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Mar 4, 2015 at 10:08 comment added Charpe Ok I think I understand. The bound gives $u_{n_{j}(t)}(t) \rightharpoonup u(t)$ in $L^2$ for a.e. $t$, but the subsequence depends on $t$. So the Theorem cannot be applied for even a subsequence, and my desired result does not hold even for a subsequence (a subsequence convergence is enough for PDE type arguments). It is strange that this theorem is used in papers without considering such issues.
Mar 4, 2015 at 9:20 comment added Jean Van Schaftingen @Charpe The uniform bound in $L^\infty (0, T; L^2)$ only implies that for almost every $t \in [0, T]$, a subsequence converges weakly; since the set $[0, T]$ is not contable, you cannot conclude by a diagonal argument.
Mar 4, 2015 at 8:06 vote accept Charpe
Mar 4, 2015 at 8:05 comment added Charpe Oh I see. Well, we certainly have the equiintegrability by Remark 1, point 2. I had thought that the uniform bound on $u_n$ in $L^\infty(0,T;L^2)$ gave $u_n(t) \rightharpoonup u(t)$ in $H=L^2$ a.e. $t$.
Mar 4, 2015 at 7:25 history edited Jean Van Schaftingen CC BY-SA 3.0
Explained L^\infty convergence
Mar 4, 2015 at 7:19 comment added Jean Van Schaftingen The theorem that you mention is quite different: it assumes that for almost every $t \in [0, T]$, the sequence $(u_n)$ converges weakly in $H$ and that it is equiintegrable in $L^2 (0, T; H)$. This is not a consequence of the weak-* $L^\infty (0, T; V)$ convergence.
Mar 3, 2015 at 19:35 comment added Charpe I am of course aware of the classical example involving the sine that weak convergence does not imply strong convergence. But I don't see why your example converges weakly in the $L^\infty(0,T;L^2)$ either.
Mar 3, 2015 at 19:34 comment added Charpe Thank you, but doesn't this contradict Theorem 1 of this paper?
Mar 3, 2015 at 7:35 history answered Jean Van Schaftingen CC BY-SA 3.0