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

