Prove that the weak convergence in $L^p \left(\mathbb{R}^N\right)$ does not imply the weak convergence of the modulus

Question

Let be $p\in [1,\infty)$ and $\{u_n\}\subset L^p\left(\mathbb{R}^N\right)$ a sequence.

How do I prove that $u_n \rightharpoonup u$ in $L^p \left(\mathbb{R}^N\right) \not\Longrightarrow|u_n|\rightharpoonup |u|$ in $L^p \left(\mathbb{R}^N\right)$?

Answer 1

Define $$g(x):=\cases{\frac{1}{x^{2N}}, & $||x||_\infty >\pi$ \\ 1, & $||x||_\infty \le \pi$ \\ } \quad u_n(x):=g(x)\prod_{j=1}^N \sin(n x_j)$$ Both $g$ and $u_n$ are in $L^p\left(\mathbb{R}^N\right)$. Riemann-Lebesgue lemma says that $$\lim_{n\to+\infty}\int_{\mathbb{R}^N} f(x)\prod_{j=1}^N \sin(n x_j)\,dx=0\quad\forall f\in L^1\left(\mathbb{R}^N\right)$$ So it is clear that $u_n\rightharpoonup 0$ in $L^p\left(\mathbb{R}^N\right)$.

Pick $h=\chi_{(-\pi,\pi)^N}\in L^{p'}\left(\mathbb{R}^N\right)$

Show that $\int |u_n| h\not\to 0$ $$\lim_{n\to+\infty}\int_\mathbb{R^N}|u_n(x)| h(x)\,dx =\lim_{n\to+\infty}\int_{(-\pi,\pi)^N}\prod_{j=1}^N |\sin(n x_j)|\,dx= \\ =\lim_{n\to+\infty}\left(\int_{-\pi}^\pi |\sin(n x)|\,dx\right)^N= \\ =\lim_{n\to+\infty}\left(\frac{2}{n}\int_{0}^{n\pi}|\sin(x)|\,dx\right)^N=\lim_{n\to+\infty}4^N=4^N\neq 0$$ Hence $|u_n|\not\rightharpoonup 0$ in $L^p\left(\mathbb{R}^N\right)$.