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# Banach limit

Consider the set $c_{0}$ of all convergent complex-valued sequences $\{x(n)\}_{{n\in\mathbb{N}}}$.
The limit operation $x\mapsto\lim_{{n\to\infty}}x(n)$
is a linear functional on $c_{0}$, by the usual limit laws.
A *Banach limit* is, loosely speaking, any linear functional that generalizes $\lim$ to apply to non-convergent sequences as well. The formal definition follows:

Let $\ell^{\infty}$ be the set of bounded complex-valued sequences $\{x(n)\}_{{n\in\mathbb{N}}}$, equipped
with the sup norm.
Then $c_{0}\subset\ell^{\infty}$, and $\lim\colon c_{0}\to\mathbb{C}$ is a linear functional.
A *Banach limit* is any continuous linear functional $\phi\in(\ell^{\infty})^{*}$ satisfying:

i. $\phi(x)=\lim_{n}x(n)$ if $x\in c_{0}$ (That is, $\phi$ extends $\lim$.)

ii. $\lVert\phi\rVert=1$.

iii. $\phi(Sx)=\phi(x)$, where $S\colon\ell^{\infty}\to\ell^{\infty}$ is the shift operator defined by $Sx(n)=x(n+1)$. (Shift invariance)

iv. If $x(n)\geq 0$ for all $n$, then $\phi(x)\geq 0$. (Positivity)

There is not necessarily a unique Banach limit. Indeed, Banach limits are often constructed by extending $\lim$ with the Hahn-Banach theorem (which in turn invokes the Axiom of Choice).

Like the limit superior and limit inferior, the Banach limit can be applied for situations where one wants to algebraically manipulate limit equations or inequalities, even when it is not assured beforehand that the limits in question exist (in the classical sense).

# 1 Some consequences of the definition

The positivity condition ensures that the Banach limit of a real-valued sequence is real-valued, and that limits can be compared: if $x\leq y$, then $\phi(x)\leq\phi(y)$. In particular, given a real-valued sequence $x$, by comparison with the sequences $y(n)=\inf_{{k\geq n}}x(k)$ and $z(n)=\sup_{{k\geq n}}x(k)$, it follows that $\liminf_{n}x(n)\leq\phi(x)\leq\limsup_{n}x(n)$.

The shift invariance allows any finite number of terms of the sequence to be neglected when taking the Banach limit, as is possible with the classical limit.

On the other hand, $\phi$ can never be multiplicative, meaning that $\phi(xy)=\phi(x)\phi(y)$ fails. For a counter-example, set $x=(0,1,0,1,\ldots)$; then we would have $\phi(0)=\phi(x\cdot Sx)=\phi(x)\phi(Sx)=\phi(x)^{2}$, so $\phi(x)=0$, but $1=\phi(1)=\phi(x+Sx)=\phi(x)+\phi(Sx)=2\phi(x)=0$.

That $\phi$ is continuous means it is compatible with limits in $\ell^{\infty}$. For example, suppose that $\{x_{k}\}_{{k\in\mathbb{N}}}\subset\ell^{\infty}$, and that $\sum_{{k=0}}^{\infty}x_{k}$ is absolutely convergent in $\ell^{\infty}$. (In other words, $\sum_{{k=0}}^{\infty}\lVert x_{k}\rVert_{\infty}<\infty$.) Then $\phi(\sum_{{k=0}}^{\infty}x_{k})=\sum_{{k=0}}^{\infty}\phi(x_{k})$ by continuity. Observe that this is just the dominated convergence theorem, specialized to the case of the counting measure on $\mathbb{N}$, in disguise.

# 2 Other definitions

In some definitions of the Banach limit, condition (i) is replaced by the seemingly weaker condition that $\phi(1)=1$ — the Banach limit of a constant sequence is that constant. In fact, the latter condition together with shift invarance implies condition (i).

If we restrict to real-valued sequences, condition (ii) is clearly redundant, in view of the other conditions.

## Mathematics Subject Classification

46E30*no label found*40A05

*no label found*

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## Comments

## Banach limit and AC

I have to questions about Banach limits. Maybe someone of you knows something about it or could provide at least a pointer, where I could find information I'm interested in.

1. I'm aware of two possible proofs of existence of Banach limits. One of them employs ultrafilters, the other one uses Hahn-Banach theorem. How much Choice is really needed for existence of Ban. limit? (Maybe this example shows better what I want to know: It's known that equivalence of Heine's and Cauchy's definition of continuity implies Countable Choice. Is some similar result known for Banach limits?)

2. Was analogous concept defined also for some more general setting, e.g. in Banach spaces?

TIA

Martin

## Re: Banach limit and AC

For question #2:

Yes, although I am not too familiar with it.

There is a notion of a "generalized limit"

in Yoshida's Functional Analysis book, which uses

nets rather than l^\infty sequences.

So I would think that the procedure

employed, for example, to extend the

Riemann integral to the Lebesgue integral.

(note Riemann integral is a limit I

st. L <= I <= U for the supremum L of

the lower sums and the infimum U of the upper sums.

The situation seems to be analogous to the liminf

and limsup of a sequence.)

If you (or anyone else) would like to elaborate

on the current article, I can give you write access.

// Steve

## Re: Banach limit and AC

I would say that nets is another type of generalized limit. What I mean is this: nets are like sequences, but instead of N they are indexed by an upwards directed set. It means, we changed the "domain" of the sequence in this generalization - the notion of sequence is replaced with the notion of net in this generalization of limit. Nets can be defined for any topological space.

In the case of Banach limit, the notion of sequence isn't changed, but this operator assigns the value also to some non-convergent sequences.

> If you (or anyone else) would like to elaborate

> on the current article, I can give you write access.

There already is entry on nets in topological spaces - http://planetmath.org/encyclopedia/Subnet.html. (Although it doesn't mention the example with the Riemann integral.)

Martin