# If $u_n$ converges weakly to $u$. Does $|u_n|$ converges weakly to $|u|$?

This question comes from the proof of Theorem 1 in E.N.Dancer's paper. (page 427 line 13-14)

Question:

Suppose $u_n$ converges weakly to $u$ in $L^2(\Omega)$.

Does $|u_n|$ converges weakly to $|u|$ in $L^2(\Omega)$?

My Attempt

From the facts that

1.) weak convergence $\Rightarrow$ norm-boundedness $\Rightarrow$ subsequence weak convergence.

2.) If every subsequence contains a weakly convergent subsubsequence, then the whole sequence converges weakly.

We know that $|u_n|$ converges weakly to some $v \in L^2(\Omega)$. But I don't know how to show $v = |u|$.

This is not true. In $\Omega = (0,1)$, consider the sequence $\{u_n\}$ given by $$u_n(t) = \operatorname{sign}(\sin(\pi \, n \, t)).$$ Then, it is easy to check that $u_n \rightharpoonup 0$ but $|u_n| \equiv 1$.
This is not true. Let $e_k:[0,2\pi]\to \mathbb C, t\mapsto e^{ikt}$ for $k\in\mathbb Z$. Then $(\frac{1}{2\pi}e_k)_{k\in\mathbb Z}$ is an orthonormal basis of $L^2[0,2\pi]$ and thus converges weakly to $0$ for $k\to\infty$, but $|e_k|\equiv \frac{1}{2\pi}$ converges in norm to $\frac{1}{2\pi}$.