Added to Favorites

Related Searches

Nearby Words

In functional analysis and related branches of mathematics, the Banach–Alaoglu theorem (also known as Alaoglu's theorem) states that the closed unit ball of the dual space of a normed vector space is compact in the weak* topology. A common proof identifies the unit ball with the weak* topology as a closed subset of a product of compact sets with the product topology. As a consequence of Tychonoff's theorem, this product, and hence the unit ball within, is compact.## Generalization: Bourbaki–Alaoglu theorem

## Proof

_{1}(X*) having the weak-* topology and D the product topology. Its inverse, defined on its range, is also continuous.## Consequences

## Notes

## See also

## Further reading

A proof of this theorem for separable normed vector spaces was published in 1932 by Stefan Banach, and the first proof for the general case was published in 1940 by the mathematician Leonidas Alaoglu.

Since the Banach–Alaoglu theorem is proven via Tychonoff's theorem, it relies on the ZFC axiomatic framework, in particular the axiom of choice. Most mainstream functional analysis also relies on ZFC.

The Bourbaki–Alaoglu theorem is a generalization by Bourbaki to dual topologies.

Given a separated locally convex space X with continuous dual X ' then the polar U^{0} of any neighbourhood U in X is compact in the weak topology σ(X ',X) on X '.

In the case of a normed vector space, the polar of a neighbourhood is closed and norm-bounded in the dual space. For example the polar of the unit ball is the closed unit ball in the dual. Consequently, for normed vector space (and hence Banach spaces) the Bourbaki–Alaoglu theorem is equivalent to the Banach–Alaoglu theorem.

For any x in X, let

- $D\_x=\{zinmathbb\{C\}:\; |z|leq\; |x|\},$

and

- $D=Pi\_\{xin\; X\}\; D\_x\; .$

Since each D_{x} is a compact subset
of the complex plane, D is also compact in the product topology by Tychonoff theorem.

We can identify the closed unit ball in X*, B_{1}(X*), as a subset of D in a natural way:

- $f\; in\; B\_1(X^*)\; mapsto\; (f(x))\_\{x\; in\; X\}\; in\; D.$

The claim will be proved if the range of the above map is closed. But this is also clear. If one has a net

- $(f\_\{alpha\}(x))\_\{x\; in\; X\}\; rightarrow\; (lambda\_x)\_\{x\; in\; X\}$

in D, then the functional defined by

- $g(x)\; =\; lambda\_x\; ,$

lies in B_{1}(X*).

If X is a reflexive Banach space, then every bounded sequence in X has a weakly convergent subsequence. (This follows by applying the Banach–Alaoglu theorem to a weakly metrizable subspace of X; or, more succinctly, by applying the Eberlein–Šmulian theorem.) For example, suppose that X=L^{p}(μ), 1<p<∞. Let f_{n} be a bounded sequence of functions in X. Then there exists a subsequence f_{n}k and an f ∈ X such that

- $int\; f\_\{n\_k\}\; g,dmu\; to\; int\; f\; g,dmu$

for all g ∈ L^{q}(μ) = X* (where 1/p+1/q=1). The corresponding result for p=1 is not true, as L^{1}(μ) is not reflexive.

- John B. Conway (1994).
*A course in functional analysis*. 2nd edition, Berlin: Springer-Verlag. ISBN 0-387-97245-5. See Chapter 5, section 3. - . See section 3.15, p.68.

Wikipedia, the free encyclopedia © 2001-2006 Wikipedia contributors (Disclaimer)

This article is licensed under the GNU Free Documentation License.

Last updated on Monday June 09, 2008 at 21:41:07 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

This article is licensed under the GNU Free Documentation License.

Last updated on Monday June 09, 2008 at 21:41:07 PDT (GMT -0700)

View this article at Wikipedia.org - Edit this article at Wikipedia.org - Donate to the Wikimedia Foundation

Copyright © 2015 Dictionary.com, LLC. All rights reserved.