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Let $\Omega$ be a bounded and smooth domain. Suppose I have a sequence of non-negative functions $u_n \in L^\infty((0,1)\times \Omega) \cap L^\infty((0,1);L^\infty(\Omega))$ with $$0 \leq u_n \leq 1 \quad \text{a.e}$$ and $$u_n \rightharpoonup^* u \quad\text{in $L^\infty((0,1)\times \Omega)$}$$ to some $u$.

Is it possible to conclude that for a subsequence, $u_{n_j}(t) \rightharpoonup^* u(t)$ in $L^\infty(\Omega)$ for a.e. $t$? The subsequence $\{n_j\}$ should not depend on the point $t$.

Additional info:

  • For each $n$, $u_n$ is a piecewise constant function, i.e., $u_n = \sum_{k=1}^n a_{kn}\chi_{I_{kn}}(t)$ holds for a partition $\{I_{kn}\}$ and $a_{kn} \in L^\infty(\Omega)$.

    • The sequence $u_n(t)$ is bounded uniformly in $L^\infty(\Omega)$ for a.e. $t$.
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    $\begingroup$ If I understood correctly the question, this is not true. If $u_n(t,x)=\sin (2\pi n\, t)f(x)$ for a fixed $f \neq 0$, then $u_n \to 0$, $w^*$ in $(t,x)$ by Riemann Lebesgue, but for fixed $t\neq 0,1/2,1$ does not converge $w^*$ in $x$. Similarly, there is no way to find a sequence $n_j$ such that $\sin (n_j t)$ converges a.e. Positivity does not help: tafe $f$ positive and add 1 to the sinus. $\endgroup$ Feb 7, 2020 at 15:02
  • $\begingroup$ @GiorgioMetafune: I think it's worth posting this as an answer. $\endgroup$
    – Nik Weaver
    Feb 7, 2020 at 17:20
  • $\begingroup$ Ok, I will do. I was not sure about the question. $\endgroup$ Feb 7, 2020 at 17:39

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If I understood correctly the question, this is not true. If $u_n(t,x)=sin(2πnt)f(x)$ for a fixed $f\neq 0$, then $u_n \to 0$, $w^*$ in $(t,x)$ by Riemann Lebesgue, but for fixed $t\neq 0,1/2,1$ does not converge $w^*$ in $x$. Similarly, there is no way to find a sequence $n_j$ such that $sin(n_jt)$ converges a.e. Positivity does not help: take f positive and add 1 to the sinus.

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