1
$\begingroup$

If $\{u_n\}$ is bounded in a real Hilbert space $H$, with inner product $(\cdot,\cdot)$, then ${\|u_n\|^2u_n}$ is also bounded.

As there is a weakly converging sub-sequence, we can WLOG assume that $\{\|u_n\|^2u_n\}$ converges weakly to $u_0\in H$.

Is it right that $u_0=\|u\|^2u$ ? Practically speaking, can we repeatedly choose a sub-sequence of $\{u_n\}$ to obtain $(\|u_n\|^2u_n,v)\rightarrow (\|u\|^2u,v), \, \forall v\in H $?

$\endgroup$

1 Answer 1

2
$\begingroup$

Assuming $u$ is supposed to be the weak limit of the original sequence $u_n$, the answer is not in general. For example, if $u \neq 0$, then $||u_n||^2u_n \rightharpoonup ||u||^2u$ iff $||u_n||^2/||u||^2 u_n \rightharpoonup u$. Since $u_n \rightharpoonup u$, this in turn occurs iff $||u_n||^2/||u||^2 \to 1$, which fails in general; the problem is that all we can conclude in general is $||u|| \leq \liminf ||u_n||$ (cf. Fatou's Lemma) but we don't in general have $||u|| = \lim ||u_n||$ (and indeed, $\lim ||u_n||$ need not exist).

$\endgroup$
1
  • $\begingroup$ As $\{u_n\}$ is bounded in $H$, that is, $\{\|u_n\|\}$ is bounded in $R^+$, which implies that $\{\|u_n\|\}$ has convergent subsequence $\|u_{n_k}\|→\|u_0\|$ in $R^+$. But $lim\|u_{n_k}\|≠\|u\|.$ Thank your answer. $\endgroup$
    – jiahua
    Mar 24, 2013 at 13:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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