Follow-up question: Cesaro mean of cesaro mean of ...

Question

In this post, Cesaro mean of Cesaro means, benny asked about a bounded sequence for which the Cesàro mean diverges, but the Cesàro mean of the Cesàro mean converges (which still isn't answered completely). To revive this topic, I'd like to ask a related but more general question:

Given a bounded sequence of real numbers $(x_{n})_{n\in\mathbb{N}}:=x$, does there always exist a positive integer $n$ such that $C^{n}(x)$ converges?

Where $C^{n}$ denotes the $n$-th iteration of the Cesàro mean $C$, which is defined to be $$C(y)= \Bigg(\sum_{k=1}^{n} \frac{y_{k}}{n}\Bigg)_{n\in\mathbb{N}} $$ for any real number sequence $y:=(y_{n})_{n\in\mathbb{N}}$.

Answer 1

No.

Lemma 1. (i) For every $N$ and every $\epsilon>0$ there exist $M>N$ and $\delta>0$ such that if $M'\ge M$, $0\le a_n\le 1$ for $1\le n\le N$ and $a_n\ge 1-\delta$ for $N<n\le M'$ then $C(a)_{M'}>1-\epsilon$. (ii) For every $N$ and every $\epsilon>0$ there exist $M>N$ and $\delta>0$ such that if $M'\ge M$, $0\le a_n\le 1$ for $1\le n\le N$ and $a_n\le\delta$ for $N<n\le M'$ then $C(a)_{M'}<\epsilon$.

By induction on $k$ this implies

Lemma 2. (i) For every $N$, every $k$ and every $\epsilon>0$ there exists $M>N$ such that if $M'\ge M$, $0\le a_n\le 1$ for $1\le n\le N$ and $a_n=1$ for $N<n\le M'$ then $C^j(a)_{M'}>1-\epsilon$ for $1\le j\le k$. (ii) For every $N$, every $k$ and every $\epsilon>0$ there exists $M>N$ such that if $M'\ge M$, $0\le a_n\le 1$ for $1\le n\le N$ and $a_n=0$ for $N<n\le M'$ then $C^j(a)_{M'}<\epsilon$ for $1\le j\le k$.

And now you can construct a counterexample. Start with $a_1=1$. Now we have a point where $C(a)_j=1$. If you add a long enough sequence of $0$'s you reach a point where $C(a)_j<1/3$ and $C^2(a)_j<1/3$. Now add an even longer sequence of $1$'s and you reach a point where $C(a)_j>2/3$, $C^2(a)_j>2/3$ and $C^3(a)_j>2/3$.

Etc.