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.
According to Lawrence Narici and Edward Beckenstein, the Alaoglu theorem is a “very important result—maybethe most important fact about theweak-* topology—[that] echos throughout functional analysis.”[2] In 1912, Helly proved that the unit ball of the continuous dual space of is countably weak-* compact.[3] In 1932,Stefan Banach proved that the closed unit ball in the continuous dual space of anyseparablenormed space is sequentially weak-* compact (Banach only consideredsequential compactness).[3] The proof for the general case was published in 1940 by the mathematicianLeonidas Alaoglu. According to Pietsch [2007], there are at least twelve mathematicians who can lay claim to this theorem or an important predecessor to it.[2]
TheBourbaki–Alaoglu theorem is a generalization[4][5] of the original theorem byBourbaki todual topologies onlocally convex spaces. This theorem is also called theBanach–Alaoglu theorem or theweak-* compactness theorem and it is commonly called simply theAlaoglu theorem.[2]
If is a vector space over the field then will denote thealgebraic dual space of and these two spaces are henceforth associated with thebilinearevaluation map defined bywhere the triple forms adual system called thecanonical dual system.
If is atopological vector space (TVS) then itscontinuous dual space will be denoted by where always holds. Denote theweak-* topology on by and denote the weak-* topology on by The weak-* topology is also called thetopology of pointwise convergence because given a map and anet 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
To start the proof, some definitions and readily verified results are recalled. When is endowed with theweak-* topology then thisHausdorfflocally convex topological vector space is denoted by The space is always acomplete TVS; however, may fail to be a complete space, which is the reason why this proof involves the space Specifically, this proof will use the fact that a subset of a complete Hausdorff space is compact if (and only if) it is closed andtotally bounded. Importantly, thesubspace topology that inherits from is equal to This can be readily verified by showing that given any anet in converges to in one of these topologies if and only if it also converges to in the other topology (the conclusion follows because two topologies are equal if and only if they have the exact same convergent nets).
The triple is adual pairing although unlike it is in general not guaranteed to be a dual system. Throughout, unless stated otherwise, all polar sets will be taken with respect to the canonicalpairing
Let be a neighborhood of the origin in and let:
be the polar of with respect to the canonical pairing;
be the polar of with respect to the canonical dual system Note that
A well known fact about polar sets is that
Show that is a-closed subset of Let and suppose that is a net in that converges to in To conclude that it is sufficient (and necessary) to show that for every Because in the scalar field and every value belongs to the closed (in) subset so too must this net's limit belong to this set. Thus
Show that and then conclude that is a closed subset of both and The inclusion holds because every continuous linear functional is (in particular) a linear functional. For the reverse inclusion let so that which states exactly that the linear functional is bounded on the neighborhood; thus is acontinuous linear functional (that is,) and so as desired. Using (1) and the fact that the intersection is closed in the subspace topology on the claim about being closed follows.
Conclude that is also a-totally bounded subset of Recall that the topology on is identical to the subspace topology that inherits from This fact, together with (3) and the definition of "totally bounded", implies that is a-totally bounded subset of
Finally, deduce that is a-compact subset of Because is acomplete TVS and is a closed (by (2)) and totally bounded (by (4)) subset of it follows that is compact.
If is anormed 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 theusual dual norm). Consequently, this theorem can be specialized to:
Banach–Alaoglu theorem—If is a normed space then the closed unit ball in the continuous dual space (endowed with its usualoperator norm) is compact with respect to theweak-* topology.
When the continuous dual space of is an infinite dimensional normed space then it isimpossible 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 doesnot imply that the weak-* topology islocally compact. This is because the closed unit ball is only a neighborhood of the origin in thestrong 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 ofWeil that alllocally compactHausdorff topological vector spaces must be finite-dimensional.
Primer on product/function spaces, nets, and pointwise convergence
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-indexedtuples 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:
Function Tuple: A function belonging to is identified with its (-indexed) "tuple of values"
Tuple Function: 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 equalityand 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 thecategory ofsets (which is a type ofinverse limit), also comes equipped with associated maps that are known as its (coordinate)projections.
Thecanonical projection of the Cartesian product at a given point is the functionwhere 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 subsetpartitions into then the linear bijectioncanonically identifies these two Cartesian products; moreover, this map is ahomeomorphism 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
Anet in is by definition a function from a non-emptydirected set Everysequence 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 forfunction 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 theproduct topology. It is well known that the product topology is identical to thetopology of pointwise convergence. This is because given and anet where and every is an element of then the netconverges 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 totopological 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 thesubspace 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 realwhere denotes thesupremum 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 theidentity 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 doesnot require the vector space to endowed with any topology.
Proposition[3]—Let be a subset of a vector space over the field (where) and for every real number endow the closed ball with itsusual topology ( need not be endowed with any topology, but has its usualEuclidean topology). Define
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 atopological vector space (TVS) and that is a neighborhood of the origin).
Proof that Banach–Alaoglu follows from the proposition above
Assume that is atopological vector space with continuous dual space and that is a neighborhood of the origin. Because is a neighborhood of the origin in it is also anabsorbing subset of so for every there exists a real number such that Thus the hypotheses of the above proposition are satisfied, and so the set is therefore compact in theweak-* topology. The proof of the Banach–Alaoglu theorem will be complete once it is shown that[note 2] where recall that was defined as
Proof that Because the conclusion is equivalent to If then which states exactly that the linear functional is bounded on the neighborhood thus is acontinuous linear functional (that is,), as desired.
Proof of Proposition
Theproduct space is compact byTychonoff's theorem (since each closed ball is aHausdorff[note 3]compact space). Because a closed subset of a compact space is compact, the proof of the proposition will be complete once it is shown thatis a closed subset of The following statements guarantee this conclusion:
For any let denote the projection to theth coordinate (as defined above). To prove that it is sufficient (and necessary) to show that for every So fix and let Because it remains to show that Recall that was defined in the proposition's statement as being any positive real number that satisfies (so for example, would be a valid choice for each), which implies Because is apositive homogeneous function that satisfies
Thus which shows that as desired.
Proof of (2):
Thealgebraic dual space is always a closed subset of (this is proved inthe lemma below for readers who are not familiar with this result). The setis closed in theproduct topology on since it is a product of closed subsets of Thus is an intersection of two closed subsets of which proves (2).[note 4]
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 aclosed 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.
Lemma ( is closed in)—Thealgebraic dual space of any vector space over a field (where is or) is a closed subset of in the topology of pointwise convergence. (The vector space need not be endowed with any topology).
Proof of lemma
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's net of values atBecause 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 iswhich 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 iswhich proves that Because also it follows that as desired.
The lemma above actually also follows from its corollary below since is a Hausdorffcomplete 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 theproduct 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 anyabsorbing subset of), then is aweak-* compact subset of
As a side note, with the help of the above elementary proof, it may be shown (see this footnote)[proof 1] that there exist-indexed non-negative real numbers such thatwhere 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 which shows that these numbers are unique; indeed, thisinfimum formula can be used to define them.
In fact, if denotes the set of all such products of closed balls containing the polar set then where denotes the intersection of all sets belonging to
This implies (among other things[note 5]) that the uniqueleast element of with respect to this may be used as an alternative definition of this (necessarilyconvex andbalanced) set. The function is aseminorm and it is unchanged if is replaced by theconvex balanced hull of (because). Similarly, because is also unchanged if is replaced by itsclosure in
A special case of the Banach–Alaoglu theorem is the sequential version of the theorem, which asserts that the closed unit ball of the dual space of aseparable normed vector space issequentially compact in the weak-* topology. In fact, the weak* topology on the closed unit ball of the dual of a separable space ismetrizable, and thus compactness and sequential compactness are equivalent.
Specifically, let be a separable normed space and the closed unit ball in Since is separable, let be a countable dense subset. Then the following defines a metric, where for anyin which denotes the duality pairing of with Sequential compactness of in this metric can be shown by adiagonalization argument similar to the one employed in the proof of theArzelà–Ascoli theorem.
Due to the constructive nature of its proof (as opposed to the general case, which is based on the axiom of choice), the sequential Banach–Alaoglu theorem is often used in the field ofpartial differential equations to construct solutions to PDE orvariational problems. For instance, if one wants to minimize a functional on the dual of a separable normed vector space one common strategy is to first construct a minimizing sequence which approaches the infimum of use the sequential Banach–Alaoglu theorem to extract a subsequence that converges in the weak* topology to a limit and then establish that is a minimizer of The last step often requires to obey a (sequential)lower semi-continuity property in the weak* topology.
When is the space of finite Radon measures on the real line (so that is the space of continuous functions vanishing at infinity, by theRiesz representation theorem), the sequential Banach–Alaoglu theorem is equivalent to theHelly selection theorem.
Proof
For every letand letbe endowed with theproduct topology. Because every is a compact subset of the complex plane,Tychonoff's theorem guarantees that their product is compact.
The closed unit ball in denoted by can be identified as a subset of in a natural way:
This map is injective and it is continuous when has theweak-* topology. This map's inverse, defined on its image, is also continuous.
It will now be shown that the image of the above map is closed, which will complete the proof of the theorem. Given a point and a net in the image of indexed by such thatthe functional defined bylies in and
Assume that is anormed space and endow its continuous dual space with the usualdual norm.
The closed unit ball in is weak-* compact.[3] So if is infinite dimensional then its closed unit ball is necessarilynot compact in the norm topology byF. Riesz's theorem (despite it being weak-* compact).
If is areflexive Banach space, then every bounded sequence in has a weakly convergent subsequence. (This follows by applying the Banach–Alaoglu theorem to a weakly metrizable subspace of; or, more succinctly, by applying theEberlein–Šmulian theorem.) For example, suppose that is the spaceLp space where and let satisfyLet be a bounded sequence of functions in Then there exists a subsequence and an such thatThe corresponding result for is not true, as is not reflexive.
In a Hilbert space, every bounded and closed set is weakly relatively compact, hence every bounded net has a weakly convergent subnet (Hilbert spaces arereflexive).
As norm-closed, convex sets are weakly closed (Hahn–Banach theorem), norm-closures of convex bounded sets in Hilbert spaces or reflexive Banach spaces are weakly compact.
Closed and bounded sets in are precompact with respect to theweak operator topology (the weak operator topology is weaker than theultraweak topology which is in turn the weak-* topology with respect to the predual of thetrace class operators). Hence bounded sequences of operators have a weak accumulation point. As a consequence, has theHeine–Borel property, if equipped with either the weak operator or the ultraweak topology.
Relation to the axiom of choice and other statements
The Banach–Alaoglu may be proven by usingTychonoff's theorem, which under theZermelo–Fraenkel set theory (ZF) axiomatic framework is equivalent to theaxiom of choice. Most mainstream functional analysis relies onZF + the axiom of choice, which is often denoted byZFC. However, the theorem doesnot rely upon the axiom of choice in theseparable case (seeabove): in this case there actually exists a constructive proof. In the general case of an arbitrary normed space, theultrafilter Lemma, which is strictly weaker than the axiom of choice and equivalent to Tychonoff's theorem for compactHausdorff spaces, suffices for the proof of the Banach–Alaoglu theorem, and is in fact equivalent to it.
The Banach–Alaoglu theorem is equivalent to theultrafilter lemma, which implies theHahn–Banach theorem forreal vector spaces (HB) but is not equivalent to it (said differently, Banach–Alaoglu is also strictly stronger thanHB). However, theHahn–Banach theorem is equivalent to the following weak version of the Banach–Alaoglu theorem for normed space[6] in which the conclusion of compactness (in theweak-* topology of the closed unit ball of the dual space) is replaced with the conclusion ofquasicompactness (also sometimes calledconvex compactness);
Weak version of Alaoglu theorem[6]—Let be a normed space and let denote the closed unit ball of itscontinuous dual space Then has the following property, which is called (weak-*)quasicompactness orconvex compactness: whenever is a cover of byconvexweak-* closed subsets of such that has thefinite intersection property, then is not empty.
^Explicitly, a subset is said to be "compact (resp. totally bounded, etc.) in the weak-* topology" if when is given theweak-* topology and the subset is given thesubspace topology inherited from then is acompact (resp.totally bounded, etc.) space.
^If denotes the topology that is (originally) endowed with, then the equality shows that the polar of is dependentonly on (and) and that the rest of the topology can be ignored. To clarify what is meant, suppose is any TVS topology on such that the set is (also) a neighborhood of the origin in Denote the continuous dual space of by and denote the polar of with respect to byso that is just the set from above. Then because both of these sets are equal to Said differently, the polar set's defining "requirement" that be a subset of thecontinuous dual space is inconsequential and can be ignored because it does not have any effect on the resulting set of linear functionals. However, if is a TVS topology on such that isnot a neighborhood of the origin in then the polar of with respect to is not guaranteed to equal and so the topology can not be ignored.
^Because every is also aHausdorff space, the conclusion that is compact only requires the so-called "Tychonoff's theorem for compact Hausdorff spaces," which is equivalent to theultrafilter lemma and strictly weaker than theaxiom of choice.
^The conclusion can be written as The set may thus equivalently be defined by Rewriting the definition in this way helps make it apparent that the set is closed in becausethis is true of
^This tuple is theleast element of with respect to natural induced pointwisepartial order defined by if and only if for every Thus, every neighborhood of the origin in can be associated with this unique (minimum) function For any if is such that then so that in particular, and for every
Proofs
^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 theleast element of with respect to (In fact, thefamily 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.
Köthe, Gottfried (1983) [1969].Topological Vector Spaces I. Grundlehren der mathematischen Wissenschaften. Vol. 159. Translated by Garling, D.J.H. New York: Springer Science & Business Media.ISBN978-3-642-64988-2.MR0248498.OCLC840293704.
Meise, Reinhold; Vogt, Dietmar (1997). "Theorem 23.5".Introduction to Functional Analysis. Oxford, England: Clarendon Press. p. 264.ISBN0-19-851485-9.
Narici, Lawrence; Beckenstein, Edward (2011).Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press.ISBN978-1584888666.OCLC144216834.
