Theorem in functional analysis
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
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.” 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 [2007], 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[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.
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
![{\displaystyle \left\langle x,f\right\rangle ~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/055a47cdd9bcc2b66ee625656c1d6b30390d861a)
where the triple
![{\displaystyle \left\langle X,X^{\#},\left\langle \cdot ,\cdot \right\rangle \right\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/80cdf2c93deba96a00211f1e565a9ac47723e21a)
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 — For any topological vector space (TVS)
(not necessarily Hausdorff or locally convex) with continuous dual space
the polar
![{\displaystyle U^{\circ }=\left\{f\in X^{\prime }~:~\sup _{u\in U}|f(u)|\leq 1\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a8086d345484f951e647f7ed3e9e4af3a2870260)
of any neighborhood
![{\displaystyle U}](https://wikimedia.org/api/rest_v1/media/math/render/svg/458a728f53b9a0274f059cd695e067c430956025)
of origin in
![{\displaystyle X}](https://wikimedia.org/api/rest_v1/media/math/render/svg/68baa052181f707c662844a465bfeeb135e82bab)
is compact in the weak-* topology
[note 1] ![{\displaystyle \sigma \left(X^{\prime },X\right)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/43ed437c13fa2a58f9603401ef4229ca41d82a03)
on
![{\displaystyle X^{\prime }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/980210d5ccf78c678264901dc7b0ce8a53d827bc)
Moreover,
![{\displaystyle U^{\circ }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ee5cf941b90e81fcba4b8524da26ad18ac92db2a)
is equal to the polar of
![{\displaystyle U}](https://wikimedia.org/api/rest_v1/media/math/render/svg/458a728f53b9a0274f059cd695e067c430956025)
with respect to the canonical system
![{\displaystyle \left\langle X,X^{\#}\right\rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/d954b04e248cfe884f12bed3bb6c887fefe7c1b9)
and it is also a compact subset of
Proof involving duality theory
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 ![{\displaystyle U^{\circ }=U^{\#}\cap X^{\prime }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/81d25a7c76cb4f8cbef8884a6dd236e5c62114fe)
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 ![{\displaystyle |f(u)|\leq 1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/681777aeb9bc30c8b644703719ed8c56410c3d16)
- 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 ![{\displaystyle X^{\#}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ece846c5aa8de19f2bede9835e0c64d979bb93c3)
- 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. ![{\displaystyle \blacksquare }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8733090f2d787d03101c3e16dc3f6404f0e7dd4c)
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
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.
Premiere 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" ![{\displaystyle s_{\bullet }~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~(s(x))_{x\in X}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/96aa53a247034f5387a30ba1e1b3dca8b09dd3e9) - Tuple
Function: A tuple in is identified with the function defined by ; this function's "tuple of values" is the original tuple ![{\displaystyle \left(s_{x}\right)_{x\in X}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f1bedcbb3e9e984308e58eb0d9faa7c654c4f98a) This is the reason why many authors write, often without comment, the equality ![{\displaystyle \mathbb {K} ^{X}=\prod _{x\in X}\mathbb {K} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c5b870eab8d6a382a354327b4e7f1f94e3a1ddd7) 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 ![{\displaystyle \Pr {}_{z}:\prod _{x\in X}\mathbb {K} \to \mathbb {K} \quad {\text{ defined by }}\quad s_{\bullet }=\left(s_{x}\right)_{x\in X}\mapsto s_{z}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b37454b0b7a44e616cb5a1dab7b8057cdc672fec) where under the above identification, sends a function to ![{\displaystyle \Pr {}_{z}(s)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~s(z).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2e67aa35120d5f7251144693d19e29cd43fb605a) 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 ![{\displaystyle {\begin{alignedat}{4}H:\;&&\prod _{x\in X}\mathbb {K} &&\;\to \;&\left(\prod _{u\in U}\mathbb {K} \right)\times \prod _{x\in X\setminus U}\mathbb {K} \\[0.3ex]&&\left(f_{x}\right)_{x\in X}&&\;\mapsto \;&\left(\left(f_{u}\right)_{u\in U},\;\left(f_{x}\right)_{x\in X\setminus U}\right)\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/27970a64f4e9095a733e3fb32257ee07b7eb7a49) 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 ![{\displaystyle {\begin{alignedat}{4}H:\;&&\mathbb {K} ^{X}&&\;\to \;&\mathbb {K} ^{U}\times \mathbb {K} ^{X\setminus U}\\[0.3ex]&&f&&\;\mapsto \;&\left(f{\big \vert }_{U},\;f{\big \vert }_{X\setminus U}\right)\\\end{alignedat}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bdc29181ec3967dcec729a1239ab0c5d2a879a10) 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 ![{\displaystyle F\left(x_{\bullet }\right)=\left(F\left(x_{i}\right)\right)_{i\in I}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~F\circ x_{\bullet },}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bf79de501876b2d5f32064e1746ee968e10c3089) 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 ![{\displaystyle \mathbb {K} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07453cff88535372e3ebb320517acf77be68529d) where because and this happens if and only if - for every
the net converges in ![{\displaystyle \mathbb {K} ,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/07453cff88535372e3ebb320517acf77be68529d) 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 ![{\displaystyle \sup _{u\in U}|f(u)|\leq r\qquad {\text{ if and only if }}\qquad f(U)\subseteq B_{r}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f6ffc9439040d547f757beb88e61df9f19a538a5) 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.
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
![{\displaystyle U^{\circ }~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\Big \{}f\in X^{\prime }~:~\sup _{u\in U}|f(u)|\leq 1{\Big \}}~=~U^{\#}\cap X^{\prime }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c165cb9705777996c096dc71db350dda40a05932)
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
![{\displaystyle U^{\#}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\Big \{}f\in X^{\#}~:~\sup _{u\in U}|f(u)|\leq 1{\Big \}}~=~\left\{f\in X^{\#}~:~f(U)\subseteq B_{1}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d36d06d533cefc5abe38f73456d164154a9157d)
is a closed subset of
![{\textstyle \prod _{x\in X}B_{r_{x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/de8d8fffbc64234b3f11ecbdca1d37a55d0381c5)
The following statements guarantee this conclusion:
![{\displaystyle U^{\#}\subseteq \prod _{x\in X}B_{r_{x}}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2db9099d1787fef20d7bd040e92746bd2fbff864)
is a closed subset of the product space ![{\displaystyle \prod _{x\in X}\mathbb {K} =\mathbb {K} ^{X}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a750c73af15790f5e61a09f3c66bd749cc58f398)
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
![{\displaystyle {\frac {1}{r_{x}}}|f(x)|=\left|{\frac {1}{r_{x}}}f(x)\right|=\left|f\left({\frac {1}{r_{x}}}x\right)\right|=\left|f\left(u_{x}\right)\right|\leq \sup _{u\in U}|f(u)|\leq 1.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8e77b1544a3a3f29a1d59dc4eb432542e9428928)
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
![{\displaystyle {\begin{alignedat}{9}U_{B_{1}}&\,{\stackrel {\scriptscriptstyle {\text{def}}}{=}}\,{\Big \{}~~\;~~\;~~\;~~f\ \in \mathbb {K} ^{X}~~\;~~:\sup _{u\in U}|f(u)|\leq 1{\Big \}}\\&={\big \{}~~\;~~\;~~\;~~f\,\in \mathbb {K} ^{X}~~\;~~:f(u)\in B_{1}{\text{ for all }}u\in U{\big \}}\\&={\Big \{}\left(f_{x}\right)_{x\in X}\in \prod _{x\in X}\mathbb {K} \,~:~\;~f_{u}~\in B_{1}{\text{ for all }}u\in U{\Big \}}\\&=\prod _{x\in X}C_{x}\quad {\text{ where }}\quad C_{x}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\begin{cases}B_{1}&{\text{ if }}x\in U\\\mathbb {K} &{\text{ if }}x\not \in U\\\end{cases}}\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e3b4638bf5f3a1f6a92b4aa86a5bf7b164bcb3d2)
is closed in the
product topology on
![{\displaystyle \prod _{x\in X}\mathbb {K} =\mathbb {K} ^{X}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10452c13f3d066a5a73758ef9d55de539a37fe43)
since it is a product of closed subsets of
![{\displaystyle \mathbb {K} .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7a85c73bd0f34d264a8b2a361a14e66ab0231834)
Thus
![{\displaystyle U_{B_{1}}\cap X^{\#}=U^{\#}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab998e725dff391eec35b3b809e36625dec1f3cf)
is an intersection of two closed subsets of
![{\displaystyle \mathbb {K} ^{X},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4d12a5d1d6a6add4b8ca2346b611a414dc1a8f6c)
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. ![{\displaystyle \blacksquare }](https://wikimedia.org/api/rest_v1/media/math/render/svg/8733090f2d787d03101c3e16dc3f6404f0e7dd4c)
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 ![{\displaystyle f_{\bullet }(z)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left(f_{i}(z)\right)_{i\in I}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7f00c2b93408afcd4da05d80d14e419b531808d5) 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 ![{\displaystyle f_{\bullet }(x)\to f(x),\quad f_{\bullet }(y)\to f(y),\quad f_{\bullet }(x+y)\to f(x+y),\quad {\text{ and }}\quad f_{\bullet }(sx)\to f(sx).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2d16d22139a12f8d52c96d2b40c58b2e31941166) 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
![{\displaystyle {\begin{alignedat}{4}M\left(f_{\bullet }(x)\right){\stackrel {\scriptscriptstyle {\text{def}}}{=}}&~M\circ f_{\bullet }(x)&&{\text{ by definition of notation }}\\=&~\left(M\left(f_{i}(x)\right)\right)_{i\in I}~~~&&{\text{ because }}f_{\bullet }(x)=\left(f_{i}(x)\right)_{i\in I}:I\to \mathbb {K} \\=&~\left(sf_{i}(x)\right)_{i\in I}&&M\left(f_{i}(x)\right)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~sf_{i}(x)\\=&~\left(f_{i}(sx)\right)_{i\in I}&&{\text{ by linearity of }}f_{i}\\=&~f_{\bullet }(sx)&&{\text{ notation }}\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ae87fde8b9bb9d2049604ce6521a75d8e1395410) 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
![{\displaystyle A\left(z_{\bullet }\right)~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~A\circ z_{\bullet }=\left(A\left(z_{i}\right)\right)_{i\in I}=\left(A\left(f_{i}(x),f_{i}(y)\right)\right)_{i\in I}=\left(f_{i}(x)+f_{i}(y)\right)_{i\in I}=\left(f_{i}(x+y)\right)_{i\in I}=f_{\bullet }(x+y)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2b030919302ab3a4c2ebe5f395128b7af10cf25a) 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.
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
![{\displaystyle {\begin{alignedat}{4}U^{\circ }&=U^{\#}&&\\&=X^{\#}&&\cap \prod _{x\in X}B_{m_{x}}\\&=X^{\prime }&&\cap \prod _{x\in X}B_{m_{x}}\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbea14635add1a3040dbf817ae28a3051a281838)
where these real numbers
![{\displaystyle m_{\bullet }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0cacfeb244fd66593091c98592831e179cfeaf4)
can also be chosen to be "minimal" in the following sense: using
![{\displaystyle P~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~U^{\circ }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/096eab9d83b46ab1cdc19215585863f077ab6b1d)
(so
![{\displaystyle P=U^{\#}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7ae6ba113cc30c4858dd4d9e2ce97b71c3aebdf1)
as in the proof) and defining the notation
![{\displaystyle \prod B_{R_{\bullet }}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\prod _{x\in X}B_{R_{x}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1a57be978cc738993f75d8d43eca5f57e89cf777)
for any
![{\displaystyle R_{\bullet }=\left(R_{x}\right)_{x\in X}\in \mathbb {R} ^{X},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/003e6a6e41478c6d6343875f184b0480dd2b124d)
if
![{\displaystyle T_{P}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{R_{\bullet }\in \mathbb {R} ^{X}~:~P\subseteq \prod B_{R_{\bullet }}\right\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3b511b213a6f1b4349bba3393a4757224d70e12e)
then
![{\displaystyle m_{\bullet }\in T_{P}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/843c050357bb366382516fd4983580f01f7dca1b)
and for every
![{\displaystyle m_{x}=\inf \left\{R_{x}:R_{\bullet }\in T_{P}\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b4730c719693394f3b8ba26fecd68dc64a72f7eb)
which shows that these numbers
![{\displaystyle m_{\bullet }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d0cacfeb244fd66593091c98592831e179cfeaf4)
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
![{\displaystyle \operatorname {Box} _{P}~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~\left\{\prod B_{R_{\bullet }}~:~R_{\bullet }\in T_{P}\right\}~=~\left\{\prod B_{R_{\bullet }}~:~P\subseteq \prod B_{R_{\bullet }}\right\},}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c6de4c515d7a792142623459fd6885122d107a4f)
then
![{\textstyle \prod B_{m_{\bullet }}=\cap \operatorname {Box} _{P}\in \operatorname {Box} _{P}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15c9cdbc49e9951ead6149ab940bc527d940548d)
where
![{\textstyle \bigcap \operatorname {Box} _{P}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2bd71acfc1e9d3b57c3b8c2a004ed42919392b0a)
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
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
![{\displaystyle \rho (x,y)=\sum _{n=1}^{\infty }\,2^{-n}\,{\frac {\left|\langle x-y,x_{n}\rangle \right|}{1+\left|\langle x-y,x_{n}\rangle \right|}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/39ea877e44cb39f35b761219ceb069a72567eb05)
in which
![{\displaystyle \langle \cdot ,\cdot \rangle }](https://wikimedia.org/api/rest_v1/media/math/render/svg/9a50080b735975d8001c9552ac2134b49ad534c0)
denotes the duality pairing of
![{\displaystyle X^{\prime }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d7a3a5819cc45f097de14b3ac5a8bedd902bc66d)
with
![{\displaystyle X.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5ba76c5a460c4a0bb1639a193bc1830f0a773e03)
Sequential compactness of
![{\displaystyle B}](https://wikimedia.org/api/rest_v1/media/math/render/svg/47136aad860d145f75f3eed3022df827cee94d7a)
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
![{\displaystyle D_{x}=\{c\in \mathbb {C} :|c|\leq \|x\|\}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3e8e917925d5a6b51e8d378663020a8f2497a142)
and let
![{\displaystyle D=\prod _{x\in X}D_{x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1e0221608b59dacdfe92a541c70481eb8bb8e09c)
be endowed with the
product topology. Because every
![{\displaystyle D_{x}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c15427ff95343d483864e9ad5f2f836de676312d)
is a compact subset of the complex plane,
Tychonoff's theorem guarantees that their product
![{\displaystyle D}](https://wikimedia.org/api/rest_v1/media/math/render/svg/f34a0c600395e5d4345287e21fb26efd386990e6)
is compact.
The closed unit ball in
denoted by
can be identified as a subset of
in a natural way:
![{\displaystyle {\begin{alignedat}{4}F:\;&&B_{1}^{\,\prime }&&\;\to \;&D\\[0.3ex]&&f&&\;\mapsto \;&(f(x))_{x\in X}.\\\end{alignedat}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/76a0397e85db7d048e9674d926da6aa724976576)
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
![{\displaystyle \lim _{i}\left(f_{i}(x)\right)_{x\in X}\to \lambda _{\bullet }\quad {\text{ in }}D,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/58ddd15d2ab84e558e49069c6fed21c6732f3fb6)
the functional
![{\displaystyle g:X\to \mathbb {C} }](https://wikimedia.org/api/rest_v1/media/math/render/svg/50143721e946f047840153b4dfedb7594c257bca)
defined by
![{\displaystyle g(x)=\lambda _{x}\qquad {\text{ for every }}x\in X,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/08d2fd0a76c22fa6e6792856cd01d7ed9e200185)
lies in
![{\displaystyle B_{1}^{\,\prime }}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc22daf78d6ef097646c120ebe6d83bb875dcb0c)
and
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. 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. - 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 ![{\displaystyle \int f_{n_{k}}g\,d\mu \to \int fg\,d\mu \qquad {\text{ for all }}g\in L^{q}(\mu )=X^{\prime }.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/023a51069990cdfdff562f93df46d0d8587a9c7d)
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 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
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);
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
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 ![{\displaystyle U^{\circ ,\sigma }~{\stackrel {\scriptscriptstyle {\text{def}}}{=}}~{\Big \{}f\in (X,\sigma )^{\prime }~:~\sup _{u\in U}|f(u)|\leq 1{\Big \}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8d7d3f5b038c052c2f5f5aca93a008cf849e7e59)
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. - ^ 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. - ^ 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
- ^ 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 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
- ^ Rudin 1991, Theorem 3.15.
- ^ Köthe 1983, Theorem (4) in §20.9.
- ^ Meise & Vogt 1997, Theorem 23.5.
- ^ 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
- 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. ISBN 978-3-642-64988-2. MR 0248498. OCLC 840293704.
- Meise, Reinhold; Vogt, Dietmar (1997). "Theorem 23.5". Introduction to Functional Analysis. Oxford, England: Clarendon Press. p. 264. ISBN 0-19-851485-9.
- Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
- Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277. See Theorem 3.15, p. 68.
- Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
- Schechter, Eric (1996). Handbook of Analysis and Its Foundations. San Diego, CA: Academic Press. ISBN 978-0-12-622760-4. OCLC 175294365.
- Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.
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