Strong dual space

Infunctional analysisand related areas ofmathematics,thestrong dual spaceof atopological vector space(TVS)$X$is thecontinuous dual space$X^{\prime }$of$X$equipped with thestrong(dual)topologyor thetopology of uniform convergence on bounded subsets of$X,$where this topology is denoted by$b\left(X^{\prime },X\right)$or$\beta \left(X^{\prime },X\right).$Thecoarsestpolar topology is calledweak topology. The strong dual space plays such an important role in modern functional analysis, that the continuous dual space is usually assumed to have the strong dual topology unless indicated otherwise. To emphasize that the continuous dual space,$X^{\prime },$has the strong dual topology,$X_{b}^{\prime }$or$X_{\beta }^{\prime }$may be written.

Strong dual topology

Throughout, all vector spaces will be assumed to be over the field$\mathbb {F}$of either thereal numbers$\mathbb {R}$orcomplex numbers$\mathbb {C}.$

Definition from a dual system

Main article:Dual system

Let$(X,Y,\langle \cdot,\cdot \rangle )$be adual pairof vector spaces over the field$\mathbb {F}$ofreal numbers$\mathbb {R}$orcomplex numbers$\mathbb {C}.$ For any$B\subseteq X$and any$y\in Y,$define $$|y|_{B}=\sup _{x\in B}|\langle x,y\rangle |.$$

Neither$X$nor$Y$has a topology so say a subset$B\subseteq X$is said to bebounded by a subset$C\subseteq Y$if$|y|_{B}<\infty$for all$y\in C.$ So a subset$B\subseteq X$is calledboundedif and only if $$\sup _{x\in B}|\langle x,y\rangle |<\infty \quad {\text{ for all }}y\in Y.$$ This is equivalent to the usual notion ofbounded subsetswhen$X$is given the weak topology induced by$Y,$which is a Hausdorfflocally convextopology.

Let${\mathcal {B}}$denote thefamilyof all subsets$B\subseteq X$bounded by elements of$Y$;that is,${\mathcal {B}}$is the set of all subsets$B\subseteq X$such that for every$y\in Y,$ $$|y|_{B}=\sup _{x\in B}|\langle x,y\rangle |<\infty.$$ Then thestrong topology$\beta (Y,X,\langle \cdot,\cdot \rangle )$on$Y,$also denoted by$b(Y,X,\langle \cdot,\cdot \rangle )$or simply$\beta (Y,X)$or$b(Y,X)$if the pairing$\langle \cdot,\cdot \rangle$is understood, is defined as thelocally convextopology on$Y$generated by the seminorms of the form $$|y|_{B}=\sup _{x\in B}|\langle x,y\rangle |,\qquad y\in Y,\qquad B\in {\mathcal {B}}.$$

The definition of the strong dual topology now proceeds as in the case of a TVS. Note that if$X$is a TVS whose continuous dual spaceseparates pointon$X,$then$X$is part of a canonical dual system$\left(X,X^{\prime },\langle \cdot,\cdot \rangle \right)$ where$\left\langle x,x^{\prime }\right\rangle:=x^{\prime }(x).$ In the special case when$X$is alocally convex space,thestrong topologyon the (continuous)dual space$X^{\prime }$(that is, on the space of all continuous linear functionals$f:X\to \mathbb {F}$) is defined as the strong topology$\beta \left(X^{\prime },X\right),$and it coincides with the topology of uniform convergence onbounded setsin$X,$i.e. with the topology on$X^{\prime }$generated by the seminorms of the form $$|f|_{B}=\sup _{x\in B}|f(x)|,\qquad {\text{ where }}f\in X^{\prime },$$ where$B$runs over the family of allbounded setsin$X.$ The space$X^{\prime }$with this topology is calledstrong dual spaceof the space$X$and is denoted by$X_{\beta }^{\prime }.$

Definition on a TVS

Suppose that$X$is atopological vector space(TVS) over the field$\mathbb {F}.$ Let${\mathcal {B}}$be any fundamental system ofbounded setsof$X$; that is,${\mathcal {B}}$is afamilyof bounded subsets of$X$such that every bounded subset of$X$is a subset of some$B\in {\mathcal {B}}$; the set of all bounded subsets of$X$forms a fundamental system of bounded sets of$X.$ A basis of closed neighborhoods of the origin in$X^{\prime }$is given by thepolars: $$B^{\circ }:=\left\{x^{\prime }\in X^{\prime }:\sup _{x\in B}\left|x^{\prime }(x)\right|\leq 1\right\}$$ as$B$ranges over${\mathcal {B}}$). This is a locally convex topology that is given by the set ofseminormson$X^{\prime }$: $\left|x^{\prime }\right|_{B}:=\sup _{x\in B}\left|x^{\prime }(x)\right|$ as$B$ranges over${\mathcal {B}}.$

If$X$isnormablethen so is$X_{b}^{\prime }$and$X_{b}^{\prime }$will in fact be aBanach space. If$X$is a normed space with norm$\|\cdot \|$then$X^{\prime }$has a canonical norm (theoperator norm) given by$\left\|x^{\prime }\right\|:=\sup _{\|x\|\leq 1}\left|x^{\prime }(x)\right|$; the topology that this norm induces on$X^{\prime }$is identical to the strong dual topology.

Bidual

See also:Banach space § Bidual,Reflexive space,Semi-reflexive space,andDouble dual

Thebidualorsecond dualof a TVS$X,$often denoted by$X^{\prime \prime },$is the strong dual of the strong dual of$X$: $$X^{\prime \prime }\,:=\,\left(X_{b}^{\prime }\right)^{\prime }$$ where$X_{b}^{\prime }$denotes$X^{\prime }$endowed with the strong dual topology$b\left(X^{\prime },X\right).$ Unless indicated otherwise, the vector space$X^{\prime \prime }$is usually assumed to be endowed with the strong dual topology induced on it by$X_{b}^{\prime },$in which case it is called thestrong bidualof$X$;that is, $$X^{\prime \prime }\,:=\,\left(X_{b}^{\prime }\right)_{b}^{\prime }$$ where the vector space$X^{\prime \prime }$is endowed with the strong dual topology$b\left(X^{\prime \prime },X_{b}^{\prime }\right).$

Properties

Let$X$be alocally convexTVS.

• A convexbalancedweakly compact subset of$X^{\prime }$is bounded in$X_{b}^{\prime }.$
• Every weakly bounded subset of$X^{\prime }$is strongly bounded.
• If$X$is abarreled spacethen$X$'s topology is identical to the strong dual topology$b\left(X,X^{\prime }\right)$and to theMackey topologyon$X.$
• If$X$is a metrizable locally convex space, then the strong dual of$X$is abornological spaceif and only if it is aninfrabarreled space,if and only if it is abarreled space.
• If$X$is Hausdorff locally convex TVS then$\left(X,b\left(X,X^{\prime }\right)\right)$ismetrizableif and only if there exists a countable set${\mathcal {B}}$of bounded subsets of$X$such that every bounded subset of$X$is contained in some element of${\mathcal {B}}.$
• If$X$is locally convex, then this topology is finer than all other${\mathcal {G}}$-topologieson$X^{\prime }$when considering only${\mathcal {G}}$'s whose sets are subsets of$X.$
• If$X$is abornological space(e.g.metrizableorLF-space) then$X_{b(X^{\prime },X)}^{\prime }$iscomplete.

If$X$is abarrelled space,then its topology coincides with the strong topology$\beta \left(X,X^{\prime }\right)$on$X$and with theMackey topologyon generated by the pairing$\left(X,X^{\prime }\right).$

Examples

If$X$is anormed vector space,then its(continuous) dual space$X^{\prime }$with the strong topology coincides with theBanach dual space$X^{\prime }$;that is, with the space$X^{\prime }$with the topology induced by theoperator norm.Conversely$\left(X,X^{\prime }\right).$-topology on$X$is identical to the topology induced by thenormon$X.$

See also

• Dual topology
• Dual system
• List of topologies– List of concrete topologies and topological spaces
• Polar topology– Dual space topology of uniform convergence on some sub-collection of bounded subsets
• Reflexive space– Locally convex topological vector space
• Semi-reflexive space
• Strong topology
• Topologies on spaces of linear maps

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