Pairings

A pairing or pair over a field ${\displaystyle \mathbb {K} }$ is a triple ${\displaystyle (X,Y,b),}$ which may also be denoted by ${\displaystyle b(X,Y),}$ consisting of two vector spaces ${\displaystyle X}$ and ${\displaystyle Y}$ over ${\displaystyle \mathbb {K} }$ and a bilinear map ${\displaystyle b:X\times Y\to \mathbb {K} }$, called the bilinear map associated with the pairing^[1] or simply the pairing's map or its bilinear form. For simplicity, this article only covers examples where ${\displaystyle \mathbb {K} }$ is either the real numbers ${\displaystyle \mathbb {R} }$ or the complex numbers ${\displaystyle \mathbb {C} }$.

For every ${\displaystyle x\in X}$, define

$${\displaystyle {\begin{alignedat}{4}b(x,\,\cdot \,):\,&Y&&\to &&\,\mathbb {K} \\&y&&\mapsto &&\,b(x,y)\end{alignedat}}}$$

and for every ${\displaystyle y\in Y,}$ define

$${\displaystyle {\begin{alignedat}{4}b(\,\cdot \,,y):\,&X&&\to &&\,\mathbb {K} \\&x&&\mapsto &&\,b(x,y).\end{alignedat}}}$$

Every ${\displaystyle b(x,\,\cdot \,)}$ is a linear functional on ${\displaystyle Y}$ and every ${\displaystyle b(\,\cdot \,,y)}$ is a linear functional on ${\displaystyle X}$ . Let

$${\displaystyle b(X,\,\cdot \,):=\{b(x,\,\cdot \,):x\in X\}\qquad {\text{ and }}\qquad b(\,\cdot \,,Y):=\{b(\,\cdot \,,y):y\in Y\}}$$

where each of these sets forms a vector space of linear functionals.

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

Dual pairings

A pairing ${\displaystyle (X,Y,b)}$ is called a dual system, a dual pair,^[2] or a duality over ${\displaystyle \mathbb {K} }$ if the bilinear form ${\displaystyle b}$ is non-degenerate, which means that it satisfies the following two separation axioms:

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

In this case ${\displaystyle b}$ is non-degenerate, and one can say that ${\displaystyle b}$ places ${\displaystyle X}$ and ${\displaystyle Y}$ in duality (or, redundantly but explicitly, in separated duality), and ${\displaystyle b}$ is called the duality pairing of the triple ${\displaystyle (X,Y,b)}$.^[1][2]

Total subsets

A subset ${\displaystyle S}$ of ${\displaystyle Y}$ is called total if for every ${\displaystyle x\in X}$,

$${\displaystyle b(x,s)=0\quad {\text{ for all }}s\in S}$$

implies ${\displaystyle x=0.}$

A total subset of ${\displaystyle X}$ is defined analogously (see footnote).^{[note 1]} Thus ${\displaystyle X}$ separates points of ${\displaystyle Y}$ if and only if ${\displaystyle X}$ is a total subset of ${\displaystyle X}$, and similarly for ${\displaystyle Y}$.

Orthogonality

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

The orthogonal complement or annihilator of a subset ${\displaystyle R\subseteq X}$ is

$${\displaystyle R^{\perp }:=\{y\in Y:R\perp y\}:=\{y\in Y:b(R,y)=\{0\}\}}$$

.

Thus ${\displaystyle R}$ is a total subset of ${\displaystyle X}$ if and only if ${\displaystyle R^{\perp }}$ equals ${\displaystyle \{0\}}$.

Polar sets

Given a triple ${\displaystyle (X,Y,b)}$ defining a pairing over ${\displaystyle \mathbb {K} }$, the absolute polar set or polar set of a subset ${\displaystyle A}$ of ${\displaystyle X}$ is the set:

$${\displaystyle A^{\circ }:=\left\{y\in Y:\sup _{x\in A}|b(x,y)|\leq 1\right\}.}$$

Symmetrically, the absolute polar set or polar set of a subset ${\displaystyle B}$ of ${\displaystyle Y}$ is denoted by ${\displaystyle B^{\circ }}$ and defined by

$${\displaystyle B^{\circ }:=\left\{x\in X:\sup _{y\in B}|b(x,y)|\leq 1\right\}}$$

To use bookkeeping that helps keep track of the antisymmetry of the two sides of the duality, the absolute polar of a subset ${\displaystyle B}$ of ${\displaystyle Y}$ may also be called the absolute prepolar or prepolar of ${\displaystyle B}$ and then may be denoted by ${\displaystyle {}^{\circ }B.}$

^[3]

The polar ${\displaystyle B^{\circ }}$ is necessarily a convex set containing ${\displaystyle 0\in Y}$ where if ${\displaystyle B}$ is balanced then so is ${\displaystyle B^{\circ }}$ and if ${\displaystyle B}$ is a vector subspace of ${\displaystyle X}$ then so too is ${\displaystyle B^{\circ }}$ a vector subspace of ${\displaystyle Y.}$^[4]

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

Dual definitions and results

Given a pairing ${\displaystyle (X,Y,b),}$ define a new pairing ${\displaystyle (Y,X,d)}$ where ${\displaystyle d(y,x):=b(x,y)}$ for all ${\displaystyle x\in X}$ and ${\displaystyle y\in Y}$.^[1]

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

Convention and Definition: Given any definition for a pairing ${\displaystyle (X,Y,b),}$ one obtains a dual definition by applying it to the pairing ${\displaystyle (Y,X,d).}$ This conventions also apply to theorems.

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

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

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

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

Restriction of a pairing

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

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

Canonical duality on a vector space

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

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

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

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

If ${\displaystyle N}$ is a vector subspace of ${\displaystyle X^{\#}}$ then ${\displaystyle X}$ distinguishes points of ${\displaystyle N}$ (or equivalently, ${\displaystyle (X,N,c)}$ is a duality) if and only if ${\displaystyle N}$ distinguishes points of ${\displaystyle X,}$ or equivalently if ${\displaystyle N}$ is total (that is, ${\displaystyle n(x)=0}$ for all ${\displaystyle n\in N}$ implies ${\displaystyle x=0}$).^[1]

Canonical duality on a topological vector space

Suppose ${\displaystyle X}$ is a topological vector space (TVS) with continuous dual space ${\displaystyle X^{\prime }.}$ Then the restriction of the canonical duality ${\displaystyle \left(X,X^{\#},c\right)}$ to ${\displaystyle X}$ × ${\displaystyle X^{\prime }}$ defines a pairing ${\displaystyle \left(X,X^{\prime },c{\big \vert }_{X\times X^{\prime }}\right)}$ for which ${\displaystyle X}$ separates points of ${\displaystyle X^{\prime }.}$ If ${\displaystyle X^{\prime }}$ separates points of ${\displaystyle X}$ (which is true if, for instance, ${\displaystyle X}$ is a Hausdorff locally convex space) then this pairing forms a duality.^[2]

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

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

• If ${\displaystyle (H,\langle \cdot ,\cdot \rangle )}$ is a real Hilbert space then ${\displaystyle (H,H,\langle \cdot ,\cdot \rangle )}$ forms a dual system.
• If ${\displaystyle (H,\langle \cdot ,\cdot \rangle )}$ is a complex Hilbert space then ${\displaystyle (H,H,\langle \cdot ,\cdot \rangle )}$ forms a dual system if and only if ${\displaystyle \operatorname {dim} H=0.}$ If ${\displaystyle H}$ is non-trivial then ${\displaystyle (H,H,\langle \cdot ,\cdot \rangle )}$ does not even form pairing since the inner product is sesquilinear rather than bilinear.^[1]

Suppose that ${\displaystyle (H,\langle \cdot ,\cdot \rangle )}$ is a complex pre-Hilbert space with scalar multiplication denoted as usual by juxtaposition or by a dot ${\displaystyle \cdot .}$ Define the map

$${\displaystyle \,\cdot \,\perp \,\cdot \,:\mathbb {C} \times H\to H\quad {\text{ by }}\quad c\perp x:={\overline {c}}x,}$$

where the right-hand side uses the scalar multiplication of ${\displaystyle H.}$ Let ${\displaystyle {\overline {H}}}$ denote the complex conjugate vector space of ${\displaystyle H,}$ where ${\displaystyle {\overline {H}}}$ denotes the additive group of ${\displaystyle (H,+)}$ (so vector addition in ${\displaystyle {\overline {H}}}$ is identical to vector addition in ${\displaystyle H}$) but with scalar multiplication in ${\displaystyle {\overline {H}}}$ being the map ${\displaystyle \,\cdot \,\perp \,\cdot \,}$ (instead of the scalar multiplication that ${\displaystyle H}$ is endowed with).

The map ${\displaystyle b:H\times {\overline {H}}\to \mathbb {C} }$ defined by ${\displaystyle b(x,y):=\langle x,y\rangle }$ is linear in both coordinates^{[note 2]} and so ${\displaystyle \left(H,{\overline {H}},\langle \cdot ,\cdot \rangle \right)}$ forms a dual pairing.

Other examples

• Suppose ${\displaystyle X=\mathbb {R} ^{2},}$ ${\displaystyle Y=\mathbb {R} ^{3},}$ and for all ${\displaystyle \left(x_{1},y_{1}\right)\in X{\text{ and }}\left(x_{2},y_{2},z_{2}\right)\in Y,}$ let
  $${\displaystyle b\left(\left(x_{1},y_{1}\right),\left(x_{2},y_{2},z_{2}\right)\right):=x_{1}x_{2}+y_{1}y_{2}.}$$

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

Bounded subsets

A subset ${\displaystyle S}$ of ${\displaystyle X}$ is ${\displaystyle \sigma (X,Y,b)}$-bounded if and only if

$${\displaystyle \sup _{}|b(S,y)|<\infty \quad {\text{ for all }}y\in Y,}$$

where ${\displaystyle |b(S,y)|:=\{b(s,y):s\in S\}.}$

Hausdorffness

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

1. ${\displaystyle X}$ distinguishes points of ${\displaystyle Y}$;
2. The map ${\displaystyle y\mapsto b(\,\cdot \,,y)}$ defines an injection from ${\displaystyle Y}$ into the algebraic dual space of ${\displaystyle X}$;^[1]
3. ${\displaystyle \sigma (Y,X,b)}$ is Hausdorff.^[1]

Weak representation theorem

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

Weak representation theorem^[1] — Let ${\displaystyle (X,Y,b)}$ be a pairing over the field ${\displaystyle \mathbb {K} .}$ Then the continuous dual space of ${\displaystyle (X,\sigma (X,Y,b))}$ is

$${\displaystyle b(\,\cdot \,,Y):=\{b(\,\cdot \,,y):y\in Y\}.}$$

Furthermore,

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

Consequently, the continuous dual space of ${\displaystyle (X,\sigma (X,Y,b))}$ is

$${\displaystyle (X,\sigma (X,Y,b))^{\prime }=b(\,\cdot \,,Y):=\left\{b(\,\cdot \,,y):y\in Y\right\}.}$$

With respect to the canonical pairing, if ${\displaystyle X}$ is a TVS whose continuous dual space ${\displaystyle X^{\prime }}$ separates points on ${\displaystyle X}$ (i.e. such that ${\displaystyle \left(X,\sigma \left(X,X^{\prime }\right)\right)}$ is Hausdorff, which implies that ${\displaystyle X}$ is also necessarily Hausdorff) then the continuous dual space of ${\displaystyle \left(X^{\prime },\sigma \left(X^{\prime },X\right)\right)}$ is equal to the set of all "evaluation at a point ${\displaystyle x}$" maps as ${\displaystyle x}$ ranges over ${\displaystyle X}$ (i.e. the map that send ${\displaystyle x^{\prime }\in X^{\prime }}$ to ${\displaystyle x^{\prime }(x)}$). This is commonly written as

$${\displaystyle \left(X^{\prime },\sigma \left(X^{\prime },X\right)\right)^{\prime }=X\qquad {\text{ or }}\qquad \left(X_{\sigma }^{\prime }\right)^{\prime }=X.}$$

This very important fact is why results for polar topologies on continuous dual spaces, such as the strong dual topology ${\displaystyle \beta \left(X^{\prime },X\right)}$ on ${\displaystyle X^{\prime }}$ for example, can also often be applied to the original TVS ${\displaystyle X}$; for instance, ${\displaystyle X}$ being identified with ${\displaystyle \left(X_{\sigma }^{\prime }\right)^{\prime }}$ means that the topology ${\displaystyle \beta \left(\left(X_{\sigma }^{\prime }\right)^{\prime },X_{\sigma }^{\prime }\right)}$ on ${\displaystyle \left(X_{\sigma }^{\prime }\right)^{\prime }}$ can instead be thought of as a topology on ${\displaystyle X.}$ Moreover, if ${\displaystyle X^{\prime }}$ is endowed with a topology that is finer than ${\displaystyle \sigma \left(X^{\prime },X\right)}$ then the continuous dual space of ${\displaystyle X^{\prime }}$ will necessarily contain ${\displaystyle \left(X_{\sigma }^{\prime }\right)^{\prime }}$ as a subset. So for instance, when ${\displaystyle X^{\prime }}$ is endowed with the strong dual topology (and so is denoted by ${\displaystyle X_{\beta }^{\prime }}$) then

$${\displaystyle \left(X_{\beta }^{\prime }\right)^{\prime }~\supseteq ~\left(X_{\sigma }^{\prime }\right)^{\prime }~=~X}$$

which (among other things) allows for ${\displaystyle X}$ to be endowed with the subspace topology induced on it by, say, the strong dual topology ${\displaystyle \beta \left(\left(X_{\beta }^{\prime }\right)^{\prime },X_{\beta }^{\prime }\right)}$ (this topology is also called the strong bidual topology and it appears in the theory of reflexive spaces: the Hausdorff locally convex TVS ${\displaystyle X}$ is said to be semi-reflexive if ${\displaystyle \left(X_{\beta }^{\prime }\right)^{\prime }=X}$ and it will be called reflexive if in addition the strong bidual topology ${\displaystyle \beta \left(\left(X_{\beta }^{\prime }\right)^{\prime },X_{\beta }^{\prime }\right)}$ on ${\displaystyle X}$ is equal to ${\displaystyle X}$'s original/starting topology).

Orthogonals, quotients, and subspaces

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

• ${\displaystyle S^{\perp }=(\operatorname {span} S)^{\perp }=\left(\operatorname {cl} _{\sigma (Y,X,b)}\operatorname {span} S\right)^{\perp }=S^{\perp \perp \perp }}$ and this set is ${\displaystyle \sigma (Y,X,b)}$-closed;^[1]
• ${\displaystyle S\subseteq S^{\perp \perp }=\left(\operatorname {cl} _{\sigma (X,Y,b)}\operatorname {span} S\right)}$;^[1]
  • Thus if ${\displaystyle S}$ is a ${\displaystyle \sigma (X,Y,b)}$-closed vector subspace of ${\displaystyle X}$ then ${\displaystyle S\subseteq S^{\perp \perp }.}$
• If ${\displaystyle \left(S_{i}\right)_{i\in I}}$ is a family of ${\displaystyle \sigma (X,Y,b)}$-closed vector subspaces of ${\displaystyle X}$ then
  $${\displaystyle \left(\bigcap _{i\in I}S_{i}\right)^{\perp }=\operatorname {cl} _{\sigma (Y,X,b)}\left(\operatorname {span} \left(\bigcup _{i\in I}S_{i}^{\perp }\right)\right).}$$

  
  ^[1]
• If ${\displaystyle \left(S_{i}\right)_{i\in I}}$ is a family of subsets of ${\displaystyle X}$ then ${\displaystyle \left(\bigcup _{i\in I}S_{i}\right)^{\perp }=\bigcap _{i\in I}S_{i}^{\perp }.}$^[1]

If ${\displaystyle X}$ is a normed space then under the canonical duality, ${\displaystyle S^{\perp }}$ is norm closed in ${\displaystyle X^{\prime }}$ and ${\displaystyle S^{\perp \perp }}$ is norm closed in ${\displaystyle X.}$^[1]

Subspaces

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

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

$${\displaystyle \left(m,y+M^{\perp }\right)\mapsto b(m,y).}$$

The topology ${\displaystyle \sigma \left(M,Y/M^{\perp },b{\big \vert }_{M}\right)}$ is equal to the subspace topology that ${\displaystyle M}$ inherits from ${\displaystyle (X,\sigma (X,Y,b)).}$^[5] Furthermore, if ${\displaystyle (X,\sigma (X,Y,b))}$ is a dual system then so is ${\displaystyle \left(M,Y/M^{\perp },b{\big \vert }_{M}\right).}$^[5]

Quotients

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

$${\displaystyle (x+M,y)\mapsto b(x,y).}$$

The topology ${\displaystyle \sigma \left(X/M,M^{\perp }\right)}$ is identical to the usual quotient topology induced by ${\displaystyle (X,\sigma (X,Y,b))}$ on ${\displaystyle X/M.}$^[5]

Polars and the weak topology

If ${\displaystyle X}$ is a locally convex space and if ${\displaystyle H}$ is a subset of the continuous dual space ${\displaystyle X^{\prime },}$ then ${\displaystyle H}$ is ${\displaystyle \sigma \left(X^{\prime },X\right)}$-bounded if and only if ${\displaystyle H\subseteq B^{\circ }}$ for some barrel ${\displaystyle B}$ in ${\displaystyle X.}$^[1]

The following results are important for defining polar topologies.

If ${\displaystyle (X,Y,b)}$ is a pairing and ${\displaystyle A\subseteq X,}$ then:^[1]

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

If ${\displaystyle (X,Y,b)}$ is a pairing and ${\displaystyle \tau }$ is a locally convex topology on ${\displaystyle X}$ that is consistent with duality, then a subset ${\displaystyle B}$ of ${\displaystyle X}$ is a barrel in ${\displaystyle (X,\tau )}$ if and only if ${\displaystyle B}$ is the polar of some ${\displaystyle \sigma (Y,X,b)}$-bounded subset of ${\displaystyle Y.}$^[6]