$L^p$ compactness for a sequence of functions from compactness of product with cut-off

Question

Fix $p \in [1,\infty)$. Let $f_n:[a,b] \to \mathbb R$, $n \in \mathbb N$, be a sequence of $C^1$ functions. For every fixed $m\in \mathbb N^*$, suppose that the sequence of functions $$\{f_{n}\psi_m(f_n)\}_{n \in \mathbb N}$$ has a strongly convergent subsequence in $L^p([a,b])$ and that every subsequence $\{f_{n_k}\psi_m(f_{n_k})\}_{n_k}$ is also compact in $L^p$ for any fixed $m$. Here $\psi_m$ is a smooth cut-off function such that $0 \le \psi_m \le 1$ and $$\psi_m(f) = \begin{cases} 1 \qquad \text{ if } |f - 1|\ge 1/m \\ 0 \qquad \text{ if } |f -1 |\le 1/(2m) \end{cases} $$

Assume also that $\Vert f_n \Vert_{L^p} \le C$ (for a constant $C>0$ that does not depend on $n,m$ for all $p \in [1,\infty]$). . If necessary, also asssume that $\Vert D_x(f_n\psi_m(f_n))\Vert_{L^1} \le C_m$, where $C_m$ is a constant that depends only on $m$.

• How can we prove (or disprove) that $\{f_n\}_{n\in \mathbb N}$ also has a strongly convergent subsequence in $L^p([a,b])$?

• If the result is not true, what additional assumption would make it so?

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This question is motivated by my previous question $L^p$ compactness for a sequence of functions from compactness of cut-off.

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Answer 1

The question is not very clear, as stated, but the following can be proved. Let $B$ be a bounded set of $L^p(0,1)$ and assume that for every $\epsilon>0$ the set $$B_\epsilon=\{f \chi_{\{|f| \ge \epsilon\}}, \ f \in B\}$$ is relatively compact in $L^p$, then $B$ is relatively compact, too. In fact, given $\epsilon >0$, a finite $\epsilon/2$-net for $B_\epsilon$ is a $\epsilon$-net for $B$, since $(0,1)$ has finite measure. Changing characteristic functions with smooth cut-off (around 1, as in the original problem), should not change the conclusion.