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Let $u_n\rightharpoonup u$ in $L^2(\Omega)$, where $\Omega$ is any domain in $\mathbb R^n$. Is there any relationship between the supports of $u_n$ and $u$? The reason I am asking is that I remember seeing somewhere an assertion that, for a weakly (star) convergent sequence of measures, the support of the limit is a subset of the Kuratowski limit inferior (or superior, I cannot remember exactly) of the supports of the sequence members.

Edit: there is an "obvious" relation, namely $\mathrm{supp}\, u\subseteq \cup_n \mathrm{supp}\, u_n$. I would be interested in something more "optimal".

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Ok, so we want something like $\operatorname{supp} u \subseteq \cap_{N=1}^\infty \cup_{n=N}^\infty \operatorname{supp} u_n$. Here we go:

$(u_n)_{n \in \mathbb{N}}$ converges weakly to $u$, and so does every restricted sequence $(u_n)_{n \geq N}$. $A_N = \{ f \in L^2; \operatorname{supp} f \subseteq \cup_{n=N}^\infty \operatorname{supp}u_n\}$ is a closed subspace of $L^2$, and closed implies weakly closed. Therefore $u$ needs to be in $A_N$ for each $N$. In particular,

$$ \operatorname{supp} u \subseteq \cap_{N=1}^\infty \cup_{n=N}^\infty \operatorname{supp} u_n.$$

Edit: A sketch of why I think the statement will be wrong if we replace the lim sup by lim inf:

Consider $L^2[0,1]$ and let the functions $f_{k,n}$ be defined by $$ f_{k,m} = 1_{\big [ \frac{k}{m},\frac{k+1}{m}\big ]}$$ for $m \in \mathbb{N}$ and $k\in \mathbb{N}, 0 \leq k < m$. Enumerate them somehow as $(f_n)_{n \in \mathbb{N}}$. Then $f_n \to 0$ in norm, hence $f_n \to 0$ weakly. Consider $$ u_n = 1 - f_n.$$ Then $u_n \to 1$ weakly. But on the other hand, for each $x \in [0,1]$ there are infinitely many $n \in \mathbb{N}$ such that there is an open interval $I(n)$ containing $x$ and $\operatorname{supp} u_n \cap I(n) = \emptyset$. Therefore $\lim \inf \operatorname{supp} u_n = \emptyset$ and $\operatorname{supp} u = [0,1]$.

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  • $\begingroup$ True, but there is a relation: $\emptyset \subset [0,1]$ :-). I am more of thinking of the case where the supports of the sequence members wander around inside $\Omega$... $\endgroup$
    – tks
    May 12, 2017 at 6:58
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    $\begingroup$ Alright, in that sense you always have the relation $\operatorname{supp} u \subseteq \cup_{n} \operatorname{supp} u_n$ ;-) $\endgroup$
    – agb
    May 12, 2017 at 7:09
  • $\begingroup$ Yes, the question is if it is possible to put a \cap somewhere there in between.... $\endgroup$
    – tks
    May 12, 2017 at 7:11
  • $\begingroup$ I've modified the question to be more precise... $\endgroup$
    – tks
    May 12, 2017 at 7:14
  • $\begingroup$ Ok, I've modified my answer. Maybe this is more what you are looking for. $\endgroup$
    – agb
    May 12, 2017 at 7:19

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