Strong Dual Topology

In functional analysis and related areas of mathematics, the strong dual space of a topological vector space (TVS) $X$ is the continuous dual space $X^$ 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 $b\left(X^, X\right)$ or $\beta\left(X^, X\right).$ 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, $X^,$ has the strong dual topology, $X^_b$ or $X^_$ may be written.

Strong dual topology

Throughout, all vector spaces will be assumed to be over the field $\mathbb$ of either the real numbers $\R$ or complex numbers $\C.$

Definition from a dual system

Let $(X, Y, \langle \cdot, \cdot \rangle)$ be a dual pair of vector spaces over the field $\mathbb$ of real numbers $\R$ or complex numbers $\C.$
For any $B \subseteq X$ and any $y \in Y,$ define
$, y, _B = \sup_, \langle x, y\rangle, .$

Neither $X$ nor $Y$ has a topology so say a subset $B \subseteq X$ is said to be if $, y, _B < \infty$ for all $y \in C.$
So a subset $B \subseteq X$ is called if and only if
$\sup_ , \langle x, y \rangle, < \infty \quad \text y \in Y.$
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 $\mathcal$ denote the family of all subsets $B \subseteq X$ bounded by elements of $Y$; that is, $\mathcal$ is the set of all subsets $B \subseteq X$ such that for every $y \in Y,$
$, y, _B = \sup_, \langle x, y\rangle, < \infty.$
Then the $\beta(Y, X, \langle \cdot, \cdot \rangle)$ 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 the locally convex topology on $Y$ generated by the seminorms of the form
$, y, _B = \sup_ , \langle x, y\rangle, ,\qquad y \in Y, \qquad B \in \mathcal.$

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

Definition on a TVS

Suppose that $X$ is a topological vector space (TVS) over the field $\mathbb.$
Let $\mathcal$ be any fundamental system of bounded sets of $X$;
that is, $\mathcal$ is a family of bounded subsets of $X$ such that every bounded subset of $X$ is a subset of some $B \in \mathcal$;
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 $X^$ is given by the polars:
$B^ := \left\$
as $B$ ranges over $\mathcal$).
This is a locally convex topology that is given by the set of seminorms on $X^$:
$\left, x^\_ := \sup_ \left, x^(x)\$
as $B$ ranges over $\mathcal.$

If $X$ is normable then so is $X^_$ and $X^_$ will in fact be a Banach space.
If $X$ is a normed space with norm $\, \cdot \,$ then $X^$ has a canonical norm (the operator norm) given by $\left\, x^ \right\, := \sup_ \left, x^(x) \$;
the topology that this norm induces on $X^$ is identical to the strong dual topology.

Bidual

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

Properties

Let $X$ be a locally convex TVS.

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

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

Examples

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

See also

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References

Bibliography

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{{DualityInLCTVSs

Functional analysis

Topology of function spaces