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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?

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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 '12 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 '12 at 20:34
    
Yes, thanks. $L$ is therefore an element of the topological dual of $\ell^\infty$. –  Liviu Nicolaescu Apr 8 '12 at 21:21
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Sequences with this property are called almost convergent 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 at Math.SE. Also the references from the Wikipedia article might be useful. –  Martin Sleziak Apr 9 '12 at 13:47

2 Answers 2

up vote 11 down vote accepted

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 $\|(x_n)\| \leq M$.

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Is there a typo in the second equation from the bottom? In the middle expression you write $x+y$, where must be $x-y$. (I believe the idea does work anyway!) Also, it is not quite obvious why every element of $X$ has Cesaro mean 0. –  kap44 Apr 9 '12 at 7:03
    
Thank you, I think that it is all correct now. I have added a bit of explanation about why the elements of $X$ have Cesaro mean zero. –  Aaron Tikuisis Apr 9 '12 at 10:40
    
Many thanks to you! –  kap44 Apr 9 '12 at 10:45
    
Nice answer! By the way? Is there a constructive example of Banach limit functional? –  Liviu Nicolaescu Apr 9 '12 at 13:43
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@Liviu: I believe there isn't. See mathoverflow.net/questions/5351/whats-an-example-of-a-space-that-needs-the-hahn-‌​banach-theorem/5366 –  Mark Meckes Apr 9 '12 at 16:23

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.

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