Dual system

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In mathematics, a dual system, dual pair or a duality over a field ${\displaystyle \mathbb {K} }$ is a triple ${\displaystyle (X,Y,b)}$ consisting of two vector spaces, ${\displaystyle X}$ and ${\displaystyle Y}$, over ${\displaystyle \mathbb {K} }$ and a non-degenerate bilinear map ${\displaystyle b:X\times Y\to \mathbb {K} }$.

In mathematics, duality is the study of dual systems and is important in functional analysis. Duality plays crucial roles in quantum mechanics because it has extensive applications to the theory of Hilbert spaces.

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 more simply called the pairing's map or its bilinear form. The examples here only describe when ${\displaystyle \mathbb {K} }$ is either the real numbers ${\displaystyle \mathbb {R} }$ or the complex numbers ${\displaystyle \mathbb {C} }$, but the mathematical theory is general.

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 ${\displaystyle X}$. Therefore both $${\displaystyle b(X,\,\cdot \,):=\{b(x,\,\cdot \,):x\in X\}\qquad {\text{ and }}\qquad b(\,\cdot \,,Y):=\{b(\,\cdot \,,y):y\in Y\},}$$ form vector spaces of linear functionals.

It is common practice to write ${\displaystyle \langle x,y\rangle }$ instead of ${\displaystyle b(x,y)}$, in which in some cases the pairing may 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.

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]

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 ${\displaystyle Y}$.

The vectors ${\displaystyle x}$ and ${\displaystyle y}$ are 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 ${\displaystyle \{0\}}$.

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 anti-symmetry 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 ${\displaystyle B^{\circ }.}$^[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 ${\displaystyle B^{\circ \circ }:=\left({}^{\circ }B\right)^{\circ }.}$

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 ${\displaystyle (Y,X,d).}$ These 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 ${\displaystyle \sigma (Y,X,d)}$.

Although it is technically incorrect and an abuse of notation, this article will adhere to the nearly ubiquitous convention of treating a pairing ${\displaystyle (X,Y,b)}$ interchangeably with ${\displaystyle (Y,X,d)}$ and also of denoting ${\displaystyle (Y,X,d)}$ by ${\displaystyle (Y,X,b).}$

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 ${\displaystyle (M,N,b).}$

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]

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 ${\displaystyle \left\langle X,X^{\prime }\right\rangle .}$

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.

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.

• 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.

Suppose that ${\displaystyle (X,Y,b)}$ is a pairing of vector spaces over ${\displaystyle \mathbb {K} .}$ If ${\displaystyle S\subseteq Y}$ then the weak topology on ${\displaystyle X}$ induced by ${\displaystyle S}$ (and ${\displaystyle b}$) is the weakest TVS topology on ${\displaystyle X,}$ denoted by ${\displaystyle \sigma (X,S,b)}$ or simply ${\displaystyle \sigma (X,S),}$ making all maps ${\displaystyle b(\,\cdot \,,y):X\to \mathbb {K} }$ continuous as ${\displaystyle y}$ ranges over ${\displaystyle S.}$^[1] If ${\displaystyle S}$ is not clear from context then it should be assumed to be all of ${\displaystyle Y,}$ in which case it is called the weak topology on ${\displaystyle X}$ (induced by ${\displaystyle Y}$). The notation ${\displaystyle X_{\sigma (X,S,b)},}$ ${\displaystyle X_{\sigma (X,S)},}$ or (if no confusion could arise) simply ${\displaystyle X_{\sigma }}$ is used to denote ${\displaystyle X}$ endowed with the weak topology ${\displaystyle \sigma (X,S,b).}$ Importantly, the weak topology depends entirely on the function ${\displaystyle b,}$ the usual topology on ${\displaystyle \mathbb {C} ,}$ and ${\displaystyle X}$'s vector space structure but not on the algebraic structures of ${\displaystyle Y.}$

Similarly, if ${\displaystyle R\subseteq X}$ then the dual definition of the weak topology on ${\displaystyle Y}$ induced by ${\displaystyle R}$ (and ${\displaystyle b}$), which is denoted by ${\displaystyle \sigma (Y,R,b)}$ or simply ${\displaystyle \sigma (Y,R)}$ (see footnote for details).^{[note 3]}

Definition and Notation: If "${\displaystyle \sigma (X,Y,b)}$" is attached to a topological definition (e.g. ${\displaystyle \sigma (X,Y,b)}$-converges, ${\displaystyle \sigma (X,Y,b)}$-bounded, ${\displaystyle \operatorname {cl} _{\sigma (X,Y,b)}(S),}$ etc.) then it means that definition when the first space (i.e. ${\displaystyle X}$) carries the ${\displaystyle \sigma (X,Y,b)}$ topology. Mention of ${\displaystyle b}$ or even ${\displaystyle X}$ and ${\displaystyle Y}$ may be omitted if no confusion arises. So, for instance, if a sequence ${\displaystyle \left(a_{i}\right)_{i=1}^{\infty }}$ in ${\displaystyle Y}$ "${\displaystyle \sigma }$-converges" or "weakly converges" then this means that it converges in ${\displaystyle (Y,\sigma (Y,X,b))}$ whereas if it were a sequence in ${\displaystyle X}$, then this would mean that it converges in ${\displaystyle (X,\sigma (X,Y,b))}$).

The topology ${\displaystyle \sigma (X,Y,b)}$ is locally convex since it is determined by the family of seminorms ${\displaystyle p_{y}:X\to \mathbb {R} }$ defined by ${\displaystyle p_{y}(x):=|b(x,y)|,}$ as ${\displaystyle y}$ ranges over ${\displaystyle Y.}$^[1] If ${\displaystyle x\in X}$ and ${\displaystyle \left(x_{i}\right)_{i\in I}}$ is a net in ${\displaystyle X,}$ then ${\displaystyle \left(x_{i}\right)_{i\in I}}$ ${\displaystyle \sigma (X,Y,b)}$-converges to ${\displaystyle x}$ if ${\displaystyle \left(x_{i}\right)_{i\in I}}$ converges to ${\displaystyle x}$ in ${\displaystyle (X,\sigma (X,Y,b)).}$^[1] A net ${\displaystyle \left(x_{i}\right)_{i\in I}}$ ${\displaystyle \sigma (X,Y,b)}$-converges to ${\displaystyle x}$ if and only if for all ${\displaystyle y\in Y,}$ ${\displaystyle b\left(x_{i},y\right)}$ converges to ${\displaystyle b(x,y).}$ If ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ is a sequence of orthonormal vectors in Hilbert space, then ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ converges weakly to 0 but does not norm-converge to 0 (or any other vector).^[1]

If ${\displaystyle (X,Y,b)}$ is a pairing and ${\displaystyle N}$ is a proper vector subspace of ${\displaystyle Y}$ such that ${\displaystyle (X,N,b)}$ is a dual pair, then ${\displaystyle \sigma (X,N,b)}$ is strictly coarser than ${\displaystyle \sigma (X,Y,b).}$^[1]

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

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]

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

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).

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]

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]

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]

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.