Dual system

Examples

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 ${\displaystyle \left(M,N,b{|}_{M\times N}\right).}$ 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 ${\displaystyle (M,N,b).}$

Canonical duality on a vector space

Suppose that ${\displaystyle X}$ is a vector space and let ${\displaystyle X^{\mathrm{\#}}}$ denote the algebraic dual space of ${\displaystyle X}$ (that is, the space of all linear functionals on ${\displaystyle X}$). There is a canonical duality ${\displaystyle \left(X,X^{\mathrm{\#}},c\right)}$ where ${\displaystyle c\left(x,x^{\mathrm{\prime }}\right)=⟨x,x^{\mathrm{\prime }}⟩=x^{\mathrm{\prime }}(x),}$ which is called the evaluation map or the natural or canonical 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 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 ${\displaystyle ⟨x,x^{\mathrm{\prime }}⟩=x^{\mathrm{\prime }}(x)}$ (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 ${\displaystyle x=0}$).

Canonical duality on a topological vector space

Suppose ${\displaystyle X}$ is a topological vector space (TVS) with continuous 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.

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 ${\displaystyle ⟨X,X^{\mathrm{\prime }}⟩.}$

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.

Theorem—Let ${\displaystyle X}$ be a TVS with algebraic dual ${\displaystyle X^{\mathrm{\#}}}$ and let ${\displaystyle \mathcal{N}}$ be a basis of neighborhoods of ${\displaystyle X}$ at the origin. Under the canonical duality ${\displaystyle ⟨X,X^{\mathrm{\#}}⟩,}$ the continuous dual space of ${\displaystyle X}$ is the union of all ${\displaystyle N^{\circ }}$ as ${\displaystyle N}$ ranges over ${\displaystyle \mathcal{N}}$ (where the polars are taken in ${\displaystyle X^{\mathrm{\#}}}$).

Inner product spaces and complex conjugate spaces

A pre-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 the sesquilinear form ${\displaystyle ⟨\cdot ,\cdot ⟩}$ is conjugate homogeneous in its second coordinate and homogeneous in its first coordinate.

• If ${\displaystyle (H,⟨\cdot ,\cdot ⟩)}$ is a real Hilbert space then ${\displaystyle (H,H,⟨\cdot ,\cdot ⟩)}$ forms a dual system.
• If ${\displaystyle (H,⟨\cdot ,\cdot ⟩)}$ is a complex Hilbert 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 is sesquilinear rather than bilinear.

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 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 ${\displaystyle b(x,y):=⟨x,y⟩}$ is linear in both coordinates and so ${\displaystyle \left(H,\overline{H},⟨\cdot ,\cdot ⟩\right)}$ 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 ${\displaystyle 0<p<\mathrm{\infty },}$ ${\displaystyle X:=L^p(\mu ),}$ ${\displaystyle Y:=L^q(\mu )}$ (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.
• A sequence space ${\displaystyle X}$ and its beta 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.

Weak topology

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 each map ${\displaystyle b(\cdot ,y):X\to \mathbb{K}}$ continuous as a function of ${\displaystyle x}$ for every ${\displaystyle y\in S}$. 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).

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 ${\displaystyle \sigma (X,Y,b)}$ 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 ${\displaystyle Y.}$ 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 ${\displaystyle (X,\sigma (X,Y,b)).}$ 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 of orthonormal 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).

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

Bounded subsets

A subset ${\displaystyle S}$ of ${\displaystyle X}$ is ${\displaystyle \sigma (X,Y,b)}$-bounded if and only if $${\displaystyle \underset{}{sup}|b(S,y)|<\mathrm{\infty }\text{ for all }y\in Y,}$$ 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 ${\displaystyle X}$;
3. ${\displaystyle \sigma (Y,X,b)}$ is Hausdorff.

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

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 the strong 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 is finer 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 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 ${\displaystyle {\left(X_{\beta }^{\mathrm{\prime }}\right)}^{\mathrm{\prime }}=X}$ and it will be called reflexive 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).

Orthogonals, quotients, and subspaces

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;
• ${\displaystyle S\subseteq S^{\perp <!--\; -->\perp }=\left({\mathrm{cl}}_{\sigma (X,Y,b)}\mathrm{span}S\right)}$;
  • 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 $${\displaystyle {\left(\bigcap _{i\in I}{S}_i\right)}^{\perp }={\mathrm{cl}}_{\sigma (Y,X,b)}\left(\mathrm{span}\left(\bigcup _{i\in I}{S}_i^{\perp }\right)\right).}$$
• If ${\displaystyle {\left(S_i\right)}_{i\in I}}$ is a family of subsets of ${\displaystyle X}$ then ${\displaystyle {\left(\bigcup _{i\in I}{S}_i\right)}^{\perp }=\bigcap _{i\in I}{S}_i^{\perp }.}$

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 ${\displaystyle X.}$

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, ${\displaystyle \left(M,Y/M^{\perp },b{|}_M\right)}$ 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 the subspace topology that ${\displaystyle M}$ inherits from ${\displaystyle (X,\sigma (X,Y,b)).}$ Furthermore, if ${\displaystyle (X,\sigma (X,Y,b))}$ is a dual system then so is ${\displaystyle \left(M,Y/M^{\perp },b{|}_M\right).}$

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 ${\displaystyle X/M.}$

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 ${\displaystyle X^{\mathrm{\prime }},}$ then ${\displaystyle H}$ is ${\displaystyle \sigma \left(X^{\mathrm{\prime }},X\right)}$-bounded if and only if ${\displaystyle H\subseteq B^{\circ }}$ for some barrel ${\displaystyle B}$ in ${\displaystyle X.}$

The following results are important for defining polar topologies.

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

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. ${\displaystyle A}$ 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 ${\displaystyle Y.}$

Transposes

Transposes of a linear map with respect to pairings

Let ${\displaystyle (X,Y,b)}$ and ${\displaystyle (W,Z,c)}$ be pairings over ${\displaystyle \mathbb{K}}$ and let ${\displaystyle F:X\to W}$ be a linear map.

For all ${\displaystyle z\in Z,}$ let ${\displaystyle c(F(\cdot ),z):X\to \mathbb{K}}$ be the map defined by ${\displaystyle x\mapsto c(F(x),z).}$ It is said that ${\displaystyle F}$'s transpose or adjoint is well-defined if the following conditions are satisfied:

1. ${\displaystyle X}$ distinguishes points of ${\displaystyle Y}$ (or equivalently, the map ${\displaystyle y\mapsto b(\cdot ,y)}$ from ${\displaystyle Y}$ into the algebraic dual ${\displaystyle X^{\mathrm{\#}}}$ is injective), and
2. ${\displaystyle c(F(\cdot ),Z)\subseteq b(\cdot ,Y),}$ where ${\displaystyle c(F(\cdot ),Z):=\{c(F(\cdot ),z):z\in Z\}}$ and ${\displaystyle b(\cdot ,Y):=\{b(\cdot ,y):y\in Y\}}$.

In this case, for any ${\displaystyle z\in Z}$ there exists (by condition 2) a unique (by condition 1) ${\displaystyle y\in Y}$ such that ${\displaystyle c(F(\cdot ),z)=b(\cdot ,y)}$), where this element of ${\displaystyle Y}$ will be denoted by ${\displaystyle {}^{t}F(z).}$ This defines a linear map $${\displaystyle {}^{t}F:Z\to Y}$$

called the transpose or adjoint of ${\displaystyle F}$ with respect to ${\displaystyle (X,Y,b)}$ and ${\displaystyle (W,Z,c)}$ (this should not be confused with the Hermitian adjoint). It is easy to see that the two conditions mentioned above (i.e. for "the transpose is well-defined") are also necessary for ${\displaystyle {}^{t}F}$ to be well-defined. For every ${\displaystyle z\in Z,}$ the defining condition for ${\displaystyle {}^{t}F(z)}$ is $${\displaystyle c(F(\cdot ),z)=b\left(\cdot ,{}^{t}F(z)\right),}$$ that is, $${\displaystyle c(F(x),z)=b\left(x,{}^{t}F(z)\right)}$$ for all ${\displaystyle x\in X.}$

By the conventions mentioned at the beginning of this article, this also defines the transpose of linear maps of the form ${\displaystyle Z\to Y,}$ ${\displaystyle X\to Z,}$ ${\displaystyle W\to Y,}$ ${\displaystyle Y\to W,}$ etc. (see footnote).

Properties of the transpose

Throughout, ${\displaystyle (X,Y,b)}$ and ${\displaystyle (W,Z,c)}$ be pairings over ${\displaystyle \mathbb{K}}$ and ${\displaystyle F:X\to W}$ will be a linear map whose transpose ${\displaystyle {}^{t}F:Z\to Y}$ is well-defined.

• ${\displaystyle {}^{t}F:Z\to Y}$ is injective (i.e. ${\displaystyle \ker {}^{t}F=\{0\}}$) if and only if the range of ${\displaystyle F}$ is dense in ${\displaystyle \left(W,\sigma \left(W,Z,c\right)\right).}$
• If in addition to ${\displaystyle {}^{t}F}$ being well-defined, the transpose of ${\displaystyle {}^{t}F}$ is also well-defined then ${\displaystyle {}^{tt}F=F.}$
• Suppose ${\displaystyle (U,V,a)}$ is a pairing over ${\displaystyle \mathbb{K}}$ and ${\displaystyle E:U\to X}$ is a linear map whose transpose ${\displaystyle {}^{t}E:Y\to V}$ is well-defined. Then the transpose of ${\displaystyle F\circ E:U\to W,}$ which is ${\displaystyle {}^t(F\circ E):Z\to V,}$ is well-defined and ${\displaystyle {}^t(F\circ E)={}^{t}E\circ {}^{t}F.}$
• If ${\displaystyle F:X\to W}$ is a vector space isomorphism then ${\displaystyle {}^{t}F:Z\to Y}$ is bijective, the transpose of ${\displaystyle F^{-1}:W\to X,}$ which is ${\displaystyle {}^t\left(F^{-1}\right):Y\to Z,}$ is well-defined, and ${\displaystyle {}^t\left(F^{-1}\right)={\left({}^{t}F\right)}^{-1}}$
• Let ${\displaystyle S\subseteq X}$ and let ${\displaystyle S^{\circ }}$ denotes the absolute polar of ${\displaystyle A,}$ then:
  1. ${\displaystyle [F(S){]}^{\circ }={\left({}^{t}F\right)}^{-1}\left(S^{\circ }\right)}$;
  2. if ${\displaystyle F(S)\subseteq T}$ for some ${\displaystyle T\subseteq W,}$ then ${\displaystyle {}^{t}F\left(T^{\circ }\right)\subseteq S^{\circ }}$;
  3. if ${\displaystyle T\subseteq W}$ is such that ${\displaystyle {}^{t}F\left(T^{\circ }\right)\subseteq S^{\circ },}$ then ${\displaystyle F(S)\subseteq T^{\circ \circ }}$;
  4. if ${\displaystyle T\subseteq W}$ and ${\displaystyle S\subseteq X}$ are weakly closed disks then ${\displaystyle {}^{t}F\left(T^{\circ }\right)\subseteq S^{\circ }}$ if and only if ${\displaystyle F(S)\subseteq T}$;
  5. ${\displaystyle \ker {}^{t}F=[F(X){]}^{\perp }.}$

These results hold when the real polar is used in place of the absolute polar.

If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed spaces under their canonical dualities and if ${\displaystyle F:X\to Y}$ is a continuous linear map, then ${\displaystyle \Vert F\Vert =\Vert {}^{t}F\Vert .}$

Weak continuity

A linear map ${\displaystyle F:X\to W}$ is weakly continuous (with respect to ${\displaystyle (X,Y,b)}$ and ${\displaystyle (W,Z,c)}$) if ${\displaystyle F:(X,\sigma (X,Y,b))\to (W,(W,Z,c))}$ is continuous.

The following result shows that the existence of the transpose map is intimately tied to the weak topology.

Proposition—Assume that ${\displaystyle X}$ distinguishes points of ${\displaystyle Y}$ and ${\displaystyle F:X\to W}$ is a linear map. Then the following are equivalent:

1. ${\displaystyle F}$ is weakly continuous (that is, ${\displaystyle F:(X,\sigma (X,Y,b))\to (W,(W,Z,c))}$ is continuous);
2. ${\displaystyle c(F(\cdot ),Z)\subseteq b(\cdot ,Y)}$;
3. the transpose of ${\displaystyle F}$ is well-defined.

If ${\displaystyle F}$ is weakly continuous then

• ${\displaystyle {}^{t}F:Z\to Y}$ is weakly continuous, meaning that ${\displaystyle {}^{t}F:(Z,\sigma (Z,W,c))\to (Y,(Y,X,b))}$ is continuous;
• the transpose of ${\displaystyle {}^{t}F}$ is well-defined if and only if ${\displaystyle Z}$ distinguishes points of ${\displaystyle W,}$ in which case ${\displaystyle {}^{tt}F=F.}$

Weak topology and the canonical duality

Suppose that ${\displaystyle X}$ is a vector space and that ${\displaystyle X^{\mathrm{\#}}}$ is its the algebraic dual. Then every ${\displaystyle \sigma \left(X,X^{\mathrm{\#}}\right)}$-bounded subset of ${\displaystyle X}$ is contained in a finite dimensional vector subspace and every vector subspace of ${\displaystyle X}$ is ${\displaystyle \sigma \left(X,X^{\mathrm{\#}}\right)}$-closed.

Weak completeness

If ${\displaystyle (X,\sigma (X,Y,b))}$ is a complete topological vector space say that ${\displaystyle X}$ is ${\displaystyle \sigma (X,Y,b)}$-complete or (if no ambiguity can arise) weakly-complete. There exist Banach spaces that are not weakly-complete (despite being complete in their norm topology).

If ${\displaystyle X}$ is a vector space then under the canonical duality, ${\displaystyle \left(X^{\mathrm{\#}},\sigma \left(X^{\mathrm{\#}},X\right)\right)}$ is complete. Conversely, if ${\displaystyle Z}$ is a Hausdorff locally convex TVS with continuous dual space ${\displaystyle Z^{\mathrm{\prime }},}$ then ${\displaystyle \left(Z,\sigma \left(Z,Z^{\mathrm{\prime }}\right)\right)}$ is complete if and only if ${\displaystyle Z={\left(Z^{\mathrm{\prime }}\right)}^{\mathrm{\#}}}$; that is, if and only if the map ${\displaystyle Z\to {\left(Z^{\mathrm{\prime }}\right)}^{\mathrm{\#}}}$ defined by sending ${\displaystyle z\in Z}$ to the evaluation map at ${\displaystyle z}$ (i.e. ${\displaystyle z^{\mathrm{\prime }}\mapsto z^{\mathrm{\prime }}(z)}$) is a bijection.

In particular, with respect to the canonical duality, if ${\displaystyle Y}$ is a vector subspace of ${\displaystyle X^{\mathrm{\#}}}$ such that ${\displaystyle Y}$ separates points of ${\displaystyle X,}$ then ${\displaystyle (Y,\sigma (Y,X))}$ is complete if and only if ${\displaystyle Y=X^{\mathrm{\#}}.}$ Said differently, there does not exist a proper vector subspace ${\displaystyle Y\ne X^{\mathrm{\#}}}$ of ${\displaystyle X^{\mathrm{\#}}}$ such that ${\displaystyle (X,\sigma (X,Y))}$ is Hausdorff and ${\displaystyle Y}$ is complete in the weak-* topology (i.e. the topology of pointwise convergence). Consequently, when the continuous dual space ${\displaystyle X^{\mathrm{\prime }}}$ of a Hausdorff locally convex TVS ${\displaystyle X}$ is endowed with the weak-* topology, then ${\displaystyle X_{\sigma }^{\mathrm{\prime }}}$ is complete if and only if ${\displaystyle X^{\mathrm{\prime }}=X^{\mathrm{\#}}}$ (that is, if and only if every linear functional on ${\displaystyle X}$ is continuous).

Identification of Y with a subspace of the algebraic dual

If ${\displaystyle X}$ distinguishes points of ${\displaystyle Y}$ and if ${\displaystyle Z}$ denotes the range of the injection ${\displaystyle y\mapsto b(\cdot ,y)}$ then ${\displaystyle Z}$ is a vector subspace of the algebraic dual space of ${\displaystyle X}$ and the pairing ${\displaystyle (X,Y,b)}$ becomes canonically identified with the canonical pairing ${\displaystyle ⟨X,Z⟩}$ (where ${\displaystyle ⟨x,x^{\mathrm{\prime }}⟩:=x^{\mathrm{\prime }}(x)}$ is the natural evaluation map). In particular, in this situation it will be assumed without loss of generality that ${\displaystyle Y}$ is a vector subspace of ${\displaystyle X}$'s algebraic dual and ${\displaystyle b}$ is the evaluation map.

Convention: Often, whenever ${\displaystyle y\mapsto b(\cdot ,y)}$ is injective (especially when ${\displaystyle (X,Y,b)}$ forms a dual pair) then it is common practice to assume without loss of generality that ${\displaystyle Y}$ is a vector subspace of the algebraic dual space of ${\displaystyle X,}$ that ${\displaystyle b}$ is the natural evaluation map, and also denote ${\displaystyle Y}$ by ${\displaystyle X^{\mathrm{\prime }}.}$

In a completely analogous manner, if ${\displaystyle Y}$ distinguishes points of ${\displaystyle X}$ then it is possible for ${\displaystyle X}$ to be identified as a vector subspace of ${\displaystyle Y}$'s algebraic dual space.

Algebraic adjoint

In the special case where the dualities are the canonical dualities ${\displaystyle ⟨X,X^{\mathrm{\#}}⟩}$ and ${\displaystyle ⟨W,W^{\mathrm{\#}}⟩,}$ the transpose of a linear map ${\displaystyle F:X\to W}$ is always well-defined. This transpose is called the algebraic adjoint of ${\displaystyle F}$ and it will be denoted by ${\displaystyle F^{\mathrm{\#}}}$; that is, ${\displaystyle F^{\mathrm{\#}}={}^{t}F:W^{\mathrm{\#}}\to X^{\mathrm{\#}}.}$ In this case, for all ${\displaystyle w^{\mathrm{\prime }}\in W^{\mathrm{\#}},}$ ${\displaystyle F^{\mathrm{\#}}\left(w^{\mathrm{\prime }}\right)=w^{\mathrm{\prime }}\circ F}$ where the defining condition for ${\displaystyle F^{\mathrm{\#}}\left(w^{\mathrm{\prime }}\right)}$ is: $${\displaystyle ⟨x,F^{\mathrm{\#}}\left(w^{\mathrm{\prime }}\right)⟩=⟨F(x),w^{\mathrm{\prime }}⟩\text{ for all }>x\in X,}$$ or equivalently, ${\displaystyle F^{\mathrm{\#}}\left(w^{\mathrm{\prime }}\right)(x)=w^{\mathrm{\prime }}(F(x))\text{ for all }x\in X.}$

If ${\displaystyle X=Y={\mathbb{K}}^n}$ for some integer ${\displaystyle n,}$ ${\displaystyle \mathcal{E}=\left\{e_1,\dots ,e_n\right\}}$ is a basis for ${\displaystyle X}$ with dual basis ${\displaystyle {\mathcal{E}}^{\mathrm{\prime }}=\left\{e_1^{\mathrm{\prime }},\dots ,e_n^{\mathrm{\prime }}\right\},}$ ${\displaystyle F:{\mathbb{K}}^n\to {\mathbb{K}}^n}$ is a linear operator, and the matrix representation of ${\displaystyle F}$ with respect to ${\displaystyle \mathcal{E}}$ is ${\displaystyle M:=\left(f_{i,j}\right),}$ then the transpose of ${\displaystyle M}$ is the matrix representation with respect to ${\displaystyle {\mathcal{E}}^{\mathrm{\prime }}}$ of ${\displaystyle F^{\mathrm{\#}}.}$

Weak continuity and openness

Suppose that ${\displaystyle ⟨X,Y⟩}$ and ${\displaystyle ⟨W,Z⟩}$ are canonical pairings (so ${\displaystyle Y\subseteq X^{\mathrm{\#}}}$ and ${\displaystyle Z\subseteq W^{\mathrm{\#}}}$) that are dual systems and let ${\displaystyle F:X\to W}$ be a linear map. Then ${\displaystyle F:X\to W}$ is weakly continuous if and only if it satisfies any of the following equivalent conditions:

1. ${\displaystyle F:(X,\sigma (X,Y))\to (W,\sigma (W,Z))}$ is continuous.
2. ${\displaystyle F^{\mathrm{\#}}(Z)\subseteq Y}$
3. the transpose of F, ${\displaystyle {}^{t}F:Z\to Y,}$ with respect to ${\displaystyle ⟨X,Y⟩}$ and ${\displaystyle ⟨W,Z⟩}$ is well-defined.