Banach–Alaoglu theorem
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.
Contents 1 History 2 Statement 2.1 Proof involving duality theory 2.2 Elementary proof 3 Sequential Banach–Alaoglu theorem 4 Consequences 4.1 Consequences for normed spaces 4.2 Consequences for Hilbert spaces 5 Relation to the axiom of choice and other statements 6 See also 7 Notes 8 Citations 9 References 10 Further reading 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 {displaystyle C([a,b])} 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 Banach-Alaoglu theorem or the weak-* compactness theorem and it is commonly called simply the Alaoglu theorem[2] Statement See also: Topological vector space § Dual space, Dual system, and Polar set If {displaystyle X} is a vector space over the field {displaystyle mathbb {K} } then {displaystyle X^{#}} will denote the algebraic dual space of {displaystyle X} and these two spaces are henceforth associated with the bilinear evaluation map {displaystyle leftlangle cdot ,cdot rightrangle :Xtimes X^{#}to mathbb {K} } defined by {displaystyle leftlangle x,frightrangle ~{stackrel {scriptscriptstyle {text{def}}}{=}}~f(x)} where the triple {displaystyle leftlangle X,X^{#}rightrangle } forms a dual system called the canonical dual system.
If {displaystyle X} is a topological vector space (TVS) then its continuous dual space will be denoted by {displaystyle X^{prime },} where {displaystyle X^{prime }subseteq X^{#}} always holds. Denote the weak-* topology on {displaystyle X^{#}} by {displaystyle sigma left(X^{#},Xright)} and denote the weak-* topology on {displaystyle X^{prime }} by {displaystyle sigma left(X^{prime },Xright).} The weak-* topology is also called the topology of pointwise convergence because given a map {displaystyle f} and a net of maps {displaystyle f_{bullet }=left(f_{i}right)_{iin I},} the net {displaystyle f_{bullet }} converges to {displaystyle f} in this topology if and only if for every point {displaystyle x} in the domain, the net of values {displaystyle left(f_{i}(x)right)_{iin I}} converges to the value {displaystyle f(x).} Alaoglu theorem[3] — For any topological vector space (TVS) {displaystyle X} (not necessarily Hausdorff or locally convex) with continuous dual space {displaystyle X^{prime },} the polar {displaystyle U^{circ }=left{fin X^{prime }~:~sup _{uin U}|f(u)|leq 1right}} of any neighborhood {displaystyle U} of origin in {displaystyle X} is compact in the weak-* topology[note 1] {displaystyle sigma left(X^{prime },Xright)} on {displaystyle X^{prime }.} Moreover, {displaystyle U^{circ }} is equal to the polar of {displaystyle U} with respect to the canonical system {displaystyle leftlangle X,X^{#}rightrangle } and it is also a compact subset of {displaystyle left(X^{#},sigma left(X^{#},Xright)right).} Proof involving duality theory Proof Denote by the underlying field of {displaystyle X} by {displaystyle mathbb {K} ,} which is either the real numbers {displaystyle mathbb {R} } or complex numbers {displaystyle mathbb {C} .} 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 {displaystyle X^{#}} is endowed with the weak-* topology {displaystyle sigma left(X^{#},Xright),} then this Hausdorff locally convex topological vector space is denoted by {displaystyle left(X^{#},sigma left(X^{#},Xright)right).} The space {displaystyle left(X^{#},sigma left(X^{#},Xright)right)} is always a complete TVS; however, {displaystyle left(X^{prime },sigma left(X^{prime },Xright)right)} may fail to be a complete space, which is the reason why this proof involves the space {displaystyle left(X^{#},sigma left(X^{#},Xright)right).} 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 {displaystyle X^{prime }} inherits from {displaystyle left(X^{#},sigma left(X^{#},Xright)right)} is equal to {displaystyle sigma left(X^{prime },Xright).} This can be readily verified by showing that given any {displaystyle fin X^{prime },} a net in {displaystyle X^{prime }} converges to {displaystyle f} in one of these topologies if and only if it also converges to {displaystyle f} 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 {displaystyle leftlangle X,X^{prime }rightrangle } is a dual pairing although unlike {displaystyle leftlangle X,X^{#}rightrangle ,} 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 {displaystyle leftlangle X,X^{prime }rightrangle .} Let {displaystyle U} be a neighborhood of the origin in {displaystyle X} and let: {displaystyle U^{circ }=left{fin X^{prime }~:~sup _{uin U}|f(u)|leq 1right}} be the polar of {displaystyle U} with respect to the canonical pairing {displaystyle leftlangle X,X^{prime }rightrangle } ; {displaystyle U^{circ circ }=left{xin X~:~sup _{fin U^{circ }}|f(x)|leq 1right}} be the bipolar of {displaystyle U} with respect to {displaystyle leftlangle X,X^{prime }rightrangle } ; {displaystyle U^{#}=left{fin X^{#}~:~sup _{uin U}|f(u)|leq 1right}} be the polar of {displaystyle U} with respect to the canonical dual system {displaystyle leftlangle X,X^{#}rightrangle .} Note that {displaystyle U^{circ }=U^{#}cap X^{prime }.} A well known fact about polar sets is that {displaystyle U^{circ circ circ }subseteq U^{circ }.} Show that {displaystyle U^{#}} is a {displaystyle sigma left(X^{#},Xright)} -closed subset of {displaystyle X^{#}:} Let {displaystyle fin X^{#}} and suppose that {displaystyle f_{bullet }=left(f_{i}right)_{iin I}} is a net in {displaystyle U^{#}} that converges to {displaystyle f} in {displaystyle left(X^{#},sigma left(X^{#},Xright)right).} To conclude that {displaystyle fin U^{#},} it is sufficient (and necessary) to show that {displaystyle |f(u)|leq 1} for every {displaystyle uin U.} Because {displaystyle f_{i}(u)to f(u)} in the scalar field {displaystyle mathbb {K} } and every value {displaystyle f_{i}(u)} belongs to the closed (in {displaystyle mathbb {K} } ) subset {displaystyle left{sin mathbb {K} :|s|leq 1right},} so too must this net's limit {displaystyle f(u)} belong to this set. Thus {displaystyle |f(u)|leq 1.} Show that {displaystyle U^{#}=U^{circ }} and then conclude that {displaystyle U^{circ }} is a closed subset of both {displaystyle left(X^{#},sigma left(X^{#},Xright)right)} and {displaystyle left(X^{prime },sigma left(X^{prime },Xright)right):} The inclusion {displaystyle U^{circ }subseteq U^{#}} holds because every continuous linear functional is (in particular) a linear functional. For the reverse inclusion {displaystyle ,U^{#}subseteq U^{circ },,} let {displaystyle fin U^{#}} so that {displaystyle ;sup _{uin U}|f(u)|leq 1,,} which states exactly that the linear functional {displaystyle f} is bounded on the neighborhood {displaystyle U} ; thus {displaystyle f} is a continuous linear functional (that is, {displaystyle fin X^{prime }} ) and so {displaystyle fin U^{circ },} as desired. Using (1) and the fact that the intersection {displaystyle U^{#}cap X^{prime }=U^{circ }cap X^{prime }=U^{circ }} is closed in the subspace topology on {displaystyle X^{prime },} the claim about {displaystyle U^{circ }} being closed follows. Show that {displaystyle U^{circ }} is a {displaystyle sigma left(X^{prime },Xright)} -totally bounded subset of {displaystyle X^{prime }:} By the bipolar theorem, {displaystyle Usubseteq U^{circ circ }} where because the neighborhood {displaystyle U} is an absorbing subset of {displaystyle X,} the same must be true of the set {displaystyle U^{circ circ };} it is possible to prove that this implies that {displaystyle U^{circ }} is a {displaystyle sigma left(X^{prime },Xright)} -bounded subset of {displaystyle X^{prime }.} Because {displaystyle X} distinguishes points of {displaystyle X^{prime },} a subset of {displaystyle X^{prime }} is {displaystyle sigma left(X^{prime },Xright)} -bounded if and only if it is {displaystyle sigma left(X^{prime },Xright)} -totally bounded. So in particular, {displaystyle U^{circ }} is also {displaystyle sigma left(X^{prime },Xright)} -totally bounded. Conclude that {displaystyle U^{circ }} is also a {displaystyle sigma left(X^{#},Xright)} -totally bounded subset of {displaystyle X^{#}:} Recall that the {displaystyle sigma left(X^{prime },Xright)} topology on {displaystyle X^{prime }} is identical to the subspace topology that {displaystyle X^{prime }} inherits from {displaystyle left(X^{#},sigma left(X^{#},Xright)right).} This fact, together with (3) and the definition of "totally bounded", implies that {displaystyle U^{circ }} is a {displaystyle sigma left(X^{#},Xright)} -totally bounded subset of {displaystyle X^{#}.} Finally, deduce that {displaystyle U^{circ }} is a {displaystyle sigma left(X^{prime },Xright)} -compact subset of {displaystyle X^{prime }:} Because {displaystyle left(X^{#},sigma left(X^{#},Xright)right)} is a complete TVS and {displaystyle U^{circ }} is a closed (by (2)) and totally bounded (by (4)) subset of {displaystyle left(X^{#},sigma left(X^{#},Xright)right),} it follows that {displaystyle U^{circ }} is compact. {displaystyle blacksquare } If {displaystyle X} is a normed vector space, then the polar of a neighborhood is closed and norm-bounded in the dual space. In particular, if {displaystyle U} is the open (or closed) unit ball in {displaystyle X} then the polar of {displaystyle U} is the closed unit ball in the continuous dual space {displaystyle X^{prime }} of {displaystyle X} (with the usual dual norm). Consequently, this theorem can be specialized to: Banach–Alaoglu theorem — If {displaystyle X} is a normed space then the closed unit ball in the continuous dual space {displaystyle X^{prime }} (endowed with its usual operator norm) is compact with respect to the weak-* topology.
When the continuous dual space {displaystyle X^{prime }} of {displaystyle X} is an infinite dimensional normed space then it is impossible for the closed unit ball in {displaystyle X^{prime }} to be a compact subset when {displaystyle X^{prime }} 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 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 and 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 (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 {displaystyle mathbb {K} ^{X}} of all functions of the form {displaystyle Xto mathbb {K} } is identified as the Cartesian product {textstyle prod _{xin X}mathbb {K} ,} and the relationship between pointwise convergence, the product topology, and subspace topologies they induce on subsets such as the algebraic dual space {displaystyle X^{#}} and products of subspaces such as {textstyle prod _{xin X}B_{r_{x}}.} An explanation of these details is now given for readers who are interested.
show Premiere on product/function spaces, nets, and pointwise convergence 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 {displaystyle X} to endowed with any topology.
Proposition[3] — Let {displaystyle U} be a subset of a vector space {displaystyle X} over the field {displaystyle mathbb {K} } (where {displaystyle mathbb {K} =mathbb {R} {text{ or }}mathbb {K} =mathbb {C} } ) and for every real number {displaystyle r,} endow the closed ball {textstyle B_{r}~{stackrel {scriptscriptstyle {text{def}}}{=}}~{sin mathbb {K} :|s|leq r}} with its usual topology ( {displaystyle X} need not be endowed with any topology, but {displaystyle mathbb {K} } has its usual Euclidean topology). Define {displaystyle U^{#}~{stackrel {scriptscriptstyle {text{def}}}{=}}~{Big {}fin X^{#}~:~sup _{uin U}|f(u)|leq 1{Big }}.} If for every {displaystyle xin X,} {displaystyle r_{x}>0} is a real number such that {displaystyle xin r_{x}U,} then {displaystyle U^{#}} is a closed and compact subspace of the product space {displaystyle prod _{xin X}B_{r_{x}}} (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 {displaystyle U^{#}} 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 {displaystyle X} is a topological vector space (TVS) and that {displaystyle U} is a neighborhood of the origin).
Proof that Banach–Alaoglu follows from the proposition above Assume that {displaystyle X} is a topological vector space with continuous dual space {displaystyle X^{prime }} and that {displaystyle U} is a neighborhood of the origin. Because {displaystyle U} is a neighborhood of the origin in {displaystyle X,} it is also an absorbing subset of {displaystyle X,} so for every {displaystyle xin X,} there exists a real number {displaystyle r_{x}>0} such that {displaystyle xin r_{x}U.} Thus the hypotheses of the above proposition are satisfied, and so the set {displaystyle U^{#}} is therefore compact in the weak-* topology. The proof of the Banach–Alaoglu theorem will be complete once it is shown that {displaystyle U^{#}=U^{circ },} [note 2] where recall that {displaystyle U^{circ }} was defined as {displaystyle U^{circ }~{stackrel {scriptscriptstyle {text{def}}}{=}}~{Big {}fin X^{prime }~:~sup _{uin U}|f(u)|leq 1{Big }}~=~U^{#}cap X^{prime }.} Proof that {displaystyle U^{circ }=U^{#}:} Because {displaystyle U^{circ }=U^{#}cap X^{prime },} the conclusion is equivalent to {displaystyle U^{#}subseteq X^{prime }.} If {displaystyle fin U^{#}} then {displaystyle ;sup _{uin U}|f(u)|leq 1,,} which states exactly that the linear functional {displaystyle f} is bounded on the neighborhood {displaystyle U;} thus {displaystyle f} is a continuous linear functional (that is, {displaystyle fin X^{prime }} ), as desired. {displaystyle blacksquare } Proof of Proposition The product space {textstyle prod _{xin X}B_{r_{x}}} is compact by Tychonoff's theorem (since each closed ball {displaystyle B_{r_{x}}~{stackrel {scriptscriptstyle {text{def}}}{=}}~{sin mathbb {K} :|s|leq r_{x}}} 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 {}fin X^{#}~:~sup _{uin U}|f(u)|leq 1{Big }}~=~left{fin X^{#}~:~f(U)subseteq B_{1}right}} is a closed subset of {textstyle prod _{xin X}B_{r_{x}}.} The following statements guarantee this conclusion: {displaystyle U^{#}subseteq prod _{xin X}B_{r_{x}}.} {displaystyle U^{#}} is a closed subset of the product space {displaystyle prod _{xin X}mathbb {K} =mathbb {K} ^{X}.} Proof of (1): For any {displaystyle zin X,} let {textstyle Pr {}_{z}:prod _{xin X}mathbb {K} to mathbb {K} } denote the projection to the {displaystyle z} th coordinate (as defined above). To prove that {textstyle U^{#}subseteq prod _{xin X}B_{r_{x}},} it is sufficient (and necessary) to show that {displaystyle Pr {}_{x}left(U^{#}right)subseteq B_{r_{x}}} for every {displaystyle xin X.} So fix {displaystyle xin X} and let {displaystyle fin U^{#}.} Because {displaystyle Pr {}_{x}(f),=,f(x),} it remains to show that {displaystyle f(x)in B_{r_{x}}.} Recall that {displaystyle r_{x}>0} was defined in the proposition's statement as being any positive real number that satisfies {displaystyle xin r_{x}U} (so for example, {displaystyle r_{u}:=1} would be a valid choice for each {displaystyle uin U} ), which implies {displaystyle ,u_{x}~{stackrel {scriptscriptstyle {text{def}}}{=}}~{frac {1}{r_{x}}},xin U.,} Because {displaystyle f} is a positive homogeneous function that satisfies {displaystyle ;sup _{uin U}|f(u)|leq 1,,} {displaystyle {frac {1}{r_{x}}}|f(x)|=left|{frac {1}{r_{x}}}f(x)right|=left|fleft({frac {1}{r_{x}}}xright)right|=left|fleft(u_{x}right)right|leq sup _{uin U}|f(u)|leq 1.} Thus {displaystyle |f(x)|leq r_{x},} which shows that {displaystyle f(x)in B_{r_{x}},} as desired.
Proof of (2): The algebraic dual space {displaystyle X^{#}} is always a closed subset of {textstyle mathbb {K} ^{X}=prod _{xin X}mathbb {K} } (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 _{uin U}|f(u)|leq 1{Big }}\&={big {}~~;~~;~~;~~f,in mathbb {K} ^{X}~~;~~:f(u)in B_{1}{text{ for all }}uin U{big }}\&={Big {}left(f_{x}right)_{xin X}in prod _{xin X}mathbb {K} ,~:~;~f_{u}~in B_{1}{text{ for all }}uin U{Big }}\&=prod _{xin X}C_{x}quad {text{ where }}quad C_{x}~{stackrel {scriptscriptstyle {text{def}}}{=}}~{begin{cases}B_{1}&{text{ if }}xin U\mathbb {K} &{text{ if }}xnot in U\end{cases}}\end{alignedat}}} is closed in the product topology on {displaystyle prod _{xin X}mathbb {K} =mathbb {K} ^{X}} since it is a product of closed subsets of {displaystyle mathbb {K} } (an alternative proof that utilizes nets to show that {displaystyle U_{B_{1}}=left{fin mathbb {K} ^{X}:f(U)subseteq B_{1}right}} is closed is also given below). Thus {displaystyle U_{B_{1}}cap X^{#}=U^{#}} is an intersection of two closed subsets of {displaystyle mathbb {K} ^{X},} which proves (2).[note 4] {displaystyle blacksquare } This conclusion that {displaystyle U_{B_{1}}} is closed can also be reached by applying the following more general result to the special case {displaystyle Y:=mathbb {K} } and {displaystyle B:=B_{1}.} Observation: If {displaystyle Usubseteq X} is any set and if {displaystyle Bsubseteq Y} is a closed subset of a topological space {displaystyle Y,} then {displaystyle U_{B}~{stackrel {scriptscriptstyle {text{def}}}{=}}~left{fin Y^{X}:f(U)subseteq Bright}} is a closed subset of {displaystyle Y^{X}} in the topology of pointwise convergence. Proof of observation: Let {displaystyle fin Y^{X}} and suppose that {displaystyle left(f_{i}right)_{iin I}} is a net in {displaystyle U_{B}} that converges pointwise to {displaystyle f.} It remains to show that {displaystyle fin U_{B},} which by definition means {displaystyle f(U)subseteq B.} For any {displaystyle uin U,} because {displaystyle left(f_{i}(u)right)_{iin I}to f(u)} in {displaystyle Y} and every value {displaystyle f_{i}(u)in f_{i}(U)subseteq B} belongs to the closed (in {displaystyle Y} ) subset {displaystyle B,} so too must this net's limit belong to this closed set; thus {displaystyle f(u)in B,} which completes the proof. {displaystyle blacksquare } Lemma ( {displaystyle X^{#}} is closed in {displaystyle mathbb {K} ^{X}} ) — The algebraic dual space {displaystyle X^{#}} of any vector space {displaystyle X} over a field {displaystyle mathbb {K} } (where {displaystyle mathbb {K} } is {displaystyle mathbb {R} } or {displaystyle mathbb {C} } ) is a closed subset of {textstyle mathbb {K} ^{X}=prod _{xin X}mathbb {K} } in the topology of pointwise convergence. (The vector space {displaystyle X} need not be endowed with any topology).
show Proof of lemma The lemma above actually also follows from its corollary below since {displaystyle prod _{xin X}mathbb {K} } is a Hausdorff complete uniform space and any subset of such a space (in particular {displaystyle X^{#}} ) is closed if and only if it is complete.
Corollary to lemma ( {displaystyle X^{#}} is weak-* complete) — When the algebraic dual space {displaystyle X^{#}} of a vector space {displaystyle X} is equipped with the topology {displaystyle sigma left(X^{#},Xright)} of pointwise convergence (also known as the weak-* topology) then the resulting topological space {displaystyle left(X^{#},sigma left(X^{#},Xright)right)} is a complete Hausdorff locally convex topological vector space.
show Proof of corollary to lemma The above elementary proof of the Banach–Alaoglu theorem actually shows that if {displaystyle Usubseteq X} is any subset that satisfies {displaystyle X=(0,infty )U~{stackrel {scriptscriptstyle {text{def}}}{=}}~{ru:r>0,uin U}} (such as any absorbing subset of {displaystyle X} ), then {displaystyle U^{#}~{stackrel {scriptscriptstyle {text{def}}}{=}}~left{fin X^{#}:f(U)subseteq B_{1}right}} is a weak-* compact subset of {displaystyle X^{#}.} As a side note, with the help of the above elementary proof, it may be shown (see this footnote)[proof 1] that there exist {displaystyle X} -indexed non-negative real numbers {displaystyle m_{bullet }=left(m_{x}right)_{xin X}} such that {displaystyle {begin{alignedat}{4}U^{circ }&=U^{#}&&\&=X^{#}&&cap prod _{xin X}B_{m_{x}}\&=X^{prime }&&cap prod _{xin X}B_{m_{x}}\end{alignedat}}} where these real numbers {displaystyle m_{bullet }} can also be chosen to be "minimal" in the following sense: using {displaystyle P~{stackrel {scriptscriptstyle {text{def}}}{=}}~U^{circ }} (so {displaystyle P=U^{#}} as in the proof) and defining the notation {displaystyle prod B_{R_{bullet }}~{stackrel {scriptscriptstyle {text{def}}}{=}}~prod _{xin X}B_{R_{x}}} for any {displaystyle R_{bullet }=left(R_{x}right)_{xin X}in mathbb {R} ^{X},} if {displaystyle T_{P}~{stackrel {scriptscriptstyle {text{def}}}{=}}~left{R_{bullet }in mathbb {R} ^{X}~:~Psubseteq prod B_{R_{bullet }}right}} then {displaystyle m_{bullet }in T_{P}} and for every {displaystyle xin X,} {displaystyle m_{x}=inf left{R_{x}:R_{bullet }in T_{P}right},} which shows that these numbers {displaystyle m_{bullet }} are unique; indeed, this infimum formula can be used to define them.
In fact, if {displaystyle {mathcal {B}}_{P}} denotes the set of all such products of closed balls containing the polar set {displaystyle P,} {displaystyle {mathcal {B}}_{P}~{stackrel {scriptscriptstyle {text{def}}}{=}}~left{prod B_{R_{bullet }}~:~R_{bullet }in T_{P}right}~=~left{prod B_{R_{bullet }}~:~Psubseteq prod B_{R_{bullet }}right},} then {textstyle prod B_{m_{bullet }}=cap {mathcal {B}}_{P}in {mathcal {B}}_{P}} where {textstyle bigcap {mathcal {B}}_{P}} denotes the intersection of all sets belonging to {displaystyle {mathcal {B}}_{P}.} This implies (among other things[note 5]) that {textstyle prod B_{m_{bullet }}=prod _{xin X}B_{m_{x}}} the unique least element of {displaystyle {mathcal {B}}_{P}} with respect to {displaystyle ,subseteq ;} this may be used as an alternative definition of this (necessarily convex and balanced) set. The function {displaystyle m_{bullet }~{stackrel {scriptscriptstyle {text{def}}}{=}}~left(m_{x}right)_{xin X}:Xto [0,infty )} is a seminorm and it is unchanged if {displaystyle U} is replaced by the convex balanced hull of {displaystyle U} (because {displaystyle U^{#}=[operatorname {cobal} U]^{#}} ). Similarly, because {displaystyle U^{circ }=left[operatorname {cl} _{X}Uright]^{circ },} {displaystyle m_{bullet }} is also unchanged if {displaystyle U} is replaced by its closure in {displaystyle X.} 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 {displaystyle X} be a separable normed space and {displaystyle B} the closed unit ball in {displaystyle X^{prime }.} Since {displaystyle X} is separable, let {displaystyle x_{bullet }=left(x_{n}right)_{n=1}^{infty }} be a countable dense subset. Then the following defines a metric, where for any {displaystyle x,yin B} {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|}}} in which {displaystyle langle cdot ,cdot rangle } denotes the duality pairing of {displaystyle X^{prime }} with {displaystyle X.} Sequential compactness of {displaystyle B} 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 {displaystyle F:X^{prime }to mathbb {R} } on the dual of a separable normed vector space {displaystyle X,} one common strategy is to first construct a minimizing sequence {displaystyle x_{1},x_{2},ldots in X^{prime }} which approaches the infimum of {displaystyle F,} use the sequential Banach–Alaoglu theorem to extract a subsequence that converges in the weak* topology to a limit {displaystyle x,} and then establish that {displaystyle x} is a minimizer of {displaystyle F.} The last step often requires {displaystyle F} to obey a (sequential) lower semi-continuity property in the weak* topology.
When {displaystyle X^{prime }} is the space of finite Radon measures on the real line (so that {displaystyle X=C_{0}(mathbb {R} )} 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 {displaystyle xin X,} let {displaystyle D_{x}={cin mathbb {C} :|c|leq |x|}} and let {displaystyle D=prod _{xin X}D_{x}} be endowed with the product topology. Because every {displaystyle D_{x}} is a compact subset of the complex plane, Tychonoff's theorem guarantees that their product {displaystyle D} is compact.
The closed unit ball in {displaystyle X^{prime },} denoted by {displaystyle B_{1}^{,prime },} can be identified as a subset of {displaystyle D} in a natural way: {displaystyle {begin{alignedat}{4}F:;&&B_{1}^{,prime }&&;to ;&D\[0.3ex]&&f&&;mapsto ;&(f(x))_{xin X}.\end{alignedat}}} This map is injective and it is continuous when {displaystyle B_{1}^{,prime }} 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 {displaystyle lambda _{bullet }=left(lambda _{x}right)_{xin X}in D} and a net {displaystyle left(f_{i}(x)right)_{xin X}} in the image of {displaystyle F} indexed by {displaystyle iin I} such that {displaystyle lim _{i}left(f_{i}(x)right)_{xin X}to lambda _{bullet }quad {text{ in }}D,} the functional {displaystyle g:Xto mathbb {C} } defined by {displaystyle g(x)=lambda _{x}qquad {text{ for every }}xin X,} lies in {displaystyle B_{1}^{,prime }} and {displaystyle F(g)=lambda _{bullet }.} {displaystyle blacksquare } Consequences Consequences for normed spaces Assume that {displaystyle X} is a normed space and endow its continuous dual space {displaystyle X^{prime }} with the usual dual norm.
The closed unit ball in {displaystyle X^{prime }} is weak-* compact.[3] So if {displaystyle X^{prime }} 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 {displaystyle sigma left(X,X^{prime }right)} -compact; this is known as James' theorem.[3] If {displaystyle X} is a reflexive Banach space, then every bounded sequence in {displaystyle X} has a weakly convergent subsequence. (This follows by applying the Banach–Alaoglu theorem to a weakly metrizable subspace of {displaystyle X} ; or, more succinctly, by applying the Eberlein–Šmulian theorem.) For example, suppose that {displaystyle X} is the space Lp space {displaystyle L^{p}(mu )} where {displaystyle 1
Si quieres conocer otros artículos parecidos a Banach–Alaoglu theorem puedes visitar la categoría Compactness theorems.
Deja una respuesta