# $L^p$ compactness for a sequence of functions from compactness of 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 $$L^p([a,b])$$. 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]

• "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. Aug 2, 2020 at 8:53
• @PietroMajer Thanks, I've edited the question.
– Zac
Aug 2, 2020 at 9:03
• 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. Aug 2, 2020 at 14:37
• @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
– Zac
Aug 2, 2020 at 15:38
• Sounds good to me. (I didn't realise you weren't the author.) Aug 2, 2020 at 16:02

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.

• @Zac I added a diagonal argument. Aug 7, 2020 at 11:31
• 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?
• 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 $m \to \infty$.) Aug 7, 2020 at 12:04