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

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.

• 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 $x\in [a,b]$. Aug 12, 2020 at 8:24
• @AlexRavsky Yes, exactly. I was just trying to use a shorter notation.
– Zac
Aug 12, 2020 at 8:25
• 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. Aug 12, 2020 at 8:30
• @AlexRavsky Thank you. What other (more explicit) assumption on $f_n$ would make this requirement hold?
– Zac
Aug 12, 2020 at 9:29
• @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?
– Zac
Aug 12, 2020 at 9:59

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.

• Thank you very much for your answer. How can one prove the claim in your last sentence?
– Zac
Aug 11, 2020 at 21:29
• 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?
– Zac
Aug 11, 2020 at 21:31
• 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 $B$. Aug 11, 2020 at 21:42
• Ok, thanks. How does compactness follow from the covering?
– Zac
Aug 11, 2020 at 21:44
• This is a standard and powerful equivalence which holds in metric spaces. Aug 11, 2020 at 21:49