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Suppose $(z_n)$ is a weakly convergent sequence in $L^p(\Omega)$, $\Omega$ is a limited open set of $\mathbb{R}^N$. Let $(\chi_n)$ be defined by $\chi_n(x)=\chi(x.n)$, where $\chi$ is the characteristic function of a rectangular block $B$. So, we can conclude that $\chi_n\stackrel{*}{\rightharpoonup}<\chi>_B$, where $<\chi>_B$ denotes the average (here the convergence is in $L^\infty(\Omega)$).

Is it possible to prove that $z_n \chi_n\rightharpoonup z<\chi>_B$? What kind of changes should I do to prove it?

Of course, I can't suppose strong convergence here.

user143090
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  • Maybe there is something wrong, but as you wrote it, it holds $\chi_n\to0$ in all $L^q(\Omega)$, $q<\infty$. – daw Mar 16 '17 at 16:49
  • Well, it's not. Since we can extend $\chi(x)$ periodically to $\mathbb{R}^N$ and define $\chi_n$. So, by the average theorem, we get what I told. – user143090 Mar 16 '17 at 16:55
  • Yes, we can. But why should we? There is nothing like that in the question. – daw Mar 16 '17 at 17:03

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