Banach–Alaoglu theorem
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.^{[1]} 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 socalled pure states.
History
According to Lawrence Narici and Edward Beckenstein, the Alaoglu theorem is a "very important result  maybe the most important fact about the weak* 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 any separable normed space is sequentially weak* compact (Banach only considered sequential compactness).^{[3]} The proof for the general case was published in 1940 by the mathematician Leonidas Alaoglu. According to Pietsch [2007], there are at least 12 mathematicians who can lay claim to this theorem or an important predecessor to it.^{[2]}
The Bourbaki–Alaoglu theorem is a generalization^{[4]}^{[5]} of the original theorem by Bourbaki to dual topologies on locally convex spaces. This theorem is also called the BanachAlaoglu theorem or the weak* compactness theorem and it is commonly called simply the Alaoglu theorem^{[2]}
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 evaluation map defined by
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
Alaoglu theorem^{[3]} — For any topological vector space (TVS) (not necessarily Hausdorff or locally convex) with continuous dual space the polar
Proof involving duality theory
Denote by the underlying field of by which is either the real numbers or complex numbers This proof will use some of the basic properties that are listed in the articles: polar set, dual system, and continuous linear operator.
To start the proof, some definitions and readily verified results are recalled. When is endowed with the weak* topology then this Hausdorff locally convex topological vector space is denoted by The space is always a complete 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 and totally bounded. Importantly, the subspace topology that inherits from is equal to This can be readily verified by showing that given any a net 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 a dual 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 canonical pairing
Let be a neighborhood of the origin in and let:
 be the polar of with respect to the canonical pairing ;
 be the bipolar of with respect to ;
 be the polar of with respect to the canonical dual system
A well known fact about polars of 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 a continuous 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.
 Show that is a totally bounded subset of By the bipolar theorem, where because the neighborhood is an absorbing subset of the same must be true of the set ; it is possible to prove that this implies that is a bounded subset of Because distinguishes points of a subset of is bounded if and only if it is totally bounded. So in particular, is also totally bounded.
 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 a complete TVS and is a closed (by (2)) and totally bounded (by (4)) subset of it follows that is compact. Q.E.D.
If is a normed vector space, then the polar of a neighborhood is closed and normbounded 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:
Banach–Alaoglu theorem — If is a normed space then the closed unit ball in the continuous dual space (endowed with its usual operator norm) is compact with respect to the weak* topology.
When the continuous dual space of is an infinite dimensional normed space then it is impossible 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 finitedimensional (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 not 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 finitedimensional. In fact, it is a result of Weil that all locally compact Hausdorff topological vector spaces must be finitedimensional.
Elementary proof
The following proof involves only elementary concepts from set theory, topology, and functional analysis. In particular, what is need from topology is a working knowledge of nets in topological spaces, the product topology, and their relationship to pointwise convergence (some details of this relationship are given in the proof). Familiarity with the fact that a linear functional is continuous if and only if it is bounded on a neighborhood of the origin is also needed (this is described in the article on sublinear functionals).
Denote by the underlying field of by which is either the real numbers or complex numbers For any real let
Because is a neighborhood of the origin in it is also an absorbing subset of so for every there exists a real number such that Let
Proof that 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 a continuous linear functional (that is, ) and so as desired. Q.E.D.
The rest of this proof requires a proper understanding how the Cartesian product is identified as the space of all functions of the form An explanation is now given for readers who are interested.
Premiere on identification of functions with tuples


The Cartesian product is usually thought of as the set of all indexed tuples but, as is now described, it can also be identified with the space of all functions having prototype
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
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) projections.
The Cartesian product's canonical projection at any given is the function
where under the above identification, sends a function to
In words, for a point and function "plugging into " is the same as "plugging into ".
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
where and Thus converges to in the product topology if and only if it converges to pointwise on Also used in this proof will be 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 
Having established that ^{[note 2]} to reduce symbol clutter, this olar set will be denoted by
The proof of the theorem will be complete once the following statements are verified:
 is a closed subset of
 Here is endowed with the topology of pointwise convergence, which is identical to the product topology.

 denotes the closed ball of radius centered at For each was defined at the start of this proof as any real that satisfies (so in particular, is a valid choice for each ).
These statements imply that is a closed subset of where this product space is compact by Tychonoff's theorem^{[note 3]} (because every closed ball is a compact space). Because a closed subset of a compact space is compact, it follows that is compact, which is the main conclusion of the Banach–Alaoglu theorem.
Proof of (1):
The algebraic dual space is always a closed subset of (this is proved in the lemma below for readers who are not familiar with this result). To prove that is closed in it suffices to show that the set defined by
As a side note, this proof can be generalized to prove the following more general result, from which the above conclusion follows as the special case and
 Proposition: If is any set and if is a closed subset of a topological space then is a closed subset of with respect to the topology of pointwise convergence.
Proof of (2):
For any let denote the projection to the ^{th} coordinate (as defined above). To prove that it is sufficient (and necessary) to show that for every So fix and let ; it remains to show that The defining condition on was that which implies that Because the linear functional satisfies and so implies
Thus which shows that as desired. Q.E.D.
The elementary proof above 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)^{[proof 1]} that
In fact,
This implies (among other things^{[note 4]}) that the unique least element of with respect to ; this may be used as an alternative definition of this (necessarily convex and balanced) set. The function is a seminorm and it is unchanged if is replaced by the convex balanced hull of (because ). Similarly, because is also unchanged if is replaced by its closure in
Lemma — The algebraic 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


A net in is by definition a function from a nonempty 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 this proof, 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 Start of proof: Let and suppose that is a net in the converges to in If then will denote 's net of values at To conclude that it must be shown that is a linear functional so let be a scalar and let The topology on is the topology of pointwise convergence so by considering the points and the convergence of in implies that each of the following nets of scalars converges in
which proves that Because also and limits in are unique, it follows that as desired.
which proves that Because also it follows that as desired.
Q.E.D.

Corollary to lemma — When the algebraic dual space of a vector space is equipped with the topology of pointwise convergence (also known as the weak* topology) then the resulting topological vector space (TVS) is a complete Hausdorff locally convex TVS.
Because the underlying field is a complete Hausdorff locally convex TVS, the same is true of the Cartesian product A closed subset of a complete space is complete, so by the lemma, the space is complete.
Sequential Banach–Alaoglu theorem
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 a separable normed vector space is sequentially compact in the weak* topology. In fact, the weak* topology on the closed unit ball of the dual of a separable space is metrizable, 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 any
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 of partial differential equations to construct solutions to PDE or variational 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 semicontinuity 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 the Riesz representation theorem), the sequential Banach–Alaoglu theorem is equivalent to the Helly selection theorem.
For every let
Because each is a compact subset of the complex plane, is also compact in the product topology by Tychonoff's theorem.
The closed unit ball in can be identified as a subset of in a natural way:
This map is injective and continuous, with having the weak* topology and the product topology. This map's inverse, defined on its range, is also continuous.
To finish proving this theorem, it will now be shown that the range of the above map is closed. Given a net
Consequences
Consequences for normed spaces
Assume that is a normed space and endow its continuous dual space with the usual dual norm.
 The closed unit ball in is weak* compact.^{[3]} So if is infinite dimensional then its closed unit ball is necessarily not compact in the norm topology by the F. Riesz theorem (despite it being weak* compact).
 A Banach space is reflexive if and only if its closed unit ball is compact.^{[3]}
 If is a reflexive 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 the Eberlein–Šmulian theorem.)
For example, suppose that is the space Lp space
Let be a bounded sequence of functions in
Then there exists a subsequence and an such that
for all where ). The corresponding result for is not true, as is not reflexive.
Consequences for Hilbert spaces
 In a Hilbert space, every bounded and closed set is weakly relatively compact, hence every bounded net has a weakly convergent subnet (Hilbert spaces are reflexive).
 As normclosed, convex sets are weakly closed (Hahn–Banach theorem), normclosures of convex bounded sets in Hilbert spaces or reflexive Banach spaces are weakly compact.
 Closed and bounded sets in are precompact with respect to the weak operator topology (the weak operator topology is weaker than the ultraweak topology which is in turn the weak* topology with respect to the predual of the trace class operators). Hence bounded sequences of operators have a weak accumulation point. As a consequence, has the Heine–Borel property, if equipped with either the weak operator or the ultraweak topology.
Relation to the axiom of choice
Since the Banach–Alaoglu theorem is usually proven via Tychonoff's theorem, it relies on the ZFC axiomatic framework, and in particular the axiom of choice. Most mainstream functional analysis also relies on ZFC. However, the theorem does not rely upon the axiom of choice in the separable case (see above): in this case one actually has a constructive proof. In the nonseparable case, the ultrafilter Lemma, which is strictly weaker than the axiom of choice, suffices for the proof of the BanachAlaoglu theorem, and is in fact equivalent to it.
See also
 Bishop–Phelps theorem
 Banach–Mazur theorem
 Deltacompactness theorem
 Eberlein–Šmulian theorem – Relates three different kinds of weak compactness in a Banach space
 Goldstine theorem
 James' theorem
 KreinMilman theorem
 Mazur's lemma – On strongly convergent combinations of a weakly convergent sequence in a Banach space
 Topological vector space – Vector space with a notion of nearness
Notes
 ^ Explicitly, a subset is said to be "compact (resp. totally bounded, etc.) in the weak* topology" if when is given the weak* topology and the subset is given the subspace topology inherited from then is a compact (resp. totally bounded, etc.) space.
 ^ If denotes the topology that is (originally) endowed with, then the equality shows that the polar of is dependent only 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 by
so 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 the continuous dual space is inconsequential and can be ignored because it does not have any affect on the resulting set of linear functionals. However, if is a TVS topology on such that is not 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 a Hausdorff space, the conclusion that is compact only requires the socalled "Tychonoff's theorem for compact Hausdorff spaces," which is equivalent to the ultrafilter lemma and strictly weaker than the axiom of choice.
 ^ This tuple is the least element of with respect to natural induced pointwise partial 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 nonempty 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 (nonnullary) arbitrary intersections and also under finite unions of at least one set). Statement (2) in the above 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 so 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. Q.E.D.
References
 ^ Rudin 1991, Theorem 3.15.
 ^ ^{a} ^{b} ^{c} Narici & Beckenstein 2011, pp. 235240.
 ^ ^{a} ^{b} ^{c} ^{d} ^{e} Narici & Beckenstein 2011, pp. 225273.
 ^ Köthe 1969, Theorem (4) in §20.9.
 ^ Meise & Vogt 1997, Theorem 23.5.
 Köthe, Gottfried (1969). Topological Vector Spaces I. New York: SpringerVerlag. See §20.9.
 Meise, Reinhold; Vogt, Dietmar (1997). Introduction to Functional Analysis. Oxford: Clarendon Press. ISBN 0198514859. See Theorem 23.5, p. 264.
 Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 9781584888666. OCLC 144216834.
 Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. 8 (Second ed.). New York, NY: McGrawHill Science/Engineering/Math. ISBN 9780070542365. OCLC 21163277. See Theorem 3.15, p. 68.
 Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 9781461271550. OCLC 840278135.
 Schechter, Eric (1997). Handbook of Analysis and its Foundations. San Diego: Academic Press.
 Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 9780486453521. OCLC 853623322.
Further reading
 John B. Conway (1994). A course in functional analysis (2nd ed.). Berlin: SpringerVerlag. ISBN 0387972455. See Chapter 5, section 3.
 Peter B. Lax (2002). Functional Analysis. WileyInterscience. pp. 120–121. ISBN 0471556041.