$L^p$ compactness for product of two sequences of functions Let $f_n:[a,b] \to \mathbb R$, $n \in \mathbb N$, be a sequence of  $L^p$ functions for some $p \in (1,\infty)$. 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 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 $0 \le \psi_m \le 1$.
Is it true that $\{f_n\}_{n\in \mathbb N}$ also has a strongly convergent subsequence in $L^p([a,b])$?

I wanted to apply a diagonal argument: [1], but I can't make it work properly.
 A: There exists a counterexample showing that it is in general not possible to conclude the sequentially compactness of ${f_n}$
Set $[a,b]=[0,1]$ and $p=2$
First, for $k\geq 1$ define $a_k = \sum_{j=1}^k 2^{-j}$
Then, define
$$g_k(x):=2^{k+1} \chi_{[a_{k},a_{k+1}]} (x)$$ .
$$h_{m}(x):=\chi_{[0,1/2]} (x)sin(4\pi m x)$$
Now set
$$f_n(x)=g_{\sigma_1(n)}(x) +  \frac{1}{\sigma_1(n)}h_{\sigma_2(n)}(x) $$
Where $(\sigma_1,\sigma_2): \mathbb{N} \to \mathbb{N}\times \mathbb{N}$ is a bijection.
Then the sequence $\{f_n\}_n$ does not admit convergent subsequences in $L^2[a,b]$. This can be shown using that

*

*$$
\|g_{k_1}+  \frac{1}{k_1}h_{m_1} - (g_{k_2}+  \frac{1}{k_2}h_{m_2})\|_2^2
 = \|g_{k_1} - g_{k_2}\|_2^2 +\|\frac{1}{k_1}h_{m_1} -\frac{1}{k_2}h_{m_2}\|_2^2 
$$
since the support of every $g$ is a subset $[1/2,1]$ while the support of every $h$ is a subset of $[0,1/2]$


*If $k_1 \neq k_2$ $$\|g_{k_1} - g_{k_2}\|_2^2= \|g_{k_1}\|_2^2+ \|g_{k_2}\|_2^2 \geq C$$ where $C$ is a positive constant independent of $k_1,k_2$


*If $k_1 = k_2$ the term
$$\|\frac{1}{k_1}h_{m_1} -\frac{1}{k_2}h_{m_2}\|_2^2 = \frac{1}{k_1^2}\| h_{m_1} - h_{m_2}\|_2^2 $$ is equal to zero for $m_1=m_2$ and it is equal to $\frac{1}{2 k_1^2}$ otherwise, using the well known properties of the trigonometric basis.
Nontheless, for every fixed $m$ the family $\{f_{n}\psi_m(f_n)\}_{n \in \mathbb N}$ admits a convergent subsequence.
In fact, the trigonometric term appearing in $f_n$ is cutted off in $f_{n}\psi_m(f_n)$ if $m>\sigma_1(n)$ and therefore if $\{ n_j\} = \{ n \mid \sigma_1(n)=m+1\}$ then subsequence $\{f_{n_j}\psi_m(f_{n_j})\}$ converges to $g_{m+1} \psi_m(g_{m+1})$
