# $L^p$ 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.

• Have you tried the Kolmogorov-Riesz theorem? math.ntnu.no/conservation/2009/037.pdf Jul 31, 2020 at 18:52
• @OliverDiaz How does that theorem help?
– Zac
Jul 31, 2020 at 19:28
• The argument using diagonal sequence does not work: By assumption, $\{ f_{n} \psi_1 (f_{n})\}$ has a convergent subsequence $\{ f_{n_k} \psi_1( f_{n_k})\}$. But starting at $m=2$, we already do not know if $\{ f_{n_k} \psi_2( f_{n_k})\}$ has a convergent subsequence. Jul 31, 2020 at 22:17
• @ArcticChar Thanks! But I think that if this is the issue we can solve it by a slightly stronger assumption: that the sequence is compact in $L^1$ because it satisfies the assumption of Helly's theorem: en.wikipedia.org/wiki/Helly%27s_selection_theorem What do you think?
– Zac
Jul 31, 2020 at 22:25
• @ArcticChar In this way you know that also the subsequence $\{f_{n_k}\psi_m(f_{n_k})\}$ has a converging subsequence because it satisfies the same assumption. Also, I'm pretty sure that with this assumption the counterexample below does not hold (because it seems that the cut-off sequence does not have derivatives uniformly bounded with respect to $n$).
– Zac
Jul 31, 2020 at 22:27

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

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

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 $$k_1,k_2$$

3. 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 $$g_{m+1} \psi_m(g_{m+1})$$

• Can you clarify what you mean? What is the counterexample?
– Zac
Jul 31, 2020 at 21:20
• Thanks for the edit. How do you prove that $\{f_n\}_n$ is not compact in $L^2[a,b]$ but, for every fixed $m$, the family $\{f_{n}\psi_m(f_n)\}_{n \in \mathbb N}$ admits a convergent subsequence?
– Zac
Jul 31, 2020 at 21:24
• In particular I think that in your example even $\{f_n \psi_m(f_n)\}$ does not have a subsequence that converges strongly in L^2.
– Zac
Jul 31, 2020 at 21:43
• @zac $\{f_{n}\psi_m(f_n)\}$ admits a convergent subsequence, roughly because one can choose $\sigma_1(n)>m$ fixed, that will kill the second terms. But the same cannot be said for $\{f_n\}$. Jul 31, 2020 at 21:44
• Nice example indeed. I guess the mistake in my answer is when constructing the diagonal sequence. After taking the first subsequence at $m=1$, I cannot use the same subsequence to take further subsequence. Jul 31, 2020 at 21:47