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](https://math.meta.stackexchange.com/questions/370/formatting-sandbox/4993#4993), 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]](https://math.stackexchange.com/questions/3775973/lp-compactness-for-product-of-two-sequences-of-functions/3776126?noredirect=1#comment7774430_3776126)



$L^p$ compactness for a sequence of functions from compactness of cut-off
Let $$g^m\_n := f\_n \psi\_m(f\_n).$$ The assumptions mean that $(f\_n)\_n$ is a bounded sequence in $L^p(a,b)$ and that $(g\_n^m)\_n$ is relatively compact in $L^p(a,b)$ for each $m$. We use the [Frechet-Kolmogorov theorem](https://en.wikipedia.org/wiki/Fr%C3%A9chet%E2%80%93Kolmogorov_theorem) characterizing compactness in $L^p$ spaces to show that this transfers to $(f\_n)\_n$. (Then not only $(f\_n)\_n$ has a convergent subsequence, but also every subsequence $(f\_{n\_k})\_k$.)


Split and estimate
\begin{align\*}\|\tau\_h f\_n - f\_n\|\_{L^p(a,b-h)} &\leq \|\tau\_h(f\_n-g\_n^m)\|\_{L^p(a,b-h)} + \|\tau\_h g\_n^m - g\_n^m\|\_{L^p(a,b-h)} \\ & \qquad + \|g\_n^m - f\_n\|\_{L^p(a,b-h)} \\ & \leq \|\tau\_h g\_n^m - g\_n^m\|\_{L^p(a,b-h)} + 2\|g\_n^m - f\_n\|\_{L^p(a,b)}.\end{align\*}


(I use $(\tau\_h g)(x) := g(x+h)$ because it is more familar to me.)


Fix $\varepsilon > 0$. Choose an $m$ large enough such that $\|g\_n^m - f\_n\|\_{L^p(a,b)} < \varepsilon/3$ for all $n$ (calculation in OP). Now, for the chosen $m$, choose $h\_0$ small enough such that $\|\tau\_h g\_n^m - g\_n^m\|\_{L^p(a,b-h)} < \varepsilon/3$ for all $h \leq h\_0$, uniformly for all $n$; this is possible by the Frechet-Kolmogorov theorem. ($(g\_n^m)\_n$ is clearly bounded in $L^p(a,b)$ if $(f\_n)\_n$ is.)


Then $\|\tau\_h f\_n - f\_n\|\_{L^p(a,b-h)} < \varepsilon$ for all $h \leq h\_0$ uniformly in $n$ and again the Frechet-Kolmogorov theorem says that $(f\_n)\_n$ is relatively compact in $L^p(a,b)$.




---


Due to interest by the OP, here also a less abstract diagonal argument. Iteratively for $m=1,2, \dots$, choose nested subsequences $(f\_{n\_m(i)})\_i$ such that $(f\_{n\_m(i)}\psi\_m(f\_{n\_m(i)}))\_i$ converges (to some $f^m$). Set $h\_j := f\_{n\_j(j)}$. Then split $$h\_k - h\_\ell = (h\_k - h\_k\psi\_m(h\_k)) + (h\_k\psi\_m(h\_k) - h\_\ell\psi\_m(h\_\ell)) + (h\_\ell\psi\_m(h\_\ell) - h\_\ell).$$


For $\varepsilon > 0$ given, the norms of the first and last summands can be made smaller than $\varepsilon/3$ uniformly in $k,\ell$ by choosing $m$ large enough (calculation in OP). For $k,\ell \geq m$, $(h\_k)$ and $(h\_\ell)$ are subsequences of $(f\_{n\_m(i)})\_i$, so $(h\_j \psi\_m(h\_j))\_j$ converges and is a Cauchy sequence. Thus, choosing $k,\ell$ large enough makes the middle summand smaller than $\varepsilon/3$. Hence $(h\_j)\_j$ is also a Cauchy sequence and, by completeness of $L^p(a,b)$, convergent.




---


This should not be a contradiction to the counterexample posted on MSE in the link in OP, since there the assumption was only that $(g\_n^m)\_n$ admits a convergent subsequence for each $m$, and it was shown that then $(f\_n)\_n$ need not admit a convergent subsequence. Here OP asked for $(g\_n^m)\_n$ relatively compact for each $m$ which the counterexample sequence is not, if I see it correctly.


