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Consider the class of bounded sequences to which every Banach limit (non-negative shift-invariant continuous functional on $l^\infty$ taking convergent sequences in the usual sense to their limits) assigns the same limit value.

Does a sequence belong to this class if its Cesaro means have a limit?

Also, is the converse true?

edited Apr 8 2012 at 20:40

Yemon Choi
8,456●1●19●44

asked Apr 8 2012 at 19:58

kap44
177●5

I like the question, but I would like to make sure I understand correctly the notion of Banach limit. From your formulation I understand that a Banach limit is a linear map $L:\ell^infty(\mathbb{R})\to\mathbb{R}$ with the property that $L(\;(x_n)_{n\geq 0}\;)=L(\;(y_n)_{n\geq 0}\;) $ if $x_n=y_n$ for all $n$ sufficiently large. Is the functional $L$ continuous in the $\ell^\infty$ topology? – Liviu Nicolaescu Apr 8 2012 at 20:29

If $L$ is a Banach limit, then $|Lx|\le\|x\|_\infty$ for any sequence $x$. Does this answer your question? – kap44 Apr 8 2012 at 20:34

Yes, thanks. $L$ is therefore an element of the topological dual of $\ell^\infty$. – Liviu Nicolaescu Apr 8 2012 at 21:21

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Sequences with this property are called [almost convergent](en.wikipedia.org/wiki/Almost_convergent_sequence) and there exists a well-known characterization of such sequences due to Lorenz; which is described in the Wikipedia article. One possibility how to show this is using Hahn-Banach theorem, see [this answer](math.stackexchange.com/a/80571/8297) at Math.SE. Also the references from the Wikipedia article might be useful. – Martin Sleziak Apr 9 2012 at 13:47

Aaron Tikuisis · Accepted Answer · 2012-04-09 10:38:05Z UTC

We can characterize Banach limits as continuous functionals on $\ell^\infty$ which vanish on $$ X := \{(x_n - x_{n+1}): (x_n) \in \ell^\infty\} $$ and which send the constant sequence $(1,1,\dots)$ to $1$.

Note that $X$ is a subspace. The Hahn-Banach Theorem tells us that we are asking: if $(y_n) \in \ell^\infty$ has Cesaro mean $0$, is it in the closure of $X$? (And the converse question is: does every element of $X$ have Cesaro mean $0$? Yes; since the $n^\text{th}$ Cesaro mean of $(x_n-x_{n+1})$ is $(x_1-x_{n+1})/n$, which converges to $0$ since $(x_n)$ is uniformly bounded.)

The answer is no. Consider the sequence $(y_n)$ that has $1$ once, followed by $-1$ three times, then $1$ five times, and so on. One can compute the Cesaro mean, and see that it approaches $0$ in the limit. But $(y_n)$ is not in the closure of $X$.

Surely, if it were, then let $(x_n) \in \ell^\infty$ be such that $$ \|(y_n) - (x_n-x_{n+1})\|_\infty < 1/2. $$ Let $M$ be a natural number, $M \geq \|(x_n)\|$. Let $n$ be an index such that $$ y_n = \cdots = y_{n+4M} = 1. $$ Then for $i=1,\dots,4M$, $$ x_{n+i} < x_{n + i-1} - y_{n + i - 1} + 1/2 = x_{n + i - 1} - 1/2, $$ and summing these up, we find $$ x_{n+4M} < x_n - 4M/2. $$ This contradicts the assumption that $\|(x_n)\| \leq M$.

Daniel Mansfield · Answer 2 · 2013-01-12 00:57:40Z UTC

If $(x_n) \in \ell^\infty$. According to Lorenz the Banch limit is unique (also known as almost convergent) iff $$\lim_{p\mapsto\infty} \frac{ x_n + x_{n+1} + \cdots + x_{n+p}}{p} = L \quad (*) $$ uniformly in $n$. Setting $n=0$ yields Cesaro summability.

As Aaron says, the converse is false. If each $x_n$ is chosen uniformly at random from $\{0,1\}$ then this sequence almost never has property $(*)$ (see Connor's appropriately named article Almost none of the sequences of 0's and 1's are almost convergent)

However the Cesaro limit of this random sequence $(x_n)$ is almost always $1/2$ by the law of large numbers.

Cesaro means and Banach limits

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