Timeline for $b_n \rightharpoonup b$ in $L^q(Q) \forall q < \infty$, $b_n \to b$ in $C^0([0,T];H^{-1})$ implies $b_n(t) \rightharpoonup b(t)$ in $L^q(\Omega)$

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Apr 13, 2017 at 12:19	history	edited	CommunityBot		replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
Jul 2, 2014 at 20:18	comment	added	Terry Tao		It looks like the $b_n$ are in fact bounded in $L^\infty$, because the $u_n$ are. In any case, from the uniform boundedness principle, the $b_n(t)$ have to be bounded in $L^q$ if one is to have weak convergence, so such a bound must already appear somewhere in the argument.
Jul 2, 2014 at 11:32	comment	added	riem		Thanks @TerryTao but I don't see any mention in the paper about $b_n(t)$ being uniformly bounded in $L^q(\Omega)$. The only pointwise claim I see is (A.7) which is what we want to show. Sorry if I miss something obvious.
Jun 30, 2014 at 17:51	comment	added	Terry Tao		I think you are omitting an important hypothesis (mentioned in the paper), namely that the $b_n(t)$ are uniformly bounded in $L^q(\Omega)$. With this uniform bound, one can obtain weak convergence through testing against $C^\infty_c(\Omega)$ functions, at which point one can use the $C^0$ strong convergence to conclude.
Jun 30, 2014 at 11:44	comment	added	riem		Sorry I meant $b_n$, not $u_n$. Typo.
Jun 30, 2014 at 11:44	history	edited	riem	CC BY-SA 3.0	edited body; edited title
Jun 30, 2014 at 11:14	comment	added	leo monsaingeon		perhaps you should tell us what is the connection between $u_n$ and $b_n$?
Jun 29, 2014 at 22:23	history	edited	riem	CC BY-SA 3.0	added 28 characters in body
Jun 29, 2014 at 22:11	history	asked	riem	CC BY-SA 3.0