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 so-called pure states.

History

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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 twelve 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 Banach–Alaoglu theorem or the weak-* compactness theorem and it is commonly called simply the Alaoglu theorem.[2]

Statement

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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   where the triple   forms a dual system called the canonical dual system.

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   of any neighborhood   of origin in   is compact in the weak-* topology[note 1]   on   Moreover,   is equal to the polar of   with respect to the canonical system   and it is also a compact subset of  

Proof involving duality theory

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Proof

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   Note that  

A well known fact about polar sets is that  

  1. 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  
  2. 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.
  3. 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.
  4. 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  
  5. 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.  

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:

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 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 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 finite-dimensional. In fact, it is a result of Weil that all locally compact Hausdorff topological vector spaces must be finite-dimensional.

Elementary proof

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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.

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  -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:

  • 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 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) projections.

The canonical projection of the Cartesian product 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 not 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 its usual topology (  need not be endowed with any topology, but   has its usual Euclidean topology). Define  

If for every     is a real number such that   then   is a closed and compact subspace of the product space   (where because this product topology is identical to the topology of pointwise convergence, which is also called the weak-* topology in functional analysis, this means that   is compact in the weak-* topology or "weak-* compact" for short).

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).

Proof that Banach–Alaoglu follows from the proposition above

Assume that   is a topological 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 an absorbing 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 the weak-* 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 a continuous linear functional (that is,  ), as desired.  

Proof of Proposition

The product space   is compact by Tychonoff's theorem (since each closed ball   is a Hausdorff[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 that   is a closed subset of   The following statements guarantee this conclusion:

  1.  
  2.   is a closed subset of the product space  

Proof of (1):

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   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 a positive homogeneous function that satisfies    

Thus   which shows that   as desired.

Proof of (2):

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). The set   is closed in the product 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 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.  

Lemma (  is closed in  ) — 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

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 at     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.

Corollary to lemma (  is weak-* complete) — 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 space   is a complete Hausdorff locally convex topological vector space.

Proof of corollary to lemma

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)[proof 1] 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     which shows that these numbers   are unique; indeed, this infimum 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 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  

Sequential Banach–Alaoglu theorem

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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     in which   denotes the duality pairing of   with   Sequential compactness of   in this metric can be shown by a diagonalization argument similar to the one employed in the proof of the Arzelà–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 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 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 the Riesz representation theorem), the sequential Banach–Alaoglu theorem is equivalent to the Helly selection theorem.

Proof

For every   let   and let   be endowed with the product 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 the weak-* 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 that   the functional   defined by   lies in   and    

Consequences

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Consequences for normed spaces

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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 F. Riesz's theorem (despite it being weak-* compact).
  • A Banach space is reflexive if and only if its closed unit ball is  -compact; this is known as James' theorem.[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   where   and let   satisfy   Let   be a bounded sequence of functions in   Then there exists a subsequence   and an   such that   The corresponding result for   is not true, as   is not reflexive.

Consequences for Hilbert spaces

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  • 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 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 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 and other statements

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The Banach–Alaoglu may be proven by using Tychonoff's theorem, which under the Zermelo–Fraenkel set theory (ZF) axiomatic framework is equivalent to the axiom of choice. Most mainstream functional analysis relies on ZF + the axiom of choice, which is often denoted by ZFC. However, the theorem does not rely upon the axiom of choice in the separable case (see above): in this case there actually exists a constructive proof. In the general case of an arbitrary normed space, the ultrafilter Lemma, which is strictly weaker than the axiom of choice and equivalent to Tychonoff's theorem for compact Hausdorff spaces, suffices for the proof of the Banach–Alaoglu theorem, and is in fact equivalent to it.

The Banach–Alaoglu theorem is equivalent to the ultrafilter lemma, which implies the Hahn–Banach theorem for real vector spaces (HB) but is not equivalent to it (said differently, Banach–Alaoglu is also strictly stronger than HB). However, the Hahn–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 the weak-* topology of the closed unit ball of the dual space) is replaced with the conclusion of quasicompactness (also sometimes called convex compactness);

Weak version of Alaoglu theorem[6] — Let   be a normed space and let   denote the closed unit ball of its continuous dual space   Then   has the following property, which is called (weak-*) quasicompactness or convex compactness: whenever   is a cover of   by convex weak-* closed subsets of   such that   has the finite intersection property, then   is not empty.

Compactness implies convex compactness because a topological space is compact if and only if every family of closed subsets having the finite intersection property (FIP) has non-empty intersection. The definition of convex compactness is similar to this characterization of compact spaces in terms of the FIP, except that it only involves those closed subsets that are also convex (rather than all closed subsets).

See also

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Notes

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  1. ^ 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.
  2. ^ 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 effect 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.
  3. ^ Because every   is also a Hausdorff space, the conclusion that   is compact only requires the so-called "Tychonoff's theorem for compact Hausdorff spaces," which is equivalent to the ultrafilter lemma and strictly weaker than the axiom of choice.
  4. ^ 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   because this is true of  
  5. ^ 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

  1. ^ 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.  

Citations

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  1. ^ Rudin 1991, Theorem 3.15.
  2. ^ a b c Narici & Beckenstein 2011, pp. 235–240.
  3. ^ a b c d e f Narici & Beckenstein 2011, pp. 225–273.
  4. ^ Köthe 1983, Theorem (4) in §20.9.
  5. ^ Meise & Vogt 1997, Theorem 23.5.
  6. ^ a b Bell, J.; Fremlin, David (1972). "A Geometric Form of the Axiom of Choice" (PDF). Fundamenta Mathematicae. 77 (2): 167–170. doi:10.4064/fm-77-2-167-170. Retrieved 26 Dec 2021.

References

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Further reading

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