0
$\begingroup$

This question stems from the proof of Theorem A.1 on page 425 of this paper.

Let $Q=(0,T)\times \Omega$. Suppose $b_n \rightharpoonup b$ in $L^q(Q)$ for any $q < \infty$ and $b_n \to b$ in $C^0([0,T];H^{-1}(\Omega))$. This strong convergence yields that $b_n(t) \rightharpoonup b(t)$ in $L^q(\Omega)$ for any $q < \infty$.

I don't understand how this conclusion holds. Does anyone know?

Coincidentally a similar question was asked on MSE, for which I placed a bounty but received no answers so I think it is OK to post here as I have had no luck.

$\endgroup$
5
  • $\begingroup$ perhaps you should tell us what is the connection between $u_n$ and $b_n$? $\endgroup$ Commented Jun 30, 2014 at 11:14
  • $\begingroup$ Sorry I meant $b_n$, not $u_n$. Typo. $\endgroup$
    – riem
    Commented Jun 30, 2014 at 11:44
  • 2
    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Commented Jun 30, 2014 at 17:51
  • $\begingroup$ 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. $\endgroup$
    – riem
    Commented Jul 2, 2014 at 11:32
  • 1
    $\begingroup$ 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. $\endgroup$
    – Terry Tao
    Commented Jul 2, 2014 at 20:18

0

You must log in to answer this question.