Ideal characterization of almost convergence

Question

$\bullet$ A real sequence $x=(x_n)_n$ is called convergent to $\alpha$ in usual sense if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$ is finite.

$\bullet$ A real sequence $x=(x_n)_n$ is called statistically convergent to $\alpha$ if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}$ has natural density $0$. The natural density $d$ of $A\subset\mathbb N$ is defined by $d(A)=\lim\limits_{n\to\infty}\frac{|A\cap\{1,2,\dots,n\}|}{n}$, (provided the limit exists) where $|A|$ denotes the cardinality of $A$.

$\bullet$ A bounded real sequences $x=(x_n)_n$ is said to be almost convergent to $\alpha$ if all the Banach limit functionals give an unique value for the sequence $x$.

A family $\mathcal I$ of subsets of $\mathbb N$ is said to be an ideal in $\mathbb N$ if

(i) $A,B\in \mathcal I$ $\implies $ $A\cup B\in \mathcal I$

(ii) $A\in \mathcal I$ and $B\subset A$ $\implies$ $B\in \mathcal I$

$\bullet$ A real sequence $x=(x_n)_n$ is called $\mathcal I$-convergent to $\alpha$ in usual sense if for any $\epsilon>0$ the set $\{n\in\mathbb N:|x_n-\alpha|\geq\epsilon\}\in \mathcal I$.

$$\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots\dots$$

$\mathcal I_f=\{A\subset\mathbb N: A \text{ is finite}\}$ and $\mathcal I_d=\{A\subset\mathbb N: d(A)=0\}$ become ideals in $\mathbb N$. Moreover, $\mathcal I_f$-convergence and $\mathcal I_d$-convergence coincide with usual convergence and statistical convergence respectively. But, what is the ideal in the case of almost convergence?

My Question : Find the ideal $\mathcal I$ for which $\mathcal I$-convergence coincides with the almost convergence. Is it available in literature?

Answer 1

Such an ideal does not exist.

Indeed, suppose the contrary, and let $I$ be such an ideal. The sequence $(x_n)=(1,0,1,0,\dots)$ is almost convergent, and therefore $I$-convergent, to $1/2$. So, $$\mathbb N=\{n\in\mathbb N\colon|x_n-1/2|\ge1/2\}\in I, $$ and hence $I$ is the powerset of $\mathbb N$. So, every sequence is $I$-convergent, and therefore almost convergent, to every real limit, which is of course absurd.

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This consideration also shows that the Cesàro convergence -- which is implied by the almost-convergence -- is also not the $I$-convergence, for any ideal $I$.

Answer 2

A slightly different argument using the sequence $x=(1,0,1,0,1,0,\dots)$.$\newcommand{\I}{\mathcal I}\newcommand{\Ilim}{\operatorname{\I-lim}}\newcommand{\Flim}{\operatorname{\mathcal F-lim}}\newcommand{\Glim}{\operatorname{\mathcal G-lim}}$

This sequence is almost convergent to $1/2$. At the same time, it is not difficult to show that if this sequence has $\I$-limit of some ideal $\I$, then the $\I$-limit can only be $0$ or $1$.

• We can use the fact that $\I$-limit of a sequence is a cluster point of that sequence. (This holds for any admissible ideal, i.e., for any ideal which contains all finite sets. If we allow also non-admissible ideals, then we can get cluster points or terms of the sequence as limits.)
• For $\I$-convergence we have multiplicativity, i.e., $\Ilim (x_ny_n)=\Ilim x_n\cdot\Ilim y_n$. In particular, for our sequence $x$ we have $x^2=x$. Consequently, if $L$ is an $\I$-limit, then we get $L^2=L$.

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This is basically just a reformulation of Lorenz's criterion for almost convergent sequence, but since you're looking at connection between almost convergent ideals, I'll mention that a sequence is almost convergent to $L$ if and only if $$\Flim_n \Glim_k \frac{x_k+\dots+x_{k+n-1}}n=L$$ for any free ultrafilters $\mathcal F$, $\mathcal G$.

You can check the section on almost convergent sequences of the paper Jerison, Meyer, The set of all generalized limits of bounded sequences, Can. J. Math. 9, 79-89 (1957). ZBL0077.31004, MR83697. I have notes on some part of the paper where the results are expressed using ultrafilters (rather than using nets in Stone-Čech compactification) available here: notes (Wayback Machine), slides (Wayback Machine).