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?



Cesaro means and Banach limits
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 $\|(x\_n)\| \leq M$.


