Suppose $\{x_n\}_{n \geq 1}$ is a real valued sequence such that for every $a, r \in \mathbb Z_+$, we have that

$$\lim_{N \to \infty} \frac{1}{N} \sum_{i = 0}^{N-1} x_{a + ir}$$ exists and equals $L$ for some $L \in \mathbb R$.

Question: Is it true that $x_n \to L$ along a subsequence of natural density $1$? That is, does there exist a subsequence $n_k$ with

$$\lim_{M \to \infty} \frac{|\{k \in \mathbb N \, | \, n_k \leq M\}|}{M} = 1$$

such that $x_{n_k} \to L$?

  • $\begingroup$ No. Just alternate $x_n$ between $+1$ and $-1$ with $+1$'s (and then $-1$'s) occurring in a batch of prime length repeatedly, where the prime increases slowly as you go on. $\endgroup$ Sep 30 at 12:35
  • $\begingroup$ Could you provide a few more details? I am not sure how to make this counterexample precise. $\endgroup$
    – Nate River
    Sep 30 at 12:36

1 Answer 1


No, just take $x_n = e^{2\pi i \alpha}$ for your favorite irrational $\alpha$. Then for any $a,r$, $x_{an+r} = e^{2\pi i r\alpha} (e^{2\pi i a\alpha)})^n$, and so $\sum_{n=0}^{N-1} x_{an+r} = e^{2\pi i r\alpha} \frac{1 - (e^{2\pi i a\alpha})^N}{1 - e^{2\pi i a\alpha}}$, with modulus bounded from above by a constant. So all of your Cesaro averages converge to 0, but the sequence itself always has modulus 1 and so can't converge to 0 along any subsequence.

In fact, more generally, for any totally ergodic system $(X, T, \mu)$ and any $f \in L^2$, almost every point $x$ will induce a sequence $x_n = f(T^n x)$ with all Cesaro averages along infinite arithmetic progressions converging to $\int f d\mu$, but very rarely would this sequence itself converge.

  • $\begingroup$ Does this work with a real valued counterexample? $\endgroup$
    – Nate River
    Sep 30 at 12:38
  • 1
    $\begingroup$ Hm, it suffices to take the real part of the sequence. $\endgroup$
    – Nate River
    Sep 30 at 12:41

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