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?

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

  • $\begingroup$ I guess the definition of the functions $\psi_m(f)$ should be something like $$\psi_m(f)(x) = \begin{cases} 1 \qquad \text{ if } |f(x) - 1|\ge 1/m \\ 0 \qquad \text{ if } |f(x) -1 |\le 1/(2m) \end{cases} $$ for each $x\in [a,b]$. $\endgroup$ Aug 12, 2020 at 8:24
  • $\begingroup$ @AlexRavsky Yes, exactly. I was just trying to use a shorter notation. $\endgroup$
    – Zac
    Aug 12, 2020 at 8:25
  • $\begingroup$ The condition “every subsequence $\{f_{n_k}\psi_m(f_{n_k})\}_{n_k}$ is also compact in $L^p$” is very strong. It can be shown that it is equivalent to $\{f_{n}\psi_m(f_{n})\}_{n\in\Bbb N}$ has finitely many values. I guess you mean that a set $\{f_{n}\psi_m(f_{n})\}_{n\in\Bbb N}$ is totally bounded instead. $\endgroup$ Aug 12, 2020 at 8:30
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    $\begingroup$ @AlexRavsky Thank you. What other (more explicit) assumption on $f_n$ would make this requirement hold? $\endgroup$
    – Zac
    Aug 12, 2020 at 9:29
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    $\begingroup$ @AlexRavsky Thanks! Is there anything that we can assume on $f_n$ (smoothness, integrability, a condition on the derivatives?) that would make this condition on $\mu(X_{n,m})$ automatically true? $\endgroup$
    – Zac
    Aug 12, 2020 at 9:59

1 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.

  • $\begingroup$ Thank you very much for your answer. How can one prove the claim in your last sentence? $\endgroup$
    – Zac
    Aug 11, 2020 at 21:29
  • $\begingroup$ Also, I'm not sure about the argument with nets (because I'm not very familiar with them). How can one make it more explicit? $\endgroup$
    – Zac
    Aug 11, 2020 at 21:31
  • $\begingroup$ Sorry for the language which was a bit too cryptic. For $B$ I want to prove it is totally bounded, that is for every $\epsilon>0$ there exists a finite number of balls $B(f_i, \epsilon)$ covering it. Do it first for $B_\epsilon$ (with $\epsilon/2$ balls), by assumption, and then note that the double balls cover $B$. $\endgroup$ Aug 11, 2020 at 21:42
  • $\begingroup$ Ok, thanks. How does compactness follow from the covering? $\endgroup$
    – Zac
    Aug 11, 2020 at 21:44
  • $\begingroup$ This is a standard and powerful equivalence which holds in metric spaces. $\endgroup$ Aug 11, 2020 at 21:49

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