Banach limit

Consider the set $c_0$ of all convergent complex-valued sequences ${\{x(n)\}}_{n\in \mathbb{N}}$.The limit operation $x\mapsto {lim}_{n\to \mathrm{\infty }}x(n)$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 non-convergent sequences as well. The formal definition follows:

Let ${\mathrm{\ell }}^{\mathrm{\infty }}$ be the set of bounded complex-valued sequences ${\{x(n)\}}_{n\in \mathbb{N}}$, equippedwith the sup norm.Then $c_0\subset {\mathrm{\ell }}^{\mathrm{\infty }}$, and $lim:c_0\to ℂ$ is a linear functional.A Banach limit is any continuous linear functional $\varphi \in {({\mathrm{\ell }}^{\mathrm{\infty }})}^{*}$ satisfying:

1. $\varphi (x)={lim}_{n}x(n)$ if $x\in c_0$ (That is, $\varphi$ extends $lim$.)

2. $\parallel \varphi \parallel =1$.

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

4. 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 constructedby 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 situationswhere one wants to algebraically manipulate limit equations or inequalities,even when it is not assured beforehandthat 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 $x\le y$, then $\varphi (x)\le \varphi (y)$.In particular, given a real-valued 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 counter-example, 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, specializedto the case of the counting measure on $\mathbb{N}$, in disguise.

2 Other definitions

In some definitions of the Banach limit, condition (i) is replacedby 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 real-valued sequences,condition (ii) is clearly redundant, in view of the other conditions.