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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 the Cesàro sense. To illustrate, we consider the following two examples. The first of these was proposed by Yuval Peres in discussion on a separate 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 the Cesàro 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 the Cesàro 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?

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  • $\begingroup$ 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" $\endgroup$ Apr 2, 2021 at 14:27
  • $\begingroup$ Right, sorry I forgot to include that. Will edit the post, thanks! $\endgroup$
    – Nate River
    Apr 2, 2021 at 14:31
  • $\begingroup$ I would state the complete assumptions, in the question as well $\endgroup$ Apr 2, 2021 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, 2021 at 14:40
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    $\begingroup$ I like the first one. Edited. $\endgroup$
    – Nate River
    Apr 5, 2021 at 13:08

1 Answer 1

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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

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  • $\begingroup$ I see, so continuity is not even necessary for the statement to hold. That’s a really cool result.. $\endgroup$
    – Nate River
    Apr 2, 2021 at 15:30
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    $\begingroup$ @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, 2021 at 9:39
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    $\begingroup$ It seems that your link goes only to [1], so I edited in links to both articles. $\endgroup$
    – LSpice
    Apr 5, 2021 at 23:58
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    $\begingroup$ @LSpice Thanks! $\endgroup$ Apr 6, 2021 at 3:14

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