$ for all $T\in W^{-1,p'}(\mathbb{R})$. Suppose that the support of $\varphi$ is contained in $[a,b]$. By one hand $$\left|\int f_1u_n\right|=\left|\int f_1(x)\varphi(x+n)\right|=\left|\int f_1(x-n)\varphi(x)\right|\le \int_a^b |f_1(x-n)\varphi(x)|.$$

The last integral must converge to zero when $n\to\infty$, because it is bounded by $$\left(\int_{a-n}^{b-n}|f_1|^{p'}\right)^{1/p'}\|\varphi\|_p.$$

On the other hand $$\left|\int f_2u_n'\right|\le \int_a^b |f_2(x-n)||\varphi'(x)|,$$

and as in the last case, this integral must converge to zero, therefore, $$|T(u_n)|\le \left|\int f_1u_n\right|+\left|\int f_2u'_n\right|,$$

converges to zero.

Remark: a characterization of $W^{-1,p'}(\mathbb{R})$ in terms of functions in $L^{p'}$ can be found in the Brezis book of Functional Analysis.