2
$\begingroup$

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 $v = |u|$.

$\endgroup$
2
$\begingroup$

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 $|u_n| \equiv 1$.

$\endgroup$
2
$\begingroup$

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 $\frac{1}{2\pi}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.