This reads like a textbook, not an encyclopedia'stone or style may not reflect theencyclopedic tone used on Wikipedia. See Wikipedia'sguide to writing better articles for suggestions.(January 2022) (Learn how and when to remove this message)

Inmathematics, adual system,dual pair or aduality over afield${\displaystyle \mathbb{K}}$ is a triple${\displaystyle (X,Y,b)}$ consisting of twovector spaces,${\displaystyle X}$ and${\displaystyle Y}$, over${\displaystyle \mathbb{K}}$ and a non-degeneratebilinear map${\displaystyle b:X\times Y\to \mathbb{K}}$.

Inmathematics,duality is the study of dual systems and is important infunctional analysis. Duality plays crucial roles inquantum mechanics because it has extensive applications to the theory ofHilbert spaces.

Apairing orpair 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 abilinear map${\displaystyle b:X\times Y\to \mathbb{K}}$ called thebilinear map associated with the pairing,^[1] or more simply called the pairing'smap or itsbilinear form. The examples here only describe when${\displaystyle \mathbb{K}}$ is either thereal numbers${\displaystyle \mathbb{R}}$ or thecomplex numbers${\displaystyle \mathbb{C}}$, but the mathematical theory is general.

For every${\displaystyle x\in X}$, defineand for every${\displaystyle y\in Y,}$ defineEvery${\displaystyle b(x,\cdot )}$ is alinear functional on${\displaystyle Y}$ and every${\displaystyle b(\cdot ,y)}$ is alinear functional on${\displaystyle X}$. Therefore both$${\displaystyle b(X,\cdot ):=\{b(x,\cdot ):x\in X\}\text{ and }b(\cdot ,Y):=\{b(\cdot ,y):y\in Y\},}$$form vector spaces oflinear functionals.

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

A pairing${\displaystyle (X,Y,b)}$ is called adual system, adual pair,^[2] or aduality over${\displaystyle \mathbb{K}}$ if thebilinear 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${\displaystyle b(x,y)\ne 0}$ 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${\displaystyle b(x,y)\ne 0}$ for each${\displaystyle y\in Y}$).

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

A subset${\displaystyle S}$ of${\displaystyle Y}$ is calledtotal if for every${\displaystyle x\in X}$,$${\displaystyle b(x,s)=0\text{ for all }s\in S}$$ 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${\displaystyle y}$ areorthogonal, written${\displaystyle x\perp y}$, if${\displaystyle b(x,y)=0}$. Two subsets${\displaystyle R\subseteq X}$ and${\displaystyle S\subseteq Y}$ areorthogonal, 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 definedanalogously.

Theorthogonal complement orannihilator 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${\displaystyle \mathbb{K}}$, theabsolute polar set orpolar set of a subset${\displaystyle A}$ of${\displaystyle X}$ is the set:$${\displaystyle A^{\circ }:=\left\{y\in Y:\underset{x\in A}{sup}|b(x,y)|\le 1\right\}.}$$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$${\displaystyle B^{\circ }:=\left\{x\in X:\underset{y\in B}{sup}|b(x,y)|\le 1\right\}.}$$

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 theabsolute prepolar orprepolar of${\displaystyle B}$ and then may be denoted by${\displaystyle {}^{\circ }B}$.^[3]

The polar${\displaystyle B^{\circ }}$ is necessarily aconvex 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 thereal polar of${\displaystyle A.}$ If${\displaystyle A\subseteq X}$ then thebipolar 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 adual 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 theMackey 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 therestriction of${\displaystyle (X,Y,b)}$ to${\displaystyle M\times N}$ is the pairing${\displaystyle \left(M,N,b{|}_{M\times N}\right).}$ If${\displaystyle (X,Y,b)}$ is a duality, then it's possible for a restriction to fail to be a duality (e.g. if${\displaystyle Y\ne \{0\}}$ and${\displaystyle N=\{0\}}$).

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

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

If${\displaystyle N}$ is a vector subspace of${\displaystyle X^{\mathrm{\#}}}$, then the restriction of${\displaystyle \left(X,X^{\mathrm{\#}},c\right)}$ to${\displaystyle X\times N}$ is called thecanonical pairing where if this pairing is a duality then it is instead called thecanonical 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${\displaystyle ⟨x,x^{\mathrm{\prime }}⟩=x^{\mathrm{\prime }}(x)}$ (rather than by${\displaystyle c}$) and${\displaystyle ⟨X,N⟩}$ 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${\displaystyle ⟨X,N⟩.}$

If${\displaystyle N}$ is a vector subspace of${\displaystyle X^{\mathrm{\#}}}$ 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 atopological vector space (TVS) withcontinuous dual space${\displaystyle X^{\mathrm{\prime }}.}$ Then the restriction of the canonical duality${\displaystyle \left(X,X^{\mathrm{\#}},c\right)}$ to${\displaystyle X}$ ×${\displaystyle X^{\mathrm{\prime }}}$ defines a pairing${\displaystyle \left(X,X^{\mathrm{\prime }},c{|}_{X\times X^{\mathrm{\prime }}}\right)}$ for which${\displaystyle X}$ separates points of${\displaystyle X^{\mathrm{\prime }}.}$ If${\displaystyle X^{\mathrm{\prime }}}$ 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${\displaystyle ⟨X,X^{\mathrm{\prime }}⟩.}$

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

Apre-Hilbert space${\displaystyle (H,⟨\cdot ,\cdot ⟩)}$ 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 thesesquilinear form${\displaystyle ⟨\cdot ,\cdot ⟩}$ isconjugate homogeneous in its second coordinate and homogeneous in its first coordinate.

• If${\displaystyle (H,⟨\cdot ,\cdot ⟩)}$ is arealHilbert space then${\displaystyle (H,H,⟨\cdot ,\cdot ⟩)}$ forms a dual system.
• If${\displaystyle (H,⟨\cdot ,\cdot ⟩)}$ is a complexHilbert space then${\displaystyle (H,H,⟨\cdot ,\cdot ⟩)}$ forms a dual system if and only if${\displaystyle \dim H=0.}$ If${\displaystyle H}$ is non-trivial then${\displaystyle (H,H,⟨\cdot ,\cdot ⟩)}$ does not even form pairing since the inner product issesquilinear rather than bilinear.^[1]

Suppose that${\displaystyle (H,⟨\cdot ,\cdot ⟩)}$ is a complex pre-Hilbert space with scalar multiplication denoted as usual by juxtaposition or by a dot${\displaystyle \cdot .}$ Define the map$${\displaystyle \cdot \perp \cdot :\mathbb{C}\times H\to H\text{ by }c\perp x:=\overline{c}x,}$$where the right-hand side uses the scalar multiplication of${\displaystyle H.}$ Let${\displaystyle \overline{H}}$ denote thecomplex 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${\displaystyle b(x,y):=⟨x,y⟩}$ is linear in both coordinates^{[note 2]} and so${\displaystyle \left(H,\overline{H},⟨\cdot ,\cdot ⟩\right)}$ 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.}$$ Then${\displaystyle (X,Y,b)}$ is a pairing such that${\displaystyle X}$ distinguishes points of${\displaystyle Y,}$ but${\displaystyle Y}$ does not distinguish points of${\displaystyle X.}$ Furthermore,${\displaystyle X^{\perp }:=\{y\in Y:X\perp y\}=\{(0,0,z):z\in \mathbb{R}\}.}$
• Let${\displaystyle X:=L^p(\mu ),}$${\displaystyle Y:=L^q(\mu )}$ (where${\displaystyle q}$ is such that${\displaystyle \frac{1}{p}+\frac{1}{q}=1}$), 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${\displaystyle b\left(x\otimes y,x^{\ast }\otimes y^{\ast }\right)=⟨x^{\mathrm{\prime }},x⟩⟨y^{\mathrm{\prime }},y⟩}$ places${\displaystyle X\times Y}$ and${\displaystyle X^{\mathrm{\#}}\times Y^{\mathrm{\#}}}$ in duality.^[2]
• Asequence space${\displaystyle X}$ and itsbeta dual${\displaystyle Y:=X^{\beta }}$ with the bilinear map defined as${\displaystyle ⟨x,y⟩:=\sum _{i=1}^{\mathrm{\infty }}{x}_i{y}_i}$ for${\displaystyle x\in X,}$${\displaystyle y\in X^{\beta }}$ forms a dual system.

Suppose that${\displaystyle (X,Y,b)}$ is a pairing ofvector spaces over${\displaystyle \mathbb{K}.}$ If${\displaystyle S\subseteq Y}$ then theweak 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 each map${\displaystyle b(\cdot ,y):X\to \mathbb{K}}$ continuous as a function of${\displaystyle x}$ for every${\displaystyle y\in S}$.^[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 theweak 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 dependsentirely on the function${\displaystyle b,}$ the usual topology on${\displaystyle \mathbb{C},}$ and${\displaystyle X}$'svector space structure butnot on thealgebraic structures of${\displaystyle Y.}$

Similarly, if${\displaystyle R\subseteq X}$ then the dual definition of theweaktopology 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,${\displaystyle {\mathrm{cl}}_{\sigma (X,Y,b)}(S),}$ 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${\displaystyle {\left(a_i\right)}_{i=1}^{\mathrm{\infty }}}$ 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${\displaystyle \sigma (X,Y,b)}$ islocally 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 anet 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${\displaystyle {\left(x_i\right)}_{i=1}^{\mathrm{\infty }}}$ is a sequence oforthonormal vectors in Hilbert space, then${\displaystyle {\left(x_i\right)}_{i=1}^{\mathrm{\infty }}}$ 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 strictlycoarser 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 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 aninjection from${\displaystyle Y}$ into the algebraic dual space of${\displaystyle X}$;^[1]
3. ${\displaystyle \sigma (Y,X,b)}$ isHausdorff.^[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, thecontinuous dual space of${\displaystyle (X,\sigma (X,Y,b))}$ is$${\displaystyle (X,\sigma (X,Y,b){)}^{\mathrm{\prime }}=b(\cdot ,Y):=\left\{b(\cdot ,y):y\in Y\right\}.}$$

With respect to the canonical pairing, if${\displaystyle X}$ is a TVS whose continuous dual space${\displaystyle X^{\mathrm{\prime }}}$ separates points on${\displaystyle X}$ (i.e. such that${\displaystyle \left(X,\sigma \left(X,X^{\mathrm{\prime }}\right)\right)}$ is Hausdorff, which implies that${\displaystyle X}$ is also necessarily Hausdorff) then the continuous dual space of${\displaystyle \left(X^{\mathrm{\prime }},\sigma \left(X^{\mathrm{\prime }},X\right)\right)}$ 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${\displaystyle x^{\mathrm{\prime }}\in X^{\mathrm{\prime }}}$ to${\displaystyle x^{\mathrm{\prime }}(x)}$). This is commonly written as$${\displaystyle {\left(X^{\mathrm{\prime }},\sigma \left(X^{\mathrm{\prime }},X\right)\right)}^{\mathrm{\prime }}=X\text{ or }{\left(X_{\sigma }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}=X.}$$This very important fact is why results for polar topologies on continuous dual spaces, such as thestrong dual topology${\displaystyle \beta \left(X^{\mathrm{\prime }},X\right)}$ on${\displaystyle X^{\mathrm{\prime }}}$ for example, can also often be applied to the original TVS${\displaystyle X}$; for instance,${\displaystyle X}$ being identified with${\displaystyle {\left(X_{\sigma }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}}$ means that the topology${\displaystyle \beta \left({\left(X_{\sigma }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }},X_{\sigma }^{\mathrm{\prime }}\right)}$ on${\displaystyle {\left(X_{\sigma }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}}$ can instead be thought of as a topology on${\displaystyle X.}$ Moreover, if${\displaystyle X^{\mathrm{\prime }}}$ is endowed with a topology that isfiner than${\displaystyle \sigma \left(X^{\mathrm{\prime }},X\right)}$ then the continuous dual space of${\displaystyle X^{\mathrm{\prime }}}$ will necessarily contain${\displaystyle {\left(X_{\sigma }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}}$ as a subset. So for instance, when${\displaystyle X^{\mathrm{\prime }}}$ is endowed with the strong dual topology (and so is denoted by${\displaystyle X_{\beta }^{\mathrm{\prime }}}$) then$${\displaystyle {\left(X_{\beta }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}\supseteq {\left(X_{\sigma }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}=X}$$which (among other things) allows for${\displaystyle X}$ to be endowed with the subspace topology induced on it by, say, the strong dual topology${\displaystyle \beta \left({\left(X_{\beta }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }},X_{\beta }^{\mathrm{\prime }}\right)}$ (this topology is also called the strongbidual topology and it appears in the theory ofreflexive spaces: the Hausdorff locally convex TVS${\displaystyle X}$ is said to besemi-reflexive if${\displaystyle {\left(X_{\beta }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}=X}$ and it will be calledreflexive if in addition the strong bidual topology${\displaystyle \beta \left({\left(X_{\beta }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }},X_{\beta }^{\mathrm{\prime }}\right)}$ 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${\displaystyle X}$:

• ${\displaystyle S^{\perp }=(\mathrm{span}S{)}^{\perp }={\left({\mathrm{cl}}_{\sigma (Y,X,b)}\mathrm{span}S\right)}^{\perp }=S^{\perp \perp \perp }}$ and this set is${\displaystyle \sigma (Y,X,b)}$-closed;^[1]
• ${\displaystyle S\subseteq S^{\perp \perp }=\left({\mathrm{cl}}_{\sigma (X,Y,b)}\mathrm{span}S\right)}$;^[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$${\displaystyle {\left(\bigcap _{i\in I}{S}_i\right)}^{\perp }={\mathrm{cl}}_{\sigma (Y,X,b)}\left(\mathrm{span}\left(\bigcup _{i\in I}{S}_i^{\perp }\right)\right).}$$^[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${\displaystyle X^{\mathrm{\prime }}}$ 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 thesubspace topology that${\displaystyle M}$ inherits from${\displaystyle (X,\sigma (X,Y,b)).}$

Also,${\displaystyle \left(M,Y/M^{\perp },b{|}_M\right)}$ is a paired space (where${\displaystyle Y/M^{\perp }}$ means${\displaystyle Y/\left(M^{\perp }\right)}$) where${\displaystyle b{|}_M:M\times Y/M^{\perp }\to \mathbb{K}}$ is defined by$${\displaystyle \left(m,y+M^{\perp }\right)\mapsto b(m,y).}$$

The topology${\displaystyle \sigma \left(M,Y/M^{\perp },b{|}_M\right)}$ is equal to thesubspace 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${\displaystyle \left(M,Y/M^{\perp },b{|}_M\right).}$^[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 usualquotient 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${\displaystyle X^{\mathrm{\prime }},}$ then${\displaystyle H}$ is${\displaystyle \sigma \left(X^{\mathrm{\prime }},X\right)}$-bounded if and only if${\displaystyle H\subseteq B^{\circ }}$ for somebarrel${\displaystyle B}$ 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) thebalanced hull of${\displaystyle A}$; (d) the${\displaystyle \sigma (X,Y,b)}$-closure of${\displaystyle A}$; (e) the${\displaystyle \sigma (X,Y,b)}$-closure of theconvex balanced hull of${\displaystyle A.}$
3. Thebipolar 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${\displaystyle A.}$
   • Thebipolar 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${\displaystyle A^{\circ }}$ isabsorbing 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.