PlanetMath (more info)
 Math for the people, by the people.
Encyclopedia | Requests | Forums | Docs | Wiki | Random | RSS  
Login
create new user
name:
pass:
forget your password?
Main Menu
sections
Encyclopædia
Papers
Books
Expositions

meta
Requests (200)
Orphanage
Unclass'd
Unproven (490)
Corrections (6)
Classification

talkback
Polls
Forums
Feedback
Bug Reports

downloads
Snapshots
PM Book

information
News
Docs
Wiki
ChangeLog
TODO List
Legalese
About
Owner confidence rating: High Entry average rating: No information on entry rating
Banach limit (Definition)

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:

  1. $ \phi(x) = \lim_n x(n)$ if $ x \in c_0$ (That is, $ \phi$ extends $ \lim$.)
  2. $ \lVert\phi\rVert = 1$.
  3. $ \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)
  4. 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).

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, \dotsc)$; 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.

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.



"Banach limit" is owned by stevecheng.
(view preamble)

View style:

See Also: almost convergent, construction of Banach limit using limit along an ultrafilter
Log in to rate this entry.
(view current ratings)

Cross-references: implies, definitions, counting measure, dominated convergence theorem, absolutely convergent, compatible, multiplicative, terms, number, finite, even, inequalities, equations, limit inferior, limit superior, axiom of choice, Hahn-Banach theorem, operator, continuous, sup norm, bounded, linear functional, operation, limit, sequences, convergent
There are 2 references to this entry.

This is version 4 of Banach limit, born on 2005-07-08, modified 2005-07-09.
Object id is 7213, canonical name is BanachLimit.
Accessed 2844 times total.

Classification:
AMS MSC: 46E30 (Functional analysis :: Linear function spaces and their duals :: Spaces of measurable functions ($L^p$-spaces, Orlicz spaces, Köthe function spaces, Lorentz spaces, rearrangement invariant)
 40A05 (Sequences, series, summability :: Convergence and divergence of infinite limiting processes :: Convergence and divergence of series and sequences)
Pending Errata and Addenda
None.
Discussion
Style: Expand: Order:
forum policy
Banach limit and AC by kompik on 2005-10-07 03:24:31
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
[ reply | up ]
Interact
post | correct | update request | add derivation | add example | add (any)