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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 $?

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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).

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  • $\begingroup$ 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 '13 at 13:15

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