\phi:\ellinfty\toC
\ellinfty
x=(xn)
y=(yn)
\ellinfty
\alpha
\phi(\alphax+y)=\alpha\phi(x)+\phi(y)
xn\geq0
n\inN
\phi(x)\geq0
\phi(x)=\phi(Sx)
S
(Sx)n=xn+1
x
\phi(x)=\limx
\phi
\lim:c\toC
c\subset\ellinfty
C
In other words, a Banach limit extends the usual limits, is linear, shift-invariant and positive. However, there exist sequences for which the values of two Banach limits do not agree. We say that the Banach limit is not uniquely determined in this case.
As a consequence of the above properties, a real-valued Banach limit also satisfies:
\liminfn\toinftyxn\le\phi(x)\le\limsupn\toinftyxn.
The existence of Banach limits is usually proved using the Hahn–Banach theorem (analyst's approach),[1] or using ultrafilters (this approach is more frequent in set-theoretical expositions).[2] These proofs necessarily use the axiom of choice (so called non-effective proof).
There are non-convergent sequences which have a uniquely determined Banach limit. For example, if
x=(1,0,1,0,\ldots)
x+S(x)=(1,1,1,\ldots)
2\phi(x)=\phi(x)+\phi(x)=\phi(x)+\phi(Sx)=\phi(x+Sx)=\phi((1,1,1,\ldots))=\lim((1,1,1,\ldots))=1
1/2
A bounded sequence
x
\phi
\phi(x)
Given a convergent sequence
x=(xn)
c\subset\ellinfty
x
\ell1
\langle\ell1,\ellinfty\rangle
\ellinfty
\ell1
\ell1
\ellinfty
\ellinfty
\ellinfty
\ell1
\ellinfty
. John B. Conway . A Course in Functional Analysis . Springer . New York . 1994 . 0-387-97245-5 . . 96 .