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 $L^p([a,b])$. Here $\psi_m$ is a smooth cut-off function such that $$\psi_m(f) = \begin{cases} 1 \qquad \text{ if } |f|\ge 1/m \\ 0 \qquad \text{ if } |f|\le 1/(2m) \end{cases} $$ and $0 \le \psi_m \le 1$.

Suppose in addition that $\Vert f_n \Vert_{L^p} \le C$ (for a constant $C>0$ that does not depend on $n,m$ and for all $p \in [1,\infty]$) and suppose that every subsequence $\{f_{n_k}\psi_m(f_{n_k})\}_{n_k}$ is also compact in $L^p$ for any fixed $m$.

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

Under these assumptions, can we prove the result following this argument, which is rewritten below?

For any $f$,

$$ f - f \psi_m (f) = \begin{cases} f & \text{if } |f| \le 1/2m,\\ 0 & \text{ if } |f| \ge 1/m.\end{cases}$$

In particular,

\begin{align*} \int |f - f \psi_m (f)|^p &= \int_{|f| <1/m} |f - f \psi_m (f)|^p \\ &\le \int_{|f|\le 1/2m} |f|^p + \int_{1/2m \le |f|<1/m} |f - f \psi_m (f)|^p \\ &\le \frac{b-a}{(2m)^p} + \frac{(b-a)}{m^p}\\ \Rightarrow \|f - f \psi_m (f)\|_{L^p} &< C/m \end{align*}

where $C$ depends on $b-a, p$ only. Note we used $|1-\psi_m|\le 1$.

Then using a diagonal argument, there is a subsequence $\{f_{n_k}\}$ of $\{f_n\}$ and $f\in L^p [a, b]$ so that for each $m$, the sequence $\{ f_{n_k} \psi_m (f_{n_k})\}$ converges to $f$ in $L^p$. Now we show that $\{f_{n_k}\}$ also converges to $f$ in $L^p$: for any $\epsilon>0$, fix one $m\in \mathbb N$ with $C/m < \epsilon/2$. Since $\{ f_{n_k} \psi_m (f_{n_k})\}$ converges to $f$ in $L^p$, there is $K$ so that $\| f_{n_k} \psi_m (f_{n_k}) - f\| _{L^p} < \epsilon/2$ for all $k\ge K$. Then

\begin{align*} \|f_{n_k} -f\|_{L^p} \le \|f_{n_k} - f_{n_k} \psi_m (f_{n_k}) \|_{L^p} + \| f_{n_k} \psi_m (f_{n_k}) -f\|_{L^p} < \epsilon/2 + \epsilon/2 \end{align*}

for all $k\ge K$.

If the above fails, you can also add the assumption $\Vert D_x(f_n\psi_m(f_n))\Vert_{L^1} \le C_m$, where $C_m$ is a constant that depends only on $m$.

This question is motivated by two previous questions on Math Stack Exchange [A]