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Define $$ U := C^0 ([0,T], W^{1,2} ) \cap C^1 ([0,T] \cap L^2 ) \cap L^\infty ([0,T] , W^{s,2} ).$$ Then how can I prove that $ \lim_{x_j \to \infty}| u |^2 = 0 $ by using the fact: $$ C^1 ([0,T] , \mathcal S ) \text{ is dense in } U?$$ Here $u : [0,T] \times \Bbb R^n \to \Bbb C^n$ , the norm for the space $U$ : $|u|_{s,T} := \sup _{t \in [0,T]} \| u(t) \|_{W^{s,2}}$, $W^{s,p}$ is the usual Sobolev space, $\mathcal S$ denotes the Schwarz class.

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@DavideGiraudo $|u|$ means that $\sqrt{|u_1|^2 + \cdots + |u_n|^2}$ for a fixed $t \in [0,T]$. – Ann Oct 29 '12 at 14:37
@DavideGiraudo Sorry, $\Bbb C$ should be fixed to $\Bbb C^n$. – Ann Oct 29 '12 at 14:39
and $u \in U$ means that all the components $u_j \in U$. – Ann Oct 29 '12 at 14:41
@DavideGiraudo $p$ is always 2, but $s$ can be $0,1,2,\cdots$. – Ann Oct 29 '12 at 14:48
Do you see why you just have to show it when $u_j$ is in the Schwartz space? – Davide Giraudo Oct 29 '12 at 14:53

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