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., Theorem 3.15. 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.
This theorem has applications in physics when one describes the set of states of an algebra of observables, namely that any state can be written as a convex linear combination of so-called pure states.
History
According to Lawrence Narici and Edward Beckenstein, the Alaoglu theorem is a "very important result—maybe most important fact about the weak-* topology—[that] echos throughout functional analysis." In 1912, Helly proved that the unit ball of the continuous dual space of is countably weak-* compact. In 1932, Stefan Banach proved that the closed unit ball in the continuous dual space of any separable normed space is sequentially weak-* compact (Banach only considered sequential compactness). The proof for the general case was published in 1940 by the mathematician Leonidas Alaoglu. According to Pietsch , there are at least twelve mathematicians who can lay claim to this theorem or an important predecessor to it.
The Bourbaki–Alaoglu theorem is a generalization, Theorem (4) in §20.9., Theorem 23.5. of the original theorem by Bourbaki to dual topologies on locally convex spaces. This theorem is also called the Banach–Alaoglu theorem or the weak-* compactness theorem and it is commonly called simply the Alaoglu theorem.
Statement
If is a vector space over the field then will denote the algebraic dual space of and these two spaces are henceforth associated with the bilinear defined by
where the triple forms a dual system called the .
If is a topological vector space (TVS) then its continuous dual space will be denoted by where always holds. Denote the weak-* topology on by and denote the weak-* topology on by The weak-* topology is also called the topology of pointwise convergence because given a map and a net of maps the net converges to in this topology if and only if for every point in the domain, the net of values converges to the value
Proof involving duality theory
If is a normed vector space, then the polar of a neighborhood is closed and norm-bounded in the dual space. In particular, if is the open (or closed) unit ball in then the polar of is the closed unit ball in the continuous dual space of (with the usual dual norm). Consequently, this theorem can be specialized to:
When the continuous dual space of is an infinite dimensional normed space then it is for the closed unit ball in to be a compact subset when has its usual norm topology. This is because the unit ball in the norm topology is compact if and only if the space is finite-dimensional (cf. F. Riesz theorem). This theorem is one example of the utility of having different topologies on the same vector space.
It should be cautioned that despite appearances, the Banach–Alaoglu theorem does imply that the weak-* topology is locally compact. This is because the closed unit ball is only a neighborhood of the origin in the strong topology, but is usually not a neighborhood of the origin in the weak-* topology, as it has empty interior in the weak* topology, unless the space is finite-dimensional. In fact, it is a result of Weil that all locally compact Hausdorff topological vector spaces must be finite-dimensional.
Elementary proof
The following elementary proof does not utilize duality theory and requires only basic concepts from set theory, topology, and functional analysis. What is needed from topology is a working knowledge of net convergence in topological spaces and familiarity with the fact that a linear functional is continuous if and only if it is bounded on a neighborhood of the origin (see the articles on continuous linear functionals and sublinear functionals for details). Also required is a proper understanding of the technical details of how the space of all functions of the form is identified as the Cartesian product and the relationship between pointwise convergence, the product topology, and subspace topologies they induce on subsets such as the algebraic dual space and products of subspaces such as An explanation of these details is now given for readers who are interested.
For every real will denote the closed ball of radius centered at and for any
Identification of functions with tuples
The Cartesian product is usually thought of as the set of all -indexed tuples but, since tuples are technically just functions from an indexing set, it can also be identified with the space of all functions having prototype as is now described:
: A function belonging to is identified with its (-indexed) "" : A tuple in is identified with the function defined by ; this function's "tuple of values" is the original tuple
This is the reason why many authors write, often without comment, the equality
and why the Cartesian product is sometimes taken as the definition of the set of maps (or conversely). However, the Cartesian product, being the (categorical) product in the category of sets (which is a type of inverse limit), also comes equipped with associated maps that are known as its (coordinate) .
The at a given point is the function
where under the above identification, sends a function to
Stated in words, for a point and function "plugging into " is the same as "plugging into ".
In particular, suppose that are non-negative real numbers. Then where under the above identification of tuples with functions, is the set of all functions such that for every
If a subset partitions into then the linear bijection
canonically identifies these two Cartesian products; moreover, this map is a homeomorphism when these products are endowed with their product topologies. In terms of function spaces, this bijection could be expressed as
Notation for nets and function composition with nets
A net in is by definition a function from a non-empty directed set Every sequence in which by definition is just a function of the form is also a net. As with sequences, the value of a net at an index is denoted by ; however, for this proof, this value may also be denoted by the usual function parentheses notation Similarly for function composition, if is any function then the net (or sequence) that results from "plugging into " is just the function although this is typically denoted by (or by if is a sequence). In the proofs below, this resulting net may be denoted by any of the following notations
depending on whichever notation is cleanest or most clearly communicates the intended information. In particular, if is continuous and in then the conclusion commonly written as may instead be written as or
Topology
The set is assumed to be endowed with the product topology. It is well known that the product topology is identical to the topology of pointwise convergence. This is because given and a net where and every is an element of then the net converges in the product topology if and only if
for every the net converges in
where because and
this happens if and only if
for every the net converges in
Thus converges to in the product topology if and only if it converges to pointwise on
This proof will also use the fact that the topology of pointwise convergence is preserved when passing to topological subspaces. This means, for example, that if for every is some (topological) subspace of then the topology of pointwise convergence (or equivalently, the product topology) on is equal to the subspace topology that the set inherits from And if is closed in for every then is a closed subset of
Characterization of
An important fact used by the proof is that for any real
where denotes the supremum and As a side note, this characterization does not hold if the closed ball is replaced with the open ball (and replacing with the strict inequality will not change this; for counter-examples, consider and the identity map on ).
The essence of the Banach–Alaoglu theorem can be found in the next proposition, from which the Banach–Alaoglu theorem follows. Unlike the Banach–Alaoglu theorem, this proposition does require the vector space to endowed with any topology.
Before proving the proposition above, it is first shown how the Banach–Alaoglu theorem follows from it (unlike the proposition, Banach–Alaoglu assumes that is a topological vector space (TVS) and that is a neighborhood of the origin).
The conclusion that the set is closed can also be reached by applying the following more general result, this time proved using nets, to the special case and
Observation: If is any set and if is a closed subset of a topological space then is a closed subset of in the topology of pointwise convergence.
Proof of observation: Let and suppose that is a net in that converges pointwise to It remains to show that which by definition means For any because in and every value belongs to the closed (in ) subset so too must this net's limit belong to this closed set; thus which completes the proof.
Let and suppose that is a net in the converges to in To conclude that it must be shown that is a linear functional. So let be a scalar and let
For any let denote
Because in which has the topology of pointwise convergence, in for every By using in place of it follows that each of the following nets of scalars converges in
Proof that Let be the "multiplication by " map defined by Because is continuous and in it follows that where the right hand side is and the left hand side is
which proves that Because also and limits in are unique, it follows that as desired.
Proof that Define a net by letting for every Because and it follows that in Let be the addition map defined by The continuity of implies that in where the right hand side is and the left hand side is
which proves that Because also it follows that as desired.
The lemma above actually also follows from its corollary below since is a Hausdorff complete uniform space and any subset of such a space (in particular ) is closed if and only if it is complete.
Because the underlying field is a complete Hausdorff locally convex topological vector space, the same is true of the product space A closed subset of a complete space is complete, so by the lemma, the space is complete.
The above elementary proof of the Banach–Alaoglu theorem actually shows that if is any subset that satisfies (such as any absorbing subset of ), then is a weak-* compact subset of
As a side note, with the help of the above elementary proof, it may be shown (see this footnote)For any non-empty subset the equality holds (the intersection on the left is a closed, rather than open, disk − possibly of radius − because it is an intersection of closed subsets of and so must itself be closed). For every let so that the previous set equality implies From it follows that and thereby making the least element of with respect to (In fact, the family is closed under (non-nullary) arbitrary intersections and also under finite unions of at least one set). The elementary proof showed that and are not empty and moreover, it also even showed that has an element that satisfies for every which implies that for every The inclusion is immediate; to prove the reverse inclusion, let By definition, if and only if so let and it remains to show that From it follows that which implies that as desired. that there exist -indexed non-negative real numbers such that
where these real numbers can also be chosen to be "minimal" in the following sense: using (so as in the proof) and defining the notation for any if
then and for every