dual system

In mathematics, a dual system, dual pair, or duality over a field $\mathbb$ is a triple $(X, Y, b)$ consisting of two vector spaces $X$ and $Y$ over $\mathbb$ and a non-degenerate bilinear map $b : X \times Y \to \mathbb$.
Duality theory, the study of dual systems, is part of functional analysis.

According to Helmut H. Schaefer, "the study of a locally convex space in terms of its dual is the central part of the modern theory of topological vector spaces, for it provides the deepest and most beautiful results of the subject."

Definition, notation, and conventions

;Pairings

A or pair over a field $\mathbb$ is a triple $(X, Y, b),$ which may also be denoted by $b(X, Y),$ consisting of two vector spaces $X$ and $Y$ over $\mathbb$ (which this article assumes is either the real numbers or the complex numbers $\Complex$) and a bilinear map $b : X \times Y \to \mathbb,$ which is called the bilinear map associated with the pairing or simply the pairing's map/bilinear form.

For every $x \in X,$ define
$\begin b(x, \,\cdot\,) : \,& Y && \to &&\, \mathbb \\ & y && \mapsto &&\, b(x, y) \end$
and for every $y \in Y,$ define
$\begin b(\,\cdot\,, y) : \,& X && \to &&\, \mathbb \\ & x && \mapsto &&\, b(x, y). \end$
Every $b(x, \,\cdot\,)$ is a linear functional on $Y$ and every $b(\,\cdot\,, y)$ is a linear functional on $X.$
Let
$b(X, \,\cdot\,) := \ \qquad \text \qquad b(\,\cdot\,, Y) := \$
where each of these sets forms a vector space of linear functionals.

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

;Dual pairings

A pairing $(X, Y, b)$ is called a , a , or a over $\mathbb$ if the bilinear form $b$ is non-degenerate, which means that it satisfies the following two separation axioms:
# $Y$ 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 $b(x, \,\cdot\,) : Y \to \mathbb$ is not identically $0$ (i.e. there exists a $y \in Y$ such that $b(x, y) \neq 0$);
# $X$ 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 $b(\,\cdot\,, y) : X \to \mathbb$ is not identically $0$ (i.e. there exists an $x \in X$ such that $b(x, y) \neq 0$).
In this case say that $b$ is non-degenerate, say that $b$ places $X$ and $Y$ in duality (or in separated duality), and $b$ is called the duality pairing of the $(X, Y, b).$

;Total subsets

A subset $S$ of $Y$ is called if for every $x \in X,$
$b(x, s) = 0 \quad \text s \in S$ implies $x = 0.$
A total subset of $X$ is defined analogously (see footnote).A subset $S$ of $X$ is total if for all $y \in Y,$
$b(s, y) = 0 \quad \text s \in S$ implies $y = 0.$

;Orthogonality

The vectors $x$ and $y$ are called , written $x \perp y,$ if $b(x, y) = 0.$
Two subsets $R \subseteq X$ and $S \subseteq Y$ are orthogonal, written $R \perp S,$ if $b(R, S) = \$; 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 defined analogously.

The orthogonal complement or annihilator of a subset $R \subseteq X$ is
$R^ := \ := \.$

Polar sets

Throughout, $(X, Y, b)$ will be a pairing over $\mathbb.$
The absolute polar or polar of a subset $A$ of $X$ is the set:
$A^ := \left\.$

Dually, the absolute polar or polar of a subset $B$ of $Y$ is denoted by $B^$ and defined by
$B^ := \left\$

In this case, the absolute polar of a subset $B$ of $Y$ is also called the absolute prepolar or prepolar of $B$ and may be denoted by $^ B.$

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

If $A \subseteq X$ then the bipolar of $A,$ denoted by $A^,$ is the set $^\left(A^\right).$
Similarly, if $B \subseteq Y$ then the bipolar of $B$ is $B^ := \left(^B\right)^.$

If $A$ is a vector subspace of $X,$ then $A^ = A^$ and this is also equal to the real polar of $A.$

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 $x \in X \quad \text y \in Y.$

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

:: Given any definition for a pairing $(X, Y, b),$ one obtains a by applying it to the pairing $(Y, X, d).$ This conventions also apply to theorems.

:: Adhering to common practice, unless clarity is needed, whenever a definition (or result) for a pairing $(X, Y, b)$ is given then this article will omit mention of the corresponding dual definition (or result) but nevertheless use it.

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 it allows us to avoid having to assign a symbol to $d.$

:: If a definition and its notation for a pairing $(X, Y, b)$ depends on the order of $X$ and $Y$ (e.g. the definition of the Mackey topology $\tau(X, Y, b)$ on $X$) then by switching the order of $X$ and $Y,$ then it is meant that definition applied to $(Y, X, d)$ (e.g. $\tau(Y, X, b)$ actually denotes the topology $\tau(Y, X, d)$).

For instance, once the weak topology on $X$ is defined, which is denoted by $\sigma(X, Y, b),$ then this definition will automatically be applied to the pairing $(Y, X, d)$ so as to obtain the definition of the weak topology on $Y,$ where this topology will 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 also adhere to the following 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 the restriction of $(X, Y, b)$ to $M \times N$ is the pairing $\left(M, N, b\big\vert_\right).$
If $(X, Y, b)$ is a duality then it's possible for a restrictions to fail to be a duality (e.g. if $Y \neq \$ and $N = \$).

This article will use the common practice of denoting the restriction $\left(M, N, b\big\vert_\right)$ by $(M, N, b).$

Canonical duality on a vector space

Suppose that $X$ is a vector space and let $X^$ denote the algebraic dual space of $X$ (that is, the space of all linear functionals on $X$).
There is a canonical duality $\left(X, X^, c\right)$ where $c\left(x, x^\right) = \left\langle x, x^ \right\rangle = x^(x),$ which is called the evaluation map or the natural or canonical bilinear functional on $X \times X^.$
Note in particular that for any $x^ \in X^,$ $c\left(\,\cdot\,, x^\right)$ is just another way of denoting $x^$; i.e. $c\left(\,\cdot\,, x^\right) = x^(\,\cdot\,) = x^.$

If $N$ is a vector subspace of $X^$ then the restriction of $\left(X, X^, c\right)$ to $X \times N$ is called the canonical pairing where if this pairing is a duality then it is instead called the canonical 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 $\left\langle x, x^ \right\rangle = x^(x)$ (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 a topological vector space (TVS) with continuous dual space $X^.$
Then the restriction of the canonical duality $\left(X, X^, c\right)$ to $X$ × $X^$ defines a pairing $\left(X, X^, c\big\vert_\right)$ for which $X$ separates points of $X^.$
If $X^$ 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 $\left\langle X, X^ \right\rangle.$

;Polars and duals of TVSs

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

Inner product spaces and complex conjugate spaces

A pre-Hilbert space $(H, \langle \cdot, \cdot \rangle)$ is a dual pairing if and only if $H$ is vector space over $\R$ or $H$ has dimension $0.$ Here it is assumed that the sesquilinear form $\langle \cdot, \cdot \rangle$ is conjugate homogeneous in its second coordinate and homogeneous in its first coordinate.

• If $(H, \langle \cdot, \cdot \rangle)$ is a Hilbert space then $(H, H, \langle \cdot, \cdot \rangle)$ forms a dual system.


• If