9
$\begingroup$

Notation: We say a sequence of real numbers diverges if it does not converge to a finite limit. We say a sequence $f_n$ of real valued functions on $[0, 1] $ are equibounded if $\sup_{n \in \mathbb N}\sup_{x \in [0, 1]} |f_n (x)| < \infty$.

Some motivation:

The Arzela Ascoli theorem for $C[0, 1]$ says that if we have an equibounded, equicontinuous sequence of functions, we have uniform convergence along a subsequence. What happens if we drop equicontinuity (but retain continuity of the functions)? What kind of convergence can we expect?

For any countable subset $S$ of $[0, 1]$, we can still diagonalize to get pointwise convergence on $S$ along a subsequence $f_{n_k}$ that depends on $S$. But conjecturally this is the best we can do in general. Indeed as far as pointwise convergence is concerned, we have:

Theorem 1: There exists an equibounded sequence of continuous functions $f_n: [0, 1] \to \mathbb R$ such that for every increasing sequence $n_k$ of naturals, $f_{n_k} (x)$ diverges for almost every $x \in [0, 1]$.

However, the examples that I am familiar with all rely on some sort of “independence” or equidistribution type argument. For such examples, one intuitively expects $f_n$ to converge in Cesaro sense. To illustrate, we consider the following two examples. The first of these was proposed by Yuval Peres in discussion on a seperate forum.

Example 1: Take $f_n (x) = \sin(nx)$.That $f_n$ satisfy the conditions in Theorem 1 can be seen by noting that by Weyl’s criterion for equidistribution, the sequence $n_{k}x \ \text{mod} \ 1$ is equidistributed for a.e. $x \in [0, 1]$. However by the same coin, we have that for any subsequence, $f_{n_k} (x)$ converges in Cesaro sense for almost every $x$.

Example 2: Consider the domain $[0, 1]$ as a probability space, and take $g_n$ to be the indicator function of independent events with probability $1/2$ each. Then an argument based on the second Borel Cantelli lemma gives us that the $g_n$ satisfy all conditions in Theorem 1 except continuity. We can then approximate the $g_n$ by a sequence $f_n$ of continuous functions, whence $f_n$ satisfy the conditions in Theorem 1. But again it can be shown that for any subsequence, $f_{n_k} (x)$ converges in Cesaro sense almost everywhere.

This suggests the following question:

Question: Does there exist an equibounded sequence of continuous functions $f_n: [0, 1] \to \mathbb R$ such that for every increasing sequence $n_k$ of naturals, $\lim_{N \to \infty} \frac{1}{N} \sum_{k = 0}^{N-1} f_{n_k}(x)$ almost everywhere fails to exist?

$\endgroup$
8
  • $\begingroup$ In theorem 1 you mean "a bounded sequence" don't you? Otherwise you may take a sequence of constant functions $f_n=n$, so it's even true with "every x" $\endgroup$ – Pietro Majer Apr 2 at 14:27
  • $\begingroup$ Right, sorry I forgot to include that. Will edit the post, thanks! $\endgroup$ – Nate River Apr 2 at 14:31
  • $\begingroup$ I would state the complete assumptions, in the question as well $\endgroup$ – Pietro Majer Apr 2 at 14:39
  • $\begingroup$ Yep, I’ve added the equibounded condition to the question statement. Is there anything else I should add? $\endgroup$ – Nate River Apr 2 at 14:40
  • 1
    $\begingroup$ I like the first one. Edited. $\endgroup$ – Nate River Apr 5 at 13:08
18
$\begingroup$

Under the stated conditions, there always exists a subsequence that Cesaro converges almost everywhere. This was a question of Steinhaus, solved by Revesz [1]. More generally, it suffices that the sequence $f_n$ be uniformly bounded in $L^1$; This is a striking Theorem of Komlos [2] which in particular implies the Kolmogorov strong law of large numbers.

[1] P. Revesz, On a problem of Steinhaus, Acta Mathematica Academiae Scientiarum Hungaricae, 16 (1965), pp. 310–318.

[2] Janos Komlos, A generalization of a problem of Steinhaus, Acta Mathematica Academiae Scientiarum Hungaricae, 18 (1967), pp. 217–229.

https://scholarship.libraries.rutgers.edu/discovery/delivery?vid=01RUT_INST:ResearchRepository&repId=12643427010004646#13643523800004646

$\endgroup$
4
  • $\begingroup$ I see, so continuity is not even necessary for the statement to hold. That’s a really cool result.. $\endgroup$ – Nate River Apr 2 at 15:30
  • 2
    $\begingroup$ @NateRiver If you think about it, for most notions of convergence weaker than uniform convergence, there is a theorem of the form "Set of functions X can be approximated by continuous functions". So it is very likely that this would allow to extend any convergence result that holds for continuous functions to functions in X via some sort of diagonal argument. $\endgroup$ – mlk Apr 3 at 9:39
  • $\begingroup$ It seems that your link goes only to [1], so I edited in links to both articles. $\endgroup$ – LSpice Apr 5 at 23:58
  • $\begingroup$ @LSpice Thanks! $\endgroup$ – Yuval Peres Apr 6 at 3:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.