Is still it weakly continuous ？

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 $（\|u_n\|^2u_n，v)\rightarrow (\|u\|^2u,v)\forall v\in H$?

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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).
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