Let $f_n: [0,1] \to \mathbb R$ be a uniformly bounded sequence in $L^p$. Then there exists a subsequence such that $f_{n_k} \to f$ weakly in $L^p([0,1])$. What is the weak limit of the sequence of functions $$g_{n_k} = f_{n_k} \ \mathrm{sign}(f_{n_k} - 1),$$ where sign is the signum function? Can we write it in terms of $f$? Note that $g_{n_k}$ is also uniformly bounded in $L^p$, hence it has a weak limit $g$ (up to subsequences). What is the relationship between $g$ and $f$?
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$\begingroup$ It depends. For example, if $f_n\to 1$, then we can obviously have $g_n\to 1$ as well, but also $g_n\to 0$ if $f_n-1$ changes sign frequently. $\endgroup$– Christian RemlingCommented Aug 15, 2020 at 16:02
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$\begingroup$ The question may be reformulated e.g. as: what is the closure of the graph of the map $f\mapsto f\mathrm{sign} (f-1)$ from $L^\infty[0,1]$ to itself, wrto a given topology $\endgroup$– Pietro MajerCommented Aug 15, 2020 at 18:09
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$\begingroup$ @ChristianRemling I see: for example, what assumptions do I need to add to obtain $f_{n_k}\mathrm{sign}(f_{n_k}-1) \rightharpoonup f \chi$, where $$\chi \in \begin{cases} 1 \text{ if } f > 1 \\ [-1,1] \text{ if } f = 1 \\ -1 \text{ if } f < 1 \end{cases},$$ weakly in $L^p$? $\endgroup$– LaoCommented Aug 15, 2020 at 23:16
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