Strong dual space

 ( Functional Analysis ) 

In functional analysis and related areas of mathematics, the strong dual space of a topological vector space (TVS) $X$ is the continuous dual space $X^\{\backslash prime\}$ of $X$ equipped with the strong ( dual) topology or the topology of uniform convergence on bounded subsets of $X,$ where this topology is denoted by $b\backslash left(X^\{\backslash prime\},\; X\backslash right)$ or $\backslash beta\backslash left(X^\{\backslash prime\},\; X\backslash right).$ The coarsest polar topology is called weak 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, $X^\{\backslash prime\},$ has the strong dual topology, $X^\{\backslash prime\}\_b$ or $X^\{\backslash prime\}\_\{\backslash beta\}$ may be written.

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Strong dual topology Throughout, all vector spaces will be assumed to be over the field $\backslash mathbb\{F\}$ of either the $\backslash R$ or $\backslash C.$

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Definition from a dual system Let $(X,\; Y,\; \backslash langle\; \backslash cdot,\; \backslash cdot\; \backslash rangle)$ be a dual pair of vector spaces over the field $\backslash mathbb\{F\}$ of $\backslash R$ or $\backslash C.$ For any $B\; \backslash subseteq\; X$ and any $y\; \backslash in\; Y,$ define $$|y|\_B\; =\; \backslash sup\_\{x\; \backslash in\; B\}|\backslash langle\; x,\; y\backslash rangle|.$$

Neither $X$ nor $Y$ has a topology so say a subset $B\; \backslash subseteq\; X$ is said to be if for all $y\; \backslash in\; C.$ So a subset $B\; \backslash subseteq\; X$ is called if and only if This is equivalent to the usual notion of bounded subsets when $X$ is given the weak topology induced by $Y,$ which is a Hausdorff locally convex topology.

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

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 space Separating set on $X,$ then $X$ is part of a canonical dual system $\backslash left(X,\; X^\{\backslash prime\},\; \backslash langle\; \backslash cdot\; ,\; \backslash cdot\; \backslash rangle\backslash right)$ where $\backslash left\backslash langle\; x,\; x^\{\backslash prime\}\; \backslash right\backslash rangle\; :=\; x^\{\backslash prime\}(x).$ In the special case when $X$ is a locally convex space, the on the (continuous) dual space $X^\{\backslash prime\}$ (that is, on the space of all continuous linear functionals $f\; :\; X\; \backslash to\; \backslash mathbb\{F\}$) is defined as the strong topology $\backslash beta\backslash left(X^\{\backslash prime\},\; X\backslash right),$ and it coincides with the topology of uniform convergence on in $X,$ i.e. with the topology on $X^\{\backslash prime\}$ generated by the seminorms of the form $$|f|\_B\; =\; \backslash sup\_\{x\; \backslash in\; B\}\; |f(x)|,\; \backslash qquad\; \backslash text\{\; where\; \}\; f\; \backslash in\; X^\{\backslash prime\},$$ where $B$ runs over the family of all bounded sets in $X.$ The space $X^\{\backslash prime\}$ with this topology is called of the space $X$ and is denoted by $X^\{\backslash prime\}\_\{\backslash beta\}.$

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Definition on a TVS Suppose that $X$ is a topological vector space (TVS) over the field $\backslash mathbb\{F\}.$ Let $\backslash mathcal\{B\}$ be any fundamental system of bounded sets of $X$; that is, $\backslash mathcal\{B\}$ is a family of bounded subsets of $X$ such that every bounded subset of $X$ is a subset of some $B\; \backslash in\; \backslash mathcal\{B\}$; 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 $X^\{\backslash prime\}$ is given by the polar set: $$B^\{\backslash circ\}\; :=\; \backslash left\backslash \{\; x^\{\backslash prime\}\; \backslash in\; X^\{\backslash prime\}\; :\; \backslash sup\_\{x\; \backslash in\; B\}\; \backslash left|x^\{\backslash prime\}(x)\backslash right|\; \backslash leq\; 1\; \backslash right\backslash \}$$ as $B$ ranges over $\backslash mathcal\{B\}$). This is a locally convex topology that is given by the set of on $X^\{\backslash prime\}$: $\backslash left|x^\{\backslash prime\}\backslash right|\_\{B\}\; :=\; \backslash sup\_\{x\; \backslash in\; B\}\; \backslash left|x^\{\backslash prime\}(x)\backslash right|$ as $B$ ranges over $\backslash mathcal\{B\}.$

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

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Bidual The bidual or second dual of a TVS $X,$ often denoted by $X^\{\backslash prime\; \backslash prime\},$ is the strong dual of the strong dual of $X$: where the vector space $X^\{\backslash prime\; \backslash prime\}$ is endowed with the strong dual topology $b\backslash left(X^\{\backslash prime\backslash prime\},\; X^\{\backslash prime\}\_b\backslash right).$

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Properties Let $X$ be a locally convex TVS.

• A convex balanced set weakly compact subset of $X^\{\backslash prime\}$ is bounded in $X^\{\backslash prime\}\_b.$
• Every weakly bounded subset of $X^\{\backslash prime\}$ is strongly bounded.
• If $X$ is a barreled space then $X$'s topology is identical to the strong dual topology $b\backslash left(X,\; X^\{\backslash prime\}\backslash right)$ and to the Mackey topology on $X.$
• If $X$ is a metrizable locally convex space, then the strong dual of $X$ is a bornological space if and only if it is an infrabarreled space, if and only if it is a barreled space.
• If $X$ is Hausdorff locally convex TVS then $\backslash left(X,\; b\backslash left(X,\; X^\{\backslash prime\}\backslash right)\backslash right)$ is metrizable if and only if there exists a countable set $\backslash mathcal\{B\}$ of bounded subsets of $X$ such that every bounded subset of $X$ is contained in some element of $\backslash mathcal\{B\}.$
• If $X$ is locally convex, then this topology is finer than all other $\backslash mathcal\{G\}$-topologies on $X^\{\backslash prime\}$ when considering only $\backslash mathcal\{G\}$'s whose sets are subsets of $X.$
• If $X$ is a bornological space (e.g. metrizable or LF-space) then $X^\{\backslash prime\}\_\{b(X^\{\backslash prime\},\; X)\}$ is complete.

If $X$ is a barrelled space, then its topology coincides with the strong topology $\backslash beta\backslash left(X,\; X^\{\backslash prime\}\backslash right)$ on $X$ and with the Mackey topology on generated by the pairing $\backslash left(X,\; X^\{\backslash prime\}\backslash right).$

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Examples If $X$ is a normed vector space, then its Continuous dual $X^\{\backslash prime\}$ with the strong topology coincides with the Banach dual space $X^\{\backslash prime\}$; that is, with the space $X^\{\backslash prime\}$ with the topology induced by the operator norm. Conversely $\backslash left(X,\; X^\{\backslash prime\}\backslash right).$-topology on $X$ is identical to the topology induced by the norm on $X.$

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See also

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Bibliography

• Wong (1979). 9783540095132, Springer-Verlag. ISBN 9783540095132

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