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Let $u \in C^0([0,T], H^{s-1}(\Bbb R^n)). $ Let $\{t_n \} \subset [0,T]$ such that $\lim_{n \to \infty} t_n = t_0$. Let $ u(t_n ) \to u(t_0) $ in the Sobolev space $H^{s-1} ( \Bbb R^n )$ for $s = 1,2, \cdots$. Assume $ \| u(t_n ) \|_{H^{s} (\Bbb R^n)} \leqslant R < \infty $ for all $n$.

Then how can we find a subsequence $\{t_k \}$ of $\{ t_n \}$ such that $u(t_k ) \to u(t_0) \;weakly$ in $H^s$ ?

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3  
Alaoglu's theorem. –  Nate Eldredge Nov 13 '12 at 5:08
    
@NateEldredge Thank you. Then the assumption "$u(t_n) \to u(t_0)$ in the Sobolev space $H^{s−1}$ for $s=1,2,⋯.$" needless? I think if just $ \| u (t_n ) \|_{H^s} \leqslant R$ then we can apply Alaoglu's theorem. –  Ann Nov 13 '12 at 5:18
    
Alaoglu will guarantee that a subsequence converges weakly to something. You'll need the $H^{s-1}$ convergence to show that the limit is in fact $u(t_0)$. –  Nate Eldredge Nov 13 '12 at 5:19
    
Actually, aren't you assuming $u : [0,T] \to H^s$ is continuous? That would imply that $u(t_n) \to u(t_0)$ strongly in $H^s$; you don't need weak convergence or a subsequence. Are you sure you wanted to assume that? –  Nate Eldredge Nov 13 '12 at 5:20
1  
Ok, that makes sense. When you have the details worked out, you might like to post your own answer. –  Nate Eldredge Nov 13 '12 at 11:43

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