Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If $\{u_n\}$ is bounded in $H$(real Hilbert space)with inner product such that $(\cdot,\cdot)$, then ${\|u_n\|^2u_n}$ is bounded also. Passing a subsequence, one has that $\{\|u_n\|^2u_n\}$ converges weakly to $u_0$. Is it right that $u_0=\|u\|^2u$ ? Practically speaking, can we choose repeatly subsequence of $\{u_n\}$ to obtain $(\|u_n\|^2u_n,v)\rightarrow (\|u\|^2u,v)\forall v\in H $?

share|improve this question
add comment

1 Answer

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).

share|improve this answer
    
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. –  jiahua Mar 24 '13 at 13:15
    
Thank you for answer, I see it. –  jiahua Apr 4 '13 at 6:38
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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