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Let $\Omega$ be a bounded and smooth domain.

Suppose we have $u_n \to u$ in $H^1_0(\Omega)$. We know that for a subsequence, $\chi_{\{u_n = 0\}} \rightharpoonup f$ to some $f$, weak-* in $L^\infty(\Omega)$. In general we cannot say that $f=\chi_{\{u=0\}}$.

But my question is, is it possible to have a strongly convergent subsequence of $\chi_{\{u_n = 0\}}$? I.e. can we obtain $$\chi_{\{u_n = 0\}} \to g$$ for some $g$ in some (probably $L^p$) space, for a subsequence?

Motivation: If I cannot identify the weak limit as something nice (the desired indicator function), then the next best thing is to show that a strong limit exists. You might expect a strong limit because the $u_n$ are nicely convergent.

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    $\begingroup$ You at least need some conditions on $u$. If $u=0$ then the zero sets of the $u_n$ can be whatever you like, if you just rescale them small enough to make their $H^1$ norms small. $\endgroup$ Commented Mar 6, 2020 at 1:10
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    $\begingroup$ I think you'll run into the same problem whenever $m(u=0)>0$. If $u$ is a.e. nonzero then since in particular $u_n \to u$ in measure, I think you must have $m(u_n = 0) \to 0$ and therefore $\chi\{u_n = 0\} \to 0$ in whatever $L^p$ you like. $\endgroup$ Commented Mar 6, 2020 at 1:14
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    $\begingroup$ So to summarize: I claim that if $m(u=0)=0$ then $\chi_{\{u_n = 0\}} \to 0$ in every $L^p$, $p < \infty$, without needing to pass to a subsequence. For every other $u$, there exists a sequence $u_n \to u$ in $H^1_0$ such that $\chi_{\{u_n = 0\}}$ has no convergent subsequence in $L^1$ (and hence not in any other $L^p$ either). Would that resolve your question, and shall I post an answer to that effect? $\endgroup$ Commented Mar 6, 2020 at 14:17
  • $\begingroup$ @NateEldredge thanks for your replies. Feel free to answer it, I'm not sure if there is a weaker space under which there is a convergence so maybe I'll put a bounty on it when I can $\endgroup$ Commented Mar 6, 2020 at 18:27

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