Banach limit
Consider the set ${c}_{0}$ of all convergent complexvalued sequences ${\{x(n)\}}_{n\in \mathbb{N}}$. The limit operation^{} $x\mapsto {lim}_{n\to \mathrm{\infty}}x(n)$ 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 nonconvergent sequences as well. The formal definition follows:
Let ${\mathrm{\ell}}^{\mathrm{\infty}}$ be the set of bounded^{} complexvalued sequences ${\{x(n)\}}_{n\in \mathbb{N}}$, equipped with the sup norm. Then ${c}_{0}\subset {\mathrm{\ell}}^{\mathrm{\infty}}$, and $lim:{c}_{0}\to \u2102$ is a linear functional. A Banach limit is any continuous^{} linear functional $\varphi \in {({\mathrm{\ell}}^{\mathrm{\infty}})}^{*}$ satisfying:

i
$\varphi (x)={lim}_{n}x(n)$ if $x\in {c}_{0}$ (That is, $\varphi $ extends $lim$.)

ii
$\parallel \varphi \parallel =1$.

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

iv
If $x(n)\ge 0$ for all $n$, then $\varphi (x)\ge 0$. (Positivity)
There is not necessarily a unique Banach limit. Indeed, Banach limits are often constructed by extending $lim$ with the HahnBanach 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 realvalued sequence is realvalued, and that limits can be compared: if $x\le y$, then $\varphi (x)\le \varphi (y)$. In particular, given a realvalued sequence $x$, by comparison with the sequences $y(n)={inf}_{k\ge n}x(k)$ and $z(n)={sup}_{k\ge n}x(k)$, it follows that ${lim\; inf}_{n}x(n)\le \varphi (x)\le {lim\; sup}_{n}x(n)$.
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, $\varphi $ can never be multiplicative, meaning that $\varphi (xy)=\varphi (x)\varphi (y)$ fails. For a counterexample, set $x=(0,1,0,1,\mathrm{\dots})$; then we would have $\varphi (0)=\varphi (x\cdot Sx)=\varphi (x)\varphi (Sx)=\varphi {(x)}^{2}$, so $\varphi (x)=0$, but $1=\varphi (1)=\varphi (x+Sx)=\varphi (x)+\varphi (Sx)=2\varphi (x)=0$.
That $\varphi $ is continuous means it is compatible with limits in ${\mathrm{\ell}}^{\mathrm{\infty}}$. For example, suppose that ${\{{x}_{k}\}}_{k\in \mathbb{N}}\subset {\mathrm{\ell}}^{\mathrm{\infty}}$, and that ${\sum}_{k=0}^{\mathrm{\infty}}{x}_{k}$ is absolutely convergent in ${\mathrm{\ell}}^{\mathrm{\infty}}$. (In other words, $$.) Then $\varphi ({\sum}_{k=0}^{\mathrm{\infty}}{x}_{k})={\sum}_{k=0}^{\mathrm{\infty}}\varphi ({x}_{k})$ 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 $\varphi (1)=1$ — 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 realvalued sequences, condition (ii) is clearly redundant, in view of the other conditions.
Title  Banach limit 

Canonical name  BanachLimit 
Date of creation  20130322 15:23:00 
Last modified on  20130322 15:23:00 
Owner  stevecheng (10074) 
Last modified by  stevecheng (10074) 
Numerical id  7 
Author  stevecheng (10074) 
Entry type  Definition 
Classification  msc 46E30 
Classification  msc 40A05 
Related topic  AlmostConvergent 
Related topic  ConstructionOfBanachLimitUsingLimitAlongAnUltrafilter 
Related topic  ConstructionOfBanachLimitUsingLimitAlongAnUltrafilter2 