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]

  • $\begingroup$ "how can we prove": we can't because it is not true. In your hypotheses , (f_n) can be unbounded in any L^p for p>1. $\endgroup$ Aug 2, 2020 at 8:53
  • $\begingroup$ @PietroMajer Thanks, I've edited the question. $\endgroup$
    – Zac
    Aug 2, 2020 at 9:03
  • $\begingroup$ I think one shouldn't rely on content remaining indefinitely in the formatting sandbox. (Remember MO and other SE sites are meant for long-term archiving.) Probably better to move it here. $\endgroup$
    – LSpice
    Aug 2, 2020 at 14:37
  • 1
    $\begingroup$ @LSpice I put the link just to give credit to the original writer of this proof, but I rewrote it in the body of the question to preserve it in case of changes to the Sandbox link $\endgroup$
    – Zac
    Aug 2, 2020 at 15:38
  • $\begingroup$ Sounds good to me. (I didn't realise you weren't the author.) $\endgroup$
    – LSpice
    Aug 2, 2020 at 16:02

1 Answer 1


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 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.

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    $\begingroup$ @Zac I added a diagonal argument. $\endgroup$
    – Hannes
    Aug 7, 2020 at 11:31
  • $\begingroup$ Thank you very much for your addition! $\endgroup$
    – Zac
    Aug 7, 2020 at 11:31
  • $\begingroup$ I have just one more curiosity: would all of this still work if we defined $$\psi_m(f) = \begin{cases} 1 \qquad \text{ if } |f - 1|\ge 1/m \\ 0 \qquad \text{ if } |f - 1|\le 1/(2m) \end{cases} $$ instead of as in the original question? $\endgroup$
    – Zac
    Aug 7, 2020 at 11:32
  • $\begingroup$ Uh, you roughly would need to make sure that the $(f_n)$ stay away from $1$ in a uniform manner, I guess. (In the sense that the measure of the set where $|f_n - 1| \leq 1/m$ goes to zero uniformly in $n$ as $m \to \infty$.) $\endgroup$
    – Hannes
    Aug 7, 2020 at 12:04
  • $\begingroup$ It seems fair. Why wasn't this needed in the previous case? $\endgroup$
    – Zac
    Aug 7, 2020 at 12:33

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