Relation between the sets of almost convergent, bounded statistical convergent and Cesaro convergent sequences

Question

$\bullet$ A sequence $a=(a_n)$ is said to be Cesaro-summable or Cesaro-convergent to $l$ if the sequence $y=(y_n)$ defined by $y_n=\frac{a_1+a_2+a_3+\dots+a_n}{n}$, converges to $l$.

$\bullet$ A sequence $a=(a_n)$ is said to be almost convergent to $l$ if $\forall n\in\mathbb N$ $$\lim\limits_{p\to\infty}\frac{a_{n+1}+a_{n+2}+a_{n+3}+\dots+a_{n+p}}{p}=l$$

$\bullet$ A sequence $a=(a_n)$ is said to be statistically convergent to $l$ if for each $\epsilon>0$ the limit $$\lim\limits_{n\to\infty}\frac{|\{k\in\mathbb N:|a_k-l|\geq\epsilon\}\cap\{1,2,\dots,n\}|}{n}$$ exists and equal to $0$.

Let $A$, $S$ and $C$ be the set of all almost convergent, bounded statistically convergent and Cesaro summable real sequences respectively. Then we have: $A\setminus S\not=\emptyset$, $S\setminus A\not=\emptyset$, $A\subset C$, $S\subset C$.

Here I have a question: Is there any bounded Cesaro-summable sequence which becomes almost non-convergent and statistically non-convergent? That is, I'm trying to find some sequences $x$, such that $x\in C$ but $x\notin A$ and $x\notin S$.

It seems to me that $$y=(~\underbrace{1,0,1,0,\dots}_{100\text{ copies}}~,~ \overbrace{1,1,\dots,1}^{10\text{ copies}}~,~ \underbrace{1,0,1,0,\dots}_{100^2\text{ copies}}~,~ \dots)$$ is my required sequence. I'm sure that $y\notin S$ only. But what is about the rest parts?

If $y$ is not our required sequence, then how can I get such sequence?

Is it available in literature? Any reference will be appreciated.

Answer 1

Unless I missed something, the example that you proposed does work.

$y$ is Cesaro summable to $\frac12$

We want to show that $\lim\limits_{n\to\infty} \frac{y_1+y_2+\dots+y_n}n=\frac12$, which is the same as $$\liminf\limits_{n\to\infty} \frac{y_1+y_2+\dots+y_n}n=\limsup\limits_{n\to\infty} \frac{y_1+y_2+\dots+y_n}n=\frac12.$$ We can see that for every $n$ we have $\frac{y_1+y_2+\dots+y_n}n \ge \frac12$ (since we have always at least so many ones as zeros on positions $1,2,\dots,n$), so it remains to check that the limit superior is at most $\frac12$.

If we look at the structure of the sequence, it consist of "alternating blocks" and "blocks of ones". It suffices to check the ends of "blocks of ones" (or at the beginnings of "alternating blocks"), where the value of the fraction $\frac{y_1+y_2+\dots+y_n}n$ is highest. At the end of the $k$-th such block we get $$\frac{y_1+y_2+\dots+y_n}n = \frac{\frac12(100+100^2+\dots+100^k)+(10+10^2+\dots+10^k)}{(100+100^2+\dots+100^k)+(10+10^2+\dots+10^k)}.$$ It's not difficult to check that this limit is $\frac12$, so we get $$\limsup_{n\to\infty}\frac{y_1+y_2+\dots+y_n}n \le \frac12$$ and we are done.

Another possibility would be to check that the set of indices where $y$ differs from the sequence $y'=(1,0,1,0,1,0,\ldots)$ has density zero. For bounded sequences, changing on such small set cannot influence Cesaro summability. (Although this argument is not really that different from what I wrote above.)

As soon as we know about a bounded sequence that it has Cesaro mean equal to $\frac12$, then

• if this sequence is almost convergent to some $L$, then $L=\frac12$;
• if this sequence is statistically convergent to some $L$, them $L=\frac12$.

$y$ is not statistically convergent

The only limit points of the sequence $y$ are $0$ and $1$, so they are the only possible candidates for the statistical limit. But at the same time, we have shown that the sequence is convergent to $\frac12$ in Cesaro sense, so it could only be statistically convergent to $\frac12$.

$y$ is not almost convergent

We know that a sequence $x$ is almost convergent to $L$ if and only if $$\lim_{k\to\infty} \frac{x_{n+1}+x_{n+2}+\dots+x_{n+k}}k = L$$ uniformly in $k$. (This is a result by Lorenz, it was mentioned in another of your posts.)

Just slightly reformulated, this condition be equivalently written as $$\lim_{k\to\infty} \sup_{n\in\mathbb N} \frac{x_{n+1}+x_{n+2}+\dots+x_{n+k}}k = \lim_{k\to\infty} \inf_{n\in\mathbb N} \frac{x_{n+1}+x_{n+2}+\dots+x_{n+k}}k = L.$$

For the sequence in question we have $$\frac{y_{n+1}+y_{n+2}+\dots+y_{n+k}}k \ge \frac12-\frac1k$$ for every $n$ and $k$, so we immediately see that $$\frac12 \le \lim_{k\to\infty} \sup_{n\in\mathbb N} \frac{y_{n+1}+y_{n+2}+\dots+y_{n+k}}k \le \lim_{k\to\infty} \inf_{n\in\mathbb N} \frac{y_{n+1}+y_{n+2}+\dots+y_{n+k}}k \le 1.$$ However, we get \begin{align*} \lim_{k\to\infty} \sup_{n\in\mathbb N} \frac{y_{n+1}+y_{n+2}+\dots+y_{n+k}}k &= 1\\ \lim_{k\to\infty} \inf_{n\in\mathbb N} \frac{y_{n+1}+y_{n+2}+\dots+y_{n+k}}k &= \frac12 \end{align*} Since there are arbitrarily long segments consisting of ones (where this fraction is equal to one) and arbitrarily long segments which alternate between zero and one (where this fraction is close to $1/2$).