4
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Have you tried the Kolmogorov-Riesz theorem? math.ntnu.no/conservation/2009/037.pdf $\endgroup$
    – Oliver Díaz
    Jul 31, 2020 at 18:52
  • $\begingroup$ @OliverDiaz How does that theorem help? $\endgroup$
    – Zac
    Jul 31, 2020 at 19:28
  • $\begingroup$ 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. $\endgroup$
    – Arctic Char
    Jul 31, 2020 at 22:17
  • $\begingroup$ @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? $\endgroup$
    – Zac
    Jul 31, 2020 at 22:25
  • $\begingroup$ @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 $n$). $\endgroup$
    – Zac
    Jul 31, 2020 at 22:27

1 Answer 1

Reset to default
1
$\begingroup$

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

$\endgroup$
17
  • $\begingroup$ Can you clarify what you mean? What is the counterexample? $\endgroup$
    – Zac
    Jul 31, 2020 at 21:20
  • $\begingroup$ 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? $\endgroup$
    – Zac
    Jul 31, 2020 at 21:24
  • $\begingroup$ 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. $\endgroup$
    – Zac
    Jul 31, 2020 at 21:43
  • $\begingroup$ @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 $\{f_n\}$. $\endgroup$
    – Arctic Char
    Jul 31, 2020 at 21:44
  • $\begingroup$ 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. $\endgroup$
    – Arctic Char
    Jul 31, 2020 at 21:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .