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Let $u_n \to u$ strongly in $C([0,T];H^{-1}(\Omega))$ and suppose that $u_n$ is uniformly bounded in $L^\infty((0,T)\times\Omega)$. Then $$u_n(t) \to u(t)$$ weakly in $L^1(\Omega)$ for each $t$?

Can someone show me what result this is? How it is gotten?

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    $\begingroup$ I have no idea. We must have $u_n(t) \to u(t)$ in $H^{-1}(\Omega)$ for each $t$ but other than that I do not know. We also have $u_n \rightharpoonup u$ weakly in $L^1(0,T;L^1)$. But I don't think this implies the result i wanted because $u(t)$ as a function of $L^1$ is only defined a.e. $\endgroup$ – maths_student_2000 Jun 24 '14 at 15:14
  • $\begingroup$ @riem equi-integrable (in the paper), or uniformly integrable.. see below $\endgroup$ – username Jul 3 '14 at 16:20
  • $\begingroup$ Thanks @AthanagorWurlitzer. Can you explain why $\lVert u_n(t) \rVert_{L^\infty(\Omega)} \leq C$ follows from assumptions? Does one not need $\lVert u_n \rVert_{L^\infty(0,T;L^\infty)} \leq C$ for this? Also I posted this question on MO where Terrence Tao wrote it folows from uniform boundedness principle but I didn't fully follow it. You may be interested: mathoverflow.net/questions/172942/… $\endgroup$ – riem Jul 4 '14 at 11:19
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I think you need $\Omega$ of bounded measure (e.g. bounded).

Fix $t\in[0,T]$. You have $u_n(t)\to u(t)$ in $H^{-1}$. Furthermore, $\sup_n \|u_n(t)\|_\infty=M<\infty$ by assumption.

Therefore your sequence $u_n$ is uniformly integrable, which means, that for every $\epsilon>0$, there exists $\delta>0$ such that for all measurable set $A\subset\Omega$, if $|A|<\delta$ then $\int_A|u_n| <\epsilon$, for all $n$.

This is the case here: $$ \int_A |u_n|\leq \sup_n \|u_n\|_\infty |A|, $$ so $\delta=\epsilon/(M+1)$ works.

This is a particular case of a more general result, namely that if the sequence is $u_n$ is bounded in $L^1(\Omega)$, uniformly integrable, and $\Omega$ has finite Lebesgue measure, then you can extract a weakly converging subsequence $\tilde{u}_n$ in $L^1$. This is proved here in these lecture notes for example, page 37 (prop 80).

By uniqueness of the weak limit, you then get rid of the subsequence (as $u_n(t)\to u(t)$ globally in $H^{-1}$).

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  • $\begingroup$ Also, isn't this true for a.e. $t$ only? Since we only have $u_n$ bounded in $L^\infty(0,T;L^\infty)$. $\endgroup$ – riem Jul 7 '14 at 10:30

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