# Does Cesaro convergence along all arithmetic progressions imply convergence on a full density subsequence?

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

• 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. Sep 30 at 12:35
• Could you provide a few more details? I am not sure how to make this counterexample precise. Sep 30 at 12:36

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.