Dual system

This article is about dual pairs of vector spaces. For dual pairs in representation theory, seeReductive dual pair.

Inmathematics,adual system,dual pairor adualityover afield$\mathbb {K}$is a triple$(X,Y,b)$consisting of twovector spaces,$X$and$Y$,over$\mathbb {K}$and a non-degeneratebilinear map$b:X\times Y\to \mathbb {K}$.

Inmathematics,dualityis the study of dual systems and is important infunctional analysis.Duality plays crucial roles inquantum mechanicsbecause it has extensive applications to the theory ofHilbert spaces.

Definition, notation, and conventions

Pairings

Apairingorpairover a field$\mathbb {K}$is a triple$(X,Y,b),$which may also be denoted by$b(X,Y),$consisting of two vector spaces$X$and$Y$over$\mathbb {K}$and abilinear map$b:X\times Y\to \mathbb {K}$called thebilinear map associated with the pairing,or more simply called the pairing'smapor itsbilinear form.The examples here only describe when$\mathbb {K}$is either thereal numbers$\mathbb {R}$or thecomplex numbers$\mathbb {C}$,but the mathematical theory is general.

For every$x\in X$,define $${\begin{alignedat}{4}b(x,\,\cdot \,):\,&Y&&\to &&\,\mathbb {K} \\&y&&\mapsto &&\,b(x,y)\end{alignedat}}$$ and for every$y\in Y,$define $${\begin{alignedat}{4}b(\,\cdot \,,y):\,&X&&\to &&\,\mathbb {K} \\&x&&\mapsto &&\,b(x,y).\end{alignedat}}$$ Every$b(x,\,\cdot \,)$is alinear functionalon$Y$and every$b(\,\cdot \,,y)$is alinear functionalon$X$.Therefore both $$b(X,\,\cdot \,):=\{b(x,\,\cdot \,):x\in X\}\qquad {\text{ and }}\qquad b(\,\cdot \,,Y):=\{b(\,\cdot \,,y):y\in Y\},$$ form vector spaces oflinear functionals.

It is common practice to write$\langle x,y\rangle$instead of$b(x,y)$,in which in some cases the pairing may be denoted by$\left\langle X,Y\right\rangle$rather than$(X,Y,\langle \cdot,\cdot \rangle )$.However, this article will reserve the use of$\langle \cdot,\cdot \rangle$for the canonicalevaluation map(defined below) so as to avoid confusion for readers not familiar with this subject.

Dual pairings

A pairing$(X,Y,b)$is called adual system,adual pair,or adualityover$\mathbb {K}$if thebilinear form$b$is non-degenerate,which means that it satisfies the following two separation axioms:

1. $Y$separates (distinguishes) points of$X$:if$x\in X$is such that$b(x,\,\cdot \,)=0$then$x=0$;or equivalently, for all non-zero$x\in X$,the map$b(x,\,\cdot \,):Y\to \mathbb {K}$is not identically$0$(i.e. there exists a$y\in Y$such that$b(x,y)\neq 0$for each$x\in X$);
2. $X$separates (distinguishes) points of$Y$:if$y\in Y$is such that$b(\,\cdot \,,y)=0$then$y=0$;or equivalently, for all non-zero$y\in Y,$the map$b(\,\cdot \,,y):X\to \mathbb {K}$is not identically$0$(i.e. there exists an$x\in X$such that$b(x,y)\neq 0$for each$y\in Y$).

In this case$b$isnon-degenerate,and one can say that$b$places$X$and$Y$in duality(or, redundantly but explicitly, inseparated duality), and$b$is called theduality pairingof the triple$(X,Y,b)$.

Total subsets

A subset$S$of$Y$is calledtotalif for every$x\in X$,$$b(x,s)=0\quad {\text{ for all }}s\in S$$implies$x=0.$ A total subset of$X$is defined analogously (see footnote).Thus$X$separates points of$Y$if and only if$X$is a total subset of$X$,and similarly for$Y$.

Orthogonality

The vectors$x$and$y$areorthogonal,written$x\perp y$,if$b(x,y)=0$.Two subsets$R\subseteq X$and$S\subseteq Y$areorthogonal,written$R\perp S$,if$b(R,S)=\{0\}$;that is, if$b(r,s)=0$for all$r\in R$and$s\in S$.The definition of a subset being orthogonal to a vector is definedanalogously.

Theorthogonal complementorannihilatorof a subset$R\subseteq X$is $$R^{\perp }:=\{y\in Y:R\perp y\}:=\{y\in Y:b(R,y)=\{0\}\}$$Thus$R$is a total subset of$X$if and only if$R^{\perp }$equals$\{0\}$.

Polar sets

Main article:Polar set

Given a triple$(X,Y,b)$defining a pairing over$\mathbb {K}$,theabsolute polar setorpolar setof a subset$A$of$X$is the set:$$A^{\circ }:=\left\{y\in Y:\sup _{x\in A}|b(x,y)|\leq 1\right\}.$$Symmetrically,the absolute polar set or polar set of a subset$B$of$Y$is denoted by$B^{\circ }$and defined by $$B^{\circ }:=\left\{x\in X:\sup _{y\in B}|b(x,y)|\leq 1\right\}.$$

To use bookkeeping that helps keep track of the anti-symmetry of the two sides of the duality, the absolute polar of a subset$B$of$Y$may also be called theabsolute prepolarorprepolarof$B$and then may be denoted by$B^{\circ }.$

The polar$B^{\circ }$is necessarily aconvexset containing$0\in Y$where if$B$is balanced then so is$B^{\circ }$and if$B$is a vector subspace of$X$then so too is$B^{\circ }$a vector subspace of$Y.$

If$A$is a vector subspace of$X,$then$A^{\circ }=A^{\perp }$and this is also equal to thereal polarof$A.$If$A\subseteq X$then thebipolarof$A$,denoted$A^{\circ \circ }$,is the polar of the orthogonal complement of$A$,i.e., the set${}^{\circ }\left(A^{\perp }\right).$Similarly, if$B\subseteq Y$then the bipolar of$B$is$B^{\circ \circ }:=\left({}^{\circ }B\right)^{\circ }.$

Dual definitions and results

Given a pairing$(X,Y,b),$define a new pairing$(Y,X,d)$where$d(y,x):=b(x,y)$for all$x\in X$and$y\in Y$.

There is a consistent theme in duality theory that any definition for a pairing$(X,Y,b)$has a corresponding dual definition for the pairing$(Y,X,d).$

Convention and Definition:Given any definition for a pairing$(X,Y,b),$one obtains adual definitionby applying it to the pairing$(Y,X,d).$These conventions also apply to theorems.

For instance, if "$X$distinguishes points of$Y$"(resp,"$S$is a total subset of$Y$") is defined as above, then this convention immediately produces the dual definition of"$Y$distinguishes points of$X$"(resp,"$S$is a total subset of$X$").

This following notation is almost ubiquitous and allows us to avoid assigning a symbol to$d.$

Convention and Notation:If a definition and its notation for a pairing$(X,Y,b)$depends on the order of$X$and$Y$(for example, the definition of theMackey topology$\tau (X,Y,b)$on$X$) then by switching the order of$X$and$Y,$then it is meant that definition applied to$(Y,X,d)$(continuing the same example, the topology$\tau (Y,X,b)$would actually denote the topology$\tau (Y,X,d)$).

For another example, once the weak topology on$X$is defined, denoted by$\sigma (X,Y,b)$,then this dual definition would automatically be applied to the pairing$(Y,X,d)$so as to obtain the definition of the weak topology on$Y$,and this topology would be denoted by$\sigma (Y,X,b)$rather than$\sigma (Y,X,d)$.

Identification of$(X,Y)$with$(Y,X)$

Although it is technically incorrect and an abuse of notation, this article will adhere to the nearly ubiquitous convention of treating a pairing$(X,Y,b)$interchangeably with$(Y,X,d)$and also of denoting$(Y,X,d)$by$(Y,X,b).$

Examples

Restriction of a pairing

Suppose that$(X,Y,b)$is a pairing,$M$is a vector subspace of$X,$and$N$is a vector subspace of$Y$.Then therestrictionof$(X,Y,b)$to$M\times N$is the pairing$\left(M,N,b{\big \vert }_{M\times N}\right).$If$(X,Y,b)$is a duality, then it's possible for a restriction to fail to be a duality (e.g. if$Y\neq \{0\}$and$N=\{0\}$).

This article will use the common practice of denoting the restriction$\left(M,N,b{\big \vert }_{M\times N}\right)$by$(M,N,b).$

Canonical duality on a vector space

Suppose that$X$is a vector space and let$X^{\#}$denote thealgebraic dual spaceof$X$(that is, the space of all linear functionals on$X$). There is a canonical duality$\left(X,X^{\#},c\right)$where$c\left(x,x^{\prime }\right)=\left\langle x,x^{\prime }\right\rangle =x^{\prime }(x),$which is called theevaluation mapor thenaturalorcanonicalbilinear functional on$X\times X^{\#}.$ Note in particular that for any$x^{\prime }\in X^{\#},$$c\left(\,\cdot \,,x^{\prime }\right)$is just another way of denoting$x^{\prime }$;i.e.$c\left(\,\cdot \,,x^{\prime }\right)=x^{\prime }(\,\cdot \,)=x^{\prime }.$

If$N$is a vector subspace of$X^{\#}$,then the restriction of$\left(X,X^{\#},c\right)$to$X\times N$is called thecanonical pairingwhere if this pairing is a duality then it is instead called thecanonical duality.Clearly,$X$always distinguishes points of$N$,so the canonical pairing is a dual system if and only if$N$separates points of$X.$ The following notation is now nearly ubiquitous in duality theory.

The evaluation map will be denoted by$\left\langle x,x^{\prime }\right\rangle =x^{\prime }(x)$(rather than by$c$) and$\langle X,N\rangle$will be written rather than$(X,N,c).$

Assumption:As is common practice, if$X$is a vector space and$N$is a vector space of linear functionals on$X,$then unless stated otherwise, it will be assumed that they are associated with the canonical pairing$\langle X,N\rangle.$

If$N$is a vector subspace of$X^{\#}$then$X$distinguishes points of$N$(or equivalently,$(X,N,c)$is a duality) if and only if$N$distinguishes points of$X,$or equivalently if$N$is total (that is,$n(x)=0$for all$n\in N$implies$x=0$).

Canonical duality on a topological vector space

Suppose$X$is atopological vector space(TVS) withcontinuous dual space$X^{\prime }.$ Then the restriction of the canonical duality$\left(X,X^{\#},c\right)$to$X$×$X^{\prime }$defines a pairing$\left(X,X^{\prime },c{\big \vert }_{X\times X^{\prime }}\right)$for which$X$separates points of$X^{\prime }.$ If$X^{\prime }$separates points of$X$(which is true if, for instance,$X$is a Hausdorff locally convex space) then this pairing forms a duality.

Assumption:As is commonly done, whenever$X$is a TVS, then unless indicated otherwise, it will be assumed without comment that it's associated with the canonical pairing$\left\langle X,X^{\prime }\right\rangle.$

Polars and duals of TVSs

The following result shows that thecontinuous linear functionalson a TVS are exactly those linear functionals that are bounded on a neighborhood of the origin.

Theorem—Let$X$be a TVS with algebraic dual $X^{\#}$and let${\mathcal {N}}$be a basis of neighborhoods of$X$at the origin. Under the canonical duality$\left\langle X,X^{\#}\right\rangle,$the continuous dual space of$X$is the union of all$N^{\circ }$as$N$ranges over${\mathcal {N}}$(where the polars are taken in $X^{\#}$).

Inner product spaces and complex conjugate spaces

Apre-Hilbert space$(H,\langle \cdot,\cdot \rangle )$is a dual pairing if and only if$H$is vector space over$\mathbb {R}$or$H$has dimension$0.$Here it is assumed that thesesquilinear form$\langle \cdot,\cdot \rangle$isconjugate homogeneousin its second coordinate and homogeneous in its first coordinate.

• If$(H,\langle \cdot,\cdot \rangle )$is arealHilbert spacethen$(H,H,\langle \cdot,\cdot \rangle )$forms a dual system.
• If$(H,\langle \cdot,\cdot \rangle )$is a complexHilbert spacethen$(H,H,\langle \cdot,\cdot \rangle )$forms a dual system if and only if$\operatorname {dim} H=0.$If$H$is non-trivial then$(H,H,\langle \cdot,\cdot \rangle )$does not even form pairing since the inner product issesquilinearrather than bilinear.

Suppose that$(H,\langle \cdot,\cdot \rangle )$is a complex pre-Hilbert space with scalar multiplication denoted as usual by juxtaposition or by a dot$\cdot.$ Define the map $$\,\cdot \,\perp \,\cdot \,:\mathbb {C} \times H\to H\quad {\text{ by }}\quad c\perp x:={\overline {c}}x,$$ where the right-hand side uses the scalar multiplication of$H.$ Let${\overline {H}}$denote thecomplex conjugate vector spaceof$H,$where${\overline {H}}$denotes the additive group of$(H,+)$(so vector addition in${\overline {H}}$is identical to vector addition in$H$) but with scalar multiplication in${\overline {H}}$being the map$\,\cdot \,\perp \,\cdot \,$(instead of the scalar multiplication that$H$is endowed with).

The map$b:H\times {\overline {H}}\to \mathbb {C}$defined by$b(x,y):=\langle x,y\rangle$is linear in both coordinatesand so$\left(H,{\overline {H}},\langle \cdot,\cdot \rangle \right)$forms a dual pairing.

Other examples

• Suppose$X=\mathbb {R} ^{2},$$Y=\mathbb {R} ^{3},$and for all$\left(x_{1},y_{1}\right)\in X{\text{ and }}\left(x_{2},y_{2},z_{2}\right)\in Y,$let$$b\left(\left(x_{1},y_{1}\right),\left(x_{2},y_{2},z_{2}\right)\right):=x_{1}x_{2}+y_{1}y_{2}.$$Then$(X,Y,b)$is a pairing such that$X$distinguishes points of$Y,$but$Y$does not distinguish points of$X.$Furthermore,$X^{\perp }:=\{y\in Y:X\perp y\}=\{(0,0,z):z\in \mathbb {R} \}.$
• Let$0<p<\infty,$$X:=L^{p}(\mu ),$$Y:=L^{q}(\mu )$(where$q$is such that${\tfrac {1}{p}}+{\tfrac {1}{q}}=1$), and$b(f,g):=\int fg\,\mathrm {d} \mu.$Then$(X,Y,b)$is a dual system.
• Let$X$and$Y$be vector spaces over the same field$\mathbb {K}.$Then the bilinear form$b\left(x\otimes y,x^{*}\otimes y^{*}\right)=\left\langle x^{\prime },x\right\rangle \left\langle y^{\prime },y\right\rangle$places$X\times Y$and$X^{\#}\times Y^{\#}$in duality.
• Asequence space$X$and itsbeta dual$Y:=X^{\beta }$with the bilinear map defined as$\langle x,y\rangle:=\sum _{i=1}^{\infty }x_{i}y_{i}$for$x\in X,$$y\in X^{\beta }$forms a dual system.

Weak topology

Main articles:Weak topologyandWeak-* topology

Suppose that$(X,Y,b)$is a pairing ofvector spacesover$\mathbb {K}.$ If$S\subseteq Y$then theweak topology on$X$induced by$S$(and$b$) is the weakest TVS topology on$X,$denoted by$\sigma (X,S,b)$or simply$\sigma (X,S),$making all maps$b(\,\cdot \,,y):X\to \mathbb {K}$continuous as$y$ranges over$S.$If$S$is not clear from context then it should be assumed to be all of$Y,$in which case it is called theweak topologyon$X$(induced by$Y$). The notation$X_{\sigma (X,S,b)},$$X_{\sigma (X,S)},$or (if no confusion could arise) simply$X_{\sigma }$is used to denote$X$endowed with the weak topology$\sigma (X,S,b).$ Importantly, the weak topology dependsentirelyon the function$b,$the usual topology on$\mathbb {C},$and$X$'svector spacestructure butnoton thealgebraic structuresof$Y.$

Similarly, if$R\subseteq X$then the dual definition of theweaktopologyon$Y$induced by$R$(and$b$), which is denoted by$\sigma (Y,R,b)$or simply$\sigma (Y,R)$(see footnote for details).

Definition and Notation:If "$\sigma (X,Y,b)$"is attached to a topological definition (e.g.$\sigma (X,Y,b)$-converges,$\sigma (X,Y,b)$-bounded,$\operatorname {cl} _{\sigma (X,Y,b)}(S),$etc.) then it means that definition when the first space (i.e.$X$) carries the$\sigma (X,Y,b)$topology. Mention of$b$or even$X$and$Y$may be omitted if no confusion arises. So, for instance, if a sequence$\left(a_{i}\right)_{i=1}^{\infty }$in$Y$"$\sigma$-converges "or" weakly converges "then this means that it converges in$(Y,\sigma (Y,X,b))$whereas if it were a sequence in$X$,then this would mean that it converges in$(X,\sigma (X,Y,b))$).

The topology$\sigma (X,Y,b)$islocally convexsince it is determined by the family of seminorms$p_{y}:X\to \mathbb {R}$defined by$p_{y}(x):=|b(x,y)|,$as$y$ranges over$Y.$ If$x\in X$and$\left(x_{i}\right)_{i\in I}$is anetin$X,$then$\left(x_{i}\right)_{i\in I}$$\sigma (X,Y,b)$-convergesto$x$if$\left(x_{i}\right)_{i\in I}$converges to$x$in$(X,\sigma (X,Y,b)).$ A net$\left(x_{i}\right)_{i\in I}$$\sigma (X,Y,b)$-converges to$x$if and only if for all$y\in Y,$$b\left(x_{i},y\right)$converges to$b(x,y).$ If$\left(x_{i}\right)_{i=1}^{\infty }$is a sequence oforthonormalvectors in Hilbert space, then$\left(x_{i}\right)_{i=1}^{\infty }$converges weakly to 0 but does not norm-converge to 0 (or any other vector).

If$(X,Y,b)$is a pairing and$N$is a proper vector subspace of$Y$such that$(X,N,b)$is a dual pair, then$\sigma (X,N,b)$is strictlycoarserthan$\sigma (X,Y,b).$

Bounded subsets

A subset$S$of$X$is$\sigma (X,Y,b)$-bounded if and only if $$\sup _{}|b(S,y)|<\infty \quad {\text{ for all }}y\in Y,$$where$|b(S,y)|:=\{b(s,y):s\in S\}.$

Hausdorffness

If$(X,Y,b)$is a pairing then the following are equivalent:

1. $X$distinguishes points of$Y$;
2. The map$y\mapsto b(\,\cdot \,,y)$defines aninjectionfrom$Y$into the algebraic dual space of$X$;
3. $\sigma (Y,X,b)$isHausdorff.

Weak representation theorem

The following theorem is of fundamental importance to duality theory because it completely characterizes the continuous dual space of$(X,\sigma (X,Y,b)).$

Weak representation theorem—Let$(X,Y,b)$be a pairing over the field$\mathbb {K}.$Then thecontinuous dual spaceof$(X,\sigma (X,Y,b))$is$$b(\,\cdot \,,Y):=\{b(\,\cdot \,,y):y\in Y\}.$$Furthermore,

1. If$f$is acontinuous linear functionalon$(X,\sigma (X,Y,b))$then there exists some$y\in Y$such that$f=b(\,\cdot \,,y)$;if such a$y$exists then it is unique if and only if$X$distinguishes points of$Y.$
   • Note that whether or not$X$distinguishes points of$Y$is not dependent on the particular choice of$y.$
2. The continuous dual space of$(X,\sigma (X,Y,b))$may be identified with the quotient space$Y/X^{\perp },$where$X^{\perp }:=\{y\in Y:b(x,y)=0{\text{ for all }}x\in X\}.$
   • This is true regardless of whether or not$X$distinguishes points of$Y$or$Y$distinguishes points of$X.$

Consequently, thecontinuous dual spaceof$(X,\sigma (X,Y,b))$is $$(X,\sigma (X,Y,b))^{\prime }=b(\,\cdot \,,Y):=\left\{b(\,\cdot \,,y):y\in Y\right\}.$$

With respect to the canonical pairing, if$X$is a TVS whose continuous dual space$X^{\prime }$separates points on$X$(i.e. such that$\left(X,\sigma \left(X,X^{\prime }\right)\right)$is Hausdorff, which implies that$X$is also necessarily Hausdorff) then the continuous dual space of$\left(X^{\prime },\sigma \left(X^{\prime },X\right)\right)$is equal to the set of all "evaluation at a point$x$"maps as$x$ranges over$X$(i.e. the map that send$x^{\prime }\in X^{\prime }$to$x^{\prime }(x)$). This is commonly written as $$\left(X^{\prime },\sigma \left(X^{\prime },X\right)\right)^{\prime }=X\qquad {\text{ or }}\qquad \left(X_{\sigma }^{\prime }\right)^{\prime }=X.$$ This very important fact is why results for polar topologies on continuous dual spaces, such as thestrong dual topology$\beta \left(X^{\prime },X\right)$on$X^{\prime }$for example, can also often be applied to the original TVS$X$;for instance,$X$being identified with$\left(X_{\sigma }^{\prime }\right)^{\prime }$means that the topology$\beta \left(\left(X_{\sigma }^{\prime }\right)^{\prime },X_{\sigma }^{\prime }\right)$on$\left(X_{\sigma }^{\prime }\right)^{\prime }$can instead be thought of as a topology on$X.$ Moreover, if$X^{\prime }$is endowed with a topology that isfinerthan$\sigma \left(X^{\prime },X\right)$then the continuous dual space of$X^{\prime }$will necessarily contain$\left(X_{\sigma }^{\prime }\right)^{\prime }$as a subset. So for instance, when$X^{\prime }$is endowed with the strong dual topology (and so is denoted by$X_{\beta }^{\prime }$) then $$\left(X_{\beta }^{\prime }\right)^{\prime }~\supseteq ~\left(X_{\sigma }^{\prime }\right)^{\prime }~=~X$$ which (among other things) allows for$X$to be endowed with the subspace topology induced on it by, say, the strong dual topology$\beta \left(\left(X_{\beta }^{\prime }\right)^{\prime },X_{\beta }^{\prime }\right)$(this topology is also called the strongbidualtopology and it appears in the theory ofreflexive spaces:the Hausdorff locally convex TVS$X$is said to besemi-reflexiveif$\left(X_{\beta }^{\prime }\right)^{\prime }=X$and it will be calledreflexiveif in addition the strong bidual topology$\beta \left(\left(X_{\beta }^{\prime }\right)^{\prime },X_{\beta }^{\prime }\right)$on$X$is equal to$X$'s original/starting topology).

Orthogonals, quotients, and subspaces

If$(X,Y,b)$is a pairing then for any subset$S$of$X$:

• $S^{\perp }=(\operatorname {span} S)^{\perp }=\left(\operatorname {cl} _{\sigma (Y,X,b)}\operatorname {span} S\right)^{\perp }=S^{\perp \perp \perp }$and this set is$\sigma (Y,X,b)$-closed;
• $S\subseteq S^{\perp \perp }=\left(\operatorname {cl} _{\sigma (X,Y,b)}\operatorname {span} S\right)$;
  • Thus if$S$is a$\sigma (X,Y,b)$-closed vector subspace of$X$then$S\subseteq S^{\perp \perp }.$
• If$\left(S_{i}\right)_{i\in I}$is a family of$\sigma (X,Y,b)$-closed vector subspaces of$X$then $$\left(\bigcap _{i\in I}S_{i}\right)^{\perp }=\operatorname {cl} _{\sigma (Y,X,b)}\left(\operatorname {span} \left(\bigcup _{i\in I}S_{i}^{\perp }\right)\right).$$
• If$\left(S_{i}\right)_{i\in I}$is a family of subsets of$X$then $\left(\bigcup _{i\in I}S_{i}\right)^{\perp }=\bigcap _{i\in I}S_{i}^{\perp }.$

If$X$is a normed space then under the canonical duality,$S^{\perp }$is norm closed in$X^{\prime }$and$S^{\perp \perp }$is norm closed in$X.$

Subspaces

Suppose that$M$is a vector subspace of$X$and let$(M,Y,b)$denote the restriction of$(X,Y,b)$to$M\times Y.$ The weak topology$\sigma (M,Y,b)$on$M$is identical to thesubspace topologythat$M$inherits from$(X,\sigma (X,Y,b)).$

Also,$\left(M,Y/M^{\perp },b{\big \vert }_{M}\right)$is a paired space (where$Y/M^{\perp }$means$Y/\left(M^{\perp }\right)$) where$b{\big \vert }_{M}:M\times Y/M^{\perp }\to \mathbb {K}$is defined by $$\left(m,y+M^{\perp }\right)\mapsto b(m,y).$$

The topology$\sigma \left(M,Y/M^{\perp },b{\big \vert }_{M}\right)$is equal to thesubspace topologythat$M$inherits from$(X,\sigma (X,Y,b)).$ Furthermore, if$(X,\sigma (X,Y,b))$is a dual system then so is$\left(M,Y/M^{\perp },b{\big \vert }_{M}\right).$

Quotients

Suppose that$M$is a vector subspace of$X.$ Then$\left(X/M,M^{\perp },b/M\right)$is a paired space where$b/M:X/M\times M^{\perp }\to \mathbb {K}$is defined by $$(x+M,y)\mapsto b(x,y).$$

The topology$\sigma \left(X/M,M^{\perp }\right)$is identical to the usualquotient topologyinduced by$(X,\sigma (X,Y,b))$on$X/M.$

Polars and the weak topology

If$X$is a locally convex space and if$H$is a subset of the continuous dual space$X^{\prime },$then$H$is$\sigma \left(X^{\prime },X\right)$-bounded if and only if$H\subseteq B^{\circ }$for somebarrel$B$in$X.$

The following results are important for defining polar topologies.

If$(X,Y,b)$is a pairing and$A\subseteq X,$then:

1. The polar$A^{\circ }$of$A$is a closed subset of$(Y,\sigma (Y,X,b)).$
2. The polars of the following sets are identical: (a)$A$;(b) the convex hull of$A$;(c) thebalanced hullof$A$;(d) the$\sigma (X,Y,b)$-closure of$A$;(e) the$\sigma (X,Y,b)$-closure of theconvex balanced hullof$A.$
3. Thebipolar theorem:The bipolar of$A,$denoted by$A^{\circ \circ },$is equal to the$\sigma (X,Y,b)$-closure of the convex balanced hull of$A.$
   • Thebipolar theoremin particular "is an indispensable tool in working with dualities."
4. $A$is$\sigma (X,Y,b)$-bounded if and only if$A^{\circ }$isabsorbingin$Y.$
5. If in addition$Y$distinguishes points of$X$then$A$is$\sigma (X,Y,b)$-boundedif and only if it is$\sigma (X,Y,b)$-totally bounded.

If$(X,Y,b)$is a pairing and$\tau$is a locally convex topology on$X$that is consistent with duality, then a subset$B$of$X$is abarrelin$(X,\tau )$if and only if$B$is thepolarof some$\sigma (Y,X,b)$-bounded subset of$Y.$

Transposes

Transposes of a linear map with respect to pairings

See also:Transpose of a linear map,Transpose,andTranspose § Transposes of linear maps and bilinear forms

Let$(X,Y,b)$and$(W,Z,c)$be pairings over$\mathbb {K}$and let$F:X\to W$be a linear map.

For all$z\in Z,$let$c(F(\,\cdot \,),z):X\to \mathbb {K}$be the map defined by$x\mapsto c(F(x),z).$ It is said that$F$'stransposeoradjoint is well-definedif the following conditions are satisfied:

1. $X$distinguishes points of$Y$(or equivalently, the map$y\mapsto b(\,\cdot \,,y)$from$Y$into the algebraic dual$X^{\#}$isinjective), and
2. $c(F(\,\cdot \,),Z)\subseteq b(\,\cdot \,,Y),$where$c(F(\,\cdot \,),Z):=\{c(F(\,\cdot \,),z):z\in Z\}$and$b(\,\cdot \,,Y):=\{b(\,\cdot \,,y):y\in Y\}$.

In this case, for any$z\in Z$there exists (by condition 2) a unique (by condition 1)$y\in Y$such that$c(F(\,\cdot \,),z)=b(\,\cdot \,,y)$), where this element of$Y$will be denoted by${}^{t}F(z).$ This defines a linear map $${}^{t}F:Z\to Y$$

called thetransposeoradjoint of$F$with respect to$(X,Y,b)$and$(W,Z,c)$(this should not be confused with theHermitian adjoint). It is easy to see that the two conditions mentioned above (i.e. for "the transpose is well-defined" ) are also necessary for${}^{t}F$to be well-defined. For every$z\in Z,$the defining condition for${}^{t}F(z)$is $$c(F(\,\cdot \,),z)=b\left(\,\cdot \,,{}^{t}F(z)\right),$$ that is, $$c(F(x),z)=b\left(x,{}^{t}F(z)\right)$$for all$x\in X.$

By the conventions mentioned at the beginning of this article, this also defines the transpose of linear maps of the form$Z\to Y,$ $X\to Z,$ $W\to Y,$ $Y\to W,$etc. (see footnote).

Properties of the transpose

Throughout,$(X,Y,b)$and$(W,Z,c)$be pairings over$\mathbb {K}$and$F:X\to W$will be a linear map whose transpose${}^{t}F:Z\to Y$is well-defined.

• ${}^{t}F:Z\to Y$isinjective(i.e.$\operatorname {ker} {}^{t}F=\{0\}$) if and only if the range of$F$is dense in$\left(W,\sigma \left(W,Z,c\right)\right).$
• If in addition to${}^{t}F$being well-defined, the transpose of${}^{t}F$is also well-defined then${}^{tt}F=F.$
• Suppose$(U,V,a)$is a pairing over$\mathbb {K}$and$E:U\to X$is a linear map whose transpose${}^{t}E:Y\to V$is well-defined. Then the transpose of$F\circ E:U\to W,$which is${}^{t}(F\circ E):Z\to V,$is well-defined and${}^{t}(F\circ E)={}^{t}E\circ {}^{t}F.$
• If$F:X\to W$is a vector space isomorphism then${}^{t}F:Z\to Y$is bijective, the transpose of$F^{-1}:W\to X,$which is${}^{t}\left(F^{-1}\right):Y\to Z,$is well-defined, and${}^{t}\left(F^{-1}\right)=\left({}^{t}F\right)^{-1}$
• Let$S\subseteq X$and let$S^{\circ }$denotes theabsolute polarof$A,$then:
  1. $[F(S)]^{\circ }=\left({}^{t}F\right)^{-1}\left(S^{\circ }\right)$;
  2. if$F(S)\subseteq T$for some$T\subseteq W,$then${}^{t}F\left(T^{\circ }\right)\subseteq S^{\circ }$;
  3. if$T\subseteq W$is such that${}^{t}F\left(T^{\circ }\right)\subseteq S^{\circ },$then$F(S)\subseteq T^{\circ \circ }$;
  4. if$T\subseteq W$and$S\subseteq X$are weakly closed disks then${}^{t}F\left(T^{\circ }\right)\subseteq S^{\circ }$if and only if$F(S)\subseteq T$;
  5. $\operatorname {ker} {}^{t}F=[F(X)]^{\perp }.$

These results hold when thereal polaris used in place of the absolute polar.

If$X$and$Y$are normed spaces under their canonical dualities and if$F:X\to Y$is a continuous linear map, then$\|F\|=\left\|{}^{t}F\right\|.$

Weak continuity

A linear map$F:X\to W$isweakly continuous(with respect to$(X,Y,b)$and$(W,Z,c)$) if$F:(X,\sigma (X,Y,b))\to (W,(W,Z,c))$is continuous.

The following result shows that the existence of the transpose map is intimately tied to the weak topology.

Proposition—Assume that$X$distinguishes points of$Y$and$F:X\to W$is a linear map. Then the following are equivalent:

1. $F$is weakly continuous (that is,$F:(X,\sigma (X,Y,b))\to (W,(W,Z,c))$is continuous);
2. $c(F(\,\cdot \,),Z)\subseteq b(\,\cdot \,,Y)$;
3. the transpose of$F$is well-defined.

If$F$is weakly continuous then

• ${}^{t}F:Z\to Y$is weakly continuous, meaning that${}^{t}F:(Z,\sigma (Z,W,c))\to (Y,(Y,X,b))$is continuous;
• the transpose of${}^{t}F$is well-defined if and only if$Z$distinguishes points of$W,$in which case${}^{tt}F=F.$

Weak topology and the canonical duality

Suppose that$X$is a vector space and that$X^{\#}$is its the algebraic dual. Then every$\sigma \left(X,X^{\#}\right)$-bounded subset of$X$is contained in a finite dimensional vector subspace and every vector subspace of$X$is$\sigma \left(X,X^{\#}\right)$-closed.

Weak completeness

If$(X,\sigma (X,Y,b))$is acomplete topological vector spacesay that$X$is$\sigma (X,Y,b)$-completeor (if no ambiguity can arise)weakly-complete. There existBanach spacesthat are not weakly-complete (despite being complete in their norm topology).

If$X$is a vector space then under the canonical duality,$\left(X^{\#},\sigma \left(X^{\#},X\right)\right)$is complete. Conversely, if$Z$is a Hausdorfflocally convexTVS with continuous dual space$Z^{\prime },$then$\left(Z,\sigma \left(Z,Z^{\prime }\right)\right)$is complete if and only if$Z=\left(Z^{\prime }\right)^{\#}$;that is, if and only if the map$Z\to \left(Z^{\prime }\right)^{\#}$defined by sending$z\in Z$to the evaluation map at$z$(i.e.$z^{\prime }\mapsto z^{\prime }(z)$) is a bijection.

In particular, with respect to the canonical duality, if$Y$is a vector subspace of$X^{\#}$such that$Y$separates points of$X,$then$(Y,\sigma (Y,X))$is complete if and only if$Y=X^{\#}.$ Said differently, there doesnotexist a proper vector subspace$Y\neq X^{\#}$of$X^{\#}$such that$(X,\sigma (X,Y))$is Hausdorff and$Y$is complete in theweak-* topology(i.e. the topology of pointwise convergence). Consequently, when thecontinuous dual space$X^{\prime }$of aHausdorfflocally convexTVS$X$is endowed with theweak-* topology,then$X_{\sigma }^{\prime }$iscompleteif and only if$X^{\prime }=X^{\#}$(that is, if and only ifeverylinear functional on$X$is continuous).

Identification ofYwith a subspace of the algebraic dual

If$X$distinguishes points of$Y$and if$Z$denotes the range of the injection$y\mapsto b(\,\cdot \,,y)$then$Z$is a vector subspace of thealgebraic dual spaceof$X$and the pairing$(X,Y,b)$becomes canonically identified with the canonical pairing$\langle X,Z\rangle$(where$\left\langle x,x^{\prime }\right\rangle:=x^{\prime }(x)$is the natural evaluation map). In particular, in this situation it will be assumedwithout loss of generalitythat$Y$is a vector subspace of$X$'s algebraic dual and$b$is the evaluation map.

Convention:Often, whenever$y\mapsto b(\,\cdot \,,y)$is injective (especially when$(X,Y,b)$forms a dual pair) then it is common practice to assumewithout loss of generalitythat$Y$is a vector subspace of the algebraic dual space of$X,$that$b$is the natural evaluation map, and also denote$Y$by$X^{\prime }.$

In a completely analogous manner, if$Y$distinguishes points of$X$then it is possible for$X$to be identified as a vector subspace of$Y$'s algebraic dual space.

Algebraic adjoint

In the special case where the dualities are the canonical dualities$\left\langle X,X^{\#}\right\rangle$and$\left\langle W,W^{\#}\right\rangle,$the transpose of a linear map$F:X\to W$is always well-defined. This transpose is called thealgebraic adjointof$F$and it will be denoted by$F^{\#}$; that is,$F^{\#}={}^{t}F:W^{\#}\to X^{\#}.$ In this case, for all$w^{\prime }\in W^{\#},$ $F^{\#}\left(w^{\prime }\right)=w^{\prime }\circ F$where the defining condition for$F^{\#}\left(w^{\prime }\right)$is: $$\left\langle x,F^{\#}\left(w^{\prime }\right)\right\rangle =\left\langle F(x),w^{\prime }\right\rangle \quad {\text{ for all }}>x\in X,$$ or equivalently,$F^{\#}\left(w^{\prime }\right)(x)=w^{\prime }(F(x))\quad {\text{ for all }}x\in X.$

If$X=Y=\mathbb {K} ^{n}$for some integer$n,$${\mathcal {E}}=\left\{e_{1},\ldots,e_{n}\right\}$is a basis for$X$withdual basis${\mathcal {E}}^{\prime }=\left\{e_{1}^{\prime },\ldots,e_{n}^{\prime }\right\},$$F:\mathbb {K} ^{n}\to \mathbb {K} ^{n}$is a linear operator, and the matrix representation of$F$with respect to${\mathcal {E}}$is$M:=\left(f_{i,j}\right),$then the transpose of$M$is the matrix representation with respect to${\mathcal {E}}^{\prime }$of$F^{\#}.$

Weak continuity and openness

Suppose that$\left\langle X,Y\right\rangle$and$\langle W,Z\rangle$are canonical pairings (so$Y\subseteq X^{\#}$and$Z\subseteq W^{\#}$) that are dual systems and let$F:X\to W$be a linear map. Then$F:X\to W$is weakly continuous if and only if it satisfies any of the following equivalent conditions:

1. $F:(X,\sigma (X,Y))\to (W,\sigma (W,Z))$is continuous.
2. $F^{\#}(Z)\subseteq Y$
3. the transpose ofF,${}^{t}F:Z\to Y,$with respect to$\left\langle X,Y\right\rangle$and$\langle W,Z\rangle$is well-defined.

If$F$is weakly continuous then${}^{t}F::(Z,\sigma (Z,W))\to (Y,\sigma (Y,X))$will be continuous and furthermore,${}^{tt}F=F$

A map$g:A\to B$between topological spaces isrelatively openif$g:A\to \operatorname {Im} g$is anopen mapping,where$\operatorname {Im} g$is the range of$g.$

Suppose that$\langle X,Y\rangle$and$\langle W,Z\rangle$are dual systems and$F:X\to W$is a weakly continuous linear map. Then the following are equivalent:

1. $F:(X,\sigma (X,Y))\to (W,\sigma (W,Z))$is relatively open.
2. The range of${}^{t}F$is$\sigma (Y,X)$-closed in$Y$;
3. $\operatorname {Im} {}^{t}F=(\operatorname {ker} F)^{\perp }$

Furthermore,

• $F:X\to W$is injective (resp. bijective) if and only if${}^{t}F$is surjective (resp. bijective);
• $F:X\to W$is surjective if and only if${}^{t}F::(Z,\sigma (Z,W))\to (Y,\sigma (Y,X))$is relatively open and injective.

Transpose of a map between TVSs

The transpose of map between two TVSs is defined if and only if$F$is weakly continuous.

If$F:X\to Y$is a linear map between two Hausdorff locally convex topological vector spaces, then:

• If$F$is continuous then it is weakly continuous and${}^{t}F$is both Mackey continuous and strongly continuous.
• If$F$is weakly continuous then it is both Mackey continuous and strongly continuous (defined below).
• If$F$is weakly continuous then it is continuous if and only if${}^{t}F:^{\prime }\to X^{\prime }$mapsequicontinuoussubsets of$Y^{\prime }$to equicontinuous subsets of$X^{\prime }.$
• If$X$and$Y$are normed spaces then$F$is continuous if and only if it is weakly continuous, in which case$\|F\|=\left\|{}^{t}F\right\|.$
• If$F$is continuous then$F:X\to Y$is relatively open if and only if$F$is weakly relatively open (i.e.$F:\left(X,\sigma \left(X,X^{\prime }\right)\right)\to \left(Y,\sigma \left(Y,Y^{\prime }\right)\right)$is relatively open) and every equicontinuous subsets of$\operatorname {Im} {}^{t}F={}^{t}F\left(Y^{\prime }\right)$is the image of some equicontinuous subsets of$Y^{\prime }.$
• If$F$is continuous injection then$F:X\to Y$is a TVS-embedding (or equivalently, atopological embedding) if and only if every equicontinuous subsets of$X^{\prime }$is the image of some equicontinuous subsets of$Y^{\prime }.$

Metrizability and separability

Let$X$be alocally convexspace with continuous dual space$X^{\prime }$and let$K\subseteq X^{\prime }.$

1. If$K$isequicontinuousor$\sigma \left(X^{\prime },X\right)$-compact, and if$D\subseteq X^{\prime }$is such that$\operatorname {span} D$is dense in$X,$then the subspace topology that$K$inherits from$\left(X^{\prime },\sigma \left(X^{\prime },D\right)\right)$is identical to the subspace topology that$K$inherits from$\left(X^{\prime },\sigma \left(X^{\prime },X\right)\right).$
2. If$X$isseparableand$K$is equicontinuous then$K,$when endowed with the subspace topology induced by$\left(X^{\prime },\sigma \left(X^{\prime },X\right)\right),$ismetrizable.
3. If$X$is separable andmetrizable,then$\left(X^{\prime },\sigma \left(X^{\prime },X\right)\right)$is separable.
4. If$X$is a normed space then$X$is separable if and only if the closed unit call the continuous dual space of$X$is metrizable when given the subspace topology induced by$\left(X^{\prime },\sigma \left(X^{\prime },X\right)\right).$
5. If$X$is a normed space whose continuous dual space is separable (when given the usual norm topology), then$X$is separable.

Polar topologies and topologies compatible with pairing

Main article:Polar topology

Starting with only the weak topology, the use ofpolar setsproduces a range of locally convex topologies. Such topologies are calledpolar topologies. The weak topology is theweakest topologyof this range.

Throughout,$(X,Y,b)$will be a pairing over$\mathbb {K}$and${\mathcal {G}}$will be a non-empty collection of$\sigma (X,Y,b)$-bounded subsets of$X.$

Polar topologies

Main article:Polar topology

Given a collection${\mathcal {G}}$of subsets of$X$,thepolar topologyon$Y$determined by${\mathcal {G}}$(and$b$) or the${\mathcal {G}}$-topologyon$Y$is the uniquetopological vector space(TVS) topology on$Y$for which $$\left\{rG^{\circ }:G\in {\mathcal {G}},r>0\right\}$$ forms asubbasisof neighborhoods at the origin. When$Y$is endowed with this${\mathcal {G}}$-topology then it is denoted byY_{${\mathcal {G}}$}. Every polar topology is necessarilylocally convex. When${\mathcal {G}}$is adirected setwith respect to subset inclusion (i.e. if for all$G,K\in {\mathcal {G}}$there exists some$K\in {\mathcal {G}}$such that$G\cup H\subseteq K$) then this neighborhood subbasis at 0 actually forms aneighborhood basisat 0.

The following table lists some of the more important polar topologies.

Notation:If$\Delta (X,Y,b)$denotes a polar topology on$Y$then$Y$endowed with this topology will be denoted by$Y_{\Delta (Y,X,b)},$$Y_{\Delta (Y,X)}$or simply$Y_{\Delta }$(e.g. for$\sigma (Y,X,b)$we'd have$\Delta =\sigma$so that$Y_{\sigma (Y,X,b)},$$Y_{\sigma (Y,X)}$and$Y_{\sigma }$all denote$Y$endowed with$\sigma (X,Y,b)$).

${\mathcal {G}}\subseteq {\mathcal {P}}X$ ( "topology of uniform convergence on..." )	Notation	Name ( "topology of..." )	Alternative name
finite subsets of$X$ (or$\sigma (X,Y,b)$-closeddisked hullsof finite subsets of$X$)	$\sigma (X,Y,b)$ $s(X,Y,b)$	pointwise/simple convergence	weak/weak* topology
$\sigma (X,Y,b)$-compactdisks	$\tau (X,Y,b)$		Mackey topology
$\sigma (X,Y,b)$-compact convex subsets	$\gamma (X,Y,b)$	compact convex convergence
$\sigma (X,Y,b)$-compact subsets (or balanced$\sigma (X,Y,b)$-compact subsets)	$c(X,Y,b)$	compact convergence
$\sigma (X,Y,b)$-bounded subsets	$b(X,Y,b)$ $\beta (X,Y,b)$	bounded convergence	strong topology Strongest polar topology

Definitions involving polar topologies

Continuity

A linear map$F:X\to W$isMackey continuous(with respect to$(X,Y,b)$and$(W,Z,c)$) if$F:(X,\tau (X,Y,b))\to (W,\tau (W,Z,c))$is continuous.

A linear map$F:X\to W$isstrongly continuous(with respect to$(X,Y,b)$and$(W,Z,c)$) if$F:(X,\beta (X,Y,b))\to (W,\beta (W,Z,c))$is continuous.

Bounded subsets

A subset of$X$isweakly bounded(resp.Mackey bounded,strongly bounded) if it is bounded in$(X,\sigma (X,Y,b))$(resp. bounded in$(X,\tau (X,Y,b)),$bounded in$(X,\beta (X,Y,b))$).

Topologies compatible with a pair

If$(X,Y,b)$is a pairing over$\mathbb {K}$and${\mathcal {T}}$is a vector topology on$X$then${\mathcal {T}}$is atopology of the pairingand that it iscompatible(orconsistent)with the pairing$(X,Y,b)$if it islocally convexand if the continuous dual space of$\left(X,{\mathcal {T}}\right)=b(\,\cdot \,,Y).$ If$X$distinguishes points of$Y$then by identifying$Y$as a vector subspace of$X$'s algebraic dual, the defining condition becomes:$\left(X,{\mathcal {T}}\right)^{\prime }=Y.$ Some authors (e.g. [Trèves 2006] and [Schaefer 1999]) require that a topology of a pair also be Hausdorff,which it would have to be if$Y$distinguishes the points of$X$(which these authors assume).

The weak topology$\sigma (X,Y,b)$is compatible with the pairing$(X,Y,b)$(as was shown in the Weak representation theorem) and it is in fact the weakest such topology. There is a strongest topology compatible with this pairing and that is theMackey topology. If$N$is a normed space that is notreflexivethen the usual norm topology on its continuous dual space isnotcompatible with the duality$\left(N^{\prime },N\right).$

Mackey–Arens theorem

Main articles:Mackey–Arens theorem,Mackey topology,andMackey space

The following is one of the most important theorems in duality theory.

Mackey–Arens theoremI—Let$(X,Y,b)$will be a pairing such that$X$distinguishes the points of$Y$and let${\mathcal {T}}$be a locally convex topology on$X$(not necessarily Hausdorff). Then${\mathcal {T}}$is compatible with the pairing$(X,Y,b)$if and only if${\mathcal {T}}$is a polar topology determined by some collection${\mathcal {G}}$of$\sigma (Y,X,b)$-compactdisksthat cover$Y.$

It follows that the Mackey topology$\tau (X,Y,b),$which recall is the polar topology generated by all$\sigma (X,Y,b)$-compact disks in$Y,$is the strongest locally convex topology on$X$that is compatible with the pairing$(X,Y,b).$ A locally convex space whose given topology is identical to the Mackey topology is called aMackey space. The following consequence of the above Mackey-Arens theorem is also called the Mackey-Arens theorem.

Mackey–Arens theorem II—Let$(X,Y,b)$will be a pairing such that$X$distinguishes the points of$Y$and let${\mathcal {T}}$be a locally convex topology on$X.$ Then${\mathcal {T}}$is compatible with the pairing if and only if$\sigma (X,Y,b)\subseteq {\mathcal {T}}\subseteq \tau (X,Y,b).$

Mackey's theorem, barrels, and closed convex sets

If$X$is a TVS (over$\mathbb {R}$or$\mathbb {C}$) then ahalf-spaceis a set of the form$\{x\in X:f(x)\leq r\}$for some real$r$and some continuousreallinear functional$f$on$X.$

Theorem—If$X$is alocally convexspace (over$\mathbb {R}$or$\mathbb {C}$) and if$C$is a non-empty closed and convex subset of$X,$then$C$is equal to the intersection of all closed half spaces containing it.

The above theorem implies that the closed and convex subsets of a locally convex space dependentirelyon the continuous dual space. Consequently, the closed and convex subsets are the same in any topology compatible with duality;that is, if${\mathcal {T}}$and${\mathcal {L}}$are any locally convex topologies on$X$with the same continuous dual spaces, then a convex subset of$X$is closed in the${\mathcal {T}}$topology if and only if it is closed in the${\mathcal {L}}$topology. This implies that the${\mathcal {T}}$-closure of any convex subset of$X$is equal to its${\mathcal {L}}$-closure and that for any${\mathcal {T}}$-closeddisk$A$in$X,$$A=A^{\circ \circ }.$ In particular, if$B$is a subset of$X$then$B$is abarrelin$(X,{\mathcal {L}})$if and only if it is a barrel in$(X,{\mathcal {L}}).$

The following theorem shows thatbarrels(i.e. closedabsorbingdisks) are exactly the polars of weakly bounded subsets.

Theorem—Let$(X,Y,b)$will be a pairing such that$X$distinguishes the points of$Y$and let${\mathcal {T}}$be a topology of the pair. Then a subset of$X$is a barrel in$X$if and only if it is equal to the polar of some$\sigma (Y,X,b)$-bounded subset of$Y.$

If$X$is a topological vector space, then:

1. A closedabsorbingandbalancedsubset$B$of$X$absorbs each convex compact subset of$X$(i.e. there exists a real$r>0$such that$rB$contains that set).
2. If$X$is Hausdorff and locally convex then every barrel in$X$absorbs every convex bounded complete subset of$X.$

All of this leads to Mackey's theorem, which is one of the central theorems in the theory of dual systems. In short, it states the bounded subsets are the same for any two Hausdorff locally convex topologies that are compatible with the same duality.

Mackey's theorem—Suppose that$(X,{\mathcal {L}})$is a Hausdorff locally convex space with continuous dual space$X^{\prime }$and consider the canonical duality$\left\langle X,X^{\prime }\right\rangle.$ If${\mathcal {L}}$is any topology on$X$that is compatible with the duality$\left\langle X,X^{\prime }\right\rangle$on$X$then the bounded subsets of$(X,{\mathcal {L}})$are the same as the bounded subsets of$(X,{\mathcal {L}}).$

Space of finite sequences

Let$X$denote the space of all sequences of scalars$r_{\bullet }=\left(r_{i}\right)_{i=1}^{\infty }$such that$r_{i}=0$for all sufficiently large$i.$ Let$Y=X$and define a bilinear map$b:X\times X\to \mathbb {K}$by $$b\left(r_{\bullet },s_{\bullet }\right):=\sum _{i=1}^{\infty }r_{i}s_{i}.$$ Then$\sigma (X,X,b)=\tau (X,X,b).$ Moreover, a subset$T\subseteq X$is$\sigma (X,X,b)$-bounded (resp.$\beta (X,X,b)$-bounded) if and only if there exists a sequence$m_{\bullet }=\left(m_{i}\right)_{i=1}^{\infty }$of positive real numbers such that$\left|t_{i}\right|\leq m_{i}$for all$t_{\bullet }=\left(t_{i}\right)_{i=1}^{\infty }\in T$and all indices$i$(resp. and$m_{\bullet }\in X$).

It follows that there are weakly bounded (that is,$\sigma (X,X,b)$-bounded) subsets of$X$that are not strongly bounded (that is, not$\beta (X,X,b)$-bounded).

See also

• Beta-dual space
• Biorthogonal system
• Dual space– In mathematics, vector space of linear forms
• Dual topology
• Duality (mathematics)– General concept and operation in mathematics
• Inner product– Generalization of the dot product; used to define Hilbert spacesPages displaying short descriptions of redirect targets
• L-semi-inner product– Generalization of inner products that applies to all normed spaces
• Pairing– Bilinear map in mathematics
• Polar set– Subset of all points that is bounded by some given point of a dual (in a dual pairing)
• Polar topology– Dual space topology of uniform convergence on some sub-collection of bounded subsets
• Reductive dual pair
• Strong dual space– Continuous dual space endowed with the topology of uniform convergence on bounded sets
• Strong topology (polar topology)– Continuous dual space endowed with the topology of uniform convergence on bounded setsPages displaying short descriptions of redirect targets
• Topologies on spaces of linear maps
• Weak topology– Mathematical term

Bibliography

• Narici, Lawrence; Beckenstein, Edward (2011).Topological Vector Spaces.Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press.ISBN978-1584888666.OCLC144216834.
• Michael Reed and Barry Simon,Methods of Modern Mathematical Physics, Vol. 1, Functional Analysis,Section III.3. Academic Press, San Diego, 1980.ISBN0-12-585050-6.
• Rudin, Walter(1991).Functional Analysis.International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY:McGraw-Hill Science/Engineering/Math.ISBN978-0-07-054236-5.OCLC21163277.
• Schaefer, Helmut H.;Wolff, Manfred P. (1999).Topological Vector Spaces.GTM.Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer.ISBN978-1-4612-7155-0.OCLC840278135.
• Schmitt, Lothar M (1992)."An Equivariant Version of the Hahn–Banach Theorem".Houston J. Of Math.18:429–447.
• Trèves, François(2006) [1967].Topological Vector Spaces, Distributions and Kernels.Mineola, N.Y.: Dover Publications.ISBN978-0-486-45352-1.OCLC853623322.

External links

• Duality Theory